Plus One

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 11 Conic Sections.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 11 Conic Sections

Plus One Maths Conic Sections 3 Marks Important Questions

Question 1.
1. Find the equation of the Hyperbola where foci (0,±8)are and the length of the latus rectum is 24.(IMP-2012)
Answer:
Since foci (0,±8)
=> ae = 8
Latus rectum = 24= \(\frac {2b² }{ a }\)
=> 12a = b²
b² =a²(e² -1)
=> b² – a²e² -a²
=>12a = 64 – a²
=>a²+12a-64 = 0
=> a = – 16,4
acceptable value is => a = 4
=> 48 = b²
Hence equation is

Question 2.
Find the equation of the circle with centre (- a,- b)and radius \(\sqrt{a^{2}+b^{2}}\) . (IMP-2012)
Answer:
We have the equation of a circle as;
(x-h)² + (y-k)² – r²
=> (x + a)² +(y + b)² = a² + b²
=> x² +2 ax + a² + y² +2 by + b² =a² +b²
=> x² +2ax + y² +2by = 0

Question 3.
Find the coordinate of the foci, the length of the major axis, minor axis, latus rectum and eccentricity of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) . (MARCH-2013)
Answer:

Question 4.
Consider the parabola y² =12x. (MARCH-2015)
i) Find the coordinate of the focus.
ii) Find the length of the latus rectum.
Answer:
i) Given; y² =12x comparing with y² = 4ax We have 4a = 12 => a = 3 Then; Focus is (3,0)
ii) Length of latus rectum = 4a = 12

Question 5.
Find the foci, vertices, the eccentricity and the length of the latus rectum of the hyperbola 16x² – 9y² =144. (SAY-2017) 
Answer:
The equation of the hyperbola is of the form

=>a² =9,b² =16
=>c² = a² +b² =9 + 16 = 25
=>c = 5
Coordinate of foci are (±5,0)
Coordinate of vertices are (±a,0) => (±3,0)

Question 6.
Directrix of the parabola x² = – 4ay is ……….. (MARCH-2014)
a) x + a = 0
b) x – a = 0
c) y – a = 0
d) y + a = 0
Find the equation of the ellipse whose length of the major axis is 20 and foci are (0 ±5)
(March-2015)
Answer:
i) y-a = 0
ii) The equation of the ellipse is of the form;

Question 7.
Find the coordinates of the focii, vertices, eccentricity and the length of the Latus Rectum of the ellipse 100x² + 25y² = 2500 (IMP-2015)
Answer:
Given: 100x² +25y² = 2500

Question 8.
Find the foci, vertices, length of the major axis and eccentricity of the ellipse: \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) (MARCH-2016)
Answer:
Since 25 > 9 the standard equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) => a² =25;b² =9
c² =a² – b² =25 – 9 = 16
=>c = 4
Coordinate of foci are (±4,0)
Coordinate of vertex are (±5,0)
Length of major axis = 2a = 2 x 5 = 10

Plus One Maths Conic Sections 4 Marks Important Questions

Question 1.
An ellipse whose major axis as x-axis and the centre (0,0) passes through (4,3) and (- 1,4). (MARCH-2010)
i) Find the equation of the ellipse.
ii) Find is eccentricity.
Answer:
i)

ii)

Question 2.
Consider the conic find 9y² -4x² = 36 (IMP-2010)
i) The foci.
ii) Eccentricity.
iii) Length of latus rectum.
Answer:

Question 3.
Find the equation of the circle with center (2,2) and passing through the point (4,5). (MARCH-2011)
Find the eccentricity and the length of latus rectum of the ellipse 4x² + 9y² =36
Answer:

Question 4.
For the hyperbola 9x² – 16y² =144 (IMP-2011)
i) find eccentricity.
ii) find the latus rectum.
Answer:
i)

ii)

Question 5.
A hyperbola whose transverse axis is x-axis, centre (0,0) and foci (±√10,0) passes through the point (3,2) (MARCH-2012)
i) Find the equation of the hyperbola.
ii) Find the eccentricity.
Answer:
i)

ii)

Question 6.
Find the centre and radius of the circle. (IMP-2013)
x² +y² – 8x + 10y – 12 = 0.
ii) Determine the eccentricity and length of latus rectum of the hyperbola —–
Answer:
i) Comparing with the general equation we have
g = – 4; f = 5; c = – 12
Centre – (- g,- f) => (4,- 5)
\(\sqrt{g^{2}+f^{2}-c} \)= \(\sqrt{16+25+12}=\sqrt{53}\)
ii)

Question 7.
Consider the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\). Find the coordinate of the foci, the length of the major axis, the length of the minor axis, latus rectum and eccentricity. (MARCH-2014)
Answer:

Question 8.
Which one of the following equations (IMP-2014)
represents a parabola which is symmetrical about the positive Y-axis?
a) y² = 4x
b) y² = – 8x
c) x² + 4y = 0
d) x² – 4y = 0
ii) Find the equation of the ellipse vertices are (±13,0) and foci are (±5,0)
Answer:

Question 9.
Match the following. (IMP-2014)

Answer:

Question 10.
i) Find the equation of the parabola with focus (6,0) and equation of the directrix is x = – 6. (MARCH-2017)
ii) Find the coordinate of the foci, vertices, the length of transverse axis, conjugate axis and eccentricity of the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)
(MARCH -2017)
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 10 Straight Lines.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 10 Straight Lines

Plus One Maths Straight Lines 3 Marks Important Questions

Question 1.
i) Find the slope of the line joining (- 2,6) and (4,8). (MARCH-2010)
ii) Find the value of x if the above line is perpendicular to the line joining (8,12) and (x,24).
Answer:
i)

ii) Slope of line through (8,12)and (x,24)

Since both are perpendicular to each other

Question 2.
i) Write the equation of y-axis. (IMP-2010)
ii) Find the distance between the lines
8x + 15y – 5 = 0 and 8x + 15y + 12 = 0
(Imp (Science) – 2010)
Answer:
i) Equation of y-axis is x = 0
ii) Both are parallel lines, we have;

Question 3.
The vertices of ∆ ABCare A(2,1), B(- 3,5) and C(4,5). (IMP-2012)
i) Write the coordinates of the midpoint of AC.
ii) Find the equation of the median through the vertex B.
Answer:
i)

ii) Equation of the median through the vertex B is the equation of a line passing through the midpoint of AC and the vertex B. ie; through (3,3)and (-3,5)

Question 4.
Find the slope of the straight lines √3x + y = 1, x + √3y = 1 (IMP-2012)
Also find the angles between them.
(Imp (Science) – 2012)
Answer:

Question 5.
The vertices of ∆ABC are A(-2,3), B(2,-3) and C(4,5). (MARCH-2012)
i) Find the slope of BC.
ii) Find the equation of the altitude of ∆ABC passing through A.
Answer:

Question 6.
i) Find the slope of the line joining the points (2,2) and (5,3). (MARCH-2013)
ii) Find the equation of the line joining the points (2,2) and (5,3).
Answer:

Question 7.
i) If two lines are perpendicular, then the product of their slopes is ______. (MARCH-2013)
ii) Find the equation of a line perpendicular to the line x – 2y + 3 = 0 and passing through the points (1 ,- 2).
Answer:

Question 8.
Consider the line joining the points P(- 4,1) and Q(0,5) (IMP-2013)
i) Write the coordinate of the line passing through the midpoint of PQ .
ii) Find the equation of the line passing through the midpoint of PQ and parallel to the line 3x – 4y + 2 = 0
Answer:
i) Midpoint of PQ =
ii) The equation of the parallel line is of the form 3x – 4y + k = 0. Since it passes through (- 2,3) we have;
3(- 2) – 4 + k = 0
=> – 6 – 12 + A: = 0
=>k = 18
Hence the equation is 3x – 4y + 18 = 0

Question 9.
i) Find the slope of the line y = 2x – 3. (MARCH-2013)
ii) Find the equation of the line which makes intercepts – 3 and 2 on the X and Y axes respectively. Find its slope.
Answer:

Question 10.
Consider the lines 2x – 3y + 9 = 0 and 2x – 3y + 7 = 0
i) Find the distance from the origin to these two lines.
ii) Find the distance between these two lines.
Answer:

Plus One Maths Straight Lines 4 Marks Important Questions

Question 1.
i) Find the slope of the line \(\frac{x}{a}+\frac{y}{b}=1\) (IMP-2010)
ii) If the lines joining the points and are perpendicular (0,0), (1,1) and (2,2), (4,y) find y.
Answer:

Question 2.
Find slope of the line through the points (5,-1) and (6,4). (IMP-2011)
ii) Find the equation of the line through (5,-1) and (6,4).
iii) Find x intercept and y intercept of this line.
Answer:

Question 3.
i) Find the slope of the line joining the points (3,- 1) and (4,- 2). (IMP-2012)
ii) Find the angle between the positive x-axis and the line joining the points (3,- 1) and (4,- 2).
iii) Find the equation of the line joining the points (3,- 1) and (4,- 2).
Answer:
i)

ii) Slope=- 1
=>tanθ = – 1
=>θ = 135°
iii) Equation of the line joining the points (3,1) and (4,- 2) is y +1 = – 1(x – 3)
=>y + 1 = – x + 3
=> x + y = 2

Question 4.
i) Find the point of intersection of the lines 2x + y – 3 = 0,3x – y – 2 = 0.(MARCH-2012)
ii) Find the equation of the line passing through the above point of intersection and parallel to the linex + y + 1 = 0
Answer:
i)
2x + y = 3 ……..(1)
3x-y = 2 …………(2)
(1) + (2)
=> 5x = 5
=>x = 1
(1)=> y = 3 – 2 =1
Intersection point is (1,1)
ii) Equation of the parallel line x + y + k = 0 Since it passes through (1,1) we have;
1 + 1 + A = 0
=>k = -2
Equation is x + y- 2 = 0

Question 5.
Consider the line x + 3y – 7 = 0 (IMP-2013)
i) The slope of the line is ……….
ii) Find the image of the point (3,8) with respect to the given line.
Answer:
i) We have; x + 3>’-7 = 0
\(y=-\frac{1}{3} x+\frac{7}{3}\)
Hence the slope = – \(\frac { 1 }{ 3 }\)

ii) The equation of the perpendicular to the given and passing through (3,8) is
(y – 8) = 3(x – 3)
=>y – 8 = 3x – 9
=> 3x – y = 1
Solving
=> 3x – y = 1 and x + 3y = 7
we get the coordinate of D, which is the midpoint of the points (3,8) and (x,y).
3x – y = 1
3x + 9y = 21
– 10y = – 20
=> y – 2;
x + 3y = 7
=> x = – 3 + 7 = 1
Hence midpoint is (1,2)
Therefore the coordinate of the image is
\(\left(\frac{x+3}{2}, \frac{y+8}{2}\right)=(1,2)\)
=>x + 3 = 2
=>x = – 1
=>y + 8 = 4
=>y = – 4
Hence (- 1,- 4)

Question 6.
Find the slope of the line 3x – 4y + 10 = 0 (MARCH-2014)
ii) Find the equation of the line passing through the points (1,3) and (5,6).
iii) Find the equation of the line parallel to x – 2y + 3 = 0 and passing through the point (1,- 2).
Answer:

Question 7.
i) Find the slope of the line passing through the points (3,- 2) and (- 1,4). (MARCH-2014)
ii) Find the distance of the point (3,- 5) from the line 3x – 4y – 26 = 0
iii) Consider the equation of the line 3x – 4y + 10 = 0
Find its
a) Slope.
b) x and y intercepts.
Answer:

Question 8.
i) Find the equation of the line passing through (4,2) with a slope 2. (IMP-2014)
ii) Convert the above equation into intercept form. Find x and y intercepts.
Answer:
i) The equation of the line is
y – 2 = 2(x – 4)
=> y – 2 = 2x-8
=> 2x – y – 6 = 0
ii) Given
=> 2x – y – 6 = 0

x – intercept = 3;
y – intercept = – 6

Question 9.
i) Find the equation of the line passing through the two points (1,-1) and (3,5). (IMP-2014)
ii) Find the angles between the lines
y – √3x – 5 = 0 and √3y – x + 6 = 0
Answer:

Question 10.
Slope of the line L : 2x + 3y + 5 = 0 is. (IMP-2015)
a) \(\frac { 2 }{ 3 }\)
b) – \(\frac { 2 }{ 3 }\)
c) – \(\frac { 3 }{ 2 }\)
d) \(\frac { 3 }{ 2 }\)
ii) Find the equation of the line L’ parallel to L and passing through (2, 2). Find the distance of the lines L and L’
from the origin. Also find the distance between the lines L and L’.
Answer:

Question 11.
The slope of the line passing through the points (3,-2) and (7,-2) is (MARCH-2017)
(a) – 1
(b) 2
(c) 0
(d) 1
ii) Reduce the equation 6x + 3y – 5 = 0 into slope intercept form and hence find it slope and y-intercept.
iii) Find the point on the x-axis which equidistant from the points (7,6) and (3,4).
Answer:

Plus One Maths Straight Lines 6 Marks Important Questions

Question 1.
Reduce the equation3x + 4y – 12 = 0 into intercept form. (MARCH-2010)
Find the distance of it from the origin. Find the distance of the above line from the line 6x + 8y – 18 = 0
Answer:

Question 2.
Consider the straight line 3x + 4y + 8 = 0 (MARCH-2011)
i) What is the slope of the line which is perpendicular to the given line?
ii) If the perpendicular line passes through (2,3) from its equation.
iii) Find the foot of the perpendicular drawn from (2,3) to the given line.
Answer:

Question 3.
i) Find the equation of the line passing through the points (3,- 2) and (- 1,4). (MARCH-2015)
ii) Reduce the equation √3x + y – 8 = 0 into normal form.
iii) If the angle between two lines is \(\frac { π }{ 4 }\) and slope of one of the lines is \(\frac { 1 }{ 2 }\), find the slope of the other line.
Answer:

Question 4.
i) The Slope of a line ‘ L¹‘ making an angle 135° with direction of the positive direction of x-axis is (IMP-2015)
(a)1
(b)- 1
(c)√3
(d)-√3
ii) Find the equation of the line L² perpendicular to L¹ and passing through the point (- 2, 3).
iii) Find the equation of a line passing
through the intersection of x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0 .
Answer:
i) tan(135°) = tan(90 + 45) = -1
ii) Slope of line L²= 1
Equation of line L² passing through (- 2,3)
is y – 3 = 1(x + 2)
=>y – 3 = x + 2
=>x – y + 5 = 0
iii) let the equation line passing through the intersecting point is
x + 2y – 3 + k( 4x – y + 7) = 0
=> (1 + 4k)x + (2 – k)y – 3 + 7k = 0

Question 5.
Which one of the following pair of straight lines are parallel? (MARCH-2016)
a) x – 2y – 4 = 0;2x – 3y – 4 = 0
b) x – 2y – 4 = 0;x – 2y – 5 = 0
c) 2x – 3y – 8 = 0,3x – 3y – 8 = 0
d) 2x – 3y – 8 = 0;3x – 2y – 8 = 0
ii) Equation of a straight line is 3x – 4y + 10=0. Convert it into the intercept form and write the x-intercept and write the x-intercept and y- intercept.
iii) Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.
Answer:

iii) The equation of the perpendicular line will be 7x + y + k = 0.
Since x – intercept is 3, the line passes through the point (3,0). So we have;
7(3) + 0+k = 0
=>21 + 0 + k = 0
=>k = – 21
Therefore the equation is 7x + y – 21 = 0.

