QUESTIONS
SECTION 150 MARKS (ANSWER ALL THE QUESTIONS)
 A stop watch reads correct to ^{1}/_{5} seconds. Two races are timed as 49^{3}/_{5} seconds and 49^{4}/_{5} seconds. Calculate the maximum percentage error in sum of these two timings (3 marks)
 Construct, using a scale of 1:100 construct a trianguiar plot ABC where AB=6m, AC7m and BC7.5m using AB as the base. Cows are allowed to graze inside the plot provided that they are at least 2 meters from AB and more than 3 meters from C. Indicate by shading the area available for grazing. (4 marks)
 Simplify without using calculators or mathematical tables, the value of(3 marks)
8  ^{√60}/_{2} + √5+√3
√5√3  Use binomial expansion to determine the value of (1½)^{5 }(3 marks)
 The mean and standard deviation of the marks scored by a group of 10 students was found to be 47 and 11 respectively. An eleventh student had a score of 58 marks. Calculate the mean and standard deviation of the 11 students. (4 marks)
 Two pipes fill a swimming bath in 12 hours. The larger pipe is 33^{1}/_{3} more efficient than the smaller pipe. How long does the larger pipe take to fill the bath? (3 marks)
 Given that S = (1  r^{n}) make n the subject of the formula.
1  r  State the amplitude and the period of the following function y=tan 3x
 In a Geometric Progression, the first term is 2 and the common ratio is 2. Given that the product of the last two terms of the GP is 8192, find the sum of the last two terms. (3 marks)
 Given that x=m+ n, and m varies directly as y while n varies directly as the square of y. If x=16, y=2 and when x=33, y=3. Find x when y=8. (3 marks)
 A curve is such that ^{dy}/_{dx }= 4x and the point (2,9) lies on the curve. Find the equation of the curve. (3 marks)
 Given that p=2i3j+k and q3i4j3k, a point R divides a line PQ externally in the ratio of 4:1. Find the coordinates of R. (3 marks)
 Given that x and y are both positive, solve the equations log(xy) = 7 and log(^{x}/_{y})=1 (3marks)
 Use the midordinate rule to estimate the area enclosed by the curve y = x^{2}9, xaxis and the lines x=2 and x5 using six strips (3marks)
 Find the value of p if ∫^{3}_{0}(px^{2} + 2x + 3)dx = 54
 Solve for x in the domain 0≤x≤2π^{c}
2cos2x= 0.7071  The table below shows the frequency distribution of marks scored by students in a test.
Marks 110 1120 2130 3140 4150 Frequency 2 4 8 4 2  On the grid provided, draw an ogive for the data.
 Use your graph to determine;
 The pass mark if only 6 students passed the exam.
 The quartile deviation
 Range of marks scored by the middle 60 % of the students
 A triangle ABC with vertices at A (1,1), B (3,1) and C (1, 3) is mapped onto triangle A'B'C' by a transformation whose matrix is(^{1}_{0} ^{0}_{1}) Triangle A'B'C' is then mapped onto A''B''C'' with vertices at A"(2, 2), B"(6,2) and C''(2,6) by a second transformation.
 Find the coordinates of A'B'C' (3 marks)
 Find the matrix which maps A'B'C' onto A''B''C''. (3 marks)
 Determine the ratio of the area of triangle A'B'C' to triangle A"B''C"(1 mark)
 Find the transformation matrix which maps A''B''C'' onto ABC (3 marks)
 The table below shows the taxation rates for income earned.
Income in ksh pm Tax rates (%) 19680 10 9681  18800 12 18801  27920 20 27921  37040 25 Excess over 37041 30
House Allowance Shs. 10,000
Medical Allowance Shs. 2400
Acting Allowance Shs. 2820.
He is entitled to a monthly personal relief of KShs. 1162 while 7.5% of his basic salary is taxexempted. Calculate Mr. Hamisi's monthly basic salary in KSh. (7 Marks)
 The following deductions also made every month.
 N.H.I.F. KSh. 320
 Cooperative society shares KSh. 6000
 Union dues KSh. 200
Calculate his net monthly salary. (3 Marks)
 The acceleration of a particle, t seconds after passing a fixedpoint P is given by a 4t7. Given that the initial velocity of the particle is 5m/s, find;
 Its acceleration when t = 4 seconds
 Its velocity when t=3 seconds
 Values of t when the particle is momentarily at rest
 Its maximum velocity
 Mr. Mairura has two lorries A and B used to transport at least 42 tons of potatoes to the market. Lorry A carries 4 tons of potatoes per trip while lorry B carries 6 tons of potatoes per trip. Lorry A uses 2 liters of fuel per trip while lorry uses 4 liters of fuel per trip. The two lorries are to less than 24 liters of fuel. The number of trips made by lorry A should be less than the number of trips made by lorry B. Lorry A should make more than 4 trips.
 Taking X to represent the number of trips made by lorry A and Y to represent the number of trips made by lorry B, write the inequalities that represent the above information (4marks)
 Plot the inequalities above in the graph provided below (4 marks)
 If Lorry A makes sh. 35,000 per trip and Lorry B makes sh.28,000 per trip, use the graph above to determine the number of trips made by lorry A and by lorry B to deliver the greatest number of potatoes and hence find the maximum profit. (2marks)
 The position of 3 cities P, Q and R are (15°N, 20°W) (50°N, 20°W) and (50°N, 60°E) respectively.
 Find the distance in nautical miles between:
 Cities P and Q (2 marks)
 Cities Q and R along a circle of latitudes
 A plane left city P at 0250h and flew to city Q where it stopped for 3 hours then flew on to city R, maintaining a ground speed of 900 knots throughout. Calculate:
 The local time at city R when the plane left city P
 The local time (to the nearest minute) at city R when the plane landed at R. (3 marks)
 Find the distance in nautical miles between:
 The figure below represents a cuboid ABCDEFGH, with AB = 7 cm, BC = 24 cm andCF = 7.2 cm. M and N are the midpoints of EF and DC respectively.
Calculate to 2 decimal places the: Angle AF makes with the plane ABCD (3 marks)
 Angle between the lines HF and AB. (2 marks)
 Angle between the planes GHEF and ABFE (2 marks)
 Angle between BM and the plane ABCD (3 marks)

