what is the remainder when 22^33 is divided by 45 ?
@PURITAN said:what is the remainder when 22^33 is divided by 45 ?
E(45)=45(4/5)(2/3)=24
22^9 mod 45
45=5*9
22^9 mod 5=2
22^9 mod 9=22^3 mod 9=4^3 mod 9=1
5a+2=9b+1
37 satisfies
hence 37
@PURITAN
Shld be 37..
E(45) = 45*(1 - 1/3)*(1 - 1/5) = 24..Since 22 n 45 are co-prime to each other..
Thus, Rem[22^24/45] = 1..Nw, our question reduces to Rem[22^9/45] = ?
=>Rem[22^9/45] = Rem[22*484^4/45] = Rem[22*34^4/45] = Rem[22*31^2/45]
=>Rem[22*16/45] = 37..
There are two qualities of milk- Amul and Sudha having different prices per litre, their volumes being 130L and 180L respectively. After equal amounts of milk was removed from both, the milk removed from Amul was added to Sudha and vice versa. The resulting two types of milk now have the same price. Find the amount of milk drawn out from each type of milk.
1) 58.66
2) 75.48
3) 81.23
4) None of these
Plz give answer alongwith explanation
1) 58.66
2) 75.48
3) 81.23
4) None of these
Plz give answer alongwith explanation
@PURITAN said:what is the remainder when 22^33 is divided by 45 ?
Remainder should be 37..
E(54) = 45*2/3*4/5 = 24
so, 22^9 /45 remainder we have to find...
22^9 = 22^8 * 22/45
from here on you can take pairs and find the remainder...
at last you get 22*16/45 remainder = 37...
@Exodia said:There are two qualities of milk- Amul and Sudha having different prices per litre, their volumes being 130L and 180L respectively. After equal amounts of milk was removed from both, the milk removed from Amul was added to Sudha and vice versa. The resulting two types of milk now have the same price. Find the amount of milk drawn out from each type of milk.1) 58.662) 75.483) 81.234) None of thesePlz give answer alongwith explanation
ans 4) None of these. the amount should be 77.5
what must be subratcted from both the terms A and B so that their ratios reverse ?
@Exodia said:There are two qualities of milk- Amul and Sudha having different prices per litre, their volumes being 130L and 180L respectively. After equal amounts of milk was removed from both, the milk removed from Amul was added to Sudha and vice versa. The resulting two types of milk now have the same price. Find the amount of milk drawn out from each type of milk.1) 58.662) 75.483) 81.234) None of thesePlz give answer alongwith explanation
75.5
@ishu1991
@ishu1991 said:The value ofnc0+ 2(nc1)+ 3(nc2)+ . . . + (n + 1)(ncn)equals(A) 2^n + n2^(nā1)(B) 2^n ā n2^(nā1),(C) 2^n,(D) 2^(n+2)
(1+x)^n = nc0+ x(nc1)+ . . . + (ncn)*x^n
multiply by x ...
it becomes x*(1+x)^n=x*nc0+ x^2 * (nc1)+ . . . + (ncn)*x^n+1
now differentiate and put x=1
we get A) as answer .. _/\_ :D
@Exodia said:There are two qualities of milk- Amul and Sudha having different prices per litre, their volumes being 130L and 180L respectively. After equal amounts of milk was removed from both, the milk removed from Amul was added to Sudha and vice versa. The resulting two types of milk now have the same price. Find the amount of milk drawn out from each type of milk.1) 58.662) 75.483) 81.234) None of thesePlz give answer alongwith explanation
A S
130ltr 180ltr
Rate
x /ltr y /ltr
After mixing k ltrs..
130x + ky - kx = 180y + kx - ky
(130-2k)x = (180-2k)y
Got this eqn. 
Thereafter 
Thereafter Plz tag me with the OA..!
@iLoveTorres said:bro can u post your approach?
Let the milk in 130 L be x%
milk in 180 L be y%
for the costs to be equal, the percentage of milk in both will be equal = k (say)
Equal volume removed = t lts.
Alligation in first mixture : (x-a)/(a-k) = t/(130 - t)
Alligation in second mixture : (x-a)/(a-k) = (180-t)/t
So, t/(130 - t) = (180-t)/t
Or, t= 2340/31 = 75.5 (approx)
milk in 180 L be y%
for the costs to be equal, the percentage of milk in both will be equal = k (say)
Equal volume removed = t lts.
Alligation in first mixture : (x-a)/(a-k) = t/(130 - t)
Alligation in second mixture : (x-a)/(a-k) = (180-t)/t
So, t/(130 - t) = (180-t)/t
Or, t= 2340/31 = 75.5 (approx)
@ishu1991 said:Let the milk in 130 L be x% milk in 180 L be y%for the costs to be equal, the percentage of milk in both will be equal = k (say)Equal volume removed = t lts.Alligation in first mixture : (x-a)/(a-k) = t/(130 - t)Alligation in second mixture : (x-a)/(a-k) = (180-t)/tSo, t/(130 - t) = (180-t)/tOr, t= 2340/31 = 75.5 (approx)
bro if you dont mind can u explain me in layman language. mujhe yeh alligations samajh nahi aata.. it would be really helpful if u could explain in detail
if a:b is the duplicate ratio of ( a+ x ) : ( b+x) . then what is the value of x ?
@meenu05 said:if a:b is the duplicate ratio of ( a+ x ) : ( b+x) . then what is the value of x ?
0?
@meenu05 said:if a:b is the duplicate ratio of ( a+ x ) : ( b+x) . then what is the value of x ?
sorry it should be ab^1/2