Question 6.
Which is the slope of the line perpendicular to the line with slope –\(\frac { 3 }{ 2 }\)?(MAY-2016)
(a) –\(\frac { 3 }{ 2 }\)
(b) –\(\frac { 2 }{ 3 }\)
(c) \(\frac { 3 }{ 2 }\)
(d) \(\frac { 2 }{ 3 }\)
ii) Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of origin with slope – 2.
iii) Assume that straight tines work as the plane mirror for a point, find the image of the point (1,2) in the line x – 3y + 4 = 0
Answer:
i) (d) \(\frac { 2 }{ 3 }\)
ii) Slope is m=- 2 and point is (- 3,0)
Equation is y – yx = mix – xx)
=> y – 0 = – 2(x + 3)
=> y = – 2x – 6
iii)

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 9 Sequences and Series.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 9 Sequences and Series

Plus One Maths Sequences and Series 3 Marks Important Questions

Question 1.
Consider the GP 3,3²,3³, _______. (IMP-2014)
i) Find the sum to n terms of this GP.
ii) Find the value of n so that the sum to n terms of this GP is 120.
Answer:
i)

ii)

Plus One Maths Sequences and Series 4 Marks Important Questions

Question 1.
Given sum of three consecutive terms in an AP is 21 and their product is 280  (IMP-2011)
i) Find the middle term of the above terms.
ii) Find the remaining two terms of the above AP.
Answer:
i) Let the three consecutive terms be
a-d, a, a + d
a-d + a + a + d = 21
=>3a = 21
=>a = 7
ii) Then the AP becomes 7 – d,7, 7 + d
Given product is 280;
(7 – d)(7)(7 + d) = 280
=> (7 – d)(7 + d) = 40
==> 49 – d² = 40
=> <d² = 9 => d= 3,- 3
Therefore the AP is 4,7,10 or 10,7,4.

Question 2.
Consider the GP 3,6,12  (IMP-2011)
i) Which term of this GP is 96?
ii) Find the value of n so that sum to n terms of this GP is 381.
Answer:

Question 3.
i) What is the sum of the first ‘n’ natural numbers?  (IMP-2012)
ii) Find the sum to ‘n’ terms of the series
3 x 8 + 6 x 11 + 9 x 14 + ______.
Answer:

Question 4.
If the sum of the first n terms of an Arithmetic progression is ——,where X and Y are constants, find  (IMP-2012)
i) S₁ and S₂
ii) The first term and common difference.
iii) The n^th term.
Answer:
i) S₁ = X
S₂ =2X + 1/2(2 – 1)Y=2X + Y
ii) First term = a, = Sx = X
S₂ =2 X + Y
=> a₁ +a₂ =2 X + Y
=> a₂ =2X + Y
=>a₂ = X+ Y
Common difference =
a₂ – a₁ =X + Y – X = Y
iii) n^thterm = a_n = a + (n-1)d = X + (n – 1)Y

Question 5.
Find the sum to n terms of the series;  (IMP-2012)
2² + 5² + 8² +_______
Answer:

Question 6.
i) Write the first four terms of the sequence whose nth term \(a_{n}=\frac{n}{n+1}\) (MARCH-2013)
ii) The sum of the first three terms of a GP is \(\frac {12}{13}\) and their product is -1. Find the common ratio and the terms.
Answer:

Question 7.
If the numbers \(\frac { 5 }{ 2 }\) x \(\frac { 5 }{ 8 }\) are three consecutive terms of a GP, then find x. (MARCH-2014)
Find the sum of the first n-terms of the series. 2 +22+222 + _____
Answer:

Question 8.
i) Find the 5^th term of the sequence whose nth term is \(a_{n}=\frac{n(n-2)}{(n+3)}\) (MARCH-2014)
ii) Write the sum of first n natural numbers.
iii) The 5^th, 8^th and 11^th terms of a GP are p, q and s respectively. Prove that q² – ps
Answer:

Question 9.
i) A man starts repaying a loan as a first instalment of Rs. 1,000. If he increases the instalment by Rs. 150 every month, what amount will he pay in the 30^th instalment?  (IMP-2014)
ii) Find the sum to n terms of the sequence:
7,77,777,7777 ______.
Answer:

Question 10.
i) Consider the AP 4,10,16,22…….. Find its common difference and the 7^th terms.  (IMP-2014)
ii) If the m^th term of an AP is \(\frac { 1 }{ n }\) and the nth term is \(\frac { 1 }{ m }\) , prove that the sum of the first ‘mn’ terms is \(\frac { 1 }{ 2 }\)(mn +1)
Answer:

Question 11.
The 6^th term of the sequence whose n^th term is \(t_{n}=\frac{2 n-3}{6}\) is _____. (MARCH-2015)
a) 3
b) \(\frac { 1 }{ 2 }\)
c) \(\frac { 3 }{ 2 }\)
d) \(\frac { 1 }{ 3 }\)
ii) Find the sum to infinity of the sequence 1,\(\frac { 1 }{ 3 }\) ,\(\frac { 1 }{ 9 }\), ………
iii) If a, b, c are in AP and \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\), prove that x, y, z are in AP.
Answer:

Plus One Maths Sequences and Series 6 Marks Important Questions

Question 1.
i) In an AP, the first term is 2 and the sum of the first five terms is one fourth the sum of the next five terms. (MARCH-2010)
a) Find the common difference.
b) Find the 20^th term.
ii) If AM and GM of two numbers are 10 and 8 respectively, find the numbers.
Answer:

Question 2.
i) In an AP if m^th term is ‘n’ and nth term is ‘m’ .find the (m + n)^th term.  (IMP-2010)
ii) If 3^rd, 8^th and 13^th terms of a GP are x,y,z respectively, prove that x,y,z are in GP.
iii) Prove that x,y,z in the above satisfies the equation \(\frac{y^{10}}{(x z)^{5}}=1\)
Answer:

Question 3.
Which of the following is the n^th term of an AP? (MARCH-2011)
a) 3 – 2n
b)n² – 3
c) 3ⁿ – 2
d) 2 – 3n²
ii) Find the 10th term of the sequence
– 6,- \(\frac { 11 }{ 2 }\), – 5,….
iii) The sum of the first three terms of a GP is \(\frac { 39 }{ 10 }\) and their product is 1. Find the common ratio and the terms.
Answer:

Question 4.
Find the 10^th term of an AP whose n^th \(\frac{2 n-3}{6}\) term is (MARCH-2012)
ii) Find the sum of the first 10 terms of the above AP.
iii) Find the sum of the first 10 terms of a GP, whose 3^rd term is 12 and 8^th term is 384.
Answer:

Question 5.
i) Find the 5^th term of the sequence whose n^th term, \(a_{n}=\frac{n^{2}-5}{4}\) (MARCH-2013)
ii) Find 7 + 77 + 777 +……. to n terms.
iii) Find the sum to n terms of the series.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………
Answer:

Question 6.
i) Find the sum of multiple of 7 between 200 and 400.  (IMP-2013)
ii) The sum of first 3 terms of a GP is \(\frac { 39 }{ 10 }\) and their product is 1. Find the terms.
Answer:

Question 7.
If ‘a’ is the first term and ‘cf is the common difference of an AP, then the nth term of the AP, a_n = ……. (MARCH-2014)
ii) In an AP, if the m^th‘ term is ‘n’ and the n^th term is ‘m’, where , prove that its p^th term is n + m – p.
iii) Find the sum to ‘n’ terms of the series:
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + _______.
Answer:

Question 8.
i) If the sum of certain number of terms of the AP 25,22,19 is 116, then find the last term.  (IMP-2014)
ii) Find the sum to n terms of the series
1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ………
(Imp (Science) – 2014)
Answer:

Question 9.
i) The 3^rd term of the sequence whose n^th term is (MARCH-2015)
ii) Insert three numbers between 1 and 256 so that the resulting sequence is a GP.
iii) If p^th term of an AP is q and q^th term is ‘p’, where p ≠ qfind r^th term.
Answer:

Question 10.
i) Geometric mean of 16 and 4 is ______.  (IMP-2015)
(a) 20
(b) 4
(c) 10
(d) 8
ii) Find the sum to n terms of the series: 5 + 55 + 555 + ________.
iii) Find the sum to n terms of the AP,
whose K^th term is a_k = 5K +1
Answer:

Question 11.
i) If the first three terms of an AP is x – 1,x + 1, 2x + 3, then x is  (IMP-2015)
(a)- 2
(b) 2
(c) 0
(d) 4
ii) Find the sum to n terms of the sequence.
1 x 2 + 2 x 3 + 3 x 4 + _______
iii) The nth term of the GP 5,- \(\frac { 5 }{ 2 }\),\(\frac { 5 }{ 4 }\),\(\frac { 5 }{ 8 }\),….. is \(\frac { 5 }{ 1024 }\) find ‘n’.
Answer:

Question 12.
The n^th term of the GP 5,25,125 (MARCH-2016)
is
(a) n⁵
(b) 5ⁿ
(c) (2n)⁵
(d) (5)²ⁿ
ii) Find the sum of .all natural numbers between 200 and 1000 which are multiples of 10.
iii) Calculate the sum of n-terms of the series whose n81 term is a_n = n(n + 3)
Answer:

Question 13.
i) Which among the following represents the sequence whose n^th terms is \(\frac { n}{ n+1 }\) ? (MAY-2017)
a) 1,2,3,4,5,6
b) 2,3,4,5,6
c) 2,\(\frac { 3 }{ 2 }\),\(\frac { 4 }{ 3 }\),\(\frac { 5 }{ 4 }\),\(\frac { 6 }{ 5 }\)
d) \(\frac { 1 }{ 2 }\),\(\frac { 2 }{ 3 }\),\(\frac { 3 }{ 4 }\),\(\frac { 4 }{ 5 }\),\(\frac { 5 }{ 6 }\)
ii) Using progression, find the sum of first five terms of the series 1 + \(\frac { 2 }{ 3 }\) + \(\frac { 4 }{ 9 }\) + …..
iii) Calculate: 0.6 + 0.66 + 0.666 + ………. n terms.
Answer:

Question 14.
The sum of the infinite series is 1, \(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 9 }\) ………is ________. (MARCH-2017)
(a) \(\frac { 3 }{ 2 }\)
(b) \(\frac { 5 }{ 2 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { 7 }{ 2 }\)
ii) Find the sum of all natural numbers between 100 and 1000 which is a multiple of 5.
iii) Find the sum to n terms of the series 8,88,888 ………
Answer:

Question 15.
The 6^th term of the GP \(\frac { 1 }{ 2 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 8 }\), ………. (MARCH-2017)
a) \(\frac { 1 }{ 32 }\)
b) \(\frac { 1 }{ 64 }\)
c) \(\frac { 1 }{ 16 }\)
d) \(\frac { 1 }{ 128 }\)
ii) The sum of 1st 3 terms of a G.P is \(\frac { 13 }{ 12 }\) and their product is – 1. Find the common ratio and terms.
iii) Find the sum to n terms of the series \(3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}\) + ………
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 8 Binomial Theorem.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 8 Binomial Theorem

Plus One Maths Binomial Theorem 3 Marks Important Questions

Question 1.
i) The number of terms in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\) is _____.(IMP-2013)
ii) Find the middle term in the above expansion.
Answer:
i) 11

Question 2.
i) Find the general term in the expansion of (x + y)ⁿ
ii) Find the middle term in the expansion of \(\left(2 x+\frac{1}{3 y}\right)^{18}\) (MARCH-2014)
Answer:

Question 3.
i) Write the general term in the expansion of (a + b)ⁿ 
ii) Find the 9th term in the expansion of \(\left(\frac{x}{2}+\frac{6}{x^{2}}\right)^{12}\) (IMP-2014)
Answer:

Plus One Maths Binomial Theorem 4 Marks Important Questions

Question 1.
i) Find the general term in the expansion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\) (MARCH-2010)
ii) Find the term independent of x in the above expansion.
Answer:

Question 2.
Consider the expansion of \(\left(x^{2}-\frac{1}{3 x}\right)^{9}\) (IMP-2010)
i) Find the coefficient of x⁹
ii) Find the term which is independent of x.
Answer:

Question 3.
Consider the expansion of \(\left(\frac{x}{9}+9 y\right)^{2 n}\) (MARCH-2011)
i) The number of terms in the expansion is _____
(a) 2n
(b) n+1
(c) 2n+1
(d) 2/7-1
ii) What is its (n+1)^th term?
iii) If n = 5, find its middle term.
Answer:

Question 4.
i) Write the general term in the expansion (1 + x)⁴⁴   
Write 21st and ,22nd terms in the expansion of (1 + x)⁴⁴
iii) If 21^st and 22^nd terms in the expansion of (1 + x)⁴⁴ are equal then find the value of x. (IMP-2011)
Answer:

Question 5.
8. Find(x + y)⁴ – (x – y)⁴. (IMP-2012)
Hence evaluate: (√5 + √6)⁴ – (√5 – √6)⁴
Answer:

Question 6.
i) How many terms are there in the expansion of (1 + x)²ⁿ (n is a positive integer)? (IMP-2012)
ii) Show that the middle term in the (1 + x)²ⁿ
expansion of is \(\frac{1.3 .5 \ldots(2 n-1)}{n !} 2^{n} x^{n}\)
Answer:

Question 7.
i) Find the general term in the expansion of \(\left(\frac{x}{2}-\frac{2}{x}\right)^{10}\) (MARCH-2012)
ii) Find the terms independent of x in the above expansion.
Answer:

Question 8.
i) Find the number of terms in the expansion of \(\left(x-\frac{1}{x}\right)^{14}\) (MARCH-2013)
ii) Find the general term in the expansion of \(\left(x-\frac{1}{x}\right)^{14}\)
iii) Find the term independent of x in the above expansion.
Answer:
i) 15

Question 9.
i)Write the number of terms in the expansion of (a -b)²ⁿ
ii) Find the general term in the expansion of \(\left(x^{2}-y x\right)^{12}, x \neq 0\) (MARCH-2014)
iii) Find the coefficient of x⁶y³ in the expansion of (x + 2y)⁹
Answer:

Question 10.
i) Write the expansion of (a + n)ⁿ, where n is any positive integer. (IMP-2014)
ii) Find the value of ‘a’ if the 17^th term and 18^th term in the expansion of (2 +a)⁵⁰ are equal.
Answer:

Question 11.
i) The number of term in the expansion of \(\left(x-\frac{1}{x}\right)^{2 n}\) is ______. (MARCH-2015)
(a) n+1
(b) n
(c) 2n+1
(d) 2n+2
ii) Find a, if the 17^th term and 18^th term of the expansion of (2 +a)⁵⁰ are equal.
Answer:

