Aptitude Answers 3

Here are the answers to the last week's aptitude questions. Read more for the answers. Refer here for the corresponding questions!!!





1.Answer 65292

      As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9) It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9) Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.

2.Answer:

          The next letters in the series are N, O, R, S, U, X. The pattern is - letters whose English names (Phonetic Pronunciations) start with vowels.

3.Answer:

        This is a series with odd positions containing position number whereas even positions containing square of the position.i.e. even position numbers are 4 16 36 64 100 and odd position numbers are 1 3 5 7 9 Hence, 6 does not fit in the series. It should be 5.

4.Answer:

           The mother is 29 years and 2 months old. Let's assume that son is X years old. Hence, father is 6X years old and mother is (70-7X) years old. It is given that the sum of their ages is 70, which will total 140 after 70/3 years. After 70/3 years, son will be (X + 70/3) years old and father will be (6X + 70/3) years old. Also, it is given that after 70/3 years, the father will be twice as old as the son. Thus, (6X + 70/3) = 2 * (X + 70/3) 6X + 70/3 = 2X + 140/3 4X = 70/3 X = 35/6 Hence, their ages are Son = X = 35/6 = 5 years and 10 months Father = 6X = 6(35/6) = 35 years Mother = (70 - 7X) = 70 - 7(35/6) = 29 years and 2 months 

5.Answer:

             We tried to find out some simple mathematical method and finally we wrote small C program to find out the answer. The answer is 3121 coins. Here is the breakup: First son = 624 coins Second son = 499 coins Third son = 399 coins Forth son = 319 coins Fifth son = 255 coins Daughters = 204 each Manager = 5 coins

6. Answer

            1. Mark the boxes with numbers 1, 2, 3, 4, ... upto 10 2. Take 1 ball from box 1, take 2 balls from box 2, take 3 balls from box 3, take 4 balls from box 4 and so on 3. Put all of them on the scale at once and take the measurement. 4. Now, subtract the measurement from 550 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10 + 6*10 + 7*10 + 8*10 + 9*10 + 10*10) 5. The result will give you the box number which has a ball of 9 Kg

7.Answer:

              224

8. Answer:

       For next number add one = 3

Next add one = 7 and so on ...

Now at 31, for next number double the number = 62

Add one for next number = 63

Answer

Next two numbers are 62 and 63.

The pattern is double the number, add one, double the number, add one...

Starting with 1, for next number double the number = 2 Again, double the number = 6

9.Answer

      The minute and the hour hand meet 11 times in 12 hours in normal watch i.e. they meet after every

= (12 * 60) / 11 minutes

= 65.45 minutes

= 65 minutes 27.16 seconds

But in our case they meet after every 65 minutes means the watch is gaining 27.16 seconds.

10.Answer

   There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in 64C3 ways.

So the sample space is = 41664

There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.

3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.

But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112

The require probability is

= 112 / 41664

= 1 / 372

= 0.002688