Problem Solving

5 pieces of wood have an average length of 124cm and median length of
140cm. what is the maximum possible length of the shortest piece of wood?
a. 90
b. 100
c. 110
d. 130
e. 140

OA is B

Akansha wrote:5 pieces of wood have an average length of 124cm and median length of
140cm. what is the maximum possible length of the shortest piece of wood?
a. 90
b. 100
c. 110
d. 130
e. 140

Hi Akansha!

Let's start by analyzing the question stem. We know that:

5 pieces
Average length 124cm, so sum = #*avg = 5*124cm = 620cm
Median length 140cm; odd number of terms, so the middle term must be 140.

Our goal: to find the LONGEST POSSIBLE length of the SHORTEST piece of wood.
We think: on the GMAT, to maximize one thing you must try to minimize everything else.

With 140 as our median, we know the set is {a, b, 140, c, d}

Nowhere does it say that the 5 pieces of wood have to be of different length, so let's start by picking minimum values for c and d: 140 each.

Now our set is {a, b, 140, 140, 140}

3*140 = 420, so that leaves 620cm - 420cm = 200cm for a and b.

Since we want to make the smallest piece as BIG as possible, let's let a=b.

So, 2a = 200cm, a = 100cm.

Accordingly, the BIGGEST possible value for the SMALLEST piece of wood is 100cm... choose (B)!

Akansha wrote:5 pieces of wood have an average length of 124cm and median length of
140cm. what is the maximum possible length of the shortest piece of wood?
a. 90
b. 100
c. 110
d. 130
e. 140

OA is B

Let the length of 5 individual pieces be = a,b,c,d,e such that a<b<c<d<e
Average=124
then a+b+c+d+e=5*124=620
Median =140 =c (Middle value)

Now our purpose is to maximize a

So we need to make sure the order becomes a<b<c=d=e
this implies c=d=e=140
there for c+d+e =140*3=420
Therefore a+b =620-420=200

Now a will be maximum only when it is equal to b. Hence order becomes a=b<c=d=e
So a+b=200; putting a=b; 2a=200 -> a=100 ---> Option B

let lengths be a,b,c,d,e (in ascending order)
As median = 140 =>c=140
also a+b+c+d+e = 5*124 = 620
To maximize a, minimize b,d,e (c is constant)
as d,e>=c thus min d,e = 140
=>a+b+140+140+140=620 => a+b = 200
now a<=b, to maximize a, a=b and thus a+a=200 => a=100
IMO B