Arithmetic Applications (Level 2): MCQ test, Set 1

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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 can finish a piece of work in $51$ days. What part of the work does A finish in $9$ days?
A $\displaystyle \frac{14}{17}$ B $\displaystyle \frac{17}{3}$ C $\displaystyle \frac{1}{51}$ D $\displaystyle \frac{3}{17}$

Answer and solution

Answer: (D) $\displaystyle \frac{3}{17}$

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

A's 9 days' work $\displaystyle = 9 \times \frac{1}{51} = \frac{9}{51} = \frac{3}{17}$
$7$ men can finish a piece of work in $24$ days. In how many days can $14$ men finish the same work?
A $48$ days B $12$ days C $17$ days D $31$ days

Answer and solution

Answer: (B) $12$ days

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

$7 \times 24 = 14 \times x$

$\displaystyle x = \frac{7 \times 24}{14} = 12$ days
A can finish a piece of work in $55$ days and B can finish it in $11$ days. In how many days can they finish it working together?
A $66$ days B $44$ days C $\displaystyle 9\frac{1}{6}$ days D $33$ days

Answer and solution

Answer: (C) $\displaystyle 9\frac{1}{6}$ days

A's 1 day's work $\displaystyle = \frac{1}{55}$

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

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{55} + \frac{1}{11} = \frac{1 + 5}{55} = \frac{6}{55}$

∴ Both together can finish the work in $\displaystyle \frac{55}{6}$ days $\displaystyle = 9\frac{1}{6}$ days
A can do a piece of work in $21$ days and B in $33$ days. (A + B)'s 1 day's work is
A $\displaystyle \frac{1}{27}$ B $\displaystyle \frac{4}{231}$ C $\displaystyle \frac{1}{54}$ D $\displaystyle \frac{6}{77}$

Answer and solution

Answer: (D) $\displaystyle \frac{6}{77}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{21} + \frac{1}{33} = \frac{11 + 7}{231} = \frac{18}{231} = \frac{6}{77}$
A and B together can finish a piece of work in $40$ days. A alone can do it in $120$ days. In how many days can B alone do it?
A $60$ days B $160$ days C $30$ days D $80$ days

Answer and solution

Answer: (A) $60$ days

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

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

⇒ B's 1 day's work $\displaystyle = \frac{1}{40} - \frac{1}{120} = \frac{3 - 1}{120} = \frac{2}{120} = \frac{1}{60}$

∴ B alone can finish the work in $60$ days.
A and B can do a piece of work in $30$ days; B and C in $36$ days; C and A in $20$ days. In how many days can A alone finish it?
A $45$ days B $36$ days C $180$ days D $18$ days

Answer and solution

Answer: (B) $36$ days

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

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

(C + A)'s 1 day's work $\displaystyle = \frac{1}{20}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{36} + \frac{1}{20} = \frac{6 + 5 + 9}{180} = \frac{20}{180} = \frac{1}{9}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{1}{18}$

∴ A, B, C together can finish the work in $18$ days.

A's 1 day's work $\displaystyle = \frac{1}{18} - \frac{1}{36} = \frac{1}{36}$ , so A alone takes $36$ days
A can do a piece of work in $40$ days and B can finish it in $28$ days. They work together for $7$ days and then B goes away. In how many days will A finish the remaining work?
A $33$ days B $23$ days C $\displaystyle 16\frac{1}{10}$ days D $\displaystyle 37\frac{4}{7}$ days

Answer and solution

Answer: (B) $23$ days

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

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{40} + \frac{1}{28} = \frac{7 + 10}{280} = \frac{17}{280}$

(A + B)'s 7 days' work $\displaystyle = 7 \times \frac{17}{280} = \frac{17}{40}$

Remaining work $\displaystyle = 1 - \frac{17}{40} = \frac{23}{40}$

This work is done by A in $\displaystyle \frac{23}{40} \times 40 = 23$ days
A can do a piece of work in $8$ days while B alone can do it in $48$ days. With the help of C they can finish it in $4$ days. If they are paid ₹3120 for the whole work, what is A's share?
A ₹1300 B ₹1040 C ₹1560 D ₹260

Answer and solution

Answer: (C) ₹1560

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

B's 1 day's work $\displaystyle = \frac{1}{48}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{8} + \frac{1}{48} = \frac{6 + 1}{48} = \frac{7}{48}$

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

C's 1 day's work $\displaystyle = \frac{1}{4} - \frac{7}{48} = \frac{12 - 7}{48} = \frac{5}{48}$

Consider $\displaystyle \frac{1}{8} : \frac{1}{48} : \frac{5}{48} = \frac{6}{48} : \frac{1}{48} : \frac{5}{48} = 6 : 1 : 5$

∴ ₹3120 must be divided in the ratio $6 : 1 : 5$ ( $6 + 1 + 5 = 12$ )

