# Can't figure out this problem

• Jan 8th 2008, 10:55 PM
Can't figure out this problem
Sorry that I wasn't more specific in the title, but I don't really know what to call this problem...

"In a two-digit number, the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the two-digit number."

a. 52
b. 63
c. 73
d. 85

I'm really confused, because unless there's something I'm missing here, 52, 63, and 85 would all work. Problem is, this isn't supposed to be a question with more than one answer, so I'm very lost.

I'd appreciate it if someone could help me figure this out.
• Jan 8th 2008, 11:16 PM
earboth
Quote:

Sorry that I wasn't more specific in the title, but I don't really know what to call this problem...

"In a two-digit number, the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the two-digit number."

a. 52
b. 63
c. 73
d. 85

I'm really confused, because unless there's something I'm missing here, 52, 63, and 85 would all work. Problem is, this isn't supposed to be a question with more than one answer, so I'm very lost.

I'd appreciate it if someone could help me figure this out.

Hello,

if x is the ten's digit $\displaystyle x \in \{1, 2, ..., 9\}$ and
if y is the one's digit $\displaystyle y \in \{0, 1, 2, ..., 9\}$ then a 2-digit-number is

$\displaystyle n = 10\cdot x+y$ . The number with reverse order of digits is
$\displaystyle n_{inverse \ order} = 10\cdot y+x$

According to your problem $\displaystyle x = y+3$

That means:

$\displaystyle \underbrace{10(y+3) + y}_{n} - \underbrace{(10y + y+3)}_{n_{inverse \ order}} = 27$ ..........Expand the brackets and collect like terms:

$\displaystyle 10y+30+y-10y-y-3 = 27~\iff~27 = 27$

That means there isn't a unique solution but every number which ten's digit is 3 greater then the one's digit satisfies the given conditions.
• Jan 9th 2008, 03:44 PM
Thank you for the help. I have another problem that I was hoping someone could help me with, and I think I may as well just put it in this topic as to not clutter up the board.

"A craftsperson wishes to make an alloy from an alloy which is 40% gold and another which is 60% gold. How many ounces of the 40% alloy need to be used to get 5 ounces that are 55% gold?"

a. 4.75
b. 3.25
c. 1.25
d. 2.75

I was leaning towards C, because... Well, lemme see if I can explain it. 55% is closer to 60 than it is 40, so it would only be logical that there would be more of the 60% alloy than the 40% alloy. The main reason I'm confused is because one of my quite smart friends told me it was D, so I have no idea.
• Jan 9th 2008, 03:49 PM
Oh, and by the way, sorry if what I had previously posted confused anyone. I edited out the non sensemaking part... No idea what I was thinking.
• Jan 9th 2008, 04:05 PM