 A jewel is guarded by three guards A, B and C in that order. On his way in, the probability of a thief getting caught by guard A is ^{2}/_{3}, by B is ^{3}/_{7} and by C is ¼. On his way out, the probability of being caught by guard C is ^{4}/_{5} , by B is ^{1}/_{3} and by guard A is ^{2}/_{5}. Calculate the probability that:
 The jewel is stolen and the thief escapes. (2 marks)
 The thief was caught by guard C
 Albert, Bonny and Charles competed in a game of chess. Their probabilities of winning the game are ^{2}/_{5}, ^{3}/_{5} and ^{1}/_{10} respectively.
 Draw a probability tree diagram to show all the possible outcomes. (2 marks)
 Calculate the probability that;
 No one loses the game. (2 marks)
 Only one of them wins the game.
 A jewel is guarded by three guards A, B and C in that order. On his way in, the probability of a thief getting caught by guard A is ^{2}/_{3}, by B is ^{3}/_{7} and by C is ¼. On his way out, the probability of being caught by guard C is ^{4}/_{5} , by B is ^{1}/_{3} and by guard A is ^{2}/_{5}. Calculate the probability that:
MARKING SCHEME
 AE =½ x ^{1}/5 = ^{1}/10
AE in sum = ^{1}/10 x 2 = 1/5
% error =^{1}/5 x 100
^{493}/5 + ^{494}/5
= 0.2012% 
 ^{√60}/2 = √15x√4
2
= 2√15
2
= √15
√5 + 3 (√5 + 3) = 8 +2√5
√5  3 √5 +√3 2
= 4 +√15
8  √15 + 4 + √15
= 12  (1)^{5} + 5(1^{4})(½) + 10(13)(½)^{2} + 10(12)(½)^{3} + 5(1)(½)^{4} + (½)^{5}
= 1 + ^{5}/2 + ^{5}/2 + ^{5}/4 + ^{5}/16 + ^{1}/32
= 7.59375  Σfx = 47 x 10 = 470
new(x) = 470 + 58
11
= 48
sd = √Σfx^{2}  (x)^{2}
Σf
11^{2} = Σfx^{2}  47^{2}
10
Σfx^{2} = 23300
new sd =√23300  48^{2}
11
= 10.95  bigger smaller
133^{1}/3%x x
1hr = ^{1}/x + ^{4}/3x = ^{1}/12
^{7}/3x = ^{1}/12
¾ x = 28
larger pipe = ¾ x 28
= 21hrs  S  Sr = 1  r^{n}
8n = 1  s + sr
nlog r = log(1  S + Sr)
n = log(1  S + Sr)
log r  a = undefined
period = 180  (ar^{n1})(ar^{n2}) = 8192
2 x 2^{n1} x 2 x 2^{n2} = 213
2^{n1} = 2^{13}
n = 7
last = 2 x 2^{6} = 128
2nd = 2 x 2^{5} = 64
sum = 128 + 64
= 192  m x y = m=ky
n x y^{2} = n=ay^{2}
x = ky + ay^{2}
16 = 2k + 4a
33 = 3k + 9a
k = 2
a = 3
x = 2y + 3y^{2}
x = 2(8) + 3(64)
x = 208  y = ∫(4  x)dx
y = 4x  x^{2} + c
2
9 = 82 + c
c = 3
y = 4x  x^{2} + 3
m:n
4:1 log x + log y = 7
log x  log y = 1
2log y = 6
log y = 3
y = 1000
x = 10000  A = 0.5(3.9375 + 1.4375 + 1.5625 + 5.0625 + 9.0625 + 13.5625)
A = 17.3125 sq umits  [Px^{3} + x^{2} + 3x]^{3}_{0} = 54
9p + 18 = 54
9p = 36
p = 4  cos 2x = 0.35355
2x = 110.71, 249.29, 470.71, 609.29
x = 55.355, 124.645, 235.355, 304.645
x = 0.3075c, 0.6923c, 1.307c, 1.6925c Marks 110 1120 2130 3140 4150 Frequency 2 4 8 4 2 cf 2 6 14 18 20 

