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Arithmetic Applications (Level 2)

MCQ test, Set 3

  • Set 3 of 10
  • 25 questions
  • 25 marks
  • 30 minutes

Download the PDF (with key sheet and solutions, 120.9 KB)

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

  1. If A can finish a piece of work in 3434 days, then A's 1 day's work is
    Answer and solution

    Answer: (C) 134\displaystyle \frac{1}{34}

    A finishes the whole work in 3434 days.

    ∴ A's 1 day's work =134\displaystyle = \frac{1}{34}

  2. 1818 men can do a piece of work in 1414 days. How many men are needed to finish it in 2121 days?
    Answer and solution

    Answer: (A) 1212 men

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

    18×14=x×2118 \times 14 = x \times 21

    x=18×1421=12\displaystyle x = \frac{18 \times 14}{21} = 12 men

  3. If the number of men to do a work is changed in the ratio 5:45 : 4, the time taken by them to finish the work changes in the ratio
    Answer and solution

    Answer: (D) 4:54 : 5

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

    T1T2=M2M1=45\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{4}{5}

    ∴ T1:T2=4:5T_1 : T_2 = 4 : 5

  4. A can do a piece of work in 2121 days and B alone can do it in 2424 days. How much time will both take to finish the work?
    Answer and solution

    Answer: (B) 1115\displaystyle 11\frac{1}{5} days

    A's 1 day's work =121\displaystyle = \frac{1}{21}

    B's 1 day's work =124\displaystyle = \frac{1}{24}

    ∴ (A + B)'s 1 day's work =121+124=8+7168=15168=556\displaystyle = \frac{1}{21} + \frac{1}{24} = \frac{8 + 7}{168} = \frac{15}{168} = \frac{5}{56}

    ∴ Both together can finish the work in 565\displaystyle \frac{56}{5} days =1115\displaystyle = 11\frac{1}{5} days

  5. A and B together can finish a piece of work in 1818 days. A alone can do it in 5454 days. In how many days can B alone do it?
    Answer and solution

    Answer: (A) 2727 days

    (A + B)'s 1 day's work =118\displaystyle = \frac{1}{18}

    A's 1 day's work =154\displaystyle = \frac{1}{54}

    ⇒ B's 1 day's work =118−154=3−154=254=127\displaystyle = \frac{1}{18} - \frac{1}{54} = \frac{3 - 1}{54} = \frac{2}{54} = \frac{1}{27}

    ∴ B alone can finish the work in 2727 days.

  6. A and B can do a piece of work in 7070 days; B and C in 8484 days; C and A in 105105 days. In how many days can C alone finish it?
    Answer and solution

    Answer: (A) 280280 days

    (A + B)'s 1 day's work =170\displaystyle = \frac{1}{70}

    (B + C)'s 1 day's work =184\displaystyle = \frac{1}{84}

    (C + A)'s 1 day's work =1105\displaystyle = \frac{1}{105}

    Adding, 2(A + B + C)'s 1 day's work =170+184+1105=6+5+4420=15420=128\displaystyle = \frac{1}{70} + \frac{1}{84} + \frac{1}{105} = \frac{6 + 5 + 4}{420} = \frac{15}{420} = \frac{1}{28}

    ⇒ (A + B + C)'s 1 day's work =156\displaystyle = \frac{1}{56}

    ∴ A, B, C together can finish the work in 5656 days.

    C's 1 day's work =156−170=1280\displaystyle = \frac{1}{56} - \frac{1}{70} = \frac{1}{280}, so C alone takes 280280 days

  7. A can do a piece of work in 1818 days and B can finish it in 1212 days. They work together for 44 days and then B goes away. In how many days will A finish the remaining work?
    Answer and solution

    Answer: (C) 88 days

    A's 1 day's work =118\displaystyle = \frac{1}{18}

    B's 1 day's work =112\displaystyle = \frac{1}{12}

    (A + B)'s 1 day's work =118+112=2+336=536\displaystyle = \frac{1}{18} + \frac{1}{12} = \frac{2 + 3}{36} = \frac{5}{36}

    (A + B)'s 4 days' work =4×536=59\displaystyle = 4 \times \frac{5}{36} = \frac{5}{9}

