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Math Help - Percentage with apples problem

  1. #1
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    Percentage with apples problem

    Dear sir,
    I would be grateful if someone can help me in the below question
    thanks

    There are more apples in box A than in box B. 30 % of the apples in box A is 45 more than 40% of the apples in box B. If 10 % of the apples in box A is transferred to box B, there will be 200 more apples in box A than box B.
    a) How many apples are there in box B?

    b) How many percent less apples are there in box B than in box A?
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  2. #2
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    Quote Originally Posted by kingman View Post
    Dear sir,
    I would be grateful if someone can help me in the below question
    thanks

    There are more apples in box A than in box B.

    A>B

    30 % of the apples in box A is 45 more than 40% of the apples in box B.

    0.3A=0.4B+45

    If 10 % of the apples in box A is transferred to box B, there will be 200 more apples in box A than box B.

    B_{new}=B+0.1A

    A_{new}=A-0.1A

    A-0.1A=B+0.1A+200


    a) How many apples are there in box B?

    b) How many percent less apples are there in box B than in box A?
    You can set up a solution by writing a few equations as above.
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  3. #3
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    Dear Sir,
    Thanks for the help . In fact I can set up the the 2 equations needed and get the answer but I
    wonder there is any particular hidden relationship between A and B which can be used to make
    the required equation look simpler and easier to visualise for the given situation.
    thanks
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  4. #4
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    Quote Originally Posted by kingman View Post
    Dear Sir,
    Thanks for the help . In fact I can set up the the 2 equations needed and get the answer but I
    wonder there is any particular hidden relationship between A and B which can be used to make
    the required equation look simpler and easier to visualise for the given situation.
    thanks
    Part (a) can be solved by using the two equations

    0.3A=0.4B+45\Rightarrow\ A=\displaystyle\frac{0.4B+45}{0.3}

    A-0.1A=B+0.1A+200\Rightarrow\ B=A-0.2A-200\Rightarrow\ B=0.8A-200

    Now, substitute for A

    B=\displaystyle\ 0.8\frac{0.4B+45}{0.3}-200\Rightarrow\ B=\frac{0.32B}{0.3}+\frac{36}{0.3}-200

    0.3B=0.32B+36-60\Rightarrow\ 60-36=24=0.32B-0.3B=0.02B

    2400=2B\Rightarrow\ B=1200
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  5. #5
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    We can clean up the equations and make things easier..

    multiplying both sides by 10

    0.3A=0.4B+45\Rightarrow\ 3A=4B+450

    A-0.1A=B+0.1A+200\Rightarrow\ 0.9A=B+0.1A+200\Rightarrow\ 9A=10B+A+2000

    \Rightarrow\ 8A=10B+2000

    Then, multiplying the 1st equation by 8 and the 2nd by 3.....

    24A=32B+3600

    24A=30B+6000

    Therefore

    32B+3600=30B+6000\Rightarrow\ 32B-30B=2B=6000-36000=2400



    I think that the method you are looking for may be the following.......

    From the first clue (30% and 40%.....)

    0.3A=45+0.4B

    therefore, dividing by 3

    0.1A=15+\frac{0.4B}{3}

    and multiplying by 3

    0.9A=135+1.2B

    We required 0.1A and 0.9A to use in the 2nd clue (if we move 10% of A to B....)

    135+1.2B=B+15+\frac{0.4B}{3}+200 allows us to solve for B..

    405+3.6B=3B+45+0.4B+600

    4050+36B=30B+450+4B+6000

    36B-34B=6450-4050\Rightarrow\ 2B=2400
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  6. #6
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    Quote Originally Posted by kingman View Post
    Dear sir,
    I would be grateful if someone can help me in the below question
    thanks

    There are more apples in box A than in box B. 30 % of the apples in box A is 45 more than 40% of the apples in box B. If 10 % of the apples in box A is transferred to box B, there will be 200 more apples in box A than box B.
    a) How many apples are there in box B?

    b) How many percent less apples are there in box B than in box A?
    For (b)

    90% of A = B+10% of A +200

    80% of A = B+200 = B+(1/6)B = (7/6)B

    B = (6/7)80% of A

    A-B as a percentage gives the percent less apples in B than in A.

    This is A(100-[6/7]80)% which is (100-68.57)% of A, 31.43% down.
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