Hi Gudda,
Sorry to miss some part of question 1:
question 1:
For the number 2450 find the sum and number of factors divisible by 245.
Usually we can do it if the number is a prime number like 5,7,.....but here it is 245..i.e,5^2*7^2*2...
How to find it...
can you solve it and tell me your answer?
and for 2nd question is the answer 62?
Hi Guys,
Could you please help me in these questions and let me know the procedure to solve them
Topic:Numbers
Question 1:
Find the sum and number of factors Divisible by 245
Question 2:
Find the Consecutive zeros at the end of following number
1!*2!*3!*...........*50!
1)prime factorization of 245=7^2*5
so factors are 6
sum of factors is 7^3-1*5^2-1/24=342
2)find the powers of 5 in 5!+10!+15!..50!=62
so 62 zeroes
btw what are the ans? 😛
Hi gudda,
Even i got the 2nd answer as 62 but the answer given is 262.
For the first question
question 1:
For the number 2450 find the sum and number of factors divisible by 245 (Arun sharma quantitavie book 4th edition---page 23...6th sum).
Here they are asking to find the sum and number of factors for 2450 which are divisible by 245..
The answer you got is by finding for 245...
Can you help me out in this question
hi gudda,
even i got the 2nd answer as 62 but the answer given is 262.
For the first question
question 1:
for the number 2450 find the sum and number of factors divisible by 245 (arun sharma quantitavie book 4th edition---page 23...6th sum).
here they are asking to find the sum and number of factors for 2450 which are divisible by 245..
The answer you got is by finding for 245...
Can you help me out in this question
wht's the right answer for 1st one.
rakeshkr579 Sayswht's the right answer for 1st one.
I dont have an answer
What i am thinking the answer will be is
(5^1+5^2)(7^2)(2^0+2^1)=4410...
If anyone can say me the correct procedure and solution are gad to show here
i dont have an answer
what i am thinking the answer will be is
(5^1+5^2)(7^2)(2^0+2^1)=4410...
If anyone can say me the correct procedure and solution are gad to show here
me too getting the same.
Factors =245,490,1225,2450.
Some more questions which i feel are lill difficult..please explain the procedure to solve them..it will be of great help to me
Topic:Numbers
Question 1:
Find the unit digit in the below
37^123*43^144*57^226*32^127*52^5!----------I Got the answer as 6(I doubt it)
Question 2:
Find the Least number that when divided by 16,18 and 20 leaves a remainder 4 in each case,but is completely divivsible by 7..........I dont know the procedure(Even though i can solve it via options 😃 )
Question 3:
Find the LCM of (x+3)(6x^2+5x+4) and (2x^2+7x+3)(x+3)......Answer given is (4x^2-1)(x+3)(3x+4)
Question 4:
If -4
The least multiple of 7 which leaves a remainder of 4 when divided by 6,9,15 and 18...ans is 364
me too getting the same.
Factors =245,490,1225,2450.
Hi Rakesh,
Hope the procedure we are following is right...
If any aberrations found by you...please let me know
Hi Puys, I need help with this particular ArunSharma problem.
A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%.Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available.
Ans:20%
Hi Puys, I need help with this particular ArunSharma problem.
A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%.Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available.
Ans:20%
let the initial amt. be 100.
and let x be the adulterated amount.
so,
x/100-x * 100 =25
so x = 20,which is 20% of the total mixture.:)
Hi Puys, I need help with this particular ArunSharma problem.
A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%.Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available.
Ans:20%
jst think 100Rs for 1 kg is the CP(100/kg)
now you want 125 Rs => 1.25 kg (100/kg)
=> 1kg already for pure butter
=> .25 kg for adulterated fat
=> .25/1.25 => 20%
Quote:
Originally Posted by arijitprepares
Hey guys, I am so sorry...the post was wrong...The actual problem is If x, y, z are in GP and a^x, b^y, c^z are equal then a, b, c are in
1. AP
2. GP
3. HP
4. None of these
5. Cannot be determined
The answer is 3 - HP...I want the procedure.....Apologies once again....
a^x=b^y=c^z
Since x,y,z are in GP put any value
x=1,y=2,z=4
a=b^2=c^4
c=2,b=4,a=16 fit
To be in H.P 1/2-1/4 should be 1/4-1/16 but it's not !
