**Question 1. The product of two consecutive integers is 56. Find the integers.Solution:**

Let the first number be π₯.

And second number be π₯ + 1.

π₯(π₯ + 1) = 56

π₯

^{2}+ π₯β 56 = 0

By the splitting the middle term,

π₯

^{2}+ 8π₯ β 7π₯β 56 = 0

π₯(π₯ + 8) β 7(π₯ + 8) = 0

(π₯ + 8) (π₯β 7) = 0

π₯ + 8 = 0 π₯ β 8 = 0

π₯ = β8 π₯ = 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 π₯.

And second number be π₯ + 1.

According to the question

π₯

^{2}+ (π₯ + 1)

^{2}= 41

π₯

^{2}+ π₯

^{2}+ 2π₯ + 1 = 41

2π₯

^{2}+ 2π₯ + 1β 41 = 0

2π₯

^{2}+ 2π₯ β 40 = 0

Take common 2 in given equation

2(π₯

^{2}+ π₯β 20) = 0

π₯

^{2}+ π₯β 20 = 0

π₯

^{2}+ 5π₯ β 4π₯β 20 = 0

π₯(π₯ + 5) β 4(π₯ + 20) = 0

(π₯ + 5) (π₯β 4) = 0

π₯ + 5 = 0 π₯ β 4 = 0

π₯ = β5 π₯ = 4

Here, π₯ 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 π₯

And second number be π₯ + 5

According to the question,

π₯

^{2}+ (π₯ + 5)

^{2}= 97

π₯

^{2}+ π₯

^{2}+ 2(π₯)(5) + (5)

^{2}= 97

2π₯

^{2}+ 10π₯ + 25 β 97 = 0

2π₯

^{2}+ 10π₯ β 72 = 0

2(π₯

^{2}+ 5π₯ β 36) = 0

π₯

^{2}+ 5π₯ β 36 = 0

By splitting middle term

π₯

^{2}+ 9π₯ β 4π₯ β 36 = 0

(π₯ + 9) (π₯ β 4) = 0

π₯ + 9 = 0 π₯ β 4 = 0

π₯ = β9 π₯ = 4

Here, π₯ 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 π₯

And reciprocal of that number is 1/π₯

According to question,

π₯ + 1/π₯ = 4.25

π₯

^{2}+1/π₯ = 4.25

π₯

^{2}+ 1 = 4.25π₯

π₯

^{2}+ 1 = (425/100) π₯

100(π₯

^{2}+ 1) = 425π₯

100π₯

^{2}+ 100 = 425π₯

100π₯

^{2}β 425π₯ + 100 = 0

25(4π₯

^{2}β 17π₯ + 4) = 0

4π₯

^{2}β 17π₯ + 4 = 0

4π₯

^{2}β 16π₯ β π₯ + 4 = 0

4π₯(π₯ β 4) β 1(π₯ β 4) = 0

(4π₯ β 1)(π₯ β 4) = 0

4π₯ β 1 = 0 π₯ β 4 = 0

π₯ = 1/4 π₯ = 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 π₯

And second number be π₯ + 3

According to question

10(2π₯ + 3) = 7(π₯^{2} + 3π₯)

10(2π₯ + 3) = 7(π₯^{2} + 3π₯)

20π₯ + 30 = 7π₯^{2} + 21π₯

7π₯^{2} + 21π₯ β 20π₯ β 30 = 0

7π₯^{2} + π₯ β 30 = 0

By splitting the middle term

7π₯^{2}+ π₯ β 30 = 0

7π₯^{2} + 15π₯ β 14π₯ β 30 = 0

π₯(7π₯ + 15) β 2(7π₯ β 15) = 0

(π₯ β 2)(7π₯ β 15) = 0

π₯ β 2 = 0 7π₯ β 15 = 0

π₯ = 2 π₯ = 15/7

Here, π₯ 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 π₯

And other number be 15 β π₯

According to question

150 = 3(15π₯ β π₯^{2})

150 = 45π₯ β 3π₯^{2}

3π₯^{2} β 45π₯ + 150 = 0

3(π₯^{2} β 15π₯ + 50) = 0

π₯^{2} β 15π₯ + 50 = 0

π₯^{2} β 10π₯ β 5π₯ + 50 = 0

π₯(π₯ β 10) β 5(π₯ β 10) = 0

(π₯ β 5)(π₯ β 10) = 0

π₯ β 5 = 0 π₯ β 10 = 0

π₯ = 5 π₯ = 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 timesthe smaller number, find the numbers.Solution:**

Let us assumed that,

First number be π₯

And second number be π¦

According to question,

π₯

^{2}+ π¦

^{2}= 208

π¦

^{2}= 18π₯

Put the value of π¦

^{2}in above equation,

π₯

^{2}+ 18π₯ = 208

π₯

^{2}+ 18π₯ β 208 = 0

π₯

^{2}+ 26π₯ β 8π₯ β 208 = 0

π₯(π₯ + 26) β 8(π₯ + 26) = 0

(π₯β 8)(π₯ + 26) = 0

π₯ β 8 = 0 π₯ + 26 = 0

π₯ = 8 π₯ = β26

The value of π₯ is 8 as π₯ canβt be negative.

Put the value of π₯.

π¦

^{2}= 18 Γ 8

π¦

^{2}= 144

π¦ = β144

π¦ = 12

Hence, the value of π₯ is 8 and π¦ 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 π₯

And second number be π₯ + 2

According to the question,

π₯

^{2}+ (π₯ + 2)

^{2}= 52

π₯

^{2}+ π₯

^{2}+ 2(π₯)(2) + (2)2 = 52

π₯

^{2}+ π₯

^{2}+ 4π₯ + 4 = 52

2π₯

^{2}+ 4π₯ + 4 = 52

2π₯

^{2}+ 4π₯ + 4 β 52 = 0

2π₯

^{2}+ 4π₯ β 48 = 0

2(π₯

^{2}+ 2π₯ β 24) = 0

π₯

^{2}+ 2π₯ β 24 = 0

By splitting the middle term

π₯

^{2}+ 6π₯ β 4π₯ β 24 = 0

π₯(π₯ + 6) β 4(π₯ + 6) = 0

(π₯ β 4)(π₯ + 6) = 0

π₯ β 4 = 0 π₯ + 6 = 0

π₯ = 4 π₯ = β6

π₯ canβt be negative so the value of π₯ 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 π₯

And second number be π₯ + 2

According to the question,

π₯

^{2}+ (π₯ + 2)

^{2}= 74

π₯

^{2}+ (π₯ + 2)

^{2}= 74

π₯

^{2}+ π₯

^{2}+ 2(π₯)(2) + (2)2 = 74

π₯

^{2}+ π₯

^{2}+ 4π₯ + 4 = 74

2π₯

^{2}+ 4π₯ + 4 = 74

2π₯

^{2}+ 4π₯ + 4 β 74 = 0

2π₯

^{2}+ 4π₯ β 70 = 0

2(π₯

^{2}+ 2π₯ β 35) = 0

π₯

^{2}+ 2π₯ β 35 = 0

By splitting the middle term

π₯

^{2}+ 7π₯ β 5π₯ β 35 = 0

π₯(π₯ + 7) β 5(π₯ + 7) = 0

(π₯ β 5)(π₯ + 7) = 0

π₯ β 5 = 0 π₯ + 7 = 0

π₯ = 5 π₯ = β7

π₯ canβt be negative so the value of π₯ 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 π₯

