Time And Work Test - 1

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

• Chetan is thrice as efficient as Mamta.
• Let, Mamta takes 3x days and Chetan takes x days to complete the work.
• ∴ 1/x + 1/3x = 1/60 ⇒ x = 80.
• ∴ Mamta will take 80 × 3 = 240 days to complete the work.

So Option C is correct

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

Let time taken by second one to overtake First one be xhrs

Total distance travelled by second one is 56x

Meanwhile distance covered by First one is 1.5(16)+1.5(0)+16(x+1)

⟹16x+40

at point of overtaking

56x=16x+40

⟹x=1

so they travel 

 56 kmbefore second overtakes the first one..

• 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

Charlie reaches B at 10.00 a.m. while Arun reaches B at 10:15 a.m.
At 10:15 a.m., Charlie is (15/60) × 5 = 1.25 km away from B.
Arun overtakes Ram 1.25/(10 – 5) = 0.25 hrs = 15 minutes after 10:15 a.m., i.e. at 10:30 a.m.
Hence, option C.

Since the second rabbit covers 7/120 of the distance in 2 hours 30 minutes
⇒ it covers 8.4 / 120 = 7% of the distance in 3 hours.

Thus, in 3 hours both rabbits together cover 15% of the distance which means 5% per hour so they will meet in 20 hours.

The ratio of speeds = 8 : 7.
⇒ the second rabbit would cover 700 ft to the meeting point in 20 hours and its speed would be 35 feet/hr.

So Option D is correct