    Remaining work =1−59=49\displaystyle = 1 - \frac{5}{9} = \frac{4}{9}

    This work is done by A in 49×18=8\displaystyle \frac{4}{9} \times 18 = 8 days

  8. A can do a piece of work in 77 days while B alone can do it in 4242 days. With the help of C they can finish it in 33 days. If they are paid ₹3780 for the whole work, what is C's share?
    Answer and solution

    Answer: (C) ₹1890

    A's 1 day's work =17\displaystyle = \frac{1}{7}

    B's 1 day's work =142\displaystyle = \frac{1}{42}

    (A + B)'s 1 day's work =17+142=6+142=742=16\displaystyle = \frac{1}{7} + \frac{1}{42} = \frac{6 + 1}{42} = \frac{7}{42} = \frac{1}{6}

    Given (A + B + C)'s 1 day's work =13\displaystyle = \frac{1}{3}

    C's 1 day's work =13−16=2−16=16\displaystyle = \frac{1}{3} - \frac{1}{6} = \frac{2 - 1}{6} = \frac{1}{6}

    Consider 17:142:16=642:142:742=6:1:7\displaystyle \frac{1}{7} : \frac{1}{42} : \frac{1}{6} = \frac{6}{42} : \frac{1}{42} : \frac{7}{42} = 6 : 1 : 7

    ∴ ₹3780 must be divided in the ratio 6:1:76 : 1 : 7 (6+1+7=146 + 1 + 7 = 14)

    C's share =3780×714=1890\displaystyle = 3780 \times \frac{7}{14} = 1890, i.e. ₹1890

  9. A is twice as good a workman as B and is therefore able to finish a piece of work in 2323 days less than B. In how many days can they do it working together?
    Answer and solution

    Answer: (D) 1513\displaystyle 15\frac{1}{3} days

    Suppose B finishes the work in xx days

    ⇒ A is able to finish it in (x−23)(x - 23) days

    Time taken by A =12\displaystyle = \frac{1}{2} (Time taken by B)

    ∴ x−23=12x\displaystyle x - 23 = \frac{1}{2}x

    ⇒ 2x−46=x2x - 46 = x ⇒ x=46x = 46

    ∴ Time taken by B =46= 46 days, time taken by A =23= 23 days

    A's 1 day's work =123\displaystyle = \frac{1}{23}, B's 1 day's work =146\displaystyle = \frac{1}{46}

    (A + B)'s 1 day's work =123+146=2+146=346\displaystyle = \frac{1}{23} + \frac{1}{46} = \frac{2 + 1}{46} = \frac{3}{46}

    A and B together can finish the work in 1513\displaystyle 15\frac{1}{3} days

  10. A and B can do a piece of work in 1212 and 5656 days respectively. They began the work together, but A leaves after some days and B finishes the remaining work in 55 days. After how many days did A leave?
    Answer and solution

    Answer: (A) 99 days

    A's 1 day's work =112\displaystyle = \frac{1}{12}, B's 1 day's work =156\displaystyle = \frac{1}{56}

    (A + B)'s 1 day's work =112+156=14+3168=17168\displaystyle = \frac{1}{12} + \frac{1}{56} = \frac{14 + 3}{168} = \frac{17}{168}

    B's 5 days' work =5×156=556\displaystyle = 5 \times \frac{1}{56} = \frac{5}{56}

    Remaining work =1−556=5156\displaystyle = 1 - \frac{5}{56} = \frac{51}{56}

    (A + B)'s xx days' work =5156\displaystyle = \frac{51}{56} ⇒ 17168×x=5156\displaystyle \frac{17}{168} \times x = \frac{51}{56}

    x=9x = 9, so A left after 99 days

  11. A can do a piece of work in 120120 days and B can do it in 3030 days. They work together for 55 days. Then B leaves and A alone continues the work. 77 days after that, C joins A and the work is completed in 5656 days more. In how many days can C do the work alone?
    Answer and solution

    Answer: (C) 210210 days

    A's 1 day's work =1120\displaystyle = \frac{1}{120}, B's 1 day's work =130\displaystyle = \frac{1}{30}

    (A + B)'s 1 day's work =1120+130=1+4120=5120=124\displaystyle = \frac{1}{120} + \frac{1}{30} = \frac{1 + 4}{120} = \frac{5}{120} = \frac{1}{24}

    (A + B)'s 5 days' work =5×124=524\displaystyle = 5 \times \frac{1}{24} = \frac{5}{24}

