Linear Equations in Two Variable | Maths | Class 10th | Chapter 3 | PYQ Level 1

 

Linear Equations in Two Variables :- PYQ :- Level 1

Q.1: The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically.

Solution: Let the cost of 1 kg of apples be ‘Rs. x’.
And, let the cost of 1 kg of grapes be ‘Rs. y’.
According to the question, the algebraic representation is
2x + y = 160
And 4x + 2y = 300

For, 2x + y = 160 or y = 160 − 2x, the solution table is;

x	50	60	70
y	60	40	20

For 4x + 2y = 300 or y = (300 – 4x)/ 2, the solution table is;

x	70	80	75
y	10	-10	0

Note: Students can also represent these two equations graphically, by using the given points of x-coordinate and y-coordinate.

Q.2: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1)
Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36
2x + 4 = 36
2x = 32
x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Q.3: On comparing the ratios a1/a2, b1/b2, and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

Solution:

(i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0
and 2x – 3y = 7 or 2x – 3y – 7 = 0

Comparing the above equations with a1x + b1y + c1=0
And a2x + b2y + c2 = 0
We get,
a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7

a1/a2 = 3/2, b1/b2 = 2/-3, c1/c2 = -5/-7 = 5/7
Since, a1/a2≠b1/b2 the lines intersect each other at a point and have only one possible solution.

Hence, the equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9

a1/a2 = 2/4 = 1/2, b1/b2 = -3/-6 = 1/2, c1/c2 = -8/-9 = 8/9

Since, a1/a2=b1/b2≠c1/c2

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Q.4: Solve the following pair of linear equations by the substitution method.

(i) x + y = 14
x – y = 4

(ii) 3x – y = 3
9x – 3y = 9

Solution:

(i) Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, put the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y)/3
Now, substitute the value of x in the given second equation to get,
9[(3 + y)/3] – 3y = 9

⇒ 3(3+y) – 3y = 9
⇒ 9 + 3y – 3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

Q.5: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(i)
2x – 4y = -24………………………… (ii)
From equation (ii), we get;
x = (11 – 3y)/2 ……….…………………………..(iii)

Putting the value of x in equation (ii), we get

2[(11 – 3y)/2] – 4y = −24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(iv)
Putting the value of y in equation (iii), we get;

x = (11 – 15)/2 = -4/2 = −2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.

Q.6: The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.
Solution:
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (i)
3x + 5y = 1750 ………………. (ii)
From (i), we get;
y = (3800 – 7x)/6 …………………… (iii)
Substituting (iii) in (ii). we get,
3x + 5[(3800 – 7x)/6] = 1750
⇒3x + (9500/3) – (35x/6) = 1750
3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500
x = 500
Putting the value of x in (iii), we get;

y = (3800 – 7 × 500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

Q.7: A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:
Let the fraction be x/y.

According to the question,
(x + 2)/(y + 2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)
(x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2), we get
6[(-4 + 9y)/11] – 5y = -3

-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get
x = (-4 + 81)/11 = 77/11 = 7

Hence, the fraction is 7/9.

Q.8 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution:
Let us assume, the present age of Nuri is x.
And the present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,
x – 3(20) = -10
x – 60 = -10
x = 50
Therefore,
Age of Nuri is 50 years
Age of Sonu is 20 years.

(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs. 15.
And the Additional charge per day is Rs. 3.

Q.9: Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4

Solution:

8x + 5y = 9 …………………..(1)
3x + 2y = 4 ……………….….(2)
From equation (2) we get;

x = (4 – 2y) / 3 ……………………. (3)
Substituting this value in equation 1, we get
8[(4 – 2y)/3] + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ……………………………….(4)
Substituting this value in equation (2), we get
3x + 10 = 4

3x = -6
x = -2
Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:
8x +5y – 9 = 0
3x + 2y – 4 = 0

x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15)
-x/2 = y/5 = 1/1
∴ x = -2 and y =5.

Q.10: Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

(i) Let us consider,
Speed of boat is still water = x km/hr
Speed of current = y km/hr
Now, speed of Ritu, during,
Downstream = x + y km/hr
Upstream = x – y km/hr
As per the question given,
2(x + y) = 20
Or x + y = 10……………………….(1)
And, 2(x – y) = 4
Or x – y = 2………………………(2)
Adding both the eq.1 and 2, we get,
2x = 12
x = 6
Putting the value of x in eq.1, we get,
y = 4
Therefore,

Speed of Ritu is still water = 6 km/hr
Speed of current = 4 km/hr

I hope this helps you in excelling in your academics. Please check out the Level 2 PYQs of this chapter. If you have any queries, please do let me know in the comments section.


Thanking You.