Mineral Extracted

Think in terms of the fraction that remains

A.

I did it by picking 1 as my initial number.

1-3/4 = 1/4 left after first pass

3/4 * 1/4 = 3/16 --> 1/4 - 3/16 = 1/16 after second pass

1/16 * 3/4 = 3/64 --> 1/16 - 3/64 = 1/64 left after third pass

how much removed in total = 1 - 1/64 = 63/64 = A

Ashish8 wrote:A.

I did it by picking 1 as my initial number.

1-3/4 = 1/4 left after first pass

3/4 * 1/4 = 3/16 --> 1/4 - 3/16 = 1/16 after second pass

1/16 * 3/4 = 3/64 --> 1/16 - 3/64 = 1/64 left after third pass

how much removed in total = 1 - 1/64 = 63/64 = A

I or got to do the last step there and was getting 1/64

thanks ashish

mchaubey wrote:If 3/4 of the mineral deposits in a reservoir of water are removed every time a water filtration unit completes a cycle, what fraction of the total minerals present in the water will have been removed after 3 complete filtration cycles?

a. 63/64
b.25/32
c. 13/16
d. 9/32
e. 1/62

Consider 100 as the initial quantity.


After three filtration the picture looks like :

100 -> 25 -> 25/4 -> 25/16

fraction left = 25/16 / 100 = 1/64.

therefore, the fraction removed = 1-1/64 = 63/64

mchaubey wrote:If 3/4 of the mineral deposits in a reservoir of water are removed every time a water filtration unit completes a cycle, what fraction of the total minerals present in the water will have been removed after 3 complete filtration cycles?

a. 63/64
b.25/32
c. 13/16
d. 9/32
e. 1/62

Since each filtration cycle removes 3/4 of the mineral deposits in the water, 1/4 of the mineral deposits will remain in the water. Thus, after 3 cycles, the amount of mineral deposits remaining in the water is 1/4 x 1/4 x 1/4 = 1/64. Thus, 1 - 1/64 = 63/64 of the mineral deposits will have been removed from the water.

Answer: A