Linear Equations In Two Variables Class 9th Mathematics Part I MHB Solution

Linear Equations In Two Variables Class 9th Mathematics Part I MHB Solution

Practice Set 5.1

1. By using variables x and y form any five linear equations in two variables.…
2. Write five solutions of the equation x + y = 7.
3. Solve the following sets of simultaneous equations. i. x + y = 4; 2x - 5y = 1 ii. 2x +…

Practice Set 5.2

1. In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of…
2. The denominator of a fraction is 1 more than twice its numerator. If 1 is added to…
3. The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6…
4. The total number of lions and peacocks in a certain zoo is 50. The total number of…
5. Sanjay gets fixed monthly income. Every year there is a certain increment in his…
6. The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is…
7. The sum of the digits in a two-digits number is 9. The number obtained by interchanging…
8. In ΔABC, the measure of angle A is equal to the sum of the measures of ∠B and ∠C. Also…
9. Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller…
10. In a competitive examination, there were 60 questions. The correct answer would carry…

Problem Set 5

1. Choose the correct alternative answers for the following questions. If 3x + 5y =9 and…
2. 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes…
3. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's…
4. Solve the following simultaneous equations. i. 2x + y = 5; 3x - y = 5 ii. x - 2y = -1;…
5. By equating coefficients of variables, solve the following equations. i. 3x - 4y = 7;…
6. Solve the following simultaneous equations. i. x/3 + y/4 = 4 x/2 - y/2 - y/4 = 1 ii.…
7. A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to this…
8. The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5…
9. The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3.…
10. If the length of a rectangle is reduced by 5 units and its breadth is increased by 3…
11. The distance between two places A and B on road is 70 kilometers. A car starts from A…
12. The sum of a two-digit number and the number obtained by interchanging its digits is…

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Practice Set 5.1

Question 1.

By using variables x and y form any five linear equations in two variables.

Answer:

a) x + y = 5

b) x + 2y = 6

c) 2x + y = 4

d) 3x + 4y = 8

e) 5x + 9y = 1

Question 2.

Write five solutions of the equation x + y = 7.

Answer:

(a)

Let x = 1,

As, x + y = 7

⇒ 1 + y = 7

⇒ y = 6

Hence, solution is x = 1 and y = 6.

(b)

Let x = 2,

As, x + y = 7

⇒ 2 + y = 7

⇒ y = 5

Hence, solution is x = 2 and y = 5.

(c)

Let x = 3,

As, x + y = 7

⇒ 3 + y = 7

⇒ y = 4

Hence, solution is x = 3 and y = 4.

(d)

Let x = 4,

As, x + y = 7

⇒ 4 + y = 7

⇒ y = 3

Hence, solution is x = 4 and y = 3.

(e)

Let x = 5,

As, x + y = 7

⇒ 5 + y = 7

⇒ y = 2

Hence, solution is x = 5 and y = 2.

Question 3.

Solve the following sets of simultaneous equations.

i. x + y = 4; 2x - 5y = 1

ii. 2x + y =5; 3x – y =5

iii. 3x – 5y =16; x - 3y = 8

iv. 2y – x =0; 10x + 15y =105

v. 2x +3y + 4 = 0; x – 5y = 11

vi. 2x – 7y =7; 3x + y =22

Answer:

(i)

x + y = 4 eq.[1]

2x - 5y = 1 eq.[2]

We can write eq.[1] as,

x = 4 - y eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 2(4 - y) - 5y = 1

⇒ 8 - 2y - 5y = 1

⇒ -7y = -7

⇒ y = 1

Substituting 'y' in eq.[3]

⇒ x = 4 - 1

⇒ x = 3

Hence, solution is x = 3 and y = 1.

