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Progressions [#permalink] New post 07 Dec 2011, 07:19
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Question Stats:

66% (00:00) correct 33% (00:00) wrong based on 1 sessions
Balls of equal size are arranged in rows to form an equilateral triangle. the top most row consists of one ball, the 2nd row of two balls and so on. If 669 balls are added, then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did. How many balls made up the triangle?

A)1540
B)2209
C)2878
D)1210
E)1560
[Reveal] Spoiler: OA
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Re: Progressions [#permalink] New post 07 Dec 2011, 08:28
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Good question.
As expected, this question boils down to 2 equation,
Consider total number of Balls in Triangle = T and number of balls in last row = x.
1+2+3+...+x = T
x(x+1)/2 = T ----(a)
As mentioned in the question, side of a square will be (x-8) and total number of Balls in square will be (T+669)
(x-8)^2 = T+669 -----(b)

Now the hardest part of the question will be to solve these 2 equations and this looks like time consuming BUT the easy way will be plug and play. Also, we've to find a value of T (from 5 optiosn given below) which can make a square of a a number. One we know this, it will be a cake walk.
We can see that option A fits this criteria in eq (b). Add - 1540+669 = 2209 = 47^2 = (x-8)^2 Hence, x = 55.
Cross check by putting in eq (a) = x(x+1)/2 = T = > 56*55=1540*2.

Hence, answer is A.
Hope it helps.

Cheers!
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Re: Progressions [#permalink] New post 07 Dec 2011, 08:32
Good question and excellent explanation by Capricorn!
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Re: Progressions [#permalink] New post 07 Dec 2011, 09:16
do we get these type of questions in GMAT...because i found this question tough?
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Re: Progressions [#permalink] New post 07 Dec 2011, 09:23
great question and explanation. I found this one to be very difficult
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Re: Progressions [#permalink] New post 07 Dec 2011, 10:48
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For those who like myself failed to recall/derive the formula of the sum of the members of a finite arithmetic progression:

1.“669 balls are added, which enabled to arrange all the balls in the shape of square” simply means that now we have a number of balls which is represented by a perfect square.

2. Adding 9 (units digit of 669) to the last digit of each answer choice will give us the units digits of our perfect square: A-9; B-8; C-7; D-9; E-9

3. One can easily check that from the digits above only 9 could be a units digit of a perfect square (just raise to power 2 all the numbers from 0 to 9 if unsure) => B and C are out.

4. Now adding 669 to each of the remaining ACs will give us 2209, 1879 and 2229

5. All numbers above are between 1600 and 2500 or 40^2 and 50^2, so we should look for a number between 40 and 50 ending with 3 or 7 to be the side of the square.

6. 43x43=1849, 47x47=2209 => answer A

Obviously, the method from Capricorn369 is much more elegant and safe.
Manager
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Re: Progressions [#permalink] New post 07 Dec 2011, 14:16
I found this one very difficult even after multiple attempts, let alone solving in 2 min.
Manager
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Re: Progressions [#permalink] New post 07 Dec 2011, 14:18
I guess the only way to solve this Q in 2 mins. is to know 47-squared and to know when to go "plug and play" mode. The rest is pretty routine.
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Re: Progressions [#permalink] New post 07 Dec 2011, 20:30
Balls of equal size are arranged in rows to form an equilateral triangle.
The top most row consists of one ball, the 2nd row of two balls and so on.
If 669 balls are added, then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did.

How many balls made up the triangle?

A)1540
B)2209
C)2878
D)1210
E)1560

The formula for the sum of a finite arithmetic progression, such as the total number of balls in the triangle, is

S(n) = n[A(1) + A(n)]/2

A(1) is the sum of first term (or top row in triangle)
A(n) is the sum of the nth term (or edge line in triangle)
A(n) is also the value of the side of the triange.

Here, A(1) = 1
A(n) = n

S(n) = n(1 + n)/2

Taking n = x

(x - 8)^2 = x(x+1)/2 + 669

x^2 - 16x + 64 = (x^2 + x)/2 + 669

2x^2 - 32x + 128 = x^2 + x + 1338

x^2 - 33x - 1210 = 0

We can verify that the above quadratic (a=1, b=-33, c=-1210) has real roots because b^2 - 4ac is positive.

(x - 55) (x + 22) is the solution,

where x > 0, x = 55

S(n) = 55(55 + 1)/2 = 1540
Re: Progressions   [#permalink] 07 Dec 2011, 20:30
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