# Aptitude Puzzles

1. A car is traveling at a uniform speed. The driver sees a milestone showing a 2-digit number. After traveling for an hour the driver sees another milestone with the same digits in reverse order. After another hour the driver sees another milestone containing the same two digits with a zero in between(0). What is the average speed of the driver?

Ans: 45 kmph**Solution** The last statement tells the third milestone have number 101-109.

This means the initial milestone have the number from 11-19.

So the distance between initial milestone and the last one ranges from 82(101-19) to 98(109-11).

Dividing by two(for the time taken) , and keeping in mind odd numbers cannt be halved we have 41-49.

Now as it is visible the speed is definitely above 40. So the second milestone will be having a number with its first digit either 5 or 6, and so will the first milestone as the second digit.

Hence, either 15 or 16 is the first milestone.

Checking which number lies in range : 51-15 = 36 and 61-16 = 45.

So 16 lies in range, hence 45 is the speed.

2. Mr. ANYMAN left ANYTOWN by car to attend a wedding at ANYCITY. He had been driving for exactly two hours when the car got punctured. It took his driver exactlyten minutes to change the wheel. In order to play safe they covered the remaining distance at a speed of 30 mph. consequently, Mr. ANYMAN was at wedding half an-hour behind schedule. Had the car got the puncture only 30 miles later , I would have been only FIFTEEN minutes late he told the driver . How Far is ANYCITY from ANYTOWN.

Ans: 120 miles**Solution** As he said, he would have covered other 15 minutes in time for 30 miles late puncture. Hence he would have covered 30 miles in 45 minutes. So original speed = 40mph. Other late 15 minutes include change of wheel in 10 minutes. So the remaining distance is covered costs 5 minutes. Hence distance remaining = (5/60) / (1/30-1/40) = 10. Distance covered in 2 hours = 80 kms.

Hence total distance = 10 (distance travelled after 30 miles late puncture) + 30 (30 miles late puncture) + 80 (first two hours)

3. There are three persons A, B, C. one day they set out in different directions and each one steals an animal. The animals are camel, horse and a mule though not inthe order. A CBI officer catches them and on interrogation he has the following statements:

A says: B has stolen a horseC says: A is telling wrong, B has stolen a mule.B says: A and C are both telling wrong. I have stolen nothing.

The person stealing a camel tells wrong while the person with a horse tells right. So tell which person steal which animal.

Ans: A: Camel

B: Mule

c: Horse**Solution**

Suppose A is telling the truth. According to him B stole the horse. So B is telling right. But the statement is contradictory. So A is telling a lie.

B anyways is telling a lie because it is said that each of them stole an animal.

So, C is telling the truth. Hence B has stolen the Mule. As he is telling right so stole a horse and A stole the camel.

4.There are 3 societies A, B, C. A Lent Tractors to B and C as many as they had already. After some time B gave as many tractors to A and C as many as they have. After sometime C did the same thing. At the end of this transaction each one of them had 24.Find the cars each originally had. This question was bit differently asked with the same thing.

Ans: A had 39 cars,

B had 21 cars

C had 12 cars

**Solution** Let’s back trace from the end to the start.

Logic: they gave the same number of tractor to the other person as that person had already with him. That means at the end of every step the number of tractors with people other than the current lender is doubled. And the number of tractor others had is the decrement in his number of tractors. So for a back trace half the number of tractors with the people other then the lender and add the decreased amount to the lender as follows

A—————B—————C

24————–24————–24

12————–12————–24+12+12=48

6—————12+6+24=42——24

6+21+12=39——21————–12