Question

The two numbers are in the ratio of 3:4. If the sum of the numbers is 133. Find the numbers.

Answer 1

Answer:

Let the no. be 3x and 4x

3x+4x= 133. (given)

7x=133

x= 19

3x= 3x19 =57

4x= 4x19=76

Hence the no. are 57 nd 76

Answer 2

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$\large{\sf{\underline{\underline\green{Solution!!}}}}$

Let the two numbers be ' x ' and ' y '

It is given that  $\sf\dfrac{x}{y}$  =  $\sf\dfrac{3}{4}$

Hence, $\sf{ x =\dfrac{3y}{4} ------( Equation\: 1 )}$

It is given that x + y = 133

So we know $\sf{ x =\dfrac{3y}{4} }$  Hence substituting in the above equation we get,

$:\implies\:\sf{\dfrac{3y}{4} +y=133}$

Taking LCM we get,

$:\implies\:\sf{3y+\dfrac{4y}{4}=133 }$

$:\implies\:\sf{\dfrac{7y}{4}=133 }$

$:\implies\:\sf{7y=133 \times 4}$

$:\implies\:\sf{7y=532}$

$:\implies\:\sf{y=\dfrac{532}{7} }$

$:\implies\:\sf{y=76 }$

Hence the number ' y ' is 76.

We know ' x ' is 3y / 4. Hence we get,

$:\implies\:\sf{x=76\times \dfrac{3}{4} }$

$:\implies\:\sf{x=57}$

$\displaystyle\underline{\textsf{Hence\: the \:two \:numbers \:' x ' \:and \:' y ' \:are \:57\: and \:76\: respectively.}}$

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