JAMB UTME - Mathematics - 1990

Find the solution of the equation $x - 8\sqrt{x} + 15 = 0$

3, 5

-3, -5

9, 25

-9, 25

Bayanin Amsa

x - 8

\sqrt{x}

+ 15 = 0
x + 15 = 8

\sqrt{x}

square both sides = (x + 15)2 = (8

\sqrt{x}

2
x2 + 225 + 30x = 64x
x2 + 225 + 30x - 64x = 0
x2 - 34x + 225 = 0
(x - 9)(x - 25) = 0
x = 9 or 25

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 2 Rahoto

If $ \cos \theta = \frac{12}{13} $. Find $ \theta + \cos^2 \theta $

\frac{169}{25}

\frac{25}{169}

\frac{169}{144}

\frac{144}{169}

Bayanin Amsa

Cos

θ

=

\frac{12}{13}

x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan

θ

=

\frac{5}{12}

and cos

θ

=

\frac{12}{13}

If cos2

θ

= 1 +

\frac{1}{t a n^{2} θ}

= 1 +

\frac{1}{\frac{(5)^{2}}{12}}

= 1 +

\frac{1}{\frac{25}{144}}

= 1 +

\frac{144}{25}

=

\frac{25 + 144}{25}

=

\frac{169}{25}

Duba Amsa

Tambaya 3 Rahoto

The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

3

6

18

38

Bayanin Amsa

The mean of ten positive numbers is given to be 16. This means that the sum of these ten numbers is 10 times 16, which is 160. When an eleventh number is added, the mean becomes 18. Let's call this number "x". Now we have a total of eleven numbers, and their sum is 11 times 18, which is 198. We can find the eleventh number by subtracting the sum of the first ten numbers (160) from the sum of all eleven numbers (198): x = 198 - 160 = 38 Therefore, the eleventh number is 38, which is.

Duba Amsa

Tambaya 4 Rahoto

The perimeter of a rectangular lawn is 24m. If the area of the lawn is 35m2; how wide is the lawn?

5cm

7m

12m

14m

Lengths And Perimeters

Duba Amsa

Tambaya 5 Rahoto

The locus of a point which moves so that it is equidistant from two intersecting straight lines is the

perpendicular bisector of the two lines

angle bisector of the two lines

bisector of the two lines

line parallel to the two lines

Loci

Circles

Coordinate Geometry

Duba Amsa

Tambaya 6 Rahoto

Divide x3 - 2x2 - 5x + 6 by (x - 1)

x2 - x - 6

x2 - 5x + 6

x2 - 7x + 6

x2 - 5x - 6

Algebraic Expressions

Duba Amsa

Tambaya 7 Rahoto

What is the equation of the quadratic function represented by the graph?

y = x2 + x - 2

y = -x2 - x + 2

y = x2 - x - 2

y = -x2 + x + 2

Bayanin Amsa

The required equation is y = x2 - x - 2

i.e. B where the graph touches the graph touches the x-axis y = 0

x2 - x - 2 = 0 = (x + 1)(x - 2) = 0

Hence roots of the equation are -1 and 2 as shown in the graph

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 8 Rahoto

In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY

8cm

6cm

4cm

3cm

Triangles And Polygons

Duba Amsa

Tambaya 9 Rahoto

The H.C.F. of a2bx + ab2x and a2b - b2 is

b

a + b

b(a

\div

b)

abx(a2 - b2)

Bayanin Amsa

To find the H.C.F. of the given terms, we need to factorize them first: a²bx + ab²x = abx(a + b) a²b - b² = b(a - b) The H.C.F. of two or more terms is the product of their common factors raised to the lowest power. Both the terms have a common factor of 'b', so the H.C.F. will have 'b' as a factor. Also, the first term has a common factor of 'abx', and the second term does not have any factor of 'abx'. So, the H.C.F. will not have any factor of 'abx'. Therefore, the H.C.F. of a²bx + ab²x and a²b - b² is 'b'. Hence, the correct answer is: b.

