Question

17. Out of three numbers the first and second are in the ratio 4: 5 are the first and third are in the ratio 3 : 2 and their average 70. Find the distance between the largest and smallest numbers. 2) 48 3) 42 4) 72 1) 90​

Answer 1

Solution :

Out of three numbers the first and second are in a ratio 4: 5.

And the first and third numbers are in the ratio 3 : 2 .

The average of the first and third numbers is 70.

We have to find the distance between the largest and smallest numbers

Let us start by assuming that the three numbers are 'p', 'q' and 'r' .

The two numbers 'p' and 'q' are in a ratio of 4:5

Let us express 'p' as 4x and 'q' as 5x

We can multiply this by a factor of 3 to express as :

'p' = 12x and 'q' = 15x

Now, the first and third are in a ratio of 3:2

If 'p' is 12x then

12x : r = 3:2

(12x)/r = 3/2

>> (4x)/r = 1/2

>> r = 4x × 2 = 8x

Thus, three numbere in terms of x are :

'p' - 12x

'q' - 15x

'r' - 8x

The average of the first and third numbers is 70

Average of 12x and 8x is 70

The average is 10x

So ,

10x = 70

x = 7

'p' = 12x = 84

'q' = 15x = 105

'r' = 8x = 56

Thus the numbers are 84, 105 and 56 respectively.

56 is the smallest number and 105 is the largest number.

The distance between them can be found by subtracting the smallest number from the larger number

This becomes :

>> 105 - 56

>> 49

Most probably the answer will be option 2, but it will be 49 instead of 48.

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