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Previous Year's Solved Paper SSC CGL Quant tier II

held on 13th September 2019

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1. N solid metallic spherical balls are melted and recast into a cylindrical rod whose radius is 3

times that of a spherical ball and height is 4 times the radius of a spherical ball. The value of N

is:

(1) 30

(2) 27

(3) 24

(4) 36

2. If x is the remainder when 361284 is divided by 5 and y is the remainder when 496 is divided by 6,

then what is the value of (2x - y)?

(1) -4

(2) 4

(3) -2

(4) 2

3. What is the area (in square units) of the triangular region enclosed by the graphs of the

equations x + y = 3, 2x + 5y = 12 and the x-axis?

(1) 2

(2) 3

(3) 4

(4) 6

4. The value of 28 10 3 7 4 3+ − − is closest to:

(1) 7.2

(2) 6.1

(3) 6.5

(4) 5.8

5. If sec + tan = p, (p > 1) then cosec 1

cosec 1

+

−=?

(1) p 1

p 1

+

−

(2) p2

(3) p 1

p 1

−

+

(4) 2p2

6. The value cosec (67 + ) – sec (23 - ) + cos15 cos35 cosec55 cos60 cosec75 is:

(1) 2

(2) 0

(3) 1

(4) 1/2

7. 35% of goods were sold at a profit of 65%, while the remaining were sold at x% loss. If the

overall loss is 12%, then what is the value of x? (correct to one decimal place)

(1) 51.8

(2) 50.6

(3) 53.5

(4) 52.4

8. In a circle with centre O, ABCD is a cyclic quadrilateral and AC is the diameter. Chords AB and

CD are produced to meet at E. If ∠CAE = 34 and ∠E = 30, then ∠CBD is equal to:

(1) 36

(2) 26

(3) 24

(4) 34

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9. ab (a - b) + bc (b - c) + ca (c - a) is equal to:

(1) (a + b) (b - c) (c - a)

(2) (a - b) (b + c) (c - a)

(3) (a - b) (b - c) (c - a)

(4) (b - a) (b - c) (c - a)

10. The radius of the base of a right circular cylinder is increased by 20%. By what per cent should

its height be reduced so that its volume remains the same as before?

(1) 25

(2) 2

309

(3) 305

9

(4) 28

Answer keys

1. 2 2. 3 3. 2 4. 3 5. 2 6. 4 7. 3 8. 2 9. 4 10. 3

Solutions

1. 2

Let the radius of each spherical ball be r units.

So, radius of rod = 3r and height = 4r

Now, volume of the rod = π(3r)2(4r) = 36πr3

Also, volume of N balls = N * 4πr3/3

So, 36πr3 = 4Nπr3/3

N = 27

2. 3

361284 = (3)4*15321 = 8115321

When 81 is divided by 5, remainder is 1

When 8115321 is divided by 5, remainder is 115321 = 1

So, x = 1

496 = 44*24

When 44 is divided by 6, remainder is 4.

Again, 424 = 44*6

When 46 = 42*3 is divided by 6, remainder is 4.

So, when 496 is divided by 6, remainder is 4.

So, y = 4

2x – y = 2 – 4 = –2

3. 2

The three sides of the triangle are formed by the lines

x + y = 3 … (i)

2x + 5y = 12 … (ii)

And y = 0 … (iii)

From (i) and (iii),

x = 3, y = 0

From (ii) and (iii),

x = 6, y = 0

From (i) and (ii),

x = 1, y = 2

So, the vertices of the triangle are (3,0), (6,0) and (1,2)

Area of the triangle = (1/2){3 *(0 – 2) + 6 * (2 – 0) + 1 * (0 – 0)} = (1/2) * 6 = 3 sq units

4. 3

Let 28 + 10√3 = (p + √q)2 = p2 + 2p√q + q = (p2 + q) + 2p√q

So, p2 + q = 28 and 2p√q = 10√3

Hence, p = 5, q = 3

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Similarly, 7 – 4√3 = (y – √z)2 = y2 – 2y√z + z = (y2 + z) – 2y√z

So, y2 + z = 7 and 2y√z = 4√3

Hence, y = 2, z = 3

2 2

28 10 3 7 4 3

(5 3) (2 3)

+ − −

= + − −

= 5 + √3 – (2 – √3)

= 3 + 2√3

= 3 + 2 * 1.73

= 6.46, i.e., it is closest to 6.5

5. 2

2

2 2

2

2

2

2 2 2

cosec 1

cosec 1

(cosec 1)(cosec 1)

(cosec 1)(cosec 1)

cosec 2cosec 1

cosec 1

cosec 2cosec 1

cot

1 2sin sin

cos cos cos

+

−

+ +=

− +

+ +=

−

+ +=

= + +

= sec2θ + 2tan θ sec θ + tan2 θ

= (sec + tan )2 = p2

6. 4

cosec (67 + ) – sec (23 - ) + cos15 cos35 cosec55 cos60 cosec75

= cosec (67 + ) – sec (90 – (23 - )) + cos15 cos35 cosec (90 – 35) (1/2) cosec(90 – 15)

= cosec (67 + ) – cosec (67 + ) + (1/2) cos15 cos35 sec35 sec15

= 1/2

7. 3

Let the total C.P. be np, where n is the total quantity of goods and p is the cost per n unit.

Total S.P. = 165% of 35% of np + (100 – x)% of (100 – 35)% of np

= 0.5775np + 0.65np – 0.0065xnp

= 1.2275np – 0.0065xnp

Also, total S.P. = (100 – 12)% of np = 0.88np

Hence, 1.2275np – 0.0065xnp = 0.88np

0.0065x = 0.3475

x = 53.46 ~ 53.5

8. 2

∠CBA = ∠CDA = 90 as they are angles formed on the diameter of the circle.

Also, exterior angle = sum of two interior angles

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So, ∠ACD = 34 + 30 = 64

So, in triangle CAD, ∠CAD = 180 – 64 – 90 = 26

Again, ∠CAD and ∠CBD are on the same arc CD.

So, ∠CBD = ∠CAD = 26

9. 4

ab (a - b) + bc (b - c) + ca (c - a)

= a2b – ab2 + b2c – bc2 + c2a – a2c

= a2b – ab2 + b2c – bc2 + c2a – a2c + abc – abc

= a2b + c2a – a2c – abc – ab2 + b2c – bc2 + abc

= a(ab + c2 – ac – bc) – b(ab – bc + c2 – ac)

= (a – b)(ab + c2 – ac – bc)

= (a – b)(b(a – c) – c(a – c))

= (a – b)(b – c)(a – c)

= (b – a)(b – c)(c – a)

10. 3

Let the original radius and height of the cylinder be r and h units respectively.

Original volume = πr2h

After increase in radius, new height be H

So, new volume = π(120r/100)2H

πr2h = π(36r/25)H

H = (25/36)h

So, new height should be reduced by (36 – 25)/36 * 100% of h = 30(5/9)% of h

11. A is as efficient as B and C together. Working together A and B can complete a work in 36 days

and C alone can complete it in 60 days. A and C work together for 10 days. B alone will complete

the remaining work in:

(1) 110 days

(2) 88 days

(3) 84 days

(4) 90 days

12. If 2cos2 + 3sin = 3, where 0 < < 90, then what is the value of sin22 + cos2 + tan22 +

cosec22?

(1) 35/12

(2) 29/3

(3) 35/6

(4) 29/6

13. The radius and the height of a right circular cone are in the ratio 5:12. Its curved surface area is

816.4 cm2. What is the volume (in cm3) of the cone? (Take π = 3.14)

(1) 2512

(2) 1256

(3) 3140

(4) 628

14. Given that (5x - 3)3 + (2x + 5)3 + 27 (4 - 3x)3 = 9 (3 - 5x) (2x + 5) (3x - 4), then the value of

(2x + 1) is:

(1) -13

(2) 15

(3) -15

(4) 13

15. The sides of a triangle are 12 cm, 35 cm and 37 cm. What is the circumradius of the triangle?

(1) 19 cm

(2) 17.5 cm

(3) 17 cm

(4) 18.5 cm

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16. The base of a right pyramid is an equilateral triangle with area 16 3 cm2. If the area of one of

its lateral faces is 30 cm2, then its height (in cm) is:

(1) 739

12

(2) 209

12

(3) 611

12

(4) 643

12

17. A vessel contains a 32 litre solution of acid and water in which the ratio of acid and water is 5:3.

If 12 litres of the solution are taken out and 7 1/2 litres of water are added to it, then what is

the ratio of acid and water in the resulting solution?

