Selina ICSE Class 10 Maths Solutions Chapter 6 Solving Problems Based On Quadratic Equations

Question 1. The product of two consecutive integers is 56. Find the integers.
Solution:
Let the first number be 𝑥.
And second number be 𝑥 + 1.
𝑥(𝑥 + 1) = 56
𝑥² + 𝑥– 56 = 0
By the splitting the middle term,
𝑥² + 8𝑥 − 7𝑥– 56 = 0
𝑥(𝑥 + 8) − 7(𝑥 + 8) = 0
(𝑥 + 8) (𝑥– 7) = 0
𝑥 + 8 = 0           𝑥 − 8 = 0
𝑥 = −8               𝑥 = 7
Hence, the required integers are 8 and 7.

Question 2. The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the first number be 𝑥.
And second number be 𝑥 + 1.
According to the question
𝑥² + (𝑥 + 1)² = 41
𝑥² + 𝑥² + 2𝑥 + 1 = 41
2𝑥² + 2𝑥 + 1– 41 = 0
2𝑥² + 2𝑥 − 40 = 0
Take common 2 in given equation
2(𝑥² + 𝑥– 20) = 0
𝑥² + 𝑥– 20 = 0
𝑥² + 5𝑥 − 4𝑥– 20 = 0
𝑥(𝑥 + 5) − 4(𝑥 + 20) = 0
(𝑥 + 5) (𝑥– 4) = 0
𝑥 + 5 = 0           𝑥 − 4 = 0
𝑥 = −5               𝑥 = 4
Here, 𝑥 is 4 as -5 is not a natural number.
Hence, the first number be 4 and second be 5.

Question 3. Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Solution:
Let the first number be 𝑥
And second number be 𝑥 + 5
According to the question,
𝑥² + (𝑥 + 5)² = 97
𝑥² + 𝑥² + 2(𝑥)(5) + (5)² = 97
2𝑥² + 10𝑥 + 25 – 97 = 0
2𝑥² + 10𝑥 − 72 = 0
2(𝑥² + 5𝑥 − 36) = 0
𝑥² + 5𝑥 − 36 = 0
By splitting middle term
𝑥² + 9𝑥 − 4𝑥 − 36 = 0
(𝑥 + 9) (𝑥 – 4) = 0
𝑥 + 9 = 0           𝑥 − 4 = 0
𝑥 = −9               𝑥 = 4
Here, 𝑥 is 4 as -9 is not a natural number.
Hence, the first number be 4 and second be 9.

Question 4. The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Let the number be 𝑥
And reciprocal of that number is 1/𝑥
According to question,
𝑥 + 1/𝑥 = 4.25
𝑥²+1/𝑥 = 4.25
𝑥² + 1 = 4.25𝑥
𝑥² + 1 = (425/100) 𝑥
100(𝑥² + 1) = 425𝑥
100𝑥² + 100 = 425𝑥
100𝑥² − 425𝑥 + 100 = 0
25(4𝑥² − 17𝑥 + 4) = 0
4𝑥² − 17𝑥 + 4 = 0
4𝑥² − 16𝑥 − 𝑥 + 4 = 0
4𝑥(𝑥 − 4) − 1(𝑥 − 4) = 0
(4𝑥 − 1)(𝑥 − 4) = 0
4𝑥 − 1 = 0            𝑥 − 4 = 0
𝑥 = 1/4                 𝑥 = 4
Hence, the required numbers are 4 and 1/4

Question 5. Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.
Solution:
Let the number be 𝑥
And second number be 𝑥 + 3
According to question

10(2𝑥 + 3) = 7(𝑥² + 3𝑥)
10(2𝑥 + 3) = 7(𝑥² + 3𝑥)
20𝑥 + 30 = 7𝑥² + 21𝑥
7𝑥² + 21𝑥 − 20𝑥 − 30 = 0
7𝑥² + 𝑥 − 30 = 0
By splitting the middle term
7𝑥²+ 𝑥 − 30 = 0
7𝑥² + 15𝑥 − 14𝑥 − 30 = 0
𝑥(7𝑥 + 15) − 2(7𝑥 − 15) = 0
(𝑥 − 2)(7𝑥 − 15) = 0
𝑥 − 2 = 0            7𝑥 − 15 = 0
𝑥 = 2                    𝑥 = 15/7
Here, 𝑥 is 2 as 15/7is not a natural number.
Hence, the first number be 2 and second be 5.

Question 6. Divide 15 into two parts such that the sum of their reciprocals is 3/10.
Solution:
Let the first number be 𝑥
And other number be 15 − 𝑥
According to question

150 = 3(15𝑥 − 𝑥²)
150 = 45𝑥 − 3𝑥²
3𝑥² − 45𝑥 + 150 = 0
3(𝑥² − 15𝑥 + 50) = 0
𝑥² − 15𝑥 + 50 = 0
𝑥² − 10𝑥 − 5𝑥 + 50 = 0
𝑥(𝑥 − 10) − 5(𝑥 − 10) = 0
(𝑥 − 5)(𝑥 − 10) = 0
𝑥 − 5 = 0 𝑥 − 10 = 0
𝑥 = 5 𝑥 = 10
Hence, the required part is 5 and 10.

Question 7. The sum of the square of two positive integers is 208. If the square of larger number is 18 times
the smaller number, find the numbers.
Solution:
Let us assumed that,
First number be 𝑥
And second number be 𝑦
According to question,
𝑥² + 𝑦² = 208
𝑦² = 18𝑥
Put the value of 𝑦² in above equation,
𝑥² + 18𝑥 = 208
𝑥² + 18𝑥 − 208 = 0
𝑥² + 26𝑥 − 8𝑥 − 208 = 0
𝑥(𝑥 + 26) − 8(𝑥 + 26) = 0
(𝑥– 8)(𝑥 + 26) = 0
𝑥 − 8 = 0            𝑥 + 26 = 0
𝑥 = 8                 𝑥 = −26
The value of 𝑥 is 8 as 𝑥 can’t be negative.
Put the value of 𝑥.
𝑦² = 18 × 8
𝑦² = 144
𝑦 = √144
𝑦 = 12
Hence, the value of 𝑥 is 8 and 𝑦 is 12.

Question 8. The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let the first number be 𝑥
And second number be 𝑥 + 2
According to the question,
𝑥² + (𝑥 + 2)² = 52
𝑥² + 𝑥² + 2(𝑥)(2) + (2)2 = 52
𝑥² + 𝑥² + 4𝑥 + 4 = 52
2𝑥² + 4𝑥 + 4 = 52
2𝑥² + 4𝑥 + 4 − 52 = 0
2𝑥² + 4𝑥 − 48 = 0
2(𝑥² + 2𝑥 − 24) = 0
𝑥² + 2𝑥 − 24 = 0
By splitting the middle term
𝑥² + 6𝑥 − 4𝑥 − 24 = 0
𝑥(𝑥 + 6) − 4(𝑥 + 6) = 0
(𝑥 − 4)(𝑥 + 6) = 0
𝑥 − 4 = 0           𝑥 + 6 = 0
𝑥 = 4                 𝑥 = −6
𝑥 can’t be negative so the value of 𝑥 is 4.
Hence, the first positive number be 4 and other be 6.

Question 9. Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let the first number be 𝑥
And second number be 𝑥 + 2
According to the question,
𝑥² + (𝑥 + 2)² = 74
𝑥² + (𝑥 + 2)² = 74
𝑥² + 𝑥² + 2(𝑥)(2) + (2)2 = 74
𝑥² + 𝑥² + 4𝑥 + 4 = 74
2𝑥² + 4𝑥 + 4 = 74
2𝑥² + 4𝑥 + 4 − 74 = 0
2𝑥² + 4𝑥 − 70 = 0
2(𝑥² + 2𝑥 − 35) = 0
𝑥² + 2𝑥 − 35 = 0
By splitting the middle term
𝑥² + 7𝑥 − 5𝑥 − 35 = 0
𝑥(𝑥 + 7) − 5(𝑥 + 7) = 0
(𝑥 − 5)(𝑥 + 7) = 0
𝑥 − 5 = 0            𝑥 + 7 = 0
𝑥 = 5                  𝑥 = −7
𝑥 can’t be negative so the value of 𝑥 is 5.
Hence, the first positive number be 5 and other be 7.