Question 12.
i) Number of terms in the expansion of \(\left(x+\frac{1}{x}\right)^{20}\) (IMP-2016)
(a) 19
(b) 20
(c) 21
(d) 22
Consider the expansion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\)
find the coefficient of x⁶ and the term independent of x.
Answer:

Question 13.
The 8th term in the expression (MARCH-2016)
of(√2 + √3)⁷ is
a) 27√2
b) 27√3
c) 72√2
d) 72√3
ii) Find the term independent of x in the expansion of \(\left(x+\frac{1}{2 x}\right)^{18} ; x>0\)
Answer:

Question 14.
Write the expansion of (a + b)^{4  (MAY-2017)}
Evaluate: (√5 + √6)4+ (√5 – √6)⁴
Answer:

Question 15.
Consider the expansion of \(\left(x+\frac{1}{x}\right)^{10}\) (MARCH-2017)
i) The number of terms in the expansion is _____.
(a) 10
(b) 9
(c)11
(d) 12
ii) Find the term which is independent of x in the above expansion.
Answer:

Plus One Maths Binomial Theorem 6 Marks Important Questions

Question 1.
i) Write the number of terms in the expansion of (a + b)ⁿ
ii) Expand \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\) (MARCH-2013)
iii) Find the general term in the expansion of \(\left(x^{2}-y\right)^{6}\)
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 7 Permutation and Combinations.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 7 Permutation and Combinations

Plus One Maths Permutation and Combinations 3 Marks Important Questions

Question 1.
i) if \({ }^{n} C_{9}={ }^{n} C_{8}\) find ‘n’ and \({ }^{n} C_{17}\) (IMP-2014)
ii) How many chords can be drawn, through 23 points on a circle?
Answer:

Plus One Maths Permutation and Combinations 4 Marks Important Questions

Question 1.
i) Simplify (MARCH-2011)

ii) In how many different ways can the letters of the word HEXAGON be permuted?
iii) In how many different ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
(March (Science) – 2011)
Answer:

Total number of ways = 10 x 4 = 40

Question 2.
i) If \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}\) (MARCH-2011)
then find x.
ii) How many 4 digit numbers are there with no digit repeated?
iii) If nC₈ = nC₂, then find nC₃ ?
Answer:

Question 3.
Consider all the letters of the word ‘FALIURE’.  (IMP-2011)
i) How many words can be formed using these letters?
ii) How many words can be formed so that the vowels being together?
iii) How many words begin with A and end with E?
(Imp (Commerce) – 2011)
Answer:
i) ‘FALIURE’ word has 7 letters in it, can be arranged in 7! Ways = 7.6.5.4.3.2.1 = 5040
ii)

A,E,I,U are vowels in the word, should be kept together, so should be treated as on block. Hence there are 4 such blocks can be arranged in 4! ways. These 4 vowels can be arranged in 4! Ways.
Hence the total words = 4! x 4! = 24 x 24 = 576
iii)

The only possible arrangement is for 5 blocks; hence total number of ways is 5! = 120.

Question 4.
i) Find the value of n if (IMP-2012)

ii) How many words, with or without meaning,
can be formed using all the letters of the word CHEMISTRY, using each letter exactly once? How many of them start with C and end with Y?
Answer:
i)

n —10,- 3
n = – 3 is not possible since negative son = 10
ii) Total number of words = 9!
Total number of words starting by C and ending by Y= 7!

Question 5.
i) If 2nC₃ : nC₃ = 12 : 1 find n. (IMP-2012)
ii) What is the total number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these four cards of the same suit?
Answer:

Question 6.
i) If nC₉ = nC₈, find nC₁₇ 
ii) A committee of 5 person is to be selected from a group of 4 men and 5 women. In how many ways can this be done? How many of these committees would consist of 2 men and 3 women? (IMP-2012)
Answer:

Number of different selection = 6 x 10 = 60

Question 7.
i) If nC₉ = nC₈, find nC₁₇ 
ii) How many three digit number can be formed using the digits 1,2,3,4,5 if repetition is not allowed? (MARCH-2013)
iii) In How many ways can a team of 4 boys and 3 girls be selected from 6 boys and 4 girls?
Answer:

Question 8.
i) If nC₅ = nC₄, find nC₈
ii) How many chords can be drawn through 20 points on a circle? (MARCH-2014)
iii) A bag contains 6 red and 5 blue balls. In how many ways can one choose 3 red and 2 blue balls from this bag?
Answer:

Question 9.
i) Find the number of permutation of the letters of the word ALLAHABAD.
ii) Find r,if \({ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}\) (IMP-2014)
Answer:
i) Total number of letters is 9.
A: 4 times; L: 2 times.

Therefore r = 3.

Plus One Maths Permutation and Combinations 6 Marks Important Questions

Question 1.
i) Find the value of n such that (MARCH-2010)
nP₅ = 42 x nP₃ for n > 4
ii) A committee of 3 persons is to be constituted from a group of 2 men and 3 women.
a) In how many ways can this be done?
b) How many of these committees would consist of 1 man and 2 women? (March (Science) – 2010)
Answer:
i) nP₅ = 42 x nP₂
=> n(n – 1)(n – 2 )(n – 3)(n – 4) = 42 x n(n -1)(n- 2)
=> (n – 3)(n – 4) = 42
=>n² – 7n + 12 = 42
=> n² – 7n – 30 = 0
=> (n -10)(n + 3) = 0
=> n = 10; n = – 3
n can’t be negative, so the acceptable value is n = 10
ii) a) 3 person can be selected from 5 in 5C₃ = 10
(b) 1 man can be selected from 2 in 2C₁ = 2 ways.
2 women can be selected from 3 in 3 C₂ = 3 ways.
Total ways = 2×3 = 6

Question 2.
i) lf \({ }^{n} C_{2}:{ }^{2 n} C_{1}=3: 2\), find n. (MARCH-2010)
ii) a) Find the number of words that can be
formed from the letters of the word MALAYALAM.
b) How many of these arrangements start with Y?
(March (Science) – 2010)
Answer:
i)

ii) a) MALAYALAM, this word has 9 letters
M- repeated 2 times.
A- repeated 4 times.
L- repeated 2 times.
Number of words formed by these 9 letters
= \(\frac{9 !}{2 ! \times 4 ! \times 2 !}\)
b) If the word starts with Y, then total number of letters that can be arranged become 8. Number of words formed which begin with Y
= \(\frac{8 !}{2 ! \times 4 ! \times 2 !}\)

Question 3.
i) if \(5 \times 4 P_{r}=6 \times 5 P_{r-1}\) find ‘r’, (IMP-2014)
ii) How many 3 digit number can be formed with the digits 0,1,2,3 and 4?
iii) In a Panchayath there are 10 Panchayath members. Ladies contested only in the 50 % reserved constituency. If the post president and vice president are reserved for ladies, in how many ways both the president and vice president can be selected?
(Imp (Science) – 2010)
Answer:
i)

ii)

Total 3 digit numbers = 4x5x5 = 100 (IMP-2010)
iii) The 10 member Panchayath has 5 men and 5 ladies. The president and vice president are to be selected from these ladies in 5C₁₂ = 10 ways.

Question 4.
i) Prove that nC_r = nC_n-r
ii) Twenty eight matches were played in a volley ball tournament. Each team playing one against each of others. How many teams were there? (IMP-2010)
iii) If the letters of the word ‘TUTOR’ be . permuted among themselves and arranged as in a dictionary, then find the position of the word ‘TUTOR’.
Answer:

Question 5.
A student is instructed to answer any 8 out of 12 questions. (IMP-2011)
i) How many different ways he can choose the questions?
ii) How many different ways he can choose the questions so that question no.1 will be included?
iii) How many different ways, he can choose the questions so that question no.1 will be included and question no.10 will be excluded?
Answer:
i) 8 out of 12 questions can be selected

ii) Since question no.1 is included, the possible is selection is from 11 questions and the number of questions to be selected becomes 7.
Hence the total selection

iii) Question no.10 will be excluded so total questions become 11. Question no.1 is included again total questions reduced to 10. Now we have to select 7 questions out of 10, can be done in 10C₇ = 10C₃ = \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) = 120.

Question 6.
Solve for the natural n; (MARCH-2012)
12.(n-1)P10C₃ =5.(n + 1)P10C₃
In how many ways seven althlets can be chosen out of 12?
iii) The English alphabets has 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed without repetition of letters?
Answer:
i)

ii) 7 athletes be chosen out of 12 in 12C₇ =12C₅ ways
iii) Two different vowels can be selected in 5C₂. Two different consonants can be selected in 21C₂.
Therefore total numbers of words

Question 7.
i) Find r if 5P_r = 6P_r-1.
ii) If there are 12 persons in a party and each of them shake hands with all others, what is the total number of handshakes? (MARCH-2012)
iii) In How many ways can a committee of 3men and 2 women be selected out of 7 men and 5 women?
Answer:

Question 8.
i) Find the value of n such that (MARCH-2013)
3.nP₄ = 5.(n -1 )P₄,n > 4
ii) In how many ways can 5 students be seated on a bench?
iii) Find the number of different 8-letter arrangements that can be made from the letters of the word, ‘DAUGHTER’ so that:
a) All vowels are occur together.
b) All vowels do not occur together.
Answer:

Question 9.
i) Determine n if 2nC₃ = 11.nC₃ 
ii) In how many ways can a cricket team of 11 of players be selected from 15 players? (MARCH-2013)
iii) A bag contains 5 white, 6 red and 4 blue balls. Determine the number of ways in which 2 white, 3red and 2 blue balls can be selected.
Answer:

Question 10.
i) The number of 3 digit numbers can be formed from the digits 1,2,3,4,5 assuming that repetition of the digits is not allowed is _______.
ii) If \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\), find x. (IMP-2013)
iii) How many words, with or without meaning, can be formed using all the letters of the word ‘FRIDAY’, using each letter exactly once? How many of them have first letter is a vowel?
Answer:

Question 11.
i) nC₇=nC₅,n =
ii) A bag contains 5 blue and 6 white balls. Determine the number of ways in which 3 blue and 4 white balls can be selected. (IMP-2013)
iii) What is number of choosing 3 cards from a pack of 52 playing cards? In how many of these 3 cards of the same colour?
Answer:

Question 12.
i) if \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}\) find x? (MARCH-2014)
ii) How many four digit numbers can be formed using the digits 4,5,6,7,8 if repetition of digits is not allowed?
iii) Find the number of arrangements that can be made from the letters of the word ‘MOTHER’ so that all vowels occur together.
Answer:

Question 13.
i) In how many ways can the letters of the word PERMUTATIONS be arranged if; (MARCH-2014)
a) the word starts with P and ends with S?
b) there are always 4 letters between P and S?
ii) In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
iii) How many chords can be drawn through 21 points?
Answer:
i) a) In the word T is repeated twice

b) P should move from 1 to 7th position and S should move from 6th to 12th position. Hence the arrangements

ii) First arrange 5 girls in 5P₅ ways. Arrange 3 boys in 6P₃.
Hence the total arrangements
= 5P₅ x 6P₃ = 14400
iii) Chord is the join of two points. Hence selection 2 points from 21, which can be done in 21C₂ =210

Question 14.
i) What is the minimum number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
a) are 4 cards of the same suit? (MARCH-2014)
b) do 4 cards belong to 4 different suits?
ii) Find the number of permutation of the letters of the word, ALLAHABAD.
iii) How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:

Question 15.
i) Find ‘n’ if \(9 \times^{(n-1)} P_{3}={ }^{n} P_{4}\) (IMP-2014)
ii) Find the number of words that can be formed from the letters of the word, COMMERCE’.
iii) In how many ways can a group of 12 students be selected from 15 students? How many of these groups would include one particular student?
Answer:

Question 16.
i) \(\frac{0 !}{1 !}\) = _____ (MARCH-2015)
a) 0
b) 1
c) 2
d) 3
ii) Find r, if \(5 \times^{4} P_{r}=6 \times^{5} P_{r-1}\)
iii) Find the number of 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.
Answer:

Question 17.
nC_{n – 1} = _____ (MARCH-2015)
(a) n-1
(b) n
(c) 0
(d) 1
If nC₉=nC₈,find nC₂
How many ways can a team of 5 persons be selected out of a group of 4 men and 7 women, if the team has at least one man and one women?
Answer:

Question 18.

i) \({ }^{7} P_{7}\) = _____ (IMP-2014)
a) 7
b) 7!
c) 1
d) 77
ii) Find the number of words that can be formed from the letters of the word “MALAYALAM”. How many of them start with Y?
iii) \({ }^{2 n} C_{3}=11 \times{ }^{n} C_{3}\) Find ’n’.
Answer:

iii)

Question 19.
i) \({ }^{29} C_{29}\) = ______ (IMP-2015)
(a) 0
(b) 1
(c) 2
(d) 3
ii) Prove that \({ }^{61} C_{57}-{ }^{60} C_{56}={ }^{60} C_{3}\)
iii) In how many ways can the letters of the word ‘ARRANGE’ be arranged such that two A’s do not occur together?
Answer:

Question 20.
Write the value of \({ }^{7} C_{5}\) (MARCH-2016)
Find the vale of n, if \(3 \times^{n} P_{4}=5 \times^{n-1} P_{4}\)
What is the number of ways of choosing 4 cards from a pack of 52 cards, provided all 4 cards belong to 4 different suits? (March (Science) – 2016)
Answer:

Question 21.
i) \({ }^{29} C_{29}\) = ______ (MARCH-2016)
a) 0
b) 1
c )2
d )3
ii) Find the value of n,
if \(12 \times^{n-1} P_{3}=5 \times^{n+1} P_{3}\)
iii) A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least one boy and one girl?
Answer:

Question 22.
i) How many 4 digit numbers can be formed using the digits 9, 8, 7, 6, 5, 4, if no digits are repeated? (MAY-2017)
(a) 630
(b) 603
(c) 306
(d) 360
ii) In how many ways a committee of 3 persons can be formed from a group of 2 men and 3 women?
iii) Find the value of n,
if \({ }^{2 n} C_{3}=11 \times{ }^{n} C_{3}\)
Answer:

Question 23.
i) \({ }^{569} \mathrm{C}_{569}\) = ______. (MAY-2017)
ii) \({ }^{2 n} C_{3}:{ }^{n} C_{3}=12: 1\) Find n.
iii) If the letters of the word EQUATION are arranged, find the number of arrangements in which no two consonants occur together?
Answer:
i) 1

Question 24.
i) if \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\) , then x is ______.
(a) 32
(b) 16
(c) 64
(d) 8
ii) Given 5 flags of different colour, how many different signals can be generated if each other.
iii) Find r, if \({ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}\)
Answer:
i) c) 64
ii) Number of ways = 5 x 4 = 20

Question 25.
i) lf nC₉=nC₈, then n = ______.
a) 9
b) 8
c) 17
d) 1
ii) How many chords can be drawn through 12 point on a circle?
iii) What is the number of way of choosing 4 cards from a pack 52 playing cards? In how many of these:
a) Four cards are of the same suit.
b) Cards are of the same colour.
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 6 Linear Inequalities.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 6 Linear Inequalities

