Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A.

Other Exercises

Question 1.
How much money will be required to buy 400, ₹ 12.50 shares at a premium of ₹ 1?
Solution:
Number of shares purchased = 400
Rate of each share = ₹ 12.50
M.V. = ₹ 1 premium = ₹ 12.50 + ₹ 1 = ₹ 13.50
Amount of in vestment = ₹ 400 x ₹ 13.50 = ₹ 5400

Question 2.
How much money will be required to buy 250, ₹ 15 shares at a discount of ₹ 1.50?
Solution:
Number of shares = 250
M.V. = at ₹ 15 at a discount of ₹ 1.50 = ₹ 15 – ₹ 1.50 = ₹ 13.50
Amount of investment = ₹ 13.50 x 250 = ₹ 3375

Question 3.
A person buys 120 shares at a nominal value of ₹ 40 each, which he sells at ₹ 42.50 each. Find his profit and profit percent.
Solution:
No. of shares = 120
Nominal value of each share = ₹ 40.00
Profit at each share = ₹ 42.50 – ₹ 40.00 = ₹ 2.50
Total profit = 2.50 x 120 = ₹ 300
Cost price of 120 shares = ₹ 40 x 120 = ₹ 4,800
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 3.1

Question 4.
Find the cost of 85 shares of Rs. 60 each when quoted at ₹ 63.25
Solution:
No. of shares = 85
Market value of cach share = ₹ 63.25
Total cost = ₹ 63.25 x 85 = ₹ 5,376.25

Question 5.
A man invests ₹ 800 in buying 75 shares and when they are selling at a premium of ₹ 1.15, he sells all the shares. Find his profit and profit percent.
Solution:
Investment = ₹ 800
In first case face value of each share = ₹ 5
and market value of each share = ₹ 5.00 + ₹ 1.15 = ₹ 6.15
Gain on each share of ₹ 5 = ₹ 1.15
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 5.1

Question 6.
Find the annual income derived from 125, ₹ 120 shares paying 5% dividend.
Solution:
Amount of investment = ?
Number of shares purchased = 125 at ₹ 120, 5% dividend
Amount of investment = ₹ 125 x 120 = ₹ 15000
His annual income = 15000 x \(\frac { 5 }{ 100 }\) = ₹ 750

Question 7.
A man invests ₹ 3,072 in a company paying 5% per annum when its ₹ 10 share can be bought for ₹ 16 each. Find:
(i) his annual income;
(ii) his percentage income on his investment.
Solution:
Total investment = ₹ 3,072
Market value of each shares = ₹ 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 7.1

Question 8.
A man invests ₹ 7,770 in a company paying 5 percent dividend when a share of nominal value of ₹ 100 sells at a premium of ₹ 5. Find :
(i) the number of shares bought;
(ii) annual income ;
(iii) percentage income ;
Solution:
Investment = ₹ 7770
Nominal value of each share = 100
Market value = 100 + 5 = 105
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 8.1

Question 9.
A man buys ₹ 50 shares of a company paying 12 percent dividend, at a premium of ₹ 10. Find :
(i) the market value of 320 shares ;
(ii) his annual income ;
(iii) his profit percent.
Solution:
(i) Market value of each share = ₹ 50 + ₹ 10 = ₹ 60
Market value of 320 shares = ₹ 60 x 320 = ₹ 19,200
(ii) Rate of dividend = 12%
Face value of 320 shares = Rs. 50 x 320 = Rs. 16,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 9.1

Question 10.
A man buys of Rs. 75 shares at a discount of Rs. 15 of a company paying 20% dividend. Find :
(i) the market value of 120 shares ;
(ii) his annual income ;
(iii) his profit percent.
Solution:
(i) Market value of one share = Rs. 75 – 15 = Rs. 60
Market value of 120 shares = Rs. 60 x 120 = Rs. 7,200
(ii) Rate of dividend = 20%
Face value of 120 shares = Rs. 75 x 120 = Rs. 9,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 10.1

Question 11.
A man has 300, ₹ 50 shares of a company paying 20% dividend. Find his net income after paying 3% income tax.
Solution:
No. of shares = 300
Face value of 50 shares = Rs. 50 x 300 = Rs. 15,000
Rate of dividend = 20%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 11.1

Question 12.
A company pays dividend of 15 % on its ten-rupee shares from which it deducts income tax at the rate of 22%. Find the annual income of a man who owns one thousand shares of this company.
Solution:
No. of shares = 1,000
Face Value of each share = Rs. 10
Rate of dividend = 15%
Rate of income tax = 22%
Face value of 1,000 shares = 1,000 x 10 = Rs. 10,000
Total dividend = Rs. 10,000 x \(\frac { 15 }{ 100 }\) = Rs. 1,500
Income tax deducted = Rs. 1500 x \(\frac { 22 }{ 100 }\) = Rs. 330
Net income = Rs.1500 – Rs. 330 = Rs. 1170

Question 13.
A man invests Rs. 8,800 in buying shares of a company of face value of rupees hundred each at a premium of 10%. If he earns Rs. 1,200 at the end of the year as dividend find:
(i) the number of shares he has in the company;
(ii) the dividend percent per share. [2001]
Solution:
Investment = Rs. 8,800
Face value of each share = Rs. 100
Market value of each share = Rs. 100 + 10 = Rs. 110
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 13.1

Question 14.
A man invests Rs. 1,680 in buying shares of nominal value Rs. 24 and selling at 12% premium. The dividend on the shares is 15% per annum. Calculate :
(i) The number of shares he buys ;
(ii) The dividend he receives annually. [1999]
Solution:
Investment = Rs. 1680
Nominal value of each share = Rs. 24
Market value of each share = Rs. 24 + 12% of 24
= Rs. 24 + 2.88 = Rs. 26.88
Rate of dividend = 15%
(i) No. of shares = \(\frac { 1680 }{ 26.88 }\) = 62.5
(ii) Face value of 62.5 shares = 62.5 x 24 = Rs. 1500
Amount of dividend = 1500 x \(\frac { 15 }{ 100 }\) = Rs. 225

Question 15.
By investing Rs. 7,500 in a company paying 10 percent dividend, an annual income of Rs. 500 is received. What price is paid for each of Rs. 100 share? [1990]
Solution:
Investment = Rs. 7,500
Rate of dividend = 10%
Total income = Rs. 500
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 15.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking (Recurring Deposit Accounts) Ex 2B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B.

Other Exercises

Question 1.
Pramod deposits ₹ 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.
Solution:
Deposit per month (P) = ₹ 600
Rate of interest (r) = 8%
Period (n) = 4 years = 48 months.
According to formula,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 1.1
Maturity value = ₹ 600 x 48 + ₹ 4,704 = ₹ 28,800 + ₹ 4,704 = ₹ 33504

Question 2.
Ritu has a Recurring Deposit Account in a bank and deposits ₹ 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of this account is ₹ 1,554.
Solution:
Let rate of interest = r%,
n = 18,
P = ₹ 80
and A is maturity value.
Using formula
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 2.1

Question 3.
The maturity value of a R.D. Account is ₹ 16,176. If the monthly installment is ₹ 400 and the rate of interest is 8%; find the time (period) of this R.D. Account.
Solution:
Here maturity value (A) = ₹ 16,176
Rate = 8%,
P = ₹ 400
Let period = n (No. of months)
Using formula :
I = A – P x n = 16,176 – 400 x n = 16,716 – 400n.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 3.1
⇒ 48,528 – 1,200n = 4n² + 4n
⇒ 4n² + 4n + 1200n – 48,528 = 0
⇒ 4n² + 1,204n – 48,528 = 0
⇒ n² + 301n — 12,132 = 0 (dividing by 4)
⇒ n² – 36n + 337n – 12,132 = 0
⇒ n (n – 36) + 337 (n – 36) = 0
⇒ (n – 36) (n + 337) = 0
Either n = 36 months or n = -337, which is not possible.
Time = 36 months = 3 years

Question 4.
Mr. Bajaj needs ₹ 30,000 after 2 years. What least money (in multiple of ₹ 5) must he deposit every month in a recurring deposit account to get required money at the end of 2 years, the rate of interest being 8% p.a. ?
Solution:
Amount of maturity = ₹ 30000
Period (n) = 2 years = 24 months
Rate = 8% p.a.
Let x be the monthly deposit
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 4.1
Amount of monthly deposit in the multiple of ₹ 5 = ₹ 1155

Question 5.
Rishabh has a recurring deposit account in a post office for 3 years at 8% p.a. simple interest. If he gets ₹ 9,990 as interest at the time of maturity, find :
(i) the monthly installment.
(ii) the amount of maturity.
Solution:
Total interest = ₹ 9990
Period (n) = 3 years = 36 months
Rate of interest (r) = 8%
(i) Let monthly installment = x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 5.1
Monthly installment = ₹ 2250
(ii) Amount of maturity = Principal + Interest
= 36 x 2250 + 9990
= ₹ 81000 + 9990 = ₹ 90990

Question 6.
Gopal has a cumulative deposit account and deposits ₹ 900 per month for a period of 4 years. If he gets ₹ 52,020 at the time of maturity, find the rate of interest.
Solution:
Maturity value = ₹ 52,020
Monthly installment (P) = ₹ 900
Total principal = ₹ 900 x 48 = ₹ 43200
Amount of interest = ₹ 52020 – ₹ 43200 = ₹ 8820
Let rate of interest = r%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 6.1

Question 7.
Deepa has a 4 year recurring deposit account in a bank and deposits ₹ 1,800 per month. If she gets ₹ 1,800 per month. If she gets ₹ 1,08,450 at the time of maturity, find the rate of interest.
Solution:
Deposit per month = ₹ 1800
Period = 4 years = 48 months
Maturity value = ₹ 108450
Total principal = ₹ 1800 x 48 = ₹ 86400
Amount of interest = ₹ 108450 – 86400 = ₹ 22050
Let r be the rate of interest
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 7.1

Question 8.
Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8,088 from the bank after 3 years, find the value of his monthly installment. (2013)
Solution:
Let monthly installment = ₹ x
Period (n) = 3 x 12 months = 36 months
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 8.1

Question 9.
Sharukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 1\(\frac { 1 }{ 2 }\) years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum. (2014)
Solution:
Money deposited per month (P) = ₹ 800
r = ?
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 9.1

Question 10.
Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is ₹ 1,000, find the :
(i) interest earned in 2 years
(ii) maturity value. (2015)
Solution:
Period (n) = 2 years = 2 x 12 = 24 months
Rate of interest (r) = 6%
Monthly installment (P) = ₹ 1000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 10.1

Question 11.
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 installment
(ii) the amount of maturity
Solution:
(i) Interest = ₹ 1200,
n = 2 x 12 = 24 months,
r = 6%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 11.1
⇒ P = ₹ 800
So the monthly installment 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 Received
= ₹ 19200 + ₹ 1200 = ₹ 20400
Hence, the amount of maturity is ₹ 20400

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking (Recurring Deposit Accounts) Ex 2A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A.