Question 10. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Let us assumed that,
Numerator will be 𝑥
And denominator will be 2𝑥 + 1
The fraction is 𝑥/2𝑥+1
According to the question,

10(5𝑥² + 4𝑥 + 1) = 29(2𝑥² + 𝑥)
50𝑥² + 40𝑥 + 10 = 58𝑥² + 29𝑥
50𝑥² − 58𝑥² + 40𝑥 − 29𝑥 + 10 = 0
−8𝑥² + 11𝑥 + 10 = 0
−(8𝑥² − 11𝑥 − 10) = 0
8𝑥² − 11𝑥 − 10 = 0
By the quadratic formula
𝑎 = 8
𝑏 = −11
𝑐 = −10

Here,
Numerator will be 2 and
Denominator = 2𝑥 + 1
Denominator = 2(2) + 1
Denominator = 4 + 1
Denominator = 5
Hence, the fraction is 2/5

Question 12. Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution:
Let the first number be 𝑥
Second number be 𝑦
According to question,
𝑥 + 𝑦 = 20
𝑦 = 20– 𝑥 _(i)
3𝑥² = (20 − 𝑥) + 10
3𝑥² = 20 − 𝑥 + 10
3𝑥² = 30 − 𝑥
3𝑥² + 𝑥 − 30 = 0
3𝑥² − 9𝑥 + 10𝑥 − 30 = 0
3𝑥(𝑥 − 3) + 10(𝑥 − 3) = 0
(3𝑥 + 10)(𝑥 − 3) = 0
3𝑥 + 10 = 0           𝑥 − 3 = 0
𝑥 = −10/3             𝑥 = 3
𝑥 can’t be negative so the value of 𝑥 is 3.
Put the value of 𝑥 in equation (i)
𝑦 = 20– 𝑥
𝑦 = 20– 3
𝑦 = 17
Hence, the value of 𝑥 is 3 and 𝑦 is 17.

Question 13. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.
Assume the middle number to be x and form a quadratic equation satisfying the above statement.
Hence; find the three numbers.
Solution:
Let us assumed that,
First number be 𝑥– 1
Second number be 𝑥
Third number be 𝑥 + 1
According to question,
𝑥² = (𝑥 + 1)2 − (𝑥 − 1)² + 60
𝑥² = 𝑥² + 2(𝑥)(1)+(1)² − [𝑥² − 2(𝑥)(1)+(1)²] + 60
𝑥² = 𝑥² + 2𝑥 + 1 − 𝑥² + 2𝑥 − 1 + 60
𝑥² = 4𝑥 + 60
𝑥² − 4𝑥 − 60 = 0
𝑥² − 10𝑥 + 6𝑥 − 60 = 0
(𝑥– 10)(𝑥 + 6) = 0
𝑥– 10 = 0          𝑥 + 6 = 0
𝑥 = 10               𝑥 = −6
𝑥 can’t be negative so the value of 𝑥 is 10.
First number = 𝑥– 1
First number = 10– 1
First number = 9
Second number = 𝑥
Second number = 10
Third number = 𝑥 + 1
Third number = 10 + 1
Third number = 11
Hence, the required numbers be 9, 10 and 11.

Question 14. Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Solution:
Let us assumed that,
First number be 𝑥– 1
Second number be 𝑥
Third number be 𝑥 + 1
According to question,
3(𝑥 + 1)² = (𝑥 − 1)² + 𝑥² + 67
3[(𝑥)² + 2(𝑥)(1) + (1)²] = (𝑥)² − 2(𝑥)(1) + (1)² + 𝑥² + 67
3(𝑥² + 2𝑥 + 1) = (𝑥)² − 2𝑥 + 1 + 𝑥² + 67
3𝑥² + 6𝑥 + 3 = 𝑥² − 2𝑥 + 1 + 𝑥² + 67
3𝑥² + 6𝑥 + 3 = 2𝑥² − 2𝑥 + 68
3𝑥² − 2𝑥² + 6𝑥 + 2𝑥 + 3 − 68 = 0
𝑥² + 8𝑥 − 65 = 0
𝑥² + 13𝑥 − 5𝑥 − 65 = 0
𝑥(𝑥 + 13) − 5(𝑥 + 13) = 0
(𝑥 − 5)(𝑥 + 13) = 0
𝑥– 5 = 0           𝑥 + 13 = 0
𝑥 = 5               𝑥 = −13
𝑥 can’t be negative so the value of 𝑥 is 5.
First number = 𝑥– 1
First number = 5– 1
First number = 4
Second number = 𝑥
Second number = 5
Third number = 𝑥 + 1
Third number = 5 + 1
Third number = 6
Hence, the required numbers be 4, 5 and 6.

Question 15. A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate ‘x’.
Solution:
It is given that,
Work done by A = 1/𝑥
Work done by B = 1/𝑥+16
According to the question
A and B together can do the work in 15 days.

15(2𝑥 + 16) = 𝑥² + 16𝑥
30𝑥 + 240 = 𝑥² + 16𝑥
𝑥² + 16𝑥 − 30𝑥 − 240 = 0
𝑥² + 14𝑥 − 240 = 0
𝑥² + 24𝑥 − 10𝑥 − 240 = 0
𝑥(𝑥 + 24) − 10(𝑥 + 24) = 0
(𝑥 − 10)(𝑥 + 24) = 0
𝑥– 10 = 0           𝑥 + 24 = 0
𝑥 = 10                𝑥 = −24
𝑥 can’t be negative so the value of 𝑥 is 10.
Hence, the value of 𝑥 is 10.

Question 16. One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Solution:
It is given that,
A pipe can fill a cistern in = 1/𝑥
B pipe can fill a cistern in = 1/𝑥−3
According to the question
A and B together can fill it in 6 hours and 40 minutes.

By cross multiply,
20(2𝑥 − 3) = 3(𝑥² − 3𝑥)
40𝑥 − 60 = 3𝑥² − 9𝑥
3𝑥² − 9𝑥 − 40𝑥 + 60 = 0
3𝑥² − 49𝑥 + 60 = 0
3𝑥² − 45𝑥 − 4𝑥 + 60 = 0
(3𝑥 − 4)(𝑥 − 15) = 0
3𝑥– 4 = 0          𝑥 − 15 = 0
𝑥 = 4/3             𝑥 = 15
𝑥 can’t be in fractions so the value of 𝑥 is 15.
Hence, one pipe fill cistern in 15 hours and other fills in (15 − 3) = 12.

Question 17. A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking 𝑥 as the smaller part of the two parts, find the number.
Solution:
Let us assumed that,
Smaller part be 𝑥
(Larger part)² = 8𝑥
Larger part = √8𝑥
According to the question,
𝑥² + (√8𝑥)² = 20
𝑥² + 8𝑥 = 20
𝑥² + 8𝑥 − 20 = 0
𝑥² − 2𝑥 + 10𝑥 − 20 = 0
𝑥(𝑥 − 2) + 10(𝑥 − 2) = 0
(𝑥 + 10)(𝑥 − 2) = 0
𝑥 + 10 = 0         𝑥 − 2 = 0
𝑥 = −10             𝑥 = 2
𝑥 can’t be in fractions so the value of 𝑥 is 2.
Smaller part = 2
Larger part = √8 × 2 = √16 = 4
Hence, the smaller part is 2 and larger part is 4.

Exercise 6B

Question 1. The sides of a right-angled triangle containing the right angle are 4𝑥 cm and (2𝑥– 1)cm. If the area of the triangle is 30𝑐𝑚²; Calculate the lengths of its sides.
Solution:
It is given that,
Height of a triangle is 4𝑥cm
Base of a triangle be (2𝑥– 1)cm
Area of triangle = 30cm²
We know that,
Area of triangle = 1/2 × ℎ𝑒𝑖𝑔ℎ𝑡 × 𝑏𝑎𝑠𝑒
30 = 1/2 × 4𝑥 × (2𝑥 − 1)
30 × 2 = 4𝑥 × (2𝑥 − 1)
60 = 8𝑥² − 4𝑥
8𝑥² − 4𝑥 − 60 = 0
4(2𝑥² − 𝑥 − 15) = 0
2𝑥² − 𝑥 − 15 = 0
2𝑥² − 6𝑥 + 5𝑥 − 15 = 0
2𝑥(𝑥 − 3) + 5(𝑥 − 3) = 0
(2𝑥 + 5)(𝑥 − 3) = 0
2𝑥 + 5 = 0          𝑥 − 3 = 0
𝑥 = −5/2            𝑥 = 3
𝑥 can’t be negative so the value of 𝑥 is 3.
Height of a triangle = 4 × 3 = 12cm
Base side = (2 × 3)– 1 = 6 − 1 = 5cm

Question 2.The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides.
Solution:
It is given that,
Hypotenuse = 26 cm
The sum of other two sides = 34 cm
Let the one side be 𝑥
Other two sides be (34– 𝑥) cm.
By Pythagoras theorem,
(Hypotenuse)² = (𝐻𝑒𝑖𝑔ℎ𝑡)² + (𝐵𝑎𝑠𝑒)²
(26)² = 𝑥² + (34– 𝑥)²
676 = 𝑥² + 𝑥² + 1156– 68𝑥
2𝑥² – 68𝑥 + 480 = 0
𝑥² – 34𝑥 + 240 = 0
𝑥² – 10𝑥 – 24𝑥 + 240 = 0
𝑥(𝑥– 10) – 24(𝑥– 10) = 0
(𝑥– 10)(𝑥– 24) = 0
𝑥 − 10 = 0         𝑥 − 24 = 0
𝑥 = 10                𝑥 = 24
If,
𝑥 = 10
(34– 𝑥) = 24
If
𝑥 = 24
(34– 𝑥) = 10
Hence, the lengths the three sides of the right-angled triangle are 10 cm, 24 cm and 26 cm.

