Sat - MC Grawhill part 36 ppsx

Digit Problems Some of the most common problems on the SAT are numerical reasoning problems, which ask you to think about what happens to numbers when you perform basic operations on the

Concept Review 2

1 It should look like this:

2 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

3 A median is the “middle” number when the

bers are listed in order If there are an even

num-ber of numnum-bers in the set, it is the average of the

two middle numbers

4 The number that appears the most frequently in

a set

5 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]

6 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 x is 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 if there 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

1 D The average of 2x, 2x, y, and 3y is

(2x + 2x + y + 3y)/4 = (4x + 4y)/4 = x + y

Substitut-ing 2x + 1 for y gives x + (2x + 1) = 3x + 1.

2 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 x is the least possible of these integers,

x+ 19 + 19 + 19 + 19 + 19 + 19 = 77

Simplify: x+ 114 = 77

Subtract 114: x= −37

3 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 k must be 4.

4 C Call the number you are looking for y The av-erage of x and y is z, so set up the equation and

solve:

(x + y)/2 = z

Multiply by 2: x + y = 2z Subtract x: y = 2z − x

5 A Just fill in the pyramid: n = 12m/3k = 4m/k.

6 B The average is 0, so (5 + 8 + 2 + k) = 0 Solving for k gives 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 howmany

sum

×

÷

÷

both of these are 3, b= 3 Therefore, statement II

is true, and you can eliminate choice (A) To find

the average, just divide the sum by 20: ((1)(5)

+ (2)(3) + (3)(3) + (4)(3) + (5)(3) + (6)(3))/20 = 3.25,

so a= 3.25 Therefore, statement I is not true, so

the answer is (B)

8 B When a 30% solution and a 50% solution are

combined, the concentration must be anywhere

between 30% and 50%, depending on how much

of each you added It can’t be 50%, though, because the 30% solution dilutes it

9 A If Set A were {0, 0, 10, 10, 10}, then its median,

m, would be 10 Set B would be {2, 2, 12, 12, 12}.

Inspection of Set B shows that it is a counterexam-ple to statements II and III, leaving (A) as a possible answer

Arithmetic Reasoning

Lesson 3: Numerical Reasoning Problems

The first statement, m < n < p < r, tells you that the

al-phabetical order is also the numerical order of the

numbers The second statement, mnpr= 0, tells you that one of the numbers must be 0 (This is the zero

product property!) The third statement, m + n + p + r

= 0, tells you that you must have at least one positive and one negative, and all the numbers must “cancel

out.” This means that m can’t be 0 because then none

of the numbers would be negative, and r can’t be 0,

because then none of the numbers would be positive

Thus, either n or p is 0 This means that both I and II

are necessarily true, so you can eliminate choices (A),

(B), and (D) The example m = −3, n = 0, p = 1, r = 2

shows that statement III is not necessarily true, so the answer is (C)

Digit Problems

Some of the most common problems on the SAT

are numerical reasoning problems, which ask

you to think about what happens to numbers

when you perform basic operations on them

You just need to know the common numerical

and arithmetic rules and think logically

Example:

If a + b is negative, which of the following CANNOT

be negative?

(A) ab (B) ab2 (C) a2b

(D) a2b2 (E) a − b

Start by thinking about what might be true about a and

b and what must be true about a and b First think of

possible values for a and b.−2 and 1 work, because

a + b = −2 + 1 = −1 Notice that this proves that (A), (B),

and (E) are incorrect, because they can be negative:

(A) ab = (−2)(1) = −2, (B) ab2 = (−2)(1)2 = −2, and

(E) a − b = (−2) − (1) = −3 But (C) a2b= (−2)2(1) = 4 is

positive, so does that mean the answer is (C)? Not so

fast! Your job is not to find which one can be positive,

but rather which cannot be negative Notice that

(C) can be negative if a and b are, say, 1 and −2

(notice that a + b is still negative, so those values

work): (C) a2b= (1)2(−2) = −2 Therefore, by process of

elimination, the answer is (D)

This question is much easier if you remember a

sim-ple fact: If x is a real number, then x2is never

nega-tive If you don’t know this already, play around with

possible values of x until you see why this is true.

Then look at choice (D) a2b2 a2can’t be negative, and

neither can b2, so a2b2can’t be negative

Example:

If m < n < p < r, mnpr = 0, and m + n + p + r = 0,

then which of the following must be true?