Other Exercises

Question 1.
Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits ₹ 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.
Solution:
Recurring Deposit per month = ₹ 600
Period (n) = 20 months
Rate of interest (r) = 10% p.a.
Total principal for 1 month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 1.1
Maturity value = ₹ 600 x 20 + ₹ 1,050 = ₹ 12,000 + ₹ 1,050 = ₹ 13,050

Question 2.
Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4\(\frac { 1 }{ 2 }\) years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Solution:
Recurring deposit per month = ₹ 640
Period (n) = 4\(\frac { 1 }{ 2 }\) years = 54 months
Rate of interest (r) = 12%
Total principal for 1 month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 2.1
Maturity value = ₹ 640 x 54 + ₹ 9,504 = ₹ 34,560 + ₹ 9,504 = ₹ 44,064

Question 3.
Each of A and B opened a recurring deposit accounts in a bank. If A deposited ₹ 1,200 per month for 3 years and B deposited ₹ 1,500 per month for 2\(\frac { 1 }{ 2 }\) years; find, on maturity, who will it get more amount and by how much ? The rate of interest paid by the bank is 10% per annum.
Solution:
A’s deposit per month (P) = ₹ 1200
Period = 3 years = 36 months
Total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 3.2
and maturity value = ₹ 1500 x 30 + Interest
= ₹ 45000 + 5812.50
= ₹ 50812.50
It is clear that B’s maturity value is greater Difference = ₹ 50812.50 – ₹ 49860 = ₹ 952.50

Question 4.
Ashish deposits a certain sum of money every month in a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did he pay every month ?
Solution:
Let Recurring deposit per month = ₹ x
Period (n) = 12 months
Rate of interest (r) = 11%
Maturity value = ₹ 12,715 ………. (i)
Total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 4.1
Recurring deposit per month ₹ 1000 p.m.

Question 5.
A man has a Recurring Deposit Account in a bank for 3\(\frac { 1 }{ 2 }\) years. If the rate of interest is 12% per annum and the man gets ₹ 10206 on maturity, find the value of monthly installments.
Solution:
Let Recurring deposit per month = ₹ x
Period (n) = 3\(\frac { 1 }{ 2 }\) years = 42 months
Rate of interest (r) = 12% p.a.
Amount of maturity = ₹ 10206 ……… (i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 5.1
Amount of each installment = ₹ 200

Question 6.
(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank.
(ii) 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. (2008)
Solution:
(i) Recurring deposit per month = ₹ 140
Period (n) = 4 years = 48 months
Let Rate of interest (r) = r % p.a.
Amount of maturity = ₹ 8,092
Total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 6.2

Question 7.
Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposits Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month ? [I.C.S.E. 2001]
Solution:
Amount of Recurring deposit = ₹ 150
Period (n) = 8 months
Rate of interest (r) = 8% p.a.
Total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 7.1
Amount of maturity value = ₹ 150 x 8 + ₹ 36 = ₹ 1,200 + ₹ 36 = ₹ 1,236

Question 8.
Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565; find the rate of interest per annum.
Solution:
Amount of recurring deposit per month = ₹ 350
Period (n) = 1 year 3 months = 15 months
Let rate of interest = r % p.a.
Amount of maturity = ₹ 5565
Amount of interest = ₹ 5,565 – ₹ 350 x 15 = ₹ 5,565 – 5,250 = ₹ 315 ….(i)
Now, total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 8.1

Question 9.
A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.
Solution:
Amount of recurring deposit per month = ₹ 1,200
Rate of interest (r) = 8% p.a.
Let period = n months
Amount of maturity = ₹ 12,440
Amount of interest = ₹ 12440 – ₹ 1200 x n ….(i)
Total principal for one month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 9.1
from (i) and (ii), we get,
4n (n + 1) = 12440 – 1200n
⇒ 4n² + 4n = 12440 – 1200n
⇒ 4n² + 1204n – 12440 = 0
Dividing by 4,
⇒ n² + 301n – 3110 = 0
⇒ n² + 311n – 10n – 3110 = 0
⇒ n (n + 311) – 10 (n + 311) = 0
⇒ (n + 311) (n – 10) = 0
Given n + 311 = 0, then n = – 311 Which is not possible,
or n – 10 = 0, then n = 10
Period = 10 months.

Question 10.
Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time (in years) of this Recurring Deposit Account.
Solution:
Amount of recurring deposit per month = ₹ 300
Let Period = n months
Rate of interest (r) = 12% p.a.
Amount of maturity = ₹ 8,100
Interest = 8,100 – 300 x n ……. (i)
Total principal for 1 month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 10.1
⇒ 3n (n + 1) = 16,200 – 600 n
⇒ 3n² + 3n = 16,200 – 600 n
⇒ 3n² + 603n – 16,200 = 0
Dividing by 3, we get,
⇒ n² + 201n – 5,400 = 0
⇒ n² + 225n – 24n – 5,400 = 0
⇒ n(n + 225) – 24 (n + 225) = 0
⇒ (n + 225) (n – 24) = 0
Either n + 225 = 0, then n = – 225 Which is not possible
or n – 24 = 0, then n = 24
Period = 24 months or 2 years.

Question 11.
Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2,500 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. (2010)
Solution:
(i) Total amount deposited by Mr. Gupta in 24 months = ₹ 2500 x 24 = ₹ 60,000
Maturity amount = ₹ 67,500
Total interest earned by Mr. Gupta = ₹ 67,500 – ₹ 60,000 = ₹ 7,500
(ii) Total principal for 1 month
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 11.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C.

Other Exercises

Question 1.
By investing 745,000 in 10% 7100 shares, Sharad gets 73,000 as divided. Find the market value of each share.
Solution:
Total investment = ₹ 45000 at 10% of ₹ 100 shares
and amount of dividend = ₹ 3000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 1.1

Question 2.
Mrs. Kulkarni invests ₹ 1,31,040 in buying ₹ 100 shares at a discount of 9%. She sells shares worth ₹ 72,000 at a premium of 10% and the rest at a discount of 5%. Find her total gain or loss on the whole.
Solution:
Total investment = ₹ 1,31,040 in ₹ 100 share at discount of 9%
Market value of each share = ₹ 100 – ₹ 9 = ₹ 91
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 2.1

Question 3.
A man invests a certain sum in buying 15% ₹ 100 shares at 20% premium. Find:
(i) his income from one share.
(ii) the number of shares bought to have an income, from the dividend, ₹ 6,480.
(iii) sum invested.
Solution:
Face value of each share = ₹ 100
Market value of each share = ₹ 100 + ₹ 20 = ₹ 120
Rate of dividend = 15%
(i) Income from one share = ₹ 15
(ii) and number of shares when amount of dividend
= \(\frac { 6480 }{ 15 }\) = 432
(iii) and sum invested = ₹ 432 x 120 = ₹ 51,840

Question 4.
Gagan invested 80% of his savings in 10% ₹ 100 shares at 20% premium and the rest of his savings in 20% ₹ 50 shares at 20% discount. If his incomes from these shares is ₹ 5,600, calculate:
(i) his investment in shares on the whole.
(ii) the number of shares of first kind that he bought
(iii) percentage return, on the shares bought, on the whole.
Solution:
(i) Total income = ₹ 5600
Let total investment = ₹ x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 4.2

Question 5.
Aishwarya bought 496, ₹ 100 shares at ₹ 132 each. Find:
(i) investment made by her.
(ii) income of Aishwarya from these shares, if the rate of dividend is 7.5%.
(iii) how much extra must Aishwarya invest in order to increase her income by ₹ 7,200?
Solution:
Number of shares = 496
Market value of each share = ₹ 132
(i) Total investment = 496 x 132 = ₹ 65472
(ii) Rate of dividend = 7.5%
Income = 496 x 7.5 = ₹ 3720
(iii) New income (increase in income) = ₹ 7200
Market value of share = ₹ 132
Rate of income = 7.5%
Exit investment
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 5.1

Question 6.
Gopal has some ₹ 100 shares of company A, paying 10% dividend. He sells a certain number of these shares at a discount of 20% and invests the proceeds in ₹ 100 shares at ₹ 60 of company B paying 20% dividend. If his income, from the shares sold, increases by ₹ 18,000, find the number of shares sold by Gopal.
Solution:
Let number of share purchased = x
Face value of these shares = ₹ 100 x x = 100x
dividend = 10%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 6.1

Question 7.
A man invests a certain sum of money in 6% hundred rupee shares at ₹ 12 premium. When the shares fell to ₹ 96, he sold out all the shares bought and invested the proceed in 10%, ten rupee shares at ₹ 8. If the change in his income is ₹ 540, find the sum invested originally.
Solution:
Let investment = ₹ x
Dividend at the rate of 6% at 12% premium
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 7.2

