JAMB UTME - Mathematics - 2015

Simplify 4 √[(390695T− 8)½)]

1/T

2T

5/T

T/5

Akọwa Nkọwa

4√(390625T − 8)½
[(390625T − 8)½]¼
[((390625)½ × T − 8)½]¼
(√ 390625 × T − 4 )¼
(625 × T −¼)¼
(54 × T −4)¼
= (54)¼ × (T − 4 )¼
= 5 × T − 1
= 5 × 1/T
= 5/T

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 2 Ripọtì

Solve the equation

\[ 5^{(x-2)}=\left(1\div125\right)^{(x+3)} \]

3/2

− 7/4

4/7

7/4

Akọwa Nkọwa

To solve the equation 5(x - 2) = (1 ÷ 125)(x + 3), we need to isolate the variable "x" on one side of the equation. First, let's simplify the right side of the equation: (1 ÷ 125)(x + 3) = (1 ÷ 125)x + (1 ÷ 125)3 = x/125 + 3/125 = (x + 3)/125 Now, let's simplify the left side of the equation: 5(x - 2) = 5x - 10 Now, we have: 5x - 10 = (x + 3)/125 To isolate "x" on one side, we'll multiply both sides by 125: 125 * 5x - 125 * 10 = 125 * (x + 3) / 125 Expanding the right side: 125 * 5x - 125 * 10 = x + 3 Combining like terms on the left side: 625x - 1250 = x + 3 Subtracting "x" from both sides: 624x - 1250 = 3 Adding 1250 to both sides: 624x = 1253 Dividing both sides by 624: x = 2 So the solution to the equation is x = 2, which is not equal to any of the options (-7/4, 4/7, or 7/4).

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Lee Azịza

Ajụjụ 3 Ripọtì

Find the area of the curved surface of a cone whose base radius is 3cm and whose height is 4cm (π = 3.14)

17.1cm2

27.2cm2

47.1cm2

37.3cm2

Akọwa Nkọwa

To find the curved surface area of a cone, we need to use the formula: `Curved Surface Area = πrs` where `r` is the radius of the base of the cone and `s` is the slant height of the cone. To find the slant height of the cone, we can use the Pythagorean theorem, which tells us that: `h^2 + r^2 = s^2` where `h` is the height of the cone. Substituting `r = 3cm` and `h = 4cm`, we get: `4^2 + 3^2 = s^2` `16 + 9 = s^2` `25 = s^2` `sqrt(25) = s` `s = 5cm` Now, substituting `r = 3cm` and `s = 5cm` into the formula for curved surface area, we get: `Curved Surface Area = πrs` `= 3.14 × 3 × 5` `= 47.1cm^2` Therefore, the curved surface area of the cone is `47.1cm^2`. Hence, option C, "47.1cm^2" is the correct answer.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 4 Ripọtì

Factorize \(a^2 - b^2 - 4a + 4\)

(a + b)(a − b)

(a − 2 + b)(a − 2 − b)

(a + 1)(a − 2 + b)

(a + b)2

Akọwa Nkọwa

The trinomial =

a^{2} - 4 a + 4

a^{2} - b^{2} - 4 a + 4 = (a^{2} - 4 a + 4) - b^{2}

(a^{2} - 2 a - 2 a + 4) - b^{2}

a (a - 2) - 2 (a - 2) - b^{2}

(a - 2)^{2} - b^{2}

(a - 2 + b) (a - 2 - b)

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 5 Ripọtì

One bag contain 3 blue and 5 red balls, another bag contain 2 blue and 4 red balls respectively. One ball is drawn for each bag. What is the probability both balls are blue

\frac{2}{15}

\frac{3}{24}

\frac{3}{21}

\frac{3}{28}

Akọwa Nkọwa

One bag with 3 blue and 5 red
Pr (H) = 3/8Pr (r) 5/8
Another bag with 2 blue and 4 red (note one ball is drawn from each bag)
i.e. prob.=8 − 1 = 7
prob. (2 ball both are blue=pr (1st blue and 2nd blue)
prob. 2 balls both blue = 3/8 × 2/7
=

