Practice Basics on time and work - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A can do a work in 6 days and B in 9 days. How many days will both take together to complete the work?
(a)
(b)
(c)
(d)
According to question,
A can finish the whole work in 6 days.
A’s one day’s work = $1/6$
Similarly, B’s one day’s work = $1/9$
(A + B)’s one day’s work
= $(1/6 + 1/9) = ({3 + 2}/18) = 5/18$
Therefore, (A + B)’s can finish the
whole work in $18/5$ days i.e., 3.6 days.
Using Rule 2If A completes a piece of work in 'x' days, and B completes the same work in 'y' days, then,Work done by A in 1 day = $1/x$, Work done by B in 1 day = $1/y$ Work done by A and B in 1 day = $1/x + 1/y = {x + y}/{xy}$Total time taken to complete the work by A and B both = $({xy}/{x + y})$
Time taken = ${6 × 9}/{9 + 6} = 54/15$ = 3.6 days
Q-2) A and B can complete a piece of work in 8 days, B and C can do it in 12 days, C and A can do it in 8 days. A, B and C together can complete it in
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/8$
(B + C)’s 1 day’s work = $1/12$
(C + A)’s 1 day’s work = $1/8$
On adding,
2 (A + B + C)’s 1 day’s work = $1/8 + 1/12 + 1/8$
${3 + 2 + 3}/24 = 8/24 = 1/3$
(A + B + C)’s 1 day’s work = $1/6$
Hence, the work will be completed in 6 days.
Method 2 :
Time = ${2xyz}/{xy + yz + zx}$
(Here, x = 8, y = 12; z = 8)
= ${2 × 8 × 12 × 8}/{96 + 96 + 64}$
${2 × 8 × 12 × 8}/256$ = 6 days.
Using Rule 5,
Time taken = ${2 × 8 × 12 × 8}/{8 × 12 + 12 × 8 + 8 × 8}$
= ${16 × 96}/{96 + 96 + 64} = {16 × 96}/256$ = 6 days
Q-3) A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in
(a)
(b)
(c)
(d)
A’s 1 day’s work = $1/24$
B’s 1 day’s work = $1/6$
C’s 1 day’s work = $1/12$
(A + B + C)’s 1 day’s work
=$1/24 + 1/6 + 1/12 = {1 + 4 + 2}/24 = 7/24$
The work will be completed by them in $24/7$ i.e., 3$3/7$ days
Using Rule 3If A can do a work in 'x' days, B can do the same work in 'y' days, C can do the same work in 'z' days then, total time taken by A, B and C to complete the work together = $1/{1/x + 1/y + 1/z} = {xyz}/{xy + yz + zx}$and If workers are more than 3 then total time taken by A, B, C ...... so on to complete the work together = $1/{1/x + 1/y + 1/z + ...}$
Time taken = ${24 × 6 × 12}/{24 × 6 + 6 × 12 + 24 × 12}$
= $1728/{144 + 72 + 288}$
= $1728/504 = 24/7 = 3{3}/7$ days
Q-4) A, B and C individually can do a work in 10 days, 12 days and 15 days respectively. If they start working together, then the number of days required to finish the work is
(a)
(b)
(c)
(d)
Work done by A, B and C in 1 day
= $1/10 + 1/12 + 1/15 = {6 + 5 + 4}/60$
= $15/60 = 1/4$
Required time = 4 days
Using Rule 3,
Time Taken = ${xyz}/{xy + yz + zx}$
= ${10 × 12 × 15}/{10 × 12 + 12 × 15 + 15 × 10}$
= $1800/{120 + 180 + 150}$
= $1800/450$ = 4 days
Q-5) A alone can complete a work in 12 days. A and B together can complete it in 8 days. How long will B alone take to complete the work ?
(a)
(b)
(c)
(d)
A’s 1 day’s work = $1/12$
(A+B)’s 1 day’s work = $1/8$
B’s 1 day’s work =$1/8 - 1/12 = {3 - 2}/24 = 1/24$
B alone can do the work in 24 days.
Using Rule 4,
Time taken by B = ${12 × 8}/{12 - 8}$ = 24 days
Q-6) If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work ?
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/15$
B’s 1 day’s work = $1/20$
A’s 1 day’s work = $1/15 - 1/20 = {4 - 3}/60 = 1/60$
A alone will do the work in 60 days.
