**A school room is to be built to accommodate 70 children, so as to allow 2.2 m**^{2}of floor and 11 m^{3}of space for each child. If the room be 14 metres long, what must be its breadth and height?- 12 & 5.5 metres
- 13 & 6 metres
- 11 & 5 metres
- 11 & 4 metres

Answer: Option C.

22 × 70 = 154 m2 for 70 students.

11 × 70 = 770 m3 for 70 students.

Matching through options answer is 3rd option.**Two pipes P and Q can fill a cistern in 3 and 6 minutes respectively, while an empty pipe R can empty the cistern in 4 minutes. All the three pipes are opened together and after 2 minutes pipe R is closed. Find when the tank will be full**- 3 minutes
- 6 minutes
- 5 minutes
- 8 minutes

Answer: Option A.

1/3 + 1/6 – 1/4 = 3/12.

Work done = 3 × 2 = 6.

Remaining work = 12 – 6 = 6

Hence in addition to 2 minutes, 1 minute more will be needed. So answer is 1st option.**There is a leak in the bottom of a cistern. Before the leak, it could be filled in 4 1/2 hours. It now takes 1/2 hour longer. If the cistern is full, in how much time would the leakage empty the full cistern?**- 23 hours
- 35 hours
- 52 hours
- 45 hours

Answer: Option D.

2/9 – 1/x = 1/5.

x = 45 hours.

So answer is 4th option.**Two filling pipes A and B can fill a tank in 30 and 20 hours respectively. Pipe B alone is kept open for half the time and both pipes are kept open for the remaining time. In how many hours, will the tank be completely full?**- 25 hours
- 40 hours
- 15 hours
- 28 hours

Answer: Option C.

1/30 + 1/20 = 5/60.

Let tank will be completely full in x hours.

x/2 × 1/20 + x/2 × 1/12 = 1.

So x = 15.

Hence, 3rd option.**A man covers a certain distance on a toy train. If the train moved 4 km/hr faster, it would take 30 minutes less. If it moved 2 km/hr slower, it would have taken 20 minutes more. Find the distance**- 60 km
- 45 km
- 30 km
- 20 km

Answer: Option A.

Distance = Speed × Time.

Distance = (Speed+4) (Time-30/60).

Distance= (Speed-2) (Time+20/60).

Solving all equations we get Time = 3 hours,

Speed = 20 km/h.

So, Distance=20 x 3 = 60km

So answer is 1st option.Godrej Preparation Links- Sample Aptitude Questions of Godrej
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**The average speed of a train is 20% less on the return journey than on the onward journey. The train halts for half an hour at the destination station before starting on the return journey. If the total time taken for the to and fro journey is 23 hours, covering a distance of 1000 km, the speed of the train on the return journey is**- 60 km/hr
- 40 km/hr
- 50 km/hr
- 55 km/hr

Answer: Option B.

Let speed of onward journey = x km/h.

Equation will be = 500/x +500/0.8x = 22.5.

By solving we can get the answer as 40 km/h which is 2nd option.**Two trains move from station A and station B towards each other at the speed of 50 km/hr and 60 km/hr. At the meeting point, the driver of the second train felt that the train has covered 120 km more. What is the distance between A and B?**- 1320 km
- 1100 km
- 1200 km
- 960 km

Answer: Option A.

Do this question through options we get the answer as 1st option.**Rohit took a loan of ` 20,000 to purchase an LCD TV set from a finance company. He promised to make the payment after three years. The company charges compound interest at the rate of 10% per annum for the same. But, suddenly the company announces the rate of interest as 15% per annum for the last one year of the loan period. What extra amount does Rohit have to pay due to this announcement of the new rate of interest?**- 1320 km
- 1100 km
- 1200 km
- 960 km

Answer: Option D

20000(1+10/100)3,

20000(1+10/100)2(1+15/100) = 27830.

So taking the difference = 1210.

So answer is 4th option.**A tree was planted three years ago. The rate of its growth is 30% per annum. If at present the height of the tree is 670 cm, what was the height of the tree when it was planted?**- 305 cm
- 500 cm
- 405 cm
- 625 cm

Answer: Option A.

Let 3 years ago the height of tree = x cm.

After 3 years = x(130/100)^{3}= 670.

⇒ x=305

So answer is 1st option**Sanju put equal amounts of money: one at 10% per annum compound interest payable half yearly and the second at a certain percent per annum compound interest payable yearly. If he gets equal amounts after 3 years, what is the value of the second percent?**- 10
^{1}⁄_{5}% - 10%
- 9
^{1}⁄_{2}% - 10
^{1}⁄_{4}%

Answer: Option D.

Let amount of principal = x.

x(1+5/100)^{6}= x(1 + r/100)^{3}.

⇒ r = 10^{1}⁄_{4}%

Hence option 4.- 10