RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B.

Other Exercises

Question 1.
Solution:
48 men can dig a trench in = 14 days
1 man will dig the trench in = 14 x 48 days (less men more days)
28 men will dig the trench m = \(\frac { 14 x 48 }{ 28 }\) (more men less days)
= 24 days

Question 2.
Solution:
In 30 days, a field is reaped by = 16 men
In 1 day, it will be reaped by = 16 x 30 (less days, more men)
and in 24 days, it will be reaped by = \(\frac { 16 x 30 }{ 24 }\) men (more days, less men)
= \(\frac { 480 }{ 24 }\)
= 20 men

Question 3.
Solution:
In 13 days, a field is grazed by = 45 cows.
In 1 day, the field will be grazed by = 45 x 13 cows (less days, more cows)
and in 9 days the field will be grazed by = \(\frac { 45 x 13 }{ 9 }\) cows (more days, less cows)
= 5 x 13 = 65 cows

Question 4.
Solution:
16 horses can consume corn in = 25 days
1 horse will consume it in = 25 x 16 days (Less horse, more days)
and 40 horses will consume it in = \(\frac { 25 x 16 }{ 40 }\) days (more horses, less days)
= 10 days

Question 5.
Solution:
By reading 18 pages a day, a book is finished in = 25 days
By reading 1 page a day, it will be finished = 25 x 18 days (Less page, more days)
and by reading 15 pages a day, it will be finished in = \(\frac { 25 x 18 }{ 15 }\) days (more pages, less days)
= 5 x 6 = 30 days

Question 6.
Solution:
Reeta types a document by typing 40 words a minute in = 24 minutes
She will type it by typing 1 word a minute in = 24 x 40 minutes (Less speed, more time)
Her friend will type it by typing 48 words 24 x 40 a minute in = \(\frac { 24 x 40 }{ 48 }\) minutes
(more speed, less time)
= 20 minutes

Question 7.
Solution:
With a speed of 45 km/h, a bus covers a distance in = 3 hours 20 minutes
= 3\(\frac { 1 }{ 3 }\) = \(\frac { 10 }{ 3 }\) hours
With a speed of 1 km/h it will cover the distance m = \(\frac { 10 x 45 }{ 3 }\) h
(Less speed, more time)
and with a speed of 36 km/h, it will cover the distance in
= \(\frac { 10 x 45 }{ 3 x 36 }\) hr (more speed, less time)
= \(\frac { 25 }{ 6 }\) h
= 4\(\frac { 1 }{ 6 }\) h
= 4 hr 10 minutes

Question 8.
Solution:
To make 240 tonnes of steel, material is sufficient in = 1 month or 30 days
To make 1 tonne of steel, it will be sufficient in = 30 x 240 days (Less steel, more days)
To make 240 + 60 = 300 tonnes of steel it will be sufficient in = \(\frac { 30 x 240 }{ 300 }\) days
= 24 days (more steel, less days)

Question 9.
Solution:
In the beginning, number of men = 210
After 12 days, more men employed = 70
Total men = 210 + 70 = 280
Total period = 60 days.
After 12 days, remaining period = 60 – 12 = 48 days
Now 210 men can build the house in = 48 days
and 1 man can build the house in = 48 x 210 days (less men, more days) .
280 men can build the house in = \(\frac { 48 x 210 }{ 280 }\) days
(more men, less days)
= 36 days

Question 10.
Solution:
In 25 days, the food is sufficient for = 630 men
In 1 day, the food will be sufficient for = 630 x 25 men (less days, more men)
and in 30 days, the food will be sufficient for = \(\frac { 630 x 25 }{ 30 }\) hr
(more days less men)
= 525 men
Number of men to be transfered = 630 – 525 = 105 men

Question 11.
Solution:
Number of men in the beginning = 120
Number of men died = 30
Remaining = 120 – 30 = 90 men
Total period = 200 days
No. of days passed = 5
Remaining period = 200 – 5 = 195
Now, The food lasts for 120 men for = 195 days
The food will last for 1 man for = 195 x 120 days (Less men, more days)
The food will last for 90 men for = \(\frac { 195 x 120 }{ 90 }\)
(more men less days)
= 65 x 4 = 260 days

Question 12.
Solution:
Period in the beginning = 28 days
No. of days passed = 4 days.
Remaining period = 28 – 4 = 24 days
The food is sufficient for 24 days for = 1200 soldiers
The food will be sufficient for 1 day for = 1200 x 24 soldiers (Less days, more men)
and the food will be sufficient for 32 days = \(\frac { 1200 x 24 }{ 32 }\)
= 900 soldiers (more days, less men)
No. of soldiers who left the fort = 1200 – 900 = 300 soldiers

Hope given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B are helpful to complete your math homework.

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