Question 8.
Mr. Gupta has a choice to invest in ten rupee shares of two firms at ₹ 13 or at 716. If the first firm pays 5% dividend and the second firm pays 6% dividend per annum, find:
(i) which firm is paying better ?
(ii) If Mr. Gupta invests equally in both the firms and the difference between the returns from them is ₹ 30, Find how much in all does he invest ?
Solution:
Face value of each share = ₹ 10
Market value of first firm’s share = ₹ 13
and market value of second firm’s share = ₹ 16
Dividend from first firm = 5%
and dividend from second firm = 6%
(i) Let investment in each firm = ₹ 13 x 16
Income from first firm’s shares
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 8.1
It is clear from the above that first firm’s shares are better.
(ii) Difference in income = ₹ 8.00 – ₹ 7.80 = ₹ 0.20
If difference is ₹ 0.20 then investment in each firm = ₹ 13 x 16
and if difference is ₹ 30, then investment
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 8.2
Total investment in both firms = ₹ 31200 x 2 = ₹ 62,400

Question 9.
Ashok invested ₹ 26,400 in 12%, ₹ 25 shares of a company. If he receives a dividend of ₹ 2,475, find the :
(i) number of shares he bought.
(ii) market value of each share.
Solution:
(i) Given,
Investment = ₹ 26400
Rate of dividend = 12%
Dividend earned = ₹ 2475
Face value of one share = ₹ 25
Total dividend earned = No. of shares x Rate of dividend x Face value of one share
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 9.1

Question 10.
A man invested ₹ 45,000 in 15% ₹ 100 shares quoted at ₹ 125. When the market value of these shares rose to Rs. 140, he sold some shares, just enough to raise ₹ 8,400. calculate :
(i) the number of shares he still holds;
(ii) the dividend due to him on these remaining shares. [2004]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 10.1

Question 11.
Mr. Tiwari invested ₹ 29,040 in 15% ₹100 shares quoted at a premium of 20%. Calculate :
(i) the number of shares bought by Mr. Tiwari.
(ii) Mr. Tiwari’s income from the investment.
(iii) the percentage return on his investment.
Solution:
Mr. Tiwari’s investment = ₹ 29040
Face value of each share = ₹ 100
Market value of each share = ₹ 100 + ₹ 20 = ₹ 120
Rate of income = 15%
(i) Number of shares purchased
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 11.1

Question 12.
A dividend of 12% was declared on ₹ 150 shares selling at a certain price. If the rate of return is 10%, calculate :
(i) the market value of the shares.
(ii) the amount to be invested to obtain an annual dividend of ₹ 1,350.
Solution:
Let market value of each share = x
Rate of return on investment = 10%
Face value of each share = ₹ 150
Dividend rate = 12%
(i) Now, rate of return x market value = Rate of dividend x Face value
⇒ 10 x x = 12 x 150
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 12.1
Amount of investment in ₹ 5 shares = ₹ 5 x ₹ 180 = ₹ 13500

Question 13.
Divide ₹ 50,760 into two parts such that if one part is invested in 8% ₹ 100 shares at 8% discount and the other in 9% ₹ 100 shares at 8% premium, the annual incomes from both the investments are equal.
Solution:
Total investment = ₹ 50,760
Let first part of investment = x
Then second part = ₹ 50,760 – x
Rate of dividend in first part = 8% ₹100 at discount = 8%
M.V. of each share = ₹ 100 – 8 = ₹ 92
Rate of dividend second part = 9% ₹ 100 at premium = 8%
M.V. of each share = 100 + 8 = ₹ 108
But annual income from both part is same
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 13.1

Question 14.
Mr. Shameem invested 33\(\frac { 1 }{ 3 }\) % of his savings in 20% ₹ 50 shares quoted at ₹ 60 and the remainder of the savings in 10% ₹ 100 shares quoted at ₹ 110. If his total income from these investments is ₹ 9,200 ; find :
(i) his total savings
(ii) the number of ₹ 50 shares.
(iii) the number of ₹ 100 shares.
Solution:
Let total investment = x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 14.2

Question 15.
Vivek invests ₹ 4500 in 8% ₹ 10 shares at ₹ 15. He sells the shares when the price rises to ₹ 30, and invests the proceeds in 12% ₹ 100 shares at ₹ 125. Calculate,
(i) the sale proceeds
(ii) the number of ₹ 125 shares he buys.
(iii) the change in his annual income from dividend.
Solution:
(i) By investing ₹ 15, share bought = ₹ 10
By investing ₹ 4500, share bought = \(\frac { 10 }{ 15 }\) x 4500 = ₹ 3000
Total face value of ₹ 10 shares = ₹ 3000, Income = 8%
= \(\frac { 8 }{ 100 }\) x 3000 = ₹ 240
By selling Rs. 10 share money received = ₹ 30
By selling Rs. 3000 shares money = \(\frac { 30 }{ 10 }\) x 3000 = ₹ 9000
(ii) By investing ₹ 125, no. of share of ₹ 100 bought = 1
By investing ₹ 9000, no. of share of ₹ 100 bought = \(\frac { 1 }{ 125 }\) x 9000 = 72
No. of ₹ 125 shares bought = 72
(iii) By investing ₹ 125 in Rs. 100 share, income = ₹ 12
By investing ₹ 9000 in ₹ 100 share, income = \(\frac { 12 }{ 125 }\) x 9000 = ₹ 864
Increase in income = ₹ 864 – ₹ 240 = ₹ 624

Question 16.
Mr. Parekh invested ₹ 52,000 on ₹ 100 shares at a discount of ₹ 20 paying 8% dividend. At the end of one year he sells the shares at a premium of ₹ 20. Find
(i) The annual dividend.
(ii) The profit earned including his dividend.
Solution:
Investment = ₹ 52000,
N.V. of 1 share = ₹ 100
M.V. of 1 share for 1 st year = ₹ 100 – 20 = ₹ 80
No. of shares = \(\frac { 52000 }{ 80 }\) = 650
(i) Annual dividend = \(\frac { 8 }{ 100 }\) x 650 x 100 = ₹ 5200
(ii) S.P of 1 share = ₹ 100 + 20 = ₹ 120
S.P. of 650 shares = ₹ 120 x 650 = ₹ 78000
C.P. of 650 shares = ₹ 100 x 650 = ₹ 65000
Profit = S.P. – C.P. = ₹ 78000 – ₹ 65000 = ₹ 13000
Profit including dividend = ₹ 13000 + ₹ 5200 = ₹ 18200

Question 17.
Salman buys 50 shares of face value ₹ 100 available at ₹ 132.
(i) What is his investment ?
(ii) If the dividend is 7.5%, what will be his annual income ?
(iii) If he wants to increase his annual income by ₹ 150, how many extra shares should he buy?
Solution:
F.V. = ₹ 100
(i) M. V. = ₹ 132,
no. of shares = 50
Investment = no. of shares x M.V. = 50 x 132 = ₹ 6600
(ii) Income per share = 7.5% of N.V.
= \(\frac { 75 }{ 10 x 100 }\) x 100 = ₹ 7.5
Annual incomes = 7.5 x 50 = ₹ 375
(iii) New annual income = 375 + 150 = ₹ 525
No. of shares = \(\frac { 525 }{ 7.5 }\) = 70
No. of extra share = 70 – 50 = 20

Question 18.
Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought
(ii) his total investment
(iii) the rate of return on his investment. (2014)
Solution:
Nominal value = ₹ 50
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 18.1

Question 19.
Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the :
(i) original number of shares.
(ii) sale proceeds.
(iii) new number of shares.
(iv) change in the two dividends. (2015)
Solution:
Investment by Rohit = ₹ 9600
Rate of dividend = 8% on 100 shares at ₹ 20 premium
Market value = ₹ 100 + ₹ 20 = ₹ 120
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 19.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B

 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B.

Other Exercises

Question 1.
A man buys 75, Rs. 100 shares paying 9 per cent dividend. He buys shares at such a price that he gets 12 per cent of his money. At what price did he buy the shares?
Solution:
No. of shares = 75
Value of each share = Rs. 100
Rate of dividend = 9%
Let the market value of each share = x
Thus. 12% of x = 9% of 100
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 1.1
Market value of each share = Rs. 75

Question 2.
By purchasing Rs. 25 gas shares for Rs. 40 each, a man gets 4 per cent profit on his investment. What rate per cent is the company paying ? What is his dividend if he buys 60 shares?
Solution:
Face value of each share = Rs. 25
Market value of each share = Rs. 40
Profit = 4% of his investment
Let the rate of dividend = x %
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 2.1

Question 3.
Hundred rupee shares of a company are available in the market at a premium of Rs. 20. Find the rate of dividend given by the company, when a man’s return on his investment is 15 percent.
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 120
Total dividend = 15% on his investment
Let the rate of dividend = x %
Then x % of 100 = 15 % of 120
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 3.1
x = 18%
Rate of dividend = 18%

Question 4.
Rs. 50 shares of a company are quoted at a discount of 10%. Find the rate of dividend given by the company, the return on the investment on these shares being 20 percent.
Solution:
Face value of each share = Rs. 50
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 4.1
Rate of dividend = 18%

Question 5.
A company declares 8 per cent dividend to the share holders. If a man receives Rs. 2,840 as his dividend, find the nominal value of his shares.
Solution:
Rate of dividend = 8%
Total dividend = Rs. 2.840
Let Nominal value of shares = x
Then 8% of x = Rs. 2840
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 5.1
Nominal value of his shares = Rs. 35,500

Question 6.
How much should a man invest in Rs. 100 shares selling at Rs. 110 to obtain an annual income of Rs. 1,680, if the dividend declared is 12% ?
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 110
Total annual income = Rs. 1,680.
Rate of dividend = 12%
Let total amount of shares = x
Then x x 12% = 1,680
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 6.1

