SSC CGL Exam  >  SSC CGL Tests  >  General Intelligence and Reasoning for SSC CGL  >  MCQ: Clock & Calendar - 2(Medium) - SSC CGL MCQ

MCQ: Clock & Calendar - 2(Medium) - SSC CGL MCQ


Test Description

15 Questions MCQ Test General Intelligence and Reasoning for SSC CGL - MCQ: Clock & Calendar - 2(Medium)

MCQ: Clock & Calendar - 2(Medium) for SSC CGL 2024 is part of General Intelligence and Reasoning for SSC CGL preparation. The MCQ: Clock & Calendar - 2(Medium) questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Clock & Calendar - 2(Medium) MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Clock & Calendar - 2(Medium) below.
Solutions of MCQ: Clock & Calendar - 2(Medium) questions in English are available as part of our General Intelligence and Reasoning for SSC CGL for SSC CGL & MCQ: Clock & Calendar - 2(Medium) solutions in Hindi for General Intelligence and Reasoning for SSC CGL course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt MCQ: Clock & Calendar - 2(Medium) | 15 questions in 15 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study General Intelligence and Reasoning for SSC CGL for SSC CGL Exam | Download free PDF with solutions
MCQ: Clock & Calendar - 2(Medium) - Question 1

Find the percentage change in angle between minute hand hour hand from 2 PM to 6 PM.

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 1

At 2 PM angle between minute hand and hour hand is 60° and that at 6 PM is 180° hence percentage change is 200%.

MCQ: Clock & Calendar - 2(Medium) - Question 2

At what time between 9 and 10 o'clock will the hands of a clock be in the same straight line but not together ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 2

9 :164/11

1 Crore+ students have signed up on EduRev. Have you? Download the App
MCQ: Clock & Calendar - 2(Medium) - Question 3

By how many degree does the minute hand move in the same time, in which the hour hand move by 28 ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 3

28 x 2 x 6 = 336 degree

MCQ: Clock & Calendar - 2(Medium) - Question 4

At what time between 1'O clock and 2'O clock the hands of the clock are opposite to each other.

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 4

The minute hand to coincide with the hour hand it should trace at first 5 minute spaces and then the hands of the clocks to be opposite to each other minute hand should trace 30 minute spaces i.e. totally it should gain 5 + 30 = 35 minute spaces to be opposite to that of hour hand.
We know that, Minute hand gains 55 minute spaces over hour hand in 1 hour. Therefore, Minute hand gain 40 minute spaces over hand in 35 x (60/55) = 38(2/11).
Hence the hand of the clock will minutes be opposite to each at 38 (2/11) past 1'O clock. Therefore, Correct option is B.

MCQ: Clock & Calendar - 2(Medium) - Question 5

How many times in a day (24 Hrs) are the hands of a clock are in a straight line ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 5

In 12 hours both hands are in straight line (either coincides or in opposite direction) 22 times. Hence in 24 hours both hands are in straight line 44 times.

MCQ: Clock & Calendar - 2(Medium) - Question 6

A watch, which gains uniformly, is 2 min, slow at noon on Saturday, and is 4 min 48 seconds fast at 3 p.m on the following Sunday when was it correct ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 6

From Sunday noon to the following Sunday at 2 p.m. there are 7 days 2 hours or 170 hours. The watch gains 2 + 44/5 min in 170 hrs. Therefor, the watch gains 2 min in 2 x 170 hrs i.e., 50 hours 64/5. Now 50 hours Sunday noon = 2 p.m on Tuesday.

MCQ: Clock & Calendar - 2(Medium) - Question 7

A watch which gains uniformly is 2 minutes slow at noon on Sunday and is 4 min. 48 sec fast at 2 p.m. on the following Sunday. When it has shown the correct time ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 7

Time from 12 p.m. on Sunday to 2 p.m. on the following Sunday = 7 days 2 hours.
= 24 x 7 + 2 = 170 hours.
The watch gain = (2 + 4 x 4/5) min = 34/5 minute in 170 hrs.
Since, 34/5 min are gained in 170 hrs.
2 min are gained in (170 x 5/34 x 2) hrs = 50 hours i.e., 2 days 2 hrs.
after 12 p.m. on Sunday i.e., it will be correct at 2 p.m. on Tuesday.