(1) 4:7

(2) 5:6

(3) 4:9

(4) 8:11

18. A sphere of maximum volume is cut out from a solid hemisphere. What is the ratio of the volume

of the sphere to that of the remaining solid?

(1) 1:4

(2) 1:2

(3) 1:3

(4) 1:1

19. If ( ) ( )3 3 2 25 5x 2 2y Ax 2y Bx 2y Cxy ,+ = + + + then the value of (A2 + B2 – C2) is:

(1) 15

(2) 20

(3) 30

(4) 40

20. The value of (1 + cot - cosec ) (1 + cos + sin ) sec = ?

(1) -2

(2) 2

(3) sec cosec

(4) sin cos

Answer keys

11. 1 12. 3 13. 1 14. 2 15. 4 16. 3 17. 2 18. 3 19. 2 20. 2

Solutions

11. 1

Let the total work be 180 units (LCM of 36 and 60).

Work done by A and B per day = 180/36 = 5 units

Work done by C per day = 180/60 = 3 units

As A = B + C

And A + B = 5

So, A = 5 – A + 3

A = 4

So, B = 1

Hence, when A and C work for 10 days, work done = 10 *(4 + 3) = 70 units

Remaining work will be done by B in = (180 – 70)/1 = 110 days

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12. 3

2cos2 + 3sin = 3

2(1 – sin2) + 3sin = 3

2 – 2sin2 + 3sin - 3 = 0

2sin2 - 3sin + 1= 0

Let sin = a, then 2a2 – 3a + 1 = 0

a = 1, 1/2

sin = 1, 1/2

As 0 < < 90, so sin = 1/2

= 30

sin22 + cos2 + tan22 + cosec22

= sin260 + cos230 + tan260 + cosec260

= (√3/2)2 + (√3/2)2 + (√3)2 + (2/√3)2

= 35/6

13. 1

Let the radius and the height of the cone be 5a and 12a respectively (in cm).

Curved surface area = π * 5a * √((12a)2 + (5a)2)

= 3.14 * 5a * 13a = 204.1a2

204.1a2 = 816.4

a2 = 4

a = 2

So, volume = (1/3)π * (5a)2 * 12a = 2512

14. 2

We know, if a3 + b3 + c3 = 3abc, then a + b + c = 0

Here, a = 5x – 3, b = 2x + 5, c = 3(4 – 3x)

So, 5x – 3 + 2x + 5 + 3(4 – 3x) = 0

14 – 2x = 0

2x = 14

So, 2x + 1 = 14 + 1 = 15

15. 4

We know, circumradius of a triangle with sides a, b and c =

+ + + − + − + −

=

+ + + − + − + −

=

abc

(a b c)(b c a)(c a b)(a b c)

12 35 37

(12 35 37)(35 37 12)(37 12 35)(12 35 37)

12 35 37

84 60 14 10

= 12 * 35 * 37/840

= 37/2 cm

= 18.5 cm

16. 3

Area of equilateral triangle BCD (base) = (3/4)a2, where a = BC = DB = CD

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So, a2 = 163/(3/4) = 64

a = 8

Inradius of BCD = OE = a/23 = 4/3

Area of the lateral face ACD = (1/2) * CD * AE, where AE = height of the face and slant height

of the pyramid.

So, 4 * AE = 30

AE = 15/2

Hence, in right AEO,

AE2 = EO2 + OA2

(15/2)2 = (4/3)2 + OA2

OA2 = 225/4 – 16/3 = 611/12

OA = (611/12) cm, where OA is the height of the pyramid.

17. 2

Quantity of acid in the vessel = 5/(5 + 3) * 32 = 20 litres

So, quantity of water = 32 – 20 = 12 litres

12 litres of solution is taken out, then quantity of acid left = 20 - (12/32) * 20 = 12.5

And quantity of water left = 12 - (12/32) * 12 = 12 - 4.5 = 7.5

Again when 7.5 litres of water is added, quantity of water = 7.5 + 7.5 = 15

Hence, ratio = 12.5: 15 = 5:6

18. 3

Let the volume of the hemisphere be (2/3)πr3, where r = radius of the hemisphere

The radius of the hemisphere will be the diameter of the new sphere formed as it should have

max volume.

So, volume of sphere formed = (4/3)π(r/2)3 = (1/6)πr3

Remaining solid = (2/3)πr3 – (1/6)πr3 = (1/2)πr3

Required ratio = (1/6):(1/2) = 1:3

19. 2

( ) ( )3 3 2 25 5x 2 2y Ax 2y Bx 2y Cxy+ = + + +

We know, a3 + b3 = (a + b)(a2 – ab + b2)

Here, a = √5x, b = √2y

(√5x)3 + (√2y)3 = (√5x + √2y)(5x2 - √10xy + 2y2)

So, (Ax + √2y)(Bx2 + Cxy + 2y2) = (√5x + √2y)(5x2 - √10xy + 2y2)

A = √5, B = 5 and C = - √10

A2 + B2 – C2 = 5 + 25 – 10 = 20

20. 2

(1 + cot - cosec ) (1 + cos + sin ) sec

= (1 + cos /sin - 1/sin ) (1 + cos + sin ) (1/cos )

= {(sin + cos - 1)/sin } (1 + cos + sin ) (1/cos )

= {(sin + cos )2 - 12)/(sin cos )

= (1 + 2sin cos - 1)/(sin cos )

= 2sin cos /sin cos

= 2

21. S is the incentre of PQR. If ∠PSR = 125, then the measure of ∠PQR is:

(1) 75

(2) 55

(3) 80

(4) 70

22. The value of 0.47 0.503 0.39 0.8+ − is:

(1) 0.615

(2) 0.615

(3) 0.625

(4) 0.625

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23. If in ABC, D and E are the points on AB and BC respectively such that DE ll AC, and AD:AB =

3:8, then (area of BDE) : (area of quadrilateral DECA) = ?

(1) 9:55

(2) 9:64

(3) 8:13

(4) 25:39

24. Monika spends 72% of her monthly income. If her income increases by 20% and savings

increase by 15%, then her expenditure increases by: (correct to 1 decimal place)

(1) 20.8%

(2) 20.2%

(3) 21.9%

(4) 19.8%

25. A, B and C started a business with their capitals in the ratio 2:3:5. A increased his capital by

50% after 4 months, B increased his capital by 33 (1/3) % after 6 months and C withdrew 50%

of his capital after 8 months, from the start of the business. If the total profit at the end of a

year was ₹ 86,800, then the difference between the shares of A and C in the profit was:

(1) ₹ 12,600

(2) ₹ 7,000

(3) ₹ 9,800

(4) ₹ 8,400

26. The graph of the equations 5x - 2y + 1 = 0 and 4y – 3x + 5 = 0, intersect at the point P(, ).

What is the value of (2 - 3)?

(1) 4

(2) 6

(3) -4

(4) -3

27. An article was sold at a profit of 14%. Had it been sold for ₹ 121 less, a loss of 8% would have

been incurred. If the same article would have been sold for ₹ 536.25, then the profit/loss per

cent would have been:

(1) Profit, 5%

(2) Loss, 5%

(3) Loss, 2.5%

(4) Profit, 2.5%

28. A shopkeeper allows 18% discount on the marked price of an article while selling and still makes

a profit of 23%. If he gains ₹ 18.40 on the sale of the article, then what is the marked price of

the article?

(1) ₹ 140

(2) ₹ 125

(3) ₹ 120

(4) ₹ 146

29. The value of ( )2 2

2 2

2 2

sec cosecsec cosec

cosec sec

+ − +

is:

(1) 0

(2) -2

(3) 2

(4) 1

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30. The given graph shows the weights of students in school on a particular day.

The number of students weighing less than 50kg is what per cent less than the number of

students weighing 55 kg or more?

(1) 44

(2) 40

(3) 55

(4) 30

Answer keys

21. 4 22. 4 23. 4 24. 3 25. 1 26. 1 27. 3 28. 3 29. 2 30. 1

Solutions

21. 4

We know when S is the incentre in PQR.

