RS Aggarwal Class 7 Solutions Unitary Method

 

QUESTION – 1: If 15 oranges cost Rs 110, what do 39 oranges cost?

Solution:

Cost of 15 oranges = Rs 110

Cost of 1 orange = Rs 110/15

Therefore, Cost of 39 oranges = \(\frac{110}{15}\times 39\) = Rs 286

QUESTION – 2: If 8 kg sugar costs Rs 260, how much sugar can be bought for Rs 877.50?

Solution:

Amount of sugar bought for Rs 260 = 8 kg

Amount of sugar bought for Re 1 = 8/260 kg

Now, amount of sugar bought for Rs 877.50 = \(\frac{8}{260}\times 877.50 kg\) = 27 kg

Therefore, 27 kg of sugar can be bought for Rs 877.50.

QUESTION – 3: The cost of 37 m of silk is Rs 6290. What length of this silk can be purchased for Rs 4420?

Solution:

Length of the silk purchased for Rs 6290 = 37 m

Length of the silk purchased for Re 1 = 37/6290 m

Now, length of the silk purchased for Rs 4,420 = \(\frac{37}{6290}\times 4420 m\) = 26 m

Therefore, 26 m of silk can be purchased for Rs 4,420.

QUESTION – 4: A worker is paid Rs 1110 for 6 days. If his total wages during a month are Rs 4625, for how many days did he work?

Solution:

Number of days for which a worker is paid Rs 1,110 = 6

Number of days for which a worker is paid Re 1 = 6/1110 days

Now, number of days for which a worker is paid Rs 4625 = \(\frac{6}{1110}\times 4625 days\) = 25 days

Therefore, The worker worked 25 days in a month.

QUESTION – 5: A car can cover a distance of 357 km on 42 litres of petrol. How far can it travel on 12 litres of petrol?

Solution:

Distance covered by the car with 42 L of petrol = 357 km

Distance covered by the car with 1 L of petrol = 357/42 km [less petrol, less distance]

Now, distance covered by the car with 12 L of petrol = \(\frac{357}{42}\times 12 days\) = 102 km [more petrol, more distance]

QUESTION – 6:Travelling 900 km by rail costs Rs 2520. What would be the fare for a journey of 360 km when a person travels by the same class?

Solution:

Cost of travelling 900 km by train = Rs 2520

Cost of travelling 1 km by train = Rs 2520/900

Now, cost of travelling 360 km by train = Rs\(\frac{2520}{900}\times 360 m\) = Rs 1008

Therefore, The train fare for a journey of distance 360 km is Rs 1,008.

QUESTION – 7: A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km?

Solution:

Time taken to cover a distance of 51 km = 45 min

Time taken to cover a distance of 1 km = 45/51 min

Time taken to cover distance of 221 km = \(\frac{45}{51}\times 221 km\).

QUESTION – 8: If 22.5 metres of a uniform iron rod weighs 85.5 kg, what will be the length of 22.8 kg of the same rod?

Solution:

Length of the iron rod that weighs 85.5 kg = 22.5 m

Length of the iron rod that weighs 1 kg = 22.5/85.5 m [less weight, less length]

Therefore, Length of the iron rod that weighs 22.8 kg = \(\frac{22.5}{85.5}\times 22.8m\) [more weight, more length]

QUESTION – 9: If the weight of 6 sheets of a paper is 162 grams, how many sheets of the same quality of paper would weigh 13.5 kg?

Solution:

Number of paper sheets that weighs 162 g = 6

Number of paper sheets that weighs 1 kg = 6/162 [less weight, less sheets]

Therefore, Number of paper sheets that weighs 13.5 kg = \(\frac{6}{162}\times 13.5\times 1000=500\) [more weight, more sheets]

QUESTION – 10: 1152 bars of soap can be packed in 8 cartons of the same size. How many such cartons will be required to pack 3888 bars?

Solution:

Number of cartons needed to pack 1152 soap bars = 8

Number of cartons needed to pack 1 soap bar = 8/3888 = 27 [less number of soaps, less number of cartons needed]

Now, number of cartons needed to pack 3888 soap bars = \(\frac{8}{1152}\times 3888=27\) [more soaps, more cartons needed]

Therefore, 27 cartons are needed to pack 3888 soap bars.

QUESTION – 11: If the thickness of a pile of 16 cardboards is 44 mm, how many cardboards will be there in a pile which is 71.5 cm thick?

