Problem Solving

A cashier mentally reversed the digits of one customer's correct amount of change and this gave the customer an incorrect amount of change. Id the cash register contained 45 cents more than it should have as a result of this error, which of the following could have been the correct amount of change in cents.

A 14
B 45
C 54
D 65
E 83

rtaha2412 wrote:A cashier mentally reversed the digits of one customer's correct amount of change and this gave the customer an incorrect amount of change. Id the cash register contained 45 cents more than it should have as a result of this error, which of the following could have been the correct amount of change in cents.

A 14
B 45
C 54
D 65
E 83

Let us say cashier has X cents in his counter.

Actually he should have X-45 cents with him, but because of mistake he has X.

Try the options (try from the larger options as he has more money then he should have)

E. 83 - 45 = 38 . (Reversed)

IMO [spoiler]E

But 83 is the miscalculated quantity, question asks correct quantity?
[/spoiler]

Nice explanation, shovan85!

One thing to add to this one - if you'd like to try to solve these algebraically, there's a method for these "reverse the digits" type problems that uses pure algebra:

You can express any two-digit number as a sum of its tens digit plus its ones digit by remembering to multiply the tens by 10. For example, 45 is 4*10 + 5.

So if we want to say that flip-flopping the digits is the same as adding 45, we'd say that:

10x + y is the initial number (x is the tens digit and y is the units)
10y + x is the new number (now y is the tens digit and x is the units)

And by flip-flopping we're adding 45, so we'd have:

10x + y = 10y + x + 45

Combine like terms to get:

9x = 9y + 45

Divide by 9 to simplify:

x = y + 5

so we know that there's a difference of 5 between the digits, so we could have:

16 & 61
27 & 72
38 & 83
49 & 94

The only option that's here in the answer choices is 83, so that's correct.