4 Mark Question - Maths STD 8 Questions - Vidyadip

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Find the amount and the compound interest on Rs. 3000 for 2 years at 10% per annum.

Answer

Here, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 3000\times\Big(1+\frac{10}{100}\Big)^2$
$=\text{Rs}.\ 3000\times\Big(\frac{110}{100}\Big)^2$
$=\text{Rs}.\ 3000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$
$=\text{Rs}.(30\times11\times11)$
$=\text{Rs}. 3630$
$\therefore$ CI = A = P = Rs. (3630 - 3000) = Rs. 630
Hence, the amount is Rs. 3630 and the CI is Rs. 630.

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Question 24 Marks

Find the amount and the compound interest on Rs. 31250 for $1\frac{1}{2}$ years at 8% per annum, compounded half-yearly.

Answer

Principal, (P) = Rs. 31250
Annual rate of interest, $\text{R}=\frac{3}{2}\%$
Rate of interest for a half year $=\frac{1}{2}\Big(\frac{3}{2}\Big)\%=\frac{3}{4}\%$
Time, $\text{n}=1\frac{1}{2}$ years = 3 half years
Then the amount with the compound interest is given by
$\text{A}=\text{P }\times \Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$= 31250 \times\Big(1+\frac{\frac{3}{4}}{100}\Big)^3$
$=31250\times\Big(1+\frac{3}{100\times4}\Big)^3$
$=31250\times\Big(\frac{400+3}{400}\Big)^3$
$=31250\times\Big(\frac{403}{400}\Big)^3$
$=31250\times\Big(\frac{403}{400}\Big)\times\Big(\frac{403}{400}\Big)\times\Big(\frac{403}{400}\Big)$
= Rs. 31250 × (1.0075 × 1.0075 × 1.0075)$$
= Rs. 31958.41
Therefore, compound interest = amount - principal
= Rs. (31958.41 - 31250)
= Rs. 708.41.

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Question 34 Marks

Manoj deposited a sum of Rs. 64000 in a post office for 3 years, compounded annually at $7\frac{1}{2}\%$ per annum. What amount will he get on maturity?

Answer

Principal (P) = Rs. 64000

Rate (r) $=7\frac{1}{2}\%=\frac{15}{2}\%$

Period (t) = 3 years

Interest for the first year $=\frac{\text{prt}}{100}$

$=\text{Rs.}\ \frac{61000\times15\times1}{100\times2}$

$ =\text{Rs.}\ 4800$

Amount at the end of first year = Rs. 64000 + Rs. 4800 = Rs. 68800

Principal to the second year = Rs. 68800

Interest for the second year $=\text{Rs.}\ \frac{6880\times15\times1}{100\times2}$

$ =\text{Rs.}\ 5160$

Amount at the end of 2nd year = Rs. 68800 + 5160 = Rs. 73960

Principal for the third year $=\frac{73960\times15\times1}{100\times2}$

$ =\text{Rs.}\ 5547$

$\therefore$ Amount at the end of 3rd year = Rs. 73960 + Rs. 5547 = Rs. 79507

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Question 44 Marks

Find the amount and the compound interest on Rs. 12800 for 1 year at $7\frac{1}{2}\%$ per annum, compounded semi-annually.

Answer

Principal (P) = Rs. 12800 Rate (R) $=7\frac{1}{2}\%\text{ p.a }\frac{15}{2}\%$ half yearly Period (n) = 1 year or 2 half years $\therefore$ Amount (A) $=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }12800\Big(1+\frac{15}{4\times100}\Big)^2$ $=\text{Rs. }12800\times\frac{83}{80}\times\frac{83}{80} =\text{Rs. } 13778$ $\therefore$ C.I. = A - P$$= Rs. 13778 - Rs. 12800
= Rs. 978.

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Question 54 Marks

Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank charges $12\frac{1}{2}\%$ per annum, compounded half-yearly, what amount will she have to pay after $1\frac{1}{2}$ years? Also, find the interest paid by her.

Answer

Principal (P) = Rs. 40960
Rate(R) $=12\frac{1}{2}=\frac{25}{2}\%\text{ p.a or }\frac{25}{4}\%$ half yearly
Period (n) $=1\frac{1}{2}$ year or 3 half years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{R}{100}\Big)^{\text{n}}$
$=\text{Rs. }40960\Big(1+\frac{25}{4\times100}\Big)^3$
$=\text{Rs. }40960\times\Big(\frac{17}{16}\Big)^3$
$=\text{Rs. }40960\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}$
$=\text{Rs. }49130$
$\therefore$ C.I. paid = A - P
= Rs. 49130 - Rs. 40960
= Rs. 8170.

