Arithmetic Applications (Level 2): MCQ test, Set 7

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A can finish a piece of work in $21$ days. What part of the work does A finish in $12$ days?
A $\displaystyle \frac{3}{7}$ B $\displaystyle \frac{7}{4}$ C $\displaystyle \frac{4}{7}$ D $\displaystyle \frac{1}{21}$

Answer and solution

Answer: (C) $\displaystyle \frac{4}{7}$

A's 1 day's work $\displaystyle = \frac{1}{21}$

A's 12 days' work $\displaystyle = 12 \times \frac{1}{21} = \frac{12}{21} = \frac{4}{7}$
$20$ men can do a piece of work in $15$ days. How many men are needed to finish it in $30$ days?
A $10$ men B $35$ men C $40$ men D $5$ men

Answer and solution

Answer: (A) $10$ men

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

$20 \times 15 = x \times 30$

$\displaystyle x = \frac{20 \times 15}{30} = 10$ men
A can do a piece of work in $9$ days and B alone can do it in $45$ days. How much time will both take to finish the work?
A $54$ days B $27$ days C $\displaystyle 7\frac{1}{2}$ days D $36$ days

Answer and solution

Answer: (C) $\displaystyle 7\frac{1}{2}$ days

A's 1 day's work $\displaystyle = \frac{1}{9}$

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

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{9} + \frac{1}{45} = \frac{5 + 1}{45} = \frac{6}{45} = \frac{2}{15}$

∴ Both together can finish the work in $\displaystyle \frac{15}{2}$ days $\displaystyle = 7\frac{1}{2}$ days
A and B together can do a piece of work in $8$ days. B alone can finish it in $12$ days. In how many days can A alone finish the work?
A $24$ days B $10$ days C $4$ days D $\displaystyle 4\frac{4}{5}$ days

Answer and solution

Answer: (A) $24$ days

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

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

⇒ A's 1 day's work $\displaystyle = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$

∴ A alone can finish the work in $24$ days.
A and B can do a piece of work in $30$ days; B and C in $55$ days; C and A in $22$ days. In how many days will they finish it working together?
A $\displaystyle 20\frac{5}{8}$ days B $\displaystyle 35\frac{2}{3}$ days C $\displaystyle 41\frac{1}{4}$ days D $107$ days

Answer and solution

Answer: (A) $\displaystyle 20\frac{5}{8}$ days

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

(B + C)'s 1 day's work $\displaystyle = \frac{1}{55}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{22}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{55} + \frac{1}{22} = \frac{11 + 6 + 15}{330} = \frac{32}{330} = \frac{16}{165}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{8}{165}$

∴ A, B, C together can finish the work in $\displaystyle 20\frac{5}{8}$ days.
A can do a piece of work in $30$ days and B can finish it in $25$ days. They work together for $12$ days and then A goes away. In how many days will B finish the remaining work?
A $13$ days B $\displaystyle 3\frac{3}{5}$ days C $3$ days D $\displaystyle 23\frac{1}{6}$ days

Answer and solution

Answer: (C) $3$ days

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

B's 1 day's work $\displaystyle = \frac{1}{25}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{25} = \frac{5 + 6}{150} = \frac{11}{150}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{11}{150} = \frac{22}{25}$

Remaining work $\displaystyle = 1 - \frac{22}{25} = \frac{3}{25}$

This work is done by B in $\displaystyle \frac{3}{25} \times 25 = 3$ days
A can do a piece of work in $15$ days while B alone can do it in $45$ days. With the help of C they can finish it in $3$ days. If they are paid ₹3600 for the whole work, what is C's share?
A ₹2640 B ₹1200 C ₹720 D ₹240

Answer and solution

Answer: (A) ₹2640

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

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{15} + \frac{1}{45} = \frac{3 + 1}{45} = \frac{4}{45}$

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

C's 1 day's work $\displaystyle = \frac{1}{3} - \frac{4}{45} = \frac{15 - 4}{45} = \frac{11}{45}$

Consider $\displaystyle \frac{1}{15} : \frac{1}{45} : \frac{11}{45} = \frac{3}{45} : \frac{1}{45} : \frac{11}{45} = 3 : 1 : 11$

∴ ₹3600 must be divided in the ratio $3 : 1 : 11$ ( $3 + 1 + 11 = 15$ )

C's share $\displaystyle = 3600 \times \frac{11}{15} = 2640$ , i.e. ₹2640
A is thrice as good a workman as B and is therefore able to finish a piece of work in $28$ days less than B. In how many days can they do it working together?
A $\displaystyle 10\frac{1}{2}$ days B $42$ days C $28$ days D $14$ days

