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Test Level 3: Clocks & Calendars - CAT MCQ


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Test Level 3: Clocks & Calendars - Question 1

A clock is set right at 5:00 a.m. and gains 16 minutes in 24 hours. What will be the right time when the clock indicates 10:00 p.m. on the fifth day?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 1

From 5:00 a.m. today to 10:00 p.m. on the 5th day, there are a total of (24 × 4 + 17) hours.
So, the clock has gained 4 x 16 minutes + ((17/24) x 16) minutes = 75.333 minutes = 75 minutes (approx.)
Thus, the clock has gained 1 hour 15 minutes.
Hence, if the clock is displaying 10:00 p.m, then the correct time must be 1 hour 15 minutes before 10:00 p.m. i.e. 8:45 p.m.

Test Level 3: Clocks & Calendars - Question 2

What was the day on 26th January, 1950, when the first Republic Day of India was celebrated?  

Detailed Solution for Test Level 3: Clocks & Calendars - Question 2

As there is 1 odd day in 1900 years
In 49 years from 1900 to 1949, there are 49 ÷ 4 = 12 leap years and 37 ordinary years. There are (12 × 2 = 24) + 37 = 61 odd days or 5 odd days
From 1st January, 1950 till 26th January, 1950, there are 5 odd days
So, in total there are 0 + 1 + 5 + 5 = 11 odd days or 4 odd days
Hence, From 1st January, 1600 till 26th January, 1950, there are 4 odd days
So 26th January 1950 is Thursday.

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Test Level 3: Clocks & Calendars - Question 3

The price of petrol (in rupees per ltr) is 100 + 0.1x on the xth day of 2007 (x = 1, 2, 3, ...., 100) and then remains constant. The price of diesel (in rupees per ltr) is 89 + 0.15x on the xth day of 2007 (x = 1, 2, 3, ..., 365). On which of the following days in 2007 will the price of these two be equal?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 3

The price of petrol (in rupees per ltr) is 100 + 0.1x on the xth day of 2007 (x = 1, 2, 3, ...., 100) and then remains constant.
Hence, price of petrol remains constant after the 100th day (x = 100).
After 100 days, its price = 100 + 0.1 × 100 = 110
The price of diesel (in rupees per ltr) is 89 + 0.15x on the xth day of 2007 (x = 1, 2, 3, ..., 365).
After x days, the price of both will be equal.
89 + 0.15x = 110
$%5CRightarrow$ x = 140 days
After 140 days, the price of both will be equal.
Since 2007 is not a leap year.
January + February + March + April + May = 31 + 28 + 31 + 30 + 20
Therefore, the prices of both will be equal on 20th May.

Test Level 3: Clocks & Calendars - Question 4

There are clocks A and B. The hands of clock A move normally as clockwise, while in clock B (due to reverse connection), they move anticlockwise. Initially, the two hands of both clocks are at mark showing 12. If after some time, the angle between the directions of two hour hands is 90º (for the first time), then at the same instant, the angle between the directions of minute hand will be

Detailed Solution for Test Level 3: Clocks & Calendars - Question 4

The angle between the two hour hands after some time is 90°.
As they both start at 12 o'clock, the hour hand of clock A makes 45° and the hour hand of clock B makes 45° from their initial position.
The hour hand of clock A moved 45° from its initial position. Therefore, number of minutes the hour hand makes is 45 × 2 = 90 minutes
In 90 minutes, the minute hand makes an angle of 90 × 6° = 540°
360° is a complete revolution ⇒ 540° = 360° + 180° from its initial position.
As both hands move 180° in either direction, the minute hands of two clocks coincide i.e. the angle between them is zero degree.

Test Level 3: Clocks & Calendars - Question 5

For a project, a few computer science students form a virtual planet called SRAM. They introduce a new calendar and time period structure where there are 8 days per week, starting Sunday to Saturday and another day called Funday. Every day is of 36 hours and each hour is of 90 minutes. Each minute is of 60 seconds. Hour hand covers the dial twice daily. What is the angle between the hour and minute hand when time is 12.18 am?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 5