Question 3. The sides of a right-angled triangle are (𝑥– 1)cm, 3𝑥cm and (3𝑥 + 1)cm. Find:
(i) the value of x,
(ii) the lengths of its sides,
(iii) its area.
Solution:
We know that,
The longer side of a triangle is hypotenuse.
Hypotenuse = (3𝑥 + 1)cm
Lengths of a sides is (𝑥– 1)cm
Length of other side 3𝑥 cm
By Pythagoras theorem,
(Hypotenuse)² = (𝐻𝑒𝑖𝑔ℎ𝑡)² + (𝐵𝑎𝑠𝑒)²
(3𝑥 + 1)² = (𝑥– 1)² + (3𝑥)²
(3𝑥)² + 2(3𝑥)(1) + (1)² = (𝑥)² − 2(𝑥)(1) + (1)² + (3𝑥)²
9𝑥² + 6𝑥 + 1 = 𝑥² − 2𝑥 + 1 + 9𝑥²
6𝑥 = 𝑥² − 2𝑥
𝑥²– 8𝑥 = 0
𝑥(𝑥 – 8) = 0
𝑥 = 0         𝑥 − 8 = 0
                  𝑥 = 8
𝑥 can’t be 0 so the value of 𝑥 is 8.
Hypotenuse = (3(8) + 1)cm
Hypotenuse = 25cm
Lengths of a sides is (8– 1)cm
Lengths of a sides is 7cm
Length of other side 3 × 8 cm
Length of other side 24cm
Area of the triangle = 1/2 × 7cm × 24cm
Area of the triangle = 84cm²
Hence, the area of triangle is 84cm².

Question 4. The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.
Solution:
Let the hypotenuse will be 𝑥
Length of one side = (𝑥– 1)cm
Length of other side = (𝑥– 18)cm
By Using Pythagoras theorem,
(Hypotenuse)² = (𝐻𝑒𝑖𝑔ℎ𝑡)² + (𝐵𝑎𝑠𝑒)²
𝑥² = (𝑥– 1)² + (𝑥– 18)²
𝑥² = 𝑥²– 2 × 𝑥 × 1 + (1)² + 𝑥²– 2 × 𝑥 × 18 + (18)² = 0
𝑥² = 𝑥²– 2𝑥 + 1 + 𝑥²– 36𝑥 + 324 = 0
𝑥² – 38𝑥 + 325 = 0
𝑥²– 13𝑥– 25𝑥 + 325 = 0
𝑥(𝑥– 13)– 25(𝑥– 13) = 0
(𝑥– 13) (𝑥– 25) = 0
𝑥– 13 = 0          𝑥 − 25 = 0
𝑥 = 13               𝑥 = 25
If,
𝑥 = 13
Hypotenuse =13cm
Length of one side = (13– 1)cm = 12cm
Length of other side = (13– 18)cm = -5cm
Length can’t be negative.
If
𝑥 = 25
Hypotenuse =25cm
Length of one side = (25– 1)cm = 24cm
Length of other side = (25– 18)cm = 7cm

Question 5. The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.
Solution:
Let us assumed that,
Shorter side be 𝑥 m
According to question,
Length of the other side = (𝑥 + 30)m
Length of hypotenuse = (𝑥 + 60) m
Using Pythagoras theorem,
(Hypotenuse)² = (𝐻𝑒𝑖𝑔ℎ𝑡)² + (𝐵𝑎𝑠𝑒)²
(𝑥 + 60)² = 𝑥² + (𝑥 + 30)²
𝑥² + 2 × 𝑥 × 60 + (60)² = 𝑥² + 𝑥² + 2 × 𝑥 × 30 + (30)² = 0
𝑥² + 120𝑥 + 3600 = 2𝑥² + 60𝑥 + 900 = 0
𝑥² + 120𝑥 + 3600 − 2𝑥² − 60𝑥 − 900 = 0
𝑥² + 120𝑥 + 3600 − 2𝑥² − 60𝑥 − 900 = 0
−𝑥² + 60𝑥 + 2700 = 0
𝑥² − 60𝑥 − 2700 = 0
𝑥² − 90𝑥 + 30𝑥 − 2700 = 0
𝑥(𝑥 − 90)30(𝑥 − 90) = 0
(𝑥 + 30)(𝑥 − 90) = 0
𝑥 + 30 = 0          𝑥 − 90 = 0
𝑥 = −30               𝑥 = 90
𝑥 can’t be negative so the value of 𝑥 is 90.
Length of shorter side = 90m.
Length of the other side = (𝑥 + 30)m = (90 + 30)m = 120m
Length of hypotenuse = (𝑥 + 60)m = (90 + 60)m = 150m
Hence, the shorter side is 90, other side is 120m and hypotenuse is 150m.

Question 6. The perimeter of a rectangle is 104 m and its area is 640 m². Find its length and breadth.
Solution:
It is given that,
Perimeter of rectangle = 104m
Area of rectangle = 640m²
Let us assumed that,
Length of rectangle be 𝑥
Breadth of rectangle be 𝑦
Perimeter = 2(x + y)m
104 = 2(𝑥 + 𝑦)
𝑥 + 𝑦 = 52
𝑦 = 52 – 𝑥
Area = L × B m²
𝑥𝑦 = 640
𝑥(52– 𝑥) = 640
𝑥²– 52𝑥 + 640 = 0
𝑥²– 32𝑥– 20𝑥 + 640 = 0
𝑥(𝑥– 32)– 20(𝑥– 32) = 0
(𝑥– 32)(𝑥– 20) = 0
𝑥 − 32 = 0            𝑥 − 20 = 0
𝑥 = 32                  𝑥 = 20
If,
𝑥 = 32
𝑦 = 52– 32 = 20
If,
𝑥 = 20
𝑦 = 52– 20 = 32
Hence, the length and breadth of the rectangle are 32 cm and 20 cm

Question 7. A footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m², find the width of the footpath.
Solution:
It is given that,
Length of outer rectangular field is 32m
Breadth of outer rectangular field is 24m
Let us assumed that,
𝑥 be the width of the footpath
Length of inner rectangular field is (32 − 2𝑥)𝑚
Breadth of outer rectangular field is (24 − 2𝑥)𝑚
We know that,
Area of the path = Area of outer rectangle – Area of inner rectangle
208 = (32)(24)– (32– 2𝑥)(24– 2𝑥)
208 = 768– [768 − 64𝑥 − 48𝑥 + 4𝑥²]
208 = 768– 768 + 64𝑥 + 48𝑥 − 4𝑥²
208 = 112𝑥 − 4𝑥²
4𝑥² − 112𝑥 + 208 = 0
4(𝑥² − 28𝑥 + 52) = 0
𝑥² − 28𝑥 + 52 = 0
𝑥² − 26𝑥 − 2𝑥 + 52 = 0
𝑥(𝑥 − 26) − 2(𝑥 − 26) = 0
(𝑥 − 2)(𝑥 − 26) = 0
𝑥 − 2 = 0            𝑥 − 26 = 0
𝑥 = 2                  𝑥 = 26
If 𝑥 = 2
Length of inner rectangular field is (32 − 2𝑥)𝑚
= (32 − 2(2))𝑚
= (32 − 4)𝑚
= 28𝑚
Breadth of outer rectangular field is (24 − 2𝑥)𝑚
= (24 − 2(2))𝑚
= (24 − 4)𝑚
= 20𝑚
If 𝑥 = 26
Length of inner rectangular field is (32 − 2𝑥)𝑚
= (32 − 2(26))𝑚
= (32 − 52)𝑚
= −20𝑚
It can’t be possible.
Hence, the width of the footpath is 2𝑚.

Question 8. Two squares have sides x cm and (x + 4)cm. The sum of their area is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Solution:
It is given that,
Side of first square = 𝑥 𝑐𝑚
Side of second square = (𝑥 + 4)𝑐𝑚.
Sum of their area = 656𝑐𝑚²
𝑥² + (𝑥 + 4)² = 656
𝑥² + 𝑥² + 2 × 4 × 𝑥 + 16 = 656
2𝑥² + 8𝑥 + 16 = 656
2𝑥² + 8𝑥 + 16– 656 = 0
2𝑥² + 8𝑥– 640 = 0
2(𝑥² + 4𝑥– 320) = 0
𝑥² + 4𝑥– 320 = 0
𝑥² + 20𝑥– 16𝑥– 320 = 0
𝑥(𝑥 + 20) – 16(𝑥 + 20) = 0
(𝑥 + 20) (𝑥 – 16) = 0
𝑥 + 20 = 0          𝑥 − 16 = 0
𝑥 = −20               𝑥 = 16
𝑥 can’t be negative so the value of 𝑥 is 16.
Hence, the sides of the two squares are 16cm and 20cm.

