Core Mathematics 2019 WASSCE-Past Questions and Answers

“Core Mathematics 2019 WASSCE-Past Questions and Answer” – is an online learning resource for students.

 

 

Thus, it has been designed to help students to pass their Core Maths WASSCE and other relevant mathematics exams. So what does our site offer? Now read on and see for yourself. Read more.

Specifically, let’s concentrate on the solution to the Core Mathematics 2019 WASSCE-Past Questions and Answers for wassce students. Very soon, we will make the wassce maths questions and answers pdf available for your download

  

General Overview – Core Mathematics 2019 WASSCE

Click here to read more   

We are here to encourage you that you can easily pass mathematics subject. Educareguide has made mathematics very easy for every student. It is actually not different from the following subjects or papers:   

 

Other Relevant Examination

These mathematics questions and answer are not only for WASSCE. Rather, they are also relevant to the following examination:   

 

1   (a)Mathematics question on standard Form

Without using mathematical tables or calculators, evaluate

$\frac{0.15\times 0.567}{0.05\times 0.189}$

   , leaving the answer in standard form

Solution:

$$\begin{gathered} \frac{0.015 x 0.567}{0.05 x 0.189} \\ =\frac{15x{10}^{–3} x 567 x{10}^{–3}}{5 x{10}^{–2} x189 x{10}^{–3}} \\ =\frac{15 x567 x{10}^{–3} x{10}^{–3}}{5 x189 x{10}^{–2} x{10}^{–3}} \\ =\frac{3x3\times {10}^{–6}}{1x1x{10}^{–5}} \\ = 9 x{10}^{–6–(–5)} \\ =9 x {10}^{4+5} \\ =9x{10}^{–1} \end{gathered}$$

1   (b) Mathematics question on Linear Equations and Inequalities

 

$$\begin{gathered} If \frac{5y–x}{8y+3x}=\frac{1}{5}, \mathrm{find}, \mathrm{correct} \mathrm{to} \mathrm{two} \mathrm{decimal} \mathrm{places} \\ \mathrm{the} \mathrm{value} \mathrm{of} \frac{x}{y} \end{gathered}$$

Solution:

This is a mathematics question on Linear Equations and Inequalities

The answer to the above question can be found below:

$$\begin{gathered} \frac{5y–x}{8y+3x}=\frac{1}{5} \\ 1\left(8y+3x\right)=5\left(5y–x\right) \\ 8y+3x=25y–8y \\ 8x=17y \\ Divide both \mathrm{si}des by 8y, that gives us \\ \frac{8x}{8y}=\frac{17y}{8y} \\ \frac{x}{y}=\frac{17}{8} \\ \frac{x}{y}=2.13 Answer corrected to two decimal places \end{gathered}$$

 

2 (a) Mathematics question on variation

Z varies directly as x and inversely as twice the cube root of y. If Z = 8, when x = 4 and y = 

$\frac{1}{8}$

find the relation for y in terms of x and Z.

Solution:

Answer to question to wassce 2019 core maths question 2a:

$$\begin{gathered} Z\infty \frac{x}{2\sqrt[3]{y}} \Rightarrow Z=\frac{kx}{2\sqrt[3]{y}} ............\left(1\right) \\ When Z=8, x=4, y=\frac{1}{8} \\ 8=\frac{k\left(4\right)}{2\sqrt[3]{{\displaystyle \frac{1}{8}}}} \Rightarrow 8=\frac{4k}{2\left({\displaystyle \frac{1}{2}}\right)} \Rightarrow 8=\frac{4k}{2\left(1\right)} \\ \Rightarrow 8=\frac{4k}{2} \Rightarrow 8=4k \Rightarrow k=\frac{8}{4}=2 \\ Putting k=2 into \left(1\right) gives you \\ Z=\frac{2x}{2\sqrt[3]{y}} \\ Z= \frac{x}{2\sqrt[3]{y}} \\ \sqrt[3]{y}=\frac{x}{Z} \\ {\left(\sqrt[3]{y}\right)}^3={\left(\frac{x}{Z}\right)}^3 \\ y=\frac{x^3}{Z^3} Ans. \end{gathered}$$

 

2   (b)A mathematics question on Algebraic Expressions

Factorize completely:

