Official Quant thread for CAT 2013

@vbhvgupta said:
Farook marks up the price of an article by 50% and then offers a discount of 20% to Shahrukh. Shahrukh sells it for र20 more than the price at which he purchased it. If Shahrukh's selling price is 30% more than the original cost price of the article, then Shahrukh's profit percentage is 10%9%6.66%8.33% ANSfor farook let the cp = 100 mp = 150 so sp = 120for shahrukh cp = 120 sp = 120+20 = 140 profit = (20*100)120 = 16.66where I am doing mistake???
see friend...had the complete transaction be done in terms of percentage..your method would have hit the gold...but unfortunately, in this case, we have an absolute value (Rs. 20)..hence the basic assumption of CP being 100 won't take you on the correct path..(this method is useful only when we are talking in terms of percentage).

here to solve it, assume CP to be x (or 100x for convenience in calculations)and proceed accordingly, you will get the required ans.

@vbhvgupta said:
The prices per litre of petrol, diesel and CNG are in the ratio 1 : 1/2 : 1/3 while the mileages (in km/litre) are in the ratio 2 : 1 : 3/2.If petrol, diesel and CNG that cost the same amount is filled in 3 similar cars respectively, the distance travelled by the vehicles on that fuel is in the ratio of 4 : 4 : 94 : 6 : 96 : 6 : 99 : 9 : 4
the prices are in the ratio 6:3:2..and the mileage is in the ratio 4:2:3

hence if they fill fuel for Rs. 60, the amount of each will be in the ratio of 10:20:30

hence they will run in the ratio 40:40:90 or 4:4:9
@vbhvgupta said:
The prices per litre of petrol, diesel and CNG are in the ratio 1 : 1/2 : 1/3 while the mileages (in km/litre) are in the ratio 2 : 1 : 3/2.If petrol, diesel and CNG that cost the same amount is filled in 3 similar cars respectively, the distance travelled by the vehicles on that fuel is in the ratio of 4 : 4 : 94 : 6 : 96 : 6 : 99 : 9 : 4
4:4:9??
Four pieces of cake A, B, C and D have weights
6 3/10 lbs, 11 1/5 lbs, 24 1/2 lbs, 15 3/4 lbs
respectively. A and B have to be cut into parts of equal weights. C and D have to be cut into parts of equal weights. Furthermore, each part is cut in such a way that each piece is as large as possible. All the pieces of A and B are served to x guests-one piece per guest and all the pieces of C and D are served to y guests-one piece per guest. What is the value of x - y?
In the sequence 1,9,7,7,4,7,5,3,9,4,1, €Ś every digit from the fifth on is the sum of the preceding 4 digits mod 10.Does one of the following set ever occur in the sequence?

a) 1,2,3,4 b) 3,2,6,9 c) 0,1,9,8 d) 7,9,5,3

1.For each natural number N, define S(N) be the sum of digits of N.

For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18
N = 25, S(N) = 2 + 5 = 7

How many natural numbers, N, exist such that

S(N) + S(N^2) = 2011

@vbhvgupta -is 0 the answer?what are the options
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?

(a) 147
(b) 76
(c) 123
(d) Cannot be determined

@joyjitpal 123
@vbhvgupta said:
Four pieces of cake A, B, C and D have weights 6 3/10 lbs, 11 1/5 lbs, 24 1/2 lbs, 15 3/4 lbs respectively. A and B have to be cut into parts of equal weights. C and D have to be cut into parts of equal weights. Furthermore, each part is cut in such a way that each piece is as large as possible. All the pieces of A and B are served to x guests-one piece per guest and all the pieces of C and D are served to y guests-one piece per guest. What is the value of x - y?
HCF of A & B=7/10
HCF of C & D =7/4
Total pieces of A,B,C & D are 9,16,14 & 9 respectively
x=25, y=23
x-y=2
@audiq7 said:
@joyjitpal 123
how?
@sparklingaubade said:
1.For each natural number N, define S(N) be the sum of digits of N.For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18 N = 25, S(N) = 2 + 5 = 7How many natural numbers, N, exist such thatS(N) + S(N^2) = 2011N+N^2 = 2011
none ? :|
@joyjitpal said:
how?
suppose t7 and t6 are the seventh and 6th terms.
so t7^2-t6^2=517
(t7+t6)(t7-t6)=47*11

so t7=29 and t6=18.
now calculate 10th term
@joyjitpal said:
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?(a) 147 (b) 76 (c) 123 (d) Cannot be determined
123?
@sparklingaubade -d??the whole sequence has 4odd numbers followed by even..in other words sequence mod 2 yields 11110..the 1st three options are not in sync with the sequence but the fourth one seems to be..just an observation..wats the OA
@joyjitpal said:
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?(a) 147 (b) 76 (c) 123 (d) Cannot be determined
Difference between the squares of seventh and sixth term is 517 ( product of 2 primes 47 and 11) if a and b are the sixth and seventh term respectively then b^2 - a^2 = ( b + a) (b-a) = 47 * 11.. Solving b = 29 and a = 18...and now keep on calculating the further terms.. I think the answer ( 10th term) is 123
@sparklingaubade said:
1.For each natural number N, define S(N) be the sum of digits of N.For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18 N = 25, S(N) = 2 + 5 = 7How many natural numbers, N, exist such thatS(N) + S(N^2) = 2011N+N^2 = 2011
0 ?

N*(N+1) = 2011 , but N*(N + 1) is always even for N being a natural number

Also, S(N) mod 9 = r , S(N^2) mod 9 = (r)^2

So, r + r^2 = 2011 mod 9 = 4
r = 1,-1,2,-2,3,-3,4,-4 don't satisfy
@sparklingaubade said:
1.For each natural number N, define S(N) be the sum of digits of N.For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18 N = 25, S(N) = 2 + 5 = 7How many natural numbers, N, exist such thatS(N) + S(N^2) = 2011N+N^2 = 2011
i don't think there exist any such (just a blind guess though have some clue)

see i got to this conclusion using this logic.

the sum of N and N^2 is 2011..so the number has to be less than 45..
secondly, since the sum of its digits is 2011, this implies that a number has to be numerous digits(dunno how many)

hence none
@joyjitpal said:
For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?(a) 147 (b) 76 (c) 123 (d) Cannot be determined
should be 123

@sparklingaubade said:
In the sequence 1,9,7,7,4,7,5,3,9,4,1, €Ś every digit from the fifth on is the sum of the preceding 4 digits mod 10.Does one of the following set ever occur in the sequence?a) 1,2,3,4 b) 3,2,6,9 c) 0,1,9,8 d) 7,9,5,3
anyone ?