Plus One Maths Linear Inequalities 4 Marks Important Questions

Question 1.
i) Draw the graphs of 2x + 3y = 24 and x + y = 9 (IMP-2011)
ii) Solve the following system of inequalities graphically;
2x + 3y ≤ 24,
x + y ≤ 9, x, y ≥ 0
Answer:

Question 2.
i) Solve 4x – 5 < 7, when x is a real number. (IMP-2012)
ii) Solve the following system of inequalities
graphically. 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Answer:
i) 4x – 5 < 7 => 4x < 12 => x < 3
ii) 3x + 4y ≤ 12, x ≥ 0, y ≥ 0

Question 3.
Solve: 4x + 3 < 3x + 7 represent the solution on the real line. (MARCH-2013)
ii) Solve the following system of inequalities graphically.
3x + 2y ≤ 12;
x,y ≥ 0
Answer:
i) 4x + 3 < 3x + 7 => 4x – 3x < 7 – 3 => x < 4
ii)

Question 4.
i) Represent the inequality x ≥ – 3 on a number line. (IMP-2014)
ii) Solve the following inequalities graphically:
x + y≥5;
x – y≤ 3
Answer:

Question 5.
The interval representing the solution of the inequality 3x-1 ≥ 5, x∈R is (MARCH-2015)
a) [5,∞) b) [2, ∞)
c) [3,∞) d) (— ∞, ∞)
ii) Solve the system of inequality graphically
x + 2y ≤ 8,2x + y ≤ 8,
x ≥ 0,y ≥ 0
Answer:

Question 6.
i) Which among the following is the interval corresponding to the inequality – 2 < x ≤ 3 . (MARCH-2016)
(a) [- 2,3]
(b) [- 2,3)
(c) (- 2,3]
(d) (- 2,3)
ii) Solve the following equation.
2x + y ≥ 4;
x + y ≤ 3;
2x – 3y ≤ 6.
Answer:

Question 7.
i) Which among the following inequality represents the intervals [2,∞)
a) x – 3 ≥ 5, x∈R
b) 3x – 3 ≥5, x∈R
c) 3x – 1≥ 3, x∈R
d) 3x – 1 ≥ 5, x∈R
ii) Solve the following system of inequalities graphically. 3x + 2y ≤ 12; x ≥ 1; y ≥ 2
Answer:

Plus One Maths Linear Inequalities 6 Marks Important Questions

Question 1.
i) Solve the inequality 3(x – 1) ≤ 2(x – 3)  (MARCH-2010)
ii) Solve the following system of inequalities graphically. 5x + 4y ≤ 20; x ≥ 1, y ≥ 2
Answer:
i) 3(x – 1) ≤ 2(x – 3) =>3x – 3 ≤2x – 6
=>3x – 2x ≤ 3 – 6
=> x ≤ – 3
ii) 5x + 4y = 20

Question 2.
i) Arathi took 3 examinations in an year. The marks obtained by her in the second and third examinations are more than 5 and 10 respectively than in the first examination. If her average mark is at least 80 find the minimum mark that she should get in the final examinations? (IMP-2010)
ii) Solve the following system of inequalities graphically 2x + y ≥6; 3x + 4y ≤12
Answer:
Let x denote the marks of arathi in first examination. then mark in second exam and third exam are x +5 and x + 10 respectively. Given average in three examinations is atleast 80.

Question 3.
i) Solve the inequality
2(2x + 3) – 10 < 6(x – 2) (MARCH-2011)
ii) Solve the following inequalities graphically. system of
x – 2y < 3;
3x + 4y ≥ 12; x,y ≥ 0
Answer:
i) 2(2x + 3) – 10 < 6(x – 2)
=> 4x + 6 – 10 ≤ 6x – 12
=> – 2x ≤ -12 + 4
=> – 2x ≤ – 8
=> x ≥ 4
ii)

Question 4.
i) Find all pairs of consecutive odd natural numbers, both of which are smaller than 10, such that their sum is more than 11. (IMP-2012)
ii) Solve 2x + y ≤ 6 graphically.
Answer:
i) Consecutive odd natural numbers be x and x+2. Then,
x + x + 2 > 11;
x + 2 < 10
=> 2x > 11 – 2;
x < 10 – 2
=> x > 9/5 = 4.5;
x < 8
5 ≤ x < 8 Therefore x can take values 5,7.
Hence the pairs are (5,7),(7,9)
ii)

Question 5.
i) Solve the inequality: 3(2 – x) ≥ 2(1 – x)  (MARCH-2013)
ii) Solve the following system of inequalities graphically.
2x + y ≥ 4;
x + y ≤ 3;
2x – 3 ≤ 6
Answer:
i) 3(2 – x) ≥ 2(1 – x)
=> 6 – 3x ≥ 2 — 2x
– 3x + 2x ≥ 2 – 6
=>- x ≥ – 4
=> x ≤ 4
ii)

Question 6.
i) Solve: 5x – 3 < 3x + l (MARCH-2014)
ii) Solve the following inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x,y ≥ 0
Answer:

Question 7.
i) Raju obtained 70 and 60 marks in first two examinations. Find the minimum mark he should get in the third examination to have an average of atleast 50 marks. (IMP-2013)
ii) Solve the following system of inequalities graphically.
3x + 2y ≤ 12;
x ≥ 1;
y ≥ 2
Answer:
i) Let x be the mark obtained by Raju in third exam. Then,

Question 8.
i) Solve: 5x + 3 < 2x + 7 represent the solution on the real line. (MARCH-2014)
ii) Solve the following system of inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x, y ≥ 0
Answer:

Question 9.
i) Solve: 7x + 3 < 5x + 9 represent the solution on the real line. (MARCH-2014)
ii) Solve the following system of inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x, y ≥ 0
Answer:
i)

ii)
15

Question 10.
i) Solve 10x – 23 < 3x + 5
ii) Solve the following system of inequalities graphically: 3x + 5y ≤ 15; 5x + 2y ≤ 10; x,y ≥ 0 (IMP-2014)
Answer:

Question 11.
i) Solve; 7x + 3 ≤ 5x + 9; x∈R . Express the solution on a number line.  (IMP-2015)
ii) Solve graphically; 3x + 4y ≤ 60;
x + 3y ≤ 30;
x,y ≥ 0.
Answer:

Question 12.
Solve the inequality \(\frac{x}{3}>\frac{x}{2}+1\) (MARCH-2017)
Solve the system of inequality graphically
2x + y > 6,3x + 4y ≤ 12
Answer:
i) 2x > 3x + 6
=> – x > 6
=> x < – 6
ii)

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 5 Complex Numbers and Quadratic Equations.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 5 Complex Numbers and Quadratic Equations

Plus One Maths Principle of Complex Numbers and Quadratic Equations 3 Marks Important Questions

Question 1.
Find the modulus and argument of the complex number \(\frac{1+i}{1-i}\) . Find its multiplicative inverse in the form a + ib (IMP-2012)
Answer:
Convert into a + ib form

Hence Modulus is 1 and argument is \(\frac{\pi}{2}\)
Multiplicative inverse of i is \(\frac{1}{i}=\frac{1}{i} \times \frac{-i}{-i}=-i\)

Plus One Maths Principle of Complex Numbers and Quadratic Equations 4 Marks Important Questions

Question 1.
i) Express the complex number \(z=\frac{5 + i}{2 + 3i}\) in the form of a + ib. (MARCH-2010)
ii) Represent z in the polar form.
Answer:

Question 2.
consider the complex number (MARCH-2011)

i) Express z in the form of a + ib
ii) Represent z in the polar form.
Answer:

The complex number lies in the first quadrant;

Question 3.
i) Express \(\frac{1}{1-i}\) in the form of a + ib  (IMP-2011)
ii) Express \(\frac{1}{1-i}\) in the polar form.
Answer:

Question 4.
Represent the complex number 1 + i√3 in the polar form.  (IMP-2012)
Express \(\frac{2 + i}{2 – i}\) in the form of a + ib.
Answer:

Question 5.
Consider the complex number  (MARCH-2012)

i) Express complex number in the form of a + ib.
ii) Express complex number in the polar form
Answer:

Question 6.
i) Express the following complex number in the form a + ib  (MARCH-2013)
(1 +i) – (1 – 6i) + (2 + i)
ii) Represent the complex number z = 1 + i in the polar form.
Answer:

Question 7.
i) Represent the complex number √3+ i in the polar form.  (MARCH-2013)
ii) Solve : √5x² + x + √5 = 0
(March (Science) – 2013)
Answer:

Question 8.
i) Express \(\frac{1+i}{1-i}\) in the form a + ib.  (IMP-2013)
ii) Represent the \(\frac{1+i}{1-i}\) in the polar form.
Answer:

Question 9.
i) Solve the quadratic Equation – x² + x – 2 = 0  (IMP-2014)
ii) Express ‘i’ in the form r(cosθ+i sinθ )
Answer:

Plus One Maths Principle of Complex Numbers and Quadratic Equations 6 Marks Important Questions

Question 1.

a) a + ib form. (IMP-2010)
b) polar form.
Answer:

Question 2.
i) If z = √3 + i , find the conjugate of Z. (IMP-2010)
ii) Write the polar form of the complex number z = √3 + i
iii) Solve 2x² + 3x + 1 = 0
Answer:

Question 3.
i) Solve: √3x² + x + √3 = 0 (MARCH-2014)
ii) Represent the complex number z = 1 + i √3 in the polar form.
Answer:

Question 4.
The conjugate of 1 – 2i is _______. (IMP-2014)
ii) Express the complex number \(\frac{2 + 3i}{1 – 2i}\) in the form a + ib .
iii) Solve x² + 3x + 5 = 0
Answer:

Question 5.
i) Represent the complex number 1 + √3i in the polar form. (MARCH-2015)
ii) Find the square root of the complex number – 7 – 24i.
Answer:

ii)

Since the product xy is negative, we have x = 3, y = – 4 or, x = – 3, y = 4
Thus, the square roots of – 7 – 24i are 3 – 4i and – 3 + 4i.

Question 6.
i) What is i ^{– 35} ? (IMP-2015)
a) i
b) -i
c) 1
d) -1
ii) Express the complex number √3 + i ’ in the polar form.
iii) Solve: √5x² + x + √5 = 0
Answer:
i) a) i
ii)

The complex number lies in the first quadrant;

iii)

Question 7.
i) Which one of the following is the real part and imaginary part of the complex number: (MARCH-2016)

a) 0 and 1
b) 0 and 2
c) 3 and 2
d) 0 and 4
ii) Express the complex number ‘ i ’ in the polar form.
iii) Solve: √5x² + x + √5 = 0
Answer:
i) b) 0 and 2
ii)

iii)

Question 8.
i) Write the real and imaginary part of the complex number – 3 + √- 7 (MAY-2017)
ii) Find the modulus and argument of the complex number 1 + i√3
iii) Solve: x² – 2x + 3 = 0
Answer:
i) Real part is – 3 and imaginary part is √7
ii)

iii)

Question 9.
i) i¹⁸= ________
a) 1
b) 0
c) – 1
d) i
ii) complex number in polar form √3 + i
iii) Find the square root of the complex number – 8 – 6i.
Answer:
i) c) – 1
ii)

The complex number lies in the first quadrant;

iii)

Since the product xy is negative, we have x = 1, y = – 3 or, x = – 1, y = 3 Thus, the square roots of – 8 – 6i are 1 – 3i and – 1 + 3i.

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 4 Principle of Mathematical Induction.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction 3 Marks Important Questions

Question 1.
Consider the statement “p(n): 9ⁿ – 1 is a multiple of 8 ”. Where n is a natural number. (MARCH-2011)
i) is p true?
ii) Assuming p(k) is true, Show that p(k+1) is true.
Answer:

Question 2.
Consider the statement “ P(n): xⁿ – yⁿ is divisible by x – y” (MAY-2017)
i) Show that P is true.
ii) Using the principle of Mathematical induction verify that P(n) is true for all natural numbers.
Answer:

Hence divisible by (x – y).
Therefore by using the principle of mathematical induction true for all n ∈ N .

Plus One Maths Principle of Mathematical Induction 4 Marks Important Questions

Question 1.
Consider the statement “ 7ⁿ – 3ⁿ is divisible by 4” (MARCH-2010)
i) Verify the result for n = 2.
ii) Prove the statement using mathematical induction.
Answer:


Therefore p(k +1) is multiple of 4. Hence p(n) is true for all natural number n.

Question 2.
i) Which among the following is the least number that will divide 7²ⁿ-4²ⁿ for every positive integer n? [4,7,11,13] (IMP-2010)
ii) Prove by mathematical induction.
(cos θ + i sin θ)ⁿ = (cosnθ + i sin nθ)
Answer:

Therefore true for p(k +1). Hence p(n) is true for all natural number n.

Question 3.
Given P(n): 3²ⁿ -1 is divisible by 8. (IMP-2011)
i) Check whether P(1) is true.
ii) If P(k) is true then prove P(k+1) is true.
iii) Is the statement P(n) true for all natural numbers? Justify your answer.
Answer:
i) P(1):3²⁽¹⁾ = 9 -1 = 8 divisible by 8, hence true.
ii) P(k) : 3^2k -1 is divisible by 8.
3^2k– 1 = 8M .
M is a positive real
P(k + 1):3^2k+1 – 1= 3^2k+2 – 1
= 3^2k3² – 1
= 3^2k x 9 – 1
=3^2k x 9 – 9 + 8
= 9(3^2k – 1) + 8
= 9(8M) + 8
Hence divisible by 8
iii) By PMI; true for all natural number n.

Question 4.
Prove that by 1.2 +2.3 + 3.4 + + n(n +1) = \(\frac{n(n+1)(n+2)}{3}\) by using the principle of mathematical induction for all n∈N. (IMP-2012)
Answer:

Hence by using the principle of mathematical induction true for all n∈N .

Question 5.
By the principle of Mathematical Induction, Prove that (IMP-2012)

Answer:


Hence by using the principle of mathematical induction true for all n∈N .

Question 6.
Consider the statement P(n): n(n+1)(2n+1) is divisible by 6 (MARCH-2012)
i) Verify the statement for n = 2.
ii) By assume that P(k) is true for a natural number k, verify that P(k+1) is true.
Answer:
P : 2(2 + 1)(2 x 2 + 1) = 2 x 3 x 5 = 30
Which is divisible by 6.
ii) Assuming that true for p(k)
p(k): k(k +1)(2k +1) is divisible by 6.
k(k +1)(2k +1) = 6M, M is a positive real
Let p(k +1): (k +1)(k + 2)(2(k +1) +1)
= (k + 1)(k + 2)(2k + 3)
= {k + 1){2k² +7k + 6)}
= {(k + 1){(2k²+k) + (6k + 6)}
= k(k + 1)(2k +1) + 6(k + 1)(k +1)
= 6M + 6(k + 1)(k +1)
Hence divisible by 6. Therefore by using the principle of mathematical induction true for all n∈N .