(ii)

2x + y = 5 eq.[1]

3x - y = 5 eq.[2]

We can write eq.[1] as,

y = 5 - 2x eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 3x - (5 - 2x) = 5

⇒ 3x - 5 + 2x = 5

⇒ 5x = 10

⇒ x = 2

Substituting 'x' in eq.[3]

⇒ y = 5 - 2(2)

⇒ y = 1

Hence, solution is x = 2 and y = 1.

(iii)

3x - 5y = 16 eq.[1]

x - 3y = 8 eq.[2]

We can write eq.[2] as,

x = 8 + 3y eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 3(8 + 3y) - 5y = 16

⇒ 24 + 9y - 5y = 16

⇒ 4y = -8

⇒ y = -2

Substituting 'y' in eq.[3]

⇒ x = 8 + 3(-2)

⇒ x = 8 - 6 = 2

Hence, solution is x = 2 and y = -2

(iv)

2y - x = 0 eq.[1]

10x + 15y = 105 eq.[2]

We can write eq.[1] as,

x = 2y eq.[3]

Substituting eq.[3] in eq.[2],

⇒ 10(2y) + 15y = 105

⇒ 20y + 15y = 105

⇒ 35y = 105

⇒ y = 3

Substituting 'y' in eq.[3]

⇒ x = 2(3)

⇒ x = 6

Hence, solution is x = 6 and y = 3.

(v)

2x + 3y + 4 = 0 eq.[1]

x - 5y = 11 eq.[2]

We can write eq.[2] as,

x = 11 + 5y eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 2(11 + 5y) + 3y + 4 = 0

⇒ 22 + 10y + 3y + 4 = 0

⇒ 13y + 26 = 0

⇒ 13y = -26

⇒ y = -2

Substituting 'y' in eq.[3]

⇒ x = 11 + 5(-2)

⇒ x = 11 - 10 = 1

Hence, solution is x = 1 and y = -2.

(vi)

2x - 7y = 7 eq.[1]

3x + y = 22 eq.[2]

We can write eq.[2] as,

y = 22 - 3x eq.[3]

Substituting eq.[3] in eq.[1],

⇒ 2x - 7(22- 3x) = 7

⇒ 2x - 154 + 21x = 7

⇒ 23x = 161

⇒ x = 7

Substituting 'x' in eq.[3]

⇒ y = 22 - 3(7)

⇒ y = 22 - 21 = 1

Hence, solution is x = 7 and y = 1.

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Practice Set 5.2

Question 1.

In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.

Answer:

Let the number of 5 rupees notes = x

Let the number of 10 rupees notes = y

Given, Total amount is 350 Rupees

⇒ 5x + 10y = 350 eq.[1]

Also,

Number of 5 rupees notes are less by 10 than number of 10 rupees note,

y = x - 10

⇒ x = y + 10 eq.[2]

Putting [2] in [1]

⇒ 5(y + 10) + 10y = 350

⇒ 5y + 50 + 10y = 350

⇒ 15y = 300

⇒ y = 20

Then, x = y + 10

⇒ x = 20 + 10

⇒ x = 30.

Answer: 30 notes of Rs 5 and 20 notes of Rs. 10.

Question 2.

The denominator of a fraction is 1 more than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is
1 : 2. Find the fraction.

Answer:

Let the numerator be 'x' and denominator be 'y'

Given,

The denominator of a fraction is 1 more than twice its numerator

⇒ y =2 x + 1 

⇒ y - 2 x = 1 ....... (1)

If 1 is added to numerator and denomination, the ratio of the numerator to denominator becomes 1:2.

⇒ 2(x + 1) = y + 1

⇒ 2x + 2 = y + 1

⇒ y - 2x = 1 ....... (2)

As (1) and (2) are the same, there can be infinitely many solutions for x and y.

One such solution is:

x = 4 and y = 9

Now,

y - 2 x = 9 - 2(4)
= 9 - 8
= 1

Question 3.

The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today's ages.

Answer:

Let the ages of Priyanka and Deepika be 'x' and 'y' respectively.

