@bs0409 said:ABCD a Convex quadrilateral . 3,4,5 and 6 points are marked on the sides AB,BC,CD,DA respectively . The No. of triangle with vertices on different side is..??
342??
@CSK23 said:bhai this is true for only even number of times, like LCM ( 222, 33 ) = 2442 (2's 3 times repeated) LCM ( 2222, 33 ) = 6666 (2's 4 times repeated) LCM ( 22, 333) = 7326 (3's 3 times repeated) LCM ( 22, 3333) = 6666 ( 3's 4 times repeated)
actually it comes from following (suppose I take your last example)
LCM of (22,3333) = (2*11, 3*1111) = LCM of (2,3) * LCM of (11,1111)
LCM of (11,1111) can be found out by observing that 11*101=1111 (or for that matter 1000001*111111=111111111111 or 11*10101=111111) So LCM is 11 repeated the no. of times the LCM of the no. of times 11 is repeated in each.
@bs0409 said:13^73 + 14^3 MOD 11 =???
2^73 + 3^3 mod 11 ==> E ( 11 ) = 10 . ==> 73 mod10 = 3 ==> 2^3 + 3^3 / 11 = 8 + 27 = 35/ 11 ==> rem = 2
@Calvin4ever said:If (pqr)^2 = (ijkpqr) where i, j, k, p, q,r E of W. , and pqr and ijkpqr are three digits and 6 digits numbers respectively. Then the value of i*j*k*p*q*r is :
square of 5 ends with 25..square of 25 ends with 625..so here qr = 25 then p = 6 ==> pqr = 625 ==> (625)^2 = 390625 ..i=3 j=9 k =0.....product = 0
@Calvin4ever said:The number of co-primes of 200 lying between 1 to 100 is ____
question is same as no.of co-primes of 100 less than 100 .because no.of primes involved in 200 is same as that of 100 ..i.e., 2^3 * 5 .
so, no.of co-primes = 100 (1-1/2 ) ( 1-1/5 ) =40
so, no.of co-primes = 100 (1-1/2 ) ( 1-1/5 ) =40
@Calvin4ever said:The greatest possible number which can always divide the sum of the cubes of any three consecutive integers is _____
(n-1)^3 + n^3 + (n+1)^3 = 3n (n^2 +2) .
Here if n is even..it can be divsible by greater multiples of 6.. if n is odd...the greatest would be 3...so ans is 3??
Here if n is even..it can be divsible by greater multiples of 6.. if n is odd...the greatest would be 3...so ans is 3??
A and B are efficient Workers and can complete a job in 12 days together . they worked for 4 days after which the work get stalled for next 15 days. after 15 days, 5/6th of the remaining work is added. therefore A works alone thrice as fast as and fininsh the work in 6 days. If B had worked alone and at thrice his speed in latter part. in how many days would B working alone finish the entire work
@abhishekj1991 said:find a remainder of 2^133 divided by 133?
133 = 19 * 7 ; 2^133 mod 19 ==> E (19) = 18 ==> 133mod18 = 7 ==> 2^7 / 19 = 128/19 ==> rem = 2 ; 19K + 2
2^133mod 7 = 8^ 44 * 2 /7 ==> rem =2 ; 7m +2 ; 19K +2 = 7m+2 satisfies at m=k=0...so rem = 2??
2^133mod 7 = 8^ 44 * 2 /7 ==> rem =2 ; 7m +2 ; 19K +2 = 7m+2 satisfies at m=k=0...so rem = 2??
@catahead said:A and B are efficient Workers and can complete a job in 12 days together . they worked for 4 days after which the work get stalled for next 15 days. after 15 days, 5/6th of the
26.5 days approx?
@catahead said:A and B are efficient Workers and can complete a job in 12 days together . they worked for 4 days after which the work get stalled for next 15 days. after 15 days, 5/6th of the remaining work is added. therefore A works alone thrice as fast as and fininsh the work in 6 days. If B had worked alone and at thrice his speed in latter part. in how many days would B working alone finish the entire work
exactly 26.4 days
A and B can do the job in 12 days so 1/A + 1/B = 1/12
4 days they worked together so 1/3 work is completed. Remaining work = 2/3
After 15 days remaining work = 2/3(1+ 5/6) = 11/9
a thrice efficient A finishes the work in 6 days => (3/A)*6 = 11/9 => 1/A = 11/(9*3*6)
putting this value in 1/A + 1/B = 1/12 we get 1/B = 30/(12*162)
a thrice efficient B will complete the work in (11/9)*(B/3) = 132/5 = 26.4 days
A and B play 12 games of chess of which 6 are won by A, 4 by B and 2 ends in a tie. They agree to play 3 more games. Find the probability that
1) A winal the 3 games
2) 2 Games end in a tie
3) A and B win alternative
4) B wins at least 1 game
No OA :D
@chandrakant.k said:A and B play 12 games of chess of which 6 are won by A, 4 by B and 2 ends in a tie. They agree to play 3 more games. Find the probability that 1) A winal the 3 games2) 2 Games end in a tie 3) A and B win alternative 4) B wins at least 1 gameNo OA
1) 1/8
2) 15/216
3) 5/ 36
4) 2/3
2) 15/216
3) 5/ 36
4) 2/3
@chandrakant.k said:explain the method pls!!! as i dont have OA
p( a ) + p(b) + p(t) =1 ;
p(a) = 6/12 = 1/2 . p(b) = 4/12 = 1/3 ; p(t) = 2/12 = 1/6 .
