QUANT TEST PAPER 2

**Explanation
to Paper II
**

1. S.I = 60,000 * 10 /100

= Rs. 6000

Rent received = 1500 * 12

= 18000

Therefore debt cleared in one year = 18000 - 6000

=12000

Therefore in 5 years.

2. 800 - 600 = 200

n = 6 months. = 1 /2 year

r = 100 I /Pn

= 100 * 15.50/ 200 * 1/2

= 1550 / 100

= 15.5%

3. A = P ( 1 + r/ 100^{2} ( 1 + r /100)

= 2400 ( 1+ 5 /100^{2 }( 1 + 2 1/2/100)

= Rs. 2712.15

Therefore C. I. = 2712.15 -150; 2400

= Rs. 312.15

4. Perimeter of square = 4a = 4 * 45

=180 m

Length of fence = 10 * 180

=1800m. with 10 lines of wire.

Therefore cost of wire = 1.50 * 1800

= Rs. 2700

5. External volume = l * b * h

= 10 * 6 * 4

= 240 cm^{3}

Internal dimensions = ( 10 -2); ( 6- 2); ( 4- 2)

= 8 , 4 , 2 cm.

Internal volume = 8 * 4 * 2

= 64 cm^{3}

Therefore volume of cube = 240 - 64

= 176 cm^{3}

6. Let radii --> 2x and 3x

Let height --> 5y and 3y

V_{1} /V _{2}= ( 22 /7 ) ( 2x ) ^{2} ( 5 y ) / ( 22
/7 ) ( 3x ) ^{2 }( 3y )

= 20 / 27

7. Diameter of a roller = 60 cm

r = 30 cm = 0.30 m

Height of roller = 1.00 m

Curved surface area = 2 ( 22 / 7 ) rh

= 2 * ( 22 / 7 ) * 0.30 * 1

= 1.32 / 7

= 0.88

Area covered by roller = 500 * 0.88

= 88 * 5

= 440 cm^{2}

Therefore cost of levelling = 0.50 * 440

=5 * 44

= 220

8. Since a+b+c = 0

a+b = -c

(a+b^{2}=(-c^{2})

a^{2}+2ab+b^{2}=c^{2}

a^{2}+b^{2}-c^{2}= -2ab

(a^{2}+b^{2}+c^{2}) ^{2}=(-2ab) ^{2}

or

a^{4}+b^{4}+c^{4}+2a^{2}b^{2}-2b^{2}c^{2}-2c

^{2}a^{2}=4a^{2}b^{2}

or

a^{4}+b^{4}+c^{4}=2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}

or

=2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})

a^{4}+b^{4}+c^{4}/a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}

= 2

9. Let x be the total value of the property.

x = x/2+x/4+1/5x+10000

= (10+5+4)x/20+10000

= 19x/20+10000

= 20*10000

= 200000

10. m/n = 2/3

3m = 2n

now, 3m+5n/6m-n = 2n+5n/4n-n = 7n/3n = 7/3

11. a+b = 27

5a+11b = 195

solving simultaneously we have,

a = 17 b = 10

12. 40% of 5/6 = 40/100*5/6 = 1/3

13. 3 leaps of the dog = 4 leaps of the hare.

1 leap of the dog = 4/3 leaps of the hare.

4 leaps of the dog = 16/3 leaps of the hare.

Speed of the dog : Speed of the hare

16: 15

14. 9 days earlier mean that the work should be completed in (42-9) =
33 days.

No. of Men = 42*55/33 = 70

Hence the no. of extra men = 70-55

=15 men

15. Total cost of the electronic buzzer = Rs.215+Rs.25 = Rs.240

Selling Price = Rs.300

Profit% = 60/240*100 = 25%

**
**16. 7500*20%=1500

No of valid votes=6000

45% of 6000=2700

Hence[1]

17. At the speed of 60 kmph, aunt will reach nashik in 4 hours.

The distance between mumbai and nashik is 240 kms.

