Question

please solve question no 4 and 5 by step by step explain ​

Answer 1

Answer:                                                  C                                   Q                                                                

1. A------------------------------------------------------------B-----------------------

                                           100Km

Let the speed of the car A be xKm/hr

Let the speed of the car B be yKm/hr.

1st case: let the cars meet at point C.

We know that $speed=\frac{distance}{time}$.

Distance travelled by CarA = 100-BC.

Distance travelled by CarB=BC.

Time=5 hours

SpeedA+speedB=100-BC+BC/5

x+y=20...................(1)

2nd case:Let the cars meet at point Q.

Again, following the same method:

Distance travelled by CarB=BQ

Distance travelled by CarA=100+BQ

Time=1 hours.

To eliminate BQ, we will now substract The speeds of the car.

SpeedA+SpeedB=100+BQ-BQ/1

x-y=100..............................(2)

Add (1) and (2)

you will get 2x=120km/hr------>x=60km/hr

Substract (1) and (2)

you will get 2y=-80

y=-40km/hr.

Since speed is a scalar quantity, direction is not involved, thus we can ignore the negative sign.

x=60km/hr

y=40km/hr

2.If lenght is x and y, area=xy

Let the area, lenght and breadth be xy, x, y respectively.

If length is reduced by 5 units and the breadth is increases by 3 units, then area is reduced by 9 square units.

∴xy−9=(x−5)(y+3)

⇒xy−9=xy+3x−5y−15

⇒3x−5y−6=0 ...(1)

When length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq. units.

∴xy+67=(x+3)(y+2)

⇒xy+67=xy+2x+3y+6

⇒2x+3y−61=0 ...(2)

lets eliminate x. for that we multiply 2 on the (1) equation and 3 on the (2) equation.

Using elimination in (1) and (2),

2(3x−5y−6=0)= 6x-10y-12=0.................(3)

3(2x+3y−61=0)=6x+9y-183=0.......................(4)

Substract (3) and (4).

You will get y=19units.

Substitute the value of y in (1) or (2) equation. you will get x=17units.

HOPE THIS HELPS :D