Thrice the time taken by P to complete a piece of work is equal to 5

Question

Thrice the time taken by P to complete a piece of work is equal to 5 times the time taken by Q and R working together to complete the same piece of work. Thrice the time taken by Q to complete a piece of work is equal to 7 times the time taken by P and R working together to complete the same piece of work. P, Q, and R working together taken 18 days to complete the same piece of work. Find the time taken (in approximate days) by R alone to complete the same piece of work?

A. 50
B. 52
C. 55
D. 57
E. 59

A. 50
B. 52
C. 55
D. 57
E. 59

kingbucky · Answer

Let the time taken by P, Q, and R to complete the work be $ T_P $, $ T_Q $, and $ T_R $ respectively. The efficiencies are given by:

E_P = \frac{1}{T_P}, \quad E_Q = \frac{1}{T_Q}, \quad E_R = \frac{1}{T_R}

From the problem, we have the following relationships:

1.   3T_P = 5(T_Q + T_R)  
2.   3T_Q = 7(T_P + T_R)

We also know that P, Q, and R together can complete the work in 18 days:

E_P + E_Q + E_R = \frac{1}{18}

### Step 1: Express $ T_Q + T_R $ and $ T_P + T_R $

From the first equation:

T_Q + T_R = \frac{3T_P}{5}

From the second equation:

T_P + T_R = \frac{3T_Q}{7}

### Step 2: Substitute $ T_R $

From $ T_Q + T_R = \frac{3T_P}{5} $:

T_R = \frac{3T_P}{5} - T_Q

Substituting $ T_R $ into $ T_P + T_R = \frac{3T_Q}{7} $:

T_P + \left(\frac{3T_P}{5} - T_Q\right) = \frac{3T_Q}{7}

### Step 3: Simplify the equation

Multiplying through by 35 to eliminate the fractions:

35T_P + 21T_P - 35T_Q = 15T_Q

Combining like terms:

56T_P = 50T_Q \implies \frac{T_P}{T_Q} = \frac{25}{28}

### Step 4: Substitute into $ T_R $

Now, substituting $ T_P = \frac{25}{28}T_Q $ into $ T_R $:

T_R = \frac{3T_P}{5} - T_Q = \frac{3 \cdot \frac{25}{28}T_Q}{5} - T_Q

= \frac{15}{28}T_Q - T_Q = \frac{15}{28}T_Q - \frac{28}{28}T_Q = -\frac{13}{28}T_Q

### Step 5: Substitute into the efficiency equation

Now substituting into the efficiency equation:

E_P + E_Q + E_R = \frac{1}{T_P} + \frac{1}{T_Q} + \frac{1}{T_R} = \frac{28}{25T_Q} + \frac{1}{T_Q} + \frac{28}{13T_Q}

Setting this equal to   \frac{1}{18}  :

\frac{81}{25T_Q} = \frac{1}{18}

### Step 6: Solve for $ T_R $

Cross-multiplying gives:

81 \cdot 18 = 25T_Q \implies T_Q = \frac{1458}{25}

Now substituting $ T_Q $ back to find $ T_R $:

T_R = \frac{3T_P}{5} - T_Q

Calculating $ T_R $:

T_R = \frac{3 \cdot \frac{25}{28} \cdot \frac{1458}{25}}{5} - \frac{1458}{25}

Thus, the time taken by R alone to complete the work is approximately:

**C. 55**

btsaami · Answer

Use the concept, Efficiency ~ 1/time.

Let P, Q and R be the efficiency.

P/(Q+R) = 3/5 
Let P=3x---1) 
Q+R=5x---2)

Similarly, Q/(P+R) = 3/7
Q= 3y---3)
P+R= 7y---4)

From 2 and 3, R= 5x-3y
From 1 and 4, R= 7y- 3x

Therefore, 5x-3y= 7y-3x
x/y= 10/8 = 5/4

R= 25-12= 13
 
Work done= eff.*time
(P+Q+R)*18= (3x+3y+5x-3y)*18

8x*18= 8*5*18= 720

720/13 ~ 55. Hence, C

Can anyone suggest shortcut, if any?

versewriter · Answer

Here is the solution -

From the question the following points are certain or true -
1. 1/p * 3 = 1/(q+r) * 5
2. 1/q * 3 = 1/(p+r) * 7
3. p+q+r = 1/18

So we can form he following equations -

From point 1 above we can know the 1/p * 3 = 1/(q+r) * 5
so, 
=> p5 = 3q+3r
we add 3*p both the sides because we know that 1/(p+q+r) = 18
=> 5p + 3p = 3q+3r+3p
=> 8p = 3(p+q+r)
=> 8p = 3 * 1/18
=> p = 1/48 .............. A

From point 2 above we can know the 1/q * 3 = 1/(p+r) * 7
so, 
=> q7 = 3p+3r
we add 3*q both the sides because we know that 1/(p+q+r) = 18
=> q7 + 3q = 3p+3r+3q
=> 10q = 3(p+q+r)
=> 7q = 3 * 1/18
=> q= 1/60 ................B

We put value of p & q in equation/point 3 i.e. p+q+r = 1/18
=> 1/48 + 1/60 + 1/r = 1/18
=> 1/r = 1/18 - 1/48- 1/60
=> 1/r = 13/720
=> r = 720/13
=> r = 55 ~

So Option c is answer

Thrice the time taken by P to complete a piece of work is equal to 5

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Thrice the time taken by P to complete a piece of work is equal to 5

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