Boats and Streams, Pipes and Cisterns for CUET Applied Maths

AnswerC · 15 km/h

1. Downstream, the stream carries the boat along, so the speeds add.

2. $12 + 3 = 15$ km/h.

AnswerB · 8 km/h

1. Downstream $= b + s$ and upstream $= b - s$, so $b$ is the average of the two speeds.

2. $b = \frac{10 + 6}{2} = 8$ km/h.

AnswerB · 14 km/h

1. Downstream the speeds add; upstream the stream's speed is taken away.

2. $18 - 4 = 14$ km/h.

AnswerB · 5.5 km/h

1. Downstream the speeds add; upstream the stream's speed is taken away.

2. $4 + \frac{3}{2} = \frac{11}{2}$ km/h.

AnswerC · 11 km/h

1. Downstream $= b + s$ and upstream $= b - s$.

2. $b$ is their average: $\frac{14 + 8}{2} = 11.0$ km/h.

AnswerC · 12 km/h

1. Downstream it covers 24 km in 2 hours.

2. $\frac{24}{2} = 12$ km/h.

AnswerA · 2 km/h

1. The current adds to the speed one way and takes it away the other.

2. $\frac{12 - 8}{2} = 2$ km/h.

AnswerA · 4 km/h

1. Downstream − upstream $= (b + s) - (b - s) = 2s$.

2. $2s = 18 - 10 = 8$, so $s = 4$ km/h.

AnswerD · 8.5 km/h

1. Downstream speed $= \frac{24}{2} = 12$ km/h; upstream speed $= \frac{15}{3} = 5$ km/h.

2. Still water: $\frac{12 + 5}{2} = 8.5$ km/h.

AnswerC · Both I and II

1. I: $b + s > b - s$ whenever $s > 0$. True.

2. II: upstream speed $= b - s = 0$ when $s = b$, so the boat stays where it is. True.

AnswerD · A-II, B-IV, C-I, D-III

1. First boat: $10 + 2 = 12$ and $10 - 2 = 8$.

2. Second boat: $\frac{14 + 6}{2} = 10$ in still water, $\frac{14 - 6}{2} = 4$ for the stream.

AnswerB · 2 : 1

1. Upstream 6 km/h, downstream 12 km/h. For a fixed distance, time is inversely proportional to speed.

2. Time up : time down $= 12 : 6 = 2 : 1$.

AnswerA · 2 hours

1. A floating log moves at the stream's speed: $\frac{12}{4} = 3$ km/h.

2. Downstream speed $10 + 3 = 13$ km/h, so $\frac{26}{13} = 2$ hours.

AnswerB · 6 km/h

1. Downstream $= b + s$ and upstream $= b - s$.

2. $b$ is their average: $\frac{\frac{15}{2} + \frac{9}{2}}{2} = 6$ km/h.

AnswerA · 4 km/h

1. Downstream $= b + s$ and upstream $= b - s$.

2. $s$ is half their difference: $\frac{22 - 14}{2} = 4.0$ km/h.

AnswerA · 3 km/h

1. Downstream $\frac{30}{2} = 15$ km/h, upstream $\frac{18}{2} = 9$ km/h.

2. Stream $= \frac{15 - 9}{2} = 3$ km/h.

AnswerC · 14 km/h

1. Downstream $\frac{48}{3} = 16$ km/h; upstream $\frac{48}{4} = 12$ km/h.

2. Still water $= \frac{16 + 12}{2} = 14$ km/h.

AnswerB · 144 minutes

1. In still water it moves at 10 km/h.

2. $\frac{24}{10} = 2.4$ hours, that is 144 minutes.

AnswerB · 10 km/h

1. Downstream 12 km/h and upstream $\frac{24}{3} = 8$ km/h.

2. Still water $= \frac{12 + 8}{2} = 10$ km/h.

AnswerB · 3 : 2

1. Upstream speed $= 4$ km/h and downstream speed $= 6$ km/h.

2. For the same distance, time is inversely proportional to speed: upstream : downstream $= 6 : 4 = 3 : 2$.

AnswerD · 3 : 1

1. Twice the time means half the speed: $b + s = 2(b - s)$.

2. So $b = 3s$, and $b : s = 3 : 1$.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. R is the rule for upstream speed. True.