Hour hand:
1 day = 36 hours and the hour hand covers the dial twice every day
So, the dial covers 36/2 = 18 hours in one complete round
One complete angle in the clock dial is 360°.
So, in 1 hour, hour hand moves 360/18 = 20°
So, for 12 hours, hand moves 12 × 20 = 240°
Also, 1 hour = 90 minutes
Now, for 1 minute, hour hand moves (20/90)°
Thus, for 18 minutes, it moves 20/90 × 18 = 4°
So, for 12:18 am, hour hand moves = 240 + 4 = 244°
Minute hand:
Now, minute hand moves 90 minutes to complete one hour.
Thus, for 1 minute, it moves 360/90 = 4°
For 18 minutes, it moves 4 × 18 = 72°
So, the angle between hour hand and minute hand at 12.18 am is 244 - 72 = 172°

Test Level 3: Clocks & Calendars - Question 6

In an year N, the 320th day of the year is a Thursday. In the year N + 1, the 206th day of the year is also a Thursday. What is the 168th day of the year N - 1?  

Detailed Solution for Test Level 3: Clocks & Calendars - Question 6

As 320th day of the Nth year and 206th day of the N + 1th year are same days i.e. Thursday, there must be 0 odd days between these 2 days (The number of days between these 2 days must be divisible by 7).
If Nth year is a non-leap year, the number of days between 2 days = (365 - 320) + 206 = 251 days
Here, number of odd days in 251 days is 6.
So, we can say that it is not an ordinary year.
If Nth year is a leap year, then number of days between 2 days
⇒ (366 - 320) + 206 = 252 days
Number of odd days in 252 days is 0.
So, we can say that Nth year is a leap year.
To find the 168th day of the year N - 1, we have to find the number of odd days between 320th day of Nth year and 168th day of N - 1th year.
⇒ (365 - 168) + 320
⇒ 517 days
The number of odd days in 517 ⇒ 6 odd days
The 168th day of N - 1 year is a Thursday - 6 = Friday

Test Level 3: Clocks & Calendars - Question 7

Ram was born on Sunday, March 20, 1992. On what day of the week was he 5 years, 2 months and 16 days old?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 7

After 5 years, the year would have been 1997 and there is a leap year (1996). So, the total days after 5 years = 4 × 365 + 366 = 1826. April has 30 days and May has 31 days. Hence, total days after 5 years 2 months and 16 days are 1903. Upon dividing by 7, remainder is 6. So, the required day is 6 days after Sunday. Thus, the correct answer is Saturday.

Test Level 3: Clocks & Calendars - Question 8

The pendulum of a clock takes 7 seconds to strike 4 o'clock. How much time will it take to strike 11 o'clock?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 8

 Please note that the more relevant thing in this case is not the number of strikes, but the number of time intervals. In other words, if a clock has to strike 4, there are three time intervals between the 4 strikes (this is so taken because the first strike happens at the 0th second). So in 7 seconds, the pendulum elapses three time intervals. To strike 11, there has to be 10 time intervals, which will take = 23.33 seconds.

Test Level 3: Clocks & Calendars - Question 9

One watch loses 3 minutes and another gains 4 minutes daily. They were set right at 3:00 p.m. What time will the slower watch indicate the next day when the faster watch shows 9:00 p.m?

Detailed Solution for Test Level 3: Clocks & Calendars - Question 9

From 3:00 p.m. today to 9:00 p.m. the next day, there are a total of 30 hours of faster clock.
In a day (i.e. in 24 × 60 = 1440 minutes), the slow watch will show 1437 minutes and the faster will show 1444 minutes.
The faster is showing 1444 minutes and the slower will show 1437 minutes.
If the faster is showing 30 hours, the slower will show 
Thus, the slow clock will show = 8 

Test Level 3: Clocks & Calendars - Question 10

An antique grandfather's clock, which gains uniformly, is 10 minutes slow at 9:00 a.m. in the morning on Monday and 10 minutes 48 seconds fast at 9:00 p.m. on the following Monday. When was it correct?  

Detailed Solution for Test Level 3: Clocks & Calendars - Question 10

From 9:00 a.m. Monday to 9:00 p.m. on the following Monday, total hours = 24  7 + 12 = 180 hours
The clock has gained in 180 hours
minutes in 180 hours
To gainminutes, the clock needs 180 hours.
To gain 10 minutes, the clock needs

= 3 days 14.53 hours
= 3 days 14 hours 32 minutes (approx.)
Thus, the clock will be correct after 3 days 14 hours 32 minutes from Monday morning.
i.e. 11:32 p.m. Thursday

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