And denominator will be 2π₯ + 1

The fraction is π₯/2π₯+1

According to the question,

10(5π₯^{2} + 4π₯ + 1) = 29(2π₯^{2} + π₯)

50π₯^{2} + 40π₯ + 10 = 58π₯^{2} + 29π₯

50π₯^{2} β 58π₯^{2} + 40π₯ β 29π₯ + 10 = 0

β8π₯^{2} + 11π₯ + 10 = 0

β(8π₯^{2} β 11π₯ β 10) = 0

8π₯^{2} β 11π₯ β 10 = 0

By the quadratic formula

π = 8

π = β11

π = β10

Here,

Numerator will be 2 and

Denominator = 2π₯ + 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 π₯

Second number be π¦

According to question,

π₯ + π¦ = 20

π¦ = 20β π₯ ** _**(i)

3π₯

^{2}= (20 β π₯) + 10

3π₯

^{2}= 20 β π₯ + 10

3π₯

^{2}= 30 β π₯

3π₯

^{2}+ π₯ β 30 = 0

3π₯

^{2}β 9π₯ + 10π₯ β 30 = 0

3π₯(π₯ β 3) + 10(π₯ β 3) = 0

(3π₯ + 10)(π₯ β 3) = 0

3π₯ + 10 = 0 π₯ β 3 = 0

π₯ = β10/3 π₯ = 3

π₯ canβt be negative so the value of π₯ is 3.

Put the value of π₯ in equation (i)

π¦ = 20β π₯

π¦ = 20β 3

π¦ = 17

Hence, the value of π₯ is 3 and π¦ 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 π₯β 1

Second number be π₯

Third number be π₯ + 1

According to question,

π₯^{2} = (π₯ + 1)2 β (π₯ β 1)^{2} + 60

π₯^{2} = π₯^{2} + 2(π₯)(1)+(1)^{2} β [π₯^{2} β 2(π₯)(1)+(1)^{2}] + 60

π₯^{2} = π₯^{2} + 2π₯ + 1 β π₯^{2} + 2π₯ β 1 + 60

π₯^{2} = 4π₯ + 60

π₯^{2} β 4π₯ β 60 = 0

π₯^{2} β 10π₯ + 6π₯ β 60 = 0

(π₯β 10)(π₯ + 6) = 0

π₯β 10 = 0 π₯ + 6 = 0

π₯ = 10 π₯ = β6

π₯ canβt be negative so the value of π₯ is 10.

First number = π₯β 1

First number = 10β 1

First number = 9

Second number = π₯

Second number = 10

Third number = π₯ + 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 π₯β 1

Second number be π₯

Third number be π₯ + 1

According to question,

3(π₯ + 1)^{2} = (π₯ β 1)^{2} + π₯^{2} + 67

3[(π₯)^{2} + 2(π₯)(1) + (1)^{2}] = (π₯)^{2} β 2(π₯)(1) + (1)^{2} + π₯^{2} + 67

3(π₯^{2} + 2π₯ + 1) = (π₯)^{2} β 2π₯ + 1 + π₯^{2} + 67

3π₯^{2} + 6π₯ + 3 = π₯^{2} β 2π₯ + 1 + π₯^{2} + 67

3π₯^{2} + 6π₯ + 3 = 2π₯^{2} β 2π₯ + 68

3π₯^{2} β 2π₯^{2} + 6π₯ + 2π₯ + 3 β 68 = 0

π₯^{2} + 8π₯ β 65 = 0

π₯^{2} + 13π₯ β 5π₯ β 65 = 0

π₯(π₯ + 13) β 5(π₯ + 13) = 0

(π₯ β 5)(π₯ + 13) = 0

π₯β 5 = 0 π₯ + 13 = 0

π₯ = 5 π₯ = β13

π₯ canβt be negative so the value of π₯ is 5.

First number = π₯β 1

First number = 5β 1

First number = 4

Second number = π₯

Second number = 5

Third number = π₯ + 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 βxβ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate βxβ.

Solution:

It is given that,

Work done by A = 1/π₯

Work done by B = 1/π₯+16

According to the question

A and B together can do the work in 15 days.

15(2π₯ + 16) = π₯^{2} + 16π₯

30π₯ + 240 = π₯^{2} + 16π₯

π₯^{2} + 16π₯ β 30π₯ β 240 = 0

π₯^{2} + 14π₯ β 240 = 0

π₯^{2} + 24π₯ β 10π₯ β 240 = 0

π₯(π₯ + 24) β 10(π₯ + 24) = 0

(π₯ β 10)(π₯ + 24) = 0

π₯β 10 = 0 π₯ + 24 = 0

π₯ = 10 π₯ = β24

π₯ canβt be negative so the value of π₯ is 10.

Hence, the value of π₯ 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/π₯

B pipe can fill a cistern in = 1/π₯β3

According to the question

A and B together can fill it in 6 hours and 40 minutes.

By cross multiply,

20(2π₯ β 3) = 3(π₯^{2} β 3π₯)

40π₯ β 60 = 3π₯^{2} β 9π₯

3π₯^{2} β 9π₯ β 40π₯ + 60 = 0

3π₯^{2} β 49π₯ + 60 = 0

3π₯^{2} β 45π₯ β 4π₯ + 60 = 0

(3π₯ β 4)(π₯ β 15) = 0

3π₯β 4 = 0 π₯ β 15 = 0

π₯ = 4/3 π₯ = 15

π₯ canβt be in fractions so the value of π₯ is 15.

Hence, one pipe fill cistern in 15 hours and other fills in (15 β 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 π₯ as the smaller part of the two parts, find the number.

Solution:

Let us assumed that,

Smaller part be π₯

(Larger part)^{2} = 8π₯

Larger part = β8π₯

According to the question,

π₯^{2} + (β8π₯)^{2} = 20

π₯^{2} + 8π₯ = 20

π₯^{2} + 8π₯ β 20 = 0

π₯^{2} β 2π₯ + 10π₯ β 20 = 0

π₯(π₯ β 2) + 10(π₯ β 2) = 0

(π₯ + 10)(π₯ β 2) = 0

π₯ + 10 = 0 π₯ β 2 = 0

π₯ = β10 π₯ = 2

π₯ canβt be in fractions so the value of π₯ is 2.

Smaller part = 2

Larger part = β8 Γ 2 = β16 = 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 4π₯ cm and (2π₯β 1)cm. If the area of the triangle is 30ππΒ²; Calculate the lengths of its sides.

Solution:

It is given that,

Height of a triangle is 4π₯cm

Base of a triangle be (2π₯β 1)cm

Area of triangle = 30cm^{2}

We know that,

Area of triangle = 1/2 Γ βπππβπ‘ Γ πππ π

30 = 1/2 Γ 4π₯ Γ (2π₯ β 1)

30 Γ 2 = 4π₯ Γ (2π₯ β 1)

60 = 8π₯^{2} β 4π₯

8π₯^{2} β 4π₯ β 60 = 0

4(2π₯^{2} β π₯ β 15) = 0

2π₯^{2} β π₯ β 15 = 0

2π₯^{2} β 6π₯ + 5π₯ β 15 = 0

2π₯(π₯ β 3) + 5(π₯ β 3) = 0

(2π₯ + 5)(π₯ β 3) = 0

2π₯ + 5 = 0 π₯ β 3 = 0

π₯ = β5/2 π₯ = 3

π₯ canβt be negative so the value of π₯ is 3.

