official quant thread for cat08

[KIMSTER]

Originally Posted by getintoiimb

puys pls solve these with steps !

3)a shopkeeper makes a profit of Q% by selling an object for Rs 24.has the cP and SP been interchanged it would have led to a loss of 62.5Q%.with the latter CP what should be the new SP to get a profit of Q%?

ans : Rs 38.8
Thanks!

Small doubt...
when cP and SP been interchanged it would have led to a loss of 62.5Q%..

its 62.5*Q% only no?? and not 62.5%...?? just confirming...

[selebratinglife]

Originally Posted by implex

New Problem!!
Quadrilateral ABCD is inscribed in a circle with diameter AD = 4, if side AB and BC each have length 1, then find CD?

angle ABD=ACD=90 degrees since angle in a semi circle ..And BD bisetc the angle ADC can be proved by alternate segment theorem and ABC being isosceles triangle..From triangles ADB and ADC it can be deduced that CD=3.5

[getintoiimb]

Originally Posted by KIMSTER

Small doubt...
when cP and SP been interchanged it would have led to a loss of 62.5Q%..

its 62.5*Q% only no?? and not 62.5%...?? just confirming...

even i was not clear on this..but i have typed the question exactly as given in Arun sharma..quant gods here pls explain this question and the solution as well

[implex]

Originally Posted by naga25french

i guess the number for the given condition is 510511... but i am afraid that this is not the right answer as its not divisible by 17 but this number satisfy the given condition

@implex

am i atleast close to the answer????

answer is wrong . the concept is simple
it should be k times lcm of(2,3,4..16) +1
N=k.lcm(2,3,4..16) +1
we will find the number k as 10

so the number is N=10.16.9.5.7.11.13+1

[implex]

Originally Posted by deep@k

CD = 1..?
is it correct?

nopes it's not

options are a) 5/2 b) 3 c) 7/2 d) 4 e) none of these

[the_hate]

Originally Posted by naga25french

New Problem:
Find the least multiple of 17 which when divided by 2,3,4..16 leaves a remainder of 1 in each case

LCM(2,3,4...16) = 2^4*3^2*5*7*11*13
=16*9*5*7*11*13

so no must be of form 16*9*5*7*11*13*k + 1 = 17*m

where k,m are integers

m = (720720k+1)/17
= 42395 + (5k+1)/17
smallest value for k such that (5k+1)/17 is integer is 10
m=42395+3 = 42398

m not confident bout calculation portion

--------------------
the calcln is wrong...must hv seen it...no ends in 8 and we need remainder 1 when divided by 5...
@implex..thnx. ur approach is better

[raghav507]

Originally Posted by implex

nopes it's not

options are a) 5/2 b) 3 c) 7/2 d) 4 e) none of these

i think CD must be 3.5...
i have used the formula AC.BD = AB.CD + AD.BC
here CD can be taken as a variable x and the other sides can be found out in terms of x(using the Pythagoras theorem).
it will result in a quadratic equation with roots 7/2 and -4.

hence the answer...

[implex]

Originally Posted by the_hate

LCM(2,3,4...16) = 2^4*3^2*5*7*11*13
=16*9*5*7*11*13

so no must be of form 16*9*5*7*11*13*k + 1 = 17*m

where k,m are integers

m = (720720k+1)/17
= 42395 + (5k+1)/17
smallest value for k such that (5k+1)/17 is integer is 10
m=42395+3 = 42398

m not confident bout calculation portion

nice attempt but do it in a simpler way
16*9*5*7*11*13 gives a remainder 5 on division by 17
5.10+1 =51 divisible by 17
so the number is 10.16*9*5*7*11*13+1

[implex]

Originally Posted by raghav507

i think CD must be 3.5...

yeah it's 3.5 can u post ur solution!!

[selebratinglife]

Originally Posted by implex

yeah it's 3.5 can u post ur solution!!

Posted my solution..You seem to have missed it ..Anyway, reposting
angle ABD=ACD=90 degrees since angle in a semi circle ..And BD bisetc the angle ADC can be proved by alternate segment theorem and ABC being isosceles triangle..From triangles ADB and ADC it can be deduced that CD=3.5