QUANT TEST PAPER 4

**Explanatory answers for paper IV**

1. Required distance = sqrt (3-3) ^{2} + (3 – (- 4) ^{2}
)

= sqrt (7) ^{2}

= 7 Units.

Hence[1]

2. The required point is

[ 2 * 8 + 1 * 5 / 2 + 1 , 2 * 5 + 1 * (- 3 ) / (2 + 1) ]

= 21 / 3, 15 / 3

= ( 7 , 5)

Hence[2]

3. A is the midpoint of yz,

The coordinates of A are [ 2+4 / 2 , - 2 + 4 / 2 , -2 + 2 / 2 ]

= ( 3 , 0)

XA = sqrt (6 – 3) ^{2 }+ ( - 2 – 0 ) ^{2}

=sqrt 3 ^{2} + 2 ^{2 }

= sqrt 9 + 4 = sqrt 13 Units

Hence[3]

4. Let the ratio be K : 1

TK + 1 * ( - 4 ) / K + 1 = 0

TK = 4

K = 4 / 7

The ratio is 4/7 : 1

= 4 : 7

Hence [4]

5. b + b + 2 / 2 = 2

2b + 2 = 4

2b = 2

b = 1

Hence[1]

6. Slope = y 2 - y1 / x2 - x1

= 3 - 6 / 5 + 4 = -3 / 9 = -1/3

Hence[2]

7. Putting y = 5 in x ^{2} + y ^{2} = 34

= x ^{2 }+ 25 = 34

x 2 = a \ x = ± 3

points are ( 3 , 5 ) and ( 3, -5)

Hence[3]

8. The series is 5^{2}, 5^{3}, 6^{2} and so on

9. Value of Machinery after 2 years = 60000 ( 1 – 15 / 100) ^{2}

60000 ( 115 / 100) ^{2}

= Rs 43350

Hence[1]

10. 8 0 / 100 P = 40 / 100 Q = 40 / 100 * X/100

8 / 10 = 4 x / 1000

X = 8000 / 40 = 200

Hence[2]

11. Population 2 years ago = 16093 / (1 + 10 / 100 ) ^{2}

= 16093 * 10000 / 12100

= 13300

Hence[1]

12. A:C = 2/3 * 3/4

=1:2

Hence[4]

13. Let the fraction be x/y

New fraction = 115% of x / 110% of y = 23x / 22y = 15 / 26

x / y = ( 15/26 * 22 / 23 ) = 330 / 598 = 165 / 299

Hence[4]

14. When y = 10, X = 20, Z = 5

x = k, and z = k _{2} / y 5 = k _{2} / 10

K 2 = 50

X = 2y and Z = 50 / y

When y = 20

Z = 50 / 20 = 2.5

Hence[1]

15. For a perfect division into whole numbers the sum of the terms of the ratio must
divide 10 Therefore the ratio cannot be 3 : 4

Hence[4]

16. SP = 10 , Q = 15 , R = x

P = x / 5, Q = x / 10 and R = x / 15

P : Q : R = x /5 : x / 10 : x / 15

= 6 : 3 : 2

Hence[1]

17. S.P at 50% discount = Rs 250

S.P. after 2 successive discounts of 5 % and 5% = 95% of ( 55% of 500) =

[ 95 / 100 * 55 / 100 * 500 ]

= Rs. 261.25

differences = Rs 11.25

Hence[2]

18. Rate unstream = 10 / 4 = 2.5 kmph

Rate downstream = 20 / 4 = 5 kmph

velocity of current = 1 / 2 ( 5 - 2.5 ) kmph

= 1.25 kmph

Hence[2]

19. Rate upstream = 50 / 10 = 5 kmph

Rate downstream = 40 / 9 = 4.44 kmph

Rate in still water = 1 / 2 ( 5 * 4.44)

= 4.72 kmph

Hence[4]

20. Rate upstream = 1

rate downstream = 7

Total time take = [ 14 / 1 + 14 / 7]

= 14 + 2 = 16 hours

Hence[1]

21. A cone is generated with radius 10 cm & vertical height = 6 cm

Volume = 1 P / 3 * 100 * 6 = 200 P

Hence [ 1 ]

22. Let him have Rs. X when he intered the Amount spent = 2 + x / 5

x – 2 – x / 5 = 0

5x - 10 – x = 0

4x = 10 , x = 2.5

When I entered the 3^{rd} shop, I had 2.5 ( 2.5 + 2 ) = 11.25

When I entered the 2^{nd} shop I had 11.25 + 2 ( 2.5) = 33.125

When I entered the 1^{st} shop I had 33.125 + 2 (2.5) = 87.8125

Hence [1]

23. Total time = 5 / 1 + 4 / 2 + 12 / 6

= 9 hours

Total distance = 21 km

Average speed = 21 / 9 = 2.33 every hr.

Hence[1]

24. It cannot be determined as it depends on the position of the path. Hence[4]

25. ( p - q ) 2 = ( x – y ) 2

p - q = ± x – y

p – q = x - y or y – x

x = p - q + y or q - p + y

Hence [3]

26. Let the sum interested at 6% be x

X * 7 * 6 / 100 = 100

= ( 13400 – x ) * 7 * 4 / 100

42 x / 10 = 93800 – 7x / 25

42 x = 93800 * 4 - 28 x

70 x = 93800 * 4 / 70 = Rs 5360

Hence[1]

27. Let ‘d ‘ be the distance and t’ be the normal time

D / 6 = t – 10 / 60

D / 5 = t + 10 / 60

D / 6 - d / 5 = - 10 / 60 - 10 / 60

5d – 6d / 30 = -20 / 60

-2d = -20

d = 10 km

Hence [ 1 ]

