akhilmisra

@akhilmisra

abhpr
abhigya1 .... @abhpr
please suggest some good books for anthro as optional? n which coaching material is best?brilliant?vaid ias?
akhilmisra
Akhil @akhilmisra
I think books are enough for anthro.... you can refer to coaching material only for those few topics for which you do not get enough theory from books.

For books you should refer the below
P Nath - Physical anthro
Indian anthro - Nadeem hasain
Anthropological theories - Makhan jha.

You should first refer these standard books, they are good enough for initial stage of preparation.

Let me know if you need any more help.
abhpr
abhigya1 .... @abhpr
can anyone tell me which optional is good between anthro n socio?i had bio in 12.... i like socio also bt kind of afraid that i wont be able to write in "flowery language" n wont be able to :lookround: score good in socio. anyone with anthro as optional?share ur strategy and plan.. ty
akhilmisra
Akhil @akhilmisra
I m going with Anthro. I find it very interesting moreover it can be very helpful for essay paper too. As mentioned by Guns.Ablaze paper one is quite a lot of biology while paper 2 is India centric.
You should check the syllabus and then decide. Best of luck.
Hi Everyone,
I believe that every regular and serious member should post a couple of questions daily, focused on prelims.... this will lead to nice quiz en route to the final exams. We still have some days to go, few questions a day will certainly do a world of good to preparations...

My first..

Q. What is the pH level of blood of normal person?
(a) 4.5-4.5 (b) 6.45-6.55
(c) 7.35-7.45 (d) 8.25-8.35
i think its none..
10(x-3)=9x+p....p=age of younger guy..
solving it,we get x=30+p..for different values of x..we get different p..
am i correct..?


here what mistake u are doin according to me is that u are considering the age of the removed older person as 'x' .... but we do not know whether his age is actually 'x' as x is just the average.

for me it should be,

10(x-3)=10x - older + younger
10(x-3)=10x - difference
difference = 30
amitshanky Says
you are correct can you please explain 2nd one:)


let the average age of the original 10 people currently be 'x+3'
so their average age three yrs before would have been 'x'

x = ((x+3)*10 - older + younger) / 10
x = ((x+3)*10 - difference) / 10

difference = 30

pls someone cross check....
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1)the average weight of n boys in a group is 36 kg.if 20 other boys whose average weight is 30 kg join the group,the avg weight of the group would be the same as what it would be if 5 boys whose average weight is 40 kg leave the group.find N:-
10,15,30,25



forming the equantions,
total weight of n boys = 36n

(36n + 20*30) / (n+20) = (36n - 40*5) / (n-5)

420n - 3000 = 520n - 4000
n=10
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A and B have some guavas divided among themselves. A says to B "If i give 25% of the guava I have, I will still have 2 more guavas than u have." To this B says, If u give me guavas equal to 70% of what I have now, I will have 4 more guavas than u have". What is the total number of guavas that they have?

80
64
36
88

A has x
B has y

I am bit confused.
Stmt1-> If i give 25% of the guava I have, I will still have 2 more guavas than u have.

3x/4 = y + 2
3x = 4y + 8

Had it been this way
Stmt1-> If i give 25% of the guava I have, I will still have 2 more guavas than u will have.

3x/4 = y + 2 + x /4
x = 2y + 4 ------

Stmt2-> If u give me guavas equal to 70% of what I have now, I will have 4 more guavas than u have

1.7y = x + 4

Now acc to me eqs are :
3x = 4y + 8 => 30x = 40y + 80 ---eq1
17y =10 x + 40=>51y = 30x + 120 ---eq2

51y = 40y + 200
11y = 200
y = 18.18 ~ 18
x = 88/3 = 29.7
my ans 48

Where am i going wrong ???


I feel the eqn 2 that u form i.e. "1.7y = x + 4" her u have added 0.7y to B's gauvas but according to me since they are being given by A so they need to be reduced from the number of gauvas A has currently. Which would give
"x- 0.7y"

so eqn should be
1.7y = x + 4 - 0.7y

P.S. => I have edited my second eqation in the original post because I wrote " 4 + x+ 7y/10" instead of 4 + x- 7y/10...... by mistake
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A locomotive engine runs at a speed of 50 kmph when no compartment is attached to it. For every new compartment that is attached to it, the speed of the train is reduced by 10% of the earlier speed. At most, how many compartments can be attached so that the train can cover a distance of 180 km in a maximum of 8 hours?
7
5
6
8



to cover 180 km in maximum 8 hrs the speed should be > (180/km/hr
i.e. speed should be greater than 22.5 km/hr

for every compartment the speed is reduced by 10% so we have,

speed after 'n' compartments are connected = 50* (1 - 1/10)^n

therefore by the above condition we have,

50* (1 - 1/10)^n > 22.5

(1 - 1/10)^n > 9/20

now checking, we get

max n=7

hope I have done it correctly.....
Found the problem confusing.

A and B have some guavas divided among themselves. A says to B "If i give 25% of the guava I have, I will still have 2 more guavas than u have." To this B says, If u give me guavas equal to 70% of what I have now, I will have 4 more guavas than u have". What is the total number of guavas that they have?

80
64
36
88


let A's be ='x' and B's be 'Y'

if A gives B 25% of his gauvas,
x - x/4 =y+ x/4 +2
2x - 4y =8 ..................(1)

if A gives 70% of what B has now i.e. y , B will have 4 more gauvas

4 + x - 7y/10 = y + 7y/10

10x - 24y =-40 .............(2)

solving (1) and (2), we get....
y = 20 and x =44
therefore total gauvas = 64.
Its you who has to change it all,
and step forward to fight,
Its you who has to reinstate faith,
and truth and the holy light;

cheers....
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