ICSE Class 10 Mathematics Previous Year Question Paper 2019

Mathematics is one of the crucial and most scoring subjects in ICSE Class 10. TopperLearning presents study materials for ICSE Class 10 Mathematics which will enable students to score well in the board examination. The syllabus includes certain challenging concepts like Value Added Tax, Ratio and Proportion, Quadric Equations among others which require effective study materials. Our study materials for ICSE Class 10 consist of video lessons, question banks, sample papers, revision notes and past year papers which improve the quality of learning.

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Q 1.

(a) Solve the following inequation and write down the solution set:

11x - 4 < 15x + 4 <13x + 14, x ∈ W

Represent the solution on a real number line.

 

(b) A man invests Rs. 4500 in shares of a company which is paying 7.5% dividend. If Rs. 100 shares are available at a discount of 10%.

Find:

(i) Number of shares he purchases.

(ii) His annual income.

 

(c) In class of 40 students, marks obtained by the students a class test (out of 10) are given below,

Marks 1 2 3 4 5 6 7 8 9 10
Number of students 1 2 3 3 6 10 5 4 3 3

Calculate the following for the given distribution:

(i) Median

(ii) Mode

 

Q 2.

(a) Using the factor theorem, show that (x - 2) is a factor of x3 + x2 – x4 – 4, Hence factorise the polynomial completely.

(b) Prove that:

(cosec θ - sin θ)(sec θ – cos θ) (tan θ + cot θ) =1

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively, Find the:

(i) first term

(ii) common difference

(iii) sum of the first 20 terms

 

Q 3.

(a) Simplify

begin mathsize 11px style sin space straight A open square brackets table row cell sin space straight A end cell cell negative cos space straight A end cell row cell cos space straight A end cell cell sin space straight A end cell end table close square brackets space plus space cos space straight A open square brackets table row cell cos space straight A end cell cell sin space straight A end cell row cell negative sin space straight A end cell cell cos space straight A end cell end table close square brackets end style

 

(b) M and N are two points on the X axis and Y axis respectively.

P(3, 2) divides the line segment MN in the ratio 2 : 3.

Find:

(i) the coordinates of M and N

(ii) slope of the line MN.

 

(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32cm. Find the

(i) radius of the cylinder

(ii) curved surface area of the cylinder

Take Π = 3.1

 

Q 4. 

(a) The following numbers, K + 3, K + 2, 3K - 7 and 2K – 3 are in proportion. Find K.

(b) Solve for x the quadratic equation x2 – 4x – 8 = 0.

Give your answer correct to three significant figures.

(c) Use ruler and compass only for answering this question.

Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre, Construct two tangents to the circle from the external point P.

Measure and write down the length of any one tangent.

 

Q 5.

(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box.

Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by 2 and 3 both.

(iii) a number less than 16.

 

(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives Rs. 441 as interest at the time of maturity.

Find the amount Rekha deposited each month.

 

(c) Use a graph sheet for this question.

Take 1 cm = 1 unit along both x and y axis.

(i) Plot the following points:

A(0,5), B(3,0), C(1,0), and D(1,-5)

(ii) Reflect the points B, C, and D on the y axis and name them as B’, C’ and D’ respectively.

(iii) Write down the coordinates of B’, C’ and D’.

(iv) Join the points A, B, C, D, D’, C’, B’, A in order and give a name to the closed figure ABCDD’C’B’.


Q 6.

(a) In the given figure PQR = PST = 90o, PQ = 5 cm and PS  = 2 cm.

(i) Prove that PQR = PST

(ii) Find Area of PQR: Area of quadrilateral SRQT.


(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:

(i) ‘n’ the number of terms of the G.P.

(ii) Sum of the n terms.

 

(c) A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows:

The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm.

Give your answer correct to the nearest whole number Take begin mathsize 11px style straight pi space equals space 22 over 7 end style


Q 1.

(a) Solve the following inequation and write down the solution set:

11x - 4 < 15x + 4 <13x + 14, x ∈ W

Represent the solution on a real number line.

 

(b) A man invests Rs. 4500 in shares of a company which is paying 7.5% dividend. If Rs. 100 shares are available at a discount of 10%.

Find:

(i) Number of shares he purchases.

(ii) His annual income.