Given,

Sum of ages is 34

⇒ x + y = 34

⇒ y = 34 - x eq.[1]

Also, Priyanka is elder to Deepika by 6 years

⇒ x = y + 6

Using eq.[1] we have

⇒ x = 34 - x + 6

⇒ 2x = 40

⇒ x = 20

Putting this value in eq.[1]

⇒ y = 34 - 20 = 14 years.

Hence, Age of Priyanka = x = 20 Years

Age of Deepika = y = 14 years.

Question 4.

The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.

Answer:

Let the number of lions be 'x' and peacocks be 'y'

Given, Total no of lions and peacocks is 50

⇒ x + y = 50

⇒ x = 50 - y eq.[1]

Also, Total no of their legs is 140, as lion has four legs and peacocks has 2 legs

⇒ 4x + 2y = 140

⇒ 4(50 - y) + 2y = 140

⇒ 200 - 4y + 2y = 140

⇒ 2y = 60

⇒ y = 30

Using this in eq.[1]

⇒ x = 50 - 30 = 20

Therefore,

No of lions, x = 20

No of peacocks, y = 30

Question 5.

Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.

Answer:

Let the original salary be 'x' and yearly increment be 'y'

After 4 years, his salary was Rs. 4500

⇒ x + 4y = 4500

⇒ x = 4500 - 4y eq.[1]

After 10 years, his salary becomes 5400

⇒ x + 10y = 5400

⇒ 4500 - 4y + 10y = 5400

⇒ 6y = 900

⇒ y = 150

Putting this in eq.[1],

⇒ x = 4500 - 4(150)

⇒ x = 4500 - 600 = 3900

Hence, his original salary was Rs. 3900 and increment per year was 150 Rs.

Question 6.

The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the price of 2 chairs and 2 tables.

Answer:

Let the price of one chair be 'x' and one table be 'y'.

Given,

Price of 3 chairs and 2 tables = 4500 Rs

⇒ 3x + 2y = 4500

Multiplying by 3 both side,

⇒ 9x + 6y = 13500

⇒ 6y = 13500 - 9x eq.[1]

Price of 5 chairs and 3 tables = 7000 Rs

⇒ 5x + 3y = 7000

Multiplying by eq.[2] both side,

⇒ 10x + 6y = 14000

⇒ 10x + 13500 - 9x = 14000 eq.[From 1]

⇒ x = 500

Putting this in eq.[1]

⇒ 6y = 13500 - 9(500)

⇒ 6y = 13500- 4500

⇒ 6y = 9000

⇒ y = 1500

Also, Price of 2 chairs and 2 tables = 2x + 2y

= 2(500) + 2(1500)

= 1000 + 3000 = 4000 Rs.

Question 7.

The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.

Answer:

Let the unit digit be 'x'

Let the digit at ten's place be 'y'

The original number will be 10y + x

Given,

Sum of digits = 9

⇒ x + y = 9

⇒ x = 9 - y eq.[1]

Also,

If the digits are interchanged,

Reversed number will be = 10x + y

As, reversed number exceeds the original number by 27,

⇒ (10x + y) - (10y + x) = 27

⇒ 10x + y - 10y - x = 27

⇒ 9x - 9y = 27

⇒ x - y = 3

⇒ 9 - y - y = 3 eq.[using 1]

⇒ -2y = -6

⇒ y = 3

Using this in eq.[1]

⇒ x = 9 - 3 = 6

Hence the original number is 10y + x = 10(3) + 6 = 30 + 6 = 36.

Question 8.

In ΔABC, the measure of angle A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.