I If m and n are negative, then p= 0

II np= 0

III m + r = 0

(A) I only (B) II only

(C) I and II only (D) I and III only

(E) I, II, and III

You may see a question on the SAT like the one

below, where letters represent digits Remember that digits can only take the values 0, 1, 2, 3, 4,

5, 6, 7, 8, and 9 Also remember that you may

have to consider “carried” digits when looking

at a sum or product Lastly, you may find it best to work from left to right rather than right to left

Example:

1BA + 8B

2 1 1

If A and B represent distinct digits in this addi-tion problem, what is the value of A − B?

(A) 9 (B) 7 (C) 2 (D) 7 (E) 9

Look at the left (hundreds) column first Since the sum has a 2 in the hundreds place, there must be a

carry of 1 from the tens place Therefore, B + 8 + (carry from ones column, if any) = 11 This means

B = 2 or 3 Trying each one shows that only

B = 2 and A = 9 works, giving 129 + 82 = 211 There-fore A − B = 9 − 2 = 7, so the answer is (D).

1 If neither a nor b is 0, what is the relationship between a ÷ b and b ÷ a?

2 What is the relationship between a − b and b − a?

Complete the following “parity rules.”

3 Odd × even = _ 4 Even × even = _ 5 Odd × odd = _

6 Even + even = _ 7 Odd + even = _ 8 Odd + odd = _

Complete the following “sign rules.”

9 If n is odd, (−1)n= _ 10 If n is even, (−1)n= _ 11 If x + y = 0 and x ≠ 0, then x/y = _.

12 Dividing by x is the same as multiplying by .

13 Subtracting (x+ 1) is the same as adding

14 When a number is multiplied by its reciprocal, the result is

15 When a number and its opposite are added, the result is

16 When a number (other than 0) is divided by its opposite, the result is

17 If a positive number is multiplied by a number greater than 1, what happens to it? _

18 If a positive number is multiplied by a number between 0 and 1, what happens to it? _

19 If a negative number is multiplied by a number greater than 1, what happens to it? _

20 Is x always bigger than −x? Explain.

21 Is x2always bigger than x? Explain.

22 Is x3always bigger than x2? Explain

23 If x is between 0 and 1, then 1/x is _.

24 If a > b > 0, then is

25 If b > a > 0, then a is

b

a b

Concept Review 3:

Numerical Reasoning Problems

SAT Practice 3: Numerical Reasoning Problems

1. If m and n are both odd integers, which of the

following must be true?

I m2+ n2is even

II m2+ n2is divisible by 4

III (m + n)2is divisible by 4

(A) none

(B) I only

(C) I and II only

(D) I and III only

(E) I, II, and III

2. 6 A A

× 8

50B4

If A and B represent distinct digits in this

cor-rectly worked multiplication problem, what is

the value of B?

(A) 2 (B) 3 (C) 5

(D) 6 (E) 8

3. If j is the number of integers between 1 and 500

that are divisible by 9 and k is the number of

in-tegers between 1 and 500 that are divisible by 7,

what is j + k?

(A) 126 (B) 127 (C) 128

(D) 129 (E) 130

4. If 60 is written as the product of four integers,

each greater than 1, then what is the sum of

those integers?

5. If n is an integer and 2 nis a factor of

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9, what is the greatest possible value of n?

(A) 5 (B) 6 (C) 7 (D) 8 (E) 9

6. If p + pq is 4 times p − pq, and pq ≠ 0, which of

the following has exactly one possible value? (A) p

(B) q

(C) pq

(D) p + pq

(E) p − pq

7. If a, b, c, d, and e are whole numbers and

a(b(c + d) + e) is odd, then which of the

follow-ing CANNOT be even?

(A) a

(B) b

(C) c

(D) d

(E) e

a + b + c = 7

c + d + e = 9

8. If each letter in the sums above represents a

different positive integer, then c= (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

ABB

+9B7

AA7C

9. If A, B, and C are distinct digits in the correctly

worked addition problem above, what is the

value of A + B + C?

(A) 4 (B) 9 (C) 14 (D) 16 (E) 17

.

