Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

Make R the subject of the fomula S = √2R+T2RT

**Answer Details**

S = √2R+T2RT
$\sqrt{\frac{2R+T}{2RT}}$

squaring both sides = S2 = 2R+T2RT
$\frac{2R+T}{2RT}$

S2(2RT) = 2R + T

2S2RT - 2R = T

R(2S2T - 2) = T

R = R = T2(TS2−2)

**Question 2**
**Report**

If -8, m, n, 19 are in arithmetic progression, find (m, n)

**Answer Details**

To solve this problem, we need to use the arithmetic progression formula: a_n = a_1 + (n-1)d where a_n is the nth term of the arithmetic progression, a_1 is the first term, n is the number of terms, and d is the common difference. We are given that -8, m, n, and 19 are in arithmetic progression. So we can set up the following equations: m = -8 + d n = -8 + 2d 19 = -8 + 3d We can solve for d by subtracting the first equation from the second equation: n - m = 2d - d n - m = d We can substitute this expression for d into the third equation: 19 = -8 + 3(n - m) Simplifying this equation gives: 27 = 3(n - m) 9 = n - m We can substitute this expression for n - m into the equation we derived earlier: n - m = d So we have: d = 9 Substituting this value of d into any of the earlier equations will allow us to solve for m and n. For example, using the equation: m = -8 + d gives: m = -8 + 9 m = 1 And using the equation: n = -8 + 2d gives: n = -8 + 2(9) n = 10 Therefore, the solution is (m, n) = (1, 10)

**Question 3**
**Report**

In a class of 30 students, the marks scored in an examination are displayed in the histogram. What percentage of the student scored more than 40%?

**Answer Details**

This histogram is transferred into this frequency table

Marks20406080100students97662
$\begin{array}{cccccc}Marks& 20& 40& 60& 80& 100\\ students& 9& 7& 6& 6& 2\end{array}$

Students who scored more than 40 = 6 + 6 + 2 = 14

i.e. 1430 $\frac{14}{30}$ x 100% = 4634 $\frac{3}{4}$%

**Question 4**
**Report**

Factorize 4a2 - 12ab - C2 + 9b2

**Answer Details**

4a2 - 12ab - C2 + 9b2

rearranges: (4a2 - 12ab + 9b2) - c2

(2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2

= (2a - 3b + c)(2a - 3b - c)

**Question 5**
**Report**

If f(x) = 2x2 - 5x + 3, find f(x + 1)

**Answer Details**

To find f(x + 1), we need to replace x in the given function f(x) with (x + 1): f(x + 1) = 2(x + 1)^2 - 5(x + 1) + 3 Expanding the brackets and simplifying: f(x + 1) = 2(x^2 + 2x + 1) - 5x - 5 + 3 f(x + 1) = 2x^2 + 4x + 2 - 5x - 2 f(x + 1) = 2x^2 - x Therefore, the answer is option A: 2x^2 - x.

**Question 6**
**Report**

Solve the pair of pair of equation for x and y respectively 2x-t - 3y-1 = 4, 4x-1 + y-1 = 1

**Answer Details**

2x-t - 3y-1 = 4, 4x-1 + y-1 = 1

Let x - 1 = a and y - 1= b

2a - 3b = 4 .......(i)

4a + b = 1 .........(ii)

(i) x 3 = 12a + 3b = 3........(iii)

2a - 3b = 4 ...........(i)

(i) + (iii) = 14a

= 7

∴ a = 714
$\frac{7}{14}$ = 12
$\frac{1}{2}$

From (i), 3b = 2a - 4

3b = 1 - 4

3b = -3

∴ x = -1

From substituting, 2-1 = x - 1

∴ x = 2

y-1 = -1, y = -1

**Question 7**
**Report**

Express 1x+1 $\frac{1}{x+1}$ - 1x?2 $\frac{1}{x?2}$ as a single algebraic fraction

**Answer Details**

To express the given expression as a single algebraic fraction, we first need to find a common denominator for all the terms. The common denominator is (x+1)(2-x). Then, we can simplify each term by multiplying the numerator and denominator by the missing factor in the denominator to obtain: 1(x+1)(2-x) / (x+1) + 1(2-x) / (2-x)(x+1) - 1x(x+1) / (2-x)(x+1) Simplifying further by combining like terms, we get: [(x+1)(2-x) + 1(1-x)(x+1) - x(x+1)] / [(2-x)(x+1)] Simplifying the numerator by distributing, we get: [-x^2 + 3x - 1] / [(2-x)(x+1)] Therefore, the expression 1x+1 / (x+1) + 1 / (2-x) - 1x?2 / (x+1)(2-x) can be simplified to -[-x^2 + 3x - 1] / [(2-x)(x+1)]. So, the answer is (a) -3(x+1)(2-x) / [-x^2 + 3x - 1].