\frac{3}{28}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Probability

Areas

Lee Azịza

Ajụjụ 6 Ripọtì

The area of a circle of radius 4cm is equal to (Take π = 3.142 )

10.3cm2

15.7cm2

50.3cm2

17.4cm2

Akọwa Nkọwa

Area of a circle (A) = πr2
Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 7 Ripọtì

Find the value of x if \( [1 \div 64^{(x+2)}] = [4^{(x-3)} \div 16^x] \)

\frac{3}{2}

\frac{2}{3}

\frac{1}{3}

- \frac{3}{2}

Akọwa Nkọwa

[1 \div 64^{(x + 2)}] = [4^{(x - 3)} \div 16^{x}]

64^{- (x + 2)} = [4^{(x - 3)}] \div [16^{x}]

Break down 4, 16 and 64 into smaller index numbers:

2^{- 6 (x + 2)} = 2^{2 (x - 3)} \div 2^{4 (x)}

2^{- 6 x - 12} = 2^{2 x - 4 x - 6}

2^{- 6 x - 12} = 2^{- 2 x - 6}

− 6x − 12 = − 2x − 6
Collect the like terms:
−6x + 2x = −6 + 12
−4x =6
x =

\frac{6}{4}

x =

\frac{- 3}{2}

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Number Bases

Algebraic Fractions

Lee Azịza

Ajụjụ 8 Ripọtì

Simplify (0.09)2 and give your answer correct to 4 significant figures

0.81

0.081

0.0081

8.0001

Akọwa Nkọwa

(0.09)2 = 0.09 × 0.09
= 0.0081
= 0.008100 to 4 significant figures
Please, start counting first from the non-zero digits i.e. 8

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 9 Ripọtì

If A = (−3,5) and B = (4,−1) find the co-ordinate of the mid point

2, ½

½ , 2

1, ½

0, 2

Lee Azịza

Ajụjụ 10 Ripọtì

A man sells his new brand car for ₦420,000 at a gain of 15%. What did it cost him?

₦410,000

₦365,217

₦157,250

₦257,000

Akọwa Nkọwa

If the man sold the car for ₦420,000 at a gain of 15%, then we can find the cost price of the car using the following formula: Selling price = Cost price + Profit We know the selling price is ₦420,000 and the profit is 15% of the cost price. Let's call the cost price "x." Profit = 15% of x = 0.15x Substituting the values in the formula: ₦420,000 = x + 0.15x ₦420,000 = 1.15x To solve for x, we need to divide both sides by 1.15: x = ₦420,000 / 1.15 x = ₦365,217.39 Therefore, the cost price of the car was ₦365,217.39. So, option B - ₦365,217 is the correct answer.

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 11 Ripọtì

Find the simple interest on ₦325 in 5years at 3% per annum.

₦48.75K

₦50.10K

₦75.50K

₦15.75K

Akọwa Nkọwa

The formula for finding the simple interest on a principal amount P for a period of t years at an interest rate of r per annum is:

simple interest = P * r * t / 100

In this case, P = 325, t = 5, and r = 3, so we can plug in the values into the formula:

simple interest = 325 * 3 * 5 / 100
simple interest = 48.75

So, the simple interest on ₦325 in 5 years at 3% per annum is ₦48.75.

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Sine, Cosine And Tangent Of An Angle

Representation Of Data

Lee Azịza

Ajụjụ 12 Ripọtì

X and Y are two sets such that nX = 15, nY = 12 and n{X ∩ Y} = 7. Find n{X ∪ Y}.