Question 7.
A company declares a dividend of 11.2% to all its share-holders. If its Rs. 60 share is available in the market at a premium of 25%, how much should Rakesh invest in buying the shares of this company in order to have an annual income of Rs. 1,680 ?
Solution:
Face value of each share = Rs. 60
Market value = Rs. 60 x \(\frac { 125 }{ 100 }\) = Rs. 75
Rate of dividend = 11.2%
Annual income = Rs. 1,680
Let the face value of shares = x
Dividend = x x 11.2%
x x 11.2% = 1680
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 7.1

Question 8.
A man buys 400, twenty rupee shares at a premium of Rs. 4 each and receives a dividend of 12%. Find :
(i) the amount invested by him.
(ii) his total income from the shares.
(iii) percentage return on his money.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 + 4 = Rs. 24
Rate of dividend = 12%
(i) Amount invested by the man = Rs. 24 x 400 = Rs. 9600
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 8.1

Question 9.
A man buys 400, twenty-rupee shares at a discount of 20% and receives a return of 12% on his money. Calculate
(i) the amount invested by him.
(ii) the rate of dividend paid by the company.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 x \(\frac { 80 }{ 100 }\) = Rs. 16
Amount of shares = Rs. 20 x 400 = Rs. 8,000
(i) Amount invested = Rs. 16 x 400 = Rs. 6,400
(ii) Total dividend = Rs. 6,400 x \(\frac { 12 }{ 100 }\) = Rs. 768
Rate of dividend = \(\frac { 768 x 100 }{ 8000 }\)= 9.6 %

Question 10.
A company with 10,000 shares of Rs. 100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company ?
(ii) What should be the annual income of a man who has 72 shares, in the company ?
(iii) If he received only 4% of his investment, find the price he paid for each share.
Solution:
No. of shares = 10000
Face value of each share = 100
Rate of dividend = 5%
Amount of shares = Rs. 100 x 10,000 = Rs. 10,00,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 10.1

Question 11.
A lady holds 1800, Rs. 100 shares of a company that pays 15 % dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what is the return she gets as percent on her investment. Give your answer to the nearest integer.
Solution:
No. of shares = 1800
Face value of each share = Rs. 100
Rate of dividend = 15 %
Market value of each share Rs. 140
Total value of shares = Rs. 1800 x 100 = Rs. 1,80,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 11.1

Question 12.
A man invests Rs. 11,200 in a company paying 6 percent per annum when its Rs. 100 shares can be bought for Rs. 140 find:
(i) his annual dividend.
(ii) his percentage return on his investment.
Solution:
Investment = Rs. 11,200
Rate of dividend = 6%
Market value of a share = Rs. 140
No. of shares = Rs. 11,200 ÷ Rs. 140 = 80
Face value of 80 shares = 80 x Rs. 100 = Rs. 8,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 12.1

Question 13.
Mr. Sharma has 60 shares of N.V. ₹ 100 and sells them when they are at a premium of 60%. He invests the proceeds in shares of nominal value ₹ 50, quoted at 4% discount, and paying 18% dividend annually. Calculate:
(i) the sale proceeds ;
(ii) the number of shares he buys and
(iii) his annual dividend from the shares.
Solution:
(i) No. of shares = 60
Face value of each share = Rs. 100
Total amount = Rs. 100 x 60 = Rs. 6,000
Market value = Rs. 160
His sale proceed = Rs. 160 x 60 = Rs. 9,600
(ii) In second case :
Nominal value of each share = Rs.50
and Market value = Rs. 50 x \(\frac { 96 }{ 100 }\) = Rs. 48
Rate of dividend = 18%
No. of shares he purchased = \(\frac { 9600 }{ 48 }\) = 200
(iii) Face value of 200 shares = 200 x Rs. 50 = Rs. 10,000
Dividend = Rs. 10000 x \(\frac { 18 }{ 100 }\) = Rs. 1,800

Question 14.
A company with 10,000 shares of nominal value ₹ 100 declares an annual dividend of 8% to shareholders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh had bought 90 shares of the company at ₹ 150 per share. Calculate the dividend he receives and the percentage of return on his investment.
Solution:
(i) No. of shares = 10,000
Nominal value of each share = Rs. 100
Dividend = 8%
Total face value of 10,000 shares = Rs. 100 x 10,000 = Rs. 10,00,000
and amount of dividend = Rs. \(\frac { 1000000 x 8 }{ 100 }\) = Rs. 8000
(ii) In second case :
Ramesh bought = 90 shares
Market value of each share = Rs. 150
His investment = Rs. 150 x 90 = Rs. 13,500
Face value of 90 shares = Rs. 100 x 90 = Rs. 9,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 14.1

Question 15.
Which is the better investment:
16% Rs. 100 shares at 80 or 20% Rs. 100 shares at 120 ?
Solution:
In first case :
Income on Rs. 80 = Rs. 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 15.1
From above, it is clear that first investment is better.

Question 16.
A man has a choice to invest in hundred rupee shares of two firms at Rs. 120 or at Rs. 132. The first firm pays r. dividend of 5% per annum and the second firm pays a dividend of 6% per annum. Find :
(i) Which company is giving a better return.
(ii) If a man invests Rs. 26,400 with each firm, how much will be the difference between the annual returns from the two firms.
Solution:
In first case :
Market value of share = Rs. 120
and dividend = 5%
Income on Rs. 120 = Rs. 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 16.1
= Rs. 1,200
Difference = Rs. 1,200 – Rs. 1,100 = Rs. 100

Question 17.
A man bought 360, ten rupee shares of a company paying 12 percent per annum. He sold the shares when their price rose to Rs. 21 per share and invested the proceeds in five rupee shares paying 4.5 per cent per annum at Rs 3.50 per share. Find the annual change in his income.
Solution:
No. of shares bought = 360
Face value of each share = Rs. 10
Dividend = 12%
Cost price of 360 shares = Rs. 360 x 10 = Rs. 3,600
Market value = Rs. 21
Selling price = Rs. 21 x 360 = Rs. 7,560
In second case :
Face value of each share = Rs. 5
Market value of each share = Rs. 3.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 17.1
Difference in his income = Rs. 486 Rs. 432 = Rs. 54

Question 18.
A man sold 400 (Rs. 20) shares of a company paying 5% at Rs. 18 and invested the proceeds in (Rs. 10) shares of another company paying 7% at Rs. 12. How many (Rs. 10) shares did he buy and what was the change in his income?
Solution:
In first case :
No. of shares sold = 400
Face value of each share = Rs. 20
Market value = Rs. 18
Income = 5%
Amount of his investment = Rs. 18 x 400 = Rs. 7,200
and amount of shares = Rs. 20 x 400 = Rs. 8000
In second case :
Market value of each share = Rs. 12
Face value of each share = Rs. 10
Rate of dividend = 7%
No. of shares purchased = \(\frac { 7200 }{ 12 }\) = 600
Face value of 600 shares = Rs. 10 x 600 = Rs. 6,000
Now, income in first case = Rs. 8000 x \(\frac { 5 }{ 100 }\) = Rs. 400
and income in second case = Rs. 6000 x \(\frac { 7 }{ 100 }\) = Rs. 420
Increase in income = Rs. 420 – 400 = Rs. 20

Question 19.
Two brothers A and B invest Rs. 16,000 each in buying shares of two companies. A buys 3% hundred-rupee shares at 80 and B buys ten rupee shares at par. If they both receive equal dividend at the end of the year, find the rate percent of the dividend received by B.
Solution:
A’s investment = Rs. 16,000
Face value of each share = Rs. 100
Market value of each share = Rs. 80
and rate of dividend = 3%
No. of shares purchased = \(\frac { 16000 }{ 80 }\) = 200
Amount of shares = 200 x Rs. 100 = Rs. 20,000
and amount of dividend = 20,000 x \(\frac { 3 }{ 100 }\) = Rs. 600
B’s investment = Rs. 16,000
Face value of each share = Rs. 10
and amount of dividend = Rs. 600
Rate of dividend = \(\frac { 600 }{ 16000 }\) x 100 = 3.75 %

Question 20.
A man invests Rs. 20,020 in buying shares of nominal value Rs. 26 at 10% premium. The dividend on the shares is 15% per annum Calculate :
(i) The number of shares he buys.
(ii) The dividend he receives annually.
(iii) The rate of interest he gets on his money [2003]
Solution:
Total investment = Rs. 20,020
Nominal value of each share = Rs. 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 20.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C.

