Official Quant thread for CAT 2013

MBA CAT 2022
2821
@milestogo3 said:
@Torque024 said:Q) Determine the remainder when the number 1+22+33+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base? for binary1for base 3(1+22)base 3 = 1+ 8=9/2 = 1for base 41+22+333 base 4,1+10+3+12+48 = 64/3 =1one in all cases.
1 + 22 +333 + 4444 +..............
1+ 2+ 18 + 3 + 27 + 243
1 + 2+2 + 3+3+3 + 4+4+4+4 +5+5+5+5+5 +6+6+6+6+6+6+ 7+7+7+7+7....
1 + 4 + 9 + 16 + 25+ 36 + 49 + 9
55 + 45+49
149 /8
yes,giving it 5.



1+22+333+4444 base 5
1+12+ 3+15+125 + 4+20+100+500
16+140+624
780/4

differnt aa rha hai ye to. :/

2822
@ScareCrow28 said:
A cube of side €˜a €™ is converted into a sphere by adding clay on all the faces of the cube. The minimum volume of clay required is ?
Volume of clay required = vol of sphere - vol of cube
=> 4/3*pi*(root(3)*a/2)^3 - a^3

-> (root(3)/2 - 1)*(a^3) ?
2823
@krum said:
@ScareCrow28 , since even number if AM are inserted, the series won't be an A.Pthe answer will depend on the missing termx and y were 2,30if we insert odd number of AMs in between, the series will be in A.Psay2 16 302 9 16 23 30;however, here out series will be something like2 9 16 302 5.5 9 16 30i.e one of the terms will skip by 2d, the particular term will decide the value of y as every term after that term will skip by d@Estallar12 sir, kaise kia apne
Bhai.. how can you say inserting even nos won't give AP? The numbers can be fraction..they are not necessarily integrs..so if we insert even nos..they can still form AP..ryt?
2824
@ScareCrow28 said:
We are given 1997 distinct positive integers, any 10 of which have the same least common multiple. Find the maximum possible number of pairwise coprime numbers among them.
Age old question 😁 answer is 9 .. bit lazy to type 😁

6th season quant thread :w00t:
2825
@naga25french said:
Age old question answer is 9 .. bit lazy to type 6th season quant thread
__/\__ Sir apke lie nahi hai ye question 😛 mere jaise baalko k lie hai.. Happy Bal Diwas waise
2826
@ScareCrow28 said:
Bhai.. how can you say inserting even nos won't give AP? The numbers can be fraction..they are not necessarily integrs..so if we insert even nos..they can still form AP..ryt?
can u give me an example ?
2827
@krum said:
can u give me an example ?
6/3...34/3...62/3...90/3?? I may be horribly wrong here..but i didn't actually get what you said :P
2828
@krum said:
can u give me an example ?
4 a1 a2 8

4 + 3d = 8
d =4/3

a1 = 16/3 a2 = 16/3 + 4/3 = 20/3


2829
@krum

The question states about the insertion of Arithmetic Mean(AM)..Hence how come the series won't be an AP??
2830
@ScareCrow28 said:
6/3...34/3...62/3...90/3?? I may be horribly wrong here..but i didn't actually get what you said
sahi kaha, main lallu jaisa soch hi rha tha
2831
Have a worksheet in my lapy, don't know the answers. You may discuss.

1. A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. Find the radius of the circle?

2. A triangle ABC has positive integer sides, angle A = 2 angle B and angle C > 90. Find the minimum length of the perimeter of ABC.

3. Given a certain hospital is opened only in last week. If 10 people travel through bus to hospital then what is the probability atleast two people travel on the same day?
2832
@krum said:
can u give me an example ?
Say x=0, y=1.
Insert 1/3, 2/3 b/w x and y.
Now, we have 0,1/3,2/3 and 1.
2833
@krum said:
sahi kaha, main lallu jaisa soch hi rha tha
, aaj bal diwas hai.
2834
@Torque024

The answer to last question is 1..Coz at the least 7 people will travel in 7 days..Hence the remaining 3 can be put on any days..Hence P(E) is 1..
2835
Q) How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
2836
@pyashraj said:
@Torque024The answer to last question is 1..Coz at the least 7 people will travel in 7 days..Hence the remaining 3 can be put on any days..Hence P(E) is 1..
yes it is 1.
2837
@Torque024 said:
Q) Determine the remainder when the number 1+22+33+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base?
@Torque024

I will go by general method i.e. base (s+1)
(s+1) mod (s-1) = 2 [Here (s-1) is greater than 2, if not do it manually]

1 mod (s-1) = 1
22 mod (s-1) = 2*(s+1) + 2 mod (s-1) = 2*(1 + 2)
333 mod (s-1) = 3*(s+1)^2 + 3*(s+1) + 3 mod (s-1) = 3*(1 + 2 + 4)

and so on

=> If I assume remainders in an AGP, general term = n*(2^n - 1)

Number of terms in that series = s

Sum of the series = S1 - S2
S1 = Sigma(n*[2^n]) = (2s-2)*(2^s) + 2
S2 = Sigma(n) = s*(s+1)/2

S mod (s-1) = (0)*(2^s) + 2 - (s*(s+1)/2) = 2 - [(s*(s+1))/2] (for s>=3)

I guess the answer will depend on s. This may be wrong, if yes, please tag me when the OA is shared.

EDIT: Calculation mistake earlier. Checked for various values, and it comes out to be correct.
2838
@Torque024 said:
Q) How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
10^ 5 numbers?

total numbers possible 9.10^5, every 9th number would be divisible by 9
so 10^5.9/9


ps: what was the source of last question....base one?
2839
@Torque024 said:
Q) How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
lengthy approach hai..koi choti nhn soch pa rha hu
2840
@Torque024

Here, _ _ _ _ _ 5_.. Thus, x1 + x2 + x3 + x4 + x5 + x6 + 5 has to be a multiple of 9..

Thus, x1x2x3x4x5x6 is a 6-digit number which can be generalized as 9y + 4.., and x1>1..

Hence, there are 9*10^5 numbers, and out of such numbers which are 9y+4 form=1/9*9*10^5

=>10^5??