CBSE Class 8

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

Find each of the following products (1-8)
Question 1.
5x² x 4x³
Solution:
5x² x 4x³ = 5 x 4 x x² x x³
= 20x^{2 + 3} = 20x^s

Question 2.
3a² x 4b⁴
Solution:
-3a² x 4b⁴ = -3 x 4 x a²b⁴
= -12a²b⁴

Question 3.
(-5xy) x (-3x²yz)
Solution:
(-5xy) x (-3x²yz)
= (-5) x (-3)xy x x²yz
= 15x^{1 + 2}xy^{1+ 1}z= 15x³y²z

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab²c) x (6abc²)
Solution:
(7ab) x (-5ab²c) x (6abc²)
= 7 x (-5) x 6 x a x a x a x b x b² x b x c x c²
=-210 x a¹⁺¹⁺¹ x b¹⁺²⁺¹x c¹⁺²
=-210 x a³b⁴c³

Question 10.
(-5a) x (-10a²) x (-2a³)
Solution:
(-5a) x (-10a²) x (-2a³)
= (-5) (-10) (-2) x a x a² x a³
= -100a^{1 + 2 + 3} = -100a⁶

Question 11.
(-4x²) x (-6xy²) x (-3yz²)
Solution:
(-4x²) x (-6xy²) x (-3yz²)
= (-4) x (-6) x (-3) x² x x x y² x y xz²
= -72x²⁺¹ x y²⁺¹ x _z²
= 72x³y³z³

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x^{1 +1} x y = 0.0368x²y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:

Question 19.

Solution:

Question 20.
(5x⁴) x (x²)³ x (2x)²
Solution:
(5x⁴) x (x²)³ x (2x)²
= 5x⁴ x x^{2 x 3} x 2x x 2x
= 5x⁴ * x⁶ x 4x² = 5 x 4 x x^{4 + 6 + 2}
= 20x¹²
Verification:
L.H.S. = (5x⁴) x (x²)³ x (2x)²
= 5 x (1)⁴ x [(1)²]³ x (2 x 1)²
= 5 x 1 x (1)^{2 x 3}_x (2)²
= 5 x 1⁶ x 2² = 5 x 1 x 4 = 20
R.H.S. = 20x¹² = 20 (1)¹² = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x²)³ x (2x) x (-4x) x 5
Solution:
(x²)³ x (2x) x (-4x) x (5)
= x^{2 x 3} X 2x X (-4x) X 5
= x⁶ x 2x x (-4x) x 5 = 2 x (-4) x 5x^{6+1 +1}
= -40x⁸
Verification
L.H.S. = (x²)³ x (2x) x (-4x) x (5)
= (1²)³ x (2 x 1) x (-4 x 1) x 5
= 1^2 x 3 x 2 x (- 4) x 5 = 1⁶ x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x⁸ = -40 x (1)⁸
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x²y⁶ and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x²y⁶ and -20xy
= (-8x²y⁶) x (-20xy)
= -8 x (-20) x² x _x x y⁶ x y = 160x^{2 + 1} x y^{6 + 1}
= 160x³y³
Verification.
L.H.S. = (-8x²y⁶) x (-20xy)
= -8 x (2.5)² x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)³ x (1)⁷
= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x⁶y³) x (2.1x²y²) when x = 1 and y = 0.5.
Solution:
3.2x⁶y³ x 2.1x²y²
= 3.2 x 2.1 x x⁶⁺² x y³⁺²
= 6.72x⁸y⁵ = 6.72 x (1)⁸ x (0.5)⁵
= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x⁶) x (-1.5x²y³) x (-12xy²) when x = 1, y = 0.5.
Solution:

Question 25.
Evaluate : (2.3a⁵b²) x (1.2a²b²) when a = 1, b = 0.5.
Solution:

Question 26.
Evaluate : (-8x²y⁶) x (-20xy) for x = 2.5 and y = 1.
Solution:

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy³) x (yx³) x (xy)
Solution:
(-xy³) x (yx³) x (xy)
= -x x x³ x x x y³ x y x y = -x^{1 + 3 + 1} x y^{3 + 1 + 1} = -x⁵y⁵
Verification:
L.H.S. = (-xy³) x (yx³) x (xy)
= (-1 x 2³) x [2 x (1)³] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x⁵y⁵  = -(1)⁵ (2)⁵
= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 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.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 12\(\frac { 1 }{ 2 }\) % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years

n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 1\(\frac { 1 }{ 2 }\) years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then

Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 6\(\frac { 2 }{ 3 }\) % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years

Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 6\(\frac { 3 }{ 4 }\) % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 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.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 7\(\frac { 1 }{ 2 }\) %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years

and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years

and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.

