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Q:

A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:

A) 4pm	B) 5pm
C) 6pm	D) 7pm

Answer & Explanation Answer: A) 4pm

Explanation:

Time from 7 am to 4.15 pm = 9 hrs 15 min = $\frac{37}{4}$ hrs

3 min. 5 sec. of this clock = 3 min. of the correct clock.

Now 3min.5sec. is $\frac{37}{720}$ hrs of this clock = 3 min.is $\frac{1}{20}$ hrs of the correct clock

$\frac{37}{4}$ hrs of clock = $(\frac{1}{20} \times \frac{720}{37} \times \frac{37}{4})$ hrs of the correct clock.

= 9 hrs of the correct clock.

$\therefore$ The correct time is 9 hrs after 7 am. i.e., 4 pm.

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Filed Under: Clocks
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

A watch which gains uniformly ,is 5 min,slow at 8 o'clock in the morning on sunday and it is 5 min 48 sec.fast at 8 p.m on following sunday. when was it correct?

A) 7pm on wednesday	B) 20 min past 7pm on wednesday
C) 15min past 7pm on wednesday	D) 8pm on wednesday

Answer & Explanation Answer: B) 20 min past 7pm on wednesday

Explanation:

This sunday morning at 8:00 AM, the watch is 5 min. Slow, and the next sunday at 8:00PM it becomes 5 min 48 sec fast. The watch gains $5 + 5 \frac{48}{60}$ min in a time of (7×24)+12 = 180 hours.

To show the correct time, it has to gain 5 min.

$\frac{54}{5} m i n \to 180 h o u r s$

5min ->

$(\frac{5}{\frac{54}{2}} \times 180)$

$83 \frac{1}{3} h r s = 72 h r s + 11 \frac{1}{3} h r s = 3 d a y s + 11 h r s + 20 m i n$

So the correct time will be shown on wednesday at 7:20 PM

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Filed Under: Clocks
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

A straight fence is to be constructed from posts 6 inches wide and separated by lengths of chain 5 feet long. If a certain fence begins and ends with a post, which of the following could be the length of the fence in feet? (12 inches = 1 foot).

A) 17	B) 18
C) `19	D) 20

Answer & Explanation Answer: A) 17

Explanation:

The fence will consist of one more post than there are chains. (e.g. P-c-P-c-P).

Therefore, a total length has to be a multiple of the length of the chain plus one post (5.5) plus one post extra.We have length = (5.5n + 0.5), where n can be any positive whole number. If n= 3, length =17

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Filed Under: Numbers
Exam Prep: GRE

Q:

Of the following, which is greater than ½ ?

Indicate ALL such fractions.

A) 2/5	B) 4/7
C) 4/9	D) 5/11

Answer & Explanation Answer: B) 4/7

Explanation:

One way to deal with fractions is to convert them all to decimals.

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7.

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Filed Under: Average
Exam Prep: GRE

Q:

Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters.

A) 56m	B) 16m
C) 14m	D) 76m

Answer & Explanation Answer: A) 56m

Explanation:

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.

Let the length of the rope be r mts

then, ${πr}^{2} = 9856 \Rightarrow r^{2} = \frac{9856 \times 7}{22} = 3136 \Rightarrow r = 56 m$

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Filed Under: Area
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Q:

A particle which is similar to electron is

A) positron	B) beta particle
C) photon	D) mason

Answer & Explanation Answer: B) beta particle

Explanation:

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Filed Under: Chemistry

Q:

A particle which is four times in mass and two times in charge that of proton is

A) helium atom	B) an alpha particle
C) deuteron	D) tritium

Answer & Explanation Answer: B) an alpha particle

Explanation:

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Filed Under: Chemistry

Q:

Wheels of diameters 7cm and 14cm start rolling simultaneously from x & y which are 1980 cm apart towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10seconds.The speed of the smaller wheel is

A) 22m/s	B) 33m/s
C) 44m/s	D) 55m/s

Answer & Explanation Answer: A) 22m/s

Explanation:

let each wheel make x revolutions per sec. Then $[(2 π * \frac{7}{2} * x) + (2 π * 7 * x)] * 10 = 1980$ $(2 * \frac{22}{7} * \frac{7}{2} * x) + (2 * \frac{22}{7} * \frac{7}{2} * x) = 198$

$66 x = 198$

distance moved by smaller wheel in 3 revolutions

= $2 * \frac{22}{7} * \frac{7}{2} * 3 = 66 c m$

speed of smaller wheel = $\frac{66}{3}$ m/s = 22m/s

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

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