Duba Amsa

Tambaya 10 Rahoto

What must be added to 4x2 - 4 to make it a perfect square?

)

\frac{- 1}{x^{2}}

\frac{1}{x^{2}}

1

-1

Simple Operations On Algebraic Expressions

Quadratic Equations

Duba Amsa

Tambaya 11 Rahoto

At what value of x is the function x2 + x + 1 minimum?

1

\frac{3}{4}

\frac{5}{3}

9

Bayanin Amsa

To find the value of `x` that minimizes the function `x^2 + x + 1`, we need to find the vertex of the parabola represented by this function. The vertex of a parabola of the form `ax^2 + bx + c` is given by `(-b/2a, f(-b/2a))`. In this case, `a = 1`, `b = 1`, and `c = 1`, so the vertex is located at `(-1/2, f(-1/2))`. To find `f(-1/2)`, we substitute `-1/2` for `x` in the function: `f(-1/2) = (-1/2)^2 + (-1/2) + 1 = 3/4` Therefore, the vertex of the parabola is `(-1/2, 3/4)`. Since the parabola opens upwards (the coefficient of `x^2` is positive), the value of `x` that minimizes the function is the x-coordinate of the vertex, which is `-1/2`. Therefore, the answer is `1`.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 12 Rahoto

Correct 241.34(3 x 10-3)2 to 4 significant figures

0.0014

0.001448

0.0022

0.002172

Bayanin Amsa

To correctly express 241.34 x 10^-3 to 4 significant figures, we need to start by determining the first significant digit, which is the first non-zero digit. In this case, it is 2. Next, we count the next three digits after the first significant digit, which are 4, 1, and 3. Therefore, the number 241.34 x 10^-3 to 4 significant figures is 0.2413. Out of the given options, the number that matches 0.2413 to 4 significant figures is: 0.002172.

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

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 13 Rahoto

4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16.

48o

84o

92o

276o

Bayanin Amsa

Given that 4, 16, 30, 20, 10, 14 and 26
Adding up = 120
nos

\geq

16 are 16 + 30 + 20 + 26 = 92
The requires sum of angles =

\frac{92}{120}

x

\frac{360^{o}}{1}

= 276o

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 14 Rahoto

Simplify 3 log69 + log612 + log664 - log672

5

7776

log631

(7776)6

Bayanin Amsa

3 log69 + log612 + log664 - log672
= log693 + log612 + log664 - log672
log6729 + log612 + log664 - log672
log6(729 x 12 x 64) = log6776
= log665 = 5 log66 = 5
N.B: log66 = 1

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 15 Rahoto

Find the curved surface area of the frustrum in the figure.

16

π \sqrt{10}

cm2

20

π \sqrt{10}

cm2

24

π \sqrt{10}

cm2

36

π \sqrt{10}

cm2

Bayanin Amsa

\frac{x}{4} = \frac{6 + x}{6}

6x = 4(6 + x) = 24 + 4x

x = 12 = c = $π R L - π L$

= $π (6) \sqrt{18^{2}} + 6^{2} - π \times 4 \times \sqrt{12^{2}} + 4^{2}$

= 6 $π \sqrt{360} - 4 π \sqrt{160}$

= 36 $π \sqrt{10} - 16 π \sqrt{10}$

= 20 $π \sqrt{10}$ cm2

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Matrices And Determinants

Duba Amsa

Tambaya 16 Rahoto

Given that $x^2 + y^2 + z^2 = 194$, calculate z if $x = 7$ and $\sqrt{y} = 3$

√10

8

12.2

13.4

Bayanin Amsa

Given that x2 + y2 + z2 = 194, calculate z if x = 7 and $\sqrt{y}$ = 3
x = 7
∴ x2 = 49
$\sqrt{y}$ = 3
∴ y2 = 81 = x2 + y2 + z2 = 194
49 + 81 + z2 = 194
130 + z2 = 194
z2 = 194 - 130
= 64
z = $\sqrt{64}$
= 8

