QUESTIONS
Answer ALL the questions in this section.
- Use logarithms to evaluate:
∛45.3 × 0.00697(3 marks)
0.534 -
- Expand (1-½x)6 to fourth term. (2 marks)
- Use the expansion above to evaluate (0.98)6 (2 marks)
- The price of a new car is shs. 800,000. If it depreciates at a constant rate to shs. 550,000 within 4 years, find the annual rate of depreciation. (3 marks)
- Object A of the area 10cm2 is mapped onto its image B of area 60cm2 by a transformation whose matrix is given by P = (x3 4 x+3). Find the positive values of x. (3 marks)
- Without using a calculator or mathematical tables, express √3 in surd form and simplify. (3 marks)
1-Cos30º - Pipe A can fill a tank in 2 hours, Pipe B and C can empty the tank in 5 hours and 6 hours respectively. How long would it take:
- To fill the tank if A and B are left open and C is closed. (2 marks)
- To fill the tank with all pipes open. (2 marks)
- The position vector of A and B are a = 4i + 4j – 6k and b = 10i + 4j + 12k. D is a point on AB such that AD:DB is 2:1. Find the co-ordinates of D. (2 marks)
- Given that y is inversely proportional to xn and k is the constant of proportionality and that x=2, when y=4.5 , and x=3, when y=4/3. Find the values of n and k. (4marks)
- The coordinates of the end points of diameter are A(2,4) B(-2,6). Find the equation of a circle in the form ax2 + by2 +cx + dy + e = 0 (3 marks)
- Given that the mean of 9, 8, 5, 5 and 8 is 7; find the standard deviation of the numbers to 2 d.p (3 marks)
- A vendor mixed grade 1 rice and grade 2 rice in the ratio 1:3 to form a mixture which she sold at sh.105 making a profit of 40%.Given that the cost price of grade 2 rice is sh.80 per kg. Find the cost price of 1kg grade 1 rice. (3marks)
- Chord WX and YZ intersect externally at Q. The secant WQ =11cm and QX =6cm while ZQ=4cm
- Calculate the length of chord YZ (2 marks)
- Find the length of the tangent SQ (2 marks)
- Given that x, y and z are integers and that 8 ≤ x ≤ 10, 5 ≤ y ≤ 7, 4≤ z ≤ 6.
Find the percentage error in x+y/z (3 marks) - Solve the equation below by completing the square. 5-9x-2x2=0 (2 marks)
- Solve the simultaneous equations
2x - y = 3
x2 –xy = -4 (3marks) - Solve the equation
Log2 (2+3x) +3log22 = 2 + log2 (2x+6) (3marks)
ANSWER ANY FIVE QUESTIONS IN THIS SECTION
- Two fair dice one a regular tetrahedron (4 faces) and the other a cube are thrown. The scores are added together.
- Draw a table to show all possible outcomes. (2 marks)
- Find the probability that:
- The sum is 6. (1 mark)
- The sum is an odd number. (1 mark)
- The sum is 6 or 9. (2 marks)
- If a player wins a game by throwing a sum of 6 or 9, draw a tree diagram and use it to find probability that he wins at least once when the dice are thrown twice. (4 marks)
- The Hire Purchase (H.P) price of a public address system was Ksh 448 000. A deposit of Ksh 112 000 was paid followed by 24 equal monthly instalments. The cash price of the public address system was 15% less than the H.P price.
- Calculate :
- The monthly instalment. (2 marks)
- The cash price. (2marks)
- A customer decided to buy the system in cash and was allowed an 8% discount on the cash price. He took a bank loan to buy the system in cash. The bank charged compound interest on the loan at rate of 16% p.a. compounded quarterly. The loan was repaid in 2years. Calculate the amount repaid to the bank by the end of the 2 years. (3 marks)
- Express as a percentage of the Hire Purchase price, the difference between the amounts repaid to the bank and the Hire Purchase price. (3 marks)
- Calculate :
- The vertices of a rectangle ABCD are ( -1, 1) , (1, 1) ,(1, 4 )and ( -1, 4 )respectively. The vertices of its image under transformation T areA’ (1,1), B’(3,1) C’ (9, 4) and D’(7, 4)
- Draw on the grid provided rectangle ABCD and its image A’B’C’D’ under the transformation T. (2 marks)
- Describe fully the transformation (2mks)
- Determine the matrix of transformation(3mks)
- On the grid as in (a), draw rectangle A”B”C”D”, the image of rectangle ABCD under a stretch with line y=1 invariant and stretch factor 2. State the coordinates ofA”B”C”D” (3mks)
- An arithmetic progression of 41 terms is such that the sum of the first five terms is 560 and the sum of the last five terms is -250. Find:
- The first term and the common difference (5marks)
- The last term (2marks)
- The sum of the progression (3 marks)
- The table below shows the masses measured to the nearest Kg of 100 people.
