Official Quant thread for CAT 2013

MBA CAT 2024
9601
@chillfactor said:
Just a small extension.Is it possible to have 17 or more terms in such a series given rest all conditions remains the same??
Why not?
It is making the following sequence..

5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13 ......
9602
@chillfactor said:
Ram purchased some items: books, pens, pencils, erasers, sharpeners, compass and geometry boxes. He bought at least seven pieces of each of above mentioned items but he didn't but no two items in same quantity. He did not buy anything else. It is given that among the seven items he bought geometry boxes are least in quantity. Number of geometry boxes he bought can be exactly determined if the total number of items bought is at most a) 70b) 72c) 74d) 76e) 78
At least 7 of each item, so 49 items are bought surely

Now, the case he buys the things in consecutive numbers , extra items he bought for each can be 0,1,2,3,4,5,6
Sum of extra = 21

So, the number of geometry boxes can surely be found out uniquely for 70

But, if I buy one more item, then also I can determine it uniquely , so we see that when we keep on adding the items, I can add up to 6 more items to 70 to keep the 0 in front undisturbed and maintain the constraints

So, 76 ?
9603
@vijay_chandola said:
Why not? It is making the following sequence..5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13 ......
Sum = -52 + 35 = -17
9604
In how many ways can 2700 be written as a product of 8 distinct integers?

a-6
b-0
c-4
d-More than 6
9605

A 3 digit no in base 7 when written in base 9 has its digits reversed. what is the sum of the digits of the number?


5
6
7
8
9606

two positive numbers a and b satisfy: (a+b)/x = HCF(a,b). Which of the following two numbers sum up to x?


13 and 52
132 and 96
18 and 126
56 and 45
9607

@tarun.g said:
In how many ways can 2700 be written as a product of 8 distinct integers?a-6 b-0 c-4 d-More than 6
2700 = 1*2*2*3*3*3*5*5
So, its not possible to have all 8 factors as positive integer. Hence we need to consider negative integers also

Now, (-1)(1)(-2)(2)(-3)(3)(-5)(5) = 900 (so one 3 is left)
Now, this last three can go to any of the 8 factors, but when it combines with 1 or -1, then we will get 3 or -3 which are already there.

So, only 6 possibilities are there
9608
@chillfactor said:
Sum = -52 + 35 = -17
:banghead:

After 16th term, sum of 11 term would not be 1 whenever the sequence will start from -13 :O
Checked it for the sum of 7 terms. So, thought it is correct. :embarrassed:

That is why they have specially mentioned it 16 term series. :-/
9609
@tarun.g said:
In how many ways can 2700 be written as a product of 8 distinct integers?a-6 b-0 c-4 d-More than 6
2700 = 1*3*3*3*2*2*5*5

But we want distinct integers, so adjusting the above [3 has to be involved in 3,2,5] , we get

Case 1: (1)*(-1)*(9)*(3)*(-2)*(2)*(-5)*(5) => 2 ways (change sign of both 9,3)

Case 2: (1)*(-1)*(6)*(3)*(-3)*(2)*(5)*(-5) => 2 ways (change sign of both 6,2)

Case 3: (1)*(-1)*(15)*(3)*(-3)*(-2)*(2)*(5) => 2 ways (change sign of both 15,5)

So, Total 6 ways ?
9610
@chillfactor said:
2700 = 1*2*2*3*3*3*5*5So, its not possible to have all 8 factors as positive integer. Hence we need to consider negative integers alsoNow, (-1)(1)(-2)(2)(-3)(3)(-5)(5) = 900 (so one 3 is left)Now, this last three can go to any of the 8 factors, but when it combines with 1 or -1, then we will get 3 or -3 which are already there.So, only 6 possibilities are there
@YouMadFellow said:
2700 = 1*3*3*3*2*2*5*5 But we want distinct integers, so adjusting the above [3 has to be involved in 3,2,5] , we get Case 1: (1)*(-1)*(9)*(3)*(-2)*(2)*(-5)*(5) => 2 ways (change sign of both 9,3)Case 2: (1)*(-1)*(6)*(3)*(-3)*(2)*(5)*(-5) => 2 ways (change sign of both 6,2)Case 3: (1)*(-1)*(15)*(3)*(-3)*(-2)*(2)*(5) => 2 ways (change sign of both 15,5)So, Total 6 ways ?
Yes OA is 6
9611
@chillfactor said:
Ram purchased some items: books, pens, pencils, erasers, sharpeners, compass and geometry boxes. He bought at least seven pieces of each of above mentioned items but he didn't but no two items in same quantity. He did not buy anything else. It is given that among the seven items he bought geometry boxes are least in quantity. Number of geometry boxes he bought can be exactly determined if the total number of items bought is at most a) 70b) 72c) 74d) 76e) 78
Can be determined for 70 and 76.
condition 'at most' is given.
hence, d) 76
9612
@tarun.g said:
A 3 digit no in base 7 when written in base 9 has its digits reversed. what is the sum of the digits of the number?5678
(abc)₇ = (cba)₉
49a + 7b + c = 81c + 9b + a
48a = 80c + 2b
24a = 40c + b

