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18 May 2018, 11:48
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C
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:
49% (02:52) correct
51%
(02:40)
wrong
based on 260
sessions
History
Date
Time
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Not Attempted Yet
There are three types of Chandeliers hung in a party hall: size A, size B and size C. All three types of chandeliers contain bulbs of same standard size only, but the difference among their sizes is due to the number of bulbs. Each chandelier of size A can contain 15 to 18 bulbs , inclusive. Each chandelier of size B can contain 22 to 25 bulbs, inclusive while each chandelier of size C can contain 28 to 30 bulbs, inclusive. If total 10 chandeliers are hung inside this hall and all bulbs are working in proper condition, and also there is at least one chandelier of each size, then how many chandeliers of size B are there?
(1) There are 247 bulbs in total, including all the chandeliers of all three sizes.
(2) There are exactly 4 chandeliers of size A.
(1) There are 247 bulbs in total, including all the chandeliers of all three sizes.
(2) There are exactly 4 chandeliers of size A.
18 May 2018, 21:00
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amanvermagmat
Let there be \(a\), \(b\) & \(c\) chandeliers of Size A, Size B & Size C respectively. We need to find \(b\)?
Given \(a+b+c=10\) -----------(1)
Statement 1: let there be \(x\), \(y\) & \(z\) bulbs in chandelier of Size A, Size B & Size C respectively. From the ranges given it implies that \(z>y>x\)
Hence we have \(ax+by+cz=247\) ----------------(2)
But this information is not helpful in finding \(b\). Insufficient
Statement 2: \(a=4 => b+c=6\), or \(c=6-b\)------------------(3)
But this information is not helpful in finding \(b\). Insufficient
Combining 1 & 2: equation (2) can be written as \(4x+by+(6-b)z=247\), Solve this to get \(b\) as
\(bz-by-6z=4x-247 => b=\frac{4x+6z-247}{(z-y)}\)-------------(4)
Now \(b\) is positive, hence \(4x+6z>247 => z>\frac{247-4x}{6}\)
the minimum value of \(z\) will occur when \(x\) is maximum. Maximum \(x\) possible is \(x=18\). Hence \(z>\frac{247-4*18}{6} =>z>29.16\)
As \(z\) is integer, hence \(z=30\). So our equation (4) becomes
\(=> b=\frac{4x+180-247}{(30-y)}=>b=\frac{4x-67}{(30-y)}\)
Now \(4x>67\), because \(b\) is positive. Hence \(x\) is either \(17\) or \(18\). So
\(b\) is either \(\frac{1}{30-y}\) or \(\frac{5}{30-y}\). Now least value of \(30-y\) can be \(5\) when \(y=25\) and as \(b\) is integer hence
\(b=\frac{5}{30-25}=1\). Sufficient
Option C
General Discussion
18 May 2018, 20:28
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Tough one...
Given:
A: 15-18 bulbs
B: 22-25 bulbs
C: 28-30 bulbs
A+B+C=10
Find B.
Statement 1
247 total bulbs. I thought to myself "how can I make the 7 in the units digit go away to leave a number that ends in 5 or 0". Obvious option is to have 17 bulbs for A. If we have one A chandelier, then that leaves 230 bulbs to split among 9 chandeliers. Let's add one C chandelier with 30 bulbs. Now there are 200 bulbs left to split among 8 chandeliers. That's a nice even number that happens to be perfectly divisible by 25. So we have eight B chandeliers with 25 bulbs. That's a total of 10 chandeliers with 247 bulbs. It can't be that easy though... there has to be some other combination hiding there...
Guessing and checking other combinations will take too long. Let's see if Statement 2 can give us a hint. Statement 2 says there are four A chandeliers, so lets try that. Again, I will try to reduce 247 total bulbs to a number that ends in 5 or 0 so that I can easily utilize 25 bulbs for B and 30 bulbs for C. Only way I can do that is by giving A 18 bulbs (18*4=72). Now we have 175 bulbs for 6 chandeliers. That's one B (with 25 bulbs) and five C (with 30 bulbs). So this is a 2nd example of 10 chandeliers with 247 bulbs.
Insufficient
Statement 2
We can have 4 A, 1 B, 5 C.
Or we can have 4 A, 2 B, 4 C... etc
Insufficient
Combining both statements
The only combo that has 247 bulbs and 4 A is:
4 A (with 18 bulbs)
1 B (with 25 bulbs)
5 C (with 30 bulbs)
We can be confident that no other combo exists because we gave A the max number of bulbs, then we maximized the remaining bulbs. So any other combo with 4 A will produce less than 247 bulbs.
Sufficient
Answer: C
Given:
A: 15-18 bulbs
B: 22-25 bulbs
C: 28-30 bulbs
A+B+C=10
Find B.
Statement 1
247 total bulbs. I thought to myself "how can I make the 7 in the units digit go away to leave a number that ends in 5 or 0". Obvious option is to have 17 bulbs for A. If we have one A chandelier, then that leaves 230 bulbs to split among 9 chandeliers. Let's add one C chandelier with 30 bulbs. Now there are 200 bulbs left to split among 8 chandeliers. That's a nice even number that happens to be perfectly divisible by 25. So we have eight B chandeliers with 25 bulbs. That's a total of 10 chandeliers with 247 bulbs. It can't be that easy though... there has to be some other combination hiding there...
Guessing and checking other combinations will take too long. Let's see if Statement 2 can give us a hint. Statement 2 says there are four A chandeliers, so lets try that. Again, I will try to reduce 247 total bulbs to a number that ends in 5 or 0 so that I can easily utilize 25 bulbs for B and 30 bulbs for C. Only way I can do that is by giving A 18 bulbs (18*4=72). Now we have 175 bulbs for 6 chandeliers. That's one B (with 25 bulbs) and five C (with 30 bulbs). So this is a 2nd example of 10 chandeliers with 247 bulbs.
Insufficient
Statement 2
We can have 4 A, 1 B, 5 C.
Or we can have 4 A, 2 B, 4 C... etc
Insufficient
Combining both statements
The only combo that has 247 bulbs and 4 A is:
4 A (with 18 bulbs)
1 B (with 25 bulbs)
5 C (with 30 bulbs)
We can be confident that no other combo exists because we gave A the max number of bulbs, then we maximized the remaining bulbs. So any other combo with 4 A will produce less than 247 bulbs.
Sufficient
Answer: C
