A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.

Answer

1st year Feb 2011 – 2012 @ 30%

= 100-30 /100 x 900,000 = N630,000

2nd yr Feb 2012 – 2013

=100-22/100 x 630,000 = 491,400

3rd yr Feb 2013 – 2014

= 100-22/100 x 491,400 = 298.967.76

Value of the car on 2015 is N299,000 (to the nearest hundred naira)

^{2}−p

^{2}x−14 passes through the point (3, 10). Find the values of p. (b) Two lines, 3y−2x = 21 and 4y + 5x = 5 intersect at the point Q. Find the coordinates of Q. Answer (a) 2px

^{2}−p

^{2}x−14 At point (3, 10), y = 10 when x = 3. ⟹10=2p(32)−p2(3)−14 10=18p−3p

^{2}−14 −10=3p

^{2}−18p+14⟹3p2−18p+24=0 3p

^{2}−12p−6p+24=0 (p−4)(3p−6)=0⟹p = 2 or 4. (b) 3y−2x=21…(1) 4y+5x=5….(2) Using elimination method, multiply (1) by 4 and (2) by 3. (1)×4:12y−8x=84…(3) (2)×3:12y+15x=15…(4) Subtracting (3) – (4), we have: −23x=69⟹x=−3 Putting x = -3 in (1), we have 3y−2(−3)=3y+6=21 3y=15⟹y=5 Hence, the coordinates of Q are (-3, 5).

(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.

(b) Given that sinx = 3/5, 0°<x<90°, find the value of 5cosx−4tanx.

Answer

(a)

Let a side of the rhombus be n.

Using Pythogoras

a^{2 }= 4.65^{2 }+ 5.1^{2} = 21.6225+26.01

a^{2} = 47.6325

a= √47.6325 = 6.902cm

Hence, the perimeter of the rhombus = 4a

= 4×6.902

= 27.608cm ≊ 27.6cm (to 1 d.p)

(b) Given that sinx = 3/5, 0°<x<90°, find the value of 5cosx−4tanx.

sinx = 3 / 5, Then , a^{2 }= 5^{2 }− 3^{2 }= 25−9 = 16

a = 4.

cosx = 4 / 5 ⟹ 5cosx =5 × 4 /5 = 4

tanx = 3/ 4 ⟹ 4tanx = 4 × 3/4 = 3

5cosx − 4tanx = 4−3 = 1.

(a)

In the diagram, Q0S is a diameter o and o .

Find : (i) the value of x (ii) <RSQ

(b) If 2N4_{seven} =15N_{nine}, find the value of N.

Answer

(a) x°+90° = (3x+15)° (Sum of 2 opposite interior angles equal to exterior angle)

Thus, 90°−15° = 3x − x ⟹ 75° = 2x

x = 37.5°

(b) 0 (sum of angles in a straight line )

0 = 180^{0}

0

0

0 ^{}

= 52.50

(b) 2N4_{seven }= 15N_{nine}

2N4seven = (2×7^{2}) + (N×7^{1}) + (4×7^{0})

= 98 + 7N + 4

= 102 + 7N

15Nnine = (1×9^{2}) + (5×9^{1}) + (N×9^{0})

= 81 + 45 + N

= 126 + N

⟹102 + 7N = 126 + N

7N−N = 126−102 = 24

6N = 24 ⟹ N = 4

Therefore, 244seven=154nine

(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n – 3), (s + 6), (p – 2) and (q + 8).

(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?

Answer

(a)

In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only – 28, Clutch and Steering – 14; Clutch, Steering and Brakes – 8; Clutch and Brakes – 20; Brakes and Steering only – 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.

(a) Draw a Venn Diagram to illustrate this information.

(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?

Answer

If 60% passed , then 40% failed

µ = 100-60 /100 X 240 = 96 car failed .

Clutch , steering and brake = 8

Clutch only = 28

Clutch and steering = 14

Clutch and Steering only = 14 -8 = 6

Clutch and brake = 20

Clutch and brakes only = 20 – 8 = 12

Brakes and steering only = 6

Let the number of cars with fault in brakes only be x

x + 28 + 2x + 12 + 6 + 6 + 8 = 96 ; 3x + 60 = 96

3x = 96 – 60 ; 3x = 36 ;

x = 12

(i) then No of cars with faulty brakes = 12 + 12 + 8 + 6 = 38

(ii) Only one fault?

= brake only + clutch only + steering only = x + 2x + 28

12 + 2(12) + 28 = 64 cars

So 64cars had only one fault

(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).

(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,

(i) Illustrate the information in a diagram.

(ii) Calculate, correct to three significant figures, the:

(1) distance between Q and R

(2) bearing of R from Q.

Answer

(a) line passing through the points (2, 5) and (-4, -7).

let (2, 5) = ( x1, y1 ) and (-4, -7) = ( x2, y2 )

Apply m = y- y_{1} /x- x_{1} ;

y_{2}– y_{1} / x_{2} -x_{1} = y – y1 / x-x_{1}

-7 – 5 /- 4 -2 = y – 5 /x – 2 ; -12 / -6 = y – 5/ x -2 = 2 /1 = y – 5 /x -2

cross multiply

2(x – 2) = 1(y – 5) ; 2x -4 = y -5

2x – y = -5 + 4 ; 2x – y = -1

2x + y + 1 = 0 or y = 2x + 1

(b)

In ΔPQR |QP|2 = |PQ|2 + |PR|2 = 52 + 82

|QP|2 = 25 + 64 = 89 ; |QP| = √89 = 9.43km

|QP| = 9.43km

The bearing of R from Q

Using Sine Rule —– Sin C /8 = Sin90/9.43 ;

Sin C = 0.8483 ; C = Sin-1 0.8483 ; C = ≅58o

Bearing of R from Q = 360 -90 -60 -58 = 152

Bearing of R from Q = 152°

^{ }

(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid N(6x+34)

“>N(6x+34)

N(6x+34) more for the book than Lamin paid, find the value of x.

