Secondary 2 Mathematics. Chapter 2 Direct and Inverse Proportion page 1

Chapter 2 Direct and Inverse Proportion


1. If q is inversely proportional to p, and q = 120 when p = 2, form an equation connecting p and q and calculate q when p = 5.

q = ^k/_p
120 = ^k/₂
k = 120 x 2 = 240
if p = 5 , so
q = ^k/_p
q = ²⁴⁰/₅
q = 48

2.         If y is inversely proportional to (2x + 1) and y = 5 when x = 3, find
(a)  y when x = 17,                  (b)  x when y = 7.

y = ^k/_{(2x + 1)}
5 = ^k/_[2(3)+1]
5 = ^k/₇
k = 5 x 7 = 35
(i) if x = 17 , so
y = ³⁵/_{[2(17) + 1]}
y = ³⁵/₃₅
y = 1
(ii) if y = 7, so
y = ³⁵/_{[2x + 1]}
7 = ³⁵/_{[2x + 1]}
35 = 7 x (2x + 1)
35 = 14x + 7
14x = 35 - 7
x = 28 / 14 = 2

3. If y is inversely proportional to the square of (3x + 2) and y = 4 when x = ²/₃  , find
(a)  y when x = 11 ¹/₃ , (b)  x when y = 16.

k = 64
(i) if x = 11¹/₃, so
y = 64 / [3(11¹/₃) + 2]²
y = 64 / 36²
y = 64 / 1296
y= 4/81
(ii) when y = 16, so
y = k / (3x + 2)²
16 = 64 / (3x + 2)²
(3x + 2)² = 64 / 16
(3x + 2) = 4^1/2
3x = 2 - 2
x = 0 / 3 = 0
or 3x = -2 - 2
x = -4 / 3 = -1¹/₃

4. Given that p is directly proportional to (2q + 1)^1/2  and p = 63 when q = 24, find
(a)  p when q = 12, (b)  q when p = 27.

p = k x (2q + 1)^1/2
63 = k x [2(24)+1]^1/2
63 = k x 49^1/2
k = 63 / 7 = 9
(i) if q = 12, so
p = 9 x (2(12) + 1)^1/2
p = 9 x 5
p = 45
(ii) when p = 27, so
p = k x (2q + 1)^1/2
27 = 9 x (2q + 1)^1/2
(2q + 1)^1/2 = 27 / 9
(2q + 1)^1/2 = 3
(2q + 1) = 3²
2q = 9 - 1
q = 8/2
q = 4

5. Given that d is directly proportional to the square root of t, copy and complete the table below.

The equation is d = k x t^1/2
8 = k x 4^1/2
k = 8 / 2
k = 4
(i) if t = 9, so we can calculate distance from that equation
d = k x t^1/2
d = 4 x 9^1/2
d = 4 x 3 d = 12
(ii) if d = 20, so the time is
d = k x t^1/2
20 = 4 x t^1/2
t^1/2= 20/4
t= 5²
t = 25
(iii) if t = 2¹/₄, so we can calculate distance from that equation
d = k x t^1/2
d = 4 x (2¹/₄)^1/2
d = 4 x (3/2)
d = 6

Need more practise, go to number 6