GMAT Club World Cup 2022 (DAY 8): Mbappe withdraws 100/x% of his money

let's say the amount of money is M, then
after the 1st withdraw he'll have M(1-1/x)
after the 2nd withdraw he'll have M(1-1/x)^2
after the 3rd withdraw he'll have M(1-1/x)^3
after the 4th withdraw he'll have M(1-1/x)^4
after the 5th withdraw he'll have M(1-1/x)^5

after 5 visits, he has less than 1xthof the initial amount in the bank then
M(1-1/x)^5 < 1/x
=>(x-1)^5<x4
possible values of x that satisfy this equation are : 2, 3,and 4
the range then is 2
our answer is B

As per the question, (1-1/x)ˆ5< 1/x;

As x >1, the minimum value will be 2. For 2,3,4,5& 6, this equation is valid. Once you put 7, (1-1/x)ˆ5>1/x.

Hence, range will be 6-2=4. Option D is the answer

withdrawal done each time is 100/x % which is 1/x
so 5 times withdrawal is 5/x
money left is 1/x
x-5/x <1/x
x^2-5-1 <0
x^2<6
given x >1
so possible value of x = 2
option B

Given: Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank, where \(x > 1\).

Asked: If after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount in the bank, what is the range of all possible value of \(x\) ?

Let the initial amount in the bank = M
Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank = M/x
Money left after 1st withdrawal = M - M/x = M(x-1)/x
Money left after 2nd withdrawal = M(x-1)^2/x^2
Similarily money left after 5th withdrawal = M(x-1)^5/x^5 < M/x
(x-1)^5/x^5 - 1/x < 0
(x^4 - (x-1)^5)/x^5 > 0
x^4 - (x-1)^5 > 0
x^4 > (x-1)^5

If x = 2 ; x^4 = 2^4 = 16; (x-1)^5 = 1; x^4 > (x-1)^5
If x =3; x^4 = 3^4 = 81 ; (x-1)^5 = 2^5 = 32; x^4 > (x-1)^5
If x =4; x^4 = 4^4 = 256; (x-1)^5 = 3^5 = 243; x^4 > (x-1)^5
If x = 5 ; 5^4 = 625 ; 4^5= 1024; x^4 < (x-1)^5

Range of values of x = 4 - 2 = 2
I am assuming x> 1 ; where x is an integer.

IMO B

I don't understand the question....
Guess C and move on

If x equals 1.001, he would withdraw 99.9% of his money on each visit, thereby leaving him with effectively 0%, which is less than 1/1.001 (99.9%) of the initial amount. 1 is the lower bound of our range.

If x equals 2, he would withdraw 50% (1/2) of his money on each visit, leaving him with 1/2, then 1/4, then 1/8, then 1/16, then 1/32, which is less than 1/2 of the initial amount.

If x equals 3, he would withdraw 33.333% (1/3) of his money on each visit, leaving him with 2/3, then 4/9, then 8/27, then 16/81, then 32/243, which is less than 1/3 of the initial amount.

If x equals 4, he would withdraw 25% (1/4) of his money on each visit, leaving him with 3^5/4^5, which is 243/1024. That is just baaaaarely less than 1/4 of the initial amount.

Range is from 1 to 4.

Answer choice C.

after 5th visit we get the inequality
1/x > (1-1/x)^5
this equality holds for x=2,3 and 4
Hence 3 values
Answer: C

I just straight up took values.
If you take x = 2 and the initial amount as 100, the statement will hold true, and the same holds true till x = 4

But if you take x = 5, then the statement doesn't hold true.

Hence, the range = 4-2 = 2.

let initial amount be y then

(y-(5y/x)<y
(x-1)<5
x<6

also as per question x>1, thus possible values of x are 2,3,4,5 hence d is the correct answer.

Correct answer : choice B

let amount = 100
for x = 2 , amount at 5th visit is less than 1/2 (100) ---> works
for x = 3, amount at 5th visit is less than 1/3(100) ---> works
for x = 4, amount at 5th visit is less than 1/4(100) ---> works
for x = 5, amount at 5th visit is not less than 1/5(100)

Given x > 1
hence range is 4-2 = 2
choice B

(1-1/x)^5 < 1/x

For x =1, 0<1
For x = 2, 1/32 < 1/2
For x = 3, 32/243 < 1/3
For x = 4, 243/1024 < 1/4
For x = 5, 1024/3125 > 1/5

Range = 4-1 = 3
Ans C

Here let amount after 5 visit A=P(1-r/100)^n [P=Original amount, r=rate %, n= Number of visits]
Thus,
P(1-(1/x))^5<P/x
Or (1-(1/x))^5<1/x [All values are positive]
Or ((x-1)^5)/(x^4) < 1

Or x < ~ 4.08
Therefore range of values of x will be given by: 1 < x < ~4.08

If we consider range of all integer values of (answers given for range are integers) x, we have x belonging to [2,4]
Hence range is 2.

