(Solved) 2019 BECE Mathematics (Maths) Past Questions and Answers Paper Two

In our earlier update, we gave you the (Solved) 2019 BECE Mathematics (Maths) Past Questions and Answers Paper one. In this update, we are bringing your the (Solved) 2019 BECE Mathematics (Maths) Past Questions and Answers Paper Two.

You can go through these questions and have a feel of how the questions and answers are. This will help you in your coming examination. Do let us know in the comment if you need further explantion.

 Paper Two

 

1. (a) Given that x = {whole numbers from 4 to 13} and y = {multiples of 3 between 2 and 20},

find .

(b) Find the Least Common Multiple (L.C.M) of the following numbers: 3, 5, and 9.

(c) If, find the value of  \frac{q}{p} ,\ where\ p\neq0 .

2. (a) Solve: \frac{4x+5}{5}+\frac{x-3}{4}=-1.

(b) The ratio of boys to girls in a school is 12:25. If there are 120 boys.

(i) how many girls are in the school?

(ii) what is the total number of boys and girls in the school?

(c) Simplify:  \left(8x^2y^3\right)\left(\frac{3}{8}xy^4\right)

 

3. (a) In an examination, 60 candidates passed Integrated Science or Mathematics. If 15 passed both subjects and 9 more passed Mathematics than Integrated Science, find the:

(i) number of candidates who passed each subject

(ii) probability that a candidate passed exactly one subject.

(b) Factorize: xy+6x+3y+18.

4. (a) Express 250% as a fraction in its lowest term.

(b)
2019 bece past questions maths

Use the diagram to find the value of x.

(c) Simplify:  2\div\left(\frac{15}{64}+\frac{6}{7}\right)

(d) if q=\left(\begin{matrix}7\\-1\\\end{matrix}\right)and\ r=\left(\begin{matrix}4\\-5\\\end{matrix}\right)and\ find (q + r )

5. (a)
https://www.passco.org/wp-content/uploads/2020/04/mappings.jpg

The mapping shows the relationship between x and y.
Using a scale of 2cm to 1unit on the x-axis and 2 cm to 2 units on the y-axis, draw two perpendicular axes 0x and 0y on a graph sheet for 1\le x\le5 and 0\le y\le14;

plot the point for each ordered pair, \left(x,\ y\right) .

Join the points with a straight line;

Using the graph, find the gradient of the line in 5 (a)(iii);

Use the graph to find the equation of the line in 5 (a)(iii).

6. The marks obtained by students in a class test were

4 8 7 6 7
2 1 7 4 7
3 7 8 4 3
7 5 2 7 2
5 4 8 3 2

(a) construct a frequency distribution table for the data.

(b) Find the:
(i) mode of the distribution;
(ii) median mark of the test;
(iii) mean mark.

 

Paper 2 Answers

 

Question 1.

(a)
x = { 4,5,6,7,8,9,10,11,12,13}
y= {3,6,9,12,15,18}

X\ \cap Y = {6,9,12}

(b) The LCM is the smallest positive number that all of the numbers divide into evenly.
We solve this question by;

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Since 3 has no factors besides 1 and 3. 3 is a prime number

Since 5 has no factors besides 1 and 5. 5 is a prime number

9 has factors of 1,3 and  3^2 .

The LCM of 3,5,9 is the result of multiplying all prime factors the greatest number of times they occur in either number.

The LCM of 3,5,9 is  3^2 \times 5 =45.

C.

\\ \frac{p+2q}{p}=\frac{7}{5}, \\ \text{simplifying }\\\ p+2q= \frac{7p}{5}, \\ \ 5p+10q= 7p,\\ \ 10q= 7p-5p \\ \ 10q = 2p \\ \frac{q}{p} = \frac{2}{10}\\ \frac{q}{p} = \frac{1}{5}\\

Question 2. (a) Find the LCM of 5 and 4 which is 20.

