Rahul pays Rs. 3 and gets to throw a dice and receives an amount equal to the number that the top face of the die shows. If Rahul keeps on playing the game, how much does he win per throw in the long run?
Rahul pays Rs. 3 and gets to throw a dice and receives an amount equal to the number that the top face of the die shows. If Rahul keeps on playing the game, how much does he win per throw in the long run?Solution.50rs1rs0paisa.50 paisaCorrect OA
On a meter scale,100 ants are sitting consecutively at a distance of 1 cm.All the ants are idle at this moment .Suddenly,all the ants start moving with a speed of 1 m/s.An ant starts moving towards left or right along the scale and it continues moving in that direction till it collides with another ant or reaches the end of scale,where it falls off.When ever two ants collide,both of them reverse their directions and keep moving with the same speed.How long will it take for all the ants to fall off the scale?
Rahul pays Rs. 3 and gets to throw a dice and receives an amount equal to the number that the top face of the die shows. If Rahul keeps on playing the game, how much does he win per throw in the long run?Solution.50rs1rs0paisa.50 paisaCorrect OA
Total = 1+2+3+4+5+6 = 21
Subtract 3*6 = 18 from it.
Now, Dividing by 6 as 6 faces => 21 - 18 / 6 = 3/6 = 0.50 Ruppee.
Rahul pays Rs. 3 and gets to throw a dice and receives an amount equal to the number that the top face of the die shows. If Rahul keeps on playing the game, how much does he win per throw in the long run?Solution.50rs1rs0paisa.50 paisaCorrect OA
On a meter scale,100 ants are sitting consecutively at a distance of 1 cm.All the ants are idle at this moment .Suddenly,all the ants start moving with a speed of 1 m/s.An ant starts moving towards left or right along the scale and it continues moving in that direction till it collides with another ant or reaches the end of scale,where it falls off.When ever two ants collide,both of them reverse their directions and keep moving with the same speed.How long will it take for all the ants to fall off the scale?1) 100 s2) 1 s3) (2^100+1)/2 s4) (2^100-1)/2 s PS: No OA ........so plz post your approach
As I can see it, here we can for three cases and I guess we need to find MAXIMUM time taken by them to fall off. In that case -
1. All ants moving in one direction => 100cm/(1m/sec) = 1 sec.
2. From the middle, right ones move towards right side and left towards left side. Time taken then would be 0.5/1 = 1/2 sec.
3. Random Movement - This will also take 1 sec overall. I checked it taking small number of Ants. :splat:
On a meter scale,100 ants are sitting consecutively at a distance of 1 cm.All the ants are idle at this moment .Suddenly,all the ants start moving with a speed of 1 m/s.An ant starts moving towards left or right along the scale and it continues moving in that direction till it collides with another ant or reaches the end of scale,where it falls off.When ever two ants collide,both of them reverse their directions and keep moving with the same speed.How long will it take for all the ants to fall off the scale?1) 100 s2) 1 s3) (2^100+1)/2 s4) (2^100-1)/2 s PS: No OA ........so plz post your approach
Another Logic -
I feel that number of ants on the scale doesn't matter here. As all ants are indistiguishable here, so their actual behaviour when they will bump into each other (collide and turn around) is equivalent to ants passing through each other.
Thus, I can assume that they are running independently. So, at the most in ONE second, all ants will fall off.
bhai so k baad waala differentiation theek nhi lag rha hai.....
woh denominator ki power -1 ho jaayegi numerator mein aa kar....toh differentiation k time first part mein -1 and second part mein -2 ho jayegi uski power...please check...
bhai so k baad waala differentiation theek nhi lag rha hai.....woh denominator ki power -1 ho jaayegi numerator mein aa kar....toh differentiation k time first part mein -1 and second part mein -2 ho jayegi uski power...please check...
nothing wrong in it. y=f/g dy/dx = (g.df/dx - f.dg/dx)/g^2 as dy/dx=0, g.df/dx = f.dg/dx
i didnt bother to write second step, just directly wrote 3rd step.
A milkman claims to sell milk at the cost price but actually mixes water and milk in the ratio 1 : 4. By selling this product, his revenue is Rs. 600 every day. The amount of milk remains the same every day. One day his revenue is Rs. 560 by selling the product at its normal fixed price of Rs. 10 per litre. What is the proportion of water in milk on that day ?
A milkman claims to sell milk at the cost price but actually mixes water and milk in the ratio 1 : 4. By selling this product, his revenue is Rs. 600 every day. The amount of milk remains the same every day. One day his revenue is Rs. 560 by selling the product at its normal fixed price of Rs. 10 per litre. What is the proportion of water in milk on that day ?a) 20% b) 14.28% c) 15% d) 12% e) 16.66%
A milkman claims to sell milk at the cost price but actually mixes water and milk in the ratio 1 : 4. By selling this product, his revenue is Rs. 600 every day. The amount of milk remains the same every day. One day his revenue is Rs. 560 by selling the product at its normal fixed price of Rs. 10 per litre. What is the proportion of water in milk on that day ?a) 20% b) 14.28% c) 15% d) 12% e) 16.66%
A milkman claims to sell milk at the cost price but actually mixes water and milk in the ratio 1 : 4. By selling this product, his revenue is Rs. 600 every day. The amount of milk remains the same every day. One day his revenue is Rs. 560 by selling the product at its normal fixed price of Rs. 10 per litre. What is the proportion of water in milk on that day ?a) 20% b) 14.28% c) 15% d) 12% e) 16.66%