A's share $\displaystyle = 3120 \times \frac{6}{12} = 1560$ , i.e. ₹1560
A is thrice as good a workman as B and is therefore able to finish a piece of work in $22$ days less than B. In how many days can they do it working together?
A $\displaystyle 8\frac{1}{4}$ days B $11$ days C $22$ days D $33$ days

Answer and solution

Answer: (A) $\displaystyle 8\frac{1}{4}$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 22)$ days

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

∴ $\displaystyle x - 22 = \frac{1}{3}x$

⇒ $3x - 66 = x$ ⇒ $2x = 66$ ⇒ $x = 33$

∴ Time taken by B $= 33$ days, time taken by A $= 11$ days

A's 1 day's work $\displaystyle = \frac{1}{11}$ , B's 1 day's work $\displaystyle = \frac{1}{33}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{11} + \frac{1}{33} = \frac{3 + 1}{33} = \frac{4}{33}$

A and B together can finish the work in $\displaystyle 8\frac{1}{4}$ days
A and B can do a piece of work in $30$ and $33$ days respectively. They began the work together, but B leaves after some days and A finishes the remaining work in $9$ days. After how many days did B leave?
A $20$ days B $21$ days C $11$ days D $\displaystyle 23\frac{1}{10}$ days

Answer and solution

Answer: (C) $11$ days

A's 1 day's work $\displaystyle = \frac{1}{30}$ , B's 1 day's work $\displaystyle = \frac{1}{33}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{33} = \frac{11 + 10}{330} = \frac{21}{330} = \frac{7}{110}$

A's 9 days' work $\displaystyle = 9 \times \frac{1}{30} = \frac{3}{10}$

Remaining work $\displaystyle = 1 - \frac{3}{10} = \frac{7}{10}$

(A + B)'s $x$ days' work $\displaystyle = \frac{7}{10}$ ⇒ $\displaystyle \frac{7}{110} \times x = \frac{7}{10}$

$x = 11$ , so B left after $11$ days
A can do a piece of work in $200$ days and B can do it in $168$ days. They work together for $18$ days. Then B leaves and A alone continues the work. $2$ days after that, C joins A and the work is completed in $74$ days more. In how many days can C do the work alone?
A $175$ days B $210$ days C $\displaystyle 93\frac{1}{3}$ days D $101$ days

Answer and solution

Answer: (A) $175$ days

A's 1 day's work $\displaystyle = \frac{1}{200}$ , B's 1 day's work $\displaystyle = \frac{1}{168}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{200} + \frac{1}{168} = \frac{21 + 25}{4200} = \frac{46}{4200} = \frac{23}{2100}$

(A + B)'s 18 days' work $\displaystyle = 18 \times \frac{23}{2100} = \frac{69}{350}$

A's 2 days' work $\displaystyle = \frac{2}{200} = \frac{1}{100}$

(A + C)'s 74 days' work $\displaystyle = 1 - \left(\frac{69}{350} + \frac{1}{100}\right) = \frac{111}{140}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{111}{140} \div 74 = \frac{3}{280}$

C's 1 day's work $\displaystyle = \frac{3}{280} - \frac{1}{200} = \frac{15 - 7}{1400} = \frac{8}{1400} = \frac{1}{175}$

C can finish the work in $175$ days
A and B together can do a piece of work in $24$ days. B and C together can do it in $40$ days. If A is twice as good a workman as C, B alone can do it in
A $16$ days B $30$ days C $120$ days D $60$ days

Answer and solution

Answer: (C) $120$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{24}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{40}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{24}$

⇒ C's 1 day's work + (B + C)'s 1 day's work $\displaystyle = \frac{1}{24}$

⇒ C's 1 day's work $\displaystyle + \frac{1}{40} = \frac{1}{24}$

C's 1 day's work $\displaystyle = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{2}{120} = \frac{1}{60}$

B's 1 day's work $\displaystyle = \frac{1}{40} - \frac{1}{60} = \frac{3 - 2}{120} = \frac{1}{120}$

∴ B alone can finish the work in $120$ days
A and B working alone can finish a piece of work in $23$ and $12$ days respectively. If they work at it alternately for a day, B beginning, in how many days will the work be finished?
A $15$ days B $\displaystyle 15\frac{2}{3}$ days C $\displaystyle 16\frac{2}{3}$ days D $16$ days

Answer and solution

Answer: (B) $\displaystyle 15\frac{2}{3}$ days

A's 1 day's work $\displaystyle = \frac{1}{23}$ , B's 1 day's work $\displaystyle = \frac{1}{12}$

Work done in the first 2 days $\displaystyle = \frac{1}{12} + \frac{1}{23} = \frac{23 + 12}{276} = \frac{35}{276}$

Work done in the first 14 days $\displaystyle = 7 \times \frac{35}{276} = \frac{245}{276}$

