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Q:

What is the least number of squares tiles required to pave the floor of a room 15.17 m long and 9.02 m broad?

A) 814	B) 714
C) 614	D) 713

Answer & Explanation Answer: A) 814

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = $(41 \times 41) c m^{2}$

Required number of tiles = $\frac{1517 \times 902}{41 \times 41}$ = 814

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

Q:

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A) 18	B) 28
C) 38	D) 48

Answer & Explanation Answer: A) 18

Explanation:

$\sqrt{(l^{2} + b^{2})} = \sqrt{41}$ (or) $l^{2} + b^{2} = 41$

Also, $l b = 20$

${(l + b)}^{2} = l^{2 +} b^{2} + 2 l b$ = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

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Filed Under: Area
Exam Prep: CAT , Bank Exams , AIEEE
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Q:

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

A) 18%	B) 28%
C) 38%	D) 48%

Answer & Explanation Answer: B) 28%

Explanation:

Let original length = x and original breadth = y.

Decrease in area = xy - $(\frac{80}{100} x \times \frac{90}{100} y)$ = (7/25)xy

Decrease % = (7/25)xy * (1/xy) * 100 = 28%

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Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: C) 3

Explanation:

Area of the park = (60 x 40) = 2400 $m^{2}$

Area of the lawn = 2109 $m^{2}$

Area of the crossroads = (2400 - 2109) = 291 $m^{2}$

Let the width of the road be x metres. Then,

$60 x + 40 x - X^{2} = 291$

$x^{2} - 100 x + 291 = 0$

(x - 97)(x - 3) = 0
x = 3.

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Filed Under: Area
Exam Prep: Bank Exams
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Q:

If each of the 8 teams in a league must play each other three times,how many games will be played?

A) 72	B) 84
C) 36	D) 79

Answer & Explanation Answer: B) 84

Explanation:

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be $8 C_{2}$ games.Since,each pairing of teams will occur three times,the answer will be triple.

Therfore, $8 C_{2} * 3$ =84

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Filed Under: Permutations and Combinations

Q:

There are 7 men and 10 women on a committee selection pool.A committee consisting of President,Vice-President, and Treasurer is to be formed.How many ways can exactly two men be on the committe ?

A) 1200	B) 1240
C) 1260	D) 1620

Answer & Explanation Answer: C) 1260

Explanation:

There are $7 C_{2}$ ways of selecting two men, and $10 C_{1}$ ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.

So, $7 C_{2} * 10 C_{1} * 3! = 1260$

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Q:

A research team of 6 people is to be formed from 10 chemists,5 politicians, 8 economists and 15 biologists.How many teams have atleast 5 chemists?

A) 7350	B) 6400
C) 6379	D) 7266

Answer & Explanation Answer: D) 7266

Explanation:

$10 C_{5} * 28 C_{1} * 10 C_{6} = 7266$

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Filed Under: Permutations and Combinations

Q:

A research team of 6 people is to be formed from 10 chemists,5 politicians, 8 economists and 15 biologists.How many teams have atleast 5 chemists?

A) 7350	B) 6400
C) 6379	D) 7266

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