Bank Clerk Questions and Answers with Explanation, Job Roles

Q:

What is the number of digits in $3^{3^{3}}$ ? Given that log3 = 0.47712?

A) 12	B) 13
C) 14	D) 15

Answer & Explanation Answer: B) 13

Explanation:

Let $L e t x = (3^{3^{3}})$ = ${(3^{3})}^{3}$

Then, $\log (x) = 3^{3} \log (3)$

= 27 x 0.47712 = 12.88224

Since the characteristic in the resultant value of log x is 12

$∴$ The number of digits in x is (12 + 1) = 13

Hence the required number of digits in $3^{3^{3}}$ is 13.

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

Find value of $\frac{\log \sqrt{27} + \log 8 + l o g \sqrt{1000}}{\log 120}$

A) 1/2	B) 3/2
C) 2	D) 2/3

Answer & Explanation Answer: B) 3/2

Explanation:

$= \frac{\log \sqrt{3^{3}} + \log 2^{3} + \log \sqrt{10^{3}}}{\log (10 \times 3 \times 2^{2})}$

$= \frac{\log (3^{3})^{\frac{1}{2}} + \log 2^{3} + \log {(10^{3})}^{\frac{1}{2}}}{\log (10 \times 3 \times 2^{2})}$

$= \frac{\frac{1}{2} \log 3^{3} + 3 \log 2 + \frac{1}{2} \log 10^{3}}{\log 10 + \log 3 + \log 2^{2}}$

$= \frac{\frac{3}{2} (\log 3 + 2 \log 2 + \log 10)}{\log 3 + 2 \log 2 + \log 10}$ = $\frac{3}{2}$

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Filed Under: Logarithms
Exam Prep: CAT , Bank Exams , AIEEE
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27 4764

Q:

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96	B) 19/90
C) 19/96	D) None of these

Answer & Explanation Answer: B) 19/90

Explanation:

$A S S I S T A N T \to A A I N S S S T T$

$S T A T I S T I C S \to A C I I S S S T T T$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{1_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{9_{C_{1}}} \times \frac{2_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing S = $\frac{3_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/10

Probability of choosing T = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$

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Filed Under: Probability
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97 21264

Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39	B) 15/39
C) 12/13	D) None of these

Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

= $1 - (\frac{16_{C_{1}} \times 14_{C_{1}} \times 12_{C_{1}} \times 10_{C_{1}}}{16_{C_{4}}}) = \frac{15}{39}$

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Filed Under: Probability
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Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42	B) 19/42
C) 7/32	D) 16/39

Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2 and P(E2) = 1/2

Now, $P (\frac{A}{E_{1}})$ = Probability of drawing a red ball when the first bag has been selected = 4/7

$P (\frac{A}{E_{2}})$ = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) = $P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}})$

= $\frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{2}{6} = \frac{19}{42}$

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Filed Under: Probability
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Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7	B) 1/5
C) 2/7	D) 2/35

Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)]

= $P [(E \cap F) \cup (F \cap E)]$

= $P (E) P (F) + P (F) P (E)$

= $\frac{1}{7} \times \frac{4}{5} + \frac{1}{5} \times \frac{6}{7} = \frac{2}{7}$

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Filed Under: Probability
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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
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Q:

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

A) 525	B) 535
C) 545	D) 555

Answer & Explanation Answer: B) 535

Explanation:

The number of points of intersection of 37 lines is $C_{2}^{37}$ . But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting $C_{2}^{13}$ points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting $C_{2}^{11}$ points, we get only one point B.

Hence the number of intersection points of the lines is $C_{2}^{37} - C_{2}^{13} - C_{2}^{11} + 2$ = 535

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Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
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