Question

What number should be subtracted from each of 50, 61, 92, 117 so that the numbers, so obtained in this order, are in proportion ?

Answer 1

Step-by-step explanation:

Let the number be x

a/q

50-x/61-x=92-x/117-x

(50-x)(117-x)=(92-x)(61-x)

5850-50x-117x-x^2=5612-92x-61x-x^2

5850-5612=167x-153x

238=14x

x=17

hence 33:44::65: 100

Answer 2

The required number is 17.

Step-by-step explanation:

To find : What number should be subtracted from each of 50, 61, 92, 117 so that the numbers, so obtained in this order, are in proportion ?

Solution :

Let the number be 'x' which should be subtracted.

So, the proportion became

$50-x:61-x::92-x:117-x$

i.e. $\frac{50-x}{61-x}=\frac{92-x}{117-x}$

Cross multiply,

$(50-x)(117-x)=(92-x)(61-x)$

$5850-50x-117x+x^2=5612-92x-61x+x^2$

$-50x-117x+92x+61x=5612-5850$

$-14x=-238$

$x=\frac{238}{14}$

$x=17$

Therefore, the required number is 17.

#Learn more

What number should be subtracted from each of the number 18 ,24,28,38 so that remainders may be in proportion?

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