Solutions

1. 25

   Use two percent tables to solve this problem. Begin with the fact that y% of 120 is 48:

   PART	48	y
   WHOLE	120	100

   

   Then, set up a percent table for the fact that x % of 40 is 10:

   PART	10	x
   WHOLE	40	100

   

   You could also set up equations with decimal equivalents to solve: (0.01y)(120) = 48, so 1.2 y = 48 or y = 40. Therefore, because you know that (0.01x)(y) = 10, you have:

   

2. $120

   Use a percent table to solve this problem. Remember that the stereo was marked down 30% from the original, so you have to solve for the original price:

   CHANGE	x	30
   ORIGINAL	$84 + x	100

   

   Therefore, the original price was 84 + 36, or $120.

   You could also solve this problem using the formula,
   

   

3. 2.82% increase

   For percent problems, the Smart Number is 100. Therefore, assume that the population of Mitannia in 1980 was 100. Then, apply the successive percents procedure to find the overall percent change:

   From 1980–1990, there was a 6% increase:	100(1 + 0.06) = 100(1.06) = 106
   From 1990–2000, there was a 3% decrease:	106(1 − 0.03) = 106(0.97) = 102.82

   Overall, the population increased from 100 to 102.82, representing a 2.82% increase.

4. 1.28y

   For percent problems, the Smart Number is 100. Therefore, assign y a value of 100. Then, apply the successive percents procedure to find the overall percentage change:

   (1) y is decreased by 20%:	100(1 − 0.20) = 100(0.8) = 80
   (2) Then, it is increased by 60%:	80(1 + 0.60) = 80(1.6) = 128

   Overall, there was a 28% increase. If the original value of y is 100, the new value is 1.28y.

5. $3,000

   Use a percent table to solve this problem, which helps you find the decimal equivalent equation:

   PART	210	7
   WHOLE	x	100

   

6. 14

   For some problems, you cannot use Smart Numbers, because the total amount can be calculated. This is one of those problems. Instead, use a percent table:

   PART	0.5x + 4	70
   WHOLE	x	100

   

   The capacity of the bowl is 20 cups. There are 14 cups of water in the bowl {70% of 20, or 0.5(20) + 4}:

   PART	4	20
   WHOLE	x	100

   Alternatively, the 4 cups of water added to the bowl represent 20% of the total capacity. Use a percent table to solve for x, the whole.
   Because x = 20, there are 14 (4 + 50% of 20) cups of water in the bowl.

7. 46%

   For this liquid mixture problem, set up a table with two columns: one for the original mixture and one for the mixture after the alcohol evaporates from the tub:

   	Original	After Evaporation
   Water	920	920
   Alcohol	1,800	0.60(1,800) = 1,080
   TOTAL	2,720	2,000

    

    

   The remaining liquid in the tub is water.

   Alternatively, you could skip the chart and solve for the new amount of alcohol using the formula:

   

   units of alcohol. Water is  of the total.

8. 50%

   Use two percent tables to solve this problem. Begin with the fact that 50% of y is 40:

   PART	40	50
   WHOLE	y	100

   

   Then, set up a percent table for the fact that x is 40% of y:

   PART	x	40
   WHOLE	80	100

   

   Finally, 16 is 50% of 32. You could alternatively set up equations with decimal equivalents to solve: x = (0.4)y.

   You also know that (0.5)y = 40, so y = 80 and x = (0.4)(80) = 32. Therefore, 16 is half, or 50%, of x.

9. 840

   Apply the successive percents procedure:

   (1) 800 is increased by 50%:	800 × 1.5 = 1,200
   (2) Then, the result is decreased by 30%:	1,200 × 0.7 = 840

10. 20

    Apply the percents in succession with two percent tables:

    PART	x	120
    WHOLE	1,600	100

    

    Then, fill in the “change” for the part (1,920 − 1,536 = 384) and the original for the whole (1,920):

    PART	384	y
    WHOLE	1,920	100

    

    Alternatively, you could solve for the new number using formulas. Because this is a successive percents problem, you need to “chain” the formula: once to reflect the initial increase in the number, then twice to reflect the subsequent decrease:

    

    

    

11. $106.12

    Interest compounded annually is just a series of successive percents:

    (1) 100.00 is increased by 2%:	100(1.02) = 102
    (2) 102.00 is increased by 2%:	102(1.02) = 104.04
    (3) 104.04 is increased by 2%:	104.04(1.02) ≅ 106.12

12. (A)

    In dealing with percents problems, you should choose 100. In this case, the original size of the image is 100. The question tells you that Steve reduces the image by 13%. Thus:

    100 − 0.13(100) = 100 − 13 = 87

    So the image is at 87 percent of its original size. Quantity A tells you that you have to reduce the image size by another 13%.

    If the image size is reduced by 13%, then 87% of the image remains. Multiply 87 (the current size of the image) by 0.87 (87% expressed as a decimal):

    87 × 0.87 = 75.69

    Quantity A	Quantity B
    The percent of the original if Steve reduces the image by another 13% = 75.69%	75%

    Therefore, Quantity A is larger.

13. (D)

    First translate the statement in the question stem into an equation:

    

    Now try to pick some easy numbers. If , which is definitely greater than 1:

    Quantity A	Quantity B
    y = 1

    However, if y = 200, then x must also equal 200:

    Quantity A	Quantity B
    y = 200	x = 200

    y can be less than x, but y can also be equal to x. You could also choose values for which y is greater than x. Therefore, you do not have enough information to answer the question.

14. (C)

    To calculate 10% of 643.38, move the decimal to the left one place: 643.38 → 64.338.

    Quantity A	Quantity B
    10% of 643.38 = 64.338	20% of 321.69

    To calculate 20% of 321.69, don’t multiply by 0.2. Instead, find 10% first by moving the decimal to the left one place: 321.69 → 32.169.

    Now multiply by 2: 32.169 × 2 = 64.338.

    Quantity A	Quantity B
    64.338	20% of 321.69 = 64.338

    Therefore, the two quantities are equal.