RS Aggarwal 2019,2020 solution class 7 chapter 9 Unitary Method Exercise 9A

Exercise 9A

page-133

Question 1:

If 15 oranges cost Rs 110,  what do 39 oranges cost?

Answer 1:

Cost of 15 oranges = Rs 110

Cost of 1 orange = Rs $\frac{110}{15}$
∴ Cost of 39 oranges = Rs $\frac{110}{15}\times 39$ = Rs 286

Question 2:

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

Answer 2:

Amount of sugar bought for Rs 260 = 8 kg
Amount of sugar bought for Re 1 = $\frac{8}{260}$ kg
Now, amount of sugar bought for Rs 877.50 = $\frac{8}{260}\times 877.50$ kg = 27 kg

∴ 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 4420?

Answer 3:

Length of the silk purchased for Rs 6290  = 37 m
Length of the silk purchased for  Re 1= $\frac{37}{6290}$ m
Now, length of the silk purchased for Rs 4,420 = $\frac{37}{6290}\times 4420$ m = 26 m
∴ 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?

Answer 4:

Number of days for which a worker is paid Rs 1,110 = 6
Number of days for which a worker is paid Re 1 = $\frac{6}{1110}$ days
Now, number of days for which a worker is paid Rs 4625=$\frac{6}{1110}\times 4625$ days = 25 days
∴ 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?

Answer 5:

Distance covered by the car with 42 L of petrol = 357 km
Distance covered by the car with 1 L of petrol = $\frac{357}{42}$ km             [less petrol, less distance]
Now, distance covered by the car with 12 L of petrol = $\frac{357}{42}\times 12$ = 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?

Answer 6:

Cost of travelling 900 km by train = Rs 2520

Cost of travelling 1 km by train = Rs $\frac{2520}{900}$
Now, cost of travelling 360 km by train =Rs $\frac{2520}{900}\times 360$ = Rs 1008
∴ 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?

Answer 7:

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

Time taken to cover a distance of 1 km = $\frac{45}{51}$ min

Time taken to cover distance of 221 km = $\frac{45}{51}\times 221$ min = 195 min = 3 h 15 min

∴ The train will take 3 h 15 min to cover a distance of 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?

Answer 8:

Length of the iron rod that weighs 85.5 kg = 22.5 m
Length of the iron rod that weighs 1 kg = $\frac{22.5}{85.5}$ m               [less weight, less length]
∴ Length of the iron rod that weighs 22.8 kg = $\frac{22.5}{85.5}\times 22.8$m = 6 m       [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?

Answer 9:

Number of paper sheets that weighs 162 g = 6
Number of paper sheets that weighs 1 g = $\frac{6}{162}$                  [less weight, less sheets]
∴ 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?

Answer 10:

Number of cartons needed to pack 1152 soap bars = 8
Number of cartons needed to pack 1 soap bar = $\frac{8}{1152}$              [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 carton needed]

∴ 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?

Answer 11:

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

Answer 12:

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 = $\frac{7}{8.2}$ 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
∴ 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?

Answer 13:

Number of men employed to built the 16.25 m long wall = 15
Number of men required to built a 1 m long wall = $\frac{15}{16.25}$                       

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

Answer 14:

Number of patients who can consume 1350 L of milk = 60
Number of patients who can consume 1 L of milk = $\frac{60}{1350}$                                 
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?

Answer 15:

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

Weight that would produce an extension of 1 cm = $\frac{150}{2.8}$ g

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

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