Question paper, Set 1

A finishes the whole work in $57$ days.

∴ A's 1 day's work $\displaystyle = \frac{1}{57}$

Answer: $\displaystyle \frac{1}{57}$

Time taken by A $\displaystyle = \frac{1}{3} \times$ (Time taken by B)

$\displaystyle 16 = \frac{1}{3} \times$ (Time taken by B)

∴ Time taken by B $= 3 \times 16 = 48$ days

Answer: $48$ days

No. of men $\displaystyle \propto \frac{1}{\text{Time taken}}$

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{7}{9}$

∴ $T_1 : T_2 = 7 : 9$

Answer: $7 : 9$

No. of men $\displaystyle \propto \frac{1}{\text{No. of days}}$

$9 \times 50 = x \times 75$

$\displaystyle x = \frac{9 \times 50}{75} = 6$ men

Answer: $6$ men

(A + B)'s 1 day's work $\displaystyle = \frac{1}{14} + \frac{1}{35} = \frac{5 + 2}{70} = \frac{7}{70} = \frac{1}{10}$

Answer: $\displaystyle \frac{1}{10}$

A's 1 day's work $\displaystyle = \frac{1}{13}$

B's 1 day's work $\displaystyle = \frac{1}{52}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{13} + \frac{1}{52} = \frac{4 + 1}{52} = \frac{5}{52}$

∴ Both together can finish the work in $\displaystyle \frac{52}{5}$ days $\displaystyle = 10\frac{2}{5}$ days

Answer: $\displaystyle 10\frac{2}{5}$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{10}$

B's 1 day's work $\displaystyle = \frac{1}{35}$

⇒ A's 1 day's work $\displaystyle = \frac{1}{10} - \frac{1}{35} = \frac{7 - 2}{70} = \frac{5}{70} = \frac{1}{14}$

∴ A alone can finish the work in $14$ days.

Answer: $14$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{50}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{100}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{60}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{50} + \frac{1}{100} + \frac{1}{60} = \frac{6 + 3 + 5}{300} = \frac{14}{300} = \frac{7}{150}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{7}{300}$

∴ A, B, C together can finish the work in $\displaystyle 42\frac{6}{7}$ days.

A's 1 day's work $\displaystyle = \frac{7}{300} - \frac{1}{100} = \frac{1}{75}$, so A alone takes $75$ days

B's 1 day's work $\displaystyle = \frac{7}{300} - \frac{1}{60} = \frac{1}{150}$, so B alone takes $150$ days

C's 1 day's work $\displaystyle = \frac{7}{300} - \frac{1}{50} = \frac{1}{300}$, so C alone takes $300$ days

Answer: Together $\displaystyle 42\frac{6}{7}$ days; A alone $75$ days, B alone $150$ days, C alone $300$ days

A's 1 day's work $\displaystyle = \frac{1}{23}$

B's 1 day's work $\displaystyle = \frac{1}{46}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{23} + \frac{1}{46} = \frac{2 + 1}{46} = \frac{3}{46}$

(A + B)'s 11 days' work $\displaystyle = 11 \times \frac{3}{46} = \frac{33}{46}$

Remaining work $\displaystyle = 1 - \frac{33}{46} = \frac{13}{46}$

This work is done by B in $\displaystyle \frac{13}{46} \times 46 = 13$ days

Answer: $13$ days

A's 1 day's work $\displaystyle = \frac{1}{72}$, B's 1 day's work $\displaystyle = \frac{1}{18}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{72} + \frac{1}{18} = \frac{1 + 4}{72} = \frac{5}{72}$

A's 12 days' work $\displaystyle = 12 \times \frac{1}{72} = \frac{1}{6}$

Remaining work $\displaystyle = 1 - \frac{1}{6} = \frac{5}{6}$

(A + B)'s $x$ days' work $\displaystyle = \frac{5}{6}$ ⇒ $\displaystyle \frac{5}{72} \times x = \frac{5}{6}$

$x = 12$, so B left after $12$ days

Answer: $12$ days

A's 1 day's work $\displaystyle = \frac{1}{135}$, B's 1 day's work $\displaystyle = \frac{1}{60}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{135} + \frac{1}{60} = \frac{4 + 9}{540} = \frac{13}{540}$

(A + B)'s 26 days' work $\displaystyle = 26 \times \frac{13}{540} = \frac{169}{270}$

A's 13 days' work $\displaystyle = \frac{13}{135}$

(A + C)'s 24 days' work $\displaystyle = 1 - \left(\frac{169}{270} + \frac{13}{135}\right) = \frac{5}{18}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{5}{18} \div 24 = \frac{5}{432}$

C's 1 day's work $\displaystyle = \frac{5}{432} - \frac{1}{135} = \frac{25 - 16}{2160} = \frac{9}{2160} = \frac{1}{240}$

C can finish the work in $240$ days

Answer: $240$ days

Let the original number of men be $x$

No. of men $\displaystyle \propto \frac{1}{\text{No. of days}}$

$x \times 26 = (x + 8) \times 18$

$26x = 18x + 144$

$8x = 144$ ⇒ $x = 18$

Answer: $18$ men

Let 1 man do $m$ and 1 woman do $w$ units of work in 1 hour.

