CDS II Exam 2018: 18th November 2018 Mathematics Q1-20

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CDS II Exam 2018: 18th November 2018 Mathematics Q1-20 with Solutions

Q1. In the figure given below, ABC is a right-angled triangle where ∠ A = 90°, AB = p cm and AC = q cm. On the three sides as diameters semicircles are drawn as shown in the figure. The area of shaded portion, in square cm, is

(a) pq (b) π(p² + q²)/2 (c) π(p² + q²) (d) pq/2 S1. Ans.(d) Sol. In ∆ABC, by phythagoras

BC = √p2 + q2 Area of the semicircle with diameter BC

= π

2(

√p2+q2

2)

2

= π

8(p2 + q2)

Area of the semicircle with diameter AB

= π

2(

p

2)

2

= πp2

8

Area of the semicircle with diameter AC

= π

2(

q

2)

2

= πq2

8

Area of the triangle ABC = 1

2pq.

Area of shaded region = πp2

8+

πq2

8+

1

2pq–

π

8(p2 + q2)

= π(p2+q2)

8+

1

2pq–

π

8(p2 + q2)

= 1

2pq.

Q2. In the figure given below, the radius of the circle is 6 cm and AT = 4 cm. The length of tangent PT is

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(a) 6 cm (b) 8 cm (c) 9 cm (d) 10 cm S2. Ans.(b) Sol. PT is tangent and OP is a radius ⇒ ∠OPT = 90° Given that OP = OA =6 cm and AT = 4 cm. In ∆OPT, by PHythagoras OT² = OP² + PT² 10² = 6² + PT² ⇒ PT = 8 cm. Q3. In the figure given below, ABCD is the diameter of a circle of radius 9 cm. The lengths AB, BC and CD are equal. Semicircles are drawn on AB and BD as diameters as shown in the figure. What is the area of the shaded region ?

(a) 9π (b) 27π (c) 36π (d) 81π S3. Ans.(b) Sol. Given radius = 9 cm ∴ diameter ABCD = 18 cm Given that AB = BC = CD = 6 cm

Area of the semicircle with diameter AD = π

2(9)2 = π

81

2

Area of the semicircle with diameter AB = π

3(3)2 = π ×

9

2

Area of the semicircle with diameter BD = π

2(6)2 =

π

236

Area of the shaded portion = 81π

2+

9π

2–

36π

2 =

54π

2 = 27π

Q4. In the figure given below, what is ∠BCD equal to ?






(a) 70° (b) 75° (c) 80° (d) 90° S4. Ans.(c)

Sol. ∠DAC = ∠DBC = 70° [All angles inscribed in a circle and subtended by the same chord are equal]

∵ ∠DAB + ∠DCB = 180° [Opposite angles in any quadrilateral inscribed in a circle are supplements of

each other]

⇒ (70° + 30°) + ∠DCB = 180°

⇒ ∠BCD = 80°

Q5. In the figure given below, AB is the diameter of the circle whose centre is at O. Given that ∠ECD = ∠EDC

= 32°, then ∠CEF and ∠COF respectively are

(a) 32°, 64°

(b) 64°, 64°

(c) 32°, 32°

(d) 64°, 32°

S5. Ans.(b)

Sol. Given,

∠ECD = ∠EDC = 32°

⇒ ∠CEF = 32° + 32° = 64° [An exterior angle of a triangle is equal to the sum of the opposite interior

angles]

Now, ∠COF = 2∠EDC

= 2 × 32° = 64° [The inscribed angle in a circle is equal to one half of the central angle subtended by the

chord]

Q6. In the figure given below, ∆ ABR ~ ∆PQR. If PQ = 3 cm, AB = 6 cm, BR = 8.2 cm and PR = 5.2 cm, then

QR and AR are respectively






(a) 8.2 cm, 10.4 cm

(b) 4.1 cm, 6 cm

(c) 2.6 cm, 5.2 cm

(d) 4.1 cm, 10.4 cm

S6. Ans.(d) Sol. Given, ∆ABR ~ ∆PQR.

⇒ AB

PQ=

BR

QR=

AR

PR [In similar triangle, corresponding sides are all in the same proportion]

= 6

3 =

8.2

QR =

AR

5.2

⇒ QR = 4.1, AR = 10.4 Q7. In the figure given below, ABC is a triangle with AB perpendicular to BC. Further BD is perpendicular to AC. If AD = 9 cm and DC = 4 cm, then what is the length of BD ?

