# Mathematics Paper 2 Questions and Answers - MECS Cluster Joint Pre Mock Exams 2021/2022

MATHEMATICS
PAPER 2

SECTION 1 (50 MARKS):
answer all questions in the section.

1. Use Logarithms correct to four significant figures to evaluate. (4marks)
24.36×0.066547/1.482
2.
1. Given the vectors a = 3i – j + 2k, b = 4i + 2j – k and p = 2a – b. Express p in terms of i, j and k. (2 marks)
2. Hence calculate |p|correct to 3 significant figures. (1 mark)
3. W varies directly as the cube of x and inversely as y. Find W in terms of x and y given that W = 80 when x = 2 and y = 5. (2 Marks)
4. A cold water tap can fill a bath in 10 minutes while a hot water tap can fill it in 8 minutes. The drainage pipe can empty it in 5 minutes. The cold water and hot water taps are opened for 4 minutes. After four minutes all the three taps are opened. Find how long it takes to fill the bath. (3 Marks)
5. Object A of area 10cm2 is mapped onto its image B of area 60cm2 by a transformation. Whose matrix is given by
p =
. Find the positive values of x (3 Marks)
6. Make P the subject of the formula in L = 2/3x² - PT /y (3 Marks)
7.
1. Expand the expression (1+ 1/2 x)in ascending order powers of x, leaving the coefficients as fractions in their simplest form. (2 Marks)
2. Use the first three terms of the expansion in (a) above to estimate the value of (1.05)5 (2 Marks)
8. By rounding each number to the nearest tens, approximate the value of 2454 × 396/66
Hence, calculate the percentage error arising from this approximation to 4 significant figures. (3 Marks)
9. Without using a calculator or mathematical tables, express √3/1-Cos 30º in surd form and simplify (3 Marks)
10. Solve for θ in the equation sin (3 θ + 120º = √3/2 for 0º ≤  θ ≤ 180º (3Marks)
11. The equation of a circle is x² + y² + 6x -10y – 2 = 0. Determine the co-ordinates of the centre of the circle and its radius. (3 Marks)
12. Kamau deposited ksh.50,000 in a financial institution in which interest is compounded quarterly. If at the end of second year he received a total amount of ksh79,692.40. Calculate the rate of interest p.a (3 Marks)
13. Chords AB and CD in the figure below intersect externally at Q. if AB = 5cm BQ = 6cm and DQ = 4cm, calculate the length of chord CD. (3 Marks)
14. Two containers have base area of 750cm2 and 120cm2 respectively. Calculate the volume of the larger container in litres given that the volume of the smaller container is 400cm3. (3 Marks)
15.
1. Find the inverse of the matrix (1 Marks)

2. Hence or otherwise solve the simultaneous equations (3marks)
3x + 2y = 4
5x + 4y = 9
3. Solve for x in the equation. (3 Marks)
Log 8 (x + 6) – Log8 (x – 3) = 2/3

SECTION II (50 MARKS)
Attempt ONLY FIVE questions from this section.

1. In triangle ABC, shown below, AB = a AC = b point M lies on AB such that AM: MB = 2:3 and point N lies on AC such that AN: NC = 5:1 line BN intersects line MC at X.
1. Express the following in terms of a and b
1. BN (1 mark)
2. CM (1 mark)
2. Given that BX = kBN and CX = rCM where k and r are scalars
1. Write two different expressions for AX in term of a, b, k and r (4 marks)
2. Find the values of k and r (4 marks)
2. A particle moves such that t seconds after passing a given point O is given by
S = t ( t – 2 ) ( t – 1)
1. Find its velocity when t = 2 second (3 marks)
2. Find its minimum velocity. (3 marks)
3. Find the time when the particles is momentarily at rest. (3 marks)
4. Find its acceleration when t = 3 seconds. (1mark)
3. The following table shows the rate at which income tax was charged during a certain year.
 Monthly taxable income in Ksh. Tax rate % 0 - 9860 9861 - 19720 19721 - 29580 29581 - 39440 39441 - 49300 49301 - 59160 over 59160 10152025303540