**Question 8**
**Report**

In xyz, y = z = 30o ans XZ = 3cm. Find YZ

**Answer Details**

Y →
$\to $ 30o

X →
$\to $ 120o

Z →
$\to $ 30o

3sin 30
$\frac{3}{\text{sin 30}}$ = YZsin 120
$\frac{YZ}{\text{sin 120}}$

YZ = 3sin 120sin 30
$\frac{3\text{sin 120}}{\text{sin 30}}$

3 x √32
$\frac{\sqrt{3}}{2}$ x 2 = 3√3

**Question 9**
**Report**

In the diagram, HK is parallel to QR, PH = 4cm and HQ = 3cm. What is the ratio of KR:PR?

**Answer Details**

Since HK is parallel to QR, angle HPQ is equal to angle KPR, and angle HQP is equal to angle KRP. Therefore, triangles HPQ and KRP are similar triangles by the angle-angle (AA) criterion. The ratio of the lengths of the corresponding sides of similar triangles is equal. Therefore: KR/PR = HQ/PQ = 3/(3+4) = 3/7 So the ratio of KR:PR is 3:7. Therefore, the correct option is 3:7.

**Question 11**
**Report**

What is the different between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal?

**Answer Details**

When we round 0.007685 to three significant figures, we get 0.00768. The last digit (5) is rounded down because it is less than 5. Therefore, the difference between 0.007685 correct to three significant figures and 0.007685 correct to four places of decimal is: 0.007685 - 0.0077 = -0.000015 So the answer is -1.5 x 10^-5. Therefore, the correct option is: - 10-5

**Question 12**
**Report**

A square tile has side 30cm. How many of these tiles will cover a rectangular floor of length 7.2m and width 4.2?

**Answer Details**

To solve this problem, we need to find the area of the rectangular floor and divide it by the area of one tile. The area of the rectangular floor is: 7.2m x 4.2m = 30.24 m² The area of one square tile is: 30cm x 30cm = 0.09 m² Now we can divide the area of the floor by the area of one tile to get the number of tiles needed to cover the floor: 30.24 m² / 0.09 m² = 336 Therefore, 336 square tiles will cover the rectangular floor. So, the answer is 336.

**Question 13**
**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 represents meat in pie chart

**Answer Details**

To find the angle of the sector which represents meat in the pie chart, we first need to find the total weight of the ingredients used: Total weight = 75g + 40g + 105g + 20g = 240g Now, we can find the percentage of meat used: Percentage of meat = (Weight of meat / Total weight) x 100% = (105g / 240g) x 100% = 43.75% To find the angle of the sector in the pie chart, we use the formula: Angle of sector = Percentage of sector / 100% x 360 degrees Substituting the percentage of meat, we get: Angle of sector = 43.75 / 100 x 360 degrees = 157.5 degrees Therefore, the answer is 157.5 degrees.

**Question 14**
**Report**

OXYZW is a pyramid with a square base such that OX = OY= OZ = OW = 5cm and XY = XW = YZ = WZ = 6cm. Find the height OT

**Answer Details**

xz2 = 62 + 62

36 + 36 = 72

xz = √72
$72$

6√2
$2$ = xT

6√22
$\frac{6\sqrt{2}}{2}$ = 3√2
$\frac{3}{\sqrt{2}}$

OT2 = 52 + (3√2
$2$)2 = 25 + 18

OT = 4√3

**Question 15**
**Report**

a cylindrical metal pipe 1m long has an outer diameter of 7.2cm and an inner diameter of 2.8cm. Find the volume of metal used for the cylinder

**Answer Details**

Volume of cylinder pip (V) = π
$\pi $h(R2
${}^{2}$ - r2
${}^{2}$)

= 100π
$\pi $(7.22
${}^{2}$ - 2.82
${}^{2}$)

= 100π
$\pi $(51.84 - 7.84)

= 100π
$\pi $ x 44

= 440π
$\pi $cm3

**Question 16**
**Report**

What value of Q will make the expression 4x2 + 5x + Q a complete square?

