C

NB: Change the mixed fractions to improper fractions

Explanation:

Multiply the whole number(A) by the denominator(c) and add the numerator(b) to the result(Axc). Divide everything by the denominator(c).

⇒ 2¹⁄₂ = ^{(2x2 + 1)}⁄₂ = ⁵⁄₂
⇒ 1¹⁄₂ = ^{(1x2 + 1)}⁄₂ = ³⁄₂

Concepts

a = a/₁
b = b/₁

Explanation:

Change the ÷ to x (multiplication). To do so:
Reciprocate the fraction at the right side of the ÷
Reciprocate means the top goes to the down and the down comes to the top

Explanation:

The numerators multiply each other over the denominators multiplying each other.

Thus the products of the numerators over the products of the denominators

Product means multiplication

Applying all the above concepts

Answer is A

3 x 9 ^{1 + x} = 27^-x

Concept

A^b x A^c = A^b+c

Thus if the bases are the same, we can add the indexes (superscripts)

3 can go into the other numbers(9,27) hence we change then to have base 3

9 = 3¹x 3¹ = 3^{1 + 1}=3²

27 = 3¹x 3¹ x 3¹ =3¹⁺¹⁺¹ = 3³

⇒ 3 x 3^{2(1 + x)} = 3^3(-x)

But 3 = 3¹, thus every number is raised to the power 1
⇒ 3¹ x 3^{2(1 + x)} = 3^3(-x)

Applying A^b x A^c = A^{b + c}

3 ¹ x 3^{2(1+ x)} = 3^{1 + 2(1 + x )}

3^{1+2(1+ x)} = 3^3(-x)

Concept
If A^b = A^c ⇒ b = c

Thus if the base at the left side of the equation, is the same as the right side of the equation, then the powers are the same

∴ 1 + 2(1 + x ) = 3(-x)

Concept

a(b + c ) = a x b + a x c

a(b - c ) = a x b - a x c

Thus the number outside the bracket multiplies each of the number in the bracket and the sign ( + or - ) is(are) maintained

-x = -1 x x

3(-x) = 3 x (-1) x x = -3x

1 + 2(1 + x ) = 1 + 2 x 1 + 2 x x

                     = 1 + 2 + 2x

                     = 3 + 2x

1 + 2(1+x) = 3(-x)

⇒ 3 + 2x = -3x

Grouping the x at the left and the rest at the right

Concept
If the sign is positive(+) and crosses to the other side of the equation the sign changes to negative ( - ) and the vice versa

2x + 3x = -3

5x = -3

Concept
Divide both sides by the number multiplying the number or variable (x)

^5x⁄₅ = ^-3⁄₅

x = -3⁄5

Answer is D

Use a venn diagram to illustrate

In a class of 39 student ⇒ n(u)=39
let F represent students who offer Fante ⇒ n(F)= 25
Let T represent students who offer Twi ⇒ n(T)= 19
Five student do not offer any of the two languages ⇒ 5 are outside the sets F and T
Let n represent number of student who offer both languages

NB: You need to substract the number of the students who offer both languaes from each of the languages to get the students who offer only that languages.

When we add all the parts in the venn diagram we are to get the total number of students n(u) = 39

(25 - n )+ n + (19 - n) + 5 =39
25 - n + n + 19 -n + 5 = 39
-n + 49 = 39
-n = 39 - 49
-n = -10
^-n⁄_-1 = ^-10⁄ _-1
n = 10
Only Twi = 19-n
⇒ Only Twi = 19 - 10 = 9

Answer is B

Concept

√a x √a = √axa = √a²
√a = a^1⁄₂
⇒ √a² = a^2x1⁄₂ = a¹ = a

Thus √a x √a = a
√2 x √2 = 2

When solving a surd(√) questions, always remeber to get rid of the surd at the denominator by
1. Multiplying both the top and the down by the surd in the case of √a,√b,√c etc
2. In the case of a+√b, a-√b, multiply the top and down by the same expression except you need to change the sign(- to + and + to -)
Thus for a+√b multiply by a-√b and a-√b multiply by a+√b

NB: (a+√b)x(a-√b) = a²-b

Applying concepts

Answer is D

Ratio of profit
Mary : Charity
4 : 5

NB: Respectively means the ratio are in the order of the names mentioned

Mary received GH¢ 5,000.00 less than charity.

To solve, we can represent amount received by charity as x

Mary received GH¢ 5,000.00 less than charity.
⇒ Mary amount = x - 5000

Ratio is the simplification of amounts expressed as one amount(number) over the other
⇒ ^{Amount of Mary}⁄_{Amount of Charity} = ratio.

^(x-5000)⁄_x = ⁴⁄₅

Now solve for x

Get ride of the fraction(5) by multiplying both sizes by 5

5x(x-5000) = x x 4

NB: a x (b - c) = a x b - a xc

5x-5x5000 = 4x
5x-25000 = 4x
5x-4x = 25000
x = 25000
⇒ Charity received GH¢25,000.00

Mary received GH¢ 5,000.00 less than charity
⇒ Mary amount = GH¢25,000.00 - GH¢ 5,000.00 = GH¢20,000.00

Total profit = Amount received by Mary + Amount received by charity
⇒ Total profit = GH¢20,000.00 + GH¢25,000.00 = GH¢45,000.00

