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Word problems on pairs of equations · 5 marks

Form a pair of linear equations for each of the following problems and find their solutions.

  1. (i) The sum of two integers is +5 and their difference is −21. Find the two numbers.
  2. (ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
  3. (iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3 bats and 5 balls for ₹4000. Find the cost of each bat and each ball.
  4. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹155 and for a journey of 15 km, the total amount paid is ₹220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
  5. (v) A fraction becomes equal to 911 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to 56. Find the fraction.
  6. (vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to 12 if we add 1 only to the denominator. What is the fraction?
  7. (vii) 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?
  8. (viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  9. (ix) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 did she receive?
  10. (x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer: (i) −8 and 13 (ii) 39 and 13 (iii) bat ₹1200, ball ₹80 (iv) fixed ₹25, ₹13 per km; 25 km costs ₹350 (v) 79 (vi) 35 (vii) Nuri 50 years, Sonu 20 years (viii) 18 (ix) 10 notes of ₹50 and 15 notes of ₹100 (x) fixed charge ₹15, ₹3 per extra day.

Step-by-step solution

Idea: For each problem: name the two unknowns, turn each sentence into one equation, then eliminate a variable (or substitute) and check the answer in the original words.

(i) The sum of two integers is +5 and their difference is −21. Find the two numbers.

  1. Let the integers be x and y: x + y = 5 … (1), x − y = −21 … (2).
  2. (1) + (2): 2x = −16 ⇒ x = −8; y = 5 − (−8) = 13. The numbers are −8 and 13.½ mark
−8 and 13

(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.

  1. Let the larger be x and the smaller y: x − y = 26 … (1), x = 3y … (2).
  2. Substitute (2) in (1): 3y − y = 26 ⇒ y = 13, x = 39. The numbers are 39 and 13.½ mark
39 and 13

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3 bats and 5 balls for ₹4000. Find the cost of each bat and each ball.

  1. Bat ₹x, ball ₹y: 7x + 6y = 8880 … (1), 3x + 5y = 4000 … (2).
  2. (1) × 5 − (2) × 6: 35x − 18x = 44400 − 24000 ⇒ 17x = 20400 ⇒ x = 1200. Then 5y = 4000 − 3600 = 400 ⇒ y = 80. Bat ₹1200, ball ₹80.½ mark
Bat ₹1200, ball ₹80

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹155 and for a journey of 15 km, the total amount paid is ₹220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

  1. Fixed charge ₹x, ₹y per km: x + 10y = 155 … (1), x + 15y = 220 … (2).
  2. (2) − (1): 5y = 65 ⇒ y = 13; x = 155 − 130 = 25. For 25 km: 25 + 25 × 13 = 25 + 325 = ₹350. Fixed ₹25, ₹13 per km, ₹350 for 25 km.½ mark
Fixed charge ₹25, ₹13 per km; ₹350 for 25 km

(v) A fraction becomes equal to 911 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to 56. Find the fraction.

  1. Let the fraction be xy. x + 2y + 2 = 911 ⇒ 11x + 22 = 9y + 18 ⇒ 11x − 9y = −4 … (1). x + 3y + 3 = 56 ⇒ 6x + 18 = 5y + 15 ⇒ 6x − 5y = −3 … (2).
  2. (1) × 5 − (2) × 9: 55x − 54x = −20 + 27 ⇒ x = 7; then 9y = 77 + 4 = 81 ⇒ y = 9. The fraction is 79.½ mark
79

(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to 12 if we add 1 only to the denominator. What is the fraction?

  1. Let the fraction be xy. x + 1y − 1 = 1 ⇒ x − y = −2 … (1). xy + 1 = 12 ⇒ 2x − y = 1 … (2).
  2. (2) − (1): x = 3; y = 3 + 2 = 5. The fraction is 35.½ mark
35

(vii) 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?

  1. Present ages: Nuri x, Sonu y. x − 5 = 3(y − 5) ⇒ x − 3y = −10 … (1). x + 10 = 2(y + 10) ⇒ x − 2y = 10 … (2).
  2. (2) − (1): y = 20; x = 10 + 40 = 50. Nuri is 50 years and Sonu 20 years old.½ mark
Nuri 50 years, Sonu 20 years

(viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

  1. Tens digit x, units digit y: x + y = 9 … (1). 9(10x + y) = 2(10y + x) ⇒ 88x = 11y ⇒ y = 8x … (2).
  2. Substitute (2) in (1): 9x = 9 ⇒ x = 1, y = 8. The number is 18.½ mark
18

(ix) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 did she receive?

  1. x notes of ₹50 and y notes of ₹100: x + y = 25 … (1), 50x + 100y = 2000 ⇒ x + 2y = 40 … (2).
  2. (2) − (1): y = 15; x = 10. 10 notes of ₹50 and 15 notes of ₹100.½ mark
10 notes of ₹50, 15 notes of ₹100

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

  1. Fixed charge ₹x (first 3 days), ₹y per extra day. 7 days = 3 + 4 extra: x + 4y = 27 … (1). 5 days = 3 + 2 extra: x + 2y = 21 … (2).
  2. (1) − (2): 2y = 6 ⇒ y = 3; x = 21 − 6 = 15. Fixed charge ₹15, ₹3 for each extra day.½ mark
Fixed charge ₹15; ₹3 per extra day
(i) −8, 13 (ii) 39, 13 (iii) bat ₹1200, ball ₹80 (iv) fixed ₹25, ₹13/km, ₹350 for 25 km (v) 7/9 (vi) 3/5 (vii) Nuri 50, Sonu 20 (viii) 18 (ix) 10 × ₹50, 15 × ₹100 (x) fixed ₹15, ₹3 per extra day.

