Question

Class 10 CBSE Subject- Math Ch 3 Pair of Linerar Equation in two variables. pg 57 Ex. 3.4 Qn 2 ii)

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Answer 1

$\huge \underline \bold \red{Given \: - }$

5 years ago, Nuri was thrice as old as Sonu.

10 years later, Nuri will be twice as old as Sonu.

$\huge \underline \bold \red{To \: find \: - }$

Both of their ages.

$\huge \underline \bold \red{Solution \: - }$

Let the present age of Nuri be = x.

and, let the present age of Sonu be = y.

$\huge \underline \bold \red{A/Q,}$

$\bold \green{(x - 5) = 3(y - 5)}$

$\bold{x - 3y = - 10..........(i)}$

$\bold \green{(x + 10y) = 2(y + 10)}$

$\bold{x - 2y = 10.........(ii)}$

Subtracting (i) from (ii),

we get,

$\bold{y = 20...........(iii)}$

Putting this value in eq. (i), we get

$\bold \green{x - 60 = - 10}$

$\bold \green{x = 50}$

Hence,

Nuri = 50 years....

Sonu = 20 years.....

_____________xxx___________

Answer 2

As per the question.

Let the present age of Nuri = x years

Let the present age of Sonu = y years

Five years ago,

Age of Nuri = (x-5) years

Age of Sonu = (y-5) years

As already given that,

Age of Nuri = 3(Age of Sonu)

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -10 ------(1)

Ten years hence,

Age of Nuri = (x + 10) years

Age of Sonu = (y + 10) years

Age of Nuri will be twice as Sonu.

x + 10 = 2(y + 10)

x - 2y = 10 ------(2)

From using equation (1) and (2), we get.

x - 3y = -10

x - 2y = 10

(-) (+) = (-)

___________

-y = -20

y = 20 years

Putting the value of "y" in equation (2)

x - 2(20) = 10

x = 50 years.

Hence,

The present age of Nuri = x = 50 years.

The present age of Sonu = y = 20 years.