 30.5 marks
 0.25 x 20 = 5th
= 18.5
0.75 x 20 = 15th
= 32.5
32.5  18.5
2
= 7
 0.2 x 20 = 4th → 16.5
0.8 x 20 = 16th → 34.5
range → 16.5 to 34.5




 ASF = 4



 gross tax = 5512 + 1162
= 6674
9680 x 0.1 = 968
9120 x 0.15 = 1368
9120 x 0.2 = 1824
9120 x 0.25 = 2280
0.3 x y = 234
y = 780
TI = 9680 + 3(9120) + 780
= 37820
BS = 37820  (10000 + 2400 + 2820)
= 22600
22600 x 100
92.5
= 24.432.43  deducstions = 320 + 6000 + 200 + 5512 = 12032
net salary = 39652.43  12032
= 27620.43
 gross tax = 5512 + 1162

 a = 4(4)  7
= 9ms^{2}  v = ∫(4t  7)dt
v = 2t^{2}  7t + c
s = c
v = 2t^{2 } 7t + 5
v = 2(9)  4(3) + 5
v = 2m/s  v = 0
2t^{2}  7t + 5 = 0
t^{2}  3.5t = 2.5
t^{2}  3.5t + (1.75)^{2} = 0.5625
t  1.75 ± 0.75
t = 1.75 ± 0.75
t = 1sec or t = 2.5sec  a = 0
0 = 4t  7
t = 1.75 sec
v = 2(3.0625)  7(1.75) + 5
v = 1.125m/s
 a = 4(4)  7


 4x + 6y ≤ 42
 2x + 4y ≤ 24
 x < y
 x > 4
 x ≥ 0

 (6,3) → (6 x 35000) + (3 x 28000)
= 294000



D = 60 x 35
= 2100nm D = 60 x 80 cos 50
D = 3085.38nm

 80 x 4 = 320min
= 5hrs 20min
0250 + 520 = 8:10
= 0810hrs  Time from P to Q
= 2100 = 2^{1}/3
900
Q to R = 3085.38 = 3hrs26min
900
total time = 3hrs26min + 2hr20min + 3hrs
= 8hrs46mins
= 16:56hrs
 80 x 4 = 320min


 AC = √24^{2} + 7^{2}
= 25cm
tan = ^{7.2}/25
= 16.07  tan x = ^{24}/7
x = 73.74º  tanθ = ^{7.2}/24
= 16.70º  tanθ =^{ 7.2}/23.74
BN = √24^{2}  3.5^{2}
θ = 16.87º
 AC = √24^{2} + 7^{2}

 ^{ }
 ^{}^{1}/_{3} x ^{4}/_{7} x ¾ x^{ 1}/_{5} x ^{2}/_{3} x ^{3}/_{5}
= ^{2}/_{175}  (^{1}/_{3} x ^{4}/_{7} x ¼) + (/_{3} x ^{4}/_{7} x ¾ x 4/5)
= ^{4}/_{84} + ^{4}/_{35}
= ^{17}/_{105}
 ^{}^{1}/_{3} x ^{4}/_{7} x ¾ x^{ 1}/_{5} x ^{2}/_{3} x ^{3}/_{5}



 P(ABC)
= ^{2}/_{5} x ^{3}/_{5} x^{ 1}/_{10} = ^{3}/_{125}  ( ^{2}/_{5} x ^{2}/_{5} x ^{9}/_{10}) + (^{3}/_{5} x ^{3}/_{5} x ^{9}/_{10}) + (^{3}/_{5} x ^{2}/_{5} x ^{1}/_{10})
=^{ 123}/_{250}
 P(ABC)

 ^{ }
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