# @gokuwow

##### Nishant Goyal 10 karma
###### Jobs & CareersBHEL ET 2012 Interview selected guys
@shailendra53 no.....but a guy who has got hyderabad got an acknowledgement mail that his acceptance letter has been received.
###### Jobs & CareersBHEL ET 2012 Interview selected guys
for Delhi people.....has anyone got acceptance letter acknowledgement mail?

###### Jobs & CareersBHEL ET 2012 Interview selected guys

for Delhi people.....has anyone got acceptance letter acknowledgement mail?

@yrathee hi ...i hav also got posting in delhi.....but how did u get to know that u got project management group...and did u get any acknowledgement for the acceptance letter...please reply asap.

Hi
I have a query related to the 60% eligibility criteria. I am from IIT-R and IITs dont give any conversion formula for converting CGPA into percentage. So what is the method followed for checking the eligibility of such candidates?

Hi
I have a query related to the 60% eligibility criteria. I am from IIT-R and IITs dont give any conversion formula for converting CGPA into percentage. So what is the method followed for checking the eligibility of such candidates?

###### CATTime speed, work, Alligations and mixtures problems
Amul and Cadbury run a 12km race around a circular track of circumference 750m. Amul can beat cadbury in a race of 2000 m by 500 m. If amul gives a headstart to cadbury by 500m then how many times amul overtake cadbury in the race?

ans is 4

let the speed of amul is "a" and cadbury is "b"
then 2000/a=1500/b
-> a/b=4/3. let a=4 and b=3 m/s. (it wont affect anything)

now cadbury starts 500 m ahead of amul so they meet at some point where amul has covered 2000m and cadbury 1500m. then remaining journey is 10000m. now they again meet when
4t-750n = 3t (n is an integer)
t = 750n
so they meet for the second time when t=750, and a covers 3000m and so on for the 3rd and 4th time. after that their 10000m completes and they cant meet again.
so ans=4.
• 1 Like
###### CATTime speed, work, Alligations and mixtures problems
Shop X prepares a mixture of curd, honey and sugar in ratio of5:4:1, Shop Y mixes them in ratio 5:6:4. A sourced the mixture in ratio 27:14 from these shops. In what ratio should she dilute it with curd , so that honey concentration is 25% finally?
3:2
3:7
7:3
5:6
2:3

sol: Ans is 2.66
volume of curd in mixture bought by A = 5*27/10 + 5*14/15 = 18.16
volume of honey in mixture = 4*27/10 + 6*14/15 = 16.4
volume of sugar in mixture = 1*27/10 + 4*14/15 = 6.43

total vol = 41

to make conc of honey 25% we have to make the volume = 16.4*4 = 65.6
Hence required amt of curd = 65.6-41 = 24.6
Hence the ratio = 65.6:24.6 = 2.66
• 1 Like
###### CATTime speed, work, Alligations and mixtures problems
i think there is some mistake in the solution, because by putting the value of y=sqrt(3)*x in eqn (1) and (2) and dividing them will give a ratio of the speed S1 and S2 and not the initial distance between A and B.

I think after getting this speed ratio using the further information of 6.66km distance we can find the value of x and y and hence will get the value of distance between A and B.

oops sorry i forgot to mention one equation ...
we also have: sqrt(x^2+y^2)=s1*t
x-6.66=s2*t
now divide these two eqns and equate the ratio s1/s2 to that obtained from the previous two eqns...u will get the same ans.
:):)
• 2 Likes
###### CATTime speed, work, Alligations and mixtures problems
Puys!

A abd B start walking towards Connaught Place from their respective positions, and move in straight roads with constant speeds. At the initial moment, A,B and Connaught place form a right triangle. After A travelled 30 km, the triangle became equilateral. When A arrived at connaught place, B still had to cover 6.66 km. to reach connaught place.
Find the initial distance between A and B.

(a)10 root 3
(b)12 root 3
(c)30 root 5
(d)20 root 3
(e) None of these

Plz give complete solution and time taken!

the ans is (d).
sol: let distance between A and B is y, and between B and CP is x. The triangle is right angled at B. Let A's speed is s1 and B's s2.
Now when A has traveled 30km , we have following relations;
30 = s1*t .......(1)
x-(sqrt(x^2+y^2)-30) = s2*t ........(2)
Also tan60 = y/x (since after A has traveled 30km, the triangle becomes an equilateral one, therefore angle ACB is 60)
we get: y=sqrt(3)*x
Put in (1) and (2) and divide them: u will get eqn; 2x^2-30x-200=0
solving it gives x=20
hance AB=y=20*sqrt(3)

• 1 Like