21

225

15

20

Akọwa Nkọwa

The formula for finding the number of elements in the union of two sets X and Y is:


n(X U Y) = nX + nY - n(X ∩ Y)

where nX is the number of elements in set X, nY is the number of elements in set Y, and n(X ∩ Y) is the number of elements in the intersection of X and Y. In this case, nX = 15, nY = 12, and n(X ∩ Y) = 7, so we can plug in the values into the formula:


n(X U Y) = 15 + 12 - 7
n(X U Y) = 20

So, the number of elements in the union of X and Y is 20.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 13 Ripọtì

Simplify \( [1 \div (x^2 + 3x + 2)] + [1 \div (x^2 + 5x + 6)] \)

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

\frac{2}{(x + 1) (x + 2}

2/((x + 2)(x + 3))

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

Akọwa Nkọwa

[1 \div (x^{2} + 3 x + 2)] + [1 \div (x^{2} + 5 x + 6)]

=

1 \div (x^{2} + 3 x + 2) + [1 \div (x^{2} + 5 x + 6)]

=

[1 \div ((x^{2} + x) + (2 x + 2))] + [1 \div ((x^{2} + 3 x) + (2 x + 6))]

= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]
= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]
=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)
Using the L.C.M
=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)
=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)
=

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

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Algebraic Fractions

Lee Azịza

Ajụjụ 14 Ripọtì

Simplify 2.04 × 3.7 (Leave your answer in 2 decimal place)

7.51

7.71

5.5

7.55

Akọwa Nkọwa

To simplify the expression 2.04 × 3.7, we can use the distributive property of multiplication over addition. That is: 2.04 × 3.7 = (2 + 0.04) × 3.7 We can now expand this expression as: (2 + 0.04) × 3.7 = 2 × 3.7 + 0.04 × 3.7 Simplifying further: 2 × 3.7 = 7.4 0.04 × 3.7 = 0.148 So, substituting these values back into the expression: 2.04 × 3.7 = 7.4 + 0.148 = 7.548 Rounding this answer to two decimal places, we get: 2.04 × 3.7 = 7.55 Therefore, the answer is option D - 7.55.

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Fractions, Decimals And Approximations

Number Bases

Lee Azịza

Ajụjụ 15 Ripọtì

If an investor invest ₦450,000 in a certain organization in order to yield X as a return of ₦25,000. Find the return on an investment of ₦700,000 by Y in the same organization.

₦14,950.50K

₦25,150.30K

₦15,000.00K

₦38,888.90K

Akọwa Nkọwa

To solve the problem, we need to use the concept of proportionality. We can set up a proportion between the return and the investment amount: Return/Investment = X/₦450,000 We can rearrange this to solve for X: X = (Return * ₦450,000) / Investment Now, we can use this equation to find the return Y on an investment of ₦700,000: Y = (X * ₦700,000) / ₦450,000 Substituting the given values, we get: X = (₦25,000 * ₦450,000) / ₦450,000 = ₦25,000 Y = (₦25,000 * ₦700,000) / ₦450,000 = ₦38,888.89 (rounded to ₦38,888.90K) Therefore, the answer is option D: ₦38,888.90K.

Gụọ akwụkwọ nkuzi na Application Of Differentiation (JAMB)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Application Of Differentiation

Financial Arithmetic

Lee Azịza

Ajụjụ 16 Ripọtì

Evaluate \( \log_{5}\left(y^{2}x^{5} \div 125b\frac{1}{2}\right) \)

2 log5y + 5log5 y2 −

log5 y2 + 5log5 x + 3

25logy 5 + 3

2log5y + 5log5x − ½ log5b −3

Logarithms

Lee Azịza

Ajụjụ 17 Ripọtì

If √(2x + 2) − √x = 1, find x.

x = 1 twice

x = 0 or 1

x = 3 or 1

x = 2 twice

Akọwa Nkọwa

To solve for x in the equation √(2x + 2) − √x = 1, we can use algebraic manipulation to isolate x on one side of the equation. Here's how: 1. Start by adding √x to both sides of the equation: √(2x + 2) = √x + 1 2. Square both sides of the equation to eliminate the square root: (√(2x + 2))^2 = (√x + 1)^2 Simplifying the left side: 2x + 2 = x + 1 + 2√x + 1 3. Rearrange terms: x - 2√x + 1 = 0 4. Factor the left side of the equation: (√x - 1)^2 = 0 5. Solve for x by taking the square root of both sides: √x - 1 = 0 √x = 1 x = 1^2 = 1 Therefore, the solution to the equation √(2x + 2) − √x = 1 is x = 1.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 18 Ripọtì

Simplify \(6\frac{1}{12} - 2\frac{3}{4} + 1\frac{1}{2}\)

3 \frac{5}{6}

4 \frac{5}{6}

216

\frac{1}{3}

Akọwa Nkọwa

6½ − 2¾ + 1½
73/12 − 11/4 + 3/2
[(73 − 33 + 18) ÷ (12)]
58/12
29/6

4 \frac{5}{6}

Lee Azịza

Ajụjụ 19 Ripọtì

Find x if 132x = 70 eight.

