0.0098 = 0.⌒0⌒0⌒9.8 = 9.8 x 10^-3

Note: the power is negative 3 because we had to move the decimal point to the right 3 times.

Express both the left and the right hand sides to be in the same base. Once the bases are the same, you can equate the powers and solve for m.

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

2^{m - 2} = 32

2^{m - 2} = 2⁵

Since the bases are the same, the powers are also the same.

m - 2 = 5

m = 5 + 2

m = 7

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = -2
Y₁ = 6
X₂ = 3
X₁ = -2

Gradient = $\frac{\mathrm{-2\; -\; 6}}{\mathrm{3\; -\; -2}}$ = $\frac{-8}{\mathrm{3\; +\; 2}}$ = -$\frac{8}{5}$

Note: - - = +

Ordered Numbers

2, 2, 3, 4, 4, 6, 8, 8, 9, 9

The middle numbers are 4 and 6. Hence sum them and divide by 2.

Median = $\frac{\mathrm{4\; +\; 6}}{2}$ = $\frac{10}{2}$ = 5

Mean ag of the 5 boys = 4

$\frac{\mathrm{Sum\; of\; ages\; of\; the\; 5\; boys}}{5}$ = 4

Sum of the ages of the 5 boys = 4 x 5 = 20

Th mean age of 4 of them is 4

$\frac{\mathrm{Sum\; of\; ages\; of\; the\; 4\; boys}}{4}$ = 4

Sum of ages of the 4 boys = 4 x 4 = 16

The age of the remaining boy = Sum of the ages of the 5 boys - Sum of the ages of the 4 boys

The age of the remaining boy = 20 - 16

The age of the remaining boy = 4

Volume of water = Area of the base of the tank x depth of water

The base area = 2 m x 2 m = 4 m²

Since the depth is given in cm, you must convert the cm to m by dividing the cm by 100

60 cm = $\frac{60}{100}$ m

Volume of water = 4 m² x $\frac{60}{100}$ m = 2.4 m³

Notes:

1. The zeros in 60 and 100 cancels each other
2. 2 divides 4, 2 times and 10, 5 times
3. 2 x 6 = 12
4. 12 divided by 5 = 2.4

a = $\left(\begin{array}{c}-2\\ -3\end{array}\right)$ and b = $\left(\begin{array}{c}-3\\ 2\end{array}\right)$

3a - b = 3$\left(\begin{array}{c}-2\\ -3\end{array}\right)$ - $\left(\begin{array}{c}-3\\ 2\end{array}\right)$

3a - b = $\left(\begin{array}{c}\mathrm{-2\; x\; 3}\\ \mathrm{-3\; x\; 3}\end{array}\right)$ - $\left(\begin{array}{c}-3\\ 2\end{array}\right)$

3a - b = $\left(\begin{array}{c}-6\\ -9\end{array}\right)$ - $\left(\begin{array}{c}-3\\ 2\end{array}\right)$

3a - b = $\left(\begin{array}{c}\mathrm{-6\; -\; -3}\\ \mathrm{-9\; -\; 2}\end{array}\right)$

Note: - - = +. - - 3 = + 3

3a - b = $\left(\begin{array}{c}\mathrm{-6\; +\; 3}\\ -11\end{array}\right)$

3a - b = $\left(\begin{array}{c}-3\\ -11\end{array}\right)$

PQ→ = OQ→ - OP→

PQ→ = $\left(\begin{array}{c}-1\\ 2\end{array}\right)$ - $\left(\begin{array}{c}-2\\ 1\end{array}\right)$

PQ→ = $\left(\begin{array}{c}\mathrm{-1\; -\; -2}\\ \mathrm{2\; -\; 1}\end{array}\right)$

Note: - - = +. - -2 = + 2

PQ→ = $\left(\begin{array}{c}\mathrm{-1\; +\; 2}\\ 1\end{array}\right)$

PQ→ = $\left(\begin{array}{c}1\\ 1\end{array}\right)$

Translation by a vector

Translation = Point + Vector

P₁ = P + V

If P(x,y) translated by the vector $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$

P₁ = $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{x\; +\; a}\\ \mathrm{y\; +\; b}\end{array}\right)$

Let $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ be the vector which translates the point (1, 3) to the point (3, 2)

$\left(\begin{array}{c}3\\ 2\end{array}\right)$ = $\left(\begin{array}{c}1\\ 3\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{1\; +\; a}\\ \mathrm{3\; +\; b}\end{array}\right)$

The top at the left equals the top at the right.

