Official Quant thread for CAT 2013

MBA CAT 2022
1981
@sauravd2001 said:
find the sum of squares and cubes of the root of eqaution x^3-2x^2+x-3?
a+b+c = 2
ab+bc+ac = 1
abc = 3
now use (a+b+c)^2 = a^2+b^2+c^2+2ab+abc+2ca
so a^2+b^2+c^2 = 4-2(1) = 2
(a^2+b^2+c^2)(a+b+c) = 4
solving we get a^3+b^3+c^3 = 11
1982
@pankaj1988 said:
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
1+2+..6 = 21
1983
@pankaj1988 bhai 21 ???????
1984
@pankaj1988 said:
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
2) 21 ?

Balls of same color are identical - Assumption

One green ball - 6 ways
Two green ball - 5 ways
Three green ball - 4 ways
and so on

6 + 5 + 4 + ..1 = 21 ?
1985
@pankaj1988 said:
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
1 green = 6C1
2 green = 5C1
3 green = 4C1
.
.
6 green = 1C1

total = 6 + 5 + 4 + 3 + 2 + 1 = 21
1986
@pankaj1988 said:
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
21 aa rha hai.
1987
@Torque024 said:
Please explain your solution, I'm not getting. :/
2 2 1

balls = A B C D E

suppose we select 1st pair as AB and 2nd pair as CD

now there will be one more case when we select 1st pair as CD and 2nd pair as AB

but here we have identical boxes , that's why divided by 2

@sujamait bhai ___/\___
1988
@jain4444 said:
2 2 1 balls = A B C D E suppose we select 1st pair as AB and 2nd pair as CD now there will be one more case when we select 1st pair as CD and 2nd pair as AB but here we have identical boxes , that's why divided by 2 @sujamait bhai ___/\___
Box ke andar toys ka arrangement matter nai karega like AB and BA?
1989
@Torque024 said:
Box ke andar toys ka arrangement matter nai karega like AB and BA?
neah..finger mein rings hoti toh ..matter kar sakti thi..box ke ander kaise arrange karoge. ? infinite possibilities hein..untill n unless kuch size bagera aisa ho..
1990
@pankaj1988 said:
There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ____(1) 5(2) 21(3) 33(4) 60(5) 27
6C1+5C1+4C1+3C1+2C1+1C1 = 21
1991

sanatta cha gaya idhar toh

1992
Find (x+y)where x^4+y^4=4721(a)13, (b)17, (c)21, (d)25, (e)None.
1993

A wire if bent into a square encloses an area of 484 cm^2. The wire is cut into 2 pieces; with the larger piece having a length 3/4th of the original wire's length. Now, if a circle and a square are formed with the bigger and smaller pieces resp, what should be the area enclosed by the 2 pieces?


1) 464
2) 576.25
3) 376.75
4) 424.25
1994
@pankaj1988 said:
Find (x+y)where x^4+y^4=4721(a)13, (b)17, (c)21, (d)25, (e)None.
x=8, y=5

13
1995
@pankaj1988 said:
Find (x+y)where x^4+y^4=4721(a)13, (b)17, (c)21, (d)25, (e)None.
13
8^4+5^4 = 4721
1996
@pankaj1988 said:
Find (x+y)where x^4+y^4=4721(a)13, (b)17, (c)21, (d)25, (e)None.
13
8,5?
1997
@pankaj1988 said:
Find (x+y)where x^4+y^4=4721(a)13, (b)17, (c)21, (d)25, (e)None.
5,8 = 13
1998
@soumitrabengeri said:
A wire if bent into a square encloses an area of 484 cm^2. The wire is cut into 2 pieces; with the larger piece having a length 3/4th of the original wire's length. Now, if a circle and a square are formed with the bigger and smaller pieces resp, what should be the area enclosed by the 2 pieces?1) 4642) 576.253) 376.754) 424.25
length = 4*root(484)

given

2*pi*r=3/4*4*root(484)
=>r=3*root(484)/2pi
==>area of circle = 9/4*484/pi=1089/pi=346.64

4*a=root(484)
==>area of square = 484/16= 30.25

so total area= 376.89


1999
@soumitrabengeri said:
A wire if bent into a square encloses an area of 484 cm^2. The wire is cut into 2 pieces; with the larger piece having a length 3/4th of the original wire's length. Now, if a circle and a square are formed with the bigger and smaller pieces resp, what should be the area enclosed by the 2 pieces?1) 4642) 576.253) 376.754) 424.25
22*22 = 484
22*4= 88 length
66 , 22

2 pie r = 66 => r = 33*7/22 = 21/2 area = pie*441/4 = ?
4x = 22 => x= 11/2 area = 121/4
...
2000
@soumitrabengeri said:
A wire if bent into a square encloses an area of 484 cm^2. The wire is cut into 2 pieces; with the larger piece having a length 3/4th of the original wire's length. Now, if a circle and a square are formed with the bigger and smaller pieces resp, what should be the area enclosed by the 2 pieces?1) 4642) 576.253) 376.754) 424.25
option 3?