5

3

6

1

Akọwa Nkọwa

132x = 708

1 \times x^{2} + 3 \times x^{1} + 2 \times x^{0}

7 \times 8^{1} + 0 \times 8^{0}

x^{2} + 3 x + 2 - 56 = 0

x^{2} + 3 x - 54 = 0

x(x + 9)−6(x + 9) = 0
(x + 9)(x − 6) = 0
Either (x + 9) = 0 or (x − 6) = 0
x = − 9 or +6
The positive values for x = 6
The base number for 132x = 1326

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Lee Azịza

Ajụjụ 20 Ripọtì

Factorize x2 + 9x + 20

(x − 5) 2

(x + 5)(x + 4)

(x – 5)(x + 3)

(x + 3) 2

Akọwa Nkọwa

The correct answer is (x + 5)(x + 4). This can be determined using the method of factoring by grouping. The general idea is to rewrite the expression as the product of two binomials (in this case, two linear expressions) that add up to the original expression. To do this, we need to find two numbers that multiply to give 20 and add up to 9. These numbers are 4 and 5. So, we write the expression as: (x + 5)(x + 4) = x^2 + 9x + 20 Therefore, (x + 5)(x + 4) is the factorized form of x^2 + 9x + 20.

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 21 Ripọtì

Find at which rate per annum simple interest ₦525 will amount to ₦588 in 3 years.

3%

2%

5%

4%

Akọwa Nkọwa

To find the rate per annum of simple interest, we can use the formula: Simple Interest = (Principal * Rate * Time) / 100 where Principal is the initial amount, Rate is the interest rate per annum, and Time is the time period in years. In this case, we know that the Principal is ₦525, the Simple Interest is ₦588 - ₦525 = ₦63, and the Time is 3 years. Substituting these values into the formula, we get: 63 = (525 * Rate * 3) / 100 Simplifying the equation, we get: Rate = (63 * 100) / (525 * 3) = 4% Therefore, the rate per annum of simple interest that will amount to ₦588 in 3 years is 4%.

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 22 Ripọtì

The volume of a cylinder whose height is 4cm and whose radius 5cm is equal to (π = 3.14)

313cm2

145cm2

314cm2

214cm2

Akọwa Nkọwa

The formula for the volume of a cylinder is V = πr^2h, where "r" is the radius of the base, "h" is the height, and π (pi) is a mathematical constant approximately equal to 3.14. In this case, the height of the cylinder is 4cm, and the radius is 5cm. So we can substitute these values into the formula to find the volume: V = πr^2h V = π(5cm)^2(4cm) V = π(25cm^2)(4cm) V = π(100cm^3) V = 100π cm^3 Therefore, the volume of the cylinder is 100π cubic centimeters. We can leave the answer in terms of π or use the approximate value of 3.14 to get an estimate: V ≈ 100(3.14) cm^3 V ≈ 314 cm^3 So the answer is option C: 314cm^2.

Lee Azịza

Ajụjụ 23 Ripọtì

Solve the equation \(3x^2 - 4x - 5 = 0\)

x = 1.75 or − 0.15

x = 2.12 or − 0.79

x = 1.5 or − 0.34

x = 2.35 or −1.23

Solution Of Linear Equations

Lee Azịza

Ajụjụ 24 Ripọtì

If \(y = (x^2 + 3x - 1) \div (3x + 4)\). Find dy/dx

(3 x^{2} + 8 x + 7) \div (3 x + 4)^{2}

x^{2} + 2 x + (1 \div (x + 1)^{2})

2 x^{2} + 4 x + (1 \div (2 x + 4)^{2})

x^{2} + 3 x + (2 \div (x + 3)^{2})

Akọwa Nkọwa

dy/dx

(4 x^{3} + 3 x^{2} + 2 x + 1)

=

12 x^{2} + 6 x + 2

⇒

12 x^{2} + 6 + 2

Divide both sides by 2

\frac{12 x^{2}}{2} - \frac{6 x}{2} - \frac{2}{2}

=

6 x^{2} + 3 x + 1

dy/dx
=

6 x^{2} + 3 x + 1

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 25 Ripọtì

In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate.