Height of a triangle = 4 Γ 3 = 12cm

Base side = (2 Γ 3)β 1 = 6 β 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 π₯

Other two sides be (34β π₯) cm.

By Pythagoras theorem,

(Hypotenuse)^{2} = (π»πππβπ‘)^{2} + (π΅ππ π)^{2}

(26)^{2} = π₯^{2 }+ (34β π₯)^{2}

676 = π₯^{2} + π₯^{2} + 1156β 68π₯

2π₯^{2} β 68π₯ + 480 = 0

π₯^{2} β 34π₯ + 240 = 0

π₯^{2} β 10π₯ β 24π₯ + 240 = 0

π₯(π₯β 10) β 24(π₯β 10) = 0

(π₯β 10)(π₯β 24) = 0

π₯ β 10 = 0 π₯ β 24 = 0

π₯ = 10 π₯ = 24

If,

π₯ = 10

(34β π₯) = 24

If

π₯ = 24

(34β π₯) = 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 (π₯β 1)cm, 3π₯cm and (3π₯ + 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 = (3π₯ + 1)cm

Lengths of a sides is (π₯β 1)cm

Length of other side 3π₯ cm

By Pythagoras theorem,

(Hypotenuse)^{2} = (π»πππβπ‘)^{2} + (π΅ππ π)^{2}

(3π₯ + 1)^{2} = (π₯β 1)^{2} + (3π₯)^{2}

(3π₯)^{2} + 2(3π₯)(1) + (1)^{2} = (π₯)^{2} β 2(π₯)(1) + (1)^{2} + (3π₯)^{2}

9π₯^{2} + 6π₯ + 1 = π₯^{2} β 2π₯ + 1 + 9π₯^{2}

6π₯ = π₯^{2} β 2π₯

π₯^{2}β 8π₯ = 0

π₯(π₯ β 8) = 0

π₯ = 0 π₯ β 8 = 0

π₯ = 8

π₯ canβt be 0 so the value of π₯ is 8.

Hypotenuse = (3(8) + 1)cm

Hypotenuse = 25cm

Lengths of a sides is (8β 1)cm

Lengths of a sides is 7cm

Length of other side 3 Γ 8 cm

Length of other side 24cm

Area of the triangle = 1/2 Γ 7cm Γ 24cm

Area of the triangle = 84cmΒ²

Hence, the area of triangle is 84cmΒ².

**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 π₯

Length of one side = (π₯β 1)cm

Length of other side = (π₯β 18)cm

By Using Pythagoras theorem,

(Hypotenuse)^{2} = (π»πππβπ‘)^{2} + (π΅ππ π)^{2}

π₯^{2} = (π₯β 1)^{2} + (π₯β 18)^{2}

π₯^{2} = π₯^{2}β 2 Γ π₯ Γ 1 + (1)^{2} + π₯^{2}β 2 Γ π₯ Γ 18 + (18)^{2} = 0

π₯^{2} = π₯^{2}β 2π₯ + 1 + π₯^{2}β 36π₯ + 324 = 0

π₯^{2} β 38π₯ + 325 = 0

π₯^{2}β 13π₯β 25π₯ + 325 = 0

π₯(π₯β 13)β 25(π₯β 13) = 0

(π₯β 13) (π₯β 25) = 0

π₯β 13 = 0 π₯ β 25 = 0

π₯ = 13 π₯ = 25

If,

π₯ = 13

Hypotenuse =13cm

Length of one side = (13β 1)cm = 12cm

Length of other side = (13β 18)cm = -5cm

Length canβt be negative.

If

π₯ = 25

Hypotenuse =25cm

Length of one side = (25β 1)cm = 24cm

Length of other side = (25β 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 π₯ m

According to question,

Length of the other side = (π₯ + 30)m

Length of hypotenuse = (π₯ + 60) m

Using Pythagoras theorem,

(Hypotenuse)^{2} = (π»πππβπ‘)^{2} + (π΅ππ π)^{2}

(π₯ + 60)^{2} = π₯^{2} + (π₯ + 30)^{2}

π₯^{2} + 2 Γ π₯ Γ 60 + (60)^{2} = π₯^{2} + π₯^{2} + 2 Γ π₯ Γ 30 + (30)^{2} = 0

π₯^{2} + 120π₯ + 3600 = 2π₯^{2} + 60π₯ + 900 = 0

π₯^{2} + 120π₯ + 3600 β 2π₯^{2} β 60π₯ β 900 = 0

π₯^{2} + 120π₯ + 3600 β 2π₯^{2} β 60π₯ β 900 = 0

βπ₯^{2} + 60π₯ + 2700 = 0

π₯^{2} β 60π₯ β 2700 = 0

π₯^{2} β 90π₯ + 30π₯ β 2700 = 0

π₯(π₯ β 90)30(π₯ β 90) = 0

(π₯ + 30)(π₯ β 90) = 0

π₯ + 30 = 0 π₯ β 90 = 0

π₯ = β30 π₯ = 90

π₯ canβt be negative so the value of π₯ is 90.

Length of shorter side = 90m.

Length of the other side = (π₯ + 30)m = (90 + 30)m = 120m

Length of hypotenuse = (π₯ + 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 mΒ². Find its length and breadth.

Solution:

It is given that,

Perimeter of rectangle = 104m

Area of rectangle = 640m^{2}

Let us assumed that,

Length of rectangle be π₯

Breadth of rectangle be π¦

Perimeter = 2(x + y)m

104 = 2(π₯ + π¦)

π₯ + π¦ = 52

π¦ = 52 β π₯

Area = L Γ B m^{2}

π₯π¦ = 640

π₯(52β π₯) = 640

π₯^{2}β 52π₯ + 640 = 0

π₯^{2}β 32π₯β 20π₯ + 640 = 0

π₯(π₯β 32)β 20(π₯β 32) = 0

(π₯β 32)(π₯β 20) = 0

π₯ β 32 = 0 π₯ β 20 = 0

π₯ = 32 π₯ = 20

If,

π₯ = 32

π¦ = 52β 32 = 20

If,

π₯ = 20

π¦ = 52β 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 mΒ², 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,

π₯ be the width of the footpath

Length of inner rectangular field is (32 β 2π₯)π

Breadth of outer rectangular field is (24 β 2π₯)π

We know that,

Area of the path = Area of outer rectangle β Area of inner rectangle

208 = (32)(24)β (32β 2π₯)(24β 2π₯)

208 = 768β [768 β 64π₯ β 48π₯ + 4π₯^{2}]

208 = 768β 768 + 64π₯ + 48π₯ β 4π₯^{2}

208 = 112π₯ β 4π₯^{2}

4π₯^{2} β 112π₯ + 208 = 0

4(π₯^{2} β 28π₯ + 52) = 0

π₯^{2} β 28π₯ + 52 = 0

π₯^{2} β 26π₯ β 2π₯ + 52 = 0

π₯(π₯ β 26) β 2(π₯ β 26) = 0

(π₯ β 2)(π₯ β 26) = 0

π₯ β 2 = 0 π₯ β 26 = 0

π₯ = 2 π₯ = 26

If π₯ = 2

Length of inner rectangular field is (32 β 2π₯)π

= (32 β 2(2))π

= (32 β 4)π

= 28π

Breadth of outer rectangular field is (24 β 2π₯)π

= (24 β 2(2))π

= (24 β 4)π

= 20π

If π₯ = 26

Length of inner rectangular field is (32 β 2π₯)π

= (32 β 2(26))π

= (32 β 52)π

= β20π

It canβt be possible.