3 = 1 + a

3 - 1 = a

2 = a

a = 2

The bottom at the left equals the bottom at the right.

2 = 3 + b

2 - 3 = b

-1 = b

b = -1

The translation vector is therefore $\left(\begin{array}{c}2\\ -1\end{array}\right)$

The right angle symbol at ∠RUT implies ∠RUT = 90°

The mark on lines PQ and QU indicates that both sides are equal.

A triangle with two sides equal (Isosceles triangle) has the base angles the same.

∠QPU = ∠PUQ (Base angles of isosceles triangles are the same)

∠QPU + ∠PUQ + 130° = 180° (Sum of interior angles of a triangle is 180°)

∠QPU + ∠PUQ = 180° - 130°

∠QPU + ∠PUQ = 50°

But ∠QPU = ∠PUQ = $\frac{\mathrm{50°}}{2}$ = 25°

∠PUQ + ∠RUQ + ∠RUT = 180° (Sum of angles on a straight line is 180°)

25° + ∠RUQ + 90° = 180°

∠RUQ + 25° + 90° = 180°

∠RUQ + 115° = 180°

∠RUQ = 180° - 115°

∠RUQ = 65°

∠PUR = ∠PUQ + ∠RUQ

∠PUR = 25° + 65° = 90°

Alternatively,

∠PUR + ∠RUT = 180°

But ∠RUT = 90°

∠PUR + 90° = 180°

∠PUR = 180° - 90°

∠PUR = 90°

x = ∠QPU + ∠PUR (The sum of the interior opposite angle is equal to the exterior angle)

x = 25° + 90° = 115°

Alternatively,

∠QRU + x = 180° (Sum of angles on a straight line is 180°)

x = 180° - ∠QRU

∠QPU + ∠PUR + ∠QRU = 180°

25° + 90° + ∠QRU = 180°

115° + ∠QRU = 180°

∠QRU = 180° - 115°

∠QRU = 65°

x = 180° - ∠QRU

x = 180° - 65°

x = 115°

Note: If the mapping has equal interval (difference) for the y and x interval of 1, you can use the general linear sequence formula to find the rule of the mapping.

U_n = U₁ + (n-1)d

In the mapping, U_n = y and n = x and U₁, the first value of y

U₁ = 4

y = 4 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 2-4 = 0-2 = -2 - 0 = -4 -(-2) = -2

We can substitute our d and simplify the expression for y

y = 4 + (x-1)d

y = 4 + (x-1)(-2)

y = 4 + -2 x x -1 x -2

y = 4 + -2x + 2

y = 4 + 2 -2x

y = 6 + -2x

y = -2x + 6

a³ = a x a x a

When a = 2, a³ = 2³ = 2 x 2 x 2 = 8

b³ = b x b x b

When b = -1, b³ = (-1)³ = -1 x -1 x -1 = -1

Note: - 1 x - 1 = + 1 and 1 x -1 = -1

b² = b x b

When b = -1, b² = (-1)² = -1 x -1 = 1

a³ + 2ab² + b³ = 8 + 2 x 2 x 1 + -1

a³ + 2ab² + b³ = 8 + 4 - 1

a³ + 2ab² + b³ = 11

Let C = {Students who read Chemistry}

P = {Students who read Chemistry}

n = number of students who read both Chemistry and Physics

n(U) = 40

n(C) = 30

n(P) = 20

Covering Chemistry (C),

30 + 20 - n = 40

50 - n = 40

50 - 40 = n

10 = n

∴ 10 students read both Chemistry and Physics.

