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Q:

A mixture contains alcohol and water in the ratio 4:3. If 5 liters of water is added to the mixture, the ratio becomes 4:5. Find the quantity of alcohol in the given mixture.

A) 8 Liters	B) 10 Liters
C) 18 Liters	D) 22 Liters

Answer & Explanation Answer: B) 10 Liters

Explanation:

Let the quantity of alcohol and water be 4x liters and 3x liters respectively. Then,

(4x)/(3x+5) = 4/5

=> 20x = 4(3x+5)

=> x = 2.5

Quantity of alcohol = (4 * 2.5) liters = 10 liters.

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Filed Under: Ratios and Proportions

Q:

Loiter : Dwandle :: Impugn : ?

A) Revenge	B) Confiscate
C) Gamble	D) Challenge

Answer & Explanation Answer: D) Challenge

Explanation:

The words in each pair are synonyms of each other.

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Filed Under: Analogy
Exam Prep: AIEEE , Bank Exams
Job Role: Analyst , Bank Clerk , Bank PO

Q:

Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds?

A) 48	B) 24
C) 38	D) 36

Answer & Explanation Answer: B) 24

Explanation:

Less Cogs $\Rightarrow$ more turns and less time $\Rightarrow$ less turns

$C o g s T i m e T u r n s A 54 45 80 B 32 8 ?$

Number of turns required=80 × 54/32 × 8/45 = 24 times

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Filed Under: Problems on Trains

Q:

A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

A) 230 m	B) 240 m
C) 260 m	D) 270 m

Answer & Explanation Answer: D) 270 m

Explanation:

Speed =[ 72 x (5/18) ]m/sec= 20 m/sec.

Time = 26 sec.

Let the length of the train be x metres.

Then,[ (x+250)/26 ]= 20

=> x + 250 = 520

=> x = 270.

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Filed Under: Problems on Trains

Q:

Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:

A) 9	B) 9.6
C) 10	D) 10.8

Answer & Explanation Answer: D) 10.8

Explanation:

Relative speed = (60 + 40) km/hr =[ 100 x ( 5 / 18 ) ]m/sec = ( 250 /9 ) m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [ 300 x ( 9/250 ) ] sec = ( 54/ 5 )sec = 10.8 sec.

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Filed Under: Problems on Trains

Q:

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:

A) 50 m	B) 150 m
C) 200 m	D) data inadequate

Answer & Explanation Answer: B) 150 m

Explanation:

Let the length of the train be x metres and its speed be y m/sec.
Then, (x/y)= 15 $\Rightarrow$ y =(x/15)

(x+100)/25 = x/15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

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Filed Under: Problems on Trains

Q:

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

A) 69.5 km/hr	B) 70 km/hr
C) 79 km/hr	D) 79.2 km/hr

Answer & Explanation Answer: D) 79.2 km/hr

Explanation:

Let the length of the train be x metres and its speed by y m/sec.

Then, x/y = 8 => x = 8y

Now, (x+264)/20 = y
$\Rightarrow$ 8y + 264 = 20y
$\Rightarrow$ y = 22.
$\therefore$ Speed = 22 m/sec = (22*18/5) km/hr = 79.2 km/hr.

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Filed Under: Problems on Trains

Q:

0.1 decates ago, Vijaya was quadrice as old as her daughter Simran. 0.6 decades hence, Vijaya’s age will beat her daughter’s age by 0.9 decades. The proportion of the current ages of Vijaya and Simran is : [ 0.1 Decates =1 Year; quadrice = 4 times]

A) 8:1	B) 10:2
C) 11:4	D) 13:4

Answer & Explanation Answer: D) 13:4

Explanation:

Let, Ages of Vijaya and Simran – 1 year ago was 4A and A years correspondingly.
Then, [(4A + 1) + 6] – [(A + 1) + 6] = 9
3A = 9 => A = 3

Needed Ratio = (4A + 1) : (A + 1) = 13 : 4

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Filed Under: Problems on Ages

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