Official Quant thread for CAT 2013

MBA CAT 2024
29586
@shattereddream said:
Total cost = 1347 x(9/8 x (20/21) = 1443.214
@shattereddream said:
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
1+7*3=22
29587
@shattereddream said:
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till 창€˜n창€™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
OA is 22 @Narci @YouMadFellow ___/ \___ @abhishek.2011
29588
@shattereddream said:
The half-life of a substance is 1 hr, i.e. at the end of 1 hr, its quantity reduces to one-half of what it was at the beginning of the hour. In how many hours will the quantity become less than 1% of the initial quantity?
100/2^n should be less than 1
n=7 satisfies
29589
@shattereddream said:
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
every n(n+1) ends with n 7's ... for 5040 nearest number in the form n(n+1) = 5112... (71*72) 5112th term is 7 and so is 5112 - 71 + 1 = 5042th term is 7 ..
hence ans is 1 + 7 + 7 + 7 = 22
29590
@shattereddream 16
29591
@abhi_14 said:
In how many ways can 10 identical presents b distributed among 6 children so tht each gets atleast one present15C516C69C56^10NOTPlease share the detail approach
as gifts are identical so
a+b+c+d+e+f=10
o m sorry didnt see that all are greater than 1 than its
a+b+c+d+e+f=4
9c5

15c5 -------
29592
@pratskool said:
every n(n+1) ends with n 7's ... for 5040 nearest number in the form n(n+1) = 5112... (71*72) 5112th term is 7 and so is 5112 - 70 + 1 = 5043 term is 7 ..hence ans is 1 + 1 + 7 + 7 = 16
no terms from 4971 to 5041 are 1 after that till 5041+71 terms are 7
so 1+7*3=22
29593
@abhishek.2011 said:
no terms from 4971 to 5041 are 1 after that till 5041+71 terms are 7so 1+7*3=22
yeah, thnks i have corrected my solution
29594

If we type all numbers from 1 to 1000 on a computer, then how many times will we press the buttons 9 and 1 consecutively (in same order) ?

29595
@nadalobama said:
@shattereddream 16
No the OA is 22 u can see the solution posted by youmadfellow
29596
@catahead said:
If we type all numbers from 1 to 1000 on a computer, then how many times will we press the buttons 9 and 1 consecutively (in same order) ?
24?
29597
@albiesriram 16(3+Root(2))
29598
@catahead said:
If we type all numbers from 1 to 1000 on a computer, then how many times will we press the buttons 9 and 1 consecutively (in same order) ?
abc is the number having unit digit 9 and (abc + 1) is the number having first digit from left as 1

When a is 0, bc = 09, 99 (two cases)
When a is 1, bc = 09, 19, 29, ......., 89 (9 cases)
When a is 2, 3, 4, ..., 8, (no such number possible)
When a = 9, bc = 99 (one case)

a91 - 10 cases

91a - 10 cases

So, 2 + 9 + 1 + 10 + 10 = 32 times
29599

@shattereddream said:
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
the terms are forming a group of 2,4,6,8 i.e, 17 1177 111777 11117777
so the terms after grouping can be expressed as a sum
n/2(2a+(n-1)d)=n(n+1)
70*71=4970
71*72=5142
5142-4970=142
it will have 71 1's and 71 7's
so only term 5040 will be 1 and the rest will be 7
net sum =1+7*3=22
29600
A = {179, 180, 181, €Ś..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed
29601
@shattereddream said:
A = {179, 180, 181, €Ś..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed
numbers are of the form
9k -- 21
9k+1 --20
9k+2--20
9k+3 -- 20
9k+4 --- 20
9k+5 --20
9k+ 6---20
9k+7 -- 20
9k+8---21
now max set 9k+8,9k+2,9k+3,9k+4 , + 1 9k member
20*3+21+1=82
29602
@shattereddream said:
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
22?

2 + 4 + 6 + 8 + .........140
2(1 + 2 + 3 + .....70) = 2(70)(71)/2 = 4970
after 140 terms....142 terms will be there....in which 71 1's will be there and 71 7's will be there..so from 4971 to 5041 will contain 1's....after that 7's
therefore T5040 + T5042 + T5044 + T5046 = 1 + 7 + 7 + 7 = 22
29603
@Vishnu.shan said:
The solution is option 1:for every x in the denominator, the numerator doubles so the initial divisor factor is 100000/1040.... for the given series it is, 99999............999996 with a common remainder of 160
@DivineSeeker said:@chillfactor cant be zero dude....check ur method again....remainder would undoubtely be zero bt nt the quotient. i ll post the answer as soon as i get it.. thanks anyways.

Try this question:-

What will be the unit digit of the quotient when 10^20000 is divided by (10^200 + 7) ??
29604
@chillfactor said:
@DivineSeekersaid:@chillfactor cant be zero dude....check ur method again....remainder would undoubtely be zero bt nt the quotient. i ll post the answer as soon as i get it.. thanks anyways.Try this question:-What will be the unit digit of the quotient when 10^20000 is divided by (10^200 + 7) ??
hemant sir according to ur approach unit digit of Q is 7?????
29605
@chillfactor said:

What will be the unit digit of the quotient when 10^20000 is divided by (10^200 + 7) ??
N = 10^20000

Divisor = 10^200 + 7

Remainder = (7)^100

10^20000 = Q*(10^200 + 7) + (7^100)

Q*(10^200 + 7) = 10^20000 - 7^100

Last digit of RHS = last digit of (10^20000 - 7^100) = [10- 1] = 9

Last digit of LHS = 0 + Last digit of (Q * 7 )

=> Last digit of Q = 7 ?