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Q:

The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is

A) 49	B) 154
C) 378	D) 1078

Answer & Explanation Answer: B) 154

Explanation:

The diameter is equal to the shortest side of the rectangle.

So radius= 14/2 = 7cm.

Therefore, $a r e a o f c i r c l e = {πr}^{2} = \frac{22}{7} \times 49 = 154 {cm}^{2}$

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Q:

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

A) (5 + 5/11 )min past 7	B) (6 + 5/11) min past 7
C) 7	D) 8

Answer & Explanation Answer: A) (5 + 5/11 )min past 7

Explanation:

When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
$\therefore$ Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.

5 min spaces are gained in $(\frac{60}{55} * 5)$ min = $5 \frac{5}{11}$ min

so, Required time = $5 \frac{5}{11}$ min past 7

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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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Q:

A clock is started at noon. By 10 minutes past 5, the hour hand has turned through

A) 145	B) 155
C) 165	D) 175

Answer & Explanation Answer: B) 155

Explanation:

Angle traced by hour hand in 12 hrs = 360º

Angle traced by hour hand in 5 hrs 10 min. i.e $\frac{31}{6}$ hrs = ${(\frac{360}{12} \times \frac{31}{6})}^{0}$ = $155^{0}$

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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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Q:

Find at what time between 8 and 9 o'clock will the hands of a clock be in the same straight line but not together ?

A) 100/11 min past 8	B) 120/11 min past 8
C) 90/11 min past 8	D) 80/11 min past 8

Answer & Explanation Answer: B) 120/11 min past 8

Explanation:

In this type of problems the formulae is $(5 x - 30) \times \frac{12}{11}$
x is replaced by the first interval of given time Here i.e 8
$(5 * 8 - 30) * \frac{12}{11}$
= $\frac{120}{11}$ min

Therefore the hands will be in the same straight line but not
together at $\frac{120}{11}$ min.past 8.

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Q:

Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to British stamps is 5 to 1, what is the ratio of US to British stamps?

A) 10 : 5	B) 15 : 2
C) 20 : 2	D) 25 : 2

Answer & Explanation Answer: D) 25 : 2

Explanation:

Indian stamps are common to both ratios. Multiply both ratios by factors such that the Indian stamps are represented by the same number.

US : Indian = 5 : 2, and Indian : British = 5 : 1. Multiply the first by 5, and the second by 2.

Now US : Indian = 25 : 10, and Indian : British = 10 : 2

Hence the two ratios can be combined and US : British = 25 : 2

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Q:

Amy has to visit towns B and C in any order. The roads connecting these towns with her home are shown on the diagram. How many different routes can she take starting from A and returning to A, going through both B and C (but not more than once through each) and not travelling any road twice on the same trip?

A) 10	B) 8
C) 6	D) 4

Answer & Explanation Answer: B) 8

Explanation:

Amy can travel clockwise or anticlockwise on the diagram.

Clockwise, she has no choice of route from A to B, a choice of one out of two routes from B to C, and a choice of one out of two routes from C back to A. This gives four possible routes.

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