Chain Rule Questions and Answers

Answer: D

224 more workers are needed. The work is 560 × 7 = 3920 worker-months. In 5 months the workers needed are 3920 / 5 = 784. The extra workers are 784 - 560 = 224.

784 is the total required, not the extra, and 400 and 408 do not come from the inverse proportion.

Answer: D

320 prisoners were transferred. The provisions equal 800 x 15 = 12,000 prisoner-days. For 25 days they feed 12000/25 = 480 prisoners, so 800 - 480 = 320 were transferred. 200, 1120 and 480 are the other quantities that appear in the working but do not answer the question. Provisions can be treated as a fixed stock of prisoner-days. A stock that lasts 15 days for 800 people can last 25 days only for fewer people, in inverse proportion: 800 x 15 / 25 = 480.

Answer: B

The provisions will last 16 days.

The total food is fixed, so the number of students times the number of days is constant:

\[ \begin{aligned} 100 \times 20 &= 125 \times d \\ d &= \frac{2000}{125} = 16 \end{aligned} \]

More students means fewer days, in inverse proportion. Subtracting 25% of 20 days gives 15, which is not offered; 18, 14 and 12 do not keep the product at 2000 student-days.

Answer: B

The work will take 13 days.

The total work in person-hours is the same in both cases:

\[ \begin{aligned} 39 \times 12 \times 5 &= 30 \times d \times 6 \\ 2340 &= 180\,d \\ d &= 13 \end{aligned} \]

Fewer people means more days, but longer working days mean fewer, and the two partly cancel. The other options do not balance: 15 days, for example, gives \(30 \times 15 \times 6 = 2700\) person-hours, more than the 2340 the job needs.

Answer: C

The food will last 84 days more.

After 20 days the provisions left are enough for 300 men for 70 more days. That stock is now shared by 300 - 50 = 250 men:

\[ \begin{aligned} 300 \times 70 &= 250 \times d \\ d &= \frac{21000}{250} = 84\ \text{days} \end{aligned} \]

80 days and 160 days come from forgetting that 20 days have already passed or from using the wrong number of men; with fewer men the food must last longer than 70 days, but only in the ratio 300 : 250.

Answer: A

Five machines make 5 parts in 5 minutes, so each machine makes 1 part in 5 minutes.

One hundred machines then make 100 parts in the same 5 minutes, option A. The answers 100, 50 and 30 minutes come from wrongly scaling the time with the number of parts.

The rate per machine is 1 part per 5 minutes, which does not change as more machines are added.

Answer: A

The 12 labourers earn Rs. 5580.

Find the wage for one labourer for one day, then scale up.

\[ \begin{aligned} \text{One day's wage} &= \frac{3100}{25 \times 4} = \text{Rs. } 31 \\ \text{Earnings} &= 12 \times 15 \times 31 \\ &= 180 \times 31 = \text{Rs. } 5580 \end{aligned} \]

Check by proportion: the work rises from 100 labourer-days to 180 labourer-days, so the earnings rise in the ratio 180 : 100, and 3100 times 1.8 is 5580. The figures 6580 and 4580 differ from this by exactly a thousand rupees, and 3810 is barely more than the original Rs. 3100 although the labourer-days nearly double.

Answer: E

Let there be x men originally.
They were to complete the work in 180 days but as the number of persons is reduced to x - 30.
∴ Work takes 20 more days. So the equation is 180x = (x - 30)200 ⇒ 20x = 6000 ⇒ x = 300.

Answer: E

Time is proportional to distance and inversely proportional to speed.

\[ t' = 100 \times 2 \times \frac{1}{1/2} = 100 \times 2 \times 2 \]

\[ = 400 \text{ days} \]

Doubling the distance doubles the time, and halving the speed doubles it again - the two effects compound.

Answer: B

Earlier dimensions of the wall = 100 × 3 × 0.30.
New dimensions = L × 1.5 × 0.3.
∴ As men, women and children are given to be equally efficient, so in the first case, the total number of persons is 100 (i.e. 30 + 20 + 50) and the same in the second case is 75 (15 + 25 + 35).
Length of wall = L = (75x100) x (2x9) x (15x20) x (100 x 3 x 3 x 0.3)/(1.5 x 0.3) ⇒ L = 25 m.