- Find all the integral values of X Which satisfy the following inequalities. 2x-7<x + 2 ≤ 2 (x-2) (4MKS)
- Using method of completing the square, find the values of x that satisfies the quadratic equation below.x
^{2}+ x – 6 = 0 (3mks) - A triangle PQR is such that <PQR =60º length QR = 600cm and length QP=540cm.Find the area of triangle PQR and leave your answer in M
^{2}(3mks) - Use logarithms tables to evaluate ∛239.5 ×0.004832

sin30º (4mks) - Juma, Ali and Choge are three masons together to complete building a stone wall together in 12 hours.Juma and Ali working together complete the work in 18 hours while Juma and Choge working together complete the work in 24 hours. For how long each would take to complete building of the stone wall while alone. (3mks)
- Find the value of X that satisfy the equation

Log (x + 5) + log (x + 2)=log 4 (3mks) - Simplify X + 3x - 2 (2mks)

x+2 X^{2}-4 - Without using a calculator or mathematical tables, express Cos 30º in surd form and Tan 45°+√3

simplify leaving your answer in the form a+b√ c where a, b and c are rational numbers. (3mks) - The figure below shows a quadrilateral ABCD which is cyclic. Solve for x. (3MKS)
- Two cyclists moving towards each other take two hours to meet. If one cycles at a speed of 25km/h and the other at Y km/hr,find the value of Y if their distance apart is 92km. (3mks)
- Using a ruler and a pair of compass only construct a triangle in which AB is 7cm and BC=9cm and angle BAC =75º measure AC.Construct an inscribed circle in triangle ABC and measure its radius 4mks
- Given that cos (x - 20)° = Sin (2x + 32)° and x is an acute angle.Find tan (x - 4)(3mks
- Given that 3 + 3√5 = a + b√5 Find the values of a and b (4mks)

3 + √5 3 - √5 - In the figure below, 0 is the centre of the circle which passes through the points T,C and D. Line TC is parallel to OD and line ATB is a tangent to the circle at T. Angle DOC = 38°. Calculate the size of angle CTB (3mks)
- In the circle below chords PQ and RS interest intersect at X .Given that RX -8cm, XS=3cm and PQ =10cm.Calculate PX (2mks)
- The length and width of a rectangular window pane measured to the nearest millimeter are 8.6cm and 5.3 respectively.

Find to four significant figures. The percentage error in the area of the windowpane. 3mk

**DO FIVE QUESTIONS IN SECTION B**

- Using a ruler and a pair of compasses only, draw a parallelogram ABCD, such that angle DAB = 750. Length AB = 6.0cm and BC = 4.0cm.From point D, drop a perpendicular to meet line AB at N. (7 Marks)
- Measure length DN (1 Mark)
- Find the area of the parallelogram. (2 Marks)

- The table below shows income tax rates in the year 2013.

A constable earned a monthly salary of ksh. 42500 , house allowance of 10,000. Travelling allowance of 5000.

**Monthly Income in Ksh****Tax rate in each shilling**Up to 9680 10% 9681-18800 15% 18801 – 27920 20% 27921 – 37040 25% Over 37040 30% - taxable income (2mks)
- Tax due (3mks)
- Payable tax (1mk)
- Net pay of the constable 4mks

- The figure below shows a circle centre O, PQRS is a cyclic quadrilateral and QOS is a straight line.

Giving reasons for your answers, find the size of- Angle PRS (2 Marks)
- Angle POQ (2 Marks)
- Angle QSR (3 Marks)
- Reflex angle POS (3 Marks)

- A group of people planned to contribute equally towards a water project which was to cost Ksh 2000,000 to complete.40 members of the group withdrew before the project started. Hence the remaining members had to contribute Ksh. Using x as the original no. of members,
- Form expressions to show the original and the new contribution for each member. 2 mks
- Form a quadratic expression to represent the information above (no. 24) (2mks)
- find the original number of members 3mks
- luckily, 45% of the value of the project was funded by CDF. Calculate amount contributed by each member (3mks)

- The figure below shows a frustum of a right pyramid whose top face is a rectangle of side 3 cm by 5 cm and the bottom face is also a rectangle of side 6cm by 10cm. The perpendicular distance between the top and bottom faces (height) is 25cm.