2. If $s > b$, then $b - s < 0$: the boat is carried back, so A is true because of R.

AnswerA · 30 minutes

1. Downstream speed $= 4 + 2 = 6$ km/h.

2. $\frac{3}{6} = 0.5$ hour, 30 minutes.

AnswerD · 90 minutes

1. Upstream speed $= 4 - 2 = 2$ km/h.

2. $\frac{3}{2} = 1.5$ hours, 90 minutes.

AnswerC · 6 hours

1. Upstream at $9 - 3 = 6$ km/h takes $\frac{24}{6} = 4$ hours; downstream at 12 km/h takes 2 hours.

2. In all, 6 hours.

AnswerB · 4 hours

1. Downstream $\frac{36}{3} = 12$ km/h, upstream $\frac{36}{6} = 6$ km/h, so still water $\frac{12 + 6}{2} = 9$ km/h.

2. $\frac{36}{9} = 4$ hours.

AnswerA · 7.5 km/h

1. For a distance $d$ each way: time $= \frac{d}{15} + \frac{d}{5} = \frac{4d}{15}$.

2. Average speed $= \frac{2d}{4d/15} = 7.5$ km/h, which is $\frac{b^2 - s^2}{b}$.

AnswerC · 2.4 hours

1. Downstream 15 km/h, upstream 10 km/h, so still water $\frac{15 + 10}{2} = 12.5$ km/h.

2. $\frac{30}{12.5} = 2.4$ hours.

AnswerD · $\frac{10}{3}$ hours

1. Downstream speed $\frac{40}{2} = 20$ km/h, so the boat's own speed is $20 - 4 = 16$ km/h and upstream it goes $16 - 4 = 12$ km/h.

2. $\frac{40}{12} = \frac{10}{3}$ hours, that is 3 hours 20 minutes.

AnswerD · I, II and IV only

1. I and II come from $b + s$ and $b - s$. IV: downstream is faster.

2. III is false: the round trip takes longer on a flowing river, since the slow leg upstream lasts longer.

AnswerC · 5 hours

1. Down at 12 km/h: $\frac{24}{12} = 2$ hours. Back at 8 km/h: $\frac{24}{8} = 3$ hours.

2. Total 5 hours.

AnswerB · 8 hours

1. Down at 15 km/h: $\frac{45}{15} = 3$ hours. Back at 9 km/h: $\frac{45}{9} = 5$ hours.

2. Total 8 hours.

AnswerB · She could not make any headway upstream

1. Upstream speed would be $4 - 4 = 0$ km/h.

2. Swimming as hard as she can, she would only hold her place, so she could never swim back.

AnswerC · 3 km/h

1. 6 km in $0.5 + 1.5 = 2$ hours.

2. Average speed $= \frac{6}{2} = 3$ km/h.

AnswerB · 5 km/h

1. $\frac{30}{15 + s} + \frac{30}{15 - s} = 4.5$, which gives $\frac{900}{225 - s^{2}} = 4.5$.

2. $225 - s^{2} = 200$, so $s^{2} = 25$ and $s = 5$ km/h.

AnswerA · 8 km

1. Downstream he makes 8 km/h, upstream 4 km/h. For a distance $d$: $\frac{d}{8} + \frac{d}{4} = 3$.

2. $\frac{3d}{8} = 3$, so $d = 8$ km.

AnswerA · $\frac{32}{3}$ km/h

1. Downstream 16 km/h, upstream 8 km/h. Over equal distances the average speed is $\frac{2 \times 16 \times 8}{16 + 8}$.

2. $= \frac{256}{24} = \frac{32}{3}$ km/h, about 10.7 km/h.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. For a distance $d$: $\frac{d}{b - s} + \frac{d}{b + s} = \frac{2db}{b^{2} - s^{2}}$, which is more than $\frac{2d}{b}$ when $s > 0$. A is true.

2. R is why: the slow leg lasts longer, so it outweighs the fast one. Both are true and R explains A.

AnswerB · 12 km/h

1. Equal times: $\frac{21}{b + 2} = \frac{15}{b - 2}$.

2. $21b - 42 = 15b + 30$, so $6b = 72$ and $b = 12$ km/h.

AnswerB · 2 km/h

1. $\frac{48}{10 + s} + \frac{48}{10 - s} = 10$, so $\frac{960}{100 - s^2} = 10$.

2. $100 - s^2 = 96$, so $s = 2$ km/h. Check: $4 + 6 = 10$ hours.

