Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

Find the gradient of the line passing through the points (-2, 0) and (0, -4)

**Answer Details**

The gradient of a line describes how steep the line is, and it can be calculated as the change in the y-coordinate divided by the change in the x-coordinate between two points on the line. In this case, the two points are (-2, 0) and (0, -4), so the change in x is 0 - (-2) = 2, and the change in y is -4 - 0 = -4. Therefore, the gradient of the line passing through these two points is -4 ÷ 2 = -2. So, the answer is -2.

**Question 2**
**Report**

A binary operation x is defined by a x b = ab ${}^{b}$. If a x 2 = 2 - a, find the possible values of a?

**Answer Details**

The equation a x 2 = 2 - a can be solved for the possible values of a. To do this, we can simplify the equation by adding a to both sides: a x 2 + a = 2 Combining like terms: a x 2 + a = 2 a(x + 1) = 2 Dividing both sides by x + 1: a = 2 / (x + 1) Since the binary operation x is defined as a x b = a, we can substitute this definition into the equation: a = 2 / (a x 1 + 1) a = 2 / (a + 1) Since a x b = a, we know that a x 1 = a, so we can substitute this into the equation: a = 2 / (a + 1) a = 2 / (a + 1) Solving for a, we get: a = 2 / (a + 1) = 2 / (2) = 1 So, the possible values of a are 1.

**Question 3**
**Report**

Express the product of 0.0014 and 0.011 in standard form

**Question 4**
**Report**

Find the simple interest rate percent annum at which N1000 accumulates to N1240 in 3 years.

**Answer Details**

Simple interest is the interest calculated on the original amount of a loan or deposit, without taking into account any compound interest. To find the simple interest rate at which N1000 accumulates to N1240 in 3 years, we can use the formula for simple interest: I = Prt where I is the interest, P is the principal (original amount), r is the interest rate, and t is the time in years. We know that I = N1240 - N1000 = N240, and t = 3 years. So, we can plug in these values into the formula and solve for r: N240 = N1000 x r x 3 Dividing both sides by N1000 x 3, we get: r = N240 / (N1000 x 3) = 0.08 So, the interest rate is 8%. So, the answer is 8%.

**Question 5**
**Report**

Four boys and ten girls can cut a field first in 5 hours. If the boys work at 54 $\frac{5}{4}$ the rate at which the girls work, how many boys will be needed to cut the field in 3 hours?

**Answer Details**

Let x rep. numbers of boys that can work at 54 $\frac{5}{4}$ the rate at

which the 10 girls work

For 1 hrs, x boys will work for 15×104 $\frac{\frac{1}{5}\times 10}{4}$

x = 54 $\frac{5}{4}$ x 10

= 8 boys

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

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

12×53 $\frac{12\times 5}{3}$ boys will be needed = 20 boys.

**Question 6**
**Report**

Reach each number to two significant figures and then evaluate 0.02174×1.20470.023789

**Answer Details**

0.021741×1.20470.023789 = 0.0255×1.20.024(to 216)

= 0.02640.024= 1.1

**Question 7**
**Report**

In this fiqure, PQ = PR = PS and SRT = 68o
${}^{o}$. Find QPS

**Answer Details**

Since PQRS is quadrilateral

2y + 2x + QPS = 360o ${}^{o}$

i.e. (y + x) + QPS = 360o ${}^{o}$

QPS = 360o ${}^{o}$ - 2 (y + x)

But x + y + 68o ${}^{o}$ = 180o ${}^{o}$

There; x + y = 180o ${}^{o}$ - 68o ${}^{o}$ = 112o ${}^{o}$

QPS = 360 - 2(112o ${}^{o}$)

= 360o ${}^{o}$ - 224 = 136o

**Question 8**
**Report**

A crate of soft drinks contains 10 bottle of Coca-Cola 8 of Fanta and 6 of Sprite. If one bottle is selected at random, what is the probability that it is Not a Coca-Cola bottle?

**Answer Details**

Coca-Cola = 10 bottles, Fanta = 8 bottles, Spirite = 6 bottles

Total = 24

P(Coca-Cola) = 1024 $\frac{10}{24}$; P(not Coca-Cola)

1 - 1024 $\frac{10}{24}$

24−1024=1424=712 $\frac{24-10}{24}=\frac{14}{24}=\frac{7}{12}$

**Question 9**
**Report**

A sector of circle of radius 7.2cm which substends an angle of 300o ${}^{o}$ at the centre is used to form a cone. What s the radius of the base of the cone?

**Question 10**
**Report**

What is the product of 275 $\frac{27}{5}$, (3)−3 $(3{)}^{-3}$ and (15)−1 $\frac{1}{5}{)}^{-1}$?

**Answer Details**

275×3−3×(1)−15 $\frac{27}{5}\times {3}^{-3}\times \frac{(1{)}^{-1}}{5}$

= 275×133×115 $\frac{27}{5}\times \frac{1}{{3}^{3}}\times \frac{1}{\frac{1}{5}}$

275 $\frac{27}{5}$ x 127 $\frac{1}{27}$ x 51 $\frac{5}{1}$ = 1

**Question 11**
**Report**

A school boy lying on the ground 30m away from the foot of a water tank towel observes that the angle of elevation of the top of the tank is 60o
${}^{o}$. Calculate the height of the tank.

**Answer Details**

h = 30 tan 60

= 303–√ $\sqrt{3}$

**Question 12**
**Report**

The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group.

