In a particular dilution technique, 10% of the solution is removed and

Question

In a particular dilution technique, 10% of the solution is removed and replaced with the diluter. If we start with pure alcohol, minimum how many times would the operation need to be performed to bring the percentage of alcohol below 65%.

A. 3
B. 4
C. 5
D. 6
E. 7

A. 3 
B. 4 
C. 5 
D. 6 
E. 7

Tapesh03 · Answer

IMO - C
Consider the solution is 100 liters of alcohol and we need to remove 10% every time
1st attempt - 10 liters of alcohol is replaced with 10 liters of diluter; alocohol left is 91 liters ie. 91%
2nd attempt - 9.1 liters of alcohol is removed because 10% of of 91 is 9.1. Alcohol left is about 81.9 liters 
3rd attempt - 8.19 liters of alcohol is removed becuase 10% of 81.9 liters is 8.19. Alcohol left is about 73.71
4th attempt - 7.37 liters of alcohol is removed becuase 10% of 73.71 liters is 7.37. Alcohol left is about 66.34
5th attempt - 6.63 liters of alcohol is removed because 10% of 66.34 liters is 6.63. Alcohol left would be less than 65 liters i.e less than 65%
it's a long process i think, pelase do share if any shortcut

yashwardhan · Answer

We know that:
Cf = Ci * (Vi/Vf)
Given :
Cf = less than 65%, assume 64.9
Ci = 1 (as we are given pure alcohol)
Vi = ( 10% is removed, assume 100L, so 90% remains)
Vf = 100 ( as 10% is replaced too)

so, 64.9/100 = 1 * (90/100)
.649 = (.9)*(.9)*(.9)*(.9)*(.9) - when you multiply (.9), 4 times, you will see a number close to .6561>65% and we need a number less than that, hence multiply with (.9) again to make it less than 65%

Hence, the operation needs to be performed 5 times.

Answer : C

ScottTargetTestPrep · Answer

We can start with 100 grams of pure alcohol and see how many times it takes the alcohol to drop below 65 grams where each time we remove 10% of the solution.

After the first time, we have 100 x 0.9 = 90 grams of alcohol left.

After the second time, we have 90 x 0.9 = 81 grams of alcohol left.

After the third time, we have 81 x 0.9 = 72.9 grams of alcohol left.

After the fourth time, we have 72 x 0.9 = 65.61 grams of alcohol left.

Without further calculation, we can see that after the fifth time, the amount of alcohol will drop below 65 grams (since the the fourth time, that amount is just above 65 grams).

Answer: C

GmatPoint · Answer

Let the total volume of the solution be 100l
The initial concentration of alcohol is 100%.
Thus, the volume of alcohol is 100l.

After first dilution, the volume of the alcohol removed = 10% of 100l = 100*10/100 = 10l
Thus, the volume of alcohol left = 100 - 10 = 90l or 0.9*100

After the second dilution, volume of the alcohol left =  *90l = 0.9*90 = 81l

Similarly, after the third dilution, the volume of alcohol left = 0.9*81 = 72.9l.

Thus, after nth dilution, the volume of alcohol left = (0.9)^n*100

Since the percentage, if alcohol should be below 65% or 65*100/100 = 65l.

(0.9)^n*100 < 65l
(0.9)^n < 0.65

When n = 5, (0.9)^5 = 0.5905

Thus, after 5 dilutions, the concentration of alcohol will be below 65%.

Thus, the correct answer is C.

glennc · Answer

Each operation removes 10% of the alcohol, so 90% remains.
After n operations, alcohol percentage = 100 × (0.9)^n
We need:
 (0.9)^n < 0.65
Check powers of 0.9:
 (0.9)^4 = 0.6561 → 65.61% (still above 65%)
 (0.9)^5 = 0.59049 → 59.05% (below 65%)
Minimum number of operations required = 5

ANS C

pappal · Answer

on every removal of 10% of the solution and replacing it with water the new concentration of the alcohol is 9/10 of the concentration before the removal.
since we start with 100% alcohol solution 
so after 
Ist- 0.9 or 90%
IInd- 0.9^2=0.81 or 81%
IIIrd - 0.9^3=0.81*0.9=0.729 or 72.9%
IVth- 0.9^4=0.729*0.9=0.6561 or 65.61%
it will be the  Vth removal and replacement  after which the value of the alcohol concentration drops below 65%.

egmat · Answer

Understanding the Problem:
 
Imagine you have a full glass of pure alcohol. Every time you do the operation:
- You pour out 10% of the liquid
- You fill it back up with water

What happens to the alcohol each time? You're left with 90% of whatever alcohol was there before.

That's it. Each operation = multiply by 0.9

The Best Method - Sequential Multiplication:
 
Start: 100% alcohol

Operation 1: 100 × 0.9 = 90%
Operation 2: 90 × 0.9 = 81%
Operation 3: 81 × 0.9 = 72.9%
Operation 4: 72.9 × 0.9 = 65.61%
Operation 5: 65.61 × 0.9 = 59.05%

Question asks: When does alcohol go BELOW 65%?

After 4 operations: 65.61% → Still ABOVE 65%
After 5 operations: 59.05% → BELOW 65% ✓

Answer: C (5)

The Most Common Wrong Answer - B (4):
 
25% of people chose 4. Let's understand why.

After 4 operations, you get 65.61%.

The Trap: This number is dangerously close to 65%. People see 65.61% and think:
- "That's basically 65%"
- "Close enough"
- Or they mentally round 65.61% down to 65%

But here's the thing: 65.61% is NOT below 65%. It's above it.

The question specifically says "below 65%" — not "approximately 65%" or "around 65%."

This is intentional. GMAT loves placing a tempting wrong answer exactly one step before the correct one.

The Gap in Thinking:
 
People who chose 4 likely:
1. Did the math correctly up to 65.61%
2. Then got lazy with the final comparison
3. Assumed "close enough" counts

They confused "close to 65%" with "below 65%."

General Principle - How to Avoid This Mistake:
 
When a question uses precise language like "below," "less than," "at least," or "more than," treat it mathematically, not approximately.

Below 65% means < 65%. Not ≤ 65%. Not ≈ 65%.

65.61 > 65, so it fails the condition. Period.

Always re-read the exact condition before selecting your answer, especially when your calculated value is suspiciously close to the boundary.

In a particular dilution technique, 10% of the solution is removed and

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In a particular dilution technique, 10% of the solution is removed and

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