    A's 7 days' work =7120\displaystyle = \frac{7}{120}

    (A + C)'s 56 days' work =1−(524+7120)=1115\displaystyle = 1 - \left(\frac{5}{24} + \frac{7}{120}\right) = \frac{11}{15}

    ⇒ (A + C)'s 1 day's work =1115÷56=11840\displaystyle = \frac{11}{15} \div 56 = \frac{11}{840}

    C's 1 day's work =11840−1120=11−7840=4840=1210\displaystyle = \frac{11}{840} - \frac{1}{120} = \frac{11 - 7}{840} = \frac{4}{840} = \frac{1}{210}

    C can finish the work in 210210 days

  12. A and B together can do a piece of work in 1515 days. B and C together can do it in 2424 days. If A is twice as good a workman as C, B alone can do it in
    Answer and solution

    Answer: (D) 6060 days

    (A + B)'s 1 day's work =115\displaystyle = \frac{1}{15}, (B + C)'s 1 day's work =124\displaystyle = \frac{1}{24}

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

    C's 2 days' work + B's 1 day's work =115\displaystyle = \frac{1}{15}

    ⇒ C's 1 day's work + (B + C)'s 1 day's work =115\displaystyle = \frac{1}{15}

    ⇒ C's 1 day's work +124=115\displaystyle + \frac{1}{24} = \frac{1}{15}

    C's 1 day's work =115−124=8−5120=3120=140\displaystyle = \frac{1}{15} - \frac{1}{24} = \frac{8 - 5}{120} = \frac{3}{120} = \frac{1}{40}

    B's 1 day's work =124−140=5−3120=2120=160\displaystyle = \frac{1}{24} - \frac{1}{40} = \frac{5 - 3}{120} = \frac{2}{120} = \frac{1}{60}

    ∴ B alone can finish the work in 6060 days

  13. A and B working alone can finish a piece of work in 77 and 2020 days respectively. If they work at it alternately for a day, B beginning, in how many days will the work be finished?
    Answer and solution

    Answer: (D) 1057\displaystyle 10\frac{5}{7} days

    A's 1 day's work =17\displaystyle = \frac{1}{7}, B's 1 day's work =120\displaystyle = \frac{1}{20}

    Work done in the first 2 days =120+17=7+20140=27140\displaystyle = \frac{1}{20} + \frac{1}{7} = \frac{7 + 20}{140} = \frac{27}{140}

    Work done in the first 10 days =5×27140=2728\displaystyle = 5 \times \frac{27}{140} = \frac{27}{28}

    Remaining work =1−2728=128\displaystyle = 1 - \frac{27}{28} = \frac{1}{28}

    On day 11 it is B's turn; B does 128\displaystyle \frac{1}{28} in 128÷120=57\displaystyle \frac{1}{28} \div \frac{1}{20} = \frac{5}{7} day

    ∴ Total no. of days =10+57=1057\displaystyle = 10 + \frac{5}{7} = 10\frac{5}{7} days

  14. A certain number of men complete a piece of work in 6363 days. If there were 55 men more, the work could be finished in 77 days less. How many men were there originally?
    Answer and solution

    Answer: (B) 4040

    Let the original number of men be xx

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

    x×63=(x+5)×56x \times 63 = (x + 5) \times 56

    63x=56x+28063x = 56x + 280

    7x=2807x = 280 ⇒ x=40x = 40

  15. If 22 men or 1010 boys can finish a piece of work in 2626 days, in how many days can 44 men and 66 boys finish it?
    Answer and solution

    Answer: (A) 1010 days

    22 men =10= 10 boys

    44 men =102×4=20\displaystyle = \frac{10}{2} \times 4 = 20 boys

    44 men + 66 boys =20= 20 boys + 66 boys =26= 26 boys

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

    10×26=26×x10 \times 26 = 26 \times x ⇒ x=10x = 10 days

  16. 88 men and 88 women working 33 hours a day finish a piece of work in 1212 days; 77 men and 22 women working 66 hours a day do the same work in 1212 days. Find the number of days in which the work is done by 88 men only, working 55 hours a day.
    Answer and solution

    Answer: (B) 1818 days

    Let 1 man do mm and 1 woman do ww units of work in 1 hour.

