2022 - JAMB Mathematics Past Questions and Answers - page 3

Here's how to solve the problem: 1. **Calculate the number of eggs sold:** * Each crate contains 30 eggs (a standard assumption). * The farmer sold 5.5 crates, so he sold 5.5 * 30 = 165 eggs. 2. **Calculate the number of eggs remaining:** * The farmer started with 200 eggs and sold 165. * Eggs remaining: 200 - 165 = 35 eggs. 3. **Calculate the percentage of eggs remaining:** * (Eggs remaining / Total eggs) * 100% * (35 / 200) * 10

sec\(^2\)θ + tan\(^2\)θ = 3

sec\(^2\)θ = 1  + tan\(^2\)θ

: Hence,

1 + tan\(^2\)θ + tan\(^2\)θ = 3

2tan\(^2\)θ = 3 - 1

2tan\(^2\)θ = 2

divide through by 2

tan\(^2\)θ = 1

tanθ =  √1

tanθ = 1

θ = tan\(^{-1}\) (1)

θ = 45º

\(\frac{x^3 + 6x - 7}{x-1}\):

When numerator is differentiated => 3x\(^2\) + 6  

When the denominator is differentiated => 1

Hence,

\(\frac{3x^2 + 6}{1}\)

Substitute x for 1

 \(\frac{3 * 1^2 + 6}{1}\) =  \(\frac{3 + 6}{1}\) 

=  \(\frac{9}{1}\)

= 9

If two lines are parallel then their slopes (gradients) will be equal.

Line 2x - y = 5

Equation of a straight line: y = mx + c

y = 2x - 5 

line 2x - y = 5 has a gradient of 2

A parallel line with a gradient of 2 and intercept of 5

=> 2x - y = -5

=> y = 2x + 5

14, 17, 10, 13, 18 and 10.

=> 10, 10, 13, 14, 17, 18

median = \(\frac{13+14}{2}\)

= \(\frac{27}{2}\) or 13.5

Apply the expansion method on (10110.011)\(_2\)

(1 * 2\(^4\)) + (0 * 2\(^3\)) + (1 * 2\(^2\)) + (1 * 2\(^1\)) (0 * 2\(^0\)) + (0 * 2\(^{-1}\)) + (1 * 2\(^{-2}\))  + (1 * 2\(^{-3}\))

(1*16) + 0 (1*4) + (1*2) + 0 + 0 + (1*\(\frac{1}{4}\)) + (1*\(\frac{1}{8}\)) 

16 + 4 + 2 + (0.25 + 0.125)

22 + 0.375

(10110.011)\(_2\) = 22.375\(_{10}\)

if 20 students = 10000

₦‎500 per student

₦‎1,000,000 ÷ 500

= 2,000 students

Given that x\(_1\) = -3,  x\(_2\) = 2,  y\(_1\) = 5,  y\(_2\) = 10

Coordinates of the midpoint of the line = (\(\frac{x_1 + x_2}{2}\) , \(\frac{y_1 + y_2}{2}\))

(\(\frac{-3 +2}{2}\) , \(\frac{5 + 10}{2}\))

=  \(\frac{-1}{2}\)) , \(\frac{15}{2}\))

\(^{3^2 + 1}\)C\(_{3+5}\)

\(^{9 + 1}\)C\(_{3+5}\) 

\(^{10}\)C\(_{8}\) = \(\frac{10!}{8! 2!}\)

\(\frac{10 * 9 * 8!}{8! 2!}\) = \(\frac{10 * 9}{2}\)

= 45

s = ut + 1/2at^2