Aptitude and Reasoning Questions

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Only son of woman's grandfather - woman's father;  Man's brother's father - Man's father 

So, man's father is woman's father i.e woman is the man's sister.

6th to the left of 17th letter to the right of 7th letter from the left = 6th to the left of (17+7 =) 24th letter from the left = (24-6 =) 18th letter from the left.

Thus, we have to count from the left and we have to find out 18th letter from the left that falls into the second section of English alphabet.

Here, both the first half and the second half of the English alphabet have been reversed and we have to count from the left end, then the required letter can be found out using the formula given below. Suppose , the first and the second reversed sections of the English alphabet are α and  βrespectively.Then,

  The required letter = 2α+β+1-(required position of letter from left side)

Therefore, The required letter = 26 + 13 + 1 - 18 = 22 = V

Let the large number be x.

Then x - 20 = 20x/100

=> x - x/5 = 20  => x = 25.

Each term of the series is the product of the preceding two terms
So, 5000 is wrong and it must be replaced by (500*50) = 25000

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways. 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280