Class 7: Maths Chapter 9 solutions. Complete Class 7 Maths Chapter 9 Notes.
Contents
RS Aggarwal Solutions for Class 7 Maths Chapter 9–Unitary Method
RS Aggarwal 7th Maths Chapter 9, Class 7 Maths Chapter 9 solutions
Ex 9A Solutions
Question 1.
Solution:
Cost of 15 oranges = Rs. 110
Cost of 1 orange = Rs. 11015
and cost of 39 oranges = Rs. 11015 x 39
= Rs. 22 x 13 = Rs. 286
Question 2.
Solution:
In Rs. 260, the sugar is bought = 8 kg
and in Re. 1, the sugar is bought = 8260 kg
Then in Rs. 877.50, the sugar will be bought = 8260 x 877.50 kg
= 8260 x 87750100
= 27 kg
Question 3.
Solution:
In Rs. 6290, silk is purchased = 37 m
and in Re. 1, silk is purchased = 376290 m
and in Rs. 4420, silk will be purchased 37
= 376290 x 4420 m = 26 m
Question 4.
Solution:
Rs. 1110 is wages for = 6 days.
Re. 1 will be wages for = 61110 days
and Rs. 4625 will be wages for
Question 5.
Solution:
In 42 litres of petrol, a car covers = 357 km
and in 1 litre, car will cover = 35742 km
and in 12 litres, car will cover = 35742 x 12 = 102 km
Question 6.
Solution:
Cost of travelling 900 km is = Rs. 2520
and cost of 1 km will be = Rs. 2520900
andcostof360kmwillbe = Rs. 2520900 x 360 = Rs. 1008
Question 7.
Solution:
To cover a distance of 51 km, time is taken = 45 minutes
Question 8.
Solution:
If weight is 85.5 kg, then length of iron rod = 22.5 m
If weight is 1 kg, then length of rod will be
Question 9.
Solution:
In 162 grams, sheets are = 6
Question 10.
Solution:
1152 bars of soap can be packed in 8 cartons
1 bar of soap coil be packed in
Question 11.
Solution:
In 44 mm of thickness, cardboards are = 16
In 1 mm of thickness, cardboards will be
Question 12.
Solution:
If length of shadow is 8.2 m, then
height of flag staff is = 7 m
If length of shadow is 1 m, then height will
Question 13.
Solution:
16.25 m long wall is build by = 15 men
Question 14.
Solution:
1350 litres of milk cm be consumed by = 60 patients
1 litres of milk can be consumed by = 601350 patients
and 1710 litres of milk can be consumed
Question 15.
Solution:
2.8 cm extension is produced by = 150 g.
1 cm extension will be produced by = 1502.8 g
and 19.6 cm extension will be produced by
Ex 9B Solutions
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 = 14×4828 (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 = 16×3024 men (more days, less men)
= 48024
= 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 = 45×139 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 = 25×1640 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 = 25×1815 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 = 24×4048 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
= 313 = 103 hours
With a speed of 1 km/h it will cover the distance m = 10×453 h
(Less speed, more time)
and with a speed of 36 km/h, it will cover the distance in
= 10x453x36 hr (more speed, less time)
= 256 h
= 416 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 = 30×240300 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 = 48×210280 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 = 630×2530 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 = 195×12090
(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 = 1200×2432
= 900 soldiers (more days, less men)
No. of soldiers who left the fort = 1200 – 900 = 300 soldiers
Ex 9C Solutions
Objective Questions.
Marks (✓) against the correct answer in each of the following :
Question 1.
Solution:
(c)
Weight of 4.5 m rod = 17.1 kg
Question 2.
Solution:
(d) None of these 0.8 cm represent the map = 8.8 km
Question 3.
Solution:
(c) In 20 minutes, Raghu covers = 5 km
in 1 minutes, he will cover = 520 km
and in 50 minutes, he will cover
Question 4.
Solution:
(d)
No. of men in the beginning = 500
More men arrived = 300
No. of total men = 500 + 300 = 800
For 500 men, provision are for = 24 days
For 1 man, provision will be = 24 x 500 days (less men, more days)
and for 800 men, provision will be = 24800 x 500 days
(more men less days)
= 15 days
Question 5.
