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Q:

At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other?

A) 11 4/11	B) 13 4/11
C) 15 4/11	D) 16 4/11

Answer & Explanation Answer: D) 16 4/11

Explanation:

At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.55 min. are gained in 60 min.
15 min. are gained in
= (60/55 x 15)min
=16+4/11
The hands are coincident at 16 + 4/11 min past 3

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Filed Under: Clocks
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Q:

How much does a watch lose per day, if its hands coincide ever 64 minutes?

A) 32 8/11	B) 33 8/11
C) 34 8/11	D) 35 8/11

Answer & Explanation Answer: A) 32 8/11

Explanation:

55min spaces are covered in 60min

60 min. spaces are covered in (60/55 x 60) min= 65+5/11 min.

loss in 64min=(65+5/11)- 64 =16/11

Loss in 24 hrs (16/11 x 1/64 x 24 x 60) min= 32 8/11.

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Q:

A room 5m 55cm long and 3m 74cm broad is to be paved with square tiles.Find the least number of square tiles required to cover the floor?

A) 156	B) 166
C) 176	D) 186

Answer & Explanation Answer: C) 176

Explanation:

Area of the room=(544 * 374)

size of largest square tile= H.C.F of 544 & 374 = 34 cm

Area of 1 tile = (34 x 34) $c m^{2}$

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

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Filed Under: Area
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Q:

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

A) 56m	B) 65m
C) 34m	D) 36m

Answer & Explanation Answer: A) 56m

Explanation:

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

Hence number of poles required = 280 / 5 = 56

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Filed Under: Area
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Q:

A typist uses a sheet measuring 20cm by 30cm lengthwise.If a margin of 2 cm is left on each side and a 3 cm margin on top and bottom, then percent of the page used for typing is

A) 64%	B) 74%
C) 84%	D) 94%

Answer & Explanation Answer: A) 64%

Explanation:

Area of the sheet = $(20 \times 30) c m^{2}$

Area used for typing = $[(20 - 4) \times (30 - 6)] c m^{2}$

required % = $\frac{384}{600} \times 100$ =64%

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Filed Under: Area
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Q:

2 metres broad pathway is to be constructed around a rectangular plot on the inside. The area of the plots is 96 sq.m. The rate of construction is Rs. 50 per square metre. Find the total cost of the construction?

A) 20	B) 30
C) 40	D) data is inadequate

Answer & Explanation Answer: D) data is inadequate

Explanation:

Given lb =96

Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b)

Which cannot be determined. so, data is inadequate.

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Filed Under: Area
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Q:

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required?

A) 44ft	B) 88ft
C) 22ft	D) 11ft

Answer & Explanation Answer: B) 88ft

Explanation:

Given that length and area, so we can find the breadth.

Length x Breadth = Area

=> 20 x Breadth = 680

=> Breadth = 34 feet

Area to be fenced = 2B + L = 2 (34) + 20 = 88 feet

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Filed Under: Area
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Q:

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is

A) 200m	B) 300m
C) 400m	D) 100m

Answer & Explanation Answer: A) 200m

Explanation:

Let the triangle and parallelogram have common base b,

let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then

Area of triangle = $\frac{1}{2} \times b \times h 1$

Area of rectangle = b*h2

As per Given,

$\frac{1}{2} * b * h 1 = b * h 2$

$\frac{1}{2} * b * h 1 = b * 100$

h1=200

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