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11 May 2022, 07:15
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The average (arithmetic mean) of 10, 30, 50, and 70 is 5 more than the average of 20, 40, and
A 35
B 40
C 45
D 50
E 60
A 35
B 40
C 45
D 50
E 60
11 May 2022, 08:44
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Bunuel
The average (arithmetic mean) of 10, 30, 50, and 70 is 5 more than the average of 20, 40, and
A 35
B 40
C 45
D 50
E 60
\(\frac{10+30+50+70}{4} = \frac{20+40+x}{3} + 5\)A 35
B 40
C 45
D 50
E 60
So, \(3(35) = 60 + x\)
Or, \(x = 45\), Answer must be (C)
11 May 2022, 09:58
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You could set up the algebra...or you could get the right answer faster and without any risk of making a silly calculation error...your choice!
If you just eyeball the first group of numbers for a second before you start writing anything, you can probably spot that the mean is 40 (halfway between 30 and 50, and also halfway between 10 and 70). If you really need to, you can sum the four elements (160) and divide by 4 to get 40.
Okay, so the average of the first group is 40. And that's 5 more than the average of the second group. So the average of the second group is 35.
There are three elements in the second group. The math is pretty straight-forward, so maybe you just crunch the numbers: sum of the three elements/3 = 35, so sum of the three elements = 105. We've already got 20 and 40, which is 60. That leaves 45 that we still need, so answer choice C.
But what if the arithmetic were a little more cumbersome on this last part? We need an average of 35 in the second set, and there are an odd number of elements, so the sum of the elements has to have a units digit of 5. This eliminates answer choices B, D, and E. There are only two answer choices remaining that would result in a units digit of 5. One of those answer choices is 35 (trap answer alert!), but that would only work if the other elements in the set averaged to also be 35. They don't, so A has to be wrong, and we are left with C...and we didn't really do any math!
If you just eyeball the first group of numbers for a second before you start writing anything, you can probably spot that the mean is 40 (halfway between 30 and 50, and also halfway between 10 and 70). If you really need to, you can sum the four elements (160) and divide by 4 to get 40.
Okay, so the average of the first group is 40. And that's 5 more than the average of the second group. So the average of the second group is 35.
There are three elements in the second group. The math is pretty straight-forward, so maybe you just crunch the numbers: sum of the three elements/3 = 35, so sum of the three elements = 105. We've already got 20 and 40, which is 60. That leaves 45 that we still need, so answer choice C.
But what if the arithmetic were a little more cumbersome on this last part? We need an average of 35 in the second set, and there are an odd number of elements, so the sum of the elements has to have a units digit of 5. This eliminates answer choices B, D, and E. There are only two answer choices remaining that would result in a units digit of 5. One of those answer choices is 35 (trap answer alert!), but that would only work if the other elements in the set averaged to also be 35. They don't, so A has to be wrong, and we are left with C...and we didn't really do any math!
