Official Quant thread for CAT 2013

MBA CAT 2024
10161
@nole said:
A three digit number on getting divided by 21 leaves the remainder of 5, and on getting divided by 25 leaves the remainder leaves the remainder 16.Find out how many such numbers are possible.
N=21*x+5
=25*y+16

=> 25*y=21*x-11
=> y=(21*x-11)/25

Now 21*x-11 would be divisible by 25 only when 21*x will end up with digits 36, 86, 11 or 61

==> check for x=11,21, 31, 41, 16, 26, 36, 46
only x=41, 16 satisfies. so, 2 numbers
10162
@nole said:
@soumitrabengeri 341 and 866 are the correct answers.If you get the answer please tell me the complete logic because i asked such questions before also and there is also a link in pagalguy by maximus but still i didm't get it @sujamait : how are u checking the values for k and m ? please elucidate.
k=11+25m /21
now k is int.. so 11 + 25m shall be divi by 21
which is nothing but => 11+4m
m=13 satisfy.next will be m = 13+21 = 34 that will satisfy..now just put and get numbers.
10163
@nole said:
A three digit number on getting divided by 21 leaves the remainder of 5, and on getting divided by 25 leaves the remainder leaves the remainder 16.Find out how many such numbers are possible.
the three digit number is of the form 21x+5 or 25y+16

equating 21x+5 = 25y+16
21x-25y=11
no of solutions of the equation 21x - 25y =11
then what ?
@chillfactor

10164
In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

(1) 20
(2) 30
(3) 25
(4) 600
(5) 480

I know its easy. But had the toys been identical would the ans be different. If true then how so? Thanks
10165
@Ak92 said:
In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?(1) 20 (2) 30 (3) 25 (4) 600 (5) 480I know its easy. But had the toys been identical would the ans be different. If true then how so? Thanks
5 different toys and 3 identical boxes => Only the grouping of the toys matter

Case 1: (3,1,1) => (5!/3!1!1!2!) = 10
Case 2: (2,2,1) => (5!/2!2!1!2!) = 15

Total = 25 ?

When the toys are identical, only the types of groupings matter i.e. (3,1,1) and (2,2,1) => only 2 ways ?
10166
@nole said:
A three digit number on getting divided by 21 leaves the remainder of 5, and on getting divided by 25 leaves the remainder leaves the remainder 16.Find out how many such numbers are possible.
N = 21k + 5 = 25n + 16

21k = 21n + (4n + 11)
(4n + 11) is multiple of 21

4n + 11 = 21a
4n = (20a - 12) + (a + 1)
(a + 1) is multiple of 4
So a = 4p + 3

4n + 11 = 21(4p + 3)
n = 21p + 13

=> N = 25(21p + 13) + 16 = 525p + 341

So, 341 and 866
@tmohan02 said:
Find all numbers of the form 517xy that are divisible by 89.Any short cut method?????
517xy is divisible by 89
51700 + xy is divisible by 89

51700 = 580*89 + 80

So, 80 + xy should be divisible by 89

xy can be (89 - 80) or (178 - 80)

so, xy can be 09 or 98
10167

@Ak92 said:
In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?(1) 20 (2) 30 (3) 25 (4) 600 (5) 480I know its easy. But had the toys been identical would the ans be different. If true then how so? Thanks
case 1 : 1 1 3

no of ways 5! /1!*1!*3!*2!
equal to 10

case2: 2 2 1

no of ways of distributing 5!/2!*2!*1!*2! equal to 15

total 25

10168
@tmohan02 said:
Find all numbers of the form 517xy that are divisible by 89.Any short cut method?????
2 nos possible with
xy as
09
98

10169
@Ak92 said:
In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?(1) 20 (2) 30 (3) 25 (4) 600 (5) 480I know its easy. But had the toys been identical would the ans be different. If true then how so? Thanks
two cases are possible (1,1,3) and (1,2,2)

so, number of ways=5!/3!*2!+5!/2!*2!*2!=25

Check what is given identical and then try to make the combinations :D
10170
@tmohan02 said:
Find all numbers of the form 517xy that are divisible by 89.Any short cut method?????
N= 517xy = 51700 + 10x + y
N mod 89 = 0

(51700 + 10x + y) mod 89 = 0
80 + [(10x + y) mod 89] = 0

-> 10x + y = 89*k - 80

But LHS is a two digit number => k = 1 or 2

Numbers are 51709 and 51798

10171
If a, b are integers then how many ordered pairs (a, b) satisfy the equation a2 + ab + b2 = 1?
10172
@amresh_maverick said:
If a, b are integers then how many ordered pairs (a, b) satisfy the equation a2 + ab + b2 = 1?
a^2 + b^2 + 2ab - ab = 1
(a+b)^2 = 1+ab

a=0,b=1
a=1,b=0
1,-1
-1,1
-1,0
0,-1
6?
10173
@amresh_maverick said:
If a, b are integers then how many ordered pairs (a, b) satisfy the equation a2 + ab + b2 = 1?
6 orderd pairs...(-1,0): (0,-1) : (1,-1) : (-1,1) : (0,1) : (1,0)
10174
@antodaya said:
6 orderd pairs...(-1,0):(0,-1)1,-1)-1,1):(0,1):(1,0)
sahi jawab sir ji

How many arrangements of the letters of the word CATASTROPHE are there in which both the 'A's appear before both the 'T's?

approach plz ?

PS: @antodaya @sujamait @chillfactor sabko mera pranam
10175
@amresh_maverick said:
If a, b are integers then how many ordered pairs (a, b) satisfy the equation a2 + ab + b2 = 1?
a^2 + ab + b^2 = 1
4a^2 + 4ab + 4b^2 = 4
(2a + b)^2 + 3b^2 = 4

Only possibilities are:-
i) (2a + b)^2 = 1 and 3b^2 = 3
ii) (2a + b)^2 = 4 and 3b^2 = 0

i) 2a + b = +1 or -1 and b = +1 or -1
(a, b) = (0, 1), (-1, 1), (0, -1), (1, -1)

ii) 2a + b = +2 or -2 and b = 0
(a, b) = (1, 0), (-1, 0)

So, 6 solutions
10176
@antodaya said:
AATT-CSROPHEfix first 4 in their place... no 8 digitno...so arrange in 8! ways..?
options are :

11!/4!
11!/4!*2!^2
11!/6
11!*2!^2/4!
10177
@amresh_maverick said:
sahi jawab sir jiHow many arrangements of the letters of the word CATASTROPHE are there in which both the 'A's appear before both the 'T's?approach plz ?PS: @antodaya@sujamait@chillfactor sabko mera pranam
- a -a- t- t-
11C4*7!
10178
@amresh_maverick said:
options are :11!/4!11!/4!*2!^211!/611!*2!^2/4!
11!/4!

select 4 from 11 and arrrange rest 7 in 7! ways...

11c4 *7!
10179
@amresh_maverick said:

How many arrangements of the letters of the word CATASTROPHE are there in which both the 'A's appear before both the 'T's?approach plz ?
CATASTROPHE

Arrange them in (11!)/(2!2!) ways, but we need only those where AA TT are arranged in a particular way

So, (11!/2!2!)/ (4!/2!2!) = 11!/4! ?

EDITED
10180
@antodaya said:
11!/4!select 4 from 11 and arrrange rest 7 in 7! ways...11c4 *7!
why do we have to select : 11C4 ?