Question 7.
Consider the statement (MARCH-2013)

i) Check whether P is true.
ii) By assume that P(k) is true, prove that P(k+1) is true.
iii) Is P(n) true for all natural number n?
justify your answer.
Answer:


iii) Hence by using the principle of mathematical induction true for all n∈ N .

Question 8.
Consider the statement (MARCH-2013)

i) Check whether P is true.
ii) If P(k)is true, prove thatP(k+1) is true.
iii) Is P(n) true for all natural number n?
Justify your answer.
Answer:


iii) By PMI; true for all natural number n.

Question 9.
Consider the statement (IMP-2014)

i) Verify the result for n = 2.
ii) Prove the statement using mathematical induction.
Answer:

Hence by using the principle of mathematical induction true for all n∈ N

Question 10.
Consider the statement (MARCH-2014)
P(n) : 1.2 + 2.3 + 3.4 +……… + n(n +1) = \(\frac{n(n+1)(n+2)}{3}\)
i) Prove that P is true.
ii) Assume that P(k) is true for a natural number k, verify that P(k+1) is true.
Answer:

Therefore p(k+1) is true.

Question 11.
Consider the statement (MARCH-2014)
P{n) = 3²ⁿ⁺² – 8n – 9 is divisible by 8
i) Verify the statement for n = 1.
ii) Prove the statement using the principle of mathematical induction for all natural numbers.
Answer:
i) P(1) = 3²⁺² – 8 – 9 = 64 is divisible by 8 . hence true
ii) Assuming true for P(k) is divisible by 8
P(k) = 3^2k+2 – 8k – 9=8M,
M is a positive real
Let P(k +1) = 3^2(k+1)+2 – 8(k) – 9
3^2k+2+2 – 8k – 8 – 9
3^2k+2 x 3² – 8k – 17
= (8M + 8k + 9)9 – 8k – 17
= 72M + 12k + 81 – 8k – 17
= 72M + 64k + 64 = 8(9M + 8k + 8)
Therefore p(k +1) is divisible by 8. Hence p(n) is true for all natural number n.

Question 12.
Consider the statement:(IMP-2014)

i) Prove that P is true.
ii) If P(k) is true, Prove that P(k+1) is true
iii) Is P(n) true for all natural number n? Why?
Answer:
11 i,ii

Question 13.
Using the principal of mathematical induction, prove that

Answer:

It is true.
Therefore P(k+1) is true when ever P(k) is true. By PMI P(n) is true for all n.

Question 14.
A statement p(n) for a natural number n is given by (MARCH-2015)

i) Verify that p(1) is true.
ii) By assuming that p(k) is true for a natural number k, show that p(k+1) is true.
Answer:

Question 15.
Consider the statement (IMP-2015)
P(n): 7ⁿ-3ⁿ is divisible by 4.
i) Show that P(1) is true.
ii) Verify, by the method of Mathematical induction, that P (n) is true for all natural numbers.
Answer:

Question 16.
Consider the following statement:

i) Prove that P is true.
ii) Hence by using the principle of mathematical induction, prove that P(n) is true for all natural numbers n. (MARCH-2016)
Answer:

Hence by PMI the result is true for all natural numbers.

Question 17.
Consider the statement “10^2n-1+1 is divisible by 11” verify that P is true and prove the statement by using Mathematical induction.
(MARCH-2017)
Answer:
p(1): 10¹ + 1 = 11 divisible by 11, hence true. Assuming that true for p(k)
p(k): 10^{2k – 1} + 1 is divisible by 11.
10^{2k – 1} + 1 = 11M
p(k + 1):10^{2(k+1) – 1} + 1
= 1o^{2k + 2 – 1} +1
=10^{2k – 1} x 10² + 1
= 10^{2k – 1} x 100 + 1
(11M – 1)100 + 1
= 1100M – 100+1
= 1100M – 99
= 11(100M – 9)
Hence p(k+1)divisible by 11. Therefore by using the principle of mathematical induction true for all n ∈ N .

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 3 Trigonometric Functions.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions 3 Marks Important Questions

Question 1.
i) Find the degree measure corresponding to \(\frac { 11 }{ 14 }\) radians.( use π = \(\frac { 22 }{ 7 }\)) (MARCH-2010)
ii) If cos x = \(\frac { -1 }{ 2 }\), x lies in the third quadrant,find sin x and tan x
Answer:
i) Degree measure corresponding to \(\frac { 11 }{ 14 }\) radians
= \(\frac{11}{14} \times \frac{180}{\pi}=\frac{11}{14} \times \frac{180 \times 7}{22}=\frac{1}{2} \times \frac{180}{2}=45^{\circ}\)
ii) since x lies in the third quadrant

sin x = –\(\frac{\sqrt{3}}{2}\)
tan x = √3

Question 2.
prove that (MARCH-2010)

Answer:

Question 3.
Showthat (cos x + cos y)² +(sin x + sin y)² =\(4 \cos ^{2}\left(\frac{x-y}{2}\right)\) (MARCH-2011)
Answer:
(cos x + cos y)² + (sin x + sin y)²
= cos² x + cos² y + 2cosx cosy + sin² x + sin² y + 2 sin x sin y
= 1+ 1 + 2(cosxcosy + sinxsiny)
= 2 + 2cos(x – y) = 2(1 + cos(x – y))
= \(4 \cos ^{2}\left(\frac{x-y}{2}\right)\)

Question 4.
prove that (MARCH-2013)

Answer:

Question 5.
Consider the trigonometric equation tan x = √3 (IMP-2013)
i) Write the general solution.
ii) Write the principal solution.
Answer:

Question 6.
i) The value of sin(π – x) is _______. (MARCH-2014)
ii) Find the principal and general solution of the equation sin x = \(\frac { √3 }{ 2 }\)
Answer:
i) sin x
ii)

Question 7.
prove that (MARCH-2014)

Answer:

Question 8.
i) 1 + tan² x = _______. (IMP-2014)
ii) If sin x = \(\frac { 3 }{ 5 }\) and x lies in the second quadrant, find the values of cosx, tan x and secx.
Answer:
i) sec² x
ii)

Plus One Maths Trigonometric Functions 4 Marks Important Questions

Question 1.
Expand cos(x + y) and hence prove (IMP-2010)
i) cos 2x = 1 – 2sin² x
ii) Solve the equation tan² θ + cot² θ = 2
Answer:
i) cos(x + y) = cos x cos y — sinx sin y
Put y = x
cos(x + x) = cos x cosx – sin x sin x
=> cos(2x) = cos² x – sin² x
=> cos(2x) = 1 – sin² x – sin² x = 1 – 2sin² x
ii)

Question 2.
Show that (IMP-2010)

Answer:

Question 3.
i) Find the value of sin(\(\frac { 31π }{ 3 }\)) (MARCH-2011)
ii) Find the principle and general solution of the equation cos x = \(\frac {-√3 }{ 2 }\)
Answer:

Question 4.
Solve: sin 2x – sin 4x + sin 6x = 0 (IMP-2012)
Answer:
sin 2x + sin 6x – sin 4x = 0
=> 2sin4xcos2x – sin4x = 0
=> sin4x(2cos2x – 1) = 0
=>sin4x = 0 or (2cos2x – 1) = 0

Question 5.
tan x tan 2x tan 3x = tan 3x – tan 2x – tan x (IMP-2012)
Answer:
We have; 3x= 2x + x
Take tan on both sides;

tan 3x(1 – tan 2x tan x) = tan 2x + tan x
tan 3x- tan 3x tan 2x tan x = tan 2x + tan x
tan x tan 2x tan 3x = tan 3x – tan 2x – tan x

Question 6.
i) Evaluate tan(\(\frac {13π }{ 6 }\)) (MARCH-2012)
ii) If tan x = \(\frac {1 }{ 2 }\) and x is in the third quadrant, find sinx and cosx.
Answer:

Question 7.
Prove that (MARCH-2012)

Answer:

Question 8.
i) \(\frac {tan x + tan y}{ 1 – tan x tan y }\) = _________. (MARCH-2013)
ii) Prove that

Answer:
i) tan(x + y)
ii)

Question 9.
Match the following: (MARCH-2013)

Answer:

Question 10.
i) Prove that (MARCH-2013)

ii) Prove that

Answer:

Question 11.
Show that (IMP-2013)
i) tan 15°=2-√3
ii) tan 15°+cot 15° = 4
Answer:
i)

ii)

Question 12.
i) Prove that (MARCH-2014)

ii)

Answer:
i)

Question 13.
i) If tan x = \(\frac {3 }{ 4 }\); x lies in the third quadrant,
find the value of cos x. (MARCH-2014)
ii) Find the principal and general solution of cos x = \(\frac {1 }{ 2 }\)
Answer:
i) Since x lies in the third quadrant

Plus One Maths Trigonometric Functions 6 Marks Important Questions

Question 1.
i) Write the value of (IMP-2011)
sin 600°; cos 330°; cos 120°; sin 150°
ii) Prove that
sin 600° cos330° +cos120° sin 150° + sin 180° cos 180° = -1
Answer:

Question 2.
i) Find the value of sin 75° (IMP-2012)
ii) Prove that

Answer:

Question 3.
i) \(\frac {2π }{ 3 }\) radians = ______ degree.(IMP-2014)
ii) cos(2π – x) = _______
iii) Find the general solution of sin 2x – sin 4x + sin 6x = 0
Answer:
i) 120°
ii) cos x
iii)

Question 4.
i) sin x cos y + cos x sin y = ______. (IMP-2014)
ii) Find sin 50° cos 10° + cos50° sin10°
iii) Prove that

Answer:
i) sin(x + y)
ii) sin 50° cos 10° + cos 50° cos 10°
= sin(50° +10°) = sin(60°) = \(\frac {√3 }{ 2 }\)
iii)

Question 5.
i) Which one of the following values of sin x is incorrect? (MARCH-2015)
a) 0
b) \(\frac {1 }{ 2 }\)
c) 3
d) 1
ii) Prove that

iii) A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Answer:
i) 3
ii)

iii)

Question 6.
i) sin 225° = _______.(MARCH-2015)

ii) Find the principle and general solutions of sin x = – \(\frac {√3 }{ 2 }\)
iii) Prove that

Answer:
i) \(\frac {- 1 }{ √2 }\)
ii)

iii)

Question 7.
i) Which of the equal to 520° ? (IMP-2015)

ii) Solve Sin 2x – Sin 4x + Sin 6x = 0.
iii) In any triangle ABC, prove that

Answer:
i) \(\frac { 26π }{ 6 }\)
ii)

iii)

Question 8.
i) The degree measure of \(\frac { 7π }{ 6 }\) radian is _____. (MARCH-2016)
(a) 120° (b) 102° (c) 201° (d) 210°
ii) Prove that

iii)A lamp post is situated at the middle point M of the side AC of a triangular plot ABC with BC = 7m, CA = 8m, AB =9m. Lamp post subtends an angle 15° at the point B. Determine the height of the lamp post.
Answer:
i) 210°
ii)

iii)

Question 9.
i) 40°20′ = ____ radians (MAY-2016)
a) \(\frac { 112π }{ 540 }\)
b) \(\frac { 211π }{ 540 }\)
c) \(\frac { 122π }{ 540 }\)
d) \(\frac { 121π }{ 540 }\)
ii) Prove that

iii) Solve sin 2x – sin 4x + sin 6x = 0
Answer:
i)
d) \(\frac { 121π }{ 540 }\)
ii)

iii)

Question 10.
i) sin 405°= _____. (MARCH-2017)

ii)
sin x = \(\frac { 3 }{ 5 }\)
x lies in the second quadrant.Find the values of cosx,secx,tanx,cotx
iii) Solve: sin 2x – sin 4x + sin 6x = 0
Answer:
i) a) \(\frac { 1 }{ 2 }\)
ii)

iii)

Question 11.
i) \(\frac { 7π }{ 6 }\) radian = ______ degree. (MARCH-2017)
a) 200
b) 300
c) 240
d) 120
ii) Find the value of tan 75°
iii) In a triangle ABC, prove that
a sin(B – C)+b sin(C -A)+c sin (A – B) = 0
Answer:
i) a) 210
ii)

iii)

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 2 Relations and Functions.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 2 Relations and Functions

Plus One Maths Relations and Functions 3 Marks Important Questions

Question 1.
Let A = {1,2,3,4} and B = {1,4,5} be two sets. (IMP-2011)
If R is the relation”<” from A to B then )
i) Write R in Roster form.
ii) Write Domain and Range of R.
iii) Find the number of relations from AtoB.
Answer:
i) R = {(1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (4,5)}
ii) Domain of R = {1,2,3,4}; Range of R = {4,5}
iii) Number of relations from A to B = 2^{4 x 3} =2¹²

Question 2.
\(\left(\frac{2 x}{5}+1, y-\frac{3}{4}\right)=\left(\frac{1}{5}, \frac{1}{4}\right)\) find xand y (IMP-2012)
Answer:

Question 3.
Let A = {1,2,3}; B = {3,4}. Write a relation from A and B having 5 elements. Write its domain, co-domain and range. (IMP-2012)
Answer:
R = {(1,3), (1,4), (2,3), (3,4), (3,3)}
Domain of R = {1,2,3}
Range of R = {3,4}
Co-domain of R = {3,4} = B

Question 4.
The function f is defined by (IMP-2012)

Draw the graph of Find f(x)
Answer:
For x < 0

Question 5.
5. Let A = {1,2,3}, B={4,5} (MARCH- 2013)
i) Find A x B and B x A
ii) Find the number of relations from A to B.
Answer:
i) A x B = {(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)} B x A = {(4,1), (4,2), (4,3), (5,1), (5,2), (5,3)}
ii) n(R) = 2^{2 x 3} = 2⁶ = 64

Question 6.
i) Let A = {7,8} and B = {2,4,5}. Find A x B (MARCH – 2013)
ii) Determine the domain and range of the relation R defined by R={(x,y):y=x+1},
x ∈ {0,1,2,3,4,5,6}
Answer:
i) A x B = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
ii) R= {(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)}
Domain = {0,1,2,3,4,5}
Range = {1,2,3,4,5,6}

Question 7.
i) If A = {2,4}, B = {1,3,5}. Then the number of relations from A to B is …………. (IMP-2013)
ii) If P={-1,1}, form the set P x P x P
Answer:
i) number of relations from A to B 2^{2 x 3} = 2⁶ = 64
ii) P x P = {-1,1} x {-1,1} – {(-1,-1),(-1,1), (1-1), (1,1)}
P x P x P={(-1,-1), (-1,1), (1,-1),(1,1)} x {-1,1}
= {(-1,-1,-1),(-1-1,1),(-1,1,-1),(-1,1,1), (1,-1,-1), (1,-1,1), (1,1-1), (1,1,1)}

Question 8.
Consider the function f :R -> R defined by f(x) = -|x| (IMP-2013)
i) Find the domain and range of f.
ii) Draw the graph of f.
Answer:
i) Domain = R; Range = (-∝,0]
ii)

Plus One Maths Relations and Functions 4 Marks Important Questions

Question 1.
Consider the functions (MARCH- 2010)
\(f(x)=\sqrt{x-2}, \quad g(x)=\frac{x+1}{x^{2}-2 x+1}\)
Find
i) Domain of ‘f.
ii) Domain of ‘g’.
iii) (f + g)(x)
iv) (fg)(x)
Answer:

