Official Quant thread for CAT 2013

MBA CAT 2024
10421
@bullseyes said:
X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?(1) The hundreds digit of XY is 6.(2) The tens digit of XY is 4.a) using 1 aloneb) using 2 alonec) using bothd) eithere) CBD
d) Either?
10422
@bullseyes said:
X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?(1) The hundreds digit of XY is 6.(2) The tens digit of XY is 4.a) using 1 aloneb) using 2 alonec) using bothd) eithere) CBD
Total No. Possible
111
112,211,121
122,212,221
222

Total Multiplication Possible
-> 122*221=26962
-> 212*212=44944
> 112*211=23632
> 121*121=14641
> 222*222=49284


BY FIRST STATEMENT We have two no's 23632 & 14641. in both these no. 's divided by 3 remainder is 1

So this statement is sufficient

Again in second statement We have two no's 44944 & 14641. in both these no. 's divided by 3 remainder is 1
So this statement is sufficient


So ans is D.Either
10423
If N = 1! 2! + 3! 4! +..+ 47! 48! + 49!, then what is the unit digit of N^N?(a) 0 (b) 9 (c) 7 (d) 1
10424
@Aman.Malhotra said:
If N = 1! 2! + 3! 4! +..+ 47! 48! + 49!, then what is the unit digit of N^N?(a) 0 (b) 9 (c) 7 (d) 1
it shud be 6 yaar,,no option..
2 + xx4 + xx0 + xx0........+xx0 = xxxxx6
and no ending with raised to any power ends in 6
10425
@bullseyes said:
X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?(1) The hundreds digit of XY is 6.(2) The tens digit of XY is 4.a) using 1 aloneb) using 2 alonec) using bothd) eithere) CBD
d) using 1 alone?
if xyz is a number
then tens digit of xyz*zyx=(x*y)+(y*z)
(x*y)+(y*z)=4 only true for 121 or 212, mod 3= 1 or 2 so can't be determined by this

hundred digit will be x^2+y^2+z^2
, should be 6

numbers can be 121, 112, 211, mod 3 for all=1
10426

@rkshtsurana ya getting same but answer is given 1
thanks
10427
@Aman.Malhotra said:
If N = 1! 2! + 3! 4! +..+ 47! 48! + 49!, then what is the unit digit of N^N?(a) 0 (b) 9 (c) 7 (d) 1
N = 1! 2! + 3! 4! +..+ 47! 48! + 49!
Nmod10=(1!*2!+2!*3!)mod10 as combinations having 5!or higher will be divided by 10
Nmod10=2+4=6, unit digit=6
10428
How many four letter word from the word "INEFFECTIVE" ?????
10429
@Aman.Malhotra said:
If N = 1! 2! + 3! 4! +..+ 47! 48! + 49!, then what is the unit digit of N^N?(a) 0 (b) 9 (c) 7 (d) 1
getting 6. but not in option
please clarify...
10430
@Cat.Aspirant123 said:
Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
75 litres is the answer
10431
@Aman.Malhotra said:
How many four letter word from the word "INEFFECTIVE" ?????
1422?
10432
@Aman.Malhotra 1422.
10433

@Aman.Malhotra

All words diff. : 7P4

All words same : 0

2 same 2 diff : 3C1* 6C2 * 4!/2!

2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}

3 same 1 diff : 6

Adding all above we get the required OA......

10434
@LeoN88 said:
@Aman.MalhotraAll words diff. : 7P4All words same : 02 same 2 diff : 3C1* 6C2 * 4!/2!2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}3 same 1 diff : 6Adding all above we get the required OA......

2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}
Here We will use 3C2 instead of 3C1*2C1

and here " 3 same 1 diff : 6" you forgot to multiply by 4!/3!
10435
@Aman.Malhotra said:

2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}
Here We will use 3C2 instead of 3C1*2C1

and here " 3 same 1 diff : 6" you forgot to multiply by 4!/3!
I guess 1st mistake you identified is ok, Yeah got it, it was a big mistake :-)
& yes I forgot to * 4!/3! in next case...........
10436

@Omkarp said:
using either?
@audiq7 said:
@bullseyes is it C?

@Aman.Malhotra said:
Total No. Possible111

So this statement is sufficientSo ans is D.Either
@ScareCrow28 said:
d) Either?
nope




@Torque024 said:
d) using 1 alone?
yes
10437
The product of the ages of some teenagers is 10584000. The sum of their ages is equal toA. 86 B. 88 C. 85 D. 89 E. 87
10438
@Aman.Malhotra said:
The product of the ages of some teenagers is 10584000. The sum of their ages is equal toA. 86 B. 88 C. 85 D. 89 E. 87
14*14*15*15*15*16 = 10584000
Sum = 89
10439
@Aman.Malhotra 89 ?
=>10584000=2^6*3^3*5^3*7^2
So we can take the ages as : 14,14,15,15,15,16
Sum=14+14+15+15+15+16=89
10440
@Aman.Malhotra said:
2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}Here We will use 3C2 instead of 3C1*2C1and here " 3 same 1 diff : 6" you forgot to multiply by 4!/3!
what is the mistake if we do : "2 same 2 same : 3C1* 2C1 * (4!)/{(2!) * (2!)}"