Model 1 Basics on Time & Work Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

A alone can complete the work in 42 days working 1 hour daily. Similarly, B will take 56 days working 1 hour daily.

A 's 1 day's work = $1/42$

B 's 1 day's work = $1/56$

(A + B) 's 1 day's work = $1/42 + 1/56 = {4 + 3}/168 = 7/168$

Time taken by (A + B) working 8 hours daily

= $168/{7 × 8}$ = 3 days

Here, $h_1$ = 7 hours, $h_2$ = 7 hours, $d_1$ = 6 days, $d_2$ = 8 days, h = 8 hours

Required Time = $[{({7 × 6}) × ({7 × 8})}/{7 × 6 + 7 × 8}] × 1/8$

= ${42 × 56}/98 × 1/8 = 2352/98 × 1/8 = 24/8$ = 3 days

Answer: (b)

(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

Answer: (a)

Work done by (A + B) in 1 day = $1/30$

Work done by (B + C) in 1 day = $1/20$

Work done by (C + A) in 1 day = $1/15$

On adding,

Work done by 2 (A +B + C) in 1 day

= $1/30 + 1/20 + 1/15 = {2 + 3 + 4}/60$

= $9/60 = 3/20$

Work done by (A + B + C) in 1 day = $3/40$

(A + B + C) will do the work in $40/3 = 13{1}/3$ days

Using Rule 5,

Time taken = ${2 × 30 × 20 × 15}/{30 × 20 + 20 × 15 + 15 × 30}$

= $18000/{600 + 300 + 450}$

= $18000/1350 = 13{1}/3$ days

Answer: (d)

Work done by (A + B + C) in 1 minute = $1/30$

Work done by (A + B) in 1 minute = $1/50$

Work done by C alone in 1 minute = $1/30 - 1/50$

= ${5 - 3}/150 = 2/150 = 1/75$

C alone will complete the work in 75 minutes.

Using Rule 4,

C alone can do in = ${xy}/{x - y}$

= ${50 × 30}/{50 - 30}$ = 75 minutes

Answer: (b)

(A + B)’s 1 day’s work = $1/10$

(B + C)’s 1 day’s work = $1/6$

(C + A)’s 1 day’s work = $1/12$

Adding all three

2 (A + B + C)’s 1 day’s work = $1/10 + 1/6 + 1/12$

${6 + 10 + 5}/60 = 21/60 = 7/20$

(A + B + C)’s 1 day’s work = $7/40$

All three together will complete the work in $40/7 = 5{5}/7$ = days

Using Rule 5,

Time taken = ${2 × 10 × 6 × 12}/{10 × 6 + 6 × 12 + 12 × 10}$

= $1440/{60 + 72 + 120}$

= $1440/252 = 40/7 = 5{5}/7$ days

Answer: (b)

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.

Time taken = ${6 × 9}/{9 + 6} = 54/15$ = 3.6 days