Question 31

Of three numbers,the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is

Solution

let the three numbers be x,y,z

second is thrice the first

y = 3x...........(1)

third number is three-fourth of the first

$$ z = \frac{3}{4}x $$

average of the three numbers is 114

sum of 3 numbers = $$ 114 \times 3 = 342 $$

that is $$ x + 3x + \frac{3}{4}x = 342 $$

solving x = 72

y = 3x = $$ 3 \times 72 = 216 $$

z = $$ \frac{3}{4}x = \frac{3}{4} \times 72 = 54 $$

thus the largest number is 216

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