broker wrote:

All,

I came upon this sample question online and I'm trying to figure out how they came up with their solution because of their solution makes no sense to me. Can anyone help?

QUESTION #1 (Fractions):

After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?

Choices:

A) 8/27

B) 1/27

C) 2/3

D) 19/27 ---> was the chosen answer

E) 27/28

QUESTION #2 (Expression Simplification):

What is the units digit of 6^15 - 7^4 - 9^3?

/

A) 8

B) 7

C) 6 ---> they those this answer, WHY?

D) 5

E) 4

**For question #2, I guess I see why it will be 6 since the unit digit of 6^15 is far greater than 7^4 and 9^3 but I'm hoping to see how you arrive at that solution

Thanks!!

Answers:

1. 19/27.

Total gasoline -- 1 unit.

First time the pump is run:

Amount of gasoline removed -- (1/3), amount remaining - 1 - (1/3) = (2/3)

Second time the pump is run:

Amount of gasoline removed -- (1/3) * (2/3), amount remaining - (2/3) - (1/3) * (2/3) -- 4/9

Third time the pump is run:

Amount of gasoline removed -- (1/3) * (4/9).

Hence total gasoline removed -- (1/3) + (2/9) + (4/27).

= (19/27) units.

2. 6

Explanation:

6^15 - 7^4 - 9^3

6^15 -- unit digit would be 6. (Unit digit repeats for all positive power of 6 i.e., 6^1, 6^2 .... is always 6.)

7^4 -- 1.

9^3 -- 9.

Hence 6^15 - (7^4 + 9^3)

= 6 - (1 + 9) == 6 - 10

6^15 --- would be something like XXX..X6

(7^4 + 9^3) -- would be something like XXX..10

And hence when you subtract the 6 with 0, the unit digit would remain in the answer.

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