MCQ: Clock & Calendar - 2(Medium) - Question 8

A clock is set at 10 a.m. The clock loses 16 minute in 24 hours. What will be the true time when the clock indicates 3 a.m. on 4th day ?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 8

Time from 10 a.m on a day to 3 a.m on 4th day = 24 x 3 + 17 = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
89 hrs of faulty clock = (24 x 15/356 x 89) hrs = 90 hrs.
So, the correct time is 11 p.m

MCQ: Clock & Calendar - 2(Medium) - Question 9

Today is Sunday what day of the week was 79 days back.

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 9

When we divide 79 by 7 we will get remainder 2 so we have 2 odd days, so required day must be 2 days back from today (i.e Sunday) and that day should be Friday.

MCQ: Clock & Calendar - 2(Medium) - Question 10

Which of the following day of the week was 15th august 1947?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 10

From Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 15 (since 15th August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1st two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6 - 1 / 5] + 47 + [47/4] + [19/4] - 2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75] - 38.
f = 15 + 15 + 47 + 11 + 4 - 38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15th August 1947 is 5 days more than Sunday, i.e Friday.

MCQ: Clock & Calendar - 2(Medium) - Question 11

If March 5, 2012 was a Wednesday, what was the day on November 5, 2014?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 11

730 ÷ 7 ⇒ remainder = 2 days

So, 5 March 2012 (Wednesday) + 2 odd days = 5 March 2014

Wednesday + 2 odd days = Friday.

Hence, 5 March 2014 = Friday

5 March 2014 to 5 November 2014 = 30 + 31 + 30 + 31 + 31 + 30 + 31 + 5 + 26 = 245 days i.e. 0  odd days.

Hence, Friday + 0 = Friday

Thus, 5 November 2014 will be Friday.

Hence, 'Friday' is the correct answer.

MCQ: Clock & Calendar - 2(Medium) - Question 12

If 2nd June 2013 is Sunday then which day was on 2nd June 2010?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 12

Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2nd June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m - 1/ 5 ] + D + [D/4] + [C/4] - 2 x C.
In this case k = 2 (since 2nd June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4 - 1 / 5] + 10 + [10/ 4] + [ 20/4] - 2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5] - 40.
f = 2 + 10 + 10 + 2 + 5 - 40 = -11
This - ve value of f can be made positive by adding multiple of 7
So f = - 11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2nd June 2010 is 3 days more than Monday, i.e Wednesday.

MCQ: Clock & Calendar - 2(Medium) - Question 13

Which of the following day could be the 18th October 2050?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 13

From Zeller's Formula
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 18 (since 18th October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1st two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8 - 1 / 5] + 50 + [50/4] + [20/4] - 2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5] - 40.
f = 18 + 20 + 50 + 12 + 5 - 40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18th October 2050 is 2 days more than Sunday, i.e Tuesday.

MCQ: Clock & Calendar - 2(Medium) - Question 14

How many weekends are there in March 2009?

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 14

From Zeller's Formula we can find that 1st March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.

MCQ: Clock & Calendar - 2(Medium) - Question 15

Which of the following two months in a particular year will have same calendar.

Detailed Solution for MCQ: Clock & Calendar - 2(Medium) - Question 15

January and August or October depends on leap year or non leap year. But if we find the number of odd days between March and November we will get number of odd days is 0 hence they will have same calendar.

177 videos|126 docs|197 tests
Information about MCQ: Clock & Calendar - 2(Medium) Page
In this test you can find the Exam questions for MCQ: Clock & Calendar - 2(Medium) solved & explained in the simplest way possible. Besides giving Questions and answers for MCQ: Clock & Calendar - 2(Medium), EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next