Answer:

Given that, In ΔABC

∠A = ∠B + ∠C eq.[1]

Let ∠B = x and ∠C = y

Then,

∠A = x + y

In ΔABC, By angle sum property of triangle

∠A + ∠B + ∠C = 180°

⇒ x + y + x + y = 180

⇒ 2x + 2y = 180

⇒ x + y = 90

⇒ x = 90 - y eq.[2]

Also, Given that

⇒ 5x = 4y

From eq.[2]

⇒ 5(90 - y) = 4y

⇒ 450 - 5y = 4y

⇒ 9y = 450

⇒ y = 50°

Putting this in eq.[2]

⇒ x = 90 - 50 = 40°

Therefore, we have

∠A = x + y = 40° + 50° = 90°

∠B = x = 40°

∠C = y = 50°

Question 9.

Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to 1/3 of the larger part. Then find the length of the larger part.

Answer:

Let the length of smaller part be 'x' cm and larger part be 'y' cm.

Length of rope = 560 cm

⇒ x + y = 560

⇒ y = 560 - x eq.[1]

Also,

Twice the length of smaller part is equal to  of the larger part

⇒ 6x = y

⇒ 6x = 560 - x

⇒ 7x = 560

⇒ x = 80

Using this in eq.[1]

⇒ y = 560 - 80 = 480

Therefore,

Length of smaller part = 'x' cm = 80 cm

Length of larger part = 'y' cm = 480 cm

Question 10.

In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?

Answer:

Let the no of questions he got wrong be 'x'

And the no of questions he got right be 'y'

As, he attempted all the questions,

⇒ x + y = 60

⇒ y = 60 - x eq.[1]

Also, he carries 2 for each corrects question and (-1) for each wrong question, also he got 90 marks

⇒ y(2) + x(-1) = 90

⇒ 2y - x = 90

⇒ 2(60 - x) - x = 90 eq.[Using 1]

⇒ 120 - 2x - x = 90

⇒ -3x = -30

⇒ x = 10

⇒ he got 10 wrong questions.

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Problem Set 5

Question 1.

Choose the correct alternative answers for the following questions.

If 3x + 5y =9 and 5x + 3y =7 then What is the value of x + y?
A. 2

B. 16

C. 9

D. 7

Answer:

3x + 5y = 9 eq.[1]

5x + 3y = 7 eq.[2]

Adding eq.[1] and eq.[2] we get

3x + 5y + 5x + 3y = 9 + 7

⇒ 8x + 8y = 16

Dividing both side by 8, we get

⇒ x + y = 2

Question 2.

'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.' What is the mathematical form of the statement?
A. x – y = 8

B. x + y = 8

C. x + y =23

D. 2x + y = 21

Answer:

Let the length be 'x' and breadth be 'y' units.

Perimeter of triangle = 2(x + y) units

If 5 is subtracted from length and breadth

Perimeter = 26 units eq.[Given]

⇒ 2( x - 5 + y - 5) = 26

⇒ 2(x + y - 10) = 26

⇒ x + y - 10 = 13

⇒ x + y = 23

Question 3.

Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age?
A. 20

B. 15

C. 10

D. 5

Answer:

Let Ajay's age be 'x' years and Vijay's age be 'y' years.

Given, Ajay is younger than Vijay by 5 years

⇒ x = y - 5 eq.[1]

Also, Sum of their ages is 25 years,

⇒ x + y = 25

From eq.[1]

⇒ y - 5 + y = 25

⇒ 2y = 30

⇒ y = 15

Putting this in eq.[1]

⇒ x = 15 - 5 = 10

Age of Ajay = x = 10 Years

Age of Vijay = y = 15 Years

Question 4.

