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:
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
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