Case (i): $(6m + 4w) \times 5 \times 4 = 120m + 80w$

Case (ii): $(4m + 4w) \times 4 \times 6 = 96m + 96w$

Same work: $120m + 80w = 96m + 96w$

⇒ $16w = 24m$ ⇒ $2w = 3m$

So $4$ women $= 6$ men, and $6m + 4w = 12$ men

$12$ men working $5$ hours a day for $4$ days do the same work as $10$ men working $8$ hours a day for $x$ days

$12 \times 5 \times 4 = 10 \times 8 \times x$ ⇒ $x = 3$ days

Answer: $3$ days

If B does $x$, A does $\displaystyle \frac{x}{2} + x = \frac{3x}{2}$ in the same time

Ratio of working of A and B $\displaystyle = \frac{3}{2} : 1 = 3 : 2$

Ratio of time taken for the same work $= 2 : 3$

Time taken by B $= 3k = 12$ ⇒ $k = 4$

Time taken by A $= 2k = 8$ days

A's 1 day's work $\displaystyle = \frac{1}{8}$, B's 1 day's work $\displaystyle = \frac{1}{12}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{8} + \frac{1}{12} = \frac{3 + 2}{24} = \frac{5}{24}$

Time taken by (A + B) to finish the work $\displaystyle = 4\frac{4}{5}$ days

Answer: $\displaystyle 4\frac{4}{5}$ days

Speed upstream $= (x - y)$ km/hr

$= 7.5 - 1 = 6.5$ km/hr

Answer: $6.5$ km/hr

Let the man's rate in still water be $x$ km/hr and the rate of the current be $y$ km/hr.

Given $x + y = 20$ and $x - y = 17$

Adding, $2x = 37$ ⇒ $x = 18.5$ km/hr

Subtracting, $2y = 3$ ⇒ $y = 1.5$ km/hr

Answer: Rate in still water $18.5$ km/hr, rate of current $1.5$ km/hr

Speed $\displaystyle = \frac{\text{Distance}}{\text{Time}}$

$\displaystyle = \frac{25}{8} = 3.125$ km/hr

Answer: $3.125$ km/hr

Let the rate of the stream be $y$ km/hr

Same distance, thrice the time, so speed down $= 3 \times$ speed up

Given $3(3 - y) = 3 + y$

⇒ $9 - 3y = 3 + y$ ⇒ $4y = 6$

⇒ $y = 1.5$ km/hr

Answer: $1.5$ km/hr

Let the required distance be $d$ km

Speed downstream $= 15 + 1 = 16$ km/hr

Speed upstream $= 15 - 1 = 14$ km/hr

Total time $=$ 1 hour 30 minutes $= 1.5$ hours

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{16} + \frac{d}{14} = 1.5$

⇒ $\displaystyle d\left(\frac{1}{16} + \frac{1}{14}\right) = 1.5$

⇒ $\displaystyle d \times \frac{15}{112} = 1.5$

⇒ $d = 11.2$ km

Answer: $11.2$ km

Let the speed of the man in still water be $x$ km/hr and the rate of the current $y$ km/hr.

Speed upstream $= (x - y)$ km/hr, speed downstream $= (x + y)$ km/hr. Put $\displaystyle p = \frac{1}{x - y}$, $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{7}{x - y} + \frac{60}{x + y} = 5$ ⇒ $7p + 60q = 5 \quad \cdots (1)$

$\displaystyle \frac{42}{x - y} + \frac{60}{x + y} = 10$ ⇒ $42p + 60q = 10 \quad \cdots (2)$

(1) $\times 6$ ⇒ $42p + 360q = 30$

(1) $\times 6$ $-$ (2) ⇒ $300q = 20$ ⇒ $\displaystyle q = \frac{1}{15}$

From (1): $7p + 4 = 5$ ⇒ $7p = 1$ ⇒ $\displaystyle p = \frac{1}{7}$

So $x - y = 7$ and $x + y = 15$

Adding, $2x = 22$ ⇒ $x = 11$; then $y = 15 - 11 = 4$

∴ Rate of the current $= 4$ km/hr, speed in still water $= 11$ km/hr

Answer: Rate of current $4$ km/hr, speed in still water $11$ km/hr