Duba Amsa

Tambaya 17 Rahoto

If $x + \frac{1}{x} = 4$, find $x^2 + \frac{1}{x^2}$

16

14

12

9

Bayanin Amsa

x +

\frac{1}{x}

= 4, find x2 +

\frac{1}{x^{2}}

= (x +

\frac{1}{x}

)2 = x2 +

\frac{1}{x^{2}}

+ 2
x2 +

\frac{1}{x^{2}}

= ( x +

\frac{1}{x^{2}}

)2 - 2
= (4)2 - 2
= 16 - 2
= 14

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 18 Rahoto

A car painter charges ₦40.00 per day for himself and ₦10.00 per day for his assistant. if a fleet of cars were painted for ₦2000.00 and the painter worked 10days more than his assistant, how much did the assistant receive?

₦32.00

₦320.00

₦420.00

₦1680.00

Bayanin Amsa

To solve this problem, we can use a system of equations. Let's call the number of days the assistant worked "x". Then the number of days the car painter worked would be "x + 10" (since he worked 10 days more than the assistant). We can then set up the following equations: 40(x+10) + 10x = 2000 (this represents the total amount the car painter and his assistant earned for painting the cars) Simplifying this equation: 50x + 400 = 2000 50x = 1600 x = 32 So the assistant worked for 32 days. To find out how much he earned, we can plug this value back into one of the original equations: 10x = 10(32) = 320 Therefore, the assistant earned ₦320.00.

Karanta bayanin darasi akan Percentages (WAEC)

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Percentages

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Areas

Duba Amsa

Tambaya 19 Rahoto

If $a = 2$, $b = -2$ and $c = -\frac{1}{2}$, evaluate $(ab^2 - bc^2)(a^2c - abc)$

2

-28

-30

-34

Bayanin Amsa

(ab2 - bc2)(a2c - abc)
[2(2)2 - (- 2x

\frac{1}{2}

)] [22(-

\frac{1}{2}

) - 2(-2)(-

\frac{1}{2}

)]
[8 =

\frac{1}{2}

][-2 - 2] =

\frac{17}{2}

x 42
= -34

Duba Amsa

Tambaya 20 Rahoto

At what rate would a sum of ₦100.00 deposited for 5 years raise an interest of ₦7.50?

1

\frac{1}{2}

%

2

\frac{1}{2}

%

1.5%

25%

Bayanin Amsa

To calculate the interest rate that would raise a sum of ₦100.00 deposited for 5 years to an interest of ₦7.50, we need to use the formula: Simple Interest = (Principal * Rate * Time) / 100 Where: - Principal = ₦100.00 - Simple Interest = ₦7.50 - Time = 5 years By substituting the values in the formula, we get: ₦7.50 = (₦100.00 * Rate * 5) / 100 Simplifying the equation, we get: Rate = (₦7.50 * 100) / (₦100.00 * 5) Rate = 1.5% Therefore, the interest rate that would raise a sum of ₦100.00 deposited for 5 years to an interest of ₦7.50 is 1.5%.

Karanta bayanin darasi akan Percentages (WAEC)

Percentages

Duba Amsa

Tambaya 21 Rahoto

Simplify $ \left(\frac{1}{x^{-1}} + \frac{1}{y^{-1}}\right)^{-1} $

\frac{x}{y}

xy

\frac{x}{y}

(xy)-1

Bayanin Amsa

Simplify (

\frac{1}{x^{- 1}} + \frac{1}{y^{- 1}}

)-1 = (

\frac{1}{x^{- 1}} + \frac{1}{y^{- 1}}

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

\frac{(x)}{y}

=

\frac{x}{y}

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Solution Of Linear Equations

Circles

Algebraic Expressions

Duba Amsa

Tambaya 22 Rahoto

A cylinder pipe, made of metal is 3cm thick.If the internal radius of the pope is 10cm.Find the volume of metal used in making 3m of the pipe.

