Rd Sharma Solutions Class 8 Chapter 11 Time and Work
Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 11, Time and Work. Here students can easily find Exercise wise solution for chapter 11, Time and Work. Students will find proper solutions for Exercise 11.1 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.
Time and Work Exercise 11.1 Solution :
Question no – (1)
Solution :
Rakesh need 20 day to complete a piece of work
∴ Rakesh need 1 day to complete a piece of work
= 1/20 part
∴ Rakesh need 4 day to complete a piece of work
= 4/20
= 1/5
Therefore, Rakesh can do 1/5 piece of work in 4 days.
Question no – (2)
Solution :
Let, Total work of painting = 1
Rohan can paint = 1/3th part
∴ Rest part of painting
= (1 – 1/3)
= 3 – 1/3
= 2/3th part
Now, Rohan can paint 1/3th part in 6 days
∴ Now, Rohan can paint 1 th part in 6/(1/3) days
= 6 × 3
= 18 Days
Therefore, Rohan can complete 1 work in 18 days.
Question no – (3)
Solution :
Anil can complete in 5 days 1 work
∴ Anil can complete in 1 days 1/5 work
Ankur can complete in 4 day 1 work
Ankur can complete in 1 days 1/4 work
∴ (Anil + Ankur) can complete in 1 day,
= 1/5 + 1/4
= 4 + 5/20
= 9/20th
∴ (Anil + Ankur) can complete in 1 days 1 work,
= 20/9
= 2 2/9 days
Therefore, it will take 2 2/9 days if they work together.
Question no – (4)
Solution :
Mohan takes to do in 9 hr = 1 work
∴ Mohan takes to do in 1 hr = 1/9 work
Mohan + Sohan takes to do in 4 hr = 1 wok
∴ Mohan + Sohan takes to do in 1 hr = 1/4 work
∴ Sohan takes to do in 1 hr,
= (1/4 – 1/9)
= 9 – 4/36
= 5/36
Therefore, Sohan will takes to do 36/5 hr in 1 work.
Question no – (5)
Solution :
Sita can finish in 9 hr = 100 page document
∴ Sita can finish in 1 hr = 100/9 page document
Mita can finish in 6 hr 100 page document
∴ Mita can finish in 1 hr = 100/6 page document
Rita can finish in 12 hr = 100 page document
∴ Rita can finish in 1 hr = 100/012 page document
∴ (Sita + Mita + Rita) can finish in 1 hr
= (100/9 + 100/6 + 100/12)
= 400+600 + 30/36
= 1300/36 work
∴ (Sita + Mita + Rita) can finish 1300/36 in 1 hr
∴ (Sita + Mita + Rita) can finish 100 in,
= 100 × 36 /1300 hr
= 36/13 hr.
Therefore, together they will take 36/13 hours to complete the work.
Question no – (6)
Solution :
(A + B + C) can do in 8 hr = 1 work.
∴ (A + b + c) can do in 8 hr = 1/8 work
A can do in 20 hr 20 hr = 1 work
∴ A can do in 1 hr = 1/20 work
B can do in 24 hr = 1 work
∴ B can do in 1 hr = 1/24 work
∴ C can do in work 1 hr,
= 1/8 – (1/20 + 1/24)
= 1/8/ – ( 6 +5/120)
= 1/8 – 11/120
= 15 – 11/120 = 4/120
= 1/30 work
Therefore, C can do 1 work in 30 hours.
Question no – (8)
Solution :
(A + B) can do 12 days = 1 work
∴ (A + B) can do 1 days = 1/12 work
(B + C) can do 15 days = 1 work
∴ (B + C) can do 1 days = 1/15 work
(A + C) can do 20 days = 1 work
∴ (A+ C) can do 1 days = 1/20 work
Now, (A + B) + (B + C) + (A + C) can do in 1 day = (1/12 + 1/15 + 1/20)
or, 2A + 2B + 2C can do in 1 day = (5 + 4 + 3/60)
or, (A + B+ C) can do in 1 day = 12/60
or, (A+ B + C) can do in 1 day = 1/5 × 2 = 1/10
∴ (A + B + C) can do 1 work = 10 days
Now, (A + B + C) – (B + C) = A = B + C – B – C
∴ A can do in 1 day
= 1/10 – 1/15
= 3 – 2/30
= 1/30
Therefore, A can do 1 work in 30 days.
