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Thread: COIN PROBLEM AND ALLEGATIONS & MIXTURES CONCEPT(2 problems)

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    COIN PROBLEM AND ALLEGATIONS & MIXTURES CONCEPT(2 problems)

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    I need help with the following two problems which are in Allegations and mixtures concept.

    1. If a man decides to travel 72 miles in 8 hours partly by foot and partly on a bicycle, his speed of foot being 8 mph and that on bicycle being 10 mph. What distance would he travel on foot?
    (A) 24 miles
    (B) 32 miles
    (C) 48 miles
    (D) 64 miles
    (E) Cannot be determined

    2. A sum of $80 is made up of 80 coins that are either dimes or quarters coins. Find out how many dimes or quarters coins are there in the total amount?
    (A) 10
    (B) 20
    (C) 30
    (D) 40
    (E) 50

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    For 1st
    let the man travel x miles on foot. he will then travel 72-x miles on bicycle. the total time taken is 8 hrs.
    Therefore [ x/8 + (72-x)/10] = 8 since time= distance / speed .
    solving the equation you will get x as 32miles.

    for 2nd question:
    the question seems to be flawed since even if u have 80 quarters it amounts to $20 . therefore the total amount given of $80 is flawed it should be something much less than that.
    does somebody have a different explanation?

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    1. Suppose he walks -

    1 hour on foot (8 mile) and 7 hour on bicycle 70 mile, total 78 miles in 8 hours which doesn't fulfill the condition.
    2 hour on foot (16 mile) and 6 hour on bicycle 60 mile, total 76 miles in 8 hours which doesn't fulfill the condition.
    3 hour on foot (24 mile) and 5 hour on bicycle 50 mile, total 74 miles in 8 hours which doesn't fulfill the condition.
    4 hour on foot (32 mile) and 4 hour on bicycle 40 mile, total 72 miles in 8 hours which fulfills the condition.
    5 hour on foot (50 mile) and 3 hour on bicycle 30 mile, total 80 miles in 8 hours which doesn't fulfill the condition.
    6 hour on foot (48 mile) and 2 hour on bicycle 20 mile, total 68 miles in 8 hours which doesn't fulfill the condition.
    7 hour on foot (56 mile) and 1 hour on bicycle 10 mile, total 66 miles in 8 hours which doesn't fulfill the condition.

    So the answer is B.

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    Thanks Achilles and Zulkarnain.

    @ Achilles: I found a solution to a similar coin problem posted above. But somehow, I can't relate this to that. Only dimes have been asked here. Please go through the following:

    Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?
    (A) 3 (B) 7 (C) 10 (D) 13 (E) 16
    Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 - D). Summarizing this information in a table yields
    Dimes Quarters Total Number D 20 - D 20 Value 10D 25(20 - D) 305
    Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:
    10D + 25(20 - D) = 305
    10D + 500 - 25D = 305
    -15D = -195
    D = 13
    Hence, there are 13 dimes, and the answer is (D).

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    This is a typical denomination question. let there be D number of dimes and Q number of quarters. therefore D+Q= 20. [since total number of coins is 20]
    now we know that 1 dime is equal to 10 cents therefore the amount of D dimes will be 10D cents . similarly 1 quarter equal to 25 cents therefore the amount Q quarter will be 25Q cents .
    It is given that the total is $3.05 = 305 cents since [$1 = 100 cents].
    therefore 10D + 25Q = 305
    and D+Q = 20
    by solving the two equations we get D= 13.

    Now if we solve the first problem in the same way we will get two equations:

    Q+D= 80
    25Q+10D=8000
    solving these we will get Q= 528. and D negative. therefore there is some problem with the question. also if we consider that there are only quarters that is 80 quarters it will amount to $20 and not $80.

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    This is a typical denomination question. let there be D number of dimes and Q number of quarters. therefore D+Q= 20. [since total number of coins is 20]
    now we know that 1 dime is equal to 10 cents therefore the amount of D dimes will be 10D cents . similarly 1 quarter equal to 25 cents therefore the amount Q quarter will be 25Q cents .
    It is given that the total is $3.05 = 305 cents since [$1 = 100 cents].
    therefore 10D + 25Q = 305
    and D+Q = 20
    by solving the two equations we get D= 13.

    Now if we solve the first problem in the same way we will get two equations:

    Q+D= 80
    25Q+10D=8000
    solving these we will get Q= 528. and D negative. therefore there is some problem with the question. also if we consider that there are only quarters that is 80 quarters it will amount to $20 and not $80.

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