∠PSR = 90 + ∠PQR/2

125 = 90 + ∠PQR/2

∠PQR = 70

22. 4

Let 'n' 0.47 0.4777777....= =

10n = 4.777777…

100n = 47.77777…

So, 100n – 10n = 47.7777… – 4.7777…

90n = 43

n = 43/90

Similarly,

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4980.503

990

360.39

90

80.8

9

So,0.47 0.503 0.39 0.8

=

=

=

+ −

= (43/90) + (498/990) – (36/90) * (8/9)

= (43/90) + (498/990) – (32/90)

= (11/90) + (498/990)

= (121 + 498)/990

= 619/990

= 0.625252525..

0.625=

23. 4

AD/AB = 3/8

So, BD/AB = 5/8

In ABC and BDE, we have

∠BAC = ∠BDE (corresponding angles)

∠BCA = ∠BED (corresponding angles)

∠B = ∠B

So, ABC is congruent to BDE (AAA)

Hence, (area of ABC)/(area of BDE) = (BD/AB)2 = (5/8)2 = 25/64

So, area of ABC = (25/64) of area of BDE

Area of quad. DECA = area of ABC – area of BDE = (64 – 25)/64 of area of BDE = 39/64 of

area of BDE

Hence, required ratio = (25/64):(39/64) = 25:39

24. 3

Let Monika’s income be I.

Expenditure = 72% of I, so savings = (100 – 72)% of I = 28% of I

When income becomes 120% of I = 1.2I

Savings becomes 115% of 28% of I = 0.323I

Hence, new expenditure = 1.2I – 0.323I = 0.878I

Increase in expenditure = (0.878 – 0.72)/0.72 * 100% = 21.9%

25. 1

Let the initial capitals of A, B and C respectively be 2k, 3k and 5k.

Ratio of their investments = (2k * 4 + (3/2) * 2k * 8):(3k * 6 + (4/3) * 3k * 6):(5k * 8 + (1/2)

* 5k * 4) = 32k:42k:50k = 16:21:25

So, at the end of the year difference between shares of A and C = (25 – 16)/(16 + 21 + 25) *

86800 = (9/62) * 86800 = 12600

26. 1

5x - 2y + 1 = 0

5x = 2y – 1

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x = (2y – 1)/5 … (i)

4y – 3x + 5 = 0 … (ii)

Putting (i) in (ii),

4y – 3 * (2y – 1)/5 + 5 = 0

20y – 6y + 3 + 25 = 0

14y = –28

y = –2

Putting y = –2 in (i), we get

x = (–4 – 1)/5 = –1

So, P(, ) = P(–1, –2)

2 - 3 = –2 + 6 = 4

27. 3

Let the Cost Price of the article be ₹ P.

Selling Price = 114% of P = 1.14P

Also, 1.14P – 121 = (100 – 8)% of P = 0.92P

1.14P – 0.92P = 121

P = 550

So, loss percentage = (550 – 536.25)/550 * 100% = 2.5%

28. 3

Let the Marked Price be ₹ p.

After discount, Selling Price = (100 – 18)% of p = 0.82p

Profit % = 23% and profit = 18.40

So, Cost Price = 0.82p * 100/(100 + 23) = 2p/3

Now, 0.82p – 2p/3 = 18.40

0.46p/3 = 18.40

p = 40 * 3 = ₹ 120

29. 2

( )

( )

2 22 2

2 2

2 22 2

2 2

2 22 2

2 2

sec cosecsec cosec

cosec sec

1 / cos 1 / sin1 tan 1 cot

1 / sin 1 / cos

sin cos(2 t an cot )

cos sin

+ − +

= + − + + +

= + − + +

= tan2θ + cot2θ – 2 – tan2θ – cot2θ

= –2

30. 1

Number of students weighing less than 50kg = 40 + 30 = 70

Number of students weighing 55 kg or more = 55 + 45 + 25 = 125

So, percentage = (125 – 70)/125 * 100% = 44% less

31. A right triangular prism has height 18 cm and its base is a triangle with sides 5 cm, 8 cm and 12

cm. What is its lateral surface area (in cm2)?

(1) 450

(2) 468

(3) 432

(4) 486

32. A can do one-third of a work in 15 days, B can do 75% of the same work in 18 days and C can

do the same work in 36 days. If B and C work together for 8 days. In how many days will A

alone complete the remaining work?

(1) 24 days

(2) 18 days

(3) 20 days

(4) 16 days

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33. A person buys 80 kg of rice and sells it at a profit of as much money as he paid for 30 kg. His

profit per cent is:

(1) 27 (3/11)

(2) 35

(3) 40

(4) 37 (1/2)

34. To cover a distance of 416 km, a train A takes 2 (2/3) hours more than train B. If the speed of A

is doubled, it would take 1 (1/3) hours less than train B. What is the speed (in km/h) of train A?

(1) 56

(2) 54

(3) 52

(4) 65

35. The value of 2 10 5 2 3

5 2 7 5 2 7 2

−− −

+ − + − is:

(1) 2 + 2

(2) 25

(3) 2

(4) 7

36. If the price of oil is increased by 20%. However, its consumption decreased by 8 (1/3) %. Then,

what is the percentage increase or decrease in the expenditure on it?

(1) Increase by 10%

(2) Decrease by 5%

(3) Decrease by 10%

(4) Decrease by 5%

37. The average age of 120 students in a group is 13.56 years. 35% of the number of students are

girls and the rest are boys. If the ratio of the average age of boys and girls is 6:5, then what is

the average age (in years) of the girls?

(1) 12

(2) 11.6

(3) 10

(4) 14.4

38. The marked price of an article is ₹ 1500. If two successive discounts, each of x%, on the marked

price is equal to a single discount of ₹ 587.40, then what will be the selling price of the article if

a single discount of x% is given on the marked price?

(1) ₹ 1.025

(2) ₹ 1,155

(3) ₹ 1,170

(4) ₹ 1,200

39. Two parallel chords on the same side of the centre of a circle are 12 cm and 20 cm long and the

radius of the circle is 5 13 cm. What is the distance (in cm) between the chords?

(1) 2

(2) 3

(3) 2.5

(4) 1.5

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40. Study the following bar graph and answer the question given.

The ratio of the total demand of motorcycles of companies A, C and E to the total production of

motorcycles of B and C is:

(1) 1:1

(2) 2:1

(3) 11:10

(4) 3:2

Answer keys

31. 1 32. 3 33. 4 34. 3 35. 3 36. 1 37. 1 38. 3 39. 1 40. 4

Solutions

31. 1

Lateral surface area of a right prism = perimeter of base * height of prism

= (5 + 8 + 12) * 18

= 25 * 18

= 450 cm2

32. 3

Let the total work be 180 units (LCM of 15, 18 and 36)

Work done by A in a day = (180/3)/15 = 4 units

Work done by B in a day = (75% of 180)/18 = 7.5 units

Work done by C in a day = 180/36 = 5 units

So, work done by B and C together in 8 days = 8 * (7.5 + 5) = 100 units

So, remaining work = 180 – 100 = 80 units

Number of days required by A = 80/4 = 20 days

33. 4

Cost Price = Cost of 80 kg

Profit = Cost of 30 kg

So, profit percent = (30/80) * 100% = 37(1/2)%

34. 3

Let the speed of train A be v km/hr and the time take by B be ‘t’ hrs.

So, time taken by A = t + 2(2/3) = t + 8/3 = (3t + 8)/3

Also, time taken by A = 416/v

416/v = (3t + 8)/3

1248 = 3tv + 8v … (i)

Again, if speed of A becomes 2v, then

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Time taken = t – 1(1/3) = (3t – 4)/3

Also, time taken = 416/2v

So, 416/2v = (3t – 4)/3

1248 = 6tv – 8v … (ii)

From (i) and (ii),

3tv + 8v = 6tv – 8v

3t = 16

t = 16/3

So, v = 1248/(8 + 3 * 16/3) = 1248/24 = 52 km/hr

35. 3

2 10 5 2 3

5 2 7 5 2 7 2

−− −

+ − + −

Rationalizing:

( )

( )

2 2 22 22

2 10( 5 2 7) ( 5 2)( 5 2) 3( 7 2)

( 5 2 7)( 5 2 7) ( 5 2)( 5 2) ( 7 2)( 7 2)

2 10( 5 2 7) 5 2 3( 7 2)

( 7) 2( 5) 25 2 ( 7)

2 10( 5 2 7) 5 2 3( 7 2)

7 45 45 2 2 5 2 7

2 10( 5 2 7) 5 2 3( 7 2)

32 10 1

+ + − − +− −

+ − + + + − − +

+ + − += − −

−−+ −

+ + − += − −

−−+ + −

+ + − += − −

= 5 + 2 + 7 - 5 + 2 - 7 – 2

= 2

36. 1

Let the initial price of oil be ₹ p and its initial consumption be ‘c’ units.