P(C and P) = $\frac{\mathrm{Number\; of\; students\; who\; read\; both\; Chemistry\; and\; Physics}}{\mathrm{Number\; of\; students}}$

P(C and P) = $\frac{10}{40}$ = $\frac{1}{4}$

V = $\frac{1}{3}$πr²h

Whenever you have a fraction, get rid of the fraction by multiplying both sides by the denominator or the L.C.M of the denominators.

Multiply both sides by the denominator, 3

3 x V = 3 x $\frac{1}{3}$πr²h

3V = πr²h

Since we are making h the subject, we divide both sides by the expressions multiplying it to cancel them out for the h to stand alone.

Divide both sides by πr²

h = $\frac{\mathrm{3V}}{\mathrm{\pi }{\mathrm{r}}^2}$

PR→ = PQ→ + QR→

Note: the letter which is not part of what you are finding must face each other as you can see with the Q.

QR→ = -RQ→

Note: to get the opposite vector, you must negative the vector.

QR→ = -$\left(\begin{array}{c}-1\\ 5\end{array}\right)$ = $\left(\begin{array}{c}\mathrm{-1\; x\; -1}\\ \mathrm{5\; x\; -1}\end{array}\right)$ = $\left(\begin{array}{c}1\\ -5\end{array}\right)$

Notes:
1. - x - = +. -1 x -1 = + 1
2. - x + = - or + x - = -. 5 x -1 = -5

PQ→ = $\left(\begin{array}{c}2\\ 4\end{array}\right)$

PR→ = $\left(\begin{array}{c}2\\ 4\end{array}\right)$ + $\left(\begin{array}{c}1\\ -5\end{array}\right)$

PR→ = $\left(\begin{array}{c}\mathrm{2\; +\; 1}\\ \mathrm{4\; +\; -5}\end{array}\right)$

PR→ = $\left(\begin{array}{c}3\\ \mathrm{4\; -\; 5}\end{array}\right)$

PR→ = $\left(\begin{array}{c}3\\ -1\end{array}\right)$

Notes:
1. The sum of the ratios represents the total amount shared
2. The highest ratio represents the highest amount

Sum of ratios = 1 + 2 + 3 = 6

The ratio for the highest amount is 3.

If 6 = GH₵ 300

3 = $\frac{\mathrm{3\; x\; GH₵\; 300}}{6}$ = GH₵ 150

Notes:
1. the left below multiplies the right and divided by the left at the top
2. 3 divides itself 1 time and 6, 2 times
3. 2 divides itself 1 time and 300, 150

Notes:
1. Bearing starts from the north clockwise
2. Apply the concept of alternate angles (the Z-shaped angles are the same)
3. Angle on a straight line is 180° (North to South)

Bearing of P from Q = 180° + 40° = 220°

Intersection (∩) are elements that can be found in both sets

Union (∪) is a set of the list of elements in both sets. If an element is found in both set, it is written only once

P = {3, 4, 5, 6}, Q = {5, 6, 7, 8} and R = {2, 3, 7, 9}

Q ∪ R = {2, 3, 5, 6, 7, 8, 9}

P ∩ (Q ∪ R) = P ∩ {2, 3, 5, 6, 7, 8, 9}

P ∩ (Q ∪ R) = {3, 5, 6}

The perimeter = 10 m + 7 m + 10 m + the semi-circle

The perimeter = 27 m + the semi-circle

The semi-circle is half of the circumference of a circle.

π = $\frac{22}{7}$

Diameter = 7 m

Circumference = $\frac{22}{7}$ x 7 m = 22 m

Note: the 7s cancel each other.

The semi-circle is half the circumference.

