Arithmetic Applications (Level 2): MCQ test, Set 5

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

A is twice as good a workman as B. If B can finish a piece of work in $58$ days, in how many days can A finish it?
A $116$ days B $\displaystyle 19\frac{1}{3}$ days C $29$ days D $\displaystyle 38\frac{2}{3}$ days

Answer and solution

Answer: (C) $29$ days

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

$\displaystyle = \frac{1}{2} \times 58 = 29$ days
A can finish a piece of work in $56$ days. What part of the work does A finish in $6$ days?
A $\displaystyle \frac{1}{56}$ B $\displaystyle \frac{3}{28}$ C $\displaystyle \frac{25}{28}$ D $\displaystyle \frac{28}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{3}{28}$

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

A's 6 days' work $\displaystyle = 6 \times \frac{1}{56} = \frac{6}{56} = \frac{3}{28}$
$12$ men can do a piece of work in $7$ days. How many men are needed to finish it in $14$ days?
A $6$ men B $5$ men C $24$ men D $19$ men

Answer and solution

Answer: (A) $6$ men

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

$12 \times 7 = x \times 14$

$\displaystyle x = \frac{12 \times 7}{14} = 6$ men
A can do a piece of work in $32$ days and B alone can do it in $64$ days. How much time will both take to finish the work?
A $32$ days B $\displaystyle 21\frac{1}{3}$ days C $96$ days D $48$ days

Answer and solution

Answer: (B) $\displaystyle 21\frac{1}{3}$ days

A's 1 day's work $\displaystyle = \frac{1}{32}$

B's 1 day's work $\displaystyle = \frac{1}{64}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{32} + \frac{1}{64} = \frac{2 + 1}{64} = \frac{3}{64}$

∴ Both together can finish the work in $\displaystyle \frac{64}{3}$ days $\displaystyle = 21\frac{1}{3}$ days
A and B together can do a piece of work in $16$ days. B alone can finish it in $80$ days. In how many days can A alone finish the work?
A $48$ days B $20$ days C $\displaystyle 13\frac{1}{3}$ days D $64$ days

Answer and solution

Answer: (B) $20$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{16}$

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

⇒ A's 1 day's work $\displaystyle = \frac{1}{16} - \frac{1}{80} = \frac{5 - 1}{80} = \frac{4}{80} = \frac{1}{20}$

∴ A alone can finish the work in $20$ days.
A and B can do a piece of work in $112$ days; B and C in $42$ days; C and A in $48$ days. In how many days can A alone finish it?
A $336$ days B $\displaystyle 37\frac{1}{3}$ days C $56$ days D $168$ days

Answer and solution

Answer: (A) $336$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{112}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{42}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{48}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{112} + \frac{1}{42} + \frac{1}{48} = \frac{3 + 8 + 7}{336} = \frac{18}{336} = \frac{3}{56}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{3}{112}$

∴ A, B, C together can finish the work in $\displaystyle 37\frac{1}{3}$ days.

A's 1 day's work $\displaystyle = \frac{3}{112} - \frac{1}{42} = \frac{1}{336}$ , so A alone takes $336$ days
A can do a piece of work in $11$ days and B can finish it in $44$ days. They work together for $5$ days and then A goes away. In how many days will B finish the remaining work?
A $39$ days B $\displaystyle 4\frac{3}{4}$ days C $\displaystyle 19\frac{1}{2}$ days D $19$ days

Answer and solution

Answer: (D) $19$ days

A's 1 day's work $\displaystyle = \frac{1}{11}$

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{11} + \frac{1}{44} = \frac{4 + 1}{44} = \frac{5}{44}$

