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Q:

A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram?

A) 3560 sec	B) 3600 sec
C) 3576 sec	D) can't be determined

Answer & Explanation Answer: C) 3576 sec

Explanation:

Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then

$\frac{L}{R - A} = 36$ ---------- (i)

and $\frac{L}{R + K} = 24$ ---------(ii)

From eq.(i) and (ii)

3(R - A ) = 2 (R + K)

$\Rightarrow$ R = 3A + 2K

In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram

= 1800 ( R - A) - 24 ( A + K)

Time required = $\frac{1800 (R - A) - 24 (A + K)}{(A + K)}$

= (3600 - 24) = 3576 s

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Filed Under: Time and Distance

Q:

Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train travels at 120 km/h

A) 90 km/h	B) 160 km/h
C) 67.5 km/h	D) None of these

Answer & Explanation Answer: B) 160 km/h

Explanation:

$\frac{s_{1}}{s_{2}} = \sqrt{\frac{t_{2}}{t_{1}}}$

$\Rightarrow \frac{120}{s_{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

$\Rightarrow$ $s_{2} = 160 k m / h$

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Q:

Akash leaves mumbai at 6 am and reaches Bangalore at 10 am . Prakash leaves Bangalore at 8 am and reaches Mumbai at 11:30 am. At what time do they cross each other?

A) 10 am	B) 8:32 am
C) 8:56 am	D) 9:20 am

Answer & Explanation Answer: C) 8:56 am

Explanation:

Time taken by Akash = 4 h

Time taken by Prakash = 3.5 h

For your convenience take the product of times taken by both as a distance.

Then the distance = 14km

Since, Akash covers half of the distance in 2 hours(i.e at 8 am)

Now, the rest half (i.e 7 km) will be coverd by both prakash and akash

Time taken by them = 7/7.5 = 56 min

Thus , they will cross each other at 8 : 56am.

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Q:

In a race of 800m Kaushik gives Piyush a start of 200m and then loses the race by 20 seconds. What is the speed of Piyush and Kaushik respectively ? If the ratio of respective speeds be 3:2.

Answer

Let the speed of Kaushik be Sk and speed of Piyush be Sp and let the time taken by Piyush be t second then,

$\frac{Sk}{Sp} = \frac{\frac{800}{t + 20}}{\frac{600}{t}} = \frac{2}{3} (since given ratio)$

t = 20 s

Speed of Piyush = Sp = 600/20 = 30 m/s

and Speed of Kaushik = 800/40 = 20 m/s

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Subject: Time and Distance Exam Prep: AIEEE , Bank Exams , CAT
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Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

A) 100 m	B) 125 m
C) 150 m	D) 175 m

Answer & Explanation Answer: C) 150 m

Explanation:

$T i m e = \frac{S u m o f l e n g t h o f t h e t w o t r a i n s}{D i f f e r e n c e i n s p e e d s}$

$20 = \frac{(100 + x)}{\frac{25}{2}}$

$\Rightarrow$ X=150m

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Filed Under: Logarithms

Q:

The distance of the college and home of Rajeev is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed (or increased) speed of Rajeev?

A) 28 km/h	B) 30 km/h
C) 40 km/h	D) 20 km/h

Answer & Explanation Answer: D) 20 km/h

Explanation:

Let the normal speed be x km/h, then

$\frac{80}{x} - \frac{80}{(x + 4)} = 1$

$\Rightarrow$ $x^{2} + 4 x - 320 = 0$

$\Rightarrow$ x (x + 20) - 16 (x + 20) = 0

(x + 20 ) (x - 16) =0

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore increased speed = 20 km/h

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Q:

For $x \in N, x > 1$ , and $p = \log_{x} (x + 1)$ , $q = \log_{x + 1} (x + 2)$ then which one of the following is correct?

A) p < q	B) p = q
C) p > q	D) can't be determined

Answer & Explanation Answer: C) p > q

Explanation:

$\frac{k}{l} > \frac{k + 1}{l + 1}$ for (k,l) > 0 and k > l

Let k = x+1 and l = x

Therefore, $\frac{x + 1}{x} > \frac{(x + 1) + 1}{(x) + 1}$

(x + 1) > x

Therefore, $\frac{\log (x + 1)}{\log (x)} > \frac{\log (x + 2)}{\log (x + 1)}$

$\Rightarrow \log_{x} (x + 1) > \log_{x + 1} (x + 2)$

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Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT
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Q:

The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is

A) -1	B) 0
C) 1/2	D) 1

Answer & Explanation Answer: B) 0

Explanation:

$= [\log \tan (1^{0}) + \log \tan (89^{0})] + [\log \tan (2^{0}) + \log \tan (88^{0})] + \dots \dots + [\log \tan (45^{0})]$

$= \log [\tan (1^{0}) \times \tan (89^{0})] + \log [\tan (2^{0}) \times \tan (88^{0})] + \dots \dots + \log (1)$

$[∵ \tan (90 - θ) = c o t θ a n d \tan (45^{0}) = 1]$

= log 1 + log 1 +.....+log 1

= 0.

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Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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