Official Quant thread for CAT 2013

MBA CAT 2024
9861
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm

(x+2)/8 - x/20 =1
Solve it and x will be 10
So 10 hour after 9 Am is 7 Pm

So ans 7 PM
9862
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
By 9 am, p can fill 2(1/8) = 1/4th of the tank
Remaining part of the tank = 3/4th
After 9 am..every hour (1/8-1/20)th = 3/40th part of the tank gets filled
So it will take 10 more hours to fill up the remaining tank
7 pm
9863
@Cat.Aspirant123 said:
steps
let capacity of tank be 40 unit (LCM of 8 and 20)
then efficiency of p and q = 5 and -2unit/hour respectively
from 7 to 9 am work done by p= 5*2=10 units
work left = 40-10=30 units.......effective efficiency of both=5-2=3units/hr
hence time taken=30/3=10hr
hence 9am +10 hr=7pm
9864
@chillfactor said:
A group of children participated in a competition. Winner got n points and the child in r-th position got (n + 2 - 2r) points. Total of all the points secured by all the children is 2009. Find the smallest possible value of n.
P(r) = n + 2 - 2r

Total students = T

(n+2)*T - T*(T + 1) = 2009

T*( n - T + 1) = 2009 = 7*7*41

T = 7 => n - 7 + 1 = 287 => n = 293
T = 41 => n - 41 + 1 = 49 => n = 89
T = 49 => n - 49 + 1 = 41 => n = 89
and so on..

So, Smallest (n) = 89 ?
9865
@chillfactor said:
680A group of children participated in a competition. Winner got n points and the child in r-th position got (n + 2 - 2r) points. Total of all the points secured by all the children is 2009. Find the smallest possible value of n.
1st position = n
2nd position = n - 2
3rd position = n - 4
4th position = n - 6
.
.
.
rth position = n + 2 - 2r

2009 = 7*7*41

sum of 1st even natural numbers for (r - 1) terms = (r - 1)*r

rn - (r - 1)*r = 2009
r*(n - r + 1) = 2009

r = 49
n = 89
9866
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
Rate of p = v/8
Rate of q = -v/20

(v/8)*(x) - (v/20)*(x-2) = v

(x/8) - x/20 = 1 - 1/10
=> 3x/40 = 9/10 => x = 12 hours

=> 7 pm
9867
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
LCM (8 and 20) = 40 = total work

till 9 am P will fill 10 units after that their relative work done is = 3 units

total hrs = 2 + 30/3 = 10

7am + 12 = 7 pm
9868
@Cat.Aspirant123 said:
pipe p can fill d tank in 8 hrs, q can empty in 20 hours, on dat day p started at 7 a.m and q started at 9 am. At what time tank will be filled completely.9 pm7 pm5 pm1 pm
2+x/8-x/20=1
x=10
therefore 9:00 am +10hrs=7 pm
9869

Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?
1) 21
2)22
3)23
4)18.

9870
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
23
Went through options and eliminated one by one
9871
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
95x + 77y = 2724
x + y = 30

Solve and get x, y
x = 23 ?

9872
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
3) 23?
95 for target
77 for no target
95x + 77(30 - x) = 2724
95x + 2310 - 77x = 2724
18x = 414
x = 23
9873
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
95*x+77(30-x)=2724
x=23
9874
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
95x + 77y = 2724
x + y = 30

y = 7 and x = 23
9875
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
77y + 95 (30-y) =2724

Y=7
=> 30-7=23 Ans
9876
@soumitrabengeri
@YouMadFellow How to solve it using just first equation? i.e if we don't have x+y=30,in TIME'S solution they solved only using first equation.I wanted to know that method,since many times we have only one equation and two variable.
9877
@nole Well, generally, we need two equations for two variables, but sometimes a case may arrive where some implicit condition is already present. Like in this case, the implicit condition is that (x, y) are integers and positive because the number of days can't be negative and non-integral.
So, let's take a small example: 2x + 3y = 5
Now, in reality there are infinite solutions possible, but when we add the positive integral part, we observe that is only one solution i.e. (1,1)
In such cases, we divide both the sides by a common divisor and equate the remainders of both sides, if I divide by 2, then 0 + 1*(y mod 2) = 1 => y mod 2 = 1 -> y is of the form 2k + 1
This is just an example, you need to change your strategy as per the question.
9878
@nole said:
@soumitrabengeri@YouMadFellow How to solve it using just first equation? i.e if we don't have x+y=30,in TIME'S solution they solved only using first equation.I wanted to know that method,since many times we have only one equation and two variable.
Use the following method

95x+77y = 2724
Divide by 77 throughout

x+18x/77+y = 35+29/77
Considering the fact that x and y both are integers,

(18x - 29)/77 = k (some integer)
x = (77k+29)/18

Substitute values for k such that x is an integer
You get x = 23 for k = 5
9879
@nole said:
Nikhil takes up an assignment.On each working day he is given a target.For each day he meets target he is paid rs95 and for each day he doesn't meet target he is paid rs18 less.By the end of the month(30 days) he is paid a total of rs 2724.For how many days did he meet the target ?1) 212)223)234)18.
Without writing: If he met target on all days he would get 95 x 30 = 2850. He received 2724 i.e. 126 less which is 18 x 7 so he missed target for 7 days. Hence achieved it on 23 days.

regards
scrabbler
9880
A shopkeeper makes a profit of Q% by selling an object for rs24.Had c.p and s.p be interchanged it would have led to a loss of 62.5Q%.With the latter c.p what should be the new s.p to make a profit of Q%?
a.34.4 b. 32.5 c.25.6 d.38.4
please tell the approach to solve?