Question 9. The dimensions of a rectangular field are 50 m and 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs. 30 and Rs. 20 per square meter, respectively, is Rs. 52,000.
Find the width of the gravel path.
Solution:
It is given that,
Length of the rectangular field = 50m
Breadth of the rectangular field = 40m
Let us assumed that,
Width of the gravel path be 𝑤m
Length of the flower bed be 𝑥m
Breadth of the flower bed be 𝑦m
Therefore, we have:
x + 2𝑤 = 50
x = 50 − 2w__________(1)
y + 2w = 40
y = 40 − 2𝑤__________(2)
Area of rectangle = Length × Breadth
Area of rectangular field = 50m × 40m
Area of rectangular field = 2000m²
Area of the flower bed = xym²
Area of gravel path = Area of rectangular field – Area of flower bed
Area of gravel path = (2000– 𝑥𝑦)𝑚²
Cost of laying flower bed + Gravel path = Area x cost of laying per sq. m
52000 = 30𝑥𝑦 + 20(2000– 𝑥𝑦)
52000 = 30𝑥𝑦 + 40000 − 20𝑥𝑦
52000 = 10𝑥𝑦 + 40000
−10𝑥𝑦 = 40000 − 52000
−10𝑥𝑦 = −12000
𝑥𝑦 = 1200
Using (1) and (2), we have:
(50– 2𝑤)(40– 2𝑤) = 1200
2000– 180𝑤 + 4𝑤² = 1200
2000– 180𝑤 + 4𝑤² = 1200 − 2000
– 180𝑤 + 4𝑤² = −800
4𝑤²– 180𝑤 + 800 = 0
4(𝑤²– 45𝑤 + 200) = 0
𝑤²– 45𝑤 + 200 = 0
𝑤²– 5𝑤– 40𝑤 + 200 = 0
𝑤(𝑤– 5)– 40(𝑤– 5) = 0
(𝑤– 5)(𝑤– 40) = 0
𝑤 − 5 =         0 𝑤 − 40 = 0
𝑤 = 5               𝑤 = 40
If, 𝑤 = 40
𝑥 = 50– 2𝑤
𝑥 = −30
It can’t be possible.
Hence, the width of the gravel path is 5m.

Question 10. An area is paved with square tiles of a certain size and the number required is 128. If the tiles had been 2 cm smaller each way, 200 tiles would have been needed to pave the same area. Find the size of the larger tiles.
Solution:
Let us assumed that,
The size of the larger tiles be 𝑥cm.
Size of smaller tiles = (𝑥– 2)cm
Area of larger tiles = 𝑥𝑐𝑚²
Area of smaller tiles = (𝑥– 2)²𝑐𝑚²
It is given that,
Number of larger tiles required = 128.
Area needed to be paved = 128𝑥²𝑐𝑚²__________(1)
Number of larger tiles required to pave an area is 200.
Area needed to be paved = 200(𝑥– 2)2𝑐𝑚²_________(2)
From equation (1) and (2),
We have,
128𝑥² = 200(𝑥– 2)²
128𝑥²
= 200𝑥²
+ 800 – 800𝑥
72𝑥²– 800𝑥 + 800 = 0
9𝑥²– 100𝑥 + 100 = 0
9𝑥²– 90𝑥– 10𝑥 + 100 = 0
9𝑥(𝑥– 10) – 10(𝑥– 10) = 0
(𝑥– 10)(9𝑥– 10) = 0
𝑥– 10 = 0           9𝑥– 10 = 0
𝑥 = 10                 𝑥 = 10/9
If,
𝑥 = 10/9

Question 11. A farmer has 70 m of fencing, with which he encloses three sides of a rectangular sheep pen; the fourth side being a wall. If the area of the pen is 600 sq. m, find the length of its shorter side.
Solution:
Let us assumed that,
The length of the rectangular sheep pen be x
The breadth of the rectangular sheep pen be y
According to the question,
𝑥 + 𝑦 + 𝑥 = 70
2𝑥 + 𝑦 = 70 __(1)
Get the value of 𝑦 from equation (1)
𝑦 = 70 − 2𝑥
It is given that,
The area of rectangle is 600.
Area = Length × Breadth
𝑥(70– 2𝑥) = 600
70𝑥– 2𝑥² = 600
2𝑥²– 70𝑥 + 600 = 0
2(𝑥²– 35𝑥 + 300) = 0
𝑥²– 35𝑥 + 300 = 0
𝑥²– 15𝑥– 20𝑥 + 300 = 0
𝑥(𝑥– 15)– 20(𝑥– 15) = 0
(𝑥– 15)(𝑥– 20) = 0
𝑥– 15 = 0            𝑥– 20 = 0
𝑥 = 15                𝑥 = 20
If,
𝑥 = 15
𝑦 = 70 – 2𝑥
𝑦 = 70 – 30
𝑦 = 40
If,
𝑥 = 20
𝑦 = 70 – 2𝑥
𝑦 = 70 – 40
𝑦 = 30
Hence, the length of the shorter side is 15 m then longer side is 40 m and length of the shorter side is 20 m then longer side is 30 m.

Question 12. A square lawn is bounded on three sides by a path 4 m wide. If the area of the path is 7/8 that of the lawn, find the dimensions of the lawn.
Solution:
Let us assumed that,
The side of the square lawn be 𝑥𝑚
Area of the square lawn = 𝑥²𝑚²
The square lawn is bounded on three sides by a path which is 4 m wide.
Area of outer rectangle = (𝑥 + 4)(𝑥 + 8)
Area of outer rectangle = 𝑥² + 8𝑥 + 4𝑥 + 32
Area of outer rectangle = 𝑥² + 12𝑥 + 32
Area of path = 𝑥² + 12𝑥 + 32– (𝑥²)
Area of path = 12𝑥 + 32
According to the question,
12𝑥 + 32 = 7/8 × 𝑥²
8(12𝑥 + 32) = 7𝑥²
96𝑥 + 256 = 7𝑥²
7𝑥² − 96𝑥 − 256 = 0
7𝑥² − 112𝑥 + 16𝑥 − 256 = 0
7𝑥(𝑥 − 16) + 16(𝑥 − 16) = 0
(7𝑥 + 16)(𝑥 − 16) = 0
7𝑥 + 16 = 0           𝑥– 16 = 0
𝑥 = −16/7            𝑥 = 16
𝑥 can’t be negative so the value of 𝑥 is 16.
Hence, sides of the square lawn are 16m.

Question 13. The area of a big rectangular room is 300 m². If the length were decreased by 5m and the breadth increased by 5m; the area would be unaltered. Find the length of the room.
Solution:
Let us assumed that,
Length of the rectangular room be 𝑥m
Breadth of the rectangular room be 𝑦m
Area of the rectangular room = 300𝑚²
𝑥𝑦 = 300
𝑦 = 300/𝑥 __(1)
New length = (𝑥– 5)𝑚
New breadth = (𝑦 + 5)𝑚
Area of new rectangular = (𝑥– 5)(𝑦 + 5)
(𝑥– 5)(𝑦 + 5) = 300

5𝑥² − 1500 − 25𝑥 = 0
5(𝑥² − 5𝑥 − 300) = 0
𝑥² − 5𝑥 − 300 = 0
𝑥² − 20𝑥 + 15𝑥 − 300 = 0
𝑥(𝑥 − 20) + 15(𝑥 − 20) = 0
(𝑥 + 15)(𝑥 − 20) = 0
𝑥 + 15 = 0            𝑥 − 20 = 0
𝑥 = −15                𝑥 = 20
𝑥 can’t be negative so the value of 𝑥 is 20.
Hence, length of the room is 20m.

Exercise 6C

Question 1. The speed of an ordinary train is x km per hrs. and that of an express train is (x + 25) km per hr.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs. more than the express train; calculate speed of the express train.
Solution:
(i) Speed of ordinary train = 𝑥km/hr
Speed of express train = (𝑥 + 25)km/hr
Distance = 300km

7500 = 2(𝑥² + 25𝑥)
7500 = 2𝑥² + 50𝑥
2𝑥² + 50𝑥 − 7500 = 0
2(𝑥² + 25𝑥 − 3750) = 0
𝑥² + 25𝑥 − 3750 = 0
𝑥² + 75𝑥 − 50𝑥 − 3750 = 0
𝑥(𝑥 + 75) − 50(𝑥 + 75) = 0
(𝑥 − 50)(𝑥 + 75) = 0
𝑥 − 50 = 0            𝑥 + 75 = 0
𝑥 = 50                 𝑥 = −75
𝑥 can’t be negative as the value of x is 50.
Speed of the express train is (x + 25) km/hr = 75 km/hr
Hence, the speed of ordinary train is 50km/hr and express train is 75km/hr.

Question 2. If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Solution:
Let the speed of the car be x km/hr.
Distance = 36 km

360(60) = 18(𝑥² + 10𝑥)
21600 = 18𝑥² + 180𝑥
18𝑥² + 180𝑥 − 21600 = 0
18(𝑥² + 10𝑥 − 1200) = 0
𝑥² + 10𝑥 − 1200 = 0
𝑥² + 40𝑥 − 30𝑥 − 1200 = 0
𝑥(𝑥 + 40) − 30(𝑥 + 40) = 0
(𝑥 − 30)(𝑥 + 40) = 0
𝑥 − 30 = 0          𝑥 + 40 = 0
𝑥 = 30                 𝑥 = −40
𝑥 can’t be negative so the value of 𝑥 is 30.
Hence, speed of the car is 30km/hr.

