CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables MCQs Set 22

Practice MCQs for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables

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Multiple Choice Questions

Question 1: The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages (in years) of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Show Answer & Explanation

Answer: (c) 6 and 36

Explanation:
1. Let son = s, father = 6s: 6s + 4 = 4(s + 4).
2. 2s = 12, so s = 6 and father = 36.

Teacher's Note:
1. Add 4 to both ages for the future condition.
2. Check: 40 = 4 × 10.

Question 2: A purse contains 25 paise and 10 paise coins. The total amount in the purse is Rs. 8.25. If the number of 25 paise coins is one-third of the number of 10 paise coins, then the total number of coins in the purse is
(a) 60
(b) 40
(c) 80
(d) 72
Show Answer & Explanation

Answer: (a) 60

Explanation:
1. Let 10 paise coins = b and 25 paise coins = b/3: 25(b/3) + 10b = 825.
2. (55/3)b = 825, so b = 45 and 25 paise coins = 15.
3. Total = 45 + 15 = 60.

Teacher's Note:
1. Work in paise to avoid decimals.
2. Check: 15 × 25 + 45 × 10 = 375 + 450 = 825.

Question 3: A man has some hens and cows. If the number of heads is 48 and the number of feet equals 140, the number of hens will be
(a) 18
(b) 26
(c) 32
(d) 40
Show Answer & Explanation

Answer: (b) 26

Explanation:
1. Hens h and cows c: h + c = 48 and 2h + 4c = 140.
2. From the first, 2h + 2c = 96; subtract: 2c = 44, so c = 22 and h = 26.

Teacher's Note:
1. Each animal has one head; hens have 2 feet, cows 4.
2. Check: 52 + 88 = 140.

Question 4: If 3 chairs and 1 table cost Rs. 1500 and 6 chairs and 1 table cost Rs. 2400, the pair of linear equations that represents this situation is (x = cost of a chair, y = cost of a table)
(a) 6x + y = 1500, 3x + y = 2400
(b) x/3 + y = 1500, x/6 + y = 2400
(c) 3x + y = 1500, 6x + y = 2400
(d) none of these
Show Answer & Explanation

Answer: (c) 3x + y = 1500, 6x + y = 2400

Explanation:
1. 3 chairs and 1 table: 3x + y = 1500.
2. 6 chairs and 1 table: 6x + y = 2400.

Teacher's Note:
1. Multiply the unit cost by the number of items.
2. (Solving gives chair Rs. 300, table Rs. 600.)

Question 5: The difference between two numbers is 26 and the larger number exceeds thrice the smaller number by 4. The numbers are
(a) 39, 13
(b) 12, 38
(c) 37, 11
(d) none of these
Show Answer & Explanation

Answer: (c) 37, 11

Explanation:
1. L − S = 26 and L = 3S + 4.
2. 3S + 4 − S = 26, so S = 11 and L = 37.

Teacher's Note:
1. Translate each sentence into an equation.
2. Check: 3 × 11 + 4 = 37.

Question 6: Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. The number of Rs. 50 and Rs. 100 notes she received is
(a) Rs. 50 : 10, Rs. 100 : 15
(b) Rs. 50 : 12, Rs. 100 : 10
(c) Rs. 50 : 15, Rs. 100 : 10
(d) none of these
Show Answer & Explanation

Answer: (a) Rs. 50 : 10, Rs. 100 : 15

Explanation:
1. a + b = 25 and 50a + 100b = 2000.
2. From the first, a = 25 − b: 1250 + 50b = 2000, so b = 15 and a = 10.

Teacher's Note:
1. One equation counts notes, the other counts money.
2. Check: 500 + 1500 = 2000.

Question 7: The two consecutive odd positive integers, the sum of whose squares is 290, are
(a) 5, 13
(b) 11, 13
(c) 13, 17
(d) none of these
Show Answer & Explanation

Answer: (b) 11, 13

Explanation:
1. Let the numbers be x and x + 2: x² + (x + 2)² = 290.
2. 2x² + 4x − 286 = 0 gives x = 11, so the numbers are 11, 13.

Teacher's Note:
1. Consecutive odd numbers differ by 2.
2. Check: 121 + 169 = 290.

Question 8: Jamila sold a table and a chair for Rs. 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair, she would have got Rs. 1065. The cost price of each is
(a) Table: Rs. 500, chair: Rs. 400
(b) Table: Rs. 400, chair: Rs. 500
(c) Table: Rs. 700, chair: Rs. 600
(d) none of these
Show Answer & Explanation

Answer: (a) Table: Rs. 500, chair: Rs. 400

Explanation:
1. Let the cost prices be T and C: 1.10T + 1.25C = 1050 and 1.25T + 1.10C = 1065.
2. Solving, T = 500 and C = 400.

