## RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B.

Other Exercises

Question 1.
Solution:
∵ x and y are inversely proportional
Then xy are equal
(i) xy = 6 x 9 = 54
= 10 x 15 = 150
= 14 x 21 = 294
= 16 x 24 = 384
∵ xy in each case is not equal
So, x and y are not inversely proportional
(ii) xy = 5 x 18 = 90
= 9 x 10 = 90
= 15 x 6 = 90
= 3 x 30 = 90
= 45 x 2 = 90
∵ xy in each case is equal
x and y are inversely proportional
(iii) xy = 9 x 4 = 36
= 3 x 12 = 36
= 6 x 9 = 54
= 36 x 1 = 36
∵ xy in each is not equal
x and y are not inversely proportional

Question 2.
Solution:
x and y are inversely proportional
xy is equal
Now,

Question 3.
Solution:
Let required number of days = x

Question 4.
Solution:
A pond is! dug in 8 days by = 12 men
It can be dug in 1 day by = 12 x 8 men (Less days, more men)
and it can be dug in 6 days by = $$\\ \frac { 12X8 }{ 6 }$$
= 16 men Ans. (more days, less men)

Question 5.
Solution:
Let 14 cows can graze in x days

Question 6.
Solution:
Let required time take = x hour

By inverse proportion
60 : x :: 75 : 5
x = $$\\ \frac { 50X5 }{ 75 }$$
Time required = 4 hours

Question 7.
Solution:
Let machines required = x

Question 8.
Solution:
Let 8 taken will fill in tank in x hour

Question 9.
Solution:
8 taps can fill tank in = 27 minutes
1 tap can fill that tank in = 27 x 8 minutes (less tap, more time)
8 – 2 = 6 taps can fill that tank in
= $$\\ \frac { 27X8 }{ 6 }$$ minutes
= 36 minutes

Question 10.
Solution:
Let total animals can be feed with food in x days

Question 11.
Solution:
Let for x day, the food provision is sufficient for 900 + 500 = 1400 men

Question 12.
Solution:
Let the food will be for x days

Question 13.
Solution:
Let each period will be of x minutes

Question 14.
Solution:
x and y are inversely
and x = 15, y = 6
Then xy = 15 x 6 = 90
Now if x = 9, then y = $$\\ \frac { 90 }{ 9 }$$
= 10

Question 15.
Solution:
x and y are inversely and x = 18, y = 8
xy = 18 x 8 = 144
Now if y = 16,
then x = $$\\ \frac { 144 }{ 16 }$$
= 9

Hope given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A.

Other Exercises

Question 1.
Solution:
(i) $$\\ \frac { x }{ y }$$ = $$\\ \frac { 3 }{ 9 }$$ = $$\\ \frac { 1 }{ 3 }$$

Question 2.
Solution:
x and y are directly proportional
$$\\ \frac { x }{ y }$$ = $$\\ \frac { 3 }{ 72 }$$ = $$\\ \frac { 1 }{ 24 }$$

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Let distance covered = x then

Question 6.
Solution:
Let no. of dolls = x, then

Question 7.
Solution:
Let x kg of sugar will be bought

Question 8.
Solution:
Let cloth bought = x m

Question 9.
Solution:
Let length of model ship = x m

Question 10.
Solution:
Let x kg dust will be picked up

Question 11.
Solution:
A speed of car = 50 km/hr
Distance travelled in 1 hr. = 5 m
Let required distance travelled in 1 hr. 12 min.

Question 12.
Solution:
Let required distance covered = x km
Speed of man = 5 km/hr

Question 13.
Solution:
Let required thickness = x mm

Question 14.
Solution:
Let men required = x

Question 15.
Solution:
Let no. of words type in 8 minutes = x

Hope given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D.

Other Exercises

Tick the correct answer in each of the following

Question 1.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 8% p.a.
Period (n) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 10% p.a.
Period (n) = 3 years

Question 3.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 12% p.a.

Question 4.
Solution:
Principal (P) = Rs. 4000
Rate (R) = 10% p.a.
Period (a) = 2 years 3 months

Question 5.
Solution:
Principal (P) = Rs. 25000
Rate (R1) = 5% for the first year
R2 = 6% for the second year
R3 = 8% for the third year

Question 6.
Solution:
Principal (P) = Rs. 6250
Rate (R) = 8% p.a. or 4% half yearly
Period (n) = 1 year or 2 half years

Question 7.
Solution:
Principal (P) = Rs. 40000
Rate (R) = 6% p.a. $$\\ \frac { 6 }{ 4 }$$ = $$\\ \frac { 3 }{ 2 }$$ % quarterly
Period (n) = 6 months = 2 quarters

Question 8.
Solution:
Present population (P) = 24000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years

Question 9.
Solution:
3 years ago, the value of machine = Rs. 60000
Rate of depreciation (R) = 10%
Period (n) = 3 years

Question 10.
Solution:
Present value = Rs. 40000
Rate of depreciation (R) = 20% p.a.
Value of machine 2 years ago

