Question

Three numbers are in ratio 3:4:5. If the sum of the largest and the smallest numbers exceeds the third number by 64, find the number.
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Answer 1

Answer:

48, 64 , & 80

Step-by-step explanation:

let x be the proportionate,

=> 3x,4x,and 5x are the no.s

also , given that ,

3x+5x = 4x+64

=> 8x-4x = 64

=> 4x = 64

=> x = 64/4

=> x=16

so, the 3 numbers are,

3x,4x & 5x

3*16,4*16 & 5*16

48, 64 , & 80

Verification:

sum of largest and smallest = 48+80 = 128

3rd no. other than smallest & largest = 64

128 = 64 +64

=> largest+smallest = 3rd no. + 64

hence verified