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

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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. A can finish a piece of work in $27$ days. How many days will B take to finish it?
A $29$ days B $\displaystyle 13\frac{1}{2}$ days C $54$ days D $81$ days

Answer and solution

Answer: (C) $54$ days

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

$\displaystyle 27 = \frac{1}{2} \times$ (Time taken by B)

∴ Time taken by B $= 2 \times 27 = 54$ days
If the number of men to do a work is changed in the ratio $3 : 4$ , the time taken by them to finish the work changes in the ratio
A $3 : 4$ B $4 : 3$ C $9 : 16$ D $16 : 9$

Answer and solution

Answer: (B) $4 : 3$

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{4}{3}$

∴ $T_1 : T_2 = 4 : 3$
A can finish a piece of work in $35$ days and B can finish it in $15$ days. In how many days can they finish it working together?
A $20$ days B $\displaystyle 10\frac{1}{2}$ days C $50$ days D $25$ days

Answer and solution

Answer: (B) $\displaystyle 10\frac{1}{2}$ days

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

B's 1 day's work $\displaystyle = \frac{1}{15}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{35} + \frac{1}{15} = \frac{3 + 7}{105} = \frac{10}{105} = \frac{2}{21}$

∴ Both together can finish the work in $\displaystyle \frac{21}{2}$ days $\displaystyle = 10\frac{1}{2}$ days
A and B together can finish a piece of work in $28$ days. A alone can do it in $77$ days. In how many days can B alone do it?
A $44$ days B $49$ days C $105$ days D $\displaystyle 52\frac{1}{2}$ days

Answer and solution

Answer: (A) $44$ days

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

A's 1 day's work $\displaystyle = \frac{1}{77}$

⇒ B's 1 day's work $\displaystyle = \frac{1}{28} - \frac{1}{77} = \frac{11 - 4}{308} = \frac{7}{308} = \frac{1}{44}$

∴ B alone can finish the work in $44$ days.
A, B and C together can finish a piece of work in $24$ days. B and C together can finish it in $60$ days. In how many days can A alone finish it?
A $36$ days B $\displaystyle 17\frac{1}{7}$ days C $42$ days D $40$ days

Answer and solution

Answer: (D) $40$ days

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

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

A's 1 day's work $=$ (A + B + C)'s 1 day's work $-$ (B + C)'s 1 day's work

$\displaystyle = \frac{1}{24} - \frac{1}{60} = \frac{5 - 2}{120} = \frac{3}{120} = \frac{1}{40}$

∴ A alone can finish the work in $40$ days
A can do a piece of work in $36$ days and B can finish it in $18$ 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 $15$ days C $29$ days D $\displaystyle 7\frac{1}{2}$ days

Answer and solution

Answer: (B) $15$ days

A's 1 day's work $\displaystyle = \frac{1}{36}$

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{36} + \frac{1}{18} = \frac{1 + 2}{36} = \frac{3}{36} = \frac{1}{12}$

(A + B)'s 7 days' work $\displaystyle = 7 \times \frac{1}{12} = \frac{7}{12}$

Remaining work $\displaystyle = 1 - \frac{7}{12} = \frac{5}{12}$

This work is done by A in $\displaystyle \frac{5}{12} \times 36 = 15$ days
A can do a piece of work in $40$ days and B in $12$ days. They work together for $5$ days. The rest of the work is finished by C in $4$ days. If they get ₹960 for the whole work, what is C's daily wage?
A ₹440 B ₹80 C ₹24 D ₹110

Answer and solution

Answer: (D) ₹110

A's 1 day's work $\displaystyle = \frac{1}{40}$ , B's 1 day's work $\displaystyle = \frac{1}{12}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{40} + \frac{1}{12} = \frac{3 + 10}{120} = \frac{13}{120}$

(A + B)'s 5 days' work $\displaystyle = \frac{13}{24}$

Remaining work $\displaystyle = 1 - \frac{13}{24} = \frac{11}{24}$ , done by C in $4$ days

