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Am I wrong if I say that there is another possible combination which is :

4 x 500 + 3 x 100 + 2 x 50 = 2400.

Moreover, it is specified in the problem that the number of denominatios of 100 is more than the number of denominations of 50.

Thus, with this reasoning, the less amount whe could have taken out of her pocket is

2 x 50 + 3 x 100 = 400.
meanup
We needed minimum possible pocket money.

Given, \(500 \times X > 100 \times Y\), and \(50\times Z > 100 \times y\)

\(X_n + Y_n + Z_n = 5\) paid for fees and we needed to maximize this value.

Possible arrangements,

\(4\times 500 + 1\times 100 + 6\times 50 = 2400 \to (1)\)

\(4\times 500 + 2\times 100 + 4\times 50 = 2400 \to (2)\)

Only two combinations are possible as mentioned in 1 and 2.

The Fees maximum amount in both cases \(= 4\times 500 + 1\times 100 = 2100\)

Pocket money \(= 2400 - 2100 = 300\)

ANS B
Hi, its the reverse. The number of 100 notes are lesser than the number of 50 notes.

Hence, its 300.
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To minimize the amount of money leftover, we must maximize the value of the of 5 bills that were used. This would mean the number of $500 bills while satisfying the conditions of Y < Z and Y < X with X being $100, Y $100, and Z $500

One possible outcome is (500)(4) + (100)(2) + (50)(4) = 2,400. Taking away the 5 largest bills would leave 100 + 200 = 300
B
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I wish there were a better clarification in the last line "If the number of $500 notes is greater than the number of $100 notes which in turn was less than the number of $50 notes?"

Because there could be confusion about when the scenario occurred—before the fee payment at the counter or later. I calculated the later scenario, which gave me a different answer.
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Let the number of $50, $100 and $500 notes with Amy be x,y and z respectively.

Since the total amount is $2400, it can be written as:-

50x + 100y + 500z = 2400

x + 2y + 10z = 48

Now, the total number of notes with Amy is x+y+z. Out of these, she gives away 5 notes as school fee such that she is left with minimum pocket money.

How to minimize pocket money?

Pocket money + school fee = $2400
Pocket money = $2400 - (school fee)
Min.(Pocket Money) = $2400 - max.(school fee)

Hence, to minimize the pocket money, we have to maximize the school fee.

How to maximize school fee?

Case I:
Maximum value of school fee would be if all the 5 notes given away as school fee are of $500 denomination i.e. the highest denomination.

However, z can not be equal to 5 as x+2y+10z = 48 (if z=5, then x+2y+10(5) > 48)

Case II:
Now, we look at the next best possible option to maximize the school fee i.e. 4 notes of $500 denomination (highest denomination) and 1 note of $100 denomination (second highest denomination).
It means y=1 and z=4.
Testing these values in the condition x+2y+10z = 48, we get x = 6.
This gives us valid values of x,y and z.

Therefore, the maximum value of school fee = 4*$500 + 1*$100 = $2100
Minimum value of pocket money = $2400 - $2100 = $300

Correct option is (B).



Bunuel
Amy had $2400, which comprises of her school fee and her pocket money. The amount was in denominations of $50, $100 and $500 notes. She gave 5 notes at her school fee counter. What is the minimum possible amount that she got as her pocket money if the number of $500 notes is greater than the number of $100 notes which in turn was less than the number of $50 notes?

A. $250
B. $300
C. $350
D. $400
E. None of these


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