Pre-KCSE Joint Maths Paper 1

In a game park \(\frac{1}{5}\) of the animals are rhinos and   \(\frac{3}{4}\) of them are zebras. Two-thirds of the remaining animals are lions and the rest are warthogs. Find the fraction of warthogs in the game park.

\(Solution\)

Let \(x\) represent the number of animals in the game park

 Rhinos 

\( = \frac{1}{5}x\)

 Zebras 

\(  =  \frac{3}{4}x\) 

 Remaining 

\( = x-(\frac{1}{5}x+ \frac{3}{4}x )\) 

\(=\frac{1}{20}x\)

 Lions 

\(  =  \frac{2}{3} of \frac{1}{20}x\) 

\(=\frac{1}{30}x \)

 Warthogs 

\(  =  \frac{1}{20}x -  \frac{1}{30}x  \)

\(=\frac{1}{60}x \) 

2. The average mark scored by the first 27 students in a mathematics test is 52. The average mark scored by the remaining 37 is 58. Calculate the mean mark for the whole class

average marks = \(\frac{27 \times 52+37\times 58}{27+37}\)

 \(=55.46875\) 

3. Use square roots, reciprocal and square tables to evaluate to 4 significant figures the expression

\(0.06458^{\frac{1}{2}} + (\frac{2}{0.4327})^{2}\)

It is advisable that we solve the first part of this expression. 

\((6.458 \times 10^{-2})^{\frac{1}{2}}\)

Applying the law of indices,  \((a^{m})^{n} = a^{m \times n}\), we multiply the powers and obtain:

\(6.458^{\frac{1}{2}} \times 10^{-1}\)

Remember, a number raised to the power of a half means the square root of that number

Now we get the square root of 6.458 from square roots table, therefore we have:

\(2.541 \times 10^{-1}\)

= 0.2541

For the second part, we need to rearrange the fraction within the bracket to be to apply table of reciprocals

Here is the rearrangement:

\(2 \times \frac{1}{0.4327}\)

Now we just use reciprocals table to solve  \(\frac{1}{0.4327} \). We need to rewrite the denominator in standard form

Therefore, we have:

\(2 \times  \frac{1}{4.327 \times 10^{-1}}\) 

This becomes   \(2 \times 0.2311 \times 10\)  which gives \(4.622\)

Remember we had the fraction inside the bracket raised to the power of 2 so we have \((4.622)^{2}\)   \(=21.36\) 

Therefore, recombining the two parts,

\(=0.2541 + 21.36\)

\(=21.6141\)

Question 11. KCSE 2015 Paper 1 

Given:

\( P = 5a - 2b \)

where:

\( a = \begin{pmatrix} 3 \\ 2 \end{pmatrix}, \quad b = \begin{pmatrix} 4 \\ 1 \end{pmatrix} \)

(a) Finding \( P \):

\( P = 5 \begin{pmatrix} 3 \\ 2 \end{pmatrix} - 2 \begin{pmatrix} 4 \\ 1 \end{pmatrix} \)

\( = \begin{pmatrix} 5 \times 3 \\ 5 \times 2 \end{pmatrix} - \begin{pmatrix} 2 \times 4 \\ 2 \times 1 \end{pmatrix} \)

\( = \begin{pmatrix} 15 \\ 10 \end{pmatrix} - \begin{pmatrix} 8 \\ 2 \end{pmatrix} \)

\( = \begin{pmatrix} 15 - 8 \\ 10 - 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 8 \end{pmatrix} \)

Thus, \( P = \begin{pmatrix} 7 \\ 8 \end{pmatrix} \).

(b) Finding \( P' \) after translation by \( \begin{pmatrix} -6 \\ 4 \end{pmatrix} \):

\( P' = P + \text{translation vector} \)

\( = \begin{pmatrix} 7 \\ 8 \end{pmatrix} + \begin{pmatrix} -6 \\ 4 \end{pmatrix} \)

\( = \begin{pmatrix} 7 - 6 \\ 8 + 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 12 \end{pmatrix} \)

Thus, \( P' = \begin{pmatrix} 1 \\ 12 \end{pmatrix} \).