153

π

cm3

207

π

cm3

15 300

π

cm3

20 700

π

cm3

Bayanin Amsa

To find the volume of metal used to make 3 meters of the pipe, we first need to find the volume of one meter of the pipe, and then multiply it by 3. The pipe is a cylinder with an internal radius of 10cm and a thickness of 3cm, which means that the external radius of the pipe is 13cm (10cm + 3cm). To find the volume of one meter of the pipe, we need to find the volume of the outer cylinder (with radius 13cm) and subtract the volume of the inner cylinder (with radius 10cm): Volume of outer cylinder = π x radius^2 x height = π x 13^2 x 100 = 16,900π cm^3 Volume of inner cylinder = π x radius^2 x height = π x 10^2 x 100 = 10,000π cm^3 Volume of metal in one meter of the pipe = Volume of outer cylinder - Volume of inner cylinder = 16,900π - 10,000π = 6,900π cm^3 Therefore, the volume of metal used to make 3 meters of the pipe is: Volume of metal in 3 meters of the pipe = Volume of metal in 1 meter x 3 = 6,900π x 3 = 20,700π cm^3 So the answer is 20,700π cubic cm.

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 23 Rahoto

Simplify $ \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} $

\frac{2 x}{(x - 2) (x + 2) (x^{2} - 4)}

\frac{2 x}{x^{2} - 4}

\frac{x}{x^{2} - 4}

\frac{4 x}{x^{2} - 4}

Bayanin Amsa

\frac{1}{x - 2}

+

\frac{1}{x + 2}

+

\frac{2 x}{x^{2} - 4}

=

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

=

\frac{4 x}{x^{2} - 4}

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

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Sine, Cosine And Tangent Of An Angle

Matrices And Determinants

Algebraic Expressions

Duba Amsa

Tambaya 24 Rahoto

Three children shared a basket of mangoes in such a way that the first child took $ \frac{1}{4} $ of the mangoes and the second $ \frac{3}{4} $ of the remainder. What fraction of the mangoes did the third child take?

\frac{3}{16}

\frac{7}{16}

\frac{9}{16}

\frac{13}{16}

Bayanin Amsa

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.
If the first child takes

\frac{1}{4}

it will remain 1 -

\frac{1}{4}

=

\frac{3}{4}

Next, the second child takes

\frac{3}{4}

of the remainder
which is

\frac{3}{4}

i.e. find

\frac{3}{4}

of

\frac{3}{4}

=

\frac{3}{4}

x

\frac{3}{4}

=

\frac{9}{16}

the fraction remaining now =

\frac{3}{4}

-

\frac{9}{16}

=

\frac{12 - 9}{16}

=

\frac{3}{16}

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 25 Rahoto

Find the sum of the first twenty terms of the progression log a, log a2, log a3.....

log a20

log a21

log a200

log a210

Sequence And Series

Logarithms

Duba Amsa

Tambaya 26 Rahoto

If f(x - 4) = x2 + 2x + 3, Find, f(2)

6

11

27

51

Bayanin Amsa

f(x - 4) = x2 + 2x + 3
To find f(2) = f(x - 4)
= f(2)
x - 4 = 2
x = 6
f(2) = 62 + 2(6) + 3
= 36 + 12 + 3
= 51

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 27 Rahoto

An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7]

70o

84o

96o

156o

168o

Bayanin Amsa

We can use the formula for the length of an arc of a circle given by: L = rθ where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the center of the circle in radians. In this problem, we are given L = 22cm and r = 15cm, and we need to find θ. We are also given that π = 22/7. Using the formula above and substituting the given values, we can solve for θ: 22 = 15θ θ = 22/15 radians To convert this to degrees, we can multiply by 180/π: θ = (22/15) * (180/22/7) θ = 84 degrees (rounded to the nearest degree) Therefore, the answer is: - 84o

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 28 Rahoto

Simplify $4\frac{3}{4} - 6\frac{1}{4}$

-7

\frac{7}{8}

\frac{- 2}{7}

\frac{- 10}{21}

\frac{10}{21}

Bayanin Amsa

4

\frac{3}{4}

- 6

\frac{1}{4}

\frac{19}{4}

-

\frac{25}{4}

............(A)

