Rate, ratio, proportion and variation
1.
If
a : b = c : d, prove that (a – b) : (a + b) = (c – d) : (c + d) .
x 2  3x  4 x 3  3x 2  2x  5
.

x 2  3x  4 x 3  3x 2  2x  5
Hence, solve the equation :
2.
If
b
2

x : a = b : y or
3.
If
 

, prove that either
x:b=y:a.
2 x  3y y  z
x  3z


, prove that each of these ratios is equal to
y  3z
x  z 2 y  3x
Hence, show that either
4.

 bx  x 2 : a 2  ay  y 2  b 2  bx  x 2 : a 2  ay  y 2
Given that
x = y or
x
.
y
x+y=z.
a
b
a
b
ab
, prove that if x  z , then each side is equal to
.



x a x a za za
xz
Hence, prove that
a
b
a
b
ab
.




x a za za x a
2a
x
y
z
 2  2
2
a p b q c r
and
x 2 y2 z2


 1 , find the possible values of x, y and z. .
a 2 b2 c2
5.
If
6.
The receipts on a railway vary as the excess of speed above
20 km/h, while the expenses vary as the
square of that excess .
(a)
Find the speed at which the profits will be greatest, if at
40 km/h the expenses are just covered.
(b) What percentage of the receipts is the profit, when the speed is 35 km/h ?
7.
The expense of publishing a book varies partly on the cost of setting up the type, which is a constant
and partly on the cost of printing, which varies as the number of copies required. If
sold, a loss of
10% is incurred and if
630 copies are
980 copies are sold, a gain of 12% is made. How many
copies must be sold just to pay expenses ?
8.
Solve the equations :
(a)
(b)
9.
If
 xyz0

 2x  y  3z  0
x 2  2 y 2  3z 2  9

(1)
(2)
(3)
 2 x  3y  7 z  0

 5x  2 y  8z  0
3x 2  4 y 2  z 2  9

ax  cy  bz  0

cx  by  az  0
bx  ay  cz  0

(1)
(2)
(3)
(1)
(2)
for all real values of x, y and z, show that
a 3  b 3  c 3  3abc .
(3)
1
Solution
1.
a c

b d

 a  kc

b  dk
a b

c d
for some constant
k.
a  b kc  kd k c  d  c  d



a  b kc  kd k c  d  c  d
x 2  3x  4 x 3  3x 2  2x  5

x 2  3x  4 x 3  3x 2  2x  5

 
 
 
 
 
 
 23x   23x  2x 


 3x x  5  3x  2x x  4
2x  4
2x  5
 x3x  2x  4  3x  5  0  x2x  12x  7  0

x
x


 3x  4  x 2  3x  4
x 3  3x 2  2x  5  x 3  3x 2  2x  5

2
 3x  4  x 2  3x  4
x 3  3x 2  2x  5  x 3  3x 2  2x  5
2
2
3
2
2
3
2
b 2  bx  x 2 a 2  ay  y 2

b 2  bx  x 2 a 2  ay  y 2


 


b
b
 xy  ab
or

2
 
 
 
 
 
 
 bx  x 2  b 2  bx  x 2
a 2  ay  y 2  a 2  ay  y 2

 bx  x 2  b 2  bx  x 2
a 2  ay  y 2  a 2  ay  y 2





 xyax  by   abax  by   xy  abax  by   0
ax  by
 x : a  b : y or
x:b  y:a
2 x  3y y  z
x  3z


k
y  3z
x  z 2 y  3x
By Equal Ratio Theorem,
Now, we have
k
2x  3y   3y  z   x  3z   3x  x .
y  3z   3x  z   2 y  3x  3y y
yz x
 k
xz y
By Equal Ratio Theorem again, k 
Then either k =
4.
2
2 b2  x 2
2 a 2  y2

 ay b 2  x 2  bx a 2  y 2  ab2 y  ax 2 y  a 2 bx  bxy 2
2bx
2ay
 ax 2 y  bxy 2  a 2 bx  ab2 y
3.
2
 3  50
2
 x  0,
2.
2
3
y  z   x  1
x  z   y
, if (x – z) + y  0 .
x
= 1 or the denominator (x – z) + y = 0 .
y
Result follows.
a
b
a
b
a x  a   bx  a  a z  a   bz  a 