3%

1.0%

2.5%

5.40%

Akọwa Nkọwa

To find the percentage birth rate, we need to divide the number of births by the total population and multiply by 100 to get the percentage. So, the birth rate would be (62/6250) x 100 = 0.992 x 100 = 0.992%. Therefore, the closest option to the birth rate would be 1.0%.

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Percentages

Lee Azịza

Ajụjụ 26 Ripọtì

Rationalize 5 ÷ (2 − √3)

3(2 + √5)

2(3 + √5)

5(2 + √3)

3 √2 + 1

Akọwa Nkọwa

To rationalize the expression 5 ÷ (2 − √3), we need to eliminate the square root in the denominator. We can do this by multiplying both the numerator and denominator by the conjugate of the denominator, which is 2 + √3. This is because when we multiply the conjugate of the denominator by the original denominator, we get a difference of squares that simplifies the expression: 5 ÷ (2 − √3) * (2 + √3) / (2 + √3) Simplifying the numerator by multiplying the terms inside the brackets, we get: 5(2 + √3) / (2^2 - (√3)^2) Simplifying further, we get: 5(2 + √3) / (4 - 3) 5(2 + √3) / 1 Therefore, the rationalized form of the expression is 5(2 + √3), which is equivalent to option C.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 27 Ripọtì

The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100))n. Find A, if P = 126, r = 4 and n = 2.

₦132.50K

₦136.30K

₦125.40K

₦257.42K

Akọwa Nkọwa

The formula for finding the amount of interest earned on a principal amount is given as A = P(1 + (r ÷ 100))n, where P is the principal amount, r is the interest rate, and n is the number of years the money is invested for. Using the given values of P = 126, r = 4 and n = 2, we can plug these values into the formula and find A: A = 126 * (1 + (4 ÷ 100))^2 = 126 * (1 + 0.04)^2 = 126 * 1.04^2 = 126 * 1.0816 = 136.30 So, the amount A to which the principal of 126 amounts at 4% compound interest for 2 years is ₦136.30K.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 28 Ripọtì

The probability of an outcome A is 1/6. The probability of the B outcome is 1/4. If the probability of A or B or both is 1/12, what is the probability of both outcomes A and B?

1/2

1/3

2/5

3/4

Akọwa Nkọwa

P(A outcome) =

\frac{1}{6}

, P(B outcome) = ¼
P(A ∪ B) = 1/12, P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
1/6 + 1/4 − 1/12
= (2 + 3 − 1) ÷ 12
= 4/12
= 1/3
P(A ∩ B) = 1/3

Lee Azịza

Ajụjụ 29 Ripọtì

Make x the subject of the equation

\(s = 2 + \frac{t}{5}(x + \frac{3}{5}y)\)

x = 5[(s − 2) ÷ t] + 3yt

x = 25[(s − 2) ÷ t] − 3ty

x = [1 ÷ (s − 2)3ty]

x = [5(s − 2) 2 ÷ 3ty] × t

Akọwa Nkọwa

2 + t/5(x + 3/5y ) = s
t/5(x + 3/5y ) = s − 2
[(t(5x + 3y) ÷ 25)] = s − 2
t(5x + 3y) = 25 (s − 2)
5x + 3y = [(25(s − 2))÷ t]
5x = [(25(s − 2))÷ t] − 3y
Divide both sides by 5
x = [25(s − 2) − 3ty)]÷ 5t

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 30 Ripọtì

In the venn diagram, the shaded region is

(P

\cap

Q)

\cup

R

(P

\cap

Q)

\cap

R

(P

\cap

Q1)

\cap

R

(P

\cap

Q1)

