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(A) Let the marks of the first student be $x$.Then,the marks of the second student are $x + 19$.The sum of their marks is $x + (x + 19) = 2x + 19$.Acc

Vedclass · Accepted Answer

$57$ and $38$

Vedclass · Answer

(A) Let the marks of the first student be $x$.Then,the marks of the second student are $x + 19$.The sum of their marks is $x + (x + 19) = 2x + 19$.According to the problem,the marks of the second student are $60 \%$ of the sum of their marks:$x + 19 = 60 \% 	imes (2x + 19)$$x + 19 = \frac{60}{100} 	imes (2x + 19)$$x + 19 = \frac{3}{5} 	imes (2x + 19)$Multiply both sides by $5$:$5(x + 19) = 3(2x + 19)$$5x + 95 = 6x + 57$$95 - 57 = 6x - 5x$$x = 38$Therefore,the marks of the first student are $38$,and the marks of the second student are $38 + 19 = 57$.The marks obtained by them are $57$ and $38$.

Two students appeared for an examination. One of them secured $19$ marks more than the other and his marks were $60 \%$ of the sum of their marks. The marks obtained by them are

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