Hence, the width of the footpath is 2π.

**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 = π₯ ππ

Side of second square = (π₯ + 4)ππ.

Sum of their area = 656ππ^{2}

π₯^{2} + (π₯ + 4)^{2} = 656

π₯^{2} + π₯^{2} + 2 Γ 4 Γ π₯ + 16 = 656

2π₯^{2} + 8π₯ + 16 = 656

2π₯^{2} + 8π₯ + 16β 656 = 0

2π₯^{2} + 8π₯β 640 = 0

2(π₯^{2} + 4π₯β 320) = 0

π₯^{2} + 4π₯β 320 = 0

π₯^{2} + 20π₯β 16π₯β 320 = 0

π₯(π₯ + 20) β 16(π₯ + 20) = 0

(π₯ + 20) (π₯ β 16) = 0

π₯ + 20 = 0 π₯ β 16 = 0

π₯ = β20 π₯ = 16

π₯ canβt be negative so the value of π₯ 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 π€m

Length of the flower bed be π₯m

Breadth of the flower bed be π¦m

Therefore, we have:

x + 2π€ = 50

x = 50 β 2w__________(1)

y + 2w = 40

y = 40 β 2π€__________(2)

Area of rectangle = Length Γ Breadth

Area of rectangular field = 50m Γ 40m

Area of rectangular field = 2000m^{2}

Area of the flower bed = xym^{2}

Area of gravel path = Area of rectangular field β Area of flower bed

Area of gravel path = (2000β π₯π¦)π^{2}

Cost of laying flower bed + Gravel path = Area x cost of laying per sq. m

52000 = 30π₯π¦ + 20(2000β π₯π¦)

52000 = 30π₯π¦ + 40000 β 20π₯π¦

52000 = 10π₯π¦ + 40000

β10π₯π¦ = 40000 β 52000

β10π₯π¦ = β12000

π₯π¦ = 1200

Using (1) and (2), we have:

(50β 2π€)(40β 2π€) = 1200

2000β 180π€ + 4π€^{2} = 1200

2000β 180π€ + 4π€^{2} = 1200 β 2000

β 180π€ + 4π€^{2} = β800

4π€^{2}β 180π€ + 800 = 0

4(π€^{2}β 45π€ + 200) = 0

π€^{2}β 45π€ + 200 = 0

π€^{2}β 5π€β 40π€ + 200 = 0

π€(π€β 5)β 40(π€β 5) = 0

(π€β 5)(π€β 40) = 0

π€ β 5 = 0 π€ β 40 = 0

π€ = 5 π€ = 40

If, π€ = 40

π₯ = 50β 2π€

π₯ = β30

It canβt 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 π₯cm.

Size of smaller tiles = (π₯β 2)cm

Area of larger tiles = π₯ππ^{2}

Area of smaller tiles = (π₯β 2)^{2}ππ^{2}

It is given that,

Number of larger tiles required = 128.

Area needed to be paved = 128π₯^{2}ππ^{2}__________(1)

Number of larger tiles required to pave an area is 200.

Area needed to be paved = 200(π₯β 2)2ππ^{2}_________(2)

From equation (1) and (2),

We have,

128π₯^{2} = 200(π₯β 2)^{2}

128π₯^{2}

= 200π₯^{2}

+ 800 β 800π₯

72π₯^{2}β 800π₯ + 800 = 0

9π₯^{2}β 100π₯ + 100 = 0

9π₯^{2}β 90π₯β 10π₯ + 100 = 0

9π₯(π₯β 10) β 10(π₯β 10) = 0

(π₯β 10)(9π₯β 10) = 0

π₯β 10 = 0 9π₯β 10 = 0

π₯ = 10 π₯ = 10/9

If,

π₯ = 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,

π₯ + π¦ + π₯ = 70

2π₯ + π¦ = 70 ** __**(1)

Get the value of π¦ from equation (1)

π¦ = 70 β 2π₯

It is given that,

The area of rectangle is 600.

Area = Length Γ Breadth

π₯(70β 2π₯) = 600

70π₯β 2π₯

^{2}= 600

2π₯

^{2}β 70π₯ + 600 = 0

2(π₯

^{2}β 35π₯ + 300) = 0

π₯

^{2}β 35π₯ + 300 = 0

π₯

^{2}β 15π₯β 20π₯ + 300 = 0

π₯(π₯β 15)β 20(π₯β 15) = 0

(π₯β 15)(π₯β 20) = 0

π₯β 15 = 0 π₯β 20 = 0

π₯ = 15 π₯ = 20

If,

π₯ = 15

π¦ = 70 β 2π₯

π¦ = 70 β 30

π¦ = 40

If,

π₯ = 20

π¦ = 70 β 2π₯

π¦ = 70 β 40

π¦ = 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 π₯π

Area of the square lawn = π₯^{2}π^{2}

The square lawn is bounded on three sides by a path which is 4 m wide.

Area of outer rectangle = (π₯ + 4)(π₯ + 8)

Area of outer rectangle = π₯^{2} + 8π₯ + 4π₯ + 32

Area of outer rectangle = π₯^{2} + 12π₯ + 32

Area of path = π₯^{2 }+ 12π₯ + 32β (π₯^{2})

Area of path = 12π₯ + 32

According to the question,

12π₯ + 32 = 7/8 Γ π₯^{2}

8(12π₯ + 32) = 7π₯^{2}

96π₯ + 256 = 7π₯^{2}

7π₯^{2} β 96π₯ β 256 = 0

7π₯^{2} β 112π₯ + 16π₯ β 256 = 0

7π₯(π₯ β 16) + 16(π₯ β 16) = 0

(7π₯ + 16)(π₯ β 16) = 0

7π₯ + 16 = 0 π₯β 16 = 0

π₯ = β16/7 π₯ = 16

π₯ canβt be negative so the value of π₯ is 16.

Hence, sides of the square lawn are 16m.

**Question **13. The area of a big rectangular room is 300 mΒ². 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 π₯m

Breadth of the rectangular room be π¦m

Area of the rectangular room = 300π^{2}

π₯π¦ = 300

π¦ = 300/π₯ ** __**(1)

New length = (π₯β 5)π

New breadth = (π¦ + 5)π

Area of new rectangular = (π₯β 5)(π¦ + 5)

(π₯β 5)(π¦ + 5) = 300

5π₯^{2} β 1500 β 25π₯ = 0

5(π₯^{2} β 5π₯ β 300) = 0

π₯^{2} β 5π₯ β 300 = 0

π₯^{2} β 20π₯ + 15π₯ β 300 = 0

π₯(π₯ β 20) + 15(π₯ β 20) = 0

(π₯ + 15)(π₯ β 20) = 0

π₯ + 15 = 0 π₯ β 20 = 0

π₯ = β15 π₯ = 20

π₯ canβt be negative so the value of π₯ 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 = π₯km/hr

Speed of express train = (π₯ + 25)km/hr

Distance = 300km

7500 = 2(π₯^{2} + 25π₯)

7500 = 2π₯^{2} + 50π₯

2π₯^{2} + 50π₯ β 7500 = 0

2(π₯^{2} + 25π₯ β 3750) = 0

π₯^{2} + 25π₯ β 3750 = 0

π₯^{2} + 75π₯ β 50π₯ β 3750 = 0

π₯(π₯ + 75) β 50(π₯ + 75) = 0

(π₯ β 50)(π₯ + 75) = 0

π₯ β 50 = 0 π₯ + 75 = 0

π₯ = 50 π₯ = β75

π₯ canβt 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(π₯^{2} + 10π₯)

21600 = 18π₯^{2} + 180π₯

18π₯^{2} + 180π₯ β 21600 = 0

18(π₯^{2} + 10π₯ β 1200) = 0

π₯^{2} + 10π₯ β 1200 = 0

π₯^{2} + 40π₯ β 30π₯ β 1200 = 0

π₯(π₯ + 40) β 30(π₯ + 40) = 0

(π₯ β 30)(π₯ + 40) = 0

π₯ β 30 = 0 π₯ + 40 = 0

π₯ = 30 π₯ = β40

π₯ canβt be negative so the value of π₯ 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 π₯km/hr.