Using Rule 4,
A alone do in = ${15 × 20}/{20 - 15}$
= ${15 × 20}/5$ = 60 days
Q-7) If A and B together can complete a work in 12 days, B and C together in 15 days and C and A together in 20 days, then B alone can complete the work in
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/12$
(B + C)’s 1 day’s work = $1/15$
(C + A)’s 1 day’s work = $1/20$
On adding,
2 (A + B + C)’s 1 day's work = $1/12 + 1/15 + 1/20$
= ${5 + 4 + 3}/60 = 1/5$
(A+B+C)’s 1 day’s work = $1/10$
B’s 1 day’s work = $1/10 - 1/20 = {2 - 1}/20 = 1/20$
B alone can do the work in 20 days.
Using Rule 19,
B alone can do in = ${2 × 12 × 15 × 20}/{-12 × 15 + 15 × 20 + 20 × 12}$
= ${24 × 300}/{-180 + 300 + 240} = {24 × 300}/360$ = 20 days
Q-8) A and B can do a piece of work in 12 days, B and C in 8 days and C and A in 6 days. How long would B take to do the same work alone ?
(a)
(b)
(c)
(d)
(A + B)'s 1 day's work = $1/12$ ...(i)
(B + C)'s 1 day's work = $1/8$ ...(ii)
(C + A)'s 1 day's work = $1/6$ ...(iii)
On adding,
2(A + B + C)'s 1 day's work = $1/12 + 1/8 + 1/6$
= ${2 + 3 + 4}/24 = 9/24$
(A+ B + C)'s 1 day’s work = $9/{24 × 2} = 9/48$ ...(iv)
On, subtracting (iii) from (iv),
B’s 1 day’s work = $9/48 - 1/6$
= ${9 - 8}/48 = 1/48$
B can complete the work in 48 days.
Using Rule 19,
B alone can do in = ${2 × 12 × 8 × 6}/{-12 × 8 + 8 × 6 + 6 × 12}$
= ${24 × 48}/{-96 + 48 + 72} = {24 × 48}/{-96 + 120}$
= ${24 × 48}/24$ = 48 days
Q-9) A and B can do a piece of work in 10 days, B and C in 15 days and C and A in 20 days. C alone can do the work in :
(a)
(b)
(c)
(d)
According to the question
Work done by A and B together in one day = $1/10$ part
Work done by B and C together in one day = $1/15$ part
Work done by C and A together in one day = $1/20$ part
So, A + B = $1/10$ ....(I)
B + C = $1/15$ ...(II)
C + A = $1/20$ ....(III)
Adding I, II, III, we get
2 (A + B + C) = $1/10 + 1/15 + 1/20$
2 (A + B + C) = ${6 + 4 + 3}/60 = 13/60$
A + B + C = $13/120$ ....(IV)
Putting the value of eqn. (I) in eqn. (IV)
$1/10 + C =13/120$
C = $13/120 - 1/10 = {13 - 12}/120 = 1/120$
Work done in 1 day by C is $1/120$ part
Hence, C will finish the whole work in 120 days
Using Rule 19A and B can do a work in 'x' days, B and C can do the same work in 'y' days. C and A can do the same work in 'z' days. Then, all can do alone the work as following:A alone can do in =${2xyz}/{xy + yz - zx}$daysB alone can do in =${2xyz}/{-xy + yz + zx}$daysC alone can do in =${2xyz}/{xy - yz + zx}$days
Time Taken by C= ${2xyz}/{xy - yz + zx}$
= ${2 × 10 × 15 × 20}/{10 × 15 - 15 × 20 + 20 × 10 }$
= $6000/{150 - 300 + 200} = 6000/50$ = 120 days
Q-10) A and B together can complete a piece of work in 18 days, B and C in 24 days and A and C in 36 days. In how many days, will all of them together complete the work ?
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/18$
(B + C)’s 1 day’s work = $1/24$
(A + C)’s 1 day’s work = $1/36$
Adding all three,
2 (A + B + C)’s 1 day’s work= $1/18 + 1/24 + 1/36$
= ${4 + 3 + 2}/72 = 1/8$
(A + B + C)’ 1 day’s work = $1/16$
A, B and C together will complete the work in 16 days.
Using Rule 5,
Total time taken = ${2 × 18 × 24 × 36}/{18 × 24 +24 × 36 + 36 × 18}$
= ${36 × 24 × 36}/{432 + 864 + 648}$
= $31104/1944$ = 16 days