Percival has a total of 26 bills, consisting of a combination of $1, $

Question

Percival has a total of 26 bills, consisting of a combination of $1, $5, $10, and $20 bills.
 
If the total value of these bills is $386, and Percival has at least one bill of each denomination, select for  $1 bills  the possible number of $1 bills Percival could have, and for  $20 bills  the possible number of $20 bills Percival could have that are jointly consistent with the information provided. Make only two selections, one in each column

Bunuel · Accepted Answer

Official Solution: Let's analyze the provided options and determine how many $20 bills Percival could have. • Percival cannot have 18 $20 bills, as that would amount to $360, leaving only $26 to be distributed among the remaining 8 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6. If there were 1 $1 bill, the remaining 7 bills of $5 and $10 denominations would exceed $25. Conversely, if there were 6 $1 bills, the remaining 2 bills of $5 and $10 denominations would total less than $20. Thus, 18 $20 bills is not possible. • He also cannot have 12 $20 bills because, in that case, the total from the $20 bills would be $240. The remaining 14 bills, even if all were $10 bills, would amount to only $140, resulting in a total of $380, which is less than $386. • Therefore, from the options, Percival must have 16 $20 bills, which amounts to $320. This leaves $66 to be distributed among the remaining 10 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6. • If there were 6 $1 bills, they would amount to $6, leaving $60 to be accounted for by the remaining 4 bills. However, even if all 4 bills were $10 bills, this would only amount to $40, which is less than $60. • Therefore, there must be exactly 1 $1 bill. Correct answer: $1 bills "1" $20 bills "16" GMAT-Club-Forum-rb3h7f5z.png

Bismuth83 · Answer

1. Let's rephrase the question using the information that Percival has 1 of each bill (there is $386 - $1 - $5 - $10 - $20 = $350 left). Find the number of possible $1 and $20 bills that match the answer choices, when the amount of money is $350 and there are 26 - 4 = 22 bills. The answer choices would then be: 0, 5, 9, 11, 15, and 17.

2. The maximum number of $20 bills we can have is 17, and the nearest answer choice to that is 17. This would make up $340 of $350 with 5 bills left, which is impossible to cover.

3. The next number of $20 bills we can have is 15. This would make up $300 of $350 with 7 bills left, which can be covered with 0 $1 bills, 3 $10 bills, and 4 $5 bills.

4. This means that altogether, we used 0 + 1 = 1 $1 bill, 4 + 1 = 5 $5 bills, 3 + 1 = 4 $10 bills, and 15 + 1 = 16 $20 bills.

5. Our answer will be:   $1 bills - 1 and $20 bills - 16  .

MegB07 · Answer

Hi, how did you get 26 - 4 = 22?

KevoK · Answer

They arrived at that number becuase the -4 is the 4 bills for each denomination ($1, $5, $10, $20)

Vivek1707 · Answer

One of the quickest way would be to test cases with numbers in a vertical tabular format

Start with 1 units for 1$ cause that’s the lowest values. Additionally take note that total number of billls are 26 and each bill has to have minimum 1 unit.

$ |Bills| Amount
1$ 1 = 1 
5$ 1 = 5 
10$ 2 = 20
20$ 18 = 360

Total = 386 the but the number of bills is 22

So to increase the number of bills remove 1 bill from 20$ i.e make it 17 and increase 4 bills from 5$ the amount will be 386 but the total number of bills would still be 25 i.e < 26.

Next remove 1 bill from 20$ and increase 2 bills from 10$ ---> gives you total values as 386 and total number of bills as 26

Adit_ · Answer

Trick to such problems is maximise the number of max denominations and minimise the number of least denominations this is because 26*20=520 and we are not too far away from 386 which is what we need

Meaning we are just 140 ish away from 386 if we ONLY take 20 OR we need a LOT of 20 denominations to cover up the gap with just 26 notes (I hope that makes sense)

Take max of 20 and then work ur way back now if 18 then we quickly notice that its a BIT too much 
Once we reduce it to 16 voila! 1 denomination of 1 sticks perfectly

The way to choose number of 1s is basically only 3 ways here:
1,6,16 This is because the 5 and 10s must cover up values ending with 5 or 0 and only 1,6,16 1s will help cover that unit digit.

16 is too much eliminated.
6 is a possibility to try out with 18 as 20 (doesnt work, here we notice that there is plenty of gap when we choose 18 still, so take 16 and reduce the number of 1s and check)

Trial and error IMO helps the best once u hit on the strat.

egmat · Answer

The  26 - 4 = 22  comes from the constraint  "at least one bill of each denomination."

Here's the logic:

We have  4  different bill types: $1, $5, $10, and $20.

Since Percival  must  have at least one of each:
-  1  bill reserved for $1
-  1  bill reserved for $5
-  1  bill reserved for $10
-  1  bill reserved for $20

That's  4  bills "locked in" from the start.

Therefore: 
- Total bills =  26 
- Bills already accounted for =  4  (one of each type)
- Remaining bills to distribute =  26 - 4 = 22

Similarly, for the value:
- Total value = $ 386 
- Value of the  4  reserved bills = $1 + $5 + $10 + $20 = $ 36 
- Remaining value = $ 386  - $ 36  = $ 350

So we need to figure out how to distribute  22  additional bills worth $ 350 .

axeofflane · Answer

Way I did this was by cutting pieces one by one:

Step 1: Account for the "at least possibility" - so one of each would mean 35$. Subtract it from the total, so remaining amount is 350$, remaining bills is 21. Now we just have to test cases for 20$

Step 2: Test cases from high to low for 20$ bills. So try 18 20$ bills, but this would mean 360$, hence out.

Step 3: Test 16 20$ bills now, comes to 320, which is less than 350, yay. You have 5 bills to play with and 30 you need to get, can you squeeze it in? Absolutely, with 2 10$ bills and 2 5$ bills.

And there you go, the answer. Took me a minute to solve, but maybe I just got lucky with this method for this question as i was able to get to the correct case quickly

DishaAgarwal12 · Answer

How i thought about it:

$20 if it is just 1 bill- then we will left with 366- (which cant be covered even if we have all 10s= 250) so out, same for 6 (20*6=120) we are left with 266 (20*10=200)

we now look at 10,12,16,18

10*$20= 200, We have 16 bills and 186 (all 16s as $10 still $160,short of $26) so out

12*$20= $240, $146 left and 14 bills left (still with all $10 not possible)

16*20= 320, $66 left (1 note each of $1,$5,$10= $16), so $50 left and 7 bills left (26-16-3=19); 7 bills- if all 7 $10= $70, you can hve 0 or 5 $1, which wont be sufficient

take 3 $10= $30, and 4 $5 bills which gives $20

therefore 1 and 16 is the ans

Percival has a total of 26 bills, consisting of a combination of $1, $

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Percival has a total of 26 bills, consisting of a combination of $1, $

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