A's 5 days' work $\displaystyle = \frac{5}{40} = \frac{1}{8}$ , B's 5 days' work $\displaystyle = \frac{5}{12}$ , C's 4 days' work $\displaystyle = \frac{11}{24}$

Consider $\displaystyle \frac{1}{8} : \frac{5}{12} : \frac{11}{24} = 3 : 10 : 11$

A's share $\displaystyle = 3 \times \frac{960}{24} = 120$ , i.e. ₹120

B's share $\displaystyle = 10 \times \frac{960}{24} = 400$ , i.e. ₹400

C's share $\displaystyle = 11 \times \frac{960}{24} = 440$ , i.e. ₹440

C's daily wage $\displaystyle = \frac{440}{4} = 110$ , i.e. ₹110
A is thrice as good a workman as B and is therefore able to finish a piece of work in $20$ days less than B. How many days does A take to finish it alone?
A $10$ days B $60$ days C $30$ days D $\displaystyle 6\frac{2}{3}$ days

Answer and solution

Answer: (A) $10$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 20)$ days

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

∴ $\displaystyle x - 20 = \frac{1}{3}x$

⇒ $3x - 60 = x$ ⇒ $2x = 60$ ⇒ $x = 30$

∴ Time taken by B $= 30$ days, time taken by A $= 10$ days
A can do a piece of work in $32$ days, B in $24$ days and C in $16$ days. All begin together, but A leaves the work after $4$ days and B leaves $4$ days before the work is finished. How long did the work last?
A $10$ days B $14$ days C $6$ days D $2$ days

Answer and solution

Answer: (A) $10$ days

A's 1 day's work $\displaystyle = \frac{1}{32}$ , B's $\displaystyle = \frac{1}{24}$ , C's $\displaystyle = \frac{1}{16}$

A, B, C together work for $4$ days; C alone works for the last $4$ days; B and C work together in between.

(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{32} + \frac{1}{24} + \frac{1}{16} = \frac{3 + 4 + 6}{96} = \frac{13}{96}$

(A + B + C)'s 4 days' work $\displaystyle = 4 \times \frac{13}{96} = \frac{13}{24}$

C's 4 days' work $\displaystyle = 4 \times \frac{1}{16} = \frac{1}{4}$

Remaining work $\displaystyle = 1 - \left(\frac{13}{24} + \frac{1}{4}\right) = \frac{5}{24}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{24} + \frac{1}{16} = \frac{2 + 3}{48} = \frac{5}{48}$

Remaining work is done by (B + C) in $\displaystyle \frac{5}{24} \div \frac{5}{48} = 2$ days

Total time $= 4 + 2 + 4 = 10$ days
A can do a piece of work in $56$ days and B can do it in $160$ days. They work together for $12$ days. Then B leaves and A alone continues the work. $17$ days after that, C joins A and the work is completed in $6$ days more. In how many days can C do the work alone?
A $14$ days B $20$ days C $76$ days D $40$ days

Answer and solution

Answer: (B) $20$ days

A's 1 day's work $\displaystyle = \frac{1}{56}$ , B's 1 day's work $\displaystyle = \frac{1}{160}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{56} + \frac{1}{160} = \frac{20 + 7}{1120} = \frac{27}{1120}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{27}{1120} = \frac{81}{280}$

A's 17 days' work $\displaystyle = \frac{17}{56}$

(A + C)'s 6 days' work $\displaystyle = 1 - \left(\frac{81}{280} + \frac{17}{56}\right) = \frac{57}{140}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{57}{140} \div 6 = \frac{19}{280}$

C's 1 day's work $\displaystyle = \frac{19}{280} - \frac{1}{56} = \frac{19 - 5}{280} = \frac{14}{280} = \frac{1}{20}$