We know that,

3(48000) = π₯^{2} β 40π₯

144000 = π₯^{2} β 40π₯

π₯^{2} β 40π₯ β 144000 = 0

π₯^{2} β 400π₯ + 360π₯ β 144000 = 0

π₯(π₯ β 400) + 360(π₯ β 400) = 0

(π₯ + 360)(π₯ β 400) = 0

π₯ + 360 = 0 π₯ β 400 = 0

π₯ = β360 π₯ = 400

π₯ as a speed canβt 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 π₯ km/h

We know that,

3(4800) = 5(π₯^{2 }+ 12π₯)

14400 = 5π₯^{2 }+ 60π₯

5π₯^{2} + 60π₯ β 14400 = 0

5(π₯^{2} + 12π₯ β 2880) = 0

π₯^{2} + 12π₯ β 2880 = 0

π₯^{2} + 60π₯ β 48π₯ β 2880 = 0

π₯(π₯^{2} + 60) β 48(π₯ + 60) = 0

(π₯ β 48)(π₯ + 60) = 0

π₯ β 48 = 0 π₯ + 60 = 0

π₯ = 48 π₯ = β60

π₯ as a speed canβt be negative as the value of x is 48.

Hence, the speed of car is 48km/hr.

**Question **5. A girl goes to her friendβs 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 βxβ.

Solution:

It is given that,

Total distance cover by a girl = 6km

Speed = π₯km/hr

2(12π₯ + 12) = 5(π₯^{2} + 2π₯)

24π₯ + 24 = 5π₯^{2} + 10π₯

5π₯^{2} + 10π₯ β 24π₯ β 24 = 0

5π₯^{2} β 14π₯ β 24 = 0

5π₯^{2} β 14π₯ β 24 = 0

5π₯^{2} β 20π₯ + 6π₯ β 24 = 0

5π₯(π₯ β 4) + 6(π₯ β 4) = 0

(5π₯ + 6)(π₯ β 4) = 0

5π₯ + 6 = 0 π₯ β 4 = 0

π₯ = β6/5 π₯ = 4

π₯ as a speed canβt 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 βxβ hours. If the speed had been 4 km/hour more, it would have taken 2 hours less for the journey. Find βxβ.

Solution:

It is given that,

Distance cover by car is 390km

Time taken by car is π₯ hrs.

Let us assumed that,

Original speed of the car be π¦km/h

780 = π¦^{2} + 4π¦

π¦^{2} + 4π¦ β 780 = 0

π¦^{2} + 30π¦ β 26π¦ β 780 = 0

π¦(π¦ + 30) β 26(π¦ + 30) = 0

(π¦ β 26)(π¦ + 30) = 0

π¦ β 26 = 0 π¦ + 30 = 0

π¦ = 26 π¦ = β30

π¦ as a speed canβt 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 π₯ km/hr

1600 Γ 5 = π₯^{2} + 20π₯

8000 = π₯^{2} + 20π₯

π₯^{2} + 20π₯ β 8000 = 0

π₯^{2} β 80π₯ + 100π₯ β 8000 = 0

π₯(π₯ β 80) + 100(π₯ β 80) = 0

(π₯ + 100)(π₯ β 80) = 0

π₯ β 80 = 0 π₯ + 100 = 0

π₯ = 80 π₯ = β100

π₯ as a speed canβt 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. π₯ 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. π₯

Selling Price of an article = Rs 16

Loss = Rs. (π₯β 16)

π₯^{2} = (π₯ β 16) Γ 100

π₯^{2} = 100π₯ β 1600

π₯^{2} β 100π₯ + 1600 = 0

π₯^{2} β 80π₯ β 20π₯ + 1600 = 0

π₯(π₯ β 80) β 20(π₯ β 80) = 0

(π₯ β 20)(π₯ β 80) = 0

π₯ β 20 = 0 π₯ β 80 = 0

π₯ = 20 π₯ = 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 β 10) percent on his outlay. Calculate the cost price.

Solution:

It is given that,

Cost Price of an article = Rs. π₯

Selling Price of an article = Rs. 52

Profit = Rs. (52 β π₯)

Profit % = Rs. (π₯ β 10)

π₯(π₯ β 10) = (52 β π₯) Γ 100

π₯^{2} β 10π₯ = 5200 β 100π₯

π₯^{2} β 10π₯ + 100π₯ β 5200 = 0

π₯^{2} + 90π₯ β 5200 = 0

π₯^{2} + 130π₯ β 40π₯ β 5200 = 0

π₯(π₯ + 130) β 40(π₯ + 130) = 0

(π₯ β 40)(π₯ + 5200) = 0

π₯ β 40 = 0 π₯ + 130 = 0

π₯ = 40 π₯ = β130

Cost Price canβt 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. π₯

S.P. of chair = Rs. 75

Profit = Rs. (75 β π₯)

π₯^{2} = (75 β π₯) Γ 100

π₯^{2} = 7500 β 100π₯

π₯^{2} + 100π₯ β 7500 = 0

π₯^{2} + 150π₯ β 50π₯ β 7500 = 0

π₯(π₯ + 150) β 50(π₯ + 150) = 0

(π₯ β 50)(π₯ + 150) = 0

π₯ β 50 = 0 π₯ + 150 = 0

π₯ = 50 π₯ = β150

Cost Price canβt 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,

π(π + 2) = 168

πΒ² + 2πβ 168 = 0

πΒ² + 14πβ 12πβ 168 = 0

π(π + 14)β 12(π + 14) = 0

(π + 14)(πβ 12) = 0

π + 14 = 0 π β 12 = 0

π = β14 π = 12

π Canβt be negative, Hence, the value of π be 12.

**Question **2. A stone is thrown vertically downwards and the formula d = 16tΒ² + 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,

16π‘^{2} + 4π‘ = 420

4π‘^{2} + π‘β 105 = 0

4π‘^{2}β 20π‘ + 21π‘β 105 = 0

4π‘(π‘β 5) + 21(π‘β 5) = 0

(4π‘ + 21)(π‘β 5) = 0

4π‘ + 21 = 0 π‘β 5 = 0

π‘ = β21/4 π‘ = 5

π‘ 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 unitβs digit exceeds twice its tenβs digit by 2; find the number.

Solution:

Let the first number be π₯

And second number be π¦

According to the question,

π₯π¦ = 24___________(1)

It is given that, unitβs digit exceeds twice its tenβs digit by 2.

