How Did The Blind Man Catch This Cheating Cashier?

Total cost of items bought by Matthew = $58

Change that should have been given to Matthew = $42

If the change was correct, then the teller could give him 4, 5 or 6 bills.

$(20 + 10 + 10 + 1 + 1)= $42 (5 bills)

$(20 + 20 + 1 + 1) = $42 (4 bills)

$(1 + 1 + 5 + 5 + 10 + 20) = $42 (6 bills)

The question says that the change given was incorrect.

Therefore, the change given to him must not be $42.

Using 3 bills we cannot total up $42.

Hence , the possibility that the number of bills that the teller handed to Matthew is $\boxed{3}$ .

you know that you are going to have two, one dollar bills no matter what.

since there is no such thing as a $40 bill then 3 bills must be the only one which is impossible. Unless he made his own $40 bill ;)

the amount Matthew had to receive back were (40 +2)$. The 2$ can be given only in two notes of 1$. The remaining balance is 40$. Since there is no note for 40$ itself it has to be given as a combination of the available denominations. Hence the minimum number of notes given cannot be 3. hence the answer. :)

I got i this way.

The total shopping was of $58 and he should have gotten a change of $42.

So this means he could have received money in he following denominations $1,$5,$10 and $20.

Receiving a change of $42 is impossible if he gets only 3 notes.

Hope this helps...cheers!!!!

$25 Spiderman $13 LEGO CLRT + $20 Despicable

$58

$100-$58=$42.

The cashier owed $42 change to Matthew.

Trying the lowest of these combinations, it's obvious, that $20, $20, and $2 combination is invalid since $2 dollars are not valid. Any other bill combination will deem invalid also.

Legal bills: $1, $5, $10, $20, $50, $100. $50 is useless, $10 too low, so 2 $20's makes sense, but nothing to top off with.

Therefore, 3 is a possibility.

A change of $42 to be tendered can be done by:

6 notes ($10 x 4 notes + $1 x 2 notes)

4 notes ($20 x 2 notes + $1 x 2 notes)

5 notes ($20 x 1 note + $10 x 2 notes + $1 x 2 notes)

But no combinations, can he tender the $42 change with 3 notes.

It is very simple. The person had to return change i.e. $42. Now we look at the minimum of the option i.e. 3. The person could have returned only $41 and he would have to give one more note to make it $42.

Hence the answer is 3

This is what the blind man bought for his nephew. Spiderman action figure $25 + Truck $13 + Despicable Me 2 $20 = For a total of $58 He handed the cashier a $100 bill. This question states that in general, bills will come in $1, $5, $10, $20, $50, and $100's. Therefor, when the teller gave him the least amount of bills, he would have received two $20's and two $1 bills, which is 4 bills.

Since that he has 100$ bill,

so the change=$10--$20-$25-$13=$42

so the probable change notes are=$1,$5,$10,$20 that is 4 of them as the change is < $50 and < $100

so by combination to fill up that $42 with bills= 4C1=4

thanks...

6 bills 10,10,10,10,1,1 or 20,10,5,5,1,1 5 bills 20,10,10,1,1 4 bills 20,20,1,1 3 bills 20,20,1 not=42 in any case

Toys has total cost of $58 so he should be changed of $42. Means, it can be $20,$20,$1 or $1 or $20,$10,$10,$1&$1 or $10,$10,$10,$10,$1,&1. Or with 4,5 or 6 number of bills.

The Teller will never return $100 or $50 bills. So we remain with $1, $5, $10, $20. Matthew gets back $(100-58)=$42. So to get wrong return he must have got * 2 $20 bills * 1 $20 bill & 2 $10 bills and so on...... but the number of bills increases a lot So the minimum no. of bills he gets back is 2 but since there is no option for 2 , I chose 3.........

Total expense = 25 + 13 + 20 = 58 Amount Mathew should receive = 100 -58 = 42 There are many ways to make up this amount But you need at least 4 bills to do it. 20+20+1+1(4 bills), 20+10+10+1+1(5 bills), 20+10+5+5+1+1(6 bills),upto (5+5+5+5)+(5+5)+(5+5)+1+1(10 bills), and then you can break up the $5 into $1 bills. This will give 5+5+5+5+5+5+5+(1+1+1+1+1)+ 1+1(14 bills) and then on you add 4 bills for every $5 you break. The sequence from 10 bills becomes 14,18, 22, 26, 30, 34, 38, and the last is 42 where everything is in $1 bills. Therefore, the wrong number is 3.

the ans is three because the minimum number of notes that the cashier can give is 4.... 2 of 20 and 2 of 2..... ;)

The total cost for the purchases are $=\$(25+20+13)=\$58$

Now, since Matthew handed a $\$100$ bill, so the cashier should give change $=\$(100-58)=\$42$

Now, the cashier can give the change at the very least using 4 bills ( $2$ $\$20$ bills and $2$ $\$1$ bills).

