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Q:

A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A) No profit, no loss	B) 5%
C) 8%	D) 10%

Answer & Explanation Answer: B) 5%

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =(80/1600*100) % = 5%

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Filed Under: Profit and Loss

Q:

The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?

A) Rs. 2000	B) Rs. 2200
C) Rs. 2400	D) Data inadequate

Answer & Explanation Answer: A) Rs. 2000

Explanation:

Let C.P. be Rs. x.

Then,= > $\frac{1920 - x}{x} * 100 = \frac{x - 1280}{x} * 100$

=> 1920 - x = x - 1280

=> 2x = 3200

=> x = 1600

Required S.P. = 125% of Rs. 1600 =Rs(125/100*1600) = Rs2000

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Filed Under: Profit and Loss

Q:

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

A) 10 min. 20 sec.	B) 11 min. 45 sec.
C) 12 min. 30 sec.	D) 14 min. 40 sec.

Answer & Explanation Answer: D) 14 min. 40 sec.

Explanation:

Part filled in 4 minutes =4(1/15+1/20) = 7/15

Remaining part =(1-7/15) = 8/15

Part filled by B in 1 minute =1/20 : 8/15 :: 1:x

x = (8/15*1*20) = $10 \frac{2}{3} m i n = 10 m i n 40 s e c$

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

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Filed Under: Pipes and Cistern

Q:

A pump can fill a tank with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the tank. The leak can drain all the water of the tank in:

A) 7 hours	B) 8 hours
C) 12 hours	D) 14 hours

Answer & Explanation Answer: D) 14 hours

Explanation:

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part

The pumop take1 hour to fill 1/2 portion

Due to lickage

The pumop take 7/3 hrs to fill 1 part

The pumop take1 hour to fill 3/7 portion

Now the difference of area = (1/2-3/7)=1/14

This 1/14 part of water drains in 1 hour

Total area=1 part of water drains in (1x14/1)hours= 14 hours

So the leak can drain all the water of the tank in 14 hours.

Leak will empty the tank in 14 hrs

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Filed Under: Pipes and Cistern
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

Q:

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

A) 3 hrs 15 min	B) 3 hrs 45 min
C) 4 hrs 15 min	D) 4 hrs 1

Answer & Explanation Answer: B) 3 hrs 45 min

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part = $(1 - \frac{1}{2}) = \frac{1}{2}$

$\frac{2}{3} : \frac{1}{2} : : 1 : x$

=> x = $(\frac{1}{2} * 1 * \frac{3}{2}) = \frac{3}{4}$

So, total time taken = 3 hrs. 45 mins.

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Filed Under: Pipes and Cistern

Q:

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

A) 10	B) 12
C) 14	D) 16

Answer & Explanation Answer: C) 14

Explanation:

Part filled in 2 hours = 2/6=1/3

Remaining part = $(1 - \frac{1}{3}) = \frac{2}{3}$

(A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

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Filed Under: Pipes and Cistern

Q:

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

A) 1	B) 1/40
C) 40	D) 80

Answer & Explanation Answer: C) 40

Explanation:

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

$c o w s 1 : 40 : 40 : x b a g s 40 : 1 : 40 : x$

1 x 40 x x = 40 x 1 x 40

=> x = 40.

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Filed Under: Chain Rule

Q:

A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is:

A) 4	B) 9
C) 12	D) 49

Answer & Explanation Answer: B) 9

Explanation:

Let the required number of revolutions made by larger wheel be x.

Then, More cogs, Less revolutions (Indirect Proportion)

14 : 6 :: 21 : x => 14 x x = 6 x 21

x=6*21/14 => x=9

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Filed Under: Chain Rule

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