Today our topic of discussion is Formation of simultaneous equations from real life problems and solution.
Formation of simultaneous equations from real life problems and solution
Formation of simultaneous equations from real life problems and solution
In everyday life, there occur some such mathematical problems which are easier to solve by forming equations. For this, from the condition or conditions of the problem, two mathematical symbols, mostly the variables x, y are assumed for two unknown expressions. Two equations are to be formed for determining the values of those unknown expressions. If the two equations thus formed are solved, values of the unknown quantities will be found.
Example 12. If 5 is added to the sum of the two digits of a number consisting of two digits, the sum will be three times the digits of the tens place. Moreover, if the places of the digits are interchanged, the number thus found will be 9 less than the original number. Find the number.
Solution: Let the digit of the tens place of the required number be z and its digits of the units place is y. Therefore, the number is 10x + y
∴ by the 1st condition, x + y + 5 =3x…(1)
and by the 2nd condition, 10y+x= (10x + y) -9…(2)
From equation (1) we get, y = 3x – x – 5 or, y = 2x – 5 (3)
Again, from equation (2) we get,
10y – y + x – 10x + 9 = 0
or, 9y – 9x + 9 = 0
or, y – x + 1 = 0
or, 2x – 5 – x + 1 = 0 [putting the value of y from equation (3) we get]
or, x = 4
putting the value of x in equation (3) we get, y = 2 * 4 – 5 = 8 – 5 = 3
∴ the number is 10x + y = 10 * 4 + 3 = 40 + 3 = 43
Example 13. 8 years ago, father’s age was eight times the age of his son. After 10 years, father’s age will be twice the age of the son. What are their present ages?
Solution: Let the present age of father be x years and age of son is y years. .. by 1st condition, x-8= 8(y – 8) …(1)
and by 2nd condition, x+10= 2(y + 10) …(2)
From (1) we get, x – 8 = 8y – 64
or, x = 8y – 64 + 8
or ,x=8y-56…(3)
From (2) we get, x + 10 = 2y + 20
or, 8y – 56 + 10 = 2y + 20 [Putting the value of x from (3)]
or, 8y – 2y = 20 + 56 – 10
or, 6y = 66
or, y = 11
(3) we get, x = 8 * 11 – 56 = 88 – 56 = 32
∴ at present, Father’s age is 32 years and son’s age is 11 years.
Example 14. Twice the breadth of a rectangular garden is 10 metre more than its length and perimeter of the garden is 100 metre. There is a path of width 2 metre around the outside boundary of the garden. To make the path by bricks, it costs TK. 110 per square metre.
1) Assuming the length of the garden to be a metre and its breadth to be y metre, form system of simultaneous equations.
2) Find the length and breadth of the garden.
3) What will be the total cost to make the path by bricks?
Solution:
1)
Length of the rectangular garden is z metre and breadth is y metre.
∴ by first condition, 2y = x + 10 (1)
and by second condition, 2(x + y) = 100 (2)
2) from equation (2) we get, 2x + 2y = 100
or, 2x + x + 10 = 100 [putting the value of 2y from (1)]
or, 3x = 90
or, x = 30
∴ from (1) we get, 2y = 30 + 10 [putting the value of x]
or, 2y = 40
or, y = 20
∴ length of the garden is 30 metre and breadth is 20 metre.
3)
Length of the garden with the path = (30 + 4) m .=34 m
and breadth of the garden with the path (204) m .=24 m.
∴ Area of the path Area of the garden with
the path – Area of the garden
overline aZ
= (34 * 24 – 30 * 20) square metre.
=(816-600) square metre.
=216 square metre.
∴ cost for making the path by bricks =Tk. (216 * 110) = Tk . 23760 the times.
Example 15. How many times will minute hand and hour hand coincide? Find
Solution: Let minute hand and hour hand coincide at the time zy. We need to remember that x( x = 0, 1 ,***11 where 0 means 12) is always an integer but y may not be. We know, minute hand runs 12 times faster than the hour hand. At time a the hour hand is exactly on the z and minute hand is on the 12. Within y minutes the hour hand passes y/12 and the minute hand passes y ticks. So,
5x + y/12 = y
or, y – y/12 = 5x
or, 11/12 * y = 5x
y = 60/11 * x
Now we put the possible values of x:
If x = 0 y = 0 minute i.e. 12:00.
If x = 1, 1 / (5 5/11) minute.
If x = 2, 2 / (10 10/11) minute.
If x = 11 , 11: 60 minute or, 12:00.
As the first and last times are same, these two hands coincides 11 times and the times are: : 60/11 * x minute.
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