population after 2 yr is ?

1) 2500 2) 10000 3) 252000 4) 10100 5) None of these

Solution: (4).

Population after 2 year

= P ( 1 + R/100 )2 = 250000 ( 1 + 2/100 )2

= 250000 x 51/50 x 51/50 = 260100

Therefore Growth = 260100 – 250000 = 10100

2. A jogger desires to run a certain course in ¼ less time than he usually takes. By what per cent must

be increase his average running speed to accomplish the goal?

1) 50% 2) 20% 3) 25% 4) 33 1/3 % 5) None of these

Solution: (4).

xt = x’ 3/4t = x = 3/4x’ = x’ = 4/3 x

Thus, he has to increase his speed by

4/3 x- x / x

* 100% i.e. 33 1/3 %

3. The prices of two articles are as 3:4 . If the price of the first article is increased by 10% and that of

the second by Rs.4, one original ratio remains the same. The original price of the second article is

1) Rs.40 2) Rs.10 3) Rs.30 4) Rs.35 5) None of these

Solution: (1).

Let cost prices of two articles be 3x and 4x , respectively . then,

(110% of 3 x) / (4x +4)

1.1x = x+1

0.1x = 1

x= 10

Thus, cost price of the second article is 4 x 10 = Rs. 40

4. An alloy of gold and silver weights 50g. It contains 80% gold. How much gold should be added to the

alloy, so that percentage of gold is increased to 90?

1) 50g 2) 60g 3) 30g 4) 40g 5) None of these

Solution: (1).

Gold in 50g of alloy

= 80 x 50/100 = 40g

Let x g gold must be added

Now, according to the question.

(40 +x) / (50 + x ) = 90/100

= 100 ( 40+x) = 90 (50+x)

= 10 ( 40 + x) = 9 ( 50+x )

= 400 + 10x = 450 + 9x

x = 450-400 = x = 50g

Thus, 50 g of gold must be added to make it 90%

5. 1 L of water is added to 5 L of alcohol and water solution containing 40% alcohol strength. The

strength of alcohol in the new solution will be

1) 30% 2) 33 1/3 % 3) 33 2/3% 4) 33% 5) None of these

Solution: (2).Quantity of alcohol in 5 L of solution

= 40/100 x 5 = 2L

Quantity of alcohol in 6 L of solution = 2L

Therefore Strength of alcohol in new solution

= ( 2/6 x 100) % = 33 1/3%

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