(Car)
QUADRATIC �EQUATIONS
x Km/hr
What do we have to find ?
Sol.
Let the speed of the car be x km/hr
∴ New speed of the car = (x + 20)km/hr
As per the given condition,
–
= 2
For example if the old time taken = 10 hrs
What will be the new time taken ?
New time taken =
= 8 hrs
10 – 2
240
x
240
x + 20
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240 Km
(x+20) Km/hr
240 Km
DISTANCE
SPEED
TIME=
DISTANCE SPEED
Old
New
x km/hr
240 km
240
x + 20 km/hr
240 km
240
(Q) A car covers a distance of 240km with some speed. If its speed is increased by 20 km/hr, it will cover the same distance in 2 hours less. Find the speed of the car.
Old time taken – New time taken = 2
x
hrs
x + 20
hrs
(Q) A car covers a distance of 240km with some speed. If its speed is increased by 20 km/hr, it will cover the same distance in 2 hours less. Find the speed of the car.
Sol.
Let the speed of the car be x km/hr
∴ New speed of the car = (x + 20)km/hr
As per the given condition,
–
= 2
240
x
240
x + 20
+ 20x
– 2400
= 0
(Q) A car covers a distance of 240km with some speed. If its speed is increased by 20 km/hr, it will cover the same distance in 2 hours less. Find the speed of the car.
Sol.
Let the speed of the car be x km/hr
∴ New speed of the car = (x + 20)km/hr
As per the given condition,
240
x
240
x + 20
–
= 2
1
240
–
1
x +20
=
2
x + 20
x (x + 20)
=
2
120
20
x2
1
240
∴
=
20 (120)
2400 = x2 + 20x
∴
∴
∴
– x2
∴
x2
x2
x
∴
∴
∴
(x + 60)
x + 60 = 0
x = – 60
∴
∴
∴
The speed of car can never be negative
∴
x ≠ – 60
Hence x = 40
The original speed of car is 40 km/hr.
∴
6
4
24
+
–
40
–
= 20
60
∴
– x
(x – 40) = 0
– 40
+ 60x
– 40x
– 2400
= 0
Since we are subtracting the factors give middle term sign to the bigger factor and the opposite sign to the smaller factor
= 1 (x2 + 20x)
0
0
00
∴
1
120
+ 20x
Find two factors of 2400 in such a way that by subtracting factors we get middle no. 20
(x + 60)
= 0
(x + 60)
or x – 40 = 0
or x = 40
– 20x
+ 2400
= 0
Multiplying throughout by – 1