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I was wondering whether there is a shortcut to solve this question without going through the tedious calculations of Pythagorean theorem.

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
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I cannot think of any faster way to do it.

However, here are some tips when dealing with phthagorus theorem and in general, problems involving sqauring numbers.

1. Finding the sqaure of a number ending in 5 (such as 15, 25,35,45, etc).

The answer has two parts, Part A and Part B

Step a: Take whatever is there to the left 5 in the original number and multiply it by one more. the product is part A.

Step b: Take 5^2 = 25 and write it down on the right hand side, this is part B.

eg: 15^2 = (1*2) 25 = 225
25^2 = (2*3) 25 = 625
35^2 = (3*4) 25 = 1225
45^2 = (4*5) 25 = 2025
etc.

95^2 = (9*10)25 = 9025

115^2 = (11*12) 25 = 13225

2. Finding sqaures of any two digit number

Note that (x+y)^2 = x^2 + 2xy + y^2

This is a very powerful algebric formula and can be effectively used to find the sqaure of any two digit number.

for example, 32^2

it has three parts - part A, part B, part C

part A - x^2 which is 3^2 = 9
part B - 2xy = 2*3*2 =12
part C - y^2 = 2^2 = 4

the answer is (9)(12)(4) = 1024

when putting together ABC, you need to add from right to left with any carry forward.

eg: 56^2 = (25)(60)(36) = 3136

eg: Small numbers are very trivial, 12^2=(1)(4)(4) = 144, 13^2 = (1)(6)(9) = 169

eg: 43^2 = (16)(24)(9) = 1849

Now combing back to the original problem:

it is good to know some basic phthagorun triplets (integers that form the sides of a right triangle), 3,4,5, 5,12,13, 7,24,25, 9,40,41 etc..

the follow a specific pattern, 3^2 = 4+5, 5^2 = 12+13, 7^2=24+25, 9^2=40+41 and so on (next one is 11,60,61).

(Note that there are other types of phthagorun triplets which do not satisfy the above condition).

-mathguru
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Quote:
it is good to know some basic phthagorun triplets (integers that form the sides of a right triangle), 3,4,5, 5,12,13, 7,24,25, 9,40,41 etc..

This is precisely how to do the above question quickly! Don't forget 8:15:17
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Excellent tips! Thanks guys. BTW, the OA is C.
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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..

Last edited by haas_mba07 on 21 Jul 2006, 08:53, edited 1 time in total.
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Great information gmatmathguru... Thanks/

Could you clarify a little more as to how to get the sum after the x^2, 2xy, y^2 terms are calculated.

gmatmathguru wrote:
the answer is (9)(12)(4) = 1024

when putting together ABC, you need to add from right to left with any carry forward.

eg: 56^2 = (25)(60)(36) = 3136

eg: Small numbers are very trivial, 12^2=(1)(4)(4) = 144, 13^2 = (1)(6)(9) = 169

eg: 43^2 = (16)(24)(9) = 1849

Now combing back to the original problem:

it is good to know some basic phthagorun triplets (integers that form the sides of a right triangle), 3,4,5, 5,12,13, 7,24,25, 9,40,41 etc..

the follow a specific pattern, 3^2 = 4+5, 5^2 = 12+13, 7^2=24+25, 9^2=40+41 and so on (next one is 11,60,61).

(Note that there are other types of phthagorun triplets which do not satisfy the above condition).

-mathguru
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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

eg: 43^2 = (16)(24)(9) = 1849

-mathguru
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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

eg: 43^2 = (16)(24)(9) = 1849

-mathguru
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You guys may want to check this out for squares:

http://www.gmatclub.com/phpbb/viewtopic.php?t=25664

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.
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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...

ps_dahiya wrote:
You guys may want to check this out for squares:

http://www.gmatclub.com/phpbb/viewtopic.php?t=25664

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.
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haas_mba07 wrote:
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...

ps_dahiya wrote:
You guys may want to check this out for squares:

http://www.gmatclub.com/phpbb/viewtopic.php?t=25664

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.
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I dont know any shorter method. But it took me less than a minute.
What you can do is memorize some important squares.

Height of wall till ladder = (25^2 - 7^2)^1/2 = (625-49)^1/2 = 576^1/2 = 24

New height = 20

New distance from the wall to the base of the ladder
= (25^2 - 20^2)^1/2
= (225)^1/2
= 15
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ps_dahiya,
Thank you very much. I have never heard about vedic mathematics before. It's very interesting to know.

gmatmathguru,
thank you for your explanations. Great shortcuts and really useful for GMAT.
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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:

20^2 + b^2 = 25^2 --> 400 + b^2 = 625
b^2 = 225 --> b=15

Now we subtract 7 from 15 and we find that the base moved 8 ft!

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