Total expenditure = Rs. pc

Now, price of oil = 120% of p = 1.2p

And consumption = (100 – 8(1/3))% of c = 275/3% of c = 2.75c/3

New expenditure = 1.2p * 2.75c/3 = 1.1pc, i.e., 110% of pc

So, the expenditure increases by 10%.

37. 1

Ratio of the number of girls and boys = 35:(100 – 35) = 7:13

Let the average age of boys and girls be 6a and 5a respectively.

By alligations,

(6a – 13.56):(13.56 – 5a) = 7:13

13(6a – 13.56) = 7(13.56 – 5a)

78a + 35a = 13.56(13 + 7)

113a = 271.2

a = 2.4

so, average age of girls = 5 * 2.4 = 12 years

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38. 3

Marked Price = 1500

After two successive discounts, Selling Price = (100 – x)2% of 1500 = ((100 – x)2/10000) *

1500 = ((100 – x)2/20) * 3

Given, 3(100 – x)2/20 = 1500 – 587.40 = 912.60

(100 – x)2 = 6084

100 – x = ±78

x = 22, 178 (invalid as discount% can’t be greater than 100)

So, x =22%

And new Selling Price = ((100 – 22)/100) of 1500 = ₹ 1170

39. 1

CD = 12 cm, AB = 20 cm, OA = OD = 5√(13) cm

PQ is the distance between the chords AB and CD, so OQ is perpendicular to CD and OP is

perpendicular to AB. So, AP = PB and CQ = QD and APO and DQO are right-angled triangles.

In APO, OA2 = OP2 + PA2

(5√(13))2 = OP2 + (AB/2)2

OP2 = 325 – 100 = 225

OP = 15

In DQO, OD2 = OQ2 + QD2

(5√(13))2 = (OP + PQ)2 + (CD/2)2

(15 + PQ)2 = 325 – 36 = 289

15 + PQ = 17

PQ = 2 cm

40. 4

Total demand of motorcycles of companies A, C and E = 50 + 60 + 55 = 165

Total production of motorcycles of B and C = 38 + 72 = 110

Ratio = 165:110 = 3:2

41. A circle touches the side BC of ABC at D and AB and AC are produced to E and F respectively

such that circle touches point E and F. If AB = 10 cm, AC = 8.6 cm and BC = 6.4 cm, then BE =

?

(1) 3.2 cm

(2) 3.5 cm

(3) 2.2 cm

(4) 2.5 cm

42. If the measure of each exterior angle of a regular polygon is 3

517

, then the ratio of the

number of its diagonals to the number of its sides is:

(1) 5:2

(2) 13:6

(3) 3:1

(4) 2:1

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43. Two numbers are in the ratio 3:5. If 13 is subtracted from each, the new numbers are in the

ratio 10:21. If 15 is added to each of the originals numbers, then the ratio becomes:

(1) 5:7

(2) 23:33

(3) 4:5

(4) 24:35

44. Pipes A and B are filling pipes while pipe C is an emptying pipe. A and B can fill a tank in 72 and

90 minutes respectively. When all the three pipes are opened together, the tank gets filled in 2

hours. A and B are opened together for 12 minutes, then closed and C is opened. The tank will

be empty after:

(1) 15 minutes

(2) 18 minutes

(3) 12 minutes

(4) 16 minutes

45. The LCM of two numbers x and y is 204 times their HCF. If their HCF is 12 and the difference

between the numbers is 60, then x + y = ?

(1) 660

(2) 426

(3) 852

(4) 348

46. In ABC, BE ⊥ AC, CD ⊥ AB and BE and CD intersect each other at O. The bisectors of ∠OBC and

∠OCB meet at P. If ∠BPC = 148, then what is the measure of ∠A?

(1) 56

(2) 28

(3) 32

(4) 64

47. The value of ( ) ( )6 6 4 4

4 4 2

2 sin cos 3 sin cos

cos sin 2cos

+ − +

− − is:

(1) -1

(2) -2

(3) 2

(4) 1

48. The value of 24 x 2 12 + 12 6 of 2 (15 8 x 4) of (28 7 of 5) is:

(1) 4 1/6

(2) 4 8/75

(3) 4 2/3

(4) 4 32/75

49. A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9

km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

(1) 9 5/8

(2) 9 2/3

(3) 9 3/8

(4) 9 1/3

50. A 15 m deep well with radius 2.8 m is dug and the earth taken out from it is spread evenly to

form a platform of breadth 8 m and height 1.5 m. What will be the length of the platform? (Take

π = 22/7)

(1) 28.4 m

(2) 28.8 m

(3) 30.2 m

(4) 30.8 m

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Answer keys

41. 4 42. 4 43. 4 44. 2 45. 4 46. 4 47. 4 48. 1 49. 3 50. 4

Solutions

41. 4

Given that, AB = 10 cm, AC = 8.6 cm and BC = 6.4 cm

We know that by tangent property,

BD = BE and CD = CF and AE = AF

Let BD = BE = x cm

So, CD = CF = (6.4 – x) cm

And, AE = AF

=> AB + BE = AC + CF

=> 10 + x = 8.6 + 6.4 - x

=> 2x = 15 – 10

=> x = 2.5 cm

So, BE = 2.5 cm

42. 4

Let number of sides of the regular polygon is n

3 36051

7 n

=

=> n = 360 * 7/360

=> n = 7

So, number of its diagonals = n(n – 3)/2 = 7 * 4/2 = 14

So, required ratio = 14: 7 = 2: 1

43. 4

Let numbers are 3x and 5x.

(3x – 13)/(5x – 13) = 10/21

=> 63x – 273 = 50x – 130

=> 13x = 143

=> x = 11

So, new ratio = (33 + 15): (55 + 15) = 48: 70 = 24: 35

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44. 2

According to the question,

1/72 + 1/90 – 1/C = 1/120

=> 1/C = 1/72 + 1/90 – 1/120

=> 1/C = 1/60

=> C = 60 minutes

Part of tank filled by A and B together in 12 minutes = 12 * (1/72 + 1/90) = 3/10

So, 3/10 part of tank empty by pipe C in = 3/10 * 60 = 18 minutes

45. 4

Let x = 12a and y = 12b

Given, x – y = 60

=> 12a – 12b = 60

=> a – b = 5 ---(1)

And, LCM of x and y = 12ab

=> 12ab = 204 * 12

=> ab = 204 ---(2)

Now, (a + b)2 = (a – b)2 + 4ab = 25 + 4 * 204

=> (a + b)2 = 841

=> a + b = 29 ---(3)

Solving equation (1) and (3), we get

a = 17 and b = 12

So, x + y = 12(17 + 12) = 12 * 29 = 348

46. 4

By orthocentre property,

∠BPC = 180° - ∠A/2

=> 148° = 180° - ∠A/2

=> ∠A/2 = 180° - 148°

=> ∠A = 64°

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47. 4

( ) ( )

( ) ( )( ) ( )

6 6 4 4

4 4 2

2 2 4 2 2 4 4 4 2 2 2 2

4 4 2 2 2 2 4 2 2 2

4 2 2

2 sin cos 3 sin cos

cos sin 2cos

2 sin cos sin sin cos cos 3 sin cos 2sin cos 2sin cos

cos sin 2sin cos 2sin cos 2sin 2cos 2sin 2sin

2 sin 2sin cos co

+ − +

− −

+ − + − + + − =

+ + − − − − +

+ +=

( ) ( )

( ) ( )

4 2 2 2 2

2 2 4 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2

s 3sin cos 3 1 2sin cos

1 2sin cos 2sin 2 2sin

2 1 3sin cos 3 1 2sin cos

1 2sin (cos sin ) 2 2sin

2 6sin cos 3 6sin cos

2sin 1 2sin

2 3

1

1

− − −

− − − +

− − − =

− + − +

− − + =

− − +

−=

−

=

48. 1

24 x 2 12 + 12 6 of 2 (15 8 x 4) of (28 7 of 5)

= 48/12 + 12/12/(15/2) of (28/35)

= 4 + 1/6

= 4(1/6)

49. 3

Let total distance = 100d

Time taken to cover 40% distance = 40d/8 = 5d hours

Time taken to cover 40% of remaining distance = 40/100 * 60d/9 = 8d/3 hours

Time taken to cover remaining distance = (60d – 24d)/12 = 3d hours

So, average speed = (total distance)/(total time) = 100d/(5d + 8d/3 + 3d) = 9(3/8) km/h

50. 4

Volume of well = volume of platform

=> 22/7 * 2.8 * 2.8 * 15 = length * 8 * 1.5

=> length = (22 * 2.8 * 2.8 * 15)/(7 * 8 * 1.5)

=> length = 30.8 m

51. In PQR, ∠Q > ∠R, PS is the bisector of ∠P and PT ⊥ PQ. If ∠SPT = 28 and ∠R = 23, then the

measure of ∠Q is:

(1) 74

(2) 79

(3) 82

(4) 89

52. 25 persons can complete a work in 60 days. They started the work. 10 persons left the work

after x days. If the remaining work was completed in 80 days, then what is the value of x?

