FN142 Marking Scheme
May/June 2013
Question 1
Simplify each expression.
a)
2  ( 4a  5a )  2a = -2a +2a = 0
b)
6a  2b
4ab
= (b-3ab) / 2ab = 1/2a – 3/2b

8
(2a  10a )
c)
8a 3b  4a 3b 2  2a 2
= 4ab + 2ab2 -1
2a
2
 ( x  1) 2 ( x  1) 5 
 = (x + 2)6 / ((x + 1)4(x - 1)10)
d) 
3
(
x

2
)


e)
4
x x=
4
x 3 or x3/4
(Total 5 marks)
Question 2
a) Indicate an order of operations; clearly write the relevant number into
7
2 5 0
 ( )  (3  8  2  4  (6  5  21 )) = 6.4
5 2
13, 11, 12, 9, 10, 14, 6, 5, 7, 4, 8, 3, 2, 1.
(2 marks)
b) Round off:
i.
To the nearest thousandth
3.249831 = 3.250
ii.
To 2 decimal places (2 d. p.)
1.31591 = 1.32
(2 marks)
c) State the reciprocal of:
 5 => - 1/5
i.
ii.
2
1
=> 4/9
4
(2 marks)
(Total 6 marks)
Question 3
a) Two friends have started a business. They have agreed to share all profits and losses in the
same ratio as they have invested their money to buy stock and premises. Matt has invested
£180,000 and Will has invested £120,000.
How would they share a profit of £70,000?
180,000 : 120,000 = 3:2 (5 parts in total)
70,000 x 3/5 = £42,000 (Matt)
70,000 x 2/5 = £28,000 (Will)
b) A third friend, Harry, has joined the partnership on the same terms about sharing profits and
losses and invested £300,000.
How would 3 of them share a loss of £18,080?
180,000 : 120,000 : 300,000 = 3 : 2 : 5 (10 parts in total)
18,080 x 3/10 = £5,424 (Matt)
18,080 x 2/10 = £3,616 (Will)
18,080 x 5/10 = £9,040 (Harry)
(Total 6 marks)
Question 4
a) There is a maximum possible mark of 60 points in an exam. You need 70% to get an A grade.
How many points do you need? 70 x 60 /100 = 42 marks
(1 mark)
b) 75 per cent of what amount is equal to £30? 30 x 100 / 75 = £40
(1 mark)
c) Express:
i.
ii.
7
as a percentage => 43.75%
16
162% as the sum of a whole number and a proper fraction. => 1 + 31/50
(2 marks)
(Total 4 marks)
Question 5
Use “<”, “=”, “>” to compare the following offers:
Insert a suitable symbol:
“<” or “=” or ”>”
a)
Original price £40.
50% OFF!
b) Bargain!
Was £10 but… take 20% off!
Was £30.
Now 1/3 cheaper!
Original price £11.
Now half-price!
a) £20 = £20
b) £8 > £5.5
(Total 4 marks)
Question 6
a) Write an appropriate equation and solve the problem below.
At what price should a dealer mark a car for which he paid £15,000 in order to offer a
discount of 10% on the marked price while still making a profit of 20% on the sale price?
Let x be the marked price of a car. Since the sale price is 90% of a marked price (i.e. 10%
discount), the sale price is 0.9x. Since the profit is 20% of the sale price, the cost is
therefore 80% of the sale price. As cost is £15,000, then 15000 = 0.8 (0.9x).
X = £20,833.33
(4 marks)
b) What is the percentage change from £15,000 to the sale price? Round your answer to 2 d. p.
(20833.33-15000) / 15000 = 0.3889 or +38.89%
(1 mark)
c) What would be the percentage change if the sale price from part (a) was marked down to
the cost of £15,000? Round your answer to 2 d. p.
(15000-20833.33) / 20833.33 = - 0.28 or -28%
(1 mark)
(Total 6 marks)
Question 7
Express each of the following statements algebraically or as an equation, whichever applicable
a) P minus twice the sum of x and y. => p – 2(x + y)
b) Twice what number added to six is equal to twelve. => 2x + 6 = 12
c) How old is the person now if ten years ago the person was twenty five. => x – 10 = 25
(Total 3 marks)
Question 8
Solve the following system of equations:
x  y  z  4

