2 and half day.
13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13
To solve this problem, we can break it down into steps:
Step 1: Determine the individual rates of work for A, B, and C.
If A needs 8 days to finish the task, then their work rate is 1/8 of the task per day.
If B needs 12 days to finish the task, then their work rate is 1/12 of the task per day.
If C needs 16 days to finish the task, then their work rate is 1/16 of the task per day.
Step 2: Calculate the combined work rate of A and B.
If A works for 2 days, their contribution will be 2 * (1/8) = 1/4 of the task completed.
If B works until 25% of the job is left for C, then they will complete 75% of the task.
Step 3: Calculate the time it takes for B to complete 75% of the task.
Since B’s work rate is 1/12 of the task per day, it will take B (75%)/(1/12) = 9 days to complete 75% of the task.
Step 4: Calculate the remaining work for C.
If B completes 75% of the task, then the remaining work for C is 100% – 75% = 25% of the task.
Step 5: Calculate the time it takes for C to complete the remaining work.
Since C’s work rate is 1/16 of the task per day, it will take C (25%)/(1/16) = 4 days to complete the remaining 25% of the task.
Step 6: Calculate the total time required.
A worked for 2 days, B worked for 9 days, and C worked for 4 days, totaling 2 + 9 + 4 = 15 days.
Therefore, it will take a total of 15 days for A to work for 2 days, B to work until 25% of the job is left, and C to complete the remaining work.
C
Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.
Total 55
Manisha is a girl name so 54 boys
1 girl
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
A can copy 50 papers in 10 hrs
that means 5 papers in 1 hour
A & B can copy 70 papers in 10 hrs
that means both of them copy 7 papers in 1hour
we know A can copy 5 papers
there fore B can copy 2papers in 1 hour
so B can copu 26 papers in x hours
2*x=26
x=26/2=13 hours