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GMAT Club Olympics 2025 (Day 8): On Monday, Elena had a certain amount

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Edwin555

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09 Jul 2025, 21:05

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Let $x$ be the amount present on Monday

Forming an equation from the given details we have,
$x + (mx/100) - nx/100(1 + m/100) = .50x$
Re-arranging,
$1 + (m/100) - n/100(1 + m/100) = .5$
$(1 + m/100)(1 - n/100) = .5$

Re-arranging and multiplying both sides by $100$,
$(100 + m)(100 - n) = 5000$
$(100 - n) = \frac{ 5000 }{ (100 + m) } $
$\frac{100 (100 + m) - 5000}{(100 + m)} = n$
$\frac{100(m + 50)}{(100 + m)} = n$

Correct option is Option A

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We can solve this by taking sample values. Initially suppose we have 100 rs and we increased it by 50% which will get us to 150. Now to get 50% of initiall amount which will be 50, we need to withdraw 200/3% So substituting in options 1st one will get us to 100*100/150=200/3. Option A

Bunuel

On Monday, Elena had a certain amount of money in her savings account. On Tuesday, she increased that amount by m percent. On Wednesday, she withdrew n percent of her Tuesday's closing balance, leaving her with exactly 50 percent of what she had on Monday. What is the value of n in terms of m?

A. $\frac{100(m + 50)}{m + 100}$

B. $ \frac{100(m - 50)}{m + 100}$

C. $ \frac{100(m + 50)}{100 - m}$

D. $ \frac{100m}{m + 100}$

E. $ \frac{100(50 - m)}{m - 100}$

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09 Jul 2025, 22:02

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Let the Monday amount be x.
Tuesday: after m% increase x(1+m/100)
Wednesday: after withdrawing n% x(1+m/100)( 1-n/100)
x(1+m/100)(1-n/100) =x/2

Answer: n=100(m+50)/m+100.
Option A.

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09 Jul 2025, 22:09

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let money on Monday be = x

so on tuesday, money in account will be: (1+m/100)x

on wednesday, money in account will be: (1-n/100)(1+m/100)x

which is equal to x/2.

(1-n/100)(1+m/100)x = x/2

on solving the equation we get, option A as the correct answer.

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09 Jul 2025, 22:19

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Let Monday = 100 (a smart, simple choice)

Tuesday: Increased by m%, so it becomes ->
100+m100 + m100+m
Wednesday: Withdraw n% of Tuesday’s amount ->
Left with:(100+m) * (1− (n /100))

But we know this equals 50 (i.e., half of 100):
(100+m)*(1−(n/100)) = 50
Now isolate n:
1− n/100 = 50 / (100+m)
=>
n/100 = 1− (50 / (100+m))
=> (100+m−50)/(100+m) => (50+m) / (100+m)

So:
n= 100 * (50+m)/(100+mn)
(A)

Bunuel

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09 Jul 2025, 22:39

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Bunuel

Considering the Monday Amount be = 100.
On Tuesday the new amount will be = 100 + m*100/100= 100+m
On Wednesday, the withdrawal Amount = n/100*(100+m) = n[1+(m/100)]
Amount balance on Tuesday = 100+m - n[1+(m/100)] = 100*50/100
100+m- n[1+(m/100)] = 50
n[1+(m/100)]= m + 50
n= (m + 50 )/ [1+(m/100)] = 100(m+50)/[100+m] A

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09 Jul 2025, 22:56

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Let's start with a 100 on Monday.
m% increase and then n% decrease on Tue and Wed respectively.
Final amount left is 50 at end of Wed.

We can use the percentage increase formula to find the final increase/decrease as below:

100 + m - n - (mn/100) = 50

Simplifying:

100m - 100n - mn = -50 *100
=> 100m + 50*100 = mn + 100n
=> 100m + 50*100 = n(m+100)
=> 100(m + 50) = n (m +100)
=> n = 100(m+50) / (m +100)

Answer: B

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09 Jul 2025, 23:38

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Let us assume initial amount on Monday be x units
Given
On Tuesday amount is increased by m%
I.e (x + m% of x) =x * $((100+m)/100$)

On Wednesday n% of Tuesday's closing balance was withdrawn
I.e Amount withdrawn = n% of x * ($\frac{(100+m)}{100}$)=($\frac{nx}{100}$) * $(\frac{(100+m)}{100}$)
Remaining amount = x * $(\frac{(100+m)}{100}$) - ($\frac{nx}{100}$) * $(\frac{(100+m)}{100}$) = x$\frac{(m+100)}{100}$ * (1-$\frac{n}{100}$)

Final amount is 50% of Monday's (x) amount
$\frac{x(m+100)}{100}$ (1-$\frac{n}{100}$) = x/2

On solving:
(1-$\frac{n}{100}$) = $\frac{1}{2}$$\frac{100}{m+100}$
==>$\frac{n}{100}$ = 1 - $\frac{50}{m+100}$
==>n = $\frac{100(m+50)}{m+100}$

Option A correct

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09 Jul 2025, 23:58

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Used numbers to solve this.
Suppose on Monday she started with 100 & m= 5
Then on Tuesday, total amount becomes 105
After withdrawing n% of 105, she has 50% of Monday's amount left = 50. Hence she withdrew the remaining 50.