\frac{21}{5}

-

\frac{5}{4}

.............(B)
Now work out the value of A and the value of B and then find the value

\frac{A}{B}

A =

\frac{19}{4}

-

\frac{25}{4}

=

\frac{- 6}{4}

B =

\frac{21}{5}

x

\frac{5}{20}

=

\frac{105}{20}

=

\frac{21}{4}

But then

\frac{A}{B}

=

\frac{- 6}{4}

\frac{21}{4}

=

\frac{- 6}{4}

\div

\frac{21}{4}

=

\frac{- 6}{4}

x

\frac{4}{21}

=

\frac{- 24}{84}

=

\frac{- 2}{7}

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Solution Of Linear Equations

Areas

Duba Amsa

Tambaya 29 Rahoto

Factorize 9(x + y)2 - 4(x - y)2

(x + y)(5x + y)

(x + y)2

(x + 5y)(5x + y)

5(x + y)2

Bayanin Amsa

9(x + y)2 - 4(x - y)2
Using difference of two squares which says
a2 - b2 = (a + b)(a - b) = 9(x + y)2 - 4(x - y)2
= [3(x + y)]2 - [2(x - y)]-2
= [3(x + y) + 2(x - y) + 2(x - y)][3(x + y) - 2(x - y)]
= [3x +3y + 2x - 2y][3x + 3y - 2x + 2y]
= (5x + y)(x + 5y)

Duba Amsa

Tambaya 30 Rahoto

Simplify and express in standard form $ \frac{0.00275 \times 0.0064}{0.025 \times 0.08} $

8.8 x 10-1

8.8 x 10-2

8.8 x 10-3

8.8 x 103

Bayanin Amsa

To simplify and express in standard form the expression: $$\frac{0.00275 \times 0.0064}{0.025 \times 0.08}$$ We can start by simplifying the numerator and denominator: $$\frac{0.00275 \times 0.0064}{0.025 \times 0.08} = \frac{0.0000176}{0.002}$$ Then, we can simplify the fraction by dividing both the numerator and denominator by the greatest common factor of the two: $$\frac{0.0000176}{0.002} = \frac{8.8}{1000} = \frac{22}{2500}$$ Finally, we can express this fraction in standard form by writing it as a number between 1 and 10 multiplied by a power of 10: $$\frac{22}{2500} = 0.0088 = 8.8 \times 10^{-3}$$ Therefore, the expression $\frac{0.00275 \times 0.0064}{0.025 \times 0.08}$ simplified and expressed in standard form is $8.8 \times 10^{-3}$.

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Modular Arithmetic

Duba Amsa

Tambaya 31 Rahoto

Find the sum of the first 18 terms of the progression 3, 6, 12......

3(217 - 1)

3(218 - 1)

3(218 + 1)

3(217 - 1)

Bayanin Amsa

3 + 6 + 12 + .....18thy term
1st term = 3, common ratio

\frac{6}{3}

= 2
n = 18, sum og GP is given by Sn = a

\frac{(r^{n} - 1)}{r - 1}

s18 = 3

\frac{(2^{18} - 1)}{2 - 1}

= 3(217 - 1)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 32 Rahoto

Simplify $ \sqrt{27} + \frac{3}{\sqrt{3}} $

4

\sqrt{3}

\frac{4}{\sqrt{3}}

3

\sqrt{3}

\frac{\sqrt{3}}{4}

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Duba Amsa

Tambaya 33 Rahoto

Find the probability that a number selected at random from 40 to 50 is a prime

\frac{3}{11}

\frac{5}{11}

\frac{3}{11}

\frac{4}{11}

Bayanin Amsa

From 40 to 50 = 11 & number are prime i.e. 41, 43, 47
prob. of selecting a prime No. is

\frac{3}{11}

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 34 Rahoto

If the heights of two circular cylinder are in the ratio 2 : 3 and their volumes?