C can finish the work in $20$ days
A and B together can do a piece of work in $27$ days. B and C together can do it in $36$ days. If A is twice as good a workman as C, A alone can do it in
A $216$ days B $54$ days C $9$ days D $108$ days

Answer and solution

Answer: (B) $54$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{27}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{36}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{27}$

⇒ C's 1 day's work + (B + C)'s 1 day's work $\displaystyle = \frac{1}{27}$

⇒ C's 1 day's work $\displaystyle + \frac{1}{36} = \frac{1}{27}$

C's 1 day's work $\displaystyle = \frac{1}{27} - \frac{1}{36} = \frac{4 - 3}{108} = \frac{1}{108}$

A's 1 day's work $\displaystyle = 2 \times \frac{1}{108} = \frac{1}{54}$ , so A alone takes $54$ days
A and B working alone can finish a piece of work in $16$ and $4$ days respectively. If they work at it alternately for a day, B beginning, in how many days will the work be finished?
A $7$ days B $\displaystyle 6\frac{2}{5}$ days C $\displaystyle 6\frac{1}{4}$ days D $\displaystyle 3\frac{1}{5}$ days

Answer and solution

Answer: (C) $\displaystyle 6\frac{1}{4}$ days

A's 1 day's work $\displaystyle = \frac{1}{16}$ , B's 1 day's work $\displaystyle = \frac{1}{4}$

Work done in the first 2 days $\displaystyle = \frac{1}{4} + \frac{1}{16} = \frac{4 + 1}{16} = \frac{5}{16}$

Work done in the first 6 days $\displaystyle = 3 \times \frac{5}{16} = \frac{15}{16}$

Remaining work $\displaystyle = 1 - \frac{15}{16} = \frac{1}{16}$

On day 7 it is B's turn; B does $\displaystyle \frac{1}{16}$ in $\displaystyle \frac{1}{16} \div \frac{1}{4} = \frac{1}{4}$ day

∴ Total no. of days $\displaystyle = 6 + \frac{1}{4} = 6\frac{1}{4}$ days
A certain number of men complete a piece of work in $52$ days. If there were $12$ men more, the work could be finished in $24$ days less. How many men were there originally?
A $38$ B $2$ C $14$ D $26$

Answer and solution

Answer: (C) $14$

Let the original number of men be $x$

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

$x \times 52 = (x + 12) \times 28$

$52x = 28x + 336$

$24x = 336$ ⇒ $x = 14$
If $2$ men or $6$ boys can finish a piece of work in $11$ days, in how many days can $3$ men and $2$ boys finish it?
A $22$ days B $\displaystyle 20\frac{1}{6}$ days C $\displaystyle 13\frac{1}{5}$ days D $6$ days

Answer and solution

Answer: (D) $6$ days

$2$ men $= 6$ boys

$3$ men $\displaystyle = \frac{6}{2} \times 3 = 9$ boys

$3$ men + $2$ boys $= 9$ boys + $2$ boys $= 11$ boys

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

$6 \times 11 = 11 \times x$ ⇒ $x = 6$ days
$6$ men and $6$ women working $5$ hours a day finish a piece of work in $8$ days; $8$ men and $6$ women working $7$ hours a day do the same work in $5$ days. Find the number of days in which the work is done by $8$ men only, working $5$ hours a day.
A $14$ days B $70$ days C $\displaystyle 10\frac{1}{2}$ days D $12$ days

Answer and solution

Answer: (A) $14$ days

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

Case (i): $(6m + 6w) \times 5 \times 8 = 240m + 240w$

Case (ii): $(8m + 6w) \times 7 \times 5 = 280m + 210w$

Same work: $240m + 240w = 280m + 210w$

⇒ $30w = 40m$ ⇒ $3w = 4m$

So $6$ women $= 8$ men, and $6m + 6w = 14$ men

$14$ men working $5$ hours a day for $8$ days do the same work as $8$ men working $5$ hours a day for $x$ days

