75 Since we’re starting off with a group of 10 numbers and then removing 2 of them, the highest and the lowest, let’s express the sum of the 10 numbers as the sum of the highest number,
Trang 1So put in your grid (And even though is the same as 1 , don’t be tempted to
put 1 into the grid, since you can’t grid in mixed numbers!)
22 9,984
If the ratio of students to teachers to parents is 11:2:3, then there might be 11
students, 2 teachers, and 3 parents, for a total of 16 people On the other hand,
there might be 10 × 11, or 110 students, 10 × 2, or 20 teachers, and 10 × 3, or 30
parents, for a total of 10 × 16, or 160 people In fact there could be any number of
people at all, as long as they are there in the ratio 11:2:3 That means that the
number of people in attendance must be a multiple of 11 + 2 + 3, or 16 So what
this question is really asking is – what’s the greatest multiple of 16 that fits on the
grid? Since the grid only has 4 columns, the largest multiple of 16 that fits will be a
4-digit number, in other words a number greater than 999 but less than 10,000 So
you need the largest multiple of 16 that is less than 10,000 The easiest way to find
that number is to use your calculator and your knowledge of arithmetic First of all,
= 625, so 10,000 is a multiple of 16 Therefore, the largest multiple of 16 that the grid can accomodate must be 10,000 – 16 = 9,984 You can check that
624× 16 = 9,984 on your calculator, but you don’t have to
23 75
Since we’re starting off with a group of 10 numbers and then removing 2 of them,
the highest and the lowest, let’s express the sum of the 10 numbers as the sum of
the highest number, the lowest number, and the other 8 numbers The sum of the
10 numbers is equal to the highest number + the lowest number + the sum of the
other 8 numbers If we call the highest number h, the lowest number , and the
sum of the other 8 numbers S, we can express the sum of the 10 numbers as
h + + S Then the average of the 10 numbers is , which we’re given is
87 The average of the eight numbers is just , and we’re given that that’s 90 We
want to know the average of the 2 grades that were removed, or
If = 90 then S = 90 × 8 = 720 And if = 87 then h + + S = 87 × 10 =
870 But we know that S = 720, so h + + 720 = 870, and h + = 150 Then the
average of h and is 150, or 75, so grid in a 75
2
h + + S
10
S
8
h +
2
S
8
h + + S
10
10,000
16
1
3
1
3
4
3
4
3
SAT
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25
Trang 224 12
Here we have a figure that is not drawn to scale, so eyeballing won’t help much The area of the top of the solid is much greater than the areas of the sides, so the solid probably doesn’t look much like the figure anyway Don’t let the figure mislead you
You know that volume is length × width × height, so label the edges accordingly Let’s say that the vertical lines represent the height of the solid and label all the vertical lines h Let’s also say that the horizontal lines represent the width and label them w, and the diagonal lines are the length, so label them So the area of face I
is w , the area of face II is hw, and the area of face III is h This means that we can write down 3 algebraic equations; w = 24, hw = 2, and h = 3 Now you have
3 equations and 3 unknowns and you should be able to solve It’s not obvious how
to solve, however, so you’ll have to play around with these equations Adding or subtracting them doesn’t get you very far, but if you multiply any 2 of them together something interesting happens For example, if you multiply the first 2 equations together you get a new equation, w× hw = 24 × 2, or w2h = 48 Notice that the
left side of the new equation is w2 times h, which is also w2times the left side of the third equation So, if you divide the new equation by the third equation you get
= , or w2= 16 That means that w must be 4, so is 6 and h is , and the volume is 4 × 6 × , or 12
25 25
Increasing 45 by P percent means adding P percent of 45 to 45, so increasing 45 by
P percent results in 45 + (P % of 45) or 45 + × 45 which equals 45 + Decreasing 75 by P percent means taking P percent of 75 away from 75 so decreasing 75 by P percent results in 75 – (P % of 75) or 75 – × 75 which equals 75 – Increasing 45 by P percent gives the same result as decreasing