k
x a x a za za
x2  a2
z2  a 2
By Equal Ratio Theorem,

k
k
a  bx  z   a  b
x 2  z2
xz
a x  a   bx  a   a z  a   bz  a 
x
2
 
 a 2  z2  a 2

, if x  z .
2
a
b
a
b
a z  a   bx  a  a x  a   bz  a 





k
x a za za x a
xz  ax  az  a 2
xz  ax  az  a 2
k
By Equal Ratio Theorem,

5.
k
xz  ax  az  a  xz  ax  az  a 
2
2
az  bx  ax  bz a  b z  x  a  b


, if
2az  2ax
2a z  x 
2a
xz.
x  ka 2 p
x
y
z

 2  2  k   y  kb 2 q
2
a p b q c r
 z  kc 2 r




k 2 a 2 p2  b2q 2  c2 r 2  1
x
Let
ka p  kb q  kc r 
x 2 y2 z2


 1 , we get
a 2 b2 c2
Substitute this equalities in
6.
a z  a   bx  a   a x  a   bz  a 
a 2 p 2  b 2q 2  c2 r 2
R = receipts, E = expenses,
, y
a2
2
b2
2
2
c2
2
1
1
k
a 2p
2
2
a p  b2q 2  c2 r 2
2
2
b 2q
a 2 p 2  b2q 2  c2 r 2
, z
c2r
a 2 p 2  b 2q 2  c 2 r 2
v = velocity
 R  v  20 
 R  k v  20 
k , m are constants .

2  
2 , where
E  mv  20 
E  v  20 
(a)
By given

when
v = 40, R = E
k(40 – 20) = m(40 – 20)2 
k = 20m
Profit = R – E

= kv  20   mv  20   20mv  20   mv  20   m v 2  40 v  400  20 v  400
2

2


 m v 2  60v  800  m v  30  100

2


The speed at which the profits will be greatest is 30 km/h .
(b) When the speed is 35 km/h,
R = 20m(35 – 20) = 300m
E = m(35 – 20)2 = 225m
Profit = 300m – 225m = 75m
Required Percentage =
7.
E  k 1  k 2 C , where
75m
100%  25%
300 m
E = expense, k1 = cost of setting up the type, C = number of copies,
and k2C = cost of printing .
Let P be the selling price of the book.
3
If
630 copies are sold,
If
980 copies are sold,
980 P  k 1  980 k 2 1  12% 
175P = 350k2
(3)(1),
700 P  k 1  630 0.5P   k 1  385 P
x copies are sold just to pay expenses, then
x
(a)
From (1) and (2),
x:y:z 
 875 P  k 1  980 k 2
…. (2)
…. (3)
xP = k1 + xk2 .
1 1
1 1 1 1
:
:
  1 3  11 : 12  1 3 : 11  2 1  2 : 5 : 3
1 3 3 2 2 1
x = 2k , y = 5k , z = 3k
Substitute in (3),

…. (1)
k1
385P

 770
P  k 2 0.5P


 700 P  k 1  630 k 2
 k 2  0.5P
(2) – (1),
If
8.
630 P  k 1  630 k 2 1  10% 
, where
k is a constant .
2k 2  25k 2  33k 2  9
x, y, z     2 ,
 3
1
k .
3
 81k2 = 9
5

 ,  1 .
3

(b) From (1) and (2),
x:y:z 
3 7 7 2 2 3
:
:
 3 8   7  2 :  7 5  2 8 : 2 2  35
 2 8 8 5 5  2
 38 : 19 : 19  2 : 1 : 1

x = 2k , y = k , z = k
Substitute in (3),

9.
x, y, z    2,
(1) + (2) +(3),

2
2
 1,  1
(a + b + c) (x + y + z) = 0

Since
, where k is a constant .
32k   4k   k   9
2
a+b+c=0

 9k2 = 9  k  1
for all real values of x, y and z .

a 3  b3  c3  3abc  a  b  c a 2  b 2  c 2  ab  bc  ca  0
a  b  c  3abc
3
3
3
4