## OP Malhotra Class 10 Maths Solutions Chapter 2 Banking Ex 2

Practicing OP Malhotra Class 10 Solutions Chapter 2 Banking Ex 2 is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 10 ICSE Maths Solutions Chapter 2 Banking Ex 2

Question 1.
Mr. Rajiv Anand has opened a recurring deposit account of 7400 per month for 20 months in a bank. Find the amount he will get at the time of maturity, if the rate of interest is 8.5% p.a., and if the interest is calculated at the end of each month.
Solution:
Deposit per month (P) = ₹ 400
Period (n) = 20 months
Rate (R) = 8.5% p.a.
Now principal for 1 month = P x $$\frac{n(n+1)}{2}$$
= $$\frac{400 \times 20 \times(20+1)}{2}$$
= ₹ $$\frac{400 \times 20 \times 21}{2}$$ = ₹ 84000
∴ Interest = $$\frac {PRT}{100}$$ = $$\frac{84000 \times 8.5 \times 1}{100 \times 12}$$
= $$\frac {1}{2}$$ = ₹ 595
∴ Maturity value = P x 20 + Interest
= ₹ 400 x 20 + 595
= ₹ 8000 + 595
= ₹ 8595

Question 2.
Mrs. Savita Khosla deposits 7900 per month in a recurring account for 2 years. If she gets 71800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.
Solution:
Monthly deposit (P) = ₹ 900
Let Rate = R%
Period (n) = 2 years = 24 months
Interest = ₹ 1800
Now principal for 1 month
= $$P \times \frac{n(n+1)}{2}=₹ 900 \times \frac{24 \times(24+1)}{2}$$
= ₹ $$\frac{900 \times 24 \times 25}{2}$$ = ₹ 270000
Interest = ₹ 1800
∴ Rate = $$\frac{\text { Interest } \times 100}{\mathrm{P} \times t}=\frac{1800 \times 100 \times 12}{270000 \times 1}$$
= 2 x 4 = 8% p.a. Question 3.
Mr. Brown deposits ₹ 1100 per month in a cumulative time deposit account in a bank for 16 months. If at the end of maturity he gets ₹ 19096, find the rate of interest if interest is calculated at the end of each month.
Solution:
Deposit per month (P) = ₹ 1100
Period (n) = 16 months
Maturity value = ₹ 19096
Let rate = R%
Interest = Maturity value – Principal
= ₹ 19096 – 1100 x 16
= ₹ 19096 – 17600
= ₹ 1496
Principal for one month
= P x $$\frac{n(n+1)}{2}$$ = ₹ 1100 x $$\frac{16 \times(16+1)}{2}$$
= ₹ $$\frac{1100 \times 16 \times 17}{2}$$ = ₹ 149600
∴ Rate = $$\frac{\text { Interest } \times 100}{P \times t}=\frac{1496 \times 100 \times 12}{149600 \times 1}$$
= 12% p.a.

Question 4.
Sandhya has a recurring deposit account in Vijaya Bank and deposits ₹ 400 per month for 3 years. If she gets ₹ 16,176 on maturity, find the rate of interest given by the bank.
Solution:
Monthly deposit (P) = ₹ 400
Period (n) = 3 years or 36 months
∴ Maturity value = ₹ 16176
Let rate = R% p.a.
Principal for 1 month = P x $$\frac{n(n+1)}{2}$$
= 400 x $$\frac{36 \times(36+1)}{2}$$
= ₹ $$\frac{400 \times 36 \times 37}{2}$$ = ₹ 266400
Interest = M.V. – Deposit
= 16176 – 400 x 36
= 16176 – 14400 = ₹ 1776
∴ Rate = $$\frac{\text { Interest } \times 100}{\mathrm{P} \times t}=\frac{1776 \times 100 \times 12}{266400 \times 1}$$
= 8% p.a.

