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Official Quant Thread for CAT 2011 [Part 8]

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find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

=> S = 1 + (4/7) + (9/7^2) + ...
=> S/7 = 1/7 + (4/7^2) + (9/7^3) + ...
Subtract them we get
=> 6S/7 = 1 + 3/7 + 5/49 + ...
=> 6S/49 = 1/7 + 3/49 + 5/343 + ...
Subtract them again
=> 36S/49 = 1 + 2/7 + 2/49 + ...
=> 36S/49 = 1 + (2/7) / (6/7)
=> S = 49/27
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find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

S= 1+(4/7)+(9/(7^2))+(16/(7^3))+.....

S/7 = 1/7 +(4/(7^2))+(9/(7^3))+.....
6S /7 = 1+3/7+5/7^2+.......

Now this series is in A.P. & G.P.

S = ab/1-r+dab/(1-r)^2

here a = first number in A.P.
B= first number in G.P.

1/1-1/7+2/(1-1/7)^2 = 70/18

so sm is: 7/6*70/18

P.S.: Kisi ko formula pata ho to chk kar lo, I am confused betn (1-r)^2 otr 1-r^2
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find the infinite sum of the foll seuence
1+(4/7)+(9/(7^2))+(16/(7^3))+.....

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TT School of Management institute involved in teaching, training and research. Currently it has 37 faculty
members. They are involved in three jobs: teaching, training and research. Each faculty member working
with TT School of Management has to be involved in at least one of the three job mentioned above:

-
A maximum number of faculty members are involved in training. Among them, a number of faculty
members are having additional involvement in the research.
-

The number of faculty members in research alone is double the number of faculty members involved
in all the three jobs.
-

17 faculty members are involved in teaching.
-

The number of faculty members involved in teaching alone is less than the number of faculty members
involved in research alone.
-

Ten faculty members involved in the teaching are also involved in at least one more job.

80. After some time, the faculty members who were involved in all the three tasks were asked to with
drawl from one task. As a result, one of the faculty members each opted out of teaching and
research, while remaining ones involved in all the three tasks opted out of training. Which one of the
following statements, then necessarily follows:
(A) The least number of faculty members is now involved in teaching.
(B) More faculty members are now associated with training as compared to research.
(C) More faculty members are now involved in teaching as compared to research.
(D) None of the above.

81. Based on the information given above, the minimum number of faculty members involved in both
training and teaching, but not in research is:
(A) 1 (B) 3 (C) 4 (D) 5

Approach plz ?

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Solve and find the pair of non negative integers (x, y) such that
(x+ y - 5)^2 = 9xy.

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rohitpal Says
hi plz solve the attached problems.

Doing Second: log(4)a-b = log(4) (_/ a-_/b)^2+log(4)4
log(base 4)a-b = log(base 4) 4*(_/ a-_/b)^2
a-b = 4*(_/ a-_/b)^2
taking sqaure root
_/ (a-b) = 2_/a-2_/b
iI think it should be 1 only keeping option in mind
Attachments
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ind the unit digit : 12^55/3^11 + 8^48/16^18

is ans 0?

(12^5/3) ^11 % 3

obviously this will be 0 .

8^48/16^18 = 2^144/ 2^72 : remainder will again 0 so my take 0
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Shyam, Gopal and Madhur are three partners in a business. Their capitals are respectively
Rs 400, Rs 8000 and Rs 6000. Shyam gets 20% of total profit for managing the business.
The remaining profit is divided among the three in the ratio of their capitals. At the end of the
year, the profit of Shyam is Rs 2200 less than the sum of the profit of Gopal and Madhur.
How much profit, Madhur will get?

could anyone expalin me that?

remaining whatever it is divided into

S:G:M = 4: 80: 60 = 1:20:15

Now let profit y then 80% of the same will be distributed and S will get y/36*80/100+20/100Y

G = 20/36*80/100y

M=15/36*80/100y

so the final ratio will be 800:1600:1200

8:16:12 = 2:4:3

2x 4x 3x

5x = 2200

x= 440

so 3x = 440 =1320 my take
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Q. A card is drawn at random from a well shuffled pack of 52 cards.

X: The card drawn is black or a king.
Y: The card drawn is a club or a heart or a jack.
Z: The card drawn is an ace or a diamond or a queen.
Then which of the following is correct?
A. P(X) > P(Y) > P(Z)
B. P(X) >= P(Y) = P(Z)
C. P(X) = P(Y) > P(Z)
D. P(X) = P(Y) = P(Z)

IIFT - 2009

p(x) = 26/52+6/52 =32/52
p(y) = 13/52 +13/52+2/52 =28/52
p(z) = 13/52+3/52+3/52 = 19/52

so p(x)> p(y) > p(z)

option A

P.S.: Pata nahi me tash k patto ko thik se samaj payi hu ya nai, I always got confused betn "Ace" and "club" or "jack"
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thanks dude..i tried the same eqn to solve,but i couldnt solve it..can u pls show the solving procedure..
x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

x^2 + y^2 = 2.5(x+y) --(i)
=> x^2 + y^2 - (3xy) = 1 -- (ii)

2.5x+2.5y = 1+3xy

let x=0 y =1/2.5
y=0 X= 1/2.5

I think soln can be more than 1
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