FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 2.19(vii) If A ∝ B and C ∝ D, then AC ∝ BD and =
A B
C D
(viii) If A ∝ BC, then B∝CA and C∝AB
SOLVED EXAMPLES
Example 25: Given c ∝ (ax + b), value of c is 3 when a = 1, b = 2 value of c is 5, when a = 2, b = 3. Find x.
Solution:
As c ∝ (ax +b), so c = k (ax + b), k is constant.
For a = 1, b = 2, we get c = k (x + 2) ..... (i)
a = 2, b = 3, we get c = k (2x + 3) ...... (ii)
Subtracting (i) from (ii), 0 = k (x + 1) or, x + 1 = 0 as k ≠ 0 ∴ x = – 1.
Example 26: If the cost price of 12 kg. of rice is ` 10, what will be the cost of 15 kg. of rice?
Solution:
Let A ( = cost) = ` 10, B ( = quantity) = 12 kg. Now A ∝ B i.e., A = KB or, 10 = K. 12 or, K=^1012. Now, we are to
find A, when B = 15 kg.
Again from A = KB, we have A=^1120 .15 = ` 12.50.
Example 27 : A man can finish a piece of work, working 8 hours a day in 5 days. If he works now 10 hours
daily, in how many days can he finish the same work?
Solution:
Let A ( = days) = 5, B ( = hours) = 8, it is clear that A∝B^1
i.e., A K. or,5 K. or,= B^1 = 81 K = 40.
To find A, when B = 10, we have A = 40. 101 = 4 days.
(Partly fixed and partly variable)
Example 28 : The publisher of a book pays author a lump sum plus an amount for every copy sold. If 500
copies are sold, the author would receive 750 and for 1350 copies 1175. How much would the author
receive if 10000 copies are sold?
Solution:
Let x = lump (i.e. fixed) sum received, y = variable amount received on sale.
n = number of copies sold, so that y ∝ n or, y = kn, k = constant.
Again, total amount (T) = x + y = x + kn ..... (i)
So we get, 750 = x + k. 500 ....... (ii)
1175 = x +k. 1350 ............ (iii)