Question paper, Set 7

B does $\displaystyle \frac{1}{15}$ of the work in 1 day.

∴ B finishes the whole work in $15$ days.

Answer: $15$ days

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

$\displaystyle = \frac{1}{2} \times 20 = 10$ days

Answer: $10$ days

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{7}{8}$

∴ $T_1 : T_2 = 7 : 8$

Answer: $7 : 8$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{42} + \frac{1}{56} = \frac{4 + 3}{168} = \frac{7}{168} = \frac{1}{24}$

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

A's 1 day's work $\displaystyle = \frac{1}{39}$

B's 1 day's work $\displaystyle = \frac{1}{26}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{39} + \frac{1}{26} = \frac{2 + 3}{78} = \frac{5}{78}$

∴ Both together can finish the work in $\displaystyle \frac{78}{5}$ days $\displaystyle = 15\frac{3}{5}$ days

Answer: $\displaystyle 15\frac{3}{5}$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{14}$

A's 1 day's work $\displaystyle = \frac{1}{63}$

⇒ B's 1 day's work $\displaystyle = \frac{1}{14} - \frac{1}{63} = \frac{9 - 2}{126} = \frac{7}{126} = \frac{1}{18}$

∴ B alone can finish the work in $18$ days.

Answer: $18$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{35}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{70}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{30}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{35} + \frac{1}{70} + \frac{1}{30} = \frac{6 + 3 + 7}{210} = \frac{16}{210} = \frac{8}{105}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{4}{105}$

∴ A, B, C together can finish the work in $\displaystyle 26\frac{1}{4}$ days.

A's 1 day's work $\displaystyle = \frac{4}{105} - \frac{1}{70} = \frac{1}{42}$, so A alone takes $42$ days

B's 1 day's work $\displaystyle = \frac{4}{105} - \frac{1}{30} = \frac{1}{210}$, so B alone takes $210$ days

C's 1 day's work $\displaystyle = \frac{4}{105} - \frac{1}{35} = \frac{1}{105}$, so C alone takes $105$ days

Answer: Together $\displaystyle 26\frac{1}{4}$ days; A alone $42$ days, B alone $210$ days, C alone $105$ days

A's 1 day's work $\displaystyle = \frac{1}{20}$

B's 1 day's work $\displaystyle = \frac{1}{40}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{20} + \frac{1}{40} = \frac{2 + 1}{40} = \frac{3}{40}$

(A + B)'s 6 days' work $\displaystyle = 6 \times \frac{3}{40} = \frac{9}{20}$

Remaining work $\displaystyle = 1 - \frac{9}{20} = \frac{11}{20}$

This work is done by A in $\displaystyle \frac{11}{20} \times 20 = 11$ days

Answer: $11$ days

A's 1 day's work $\displaystyle = \frac{1}{22}$, B's 1 day's work $\displaystyle = \frac{1}{44}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{22} + \frac{1}{44} = \frac{2 + 1}{44} = \frac{3}{44}$

B's 35 days' work $\displaystyle = 35 \times \frac{1}{44} = \frac{35}{44}$

Remaining work $\displaystyle = 1 - \frac{35}{44} = \frac{9}{44}$

(A + B)'s $x$ days' work $\displaystyle = \frac{9}{44}$ ⇒ $\displaystyle \frac{3}{44} \times x = \frac{9}{44}$

$x = 3$, so A left after $3$ days

Answer: $3$ days

A's 1 day's work $\displaystyle = \frac{1}{180}$, B's 1 day's work $\displaystyle = \frac{1}{90}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{180} + \frac{1}{90} = \frac{1 + 2}{180} = \frac{3}{180} = \frac{1}{60}$

(A + B)'s 7 days' work $\displaystyle = 7 \times \frac{1}{60} = \frac{7}{60}$

A's 3 days' work $\displaystyle = \frac{3}{180} = \frac{1}{60}$

(A + C)'s 66 days' work $\displaystyle = 1 - \left(\frac{7}{60} + \frac{1}{60}\right) = \frac{13}{15}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{13}{15} \div 66 = \frac{13}{990}$