Solution:

Number of cardboards in a pile of thickness 44 mm = 16

Number of cardboards in a pile of thickness 1 mm = 16/44

Number of cardboards in a pile of thickness 71.5 cm = \(\frac{16}{44}\times 71.5\times 10=260\) [1 cm = 10 mm]

Therefore, 260 cardboards will be there in a pile of thickness 71.5 cm.

QUESTION – 12: At a particular time of a day, a 7 m high flagstaff casts a shadow which is 8.2 m long. What is the height of the building which casts a shadow 20.5 metres in length at the same moment?

Solution:

Height of the flagstaff that casts a shadow of length 8.2 m = 7 m

Height of the building that casts a shadow of length 1 m = 7/8.2 m

Height of the building that casts a shadow of length 20.5 m = \(\frac{7}{8.2}\times 20.5 m=17.5 m\)

Therefore, The height of the required building is 17.5 m.

QUESTION – 13: 15 men can build a 16.25 m long wall up to a certain height in one day. How many men should be employed to build a wall of the same height but of length 26 metres in one day?

Solution:

Number of men employed to built the 16.25 m long wall = 15

Number of men required to built a 1 m long wall = 15/16.25

Number of men that should be employed to built a 26 m long wall = \(\frac{15}{16.25}\times 26 =24\)

Therefore, 24 men should be employed to build a wall of length 26 m in a day.

QUESTION – 14: In a hospital, the monthly consumption of milk of 60 patients is 1350 litres. How many patients can be accommodated in the hospital if the monthly ration of milk is raised to 1710 litres, assuming that the quota per head remains the same?

Solution:

Number of patients who can consume 1350 L of milk = 60

Number of patients who can consume 1 L of milk = 60/1350

Now, number of patients who can consume 1710 L of milk = \(\frac{60}{1350}\times 1710 =76\)

Hence, 76 patients can be accommodated in the hospital if the monthly ration of milk is raised to 1710 L.

QUESTION – 15: The extension in an elastic string varies directly as the weight hung on it. If a weight of 150 g produces an extension of 2.8 cm, what weight would produce an extension of 19.6 cm?

Solution:

Weight that would produce an extension of 2.8 cm = 150 g

Weight that would produce an extension of 1 cm = 150/2.8 g

Weight that would produce an extension of 19.6 cm = \(\frac{150}{2.8}\times 19.6\) = 1050 g = 1 kg 50 g [I kg = 1000 g]

Therefore, A weight of 1 kg 50 g would produce an extension of 19.6 cm.

EXERCISE – 9B

QUESTION – 1: If 48 men can dig a trench in 14 days, how long will 28 men to dig a similar trench?

Solution:

48 men can dig a trench in 14 days.

1 man can dig the trench in 14 x 48 days. [less men, more days]

Therefore, 28 men can dig the trench in \(\frac{14\times 14}{28}\) days = 24 days [ more men, less days]

Hence, 28 men will take 24 days to dig a similar trench.

QUESTION – 2: 16 men can reap a field in 30 days. How many men must be engaged to reap the same field in 24 days?

Solution:

No. of men required to reap the field in 30 days = 16

No. of men required to reap the field in 1 day = 16 x 30 [less days, more men]

Now, no. of men required to reap the field in 24 days = \(\frac{16\times 30}{24}\) = 20 [more days, less men]

Therefore, 20 men are required to reap the field in 24 days.

QUESTION – 3: 45 cows can graze a field in 13 days. How many cows will graze the same field in 9 days?

Solution:

Number of cows that can graze the field in 13 days = 45

Number of cows that can graze the field in 1 day = 45 x 13 [less days, more cows]

Therefore, number of cows that can graze the field in 9 days = \(\frac{45\times 13}{9}\) = 65 [More days, less cows]

Hence, 65 cows can graze the field in 9 days.

QUESTION – 4: 16 horses can consume a certain quantity of corn in 25 days. In how many days would the same quantity be consumed by 40 horses?

Solution:

Time taken by 16 horses to consume the corn = 25 days

Time taken by 1 horse to consume the corn = 25 x 16 [less horses, more time taken]

Time taken by 40 horses to consume the corn = \(\frac{25\times 10}{40}\) = 10 days [more horses, less time taken]

Hence, 40 horses would consume the same quality of corn in 10 days.

QUESTION – 5: A girl can finish a book in 25 days if she reads 18 pages of it every day. How many days will she take to finish it, if she reads 15 pages every day?

