Time and Work: CAT Previous Year Questions

Ans: A

Assume the total effort required for the task is a unit of 1.

If each individual were to complete the entire task independently, their respective total expenditures would be:

* Arun: $2160 \text{ (per day)} \times 24 \text{ (days)} = 51840$
* Varun: $2400 \text{ (per day)} \times 21 \text{ (days)} = 50400$
* Tarun: $2160 \text{ (per day)} \times 15 \text{ (days)} = 32400$

Tarun presents the most economical option for completing the task, as he can finish it in 15 days. However, the requirement is to have the task completed within a 10-day timeframe.

Tarun will complete a portion of the work equal to:
$10 \text{ (days)} \times \dfrac{1}{15} \text{ (work per day)} = \dfrac{2}{3}$

The remaining portion of the work is:
$1 – \dfrac{2}{3} = \dfrac{1}{3}$

This remaining $\dfrac{1}{3}$ of the work will be undertaken by the next most cost-effective individual, Varun.

The number of days Varun will need to complete this portion is calculated as:
$\dfrac{\text{Work to be done}}{\text{Varun’s work rate}} = \dfrac{1/3 \text{ (work)}}{1/21 \text{ (work per day)}} = \dfrac{1}{3} \times 21 = 7$ days.

The total expenditure will be the sum of the payments made to Tarun and Varun:
(Tarun’s payment) + (Varun’s payment) = $(10 \text{ days} \times 2160 \text{ per day}) + (7 \text{ days} \times 2400 \text{ per day})$
$= 21600 + 16800 = 38400$

Ans: 340

Explanation:Let the efficiency of Chandan be $x$ units per day.
Then, the efficiency of Bipin is $2x$ units per day.
And the efficiency of Ankita is $2 \times (2x) = 4x$ units per day.

Let the total work to be done be $W$.
The total work done by all three together in one day is the sum of their efficiencies: $4x + 2x + x = 7x$ units per day.

Bipin leaves the job after 20 days.
In the first 20 days, all three worked together.
Work done in the first 20 days = (efficiency of Ankita + efficiency of Bipin + efficiency of Chandan) $\times$ 20
Work done in the first 20 days = $(4x + 2x + x) \times 20 = 7x \times 20 = 140x$ units.

The job was completed in 60 days.
This means that after Bipin left, the remaining work was completed by Ankita and Chandan in $60 – 20 = 40$ days.
The combined efficiency of Ankita and Chandan is $4x + x = 5x$ units per day.
Work done by Ankita and Chandan in the remaining 40 days = $(4x + x) \times 40 = 5x \times 40 = 200x$ units.

The total work $W$ is the sum of the work done in the first 20 days and the work done in the remaining 40 days.
$W = 140x + 200x = 340x$ units.

We need to find the number of days needed by Chandan to complete the job alone.
Let $D$ be the number of days Chandan needs to complete the job alone.
Work done by Chandan alone = Chandan’s efficiency $\times$ Number of days
$W = x \times D$

We have $W = 340x$.
So, $340x = x \times D$.
Dividing both sides by $x$ (since $x$ is an efficiency, $x > 0$), we get $D = 340$.

Therefore, Chandan needs 340 days to complete the job alone.

Final_Answer:340

Ans: B

Explanation:
Let the productivity of each member in team A, B, and C be $p_A$, $p_B$, and $p_C$ respectively.
The number of members in team A, B, and C are 5, 8, and 10 respectively.

The total productivity of team A is $5p_A$.
The total productivity of team B is $8p_B$.
The total productivity of team C is $10p_C$.

Team A can complete the job in 40 hours, so the total work W = $5p_A \times 40 = 200p_A$.
Team B can complete the job in 50 hours, so the total work W = $8p_B \times 50 = 400p_B$.
Team C can complete the job in 4 hours, so the total work W = $10p_C \times 4 = 40p_C$.

Equating the work, we get:
$200p_A = 400p_B = 40p_C$

From $200p_A = 400p_B$, we get $p_A = 2p_B$.
From $400p_B = 40p_C$, we get $10p_B = p_C$.

Let’s express the productivity of each member in terms of $p_B$.
$p_A = 2p_B$
$p_B = p_B$
$p_C = 10p_B$

The total work W can be expressed in terms of $p_B$. Let’s use the work done by team B.
W = $400p_B$.

Two members from team A, three members from team B, and one member from team C together start the job.
The combined productivity of these members is $2p_A + 3p_B + 1p_C$.
Substitute the equivalent productivities in terms of $p_B$:
Combined productivity = $2(2p_B) + 3p_B + 1(10p_B) = 4p_B + 3p_B + 10p_B = 17p_B$.

These members work for 23 hours.
Work done in the first 23 hours = $17p_B \times 23 = 391p_B$.

The remaining work = W – Work done = $400p_B – 391p_B = 9p_B$.

After 23 hours, the member from team C leaves.
The remaining members are two from team A and three from team B.
Their combined productivity is $2p_A + 3p_B = 2(2p_B) + 3p_B = 4p_B + 3p_B = 7p_B$.

The job needs to be finished in the next 1 hour.
Let $x$ be the number of additional members from team B required to finish the remaining work in 1 hour.
The new productivity will be $7p_B + xp_B = (7+x)p_B$.

The work to be done in the next 1 hour is $9p_B$.
So, $(7+x)p_B \times 1 = 9p_B$.
$7+x = 9$
$x = 9 – 7$
$x = 2$.

Therefore, 2 additional members from team B are required.

Correct_Option: B

Ans: A

Explanation:Let the rate of water flow through pipes A, B, and C be $4x$, $9x$, and $36x$ respectively, where $x$ is a constant.
The rate of pipe A is $4x$.
Pipe A can fill the empty tank completely in 15 hours.
The volume of the tank can be represented as the rate multiplied by the time.
Volume of the tank = Rate of pipe A * Time taken by pipe A
Volume of the tank = $(4x) * 15$ hours = $60x$ tank-units-hours.

If all three pipes A, B, and C are used simultaneously, their combined rate is the sum of their individual rates:
Combined rate = Rate of A + Rate of B + Rate of C
Combined rate = $4x + 9x + 36x = 49x$ tank-units-hours.

Let $T$ be the time required to fill the tank completely when all three pipes are used simultaneously.
Volume of the tank = Combined rate * $T$
$60x = (49x) * T$

To find $T$, we can divide both sides by $49x$:
$T = \frac{60x}{49x} = \frac{60}{49}$ hours.

The question asks for the time in minutes. To convert hours to minutes, we multiply by 60:
Time in minutes = $\frac{60}{49} * 60$ minutes
Time in minutes = $\frac{3600}{49}$ minutes.

Now, we need to calculate the value of $\frac{3600}{49}$ and find the nearest option.
$\frac{3600}{49} \approx 73.469$ minutes.

Let’s check the given options:
Option A: 73
Option B: 78
Option C: 76
Option D: 71

The calculated time is approximately 73.469 minutes. The nearest option is 73 minutes.

Correct_Option:A

Ans: 6

Let R represent the quantity of work Renu completes in one hour.
Let S represent the quantity of work Seema completes in one hour.

We are given that a specific job requires a certain amount of work.
Renu can complete the job in 15 days, working 4 hours per day. The total work Renu contributes is therefore $15 \times 4 \times R = 60R$.
Seema can complete the job in 8 days, working 5 hours per day. The total work Seema contributes is therefore $8 \times 5 \times S = 40S$.

Since the total work for the job is the same in both cases, we have $60R = 40S$.
This simplifies to $S = \frac{60}{40}R = 1.5R$.

Now, consider a scenario where Seema works for X days and Renu works for Y hours each day to complete this same job.
The problem states that Seema works for 2Y hours per day, and Renu works for 2X days.
The work contributed by Seema in this new scenario is $X \times (2Y) \times S = 2XYS$.
The work contributed by Renu in this new scenario is $(2X) \times Y \times R = 2XYR$.

We are also given that Y = 2. Substituting this into the work expressions:
Work by Seema = $2X(2)S = 4XS$.
Work by Renu = $2X(2)R = 4XR$.

The total work for the job remains $60R$.
So, the sum of the work done by Seema and Renu in the new scenario equals the total work:
$4XR + 4XS = 60R$.

Substitute $S = 1.5R$ into the equation:
$4XR + 4X(1.5R) = 60R$.
$4XR + 6XR = 60R$.
$10XR = 60R$.

Dividing both sides by 10R (assuming R is not zero), we get:
$X = 6$.

Therefore, X, which represents the number of days Seema will work, is 6.

Ans: 139

Let the individual work outputs of Amal, Vimal, and Sunil be represented by A, V, and S, respectively.
Assume the total task to be completed is denoted by T.
The problem provides the following relationships:
150A + 150V = T      …(1)
100V + 100S = T      …(2)
75A + 135V + 45S = T    …(3)

Summing equations (1) and (2) yields: 150A + 250V + 100S = 2T   …(4)
Multiplying equation (3) by 2 gives: 150A + 270V + 100S = 2T  …(5)

Subtracting equation (5) from equation (4) results in 10S = 20V, which simplifies to S = 2V.
Substituting S = 2V into equation (2) allows us to express the total work T in terms of V: T = 100V + 100(2V) = 300V.
Now, substituting T = 300V into equation (1) gives 150A + 150V = 300V, leading to 150A = 150V, or A = V.
Thus, the daily work rates of Amal, Vimal, and Sunil are equivalent to V, V, and 2V units per day, respectively.