Multiply all terms by 20 and simplify

 \frac{4x+5}{5}+\frac{x-3}{4}=-1. \\ 20 \times \frac{4x+5}{5}+20 \times \frac{x-3}{4}=20 \times-1\\ 4 \times {(4x+5)}+5 \times {(x-3)}=20 \times-1\\ 16x+20+5x-15=-20\\ 21x+5=-20\\ 21x=-20-5\\ 21x=-25\\ x = -\frac{25}{21}

(b)(i)

Boys : Girls = 12:25,

If there are 120 boys, let x be the number of girls

120 boys: x girls

12:25 = 120: x

If more less divides

Therefore

 x = \frac{25}{12} \times 120 = 250

Hence there are 250 girls in the class

(ii) the total number of people in the school = Number of boys + number of girls

=250 + 120
=370

(c) Simplify:  \left(8x^2y^3\right)\left(\frac{3}{8}xy^4\right).

To solve, start by grouping exponential terms

(8\times\frac{3}{8})\cdot(x^2\times x) \cdot (y^3 \times y^4)\\

Simplifying..

3x^3y^7

Question 3

(a) Let U be the universal set

X – Number of students who passed integrated science

Y – Number of students who passed Mathematics

Solving this question requires the use of Venn diagrams.

From the information on given

X n Y = 15

9 more students passed mathematics than integrated science
Hence Y = x+9

2019 bece maths

Solving for x

x+9-15+15+x-15=60
2x-6=60
2x=66
X = 33

Number of students who passed math only = x+9-15 = 27

Number of students who passed science = 33-15 = 18

Probability of passing only one subject = (18+27)/60 = 3/4

(b) Factorize: xy+6x+3y+18.

xy+6x+3y+18
x(y+6)+3(y+6)
(x+3)(y+6)

 

Question. 4

a. 250% =  \frac{250}{100} = \frac{25}{10} = \frac{5}{2}

b. Using the knowledge that angles on a straight-line sum to 180, write an equation for the missing internal angles

2019 bece maths past questions

We solve this question by using the least known property of triangles shown in the image below.

 

Hence

 3x = (180-4x)+107\\ 7x = 287 \\ \frac{7x}{7} = \frac{287}{7}\\x= 41 \\

C.
 2\div\left(\frac{15}{64}+\frac{6}{7}\right) \\ \text{simplifying the term in the brackets,by first finding the LCM of denominators} \\ 2\div\left(\frac{7(15)+64(6)}{448}\right)\\ 2\div\left(\frac{105+384}{448}\right) \\ 2\div\left(\frac{489}{448}\right)\\ \text{applying the fraction rule} \\ 2\times\left(\frac{448}{489}\right)\\ \left(\frac{896}{489}\right)\\

D.  q=\left(\begin{matrix}7\\-1\\\end{matrix}\right) r=\left(\begin{matrix}4\\-5\\\end{matrix}\right)and\ \\ (q + r ) = \left(\begin{matrix}7+4\\-1+(-5)\\\end{matrix}\right) \\ (q + r ) = \left(\begin{matrix}11\\-6)\\\end{matrix}\right)

 Question. 5

The relation is plotted on a graph sheet as shown below.

2019 bece maths graph

(iv) gradient

The gradient can be found from the graph using the formula below, choosing any 2 points on the line

 =\frac{y_2-y_1}{x_2-x_1}\\ =\frac{12-0}{5-1}=\frac{12}{4}=3

Equation of a line is given as

Y = mx+Cw

Where m is the gradient
C is y-intercept

From the graph, the y-intercept is -3

So the equation of the line

Y=3x-3

(b) Simplify:  32\ \times8\times4\times2 , leaving your answer in the form 2^n.

 32 = 2^5\\ 8 = 2^3\\ 4 = 2^2\\ 2 = 2^1\\

 32\ \times8\times4\times2 = 2^{5+3+2+1} = 2^11

 Question. 6 (a).

Marks frequency cumulative frequency
1 1 1
2 4 5
3 3 8
4 4 12
5 2 14
6 1 15
7 7 22
8 3 25

 

(b) Find the:

(i) mode of the distribution = 7

(ii) median mark of the test = 5

(iii) mean mark = 4.92

In the Venn diagram M and N are the subsets of the universal set U. Use this information to answer questions 39 and 40.

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