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25 Jun 2021, 03:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (02:20) correct
38%
(02:20)
wrong
based on 55
sessions
History
Date
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Not Attempted Yet
How many parallelograms each with a base of 15 cm and heights of 5 cm, 10 cm, 15 cm and so on, will have a total area of 4,125 cm^2?
(A) 6
(B) 9
(C) 10
(D) 12
(E) 14
(A) 6
(B) 9
(C) 10
(D) 12
(E) 14
25 Jun 2021, 04:02
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=15*5+15*10+15*15+.....
=15*5(1+2+3+...)
[4125/75=55 and (10*(10+1))/2=55]
So,
=15*5(1+2+3...+10)
=4125 so number of parallelograms is 10
Posted from my mobile device
=15*5(1+2+3+...)
[4125/75=55 and (10*(10+1))/2=55]
So,
=15*5(1+2+3...+10)
=4125 so number of parallelograms is 10
Posted from my mobile device
25 Jun 2021, 04:05
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Bunuel
These NMAT questions are so badly worded that it's often almost impossible to figure out what they mean. Here, I think it's reasonable to think the question is saying "how many different parallelograms could you draw with a base of 15 and an area of 4125, if the height must be a multiple of 5?" The answer to that question is "infinitely many", so that's clearly not the intention of the question. But it's a reasonable way to interpret the wording.
Instead, I think they want to ask (phrasing this casually rather than precisely) : If you add up the areas of several parallelograms, each with a base of 15, but with heights of 5, 10, 15, 20, etc, how many parallelograms are there if the sum of those areas is 4125?
Since the area of a parallelogram is bh, the sum of those areas (if we have n parallelograms) will be 15*5 + 15*10 + 15*15 + 15*20 + ... + 15*5n = 15(5 + 10 + 15 + 20 + ...+ 5n) = 75(1 + 2 + 3 + ...n)
That sum equals 4125, so
75(1 + 2 + 3 + ...+ n) = 4125
1 + 2 + 3 + ... + n = 55
and the sum of an equally spaced list is just the number of terms, which here is n, times the average term, which is the average of the first and last, so here is (n + 1)/2, so
n(n+1)/2 = 55
n(n+1) = 110
so we want two consecutive positive integers which multiply to 110, and those are 10 and 11, so n = 10.