If 120 km are to be covered in 3 hours then aunt should travel at the uniform speed of

120/3 = 40kmph

18. If x and y are the ages of Nitya and Nitya's aunt,

x+y = 63

y-x = 4(x-4)

we have Nitya's age = 15 years

her aunt's age = 48 years

19. First watch ticks after 95/90 seconds.

second watch after 323/315 seconds.

LCM = 19*5*17/45 seconds.

No. of times they will tick in first 3600 seconds = 3600/(19*5*17)/45

Once they have ticked together in the begining , 50 in 1 hour they tick 100+1 = 101 times

20. If area = S, Sides = sqrt(S)

If side is doubled i.e. 2sqrt(S)

area = {2sqrt(S)}

^{2}

= 4S

21. Let a and b be the numbers where b is the bigger number.

b =3/2a

a-b = 2400(given)

3/2a

^{2}= 2400

a

^{2}= 1600

a = 40

22. 6 M = 8W, 2W = 3B, 4B = 5G

1 G = Rs. 50 a day.

Now, 1B = 5/4G 1 W = 3/2 * 5/4G

1 M= 8/6* 3/2* 5/4G = 5/2G

1 M = 5/2 * 50 = 125/-Rs.

23. The distance covered by the train in 30 seconds = 6000/3600* 30 metres

= 500 metres.

Length of the Bridge = 500 - 150

= 350m.

24. If we assume that the sum = Rs. 100

Then, 1st rate of interest = 100*100/100*5 = 20

2nd rate of interest = 200*100/100*12 = 16 2/3

76n

_{1}=5, x

_{1}= 42.2 n

_{2}= 4, x

_{2}= 35.5

x

_{12}= n

_{1}x

_{1}+ n

_{2}x

_{2 }/ n

_{1}+n

_{2}

= 5 * 42.2 + 4 * 35.5 /9

= 211.0 + 142.0 /9

= 353 /9

= 39 2/9

25.*6, /3

Hence[1]

26. Area of the four walls.

= 2(length + breadth ) * height

= 2 (30 + 20 ) *12 sq. meteres

= 1200 sq. metres

Total cost of painting = Rs. 1200*25/100

= Rs.300

27.Product of 3 sqrt (4), 6 sqrt (6), 2 sqrt (5) is

4 ^{1/3} * 6 ^{1/6} *5 ^{1/2}

^{=}4^{2/6 }* 6^{1/6 }* 5^{3/6=}(4^{2 }* 6 * 5^{3})^{1/6}

=( 16 *6 *125)^{1/6}

^{= 6 sqrt.12000}

28. A takes 6 hours to complete a work.

He can complete 1/6th of the work in one hour.(taking complete work as 1.)

B takes 8 hours to complete it.

He can complete 1/8th of the work in one hour.

Together, A and B can complete,

1/6+1/8 = 14/48 = 7/24 of the work.

Hour Work

1 7/24

?
24/24

=24/7 = 3 3/7

Hence, [1]

29. In one hour, C can complete 1/12th of the work.

In two hours, he can complete 1/12*2 = 2/12th of the work.

A and B start work simultaneously,

together work completed by A and B,

1/6+1/8 = 14/48 work = 7/24 work.

In 2 hours they complete 7/24*2 = 7/12 work

Total work done by A,B,C = 7/12+2/12 = 9/12

Total work to be completed = 12/12 or 1

Remaining work = 3/12

Hour work done by A

1
1/6

?
3/12

3/12*6 = 1.5 hours

Hence[2]

30./4, *7

Hence[1]

31. In 1 hour C can empty 1/8th of tank

In 1 hour amount of water filled in the tank = A+B-C

= 1/6+1/10-1/8

68/480 = 17/120

Hour
Water filled in tank

1
17/120

?
120/120

= 120/17 = 7.05 hours

32. Let the present ages of Ram and Shyam be R and S resp.

3R-2S = 10 -- I

4S-5R = 10 -- II

15R-10S = 50

-15R+12S = 30

2S = 80

S = 40

R = 30

Hence,[1]

