If 3/4 of the mineral deposits in a reservoir of water are

Brunel, the ans doesnt come by going the straight way..i.e. if 3/4 of the minerals are removed in one cycle, then in 3 cycles = (3/4)^3 would be removed..now we know this is not the right ans...so what exactly went wrong here?

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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1. Assume a convenient amount of deposits. Since the fraction used is 3/4 assume a multiple of 4, let's say 32 units of deposits
2. 3/4 of 32 units removed in the first cycle. So remaining is 8 units
3. 3/4 of 8 units removed in the second cycle. So remaining is 2 units
4. 3/4 of 2 units removed in the third cycle. So remaining is 0.5 units
5. So 0.5 units out of 32 units remain or 31.5 units have been removed
6. Therefore the fraction of units removed is 31.5/32 = 63/64.

Here we can plug in as well but the nos should be chosen carefully. All answer choices are multiple of 4 in denominator and hence we can assume total no of minerals as multiple of 4

Let us take no of minerals to be 64 (Did not arrive at this in 1st try)

then after 1 filteration we have = 48 (removed) and 16 remaining
2nd filteration we have, 12 removed and 4 remaining
3rd filteration, 3 removed 1 remaining

total removed = 63

Total fraction of mineral removed = 63/64

I was trying for alt explanation and ended up choosing 64 after trying with 40 and other multiple of 4.
Another point to note is that it is better to look at answer choices to get some plug in options

Thanks

Just to check my understanding. If we modify the question to say that 3/5 of mineral deposits were removed and asked to calculate the removed fraction after three cycles, then we should calculate by \(1 - (2/5)^3\) ?

Yes, that's correct.

This question was done by the exponential growth formula?

Yes, there is an exponential decrease here - \((\frac{1}{4})^3\)

after 1st round (a)= 3/4
remaining 1-3/4=1/4
in the second round removed fraction will be (b)=(1/4)*3/4 =3/16
so total removed after second round=(3/4)+(3/16)= 15/16
remaining fraction =1-(15/16)=1/16

after 3rd round filtration (c)=3/4*(1/16)=3/64
so, total fraction removed=a+b+c=63/64
ANS (A)

Let the Total mineral deposit be \(= 64\)

Minerals removed in first cycle \(= \frac{3}{4}* 64 = 48\)

Minerals left after first cycle \(= 64 - 48 = 16\)

Minerals removed in second cycle \(= \frac{3}{4} * 16 = 12\)

Minerals left after second cycle \(= 16 - 12 = 4\)

Minerals removed in third cycle \(= \frac{3}{4} * 4 = 3\)

Minerals left after third cycle \(= 4 - 3 = 1\)

Total Minerals removed in 3 cycles \(= 48 + 12 + 3 = 63\)

Fraction of Minerals removed to Total mineral \(= \frac{63}{64}\) . Answer (A) ...

What will be left behind after 3 cycles = \(\frac{1}{4} * \frac{1}{4} * \frac{1}{4} = 1/64\), so what's removed = \(1 - \frac{1}{64} = \frac{63}{64}\).

So to solve this problem, lets start checking how much mineral is filtered in each cycle,

1st cycle : filtered mineral : 3/4 left mineral : 1/4
2nd cycle filtered mineral : 3/4*1/4 = 3/16 , left Mineral = 1/4 - 3/16 = 4/16 - 3/16= 1/16
3rd cycle : Filtered Mineral : 3/4*1/16 = 3/64

So , Total mineral removed = 3/4 + 3/16 + 3/64 =( 48 + 12 + 3 )/64 = 63/64

Answer A.

3/4 + 3/4 of remaining 1/4 (3/16) + 3/4 of 3/16 == 63/64

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

after 1 cycle, 1/4 of the mineral deposits will remain => after 3 cycles, we have (1/4) * (1/4) * (1/4) = 1/64 of the mineral deposits remain.
Thus, the fraction of the total minerals present in the water have been removed after 3 cycles = 1-1/64 = 63/64

Hence the answer is A

for 1 unit of mineral ; 3/4 is removed
so we have 1/4 remaining which in 3 cycles; ( 1/4) ^3 ; 1/64
left with 1-1/64; 63/64
IMO A:

Minerals filtered in the first cycle = 3/4 [remaining in the tank =1/4]
minerals filtered in the second cycle = (3/4) (1/4)
Minerals filtered in the third cycle= (3/4)(1/4) (1/4)

in order to get the fraction of total minerals removed , add all the above results which will be = 63/64