2023 - JAMB Mathematics Past Questions and Answers - page 3

The mean deviation (or average deviation) of a set of values from their mean is calculated using the formula:

$$\text{Mean Deviation}=\frac{\sum _{i=1}^n|X_i-\overline{X}|}{n},$$

where $X_i$ is each individual value, $\overline{X}$ is the mean of the values, and $n$ is the number of values.

The first five prime numbers are 2, 3, 5, 7, and 11. Let's calculate the mean deviation:

$$\overline{X}=\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6.$$

Now, calculate the mean deviation:

$$\text{Mean Deviation}=\frac{|2-5.6|+|3-5.6|+|5-5.6|+|7-5.6|+|11-5.6|}{5}$$

$$=\frac{3.6+2.6+0.6+1.4+5.4}{5}=\frac{13.2}{5}=2.64.$$

Therefore, the correct answer is approximately 2.72

A binary operation '*' is said to be commutative if $a\ast b=b\ast a$ for all elements $a,b$ in the set $A$.

Therefore, the correct option is  $a\ast b=b\ast a$.

Let $F$ be the set of people who can speak French, and $E$ be the set of people who can speak English.

Given that:

$$n(F)=400$$

$$n(E)=350$$

$$n(F\cup E)=500$$

We want to find $n(F\cap E)$ (the number of people who can speak both languages).

The inclusion-exclusion principle states:

$$n(F\cup E)=n(F)+n(E)-n(F\cap E)$$

Substitute the given values:

$$500=400+350-n(F\cap E)$$

Solve for $n(F\cap E)$:

$$n(F\cap E)=750-500=250$$

Therefore, $250$ people can speak both languages.

So, the correct answer is 250.

Let the sides of the triangle be $2x,3x,$ and $4x$, where $x$ is a positive constant.

The perimeter of the triangle is given as:

$$2x+3x+4x=180$$

Combine like terms:

$$9x=180$$

Solve for $x$:

$$x=\frac{180}{9}=20$$

Now, the sides of the triangle are:

$$2x=2\times 20=40$$

$$3x=3\times 20=60$$

$$4x=4\times 20=80$$

The semi-perimeter ($s$) is half of the perimeter:

$$s=\frac{40+60+80}{2}=\frac{180}{2}=90$$

Now, use Heron's formula to find the area ($A$):

$$A=\sqrt{s(s-a)(s-b)(s-c)}$$

Substitute the values:

$$A=\sqrt{90\cdot (90-40)\cdot (90-60)\cdot (90-80)}$$

$$A=\sqrt{90\cdot 50\cdot 30\cdot 10}=\sqrt{1350000}$$

Therefore, $A\approx 1162$ cm${}^2$ (to the nearest cm${}^2$).

So, the correct answer is 1162 cm${}^2$.

To find the area of the region bounded by the curves $y=2x^2+10$ and $y=4x+16$, we need to set them equal to each other and find the points of intersection.

$$2x^2+10=4x+16$$

Rearrange it to a quadratic equation:

$$2x^2-4x-6=0$$

Now, factor the quadratic equation:

$$2(x^2-2x-3)=0$$

$$(x-3)(x+1)=0$$

So, the points of intersection are $x=3$ and $x=-1$.

Now, integrate to find the area:

$$A={\int }_{-1}^3(4x+16-(2x^2+10))dx$$

Simplify the integrand:

$$A={\int }_{-1}^3(-2x^2+4x+6)dx$$

Now integrate with respect to $x$:

$$A={[-\frac{2}{3}{x}^3+2x^2+6x]}_{-1}^3$$

Evaluate the expression at the upper and lower limits:

$$A=[-\frac{2}{3}(3{)}^3+2(3{)}^2+6(3)]-[-\frac{2}{3}(-1{)}^3+2(-1{)}^2+6(-1)]$$

$$A=[-18+18+18]-[\frac{2}{3}+2-6]$$

$$A=18-[-\frac{4}{3}]$$

$$A=18+\frac{4}{3}$$

$$A=\frac{54}{3}+\frac{4}{3}=\frac{58}{3}$$

Therefore, the correct answer is $\frac{64}{3}$.

To estimate the mean, we need to find the sum of the products of the class marks and frequencies, and then divide by the sum of frequencies.

$$\text{Class Interval}|\text{Class Mark}|\text{Frequency (f)}|fx$$
$$\text{5 - 6}|5.5|29|5.5\times 29=159.5$$
$$\text{7 - 8}|7.5|40|7.5\times 40=300$$
$$\text{9 - 10}|9.5|38|9.5\times 38=361$$
$$\sum f=107|\sum fx=820.5$$

Now, calculate the mean:

$$\text{Mean}=\frac{\sum fx}{\sum f}=\frac{820.5}{107}\approx 7.663$$

Rounded to one decimal place, the mean is approximately 7.7.

Therefore, the correct answer is 7.7.

Let's solve the given logarithmic equation:

$$\log (y+8)+\log (y-8)=2\log 3+2\log 5$$

We can use logarithmic properties to simplify the equation. The sum of logarithms is equal to the logarithm of their product:

$$\log ((y+8)(y-8))=\log (3^2\cdot 5^2)$$

Now, we can set the arguments equal to each other:

$$(y+8)(y-8)=3^2\cdot 5^2$$

Expand and simplify the equation:

$$y^2-64=9\cdot 25$$

$$y^2=9\cdot 25+64$$

$$y^2=225+64$$

$$y^2=289$$

Now, take the square root of both sides:

$$y=\pm \sqrt{289}$$

So, $y=\pm 17$.

Therefore, the correct answer is $y=\pm 17$

We can use the formula $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ to find the speed of the car. Once we have the speed, we can use it to find the time required to cover a different distance.

Given that the car takes 4.5 hours to cover 225 km, we can find the speed:

$$\text{Speed}=\frac{225\text{km}}{4.5\text{hrs}}=50\text{km/hr}$$

Now that we know the speed, we can find the time required to cover 150 km:

$$\text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{150\text{km}}{50\text{km/hr}}=3\text{hrs}$$

Therefore, the correct answer is 3 hrs

Let's find $(A-B)\cup (B-A)$, where $(A-B)$ represents the elements that are in A but not in B, and $(B-A)$ represents the elements that are in B but not in A.

$$A-B=\{1,3,5\}$$
$$B-A=\{8\}$$

Now, take the union of these two sets:
$$(A-B)\cup (B-A)=\{1,3,5\}\cup \{8\}$$

The resulting set is \{1, 3, 5, 8\}.

Therefore, the correct answer is {1, 3, 5, 8}

Let's simplify the expression:

$${16}^{0.16}\times {16}^{0.04}\times 2^{0.2}$$

First, observe that ${16}^{0.16}$ and ${16}^{0.04}$ are both powers of 16, so we can combine them:

$${16}^{0.16}\times {16}^{0.04}={16}^{0.2}$$

Now, we have:

$${16}^{0.2}\times 2^{0.2}$$

Since ${16}^{0.2}$ is the same as $(2^4{)}^{0.2}$ (which is equivalent to $2^{0.8}$), we can rewrite the expression as:

$$2^{0.8}\times 2^{0.2}$$

Now, use the property that $a^m\times a^n=a^{m+n}$:

$$2^{0.8+0.2}=2^{1.0}=2$$