Solution:

Days taken to finish the book if 18 pages are read everyday = 25

Days taken to finish the book if 1 page is read everyday = 18 x 25 [less pages, more days]

Now, days taken to finish the book if 15 pages are read everyday = \(\frac{18\times 25}{15}\) = 30 [more pages, less days]

Hence, the girl will take 30 days to finish the book if she reads 15 pages everyday.

QUESTION – 6: Reeta types 40 words per minute and takes 24 minutes to type a certain document. Her friend Geeta has a typing speed of 48 words per minute. In how much time, will she be able to type the same document?

Solution:

Time taken to type 40 words per minute = 24 min

Time taken to type a word per minute = 24 x 40 min

Now, time taken to type 48 words per minute = \(\frac{24\times 40}{48}\) = 20 min

Hence, Geeta will take 20 minutes to type the same document if her typing speed is 48 words/min.

QUESTION – 7: A bus covers a certain distance in 3 hours 20 minutes at an average speed of 45 km/h. How long will it take to cover the same distance at a speed of 36 km/h?

Solution:

Time taken to cover the distance at a speed of 45 km/h = 3 h 20 min = 200 min

Time taken to cover the distance at a speed of 1 km/h = 45 x 3.33 min [less speed, more time]

Time taken to cover the distance at a speed of 36 km/h = \(\frac{45\times 3.33}{36}=4.1625:happrox 4:h:10:min\)

Hence, the bus will take 4 h 10 min to cover the distance if its speed is 36 km/h.

QUESTION – 8: At the beginning of a month, a factory has enough materials to make 240 tonnes of steel in a month. If 60 more tonnes of steel is to be made that month, how long will the materials last?

Solution:

Time taken to make 240 tonnes of steel = 30 days

Time taken to make 1 tonne of steel = 30 x 240 days

Now, time taken to make 300 or (240 + 60) tonnes of steel = \(\frac{30\times 240}{300}\) = 24 days

Therefore, The materials will last for 24 days if 60 more tonnes of steel is to be made that month.

QUESTION – 9: A contractor employed 210 men to build a house in 60 days. After 12 days, he was joined by 70 more men. In how many days will the remaining work be finished?

Hint. The remaining food is sufficient for 120 men for 195 days. Find out how many days, the food will be sufficient for 90 men.

Solution:

Initially, the contractor had 210 men for 60 days. After 12 days, 70 more men joined.

210 men can finish the work in 48 days

1 man can finish the work in 210 x 48 days

Now, 280 men can finish the work in \(\frac{210\times 48}{280}\) days = 36 days.

Hence, it will take 36 days to finish the remaining work.

QUESTION – 10: A military camp has provisions for 630 men to last for 25 days. How many men must be transferred to another camp so that the food lasts for 30 days?

Solution:

No. of men for which the provision will last for 25 days = 360

No. of men for which the provision will last for 1 day = 360 x 25

Now, no. of men for which the provision will last for 30 days = \(\frac{360\times 25}{30}\) = 300

Therefore, 60 men, i.e., (360 – 300), must be transferred to another camp so that the provision lasts for 30 days.

QUESTION – 11: A group of 120 men had provisions for 200 days. After 5 days, 30 men died due to an epidemic. How long will the remaining food last?

Hint. The remaining food is sufficient for 120 men for 195 days. Find out for how many days, the food will be sufficient for 90 men.

Solution:

Number of days for which the food is sufficient for 120 men = 195

Number of days for which food is sufficient for 1 man = 120 x 195

Number of days for which food is sufficient for 90 men = \(\frac{120\times 195}{90}\) = 260

Hence, the food will last for 260 days.

QUESTION – 12: 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted for an extra 32 days. How many soldiers left the fort?

Solution:

We are given that in a fort, 1200 soldiers had enough food for 28 days.

Let x soldiers left after 4 days, thus, remaining soldiers = 1200 – x

Now, for these remaining soldiers food lasts for 32 days.

As number of soldiers decrease, food lasts long.

Thus, situation after 4 days is

1200 x 24 = (1200 – x) x 32

=> (1200 – x) = \(\frac{1200\times 24}{32}\)

=> 1200 – x = 900

=> x = 1200 – 900

=> x = 300

Thus, 300 soldiers left the fort after 4 days.


Practise This Question

Which of the following option correctly defines wind?