33. Let their present ages be A and B.

3A-B = 100 -- I

2A-B = 50 -- II

3A-B = 100

-2A-B = 50

A = 50

B = 50

Sum of their ages = A+B

=100

Hence[2]

34. Let their present ages be X and Y.

X+Y = 105 --I

1.5X-2Y = 0 --II

X = 60

Y = 45

Hence[4]

35. M
D H L

old 10
20 5 20

new x
5 10 100

10/x = 5/20*10/5*20/100

x = 100

36. a^{4}+b^{4} = (a^{2}+ab+b^{2})(a^{2}-ab+b^{2})

=(a^{2}+4a+8)(a^{2}-4a+8)

=(a^{2}+8)2-(4a)^{2}

=a^{2}+64

Hence[3]

37. x = 10*2 = 20 days

100 men would require 20 days from 46 days

46+20 = 66 days

But we want to complete the work in 56 days.

Hence[1]

38. Exam 1 = 30/50*100 = 60%

Exam 2 = 40/50*100 = 80%

Increase in percent =20%

percent in increase = 20/60*100 = 33 1/3%

39. a)30/100*60/100 = 18/100 = 18%

b)30% is 50% of 60%

c)60% is 200% of 30%

40. a)Old price of sugar = Rs.10/-

New price of sugar = Rs.12.5/-

Old consumption = (10*100) = 1000

New consumption = (12.5*x)= 1000

x = 1000/12.5 = Rs.80/-

Decrease in consumption = 20/100*100 = 20%

b) old price = Rs.10

New price=Rs.15

10*100 = 1000

15*x = 1000

x = 1000/15 = 66.66Rs.

Decrease in consumption = 33.33%

41. a)Let my cost price be 100

Selling price = 90

Profit for Mr. X = 10

Profit % = 10/90*100 = 11.11%

b) Let cost price be 100

Selling Price = 900

My loss = 10 = 10/100*100 = 10%

42. Let cost price be 100 of both horses.

horse1 = S.P. = 110

horse 2 = S.P.= 90

Total cost price = 100+100=200

total selling price= 110+90 = 200

Hence[3]

43. Horse 1- C.P
S.P.

x 100

?
120

S.P. = C.P.(100+profit%/100)

x = 83.33

Horse 2- C.P
S.P.

y 100

?
80

y = 125

Total cost price = cost price of Horse1+cost price of Horse2

= 125 + 83.33 = 208.33

Total S.P.= 100+100 = 200

loss = 8.33

% of loss = 4%

Hence[2]

44. 3 years - 2 times

6 years - 4 times

9 years - 6 times

12 years - 8 times

45. In 60 min - box is full.

In 59 min - box is half filled.

In 58 min - box is 1/4 filled.

hence[3]

2

46. A = 100(1+10/100)

= 100(110/100*110/100)

= 121

47 . Income = Rs.10/-

20%*50 = 10

capital gain = 55-45 = 10

gain = 10+10 = 20

% gain = 20/45*100 = 44.45%

48. Suppose goods cost him Rs.1/kg, then he sells goods costing Rs. 0.95 for Rs.1/-

gain = 1-0.95 = 0.05

% gain = 0.05/0.95*100 = 500/95 = 5%

Hence[1]

49. Let r be the rate of interest.

600*2*r/100 + 150*4*r/100 = 90

1200r/100 + 600r/100 = 90

r = 5

Hence[2]

50. p = Rs.455, n = 4months = 1/3 year, rate = 5%

Amount of Rs.100 = (455*1/3*5)/100 = 107.58

455
?

107.58 100

= 423

q = Rs.450, n = 2months = 1/6 year, rate = 5%

Amount of Rs.100 = (450*1/6*5)/100 = 103.75

450
?

103.75 100

= 434

Sum to be repaid to p = 434-423 = Rs.11

Hence [3]