admin@kcseforecast.com

+2541026301456

MATHEMATICS KCSE 2021 PAPER 1 MARKING SCHEME
5. The size of two interior angles of an irregular polygon each measures \(90^\text{o}.\) All the other remaining angles each measures \(150^\text{o}\). Determine the number of sides of the polygon. $$ Solution $$ \[\begin{align*} (n-2)180 = 90 + 90+150(n-2) \\ 180n-360 = 180+150n-300 \\ 180n-150n = 180-300+360 \\ 30n=240 \\ n=8 \end{align*} \] 7. Simplify \((4+2y)^2 - (2y-4)^2\)
$$ Solution $$ \[\begin{align*} =[(4+2y)-(2y-4)] [(4+2y)+(2y-4)] \\ = (4+2y-2y+4)(4y) \\ = 8(4y)\\ =32y \end{align*} \]
11. Solve for \(\theta\) \[\begin{align*} sin(2\theta-15)= cos 3\theta \\ 2\theta - 15+3\theta = 90 \\ 5\theta = 105 \\ \theta = 21 \end{align*} \] 19. Elimu School bought 25 textbooks and 35 exercise books for Kshs. 13500 from bookshop A. From the same bookshop Soma School bought 21 textbooks and 38 exercise books and spent Kshs. 1300 less than Elimu School. Take \(x\) to represent the price of a textbook and \(y\) to represent the price of an exercise book.
(a) Form two equations representing the above infromation (2mks) $$ Solution $$ \[\begin{align*} 25x+35y=13500 \\ 21x+38y=12200 \\ \end{align*} \] (b)Use matrix method to determine the price of each item.
$$ Solution $$ \begin{align*} \begin{bmatrix}5 & 7 \\21 & 38 \end{bmatrix}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}2700\\ 12200\end{array}\right) \end{align*} Determinant of the co-efficient matrix \begin{align*} 5\times 38 - 7\times 21 \\ =43 \end{align*} Inverse of the co-efficient matrix \begin{align*} =\frac{1}{43}\begin{bmatrix}38 & -7 \\-21 & 5 \end{bmatrix} \end{align*} \begin{align*} \left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{43}\begin{bmatrix}38 & -7 \\-21 & 5 \end{bmatrix}\left(\begin{array}{c}2700\\ 2200\end{array}\right) \\ =\frac{1}{43}\left(\begin{array}{c}17200\\ 4300\end{array}\right)\\ =\left(\begin{array}{c}400\\ 100\end{array}\right) \\ textbook = sh. 400 \\ \text{Exercise book} = sh. 100 \end{align*} (c) In bookshop B the cost of a textbook was 5% less and that of an excercise book is 5% more than in bookshop A. Kasuku school bought the same number of exercise books and textbooks as Elimu school
Calculate the difference in the amount spent by Kasuku school and Elimu school. $$ Solution $$ \begin{align*} \small{\text{Amount spent by kasuku school}} \\ =\small{25\times\frac{95}{100}\times400+35\times\frac{105}{100}\times100} \\ =\small{Kshs.13175} \\ \small{\text{Difference}=13500-13175 }\\ =\small{Kshs. 325} \end{align*} 21. (a) Solve for \(x\) \begin{align*} (x-4)^2 = (x-8)(2x+7) \end{align*} $$ Solution $$ \begin{align*} \small{ x^2-8x+16 =2x^2+7x-16x-56 }\\ -x^2+x+72=0\\ x^2-x-72=0\\ (x-9)(x+8)=0\\ x=9 \\ \text{or}\\ x=-8 \end{align*} 9. A translation T maps A (-6, 2) onto A'(3,5) $$Solution$$ (a)Determine the translation vector \begin{align*} T = \left(\begin{array}{c}3\\ 5\end{array}\right)-\left(\begin{array}{c}-6\\ 2\end{array}\right)=\left(\begin{array}{c}9\\ 3\end{array}\right)\\ \end{align*} (b) A point P'(-4,2) is the image of P under T. Determine the co-ordinates of P (2mks) \begin{align*} \left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}-4\\ 2\end{array}\right)-\left(\begin{array}{c}9\\ 3\end{array}\right) \\ a= -13, b=-1 \\ P(-13,-1) \end{align*}

6.In a race Kipsang maintained an average speed of \(5m/s\).When he was \(310 m\) to the finishing line, Mutunga was \(50 m\) behind him. However Mutunga finished the race \(10 m\) ahead of kipsang. Determine Mutunga's average speed

\begin{align*} \text{Time taken by Kipsang to cover 300 m} =\frac{300}{5}\\ =60s\\ \text{Mutunga's speed}= \frac{(310+50)}{60}\\ =6m/s \end{align*}

A table is sold at Kshs. 4500 and a chair at Kshs.2000 . A saleman earns a commission of 8\% on every table and 5\% on every chair sold. On a certain week he sold 3 more chairs than tables and his total earnings were Kshs. 3980. Determine the number chairs he sold that week. 3mks