28. Each number in the series in the precious numbers added to the sum
of its digits.

the last no. = 1 + 4 + 7 + 147

= 159

Hence [ 3 ]

29. Both the digits must be even and odd multiples of 6 will not be
divisibleby 12

The number is 42

And its reverse is 24

The difference = 18

Hence [ 1 ]

30. Let there be x tractors

Area of farm = 2x 2

When one tractor is stolen x – 1 will

Remain 2x ^{2} - 2(x – 1) ^{2 }= 62

Solving

2x ^{2 }- 2 (x ^{2 }- 2x + 1) = 62

2x ^{2} - 2x ^{2} + 4x - 2 = 62

4x = 60

x = 15

No. of tractors = 15 - 1 = 14

Hence [ 1 ]

31. All seven digits with have to be used to make a number greater than
a million.

Since there are 3 6’s and 2 7’s the number of distinct persutations = 7! / 2! 3!

But all persutations starting with zero should let be counted

7! / 2! 3! - 5! / 2! 3!

= 410

Hence [1]

32. Let the speed of the burglar and the guard be ‘x’ min /sec
and ‘y’ min / sec

The guard covered the distance in 3min 10 sec for which the thief took 5 minutes

Therefore 300 x = 190 Y

30x = 19y

Also given that

356 * x = 3560 m

x = 10 m / sec

and y = 300 * 10 / 190 = 15.07 m / sec

Hence [ 1 ]

33. Let profit be Rs p

Then x gets 0.25 p. and y gets 0.75 p. in the ratio of their investement.

Y pay Rs. 10, 000 per month for 8 months

= 10000 * 8 = Rs 80, 000

= 0.25 p + 80000 = 0.75 p - 80000

0. 5 p = 160000

P = 320, 000

Hence [2]

34. The series is 3*1+1, 4*3+1 and so on

35. For any positions of A B & C there are 2 ways of completing the
queue either D willl be

ahead or behind E since of the total combinations of forming a queue half will have D
ahead

of E.

Total No. of ways = 5! = 120

But in this case = 60

Hence[ 1 ]

36. One task = 5 men 10 hrs 1 day = 50 - man hours

same task = 4 women 8 hrs 2 days = 64 - women hours

same tassk = 5 boys 5 hrs 4 days = 100 - boy hours

each day total labour available = 20 men = (200 man hours) + 9 women (72 women hours ) +
10 boys = 50 boy hours

200 * 100 / 50 + 72 * 100 / 64 + 50

= 400 + 150 + 50

= 600 boy hours

boy hours task days

100 1
1

600 1000
?

= 1000 * 100 / 600 = 167 days

from 1^{st} march 167 days = 14 th August.

Hence [ 4 ]

37. Let y’s salary be x

x’s salary = 150 x

z’s salary = 75 x / 100

x + 75 x / 100 + 150x / 100 = 325 x / 100

325x / 100 = 3, 25, 000

x = 325000 * 100 / 325 = 1, 00, 000.

[ Hence ]

38. Distance Time Speed Days

x 24 - 18 = 6 hrs y 200

2x 24 - 9 = 15 hrs 2y ?

Days = 200 * 2x / x * 6 / 15 * y / 2y = 80 days.

Hence [ 1 ]

39. In 1 minute, tyre flat = 1 / 10 -------- Puncture (I)

In 1 minute, tyre flat = 1 / 5 -------- puncture (II)

Together 1 / 10 + 1 / 5 = 3 / 10 in one minute

Remaining = 7 / 10

= 7 / 10 * 10 / 3 = 7 / 3

= 2 1/ 3 minutes

Hence[1]

40. ( 1 ) ( 1 ) / 20 – 4 = 1 / 16

Hence [1]

41. Two numbers whose sum is 11 and product of their squares is 30 are
sqrt 5 & sqrt 6

sqrt 11 + 2 sqrt 30

= sqrt ( sqrt 5 + sqrt 6 ) ^{2}

= ( sqrt5 + sqrt 6 )

Hence [ 1 ]

42. 17164 - 3 = 17161

sqrt 17161 = 131

No. of men in the last row was 131

Hence [ 2 ]

43. Let the term be x : x + 50

X / x + 50 = 3 / 5

5x = 3x + 150 = 2x = 150

x = 75

Hence [4]

44. The ratio of coins = 3/1 : 2/2 : 4/4

= 3 : 1 : 1

The amount of 25 paise coins is Rs. 60

No of coins = 60 / 0.25

= 240 coins

Hence [1]

45. J + A + V = 11250

J = ¼ ( A + V )

4 J = A + V

5 J = 11250

J = 11250 / 5 = 2250

Hence [ 3 ]

46. Let x invest Rs a for 12 months Y invest Rs 4a for 4 months Z invest Rs 3a for 2 months

The ratio is 12a : 16a : 6a

Z’s share = Rs 8500 * 3 / 17 = Rs 1500

Hence [ 4 ]

47. S. P of 75 toys = C P of 75 toys

Let CP of each toy = Rs 1

CP of 50 toys = Rs 50

SP of 50 toys = Rs 75

= 25 * 100 / 50

= 50%

Hence [ 3 ]

48.Let Q’s contribution = Rs x

8000 * 6 : 3x

16000 : x

ratio of profit ½ : ½

= 1 : 1

16000 / x = 1 / 1

= Rs 16000

Hence [ 4 ]

49. Let the marked price be Rs. 100

Final SP after 2 discounts = 15%, 75% of Rs 100

= Rs. 63.75

Single discount = 100 - 63.75 = 36.25 %

Hence [ 3 ]

50. Let C.P. of 1 car=x

C.P. of 4 cars=4x

% profit=2x/4x*100

=50%

Hence[1]