Question 3. If the speed of an aero-plane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200km. Find the speed of the aero-plane.
Solution:
It is given that,
Total distance cover = 1200km
Let us assumed that,
Original speed of the aero-plane is 𝑥km/hr.
We know that,

3(48000) = 𝑥² − 40𝑥
144000 = 𝑥² − 40𝑥
𝑥² − 40𝑥 − 144000 = 0
𝑥² − 400𝑥 + 360𝑥 − 144000 = 0
𝑥(𝑥 − 400) + 360(𝑥 − 400) = 0
(𝑥 + 360)(𝑥 − 400) = 0
𝑥 + 360 = 0          𝑥 − 400 = 0
𝑥 = −360              𝑥 = 400
𝑥 as a speed can’t be negative as the value of x is 400.
Speed of the express train is (x + 25) km/hr = 75 km/hr
Hence, the speed of aero-plane is 400km/hr.

Question 4. A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
It is given that,
Total distance cover = 400km
Let us assumed that,
The original speed of the car is 𝑥 km/h
We know that,

3(4800) = 5(𝑥² + 12𝑥)
14400 = 5𝑥² + 60𝑥
5𝑥² + 60𝑥 − 14400 = 0
5(𝑥² + 12𝑥 − 2880) = 0
𝑥² + 12𝑥 − 2880 = 0
𝑥² + 60𝑥 − 48𝑥 − 2880 = 0
𝑥(𝑥² + 60) − 48(𝑥 + 60) = 0
(𝑥 − 48)(𝑥 + 60) = 0
𝑥 − 48 = 0            𝑥 + 60 = 0
𝑥 = 48                 𝑥 = −60
𝑥 as a speed can’t be negative as the value of x is 48.
Hence, the speed of car is 48km/hr.

Question 5. A girl goes to her friend’s house, which is at a distance of 12km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2)km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’.
Solution:
It is given that,
Total distance cover by a girl = 6km
Speed = 𝑥km/hr

2(12𝑥 + 12) = 5(𝑥² + 2𝑥)
24𝑥 + 24 = 5𝑥² + 10𝑥
5𝑥² + 10𝑥 − 24𝑥 − 24 = 0
5𝑥² − 14𝑥 − 24 = 0
5𝑥² − 14𝑥 − 24 = 0
5𝑥² − 20𝑥 + 6𝑥 − 24 = 0
5𝑥(𝑥 − 4) + 6(𝑥 − 4) = 0
(5𝑥 + 6)(𝑥 − 4) = 0
5𝑥 + 6 = 0          𝑥 − 4 = 0
𝑥 = −6/5             𝑥 = 4
𝑥 as a speed can’t be negative as the value of x is 4.
Hence, the speed is 4km/hr.

Question 6. A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km/hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Solution:
It is given that,
Distance cover by car is 390km
Time taken by car is 𝑥 hrs.
Let us assumed that,
Original speed of the car be 𝑦km/h

780 = 𝑦² + 4𝑦
𝑦² + 4𝑦 − 780 = 0
𝑦² + 30𝑦 − 26𝑦 − 780 = 0
𝑦(𝑦 + 30) − 26(𝑦 + 30) = 0
(𝑦 − 26)(𝑦 + 30) = 0
𝑦 − 26 = 0          𝑦 + 30 = 0
𝑦 = 26                𝑦 = −30
𝑦 as a speed can’t be negative as the value of y is 26.
Hence, the time taken by car is 390/26 = 15.

Question 7. A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.
Solution:
It is given that,
Distance covered by the train = 1040km
Let us assumed that,
The speed of goods train be 𝑥 km/hr

1600 × 5 = 𝑥² + 20𝑥
8000 = 𝑥² + 20𝑥
𝑥² + 20𝑥 − 8000 = 0
𝑥² − 80𝑥 + 100𝑥 − 8000 = 0
𝑥(𝑥 − 80) + 100(𝑥 − 80) = 0
(𝑥 + 100)(𝑥 − 80) = 0
𝑥 − 80 = 0            𝑥 + 100 = 0
𝑥 = 80                  𝑥 = −100
𝑥 as a speed can’t be negative as the value of x is 80.
Hence, speed of goods train and express train is 80km/hr and 100km/hr respectively.

Question 8. A man bought an article for Rs. 𝑥 and sold it for Rs. 16. If his loss was x per cent, find the cost price of the article.
Solution:
It is given that,
Cost Price of an article = Rs. 𝑥
Selling Price of an article = Rs 16
Loss = Rs. (𝑥– 16)

𝑥² = (𝑥 − 16) × 100
𝑥² = 100𝑥 − 1600
𝑥² − 100𝑥 + 1600 = 0
𝑥² − 80𝑥 − 20𝑥 + 1600 = 0
𝑥(𝑥 − 80) − 20(𝑥 − 80) = 0
(𝑥 − 20)(𝑥 − 80) = 0
𝑥 − 20 = 0           𝑥 − 80 = 0
𝑥 = 20                 𝑥 = 80
Hence, the cost price of the article is Rs. 20 or Rs. 80.

Question 9. A trader bought an article for Rs x and sold it for Rs 52, thereby making a profit of (x – 10) percent on his outlay. Calculate the cost price.
Solution:
It is given that,
Cost Price of an article = Rs. 𝑥
Selling Price of an article = Rs. 52
Profit = Rs. (52 – 𝑥)
Profit % = Rs. (𝑥 − 10)

𝑥(𝑥 − 10) = (52 − 𝑥) × 100
𝑥² − 10𝑥 = 5200 − 100𝑥
𝑥² − 10𝑥 + 100𝑥 − 5200 = 0
𝑥² + 90𝑥 − 5200 = 0
𝑥² + 130𝑥 − 40𝑥 − 5200 = 0
𝑥(𝑥 + 130) − 40(𝑥 + 130) = 0
(𝑥 − 40)(𝑥 + 5200) = 0
𝑥 − 40 = 0            𝑥 + 130 = 0
𝑥 = 40                  𝑥 = −130
Cost Price can’t be negative.
Hence, cost price of the article is Rs. 40.

Question 10. By selling a chair for Rs 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let us assumed that,
Cost Price of the chair be Rs. 𝑥
S.P. of chair = Rs. 75
Profit = Rs. (75 – 𝑥)

𝑥² = (75 − 𝑥) × 100
𝑥² = 7500 − 100𝑥
𝑥² + 100𝑥 − 7500 = 0
𝑥² + 150𝑥 − 50𝑥 − 7500 = 0
𝑥(𝑥 + 150) − 50(𝑥 + 150) = 0
(𝑥 − 50)(𝑥 + 150) = 0
𝑥 − 50 = 0          𝑥 + 150 = 0
𝑥 = 50                  𝑥 = −150
Cost Price can’t be negative.
Hence, cost of the chair is Rs. 50.

Exercise 6D

Question 1. The sum S of n successive odd numbers starting from 3 is given by the relation n(n + 2). Determine n, if the sum is 168.
Solution:
According to the question,
𝑛(𝑛 + 2) = 168
𝑛² + 2𝑛– 168 = 0
𝑛² + 14𝑛– 12𝑛– 168 = 0
𝑛(𝑛 + 14)– 12(𝑛 + 14) = 0
(𝑛 + 14)(𝑛– 12) = 0
𝑛 + 14 = 0            𝑛 − 12 = 0
𝑛 = −14               𝑛 = 12
𝑛 Can’t be negative, Hence, the value of 𝑛 be 12.

Question 2. A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d meters, that it falls in t seconds. How long does it take to fall 420 meters?
Solution:
According to the question,
16𝑡² + 4𝑡 = 420
4𝑡² + 𝑡– 105 = 0
4𝑡²– 20𝑡 + 21𝑡– 105 = 0
4𝑡(𝑡– 5) + 21(𝑡– 5) = 0
(4𝑡 + 21)(𝑡– 5) = 0
4𝑡 + 21 = 0            𝑡– 5 = 0
𝑡 = −21/4              𝑡 = 5
𝑡 as time cannot be negative.
Hence, the required time taken is 5 seconds.

Question 3. The product of the digits of a two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
Let the first number be 𝑥
And second number be 𝑦
According to the question,
𝑥𝑦 = 24___________(1)
It is given that, unit’s digit exceeds twice its ten’s digit by 2.
𝑦 = 2𝑥 + 2 __(2)
From equation (2) we get the value of 𝑦
𝑦 = 2𝑥 + 2
Put the value of 𝑦 in equation (1)
𝑥(2𝑥 + 2) = 24
2𝑥² + 2𝑥 = 24
2𝑥² + 2𝑥 − 24 = 0
2(𝑥² + 𝑥 − 12) = 0
𝑥² + 𝑥 − 12 = 0
𝑥² + 4𝑥 − 3𝑥 − 12 = 0
𝑥(𝑥 + 4) − 3(𝑥 + 4) = 0
(𝑥 − 3)(𝑥 + 4) = 0
𝑥 − 3 = 0            𝑥 + 4 = 0
𝑥 = 3                  𝑥 = −4
𝑥 Can’t be negative, the value of 𝑥 is 3.
Put the value of x in equation no (2).
𝑦 = 2(3) + 2
𝑦 = 6 + 2
𝑦 = 8
Hence, the required number is 38.

