Regina's age now = 34.

In n years time, Regina will be (34 + n) years.

Daughter's age now = 5.

In n years time, the daughter will be (5 + n) years.

In n years, Regina will be twice as old as her daughter

Twice means 2 times (2 x)

Regina's age in n years time = 2 x daughter's age in n years time.

34 + n = 2 x (5 + n)

34 + n = 2 x 5 + 2 x n

34 + n = 10 + 2n

34 - 10 = 2n - n

24 = n

∴ n = 24

Mean = 16

Number of numbers = 10

$\frac{\mathrm{Sum\; of\; numbers}}{10}$ = 16

Get rid of the fraction by multiplying both sides by the denominator 10.

Sum of numbers = 16 x 10 = 160

When k is added, the new sum = 160 + k and the number of numbers become 10 + 1 = 11

Mean for the eleven (11) numbers is 18

$\frac{\mathrm{160\; +\; k}}{11}$ = 18

Get rid of the fraction by multiplying both sides by the denominator 11

160 + k = 18 x 11

160 + k = 198

k = 198 - 160

k = 38

Let y = Juliet's age.

Badu is four times as old as Juliet

Badu's age = 4 x Juliet's age

Badu's age = 4 x y

Badu's age = 4y

Badu's age 10 years time = Badu's age now + 10 years.

Badu's age 10 years time = 4y + 10

Juliet's age in 10 years time = Juliet's age now + 10

Juliet's age in 10 years time = y + 10

In 10 years Badu will be twice as old as Juliet

Badu's age in 10 years will be 2 x Juliet's age

4y + 10 = 2 x (y + 10)

4y + 10 = 2 x y + 2 x 10

4y + 10 = 2y + 20

4y - 2y = 20 - 10

2y = 10

Divide both sides by 2

y = $\frac{10}{2}$ = 5

∴ Juliet is 5 years old now

Let x = Fafa's present age.

Metuh is nine times as old as Fafa → Metuh's present age = 9 x Fafa's present age.

Metuh's present age = 9 x x

In six year's time, Metuh's age will be 9x + 6

In six year's time, Fafa's age will be x + 6

In six years time, he will be five times as old as Fafa

9x + 6 = 5 (x + 6)
9x + 6 = 5x + 30
9x - 5x = 30 - 6
4x = 24

Divide both sides by 4

x = $\frac{24}{4}$

x = 6

Fafa's present age is 6 years

Change the mixed fractions to improper fractions.

2$\frac{1}{6}$ = $\frac{\mathrm{2\; x\; 6\; +\; 1}}{6}$ = $\frac{\mathrm{12\; +\; 1}}{6}$ = $\frac{13}{6}$

2$\frac{7}{12}$ = $\frac{\mathrm{2\; x\; 12\; +\; 7}}{12}$ = $\frac{\mathrm{24\; +\; 7}}{12}$ = $\frac{31}{12}$

3$\frac{1}{4}$ = $\frac{\mathrm{4\; x\; 3\; +\; 1}}{4}$ = $\frac{\mathrm{12\; +\; 1}}{4}$ = $\frac{13}{4}$

Let x the number to be subtracted.

Sum is addition.

$\frac{13}{6}$ + $\frac{31}{12}$ - x = $\frac{13}{4}$

$\frac{\mathrm{2\; x\; 13\; +\; 1\; x\; 31}}{12}$ - x = $\frac{13}{4}$

$\frac{\mathrm{26\; +\; 31}}{12}$ - x = $\frac{13}{4}$

$\frac{57}{12}$ - x = $\frac{13}{4}$

$\frac{57}{12}$ - $\frac{13}{4}$ = x

$\frac{\mathrm{1\; x\; 57\; -\; 3\; x\; 13}}{12}$ = x

$\frac{\mathrm{57\; -\; 39}}{12}$ = x

$\frac{18}{12}$ = x

$\frac{3}{2}$ = x

x = $\frac{3}{2}$ = 1$\frac{1}{2}$

Let n = number of adults sold to and m = number of children sold to.

Five hundred tickets were sold for a concert implies that when you add the number of adults and children, you should get 500

n + m = 500 ------- eqn 1

Total tickets sold = number of adults x cost for adult + number of children x cost for children.

Total tickets sold = n x 4.5 + m x 3

Total tickets sold = 4.5n + 3m

But total tickets sold = $1987.50

4.5n + 3m = 1987.50 ------ eqn 2

Get rid of m.

eqn 1 x 3

3 x n + 3 x m = 3 x 500

3n + 3m = 1500 ------ eqn 3

eqn 2 - eqn 3

1.5n = 487.5

Divide both sides by 1.5

n = $\frac{487.5}{1.5}$ = 325

∴ 325 tickets were sold to adults.

More than 18 km means the distance d is greater than 18 km and hence 18 km is less than the distance (18 < d)

Not more than 23 km means it can be 23 or less hence the less than or equal to sign is used (d ≤ 23)

When you combine the two, the inequality will be 18 < d ≤ 23 which is read as 18 is less than d and d is also less than or equal to 23.

