QR question

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Dental2000

Full Member
10+ Year Member
Joined
Jan 12, 2011
Messages
545
Reaction score
155
Points
5,266
Location
Chicago
  1. Dental Student
Advertisement - Members don't see this ad
DAT destroyer 2010 question 113 for pl who want the real question infront of them.

Picture shows a Rectangle with L=20 and B=10. If L and B of this rectangle were increased by 20% what is the % increase in area?

1) 44%
2) 30%
3) 55%
4)17%
5) 20%




Ans is 44% but i need to know how you get it with a good explanation
 
DAT destroyer 2010 question 113 for pl who want the real question infront of them.

Picture shows a Rectangle with L=20 and B=10. If L and B of this rectangle were increased by 20% what is the % increase in area?

1) 44%
2) 30%
3) 55%
4)17%
5) 20%

Ans is 44% but i need to know how you get it with a good explanation

Area of Original rectangle:
L X B = LB

Area of new rectangle:
1.2L X 1.2B = 1.44LB

New/Original = 1.44LB/LB = 1.44

Increase = 44%
 
DAT destroyer 2010 question 113 for pl who want the real question infront of them.

Picture shows a Rectangle with L=20 and B=10. If L and B of this rectangle were increased by 20% what is the % increase in area?

1) 44%
2) 30%
3) 55%
4)17%
5) 20%




Ans is 44% but i need to know how you get it with a good explanation

This question is very easy and most of it can be done in your head in about 25 seconds if you work fast. 1. First thing you do is calculate the original area which is 200: 10x20
2. Next you increase each side by 20%: 10x.2= 12 (10+2) 20x.2=24 (20+4)
3. You calculate the new area: 12x24= 288 (area increased by 88)
4. You compare the 2 areas: Hint: since 50% of 200 is 100, and since 88 is a little less than 100 your answer will be a little less than 50%== 44%
 
Area of Original rectangle:
L X B = LB

Area of new rectangle:
1.2L X 1.2B = 1.44LB

New/Original = 1.44LB/LB = 1.44

Increase = 44%

Yeah that's actually a way better way than mine since you can apply this to large numbers since it deals with % increase/decrease type of problem
 
Good thing I refreshed I was about to give a detailed explanation. Good job guys I like it how we all spring into action once someone poses a problem haha

Yeah Albino I agree with above poster thats good thinking on your feet. What you get on QR?
 
Last edited:
This question is very easy and most of it can be done in your head in about 25 seconds if you work fast. 1. First thing you do is calculate the original area which is 200: 10x20
2. Next you increase each side by 20%: 10x.2= 12 (10+2) 20x.2=24 (20+4)
3. You calculate the new area: 12x24= 288 (area increased by 88)
4. You compare the 2 areas: Hint: since 50% of 200 is 100, and since 88 is a little less than 100 your answer will be a little less than 50%== 44%

That's not fast at all. I used like 2 seconds. 😛

got 23, and was pretty disappointed in my score. I have always been a math whiz though. Always been a 95+ student in any math course.
 
Top Bottom