@vbhvgupta said:Remainder 32^32^32/7 ??? 50^51^52/11 30^72^87/11 33^34^35/7Approach?
4,10,5,1 ??
@Budokai001 said:For a trapezium , S1 denotes the sum of squares of of the sides and S2 denotes the sum of squares of the diagonals . S1-S2=576If the longer parallel side is 50 cm,Find the shorter parallel side
26 hoga na?
3 whole numbers are taken at random and multiplied .What is the probability that the units place of the resultant number is 1 or 3 or 7 or 9 ?
@Budokai001 said:3 whole numbers are taken at random and multiplied .What is the probability that the units place of the resultant number is 1 or 3 or 7 or 9 ?
8/125 ??
@allan89 said:approach dalo bhai... geo..
Try using the figure as below bro...
This way u can represent both diagonals using the sides itself
@Budokai001 said:3 whole numbers are taken at random and multiplied .What is the probability that the units place of the resultant number is 1 or 3 or 7 or 9 ?
u cannot select any of 0,2,4,6,8 or 5..u can only choose among 1,3,7,9 ==> 4/10 * 4/10 * 4/ 10 = 8/125
@Budokai001
Let x cm be the smaller side, a be the two adjacent sides and h cm be the height..Thus, two lngt of the sides that lie on the other longer side n meet the two vertices = (25 - x/2) cm..
Given, S1 = x^2 + 2a^2 + 2500, and, S2 = 2[h^2 + (25 + x/2)^2]
We Know, a^2 = h^2 + (25 - x/2)^2...(i)
Thus, S1 - S2 = 576
=>x^2 + 2[h^2 + (25-x/2)^2] + 2500 - 2[h^2 + (25 + x/2)^2] = 576
=> x^2 - 100x + 1924 = 0
=>Thus, x = 26, 74
Since, 74 cannot be..26 should be the answer..
In a temple there are some magical bells which tolls 18 times in a day,simultaneously.But every bell tolls at a different interval of time,but not in fraction of minutes.The maximum no. of bells in the temple can be
a)18
b)10
c)24
d)6
a)18
b)10
c)24
d)6
Approach For remainder of X ^Y^ Z/A?
Suppose you get R as remainder when X/A, therefore by remainder theorem
X^Y^Z/A is similar to R^Y^Z/A.
Now, find the power of R that gives remainder +1 when divided by A. If -1 is the remainder you are getting for a particular power , twice of that power will give the remainder 1.
Say that power is k.
Now find the remainder when Y^Z is divided by k. say the remainder is x.
Therefore, Y^Z is of the form (kt+x)
Therefore, R^Y^Z is similar to R^(kt+x)/A. This will be similar to R^x/A. Find the remainder for this you , will get the answer
Consider for 33^34^35/7 by remainder theorem this will be similar to 2^33^34/7
Now, in this case our R=2 and k=3, since 2^3/7 gives remainder 1. Now find the remainder when 34^35 is divided by 3. The remainder is (x=1) .
Therefore, 34^35 is of the form (3t+1)
Therefore, 33^34^35 is similar to 2^(3t+1)/7. Now (2^3t)/7 gives remainder 1.
Therfore, 2^(3t+1)/7 is similar to (2^1)/7. and hence this gives remainder as 2.
@umair2 said:In a temple there are some magical bells which tolls 18 times in a day,simultaneously.But every bell tolls at a different interval of time,but not in fraction of minutes.The maximum no. of bells in the temple can be a)18 b)10 c)24 d)6
The bells toll every 24/18=4/3 hours or 80 mins
The maximum no. of bells wd be the no. of factors of 80=10
The maximum no. of bells wd be the no. of factors of 80=10
@umair2
Should be 10..
Given, In a period of 24 hrs..the bells rings simultaneously 18 times..
Thus, after time period in which the bell rang simultaneously for the 1st time = 24*60/18 = 80 Mins..
Nw, LCM(n bells) = 80 Mins...
=>This can be maximized as LCM of 1,2,3,4,5,8,10,16,20,40 and 80...[Which is basically nothing but the Number of factors of 80]
Thus, a total of 10 bells..
@allan89 said:approach dalo bhai... geo..
a^2+b^2+c^2+d^2 - d1^2-d2^2 = 576
==> a^2+b^2+c^2+d^2 - b^2-d^2 - 2ac = 576
==> a-c=24
so c=a-24 = 50-24 = 26
==> a^2+b^2+c^2+d^2 - b^2-d^2 - 2ac = 576
==> a-c=24
so c=a-24 = 50-24 = 26
@vbhvgupta said:Remainder 32^32^32/7 ??? 50^51^52/11 30^72^87/11 33^34^35/7Approach?
Lets try 32^32^32 mod 7
E(7)=6
so basically we need to find 32^32mod6
=2^32 mod6
=2^31mod3 [cancelling 2]
=2
So, 32^32 mod6=2*2=4
Our problem reduces to 32^(6k+4) mod 7
=32^4 mod7
=4^4mod7
=(-3)^4 mod7
=81 mod7=4
ANS=4
E(7)=6
so basically we need to find 32^32mod6
=2^32 mod6
=2^31mod3 [cancelling 2]
=2
So, 32^32 mod6=2*2=4
Our problem reduces to 32^(6k+4) mod 7
=32^4 mod7
=4^4mod7
=(-3)^4 mod7
=81 mod7=4
ANS=4
@umair2 said:In a temple there are some magical bells which tolls 18 times in a day,simultaneously.But every bell tolls at a different interval of time,but not in fraction of minutes.The maximum no. of bells in the temple can be a)18 b)10 c)24 d)6
18 times in 24 hrs
so 1 time in 24/18 * 60 = 80 mins
total no. of factors of 80=2^4*5=(4+1)(1+1)=5*2=10?
If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?
(a)14, (b)8, (c)15, (d)7,(e)16
OA-(c)15
(a)14, (b)8, (c)15, (d)7,(e)16
OA-(c)15
@bs0409 said:If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?(a)14, (b)8, (c)15, (d)7,(e)16OA-(c)15
(1/r + 1/k) = 1/12
==> (k-12)(r-12) = 144
15 solution
k=13, r=156
k=14, r=94
...
k=156, r=13
==> (k-12)(r-12) = 144
15 solution
k=13, r=156
k=14, r=94
...
k=156, r=13
@bs0409 said:If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?(a)14, (b)8, (c)15, (d)7,(e)16OA-(c)15
(1/R + 1/K ) + (1/R +1/K) + ....12 times =1
12/R + 12/K =1
12K +12R =RK
(K-12)(R-12) =144
144 = 2^4 * 3^2
TOTAL WAYS TO EXPRESS 144 AS PRODUCT OF 2 ORDERED FACTORS = 5*3 =15.