Other Exercises

[In this exercise, all the prices are excluding tax/VAT unless specified].
Question 1.
Madan purchases a compact computer system for ₹ 47,700 which includes 10% rebate on the marked price and then 6% Sales Tax on the remaining price. Find the marked price of the computer.
Solution:
Sales price of the computer = ₹ 47700
Rate of Sales Tax = 6%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 1.1
Hence marked price = ₹ 50,000

Question 2.
An article is marked at ₹ 500. The wholesaler sells it to a retailer at 20% discount and charges sales-tax on the remaining price at 12.5%. The retailer, in turn, sells the article to a customer at its marked price and charges sales-tax at the same rate. Calculate :
(i) the price paid by the customer.
(ii) the amount of VAT paid by the retailer.
Solution:
Marked price of an article = ₹ 500
Rate of discount = 20%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 2.1
(i) Price paid by-the customer = ₹ 500 + ₹ 62.50 = ₹ 562.50
(ii) VAt paid by the retailer = ₹ 62.50 – ₹ 50.00 = ₹ 12.50

Question 3.
An article is marked at ₹ 4,500 and the rate of sales-tax on it is 6%. A trader buys this article at some discount and sells it to a customer at the marked price. If the trader pays ₹ 81 as VAT; find :
(i) how much per cent discount does the trader get ?
(ii) the total money paid by the trader, including tax, to buy the article.
Solution:
Marked price = ₹ 4500
Rate of S.T. = 6%
VAT paid by the trader = ₹ 81
C.P. for the customer = ₹ 4500
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 3.1

Question 4.
A retailer sells an article for ₹ 5,350 including 7% Sales Tax on the listed price. If he had bought it at a discount and has made a profit of 25% on the whole, And the rate of discount he gets.
Solution:
S.P. of an article of the retailer (including S.T.) = ₹ 5350
Rate of S.T. = 7%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 4.1

Question 5.
A shopkeeper buys a camera at a discount of 20% from the wholesaler, the printed price of the camera being ₹ 1600 and the rate of sales tax is 6%. The shopkeeper sells it to the buyer at the printed price and charges tax at the same rate. Find :
(i) The price at which the camera can be bought from the shopkeeper.
(ii) The VAT (Value Added Tax) paid by the shopkeeper. (2008)
Solution:
Printed price (M.P) = ₹ 1600
Rate of discount = 20%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 5.1
(i) The buyers bought the camera = ₹ 1600 + ₹ 96 = ₹ 1696
(ii) VAT paid by the shopkeeper = ₹ 96 – ₹ 76.80 = ₹ 19.20

Question 6.
Tarun bought an article for ₹ 8000 and spent ₹ 1000 for transportation. He marked the article at ₹ 11,700 and sold it to a customer. If the customer had to pay 10% sales tax, find:
(i) The customer’s price
(ii) Tarun’s profit percent.
Solution:
Tarun’s cost price = ? 8000
Transportation charges = ₹ 1000
Tarun’s total cost = (₹ 8000 + ₹ 1000) = ₹ 9000
Tarun’s selling price = ₹ 11,700
Rate of sales tax = 10%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 6.1

Question 7.
A shopkeeper sells an article at the listed price of ₹ 1,500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ₹ 36 to the Government, what was the price, inclusive of Tax, at which the shopkeeper purchased the articles from the wholesaler?
Solution:
Tax charged by shopkeeper
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 7.1

Question 8.
A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18,000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax under is 8%, find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine. (2014)
Solution:
(i) S.P. of washing machine
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 8.1
VAT paid by shopkeeper = Tax charged – Tax paid = ₹ 1296 – ₹ 1152 = ₹144
(ii) Price paid by customer = ₹ 16200 + ₹ 1296 = ₹ 17496

Question 9.
Mohit, a dealer in electronic goods, buys a high class TV set for ₹ 61,200. He sells this TV set to Geeta, Geeta to Rohan and Rohan sells it to Manoj. If the profit at each stage is ₹ 2,000 and the rate of VAT at each stage is 12.5%, find :
(i) total amount of tax (under VAT) paid to the Government.
(ii) Money paid by Manoj to buy the TV set.
Solution:
For Mohit,
C.P. of electronic goods = ₹ 61200
Amount of profit in each case = ₹ 2000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 9.1
Net VAT = ₹ 8400 – ₹ 8150 = ₹ 250
(i) Total amount of VAT paid to govt. = ₹ 7650 + ₹ 250 + ₹ 250 + ₹ 250 = ₹ 8400
(ii) Money paid by Manoj to buy the T.V. set = ₹ 67200 + ₹ 8400 = ₹ 75600

Question 10.
A shopkeeper buys an article at a discount of 30% of the list price which is ₹ 48,000. In turn, the shopkeeper sells the article at 10% discount. If the rate of VAT is 10%, find the VAT to be paid by the shopkeeper.
Solution:
Market value (M.P.) of an article = ₹ 48000
Rate of discount = 30%
Amount of discount = ₹ 48000 x \(\frac { 30 }{ 100 }\) = ₹ 14400
(i) Cost price of shopkeeper = ₹ 48000 – ₹ 14400 = ₹ 33600
(ii) C.P. for shopkeeper = ₹ 33600
Rate of VAT = 10%
Amount of VAT paid by shopkeeper
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 10.1
Net VAT paid by the shopkeeper = VAT recovered from customer – VAT paid to dealer
= ₹ 4320 – ₹ 3360 = ₹ 960

Question 11.
A company sells an article to a dealer for ₹ 40,500 including VAT (sales-tax). The dealer sells it to some other dealer for ₹ 42,500 plus tax. The second dealer sells it to a customer at a profit of ₹ 3,000. If the rate of sales-tax under VAT is 8%, find :
(i) the cost of the article (excluding tax) to the first dealer.
(ii) the total tax (under VAT) received by the Government.
(iii) the amount that a customer pays for the article.
Solution:
For a company,
S.P. of an article including VAT = ₹ 40500
Rate of VAT = 8%
(i) Amount of article excluding VAT
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 11.1
= ₹ 425 x 108 = ₹ 45900
Total tax (VAT) = ₹ 45900 – ₹ 42500 = ₹ 3000
S.P. for the second dealer = ₹ 42500 + ₹ 3000 = ₹ 45500
(ii) VAT = ₹ \(\frac { 45500 x 8 }{ 100 }\) = ₹ 3640
(iii) Total amount paid by customer = ₹ 45500 + ₹ 3640 = ₹ 49140

Question 12.
A wholesaler buys a TV from the manufacturer for ₹ 25,000. He marks the price of the TV 20% above his cost price and sells it to a retailer at 10% discount on the marked price. If the rate of VAT is 8%, find the :
(i) marked price.
(ii) retailer’s cost price inclusive of tax.
(iii) VAT paid by the wholesaler. (2015)
Solution:
S.P. for the manufactures = ₹ 25000
or C.P. for a whole seller a T.V. = ₹ 25000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 12.1
C.P. for the retailer = ₹ 27000 + ₹ 2160 = ₹ 29160
(iii) VAT paid by the whole seller = ₹ 2160 – ₹ 2000 = ₹ 160

Question 13.
A dealer buys an article at a discount of 30 % from the wholesaler, the marked price being ₹ 6,000. The dealer sells it to a shopkeeper at a discount of 10% on the marked price. If the rate of VAT is 6%, find :
(i) the price paid by the shopkeeper including the tax.
(ii) the VAT paid by the dealer. (2016)
Solution:
Market price of the article = ₹ 6000
A dealer buys an article at a discount of 30% from the wholesaler.
Price of the article which the dealer paid to the wholesaler = 6000 – 30% of 6000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 13.1
Amount of the article inclusive of sales tax at which the dealer bought it
= ₹ 4200 + ₹ 252 = ₹ 4452
Dealer sells the article at a discount of 10% to the shopkeeper.
Price of the article which the shopkeeper paid to the dealer = ₹ 6000 – 10% of 6000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 13.2
Amount of the article inclusive of sales tax at which the shopkeeper bought it
= ₹ 5400 + ₹ 324 = ₹ 5724
The value added by dealer = ₹ 5400 – ₹ 4200 = ₹ 1200
Amount of VAT paid by dealer = 6% of 1200
= \(\frac { 6 }{ 100 }\) x 1200 = ₹ 72
Price paid by shopkeeper including tax is ₹ 5724
VAT paid by dealer is ₹ 72

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B

 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B.

Other Exercises

[In this exercise, all the prices are excluding tax/VAT unless specified].
Question 1.
A shopkeeper purchases an article for ₹ 6,200 and sells it to a customer for ₹ 8,500. If the sale tax (under VAT) is 8%; find the VAT paid by the shopkeeper.
Solution:
C.P. of article = ₹ 6200
Rate of VAT = 8%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 1.1
Amount of VAT paid by the shopkeeper = ₹ 680 – ₹ 496 = ₹ 184

Question 2.
A purchases an article for ₹ 3,600 and sells it to B for ₹ 4,800. B, in turn, sells the article to C for ₹ 5,500. If the sale tax (under VAT) is 10%, find the VAT levied on A and B.
Solution:
C.P. of an article for A = ₹ 3600
C.P. of the article for B = ₹ 4800
and C.P. for C = ₹ 5500
Rate of VAT in each case = 10%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 2.1
Now VAT levied on A = ₹ 480 – ₹ 360 = ₹ 120
and VAT levied on B = ₹ 550 – ₹ 480 = ₹ 70

Question 3.
A manufacturer buys raw material for ₹ 60,000 and pays 4% tax. He sells the ready stock for ₹ 92,000 and charges 12.5% tax. Find the VAT paid by the manufacturer.
Solution:
C.P. of raw material = ₹ 60000
Rate of tax = 4%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 3.1
VAT paid by the manufacturer = ₹ 11500 – ₹ 2400 = ₹ 9100

Question 4.
The cost of an article is ₹ 6,000 to a distributor. He sells it to a trader for ₹ 7,500 and the trader sells it to a customer for ₹ 8,000. If the VAT rate is 12.5% ; find the VAT paid by the :
(i) distributor
(ii) trader.
Solution:
Cost price of an article to a distributor = ₹ 6000
and selling price of distributor = ₹ 7500
and selling price of trader = ₹ 8000
Rate of VAT = 12.5% = \(\frac { 25 }{ 2 }\) %
Now, VAT for two distributor
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 4.1
(i) Now VAT paid by distributor = ₹ 937.50 – ₹ 750 = ₹ 187.50
(ii) and VAT paid by trader = ₹ 1000 – ₹ 937.50 = ₹ 62.50

Question 5.
The printed price of an article is ₹ 2,500. A wholesaler sells it to a retailer at 20% discount and charges sales-tax at the rate of 10%. Now the retailer, in turn, sells the article to a customer at its list price and charges the sales-tax at the same rate. Find :
(i) the amount that retailer pays to the wholesaler.
(ii) the VAT paid by the retailer.
Solution:
M.P. of an article = ₹ 2500
Rebate = 20%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 5.1