Teacher's Note:
1. Selling price = cost × (1 + profit%).
2. Check: 550 + 500 = 1050 and 625 + 440 = 1065.

Question 9: A part of the monthly hostel charge is fixed and the remaining depends on the number of days one has taken food in the mess. When Swati takes food for 20 days she has to pay Rs. 3000, whereas Mansi who takes food for 25 days pays Rs. 3500. The fixed charge and the cost of food per day are, respectively
(a) Rs. 400, Rs. 300
(b) Rs. 300, Rs. 200
(c) Rs. 1000, Rs. 100
(d) Rs. 500, Rs. 100
Show Answer & Explanation

Answer: (c) Rs. 1000, Rs. 100

Explanation:
1. F + 20r = 3000 and F + 25r = 3500.
2. Subtract: 5r = 500, so r = 100 and F = 1000.

Teacher's Note:
1. Fixed charge plus per-day cost times days.
2. Check: 1000 + 2500 = 3500.

Question 10: Children were fallen in for drill. If each row contained 4 children less, 10 more rows would have been made. But if 5 more children were fallen in each row, the number of rows would have reduced by 5. The number of children in the school is
(a) 150
(b) 160
(c) 170
(d) 120
Show Answer & Explanation

Answer: (a) 150

Explanation:
1. Let r rows and c children per row: (r + 10)(c − 4) = rc and (r − 5)(c + 5) = rc.
2. These simplify to 10c − 4r = 40 and r − c = 5, giving r = 15 and c = 10.
3. Total children = 15 × 10 = 150.

Teacher's Note:
1. Both conditions keep the total the same.
2. Expand and cancel the rc term.

Question 11: A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5, or by multiplying the difference of the digits by 16 and adding 3. The number is
(a) 23
(b) 34
(c) 83
(d) 119
Show Answer & Explanation

Answer: (c) 83

Explanation:
1. Let the digits be a and b: 10a + b = 8(a + b) − 5 and 10a + b = 16(a − b) + 3.
2. These give 2a − 7b = −5 and 6a − 17b = −3, so a = 8 and b = 3.
3. The number is 83.

Teacher's Note:
1. Write the number as 10a + b.
2. Check: 8 × 11 − 5 = 83 and 16 × 5 + 3 = 83.

Question 12: The sum of a two-digit number and the number obtained by interchanging its digits is 132. If the two digits differ by 2, the number is
(a) 45
(b) 75
(c) 85
(d) 115
Show Answer & Explanation

Answer: (b) 75

Explanation:
1. Sum of the two numbers = 11(a + b) = 132, so a + b = 12.
2. With a difference of 2 the digits are 7 and 5; among the options only 75 fits (57 is not offered).

Teacher's Note:
1. Original plus reversed is 11 times the digit sum.
2. Check: 75 + 57 = 132.

Question 13: Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. The present ages of father and son, respectively, are
(a) 40 years, 12 years
(b) 30 years, 6 years
(c) 32 years, 8 years
(d) 42 years, 10 years
Show Answer & Explanation

Answer: (d) 42 years, 10 years

Explanation:
1. F − 2 = 5(S − 2) and F + 2 = 3(S + 2) + 8.
2. The equations are F = 5S − 8 and F = 3S + 12; subtracting gives 2S = 20, so S = 10 and F = 42.

Teacher's Note:
1. Handle past and future conditions separately.
2. Check: 40 = 5 × 8 and 44 = 3 × 12 + 8.

Question 14: A boat travels for 7 hours. If it travels 4 hours downstream and 3 hours upstream, it covers 116 km. But if it travels 3 hours downstream and 4 hours upstream, it covers 108 km. The speed of the boat in still water is
(a) 16 km/h
(b) 22 km/h
(c) 18 km/h
(d) none of these
Show Answer & Explanation

Answer: (a) 16 km/h

Explanation:
1. With downstream speed d and upstream speed u: 4d + 3u = 116 and 3d + 4u = 108.
2. Solving, d = 20 and u = 12.
3. Speed in still water = (20 + 12) ÷ 2 = 16 km/h.

Teacher's Note:
1. Still-water speed is the average of downstream and upstream speeds.
2. Check: 80 + 36 = 116 and 60 + 48 = 108.

Free study material for Mathematics

Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables

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FAQs

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