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CAT 2025 Lesson : Time & Work - Relative Efficiencies

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4. Efficiencies provided in relative terms

These questions would provide efficiencies in terms of relative terms. For example, “Pipe A pumps twice as much water as Pipe B” or “A works at twice the speed of B”, “C completes thrice the amount of work in a day” and “E is 20%20 \% more efficient as F”. The efficiency or rate of completion over here needs to be compared.

For example, if A takes 4040 days to complete a piece of work and B is 20%20 \% more efficient than A, then rate of completion of work for A is 140\dfrac{1}{40} in a day. So, rate of completion for B is 1.2×140=1.2401.2 \times \dfrac{1}{40} = \dfrac{1.2}{40} in a day.

Note: We cannot multiply the time taken with the increase/decrease in efficiency. Only the rate of completion of work or efficiency can be multiplied with the increase/decrease in efficiency. And, these questions are best solved using the standard unitary method.

Example 15

Jack is 40%40 \% more efficient than Jill. Jack and Jill together can complete a piece of work in 55 days. How long would Jill take to the complete the work while working alone?

Solution

Let Jill take xx days to complete the work.
∴ Portion of work completed by Jill in 11 day =1x= \dfrac{1}{x}

Jack is 40%40 \% more efficient than Jill
∴ Portion of work completed by Jack in 11 day=1.4×1x=1.4x = 1.4 \times \dfrac{1}{x} = \dfrac{1.4}{x}

Work completed by Jack and Jill in 11 day =1x+1.4x=15= \dfrac{1}{x} + \dfrac{1.4}{x} = \dfrac{1}{5}

⇒ x=12 x = 12 days

Answer: 1212 days


If the efficiency becomes x\bm{x} times initial efficiency, then the time taken becomes 1x\dfrac{\bm{1}}{\bm{x}} times initial time taken.

For instance, let's say Raj takes 99 days to complete a piece of work. If Mithali is 50%50 \% more efficient than Raj, then the multiplication factor is 1+50%=1.51 + 50 \% = 1.5 and the following is true.

Mithali's efficiency =1.5×Raj’s Efficiency=1.5×19=16= 1.5 \times \text{Raj's Efficiency} = 1.5 \times \dfrac{1}{9} = \dfrac{1}{6}

Mithali's time taken =Time taken by Raj1.5=91.5=6= \dfrac{\text{Time taken by Raj}}{1.5} = \dfrac{9}{1.5} = 6

Note that to compute efficiency we multiplied the multiplication factor. And, to compute the time taken we divided the multiplication factor.

Example 16

At work, Dipika is twice as fast as Sree and Sree is twice as fast as Karan. Dipika takes 3030 hours less than Karan to complete a piece of work. Working together, if it took Dipika and Sree a total of xx hours to complete the work, then

(1) x<4x \lt 4            (2) 4≤x<64 \leq x \lt 6            (3) 6≤x<86 \leq x \lt 8            (4) x≥8x \geq 8           

Solution

Let k,sk, s and dd be the hours taken by Karan, Sree and Dipika respectively to complete the work alone,

Dipika's efficiency is twice that of Sree (s=2d)(s = 2d), and Sree's efficiency is twice that of Karan (k=2s)(k = 2s). ∴ Sree takes twice the time as Dipika (s=2d)(\bm{s = 2d}) and Karan takes twice the time as Sree (k=2s)(\bm{k = 2s}).
⇒ k=4d k = 4d

As Dipika takes 3030 hours less than Karan,
k−d=30k - d = 30
⇒ 4d–d=30 4d – d = 30 ⇒ d=10 d = 10

∴s=20\therefore s = 20 and k=40k = 40

Portion of work completed by Dipika and Sree in 11 hour =110+120=320= \dfrac{1}{10} + \dfrac{1}{20} = \dfrac{3}{20}

Time taken by Dipika and Sree =x=203=6.67= x = \dfrac{20}{3} = 6.67 hours

Answer: (3)6≤(3) 6 \leq x < 8
If the ratio of efficiencies of A and B is a:ba : b, then the ratio of time taken by them is 1a:1b=b:a\dfrac{1}{a} : \dfrac{1}{b} = b : a

Example 17

The ratio of time taken by Pepe and Quico to build a wall is 3:23 : 2 and the ratio of efficiencies of Quico and Ricardo is 4:34 : 3. Working alone, Pepe takes 55 days more than what Ricardo takes to build the wall. How many days (rounded to the nearest integer) do the three of them take to build the same wall?

