dark_knight89 Saysthe answer is 12,since now he gets 1 dozen of apple for 11 rs,n initial price was rs12,he have 1 more ruppee to buy one more apple,since there is no discount given
i get you but as per the the book answer is 15 .
i get you but as per the the book answer is 15 .
abhi108890 Saysi get you but as per the the book answer is 15 .
abhi,our logic is right,and let me explain u if the initial price is 15,then the cost of 1 apple would be more than 1 ruppee,so mr x will not get anything in 1ruppee he saved even with discount n question has given that he get another one apple without discount.i believe it is a error in the book.so dont worry,there are lot of mistake in this book
LOD 3 NUMBER SYSTEM QUESTION 54 find the remainder when 50^51^52 is divided by 11. a)6 b)4 c)7 d)3... ans(a) PLEASE GIVE DETAILED ANSWER
dark_knight89 SaysLOD 3 NUMBER SYSTEM QUESTION 54 find the remainder when 50^51^52 is divided by 11. a)6 b)4 c)7 d)3... ans(a) PLEASE GIVE DETAILED ANSWER
50^51^52 is divided by 11
=>6^51^52 divided by 11
=>(3x2)^51^52 divided by 11
=>(3)^51^52 divided by 11 x (2)^51^52 divided by 11
now use format little theorem
from format little theorem wkt 3^10/11=1 and 2^10/11=1
51^52/10=1 so 51^52 can be put in the form 10k+1
so 3^(10k+1)/11 x 2^(10k+1)/11
3^10k=1 and 2^10k=1
so remainder is 3x2/11=>6
you can also find out the individual remainders.50 gives 6,51 gives 7 and 52 gives 8 as remainder when divided by 11.now 7^8=56 which gives 1 as remainder when divided by 11.Thus 6 is the remainder when divided by 11.
Question : (50^51^52 )/ 11
Step 1: 50/11 remainder = 6
Step 2: 6^51^52/11
Step 3 : Let x= 6^51/11 which is equal to (6*(6^50))/11
Step 4 : (2*3)^50/11 remainder is 1 {As 2^5/11 is -1 and 3^5/11 is 1}
Step 5 : so 6^51/11 which is (6*6^50)/11 has remainder 6
Step 6: 6^52/11 is 6*6^51/11 which is 6*6/11 which is 3
tahagata ,
do your process of solving work in every case?
50^51^52 is divided by 11
=>6^51^52 divided by 11
=>(3x2)^51^52 divided by 11
=>(3)^51^52 divided by 11 x (2)^51^52 divided by 11
now use format little theorem
from format little theorem wkt 3^10/11=1 and 2^10/11=1
51^52/10=1 so 51^52 can be put in the form 10k+1
so 3^(10k+1)/11 x 2^(10k+1)/11
3^10k=1 and 2^10k=1
so remainder is 3x2/11=>6
can you please explain the format little theorem
in a certain examination paper there are n questions. for j=1,2...n there are 2^n-1 students who answered j or more questions wrongly. If total no of wrong answers is 4095.then value of n is:
a.12b.11c.10d.9
ans= d
Please help me solve this!
50^51^52 is divided by 11
=>6^51^52 divided by 11
=>(3x2)^51^52 divided by 11
=>(3)^51^52 divided by 11 x (2)^51^52 divided by 11
now use format little theorem
from format little theorem wkt 3^10/11=1 and 2^10/11=1
51^52/10=1 so 51^52 can be put in the form 10k+1
so 3^(10k+1)/11 x 2^(10k+1)/11
3^10k=1 and 2^10k=1
so remainder is 3x2/11=>6
can you repeat your solving process in question what will be remainder when 30^72^87/7?
dark_knight89 Sayscan you repeat your solving process in question what will be remainder when 30^72^87/7?
-----------------------------------------
Its 1...