Question 12 KCSE Paper 1 2015:

Given:

\( a = 3, \quad b = 5, \quad c = -\frac{1}{2} \)

Evaluate:

\( \frac{4a^2 + 2b - 4c}{\frac{1}{4} (b^2 - 3a)} \)

Step 1: Compute the numerator

\( 4a^2 + 2b - 4c \)

\( = 4(3^2) + 2(5) - 4\left(-\frac{1}{2}\right) \)

\( = 4(9) + 10 + 2 \)

\( = 36 + 10 + 2 = 48 \)

Step 2: Compute the denominator

\( \frac{1}{4} (b^2 - 3a) \)

\( = \frac{1}{4} (5^2 - 3(3)) \)

\( = \frac{1}{4} (25 - 9) \)

\( = \frac{1}{4} \times 16 = 4 \)

Step 3: Compute the final value

\( \frac{48}{4} = 12 \)

Final Answer: \( 12\)

KCSE 2015 Maths Paper 1  Q14

14. The cost of 2 jackets and 3 shirts was Ksh 1 800. After the cost of a jacket and that of a shirt were increased by 20%, the cost of 6 jackets and 2 shirts was Ksh 4 800. Calculate the new cost of a jacket and that of a shirt. (4 marks)

\(2j + 3s = 1800 -- (i) \)

\((6 \times 1.2)j + (2 \times 1.2)s = 4800 \)

\(7.2j + 2.4s = 4800 \)

\(72j + 24s = 48000 \)

\(3j + s = 2000 -- (ii) \)

\(2j + 3s = 1800 \)

\(3j + s = 2000\)

Solving equation (1) and (2) simultaneously by elimination method we have:

\(2j + 3s = 1800 \)

\( \underline{ 9j + 3s = 6000 }\)

\(-7j = -4200 \)

\(j = 600\)

\(2(600) + 3s = 1800 \)

\(1200 + 3s = 1800 \)

\(3s = 1800 - 1200 \)

\(3s = 600 \)

\(s = 200\)

\(2j + 3s = 1800\)

 New price of a shirt \(= (1.2 \times 200) = 240\) 

 New price of jacket \(= (1.2 \times 600) = 720\) 

 KCSE 2015 Maths Paper 1 Q15 

15. A tailor had a piece of cloth in the shape of a trapezium. The perpendicular distance between the two parallel edges was 30cm. The lengths of the two parallel edges were 36 cm and 60cm. The tailor cut off a semi-circular piece of the cloth of radius 14cm from the 60cm edge. Calculate the area of the remaining piece of cloth. \((Take  \quad \pi = \frac{22}{7})\) (3 marks)

KCSE 2015 Maths Paper 1  Q17

17. Three partners Amina, Bosire and Karuri contributed a total of Ksh 4800000 in the ratio 4:5:7 to buy an 8 hectares piece of land. The partners set aside 14\frac{1}{4} of the land for social amenities and sub-divided the rest into 15 m by 25 m plots.

(a) Find:
 (i) the amount of money contributed by Karuri; (2 marks)
 (ii) the number of plots that were obtained. (3 marks)

(b) The partners sold the plots at Ksh 50000 each and spent 30% of the profit realised to pay for administrative costs. They shared the rest of the profit in the ratio of their contributions.
 (i) Calculate the net profit realised. (3 marks)
 (ii) Find the difference in the amount of the profit earned by Amina and Bosire. (2 marks)

KCSE 2015 Maths Paper 1 Q18

18. Two shopkeepers, Juma and Wanjiku bought some items from a wholesaler. Juma bought 18 loaves of bread, 40 packets of milk and 5 bars of soap while Wanjiku bought 15 loaves of bread, 30 packets of milk and 6 bars of soap. The prices of a loaf of bread, a packet of milk and a bar of soap were Ksh 45, Ksh 50 and Ksh 150 respectively.

(a) Represent:
 (i) the number of items bought by Juma and Wanjiku using a  \(2 \times 3 \) matrix. (1 mark)
 (ii) the prices of the items bought using a \(3 \times 1 \) matrix. (1 mark)

(b) Use the matrices in (a) above to determine the total expenditure incurred by each person and hence the difference in their expenditure. (3 marks)

(c) Juma and Wanjiku also bought rice and sugar. Juma bought 36 kg of rice and 23 kg of sugar and paid Ksh 8 160. Wanjiku bought 50kg of rice and 32kg of sugar and paid Ksh 11 340. Use the matrix method to determine the price of one kilogram of rice and one kilogram of sugar. (5 marks)