125 ltr mixture of wine n water contains 20% water, how much more water b added to result in 25% concentration of water
pls explain how alligation technique can be used here in arun sharma ( straight line method)
(25/3) ltrs of water....
(25/3) ltrs of water....
Thanks 4 d ans, but I too knw d sol like: (25+x)/(125+x)=1/4
thing is arun sharma explained a method in form of A1,A2,Aw,n1,n2
can u give values of these variables in ref to this ques..
Thanks 4 d ans, but I too knw d sol like: (25+x)/(125+x)=1/4
thing is arun sharma explained a method in form of A1,A2,Aw,n1,n2
can u give values of these variables in ref to this ques..
Hey Guyz,
Question from AS:
Concentrations of three wines A, B and C are 10%,20%,30% respectively. They are mixed in the ratio 2:3:x resulting in a 23% concentration solution.
Find x.
Guyz plz post the solution to this with a detailed explanation ...
Hey Guyz,
Question from AS:
Concentrations of three wines A, B and C are 10%,20%,30% respectively. They are mixed in the ratio 2:3:x resulting in a 23% concentration solution.
Find x.
Guyz plz post the solution to this with a detailed explanation ...
Total concentration is 23% of ( 2 + 3 + x )
It's also the sum of the individual concentration and ratios:
10% of 2 + 20% of 3 + 30% of x
So:
0.2 + 0.6 + 0.3*x = 0.23 * ( 5 + x ), or
0.8 + 0.3*x = 1.15 + 0.23*x, or
0.07*x = 0.35, or
x = 5
thanks dost!!!!!
Suppose x is the product of n consecutive integers .
(x+1)(x+2)(x+3)..(x+(n-1))=1000 then which of the following cannot be true about number of terms n
a)The number of terms can be 16
b)The number of terms can be 5
c)The number of terms can be 25
d)The number of terms can be 20
can anyone help me out with this one ?
page no 51 ,2008 reprint. number systems lod -3 ,no. 42
Suppose x is the product of n consecutive integers .
(x+1)(x+2)(x+3)..(x+(n-1))=1000 then which of the following cannot be true about number of terms n
a)The number of terms can be 16
b)The number of terms can be 5
c)The number of terms can be 25
d)The number of terms can be 20
can anyone help me out with this one ?
page no 51 ,2008 reprint. number systems lod -3 ,no. 42
Product of n consecutive integers is always divisible by n!. So for n = 16, 5, 20 and 25, product should be divisible by 16!, 5!, 20!, 25! respectively. But 1000 is not divisible by any of these.
So, I think its a misprint in the book. The correct question should be like:-
If x is positive integer and
x + (x + 1) + (x + 2) + (x + 3) + .... + (x + (n - 1)) = 1000 then which of the following cannot be true about number of terms n
a)The number of terms can be 16
b)The number of terms can be 5
c)The number of terms can be 25
d)The number of terms can be 20
x + (x + 1) + (x + 2) + (x + 3) + .... + (x + (n - 1)) = n(n + 2x - 1)/2 = 1000
=> n(n + 2x - 1) = 2000
Now, we can notice that one of n and (n + 2x - 1) is odd and other will be even and n
2000 = 1*2000 = 5*400 = 16*125 = 25*80
=> n can be 5, 16 and 25
So, 20 should be the answer.
thanks a LOT man , almost lost sleep over it 
dis one ? cant find any pattern 
the remainder en 2^2 +22^2 +222^2+....(222...49 twos's)^2 is divied by 9 is
a)2 b)5 c)6 d)7
problem no 43 lod-3 number systems (2008 reprint)
Bad Post...
dis one ? cant find any pattern
the remainder en 2^2 +22^2 +222^2+....(222...49 twos's)^2 is divied by 9 is
a)2 b)5 c)6 d)7
problem no 43 lod-3 number systems (2008 reprint)
On dividing by 9,
2^2 leaves a remainder of 4.
Similarly, 22^2 will leave a remainder of 7
222^2 leaves 0
2222^2 leaves 1
22222^2 leaves 1
222222^2 leaves 0.
Sorry not finding the pattern ..
dis one ? cant find any pattern
the remainder en 2^2 +22^2 +222^2+....(222...49 twos's)^2 is divied by 9 is
a)2 b)5 c)6 d)7
problem no 43 lod-3 number systems (2008 reprint)
2^2 + 22^2 + 222^2 + 2222^2 + ...... + (222...49 2's)^2
=2^2 * (1^2 + 11^2 + 111^2 + 1111^2 +..... + (1111...49 1's)^2)
Considering 1^2 + 11^2 + .... + (11...9 1's)^2
This when divided by 9 can be reduced to the form -
(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 0)
=> 2 * (1^2 + 2^2 + 3^2 + 4^2)
=> 2 * 30 = 60
Hence the remainder for this part is 6.