Question 11.
Solution:
Rate of growth in population (R) = 10%
Present population = 33275
Population 3 years ago = A

Question 12.
Solution:
S.I. = Rs. 1200
Rate (R) = 5%
Period (T) = 3 years

Question 13.
Solution:
C.I. on a sum = Rs. 510
Rate (R) = $$12\frac { 1 }{ 2 }$$ % = $$\\ \frac { 25 }{ 2 }$$ % p.a.
Period (n) = 2 years

Question 14.
Solution:
Amount = Rs. 4913
Rate (R) = $$6\frac { 1 }{ 4 }$$ = $$\\ \frac { 25 }{ 4 }$$ %
Period (n) = 3 years

Question 15.
Solution:
Sum (P) = Rs. 7500
Amount (A) = 8427
Period = 2 years
Let R be the rate of p.a., then

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B.

Other Exercises

By using the formula, find the amount and compound interest on :

Question 1.
Solution:
Principal (P) = Rs. 6000
Rate (R) = 9% p.a.
Period (n) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 10000
Rate (R) = 11% p.a.
Period (n) = 2 years

Question 3.
Solution:
Principal (P) = Rs. 31250
Rate (R) = 8% p.a.
Period (n) = 3 years

Question 4.
Solution:
Principal (P) = Rs. 10240
Rate (R) = $$12\frac { 1 }{ 2 }$$% = $$\\ \frac { 25 }{ 2 }$$% p.a.
Period (n) = 3 years

Question 5.
Solution:
Principal (P) = Rs. 62500
Rate (R) = 12% p.a.
Period (n) = 2 years 6 months

Question 6.
Solution:
Principal (P) = Rs. 9000
Rate (R) = 10% p.a.
Period (n) = 2 years 4 months

Question 7.
Solution:
Principal (P) = Rs. 8000
Period (n) = 2 years

Question 8.
Solution:
Principal (p) = Rs. 1, 25,000
Rate of interest (r) = 8% p.a.
Period (n) = 3 years

Question 9.
Solution:
Price of a buffalo (P) = Rs. 11000
Rate of interest (R) = 10% p.a.
Period (n) = 3 years
Price of buffalo at present

Question 10.
Solution:
Amount of loan taken (P)
= Rs. 18000

Question 11.
Solution:
Amount borrowed from Bank (P) = Rs. 24000
Rate (R) = 10% p.a.
Period (n) = 2 years 3 months

Question 12.
Solution:
In case of Abhay,
Principal (p) = Rs. 16000

Question 13.
Solution:
Simple interest (S.I.) = Rs. 2400
Rate (R) = 8% p.a.
Period (T) = 2 years

Question 14.
Solution:
Difference between C.I. and S.I.
= Rs. 90
Rate (R) = 6% p.a.
Period (n) = 2 years
Let principal (P) = Rs. 100

Question 15.
Solution:
Let sum (p) = Rs. 100
Rate (r) 10% p.a.
Period (t) = 3 years.

Question 16.
Solution:
Amount (A) = Rs. 10240
Rate (r) = $$6\frac { 2 }{ 3 }$$% = $$\\ \frac { 20 }{ 3 }$$% p.a.
Period (n) = 2 years
Let sum = P, then

Question 17.
Solution:
Amount (A) = Rs. 21296
Rate (r) = 10% p.a.
Period (n) = 3 years.
Let P be the sum, Then

Question 18.
Solution:
Principal (P) = 4000
Amount (A) = Rs. 4410
Period (n) = 2 years
Let r be the rate per cent per annum
We know that,

Question 19.
Solution:
Principal (P) = Rs. 640
Amount (A) = Rs. 774.40
Period (n) = 2 years
Let r be the rate per cent per annum.
We know that

Question 20.
Solution:
Principal (P) = Rs. 1800
Amount (A) = Rs. 2178
Rate (r) = 10% p.a.
Let n be the number of years,
We know that

Question 21.
Solution:
Principal (P) = Rs. 6250
Amount (A) = Rs. 7290
Rate (R) = 8% p.a.
Let n be the time, then

Question 22.
Solution:
Present population (P) = 125000
Rate of increasing (R) = 2% p.a.
Period (n) = 3 years

Question 23.
Solution:
3 years ago, the population was = 50000
Rate of increase successively (r1, r2, r3) = 4%, 5% and 3% p.a.
Period (n) = 3 years
Present Population

Question 24.
Solution:
Population of a city in 2013 = 120000
Increase in next year = 6%
and decrease in the following year = 5%
Population in 2015

Question 25.
Solution:
Initially bacteria = 500000
Increase in bacteria = 2% per hour
Period (n) = 2 hours

Question 26.
Solution:
Growth of bacteria in a culture (R1) = 10% in first hour
Decrease in next hour (R2) = 10%
Increase in the third hour (R3) = 10%
Bacteria in the beginning = 20000
Bacteria after 3 hours

Question 27.
Solution:
Value of machine (P) = Rs. 625000
Rate of depreciation (R) = 8% p.a.
Period (n) = 2 years

Question 28.
Solution:
Value of scooter (P) = Rs. 56000
Rate of depreciation (R) = 10% p.a.
Period = 3 years
Value of scooter after 3 years

Question 29.
Solution:
Cost of car = Rs. 34800
Rate of depreciation (R1) = 10% p.a. for first year

Question 30.
Solution:
Rate of depreciation (R) = 10% p.a.
Period (n) = 3 years
Present value (A) = Rs. 291600
Value of machine 3 years ago

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A.