Solve the following simultaneous equations.
i. 2x + y = 5; 3x - y = 5

ii. x - 2y = -1; 2x - y = 7

iii. x + y = 11; 2x - 3y = 7

iv. 2x + y = -2; 3x - y = 7

v. 2x - y = 5; 3x + 2y = 11

vi. x - 2y = -2; x + 2y = 10

Answer:

(i)

2x + y = 5

⇒ y = 5 - 2x eq.[1]

3x - y = 5

Using eq.[1] we have

⇒ 3x - (5 - 2x) = 5

⇒ 3x - 5 + 2x = 5

⇒ 5x = 10

⇒ x = 2

Using 'x' in eq.[1]

⇒ y = 5 - 2(2)

⇒ y = 5 - 4 = 1 cm

(ii)

x - 2y = -1

⇒ x = 2y - 1 eq.[1]

2x - y = 7

Using eq.[1], we have

⇒ 2(2y - 1) - y = 7

⇒ 4y - 2 - y = 7

⇒ 3y = 9

⇒ y = 3

Using this value in eq.[1]

⇒ x = 2(3) - 1

⇒ x = 5

(iii)

x + y = 11

⇒ y = 11 - x eq.[1]

2x - 3y = 7

Using eq.[1], we have

⇒ 2x - 3(11 - x) = 7

⇒ 2x - 33 + 3x = 7

⇒ 5x = 40

⇒ x = 8

Using this in eq.[1]

⇒ y = 11 - 8

⇒ y = 3

(iv)

2x + y = -2

⇒ y = -2x - 2 eq.[1]

3x - y = 7

Using eq.[1]

3x - (-2x - 2) = 7

⇒ 3x + 2x + 2 =7

⇒ 5x = 5

⇒ x = 1

Using this in eq.[1]

⇒ y = -2(1) - 2

⇒ y = -2 - 2 = -4

(v)

2x - y = 5

⇒ y = 2x - 5 eq.[1]

3x + 2y = 11

Using eq.[1]

⇒ 3x + 2(2x - 5) = 11

⇒ 3x + 4x - 10 = 11

⇒ 7x = 21

⇒ x = 3

Using this in eq.[1]

⇒ y = 2(3) - 5

⇒ y = 1

(vi)

x - 2y = -2

x = 2y - 2 eq.[1]

x + 2y = 10

using eq.[1], we have

⇒ 2y - 2 + 2y = 10

⇒ 4y = 12

⇒ y = 3

Using this in eq.[1]

⇒ x = 2(3) - 2

⇒ x = 4

Question 5.

By equating coefficients of variables, solve the following equations.

i. 3x - 4y = 7; 5x + 2y = 3

ii. 5x + 7y = 17; 3x - 2y = 4

iii. x - 2y = -10; 3x - 5y = -12

iv. 4x + y = 34; x + 4y = 16

Answer:

(i)

3x - 4y = 7 eq.[1]

5x + 2y = 3 eq.[2]

Multiplying eq.[2] by 2 both side, we get

10x + 4y = 6 eq.[3]

Adding eq.[1] and eq.[3], we get

3x - 4y + 10x + 4y = 7 + 6

⇒ 13x = 13

⇒ x = 1

Putting this in eq.[1], we get

3(1) - 4y = 7

⇒ -4y = 7 - 3

⇒ -4y = 4

⇒ y = -1

(ii)

5x + 7y = 17 eq.[1]

3x - 2y = 4 eq.[2]

Multiplying eq.[1] by 3 both side and Multiplying eq.[2] by 5 both side we get,

15x + 21y = 51 eq.[3]

15x - 10y = 20 eq.[4]

Subtracting eq.[4] from eq.[3], we get

15x + 21y - 15x + 10y = 51 - 20

⇒ 31y = 31

⇒ y = 1

Putting this in eq.[1], we get

5x + 7(1) = 17

⇒ 5x = 10

⇒ x = 2

(iii)

x - 2y = -10 eq.[1]

3x - 5y = -12 eq.[2]

Multiplying eq.[1] by 3

3x - 6y = -30 eq.[3]

Subtracting eq.[2] from eq.[3], we get

3x - 6y - 3x + 5y = -30 + 12

⇒ -y = -18

⇒ y = 18

Putting this in eq.[1], we get

x - 2(18) = -10

⇒ x - 36 = -10

⇒ x = 26

(iv)

4x + y = 34 eq.[1]

x + 4y = 16 eq.[2]

Multiplying eq.[2] by 4 both side, we get

4x + 16y = 64 eq.[3]

Subtracting eq.[3] from eq.[1], we get

4x + 16y - 4x - y = 64 - 34

⇒ 15y = 30

⇒ y = 2

Putting this in eq.[2], we get

x + 4(2) = 16

⇒ x + 8 = 16

⇒ x = 8

Question 6.