75 by P percent, so we have the equation 45 + = 75 – Now let’s solve this equation for P Subtracting 45 from both sides results in = 30 – Adding to both sides gives us = 30 Multiplying both sides by results in P = × 30 = Cancelling a factor of 30 from the top and bottom of gives us 100× 1 which equals 25 So P = 25
4
100 × 30
120
100 × 30
120
100
120
100
120
120P
100
75P
100
75P
100
45P
100
75P
100
45P
100
75P
100
P
100
45P
100
P
100
1
2
1
2
48
3
w2h
h
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Trang 3Section 4 (Math)
1 B Instead of trying to multiply the fractions in Column A, try canceling as much as you
can:
Now compare the columns Since is greater than 1 and is less than 1, is
greater than , and Column B is greater than Column A
2 B This problem is relatively easy to calculate Since 4 quarters make up one dollar and
10 dimes make up one dollar, simply multiply each of these numbers by 5 and
compare them 20 quarters can be changed for $5.00, while 50 dimes can be
changed for the same amount Column B is greater
An even faster way to do this problem is to simply compare and skip the calculations
altogether Since a quarter is worth more than a dime, it takes fewer quarters than
dimes to make up $5.00 Therefore, the number of dimes that are exchanged for
$5.00 is greater than the number of quarters exchanged for $5.00 and the answer is
(B)
3 A Solve the centered equation for x :
Now plug 3 in for the x in Column A 3 + 2 = 5, which is greater than 4, so Column A
is greater than Column B
4 C The easiest way to do this problem is to list all the possible values of a + b :
Now count up how many DIFFERENT sums there are There are 6 combinations
total, but 4 + 6 and 7 + 3 both give you 10, so there are only 5 different sums for
a + b The columns are equal and the answer is (C).
4+ 3 = 7
4+ 6 = 10
7+ 3 = 10
7+ 6 = 13
8+ 3 = 11
8+ 6 = 14
3x − 12 = 3 − 2x 5x = 15
x = 3
4 5
5 4
4 5
5 4
4
9 ×18
6 × 12
20 = 1
1× 2
1× 2
5 = 4 5
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Trang 45 A The easiest way to deal with this problem is to get 2 out of the denominator of
Column B To do this, multiply each column by 2, yielding 9p in Column A and p in Column B Multiplying p by 9 brings its value closer to 0, and since p is negative, it increases in value Therefore, Column A is greater than Column B
If this is too abstract for you, pick numbers Choose a number for p that’s easy to work with and meets the criteria of the centered information, like –.2 Multiplying –.2
by 4.5 gives you –.9, while dividing –.2 by 2 gives you –1 Since –.9 is greater than –1, Column A is greater than Column B
6 C First, try to make the expressions in each column look more like the centered
information You can do this by breaking each expression into its factors, rewriting Column A as (6)2(x – y )2and Column B as (6)2(y – x )2 Now you can compare the columns piece by piece Since both columns contain (6)2, ignore it Now compare the second term of each expression You know that x – y = 6, so (x – y )2 = 36 What does y – x equal? Well, y – x = (–1)(x – y), which equals –6 Since (–6)2= 36, the two columns are equal
7 B Compare—don’t calculate! Use common sense to find the relationship between the
two columns In Column A, the average price of 2 dozen grade A eggs and 2 dozen grade B eggs is the same as the average of 1 dozen of each In Column B, however, the average of 2 dozen grade A eggs and only 1 dozen grade B eggs is closer to the price of grade A eggs than it is to the price of grade B eggs Since grade A eggs are more expensive than grade B eggs, Column B is greater than Column A
8 B You don’t have to know the values of a and b to make the comparison Since SRU
and TRV share TRU, the only difference between the two angles is the difference
between a and b TRVis larger than SRU, which means that b must be larger than a
and Column B must be greater than Column A
9 D First solve the equation in the centered information for x2:
Your first instinct may be to jump to the conclusion that x = 3, which makes Column A greater than Column B However, don’t forget that x could also equal –3, which is less than 2 In that case, Column B would be greater than Column A Since there are two possible relationships between the columns, the answer is (D)
5x2 + 1= 46
5x2 = 45
x2 = 9
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Trang 510 C At first you may look at this QC and decide that there’s not enough information to
