## GMAT - Problem Solving - Time and Work

LEAP Administrator
6 months ago
**Problem Solving** |
**Time and Work**

### An empty pool being filled with water at a constant rate takes 8hours to fill to 3/5 of its capacity.how much more time will it take to finish filling the pool? (more)

An empty pool being filled with water at a constant rate takes 8hours to fill to 3/5 of its capacity.how much more time will it take to finish filling the pool?

A. 5hr 30min

B. 5hr 20min

C. 4hr 48min

D. 3 hr 12min

E. 2hr 40 min

Time taken to fill 3/5th of the pool = 8 hours = 480 minutes

Rate of filling the pool per minute = volume filled / time taken = (3/5) / 480 = 1/800 per minute

Time taken to fill the complete pool = 1/rate = 800 minute

time left = 800 - 480 = 320 minutes

= 5 hours 20 minutes

B is correct

LEAP Administrator
6 months ago
**Problem Solving** |
**Time and Work**

### Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alon (more)

Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x y)

(B) y/(x y)

(C) xy/(x y)

(D) xy/(x-y)

(E) xy/(y-x)

Assume the time taken by B to produce 800 nails be k

Rate at which B produces nails = 800/k

Rate at which A produces nails = Number of nails produced / time taken = 800/y

Combined rate = 800/k + 800/y

Combined time taken for production of 800 nails = x = 800 / (800/k + 800/y)

Simplifying,

1/k + 1/y = 1/x

1/k = 1/x - 1/y = (y - x)/xy

k = xy / (y - x)

Ans. xy / (y - x)

E.

LEAP Administrator
6 months ago
**Problem Solving** |
**Time and Work**

### Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneou (more)

Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25

B. 3.6

C. 4.2

D. 4.4

E. 5.5

Let the total capacity of the tank be V

pipe 1 fills the V/2 in 3 hours

Rate of flow through pipe 1 = capacity filled / time = (V/2) / 3 = V/6

pipe 2 fills 2V/3 in 6 hours

Rate of flow through pipe 2 = (2V/3) / 6 = V/9

Combined flow rate = V/6 + V/9 = 5V/18

time taken by both the pipes to fill the tank = total capacity / flow rate

= V / (5V/18) = 18/5 hours = 3.6 hours

Option B is correct.