Answer and solution

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

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 28)$ days

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

∴ $\displaystyle x - 28 = \frac{1}{3}x$

⇒ $3x - 84 = x$ ⇒ $2x = 84$ ⇒ $x = 42$

∴ Time taken by B $= 42$ days, time taken by A $= 14$ days

A's 1 day's work $\displaystyle = \frac{1}{14}$ , B's 1 day's work $\displaystyle = \frac{1}{42}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{14} + \frac{1}{42} = \frac{3 + 1}{42} = \frac{4}{42} = \frac{2}{21}$

A and B together can finish the work in $\displaystyle 10\frac{1}{2}$ days
A and B can do a piece of work in $65$ and $60$ days respectively. They began the work together, but A leaves after some days and B finishes the remaining work in $35$ days. After how many days did A leave?
A $\displaystyle 27\frac{1}{12}$ days B $48$ days C $25$ days D $13$ days

Answer and solution

Answer: (D) $13$ days

A's 1 day's work $\displaystyle = \frac{1}{65}$ , B's 1 day's work $\displaystyle = \frac{1}{60}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{65} + \frac{1}{60} = \frac{12 + 13}{780} = \frac{25}{780} = \frac{5}{156}$

B's 35 days' work $\displaystyle = 35 \times \frac{1}{60} = \frac{7}{12}$

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

(A + B)'s $x$ days' work $\displaystyle = \frac{5}{12}$ ⇒ $\displaystyle \frac{5}{156} \times x = \frac{5}{12}$

$x = 13$ , so A left after $13$ days
A can do a piece of work in $55$ days and B can do it in $36$ days. They work together for $12$ days. Then B leaves and A alone continues the work. $2$ days after that, C joins A and the work is completed in $16$ days more. In how many days can C do the work alone?
A $132$ days B $264$ days C $116$ days D $187$ days

Answer and solution

Answer: (A) $132$ days

A's 1 day's work $\displaystyle = \frac{1}{55}$ , B's 1 day's work $\displaystyle = \frac{1}{36}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{55} + \frac{1}{36} = \frac{36 + 55}{1980} = \frac{91}{1980}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{91}{1980} = \frac{91}{165}$

A's 2 days' work $\displaystyle = \frac{2}{55}$

(A + C)'s 16 days' work $\displaystyle = 1 - \left(\frac{91}{165} + \frac{2}{55}\right) = \frac{68}{165}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{68}{165} \div 16 = \frac{17}{660}$

C's 1 day's work $\displaystyle = \frac{17}{660} - \frac{1}{55} = \frac{17 - 12}{660} = \frac{5}{660} = \frac{1}{132}$

C can finish the work in $132$ days
A and B together can do a piece of work in $42$ days. B and C together can do it in $48$ days. If A is twice as good a workman as C, A alone can do it in
A $336$ days B $168$ days C $56$ days D $672$ days

Answer and solution

Answer: (B) $168$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{42}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{48}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{42}$

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

⇒ C's 1 day's work $\displaystyle + \frac{1}{48} = \frac{1}{42}$

C's 1 day's work $\displaystyle = \frac{1}{42} - \frac{1}{48} = \frac{8 - 7}{336} = \frac{1}{336}$

A's 1 day's work $\displaystyle = 2 \times \frac{1}{336} = \frac{1}{168}$ , so A alone takes $168$ days
A and B working alone can finish a piece of work in $10$ and $11$ days respectively. If they work at it alternately for a day, B beginning, in how many days will the work be finished?
A $10$ days B $11$ days C $\displaystyle 11\frac{1}{2}$ days D $\displaystyle 10\frac{1}{2}$ days

Answer and solution

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

A's 1 day's work $\displaystyle = \frac{1}{10}$ , B's 1 day's work $\displaystyle = \frac{1}{11}$

Work done in the first 2 days $\displaystyle = \frac{1}{11} + \frac{1}{10} = \frac{10 + 11}{110} = \frac{21}{110}$

Work done in the first 10 days $\displaystyle = 5 \times \frac{21}{110} = \frac{21}{22}$

Remaining work $\displaystyle = 1 - \frac{21}{22} = \frac{1}{22}$

On day 11 it is B's turn; B does $\displaystyle \frac{1}{22}$ in $\displaystyle \frac{1}{22} \div \frac{1}{11} = \frac{1}{2}$ day

∴ Total no. of days $\displaystyle = 10 + \frac{1}{2} = 10\frac{1}{2}$ days
A certain number of men complete a piece of work in $27$ days. If there were $21$ men more, the work could be finished in $9$ days less. How many men were there originally?
A $84$ B $42$ C $21$ D $63$

Answer and solution

Answer: (B) $42$

Let the original number of men be $x$

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

$x \times 27 = (x + 21) \times 18$

$27x = 18x + 378$

$9x = 378$ ⇒ $x = 42$
If $4$ men or $8$ boys can finish a piece of work in $22$ days, in how many days can $6$ men and $10$ boys finish it?
A $\displaystyle 60\frac{1}{2}$ days B $\displaystyle 5\frac{1}{2}$ days C $8$ days D $11$ days

