Konda bhai is the answer for this 4752....

i mean f(94)=4752

JUs check if you are getting the same...



we have f(20)+f(19) = 20^2
f(21)+f(20) = 21^2 and so on
f(94)+f(93) = 94^2


2-1 = f(21)-f(19) = 21^2-20^2
3-(2-1) = f(22) + f(19) = 22^2-21^2+20^2

hence f(94)+f(19) = 94^2-93^2+92^2........+20^2

f(94) = 94^2-93^2+92^2........+20^2 - 94

deoarshi Says

its simple put x= (x+503)/(x-1) , and two equations will be formed ..the answer wud be 501

yes the answer is 501...

could be done saisly by putting x=2 first
and then x=505
whixh will result in 2 equations in terms of f(2) and f(505) with interchanges coeficient of 1 and 2

solving it gives f(505)= 501

i am getting very big answer for this . answer wud be something like (94^2)-(93^2)+(92^2)+ ..........-f(0)
value of f(0) is (19^2)-(18^2) +(17^2) -(16^2) +.......1^2

Konda bhai is the answer for this 4752....

i mean f(94)=4752

answer wud be -250/3 start substituting N=0,1,2,3 and add all we will get
3= -250

ok..

its correct ..the answer is -250/3

its simple put x= (x+503)/(x-1) , and two equations will be formed ..the answer wud be 501

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i am getting very big answer for this . answer wud be something like (94^2)-(93^2)+(92^2)+ ..........-f(0)
value of f(0) is (19^2)-(18^2) +(17^2) -(16^2) +.......1^2

=====

answer wud be -250/3 start substituting N=0,1,2,3 and add all we will get
3= -250

===





Hello bhai, I solved the fifth one because that seemed to be pretty easy. I think the number a= 12. So you have the sum of digits as 3. Cross check, f(12)=11, f(11)=22,f(22)=21. So substitute in the original equation, It satisfies. Hope the answer is correct.

ques

in a circle with chord ab we extend it to points c and d such that ac=bd.
frm c and d tangents are drawn on opposit side of chord ab touching at p and q.
in wat ratio pq divides ab.
1:1
1:2
2:1
2:3

If we consider the chord ab as the diameter and draw the figure, then we see that pq is also another diameter of the circle. Hence the ratio has to be 1:1. Whats the answer anyways?

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COPY PASTING FROM ANOTHER SITE>>>>>>>>>>>>>>>There nobody solved it...

QUANT GODS SOLVE THIS...............


A,B,C,D ,E are Sections of a class

Xstandard A(55) B (26) C (38 D (37) E(48

what is the max no. ofstudents who expected exactly 4 out of 5 sections?????????

X11 standard A(47) B (15) C (41) D (28 E(25)

how many in X11 standard expected a minimum of two of the five sections ????

**********Maximum no. of students in a section is 100 and one student can be a part of more than one section...
Waiting for solution..........

Please gve the detailed approach.........



ESH BHAI EACH SECTION HAS MAXIMUM of 100 students but I think U Misunderstood as collectively they have 100 students

TRY AGAIN Sir ji.............

issue clarified

AC and BD are diagonals
AC can be found ut by considering triangle ABC which is right angled..
so AC=60
similarly BD can befound out by applying pyth triplet in triangle BDC
so BD=52
now apply ptolemy's theorem
AB.DC+AD.BC=AC.BD
we get AD=33

is the bold part true for any quadrilateral???
pls let me know puys

@OperationBLACKI, (is it okay..or what shd I call u ) If b is B schools evaluated on exactly 5 parameters, then dont you think the total no of b-schools which fulfill atleast 2 parameters should be 5b and not b. Coz all 5b schools satisfy "fulfill atleast 2 parameters" and hence all of them should be counted.

Same logic holds for all the cases that u have worked out. Pls chk and see.

