Advertisements
Advertisements
Question
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,
`"speed"="distance travelled"/"time taken to tavelled that distance"`
`x = d/t`
Or, d = xt (i)
Using the information given in the question, we obtain
`(x + 10) = d/(t-2)`
(x + 1)(t - 2) = d
xt + 10t - 2x - 20 = d
By using equation (i), we obtain
− 2x + 10t = 20 (ii)
`(x- 10) = d/(t+3)`
(x-10)(t+3) = d
xt - 10t + 3x - 30 = d
By using equation (i), we obtain
3x − 10t = 30 (iii)
Adding equations (ii) and (iii), we obtain
x = 50
Using equation (ii), we obtain
(−2) × (50) + 10t = 20
−100 + 10t = 20
10t = 120
t = 12 hours
From equation (i), we obtain
Distance to travel = d = xt
= 50 × 12
= 600 km
Hence, the distance covered by the train is 600 km.
APPEARS IN
RELATED QUESTIONS
Solve the following pairs of equations by reducing them to a pair of linear equations
`1/(2x) + 1/(3y) = 2`
`1/(3x) + 1/(2y) = 13/6`
Solve the following pairs of equations by reducing them to a pair of linear equations
`10/(x+y) + 2/(x-y) = 4`
`15/(x+y) - 5/(x-y) = -2`
Formulate the following problems as a pair of equations, and hence find their solutions:
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
One says, "Give me a hundred, friend! I shall then become twice as rich as you". The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II)
[Hint: x + 100 = 2 (y − 100), y + 10 = 6(x − 10)]
Solve the following pair of linear equations: px + qy = p − q, qx − py = p + q
Solve the following for x:
`1/(2a+b+2x)=1/(2a)+1/b+1/(2x)`
A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you." B replies, "if you give me 10, I will have thrice as many as left with you." How many mangoes does each have?
A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes `6/5`. Thus, we have
`(2x)/(y-5)=6/5`
`⇒ 10x=6(y-5)`
`⇒ 10x=6y-30`
`⇒ 10x-6y+30 =0`
`⇒ 2(5x-3y+15)=0`
`⇒ 5x - 3y+15=0`
If the denominator is doubled and the numerator is increased by 8, the fraction becomes `2/5`. Thus, we have
`(x+8)/(2y)=2/5`
`⇒ 5(x+8)=4y`
`⇒ 5x+40=4y`
`⇒ 5x-4y+40=0`
So, we have two equations
`5x-3y+15=0`
`5x-4y+40=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-3)xx40-(-4)xx15)=-y/(5xx40-5xx15)=1/(5xx(-4)-5xx(-3))`
`⇒ x/(-120+60)=(-y)/(200-75)=1/(-20+15)`
`⇒x/(-60)=-y/125``=1/-5`
`⇒ x= 60/5,y=125/5`
`⇒ x=12,y=25`
Hence, the fraction is `12/25`
Asha has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively.