Find;- the height of the chopped off pyramid (2marks)
- the volume of the frustrum (4marks)
- The surface area of the frustum. (4marks)

- The figure below shows the position of a boat Q which is observed sailing directly towards the pier P at the base of a vertical cliff PT. The angle of elevation of the top of the cliff from Q is 25.4º. After 14 seconds the boat is at point R, and the angle for elevation of T is now 64.7º.

If the cliff is 50m high, calculate- The distance PQ (2 Marks)
- The distance QR (4 Marks)
- The speed of the boat in km/h (4 Marks)

- A surveyor recorded the measurements of a field book using XY=400m as the base line as shown below.

To E 200

To F 250

Y

320

210

170

50

X

150 To D

150 To C

225 To B

100 To A- Use a scale of 1cm to represent 50m to draw the map of the field. (5mks)
- Find the area of the field in hectares (5mks)

- In the figure below PQ = 2500m, U T= 1000m and TS =2350m. PQR is a straight line. Parallel to UT and angle UPQ =22.5º.
- Calculate to the nearest meter
- U Q (2marks)
- T V (2marks)
- V S (2marks)
- P U (2marks)

- Find the perimeter of the figure. (2marks)

- Calculate to the nearest meter

## MARKING SCHEME

- Find all the integral values of X Which satisfy the following inequalities. 2x-7<x + 2 ≤ 2 (x-2) (4MKS)

2x - 7 < x + 2

2x - x <9

x < 9

x + 2 ≤ 2(x - 2)

x + 2 ≤ 2x -4

x - 2x ≤ -6

-x ≤ -6

x ≥ 6 - Using method of completing the square, find the values of x that satisfies the quadratic equation below.x
^{2}+ x – 6 = 0 (3mks)

x^{2}+ x = 6

x^{2}+ x + (½)^{2}= 6 + (½)^{2 }

(x^{2}+ ½)^{2}= 6 + 0.25

(x + ½) = ±√6.25

Either x = 2.5 - 0.5

= 2

or

(x + 0.5) = -2.5

x = -2.5

x = -3 - A triangle PQR is such that <PQR =60º length QR = 600cm and length QP=540cm.Find the area of triangle PQR and leave your answer in M
^{2}(3mks)

A = ½abSinθ

= ½ x 540 x 600Sin60º

= 16933.61

= 1.6933m^{2} - Use logarithms tables to evaluate ∛239.5 ×0.004832

sin30º (4mks)

No Stand Log 239.5

42.395 x 10 ^{2 }

4.832 x 10^{-3}2.3793

- 3.6841Sin 30

0.55x 10 ^{-1}0.0634

1.6990

- 0.3644

3 - Juma, Ali and Choge are three masons together to complete building a stone wall together in 12 hours.Juma and Ali working together complete the work in 18 hours while Juma and Choge working together complete the work in 24 hours. For how long each would take to complete building of the stone wall while alone. (3mks)

In 1 hour Juma and Ali complete^{1}/_{18}of the job hence work done by Choge^{1}/_{12}-^{1}/_{18}= 6 - 4

72

=^{2}/_{72}

Hence Choge takes 36hrs

Amount of work done by Juma^{1}/_{24}-^{1}/_{36}=3 - 2 =^{1}/_{72 }

72

Juma takes 72 hrs - Find the value of X that satisfy the equation

Log (x + 5) + log (x + 2)=log 4 (3mks)

log(x + 5)(x + 2) = log4

x^{2}+ 7x + 10 = 4

x^{2}+ 7x + 6 = 0

(x^{2}+ x) + (6x + 6)

x(x + 1) + 6(x + 1)