$4b^2 — ab + {(a + 9b)}^2 — a^2$

Solution:

$$\begin{gathered} 4b^2 —ab + {(a +9b)}^2 —a^2 = 4b^2 —ab+ (a +9b)(a +9b)— a^2 \\ = 4b^2 —ab+a^2 +9ab+9ab+81b^2 —a^2 \\ = 4b^2 —ab+ a^2 +18ab +81b^2 — a^2 \\ =4b^2 +81b^2 — ab+18ab+ a^2 —a^2 \\ = 85b^2 +17 ab \\ =17b(5b+ a) \end{gathered}$$

 

3   (a)A mathematics question on Algebraic Expressions

Solve 

$\frac{5x–7}{6}+\frac{2x–3}{4}=\frac{2}{3}$

Solution:

$$\begin{gathered} \frac{5x–7}{6}+\frac{2x–3}{4}=–\frac{2}{3} \\ \mathrm{Multiplying} \mathrm{through} \mathrm{by} 12, \mathrm{we} \mathrm{will} \mathrm{get} \\ 12\left(\frac{5x–7}{6}\right)+12\left(\frac{2x–3}{4}\right)=\left(–\frac{2}{3}\right) \\ 2\left(5x–7\right)+3\left(2x–3\right)=–8 \\ 10x–14+6x–9=–8 \\ 10x+6x=8+14+9 \\ 16x=15 \\ x=\frac{15}{16} \end{gathered}$$

 

3 (b) A mathematics question on Surds

Evaluate:

$\frac{\sqrt{\sqrt{28}+\sqrt{343}}}{2\sqrt{63}}+\frac{5}{3}$

Solution:

Below is the answer to question to wassce 2019 core maths question 3b:

$$\begin{gathered} \frac{\sqrt{28+\sqrt{343}}}{2\sqrt{63}}+\frac{3}{5} \\ =\frac{\sqrt{4\times 7}+\sqrt{49\times 7}}{2\sqrt{9\times 7}}+\frac{5}{3} \\ =\frac{2\sqrt{7}+7\sqrt{7}}{2\left(3\sqrt{7}\right)}+\frac{5}{3} \\ =\frac{9\sqrt{7}}{6\sqrt{7}}+\frac{5}{3} \\ =\frac{3}{2}+\frac{5}{3} \\ =\frac{19}{6} \\ 3\frac{1}{6} Ans. \end{gathered}$$

 

4 Mathematics question on Percentages

A car dealer made a profit of 22.5% by selling a car for GH¢ 58,000.00. Find, correct to two decimal places, the percentage profit if the car had been sold for GH¢ 61,200.00

Find the answer 2019 core mathematics, question 4

$$\begin{gathered} At the first ins\tan ce, \\ Let P\% represent profit percentage and \\ SP represent selling price \\ CP=\frac{100}{100+P\%}\times SP \\ CP=\frac{100}{100+22.5}58,000 \\ CP=\frac{100}{122.5}\times 58,000 \\ CP=47,346.94 \\ At the second ins\tan ce: \\ CP=47,346.94, SP=61,200 \\ P=SP–CP \\ P=61200–47346.94 \\ P=13,853.06 \\ P\%=\frac{P}{CP}\times 100\% \\ P\%=\frac{13,853.06}{47,346.06}\times 100\% \\ P\%29.2586\% \\ P\%=29.26\% Ans. (remember the question said to 2 decimal places. \end{gathered}$$

 

5(a) Mathematics question on Probability

A number is chosen at random from Q = { 1, 2, 3, …, 10). Find the probability that the chosen number is either a prime factor of 42 or a multiple of 3.

Solution:

Solution to wassce 2019 core maths quuestion 5a

$$\begin{gathered} Q= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\ n\left(Q\right) =10 \\ Let A = \{Prime factors of 42 in set Q\} = \{2, 3, 7\} \\ n\left(A\right)= 3 \\ Let B = \{Multiples of 3 in set Q) = \{3, 6, 9\} \\ n\left(B\right)= 3 \\ AnB= \left\{3\right) \\ n(A n B) = 1 \\ P(A or B) = P\left(A\right) + P\left(B\right) — P(A n B) \\ \frac{3}{10}+\frac{3}{10}–\frac{1}{10} \\ \frac{5}{10} \\ \frac{1}{2} \end{gathered}$$

 

5   (b)Mathematics question Number Bases

If

${110}_x ={1020}_4$

, find the value of x.