 

(c) In class of 40 students, marks obtained by the students a class test (out of 10) are given below,

Marks 1 2 3 4 5 6 7 8 9 10
Number of students 1 2 3 3 6 10 5 4 3 3

Calculate the following for the given distribution:

(i) Median

(ii) Mode


Solution:

(a) We have,

11x - 4 < 15x + 4 13x + 14

So, we can say

11x - 4 < 15x + 4

begin mathsize 11px style rightwards double arrow 11 straight x minus 4 less than 15 straight x plus 4
rightwards double arrow 15 straight x minus 11 straight x greater than negative 4 minus 4
rightwards double arrow 4 straight x greater than negative 8
rightwards double arrow straight x greater than negative 2

15 straight x plus 4 less or equal than 13 straight x plus 14
rightwards double arrow 15 straight x minus 13 straight x less or equal than 10
rightwards double arrow 2 straight x less or equal than 10
rightwards double arrow straight x less or equal than 5

straight x greater than negative 2 space and space straight x space less or equal than 5
rightwards double arrow negative 2 less than straight x less or equal than 5

straight x element of straight W
rightwards double arrow straight x equals open curly brackets 0 comma 1 comma 2 comma 3 comma 4 comma 5 close curly brackets end style

 

(b) Total investment = Rs. 4500

Dividend percentage = 7.5%

Nominal value = Rs. 100

Discount % = 10%

begin mathsize 11px style Market space value equals Nominal space value space minus Discount
equals 100 minus 10 over 100 cross times 100 equals Rs.90
Number space of space shares equals fraction numerator Investment over denominator Market space value end fraction equals 4500 over 90 equals 50 space shares
Divident space on space 1 space share space equals space 7.5 percent sign space cross times 100 equals Rs. space 7.5
Divident space on space 50 space share space equals space 7.5 percent sign space cross times 50 equals Rs.375
Annual space income equals Rs.375
end style

(c)

Marks

xi

No. of
Students

fi
xifi
1 1 1
2 2 4
3 3 9
4 3 12
5 6 30
6 10 60
7 5 35
8 4 32
9 3 27
10 3 30
Total 40 240

 

begin mathsize 11px style open parentheses straight i close parentheses Mean equals fraction numerator sum straight x subscript straight i straight f subscript straight i over denominator sum straight f subscript straight i end fraction equals 240 over 40 equals 6
left parenthesis ii right parenthesis space Mode space equals space Class space with space highest space frequency space that space is space 10
Mode equals 6 end style

 

Q 2.

(a) Using the factor theorem, show that (x - 2) is a factor of x3 + x2 – x4 – 4, Hence factorise the polynomial completely.

(b) Prove that:

(cosec θ - sin θ)(sec θ – cos θ) (tan θ + cot θ) =1

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively, Find the:

(i) first term

(ii) common difference

(iii) sum of the first 20 terms

Solution:

(a) The given polynomial is p(x) = x3 + x2 - 4x - 4

Using factor theorem, we know that

(x - a) is a factor of p(x) if p(a) = 0.

So to prove that (x - 2) is a factor of p(x), we need to prove that p(2) = 0

Consider, p(2) = 23 + 22 - 4 × 2 - 4

= 8 + 4 - 8 - 4

= 0

Therefore, (x - 2) is a factor of p(x).

Now,

x3 + x2 - 4x - 4

= x3 - 4x +  x2 - 4

= x(x2 - 4) + 1(x2 - 4)

= (x2 - 4) (x + 1)

= (x- 22) (x + 1)

= (x - 2) (x + 2) (x + 1)


(b) 

Consider

Error converting from MathML to accessible text.


(c) Let  be the nth term of A.P.

Given: Fourth and sixth terms are 8 and 14

i.e. t4 = 8 and t6 = 14

begin mathsize 11px style rightwards double arrow straight a plus open parentheses 4 minus 1 close parentheses straight d space equals space 8 space and space straight a space plus space left parenthesis 6 space minus space 1 right parenthesis straight d space equals space 14....... open parentheses 1 close parentheses end style

Where, a and d are first term and common difference of an A.P respectively.

i. Using (1), we get

a + 3d = 8   ....(2) and a + 5d = 14  ......(3)

Solving (2) and (3) simultaneously, we get

a = -1

ii. Substituting the value of 'a' in equation (2), we get

d = 3

iii. Sum of first n terms of an A.P. is given by

begin mathsize 11px style straight S subscript straight n equals straight n over 2 open square brackets 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close square brackets
Therefore comma space sum space of space first space 20 space term space is colon
straight S subscript 20 equals 20 over 2 open square brackets 2 straight a space plus space open parentheses 20 minus 1 close parentheses straight d close square brackets
rightwards double arrow straight S subscript 20 equals 10 open square brackets 2 cross times open parentheses negative 1 close parentheses plus 19 cross times 3 close square brackets
rightwards double arrow straight S subscript 20 equals 10 open parentheses negative 2 plus 57 close parentheses
rightwards double arrow straight S subscript 20 equals 10 open parentheses 55 close parentheses
rightwards double arrow straight S subscript 20 equals 550 end style


Q 3.