Question 6.
A retailer buys an article for ₹ 800 and pays the sales-tax at the rate of 8%. The retailer sells the same article to a customer for ₹ 1,000 and charges sales- tax at the same rate. Find :
(i) the price paid by a customer to buy this article.
(ii) the amount of VAT paid by the retailer.
Solution:
C.P. of an article for retailer = ₹ 800
Rate of S.T. = 8%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 6.1
(i) Cost price of the customer = ₹ 1000 + ₹ 80 = ₹ 1080
(ii) VAT paid by the retailer = ₹ 80 – ₹ 64 = ₹ 16

Question 7.
A shopkeeper buys 15 identical articles for ₹ 840 and pays sales-tax at the rate of 8%. He sells 6 of these articles at ₹ 65 each and charges sales-tax at the same rate. Calculate the VAT paid by the shopkeeper against the sale of these six articles.
Solution:
C.P. of 15 articles = ₹ 840
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 7.1
VAT paid by the retailer = ₹ 31.20 – ₹ 26.88 = ₹ 4.32

Question 8.
The marked price of an article is ₹ 900 and the rate of sales-tax on it is 6%. If on selling the article at its marked price, a retailer has to pay VAT = ₹ 4.80; find the money paid by him (including sales-tax) for purchasing this article.
Solution:
M.P. an article = ₹ 900
Rate of S.T. = 6%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 8.1
Amount paid including S.T. = ₹ 820 + ₹ 49.20 = ₹ 869.20

Question 9.
A manufacturer marks an article at ₹ 5,000. He sells this article to a wholesaler at a discount of 25% on the marked price and the wholesaler sells it to a retailer at a discount of 15% on its marked price. If the retailer sells the article without any discount and at each stage the sales-tax is 8%, calculate the amount of VAT paid by :
(i) the wholesaler.
(ii) the retailer.
Solution:
M.P. of an article = ₹ 5000
Rate of discount = 25%
S.R of the manufacturer or C.P of
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 9.2
Now VAT paid by
(i) The wholesaler = ₹ 340 – ₹ 300 = ₹ 40
(ii) Retailer = ₹ 400 – ₹ 340 = ₹ 60

Question 10.
A shopkeeper buys an article at a discount of 30% and pays sales-tax at the rate of 8%. The shopkeeper, in turn, sells the article to a customer at the printed price and charges sales tax at the same rate. If the printed price of the article is ₹ 2,500; find :
(i) the price paid by the shopkeeper.
(ii) the price paid by the customer.
(iii) The VAT (Value Added Tax) paid by the shopkeeper.
Solution:
Printed price of an article = ₹ 2500
Discount = 30%
C.P of the article for shopkeeper
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 10.1
(i) Price paid by the shopkeeper = ₹ 1750 + ₹ 140 = ₹ 1890
(ii) Price paid by the customer = ₹ 2500 + ₹ 200 = ₹ 2700
(iii) VAT paid by the shopkeeper = ₹ 200 – ₹ 140 = ₹ 60

Question 11.
A shopkeeper sells an article at its list price (₹ 3,000) and charges sales-tax at the rate of 12%. If the VAT paid by the shopkeeper is ₹ 72, at what price did the shopkeeper buy the article inclusive of sales-tax ?
Solution:
S.P. or list price of an article = ₹ 3000
Rate of sales tax = 12%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 11.1
C.P paid by the shopkeeper including S.T. = ₹ 2400 + ₹ 288 = ₹ 2688

Question 12.
A manufacturer marks an article for ₹ 10,000. He sells it to a wholesaler at 40% discount. The wholesaler sells this article to a retailer at 20% discount on the marked price of the article. If retailer sells the article to a customer at 10% discount and the rate of sales-tax is 12% at each stage; find the amount of VAT paid by the :
(i) wholesaler
(ii) retailer
Solution:
Marked price of an article = ₹ 10000
Rate of discount = 40%
C.P of the wholesaler
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 12.2
Now
(i) Vat paid by wholesaler = ₹ 960 – ₹ 720 = ₹ 240
(ii) Vat paid by retailer = ₹ 1080 – ₹ 960 = ₹ 120

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1B 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A.

Other Exercises

Question 1.
Rajat purchases a wrist-watch costing ₹ 540. The rate of Sales Tax is 8%. Find the total amount paid by Rajat for the watch.
Solution:
Cost of watch = ₹ 540
Rate of Sales Tax = 8%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 1.1
Total Amount of Watch = ₹ 540 + ₹ 43.20 = ₹ 583.20

Question 2.
Ramesh paid ₹ 345.60 as Sales Tax on a purchase of ₹ 3,840. Find the rate of Sales Tax.
Solution:
On ₹ 3840, sales-tax is = ₹ 345.60
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 2.1

Question 3.
The price of a washing machine, inclusive of sales tax is ₹ 13,530/-. If the Sales Tax is 10%, find its basic (cost) price. [2003]
Solution:
Selling price of washing machine = ₹ 13,530.
Rate of Sales Tax = 10%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 3.1

Question 4.
Sarita purchases biscuits costing ₹ 158 on which the rate of Sales Tax is 6%. She also purchases some cosmetic goods costing ₹ 354 on which the rate of Sales Tax is 9%. Find the total amount to be paid by Sarita.
Solution:
Cost of biscuits = ₹ 158.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 4.1
Total cost of cosmetic goods = ₹ 354 + ₹ 31.86 = ₹ 385.86
Total amount paid by Sarita = ₹ 167. 48 + ₹ 385.86 = ₹ 553. 34

Question 5.
The marked price of two articles A and B together is ₹ 6,000. The sales tax on article A is 8% and that on article B is 10%. If on selling both the articles, the total sales tax collected is ₹ 552, find the marked price of each of the articles A and B.
Solution:
M.P of two articles A and B = ₹ 6000
Rate of S.T. on A = 8%
and rate of S.I. on B = 10%
Total sales tax on A and B = ₹ 552
Let MP of A be = ₹ x
and then of MP of B = ₹ (6000 – x)
according to the condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 5.1
M.P. of article A = ₹ 2400
and MP of article B = ₹ 6000 – ₹ 2400 = ₹ 3600

Question 6.
The price of a T.V. set inclusive of sales tax of 9% is ₹ 13,407. Find its marked price. If Sales Tax is increased to 13%, how much more does the customer has to pay for the T.V. ? [2002]
Solution:
Sale price of T.V. set = ₹ 13,407
Rate of sales tax = 9%
Let marked price of T.V. = x
Then sale price
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 6.1
Sale price = ₹ 12,300 + ₹ 1,599 = ₹ 13,899
Difference between the two sales price = ₹ 13,899 – ₹ 13,407 = ₹ 492

Question 7.
The price of an article is ₹ 8,250 which includes Sales Tax at 10%. Find how much more or less does a customer pay for the article, if the Sales Tax on the article:
(i) increases to 15%
(ii) decreases to 6%
(iii) increases by 2%
(iv) decreases by 3%
Solution:
Price of an article = ₹ 8,250
Rate of Sales Tax = 10%
Let the list price = x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 7.2
The customer will have to pay less = ₹ 8,250 – ₹ 8,025 = ₹ 225

Question 8.
A bicycle is available for ₹ 1,664 including Sales Tax. If the list price of the bicycle is ₹ 1,600, find :
(i) the rate of Sales Tax
(ii) the price a customer will pay for the bicycle if the Sales Tax is increased by 6%.
Solution:
Sale price of bicycle = ₹ 1,664.
List price = ₹ 1,600
Amount of Sales Tax = ₹ 1,664 – ₹ 1,600 = ₹ 64.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 8.1
Sales price = ₹ 1,600 + ₹ 160 = ₹ 1,760

Question 9.
When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V.; Mrs Geeta will save ₹ 780 in buying this T.V. Find the list price of the T.V.
Solution:
Rate of sales tax in the beginning = 9%
and Reduced rate = 6%
Diff. = 9 – 6 = 3%
Total saving = ₹ 780
List, price of TV = Total saving x \(\frac { 100 }{ 3 }\)
= \(\frac { 780 x 100 }{ 3 }\) = ₹ 26000

Question 10.
A trader buys an unfinished article for ₹ 1,800 and spends ₹ 600 on its finishing, packing, transportation, etc. He marks the article at such a price that will give him 20% profit. How much will a customer pay for the article including 12% sales tax.
Solution:
Cost price of an unfinished article = ₹ 1800
Cost on finishing, packing etc or over head charges = ₹ 600
Total C.P. = ₹ 1800 + ₹ 600 = ₹ 2400
Gain = 20%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 10.1
Total sale price = ₹ 2880 + ₹ 345.60 = ₹ 3225.60

Question 11.
A shopkeeper buys an article for ₹ 800 and spends ₹ 100 on its transportation, etc. He marks the article at a certain price and then sells it Tor ₹ 1,287 including 10% sales tax. Find this profit as per cent.
Solution:
C.P. of an article = ₹ 800
Over head charges = ₹ 100
Total C.P. = ₹ 800 + ₹ 100 = ₹ 900
Sale price = ₹ 1287 including S.T.
Rate of S.T. = 10%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 11.1

Question 12.
A shopkeeper announces a discount of 15% on his goods. If the marked price of an article, in his shop, is ₹ 6000 ; how much a customer has to pay for it, if the rate of Sales Tax is 10% ?
Solution:
Marked price of an article = ₹ 6000
Rate of discount = 15%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 12.1

Question 13.
The catalogue price of a music system is ₹ 24,000. The shopkeeper gives a discount of 8% on the list price. He gives a further off-season discount of 5% on the balance. But Sales Tax at 10% is charged on the remaining amount. Find:
(a) the Sales Tax a customer has to pay.
(b) the final price he has to pay for the music system. (2001)
Solution:
List price of colour music system = ₹ 24,000
First rate of discount = 8%
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 13.1
(b) Final price of music system = ₹ 20,976 + ₹ 2,097.60 = ₹ 23,073.60

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1A 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A.