Solution to Mathematics question wassce 2019 5b

$$\begin{gathered} 110, =10204 \\ \left(1\mathrm{x}{x}^2\right)+\left(l\mathrm{x}{x}^1\right)+\left(0x{x}^0\right)=\left(1\mathrm{x}{4}^3\right)+\left(1\mathrm{x}{4}^2\right)+ \left(1\mathrm{x}{4}^1\right)+(l\mathrm{x} 4^0) \\ {X}^2 +X+ 0 =64+0+8+0 \\ x^2–X = 72 \\ x2+ x–72=0 \\ (x–8)(x+ 9) = 0 \\ x = 8, x = —9 \\ x = 8 \end{gathered}$$

 

6(a) Mathematics question on vectors

$$\begin{gathered} If a=\left(\begin{array}{c}2\\ 3\end{array}\right), b=\left(\begin{array}{c}4\\ 5\end{array}\right), and r=a+\frac{1}{2}\left(a–b\right), find \\ \left(i\right) r \\ \left(ii\right) \left|r\right| \end{gathered}$$

Solution:

The answer to question to wassce 2019 core maths question 6a(i)

$$\begin{gathered} r=a+\frac{1}{2}\left(a–b\right) \\ r=\left(\begin{array}{c}2\\ 3\end{array}\right)+\frac{1}{2}\left[\left(\begin{array}{c}2\\ 3\end{array}\right)–\left(\begin{array}{c}4\\ 5\end{array}\right)\right] \\ r=\left(\begin{array}{c}2\\ 3\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}–2\\ –2\end{array}\right) \\ r=\left(\begin{array}{c}2\\ 3\end{array}\right)+\left(\begin{array}{c}–1\\ –1\end{array}\right) \\ r=\left(\begin{array}{c}1\\ 2\end{array}\right) \end{gathered}$$

The answer to question to wassce 2019 core maths question 6a(ii)

$$\begin{gathered} \left|r\right|=\sqrt{1^2+2^2} \\ \left|r\right|=\sqrt{1+4} \\ \left|r\right|=\sqrt{5units} \end{gathered}$$

 

6   (b) A mathematics question on Change of Subject and Substitution into Formulae

$$\begin{gathered} Given that a=bc and n=\frac{mk}{ec}, \\ \left(i\right) express k in terms of a,b,e,m and n; \\ \left(ii\right) find, correct to three significant figures, the value of k, when \\ a=\frac{1}{2}, b=–4, e=3, m=7, n=–5. \end{gathered}$$

Solution:

Below is the answer to question to wassce 2019 core maths question 6b:

(i) Find the answer 2019 core mathematics, question, 6b(i)

$$\begin{gathered} a=bc ..............\left(1\right) \\ n=\frac{mk}{ec} ............\left(2\right) \\ From \left(1\right) \\ c=\frac{a}{b} ...............\left(3\right) \\ Putting \left(3\right) into \left(2\right), we will get \\ n=\left(\frac{a}{b}\right) \\ n=\frac{{\displaystyle \frac{mk}{ae}}}{b} \\ n=mk\times \frac{a}{ae} \\ n=\frac{bmk}{ae} \\ bmk=\frac{aen}{bm} \end{gathered}$$

 

(ii)Find the answer 2019 core mathematics, question, 6b(ii)

 

$$\begin{gathered} When a=\frac{1}{2}, b=–4, c=3, m=7, n=–5 \\ k=\frac{{\displaystyle \frac{1}{2}}\left(3\right)\left(–5\right)}{–4\left(7\right)} \\ =\frac{–{\displaystyle \frac{15}{2}}}{–28} \\ =\frac{15}{2}\times \frac{1}{28} \\ =\frac{15}{56} Ans. \end{gathered}$$

 

7  Mathematics question Trigonometry II

(a) Copy and complete the table of values for y =5sinx+9cosx for 

$0\le x\le {150}^o$

x	$0^o$	${30}^o$	${60}^o$	${90}^o$	${120}^o$	${150}^o$
y		10.3			-0.2

(b) Using a scale of 2 cm to 30° on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = 5 sin x + 9cos x for 0° x 150°.
(c) Use the graph to solve the equations:
(i) 5sinx+9cosx= 0 ;
(ii) 5sinx+9cosx= 2.
(d) Using the graph, find, the value of y when x= 45°

