Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 9 Unitary Method are provided here with simple step-by-step explanations. These solutions for Unitary Method are extremely popular among Class 7 students for Math Unitary Method Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 133:

Cost of 15 oranges = Rs 110

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

#### Page No 133:

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 = $\frac{8}{260}×877.50$ kg = 27 kg

∴ 27 kg of sugar can be bought for Rs 877.50.

#### Page No 133:

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 = $\frac{37}{6290}×4420$ m = 26 m
∴ 26 m of silk can be purchased for Rs 4,420.

#### Page No 133:

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=$\frac{6}{1110}×4625$ days = 25 days
∴ The worker worked 25 days in a month.

#### Page No 133:

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 = $\frac{357}{42}×12$ = 102 km   [more petrol, more distance]

#### Page No 133:

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 $\frac{2520}{900}×360$ = Rs 1008
∴ The train fare for a journey of distance 360 km is Rs 1,008.

#### Page No 133:

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 = $\frac{45}{51}×221$ min = 195 min = 3 h 15 min

∴ The train will take 3 h 15 min to cover a distance of 221 km.

#### Page No 133:

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 = m = 6 m       [more weight, more length]

#### Page No 133:

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 = $\frac{6}{162}×13.5×1000$ = 500    [more weight, more sheets]

#### Page No 133:

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 = $\frac{8}{1152}×3888$ = 27        [more soaps, more carton needed]

∴ 27 cartons are needed to pack 3888 soap bars.

#### Page No 133:

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 = = 260         [1 cm=10 mm]
∴ 260 cardboards will be there in a pile of thickness 71.5 cm.

#### Page No 133:

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 = $\frac{7}{8.2}×20.5$ m = 17.5 m
∴ The height of the required building is 17.5 m.

#### Page No 133:

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 = $\frac{15}{16.25}×26$ = 24
∴ 24 men should be employed to build a wall of length 26 m in a day.

#### Page No 133:

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  = $\frac{60}{1350}×1710$ = 76

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

#### Page No 133:

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}$$×$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.

#### Page No 134:

48 men can dig a trench in 14 days.
1 man can dig the trench in 14 $×$ 48 days.                                               [less men, more days]
Therefore, 28 men can dig the trench in $\frac{14×48}{28}$ days = 24 days              [more men, less days]
Hence, 28 men will take 24 days to dig a similar trench.

#### Page No 134:

No. of men required to reap the field in 30 days = 16
No. of men required to reap the field in 1 day = 16 $×$30                        (less days, more men)
Now, no. of men required to reap the field in 24 days = $\frac{16×30}{24}$ = 20    (more days, less men)

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

#### Page No 135:

Number of cows that can graze the field in 13 days = 45
Number of cows that can graze the field in 1 day = 45 $×$13           [Less days, more cows]
Therefore, number of cows that can graze the field in 9 days = $\frac{45×13}{9}$ = 65   [More days, less cows]

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

#### Page No 135:

Time taken by 16 horses to consume the corn = 25 days
Time taken by 1 horse to consume the corn = 25$×$16                    [less horses, more time taken]
Time taken by 40 horses to consume the corn = $\frac{25×16}{40}$= 10 days     [more horses, less time taken]

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

#### Page No 135:

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 $×$25                          [less pages, more days]
Now, days taken to finish the book if 15 pages are read everyday = $\frac{18×25}{15}$ = 30     [more pages, less days]
Hence, the girl will take 30 days to finish the book if she reads 15 pages everyday.

#### Page No 135:

Time taken to type 40 words per minute = 24 min
Time taken to type a word per minute = 24$×$40 min
Now, time taken to type 48 words per minute = $\frac{24×40}{48}$ = 20 min

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

#### Page No 135:

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 $×$ 3.33 min             [less speed, more time]
Time taken to cover the distance at a speed of 36 km/h = $\frac{45×3.33}{36}$ = 4.1625 h ≈ 4 h 10 min

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

#### Page No 135:

Time taken to make 240 tonnes of steel = 30 days
Time taken to make 1 tonne of steel = 30 $×$ 240 days
Now, time taken to make 300 or (240 + 60) tonnes of steel = $\frac{30×240}{300}$ = 24 days

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

#### Page No 135:

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$×$48 days
Now, 280 men can finish the work in $\frac{210×48}{280}$ days = 36 days.

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

#### Page No 135:

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

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

#### Page No 135:

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$×$195
Number of days for which food is sufficient for 90 men = $\frac{120×195}{90}$ = 260

Hence, the food will last for 260 days.

#### Page No 135:

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

Thus, 300 soldiers left the fort after 4 days.

#### Page No 135:

(c) 45.6 kg

Weight of the rod of length 4.5 m = 17.1 kg
Weight of the rod of length 1 m = $\frac{17.1}{4.5}$ kg     [less length, less weight]

∴ Weight of the rod of length 12 m = $\frac{17.1}{4.5}×12$ = 45.6 kg       [more length, more weight]

#### Page No 135:

(d) none of these

0.8 cm represents 8.8 km.