Question 2.
The Cartesian product P*P has 9 elements among which are found (- a, 0) and (a, 0). A relation from P to P is defined as (IMP- 2010)
R = {(x,y):x + y = 0)
i) Find P.
ii) Depict the relation using an arrow
diagram.
iii) Write down the domain and range of R,
iv) How many relations are possible from P to P?
Answer:
i) (-a,0), (a,0) are elements in P*P and P*P have 9 elements, implies P has 3 elements Therefore; P = {- a,a,0}

ii) Given; R = {(x,.y) : x + y
x = -a=>y = a=> (-a, a)
x = a =>y = -a => (a,-a)
x = 0=>y = 0 => (0,0)
iii) R = {{-a, a), (a-a), (0,0)}
Domain of R = {-a,a,0}; Range of R = {-a,a,0}
iv) Number of relation from P to P = 2₉

Question 3.
Consider the real function (MARCH-2012)
\(f(x)=\frac{x+2}{x-2}\)
i) Find the domain and range of thefunction.
ii) Prove that f(x)f(-x) + f(0) = 0
Answer:

Question 4.
i) Let P = {1,2}. Find P x P x P (MARCH-2014)
ii) Let A = {1,2,3,……..13,14},
R is the relation on A defined by R={(x,y):3x-y=0,x,y∈A}
a) Write R in a tabular form.
b) Find the domain and range of R.
Answer:
i) P x P x P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}
ii) R = {(1,3), (2,6), (3,9), (4,12)}
Domain = {1,2,3,4}
Range = {3,6,9,12}

Question 5.
i) Write the domain and range of the relation (IMP-2014)
R = {(2,5),(3,10),(4,17),(5,26)}
ii) If f{x) – x² -3x and g(x) = x + 2 find (f + g)(x), (f – g)(x) and (fg)(x)
Answer:
i) Domain = {2,3,4,5}; Range = {5,10,17,26}
ii) (f – g)(x) = f(x) – g(x)
= x² – 3x + x + 2
=x²—2x + 2
(f – g)(x) = f(x) – g(x)
= x² – 3x – x – 2 =x²-4x – 2
(fg)(x) = f(x) x g(x)
= (x² —3x) x (x + 2) = x³ – x² – 6x

Question 6.
i) If P = {m,n}, Q = {n,m} ; state whether the following is TRUE or FALSE (MAY-2016)
P x Q – {(m,n),(n,m)}
ii) Write the relation R= {(x, x³), x is a prime number less than 10}, in roster form.
iii) Let A = {1,2,3,4} B = {1,5,9,11,15,16} and
f = {(1,5),(2,9),(3,1),(4,5),(2,1)}. State with the reason whether f is a relation or a function.
Answer:
i) False
ii) R = {(2,2³)(3,3³),(5,5³),(7,7³)}
= {(2,8)(3,27), (5,125), (7,343)}
iii) F is not a function since the element 2 has two images.

Plus One Maths Relations and Functions 6 Marks Important Questions

Question 1.
Let R be the set of Reals. Define a function (IMP-2011)
f : R→R by f(x) = 2x²-1
i) Find \(\frac{f(-1)+f(1)}{2}\)
ii) Find f[f(x)]
iii) Draw the graph of f(x)
Answer:

Question 2.
i) If A = {-1,1}, Find A x A. (MARCH-2014)
ii) Consider the relation R defined by R = {(x,x + l) : x ∈ {-1,1}} Write R in the roster form. Also find the range.
iii) Draw the graph of the function. y = x, x ∈ R
Answer:
i) A x A ={-1,1} x {-1,1}
= {(-1,-1), (-1,1), (1,-1), (1,1)}
ii) R = {(-1,0),(1,2)}; Range ={0,2}
iii)

Question 3.
Let A = {1,2,3,4,5,6} be a set. Defined a relation R from A to A by R = {(x,y)/y = x+1} (IMP-2014)
i) Express R in the roster form.
ii) Represent the relation R using an arrow diagram.
iii) Write the domain and range of R.
Answer:
i) R = {(1,2),(2,3),(3,4),(4,5),(5,6)}
ii) Given; y = x +1
x = l =>j = 1 + 1 = 2; x = 2=> y = 3
x = 3 =>j = 4; x = 4 =>y = 5
x = 5=> y = 6
R = {(1,2), (2,3), (3,4), (4,5), (5,6)}

iii) Domain of R = {1,2,3,4,5}
Range of R = {2,3,4,5,6}
Codomain of R = {1,2,3,4,5,6}

Question 4.
i) Find the domain of the function. (MARCH-2015)
\(f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)
ii) Sketch the graph of the function
f(x) = |x + 1|
iii) Consider A = {1,2,3,5} and B = {4,6,9}. Define a relation
R : A → B by R = {(x,y): x – y is odd,x ∈ A,y ∈ B}.
Write R in roster form and find range of R.
Answer:
i) x² – 5x + 4 = 0 => (x-4)(x-l) = 0
Therefore f(x) is defined for all x∈R? , except x = 4 and x = 1. Hence the domain is R – {1,4}.
ii)

iii) R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
Range = {4,6,9}

Question 5.
i) Match the following  (IMP-2015)

ii) A = {1,2,3 ………. 14}. R is a relation from
A to A defined by R = {(x,y): 3x – y = 0,x,y ∈A}. Write the domain, range, co-domain of ,R.
Answer:
i) 1- Identity function, f :R →R : f(x) = x
2- Modulus function, f :R →R : f(x) = |x|
3 – Signum function f :R → R

Question 6.
i) If (x+1, y-2) = (3,1), write the values of x and y. (MARCH-2016)
ii) Let A = {1,2,3,4,5,} and B = {4,6,9} be two sets. Define a relation R from A to B by R = {(x,y): x – y} is a positive integer}
iii) Define the modulus function. What is its domain? Draw a rough sketch.
Answer:
i) Given; (x + l,y-2) = (3,1)
=> x + l = 3;
y – 2 = 1
=>x = 2; y = 3
ii) A x B= {(1,4),(1,6),(1,9),(2,4),(2,6),(2,9), (3,4),(3,6),(3,9),(4,4),(4,6)(4,9),(5,4),(5,6),(5,9)}
Then R = {(5,4)}

Question 7.
The domain of the function (MARCH-2017)
\(f(x)=\frac{1}{x-1}\)
(a) {1}
(b) R
(c) R – {1}
(d) R – {0}
ii) A relation R on set natural numbers is defined by R = {(x,y):y = x+5,x is a natural number less than 4, x, y ∈ N}
a) Write the relation in roster form.
b) Write the domain and range of the relation.
iii) Draw the graph of the relation
f(x) = |x|,x∈R
Answer:
i) c) R – {1}
ii) a) R = {(1,6), (2,7), (3,8)}
b) Domain = {1,2,3,}
Range of R = {6,7,8}

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 1 Sets.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 1 Sets

Plus One Maths Sets 3 Marks Important Questions

Question 1.
Consider the sets A and B given by (MARCH-2010)
A = {x\ x is a natural number, 2 ≤ x ≤ 6 }
B = {x: x is a prime number x ≤ 7 }
C = {x: x² – 5x + 6 = 0 }
i) Write A, B and C in the roster form.
ii) Verify that(A∪B)∪C = A∪(B∪C)
Answer:
i) A = {2,3,4,5,6}; B = {2,3, 5,7};
x² – 5x + 6 = 0
=> (x – 2)(x – 3) = 0
x = 2,3
C = {2,3}

ii) A∪B = {2,3,4,5,6,7};
(A∪B)∪C = {2,3,4,5,6,7}
B∪C = {2,3,5,7};
A∪(B∪C )= {2,3,4,5,6,7}
Hence;
(A∪B)∪C = A∪(B∪C)

Question 2.
Consider the sets A and B given by (MARCH-2011)
A = {x: x is a prime number <10}
B = {y. x is a natural number which divides 12}
i) Write A and B in the roster form.
ii) Find A∪BandB-A.
iii) Verify that(A∪B) – A = B – A.
Answer:
i) A = (2,3,5,7};B = {1,2,3,4,6,12}
ii) A∪B = {1,2,3,4,5,6,7,12};
B-A = {1,4,6,12}
iii) (A∪B)-A = {1,4,6,12} = B – A

Plus One Maths Sets 4 Marks Important Questions

Question 1.
If A = {1},
B = {{1},2}
C = {{1}, 3} and ∪ = {{1}, {2}, {3},1,2,3}, then find (IMP-2010)
i) A∩B
ii) B∩C
iii) n{{A∩B)’∪(B∩C)’)
Answer:
i) A∩B = ????
ii) B∩C={{1}}
iii) (A∩B)’ = ∪ ;
(B∩C)’ = {{2},{3},1,2,3}
(A∩B)’∪(B∩C)’ = U∪{ {2}, {3} ,1,2,3} =U
n((A∩B)'{B∩C)’)= 6

Question 2.
Given U = {1,2,3,4,5,6,7,8,9,10},A= {1,2,3,4,5} and B = {3,4,5,6} (IMP-2011)
i) Write A∪B
ii) Verify whether(A∪B) ‘ = A’∩B’
iii) Verify whether
n(A∪B)=n(A-B)+n(A∩B)+n(B – A)
Answer:
i) A∪B = {1,2,3,4,5,6}
ii) A’ = {6,7,8,9,10} ,B’-= {1,2,7,8,9,10}
A’∩B’ = {7,8,9,10}
(A∪ B)’ = {7,8,9,10}
Hence (A∪B)’ = A’∩B’
iii) (A – B)={1,2};B – A = {6}; A∩B = {3,4,5}
n(A∪B) = 6,n(A – B) = 2,
n{B-A) = 1,n(A∩B) = 3
n(A -B) + n(A ∩ B) + n(B -A)
= 2 + 3 +1 = 6
= n(A ∪ B)
Hence verified.

Question 3.
Let A = {x:x is an integer \(\frac{1}{2}<x<\frac{7}{2}\) } and B = {2,3,4} (MARCH-2012)
i) Write A in the roster form.
ii) Find the power set of (A∪ B).
iii) Verify that(A-B)∪(A∩B) = A
Answer:
i) A = {1,2,3}
ii) A ∪ B= {1,2,3,4}
P(A) = {????, {1} ,{2}, {3}, {4}, {1,2}, {1,3}, {1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4}, {2,3,4}, {1,2,3,4}}
iii) (A – B)∪(A ∩ B)
= {1} ∪ {2,3} = {1,2,3} = A

Question 4.
i) How many elements has P(A) if A = {1,2,3}? (IMP-2012)
ii) U = {1,2,3,4,5,6,7}; A = {1,4,6,7}; B = {1,2,3}. Find
A’,B’,A’∩B’,A∪B and
hence show that A’ ∩ B’ = (A ∪ B)’.
iii) If A and B are two sets such that A⊂B, then what is A ∩ B ?
Answer:
i) n(P(A)) = 2³ = 8
ii) A’ = {2,3,5},B’= {4,5,6,7}
A’ ∩B’ = {5}
A∪B = {1,2,3,4,6,7} =>(A∪B)’={5}
Hence A’∩B’ = (A∪ B)’
iii) A∩B = A

Question 5.
i) Let A = {1,2,3,5,6} and B = {1,3,4,6,7} (MARCH-2013)
a) Find A ∪ B? and A ∩B?
b) Find A -(A ∩B)
ii) If X and Y are two sets such that n(X) = 17, n{Y) = 23 and n(X∪Y)=38, find n(X∩Y).
Answer:
i) a) (A∪B) = {1,2,3,4,5 6,7} ,A∩B = {1,3,6}
b) A – (A∩B) = A – {1,3,6} = {2,5}
ii) We have; n(X∪Y)=n(X) + n(Y) – n(X∩Y)
38 = 17 + 23 – n(X∩Y)
n(X∩Y)= 40 – 38 = 2

Question 6.
Let ∪ = {1,2,3,4,5,6,7,8,9}, A = {1,2,4,7} and B = {1,3,5,7} (MARCH-2013)
i) Find A∪B.
ii) Find A’, B’ and hence show that
(A∪B)’ = A’∩B’
iii) The power set P(A) has …………. elements.
Answer:
i) A∪B={l,2,3,4,5,7}
ii) A’ = {3,5,6,8,9} ,B’ = {2,4,6,8,9}
(A∪B)’ = {6,8,9} = A’ ∩B’
iii) 2⁴= 16

Question 7.
i) If two sets A and B are disjoint, which none among the following is true? (IMP-2014)
a) A∪B = A
b)A∪B = B
c) A∩B = A
d )A∩B = ????
ii) Find the set of the equation x² + x – 2 = 0 in roster form.
iii) In a group of students, 100 students know Hindi, 50 know English and 33 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Answer:
i) d)A∩B = ????
ii) x² + x – 2 = 0
=> (x – 1)(x + 2) = 0 =>
x = 1;-2
Solution set is {1,-2}
iii) Let H-Hindi; E- English
n(H) = 100;
n(E) = 50;
n(H ∩E) = 33
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50-33 = 117

Question 8.
Let A = {x: x ∈ W ,x<5} and B = (x : x is a prime number less than 5} and U = {x: x is an integer O ≤ x ≤ 6} (MARCH-2015)
i) Write A,B in roster form.
ii) Find (A-B)∪(B-A)
iii) Verify (A ∪ B)’ = A’ ∩ B’
Answer:
i) A= {0,1,2,3,4}; B = {2,3}
ii) A – B = {0,1,4}; B – A ={ }
iii) U= {0,1,2,3,4,5,6}
A’ = {5,6}, B’ = {0,1,4,5,6}
(A∪B)’ = {5,6}
A’ ∩ B’ = {5,6}

Question 9.
If U = {1,2,3,4,5,6,7,8}, A = {2,4,6,8} and B = {2,4,8} (MAY-2016)
then:
i) Write A’, B’.
ii) For the above sets A and B prove that (A∪B)’ = A’∩B’
iii) Check whether (A∩B)’ = A’∪B’
Answer:
i) A’ = {1,3,5,7};B’ = {1,3,5,6,7}
ii) A ∪ B = {2,4,6,8}, (A ∪ B)’ = {1,3,5,7}
A’∩B’ = {l,3,5,7} => (A∪B)’ = A’∩B’
iii) A ∩ B = {2,4,8}; (A ∩ B)’ = {1,3,5,6,7}
A’∪B’ = {1,3,5,6,7} => (A∩B)’ = A’∪B’

Plus One Maths Sets 6 Marks Important Questions

Question 1.
i) What is A∪A’? (IMP-2012)
ii) If A and B are two sets such that n(A∪B) = 17 and n(A∩B) = 5,
find n(A) + n(B)
iii) In a group of 100 peoples, 40 people like cinema, 10 like both drama and cinema. How many like drama? How many like drama only not cinema?
Answer:
i) A∪A’ = ∪
ii) We have; A(A∪B)=n(A) + n(B) – n(A∩B)
=> 17 = n(A) + n(B) – 5
=> n(A) + n(B) = 17 + 5 = 22
iii) Let D – is the set of Drama, C is the set of Cinema.
Given;
n(D∪C) = 100, n{C) = 40, n(D∩C) = 10,
n(D ∪ C) = n(D) + n(C) – n(D ∩ C)
=>100 = n(D) +40 – 10
=> n(D) = 100-40 +10 = 70
n(Drama not Cinema) = n(D ∩ C’)
= n(D) – n(D∩C) = 70-10 = 60