make a comparison This problem is doable, however, if you realize that the two
triangles share the unmarked angle on the right We will call the measure of that
shared angle c for purposes of discussion The sum of the angles of the larger
triangle can be written as x + y + c = 180, while the sum of the angles of the smaller
triangle can be written as a + b + c = 180 Because both sums equal 180, set them
equal to each other:
x + y + c = a + b + c
Subtract c from both sides:
x + y = a + b
Now that we have the relationship between the measures of the angles in this form,
we can rearrange the variables to match the expressions in the columns:
x – b = a – y
The answer is (C)
11.D Start this problem by multiplying out the expressions using FOIL This gives you
x2– 2xy + y2 = x2+ 2xy + y2 You can subtract x2+ y2from each side of the equation,
leaving –2xy = 2xy Getting all the variables over to one side leaves –4xy = 0, which
can be simplified to xy = 0 Because you aren’t given any information about the value
of either variable, x or y must equal 0, but either one could In fact, if x = 0, y could be
any value, positive, negative or zero Therefore, the answer is (D)
12.A If the 1990 tuition was 20 percent greater than the 1980 tuition, then it was equal to
the 1980 tuition times 120 percent Knowing this allows you to set up and solve an
equation for the 1980 tuition:
Column A is greater than Column B
13.A This problem is most easily solved by making each side look more like the other To
do this, break each quantity into components In Column A, 25100can be broken into
(2550)(2550) The value in Column B can be rewritten as (2 × 25)50
, or (250)(2550)
Since both columns have one factor of 2550, it can be canceled out What’s left is
2550 in Column A and 250in Column B Since 2550 is greater than 250, Column A is
greater than Column B
3,000= 120%(1980 tuition) 3,000= 1.2(1980 tuition) 3,000÷ 1.2 = 1980 tuition
1980 tuition= $2,500
SAT
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Trang 614 D First solve for x by subtracting 1 from both sides of the equation in the centered
information This gives you x = 2 Now you must find the range of values for y, but be careful If y2is greater than 9, it may be true that y > 3, but it may also be true that y
< –3 Substitute these values into Column A and see how the results compare to 5 If
x = 2 and y > 3, then x + y > 5 However, if y < –3, then x + y < –1, which is less than
5 Since more than one relationship exists between the two columns, the answer is (D)
15.B There are no numbers provided, so you might think that there’s not enough
information to compare the areas and that the answer is (D) That, however, would be too easy for the last QC Manipulate the diagram a bit and you can see how the area
of the triangle compares to the area of the rest of the square First, imagine how they’d compare if AB were coincident with DE:
If this were the case, the area of the triangle would be exactly half that of the square, and therefore equal to the area of the shaded regions Now, if you move A and B a bit
to the right, the area of the triangle decreases:
The more you moveA and B to the right, the smaller the area of the triangle:
So you can see that the area of the triangle is less than half that of the square
Therefore the sum of the areas of the shaded regions is greater than the area of the triangle, and Column B is greater
A
B
C A
B
C
A
B
C
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Trang 716 12 First multiply both sides of the equation by x + 2:
Now solve for x:
The question asks for the value of 3x, not x, so multiply 4 by 3 Your answer is 12
17.192, 480, or 960
The best way to do this problem is to use trial and error Just choose three
consecutive even integers and see if their product is a 3-digit number Start with 2, 4,
and 6 and go from there:
2 × 4 × 6 = 48 That doesn’t work, so now try 4, 6, and 8:
4 × 6 × 8 = 192 That combination works, so grid in 192 Other possible combinations are 6 × 8 × 10 =
480 and 8 × 10 × 12 = 960
18.6.5 or 13/2
An equilateral triangle with side x has 3 sides of length x, which means it has a
perimeter of 3x The second triangle has sides of length 5, 8, and x, which means its
perimeter is equal to 5 + 8 + x, or 13+x Since the two perimeters have the same
value, you can set 3x equal to 13 + x and solve for x:
