Reena has pend and pencils which together are 40 in number.
Question:

Reena has pend and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.

Solution:

Given:

(i) Total numbers of pens and pencils = 40.

(ii) If she has 5 more pencil and 5 less pens, the number of pencils would be 4 times the number of pen.

To find: Original number of pens and pencils.

Suppose original number of pencil = x

And original number of pen = y

According the given conditions, we have,

$x+y=40$

$x+y-40=0$…(1)

$5+x=4(y-5)$

$5+x=4 y-20$

$x-4 y+5+20=0$

$x-4 y+25=0$….(2)

Thus we got the following system of linear equations

$x+y-40=0$$\ldots \ldots(1)$

$x-4 y+25=0$$\ldots \ldots(2)$

Substituting the value of y from equation 1 in equation 2 we get

$x-4(40-x)+25=0 \quad[y=(40-x)$ from equation 1$]$

$x-160+4 x+25=0$

$5 x-135=0$

$x=\frac{135}{5}$

$x=27$

Substituting the value of in equation 1 we get

$27+y=40$

$y=40-27$

$y=13$

Hence we got the result number of pencils is $x=27$ and number of pens are $y=13$

 

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