Question paper, Set 10

A finishes the whole work in $24$ days.

∴ A's 1 day's work $\displaystyle = \frac{1}{24}$

Answer: $\displaystyle \frac{1}{24}$

Time taken by A $\displaystyle = \frac{1}{2} \times$ (Time taken by B)

$\displaystyle 41 = \frac{1}{2} \times$ (Time taken by B)

∴ Time taken by B $= 2 \times 41 = 82$ days

Answer: $82$ days

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{5}{6}$

∴ $T_1 : T_2 = 5 : 6$

Answer: $5 : 6$

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

$8 \times 26 = 52 \times x$

$\displaystyle x = \frac{8 \times 26}{52} = 4$ hours

Answer: $4$ hours

(A + B)'s 1 day's work $\displaystyle = \frac{1}{10} + \frac{1}{70} = \frac{7 + 1}{70} = \frac{8}{70} = \frac{4}{35}$

Answer: $\displaystyle \frac{4}{35}$

A's 1 day's work $\displaystyle = \frac{1}{17}$

B's 1 day's work $\displaystyle = \frac{1}{51}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{17} + \frac{1}{51} = \frac{3 + 1}{51} = \frac{4}{51}$

∴ Both together can finish the work in $\displaystyle \frac{51}{4}$ days $\displaystyle = 12\frac{3}{4}$ days

Answer: $\displaystyle 12\frac{3}{4}$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{9}$

B's 1 day's work $\displaystyle = \frac{1}{90}$

⇒ A's 1 day's work $\displaystyle = \frac{1}{9} - \frac{1}{90} = \frac{10 - 1}{90} = \frac{9}{90} = \frac{1}{10}$

∴ A alone can finish the work in $10$ days.

Answer: $10$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{63}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{28}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{36}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{63} + \frac{1}{28} + \frac{1}{36} = \frac{4 + 9 + 7}{252} = \frac{20}{252} = \frac{5}{63}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{5}{126}$

∴ A, B, C together can finish the work in $\displaystyle 25\frac{1}{5}$ days.

A's 1 day's work $\displaystyle = \frac{5}{126} - \frac{1}{28} = \frac{1}{252}$, so A alone takes $252$ days

B's 1 day's work $\displaystyle = \frac{5}{126} - \frac{1}{36} = \frac{1}{84}$, so B alone takes $84$ days

C's 1 day's work $\displaystyle = \frac{5}{126} - \frac{1}{63} = \frac{1}{42}$, so C alone takes $42$ days

Answer: Together $\displaystyle 25\frac{1}{5}$ days; A alone $252$ days, B alone $84$ days, C alone $42$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 38)$ days

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

∴ $\displaystyle x - 38 = \frac{1}{3}x$

⇒ $3x - 114 = x$ ⇒ $2x = 114$ ⇒ $x = 57$

∴ Time taken by B $= 57$ days, time taken by A $= 19$ days

A's 1 day's work $\displaystyle = \frac{1}{19}$, B's 1 day's work $\displaystyle = \frac{1}{57}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{19} + \frac{1}{57} = \frac{3 + 1}{57} = \frac{4}{57}$

A and B together can finish the work in $\displaystyle 14\frac{1}{4}$ days

Answer: $\displaystyle 14\frac{1}{4}$ days

A's 1 day's work $\displaystyle = \frac{1}{56}$, B's 1 day's work $\displaystyle = \frac{1}{28}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{56} + \frac{1}{28} = \frac{1 + 2}{56} = \frac{3}{56}$

A's 44 days' work $\displaystyle = 44 \times \frac{1}{56} = \frac{11}{14}$

Remaining work $\displaystyle = 1 - \frac{11}{14} = \frac{3}{14}$

(A + B)'s $x$ days' work $\displaystyle = \frac{3}{14}$ ⇒ $\displaystyle \frac{3}{56} \times x = \frac{3}{14}$

$x = 4$, so B left after $4$ days

Answer: $4$ days

A's 1 day's work $\displaystyle = \frac{1}{120}$, B's 1 day's work $\displaystyle = \frac{1}{80}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{120} + \frac{1}{80} = \frac{2 + 3}{240} = \frac{5}{240} = \frac{1}{48}$