Question no – (9)
Solution :
(A + B + C) can do 1 work = 15 3/4 = 63/4 days
∴ (A + B+ C) can do in 1 days = 4/63 work
(B + C + D) can do in 14 days = 1 work
∴ (B + C + D) can do in 1 work = 1/14 work
(C + D + A) can do in 18 days = 1 work
∴ (C + D + A) can do in 1 days = 1/18 work
(D + A + B) can do in 21 days = 1 work
∴ (D + A + B) can do in 1 days = 1/21 work
∴ (A + B+ C) (B + C + D) + (C + D + A) + (D + A + B) can do 1 day
= (4/63 + 1/14 + 1/18 + 1/21)
or, A + B + C + B + C + D + A + D + A + B = 8 + 9 + 7 + 6/126
or, 3A + 3B + 3C + 3D = 30/126
or, 3 (A + B + C + D) = can do in 1 day = 30/126
or, (A + B + C + D) can do in 1 day
= 30/126 × 3
= 5/63
∴ (A + B + C + D) can do 1 work
= 63/5
= 12 3/5 Days
Question no – (10)
Solution :
Let, Total work = 1
A can do = 1/4th
The rest part,
= (1 – 1/4)
= 4 – 1/4
= 3/4th
∴ A can do 1/4th in 12 days
∴ A can do 1 work in = 12/1/4 = 48 days
∴ A can do 1 days = 1/48 work
∴ (A + B) can do 1 days in = 10 days
∴ (A + B) can do in 1 days = 1/10 work
∴ B can do in 1 days (10/-1/48)
= 24 – 5/240 = 19/240
∴ B can do in 1 work
= 240/19
= 12 12/19 days
Therefore, B can polish the floor alone in 12 12/19 days.
Question no – (11)
Solution :
(A + B) can do in 20 days = 1 work
∴ (A + B) can do in 1 days = 1/20 work
A alone can do in 1/5th = 12 days
∴ A alone can do in 1 work
= 12 /1/5
= 12 × 5
= 60 days
∴ A alone can do in 60 days = 1 work
∴ A alone can do in 1 days = 1/60 work
∴ B alone can do in 1 days
= (1/20 – 1/60)
= ( 3 – 1/60)
= 2/60
= 1/30 work
Therefore, B alone can do the work in 1/30 days.
Question no – (12)
Solution :
(A + B) can do in 20 days = 1 work
(A + B) can do in 1 days = 1/20 work
B can do in 15 days = 1 work
∴ B can do in 1 days = 1/15 work
∴ (A + B) – B = A can do in 1 days
= (1/20 – 1/15)
= 3 – 4/60
= 1/60 days
∴ A can do in 1 work = 60 days
(A + B) can do it 2 days
= 2/20
= 1/10
Total work = 1
∴ Work remaining,
= 1 – 1/10
= 10 – 1/10
= 9/10
∴ B can finish 1 work = 15 days
∴ B can finish 9/10 work
= 15 × 9/10
= 27/2
= 13 1/2 days
Therefore, B can finish the remaining work in 13 1/2 days.
Question no – (13)
Solution :
A can do in 40 days = 1 work
∴ A can do in 1 days = 1/40 work
B can do in 45 days = 1 work
∴ B can do it 1 days = 1/45 work
∴ (A + B) can do in 1 work
= (1/40 + 1/45)
= 9 + 8/360
= 17/360 work
∴ (A + B) can do in 10 days,
= 17/ 360 × 10
= 17/36 work
Total work = 1
∴ Remaining work,
= 1 – 17/36
= 36 -17/36
= 19/36 work
A can do in 1 work = 40 days
∴ A can do in 19/36 work,
= 19/36 × 40
= 190/9 days
= 21 1/9 days.
Therefore, ‘A’ will finish the remaining work in 21 1/9 days.
Question no – (14)
Solution :
Aasheesh can paint 1 work = 20 min
∴ Aasheesh can paint in 1 min = 1//20 work
Chinki can paint in 25 min = 1 work
∴ Chinki can paint 1 min = 1/25 work
(Aasheesh + Chinki) can paint 1 min = 1 work
= (1/20 + 1/25+
= 5 + 4/100
= 9/100
∴ (Aasheesh + Chinki) can paint 5 min
= 9/100 × 5
= 90/20
∴ Total work = 1
∴ Remaining work = 1 – 9/20
= 20 – 9/20
= 11/20
∴ Aasheesh complete 11/20 work
= 11/20 ×
= 11 min
Therefore, Aasheesh finish painting of the remaining doll in 11 minutes.
Question no – (16)
Solution :
6 men can complete the electric fitting = 7 days
∴ 1 men can complete the electric fitting = 6 × 7 days
Let, men can complete the electric fitting = x days
∴ 1 men can complete the electric fitting = 21x days
Now, according to question,
6 × 7 = 21 × x
or, x = 6 × 7/21 = 2 days
Therefore, if 21 men do the job it will taker 2 days.
Question no – (17)
Solution :
8 men can do 1 work = 9 days
∴ 1 men can do 11 work = 8 × 9 days
Let, 6 men can do 11 work = x days
∴ 1 men can do 11 work = x × 6 days
Now, according to question,
= 8 × 9 = x × 6
or, x = 48 × 9 / 6
= x = 12 days
Therefore, the 6 men will do it in 612 days.
Question no – (18)
Solution :
As per the given question,
35 baskets weaves = 25 days
∴ 1 baskets weaves = 25/35 days
∴ 55 baskets weaves,
= 25/35 × 11/55
= 275/7
= 39 2/7 days
Therefore, Reema will weave 55 baskets in 39 2/7 days.
Question no – (19)
Solution :
14 hr need to types = 75 pages
∴ 1 hr need to types = 75/14 pages
∴ 20 hr need to types,
= 75 × 20/14
= 750/7
= 107 1/7 pages
Therefore, Neha will type 107 1/7 pages.