Answer and solution

Answer: (C) $8$ days

$4$ men $= 8$ boys

$6$ men $\displaystyle = \frac{8}{4} \times 6 = 12$ boys

$6$ men + $10$ boys $= 12$ boys + $10$ boys $= 22$ boys

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

$8 \times 22 = 22 \times x$ ⇒ $x = 8$ days
$3$ men and $6$ women working $10$ hours a day finish a piece of work in $7$ days; $2$ men and $6$ women working $7$ hours a day do the same work in $11$ days. Find the number of days in which the work is done by $10$ men only, working $7$ hours a day.
A $\displaystyle 7\frac{7}{10}$ days B $\displaystyle 7\frac{1}{2}$ days C $11$ days D $9$ days

Answer and solution

Answer: (C) $11$ days

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

Case (i): $(3m + 6w) \times 10 \times 7 = 210m + 420w$

Case (ii): $(2m + 6w) \times 7 \times 11 = 154m + 462w$

Same work: $210m + 420w = 154m + 462w$

⇒ $42w = 56m$ ⇒ $3w = 4m$

So $6$ women $= 8$ men, and $3m + 6w = 11$ men

$11$ men working $10$ hours a day for $7$ days do the same work as $10$ men working $7$ hours a day for $x$ days

$11 \times 10 \times 7 = 10 \times 7 \times x$ ⇒ $x = 11$ days
If $5$ men with $8$ boys can earn ₹3210 in $6$ days and $10$ men with $9$ boys can earn ₹4650 in $5$ days, how much does 1 man earn in a day?
A ₹107 B ₹75 C ₹20 D ₹95

Answer and solution

Answer: (B) ₹75

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

$(5x + 8y) \times 6 = 3210$ ⇒ $5x + 8y = 535 \quad \cdots (1)$

$(10x + 9y) \times 5 = 4650$ ⇒ $10x + 9y = 930 \quad \cdots (2)$

(1) $\times 2$ ⇒ $10x + 16y = 1070$

(1) $\times 2$ $-$ (2) ⇒ $7y = 140$ ⇒ $y = 20$

From (1): $5x = 535 - 160 = 375$ ⇒ $x = 75$
$42$ men are employed to do a piece of work which they could finish in $33$ days, but the men drop off by $6$ at the end of every $11$ days. In what time will the work be completed?
A $33$ days B $\displaystyle 38\frac{1}{2}$ days C $44$ days D $\displaystyle 41\frac{1}{4}$ days

Answer and solution

Answer: (D) $\displaystyle 41\frac{1}{4}$ days

Work done by 42 men in 11 days $\displaystyle = 11 \times \frac{1}{33} = \frac{1}{3}$

So 1 man's work in 11 days $\displaystyle = \frac{1}{3} \div 42 = \frac{1}{126}$

Remaining work $\displaystyle = \frac{2}{3}$ , remaining men $= 36$ ; work done by 36 men in 11 days $\displaystyle = 36 \times \frac{1}{126} = \frac{2}{7}$

Remaining work $\displaystyle = \frac{8}{21}$ , remaining men $= 30$ ; work done by 30 men in 11 days $\displaystyle = 30 \times \frac{1}{126} = \frac{5}{21}$

Remaining work $\displaystyle = \frac{1}{7}$ , remaining men $= 24$ ; work done by 24 men in 11 days $\displaystyle = \frac{4}{21}$

$\displaystyle \frac{1}{7}$ work is done by 24 men in $\displaystyle \frac{1}{7} \div \frac{4}{21} \times 11 = 8\frac{1}{4}$ days

Total time $\displaystyle = 11 + 11 + 11 + 8\frac{1}{4} = 41\frac{1}{4}$ 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 $58$ days. In how many days can B alone finish it?
A $87$ days B $\displaystyle 38\frac{2}{3}$ days C $116$ days D $29$ days

Answer and solution

Answer: (A) $87$ 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 $= 2k = 58$ ⇒ $k = 29$

Time taken by B $= 3k = 87$ 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 $22$ days. In how many days can both together finish it?
A $\displaystyle 27\frac{1}{2}$ days B $\displaystyle 13\frac{1}{5}$ days C $\displaystyle 16\frac{1}{2}$ days D $\displaystyle 19\frac{4}{5}$ days

Answer and solution

Answer: (B) $\displaystyle 13\frac{1}{5}$ 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 $= 2k = 22$ ⇒ $k = 11$

Time taken by B $= 3k = 33$ days

A's 1 day's work $\displaystyle = \frac{1}{22}$ , B's 1 day's work $\displaystyle = \frac{1}{33}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{22} + \frac{1}{33} = \frac{3 + 2}{66} = \frac{5}{66}$