1

2

3

4

5

7

8

6

1

0

2

3

4

5

7

8

6

1 0

2 3 4 5

7 8 6

1 0

2 3 4 5

7 8 6

SAT Practice 3

1 D Start with the simplest odd values for m and n: m

= n = 1 (There’s no reason why m and n can’t equal the

same number!) Notice that m2+ n2= 12+ 12= 2, which

isn’t divisible by 4, so statement II is not necessarily

true, and you can eliminate choices (C) and (E) Next,

notice that m2and n2must both be odd, so m2+n2must

be even, so statement I is necessarily true, and you can

eliminate choice (A) (m + n)2must be a multiple of 4

because m + n must be even (odd + odd = even), so it is

a multiple of 2 When it is squared, it becomes a

multi-ple of 4 So III is true, and the answer is (D)

2 D Trial and error should show that A = 3 If A is less

than 3, the product is too small If A is greater than 3,

the product is too large Since 633 × 8 = 5,064, B = 6

3 A 500 ÷ 9 = 55.55, so there are 55 multiples of 9

be-tween 1 and 500 500 ÷ 7 = 71.43, so there are 71

mul-tiples of 7 between 1 and 500 So j + k = 55 + 71 = 126.

4 12 Trial and error shows that the only way to

write 60 as the product of four integers, each

greater than 1, is 2 × 2 × 3 × 5 Their sum is

2 + 2 + 3 + 5 = 12

5 C Do the prime factorization:

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 1 × 2 × 3 × (2 × 2)

× 5 × (2 × 3) × 7 × (2 × 2 × 2) × (3 × 3)

Since there are seven factors of 2, the greatest power of 2 that is a factor is 27

6 B p + pq = 4(p − pq)

Distribute: p + pq = 4p − 4pq Divide by p: 1 + q = 4 − 4q (This is okay as long as p is anything but 0.) Add 4q: 1 + 5q = 4

Subtract 1: 5q= 3 Divide by 5: q= 3/5

Because p can have many possible values but q

can only equal 3/5, (B) q is the only expression that has only one possible value

7 A a cannot be even, because an even number

times any other integer yields an even number,

but a(b(c + d) + e) is odd.

8 A The only three different positive integers that have a sum of 7 are 1, 2, and 4 The only three dif-ferent positive integers that have a sum of 9 are 1,

3, and 5 or 1, 2, and 6 But 1 + 2 + 6 doesn’t work,

since that would have two numbers in common

with the first set, but it may only have one (C) Since (C) is the only number they may have in common, it must be 1

9 C The only solution is 188 + 987 = 1,175, so

A + B + C = 1 + 8 + 5 = 14

Concept Review 3

1 They are reciprocals, so their product is 1

2 They are opposites, so their sum is 0

3 Odd × even = even

4 Even × even = even

5 Odd × odd = odd

6 Even + even = even

7 Odd + even = odd

8 Odd + odd = even

9 If n is odd, (−1)n= −1

10 If n is even, (−1)n= 1

11 If x + y = 0 and x ≠ 0, then x/y = −1.

12 Dividing by x is the same as multiplying by 1/x.

13 Subtracting (x + 1) is the same as adding −x − 1.

14 When a number is multiplied by its reciprocal,

the result is 1

15 When a number and its opposite are added, the

result is 0

16 When a number (other than 0) is divided by its opposite, the result is −1

17 It gets bigger

18 It gets smaller

19 It gets smaller (more negative)

20 No If x is 0, then −x is equal to x, and if x is

neg-ative, then −x is greater than x.

21 No If x is between 0 and 1, then x2is smaller than

x And if x is 0 or 1, then they are the same If x is

negative, then x2is positive, and therefore greater

than x.

22 No If x is between 0 and 1, then x3is smaller than

x2 And if x is 0 or 1, then they are the same If x

is negative, then x2 is positive, and therefore

greater than x3

23 greater than 1

24 greater than 1

25 between 0 and 1

Answer Key 3: Numerical Reasoning Problems

What Are Rates?

The word rate comes from the same Latin root as the

word ratio All rates are ratios The most common type

of rate is speed, which is a ratio with respect to time, as

in miles per hour or words per minute, but some rates

don’t involve time at all, as in miles per gallon Rate

units always have per in their names: miles per gallon,

meters per second, etc Per, remember, means divided

by, and is like the colon (:) or fraction bar in a ratio.

The Rate Pyramid

Watch Your Units

Whenever you work with formulas, you can check your work by paying attention to units.

For instance, the problem above asks how

long, so the calculation has to produce a time

unit Check the units in the calculation:

miles miles hours

Lesson 4: Rate Problems

The name of any rate is equivalent to its formula.