**Answer Details**

4x2 + 5x + Q

To make a complete square, the coefficient of x2 must be 1

= x2 + 5x4
$\frac{5x}{4}$ + Q4
$\frac{Q}{4}$

Then (half the coefficient of x2) should be added

i.e. x2 + 5x4
$\frac{5x}{4}$ + 2564
$\frac{25}{64}$

? Q4
$\frac{Q}{4}$ = 2564
$\frac{25}{64}$

Q = 4×2564
$\frac{4\times 25}{64}$

= 2516

**Question 17**
**Report**

The prime factors of 2520 are

**Answer Details**

To find the prime factors of 2520, we need to divide it by its prime factors until we are left with prime numbers only. We start by dividing 2520 by 2, which gives 1260. 1260 is still an even number, so we divide it by 2 again to get 630. 630 is divisible by 2 and 3, so we divide it by 2 to get 315, and then by 3 to get 105. 105 is divisible by 3 and 5, so we divide it by 3 to get 35, and then by 5 to get 7. 7 is a prime number, so we can't divide it any further. Therefore, the prime factors of 2520 are 2, 2, 2, 3, 3, 5, and 7. So the answer is 2, 3, 5, 7.

**Question 18**
**Report**

PQRS is a rhombus. If PR2 + QS2 = KPQ2, determine k

**Answer Details**

PR2 + QS2 = KPQ2

SQ2 = SR2 + RQ2

PR2 + SQ2 = PQ2 + SR2 + 2RQ2

= 2PQ2 + 2RQ2

= 4PQ2

∴ K = 4

**Question 19**
**Report**

A rectangular polygon of (2k + 1) sides has 140o as the size of each interior angle. Find k

**Answer Details**

A rectangular has all sides and all angles equal. If each interior angle is 140o each exterior angle must be

180o - 140o = 40o

The number of sides must be 360o40o
$\frac{{360}^{o}}{{40}^{o}}$ = 9 sides

hence 2k + 1 = 9

2k = 9 - 1

8 = 2k

k = 82
$\frac{8}{2}$

= 4

**Question 20**
**Report**

If n is the median and m is the mode of the following set of numbers, 2.4, 2.1, 1.6, 2.6, 2.6, 3.7, 2.1, 2.6, then (n, m) is

**Answer Details**

Arrange the numbers in order, 1.6, 2.1, 2.1| 2.4, 2.6| 2.6, 2.6, 3.7

n = median = 2.4+2.62
$\frac{2.4+2.6}{2}$

= 2.5

m = mode = 2.6

∴ (n, m) = (2.5, 2.6)

**Question 21**
**Report**

The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is ₦74.00 when the number of is 20 and ₦96.00 when the number is 30, find the cost when there are 15 students

**Answer Details**

We know that the cost of dinner for the group of students is partly constant and partly varies directly as the number of students. Let the constant cost be c and the cost per student be k. Therefore, the cost of dinner for 20 students is: c + 20k = ₦74.00 ...(1) Similarly, the cost of dinner for 30 students is: c + 30k = ₦96.00 ...(2) Now, we need to find the cost of dinner when there are 15 students. We can solve for the values of c and k using simultaneous equations: Subtracting equation (1) from (2) gives: 10k = ₦22.00 Therefore, k = ₦2.20 Substituting the value of k in equation (1) gives: c + 20(2.20) = ₦74.00 Therefore, c = ₦30.00 So, the cost of dinner when there are 15 students is: c + 15k = ₦30.00 + 15(2.20) = ₦63.00 Therefore, the answer is ₦63.00.

**Question 22**
**Report**

Simplify x(x+1)12−(x+1)12(x+1)12

**Answer Details**

x(x+1)
$\frac{x}{(x+1)}$ - √(x+1)√x+1
$\frac{\sqrt{(x+1)}}{\sqrt{x+1}}$

= xx+1
$\frac{x}{x+1}$ - 1

x−x−1x+1
$\frac{x-x-1}{x+1}$ = −1x+1

**Question 23**
**Report**

If PST is a straight line and PQ = QS = SR in the diagram, find y.

**Answer Details**

< PSQ = < SQR = < SRQ = 24∘ $\circ $

< QSR = 180∘ $\circ $ - 48∘ $\circ $ = 132∘ $\circ $

< PSQ + < QSR + y + 180 (angle on a straight lines)

24 + 132 + y = 180∘ $\circ $ = 156∘ $\circ $ + y = 180

y = 180∘ $\circ $ - 156∘ $\circ $

= 24∘

**Question 24**
**Report**

Oke deposited ₦800.00 in the bank at the rate of 1212 $\frac{1}{2}$% simple interest. After some time the total amount was one and half times the principal. For how many years was the money left in the bank?

