Official Quant thread for CAT 2013

MBA CAT 2024
23647
@iLoveTorres said:
i need a simpler solution to the question.. the diagram by chillfactor is incomprehensible
See the attached pic. I have rotated Triangle PCB by 60 degrees about C such that B coincides with A and M is the image of P. PCM becomes equilateral triangle. The whole idea is to get the value of angle APC
23648
@jain4444 said:
Out of 200 in an aquarium, 99% are red. How many red fish must be removed in order to reduce the percentage of red fish to 98%?
200.. 198 are red.

say x are removed.

200-x ... new total.

(198-x)*100 / 200-x = 98/100
or x=100.
23649
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
23650
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
12?
Edited 102?
23651

In triangle ABC, AB=AC and D is any point on BC. Find BD, if AB=65cm AD=63cm and CD=8cm
a) 8cm
b) 24cm
c) 32cm
d) 16cm

23652
A set S consists of first n natural numbers such that S = {1, 2, 3, 4, ....., n}. For each of its non-zero subset, alternating sum is defined as :-
Arrange the elements in subsets in decreasing order and then alternately add and subtract successive terms such that largest term is always added. For eg:- alternating sum for (1, 3, 5, 7} is 7 - 5 + 3 - 1 = 4 and for {3}, its just 3

Find the sum of all such alternating sums for n = 8
23653
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
57?
23654
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
if repetition not allowed (x y and z are different)
2even, one 5
4c2*3!
case 2
(4,8), (5),(1,3,7,9)
2c1*4c1*3!
18+48
66
23655
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
if repetition is allowed
66 as above post
case 3 (4,8),(5)(5)
2c1*3!/2!
=6
case 4
2 same even numbers, one 5
4c1*3!/2!
=4*3
sum
=18
total =66 +18=84
ordered so combination matters
23656
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
Is it 74?
23657
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
Intelligent question :mg:

Take two number whose product ends in 0

40 --> 8,5 ---> third number can take 7 values --> 3! * 7 = 42

20 --> 4,5 -- > third number can take 7 values --> 3! * 7 = 42

30 --> 6,5 --> third number can take 2 values (2,6) -- > 3 + 6 = 9

10 --> 2.5 ---> third number can take 2 values (2,6) -- > 3 + 6 = 9

so intotal 102 ways




23658
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
Consider following cases:-
i) 5(2a)(2b) , so 3*4*4 = 48 cases

ii) 5*4s*t===> 6*2*5 - 3*2 = 54 cases
(3*2 is subtracted as 5*4*5 or 5*8*5 are counted 6 times instead of 3 times)

So, total = 48 + 54 = 102
@ravi.theja said:
In triangle ABC, AB=AC and D is any point on BC. Find BD, if AB=65cm AD=63cm and CD=8cma) 8cmb) 24cmc) 32cmd) 16cm
Isosceles triangle and ADC is rt angle triangle
=> ADC and ADB are congruent
=> BD = CD = 8

Oops calculation mistake!!

We can use cosine rule twice and then we will get BD
23659
@ravi.theja said:
In triangle ABC, AB=AC and D is any point on BC. Find BD, if AB=65cm AD=63cm and CD=8cma) 8cmb) 24cmc) 32cmd) 16cm
32 ?
23660
@jain4444 said:
How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
5*4*z ==> 9 values ; total ways 7*6 +2*3 = 48 ;

5*2*z ==> 3 values ; z=2,6,8 ==> 15 ways ;

5*8*z ==> z= 1,3,5,6,7,8,9 ==> 5*6+2*3 = 36 ways ;

total = 102 ??
23661
@naga25french said:
Intelligent question

Take two number whose product ends in 0

40 --> 8,5 ---> third number can take 7 values --> 3! * 7 = 42

20 --> 4,5 -- > third number can take 7 values --> 3! * 7 = 42

30 --> 6,5 --> third number can take 2 values (2,6) -- > 3 + 6 = 9

10 --> 2.5 ---> third number can take 2 values (2,6) -- > 3 + 6 = 9

so intotal 102 ways

i might be wrong but isnt their repetiton in first 2 cases
(4,8,5)
and
(8,4,5)
pardon my ignorance, bad at pnc
23662
@ravi.theja said:
In triangle ABC, AB=AC and D is any point on BC. Find BD, if AB=65cm AD=63cm and CD=8cma) 8cmb) 24cmc) 32cmd) 16cm
23663
@techsurge said:
i might be wrong but isnt their repetiton in first 2 cases(4,8,5)and (8,4,5)pardon my ignorance, bad at pnc
Ordered triplets pucha hai :)
23664
@ScareCrow28 said:
Ordered triplets pucha hai
but in both cases naga bhai hai use 3! for ordering it
1st case he used 8,5 and 3rd as one among (1,2,3,4,6,7,9)
selecting in 7 ways ordering it in 3!
2nd case 4,5 and 3rd among (1,2,3,6,7,8,9)
here ordering in 3! as above
1n 1st case 1 selected 8,5 with 4 ordering in 3! ways
2nd case i selected 4,5 with 8 among 7
ordering in 3! ways
isnt it repetition
23665
@chillfactor said:
Consider following cases:-i) 5(2a)(2b) , so 3*4*4 = 48 casesii) 5*4s*t===> 6*2*5 - 3*2 = 54 cases(3*2 is subtracted as 5*4*5 or 5*8*5 are counted 6 times instead of 3 times)So, total = 48 + 54 = 102Isosceles triangle and ADC is rt angle triangle=> ADC and ADB are congruent=> BD = CD = 8Oops calculation mistake!!We can use cosine rule twice and then we will get BD
sir can u please explain me how to apply cosine rule..am not able to proceed .tanx in advance sir
23666
@chillfactor
I think i have misinterpreted the question, please clarify if this is what the question is asking:
sum of natural numbers from 1-8= 36 . so take the negative component to x and positive one to be 36-x. hence the alternating sum will be of the form (36-x)-x = 36 -2x, and since 36-2x>0 because of largest number being placed first x can range from 1-18.
so the sum will be 36*9 - 2(18*19) = 306. I haven't filtered out the invalidate cases and such yet, just want to confirm if my interpretation of the question is correct?