The man's age now = Age in 8 years time - 8

The man's age now = (x + 10) - 8

The man's age now = x + 10 - 8

The man's age now = x + 2

2 years ago he was 63 years.

The man's age now = 63 + 2 = 65 years

But the man's age now = x + 2

x + 2 = 65

x = 65 - 2

x = 63

The man is five times as old as his son means when you multiply the son's age by 5, you should get the man's age.

The son is y years now.

Man's age now = 5 x y = 5y

In four years time,

Son's age = y + 4

Man's age = 5y + 4

The product of their ages would be 340

Product means multiplication.

Man's age x Son's age in four years time = 340

(5y + 4) x (y + 4) = 340

5y(y + 4) + 4(y + 4) = 340

5y² + 20y + 4y + 16 = 340

5y² + 24y + 16 - 340 = 0

5y² + 24y + 16 - 340 = 0

5y² + 24y - 324 = 0

The amount of money(f) which Dede has was not enough to buy the car
Which implies that Dede's amount is less than the cost of 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

This implies Kofi has either the exact amount or more than the cost of the car

⇒k ≥ ₦3,000,000.00 and the cost of the car is less than or equal (≤) to Kofi's amount
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)

Wednesday → Went, day 1

Thursday → No walk, day 2

Friday → No walk, day 3

Saturday → No walk, day 4

Sunday → No walk, day 5

Monday → Will go for a walk, day 6

Let the number = n

Two times = 2 x n = 2n

one-third = $\frac{1}{3}$

one-third of the number = $\frac{1}{3}$ x n = $\frac{\mathrm{n}}{3}$

Two times a number added to one-third of the number = 2n + $\frac{\mathrm{n}}{3}$

Gives means =

2n + $\frac{\mathrm{n}}{3}$ = 5$\frac{1}{6}$

Change the mixed fraction to improper fraction.

5$\frac{1}{6}$ = $\frac{\mathrm{6\; x\; 5\; +\; 1}}{6}$ = $\frac{\mathrm{30\; +\; 1}}{6}$ = $\frac{31}{6}$

2n + $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3 and 6. The L.C.M is 6

6 x 2n + 6 x $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$ x 6

12n + 2 x n = 31

12n + 2n = 31

14n = 31

Divide both sides by 14

n = $\frac{31}{14}$ = 2$\frac{3}{14}$

∴ the number is 2$\frac{3}{14}$

Mean = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$

The mean is not greater than 5 means the mean can be less than 5 or equal to 5.

$\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$ ≤ 5

$\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5

Get rid of the fraction by multiplying both sides by the denominator, 4

4 x $\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5 x 4

14 + 2x ≤ 20

2x ≤ 20 - 14

2x ≤ 6

Divide both sides by 2.

x ≤ $\frac{6}{2}$

x ≤ 3

When the daughter was x years old, the woman is y + x years old. Thus if you add the daughters age to the mother's age when she gave birth to the daughtr, you will get the mother's current age.

When the daughter was x years old, the mother was n times as old as her daughter.

Times means multiplication (x)

y + x = n x x

y + x = nx

y = nx - x

Factorize x out.

y = x(n - 1)

Since the difference (subtraction) is 6 years, when you add 6 years to the younger age, you should get the eldest's age.

Let y = Esi's age

Amina's age = Esi's age + 6

Amina's age = y + 6

Amina's age + Esi's age = 24

(y + 6) + y = 24

y + 6 + y = 24

y + y + 6 = 24

2y = 24 - 6

2y = 18

Divide both sides by 2

y = $\frac{18}{2}$

y = 9

Esi's age is 9 years

Let g = Gifty's current age

Gifty is 19 years younger than her mother means if you add 19 years to Gifty's age, you will get that of the mother's age.

Mother's age = g + 19

In three years time,

Gifty's age = g + 3

Mother's age = (g + 19) + 3 = g + 22

The sum of their ages in three years will be:

(g + 3) + (g + 22) = 67

Simplifying:

g + 3 + g + 22 = 67

2g + 25 = 67

2g= 42

Divide both sides by 2

$\frac{\mathrm{2g}}{2}$ = $\frac{42}{2}$

g= 21.

Gifty is currently 21 years old.

Mother's current age = g + 19 = 21 + 19 = 40

The sum of their ages now is: 21 + 40 = 61

Let y = Juliet's age.

Badu is four times as old as Juliet

Badu's age = 4 x Juliet's age

Badu's age = 4 x y

Badu's age = 4y

Badu's age 10 years time = Badu's age now + 10 years.

Badu's age 10 years time = 4y + 10

Juliet's age in 10 years time = Juliet's age now + 10

Juliet's age in 10 years time = y + 10

In 10 years Badu will be twice as old as Juliet

Badu's age in 10 years will be 2 x Juliet's age

4y + 10 = 2 x (y + 10)

4y + 10 = 2 x y + 2 x 10

4y + 10 = 2y + 20

4y - 2y = 20 - 10

2y = 10

Divide both sides by 2

y = $\frac{10}{2}$ = 5

∴ Juliet is 5 years old now

Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Number of items = 3 + 9
Number of items = 12

Average cost = $\frac{\mathrm{Sum\; of\; cost}}{12}$

Sum of cost = 3 x 0.4 + 9 x 0.2
Sum of cost = 1.2 + 1.8
Sum of cost = GH₵3

Average cost = $\frac{\mathrm{GH₵3}}{12}$

Average cost = GH₵ 0.25

Let x = Fafa's present age.