Check: (iii) 7(1200) + 6(80) = 8400 + 480 = 8880 ✓. (v) 911 ✓ and 1012 = 56 ✓. (vii) 5 years ago 45 = 3 × 15 ✓; in 10 years 60 = 2 × 30 ✓. (viii) 9 × 18 = 162 = 2 × 81 ✓. (x) 15 + 4 × 3 = 27 ✓.

Answer to write in the exam

(i)

x + y = 5 … (1); x − y = −21 … (2)

(1) + (2): 2x = −16 ⇒ x = −8

y = 13

∴ The integers are −8 and 13

(ii)

x − y = 26 … (1); x = 3y … (2)

3y − y = 26 ⇒ y = 13

x = 39

∴ The numbers are 39 and 13

(iii)

7x + 6y = 8880 … (1); 3x + 5y = 4000 … (2)

(1) × 5: 35x + 30y = 44400; (2) × 6: 18x + 30y = 24000

Subtract: 17x = 20400 ⇒ x = 1200

3(1200) + 5y = 4000 ⇒ y = 80

∴ Bat = ₹1200, ball = ₹80

(iv)

x + 10y = 155 … (1); x + 15y = 220 … (2)

(2) − (1): 5y = 65 ⇒ y = 13

x = 155 − 130 = 25

Fare for 25 km = 25 + 25 × 13 = 350

∴ Fixed charge ₹25, ₹13 per km; ₹350 for 25 km

(v)

Let the fraction = xy

x + 2y + 2 = 911 ⇒ 11x − 9y = −4 … (1)

x + 3y + 3 = 56 ⇒ 6x − 5y = −3 … (2)

(1) × 5 − (2) × 9: x = 7

11(7) − 9y = −4 ⇒ y = 9

∴ Fraction = 79

(vi)

Let the fraction = xy

x + 1y − 1 = 1 ⇒ x − y = −2 … (1)

xy + 1 = 12 ⇒ 2x − y = 1 … (2)

(2) − (1): x = 3; y = 5

∴ Fraction = 35

(vii)

Let Nuri = x years, Sonu = y years (now)

x − 5 = 3(y − 5) ⇒ x − 3y = −10 … (1)

x + 10 = 2(y + 10) ⇒ x − 2y = 10 … (2)

(2) − (1): y = 20; x = 50

∴ Nuri is 50 years old and Sonu is 20 years old

(viii)

Let number = 10x + y

x + y = 9 … (1)

9(10x + y) = 2(10y + x) ⇒ 88x = 11y ⇒ y = 8x … (2)

(2) in (1): 9x = 9 ⇒ x = 1, y = 8

∴ The number is 18

(ix)

x + y = 25 … (1)

50x + 100y = 2000 ⇒ x + 2y = 40 … (2)

(2) − (1): y = 15; x = 10

∴ 10 notes of ₹50 and 15 notes of ₹100

(x)

x + 4y = 27 (7 days = 3 + 4 extra) … (1)

x + 2y = 21 (5 days = 3 + 2 extra) … (2)

(1) − (2): 2y = 6 ⇒ y = 3; x = 15

∴ Fixed charge ₹15, extra ₹3 per day

Common mistakes that cost marks

  • In (x), multiplying the daily charge by 7 days instead of by the 4 days after the first three.
  • In (vii), using present ages in both conditions; five years ago both ages were 5 less.
  • In (v) and (vi), adding the 2 or 3 to only the numerator, or cross-multiplying incorrectly.

How this can come in the exam

MCQ (1 mark)

A two-digit number is 4 times the sum of its digits and the digits differ by 3 (units digit larger). The number is

  1. 36
  2. 63
  3. 14
  4. 47
Show answer

(A) 36
10x + y = 4(x + y) ⇒ 2x = y; y − x = 3 ⇒ x = 3, y = 6: 36.

Case-based (4 marks)

A canteen sells veg rolls and paneer rolls. On Monday it sold 30 veg and 20 paneer rolls for ₹2300; on Tuesday 25 veg and 30 paneer rolls for ₹2850.
(i) Write the two equations. (ii) Find the price of each roll. (iii) What would 10 of each cost?

Show answer(i) 30x + 20y = 2300 ⇒ 3x + 2y = 230; 25x + 30y = 2850 ⇒ 5x + 6y = 570 (1 mark). (ii) 3 × first − second: 4x = 120 ⇒ x = 30, y = 70 (2 marks). (iii) 10 × 100 = ₹1000 (1 mark).

Try one yourself

The sum of two numbers is 1000 and their difference is 212. Find them.

Show answer

2x = 1212 ⇒ 606 and 394.

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