(a) Simplify

begin mathsize 11px style sin space straight A open square brackets table row cell sin space straight A end cell cell negative cos space straight A end cell row cell cos space straight A end cell cell sin space straight A end cell end table close square brackets space plus space cos space straight A open square brackets table row cell cos space straight A end cell cell sin space straight A end cell row cell negative sin space straight A end cell cell cos space straight A end cell end table close square brackets end style

 

(b) M and N are two points on the X axis and Y axis respectively.

P(3, 2) divides the line segment MN in the ratio 2 : 3.

Find:

(i) the coordinates of M and N

(ii) slope of the line MN.

 

(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32cm. Find the

(i) radius of the cylinder

(ii) curved surface area of the cylinder

Take Π = 3.1

Solution:

(a) 

begin mathsize 11px style sin space straight A open square brackets table row sinA cell negative cosA end cell row cosA sinA end table close square brackets space plus space cosA open square brackets table row cosA sinA row cell negative sinA end cell cosA end table close square brackets
equals open square brackets table row cell sin squared straight A end cell cell negative sinA space cosA end cell row cell sinA space cosA end cell cell sin squared straight A end cell end table close square brackets space plus space open square brackets table row cell cos squared straight A end cell cell sinA space cosA end cell row cell negative sinA space cosA end cell cell cos squared straight A end cell end table close square brackets
equals open square brackets table row cell sin squared straight A space plus space cos squared straight A end cell cell space space space space space space minus sinA space cosA space plus space sinA space cosA end cell row cell sinA space cosA space minus space sinA space cosA end cell cell sin squared straight A space plus space cos squared straight A end cell end table close square brackets
equals open square brackets table row 1 0 row 0 1 end table close square brackets
end style


(b)

i. Let M and N be the points be (x, 0) and (0, y) on the x – axis and y - axis respectively.

P(3, 2) divides MN in the ratio 2:3.

begin mathsize 11px style therefore open parentheses fraction numerator 3 straight x over denominator 5 end fraction comma space fraction numerator 2 straight y over denominator 5 end fraction close parentheses equals open parentheses 3 comma 2 close parentheses
rightwards double arrow fraction numerator 3 straight x over denominator 5 end fraction equals 3 comma space fraction numerator 2 straight y over denominator 5 end fraction equals 2
rightwards double arrow straight x space equals space 5 comma space straight y space equals space 5
therefore The space coordinates space of space straight M space and space straight N space are space open parentheses 5 comma space 0 close parentheses space and space left parenthesis 0 comma space 5 right parenthesis end style

ii. Slope of MN = begin mathsize 11px style fraction numerator 5 space minus space 0 over denominator 0 space minus space 5 end fraction equals fraction numerator 5 over denominator negative 5 end fraction space equals space minus 1 end style

 

(c)

i. Let r and R be the radius of sphere and cylinder respectively and h be the height of the cylinder.
r = 6 cm, h = 32 cm
According to the question,
Volume of sphere = Volume of cylinder

begin mathsize 11px style rightwards double arrow 4 over 3 πr cubed equals πR squared straight h
rightwards double arrow 4 over 3 cross times space 6 cubed space equals space straight R squared space cross times space 32
rightwards double arrow straight R squared equals 9 end style

R = 3 cm

The radius of the sphere is 3 cm.

ii. For cylinder, h = 32 cm and R = 9 cm

Curved surface area of a cylinder = ΠR2h = 3.1 × 32 × 32 = 892.8 cm2


Q 4. 

(a) The following numbers, K + 3, K + 2, 3K - 7 and 2K – 3 are in proportion. Find K.

(b) Solve for x the quadratic equation x2 – 4x – 8 = 0.

Give your answer correct to three significant figures.

(c) Use ruler and compass only for answering this question.

Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre, Construct two tangents to the circle from the external point P.

Measure and write down the length of any one tangent.

Solution:

(a) K + 3, K + 2, 3K - 7 and 2K – 3 are in proportion.

begin mathsize 11px style therefore space fraction numerator straight K space plus space 3 over denominator straight K space plus space 2 end fraction equals fraction numerator 3 straight K space minus space 7 over denominator 2 straight K space minus space 3 end fraction
therefore space open parentheses straight K space plus space 3 close parentheses space open parentheses 2 straight K space minus space 3 close parentheses space equals space open parentheses straight K space plus space 2 close parentheses open parentheses 2 straight K space minus space 3 close parentheses
therefore space 2 straight K squared space minus space 3 straight K space plus space 6 straight K space minus space 9 space equals space 2 straight K squared space minus space 3 straight K space plus space 4 straight K space minus space 6
therefore space 2 straight K space equals space 3
therefore space straight K space equals space 3 over 2 end style

 

(b) Comparing with ax2 + bx + c = 0 we get

a = 1, b = -4 and c = -8

b2 – 4ac = (-4)2 – 4 × 1 × (-8)