Question 4. The ages of two sisters are 11 years and 14 years. In how many years’ time will the product of their ages be 304?
Solution:
It is given that,
Ages of two sisters are 11 years and 14 years.
Let us assumed that,
𝑥 Number of years the product of their ages be 304.
(11 + 𝑥)(14 + 𝑥) = 304
154 + 11𝑥 + 14𝑥 + 𝑥² = 304
154 + 11𝑥 + 14𝑥 + 𝑥² − 304 = 0
𝑥² + 25𝑥 − 150 = 0
𝑥² + 30𝑥 − 5𝑥 − 150 = 0
𝑥(𝑥 + 30) − 5(𝑥 + 30) = 0
(𝑥 − 5)(𝑥 + 30) = 0
𝑥 − 5 = 0            𝑥 + 30 = 0
𝑥 = 5                  𝑥 = −30
The number of years cannot be negative so, the value of 𝑥 is 5.
Hence, the required number of years is 5 years.

Question 5. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Solution:
Let the present age of the son be 𝑥years
Present age of man = 𝑥²years
According to question, One year ago
Son’s age = (𝑥 – 1)years
Man’s age = (𝑥²– 1) years
It is given that one year ago; a man was 8 times as old as his son.
(𝑥²– 1) = 8(𝑥 – 1)
𝑥²– 1 = 8𝑥– 8
𝑥²– 8𝑥 – 1 + 8 = 0
𝑥²– 8𝑥 + 7 = 0
𝑥²– 7𝑥 − 𝑥 + 7 = 0
𝑥(𝑥– 7) − 1(𝑥 − 7) = 0
(𝑥– 7)(𝑥– 1) = 0
𝑥 − 7 = 0            𝑥 − 1 = 0
𝑥 = 7                   𝑥 = 1
If,
𝑥 = 1
𝑥² = 1,
It is not possible as father’s age can’t be equal to son’s age.
If,
𝑥 = 7
Present age of son = 𝑥 years = 7 years
Present age of man = 𝑥²years = 49 years
Hence, the son’s age is 7 years and father’s age 49 years.

Question 6. The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let us assumed that,
The present age of the son be x years.
According to the question,
Present age of father = 2𝑥²years
Eight years hence,
Son’s age = (𝑥 + 8)years
Father’s age = (2𝑥² + 8)years
The age of the father will be 4 years more than three times the age of the son.
2𝑥² + 8 = 3(𝑥 + 8) + 4
2𝑥² + 8 = 3𝑥 + 24 + 4
2𝑥² + 8 = 3𝑥 + 28
2𝑥² − 3𝑥 + 8 − 28 = 0
2𝑥²– 3𝑥– 20 = 0
2𝑥²– 8𝑥 + 5𝑥– 20 = 0
2𝑥(𝑥– 4) + 5(𝑥– 4) = 0
(𝑥– 4)(2𝑥 + 5) = 0
𝑥 − 4 = 0          2𝑥 + 5 = 0
𝑥 = 4                  𝑥 = − 5/2
Age can’t be negative, so the value of 𝑥 is 4.
Present age of son = 4 years
Present age of father = 2(42)years = 32 years
Hence, the present age of the son is 4years and father’s age is 32.

Question 7. The speed of a boat in still water is 15 km/hr. It can go 30km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let us assumed that,
The speed of the stream be 𝑥km/hr
Speed of the boat downstream = (15 + 𝑥)km/hr
Speed of the boat upstream = (15– 𝑥)km/hr
Distance cover by board = 30km

900 × 2 = 9(225 − 𝑥²)
1800 = 2025 − 9𝑥²
9𝑥² = 2025 − 1800
9𝑥² = 225
𝑥² = 225/9
𝑥² = 25
𝑥 = ±√25
𝑥 = ±5
𝑥 can’t be negative, so the value of 𝑥 is 5.
Hence, the speed of the stream is 5km/hr.

Question 8. Mr. Mehra sends his servant to the market to buy oranges worth Rs. 15. The servant having eaten three oranges on the way. Mr. Mehra pays Rs. 25paise per orange more than the market price.
Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
Let Number of oranges = y
Cost of one orange = Rs. 15/𝑦
The servant ate 3 oranges, so Mr. Mehra received (𝑦 − 3) oranges.
𝑥 = 𝑦 − 3
𝑦 = 3 + 𝑥

63𝑥 + 𝑥² = 15(4𝑥 + 12)
63𝑥 + 𝑥² = 60𝑥 + 180
63𝑥 + 𝑥² − 60𝑥 − 180 = 0
𝑥² − 3𝑥 − 180 = 0
𝑥² + 15𝑥 − 12𝑥 − 180 = 0
𝑥(𝑥 + 15) − 12(𝑥 + 15) = 0
(𝑥 − 12)(𝑥 + 15) = 0
𝑥 − 12 = 0            𝑥 + 15 = 0
𝑥 = 12                   𝑥 = −15
𝑥 can’t be negative so, the value of 𝑥 is 12.
Hence, the number of oranges is 12.

Question 9. Rs. 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50paise less. Find the number of children.
Solution:
Let us assumed that,
The number of children be x.
According to the question,
It is given that Rs. 250 is divided amongst x students.
Money received by each child = 250/𝑥
As per question if we increase 25 children,

6250 × 2 = 𝑥² + 25𝑥
12500 = 𝑥² + 25𝑥
𝑥² + 25𝑥 − 12500 = 0
𝑥² + 125𝑥 − 100𝑥 − 12500 = 0
𝑥(𝑥 + 125) − 100(𝑥 + 125) = 0
(𝑥 − 100)(𝑥 + 125) = 0
𝑥 − 100 = 0 𝑥 + 125 = 0
𝑥 = 100 𝑥 = −125
𝑥 as the number of students can’t be negative, so, the value of 𝑥 is 100.
Hence, the number of students is 100.

Question 10. An employer finds that if he increased the weekly wages of each worker by Rs. 5 and employs five workers less, he increases his weekly wage bill from Rs. 3,150 to Rs. 3,250. Taking the original weekly wage of each worker as Rs x; obtain an equation in x and then solve it to find the weekly wages of each worker.
Solution:
Original weekly wage of each worker = Rs. 𝑥
Original weekly wage bill of employer = Rs. 3150
Number of workers = 3150/𝑥
New weekly wage of each worker = Rs. (𝑥 + 5)
New weekly wage of each worker = Rs. 3250

(3150 − 5𝑥)(𝑥 + 5) = 3250𝑥
3150𝑥 + 15750 − 5𝑥² − 25𝑥 = 3250𝑥
3150𝑥 − 3250𝑥 + 15750 − 5𝑥² − 25𝑥 = 0
15750 − 5𝑥² − 125𝑥 = 0
−5(−3150 + 𝑥² + 25𝑥) = 0
𝑥² + 25𝑥 − 3150 = 0
𝑥² + 70𝑥 − 45𝑥 − 3150 = 0
𝑥(𝑥 + 70) − 45(𝑥 + 70) = 0
(𝑥 − 45)(𝑥 + 70) = 0
𝑥 − 45 = 0            𝑥 + 70 = 0
𝑥 = 45                  𝑥 = −70
𝑥 can’t be negative so, the value of 𝑥 is 45.
Hence, the weekly wages of each worker is Rs. 45.

Question 11. A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on whole
transaction.
Taking the number of articles he bought as x, form an equation in x and solve it.
Solution:
It is given that,
Total Cost of the articles is Rs. 1200.
Let us assumed that,
Articles bought by the trader = 𝑥
Cost per article = 1200/𝑥
According to question,
Ten articles were damaged then remaining articles are 𝑥 − 10

2𝑥² − 12000 − 20𝑥 = 60𝑥
2𝑥² − 12000 − 20𝑥 − 60𝑥 = 0
2𝑥² − 12000 − 80𝑥 = 0
2𝑥² − 80𝑥 − 12000 = 0
2(𝑥² − 40𝑥 − 6000) = 0
𝑥² − 40𝑥 − 6000 = 0
𝑥² − 100𝑥 + 60𝑥 − 6000 = 0
𝑥(𝑥 − 100) + 60(𝑥 − 100) = 0
(𝑥 + 60)(𝑥 − 100) = 0
𝑥 − 100 = 0 𝑥 + 60 = 0
𝑥 = 100 𝑥 = −60
𝑥 can’t be negative. So, the value of 𝑥 is 100.
Hence, the number of articles will be 100.

