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Test Level 2: Clocks & Calendars - 2 (August 21) - CAT MCQ


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10 Questions MCQ Test - Test Level 2: Clocks & Calendars - 2 (August 21)

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Test Level 2: Clocks & Calendars - 2 (August 21) - Question 1

At what time between 5 and 6 o'clock will the hands of a clock be 3 minutes apart?  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 1

In this type of problem, the formula is"

Here, H is replaced by the first interval of given time and t is spaces apart.
Given that H is 5.

 mins or 24 mins
Therefore, the hands will by 3 minutes apart at 336/11 minutes or 24 minutes past 5.

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 2

In the year 1648, if February had 5 Sundays, then what was the day on February 13, 1750?  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 2

Since February had 5 Sundays in 1648, so February 1, 1648 was a Sunday.
So, number of odd days up to February 1, 1748 = 5
Number of odd days up to February 1, 1750 = (2 + 1) = 3
Number of odd days up to February 13, 1750 = 12
So, total number of odd days = 5 + 3 + 12 = 20
So, number of odd days from February 1, 1648 to February 13, 1750 = 6
Hence, the answer is 'Saturday'.

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Test Level 2: Clocks & Calendars - 2 (August 21) - Question 3

At what time between 4 o'clock and 5 o'clock will the hands of a clock be at a right angle?

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 3

Using formula, [where θ = angle between the hour-hand and the minute-hand, m - minutes, h - hours]

There will be 2 values.
 minutes past 4 or m = minutes past 4.

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 4

At what time between 4:15 a.m. and 5:05 a.m. will the angle between the hour hand and the minute hand of a clock be the same as the angle between the hands at 8:45 p.m.?

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 4

Hour hand rotates 360° in 12 hrs, so it rotates 30° in 1 hour.
So, in 1 min, it rotates 0.5°.
Minute hand rotates 6° in 1 min.
So, angle at 8:45 = 0.25 × 30° = 7.5°
At 4:15, angle will be 0.25 × 30° + 30° = 7.5° + 30° = 37.5°
So, 37.5° + 0.5t + 7.5° = 6t
45 = 5.5t

Hence clock makes same angle at 23minutes past 4 o'clock

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 5

Alex turned a clock on at 3:00 pm. But the clock is defective, due to which it lags behind by 9 minutes after each day (24 hours). What will be the real time when the clock indicates 6:00 am on the 4th day of it's successive working?  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 5

Time from 3 pm on a day to 6 am after 4 days is 87 hours. Now, 23 hr 51 min on this clock are the same as 24 hr on the correct clock.
That is, on this clock = 24 hr on the correct clock
∴ 87 hr on this clock = hr on the correct clock
= 87 hr and 33 minutes approx. on the correct clock
So, the correct time will be 6:33 am.

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 6

If 15th March, 2013 was a Friday, then 10th July, 2013 will be a  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 6

Number of days from March 15th, 2013 to July 10th, 2013 = 16 + 30 + 31 + 30 + 10 = 117
On dividing 117 by 7, remainder is 5.
Now, counting 5 days after Friday, we get Wednesday.

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 7

At what time between 7 o'clock and 8 o'clock will the hands of a clock be in a straight line but not together?

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 7

Using formula: 

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 8

How many days will there be from 23rd January, 2011 to 31st July, 2013 (both days included)?  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 8

As 2012 was a leap year, so the number of days in each month from 23rd January, 2011 to 31st July, 2013 is given as follows.
Since 2011 is not a leap year = 365 - 22 = 343 days
Number of days in 2012 = 366 days (leap year)
Number of days in 2013:
January = 31 days
February = 28 days
March = 31 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Total number of days in 2013 = 212
Therefore, total number of days from 23rd January, 2011 to 31st July, 2013 = 343 + 366 + 212 = 921

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 9

Two clocks are set correctly at 10 a.m. on Friday. The first clock gains 2 minutes per hour, which is twice as much as gained by the second clock. What time will the second clock register when the correct time is 2 p.m. on the following Monday?  

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 9

The time duration from 10 a.m. on Friday to 2 p.m. on the following Monday is 76 hours.
The clock gains 1 minute per hour.
∴ Time gained in 76 hours = 76 minutes = 1 hour 16 minutes
∴ Time shown by the second clock will be 3:16 p.m.

Test Level 2: Clocks & Calendars - 2 (August 21) - Question 10

A man went outside between 7 o'clock and 9 o'clock at such a time that the minute hand and the hour hand were found to be coinciding before 8 o'clock; and when he returned, again he found both the hands to be coinciding, but after 8 o'clock. What was the time when he returned to the house?

Detailed Solution for Test Level 2: Clocks & Calendars - 2 (August 21) - Question 10

He returned after 8 o'clock but before 9 o'clock, at such a time when both the minute hand and the hour hand were found to be coinciding.
In a simple way, we need to calculate the time when the angle between the minute hand and the hour hand is zero.
Angle between the minute hand and the hour hand at 8 o'clock = 8 × 30 = 240°
We can reduce the difference of 5.5 degree in 1 min,
Required time = 8 hrs + 240 × 2 ÷ 11

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