Question

7. The numerator of a fraction is 2 less than the denominator. If 1 is added to its denominator, it becomes . Find
fraction.
(3M)
8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest or
drinking water from the pond. Find the number of deer in the herd.
(4M)​

Answer 1

Solution 1 :

Given :

• The numerator of a fraction is 2 less than the denominator.
• If 1 is added to the denominator, the fraction becomes ½

To Find :

• The Fraction.

Solution :

Let the numerator of the fraction be x.

Let the denominator of the fraction be y.

Fraction → $\sf{\dfrac{x}{y}}$

Case 1 :

The numerator (x) is 2 less than denominator (y).

Equation :

$\longrightarrow$ $\sf{x=y-2}$

$\sf{x-y=-2\:\:(i)}$

Case 2 :

1 is added to the denominator.

Numerator → x

Denominator → (y+1)

Equation :

$\longrightarrow$ $\sf{\dfrac{x}{(y+1)}=\dfrac{1}{2}}$

$\longrightarrow$ $\sf{2x=y+1}$

$\sf{2x-y=1\:\:\:\:(ii)}$

Multiply equation (i) by 2,

$\longrightarrow$ $\sf{2\:\times\:x-2\:\times\:y\:=\:2\:\times\:-2}$

$\sf{2x-2y=-4\:\:\:\:(iii)}$

Solve equation (ii) and (iii) simultaneously.

Subtract equation (ii) from (iii)

$\longrightarrow$ $\sf{2x-y-(2x-2y) =1-(-4)}$

$\longrightarrow$ $\sf{2x-y-2x+2y=1+4}$

$\longrightarrow$ $\sf{-y+2y=5}$

$\longrightarrow$ $\sf{y=5}$

Substitute, y = 5 in equation (i),

$\longrightarrow$ $\sf{x-y=-2}$

$\longrightarrow$ $\sf{x-5=-2}$

$\longrightarrow$ $\sf{x=-2+5}$

$\longrightarrow$ $\sf{x=3}$

$\large{\boxed{\sf{\purple{Numerator\:=\:x=3}}}}$

$\large{\boxed{\sf{\purple{Denominator\:=\:y=5}}}}$

$\large{\boxed{\sf{\purple{Fraction\:=\:\dfrac{x}{y}\:=\:\dfrac{3}{5}}}}}$

Solution 2 :

Given :

• Half of a herd of deer are grazing in the field.
• Three fourth of the remaining herd are playing nearby.
• Rest 9 deers are drinking water from the pond.

To Find :

• The number of deer in the herd.

Solution :

Let the number of deer in the herd be x.

Case 1 :

Half of the deer are grazing in the field.

•°• $\sf{Number\:of\:deer\:grazing\:in\:field\:=\:\dfrac{x}{2}\:\:\:\:(i)}$

Case 2 :

The remaining ¾ of deer are playing nearby.

$\sf{Remaining\:deer\:=\:\dfrac{3}{4}\:\times\:\big(x-\dfrac{x}{2}\big)}$

$\longrightarrow$ $\sf{\dfrac{3}{4}\:\times\:\big(\dfrac{2x-x}{2}\big)}$

$\longrightarrow$ $\sf{\dfrac{3}{4}\:\times\:\dfrac{x}{2}}$

$\longrightarrow$ $\sf{\dfrac{3x}{2\:\times\:4}}$

$\longrightarrow$ $\sf{\dfrac{3x}{8}\:\:\:\:(ii)}$

Case 3 :

The number of deer drinking water in the pond is 9.

Equation :

The total number of deer will be sum of (i), (ii) and the deer drinking water in the pond.

$\longrightarrow$ $\sf{\dfrac{x}{2}\:+\:\dfrac{3x}{8}\:+\:9=x}$

$\longrightarrow$ $\sf{\dfrac{8x+6x}{16}+9\:=\:x}$

$\longrightarrow$ $\sf{\dfrac{14x}{16}+9=x}$

$\longrightarrow$ $\sf{\dfrac{14x+144}{16}=x}$

$\longrightarrow$ $\sf{14x+144=16x}$

$\longrightarrow$ $\sf{14x-16x=-144}$

$\longrightarrow$ $\sf{\cancel{-}2x=\:\cancel{-}144}$

$\longrightarrow$ $\sf{x=\dfrac{-144}{-2}}$

$\longrightarrow$ $\sf{x=72}$

$\large{\boxed{\sf{\red{Number\:of\:deer\:in\:the\:herd=72}}}}$