(1) 9

(2) 8

(3) 12

(4) 15

53. The value of sin264 + cos64 sin26 + 2cos43 cosec47 is:

(1) 4

(2) 1

(3) 2

(4) 3

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54. A tank is in the form, of a cuboid with length 12 m. If 18 kilolitre of water is removed from it,

the water level goes down by 30 cm. What is the width (in m) of the tank?

(1) 4

(2) 5

(3) 5.5

(4) 4.5

55. In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients

are 1, 11 and 2 respectively. What is sum of the two numbers?

(1) 833

(2) 867

(3) 816

(4) 901

56. A person invested one-fourth of the sum of ₹ 25,000 at a certain rate of simple interest and the

rest at 4% p.a. higher rate. If the total interest received for 2 years is ₹ 4,125, what is the rate

at which the second sum was invested?

(1) 9.5%

(2) 9.25%

(3) 5.25%

(4) 7.5%

57. The radius of the base of a right circular cylinder is 3 cm and its curved surface area is 60π cm2.

The volume of the cylinder (in cm3) is:

(1) 90π

(2) 72π

(3) 60π

(4) 81π

58. If ( )23 x 1 7x

6,x 0,3x

+ −= then the value of

1x

x+ is:

(1) 25

3

(2) 11

3

(3) 35

3

(4) 31

3

59. Basir’s working hours per day were increased by 15% and his wages per hour were increased by

20%. By how much per cent did his daily earnings increase?

(1) 40

(2) 38

(3) 35

(4) 36

60. A student was asked to find the value of4 1 1 1 4 3 1 2 4 2

9 11 of 1 1 2 of of .9 3 6 3 5 5 6 3 3 3

+

His

answer was 1

194

. What is the difference between his answer and the correct answer?

(1) 7 3/4

(2) 6 2/3

(3) 7 1/2

(4) 6 1/3

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Answer keys

51. 2 52. 3 53. 4 54. 2 55. 3 56. 2 57. 1 58. 4 59. 2 60. 1

Solutions

51. 2

In ∆PST

∠RPT = 180° – 90° – 23° = 67°

So, ∠RPS = 67° – 28° = 39°

Since, PS is the bisector of ∠P

So, ∠P = 2 * ∠RPS = 2 * 39 = 78°

Hence, ∠Q = 180° – 23° – 78° = 79°

52. 3

Part of work done by 25 persons in 1 day = 1/60

And, part of work done by 1 person in 1 day = 1/(25 * 60)

And part of work done by 15 persons in 1 day = 15/(25 * 60)

According to the question,

x/60 + 80 * 15/(25 * 60) = 1

=> x/60 + 4/5 = 1

=> x/60 = 1/5

=> x = 12

53. 4

sin264 + cos64 sin26 + 2cos43 cosec47

= sin264 + cos64 cos(90° – 26°) + 2cos43° sec(90° - 47°)

= sin264 + cos264 + 2cos43° sec43°

= 1 + 2

= 3

54. 2

Let breadth of the tank is b m

Length of the tank = 12 m

So, volume of water = 18 kilolitre

=> 12 * b * 0.30 = 18 m3

=> b = 18/(12 * 0.30)

=> b = 5 m

55. 3

In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients

are 1, 11 and 2 respectively.

Since last divisor is 17 and quotient is 2. So, dividend = 17 * 2 + 0 = 34

Now, 34 is the second last divisor and quotient is 11. So, dividend = 34 * 11 + 17 = 391

Now, 391 is the first divisor and 1 is quotient. So, dividend = 391 * 1 + 34 = 425

So, two numbers are 391 and 425

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Hence, sum of two numbers = 391 + 425 = 816

56. 2

Let rate of interest for first and second sum is x% and (x + 4)% respectively

according to the question,

1/4 * 25000 * x * 2/100 + 3/4 * 25000 * (x + 4) * 2/100 = 4125

=> 125 x + 375 (x + 4) = 4125

=> x = 5.25%

So, rate of interest for second sum of interest = 5.25 + 4 = 9.25%

57. 1

Curved surface area of the cylinder = 60π

=> 2πrh = 60π

=> 3 * h = 30

=> h = 10 cm

So, volume of the cylinder = πr2h = π * 3 * 3 * 10 = 90π cm3

58. 4

( )2

2

3 x 1 7x6

3x

1 7x 6

x 3

1 7x 6

x 3

1 25x 2 2

x 3

1 31x

3x

1 31x

3x

+ −=

+ − =

+ = +

+ + = +

+ =

+ =

59. 2

his daily earnings increase by = 100 - 100 * 115/100 * 120/100 = 100 – 138 = 38%

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60. 1

4 1 1 1 4 3 1 2 4 29 11 of 1 1 2 of of

9 3 6 3 5 5 6 3 3 3

85 34 1 4 9 5 13 2 4 2

9 3 6 3 5 3 6 3 3 3

85 9 13 94

9 17 9 8

15 6

2

111

2

+

= +

= +

= +

=

So, required difference =1 1 (77 46) 31 3

19 11 74 2 4 4 4

−− = = =

61. If a 10-digit number 5 4 3 2 y 1 7 4 9 x is divisible by 72, then what is the value of (5x - 4y)?

(1) 14

(2) 15

(3) 10

(4) 9

62. What is the remainder when (12797 + 9797) is divided by 32?

(1) 4

(2) 2

(3) 7

(4) 0

63. The value of ( ) ( )sin cos 1 tan cot

?1 sin cos

− + + =

+

(1) sec - cosec

(2) cosec - sec

(3) sin + cos

(4) tan - cot

64. A, B and C spend 80%, 85% and 75% of their incomes respectively. If their savings are in the

ratio 8:9:20 and the difference between the incomes of A and C is ₹ 18,000, then the income of

B is:

(1) ₹ 24,000

(2) ₹ 27,000

(3) ₹ 30,000

(4) ₹ 36,000

65. If 25% of half of x is equal to 2.5 times the value of 30% of one-fourth of y, then x is what per

cent more or less than y?

(1) 33 1/3% more

(2) 50% more

(3) 33 1/3% less

(4) 50% less

66. The value of ( )2 2tan cosec 1sin cos 1

sin cos 1 sec tan

− + −

− + − is:

(1) 0

(2) -1

(3) 1

(4) 1/2

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67. In an examination, A obtained 10% more marks than B, B obtained 20% more marks than C and

C obtained 32% less marks than D. If A obtained 272 more marks than C, then the marks

obtained by B is:

(1) 850

(2) 816

(3) 1020

(4) 952

68. In quadrilateral ABCD, ∠C = 72 and ∠D = 28. The bisectors of ∠A and ∠B meet in O. What is

the measure of ∠AOB?

(1) 48

(2) 54

(3) 50

(4) 36

69. a, b and c are three fractions such that a < b < c. If c is divided by a, the result is 9/2, which

exceeds b by 23/6. The sum of a, b and c is 19/12. What is the value of (2a + b - c)?

(1) 1/2

(2) 1/3

(3) 1/12

(4) 1/4

70. How many kg of salt costing ₹ 28 per kg must be mixed with 39.6 kg of salt costing ₹ 16 per kg,

so that selling the mixture at ₹ 29.90, there is gain of 15%?

