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If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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15 Jul 2010, 12:40

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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?

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. A – 63/64 b. B – 25/32 c. C – 13/16 d. D – 9/32 e. E – 1/64

"3/4th of the mineral deposits are removed after 1 cycle" means that 1/4th of mineral deposits are remained after 1 cycle.

After 3 cycles \((\frac{1}{4})^3=\frac{1}{64}\)th of the minerals will remain thus\(1-\frac{1}{64}=\frac{63}{64}\) will be removed.

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. A – 63/64 b. B – 25/32 c. C – 13/16 d. D – 9/32 e. E – 1/64

"3/4th of the mineral deposits are removed after 1 cycle" means that 1/4th of mineral deposits are remained after 1 cycle.

After 3 cycles \((\frac{1}{4})^3=\frac{1}{64}\)th of the minerals will remain thus\(1-\frac{1}{64}=\frac{63}{64}\) will be removed.

Answer: A.

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?
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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. A – 63/64 b. B – 25/32 c. C – 13/16 d. D – 9/32 e. E – 1/64

"3/4th of the mineral deposits are removed after 1 cycle" means that 1/4th of mineral deposits are remained after 1 cycle.

After 3 cycles \((\frac{1}{4})^3=\frac{1}{64}\)th of the minerals will remain thus\(1-\frac{1}{64}=\frac{63}{64}\) will be removed.

Answer: A.

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?

When you say that for example after 2 cycles we removed (3/4)^2 of the total that is not true as (3/4)^2<3/4 which means that after 2 cycles we removed less part of the total then after 1 cycle. Whereas when we say that after 1 cycle 1/4 of the total is left and after 2 cycles 1/16 it's true as the remaining part is always 1/4 of the previous.

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.
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Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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25 Jun 2013, 11:46

aimingformba 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. 1/64

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
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When you say that for example after 2 cycles we removed (3/4)^2 of the total that is not true as (3/4)^2<3/4 which means that after 2 cycles we removed less part of the total then after 1 cycle. Whereas when we say that after 1 cycle 1/4 of the total is left and after 2 cycles 1/16 it's true as the remaining part is always 1/4 of the previous.

Hope it's clear.

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 \(1 - (2/5)^3\) ?

When you say that for example after 2 cycles we removed (3/4)^2 of the total that is not true as (3/4)^2<3/4 which means that after 2 cycles we removed less part of the total then after 1 cycle. Whereas when we say that after 1 cycle 1/4 of the total is left and after 2 cycles 1/16 it's true as the remaining part is always 1/4 of the previous.

Hope it's clear.

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 \(1 - (2/5)^3\) ?

Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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08 Jul 2016, 07:29

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)

Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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24 Jun 2017, 00:57

aimingformba 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. 1/64

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) ...

Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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24 Jun 2017, 03:49

aimingformba 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. 1/64

What will be left behind after 3 cycles = \(\frac{1}{4} * \frac{1}{4} * \frac{1}{4} = 1/64\), so what's removed = \(1 - \frac{1}{64} = \frac{63}{64}\).
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Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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21 Jul 2017, 06:20

aimingformba 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. 1/64

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

Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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30 Oct 2017, 21:14

krishnasty wrote:

Bunuel wrote:

aimingformba 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. A – 63/64 b. B – 25/32 c. C – 13/16 d. D – 9/32 e. E – 1/64

"3/4th of the mineral deposits are removed after 1 cycle" means that 1/4th of mineral deposits are remained after 1 cycle.

After 3 cycles \((\frac{1}{4})^3=\frac{1}{64}\)th of the minerals will remain thus\(1-\frac{1}{64}=\frac{63}{64}\) will be removed.

Answer: A.

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?

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

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. A – 63/64 b. B – 25/32 c. C – 13/16 d. D – 9/32 e. E – 1/64

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
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Re: If 3/4 of the mineral deposits in a reservoir of water are [#permalink]

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02 Nov 2017, 19:12

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