Question 5.
A man deposits ₹ 600 per month in a bank for 12 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits if the rate of interest is 8% p.a., and interest is calculated at the end of every month?
Solution:
Deposit per month (P) = ₹ 600
Period (n) = 12 months
Rate (R) = 8% p.a.
Principal for 1 month = P x $$\frac{n(n+1)}{2}$$
= ₹ 600 x $$\frac{12 \times(12+1)}{2}$$
= ₹ $$\frac{600 \times 12 \times 13}{2}$$ = ₹ 46800
∴ Interest = $$\frac {PRT}{100}$$ = $$\frac{46800 \times 8 \times 1}{100 \times 12}$$
∴Maturity value = P x n + Interest
= ₹ 600 x 12 + 312
= ₹ 7200 + 312
= ₹ 7512

Question 6.
Anil deposits ₹ 300 per month in a recurring deposit account for 2 years. If the rate of interest is 10% per year, calculate the amount that Anil will receive at the end of 2 years, i.e., at the time of maturity.
Solution:
Deposit per month (P) = ₹ 300
Period (n) = 2 years = 24 months
Rate (R) = 10% p.a.
Principal for one month = P x $$\frac{n(n+1)}{2}$$
= ₹ $$\frac{300 \times 24 \times(24+1)}{2}$$
= ₹ $$\frac{300 \times 24 \times 25}{2}$$ = ₹ 90000
∴ Interest = $$\frac {PRT}{100}$$ = $$\frac{90000 \times 10 \times 1}{100 \times 12}$$ = ₹ 750
∴Maturity value = P x n + Interest
= ₹ 300 x 24 + 750
= ₹ 7200 + 750
= ₹ 7950 Question 7.
Sudhir opened a recurring deposit account with a bank for 1$$\frac {1}{2}$$ years. If the rate of interest is 10% and the bank pays ₹ 1554 on maturity, find how much did Sudhir deposit per month ?
Solution:
Maturity value = ₹ 1554
Rate (R) = 10%
Period (n) = 1$$\frac {1}{2}$$ years = 18 months
Let P be the monthly deposit, then
principal for one month = $$\frac{\mathrm{P}(n)(n+1)}{2}$$
= $$\frac{P \times 18 \times(18+1)}{2}=\frac{P \times 18 \times 19}{2}$$ = 171 P
Interest = $$\frac{\mathrm{PRT}}{100}=\frac{171 \mathrm{P} \times 10 \times 1}{100 \times 12}=\frac{171}{120}$$P
Now Maturity value = P x n + Interest
1554 = P x 18 + $$\frac {171}{120}$$P
⇒ 1554 = $$\frac{2160 \mathrm{P}+171 \mathrm{P}}{120}=\frac{2331}{120}$$P
⇒ p = $$\frac{1554 \times 120}{2331}$$ = 80
∴ Monthly deposit = ₹ 80

Question 8.
Renu has a cumulative deposit account of ₹ 200 per month at 10% per annum. If she gets ₹ 6775 at the time of maturity, find the total time for which the account was held.
Solution:
Deposit per month (P) = ₹ 200
Rate (R) = 10% p.a.
Maturity value = ₹ 6775
Let period = n months
∴ Principal for 1 month = P x $$\frac{n(n+1)}{2}$$
= $$\frac{n(n+1)}{2}$$ x 200
= ₹ 100 n(n + 1)
Interest = $$\frac{\mathrm{PRT}}{100}=\frac{100 n(n+1) \times 10 \times 1}{100 \times 12}$$
= $$\frac{5n(n+1)}{6}$$ … (i)
But Interest = M.V. – Deposit
= 6775 – 200 x n
= 6775 – 200n
From (i) and (ii)
$$\frac{5n(n+1)}{6}$$ = 6775 – 200n
⇒ 5n(n+ 1) = 40650 – 1200n
⇒ 5n² + 5n + 1200n – 40650 = 0
⇒ 5n² + 1205n – 40650 = 0
⇒ n² + 241n – 8130 = 0
⇒ n² – 30n + 271n – 8130 = 0 {∵ \begin{aligned} & 8130=-30 \times 271 \\ & 241=-30+270 \end{aligned}
= n(n- 30) + 271 (n – 30)
⇒ (n – 30) (n + 271) = 0
Either n – 30 = 0, then n = 30
or n + 271 = 0, n = – 271 which is not possible
∴ Period = n = 30 months
= 2$$\frac { 1 }{ 2 }$$ years

