The answer is a.
First calculate how many paperbacks could fit on the seven shelves if they only contained paperbacks.
7*45 = 315 paper
285 paper + 15 hard = 285 + 2*15 = 315 paper
And the could be shelved as
3 shelves = 35 paper + 5 hard
4 shelves = 45 paper
Total = (3*35 + 4*45) paper + 3*5 hard = 285 paper + 15 hard
The answer is a.
2007-12-14 19:44:06
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answer #1
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answered by Northstar 7
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a. 285 paper/15 hard explained as under.
7 shelves can hold a total of 45 x 7 = 315 paperback books.
285 paperback + 15 hardback = 285 + 2x15 = 315 paperback books.
Other options either exceed 315 or add up to less than 315 and hence are ruled out.
2007-12-14 17:27:04
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answer #2
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answered by Madhukar 7
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1 hardbook = 2 paperback
1 shelf = 45 paperback
7 shelf = 7*45 paperback = 315 paperback
The answer is a because
285 * 1 + 15 * 2 = 315
Are we filling up 7 shelves or 1 shelf?
Is this question from the math or the critical reading section?
@ mitch,
15 paper + 15 hard = 15 *1 + 15*2 = 45
45 = 1 shelf. Technically you are correct, but the poor wording of the problem threw you off.
2007-12-14 17:23:52
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answer #3
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answered by Axis Flip 3
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We would have to see how many paperback books can fit into the bookshelf.
1 shelf can hold 45 paperback books
The whole bookshelf can hold 45*7 = 315
Taking the first choice
285 paperback books plus 15 hard books which is equivalent to 30 paperback books equal 315 paperback books which fill the self.
Answer is a
2007-12-14 17:28:25
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answer #4
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answered by Loric M 2
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Convert everything to paperbacks:
7 shelves X 45 paperbacks = 315 total capacity
a = 285 + 30 = 315 (you don't even have to convert the rest - you already have your answer: "A"
2007-12-14 17:24:36
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answer #5
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answered by joes_shmoe 3
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you be attentive to the thank you to get remainders once you do branch, proper? it would have been an prolonged time, in view which you have probable been working with decimals and fractions for a jointly as. So purely a speedy overview - in case you divide 13 by potential of three, then what share situations does 3 pass into 13 frivolously? 4, proper? So 4x3 = 12, and 13 - 12 aspects something of a million. in case you probably did the long branch, it would appear like .__4 r a million 3)13 . -12 ------- ... a million ok. one thank you to get the respond is to locate an n that satisfies each and every of the circumstances you're given interior the difficulty. that's definitely extraordinarily complication-free, in view that there is entirely one concern - n/9 provides a the remainder of 7. So, while you're taking n = 7, then n/9 = 0 r 7. ok, now, 5n = 5*7 = 35, and 35/9 = 3 r 8. So, something is 8. in case you like a greater rigorous answer, you ought to do it like this: valuable integer n divided by potential of 9 is comparable to a favorable integer x with a the remainder of 7. So, n / 9 = x r 7 n = 9 * (x r 7) n = 9x + 7 5n = 5*(9x + 7) 5n = 45x + 35 5n / 9 = (45x + 35) / 9 5n / 9 = 45x / 9 + 35 / 9 5n / 9 = 5x + 35 / 9 5n / 9 = 5x + (27 + 8) / 9 5n / 9 = 5x + 27/9 + 8/9 5n / 9 = 5x + 3 + 8/9 Then, in view that x is an integer, 5x + 3 is an integer. So, in the rest sort, 5n / 9 = (5x + 3) r 8. so which you land up with an identical answer. desire this helps!
2016-11-27 01:34:15
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answer #6
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answered by Anonymous
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a total of 45*7 = 315 paperbook books could be shelved. Since 1 hb = 2 pb, a) is a possibility
2007-12-14 17:48:27
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answer #7
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answered by jimbob 6
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alright... so total number of paperbacks allowed is found by:
7 x 45
Take that answer and start checking each of the possible answers to see which one is equal or less take the paperback number and add it to 2x the hardback number because hardbacks take up the same space as 2 paperbacks
Or you could take the number of possible paperbacks and subtract 2x the hardbacks in each answer then check to see if the number you get is the same as the paperbacks for that answer and if it is then that's the correct answer
You can do the math.
2007-12-14 17:28:14
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answer #8
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answered by Crystal 2
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c
2007-12-14 17:22:05
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answer #9
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answered by mitch 5
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