(A + B)'s 5 days' work $\displaystyle = 5 \times \frac{5}{44} = \frac{25}{44}$

Remaining work $\displaystyle = 1 - \frac{25}{44} = \frac{19}{44}$

This work is done by B in $\displaystyle \frac{19}{44} \times 44 = 19$ days
A can do a piece of work in $24$ days while B alone can do it in $30$ days. With the help of C they can finish it in $6$ days. If they are paid ₹9000 for the whole work, what is B's share?
A ₹2250 B ₹3000 C ₹4950 D ₹1800

Answer and solution

Answer: (D) ₹1800

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

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{24} + \frac{1}{30} = \frac{5 + 4}{120} = \frac{9}{120} = \frac{3}{40}$

Given (A + B + C)'s 1 day's work $\displaystyle = \frac{1}{6}$

C's 1 day's work $\displaystyle = \frac{1}{6} - \frac{3}{40} = \frac{20 - 9}{120} = \frac{11}{120}$

Consider $\displaystyle \frac{1}{24} : \frac{1}{30} : \frac{11}{120} = \frac{5}{120} : \frac{4}{120} : \frac{11}{120} = 5 : 4 : 11$

∴ ₹9000 must be divided in the ratio $5 : 4 : 11$ ( $5 + 4 + 11 = 20$ )

B's share $\displaystyle = 9000 \times \frac{4}{20} = 1800$ , i.e. ₹1800
A is twice as good a workman as B and is therefore able to finish a piece of work in $56$ days less than B. In how many days can they do it working together?
A $84$ days B $112$ days C $\displaystyle 37\frac{1}{3}$ days D $56$ days

Answer and solution

Answer: (C) $\displaystyle 37\frac{1}{3}$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 56)$ days

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

∴ $\displaystyle x - 56 = \frac{1}{2}x$

⇒ $2x - 112 = x$ ⇒ $x = 112$

∴ Time taken by B $= 112$ days, time taken by A $= 56$ days

A's 1 day's work $\displaystyle = \frac{1}{56}$ , B's 1 day's work $\displaystyle = \frac{1}{112}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{56} + \frac{1}{112} = \frac{2 + 1}{112} = \frac{3}{112}$

A and B together can finish the work in $\displaystyle 37\frac{1}{3}$ days
A and B can do a piece of work in $64$ and $32$ days respectively. They began the work together, but A leaves after some days and B finishes the remaining work in $5$ days. After how many days did A leave?
A $18$ days B $54$ days C $23$ days D $27$ days

Answer and solution

Answer: (A) $18$ days

A's 1 day's work $\displaystyle = \frac{1}{64}$ , B's 1 day's work $\displaystyle = \frac{1}{32}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{64} + \frac{1}{32} = \frac{1 + 2}{64} = \frac{3}{64}$

B's 5 days' work $\displaystyle = 5 \times \frac{1}{32} = \frac{5}{32}$

Remaining work $\displaystyle = 1 - \frac{5}{32} = \frac{27}{32}$

(A + B)'s $x$ days' work $\displaystyle = \frac{27}{32}$ ⇒ $\displaystyle \frac{3}{64} \times x = \frac{27}{32}$

$x = 18$ , so A left after $18$ days
A can do a piece of work in $208$ days and B can do it in $104$ days. They work together for $14$ days. Then B leaves and A alone continues the work. $5$ days after that, C joins A and the work is completed in $69$ days more. In how many days can C do the work alone?
A $87$ days B $624$ days C $\displaystyle 89\frac{1}{7}$ days D $156$ days

Answer and solution

Answer: (D) $156$ days

A's 1 day's work $\displaystyle = \frac{1}{208}$ , B's 1 day's work $\displaystyle = \frac{1}{104}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{208} + \frac{1}{104} = \frac{1 + 2}{208} = \frac{3}{208}$

(A + B)'s 14 days' work $\displaystyle = 14 \times \frac{3}{208} = \frac{21}{104}$

A's 5 days' work $\displaystyle = \frac{5}{208}$

(A + C)'s 69 days' work $\displaystyle = 1 - \left(\frac{21}{104} + \frac{5}{208}\right) = \frac{161}{208}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{161}{208} \div 69 = \frac{7}{624}$