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and the interest is calculated at the end of every month. (ICSE 2001)
Solution:
Deposit per month (P) = ₹ 150
Rate (R) = 8%
Period (n) = 8 months
∴ Principal for 1 month = $$\frac{\mathrm{P} \times n(n+1)}{2}$$
= $$\frac{150 \times 8 \times(8+1)}{2}=₹ \frac{150 \times 8 \times 9}{2}$$
= ₹ 5400
Interest = $$\frac{\text { PRT }}{100}=₹ \frac{5400 \times 8 \times 1}{100 \times 12}$$ = ₹ 36
∴ Maturity value = Deposit + Interest
= ₹ 150 x 8 + 36
= ₹ 1200 + 36 = ₹ 1236

Question 2.
Mr. R.K. Nair gets ₹ 6455 at the end of one year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment. (ICSE 2005)
Solution:
Let deposit per month = ₹ x
Rate (R) = 14% p.a.
Period (n) = 1 year =12 months
Maturity value = ₹ 6455
Principal for 1 month = P x $$\frac{n(n+1)}{2}$$
= x × $$\frac{12 \times(12+1)}{2}=\frac{x+12 \times 13}{2}$$ = 78 x
and interest = $$\frac{\mathrm{PRT}}{100}=\frac{78 x \times 14 \times 1}{100 \times 12}$$
= $$\frac { 91x }{ 100 }$$ … (i)
Interest = Maturity value – Total deposit
= ₹ 6455 – x × 12 = 6455 – 12x … (ii)
From (i) and (ii)
$$\frac { 91x }{100}$$ = 6455 – 12x
⇒ 91x = 645500 – 1200x
⇒ 91x + 1200x = 645500
⇒ 1291x = 645500
⇒ x = $$\frac { 645500 }{ 1291 }$$ = 500
∴ Deposit per month = ₹ 500

Question 3.
Mohan deposits ₹ 80 per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum.
Solution:
Deposit per month (P) = ₹ 80
Rate (R) = 6% p.a.
Period (n) = 6 years = 72 months
Now principal for 1 month = $$\frac{n(n+1)}{2}$$
= $$\frac{80 \times 72 \times(72+1)}{2}$$
= ₹ $$\frac{80 \times 72 \times 73}{2}$$ = ₹ 210240
Interest = $$\frac{P R T}{100}=\frac{210240 \times 6 \times 1}{100 \times 12}$$
= $$\frac { 105120 }{ 100 }$$ = ₹ 1051.20
∴ Maturity value = Deposit + Interest
= 72 x 80+ 1051.20
= ₹ 5760 + 1051.20 = ₹ 6811.20 Question 4.
Saloni deposited ₹ 150 per month in her bank for eight months under the Recurring Deposit Scheme. What will the maturity value of her deposit, if the rate of interest is 8% per annum and the interest is calculated at the end of every month.
Solution:
Deposit per month = ₹ 150
Rate (R) = 8%
Period (n) = 8 months
∴ Principal for 1 month = $$\frac{P×n(n+1)}{2}$$
= ₹$$\frac{150 \times 8 \times(8+1)}{2}$$
= $$\frac{150 \times 8 \times 9}{2}$$
= ₹ 5400
∴ Interest = $$\frac{\mathrm{PRT}}{100}=\frac{5400 \times 8 \times 1}{100 \times 12}$$ = ₹ 36
∴ Maturity value = P x n + Interest
= ₹ 150 x 8 + 36
= ₹ 1200 + 36
= ₹ 1236