1) a wining all three = 1/2 *1/2 * 1/2 = 1/8 .
2) 2 games end in a tie = 3c2 * (1/6)^2 * 5/6 = 15/216 [ 5/6 is non tie case ]
3) ABA and BAB are the cases = 1/2 * 1/3 * 1/2 + 1/3 * 1/2 * 1/3 = 1/12 + 1/18 = 5/36.
4) 3c1 *(1/3)*(2/3)^2 + 3c1 (1/3)^2 * 2/3 + 3c3 (1/3)^3 = 18/27 = 2/3 { cases of B wing 1 ,2,3 }
p(a) = 6/12 = 1/2 . p(b) = 4/12 = 1/3 ; p(t) = 2/12 = 1/6 .
1) a wining all three = 1/2 *1/2 * 1/2 = 1/8 .
2) 2 games end in a tie = 3c2 * (1/6)^2 * 5/6 = 15/216 [ 5/6 is non tie case ]
3) ABA and BAB are the cases = 1/2 * 1/3 * 1/2 + 1/3 * 1/2 * 1/3 = 1/12 + 1/18 = 5/36.
4) 3c1 *(1/3)*(2/3)^2 + 3c1 (1/3)^2 * 2/3 + 3c3 (1/3)^3 = 18/27 = 2/3 { cases of B wing 1 ,2,3 }
@toshalimitra said:P,Q & R start from the same point and they need to cover a distance of 100 km. They have only one bike with them. Only two person can sit on the bike at a time. P & Q start on the bike and they are moving at the speed of 25 kmph. P drops Q at a place on the way and returns to pick R who is already moving at a speed of 5 kmph. In the mean time, Q is also moving at a speed of 5 kmph towards the destination. Three of them reach the destination at the same time. What is the total time take?(with approach)
since both the walkers have same starting point and same ending point...and they reach at the same time hence their average speed of the entire journey must be the same, which in turn implies that for both of them they should have walked equal distance and covered equal distance in car. The only difference being that one person first walks and then takes the bike ride, while other person first rides the bike and then walks.
So let the distance walked by each of them = d.
=> distance covered on bike by each = 100-d
|--------d----------|-----------------(100-2d)--------------|---------d---------|
=>total time taken by each = d/5 + (100-d)/25
since car covers a distance of 100 + going back and forth (100-2d) distance
hence time taken by car = {100 + 2*(100-2d)}/25
now equate the time
=>d/5 + (100-d)/25 = {100+2(100-2d)}/25
=>5d + 100 - d = 100 + 2(100-2d)
=>8d = 200
=>d = 25
hence the time = 25/5 + (100-25)/25 = 5+3 = 8 hrs
ATDH
@meenu05 said:question : a test has 90 questions . Each correct answer is awarded 1 mrk each wrong answer is awarded -1/2 and an unattempted question is awareded -1/4 mark . a candidate scored 5 marks in the test . the number of wrongly answered question by him cannot be more than ?
a+b+c = 90 ; a-1/2 *b -1/4 *c = 5....to hav max value for b ..minimise the value of c..say zero.
==> a+b = 50 and a-b/2 = 5 ==> 2a-b = 10
on solving a = 100/3 = 33 (approx).. so now a=33 and to get total 5..c can take value of 2..==> b = 55
==> a+b = 50 and a-b/2 = 5 ==> 2a-b = 10
on solving a = 100/3 = 33 (approx).. so now a=33 and to get total 5..c can take value of 2..==> b = 55
@meenu05 said:question : a test has 90 questions . Each correct answer is awarded 1 mrk each wrong answer is awarded -1/2 and an unattempted question is awareded -1/4 mark . a candidate scored 5 marks in the test . the number of wrongly answered question by him cannot be more than ?
x - y/2 - z/4 = 5
x + y + z = 90
subtracting -> -3y/2 -5z/4 = -85
3y/2 + 5z/4 = 85
6y+5z = 340
y=55 and z=2...
@swapnil4ever2u said:how many distinct rectangles of perimeter not more than 258 cm can have both length and breadth as positive integers when expressed in cms ??
Is it 4160??
2(l+b) =
l+b =
l+b = 2 -> 1 solution
l+b = 3 -> 1 solution
l+b = 4 -> 2 solutions
l+b = 5 -> 2 solutions
.
.
.
l+b = 128 -> 64 solutions
l+b = 129 -> 64 solutions
hence total = 2(1+2+3+...+64) = 64*65 = 4160 solutions....
Solve for x:
(a/ax-1)+(b/bx-1)=a+b
(a/ax-1)+(b/bx-1)=a+b