Gerry had a total of 30 pens and pencils. He decided to trade with his friends all of his pens and pencils. If he traded every pen for 2 pencils, he would have 48 pencils in all. How many pens and how many pencils did he have before the trade?
12 pencils and 18 pens
first,you have to say in your head ok he had 30 pens and pencils.you have to say hmm what is half of 30?half of 30 is 15.so,he most likely had 15 pens.So i believe he Gerry had 15 pens and 15 pencils.Also,i am a 5th grader and i hope this helps everyone who needs it! since it is Halloween and i got lots of candy,i decided to do this! i hope this helps! Also i had this homework from my amazing math teacher and i decided to solve it then post it for people who need it!Have to go eat candy now!
Anasiah from F.a.s out!
Anasiah from f.a.s out!
Anasiah from F.a.s out!
To solve this problem, let's assume Gerry had x pens before the trade. Since Gerry traded every pen for 2 pencils, he would have received 2x pencils in exchange.
After the trade, Gerry had a total of 30 pens and pencils. So, the equation to represent this situation is:
x (pens) + 2x (pencils) = 30
Now, we are given that if Gerry traded every pen for 2 pencils, he would have 48 pencils in total. So, we can set up another equation to represent this situation:
x (pens) + 2(2x) (pencils) = 48
Let's solve the first equation for x:
x + 2x = 30
Combine like terms:
3x = 30
Divide both sides by 3:
x = 10
Now that we know x = 10, we can substitute it into the second equation:
10 (pens) + 2(2 * 10) (pencils) = 48
10 + 40 = 48
Simplify:
50 = 48
Since this is not a true statement, there is no solution that satisfies both equations. It seems there might be an error in the problem statement or the given information. Please double-check the problem and try again.