AnswerA · 12 km

1. Downstream 6 km/h, upstream 4 km/h: $\frac{d}{6} + \frac{d}{4} = 5$.

2. $\frac{5d}{12} = 5$, so $d = 12$ km.

AnswerA · 2 hours

1. In one hour A fills $\frac{1}{6}$ and B fills $\frac{1}{3}$ of the tank: together $\frac{1}{6} + \frac{2}{6} = \frac{1}{2}$.

2. Half a tank an hour fills it in 2 hours.

AnswerC · 8 hours

1. A third in 4 hours means the whole tank in 12 hours.

2. The rest is $\frac{2}{3}$ of the tank: $\frac{2}{3} \times 12 = 8$ hours.

AnswerA · Only I

1. I: the slower pipe still adds some water each hour, so the tank fills sooner than with the faster pipe alone. True.

2. II: the rates add, not the times; the average lies between the two times, so it cannot be right. False.

AnswerA · 6 hours

1. Rates add: $\frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ of the tank per hour.

2. Together: $6$ hours. (Shortcut: $\frac{ab}{a + b} = \frac{150}{25}$.)

AnswerA · 3 hours

1. Together: $\frac{1}{6} + \frac{1}{9} = \frac{5}{18}$ of the tank an hour.

2. $\frac{5}{6} \div \frac{5}{18} = 3$ hours.

AnswerB · 12 minutes

1. Rates add: $\frac{1}{20} + \frac{1}{30} = \frac{1}{12}$ of the tank per minute.

2. Together: $12$ minutes. (Shortcut: $\frac{ab}{a + b} = \frac{600}{50}$.)

AnswerA · 3 hours

1. Rates add: $\frac{1}{4} + \frac{1}{12} = \frac{1}{3}$ of the tank per hour.

2. Together: $3$ hours. (Shortcut: $\frac{ab}{a + b} = \frac{48}{16}$.)

AnswerA · 7.2 hours

1. $\frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}$ of the tank an hour.

2. $\frac{36}{5} = 7.2$ hours.

AnswerC · 4 hours

1. Per hour: $\frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6 + 5 + 4}{60} = \frac{15}{60} = \frac{1}{4}$ of the tank.

2. So together they take 4 hours.

AnswerD · 30 minutes

1. The other pipe's rate $= \frac{1}{12} - \frac{1}{20} = \frac{5 - 3}{60} = \frac{1}{30}$ of the tank a minute.

2. So it takes 30 minutes alone.

AnswerD · 5.6 hours

1. In 2 hours A fills $\frac{2}{8} = \frac{1}{4}$. The rest, $\frac{3}{4}$, fills at $\frac{1}{8} + \frac{1}{12} = \frac{5}{24}$ an hour.

2. That takes $\frac{3}{4} \div \frac{5}{24} = \frac{18}{5} = 3.6$ hours, so $2 + 3.6 = 5.6$ hours in all.

AnswerD · $\frac{5}{9}$

1. Together per hour: $\frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}$.

2. In 4 hours: $\frac{20}{36} = \frac{5}{9}$.

AnswerB · 8 minutes

1. In 4 minutes both fill $4\left(\frac{1}{12} + \frac{1}{18}\right) = \frac{5}{9}$.

2. B fills the remaining $\frac{4}{9}$ in $\frac{4}{9} \times 18 = 8$ minutes.

AnswerB · 18 hours

1. A does 2 parts of the work for every 1 of B, so A does $\frac{2}{3}$ of the combined rate: $\frac{2}{3} \times \frac{1}{12} = \frac{1}{18}$.

2. A alone takes 18 hours (and B 36).

AnswerC · 3 hours

1. $\frac{1}{6} + \frac{1}{8} + \frac{1}{24} = \frac{4 + 3 + 1}{24} = \frac{1}{3}$ of the tank an hour.

2. So 3 hours together.

AnswerD · 24 hours

1. The other's rate $= \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$.

2. So it takes 24 hours alone.

AnswerC · 4.8 hours

1. In 2 hours A fills $\frac{2}{10} = \frac{1}{5}$, leaving $\frac{4}{5}$.

2. Together they fill $\frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ an hour, so $\frac{4}{5} \times 6 = 4.8$ hours.

AnswerD · 36 hours

1. Net rate $= \frac{1}{12} + \frac{1}{18} - \frac{1}{9} = \frac{3 + 2 - 4}{36} = \frac{1}{36}$.

2. So 36 hours.