hey i didn't know there was a thread just for A.Sharma....nice....
plz someone tell me how to apply allegation method(according to what arun sharma has explained in his book) for the below problem
there is a 60litre solution of milk forms 84%.how much water must be added to this solution to make it a solutio in which milk forms 64%??
which one is A1 A2 n1 n2:oops::oops:....am doing for the first time this chapter in arun sharma
here A1 = 84% A2 = 0%(As its pure water) A = 64
n1 = 60 and u gotta find n2
hence applying allegations here we get
n1(84-64) = n2(64 - 0)
hence n2 = 60 * 10 / 64 = 9.375 ltrs
is this the ans?
here A1 = 84% A2 = 0%(As its pure water) A = 64
n1 = 60 and u gotta find n2
hence applying allegations here we get
n1(84-64) = n2(64 - 0)
hence n2 = 60 * 10 / 64 = 9.375 ltrs
is this the ans?
bro answer is 18.75 litres of water to added...
msabhi Saysbro answer is 18.75 litres of water to added...
hey srry made a mistake here 84-64 = 20 not 10. thn u'll get 18.75. 😃
here A1 = 84% A2 = 0%(As its pure water) A = 64
n1 = 60 and u gotta find n2
hence applying allegations here we get
n1(84-64) = n2(64 - 0)
hence n2 = 60 * 10 / 64 = 9.375 ltrs
is this the ans?
thanks a lot bro...can u explain me as to how n1 = 60 ? bro...
msabhi Saysthanks a lot bro...can u explain me as to how n1 = 60 ? bro...
here n1 and n2 are the ratio in which the two quantities are mixed. hence i have taken n1 = 60 litres:)
plz someone tell me how to apply allegation method(according to what arun sharma has explained in his book) for the below problem
there is a 60litre solution of milk forms 84%.how much water must be added to this solution to make it a solutio in which milk forms 64%??
which one is A1 A2 n1 n2
Apart from the method mentioned, we can also do it in simple unitary method. Since, it's mentioned that milk forms 84% of 60 litres, thus, the solution contains (60*0.84) = 50.4 Litres of milk.
Now, the amount of milk will remain same even though it's %age will come down to 64%. Thus, the total amount of solution for 64% of milk would be 50.4/64 = 78.75. Thus, (78.75-60) = 18.75 Litres of water need to be added.
I've always found the alligation method a bit off track for me, thus I just used my common sense for such problems and well, it worked for me 😃
The series is of the form a1, a2, a3, ..... an.
The second term is 2 more than the third. If a5-a2 = 12, then what is the first term ?
I got 4 but it's not the correct answer. Could someone explain ?
Thanks.
Could someone tell me the answer to this questions ?
If n is an even number, which one the following has to be odd?
1) 3n
2) 3xsquare/2 +1
3) 5(n)+4
4) 3x/2 + 1
The answer is the second choice. But, I think choice number 4 is also an option . Please explain.
Take n to be 2a where a is any odd number(worst case scenario..)
3n/2 will be an odd number and odd+odd will be even..
3(n^2)=3*4(a^2) and when you will divide it by 2,u will get an even number and hence,even+odd Will necessarily be odd...
hey guyz can anyone solve this question lod-3 number system ,question 43
The remainder when 2^2 + 22^2 + 222^2 + 2222^2 .....til 222 (49 times)^2???
hey guyz can anyone solve this question lod-3 number system ,question 43
The remainder when 2^2 + 22^2 + 222^2 + 2222^2 .....til 222 (49 times)^2???
what is the divisor here?
Could someone tell me the answer to this questions ?
If n is an even number, which one the following has to be odd?
1) 3n
2) 3xsquare/2 +1
3) 5(n)+4
4) 3x/2 + 1
The answer is the second choice. But, I think choice number 4 is also an option . Please explain.
You can easily solve such questions using examples... For the 4th option, try using 18 in place of x. You'll get a value of 28, which is even. The point in option 4 is to use any no. which, when divided by 2, gives an odd no.
Just checking the options, option 2 has to give you an odd number if x is taken as even. As x^2 will be even, any no. multiplied by even gives you even, and then adding 1 would obviously give you an odd no.
I have some questions on P&C; LOD 3 from Arun Sharma for which I need your help.
31) How many numbers smaller than 2*10^8 (2 multiply by 10 to the power )
and are divisible by 3 can be written by means of digit 0, 1 and 2 (exclude single digit and double digit numbers).
a) 4536
b) 294840
c) 191520
30) How many six digit numbers contain exactly 4 different digits?