Solve the following simultaneous equations.
i.

ii. 

iii. 

Answer:

(i)

⇒ 4x + 3y = 48 eq.[1]

⇒ 2x - 3y = 4 eq.[2]

Adding eq.[1] and eq.[2], we get

⇒ 4x + 3y + 2x - 3y = 48 + 4

⇒ 6x = 52

Using this in eq.[1], we have

⇒ 104 + 9y = 144

⇒ 9y = 40

(ii)

⇒ x + 15y = 39

⇒ x = 39 - 15y eq.[1]

⇒ 4x + y = 38

Using eq.[1], we have

⇒ 4(39 - 15y) + y = 38

⇒ 156 - 60y + y = 38

⇒ 59y = 118

⇒ y = 2

Putting this value in eq.[2]

⇒ x = 39 - 15(2)

⇒ x = 39 - 30

⇒ x = 9

(iii)

⇒ 2y + 3x = 13xy eq.[1]

⇒ 5y - 4x = -2xy eq.[2]

Multiplying eq.[1] by 4 both side, and Multiplying eq.[2] by 3 both side, we get

8y + 12x = 52xy eq.[3]

15y - 12x = -6xy eq.[4]

Adding eq.[3] and eq.[4]

⇒ 8y + 12x + 15y - 12x = 52xy - 6xy

⇒ 23y = 46xy

⇒ 1 = 2x

Putting this in eq.[1]

Question 7.

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.

Answer:

Let the unit digit be 'x'

Let the digit at ten's place be 'y'

The original number will be 10y + x

Given, number is 3 more than 4 times the sum of its digits

⇒ 10y + x = 4(x + y) + 3

⇒ 10y + x = 4x + 4y + 3

⇒ 6y - 3x = 3

⇒ 2y - x = 1

⇒ x = 2y - 1 eq.[1]

Also,

If the digits are interchanged,

Reversed number will be = 10x + y

As, reversed number exceeds the original number by 18,

⇒ (10x + y) - (10y + x) = 18

⇒ 10x + y - 10y - x = 18

⇒ 9x - 9y = 18

⇒ x - y = 2

⇒ 2y - 1 - y = 2 eq.[using 1]

⇒ y = 3

Using this in eq.[1]

⇒ x = 2(3) - 1 = 5

Hence the original number is 10y + x = 10(3) + 5 = 30 + 5 = 35.

Question 8.

The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5 pens is 77 ruppees. Find the cost of 1 book and 2 pens.

Answer:

Let the cost of one book be 'x' rupees and cost of one pen be 'y' rupees.

Cost of 6 books and 7 pens = 79 Rs

⇒ 6x + 7y = 79 eq.[1]

Cost of 7 books and 5 pens = 77 Rs

⇒ 7x + 5y = 77 eq.[2]

Multiplying eq.[1] by 5 both side, and Multiplying eq.[2] by 7 both side, we get

⇒ 30x + 35y = 395 eq.[3]

⇒ 49x + 35y = 539 eq.[4]

Subtracting eq.[3] from eq.[4], we get

⇒ 49x + 35y - 30x - 35y = 539 - 395

⇒ 19x = 144

Using this in eq.[1]

⇒ 864 + 19×7y = 79×19

⇒ 19×7y = 79×19 – 864

⇒ 

⇒ y = 5

& 6x + 7y = 79

⇒ 6x + 35 = 79

⇒ 6x = 44

⇒ x = 7

Hence, the cost of 1 pen & 2 books = Rs 1(y) + 2x

= 5 + 14 = Rs 19.

