CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

Question 1.
Solution:
We have :

Question 2.
Solution:
(i) Let x% of 1 kg be 125 g

Question 3.
Solution:

Question 4.
Solution:
(i) Let x be the required number.

Question 5.
Solution:
Let x be the number
The number is increased by 10%
Increased number = 110% of x

Question 6.
Solution:
The present value of the machine = ₹ 10000
The decrease in its value after the 1 st year = 10% of ₹ 10000
\(\frac { 10 }{ 100 }\) x 10000 = ₹ 1000
The depreciated value of the machine after the 1 st year = ₹ (10000 – 1000) = ₹ 9000
The decrease in its value after th 2nd year = 10% of ₹ 9000
\(\frac { 10 }{ 100 }\) x 9000 = 900
The depreciated value of the machine after the 2nd year = ₹ (9000 – 900) = ₹ 8100
Hence, the value of the machine after two years will be ₹ 8100.

Question 7.
Solution:
The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
= (\(\frac { 5 }{ 100 }\) x 16000) = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
= \(\frac { 5 }{ 100 }\) x 16800 = 840
Population after two years = 16800 + 840 = 17640
Hence, the population of the town after two years will be 17,640.

Question 8.
Solution:
Let us assume that the original price of the tea set is Increase in price = 5%
So, value increased on the tea set = 5% of ₹ x

Hence, the original price of the tea set is ₹ 420.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:

Question 10.
Solution:
(c) 12
Given that x% of 75 = 12

Question 11.
Solution:
(c) 25
Let the number be x. Then, we have:
120% of x = increased number

Question 12.
Solution:
(d) 180
Let the required number be x.
Then, we have :
5% of x = 9

Question 13.
Solution:
(a) 60
Let the number be x According to question, we get:
(35 % of x) + 39 = x

Question 14.
Solution:
(c) 500
Let x be the maximum marks Pass marks = (160 + 20) = 180
36 % of x = 180

Question 15.
Solution:
(i) 3 : 4 = (75) %
Explanation:
3 : 4 = \(\frac { 3 }{ 4 }\)
= (\(\frac { 3 }{ 4 }\) x 100) %
= (3 x 25)% = 75%
(ii) 0.75 = (75)%
Explanation : (0.75 x 100)% = 75%
(iii) 6% = 0.06 (express in decimals)
Explanation :
6% = \(\frac { 6 }{ 100 }\) = 0.06
(iv) If x decreased by 40% gives 135, then x = 225
Explanation :
Let the number be x.
According to question, we have :
x – 40% of x = 135

Question 16.
Solution:
(i) True (T)

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b) \(\frac { 3 }{ 4 }\) = \(\frac { 3 }{ 4 }\) x 100 = 75 %

Question 2.
Solution:
(c)
2 : 5 = \(\frac { 2 }{ 5 }\) = \(\frac { 2 }{ 5 }\) x 100 = 40%

Question 3.
Solution:
(c)

Question 4.
Solution:
(c) x% of 75 = 9

Question 5.
Solution:
(d)

Question 6.
Solution:
(b)
Let x% of 1 day = 36 minutes

Question 7.
Solution:
(a)
Let x be the required number

Question 8.
Solution:
(b)
Let x be the required number, then

Question 9.
Solution:
(d)
Let ore = x, then
5% of x = 400g
\(\frac { x x 5 }{ 100 }\) 400

Question 10.
Solution:
(b)
Let gross value of T.V = x
Commission = 10%
After deducting commission, the value

Question 11.
Solution:
(b)
Increase in salary = 25%
Let original salary = x

Question 12.
Solution:
(c)
Let x be the number of total examinees
No. of examinees passed

Question 13.
Solution:
(c)
Let total number of apples = x

Question 14.
Solution:
(b)
Present value of machine = Rs. 25000
Rate of depreciation = 10% p.a.
Value of machine after one year

Question 15.
Solution:
(c) Let x be numbers

Question 16.
Solution:
(c) 60% of 450
= \(\frac { 60 }{ 100 }\) x 450 = 270

Question 17.
Solution:
(d) Rate of reduction = 6%
Price after reduction = Rs. 658

Question 18.
Solution:
(b) Boys = 70% of students
No. of girls = 240
Girls percentage = 100 – 70 = 30%

Question 19.
Solution:
(c) Let number = x
11% of x – 7% of x = 18

Question 20.
Solution:
(a) Let number = x
35% of x + 39 = x

Question 21.
Solution:
(c) Pass marks = 36%
A students get =145 marks
But failed by 3 5 marks
Then pass marks = 145 + 35 = 180
Maximum marks = \(\frac { 180 x 100 }{ 36 }\) = 500

Question 22.
Solution:
(d) Let number be = x
Then decreasing by 40%

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

Question 1.
Solution:
Rupesh seemed 495 marks out of 750
Percentage of marks = \(\frac { 495 }{ 750 }\) x 100 = 66%

