# 2 questions of Maths.

1. Nov 25, 2006

### ritwik06

1. Parag arrives at Pimpri Railway station from office every day at 6:00 p.m.
where Prajakta picks him up in the car to take him home. One day he arrives
at Pimpri one hour early and starts walking towards home. On the way he
meets Prajakta and they return home together in the car. If they arrive the
home 10 minutes earlier then what is the ratio of Parag's walking speed to
Prajakta's driving speed???

Work done by me on this question:
Let distance from station to home= d metres
Driving speed=v1 m/s
Walking speed=v2 m/s
Time taken by car usually=d/v1
according question (if x is the distance travelled by walking),
d/v1={(d-x)/v2}+(x/v1)+600
On solving,
{(d-x)/v1}-{(d-x)/v2}=600
But I cant find the ratio??????????????????????????????????

2.If a/(b+c) = b/(a+c) = c/(a+b) and a+b+c is not 0 then find value of each
ratio.
Work done:
The answer given in my book is -1. But I dont think that a+b+c will not be
equal to zero in that case???

This is all I could do. Please dont give me answers like show me your work etc. If you know it tell me otherwise dont.

2. Nov 25, 2006

### Office_Shredder

Staff Emeritus
For question one: If he walked a distance of d before getting picked up, then the car coming to pick him up didn't need to drive that distance d. It would have taken (presumably), the same time for the car to drive both ways, so it would have taken 5 minutes for the car to go that distance, since they got home a total of 10 minutes earlier (five minutes for the car to drive each way d meters)

3. Nov 25, 2006

### ritwik06

Thanx a lot. But please can you let me out while I myself am thinking about this. Please let me have a solution. Thnak you very very much!

4. Nov 25, 2006

### HallsofIvy

Where does that "600" come from? Since he arrived at the station 1 hour= 60 minutes earlier than usual but arrived at home 10 minutes earlier, the "walking and driving" time is 60-10= 50 minutes longer than the "driving only" time. In your terms,
$$\frac{d}{v_1}+ 50= \frac{d-x}{v_1}+ \frac{x}{v_2}$$
$$\frac{d}{v_1}+ 50= \frac{d}{v_1}- \frax{x}{v_1}+ \frac{x}{v_2}$$
$$\frac{x}{v_1}+ 50= \frac{x}{v_2}$$
Multiply both sides by $v_1$ and divide by x:
$$\frac{v_2}{v_1}= 1+ \frac{50v_1}{x}$$

Now use Office Shredder's point: Since the car arrived at home 10 minutes earlier than normal, it must have saved 5 minutes each way by not having to go that distance x:
$$\frac{x}{v_1}= 5$$
so
$$\frac{50v_1}{x}= \frac{50}{5}= 10[/itex] As for the second question: You are exactly correct. If each of the ratio's is equal to -1, then, for example, [tex]\frac{a}{b+c}= -1$$
so a= -b- c, a+ b+ c= 0. From
$$\frac{a}{b+c}= \frac{c}{a+b}$$
"cross multiplying" gives $a^2+ ab= bc+ c^2$
and from
$$\frac{a}{b+c}= \frac{b}{a+ c}$$
we get $a^2+ ac= b^2+ bc$
Subtracting $ab- ac= c^2- b^2$ or a(b-c)= (c-b)(c+b).
If b- c is not 0 we must have a= -(b+c) which would make each ratio -1, but that is exactly the "forbidden" case that a+ b+ c= 0. If b- c = 0, then, using the other equality, we get a= b= c, in which case each ratio is 1/2.

5. Nov 25, 2006

### ritwik06

I really cant find words to thank you. Thanks a lot!