12 Days of Christmas GMAT Competition - Day 7: An investor has two : Two-part Analysis (TPA)

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24 Dec 2024, 09:43

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Bunuel

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Let maximum capacity be T for both main and backup wallets.

Then, initial occupied capacity of main wallet is T*m/100 and of backup wallet is T*b/100

Let x be the storage transferred from backup wallet to main wallet.

Then,

Tm/100 + x = T --- (1)
Tb/100 - x = (T*(b/2))/100 --- (2) as backup wallet now contains half of its previous percentage

From (2) => Tb/200=x

Replacing x in (1)

Tm/100 + Tb/200 = T
m/100 + b/200 = 1
2m + b = 200

Trying out option pairs for value of m,

m=30, b=140 --- Not given
m=45, b=110 --- Not given
m=50, b=100 --- Not given
m=55, b=90 => Only possible option choice
m=60, b=80 --- Not given
m=90, b=20 --- Not given

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An investor has two cryptocurrency wallets, Main and Backup. The Main wallet holds m% of its maximum capacity, while the Backup wallet holds b% of its maximum capacity. Both wallets have identical maximum capacities.

Due to new regulations, the investor must optimize storage by transferring tokens from the Backup wallet to the Main wallet until the Main wallet reaches its maximum capacity. After this transfer, the Backup wallet contains exactly half of its previous percentage.

If the Backup wallet initially held a higher percentage than the Main wallet, select for m the possible value of m, and select for b the possible value of b that would be jointly consistent with the given information. Make only two selections, one in each column.

Let the maximum capacity of both main and backup wallets be M each.

Before regulations: -
Main wallet = Mm%
Backup wallet = Mb%
Total = M(m%+b%)

After regulations: -
Main wallet = M
Backup wallet = M(m%+b%) - M = M(m%+b%-100%) = Mb%/2 ;
m%+b%/2 =100%
Total = M(m%+b%)

Since backup wallet initially held a higher percentage than the Main wallet
b% > m%

b	90	60	80	100	110	140
m	55	70	60	50	45	30

Only possible option with given data
b% = 90%
m% = 55%

m	b
55	90

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24 Dec 2024, 10:04

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Let max capacity = x

From given information: b > m

After transferring half of the tokens from backup wallet

mx/100 + bx/200 = x
=> 2m + b = 200

Satisfying from the given options we get,

m = 55
b = 90

Answer: (55, 90)

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24 Dec 2024, 10:05

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Let's assume max = 100, b has transferred half of his amount. Let's assume x is the amount being transferred.

x = b/2 -- 1
m+x = 100 -- 2

Using this, we have 2 constraints:
m+b/2 = 100
and b> m

checking values, m = 55 and b = 90, fits the bill

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24 Dec 2024, 10:07

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Bunuel

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Equation: b + 2m = 200 can be derived by solving the eqns mentioned in problem
Constraint: b > m
Test given values:
If m = 30: b = 200 - 2(30) = 140. 140 > 30, so this works.
if m = 45: b = 200 - 2(45) = 110. 110 > 45, so this works.
If m = 50: b = 200 - 2(50) = 100. 100 > 50, so this works.
If m = 55: b = 200 - 2(55) = 90. 90 > 55, so this works.
If m = 60: b = 200 - 2(60) = 80. 80 > 60, so this works.
If m = 90: b = 200 - 2(90) = 20. 20 is not greater than 90, so this does not work.
the possible pairs from the given options are (30, 140), (45, 110), (50, 100), (55,90), and (60,80). Since only 90 is an option for b, the correct pair from the provided options is m = 55 and b = 90.

IMO 55 and 90

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24 Dec 2024, 10:23

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We need to determine values ofm (Main wallet percentage) and b (Backup wallet percentage) consistent with the problem's conditions:

The Main wallet starts at m% of its maximum capacity.
The Backup wallet starts at b% of its maximum capacity.
Tokens are transferred from the Backup wallet to the Main wallet until the Main wallet reaches 100%.
After the transfer, the Backup wallet contains half of its previous percentage (i.e.,
b/12%).
Initially,
b>m (the Backup wallet held a higher percentage than the Main wallet).
Expression for the transfer amount
The transfer amount is the difference between the Main wallet's maximum capacity (
100
100%) and its initial capacity (m%):

Transfer amount=100−m.

Step 2: Post-transfer condition for the Backup wallet
After the transfer, the Backup wallet retains half of its previous percentage. Thus, the transfer amount also equals the percentage removed from the Backup wallet:

Transfer amount
= b − b/2= b2

Equating the two expressions for the transfer amount:
100−m= b/

Step 3: Solve for b in terms of m
Multiply through by 2:
200−2m=b.

Step 4: Check valid pairs of m and b
Given the condition
b>m, we substitute possible values for
m and compute b.

