Question

to cover a distance of 680 km Rakesh travelled for 4hours by car and five and half hours by train Raja covers the same distance by travelling 3 hour 20 minutes by car and 6 hours by train assuming a car has a speed of X kilometre by hour and a train has a speed of y kilometre by hour find x and y by the method of elimination​

Answer 1

Given :-

• Rakesh = 4 Hours by car + 5(1/2) Hours By Train
• Raja = 3 Hours , 20 Min. By car + 6 Hours By Train .
• Total Distance = 680km.
• Speed of Car = x km/h.
• Speed of Train = y km/h.

To Find :-

• value of x & y ?

Formula used :-

• Distance = Speed * Time .

Solution :-

Case ❶ :-

➻ Rakesh Time by Car = 4 Hours.

➻ Speed of Car = x km/h.

➻ Distance Cover by Rakesh by car= S * T = 4 * x = 4x km.

And,

➻ Rakesh Time by Train = 5(1/2) Hours = (11/2) Hours.

➻ Speed of Train = y km/h.

➻ Distance cover by Rakesh Train = S * T = (11/2) * y = (11y/2) km .

So,

➻ Total Distance = Distance by car + Distance by Train.

➻ 4x + (11y/2) = 680

Taking LCM

➻ 8x + 11y = 680 * 2

➻ 8x + 11y = 1360 -------------------- Equation (1).

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Case ❷ :-

➻ Raja Time by Car = 3 Hours + 20min = 3(20/60) = (10/3) Hours.

➻ Speed of Car = x km/h.

➻ Distance Cover by Rakesh by car= S * T = (10/3) * x = (10x/3) km.

And,

➻ Raja Time by Train = 6 Hours.

➻ Speed of Train = y km/h.

➻ Distance cover by Rakesh Train = S * T = 6 * y = 6y km .

So,

➻ Total Distance = Distance by car + Distance by Train.

➻ (10x/3) + 6y = 680

Taking LCM

➻ 10x + 18y = 680 * 3

➻ 10x + 18y = 680 * 3

➻ 2(5x + 9y) = 680 * 3

➻ 5x + 9y = 1020 -------------------- Equation (2).

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Now, Solving Equation (1) & (2) By Elimination Method:-

➼ 8x + 11y = 1360 -------------------- Equation (1).

➼ 5x + 9y = 1020 -------------------- Equation (2).

Multiply Equation (1) By 5 and Equation (2) by 8 and Than Subtracting Equation (2) From Equation (1) , we get,

➼ 5(8x + 11y) - 8(5x + 9y) = 5*1360 - 8*1020

➼ 40x - 40x + 55y - 72y = 6800 - 8160

➼ - 17y = - 1360

➼ y = (-1360)/(-17)

☛ y = 80 km/h. (Ans).

Putting This value of y in Equation (1) now, we get,

➼ 8x + 11*80 = 1360

➼ 8x = 1360 - 880

➼ 8x = 480

➼ x = (480/8)

☛ x = 60km/h. (Ans.)

Hence, Speed of Car is 60km/h & speed of Train is 80km/h.

Answer 2

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$\huge\tt{GIVEN:}$

• Rakesh travelled 680 km by different modes (4 hours by car,5½hours by train)
• Raja also travelled 680 km by different modes(3hour 20 min by car, 6 hours by train)
• Assuming car have x speed/hr
• Train have speed y/hr

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$\huge\tt{TO~FIND:}$

• Value of x and by (by elimination method)

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$\huge\tt{SOLUTION:}$

• Distance= Speed × Time

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In the case of Rakesh,

Time of car = 4 Hours

Speed of car = x km/hr

Time of train = 5½ Hours

Speed of train = ykm/hr

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Distance via car by Rakesh = S×T

↪4×X = 4x km

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Distance via train by Rakesh = S×T

↪11/2×y = (11y/2)km

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➡Total Distance = Distance covered by car + Distance covered by train

➡4x + (11y/2) = 680

➡8x+11y = 680×2

➡8x+11y = 1360______(EQ.1)

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In the case of Raja,

Time of car = (10/3) hours

Speed of car = x km/hr

Time of train = 6 hours

Speed of train = y km/hr

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Distance via car by Raja= S×T

↪(10/3)× x = (10x/3)km

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Distance via train by Raja= S×T

↪6×y = 6y km

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➡Total Distance = Distance covered by car + Distance covered by train

➡(10x/3)+6y = 680

➡10x + 18y = 680× 3

➡5 x + 9y = 1020 _______(EQ.2)

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Now,

Let's compare both equations-

↪5(8x+11y)-8(5x+9y)=5×1360-8×1020

↪40x-40x+55y-72y = 6800-8160

↪-17y = - 1360

↪y = (-1360)/(-17)

↪y = 80 km/hr

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Again,

Let's compare both equations-

↪8x+11×80 = 1360

↪8x = 1360-880

↪x = 480/8

↪x = 60km /hr

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$\huge\tt\purple{X= 60 km/hr}$

$\huge\tt\green{Y= 80 km/hr}$

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