Question 2.
i) If A and B are two sets such that A ⊂ B, (IMP-2013)
a) A ∪B is
b) Draw the Venn diagram of B – A
ii) In a committee, 60 people speak English,30 speak Hindi and 15 speak both English and Hindi. How many speak atleast one of these two languages?
Answer:
i) a) B
b)

ii) Let the events be E – English and H – Hindi
n(E) = 60; n(H) = 30; n(E∩H) = 15
People speak atleast one of these two languages = n(E∪H)
= n(E) + n(H)-n(E∩H)
= 60 + 30 – 15 = 75

Question 3.
Consider the set A = {2,3,5,7} and B = {1,2,3,4,6,12} (MARCH-2014)
i) Find A∩B.
ii) Find A – B,B – A and hence show that (A∩B)∪(A – B)∪(B – A)=A∪B
iii) Write the power set of A ∩ B
Answer:
i) A ∩ B = {2,3}
ii) A – B = {5,7} ,B-A = {1,4,6,12}
(A∪ B) = {1,2,3,4,5,6,712}
(A ∩B) ∪(A – B) ∪ (B – A) = {1,2,3,4,5,6,7,12}
iii) P(A∩B) = {????,{2},{3},{2,3}}

Question 4.
i) If A = {1,2,3,4} and B = {3,4,5,6} then (MARCH-2014)
A – B= ………
(ii) In a group of 70 people, 37 like coffee, 52 like tea and each person likes atleast one of the two drinks. How many people like both coffee and tea?
iii) Let U = {1,2,3,4,5,6}, A= {2,3} and B = {3,4,5}. Find A’,B’and hence show that (A ∪ B)’ = A’ ∩ B’
Answer:
i) A – B = {1,2}
ii) n(C∪T) = 70,n(C) = 37,n(t) = 52 ,n(C∪T) = n(C) + n(T) – n(C∩T)
=> 70 = 37 + 52 – n(C ∩ T)
=> n(C ∩ T) = 89 – 70 = 19
iii) A= {1,4,5,6},B’ = {1,2,6}
A’∩B’ = {1,6}
{A∪B)’ = ({2,3,4,5})’ = {1,6}
Hence; (A∪ B)’= A’ ∩ B’

Question 5.
Let A= {1,2,3,4,5}, B = {1,2,6} and C = {1,6,7,8} (IMP-2014)
i) Find A∪B and B∩C
ii) Show that
A∪(B∩C) = (A ∪ B)∩(A ∪ C)
iii) Find A – (B∩C) and (A ∪ B) – (B∩C)
Answer:
i) A∪B = {1,2,3,4,5,6}
(B∩C) = {1,6}
ii) A∪ (B∩C) = {1,2,3,4,5} ∪ {1,6}
= {1,2,3,4,5,6}
A∪C = {1,2,3,4,5,6,7,8}
(A ∪ B) ∩ (A ∪ C) = {1,2,3,4,5,6}
Hence A∪(B∩C) = (A∪B)∩(A∪C)
iii) A- (B∩C) = {1,2,3,4,5} – {1,6} = {2,3,4,5}
(A∪B) – (B∩C) = {2,3,4,5}

Question 6.
i) A: {x : x is a prime number x ≤ 6} (IMP-2015)
a) Represent A in Roster form.
b) Write the Power set of A.
ii) Out of 25 members in an office 17 like to take tea, 16 like to take coffee. Assume that each takes at least one of the two drinks. How many like (a) Both coffee and tea? (b) Only tea and not coffee?
Answer:
i) A ={2,3,5}
ii) P(A) = {????,{2},{3},{5},{2,3},{2,5},{3,5},{2,3,5}}
iii) Let the sets be defined as follows;
T- Tea; C – Coffee.
n(T) = 17; n(C) = 16; n(T ∪ C) = 25
n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
Coffee and Tea = n(T∩C) = 17 + 16 – 25 = 8
Only tea and not coffee =
n(T ∩C’) = n(T) – n(T∩C) = 17 – 8 = 9

Question 7.
i) If A is a subset of the set B, then (MARCH-2016)
A∩B = _____
ii) Represent the above set A∩B by Venn diagram.
iii) In a school, there are 20 teachers who teach Mathematics or Physics. Of These, 12 teach Mathematics, 12 teach Physics. How many teach both the subject?
Answer:
i) A
ii)

iii) Let the sets be defined as follows;
M- Mathematics; P – Physics.
n(M) = 12
n(P) = 12;
n(M∩P) = 20
n(M ∪ P) = n(M) + n(P) – n(M ∩ P)
n(M ∪ P) = 12 +12- 20 = 4

Question 8.
i) If U is the universal set and A is any set U∩A (MARCH-2017)
(a) U
(b) A
(c)????
(d) A’
ii) Consider the sets
U = {a,b,c,d,e,f,g}, A = {b,c,d,e}, B = {a,c,g). Find A’and B’ the verify (A∪B)’ = A’∩B’
iii) In a group of 400 people, 250 speaks Hindi, 200 can speak Malayalam. How many people can speak both Malayalam and Hindi?
Answer:
i) b) A
ii) A∪B = {a,b,c,d,e,g}
A’ = {a,f,g} ;
B’ = {b,d,e,f};
(A∪B)’ = {f}; A’∩B’ = {f}
iii) n(A∩B) = 400,n(A) = 250
n(B) = 200
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 250 + 200 – 400
= 50

Kerala State Board New Syllabus Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 12 Accounting System Using Database Management System.

Kerala Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 12 Accounting System Using Database Management System

Question 1.
Krishna intends to create a table in MS-Access. Suggest suitable data types for the following fields. (March 2015)

Answer:
Field Name – Data Type
a) Job Id – Number/Text
b) Job Name – Text
c) Job Type – Text
d) Job Department – Text
e) Job Unit – Number
f) Job Time – Date/Time

Question 2.
A company has supplied certain information required for creating a table under MS Access. (Say 2015)

Fill in the blanks with suitable data types wherever you find a question mark.
Answer:
Field Name – Data type
a) Account code – Number/Text
b) Account name – Text
c) Debit/Credit – Number/Text
d) Amount – Number
c) Narration – Text

Question 3.
Sorting or filtering of records is not possible in _____ data type. (March 2016)
a) text
b) number
c) memo
d) date
Answer:
c) Memo

Question 4.
Fill in the blank with a suitable data type on the basis of the hint given: (March 2016)
a) Name : Text
b) Birthdate : ________
Answer:
b) Birthdate : Date/Time/Text

Question 5.
List any two ways in which queries may be created in access. (March 2016)
Answer:
Design view, SQL, wizard

Question 6.
Consider the following table and answer the questions given below. (Say 2016)

a) Name of the entity.
b) Primary key.
c) Generate a meaningful record by writing an example in the above boxes with a question mark.
Answer:
a) Student
b) Student Adm. No.
c)

Question 7.
Mention the steps for creating a simple table in MS Access. (Say 2016)
Answer

• MS Access
• Table/Table design
• Data Type
• Set primary key
• Save the table

Question 8.
Entity in MS Access is a ________ (March 2017)
a) object of relation
b) present a working model
c) thing in the real world
d) model of the relation
Answer:
c) Things in the real world

Question 9.
What is the maximum length a text field can be in MS Access? (March 2017)
a) 120
b) 255
c) 265
d) 75
Answer:
b) 255

Question 10.
Write any two data types in MS-Access with suitable examples. (March 2017)
Answer:
The followings are the different data types:

• Text → Example: Job name, Job type, etc.
• Number → Example: Job unit, Job ID
• Date/Time → Example: Job time
• Currency → Example: Dollars, Rupees, etc.
• Yes/No

Kerala State Board New Syllabus Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 11 Structuring Database for Accounting.

Kerala Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 11 Structuring Database for Accounting

Question 1.
The formal blueprint or pictorial representation of accounting reality using the entity-relationship model concept is called ________ (March 2015)
a) E R design
b) Voucher
c) Entity
d) Database structure
Answer:
a) E R design

Question 2.
The most suitable ‘data-type’ for storing ‘name’ in a database is a _________ (March 2015)
a) number
b) text
c) date
d) memo
Answer:
b) Text

Question 3.
Consider the following fields of MS-Access and answer the questions. (March 2015)

1. Here, ‘Voucher’ stands for _______
2. ‘Voucher Name’ is termed as _________

Answer:

1. Entity
2. Attribute

Question 4.
State the entity and its attributes for the given information wherever you find a question mark. Entity:? (Say 2015)

(Hint: The above information is taken from the payroll records of a company.)
Answer:
Entity → Employee
Attributes → ID, Name, Age, Designation

Question 5.
________ are some properties or characteristics that give more description of an entity. (March 2016)
a) Objects
b) Attributes
c) Entity types
d) Values
Answer:
b) Attributes

Question 6.
Observe the following table. (March 2016)

a) Give a suitable entity name.
b) Write the relevant attributes relating to the entity.
Answer:
Entity → Employee/student
Attribute → Job ID, Name, Sex, Date of Birth

Question 7.
Arrange the following items given in the brackets into their order of occurrence in the transaction processing cycle. (Say 2016)
(Storage, Data Entry, Processing, validation)

Answer:

1. Data Entry
2. Validation
3. Processing
4. Storage

Question 8.
a) Write a suitable entity for the attributes given below. (March 2017)
Account Code, Account Name
b) Also identify the key attribute to that entity.
Answer:
a) Entity – Accounts
b) Key attributes – Account code, Account name.

Kerala State Board New Syllabus Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 10 Applications of Computers in Accounting.

Kerala Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 1.
The physical components of a computer are collectively called as _________ (March 2010)
a) software
b) hardware
c) live-wave
d) None of these
Answer:
b) Hardware

Question 2.
The use of computers in accounting is called computerized accounting.” Explain four points of difference between manual accounting and computerized accounting. (March 2010)
Answer:
Computerised Accounting

1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.
2. Mass data can be stored in a very small space and brought back very easily.
3. Coding is essential in computerised accounting.
4. Closing entries are not necessary.
5. The possibility of errors is less in computerised accounting.

Manual Accounting

1. In manual accounting financial statements cannot be prepared with such speed and accuracy.
2. Data are stored in a large number of books and retrieval of data is a very tedious job.
3. Coding is not essential.
4. Closing entries are necessary.
5. The possibility of errors is more.

Question 3.
“Even though computers possess many capabilities, they suffer from a number of limitations. Mention any four limitations of computers. (March 2010)
Answer:

1. Lack of commonsense
2. Computers have no intelligence
3. Lack of decision making skill
4. No feeling

Question 4.
What is a computer? Explain its capabilities and limitations. (March 2011)
Answer:
A computer is an electronic device which accepts data and instructions input, stores them, processes the data according to the instructions, and communicates the results as output.
Capabilities of a computer:

• High-speed
• A large volume of data can be stored.
• Accuracy is very high.
• Computers are multi-purpose ¡nformaon machine i.e. versatility.

Limitations of a computer:

• Computers lack common sense.
• Lack of decision-making skills
• Computers have no intelligence.
• Computers cannot make judgments based on feelings.

Question 5.
How is computerized accounting more helpful to business decision making than manual accounting? (March 2011)
Answer:

• Speed and accuracy
• Storage and retrieval of data
• Automatic and instant processing
• Ease in alteration
• Periodic availability of results

Question 6.
Processed data becomes _________ (March 2012)
a) data based
b) information N
c) record
d) data
Answer:
b) Information

Question 7.
What is computerized accounting? What are the differences between computerized accounting and manual accounting? (March 2012)
Answer:
Computerised accounting is different from manual accounting, the following are the main difference between these two:
Computerised Accounting

1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.
2. Mass data can be stored in a very small space and brought back very easily.
3. Coding is essential in computerised accounting.
4. Closing entries are not necessary.
5. The possibility of errors is less in computerised accounting.

Manual Accounting

1. In manual accounting financial statements cannot be prepared with such speed and accuracy.
2. Data are stored in a large number of books and retrieval of data is a very tedious job.
3. Coding is not essential.
4. Closing entries are necessary.
5. The possibility of errors are more.

Question 8.
Find the odd one out from the following and state the reasons. (March 2012)
a) Printer
b) Keyboard
c) Monitor
d) Plotter
Answer:
b) Keyboard – All others are output devices, the keyboard is an input device.

Question 9.
“Even though computers possess many capabilities it suffers from a number of limitations. Mention any four limitations of computers. (Say 2012)
Answer:

• Computers lack common sense.
• Lack of decision-making skills
• Computers have no intelligence.
• Computers cannot make judgments based on feelings.

Question 10.
Find out the odd one from the following. (Say 2012)
a) Monitor
b) Printer
c) Barcode reader
d) Plotter
Answer:
c) Barcode reader

Question 11.
Mr. A maintains his accounts manually. Mr. B uses computers for the preparation of accounts. State any two differences between these two systems. (Say 2012)
Answer:
Computerised Accounting

1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.
2. Mass data can be stored in a very small space and brought back very easily.
3. Coding is essential in computerised accounting.
4. Closing entries are not necessary.
5. The possibility of errors is less in computerised accounting.

Manual Accounting

1. In manual accounting financial statements cannot be prepared with such speed and accuracy.
2. Data are stored in a large number of books and retrieval of data is a very tedious job.
3. Coding is not essential.
4. Closing entries are necessary.
5. The possibility of errors is more.

Question 12.
Tally ERP-9 is a/an _______ software. (March 2013)
a) Operating
b) System
c) Application
d) Utility
Answer:
c) Application software

Question 13.
You are given certain components of a computer. Classify them into software and hardware. (March 2013)
a) Keyboard
b) MS-Office
c) Windows
d) Processor
Answer:
Software – MS Office, Windows
Hardware – Keyboard, Processor

Question 14.
A series of operations in a certain specific order to achieve the desired result in data processing is termed as ________ (March 2013)
a) Processor
b) procedure
c) DBMS
d) hardware
Answer:
b) Procedures

Question 15.
“Computerized accounting is more helpful in decision making when compared to manual accounting”. Give any two points favoring this argument. (March 2013)
Answer:
Computerised Accounting

1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.
2. Mass data can be stored in a very small space and brought back very easily.
3. Coding is essential in computerised accounting.
4. Closing entries are not necessary.
5. The possibility of errors is less in computerised accounting.

Question 16.
VDU stands for ________ (March 2014)
Answer:
Visual Display Unit (VDU)

Question 17.
Which item is different from others? Why? (March 2014)
a) Mouse
b) Printer
c) Scanner
d) Joystick
Answer:
b) Printer – is an output device, others are input devices.

