Official Quant thread for CAT 2013

@sharmaa.abhay said:
Three people A, B, C participate in a cycling race. They start simultaneously from point P, reach another point Q, turn back without changing their speeds and return to P. In the process, each person meets each of the other 2 at either of the points R or S(b/w P and Q). RQ=10 km, SQ=18 km.What is the ratio of the fastest to the slowest? Assume P and Qare in a straight line.(a) 2:1 (b)3:2 (c) 4:1 (d) 5:2 (e) Can't be determinedOA is (c). Please explain your approach.
A=fastest
B=second fastest
C=slowest
A/C=36+x/x
B/C=28+x/8+x
A/B=x+28/x+8

A/C*C/B=A/B
36+x/x*8+x/28+x=28+x/x+8
x=12
A/C=36+12/12=4/1
@gautam22 get it .. thanks

@sharmaa.abhay said:
Ajit invested a certain amount of money in a FD. It pays at r% interest p.a. compounded annually If interest credited for 4th and 6th year is Rs. 12000 and Rs. 14520 respectively, what is the amount invested by ajit?(a) 90157 (b) 88157 (c) 92157 (d) 93517 (e) 94157OA is (a). Please explain your approach.
Let A be d amt invested initially and R be the interest,
So interest for 4th yr=AR/100*(1+R/100)^3=12000
n for 6th yr= AR/100*(1+R/100)^5=14520
on solving above 2 equations,

R=10 and A~90000
@catahead said:
The interetst for the 6th year is the CI on the interest in 4th year(i.e 12000 ) for two yearsSo 14520 = 12000(1+r/100)^2 => r=10%Now, int in 4th year is the SI on the Amount after 3 years. So 12000 = P [(1+r/100)^3 ] (r/100) => 12000 = P [ (11/10)^3 ] (1/10) => P = 90157
bhai...interest for 6 th year should b CI ( P + 4 th year interest ) for 2 years na ...

whereas u hv considered on the CI of "interest part" ,,,

or m missing sumthing ....

@chandrakant.k said:
Starting off with an easy questionThe numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x
@chandrakant.k wht is the right ans?
@Estallar12 said:
Equation will be -(3x/4)*100*(1/10) +(x/4)*200*1/12 = 600Solving, x = 51 kg (Approx.)
cn u xplain it ... i m nt able to get ur xplaination 4 dis ques
@priyadhand said:
@chandrakant.k wht is the right ans?
I think that the answer should be 3.
Ist statement cannot be true for the following case..suppose units digit of x and y are 5 and 6 resp..then the units digit of x+y will be 1 which is not greater than 5 or 6
IInd statement cannot be true as well. If there is a carry forward from the units digit of the sum of x and y, then the tens place will increase by 1..which will make the tens digit 3 and not 2
IIIrd statement has to be true since the addition of x and y is a 4 digit number and x
Consider any possible case of 2 numbers such that hundreds place of y is less than 5 and x
@priyadhand
@soumitrabengeri thanks ... get it

Mini and vinay are quiz masters preparing for a quiz. in x minutes, Mini makes y questions more than vinay. If it were possible to reduce the time needed by each to make a question by two minutes, then in x minutes Mini would make 2y questions more than vinay. How many questions does Mini make in x minutes?
a. 1/4[2(x+y)-(2x^2 + 4y^2)^1/2]

b.1/4[2(x-y)-(2x^2 + 4y^2)^1/2]

c. either a or b

d. 1/4[2(x-y)-(2x^2 - 4y^2)^1/2]

please suggest .

at a bookstore "MODERN BOOK STORE" is flashed using neon lights.The words are individually flashed at long intervals of 5/2,17/4,41/8 seconds respectively,and each word is put of after a second.The least time after which the full name of the bookstore can be read again is..ďťż
@priyadhand said:
@chandrakant.k wht is the right ans?
ONly 3 is correct
For detailed explanation please look at the estallar12 sir's explanation 😃
@nole said:
at a bookstore "MODERN BOOK STORE" is flashed using neon lights.The words are individually flashed at long intervals of 5/2,17/4,41/8 seconds respectively,and each word is put of after a second.The least time after which the full name of the bookstore can be read again is..
LCM(7/2, 21/4, 49/8) = 147/2 = 73.5 seconds ?

@grkkrg yeah correct answer. how to solve this ? "each word is put of after a second " . how to use that part while solving the question ?
@nole said:
@grkkrg yeah correct answer. how to solve this ? "each word is put of after a second " . how to use that part while solving the question ?
Time taken by first word to repeat itself = 5/2 + 1 (ON cycle and OFF cycle)
Time taken by second word to repeat itself = 17/4 + 1
Time taken by third = 41/8 + 1

Take the lcm to find the next time they coincide.

This is similar to 3 people running around a circular track taking different times to complete one round. The time taken for them to meet at the starting point is the LCM of their individual times.

@grkkrg first one will flash at 5/2 seconds and will remain on till (5/2 +1 ) seconds.similarly
for second 17/4 to 17/4+1 second and so on.How do we know that if we add 1 second to all,we will get minimum possible time at which they all will flash together? can't it be 5/2+x seconds,
17/4+y seconds,....?? I was not able to get the logic of adding 1 second to all

@grkkrg ok got it thanks :)
@nole said:
@grkkrg first one will flash at 5/2 seconds and will remain on till (5/2 +1 ) seconds.similarly for second 17/4 to 17/4+1 second and so on.How do we know that if we add 1 second to all,we will get minimum possible time at which they all will flash together? can't it be 5/2+x seconds,17/4+y seconds,....??
No no
It flashes for 5/2 seconds then it is off for 1 second.
It is given that "each word is put of after a second".
There's a typo here. It should be put "off". :D
A manufacturer produces two types of products-A and B, which are subjected to two types of
operations, viz. grinding and polishing. Each unit of product A takes 2 hours of grinding and 3 hours of polishing whereas product B takes 3 hours of grinding and 2 hours of polishing. The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours/day and each
polisher 10 hours/day. The profit margin per unit of A and B are Rs. 5/- and Rs. 7/- respectively. If
the manufacturer utilises all his resources for producing these two types of items, what is the
maximum profit that the manufacturer can earn?
A. Rs.280/- B. Rs.294/- C. Rs.515/- D. Rs.550/- E. None of the above
@choosen04 said:
A manufacturer produces two types of products-A and B, which are subjected to two types ofoperations, viz. grinding and polishing. Each unit of product A takes 2 hours of grinding and 3 hours of polishing whereas product B takes 3 hours of grinding and 2 hours of polishing. The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours/day and eachpolisher 10 hours/day. The profit margin per unit of A and B are Rs. 5/- and Rs. 7/- respectively. Ifthe manufacturer utilises all his resources for producing these two types of items, what is themaximum profit that the manufacturer can earn?A. Rs.280/- B. Rs.294/- C. Rs.515/- D. Rs.550/- E. None of the above
294?
@choosen04 said:
A manufacturer produces two types of products-A and B, which are subjected to two
294 ??
@choosen04 said:
A manufacturer produces two types of products-A and B, which are subjected to two types ofoperations, viz. grinding and polishing. Each unit of product A takes 2 hours of grinding and 3 hours of polishing whereas product B takes 3 hours of grinding and 2 hours of polishing. The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours/day and eachpolisher 10 hours/day. The profit margin per unit of A and B are Rs. 5/- and Rs. 7/- respectively. Ifthe manufacturer utilises all his resources for producing these two types of items, what is themaximum profit that the manufacturer can earn?A. Rs.280/- B. Rs.294/- C. Rs.515/- D. Rs.550/- E. None of the above
B. Rs.294/-