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Jonathan buys c chairs and t tables for his newly set up res

This topic has 7 expert replies and 1 member reply
duahsolo Newbie | Next Rank: 10 Posts Default Avatar
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Jonathan buys c chairs and t tables for his newly set up res

Post Thu Jul 27, 2017 2:42 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Jonathan buys c chairs and t tables for his newly set up restaurant, where c and t are positive integers. The chairs regularly sell at $50 apiece and the tables regularly sell at $80 apiece. Jonathan gets a discount of 20 percent on the price of each chair and 15 percent on the price of each table that he buys. What percentage of the total amount spent by him in buying the chairs and the tables at discounted prices was spent in buying chairs? (source: e-gmat)

    (1) He saved a total of $100 by purchasing the chairs and the tables at their discounted prices instead of their regular prices
    (2) Due to the discounts, he paid only five-sixths of the amount he would have paid in purchasing the chairs and the tables at their regular prices

    Correct answer: D

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    GMAT/MBA Expert

    Ian Stewart GMAT Instructor
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    Post Fri Jul 28, 2017 4:22 am
    He saves$10 on each chair, and $12 on each table. So using Statement 1, we know

    10c + 12t = 100

    Dividing by 2,

    5c + 6t = 50

    If you have an equation where you are adding or subtracting two integers on one side, and getting some result on the other side, and two of the terms share some common divisor, then the third term must also be divisible by that divisor. Here two of our terms are divisible by 5, so the third term must be as well. That is, 6t must be divisible by 5, so t must be divisible by 5. Since t cannot be 10 or more (then c would need to be negative), t must be 5, and the equation has only one solution. t cannot be 0, because the question tells us c and t are positive.

    If it's not clear why t must be divisible by 5, you can rearrange the equation with the multiples of 5 on one side:

    6t = 50 - 5c
    6t = 5 * (10 - c)

    and now, since 6t is the same number as 5 * (10 - c), it must be true that 6t has the same divisors as 5 * (10 - c), and since 5 is clearly a divisor of 5 * (10 - c), it must also be a divisor of 6t.

    Statement 2 tests weighted average principles. If you get different percent discounts on two items, your overall percent discount is a weighted average of the two percent discounts, weighted by the original prices of the two items. If you know the 'alligation' method (or 'number line method') for weighted average problems, and if you recognize this as a weighted average problem, it will be immediately clear we can use Statement 2 to find the ratio of the original cost of the chairs to the original cost of the tables, and from there find the ratio of c to t and answer the question. If you are not familiar with that method, we can also use algebra. The amount he spent, after discount, was 40c + 68t. The amount he would have spent, with no discounts, was 50c + 80t. We know the first number is 5/6 of the second:

    40c + 68t = (5/6) (50c + 80t)
    20c + 34t = (5/6) (25c + 40t)
    120c + 204t = 125c + 200t
    4t = 5c
    t = (5/4)c

    So we can find the ratio of t to c, and with that information we can answer the question, since the question asks us to find 40c / (40c + 68t), and if we substitute now for t, the letters will all cancel out.

    It could be realistic to see either one of these two statements in a real GMAT question, but not both in the same question, since for most test takers it would take too long to analyze each of these statements individually.

    _________________
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    Thanked by: duahsolo

    GMAT/MBA Expert

    Post Sun Aug 06, 2017 10:19 pm
    Here's another approach that I think gets through S1 more quickly. (It isn't very elegant, but it's great for timed test conditions.)

    Let's start by sorting out the prompt. We know that he spent $40c on chairs and $68t on tables, and we're asked for 40c / (40c + 68t). OK!

    S1:

    He saved $10c + $12t. We're told that 10c + 12t = 100, or that 5c + 6t = 50.

    From here we can cheat a bit. Suppose that he only bought chairs. If that's the case, then 5c = 50, and c = 10. But we know that he bought at least one table, so c + t < 10. (Tables cost more than chairs, we'd get fewer items for our money.)

    Now suppose he only bought tables. Then 6t = 50, or t > 8.333.... But we know that he bought at least one chair, so c + t > 8.3333. (Chairs cost less than tables, so we'd get more items for our money.)

    Those two deductions give us 10 > c + t > 8.3333... Since c and t are integers, we MUST have c + t = 9.