A civil servant earns a basic salary of Ksh.35750 and a monthly house allowance of sh.12500. The civil servant is entitled to a personal relief of sh.1062 per month. Calculate:
1. Taxable income (2 marks)
2. Calculate his net monthly tax (5 marks)
3. Apart from the salary the following deduction are also made from his monthly income.
WCPS at 2% of the basic salary
Loan repayment Ksh.1325
NHIF sh.480
Calculate his net monthly earning. (3 marks)
4.
1.
1. Taking the radius of the earth, R= 6370km and π = 22/7, calculate the shortest distance between the two cities P(60oN, 29oW) and Q(60oN, 31oE) along the parallel of latitude. (3marks)
2. If it is 1200hrs at P, what is the local time at Q. (3marks)
2. An aeroplane flew due south from a point A (60oN, 45oE) to a point B. the distance covered by the aeroplane was 8000km. determine the position of B. (4marks)
5. A married couple intends to have 3 children. They consult an expert who tells them that the probability of a male birth is 0.55.
1. Draw a tree diagram to represent this occurrence. (2 marks)
2. Find the probability that
1. All the three children will be female. (2 marks)
2. At least a male is born. ( 2 marks)
3. At least 2 will be females, giving your answer to 3 s.f. ( 4 marks )
6. An arithmetic progression (AP) has the first term a and the common difference d.
1. Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (1mark)
2. The AP above is increasing and the third, ninth and twenty fifth terms form the first three consecutive terms of a Geometric Progression (G.P) The sum of the seventh and twice the sixth terms of the AP is 78. Calculate:
1. the first term and common difference of the AP. (5marks)
2. the sum of the first nine terms of the AP. (2marks)
3. The difference between the fourth and the seventh terms of an increasing AP. (2marks)
7.
1. Draw ∆PQR whose vertices are P(1,1)Q(-3,2) and R(0,3) on the grid provided (2marks)
2. Find and draw the image of ∆PQR under the transformation whose matrix is     and label the image P’Q’R’ (2mks)
3. P’Q’R’ is then transformed into P11 Q11 R11 by the transformation with the matrix Find the co-ordinates of P11Q11R11 and draw P11Q11R11 (3marks)
4. Describe fully the single transformation which maps PQR onto P11Q11R11 and find the matrix of this transformation (3marks)
8. For a sample of 100 bulbs, the time taken for each bulb to burn was recorded. The table below shows the result of the measurements.
 Time (in hours) 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Number of bulbs 6 10 9 5 7 11 15 13 8 7 5 4
1. Using an assumed mean of 42, calculate
1. the actual mean of distribution (4 Marks)
2. the standard deviation of the distribution (3 Marks)
2. Calculate the quartile deviation (3 Marks)

MARKING SCHEME

1.
 No. Log 24.360.0665471.4810-1 x 9.0450.9045 1.3867   -2.82310.20980.1703X      20.34060.20980.3406     -1.8692 x 1/33 /3 + -2.8692 3=-1.9564
= 0.9045
2.
1. P = 2 (3i – j + 2k) – 4i + 2j – k)
= 6i – 2j + 4k – 4i – 2j + k
= 2i – 4j + 5k

2. /p/ = √22 + -42 + 52
= 6.708
=6.71
3. W =   Kx3
y
80 = 8/5 k
K = 50
W = 50 x3
y

4. (1/10+ 1/8) x 4 = 9/40 x 4 = 9/10
1/10 + 1/8 - 1/5 = (4+5-8)/40 = 1/40
∴Time taken = 1/10 ÷  1/40
= 40/10 = 4 Min

5. ASF = determinant of the matrix
60/10 = x(x + 3) – 12
6 = x2 + 3x – 12
(x2 – 3x) + (6x – 18) = 0
x(x-3) + 6 (x – 3) = 0
(x + 6) (x-3) = 0
x = 6 or 3

6. 4/9 (X²-PT)/y = L2
x2-PT = 9/4 L2y
-PT = 9/4 L2 - x2

P = x² - ⁹/₄L²y
T
7.
1. 1.15 [1/2 x]+ 5.14 [1/2 x]1
+ 10.13 [1/2 x]2 + 10.12 [1/2 x]3 + 5.11 + [1/2 x]4+ 1.10+[1/2 x]5
1 + 5/2 x + 5/2 x2 + 5/4 x3 + 5/16 x4 + 1/32 x5

2. [1.05]5 = 1 + 5/2  × 0.1 + 5/2× 0.12
= 1.275

8. 2450 × 400 = 14,000
70
2454×396= 14724
66
% error = (14724-14000)/14724 x 100
≈ 4.917

9.           √3                  =       √3
1-Cos30º                 1-√3/2
(2√3   -    (2+√3)
(2-√3)  -  (2+√3)
= 2√3 (2+ √3)
4-3
= 4√3+6

10. Sin (3θ+120)=√3/2
3θ+120 = 60,120,420,480
3θ+120 = 120,420,480
θ=0º,100º,120º

11. x2 + 6x + y2 – 10y = 2
complete the square
x2 + 6x + (6/2)2 + y2 – 10y +(-10/2)2 = 2 + (6/2)2 +(-10/2)2
x2 + 6x + 9 + y2 – 10y + 25 = 36
(x + 3)2 + (y – 5)2 = 36
(x + 3)2 + (y – 5)2 = 62
(x – h)2 + (y – k) = r2
Center (-3,5)

12.  50,000 (1 + r/100)8. =79, 692.4
50,000                         50 000
= ³√(1 + r/100)8   = ⁸√1.593848
(1 + r/100 )= 1.06
r/100= 0.06
r = 6% per interest period
per annum = 6 x 4 = 24%

13. AQ x BQ = CQ x DQ
11 x 6 = (x + 4) 4
66 = 4x + 16
4x/4 = 50/4
x = 12.5

14. A.S.F = 750/120 = 25/4
L.S.F = √25/4= 5/2
V.S.F = (5/2)3= 125/8
V.S.F = 125/8 = V/400cm³
V = 125/8 x 400cm3
Volume = 6250/1000
= 6.25 litres
15.

