Official Quant thread for CAT 2013

MBA CAT 2024
10321
@jain4444 said:
Determine the exact value:The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).
@jain4444 said:
Determine the exact value:The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).
[ 1/n - 1/(n+1)]^3
1/n^3 - 1/(n+1)^3 -3[(1/n-1/(n+1)]^2
1-3(1)^2
-2???
ye to wierd lag rha h...:P
10322
@jain4444 said:
Determine the exact value:The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).
Some special series is there or what ? :banghead:

I am stuck here : Sum = 3S - 1/2

Where T(n) of S = ( 1/[(n)*(n+1)^2 *(n+2)] ) 😐
10323
@jain4444 said:
Determine the exact value:The summation of 1/[n(n+1)]^3 as n takes integral values fr. [1,+inf).
finally got the answer
betting on 4 heres my approach
the expression can be written as [1/n-1/(n+1)]^3
it can be written as 1/n^3-1/(n+1)^3 -3[(1/n)(1/n+1)][1/n-1/(n+1)]=
1/n^3-1/(n+1)^3 -3[[1/n-1/(n+1)]^2
summate first two terms to get 1
next 2 terms -3[[1/n-1/(n+1)]^2
it can be written as -3[1/n^2-1/(n+1)^2 -2(1/n)(1/n+1)]
=-3[1/n^2-1/(n+1)^2 +6[1/n-1/(n+1)]
sum the first two terms to get -3
remianing term +6[1/n-1/(n+1)]v =6 till infinity
so sum = 1-3+6 =4
@jain4444
10324
@techsurge said:
finally got the answerbetting on 4 heres my approachthe expression can be written as [1/n-1/(n+1)]^3it can be written as 1/n^3-1/(n+1)^3 -3[(1/n)(1/n+1)][1/n-1/(n+1)]=1/n^3-1/(n+1)^3 -3[[1/n-1/(n+1)]^2summate first two terms to get 1next 2 terms -3[[1/n-1/(n+1)]^2it can be written as -3[1/n^2-1/(n+1)^2 -2(1/n)(1/n+1)]=-3[1/n^2-1/(n+1)^2 +6[1/n-1/(n+1)]sum the first two terms to get -3remianing term +6[1/n-1/(n+1)]v =6 till infinityso sum = 1-3+6 =4@jain4444
i also got answer as 4 , same as your approach

but given solution says

Perform partial fraction decomposition to get

1/n^3 - 1/(n+1)^3 - 3[1/(n^2) + 1/(n+1)^2] + 6[1/n - 1/(n+1)]

Sum from n = 1 to infinity:

sum = 1 - 3[(pi^2)/6 + (pi^2)/6 - 1] + 6(1)

sum = 10 - pi^2
10325
@hiteshkhurana82 said:
The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
40 ?....
10326
@jain4444 said:
i also got answer as 4 , same as your approach

but given solution says

Perform partial fraction decomposition to get

1/n^3 - 1/(n+1)^3 - 3[1/(n^2) + 1/(n+1)^2] + 6[1/n - 1/(n+1)]

Sum from n = 1 to infinity:

sum = 1 - 3[(pi^2)/6 + (pi^2)/6 - 1] + 6(1)

sum = 10 - pi^2
bhai ye sab to higher maths mein divergent, convergent series mein pada tha
Partial fraction decomposition tak to samajh aa gaya
uske baad jo inhone high fundo maths lagayi hai
wil that count as CAT level let alone aptitude level.....
PS: I hated higher maths that was taught in engg :P
10327
@YouMadFellow : ah yeah thamks for pointing that out... such silly mistakes i often have to pay dearly for ...
10328
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?
10329
@jain4444 said:
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?
4?
10330
@jain4444 said:
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?

3 children
3rd born in present
a = sum of father and mother age
b=sum of 2 children age
a=2b
a-10=4b-40
b=15, a=30
3rd child is born ,after 15 years
a+30 =b+45
60=60
????
10331
@jain4444 said:
question says "first students" it doesn't say any particular should get baton after 1 operation I think we need to multiply nC2 by 2!
:thumbsup:

He chooses 2 people by C(N, 2) ways.
then the teacher can pass the baton to either of the students, ultimately the first one will get the baton. :)
10332
@jain4444 said:
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?
p=sum of ages of parents
c=sum of ages of children
n=number of children
p=2c
p-10=4(c-5n)
p+30=c+15n
Solving
n=5?
10333

@jain4444 another case

and valid one 5

a = sum of father and mother age
b=sum of 5 children age
a=2b
a-10 =4b-100
b=45
a=90
after 15 years
a+30 =b+75
120=120
???
so betting on 5 , unless stated no nnew child was born in this period

@jain4444
@jain4444 said:
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?