The question involves work cycles. To simplify calculations, we can consider a time frame that is the least common multiple (LCM) of the periods mentioned. In this scenario, the LCM of the given periods is 6.

Over a 6-day period:
Amal completes 6 * V = 6V units of work.
Vimal completes 3 * V = 3V units of work.
Sunil completes 2 * 2V = 4V units of work.

Therefore, in one 6-day cycle, a total of 6V + 3V + 4V = 13V units of work are accomplished.
To determine the number of full cycles to complete the total work T (which is 300V), we divide the total work by the work done per cycle: 300V / 13V = 23 with a remainder.
This means that 23 full cycles are completed, accounting for 23 * 13V = 299V units of work.
The duration of these 23 cycles is 23 * 6 days = 138 days.

The remaining 1V unit of work will be completed on the subsequent day.
Consequently, the total time required is 138 days + 1 day = 139 days.

Ans: B

Suppose an individual, Sam, can finish a task in 20 days. Another individual, Mohit, works at double Sam’s efficiency, implying Mohit would complete the task in 10 days. Mohit’s efficiency is thrice that of Ayna, meaning Ayna would take 30 days to complete the same task.

Let’s represent the total effort required for the task as 60 units. This assignment allows us to calculate the daily output of Mohit, Sam, and Ayna as 6, 3, and 2 units, respectively.

In the first day, Sam and Mohit collaborate, accomplishing 9 units.
On the second day, Sam and Ayna work together, completing 5 units.
On the third day, Mohit and Ayna team up, finishing 8 units.

This pattern results in 22 units of work completed over a 3-day period.
After two such 3-day cycles, 44 units of work are done, leaving 16 units remaining (60 – 44 = 16).

On the seventh day (the start of the next cycle), Sam and Mohit work, contributing 9 units, leaving 7 units.
On the eighth day, Sam and Ayna work, completing 5 units, leaving 2 units.
On the ninth day, Ayna and Mohit will finish the remaining 2 units.

Considering Sam’s contribution: Sam participated on days 1, 2, 4, 5, 7, and 8. He worked for a total of 6 days, contributing 3 units per day, for a cumulative output of 18 units of work (6 days * 3 units/day).

The proportion of the total work completed by Sam is calculated as $\frac{18}{60}$, which simplifies to $\frac{3}{10}$.
Consequently, Option B represents the correct solution.

Ans: 27

Let us denote the work rates of Amal, Sunil, and Kamal as ‘a’, ‘s’, and ‘k’, respectively.
We are given that their work rates are in Harmonic Progression (H.P.).
This implies:
$ \dfrac{2}{s} = \dfrac{1}{a} + \dfrac{1}{k} $ —(1)

We are also informed that Kamal takes twice the time Amal takes to complete an equivalent task. This means Kamal’s work rate is half of Amal’s work rate:
$ k = \dfrac{1}{2}a \quad \text{or} \quad a = 2k $

Next, we are told that if Amal works for 4 days, Sunil for 9 days, and Kamal for 16 days, the entire job is completed. Let ‘W’ represent the total amount of work.
$ 4a + 9s + 16k = W $

Substitute the relationship from H.P. (1) with the information $a=2k$ (which means $k = a/2$):
$ \dfrac{2}{s} = \dfrac{1}{a} + \dfrac{1}{a/2} $
$ \dfrac{2}{s} = \dfrac{1}{a} + \dfrac{2}{a} $
$ \dfrac{2}{s} = \dfrac{3}{a} $
From this, we get:
$ a = \dfrac{3}{2}s $
And since $a = 2k$, we have:
$ k = \dfrac{1}{2}a = \dfrac{1}{2} \left(\dfrac{3}{2}s\right) = \dfrac{3}{4}s $

Now, substitute these relationships for ‘a’ and ‘k’ into the total work equation:
$ 4\left(\dfrac{3s}{2}\right) + 9s + 16\left(\dfrac{3s}{4}\right) = W $
Simplify the equation:
$ 6s + 9s + 12s = W $
$ 27s = W $
This means:
$ s = \dfrac{W}{27} $
The work rate ‘s’ represents the fraction of the total work Sunil can complete in one day. Therefore, Sunil would take 27 days to complete the entire work if he were to work alone.

Ans: A

Let the duration for A to fill the tank independently be $x$ hours. Consequently, the duration for B to drain the tank independently is $(x-1)$ hours, as B functions as a drainage pipe. Let the duration for C to fill the tank independently be $y$ hours.

The problem states that when pipes A, B, and C operate simultaneously, the empty tank is filled in 2 hours.
This leads to the equation:
$\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}$ …. Eq(1)

Another condition is provided: if pipes B and C are opened together with an empty tank, and pipe B is closed after 1 hour, pipe C then takes an additional 1 hour and 15 minutes to complete the filling of the tank.
The total time pipe C was open is $1 + (1 \text{ hour } 15 \text{ minutes}) = 2 \text{ hours } 15 \text{ minutes}$, which converts to $\frac{9}{4}$ hours.
In 1 hour, pipe B’s contribution to emptying is $-\frac{1}{x-1}$ of the tank.
In $\frac{9}{4}$ hours, pipe C’s contribution to filling is $\frac{9}{4y}$ of the tank.
The net effect of these operations is the filling of the entire tank (1 unit).
Thus, we can write the equation:
$\frac{9}{4y}-\frac{1}{x-1}=1$ …. Eq(2)

Solving these two equations simultaneously yields $y=\frac{3}{2}$ and $x=3$.
Therefore, the time taken by C to fill the tank is $\frac{3}{2}$ hours, which is equivalent to 90 minutes.
The correct option is A.

Ans: D

Let the daily output of Rahul, Rakshita, and Gurmeet be represented by ‘r’, ‘x’, and ‘g’ units, respectively, with ‘T’ denoting the total work units.

From the problem statement, we can infer the following:
Working together, Rahul, Rakshita, and Gurmeet would take more than 7 days to complete the task. This translates to:
7(r + x + g) < T

Similarly, Rahul and Gurmeet, working collaboratively, would complete the task in fewer than 15 days. This can be expressed as:
15(r + g) > T

Combining these inequalities, we establish a range for the total work:
7(r + x + g) < T < 15(r + g)

We are also given that they all worked jointly for 6 days, and then Rakshita continued alone for an additional 3 days to complete the job. The total work done can thus be expressed as:
T = 6(r + x + g) + 3x

Now, substituting this expression for T into our established inequality:
7(r + x + g) < 6(r + x + g) + 3x < 15(r + g)

This leads to two separate inequalities:

1. 7(r + x + g) < 6(r + x + g) + 3x
Simplifying this, we get:
(r + x + g) < 3x
Which further simplifies to:
r + g < 2x

2. 6(r + x + g) + 3x < 15(r + g)
Combining terms and rearranging:
6r + 6x + 6g + 3x < 15r + 15g
9x < 9r + 9g
Dividing by 9:
x < r + g

Therefore, we have established that r + g < 2x and x < r + g.

Now let’s consider the time Rakshita would take to complete the work alone. This is T/x.
From our earlier inequalities involving T:
7(r + x + g) < T => T/x > 7(r + x + g) / x = 7(r/x + 1 + g/x)
T < 15(r + g) => T/x < 15(r + g) / x = 15(r/x + g/x)

From r + g < 2x, we can infer r/x + g/x < 2.
From x < r + g, we can infer 1 < r/x + g/x.

So, we have 1 < r/x + g/x < 2.

Substituting these into the expressions for T/x:
Lower bound for T/x: 7(r/x + 1 + g/x) > 7(1 + 1) = 14. However, this is a loose lower bound.
A more direct approach:
From 7(a+b+c) < W < 15(a+c) and W = 6(a+b+c)+3b, we have:
7(a+b+c) < 6(a+b+c)+3b => (a+b+c) < 3b => a+c < 2b
6(a+b+c)+3b < 15(a+c) => 6a+6b+6c+3b < 15a+15c => 9b < 9a+9c => b < a+c

So we have 2b > a+c and b < a+c. This means a+c is between b and 2b.

The number of days Rakshita takes alone is W/b.
We have 7(a+b+c) < W and W < 15(a+c).
Also W = 6(a+b+c) + 3b.

From 7(a+b+c) < 6(a+b+c) + 3b, we get a+b+c < 3b, which means a+c < 2b.
From 6(a+b+c) + 3b < 15(a+c), we get 6a+6b+6c+3b < 15a+15c, which simplifies to 9b < 9a+9c, or b < a+c.

So, b < a+c < 2b.

Now consider the bounds for W/b:
W = 6(a+b+c) + 3b.
W/b = 6(a+b+c)/b + 3 = 6(a/b + 1 + c/b) + 3 = 6(a/b + c/b) + 9.

Since b < a+c, we have 1 < (a+c)/b.
Since a+c < 2b, we have (a+c)/b < 2.
So, 1 < (a+c)/b < 2.

Therefore, the lower bound for W/b is 6 * (lower bound of (a+c)/b) + 9 = 6 * 1 + 9 = 15.
The upper bound for W/b is 6 * (upper bound of (a+c)/b) + 9 = 6 * 2 + 9 = 12 + 9 = 21.

Thus, the number of days required for Rakshita working alone is strictly between 15 and 21.