@ harshavardhanab .......I think we need to consider b as exactly 5 parameters . Simply because in the above case we have got the value of b as 26 and as per ur understanding if we were to take 5b as exactly 5 parameters we would get more than 100 which is not possible . We use the coefficient 5 because we end up adding the same value 'b' 5 times when we sum up all the values . So a would stand for exactly one and b would mean exactly 5 . Hope I have reduced ur confusion

consider the diameter as chord(ab)
& o as centre
then op is perpendicular to pc & oq is perpendicular to dq
cop is congruent to doq (sas)
hence pc parrallel to qd
op is perpendicular to cd & oq is perpendicular to dq
this can only possible when op & oq form a straight line &
ie o is the centre which bisect the diameter ab
hence answer 1:1
plz. confirm

This is what even I had got 1:1 . Experts @ PG please clarify

Please help me solving the below mentioned problem.

1) on their birthday , Ram and shyam , a pair of twins , were gifted a large packet of chocolates containing chocolates of 4 different flavours . they started eating the chocolates alternately . one choco at a time. what should be the mimimum total number of chocolates that both of them together should have eaten before you can say that at least one of them has certainly eaten atleast 4 chocolates of a single flavour ?

My take

The worst case scenario would require them to eat 25 chocolates together
The best case scenario would require them to eat 7 chocolate together ..

Answer for the second would be as follows :
Maximum value of x^3*y^4 would occur when x/3=y/4 i.e x= 3y/4 .

7(3y/4) +5y = 7 , So 41y =28 which means y= 28/41 and x= 21/41

So x^3.y^4 = 28^3*21^4/41^7

hey operation bhai
how did u arrive to this approach
can elaborate? how x/3=y/4 for max value

ques

in a circle with chord ab we extend it to points c and d such that ac=bd.
frm c and d tangents are drawn on opposit side of chord ab touching at p and q.
in wat ratio pq divides ab.
1:1
1:2
2:1
2:3

consider the diameter as chord(ab)
& o as centre
then op is perpendicular to pc & oq is perpendicular to dq
cop is congruent to doq (sas)
hence pc parrallel to qd
op is perpendicular to cd & oq is perpendicular to dq
this can only possible when op & oq form a straight line &
ie o is the centre which bisect the diameter ab
hence answer 1:1
plz. confirm

ques

in a circle with chord ab we extend it to points c and d such that ac=bd.
frm c and d tangents are drawn on opposit side of chord ab touching at p and q.
in wat ratio pq divides ab.
1:1
1:2
2:1
2:3

can you give the figure??

ques

in a circle with chord ab we extend it to points c and d such that ac=bd.
frm c and d tangents are drawn on opposit side of chord ab touching at p and q.
in wat ratio pq divides ab.
1:1
1:2
2:1
2:3

What we need is Minimum 4 chocolates of 1 flavor.
So, worst case scenario : 3 chocolates of each flavour = 3 * 4 = 12.
2 people eating one-by-one = 2 * 12 = 24.
Then, when the 25th chocolate is consumed, we can say for sure that one of them has had 4 chocolates of a particular flavor. Krsh.vik's answer is correct.

Please help me solving the below mentioned problem.

1) on their birthday , Ram and shyam , a pair of twins , were gifted a large packet of chocolates containing chocolates of 4 different flavours . they started eating the chocolates alternately . one choco at a time. what should be the mimimum total number of chocolates that both of them together should have eaten before you can say that at least one of them has certainly eaten atleast 4 chocolates of a single flavour ?

uff.. and all this time i misunderstood the question
i think the final answer is 7

We need not know the number of chocolates ... The worst possible case in which a person will not eat four of any variety is 3 of each type .. the moment he eats the next one we are pretty sure that 4 of atleast one type are eaten ...
Hence 13 should be eaten by the first person ... meanwhile the second person also ate 12 chocs ...

But buddy thats the maximum choco case. The min should be 3 + 4 only.(When both of then eat the same type of choco). Otherwise the ques needs to be re framed.
Please correct me if I am wrong

@krish
hey u r considering thw worst case.. but the question is wat is the minimum number of total chocolates.. so u must consider the best case..
so in my view the answer shud be:3+4=7

out of the four chocs ... the person can also eat one of each type ... we can't be certain in ur case

@krish
hey u r considering thw worst case.. but the question is wat is the minimum number of total chocolates.. so u must consider the best case..
so in my view the answer shud be:3+4=7

krsh bhai.. without knowing the total no of chocolates, how can we do this problem? if there are some 1000 chocolates with 250 of each type?
I am not clear.. pls enlighten...

