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Q:

A shopkeeper cheats to the extent of 10% while buying and selling, by using false weights. His total gain is.

A) 20%	B) 21%
C) 22%	D) 23%

Answer & Explanation Answer: B) 21%

Explanation:

Gain % = $(\frac{{(100 + c o m m o n g a i n %)}^{2}}{100} - 100)$ %

= $(\frac{{(100 + 10)}^{2}}{100} - 100)$

= 21%

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Filed Under: Profit and Loss

Q:

How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?

A) 36 Kg	B) 42 Kg
C) 54 Kg	D) 63 Kg

Answer & Explanation Answer: D) 63 Kg

Explanation:

By the rule of alligation:

C.P. of 1 kg sugar of 1st kind C.P. of 1 kg sugar of 2nd kind

Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind.

Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

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Filed Under: Alligation or Mixture

Q:

A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A) 10	B) 20
C) 21	D) 25

Answer & Explanation Answer: C) 21

Explanation:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left = $(7 x - \frac{7}{12} \times 9) = (7 x - \frac{21}{4})$ litres.

Quantity of B in mixture left = $(5 x - \frac{5}{12} \times 9) = (5 x - \frac{15}{4})$ litres.

$\frac{(7 x - \frac{21}{4})}{(5 x - \frac{15}{4}) + 9} = \frac{7}{9}$

$\Rightarrow \frac{28 x - 21}{20 x + 21} = \frac{7}{9}$

$\Rightarrow x = 3$

So, the can contained 21 litres of A.

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Filed Under: Alligation or Mixture

Q:

Which is smallest state by area of India ?

Answer

Goa

Workspace

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Subject: Longest, Shortest, Deepest

Q:

The proportion of forest to the total national geographical area of India as envisaged by National Forest Policy is

A) 30.3%	B) 33.3%
C) 38.3%	D) 42.3%

Answer & Explanation Answer: B) 33.3%

Explanation:

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Filed Under: Indian Geography

Q:

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

A) 100 m	B) 150 m
C) 190 m	D) 200 m

Answer & Explanation Answer: A) 100 m

Explanation:

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

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Filed Under: Time and Distance

Q:

The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

A) 456 kms	B) 482 kms
C) 552 kms	D) 556 kms

Answer & Explanation Answer: C) 552 kms

Explanation:

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.

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Filed Under: Time and Distance

Q:

Atomic theory was devised by

A) Mosley	B) John Dalton
C) Daimler	D) None of these

Answer & Explanation Answer: B) John Dalton

Explanation:

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Filed Under: Inventions

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