BODMAS

Algebra

Student’s Name

Institution of Affiliation

 

 

 

 

 

 

 

Question 1

a). i) To work out  6.38–3.560.324+0.412x 2.35

 ,    we use BODMAS formula. The first step is working out the bracket in the denominator and finding the difference in the numerator to obtain;

2.820.133488 x 2.35

  Working out the denominator we have;    

= 2.820.313697

  We divide to obtain;

= 8.99 (3 significant figures). The correct significant figures equal the smallest significant figures of the numbers used in the above expression.

 

ii).  2π x 3.1232 x (0.74+5.2)

 

We begin by working out the bracket and the square under the square root to obtain;

=2π x 9.753 x (5.94)

 Substituting the value of π =227

 and multiplying by the coefficient (2) yields;

= 6.286 x 9.753 x 5.94

   We now multiply the values under the square root and express the result to 3 significant figures.

= 364

 We find the square root as;

= 19.08 (4 significant figures

 

iii).  4.32 x 10-322.45 x 103

   in this expression, 10^-3 has only one significant figure, thus the result of working out the bracket is presented in 1 significant fgure.

= 0.00422450

= 0.000022450

 The numerator is expressed in 1 significant figure and the resulting solution is also expressed in 1 significant figure as indicated below;

= 0.000000008 (1 significant figure)

 

b).  5(4 ÷ 2) + (7 – 4 x 3) + 11.  For this expression, we first work out what is in the brackets to get;

= 5(2) + (7- 12) + 11

= 10 – 5 + 11

= 16

 

c).

 i).

The equation is h = h*p1p

  ……………… (i)      Or

p = h1p*h

  ………………………….. (ii) where p1 is for h^* in equation (i) and h is for p* in equation (ii). h₁ ≥ h, h^* and p^* are known.

ii).

h = 24
x 45

  Formed by substituting the values of p₁, p and h^*.

h = 965

h = 19.2 hours

iii).

P = 7.5
x 244

  Formed by substituting the values of

   = 7.5 x 6

  = 45 people

 

Question 2

a).  i). Expanding the expression 3n² − 2n(3n − 3) − 7 + 4n

= 3n² – 6n² + 6n + 4n – 7

= – 3n² + 10n  – 7 multiply all through by -1

= 3n² – 10n + 7 using the completing square method

= 3n² – 3n – 7n + 7

= 3n (n – 1) – 7 (n – 1)

= (3n – 7) (n – 1)

 

ii) -2 (3p – 2) (-2p + 4)

 = -2 [3p (-2p + 4) – 2 (-2p + 4)] Expanding the brackets

= -2 [-6p² + 12p + 4p – 8] Collecting the like terms

= -2 [-6p² + 16p – 8]

= 12p² – 32p + 16

 

 

b).

i).  32a32 x b223a4b3

 

= 12a6 b223a4b3

  Working out the bracket in the numerator and the known values

= 18a6 b2a4b3

  Cancelling the like terms

= 18a2b

    or

= 18a²b^-1

ii).  12x–3x-3

 

=  x -3– 6x 2x(x-3)

  Expanding the brackets

= -5x-3 2x(x-3)

 

= -5x-3 2x2–
6x

  Multiplying both the numerator and the denominator by -1 we get;

= 5x+3 6x-2x2

 

c).  3 (5 + x) = 7 – 5x

15 + 3x = 7 – 5x

3x + 5x = 7 – 15

8x = 8

x = 1

 

d).  

3x + y = 3 ……………………………………………….. (i)

4x + 3y = -1……………………………………………… (ii)

Making y the subject of equation (i) we have;

y = 3 – 3x ……………… (iii) Substituting the value of y into equation (ii) we get;

4x + 3 (3 – 3x) = -1

4x + 9 – 9x = -1

-5x = -10

x = 2

We substitute the value of x in equation (iii) to obtain the value of y as follows.

y = 3 – 3 (2)

y = 3 – 6

y = -3

e).   p=3q+9m2q

  ………………………………………….. (i)

Multiplying all through by 2q we have;

2pq=6q2+9m

  Taking 6q² to the other side of the equal sign yields;

9m = 2pq – 6q²   Dividing all through by 9s

m= 2pq-6q29

    ………………………………………….. (ii) 

Given that p = 12 and q = 3, we substitute these values in equation (ii) to find the value of m

m= 212(3)-6(3)29

m= 72-549

 

m= 189

m = 2

 

Question 3

a).

i).