We need not know the number of chocolates ... The worst possible case in which a person will not eat four of any variety is 3 of each type .. the moment he eats the next one we are pretty sure that 4 of atleast one type are eaten ...
Hence 13 should be eaten by the first person ... meanwhile the second person also ate 12 chocs ...

krsh.vik Says

13+12 = 25

krsh bhai.. without knowing the total no of chocolates, how can we do this problem? if there are some 1000 chocolates with 250 of each type?
I am not clear.. pls enlighten...

krsh.vik Says

13+12 = 25

krish bhai..
pllzzz explain

Please help me solving the below mentioned problem.

1) on their birthday , Ram and shyam , a pair of twins , were gifted a large packet of chocolates containing chocolates of 4 different flavours . they started eating the chocolates alternately . one choco at a time. what should be the mimimum total number of chocolates that both of them together should have eaten before you can say that at least one of them has certainly eaten atleast 4 chocolates of a single flavour ?

13+12 = 25

Please help me solving the below mentioned problem.

1) on their birthday , Ram and shyam , a pair of twins , were gifted a large packet of chocolates containing chocolates of 4 different flavours . they started eating the chocolates alternately . one choco at a time. what should be the mimimum total number of chocolates that both of them together should have eaten before you can say that at least one of them has certainly eaten atleast 4 chocolates of a single flavour ?

Kaise karte ho bhai? Baat to sach hai aapki. Approach? Or trial and error?



Check my answers a few posts before your's. Aapka answer galat hai. Use, A.M. >= G.M. You will get a still greater value than this. :)

I have very basic(or call it stupid) question...did we prove that x and y are positive numbers..before applying A.M. >= G.M..that is a necessary condition right?? can we assume it and go ahead in these type of problems..

Apologies Esh bhai,

Maximum no. of students in a section is 100 and one student be part of more than one section...
Waiting for solution..........

@junoon bhai... I will explain in more detail why I took all those a,b,c,d,e with a small example.. consider the diagram where there are just 2 parameters.
7130
Now as per the definition,
let a be the number of students satisfying one criteria
let b be the number of students satisfying 2 criteria
now from the diagram, a = (x-z)+(y-z)
b = z
what we know is X, Y (similar to section A, section B in the problem)
also we know the total sample space(i.e. no of students), which here is
X+Y-2z+z = X+Y-z = a+b (as per our defn)...(1)
Now consider X+Y = x-z+y-z+2z = a+2b ......(2)
These are the 2 eqns for 2 different attributes which I had written for the 5 attributes...

I have given the soln for the first part, for the second one,

a+b+c+d+e =100
a+2b+3c+4d+5e = 176
b+2c+3d+4e = 76
we need to minimise b+2c+3d+4e = 76
so b=c=d=0, 4e=76, minimum no of students = 19
Hope this is correct..

Qns are always welcome... but i cant assure a ans for all .. but fellow puys will def correct us..

OperationBLACKI Says

Monil bhai there's another combination 7!*6!= 10! ( 7! *8*9*10) gives 10!

Kaise karte ho bhai? Baat to sach hai aapki. Approach? Or trial and error?

Answer for the second would be as follows :
Maximum value of x^3*y^4 would occur when x/3=y/4 i.e x= 3y/4 .

7(3y/4) +5y = 7 , So 41y =28 which means y= 28/41 and x= 21/41

So x^3.y^4 = 28^3*21^4/41^7

Check my answers a few posts before your's. Aapka answer galat hai. Use, A.M. >= G.M. You will get a still greater value than this.

Hi Harsh, esh.nil....

Thanks for the answer given. But still, I cannot understand the logic followed. Can any of you please explain it in a more lucid manner...Plse

@abhrabumba, plz go thru previous posts and let me know which part u didnt understand

Hey guys,

Can someone here please guide me about the topics i need to cover from 11th standard?

