Please Help..
Thanks.
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GMAT Prep2?? (-x)
Source: Beat The GMAT — Problem Solving |
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AleksandrM
- Legendary Member
- Posts: 566
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
The mistake you made is in your answer choice.
When you plug in -1 for x, you end up with 1 as the answer.
sqroot -(-1)|-1| = sqroot 1 = 1
When you are looking at the answer choices you plug the -1 back into the answer choices:
eliminate B and C.
A = -(-1) = 1
D = -1
E = 1i, which is an imaginary number. Answer is A.
When you plug in -1 for x, you end up with 1 as the answer.
sqroot -(-1)|-1| = sqroot 1 = 1
When you are looking at the answer choices you plug the -1 back into the answer choices:
eliminate B and C.
A = -(-1) = 1
D = -1
E = 1i, which is an imaginary number. Answer is A.
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AleksandrM
- Legendary Member
- Posts: 566
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
For the second one, the best way is to redraw the picture and plug in the numbers for r and s. I plugged in 5 for r and 2 for s. Then, just solve this problem, and plug the numbers into the answer choices to get the answer. This all took me 2 minutes and 23 seconds.
1) Get the area of the entire circle: 5^2pi = 25pi
2) Get the area of the circle without the shaded region around it:
5 - 2 = 3 ----> 3^2pi = 9pi
3) Get the area of the shaded region: 25pi - 9pi = 16pi
A = 9pi eliminate
B = 21pi eliminate
C = 6pi eliminate
D = 40pi eliminate
E = 16pi choose E.
1) Get the area of the entire circle: 5^2pi = 25pi
2) Get the area of the circle without the shaded region around it:
5 - 2 = 3 ----> 3^2pi = 9pi
3) Get the area of the shaded region: 25pi - 9pi = 16pi
A = 9pi eliminate
B = 21pi eliminate
C = 6pi eliminate
D = 40pi eliminate
E = 16pi choose E.
Just a friendly request.. can you please post each question in a separate thread... helps everyone to keep focused..
Thanks in advance
Thanks in advance
The area of the metal frame is pie(r+s)^2 - pie(r)^2
which will be pie(r^2+2rs+s^2) - pie(r)^2
= pie s (2r+s)
What am I missing here?
which will be pie(r^2+2rs+s^2) - pie(r)^2
= pie s (2r+s)
What am I missing here?
Maxx
Im sorry..The area of the metal frame would be
pie(r)^2 - pie(r-s)^2
pie(r)^2-pie(r^2-2sr+s^2)
pie(r)^2-pie(r)^2+pie(2sr)-pie(s)^2
pie(2sr)-pie(s)^2
pie(s) [2r-s]
This should not take more than a minute
pie(r)^2 - pie(r-s)^2
pie(r)^2-pie(r^2-2sr+s^2)
pie(r)^2-pie(r)^2+pie(2sr)-pie(s)^2
pie(2sr)-pie(s)^2
pie(s) [2r-s]
This should not take more than a minute
Maxx
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AleksandrM
- Legendary Member
- Posts: 566
- Location: Philadelphia
- Thanked: 31 times
- GMAT Score:640
There is nothing wrong with the approach that moneyman used. Algebra should be applied with some problems. However, keep in mind: during the exam, you are more likely to make a mistake with applied algebra than you are doing arithmetic.