Question 5.
David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7725 at the time of maturity, find the rate of interest per annum.
Solution:
Deposit per month (P) = ₹ 300
Period (n) = 2 years or 24 months
Let R be the rate % per annum
Maturity value = ₹ 7725
Total principal for 1 month = $$\frac{P×n(n+1)}{2}$$
= $$\frac{300 \times 24 \times(24+1)}{2}$$
= $$\frac{300 \times 24 \times 25}{2}$$ = ₹ 90000
∴ Interest = $$\frac{\text { PRT }}{100}=\frac{90000 \times R \times 1}{100 \times 12}$$ = ₹ 75R … (i)
Now maturity value = P x n + Interest
7725 = 300 x 24 + 75R
⇒ 7725 = 7200 + 75R
⇒ 75R = 7725 – 7200 = 525
⇒ R = $$\frac { 525 }{ 75 }$$ = 7
Rate of interest = 7% p.a.

Question 6.
Mrs. Goswami deposits ₹ 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.
Solution:
P = $$\frac{36(36+1)}{2}$$ x 100
∴ Interest = $$\frac{36 \times 37 \times 1000 \times 8}{2 \times 12 \times 100}$$
= 12 x 37 x 10 = 4440
Matured value = 36000 + 4440 = ₹ 40440

Question 7.
Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2500 per month for two years. At the time of maturity he got ₹ 67,500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.
Solution:
Total amount deposited by Mr. Gupta in 24 months = ₹ 2500 x 24 = ₹ 60000
Total principal for 1 month
= ₹ $$\frac{2500 \times 24 \times(24+1)}{2}$$
= ₹ 2500 x 12 x 25
= ₹ 750000
Let rate of interest be r% p.a.
Now interest on ₹ 750000 for 1 month
= 750000 x $$\frac { 1 }{ 12 }$$ x $$\frac { r }{ 100 }$$ = 625 r
Maturity amount = ₹ 67500
⇒ 60000 + 625r = 67500
625r = 67500 – 60000 = 7500
r = $$\frac { 7500 }{ 625 }$$ = 12 % p.a.

Question 8.
Ahmed has a recurring deposit account in a bank. He deposits ₹ 2,500 per month for 2 years. If he get ₹ 66,250 at the time of maturity, find
(i) The interest paid by the bank.
(ii) The rate of interest.
Solution:
Monthly deposit (M.D.) = ₹ 2500
n = 2 x 12 = 24 months
Total deposited amount = ₹ 2500 x 24 = ₹ 60000
Matured amount = ₹ 66250
(i) Interest paid by the bank
= Rs. (66250 – 60000) = ₹ 6250

(ii) Equivalent principle for 1 month
= $$\frac{\text { M.D. } \times n(n+1)}{2}$$
= $$\frac{2500 \times 24 \times 25}{2}$$ = ₹ 750000
R = $$\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{6250 \times 100 \times 12}{750000 \times 1}$$ = 10 % p.a.

Question 9.
Kiran deposited ₹ 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at the rate of 11 % per annum, find the amount she gets on maturity.
Solution:
P (Principal) = ₹ 200, n (Time) = 36 months,
R (Rate) = 11% p.a.
Amount deposited in 36 months
= ₹ 200 x 36 = ₹ 7200
S.I. = P x $$\frac{n(n+1)}{2} \times \frac{1}{12} \times \frac{\mathrm{R}}{100}$$
= ₹ $$\frac{200 \times 36 \times 37 \times 11}{2 \times 12 \times 100}$$ = ₹ 1221
Amount Kiran will get on maturity
= ₹ (7200 + ₹ 1221) = ₹ 8421