Since i had maths in 11th standard, but have lost touch with it after 6 years.......

So if someone can just guide me as

1) Which topics need full practise from 11th Standard R.D.Sharma?

2) Which topics just require me to mug up the formulae.....?
(I am specially confused about Binomial theorem,logarithmic series & Exponential Series....though Special Sequences now can't be left since in the last 2 yrs. CAT papers we had questions from this part)

3) Do i need to cover Ellipses,Hyperbola...etc......?Do we get questions from these topics?
How much Trigo....sould i cover.....?

I think as far as CAT is concerned, you need not bother about conic sections and higher maths. The CAT paper is designed for anyone who has done basic mathematics.

Sequences and series will form an important part of your preparation.Make sure you are comfortable with AP GP and binomial theorem. You can leave logarithmic and exponential series safely.

Also if your query is concerning JMET, please post it in the JMET thread

Cheers

Hey guys,

Can someone here please guide me about the topics i need to cover from 11th standard?

Since i had maths in 11th standard, but have lost touch with it after 6 years.......

So if someone can just guide me as

1) Which topics need full practise from 11th Standard R.D.Sharma?

2) Which topics just require me to mug up the formulae.....?
(I am specially confused about Binomial theorem,logarithmic series & Exponential Series....though Special Sequences now can't be left since in the last 2 yrs. CAT papers we had questions from this part)

3) Do i need to cover Ellipses,Hyperbola...etc......?Do we get questions from these topics?
How much Trigo....sould i cover.....?

Esh,

the formula is (2p+1)(2q+1)(2r+1) ...so on...

DIdn't you see it?.....I already posted before somewhere...

Yup, u r right. we can actually get that ans from P&C.;
Let LCM be a^p*b^q*c^r. For 2 numbers to have LCM as this then atleast one of them should have highest powers of a,b,c. Let 2 nos be k1 and k2.

k1 = a^p1*b^q1*c^r1..... p1<=p, q1<=q, r1<=r
k2 = a^p2*b^q2*c^r2..... p2<=p, q2<=q, r2<=r.

In order for these nos to have required LCM we need atleast p1 or p2 to be p, q1 or q2 to be q and so on.

if p1 = p, then p2 can take p values .i.e., we can have combinations, (a^p,a^0), (a^p,a^1)..so on till (a^p,a^p-1)
possibilities = p.
Similarly, if p2 = p, then p1 can take p values.
possibilities = p.
If p1=p and p2=p, then possibilities = 1.
Total possibilities is 2p+1.

Similarly for b,c we get 2q+1 and 2r+1.

Now grand total = (2p+1)(2q+1)(2r+1)

Esh,

the formula is (2p+1)(2q+1)(2r+1) ...so on...

DIdn't you see it?.....I already posted before somewhere...

@konda bhai... Missed the post.. thanks for the formula,
but 2((p+1)(q+1)(r+1)+(p+q+r))-1 = (2p+1)(2q+1)(2r+1)

@junnon bhai.. AIMCAT famous problem..
let a-- no of students in one section
b- no of students in 2 sections
c - 3 sections
d - 4 sections and
e - 5 sections
given a+b+c+d+e = 100
a+2b+3c+4d+5e = 204
subtracting, b+2c+3d+4e = 104
max no of students in exactly 4 sections,
to maximise d, we will have b=c=e = 0,
so the answer is max no = 104/4 = 26 students.
Is that right konda bhai???
if this is right then u can proceed in similar lines for the other problems as well...
pls revert in case of qns..

Is the answer for this question is 3^3.4^4/7^3.5^4??

As AM>=GM so,

/3+4>=(7/3)^3.(5/4)^4x^3.y^4
And ,LHS will be 7 thus giving max value of x^3.y^4 equals 3^3.4^4/7^3.5^4

correct me if i m wrong.

yep sachin
u r rite its the correct answer...