**Answer Details**

This problem is asking us to find the number of years that Oke's ₦800.00 deposit was left in the bank at a certain interest rate, given that the total amount he received was one and a half times the original deposit amount. We can start by using the simple interest formula, which is: Simple Interest = Principal * Rate * Time Here, we know the principal is ₦800.00 and the rate is 12.12%. We don't know the time, which is what we're trying to find. Let's call it "t". We also know that the total amount Oke received after some time was one and a half times the original deposit amount, or: Total Amount = 1.5 * Principal Substituting the values we know, we get: Total Amount = 1.5 * ₦800.00 = ₦1200.00 We can use this equation to solve for "t" by first finding the simple interest: Simple Interest = Total Amount - Principal Simple Interest = ₦1200.00 - ₦800.00 = ₦400.00 Then, we can rearrange the simple interest formula to solve for "t": Time = Simple Interest / (Principal * Rate) Substituting the values we know, we get: Time = ₦400.00 / (₦800.00 * 0.1212) = 4 years (rounded to the nearest whole number) Therefore, Oke's money was left in the bank for 4 years.

**Question 25**
**Report**

Factorize completely y3 -4xy + xy3 - 4y

**Answer Details**

y3 -4xy + xy3 - 4y = y3(1 + x) - 4y(1 + x)

(y3 - 4y)(1 + y) = (y3(1 + x) - 4y(1 + x))

∴ = y(1 + x)(y + 2)(y - 2)

**Question 26**
**Report**

In a family of 21 people, the average age is 14years. If the age of the grandfather is not counted, the average age drops to 12 years. What is the age of the grandfather?

**Answer Details**

Let's call the grandfather's age "G". We know that the family consists of 21 people, so if we exclude the grandfather, there are 20 people remaining. We're given that the average age of the family is 14 years, so we can write: Total age of the family = 21 x 14 We're also given that if we exclude the grandfather, the average age drops to 12 years, so we can write: Total age of the family (excluding grandfather) = 20 x 12 We can set up an equation with these two expressions and solve for G: 21 x 14 - G = 20 x 12 294 - G = 240 G = 54 Therefore, the age of the grandfather is 54 years. So the correct option is: 54 years.

**Question 27**
**Report**

If the surface area of a sphere increased by 44%, find the percentage increase in diameter

**Answer Details**

Surface Area of Sphere A = 4πr2
$\pi {r}^{2}$

∴ A = 4π
$\pi $(D)22
$\frac{(D{)}^{2}}{2}$

= (D)22
$\frac{(D{)}^{2}}{2}$

= π
$\pi $D2

When increased by 44% A = 144πD2100
$\frac{144\pi {D}^{2}}{100}$

π
$\pi $(12D)210
$\frac{(12D{)}^{2}}{10}$ = π
$\pi $(6D)25
$\frac{(6D{)}^{2}}{5}$

Increase in diameter = 6D5
$\frac{6D}{5}$ - D = 15
$\frac{1}{5}$D

Percentage increase = 15
$\frac{1}{5}$ x 1100
$\frac{1}{100}$%

= 20%

**Question 28**
**Report**

On the curve, the points at which the gradient of the curve is equal to zero are

**Answer Details**

The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m

**Question 29**
**Report**

What are K and L respectively if 12 $\frac{1}{2}$(3y - 4x)2 = (8x2 + kxy + Ly2)

**Answer Details**

12
$\frac{1}{2}$(3y - 4x)2 = (8x2 + kxy + Ly2)

12
$\frac{1}{2}$(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2

92
$\frac{9}{2}$y2 - 12xy) = kxy + Ly2

k = -12 ∴ L = 92

**Question 30**
**Report**

Simplify 3√(64r6)12

**Answer Details**

√(64r6)12
$\sqrt[3]{(64{r}^{6}{)}^{\frac{1}{2}}}$

=(64r6)12⋅13
$=(64{r}^{6}{)}^{\frac{1}{2}\cdot \frac{1}{3}}$

=64r6⋅16
$=64{r}^{6\cdot \frac{1}{6}}$

=64r

**Question 31**
**Report**

If a : b = 5 : 8, x : y = 25 : 16; evaluate ax $\frac{a}{x}$ : by

**Answer Details**

a : b = 5 : 8 = 2.5 : 40

x : y = 25 : 16

ax
$\frac{a}{x}$ : by
$\frac{b}{y}$ = 2525
$\frac{25}{25}$ : 4016
$\frac{40}{16}$

= 1 : 4016
$\frac{40}{16}$

= 16 : 40

= 2 : 5

**Question 32**
**Report**

Find P in terms of q if log3P + 3log3q = 3

**Answer Details**

log3P + 3log3q = 3

log3(Pq3) = 3

Pq3 = 33

P = (3P
$\frac{3}{P}$)3

**Question 33**
**Report**

Simplify 4 -