-
gmattesttaker2
- Legendary Member
- Posts: 641
- Joined: Tue Feb 14, 2012 3:52 pm
- Thanked: 11 times
- Followed by:8 members
Hello,
For the following:
Right triangle ABC's hypotenuse measures 20 inches. If AB and BC are its
legs, what is the length of AB?
(1) AB - 4 = BC.
(2) The ratio of AB to BC is 4:3.
OA: D
I was trying to solve as follows:
AB^2 + BC^2 = AC^2
1) AB - 4 = BC
=> ( 4 + BC )^2 + BC^2 = AC^2
=> ( 4 + BC )^2 + BC^2 = 20^2
Now upon solving this equation I got:
( BC + 16 ) ( BC - 12 ) = 0
=> BC = 12. Once I know BC I can solve for AB.
My question is if I have such a quadratic equation do I need to solve it in a DS question or can I safely assume that only 1 value will be positive and therefore conclude it to be Sufficient?
2) (4x)^2 + (3x)^2 = 20^2
=> 16x^2 + 9x^2 = 400
=> 25x^2 = 400
=> x^2 = 16
=> x = 4
Hence, AB = 4(4) = 16 - Suff.
Hence D
Thanks for your help,
Sri
For the following:
Right triangle ABC's hypotenuse measures 20 inches. If AB and BC are its
legs, what is the length of AB?
(1) AB - 4 = BC.
(2) The ratio of AB to BC is 4:3.
OA: D
I was trying to solve as follows:
AB^2 + BC^2 = AC^2
1) AB - 4 = BC
=> ( 4 + BC )^2 + BC^2 = AC^2
=> ( 4 + BC )^2 + BC^2 = 20^2
Now upon solving this equation I got:
( BC + 16 ) ( BC - 12 ) = 0
=> BC = 12. Once I know BC I can solve for AB.
My question is if I have such a quadratic equation do I need to solve it in a DS question or can I safely assume that only 1 value will be positive and therefore conclude it to be Sufficient?
2) (4x)^2 + (3x)^2 = 20^2
=> 16x^2 + 9x^2 = 400
=> 25x^2 = 400
=> x^2 = 16
=> x = 4
Hence, AB = 4(4) = 16 - Suff.
Hence D
Thanks for your help,
Sri












