JAMB UTME - Mathematics - 2015

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}

Bayanin Amsa

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

4 \frac{5}{6}

Duba Amsa

Tambaya 2 Rahoto

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

3%

1.0%

2.5%

5.40%

Bayanin Amsa

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%.

Karanta bayanin darasi akan Percentages (WAEC)

Percentages

Duba Amsa

Tambaya 3 Rahoto

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

Duba Amsa

Tambaya 4 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 5 Rahoto

Factorize x2 + 9x + 20

(x − 5) 2

(x + 5)(x + 4)

(x – 5)(x + 3)

(x + 3) 2

Bayanin Amsa

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.

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 6 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Algebraic Expressions

Coordinate Geometry

Duba Amsa

Tambaya 7 Rahoto

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

(x + 3)2

(x + 5)(x − 3)

(x − 5)2

(x − 5)(x + 3)

Bayanin Amsa

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).

Karanta bayanin darasi akan Polynomials (JAMB)

Polynomials

Duba Amsa

Tambaya 8 Rahoto

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

10.3cm2

15.7cm2

50.3cm2

17.4cm2

Bayanin Amsa

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)

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 9 Rahoto

Simplify 4 √[(390695T− 8)½)]

1/T

2T

5/T

T/5

Bayanin Amsa

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

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Surds (radicals)

Duba Amsa

Tambaya 10 Rahoto

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}

Bayanin Amsa

[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}

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Number Bases

Algebraic Fractions

Duba Amsa

Tambaya 11 Rahoto

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

Bayanin Amsa

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

Duba Amsa

Tambaya 12 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Probability

Areas

Duba Amsa

Tambaya 13 Rahoto

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

7.51

7.71

5.5

7.55

Bayanin Amsa

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.

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Number Bases (JAMB)

Fractions, Decimals And Approximations

Number Bases

Duba Amsa

Tambaya 14 Rahoto

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})

Bayanin Amsa

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

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 15 Rahoto

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

x6 = 123.346

x6 = 424.56

x6 = 122.436

x6 = 124.456

Bayanin Amsa

(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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 16 Rahoto

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

Duba Amsa

Tambaya 17 Rahoto

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

157cm3

155cm3

175cm3

145cm3

Bayanin Amsa

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

Duba Amsa

Tambaya 18 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Modular Arithmetic

Sequence And Series

Duba Amsa

Tambaya 19 Rahoto

Simplify √30 × √40

10 √3

5 √3

20 √3

15 √3

Bayanin Amsa

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

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Duba Amsa

Tambaya 20 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 21 Rahoto

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

2, ½

½ , 2

1, ½

0, 2

Duba Amsa

Tambaya 22 Rahoto

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

0.81

0.081

0.0081

8.0001

Bayanin Amsa

(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

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 23 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 24 Rahoto

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}

Bayanin Amsa

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.

Duba Amsa

Tambaya 25 Rahoto

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

₦100

₦20

₦80

₦150

Bayanin Amsa

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.

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Financial Arithmetic

Duba Amsa

Tambaya 26 Rahoto

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

₦48.75K

₦50.10K

₦75.50K

₦15.75K

Bayanin Amsa

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.

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Representation Of Data (JAMB)

Sine, Cosine And Tangent Of An Angle

Representation Of Data

Duba Amsa

Tambaya 27 Rahoto

If √(2x + 2) − √x = 1, find x.

x = 1 twice

x = 0 or 1

x = 3 or 1

x = 2 twice

Bayanin Amsa

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.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Duba Amsa

Tambaya 28 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Financial Arithmetic

Duba Amsa

Tambaya 29 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Duba Amsa

Tambaya 30 Rahoto

Rationalize 5 ÷ (2 − √3)

3(2 + √5)

2(3 + √5)

5(2 + √3)

3 √2 + 1

Bayanin Amsa

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.

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Duba Amsa

Tambaya 31 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 32 Rahoto

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

16

8

10

12

Bayanin Amsa

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.

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 33 Rahoto

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}

Bayanin Amsa

[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)}

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Algebraic Fractions

Duba Amsa

Tambaya 34 Rahoto

Evaluate log717

1.35

1.353

1.455

0.455

Bayanin Amsa

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.

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 35 Rahoto

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

Duba Amsa

Tambaya 36 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Application Of Differentiation (JAMB)

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Application Of Differentiation

Financial Arithmetic

Duba Amsa

Tambaya 37 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 38 Rahoto

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

Bayanin Amsa

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)

Karanta bayanin darasi akan Polynomials (JAMB)

Polynomials

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Tambaya 39 Rahoto

Simplify 1¼ ÷ (2 ÷ ¼ of 28)

1

\frac{3}{8}

2

\frac{3}{4}

4

\frac{3}{8}

3

\frac{1}{5}

Bayanin Amsa

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.

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Fractions, Decimals And Approximations

Sets

Modular Arithmetic

Duba Amsa

Tambaya 40 Rahoto

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

313cm2

145cm2

314cm2

214cm2

Bayanin Amsa

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.

Duba Amsa

Tambaya 41 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 42 Rahoto

Find x if 132x = 70 eight.

5

3

6

1

Bayanin Amsa

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

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 43 Rahoto

Simplify log101.5 + 3 log102 − log100.3

log104

log1040

log10-40

log104-

Bayanin Amsa

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

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 44 Rahoto

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

Bayanin Amsa

[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)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 45 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Positive And Negative Integers, Rational Numbers

Duba Amsa

Tambaya 46 Rahoto

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

1458

1485

1345

1258

Bayanin Amsa

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

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 47 Rahoto

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

4

5

6

3

Bayanin Amsa

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.

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 48 Rahoto

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

3m

2.4m

3.2m

5.1m

Duba Amsa

Tambaya 49 Rahoto

Solve the equation

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

3/2

− 7/4

4/7

7/4

Bayanin Amsa

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).

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 50 Rahoto

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

3%

2%

5%

4%

Bayanin Amsa

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%.

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Ratio, Proportions And Rates (WAEC)

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

Duba Amsa






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