Solution mathematics question9 wassce 2019 

(a) y=5sinx+9cosx

x	$0^o$	${30}^o$	${60}^o$	${90}^o$	${120}^o$	${150}^o$
y	9.0	10.3	8.8	5.0	-0.2	-5.3

Refer to the graph

(c) (i) 5 sin x+9 cos x …………….(1)
5 sin x+9 cos x= y (2)
Comparing (1) and (2): y=o
The line y = 0 cuts the curve at x =119°
x =119°
.
(ii) 5 sin x+9 cos x = 2 (1)
5 sin x+9 cos x= y (2)
Comparing (1) and (2):
y = 2
The line y = 2 cuts the curve at x =107.5° x =107.5°
(d) From the graph, when x = 45°, y = 9.9

 

8 Mathematics question on Construction and Loci

(a) Using ruler and a pair of compasses only, construct;
(i) the quadrilateral ABCD such that

$\left|AB\right|= 6.5 cm, \left|BC\right| = 9 cm, \left|AD\right| = 4 cm, \angle ABC = 60° and \angle BAD =120°$

(ii) the perpendicular bisectors of 

$\overline{BC} and \overline{CD}$

(b) Locate the point of intersection, T, of the two bisectors in (a)(ii).
(c) With the point T in (b) as centre, draw a circle to pass through the vertices B, C and D.
(d) Measure:
(i) 

$\left|BT\right|$

(ii) 

$\left|CD\right|$

Solution:

Answer – wassce 2019 core mathematics question 8

For answer a, b, and c, refer to the diagram below:

 

 

 

 

 

 

 

d(i) 

$\left|BT\right|=4.9cm$

d(ii) 

$\left|CD\right|=5.9cm$

 

9 Mathematics question on Rigid Motion (Transformation)

(a) Using a scale of 2 cm to 1 unit on both axes, draw on a sheet of graph paper, two perpendicular axes Ox and Oy for 

$–5\le y\le 5.$

(b) Draw on the same graph sheet, indicating clearly all vertices and their coordinates:
(i)

$∆$

ABC with vertices A(2, 1), B(1, 4) and C( —1, 2);
(ii) the image

$∆A_1{B}_1{C}_1 of ∆ABC$

under a reflection in the line y = 0, where

$A \to A_1 , B \to B_1, and C \to C_1$

 ;

(iii) the image 

$∆A_2{B}_2{C}_2 of ABC$

under a translation by the vector

$\left(\begin{array}{c}–2\\ 1\end{array}\right)$

, where

$A \to A_2, B \to B_2 and C \to C_2 ;$

(iv) the image 

$∆A_3{B}_3{C}_3 of ABC$

under an anticlockwise rotation of 90° about the origin, where 

$A \to A_3, B \to B_3 and C \to C_3$

.

(c) What single transformation maps 

$∆A_1{B}_1{C}_1 onto ∆A_3{B}_3{C}_3, where A_1 \to A_3, B_{1\to } B_3 and C_1 —› C_3 ?$

Solution:

The answer to question to wassce 2019 core maths question 9

(a) Refer to the graph below:

(b) (i) Refer to the graph below:

(ii) Reflection  in the line y=0 or the x-axis

$$\begin{gathered} \left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}x\\ –y\end{array}\right) \\ A\left(\begin{array}{c}2\\ 1\end{array}\right)\to A_1\left(\begin{array}{c}2\\ –1\end{array}\right) \\ B\left(\begin{array}{c}1\\ 4\end{array}\right)\to B_1\left(\begin{array}{c}1\\ –4\end{array}\right) \\ C\left(\begin{array}{c}\\ \end{array}\right)\to C_1\left(\begin{array}{c}–1\\ –2\end{array}\right) \end{gathered}$$

(iii) Point +Translation Vector = image

$$\begin{gathered} A\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{c}–2\\ 1\end{array}\right)=A_2\left(\begin{array}{c}0\\ 2\end{array}\right) \\ B\left(\begin{array}{c}1\\ 4\end{array}\right)+\left(\begin{array}{c}–2\\ 1\end{array}\right)=B_2\left(\begin{array}{c}–1\\ 5\end{array}\right) \\ C\left(\begin{array}{c}–1\\ 2\end{array}\right)+\left(\begin{array}{c}2\\ 1\end{array}\right)=C_2\left(\begin{array}{c}–3\\ 3\end{array}\right) \end{gathered}$$