@Konda and Monil... for the problems where we are needed to find all sets of numbers which give a lcm of a certain number, i tried some approach and looks like this can be generalised and used until there is some better approach posted..
now let N = 12 = 2^2*3
now the no of factors of 12 = 3*2 = 6
all these 6 factors will give the LCM 12 with the number 12,
apart from these we have the co-primes, we will take one at a time, also to get the LCM, we need the highest power of the factor in the number,
so (2^2,3) also (2^2,6)
so the total un-ordered pairs of nos = 6+2 = 8
If the ordered pairs are asked then = 8*2-1 =15 {(12,12) counted twice)

let us consider for some bigger no, n =60 = 2^2*3*5
no that have lcm 60 wit 60 = 3*2*2 = 12
co-primes: (2^2*15), -- 2 terms
(12,5) -- 1 term
(20,3) -- 1 term
no of un-ordered terms --- 16
no of ordered ----- 31 (excluding (60,60) which is counted twice)

for the given problem,2^3.5^7.11^13
no of factors that have lcm as 2^3.5^7.11^13 with 2^3.5^7.11^13 = 4*8*14 = 448
co-primes: (2^3, 5^7.11^13) -- 3 terms
(5^7, 2^3*11^13) ---- 7 terms
(11^13, 2^3*5^7) ---- 11 terms
so un-ordered terms = 448+21 = 469 terms...
ordered pairs = 937 pairs
Is this answer correct?? I think it is right..

I think u can identify a pattern in all these questions, and if u need to generalise using a formula, then it is, (for n=a^p.b^q.c^r)
(p+1)(q+1)(r+1) +(p+q+r)...for un-ordered pairs
I hate formulas...The more I know the less confident I am... (may be my next signature.. )

Esh,

the formula is (2p+1)(2q+1)(2r+1) ...so on...

DIdn't you see it?.....I already posted before somewhere...

@Konda and Monil... for the problems where we are needed to find all sets of numbers which give a lcm of a certain number, i tried some approach and looks like this can be generalised and used until there is some better approach posted..
now let N = 12 = 2^2*3
now the no of factors of 12 = 3*2 = 6
all these 6 factors will give the LCM 12 with the number 12,
apart from these we have the co-primes, we will take one at a time, also to get the LCM, we need the highest power of the factor in the number,
so (2^2,3) also (2^2,6)
so the total un-ordered pairs of nos = 6+2 = 8
If the ordered pairs are asked then = 8*2-1 =15 {(12,12) counted twice)

let us consider for some bigger no, n =60 = 2^2*3*5
no that have lcm 60 wit 60 = 3*2*2 = 12
co-primes: (2^2*15), -- 2 terms
(12,5) -- 1 term
(20,3) -- 1 term
no of un-ordered terms --- 16
no of ordered ----- 31 (excluding (60,60) which is counted twice)

for the given problem,2^3.5^7.11^13
no of factors that have lcm as 2^3.5^7.11^13 with 2^3.5^7.11^13 = 4*8*14 = 448
co-primes: (2^3, 5^7.11^13) -- 3 terms
(5^7, 2^3*11^13) ---- 7 terms
(11^13, 2^3*5^7) ---- 11 terms
so un-ordered terms = 448+21 = 469 terms...
ordered pairs = 937 pairs
Is this answer correct?? I think it is right..

I think u can identify a pattern in all these questions, and if u need to generalise using a formula, then it is, (for n=a^p.b^q.c^r)
(p+1)(q+1)(r+1) +(p+q+r)...for un-ordered pairs
I hate formulas...The more I know the less confident I am... (may be my next signature.. )

Answer :

1 ) Minimum of atleast 2 would be calculated as follows : a+5b= 152 and a+b=100 , where a = B schools evaluated on exactly one parameter and b = B schools evaluated on exactly 5 parameters . So 4b =52 , b =13 . The minimum number of B schools which fulfill atleast 2 parameters is 13

To maximise we use a+2b = 152 and a+b = 100 where b = B schools evaluated on exactly 2 parameters . b = 52 . The MAXIMUM number of B schools which fulfill atleast 2 parameters is 52


2) Minimum of atleast three of the five parameters would be 13 as done above
3) Maximum of atleast three of the five parameters would be a+3b=152 , a+b= 100
2b=52 , So b be would be 26
Correct me if I am wrong

@OperationBLACKI, (is it okay..or what shd I call u ) If b is B schools evaluated on exactly 5 parameters, then dont you think the total no of b-schools which fulfill atleast 2 parameters should be 5b and not b. Coz all 5b schools satisfy "fulfill atleast 2 parameters" and hence all of them should be counted.