Hence n% of 105 = 50
n = 5000/105

Now putting m = 5 in options to find which gives this value of n.

(D) 100m/(m+100) = 5000/105 hence correct option is (D)

You can put m=5 in other options as well to check, only D gives the value of n.

Ans D

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10 Jul 2025, 00:39

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suppose she has x on Monday

Tuesday it increased by m% = x+(m/100)*x = (100+m)*x/100 ----- eq (1)
Wednesday she spend n% of the Tuesday balance = (100-n)*(100+m)*x/100*100 ----eq(2)

remaining balance = 50*x/100 ----eq(3)

As per the equation eq(2) and eq(3) are same
(100-n)*(100+m)*x/100*100 = 50*x/100

on simplifying
n = 100*(50+m)/(100+m)
Option A

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10 Jul 2025, 00:51

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Let Monday's balance= M

Tuesday's balance = M *(1+(m/100))

Wednesday balance

M *(1+(m/100)) * (1 - (n/100)) = M/2

(1+(m/100)) * (1 - (n/100))=1/2

(1 - (n/100))= 1/(2*(1+(m/100)))

n/100= 1-(1/(2*(1+m/100))

n= 100 *(1-(1/(2*(1+m/100)))

n= 100 ((2-1+m/100))/(2+m/100)
n=100(1+m/100)/(2+m/100)
n= 100+m/(2+m/100)
n=100(100+m)/100(2+m)
n=100(100+m)/(m +200)

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10 Jul 2025, 01:34

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	mon	tue	wed
Elena	let, x	x*(100+m/100) after m% increment	x(100+m/100)(100-n/100), after n% decrement

so finally,
x*(100+m/100)*(100-n/100) =x*50%= x/2
after solving we will get n=100(m+50)/100+m
option A

Bunuel

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We can sole this with fair assumption.
Lets take Monday Amt be $100 -- increased by 50% (m) --> $150 ---- Now we need to decrease to get $50 so n = 60%.

Now we know if we take m = 50 ; then n= 60%, plug m value to check which option give n- 60% its A

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10 Jul 2025, 04:14

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So best way to solve this is by choosing smart numbers

On Monday lets assume she deposit 100 in bank
on Tuesday she increased m% we can assume this to be 50%. So now the bank has 150.

We know on wednesday, after withdrawing her money was exactly 50% so she will have 50 in her account.
Money withdrew on wednesday is 150-50=100.
So, N will be \frac{100}{150}=\frac{2}{3}or \frac{200}{3}%

m=50 n=200/3

So lets see the answers:
1. \frac{100(50+50)}{(50+100)}=\frac{100(100)}{150}=\frac{200}{3 }

Yes this is the answer

Answer A

Bunuel

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10 Jul 2025, 04:15

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increase by m percent -> Money is (1+m/100) greater
withdraw n percent -> Money is (1-n/100) less

Money*(1+m/100)*(1-n/100)=Money/2
(1+m/100)*(1-n/100)=1/2
(m+100)*(1-n/100)=50
1-n/100=50/(m+100)
n=100*(1-50/(m+100))=100*((m+100-50)/(m+100))=100*((m+50)/(m+100))

Answer A

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10 Jul 2025, 04:22

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Given,
Monday, Elena has money in her account = x
Tuesday, after adding m %, the money in the account will become
= x + xm/100
Wednesday, after withdrawing n%, the money in the account will become
= x + xm/100 – (x + xm/100)*n/100
In other words, = x/2
To find
the value of n in terms of m?

Solve:
x + xm/100 – (x + xm/100)*n/100 = x/2
taking x common & cancelling from both sides
1+ m/100 – n/100 – mn/10000 = 1/2
1⁄2 + (m – n)/100 = mn/10000

Multiplying 100 on both sides,

100/2 + m = n + mn/100
50 + m = n (100 +m)/100
n = 100(m+50)/(m+100)

Ans: A

Bunuel

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10 Jul 2025, 04:42

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a = amount

a(1+m/100)(1-n/100) = a/2
(100+m)/100 * (1-n/100) = 1/2
1-n/100 = 100/(2*(100+m)) = 50/(100+m)
n/100 = 1-50/(100+m) = (100+m-50)/(100+m) = (m+50)/(m+100)
n = 100(m+50)/(m+100)

IMO A

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10 Jul 2025, 04:42

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Successive percent calculation : if a % increase then b% decrease = (a - b - (ab/100))%

Using this, m - n - (mn/100) = 50

=> n = 100(m-50)/(100+m)

B

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10 Jul 2025, 04:43

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let amount of money on Monday be = $A$

On Tuesday, amount increases to =$ A(1 + \frac{m} {100} )$

On Wednesday, amount decreases to = $ A(1 + \frac{m} {100} )(1 - \frac {n} {100})$ = 50% of amount on Monday =$ \frac{50}{100}A$

>> $ A(1 + \frac{m} {100} )(1 - \frac {n} {100})$ = $ \frac{50}{100}A$

>> $ (1 - \frac {n} {100}) $ = $ \frac {50} {100+m} $

>> $\frac{n}{100}= \frac {(100 +m )- (50)} {100+m}$

>>$ n = \frac {100(50+m)} {100 + m}$

Answer is A : $\frac{100(m + 50)}{m + 100}$