Metuh is nine times as old as Fafa → Metuh's present age = 9 x Fafa's present age.

Metuh's present age = 9 x x

In six year's time, Metuh's age will be 9x + 6

In six year's time, Fafa's age will be x + 6

In six years time, he will be five times as old as Fafa

9x + 6 = 5 (x + 6)
9x + 6 = 5x + 30
9x - 5x = 30 - 6
4x = 24

Divide both sides by 4

x = $\frac{24}{4}$

x = 6

Fafa's present age is 6 years

Mensah's age now = 5

Thrice means 3 times (3 x ).

Joyce's age now = 3 x Mensah's age

Joyce's age now = 3 x 5

Joyce's age now = 15

Let y years in the future that Joyce will be twice as old as Mensah.

In y years, Mensah will be (5 + y) years.

In y years, Joyce will be (15 + y) years.

In y years, Joyce will be twice as old as Mensah.

Twice means 2 times (2 x )

(15 + y) = 2 x (5 + y)

Now expand and solve for y

15 + y = 2 x 5 + 2 x y

15 + y = 10 + 2y

15 - 10 = 2y - y

5 = y

∴ Joyce will be twice as old as Mensah in 5 years time.

Let n = number of adults sold to and m = number of children sold to.

Five hundred tickets were sold for a concert implies that when you add the number of adults and children, you should get 500

n + m = 500 ------- eqn 1

Total tickets sold = number of adults x cost for adult + number of children x cost for children.

Total tickets sold = n x 4.5 + m x 3

Total tickets sold = 4.5n + 3m

But total tickets sold = $1987.50

4.5n + 3m = 1987.50 ------ eqn 2

Get rid of m.

eqn 1 x 3

3 x n + 3 x m = 3 x 500

3n + 3m = 1500 ------ eqn 3

eqn 2 - eqn 3

1.5n = 487.5

Divide both sides by 1.5

n = $\frac{487.5}{1.5}$ = 325

∴ 325 tickets were sold to adults.

The man is five times as old as his son means when you multiply the son's age by 5, you should get the man's age.

The son is y years now.

Man's age now = 5 x y = 5y

In four years time,

Son's age = y + 4

Man's age = 5y + 4

The product of their ages would be 340

Product means multiplication.

Man's age x Son's age in four years time = 340

(5y + 4) x (y + 4) = 340

5y(y + 4) + 4(y + 4) = 340

5y² + 20y + 4y + 16 = 340

5y² + 24y + 16 - 340 = 0

5y² + 24y + 16 - 340 = 0

5y² + 24y - 324 = 0

Selling Price = Cost Price + Profit

Selling Price of each = ₦ (x + y)

For the 10 tins, selling price = 10 x ₦ (x + y)

Amount received for the 10 tins = ₦ 10(x + y)

Wednesday → Went, day 1

Thursday → No walk, day 2

Friday → No walk, day 3

Saturday → No walk, day 4

Sunday → No walk, day 5

Monday → Will go for a walk, day 6

Let the number = n

Two times = 2 x n = 2n

one-third = $\frac{1}{3}$

one-third of the number = $\frac{1}{3}$ x n = $\frac{\mathrm{n}}{3}$

Two times a number added to one-third of the number = 2n + $\frac{\mathrm{n}}{3}$

Gives means =

2n + $\frac{\mathrm{n}}{3}$ = 5$\frac{1}{6}$

Change the mixed fraction to improper fraction.

5$\frac{1}{6}$ = $\frac{\mathrm{6\; x\; 5\; +\; 1}}{6}$ = $\frac{\mathrm{30\; +\; 1}}{6}$ = $\frac{31}{6}$

2n + $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3 and 6. The L.C.M is 6

6 x 2n + 6 x $\frac{\mathrm{n}}{3}$ = $\frac{31}{6}$ x 6

12n + 2 x n = 31

12n + 2n = 31

14n = 31

Divide both sides by 14

n = $\frac{31}{14}$ = 2$\frac{3}{14}$

∴ the number is 2$\frac{3}{14}$

When the daughter was x years old, the woman is y + x years old. Thus if you add the daughters age to the mother's age when she gave birth to the daughtr, you will get the mother's current age.

When the daughter was x years old, the mother was n times as old as her daughter.

Times means multiplication (x)

y + x = n x x

y + x = nx

y = nx - x

Factorize x out.

y = x(n - 1)

1. Find the number of times the cyclist complete one cycle by dividing the distance overed by the circumference.

Number of completed cycles = $\frac{\mathrm{302\; km}}{\mathrm{9\; km}}$ = 33.555

Number of completed cycles = 33

Note: completed cycle is when the cyclist reaches the starting point.

2. Distance from starting point = Distance covered - Number of completed cycles x the circumference

Distance from starting point = 302 km - 33 x 9 km

Distance from starting point = 302 km - 297 km

Distance from starting point = 5 km