The correct option is D.

Ans: 36

Let ‘G’ represent Gautam’s work rate and ‘S’ represent Suhani’s work rate. Let ‘T’ be the total amount of work.

When working together, they complete the work in 20 days, meaning their combined daily output is T/20:
G + S = T/20 —-(1)

In the second scenario, Gautam works at 60% of his usual rate (0.6G or 3G/5) and Suhani works at 150% of her usual rate (1.5S or 3S/2). They still complete the same total work in 20 days:
3G/5 + 3S/2 = T/20

Equating the daily work from both scenarios:
G + S = 3G/5 + 3S/2

To find the ratio of their efficiencies, we can rearrange this equation:
S – 3S/2 = 3G/5 – G
-S/2 = -2G/5
S/G = 4/5

This ratio indicates that Gautam (G) is more efficient than Suhani (S).

Now, substitute this ratio back into the first equation (Equation 1). We can express S in terms of G: S = (4/5)G.
G + (4/5)G = T/20
(9/5)G = T/20

Solving for G (Gautam’s daily work rate):
G = (5/9) * (T/20)
G = T/36

This result means Gautam can complete the entire work ‘T’ on his own in 36 days.

Ans: 6

Let the work rates of Anu, Tanu, and Manu be represented by $5k$, $8k$, and $10k$ units per hour, respectively.
The total amount of work to be completed is the least common multiple of these rates, which is $40k$ units.

Based on this total work and their individual rates:
Anu completes $40k / 5k = 8$ units per hour.
Tanu completes $40k / 8k = 5$ units per hour.
Manu completes $40k / 10k = 4$ units per hour.

It’s stated that when all three work together, they complete the task in 32 hours.
The combined work rate is $8 + 5 + 4 = 17$ units per hour.
So, total work = $17 \text{ units/hour} \times 32 \text{ hours} = 544$ units.
Thus, $40k = 544$, which means $k = \frac{544}{40} = \frac{68}{5}$.

The problem then describes a scenario where Anu and Tanu work together for a certain period.
They work for 6 days, and each day they work for 6 hours and 40 minutes, which is equal to $6 \frac{2}{3}$ hours or $\frac{20}{3}$ hours per day.
The total time Anu and Tanu work together is $6 \text{ days} \times \frac{20}{3} \text{ hours/day} = 40$ hours.

In these 40 hours, the work done by Anu and Tanu together is:
Work done by Anu = $8 \text{ units/hour} \times 40 \text{ hours} = 320$ units.
Work done by Tanu = $5 \text{ units/hour} \times 40 \text{ hours} = 200$ units.
Total work by Anu and Tanu = $320 + 200 = 520$ units.

The remaining work is $544 \text{ units} – 520 \text{ units} = 24$ units.

It is given that Manu completes this remaining work alone in 6 hours.
Manu’s work rate is 4 units per hour.
Work done by Manu = $4 \text{ units/hour} \times 6 \text{ hours} = 24$ units.

This confirms that Manu completes the remaining 24 units of work in 6 hours.

Ans: D

Explanation:Let N be the number of people in the group initially.
The total work done is 100%.
The work done in the first phase is 35%.
The work done per day by N people is proportional to the number of people and the hours worked per day.
Let W be the total work.
In the first phase:
Work done = 35% of W
Number of people = N
Hours per day = 7 hours
Number of days = 10 days
So, the total man-hours in the first phase = N * 7 * 10 = 70N

The remaining work is 100% – 35% = 65% of W.
In the second phase:
Number of people = N – 10
Hours per day = 10 hours
Number of days = 14 days
So, the total man-hours in the second phase = (N – 10) * 10 * 14 = 140(N – 10)

Assuming that the work done is directly proportional to the man-hours, we can set up a ratio:
(Work done in phase 1) / (Man-hours in phase 1) = (Work done in phase 2) / (Man-hours in phase 2)
0.35W / 70N = 0.65W / 140(N – 10)

We can cancel out W from both sides:
0.35 / 70N = 0.65 / 140(N – 10)

Multiply both sides by 140N(N – 10) to clear the denominators:
0.35 * 140(N – 10) = 0.65 * 70N
49(N – 10) = 45.5N
49N – 490 = 45.5N

Now, rearrange the equation to solve for N:
49N – 45.5N = 490
3.5N = 490
N = 490 / 3.5
N = 4900 / 35
N = 140

Let’s check the options.
Option A: N = 150
Option B: N = 23
Option C: N = 36
Option D: N = 140

Our calculated value of N is 140, which matches Option D.

Let’s verify with N = 140:
Phase 1: 140 people, 7 hours/day, 10 days. Total man-hours = 140 * 7 * 10 = 9800. Work done = 35%.
Phase 2: (140 – 10) = 130 people, 10 hours/day, 14 days. Total man-hours = 130 * 10 * 14 = 18200. Work done = 65%.

Ratio of work to man-hours:
Phase 1: 0.35 / 9800
Phase 2: 0.65 / 18200

Let’s see if these ratios are equal.
0.35 / 9800 = 35 / 980000 = 1 / 28000
0.65 / 18200 = 65 / 1820000 = 13 / 364000 = 1 / 28000

The ratios are equal, so our value of N = 140 is correct.

Correct_Option:D

Ans: 11

Let the output of Bob be set to 3 units per day. Consequently, the output of Alex will be 6 units per day, and the output of Cole will be 2 units per day.
Given that Bob completes the task in 40 days, the entire workload amounts to 40 multiplied by 3, which equals 120 units.
On the first day, with Alex and Bob collaborating, the combined output is 3 + 6 = 9 units.
In a similar fashion, on days 2 and 3, the respective outputs are 5 and 8 units.
Therefore, the aggregate output over the initial 3 days is 9 + 5 + 8 = 22 units.
The work accomplished within the first 15 days is 22 multiplied by 5, resulting in 110 units.
Thus, the task will be concluded on the 17th day, as the combined output of 9 + 5 = 14 units surpasses the remaining workload.
Considering that Alex contributes on two out of every three days, he will be involved for 10 days within the first 15.
He will also participate on the 16th day.
The total duration of his involvement is 11 days.

Ans: D

Explanation:Let the total work be the least common multiple (LCM) of 15, 12, and 20.
LCM(15, 12, 20) = 60 units of work.

Individual work rates:
Anu’s rate = 60 units / 15 days = 4 units/day
Vinu’s rate = 60 units / 12 days = 5 units/day
Manu’s rate = 60 units / 20 days = 3 units/day

Working pattern:
Vinu works every day.
Anu works on days 1, 3, 5, … (alternate days starting from the first day).
Manu works on days 2, 4, 6, … (alternate days starting from the second day).

Let’s analyze the work done in cycles of two days:
Day 1: Anu and Vinu work.
Work done on Day 1 = Anu’s rate + Vinu’s rate = 4 + 5 = 9 units.

Day 2: Vinu and Manu work.
Work done on Day 2 = Vinu’s rate + Manu’s rate = 5 + 3 = 8 units.

Work done in a 2-day cycle (Day 1 and Day 2) = 9 + 8 = 17 units.

We need to complete 60 units of work.

Let’s see how many full 2-day cycles are needed.
Number of full cycles = 60 units / 17 units/cycle $\approx$ 3.53 cycles.
This means we will have 3 full cycles and then some remaining work.

Work done in 3 full cycles (6 days) = 3 cycles * 17 units/cycle = 51 units.

After 6 days, 51 units of work are completed. Remaining work = 60 – 51 = 9 units.

Now, we start the 7th day.
On the 7th day, Anu and Vinu will work (since it’s an odd-numbered day, Anu works, and Vinu works every day).
Work done on Day 7 = Anu’s rate + Vinu’s rate = 4 + 5 = 9 units.

The remaining 9 units of work will be completed on the 7th day.

Therefore, the total number of days needed to complete the work is 6 days (for 3 cycles) + 1 day (for the remaining work) = 7 days.

Let’s verify:
Day 1: Anu + Vinu = 4 + 5 = 9
Day 2: Vinu + Manu = 5 + 3 = 8 (Total = 9 + 8 = 17)
Day 3: Anu + Vinu = 4 + 5 = 9 (Total = 17 + 9 = 26)
Day 4: Vinu + Manu = 5 + 3 = 8 (Total = 26 + 8 = 34)
Day 5: Anu + Vinu = 4 + 5 = 9 (Total = 34 + 9 = 43)
Day 6: Vinu + Manu = 5 + 3 = 8 (Total = 43 + 8 = 51)
Day 7: Anu + Vinu = 4 + 5 = 9 (Total = 51 + 9 = 60)

The work is completed on the 7th day.

Correct_Option:D

Ans: C

Explanation:Let N, G, and S be the daily earnings of Neeta, Geeta, and Sita, respectively.

From the first statement: “The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days.”
This can be written as an equation:
N + G = 6S (Equation 1)

From the second statement: “The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days.”
This can be written as an equation:
S + N = 2G (Equation 2)

We want to find the ratio of the daily earnings of the one who earns the most to that of the one who earns the least. To do this, we need to find the individual daily earnings of Neeta, Geeta, and Sita.

We have a system of two linear equations with three variables. We can express two variables in terms of the third. Let’s try to express N and G in terms of S.

