2024 - JAMB Mathematics Past Questions and Answers - page 4

The diagram shows a pair of intersecting lines forming angles at the intersection. To find the value of x, we use the property of vertically opposite angles. Since the angles are equal, the value of x is 40^o.

In the diagram above, < CDE = < CED = 15^o

(base < s of isos. \(\bigtriangleup\))

< ECD = 180^o - (15 + 15)^o

= 180^o - 30^o = 150^o

But x + 110^o = 150^o

(Sum of opp. interior < s of a\(\bigtriangleup\) = opp. exterior < )

x = 150^o - 110^o = 40^o

0.04945 to 2s.f is 0.05

For the committee to include 2 females, we must have 3 males, so that there should be a total of 5 members.

thus, \(^4C_3 \times ^3C_2\)

= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)

= 4 × 3

= 12 ways

We are given that the capacity (volume) of the cylindrical tank is 3080 m³ and the diameter of its base is 14 m. We are asked to find the depth (height) of the tank.

The formula for the volume of a cylinder is:

\[ V = \pi r^2 h \]

Where:

• \( V \) is the volume of the cylinder (3080 m³),
• \( r \) is the radius of the base,
• \( h \) is the height (depth) of the tank.

The diameter of the base is given as 14 m, so the radius \( r \) is half of the diameter:

\[ r = \frac{14}{2} = 7 \, \text{m} \]

Substituting the given values into the volume formula:

\[ 3080 = \pi \times (7)^2 \times h \]

This simplifies to:

\[ 3080 = \pi \times 49 \times h \]

Now, solve for \( h \) by dividing both sides by \( \pi \times 49 \):

\[ h = \frac{3080}{\pi \times 49} \]

Using \( \pi \approx 3.1416 \), we can calculate \( h \):

\[ h = \frac{3080}{3.1416 \times 49} \approx \frac{3080}{153.938} \approx 20 \, \text{m} \]

The locus of a point that is equidistant from two intersecting straight lines is the angle bisector of the two lines. This is the line that divides the angle between the two lines into two equal parts.

The given sequence is a geometric progression (G.P.) where each term after the first is obtained by multiplying the previous term by a constant called the common ratio.

Let's first express the terms more clearly:

• The first term \( T_1 = \sqrt{10} + \sqrt{5} \),
• The second term \( T_2 = \sqrt{10} + 2\sqrt{5} \),
• And so on.

The common ratio \( r \) of a G.P. can be found by dividing the second term by the first term:

\[ r = \frac{T_2}{T_1} = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \]

To simplify this, we multiply the numerator and denominator by \( \sqrt{10} - \sqrt{5} \) to rationalize the denominator:

\[ r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}} \]

Expanding the numerator and denominator:

Denominator:

\[ (\sqrt{10})^2 - (\sqrt{5})^2 = 10 - 5 = 5 \]

Numerator:

\[ (\sqrt{10})(\sqrt{10}) + (\sqrt{10})(-\sqrt{5}) + (2\sqrt{5})(\sqrt{10}) + (2\sqrt{5})(-\sqrt{5}) = 10 - \sqrt{50} + 2\sqrt{50} - 10 = \sqrt{50} \]

Now, simplifying:

\[ r = \frac{\sqrt{50}}{5} \]

\[ r = \frac{\sqrt{25 \times 2}}{5} = \frac{5\sqrt{2}}{5} = \sqrt{2} \]

The common ratio is \( \sqrt{2} \).

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

\(\sum fx\) = 91

To simplify this expression, we will apply the logarithmic properties:

• Product Rule: \( \log_b x + \log_b y = \log_b (x \cdot y) \)
• Quotient Rule: \( \log_b x - \log_b y = \log_b \left(\frac{x}{y}\right) \)
• Power Rule: \( n \log_b x = \log_b (x^n) \)

We can start by simplifying the expression using the Power Rule on the second term:

\[ 3 \log_{10} 2 = \log_{10} (2^3) = \log_{10} 8 \]

The expression now becomes:

\[ \log_{10} 1.5 + \log_{10} 8 - \log_{10} 0.3 \]

Next, we can apply the Product Rule to \( \log_{10} 1.5 + \log_{10} 8 \):

\[ \log_{10} 1.5 + \log_{10} 8 = \log_{10} (1.5 \times 8) = \log_{10} 12 \]

So the expression is now:

\[ \log_{10} 12 - \log_{10} 0.3 \]

Now, applying the Quotient Rule:

\[ \log_{10} 12 - \log_{10} 0.3 = \log_{10} \left(\frac{12}{0.3}\right) \]

We simplify the fraction:

\[ \frac{12}{0.3} = \frac{12}{\frac{3}{10}} = 12 \times \frac{10}{3} = 40 \]

The final expression is:

\[ \log_{10} 40 \]

The integral ∫(cos4x + sin3x) dx can be solved by integrating each term separately. The integral of cos(4x) is sin(4x)/4, and the integral of sin(3x) is -cos(3x)/3. Therefore, the result is:

∫(cos4x + sin3x) dx = (1/4)sin4x - (1/3)cos3x + C

The mean of the data is given as 3. To find the value of y, we first need to calculate the total sum of the frequencies, which is:

Sum of frequencies = (2y - 2) + (y - 1) + (3y - 4) + (3 - y) + (6 - 2y) = 3y - 4. The mean is the sum of the products of frequency and marks divided by the total frequency:

Mean = (1 * (2y - 2) + 2 * (y - 1) + 3 * (3y - 4) + 4 * (3 - y) + 5 * (6 - 2y)) / (3y - 4).

We are told the mean is 3, so solving for y will give us the value of 2. Therefore, y = 2.