• MATHEMATICS

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(x^x) when x = 2 is:

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

Solution:

x + x(x^x)

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

Answer: (a)

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 centsSolution:

20 % profit on $ 1.20

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24

Similarly, 20 % loss on $ 1.20

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24

Therefore, in one pipe his profit is $ 0.24 and in the other pipe his loss is $ 0.24.

Since both profit and loss amount is same so, it’s broke even.

Answer: (a)

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 × 10¹²miles, (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.

Answer: (d)

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.

Answer: (e)



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.

Answer: (b)

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)

Answer: (c)



8. If 10^2y = 25, then 10^-y equals:

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

Solution:

10^2y = 25

(10^y)² = 5²

10^y = 5

¹/_10^y = ¹/₅ 

10^-y = ¹/₅

Answer: (e)