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Various types of Math Problem Answers are solved here.

1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck. Solution: Let the 1st paycheck be x (integer). Mrs. Rodger got a weekly raise of$ 145.

So after completing the 1st week she will get $(x+145). Similarly after completing the 2nd week she will get$ (x + 145) + $145. =$ (x + 145 + 145)

= $(x + 290) So in this way end of every week her salary will increase by$ 145.

2. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,

2 + 2(22)

= 2 + 2(2 × 2)

= 2 + 2(4)

= 2 + 8

= 10

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he: (a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents Solution: Selling price of the first pipe =$1.20

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price - Profit

CP = 1.20 - 20% of CP

CP = 1.20 - 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = $$\frac{1.20}{1.20}$$

CP = $1 Selling price of the Second pipe =$1.20

Loss = 20%

Let’s try to find the cost price of the second pipe

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP - 0.20CP = 1.20

0.80CP = 1.20

CP = $$\frac{1.20}{0.80}$$

CP = $1.50 Therefore, total cost price of the two pipes =$1.00 + $1.50 =$2.50

And total selling price of the two pipes = $1.20 +$1.20 = $2.40 Loss =$2.50 – $2.40 =$0.10

Therefore, Mr. Jones loss 10 cents.

4. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is:

(a) 587 × 108 miles, (b) 587 × 1010 miles, (c) 587 × 10-10 miles, (d) 587 × 1012 miles, (e) 587 × 10-12 miles

Solution:

The distance of the light travels in 100 years is:

5,870,000,000,000 × 100 miles.

= 587,000,000,000,000 miles.

= 587 × 1012 miles.

5. A man has $10,000 to invest. He invests$ 4000 at 5 % and $3500 at 4 %. In order to have a yearly income of$ 500, he must invest the remainder at:

(a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 %

Solution:

Income from $4000 at 5 % in one year =$ 4000 of 5 %.

= $4000 × 5/100. =$ 4000 × 0.05.

= $200. Income from$ 3500 at 4 % in one year = $3500 of 4 %. =$ 3500 × 4/100.

= $3500 × 0.04. =$ 140.

Total income from 4000 at 5 % and 3500 at 4 % = $200 +$ 140 = $340. Remaining income amount in order to have a yearly income of$ 500 = $500 -$ 340.

= $160. Total invested amount =$ 4000 + $3500 =$7500.

Remaining invest amount = $10000 -$ 7500 = $2500. We know that, Interest = Principal × Rate × Time Interest =$ 160,

Principal = $2500, Rate = r [we need to find the value of r], Time = 1 year. 160 = 2500 × r × 1. 160 = 2500r 160/2500 = 2500r/2500 [divide both sides by 2500] 0.064 = r r = 0.064 Change it to a percent by moving the decimal to the right two places r = 6.4 % Therefore, he invested the remaining amount$ 2500 at 6.4 % in order to get \$ 500 income every year.

6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:

(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

Solution:

Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.

We know that

Speed = Distance/Time.

Or, Time = Distance/Speed.

So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.

And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.

= 100/x hr.

So we can clearly see that his new time compared with the old time was: twice as much.

7. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:

(a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0

Solution:

(0.2)x = 2.

Taking log on both sides

log (0.2)x = log 2.

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 - log 10] = 0.3010.

x [log 2 - 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4306….

x = -0.4 (nearest tenth)

8. If 102y = 25, then 10-y equals:

(a) -1/5, (b) 1/625, (c) 1/50, (d) 1/25, (e) 1/5

Solution:

102y = 25

(10y)2 = 52

10y = 5

1/10y = 1/5

10-y = 1/5

9. The fraction (5x-11)/(2x2 + x - 6) was obtained by adding the two fractions A/(x + 2) and B/(2x - 3). The values of A and B must be, respectively:

(a) 5x, -11, (b) -11, 5x, (c) -1, 3, (d) 3, -1, (e) 5, -11

Solution:

10. The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

Solution:

Let the three numbers be x, y and z.

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.

Unsolved Questions:

1. Fahrenheit temperature F is a linear function of Celsius temperature C. The ordered pair (0, 32) is an ordered pair of this function because 0°C is equivalent to 32°F, the freezing point of water. The ordered pair (100, 212) is also an ordered pair of this function because 100°C is equivalent to 212° F, the boiling point of water.

2. A sports field is 300 feet long. Write a formula that gives the length of x sports fields in feet. Then use this formula to determine the number of sports fields in 720 feet.

3. A recipe calls for 2 1/2 cups and I want to make 1 1/2 recipes. How many cups do I need?

4. Mario answered 30% of the questions correctly. The test contained a total of 80 questions. How many questions did Mario answer correctly?