32) Six white and six black balls of the same size are distributed among 10 urns so that there is at least
one ball in each urn. What is the number of different distributions of the balls?
35) How many distinct 6 digit numbers are there having 3 odd and 3 even digits?
You may not need to show very elaborated but would be helpful if can break in some steps that I can understand.
Thanks
Regards
Sandeep Mittal
shanks4mba Sayswhat is the divisor here?
divisor is 9
divisor is 9
hey guyz can anyone solve this question lod-3 number system ,question 43
The remainder when 2^2 + 22^2 + 222^2 + 2222^2 .....til 222 (49 times)^2???
2^2(1^2 + 11^2 +111^2....................)
Each term no is the sum of its digits. For eg 8th term 11111111^2 will give 8^2 remainder with 9.
so all the reaminders add up to (1^2 + 2^2 + 3^2--------8^2 + 0) upto 111,111,111^2. After this will repeat. five complete cycle upto (111...45 times ^2) + 1^2 + 2^2 + 3^2 +4^2.
Now (1^2 + 2^2 + 3^2--------8^2 + 0) Gives 3 remainder with 9.
So 3*5 (upto 1111....45 times) + 1^2 + 2^2 + 3^2 +4^2.
15+3= 18 =0 Rem
Seems answer in Book as 6 is wrong.
I have some questions on P&C; LOD 3 from Arun Sharma for which I need your help.
31) How many numbers smaller than 2*10^8 (2 multiply by 10 to the power )
and are divisible by 3 can be written by means of digit 0, 1 and 2 (exclude single digit and double digit numbers).
a) 4536
b) 294840
c) 191520
30) How many six digit numbers contain exactly 4 different digits?
32) Six white and six black balls of the same size are distributed among 10 urns so that there is at least
one ball in each urn. What is the number of different distributions of the balls?
35) How many distinct 6 digit numbers are there having 3 odd and 3 even digits?
You may not need to show very elaborated but would be helpful if can break in some steps that I can understand.
Thanks
Regards
Sandeep Mittal
I have some questions on P&C; LOD 3 from Arun Sharma for which I need your help.
31) How many numbers smaller than 2*10^8 (2 multiply by 10 to the power )
and are divisible by 3 can be written by means of digit 0, 1 and 2 (exclude single digit and double digit numbers).
You may not need to show very elaborated but would be helpful if can break in some steps that I can understand.
Thanks
Regards
Sandeep Mittal
we can consider one digit nos. then two digit nos. then threee digits and so on till nine digits.
for one digit nos. we have two ways i.e, 1,2 so no of ways 2^1
for two digit nos. we have four ways i.e, 11,12,21,22 so no of ways 2^2
similarly doing,
.
.
.
for eight digit nos. we have 2^8 ways becoz the eight digits may be either 1 or 2.
for nine digit nos. we have 2^8 ways as the first digit can only be 1.
so total no. of numbers = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^8
= 766 .................(answer)
cheers....
2^2(1^2 + 11^2 +111^2....................)
Each term no is the sum of its digits. For eg 8th term 11111111^2 will give 8^2 remainder with 9.
so all the reaminders add up to (1^2 + 2^2 + 3^2--------8^2 + 0) upto 111,111,111^2. After this will repeat. five complete cycle upto (111...45 times ^2) + 1^2 + 2^2 + 3^2 +4^2.
Now (1^2 + 2^2 + 3^2--------8^2 + 0) Gives 3 remainder with 9.
So 3*5 (upto 1111....45 times) + 1^2 + 2^2 + 3^2 +4^2.
15+3= 18 =0 Rem
Seems answer in Book as 6 is wrong.
dude, i think the answer in the book is correct.
2^2 divided by 9 gives remainder 2^2
remainder of 22^2 divided by 9 => 4^2 divided by 9 , ( 22 by 9 gives remainder 4)
remainder of 222^2 divided by 9 => 6^2 divided by 9 , ( 222 by 9 gives remainder 6)
remainder of 2222^2 divided by 9 => 8^2 divided by 9 , ( 2222 by 9 gives remainder
similarly doing, for ( 22....49times)^2 by 9 => (2*49)^2
so adding these, we get
= remainder on dividing (2^2 + 4^2 + 6^2 +..........+ 98^2) by 9
= remainder on dividing 2^2 ( 1^2 + 2^2 + 3^2 +.......+ 49^2) by 9
= remainder on dividing 2^2 * (using the sum of squares formula)
= remainder on dividing (4 * 49 * 25 * 33) by 9
= remainder on dividing (4 *4 * 7 * 6) by 9
=remainder on dividing (7*7*6) by 9
= remainder on dividing 24 by 9
= 6