Other Exercises

Question 1.
Solution:
Principal (p) = Rs. 2500
Rate (r) = 10% p.a.
Period (t) = 2 years
Interest for the first year = $$\\ \frac { prt }{ 100 }$$
= $$\\ \frac { 2500X10X1 }{ 100 }$$
= Rs. 250
Amount at the end of first year = Rs. 2500 + Rs. 250
= Rs 2750
Principal for the second year = Rs 2750
Interest for the second year = Rs $$\\ \frac { 2750X10X1 }{ 100 }$$
= Rs. 275
Amount at the end of second year = Rs 2750 + Rs. 275
= Rs. 305
and compound interest for the 2 years = Rs. 3025 – Rs. 2500
= Rs 525 Ans.

Question 2.
Solution:
Principal (P) = Rs. 15625 Rate
(R) = 12% p.a.
Period (n) = 3 years

Question 3.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 9% p.a.
Time (n) = 2 years

Question 4.
Solution:
Amount of loan (p) = Rs. 25000
Rate of interest (r) = 8% p.a.
Period (t) = 2 years

Question 5.
Solution:
In case of Harpreet :
Amount borrowed by Harpreet (P) = Rs. 20000
Rate (r) = 12%
Period (t) = 2 Years

Question 6.
Solution:
Principal (p) = Rs. 64000
Rate (r) = $$7\frac { 1 }{ 2 }$$ = $$\\ \frac { 15 }{ 2 }$$%
Period (t) = 3 years
Interest for the first year = $$\\ \frac { prt }{ 100 }$$
= Rs $$\\ \frac { 64000X15X1 }{ 100X2 }$$
= Rs 4800

= Rs 79507

Question 7.
Solution:
Principal (p) = Rs 6250
Rate (r) 8% p.a. or 4% half yearly
Period (t) = 1 year = 2 half years

Question 8.
Solution:
Principal (p) = Rs. = 16000
Rate (r) = 10% p.a. or 5% half yearly

Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D.

Other Exercises

Question 1.
Solution:
C.P. of toy Rs. = 75
S.P. = Rs. 100

Question 2.
Solution:
C.P. of bat = Rs. 120
S.P. = Rs. 105
Loss = Rs. 120 – Rs. 105 = Rs. 15

Question 3.
Solution:
S.P. of book = Rs. 100
gain = Rs. 20
C.P. = 100 – 20 = Rs. 80

Question 4.
Solution:
SP of an article = Rs. 48
Loss = 20%

Question 5.
Solution:
First time gain = 10%
Let SP – Rs. 100

Question 6.
Solution:
Let no. of bananas bought = 6
Now C.P. of bananas at the sale of 3

Question 7.
Solution:
SP of 10 pens = CP of 12 pens
= Rs. 100 (suppose)

Question 8.
Solution:
Gain on 100 pencils = SP of 20 pencils
SP of 100 pencils gains = CP of 100 pencils
=> SP of 100 pencils – SP of 20 pencils = CP of 100 pencils
=> SP of 80 pencils – CP of 100 pencils = Rs 100 (suppose)

Question 9.
Solution:
SP of 5 toffees = Re. 1
SP of 2 toffees = Re. 1
Now CP of 1 toffee = Rs. $$\\ \frac { 1 }{ 5 }$$
and SP of 1 toffee = Rs. $$\\ \frac { 1 }{ 2 }$$

Question 10.
Solution:
CP of 5 oranges = Rs. 10
SP of 6 oranges = Rs. 15

Question 11.
Solution:
SP of a radio = Rs. 950
Loss = 5%

Question 12.
Solution:
Let CP of an article = Rs. 100
SP = $$\\ \frac { 6 }{ 5 }$$ of CP = $$\\ \frac { 6 }{ 5 }$$ x 100 = Rs. 120

Question 13.
Solution:
SP of a chair = Rs. 720
Loss = 25%

Question 14.
Solution:
Ratio in CP and SP = 20 : 21
Let CP = Rs. 20
and SP = Rs. 21
Gain = SP – CP = Rs. 21 – 20 = Re. 1

Question 15.
Solution:
SP of first chair = Rs. 500
Gain = 20%

Question 16.
Solution:
Gain % SP of Rs. 625 = Loss on SP of Rs. 435
CP of an article = x

Question 17.
Solution:
CP of an article = Rs. 150
Overhead expenses = 10% of CP

Question 18.
Solution:
In first case, gain = 5%
and in second case, loss = 5%
and difference = Rs. 5 more
But difference in % = 5 + 5 = 10%

Question 19.
Solution:
Let CP of an article = Rs. 100
List price = Rs. 100 + 20% of Rs. 100
= Rs. 100 + 20 = Rs. 120
Discount = 10%

Question 20.
Solution:
Let CP of an article = Rs. 100
Then Marked price
= Rs. 100 + 10% of 100
= Rs. 100 + 10 = Rs. 110
Discount = 10%

Question 21.
Solution:
Price of watch including VAT = Rs. 825
VAT% = 10%

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.

