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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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Filed Under: Time and Distance

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:

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
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Q:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

A) 1/4	B) 1/2
C) 3/4	D) 7/12

Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and $A, B, C$ be the respective events of not solving the problem. Then A, B, C are independent event

$∴ A, B, C$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P (A) = \frac{1}{2}, P (B) = \frac{2}{3}, P (C) = \frac{3}{4}$

$∴$ P( none solves the problem) = P(not A) and (not B) and (not C)

= $P (A \cap B \cap C)$

= $P (A) P (B) P (C)$ $[∵ A, B, C a r e I n d e p e n d e n t]$

= $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1 - \frac{1}{4}$ = 3/4

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Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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Q:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204	B) 1/204
C) 13/204	D) None of these

Answer & Explanation Answer: B) 1/204

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then

Required probability = $P (A \cap B \cap C)$

= $P (A) P (\frac{B}{A}) P (\frac{C}{A \cap B})$

Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

$∴ P (\frac{B}{A}) = \frac{3}{17}$

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

$∴ P (\frac{C}{A \cap B}) = \frac{2}{16} = \frac{1}{8}$

Hence the required probability = $\frac{2}{9} \times \frac{3}{17} \times \frac{1}{8} = \frac{1}{204}$

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Filed Under: Probability

Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

A) 7/15	B) 5/18
C) 13/18	D) 3/16

Answer & Explanation Answer: B) 5/18

Explanation:

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

$n (A) = 6, n (B) = 5, n (A \cap B) = 1$

$∴$ Required probability = $P (A \cup B)$

= $P (A) + P (B) - P (A \cap B)$

= $\frac{6}{36} + \frac{5}{36} - \frac{1}{36}$ = $\frac{5}{18}$

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Filed Under: Probability

Q:

Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

A) 4/9	B) 5/9
C) 11/18	D) 7/9

Answer & Explanation Answer: B) 5/9

Explanation:

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in ${}^{6}C_{1}$ , ways.

Now select two distinct number out of remaining 5 numbers which can be done in ${}^{5}C_{2}$ ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is

${}^{6}C_{1} \times {}^{5}C_{2} \times \frac{4!}{2!} = 720$

=> n(E) = 720

$∴ P (E) = \frac{720}{6^{4}} = \frac{5}{9}$

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Filed Under: Probability

Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

A) 0	B) 9/13
C) 1/2	D) 4/13

Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus n(E) = 13 + 3 = 16

Hence $n (\bar{E}) = 52 - 16 = 36$

Therefore, $P (E) = \frac{36}{52} = \frac{9}{13}$

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Filed Under: Probability

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