(x + 6) (x + 1)

x = -6 or x = -1 - Simplify X + 3x - 2 (2mks)

x+2 X^{2}- 4

x^{2 }(x - 2) + 3x - 2 = x^{2 }- 2x + 3x - 2

(x - 2) (x + 2) (x - 2) (x + 2)

x + x - 2 = (x + 2)(x - 1) = (x - 1)

(x - 2) (x + 2) (x - 2) (x + 2) (x - 2) - A trader marks his goods 20% above the cost price, He gives 10% discount to a customer and sells goods at Ksh486. Calculate the cost price of the goods(3mks)

120C x 90 = 486

100 100

C = 486 x 100

108

= Ksh450 - The figure below shows a quadrilateral ABCD which is cyclic. Solve for x. (3MKS)

2x + 60 = 180

2x = 120

x = 60 - Two cyclists moving towards each other take two hours to meet. If one cycles at a speed of 25km/h and the other at Y km/hr,find the value of Y if their distance apart is 92km. (3mks)

In one hour both travel (25 + y)km

92 = 2

25 + y

92 = 50 + 2y

42 = 2y

y = 21Km/h - Using a ruler and a pair of compass only construct a triangle in which AB is 7cm and BC=9cm and angle BAC =75º measure AC.Construct an inscribed circle in triangle ABC and measure its radius (4mks)
- Given that cos (x - 20)° = Sin (2x + 32)° and x is an acute angle.Find tan (x - 4)(3mks)

(x - 20) + (2x + 32) = 90

3x + 12 = 90

3x = 78

= 26 - Given that 3 + 3√5 = a + b√5 Find the values of a and b (4mks)

3 + √5 3 - √5

Numerator

9 - 3√5 + 9√5 - 3√5

3(3√5) + 3√5(3√5)

9 -3√5 + 9√5 + 3√25

9 + 6√5 + 15

24 + 6√5

4

Denominator

(3 + √5) (3 - √5)

3(3√5) + √5(3 - √5)

9 - 3√5 + 3√5 - √25

= 4

6 +^{3}/_{2}√5 - Using a ruler and a pair of compasses only, draw a parallelogram ABCD, such that angle DAB = 750. Length AB = 6.0cm and BC = 4.0cm.From point D, drop a perpendicular to meet line AB at N. (7 Marks)
- Measure length DN (1 Mark)

3.9 ± 0.1 - Find the area of the parallelogram. (2 Marks)

(3.9 ± x 6) = 23.4cm^{2}

- Measure length DN (1 Mark)
- The table below shows income tax rates in the year 2013.

A constable earned a monthly salary of ksh. 42500 , house allowance of 10,000. Travelling allowance of 5000.

**Monthly Income in Ksh****Tax rate in each shilling**Up to 9680 10% = 968 9681-18800 15% = 1.368 18801 – 27920 20% = 1824 27921 – 37040 25% = 2280 Over 37040 30% = 6138 - taxable income (2mks)

Salary + allowances

42500 + 10000 + 5000 = Ksh57,500 - Tax due (3mks)

968 + 1368 + 1824 + 2280 + 6138 = Ksh 12578 - Payable tax (1mk)

12578 - 1056 = Ksh11522 - Net pay of the constable (4mks)

Gross Pay - Deductions

57500 - 1056 = Ksh56444

- taxable income (2mks)
- The figure below shows a circle centre O, PQRS is a cyclic quadrilateral and QOS is a straight line.