Semi-circle = $\frac{\mathrm{22\; m}}{2}$ = 11 m

The perimeter = 27 m + the semi-circle

The perimeter = 27 m + 11 m = 38 m

To simplify surds, express $\sqrt{a}$ as $\sqrt{\mathrm{Perfect\; Square\; x\; b}}$ which equals $\sqrt{\mathrm{Perfect\; Square}}$ x $\sqrt{b}$

$\sqrt{\mathrm{500}}$ = $\sqrt{\mathrm{100\; x\; 5}}$ = $\sqrt{\mathrm{100}}$ x $\sqrt{5}$ = 10$\sqrt{5}$

Note: $\sqrt{\mathrm{100}}$ = 10 because 10 x 10 = 100

$\sqrt{\mathrm{125}}$ = $\sqrt{\mathrm{25\; x\; 5}}$ = $\sqrt{\mathrm{25}}$ x $\sqrt{5}$ = 5$\sqrt{5}$

Note: $\sqrt{\mathrm{25}}$ = 5 because 5 x 5 = 25

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 3 x 10$\sqrt{5}$ - 2 x 5$\sqrt{5}$

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 30$\sqrt{5}$ - 10$\sqrt{5}$

3$\sqrt{\mathrm{500}}$ - 2$\sqrt{\mathrm{125}}$ = 20$\sqrt{5}$

Length of vector $\left(\begin{array}{c}-5\\ 10\end{array}\right)$ = $\sqrt{{\mathrm{(-5)}}^2+{10}^2}$

Length of vector $\left(\begin{array}{c}-5\\ 10\end{array}\right)$ = $\sqrt{\mathrm{25\; +\; 100}}$

Length of vector $\left(\begin{array}{c}-5\\ 10\end{array}\right)$ = $\sqrt{\mathrm{125}}$

To simplify surds, express $\sqrt{a}$ as $\sqrt{\mathrm{Perfect\; Square\; x\; b}}$ which equals $\sqrt{\mathrm{Perfect\; Square}}$ x $\sqrt{b}$

$\sqrt{\mathrm{125}}$ = $\sqrt{\mathrm{25\; x\; 5}}$ = $\sqrt{\mathrm{25}}$ x $\sqrt{5}$ = 5$\sqrt{5}$

Note: $\sqrt{\mathrm{25}}$ = 5 because 5 x 5 = 25

The mark on lines XY and XZ indicates that both lengths are the same. Hence ∆XYZ is an isosceles.

Isosceles triangle has base angles the same.

∠XYZ = ∠XZY

∠XYZ + ∠XZY + ∠YXZ = 180°(Sum of angles in a triangle)

∠YXZ = 40°

∠XYZ + ∠XZY + 40° = 180°

∠XYZ + ∠XZY = 180° - 40°

∠XYZ + ∠XZY = 140°

But ∠XYZ = ∠XZY

∠XYZ = ∠XZY = $\frac{\mathrm{140°}}{2}$ = 70°

∠XZY + ∠XZT = 180°(Sum of angles on a straight line)

∠XZY = 70°

∠XZT = (13x - 20)°

70° + (13x - 20)° = 180°

70 + 13x - 20 = 180

13x + 70 - 20 = 180

13x + 50 = 180

13x = 180 - 50

13x = 130

Divide both sides by 13

x = $\frac{130}{13}$ = 10

$\frac{\mathrm{3x}}{2}$ - $\frac{\mathrm{5x}}{3}$ = 1

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators.

The Least Common Multiples of the denominators 2 and 3 is 6

6 x $\frac{\mathrm{3x}}{2}$ - $\frac{\mathrm{5x}}{3}$ x 6 = 1 x 6

3 x 3x - 5x x 2 = 6

9x - 10x = 6

-x = 6

Divide both sides by -1

x = $\frac{6}{-1}$ = -6

6x : 64 = 18 : 32

Change the ratios to fractions.

$\frac{\mathrm{6x}}{64}$ = $\frac{18}{32}$

Note: 2 divides 6, 3 times and 64, 32 times.

$\frac{\mathrm{3x}}{32}$ = $\frac{18}{32}$

Since the denominators at both sides are equal (32), the numerators at both sides are also equal.