From Equation 2, we can express N in terms of G and S:
N = 2G – S

Substitute this expression for N into Equation 1:
(2G – S) + G = 6S
3G – S = 6S
3G = 7S
G = (7/3)S

Now that we have G in terms of S, we can substitute this back into the expression for N:
N = 2G – S
N = 2 * (7/3)S – S
N = (14/3)S – S
N = (14/3)S – (3/3)S
N = (11/3)S

So, we have the daily earnings of each person in terms of S:
Sita’s earning = S
Geeta’s earning = G = (7/3)S
Neeta’s earning = N = (11/3)S

Now, let’s compare the earnings to find who earns the most and who earns the least.
To compare, we can express them with a common denominator, which is 3:
Sita: S = (3/3)S
Geeta: G = (7/3)S
Neeta: N = (11/3)S

Comparing the coefficients of S: 3/3, 7/3, 11/3.
The largest coefficient is 11/3, so Neeta earns the most.
The smallest coefficient is 3/3, so Sita earns the least.

The ratio of the daily earnings of the one who earns the most (Neeta) to that of the one who earns the least (Sita) is:
Ratio = N : S
Ratio = (11/3)S : S

To simplify the ratio, we can divide both sides by S (assuming S is not zero, which is a reasonable assumption for earnings):
Ratio = (11/3) : 1

To get rid of the fraction, multiply both sides by 3:
Ratio = 11 : 3

Let’s check the options.
Option A: 3:2
Option B: 11:7
Option C: 11:3
Option D: 7:3

Our calculated ratio is 11:3, which matches Option C.

Let’s also find the earnings in terms of a common value, for example, if S = 3 units.
Then Sita earns 3 units.
Geeta earns (7/3) * 3 = 7 units.
Neeta earns (11/3) * 3 = 11 units.

Earnings: Sita = 3, Geeta = 7, Neeta = 11.
The one who earns the most is Neeta (11).
The one who earns the least is Sita (3).
The ratio of the highest earner to the lowest earner is 11:3.

Correct_Option:C

Ans: 32

Let the total effort required for the project be 48 units. Assume Amar’s monthly contribution is ‘a’ units, Akbar’s is ‘b’ units, and Anthony’s is ‘c’ units.

Amar and Akbar together finish the project in 12 months. Therefore, their combined monthly output is $\frac{48}{12} = 4$ units.
So, $a + b = 4$.

Similarly, Akbar and Anthony complete the project in 16 months, meaning their combined monthly output is $\frac{48}{16} = 3$ units.
Thus, $b + c = 3$.

And, Amar and Anthony complete the project in 24 months, implying their combined monthly output is $\frac{48}{24} = 2$ units.
Hence, $a + c = 2$.

Solving these three linear equations:
1) $a + b = 4$
2) $b + c = 3$
3) $a + c = 2$

Adding all three equations: $2(a + b + c) = 4 + 3 + 2 = 9$, so $a + b + c = 4.5$.
Substituting (2) into the sum: $a + 3 = 4.5 \implies a = 1.5$.
Substituting (3) into the sum: $b + 2 = 4.5 \implies b = 2.5$.
Substituting (1) into the sum: $c + 4 = 4.5 \implies c = 0.5$.

So, Amar’s monthly output is 1.5 units, Akbar’s is 2.5 units, and Anthony’s is 0.5 units.
Amar’s work rate (1.5 units/month) is between the fastest (Akbar, 2.5 units/month) and the slowest (Anthony, 0.5 units/month).

To complete the entire project of 48 units, Amar alone would take $\frac{48}{1.5} = 32$ months.

Ans: C

Let the total task be represented by T.

Anil worked for 24 days.
Bimal worked for 14 days.
Charu worked for 14 days.

Anil can complete the entire task T in 60 days.
In 24 days, Anil completed a fraction of the task equal to $ \frac{24}{60} = \frac{2}{5} $ of T, which is $ 0.4T $.

Bimal can complete the entire task T in 84 days.
In 14 days, Bimal completed a fraction of the task equal to $ \frac{14}{84} = \frac{1}{6} $ of T.

The remaining portion of the task was completed by Charu.
Work done by Charu = Total task – Work done by Anil – Work done by Bimal
Work done by Charu = $ T – \frac{2}{5}T – \frac{1}{6}T $
Work done by Charu = $ T – \frac{4}{10}T – \frac{1}{6}T $
Work done by Charu = $ \frac{30T – 12T – 5T}{30} $
Work done by Charu = $ \frac{13T}{30} $

The total remuneration for the task is $21000.
Charu’s share of the remuneration is proportional to the work done by Charu.
Charu’s share = $ \frac{\text{Work done by Charu}}{\text{Total task}} \times \text{Total remuneration} $
Charu’s share = $ \frac{\frac{13T}{30}}{T} \times 21000 $
Charu’s share = $ \frac{13}{30} \times 21000 $
Charu’s share = $ 13 \times 700 $
Charu’s share = $9100

Ans: A

Explanation:Let the capacity of the tank be $V$ units.
Let the rate at which pipe A fills the tank be $r_A$ units per hour.
Let the rate at which pipe B drains the tank be $r_B$ units per hour.

Case 1: Pipe A is opened at 2 pm and pipe B is opened at 3 pm. The tank is full at 10 pm.
Pipe A works for $10 \text{ pm} – 2 \text{ pm} = 8$ hours.
Pipe B works for $10 \text{ pm} – 3 \text{ pm} = 7$ hours.
The net filling in this case is given by:
$8 \times r_A – 7 \times r_B = V$ (Equation 1)

Case 2: Pipe A is opened at 2 pm and pipe B is opened at 4 pm. The tank is full at 6 pm.
Pipe A works for $6 \text{ pm} – 2 \text{ pm} = 4$ hours.
Pipe B works for $6 \text{ pm} – 4 \text{ pm} = 2$ hours.
The net filling in this case is given by:
$4 \times r_A – 2 \times r_B = V$ (Equation 2)

Now we have a system of two linear equations with two variables $r_A$ and $r_B$ (assuming $V=1$ unit for simplicity, and we can find the rates as fractions of the tank capacity per hour).

From Equation 2, we can express $2 \times r_B$ in terms of $r_A$:
$2 \times r_B = 4 \times r_A – V$

Substitute this into Equation 1:
$8 \times r_A – (7/2) \times (2 \times r_B) = V$
$8 \times r_A – (7/2) \times (4 \times r_A – V) = V$
$8 \times r_A – 14 \times r_A + (7/2) \times V = V$
$-6 \times r_A = V – (7/2) \times V$
$-6 \times r_A = -\frac{5}{2} \times V$
$r_A = \frac{5}{12} \times V$ units per hour.

Now, substitute the value of $r_A$ back into Equation 2 to find $r_B$:
$4 \times (\frac{5}{12} \times V) – 2 \times r_B = V$
$\frac{20}{12} \times V – 2 \times r_B = V$
$\frac{5}{3} \times V – 2 \times r_B = V$
$2 \times r_B = \frac{5}{3} \times V – V$
$2 \times r_B = \frac{2}{3} \times V$
$r_B = \frac{1}{3} \times V$ units per hour.

We need to find the time taken to fill the tank if pipe B is not opened at all. This means only pipe A is working.
Time taken = $\frac{\text{Capacity of the tank}}{\text{Rate of pipe A}}$
Time taken = $\frac{V}{r_A} = \frac{V}{\frac{5}{12} \times V} = \frac{12}{5}$ hours.

We need to convert this time into minutes:
Time in minutes = $\frac{12}{5} \times 60 = 12 \times 12 = 144$ minutes.

Correct_Option:A

Ans: B

Explanation:Let R be the time taken by Rahul to complete the work alone, and G be the time taken by Gautam to complete the work alone.
In one hour, Rahul completes 1/R of the work, and Gautam completes 1/G of the work.

Rahul started at 9 AM and worked until 5 PM, which is 8 hours.
Gautam joined 2 hours later, so he started at 11 AM and worked until 5 PM, which is 6 hours.

In the actual scenario, the work done by Rahul is R_work = 8 * (1/R) and the work done by Gautam is G_work = 6 * (1/G).
Since they completed the work together, the total work done is 1:
8/R + 6/G = 1 (Equation 1)

In the hypothetical scenario, both started at 9 AM and worked together.
If they had completed the work 30 minutes earlier than 5 PM, they would have finished at 4:30 PM.
The total time they would have worked together is from 9 AM to 4:30 PM, which is 7.5 hours.

In this hypothetical scenario, the total work done is:
7.5 * (1/R + 1/G) = 1
7.5/R + 7.5/G = 1 (Equation 2)

Now we have a system of two equations with two variables:
1) 8/R + 6/G = 1
2) 7.5/R + 7.5/G = 1

Let x = 1/R and y = 1/G. The equations become:
1) 8x + 6y = 1
2) 7.5x + 7.5y = 1

Multiply Equation 2 by 2 to get rid of the decimal:
15x + 15y = 2 (Equation 3)

Now we can solve the system of equations 1) and 3):
1) 8x + 6y = 1
3) 15x + 15y = 2

Multiply Equation 1 by 5 and Equation 3 by 2 to eliminate y:
5 * (8x + 6y) = 5 * 1 => 40x + 30y = 5
2 * (15x + 15y) = 2 * 2 => 30x + 30y = 4

Subtract the second new equation from the first new equation:
(40x + 30y) – (30x + 30y) = 5 – 4
10x = 1
x = 1/10

Since x = 1/R, we have 1/R = 1/10.
Therefore, R = 10 hours.

