QR age problems

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diene

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Can anyone provide me some tips for the age problems? I know Chad's videos provide a good technique, but my subscription has expired and I can't quite understand what I wrote in my notes. Thanks!
 
Make a table with everyone's name on them. Then rows, one for "now" and one for "later." It's basically just a guess and check with the answer choices. Choose one and put that in the "now" column for whoever the choices are about. Then based off the problem you fill out the chart to see if it fulfills the requirement...(ex... In 4 years Jane will be twice as old as Peter).
 
i liked setting up equations.
diene= x
junebug is 5 years older (make sure this part is set up correctly, because i would initially make mistakes with adding and subtracting
junebug=x+5

in 10 year junebug is 2 times as old as diene

so, lets add 10 years..
diene=x+10
junebug=x+5+10
now how do we denote june is 2 times as old?
we know 2 (x+10) = x+5+10
2x+20=x+15
x=-5
and then you plug back in, know you have X. X is not the answer per say, just whatever you set X equal to. It can be diene, or junebug, and it can be in the past or the future. They may ask for Junebugs age now (post the 10 years), which would be junebug=x+5+10.

this equation didnt work since i made up the problem, give me a few destroyer problems and i can do a walkthrough..
 
Here is one from destroyer

Donna says that she was 21 when her daughter Trish was born. But when asked how old her daughter is today, donna said that in 6 years she will be 5 more than twice her daughter's age. How old is donna's daughter, trish?
 
Here is one from destroyer

Donna says that she was 21 when her daughter Trish was born. But when asked how old her daughter is today, donna said that in 6 years she will be 5 more than twice her daughter's age. How old is donna's daughter, trish?

it would be easier for us to explain if you can give us the answer choices.
after all, chad's technique uses the answer choices as the base for answering this type of question.
 
Okay, so Chad’s method to this type of problem involves guessing and checking with the given answer choices.

First, make a table with “Now” and “Later” in separate columns, and “Donna” and “Trish” in separate rows. It will pretty much look like Punnett square.
In the “Now” column for “Trish”, you will plug in the given answer choices, since you’re answering how old Trish is NOW.
First, plug in the first choice “21” in that column.
Then we move onto the “Later” column. Since we’re talking about 6 years later, you add 6 to Trish’s possible age. 21 + 6 = 27. Now, to calculate Donna’s age for “Later” column, you multiply Trish’s possible age by 2 and add 5, since Donna will be 5 more than twice her daughter’s age. That would be 2(27)+5 = 57.
Since you now have the answer to the “Later” column for both Donna (57) and Trish (27), you look at the difference between the two numbers.
You know that you will have the right number if the difference is 20, because Donna says that she was 21 when her daughter Trish was born. This means that when Trish was 1, Donna was 21. 21-1 = 20 years of age difference.
57-27 = 30, so the first choice is incorrect.
You do the same for each choice, and eventually you get to choice D, 9.
9 would be [Trish] “Now”, and 9+6 = 15 for [Trish] “Later”.
[Donna] “Later” would then be 2(15)+5 = 35.
You see that the difference is now 35-15 = 20.
Thus, you can conclude that the answer is d.

I hope this isn’t too confusing.
It seems time-consuming judging by how long my explanation is, but once you know how to set up the table and everything, it’s pretty quick.

I hope this helps, and good luck on your journey to dental school!
 
why did you add the 10 to Junebug? it should be J=d+5 and J= 2(d+10)
now solve for diene
d+5=2d+20 ___> d= 15 and Junebug= 20

Correct me if i'm wrong!
 
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