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Originally Posted by grimreaper i see your point... but then how did u arrive at this... i mean... u take the condition of each day 6 men more to the initial problem itself right...
but even then... i can only think of trial and error to get at this table....
even then i wnt be sure till my total comes upto 545....
i mean yes i would have started with only one man on the first day... but nonetheless... i still wld have to calculate all the way till day 5 to be sure....
but yes the table is right
@ mohsan  post your method for that circular Question..... lets see what you did , and try the previous Qs posted in this forum... @GrimSaar:
I need to learn such shortcuts and quckmethods...please do help me with tips and suggestions...
I'm pretty old fashioned
Assumed x as no of men on 1st day and N as no. of crates/men on 1st day
x.N + (x+6)(N5) + (x+12)(N10) + (x+18)(N15) + (x+24)(N20) = 545
(x+12)(N10)=169
Clearly N>20
So, X=1 and N=23...
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by grimreaper yup.... an easy prob ... ehmmm bt x=2006 :P not 1003.... oops !!.... yeah its 2006 !


Originally Posted by shaunak_87 @GrimSaar:
I need to learn such shortcuts and quckmethods...please do help me with tips and suggestions...
I'm pretty old fashioned
Assumed x as no of men on 1st day and N as no. of crates/men on 1st day
x.N + (x+6)(N5) + (x+12)(N10) + (x+18)(N15) + (x+24)(N20) = 545
(x+12)(N10)=169
Clearly N>20
So, X=1 and N=23... I guess, with nothing else mentioned, the only way thru this question was the long way.... however, calculation can b eased a bit using AP summation.


Read the information given below and answer the question that follows:
Kallu kallan kalia's night club number is a 3 digit perfect square.This number when written in the reverse order also gives a perfect square and is telephone extension of kallu's office.His Mercedes registration number is a four digit perfect square formed by repeating the right most digit of his nightclub number.
The sum of the digits of kallu's night club number is
(A) a perfect square
(B) an even number
(C) a prime number
(D) a perfect number


Originally Posted by mohasan02 I guess, with nothing else mentioned, the only way thru this question was the long way.... however, calculation can b eased a bit using AP summation. Somehow i feel that the method is not too long and can u elaborate on AP summation for this, causse there are 5 products and can be easily reduced
I feel the sum was made for the part
(x+12)(N10)=13*13 part...means the factorization part....
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by anjalig Read the information given below and answer the question that follows:
Kallu kallan kalia's night club number is a 3 digit perfect square.This number when written in the reverse order also gives a perfect square and is telephone extension of kallu's office.His Mercedes registration number is a four digit perfect square formed by repeating the right most digit of his nightclub number.
The sum of the digits of kallu's night club number is
(A) a perfect square
(B) an even number
(C) a prime number
(D) a perfect number hmmmm.... well i assume that when you say repeating the last most digit it means adding the digits once again at the end.... rather than repeatin the same digit four times cause else such a number can never be a perfect square.... since "aaaa" means 1111*a which cant be a square.
so said that... your number would be 144
when read backwards its 441 = 21^2
and the last digit repeated gives 1444 = 38^2
so well sum is 9 and its a perfect square....
though i dnt knw wht a perfect number means
so going with option a


Hedonist


Originally Posted by anjalig Read the information given below and answer the question that follows:
Kallu kallan kalia's night club number is a 3 digit perfect square.This number when written in the reverse order also gives a perfect square and is telephone extension of kallu's office.His Mercedes registration number is a four digit perfect square formed by repeating the right most digit of his nightclub number.
The sum of the digits of kallu's night club number is
(A) a perfect square
(B) an even number
(C) a prime number
(D) a perfect number four digit perfect square formed by same digit can be 0000
night club number = 400
Not sure
4000040 being right most digit of 400....Car no being 0000
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by grimreaper hmmmm.... well i assume that when you say repeating the last most digit it means adding the digits once again at the end.... rather than repeatin the same digit four times cause else such a number can never be a perfect square.... since "aaaa" means 1111*a which cant be a square.
so said that... your number would be 144
when read backwards its 441 = 21^2
and the last digit repeated gives 1444 = 38^2
so well sum is 9 and its a perfect square....
though i dnt knw wht a perfect number means
so going with option a Exactly,
aaaa can't be a perfect sqr...so the qstn seems ambiguous
thanks Grim Saar CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Hedonist
Originally Posted by shaunak_87 four digit perfect square formed by same digit can be 0000
night club number = 400
Not sure
4000040 being right most digit of 400....Car no being 0000 no no... i think its only last digit repeated once...
so number wld be 144 , 441 and 1444 else Question wnt hold....


Originally Posted by grimreaper no no... i think its only last digit repeated once...
so number wld be 144 , 441 and 1444 else Question wnt hold.... Thank U Saar
20^2,12^2,21^2,13^2,31^2,22^2 all have this PROP...
tHANK U SARR
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