Question no – (20)
Solution :
Earning of 12 boys in 7 days = 840
∴ Earning of 12 boys in 1 days
= 840/7
= 120
∴ Earning of 1 boys in 1 days
= 120/12
= 10
∴ Earning of 1 boys in 6 days
= 10 × 6
= 60
∴ Earning of 15 boys in 6 days
= 60 × 15
= 900
Therefore, 15 boys will earn 900 in 6 days.
Question no – (21)
Solution :
Earning of 25 men in 10 days = 1000
∴ Earning of 25 men in 10 days
= 1000/10
= 100
∴ Earning of 1 men in 1 days
= 100/25
= 4
∴ Earning of 1 man in 15 days
= 4 × 14
= 60
∴ Earning of 15 men in 15 days
= 15 × 60
= 900
Therefore, 15 men will earn 900 in 15 days.
Question no – (22)
Solution :
Ashu can copy 1 book = 18 days
∴ Ashu can worked 1 day = 8 hr
∴ Ashu can worked 18 hr = 18 × 8
∴ Ashu can worked 1 book = 12 days
Let, Ashu can worked 1 day = x hr
∴ Ashu can worked 12 day = 12 hr
Now, according to question,
12 x = 144
or, x = 144/12
= x = 12 hr
Therefore, Ashu should work 12 hr to finish the work in 12 days.
Question no – (23)
Solution :
Let, m₁ = 9, t₁ = 3 hr, w₁= 135
And, m₂ = ?, t₂ = 1 hr, w₂ = 270
∴ According to question,
m₁t₁/w₁ = m₂ t₂ / w₂
or, m₂ = m₁t₁w₁/t₂w₁
= 9 × 3× 270/1 × 135
= 1 × 135
= 54 girls
Therefore, 54 girls are needed to prepare 270 garlands in 1 hour.
Question no – (24)
Solution :
A cistern can be filled by 8 hr by 1 tap
A cistern can be filled by 1 hr in = 1/8
Another A cistern can be filled by 4 hr in = 1 hr
∴ A cistern can be filled by 1 hr in = 1/4 hr
∴ Both cistern filled 1 hr
= (1/8 + 1/4)
= 1 +2/8 = 3/8
∴ Both cistern filled the taps,
= 8/3 hr
= 2 2/3 hr
Therefore, it will take 2 2/3 hr to fill the tap.
Question no – (25)
Solution :
A can fill 1 tank = 10 hr
∴ A can fill 1 hr = 1/10
B can fill in 15 hr = 1 tank
∴ B can fill in 1 hr = 1/15 tank
∴ (A + B) can fill 1 tank,
= (1/10 + 1/15)
= (3 + 2/30)
= 5/30
= 1/6
Let, (A + B) can be fill 1 tank = x hr
∴ (A + B) can be fill 1 hr = 1/x
Now, according to question,
x/1 = 1/6
or, x = 6
Now, Both tap filled in 4 hr
= 4 × 1/6
= 2/3
Total tank = 1
∴ Remaining,
= 1 – 2/3
= 3 – 2/3
= 1/3
∴ A can filled 1 tank = 10 hr
∴ A can filled 1/3 tan
= 10 × 1/3
= 10/3 hr
= 3 1/3 hr
Therefore, A will take 3 1/3 hr to fill the remaining tank.
Question no – (26)
Solution :
Without leakage a pipe can fill the cistern = 10 hr
∴ Without leakage a pipe can fill in 1 hr = 1/10
With leakage a pipe can fill in 12 hr = 1 cistern
∴ With leakage a pipe can fill in = 1/12 cistern
∴ Due to leakage a pipe can fill in 1 hr,
= 1/10 – 1/12
= 12 – 10/120
= 120/2
= 60 hours
Therefore, it will emptied by the leak in 60 hours.
Question no – (27)
Solution :
A can fill in 12 hr = 1 work
∴ A can fill in 1 hr = 1/12 work
B can fill in 15 hr = 1 work
∴ B can fill in 1 hr = 1/15 work
C can be fill in 10 hr = 1 work
∴ C can be fill in 1 hr = -1 1/0 work [due to inlet emptying take it as negative]
∴ Now (A + B + C) can fill in 1 hr
= (1/12 + 1/15 – 1/10)
= (5 + 4 – 6/60)
= 1/20
∴ Now (A + B + C) can fill a cistern in 20 hr.
Therefore, they will take 20 hours to fill the cistern completely.
Question no – (28)
Solution :
A cistern can fill in 4 hr = 1 tap
∴ A cistern can fill in 1 hr = 1/4
A cistern can fill in 6 hr = 1 pipe
∴ A cistern can fill 1 hr = 1/6
Both (tap and pipe) are opened
∴ Both (tap and pipe) can filled in 1 hr
= (1/4 – 1/6)
= 3 – 6/12
= 1/12
Therefore, Both (tap and pipe) can be filled in 12 hours.
Next Chapter Solution :
👉 Chapter 12 👈