Giving reasons for your answers, find the size of- Angle PRS (2 Marks)

55º - opp angles in cyclic add up to 180 - Angle POQ (2 Marks)

140º Angle at centre is twice the angle at the circumfrence - Angle QSR (3 Marks)

55º opp angles in cyclic

add up to 180º - Reflex angle POS (3 Marks)

360 - 40 = 320º

angle of a point add up to 360

- Angle PRS (2 Marks)
- A group of people planned to contribute equally towards a water project which was to cost Ksh 2000,000 to complete.40 members of the group withdrew before the project started. Hence the remaining members had to contribute Ksh. Using x as the original no. of members,
- Form expressions to show the original and the new contribution for each member. (2 mks)

New contribution = 2000000

x - 40

Original Contribution = 2000000

x - Form a quadratic expression to represent the information above (no. 24) (2mks)

2000000 - 2000000 = 2500

x - 40 x

2000000x - 2000000x - 80000 = 0

x^{2}- 200x + 160x - 32000 = 0

x^{2}- 40x - 3200 = 0 - find the original number of members (3mks)

(x + 160)(x -200) = 0

x = 200members - luckily, 45% of the value of the project was funded by CDF. Calculate amount contributed by each member (3mks)

55 x 2000000 x 1 = 6.875shillings

100 160

- Form expressions to show the original and the new contribution for each member. (2 mks)
- The figure below shows a frustum of a right pyramid whose top face is a rectangle of side 3 cm by 5 cm and the bottom face is also a rectangle of side 6cm by 10cm. The perpendicular distance between the top and bottom faces (height) is 25cm.

Find;- the height of the chopped off pyramid (2marks)

25 + x = 10

x 5

125 + 5x = 10x

125 = 5x

x = 25

Height = 25cm - the volume of the frustrum (4marks)
^{1}/_{3}AH -^{1}/_{3}ah

50x^{1}/_{3 }(6 x 10) -^{1}/_{3 }x 3 x 5 x 25

1000 - 41.67 = 958.33cm^{2} - The surface area of the frustum. (4marks)

- the height of the chopped off pyramid (2marks)
- The figure below shows the position of a boat Q which is observed sailing directly towards the pier P at the base of a vertical cliff PT. The angle of elevation of the top of the cliff from Q is 25.4º. After 14 seconds the boat is at point R, and the angle for elevation of T is now 64.7º.

If the cliff is 50m high, calculate- The distance PQ (2 Marks)

PQ = 50

Tan25.4

PQ = 105.3 - The distance QR (4 Marks)

QR = 50

tan64.7

= 23.63

QR = 105.3 - 23.63

= 81.67 - The speed of the boat in km/h (4 Marks)

81.67 ÷ 14 = 5.834 m/sec

km/h = 5.834 x 1000

3600

= 1.620km/hr

- The distance PQ (2 Marks)
- A surveyor recorded the measurements of a field book using XY=400m as the base line as shown below.

To E 200

To F 250

Y

320

210

170

50

X

150 To D

150 To C

225 To B

100 To A- Use a scale of 1cm to represent 50m to draw the map of the field. (5mks)
- Find the area of the field in hectares (5mks)
- = 100 + 225 x 50 = 8.125

2 - = 225 + 150 x 120 = 22500

2 - =40 x 150 = 6000
- ½ x 150 x 190 = 14250
- ½ x 200 x 80 = 8000
- 250 + 200 x 150 = 33750

2

½ x 170 x 250 = 21250

TOTAL**:**113875m^{2}

10000

11.3875 Hectares

- = 100 + 225 x 50 = 8.125

- Use a scale of 1cm to represent 50m to draw the map of the field. (5mks)
- In the figure below PQ = 2500m, U T= 1000m and TS =2350m. PQR is a straight line. Parallel to UT and angle UPQ =22.5º.
- Calculate to the nearest meter
- U Q (2marks)

Cos22.5 = TV

2350 - T V (2marks)

2350 Cos22.5 - V S (2marks)

Sin 22.5 = VS

TV

VS = 2171 Cos22.5

= 899.30 - P U (2marks)

Cos25 = PQ

PU

PU = PQ

Cos25

=1035.53

Cos22.5

= 2706m

- U Q (2marks)
- Find the perimeter of the figure. (2marks)

= 2706 + 1000 + 2350 + 899 + 1036 + 3171 + 2500

= 13662m

- Calculate to the nearest meter

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