(a) 13/36 cm (b) 36/13 cm (c) 13/2 cm (d) 6 cm S7. Ans.(d) Sol. Let BD = h cm. In ∆BDC,

BC = √h2 + 16 In ∆BDA

AB = √h2 + 81 …(i) In ∆ABC

AB = √169– (h2 + 16)

= √169– h2– 16

= √153– h2 …(ii) From (i) and (ii)

√h2 + 81 = √153– h2 h2 + h2 = 153 – 81 2h² = 72 h² = 36 h=6 cm Q8. In the figure given below, the diameter of bigger semicircle is 108 cm. What is the area of shaded region ?






(a) 201π cm²

(b) 186.3π cm²

(c) 405π cm²

(d) 769.5π cm²

S8. Ans.(c) Sol.

Draw a perpendicular OC on the line CD. Let OA = r ⇒ OC = 54 – r OD = 27 + r In ∆OCD, (27 + r)² = (54 – r)² + (27)² ⇒ r = 18. Area of the semi circle with diameter EF

= π

2(54)2

Area of the semi circle with diameter EC = Area of the semicircle with diameter CF = π

2(27)2

Area of the circle with diameter AB = π(18)²

Area of the shaded portion = π

2(54)2– [

π

2(27)2 +

π

2(27)2 + π(18)2]

= 405π cm² Q9. In the figure given below, ABC is an equilateral triangle with each side of length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. If XY + XP + YQ is 40 cm, then the value of PQ is

(a) 5 cm (b) 12 cm (c) 15 cm (d) 10 cm S9. Ans.(d) Sol.






Clearly XP = YQ

[∵ XY ∥ BC

PX ∥ ACYQ ∥ AB

]

Let XP = x and XY = y ⇒ 2x + y = 40 ...(i) ∵ XP ∥ AC. In ∆XBP ⇒ ∠X = ∠P = ∠B = 60° ( ∵ ∠BXP = ∠BAC

∥ly ∠BPX = ∠BCA)

⇒ ∆XBP is an equilateral triangle ⇒ BX = XP = x Similarly, ∆XAY is an equilateral triangle. ⇒ AX = XY =y Given = AB = AX + XB = 30° x + y = 30° …(ii) Solving (i) & (ii), we get x = 10 ∵ BP = QC = x = 10 Given BC = BP + PQ + QC = 30° ⇒ PQ = 10° Q10. In the figure given below, ABCD is a square of side 4 cm. Quadrants of a circle of diameter 2 cm are removed from the four corners and a circle of diameter 2 cm is also removed. What is the area of shaded region ?

(a) 5

7

9 cm²

(b) 7 7

9 cm²

(c) 9 5

7 cm²

(d) 95

6 cm²

S10. Ans.(c) Sol. Area of the square = 16 cm² Area of the 4 quadrants of a circle = π(1)² = π Area of the central circle = π (1)² = π






Area of the shaded portion = 16 – (π + π) = 16 – 2π

= 16– 2 ×22

7

= 112–44

7

= 68

7

= 95

7 cm²

Q11. The sum of ages of a father, a mother, a son Sonu and daughters Savita and Sonia is 96 years. Sonu is

the youngest member of the family. The year Sonu was born, the sum of the ages of all the members of the

family was 66 years. If the father’s age now is 6 times that of Sonu’s present age, then 12 years hence, the

father’s age will be

(a) 44 years

(b) 45 years

(c) 46 years

(d) 48 years

S11. Ans.(d)

Sol. ATQ,

At present,

F + M + Sonu + 2 daughters = 96 years. _______(1)

When Sonu was born,

F + M + 2 Daughters = 66 years ________(2)

Difference between eq. (1) and (2) is 30 years. It means 30 years is the age of 5 persons during the given

period.

So, each person’s age is increase by 6 years.

So, present age of Sonu = 6 years.

And, present age of Sonu’s father = 6 × 6

= 36 years

12 years hence, the father’s age will

= 36 + 12 = 48 years

Q12. ‘A’ is thrice as good a workman as ‘B’ and takes 10 days less to do a piece of work than ‘B’ takes. The

number of days taken by ‘B’ alone to finish the work is

(a) 12

(b) 15

(c) 20

(d) 30

S12. Ans.(b)

Sol. A B






Efficiency 3 : 1

Thus, time taken by B is 15 days

Q13. Out of 85 children playing badminton or table tennis or both, the total number of girls in the group is

70% of the total number of boys in the group. The number of boys playing only badminton is 50% of the

number of boys and total number of boys playing badminton is 60% of the total number of boys. The

number of children playing only table tennis is 40% of the total number of children and a total of 12

children play badminton and table tennis both. The number of girls playing only badminton is

(a) 14

(b) 16

(c) 17

(d) 35

S13. Ans.(a)

Sol. Total Children = 85

Let, boys are x, then girls are (85 – x)

ATQ,

(85 – x) = 70

100× x

x = 50

Boys = 50 and Girls = 35

Now,

No. of boys playing badminton only = 25

Total no. of boys playing badminton = 30

And,

No. of children playing only table tennis = 34

Children play both badminton and table tennis are 12

Thus, the numbers of girls playing badminton only are 14.