C's 1 day's work $\displaystyle = \frac{7}{624} - \frac{1}{208} = \frac{7 - 3}{624} = \frac{4}{624} = \frac{1}{156}$

C can finish the work in $156$ days
A and B together can do a piece of work in $9$ days. B and C together can do it in $12$ days. If A is twice as good a workman as C, B alone can do it in
A $72$ days B $3$ days C $36$ days D $18$ days

Answer and solution

Answer: (D) $18$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{9}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{12}$

A's 1 day's work $=$ C's 2 days' work

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{9}$

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

⇒ C's 1 day's work $\displaystyle + \frac{1}{12} = \frac{1}{9}$

C's 1 day's work $\displaystyle = \frac{1}{9} - \frac{1}{12} = \frac{4 - 3}{36} = \frac{1}{36}$

B's 1 day's work $\displaystyle = \frac{1}{12} - \frac{1}{36} = \frac{3 - 1}{36} = \frac{2}{36} = \frac{1}{18}$

∴ B alone can finish the work in $18$ days
A and B working alone can finish a piece of work in $11$ and $24$ days respectively. If they work at it alternately for a day, B beginning, in how many days will the work be finished?
A $\displaystyle 15\frac{1}{3}$ days B $15$ days C $\displaystyle 16\frac{1}{3}$ days D $16$ days

Answer and solution

Answer: (A) $\displaystyle 15\frac{1}{3}$ days

A's 1 day's work $\displaystyle = \frac{1}{11}$ , B's 1 day's work $\displaystyle = \frac{1}{24}$

Work done in the first 2 days $\displaystyle = \frac{1}{24} + \frac{1}{11} = \frac{11 + 24}{264} = \frac{35}{264}$

Work done in the first 14 days $\displaystyle = 7 \times \frac{35}{264} = \frac{245}{264}$

On day 15 it is B's turn: work done in the first 15 days $\displaystyle = \frac{245}{264} + \frac{1}{24} = \frac{32}{33}$

Remaining work $\displaystyle = 1 - \frac{32}{33} = \frac{1}{33}$

A does it in $\displaystyle \frac{1}{33} \div \frac{1}{11} = \frac{1}{3}$ day

∴ Total no. of days $\displaystyle = 15 + \frac{1}{3} = 15\frac{1}{3}$ days
A certain number of men complete a piece of work in $26$ days. If there were $2$ men more, the work could be finished in $2$ days less. How many men were there originally?
A $22$ B $26$ C $24$ D $28$

Answer and solution

Answer: (C) $24$

Let the original number of men be $x$

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

$x \times 26 = (x + 2) \times 24$

$26x = 24x + 48$

$2x = 48$ ⇒ $x = 24$
If $8$ men or $12$ boys can finish a piece of work in $38$ days, in how many days can $12$ men and $6$ boys finish it?
A $19$ days B $\displaystyle 32\frac{4}{7}$ days C $76$ days D $\displaystyle 25\frac{1}{3}$ days

Answer and solution

Answer: (A) $19$ days

$8$ men $= 12$ boys

$12$ men $\displaystyle = \frac{12}{8} \times 12 = 18$ boys

$12$ men + $6$ boys $= 18$ boys + $6$ boys $= 24$ boys

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

$12 \times 38 = 24 \times x$ ⇒ $x = 19$ days
$8$ men and $4$ women working $3$ hours a day finish a piece of work in $8$ days; $4$ men and $8$ women working $6$ hours a day do the same work in $5$ days. Find the number of days in which the work is done by $4$ men only, working $4$ hours a day.
A $15$ days B $20$ days C $18$ days D $24$ days