For instance, speed is miles per hour

can be translated as

or

Speed = distance

time

Speed = number of miles

number of hours

Since this formula is similar to the “average” formula,

you can make a rate pyramid.

This can be a great tool for

solving rate problems If a

problem gives you two of the

quantities, just put them in

their places in the pyramid,

and do the operation

be-tween them to find

the missing quantity

Example:

How long will it take a car to travel 20 miles at

60 miles per hour?

Simply fill the quantities into the pyramid: 20 miles

goes in the distance spot, and 60 miles an hour goes

in the speed spot Now what? Just do the division the

way the diagram says: 20 miles ÷ 60 miles per hour =

1/3 hour

Two-Part Rate Problems

Rate problems are tougher when they involve two parts When a problem involves, say, two people working together at different rates and times, or a two-part trip, you have to analyze the problem more carefully

Example:

Toni bicycles to work at a rate of 20 miles per

hour, then takes the bus home along the same

route at a rate of 40 miles per hour What is her

average speed for the entire trip?

At first glance, it might seem that you can just aver-age the two rates: (20 + 40)/2 = 30 miles per hour, since she is traveling the same distance at each of the two speeds But this won’t work, because she isn’t

spending the same time at each speed, and that is

what’s important But if that’s true, you might notice

that she spends twice as much time going 20 miles per

hour as 40 miles per hour (since it’s half as fast), so instead of taking the average of 20 and 40, you can

take the average of two 20s and a 40:

(20 + 20 + 40)/3 = 26.67 miles per hour Simple! But if that doesn’t make sense to you, think of it this way: Imagine, for simplicity’s sake, that her trip to work is

40 miles (It doesn’t matter what number you pick, and 40 is an easy number to work with here.) Now the average speed is simply the total distance divided by the total time (as the pyramid says) The total distance, there and back, is 80 miles The total time is in two parts Getting to work takes her 40 miles ÷ 20 miles per hour = 2 hours Getting home takes her 40 miles ÷ 40 miles per hour = 1 hour So the total time of the trip

is 3 hours The average speed, then, must be 80 miles

÷ 3 hours = 26.67 miles per hour!

speed time

distance

×

÷

÷

For each of the following rates, write the formula of the rate and the corresponding “rate pyramid.”

1 Speed is miles per hour

2 Efficiency is miles per gallon of fuel

3 Typing speed is pages per minute

Find the missing quantity, including the units, in each of these rate situations

4 A train travels for 375 miles at 75 mph

5 A car that gets 28 miles per gallon uses 4.5 gallons of fuel

6 Harold can type 600 words in 5 minutes

7 A landscaper who cuts 1.6 acres of grass per hour cuts an 8-acre lot

8 A train leaves New York at 1:00 pm, going 50 mph, bound for Philadelphia, which is 90 miles away If it makes no stops, at what time should it be expected to arrive?

9 Anne can paint a room in 2 hours, and Barbara can paint a room in 3 hours When they work together, their work rate is the sum of their rates working separately How long should it take them to paint a room if they work together?

Concept Review 4: Rate Problems

1. Janice and Edward are

editors at a newspaper

Janice can edit 700 words

per minute and Edward

can edit 500 words per

minute If each page of

text contains 800 words,

how many pages can they

edit, working together, in

20 minutes?

2. Two cars leave the same

point simultaneously,

going in the same

direc-tion along a straight, flat

road, one at 35 mph and

one at 50 mph After how

many minutes will the

cars be 5 miles apart?

3. What is the average speed, in miles per hour,

of a sprinter who runs 1⁄4mile in 45 seconds?

(1 hour = 60 minutes)

(A) 11.25 mph (B) 13.5 mph

(C) 20 mph (D) 22 mph

(E) 25 mph

4. A car travels d miles in t hours and arrives at

its destination 3 hours late At what average

speed, in miles per hour, should the car have

gone in order to have arrived on time?

(A) t – 3 (B) (C)

(D) (E) t

d− 3

d

t − 3

d

t− 3

t d

− 3

5. If x > 1, how many hours does it take a train traveling at x − 1 miles per hour to travel

x2− 1 miles?