\cup

R

Functions And Relations

Lee Azịza

Ajụjụ 31 Ripọtì

The volume of a cone (s) of height 6cm and base radius 5cm is

157cm3

155cm3

175cm3

145cm3

Akọwa Nkọwa

The formula for the volume of a cone is given by: V = (1/3)πr²h where V is the volume, r is the radius of the base of the cone, h is the height of the cone, and π is a constant value of approximately 3.14. Substituting the given values into the formula, we have: V = (1/3)π(5cm)²(6cm) Simplifying this expression, we get: V = (1/3)π(25cm²)(6cm) V = (1/3)π(150cm³) V = 50π cm³ Using a calculator to approximate π to two decimal places (π ≈ 3.14), we have: V ≈ 157 cm³ Therefore, the correct option is: 157cm3

Lee Azịza

Ajụjụ 32 Ripọtì

Given that Z = {1,2,4,5} what is the power of set Z?

16

8

10

12

Akọwa Nkọwa

The power set of a set is the set of all possible subsets of that set, including the empty set and the set itself. The formula for finding the power set of a set is 2^n, where n is the number of elements in the original set. In this case, Z has four elements, so the number of subsets is 2^4 = 16. Therefore, the power set of Z has 16 elements, which is the answer.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 33 Ripọtì

Find the equation of a line which is from the origin and passes through the point (−3, −4)

y =

\frac{3 x}{4}

y =

\frac{4 x}{3}

y =

\frac{2 x}{3}

y =

\frac{x}{2}

Akọwa Nkọwa

Equation of a straight line is given by

y - y_{1} = m (x - x_{1})

Where m is the slope or gradient
Don't forget, the line passes through the origin. That means it has a coordinates of

{0, 0}

and

{- 3, - 4}

. So to calculate m

m = \frac{Δ y}{Δ x} = \frac{0 - (- 4)}{0 - (- 3)} = \frac{4}{3}

Equation of straight line is therefore given as:

y - (- 4) = \frac{4}{3} [x - (- 3)]

3 (y + 4) = 4 (x + 3)

3 y + 12 = 4 x + 12

3 y = 4 x

∴ y = \frac{4 x}{3}

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Algebraic Expressions

Coordinate Geometry

Lee Azịza

Ajụjụ 34 Ripọtì

Given that A = {1, 5, 7}

B = {3, 9, 12, 15}

C = {2, 4, 6, 8}

Find (A ∪ B) ∪ C

{1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

{1, 2, 3, 5, 6, 8, 12, 15}

{2, 4, 5, 9, 12, 15}

{1, 5, 6, 7, 8, 9, 12, 15}

Akọwa Nkọwa

The answer is {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}. (A ∪ B) means the union of sets A and B, which is the set of all elements that are in either set A or set B or both. Therefore, (A ∪ B) is {1, 3, 5, 7, 9, 12, 15}. Then, we take the union of (A ∪ B) and set C, which is the set of all elements that are in either (A ∪ B) or set C or both. Therefore, the final answer is {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}. This is because all the elements in A, B, and C are present in the final set.

Lee Azịza

Ajụjụ 35 Ripọtì

A pipe made of metal 10cm thick has an external radius of 11cm. find the area of metal in making 2.4cm of pipe

21πcm2

15πcm2

17πcm2

Akọwa Nkọwa

The external radius = 11cm
The internal radius = 10cm
The area of cross section = π(π2 − 102)
= π (11 + 10)(11 − 10)
= π(21)(1)
= 21πcm2

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 36 Ripọtì

Given that \( \tan x = \frac{2}{3} \), where \(0^\circ < x < 90^\circ\), Find the value of \(2\sin x\).

