Out of three numbers, the first is twice the second and

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Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then difference of first and third number is

1. 12
2. 20
3. 24
4. 48

Correct Option: D

Let the numbers be 2k, k and 4k respectively

∴ Average =	2k + k + 4k
	3

⇒	7k	= 56
	3

⇒ k =	3 × 56	= 24
	7

∴ First number = 2k = 2 × 24 = 48
Third number = 4k = 4 × 24 = 96
∴ Required difference = Third number - First number = 96 – 48 = 48
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 1/4 , X = 56

First number =	3ab	× X
	1 + b + ab

First number =		3 × 2 ×	1		× 56
			4

	1 +	1	+ 2 ×	1
		4		4

First number =	3	× 4	× 56 = 48
	2

	4 + 1 + 2

Third number =	3	× X
	1 + b + ab

Third number =		3			× 56

	1 +	1	+ 2 ×	1
		4		4

Third number =	3 × 4	× 56 = 96
	4 + 4 + 2

Required difference = 96–48 = 48