3x = 18

Divide both sides by 3

x = $\frac{18}{3}$ = 6

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = 1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$)

Change the mixed fractions to improper fractions

1$\frac{2}{3}$ = $\frac{\mathrm{1\; x\; 3\; +\; 2}}{3}$ = $\frac{\mathrm{3\; +\; 2}}{3}$ = $\frac{5}{3}$

1$\frac{3}{4}$ = $\frac{\mathrm{1\; x\; 4\; +\; 3}}{4}$ = $\frac{\mathrm{4\; +\; 3}}{4}$ = $\frac{7}{4}$

2$\frac{5}{8}$ = $\frac{\mathrm{2\; x\; 8\; +\; 5}}{8}$ = $\frac{\mathrm{16\; +\; 5}}{8}$ = $\frac{21}{8}$

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = $\frac{5}{3}$ - ($\frac{7}{4}$ ÷ $\frac{21}{8}$)

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = $\frac{5}{3}$ - ($\frac{7}{4}$ x $\frac{8}{21}$)

Notes:
1. 7 divides itself 1 time and 21, 3 times
2. 4 divides itself 1 time and 8, 2 times

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = $\frac{5}{3}$ - ($\frac{1}{1}$ x $\frac{2}{3}$)

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = $\frac{5}{3}$ - $\frac{2}{3}$

1$\frac{2}{3}$ - (1$\frac{3}{4}$ ÷ 2$\frac{5}{8}$) = $\frac{\mathrm{5\; -\; 2}}{3}$ = $\frac{3}{3}$ = 1

48 < $\frac{1}{2}$(1 + x)

Get rid of the fraction by multiplying both sides by the denominator, 2

2 x 48 < 2 x $\frac{1}{2}$(1 + x)

96 < 1 x (1 + x)

96 < 1 + x

96 - 1 < x

95 < x

It is read as 95 is less than x. Since 95 is less than x then x is greater than 95.

When the x is brought to the left, the inequality sign is reversed.

x > 95

P(Neither green nor red) = $\frac{\mathrm{Number\; of\; pens\; that\; are\; neither\; green\; nor\; red}}{\mathrm{Number\; of\; pens}}$

Number of pens that are neither green nor red = Total number of pens - (Number of green pens + Number of red pens)

Total number of pens = 40

Number of green pens = 10

Number of red pens = 18

Number of pens that are neither green nor red = 40 - (10 + 18)

Number of pens that are neither green nor red = 40 - 28

Number of pens that are neither green nor red = 12

P(Neither green nor red) = $\frac{12}{40}$ = $\frac{3}{10}$

Note: 4 divides 12, 3 and 4 divides 40, 10

From Pythagorean theorem, a² + 4² = 5²

4² = 4 x 4 = 16

5² = 5 x 5 = 25

a² + 16 = 25

a² = 25 - 16

a² = 9

9 = 3 x 3 = 3²

a² = 3²

Since the powers are the same, the bases are also the same.

a = 3

The length of the base = a + a

The length of the base = 3 cm + 3 cm = 6 cm

Express 64 as a base to the power 6 so that the power cancels the denominator 6 of the power fraction of 64.

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

${\mathrm{(64)}}^{\frac{5}{6}}$ = $2^{\mathrm{6\; x}\frac{5}{6}}$

The 6s cancel out.

${\mathrm{(64)}}^{\frac{5}{6}}$ = 2⁵

${\mathrm{(64)}}^{\frac{5}{6}}$ = 2 x 2 x 2 x 2 x 2 = 32

2(kx + 6) = 6 + 8x

Substitute x = 3

2(k x 3 + 6) = 6 + 8 x 3

2(3k + 6) = 6 + 24

2 x 3k + 2 x 6 = 6 + 24

6k + 12 = 30

6k = 30 - 12

6k = 18

Divide both sides by 6

k = $\frac{18}{6}$ = 3

$\frac{\mathrm{4x}}{7}$ - 2 = $\frac{\mathrm{x\; -\; 1}}{3}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 7 and 3.

The L.C.M of 7 and 3 is 21.

