Practice Test: Time & Work -1 - UPSC MCQ

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

∴ A can complete 60% of the work working alone in 80 days.

Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins

Thus, 25 km/h is the correct answer. 

So Option A is correct

Correct option is C
Time taken by first sprinter 
= 100/10 = 10sec
Time taken by second sprinter 
= 100/80 = 12.5sec
Difference = 12.5 - 10 = 2.5 sec

• X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
• This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.  
• X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

So Option C is correct

Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.

So Option D is correct

Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,

Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 

∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t² + 15t
⇒ 3t² − 5t − 50 = 0
⇒ 3t² + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 

Thus, Total time taken by slower horse = 5 + 5 = 10 hr

So Option B is correct

Time take for reaching B for ram is T = 5/5 = 1hr
Time taken for reaching B for Shyam is T = 5/10 = 1/2 hr

Its 10 am when ram reaches B and 10: 15 when Shyam reaches B , so they must meet each after 10 and before 10:15 for sure as ram starts back after reaching B

if we see for options 10: 10 am is the only answer

Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)

Work done by pipe 1 in 1 hour = 24/3 = 8 units.

Work done by pipe 2 in 1 hour = 24/4 = 6 units.

Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units

Total work done in 1 hour = 8 + 6 – 3 = 11 units

The time required to complete 11/12^th of the work = 11/12 × 24/ 11 = 2 hours

∴ The correct answer is 2 hours.