Solution

Let p,qp, q and rr be the days taken by Pepe, Quico and Ricardo to build the wall working alone.

p:q=3:2p : q = 3 : 2
q:r=14:13=3:4q : r = \dfrac{1}{4} : \dfrac{1}{3} = 3 : 4

Combining the two ratios we get p:q:r=9:6:8p : q : r = 9 : 6 : 8
(Refer Ratio & Partnership lesson for combining ratios)

∴∴ Let the number of days taken by Pepe, Quico and Ricardo be 9x\bm{x}, 6x\bm{x} and 8x\bm{x} respectively.
9x−8x=59x - 8x = 5 ⇒ x=5 x=5

Pepe, Quico and Ricardo take 45, 30 and 40 days respectively.

Portion of work completed by the three in 11 day =145+130+140= \dfrac{1}{45} + \dfrac{1}{30} + \dfrac{1}{40}

=8+12+9360=29360 = \dfrac{8 + 12 + 9}{360} = \dfrac{29}{360}

Days taken by the three to complete the work =36029∽12.4= \dfrac{360}{29} \backsim 12.4

Rounded to the nearest integer, they take 12 days.

Answer: 1212


Let's look at a complex problem that requires application of multiple concepts covered in this lesson so far.

Example 18

A and C together take 50%50 \% less time than what B would take for building a wall. B and C working together complete 50%50 \% more work than what A does working alone. All three of them, working together, would have taken 88 days to build a wall. A, B and C started building the wall together. B left after working for 44 days, while A took 44 more days to leave. C continued to work till the wall was fully built. For how many days did C work alone?

Solution

Let a,ba, b and cc be the number of days taken to complete the work by A, B and C respectively working alone.

As A and C together take 12\dfrac{1}{2} the time that B takes, A and C together are 22 times as efficient as B.

1a+1c=2b⟶(1) \dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b} \longrightarrow (1)

As B and C together complete 50%50 \% more work that A,

1b+1c=1.5a⟶(2) \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1.5}{a} \longrightarrow (2)

(1)−(2)(1) - (2) ⇒ 1a−1b=2b−1.5a\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{2}{b} - \dfrac{1.5}{a} ⇒ 2.5a=3b \dfrac{2.5}{a} = \dfrac{3}{b}

⇒ ab=56⟶(3) \dfrac{a}{b} = \dfrac{5}{6} \longrightarrow (3)

Substituting a=5b6a = \dfrac{5b}{6} in (1)(1), we get bc=45⟶(4)\dfrac{b}{c} = \dfrac{4}{5} \longrightarrow (4)

Combining the ratios in (3)(3) and (4)(4), we get a:b:c=10:12:15a : b : c = 10 : 12 : 15

Let the time taken by A, B and C be 10x,12x10x, 12x and 15x15x respectively. All three working together take 8 days.

∴110x+112x+115x=18\therefore \dfrac{1}{10x} + \dfrac{1}{12x} + \dfrac{1}{15x} = \dfrac{1}{8} ⇒ 6+5+460x=18x=2 \dfrac{6 + 5 + 4}{60x} = \dfrac{1}{8} x = 2

∴∴ Time taken by A, B and C are 20,2420, 24 and 3030 days respectively. It is given that B worked for 44 days and A worked for 44 more days, which is a total of 88 days. Let yy be the total number of days that C worked.

424+820+y30=1\dfrac{4}{24} + \dfrac{8}{20} + \dfrac{y}{30} = 1 ⇒y30=1−16−25 \dfrac{y}{30} = 1 - \dfrac{1}{6} - \dfrac{2}{5} ⇒ y30=1330y=13 \dfrac{y}{30} = \dfrac{13}{30} y = 13 days

Therefore, C worked alone for 13−8=513 - 8 = 5 days.

Answer: 55

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