My take would be 4.
Any corrections are welcomed !
the answer given to this in arun sharma is (b) GP not (c) HP ... but i think it is incorrect too !!! (d) none of these seems to be the correct option !!!
spectramind07 SaysHow are you getting 1024???I got 2048 :shocked:
2048 is the correct answer !!! ... Arun Sharma has maximum wrong answers in LOD-III !!!!
Your methodology is spot on but the answer I am afraid is not 4th..it's 7th.
Pg 77 question 25.
The answer should be 8th !!
a + b + c = 24 then what is the maximum value for a^2 b^3 c?
i am getting ans 2^14 3^3 but ans should be 2^3 3^10.
a + b + c = 24 then what is the maximum value for a^2 b^3 c?
i am getting ans 2^14 3^3 but ans should be 2^3 3^10.
i am also getting same answer i think you are correct...
there are some mistakes in the answers of the book
somebody please correct if i am wrong
Quote:
Originally Posted by arijitprepares
Hey guys, I am so sorry...the post was wrong...The actual problem is If x, y, z are in GP and a^x, b^y, c^z are equal then a, b, c are in
1. AP
2. GP
3. HP
4. None of these
5. Cannot be determined
The answer is 3 - HP...I want the procedure.....Apologies once again....
a^x=b^y=c^z
Since x,y,z are in GP put any value
x=1,y=2,z=4
a=b^2=c^4
c=2,b=4,a=16 fit
To be in H.P 1/2-1/4 should be 1/4-1/16 but it's not !
My take would be 4.
Any corrections are welcomed !
the answer given to this in arun sharma is (b) GP not (c) HP ... but i think it is incorrect too !!! (d) none of these seems to be the correct option !!!
My Take is also 4.
a^x=b^y=c^z
Since x,y,z are in GP put any value
x=1,y=2,z=4
a=b^2=c^4
c=2,b=4,a=16 fit
To be in H.P 1/2-1/4 should be 1/4-1/16 but it's not !
P.S. --> Arun Sharma might have the answer given Wrong. So relax.. You are right Buddy..
83^261/17=(85-2)^261/17=(-2)^261/17 since{85 is divisible by 17}=(-2)(16)^65/17=(-2)(17-1)^65/17=(-2)(-1)^65/17=2/17
remainder is 2
cheers!!!
I want views of puys for this a.sharma question
Q) A boy took a 7 digit no. ending in 9 and raised it to a even power greater than 2000. He then took number 17 and raised it to a power which leaves the remainder 1 when divided by 4. If he now multiplies both the no.s , what will be the unit's digit of the no. he so obtains?
a) 7 b)9 c)3 d) cannot be determined
now my doubt is if a no. ending with 9 is raised to even power then the units digit will always be 1 and also 17 raised to any power(whether even or odd) when divided by 4 leaves remainder 1 .Therefore if those 2 digits are multiplied then the 7 digit no. ending with 1 will get multiplied to 7 or 9 or 3 or 1
Therefore the unit's digit could be 7 or 9 or 3 or 1. At the back no explanation is given and the answer is (a)
sometimes arun sharma's book brings huge disappointment , its a pity that i share the same surname 😞
Ok guys, I have been struggling with this "permutation & combination" problem for quite a while.
"The crew of an 8 member rowing team is to be chosen from 12 men, of which 3 can row on one side only & 2 can row on the other side only.
Find the number of ways of arranging the crew with 4 on each side"
The answer provided is 60480, which I guess is wrong because , the answer assumes those 3 & 2 guys who can row on either side are always chosen, but is nowhere explictly mentioned in the problem
I guess this question needs to be tackled in whole different view.
please provide your opinions, whether I am treading the wrong path?