Solution:
(b) Total cistern = 1
Filled in 1 minute = 45
Unfilled = 1 – 45 = 15
45 of cistern is filled in = 1 minutes = 60 seconds
1 full cistern can be filled in = 60×54 = 75 seconds
More time = 75 – 60 = 15 seconds
Question 6.
Solution:
(a)
15 buffaloes can eat as much as = 21 cows
1 buffalo will eat as much as = 2115 cows
35 buffaloes will eat as much as
= 21×3515 cm = 49 cows
Question 7.
Solution:
(b) 4 m long shadow is of a tree of height = 6 m
1 m long shadow of flagpole will of height = 64 m
50 m long shadow, the height of pole 6 will be = 64 x 50 = 75 m
Question 8.
Solution:
(b) 8 men can finish the work in = 40 days
1 man will finish it in=40 x 8 days (less men, more days)
8 + 2 = 10 men will finish it in = 40×810 days
(more men, less days)
= 32 days
Question 9.
Solution:
(b)
16 men can reap a field in = 30 days
1 man will reap the field in = 30 x 16 days
and 20 men will reap the field in = 30×1620 = 24 days
Question 10.
Solution:
(c) 10 pipe can fill tank in = 24 minutes
1 pipe will fill it in = 24 x 10 minutes (less pipe, more time)
and 10 – 2 = 8 pipes will fill the tank in
= 24×108 = 30 minutes
Question 11.
Solution:
(d) 6 dozen or 6 x 12 = 72 eggs
Cost of 72 eggs is = Rs. 108
Cost of 1 egg will be = Rs. 10872
and cost of 132 eggs will be 108
= Rs. 10872 x 132 = Rs. 198
Question 12.
Solution:
(b) 12 workers take to complete the work = 4 hrs.
1 worker will take = 4 x 12 hrs. (less worker, more time)
15 workers will take = 4×1215 hrs. (more workers, less time)
= 165 hr. = 3 hrs. 12 min
Question 13.
Solution:
(a) 27 days – 3 days = 24 days
Men = 500 + 300 = 800
For 500 men, provision is sufficient = 24 days
For 1 man, provision will be = 24 x 500 (less man, more days)
and for 500 + 300 = 800 men provision
will be sufficient = 24×500800 = 15 days
(more men, less days)
Question 14.
Solution:
(c) No. of rounds of rope = 140
Radius of base of cylinder = 14 cm
Radius of second cylinder of cylinder = 20 cm
If radius is 14 cm, then rounds of rope are = 140
If radius is 1 cm, then round = 140 x 14 (less radius more rounds)
and if radius is 20 cm, then rounds will
be = 140×1420 = 98 (more radius less rounds)
Question 15.
Solution:
(d) A worker makes toy in 23 hr= 1
He will make toys in 1 hr = 1 x 32
and will make toys in 223 hrs. = 1 x 32 x 223
= 11 (more time more toys)
Question 16.
Solution:
(d) A wall is constructed in 8 days by = 10 men
It will be constructed in 1 day by = 10 x 8 men (less time, more men)
10 x 8
More men required = 160 – 10 = 150
RS Aggarwal Solutions for Class 7 Maths Chapter 9: Download PDF
RS Aggarwal Solutions for Class 7 Maths Chapter 9–Unitary Method
Download PDF: RS Aggarwal Solutions for Class 7 Maths Chapter 9–Unitary Method PDF
Chapterwise RS Aggarwal Solutions for Class 7 Maths :
- Chapter 1–Integers
- Chapter 2–Fractions
- Chapter 3–Decimals
- Chapter 4–Rational Numbers
- Chapter 5–Exponents
- Chapter 6–Algebraic Expressions
- Chapter 7–Linear Equations in One Variable
- Chapter 8–Ratio and Proportion
- Chapter 9–Unitary Method
- Chapter 10–Percentage
- Chapter 11–Profit and Loss
- Chapter 12–Simple Interest
- Chapter 13–Lines and Angles
- Chapter 14–Properties of Parallel Lines
- Chapter 15–Properties of Triangles
- Chapter 16–Congruence
- Chapter 17–Constructions
- Chapter 18–Reflection and Rotational Symmetry
- Chapter 19–Three-Dimensional Shapes
- Chapter 20–Mensuration
- Chapter 21–Collection and Organisation of Data (Mean, Median and Mode)
- Chapter 22–Bar Graphs
- Chapter 23–Probability
About RS Aggarwal Class 7 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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