Question 9.

The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each.

Answer:

As the ratio of incomes is 9 : 7,

Let income of first person = 9x

Income of second person = 7x

Also, ratio of incomes is 4 : 3,

Let expenses of first person = 4y

Expenses of second person = 3y

Each person saves 200 Rs,

⇒ 9x - 4y = 200 eq.[1]

⇒ 7x - 3y = 200 eq.[2]

Multiplying eq.[1] by 3 both side and Multiplying eq.[2] by 4 both side, we get

⇒ 27x - 12y = 600 eq.[3]

⇒ 28x - 12y = 800 eq.[4]

Subtracting eq.[3] from eq.[4], we get

⇒ 28x - 12y - (27x - 12y) = 800 - 600

⇒ 28x - 12y - 27x + 12y = 200

⇒ x = 200

Income of first person = 9x = 9(200) = 1800 Rs

Income of second person = 7x = 7(200) = 1400 Rs

Question 10.

If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.

Answer:

Let the length be 'x' and breadth be 'y'

Area of rectangle = length × breadth

Area of rectangle = xy

First case:

Length = x - 5

Breadth = y + 3

As, area is reduced by 8 sq. units

⇒ xy - (x - 5)(y + 3) = 8

⇒ xy - (xy + 3x - 5y - 15) = 8

⇒ xy - xy - 3x + 5y + 15 = 8

⇒ 3x - 5y = 7 eq.[1]

Second case:

Length = x - 3

Breadth = y + 2

As, the area is increased by 67 units

⇒ (x - 3)(y + 2) - xy = 67

⇒ xy + 2x - 3y - 6 - xy = 67

⇒ 2x - 3y = 73 eq.[2]

Multiplying eq.[1] by 2 both side, and Multiplying eq.[2] by 3 both side, we get

⇒ 6x - 10y = 14 eq.[3]

⇒ 6x - 9y = 219 eq.[4]

Subtracting eq.[3] from eq.[4]

⇒ 6x - 9y - 6x + 10y = 219 - 14

⇒ y = 205

Using this in eq.[1]

⇒ 3x - 5(205) = 7

⇒ 3x - 1025 = 7

⇒ 3x = 1032

⇒ x = 344

Hence, length = x = 344 units

Breadth = y = 219 units.

Question 11.

The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.

Answer:

Let the speed of car at place A is x km/h and that of car at place B is y km/h

If they travel in same direction, they will meet after 7 hours, i.e. the difference of distance covered by them in 7 hours will be equal to distance b/w A and B.

As, distance = speed × time, and distance from A to B is 70 km

⇒ 7x - 7y = 70

⇒ x - y = 10

⇒ x = y + 10 eq.[1]

If they, travel in opposite direction, they will meet after 1 hour i.e. sum of distance travelled by both cars will be equal to the distance b/w A and B.

⇒ x + y = 70

Using eq.[1], we have

⇒ y + 10 + y = 70

⇒ 2y = 60

⇒ y = 30

Using this in eq.[1], we have

x = 30 + 10 = 40

Hence,

Speed of car at A = x = 40 km/h

Speed of car at B = y = 30 km/h

Question 12.

The sum of a two-digit number and the number obtained by interchanging its digits is 99. Find the number.

Answer:

Let the unit digit be 'x' and digit at ten's place be 'y'

Original Number = 10y + x

Number obtained by interchanging digits = 10x + y

Given,

10y + x + 10x + y = 99

⇒ 11x + 11y = 99

⇒ x + y = 9

If x = 1, y = 8 and number is 18

If x = 2, y = 7 and number is 27

If x = 3, y = 6 and number is 36

If x = 4, y = 5 and number is 45

If x = 5, y = 4 and number is 54

If x = 6, y = 3 and number is 63

If x = 7, y = 2 and number is 72

If x = 8, y = 1 and number is 81