Question 12. The total cost price of a certain number of identical articles is Rs 4800. By selling the articles at Rs 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Solution:
Let the number of articles bought be x.
Total cost price of x articles = Rs 4800
Cost price of one article = Rs. 4800/𝑥
Selling price of each article = Rs 100
Selling price of x articles = Rs 100x
It is given that,
Profit = C.P. of 15 articles
100𝑥– 4800 = 15 × (4800/𝑥) 100
𝑥² – 4800𝑥 = 15 × 4800
𝑥² – 48𝑥 – 720 = 0
𝑥² – 60𝑥 + 12𝑥 – 720 = 0
𝑥(𝑥 – 60) + 12(𝑥 – 60) = 0
(𝑥 – 60) (𝑥 + 12) = 0
𝑥 + 12 = 0            𝑥 − 60 = 0
𝑥 = −12                𝑥 = +60
𝑥 can’t be negative. So, the value of 𝑥 is 40.
Hence, the number of articles will be 40.

Exercise 6E

Question 1. The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car.
Calculate:
(i) the time taken by the car to reach town B from A, in terms of x;
(ii) the time taken by the train to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.
(iv) Hence, find the speed of the train.
Solution:
It is given that,
Speed of car = 𝑥 km/h
Speed of train = (𝑥 + 16)km/h

8𝑥 + 3456 = 2(𝑥² + 16𝑥)
8𝑥 + 3456 = 2𝑥² + 32𝑥
2𝑥² + 32𝑥 − 8𝑥 − 3456 = 0
2𝑥² + 24𝑥 − 3456 = 0
2(𝑥² + 12𝑥 − 1728) = 0
𝑥² + 12𝑥 − 1728 = 0
𝑥² + 48𝑥 − 36𝑥 − 1728 = 0
𝑥(𝑥 + 48) − 36(𝑥 + 48) = 0
(𝑥 − 36)(𝑥 + 48) = 0
𝑥 − 36 = 0          𝑥 + 48 = 0
𝑥 = 36                 𝑥 = −48
𝑥 as speed can’t be negative, so x is 36km/hr.
(iv) Speed of train is = (𝑥 + 16)km/hr
Speed of train is = (36 + 16)km/hr
Speed of train is = 52km/hr

Question 2. A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x.
If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x.
Solution:
It is given that,
Number of articles = 𝑥
Total cost of article = Rs. 600

480 = 𝑥² − 4𝑥
𝑥² − 4𝑥 − 480 = 0
𝑥² − 24𝑥 + 20𝑥 − 480 = 0
𝑥(𝑥 − 24) + 20(𝑥 − 24) = 0
(𝑥 + 20)(𝑥 − 24) = 0
𝑥 − 24 = 0 𝑥 + 20 = 0
𝑥 = 24 𝑥 = −20
𝑥 can’t be negative, so 𝑥 is 24.

Question 3. A hotel bill for a number of people for overnight stay is Rs. 4800. If there were 4 people more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight.
Solution:
Let us assumed that,
The number of people staying overnight be 𝑥.
Total Hotel bill = Rs. 4800
Hotel bill each person = Rs. 4800/𝑥
According to question,

19200 = 200(𝑥² + 4𝑥)
19200 = 200𝑥² + 800𝑥
200𝑥² + 800𝑥 − 19200 = 0
𝑥² + 4𝑥 − 96 = 0
𝑥² + 12𝑥 − 8𝑥 − 96 = 0
𝑥(𝑥 + 12) − 8(𝑥 + 12) = 0
(𝑥 − 8)(𝑥 + 12) = 0
𝑥 − 8 = 0            𝑥 + 12 = 0
𝑥 = 8                  𝑥 = −12
𝑥 can’t be negative, so 𝑥 is 8.
Hence, the number of people staying overnight is 8.

Question 4. An Aero plane travelled a distance of 400 km at an average speed of 𝑥 km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for:
(i) the onward journey;
(ii) the return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Solution:
It is given that,
Distance = 400 km
Average speed of the aero plane = 𝑥km/hr
Speed while returning = (𝑥 + 40)km/hr

2(16000) = 𝑥² + 40𝑥
32000 = 𝑥² + 40𝑥
𝑥² + 40𝑥 − 32000 = 0
𝑥² + 200𝑥 − 160𝑥 − 32000 = 0
𝑥(𝑥 + 200) − 160(𝑥 + 200) = 0
(𝑥 − 160)(𝑥 + 200) = 0
𝑥 − 160 = 0           𝑥 + 200 = 0
𝑥 = 160                  𝑥 = −200
𝑥 as speed can’t be negative, so 𝑥 is 160.
Hence, the average speed of the aero plane is 160.

Question 5. Rs. 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs. 30 less. Find the original number of persons.
Solution:
Let us assumed that,
The original number of persons be 𝑥.
Total amount was divided = Rs. 6500

97500 = 30(𝑥² + 15𝑥)
3250 = 𝑥² + 15𝑥
𝑥² + 15𝑥 − 3250 = 0
𝑥² + 65𝑥 − 50𝑥 − 3250 = 0
𝑥(𝑥 + 65) − 50(𝑥 + 65) = 0
(𝑥 − 50)(𝑥 + 65) = 0
𝑥 − 50 = 0            𝑥 + 65 = 0
𝑥 = 50                   𝑥 = −65
𝑥 as number of persons can’t be negative, so 𝑥 is 50.
Hence, the original number of persons is 50.

Question 6. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500km away in time, it has to increase its speed by 250km/hr from its usual speed. Find its usual speed.
Solution:
It is given that,
The distance = 1500km
Let us assumed that,
The speed of plane be 𝑥km/hr
According to the question,

2 × 375000 = 𝑥² + 250𝑥
750000 = 𝑥² + 250𝑥
𝑥² + 250𝑥 − 750000 = 0
𝑥² + 1000𝑥 − 750𝑥 − 750000 = 0
𝑥(𝑥 + 1000) − 750(𝑥 + 1000) = 0
𝑥 − 750 = 0            𝑥 + 1000 = 0
𝑥 = 750                   𝑥 = −1000
𝑥 as speed can’t be negative, so 𝑥 is 750.
Hence, the speed of plane is 750.

Question 7. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours,
they are 50 km apart, find the speed of each train.
Solution:
Let us assumed that,
The speed of the second train be 𝑥km/hr
Speed of the first train is (𝑥 + 5)km/hr
Let O be the position of the railway station from which the two trains leave.
Distance travelled by the first train in 2 hours = OA = speed × Time = 2(𝑥 + 5)km
Distance travelled by the first train in 2 hours in OB = speed × Time = 2𝑥km
By Pythagoras Theorem, we have
(𝐴𝐵)² = (𝑂𝐴)² + (𝑂𝐵)²
(50)² = [2(𝑥 + 5)]² + (2𝑥)²
2500 = 4(𝑥 + 5)² + 4𝑥²
2500 = 4(𝑥² + 25 + 10𝑥) + 4𝑥²
8𝑥² + 40𝑥– 2400 = 0
𝑥² + 5𝑥– 300 = 0
𝑥² + 20𝑥– 15𝑥– 300 = 0
(𝑥 + 20)(𝑥 − 15) = 0
𝑥 + 20 = 0             𝑥 − 15 = 0
𝑥 = −20                  𝑥 = 15
𝑥 as a speed can’t be negative so 𝑥 is 15.
Hence, the speed of the first train is 20 km/hrs. and second train is 15 km/hrs.

Question 8. The sum S of first n even natural numbers is given by the relation 𝑆 = 𝑛(𝑛 + 1). Find n, if the sum is 420.
Solution:
𝑆 = 𝑛(𝑛 + 1)
It is given,
𝑆 = 420 𝑛(𝑛 + 1)
𝑆 = 420𝑛² + 𝑛– 420 = 0
𝑛² + 21𝑛– 20𝑛– 420 = 0
𝑛(𝑛 + 21)– 20(𝑛 + 21) = 0
(𝑛 + 21) (𝑛 – 20) = 0
𝑛 + 21 = 0            𝑛 − 20 = 0
𝑛 = −21               𝑛 = 20
𝑛 can’t be negative. Hence, the value of n is 20.

Question 9. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
Solution:
Let the present ages of father and his son be x years and (45 – x) years respectively.
Five years ago,
Father’s age = (𝑥 – 5)years
Son’s age = (45– 𝑥– 5)years
Son’s age = (40– 𝑥)years
According to question,
(𝑥– 5)(40– 𝑥) = 124
40𝑥 − 𝑥²– 200 + 5x = 124
𝑥²– 45𝑥 + 324 = 0
𝑥²– 36𝑥– 9𝑥 + 324 = 0
𝑥(𝑥– 36) – 9(𝑥– 36) = 0
(𝑥– 36) (𝑥– 9) = 0
𝑥 − 36 = 0            𝑥 − 9 = 0
𝑥 = 36                  𝑥 = 9
If,
𝑥 = 9
Father’s age = 9 years
Son’s age = (45– 𝑥)
Son’s age = (45 − 9)
Son’s age = 36 years
Father’s age can’t be less than son’s age.
If 𝑥 = 36
Father’s age = 36 years
Son’s age = (45 – 36) years
Son’s age = 9 years
Hence, father’s age is 36 years and son’s age is 9 years.

