Calculating using the mean

Number Theory Level 2

There are two baskets, one of each contains only apples and the other contains only oranges.

Assume that the means of the weight of the apples and the oranges are a a and b b , respectively; and the mean of the weight of both apples and oranges is c c .

It's given that:

  • a ≠ b a \ne b

  • a + b 2 = c \frac{a+b}{2} = c

Which of the following is true about the number of apples and oranges, using the information given?

They are equal There are more oranges than apples. There are no such numbers. Insufficient information There are more apples than oranges.

Tin Le by Tin Le , 15, Vietnam

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2 solutions

Chris Lewis
Oct 30, 2020

Say there are A A apples and B B oranges. Using the given mean weights, we can express the total weight of fruit in two different ways: a A + b B = c ( A + B ) aA+bB=c(A+B)

Substituting for c c , a A + b B = a + b 2 ⋅ ( A + B ) 2 a A + 2 b B = ( a + b ) ( A + B ) 2 a A + 2 b B = a A + a B + b A + b B 0 = a A − a B − b A + b B 0 = ( a − b ) ( A − B ) \begin{aligned} aA+bB &= \frac{a+b}{2} \cdot (A+B) \\ 2aA+2bB &=(a+b)(A+B) \\ 2aA+2bB &=aA+aB+bA+bB \\ 0 &=aA-aB-bA+bB \\ 0 &=(a-b)(A-B) \end{aligned}

From this last factorisation, we conclude that either a = b a=b or A = B A=B . Since we're told a ≠ b a \neq b , it must be the case that A = B A=B ; that is there are the same number of apples as oranges.

2 Helpful 1 Interesting 2 Brilliant 0 Confused

An orange and an apple might have different weights, so the respective average of them can vary,

The average weight of n n apples

a 1 + a 2 + a 3 + . . . . a n n = a \dfrac{a_1+a_2+a_3+....a_n}{n}=a

The average weight of m m oranges

o 1 + o 2 + o 3 + . . . . o m m = b \dfrac{o_1+o_2+o_3+....o_m}{m}=b

Given that, a + b 2 = c \dfrac{a+b}{2}= c and ( a 1 + a 2 + a 3 + . . . . a n ) + ( o 1 + o 2 + o 3 + . . . . o m ) m + n = c \dfrac{(a_1+a_2+a_3+....a_n)+(o_1+o_2+o_3+....o_m)}{m+n}=c

( a 1 + a 2 + a 3 + . . . . a n ) + ( o 1 + o 2 + o 3 + . . . . o m ) m + n = a + b 2 \dfrac{(a_1+a_2+a_3+....a_n)+(o_1+o_2+o_3+....o_m)}{m+n}=\dfrac{a+b}{2}

a n + b m m + n = a + b 2 \dfrac{an+bm}{m+n}=\dfrac{a+b}{2}

2 ( a n + b m ) = ( a + b ) ( m + n ) 2(an+bm)= (a+b)(m+n)

If m = n m=n

2 m ( a + b ) = 2 m ( a + b ) 2m(a+b)= 2m(a+b) , the equation satisfies.

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