27 : 32

27 : 23

23 : 32

32 : 27

Bayanin Amsa

$\frac{h_{1}}{h_{2}}$ = $\frac{2}{3}$
h2 = $\frac{2 h_{1}}{3}$
$\frac{r_{1}}{r_{2}}$ = $\frac{9}{8}$
r2 = $\frac{9 r_{1}}{8}$
v1 = $π$ ( $\frac{9 r_{1}}{8}$ )2( $\frac{2 h_{1}}{3}$ )
= $π$ r1 2h1 x $\frac{27}{32}$
v = $\frac{π r_{1} 2 h_{1} \times \frac{27}{32}}{π r_{1} 2 h_{1}}$ = $\frac{27}{32}$
v2 : v1 = 27 : 32

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Matrices And Determinants

Duba Amsa

Tambaya 35 Rahoto

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Bayanin Amsa

To solve this problem, we need to use the fact that the sine and cosine functions are related to each other through the unit circle. First, we need to find the angle whose cosine is equal to cos 50°. We can do this by drawing a unit circle and marking the angle 50° on the circle. Then, we draw a line from the center of the circle to the point on the circumference that corresponds to the angle 50°. This line will intersect the x-axis at a point that has the same x-coordinate as the cosine of 50°. Next, we need to find the angle whose sine is equal to this x-coordinate. We can do this by drawing a perpendicular line from the point on the circumference to the y-axis. This line will intersect the y-axis at a point that has the same y-coordinate as the sine of the angle we are looking for. Finally, we can find the angle by using the inverse sine function (also known as arcsine) to find the angle whose sine is equal to this y-coordinate. Using this method, we find that the angle x is approximately 40.5°, which is closest to option (A) 40°. Therefore, the answer is (A) 40°.

Karanta bayanin darasi akan Trigonometry (JAMB)

Trigonometry

Duba Amsa

Tambaya 36 Rahoto

In the figure, PQR is a semicircle. Calculate the area of the shaded region.

125

\frac{2}{7}

cm2

149

\frac{2}{7}

cm2

234

\frac{1}{7}

cm2

267

\frac{1}{2}

cm2

Areas

Duba Amsa

Tambaya 37 Rahoto

Find the length of a side of a rhombus whose diagonals are 6cm and 8cm

8cm

5cm

4cm

3cm

Bayanin Amsa

In a rhombus, the diagonals intersect at a right angle and bisect each other. This means that the diagonals divide the rhombus into four congruent right triangles. Let's call the length of the side we are looking for "x". Using the Pythagorean theorem, we can find the length of each of the two right triangles that are formed by the diagonals: $$(\frac{6}{2})^2 + (\frac{8}{2})^2 = x^2$$ Simplifying this equation gives: $$9+16=x^2$$ Adding 9 and 16 gives us 25, so: $$25=x^2$$ To solve for x, we take the square root of both sides: $$\sqrt{25}=\sqrt{x^2}$$ Which simplifies to: $$5=x$$ Therefore, the length of the side of the rhombus is 5cm.

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Lengths And Perimeters

Duba Amsa

Tambaya 38 Rahoto

Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

\frac{2}{3}

\frac{2}{15}

\frac{1}{2}

\frac{1}{4}

Bayanin Amsa

\begin{array}{cc} 1 & 2 & 3 & 4 \\ 1 (1, 1) & (1, 2) & (1, 3) & (1, 4) \\ 2 (2, 1) & (2, 2) & (2, 3) & (2, 4) \\ 3 (3, 1) & (3, 2) & (3, 3) & (3, 4) \\ 4 (4, 1) & (4, 2) & (4, 3) & (4, 4) \end{array}

sample space = 16
sum of nos. removed are (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even nos. = 8 of them
Pr(even sum) =

