LOD2 TIME,SPEED AND DISTANCE QUESTION
33)
ravi,who lives in the countryside,caught a train for home earlier than usual yesterday.his wife normally drive to the station to meet him.but yesterday he set out from the station to meet his wife on the way.he reached home 12 minutes earlier than he would have done had he waited at the station for his wife.the car travels at the uniform speed,which is five times the ravis speed on the foot.ravi reached home at exactly 6"0 clock.at what time would he have reached home if his wife ,forewarned of his plan,had met him at the station?
a)5:48 b)5:24 c)5:00 d)5:36
One more...and this time I am sure Arun Sharma is gonna be on the backfoot :p
Q45 Lod 3 pg 51
Twenty five sets of problems for 25 papers. one set has 50 q. out of 50, 35 are unique and 15 are repeated in some other paper.What is the max possibl no of qs
My answer-1055
35*25 + 24*15
any one for this one?
Regards
Meet_the_man
My take on this one:
Answer: 1062
Explanation:
Number of unique questions = 35*15. This is very easy to understand.
Now, 15 questions remain for each set.
To get maximum number of questions, each repeated question should be shared by only 2 papers.
However, we have 15*25 questions. So, we have off number of questions. So, one question needs to be shared by 3 papers.
So, number of remaining questions = (25*15 - 1) / 2 = 187
Possible arrangement:
Say, first 24 papers form groups of two.
5 groups i.e. 10 papers share all 15 questions.
Remaining 7 groups i.e 14 papers share 14 questions. And each paper shares remaining question with 25th paper, ging 25th paper 14 questions. 15th question of 15th paper will be the only question which is repeated for 3rd time and can be taken form any repeated question.
LOD2 TIME,SPEED AND DISTANCE QUESTION
33)
ravi,who lives in the countryside,caught a train for home earlier than usual yesterday.his wife normally drive to the station to meet him.but yesterday he set out from the station to meet his wife on the way.he reached home 12 minutes earlier than he would have done had he waited at the station for his wife.the car travels at the uniform speed,which is five times the ravis speed on the foot.ravi reached home at exactly 6"0 clock.at what time would he have reached home if his wife ,forewarned of his plan,had met him at the station?
a)5:48 b)5:24 c)5:00 d)5:36
Answer: (d) 5:36
Explanation:
Car travels 5 times faster than Ravi.
Say, Ravi started walking and met car at point A.
If Ravi has stayed at station, he had taken 12 more mins to reach home. This extra time would have been for car coming to station from point and back to point A again. So, from point to station takes 12/2 = 6 mins for car.
For Ravi it will take 6*5 = 30 mins.
It means Ravi walked for 30 mins.
Now, if wife had already been present at station, Ravi would not have walked for 30 mins. That time they would have covered by car in 6 mins. Thus, they would have saved 24 mins and reached at 5:36.
My take on this one:
Answer: 1062
Explanation:
Number of unique questions = 35*15. This is very easy to understand.
Now, 15 questions remain for each set.
To get maximum number of questions, each repeated question should be shared by only 2 papers.
However, we have 15*25 questions. So, we have off number of questions. So, one question needs to be shared by 3 papers.
So, number of remaining questions = (25*15 - 1) / 2 = 187
Possible arrangement:
Say, first 24 papers form groups of two.
5 groups i.e. 10 papers share all 15 questions.
Remaining 7 groups i.e 14 papers share 14 questions. And each paper shares remaining question with 25th paper, ging 25th paper 14 questions. 15th question of 15th paper will be the only question which is repeated for 3rd time and can be taken form any repeated question.
my answer is also 1062 but if look at the option given in the book 1055 is max questions possible.
I agree with everything you have done, except the part where you say that the qsns appear in three papers. The question reads ' .. question appears in some other paper' and not papers. Therefore, it can't be three papers.Please do correct me if i am wrong .
:)
Thanks!
hi everybody
i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15
the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41
isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help
hi everybody
i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15
the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41
isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help
you are perfectly correct, it should be 123:cheerio:
guys according to arun sharma's book... what is the reminder of 18/4??
is it 2 or 1??
hi everybody
i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15
the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41
isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help
Given ans is ryt...req no is Hcf of diff b/w ny two nos and not 3 as u did...:-)
No...it
whats the remainder of
50^51^52 when divided by 11
p.s.-plz elaborate the approach
guys according to arun sharma's book... what is the reminder of 18/4??
is it 2 or 1??
its 2
when we divide the num and denm by 2 thn the remainder is multiplied by 2(1*2)
guys according to arun sharma's book... what is the reminder of 18/4??
is it 2 or 1??
it is 2 and not 1; go by the basic remainder theorem, rather divide the conventional way you will get the answer
:thumbsup:
whats the remainder of
50^51^52 when divided by 11
p.s.-plz elaborate the approach
50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.
50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.
couldnt get kindly elaborate
50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.
thanks
could u tell abt the eulers theorem
mbafrmfms SaysFind the maximum value of the function 1/(x^2-3x+2). Ans - N.O.T
1/(x-3x+2)= 1/(x-2)(x-1)
Now at x=2 and x=1 the value reaches infinity hence a maximum value cannot be found
The domain of definition of y=sqrt{log(5x-x^2/4} Ans:For the function log x, x>0
Hence for this case (5x-x)/4>0
Also for the square root function to be defined the value present inside the root should be positive.
hence log{(5x-x)/4}>0
=> (5x-x)/4>1 ; because log function is always positive when it is taken >1 (You can see the log graph for this)
Hence x-5x+4=>Value ranges between 1 and 4 (both included)
f(x).f(1/x)=f(x)+f(1/x) and f(4)=65 ........find f(6) Ans:217Substituting x=4 we get the value of f(1/4)=65/64
Also for x=1, f(1)=0 or 2
Now check the pattern
f(4)= 4^3+1
f(1/4)=(1/4)^3+1
f(1)= 1^3+1 (The other value is not considered as it does not follow the same pattern)
Hence,
f(x)= x^3+1
So f(6)= 216+1=217
At what integral value of x will the function (((x^2+3x+1))/((x^2-3x+1))) Ans:-3I think the question is incomplete
thanks
could u tell abt the eulers theorem
euler's theorem
example:
51^8/7
now 51 and 7 are co-primes to each other.
and 7 = 7 *1
therefore, 7 * ( 1- 1/7) = 6
according to euler's theorem 51^6/7 will give remainder as 1.
therefore, (51^6 * 50^2)/7
= 51^2/7 as 51^6 gives 1 as remainder.
= 2*2/7
= 4 is remainder.
this is euler's theorem.
hope u understand.
option (e)
My answer is 4798, below is solution how i got it:
(7! -2) - (5!+5!)
please help if wrong
Hey man.. I got 4800... by
(7!) - (5!+5!)
I couldn't make out why you did with 7!... thnx..;)
I'm getting 45 can u gelp me with that
it"s 20
getting 45 as an answer kindly help