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

Answer: (c)

Let A and C complete the work in x days

(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 = $1/x$

Then (A + B + B + C + C + A)’s 1 day’s work = $1/8 + 1/12 + 1/x$

2(A + B + C)’s 1 day’s work = ${3x + 2x + 24}/{24x}$

(A + B + C)’s 1 day’s work = ${5x + 24}/{24x × 2}$

According to the question,

(A + B + C)’s 1 day’s work = $1/6$

$1/6 = {5x + 24}/{48x}$

30x + 144 = 48x ⇒ x = $144/18$ = 8 days.

Using Rule 5,

Let the time taken by A and C bc x days

Total time taken = ${2 × 8 × 12 × x}/{8 × 12 + 12 × x + 8 × x}$

6 = ${192x}/{96 +20x}$

576 + 120x = 192x

72x = 576 ⇒ x = 8

Time taken by A and C is 8 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

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: (c)

(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

Answer: (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

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

Answer: (d)

(A + B)’s 1 hour’s work = $2/9$ .....(i)

(B + C)’s 1 hour’s work = $1/3$ ......(ii)

(C + A)’s 1 hour’s work = $4/9$ .....(iii)

Adding all three equations,

2 (A + B + C)’s 1 hour’s work = $2/9 + 1/3 + 4/9$

${2 + 3 + 4}/9$ = 1

A, B and C together will complete the work in 2 hours.

Using Rule 5,

Time taken = ${2 × 9/2 × 3 × 9/4}/{9/2 × 3 + 3 × 9/4 + 9/2 × 9/4}$

= ${{18 × 27}/8}/{27/2 + 27/4 + 81/8}$

= ${18 × 27}/8 × 8/({108 + 54 + 81})$

= ${18 × 27}/243$ = 2 hours