Today our topic of discussion is Forming and applying algebraic formulae in solving real life problems.
Forming and applying algebraic formulae in solving real life problems
Forming and applying algebraic formulae in solving real life problems
In our daily business we face the realistic problems in different time and in different ways. These problems are described linguistically. In this section, we shall discuss the formation of algebraic formulae and their applications in solving different problems of real surroundings which are described linguistically.
As a result of this discussion the students on the one hand, will get the conception about the application of mathematics in real surroundings, on the other hand, they will be eager to learn mathematics for their understanding of the involvement of mathematics with their surroundings.
Methods of solving the problems:
- At first the problem will have to be observed carefully and to read attentively and then to identify which are unknown and which are to be determined.
- One of the unknown quantities is to be denoted with any variable (say x). Then realising the problem well, express other unknown quantities in terms of the same variable z.
- The problem will have to be splitted into small parts and express them by algebraic expressions.
- Using the given conditions, the small parts together are to be expressed by an equation.
- The value of the unknown quantity is to be found by solving the equation.
Different formulae are used in solving the problems based on real life. The formulae are mentioned below:
Related to Payable or Attainable
Suppose, q = amount of money payable or attainable per person n = number of person
∴ Amount of payable or attainable money, A = qn
Related to Time and Work
Suppose, q = portion of a work performed by every one in unit of time
n = number of performers of work
x = total time of doing work
W = portion of a work done by n persons in time
∴ W = qnx
Related to Time and Distance
Suppose, v = speed per hour d = total distance
t = total time
∴ d = vt
Related to pipe and water tank
Suppose, Q_{0} = amount of stored water in a tank at the time of opening the pipe
q = amount of water flowing in or flowing out by the pipe in a unit time.
t = time taken.
Q(t) = amount of water in the tank in time t
∴ Q(t) = Q_{0} plus/minus q * t
‘ +^ prime sign at the time of flowing of water in and ‘-‘ sign, at the time of flowing of water out are to be used.
Related to percentage
Suppose, b = 1 total quantity
r = of percentage = s/100 = s%
p = part of percentage s% of = b
∴ p = br
Related to profit and loss
Suppose, C = cost price
r = rate of profit or loss
∴ selling price S = C(1 plus/minus r)
in case of profit, S = C(1 + r)
in case of loss, S = C(1 – r)
Related to investment and profit
Suppose, I = profit after time n
n = specific time
P = principal
r = profit of unit principal at unit time
A = principal with profit after time n
In the case of simple profit,
I = Pr
A = P + I = P + Pnr = P(1 + nr)
In the case of compound profit,
A = P * (1 + r) ^ n
Example 34. For a function of Annual Sports, members of an association made a budget of Tk. 45000 and decided that every member would subscribe equally. But 5 members refused to subscribe. As a result, amount of subscription of each member increased by Tk. 15 per head. How many members were in the association?
Solution: Let the number of members of the association be x and amount of subscription per head be Tk. q.
Then total amount of subscription, Tk A = qx = 45000 .
Actually numbers of members were (x – 5) and amount of subscription per head became Tk. (q + 15) .
Then, total amount of subscription (x – 5)(q + 15)
By the question,
qx= (x – 5)(q + 15) ******(1)
qx =45000*****(2)
From equation (1) we get,
qx = (x – 5)(q + 15)
or, qx = qx – 5q + 15x – 75 or, 5q = 15x – 75 = 5(3x – 15)
q = 3x – 15
Putting the value of q in (2), we get,
(3x – 15) * x = 45000
or, 3x ² – 15x = 45000
or, x ² – 5x = 15000 [dividing both sides by 3]
or, x ² – 5x – 15000 = 0
or, x ² – 125x + 120x – 15000 = 0
or, x(x – 125) + 120(x – 125) = 0 or, (x – 125)(x + 120) = 0
Therefore, (x – 125) = 0 or (x + 120) = 0
or, x = 125 or, x = – 120
Since the number of members cannot be negative, hence the value of x as -120 is not acceptable.
Therefore, number of members of the association is 125
Example 35. Rafiq can do a work in 10 days and Shafiq can do that work in 15 days. In how many days do they together finish the work?
Solution: Let, they can finish the work in d days.
As per question, d/10 + d/15 = 1 ar ,d( 1 10 + 1 15 )=1
or, d((3 + 2)/30) = 1or, (5d)/30 = 1
or, d = 30/5 = 6
Therefore, they together can finish the work in 6 days.
Example 36. A boatman can row 1 km in time t_{1} hour against the current. To cover that distance along the current he takes t_{2} hour. How much is the speed of the boat and the current?
Solution: Let the speed of the boat in still water be u km per hour and that of the current be v km per hour.
Then, along the current, the effective speed of boat is (u + v) km per hour and against the current, the effective speed of boat is (u – v) km per hour.
We know, speed = distance traversed/time
According to the question, u + v = x t 2 *****(1)
and u – v = x t 1 ******(2) Adding equations (1) and (2) we get,
2u= dot x t_{2} + x/t_{1} = x(1/t_{1} + 1/t_{2})
or, u = x/2 * (1/t_{1} + 1/t_{2})
Subtracting equation (2) from equation (1) we get, 2v = x/t_{2} – x/t_{1} = x(1/t_{2} – 1/t_{1}) or, v = x/2 * (1/t_{2} – 1/t_{1})
Hence speed of current is x 2 (1/t_{2} – 1/t_{1}) rfloor km per hour and speed of boat is x/2 * (1/t_{1} + 1/t_{2}) km per hour.