    Case (i): (8m+8w)×3×12=288m+288w(8m + 8w) \times 3 \times 12 = 288m + 288w

    Case (ii): (7m+2w)×6×12=504m+144w(7m + 2w) \times 6 \times 12 = 504m + 144w

    Same work: 288m+288w=504m+144w288m + 288w = 504m + 144w

    ⇒ 144w=216m144w = 216m ⇒ 2w=3m2w = 3m

    So 88 women =12= 12 men, and 8m+8w=208m + 8w = 20 men

    2020 men working 33 hours a day for 1212 days do the same work as 88 men working 55 hours a day for xx days

    20×3×12=8×5×x20 \times 3 \times 12 = 8 \times 5 \times x ⇒ x=18x = 18 days

  17. If 22 men with 33 boys can earn ₹1715 in 77 days and 66 men with 55 boys can earn ₹4760 in 88 days, how much does 1 man earn in a day?
    Answer and solution

    Answer: (B) ₹70

    Let 1 man earn ₹xx and 1 boy earn ₹yy in 1 day.

    (2x+3y)×7=1715(2x + 3y) \times 7 = 1715 ⇒ 2x+3y=245⋯(1)2x + 3y = 245 \quad \cdots (1)

    (6x+5y)×8=4760(6x + 5y) \times 8 = 4760 ⇒ 6x+5y=595⋯(2)6x + 5y = 595 \quad \cdots (2)

    (1) ×3\times 3 ⇒ 6x+9y=7356x + 9y = 735

    (1) ×3\times 3 −- (2) ⇒ 4y=1404y = 140 ⇒ y=35y = 35

    From (1): 2x=245−105=1402x = 245 - 105 = 140 ⇒ x=70x = 70

  18. 1414 men are employed to do a piece of work which they could finish in 1515 days, but the men drop off by 22 at the end of every 66 days. In what time will the work be completed?
    Answer and solution

    Answer: (D) 1725\displaystyle 17\frac{2}{5} days

    Work done by 14 men in 6 days =6×115=25\displaystyle = 6 \times \frac{1}{15} = \frac{2}{5}

    So 1 man's work in 6 days =25÷14=135\displaystyle = \frac{2}{5} \div 14 = \frac{1}{35}

    Remaining work =35\displaystyle = \frac{3}{5}, remaining men =12= 12; work done by 12 men in 6 days =12×135=1235\displaystyle = 12 \times \frac{1}{35} = \frac{12}{35}

    Remaining work =935\displaystyle = \frac{9}{35}, remaining men =10= 10; work done by 10 men in 6 days =27\displaystyle = \frac{2}{7}

    935\displaystyle \frac{9}{35} work is done by 10 men in 935÷27×6=525\displaystyle \frac{9}{35} \div \frac{2}{7} \times 6 = 5\frac{2}{5} days

    Total time =6+6+525=1725\displaystyle = 6 + 6 + 5\frac{2}{5} = 17\frac{2}{5} days

  19. A can do half as much again as B can do in the same time. A alone can do a piece of work in 4242 days. In how many days can B alone finish it?
    Answer and solution

    Answer: (A) 6363 days

    If B does xx, A does x2+x=3x2\displaystyle \frac{x}{2} + x = \frac{3x}{2} in the same time

    Ratio of working of A and B =32:1=3:2\displaystyle = \frac{3}{2} : 1 = 3 : 2

    Ratio of time taken for the same work =2:3= 2 : 3

    Time taken by A =2k=42= 2k = 42 ⇒ k=21k = 21

    Time taken by B =3k=63= 3k = 63 days

  20. A man can row downstream at the rate of 2424 km/hr and upstream at 1313 km/hr. The rate of the current is
    Answer and solution

    Answer: (C) 5.55.5 km/hr

    Given x+y=24x + y = 24 and x−y=13x - y = 13

    Subtracting, 2y=112y = 11 ⇒ y=5.5y = 5.5 km/hr

  21. A man rows upstream 3131 km and downstream 6262 km, taking 55 hours each time. The man's speed in still water is
    Answer and solution

