Assume that they plan to complete the work in n days.
So the amount of ore that they plan to mine on a single day is 1800/n tonnes but on n/3 number of days they mine only ( 1800/n - 20) tonnes So the amount of ore mined in n/3 days = ( 1800/n - 20) * n/3 ---(1) By completing it one day before the scheduled time they complete the remaining it in (2n/3 -1) days. i.e. for 2n/3 -1 days they mine ( 1800/n + 20 ) tonnes each day So the amount of ore mined in 2n/3 -1 days = ( 1800/n + 20 ) * (2n/3 -1)
i guess u hve made a mistake here....c d highlighted portion..
if per day dey mine 1800/n tonnes in n/3 days dey shud mine 600 tonnes....but they actually mine 580 tonnes...
in d latter part f d solution u again assume dat its 20 tonnes makeup on per day basis...d sum says smethin else...20 tonnes are made up over (2n/3-1) days.
so rem tonnes = 1800-580 =1220 per day dey mine 1220/(2n/3 -1) tonnes.
u have framed d equations wrongly
now u can equate n get ur ans... still wondering how u got the correct ans though!!!
My understanding of this is that the 20 tonnes that they overachieved as well as under achieved on per day basis. Lemme try and approach this answer using the options, for both per day basis as well as overall basis.
If the overachieve/ underachieve 20 tonnes per day :-
Option I : tonnes/day 50 days = 36 amt mined in 1/3 days = 12*30 amt mined in 2/3rd - 1 dz = 70*23 total:1870 Option II : tonnes/day 100 days =18 amt mined in 1/3 days = 6*80 amt mined in 2/3rd - 1 dz = 120* 11 total:1800 Option III: tonnes/day 250 days = 12 amt mined in 1/3 days = 4*130 amt mined in 2/3rd - 1 dz = 170*7 total:1710 Option IV: tonnes/day 200 days = 9 amt mined in 1/3 days = 3*180 amt mined in 2/3rd - 1 dz = 220*5 total:1640
If the overachieve/ underachieve 20 tonnes in 1/3rd of the days :- Option I : tonnes/day : 50 no of days: 36 amt mined in 1/3 days = 12*50 -20 amt mined in 2/3rd - 1 dz = 23*50 +20 total:1750 Option II : tonnes/day:100 no of days:18 amt mined in 1/3 days = 6*100 -20 amt mined in 2/3rd - 1 dz = 100* 11 +20 total:1700 Option III: tonnes/day: 150 no of days = 12 amt mined in 1/3 days = 4*150 -20 amt mined in 2/3rd - 1 dz = 150*7 +20 total:1650 Option IV: tonnes/day:200 no of days =9 amt mined in 1/3 days = 3*200 -20 amt mined in 2/3rd - 1 dz = 200*5 +20
total:1600
so you see, if you under achieve 20 tonnes on a whole in the first 1/3 days and over achieve 20 tonnes on a whole in the nxt 2/3 days. And also work one day less then the work will never be completed in one day less. That is why in all the options the total work done is less than 1800 by the amt of ore mined in one day.
If the overachieve/ underachieve 20 tonnes in 1/3rd of the days :- Option II : tonnes/day:100 no of days:18 amt mined in 1/3 days = 6*100 -20 amt mined in 2/3rd - 1 dz = 100* 11 +20 total:1700
yaar ders somethin seriously wrong with the way ur approachin dis prob...
c d highlighted value...wen u need to cover up 20 tonnes in a certain no of days how can u consider that the efficiency wil remain same i.e 100/day and not increase...and den jus add 20 to 1100??? goin by options...ur 4gettin one basic thing...ur tryin to show that its not possible with 100tonne/day and 20 overachieved to complete d rem task in 2/3rd - 1 no f days
whereas the sum says...he does complete it in 2/3rd-1 days...so thing is the sum states that he does mine 1220 tonnes in 11days (going by the answer as 18days) so efficiency is 1220/11 tonnes /day for 2/3rd -1 dayz.
m i makin myslf clr...plz read d sum again...
nyhow u don work by options in such sums...i rest my case.
yaar ders somethin seriously wrong with the way ur approachin dis prob...
c d highlighted value...wen u need to cover up 20 tonnes in a certain no of days how can u consider that the efficiency wil remain same i.e 100/day and not increase...and den jus add 20 to 1100??? goin by options...ur 4gettin one basic thing...ur tryin to show that its not possible with 100tonne/day and 20 overachieved to complete d rem task in 2/3rd - 1 no f days
whereas the sum says...he does complete it in 2/3rd-1 days...so thing is the sum states that he does mine 1220 tonnes in 11days (going by the answer as 18days) so efficiency is 1220/11 tonnes /day for 2/3rd -1 dayz.
m i makin myslf clr...plz read d sum again...
nyhow u don work by options in such sums...i rest my case.
true : ankita plz read qs once agian..some part is missing in this qs :
Ok if we consider that they achieved 20 tonnes less during the first n/3 days => in n/3 days they mine 580 tonnes and they overachieve to mine the remaining 1220 tonnes of ore in 2n/3 -1 days => in (2n/3 -1) days they mine 1220 tonnes and since their efficiency is the 2nd time is more so any number of days would suffice.
Lets take some arbitrary value for the number of days planned(n), say 6, then it means in the first 2(n/3) days they mine 580 tonnes and initially and then increased their efficiency to to complete 1220 tonnes in 3(2n/3 -1) days. This value does not defy any conditions given in the question because their efficiency is more 2nd time.