10334
@jain4444 said:
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?
Father =F Mother= M
sum of children's age =T
number of children = n
F+M=2T...(i)
F-5+M-5=4(T-5n)...(ii)
F+15+M+15=T+15n....(iii)

solving these three eqn
2T-10=4T-20n
10n-5=T

F+M+30=T+15n
2T+30=T+15n
2(10n-5)+30=10n-5+15n
20n-10+30=25n-5
20+5=5n
n=25/5
n=5??


10335
@Torque024 said:
p=sum of ages of parentsc=sum of ages of childrenn=number of childrenp=2cp-10=4(c-5n)p+30=c+15nSolvingn=5?
@techsurge said:
@jain4444 another caseand valid one 5a = sum of father and mother ageb=sum of 5 children agea=2ba-10 =4b-100b=45a=90after 15 years a+30 =b+75120=120???so betting on 5 , unless stated no nnew child was born in this period@jain4444


@techsurge nice bet

let sum of parent's ages = N
average age of children's = x
number of children's = y

N = 2*xy --------1

(N - 10) = 4*(xy - 5y)---------2

(N + 30) = xy + 15y ------3

put 1 in (2) and (3)

2xy - 10 = 4xy - 20y

20y - 2xy = 10
15y - xy = 30

=> y = 5

The polynomial x^2-2x+7 divides the polynomial x^4+px^2+q. What is the value of q?
10336
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?
let age of parents be p and that of child be c.
Number of children=n
first condition, p=2*c

second condition, p-2*5=4*(c- 5*n)---[1]
third condition, p+2*15=c+15*n -------[2]

put p=2*c in [1] equation,
10*n-c=10---[3]

put p=2*c in [2] equation,
15*n-c=30--[4]

solve [3] and [4], n=4
10337
@jain4444 said:
@techsurge nice bet let sum of parent's ages = N average age of children's = x number of children's = yN = 2*xy --------1(N - 10) = 4*(xy - 5y)---------2(N + 30) = xy + 15y ------3put 1 in (2) and (3)2xy - 10 = 4xy - 20y20y - 2xy = 10 15y - xy = 30=> y = 5

Zara dekhna mere wale solution mai kya galati h :neutral:

at present, parents' age=60, children's age=30, number of children =4

=> 60=30*2
5 years back, parents' age=50, children's age=10 => 50=10*5

15 years later, parents' age=90, children's age=90, 90=90
10338
Ratio of incomes of a and b is 3:4,ratio of expenditure is 4:5.The ratio of savings can be.
1)4:5
2)6:7
3)8:13
4)7:9

I did it like this ratio of savings =(3x-4y)/(4x-5y). then i equated the ratio with each of the options to get the answer. the correct answer is 8/13. But is there any shorter approach?

q2) If a-b varies directly with a+b,then a^2 - b^2 will vary directly with
1) a^2 + b^2
2)ab
3)a^2 + b^2 + 3ab
4) more than one of the above

I did it like this:

a-b=k(a+b)
multiplying both sides by a+b
a^2-b^2=k(a+b)^2

What to do next to reach the answer ? Answer is 4) more than one of the above.
10339
@vijay_chandola said:
Zara dekhna mere wale solution mai kya galati h at present, parents' age=60, children's age=30, number of children =4=> 60=30*25 years back, parents' age=50, children's age=10 => 50=10*515 years later, parents' age=90, children's age=90, 90=90
The sum of the parent's ages is twice the sum of their children's ages. Five years ago, the sum of the parent's ages is four times the sum of their children's ages. In fifteen years, the sum of the parent's ages will be equal to the sum of their children's ages. How many children were in the family?

bhai question says 4 times you took 5 times in your solution
10340
@nole
q2) If a-b varies directly with a+b,then a^2 - b^2 will vary directly with
1) a^2 + b^2
2)ab
3)a^2 + b^2 + 3ab
4) more than one of the above

I did it like this:

a-b=k(a+b)
multiplying both sides by a+b
a^2-b^2=k(a+b)^2

with my understanding, u did correctly formed the equation. Now all u need to do is to open up the RHS to get the final ans.
it will be
a^2-b^2= k( a^2+b^2+2ab)
so its dependent on both,
1) a^2+b^2
2) ab