Answer and solution

Answer: (A) $15$ days

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

Case (i): $(8m + 4w) \times 3 \times 8 = 192m + 96w$

Case (ii): $(4m + 8w) \times 6 \times 5 = 120m + 240w$

Same work: $192m + 96w = 120m + 240w$

⇒ $144w = 72m$ ⇒ $2w = m$

So $4$ women $= 2$ men, and $8m + 4w = 10$ men

$10$ men working $3$ hours a day for $8$ days do the same work as $4$ men working $4$ hours a day for $x$ days

$10 \times 3 \times 8 = 4 \times 4 \times x$ ⇒ $x = 15$ days
If $4$ men with $5$ boys can earn ₹8155 in $7$ days and $12$ men with $3$ boys can earn ₹19950 in $10$ days, in what time will $7$ men with $5$ boys earn ₹18840?
A $10$ days B $14$ days C $6$ days D $12$ days

Answer and solution

Answer: (D) $12$ days

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

$(4x + 5y) \times 7 = 8155$ ⇒ $4x + 5y = 1165 \quad \cdots (1)$

$(12x + 3y) \times 10 = 19950$ ⇒ $12x + 3y = 1995 \quad \cdots (2)$

(1) $\times 3$ ⇒ $12x + 15y = 3495$

(1) $\times 3$ $-$ (2) ⇒ $12y = 1500$ ⇒ $y = 125$

From (1): $4x = 1165 - 625 = 540$ ⇒ $x = 135$

$(7x + 5y)n = 18840$

⇒ $(945 + 625)n = 18840$ ⇒ $1570n = 18840$

⇒ $n = 12$ days
$48$ men are employed to do a piece of work which they could finish in $9$ days, but the men drop off by $6$ at the end of every $3$ days. In what time will the work be completed?
A $\displaystyle 10\frac{2}{7}$ days B $12$ days C $\displaystyle 10\frac{4}{5}$ days D $9$ days

Answer and solution

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

Work done by 48 men in 3 days $\displaystyle = 3 \times \frac{1}{9} = \frac{1}{3}$

So 1 man's work in 3 days $\displaystyle = \frac{1}{3} \div 48 = \frac{1}{144}$

Remaining work $\displaystyle = \frac{2}{3}$ , remaining men $= 42$ ; work done by 42 men in 3 days $\displaystyle = 42 \times \frac{1}{144} = \frac{7}{24}$

Remaining work $\displaystyle = \frac{3}{8}$ , remaining men $= 36$ ; work done by 36 men in 3 days $\displaystyle = 36 \times \frac{1}{144} = \frac{1}{4}$

Remaining work $\displaystyle = \frac{1}{8}$ , remaining men $= 30$ ; work done by 30 men in 3 days $\displaystyle = \frac{5}{24}$

$\displaystyle \frac{1}{8}$ work is done by 30 men in $\displaystyle \frac{1}{8} \div \frac{5}{24} \times 3 = 1\frac{4}{5}$ days

Total time $\displaystyle = 3 + 3 + 3 + 1\frac{4}{5} = 10\frac{4}{5}$ days
A can do half as much again as B can do in the same time. B alone can do a piece of work in $54$ days. In how many days can A alone finish it?
A $81$ days B $18$ days C $27$ days D $36$ days

Answer and solution

Answer: (D) $36$ 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 $= 3k = 54$ ⇒ $k = 18$

Time taken by A $= 2k = 36$ days
A man can row downstream at the rate of $22$ km/hr and upstream at $16$ km/hr. The man's rate in still water is
A $3$ km/hr B $19$ km/hr C $6$ km/hr D $38$ km/hr

Answer and solution

Answer: (B) $19$ km/hr

Given $x + y = 22$ and $x - y = 16$

Adding, $2x = 38$ ⇒ $x = 19$ km/hr
A man rows upstream $12$ km and downstream $21$ km, taking $2$ hours each time. The man's speed in still water is
A $10.5$ km/hr B $2.25$ km/hr C $8.25$ km/hr D $16.5$ km/hr