5a-54=2a+74

  Multiplying all through by 4;

20a – 5 = 8a + 7 Collecting the like terms

20a – 8a = 7 + 5

12a = 12 Dividing all through by 12

a = 1

ii)

y2–3y5=15

  Multiplying all through by 10 we have;

5y – 6y = 2

-y = 2 Multiplying both sides by -1

y = 2

 

iii)

4x-3

+ 5 = 8 squaring all through;

4x -3 + 25 = 64 collecting the like terms;

4x = 67 – 25

4x = 42 Dividing all through by 4 we have;

x = 10.5

iv)

m² + 4 = 40

m² = 40 – 4

m² = 36 Taking square root in both sides

m = 6

v)

x+23x=x+12x

  Multiplying all through by 6x we have;

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

2x + 4 = 3x + 3 Collecting the like terms;

3x – 2x = 4 – 3

x = 1

b).

3x² – x – 2 = 3x² – 3x + 2x – 2 we find two numbers that when multiplied yield -6 and when summed yield -1. The two numbers are -3 and 2.

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

                 = (3x + 2) (x – 1)

c).

i)  (3x + 2)² – 2 = 9x² + 8x + 10 Working out the brackets

3x (3x + 2) + 2 (3x + 2) = 9x² + 8x + 10

9x² + 6x + 6x + 4 = 9x² + 8x + 10 Collecting the like terms

9x² – 9x² + 12x – 8x = 10 – 4

4x = 6

x = 1.5

 

ii).   2x² – 9x + 9 = 0 we find two numbers that when multiplied yield 18 and when summed yield -9. The two numbers are -6 and -3

2x² – 6x – 3x + 9 = 0

2x (x – 3) – 3(x – 3) = 0

(2x – 3) (x – 3) = 0

Thus, 2x – 3 = 0 or x – 3 = 0

2x = 3 or x = 3

x = 1.5 and/or x = 3

 

Question 4

a). To convert an angle from degrees to radians, we multiply the degrees provided by π180

. But     π = 227

, thus π180=227
x 180=221260=0.0175 (3 SF)

.

Therefore, we multiply the provided angle in degrees by 0.0175. 115⁰ will be;

115⁰ = (115 x 0.0175) radian

        = 2.007 radian. (3 decimal places)

 

b).  To convert m³ into liters (L), 1 m³ = 1000 L

To convert 15.75 m³ to L, we multiply by 1000 to obtain;

15.75 m³ = (15.75 x 1000) L

                = 15750 L

 

c).   The relation between kg/m³ and g/l is that, 1 kg/m³ = 1 g/l.

9550 kg/m³ = (9550 x 1) g/l

                    = 9550 g/l

 

d).  Given that 1 mi (mile) = 1.6093 km, then dividing both sides by 1.6093 gives;

1 km = 0.6214 mi . The result shows that, 1 km/hr = 0.6214 mi/hr.

But 1 km = 1000 m indicating that 1000 m = 0.6214 mi. dividing both sides by 0.6214 yields;

1 mi = 1609.3 m

Therefore, 1 mi/hr = (1609.33600)

 m/s

                              = 0.447 m/s

Therefore,

25 m/s = 250.447

 mi/hr

            = 55.93 mi/hr.  (4 sf)

 

Question 5

a). Volume of air in the water tank = Volume of the water tank

                        =  πr²h +23πr3

 

h = 5 m and r = 2 ms

V = 22722(5)+2322723

   = 4407 +35221

  = 1320+35221

  = 167221

   = 79.619 m³ (3 dp).

b).

Surface Area of Water Tank = πr² + 2πrh + 2πr²

But h = 5 and r = 2

= 22722+
222725+
222722

= 887+4407+1767

=  7047

 

= 100.57 m² (2 decimal places)