\frac{8}{16}

=

\frac{1}{2}

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Representation Of Data

Probability

Permutation And Combination

Duba Amsa

Tambaya 39 Rahoto

In the figure, YXZ = $30^\circ$, XYZ = $105^\circ$ and XY = 8cm. Calculate YZ

16

\sqrt{2}

cm

8

\sqrt{2}

cm

4

\sqrt{2}

cm

22cm

Bayanin Amsa

To find the length of YZ, we can use the sine rule which relates the sides and angles of a triangle. The sine rule states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in the triangle. In other words: a b c --- --- --- = 1 sinA sinB sinC where a, b, and c are the lengths of the sides of the triangle opposite the angles A, B, and C, respectively. Using the sine rule, we can find the length of YZ as follows: YZ XY XY -- = --- = --- sinX sinZ sinY Substituting the given values, we get: YZ 8 8 -- = --- = --- sin30 sin105 sin45 Using a calculator to evaluate the sines, we get: YZ 8 8 -- = --- = --- × (2 + √2) ≈ 8.485 0.5 0.966 0.707 Therefore, YZ is approximately equal to 8.485 cm, which is equivalent to 4√2 cm to two decimal places. Therefore, the correct option is: 4√2 cm.

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Angles Of Elevation And Depression (WAEC)

Karanta bayanin darasi akan Angles (WAEC)

Introductory Calculus

Angles Of Elevation And Depression

Angles

Duba Amsa

Tambaya 40 Rahoto

The angle of a sector of s circle, radius 10.5cm, is 48 o, Calculate the perimeter of the sector

8.8cm

25.4cm

25.6cm

28.8cm

Bayanin Amsa

To find the perimeter of the sector, we need to find the arc length and add the radii. The formula for the arc length of a sector is: Arc length = (angle/360) x 2πr where r is the radius of the circle, and angle is the angle of the sector in degrees. In this case, the radius is given as 10.5 cm, and the angle is 48 degrees. So, the arc length is: Arc length = (48/360) x 2π x 10.5 Arc length = 0.133 x 2 x 3.14 x 10.5 Arc length = 8.82 cm (rounded to two decimal places) Now, we add the radii to the arc length to get the perimeter of the sector: Perimeter = 2 x 10.5 + 8.82 Perimeter = 28.82 cm (rounded to two decimal places) Therefore, the answer is 28.8cm.

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Coordinate Geometry

Duba Amsa

Tambaya 41 Rahoto

A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]

6o

26o

52o

103o

206o

Bayanin Amsa

The area of a sector of a circle is given by the formula A = (θ/360)πr^2, where A is the area of the sector, r is the radius of the circle, and θ is the central angle of the sector in degrees. We are given that the radius of the circle is 7cm and the area of the sector is 44cm^2, so we can plug these values into the formula: 44 = (θ/360)(22/7)(7)^2 Simplifying this expression, we get: 44 = (θ/360)(22)(7) 44 = (77θ/180) θ = (180/77)(44) θ ≈ 102.6 degrees Therefore, the angle of the sector is approximately 103 degrees to the nearest degree, which is.

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Indices, Logarithms And Surds (JAMB)

Circles

Indices, Logarithms And Surds

Duba Amsa

Tambaya 42 Rahoto

Each of the interior angles of a regular polygon is 140o. How many sides has the polygon?

9

8

7

5

Bayanin Amsa

To find out the number of sides of a regular polygon when given the measure of its interior angles, we can use the formula: Interior angle = (n-2) x 180 / n Where 'n' is the number of sides of the polygon. In this case, we know that each interior angle of the polygon measures 140 degrees. Substituting this value into the formula, we get: 140 = (n-2) x 180 / n Simplifying this equation, we get: 140n = (n-2) x 180 140n = 180n - 360 360 = 40n n = 9 Therefore, the regular polygon has 9 sides. Thus, the correct answer is 9.

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 43 Rahoto

If the shadow of a pole 7m high is 1/2 its length what is the angle of elevation of the sun, correct to the nearest degree?

90o

63o

60o

26o

0o

Bayanin Amsa

tan θ - 7/3.5 = 2
tan θ= 2.0000
θ = 63o