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A ladder 25 feet long is leaning against a wall that is perp [#permalink]
21 Jul 2006, 02:57
3
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
55% (hard)
Question Stats:
60% (02:46) correct
40% (01:39) wrong based on 96 sessions
A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?
Not really.. there are no tricks when it comes to unusual right angle triangles. The only thing that you can do is remember the usual ones
3-4-5 (and its multiples) etc. And use the quick computation techniques above...
As for this, it should be pretty quick to calculate.
625-49 = 576 (has to be 24 or 16.. ends in 6).
which gives 625-400 = 225 = 15 x 15.
15 = 7+8..
Answer : (C). 8
Last edited by haas_mba07 on 21 Jul 2006, 08:53, edited 1 time in total.
You write down (x^2)(2xy)(y^2) terms individually.
in this case, (9), (12), (4) respectively.
Once, you have this, the process is very simple. You add them from right to left with anything above 9 (10 or more) as a carry (that you do in addition).
Step1:
Start with the (y^2) term which is 4 to begin with. (9), (12), (4)
So, I start with 4 first. my answer looks like XXX4. Because 4 is less than 10, I do not have a carry.
Step2:
(9), (12), (4)
Next, take the 2xy term which is 12. Because 12 is more than 10, you put down 2 and take 1 as the carry.
So, up to now, you have XX24 and 1 as the carry to the next step.
(9), (1 2), (4)
Step 3:
I have a 9 and 1 as the carry from step 2. I add 9 and 1 to get 10.
(9), (1 2), (4)
PUtting all three steps together, I have 1024 as the answer.
when putting together ABC, you need to add from right to left with any carry forward.
eg: 56^2 = (25)(60)(36) = 3136
Step 1: (25)(60)(3 6) and 3 is the carry
Step 2: (25) (60+3) 6
which is (25) (6 3) 6 and 6 is the carry
Step 3: (25+6) 36 = 3136
eg: Small numbers are very trivial, 12^2=(1)(4)(4) = 144, 13^2 = (1)(6)(9) = 169
Got it!! Thanks so much... This is a great way for quick squared computations...
gmatmathguru wrote:
sure.
eg: 32^2
You write down (x^2)(2xy)(y^2) terms individually.
in this case, (9), (12), (4) respectively.
Once, you have this, the process is very simple. You add them from right to left with anything above 9 (10 or more) as a carry (that you do in addition).
Step1:
Start with the (y^2) term which is 4 to begin with. (9), (12), (4)
So, I start with 4 first. my answer looks like XXX4. Because 4 is less than 10, I do not have a carry.
Step2:
(9), (12), (4)
Next, take the 2xy term which is 12. Because 12 is more than 10, you put down 2 and take 1 as the carry.
So, up to now, you have XX24 and 1 as the carry to the next step.
(9), (1 2), (4)
Step 3:
I have a 9 and 1 as the carry from step 2. I add 9 and 1 to get 10.
(9), (1 2), (4)
PUtting all three steps together, I have 1024 as the answer.
when putting together ABC, you need to add from right to left with any carry forward.
eg: 56^2 = (25)(60)(36) = 3136
Step 1: (25)(60)(3 6) and 3 is the carry
Step 2: (25) (60+3) 6
which is (25) (6 3) 6 and 6 is the carry
Step 3: (25+6) 36 = 3136
eg: Small numbers are very trivial, 12^2=(1)(4)(4) = 144, 13^2 = (1)(6)(9) = 169
This is vedic mathematics. For who don't know about Vedic mathematics: Vedas are the ancient Indian books written thousands of years ago. These books contain many things that include quick mathematics tricks. The tricks in above link are just the tip of the iceberg. _________________
I am actually reading a book on Vedic Mathematics. Recently brought it back on my trip to India... I am not sure how much it will help me on my GMAT but a great read on different techniques they devised...
This is vedic mathematics. For who don't know about Vedic mathematics: Vedas are the ancient Indian books written thousands of years ago. These books contain many things that include quick mathematics tricks. The tricks in above link are just the tip of the iceberg.
I am actually reading a book on Vedic Mathematics. Recently brought it back on my trip to India... I am not sure how much it will help me on my GMAT but a great read on different techniques they devised...
This is vedic mathematics. For who don't know about Vedic mathematics: Vedas are the ancient Indian books written thousands of years ago. These books contain many things that include quick mathematics tricks. The tricks in above link are just the tip of the iceberg.
They will help you only if you practice practice and practice using those tricks. During my graduation, I used to read vedic mathematics in my spare time and I became so proficient that I was able to calculate A/B in few seconds accurate to 3 decimal places, where A is any number upto 1000 and B any number upto 30.
They may help on GMAT a little but not that much. GMAT is not about tedious calculations but its all about tricks and techniques. _________________
I am reviving this message because I think it has a lot of useful info. Also, it is really simple if you know your pythagorean triplets.
The triplet called into play here is a 7:24:25 triangle.
The base is 7, the height up the wall is 24 and the length of the ladder is 25. Now, decrease the height by 4 feet and you get the new height of 20. The ladder length stays the same so you can set up and solve the follwing equation for the base:
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: A ladder 25 feet long is leaning against a wall that is perp [#permalink]
17 Jun 2014, 23:56
Expert's post
1
This post was BOOKMARKED
A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall. If the top of the ladder slips down 4 feet, how many feet will the bottom of the ladder slip?
(A) 4 (B) 5 (C) 8 (D) 9 (E) 15
We have a right triangle with the hypotenuse of 25 feet and the base of 7 feet, hence its the height is \(\sqrt{25^2-7^2}=24\) feet;
Since the top of the ladder slips down 4 feet then the new height becomes 24-2=20 feet and the new base becomes \(\sqrt{25^2-20^2}=15\) feet;
Question basically asks about the difference between the old base and the new base, which is 15-7=8 feet.
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