**Answer Details**

Let's call the number of students in the group "n". The total of the ages of all students is 15n. When the teacher's age is added, the total becomes 15n + 45. The mean of the ages becomes 18, which means the total of the ages is divided by the number of people, which is "n + 1". We can set up an equation to represent the situation: (15n + 45) / (n + 1) = 18 Expanding the equation: 15n + 45 = 18 * (n + 1) 15n + 45 = 18n + 18 Subtracting 18n from both sides: -3n + 45 = 18 Subtracting 45 from both sides: -3n = -27 Dividing both sides by -3: n = 9 So, there are 9 students in the group.

**Question 13**
**Report**

Three consecutive terms of a geometric progression are give as n - 2, n and n + 3. Find the common ratio

**Answer Details**

hn−2=n+3n $\frac{h}{n-2}=\frac{n+3}{n}$

n2 ${}^{2}$ = (n + 3) (n - 2)

n2 ${}^{2}$ = n2 ${}^{2}$ + n - 6

n2 ${}^{2}$ + n - 6 - n2 ${}^{2}$ = 0

n - 6 = 0

n = 6

Common ratio: nn−2=66−2=64 $\frac{n}{n-2}=\frac{6}{6-2}=\frac{6}{4}$ = 32 $\frac{3}{2}$

**Question 14**
**Report**

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

**Answer Details**

ABCD is the floor. By pathagoras

AC2 ${}^{2}$ = 144 + 81 = 225−−−√ $\sqrt{225}$

AC = 15cm

Height of room 8m, diagonal of floor is 15m

Therefore, the cosine of the angle which a diagonal of the room makes with the floor is

EC2 ${}^{2}$ = 152 ${}^{2}$ + 82 ${}^{2}$ cosine

adjHyp=1517 $\frac{adj}{Hyp}=\frac{15}{17}$

EC2 ${}^{2}$ = 225+64−−−−−−−√ $\sqrt{225+64}$

EC = 289−−−√ $\sqrt{289}$

= 17

**Question 15**
**Report**

The pie chart shows the income of a civil servant in month. If his monthly income is N6,000. Find his monthly basic salary.

**Answer Details**

360o ${}^{o}$ - (60o ${}^{o}$ + 60o ${}^{o}$ + 67 + 50 = 237o ${}^{o}$)

360o ${}^{o}$ - 237 = 130o ${}^{o}$

B. Salary = 123360XN60001 $\frac{123}{360}X\frac{N6000}{1}$

= N2,050

**Question 16**
**Report**

In preparing rice cutlets, a cook used 75g of rice, 40g of margarine, 105g of meat and 20g of bread crumbs. Find the angle of the sector which represent meat in a pie chart?

**Answer Details**

Rice = 75g, Margarine = 40g, Meat = 105g

Bread = 20g

Total = 240

Angle of sector represented by meat

= 105240×360o1 $\frac{105}{240}\times \frac{{360}^{o}}{1}$

= 157.5

**Question 17**
**Report**

A surveyor walks 500m up a hill which slopes at an angle of 30o ${}^{o}$. Calculate the vertical height through which he rises

**Answer Details**

The vertical height through which the surveyor rises can be calculated using trigonometry. When a person walks up a hill, they are moving in two dimensions: horizontally and vertically. We can use a right triangle to represent this movement, where the horizontal distance is one side, the vertical distance is another side, and the slope of the hill is the angle between these sides. In this case, the horizontal distance is 500m and the angle is 30 degrees, so we can use the trigonometric function sine to find the vertical distance: sin(30) = vertical distance / 500 Multiplying both sides by 500, we get: 500 * sin(30) = vertical distance Using a calculator, we can find that sin(30) = 0.5, so: 500 * 0.5 = 250 Therefore, the vertical height through which the surveyor rises is 250m.

**Question 18**
**Report**

Find all median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119

**Answer Details**

Arrange in ascending order

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

Median = 120+1302 $\frac{120+130}{2}$ = 125

**Question 19**
**Report**

If log10 ${}_{10}$2 = 0.3010 and log10 ${}_{10}$3 = 0.4771, eventually without using the logarithm tables, log10 ${}_{10}$4.5

**Answer Details**

log10 ${}_{10}$2 = 0.3010 and log10 ${}_{10}$3 = 04771

log104.5=log10 ${}_{10}4.5=lo{g}_{10}$ (3×32 $\frac{3\times 3}{2}$)

log10 ${}_{10}$ 3 + log10 ${}_{10}$ 3 - log10 ${}_{10}$2 = 0.4471 + 0.771 - 0.3010

= 0.6532

**Question 20**
**Report**

Find the equation of the line through the points (5, 7) parallel to the line 7x + 5y = 12.

**Answer Details**

The equation of a line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept. To find the equation of a line parallel to another line, we need to use the same slope. The line 7x + 5y = 12 can be rewritten in the form y = -(7/5)x + 12/5. The slope of this line is -(7/5). To find a line through the point (5, 7) with the same slope, we can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope. Plugging in the values for (x1, y1) and m, we get: y - 7 = -(7/5)(x - 5) Expanding and simplifying, we get: y = -(7/5)x + 7 + (35/5) = -(7/5)x + 70/5 = 14x + 7. So the equation of the line through the point (5, 7) parallel to the line 7x + 5y = 12 is 7x + 5y = 70.

**Question 21**
**Report**

The acres for rice, pineapple, cassava, cocoa and palm oil in a certain district are given respectively as 2, 5, 3, 11 and 9. What is the angle of the sector of cassava in a pie chart?