= 16 + 32

= 48

begin mathsize 11px style therefore space straight x equals fraction numerator negative straight b space plus-or-minus square root of straight b squared space minus space 4 ac end root over denominator 2 straight a end fraction
therefore space straight x equals fraction numerator negative open parentheses negative 4 close parentheses space plus-or-minus space square root of 48 over denominator 2 end fraction
therefore space straight x space equals space fraction numerator 4 space plus-or-minus space 4 square root of 3 over denominator 2 end fraction
therefore space straight x space equals 2 space plus-or-minus space 2 square root of 3 end style

∴ x = 2 ± 2 × 1.732

∴ x = 2 ± 3.464

∴ x = 5.464 or -1.464


(c) 

  1. Take measure 4 cm in compass and draw a circle, with center as O.
  2. Draw a straight line from O to P, such that OP = 7cm
  3. Now find the midpoint of OP by drawing a perpendicular bisector
  4. Mark the midpoint as X
  5. Take measure of XO in the compass and cut arcs at S and T on the Circle
  6. Join PS and PT
  7. Measure of PS comes out to be 5.74 cm
 
 

Q 5.

(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box.

Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by 2 and 3 both.

(iii) a number less than 16.

 

(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives Rs. 441 as interest at the time of maturity.

Find the amount Rekha deposited each month.

 

(c) Use a graph sheet for this question.

Take 1 cm = 1 unit along both x and y axis.

(i) Plot the following points:

A(0,5), B(3,0), C(1,0), and D(1,-5)

(ii) Reflect the points B, C, and D on the y axis and name them as B’, C’ and D’ respectively.

(iii) Write down the coordinates of B’, C’ and D’.

(iv) Join the points A, B, C, D, D’, C’, B’, A in order and give a name to the closed figure ABCDD’C’B’.

Solution:

(a) There are 25 discs numbered from 1 to 25.
Hence, the number of possible outcomes in the sample space is n(S) = 25

i. Let A be the event of getting an odd number

A = {1, 3, 5, 7, …. 25}

n(A) = 13

Thus, the probability of getting an odd number is begin mathsize 11px style 13 over 25 end style.


ii. Let B be the event of getting a number divisible by 2 and 3.
To find the numbers which are divisible by 2 and 3 both, we need to find the number which are divisible by 6.

B = {6, 12, 18, 24}

n(B) = 4

Thus, the probability of getting an odd number is begin mathsize 11px style 4 over 25 end style.


iii. Let C be the event of getting a number less than 16.

C = {1, 2, 3, 4, 5, 6, …..15}

n(C) = 15

Thus, the probability of getting a number less than 16 is begin mathsize 11px style 15 over 25 equals 3 over 5 end style.


(b) Given that n = 20 months, r = 9% per annum, Interest = Rs. 441

According to the question,

begin mathsize 11px style straight P fraction numerator straight n open parentheses straight n plus 1 close parentheses over denominator 2 end fraction cross times fraction numerator straight r over denominator 12 space cross times space 100 end fraction equals 441
therefore straight P fraction numerator 20 space cross times space 21 over denominator 2 end fraction space cross times space fraction numerator 9 over denominator 12 space cross times space 100 end fraction space equals space 441
therefore space straight P space cross times space 1.575 space equals space 441
therefore space straight P space equals space Rs. space 280
end style

Hence, the amount Rekha deposited each month is Rs. 280.

 

(c)

i. Reflected points of B, C and D on the y-axis are B’(-3, 0), C’(-1, 0) and D’(-1, -5) respectively.

ii. After joining the points A, B, C, D, D’, C’, B’, A in order gives us the closed figure as arrow.

 

Q 6.

(a) In the given figure PQR = PST = 90o, PQ = 5 cm and PS  = 2 cm.

(i) Prove that PQR = PST

(ii) Find Area of PQR: Area of quadrilateral SRQT.


(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find:

(i) ‘n’ the number of terms of the G.P.

(ii) Sum of the n terms.

 

(c) A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows:

The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm.

Give your answer correct to the nearest whole number Take begin mathsize 11px style straight pi space equals space 22 over 7 end style

Solution:

(a) In ΔPQR and ΔPST,


begin mathsize 11px style angle PQR space equals space angle PST space equals space 90 degree
angle RPQ space equals space angle TPS space space space space space space space space space because common space angle
therefore space triangle PQR space tilde triangle PST
therefore PQ over PS equals QR over ST equals PR over PT equals 5 over 2
therefore fraction numerator ar open parentheses triangle PQR close parentheses over denominator ar open parentheses triangle PST close parentheses end fraction equals 5 squared over 2 squared equals 25 over 4
therefore space ar open parentheses triangle PST close parentheses equals 4 over 25 ar open parentheses triangle PQR close parentheses space.......... open parentheses straight i close parentheses
end style

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