Maths Class 7 RS Aggarwal Solutions

RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss CCE Test Paper

Question 1.
Solution:
SP of the chair = ₹ 1375
Gain% = 10
CP of the chair

Question 2.
Solution:
Let the cost of each pen be ₹ 1
CP of 10 pens = ₹ 10
SP of 10 pens = CP of 14 pens = ₹ 14
Gain = SP – CP = 14 – 10 = ₹ 4

Question 3.
Solution:
Let the cost price of the fan be ₹ x

Question 4.
Solution:
Cost price of six lemons = ₹ 10

Question 5.
Solution:
SP of the bat = ₹ 486
Loss = 10%
CP of the bat

Mark (✓) against the correct answer in each of the following :
Question 6.
Solution:
(b) ₹ 85
SP of a football = ₹ 100
Gain = ₹ 15
Gain = SP – CP
⇒ 15 = 100 – CP
⇒ CP = ₹ (100 – 15)
⇒ CP = ₹ 85

Question 7.
Solution:
(c) 15.2%
Cost price of 12 bananas = ₹ 25
Cost price of one banana = ₹ \(\frac { 25 }{ 12 }\)
Cost price of five bananas = 5 x \(\frac { 25 }{ 12 }\)
= \(\frac { 125 }{ 12 }\) = 10.42
He shells five bananas at the cost (SP) of ₹ 12.
Gain = SP – CP = ₹ (12 – 10.42) = ₹ 1.58

Question 8.
Solution:
(c) ₹ 196
Let the cost price of the jug be ₹ x

Question 9.
Solution:
(c) 66\(\frac { 2 }{ 3 }\) %
Let the cost price of each banana be ₹ 1
Cost price of three bananas = ₹ 3
SP of three bananas = CP of five bananas = ₹ 5
Gain = SP – CP = ₹ (5 – 3) = ₹ 2

Question 10.
Solution:
(i) Loss = (CP) – (SP).

Question 11.
Solution:
(i) False
Gain or loss is always reckoned on the cost price.
(ii) True
(iii) True
(iv) True

Hope given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss 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 11 Profit and Loss Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss CCE Test Paper

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b)
C.P. = Rs. 80
S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 80 = Rs. 20

Question 2.
Solution:
(a)
C.P. = Rs. 120
S.P. = Rs. 105
Loss = C.P. – S.P. = Rs. 120 – 105 = Rs. 15

Question 3.
Solution:
(b)
S.P. = Rs. 100
Gain = Rs. 20
C.P. = S.P. – gain = Rs. 100 – 20 = Rs. 80

Question 4.
Solution:
(a)
S.P. = Rs. 198
Gain = 10%

Question 5.
Solution:
(a)
First S.P. of jug = Rs. 144
Loss = \(\frac { 1 }{ 7 }\) of C.P.
Let C.P. = x

Question 6.
Solution:
(d)
First S.P. of a pen = Rs. 48
Loss = 20%

Question 7.
Solution:
(a)
C.P. of 12 pencils = S.P. of 15 pencils = Re 1 (suppose)

Question 8.
Solution:
(d)
Let CP of 4 toffees = SP of 3 toffee = Rs. 12
(LCM of 4, 3 = 12)
CP of 1 toffee = Rs. \(\frac { 12 }{ 4 }\) = Rs. 3
SP of 1 toffee = Rs. \(\frac { 12 }{ 3 }\) = 4
Gain = Rs. 4 – 3 = Re 1

Question 9.
Solution:
(c) SP of an article = Rs. 144
Loss% = 10%

Question 10.
Solution:
(a)
CP of 6 lemons = Re 1
CP of 1 lemon = Re \(\frac { 1 }{ 6 }\)
SP of 4 lemons = Re 1
SP of 1 lemon = Re \(\frac { 1 }{ 4 }\)

Question 11.
Solution:
(d)
SP of chair = Rs. 720
Gain% = 20%

Question 12.
Solution:
(c)
SP of stool = Rs. 630
Loss% = 10%

Hope given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B 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 11 Profit and Loss Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B
• RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss CCE Test Paper