Same logic holds for all the cases that u have worked out. Pls chk and see.

esh.nil Says

@junoon bhai... what is the total number of students in the class in class X and XII?? and can one student be part of more than one section? that is also not mentioned.. does something need to be added to the qn or am I .. the latter I am sure, but please confirm..

Apologies Esh bhai,

Maximum no. of students in a section is 100 and one student be part of more than one section...
Waiting for solution..........

solve this
A,B,C,D ,E are Sections of a class

Xstandard A(55) B (26) C (3 D (37) E(4

what is the max no. ofstudents who expected exactly 4 out of 5 sections?????????

X11 standard A(47) B (15) C (41) D (2 E(25)

how many in X11 standard expected a minimum of two of the five sections ????

Please gve the detailed approach.............

Juno Bhai,

Is that all given in the question? Question is not clear...plz make it clearer.

Hi puys,
Yup I too think the answer is 20s

Every 2s, the Car1 moves 100 cm and car 2 moves 50cm behind, the gap being 500 cm to start of with, at the end of time

2s 450 cm
4s 400 cm
.
.
.
20s 0cm.

Hence 20s is the answer.
@kushal_gupta, @sachin2807 and @shradhagarg...I concur.

Regards,
Naren.

Sorry to post late...But the answer should be 20 only, because in 4 secs the cars cover 100 cm. So they ll cover 500 cm in 20 sec.

this question have been posted in other thread .
There are 5 parameters A,B,C,D,E on which B.schools are being evaluated
There are 100 business schools in all.
the following additional info is given
parameter --------------no. of B.schools fullfilling the paramter
A----------------------------------------------34
B----------------------------------------------25
C----------------------------------------------40
D----------------------------------------------33
E----------------------------------------------20
Find the minimum number of B.schools which fullfill atleast two of the five parameters?
1)Find the minimum(MAXIMUM) number of B.schools which fullfill atleast two of the five parameters?
2)Find the minimum number of B.schools which fullfill atleast three of the five parameters?
3)Find the (MAXIMUM) number of B.schools which fullfill atleast three of the five parameters?
plz expalin it

Answer :

1 ) Minimum of atleast 2 would be calculated as follows : a+5b= 152 and a+b=100 , where a = B schools evaluated on exactly one parameter and b = B schools evaluated on exactly 5 parameters . So 4b =52 , b =13 . The minimum number of B schools which fulfill atleast 2 parameters is 13

To maximise we use a+2b = 152 and a+b = 100 where b = B schools evaluated on exactly 2 parameters . b = 52 . The MAXIMUM number of B schools which fulfill atleast 2 parameters is 52


2) Minimum of atleast three of the five parameters would be 13 as done above
3) Maximum of atleast three of the five parameters would be a+3b=152 , a+b= 100
2b=52 , So b be would be 26
Correct me if I am wrong

For example we can count the no of lines that can be drawn from the figure attached. they are :
ABD,AEF,EBC,CDF,AC,ED,FB. Total =7. now if we use the formula i have given, n + 1/2*nC2* we get it as 7. If we use 6C2-3C2 we get 12 which is incorrect.

Hope u are convinced.

Hi Harsh, esh.nil....

Thanks for the answer given. But still, I cannot understand the logic followed. Can any of you please explain it in a more lucid manner...Plse

Find remainder when
1^1 + 11^11 + 111^111 +.... 10 times
is divided by 100

The point here is 11^11 would always end in xxxxxx11 , 11^8 would always end in xxxxxxxxx8 and so on .