π¦ = 2π₯ + 2 **__**(2)

From equation (2) we get the value of π¦

π¦ = 2π₯ + 2

Put the value of π¦ in equation (1)

π₯(2π₯ + 2) = 24

2π₯

^{2}+ 2π₯ = 24

2π₯

^{2}+ 2π₯ β 24 = 0

2(π₯

^{2}+ π₯ β 12) = 0

π₯

^{2}+ π₯ β 12 = 0

π₯

^{2}+ 4π₯ β 3π₯ β 12 = 0

π₯(π₯ + 4) β 3(π₯ + 4) = 0

(π₯ β 3)(π₯ + 4) = 0

π₯ β 3 = 0 π₯ + 4 = 0

π₯ = 3 π₯ = β4

π₯ Canβt be negative, the value of π₯ is 3.

Put the value of x in equation no (2).

π¦ = 2(3) + 2

π¦ = 6 + 2

π¦ = 8

Hence, the required number is 38.

**Question **4. The ages of two sisters are 11 years and 14 years. In how many yearsβ 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,

π₯ Number of years the product of their ages be 304.

(11 + π₯)(14 + π₯) = 304

154 + 11π₯ + 14π₯ + π₯^{2} = 304

154 + 11π₯ + 14π₯ + π₯^{2} β 304 = 0

π₯^{2} + 25π₯ β 150 = 0

π₯^{2} + 30π₯ β 5π₯ β 150 = 0

π₯(π₯ + 30) β 5(π₯ + 30) = 0

(π₯ β 5)(π₯ + 30) = 0

π₯ β 5 = 0 π₯ + 30 = 0

π₯ = 5 π₯ = β30

The number of years cannot be negative so, the value of π₯ 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 sonβs age. Find their present ages.

Solution:

Let the present age of the son be π₯years

Present age of man = π₯^{2}years

According to question, One year ago

Sonβs age = (π₯ β 1)years

Manβs age = (π₯^{2}β 1) years

It is given that one year ago; a man was 8 times as old as his son.

(π₯^{2}β 1) = 8(π₯ β 1)

π₯^{2}β 1 = 8π₯β 8

π₯^{2}β 8π₯ β 1 + 8 = 0

π₯^{2}β 8π₯ + 7 = 0

π₯^{2}β 7π₯ β π₯ + 7 = 0

π₯(π₯β 7) β 1(π₯ β 7) = 0

(π₯β 7)(π₯β 1) = 0

π₯ β 7 = 0 π₯ β 1 = 0

π₯ = 7 π₯ = 1

If,

π₯ = 1

π₯^{2} = 1,

It is not possible as fatherβs age canβt be equal to sonβs age.

If,

π₯ = 7

Present age of son = π₯ years = 7 years

Present age of man = π₯^{2}years = 49 years

Hence, the sonβs age is 7 years and fatherβs 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 = 2π₯^{2}years

Eight years hence,

Sonβs age = (π₯ + 8)years

Fatherβs age = (2π₯^{2} + 8)years

The age of the father will be 4 years more than three times the age of the son.

2π₯^{2} + 8 = 3(π₯ + 8) + 4

2π₯^{2} + 8 = 3π₯ + 24 + 4

2π₯^{2} + 8 = 3π₯ + 28

2π₯^{2} β 3π₯ + 8 β 28 = 0

2π₯^{2}β 3π₯β 20 = 0

2π₯^{2}β 8π₯ + 5π₯β 20 = 0

2π₯(π₯β 4) + 5(π₯β 4) = 0

(π₯β 4)(2π₯ + 5) = 0

π₯ β 4 = 0 2π₯ + 5 = 0

π₯ = 4 π₯ = β 5/2

Age canβt be negative, so the value of π₯ 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 fatherβs 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 π₯km/hr

Speed of the boat downstream = (15 + π₯)km/hr

Speed of the boat upstream = (15β π₯)km/hr

Distance cover by board = 30km

900 Γ 2 = 9(225 β π₯^{2})

1800 = 2025 β 9π₯^{2}

9π₯^{2} = 2025 β 1800

9π₯^{2} = 225

π₯^{2 }= 225/9

π₯^{2} = 25

π₯ = Β±β25

π₯ = Β±5

π₯ canβt be negative, so the value of π₯ 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/π¦

The servant ate 3 oranges, so Mr. Mehra received (π¦ β 3) oranges.

π₯ = π¦ β 3

π¦ = 3 + π₯

63π₯ + π₯^{2} = 15(4π₯ + 12)

63π₯ + π₯^{2} = 60π₯ + 180

63π₯ + π₯^{2} β 60π₯ β 180 = 0

π₯^{2} β 3π₯ β 180 = 0

π₯^{2} + 15π₯ β 12π₯ β 180 = 0

π₯(π₯ + 15) β 12(π₯ + 15) = 0

(π₯ β 12)(π₯ + 15) = 0

π₯ β 12 = 0 π₯ + 15 = 0

π₯ = 12 π₯ = β15

π₯ canβt be negative so, the value of π₯ 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/π₯

As per question if we increase 25 children,

6250 Γ 2 = π₯^{2} + 25π₯

12500 = π₯^{2} + 25π₯

π₯^{2} + 25π₯ β 12500 = 0

π₯^{2} + 125π₯ β 100π₯ β 12500 = 0

π₯(π₯ + 125) β 100(π₯ + 125) = 0

(π₯ β 100)(π₯ + 125) = 0

π₯ β 100 = 0 π₯ + 125 = 0

π₯ = 100 π₯ = β125

π₯ as the number of students canβt be negative, so, the value of π₯ 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. π₯

Original weekly wage bill of employer = Rs. 3150

Number of workers = 3150/π₯

New weekly wage of each worker = Rs. (π₯ + 5)

New weekly wage of each worker = Rs. 3250

(3150 β 5π₯)(π₯ + 5) = 3250π₯

3150π₯ + 15750 β 5π₯^{2} β 25π₯ = 3250π₯

3150π₯ β 3250π₯ + 15750 β 5π₯^{2} β 25π₯ = 0

15750 β 5π₯^{2} β 125π₯ = 0

β5(β3150 + π₯^{2} + 25π₯) = 0

π₯^{2} + 25π₯ β 3150 = 0

π₯^{2} + 70π₯ β 45π₯ β 3150 = 0

π₯(π₯ + 70) β 45(π₯ + 70) = 0

(π₯ β 45)(π₯ + 70) = 0

π₯ β 45 = 0 π₯ + 70 = 0

π₯ = 45 π₯ = β70

π₯ canβt be negative so, the value of π₯ 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 = π₯

Cost per article = 1200/π₯

According to question,

Ten articles were damaged then remaining articles are π₯ β 10

2π₯^{2} β 12000 β 20π₯ = 60π₯

2π₯^{2} β 12000 β 20π₯ β 60π₯ = 0

2π₯^{2} β 12000 β 80π₯ = 0

2π₯^{2} β 80π₯ β 12000 = 0

2(π₯^{2} β 40π₯ β 6000) = 0

π₯^{2} β 40π₯ β 6000 = 0

π₯^{2} β 100π₯ + 60π₯ β 6000 = 0

π₯(π₯ β 100) + 60(π₯ β 100) = 0

(π₯ + 60)(π₯ β 100) = 0

π₯ β 100 = 0 π₯ + 60 = 0

π₯ = 100 π₯ = β60

π₯ canβt be negative. So, the value of π₯ 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/π₯