Question 18.
Mention two capabilities of a computer over human beings. (March 2014)
Answer:
The capabilities of a computer system are as follows:

• Speed
• Accuracy
• Storage
• Versatility

Question 19.
Tailored accounting software is usually suitable for a _______ business. (March 2015)
a) small
b) medium
c) large
d) all of these
Answer:
c) Large business

Question 20.
Write any two limitations of a computer. (March 2015)
Answer:

• Lack of common sense
• Computers have no intelligence

Question 21.
Which among the following is NOT true about computers? (March 2015)
a) Computers have no common sense.
b) Computers cannot take decisions on their own.
c) Computers have zero IQ.
d) Computers have no storage capacity.
Answer:
d) Computer have no storage capacity

Question 22.
Computerized accounting has some advantages over manual accounting. List out any four such advantages. (March 2015)
Answer:
Advantages of computerised Accounting

• Speed
• Accuracy
• Efficiency
• Storage and retrieval
• Reliability
• Versatility

Question 23.
Classify the following into input devices and output devices. (March 2015)

1. Optical pen
2. VDU
3. Printer
4. Smart card reader

Answer:

1. Optical pen – Input device
2. VDU – Output device
3. Printer – Output device
4. Smart card reader – Input device

Question 24.
Find out the odd man out. (Say 2015)
a) MICR
b) OCR
c) Optical Scanner
d) VDU
Answer:
d) VDU

Question 25.
State any four elements of a computer system. (Say 2015)
Answer:
Hardware, Software, People, Procedure

Question 26.
Computerized accounting is not free from limitations. What points would you stress on to support this statement? (Say 2015)
Answer:
The following are the limitations of computerised accounting:

• Huge training cost
• System failure
• Staff opposition
• Inability to check unanticipated errors

Question 27.
Which one of the following is NOT an advantage of computerized accounting? (March 2016)
a) Quality report
b) Up to date
c) Accurate
d) Cost of training
Answer:
d) Cost of training

Question 28.
Complete the diagram. (March 2016)

Answer:

Question 29.
Find the pair which is NOT matched. (March 2016)
a) Operating system – System Software
b) Keyboard – Mouse
c) Scanner – Output device
d) Microwave transmission – Satellite link
Answer:
c) Scanner – output device

Question 30.
Computerized accounting has many advantages over manual accounting. (March 2016)

1. List any two advantages of computerized accounting.
2. Write any one condition that must fulfill by an accounting report.
3. The user-oriented programme designed and developed for performing certain specific tasks are called as ________

Answer:

1. High speed, Accuracy is high, efficiency
2. Relevant, timeliness, accuracy, completeness
3. Application software

Question 31.
Give suitable names to the following: (Say 2016)

Answer:

1. Ready-to-use or customized software
2. Tailored

Question 32.
Find the odd one out and justify your answer. (March 2017)
a) Ready-to-use
b) Customized
c) Tailored
d) Database
Answer:
d) Database, others are accounting packages

Question 33.
Complete the following chart. (March 2017)

Answer:

Question 34.
Which of the following is NOT a feature of Ready-to-use accounting software? (March 2017)
a) It is suited to small organizations.
b) Volume of accounting transactions is very low.
c) Cost of installation is high
d) Number of users is limited.
Answer:
c) Cost of installation is high

Question 35.
Classify the following into input devices and output devices. (March 2017)
a) Keyboard
b) Printer
c) Mouse
d) Monitor
Answer:
Input devices – Keyboard, Mouse
Output devices – Printer, Monitor

Question 36.
Give any two limitations of computerized accounting systems. (March 2017)
Answer:

• Huge Training cost
• Staff opposition
• System failure
• Breaches of security

Kerala State Board New Syllabus Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 9 Accounts from Incomplete Records.

Kerala Plus One Accountancy Chapter Wise Previous Questions and Answers Chapter 9 Accounts from Incomplete Records

Question 1.
Joseph started his textile shop on 1 st January, 2005 with a capital of Rs. 1,00,000. During the year he introduced Rs. 20,000 as additional capital and his withdrawal was of Rs. 1000 per month.
On 31st December, 2005, his position is as follows: Cash in hand Rs. 20,000; Debtors Rs. 75,000; Closing stock Rs. 1,00,000; Building Rs. 50,000; Creditors Rs. 25,000; Bills payable Rs. 20,000. Calculate the profit earned by Joseph in the year 2005. (March 2010)
Answer:
Statement of Affairs as on 31.12.2005

Statement of Profit/Loss

Question 2.
The closing balance of creditors can be ascertained from the ________ (March 2010)
a) Cash account
b) Balance Sheet at the end of the year.
c) Total creditors account
Answer:
c) Total Creditors Account

Question 3.
The amount of credit sale is determined by preparing a total creditors account. (March 2010)
Answer:
False, The amount of credit sale is determined by preparing a total Debtor’s Account.

Question 4.
From the following information find out the credit sales: (March 2010)

Answer:
Total Debtors Account

Question 5.
A retail trader had not kept proper books of account but from the following details, you are required to ascertain the profit or loss for the year ended 30th June, 2009 and also to prepare a statement of affairs as on that date. (March 2010)

The drawings during the year amounted to Rs. 2,600. Depreciate furniture by 10% and write off Rs. 300 on motor van. As regards to the debtors, it is ascertained that Rs. 500 are irrecoverable and create a further reserve of 5% for bad and doubtful debts. Also create a reserve of Rs. 700 with respect to bills receivable. (March 2010)
Answer:
Statement of Affairs as on 1/7/2008 & 30/6/2009

Statement of Profit or Loss for the year ended 30/6/2009

Question 6.
Ascertain the credit sales and purchase from the following figures: (March 2011)

Answer:
Total Debtors A/c

Total Creditors A/c

Question 7.
Statement of affairs is prepared to find out the ________ (March 2011)
a) assets and liabilities
b) capital
c) purchases and sales
Answer:
a) Assets and liabilities OR b) capital

Question 8.
Pooja started her business with Rs. 75,000 and she further introduced Rs. 10,000 while she withdrew Rs. 23,000 for her personal use. The capital at present will be Rs. _______ (March 2011)
Answer:
Rs. 62,000 ie (75,000 + 10,000 – 23,000)

Question 9.
Mr. Kumar commenced business on 1-1-2009 with a capital of Rs. 30,000. On the same day, he bought furniture for Rs. 6,000. During the year, he borrowed Rs. 5,000 from his father for business purposes and introduced further capital of Rs. 3,000. He had withdrawn Rs. 300 at the end of every month for domestic use. From the following particulars obtained from his books, prepare the Trading and Profit & Loss account for the year ending 31-12-2009 and the Balance Sheet on that date. (March 2011)

Mr. Kumar has used goods worth Rs. 1,400 for his private purpose and paid Rs. 1,000 to his son, but not recorded it in the books. On 31-12-2009, his debtors were Rs. 19,000 and creditors were Rs. 16,000. Stock on that date was valued at Rs. 8,000. Furniture is to be depreciated at 15% p.a.
Answer:
Total Debtors A/c

Total Creditors A/c

Cash Account

Trading and Profit & Loss A/c for the year ended 31-12-2009

Balance Sheet as on 31/12/2009

Question 10.
From the following information, find out: (March 2012)
a) Credit purchase by preparing the total Creditors A/c.
b) Total purchase.

Answer:
Total Creditors A/c

Bill Payable A/c

Total purchase = Cash Purchases + Credit Purchases
= 37,000 + 1,25,000
= 1,62,000

Question 11.
If stock at the beginning or at the end is missing, the same can be ascertained with the help of _________ (March 2012)
a) debtors account
b) creditors account
c) statement of affairs
d) Memorandum of Trading Account
Answer:
c) Statement of affairs
or
d) Memorandum of Trading Account

Question 12.
Mr. Sumesh is a retailer whose records are incomplete. You are required to ascertain his profit or loss for the year ended 31S1 December 2011 from the following details. (March 2012)

Answer:
Statement of Profit or Loss for the year ended 31-12-2011

Question 13.
From the following particulars prepare: (Say 2012)
A) Total debtors account and find out credit sales.
B) Total creditors account and find out credit purchases.
C) Bills receivable account.
D) Bills payable account.

Transactions during the year
Cash received from Debtors ₹ 28000, Cash received against bills receivables ₹ 9,000.
Cash paid to suppliers ₹ 13000, Cash sales ₹ 18000, Cash purchases ₹ 12000.
Cash paid against bills payable ₹ 3150, Discount allowed to debtors ₹ 3500, Bad debts are written off ₹ 1,750.
Return inwards ₹ 3250, Discount received from suppliers ₹ 3150, Return outwards ₹ 1,900, Bills payable dishonoured ₹ 750.
Answer:
Total Debtors a/c

Bill Receivable A/c

Total Creditors A/c

Bill Payable A/c

Question 14.
Calculate the Capital at the beginning. (Say 2012)

Answer:
Capital at the beginning = (75,000 + 12,000) – 20,000 = 67,000/-

Question 15.
Mr. Kiran, a sole trader does not keep full records of books of accounts. However, the following information is available. You are required to calculate Credit Sales and Credit Purchases by preparing the Total Debtors Account and Total Creditors Account. Also, show the Bills Receivable A/c and Bills Payable A/c. (Say 2012)
Balance on 1st January 2010

Balance on 31st December, 2010

Answer:
Bills Receivable A/c

Total Debtors A/c

Bills Payable A/c

Total Creditors A/c

Question 16.
Ascertain the amount of total credit purchases from the following. (March 2013)

Answer:
Total Creditors A/c

Question 17.
Mr. Sankaran commenced business on 1st April 2011 with cash Rs. 20,000, furniture Rs. 10,000 and machinery Rs. 50,000. On 1st January 2011, he introduced Rs. 20,000 into the business. He withdrew @ Rs. 1,000 per month. His financial position as of 31.3.2012 was as follows. Ascertain the profit for the year ending 31.3.2012. (March 2013)
Assets: Stock Rs. 25,000; sundry debtors Rs. 20000; mahcinery Rs.44,000; cash at bank Rs. 20,000; cash in hand Rs. 10,000; bills receivable Rs. 8,000 and furniture Rs. 9,500.
Liabilities: Sundry creditors Rs. 12,500; loan from bank Rs. 20,000; bills payable Rs. 5000.
Answer:
Statement of Affairs as on 01/04/2011 and 31/3/2012.

Statement of Profit and Loss for the year ended 31.3.2012

Question 18.
Prasad kept his books under the single entry system. His position as on 31-12-2012 was as follows. (March 2013)

His capital account shows a credit balance of Rs. 54,000 as on 01-01-2012. He introduced an additional capital of Rs. 5,000 during the financial year. He withdrew Rs. 10,000 for personal purposes. As certain his profit for the year ended 31-12-2012.
Answer:
Statement of Affairs as on 31 -12-2012

Statement of Profit or Loss for the year ended 31/12/2012

Question 19.
If the opening capital is less than the closing capital, provided there are no adjustment, the result is _________ (March 2014)
a) a profit
b) a loss
c) drawings
d) expenses
Answer:
a) Profit

Question 20.
From the following information, find the amount of credit purchase. (March 2014)

Answer:
Total Creditors A/c

Question 21.
The profit or loss under the single entry system is found by/on _______ basis. (March 2014)
a) An estimate
b) Drawing the profit and loss account
c) Drawing the balance sheet
d) the double-entry system
Answer:
a) An estimate

Question 22.
Jaya brothers ¡s a sole trader who invested Rs. 10,000 on 01 .01 .2012 in his business and the same was found at Rs. 18000 at the end of the accounting year. He also withdrew Rs. 6,000 for his private use. Ascertain his profit for the year. (March 2014)
Answer:
Statement of profit/loss for the year ended _______

Question 23.
Mr. John, who does not follow the double-entry system gives you the following details for the period ended 31.12.2012. (March 2014)

Other information:
a) Depreciate plant by 10%.
b) Salary outstanding Rs. 1000.
You are asked to prepare the Profit and Loss Account and the Balance Sheet as of 31.12.2012.
Answer:
Statement of Affairs as on 01.01.2012

Cash Book

Total Debtors A/c

Total Creditors A/c

Trading and Profit & Loss A/c for the year ended 31.12.2012

Balance Sheet as on 31.12.12

Question 24.
From the given information find the following: (March 2015)
a) Credit sales by preparing the total Debtors account
b) Ascertain the total sales.

Answer:
Total Debtors Account

b) Total Sales = Cash sales + Credit sales
= 25,000 + 22,900
= 47,900

Question 25.
Mr. Thomas keeps his books under the single entry system. On 1 st April 2013, his records indicated the following assets and liabilities. (March 2015)

During the year, he introduced Rs. 15000 as further capital and withdrew Rs. 400 every month. Calculate the profit or loss made by him during the year by using the statement of affairs method.
Answer:
Statement of Affairs as of 1/04/2013 & 31/03/2014

Statement of Profit or Loss for the year ended 31/3/2014

Question 26.
Ascertain the profit from the following: (March 2015)

Answer:
Statement of profit/loss

Question 27.
Credit purchases are calculated by preparing ________ (March 2015)
a) Total Debtors account
b) Total Creditors account
c) Bills Receivable account
d) Bills Payable account
Answer:
b) Total Creditors Account

Question 28.
Which of the following cannot be a posting on the credit side of the debtor’s account? (Say 2015)
a) Discount allowed
b) Bad debts
c) Credit sales
d) Cash received
Answer:
c) Credit sales

Question 29.
Abu does not follow any perfect system of accounting. He gives you the following information. (Say 2015)

a) Ascertain the net profit
b) What method have you employed for profit computation?
Answer:
a) Statement of Profit or Loss

b) Single entry system

Question 30.
The following information was extracted from the book of Mr. Raveendran, who maintains a single entry system, for the year ending 31st December 2015. (March 2016)

a) Calculate total purchase made by Mr. Raveendran during the year 31st December 2015.
b) While preparing the final account, from where did he get the value of opening capital and expenses made during the year?
Answer:
a) Total creditors A/c

Total purchase = 75,000 + 12,000 = 1,87,000
b) i) Opening capital is calculated by preparing statement of affairs.
ii) Expensens from cash book or cash summary.

Question 31.
Consider the following information from the books of Mr. Sidique. (March 2016)
a) Profit for the year Rs.60,000
b) Additional capital introduced by him Rs. 50,000
c) Drawings by him this year Rs. 30,000
d) Capital at the end of the year Rs. 2,00,000
Find the capital contributed by him at the beginning of the year.
Answer:

Question 32.
Find the capital at the beginning, from the following information. (Say 2016)

Answer:
Capital at the beginning = Capital at the end + Drawings + Loss during the year – Additional capital
= 1,56,000 + 14,000 + 20,000 – 40,000
= 1,50,000

Question 33.
Vimal started a business on 1st January 2014 with an investment of Rs. 1,60,000. He does not maintain proper books of accounts for his business. During the year 2014, he withdrew Rs. 20,000 for personal use and introduced Rs. 10,000 as fresh capital. His position of assets and liabilities as of 31st December 2014 stood as follows. (March 2017)

Prepare a statement of Profit or Loss for the year ended 31st December. 2014, provided that depreciation on furniture is to be charged at 10%.
Answer:
Statement of Affairs as on 31/12/2014

Statement of profit or loss for the year ended 31/12/14