1^1 MOD 100 -------> 1
11^11 MOD 100------>11
111^111 MOD 100 ----->11
1111^1111 MOD 100----->11
11111^11111 MOD 100----->11

........and so on

So finally we have 11*9 + 1 MOD 100 which is 100 MOD 100 gives remainder 0 Correct me if I am wrong

solve this
A,B,C,D ,E are Sections of a class

Xstandard A(55) B (26) C (3 D (37) E(4

what is the max no. ofstudents who expected exactly 4 out of 5 sections?????????

X11 standard A(47) B (15) C (41) D (2 E(25)

how many in X11 standard expected a minimum of two of the five sections ????

Please gve the detailed approach.............

@junoon bhai... what is the total number of students in the class in class X and XII?? and can one student be part of more than one section? that is also not mentioned.. does something need to be added to the qn or am I .. the latter I am sure, but please confirm..

are u sure those nos in bold are multiple of 6... it is not an even number..
So i think u will end up with my answer eventually..

Hey esh.nil...thanks a lot man for pointing out the mistake. I totally missed that divisibility by 6 criteria :(......

yup sure, then we two shall arrive at the same answer.

m! = k! * n!
Maximise k? n should be minimum.
6! = 5! * 3!
k = 5 :)

Monil bhai there's another combination 7!*6!= 10! ( 7! *8*9*10) gives 10!

1)there are 3 distinct integers m,n and k find the max. value of k such that k!=m!/n! and 0=<10.

The answer I got for this was : the largest value of k! would be 7!
where 7!*6!=10!

Harsh ur formula is right according to them...pls cite me how did u get to it.....
45 is the answer according to it

Kavish, pls refer to
http://www.pagalguy.com/forum/quantitative-questions-and-answers/20838-official-cat-2007-quant-thread-392.html#post862166

solve this
A,B,C,D ,E are Sections of a class

Xstandard A(55) B (26) C (3 D (37) E(4

what is the max no. ofstudents who expected exactly 4 out of 5 sections?????????

X11 standard A(47) B (15) C (41) D (2 E(25)

how many in X11 standard expected a minimum of two of the five sections ????

Please gve the detailed approach.............

Solve these..

1)there are 3 distinct integers m,n and k find the max. value of k such that k!=m!/n! and 0=<10.

2)7x+5y=7 what will be the greatest value of x^3.y^4

Answer for the second would be as follows :
Maximum value of x^3*y^4 would occur when x/3=y/4 i.e x= 3y/4 .

7(3y/4) +5y = 7 , So 41y =28 which means y= 28/41 and x= 21/41

So x^3.y^4 = 28^3*21^4/41^7

Hi puys,
Yup I too think the answer is 20s

Every 2s, the Car1 moves 100 cm and car 2 moves 50cm behind, the gap being 500 cm to start of with, at the end of time

2s 450 cm
4s 400 cm
.
.
.
20s 0cm.

Hence 20s is the answer.
@kushal_gupta, @sachin2807 and @shradhagarg...I concur.

Regards,
Naren.

u r right shradha.....the consider one car first.....its moving with 50cm/s so after a second it will move forward 50 cm( so total distance covered by both cars in forward direction is 100 cm) now this car will move back 25cm as it is moving with 25cm/s...so after 2 seconds the net distance covered by this car is 25cm....similarly the distance covered by other car is 25 cm....so the total distance covered by both the cars after 2 seconds is 50 cm in forward direction....so obviously it will take 20 seconds to cover 500 cm....wat u say?

Hi....:)

Two cars are 500cm apart each is moving forward for 100cm at a velocity of 50cm/s and receeding back for 50 cm at 25cm/s...At wat time they will collide each other..


Thanxx in advance...

Answer should be 20 seconds...both the cars have to travel 250 mts each and in fst 4 seconds each car travels a distance of 50 mts so it will take 20 seconds for both the car to collide.

hello mr sk abhi.. cud u pls explain cause i thought the answer shud be 20 sec..
thanks
shradha

NEVER... i mean they will never collied..as the distance between them will remain constant as 500cm always