Selling price of each article = Rs 100

Selling price of x articles = Rs 100x

It is given that,

Profit = C.P. of 15 articles

100π₯β 4800 = 15 Γ (4800/π₯) 100

π₯^{2} β 4800π₯ = 15 Γ 4800

π₯^{2} β 48π₯ β 720 = 0

π₯^{2} β 60π₯ + 12π₯ β 720 = 0

π₯(π₯ β 60) + 12(π₯ β 60) = 0

(π₯ β 60) (π₯ + 12) = 0

π₯ + 12 = 0 π₯ β 60 = 0

π₯ = β12 π₯ = +60

π₯ canβt be negative. So, the value of π₯ 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 = π₯ km/h

Speed of train = (π₯ + 16)km/h

8π₯ + 3456 = 2(π₯^{2} + 16π₯)

8π₯ + 3456 = 2π₯^{2} + 32π₯

2π₯^{2} + 32π₯ β 8π₯ β 3456 = 0

2π₯^{2} + 24π₯ β 3456 = 0

2(π₯^{2} + 12π₯ β 1728) = 0

π₯^{2} + 12π₯ β 1728 = 0

π₯^{2} + 48π₯ β 36π₯ β 1728 = 0

π₯(π₯ + 48) β 36(π₯ + 48) = 0

(π₯ β 36)(π₯ + 48) = 0

π₯ β 36 = 0 π₯ + 48 = 0

π₯ = 36 π₯ = β48

π₯ as speed canβt be negative, so x is 36km/hr.

(iv) Speed of train is = (π₯ + 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 = π₯

Total cost of article = Rs. 600

480 = π₯^{2} β 4π₯

π₯^{2} β 4π₯ β 480 = 0

π₯^{2} β 24π₯ + 20π₯ β 480 = 0

π₯(π₯ β 24) + 20(π₯ β 24) = 0

(π₯ + 20)(π₯ β 24) = 0

π₯ β 24 = 0 π₯ + 20 = 0

π₯ = 24 π₯ = β20

π₯ canβt be negative, so π₯ 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 π₯.

Total Hotel bill = Rs. 4800

Hotel bill each person = Rs. 4800/π₯

According to question,

19200 = 200(π₯^{2} + 4π₯)

19200 = 200π₯^{2} + 800π₯

200π₯^{2} + 800π₯ β 19200 = 0

π₯^{2} + 4π₯ β 96 = 0

π₯^{2} + 12π₯ β 8π₯ β 96 = 0

π₯(π₯ + 12) β 8(π₯ + 12) = 0

(π₯ β 8)(π₯ + 12) = 0

π₯ β 8 = 0 π₯ + 12 = 0

π₯ = 8 π₯ = β12

π₯ canβt be negative, so π₯ 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 π₯ 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 = π₯km/hr

Speed while returning = (π₯ + 40)km/hr

2(16000) = π₯^{2} + 40π₯

32000 = π₯^{2} + 40π₯

π₯^{2} + 40π₯ β 32000 = 0

π₯^{2} + 200π₯ β 160π₯ β 32000 = 0

π₯(π₯ + 200) β 160(π₯ + 200) = 0

(π₯ β 160)(π₯ + 200) = 0

π₯ β 160 = 0 π₯ + 200 = 0

π₯ = 160 π₯ = β200

π₯ as speed canβt be negative, so π₯ 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 π₯.

Total amount was divided = Rs. 6500

97500 = 30(π₯^{2} + 15π₯)

3250 = π₯^{2 }+ 15π₯

π₯^{2} + 15π₯ β 3250 = 0

π₯^{2} + 65π₯ β 50π₯ β 3250 = 0

π₯(π₯ + 65) β 50(π₯ + 65) = 0

(π₯ β 50)(π₯ + 65) = 0

π₯ β 50 = 0 π₯ + 65 = 0

π₯ = 50 π₯ = β65

π₯ as number of persons canβt be negative, so π₯ 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 π₯km/hr

According to the question,

2 Γ 375000 = π₯^{2} + 250π₯

750000 = π₯^{2} + 250π₯

π₯^{2} + 250π₯ β 750000 = 0

π₯^{2} + 1000π₯ β 750π₯ β 750000 = 0

π₯(π₯ + 1000) β 750(π₯ + 1000) = 0

π₯ β 750 = 0 π₯ + 1000 = 0

π₯ = 750 π₯ = β1000

π₯ as speed canβt be negative, so π₯ 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 π₯km/hr

Speed of the first train is (π₯ + 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 Γ Time = 2(π₯ + 5)km

Distance travelled by the first train in 2 hours in OB = speed Γ Time = 2π₯km

By Pythagoras Theorem, we have

(π΄π΅)^{2} = (ππ΄)^{2} + (ππ΅)^{2}

(50)^{2} = [2(π₯ + 5)]^{2} + (2π₯)^{2}

2500 = 4(π₯ + 5)^{2} + 4π₯^{2}

2500 = 4(π₯^{2} + 25 + 10π₯) + 4π₯^{2}

8π₯^{2} + 40π₯β 2400 = 0

π₯^{2} + 5π₯β 300 = 0

π₯^{2} + 20π₯β 15π₯β 300 = 0

(π₯ + 20)(π₯ β 15) = 0

π₯ + 20 = 0 π₯ β 15 = 0

π₯ = β20 π₯ = 15

π₯ as a speed canβt be negative so π₯ 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 π = π(π + 1). Find n, if the sum is 420.

Solution:

π = π(π + 1)

It is given,

π = 420 π(π + 1)

π = 420π^{2 }+ πβ 420 = 0

π^{2} + 21πβ 20πβ 420 = 0

π(π + 21)β 20(π + 21) = 0

(π + 21) (π β 20) = 0

π + 21 = 0 π β 20 = 0

π = β21 π = 20

π canβt 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 β x) years respectively.

Five years ago,

Fatherβs age = (π₯ β 5)years

Sonβs age = (45β π₯β 5)years

Sonβs age = (40β π₯)years

According to question,

(π₯β 5)(40β π₯) = 124

40π₯ β π₯^{2}β 200 + 5x = 124

π₯^{2}β 45π₯ + 324 = 0

π₯^{2}β 36π₯β 9π₯ + 324 = 0

π₯(π₯β 36) β 9(π₯β 36) = 0

(π₯β 36) (π₯β 9) = 0

π₯ β 36 = 0 π₯ β 9 = 0

π₯ = 36 π₯ = 9

If,

π₯ = 9

Fatherβs age = 9 years

Sonβs age = (45β π₯)

Sonβs age = (45 β 9)

Sonβs age = 36 years

Fatherβs age canβt be less than sonβs age.

If π₯ = 36

Fatherβs age = 36 years

Sonβs age = (45 β 36) years

Sonβs age = 9 years

Hence, fatherβs age is 36 years and sonβs 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 π₯.

The number of seats in each row in original arrangement = π₯

Total number of seats = π₯ Γ π₯ = π₯Β²

According to question,

2π₯(π₯β 10) = π₯^{2} + 300

2π₯^{2}β 20π₯ = π₯^{2} + 300

π₯^{2}β 20π₯β 300 = 0

π₯^{2}β 30π₯ + 10π₯β 300 = 0

π₯(π₯β 30) + 10(π₯β 30) = 0

(π₯ β 30)(π₯ + 10) = 0

π₯ β 30 = 0 π₯ + 10 = 0

π₯ = 30 π₯ = β10

π₯ as rows of seats cannot be negative. So, the value of π₯ is 30.

