Suppose that the average (arithmetic mean) of ?, ?, and ? is ℎ, the

Question

Suppose that the average (arithmetic mean) of 𝑎, 𝑏, and 𝑐 is ℎ, the average of 𝑏, 𝑐, and 𝑑 is 𝑗, and the average of 𝑑 and 𝑒 is 𝑘. What is the average of 𝑎 and 𝑒?

A.  h-j+k

B.  \frac{3h+3j-2k}{2}

C.  \frac{3h-3j+2k}{2}

D.  \frac{3h-3j+2k}{5}

E.  \frac{3h-3j+2k}{8}

Kinshook · Answer

Given: Suppose that the average (arithmetic mean) of 𝑎, 𝑏, and 𝑐 is ℎ, the average of 𝑏, 𝑐, and 𝑑 is 𝑗, and the average of 𝑑 and 𝑒 is 𝑘. 
Asked: What is the average of 𝑎 and 𝑒?

a + b + c = 3h
b + c + d = 3j
d + e = 2k

a - d = (a + b + c) - (b + c + d) = 3h - 3j
d + e = 2k
a + e = (a-d) + (d+e) = 3h - 3j + 2k
Average of a & e = (a + e)/2 = (3h - 3j + 2k)/2

IMO C

chetan2u · Answer

Another way would be to take some values for the variable.

Let a, b, c, d, e to be 1, 2, 3, 4 and 5 respectively. 
So h=2, j=3 and k=4.5

So average of a and e will be (1+5)/2=3

Check the choices for value of 3.
A.  h-j+k  or 2-3+4.5=3.5.......No

B.  \frac{3h+3j-2k}{2}=\frac{3*2+3*3-2*4.5}{2}=\frac{15-9}{2}=3 ..Yes

All choices other than have same numerator 3h-3j+2k=3*2-3*3+2*4.5=6-9+9=6. So, the denominator should be 2 to get 3 as the answer. 
And C has denominator 2.

Thus we have 2 options fitting in.

Change one value to eliminate wrong option.

Let a, b, c, d, e to be 1, 2, 3, 4 and 6 respectively. 
So h=2, j=3 and k=5

So average of a and e will be (1+6)/2=3.5

Check the choices for value of 3.5

B.  \frac{3h+3j-2k}{2}=\frac{3*2+3*3-2*5}{2}=\frac{15-10}{2}=2.5 ..No

C.  \frac{3h-3j+2k}{2}=\frac{3*2-3*3+2*5}{2}=\frac{16-9}{2}=3.5 ..Yes

C

MathRevolution · Answer

Average * Total = sum

Average (arithmetic mean) of 𝑎, 𝑏, and 𝑐 is ℎ: a + b + c = 3h ------ (1)

Average (arithmetic mean) of 𝑏, 𝑐, and 𝑑 is 𝑗: b + c + d = 3j : b + c = 3j - d ------ (2)

Average of 𝑑 and 𝑒 is 𝑘: d + e = 2k: d = 2k - e ----------(3)

From (1) and (2), we get a + 3j - d = 3h ------(4)

Plugging d = 2k - e in (4)

=> a + 3j - (2k - e) = 3h

=> a + e + 3j - 2k = 3h

=> a + e = 3h - 3j + 2k

Average of a and e :  \frac{a + e}{ 2} = \frac{3h - 3j + 2k }{ 2}

Answer C

Ved3 · Answer

Given: i) (a+b+c)/3 = h, ii) (b+c+d)/3 = j, iii) (d+e)/2 = k
From iii) d + e = 2k
 d = 2k - e -> iv)
From ii) b + c + d = 3j
Put iv) here
 b + c + 2k -e = 3j
 b + c = 3j + e -2k -> v)
From i) a + b + c = 3h
Put v) here
 a + 3j +e - 2k = 3h
 a+e = 3h -3j + 2k
 (a + e)/2 = (3h -3j + 2k)/2
Therefore ans is option c

bumpbot · Answer

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Suppose that the average (arithmetic mean) of ?, ?, and ? is ℎ, the

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