Q14. A person bought two articles X and Y from a departmental store. The sum of prices before sales tax

was Rs. 130. There was no sales tax on the article X and 9% sales tax on the article Y. The total amount the

person paid, including the sales tax was Rs. 136.75. What was the price of the article Y before sales tax?

(a) Rs. 75

(b) Rs. 85

(c) Rs. 122

(d) Rs. 125

S14. Ans.(a)

Sol. ATQ,

Let cost of X = x

Then, cost of Y = 130 – x

Now,

x + (130 –x) 109

100= 136.75

100x + 14170 – 109x = 13675

9x = 495

x = 55

Thus, rate of article y = (130 – x)

= 130 – 55 = Rs. 75

Q15. According to Mr. Sharma’s will, half of his property goes to his wife and the rest is equally divided

between his two sons, Ravi and Raj. Some years later, Ravi dies and leaves half of his property to his widow

and rest to his brother Raj. When Raj dies he leaves half of his property to his widow and remaining to his

mother, who is still alive. The mother now owns Rs. 88,000 worth of the property. The total worth of the

property of Mr. Sharma was

(a) Rs. 1,00,000

(b) Rs. 1,24,000

(c) Rs. 1,28,000

(d) Rs. 1,32,000

S15. Ans.(c)

Sol. Let Mr. Sharma’s property = 1000 units

From his will chart,






Now, property of Mrs. Sharma = 500 + 187.5

= 687.5 units

687.5 units → Rs. 88,000

Then,

1000 units → 88000

687.5× 1000

= Rs. 1,28,000

Q16. X bought 4 bottles of lemon juice and Y bought one bottle of orange juice. Orange juice per bottle costs twice the cost of lemon juice per bottle. Z bought nothing but contributed Rs. 50 for his share of the drink which they mixed together and shared the cost equally. If Z’s Rs. 50 is covered from his share, then what is the cost of one bottle of orange juice ? (a) Rs. 75 (b) Rs. 50 (c) Rs. 46 (d) Rs. 30 S16. Ans.(b) Sol. Let, cost of lemon Juice bottle = x cost of orange Juice bottle = 2x ATQ, It is shared among X, Y and Z. 4×x+2x

3=

6x

3= 2x

Now, 2x = 50 ⇒ x = 25 So, cost of one bottle of orange juice = 2 × 25 = Rs. 50 Q17. Ten (10) years before, the ages of a mother and her daughter were in the ratio 3 : 1. In another 10 years from now, the ratio of their ages will be 13 : 7. What are their present ages ? (a) 39 years, 21 years (b) 55 years, 25 years (c) 75 years, 25 years (d) 49 years, 32 years S17. Ans.(b) Sol. ATQ,






M

D=

3x+20

x+20=

13

7

21x + 140 = 13x + 260 8x = 120 ⇒ x = 15 So, The age of mother = 15 × 3 + 10 = 55 years The age of daughter = 15 × 1 + 10 = 25 years Q18. In a class of 60 boys, there are 45 boys who play chess and 30 boys who play carrom. If every boy of the class plays at least one of the two games, then how many boys play carrom only ? (a) 30 (b) 20 (c) 15 (d) 10 S18. Ans.(c) Sol. Use the direct formula, Boys play only carrom = (Boys play carrom + boys play chess) – total boys = (45 + 30) – 60 = 75 – 60 = 15 Q19. Two equal amounts were borrowed at 5% and 4% simple

interest. The total interest after 4 years amounted to Rs. 405. What

was the total amount borrowed ?

(a) Rs. 1075

(b) Rs. 1100

(c) Rs. 1125

(d) Rs. 1150

S19. Ans.(c)

Sol. Let amount borrowed is ‘x’

ATQ, x×5×4

100+

x×4×4

100= 405

20x

100+

16x

100= 405

36x = 40500

⇒ x = 1125

Q20. Twelve (12) men work 8 hours per day and require 10 days to build a wall. If 8 men are available, how

many hours per day must they work to finish the work in 8 days ?

(a) 10 hours

(b) 12 hours

(c) 15 hours

(d) 18 hours

S20. Ans.(c)

Sol. Using formula, M1D1H1

W1=

M2D2H2

W2

Here work done is same in both cases.






12 × 10 × 8 = 8 × 8 × H2

H2 =12×10×8

8×8

⇒ H2 = 15 hours





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CDS II Exam 2018: 18th November 2018 Mathematics Q1-20