BECE 2019 Mathematics (Maths) Past Question 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

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

 = {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

2

 (a)Solve:

(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:

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

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.

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

End Of Paper

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