**In a race of 600 metres, A can beat B by 60 metres and in a race of 500 metres, B can beat C by 50 metres. By how many metres will A beat C in a race of 400 metres?**- 76 metres
- 80 metres
- 70 metres
- 84 metres

Answer: Option A

A runs B runs C runs

600 metres race 600m 540 m

500 metres race 500 m 450m

Combing ratio A runs B runs C runs

300metres - 2700meters - 2430metres

Unitary A runs B runs C runs

Method 400mtres - 360 metres - 324 metres

∴ A beats C by 400-324 = 76 metres.**A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km per hour. Find the distance travelled on foot.**- 26 km
- 32 km
- 30 km
- 28 km

Answer: Option B

Total Time = 7hrs

Let the distance travelled by foot @ 8kmph be x kms

∴ distance travlled by bicycle @ 16kmph be (80-x)kms

ATQ. 7hr = x/8 + (80-x)/16

∴ 7 = (2x + 80 - x)/16

∴ x = 32kms**If 20 men can build a wall 112 metres long in 6 days, what length of a similar wall can be built by 25 men in 3 days?**- 65mtr.
- 52mtr.
- 70mtr.
- 78mtr.

Answer: Option C

20 men is 6 days can build 112 metres

25 men in 3 days can build = 112*(25/20)x(3/6)

= 70 meters

**If the compound interest on a certain sum of money for 3 years at 10% per annum be Rs. 993, what would be the simple interest?**- Rs. 880
- Rs. 890
- Rs. 895
- Rs. 900

Answer: Option D

Let P = Principal

A - Amount

We have a = P(1 + R/100)^{3}and CI = A - P

ATQ 993 = P(1 + R/100)^{3}- P

∴ P = Rs 3000/-

Now SI @ 10% on Rs 3000/- for 3 yrs = (3000 x 10 x 3)/100

= Rs 900/-**What annual installment will discharge a debt of Rs. 4600 due in 4 years at 10% simple interest?**- 1000
- 1030
- 1100
- None of these

Answer: Option A

Let the annual instalment be Rs. 100. The first instalment will be paid one year from now i.e. 3 years before it is actually due. The second instalment will be paid two years from now i.e. 2 years before it is actually due.

The third instalment will be paid 1 year before it is actually due.

The fourth instalment will be paid on the day the amount is actually due.

On the first instalment the interest will be paid for 3 years, on the second for 2 years, on the third for 1 year, on the fourth for 0 year. In total an interest for 6 years will be paid (3 + 2 + 1 + 0) on Rs. 100 @ 10%. Interest = (100 × 6 × 10)/100 = Rs. 60 and the principal is Rs 100 × 4 = Rs 400. The total loan that can be discharged is Rs. 400 + 60 = Rs. 460. Here the technique of Chain Rule will be applied. I.e. for Rs. 460 the instalment required is Rs. 100, for Rs. 4600 the instalment required is 4600 × 100/460 = Rs. 1000.**A number whose fifth part increased by 5 is equal to its fourth part diminished by 5, is**- 160
- 180
- 200
- 220

Answer: Option C

x/5 + 5 = x/4 - 5

⇒ x/5 - x/4 = 10

x/20 = 10

⇒ x = 200

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**Two numbers are such that the ratio between them is 3 : 5, but if each is increased by 10, the ratio between them becomes 5:7. The numbers are**- 3, 5
- 7, 9
- 13, 22
- 15, 25

Answer: Option D

No's are in the ratio 3:5

Le the No's be 3x and 5x

ATQ. (3x+10) : (5x+10) = 5:7

∴ x = 5

No's are (15,25)**A man rows downstream 30 km and upstream 18 km, taking 5 hours each time. What is the velocity of the stream (current)?**- 1.2 km/hr
- 1.5 km/hr
- 2.5 km/hr
- 1.8 km/hr

Answer: Option A

LET X =SPEED OF BOAT AND Y = SPEED OF CURRENT.

⇒ 30/(X+Y)=18/(X-Y)=5 BY SOLVING Y = 1.2**A train 125 metre long is running at 50 km/hr. In what time will it pass a man running at 5 km/hr in the same direction in which the train is going?**- 25 sec
- 10 sec
- 20 sec
- 15 sec

Answer: Option B

Distance = 125 metres Speed = 50-5 = 45kmph => 45 * (5/18) = 12.5 m/s

Time = 125 / 12.5 = 10 sec.**A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in 42 minutes, will be covered by A in**- 11 min
- 14 min
- 7 min
- 17 min

Answer: Option C

B is thrice as fast as C

C covers in 42min

∴ B covers in 42/3 = 14 min

A is twice as fast as B

∴ A covers in 14 * (1/2) = 7 min