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Nums99
Joined: 12 Jan 2019
Last visit: 18 Mar 2022
Posts: 88
Given Kudos: 211
Location: India
Concentration: Finance, Technology
Schools: Rotman '23 Desautels '23 Alberta University
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
57% (02:26) correct
43%
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wrong
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Rosy and Daisy bought some books, pens, and pencils. Out of the total amount spent by Rosy, what percentage account for the amount spent on buying pencils, if it is known that Daisy spends 1.5 times the amount spent by Rosy?
(1) Rosy bought 4 books, 14 pens, and 10pencils
(2) Daisy bought 8 books, 28 pens, and 12 pencils
(1) Rosy bought 4 books, 14 pens, and 10pencils
(2) Daisy bought 8 books, 28 pens, and 12 pencils
21 Sep 2020, 07:59
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The answer is clearly C or E. Using both Statements, Daisy bought twice as many books and twice as many pens as Rosy. So if we think of books and pens as a single thing, let's call it 'supplies', Daisy spent twice as much on supplies as Rosy. Daisy also bought 1.2 times as many pencils as Rosy, so spent 1.2 times as much on pencils. Overall, she spent 1.5 times as much as Rosy. So we now have a weighted average situation, with two groups, one with an average of 1.2, and the other with an average of 2, and an overall average of 1.5. If you know the alligation method, you really don't need to do anything more - it's clear we can answer the question. If one wanted to solve, we can draw a number line:
---1.2-------1.5------------2---
and since the ratio of the distances to the average in the middle is 3 to 5, that's the ratio of the two groups. Since Rosy must have spent more on pencils (since 1.5 is closer to 1.2 than to 2), then 5/8 of her expense was on pencils. I'm using alligation here, and if that method is unfamiliar to anyone reading, this solution might not make a lot of sense, but you can find an explanation of the method by googling the term.
You could also solve algebraically: if Rosy spends $b on books and pens combined, then Daisy spends $2b on those things. If Rosy spends $p on pencils, then Daisy spends $1.2p on pencils. We know Daisy's overall expenditure is 1.5 times that of Rosy, so we'd get this equation:
2b + 1.2p = 1.5(b + p)
which can be solved for the ratio of b to p, from which we can answer the question.
---1.2-------1.5------------2---
and since the ratio of the distances to the average in the middle is 3 to 5, that's the ratio of the two groups. Since Rosy must have spent more on pencils (since 1.5 is closer to 1.2 than to 2), then 5/8 of her expense was on pencils. I'm using alligation here, and if that method is unfamiliar to anyone reading, this solution might not make a lot of sense, but you can find an explanation of the method by googling the term.
You could also solve algebraically: if Rosy spends $b on books and pens combined, then Daisy spends $2b on those things. If Rosy spends $p on pencils, then Daisy spends $1.2p on pencils. We know Daisy's overall expenditure is 1.5 times that of Rosy, so we'd get this equation:
2b + 1.2p = 1.5(b + p)
which can be solved for the ratio of b to p, from which we can answer the question.
23 Sep 2020, 14:08
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Nums99
Clearly, each statement alone is insufficient, so you need to analyze both statements together.
You can set up equations using the statements and the fact that "Daisy spends 1.5 times the amount spent by Rosy".
> Rosy: 4a + 14b + 10c = T
> Daisy: 8a + 28b + 12c = 1.5T
Since the question is about pencils out of total, we need to isolate c and T. You can multiply the first equation by 2 and then subtract. This works because the coefficients of a and b in Rosy's equation are half of the coefficients in Daisy's -- that is the key fact in this problem.
> 8a + 28b + 20c = 2T
> 8a + 28b + 12c = 1.5T
----------------------
> 8c = 0.5T
That means that 16c = T, so Rosy's total is equal to 16 times the price of one pencil. Thus, Rosy's 10 pencils are 10/16 of the total.
Answer: (c) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