Take another value for the number of days, say 9 , that would mean that during the first 3 (n/3) days they mine 580 tonnes and then for the next 5(2n/3 -1) days they mine the remaining 1220 tonnes. Which is again fine.
So my final take would be that they increase and decrease their efficiency by 20 tonnes per day, than what was initially planned.
I have a small problem with Arun sharma and was wondering if anyone could clarify.
In the number systems chapter and LOD III type questions , only slight hints and direct answers are given and there are no detailed explanations and some of the complex questions which involve greater application are hard to understand without a detailed explanation.
What I want to know is that is there a separate answer book or these are the only answers given.
P.S. - sorry to disrupt the continuity of the thread with my first post , apologies for that
at a fast food centre,burgers are made only on an automatic buger making machine. the machine continuously makes different sort of burgers by by adding different sorts of fillings on a common bread. the machine makes the burgers at the rate of 1 burger per half minute. the various fillings are added to the burgers in the following manner. the 1st,5th,9th ... burgers are filled with chicken patty; the 2nd, 9th, 16th are filled with vegetable patty; the 1st , 5th ,9th burgers with mushroom patty; the rest with plain cheese and tomato fillings. the machine makes exactly 660 burgers per day.
1. how many burgers per day are made with cheese and tomato fillings? (a)424 (b)236 (c)237 (d)none of these 2. how many burgers are made with all three fillings: chicken, vegetable, mushroom? (a)23 (b)24 (c)25 (d)26
1st 5th n 9th burgers are filled twice??/Is it so????
at a fast food centre,burgers are made only on an automatic buger making machine. the machine continuously makes different sort of burgers by by adding different sorts of fillings on a common bread. the machine makes the burgers at the rate of 1 burger per half minute. the various fillings are added to the burgers in the following manner. the 1st,5th,9th ... burgers are filled with chicken patty; the 2nd, 9th, 16th are filled with vegetable patty; the 1st , 5th ,9th burgers with mushroom patty; the rest with plain cheese and tomato fillings. the machine makes exactly 660 burgers per day.
1. how many burgers per day are made with cheese and tomato fillings? (a)424 (b)236 (c)237 (d)none of these 2. how many burgers are made with all three fillings: chicken, vegetable, mushroom? (a)23 (b)24 (c)25 (d)26
pls check the question once again man ..... manner. the 1st,5th,9th ... burgers are filled with chicken patty; the 2nd, 9th, 16th are filled with vegetable patty; the 1st , 5th ,9th burgers with mushroom patty; the rest with plain cheese and tomato fillings.
does the 1st,5th and 9th burgers get filled twice?
at a fast food centre,burgers are made only on an automatic buger making machine. the machine continuously makes different sort of burgers by by adding different sorts of fillings on a common bread. the machine makes the burgers at the rate of 1 burger per half minute. the various fillings are added to the burgers in the following manner. the 1st,5th,9th ... burgers are filled with chicken patty; the 2nd, 9th, 16th are filled with vegetable patty; the 1st , 5th ,9th burgers with mushroom patty; the rest with plain cheese and tomato fillings. the machine makes exactly 660 burgers per day.
1. how many burgers per day are made with cheese and tomato fillings? (a)424 (b)236 (c)237 (d)none of these 2. how many burgers are made with all three fillings: chicken, vegetable, mushroom? (a)23 (b)24 (c)25 (d)26
if u writing CAT or any entrance...try to avoid such questions...as these type of questions eat ur time a lot 😃
ans for fst qs is
chicken patty series ..1,5,9.....657. no f terms=165 veg patty...2,9,16,,.....660 number of terms=95 mushroom same as chicken....terms=165 so now find out common terms among the 3 series ... dat is 9,37,.....653...no of terms=17 thus nly chicken and mshroom patty=165-17=148 only veg patty=95-17=78 so total=148+78+17=243
thus reqd figure=660-243=417...thus d)none of these...
There are three galleries in a coal mine . On the first day , two galleries are operative and after some time , the third gallery is made operative. With this , the output of the mine became half as large again. What is the capacity of the second gallery as a percentage of the first, if it is given that a four-month output of the first and third galleries was the same as the annual output of the second galllery?
hey plz tell me how to solve such kind of remainder questions??
50^51^52 remainder from 11
33^34^35 remainder from 7
b4 solving these kinda questions fst i want to thnk my frnd implex who taught me an important concept to solve reminder prbs 😃 i.e euler method :) haa i got correct ans for fst qs 😁 but couldnt get for 2nd so i asked my frnd....he told me how to solve 2nd one
1st is easy use euler method u will get remainder 6 here is the method 50 and 11 r co primes so euler number will be 10 now we know 50^10=1mod 11 now 51^52 divided by 10 again 51 and 10 are co primes so euler number is 9 51^9=1 mod 10 50^1 divided by 11 will give us remainder 6 :)
for 2nd i used again euler but wasnt getting correct ans 😁 so MFI aka harsh helped me here is the solution 33^34^35 divided by 7 we know 33^6 divided 7 =1(according to euler method) so now we have to make 33^34^35 in the form of 33^ 6k+something & to find that something 34^35 divided by 6 since 34 and 6 are not co-prime so u cant use eulers so just cancel all the terms 34^34 *34 34^34 *17 divded by 3 34%3 =1 1^34 *17 divided by 3 17 divided 3 we will get remainder 2 but we had divided the numerator by 2 so multiply it by 2.....it gives us 4 now 33^4 divded by 7 we will get remainder 2 hence ans is 2
most of the time i found answers worng 😐