## QSIP-5

Forum

### QSIP-5

QSIP5

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-5 -Part-1

Q22.
Assume that Tap B is closed after x hours; Tap A works for 8 hours, while Tap B works for x hours

Total work = Wa + Wb
1 = Ra * 8 + Rb* (x)

=> 1 = (1/10) *8 + (1/15)*x or => x = 3 hours

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-5- Part-1

Q 39:

Work = Rate * Time

1 = R *30* 33 OR R = 1/(30*33)

Work done by 44 men on day 1 = 44 [1/(30*33)]

Work Done by 43 men on day 2 =43 [1/(30*33)]
Work Done by 42 men on day 2 =42 [1/(30*33)]

If the work gets finished on the xth day , then the number of people that must be working on xth day = 44-(x-1)

Total work = work done by 44 men on day1 + Work done by 43 men on day 2 +..... work done by 44-(x-1) men on the xth day

1 = 44 [1/(30*33)] + 43 [1/(30*33)] +42[1/(30*33)]+...+[44-(x-1)] [1/(30*33)]

Now check back from the options by putting different values of x . and You will see that none of the first three options satisfies. Hence the answer is d

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-Part-1

Q 40

10 MEN work for 9 hours and 6 women work for 7.5 hours and complete the work in 18 days.
If the number of men remains the same and the number of women rises, the number of days should come down. The only possible answer is 16.

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-5- Part-1

Q43

A garrison of 2000 men has provisions for 45 days:

Total work that can be done = R X T = R (2000 X 45 )

For the first 15 days, the entire garrison eats the food thus the work done = R X 2000 X 15 ...........i

For the remaining number of days a reinforcement arrives of n men and the food lasts for 20 days more
Total men = m = 2000 + n ................ii

R x 2000 x 45 = R x 2000 x 45 + m x 20 x R

On solving for m we get m= 3000

The number of men that arrived = 3000 - 2000 =1000

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-5 -Part-1

Q44

Assume, n men have a provision for d days:

Total work = work done in the first 15 days =by n men + work done by 3n/4 men in the d days more

ndR = n x 15 x R + 3 n x d /4 x R

=> nd/4 = 15n or d/4 = 15 or d = 60 days

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QCS-Part-II

Q 13

Assume the length of the train = l m

The distance covered by the train in crossing a 200 m long platform = length of the train + length of the platform

=> l + 200

(l + 200)/s = 10 [ Where s is the speed of the train in m/s]

=> s =(l+200)/10 .............i

Similarly, the distance travelled by the train in crossing a man = lenght of the train

=> l/s = 6 or s =l/6 ............ii
equating i and ii we get

=> (l+200)/10 = l/6

=> 6 l + 1200 = 10 l => l = 300 m

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Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QSIP-5-Part-I

q48

Work = R x T
Here total cost is analogous to total work , cost per unit is analogous to rate, while the rest is analogous to time

19 = R x 320 x 21 x 11 ................i

Similarly, for the other book cost is not known:

C = R x 297 x 28 x 10 ...............ii

Divide equation ii by i we get

C /19 = (R x 297 x 28 x 10)/(R x 320 x 21 x 11)

C= Rs 21 3/8

Posts: 228
Joined: Wed Feb 13, 2013 5:51 am

### Re: QSIP-5

QCS-5-Part-II

Q 2

Average Speed = Total distance / total time

Total distance = 320 km

total time = Time for the first 160 km + time for the next 160 km

Total time = 160/64 + 160 /80 =4.5

Average Speed = 320 / 4.5 =71.11 kmph