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please help me for this question...

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please help me for this question...

by Alpturk » Thu Feb 19, 2009 8:51 am
In a certain brick wall, each row of bricks above the bottom row contains one less brick than the row just below it. if there are 5 rows in all and a total of 15 bricks in the wall, how many bricks does the bottom row contain?

a.14
b.15
c.16
d.17
e.18
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by DanaJ » Fri Feb 20, 2009 7:03 am
Say that the number of bricks in the bottom row is n. Then you have that there are n-1 bricks in the second row, n-2 in the third row, n-3 in the fourth row and n-4 in the fifth row. This means that you have a total of:
n + n - 1 + n - 2 + n - 3 + n - 4 = 5n - 10 bricks in your wall. We know that this is equal to 15, so you have the following equation:
5n - 10 = 15
5n = 25
n = 5.

I think you should check the answers... Or maybe I'm wrong....
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by gabriel16 » Fri Feb 20, 2009 8:22 am
I did it the exact same way. So if you are wrong then we both are wrong.

I am interested to hear if those are in fact the official answers.
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by Alpturk » Fri Feb 20, 2009 8:36 am
i checked that the correct answer is D.
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by DanaJ » Fri Feb 20, 2009 11:31 am
The question is wrong in wording or smth like this. If there are 15 bricks in the wall, how can the bottom row have 17 bricks? This means that the bottom row has more bricks than the entire wall, which is absurd.
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by x2suresh » Fri Feb 20, 2009 8:10 pm
I believe ther is typo in the question it should be 75 bricks instead of 15
in that case.

...13
......14
.........15
............16
...............17

Sum = 75.

Ans 17
Last edited by x2suresh on Fri Feb 20, 2009 8:17 pm, edited 1 time in total.
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by cramya » Fri Feb 20, 2009 8:15 pm
75 makes more sense. Please check for any typos when posting questions. It may be unintentonal on our part so its always good to double check. Just a friendly suggestion...



Regards,
CR
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