# algebra and algebraic problem solving!!!!!!QUESTION 1: Simplify the following A) 6ag^2 - 4a + 5 g^2 a B) 2x(5X-8)-(6x-3) QUESTION 2: A cylindrical drum has height of 1 metre and a base radius of...

algebra and algebraic problem solving!!!!!!

QUESTION 1: Simplify the following

A) 6ag^2 - 4a + 5 g^2 a

B) 2x(5X-8)-(6x-3)

QUESTION 2:

A cylindrical drum has height of 1 metre and a base radius of cm. The drum contains water but is not full. Calculate the possible values for the depth water if it contains less than 25 litres. Answer in cm's to 1 decimal place. (note: 1000 ml=1 L and 1 ml=1cm^3).

QUESTION 3:

A farmer leaves her property to drive to an appointment in the city, 330 km away. Let her initial speed be x km/h.

She drives for one hour on a dirt road and then joins the main highway where she is able to increase her speed by 30 km/h. She remains on the highway for 3 hours and arrives in the city just in time for her appointment.

Write an equation for the distance travelled by the farmer and solve it to find the speed

a) on the dirt road

b) on the highway

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### 2 Answers

Question 1A.

The given expression, particularly its last term, may be interpreted in different ways. To avoid the confusion it is rewritten in the following form.

6ag^2 - 4a + (5g^2)a

= 6ag^2 - 4a + 5ag^2 ... [As ag^2 is equal to (g^2)a]

= 11ag^2 - 4a

= a(11g^2 - 4)

Question 1B.

2x(5x-8)-(6x-3)

= 10x^2 - 16x - 6x + 3

= 10x^2 - 22x + 3

Question 2

Volume of drum in litres is given by the formula:

Volume = [(Pi)*(Radius)^2*(Height)]/1000

when both radius and height are measured in cm.

In the question the value of radius has not been given, whereas that of the height of drum is given which has no bearing on the problem. Assuming height of drum mentioned is actually meant to be radius, the given values are:

Radius = 1m = 100cm

Volume = 25 litre

Substituting these values in the equation for volume:

25 = [3.14159*100^2*(Height)]/1000 = 31.4159*Height

Therefor:

Height = 25/31.4159 = 0.79 = 0.8 cm (rounded off to nearest first decimal place)

Thus if water contains less than 25 litre of water the depth of water column will be less than 0.8 cm.

Question 3.

Let the speed on the dirt road be equal to x.

Therefor speed on the high is equal to (x + 30)

Distance travelled on dirt road

= (Speed on dirt road)*(Time on dirt road) = x*1 = x

Distance travelled on highway

= (Speed on highway)*(Time on highway) = (x + 30)*3 = 3x + 90

Total distance travelled on dirt road plus on highway

= x + 3x + 90 = 4x + 90

This is given to be equal to 330 km

Therefor

4x + 90 = 330

4x = 330 - 90 = 240

and:

x = 240/4 = 60 km/h = Speed on the dirt road.

Speed on highway = x + 30 = 60 + 30 = 90 km/h

a)

To simplify:

6ag^2 - 4a + 5 g^2 a = 11ag^2-4a , as ag^2 = ga^2.

b)To simplify:

2x(5X-8)-(6x-3) .

=2x*5x-2x*8 -6x-(-3)

=10x^2-16x-6x+3

=10x^2-22x+3

2) The volume of water = pr^2*h, where r = radius of cylinder = 1cm and h = height of water to be determined.So

25litres = 25*1000cm^3 = p*(r cm)^2*h . Therefore

h = 25000cm^3/(Pr^2cm^2) .

= 25000/(pr^2) cm .

= 25/(pr^2) meter.Substitute the value of r which is not given.

3)

Distance travelled in the 1st hr = speed *time = 1*x km = xkm

Distance travelled in next 3 hr = speed *time =( x+30)*3 lm

Total distance travelled agebraically = x km+(x+30)3 km = 4x+ 90 which is equal to 330km .Or

4x +90 =330 . Or 4x = 330-90 = 240 . Or

a)

x = 240/4 = 60 KmBut x according to the data is thedirt road distance .

b)

So, the highway distance = (x+30)*3 = (60+30)3 = 270 Kms.

Therefore