In an industry, the raw materials and the finished goods are transport

Question

In an industry, the raw materials and the finished goods are transported on the conveyor belt. There are two conveyor belt, one for carrying parts from P to point Q and another for carrying parts from R to point Q. P, Q and R in that order are in a straight line. Sometimes, the belt serves to transport cart, which can themselves move with respect to the belts. The two belts move at a speed of 0.5 m/s and the cart move at a uniform speed of 2 m/s with respect to the belts. A cart goes from point P to R and back to P taking a total of 64 s. Find the distance PR in meters. Assume that the time taken by the cart to turn back at R is negligible?

A. 48
B. 54
C. 60
D. 64
E. 72

A. 48
B. 54
C. 60
D. 64
E. 72

nick1816 · Answer

Distance between PR=x

Cart will travel one trip against the conveyor belt and one trip in the direction of conveyor belt.

\frac{x}{2+0.5}+ \frac{x}{2-0.5}= 64

x=60m

shivi23 · Answer

Can someone please explain this question.
from P to Q speed is 2.5 from Q to R speed is 1.5 while going P to R. On turning back R to Q speed is 2.5 and Q to P is 1.5. What do we do after this?

shah15 · Answer

Let  PQ = x  and  QR = y

Then,

PR = x + y

Equate it,

\frac{x}{2.5} + \frac{y}{2.5} + \frac{x}{1.5} + \frac{y}{1.5} = 64

1.5(x+y)  +  2.5(x+y)  =  64 (2.5 * 1.5)

4(x+y)  = 240
 x+y  = 60

PR = 60m

AbhinavSV · Answer

shivi23  You have already figured out the major part of the solution.
Let's try to visualize the problem

p.........................Q.....................R

Let's assume the distance PQ=QR=x

So from time=distance/speed
 (x/2.5+x/1.5)+(x/2.5+x/1.5)=64  (first part for going , second part for coming back)
after solving x=30m

So,  PR=PQ+QR=x+x=30+30=60

Fdambro294 · Answer

Conveyor belt is laid out as follows:

P———->>> Q <<<————R

In 64 seconds the cart travels from:

P to Q —- and Q to R

turns around and then from

R to Q—— and Q to P

Let distance from P to Q = D1

Let distance from Q to R = D2

Speed of cart = 2 m/s

Speed of belt = .5 m/s

When the Cart is moving with the Belt, the Belt is adding to the Cart’s Speed as a stream does for a boat.

Relative Speed of Cart WITH Belt = 2.5 m/s

When the Cart is moving against the Belt, the Belt is hindering the Cart’s Speed similar to when a Boat travels Upstream Against the current

Relative Speed of Cart AGAINST Belt = 1.5 m/s

Time = Distance / Relative Speed

[ (D1 / 2.5) + (D2 / 1.5) ] + [ (D2 / 2.5) + (D1 / 1.5) ] = 64 seconds

—Multiply both sides of equation by *(7.5) to remove Denominators—

3*(D1) + 5*(D2) + 3*(D2) + 5*(D1) = 480

—collecting like terms and diving each side by 8—

D1 + D2 = 60 meters = Distance from Point P to Point R

-C-

Posted from my mobile device

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In an industry, the raw materials and the finished goods are transport

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