Remember that 6 is not a griddable response and the answer should either be
gridded as an improper fraction or a decimal
1 2
3x = 13 + x 2x = 13
x =13
2 = 6.5
30= 5x + 10
20= 5x
4= x
30
x + 2 = 5
30= 5(x + 2)
30= 5x + 10
SAT
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Trang 819 3 To find how many peaches are left over, divide 675 by 48 The result of this
calculation is 14 remainder 3 The remainder is the number of peaches that do not fit into a crate, which in this case is 3
20.1 < AE ≤ 1.41
Although you have no way of finding AE, you can find a range of values that AE must
be within Since the perimeter of the square is 4, each side of the square must have
a length of 1 AE forms the hypotenuse of right triangle ABE, which means that it is longer than side AB Therefore, AE must be greater than 1 Since AE is not a diagonal of the square, it must be shorter than the length of AC, one of the square’s diagonals Draw in AC:
Look at triangle ADC It is an isosceles right triangle, so you can use the ratio of side lengths to find the length of AC The ratio of the sides of an isosceles right triangle is 1:1: , and since the legs of this triangle equal 1, the length of ACis , which is approximately 1.414 Therefore, AE is somewhere between 1 and 1.414, so choose a value between them
21.996
It would take too much time to write out the entire series until you reach 1,000
Notice, however, that each term in the series is 2 plus a multiple of 7 So start by finding the highest 3-digit number that is divisible by 7 and add 2 to it The best way
to find the biggest 3-digit number divisible by 7 is to divide 999 by 7 This yields 142 remainder 5 To find a number that is evenly divisible by 7, subtract 5 from 999, leaving you 994 This is the largest 3-digit number divisible by 7, but it is not the answer Remember that the terms in the series equaled 2 plus a multiple of 7 and you must add 2 to 994, giving you 996
22.7 The formula for finding the slope given two points is:
Slope = Plug the given x’s and y’s, most of which are in terms of a, into the formula and set the fraction equal to the value you were given, 9 Then solve for a:
rise run = difference in y ’s
difference in x ’s
2 2
E C D
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Trang 923 4.80
You only have to find the amount of money saved by purchasing 2 dozen roses, so
don’t spend your time finding out how much 2 dozen roses cost Instead, since you
are given that 1 rose is free with every 5 purchased, and since 1 rose costs $1.20,
you only need to know how many roses are free of charge when you buy 2 dozen
One out of every 6 roses is free, and there are 24 roses in 2 dozen, so you would
receive 24 ÷ 6 = 4 free roses in 2 dozen 4 × $1.20 = $4.80, which is the answer
24.27.5 or 55/2
If you enlarge something by 150 percent, the new dimensions are actually 250 percent
of the old ones Therefore, when trying to determine the photograph’s new
dimensions, multiply the original ones by 2.5, not 1.5:
The perimeter of a rectangle equals 2 × (length + width), so the perimeter of the
enlarged picture equals 2(8.75 + 5), which is 2(13.75), or 27.5
25 21 To find the area of a triangle, you need an altitude Drop an altitude from vertex E
(For purposes of discussion, let’s call the point where this altitude hits the base
point F.)
The area of ∆A CEis equal to (base × height), or (AC)(EF) You’re not given
enough information to find the numerical values of ACand EF, but you can find their
values in terms of BC and BD First, it’s given that BC = AC, so you can say that
AC = 3BC Second, because ∆EFC is similar to ∆BDC (they’re both right triangles
1 3
1 2
1 2
A
E
D
C B
F
3.5× 2.5 = 8.75
2× 2.5 = 5.0
Slope = Difference in y ' s
Difference in x ' s
9= 2a− 5
(a + 1) − a
9= 2a− 5
1 = 2a − 5
14= 2a
a= 7
SAT
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33
Trang 10and they both include angle C) and DC = EC, you know that BD = EF, or, in
other words, EF = 2BD Now you can express the area of ∆A CEthis way:
Because it’s given that the area of ∆BDC is , you know that (BC)(BD) = , and therefore that (BC)(BD) = 7 Plug that into the above expression for the area of
∆A CEand you get:
Area of ∆ACE = 3(BC)(BD) = 3(7) = 21
7 2
1 2
7 2
Area of ∆ACE = 1
2(AC )(EF )= 1
2(3BC )(2BD) = 3(BC)(BD)
1 2
1 2
SAT
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