Answer and solution

Answer: (C) $8.25$ km/hr

Speed downstream: $\displaystyle x + y = \frac{21}{2}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{12}{2}$ km/hr

Adding, $\displaystyle 2x = \frac{21}{2} + \frac{12}{2} = \frac{33}{2}$

$\displaystyle x = \frac{33}{4} = 8.25$ km/hr
A man can row $2$ km/hr in still water and he finds that it takes him thrice as long to row up as to row down the river. The rate of the stream is
A $1.5$ km/hr B $1$ km/hr C $0.5$ km/hr D $2$ km/hr

Answer and solution

Answer: (B) $1$ 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 $3(2 - y) = 2 + y$

⇒ $6 - 3y = 2 + y$ ⇒ $4y = 4$

⇒ $y = 1$ km/hr
A man can row $8$ km/hr in still water. If the river is running at $4$ km/hr, it takes him 2 hours to row to a place and back. The place is at a distance of
A $6$ km B $12$ km C $7$ km D $8$ km

Answer and solution

Answer: (A) $6$ km

Let the required distance be $d$ km

Speed downstream $= 8 + 4 = 12$ km/hr

Speed upstream $= 8 - 4 = 4$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{12} + \frac{d}{4} = 2$

⇒ $\displaystyle d\left(\frac{1}{12} + \frac{1}{4}\right) = 2$

⇒ $\displaystyle d \times \frac{1}{3} = 2$

⇒ $d = 6$ km
The current of a stream runs at the rate of $5$ km/hr. A motor boat goes $40$ km upstream and back again to the starting point in 3 hours 20 minutes. The speed of the motor boat in still water is
A $35$ km/hr B $30$ km/hr C $25$ km/hr D $20$ km/hr

Answer and solution

Answer: (C) $25$ km/hr

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

Speed upstream $= (x - 5)$ km/hr, speed downstream $= (x + 5)$ km/hr

Total time $= 200$ min $\displaystyle = \frac{200}{60}$ hr $\displaystyle = \frac{10}{3}$ hr

$\displaystyle \frac{40}{x - 5} + \frac{40}{x + 5} = \frac{10}{3}$

⇒ $\displaystyle 40\left(\frac{(x + 5) + (x - 5)}{(x - 5)(x + 5)}\right) = \frac{10}{3}$

⇒ $\displaystyle \frac{80x}{x^2 - 25} = \frac{10}{3}$

⇒ $240x = 10x^2 - 250$

⇒ $x^{2} - 24 x - 25 = 0$

⇒ $\left(x - 25\right) \left(x + 1\right) = 0$

⇒ $x = 25$ (or) $x = -1$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 25$ km/hr
A man can row $30$ km upstream and $84$ km downstream in $9$ hours, while he can row $70$ km upstream and $14$ km downstream in $8$ hours. The speed of the man in still water is
A $10$ km/hr B $12$ km/hr C $2$ km/hr D $14$ km/hr

Answer and solution

Answer: (B) $12$ km/hr

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

$\displaystyle \frac{30}{x - y} + \frac{84}{x + y} = 9$ ⇒ $30p + 84q = 9 \quad \cdots (1)$

$\displaystyle \frac{70}{x - y} + \frac{14}{x + y} = 8$ ⇒ $70p + 14q = 8 \quad \cdots (2)$

(1) $\times 7$ ⇒ $210p + 588q = 63$ ; (2) $\times 3$ ⇒ $210p + 42q = 24$

(1) $\times 7$ $-$ (2) $\times 3$ ⇒ $546q = 39$ ⇒ $\displaystyle q = \frac{1}{14}$

From (1): $30p + 6 = 9$ ⇒ $30p = 3$ ⇒ $\displaystyle p = \frac{1}{10}$

So $x - y = 10$ and $x + y = 14$

Adding, $2x = 24$ ⇒ $x = 12$ ; then $y = 14 - 12 = 2$

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MCQ test, Set 5