Official Quant thread for CAT 2013

MBA CAT 2024
5741
@vijay_chandola said:
OA of the previous question: a) 0

Two different solutions of honey, milk and water are mixed with each other three times in varying proportions. The concentration of honey and milk in the three resulting solutions are found to be (10%, 16%), (12%, 12%) and (16%, x%) respectively. What is the value of x?
(a) 4 (b) 7 (c) 8 (d) 10

bhai whats the OA
and pls share approach.... bilkul logic smajh nahi aa raha
5742
@vijay_chandola said:
OA of the previous question: a) 0Two different solutions of honey, milk and water are mixed with each other three times in varying proportions. The concentration of honey and milk in the three resulting solutions are found to be (10%, 16%), (12%, 12%) and (16%, x%) respectively. What is the value of x?(a) 4 (b) 7 (c) 8 (d) 10
solution btao...
5743

Fid how many positive integers a less than or equal to 50 is (n-1)! divisible byn?

a 16
b 33
c 32
d non of these
5744

@bhatkushal said:
Fid how many positive integers a less than or equal to 50 is (n-1)! divisible byn?a 16b 33c 32d non of these
please share the approach for this solution I beliieve the answer should be 34
5745
@REVEALED said:
a=1.7*3b=1.7*4ryt angle trianglep=12*1.7/5=4.08
D=_/(5.1^2+6.8^2)=8.5
max distance = r = 8.5/2=4.25
5746
@bhatkushal said:
Fid how many positive integers a less than or equal to 50 is (n-1)! divisible byn?
a 16
b 33
c 32
d non of these
bhai question thoda reframe kar do, samajh nahi aa rha
5747
@vijay_chandola said:
2007^2007 mod 100=7^2007 mod 100=7^7 mod 100= 43 mod 100==> tens digit=4
tens digit
2007^2007
(2000+7)^2007
7^2007
2007mod4=3
7^3=343
4 is the tens digit
5748
@techsurge said:
bhai whats the OAand pls share approach.... bilkul logic smajh nahi aa raha
OA: 4 %

Let the %age of honey and milk in the two solutions be (a%, b%) and (c%, d%) respectively. According to the question:
(a – 10)/(b – 16)=(10 – c)/(16 – d)

Solving, we get,
16a – ad – 160 + 10d = 10b – 160 – bc + 16c
So, ad – bc = 16a – 10b – 16c + 10d ...(1)

Similarly from the second and third proportions we can say that

ad – bc = 12a – 12b – 12c + 12d. ...(2)
and ad – bc = xa – 16b – xc + 16d ...(3)
From (1) and (2), we get
(a – c)/1=(d – b)/2

Also from (2) and (3), we get
(a – c)/4=(d – b)/(12 – x)

Hence,
4/1=(12 – x)/2
8 = 12 – x and x = 4.

Quite lengthy it is. Posted it here in search of some short cut method

5749
For how many positive integers n,less than or equal to 50 is (n-1)! divisible by n?
a 16
b 33
c 32
d non of these
5750
@krum maxm distance from the centre of circle to p(intersection point)
1/5.1^2+1/6.8^2=1/d^2
d is in the hypo and centre is the mid point of the hypo
5751
@bhatkushal said:
For how many positive integers n,less than or equal to 50 is (n-1)! divisible by n?a 16b 33c 32d non of these
50 - all primes
=> 50 - 15 = 35
5752
@bhatkushal said:
For how many positive integers n,less than or equal to 50 is (n-1)! divisible by n?a 16b 33c 32d non of these
34? all prime no + 2
5753
@krum said:
50 - all primes => 50 - 15 = 35
wat abt 2
5754
@bhatkushal all prime numbers + a number 4
16primes+1[4]=17 numbers
5755
@REVEALED said:
@krum maxm distance from the centre of circle to p(intersection point)1/5.1^2+1/6.8^2=1/d^2d is in the hypo and centre is the mid point of the hypo
5756
@bhatkushal said:
For how many positive integers n,less than or equal to 50 is (n-1)! divisible by n?a 16b 33c 32d non of these
c ?
5757
@bhatkushal said:
For how many positive integers n,less than or equal to 50 is (n-1)! divisible by n?
a 16
b 33
c 32
d non of these
35 removing all primes till 50, not sure but
5758
@REVEALED said:
@bhatkushal all prime numbers + a number 416primes+1[4]=17 numbers
even ifigured the 16 primeshow could you point the extra 4
5759
@bhatkushal said:
even ifigured the 16 primeshow could you point the extra 4
also till 50 we will have 15 primes
5760
@maddy2807 said:
wat abt 2
3! mod 4=2
4 bhi aayega na?