(ii) A train covered a certain distance at a uniform speed. If the speed of the train would
have been 15 km/h more, it would have taken 1.6 hours less than the scheduled
time. If the speed of the train would have been 12 km/h less, it would have taken
2 hours more than the scheduled time. Find the distance covered by the train.
It is stated that a train covered a certain distance at a uniform speed. If the speed of the train would have been 15 km/h more, it would have taken 1.6 hours less than the scheduled time. If the speed of the train would have been 12 km/h less, it would have taken 2 hours more than the scheduled time. So, here it has asked to find the distance covered by the train.
So ,first we need to find out the speed and time taken by the train. Thereafter, the distance covered.
For that,
Let us we consider the speed of the train be ” x km/hr “
Time taken by the train be ” ‘y’ hrs ”
So,
Let’s solve it !
We know ::
[tex]\bullet \: \: {\boxed{\bf{Distance = Speed \times Time}}} \\ \\ [/tex]
That is Distance covered by the train is :-
[tex]\bullet \: \: {\boxed{\bf{ x \times y = xy}}} \\ \\ [/tex]
Now, According to the question ::
Case I
If the train would have been 15 km/hr more, then it would have taken 1.6 hours less than the scheduled time.
Making Equation ::
[tex]\dashrightarrow\:\:\sf (x + 15) (y – 1.6) = xy \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf xy – 1.6x + 15y – 24.0 = xy \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf 15y – 1.6x = 24 \\ \\ [/tex]..(1)
Case II
If the speed of the train would have been 12 km/h less, it would have taken 2 hours more than the scheduled time.
→ (x – 12) (y + 2) = xy
[tex]\dashrightarrow\:\:\sf xy + 2x – 12y – 24 = xy \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf – 12y + 2x = 24 \\ \\ [/tex]..(2)
Taking (1) and (2)
Using Elimination Method by equating Coefficient
Multiplying eq (1) by (4)
So, we get ;;
[tex]\dashrightarrow\:\:\sf 15y – 1.6x = 24 \: \: \times (4) \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf 60y – 6.4x = 96 \\ \\ [/tex]..(3)
Similarly, Multiplying eq (2) by (5)
[tex]\dashrightarrow\:\:\sf – 12y + 2x = 24 \\ \\ [/tex] ..(2) × 5
[tex]\dashrightarrow\:\:\sf – 60y + 10x = 120 \\ \\ [/tex] ..(4)
Now, Taking (3) and (4)
By Eliminating y from the equation , making it linear equation.
60y – 6.4x = 96 ..(3)
– 60y + 10x = 120 ..(4)
_______________ By Addition ::
[tex]\dashrightarrow\:\:\sf 3.6x = 216 \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf x = \dfrac{216}{3.6} \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf x = \dfrac{\cancel{2160}^{\: \: 60}}{\cancel{36}^{\: \: 1}} \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf x = {\red{60}} \\ \\ [/tex]
Therefore, The speed of the train is 60 km/hr.
Substituting the value of x in equation (4)
[tex]\dashrightarrow\:\:\sf – 60y + 10(60) = 120 \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf – 60y + 600 = 120 \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf – 60y = 120 – 600 \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf – 60 y = – 480 \\ \\ [/tex]
Negative Sign cancel out ::
[tex]\dashrightarrow\:\:\sf y = \dfrac{\not{480}^{ \: \: \: 8} }{\not{60}^{ \: \: 1} } \\ \\ [/tex]
[tex]\dashrightarrow\:\:\sf y = {\red{ 8}} \\ \\ [/tex]
Hence, The time taken by the train is 8 hours.
Now, we know distance is :-
[tex]\dashrightarrow\:\:\sf xy = 60 \times 8 \\ \\ [/tex]
[tex]\dashrightarrow\:\: \underline{ \boxed{\sf Distance = {\purple{480 \: km/hr}} }} \\ \\ [/tex]
What we have done ?
Here, we used the method of elimination by equating coefficient. As first we made the coefficient equal then we simply eliminated the y from the equation and got the value of x and thereafter we substituted the value of x in the other equation to get the value of y. That’s how we got the values of x and y respectively.
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[tex]{\boxed{\boxed{\sf{\blue{1. \: Method \: of \: Elimination \: by \: equating \: the \: coefficient.}}}}} \\ [/tex]
In this method , we eliminate one of the two variables to obtain an equation in one variable which can easily be solved. By putting the value of this variable in any one of the given equations, the value of the other variable wil comes out.
[tex]\bigstar \: \large{\sf{\purple{Algorithm ( Steps )}}} \\ [/tex]
[tex]\bullet \: \: [/tex] Step 1 – Obtain the two equations.
[tex]\bullet \: \: [/tex] Step 2 – Multiply the equations so as to make the coefficients of the variables to be eliminated equal.
[tex]\bullet \: \: [/tex] Step 3 – Add or subtract the equations obtained in step 2 according as the terms having the same coefficients are of opposite or of the same sign.
[tex]\bullet \: \: [/tex] Step 4 – Solve the equation in one variable obtained in step 3.
[tex]\bullet \: \: [/tex] Step 5 – Substitute the value found in step 4 in any one of the given equations and find the value of the other variable.
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