(1) 198

(2) 192

(3) 188

(4) 172

Answer keys

61. 1 62. 4 63. 1 64. 2 65. 2 66. 3 67. 3 68. 3 69. 4 70. 1

Solutions

61. 1

5 4 3 2 y 1 7 4 9 x is divisible by 72

So, number must be divisible by both 8 and 9

Divisibility rule of 8 is: last 3 digit must be divisible by 8;

So, 49x must be divisible by 8 so, x = 6

Now, divisibility rule of 9 is: sum of digits must be divisible by 9

5 + 4 + 3 + 2 + y + 1 + 7 + 4 + 9 + 6 = 41 + y

Thus, y = 4

Hence, 5x - 4y = 5 * 6 - 4 * 4 = 14

62. 4

One of the factors of (12797 + 9797) = (127 + 97) = 224

Since, 224 is divisible by 32.

So, remainder when (12797 + 9797) is divided by 32 is 0.

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63. 1

( ) ( )

( ) ( )

( )

( ) ( )2 2

sin cos 1 tan cot

1 sin cos

sin cos 1 tan cot sin cos

(1 sin cos )sin cos

sin cossin cos sin cos sin cos sin cos

cos sin

(1 sin cos )sin cos

sin cos sin cos sin cos

(1 sin cos )s

− + +

+

− + + =

+

− + +

=

+

− + + =

+

( ) ( )

in cos

sin cos sin cos 1

(1 sin cos )sin cos

sec cosec

− +=

+

= −

64. 2

Let income of A and C are 100x and 100y respectively and saving of A and C are 20x and 25y

respectively.

And, 20x/25y = 8/20

=> x/y = ½

Let x = k and y = 2k

According to the question

100y – 100x = 18000

=> 200k – 100k = 18000

=> k = 180

So, saving of A = 20 * 180

Thus, saving of B = 20 * 180 * 9/8 = 4050

Hence, income of B = 4050 * 100/15 = Rs. 27000

65. 2

According to the question,

25/100 * x/2 = 2.5 * 30/100 * 1/4 * y

=> x = 3/2 * y

=> x/y = 3/2

So, percentage change = (3 – 2)/2 * 100 = 50%

Thus, x is 50% more than y.

66. 3

( )

( )

2 2

2 2

2 2

tan cosec 1sin cos 1

sin cos 1 sec tan

sin cos 1tan cot

cossin cos 1 sec tan

cos

tan 1 sec 1

tan 1 sec sec tan

sec tan tan sec 1

tan 1 sec sec tan

(sec tan )(sec tan ) (

− + −

− + −

+ −

= − + −

+ − =

− + −

− + − =

− + −

− + −=

sec tan ) 1

tan 1 sec sec tan

(sec tan )(sec tan 1) 1

tan 1 sec sec tan

1

−

− + −

− + −=

− + −

=

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67. 3

Let D obtained 100x marked in the examination.

So, C obtained = 100x * 68/100 = 68x

And, B obtained = 68x * 120/100 = 81.6x

And, A obtained = 81.6x * 110/100 = 89.76x

According to the question,

89.76x – 68x = 272

=> x = 272/21.76

=> x = 12.5

So, marks obtained by B = 81.6 * 12.5 = 1020

68. 3

By bisector property,

∠AOB = 1/2 * (∠C + ∠D)

=> ∠AOB = 1/2 * (72 + 28)

=> ∠AOB = ½ * 100

=> ∠AOB = 50°

69. 4

c/a = 9/2

So, let c = 9k and a = 2k

and b + 23/6 = 9/2

=> b = 9/2 – 23/6

=> b = 4/6 = 2/3

Given that, a + b + c = 19/12

=> 2k + 2/3 + 9k = 19/12

=> 11k = 19/12 – 2/3

=> 11k = 11/12

=> k = 1/12

So, 2a + b – c = 2 * 2/12 + 2/3 – 9 * 1/12 = 1/3 + 2/3 – 3/4 = 1/4

70. 1

Let x kg of salt costing Rs. 28 per kg mixed with another salt

According to the question,

28x + 39.6 * 16 = (39.6 + x) * 29.9 * 100/115

=> 28 x + 633.6 = 26 (x + 39.6)

=> 2x = 1029.6 – 633.6

=> x = 198 kg

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71. Study the following bar graph and answer the question given.

The total production of motorcycles of companies C, D and E is what per cent less than the total

demand of motorcycles of all the companies for five years?

(1) 43

(2) 32

(3) 38

(4) 47

72. A, B and C started a business. Thrice the investment of A is equal to twice the investment of B

and also equal to four times the investment of C. If C’s share out of the total profit is ₹ 4,863,

then the share of A in the profit is:

(1) ₹ 7,272

(2) ₹ 6,484

(3) ₹ 9,726

(4) ₹ 8,105

73. Two positive numbers differ by 2001. When the larger number is divided by the smaller number,

the quotient is 9 and the remainder is 41. The sum of the digits of the larger number is:

(1) 15

(2) 11

(3) 10

(4) 14

74. Let 6 3 45 605 245x 27 6 andy ,

4 80 125

+ += − =

+ then the value of x2 + y2 is:

(1) 223/36

(2) 221/36

(3) 221/9

(4) 227/9

75. If (5x + 2y):(10x + 3y) = 5:9, then (2x2 + 3y2):(4x2 + 9y2) = ?

(1) 31:87

(2) 10:27

(3) 16:47

(4) 1:3

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76. The average of 18 numbers is 37.5. If six numbers of average x are added to them, then the

average of all the numbers increases by one. The value of x is:

(1) 40

(2) 41.5

(3) 42

(4) 38.5

77. In an office, 5/8 of the total number of employees are males and the rest are females. 2/5 of the

number of males are non-technical workers while 2/3 of the number of females are technical

workers. What fraction of the total number of employees are technical workers?

(1) 5/8

(2) 2/5

(3) 1/2

(4) 3/8

78. A solid cylinder of base radius 12 cm and height 15 cm is melted and recast into n toys each in

the shape of a right circular cone of height 9 cm mounted on a hemisphere of radius 3 cm. The

value of n is:

(1) 27

(2) 64

(3) 48

(4) 54

79. In ABC, D and E are the points on AB and AC respectively such that AD x AC = AB x AE. If

∠ADE = ∠ACB +30 and ∠ABC = 78, then ∠A = ?

(1) 56

(2) 54

(3) 68

(4) 48

80. P and Q are two points on the ground on either side of a pole. The angles of elevation of the top

of the pole as observed from P and Q are 60 and 30, respectively and the distance between

them is 84 3 m. What is the height (in m) of the pole?

(1) 63

(2) 73.5

(3) 52.5

(4) 60

Answer keys

71. 2 72. 2 73. 4 74. 1 75. 1 76. 2 77. 1 78. 3 79. 2 80. 1

Solutions

71. 2

The total production of motorcycles of companies C, D and E together

= 72 + 75 + 40 = 187

And, total demand of motorcycles of all the companies for five years

= 50 + 45 + 60 + 65 + 55 = 275

Difference = 275 – 187 = 88

Therefore, percentage = (88/275) * 100 = 32%

72. 2

Let investment of A = a

Investment of B = 3a/2

Investment of C = 3a/4

Then, profit ratio, A: B: C = a: (3a/2): (3a/4) = 4: 6: 3

Share of C in profit = 4863

Then, total profit = 4863 * (4 + 6 + 3)/3

And, share of A in profit = 4863 * ((4 + 6 + 3)/3) * 4/(4 + 6 + 3) = Rs.6484

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73. 4

Let two positive numbers are a (larger) and b (smaller). Then,

a = b + 2001

a = b * 9 + 41

Then, b + 2001 = 9b + 41

b = 245

And, a = 245 + 2001 = 2246

Therefore, sum of the digits of the larger number = 2 + 2 + 4 + 6 = 14

74. 1

6

2

2

3 27x 27 6 3

4 4

3 33 1

2 2

3 3x

2 4

= − = −

= − = −

= − =

And,

2

45 605 245 3 5 11 5 7 5y

80 125 4 5 5 5

21 5 7

39 5

49y

9

+ + + += =

+ +

= =

=

Therefore, x2 + y2 = (3/4) + (49/9) = 223/36

75. 1

(5x + 2y):(10x + 3y) = 5:9

9(5x + 2y) = 5(10x + 3y)

45x + 18y = 50x + 15y

x = 3y/5

Therefore, (2x2 + 3y2):(4x2 + 9y2)

= (2(3y/5)2 + 3y2):(4(3y/5)2 + 9y2)

= 31: 87

76. 2

According to the question:

