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#60 Official Guide

This topic has 3 expert replies and 1 member reply
RahilM824 Newbie | Next Rank: 10 Posts Default Avatar
Joined
02 Feb 2015
Posted:
1 messages

#60 Official Guide

Post Mon Feb 02, 2015 8:19 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Struggling with this one:

    On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $.60 a glass on the first day, what was the price per glass on the second day?

    A) $.15
    B) $.20
    C) $.30
    D) $.40
    E) $.45

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    Post Mon Feb 02, 2015 8:24 am
    Initially, you have a certain amount of orangeade and the same amount of water. So let's call each 'x.'

    Water: x
    Orangeade: x
    Total: 2x

    Well, let's say that this 2x volume of juice translates to 100 servings. (I'm picking this number simply because it's nice and round. Any number will work.)

    Well my total revenue will be

    100 * .60 = 60



    The next day we're doubling the water, so we'll have
    Water: 2x
    Orangeade: x
    Total: 3x

    Note that we have 50% more juice volume, because 3x is 50% greater than 2x. So now, if we initially had 100 servings, we'll have 150 servings. The revenue remains the same so we need to solve for the price, giving us the following equation:

    150 * P = 60
    P = 60/150 = .40.

    Answer is D.

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    Post Mon Feb 02, 2015 8:28 am
    Also note that anytime we're working with the following equation: quantity*price = revenue, if we hold the revenue constant, price and quantity will have an inverse relationship. In other words, if the price is doubled, the quantity will be halved, etc. In this case, we're multiplying the quantity by 3/2, which means that if we're holding the revenue constant, we'd need to multiply the price by 2/3. .60 *2/3 = .40.

    [And the same logic would apply to the relationship between rate and time if we hold distance constant, so this is a useful piece of logic that can be applied in multiple domains.]

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    Post Mon Feb 02, 2015 9:37 am
    RahilM824 wrote:
    Struggling with this one:

    On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $.60 a glass on the first day, what was the price per glass on the second day?

    A) $.15
    B) $.20
    C) $.30
    D) $.40
    E) $.45
    First day:
    Here, equal amounts of juice and water are used.
    Let amount of orangeade = 1 glass juice + 1 glass water = 2 glasses.
    Since each glass earns 60 cents, revenue = 2*60 = 120 cents.

    Second day:
    Revenue = 120 cents (same as on the first day).
    Since twice as much water is used, amount of orangeade = 1 glass juice + 2 glasses water = 3 glasses.
    Price per glass = revenue/glasses = 120/3 = 40 cents.

    The correct answer is D.

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    Post Fri May 01, 2015 9:30 am
    RahilM824 wrote:
    Struggling with this one:

    On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $.60 a glass on the first day, what was the price per glass on the second day?

    A) $.15
    B) $.20
    C) $.30
    D) $.40
    E) $.45
    We are given that orangeade is made on Day 1 with an EQUAL AMOUNT of water and orange juice. We can set this information up into a ratio using a variable multiplier:

    W : OJ = x : x

    Thus, orangeade quantity = amount of water + amount of OJ = x + x = 2x

    We are next given that orangeade on Day 2 was made by mixing the SAME AMOUNT of orange juice with TWICE THE AMOUNT of water. We can set this information up into a ratio using a variable multiplier:

    W : OJ = 2x : x

    Thus, orangeade quantity = amount of water + amount of OJ = 2x + x = 3x

    We also know that all orangeade made was sold and that the revenue on both days was the same. We can therefore set up the following equation:

    Day 1 Revenue = Day 2 Revenue

    That is,

    (quantity sold Day 1)(price per glass Day 1) = (quantity sold Day 2)(price per glass Day 2)

    Let a be the amount of orangeade a glass can hold, then

    quantity sold Day 1 = the number of glasses of orangeade sold on Day 1 = 2x/a

    Similarly,

    quantity sold Day 2 = the number of glasses of orangeade sold on Day 2 = 3x/a

    We also know that the price per glass on day 1 = $0.6

    But we don’t know the price per glass on day 2, so let’s label it as variable p.

    We now have:

    (2x/a)(0.6) = (3x/a)(p)

    Multiplying both sides by a, we have:

    (2x)(0.6) = (3x)(p)

    1.2x = 3xp

    1.2 = 3p

    p = 0.4

    Thus, each glass of orangeade was sold for $0.40 on Day 2.

    The answer is D

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