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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

Q:

The average monthly salary of the employees of a firm is Rs.60 out of the employees 12 officers get an average of Rs.400 per month and the rest an average of Rs. 56 per month , Find the number of emp

Answer

Actual average salary of 12 officers = 400

$\inline \fn_jvn \therefore$ Total salary of 12 officers = 400 x 12 = 4800

Average of these 12 officers as a whole =60

Total salary of 12 officers as a whole = 60 x 12 = 720

The total salary of rest = 4800 - 720 = 4080

Difference in average of the rest of officers = 60 - 56 = 4

Rest of the officers = $\inline \fn_jvn \frac{4080}{4}=1020$

Total number of employees = 1020 + 12 = 1032

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Subject: Average

Q:

The average of 9 results is 50. The average of first four results is 52 and average of last four results is 49 . what is the fifth result?

Answer

Total of results = 50 x 9 = 450

Total of first four results = 52 x 4 = 208

Total of Last four results = 49 x 4 = 196

Hence the fifth result = 450 - (208 + 196) = 46

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Subject: Average

Q:

What smallest number of 6 digit is divisible by 111?

Answer

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder

$\inline \fn_jvn \therefore$ Number to be added = (111 -100) = 11

$\inline \fn_jvn \therefore$ Required Number = 100011

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Subject: Numbers

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

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

Q:

The sum of first 45 natural numbers is:

Answer

The sum of n natural numbers $\frac{n(n+1)}{2}=\frac{45\times 46}{2}=1035$

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Subject: Numbers

Q:

Simplify the following equation.

$\frac{19}{43} + \frac{1}{(2 + \frac{1}{3 + \frac{1}{1 + \frac{1}{4}}})}$

A) 43/19	B) 1
C) 38/19	D) None of these

Answer & Explanation Answer: D) None of these

Explanation:

After solving we get,

62/931.

None of the options satisfy.

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Filed Under: Simplification
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

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