Working alone, the time Rahul would have taken to complete the work is 10 hours.

Let’s verify the value of G.
Substitute x = 1/10 into Equation 1:
8(1/10) + 6y = 1
8/10 + 6y = 1
4/5 + 6y = 1
6y = 1 – 4/5
6y = 1/5
y = 1/30
So, G = 30 hours.

Check with Equation 2:
7.5/R + 7.5/G = 7.5/10 + 7.5/30 = 0.75 + 0.25 = 1. This is correct.

The question asks for the time Rahul would have taken, in hours, to complete the work alone, which is R.
R = 10 hours.

Correct_Option:B

Ans: 3000

Explanation:Anil can paint a house in 12 days, so his work rate is 1/12 of the house per day.
Barun can paint a house in 16 days, so his work rate is 1/16 of the house per day.
Let Chandu’s work rate be C (house per day).
Together, they complete the painting in 6 days. This means their combined work rate is 1/6 of the house per day.

So, we have the equation:
Anil’s rate + Barun’s rate + Chandu’s rate = Combined rate
1/12 + 1/16 + C = 1/6

To solve for C, we first find a common denominator for the fractions. The least common multiple (LCM) of 12, 16, and 6 is 48.
(4/48) + (3/48) + C = 8/48
7/48 + C = 8/48
C = 8/48 – 7/48
C = 1/48

So, Chandu’s work rate is 1/48 of the house per day.

Now, we need to find the proportion of work done by Chandu out of the total work done by all three.
In 6 days, the work done by Anil is (1/12) * 6 = 6/12 = 1/2 of the house.
In 6 days, the work done by Barun is (1/16) * 6 = 6/16 = 3/8 of the house.
In 6 days, the work done by Chandu is (1/48) * 6 = 6/48 = 1/8 of the house.

Let’s check if the sum of their work equals 1 (the whole house):
1/2 + 3/8 + 1/8 = 4/8 + 3/8 + 1/8 = 8/8 = 1. This is correct.

The total payment is ₹ 24000. Chandu is paid in proportion to the work done by him.
The proportion of work done by Chandu is his work / total work = (1/8) / 1 = 1/8.

Therefore, Chandu’s share of the payment is (1/8) of the total payment.
Chandu’s payment = (1/8) * ₹ 24000
Chandu’s payment = ₹ 3000

Final_Answer:3000

Ans: 4

Suppose Jack requires “t” days to finish a task, then John would need “2t” days for the same task. Consequently, Jack’s daily work output is (1/t), and John’s is (1/2t).

Let Jim require “m” days to complete the task. As per the problem statement, the combined daily work of Jack and Jim equals John’s daily work. This can be represented as:
$\frac{1}{t} + \frac{1}{m} = \frac{1}{2t}$
Rearranging the equation to find Jim’s work rate:
$\frac{1}{m} = \frac{1}{2t} – \frac{1}{t} = \frac{1-2}{2t} = -\frac{1}{2t}$
This indicates an inconsistency or error in the problem statement as presented. Assuming the intended relationship was that the combined work of Jack and Jim equals *some fraction* of John’s work, or that the rates are related differently.

Let’s re-evaluate based on a potential interpretation where the *sum of rates* of Jack and Jim equals John’s rate, which is generally how such problems are set up if not explicitly stated otherwise, or if there’s a typo. However, the equation provided is $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}$, which leads to $\frac{1}{m} = \frac{3}{2t} – \frac{1}{t} = \frac{3-2}{2t} = \frac{1}{2t}$, so m = 2t.

Hence, Jim also takes “2t” days to complete the work.

Now, let’s consider the scenario where the three of them complete the work together in “p” days. The problem statement then introduces a condition: “John takes ‘p+3’ days to complete the work.” This appears to be a separate condition or a misunderstanding of the initial setup, as “p” is defined as the time taken by all three working together.

If we assume the statement meant: “If John and Jim work together for ‘m+3’ days, it is equivalent to doing (4/2t) times the work Jim does in ‘m’ days.” This is an unusual phrasing.

Let’s consider another interpretation based on the provided equation:
$\frac{1}{2t} (m+3) = \left(\frac{4}{2t}\right) m$
This equation suggests a relationship between Jim’s time (“m”), John’s time (“2t”), and a factor of 4.

If we simplify the equation:
$\frac{m+3}{2t} = \frac{4m}{2t}$
Multiplying both sides by 2t:
$m+3 = 4m$
$3 = 3m$
$m = 1$
Therefore, Jim takes 1 day to complete the work.

Following this, if Jim takes 1 day (m=1), and m=2t, then 1=2t, which means t=1/2.
So, Jack takes t = 1/2 days.
John takes 2t = 1 day.

Now, let’s check the original condition: “John takes ‘p+3’ days to complete the work.” If John takes 1 day, this implies p+3 = 1, so p = -2, which is not possible for time.

Let’s re-examine the statement “Similarly John will also take 4 days to complete the work”. This statement seems to be a consequence derived from m=1.
If m=1, and we previously found m=2t, then 1=2t, so t=1/2.
Jack’s time = t = 1/2 days.
John’s time = 2t = 1 day.
Jim’s time = m = 1 day.

The statement “Hence Jim will take (1+3)=4 days to complete the work” is confusing as it uses m=1 from the calculation, but then adds 3 to get 4. If m=1 is Jim’s time, then Jim takes 1 day.

If we assume the statement “Similarly John will also take 4 days to complete the work” is a separate outcome from the calculation of m=1, and it refers to John’s time, then John takes 4 days.
If John takes 4 days, then 2t = 4, so t = 2 days.
Jack takes t = 2 days.
If m=1, then Jim takes 1 day.

The interpretation that leads to John taking 4 days and Jim taking 4 days implies a different setup. The initial calculation of m=1 leads to Jim taking 1 day. The subsequent statement about John taking 4 days and Jim taking 4 days seems to be a separate, perhaps intended, result not directly derived from the m=1 calculation as presented.

Let’s assume the provided equation and its resolution m=1 is correct.
Then Jim takes 1 day.
From m=2t, we get 1=2t, so t=1/2.
Jack takes 1/2 day.
John takes 1 day.

If the question implies that John takes 4 days, then:
John’s time = 4 days.
Since John’s time is 2t, 2t = 4, so t = 2 days.
Jack’s time = t = 2 days.

The statement “Hence Jim will take (1+3)=4 days to complete the work” suggests Jim takes 4 days.
If Jim takes 4 days, and m=2t, then 4=2t, so t=2 days.
Jack takes t = 2 days.
John takes 2t = 4 days.

This aligns with the final conclusion that Jim takes 4 days and John takes 4 days. This implies the initial relationships stated might be different from how they were used in the derivation of m=1. However, strictly following the provided derivation:

From $\frac{1}{2t}(m+3) = \left(\frac{4}{2t}\right)m$, we get m=1.
If m=1, then Jim takes 1 day.
The concluding statements “Hence Jim will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work” suggest a different outcome. If Jim takes 4 days, then m=4. If John takes 4 days, then 2t=4, so t=2. If m=4 and t=2, then m=2t holds (4 = 2*2). This implies a potential misstep in the provided derivation leading to m=1. Assuming the intended outcome is that Jim and John both take 4 days:

Let John take 4 days. Since John takes twice as long as Jack, Jack takes 2 days.
Let Jim take 4 days.
In this scenario:
Jack’s daily work = 1/2
John’s daily work = 1/4
Jim’s daily work = 1/4
Combined daily work of Jack and Jim = 1/2 + 1/4 = 3/4. This is not equal to John’s work.

Let’s revert to the derived m=1.
If m=1, Jim takes 1 day.
If m=2t, then 1=2t, t=1/2. Jack takes 1/2 day.
John takes 2t=1 day.
Jack’s work/day = 2
John’s work/day = 1
Jim’s work/day = 1
Combined work of Jack and Jim = 2+1 = 3.
John’s work = 1.
Here, the combined work of Jack and Jim is 3 times John’s work.

The phrasing “Let the three of them complete the work in “p” days. Hence John takes “p+3″ days to complete the work.” is contradictory as ‘p’ is the time for all three working together. If John takes p+3 days, it means John is slower.

If the statement “Hence Jim will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work” is the intended conclusion, it means Jim takes 4 days and John takes 4 days.
If John takes 4 days, then 2t = 4, so t = 2 days. Jack takes 2 days.
If Jim takes 4 days, then m = 4.
This satisfies m = 2t (4 = 2 * 2).
In this case:
Jack’s daily output = 1/2
John’s daily output = 1/4
Jim’s daily output = 1/4
Combined daily output of Jack and Jim = 1/2 + 1/4 = 3/4.
This combined output of Jack and Jim is 3 times John’s output (3/4 is 3 times 1/4). This matches the relation $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}$ if Jim’s rate was $\frac{1}{4}$ (m=4) and Jack’s rate was $\frac{1}{2}$ (t=2).

The steps leading to m=1 seem to be an intermediate calculation that does not align with the final conclusion. However, adhering to the provided sequence of calculations:

Given Jack takes “t” days and John takes “2t” days.
Jack’s daily rate = 1/t. John’s daily rate = 1/2t.
Let Jim take “m” days.
From the equation $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}$, we solve for m:
$\frac{1}{m} = \frac{3}{2t} – \frac{1}{t} = \frac{3-2}{2t} = \frac{1}{2t}$
So, m = 2t. Jim takes “2t” days.

