BECE 2010 Mathematics Past Questions and Answers

The BECE 2010 Mathematics paper has 40 objective and 6 theory questions. Here are sample objective questions with the correct answers and explanations. Sit the full paper with the real time limit on ePrep.

Sample questions with answers

1. Which of the following sets is well defined?

A. {Man, Kofi, Red, 14}
B. {Ink, Mango, Green, Nail}
C. {Car, Road, Glass, Book}
D. {Seth, Mary, Jacob, Evelyn}

Answer: D. {Seth, Mary, Jacob, Evelyn}
All the sets except {Seth, Mary, Jacob, Evelyn} contains different categories of items. Only the set {Seth, Mary, Jacob, Evelyn} contains only one category of item, names of people.

2. If set B is a subset of set A, then

A. sets A and B have the same number of elements.
B. some members of set B can be found in set A.
C. no member of set B is in set A.
D. all the members of set B are in set A.

Answer: D. all the members of set B are in set A.
All members of a subset can be found in the set it is subset to.

3. The least common multiple (L.C.M) of 16, 30 and 36 is

A. 3
B. 6
C. 240
D. 720

Answer: D. 720
The easiest method is the prime factor method. If you decide to list the multiples of each and pick the least, its going to be cumbersome. Thus you have to list so many before reaching the common multiple in each.

Prime factors of 16

$\begin{matrix} & 16 \\ 2 & 8 \\ 2 & 4 \\ 2 & 2 \\ 2 & 1\end{matrix}$

Prime factors of 16 = 2 x 2 x 2 x 2

Prime factors of 16 = 2$^{4}$

Prime factors of 30

$\begin{matrix} & 30 \\ 2 & 15 \\ 3 & 5 \\ 5 & 1\end{matrix}$

Prime factors of 30 = 2 x 3 x 5

Prime factors of 36

$\begin{matrix} & 36 \\ 2 & 18 \\ 2 & 9 \\ 3 & 3 \\ 3 & 1\end{matrix}$

Prime factors of 16 = 2 x 2 x 3 x 3

Prime factors of 16 = 2$^{2}$ x 3$^{2}$

The list common multiple is the product (multiplication) of each of the prime factors in their highest powers

L.C.M of 16, 30 and 36 = 2$^{4}$ x 3$^{2}$ x 5

L.C.M of 16, 30 and 36 = 16 x 9 x 5
L.C.M of 16, 30 and 36 = 720

4. The sum of 5 and x divided by 4 is equal to 3.25. Find the value of x.

A. 8
B. 7
C. 2$\frac{1}{4}$
D. -3$\frac{4}{13}$

Answer: A. 8
Sum means addition.

$\frac{5 + x}{4}$ = 3.25

3.25 = $\frac{325}{100}$

$\frac{5 + x}{4}$ = $\frac{325}{100}$

Cross multiply and solve for x

100 (5 + x) = 4 x 325

100 x 5 + 100 x x = 1300
500 + 100*x* = 1300
100*x* = 1300 - 500
100*x* = 800

Divide both sides by 100.

x = $\frac{800}{100}$

x = 8

5. The numbers 32, 33, 34, ..., ..., 42 form a sequence in base 5. Find the missing numbers.

A. 35, 36
B. 30,31
C. 40, 41
D. 31, 41

Answer: C. 40, 41
Convert the numbers from base 5 to base 10 so you can know the order of the numbers to find the missing numbers in base ten and change back to base 5

32$_{five = 3 x 5$^{1}$ + 2 x 5$^{0}$}$

Any number/letter raised to the power 0 is 1

32$_{five = 3 x 5 + 2 x 1}$

32$_{five = 15 + 2}$

32$_{five = 17}$

33$_{five = 3x 5$^{1}$ + 3 x 5$^{0}$}$

33$_{five = 3 x 5 + 3 x 1}$

33$_{five = 15 + 3}$

33$_{five = 18}$

34$_{five = 3x 5$^{1}$ + 4 x 5$^{0}$}$

34$_{five = 3 x 5 + 4 x 1}$

34$_{five = 15 + 4}$

34$_{five = 19}$

The next sequence order would be 20 and 21 so you can convert the 20 and 21 to base 5

20 base 10 to base 5

$\begin{matrix} & 20 \\ 5 & 4 r 0 \\ 5 & 0 r 4\end{matrix}$

You list the remainders from the bottom to the top.

20 = 40$_{five}$

21 base 10 to base 5

$\begin{matrix} & 21 \\ 5 & 4 r 1 \\ 5 & 0 r 4\end{matrix}$

You list the remainders from the bottom to the top.

21 = 41$_{five}$

The missing numbers are 40 and 41.

Note: There are no numbers ≥ 5 in base 5

Alternatively, when the number gets to 35 since there is no 5 in base 5, 5 is 10 in base 5 so you write the 0 and add the 1 to 3 making 40, then you continue adding 1 so the next sequence would be 41.

6. Write down all the integers in the set A = {-10, -4, 0, $\frac{1}{4}$, 2$\frac{1}{4}$, 45, 100}

A. {-10, -4, 0, 45, 100}
B. {-10, -4}
C. {0, 45, 100}
D. {$\frac{1}{4}$, 2$\frac{1}{4}$}