C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years

C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800

C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years

C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.

C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 12\(\frac { 1 }{ 2 }\) % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.

Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 17\(\frac { 1 }{ 2 }\) % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000


C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096

Question 8.
Find the amount and the compound interest on Rs 8,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly

and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 12\(\frac { 1 }{ 2 }\) % per annum to build a house. Find the amount that she pays LIC after 1\(\frac { 1 }{ 2 }\) years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000

Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years


C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760


C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters

Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters

C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 2\(\frac { 1 }{ 4 }\) years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years

= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years

C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,

Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum

Question 21.
Simple interest on a sum of money for 2 years at 6\(\frac { 1 }{ 2 }\) % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200


Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years

In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years

= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 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.

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6
• RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7

Question 1.
Add the following algebraic expressions


Solution:

Question 2.
Subtract:
(i) -5xy from 12xy
(ii) 2a² from -7a²

Solution:

Question 3.
Take away :


Solution:

Question 4.
Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and – 4X + 9y- 11z.
Solution:
Sum of x – 3y + 2z and – 4x + 9y – 11z
= x – 3y + 2z + (- 4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y – 11z
= x – 4x – 3y + 9y + 2z – 11z
= – 3x + 6y – 9z
Now (-3x + 6y – 9z) – (3x – 4y – 7z)
= -3x + 6y – 9z – 3x + 4y + 7z
= -3x – 3x + 6y + 4y -9z +7z
= -6x + 10y – 2z

Question 5.
Subtract the sum of 3l- 4m – 7n² and 2l + 3m – 4n² from the sum of 9l + 2m – 3n² and -3l + m + 4n².
Solution:
Sum of 9l + 2m – 3n² and -3l + m + 4n²
= 9l + 2m – 3 n² + (-3l) + m + 4n²
= 9l + 2m – 3n² – 3l + m + 4n²
= 9l- 3l+ 2m + m – 3 n² + 4n²
= 6l + 3m + n²
and sum of 3l – 4m – 7n² and 2l +3m- 4n²
= 3l- 4m – 7n² + 2l+ 3m- 4n²
= 3l + 2l – 4m + 3m- 7n² – 4n²
= 5l -m- 11n²
Now (6l + 3m + n²) – (5l – m – 11n²)
= 6l + 3m + n² – 5l + m + 11n²
= 6l – 5l + 3m + m + n² + 11n²
= l + 4m+ 12n²

Question 6.
Subtract the sum of 2x – x² + 5 and -4x – 3 + 7x² from 5.
Solution:
5 – (2x-x² + 5-4x-3 + 7x²)
= 5 – (2x – 4x- x² + 7x² + 5-3)
= 5 – (-2x + 6x² + 2)
= 5 + 2x – 6x² – 2
= – 6x²+2x+3
= 3 + 2x – 6x²

Question 7.

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 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.

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

Other Exercises

• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4
• RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years

Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year

Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years

Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year

Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years

Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year

Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year

Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 1\(\frac { 1 }{ 2 }\) years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 1\(\frac { 1 }{ 2 }\) years = 3 half-years

Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year

Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year

Compound interest for the third half-years or 1\(\frac { 1 }{ 2 }\) years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters

Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter

Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter

Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter

Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 12\(\frac { 1 }{ 2 }\) % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000

Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year

Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250

Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 1\(\frac { 1 }{ 2 }\) years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 1\(\frac { 1 }{ 2 }\) years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 1\(\frac { 1 }{ 2 }\) years or 3 half-years

Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year

Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year

Total compound interest for 3 half-years
or 1\(\frac { 1 }{ 2 }\) years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years

In second case
Interest for the first year

Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year

Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly

Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter

Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter

Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.

Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.

Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100

Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or \(\frac { 5 }{ 2 }\) % quarterly
Period (T) = 1 year = 4 quarters

Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter

Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240

= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter

Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters

Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter

Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter

Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 5\(\frac { 1 }{ 2 }\) % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 5\(\frac { 1 }{ 2 }\) % = \(\frac { 11 }{ 2 }\) % p.a.
Period (T) = 3 years.

Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year

Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year

Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years

Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year

Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year

Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years

Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year

Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 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.