The next set of equations:
$\frac{1}{2t}(m+3) = \left(\frac{4}{2t}\right)m$
Substituting m=2t into the equation:
$\frac{1}{2t}(2t+3) = \left(\frac{4}{2t}\right)(2t)$
$\frac{2t+3}{2t} = 4$
$2t+3 = 8t$
$3 = 6t$
$t = \frac{1}{2}$
If t = 1/2, then:
Jack takes t = 1/2 day.
John takes 2t = 1 day.
Jim takes m = 2t = 1 day.

The subsequent lines “or m=1. Hence Jim will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work” indicate a different path or an error in the provided explanation’s consistency.

If we follow the derivation that led to m=1:
$\frac{1}{2t}(m+3) = \left(\frac{4}{2t}\right)m \implies m+3 = 4m \implies 3m = 3 \implies m=1$
If m=1, Jim takes 1 day.
If m=2t, then 1=2t, so t=1/2. Jack takes 1/2 day. John takes 1 day.
This contradicts the concluding statements about 4 days.

Assuming the final conclusion (Jim and John take 4 days) is correct and the intermediate steps were meant to lead there, the initial setup might have been intended differently. However, strictly following the given explanation: the derivation of m=1 is shown. Then, a conclusion of 4 days is stated for Jim and John, which does not follow from m=1 in the context of the initial relations.

Ans: 40

Explanation:The contractor agreed to construct a 6 km road in 200 days, employing 140 persons.
After 60 days, only 1.5 km of road has been completed.
Remaining work = 6 km – 1.5 km = 4.5 km.
Remaining days = 200 days – 60 days = 140 days.

Let the number of persons required to complete the remaining work be P.
We can use the concept of work, people, and days. The total work can be represented as the product of the number of people and the number of days they work, assuming each person works at the same rate.

Work done in the first 60 days:
1.5 km road completed by 140 persons in 60 days.
Let the work rate of one person be ‘r’ km per day.
Total work done = 140 persons * 60 days * r km/(person*day) = 1.5 km.
So, 8400 * r = 1.5

Remaining work to be done is 4.5 km.
This remaining work needs to be completed in 140 days.
Let the additional number of people needed be ‘x’.
The total number of people will be 140 + x.
So, (140 + x) persons * 140 days * r km/(person*day) = 4.5 km.
(140 + x) * 140 * r = 4.5

We have two equations:
1) 8400 * r = 1.5
2) (140 + x) * 140 * r = 4.5

From equation (1), r = 1.5 / 8400.
Substitute this value of ‘r’ into equation (2):
(140 + x) * 140 * (1.5 / 8400) = 4.5
(140 + x) * (140 * 1.5 / 8400) = 4.5
(140 + x) * (210 / 8400) = 4.5
(140 + x) * (1 / 40) = 4.5
140 + x = 4.5 * 40
140 + x = 180
x = 180 – 140
x = 40

Alternatively, we can use proportionality.
The amount of work done is proportional to the number of people and the number of days.
(Work done) / (Number of people * Number of days) = Constant

For the first phase:
Work1 = 1.5 km
People1 = 140
Days1 = 60
1.5 / (140 * 60) = Constant

For the remaining work:
Work2 = 4.5 km
People2 = 140 + x (where x is the additional people)
Days2 = 140
4.5 / ((140 + x) * 140) = Constant

Equating the constants:
1.5 / (140 * 60) = 4.5 / ((140 + x) * 140)
1.5 / 8400 = 4.5 / ((140 + x) * 140)

Cross-multiply:
1.5 * (140 + x) * 140 = 4.5 * 8400
(140 + x) * 140 = (4.5 * 8400) / 1.5
(140 + x) * 140 = 3 * 8400
(140 + x) * 140 = 25200
140 + x = 25200 / 140
140 + x = 180
x = 180 – 140
x = 40

So, 40 additional people would need to be employed.

Final_Answer:40

Ans: C

Explanation:Let $E_A$ be the usual efficiency of A and $E_B$ be the usual efficiency of B. Let the total work be W.
When A and B work together at their usual efficiency, they finish the task in 12 days.
So, the combined usual efficiency is $E_A + E_B$.
Work done = Efficiency × Time
$W = (E_A + E_B) \times 12$ (Equation 1)

If A had worked half as efficiently as she usually does, her efficiency would be $E_A/2$.
If B had worked thrice as efficiently as he usually does, his efficiency would be $3E_B$.
In this case, the task would have been completed in 9 days.
So, the new combined efficiency is $E_A/2 + 3E_B$.
$W = (E_A/2 + 3E_B) \times 9$ (Equation 2)

From Equation 1, $W = 12E_A + 12E_B$.
From Equation 2, $W = 9(E_A/2) + 9(3E_B) = 4.5E_A + 27E_B$.

Since the work is the same, we can equate the two expressions for W:
$12E_A + 12E_B = 4.5E_A + 27E_B$
Rearrange the terms to group $E_A$ and $E_B$:
$12E_A – 4.5E_A = 27E_B – 12E_B$
$7.5E_A = 15E_B$
Divide both sides by 7.5:
$E_A = \frac{15}{7.5} E_B$
$E_A = 2E_B$

This means A’s usual efficiency is twice B’s usual efficiency.

We need to find how many days A would take to finish the task alone at her usual efficiency.
Let $T_A$ be the time A takes to finish the task alone.
$W = E_A \times T_A$

From Equation 1, we know $W = 12(E_A + E_B)$.
Substitute $E_A = 2E_B$ into this equation:
$W = 12(2E_B + E_B)$
$W = 12(3E_B)$
$W = 36E_B$

Now we can use the formula $W = E_A \times T_A$.
Substitute $W = 36E_B$ and $E_A = 2E_B$:
$36E_B = (2E_B) \times T_A$
Divide both sides by $2E_B$:
$T_A = \frac{36E_B}{2E_B}$
$T_A = 18$

So, A would take 18 days to finish the task alone at her usual efficiency.

Correct_Option:C

Ans: 13

Explanation:Let $M$ be the work rate of one man (jobs per day) and $K$ be the work rate of one machine (jobs per day).
Let $W$ be the total amount of work required to finish the job.
The work rate of $n$ men is $nM$ and the work rate of $m$ machines is $mK$.
The time taken to finish a job is given by $Time = \frac{Work}{Work Rate}$.

From the problem statement, we have two scenarios:
Scenario 1: Three men and eight machines finish the job in time $T_1$.
Work rate $R_1 = 3M + 8K$.
$T_1 = \frac{W}{3M + 8K}$

Scenario 2: Three machines and eight men finish the same job in time $T_2$.
Work rate $R_2 = 8M + 3K$.
$T_2 = \frac{W}{8M + 3K}$

We are given that $T_1 = \frac{1}{2} T_2$.
So, $\frac{W}{3M + 8K} = \frac{1}{2} \times \frac{W}{8M + 3K}$.
Since $W \neq 0$, we can cancel $W$ from both sides:
$\frac{1}{3M + 8K} = \frac{1}{2(8M + 3K)}$.
$2(8M + 3K) = 3M + 8K$.
$16M + 6K = 3M + 8K$.
$16M – 3M = 8K – 6K$.
$13M = 2K$.
This gives us a relationship between the work rate of a man and a machine: $K = \frac{13}{2} M$.
This means one machine does the work of $\frac{13}{2} = 6.5$ men.

Now consider the third piece of information: two machines can finish the job in 13 days.
Let the work rate of two machines be $R_3$.
$R_3 = 2K$.
Time taken $T_3 = 13$ days.
$13 = \frac{W}{2K}$.
$W = 2K \times 13 = 26K$.

We need to find how many men can finish the job in 13 days. Let the number of men be $x$.
The work rate of $x$ men is $R_4 = xM$.
The time taken is $T_4 = 13$ days.
$13 = \frac{W}{xM}$.
$W = 13xM$.

Now we equate the expressions for $W$:
$26K = 13xM$.
We know from the first part that $K = \frac{13}{2} M$. Substitute this into the equation:
$26 \left(\frac{13}{2} M\right) = 13xM$.
$13 \times 13 M = 13xM$.
$169M = 13xM$.
Since $M \neq 0$, we can cancel $M$ from both sides:
$169 = 13x$.
$x = \frac{169}{13}$.
$x = 13$.

So, 13 men can finish the job in 13 days.

Final_Answer:13

Ans: 12

Explanation:Let R be the number of regular hours John worked and O be the number of overtime hours John worked.
We are given the following information:
1. John gets Rs 57 per hour of regular work.
2. John gets Rs 114 per hour of overtime work.
3. He works altogether 172 hours, so R + O = 172.
4. His income from overtime hours is 15% of his income from regular hours.

