NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations will assist students in better understanding the fundamental principles. We get a quadratic equation when we equate the quadratic polynomial to zero. Students will be able to strategize their preparation with the help of NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. Students will get NCERT solutions for Science and Math from Class 6 to 12. They must complete the NCERT Class 10 Maths syllabus as soon as possible so that they can revise strategically. The NCERT solutions for class 10 Maths chapter 4 Quadratic Equations provide a detailed and step-by-step solution to each question. Each Quadratic Equation Class 10 exercise is solved here.

NCERT solutions for Class 10 Maths chapter 4 Quadratic Equations Excercise: 4.1

Q1 (i) Check whether the following are quadratic equations :

Answer:

We have L.H.S.

Therefore, can be written as:

i.e.,

This equation is of type: .

Hence, the given equation is a quadratic equation.

Q1 (ii) Check whether the following are quadratic equations :

Answer:

Given equation can be written as:

i.e.,

This equation is of type: .

Hence, the given equation is a quadratic equation.

Q1 (iii) Check whether the following are quadratic equations :

Answer:

L.H.S. can be written as:

and R.H.S can be written as:

i.e.,

The equation is of the type: $ax^2+bx+c = 0,a\neq0$ .

Hence, the given equation is not a quadratic equation since a=0.

Q1 (iv) Check whether the following are quadratic equations :

Answer:

L.H.S. can be written as:

and R.H.S can be written as:

i.e.,

This equation is of type: $ax^2+bx+c = 0,a\neq0$ .

Hence, the given equation is a quadratic equation.

Q1 (v) Check whether the following are quadratic equations :

Answer:

L.H.S. can be written as:

and R.H.S can be written as:

i.e.,

This equation is of type: $ax^2+bx+c = 0,a \neq 0$ .

Hence, the given equation is a quadratic equation.

Q1 (vi) Check whether the following are quadratic equations :

Answer:

L.H.S.

and R.H.S (x-2)^2 can be written as:

i.e.,

This equation is NOT of type: $ax^2+bx+c = 0 , a\neq0$ .

Here a=0, hence, the given equation is not a quadratic equation.

Q1 (vii) Check whether the following are quadratic equations :

Answer:

L.H.S. (x+2)^3 can be written as:

and R.H.S can be written as:

i.e.,

This equation is NOT of type: .

Hence, the given equation is not a quadratic equation

Q1 (viii) Check whether the following are quadratic equations :

Answer:

L.H.S. ,

and R.H.S (x-2)^3 can be written as:

\Rightarrow x^3-4x^2-x+1 = x^3-6x^2+12x-8

i.e.,

This equation is of type: .

Hence, the given equation is a quadratic equation.

Q2 (i) Represent the following situations in the form of quadratic equations : The area of a rectangular plot is . The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Answer:

Given the area of a rectangular plot is 528m^2 .

Let the breadth of the plot be 'b' .

Then, the length of the plot will be: = 2b +1 .

Therefore the area will be:

$=b(2b+1)\ m^2$ which is equal to the given plot area 528m^2 .

Hence, the length and breadth of the plot will satisfy the equation

Q2 (ii) Represent the following situations in the form of quadratic equations : The product of two consecutive positive integers is 306. We need to find the integers.

Answer:

Given the product of two consecutive integers is 306.

Let two consecutive integers be 'x' and 'x+1' .

Then, their product will be:

Or .

Hence, the two consecutive integers will satisfy this quadratic equation .

Answer:

Given the area of a rectangular plot is 528m^2 .

Let the breadth of the plot be 'b' .

Then, the length of the plot will be: = 2b +1 .

Therefore the area will be:

$=b(2b+1)\ m^2$ which is equal to the given plot area 528m^2 .

Hence, the length and breadth of the plot will satisfy the equation

Q2 (ii) Represent the following situations in the form of quadratic equations : The product of two consecutive positive integers is 306. We need to find the integers.

Answer:

Given the product of two consecutive integers is 306.

Let two consecutive integers be 'x' and 'x+1' .

Then, their product will be:

Or .

Hence, the two consecutive integers will satisfy this quadratic equation .