C's 1 day's work $\displaystyle = \frac{13}{990} - \frac{1}{180} = \frac{26 - 11}{1980} = \frac{15}{1980} = \frac{1}{132}$

C can finish the work in $132$ days

Answer: $132$ days

Let the original number of men be $x$

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

$x \times 34 = (x + 14) \times 27$

$34x = 27x + 378$

$7x = 378$ ⇒ $x = 54$

Answer: $54$ men

$5$ men $= 7$ boys

$10$ men $\displaystyle = \frac{7}{5} \times 10 = 14$ boys

$10$ men + $3$ boys $= 14$ boys + $3$ boys $= 17$ boys

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

$7 \times 17 = 17 \times x$ ⇒ $x = 7$ days

Answer: $7$ days

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

Case (i): $(2m + 6w) \times 5 \times 10 = 100m + 300w$

Case (ii): $(3m + 3w) \times 5 \times 12 = 180m + 180w$

Same work: $100m + 300w = 180m + 180w$

⇒ $120w = 80m$ ⇒ $3w = 2m$

So $6$ women $= 4$ men, and $2m + 6w = 6$ men

$6$ men working $5$ hours a day for $10$ days do the same work as $4$ men working $3$ hours a day for $x$ days

$6 \times 5 \times 10 = 4 \times 3 \times x$ ⇒ $x = 25$ days

Answer: $25$ 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 $= 3k = 27$ ⇒ $k = 9$

Time taken by A $= 2k = 18$ days

A's 1 day's work $\displaystyle = \frac{1}{18}$, B's 1 day's work $\displaystyle = \frac{1}{27}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{18} + \frac{1}{27} = \frac{3 + 2}{54} = \frac{5}{54}$

Time taken by (A + B) to finish the work $\displaystyle = 10\frac{4}{5}$ days

Answer: $\displaystyle 10\frac{4}{5}$ days

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

$= 8 + 4 = 12$ km/hr

Answer: $12$ 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 $x + y = 21$ and $x - y = 9$

Adding, $2x = 30$ ⇒ $x = 15$ km/hr

Subtracting, $2y = 12$ ⇒ $y = 6$ km/hr

Answer: Rate in still water $15$ km/hr, rate of current $6$ km/hr

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

$\displaystyle = \frac{33}{4} = 8.25$ km/hr

Answer: $8.25$ km/hr

Let the man's rate in still water be $x$ km/hr

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

Given $3(x - 2.5) = x + 2.5$

⇒ $3x - 7.5 = x + 2.5$ ⇒ $2x = 10$

⇒ $x = 5$ km/hr

Answer: $5$ km/hr

Let the required distance be $d$ km

Speed downstream $= 16.5 + 5.5 = 22$ km/hr

Speed upstream $= 16.5 - 5.5 = 11$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{22} + \frac{d}{11} = 3$

⇒ $\displaystyle d\left(\frac{1}{22} + \frac{1}{11}\right) = 3$

⇒ $\displaystyle d \times \frac{3}{22} = 3$

⇒ $d = 22$ km

Answer: $22$ 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 $\displaystyle p = \frac{1}{x - y}$, $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{32}{x - y} + \frac{70}{x + y} = 11$ ⇒ $32p + 70q = 11 \quad \cdots (1)$

$\displaystyle \frac{40}{x - y} + \frac{20}{x + y} = 7$ ⇒ $40p + 20q = 7 \quad \cdots (2)$

(1) $\times 5$ ⇒ $160p + 350q = 55$; (2) $\times 4$ ⇒ $160p + 80q = 28$

(1) $\times 5$ $-$ (2) $\times 4$ ⇒ $270q = 27$ ⇒ $\displaystyle q = \frac{1}{10}$

From (1): $32p + 7 = 11$ ⇒ $32p = 4$ ⇒ $\displaystyle p = \frac{1}{8}$

So $x - y = 8$ and $x + y = 10$

Adding, $2x = 18$ ⇒ $x = 9$; then $y = 10 - 9 = 1$

∴ Rate of the current $= 1$ km/hr, speed in still water $= 9$ km/hr

Answer: Rate of current $1$ km/hr, speed in still water $9$ km/hr