CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set C

Question. Solution of the simultaneous linear equations: 2x/y – y/2 = - 1/6 and x/2 + 2y/3 = 3 is
(a) x = 2, y = – 3
(b) x = – 2, y = 3
(c) x = 2, y = 3
(d) x = – 2, y = – 3
Answer : C

Question. The value of x satisfying both the equations 4x – 5 = y and 2x – y = 3, when y = –1 is
(a) 1
(b) –1
(c) 2
(d) –2
Answer : A

Question. Which of the following is not a solution of the pair of equations 3x – 2y = 4 and 6x – 4y = 8?
(a) x = 2, y = 1
(b) x = 4, y = 4
(c) x = 6, y = 7
(d) x = 5, y = 3
Answer : D

Answer the questions (Q 4 to Q 9) using best suitable algebric method.

Question. If 2x + 5y – 1 = 0, 2x + 3y – 3 = 0, then
(a) x = 1, y = – 3
(b) x = 3, y = –1
(c) x = 2, y = 5
(d) x = 5, y = – 3
Answer : B

Question. If x + 2y – 3 = 0, 3x – 2y + 7 = 0, then
(a) x = –1, y = 2
(b) x = 1, y = 2
(c) x = 2, y = 3
(d) x = – 2, y = – 3
Answer : A

Question. If 2x = 5y + 4, 3x – 2y + 16 = 0, then
(a) x = 2, y = –2
(b) x = 3, y = –3
(c) x = 4, y = 5
(d) x = – 8, y = – 4
Answer : D

Question. If 4/x + 3y = 8, 6/x - 4y = - 5, then
(a) x = 2, y = 2
(b) x = 1, y = –1
(c) x = 2, y = –2
(d) x = 3, y = – 3
Answer : A

Question. If 2x + 3y = 11 and 2x – 4y = –24, then the value of ‘m’ for which y = mx + 3 is
(a) 0
(b) 1
(c) –1
(d) – 2
Answer : C

Question. If 2x + 3y = 11 and x – 2y = –12, then the value of ‘m’ for which y = mx + 3 is
(a) 1
(b) –1
(c) 2
(d) – 2
Answer : B

Question. If 7(y + 3) – 2(x + 2) = 14, 4(y – 2) + 3(x – 3) = 2, then
(a) x = 1, y = 4
(b) x = 3, y = 5
(c) x = 5, y = 1
(d) None of these

Answer : C

Question. If 4/x + 5y = 7, 3/x + 4y = 5, then
(a) x = 1/3, y = –1
(b) x = 8, y = 3
(c) x = 4, y = 7
(d) x = 5, y = 9

Answer : A

Question. If 3 – (x – 5) = y + 2, 2(x + y) = 4 – 3y, then
(a) x = 13/4, y = 9/10
(b) x = 7/16, y = 5/8
(c) x = 4/9, y = 9/12
(d) x = 26/3, y = −8/3

Answer : D

DIRECTIONS : Study the given Case/Passage and answer the following questions.

Case/Passage-I

A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while 1/4 mark is deducted for every wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.

Question. If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?
Answer : No. of correct questions are 96

Question. How many questions did he guess?
Answer : He guessed 24 questions

Question. If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?
Answer : Marks = 80 – 1/4 of 40 = 70

Question. If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?
Answer : Here, x + y = 120                   ...(i)
x - (1/4) y = 95                   ...(ii)
On solving (i) & (ii) x = 100

Case/Passage-II

Amit is planning to buy a house and the layout is given below.
The design and the measurement has been made such that areas of two bedrooms and kitchen together is 95 sq.m.

Based on the above information, answer the following questions:

Question. Form the pair of linear equations in two variables from this situation.
Answer : Given area of two bedrooms and a kitchen is 95 sq m.
2 × Area of bedroom + Area of kitchen = 95
2 × 5 x + 5y = 95
or 2x + y = 19 ...(i)
and x + 2 + y = 15
or x + y = 13 ...(ii)

Question. Find the length of the outer boundary of the layout.
Answer : Length of outer boundary = 12 + 15 + 12 +15 = 54 m

Question. Find the area of each bedroom and kitchen in the layout.
Answer : On solving x + y = 13
2x + y = 19
x = 6m, y = 7m
Area of a bedroom = 5x = 5 × 6 = 30 sq m
Area of kitchen = 5y = 5 × 7 = 35 sq m

Question. Find the area of living room in the layout.
Answer : Area of living room = 9 × 5 + 2 × 15 = 75 sq m

Question. Find the cost of laying tiles in kitchen at the rate of ₹ 50 per sq.m
Answer : Total cost of laying tiles in the kitchen = ₹ 50 × 35 = ₹1750

Case/Passage-III

It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, rickshaws, taxis, radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations.

Situation 1: In city A, for a journey of 10 km, the charge paid is ₹ 75 and for a journey of 15 km, the charge paid is ₹ 110. 
Situation 2: In a city B, for a journey of 8 km, the charge paid is ₹ 91 and for a journey of 14km, the charge paid is ₹ 145.

Question. If the fixed charges of auto rickshaw be ₹ x and the running charges be ₹ y km/hr, the pair of linear equations representing the situation is
(a) x + 10y = 110, x + 15y = 75
(b) x + 10y = 75, x + 15y = 110
(c) 10x + y = 110, 15x + y = 75
(d) 10x + y = 75, 15x + y = 110

Answer : B

Question. A person travels a distance of 50km. The amount he has to pay is
(a) ₹ 155
(b) ₹ 255
(c) ₹ 355
(d) ₹ 455

Answer : C

Question. What will a person have to pay for travelling a distance of 30km?
(a) ₹ 185
(b) ₹ 289
(c) ₹ 275
(d) ₹ 305

Answer : B