2020 - JAMB Mathematics Past Questions and Answers - page 1

When working in base 3, the largest number is 2. Just like in base 10, whenever you get a sum of values up to 3, you write 0 and carry over 1 and added to the next digit.

Hence,

    212
-  121
+ 222
= 1020

Ans: (1020)\(_3\)

The goal of factorization is to breakdown the given number into a simpler one. The result is usually two or more expressions that when multiplied gives the original number.

Recall that a\(^2\) - b\(^2\) = (a + b) (a - b)

Hence,

(4a + 3) \(^2\) - (3a - 2)\(^2\) => (a + b) (a - b) 

= ([4a + 3] + [3a - 2])([4a + 3] + [3a - 2])

= (4a + 3 + 3a - 2)(4a + 3 - 3a + 2)

= (7a + 1)(a + 5)

= (a + 5) (7a + 1) 

Ans: (a + 5) (7a + 1) 

To find the median, we first arrange in ascending order:

89, 100, 108, 119, 120,130, 131, 131, 141, 161

Since there are even number of items, we take the avarage of the two middle items:

Median = \(\frac{120 + 130}{2}\) = 125

What we need is a solution set that satisfies the given inequality. Each value in the solution set will satisfy the inequality and no other value will satisfy the inequality.

\(\frac{1}{3}\) (x + 1) - 1 > \(\frac{1}{5}\)(x + 4)  = \(\frac{x + 1}{3} - 1\) > \(\frac{x + 4}{5}\) 

\(\frac{x + 1}{3} - \frac{x + 4}{5} -1\) > 0 

= \(\frac{5x + 5 - 3x - 12}{15}\) 

2x - 7 > 15

2x > 22

Divide through by 2:

x > 11

\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.0255 \times 1.2}{0.024}\) 

= \(\frac{0.0264}{0.024}\)

= 1.1

Let x = numbers of boys that can work at \(\frac{5}{4}\) the rate as 10 girls 

Then,

1 hrs, x boys will work for \(\frac{\frac{1}{5} \times 10}{4}\) 

x = \(\frac{5}{4}\) x 10 

= 8 boys 

This means that 8 boys will do the work of ten girls at the same rate 

4 + 8 = 12 boys cut the field in 5 hrs for 3 hrs, 

\(\frac{12 \times 5}{3}\) boys will be needed

= 20 boys.

The circumference of latitude 0\(^o\)s = 2\(\pi\)r cos \(\theta\)

Where R is the radius of the earth, as given in the question.

 

• Draw a rectangle with edges ABCD to represent the floor.
• Draw a line from A to C, representing the diagonal of the floor

Use Pythagoras theorem to calculate the diagonal (AC):

AC\(^2\) = 144 + 81 = \(\sqrt{225}\) 

AC = 15cm

Given that height(h) of room 8m, let's find the cosine of the angle which a diagonal of the room (EC) makes with the floor.

EC\(^2\) = AC\(^2\) + h\(^2\)

EC\(^2\) = 15\(^2\) + 8\(^2\)

\(\frac{adj}{Hyp} = \frac{15}{17}\) 

EC\(^2\) = \(\sqrt{225 + 64}\)

EC = \(\sqrt{289}\)

EC = 17 

\(Cos\theta\) = Adjacent/Hypoteneous

= 15/17

 \(\begin{array}{c|c|} x & F & Frequency(fx)\\ 0 & 4 & 0 \\ 1 & 3 & 3\\ 2 & 15 & 30 \\ 3 & 16 & 48\\ 4 & 1 & 4 \\ 5 & 0 & 0 \\ 6 & 1 & 6 \\ & \sum Fx = 91&  \end{array}\)

\(\sum fx\) = 91

Number of students offering physics are 

\(\frac{12}{360}\) x 150

= 5