Numerical problem of Average Speed based on Elimination method of linear equation of two variable.
A man travels 35 km partly at 4 km/hrs. and at 5 km/hrs. If he covers former distance at 5 km/hrs. and later distance at 4 km/hrs., he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is?
You can also watch the descriptive video on this solution
Solution.
From the question we have
1. Speed = 4 km/hrs and 5 km/hrs
2. Total Distance = 35 km
3. Time = ? (We have to find)
We know that
Average Speed = Total Distance / Total Time,
So, Total Distance = Speed x Time
Suppose the man covers first distance in X Hrs and Second distance in Y Hrs. then
4X + 5Y = 35 …. Equation 1 here 4 is Speed (4 km/hrs) and X is Time for 1st part of Distance and 5 is Speed (5 km/hrs) and Y is time for 2nd part of Distance. 35 is Total Distance.
and
5X + 4Y = 37 …. Equation 2 here 5 is Speed (5 km/hrs) and X is Time for 1st part of Distance and 4 is Speed (4 km/hrs) and Y is time for 2nd part of Distance. 37 is Total Distance (35+2=37).
From Solving this equation we can get the value of X and Y. Now we will do cross multiplication by using Elimination method of linear equation to equal the value of X.
( 4X + 5Y = 35 ) x 5 = 20X + 25Y = 175 …. Equation 3
( 5X + 4Y = 37 ) x 5 = 20X + 16Y = 148 …. Equation 4
Now we will substract the equation 4 from equation 3
20X + 25Y = 175
20X + 16Y = 148
_ _ _
_____________
= 9Y = 27
And Y = 27/9 So. Y = 3
Now put the value of Y in 2nd Equation 5X + 4Y = 37 we will get the actual value of X, So
5X + 4 x 3 = 37
5X + 12 = 37
5X = 37 – 12
5X = 25
X = 25/5 = 5 So, X = 5
So Total Time taken to cover the whole distance at original rate is
(X+Y)Hrs = (5+3) Hrs = 8 Hrs.
For Class 8, 9,10,11,12 and for Competitive Examination
***********