Clocks-Calender-Age test

1.

Calculate the angle between the two hands of clock when the clock shows 5 : 25 p.m.

Answer: Option B

Explanation:

Since at 5 : 25 the minute hand will be at 5 and the angle between them will be same as the distance covered in degree by the hour hand in 25 minutes

Required angle = distance of hour hand = speed × time = 1/2*25 =12.5 degrees
2.

How many times in a day, are the hands of a clock in straight line but opposite in direction?

Answer: Option B

Explanation:

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clcok only). So, in a day, the hands point in the opposite directions 22 times.

3.

How many non-leap years are there from 1801 to 1900?

Answer: Option C

Explanation:

100 years consists  24 leap years and 76 non-leap years (100th year is not a leap year if it cannot be divided  with 400)

4.

Father is aged three times more than his son Raj. After 8 years, he would be two and a half times of Raj’s age. After further 8 years, how many times would he be of Raj’s age?

Answer: Option A

Explanation:

Let Raj’s present age be x years.
Then, father’s present age = (x + 3x) years = 4x years
∴ (4x +8) = 5/2(x + 8)
8x + 16 = 5x + 40
3x = 24 => x = 8
Hence, required ratio = (4x + 16)/(x +16) = 48/24 = 2

5.

When Rajeev was born, his father was 32 years older than his brother and his mother was 25 years older than his sister. If Rajeev’s brother is 6 years older than Rajeev and his mother is 3 years younger than his father, how old was Rajeev’s sister when Rajeev was born?

Answer: Option D

Explanation:

fathers age is 6+32 = 38 yrs old.
mother’s age is 38-3 = 35 years old,
so,sister’s age is  35-25= 10 year old

6.

The ages of the two persons differ by 20 years. If 5 year ago, the older one be 5 times as old as the younger one, then their present ages, in year are

Answer: Option B

Explanation:

Let the age be x and y years now.

Then, x – y =20……(i)
and (x-5) = 5(y-5)….(ii)
On solving both equation we get x= 30 and y =10.
7.

A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is

Answer: Option A

Explanation:

A hours + C hours = B hours … (i)

A, C and B cannot have values greater than or equal to 24 B minutes + A minutes = C minutes … (ii)
Looking at the two equations, we get that no value of A satisfies both the equations.
Hence, option A.
8.

A watch which gains 5 seconds in 3 minutes was set right at 6 a.m. In the afternoon of the same day, when the watch indicated quarter past 3 o’ clock, the true time is :

Answer: Option B

Explanation:

Error Clock :: Original Clock
3 min 5 sec :: 3 min
185 sec :: 180 sec
[9 hr 15 min = 555 min = 33300 sec]
33300 sec :: (180/185 x 33300) sec
33300 sec :: 32400 sec

33300 – 32400 = 900 sec = 15 min

Time in  error clock is 3.15 p.m.

So time in Original clock = 3 p.m.

9.

If a light flashes every 6 seconds, how many times will it flash in ¾ of an hour?

Answer: Option B

Explanation:

There are 60 minutes in an hour.

In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.

In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.

Light flashed for every 6 seconds.

In 2700 seconds 2700/6 = 450 times.

The count start after the first flash, the light will flashes 451 times in ¾ of an hour.

10.

A father tells his son, “I was of your present age when you were born”. If the father is 36 now, how old was the boy 5 years back ?

Answer: Option A

Explanation:

: Let the father’s age be x and the son’s age be y. Then, x-y = y or x = 2y
Now, x = 36. So, 2y = 36 or y = 18.
Therefore son’s present age = 18 years.
So, son’s age 5 years ago = 13 years.

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