Time taken by (A + B) to finish the work $\displaystyle = 13\frac{1}{5}$ days
A man can row downstream at the rate of $14$ km/hr and upstream at $7$ km/hr. The man's rate in still water is
A $7$ km/hr B $10.5$ km/hr C $3.5$ km/hr D $21$ km/hr

Answer and solution

Answer: (B) $10.5$ km/hr

Given $x + y = 14$ and $x - y = 7$

Adding, $2x = 21$ ⇒ $x = 10.5$ km/hr
A man rows upstream $36$ km and downstream $79$ km, taking $5$ hours each time. The velocity of the current is
A $11.5$ km/hr B $8.6$ km/hr C $9.6$ km/hr D $4.3$ km/hr

Answer and solution

Answer: (D) $4.3$ km/hr

Speed downstream: $\displaystyle x + y = \frac{79}{5}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{36}{5}$ km/hr

Subtracting, $\displaystyle 2y = \frac{79}{5} - \frac{36}{5} = \frac{43}{5}$

$\displaystyle y = \frac{43}{10} = 4.3$ km/hr
A man can row $7.5$ km/hr in still water and he finds that it takes him twice as long to row up as to row down the river. The rate of the stream is
A $3.75$ km/hr B $5$ km/hr C $5.5$ km/hr D $2.5$ km/hr

Answer and solution

Answer: (D) $2.5$ km/hr

Let the rate of the stream be $y$ km/hr

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

Given $2(7.5 - y) = 7.5 + y$

⇒ $15 - 2y = 7.5 + y$ ⇒ $3y = 7.5$

⇒ $y = 2.5$ km/hr
A man can row $9$ km/hr in still water. If the river is running at $1.5$ km/hr, it takes him 2 hours to row to a place and back. The place is at a distance of
A $9$ km B $8.75$ km C $15$ km D $17.5$ km

Answer and solution

Answer: (B) $8.75$ km

Let the required distance be $d$ km

Speed downstream $= 9 + 1.5 = 10.5$ km/hr

Speed upstream $= 9 - 1.5 = 7.5$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{10.5} + \frac{d}{7.5} = 2$

⇒ $\displaystyle d\left(\frac{2}{21} + \frac{2}{15}\right) = 2$

⇒ $\displaystyle d \times \frac{8}{35} = 2$

⇒ $d = 8.75$ km
The current of a stream runs at the rate of $4$ km/hr. A motor boat goes $20$ km upstream and back again to the starting point in 3 hours 45 minutes. The speed of the motor boat in still water is
A $16$ km/hr B $8$ km/hr C $12$ km/hr D $20$ km/hr

Answer and solution

Answer: (C) $12$ km/hr

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

Speed upstream $= (x - 4)$ km/hr, speed downstream $= (x + 4)$ km/hr

Total time $= 225$ min $\displaystyle = \frac{225}{60}$ hr $\displaystyle = \frac{15}{4}$ hr

$\displaystyle \frac{20}{x - 4} + \frac{20}{x + 4} = \frac{15}{4}$

⇒ $\displaystyle 20\left(\frac{(x + 4) + (x - 4)}{(x - 4)(x + 4)}\right) = \frac{15}{4}$

⇒ $\displaystyle \frac{40x}{x^2 - 16} = \frac{15}{4}$

⇒ $160x = 15x^2 - 240$

⇒ $3 x^{2} - 32 x - 48 = 0$

⇒ $\left(x - 12\right) \left(3 x + 4\right) = 0$

⇒ $x = 12$ (or) $\displaystyle x = -\frac{4}{3}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 12$ km/hr
A man can row $5$ km upstream and $18$ km downstream in $3$ hours, while he can row $10$ km upstream and $45$ km downstream in $7$ hours. The rate of the current is
A $7$ km/hr B $5$ km/hr C $4$ km/hr D $2$ km/hr

Answer and solution

Answer: (D) $2$ 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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{5}{x - y} + \frac{18}{x + y} = 3$ ⇒ $5p + 18q = 3 \quad \cdots (1)$

$\displaystyle \frac{10}{x - y} + \frac{45}{x + y} = 7$ ⇒ $10p + 45q = 7 \quad \cdots (2)$

(1) $\times 2$ ⇒ $10p + 36q = 6$

(2) $-$ (1) $\times 2$ ⇒ $9q = 1$ ⇒ $\displaystyle q = \frac{1}{9}$

From (1): $5p + 2 = 3$ ⇒ $5p = 1$ ⇒ $\displaystyle p = \frac{1}{5}$

So $x - y = 5$ and $x + y = 9$

Adding, $2x = 14$ ⇒ $x = 7$ ; then $y = 9 - 7 = 2$

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MCQ test, Set 7