SAT Math : Whole and Part

Study concepts, example questions & explanations for SAT Math

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Example Questions

Example Question #1 : Whole And Part

If a TV show is 45 minutes long, what fraction of the show is completed 5 minutes after it begins?

Possible Answers:

1/3

1/4

1/9

1/2

1/10

Correct answer:

1/9

Explanation:

5 minutes out of 45 have been completed. We can represent this as a fraction: 5/45. When simplified, this yields 1/9. 

Example Question #2 : Whole And Part

A birthday cake is cut into 8 pieces.  On the first night three pieces are eaten.  On the second night, 20% of the remaining cake was eaten.  On the third night, half of the remaining cake was eaten.  How many slices were left to eat on the fourth night?

Possible Answers:

2

3

1/2

4.5

1

Correct answer:

2

Explanation:

First night: 8 – 3 = 5 remaining.

Second night: 20% of 5 remaining = .2 * 5 = 1 slice Therefore 4 remaining.

Third night: ½ of the 4 remaining = ½*4 = 2 slices remaining

Example Question #3 : Whole And Part

There was a malfunction at a candy factory and the machines only put red, purple, and green candy into the bags.  If 1/4 of the candies were red and 3/8 were green, how many purple candies were in the bag, if each bag can hold 16 pieces?

Possible Answers:

8

3

10

4

6

Correct answer:

6

Explanation:

Since 1/4 of the 16 are red: 16 * 1/4 = 4 red candies.  With 3/8 being green:

16 X 3/8 = 6 green candies.   16 – 4 – 6 = 6 candies remain, which must be purple.

Example Question #1 : How To Find The Part From The Whole

In the 30-day month of January, for every three days it snowed, there were seven days it did not snow. The number of days in January on which it did not snow was how much greater than the number of days in January on which it snowed?

Possible Answers:

13

12

11

10

14

Correct answer:

12

Explanation:

The question tells us that for every ten-day period in January (a three-day period plus a seven-day period), it snowed on 3 of those days and did not snow on 7 of those days. Since January has 30 days, it has 3 ten-day periods, so we multiply the numbers given for the 10-day period by 3 to find the  number of days with and without snow during the 30-day period. Doing this, we see that it snowed 3 * 3 = 9 days and did not snow 7 * 3 = 21 days during the 30-day period. Since the question asks how much greater the number of days on which it did not snow is than the number of days on which it snowed, we subtract as follows: number of days it did not snow - number of days it snowed = 21 – 9 = 12.

Example Question #5 : Whole And Part

Mikey has one full box of cereal.  On Saturday, he eats 1/3 of the the box.  On Sunday he eats 2/3 of what is left.  How much of the box is still left?

Possible Answers:

2/9

4/9

2/3

5/9

none

Correct answer:

2/9

Explanation:

If Mikey eats 1/3 of the box, he is left with 1 – 1/3 = 2/3.

When he eats 2/3 of what remains, he is eating 2/3 of 2/3, or 4/9 of the box.

2/3 – 4/9 = 2/9

Example Question #6 : Whole And Part

What is the remainder when 27 is divided by 6?

Possible Answers:

4.5

0.5

3

4

8

Correct answer:

3

Explanation:

Long division is the fastest way: 6 goes into 27 four times. 6 times four is 24. 27 – 24 = 3.

Example Question #7 : Whole And Part

A pitcher is currently \dpi{80} \frac{4}{5} of the way full of water.  If \dpi{80} 3 ounces are poured out, then the pitcher would be  of the way full.  How many ounces of water are in the pitcher before the \dpi{80} 3 ounces are poured out?

Possible Answers:

Correct answer:

Explanation:

Let p represent the total capacity of the pitcher.  The current volume in the pitcher is \dpi{80} \frac{4}{5}p.  If this volume is depleted by \dpi{80} 3 ounces, it becomes \dpi{80} \frac{3}{4}p.  In other words:

\dpi{80} \frac{4}{5}p - 3 = \frac{3}{4}p

Solve this equation for \dpi{80} p and get:

\dpi{80} \frac{4}{5}p-\frac{3}{4}p=3

\dpi{80} \frac{16}{20}p-\frac{15}{20}p=3

\dpi{80} \frac{1}{20}p=3

\dpi{80} p=60

Remember, \dpi{80} p represents the total capacity of the pitcher, or 60 ounces.  The question asks how many ounces are in the pitcher at the beginning of the problem, so evaluate\dpi{80} \frac{4}{5}p=\frac{4}{5}\times 60=48.

Example Question #8 : Whole And Part

Mr. Owens spent $7.50 for a dinner buffet. The amount he paid accounted for 3/4 of the money in his wallet. How much money is left in his wallet for other expenses?

Possible Answers:

$2.50

$1.00

$4.00

$10.00

$6.50

Correct answer:

$2.50

Explanation:

If $7.50 is 3/4 of the total, 7.50/3 gives us what 1/4 of his total money would be. This equals $2.50, the remaining unspent quarter.

Example Question #9 : Whole And Part

A certain ball that is dropped will bounce back to 3/5 of the height it was initially dropped from.  If after the 2nd bounce the ball reaches  39.96 ft, what was the initial height the ball was dropped from? 

Possible Answers:

100 ft

66 ft

111 ft

135 ft

150 ft

Correct answer:

111 ft

Explanation:

We know the height of the initial bounce, so work backwards to find the initial height.  39.96/0.6 = 66.6 = height of ball after first bounce

66.6/0.6 = 111 ft

Example Question #10 : Whole And Part

A pitcher of water is filled \dpi{100} \small \frac{2}{5} of full.  An additional 27 ounces of water is added.  Now the pitcher of water is completely full.  How much water does the pitcher hold?

Possible Answers:

30

35

45

40

50

Correct answer:

45

Explanation:

If \dpi{100} \small 27 ounces fills the pitcher, then it must equal the volume of \dpi{100} \small \frac{3}{5} of the pitcher.  If \dpi{100} \small \frac{3}{5} of a pitcher equals 27 ounces, then \dpi{100} \small \frac{1}{5} of a pitcher equals \dpi{100} \small 27\div 3=9ounces.  Since there are \dpi{100} \small 5 fifths in the pitcher, it must hold \dpi{100} \small 9\times 5=45 ounces total.

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