Question 2.
Solution:
Monthly salary = Rs. 15625
Increase = 12%
Amount of increase = \(\frac { 15625 x 12 }{ 100 }\) = Rs. 1875
New salary = Rs. 15625 + Rs. 1875 = Rs. 17500

Question 3.
Solution:
Excise duty in the beginning = Rs. 950
Reduced duty = Rs. 760
Reduction = Rs. 950 – Rs. 760 = Rs. 190
Reduction percent = \(\frac { 190 x 100 }{ 950 }\) = 20%

Question 4.
Solution:
Let x be the total cost of the T.V
96% of x = 10464

Total cost of T.V = Rs. 10900

Question 5.
Solution:
Let number of students = x
In a school boys = 70%
girls = 100 – 70 = 30%
Now 30% of x = 504

Number of boys = 1680 – 504 = 1176
and number of total students = 1680

Question 6.
Solution:
Copper required = 69 kg
copper in ore = 12%
Let quantity of ore = x kg
12% of x = 69

Quantity of ore = 575kg

Question 7.
Solution:
Pass marks = 36%
A students gets marks = 123
But failed by 39 marks
Pass marks = 123 + 39 = 162
Now, 36% of maximum marks = 162
Maximum marks = \(\frac { 162 x 100 }{ 36 }\) = 450 marks

Question 8.
Solution:
Let number of apples = x
Number of apples sold = 40% of x

Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= \(\frac { x x 72 }{ 100 }\)
= \(\frac { 18x }{ 25 }\)

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x

Question 11.
Solution:
Total gunpowder = 8 kg
Amount of nitre = 75%
amount of sulphur = 10%
Rest of powder which is charcoal = 100 – (75 + 10) = 100 – 85 = 15 = 15%
Amount of charcoal = 8 x \(\frac { 15 }{ 100 }\) = \(\frac { 120 }{ 100 }\)
= \(\frac { 6 }{ 5 }\) kg = 1 kg 200 grams = 1.2 kg

Question 12.
Solution:
Quantity of chalk = 1 kg or 1000 g

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219

No. of days on which was school open = 292 days

Question 14.
Solution:
Rate of commission = 3%
Amount of commission = Rs. 42660
Let value of property = x
then 3% of x = 42660

Question 15.
Solution:
Total votes of the constituency = 60000
Votes polled = 80% of total votes
= \(\frac { 80 }{ 100 }\) x 60000 = 48000
Votes polled in favour of A = 60% of polled votes
= \(\frac { 60 }{ 100 }\) x 48000 = 28800
Votes polled in favour of B = 48000 – 28800 = 19200

Question 16.
Solution:
Let original price of shirt = Rs. x
Discount = 12%

Original price of shirt = Rs. 1350

Question 17.
Solution:
Let original price of sweater = x
Rate of increase = 8%
Increased price = x + 8% of x

Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%

Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%

Question 21.
Solution:
Rate of commission = 2% on first Rs. 200000
1 % on next Rs. 200000 and 0.5% on remaining price
Sale price of property = Rs.200000 + 200000 +140000 = Rs. 540000
Now commission earned by the
= Rs. 200000 x 2% + Rs. 200000 x 1% + 140000 x 0.5%

Question 22.
Solution:
Let Akhil’s income = Rs. 100
Then income of Nikhil’s will be = Rs. 100 – 20 = Rs. 80
Amount which is more than that of Akhil’s = 100 – 80 = Rs. 20
% age = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 23.
Solution:
Let income of Mr Thomas = Rs. 100
then income of John = Rs. 100 + 20 = Rs. 120
Income of Mr Thomas is less than John = Rs. 120 – 100 = Rs. 20
% age = \(\frac { 20 x 100 }{ 120 }\)
= \(\frac { 50 }{ 3 }\) = 16\(\frac { 2 }{ 3 }\) %

Question 24.
Solution:
Present value of machine = Rs. 387000
Rate of depreciation = 10%
Let 1 year ago the value of machine was = x

1 year ago, value of machine = Rs. 430000

Question 25.
Solution:
Present value of car = Rs. 450000
Rate of decreasing of value = 20%
Value after 2 years

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years

Question 27.
Solution:
Let the price of sugar = Rs. 100
and consumption = 100 kg.
Increase price of 100 kg = Rs. 100 + 25 = Rs. 125
Now increased amount on 100 kg = Rs. 125

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10B
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C
• RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:
Let x is the required number

Question 10.
Solution:
Let x be the required number

Question 11.
Solution:
10 % of Rs. 90 = 90 x \(\frac { 10 }{ 100 }\) = Rs. 9
Required amount = Rs. 90 + Rs. 9 = Rs. 99

Question 12.
Solution:
20 % of Rs. 60 = \(\frac { 60 x 20 }{ 100 }\) = 12
Required amount = Rs. 60 – 12 = Rs. 48