Mass kg 40-49 50-59 60-69 70-79 80-89 90-99 100-109 No of people 9 19 22 18 13 11 8 - Draw a cumulative frequency curve for the data above. (4 marks)
- Use your graph to estimate
- The median mass. (1 mark)
- The number of people whose mass lies between 70.5 kg and 75.5 kg (1 mark)
- From your graph find
- The lower quartile (1 mark)
- the upper quartile (1 mark)
- the interquartile range (2 marks)
- In the triangle PQR below L and M are points on PQ and QR respectively such that PL: LQ = 1:3 and QM: MR = 1:2, PM and RL intersect at X. Given that PQ = b and PR=c,
- Express the following vectors in terms of b and c.
- QR (1mk)
- PM (1mk)
- RL (1mk)
- By taking PX = hPM and RX = kRL where handk are constants find two expressions of PX in terms of h, k, b and c. Hence determine the values of the constantsh and k. (6mks)
- Determine the ratio LX : XR (1mk)
- Express the following vectors in terms of b and c.
- In the figure below, K L M and N are points on the circumference of a circle centre O. The points K, O, M and P are on a straight line. PQ is a tangent to the circle at N.Angle KOL = 130º and angle MKN = 40º
Find the values of the following angles, stating the reasons in each case:- <MLN (2mks)
- <OLN (2mks)
- < LNP (2mks)
- <MPQ (2mks)
- <KNQ (2mks)
- Three quantities x, y and z are such that x varies directly as the square of y and inversely as the square root of z.
-
- Given that x = 12, and y= 24, and z = 36, find x when y = 27 and z = 121 (3 marks)
- If y increases by 5% and z decreases by 19%,find the percentage increase in x (4 marks)
- If y is inversely proportional to the square root of x and that x = 4 when y = 3, calculate the value of x when y=8. (3 marks)
-
MARKING SCHEME
No Log 4.53 x 101
6.97 x 10-3
5.34 x 10-11.6561 +
3.8432
T 4993 -
T.7275
T 7718 ÷ 3-
- 16 + 6(1)5 (-x) + 15(1)4(x)2 + 20(1)2(x)3
1-3x + 15/4 x2 -5/2x3 - x=0.04
1 - 3 x 0.04 + 15/4 x 0.042 - 5/2 x 0.043
= 0.88584
- 16 + 6(1)5 (-x) + 15(1)4(x)2 + 20(1)2(x)3
- 550000 = 800000 (1 - r/100)4
0.6875 = (1-r/100)4
0.91058 =
r/100 = 0.08942
r = 8.942% - x(x+3) - 12 = 6
x2 + 3x -18 = 0
x2 - 3x + 6x - 18 = 0
x(x -3) +6(x-3)=0
(x + 6)(x - 3)=0
x = -6 or x = 3
√3
1-√3/2
√3 x 1 +√3/2
1-√3/2 1+√3/2
√3 + 1.5
1 - ¾
√3 + 1.5
1 - ¼
4√3 + 6-
- ½ - 1/5 = 3/10
1hr = 3/10
10/3 hrs = 31/3hrs - ½ - 1/5 - 1/6 = 2/15
it takes 7.5hrs
- ½ - 1/5 = 3/10
-
- y = k/xn
4.5 = k/2n
4.5(2n) = k
= k/3n
(3n) = k
4.5(2n) = (3n)
3.375 x 2n = 3n
log 3.375 + nlog2 = nlog3
0.52823777 = 0.17607125
n = 3
k = 36
OR
4 x 2n = 11/3 x 3n
(2/3)n = (2/3)3
n = 3