Other Exercises

Question 1.
Solution:
x2 + 8x + 16
= (x)2 + 2 × x × 4 + (4)2
= (x + 4)2

Question 2.
Solution:
x2 + 14x + 49
= (x)2 + 2 × x × 7 + (7)2
= (x + 7)2 Ans.

Question 3.
Solution:
1 + 2x + x2
= (1)2 + 2 × 1 × x + (x)2
= (1 + x)2 Ans.

Question 4.
Solution:
9 + 6z + z2
= (3)2 + 2 x 3 x z + (z)2
= (3 + z)2 Ans.

Question 5.
Solution:
x2 + 6ax + 9a2
= (x)2 + 2 × x × 3a + (3a)2
= (x + 3a)2 Ans.

Question 6.
Solution:
4y2 + 20y + 25
= (2y)2 + 2 x 2y x 5 + (5)2
= (2y2 + 5)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 7.
Solution:
36a2 + 36a + 9
= 9 [4a2 + 4a + 1]
= 9 [(2a)2 + 2 x 2a x 1 + (1)2]
= 9 [2a + 1]2

Question 8.
Solution:
9m2 + 24m + 16
= (3m)2 + 2 x 3m x 4 + (4)2
= (3m + 4)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 9.
Solution:
z2 + z + $$\\ \frac { 1 }{ 4 }$$
= (z)2 + 2 x z x $$\\ \frac { 1 }{ 2 }$$ + $${ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$$
= $${ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }$$

Question 10.
Solution:
49a2 + 84ab + 36b2
= (7a)2 + 2 x 7a x 6b + (6b)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
= (7a + 6b)2

Question 11.
Solution:
p2 – 10p + 25
= (p)2 – 2 x p x 5 + (5)2
= (p – 5)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 12.
Solution:
121a2 – 88ab + 16b2
= (11a)2 – 2 x 11a x 4b + 4(b)2
= (11a – 4b)2

Question 13.
Solution:
1 – 6x + 9x2
= (1)2 – 2 x 1 x 3x + (3x)2
= (1 – 3x)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 14.
Solution:
9y2 – 12y + 4
= (3y)2 – 2 x 3y x 2 + (2)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
= (3y – 2)2

Question 15.
Solution:
16x2 – 24x + 9
= (4x)2 – 2 x 4x x 3 + (3)2
= (4x – 3)2 Ans.

Question 16.
Solution:
m2 – 4mn + 4n2
= (m)2 -2 x m x 2n + (2n)2
= (m – 2n)2 Ans.

Question 17.
Solution:
a2b2 – 6abc + 9c2
= (ab)2 – 2 x ab x 3c + (3c)2
= (ab – 3c)2 Ans.

Question 18.
Solution:
m4 + 2m2n2 + n4
= (m2)2 + 2m2n2 + (n2)2
= (m2 + n2)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 19.
Solution:
(l + m)2 – 4lm
= l2 + m2 + 2lm – 4lm
= l2 + m2 – 2lm
= l2 – 2lm + m2
= (l – m)2
{ ∵ a2 – 2ab + b2 = (a – b)}

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
$$\\ \frac { 3 }{ 5 }$$
= $$\\ \frac { 3X20 }{ 5X20 }$$
= $$\\ \frac { 60 }{ 100 }$$
= 60% (d)

Question 2.
Solution:
0.8%
= $$\\ \frac { 0.8 }{ 100 }$$
= $$\\ \frac { 8 }{ 10X100 }$$
= $$\\ \frac { 8 }{ 1000 }$$
= 0.008 (b)

Question 3.
Solution:
6 : 5
= $$\\ \frac { 6 }{ 5 }$$
= $$\\ \frac { 6X20 }{ 5X20 }$$
= $$\\ \frac { 120 }{ 100 }$$
= 120% (c)

Question 4.
Solution:
5% of a number is 9
Number = $$\\ \frac { 9X100 }{ 5 }$$
= 180 (d)

Question 5.
Solution:
Let x% of 90 = 120
=> $$\\ \frac { x }{ 100 }$$ x 120 = 90
=> x% = $$\\ \frac { 120X100 }{ 90 }$$
= $$133\frac { 1 }{ 3 } %$$ (c)

Question 6.
Solution:
Let x% 10 kg = 250 g
$$\\ \frac { x }{ 100 }$$ x 10 kg
= $$\\ \frac { 250 }{ 1000 }$$ kg
x = $$\\ \frac { 250X100 }{ 1000X10 }$$
= $$\\ \frac { 25 }{ 10 }$$%
= 2.5% (d)

