Paper 4: Fundamentals of Business Mathematics & Statistic

2.20 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

Solving (ii), (iii), k ,=^12 x = 500. From (i) we get T = 500 + 21 .n

For n = 1,000, T = 500 +^12 ×^1000 = ` 5,500.

Example 29: The expenses of a boarding house are partly fixed and partly varies with the number of boarders.

The charge is ` 70 per head when there are 25 boarders and ` 60 per head when there are 50 boarders. Find

the charge per head when there are 100 boarders.

Solution:

Let x = fixed monthly expense, y = variable expense, n = no. of boarders.

Now y ∝ n or, y = k n, k is constant. The monthly expenses for 25 and 50 boarders are respectively

` 1,750 and ` 3,000.

Now total expense = fixed expenses + vairable expenses.

i.e., T = x + y = x + kn, where T = total expenses.

So, from T = x + kn, we get,

Hence, 1,750 = x + 25k ..... (1)

3,000 = x + 50k....... (2)

Subtracting (1) from (2), 25k = 1,250, or k = 50.

Again, putting the value of k in (1), we find x = ` 500.

Now, charge for 100 boarders = x + 100 k = 500 + 100 x 50 = ` 5,500

∴ Charge per head = 5,500/100 = ` 55.

Example 30 : As the number of units manufactured in a factory is increased from 200 to 300, the total cost

of production increases from ` 16,000 to ` 20,000. If the total cost of production is partly fixed and other part

varies as number of units produced, find the total cost of for production 500 units.

Solution:

Total cost (T) = fixed cost + variable cost, fixed cost = a (say) variable cost ∝ no. of units (n) i.e. variable cost

= kn, k = constant

or, T = a + kn ........ (i)

20,000 a 300k......(i)

16,000 a 200k

4,000 100k,k 40,

= +

= +

= =

From (ii) 20000 = a + 12000, a = 8000

Again for 500 units ; Total cost = 8000 + 500 × 40 = ` 28,000.

Example 31 : An engine without any wagons can run 24 km/hr. and its speed is diminished by a quantity

varying as the square root of the number of wagons attached to it. With 4 wagons its speed becomes 20

km/hr. Find the maximum number of wagons with which the engine can move.

Solution :

Let n be number of wagons. So speed = 24 k n,− k = constant ...... (i)

Again 20 = 24 k. n− or, 2k = 4 or, k = 2