x  2 y  z  1
2 x  y  2 z  1

Multiply 1st equation by -1 and add it to the 2nd equation to eliminate x. => -3y – 2z = -3
Multiply 1st equation be -2 and add to the 3rd equation to eliminate x. => -3y -4z = -9
Eliminate y from the 3rd equation by multiplying 2nd equation by -1 and adding it to the 3rd
equation. => -2z = -6. Solve for z: z = 3
Substitute 3 for z in the 2nd equation and solve for y. => -3y -6 = -3. y = -1.
Substitute -1 for y and 3 for z in the 1st equation and solve for x. => x +(-1) + 3 = 4. X=2
Answer: (2, -1, 3).
(Total 6 marks)
Question 9
a) Graph a line 3x + 2y = 6 by finding its x- and y- intercepts
If Y =0, x= 2; if x=0, y = 3. Therefore, this line passes through (2,0) and (0,3)
Y
(0, 3)
(2, 0)
X
(5 marks)
b) State the gradient of the line above
Y = - 3/2x + 3, therefore gradient = -3/2
(2 marks)
c) Find the equation and the gradient of the line that passes through the points:
(-2, -3) and (2, -1)
 3  2a  b
=> b = -2; a = 0.5 (gradient);

  1  2a  b
Y = 0.5x -2
(6 marks)
d) Algebraically find a point of interception of lines from part (a) and (c).
0.5x – 2 = -1.5x + 3, x= 2.5 and y = -0.75
(4 marks)
(Total 17 marks)
Question 10
a) £1,500 is borrowed at 5% per month simple interest for a period of 6 months. Calculate the
total amount that the borrower will have to pay back by the end of the period.
1500 + 1500 x 5 x 7 / 100 = £2,025
(4 marks)
b) At what rate of simple interest will £3,500 amount to £3,800 in 2 years?
3800 = 3500 + (3500 x R x 2) / 100; R=4.29%
(4 marks)
c) How long will it take for £4,000 to triple itself at 7% simple interest?
12000 = 4000 + (4000 x 7 x T) / 100. T= 28.57
(4 marks)
d) The monthly payment on a £100,000, 25-year mortgage at 8.5% is £805.30.
Interest is always charged on the outstanding balance. How much of the first 2 payments go
to interest and how much to principal.
100000 x 0.085 x 1/12 = 708.33 (interest)
805.30 – 708.33 = 96.97 (payment to principal)
100000-96.97 = 99903.03 (balance owed on the loan after 1st month)
99903.03 x 0.085 x 1/12 = 707.65 (interest)
805.30 - 707.65 = 97.65 (payment to principal)
Therefore, the first two payments totalling £1,610.60 buy £194.62 worth of the property
while the remaining £1,415.98 is interest.
(5 marks)
(Total 17 marks)
Question 11
a) A piece of property can be purchased for £28,500 cash or for £30,000 in 12 months.
i.
Which is better plan for the buyer if money is worth 7% compounded quarterly?
30000 / (1.07/4)4 = 27,988.76 present value of 30,000
27,988.76 is less than 28,500 cash payment, so it is better to pay later
(5 marks)
ii.
Give the present value equivalent of the savings made by adopting the better plan
28,500 – 27,988.76 = £511.24
The difference is relatively small, thus an increase in the interest rate can lead to
an opposite decision.
(3 marks)
b) Equipment costs £5,000. The depreciation any year is estimated to be 12% of the value at
the beginning of that year. What is the estimated value of the equipment after 3 years of
use? 5000 (1-0.12)3 = £3,407.36
(4 marks)
(Total 12 marks)
Question 12
a) Choose a number at random from 1 to 5.
i.
What is the probability of each outcome? 1/5
ii.
What is the probability that the number chosen is even? 2/5
iii.
What is the probability that the number chosen is odd? 3/5
(3 marks)
b) A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen
at random from the jar,
i.
What is the probability of choosing a red marble? 6/22 or 3/11
ii.
What is the probability of NOT choosing a blue marble? (22-8) / 22 = 7/11
(2 marks)
(Total 5 marks)
Question 13
Investment portfolio, 2011
Share
Price (£)
Investment portfolio, 2012
Quantity
Price (£)
Quantity
A
5.06
231
6.11
316
B
3.22
312
5.01
276
C
27.19
100
22.32
50
D
11.58
358
10.05
311
E
2.40
296
4.30
412
F
15.88
250
15.21
260
Using data above, calculate the Laspeyre (a) and Paashe (b) index values for 2012.
a) 13879.73 / 13718.54 x 100 = 101.17
b) 13281.27 / 12566.16 x 100 = 105.69
(Total 6 marks)
Question 14
Using information from the question 13 above, compute:
a. MEAN and RANGE for share prices in 2011 and 2012.
Mean (2011): 65.33 / 6 =10.89; Range (2011): 27.19 – 2.4 = 24.79
Mean (2012): 63 / 6 = 10.5; Range (2012): 22.32 – 4.30 = 18.02
b. Comment on your findings.
(Total 3 marks)