gokuwow

@gokuwow

@shailendra53 no.....but a guy who has got hyderabad got an acknowledgement mail that his acceptance letter has been received.
@shailendra53 no.....but a guy who has got hyderabad got an acknowledgement mail that his acceptance letter has been received.
for Delhi people.....has anyone got acceptance letter acknowledgement mail?
for Delhi people.....has anyone got acceptance letter acknowledgement mail?

for Delhi people.....has anyone got acceptance letter acknowledgement mail?
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.
@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?
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?
> 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 please help me in solving this question ...
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

please help me in solving this question

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.
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> 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 ...
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
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> 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 6.6...
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.

Please let me know your opinion....

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.
:):)
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> 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 ...
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)

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