Example 37. A pipe can fill up an empty tank in 12 minutes. Another pipe flows out 14 litre of water per minute. If the two pipes are opened together and the empty tank is filled up in 96 minutes, how much water does the tank contain?.
Solution: Let a litre of water flows in per minute by the first pipe and the tank can contain y litre of water.
According to the question, the tank is filled up by first pipe in 12 minutes
y= 12x *…..(1)
Again, the empty tank is filled up by the two pipes together in 96 minutes.
y= 96x -96*14*****(2)
From equation (1) we get, x = y/12
putting the value of r in equation (2) we get,
y = 96 * y/12 – 96 * 14
or, y = 8y – 96 * 14
or, 7y = 96 * 14
or, y = (96 * 14)/7 = 192
Hence, total 192 litre of water is contained in the tank.
Work:
1) For a picnic, a bus was hired at Tk. 2400 and it was decided that every passenger would have to give equal fare. But due to the absence of 10 passengers, fare per head was increased by Tk. 8. How many passengers did go by the bus and how much money did each of the passengers give as fare?
2) A and B together can do a work in p days. A alone can do that work in g days. In how many days can B alone do the work?
3) A person rowing against the current can go 2 km per hour. If the speed of the current is 3 km per hour, how much time will he take to cover 32 km, rowing along the current ?
Example 38. Price of a book is Tk. 24. This price is 80% of the actual price. The Government subsidize the due price. How much money does the Government subsidize for each book?
Solution: Market price = 80% of the actual price
We know, p = br
Here, p = Tk * 0.24 and r = 80% = 80/100
24 = b * 80/100 r, b = (24 * 100)/80
b = 30Tk
Hence, the actual price of the book is Tk. 30.
∴ Amount of subsidized money Tk.= (30 – 24) or Tk .=6
Hence, subsidized money for each book is Tk. 6.
Example 39. The loss is r% when n oranges are sold per taka. How many oranges are to be sold per taka to make a profit of s%?
Solution: If the cost price is Tk. 100, the selling price at the loss of r% is Tk (100 – r)
If selling price is Tk. (100-r), cost price is Tk. 100
∴ If selling price is Tk. 1, cost price is
Again, if cost price is Tk. 100, selling price at the profit of s% is Tk. (100 + s) .
∴ if cost price is Tk.
100/(100 – r) selling price at the profit of s% is Tk. ( (100 + s)/100 *
100/(100 – r) )
=Tk. (100 + s)/(100 – r)
Therefore, in Tk. (100 + s)/(100 – r) * 1 number of oranges is to be sold is n.
∴ In tk. 1 number of oranges is to be sold is n((100 – r)/(100 + s))
Hence, (n(100 – r))/(100 + s) oranges are to be sold per taka.
Example 40. What is the profit of Tk. 650 in 6 years at the rate of profit Tk. 7 percent per annum ?
Solution: We know, I = Pnr.
Here, P = Tk * 0.65 n = 6 year, percent per annum s = Tk
r = s/100 = 7/100
∴ I = 650 * 6 * 7/100 = 273
Hence, profit is Tk. 273.
Example 41. Find the compound principal and compound profit of Tk. 15000 in 3 years at the profit of 6 percent per annum.
Solution: We know, C = P * (1 + r) ^ n [where C is the profit principal in the case of compound profit]
Given, P = 15000Tk , r = 6% = 6/100 n = 3 year
∴ C = 15000 * (1 + 6/100) ^ 3 = 15000 * (1 + 3/50) ^ 3 = 15000 * (53/50) ^ 3
= 15000 * 53/50 * 53/50 * 53/50 = 446631/25 = 17865.24
∴ Compound principal is Tk 17865.24 .. Compound profit is Tk. (17865.24-15000) Tk.
Work:
1) The loss is 50% when 10 lemons are sold 50 taka. What will be the profit or loss if 6 lemons are sold 50 taka?
2) What will be the profit of principal of Tk. 750 in 4 years at simple interest 6 1/2 percent per annum? 3) Find the compound interest of Tk. 2000 in 3 years at compound interest of Tk. 4 percent per annum.
Example 42. The loss is % when 10 sticks of ice creams are sold per taka. How many sticks of ice creams are to be sold per taka to make the profit of z% ?
Solution: If the cost price is Tk. 100, the selling price at the loss of x% is Tk.
= (100 – x)
If selling price is Tk. (100), cost price is Tk. 100
∴If selling price is Tk. 1, cost price is Tk. 100/(100 – x)
Hence, the cost price of 10 sticks is Tk. 100/(100 – x)
∴The cost price of 1 stick is Tk. 100/((100 – x) * 10)
Again if the cost price is Tk. 100, selling price at the profit of z% is Tk. (100 + z)
If the cost price is Tk. 100, selling price is Tk. (100 + z)
If the cost price is Tk. 1, selling price is Tk. (100 + z)/100
∴ If the cost price is Tk.
100/((100 – x) * 10)
(100 + z)/100 * 100/((100 – x) * 10) = (100 + z)/((100 – x) * 10) The selling price of 1 stick ice cream is Tk. (100 + z)/((100 – x) * 10) = (100 + z)/(1000 – 10x)
Hence (1000 – 10x)/(100 + z) stick have to be sold per taka.
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