Question 42

Of the four numbers whose average is 91, the first is 3/10 times the sum of other three. The first number is:

Solution

Let the four numbers be $$w,x,y,z$$

Average of 4 numbers = $$\frac{(w+x+y+z)}{4}=91$$

=> $$w+x + y + z = 91 \times 4 = 364$$

According to ques, => $$w = \frac{3}{10} \times (x+y + z)$$

=> $$w = \frac{3}{10} \times (364 - w)$$

=> $$10w = 1092 - 3w$$

=> $$10w + 3w = 13w = 1092$$

=> $$w = \frac{1092}{13} = 84$$

=> Ans - (D)

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