Answer is B

Cost of car = ₦5,000,000.00

Deposite = ₦3,000.000.00

Rate = 8% compound intesrest per annum = ⁸⁄₁₀₀ = 0.08

Balance = ₦5,000,000.00 - ₦3,000,000.00 = ₦2,000,000.00

Hence the interest will only be charged on the ₦2,000,000.00 balance after the first year

Fisrt year

Interest = Principal x Rate x Time

Time = 1 (Compound interest is calculated year by year)

Interest = 0.08 x ₦2,000,000 = ₦160,000

Total amount to be paid by the end of the first year
⇒ ₦2,000,0000 + ₦160,000 = ₦2,160,000

Since he pays ₦1000,000 at each year

Amount left to pay the subsequent year
⇒ ₦2,160,000 - ₦1,000,000 = ₦1,160,000

Second Year

Interest = 0.08 x ₦1,160,000 = ₦92,800

Amount to pay
⇒ ₦92,800 + ₦1,160,000 = ₦1,250,8000

He pays another ₦1,000,000 by the end of the second year.

Hence amount to pay after the second year
⇒ ₦1,252,800 - ₦1,000,000.00 = ₦252,800

NB: For compound interest, the new principal amount is the interest + the previous amount to be paid

y ∝ ¹⁄_x²

NB: ∝ is the symbol for proportional
inversely proportional is 1 over the expression
directly proportional is just the expression (It would had been y ∝ x²)

a constant(=k) is introduced in place of the ∝
y = kx¹⁄_x²
y = ^k⁄_x²

Use the known values to calculate the constant(k) to get the general expression

y = 100 and x = 3
⇒ 100 = ^k⁄_3²
100 = ^k⁄₉
100x9 = k
⇒ k = 900
∴ y = ⁹⁰⁰⁄_x²

when y = 25
⇒ 25 = ⁹⁰⁰⁄_x²
25 x x² = 900
x² = ⁹⁰⁰⁄₂₅
x² = 36
√x² = √36
x = 6

NB: The square root of a squared number or variable is equal to the number or variable

Change the mixed fraction to an improper fraction.

Explanation:
Multiply the whole number(A) by the denominator(c) and add the numerator(b). Divide everything by the denominator(c).

⇒ 1¹⁄₁₅ = ^(1x15+1)⁄₁₅ = ¹⁶⁄₁₅

⇒ t - ⁹⁄₅ = -¹⁶⁄₁₅

Get ride of the denominators by multiplying both sides by the least common multiple(LCM)
The denominators are 5 and 15,5 can go into 15, hence the LCM is 15

tx15 -15x⁹⁄₅ = 15x-¹⁶⁄₁₅
15t - 27 = -16
15t = -16+27
15t = 11
^15t⁄₁₅ = ¹¹⁄₁₅
t = ¹¹⁄₁₅

Concept

(a+b)² = a²+2xaxb + b² = a²+2ab + b²
(a-b)² = a²-2xaxb + b² = a²-2ab + b²

Alternatively
(a+b)² = (a+b)x(a+b) [Expand and simplify]
(a-b)² = (a-b)x(a-b) [Expand and simplify]

⇒ (x-2)² = x²-2xxx2+2² = x²-4x+4
(x+1)² = x²+2xxx1+1² = x²+2x+1

(x-2)²+3 = (x+1)² - 6
x²-4x+4+3 = x²+2x+1 - 6
x²-4x+7 = x²+2x-5
x²-x²-4x-2x = -5 - 7
-6x = -12
^-6x⁄_-6 = ^-12⁄_-6
x = 2

The amount of money(f) which Dede has was not enough to buy the car
⇒ f(Dede's amount) < (less than) the cost of car (₦3,000,000.00)
but Kofi had enough money(k) to buy the car
⇒ Kofi can have the exact amount of the car or even more than the cost of the car
⇒k≥₦3,000,000.00
⇒ f< ₦3,000,000.00 ≤ k
Read as Dede's amount(f) is less than ₦3,000,000.00 and ₦3,000,000.00 is less than or equal to Kofi's amount(k)

Notes

Arithmetic Progression

Is a sequence of numbers such that the difference between the consecutive terms is constant.

E.g The sequence 1,3,5,7,9.. is an arithmetic progression with a common difference of 2

a_n = a₁ + (n - 1)d

a_n = The value of the sequence number of on the n^th position.

a₁ = the first number in the sequence

d = the common difference

d = a_{n + 1} - a_n (The difference between two consecutive sequence numbers)

d is the same for all consecutive numbers used for the calculator.

Thus a_{n + 2} - a_{n + 1} = a_{n + 1} - a_n

For instance, the common difference for the sequence a₁, a₂, a₃, a₄, a₅, a₆ .... will be calculated as:
a₂ - a₁ = a₃ - a₂ = a₅ - a₄ = a₆ - a₅

NB: The common difference (d) is the higher sequence number value - lower previos sequence number.

For the sequence 1,3,5,7,9

⇒ d = 3 - 1 = 5 - 3 = 7 - 5 = 9 - 7 = 2

Summation of Arithmetic Progression

S_n = ⁿ/₂ (a₁ + a_n)

S_n = ⁿ/₂ [2a₁ + (n - 1)d]

If the last number is know the summation can also be writen as

S_n = ⁿ/₂ [a₁ + l]

where l = last term

From the above diagram, the length of the arc becomes the circumference of the base of the cone.

radius of the circle = 22cm
θ = 60^o

NB: Unit of area = unit square (e.g cm²,m²) and unit of volume = unit cubic (e.g cm³, m³)