Question

The difference of the present age of two brothers is 4 years. 5 years ago if the product of their ages was 96 , find their present ages.​

Answer 1

Answer

Present age of one (first) brother is 17 years and of another (second) brother is 13 years

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Explanation

$\underline{\underline{\bold{\sf{Case\:1)}}}}$

The difference of the present ages of two brothers is 4 years.

Let the present age of one brother be x years and of another brother be y years.

Difference in ages of one brother and another brother is 4 years.

$\implies\:\sf{x\:-\:y\:=\:4}$

$\implies\:\sf{x\:=\:4\:+\:y}$ ....(eq 1)

$\underline{\underline{\bold{\sf{Case\:2)}}}}$

Five years ago, the product of their ages was 96 years.

Five years ago, (means five years back) will be (x - 5) years and (y - 5) years. [5 will be decreased from both of their ages.]

i.e.

• Age of one brother = (x - 5) years
• Age of another brother = (y - 5) years

According to question,

$\implies\:\sf{(x\:-\:5)(y\:-\:5)\:=\:96}$

$\implies\:\sf{xy\:-\:5x\:-\:5y\:+\:25\:=\:96}$

$\implies\:\sf{xy\:-\:5x\:-\:5y\:=\:96\:-\:25}$

$\implies\:\sf{(4\:+\:y)y\:-\:5(4\:+\:y)\:-\:5y\:=\:71}$

$\implies\:\sf{4y\:+\:y^2\:-\:20\:-\:5y\:-\:5y\:=\:71}$

$\implies\:\sf{y^2\:-\:6y\:-\:20\:-\:71\:=\:0}$

$\implies\:\sf{y^2\:-\:6y\:-\:91\:=\:0}$

The above equation is in the form ax² + bx + c = 0.

So, apply quadratic formula on it.

Here, a = 1, b = -6 and c = - 91

$\boxed{\bold{\sf{y \: = \frac{ \: - b \pm \sqrt{ {b}^{2} \: - \: 4ac }}{2a}}}}$

$\implies\:\sf{y\: = \frac{ \: - ( - 6) \pm \sqrt{ { (- 6)}^{2} \: - \: 4(1)( - 91)}}{2(1)}}$

$\implies\:\sf{y \: = \frac{ \: 6 \pm \sqrt{ 36 \: + \: 364}}{2}}$

$\implies\:\sf{y \: = \frac{ \: 6 \pm \sqrt{ 400}}{2}}$

$\implies\:\sf{y\:=\:\frac{6\:\pm\:20}{2}}$

$\implies\:\sf{y\:=\:\frac{6\:+\:20}{2}}$

$\implies\:\sf{y\:=\:\frac{26}{2}}$

$\implies\:\sf{y\:=\:13}$

•°• Present age of another brother is 13 years.

As, age can't be negative. So, negative one is neglected.

Now, substitute value of x = 13 in equation (1)

$\implies\:\sf{x\:=\:4\:+\:13}$

$\implies\:\sf{x\:=\:17}$

•°• Present age of one (first) brother is 17 years.

Answer 2

$\Large\underline{\underline{\sf{Given}:}}$

• Difference of Present age of two brother = 4 years .
• 5 years ago , their Product of ages was = 96 .

• Find their Present age ?

$\Large\underline{\underline{\sf{Solution}:}}$

$\textbf{Let Present age of Younger brother} \\ \textbf{x years.} \\ \\ \textbf{Than Elder age is (x+4)yrs.} \:$

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→ 5 years before younger brother was = (x-5) years old.

→ 5 years before elder brother was = (x+4-5) = (x-1) years old.

As, given 5 years ago their product of ages was 96.

so,

$(x - 5)(x - 1) = 96 \\ \\ \red\leadsto {x}^{2} - x - 5x + 5 = 96 \\ \\ \red\leadsto {x}^{2} - 6x - 91 = 0 \\ \\ \green{\textbf{Splitting the middle term Now,}} \\ \\ \red\leadsto \: {x}^{2} - 13x + 7x - 91 = 0 \\ \\ \red\leadsto \: x(x - 13) + 7(x - 13) = 0 \\ \\ taking \: (x - 13) \: \: common \: Now, \\ \\ \red\leadsto \: (x - 13)(x + 7) = 0 \\ \\ \pink{\textbf{Putting both Equal to 0}} \\ we \: get \\ \\ x = 13 \\ x = - 7$

As now, Age cant be in negative ,,,

so,

$\textbf{Present Age of Younger Brother} \\ is \: \red{13 \: years} .. \\ \\\textbf{Present Age of elder Brother} \\ is \: (13 + 4) = \red{17\: years} ..$

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$\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

Factorization of Quadratic polynomials of the form x² + bx + c by splitting Middle term :-------

(i) In order to factorize x² + bx + c we have to find numbers p and q such that p + q = b and pq = c.

(ii) After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

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$\star\large\underline\textbf{Hope it Helps You.} \star$