Solutions to practice questions
Clement Radcliffe, Contributor

I hope you have experienced little, if any, difficulty with the practice questions presented the last time around. The following are the solutions which I hope will assist you with your review process.

1. 0.625 written as a common factor is

(A) 3/5
(B) 5/8
(C) 11/16
(D) 7/8

Solution: 0.625 = 625/1000 = 5/8 .Answer, therefore is (B)

2. Expressed in standard form, 0.003 68 =

(A) 3.68 x 10-3
(B) 3.68 x 10-²
(C) 3.68 x 10²
(D) 3.68 x 10³

Solution: 0.003 68 = 3.68 x 10-3. Answer is (A)

3. A quadrilateral whose diagonals are perpendicular to each other, but unequal is a

(A) rhombus
(B) trapezium
(C) rectangle
(D) square

Solution: Answer is (A)

4. 6p/5q + 4r/3s =

(A) 10pr/8qs
(B) 24pr/8qs
(C) 6ps + 4 qr/15qs
(D) 18ps + 20 qr/15qs

Solution: 6p/5q + 4r/3s = 18ps + 20qr/15qs Answer is (D)

5. In the figure below, angle BAC = 90º, angle ACB = 37º and BC = 10cm. The length of AC, in centimetres, is

(A) 10 sin 37º
(B) 10 cos 37º
(C) 10 tan 63º
(D) 10 cos 63º

Solution: Using the cosine ratio, cos 37º = AC/10 ... AC = 10 cos 37º. Answer is (B)

6. (-3)² + (-2)³ =

(A) -17
(B)0
(C)1
(D)12

Solution: (-3)² + (-2)³ = 9 -8 = 1. Answer is (C)

7. The original price of an article was $240. The price is increased by 121/2 %. The new price of the article is

(A) $210.00
(B) $228.50
(C) $252.50
(D) $270.00

Solution: Since cost price is $240 and the price is increased by 121/2 %, 112.5% of $240 = 225/200 x 240 = $270. Answer is (D)

8. 5(2x - y) - 2(3y - 5x) =

(A) -11y
(B) 2x-6y
(C) 5x-7y
(D) 20x -11y

Solution: 5(2x - y) - 2(3y - 5x) = 10x - 5y - 6y + 10x = 20x -11y Answer is (D)

9. (a) Given a = 2, b = -3 and c = 0, evaluate

(i) 4a - 2b + 3c (ii) ac

Solution: (i) 4a - 2b + 3c = 4 x 2 - 2 x -3 + 3 x 0 = 8 + 6 = 14

(ii) ac = 2º = 1

(b) Factorise completely,

(i) 7mp² + 14m²p (ii) 2y² - 11y + 15

Solution: (i) 7mp² + 14m²p = 7mp(p + 2m)

(ii) 2y² - 11y + 15 = (2y - 5)( y - 3)

(c) Write as a simple fraction in its lowest terms

2/a - 3 + 3/a

Solution: Using the LCM of the denominators

2a + 3( a - 3)/a(a - 3) = 2a + 3a - 9/a(a - 3) = 5a - 9/a(a - 3)

(d) (i) Solve for x

12 < 3x + 5

Solution: 12 < 3x + 5

3x > 12 - 5, x/3 > 7

(ii) If x is a member of the set of whole numbers, state the smallest value of x which satisfies the inequality in (d) (i) above.

Solution: x/3 > 7

Since x is the set of whole numbers, then x is 3.

10. P is the point (2, 4) and Q is the point (6, 10)

Calculate: (i) the gradient of PQ (ii) the mid-point of PQ

Solution: Given P (2, 4) and Q (6, 10)

(i) the gradient of PQ = y2 - y1/x2 - x1 Substituting

m = 10 - 4/6 - 2 = 6/4 = 3/2

(ii) the mid-point of M = x2 + x1/2, y2 + y1/2 Substituting

= 6 + 2/2 , 10 + 4/2 = ( 4 , 7 )

11. f and g are functions defined as follows:

f: x 7x + 4

g: x 1/2x

Calculate: (i)g(3) (ii)f(-2) (iii) f-1(11)

Solution: (i)Since g: x 1/2x , g(x) = 1/2x

ie. g (3) = 1/6

(ii) Since f: x- 7x + 4 ie. f(x) = 7x + 4

... f(-2) = 7 x - 2 + 4 = -10 ... f(-4) = -10

(iii) Since y = f(x) = 7x + 4 y = 7x + 4

y - 4 = 7x; x = y - 4/7 Interchanging y for x

ie. y = x - 4/7

ie. f -1(x) = x - 4/7

ie. f -1(11) = 11 - 4/7 = 1 f -1(11) = 1

12. Mr. Mitchell deposited $40,000 into a bank and earned simple interest at 7 % per annum for two years.

(i) Calculate the amount he will achieve at the end of the two-year period.

(ii) Mr. Williams bought a plot of land for $40,000. The value of the land appreciated by 7 % each year.

Calculate the value of the land after a period of two years.

Solution: (i) Principal = $ 40,000, at rate at 7 % for2 years

Simple interest = 40,000 x 2 x 7/100 = $ 5, 600

ie. The amount earned = principal + simple interest = $40,000 + $5, 600 = $45, 6000

(ii) Cost of the land is $40,000.

Since it appreciated by 7 % each year, the value at the end of the first year is 40,000 x 107/100 = $ 42,800.

(Alternatively, you could find 7% of $40,000 = $2,800 and add it to $40,000)

Since the value at the start of the second year is $ 42,800

Value at the end of the year = 42, 800 x 107/100 = $45,796

Have you noted the difference between simple interest 6

(i) and compound interest in 6(ii).

13. If h(x) = 1 + 3x and k(x) = x + 2, calculate

(i) hk(x) (ii) hk(4) (iii) (hk)-1(x) (iv) the value of x, when hk(x) = 0

Solution: Since h(x) = 1 + 3x and k(x) = x + 2

(i)ie. hk(x) = h(x + 2) NB x + 2 replaces x in h(x)

ie. hk(x) = 1 + 3(x + 2) = 1 + 3x + 6 = 3x + 7

ie. hk(x) = 3x + 7

(ii) hk(4) = 3 x 4 + 7 = 19

(iii) Since hk(x) = y = 3x + 7

ie. 3x = y - 7

ie. x = y - 7/3 Interchanging y for x

ie.y = x - 7/3

(hk)-1(x) = x - 7/3

(iv) Given that hk(x) = 0

ie. 3x + 7 = 0 ie. x = -7/3

14. Solve the simultaneous equations:

5x + 6y = 37
2x - 3y = 4

Solution: Given the equations 5x + 6y = 37 . . . . . (1)
2x - 3y = 4 . . . . . . (2)

Multiply equation (2) by 2.

4x - 6y = 8 . . . . . . (3) Add (1) and (3)
9x = 45
x = 5 Substituting in equation (2)

ie. 10 - 3y = 4
ie. 3y = 10 -4 = 6
ie. y = 2
Answer: x = 5, y = 2

15. Given the formula s = 1/2 (u + v)t, express u in terms of v, s and t.

Solution: ...2s = ut+ vt Multiply both sides by 2

ie. u = 2s - vt/t

Now, it is for you now to do additional work from your textbooks and past papers. This practice will ensure that you are adequately prepared for the external examination.

Clement Radcliffe is the principal of Glenmuir High School in May Pen.