Q2 (iii) Represent the following situations in the form of quadratic equations: Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

Answer:

Let the age of Rohan be 'x' years.

Then his mother age will be: 'x+26' years.

After three years,

Rohan’s age will be 'x+3' years and his mother age will be 'x+29' years.

Then according to question,

The product of their ages 3 years from now will be:

$\Rightarrow x^2+3x+29x+87 = 360$ Or

Hence, the age of Rohan satisfies the quadratic equation .

Q2 (iv) Represent the following situations in the form of quadratic equations : A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Answer:

Let the speed of the train be 's' km/h.

The distance to be covered by the train is $480\ km$ .

$\therefore$ The time taken will be

If the speed had been $8\ km/h$ less, the time taken would be: $\frac{480}{s-8}\ hours$ .

Now, according to question

\Rightarrow \frac{480x - 480(x-8)}{(x-8)x} = 3

Dividing by 3 on both the side

Hence, the speed of the train satisfies the quadratic equation

Quadratic Equation Class 10 Excercise: 4.2

Q1 (i) Find the roots of the following quadratic equations by factorization:

Answer:

Given the quadratic equation:

Factorization gives,

Hence, the roots of the given quadratic equation are $5\ and\ -2$ .

Q1 (ii) Find the roots of the following quadratic equations by factorization:

Answer:

Given the quadratic equation:

Factorisation gives,

Hence, the roots of the given quadratic equation are

Q1 (iii) Find the roots of the following quadratic equations by factorization:

Answer:

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0

\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0

\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2

Hence, the roots of the given quadratic equation are

Q1 (iv) Find the roots of the following quadratic equations by factorization:

Answer:

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}

Hence, the roots of the given quadratic equation are

Q1 (v) Find the roots of the following quadratic equations by factorization:

Answer:

Given the quadratic equation:

Factorization gives,

\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

Q2 Solve the problems given in Example 1. (i) (ii)

Answer:

From Example 1 we get:

Equations:

(i)

Solving by factorization method:

Given the quadratic equation:

Factorization gives,

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$ .

Therefore, John and Jivanti have 36 and 9 marbles respectively in the beginning.

(ii)

Solving by factorization method:

Given the quadratic equation:

Factorization gives,

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$ .

Therefore, the number of toys on that day was $30\ or\ 25.$

Q3 Find two numbers whose sum is 27 and the product is 182.

Answer:

Let two numbers be x and y .

Then, their sum will be equal to 27 and the product equals 182.

………………………….(1)

……………………………(2)

From equation (2) we have:

Then putting the value of y in equation (1), we get

Solving this equation:

Hence, the two required numbers are $13\ and \ 14$ .

Q4 Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

Hence, the two consecutive integers are .

Q5 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:

Let the length of the base of the triangle be $b\ cm$ .

Then, the altitude length will be: $b-7\ cm$ .

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem; we get

So,

$\Rightarrow 2b^2-14b-120 = 0$ Or

But, the length of the base cannot be negative.

Hence the base length will be $12\ cm$ .

Therefore, we have

Altitude length and Base length

Q6 A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Answer:

Let the number of articles produced in a day = x

The cost of production of each article will be =2x+3

Given the total production on that day was Rs.90 .

Hence we have the equation;

But, x cannot be negative as it is the number of articles.

Therefore, x=6 and the cost of each article

Hence, the number of articles is 6 and the cost of each article is Rs.15.

C lass 10 Maths Chapter 4 Quadratic Equations Excercise: 4.3

Q1 (i) Find the roots of the following quadratic equations, if they exist, by the method of completing the square

Answer:

Given equation:

On dividing both sides of the equation by 2, we obtain

\Rightarrow x^2-\frac{7}{2}x+\frac{3}{2} = 0

\Rightarrow (x-\frac{7}{4})^2 + \frac{3}{2} - \frac{49}{16} = 0

\Rightarrow (x-\frac{7}{4})^2 = \frac{49}{16} - \frac{3}{2}

\Rightarrow (x-\frac{7}{4})^2 =\frac{25}{16}

\Rightarrow (x-\frac{7}{4}) =\pm \frac{5}{4}

\Rightarrow x = \frac{7}{4}+\frac{5}{4}\ or\ x = \frac{7}{4} - \frac{5}{4}

Q1 (ii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square

Answer:

Given equation:

On dividing both sides of the equation by 2, we obtain

Adding and subtracting $\frac{1}{16}$ in the equation, we get

\Rightarrow (x+\frac{1}{4})^2 -2 - \frac{1}{16} = 0

\Rightarrow (x+\frac{1}{4})^2 =2+\frac{1}{16}

\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}

\Rightarrow (x+\frac{1}{4}) =\pm \frac{\sqrt{33}}{4}

\Rightarrow x =\pm \frac{\sqrt{33}}{4} -\frac{1}{4}

\Rightarrow x = \frac{\pm \sqrt{33} - 1}{4}

\Rightarrow x = \frac{ \sqrt{33} - 1}{4}\ or\ x = \frac{ -\sqrt{33} - 1}{4}

Q1 (iii) Find the roots of the following quadratic equations, if they exist, by the method of completing the square

Answer:

Given equation: $4x^2 + 4\sqrt3 + 3 = 0$

On dividing both sides of the equation by 4, we obtain

Adding and subtracting $(\frac{\sqrt3}{2})^2$ in the equation, we get

\Rightarrow (x+\frac{\sqrt3}{2})^2 +\frac{3}{4} - (\frac{\sqrt3}{2})^2 = 0

\Rightarrow (x+\frac{\sqrt3}{2})^2 = \frac{3}{4} - \frac{3}{4} = 0

\Rightarrow (x+\frac{\sqrt3}{2}) = 0\ or\ (x+\frac{\sqrt3}{2}) = 0

Hence there are the same roots and equal:

\Rightarrow x = \frac{-\sqrt3}{2}\ or\ \frac{-\sqrt3}{2}

Q2 (iv) Find the roots of the following quadratic equations, if they exist, by the method of completing the square

Answer:

Given equation:

On dividing both sides of the equation by 2, we obtain

Adding and subtracting $(\frac{1}{4})^2$ in the equation, we get

\Rightarrow (x+\frac{1}{4})^2 +2- (\frac{1}{4})^2 = 0

\Rightarrow (x+\frac{1}{4})^2 = \frac{1}{16} -2 = \frac{-31}{16}

\Rightarrow (x+\frac{1}{4}) = \pm \frac{\sqrt{-31}}{4}

\Rightarrow x = \pm \frac{\sqrt{-31}}{4} - \frac{1}{4}

\Rightarrow x = \frac{\sqrt{-31}-1}{4} \ or\ x = \frac{-\sqrt{-31}-1}{4}

Here the real roots do not exist (in the higher studies we will study how to find the root of such equations).

Q2 Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer:

(i)

The general form of a quadratic equation is : , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

And the quadratic formula for finding the roots is:

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{7 \pm \sqrt{49-24}}{4}

\Rightarrow x= \frac{7 + 5}{4} = 3\ or\ x= \frac{7 - 5}{4} = \frac{1}{2}

Therefore, the real roots are: $x =3,\ \frac{1}{2}$

(ii)

The general form of a quadratic equation is : , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

And the quadratic formula for finding the roots is:

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-1 \pm \sqrt{1+32}}{4}

\Rightarrow x= \frac{-1 \pm \sqrt{33}}{4}

\Rightarrow x= \frac{-1 + \sqrt{33}}{4} \ or\ x= \frac{-1 - \sqrt{33}}{4}

Therefore, the real roots are:

(iii)

The general form of a quadratic equation is : , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

And the quadratic formula for finding the roots is:

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-4\sqrt{3} \pm \sqrt{48-48}}{8}

\Rightarrow x= \frac{-4\sqrt{3} \pm 0}{8}

Therefore, the real roots are:

(iv)

The general form of a quadratic equation is : , where a, b, and c are arbitrary constants.

Hence on comparing the given equation with the general form, we get

And the quadratic formula for finding the roots is:

Substituting the values in the quadratic formula, we obtain

\Rightarrow x= \frac{-1 \pm \sqrt{-31}}{4}

Here the term inside the root is negative

Therefore there are no real roots for the given equation.