Question 10.
Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly instalment.
Solution:
Let monthly installment = ₹ x
n = 3 x 12 months = 36 months
Rate = 8% p.a.
∴ I = P x $$\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$$
= x × $$\frac{36 \times 37}{2 \times 12} \times \frac{8}{100}=\frac{444}{100}$$ x
Maturity value = ₹ 8088
According to the question,
I + 36 × x = 8088
⇒ $$\frac { 444 }{ 100 }$$ + 36x = 8088
4.44x + 36x = 8088
40.44x = 8088
x = $$\frac{8088}{40.44}=\frac{8080 \times 100}{4044}$$ = ₹ 200
∴ Value of monthly installment = ₹ 200 Question 11.
Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 11/2 years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum.
Solution:
Money deposited per month (P) = ₹ 800 r = ?
No.of months (n) = 1 $$\frac { 1 }{ 2 }$$ = $$\frac { 3 }{ 2 }$$ x 12 = 18
∴ Interest = P x $$\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$$
= 800 x $$\frac{18(18+1)}{2 \times 12} \times \frac{r}{100}$$
= 800 x $$\frac{18 \times 19}{2 \times 12} \times \frac{r}{100}$$ = 114r
∴ Maturity amount = 114r + 800 x 18
According to the question,
15084 = WAr + 14400
⇒ 15084 – 14400 = 114r
⇒ 684 = WAr
r = $$\frac { 684 }{ 114 }$$ = 6%

Question 12.
Katrina opened a recurring deposit account with a National Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly instalment is ₹ 1,000, find the:
(i) interest earned in 2 years.
(ii) matured value.
Solution:
Given,
p = ₹ 1000
n = 2 years = 24 months
r = 6%
(i) Interest = p x $$\frac{n(n+1)}{2} \times \frac{r}{12 \times 100}$$
= 1000 x $$\frac{24(25+1)}{2}=\frac{6}{12 \times 100}$$
= ₹ 1000 x $$\frac{24(25)}{2} \times \frac{6}{12 \times 100}$$
Thus, the interest earned in 2 years is ₹ 1500

(ii) Sum deposited in two years = 24 x ₹ 1000 = ₹ 24,000
Maturity value = Total sum deposited in two years + Interest
= ₹ 24,000 + ₹ 1,500 = ₹ 25,500
Thus, the maturity value is ₹ 25,500

Question 13.
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find :
(i) the monthly instalment
(ii) the amount of maturity
Solution:
(i) I = ₹ 1200,
n = 2 x 12 = 24 months,
r = 6%
I = P x $$\frac{n(n+1)}{2} \times \frac{r}{12 \times 100}$$
⇒ 1200 = $$\mathrm{P} \frac{24(24+1)}{2} \times \frac{1}{12 \times 100}$$
⇒ 1200 = $$P \times \frac{24(25)}{2} \times \frac{6}{12 \times 100}$$
⇒ 1200 = P x $$\frac { 3 }{ 2 }$$
⇒ P = $$\frac { 1200×2 }{ 3 }$$
⇒ P = ₹ 800
So the monthly instalment is ₹ 800

(ii) Total sum deposited = P x n = ₹ 800 x 24 = ₹ 19200
∴ The amount that Mohan will get at the time of maturity
= Total sum deposited + Interest on it
= ₹ 19200 + ₹ 1200 = ₹ 20400
Hence, the amount of maturity is ₹ 20400

## Class 10 ICSE Maths Solutions S Chand – OP Malhotra Class 10 ICSE Solutions

Unit I Commercial Mathematics

OP Malhotra Class 10 Solutions Chapter 1 GST (Goods and Service Tax)

S Chand Class 10 Maths Solutions ICSE Chapter 2 Banking

Class 10 ICSE Maths Solutions S Chand Chapter 3 Shares and Dividends

Unit II Algebra

ICSE Class 10 Maths Solutions S Chand Chapter 4 Linear Inequations in One Variable