(A + B)'s 7 days' work $\displaystyle = 7 \times \frac{1}{48} = \frac{7}{48}$

A's 29 days' work $\displaystyle = \frac{29}{120}$

(A + C)'s 21 days' work $\displaystyle = 1 - \left(\frac{7}{48} + \frac{29}{120}\right) = \frac{49}{80}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{49}{80} \div 21 = \frac{7}{240}$

C's 1 day's work $\displaystyle = \frac{7}{240} - \frac{1}{120} = \frac{7 - 2}{240} = \frac{5}{240} = \frac{1}{48}$

C can finish the work in $48$ days

Answer: $48$ days

Let 1 man earn ₹$x$ and 1 boy earn ₹$y$ in 1 day.

$(3x + 7y) \times 9 = 9990$ ⇒ $3x + 7y = 1110 \quad \cdots (1)$

$(6x + 3y) \times 11 = 11715$ ⇒ $6x + 3y = 1065 \quad \cdots (2)$

(1) $\times 2$ ⇒ $6x + 14y = 2220$

(1) $\times 2$ $-$ (2) ⇒ $11y = 1155$ ⇒ $y = 105$

From (1): $3x = 1110 - 735 = 375$ ⇒ $x = 125$

$(7x + 10y)n = 28875$

⇒ $(875 + 1050)n = 28875$ ⇒ $1925n = 28875$

⇒ $n = 15$ days

Answer: $15$ 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 $= 3k = 45$ ⇒ $k = 15$

Time taken by A $= 2k = 30$ days

A's 1 day's work $\displaystyle = \frac{1}{30}$, B's 1 day's work $\displaystyle = \frac{1}{45}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{45} = \frac{3 + 2}{90} = \frac{5}{90} = \frac{1}{18}$

Time taken by (A + B) to finish the work $= 18$ days

Answer: $18$ days

Speed downstream $= (x + y)$ km/hr

$= 7 + 1.5 = 8.5$ km/hr

Answer: $8.5$ 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 $x + y = 13$ and $x - y = 11$

Adding, $2x = 24$ ⇒ $x = 12$ km/hr

Subtracting, $2y = 2$ ⇒ $y = 1$ km/hr

Answer: Rate in still water $12$ km/hr, rate of current $1$ km/hr

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

$\displaystyle = \frac{23}{8} = 2.875$ km/hr

Answer: $2.875$ km/hr

Speed downstream: $\displaystyle x + y = \frac{63}{6}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{42}{6}$ km/hr

Subtracting, $\displaystyle 2y = \frac{63}{6} - \frac{42}{6} = \frac{21}{6}$

$\displaystyle y = \frac{21}{12} = 1.75$ km/hr

Answer: $1.75$ 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 $3(5 - y) = 5 + y$

⇒ $15 - 3y = 5 + y$ ⇒ $4y = 10$

⇒ $y = 2.5$ km/hr

Answer: $2.5$ km/hr

Let the required distance be $d$ km

Speed downstream $= 15 + 6 = 21$ km/hr

Speed upstream $= 15 - 6 = 9$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{21} + \frac{d}{9} = 1$

⇒ $\displaystyle d\left(\frac{1}{21} + \frac{1}{9}\right) = 1$

⇒ $\displaystyle d \times \frac{10}{63} = 1$

⇒ $d = 6.3$ km

Answer: $6.3$ km

Let the speed of the boat in still water be $x$ km/hr

Speed upstream $= (x - 1)$ km/hr, speed downstream $= (x + 1)$ km/hr

Total time $= 175$ min $\displaystyle = \frac{175}{60}$ hr $\displaystyle = \frac{35}{12}$ hr

$\displaystyle \frac{7}{x - 1} + \frac{7}{x + 1} = \frac{35}{12}$

⇒ $\displaystyle 7\left(\frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)}\right) = \frac{35}{12}$

⇒ $\displaystyle \frac{14x}{x^2 - 1} = \frac{35}{12}$

⇒ $168x = 35x^2 - 35$

⇒ $5 x^{2} - 24 x - 5 = 0$

⇒ $\left(x - 5\right) \left(5 x + 1\right) = 0$

⇒ $x = 5$ (or) $\displaystyle x = -\frac{1}{5}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 5$ km/hr

Answer: $5$ km/hr