Question 7.
Solution:
40% of x = 240
=> x = $$\\ \frac { 240 }{ 40 }$$ x 100
= 600 (b)

Question 8.
Solution:
?% of 400 = 60
=> x% of 400 = 60
$$\\ \frac { x }{ 100 }$$ x 400 = 60
x = $$\\ \frac { 60X100 }{ 400 }$$
= 15 (c)

Question 9.
Solution:
(180% of ?)÷2 = 504
$$\left( \frac { 180 }{ 100 } \times x \right) \div 2$$ = 504
$$\frac { 180 }{ 100 } x$$ = 504 x 2
$$x=\frac { 504\times 2\times 100 }{ 180 }$$
= 560 (d)

Question 10.
Solution:
20% of Rs. 800
= $$\\ \frac { 20 }{ 100 }$$ x 800
= Rs 160 (a)

Question 11.
Solution:
Nitin gets = 98 marks
and it is 56% of total marks
Total-marks = $$\\ \frac { 98X100 }{ 56 }$$
= 175 (c)

Question 12.
Solution:
Let a number be = 1000
Then increase = 10%
Increased number = $$\\ \frac { 100X110 }{ 100 }$$ = 110
Now decrease = 1%
Decreased number = $$\\ \frac { 110X90 }{ 100 }$$ = 99
Difference = 100 – 99 = 1
% decrease = 1% (b)

Question 13.
Solution:
4 hours 30 min = $$4\frac { 1 }{ 2 } %$$
= $$\\ \frac { 9 }{ 2 }$$ hours
% of a day = $$\\ \frac { 9 }{ 2 }$$ x $$\\ \frac { 100 }{ 24 }$$%
= $$\\ \frac { 75 }{ 4 }$$%
= $$18\frac { 3 }{ 4 } %$$ (a)

Question 14.
Solution:
Let total number of examinees = 100
Passed = 65
Failed = 100 – 65 = 35
Now 35% of total examinees = 420
Total examinees = $$\\ \frac { 420X100 }{ 35 }$$
= 1200 (c)

Question 15.
Solution:
Let number = x
Then x = $$\\ \frac { xX20 }{ 100 }$$ = 40
=> 100x – 200x = 4000
80x = 4000
=> x = $$\\ \frac { 4000 }{ 80 }$$ = 50
Number = 50 (a)

Question 16.
Solution:
Rate of decrease = $$27\frac { 1 }{ 2 } %$$
Let number = x

Question 17.
Solution:
Let x% of 20 = 0.05
$$\\ \frac { x }{ 100 }$$ x 20 = 0.05
x = $$\\ \frac { 0.05X100 }{ 20 }$$
= 0.25% (c)

Question 18.
Solution:
$$\\ \frac { 1 }{ 3 }$$ = 1206 = 402

Question 19.
Solution:
x% of y is y% = $$\\ \frac { xy }{ 100 }$$
$$\\ \frac { y }{ 100 }$$ × x
= y% of x (a)

Question 20.
Solution:
Let x% of $$\\ \frac { 2 }{ 7 }$$ = $$\\ \frac { 1 }{ 35 }$$
=>$$\\ \frac { x }{ 100 }$$ x $$\\ \frac { 2 }{ 7 }$$ = $$\\ \frac { 1 }{ 35 }$$
x = $$\\ \frac { 1X100X7 }{ 35X2 }$$
= 10% (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A.

Other Exercises

Question 1.
Solution:
(i) C.P. = Rs. 620
S.P. = Rs. 713
Gain = S.P. – C.P. = Rs. 713 – Rs. 620
= Rs. 93

Question 2.
Solution:
(i) C.P. = Rs. 1650
Gain% = 4%

Question 3.
Solution:
(i)S.P.= Rs 1596
Gain % = 12%

Question 4.
Solution:
C.P. of iron safe = Rs 12160
Paid for its transportation = Rs 340
Total cost = Rs 12160 + Rs 340
= Rs 12500

Question 5.
Solution:
C.P. of car = Rs 73500
Overhead charges = Rs 10300 + 2600
= Rs 12900

Question 6.
Solution:
C.P. of 20 kg @ Rs 36 per kg.
= 20 x 36 = Rs 720
C.P. of 25 kg @ Rs32 per kg.