ICSE Class 10 Maths Solutions OP Malhotra Chapter 5 Quadratic Equations

ICSE Class 10 S Chand Maths Solution Chapter 6 Ratio and Proportion

ICSE S Chand Maths Class 10 Solutions Chapter 7 Factor Theorem – Factorization

S Chand Class 10 ICSE Maths Solutions Chapter 8 Matrices

OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression

Class 10 ICSE Maths S Chand Solutions Chapter 10 Reflection

S Chand Maths Class 10 ICSE Solutions Pdf Chapter 11 Coordinate Geometry

Unit III Geometry

ICSE Class 10 Maths S Chand Solutions Chapter 12 Similar Triangles

S Chand ICSE Class 10 Maths Solutions Chapter 13 Loci

‘S Chand Maths Class 10 Solutions Pdf ICSE Chapter 14 Circle

Unit IV Mensuration

OP Malhotra Class 10 ICSE Solutions Chapter 15 Three Dimensional Solids

Unit V Trigonometry

S Chand ICSE Maths Class 10 Solutions Chapter 16 Trigonometrical Identities and Tables

S Chand Maths Class 10 Solutions ICSE Pdf Chapter 17 Heights and Distances

Unit VI Statistics

ICSE Maths Solutions Class 10 S Chand Chapter 18 Arithmetic Mean, Median, Mode and Quartiles

S Chand Class 10 Maths ICSE Solutions Chapter 19 Histogram and Ogive

Class 10 ICSE Maths Solutions OP Malhotra Chapter 20 Probability

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These ISC 12th Biology Previous Year Question Papers Solved with Answers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ISC Class 12 Board Exam and Score More marks.

Board – Indian Certificate of Secondary Education (CISCE), www.cisce.org
Class – Class 12
Subject – Biology
Year of Examination – 2020, 2019, 2018, 2017.

## Last 10 Years ISC Question Papers Class 12 Biology Solved

We hope the ISC Class 12 Biology Previous Year Question Papers Solved Pdf of Last 10 Years with Solutions, help you. If you have any query regarding Last 10 Years ISC Question Papers Class 12 Biology Solved with Answers, drop a comment below and we will get back to you at the earliest.

## ISC Class 12 Hindi Previous Year Question Papers Solved Last 10 Years

Download ISC Class 12 Hindi Previous Year Question Papers Solved Pdf of Last 10 Years with Solutions and Marking Scheme. Here we have given Last 10 Years ISC Question Papers Class 12 Hindi Solved with Answers. Students can view or download the ISC Board Previous Year Question Papers Class 12 Hindi for their Class 12 upcoming examination.

These ISC 12th Hindi Previous Year Question Papers Solved with Answers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ISC Class 12 Board Exam and Score More marks.

Board – Indian Certificate of Secondary Education (CISCE), www.cisce.org
Class – Class 12
Subject – Hindi
Year of Examination – 2020, 2019, 2018, 2017.

## Last 10 Years ISC Question Papers Class 12 Hindi Solved

We hope the ISC Class 12 Hindi Previous Year Question Papers Solved Pdf of Last 10 Years with Solutions, help you. If you have any query regarding Last 10 Years ISC Question Papers Class 12 Hindi Solved with Answers, drop a comment below and we will get back to you at the earliest.

## ISC Class 12 English Literature Previous Year Question Papers Solved Last 10 Years

Download ISC Class 12 English Literature Previous Year Question Papers Solved Pdf of Last 10 Years with Solutions and Marking Scheme. Here we have given Last 10 Years ISC Question Papers Class 12 English Literature Solved with Answers. Students can view or download the ISC Board Previous Year Question Papers Class 12 English Literature for their Class 12 upcoming examination.

These ISC 12th English Literature Previous Year Question Papers Solved with Answers are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for ISC Class 12 Board Exam and Score More marks.

Board – Indian Certificate of Secondary Education (CISCE), www.cisce.org
Class – Class 12
Subject – English Literature
Year of Examination – 2020, 2019, 2018, 2017.

## Last 10 Years ISC Question Papers Class 12 English Literature Solved

We hope the ISC Class 12 English Literature Previous Year Question Papers Solved Pdf of Last 10 Years with Solutions, help you. If you have any query regarding Last 10 Years ISC Question Papers Class 12 English Literature Solved with Answers, drop a comment below and we will get back to you at the earliest.