(iv) 

${90}^o$

anticlockwise rotation about the origin

$$\begin{gathered} \left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}–y\\ x\end{array}\right) \\ A\left(\begin{array}{c}2\\ 1\end{array}\right)\to A_3\left(\begin{array}{c}–1\\ 2\end{array}\right) \\ B\left(\begin{array}{c}1\\ 4\end{array}\right)\to B_3\left(\begin{array}{c}–4\\ 1\end{array}\right) \\ C\left(\begin{array}{c}–1\\ 2\end{array}\right)\to C_3\left(\begin{array}{c}–2\\ –1\end{array}\right) \end{gathered}$$

(c) 

$$\begin{gathered} A_1\left(\begin{array}{c}2\\ 1\end{array}\right)\to A_3\left(\begin{array}{c}–1\\ 2\end{array}\right) \\ {B}_1\left(\begin{array}{c}1\\ –4\end{array}\right)\to B_3\left(\begin{array}{c}–4\\ 1\end{array}\right) \\ {C}_1\left(\begin{array}{c}–1\\ –2\end{array}\right)\to C_3\left(\begin{array}{c}–2\\ –1\end{array}\right) \\ From the mappings above, it can be seen that: \\ \left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}y\\ x\end{array}\right) \end{gathered}$$

Graph work- the answer to wassce 2019 core maths question 9

 

 

 

 

 

 

 

10(a) Mathematics question on Sets and Operations on Sets

In a class of 50 students. 24 like football, 21 basketball and 18 cricket. Six like football and basketball only, 3 like basketball only. 5 like all the three. games and 14 did not like any of the three games.

1. Illustrate this information on a Venn diagram.
2. Find the number of students who like:
   1. football and cricket only;
   2. exactly one of the games.

Solution to wassce core maths 2019 question 10a

n(U)= 50 , n(F) = 24, n(B) = 21, n(C) =18,

n(F n B only) = 25, n(B only) = 3, n(F nB nC) = 5, n(F u B u C)’ = 14

Let n(F n C only) = x, n(B nC only) = y n(F only) = m, n(C only) = b

 

(i) The Venn diagram is shown below.

U (50)

 

 

 

 

 

 

(ii) From set B: 6+5+3+y= 21

14+ y = 21

y =7

 

n(C)+m+6+3+14 =50

18+m+6+3+14=50 nz + 41= 50

m =9

 

From set F:

m + 6+5 = 24

9+x+6+5 = 24 x+ 20 = 24 x =4

 

From set C: x+5+ y+n =18.

4+5+7+b=18 b+16=18 b = 2

 

(a) The number of students who like football and cricket only

From the above calculations the answer = 4

(b) The number of students who like Exactly one of the games

= 9+3+2 = 14

 

10(b) Mathematics question on Sequence

If (3-a), 6, (7 – 5a) are consecutive terms of a
Geometric Progression (G.P) with common ratio r > 0 , find the values of a.

Solution:

Solution for wassce core maths 2019 question 10b

$$\begin{gathered} \frac{U_2}{U_1}=\frac{U_3}{U_2} \\ \frac{6}{3–a}=\frac{7–5a}{6} \\ \left(3–a\right)\left(7–5a\right)=6\left(6\right) \\ 21–15a–7a+5a^2=36 \\ 5a^2–22a+21–36=0 \\ 5a^{2}22a–15=0 \\ a=5, a=–\frac{3}{5} \\ sice r>0 \\ a=–\frac{3}{5} \end{gathered}$$

 

11  (a) Mathematics question on Ratio, Proportion and Rates

Two passenger trains, A and B, 450 km apart, start to moves towards each other at the same
time and meet after 2 hours. If train B. travels 

$\frac{8}{7}$

as fast as train A, find the speed of each train.