Other Exercises

Question 1.
The product of two consecutive integers is 56. Find the integers.
Solution:
Let the first integer = x
Then second integer = x + 1
Now according to the condition given
x (x + 1) = 56
⇒ x² + x – 56 = 0
⇒ x² + 8x – 7x – 56 = 0
⇒ x (x + 8) – 7 (x + 8) = 0
⇒ (x + 8) (x – 7) = 0
Either x + 8 = 0, then x = – 8
or x – 7 = 0, then x = 7
(i) If x = -8, then
first integer = -8
and second integer = – 8 + 1 = – 7
(ii) If x = 7, then
first integer = 7
and second integer = 7 + 1 = 8
Integers are 7, 8 or -8, -7

Question 2.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the first natural number = x
Then second natural number = x + 1
Now according to the condition given,
(x)² + (x + 1)² = 41
⇒ x² + x² + 2x + 1 = 41
⇒ 2x² + 2x + 1 – 41 = 0
⇒ 2x² + 2x – 40 = 0
⇒ x² + x – 20 = 0 (Dividing by 2)
⇒ x² + 5x – 4x – 20 = 0
⇒ x (x + 5) – 4 (x + 5) = 0
⇒ (x + 5) (x – 4) = 0
Either x + 5 = 0 then x = – 5 But it is not a natural number
or x – 4 = 0, Then x = 4
Numbers are 4 and 5

Question 3.
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Solution:
Let the first natural number = x
Then second natural number = x + 5
Now according to the given condition,
(x)² + (x + 5)² = 97
⇒ x² + x² + 10x + 25 – 97 = 0
⇒ 2x² + 10x – 72 = 0
⇒ x² + 5x – 36 = 0 (Dividing by 2)
⇒ x² + 9x – 4x – 36 = 0
⇒ x (x + 9) – 4 (x + 9) = 0
⇒ (x + 9) (x – 4) = 0
Either x + 9 = 0, then x = – 9 But it is not a natural number
or x – 4 = 0, then x = 4
First number = 4
and second number = 4 + 5 = 9

Question 4.
The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Let the number = x
Now according to the condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q4.1
⇒ 4x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (4x – 1) = 0
Either x – 4 = 0, then x = 4
or 4x – 1 = 0, 4x = 1 then x = \(\frac { 1 }{ 4 }\)
Number is 4 or \(\frac { 1 }{ 4 }\)

Question 5.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is \(\frac { 7 }{ 10 }\)
Solution:
Let the first natural number = x
Then second natural number = x + 3
Now according to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q5.1
But it is not a natural number
or x – 2 = 0, then x = 2
First number = 2
and second number = 2 + 3 = 5

Question 6.
Divide 15 into two parts such that the sum of their reciprocals is \(\frac { 3 }{ 10 }\).
Solution:
Let first part = x
Then second part = 15 – x (sum = 15)
Now according to the given condition,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q6.1
⇒ x (x – 5) – 10 (x – 5) = 0
⇒ (x – 5) (x – 10) = 0
Either x – 5 = 0, then x = 5
or x – 10 = 0, then x = 10
If x = 5, then first part = 15 – 5 = 10
If x = 10, then second part = 15 – 10 = 5
Parts are 5, 10

Question 7.
The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Solution:
Let x be the larger number and y be the smaller number, then
According to the conditions x2 = 18 ….(i)
and x² + y² = 208 ….(ii)
⇒ 18y + y² = 208 [From (i)]
⇒ y² + 18y – 208 = 0
⇒ y² + 26y – 8y – 208 = 0
⇒ y (y + 26) – 8 (y + 26) = 0
⇒ (y + 26) (y – 8) = 0
Either y + 26 = 0, then y = -26
But it is not possible as it is not positive
or y – 8 = 0, then y = 8
Then x² = 18y ⇒ y² = 18 x 8 ⇒x² = 144 = (12)² ⇒ x = 12
Number are 12, 8

Question 8.
The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let first even number = 2x
and second even number = 2x + 2
Now according to the given condition,
(2x)² + (2x + 2)² = 52
⇒ 4x² + 4x² + 8x + 4 = 52
⇒ 4x² + 4x² + 8x + 4 – 52 = 0
⇒ 8x² + 8x – 48 = 0
⇒ x² + x – 6 = 0 (Dividing by 8)
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x + 3) (x – 2) = 0
Either x + 3 = 0, then x = – 3 But it is not possible because it is not positive.
or x – 2 = 0, then x = 2
First even number = 2 x 2 = 4
Second number = 4 + 2 = 6
Hence numbers are 4, 6

Question 9.
Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let first odd number = 2x -1
Second odd number = 2x – 1 + 2 = 2x + 1
Now according to the given condition,
(2x – 1)² + (2x + 1)² = 74
⇒ 4x² – 4x + 1 + 4x² + 4x + 1 = 74
⇒ 8x² + 2 – 74 = 0
⇒ 8x² – 72 = 0
⇒ x² – 9 = 0 (Dividing by 8)
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = -3 But it is not possible because it is not positive.
or x – 3 = 0, then x = 3
First odd number = 2x – 1 = 2 x 3 – 1 = 6 – 1 = 5
and second odd number = 5 + 2 = 7
Number are 5, 7

Question 10.
The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Let numerator of a fraction = x
Then denominator = 2x + 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q10.2

Question 11.
Three positive numbers are in the ratio \(\frac { 1 }{ 2 }\) : \(\frac { 1 }{ 3 }\) : \(\frac { 1 }{ 4 }\) Find the numbers; if the sum of their squares is 244.
Solution:
Multiply each ratio by L.C.M. of de-nominators i.e. by 12.
We get, 6 : 4 : 3
Let first positive number = 6x
Then second number = 4x
and third number = 3x
According to the given condition,
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² – 4 = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = -2 But it is not possible because it is not positive
or x – 2 = 0, then x = 2
First number = 6x = 6 x 2 = 12
Second number = 4x = 4 x 2 = 8
and third number = 3x = 3 x 2 = 6
Numbers are 12, 8, 6

Question 12.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution:
Let first part = x
Then second part = 20 – x (Sum = 20)
Now, according to the given condition,
3(x)² – (20 – x) = 10
⇒ 3x² – 20 + x – 10 = 0
⇒ 3x² + x – 30 = 0
⇒ 3x² + 10x – 9x – 30 = 0
⇒ x (3x + 10) – 3 (3x + 10) = 0
⇒ (3x + 10) (x – 3) = 0
Either 3x + 10 = 0, then 3x = -10 ⇒ x = \(\frac { -10 }{ 3 }\)
But it is not possible.
or x – 3 = 0 then x = 3
Then first part = 3
and second part = 20 – 3 = 17

Question 13.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement Hence ; find the three numbers.
Solution:
Let middle number = x
Then, first number = x – 1
and third number = x + 1
Now according to the condition,
(x)²= [(x + 1)² – (x – 1)²] + 60
⇒ x² = [x² + 2x + 1 – x² + 2x – 1 ] + 60
⇒ x² = 4x + 60
⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0
⇒ x (x – 10) + 6 (x – 10) = 0
⇒ (x – 10) (x + 6) = 0
Either x – 10 = 0 then x = 10
or x + 6 = 0 then x = -6. But it is not a natural number.
Middle number = 10
First number = 10 – 1 = 9
and third number = 10 + 1 = 11
Hence numbers are 9, 10, 11

Question 14.
Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Solution:
Middle number = p
then smallest number = p – 1
and greatest number = p + 1
Now, according to the condition.
3 (p + 1)² – (p – 1)² – p² = 67
⇒ 3 (p² + 2p + 1) – (p² – 2p + 1) – p² = 67
⇒ 3p² + 6p + 3 – p² + 2p – 1 – p² – 67 = 0
⇒ p² + 8p + 2 – 67 = 0
⇒ p² + 8p – 65 = 0
⇒ p² + 13p – 5p – 65 = 0
⇒ p (p + 13) – 5 (p + 13) = 0
⇒ (p + 13) (p – 5) = 0
Either p + 13 = 0, then p = -13 But it is not possible.
or p – 5 = 0, then p = 5
p = 5

Question 15.
A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days. Calculate ‘x’.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q15.2

Question 16.
One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Solution:
Let first pipe can fill the cistern in = x hrs.
Second pipe will fill the cistern in = x – 3 hrs.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A Q16.1
But it is not possible.
x = 15
First pipe can fill in 15 hrs.
and second pipe in 15 – 3 = 12 hrs.

Question 17.
A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking A as the smaller part of the two parts. Find the number. (2010)
Solution:
Let larger part = y
and smaller part = x
x² + y² = 20 ….(i)
and y² = 8x ….(ii)
Substituting the value of y² in „
x² + 8x = 20
⇒ x² + 8x – 20 = 0
⇒ x² + 10x – 1x – 20 = 0 {-20 = 10 x (-2), 8 = 10 – 2}
⇒ x (x + 10) – 2(x + 10) = 0
⇒ (x + 10) (x – 2) = 0
Either x + 10 = 0, then x = – 10 which is not possible because it is negative
or x – 2 = 0, then x = 2
Smaller number = 2
and larger number = y² = 8x = 8 x 2 = 16
⇒ y² = 16 = (4)²
⇒ y = 4
Number = x + y = 2 + 4 = 6

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24A.

