Math Digit Problems

[Jul18]

Are these questions on the DAT? I don't know why but it seems so confusing and hard for me?... anyone already took the DAt, please help us our..


In a three-digit number, the ten's digit and the one's digit are the same and the hundred's digit is three less than the ten's digit. If the sum of the digits is 9, find the number.

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[Jake8]

x+y+z = 9
y=z
x=y-3

Substitute values in I will use y

(y-3)+y+y = 9

y=4
z=4
x=1


Yes they will put these on the test. Just remember to set up your equations and write it out and it shouldn't be too hard for you.

 

[dleodyddlek]

144

 

[Electrons]

Ugh I hope not. I hate these problems. They're very easily get mixed up.

Here's another one.

The sum of the digits of a two-digit num is 15. if the digits are reversed, the new num is 9 more than the original num. find the num.

 

[Jul18]

hey electrons, are we using the same book? lol

 

[Jul18]

Thanks jake, I got it now🙂

 

[Electrons]

hey electrons, are we using the same book? lol

maybe...by Bluman?

 

[Streetwolf]

The sum of the digits of a two-digit num is 15. if the digits are reversed, the new num is 9 more than the original num. find the num.

The key to solving any word problem is to recognize what they want and translate english into math. You need to go through this EXACTLY as they do and write down equations as you are given them.

"The sum"
You are adding something.
"of the digits"
These are single digits 0-9, possibly of a 2 digit or larger number.
"of a two-digit number"
Two digits. You want to figure out how to write this out. Picture a 2 digit number. You have the tens digit and the units digit. It makes sense to write the former as 't' and the latter as 'u'. With what you have read so far, you have t + u.
"is 15."
Equals 15. So you have t + u = 15

"If the digits are reversed"
Write out the variables as a number. Normally you'd have 'tu'. For instance if t = 4 and u = 6 then 'tu' represents '46'. If the digits are reversed you'd have '64' or in terms of the variables, you'd have 'ut'.
"the new number"
This would be 'ut'. If you want to write it as the actual NUMBER itself instead of (in the example above) a 6 followed by a 4, you need to remember that the 6 represents the tens position. So it's really 60, not 6. This is written as 10u + t. The original number (46) was 10t + u. In general that's how you write it out. If you had a 3 digit number I would make 'h' be the hundreds place. Then I would write the number VALUE as 100h + 10t + u. The actual 3 digit representation of the number is 'htu'. This is important in these problems because doing t + u would not give you the value of the 2 digit number.
"is 9 more than"
Equals 9 plus. So you have the new number which is 10u + t, is 9 more than something. We write this as 10u + t = 9 + *something*
"the original number."
The original number was 10t + u. So your equation is 10u + t = 9 + 10t + u.

"Find the number."
Solve.


===

You have equations. You have extracted everything possible from the problem. Either there's enough info or there isn't. Try to solve it. You have two equations and two variables. That sounds good to me.

t + u = 15
10u + t = 9 + 10t + u

Simplify the second equation.

t + u = 15
-9t + 9u = 9

Again, simplify.

t + u = 15
-t + u = 1

Now add them.

2u = 16
u = 8
t = 7

The number is 78. The digits add to 15. When you add 9, you get 87. This is the original number with the digits swapped.

 

[Electrons]

Streetwolf, the math king! Once again thanks. That explanation tip was really useful in decomplicating things.

 

[Jul18]

😎
can u come w/ me to my DAT?

The key to solving any word problem is to recognize what they want and translate english into math. You need to go through this EXACTLY as they do and write down equations as you are given them.

"The sum"
You are adding something.
"of the digits"
These are single digits 0-9, possibly of a 2 digit or larger number.
"of a two-digit number"
Two digits. You want to figure out how to write this out. Picture a 2 digit number. You have the tens digit and the units digit. It makes sense to write the former as 't' and the latter as 'u'. With what you have read so far, you have t + u.
"is 15."
Equals 15. So you have t + u = 15

"If the digits are reversed"
Write out the variables as a number. Normally you'd have 'tu'. For instance if t = 4 and u = 6 then 'tu' represents '46'. If the digits are reversed you'd have '64' or in terms of the variables, you'd have 'ut'.
"the new number"
This would be 'ut'. If you want to write it as the actual NUMBER itself instead of (in the example above) a 6 followed by a 4, you need to remember that the 6 represents the tens position. So it's really 60, not 6. This is written as 10u + t. The original number (46) was 10t + u. In general that's how you write it out. If you had a 3 digit number I would make 'h' be the hundreds place. Then I would write the number VALUE as 100h + 10t + u. The actual 3 digit representation of the number is 'htu'. This is important in these problems because doing t + u would not give you the value of the 2 digit number.
"is 9 more than"
Equals 9 plus. So you have the new number which is 10u + t, is 9 more than something. We write this as 10u + t = 9 + *something*
"the original number."
The original number was 10t + u. So your equation is 10u + t = 9 + 10t + u.

"Find the number."
Solve.


===

You have equations. You have extracted everything possible from the problem. Either there's enough info or there isn't. Try to solve it. You have two equations and two variables. That sounds good to me.

t + u = 15
10u + t = 9 + 10t + u

Simplify the second equation.

t + u = 15
-9t + 9u = 9

Again, simplify.

t + u = 15
-t + u = 1

Now add them.

2u = 16
u = 8
t = 7

The number is 78. The digits add to 15. When you add 9, you get 87. This is the original number with the digits swapped.

 

[Streetwolf]

I just want to point out that 'Solve' was the very very very last part of this and it took me maybe 3 or 4 lines to do the math. There is not much math in these word problems. The worst part is figuring out how to START the math part. Knowing how to do systems of equations will not help you at all if you can't interpret the problems. You have to practice these, especially the ones with really long paragraphs, and come up with your own best way of CORRECTLY writing out the actual math.

 

[klutzy1987]

I just want to point out that 'Solve' was the very very very last part of this and it took me maybe 3 or 4 lines to do the math. There is not much math in these word problems. The worst part is figuring out how to START the math part. Knowing how to do systems of equations will not help you at all if you can't interpret the problems. You have to practice these, especially the ones with really long paragraphs, and come up with your own best way of CORRECTLY writing out the actual math.

Therefore the only way to get good at these problems is through practice and alot of it too. The more of these problems you see in different variations the better you will get at solving them.