(i) The number of rows in the original arrangement = x = 30

(ii) The number of seats after re-arrangement = π₯^{2} + 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/π₯ part of work.

In one day, Manoj completes 1/π₯ + 16 part of work.

It is given that they both can do the work in 15 days.

= 30π₯ + 240 = π₯^{2} + 16π₯

= π₯^{2}β 14π₯β 240 = 0

= π₯^{2}β 24π₯ + 10π₯β 240 = 0

= π₯(π₯ β 24) + 10(π₯ β 24) = 0

= (π₯ β 24)(π₯ + 10) = 0

π₯ β 24 = 0 π₯ + 10 = 0

π₯ = 24 π₯ = β10

π₯ as the number of days canβt be negative, So the value of π₯ is 24.

Hence, the work is completed by Mohan in 24 days.

**Question **12. Two years ago, a manβs age was three times the square of his sonβs age. In three yearsβ time, his age will be four times his sonβs age. Find their present ages.

Solution:

Let us assumed that,

The age of son 2years ago be π₯ years

Then, fatherβs age 2 years ago = 3π₯^{2} years

Present age of son = (π₯ + 2) years

Present age of father = (3π₯^{2} + 2) years

3 years hence,

Sonβs age = (π₯ + 2 + 3)years

Sonβs age = (π₯ + 5)years

Fatherβs age = (3π₯^{2}+ 2 + 3)years

Fatherβs age = (3π₯^{2} + 5)years

According to question,

3π₯^{2} + 5 = 4(π₯ + 5)

3π₯^{2} β 4π₯β 15 = 0

3π₯^{2}β 9π₯ + 5π₯β 15 = 0

3π₯(π₯β 3) + 5(π₯β 3) = 0

(π₯β 3)(3π₯ + 5) = 0

π₯ β 3 = 0 3π₯ + 5 = 0

π₯ = 3 π₯ = β 5/3

π₯ as age canβt be negative, so the value of π₯ is 3.

Present age of son = (π₯ + 2)years

Present age of son = (3 + 2)years

Present age of son = 5 years

Present age of father = (3π₯^{2} + 2)years

Present age of father = (3(32) + 2)years

Present age of father = (3 Γ 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 π₯

And the denominator be π₯ + 3

If 1 is subtracted from both numerator and denominator = π₯β1/π₯+2

According to the question,

(13π₯ β 3)(π₯ + 2) = (π₯ β 1)(14π₯ + 42)

13π₯^{2} + 26π₯ β 3π₯ β 6 = 14π₯^{2} + 42π₯ β 14π₯ β 42

13π₯^{2} + 23π₯ β 6 = 14π₯^{2} + 28π₯ β 42

14π₯^{2} β 13π₯^{2} + 28π₯ β 23π₯ β 42 + 6 = 0

π₯^{2} + 5π₯ β 36 = 0

π₯^{2} + 9π₯ β 4π₯ β 36 = 0

π₯(π₯ + 9) β 4(π₯ + 9) = 0

(π₯ β 4)(π₯ + 9) = 0

π₯ β 4 = 0 π₯ + 9 = 0

π₯ = 4 π₯ = β9

π₯ canβt be negative, so π₯ is 4

Hence, the required fraction be π₯/π₯+3 = 4/7.

**Question **14. In a two digit number, the tenβs 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 tenβs digit is bigger than the unitβs digit.

So, let the unitβs digit be x and tenβs digit be (π₯ + 6).

According to question,

π₯(π₯ + 6) = 27

π₯Β² + 6π₯ β 27 = 0

π₯Β² + 9π₯ β 3π₯β 27 = 0

π₯(π₯ + 9) β 3(π₯ + 9) = 0

(π₯ + 9)(π₯β 3) = 0

π₯ + 9 = 0 π₯ β 3 = 0

π₯ = β9 π₯ = 3

π₯, the digits of a number cannot be negative, so, π₯ is 3.

Unitβs digit = 3

Tenβs 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 π₯ km/hr

Speed of the bus when returning = (π₯ β 5)km/hr

Distance = 300km

According to the question,

1500 = 2(π₯^{2} β 5π₯)

1500 = 2π₯^{2} β 10π₯

2π₯^{2} β 10π₯ β 1500 = 0

2(π₯^{2} β 5π₯ β 750) = 0

π₯^{2} β 5π₯ β 750 = 0

π₯^{2} β 30π₯ + 25π₯ β 750 = 0

π₯(π₯ β 30) + 25(π₯ β 30) = 0

(π₯ + 25)(π₯ β 30) = 0

π₯ + 25 = 0 π₯ β 30 = 0

π₯ = β25 π₯ = 30

π₯ as speed canβt be negative so, π₯ is 30.

Speed of the bus when returning = (30 β 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 βxβ children. If the number of children were 20 more, then each would have got Rs.12 less. Find βxβ.

Solution:

It is given that,

Total Amount be Rs. 480.

Let us assumed that,

The number of children be π₯.

The amount each children got = 480/π₯

If the number of children were 20 more,

Amount each children got = 480/π₯+20

According to question,

9600 = 12(π₯^{2} + 20π₯)

12π₯^{2} + 240π₯ β 9600 = 0

12(π₯^{2} + 20π₯ β 800) = 0

π₯^{2} + 20π₯ β 800 = 0

π₯^{2} + 40π₯ β 20π₯ β 800 = 0

π₯(π₯ + 40) β 20(π₯ + 40) = 0

(π₯ β 20)(π₯ + 40) = 0

π₯ β 20 = 0 π₯ + 40 = 0

π₯ = 20 π₯ = β40

π₯ as number of childrenβs canβt be negative, so the value of π₯ is 20.

Hence, the number of childrenβs 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 βxβ km/h, form an equation and solve it to evaluate βxβ.

Solution:

It is given that,

Total distance covered by bus is 240.

Let us assumed that,

The speed of bus be π₯.

Time taken by bus = 240/π₯ hrs

Time taken by bus to cover total distance with speed (π₯ β 10)km/hrs = 240/π₯β10 hrs

According to the question,

2400 = 2(π₯^{2} β 10π₯)

2400 = 2π₯^{2} β 20π₯

2π₯^{2} β 20π₯ β 2400 = 0

2(π₯^{2} β 10π₯ β 1200) = 0

π₯^{2} β 10π₯ β 1200 = 0

π₯^{2} β 40π₯ + 30π₯ β 1200 = 0

π₯(π₯ β 40) + 30(π₯ β 40) = 0

(π₯ + 30)(π₯ β 40) = 0

π₯ + 30 = 0 π₯ β 40 = 0

π₯ = β30 π₯ = 40

π₯ as the speed canβt be negative, so π₯ 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 = π₯

The age of Amit = 47β π₯

The product of their ages in years is 550

According to question,

π₯(47β π₯) = 550

47π₯β π₯^{2} = 550

π₯^{2}β 47π₯ + 550 = 0

π₯^{2}β 25π₯β 22π₯ + 550 = 0

π₯(π₯β 25)β 22(π₯β 25) = 0

(π₯β 25)(π₯β 22) = 0

π₯ β 25 = 0 π₯ β 22 = 0

π₯ = 25 π₯ = 22

Hence, the age of Vivek is 25 and age of Amit is 22 years.