(b) Find the range of values of x which satisfies the inequality 3x−2<10+x<2+5x

Answer

(a)

(b) 3x -2 < 10 + x <2 + 5x ; 3x -2 < 10 + x

3x – x , 10 + 2 ; 2x < 12 ; x < 12 /2 = 6

10 + x < 2 + 5x ; 10 – 2 < 5x -x ; 8 < 4x ; 8/4 < x

2 < x ; 2 < x < 6

(a) In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:

(a) |SR| ;

(b) area of TQRS

Answer

(a) Using similar triangle

4 /6+4 = 6 /QR +6

Cross multiply 4QR + 24 = 60

4QR = 60 -24 = 36 ; QR = 36/4

QR = 9cm

Using Cosine rule to get |SR|

|SR|^{2} = |ST|2 + |RT|2 – 2|ST| |RT|2 Cos T

|SR|2 = 102 + 152 – 2(10)(15)cos30°

|SR|2 = 100 + 225 -300cos30o

|SR|2 = 325 – 300(0.866025)

|SR|2 |= 325 – 259.801 ; SR = 166 ; SR = 8.12404 = ≅ 8cm

(b) |TQ|2 = 42 +62 −2 × 4 × 6 × cos30°

= 16 + 36 −48 × 0.8660

= 52 − 41.568

=10.432

∴|TQ| = √10.432 ≊ 3.23cm

continues

(a)

In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.

(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.

(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.

Answer

(a)

In triangle SPQ, |SQ|2= 52 + 122 (Pythagoras theorem)

= 25 +144 = 169

|SQ| = √169 = 13cm

Angle b is common to triangles SPQ and PRS are similar.

Using sinb=12 / |SQ| = |PR| / 5

sinb=12 / 13 = |PR| / 5

|PR|=12×5 / 13 ≊ 4.62cm (to 3 s.f)

(b)

h2 =102 − 82 = 36

h = √36 = 6cm

(ii) In the smaller triangle, cosx = 8 / 10 = 0.8

cos−1(0.8) = 36.87°

Since these are corresponding angles, x = x in the bigger triangle.

sinx = y / 12

y =12sinx = 12 sin 36.87

= 12 × 0.6

= 7.20 m

The ladder M touches the wall at a height 7.2 m above the ground.

(a) Copy and complete the table of values for y = 2x^{2} + x – 10 for -5 ≤ x ≤ 4.

(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of y = 2x2 + x – 10 for -5 ≤ x ≤ 4.

(c) Use the graph to find the solution of :

(i) 2x2 + x = 10

(ii) 2x2 + x – 10 = 2x

Answer

(a)

(c) Using the graph to find the solution

(i) 2x2 + x = 10 ; 2x2 + x – 10 = 0

Thus, the solution of 2x2 + x – 10 = 0 are the values of x at which the curve cuts the x- axis.

x = -2.5 or x = 2.

(ii) 2x2 + x – 10 = 2x

To solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.

The values of x at which y = 2x2 + x – 10 and y = 2x intersect give the solution of 2x2 + x – 10 = 2x.

These are x = -2 or x = 2.5.

(b) (i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for −5≤x≤5,−5≤y≤5 respectively.

(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,

(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).

(2) the image W_{1}X_{1}Y_{1}Z_{1} of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where

W → W_{1},X→X_{1},Y→Y_{1} and Z→Z_{1}.

Answer

(a) px + qy = z

then

2p+5q=−4…..(1)

3p−2q=13……(2)

Solving the equations simultaneously , we have p = 3 and q = -2.

(b)

(ii) See graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2).

(iii) See image for the quadrilateral WXYZ under an anticlockwise rotation of 90° is

w(2,3)————W1(−3,2),

x(4,-1)———— X1(1,4),

y(-3,-4)———–Y1(4,−3)

z(-3,2)———– Z1(−2,−3).

The frequency distribution shows the marks distribution of a class of 30 students in an examination.

The mean mark of the distribution is 52.

(a) Find the values of x and y.

(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.

(c) Draw a histogram for the distribution

(d) Use the histogram to estimate the mode.

Answer

(a)

Mean (x) = ∑fx / ∑f = 900 + 30x + 50y / 16 + x + y

But (x) = 52

52 = 900 + 30x + 50y / 16 + x + y

cross multiply

52 (16 + x + y) = 900 + 30x + 50y

832 + 52x + 52y = 900 + 30x + 50y

52x-30x + 52y – 50y + 900 – 832

22x + 2y = 68

11x + y = 34 ——-(i)

but total frequency = 30 (given in d question)

16 + x + y = 30

x + y = 14———(ii)

solv (i) and (ii) simultaneously

10x = 34- 14 ; x = 2

sub x=2 in equ (ii)

x + y = 14 ; 2 + y = 14 ; y = 12

hence x = 2 , y = 12

(b)

(c) Draw a histogram

(d) from the histogram

Mode = 44