Q3 (i) Find the roots of the following equations:

Answer:

Given equation: $x - \frac{1}{x} = 3, x\neq 0$

So, simplifying it,

Comparing with the general form of the quadratic equation: , we get

Now, applying the quadratic formula to find the roots:

\Rightarrow x= \frac{3 \pm \sqrt{9+4}}{2}

\Rightarrow x= \frac{3 \pm \sqrt{13}}{2}

Therefore, the roots are

\Rightarrow x = \frac{3+\sqrt{13}}{2}\ or\ \frac{3 - \sqrt{13}}{2}

Q3 (ii) Find the roots of the following equations:

Answer:

Given equation: $\frac{1}{x+4} - \frac{1}{x- 7} = \frac{11}{30},\ x\neq -4,7$

So, simplifying it,

\Rightarrow \frac{x-7-x-4}{(x+4)(x-7)} = \frac{11}{30}

\Rightarrow \frac{-11}{(x+4)(x-7)} = \frac{11}{30}

$\Rightarrow x^2-3x-28 = -30$ or $\Rightarrow x^2-3x+2 = 0$

Can be written as:

Hence the roots of the given equation are:

Q4 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is . Find the present age.

Answer:

Let the present age of Rehman be 'x' years.

Then, 3 years ago, his age was (x-3) years.

and 5 years later, his age will be (x+5) years.

Then according to the question we have,

\frac{1}{(x-3)}+\frac{1}{(x+5)} = \frac{1}{3}

Simplifying it to get the quadratic equation:

\Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3}

\Rightarrow \frac{2x+2}{(x-3)(x+5)} = \frac{1}{3}

Hence the roots are: $\Rightarrow x = 7,\ -3$

However, age cannot be negative

Therefore, Rehman is 7 years old in the present.

Q5 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Answer:

Let the marks obtained in Mathematics be ‘m’ then, the marks obtain in English will be ’30-m’.

Then according to the question:

Simplifying to get the quadratic equation:

Solving by the factorizing method:

We have two situations when,

The marks obtained in Mathematics is 12, then marks in English will be 30-12 = 18.

Or,

The marks obtained in Mathematics is 13, then marks in English will be 30-13 = 17.

Q6 The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Answer:

Let the shorter side of the rectangle be x m.

Then, the larger side of the rectangle wil be $= (x+30)\ m$ .

Diagonal of the rectangle:

It is given that the diagonal of the rectangle is 60m more than the shorter side.

Therefore,

\Rightarrow x^2+x^2+900+60x = x^2+3600+120x

Solving by the factorizing method:

Hence, the roots are: $x = 90,\ -30$

But the side cannot be negative.

Hence the length of the shorter side will be: 90 m

and the length of the larger side will be $(90+30)\ m =120\ m$

Q7 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Answer:

Given the difference of squares of two numbers is 180.

Let the larger number be ‘x’ and the smaller number be ‘y’.

Then, according to the question:

and

On solving these two equations:

Solving by the factorizing method:

As the negative value of x is not satisfied in the equation:

Hence, the larger number will be 18 and a smaller number can be found by,

putting x = 18, we obtain

$y^2 = 144\ or\ y = \pm 12$ .

Therefore, the numbers are or .

Q8 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer:

Let the speed of the train be $x\ km/hr.$

Then, time taken to cover 360km will be:

According to the question,

\Rightarrow (x+5)\left ( \frac{360}{x}-1 \right ) = 360

\Rightarrow 360-x+\frac{1800}{x} - 5 = 360

Making it a quadratic equation.

Now, solving by the factorizing method:

However, the speed cannot be negative hence,

The speed of the train is .

Q9 Two water taps together can fill a tank in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer:

Let the time taken by the smaller pipe to fill the tank be $x\ hr.$

Then, the time taken by the larger pipe will be: $(x-10)\ hr$ .

The fraction of the tank filled by a smaller pipe in 1 hour:

The fraction of the tank filled by the larger pipe in 1 hour.

$= \frac{1}{x-10}$
Given that two water taps together can fill a tank in $9\frac{3}{8} = \frac{75}{8}$ hours.