\frac{2 \sqrt{13}}{13}

\frac{3 \sqrt{13}}{13}

\frac{4 \sqrt{13}}{13}

\frac{6 \sqrt{13}}{13}

Akọwa Nkọwa

tan x =

\frac{2}{3}

(given), is illustrated in a right-angled

Δ

thus m2 = 22 + 32
= 4 + 9 = 13
m =

\sqrt{13}

Hence, 2sin x = 2 x

\frac{2}{m}

2 x

\frac{2}{\sqrt{13}}

=

\frac{4}{\sqrt{13}}

=

\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}

=

\frac{4 \sqrt{13}}{13}

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Lee Azịza

Ajụjụ 37 Ripọtì

Find the distance between the points (-2,-3) and (-2,4)

3m

2.4m

3.2m

5.1m

Lee Azịza

Ajụjụ 38 Ripọtì

If (159.75)10 = (y)6. Find x

x6 = 123.346

x6 = 424.56

x6 = 122.436

x6 = 124.456

Akọwa Nkọwa

(159.75)10 = (y)6

\begin{matrix} 6 & 159 \\ 6 & 26 R 3 \\ 6 & 4 R 2 \\ 0 R 4 \end{matrix} ↑

15910 = 4236
Then Convert 0.75 to base 6
0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5
0.5 x 6 = 3.00 423 + 1.5 =424.5
159.75 = (424.5)6
(159)10 = (424.5)6

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 39 Ripọtì

Simplify log101.5 + 3 log102 − log100.3

log104

log1040

log10-40

log104-

Akọwa Nkọwa

log101.5 + 3 log102 − log100.3
log101.5 + log1023 − log100.3
[log10(1.5 × 23) ÷ log100.3]
[log10(15/10 × 8 ) ÷ log10( 3/10)]
log10(15/10 × 8 × 10/3 )
log1040

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 40 Ripọtì

Determine the third term of a geometrical progression whose first and second term are 2 and 14 respectively

1458

1485

1345

1258

Akọwa Nkọwa

1st G.P. = a =2
2nd G.P. = ar − 1 = 54
2(r) = 54
r = 54/2 = 27
r = 27
3rd term = ar2 = (2) (27)2
2 × 27 × 27
= 1458

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 41 Ripọtì

Factorize \(x^2 - 2x - 15\)

(x + 3)2

(x + 5)(x − 3)

(x − 5)2

(x − 5)(x + 3)

Akọwa Nkọwa

The expression x^2 - 2x - 15 can be factored into the product of two binomials. To find these binomials, we can use the "ac method," where we try to find two numbers that multiply to -15 and add up to -2. These numbers are -5 and 3. We can then write the expression as:

(x - 5)(x + 3)

So, the expression x^2 - 2x - 15 can be factored as (x - 5)(x + 3).

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 42 Ripọtì

Evaluate log717

1.35

1.353

1.455

0.455

Akọwa Nkọwa

To evaluate log base 7 of 17, we need to find the exponent that 7 must be raised to in order to obtain 17. In other words, we need to solve the equation 7^x = 17. This cannot be done algebraically, so we must use numerical methods or a calculator to approximate the solution. Using a calculator, we find that 7^1.455 ≈ 16.999, and 7^1.456 ≈ 17.035. Therefore, log base 7 of 17 is approximately 1.455. So, the answer is: 1.455.

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 43 Ripọtì

Simplify \( \frac{1}{(x+1)} + \frac{1}{(x-1)} \)

2x/(x + 1)(x−3)

2/(x + 1)(x−1)

2x/(x + 1)2

2x(x+1)2

Akọwa Nkọwa

[1 ÷ (x+1)] + [1 ÷ (x − 1)]
= ((x − 1) + [(x + 1)) ÷ (x+1)(x − 1)]
Using the L.C.M.
= (x − 1 + x + 1) ÷ (x + 1)(x − 1)
= (x + 2 − 1 + 1) ÷ (x + 1)(x − 1)
= 2x ÷ (x + 1)(x − 1) =2x ÷ (x + 1)(x − 1)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 44 Ripọtì

Simplify √30 × √40

10 √3

5 √3

20 √3

15 √3

Akọwa Nkọwa

√30 × √40
= √(30 × 40)
= √(3 × 10 × 4 × 10)
= √(400 × 3)
= √400 × √3
= 20 × √3
= 20√3

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 45 Ripọtì

If three staff of Chukwudi and Sons Ltd. agreed to share their salary arrears in the ration of their ages, which are 18 years, 20 years, 22 years respectively. If the sum of the money collected is ₦120,000.00K, How much does the second staff receive?

