Question

Q29. The numerator of a fraction is six less than its denominator. If the numerator is
decreased by 4 and the denominator is decreased by 1, the fraction becomes 5/8 Find the Fraction
f​

Answer 1

Answer:-

Let the fraction be x/y.

Given:

Numerator is 6 less than the denominator.

⟶ Numerator = Denominator - 6

⟶ x = y - 6 -- equation (1)

And,

If numerator is decreased by 4 and denominator is decreased by 1 , the fraction becomes 5/8.

$\longrightarrow \sf \: \frac{x - 4}{y - 1} = \frac{5}{8}$

Substitute the value of x from equation (1).

$\longrightarrow \sf \: \frac{y - 6 - 4}{y - 1} = \frac{5}{8} \\ \\ \longrightarrow \sf \:8(y - 10) = 5(y - 1) \\ \\ \longrightarrow \sf \:8y - 80 = 5y - 5 \\ \\ \longrightarrow \sf \:8y - 5y = - 5 + 80 \\ \\ \longrightarrow \sf \:3y = 75 \\ \\ \longrightarrow \sf \:y = \frac{75}{3} \\ \\ \longrightarrow \boxed{ \sf \:y = 25}$

Substitute the value of y in equation (1).

⟶ x = y - 6

⟶ x = 25 - 6

⟶ x = 19

∴ The required fraction x/y is 19/25.

Answer 2

Answer :-

$\: \: \underline{\boxed{\bf{\purple{\mapsto \: \: \: Let's \: analyse \: the \: Question \: first}}}}$

We here see that in this question, we have to find two unknown values. The only given thing is that, those values depend on the numerical values. How will we find that ? The easiest way to find this is using Linear Equations in Two Variables. We can make the value of one variable depend on other and find the values of both.

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★ Question :-

The numerator of a fraction is six less than its denominator. If the numerator is

decreased by 4 and the denominator is decreased by 1, the fraction becomes 5/8. Find the fraction .

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★ Solution :-

Given,

» The denominator of fraction = 6 + The numerator of the fraction.

» (Numerator - 4) and (denominator - 1) = ⅝

• Let the numerator of the fraction be 'x'

• Let the denominator of the fraction be 'y'.

$\: \: \: \bf{\purple{\longrightarrow \: \: The \: required \: fraction\: will \: be \: \dfrac{x}{y}}}$

Then according to the question,

~ Case I :-

$\: \: \huge{\bf{\blue{\Longrightarrow \: \: \: y \: = \: (6 \: + \: x)}}}....(i)$

~ Case II :-

$\: \: \huge{\bf{\blue{\Longrightarrow \: \: \: \dfrac{x \: - \: 4}{y \: - \: 1} \: = \: \dfrac{5}{8}}}}$

On cross multiplication, we get,

⌬ 8(x - 4) = 5(y - 1)

⌬ 8x - 32 = 5y - 5

⌬ 8x - 5y = - 5 + 32

⌬ 8x - 5y = 27 ... (ii)

From equation (i) and equation (ii), we get,

⌬ 8x - 5(6 + x) = 27

⌬ 8x - 30 - 5x = 27

⌬ 8x - 5x = 27 + 30

⌬ 3x = 57

$\: \: \huge{\boxed{\bf{\orange{\longrightarrow \: \: x \: = \: \dfrac{\not57}{\not3} \: = \red{19}}}}}$

→ Hence, the value of numerator = x = 19

Fron equation (i) and the value of y, we get,

✏ y = 6 + x

✏ y = 6 + 19

✏ y = 25

→ Hence, the value of denominator of the fraction is = y = 25.

Now the fraction is numerator divided by denominator.

$\: \: \underline{\boxed{\sf{\green{\mapsto \: \: \: Thus, \: the \: required \: fraction \: is \: \dfrac{19}{25}}}}}$

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$\: \: \: \underline{\boxed{\rm{\pink{\mapsto \: \: \: Confused? \: Don't \: worry \: let's \: verify \: it}}}}$

For verification, we need to simply apply the values we got into the equations we formed.

~ Case I :-

=> y = 6 + x

=> 25 = 6 + 19

=> 25 = 25

Clearly, LHS = RHS.

~ Case II :-

$\: \: \bf{\purple{\Longrightarrow \: \: \: \dfrac{x \: - \: 4}{y \: - \: 1} \: = \: \dfrac{5}{8}}}$

$\: \: \bf{\purple{\Longrightarrow \: \: \: \dfrac{19 \: - \: 4}{25 \: - \: 1} \: = \: \dfrac{5}{8}}}$

$\: \: \bf{\purple{\Longrightarrow \: \: \: \dfrac{15}{24} \: = \: \dfrac{5}{8}}}$

Dividing, the LHS by 3, we get,

=> ⅝ = ⅝

Clearly LHS = RHS.

Here both the conditions satisfy, so our answer is correct .

Hence, Verified.

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$\: \: \: \underline{\boxed{\rm{\pink{\mapsto \: \: \: Let's \: understand \: more}}}}$

• Linear Equation are the equations formed using constant and variable terms but terms are of only single degree.

• Polynomials are the equations formed using both constant and variable terms but can be of many degrees.

They are classified as :-

• Linear Polynomial
• Quadratic Polynomial
• Cubic Polynomial
• Bi - Quadratic Polynomial

• If we draw the graph of this question we can see that both lines of equations intersect each other at coordinate (19,25).