4 Numbers Average

The sum of the 4 numbers is $44 \times 4 = 176$ .
The sum of the first 3 numbers is $33 \times 3 = 99$ .
Hence the 4th number is $176 - 99 = 77$ .

Slightly surprising that it is so large right?

$33 \times 3 + D = 44 \times 4$ .

$D = 176 - 99$ .

$D = 77$ .

The first three are 11 less than the final average. So the fourth one is (44+11*3)

Given $(x_1 + x_2 + x_3 + x_4) / 4 = 44$ ,

it implies $\sum_{i=1}^4 x_i = 176$

Given $(x_1 + x_2 + x_3) /3 = 33$ ,

it also implies $\sum_{i=1}^3 x_i = 99$

But $\sum_{i=1}^4 x_i = ( \sum_{i=1}^3 x_i ) + x_4 = 176$

Therefore, $x_4 = 176 - 99 = 77$

The sum of the first three numbers is $3(33)=99$ . Let $x$ be the fourth number. So the sum of the first $4$ numbers is $99+x$ . and that is also equal to $4(44)$ . So we have

$99+x=4(44)$

$\color{#D61F06}\boxed{\large x=77}$

Let 'x' be the sum of 1st three numbers and 'y' be the 4th number. So, x/3=33 which makes x=99. then, (x+y)/4=44 or (99+y)/4=44. This makes y=77

A+B+C = 99

A+B+C+D = 176

D = 176 - 99 = 77

33*3=99 99+77=176 now, 176/4=44 Hence, 77 is the correct option

$\frac{A+B+C+D}{4}$ = $\frac{A+B+C}{3}$ + 11

3D = A + B + C + 132

D = $\frac{A+B+C}{3}$ + 132/3

D = 77

• 1/4(X+Y+Z+Q)=44 , X+Y+Z+Q= 176
• 1/3(X+Y+Z)=33, X+Y+Z=99
• 176-99=77

Average= the sum / the numer

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44 = the sum of the four numbers / 4

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The sum of 4 = 176

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33 = the sum of the three numbers /3

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The sum of 3 = 99

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The 4th num = 176 - 99 = 77