MATHS CLASS 10TH QUESTION PAPER 2018 (ICSE)

MATHEMATICS 

SECTION A

Q.  1 (a) Find the value of ‘x’ and ‘y’ if: 

(b) Sonia had a recurring deposit account in a bank and deposited `600 per month for 2½ years. If the rate of interest was 10% p.a., find the maturity value of this account. 

(c) Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is: 

    (i) a prime number. 

    (ii) a number divisible by 4. 

   (iii) a number that is a multiple of 6. 

   (iv) an odd number. 

Answer:

⇒ 2 JC + 6 = 10 ⇒ 2x = 4 ⇒ x = 2

And 2y – 5 = 15 ⇒ 2y = 20 ⇒ y = 10

Hence, the values of x and y are x = 2 and y = 10

(b) Here, amount deposited per month = ₹ 600

       Number of months = 2 x 12 + 6 = 30  [∵ v T = 2.5years]

        Rate of interest = 10% p.a.

Hence, the amount received by Sonia on maturity is ₹ 20325.

(c) Total number of cards in the bag = 10

(i) Total prime numbers = 1 i.e., 2

∴ Required Probability = 110

(ii) Total numbers divisible by 4 = 5 (i.e., 4, 8, 12, 16, 20]

Required Probability = 510=12

(iii) Total numbers divisible by 6 or multiple of 6 = 3 [i.e., 6, 12, 18]

∴ Required Probability = 310

(iv) Total odd number = 0

∴ Required Probability =010 = 0.

 

Q.  2 (a) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the  

    (i) radius of the cylinder 

    (ii) volume of cylinder. (use π =227) 

(b) If (k –3), (2k + 1) and (4k + 3) are three consecutive terms of an A.P., find the  value of k. 

(c) PQRS is a cyclic quadrilateral. Given ∠QPS = 73^o, ∠PQS = 55^oand ∠PSR = 82^o,calculate: 

(i) ∠QRS 

(ii) ∠RQS 

(iii) ∠PRQ 

Answer :

(a) Let r be the radius of the base of cylindrical vessel and ft = 25 cm be its height.

Now, circumference of the base = 132 cm

2πr = 132

Hence, the radius of the cylinder is 21 cm and volume of the cylinder is 34650 cm³

(b) Here, ft – 3, 2k + 1 and 4k + 3 are three consecutive terms of an A.P.

∴ 2k + 1 – (k – 3) = 4k + 3 – (2k + 1)

⇒ 2k + 1 – k + 3 = 4k + 3 – 2k – 1

⇒ k + 4 = 2k + 2

⇒ 2k – k =4 – 2

⇒ k = 2

Hence, the value of ft is 2.

(c)(i) Since PQRS is a cyclic quadrilateral

∠QPS + ∠QRS – 180°

⇒ 73° + ∠QRS = 180°

⇒ ∠QRS = 180° – 73°

∠QRS = 107°

(ii) Again, ∠PQR + ∠PSR = 180°

∠PQS + ∠RQS + ∠PSR = 180°

55° – ∠RQS + 82° = 180°

∠RQS = 180° – 82° – 55° = 43°

(iii) In ∆PQS, by using angles sum property of a ∆.

∠PSQ + ∠SQP + ∠QPS = 180°

∠PSQ + 55° + 73° = 180°

∠PSQ = 180° – 55° – 73°

∠PSQ = 52°

Now, ∠PRQ = ∠PSQ = 52° [∠s of the same segment]

Hence, ∠QRS = 107°, ∠RQS = 43° and  ∠PRQ = 52°

Q.  3 (a) If (x+ 2) and (x + 3) are factors of x³ + ax + b, find the values of ‘a’ and ‘b’. 

(b) Prove that sec²θ+cosec²θ  = tanθ + cot θ

(c) Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data: 

Runs  scored	3000- 4000	4000- 5000	5000- 6000	6000- 7000	7000- 8000	8000- 9000	9000- 10000
No. of  batsmen	4	18	9	6	7	2	4

Answer :

(a) Given that (x + 2) and (x + 3) are factors of p(x) = x³ + ax + b.

