amanvermagmat wrote:
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.
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 \(b\).
InsufficientStatement 2: \(a=4 => b+c=6\), or \(c=6-b\)------------------(3)
But this information is not helpful in finding \(b\).
InsufficientCombining 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
\(b=\frac{5}{30-25}=1\).
SufficientOption
C