Question 7.
Solution:
Ratio of the mixture = 5 : 2
Let 5 kg of coffee is mixed with 2 kg of chicory

Question 8.
Solution:
Let CR of 17 water bottles = Rs. 17
Then C.P. of one bottle

Question 9.
Solution:
Let C.P. of 12 candIes =Rs. 12
C.P of 1 candle = Re. 1

Question 10.
Solution:
S.P. of 125 cassettes – gain = C.P. of 125 cassettes
=> S.P. of 125 cassettes – S.P. of 5 cassettes = C.P. of 125 cassettes
=> S.P. of 120 cassettes = C.P. of 125 cassettes

Question 11.
Solution:
S.P. of 45 lemons = C.P. of 45 lemons – loss

Question 12.
Solution:
CP. of 6 oranges = Rs 20
and C.P of 1 orange = Rs $$\\ \frac { 20 }{ 6 }$$ = Rs $$\\ \frac { 10 }{ 3 }$$

Question 13.
Solution:
C.P. of 12 bananas = Rs 40
and C.P of 1 banana = Rs $$\\ \frac { 40 }{ 12 }$$ = Rs $$\\ \frac { 10 }{ 3 }$$

Question 14.
Solution:
C.P. of 10 apples = Rs 75
and C.P of 1 apple = Rs $$\\ \frac { 75 }{ 10 }$$ = Rs $$\\ \frac { 15 }{ 2 }$$

Question 15.
Solution:
Let eggs purchased were = 3 x 16 = 48

Question 16.
Solution:
S.P. of camera = Rs 1080
C.P. of camera + gain = S.P. of camera
=> C.P. of camera + $$\\ \frac { 1 }{ 8 }$$ of C.P. = S.P. of camera
= $$\\ \frac { 9 }{ 8 }$$ x C.P. of camera = S.P. of camera = Rs 1080

Question 17.
Solution:
S.P. of a pen = Rs 54
Loss = $$\\ \frac { 1 }{ 10 }$$ of her outlay

Question 18.
Solution:
Let C.P. of table = Rs 100
In first case, loss = 10%
S.P. = Rs 100 – 10 = Rs 90
and in second case, gain = 10%

Question 19.
Solution:
Let C.P. of chair = Rs 100
In first case, gain = 15%
then S.P. = Rs 100 + 15 = Rs 115
and in second case, gain = 8%

Question 20.
Solution:
Let the C.P. of cycle = Rs 100
In first case, gain = 10%
then S.P. = Rs 100 + 10 = Rs 110
In second case, gain = 14%
Difference between their S.P.s = Rs 114 – Rs 110 = Rs 4

Question 21.
Solution:
Cost price of 40 kg @ Rs 12.50 per kg.
= 40 x 12.50 = Rs 500
and cost price of 30 kg @ of Rs 14 per kg
= 30 x 14 = Rs 420

Question 22.
Solution:
C.P. of first bat = Rs 840
Gain% = 15%

Question 23.
Solution:
C.P. of first jean = Rs 1450
Gain% = 8%

Question 24.
Solution:
Total quantity of rice = 200 kg.
C.P. of 200 kg @ Rs 25 per kg.
= Rs 200 x 25 = Rs 5000
Gain% on total = 8%

Question 25.
Solution:
Let C.P. = Rs 100
then S.P. = $$\\ \frac { 6 }{ 5 }$$ of Rs 100

Question 26.
Solution:
Let C.P. of flower vase = Rs 100
then S.P. $$\\ \frac { 5 }{ 6 }$$ of C.P. = $$\\ \frac { 5 }{ 6 }$$ of Rs 100
= Rs $$\\ \frac { 500 }{ 6 }$$

Question 27.
Solution:
S.P. of a bouquet = Rs 322
Gain = 15%

Question 28.
Solution:
In first case,
S.P. of an umbrella = Rs 336
Loss% = 4%

Question 29.
Solution:
S.P. of Radio = Rs 3120
Loss% = 4%

Question 30.
Solution:
S.P. Of first sarees = Rs 1980
Loss% = 10%

Question 31.
Solution:
S.P. of first fan = Rs 1140
Gain% = 14%

Question 32.
Solution:
C.P. for Manoj = Rs 3990
or S.P. for Arun = Rs 3990
loss% = 5%
C.P for Arun
$$=\frac { S.P\times 100 }{ 100-losspercent} =\frac { 3990\times 100 }{ 100-5 }$$

Question 33.
Solution:
C.P. of plot of land = Rs 480000
C.P. of $$\\ \frac { 2 }{ 5 }$$th part = Rs 480000 x $$\\ \frac { 2 }{ 5 }$$
= Rs 192000

Question 34.
Solution:
C.P. of sugar = Rs 4500
C.P. of $$\\ \frac { 1 }{ 3 }$$ of sugar

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss 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 8 Solutions Chapter 8 Linear Equations Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A.

Other Exercises

Solve :

Question 1.
Solution:
8x + 3 = 27 + 2x
=> 8x – 2x
=> 27 – 3
=> 6x = 24
=> x = $$\\ \frac { 24 }{ 6 }$$ = 4
x = 4

Question 2.
Solution:
5x + 7 = 2x – 8
=> 5x – 2x = – 8 – 7
=> 3x = – 15
=> x = $$\\ \frac { -15 }{ 3 }$$ = – 5
x = – 5

Question 3.
Solution:
2z – 1 = 14 – z
=> 2z + z = 14 + 1
=> 3z = 15
=> z = $$\\ \frac { 15 }{ 3 }$$ = 5
z = 5

Question 4.
Solution:
9x + 5 = 4(x – 2) +8
=> 9x + 5 = 4x – 8 + 8
=> 9x – 4x = – 8 + 8 – 5
=> 5x = – 5
=> x = $$\\ \frac { -5 }{ 5 }$$ = – 1
x = – 1

Question 5.
Solution:
$$\\ \frac { 7y }{ 5 }$$ = y – 4
Multiplying both sides by 5,
$$\\ \frac { 7y }{ 5 }$$ x 5 = 5(y – 4)
=> 5 (y-4)
=> 7y = 5y – 20
=> 7y – 5y = – 20
=> 2y = – 20
=> y = $$\\ \frac { -20 }{ 2 }$$ = – 10
Hence y = – 10 Ans.