Therefore,

\Rightarrow \frac{1}{x}+\frac{1}{x-10} = \frac{8}{75}

\Rightarrow \frac{x-10+x}{x(x-10)} = \frac{8}{75}

\Rightarrow \frac{2x-10}{x(x-10)} = \frac{8}{75}

Making it a quadratic equation:

Hence the roots are $\Rightarrow x = 25,\ \frac{-30}{8}$

As time is taken cannot be negative:

Therefore, time is taken individually by the smaller pipe and the larger pipe will be and hours respectively.

Q10 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer:

Let the average speed of the passenger train be $x\ km/hr$ .

Given the average speed of the express train $= (x+11)\ km/hr$

also given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Therefore,

\Rightarrow \frac{132}{x} - \frac{132}{x+11} = 1

\Rightarrow 132\left [ \frac{x+11-x}{x(x+11)} \right ] = 1

\Rightarrow \frac{132\times11}{x(x+11)} = 1

Can be written as quadratic form:

Roots are: $\Rightarrow x = -44,\ 33$

As the speed cannot be negative.

Therefore, the speed of the passenger train will be $33\ km/hr$ and

The speed of the express train will be $33+11 = 44\ km/hr$ .

Q11 Sum of the areas of two squares is 468 m ². If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer:

Let the sides of the squares be $'x'\ and\ 'y'$ . (NOTE: length are in meters)

And the perimeters will be: $4x\ and\ 4y$ respectively.

Areas $x^2\ and\ y^2$ respectively.

It is given that,

……………………………(1)

……………………………(2)

Solving both equations:

x-y = 6 or x= y+6 putting in equation (1), we obtain

Solving by the factorizing method:

Here the roots are: $\Rightarrow y = -18,\ 12$

As the sides of a square cannot be negative.

Therefore, the sides of the squares are 12m and $(12\ m+6\ m) = 18\ m$ .

Quadratic Equation Class 10 Excercise: 4.4

Q1 (i) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Answer:

For a quadratic equation, the value of discriminant determines the nature of roots and is equal to:

If D>0 then roots are distinct and real.

If D<0 then no real roots.

If D= 0 then there exists two equal real roots.

Given the quadratic equation, .

Comparing with general to get the values of a,b,c.

Finding the discriminant:

Here D is negative hence there are no real roots possible for the given equation.

Q1 (ii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Answer:

b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0

Here the value of discriminant =0, which implies that roots exist and the roots are equal.

The roots are given by the formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}

So the roots are

Q1 (iii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Answer:

The value of the discriminant

The discriminant > 0. Therefore the given quadratic equation has two distinct real root

roots are

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}

So the roots are

\frac{3}{2}+\frac{\sqrt{3}}{2}, \frac{3}{2}-\frac{\sqrt{3}}{2}

Q2 (i) Find the values of k for each of the following quadratic equations so that they have two equal roots.

Answer:

For two equal roots for the quadratic equation:

The value of the discriminant D= 0 .

Given equation:

Comparing and getting the values of a,b, and, c.

The value of

Or, $\Rightarrow k=\pm \sqrt{24} = \pm 2\sqrt{6}$

Q2 (ii) Find the values of k for each of the following quadratic equations so that they have two equal roots

Answer:

For two equal roots for the quadratic equation:

The value of the discriminant D= 0 .

Given equation:

Can be written as:

Comparing and getting the values of a,b, and, c.

The value of

But k= 0 is NOT possible because it will not satisfy the given equation.

Hence the only value of is 6 to get two equal roots.

Q3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m ²? If so, find its length and breadth.

Answer:

Let the breadth of mango grove be 'b' .

Then the length of mango grove will be '2b' .

And the area will be:

Which will be equal to 800m^2 according to question.

Comparing to get the values of a,b,c .

Finding the discriminant value:

Here, D><noscript><img alt= 0″ height=”auto” src=”https://entrancecorner.codecogs.com/gif.latex?D%3E0″ width=”100%”>

Therefore, the equation will have real roots.

And hence finding the dimensions:

As negative value is not possible, hence the value of breadth of mango grove will be 20m.

And the length of mango grove will be: $= 2\times10 = 40m$

Q4 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.