1 cm represents  $\frac{8.8}{0.8}$ km.

80.5 cm represents $\frac{8.8}{0.8}×80.5$ = 885.5 km.

#### Page No 135:

Distance covered in 20 min = 5 km
Distance covered in 1 min = $\frac{5}{20}$ km                                [less time, less distance covered]
Distance covered in 50 min = $\frac{5}{20}×50$ = 12.5 km          [more time, more distance covered]
Hence, Raghu will cover a distance of 12.5 km in 50 minutes.

Thus, the correct option is (c).

#### Page No 135:

Number of days for which 500 men have enough food = 24
Number of days for which 1 man has enough food = 24$×$ 500                     [less men, more food]
Number of days for which 800 men have enough food = $\frac{24×500}{800}$ = 15        [more men, less food]
Hence, the food will last for 15 days after the reinforcement of 300 men.

Thus, the correct option is (d).

#### Page No 135:

Time taken to fill $\frac{4}{5}$ of a cistern = 1 min
Time taken to fill 1 cistern $=\frac{5}{4}$ min
Time taken to fill $\frac{1}{5}$ of a cistern = $\frac{5}{4}×\frac{1}{5}=\frac{1}{4}$ min = 15 seconds

Hence, it will take 15 seconds to fill the rest of the cistern.

Thus, the correct option is (b).

#### Page No 136:

Number of cows that eat as much as 15 buffaloes = 21
Number of cows that eat as much as 1 Buffalo = $\frac{21}{15}$
Number of cows that eat as much as 35 buffaloes = $\frac{21}{15}×35$ = 49
Hence, 49 cows will eat as much as 35 buffaloes.

Thus, the correct option is (a).

#### Page No 136:

(b) 75 m

Height of the tree that casts a 4 m long shadow = 6 m
Height of the tree that casts a 1 m long shadow = m
∴ Height of the flag pole that casts a 50 m long shadow = $\frac{6}{4}×50$ = 75 m

#### Page No 136:

8 men finish the work in 40 days.
1 man can finish the work in 8$×$40 days.                                                 [Less men, more days]
10 men can finish the work in $\frac{8×40}{10}$ = 32 days.                                      [More men, less days]
∴ If 2 more men join them, the work will be completed in 32 days.

The correct option is (b).

#### Page No 136:

Number of days taken to reap the field by 16 men = 30 days
Number of days taken to reap the field by 1 man = 30$×$16 days                            [Less men, more days]
Number of days taken to reap the field by 20 men = $\frac{30×16}{20}$ = 24 days                [More men, less days]

Hence, 20 men will take 24 days to reap the field.

The correct option is (b).

#### Page No 136:

Time taken to fill the tank by 10 pipes = 24 min
Time taken to fill the tank by 1 pipe = 24 $×$10 min                       [Less pipes, more time taken]
Time taken to fill the tank by 8 pipes = $\frac{24×10}{8}$ min = 30 min        [More pipes, less time taken]
Hence, it will take 30 minutes to fill the tank if two pipes go out of order.

The correct option is (c).

#### Page No 136:

Cost of 72 eggs = Rs 108
Cost of 1 egg = Rs $\frac{108}{72}$
Cost of 132 eggs = RS $\frac{108}{72}×132$ = Rs 198

Hence, 132 eggs will cost Rs 198.

The correct option is (d).

#### Page No 136:

Time taken by 12 workers to complete the job = 4 h
Time taken by 1 worker to complete the job = $4×12$ h
Time taken by 15 workers to complete the job = $\frac{4×12}{15}$ = 3 h 12 min

Hence, 15 workers will complete the job in 3 h 12 min.

The correct option is (b).

#### Page No 136:

500 men had enough food for 24 days.
1 man had enough food for 24$×$500 days.                          [Less men, more days]
800 men had enough food for $\frac{24×500}{800}$ = 15 days             [More men , less days]

Hence, the food will now last for 15 days after the reinforcement of 300 men.

The correct option is (a).

#### Page No 136:

(c) 98

No. of rounds around the cylinder of radius 14 cm = 140
No. of rounds around the cylinder of radius 1 cm = $140×14$             [Less radius, more rounds]
No. of rounds around the cylinder of radius 20 cm = $\frac{140×14}{20}$ = 98      [More radius, less rounds]

Hence, the rope makes 98 rounds around the circumference of the cylinder of radius 20 cm.

#### Page No 136:

No. of toys made in $\frac{2}{3}$ h = 1
No. of toys made in 1 h = $\frac{3}{2}$
No. of toys made in $7\frac{1}{3}$ h = $\frac{3}{2}×\frac{22}{3}=11$

Hence, the worker will make 11 toys in $7\frac{1}{3}$ h.

The correct option is (d).

#### Page No 136:

Men required to finish the work in 8 days = 10
Men required to finish the work in 1 day = 10$×$8                                    [More day, less men]
Men required to finish the work in half a day = = 160     [Less days, more men]
Hence, 150 (i.e., 160 − 10) men are added to finish the work in half a day.

