The average of the 4 numbers is 44.

The average of the first 3 numbers is 33.

What is the value of the 4 $^\text{th}$ number?

11
33
55
77

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Is it just a coincidence that the 4th number equals the sum of the two averages shown?

Ron Brumleve
- 6 years, 7 months ago

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That's an observation!

It is a coincidence, caused by the numbers (ratios) that were chosen. The underlying math is that if we have $n+1$ numbers with an average of $(n+1) X$ , and $n$ numbers with an average of $nX$ , then the last number would be $(n+1) \times (n+1) X - n \times nX = (2n+1) X = (n+1) X + n X$ .

No, as for eg: let any two numbers be 4 and 6 their average =(4+6)/2=5 so 2 is number of numbers(4,6) 5*2=10 that is sum of the two numbers taken

Yash Prakash
- 6 years, 7 months ago

O thank you. How i just got the answer it could be beginners luck but what i did was kind simple i just added the 2 averages 44+33 n got 77 n reciveved a corrtect response so can we do it that way to or is that beginners luck and not a good or effective way to do so? Thanl you for any feedback on this comment

Lawrence Webb
- 3 years, 6 months ago

yes i agree

Caleb Albuquerque
- 3 years, 4 months ago

yes I also agree

Felipe Coletti
- 1 year, 9 months ago

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

$D = 176 - 99$ .

$D = 77$ .

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

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Its only possible when all are equal... Which is not given.

URVISH SOLANKI
- 6 years, 7 months ago

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, $x_4 = 176 - 99 = 77$

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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

$99+x=4(44)$

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

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Nice solution.

Marvin Kalngan
- 2 months, 3 weeks ago

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A+B+C = 99

A+B+C+D = 176

D = 176 - 99 = 77

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

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

3D = A + B + C + 132

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

D = 77

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• 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

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Average= the sum / the numer

44 = the sum of the four numbers / 4

The sum of 4 = 176

33 = the sum of the three numbers /3

The sum of 3 = 99

The 4th num = 176 - 99 = 77

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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?