Income from regular hours = 57 * R
Income from overtime hours = 114 * O

From condition 4, we have:
114 * O = 0.15 * (57 * R)
114 * O = 8.55 * R

We have a system of two linear equations:
Equation 1: R + O = 172
Equation 2: 114 * O = 8.55 * R

From Equation 1, we can express R in terms of O:
R = 172 – O

Substitute this expression for R into Equation 2:
114 * O = 8.55 * (172 – O)
114 * O = 8.55 * 172 – 8.55 * O
114 * O = 1470.6 – 8.55 * O

Now, we need to isolate O. Add 8.55 * O to both sides of the equation:
114 * O + 8.55 * O = 1470.6
122.55 * O = 1470.6

Now, divide by 122.55 to find O:
O = 1470.6 / 122.55
O = 12

Let’s verify this answer.
If O = 12 hours, then R = 172 – 12 = 160 hours.
Income from regular hours = 57 * 160 = 9120
Income from overtime hours = 114 * 12 = 1368

Check if income from overtime hours is 15% of income from regular hours:
15% of 9120 = 0.15 * 9120 = 1368

The condition is satisfied.

Final_Answer:12

Ans: B

Let the total workload be the least common multiple of 20 and 40, which is 40 units.
Anil’s and Sunil’s daily work output are 2 units and 1 unit, respectively.
Anil independently completes 6 units of work over 3 days.
Bimal contributes to 10% of the total effort while collaborating with Anil and Sunil.
Therefore, Bimal’s contribution amounts to 4 units of work.
The remaining 30 units are accomplished by Anil and Sunil working together.
The duration for this joint effort is calculated as 30 units divided by their combined daily output of 3 units, resulting in 10 days.
The entire task is finalized within a total of 3 + 10 = 13 days.

Ans: C

Explanation:Let H be the amount of work a human can do in one day, and R be the amount of work a robot can do in one day.
Let the total amount of work to be done be W.

From the first condition:
Fifteen humans and five robots working together take thirty days to finish the job.
So, the total work done by 15 humans and 5 robots in 30 days is W.
Work done in one day by 15 humans and 5 robots = 15H + 5R
Total work = (15H + 5R) * 30
W = 30(15H + 5R) (Equation 1)

From the second condition:
Five humans and fifteen robots working together take sixty days to finish it.
Work done in one day by 5 humans and 15 robots = 5H + 15R
Total work = (5H + 15R) * 60
W = 60(5H + 15R) (Equation 2)

Now, we equate Equation 1 and Equation 2 as the total work is the same:
30(15H + 5R) = 60(5H + 15R)
Divide both sides by 30:
15H + 5R = 2(5H + 15R)
15H + 5R = 10H + 30R
Subtract 10H from both sides:
5H + 5R = 30R
Subtract 5R from both sides:
5H = 25R
Divide by 5:
H = 5R

This means a human is 5 times more efficient than a robot.

Now we need to find how many days will fifteen humans working together (without any robot) take to finish it.
Let this be D days.
Work done by 15 humans in D days = 15H * D
Total work W = 15HD

We can use either Equation 1 or Equation 2 to find the total work W. Let’s use Equation 1 and substitute H = 5R:
W = 30(15H + 5R)
W = 30(15(5R) + 5R)
W = 30(75R + 5R)
W = 30(80R)
W = 2400R

Now we set the work done by 15 humans in D days equal to W:
15HD = W
Substitute H = 5R and W = 2400R:
15(5R)D = 2400R
75RD = 2400R
Divide both sides by R (assuming R is not zero, which it cannot be if work is done):
75D = 2400
D = 2400 / 75

To simplify the division:
D = 2400 / 75
D = (24 * 100) / 75
D = 24 * (100 / 75)
D = 24 * (4 / 3)
D = (24 / 3) * 4
D = 8 * 4
D = 32

So, fifteen humans working together will take 32 days to finish the job.

Let’s verify with Equation 2:
W = 60(5H + 15R)
Substitute H = 5R:
W = 60(5(5R) + 15R)
W = 60(25R + 15R)
W = 60(40R)
W = 2400R
This matches our previous calculation of W.

The question asks how many days will fifteen humans working together take to finish it.
Work done by 15 humans in D days is 15H * D.
We found H = 5R.
So, work done by 15 humans in D days = 15 * (5R) * D = 75RD.
We know the total work W = 2400R.
Therefore, 75RD = 2400R.
D = 2400R / 75R = 32.

Correct_Option:C

Ans: A

Suppose individual X completes ‘x’ units of task per day and individual Y completes ‘y’ units of task per day.
X works independently until half of the total task is accomplished.
Subsequently, X and Y collaborate for a duration of 4 days.
Finally, Y works alone to finish the remaining 5% of the task.

This implies that X and Y together accomplished 45% of the task within their 4 days of joint effort.
Let’s set the total task to be 100 units.
The work done by X alone for the first half plus the work done by both X and Y over 4 days plus the work done by Y for the final 5% equals the total work.
The equation derived from the joint work is: 4x + 4y = 45 ———(Equation 1)

We are also given that Y requires 25% more time than X to complete an equivalent amount of work.
This translates to: 1.25*y = x ———-(Equation 2)

By substituting Equation 2 into Equation 1, we obtain:
5y + 4y = 45
9y = 45
y = 5 units/day

Therefore, individual Y can complete the entire task alone in 100/5 = 20 days.
This corresponds to option A being the correct choice.

Ans: 10

Explanation:Let $F$ be the rate at which each filling pipe fills the tank, and $D$ be the rate at which each draining pipe drains the tank. Let $V$ be the volume of the tank.

When 6 filling pipes and 5 draining pipes are on, the net filling rate is $6F – 5D$. The time taken to fill the tank is 6 hours. So, the volume of the tank can be expressed as:
$V = (6F – 5D) \times 6$ (Equation 1)

When 5 filling pipes and 6 draining pipes are on, the net filling rate is $5F – 6D$. The time taken to fill the tank is 60 hours. So, the volume of the tank can be expressed as:
$V = (5F – 6D) \times 60$ (Equation 2)

Equating the two expressions for $V$:
$6(6F – 5D) = 60(5F – 6D)$
$36F – 30D = 300F – 360D$
$360D – 30D = 300F – 36F$
$330D = 264F$

We can simplify this ratio:
Divide by 6: $55D = 44F$
Divide by 11: $5D = 4F$
This means $F = \frac{5}{4}D$.

Now substitute this relationship into Equation 1 to find the volume of the tank in terms of $D$:
$V = (6F – 5D) \times 6$
$V = (6 \times \frac{5}{4}D – 5D) \times 6$
$V = (\frac{30}{4}D – 5D) \times 6$
$V = (\frac{15}{2}D – \frac{10}{2}D) \times 6$
$V = (\frac{5}{2}D) \times 6$
$V = 15D$

So, the volume of the tank is equivalent to what 15 draining pipes can drain in one hour, or what 15/D hours of draining would do. Alternatively, we can express the volume in terms of F:
Since $D = \frac{4}{5}F$,
$V = 15D = 15 \times \frac{4}{5}F = 12F$.
The volume of the tank is equivalent to what 12 filling pipes can fill in one hour.

Now we need to find the time it takes to fill the empty tank when one draining and two filling pipes are on.
The net filling rate is $2F – 1D$.
We can express this rate in terms of $D$:
$2F – D = 2(\frac{5}{4}D) – D = \frac{10}{4}D – D = \frac{5}{2}D – D = \frac{3}{2}D$.

Let $T$ be the time taken to fill the tank in this case.
$V = (\frac{3}{2}D) \times T$
We know $V = 15D$.
So, $15D = \frac{3}{2}D \times T$
$15 = \frac{3}{2}T$
$T = 15 \times \frac{2}{3}$
$T = 5 \times 2$
$T = 10$ hours.

Alternatively, we can express the net rate in terms of $F$:
$2F – D = 2F – \frac{4}{5}F = \frac{10}{5}F – \frac{4}{5}F = \frac{6}{5}F$.
We know $V = 12F$.
So, $12F = (\frac{6}{5}F) \times T$
$12 = \frac{6}{5}T$
$T = 12 \times \frac{5}{6}$
$T = 2 \times 5$
$T = 10$ hours.

Final_Answer:10

Ans: 48

Explanation:Let $R_A$ be the rate at which water is brought in by one inlet of type A, and $R_B$ be the rate at which water is brought in by one inlet of type B.
Let $V$ be the volume of the tank.

From the problem statement, we have two scenarios:

Scenario 1: 10 inlets of type A and 45 inlets of type B fill the tank in 30 minutes.
The total rate of water inflow is $10R_A + 45R_B$.
The volume of water filled is $(10R_A + 45R_B) \times 30$.
So, $V = (10R_A + 45R_B) \times 30$
$V = 300R_A + 1350R_B$ (Equation 1)

Scenario 2: 8 inlets of type A and 18 inlets of type B fill the tank in 1 hour (60 minutes).
The total rate of water inflow is $8R_A + 18R_B$.
The volume of water filled is $(8R_A + 18R_B) \times 60$.
So, $V = (8R_A + 18R_B) \times 60$
$V = 480R_A + 1080R_B$ (Equation 2)

Now we have a system of two linear equations with two variables ($R_A$ and $R_B$) representing the rates. We can equate Equation 1 and Equation 2 since the volume of the tank $V$ is the same.

$300R_A + 1350R_B = 480R_A + 1080R_B$

Let’s rearrange the terms to solve for the relationship between $R_A$ and $R_B$.
$1350R_B – 1080R_B = 480R_A – 300R_A$
$270R_B = 180R_A$

Divide both sides by 90:
$3R_B = 2R_A$
This implies $R_A = \frac{3}{2}R_B$.

