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A contract is to be completed in 52 days and 125 identical robots were employed, each operational for 7 hours a day. After 39 days, five-seventh of the work was completed. How many additional robots would be required to complete the work on time, if each robot is now operational for 8 hours a day?

  1. 50
  2. 89
  3. 146
  4. 175
in Numerical Ability by Veteran (423k points)
edited by | 946 views
Migrated from GO Mechanical 5 months ago by Arjun

1 Answer

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Best answer
Let $y$ be the speed of work of a robot per hour and $S$ be the total amount of work done.

$\frac{5}{7}$ of work is completed in $39$ days with $7$ hours a day. So,

$\frac{5}{7}S = 125 \times 39 \times y \times 7  \to(1)$

Remaining work = $\frac{2}{7}S$ and now we have $13$ more days and $8$ hours per day. So,

$\frac{2}{7}S = x \times 13\times y \times 8 \to (2)$

where $x$ is the number of robots required.

$(1) /(2) \implies \frac{5}{2} x \times 8=125\times 3\times7  .$

$x = \frac{1050}{8} = 131.25.$

So number of robots required $=\lceil 131.25 \rceil = 132$. We have $125$ already and so the no. of additional robots required $= 132-125 = 7.$
by Veteran (423k points)
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This question will be challanged so you should give everyone 2 marks in paper set 2.
0
It is already given -- not even waiting for the challenge :)
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Okok , cool work guys.
0
you have given 2 marks for this question only to those candidates who have selected any of the given option.

those who did not choose any of the given option because of the risk of negative marking did not get 2 marks according to your website. So please take it into the considerration and correct it accordingly.

Thank you
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Not really. This 2 mark is given to everyone. Though it may not be shown with a tick mark.
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i have counted each mark and every question and that wrong question's 2 marks are still not included in my total marks. And before using this website's service i had calculated my marks by hand in a paper therfore i am sure i did'nt get 2 mark of the wrong question. Please take a look at your program software.
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What is your candidate ID of GATE? I'll check.
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No need; there was a small typo for Mech. calculator. Fixed now.
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thank you sir
0
How the avg marks of top 0.1% of each session is less than that of both session combined?
Please check it.
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That was a mistake as the number crossed a threshold. Now fixed.
0
I think there is some error while calculating average of top 0.1% candidate in a session.
As I checked just now and found that average of top.1% was 84.00  for set2 and total candidates who have checked till now for session 2 are 22527 which means in 0.1% there will be 23 candidates. Now from 90-95 there are 5 candidates while from 85-90 there are 27 candidates so if the average of top 23 candidates will be taken then it will be more than 85 but there it is 84.00
Please check it.
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Top 0.1% is not of the sample, but estimated for the actual population.
0

@Arjun sir what should we do in exam for questions like this..I mean should we mark some option to be eligible for marks given to all or should we leave it without marking anything??

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