Question 10. In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find:
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Solution:
Let us assumed that,
The number of rows in the original arrangement is 𝑥.
The number of seats in each row in original arrangement = 𝑥
Total number of seats = 𝑥 × 𝑥 = 𝑥²
According to question,
2𝑥(𝑥– 10) = 𝑥² + 300
2𝑥²– 20𝑥 = 𝑥² + 300
𝑥²– 20𝑥– 300 = 0
𝑥²– 30𝑥 + 10𝑥– 300 = 0
𝑥(𝑥– 30) + 10(𝑥– 30) = 0
(𝑥 − 30)(𝑥 + 10) = 0
𝑥 − 30 = 0            𝑥 + 10 = 0
𝑥 = 30                  𝑥 = −10
𝑥 as rows of seats cannot be negative. So, the value of 𝑥 is 30.
(i) The number of rows in the original arrangement = x = 30
(ii) The number of seats after re-arrangement = 𝑥² + 300
The number of seats after re-arrangement = 900 + 300 = 1200
The number of seats after re-arrangement = 1200.

Question 11. Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, in how many days will Mohan alone complete the work?
Solution:
Let us assumed that,
The number of days in which Mohan completes the work be x.
Number of days in which Manoj completes the work = x + 16
In one day, Mohan completes 1/𝑥 part of work.
In one day, Manoj completes 1/𝑥 + 16 part of work.
It is given that they both can do the work in 15 days.

= 30𝑥 + 240 = 𝑥² + 16𝑥
= 𝑥²– 14𝑥– 240 = 0
= 𝑥²– 24𝑥 + 10𝑥– 240 = 0
= 𝑥(𝑥 − 24) + 10(𝑥 − 24) = 0
= (𝑥 − 24)(𝑥 + 10) = 0
𝑥 − 24 = 0          𝑥 + 10 = 0
𝑥 = 24                 𝑥 = −10
𝑥 as the number of days can’t be negative, So the value of 𝑥 is 24.
Hence, the work is completed by Mohan in 24 days.

Question 12. Two years ago, a man’s age was three times the square of his son’s age. In three years’ time, his age will be four times his son’s age. Find their present ages.
Solution:
Let us assumed that,
The age of son 2years ago be 𝑥 years
Then, father’s age 2 years ago = 3𝑥² years
Present age of son = (𝑥 + 2) years
Present age of father = (3𝑥² + 2) years
3 years hence,
Son’s age = (𝑥 + 2 + 3)years
Son’s age = (𝑥 + 5)years
Father’s age = (3𝑥²+ 2 + 3)years
Father’s age = (3𝑥² + 5)years
According to question,
3𝑥² + 5 = 4(𝑥 + 5)
3𝑥² – 4𝑥– 15 = 0
3𝑥²– 9𝑥 + 5𝑥– 15 = 0
3𝑥(𝑥– 3) + 5(𝑥– 3) = 0
(𝑥– 3)(3𝑥 + 5) = 0
𝑥 − 3 = 0            3𝑥 + 5 = 0
𝑥 = 3                    𝑥 = − 5/3
𝑥 as age can’t be negative, so the value of 𝑥 is 3.
Present age of son = (𝑥 + 2)years
Present age of son = (3 + 2)years
Present age of son = 5 years
Present age of father = (3𝑥² + 2)years
Present age of father = (3(32) + 2)years
Present age of father = (3 × 9 + 2)years
Present age of father = 29 years

Question 13. In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by. Find the fraction.
Solution:
Let us assumed that,
The numerator be 𝑥
And the denominator be 𝑥 + 3
If 1 is subtracted from both numerator and denominator = 𝑥−1/𝑥+2
According to the question,

(13𝑥 − 3)(𝑥 + 2) = (𝑥 − 1)(14𝑥 + 42)
13𝑥² + 26𝑥 − 3𝑥 − 6 = 14𝑥² + 42𝑥 − 14𝑥 − 42
13𝑥² + 23𝑥 − 6 = 14𝑥² + 28𝑥 − 42
14𝑥² − 13𝑥² + 28𝑥 − 23𝑥 − 42 + 6 = 0
𝑥² + 5𝑥 − 36 = 0
𝑥² + 9𝑥 − 4𝑥 − 36 = 0
𝑥(𝑥 + 9) − 4(𝑥 + 9) = 0
(𝑥 − 4)(𝑥 + 9) = 0
𝑥 − 4 = 0            𝑥 + 9 = 0
𝑥 = 4                  𝑥 = −9
𝑥 can’t be negative, so 𝑥 is 4
Hence, the required fraction be 𝑥/𝑥+3 = 4/7.

Question 14. In a two digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
It is given that,
The difference between two digits is 6
The ten’s digit is bigger than the unit’s digit.
So, let the unit’s digit be x and ten’s digit be (𝑥 + 6).
According to question,
𝑥(𝑥 + 6) = 27
𝑥² + 6𝑥 – 27 = 0
𝑥² + 9𝑥 – 3𝑥– 27 = 0
𝑥(𝑥 + 9) – 3(𝑥 + 9) = 0
(𝑥 + 9)(𝑥– 3) = 0
𝑥 + 9 = 0            𝑥 − 3 = 0
𝑥 = −9                𝑥 = 3
𝑥, the digits of a number cannot be negative, so, 𝑥 is 3.
Unit’s digit = 3
Ten’s digit = 9
Hence, the number is 93.

Question 15. Some school children went on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/hr and it took two hours longer for returning. Find the time taken to return.
Solution:
Let us assume that,
The original speed of the bus is 𝑥 km/hr
Speed of the bus when returning = (𝑥 − 5)km/hr
Distance = 300km
According to the question,

1500 = 2(𝑥² − 5𝑥)
1500 = 2𝑥² − 10𝑥
2𝑥² − 10𝑥 − 1500 = 0
2(𝑥² − 5𝑥 − 750) = 0
𝑥² − 5𝑥 − 750 = 0
𝑥² − 30𝑥 + 25𝑥 − 750 = 0
𝑥(𝑥 − 30) + 25(𝑥 − 30) = 0
(𝑥 + 25)(𝑥 − 30) = 0
𝑥 + 25 = 0            𝑥 − 30 = 0
𝑥 = −25                𝑥 = 30
𝑥 as speed can’t be negative so, 𝑥 is 30.
Speed of the bus when returning = (30 − 5)km/hr
Speed of the bus when returning = 25km/hr
Time = 300/25 = 12hrs
Hence, time taken by the bus to return is 12hrs.

Question 16. Rs. 480 is divided equally among ‘x’ children. If the number of children were 20 more, then each would have got Rs.12 less. Find ‘x’.
Solution:
It is given that,
Total Amount be Rs. 480.
Let us assumed that,
The number of children be 𝑥.
The amount each children got = 480/𝑥
If the number of children were 20 more,
Amount each children got = 480/𝑥+20
According to question,

9600 = 12(𝑥² + 20𝑥)
12𝑥² + 240𝑥 − 9600 = 0
12(𝑥² + 20𝑥 − 800) = 0
𝑥² + 20𝑥 − 800 = 0
𝑥² + 40𝑥 − 20𝑥 − 800 = 0
𝑥(𝑥 + 40) − 20(𝑥 + 40) = 0
(𝑥 − 20)(𝑥 + 40) = 0
𝑥 − 20 = 0            𝑥 + 40 = 0
𝑥 = 20                  𝑥 = −40
𝑥 as number of children’s can’t be negative, so the value of 𝑥 is 20.
Hence, the number of children’s is 20.

Question 17. A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance.
Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’.
Solution:
It is given that,
Total distance covered by bus is 240.
Let us assumed that,
The speed of bus be 𝑥.
Time taken by bus = 240/𝑥 hrs
Time taken by bus to cover total distance with speed (𝑥 − 10)km/hrs = 240/𝑥−10 hrs
According to the question,

2400 = 2(𝑥² − 10𝑥)
2400 = 2𝑥² − 20𝑥
2𝑥² − 20𝑥 − 2400 = 0
2(𝑥² − 10𝑥 − 1200) = 0
𝑥² − 10𝑥 − 1200 = 0
𝑥² − 40𝑥 + 30𝑥 − 1200 = 0
𝑥(𝑥 − 40) + 30(𝑥 − 40) = 0
(𝑥 + 30)(𝑥 − 40) = 0
𝑥 + 30 = 0            𝑥 − 40 = 0
𝑥 = −30                𝑥 = 40
𝑥 as the speed can’t be negative, so 𝑥 is 40.

Question 18. The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.
Solution:
It is given that,
The sum of the ages of Vivek and his younger brother Amit is 47 years.
Let the age of Vivek = 𝑥
The age of Amit = 47– 𝑥
The product of their ages in years is 550
According to question,
𝑥(47– 𝑥) = 550
47𝑥– 𝑥² = 550
𝑥²– 47𝑥 + 550 = 0
𝑥²– 25𝑥– 22𝑥 + 550 = 0
𝑥(𝑥– 25)– 22(𝑥– 25) = 0
(𝑥– 25)(𝑥– 22) = 0
𝑥 − 25 = 0            𝑥 − 22 = 0
𝑥 = 25                  𝑥 = 22
Hence, the age of Vivek is 25 and age of Amit is 22 years.