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Q:

Two trains 85 m and 75 m long are running in same direction with speeds of 62 km/hr and 44 km/hr respectively. In what time will the first train cross the second train ?

A) 22 sec	B) 32 sec
C) 42 sec	D) 52 sec

Answer & Explanation Answer: B) 32 sec

Explanation:

Given that,

1st train speed = 62 km/hr

2nd train speed = 44 km/hr

Hence the relative speed of two trains is : 62-44 = 18 km/hr = $18 \times \frac{5}{18}$ m/s = 5 m/s.

There fore, the first train crosses the second train i.e (85+75)mts of distance with a speed of 5m/s in,

time = $\frac{d i s \tan c e}{s p e e d} = \frac{85 + 75}{5}$ = 32 sec.

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Filed Under: Time and Distance
Exam Prep: CAT
Job Role: Bank Clerk

Q:

An industrial loom weaves 1.14 meters of cloth every second. Approximately, how much time will it take to weave 52 meters of cloth ?

A) 29.32 sec	B) 42.51 sec
C) 39.25 sec	D) 45.61 sec

Answer & Explanation Answer: D) 45.61 sec

Explanation:

Given loom weaves 1.14 mts of cloth in one second then 52 mts of cloth can be weaved by loom in,

1.14 ----- 1

52.0 ------?

$\Rightarrow \frac{52}{1.14}$ = 45.61 sec

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Filed Under: Chain Rule
Exam Prep: CAT
Job Role: Bank PO

Q:

A sales boy is allowed $4 \frac{1}{2}$ % discount on the total sales made by him plus a bonus of $\frac{1}{4}$ % on the sales over Rs.12,000.If his total earnings were Rs.2400, then his total sales(in Rs.) are ?

A) Rs.60,145.23	B) Rs.54,285.54
C) Rs.52,147.85	D) Rs.50,589.47

Answer & Explanation Answer: D) Rs.50,589.47

Explanation:

Let the total earnings be Rs.x

Then $4 \frac{1}{2}$ % of x + $\frac{1}{4}$ % of (x-12000) =2400

$\Rightarrow \frac{9}{2} \times \frac{x}{100} + \frac{1}{4} \times \frac{x - 12000}{100} = 2400$

=> x=50,589.47

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Filed Under: Percentage
Exam Prep: CAT
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Q:

A group consists of 4 couples in which each of the 4 boys have one girl friend.In how many ways they can be arranged in a straight line such that boys and girls occupies alternate positions?

Answer

Answer : 1152

Total positions are 8.

In that boys can be arranged in 4 places and girls can be arranged in 4 places and hence this can be done in 2 ways.

i.e => 4! x 4! x 2 = 24 x 24 x 2=1152.

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Subject: Permutations and Combinations Exam Prep: CAT
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Q:

The population of a city is 415600.It increased by 25% in the first year and decreased by 30% in the second year.What is the population of the city at the end of second year?

Answer

As per given data, P = 415600, R1= 25% incresed, R2 = 30% decreased

Population of the city at the end of the second year= $p\left ( 1+\frac{R1}{100} \right )\left ( 1-\frac{R2}{100} \right )$

$\Rightarrow$ $415600\left ( 1+\frac{25}{100} \right )\left ( 1-\frac{30}{100} \right )$

$\Rightarrow$ 363650.

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Subject: Percentage Exam Prep: CAT
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Q:

Guess which number should be replaced by the "?" in the given series 0,10,1110,3110,132110,?

Answer

1113122110

In given series 2nd number gives number of which digits present in the 1st number and so on continues for 3rd ,4th,...

Here 1st no. has only one zero, so 2nd no. is 10 and 2nd no. has one 1 and one 0,so 3rd no. is 1110, so like that 5th no. has one 1,one 3,one 2,two 1's and one 0.So the next no. is 1113122110.

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Subject: Math Puzzles Exam Prep: CAT
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Q:

Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

(1) $K_{2} O < N a_{2} O < A l_{2} O_{3} < M g O$

(2) $A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

(3) $M g O < K_{2} O < A l_{2} O_{3} < N a_{2} O$

(4) $N a_{2} O < K_{2} O < M g O < A l_{2} O_{3}$

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

Answer & Explanation Answer: B) Option 2

Explanation:

While moving from left to right in periodic table basic character of oxide of elements will decrease.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{A l_{2} O_{3} < M g O < N a_{2} O}$

and while descending in the group basic character of corresponding oxides increases.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{N a_{2} O < K_{2} O}$

$∴ C o r r e c t O r d e r i s A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

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Filed Under: Chemistry
Exam Prep: AIEEE

Q:

Boron cannot form which one of the following anions?

(1) $B O_{2}^{-}$

(2) $B F_{6}^{3 -}$

(3) $B H_{4}^{-}$

(4) $B (O H)_{4}^{-}$

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

Answer & Explanation Answer: B) Option 2

Explanation:

Due to absence of low lying vacant d orbital in B, $s p^{3} d^{2}$ hybridization is not possible hence $B F_{6}^{3 -}$ will not formed.

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Filed Under: Chemistry
Exam Prep: AIEEE

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