18 * 37.5 + 6x = (18 + 6) * (37.5 + 1)

x = 41.5

77. 1

Let total number of employees in office = n

Number of male employees = 5n/8

Number of female employees = n – 5n/8 = 3n/8

Number of male technical workers = (1 – 2/5) * 5n/8 = 3n/8

Number of female technical workers = (2/3) * 3n/8 = n/4

Then, total technical workers = (3n/8) + (n/4) = 5n/8

Therefore, fraction = (5n/8)/n = 5/8

78. 3

Radius of cylinder = 12 cm

Height of cylinder = 15 cm

Height of conical part of each toy = 9 cm

Radius of hemispherical part of each toy = 3 cm

Volume of cylinder = π * 122 * 15 = 2160π

Volume of each toy = (1/3) * π * 32 * 9 + (2/3) * π * 33 = 45π

Then, volume of cylinder = n * volume of each toy

2160π = n * 45π

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n = 48

79. 2

Here, AD x AC = AB x AE

AD/AB = AE/AC

So, DE is parallel to BC. Then,

∠ADE = ∠ABC = ∠ACB +30 and ∠ABC = 78

Then, 78˚ = ∠ACB +30

∠ACB = 48˚

Now, ∠A + ∠ACB + ∠ABC = 180˚

∠A + 48˚ + 78˚ = 180˚

∠A = 54˚

80. 1

According to the question:

Here, PQ = 84√3 m

tan 60˚ = AB/PB = √3

AB = √3PB

And, tan 30˚ = AB/BQ = 1/√3

AB/(84√3 – PB) = 1/√3

AB = (84√3 – PB)/√3

Then, √3PB = (84√3 – PB)/√3

PB = 21√3

Therefore, AB = √3 * 21√3 = 63 m

81. If in PQR, ∠P = 120, PS ⊥ QR at S and PQ + QS = SR, then the measure of ∠Q is:

(1) 20

(2) 50

(3) 40

(4) 30

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82. The given pie-chart shows the break-up of total marks obtained by a student in five subjects A,

B, C, D and E. The maximum marks in each subject is 150 and he obtained a total of 600 marks.

In how many subjects did the student obtain more than his average score?

(1) 3

(2) 2

(3) 4

(4) 1

83. Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late. His

usual time (in hours) to reach the destination is:

(1) 2 1/2

(2) 2 1/4

(3) 3 1/8

(4) 3 1/4

84. A man can row a distance of 900 metres against the stream in 12 minutes and returns to the

starting point in 9 minutes. What is the speed (in km/h) of the man in still water?

(1) 4 1/2

(2) 6

(3) 5 1/4

(4) 5

85. If x + y + z = 6, xyz = -10 and x2 + y2 + z2 = 30, then what is the value of (x3 + y3 + z3)?

(1) 132

(2) 135

(3) 130

(4) 127

86. The value of ( ) ( ) ( )

( ) ( )

4 4 2

2 2

4.6 5.4 24.84

4.6 5.4 24.84

+ +

+ + is:

(1) 24.42

(2) 24.24

(3) 25

(4) 25.48

87. If sin 1 cos 4

,1 cos sin 3

+ + =

+ 0<< 90, then the value of (tan + sec)-1 is:

(1) 2 - 3

(2) 3 - 2

(3) 2 + 3

(4) 3 + 2

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88. Sudha bought 80 articles at the same price. She sold some of them at 8% profit and the

remaining at 12% loss resulting in overall profit of 6%. The number of items sold at 8% profit is:

(1) 64

(2) 60

(3) 72

(4) 70

89. The given pie-chart shows the break-up of total marks obtained by a student in five subjects A,

B, C, D and E. The maximum marks in each subject is 150 and he obtained a total of 600 marks

The total marks obtained by the student in subjects C and E is approximately how much per cent

more than what he obtained in A and D together?

(1) 9.09%

(2) 10.25%

(3) 8.33%

(4) 7.26%

90. If the selling price of an article is 32% more than its cost price and the discount offered on its

marked price is 12%, then what is the ratio of its cost price to the marked price?

(1) 4:5

(2) 3:8

(3) 2:3

(4) 1:2

Answer keys

81. 3 82. 2 83. 1 84. 3 85. 1 86. 4 87. 1 88. 3 89. 1 90. 3

Solutions

81. 3

According to the questions:

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Here, ∠P = 120

PQ + QS = SR

Let PQ = x, QS = y, then SR = x + y

And, let ∠Q = a

Then, ∠QPS = 90˚ - a

∠SPR = 120˚ - (90˚ - a) = 30˚ + a

∠SRP = 90˚ - (30˚ + a) = 60˚ - a

Now, sin a = PS/x…

PS = x sin a…(i)

Sin (90˚ - a) = y/x…(ii)

Sin (30˚ + a) = (x + y)/PR…(iii)

Sin (60˚ - a) = PS/PR…(iv)

Then, from (iii) and (iv), we get

Sin (30˚ + a)/Sin (60˚ - a) = (x + y)/PS…(v)

Now, from (i) and (v), we get

Sin (30˚ + a)/Sin (60˚ - a) = (x + y)/(x sin a)

Sin a sin (30˚ + a) = (1 + y/x)(sin (60˚ - a))

Sin a sin (30˚ + a) = (1 + sin (90˚ - a))(sin (60˚ - a)) (from (ii))

Sin a sin (30˚ + a) = sin(60˚ - a) + sin (90˚ - a)(sin (60˚ - a))

Here, sin (30˚ + a) = cos (90˚ – 30˚ – a) = cos (60˚ - a)

Then,

Sin a cos (60˚- a) – cos a (sin (60˚ - a)) = sin(60˚ - a)

Sin (a – (60˚-a)) = sin(60˚ - a)

(a – (60˚- a)) = (60˚ - a)

a = 40˚

82. 2

Total marks obtained by student = 600

Average marks obtained by student = 600/5 = 120

Marks obtained in A = (60/360) * 600 = 100

Marks obtained in B = (84/360) * 600 = 140(more than 120)

Marks obtained in C = (68/360) * 600 = 113.33

Marks obtained in D = (72/360) * 600 = 120

Marks obtained in E = (76/360) * 600 = 126.67 (more than 120)

Therefore, number of subjects in which student got more marks than his average marks = 2

83. 1

Let speed of man = s km/hr

Let actual time taken = t hours

Then, distance covered = s * t

Also, distance covered = (60% of s) * (t + 1+ 40/60) = (3t + 5)s/5

Then, s * t = (3t + 5)s/5

t = 2(1/2) hours

84. 3

Let speed of man in still water = a km/hr

Speed of stream = s km/hr

Upstream speed = 900/(12 * 60) = 5/4 m/sec = (5/4) * (18/5) = 9/2 km/hr

a – s = 9/2…(i)

Downstream speed = 900/(9 * 60) = 5/3 m/sec = (5/3) * (18/5) = 6 km/hr

a + s = 6…(ii)

Now, from (i) and (ii), we get

s = a – 9/2 = 6 – a

a = 5(1/4) km/hr

85. 1

We know, (x + y + z)2 = x2 + y2 + z2+ 2xy + 2yz + 2xz

And, x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2– xy – yz – xz)

Here, x + y + z = 6, xyz = -10 and x2 + y2 + z2 = 30

Then, 62 = 30 + 2xy + 2yz + 2xz

xy + yz + xz = 3

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And, x3 + y3 + z3 – 3(-10) = (6)(30 – 3)

x3 + y3 + z3= 132

86. 4

( ) ( ) ( )

( ) ( )

( )( ) ( )( ) ( )( )( ) ( )

3 3 2 2

3 32 2

3 3 2 2

3 32 2

4 4 2

2 2

2 22 2 2

2 2

6 6

2 2

3

Since,a b (a b)(a ab b )

a bAnd,(a ab b )

a b

Also,a b (a b)(a ab b )

a bAnd,(a ab b )

a b

4.6 5.4 24.84Then,

4.6 5.4 24.84

4.6 5.4 24.84

4.6 5.4 24.84

(4.6 5.4 )

(4.6 5.4 )

(4.6 5.