On day 15 it is B's turn: work done in the first 15 days $\displaystyle = \frac{245}{276} + \frac{1}{12} = \frac{67}{69}$

Remaining work $\displaystyle = 1 - \frac{67}{69} = \frac{2}{69}$

A does it in $\displaystyle \frac{2}{69} \div \frac{1}{23} = \frac{2}{3}$ day

∴ Total no. of days $\displaystyle = 15 + \frac{2}{3} = 15\frac{2}{3}$ days
A certain number of men complete a piece of work in $52$ days. If there were $4$ men more, the work could be finished in $4$ days less. How many men were there originally?
A $52$ B $56$ C $48$ D $44$

Answer and solution

Answer: (C) $48$

Let the original number of men be $x$

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

$x \times 52 = (x + 4) \times 48$

$52x = 48x + 192$

$4x = 192$ ⇒ $x = 48$
If $3$ men or $12$ boys can finish a piece of work in $9$ days, in how many days can $4$ men and $11$ boys finish it?
A $4$ days B $9$ days C $\displaystyle 7\frac{1}{5}$ days D $\displaystyle 20\frac{1}{4}$ days

Answer and solution

Answer: (A) $4$ days

$3$ men $= 12$ boys

$4$ men $\displaystyle = \frac{12}{3} \times 4 = 16$ boys

$4$ men + $11$ boys $= 16$ boys + $11$ boys $= 27$ boys

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

$12 \times 9 = 27 \times x$ ⇒ $x = 4$ days
$5$ men and $3$ women working $3$ hours a day finish a piece of work in $12$ days; $8$ men and $3$ women working $4$ hours a day do the same work in $6$ days. Find the number of days in which the work is done by $6$ men only, working $3$ hours a day.
A $28$ days B $16$ days C $12$ days D $36$ days

Answer and solution

Answer: (C) $12$ days

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

Case (i): $(5m + 3w) \times 3 \times 12 = 180m + 108w$

Case (ii): $(8m + 3w) \times 4 \times 6 = 192m + 72w$

Same work: $180m + 108w = 192m + 72w$

⇒ $36w = 12m$ ⇒ $3w = m$

So $3$ women $= 1$ men, and $5m + 3w = 6$ men

$6$ men working $3$ hours a day for $12$ days do the same work as $6$ men working $3$ hours a day for $x$ days

$6 \times 3 \times 12 = 6 \times 3 \times x$ ⇒ $x = 12$ days
If $4$ men with $5$ boys can earn ₹5720 in $8$ days and $8$ men with $12$ boys can earn ₹13500 in $9$ days, in what time will $4$ men with $2$ boys earn ₹3660?
A $\displaystyle 6\frac{7}{9}$ days B $8$ days C $3$ days D $6$ days

Answer and solution

Answer: (D) $6$ days

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

$(4x + 5y) \times 8 = 5720$ ⇒ $4x + 5y = 715 \quad \cdots (1)$

$(8x + 12y) \times 9 = 13500$ ⇒ $8x + 12y = 1500 \quad \cdots (2)$

(1) $\times 2$ ⇒ $8x + 10y = 1430$

(2) $-$ (1) $\times 2$ ⇒ $2y = 70$ ⇒ $y = 35$

From (1): $4x = 715 - 175 = 540$ ⇒ $x = 135$

$(4x + 2y)n = 3660$

⇒ $(540 + 70)n = 3660$ ⇒ $610n = 3660$

⇒ $n = 6$ days
$20$ men are employed to do a piece of work which they could finish in $15$ days, but the men drop off by $4$ at the end of every $6$ days. In what time will the work be completed?
A $\displaystyle 19\frac{1}{2}$ days B $15$ days C $24$ days D $\displaystyle 18\frac{3}{4}$ days

Answer and solution

Answer: (A) $\displaystyle 19\frac{1}{2}$ days

Work done by 20 men in 6 days $\displaystyle = 6 \times \frac{1}{15} = \frac{2}{5}$

So 1 man's work in 6 days $\displaystyle = \frac{2}{5} \div 20 = \frac{1}{50}$

Remaining work $\displaystyle = \frac{3}{5}$ , remaining men $= 16$ ; work done by 16 men in 6 days $\displaystyle = 16 \times \frac{1}{50} = \frac{8}{25}$

Remaining work $\displaystyle = \frac{7}{25}$ , remaining men $= 12$ ; work done by 12 men in 6 days $\displaystyle = 12 \times \frac{1}{50} = \frac{6}{25}$

Remaining work $\displaystyle = \frac{1}{25}$ , remaining men $= 8$ ; work done by 8 men in 6 days $\displaystyle = \frac{4}{25}$

$\displaystyle \frac{1}{25}$ work is done by 8 men in $\displaystyle \frac{1}{25} \div \frac{4}{25} \times 6 = 1\frac{1}{2}$ days

