Official Quant thread for CAT 2013

MBA CAT 2024
21606
@Logrhythm said:
(-1)^odd strike nahi kara....meine sum nikala...4 se uska remainder liya...fir answer nikala...bad manager i am...wasting time...
Think first, then write. Best time-saving technique I know...look for the lazy way out :D

regards
scrabbler
21607
@scrabbler said:
24 * 24 = 576. Now next power will be 76*24 ~ -24 *24 ~-76 ~ 24.So last 2 digits will repeat as 24, 76, 24, 76, 24, 76...so for odd power 257, it will be 24.regardsscrabbler
kya baat hai.... neend tutt gayeee
21608
@Calvin4ever said:
1> felt tough finding the series itself

2 +10 + 30 + 68 + 130
Looks like a multiplicative or exponential series (fast growing) also 5th term divisible by 5 so I tried to see what happens....

1 * 2 + 2 * 5 + 3 * 10 + 4 * 17 + 5 * 26...

So now have to see the series 2, 5, 10, 17, 26....which is 1+1, 4+1, 9+1, 16+1, 25+1...

So series is nothing but n*(n^2+1)

regards
scrabbler
21609
@iLoveTorres said:
bro 1st wale ka third order difference is same.. how to frame a quadratic equation if the third order difference is same?
third is same na...

so -> Tn = an^3 + bn^2 + cn + d

now find a,b,c and d...subs in Tn...

^^ NCERT se sikha hai ye method...

warna u can observe...

2, 10, 20, 68, 130...

1(1+1), 2(4+1), 3(9+1), 4(16+1), 5(25+1)..

so n(n^2+1)
21610
@iLoveTorres said:
bro 1st wale ka third order difference is same.. how to frame a quadratic equation if the third order difference is same?
Cubic then it becomes. But I did something else - see previous post.

regards
scrabbler
21611

N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?

PS: dozing off

21612
@amresh_maverick said:
PS: dozing off
Good idea...me too!

regards
scrabbler
21613
@amresh_maverick said:
N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?PS: dozing off
0?
21614
@amresh_maverick said:
N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?PS: dozing off
three factors mean square of a number...

n^2 + 24n + 21 = k^2
(n+12)^2 - 123 = k^2
(n+12)^12 - k^2 = 123
(n+12+k)(n+12-k) = 123

now, 123 = 1*123 or 3*41..

(n+12+k)(n+12-k) = 1*123
here n=50 and k=61

(n+12+k)(n+12-k) = 3*41
here n=10 and k=19...

hence, 2 values of n ie 10 and 50...

@scrabbler kuch galti ho toh pls batao...
21615

@amresh_maverick said:
N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?PS: dozing off
2
21616
n^2+24n+21=n^2+2*12*n+144-123=k^2
=n+12)^2-k^2=123
=(n+12+k)(n+12-k)=123=1*123=3*41
n-k=1
n+k=123

n-k=3
n+k=41


21617
Two small squares on a chessboard are chosen at random. What is the probability that they have a common side?
21618
@MANJULNEOGI said:
Two small squares on a chessboard are chosen at random. What is the probability that they have a common side?
1/18?
21619
@MANJULNEOGI said:
Two small squares on a chessboard are chosen at random. What is the probability that they have a common side?
Favourable Events = 8*7 + 7*8 = 112.

Sample Space = 64C2.

Probability = 112/64C2 = 1/18.
21620
@Estallar12 said:
Favourable Events = 8*7 + 7*8 = 112.Sample Space = 64C2.Probability = 112/64C2 = 1/18.
bro i took the cases as 4*2(corner squares)+24*3(edge squares)+36*4(centre squares) = 224
Total outcomes possible 64c2
Prob = 224/64c2..
where am i going wrong?
21621
@iLoveTorres said:
bro i took the cases as 4*2(corner squares)+24*3(edge squares)+36*4(centre squares) = 224Total outcomes possible 64c2Prob = 224/64c2.. where am i going wrong?
Total Outcomes here won't be 64C2 as say for centre squares, you are selecting any one square out of the 36 squares, so probability for that = 36/64. Now, you will select any one out of the 4 squares surrounding it. So, probability for this will be 4/63.
Thus, Overall Probability for this case will be (36/64)*(4/63) and not 36*4/64C2.

Same for the other cases!
21622
@Estallar12 said:
Total Outcomes here won't be 64C2 as say for centre squares, you are selecting any one square out of the 36 squares, so probability for that = 36/64. Now, you will select any one out of the 4 squares surrounding it. So, probability for this will be 4/63.Thus, Overall Probability for this case will be (36/64)*(4/63) and not 36*4/64C2.Same for the other cases!
Got it.. Thanks for the timely help :)
21623
@MANJULNEOGI said:
Two small squares on a chessboard are chosen at random. What is the probability that they have a common side?
1/18
21624
@amresh_maverick said:
N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?PS: dozing off
n can take 10 and 50
21625
@amresh_maverick said:
N is a natural no . How many values of N exist such that N^2 + 24N + 21 has exactly 3 factors ?PS: dozing off
For N^2+ 24N + 21 to have exactly 3 factors , it should be the square of the prime number.

Let N^2 + 24 N + 21= k^2 where k is the prime number

(n+12)^2 - 123 = k^2
(n+12-k)(n+12+k) = 123 = 3*41 or 1*123
there will be 2 values of n and k
n+12+k = 41
n+12-k = 3
n = 10 and k = 19, k is prime so satisfies
n+12+k = 123
n+12-k = 1
n = 50, k = 61 satisfies
so n = 10 and 50 satisfies
so 2 values