Question 6.
Solution:
3x + $$\\ \frac { 2 }{ 3 }$$ = 2x + 1
=> 3x – 2x = 1 – $$\\ \frac { 2 }{ 3 }$$
=> x = $$\\ \frac { 3-2 }{ 3 }$$ = $$\\ \frac { 1 }{ 3 }$$
Hence x = $$\\ \frac { 1 }{ 3 }$$ Ans.

Question 7.
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
=> 15y – 2y + 5y = 60 – 18 – 30
=> 18y = 12
=> y = $$\\ \frac { 12 }{ 18 }$$ = $$\\ \frac { 2 }{ 3 }$$
=> y = $$\\ \frac { 2 }{ 3 }$$

Question 8.
Solution:
3(5x – 7) – 2(9x – 11) = 4(8x – 13) – 17
=> 15x – 21 – 18x + 22 = 32x – 52 – 17
=> 15x – 18x – 32x = – 52 – 17 + 21 – 22
=> 15x – 50x = – 70
=> – 35x = – 70
=> x = $$\\ \frac { -70 }{ -35 }$$ = 2
x = 2

Question 9.
Solution:
$$\\ \frac { x-5 }{ 2 }$$ – $$\\ \frac { x-3 }{ 5 }$$ = $$\\ \frac { 1 }{ 2 }$$
Multiplying each term by 10, the L.C.M. of 2 and 5

Question 10.
Solution:
$$\\ \frac { 3t-2 }{ 4 }$$ – $$\\ \frac { 2t+3 }{ 3 }$$ = $$\\ \frac { 2 }{ 3 }$$ – t

Question 11.
Solution:
$$\frac { 2x+7 }{ 5 } +\frac { 3x+11 }{ 2 } =\frac { 2x+8 }{ 3 } -5$$
Multiplying by 30, the L.C.M. of 5, 2 and 3.

Question 12.
Solution:
$$\frac { 5x-4 }{ 6 } =4x+1-\frac { 3x+10 }{ 2 }$$
Multiplying by 6, the L.C.M. of 6 and 2

Question 13.
Solution:
$$5x-\frac { 1 }{ 3 } \left( x+1 \right) =6\left( x+\frac { 1 }{ 30 } \right)$$
=> 5x – $$\\ \frac { x+1 }{ 3 }$$ = 6x + $$\\ \frac { 1 }{ 5 }$$
Multiplying by 15, the L.C.M. of 3 and 5

Question 14.
Solution:
$$4-\frac { 2\left( z-4 \right) }{ 3 } =\frac { 1 }{ 2 } \left( 2z+5 \right)$$

Question 15.
Solution:
$$\frac { 3\left( y-5 \right) }{ 4 } -4y=3-\frac { \left( y-3 \right) }{ 2 }$$
Multiplying by 4, the L.C.M. of 4 and 2

Question 16.
Solution:
$$\\ \frac { 8x-3 }{ 3x }$$ = $$\\ \frac { 2 }{ 1 }$$
By cross multiplication,
8x – 3 = 6x
=> 8x – 6x = 3
=> 2x = 3
=> x = $$\\ \frac { 3 }{ 2 }$$
x = $$\\ \frac { 3 }{ 2 }$$

Question 17.
Solution:
$$\\ \frac { 9x }{ 7-6x }$$ = $$\\ \frac { 15 }{ 1 }$$
By cross multiplication,
9x = 105 – 90x
=> 9x + 90x = 105
=> 99x = 105
=> x = $$\\ \frac { 105 }{ 99 }$$ = $$\\ \frac { 35 }{ 33 }$$
x = $$\\ \frac { 35 }{ 33 }$$

Question 18.
Solution:
$$\\ \frac { 3x }{ 5x+2 }$$ = $$\\ \frac { -4 }{ 1 }$$
By cross multiplication,
3x × 1 = – 4×(5x + 2)
=> 3x = – 20x – 8
=> 3x + 20x = – 8
=> 23x = – 8
=> x = $$\\ \frac { -8 }{ 23 }$$
Hence x = $$\\ \frac { -8 }{ 23 }$$

Question 19.
Solution:
$$\\ \frac { 6y-5 }{ 2y }$$ = $$\\ \frac { 7 }{ 9 }$$
By cross multiplication,
9(6y – 5) = 7 × 2y
=> 54y – 45 = 14y
=> 54y – 14y = 45
=> 40y = 45
=> y = $$\\ \frac { 45 }{ 40 }$$ = $$\\ \frac { 9 }{ 8 }$$
Hence y = $$\\ \frac { 9 }{ 8 }$$ Ans.

