Exam Question Answers



January 2005 C1 OCR MEI


Firstly, it's usually easier to get the variable you want as the subject on the left-hand side, so we're just going to swap the sides around.

43πr3 = V

Now we're going to get rid of the fraction. You can do this by turning the fraction upside down and carrying it to the other side.

πr3 = 3V4

Next, we just need to take the π and take it over to the other side. Remember, it is multiplying the r3, so it will divide when it goes over to the other side.

r3 = 3V



Finally, as the 3 is only on the r - what we need to have last, we need to get rid of it. So we need to find the cubic root of both sides.

r = 3√(3V4π)





June 2005 C1 OCR MEI


Similar to the above, we need to get x the subject.
To begin, we're going to get all the x's on one side.

3x + mx = y + 5y
3x + mx = 6y


Next, we're going to factorise the left-hand side.

x(3 + m) = 6y

All we need to do now, is carry the (3 + m) over to the other side. It needs to divide the 6y as it's multiplying the x.

x = 6y(3 + m)



January 2006 C1 OCR MEI



As we want all the C's on the left hand side, we're going to take the C - 4 and take it over to the other side.

P(C - 4) = C
PC - 4P = C


Next we're going to get all the C's on one side.

PC - C = 4P


As we can't have the C's cancelling out, we need to factorise them.

C(1 - P) = 4P
Now all we need to do is take the brackets out.

C = 4P1 - P

June 2006 C1 OCR MEI



The first thing to do is get rid of the fraction.

3V = πr2h


Next we're going to collect the factors of r2 and take them over to the other side.

3V = r2h)
3Vπh = r2


Now all we need to do is get rid of the 2 and we can do this by taking the square root of both sides.

±√3Vπh = r


Remember to put in the ± when you square root something!


January 2007 C1 OCR MEI



First, we're going to get all of the a's over to one side.

2a - af = 7c - 5c


As we need to get a on it's own, we need to get the factors of a to get rid of that f.

a(2 - f) = 2c


Next, we're going to take the brackets over to the other side to complete.

a = 2c2 - f



January 2007 C1 OCR MEI



First we're going to split this equation up so that it looks a lot simpler.

3 × 4 × a3 × a2 × b × b2


Remember, if the letter doesn't have a power, just think of it being to the power of 1. When you're multiplying, you have to add the powers together.

12 × a5 × b3


To finish, we just need to get rid of those × symbols.

12a5b3



June 2007 C1 OCR MEI



As usual, we get rid of the fraction first. You can do that by multiplying both sides by 2.

2s = at2


As both a and t are squared, we're going to square root both sides.

±√2s = at


All that's left to do now is take away the a from the right-hand side.

±√2sa = t
t = ±√2s
a



June 2007 C1 OCR MEI



This is a quite basic question to solve. First we're going to take the 2x from underneath the left-hand side and take it over.

4x + 5 = -3 × 2x
4x + 5 = -6x


Now it's just a matter of swapping over the -6x and the +5.

4x + 6x = -5
10x = -5
x = -510



January 2008 C1 OCR MEI



As we've seen previously, we can start by getting rid of the fraction - multiplying both sides by 2.

2E = mv2


Next, we're going to get the square root of both sides as both m and v are squared.

±√2E = mv
All that's left to do is get v on it's own.

±√2Em = v

January 2008 C1 OCR MEI




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