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Thread: 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.

    $\displaystyle A>B$

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

    $\displaystyle 0.3A=0.4B+45$

    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.

    $\displaystyle B_{new}=B+0.1A$

    $\displaystyle A_{new}=A-0.1A$

    $\displaystyle A-0.1A=B+0.1A+200$


    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

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

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

    Now, substitute for A

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

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

    $\displaystyle 2400=2B\Rightarrow\ B=1200$
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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

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

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

    $\displaystyle \Rightarrow\ 8A=10B+2000$

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

    $\displaystyle 24A=32B+3600$

    $\displaystyle 24A=30B+6000$

    Therefore

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



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

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

    $\displaystyle 0.3A=45+0.4B$

    therefore, dividing by 3

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

    and multiplying by 3

    $\displaystyle 0.9A=135+1.2B$

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

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

    $\displaystyle 405+3.6B=3B+45+0.4B+600$

    $\displaystyle 4050+36B=30B+450+4B+6000$

    $\displaystyle 36B-34B=6450-4050\Rightarrow\ 2B=2400$
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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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