yea may be they are brothers only by father or by mother
23.5 days
16
400
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.
1208.9
Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.
( b ) 40 sq cm
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
Time taken to fill repaired cistern = 12 mins
In 1 min, pipe will fill = 1/12
Time taken to fill cistern with leak = 12 + 18 = 30 mins
In 1 min, it will fill = 1/30
Consider the leak as an outlet pipe, so
1/30 = 1/12 – [ in 1 min how much the outlet pipe leaks ]
= 1/12 – 1/30
= 5/60 – 2/60 = 3/60 = 1/20
Therefore, it will take 20 mins for the leak to empty the cistern