Question

que is from linear equation class 10 word problem

Answer 1

250/x + 120/y =4 
substitue 1/x with a & 1/y with b
250a+120b=4 
130/x + 140/y = 4.3 
130a+140b=4.3 
250 a + 120 b = 4 .............1
130 a + 140 b = 4.3 .............2
Eliminate y
multiply (1)by -7
Multiply (2) by 6
-1750 a -840 b = -28
780 a + 840 b = 25.8
Add the two equations
-970 a = -2.2
/ -970
a = 0.00227
plug value of a in (1)
250 a + 120 b = 4
0.57 + 120 b = 4
120 b = 4 -0.57
120 b = 3.43
b = 0.02861
x=1/a= 1/0.00227= 440 km/h-------------Train
y=1/b=1/0.02861= 35 km/h-------------- Car 

Answer 2

Let the speed of train and car be x km/h and y km/h.

A/q , Man travels total 370 km

For case 1 :- When he travels 250 km by train and rest ( 370-250=120 km ) by car , it takes 4 hours

$\sf \mapsto \: Total \: time \: taken \: = \dfrac{250}{x} + \dfrac{120}{y}$

$\sf \mapsto \dfrac{250}{x} + \dfrac{120}{y} = 4$

$\sf \mapsto \dfrac{250y + 120x}{xy} = 4$

$\sf \mapsto {250y + 120x} = 4xy \: \: \: \: ...(1)$

Case 2:- When he travels 130 km by train and rest ( 370-130=240) by car it takes 18 minutes more or 0.3 hours more , thus total hrs = 4.3

$\sf \mapsto \: Total \: time \: taken \: = \dfrac{130}{x} + \dfrac{240}{y}$

$\sf \mapsto \dfrac{130}{x} + \dfrac{240}{y} = 4.3$

$\sf \mapsto \dfrac{130y + 240x}{xy} = 4 .3$

$\sf \mapsto \: {130y + 240x} = {0.3xy} \: \: \: \: \: ...(2)$

Multiplying by 2 in (1) we get

$\maltese \: \sf \mapsto500y + 240x = 8xy...(3)$

Subtracting (2) from (3) we get ,

⇒370 y =3.7xy

⇒x= 100 km/h

Putting x= 100 in (1) we get

y = 80 km/hr

Thus, SPEED OF TRAIN = 100KM/H

AND SPEED OF CAR = 80KM/H

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