₦36,000

₦44,000

₦40,000

₦15,000

Akọwa Nkọwa

To determine how much the second staff receives, we need to find out the total ratio of the ages of the three staff, which is 18 + 20 + 22 = 60. Next, we need to find the ratio of the second staff's age to the total age, which is 20/60 = 1/3. Then, we multiply the ratio by the total amount collected to find how much the second staff receives. So, 1/3 x ₦120,000 = ₦40,000. Therefore, the second staff receives ₦40,000. Answer (c) is correct.

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Modular Arithmetic

Sequence And Series

Lee Azịza

Ajụjụ 46 Ripọtì

If duty is levied at 25%, find the duty to be added to a bill of ₦80.

₦100

₦20

₦80

₦150

Akọwa Nkọwa

The correct answer is ₦20. To find the duty to be added to a bill of ₦80, we need to calculate 25% of ₦80. 25% means 25 out of 100, so we can calculate 25% of ₦80 by multiplying ₦80 by 25/100. ₦80 * 25/100 = ₦20 So the duty to be added to a bill of ₦80 is ₦20.

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Financial Arithmetic

Lee Azịza

Ajụjụ 47 Ripọtì

How many sides has a regular polygon whose interior angles are 120° each?

4

5

6

3

Akọwa Nkọwa

A regular polygon is a polygon with equal sides and equal angles. The formula to find the interior angles of a regular polygon is: Interior Angle = (n - 2) x 180 / n where n is the number of sides of the polygon. We are given that the interior angles of the regular polygon are 120°. Substituting this value into the formula, we get: 120 = (n - 2) x 180 / n Simplifying this equation, we get: 120n = 180n - 360 360 = 60n n = 6 Therefore, the regular polygon has 6 sides.

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 48 Ripọtì

Simplify 1¼ ÷ (2 ÷ ¼ of 28)

1

\frac{3}{8}

2

\frac{3}{4}

4

\frac{3}{8}

3

\frac{1}{5}

Akọwa Nkọwa

The correct answer is 3 15/38. To simplify the expression 1 1/4 ÷ (2 ÷ 1/4 of 28), we need to perform the operations in the correct order. First, we need to calculate 2 ÷ 1/4 of 28: 2 ÷ 1/4 of 28 = 2 ÷ (1/4 * 28) = 2 ÷ 7 = 4/7 Next, we can divide 1 1/4 by 4/7: 1 1/4 ÷ 4/7 = (5/4) ÷ (4/7) = (5/4) * (7/4) = 35/4 = 8 3/4 So the answer is 8 3/4, which can be simplified to 3 15/38 by dividing both the numerator and denominator by the greatest common divisor of 15 and 38, which is 1. So the final answer is 3 15/38.

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Sets

Modular Arithmetic

Lee Azịza

Ajụjụ 49 Ripọtì

Find the total surface area of a cylinder of base radius 5cm and length 7cm (π = 3.14)

219.9cm2

157.1cm2

377.0cm2

298.7cm2

Akọwa Nkọwa

The correct answer is 377.0cm^2. The formula for the surface area of a cylinder is given by: 2πr^2 + 2πrh where r is the radius of the base, h is the height of the cylinder, and π (pi) is a mathematical constant approximately equal to 3.14. In this case, r = 5cm and h = 7cm, so we can plug these values into the formula: 2π * 5^2 + 2π * 5 * 7 = 2π * 25 + 2π * 35 = 50π + 70π = 120π = 377.0cm^2 (rounded to one decimal place) So the total surface area of the cylinder is 377.0cm^2.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 50 Ripọtì

A man with an annual salary of ₦2000, has allowances of ₦600. If Income Tax is 5%, how much tax does he pay each year?

15

20

30

25

Akọwa Nkọwa

His annual salary = ₦2000
His allowances = ₦600
So his taxable income = Annual salary − allowance
= ₦2000 − ₦600
= ₦1400
He pays at 5%
Then, his allowance income tax 5/100 × 600 = ₦30

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Financial Arithmetic

Lee Azịza






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