Question

divide rs413 among a b c such that 2a=5b=7c​

Answer 1

let, 2a= 5b =7c = k

=> a= k/2

=> b= k/5

=> c= k/7

so, k/2 + k/5 + k/7 = 413

=> 35k/70 + 14k/70 + 10k/70 = 413

=> 59k/70 = 413

=> 59k = 413*70

=> k = (413*70)/59

=> k= 70*7 = 490

so, 413 can be divided as :

490/2, 490/5, 490/7

= 245, 98, 70

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Answer 2

Answer:

here , we have to divide 581 in A,B,C such that

A + B + C = 581_(i)

let 4A = 5B = 7C = x

=> A = x / 4

=> B = x / 5

=> C = x / 7

on placing values of A,B,C in eq(i)

we get

(x/4) + (x/5) + (x/7) = 581

=> x * ( (1/4) + (1/5) + (1/7) ) = 581

=> x * ( (35+28+20) / 140 ) = 581 _.LCM (4,5,7) = 140

=> x * (83 / 140) = 581

=> x = 581 * 140 / 83

=> x = 140 * 7 ..(ii)

now , from eq(ii)

we get

A = x / 4 = 140 * 7 / 4 = 35 * 7 = 245

B = x / 5 = 140 * 7 / 5 = 28 * 7 = 196

C = x / 7 = 140 * 7 / 7 = 140