Other Exercises

Question 1.
Find the mean of following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q1.1

Question 2.
Marks obtained (in mathematics) by a students are given below :
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) Find the arithmetic mean
(b) If marks of each student be increased by 4;
what will be the new value of arithmetic mean.
Solution:
(a) Hence x = 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q2.1
(b) If marks of each students be increased by 4 then new mean will be = 59 + 4 = 63

Question 3.
Find the mean of natural numbers from 3 to 12.
Solution:
Numbers betweeen 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Here n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q3.1

Question 4.
(a) Find the means of 7, 11, 6, 5 and 6. (b) If each number given in (a) is diminished by 2; find the new value of mean.
Solution:
(a) The mean of 7, 11, 6, 5 and 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q4.1
(b) If we subtract 2 from each number, then the mean will be 7 – 2 = 5

Question 5.
If the mean of 6, 4, 7, a and 10 is 8. Find the value of ‘a’.
Solution:
No. of terms = 5
Mean = 8
∴ Sum of number (Σxi) = 5 x 8 = 40 …(i)
But Σxi = 6 + 4 + 7 + a+10 = 27 + a ….(ii)
From (i) and (ii)
27 + a = 40 ⇒ a = 40 – 27
∴ a = 13

Question 6.
The mean of the number 6, y, 7, x and 14 is 8. Express y in terms of x.
Solution:
No. of terms = 5 and
mean = 8
∴ Sum of numbers (Σxi) = 5 x 8 = 40 ….(i)
But sum of numbers given = 6 + y + 7 + x + 14
= 21 + y + x + ….(ii)
From (i) and (ii)
27 + y + x = 40
⇒ y = 40 – 27 – x
⇒ y= 13 – x

Question 7.
The ages of 40 students are given in the following table :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q7.1
Find the arithmetic mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q7.2

Question 8.
If 69.5 is the mean of 72, 70, x, 62, 50, 71, 90, 64, 58 and 82, find the value of x.
Solution:
No. of terms = 10
Mean = 69.5
∴ Sum of numbers = 69,5 x 10 = 695 ….(i)
But sum of given number = 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619+x ….(ii)
From (i) and (ii)
619 + x = 695
⇒ x = 695 – 619 = 76

Question 9.
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q9.3

Question 10.
From the data given below, calculate the mean wage, correct to the nearest rupee.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q10.1
(i) If the number of workers in each category is doubled, what would be the new mean wage? [1995]
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?
(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%. What would be the new mean wage?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q10.2
(i) Mean remains the same if the number of workers in each catagory is doubled.
∴ Mean = 80.
(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%.
∴ New mean = 80 x \(\frac { 160 }{ 100 }\)= 128
(iii) No change in the mean if the number of worker is doubled but if wages per worker is reduced by 40%, then
New mean = 80 x \(\frac { 60 }{ 100 }\)= 48

Question 11.
The contents of 100 match boxes were checked to determine the number of matches they contained.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q11.1
(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches. [1997]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q11.2
(ii) In the second case,
New mean = 39 matches
∴ Total contents = 39 x 100 = 3900
But total no of matches already given = 3813
∴ Number of new matches to be added = 3900 – 3813 = 87

Question 12.
If the mean of the following distribution is 3, find the value of p.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.1
Solution:
Mean = 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q12.3

Question 13.
In the following table, Σf= 200 and mean = 73. Find the missing frequencies f1 and f2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q13.3

Question 14.
Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.1
Solution:
Let the Assumed mean = 30
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q14.3

Question 15.
Find the mean (correct to one place of decimal) by using short-cut method.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.1
Solution:
Let the Assumed mean A = 45
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A Q15.3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D.

Other Exercises

Solve each of the following equations :
Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q1.2

Question 2.
(2x + 3)² = 81
Solution:
(2x + 3)2 = 81
⇒ 4x² + 12x + 9 = 81
⇒ 4x² + 12x + 9 – 81 = 0
⇒ 4x² + 12x – 72 = 0
⇒ x² + 3x – 18 = 0 (Dividing by 4)
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3 (x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
x = 3, – 6

Question 3.
a² x² – b² = 0
Solution:
a² x² – b² = 0
⇒ (ax)² – (b)² =0
⇒ (ax + b) (ax – b) =
Either ax + b = 0, then x = \(\frac { -b }{ a }\)
or ax – b = 0. then x = \(\frac { b }{ a }\)
x = \(\frac { b }{ a }\) , \(\frac { -b }{ a }\)

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q4.2

Question 5.
x + \(\frac { 4 }{ x }\) = – 4; x ≠ 0
Solution:
x + \(\frac { 4 }{ x }\) = -4
⇒ x² + 4 = -4x
⇒ x² + 4x + 4 = 0
⇒ (x + 2)² = 0
⇒ x + 2 = 0
⇒ x = – 2

Question 6.
2x4 – 5x² + 3 = 0
Solution:
2x4 – 5x² + 3 = 0
⇒ 2(x²)² – 5x² + 3 = 0
⇒ 2(x²)² – 3x² – 2x² + 3 = 0
⇒ 2x4 – 3x² – 2x² + 3 = 0
⇒ x² (2x² – 3) – 1 (2x² – 3) = 0
⇒ (2x² – 3) (x² – 1) = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q6.1

Question 7.
x4 – 2x² – 3 = 0
Solution:
x4 – 2x² – 3 = 0
⇒ (x²)² – 2x² – 3 = 0
⇒ (x²)² – 3x² + x² – 3 = 0
⇒ x² (x² – 3) + 1 (x² -3) = 0
⇒ (x² – 3) (x² + 1) = 0
Either x² – 3 = 0, then x² = 3 ⇒ x = √3
or x² + 1 = 0, then x² = – 1 In this case roots are not real
x = ±√3 or √3 , – √3

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q8.3

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q9.4

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q10.3

Question 11.
(x² + 5x + 4)(x² + 5x + 6) = 120
Solution:
Let x² + 5x + 4 = y then x² + 5x + 6 = y + 2
Now (x² + 5x + 4) (x² + 5x + 6) = 120
⇒ y (y + 2) – 120 = 0
⇒ y² + 2y – 120 = 0
⇒ y² + 12y – 10y – 120 = 0
⇒ y (y + 12) – 10 (y + 12) = 0
⇒ (y + 12) (y – 10) = 0
Either y + 12 = 0, then y = – 12
or y – 10 = 0, then y = 10
(i) when y = -12, then x² + 5x + 4 = -12
⇒ x² + 5x + 4 + 12 = 0
⇒ x² + 5x + 16 = 0
Here a = 1, b = 5, c = 16
D = b² – 4ac = (5)² – 4 x 1 x 16 = 25 – 64 = -39
D < 0, then roots are not real
(ii) When y = 10, then x² + 5x + 4 = 10
⇒ x² + 5x + 4 – 10 = 0
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x + 6) (x – 1) = 0
Either x + 6 = 0, then x = – 6
or x – 1 = 0, then x = 1
x = 1, -6

Question 12.
Solve each of the following equations, giving answer upto two decimal places:
(i) x² – 5x – 10 = 0 [2005]
(ii) 3x² – x – 7 = 0 [2004]
Solution:
(i) Given Equation is : x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0
a = 1, b = -5 , c = -10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q12.3

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q13.3

Question 14.
Solve:
(i) x² – 11x – 12 = 0; when x ∈ N
(ii) x² – 4x – 12 = 0; when x ∈ I
(iii) 2x² – 9x + 10 = 0; when x ∈ Q.
Solution:
(i) x² – 11x – 12 = 0
⇒ x² – 12x + x – 12 = 0
⇒ x (x – 12) + 1 (x – 12) = 0
⇒ (x – 12) (x + 1) = 0
Either x – 12 = 0, then x = 12
or x + 1 = 0, then x = -1
x ∈ N
x = 12
(ii) x² – 4x – 12 = 0
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6)=0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = -2
x ∈ I
x = 6, -2
(iii) 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = \(\frac { 1 }{ 2 }\)
x ∈ Q
x = 2, \(\frac { 5 }{ 2 }\) or 2, 2.5

Question 15.
Solve: (a + b)² x² – (a + b) x – 6 = 0, a + b ≠ 0.
Solution:
(a + b)² x² – (a + b) x – 6 = 0
Let (a + b) x = y, then y² – y – 6 = 0
⇒ y² – 3y + 2y – 6 = 0
⇒ y (y – 3) + 2 (y – 3) = 0
⇒ (y – 3) (y + 2) = 0
Either y – 3 = 0, then y = 3
or y + 2 = 0, then y = – 2
(i) If y = 3, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q15.1

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.1
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q16.3
Either x + p = 0, then x = -p
or x + q = 0, then x = -q
Hence x = -p, -q

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.1
Solution:
(i) x (x + 1) + (x + 2) (x + 3) = 42
⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0
⇒ 2x² + 6x – 36 = 0
⇒ x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3(x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
Hence x = 3, -6
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q17.2

Question 18.
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation:
(i) (m – 3) x² – 4x + 1 = 0
(ii) 3x² + 12x + (m + 7) = 0
(iii) x² – (m + 2) x + (m + 5) = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D Q18.3

Question 19.
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.
Solution:
px² – 4x + 3 = 0 …..(i)
Compare (i) with ax² + bx + c = 0
Here a = p, b = -4, c = 3
D = b² – 4ac = (-4)² – 4.p.(3) = 16 – 12p
As roots are equal, D = 0
16 – 12p = 0
⇒ \(\frac { 16 }{ 12 }\) = p
⇒ p = \(\frac { 4 }{ 3 }\)

Question 20.
Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots : x² + 2 (m – 1) x + (m + 5) = 0.
Solution:
x² + 2 (m – 1) x + (m + 5) = 0.
Here, a = 1, b = 2 (m – 1), c = m + 5
So, discriminant, D = b² – 4ac
= 4(m – 1)² – 4 x 1 (m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
For real and equal roots D = 0
So, 4m² – 12m – 16 = 0
⇒ m² – 3m – 4 = 0 (Dividingby4)
⇒ m² – 4m + m – 4 = 0
⇒ m (m – 4) + 1 (m – 4) = 0
⇒ (m – 4) (m + 1) = 0
⇒ m = 4 or m = -1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D 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.