Both m and b must be in the list of available options:
b=200−2m
Check
b>m? 30
200−2(30)=140
b>m, but
b∉{45,50,55,60,90}

200−2(45)=110
b>m, but

b ∈
{45,50,55,60,90}
200−2(50)=100
b>m, but
b ∈ {45,50,55,60,90}

200−2(55)=90
Valid: b>m, and b=96/60

200−2(60)=80
b>m, but
b ∉ b ∈ {45,50,55,60,90}

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24 Dec 2024, 10:27

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Given b>m

let x be the capacity of each of the wallet.

Wallet M: xm/100 => to reach full capacity need x(100-m)/100
Wallet B: xb/100

So xb/100 - x(100-m)/100 = (1/2) * xb/100

on simplification we get b+2m=200
since b>m option pair 90, 55 satisfy for b,m

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24 Dec 2024, 10:27

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Let X be the max amount that can be hold in the accounts.
Main account amount. = m/100* X
Backup account amount = b/100 * X
After the transfer,
X - (mX/100) = (b/(2*100)) => 2m+b = 200
Also remember, m<b. Checking the values from the options provided. Only m = 55,b=90 hold the equation true.

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24 Dec 2024, 10:34

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let the total capacity for both Main and Backup wallets be 100 units each;
m% < b%

let m = 55%;

To fill out the max capacity, 45 units are needed from backup wallet.
As a result, b = 90 because 90 - 45 = 45/100(45%) which is half of the b percentage.
Option D for m
Option F for b

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Bunuel

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Given,
b > m
m + b/2 = 100

From the options its clear that b would be multiple of 2, since m, b are integers from the options.
b = 30 => m = 85 wrong
b = 50 => m = 75 wrong
b = 60 => m = 70 wrong
b = 90 => m = 55 correct

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24 Dec 2024, 10:37

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Based on the above question, we get the equation:
2m + b = 200
{ (m/100*c) + (b/200*c) = c} => if 1/2 of b% was added to m% then that made it full capacity
Plugging combos into the above equation, only possible values are m=55, b= 90. Any values of b ending with 5 don't work for the above because 2*something can never end with a 5.

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24 Dec 2024, 10:43

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Based on the given data, we can set up the equation:
(m / 100) + (b / 200) = 1,
which simplifies to 2m + b = 200.

Since there are no three-digit values for b, we know b must be greater than 50 for the equation to hold. (Ex: For m = 50, b would be 100, which isn’t an option)

Thus, we check the next possible value for m:
For m = 55, b=200−2(55)=90

This satisfies the equation.

Answer: m = 55, b = 90.

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24 Dec 2024, 10:53

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Lets take their both capacity as C

given that
m% of C, b% of C in both the wallets

after the transfer
Main wallet is filled
so
m% * C + (b/2)% * C = C
m +b/2 = 100

From the options

the values that satisfy m and b are m = 55, b = 90

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24 Dec 2024, 10:58

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Bunuel

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Let's assume that the maximum capacity = x

$\frac{bx}{100} - \frac{(100 - m)*x}{100} * \frac{1}{x} = \frac{b}{200}$

$b - 100 + m = \frac{b}{2}$

2m + b = 200

m = 30 ; 2m = 60 ; b = 120 -- No combination exists
m = 45 ; 2m = 90 ; b = 110 -- No combination exists
m = 50 ; 2m = 100 ; b = 100 -- No combination exists
m = 55; 2m = 110 ; b = 90 -- Correct

m = 55
b = 90

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24 Dec 2024, 11:00

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After the transfer
The Main wallet will be full (100% of its capacity)
The Backup wallet will contain half of its previous percentage of capacity = b/2%

Backup wallet

	30	45	50	55	60	90
amount in Backup after transfer	15	22.5	25	27.5	30	45
amount transferred to Main	15	22.5	25	27.5	30	45

Main wallet

	30	45	50	55	60	90
previous amount in Main	30	45	50	55	60	90
Required amount for 100%	70	55	50	45	40	10

Clearly m% = 55%
and b% = 90%
45% of backup went to main making main 100%

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24 Dec 2024, 11:02

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let the amount transferred be x
so, m + x = 2(b-x)
2b-m=3x
2b-m is a multiple of 3. hence on checking with options, if b = 45 and m = 30, we get 2b-m = 60 which is a multiple of 3 and also satisfies the condition that b>m.

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Let the capacity of both wallets be 100.
b - (100 - m) = 0.5b---> 0.5b = 100 - m---> m=100-0.5b

Start with multiple of 10 for b from the options, say 50, then m=75 and b=50 , but we know that there is a constraint b>m
take b=60, then m=70 still b<m
Take b=90, then m=55---> b>m

Therefore m=55 and b=90.