$14 \times 5 \times 8 = 8 \times 5 \times x$ ⇒ $x = 14$ days
If $3$ men with $2$ boys can earn ₹2240 in $7$ days and $6$ men with $7$ boys can earn ₹8360 in $11$ days, in what time will $4$ men with $3$ boys earn ₹5280?
A $\displaystyle 9\frac{3}{7}$ days B $12$ days C $\displaystyle 16\frac{1}{2}$ days D $14$ days

Answer and solution

Answer: (B) $12$ days

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

$(3x + 2y) \times 7 = 2240$ ⇒ $3x + 2y = 320 \quad \cdots (1)$

$(6x + 7y) \times 11 = 8360$ ⇒ $6x + 7y = 760 \quad \cdots (2)$

(1) $\times 2$ ⇒ $6x + 4y = 640$

(2) $-$ (1) $\times 2$ ⇒ $3y = 120$ ⇒ $y = 40$

From (1): $3x = 320 - 80 = 240$ ⇒ $x = 80$

$(4x + 3y)n = 5280$

⇒ $(320 + 120)n = 5280$ ⇒ $440n = 5280$

⇒ $n = 12$ days
$40$ men are employed to do a piece of work which they could finish in $18$ days, but the men drop off by $5$ at the end of every $6$ days. In what time will the work be completed?
A $\displaystyle 21\frac{3}{5}$ days B $24$ days C $\displaystyle 20\frac{4}{7}$ days D $18$ days

Answer and solution

Answer: (A) $\displaystyle 21\frac{3}{5}$ days

Work done by 40 men in 6 days $\displaystyle = 6 \times \frac{1}{18} = \frac{1}{3}$

So 1 man's work in 6 days $\displaystyle = \frac{1}{3} \div 40 = \frac{1}{120}$

Remaining work $\displaystyle = \frac{2}{3}$ , remaining men $= 35$ ; work done by 35 men in 6 days $\displaystyle = 35 \times \frac{1}{120} = \frac{7}{24}$

Remaining work $\displaystyle = \frac{3}{8}$ , remaining men $= 30$ ; work done by 30 men in 6 days $\displaystyle = 30 \times \frac{1}{120} = \frac{1}{4}$

Remaining work $\displaystyle = \frac{1}{8}$ , remaining men $= 25$ ; work done by 25 men in 6 days $\displaystyle = \frac{5}{24}$

$\displaystyle \frac{1}{8}$ work is done by 25 men in $\displaystyle \frac{1}{8} \div \frac{5}{24} \times 6 = 3\frac{3}{5}$ days

Total time $\displaystyle = 6 + 6 + 6 + 3\frac{3}{5} = 21\frac{3}{5}$ 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 $30$ days. In how many days can A alone finish it?
A $10$ days B $15$ days C $45$ days D $20$ days

Answer and solution

Answer: (D) $20$ 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 $= 3k = 30$ ⇒ $k = 10$

Time taken by A $= 2k = 20$ days
The speed of a boat in still water is $10.5$ km/hr and the speed of the stream is $0.5$ km/hr. The speed of the boat downstream is
A $5$ km/hr B $10.5$ km/hr C $11$ km/hr D $10$ km/hr

Answer and solution

Answer: (C) $11$ km/hr

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

$= 10.5 + 0.5 = 11$ km/hr
A man can row downstream at the rate of $9$ km/hr and upstream at $3$ km/hr. The man's rate in still water is
A $4.5$ km/hr B $12$ km/hr C $3$ km/hr D $6$ km/hr

Answer and solution

Answer: (D) $6$ km/hr

Given $x + y = 9$ and $x - y = 3$

Adding, $2x = 12$ ⇒ $x = 6$ km/hr
A man rows upstream $32$ km and downstream $62$ km, taking $4$ hours each time. The man's speed in still water is
A $3.75$ km/hr B $23.5$ km/hr C $15.5$ km/hr D $11.75$ km/hr

