Chain Rule Questions and Answers

Answer: A

Work in man-days, the quantity that stays fixed however the numbers change. After 3 days the provision left is what 500 men would eat in the remaining \( 27 - 3 = 24 \) days:

\[ \text{food remaining} = 500 \times 24 = 12{,}000 \text{ man-days} \]

The reinforcement brings the strength to \( 500 + 300 = 800 \), so

\[ \text{days} = \frac{12{,}000}{800} = 15 \]

Note the two traps: the 27 must be reduced to 24 before anything else, and the 300 is an addition to the garrison rather than the new total.

Answer: D

The price is directly proportional to the number of toys, so scale by the ratio:

\[ \text{price of }5 = 264.37 \times \frac{5}{6} = 220.31 \]

which is about Rs. 220.

For an approximation the arithmetic can be done in the head: 6 toys at roughly Rs. 264 means about Rs. 44 each, and \( 5 \times 44 = 220 \).

Answer: B

Divide the length required by the rate of weaving:

\[ t = \frac{25}{0.128} = 195.3 \text{ seconds} \]

so about 195 seconds.

A quick check on the size of the answer: at roughly 0.128 m per second the loom weaves a little under 8 metres per minute, so 25 metres must take somewhat over three minutes - which rules out both 250 and 844 immediately.

Answer: B

Find the daily rate first, then apply it to what is left. Completing \( \tfrac{5}{8} \) in 10 days means

\[ \text{rate} = \frac{5/8}{10} = \frac{1}{16} \text{ of the job per day} \]

The remaining fraction is \( 1 - \tfrac{5}{8} = \tfrac{3}{8} \), so

\[ \text{days} = \frac{3/8}{1/16} = \frac{3}{8} \times 16 = 6 \]

The question asks how many more days, so 6 is the answer; 16 would be the total time for the whole job from the start.

Answer: C

Count the food in man-days, which does not change when men leave. After 10 days there remain \( 45 - 10 = 35 \) days of provision for 150 men:

\[ \text{food remaining} = 150 \times 35 = 5{,}250 \text{ man-days} \]

With 25 men gone the strength is \( 150 - 25 = 125 \), so

\[ \text{days} = \frac{5{,}250}{125} = 42 \]

Fewer men must make the food last longer, so any answer below 35 is wrong on inspection - which disposes of 12 at once.

Answer: B

The meal is one fixed quantity, described two ways, so the two descriptions are equivalent:

\[ 120 \text{ men} \equiv 200 \text{ children} \]

After 150 children have eaten, what is left would feed \( 200 - 150 = 50 \) children. Converting that share to men:

\[ 50 \text{ children} \times \frac{120 \text{ men}}{200 \text{ children}} = 30 \text{ men} \]

The step that matters is doing the subtraction in children and converting only at the end; converting first and then subtracting mixes the two units.

Answer: D

Reduce to a single weaver before scaling up. Four weavers making 4 mats in 4 days means

\[ 1 \text{ weaver} \rightarrow 1 \text{ mat in } 4 \text{ days} \]

Output is proportional to both the number of weavers and the time given, so

\[ \text{mats} = 8 \text{ weavers} \times \frac{8 \text{ days}}{4 \text{ days per mat}} = 8 \times 2 = 16 \]

The intuitive answer of 8 comes from doubling only one of the two factors. Both the workforce and the time double here, so the output goes up fourfold.

Answer: C

Read the rate per cow rather than for the herd. If 40 cows finish 40 bags in 40 days, then in those same 40 days

\[ \text{each cow eats } \frac{40 \text{ bags}}{40 \text{ cows}} = 1 \text{ bag} \]

so one cow takes 40 days for one bag.

The number of cows never enters the answer: each is eating alongside the others, not sharing a single bag. Dividing 40 by 40 to get 1 day is the classic slip, and it would mean the herd polished off 40 bags in a single day.

Answer: D

The total work in man-days is fixed.

\[ 100M = 110(M - 10) \]

\[ 100M = 110M - 1100 \quad\Longrightarrow\quad 10M = 1100 \]

\[ M = 110 \]

Answer: D

Let the required length be x meters

 More men, More length built     (Direct proportion) 

Less days, Less length built      (Direct Proportion)

Men20:35Days6 : 3⋮⋮ 56 :x 

=> (20 x 6 x X)=(35 x 3 x 56) 

=> x = 49 

Hence, the required length is 49 m.