Now, let’s substitute this relationship back into either Equation 1 or Equation 2 to express $V$ in terms of $R_B$. Using Equation 1:
$V = 300R_A + 1350R_B$
$V = 300(\frac{3}{2}R_B) + 1350R_B$
$V = 450R_B + 1350R_B$
$V = 1800R_B$

Now we need to find the time it takes to fill the empty tank if 7 inlets of type A and 27 inlets of type B are open.
Let $T$ be the time in minutes.
The total rate of water inflow is $7R_A + 27R_B$.
The volume of water filled is $(7R_A + 27R_B) \times T$.
So, $V = (7R_A + 27R_B) \times T$.

Substitute $R_A = \frac{3}{2}R_B$ into the rate:
$7R_A + 27R_B = 7(\frac{3}{2}R_B) + 27R_B$
$= \frac{21}{2}R_B + 27R_B$
To add these, find a common denominator:
$= \frac{21}{2}R_B + \frac{54}{2}R_B$
$= \frac{75}{2}R_B$

Now, substitute this rate and the expression for $V$ into the equation $V = (7R_A + 27R_B) \times T$:
$1800R_B = (\frac{75}{2}R_B) \times T$

We can cancel out $R_B$ from both sides (assuming $R_B \neq 0$, which is true for water to flow in).
$1800 = \frac{75}{2} \times T$

Now, solve for $T$:
$T = \frac{1800 \times 2}{75}$
$T = \frac{3600}{75}$

To simplify the division:
$T = \frac{3600}{75} = \frac{3600}{3 \times 25} = \frac{1200}{25} = 12 \times \frac{100}{25} = 12 \times 4 = 48$.

So, the empty tank will get completely filled in 48 minutes.

Final_Answer:48

Ans: C

Let ‘t’ hours past noon represent the time when the tank is emptied each day. Let ‘a’ and ‘b’ denote the filling rates in liters per hour for pump A and pump B, respectively.

On Monday, pump A alone finished filling the tank at 8 pm. This implies that pump A operated for (8 – t) hours. Consequently, the tank’s capacity can be expressed as a * (8 – t) liters.

Similarly, on Tuesday, pump B alone completed filling the tank at 6 pm. This indicates that pump B worked for (6 – t) hours. Therefore, the tank’s capacity is b * (6 – t) liters.

On Wednesday, pump A operated alone until 5 pm, after which pump B worked alone from 5 pm to 7 pm to fill the tank. This means pump A worked for (5 – t) hours, and pump B worked for 2 hours. The tank’s capacity is thus a * (5 – t) + 2b liters.

We can establish the following equalities based on the tank’s capacity:
a * (8 – t) = b * (6 – t) = a * (5 – t) + 2b

From the equality:
a * (8 – t) = a * (5 – t) + 2b
This simplifies to:
3a = 2b … (1)

Now, using the equality:
a * (8 – t) = b * (6 – t)
Substitute equation (1) into this equation:
a * (8 – t) = (3a / 2) * (6 – t)
This yields:
t = 2

Therefore, the tank is emptied at 2 pm daily. We can deduce that pump A takes 6 hours to fill the tank alone, and pump B takes 4 hours to fill the tank alone.

When working together, both pumps can fill the tank in:
(6 * 4) / (6 + 4) = 2.4 hours, which is 2 hours and 24 minutes.

Since the pumps began filling the tank at 2:00 pm, the tank will be completely filled by 4:24 pm.

Ans: C

Explanation:Let R be the efficiency of Ramesh and G be the efficiency of Ganesh.
Let W be the total work.
Together, Ramesh and Ganesh complete the work in 16 days. So, their combined efficiency is R + G.
Total work W = (R + G) * 16 —(1)

They worked together for 7 days.
Work done in 7 days = (R + G) * 7

After 7 days, Ramesh’s efficiency fell by 30%.
New efficiency of Ramesh = R – 0.30R = 0.70R

The remaining work was completed by Ramesh (with reduced efficiency) and Ganesh.
The total time taken to complete the work was 17 days.
They worked together for 7 days. So, the remaining work was done in 17 – 7 = 10 days.
Work done in the remaining 10 days = (0.70R + G) * 10

The total work is the sum of work done in the first 7 days and the remaining 10 days.
W = (R + G) * 7 + (0.70R + G) * 10

Now we have two expressions for W:
From (1): W = 16R + 16G
From the work done: W = 7R + 7G + 7R + 10G
W = 14R + 17G

Equating the two expressions for W:
16R + 16G = 14R + 17G
16R – 14R = 17G – 16G
2R = G

This means Ganesh’s efficiency is twice Ramesh’s efficiency.

Now we need to find the number of days Ganesh would have completed the remaining work if he had worked alone after Ramesh got sick.
The remaining work is the total work minus the work done in the first 7 days.
Remaining work = W – (R + G) * 7
Substitute G = 2R into the expression for W:
W = 16R + 16(2R) = 16R + 32R = 48R

Work done in the first 7 days = (R + G) * 7 = (R + 2R) * 7 = 3R * 7 = 21R
Remaining work = 48R – 21R = 27R

After Ramesh got sick, his efficiency became 0.70R.
The work done in the next 10 days was (0.70R + G) * 10 = (0.70R + 2R) * 10 = 2.70R * 10 = 27R.
This confirms our calculation for the remaining work.

The question asks: “If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?”
The remaining work is 27R.
Ganesh’s efficiency is G = 2R.
Number of days for Ganesh to complete the remaining work = Remaining work / Ganesh’s efficiency
Number of days = 27R / (2R) = 27 / 2 = 13.5 days.

Correct_Option:C

Ans: A

Explanation:Let V be the volume of the tank.
Let R_in be the rate at which the inlet pipe fills the tank (volume per hour).
Let R_out be the rate at which the outlet pipe empties the tank (volume per hour).

When the outlet pipe is closed, the inlet pipe fills the empty tank in 8 hours.
So, V = R_in * 8
This means R_in = V/8

When the outlet pipe is open, the inlet pipe fills the empty tank in 10 hours.
In this case, the net rate of filling is (R_in – R_out).
So, V = (R_in – R_out) * 10
Substitute R_in = V/8 into the equation:
V = (V/8 – R_out) * 10
Divide both sides by V (assuming V is not zero):
1 = (1/8 – R_out/V) * 10
1/10 = 1/8 – R_out/V
R_out/V = 1/8 – 1/10
To subtract the fractions, find a common denominator, which is 40:
R_out/V = (5/40) – (4/40)
R_out/V = 1/40

This means the outlet pipe empties 1/40th of the tank’s volume per hour.

We need to find out in how many hours the full tank becomes half-full if only the outlet pipe is open.
This means we need to empty half of the tank’s volume (V/2) using the outlet pipe.
Let T be the time taken.
Volume to be emptied = V/2
Rate of emptying = R_out
So, V/2 = R_out * T
We know R_out = V/40. Substitute this into the equation:
V/2 = (V/40) * T
Divide both sides by V:
1/2 = (1/40) * T
Multiply both sides by 40:
T = 40 * (1/2)
T = 20 hours.

Correct_Option:A

Ans: A

Explanation:
Let the total work be W.
Amal can complete the job in 10 days, so Amal’s work rate is W/10 per day.
Bimal can complete the job in 8 days, so Bimal’s work rate is W/8 per day.
Amal, Bimal, and Kamal together complete the job in 4 days. Let Kamal’s work rate be R_K per day.
The combined work rate of Amal, Bimal, and Kamal is (W/10) + (W/8) + R_K.
Since they complete the job in 4 days, their combined work rate is W/4.
So, (W/10) + (W/8) + R_K = W/4.

To find Kamal’s work rate, we can solve for R_K:
R_K = W/4 – W/10 – W/8
Find a common denominator for 4, 10, and 8, which is 40.
R_K = (10W/40) – (4W/40) – (5W/40)
R_K = (10W – 4W – 5W) / 40
R_K = W/40.

This means Kamal can complete the job alone in 40 days.

Now, we need to find the proportion of work done by each person. Assuming W=1 unit of work for simplicity:
Amal’s work rate = 1/10
Bimal’s work rate = 1/8
Kamal’s work rate = 1/40

The proportion of work done by each person is equal to their work rate.
Proportion of Amal’s work = (1/10) / (1/10 + 1/8 + 1/40)
Proportion of Bimal’s work = (1/8) / (1/10 + 1/8 + 1/40)
Proportion of Kamal’s work = (1/40) / (1/10 + 1/8 + 1/40)

Let’s calculate the sum of their work rates:
1/10 + 1/8 + 1/40 = 4/40 + 5/40 + 1/40 = 10/40 = 1/4.
This confirms that together they complete 1/4 of the work per day.

Now, calculate the proportion of work for each:
Proportion of Amal’s work = (1/10) / (1/4) = 1/10 * 4/1 = 4/10 = 2/5
Proportion of Bimal’s work = (1/8) / (1/4) = 1/8 * 4/1 = 4/8 = 1/2
Proportion of Kamal’s work = (1/40) / (1/4) = 1/40 * 4/1 = 4/40 = 1/10

Let’s check if the proportions add up to 1:
2/5 + 1/2 + 1/10 = 4/10 + 5/10 + 1/10 = 10/10 = 1.

The total remuneration is Rs 1000. The amount is shared in proportion to their work.
Kamal’s share = Proportion of Kamal’s work * Total remuneration
Kamal’s share = (1/10) * 1000
Kamal’s share = Rs 100.

Correct_Option:A