Linear Equations In Two Variables Class 9th Mathematics Part I MHB Solution
- By using variables x and y form any five linear equations in two variables.…
- Write five solutions of the equation x + y = 7.
- Solve the following sets of simultaneous equations. i. x + y = 4; 2x - 5y = 1 ii. 2x +…
- In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of…
- The denominator of a fraction is 1 more than twice its numerator. If 1 is added to…
- The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6…
- The total number of lions and peacocks in a certain zoo is 50. The total number of…
- Sanjay gets fixed monthly income. Every year there is a certain increment in his…
- The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is…
- The sum of the digits in a two-digits number is 9. The number obtained by interchanging…
- In ΔABC, the measure of angle A is equal to the sum of the measures of ∠B and ∠C. Also…
- Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller…
- In a competitive examination, there were 60 questions. The correct answer would carry…
- Choose the correct alternative answers for the following questions. If 3x + 5y =9 and…
- 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes…
- Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's…
- Solve the following simultaneous equations. i. 2x + y = 5; 3x - y = 5 ii. x - 2y = -1;…
- By equating coefficients of variables, solve the following equations. i. 3x - 4y = 7;…
- Solve the following simultaneous equations. i. x/3 + y/4 = 4 x/2 - y/2 - y/4 = 1 ii.…
- A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to this…
- The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5…
- The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3.…
- If the length of a rectangle is reduced by 5 units and its breadth is increased by 3…
- The distance between two places A and B on road is 70 kilometers. A car starts from A…
- The sum of a two-digit number and the number obtained by interchanging its digits is…
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.
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.
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
