Model 1 Basics On Time & Work Practice Questions Answers Test With Solutions & More Shortcuts

Question : 26 [SSC Constable (GD) 2012]

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) 2 days

b) 4 days

c) 16 days

d) 8 days

Answer: (b)

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

Question : 27 [SSC GL Tier-I 2013]

A and B can do a work in 12 days, B and C in 15 days and C and A in 20 days. If A, B and C work together, they will complete the work in :

a) 15$2/3$ days

b) 10 days

c) 5 days

d) 7$5/6$ days

Answer: (b)

According to question,

A and B can do a work in 12 days

(A + B)'s one day's work = $1/12$

Similarly, (B + C)'s one day's work = $1/15$

and (C + A)'s one day's work = $1/20$

On adding all three,

2 (A + B + C)'s one days's work

= $1/12 + 1/15 + 1/20 = {10 + 8 + 6}/120 = 1/5$

(A + B + C)'s one days's work = $1/10$

A, B and C together can finish the whole work in 10 days.

Using Rule 5
If A 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 total time taken, when A, B and C
work together =$2/{(1/x + 1/y + 1/z)}$ OR ${2xyz}/{xy + yz + zx}$ days

Time taken = ${2 × 12 × 15 × 20}/{12 × 15 + 15 × 20 + 20 × 12}$

= ${24 × 300}/{180 + 300 + 240} = 7200/720$ = 10 days.

Question : 28 [SSC GL Tier-I 2013]

A work can be completed by P and Q in 12 days, Q and R in 15 days, R and P in 20 days. In how many days P alone can finish the work?

a) 60 days

b) 30 days

c) 10 days

d) 20 days

Answer: (b)

(P + Q)’s 1 day’s work = $1/12$ ...(i)

(Q + R)’s 1 day’s work = $1/15$ ...(ii)

(R + P)’s 1 day’s work = $1/20$ ...(iii)

Adding all three equations,

2 (P + Q + R)’s 1 day’s work = $1/12 + 1/15 + 1/20$

${5 + 4 + 3}/60 = 12/60 = 1/5$

(P + Q + R)’s 1 day’s work = $1/10$ ...(iv)

P’s 1 day’s work = $1/10 - 1/15 = {3 - 2}/30 = 1/30$

P alone will complete the work in 30 days.

Question : 29 [SSC CGL Prelim 2000]

A particular job can be completed by a team of 10 men in 12 days. The same job can be completed by a team of 10 women in 6 days. How many days are needed to complete the job if the two teams work together?

a) 18 days

b) 9 days

c) 4 days

d) 6 days

Answer: (c)

According to question,

10 men’s one day’s work = $1/12$

1 man one day’s work =$1/{12 × 10} = 1/120$

Similarly, 1 woman one day’s work = $1/{6 × 10} = 1/60$

(1 man + 1 woman)’s one day’s work = $1/120 + 1/60$

= ${1 + 2}/120 = 3/120 = 1/40$

(10 men + 10 women)’s one day’s work = $10/40 = 1/4$

Therefore, both the teams can finish the whole work in 4 days.

Question : 30 [SSC CPO S.I.2009]

A and B can do a piece of work in 10 days. B and C can do it in 12 days. C and A in 15 days. In how many days will C finish it alone ?

a) 60 days

b) 40 days

c) 24 days

d) 30 days

Answer: (b)

(A + B)'s 1 day's work = $1/10$ ... (i)

(B + C)'s 1 day’s work = $1/12$ ... (ii)

(C + A)’s 1 day’s work = $1/15$ ... (iii)

On adding all these,

2(A + B + C)’s 1 day’s work = $1/10 + 1/12 + 1/15$

= ${6 + 5 + 4}/60 = 1/4$

(A + B + C)’s 1 day’s work = $1/8$ … (iv)

C’s 1 day’s work = $1/8 - 1/10 = {5 - 4}/40 = 1/40$

C will finish the work in 40 days.

Using Rule 19,

C alone can do in = ${2 × 10 × 12 × 15}/{10 × 12 - 12 × 15 + 10 × 15}$

= ${240 × 15}/{120 - 180 + 150} = {240 × 15}/90$ = 40 days

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