16. Log8 (x+6/x-3) = Log88²/₃
(x+6)
/(x-3) = 4
x + 6 = 4x – 12
3x = 18
x = 6

17.
1.
1. BN = BA + AN
= -a +5/6 b
= 5/6 b – a

2. CM = -b + r CM
= b + r (2/5 a - b)

2.
1. AX = b + rCM
= b – rb +2/3 ra
= ( 1 – r)b + 2/5 ra
AX = a + k (5/6 b – a)
= a – ka +5/6 kb
= (1 – k)a +5/6 kb

2. ( 1 – r)b + 2/5 ra= ( 1- k)a + 5/6 kb
5/6 k = 1 – r – ( i)
2/5 r = 1 – k -- (ii)
K =6/5  – 6/5 r
2/5 = 1- (6/56/5 r)= 1 – 6/5 + 6/5 r
-4r = -1 → r = ¼
5/6 k = ¾
k = 9/10

18.
1. S = t (t2 – t – 2t + 2)
= t3 – 3t2 + 2t
ds/dt = 3t2 – 6 t + 2
v = 3 ( 4) - 6 ( 2) + 2
v = 2m/s

2. dv/dt=0
dv/dt=6t-6
6t – 6 = 0
t = 1
v= 3(1)2 – 6(1) + 2
= -1m/s

3. 3t2 – 6t + 2 = 0
t =( 6±√(36-4×3×2))/(2×3)
t = (6± 3.464)/6
t= 1.577 or 0.4227

4. a = 6 ( 3) – 6 = 12m/s2

19.
1. taxable income
35750 + 12500 = 48250= sh.48250

2. 9860 × 10/100 = sh.986
9860× 15/100 = sh.1479
9860× 20/100 = sh.1972
9860× 25/100 = sh.2465
8810× 30/100 = sh.2643
Total= 9545
9545-1062
=sh.8483pm

3. WCPS =2/100 × 35750 = 715
Total deduction=(8483 + 715 + 1325 + 480)
= 11003
Net salary = 48250- 11003
= sh.37247 p.m M1

20.
1.
1. P(60º N,29º W) and Q (60º ,31º E)
longitudinal diff = 60º
Shorter distance = 60/360 × 2 × 22/× 6370 cos⁡60
= 3335.3 km
2. Time difference = 4 × 60 = 240 min
= 4hrs
Time at Q = 1200 + 0400 = 1600hrs

3. (θ+60)/360 × 2 × 22/7 × 6370=8000
θ+60=71.92
θ=11.92
B(11.92ºS,45ºE)
21.
1.

2.
1. P(FFF)= 45/100 × 45/100 × 45/100
=91125/1000000
= 0.091125

2. 1-P(FFF)
1 – 0.091125
= 0.908875

3. P(MFF)or P(FMF)or P(FFM)
=(55/100 × 45/100 × 45/100 ) +(45/100 × 55/100 × 45/100) + 45/100 × 45/100 × 55/100 )
= 0.111375 + 0.111375 + 0.111375 = 0.334125
=0.334
22.
1.  a+2d, a+8d, a+24d
2.
1. a + 8d = a + 24d
a + 2d    a + 8d         I

a+6d+2(a+5d)=78      II
16d=10aI
3a+16d=78          II

3a+10a=78
3a=78
a=6

16d=10×6
d=3.75

2. S_q=9/2 (2×6+(9-1)3.75) =189
3. ⇒[6+6(3.75)]-[6+3(3.75)] =11.25
23.

1.

2.

3. It is an enlargement centre origin (0,0) scale factor 2
24.
 Class Mid point X f t = X-A/C ft t2 ft2 cf 15-19 17 6 -5 -30 25 150 6 20-24 22 10 -4 -40 16 160 16 25-29 27 9 -3 -27 9 81 25 30-34 32 5 -2 -10 4 20 30 35-39 37 7 -1 -7 1 7 37 40-44 42 11 0 0 0 0 48 45-49 47 15 1 15 1 15 63 50-54 52 13 2 26 4 52 76 55-59 57 8 3 24 9 72 84 60-64 62 7 4 28 16 112 91 65-69 67 5 5 25 25 125 96 70-74 74 4 6 24 36 144 100 Σf=100 Σft=28 Σft2=873

1.
1. x = 42 + (28/100 ×5) = 43.4

2. 873/100 - (28/100)2 = 0.7856
2. [49.5+ (75-63)/13)5.29.5 ]× 1/2
= 12.31

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

### Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp