c-> structured oriented lang
Pascal-> Procedure oriented lang
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.
(3)^ 7.5 ÷ (27)^1.5 x (9)^2 = 3?
⇒(3)^7.5 ÷ {(3)^(3 x 1.5)} x {(3)^(2 x 2)} = 3 ?
⇒ 3^(7.5 – 4.5 + 4) = 3?
⇒ 3^7 = 3?
⇒ ? = 7
A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m
13km
451 times.
Explanation: There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will
flashes 451 times in ¾ of an hour.
10 a.m.
127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)
DISTANCE=TIME *SPEED
D=?
48*(5/18)=13.33m/s
D=9*13.333=120m