This continues for 5 times in the original series. (Till 45 1's)
Hence remainder is
4 * (6*5 + (46 1's)^2 + (47 1's)^2 + (48 1's)^2 + (49 1's)^2 )
4 * (6*5 + 1^2 + 2^2 + 3^2 + 4^2)
4 * (6*5 + 30)
=> 4* (60)
=> 4 * 6
=> 24
Remainder = 6
dis one ? cant find any pattern
the remainder en 2^2 +22^2 +222^2+....(222...49 twos's)^2 is divied by 9 is
a)2 b)5 c)6 d)7
problem no 43 lod-3 number systems (2008 reprint)
The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9
so rem of 22*22/9 = rem of 4*4/9
rem of 222*222/9 = rem of 6*6/9
rem of 2^2+ 4^2+6^2.............98^2/9
rem of 2^2(1+2^2+3^2............+49^2)/9
1+2^2+3^2............+49^2=49(49+1)(98+1)/6=40,425
rem of 4*40425/9 = 161700/9= 1+6+1+7=15
Rem=1+5=6
2^2 + 22^2 + 222^2 + 2222^2 + ...... + (222...49 2's)^2
=2^2 * (1^2 + 11^2 + 111^2 + 1111^2 +..... + (1111...49 1's)^2)
Considering 1^2 + 11^2 + .... + (11...9 1's)^2
This when divided by 9 can be reduced to the form -
(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 0)
=> 2 * (1^2 + 2^2 + 3^2 + 4^2)
=> 2 * 30 = 60
Hence the remainder for this part is 6.
This continues for 5 times in the original series. (Till 45 1's)
Hence remainder is
4 * (6*5 + (46 1's)^2 + (47 1's)^2 + (48 1's)^2 + (49 1's)^2 )
4 * (6*5 + 1^2 + 2^2 + 3^2 + 4^2)
4 * (6*5 + 30)
=> 4* (60)
=> 4 * 6
=> 24
Remainder = 6
Thanks a ton 😁 great solution
The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9
so rem of 22*22/9 = rem of 4*4/9
rem of 222*222/9 = rem of 6*6/9
rem of 2^2+ 4^2+6^2.............98^2/9
rem of 2^2(1+2^2+3^2............+49^2)/9
1+2^2+3^2............+49^2=49(49+1)(98+1)/6=40,425
rem of 4*40425/9 = 161700/9= 1+6+1+7=15
Rem=1+5=6
lajawaab solution
dho dala 
thanks plenny mate .
202X20002X200000002X20000000000000002X2000000...2(31 zeros)
*the sum of digits in this multiplication will be?
a)112 b)160 c)144 d)CBD*
ARUN SHARMA
LOD 3, number system no. 44 (8th reprint)
202X20002X200000002X20000000000000002X2000000...2(31 zeros)
*the sum of digits in this multiplication will be?
a)112 b)160 c)144 d)CBD*
ARUN SHARMA
LOD 3, number system no. 44 (8th reprint)
I could find a pattern here
202 = 202 (there are two 2's)
202 * 20002 = 4040404 (there are four 4's)
So, if you see, there are four 4's in the above multiplication, i.e., (number of zeroes + 1 ) number of fours.
Similarly, for 202 * 20002 * 200000002 = there will be eight 8's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) = there will be sixteen 16's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) * 2000... 2 (31 zeros) = there will be thirty two 32's.
Thirty two 32's contains thrity two 3's and thirty two 2's.
So, 32 * 3 + 32 * 2 = 160
What's the OA .. ?
I could find a pattern here
202 = 202 (there are two 2's)
202 * 20002 = 4040404 (there are four 4's)
So, if you see, there are four 4's in the above multiplication, i.e., (number of zeroes + 1 ) number of fours.
Similarly, for 202 * 20002 * 200000002 = there will be eight 8's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) = there will be sixteen 16's.
For 202 * 20002 * 200000002 * 20000 ...2 (15 zeroes) * 2000... 2 (31 zeros) = there will be thirty two 32's.
Thirty two 32's contains thrity two 3's and thirty two 2's.
So, 32 * 3 + 32 * 2 = 160
What's the OA .. ?
dats the answer BINGO!! 😁 one heck of a pattern must say
arijitprepares SaysHey guys anyone got to say anything on this one? Should I presume that the answer given in Arun Sharma was wrong??:splat:
hey arijit,
i have another way but answer comes as GP..it's like this
let a^x = b^y = c^ z = k
take log both side
so now log(a^x)=logk
similarly log(a^x)=log(b^y)=log(c^z)=logk
hence xloga=ylogb=zlogc=logk------(equation 1)
now y*y=x*z or y/x=z/y;(given) ----------(equation 2)
putting equation 1 in 2
we get log (ac)= log (b^2)
so b^2=ac.........hence a,b,c is in GP.....please correct if wrong...