The correct option is (d).

#### Page No 137:

Cost of 8 toys = Rs 192
Cost of 1 toy = Rs $\frac{192}{8}$ = Rs 24

∴ Cost of 14 toys = $24×14$ = Rs 336

#### Page No 137:

Distance covered with 15 L of petrol= 270 km
Distance covered with 1 L of petrol = $\frac{270}{15}$ km

Distance covered with 8 L of petrol= $\frac{270}{15}×8$ km = 144 km

#### Page No 137:

Cost of 15 envelopes = Rs 11.25
Cost of 1 envelope = Rs $\frac{11.25}{15}$

∴ Cost of 20 envelopes = $\frac{11.25}{15}×20$ = Rs 15

#### Page No 137:

Number of cows that graze the field in 20 days = 24
Number of cows that graze the field in 1 day = 24$×$20                             [Less days, more cows]
∴ Number of cows that graze the field in 15 days = $\frac{24×20}{15}$ = 32 cows        [More days, less cows]

#### Page No 137:

Time taken to finish the work by 8 men = 15 h
Time taken to finish the work by 1 man = 8$×$15 h                             [Less men, more time taken]
∴ Time taken to finish the work by 20 men = $\frac{8×15}{20}$ h = 6 h    [More men, less time taken]

#### Page No 137:

Time taken to fill $\frac{4}{5}$ of the cistern = 1 min

Time taken to fill 1 cistern = $\frac{1}{\frac{4}{5}}=\frac{5}{4}$= 1.25 min = 1 min 15 sec
Hence, it will take 1 min 15 sec to fill the empty cistern.

#### Page No 137:

Time taken to cover the distance at a speed of 45 km/h = 3 h 20 min
Time taken to cover the distance at a speed of 1 km/h = 45 $×$ 3.33 h    [Less speed, more time taken]
(20 min = 0.33 hour)
∴ Time taken to cover the distance at a speed of 50 km/h = $\frac{45×3.33}{50}$h ≈ 3 h     [More speed, less time taken]

#### Page No 137:

Number of days with enough food for 120 men = 30
Number of days with enough food for 1 man = $30×120$                [Less men, more days]
∴ Number of days with enough food for 100 men = $\frac{30×120}{100}$ = 36         [More men, less days]

#### Page No 137:

(c) 644 km
1 cm represents 8 km.
∴ 80.5 cm  represents $8×80.5$ = 644 km.

#### Page No 137:

(a) 24 days

16 men can reap the field in 30 days.
1 man can reap the field in 30$×$16 days.                        [Less men, more days]
∴ 20 men can reap the field in $\frac{30×16}{20}$ = 24 days            [More men, less days]

#### Page No 137:

(b) 49

Number of cows that eat as much as 15 buffaloes = 21
Number of cows that eat as much as 1 buffalo = $\frac{21}{15}$

∴ Number of cows that eat as much as 35 buffaloes = $\frac{21}{15}×35$ = 49

#### Page No 137:

Number of cows that graze the field in 12 days = 45

Number of cows that graze the field in 1 day =

∴ Number of cows that graze the field in 9 days = = 60

#### Page No 137:

(b) Rs 162

Cost of 72 eggs = Rs 108
Cost of 1 egg =Rs $\frac{108}{72}$

Cost of 108 eggs = Rs $\frac{108×108}{72}$ = Rs 162

#### Page No 137:

(i) 588 days
42 men can dig the trench in 14 days.
1 man can dig the trench in 14 $×$ 42 days = 588 days

(ii) Rs. 48
15 oranges cost Rs 60.
12 oranges will cost Rs  $\frac{60}{15}×12$ = Rs 48

(iii) 10.8 kg
A rod of length 10 m weighs 18 kg.
A rod of length 6 m will weigh $\frac{18}{10}×6$ = 10.8 kg

(iv) 3 h 12 min
12 workers finish the work in 4 h.
15 workers will finish the work in $\frac{4×12}{15}$ = 3.2 h = 3 h 12 min

#### Page No 137:

(i) F
10 pipes fill the tank in 24 min.
1 pipe will fill the tank in 24 $×$10 min.                (Less pipes, more time taken)
8 pipes will fill the tank in $\frac{24×10}{8}$ = 30 min      (More pipes, less time taken)

(ii) T
8 men finish the work in 40 days.
1 man finishes the work in 8$×$40 days.                  (Less men, more days taken)
10 men will finish the work in $\frac{8×40}{10}$ = 32 days      (More men, less days taken)

(iii) T
A 6 m tall tree casts a shadow of length 4 m.
A 1 m tall tree cast a shadow of length $\frac{4}{6}$ m.
A 75 m tall pole will cast a shadow of length $\frac{4}{6}×75$ =50 m

(iv) T
1 toy is made in $\frac{2}{3}$ h.                              (Less toys, less time taken)
12 toys can be made in $\frac{2}{3}×12$ = 8 h      (More toys, more time taken)

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