Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jan 2611:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
Jan 2410:00 AM EST
-11:59 PM EST
Through Friday, January 31, get an extra 25% off Target Test Prep GMAT plans. Just use the coupon code FLASH25 at checkout to save up to $400.
Jan 3110:00 AM EST
-11:59 PM EST
Get a massive $250 off the TTP GMAT OnDemand Course. Just enter the coupon code FLASH25 at checkout to save a massive $250. If you prefer learning through engaging video lessons, OnDemand is exactly what you need.- Feb 01
11:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment - Feb 02
11:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions
17 Aug 2012, 03:56
Kudos
Bookmarks
1)In the figure above, AC=6 and BC=3. Point P lies on AB between A and B such that CP is perpendicular to AB. Which of the following could be the length of CP?
A) 2
B) 4
C) 5
D) 7
E) 8
Picture Reference:https://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg
2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?
A) 25
B) 24
C) 20
D) 16
E) none
Picture Reference: https://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg
A) 2
B) 4
C) 5
D) 7
E) 8
Picture Reference:https://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg
2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?
A) 25
B) 24
C) 20
D) 16
E) none
Picture Reference: https://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg
17 Aug 2012, 04:50
Kudos
Bookmarks
arifaisal
1)In the figure above, AC=6 and BC=3. Point P lies on AB between A and B such that CP is perpendicular to AB. Which of the following could be the length of CP?
A) 2
B) 4
C) 5
D) 7
E) 8
Picture Reference:https://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg
Triangle PBC is a right triangle (PC is perpendicular to AB), therefore CP being a leg is shorter than the hypotenuse, which is BC.
Only 2 is acceptable from the given list of answers.
Answer A
2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?
A) 25
B) 24
C) 20
D) 16
E) none
Picture Reference: https://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg
A) 2
B) 4
C) 5
D) 7
E) 8
Picture Reference:https://24.media.tumblr.com/tumblr_m8wbaeMGRS1rd5jmyo1_1280.jpg
Triangle PBC is a right triangle (PC is perpendicular to AB), therefore CP being a leg is shorter than the hypotenuse, which is BC.
Only 2 is acceptable from the given list of answers.
Answer A
2) In the figure below, E is the midpoint of AC. AC is perpendicular to AB, and AD=DB. If BC=4 cm, what is the value of BEsquare+CDsquare?
A) 25
B) 24
C) 20
D) 16
E) none
Picture Reference: https://24.media.tumblr.com/tumblr_m8wbh ... 1_1280.jpg
Using Pythagoras's theorem, \(BE^2+CD^2=AB^2+(\frac{AC}{2})^2+AC^2+(\frac{AB}{2})^2=\frac{5}{4}(AB^2+AC^2)=\frac{5}{4}BC^2=\frac{5}{4}*4^2=20.\)
Answer C
17 Aug 2012, 05:55
Kudos
Bookmarks
Q1
IN Triangle CPB
CP is the base and CB is the hypotenuse. And Base will always be less than Hypotenuse.
So CP < 3
Only possible answer is A
Q2
In Right Triangle BAE
BE^2 = AB^2 + AE^2
= AB^2 + (AC/2)^2 [because AE = AC/2]
BE^2 = AB^2 + (AC^2)/4 ... (1)
In Right Triangle CAD
CD^2 = AC^2 + AD^2
= AC^2 + (AB/2)^2 [as AD= DB = AB/2]
CD^2 = AC^2 + (AB^2)/4 ...(2)
(1) + (2) gives us
BE^2 + CD^2 = AB^2 + (AC^2)/4 + AC^2 + (AB^2)/4 = (5/4) (AB^2 + AC^2)
= (5/4)* (BC^2) [ As in RIght Triangle BAC AB^2 + AC^2 = BC^2]
= (5/4) * 4^2 = 20
So, Answer is C
Hope it helps!
IN Triangle CPB
CP is the base and CB is the hypotenuse. And Base will always be less than Hypotenuse.
So CP < 3
Only possible answer is A
Q2
In Right Triangle BAE
BE^2 = AB^2 + AE^2
= AB^2 + (AC/2)^2 [because AE = AC/2]
BE^2 = AB^2 + (AC^2)/4 ... (1)
In Right Triangle CAD
CD^2 = AC^2 + AD^2
= AC^2 + (AB/2)^2 [as AD= DB = AB/2]
CD^2 = AC^2 + (AB^2)/4 ...(2)
(1) + (2) gives us
BE^2 + CD^2 = AB^2 + (AC^2)/4 + AC^2 + (AB^2)/4 = (5/4) (AB^2 + AC^2)
= (5/4)* (BC^2) [ As in RIght Triangle BAC AB^2 + AC^2 = BC^2]
= (5/4) * 4^2 = 20
So, Answer is C
Hope it helps!