    That leaves us two equations: 5c + 6t = 50 and c + t = 9. We can solve those, so S1 is SUFFICIENT.

    S2:

    discount = (1/6) of total

    10c + 12t = (1/6) * (50c + 80t)

    10c + 12t = 25c/3 + 40t/3

    30c + 36t = 25c + 40t

    5c = 4t

    c = (4/5)t

    We were asked to find 40c / (40c + 68t). Since we can replace c with (4/5)t, all the t's will cancel out and we'll get a number. SUFFICIENT!

    Thanked by: duahsolo
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    Post Sun Aug 06, 2017 10:27 pm
    A light bulb just flickered and here's another quick and dirty approach to S1:

    Start with 5c + 6t = 50. That gives us:

    6t = 50 - 5c

    6t = 5 * (10 - c)

    Since 6t = 5 * something = some multiple of 5, we know that either 6 or t is a multiple of 5. 6 obviously isn't, so t must be.

    Now notice that, since c > 0, we MUST have 0 < 6t < 50. Since t is a multiple of 5, it can now ONLY be 5 itself, since any other multiple of 5 will make 6t ≤ 0 or 6t ≥ 60.

    This forces t = 5, from which 5c + 6t = 50 is easy to solve for c, and we're done!

    Thanked by: duahsolo
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    Post Sun Aug 06, 2017 10:31 pm
    One last, very crude approach to S1:

    5c + 6t = 50

    c + 1.2t = 10

    We know that c is an integer, so 1.2t must also be an integer. 1.2t will only be an integer if t is a multiple of 5, since .2 * 5 = 1. But if t > 5, then 1.2t > 10, forcing c to be negative, which is impossible.

    So we must have t = 5, as above, and we're good to go.

    Thanked by: duahsolo
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    Post Sun Aug 06, 2017 10:34 pm
    By the way, if you like S1, the fancy name for these equations - algebraic problems that only consider integer solutions - is Diophantine equations. This is a rich and ancient topic in math, definitely worth Googling and exploring should it strike your fancy.

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    Balla Newbie | Next Rank: 10 Posts
    Joined
    28 Jul 2017
    Posted:
    8 messages
    Post Wed Aug 16, 2017 2:51 pm
    Jonathan buys c chairs and t tables for his newly set up restaurant, where c and t are positive integers. The chairs regularly sell at $50 apiece and the tables regularly sell at $80 apiece. Jonathan gets a discount of 20 percent on the price of each chair and 15 percent on the price of each table that he bu=[ ys. What percentage of the total amount spent by him in buying the chairs and the tables at discounted prices was spent in buying chairs? (source: e-gmat)

    (1) He saved a total of $100 by purchasing the chairs and the tables at their discounted prices instead of their regular prices
    (2) Due to the discounts, he paid only five-sixths of the amount he would have paid in purchasing the chairs and the tables at their regular prices

    50c+80t=40c+68t+100

    10c+12t=100

    c= (100-12t)/(10)

    c=(100-10t-2t)/(10)

    c=10-t-t/5

    t/5 must be an integer, so t must be divisible by 5. let t = 5p where p is an integer.

    c=10-5p-p=10-6p

    c must be a positive integer so p must be 1 which makes c=4 . this makes t=5 which is enough to solve the problem.

    Statement 2.

    in general ax+by=cx+dy where a, b, c, d are integers (or even numbers in general) will simplify to give the ratio between x and y. c will be given in terms of d in this example which, when substituted, will have the variable cancel out.

    GMAT/MBA Expert

    Post Fri Aug 18, 2017 12:57 pm
    Balla wrote:
    Statement 2.

    in general ax+by=cx+dy where a, b, c, d are integers (or even numbers in general) will simplify to give the ratio between x and y. c will be given in terms of d in this example which, when substituted, will have the variable cancel out.
    This is a neat point. If we have ax + by = cx + dy, we can say that:

    ax - cx = dy - by

    x * (a - c) = y * (d - b)

    x/y = (d - b)/(a - c)

    Exactly as Balla told us! Smile

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    Post Fri Aug 18, 2017 12:58 pm
    One nitty footnote to my last post, though (I didn't want to be a wet blanket, but I must): we're assuming that (a - c) doesn't equal 0, or we can't divide to set up that ratio.

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