340 MCGRAW-HILL’S SAT
Concept Review 2
- It should look like this:
- When two of the three values are given in a prob-
lem, write them in the pyramid and perform the
operation between them. The result is the other
value in the pyramid. - A median is the “middle” number when the num-
bers are listed in order. If there are an even num-
ber of numbers in the set, it is the average of the
two middle numbers. - The number that appears the most frequently in
a set. - When the numbers are evenly spaced, the mean
is always equal to the median. This is true
more generally if the numbers are distributed
“symmetrically” about the mean, as in [−10, −7,
0, 7, 10]. - If the average of four numbers is 15, then their
sum must be (4)(15) =60. If one of the numbers is
18, then the sum of the other three is 60 − 18 =42.
So the average of the other three is 42/3 =14.
7. You need to read this problem super-carefully. If
the average of the five numbers is 25, then their
sum is (5)(25) =125. If none of the numbers is
less than 10, and since they are all different inte-
gers,the least that four of them can be is 10, 11,
12, and 13. Therefore, if xis the largest possible
number in the set,
x+ 10 + 11 + 12 + 13 = 125
Simplify: x+ 46 = 125
Subtract 46: x= 79
8. If the 20 students in Ms. Appel’s class averaged 90%,
then they must have scored a total of (20)(90) =
1,800 points. Similarly, Mr. Bandera’s class scored
a total of (30)(80) =2,400. The combined average is
just the sum of all the scores divided by the number
of scores: (1,800 +2,400)/50 =84.
Notice, too, that you can get a good estimate by
just noticing that ifthere were an equal number
of students in each class, the overall average
would simply be the average of 80 and 90, which
is 85. But since there are more students in
Mr. Bandera’s class, the average must be weighted
more heavily toward 80.
Answer Key 2: Mean/Median/Mode Problems
SAT Practice 2
- D The average of 2x, 2x, y,and 3yis
(2x+ 2 x+y+ 3 y)/4 =(4x+ 4 y)/4 =x+y.Substitut-
ing 2x+1 for ygives x+(2x+1) = 3 x+1. - C If the average of seven integers is 11, their
sum is (7)(11) =77. If each of these integers is less
than 20, then the greatest any can be is 19. The
question doesn’t say that the integers must be dif-
ferent, so if xis the least possible of these integers,
x+ 19 + 19 + 19 + 19 + 19 + 19 =77.
Simplify: x+ 114 = 77
Subtract 114: x=− 37 - 4 The median is the average of the two middle
numbers. A little trial and error shows that 1, 4, 6,
and 8 have a median of 5, so kmust be 4.
4. C Call the number you are looking for y. The av-
erage of xand yis z,so set up the equation and
solve:
(x+y)/2 =z
Multiply by 2: x+y= 2 z
Subtract x: y= 2 z−x
5. A Just fill in the pyramid: n= 12 m/3k= 4 m/k.
6. B The average is 0, so (5 + 8 + 2 +k) =0. Solving
for kgives us k=−15. So we put the numbers in
order: −15, 2, 5, 8. Since there are an even number
of numbers, the median is the average of the two
middle numbers: (2+5)/2 =3.5.
7. B The most frequent number is 1, so c=1. This
means that statement III is untrue, and you can
eliminate choices (D) and (E). To find the median,
you need to find the average of the 10th and 11th
numbers, when you arrange them in order. Since
average
how
manysum×÷ ÷