24a is divisible by 8, so only possible values of b are 0 and 8 (8 is not possible as abc is a number in base 7)
b = 0,
a/c = 5/3
So, (503)₇ = 5*49 + 3 = 248

Sum of digits = 12
I think it is asking for a a + b + c = 8
@tarun.g said:
two positive numbers a and b satisfy: (a+b)/x = HCF(a,b). Which of the following two numbers sum up to x?13 and 52132 and 9618 and 12656 and 45
a = km
b = kn, where m and n are coprime

k(m + n)/x = k
So, x = m + n

Hence, the two numbers should be coprime

Only 56 and 45 are coprime
9613
@chillfactor said:
Ram purchased some items: books, pens, pencils, erasers, sharpeners, compass and geometry boxes. He bought at least seven pieces of each of above mentioned items but he didn't but no two items in same quantity. He did not buy anything else. It is given that among the seven items he bought geometry boxes are least in quantity. Number of geometry boxes he bought can be exactly determined if the total number of items bought is at most a) 70b) 72c) 74d) 76e) 78
Is it 70/7=10
7,8,9,10,11,12,13 ?

With 72,74,76,78 we can not determine minimum value exactly
9614
@gnehagarg said:
Is it 70/7=107,8,9,10,11,12,13 ?With 72,74,76,78 we can not determine minimum value exactly
We can determine it exactly for 72,74, and 76 too

Because the additional items can be added only in a particular fashion adhering to the constraints

Ex: If 72, then possible combinations are 7,8,9,10,11,12,15 or 7,8,9,10,11,13,14 .. so the minimum value remains at 7
9615
@chillfactor said:
(abc)₇ = (cba)₉49a + 7b + c = 81c + 9b + a48a = 80c + 2b24a = 40c + b24a is divisible by 8, so only possible values of b are 0 and 8 (8 is not possible as abc is a number in base 7)b = 0,a/c = 5/3So, (503)₇ = 5*49 + 3 = 248Sum of digits = 12I think it is asking for a a + b + c = 8a = kmb = kn, where m and n are coprimek(m + n)/x = kSo, x = m + nHence, the two numbers should be coprimeOnly 56 and 45 are coprime
Both answers correct as usual, and yes sir in the first question sum would be 5+0+3 = 8

I didn't understand how you got the value of b as 0 or 8
9616
@tarun.g said:
Both answers correct as usual, and yes sir in the first question sum would be 5+0+3 = 8I didn't understand how you got the value of b as 0 or 8
Both 24 and 40 are divisible by 8. so b has to be divisible by 8 in order to get answer in integer values. :D
9617
@vijay_chandola said:
Can be determined for 70 and 76.condition 'at most' is given.hence, d) 76
@gnehagarg said:
Is it 70/7=107,8,9,10,11,12,13 ?With 72,74,76,78 we can not determine minimum value exactly
Read the last line carefully:-
Number of geometry boxes he bought can be exactly determined if the total number of items bought is at most

It means we have to find the maximum number of items that he can buy and still we can figure out the number of geometry boxes he bought.

Since he bought atleast 7 items of each kind, we just have to find that what will will be minimum no of items one can buy if he buys 8 geometry boxes (given that geometry boxes has least bought quantity among the 7 items)

8 + 9 + 10 + 11 + 12 + 13 + 14 = 77

So, we can say that for any value less than 77, we can always say that number of geometry boxes bought are 7 (as for 8, total items will be 77)

Hence, at most 76 items can be bought

@tarun.g 24a = 40c + b
b = 24a - 40c
b = 8(3a - 5c)
Clearly b is a multiple of 8

But b is a digit in base 7, and only possible digit which is a multiple of 8 is 0

So, b = 0
9618

Very simple sitter:


If a 3 digit number is made by using 3 digits 1,2,3 ..What is the probablity of getting the Sum as 6..

A)1/9 B) 1/3 C) 4/27 D) 7/27 E) None of these
9619
@aimiift2012 said:
Very simple sitter: If a 3 digit number is made by using 3 digits 1,2,3 ..What is the probablity of getting the Sum as 6..A)1/9 B) 1/3 C) 4/27 D) 7/27 E) None of these
7/27 ?
9620
@aimiift2012 said:
Very simple sitter:

Redundancy! Sitter already implies 'very simple' :)
@aimiift2012 said:
If a 3 digit number is made by using 3 digits 1,2,3 ..What is the probablity of getting the Sum as 6..A)1/9 B) 1/3 C) 4/27 D) 7/27 E) None of these

7/27...123 (6 arrangements) or 222 out of 3^3 =27 cases.

regards
scrabbler