∴ p(- 2) = (- 2)³ + o(- 2) + b = 0

⇒ – 8 – 2a + b = 0 => – 2a + b = 8 …….(i)

And p(- 3) = (- 3)³ + a(- 3) + b = 0

⇒ – 27 – 3a + b = 0 => – 3a + b = 27 ……..(ii)

Subtracting (i) from (ii), we obtain

(- 3a 4 – b) – (- 2a + b) = 27 – 8

– 3a + b + 2a – b = 19

-a = 19

⇒ a = 19

From (i), we obtain

– 2(19) + b = 8

– 38 + b = 8

⇒ b = 8 + 38

⇒ b = 46

Hence, the values of a and b are a = 19 and b = 46.

Q.  4 (a) Solve the following inequation, write down the solution set and represent it on the real number line: 

    −2 + 10x ≤ 13x + 10 < 24 + 10x, x ∈ Z

(b) If the straight lines 3x − 5y = 7 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a. 

(c) Solve x² + 7x = 7 and give your answer correct to two decimal places. 

Answer.

(a)

 Given that :

Thus, the required solution set is :

Using number line, we have

(b)Given lines are

3x – 5y = 1 ……….(i) and 4x + ay + 9 = 0  …………(ii)

Slope of line (i) (m1) =  −(3-5)=35

Slope of line (ii) (m2) = −(4a)

Also, given that two lines are perpendicular to one and another

∴ (m1) (m2) = – 1

⇒

Hence, the value of a = 125 .

(c) Here, x² + 7x = 7

⇒ x² + 7x – 7 = 0

SECTION B 

Q.  5 (a) The 4^th term of a G.P. is 16 and the 7^th term is 128. Find the first term and common ratio of the series. 

(b) A man invests `22,500 in `50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate: 

    (i) The number of shares purchased 

    (ii) The annual dividend received. 

    (iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number. 

(c) Use graph paper for this Q.  (Take 2cm = 1unit along both x and y axis).  

     ABCD is a quadrilateral whose vertices are A(2,2), B(2,–2), C(0,–1) and D(0,1). 

    (i) Reflect quadrilateral ABCD on the y-axis and name it as A’B’CD.  

    (ii) Write down the coordinates of A’ and B’. 

   (iii) Name two points which are invariant under the above reflection. 

   (iv) Name the polygon A’B’CD. 

Answer.

(a) Let a and r be the first term and common ratio of given G.P.

∴ a⁴ = 16

⇒ ar³ = 16

and a⁷= 128

⇒ a⁶ = 128

Dividing (ii) and (i), we obtain

a³ = 3

a³= 23

a = 2

⇒

From (i), we have

2(r³) = 16

r³ = 8

r³ = 23

⇒ r = 2

Hence, the first term and common ratio of the given series is 2 and 2.

(b) Total investment = ₹ 22,500

Face value of a share = ₹ 50

Market value of a share = ₹ (50 – 10% of 50) = ₹ (50 – 5) = ₹ 45

∴ No. of shares purchased = 2250045= 500

Annual dividend per share = 12 % of 50

Total annual dividend = ₹ 6 × 500 = ₹ 3000

Rate of return =

= 13.3 %

= 13% (Nearest whole number)

Hence, number of shares purchased are 500, total annual dividend is ₹ 3000 and rate of return on investment is nearly 13 % p. a.

(c) Scale used is : 2 cm = 1 unit along both x and y axis.

(i) Here, vertices of the quadrilateral ABCD are A(2, 2), B(2, -2), C(0, -1) and D(0, 1)

(iii) Two points which are invariant are C and D.

(iv) A’B’CD is a trapezium.

Q.  6 (a) Using properties of proportion, solve for x. Given that x is positive: 

(c) Prove that (1 + cotθ−cosecθ)(1 + tanθ+ secθ)= 2

Answer.

(a)

By componendo and Dividendo, we have

Squaring both sides, we have

Hence the value of x is 58

(b) Given that

(c) L.H.S. = (1 +cot θ – cosec θ) (1 + tan θ + sec θ)

Q.  7 (a) Find the value of k for which the following equation has equal roots. 

                         x² + 4kx + (k² − k + 2) = 0 

(b) On a map drawn to a scale of 1 : 50,000, a rectangular plot of land ABCD has the following dimensions. AB = 6cm; BC = 8cm and all angles are right angles. Find: 

   (i) the actual length of the diagonal distance AC of the plot in km. 

  (ii) the actual area of the plot in sq km. 