Question 1.
Solution:
(i) C.P. = Rs. 950

Question 2.
Solution:
(i) CP = Rs. 2400
SP = Rs. 2592
Gain = SP – CP = Rs. 2592 – Rs. 2400 = Rs. 192

Question 3.
Solution:
(i) S.P. = Rs. 924
Gain% = 10%

Question 4.
Solution:
C.P. of an almirah = Rs. 13600
Amount spent on transportation = Rs. 400
Total cost price (CP) = Rs. 13600 + Rs. 400 = Rs. 14000
SP = Rs. 16800
Total gain = SP – CP
= Rs. 16800 – Rs. 14000 = Rs. 2800

Question 5.
Solution:
CP of an old house = Rs. 765000
Amount spent on repairs = Rs. 115000
Total cost price (CP) = Rs. 765000 + Rs. 115000 = Rs. 880000
Gain = 5%

Question 6.
Solution:
CP of 1 dozen or 12 lemons = Rs. 25
SP of 5 lemons = Rs. 12

Question 7.
Solution:
SP of 12 pens = CP of 15 pens
Let CP of 1 pen = Re. 1
The CP of 15 pens = Rs. 15
and SP of 12 pens = Rs. 12

Question 8.
Solution:
SP of 16 spoons = CP of 15 spoons
Let CP of 1 spoon = Re. 1
Then CP of 15 spoons = Rs. 15
and SP of 16 spoons = Rs. 15

Question 9.
Solution:
For Manoj
CP of video = Rs. 12000
Gain = 10%

Question 10.
Solution:
SP of sofa set = Rs. 21600
Gain = 8%

Question 11.
Solution:
SP of a watch = Rs. 11400
Loss = 5%

Question 12.
Solution:
SP of calculator = Rs. 1325
Gain = 6%

Question 13.
Solution:
SP of computer = Rs. 24480
Loss = 4%

Question 14.
Solution:
In first case gain =15%
and in second case, gain = 20%
Difference = 20 – 15 = 5%
Now 5% of the CP = Rs. 108
CP = \(\frac { 108 x 100 }{ 5 }\) = Rs. 2160

Question 15.
Solution:
In first case, loss = 8%
and in second case gain = 6%
Difference = 8 + 6 = 14%
Now 14% of CP = Rs. 3360
CP = Rs. \(\frac { 3360 x 100 }{ 14 }\) = Rs. 24000

Question 16.
Solution:
SP of first cycle = Rs. 2376
Gain = 10%

Question 17.
Solution:
SP of an exhaust fan = Rs. 7350

Question 18.
Solution:
Let CP of watch for Mohit = Rs. 100
Gain =10%
SP for Mohit = Rs. 100 + 10 = Rs. 110
CP for Karim = Rs. 110
Gain = 4%

Question 19.
Solution:
Retail price or S.P. of the machine for retailer = Rs. 37950
Gain = 25%

Question 20.
Solution:
CP of video = Rs. 20000
and CP of television = Rs. 30000
Loss on video = 5%
and gain on TV = 8%

Total CP of video and TV = Rs. 20000 + 30000 = Rs. 50000
and SP of video and TV = Rs. 190000 + 32400 = RS. 51400
Gain = SP – CP = Rs. 51400 – 50000 = Rs. 1400

Question 21.
Solution:
SP of 36 oranges = CP of 36 oranges – loss = CP of 36 oranges – SP of 4 oranges
⇒ SP of 36 oranges + SP of 4 oranges = CP of 3 6 oranges
⇒ SP of 40 oranges = CP of 36 oranges
Let CP of each orange = Re. 1
CP of 36 oranges = Rs. 36
and SP of 40 of 40 oranges = Rs. 36

Question 22.
Solution:
We know that C.P. = S.P. – gain
C.P. of 8 dozen pencils = S.P. of 8 dozen pencils – S.P. of 1 dozen pencils
= S.P. of 7 dozen pencils = Rs. 100 (suppose)
C.P. of 1 dozen = Rs.\(\frac { 100 }{ 8 }\)

Hope given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A 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 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.