Job Roles Questions and Answers

Q:

Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively . If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

Answer

As the pipes are operating alternatively, thus their 2 minutes job is = $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$

In the next 2 minutes the pipes can fill another $\frac{5}{12}$ part of cistern.

Therefore, In 4 minutes the two pipes which are operating alternatively will fill $\frac{5}{12} + \frac{5}{12} = \frac{5}{6}$ Remaining part = $1 - \frac{5}{6} = \frac{1}{6}$

Pipe A can fill $\frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\frac{1}{6}$ of the cistern in = $\frac{2}{3}$ min

Therefore, Total time taken to fill the Cistern

4 + $\frac{2}{3}$ minutes.

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Subject: Time and Work Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

28 14527

Q:

A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the work?

Answer

150 men in 25 days do = $\frac{1}{4}$ work

Let 1 man in 1 day does = x work

Total work done by 150 men in 25 days = 150x * 25 = $\frac{1}{4}$ work => x = $\frac{1}{15000}$

Therefore, 100 men in 60 days do = 100 * 60x = 6000x work = 6/15 = 2/5

Total work done = $\frac{1}{4} + \frac{2}{5} = \frac{13}{20}$

Therefore, Remaining work = 1 - $\frac{13}{20}$ = $\frac{7}{20}$

Remaining time = 130 - (25+60+10) = 35 days

Therefore, work is done in 25 days by 150 men.

Therefore, Work is done in 35 days by 150 men.

Hence, he should employ 50 more men.

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Subject: Time and Work Exam Prep: Bank Exams , AIEEE
Job Role: Bank PO

47 18723

Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

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Subject: Time and Work
Job Role: Bank PO

72 28831

Q:

Divide Rs.6500 among A,B and C so that after spending 90% , 75% and 60% of their respective saving were in the ratio of 3: 5: 6

Answer

A's spending 90% $\inline \therefore$ saving = 10%

B's spending 75% $\therefore$ saving = 25%

C's spending 60% $\inline \therefore$ saving = 40%

Let us suppose A, B and C saves Rs. 3.5 and 6 respectively.

$\inline \therefore$ 10% of A's saving = Rs.3

100% of A's saving = $\inline \frac{3}{10}\times 100$ = Rs. 30

25% of B's saving = Rs. 5

100% of B's saving = $\inline \frac{5}{25}\times 100$ = Rs. 20

40% of C's saving = Rs.6

100% of C's saving = $\inline \frac{6}{40}\times 100$ = Rs. 15

Divide Rs. 6500 in the ratio of 30 : 20 : 15 as

A's Share = $\inline \frac{30}{65}\times 6500$ = Rs. 3000

B's Share = $\inline \frac{20}{65}\times 6500$ = Rs. 2000

C's Share = $\inline \frac{15}{65}\times 6500$ = Rs. 1500

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Subject: Ratios and Proportions
Job Role: Bank PO

66 9931

Q:

A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning?

Answer

Let the average after 7th inning = x

Then average after 16th inning = x - 3

16(x-3)+87 = 17x

x = 87 - 48 = 39

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Subject: Average Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

677 94620

Q:

The value of

$\left [ \frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}} \right ]$ is:

Answer

$\frac{2^{n} + \frac{2^{n}}{2}}{2 x 2^{n} - 2^{n}} = \frac{2^{n} (1 + \frac{1}{2})}{2^{n} (2 - 1)} = \frac{3}{2}$

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Analyst , Bank Clerk , Bank PO

20 3382

Q:

(51 + 52 + 53 + .........+100) is equal to

Answer

51 + 52 + 53 + ...........+ 100

= (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)

= It is in the form of $\frac{n (n + 1)}{2} s e r i e s s u m m a t i o n$

= n1 = 100 , n2 = 50

= $[\frac{100 (100 + 1)}{2}] - [\frac{50 (50 + 1)}{2}]$

= (5050 - 1275) = 3775

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

300 61534

Q:

The total number of prime factors in the expansion

$4^{11}\times 7^{5}\times11^{2}$

Answer

$2^{2^{11}} x 7^{5} x 11^{2} = > 2^{22} x 7^{5} x 11^{2}$

Total number of prime factors is (22+5+2) = 29

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Subject: Numbers Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Analyst , Bank Clerk , Bank PO

4 5492

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