Let the total initial Money be M
Fraction of Money left after each withdrawal is [100-(100/x)]/100 -->1-1/x
Money left after 5 withdraw = M*(1-1/x)^5

hence M*(1-1/x)^5 < M*1/x

Solving [(x-1)/x]^5 < 1/x

We get range of x = 4

[quote="Bunuel"]Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank, where \(x > 1\). If after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount in the bank, what is the range of all possible value of \(x\) ?

A. 1
B. 2
C. 3
D. 4
E. 5

As per the statement Mbappe withdraws (100/x) % of his money each time.
If The initial amount of money =P
then P*(1-((100/x)*1/100)^5 =P*(1-1/x)^5 =P*(x-1)^5/x^5
Now P*(x-1)^5/x^5<P*1/x
=> (x-1)^5<x^4
Now this is only possible for x=1,2,3,4
and given x>1
Thus the range of all possible values of x=2

Answer B

Say initially he had amount m.
So after 1st withdrawal he had = m - (m/x)
After 2nd withdrawal he had = m - (2m/x) + (m/x^2)
And similarly after the 5th visit he would have
m-(5m/x)+(10m/x^2)-(10m/x^3)+(5m/x^4)-(m/x^5). This would be < 1/x
Now it's given that x > 1
So substituting the above equation:
For x = 2, we have the value of the relationship as,
-1.0625 < 1/2
i.e. -1.0625 < 0.5, which is true
Similarly for x = 3 we have,
0.008 < 0.333, which is true.
Similarly for x = 4 we have 0.55 > 0.25
So the equation does not satisfy.
For x = 5 also we find the equation does not satisfy.
Hence there are 2 possible values of x.
Hence answer choice (B)

We know that

P\([1- \frac{1}{x}]^5\) < \(\frac{P}{x}\)

Cancelling P from both ends

\([1- \frac{1}{x}]^5\) < \(\frac{1}{x}\)

\([\frac{(x-1)}{x}]^5\) < \(\frac{1}{x}\)

\(\frac{(x-1)^5}{x^5}\) < \(\frac{1}{x}\)

\((x-1)^5 < x^4\)

The above values are valid when x < 4.XX

We know that x > 1

Since x > 1 and x < 4

The range = 3

IMO C
_________________

Range=Highest value-Lowest value

Questions states that Mbappe went to bank 5 times. So, x cannot be 4 or less, since he withdraws \(\frac{100}{x}%\) each of the 5 times. If x=4, then he would have withdrawn 25% during each of his first 4 visits, totaling to 100% in 4 visits only and nothing left for the 5th visit.

If x=5, then he would have withdrawn 20% during each of his 5 visits, totaling to 100% in 5 visits, leaving 0% which is less than 20% of the initial amount in the bank. So, x=5 is a possible value.

If x=6, then he would have withdrawn 16.66% during each of his 5 visits, totaling to 83.33% in 5 visits, leaving 16.66% which is less than 16.67% of the initial amount in the bank. So, x=6 is a possible value.

If x=7, then he would have withdrawn 14.2857% during each of his 5 visits, totaling to 71.4286% in 5 visits, leaving 28.5714% which is not less than 14.2857% of the initial amount in the bank. So, x=7 and beyond is not a possible value.

Range = 6-5 = 1. Answer A.

"Mbappe withdraws \(\frac{100}{x}\)% of his money".

Let us understand this statement and what happens when Mbappe withdraws \(\frac{100}{x}\)% first time.

Say Mbappe had $100 in his account and he withdraws 25% (i.e. \(\frac{100}{4}\))% - \(He withdraws = $100 * (\frac{25}{100})\)

He has left = \($100 * (1-\frac{25}{100})\) or \($100 * (1-\frac{1}{4})\). You notice how simply we can denote a one time withdrawal of \(x\)% as \($100 (1-\frac{1}{x})\).

What happens when Mbappe withdraws second time? He has \($100 * (1-\frac{1}{x})\) in his account left, and we withdraw a second time, we simply multiply \((1-\frac{1}{x})\) again.

2nd time after withdrawal, money left is = \($100 * (1-\frac{1}{x})^2\)

After 5 withdrawals, money left is = \($100 * (1-\frac{1}{x})^5\)

The problem says that this amount is less than \(\frac{1}{x}\) of original amount, so -

\($100 * (1-\frac{1}{x})^5 < $100 * \frac{1}{x}\)

\(\frac{(x-1)^5}{x^5} < \frac{1}{x}\)

Or \((x-1)^5 < x^4\)

As we know \(x>1\), if you plug in values 2, 3, 4 and 5 for \(x\), you will see that \(x=5\) doesn't satisfy this equation.

Hence the satisfactory values of \(x\) is {2, 3, 4}, hence the range is 4-2 = 2 Answer B

Ans - E

Solution attached

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