But there is no way to give change using only 3 bills, so the possible number of bills that the cashier gave to Matthew $=\boxed{3}$

The blind man's purchase add up to $58. So, the cashier must return him $42. Now, whatever be combination the cashier uses, he has to give two $1 notes to the blind man. So, there is no possibility of adding up to $42 using two $1 notes and any other denominations if only 1 more note is to be used. So, the cashier gave him 3 notes.

Matthew purchased for a sum of 58$(25+13+20) and should receive 42$ from the keeper. the minimum number of bills that make up 42$ is 4 i.e 20+20+1+1. if Matthew is sure that, what he received is wrong, then the number of bills he has is less than what he must have had. so the only possibility is that the store keeper gave him only 3 bills.

Total cost is $58, Therefore 100-58=42 Possible combination of teller returning exact bills 1) 1+1+20+20 = 4 notes ; or 2) 1+1+10+10+20 = 5 notes ; or 3) 1+1+10+10+10+10 = 6 notes since Matthew knew this change was incorrect only possibility of teller giving 3 notes remain

total amount of the purchases is 58 the remaining money is 42 so the cashier can give him [ 4 notes of 10$ and 2 notes of 1$ that makes total of 6 notes and 42$]...or....[1 note of 20$ 2 notes of 10$ and 2 notes of 1$ that makes 5 notes and 42$]...or...[2 notes of 20$ and 2 notes of 1$ that makes 4 notes and 42$] but he cannot give 3 notes

6 for $10+$10+$10+$10+$1+$1; 4 for $20+$20+$1+$1; 5 for $20+$10+$10+$1+$1; but none of the way is suitable for 3

Total cost is $58. So Matthew should get back $42.

Since U.S. dollar bills are $1, $5, $10, $20, $50, $100, two of the returned bills must be $1. To return remaining $40, the teller needs at least two bills of $20.

Therefore the minimum number of bills must be four.

So, there is a possibility that the teller handed 3 bills to Matthew.

Balance Amount Left: 42$ Minimum number of notes required = 4 i.e.(20+20+1+1) Hence 3 is the incorrect number of notes.

Total spending = 25+13+20 = 58, Amount paid = 100, balance = 100 - 58 = 42, balance is incorrect. Bills Might have given are $1, $5, $10, $20, he was not tendering exact change. it might be $1, $10, $20 or $5, $10, $20 or $1, $5, $10 or $1, $5, $20, The number of bills = 3.

3

minimum number of bills teller handed to the Matthew should be 4

• Total Purchase Amount = $ 58 ;

• Teller Should return = $ 42 ;

• Possible combination of return amount : $1, $5, $10, $20, $50, $100

• $1 + $1 + $20 + $20 = $42( possible) ; Option 4 is possible

• $1 + $1 + $20 + $10 + $10 = $42 (possible) ; Option 5 is possible

• $1 + $1 + $10 + $10 + $10 + $10 = $42 (possible) ; Option 6 is possible

• but Teller cannot return $42 with sum of 3 combination. ;

• Option 3 is not possible

teller needs to return bill 40, considering the available denominations it is not possible to return 3 bills which would amount 40 in total.

3, because it is only with 3 bills that there is no permutation of making the amount equal to the change,and since he was blind it is the only way he could have known.....

according to the currency denomination, having 3notes doesn't sumup to the left over amount of $42

Option 1 is correct: $1,$1,$10,$10,$20

Option 2 is correct: $1,$1,$10,$10,$10,$10

Option 3 is correct: $1,$1,$20,$20

We have spent 58 out of 100 so we must get in return 42. Options are 3,4,5,6 bills. One of these is incorrect combination of bills. ---for 4 ->2 x $1 + 2 x $20---- ---for 5 ->2 x $1 + 1 x $20 + 2 x $10---- ---for 6 ->2 x $1 + 4 x $10---- and with only 3 bills Matthew cannot get $42 in change. Hence answer is 3.

the balance is $42... which means two of the bills will be $1 bills... this leaves us with $40... it's easy to figure out combinations that will give us this sum... (remember that we only have [n-2] possibilities remaining).

as the mathew has purchase only three items.it is obvious that he will get three bills

The change to be returned is 42$. The change can be in denomination of 20 or lesser and in all the sets of 3 notes, it is impossible to make an exact of 42$!

cost 58 , change = 42

6) 10 + 10 + 10 + 10+ 1 + 1

4) 20+ 20 + 1 + 1

5) 20 + 10 + 1 + 1

3 bills can never make $42 .

3

Matthew made a total purchase of $\$20 + \$25 + \$13 = \$58$

The teller was supposed to return him $\$100 - \$58 = \$42$

Now, with the available denominations it is only possible to form $42 in three ways :

1. $\$20 \times 2 + \$1 \times 2$ (4 bills)
2. $\$10 \times 4 + \$1 \times 2$ (6 bills)
3. $\$5 \times 8 + \$1 \times 2$ (10 bills)

Any number of bills handed to Matthew other than 4, 6 or 10 will certainly be an incorrect change.

Hence, 3 and 5 are the incorrect changes (and correct answers).

As he purchased three items only, bill person at counter will give 3 bills only although he may give incorrect change

20+20+1=41 that is wrong so this is seller's option