k = 36 - Centre = (2t - 2 , 4 + 6)
2 2
(0,5)
Radius = √(2 - 0)2 + (4 - 5)2
= √4 + 1 = 5
(x - 0)2 + (y - 5)2 = 5
x2 + y2 - log + 25 = 5
x2 + y2 - log + 20 = 0
x f d fd2 9
8
51
2
22
1
-24
2
8
14
= 1.67- Cost of mixture = 105 x 100/140 = sh75
x + 80 x 3 = 75
4
x = 60 -
- (y + 4)4 = 11 x 6
4y = 66-16
YZ = 12.5cm - Q = √6 x 11
= 8.124cm
- (y + 4)4 = 11 x 6
- Max quot. = 4.25
Min quot. 8+5 =21/6
6
E = 4.25 - 21/6 = 11/24
2
Actual quot. = 9+5 = 3
5
25/24 x 100
3
= 34.7222% - x2 + 4.5 = 2.5
(x + 2.25)2 = 2.5 + 5.0625
x = 2.75 - 2.25 = 0.5
or x = -5
x = 0.5
x = -5 - x2 - x(2 x - 3) = -4
x2 - 2x2 + 3x = -4
-x + 3x + 4 = 0
-x + 4x - x + 4 = 0
-x(x - 4) - 1(x - 4) = 0
(x - 4)(-x - 1) = 0
x = 4
x = -1
when x = 4
y = 5
when x = -1
y = -5 - log2(4 + 6x) = log2 2(2x + 6)
4 + 6x = 4x + 12
2x = 8
x = 4 -
-
1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 -
- 4/24 = 1/6
- 12/24 = ½
- 4/24 + 2/24 = 6/24= ¼
1 - ¾ x ¾ = 7/16
-
-
-
- 448000 - 112000 = sh14,000
245 - 85/100 x 448000 = sh380000
- 448000 - 112000 = sh14,000
- CP 92/100 x 380000 = 350336
A = 350336(1 + 16/100)4
= sh409843.5685 - (448000 - 409843.5685) x 100
448000
= 8.517%
-
-
-
- It is a shear x - axis invariant
D(-1, 4) → D1(7, 4) - (10 k1) (-14) = (74)
-1 + 4k = 7
k = 2
(10 21) - A11(-1, 1) B11(1, 1) C11(1, 7) D11(-1, 7)
-
-
- 560 = 5/2(2a + 4d)
1120 = 10a + 20d
a + 2d = 112
a + 40d + a + 39d + a + 38d + a + 37d + a + 36d = -250
5a + 190d = -250
a + 38d = -50
a + 2d = 112
36d = -162
d = -4.5
a = 121 - 121 + 40 x -4.5
= -59 - S41 = 41/2(121 - 59) = 1271
OR
41/2(2 x 121 + 40x - 4.5) = 1271
- 560 = 5/2(2a + 4d)
C.F 9 28 50 68 81 92 100 Mass kg 40-49 50-59 60-69 70-79 80-89 90-99 100-109 No of people 9 19 22 18 13 11 8 -
-
- 69.5
- 61 - 52 = 9
-
- 57.5
- 84.5
- 84.5 - 57.5
= 27
-
-
- -b + c
- 2/3b + 1/3c
- -c + 1/3b
- px = h(2/3b + 1/3c)
= c + k(-c + 1/3b)
c - kc + 1/3kb
2/3hb + 1/3hc = c - kc + 1/3kb
1/3h = 1 - k
1/3h= 1-2h
21/3h = 1
h = 3/7
k = 6/7 - 1:6
-
-
-
- 40º angles subtended by same arc MN
- LOM = 50º Angles on a straight line
LMK = 65º Angles on circumfrence a chord is half that at the centre - OLN = 25º
NKL = LNP = 65º Angles in alternate segment are equal - 10º sum of angles in a triangle
- 50º angles on a straight line add up to 180º
-
-
-
- x = ky2
√z
12 = 242k
6
12 = 96k
k = 1/8
x = 0.125y2
√z
x = 0.125 x 272
√121
x = 8 25/88 - x = k(1.05y)2
√0.81z
x = 1.1025ky2
0.9√z
x = 1.225ky2
√z
%increase = 1.225 - 1 x 100
1
= 22.5%
- x = ky2
- y = k/√x
3 = k/2
k = 6
8 = 6/√x
64 = 36/x
x = 36/64
x = 0.5625
-
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