− = − + +

−+ + =

−

+ = + − +

+− + =

+

+ +

+ +

+ +=

+ +

−

−=

− 3

3 3 3 3

3 3

3 3

2 2

4 )

(4.6 5.4)

(4.6 5.4 )(4.6 5.4 ) (4.6 5.4)

(4.6 5.4)(4.6 5.4) (4.6 5.4 )

(4.6 5.4 )

(4.6 5.4)

4.6 4.6 5.4 5.4

25.48

−

− + −=

− + −

+=

+

= − +

=

87. 1

2 2

1

1

1

sin 1 cos 4

1 cos sin 3

sin 1 cos 2cos 4

sin (1 cos ) 3

1 1 2cos 4

sin (1 cos ) 3

2(1 cos ) 4

sin (1 cos ) 3

3sin

2

sin sin60

60

Then,(tan sec )

(tan60 sec60 )

( 3 2)

1 (2 3)

(2 3) (2 3)

2 3

4 3

2

−

−

−

+ + =

+

+ + + =

+

+ + =

+

+ =

+

=

=

=

+

= +

= +

−=

+ −

−=

−

= 3−

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88. 3

Total number of articles = 80

By alligation method:

Then, ratio = 18/2 = 9: 1

Therefore, number of articles sold at 8% profit = 80 * 9/(9 + 1) = 72

89. 1

Total marks obtained by student = 600

Marks obtained in C and E together = ((68 + 76)/360) * 600 = 240

Marks obtained in A and D together = ((60 + 72)/360) * 600 = 220

Difference = 240 – 220 = 20

Therefore, percentage = (20/220) * 100 = 9.09%

90. 3

Let CP of article = Rs. a

SP of article = (100 + 32)% of a = 33a/25

Discount% = 12%

Then, MP of article = (33a/25) * 100/(100 – 12) = 3a/2

Therefore, the ratio of cost price to the marked price of article

= a: (3a/2)

= 2: 3

Alternative method

CP : MP

(100-Discount%) : ( 100 + profit%)

(100-12) : (100+32)

88 : 132

2 : 3

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91. Study the following bar graph and answer the question given

The number of companies whose production of motorcycles is equal to or more than the average

demand of motorcycles (per year) over five years is:

(1) 4

(2) 2

(3) 1

(4) 3

92. The internal diameter of a hollow hemispherical vessel is 24 cm. It is made of a steel sheet

which is 0.5 cm thick. What is the total surface area (in cm2) of the vessel?

(1) 612.75π

(2) 468.75π

(3) 600.2π

(4) 600.5π

93. The bisector of∠A in ABC meets BC in D. If AB = 15 cm, AC = 13 cm and BC = 14 cm, then DC

= ?

(1) 8.5 cm

(2) 7.5 cm

(3) 6.5 cm

(4) 8 cm

94. A certain loan was returned in two equal half yearly instalments each of ₹ 6,760. If the rate of

interest was 8% p.a. compounded yearly, how much was the interest paid on the loan?

(1) ₹ 750

(2) ₹ 810

(3) ₹ 790

(4) ₹ 770

95. A sum is divided among A, B, C and D such that the ratio of the shares of A and B is 2:3, that of

B and C is 1:2 and that of C and D is 3:4. If the difference between the shares of A and D is ₹

648, then the sum of their shares is:

(1) ₹ 2,052

(2) ₹ 2,160

(3) ₹ 2,484

(4) ₹ 1,944

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96. The given pie-chart shows the break-up of total marks obtained by a student in five subjects A,

B, C, D and E. The maximum marks in each subject is 150 and he obtained a total of600 marks.

What is the difference between the marks obtained by the student in subjects B and D?

(1) 20

(2) 27

(3) 30

(4) 12

97. A sector of radius 10.5 cm with the central angle 120is folded to form a cone by joining the two

bounding radii of the sector. What is the volume (in cm3) of the cone so formed?

(1) 343 2

6

(2) 343 3

6

(3) 343 3

12

(4) 343 2

12

98. A certain sum amounts to ₹ 4,205.55 at 15% p.a. in 2 2/5 years, interest compounded yearly.

The sum is:

(1) ₹ 3,200

(2) ₹ 3,500

(3) ₹ 2,700

(4) ₹ 3,000

99. In ABD, C is the midpoint of BD. If AB = 10 cm, AD = 12 cm and AC = 9 cm, then BD = ?

(1) 2 41 cm

(2) 2 10 cm

(3) 41 cm

(4) 10 cm

100. A sum of ₹ 10,500 amounts to ₹ 13,825 in 3 (4/5) years at a certain rate per cent per

annum simple interest. What will be the simple interest on the same sum for 5 years at double

the earlier rate?

(1) ₹ 8,470

(2) ₹ 8,750

(3) ₹ 8,670

(4) ₹ 8,560

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Answer keys

91. 4 92. 1 93. 3 94. 4 95. 1 96. 1 97. 4 98. 4 99. 1 100. 2

Solutions

91. 4

Here, average demand of motorcycles (per year) over five years

= (50 + 45 + 60 + 65 + 55)/5

= 55 lakhs

Then, production of motorcycle in company A, C and D are 55 lakhs, 72 lakhs and 75 lakhs

respectively which is more than the average demand of motorcycles (per year) over five years.

Therefore, the number of companies whose production of motorcycles is equal to or more than

the average demand of motorcycles (per year) over five years = 3

92. 1

Internal radius = 24/2 = 12 cm

External radius = 12 + 0.5 = 12.5 cm

Therefore, total surface area of hemispherical vessel

= 2 * π * 12.52 + 2 * π * 122 + π * (12.52 – 122)

= 612.75π cm2

93. 3

According to the question:

AB = 15 cm, AC = 13 cm, BC = 14 cm

Here, AB/BD = AC/DC

AB/(BC – DC) = AC/DC

15/(14 – DC) = 13/DC

DC = 6.5 cm

94. 4

Let the amount to be paid = Rs. a

Each instalment = 6760

Rate% = 8%

Then, a = (6760/(1 + 8/200)) + (6760/(1 + 8/200)2)

a = 12750

Therefore, interest paid = 2 * 6760 – 12750

= Rs.770

95. 1

A: B = 2: 3

A = 2B/3

B: C = 1: 2

C = 2B

C: D = 3: 4

D = 4C/3 = (4/3) * 2B = 8B/3

D – A = (8B/3) – (2B/3) = 648

B = 324

A = 2 * 324/3 = 216

C = 2 * 324 = 648

D = 648 + 216 = 864

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Therefore, sum = A + B + C + D

= 216 + 324 + 648 + 864

= Rs.2052

96. 1

Total marks obtained by student = 600

Marks obtained in B = (84/360) * 600 = 140

Marks obtained in D = (72/360) * 600 = 120

Therefore, difference = 140 – 120 = 20

97. 4

Radius of sector = 10.5 cm

Central angle = 120˚

Then, circumference of the base of cone = (120/360) * 2 * * 10.5 = 7 cm

Now, circumference of base of cone = 2 * * radius of cone = 7

Radius of cone = 3.5 cm

And, slant height of cone = length of sector = 10.5 cm

Then, height of cone = √(10.52 – 3.52) = 7√2 cm

Therefore, volume of cone = (1/3) * * 3.52 * 7√2

= 343 2

12 cm3

98. 4

Rate = 15%, time = 2(2/5) years

Sum after 2(2/5) years = 4205.55

Then, 4205.55 = sum * (1 + 15/100)2 * (1 + (2/5) * 15/100)

Sum = Rs.3000

99. 1

According to the question:

Here, AB = 10 cm, AD = 12 cm and AC = 9 cm

BC = CD = BD/2

Let BC = CD = a

Since, AC is median. So,

Area of ΔABC = area of ΔACD = (1/2) * area of ΔABD…(i)

In ΔABD:

Half of Perimeter = (AB + AD + BC)/2 = (10 + 12 + 2a)/2 = 11 + a

Then, area of ΔABD

= √((11 + a)(11 + a – 10)(11 + a – 12)(11 + a – 2a))

= √((121 – a2)(a2 – 1))

In ΔABC:

Half of perimeter = (AB + AC + BC)/2 = (10 + 9 + a)/2 = (19 + a)/2

Then, area of ΔABC

= √(((19 + a)/2)((19 + a)/2 – 10)((19 + a)/2 – 9)((19 + a)/2 – a))

= (√((361 – a2)(a2 – 1)))/4

Then, from (i), we get

(√((361 – a2)(a2 – 1)))/4 = (1/2) * √((121 – a2)(a2 – 1))

(361 – a2)(a2 – 1) = 4 * (121 – a2)(a2 – 1)

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a = √41 cm

100. 2

Sum = 10500

Sum after 3(4/5) years at simple interest = 13825

Then, 13825 – 10500 = 10500 * rate * (19/5)/100

Rate% = (25/3)%

Therefore, simple interest on the same sum for 5 years at double the earlier rate

= 10500 * (2 * 25/3) * 5/100

= Rs.8750

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