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02 Dec 2020, 10:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (03:18) correct
61%
(03:04)
wrong
based on 94
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Abhishek starts to paint a fence on one day. On the second day, two more friend of Abhishek join him. On the third day 3 more friends of him join him and so on. If the fence is completely painted this way in exactly 20 days, then find the number of days in which 10 girls painting together can paint the fence completely, given that every girl can paint twice as fast as Abhishek and his friends (boys)? (Assume that the friends of Abhishek are all boys).
A. 20
B. 40
C. 45
D. 77
E. 80
Are You Up For the Challenge: 700 Level Questions
A. 20
B. 40
C. 45
D. 77
E. 80
Are You Up For the Challenge: 700 Level Questions
02 Dec 2020, 12:53
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Great question! Spent close to 10 minutes to find an efficient way to add terms.
Let's say Abhishek and each of his friends can complete "x" work in 1 day
Here's the tough part: computing work done by Abhishek and his friends; let's break it down:
Day 1: x work
Day 2: x + 2x
Day 3: x + 2x + 3x
Day 4: x + 2x + 3x + 4x and so on...
Day 20: x + 2x +...20x
This work has:
1x = 20 times
2x = 19 times
3x = 18 times
4x = 17 times...
10x = 11 times
11x = 10 times
12x = 9 times...
18x = 3 times
19x = 2 times
20x = 1 time
Notice that the first 10 terms are exactly the same as the last 10 terms; therefore, we can sum the first 10 terms and multiply by 2
Work = 2x {1*20 + 2*19 + 3*18... 10*11}
Now, each girl can do "2x" work in 1 day
So, 10 girls in "y" days can do "20xy" work
This is the same work that the boys did, so equating both values:
Work = 2x {1*20 + 2*19 + 3*18... 10*11} = 20xy
At this point I was able to narrow down to D and E,
y = 77 days (after calculating/estimating the sum)
Phew! Eager to see a shorter method!
Let's say Abhishek and each of his friends can complete "x" work in 1 day
Here's the tough part: computing work done by Abhishek and his friends; let's break it down:
Day 1: x work
Day 2: x + 2x
Day 3: x + 2x + 3x
Day 4: x + 2x + 3x + 4x and so on...
Day 20: x + 2x +...20x
This work has:
1x = 20 times
2x = 19 times
3x = 18 times
4x = 17 times...
10x = 11 times
11x = 10 times
12x = 9 times...
18x = 3 times
19x = 2 times
20x = 1 time
Notice that the first 10 terms are exactly the same as the last 10 terms; therefore, we can sum the first 10 terms and multiply by 2
Work = 2x {1*20 + 2*19 + 3*18... 10*11}
Now, each girl can do "2x" work in 1 day
So, 10 girls in "y" days can do "20xy" work
This is the same work that the boys did, so equating both values:
Work = 2x {1*20 + 2*19 + 3*18... 10*11} = 20xy
At this point I was able to narrow down to D and E,
y = 77 days (after calculating/estimating the sum)
Phew! Eager to see a shorter method!
13 Dec 2020, 04:21
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Assume Abhishek or any of his friend can paint 1 unit each day
Total work = 1+3+6.....20 terms = \(\frac{20*21*22}{6 }\)
1+3+6+10.....n terms = \(\frac{n*(n+1)*(n+2)}{6 }\)
Number of units painted by 10 girls each day = 2*10 = 20 units
Number of days taken by 10 girls to finish the whole work = \(\frac{20*21*22}{6*20 } = 77\)
Total work = 1+3+6.....20 terms = \(\frac{20*21*22}{6 }\)
1+3+6+10.....n terms = \(\frac{n*(n+1)*(n+2)}{6 }\)
Number of units painted by 10 girls each day = 2*10 = 20 units
Number of days taken by 10 girls to finish the whole work = \(\frac{20*21*22}{6*20 } = 77\)
Bunuel
