Secondary 2 Mathematics. Chapter 2 Direct and Inverse Proportion page 2

6.Two quantities P and Q are related by the formula  P = A - ^B/_Q², where A and B are constants. Given that P = 1 when Q = 2 and P = 6 when Q = 3,
(a) write down two equations in A and B,
(b) solve these equations to find the value of A and the value of B,
(c) find the positive value of Q when P = 7 ³/₄

(a) we know the equation is
Given that P = 1 when Q = 2 and P = 6 when Q = 3,
P = A - B/Q2
A = P + B/Q2
B/Q2
A = 1 + B/22
A = 1 + B/4
A = 4/4 + B/4
4A = 4 + B
4A - B = 4
and
P = A - B/Q2
A = P + B/Q2
A = 6 + B/32
A = 6 + B/9
A = 54/9 + B/9
9A = 54 + B
9A - B = 54
(b)
4A - B = 4-----first equation
9A - B = 54-----second equation
Subtract equation I and equation II, so we get
A = 10 and B = 36.
(c)
value of Q when P = 7 3/4
10 = 73/4 + 36/Q2
40/4 - 31/4 = 36/Q2
Q2 = 36 : 9/4
Q2 = 16
Q = 4

7. The intensity of illumination, I lumens/m², at a point on a screen is inversely proportional to the square of the distance, d m, of the light source from the point. Given that d = 2.5 m when I = 0.8 lumens/m², find
(a)  I, when d = 1.25 m, 

(b)  d, when I = 0.05 lumens/m².

Given that d = 2.5 m when, I = 0.8 lumens/m², find I = ^k/_d²
0.8 = ^k/_2.5²
k = 0.8 x 2.5² = 5
(a) I = ⁵/_1.25²
I = 3.2
(b) 0.05 = ⁵/_d²
d² = 5 / 0.05
d = 100^1/2 = 10

8. Water flows from a container such that the depth of water x cm at any instant, is inversely proportional to the square root of the time t s, for which the water has been flowing. After 25 s, the depth is 450 cm. Calculate the depth after

(a)  100 s,
(b)  3 min 45 s.

Given that After 25 s, the depth is 450 cm. x = ^k/_t^1/2
450 = ^k/_25^1/2
k = 450 x 5² = 2250
(a) x = ²²⁵⁰/_t^1/2
if t = 100 s,
x = ²²⁵⁰/_100^1/2
x = 2250 / 10 = 225 cm
(b) x = ²²⁵⁰/_t^1/2
if t = 225 s,
x = ²²⁵⁰/_225^1/2
x = 2250 / 15 = 150 cm

9. The resistance R, to the motion of a car is directly proportional to the speed v of the car. Given that the resistance is 2 688 newtons when the speed is 16 m/s, find

(a)  the resistance when the speed is 28 m/s,

(b)  the speed when the resistance is 16 800 newtons.

Given that the resistance is 2 688 newtons when the speed is 16 m/s. R = k.v
2688 = k.16
k = 2688/16 = 168
(a) if the velocity is 28 m/s,
R = 168.v
R = 168 x 28
R = 4704 Newton
(b) if the resistance is 16,800 N,
R = 168 x v
16800 = 168.v
v = 100 m/s

10.The total cost, $ c, of owning and operating a car is given by the formula c = a + bx, where x is the distance driven in km and a and b are constants. When the car is driven 2 500 km during it lifetime, the total cost is $ 19 000 and when the car is driven 6 000 km during its lifetime, the total cost is $ 20 400.
(a) Write down two equations in a and b.
(b) Solve these equations to find the value of a and of b.
(c) Find the total cost, if the car is driven a total distance of
(i) 50 000 km, (ii) 100 000 km.

(a)
c = a + bx
first equation
19000 = a + b(2500)
second equation
20400 = a + b(6000)
(b)
Eliminate first and second equation
19000 = a + b(2500)
20400 = a + b(6000)
__________________________ _
a = 18000 ; b = 2/5 = 0.4
(c)
The cost if the distance is 50000 km
c = 18000 + 0.4x
c = 18000 + 0.4 x 50000
c = 18000 + 20000
c = $38 000
The cost if the distance is 100 000 km
c = 18000 + 0.4x
c = 18000 + 0.4 x 100 000
c = 18000 + 40000
c = $58 000

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