Question 20.
Solution:
$$\\ \frac { 2-9z }{ 17-4z }$$ = $$\\ \frac { 4 }{ 5 }$$
By cross multiplication,
5 (2 – 9z) = 4(17 – 4z)
=> 10 – 45z = 68 – 16z
=> – 45z + 16z = 68 – 10
=> – 29 = 58
=> z = $$\\ \frac { 58 }{ -29 }$$ = – 2
Hence z = – 2 Ans.

Question 21.
Solution:
$$\\ \frac { 4x+7}{ 9-3x }$$ = $$\\ \frac { 1 }{ 4 }$$
By cross multiplication,
4(4x + 7) = 1 (9 – 3x)
=> 16x + 28 = 9 – 3x
=> 16x + 3x = 9 -28
=> 19x = – 19
=> x = $$\\ \frac { -19 }{ 19 }$$ = – 1
Hence x = – 1 Ans.

Question 22.
Solution:
$$\\ \frac { 7y+4}{ y+2 }$$ = $$\\ \frac { -4 }{ 3 }$$
By cross multiplication,
3 (7y + 4) = – 4 (y + 2)
=> 21y + 12 = – 4y – 8
=> 21y + 4y = – 8 – 12
=> 25y = – 20
=> y = $$\\ \frac { -20 }{ 25 }$$ = $$\\ \frac { -4 }{ 5 }$$
y = $$\\ \frac { -4 }{ 5 }$$

Question 23.
Solution:
$$\\ \frac { 15(2-y)-5(y+6) }{ 1-3y }$$ = $$\\ \frac { 10 }{ 1 }$$
By cross multiplication,
15 (2 – y) – 5(y + 6) = 10 (1 – 3y)
=> 30 – 15y – 5y – 30 = 10 – 30y
=> – 15y – 5y + 30y = 10 – 30 + 30
=> 30y – 20y = 10
=> 10y = 10
y = $$\\ \frac { 10 }{ 10 }$$ = 1
Hence y = 1 Ans.

Question 24.
Solution:
$$\\ \frac { 2x-(7-5x) }{ 9x-(3x+4x) }$$ = $$\\ \frac { 7 }{ 6 }$$

Question 25.
Solution:
$$m-\frac { \left( m-1 \right) }{ 2 } =1-\frac { \left( m-2 \right) }{ 3 }$$

Question 26.
Solution:
$$\\ \frac { 3x+5 }{ 4x+2 }$$ = $$\\ \frac { 3x+4 }{ 4x+7 }$$
By cross multiplication,
(3x + 5)(4x + 7) = (3x + 4)(4x + 2)
=> 12x² + 21x + 20x + 35

Question 27.
Solution:
$$\\ \frac { 9x-7 }{ 3x+5 }$$ = $$\\ \frac { 3x-4 }{ x+6 }$$

Question 28.
Solution:
$$\\ \frac { 2-7x }{ 1-5x }$$ = $$\\ \frac { 3+7x }{ 4+5x }$$

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A 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 8 Solutions Chapter 10 Profit and Loss Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B.

Other Exercises

Question 1.
Solution:
Marked price of cooler = Rs 4650
Rate of discount = 18%
Selling price

Question 2.
Solution:
Marked price = Rs 960
Selling price = Rs 816
Total Discount = M.P. – S.P.
= Rs 960 – Rs 816
= Rs 144

Question 3.
Solution:
S.P. of shirt = Rs 1092
Discount = Rs 208
M.P. of shirt = S.P. + discount

Question 4.
Solution:
S.P. of toy = Rs 216.20
Discount = 8%

Question 5.
Solution:
S.P. of tea set = Rs 528
Rate of discount = 12%

Question 6.
Solution:
Let C.P. of goods = Rs 100
Marked price = Rs 100 + 35
= Rs 135
Rate of discount = 20%

Question 7.
Solution:
Let C.P. of phone = Rs 100
.’. Marked price = Rs 100 + 40
= Rs 140
Rate of discount = 30%

Question 8.
Solution:
C.P. of fan = Rs. 1080
Gain = 25%

Question 9.
Solution:
C.P of refrigerator = Rs. 11515
and gain % = 20%
S.P. of refrigerator

Question 10.
Solution:
C.P. of ring = Rs. 1190
Gain = 20%

Question 11.
Solution:
Let Marked price = Rs. 100
Discount = 10%

Question 12.
Solution:
Let C.P. = Rs. 100
Gain = 8%

Question 13.
Solution:
Marked price of TV = Rs. 18500
Series of two successive discounts = 20% and 5%

Question 14.
Solution:
Let M.P. = Rs. 100
First discount = 20%
and second discount = 5%

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss 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.