(c) A(2, 5), B(–1, 2) and C(5, 8) are the vertices of a triangle ABC, ‘M’ is a point on AB such that AM : MB = 1 : 2. Find the co-ordinates of ‘M’. Hence find the  equation of the line passing through the points C and M. 

Answer.

(a) Given quadratic equation is :

x² + 4kx + (k² – k + 2) = 0

For equal roots, we have

b²– 4 ac = 0

⇒ (4k)² – 4(1) (k²-k + 2) = 0

⇒ 16k² – 4k² + 4k – 8 = 0

⇒ 12k²+ 4k – 8 = 0

or 3k² + k – 2 = 0

⇒ 3k² + 3k – 2k -2 = 0

⇒ 3k(k + 1) – 2(k + 1) = 0

⇒ (k + 1)(3k – 2) = 0

⇒ k + 1=0 or 3k – 2 = 0

k = – 1 or k = 23

(b) Scale used on the map is 1 : 50,000

Dimensions of a rectangular plot ABCD are AB = 6 cm, BC = 8 cm Since each angle is right angle

∴ By using Pythagoras theorem, we have

(i) Actual length of the diagonal AC = 10 × 50000 cm

= 500000100000km

= 5 km

(ii) Area of the rectangular field ABCD on map

= 6 × 8 = 48 cm2

Actual area of the field = 48 × 500000 × 500000

= 12(10)10 sq. cm.

= 12 sq. km.

(c) Coordinates of the vertices of a ∆ ABC are A(2, 5), B(- 1, 2) and C (5, 8). Since M is a point on AB such that AM : MB = 1 : 2

Coordinates M are

Now, equation of the line CM is given as :

Q.  8 (a) ₹ 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹100 more. Find the original number of  children. 

(b) If the mean of the following distribution is 24, find the value of ‘a’. 

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of  students	7	a	8	10	5

(c) Using ruler and compass only, construct a ∆ABC such that BC = 5 cm and AB = 6.5

      cm and ∠ABC = 120^° 

   (i) Construct a circum-circle of ∆ABC 

  (ii) Construct a cyclic quadrilateral ABCD, such that D is equidistant from AB and  BC. 

Answer.

(a) Total amount = ₹ 7500

Let the number of children be x

∴ Share of each child = ₹7500x 

According to the statement

(x – 20) (7500 + 100x) = 7500 x

7500x + 100x²– 150000 – 2000x – 7500x = 0

100x² – 200x – 150000 = 0

x² – 20x – 1500 = 0

x² – 50x + 30x – 1500 = 0

x(x – 50) + 30(x – 50) = 0

(x – 50) (x + 30) = 0

⇒ x = 50 or x = – 30

Rejecting -ve value, because number of children cannot be negative.

∴ x = 50

Hence, the original number of children is 50.

Mean = 24 (given)

∴ (15a+810)/(30+a)= 24

15a + 810 = 720 + 24a

⇒ 24a – 15a = 810 – 720

⇒ 9a = 90

⇒ a = 10

Hence, the value of a is 10.

Steps of Construction :

1. Draw a line segment AB = 6.5 cm.
2. At B, construct an angle of 120° and cut off BC = 5 cm.
3. Join AC, to have ∆ABC.
4. Draw the perpendicular bisectors of line segments AB and BC.
5. Let they intersect each other in 0.
6. With 0 as centre and radius OA or OB or OC, draw the circumcircle of ∆ABC.
7. Produce perpendicular bisector of line segment AB and let it intersect the circumcircle of ∆ABC at D.
8. Join AD and CD.
   Thus, quad. ABCD is the required quadrilateral.

Q.  9 (a) Priyanka has a recurring deposit account of ₹1000 per month at 10% per annum. If she gets ₹5550 as interest at the time of maturity, find the total time for which the account was held. 

(b) In ΔPQR, MN is parallel to QR and  

  (ii) Prove that ΔOMN and ΔORQ are similar. 

  (iii) Find, Area of ΔOMN : Area of ΔORQ 

  (c) The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find the volume of the solid. 

Answer:

1. Amount deposited per month = ₹ 1000

           Rate of interest = 10% p.a.

            Interest = ₹ 5550

n² + n = 1332

n² + n – 1332 = 0

n² + 37n – 36n – 1332 = 0

n(n + 37) – 36(n + 37) = 0

(n – 36) (n + 37) = 0

n = 36 or n = – 37

Rejecting – ve value of n, we have n = 36

Hence, the total time for which the account was held, was 36 month or 3 years.