Total time $\displaystyle = 6 + 6 + 6 + 1\frac{1}{2} = 19\frac{1}{2}$ days
A can do half as much again as B can do in the same time. A alone can do a piece of work in $32$ days. In how many days can B alone finish it?
A $64$ days B $\displaystyle 21\frac{1}{3}$ days C $16$ days D $48$ days

Answer and solution

Answer: (D) $48$ 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 A $= 2k = 32$ ⇒ $k = 16$

Time taken by B $= 3k = 48$ days
A man can row downstream at the rate of $15$ km/hr and upstream at $5$ km/hr. The man's rate in still water is
A $7.5$ km/hr B $5$ km/hr C $20$ km/hr D $10$ km/hr

Answer and solution

Answer: (D) $10$ km/hr

Given $x + y = 15$ and $x - y = 5$

Adding, $2x = 20$ ⇒ $x = 10$ km/hr
A man rows upstream $29$ km and downstream $59$ km, taking $5$ hours each time. The man's speed in still water is
A $8.8$ km/hr B $17.6$ km/hr C $11.8$ km/hr D $3$ km/hr

Answer and solution

Answer: (A) $8.8$ km/hr

Speed downstream: $\displaystyle x + y = \frac{59}{5}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{29}{5}$ km/hr

Adding, $\displaystyle 2x = \frac{59}{5} + \frac{29}{5} = \frac{88}{5}$

$\displaystyle x = \frac{88}{10} = 8.8$ km/hr
A man can row $8$ 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 $6$ km/hr B $5$ km/hr C $2$ km/hr D $4$ km/hr

Answer and solution

Answer: (D) $4$ 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 $3(8 - y) = 8 + y$

⇒ $24 - 3y = 8 + y$ ⇒ $4y = 16$

⇒ $y = 4$ km/hr
A man can row $11$ km/hr in still water. If the river is running at $5.5$ km/hr, it takes him 1 hour to row to a place and back. The place is at a distance of
A $5.5$ km B $4.125$ km C $8.25$ km D $5.125$ km

Answer and solution

Answer: (B) $4.125$ km

Let the required distance be $d$ km

Speed downstream $= 11 + 5.5 = 16.5$ km/hr

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

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{16.5} + \frac{d}{5.5} = 1$

⇒ $\displaystyle d\left(\frac{2}{33} + \frac{2}{11}\right) = 1$

⇒ $\displaystyle d \times \frac{8}{33} = 1$

⇒ $d = 4.125$ km
The current of a stream runs at the rate of $3$ km/hr. A motor boat goes $10$ km upstream and back again to the starting point in 2 hours 30 minutes. The speed of the motor boat in still water is
A $6$ km/hr B $9$ km/hr C $15$ km/hr D $12$ km/hr

Answer and solution

Answer: (B) $9$ km/hr

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

Speed upstream $= (x - 3)$ km/hr, speed downstream $= (x + 3)$ km/hr

Total time $= 150$ min $\displaystyle = \frac{150}{60}$ hr $\displaystyle = \frac{5}{2}$ hr

$\displaystyle \frac{10}{x - 3} + \frac{10}{x + 3} = \frac{5}{2}$

⇒ $\displaystyle 10\left(\frac{(x + 3) + (x - 3)}{(x - 3)(x + 3)}\right) = \frac{5}{2}$

⇒ $\displaystyle \frac{20x}{x^2 - 9} = \frac{5}{2}$

⇒ $40x = 5x^2 - 45$

⇒ $x^{2} - 8 x - 9 = 0$

⇒ $\left(x - 9\right) \left(x + 1\right) = 0$

⇒ $x = 9$ (or) $x = -1$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 9$ km/hr
A man can row $6$ km upstream and $84$ km downstream in $7$ hours, while he can row $24$ km upstream and $84$ km downstream in $10$ hours. The speed of the man in still water is
A $6$ km/hr B $4$ km/hr C $10$ km/hr D $14$ km/hr

Answer and solution

Answer: (C) $10$ 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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{6}{x - y} + \frac{84}{x + y} = 7$ ⇒ $6p + 84q = 7 \quad \cdots (1)$

$\displaystyle \frac{24}{x - y} + \frac{84}{x + y} = 10$ ⇒ $24p + 84q = 10 \quad \cdots (2)$

(1) $\times 4$ ⇒ $24p + 336q = 28$

(1) $\times 4$ $-$ (2) ⇒ $252q = 18$ ⇒ $\displaystyle q = \frac{1}{14}$

From (1): $6p + 6 = 7$ ⇒ $6p = 1$ ⇒ $\displaystyle p = \frac{1}{6}$

So $x - y = 6$ and $x + y = 14$

Adding, $2x = 20$ ⇒ $x = 10$ ; then $y = 14 - 10 = 4$

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MCQ test, Set 1