(A) (B) (C) x

(D) x – 1 (E) x + 1

6. In three separate 1-mile races, Ellen finishes

with times of x minutes, y minutes, and z

min-utes What was her average speed, in miles per

hour, for all three races? (1 hour = 60 minutes) (A) (B)

(C) (D)

(E)

7. A hare runs at a constant rate of a mph, a tor-toise runs at a constant rate of b mph, and

0 < b < a If they race each other for d miles, how many more hours, in terms of a, b, and d, will it

take the tortoise to finish than the hare?

(A) (B) (C) (D) ad – bd (E) a – b

8. Sylvia drives 315 miles and arrives at her desti-nation in 9 hours If she had driven at an average rate that was 10 mph faster than her actual rate, how many hours sooner would she have arrived? (A) 1.75 (B) 2.00 (C) 2.25 (D) 2.50 (E) 2.75

b d

a d

−

d b

d a

−

a b d

+ 2

x+ +y z

20

180

x+ +y z

x+ +y z

180

3

x+ +y z

x+ +y z

3

1 1

x+

1 1

x−

SAT Practice 4: Rate Problems

.

1 2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

.

1 2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

1 0

2 3 4 5

7 8 9 6

Concept Review 4

1 Speed = #miles ÷ #hours

2 Efficiency = #miles ÷ #gallons

3 Typing speed = #pages ÷ #minutes

4 375 miles ÷ 75 mph = 5 hours for the trip

5 28 miles per gallon × 4.5 gallons = 126 miles the car can go before it runs out of fuel

6 600 words ÷ 5 minutes = 120 words per minute is Harold’s typing speed

7 8 acres ÷ 1.6 acres per hour = 5 hours for the job

8 90 miles ÷ 50 mph = 1.8 hours, or 1 hour 48 min-utes for the entire trip At 1 hour and 48 minmin-utes after 1:00 pm, it is 2:48 pm

9 Anne can paint one room in 2 hours, so her rate is

1⁄2room per hour Barbara can paint one room in

3 hours, so her rate is 1⁄3room per hour When they work together, their rate is 1⁄2+1⁄3=5⁄6room per hour

So to paint one room would take one room ÷ 5⁄6

room per hour =6⁄5hours, or 1.2 hours, or 1 hour

12 minutes

Answer Key 4: Rate Problems

SAT Practice 4

1 30 Working together, they edit 700 + 500 = 1,200

words per minute Since each page is 800 words,

that’s 1,200 words per minute ÷ 800 words per

page = 1.5 pages per minute In 20 minutes, then,

they can edit 1.5 × 20 = 30 pages

2 20 Since the two cars are traveling in the same

di-rection, their relative speed (that is, the speed at which

they are moving away from each other) is

50 − 35 = 15 mph In other words, they will be 15 miles

farther apart each hour Therefore, the time it takes

them to get 5 miles apart is 5 miles ÷ 15 miles per hour

= 1/3 hour, which is equivalent to 20 minutes

3 C Since there are (60)(60) = 3,600 seconds in an

hour, 45 seconds = 45/3,600 hour Speed =

dis-tance ÷ time = 1/4 mile ÷ 45/3,600 hour = 3,600/180

= 20 miles per hour

4 C To arrive on time, the car must take t − 3

hours for the whole trip To travel d miles in t− 3

hours, the car must go d/(t− 3) miles per hour

5 E According to the rate pyramid, time = distance

÷ speed = (x2− 1) miles ÷ (x − 1) miles per hour =

hours Or you can pick

a simple value for x, like 2, and solve numerically.

6 D Speed = miles ÷ hours Her total time for the

three races is x + y + z minutes, which we must

convert to hours by multiplying by the conver-sion factor (1 hour/60 minutes), which gives us

(x + y + z)/ 60 hours Since her total distance is 3

miles, her overall speed is 3 miles ÷ (x + y + z)/60

hours = 180/(x + y + z) miles per hour.

7 B If the hare’s rate is a mph, then he covers d miles in d/a hours Similarly, the tortoise covers d miles in d/b hours The difference in their finish-ing times, then, is d/b − d/a.

8 B Sylvia’s speed is 315 miles ÷ 9 hours = 35 mph

If she were to go 10 mph faster, then her speed would be 45 mph, so her time would be 315 miles

÷ 45 mph = 7 hours, which is 2 hours sooner

x x

2 1 1

−

−

( ) ( )+

speed time

distance

×

÷

÷

speed minutes

pages

×

÷

÷

efficiency gallons

miles

×

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÷