21 x $\frac{\mathrm{4x}}{7}$ - 2 x 21 = 21 x $\frac{\mathrm{x\; -\; 1}}{3}$

3 x 4x - 42 = 7 x (x - 1)

12x - 42 = 7 x x - 1 x 7

12x - 42 = 7x - 7

12x - 7x = 42 - 7

5x = 35

Divide both sides by 5

x = $\frac{35}{5}$ = 7

Any number raised to the power 0 is 1.

Total number of oranges = 12 + 9 = 21

Number of good oranges = 12

P(Good oranges) = $\frac{12}{21}$ = $\frac{4}{7}$

Note: 3 divides 12, 4 times and 21, 7 times.

Let the two numbers be a and b. Where a is the larger number and b is the smaller number.

Sum means addition (+)

The sum of two numbers is 72

a + b = 72 ----- eqn 1

Difference means subtraction (-).

The difference between them is 26.

a - b = 26 ----- eqn 2

Get rid of b by adding ----- eqn 1 and ----- eqn 2.

2a = 98

Divide both sides by 2.

a = $\frac{98}{2}$ = 49

Substitute a = 49 into ------ eqn 1

49 + b = 72

b = 72 - 49

b = 23

∴ the smaller number is 23.

The ratio 2 represents sheep and 3 represents goats.

There are 150 goats.

If 3 = 150

2 = $\frac{\mathrm{2\; x\; 150}}{3}$ = 2 x 50 = 100

Two sets A and B are equal if both have the same number of elements and all the elements in A are found in B.

Intersection (∩) are elements that can be found in both sets

On a Venn diagram, it is the region where both sets intersect/join/overlap.

Express (2.3 × 82) × 7.9 in the form (23 × 82) × 79

Change the decimal to whole number x 10^-n.

Where n is the number of times you have to move the decimal point to the right to obtain a whole number.

2.3 = 2.⌒3. = 23 x 10^-1

7.9 = 7.⌒9. = 79 x 10^-1

(2.3 × 82) × 7.9 = (23 x 10^-1 x 82) x 79 x 10^-1

The order in multiplication doesn't matter.

(2.3 × 82) × 7.9 = (23 x 82) x 79 x 10^-1 x 10^-1

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

(2.3 × 82) × 7.9 = (23 x 82) x 79 x 10^{-1 + -1}

(2.3 × 82) × 7.9 = (23 x 82) x 79 x 10^{-1 - 1}

(2.3 × 82) × 7.9 = (23 x 82) x 79 x 10^-2

But (23 x 82) x 79 = 148,994

(2.3 × 82) × 7.9 = 148,994 x 10^-2

Change the whole number x 10^-n to decimal by moving the decimal point to the left n times.

Every whole number has .0 at the end which is not written.

148994 = 148994.0

n is 2 in 10^-2 hence we have to move the decimal point 2 times to the left.

148,994 x 10^-2 = 1489.⌒9⌒4.0 = 1489.94

(2.3 × 82) × 7.9 = 1489.94

If the solid is shaded then it is either ≤ or ≥ but the solid is not shaded. Hence the inequality sign is either < or > . The inequality sign should be in the direction of the arrow. The arrow is pointed to the right hence the inequality is >

∴ x > -3

The more the workers, the lesser the time.

Hence time (t) is inversely proportional to the number of workers (w).

t ∝ $\frac{1}{\mathrm{w}}$

Remove the proportionality by introducing a constant (k).

t = k x $\frac{1}{\mathrm{w}}$

Use the known information to solve for k

15 men used 48 days.

48 = k x $\frac{1}{15}$

k = 48 x 15

The general equation is:

t = 48 x 15 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{48\; x\; 15}}{\mathrm{w}}$

Calculate how many men would be needed for 16 days.

time = 16 days

16 = $\frac{\mathrm{48\; x\; 15}}{\mathrm{w}}$

16 x w = 48 x 15

Divide both sides by 16

w = $\frac{\mathrm{48\; x\; 15}}{16}$ = 3 x 15 = 45

Note: 16 divides itself 1 time and 48, 3 times.

∴ it will take 45 men to weed the same plot of land in 16 days.