Question

- A, B, C are three numbers such that 60% ofA, 90% of B and 80% of C are all equal. If the average of the numbers is 116, find the sum of A and B.​

Answer 1

Answer:

A+B=240

Step-by-step explanation:

A=144

B=96

C=108

Answer 2

Answer:

Sum of A and B is 240.

Step-by-step explanation:

Given:

A, B, C are three numbers such that 60% of A, 90% of B and 80% of C are all equal.

Average of the numbers is 116.

Need To Find:

Sum of A and B

Solution:

According to question,

A, B, C are three numbers such that 60% of A, 90% of B and 80% of C are all equal.

So, 1st equation is

$\frac{60}{100} \times A = \frac{80 }{100} \times B = \frac{90}{100 } \times C$

6A = 8B = 9C

Also,

Average of the numbers is 116

So, 2nd equation is

$\frac{A + B + C}{3} = 116$

Now using equation 1, substituting the values of B and C in terms of A in equation 2.

$A + \frac{6A}{8} + \frac{6A}{9} = 116\times3$

$A( 1 + \frac{3}{4} + \frac{2}{3}) = 348$

$A \times \frac{29}{12}= 348$

$A = 348 \times \frac{12}{29}$

A = 144

B = $\frac{6A}{9}$= 96

Sum of A and B = 144 + 96 = 240

Hence, Sum of A and B is 240.