Answer and solution

Answer: (D) $11.75$ km/hr

Speed downstream: $\displaystyle x + y = \frac{62}{4}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{32}{4}$ km/hr

Adding, $\displaystyle 2x = \frac{62}{4} + \frac{32}{4} = \frac{94}{4}$

$\displaystyle x = \frac{94}{8} = 11.75$ km/hr
The rate of a stream is $3.5$ km/hr. A man finds that it takes him twice as long to row up the river as to row down. The man's rate in still water is
A $14$ km/hr B $10.5$ km/hr C $5.5$ km/hr D $7$ km/hr

Answer and solution

Answer: (B) $10.5$ km/hr

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

Same distance, twice the time, so speed down $= 2 \times$ speed up

Given $2(x - 3.5) = x + 3.5$

⇒ $2x - 7 = x + 3.5$ ⇒ $x = 10.5$ km/hr
A man can row $20$ km/hr in still water. If the river is running at $4$ km/hr, it takes him 1 hour 30 minutes to row to a place and back. The place is at a distance of
A $28.8$ km B $24$ km C $14.4$ km D $15$ km

Answer and solution

Answer: (C) $14.4$ km

Let the required distance be $d$ km

Speed downstream $= 20 + 4 = 24$ km/hr

Speed upstream $= 20 - 4 = 16$ km/hr

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

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{24} + \frac{d}{16} = 1.5$

⇒ $\displaystyle d\left(\frac{1}{24} + \frac{1}{16}\right) = 1.5$

⇒ $\displaystyle d \times \frac{5}{48} = 1.5$

⇒ $d = 14.4$ km
The current of a stream runs at the rate of $5$ km/hr. A motor boat goes $14$ km upstream and back again to the starting point in 1 hour 35 minutes. The speed of the motor boat in still water is
A $29$ km/hr B $24$ km/hr C $19$ km/hr D $14$ km/hr

Answer and solution

Answer: (C) $19$ 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 $= 95$ min $\displaystyle = \frac{95}{60}$ hr $\displaystyle = \frac{19}{12}$ hr

$\displaystyle \frac{14}{x - 5} + \frac{14}{x + 5} = \frac{19}{12}$

⇒ $\displaystyle 14\left(\frac{(x + 5) + (x - 5)}{(x - 5)(x + 5)}\right) = \frac{19}{12}$

⇒ $\displaystyle \frac{28x}{x^2 - 25} = \frac{19}{12}$

⇒ $336x = 19x^2 - 475$

⇒ $19 x^{2} - 336 x - 475 = 0$

⇒ $\left(x - 19\right) \left(19 x + 25\right) = 0$

⇒ $x = 19$ (or) $\displaystyle x = -\frac{25}{19}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 19$ km/hr
A man can row $6$ km upstream and $64$ km downstream in $5$ hours, while he can row $30$ km upstream and $16$ km downstream in $6$ hours. The rate of the current is
A $5$ km/hr B $10$ km/hr C $6$ km/hr D $11$ km/hr

Answer and solution

Answer: (A) $5$ 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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{6}{x - y} + \frac{64}{x + y} = 5$ ⇒ $6p + 64q = 5 \quad \cdots (1)$

$\displaystyle \frac{30}{x - y} + \frac{16}{x + y} = 6$ ⇒ $30p + 16q = 6 \quad \cdots (2)$

(1) $\times 5$ ⇒ $30p + 320q = 25$

(1) $\times 5$ $-$ (2) ⇒ $304q = 19$ ⇒ $\displaystyle q = \frac{1}{16}$

From (1): $6p + 4 = 5$ ⇒ $6p = 1$ ⇒ $\displaystyle p = \frac{1}{6}$

So $x - y = 6$ and $x + y = 16$

Adding, $2x = 22$ ⇒ $x = 11$ ; then $y = 16 - 11 = 5$

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MCQ test, Set 2