72 is perfectly by 6 => 30^0 mod 7 = 1 mod 7...!!!..
in a certain examination paper there are n questions. for j=1,2...n there are 2^n-1 students who answered j or more questions wrongly. If total no of wrong answers is 4095.then value of n is:
a.12b.11c.10d.9
ans= d
Please help me solve this!
is it 2^(n-1) or (2^n)-1 ??????
dark_knight89 Sayscan you please explain the format little theorem
If a and b are co-prime to each other and b is a prime number than a^(b-1)/b=1
ex: 32^60 divided by 31, here 32 and 31 are co-prime and 31 is a prime no, so u can apply format little theorem
32^30/31=1, therefore 32^60/31=1
dark_knight89 Sayscan you repeat your solving process in question what will be remainder when 30^72^87/7?
For 30^72^87/7
2^72^87/7,
now from format little theorem u know that 2^6/7=1
72^87/6 = 0
so 72^87 =6k
2^6k/7=1
is it 2^(n-1) or (2^n)-1 ??????
If a and b are co-prime to each other and b is a prime number than a^(b-1)/b=1
ex: 32^60 divided by 31, here 32 and 31 are co-prime and 31 is a prime no, so u can apply format little theorem
32^30/31=1, therefore 32^60/31=1
For 30^72^87/7
2^72^87/7,
now from format little theorem u know that 2^6/7=1
72^87/6 = 0
so 72^87 =6k
2^6k/7=1
Please help to solve the problem on Progressions:LODII
Q33.Find infinite some of series1/1 + 1/3 + 1/6 +1/10 +1/15.................
Please help to solve the problem on Progressions:LODII
Q33.Find infinite some of series1/1 + 1/3 + 1/6 +1/10 +1/15.................
nth term of the series is given as 2/n(n+1).
Therefore we need to find sigma(2/n(n+1))=2sigma(1/n(n+1))
=2sigma(1/n-1/(n+1))
So, if u observe carefully all, terms will get canceled except for 1/1 and 1/k {k is the number of terms of series}. But here as the series is infinite, 1/n tends to zero.So, required sum is 2*(1/1)=2
Please help to solve the problem on Progressions:LODII
Q33.Find infinite some of series1/1 + 1/3 + 1/6 +1/10 +1/15.................
here the denominator is increasing in terms of n(n+1)/2.
Consider so tn=2/n(n+1)
tn=2
t1=2
t2=2
t3=2
and so on
are u able to observe the trend? When u add to get the sum all the terms except 1/1 is getting canceled, so 2 is the sum...
How many 6-digit numbers have at least 1 even digit?
Permutations and combinations lod 2 problem no. 35
Please post detailed solution for this
How many 6-digit numbers have at least 1 even digit?Total Number of 6 digit numbers = 10^5 to 999999 =900000
Permutations and combinations lod 2 problem no. 35
Please post detailed solution for this
Odd digits possible=1,3,5,7,9
Numbers without even digits=>Numbers with only odd digits= 5^6
So numbers with atleast 1 even digit = 900000 - 5^6 =900000-15625 = 884375
How many 6-digit numbers have at least 1 even digit?
Permutations and combinations lod 2 problem no. 35
Please post detailed solution for this
total no of even digits are 02468 and 5 odd digits 13579...
Find a number with 0 even digits.
u can fill the odd digits in 5x5x5x5x5x5
and total 6 digits no`s are 9x10x10x10x10x10
total 6 digit no-6 digit odd no will give u the answer that is
9x10x10x10x10x10-5x5x5x5x5x5
tahagata ,
do your process of solving work in every case?
YES MATE.The basic thing here in this case is the remainder theorem.
you and i are doing just the same thing in the problem,but just in a different way.
LOD2 TIME,SPEED AND DISTANCE QUESTION33 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 ans(d)
QUESTION34 hemant and ajay start the two length swimming race at the same moment but from the opposite end of the pool.they swim in the lane and with uniform speeds,but hemant is faster than ajay.they first passed at point 18.5m from the deep end and having comleted the one length.each one is allowed to take a rest on edgr for exactly 45 seconds.after setting off on the return length,the swimmers pass for the second time just 10.5 m fro the shallow end.how long is the pool?
(a)55.5m (b)45m (c)66m (d)49m ans(b)
KINDLY SENT THE DETAILED SOLUTION