The value of A + B that satisfies (6^30 + 6^30) (6^30  6 ^30) =
3^A8^B3^A 8^B is
A. 20
B. 60
C. 80
D. 40


plz solve ths..
king nandan is the king divyagarh.One day he decides to sell all the animals living in his big kingdom.Now,there are five states in his kingdom and the only animals which his kingdom is having are horses ,cows and sheeps.King sells all the animals to eight neighbouring states.Each of the neighbouring states purchases same number of animals at the following rate.
horse1700 each
cow200 each
sheep200 each
king nandan receives rs 28,500 in all.
what can be the maximum number of animals?
Options
1) 80
2) 130
3) 160
4) none of these
what are the respective numbers of horses,cows and sheeps in any state?
Options
1) (3,109,8)
2) (1,37,3)
3) (2,74,6)
4) cannot be determined


these profit loss questns r killin me !
machine X produces articles at the rate of 50 units/h.SP of each article is rs 100
and the cost of production is rs 40 each.However 20% of the articles are defective and hence cannot be sold.The rate of production of machine can be increased,but with every increase of N units/h production cost would increase by 2x% and the number of defectives would become (20+1.5x)%.
1)find the maximum increase in production that can be undertaken without incurring losses?
a)14 units
b)15 units
c)16 units
d)17 units
2)What will be the profit in 1hr if the machine produces 58 units/hr?
a)Rs 1839
b)Rs 1178
c)Rs 1253
d)Rs 1624


Originally Posted by anjalig The value of A + B that satisfies (6^30 + 6^30) (6^30  6 ^30) =
3^A8^B3^A 8^B is
A. 20
B. 60
C. 80
D. 40
option C. This can be worked out orally, but is a very good question, owing to its intimidating form when faced in CAT like time crunch !
Solution: (6^{30}+6^{30})(6^{30}6^{30}) = (6^{60}6^{60})
= 8^{20}.3^{60}8^{20}.3^{60}...(i)
Comparing powers of (i) & the given RHS, we get B=20 ; A=60.
*Note It would be gr8 to follow question numbering, as suggested by grimreaper. It wud help us keep a track !


Originally Posted by quixotic_me these profit loss questns r killin me !
machine X produces articles at the rate of 50 units/h.SP of each article is rs 100
and the cost of production is rs 40 each.However 20% of the articles are defective and hence cannot be sold.The rate of production of machine can be increased,but with every increase of N units/h production cost would increase by 2x% and the number of defectives would become (20+1.5x)%.
1)find the maximum increase in production that can be undertaken without incurring losses?
a)14 units
b)15 units
c)16 units
d)17 units
2)What will be the profit in 1hr if the machine produces 58 units/hr?
a)Rs 1839
b)Rs 1178
c)Rs 1253
d)Rs 1624 Juz had a glance thru ur question while in office (during lunch time), so not enuff time to post an entire solution... Though i may suggest a hint for you to try 
Taking 1 hr to be unit time :
Max profit = 2000/. Now an increase of N would increase CP by (x/50) & unused good by (20+1.5x)/100 [which means sales reduce in proportion of (20+1.5x)/100 ] Form the equation : SP = CP in terms of N & x and may be u express N interms of x  from where u may get max. N by putting some hit & trial value of x
I am sorry for not being able to write the entire solution due to lack of time ... kindly post, if the above hint helps u 

Originally Posted by mohasan02 Juz had a glance thru ur question while in office (during lunch time), so not enuff time to post an entire solution... Though i may suggest a hint for you to try 
Taking 1 hr to be unit time :
Max profit = 2000/. Now an increase of N would increase CP by (x/50) & unused good by (20+1.5x)/100 [which means sales reduce in proportion of (20+1.5x)/100 ] Form the equation : SP = CP in terms of N & x and may be u express N interms of x  from where u may get max. N by putting some hit & trial value of x
I am sorry for not being able to write the entire solution due to lack of time ... kindly post, if the above hint helps u
thanx fr d hint.. tried..bt nt getting d answer.. plz tell me whr m wrong.. got following two eqns
(50+n)40  2000=x/50 *2000
so ,n= x
whr 2000 is the cost price of original 50 items.
now sp=cp
100(50+n)  [(20+1.5x)/100 ](50+n)100=2000+ x/50 * 2000
since x=n
3n^2+70n4000=0


Hedonist
Originally Posted by anjalig The value of A + B that satisfies (6^30 + 6^30) (6^30  6 ^30) =
3^A8^B3^A 8^B is
A. 20
B. 60
C. 80
D. 40 ok.... well answer would be 80
when you see the lhs...realise it is nothing but a^2b^2 form..
hence u knw the only way 2^n*3^b = some 6^t is when n=b=t
hence directly a=3b here
also... a/2=30 as can be seen directly...
so a=60 b=20
so answer is 80


Hedonist
Originally Posted by quixotic_me thanx fr d hint.. tried..bt nt getting d answer.. plz tell me whr m wrong.. got following two eqns
(50+n)40  2000=x/50 *2000
so ,n= x
whr 2000 is the cost price of original 50 items.
now sp=cp
100(50+n)  [(20+1.5x)/100 ](50+n)100=2000+ x/50 * 2000
since x=n
3n^2+70n4000=0 huh?? why are you bringing that 2000 into the equation... i mean that value isnt constant....
you only need to equate the S.P for a giving production = the C.P for a given x
S.P=100*(50+x)( 1 defect %)
C.P= (50+x)(40 + 40*2x/100)
but for some reason i am getting 17.3 as answer....
and the profit as 1252.8 as the answer...
u gt the answer with you ??

The Following User Says Thank You to grimreaper For This Useful Post:

Originally Posted by grimreaper huh?? why are you bringing that 2000 into the equation... i mean that value isnt constant....
you only need to equate the S.P for a giving production = the C.P for a given x
S.P=100*(50+x)( 1 defect %)
C.P= (50+x)(40 + 40*2x/100)
but for some reason i am getting 17.3 as answer....
and the profit as 1252.8 as the answer...
u gt the answer with you ?? oh yeah !! u got it right !!answer is 17 units and 1253
thanx a tonnnnn !!

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