 

$$\begin{gathered} Let the speed of train A = x \\ Speed of train B =\frac{8}{7}x \\ Dis\tan ce =Speed \times Time \\ Therefore, the dis\tan ce covered by train A=x\times 2=2x \\ Also, dis\tan ce covered by train B=\frac{8}{7}x\times 2=\frac{16}{7}x \\ Dis\tan ce covered by A+Dis\tan ce covered by B=450km \\ Therefore: \\ 2x+\frac{16}{7}x=450 \\ 14x+16x=3150 \\ 30x=3150 \\ x=\frac{3150}{30}=105 \\ Therefore, Speed of train A=105k{m}^1 Ans. \\ Speed of train B=\frac{8}{7}\times 105km{h}^1=120km{h}^1 \end{gathered}$$

11(b) Mathematics question on Mensuration II

A solid cube of side 8 cm was melted to form a solid circular cone. The base radius of the cone is 4 cm. Calculate, correct to one decimal place, the height of the cone.

$[Take \mathrm{\pi } =\frac{22}{7}$

Solution:

Answer to question to wassce 2019 core maths question 11b:

$$\begin{gathered} Volume of cube = l^3=8^3=512c{m}^3 \\ Volume of cone = \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{1}{3}\times \frac{22}{7}\times 4^2\times \mathrm{h} \\ =16.762\mathrm{h} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cone} = \mathrm{volume} \mathrm{of} \mathrm{cube} \\ \Rightarrow 16.762\mathrm{h}=512 \\ \mathrm{h}=\frac{512}{16.762} \\ =30.545 \mathrm{Ans}. \end{gathered}$$

 

 

12(a) Mathematics question on Plane Geometry II (Circle Theorems)

 

 

 

 

 

 

 

Pleas do note that the above diagram is not drawn to scale

The diagram shows a circle ABCD with centre E. Quadrilateral EADC is a rhombus,

$\angle$

BAE =

$\angle$

ECB = n and

$\angle$

ABC = m. Find:
(i) m;
(ii) n.

Solution

Answer to question to wassce 2019 core maths question 12a:

(i) Angle at the centre of a circle is twice the
angle at the circumference.

$\angle$

AEC =2m
Sum of angles at the circumference of opposite
segments of a circle is 180°.

$\angle$

ADC = 180° — m
Opposite angles of a rhombus are equal.

$\angle$

AEC =

$\angle$

ADC

$$\begin{gathered} 2m =180—m \\ 2m + m =180° \\ 3m =180° \\ m = {60}^o \end{gathered}$$

(ii)

$\angle$

AEC = 2m = 2(60°) =120°

$∆$

AEC is isosceles. For this reason,

$\angle$

EAC =

$\angle$

ECA
Let

$\angle$

EAC =

$\angle$

ECA= x

$$\begin{gathered} From ∆AEC: \\ \angle EAC + \angle ECA+ \angle AEC =180° \\ x+x+120° =180° \\ 2x =180°–120° \\ 2x = 60° x = 30° \\ Therefore, \angle EAC = \angle ECA= 30° \\ From \angle ABC: \\ \angle ABC + \angle BAC + \angle BCA =180° \\ m+ (n+ 30°) + (n + 30°) =180° \\ 60°+n+30°+n+30°=180° \\ 2n +120° =180° \\ 2n =180° —120° \\ 2n = 60° \\ n = 30° \end{gathered}$$

 

12(b) Mathematics question on Relations, Mappings and Functions

The functions f and g are defined as

$$\begin{gathered} f:x\to 2–x^2 and g:x\to \frac{1}{x–1} \\ \left(i\right) g\left(–\frac{1}{4}\right); \\ \left(ii\right)\hspace{0.17em}\frac{f\left(2\right)}{g\left(3\right)} \end{gathered}$$

 

Answer to wassce 2019 core mathematics quaestion 12(b)

$$\begin{gathered} g\left(x\right)=\frac{1}{x–1} \\ g\left(–\frac{1}{4}\right)=\frac{1}{–{\displaystyle \frac{1}{4}}–1} \\ =\frac{1}{–{\displaystyle \frac{5}{4}}} \\ =\frac{4}{5} \\ \left(ii\right) f\left(2\right)=2–2^2 \\ =2–4 \\ =–2 \\ g\left(3\right)=\frac{1}{3–1}=\frac{1}{2} \\ Therefore, \frac{f\left(2\right)}{g\left(3\right)}=\frac{–2}{{\displaystyle \frac{1}{2}}}–2\times \frac{2}{1}=–4 \end{gathered}$$

 

13(a) Mathematics question on Sequence

The fourth term of an Arithmetic Progression (A.P) is one less than twice the second term. If the sixth term is 7, find the first term.

