Question

what must be subtracted from the numerator and denominator of the fraction 7 upon 8 so that the fraction takes the new value 4 upon 5​

Answer 1

Answer:

Required number =3

Step-by-step explanation:

$Let \: x \: should \: be \\subtracted \: from\: the \: numerator\:and \\ denominator \:\frac{7}{8}\:so \:that \\fraction \:takes \:the \\new \:value \:\frac{4}{5}$

$\implies \frac{7-x}{8-x}=\frac{4}{5}$

$\implies 5(7-x)=4(8-x)$

$\implies 35-5x = 32-4x$

$\implies -5x + 4x= 32-35$

$\implies -x = -3$

$\implies x = 3$

Therefore,

Required number =3

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Answer 2

Answer = 3

Explanation-

Let the no. be x

for numerator, for denominator,

7-x=4 8-x=5

-x=4-7 -x=5-8

x=3. x=3.