समय और कार्य प्रश्नावली 3

In 8 days, Anil does = ⅓rd work.
∴ in 1 day, he does = ¹⁄₂₄th work
∴ Rakesh’s one day’s work
= 60% of ¹⁄₂₄ = ¹⁄₄₀ th work.
Remaining work

(Anil and Rakesh)’s one day’s work
=
= ¹⁄₁₅th work
Now, ¹⁄₁₅th work is done by them in one day.
∴ ⅔rd work is done by them in 15 × ⅔ = 10 days

A’s one day’s work = ¹⁄₃₂

B’s one day’s work = ¹⁄₂₀

(B + C)’s one day’s work =¹⁄₁₂

∴ C’s one day’s work

=

D’s one day’s work = ¹⁄₂₄

∴ (A + B + C + D)’s one day’s work

=

=

∴ Out of ⁵⁄₃₂ of work done, ¹⁄₃₀ of the work is done by C.

⇒ Out of ₹25 paid for the work, C will receive

₹

i.e.

i.e. ₹.¹⁶⁄₃

A’s one day’s work = ¹⁄₁₅th work.

B’s one day’s work = ¹⁄₁₀th work.

(A + B)’s one day’s work

work.

Let A left after x days.

∴ (A +B)’s x days’ work = ¹⁄₆th work.

Remaining work

work.

Now, in 5 days, work done by

.

∴ In 1 day work done by B

and

∴ x = 3 days

Let Suresh undertakes a tour of x days.

Then, expenses for each day = 360/x

Now,

⇒ = 3

⇒ x² + 4x – 480 = 0

⇒ x = – 24 or x = 20

Since, x ≠ -24 we have x = 20

Let B can finish the work in x days.

Then A can finish the work in (x – 3) days.

B’s one day’s work

A’s one day’s work work

A’s 4 days’ work =

Remaining work

The remaining work done by B in 14 – 4

= 10 days.

Now, in 10 days, work done by

th work

∴ In 1 day, work done by B

= work

and

⇒ x = 15 days

∴ B will finish in 15 days and A will finish in 12 days

(A + B)’s one day’s work = ⅕th work

Let A can do job in x days. Then,

A’s one day’s work

and B’s one day’s work

Now ,

Let 1 man’s 1 day’s work = x and

1 boy’s 1 day’s work = y.

Then, 6x + 8y = ¹⁄₁₀

and 26x + 48y = ½

Solving these two equations, we get :

∴ (15 men + 20 boys)’s 1 day’s work

=

∴ 15 men and 20 boys can do the work in 4 days.

Let the required number of days be x.

Less persons, More days (Indirect Proportion)

More working hrs per day, Less days (Indirect Proportion)

Let the required length be x metres.

More breadth, Less length (Indirect Proportion)

More depth, Less length (Indirect Proportion)

More days, More length (Direct Proportion)

∴ 20 × 15 × 10 × x = 50 × 10 × 30 × 100

Let the required number of days be x.

8 men = 17 boys

⇒ 4 men

∴ 4 men and 24 boys

Now, More boys , less days (Indirect Proportion)

But work → 50 × 0.9 times

∴ Required days

Let the required men be x.

More hours, less men (Indirect proportion)

More days, less men (Indirect proportion)

∴ 5 × 8 × 18 = 8 × 6 × x

Let the required number of mats be x.

More weavers, More mats (Direct Proportion)

More days, More mats (Direct Proportion)

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

∴ 1 × 40 × x = 40 × 1 × 40

⇒ x = 40

More machines, less hours (Indirect Proportion)

Less days, more hours (Indirect Proportion)

More amount of coal, more hours (Direct Proportion)

Less efficiency, more hours (Indirect Proportion)

⇒ 3 × 6 × 9,000 × 0.8 × x

= 2 × 8 × 12,000 × 0.9× 12

⇒ x = 16 hrs

Let M denotes man and B denotes boy.

(M + B)’s 1 day’s work = ¹⁄₄₀

i.e.

Ratio of their skill = ⁸⁄₅

i.e.

Let efficiency of a man of 1 days work = x

i.e.

Now,

Now,

⇒ M = 65 and

(100 × 35+ 200 × 5) men can finish the work in 1 day.

i.e., 4500 men can finish the work in 1 day

∴ 100 men can finish the work in 45 days

∴ The work would be 5 days behind the schedule.

Remaining work

Remaining time = 56 – 30 = 26 days

More work, more men (Direct Proportion)

Less days, more men (Indirect Proportion)

More hours, Less men (Indirect Proportion)

⇒ x = 160

∴ Additional men to be employed = 160 – 104 = 56 men

1 minute’s work of both the punctures

=

So, both the punctures will make the tyre flat in