Question 13.
Solution:
3 % of x = 9

Question 14.
Solution:
12.5 % of x = 6
⇒ x x \(\frac { 12.5 }{ 100 }\) = 6

Question 15.
Solution:
Let x % of 84 = 14

Question 16.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9C
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper

Question 1.
Solution:
Cost of 8 toys = ₹ 192
Cost of 1 toys = ₹ \(\frac { 192 }{ 8 }\) = ₹ 24
Cost of 14 toys = 24 x 14 = ₹ 336

Question 2.
Solution:
Distance covered with 15L of petrol = 270 km
Distance covered with 1L of petrol 270 = \(\frac { 270 }{ 15 }\) km
Distance covered with 8L of petrol 270 = \(\frac { 270 }{ 15 }\) x 8 km = 144 km

Question 3.
Solution:
Cost of 15 envelopes = ₹ 11.25
Cost of 1 envelope = ₹ \(\frac { 11.25 }{ 15 }\)
Cost of 20 envelopes = \(\frac { 11.25 }{ 15 }\) x 20 = ₹ 15

Question 4.
Solution:
24 cows can graze a field in = 20 days
1 cow can graze a field in = 20 x 24 days
15 cows can graze a field in = \(\frac { 20 x 24 }{ 15 }\)
= 32 days

Question 5.
Solution:
Time taken to finish the work by 8 men = 15 h
Time taken to finish the work by 1 man = 8 x 15 h
Time taken to finish the work by 20 men = \(\frac { 8 x 15 }{ 20 }\) h = 6 h
[More men, less time taken]

Question 6.
Solution:
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.

Question 7.
Solution:
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 x 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
= \(\frac { 45 x 3.33 }{ 50 }\) h
= 3h [More speed, less time taken]

Question 8.
Solution:
Number of days with enough food for 120 men = 30
Number of days with enough food for 1 man = 30 x 120 [Less men, more days]
Number of days with enough food for 100 men = \(\frac { 30 x 120 }{ 100 }\)
= 36 [More men, less days]

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(c) 644 km
1 cm represents km 80.5 cm represents = 8 x 80.5 = 644 km

Question 10.
Solution:
(a) 24 days
16 men can reap the field in 30 days.
1 man can reap the field in 30 x 16 days [Less men, more days]
20 men can reap the field in \(\frac { 30 x 16 }{ 20 }\)
= 24 days [More men, less days]

Question 11.
Solution:
(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 = \(\frac { 21 }{ 15 }\) x 35 = 49

Question 12.
Solution:
Number of cows that graze the field in 12 days = 45
Number of cows that graze the field in 1 day = 45 x 12
Number of cows that graze the field in 9 days = \(\frac { 45 x 12 }{ 2 }\) = 60

Question 13.
Solution:
(b) ₹ 162
Cost of 72 eggs = ₹ 108
Cost of 1 egg = ₹ \(\frac { 108 }{ 72 }\)
Cost of 108 eggs = ₹ \(\frac { 108 x 108 }{ 72 }\)
= ₹ 162

Question 14.
Solution:
(i) 588 days
42 men can dig the trench in 14 days 1 men can dig the trench in = 42 x 14 = 588 days
(ii) ₹48
15 oranges cost ₹ 60
12 oranges will cost ₹ \(\frac { 60 }{ 15 }\) x 12 = ₹ 48
(iii) 10.8 kg
A rod of length 10m weighs 18 kg
A rod of length 6 m will weigh = \(\frac { 18 }{ 10 }\) x 6 = 10.8 kg
(iv) 3 h 12 min
12 workers finish the work in 4 h
15 workers will finish the work in \(\frac { 4 x 12 }{ 15 }\) = 3.2 h = 3h 12 min

Question 15.
Solution:
(i) False
10 pipes fill the tank in = 24 min.
1 pipe will fill the tank in = 24 x 10 min. [Less pipes, more time taken]
8 pipes will fill the tank in = \(\frac { 24 x 10 }{ 8 }\)
= 30 min. [More pipes, less time taken]
(ii) True
8 men finish the work in = 40 days
1 man finishes the work in = 8 x 40 days [Less men, more days taken]
10 men will finish the work in = \(\frac { 8 x 40 }{ 10 }\)
= 32 days [More men, less days taken]
(iii) True
A 6 m tall tree casts a shadow of length = 4 m.
Aim tall tree cast a shadow of length = \(\frac { 4 }{ 6 }\) m
A 75 m tall pole will cast a shadow of length = \(\frac { 4 }{ 6 }\) x 75 = 50 m
(iv) True
1 toy is made in = \(\frac { 2 }{ 3 }\) h. [Less toys, less time taken]
12 toys can be made in = \(\frac { 2 }{ 3 }\) x 12
= 8h [More toys, more time taken]

Hope given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.