    Answer: (B) 9.39.3 km/hr

    Speed downstream: x+y=625\displaystyle x + y = \frac{62}{5} km/hr

    Speed upstream: x−y=315\displaystyle x - y = \frac{31}{5} km/hr

    Adding, 2x=625+315=935\displaystyle 2x = \frac{62}{5} + \frac{31}{5} = \frac{93}{5}

    x=9310=9.3\displaystyle x = \frac{93}{10} = 9.3 km/hr

  22. A man can row 1313 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
    Answer and solution

    Answer: (D) 6.56.5 km/hr

    Let the rate of the stream be yy km/hr

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

    Given 3(13−y)=13+y3(13 - y) = 13 + y

    ⇒ 39−3y=13+y39 - 3y = 13 + y ⇒ 4y=264y = 26

    ⇒ y=6.5y = 6.5 km/hr

  23. A man can row 1818 km/hr in still water. If the river is running at 66 km/hr, it takes him 3 hours to row to a place and back. The place is at a distance of
    Answer and solution

    Answer: (B) 2424 km

    Let the required distance be dd km

    Speed downstream =18+6=24= 18 + 6 = 24 km/hr

    Speed upstream =18−6=12= 18 - 6 = 12 km/hr

    Time =DistanceSpeed\displaystyle = \frac{\text{Distance}}{\text{Speed}}, so d24+d12=3\displaystyle \frac{d}{24} + \frac{d}{12} = 3

    ⇒ d(124+112)=3\displaystyle d\left(\frac{1}{24} + \frac{1}{12}\right) = 3

    ⇒ d×18=3\displaystyle d \times \frac{1}{8} = 3

    ⇒ d=24d = 24 km

  24. The current of a stream runs at the rate of 44 km/hr. A motor boat goes 2424 km upstream and back again to the starting point in 1 hour 45 minutes. The speed of the motor boat in still water is
    Answer and solution

    Answer: (D) 2828 km/hr

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

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

    Total time =105= 105 min =10560\displaystyle = \frac{105}{60} hr =74\displaystyle = \frac{7}{4} hr

    24x−4+24x+4=74\displaystyle \frac{24}{x - 4} + \frac{24}{x + 4} = \frac{7}{4}

    ⇒ 24((x+4)+(x−4)(x−4)(x+4))=74\displaystyle 24\left(\frac{(x + 4) + (x - 4)}{(x - 4)(x + 4)}\right) = \frac{7}{4}

    ⇒ 48xx2−16=74\displaystyle \frac{48x}{x^2 - 16} = \frac{7}{4}

    ⇒ 192x=7x2−112192x = 7x^2 - 112

    ⇒ 7x2−192x−112=07 x^{2} - 192 x - 112 = 0

    ⇒ (x−28)(7x+4)=0\left(x - 28\right) \left(7 x + 4\right) = 0

    ⇒ x=28x = 28 (or) x=−47\displaystyle x = -\frac{4}{7} (rejected, a speed cannot be negative)

    ∴ Speed of the boat in still water =28= 28 km/hr

  25. A man can row 1515 km upstream and 77 km downstream in 44 hours, while he can row 2525 km upstream and 77 km downstream in 66 hours. The rate of the current is
    Answer and solution

    Answer: (C) 11 km/hr

    Let the speed of the man in still water be xx km/hr and the rate of the current yy km/hr.

    Speed upstream =(x−y)= (x - y) km/hr, speed downstream =(x+y)= (x + y) km/hr. Put p=1x−y\displaystyle p = \frac{1}{x - y}, q=1x+y\displaystyle q = \frac{1}{x + y}

    15x−y+7x+y=4\displaystyle \frac{15}{x - y} + \frac{7}{x + y} = 4 ⇒ 15p+7q=4⋯(1)15p + 7q = 4 \quad \cdots (1)

    25x−y+7x+y=6\displaystyle \frac{25}{x - y} + \frac{7}{x + y} = 6 ⇒ 25p+7q=6⋯(2)25p + 7q = 6 \quad \cdots (2)

    (1) ×5\times 5 ⇒ 75p+35q=2075p + 35q = 20; (2) ×3\times 3 ⇒ 75p+21q=1875p + 21q = 18

    (1) ×5\times 5 −- (2) ×3\times 3 ⇒ 14q=214q = 2 ⇒ q=17\displaystyle q = \frac{1}{7}

    From (1): 15p+1=415p + 1 = 4 ⇒ 15p=315p = 3 ⇒ p=15\displaystyle p = \frac{1}{5}

    So x−y=5x - y = 5 and x+y=7x + y = 7

    Adding, 2x=122x = 12 ⇒ x=6x = 6; then y=7−6=1y = 7 - 6 = 1