# Thread: Percentage with apples problem

1. ## 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?

2. Originally Posted by kingman
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.

3. 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

4. Originally Posted by kingman
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

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

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

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

5. 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..

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

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

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

6. Originally Posted by kingman
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.