## CTS Sample Time Equation Problems

Below are three problems where one has to form equations with given data in units of minutes.

Question 1

Rahman was in a hurry to meet his friend Ismail who has come from abroad.He started from his home at 3 O’clock.When he looked at his watch he found that one hour ago it was four times as many minutes before eight o’clock . How many minutes is it until eight o’clock when Rahman looked at his watch?

a) 48 minutes b) 30 minutes c) 50 minutes d) none of these.

Solution :

Minutes between 3 o’clock and 8 o’ clock is 60 x 5 = 300 minutes
Let Rahman look at his watch x minutes before 8 o’clock.
Therefore, The time now is (300-x) minutes after 3 o’clock.
The time one hour ago would be (300 –x – 60) = 240-x minutes past 3 o’clock.
This is given to be 4 times as many minutes to 8 o' clock since he looked his watch.
Therefore, 240 - x = 4x
240 = 4x + x = 5x
So x = 48
Rahman would have looked at his watch 48 minutes before 8 o' clock which is 7.12.

Question 2

Rajesh Khanna had an appointment with his doctor for health check up that day. He started from his house at 1 pm after having his lunch. The clinic was quite far away from his residence and he was keen that he should reach it well in time since the doctor’s appointment was at 5 pm. After reaching the clinic he looked at the wall clock and learnt he had come early. How many minutes is it until seven o’ clock if one hour ago it was five times as many minutes to 5pm from his arrival time?

a) 20 minutes b) 150 minutes c) 40 minutes d) 15 minutes

Solution :

The total time between 1 pm and 5 pm is 240 minutes.
Let Rajesh Khanna arrived x minutes before 5 pm
Therefore the time now is 240-x minutes after 1 pm.
The time one hour ago was (240 –x – 60) = 180-x
As the time 60 minutes ago was 5 times as many minutes to 5 pm from his arrival time.
Then (180 - x) = 5x
180 -x = 5x
180 = 5x + x = 6x
So x = 30
He arrived 30 minutes before 5 pm.
But question is on the number of minutes to 7pm. Since it is 30 minutes to 5pm, it should be 2 x 60 + 30 = 150 minutes before 7pm

Question 3

Rajanikanth started one day from Mahabalipuram at 12 noon towards Chennai airport. He reached the airport a few minutes earlier to his plane's departure time of 3 pm. He found that forty minutes before that instance it was four times as many minutes before 3 pm. At what time Rajanikanth reached the airport?

a) 2..32 pm b) 2.12 pm c) 2.22 pm d) 2.42 pm

Solution :

Minutes between 12 noon and 3 pm = 180 minutes.
Let us assume he reaches airport x minutes earlier than 3 pm. This time should be (180-x) minutes past 12 noon.
It is given that forty minutes earlier to that time it was four times as many minutes past 12 noon.
Forty minutes earlier to (180-x) = 180-x - 40 = 140 - x
This was equal to 4 times as many minutes before 3 pm.
Therefore, 140 - x = 4x
140 = 4x + x = 5x
So x = 28
That means he has reached 28 minutes before 3 pm – 2.32 pm.

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