Solution

Answer to wassce 2019 core mathematics quaestion 13(a)

$$\begin{gathered} U_4 = 2U_2 –1 \\ a +3d =2(a + d) –1 \\ a +3d =2a +2d –1 \\ a –2a +3d –2d = –1 \\ –a + d = –1 .............\left(1\right) \\ {U}_6 =7 \\ –a+5d =7 .............\left(2\right) \\ \left(1\right) x 5: \\ –5a +5d = –5 \left(3\right) \\ \left(2\right) – \left(3\right): \\ 6a =12 a \\ = 2 \\ Therefore, the first term is 2. \end{gathered}$$

 

13 (b) Mathematics question on percentages

A clerk spends 

$\frac{1}{5},\frac{1}{3} and \frac{1}{8}$

of of his annual salary on rent, transport and entertainment respectively. If after all these expenses he had GH¢ 4,100.00 left, find how much he earns per annum.

Solution:

Answer to wassce 2019 core mathematics quaestion 13(b)

$$\begin{gathered} Fraction spent =\frac{1}{5}+\frac{1}{3}+\frac{1}{8}=\frac{79}{120} \\ Fraction left=1–\frac{79}{120}=\frac{41}{120} \\ Let the clerk‘s annual salary be x \\ Because the amount left was GH¢4,100, we will get \\ \Rightarrow \frac{41}{120}\times x=4,100 \\ 41x=120\times 4,100 \\ 41x=492,000 \\ x=\frac{492,000}{41} \\ x=12,000 \\ Therefore, the clerk‘s annual salary is GH¢12,000.00 \end{gathered}$$

 

13 (c) A mathematics question on Relations, Mappings and Functions

Given that   

$f\mathit{ }\mathit{:}\mathit{ }x\mathit{ }\mathit{\to }\mathit{ }\mathit{2}{x}^{\mathit{2}}\mathit{ }\mathit{—}\mathit{ }\mathit{8}x\mathit{ }\mathit{+}\mathit{5}\mathit{,}\mathit{ }g\mathit{ }\mathit{:}\mathit{ }x\mathit{ }\mathit{\to }\mathit{ }x\mathit{ }\mathit{–}\mathit{2}\mathit{;}$

find:
(i) f (-3) ;
(ii) the value of x such that f (x) = g(x) .

Solution:

Below is the answer to question to wassce 2019 core maths question 13c:

Find the answer  2019 core maths question 13c(i):

$$\begin{gathered} f \left(x\right)= 2x^2 – 8x + 5 \\ f (–3) = 2{(–3)}^2 – 8(–3) +5 \\ =18+24+5 \\ = 47 \end{gathered}$$

Find the answer  2019 core maths question 13c(ii):

$$\begin{gathered} f \left(x\right) = g\left(x\right) \\ 2x^2 – 8x + 5 = x–2 \\ 2x^2 – 9x 4– 7 = 0 \\ (x–1)(2x–7)=O \\ x=1 and x=\frac{7}{2} \end{gathered}$$

 

You Might As Well Have Interest In The Following:

• Wassce Core Mathematics 1993 Paper 2 Question 21 – Answer
• Business Management 2007 Past Questions and Answers
• Business Management 2006 Past Questions and Answers
• Business Management 2005 Past Questions and Answers
• Business Management 2004 Past Questions and Answers
• Business Management 2003 Past Questions and Answers
• Business Management 2002 Past Questions and Answers
• Business Management 2001 Past Questions and Answers
• Business Management 2000 Past Questions and Answers
• Business Management 1999 Past Questions and Answers
• Business Management 1998 Past Questions and Answers
• Business Management 1997 Past Questions and Answers
• Business Management 1996 Past Questions and Answers
• Business Management 1995 Past Questions and Answers
• Business Management 1994 Past Questions and Answers
• Business Management 1993 Past Questions and Answers

 

Conclusion:

I believe Educareguide has been of help to you with regards to your subject of concern. Also there are many other contents we have available to help you in your education.

Furthermore, if there is any contribution/comment/concern that you would want to make, it is warmly welcome on our site. Simply proceed to Login/Register to submit your post.

Now, please, subscribe to Educareguide and contact us for further assistance for your education. Finally, fill the contact form on the side bar to reach us. Nevertheless, do not forget to pass a comment in the comment section below. Indeed, we will gladly appreciate to know how you think about this article. Thanks.