(b) Given that:

In ∆PQR, MN is parallel to QR

∴ By using Basic Proportionality theorem, we have

(c) Here, radius of cone = radius of cylinder = radius of hemisphere = 7 cm

Height of cone = 4 cm

Height of cylinder = 4 cm

Q.  10 (a) Use Remainder theorem to factorize the following polynomial: 

              2x³ + 3x² − 9x − 10. 

(b) In the figure given below ‘O’ is the centre of the circle. If QR = OP and ∠ORP = 20^°. 

      Find the value of ‘x’ giving reasons. 

(c) The angle of elevation from a point P of the top of a tower QR, 50m high is 60^o and that of the tower PT from a point Q is 30^o. Find the height of the tower PT, correct to the nearest meter.

Answer :

(a)  Let p(x) = 2x³ + 3x² – 9x – 10

Factors of constant term 10 are ± 1, ± 2, ± 5

Put x = 2, we have

p(2) =2(2)³ + 3(2)² – 9(2) – 10

= 16 + 12 – 18 – 10

= 0

∴ (x – 2) is a factor of p(x)

Put x = – 1, we have

P(-1) =2(-1)³ + 3(-1)² – 9 (-1) – 10

= – 2 + 3 + 9 – 10 = 0

∴ (x + 1) is a factor of p(x)

Thus, (x + 1) (x – 2) i.e.,x2 – x – 2 is a factor of p(x)

Hence, (x + 1), (x – 2) and (2x + 5) are the factors of given polynomial 2x³ + 3x² – 9x – 10.

(b) 

Here, in ∆OPQ

OP = OQ = r

Also, OP = QR [Given]

OP = OQ = QR = r

In ∆OQR, OQ = QR

∠QOR = ∠ORP = 20°

And ∠OQP = ∠QOR + ∠ORQ

= 20° + 20°

= 40°

Again, in ∆ OPQ

∠POQ = 180° – ∠OPQ – ∠OQP

= 180°- 40° – 40°

= 100°

Now, x° + ∠POQ + ∠QOR = 180° [a straight angle]

x° + 100° + 20° = 180°

x° = 180° – 120° = 60°

Hence, the value of x is 60.

(c) 

Here, Height of the tower (QR) = 50 m

Height of the tower (PT) = h m

Inrt. ∠ed ∆ PQR, ∠RPQ = 60°

Also, inrt. ∠ed ∆ QPT, ∠TQP = 30°

Hence, the required height of tower PT is 17 m (nearest to metre). 

   Q.  11 (a) The 4^th term of an A.P. is 22 and 15^th term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms. 

(b) Use Graph paper for this Q. . 

    A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school Was conducted. The following data was recorded: 

Height in cm	135 – 140	140 – 145	145 – 150	150 – 155	155 – 160	160 – 165	165 – 170
No. of  boys	4	8	20	14	7	6	1

Taking 2cm = height of 10 cm along one axis and 2 cm = 10 boys along the other  axis draw an ogive of the above distribution. Use the graph to estimate the  following: 

 (i) the median 

 (ii) lower Quartile 

 (iii) if above 158 cm is considered as the tall boys of the class. Find the number of  boys in the class who are tall. 

Answer.

(a) Let a and d be the first term and common difference of the required A.P.

∴ a4 = 22

⇒ a + 3d = 22 ………(i)

And a15 = 66

⇒ a + 14d = 66 ………..(ii)

Subtracting (i) from (ii), we have

(14d – 3d) = 66 – 22

11d = 44

d = 4

From (i), we have

a + 3(4) = 22

a = 22 – 12 = 10

Thus, a = 10 and d = 4

Now, Sn = n/2 [2a + (n – 1)d]

⇒ S8= 8/2[2(10) + (8-1)4]

S8 = 4 [20 + 28]

S8 = 4 x 48 S8 = 192

(b) Given data was recorded as :

Plot the points (140,4), (145,12), (150,32), (155,46), (160,53), (165,59) and (170,60). Join them free hand to get the required ogive.

Now, from the graph, we obtain :

(i) Median height (in cm) = 149.5 cm

(ii) Lower Quartile = 146 cm ‘

(iii) Number of boys who are tall e., height above 158 cm = 60 – 51 = 9.