Word problems on pairs of equations · 3 marks

At a certain time, Jacob notices that his digital watch reads ‘a’ minutes after two o’clock. Fifteen minutes later, it reads ‘b’ minutes after three o’ clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?

Answer: a − b = 45 and a = 6b give b = 9, a = 54. The second time was 9 minutes past 3 (3:09).

Step-by-step solution

Idea: Count minutes after 2:00. The first reading is a minutes after 2:00; the second is 60 + b minutes after 2:00, and it is 15 minutes later.

First reading: 2:a, i.e. a minutes after 2:00. Second: 3:b, i.e. 60 + b minutes after 2:00. Fifteen minutes later: a + 15 = 60 + b ⇒ a − b = 45 … (1).1 mark
“a is six times b“: a = 6b … (2).½ mark
(2) in (1): 6b − b = 45 ⇒ b = 9, a = 54.1 mark
So he first looked at 2:54 and the second time at 3:09.½ mark

3:09 (9 minutes past three); the first reading was 2:54.

Check: 2:54 + 15 minutes = 3:09 ✓, and 54 = 6 × 9 ✓.

a + 15 = 60 + b ⇒ a − b = 45 … (1)

a = 6b … (2)

(2) in (1): 5b = 45 ⇒ b = 9, a = 54

∴ Second time = 3:09

Writing b = a + 15, forgetting that the hour changed (from 2 to 3).
Reading “six times greater” as a = 7b; here it means a = 6b, which is the reading that gives whole minutes.

Short answer (3 marks)

A clock shows p minutes past 4. Twenty-five minutes later it shows q minutes past 5, and p = 8q. Find both times.

Show answer

p + 25 = 60 + q ⇒ p − q = 35 (1 mark). 8q − q = 35 ⇒ q = 5, p = 40 (1 mark). Times 4:40 and 5:05 (1 mark).

A watch reads x minutes past 1. Twenty minutes later it reads y minutes past 2, and x = 5y. Find the second time.

Show answer

x + 20 = 60 + y ⇒ x − y = 40; 5y − y = 40 ⇒ y = 10: 2:10 (first reading 1:50).