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Question 64 Marks

Find the difference between the simple interest and the compound interest on Rs. 5000 for 2 years at 9% per annum.

Answer

Principal (p) = Rs. 5000 Rate (R) = 9% p.a. Time (n) = 2 years$\therefore\text{S.I.}=\frac{\text{PRT}}{100}$
$=\frac{5000\times9\times2}{100}$
$=\text{Rs.}\ 900$
New amount at C.I. $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$ $=\text{Rs.}\ 5000\Big(1+\frac{9}{100}\Big)^\text{2}$ $=\text{Rs.}\ 5000\times\frac{109}{100}\times\frac{109}{100}$ $=\text{Rs.}\ \frac{11881}{2}$ $=\text{Rs.}\ 5940.50$ $\therefore$ C.I = A - P = Rs. 5940.50 - Rs. 5000 = Rs. 940.50 $\therefore$ Difference between C.I. and S.I. = Rs. 940.50 - Rs. 900 = Rs. 40.50

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Question 74 Marks

Abhay borrowed Rs. 16000 at $7\frac{1}{2}\%$ per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years?

Answer

In case of Abhay,
Principal (p) = Rs. 16000
Rate (r) $7\frac{1}{2}\%=\frac{15}{2}\%\text{p.a}$
Period (t) = 2 years
$\therefore$ S.Interest $=\frac{\text{prt}}{100}$
$=\text{Rs. }\frac{16000\times15\times2}{100\times2}=\text{Rs. }2400$
In case of Gurmeet,
Principal (p) = Rs. 16000
Rate (r) $7\frac{1}{2}\%=\frac{15}{2}\%\text{p.a}$
Period (t) = 2 years
Amount (A) $=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$=\text{Rs. }16000\times\Big(1+\frac{15}{2\times100}\Big)^2$
$=\text{Rs. }16000\Big(\frac{43}{40}\Big)^2$
$=\text{Rs. }16000\times\frac{43}{40}\times\frac{43}{40}$
$=\text{Rs. }18490$
C.I = A - P = Rs. 18490 - Rs. 16000
= Rs. 2490
gain to Abhay = Rs. 2490 - Rs. 2400
= Rs. 90

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Question 84 Marks

A scooter is bought for Rs. 32000. Its value depreciates at 10% per annum. What will be its value after 2 years?

Answer

Let the principal amount be P = Rs. 32000.
Rate of interest (R) = 10%
Time (n) = 2 years
Now, $\text{A}=\text{Rs}. \text{P}\times\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 32000\times\Big(1-\frac{10}{100}\Big)^2$
$=\text{Rs}.\ 32000\times\Big(\frac{90}{100}\Big)^2$
$=\text{Rs}.\ 32000\times\Big(\frac{9}{10}\Big)\times\Big(\frac{9}{10}\Big)$
$=\text{Rs}.\ (320\times9\times9)$
$= \text{Rs}. 25920$
$\therefore$ The value of the scooter after 2 years is Rs. 25920

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Question 94 Marks

Michael borrowed Rs. 16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after $1\frac{1}{2}$ years?

Answer

Principal (p) = Rs. 16000

Rate (r) = 10% p.a. or 5% half yearly

Period (t) = $1\frac{1}{2}$ year or 3 half years

Interest for the first year $=\frac{\text{prt}}{100}$

$=\text{Rs.}\ \frac{16000\times5\times1}{100}$

$=\text{Rs.}\ 800$

Amount at the end of first half year = Rs. 16000 + Rs. 800 = Rs. 16800

Principal for the second half year = Rs. 16800

Interest for the second half year $=\text{Rs.}\ \frac{16800\times5\times1}{100}$

$=\text{Rs.}\ 840$

Amount at the end of second half year = Rs. 16800 + Rs. 840 = Rs. 17640

Interest for the third half year $=\text{Rs.}\ \frac{17640\times5\times1}{100}$

$=\text{Rs.}\ 882$

Amount at the end of 3rd half year = Rs. 17640 + 882 = Rs. 18522

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Question 104 Marks

Find the amount of Rs. 10000 after 2 years compounded annually; the rate of interest being 10% per annum during the first year and 12% per annum during the second year. Also, find the compound interest.

Answer

Here, $\text{A}=\text{P}\times\Big(1+\frac{\text{P}}{100}\Big)\times\Big(1+\frac{\text{r}}{100}\Big)$
$=\text{Rs.}\ 10000\times\Big(1+\frac{10}{100}\Big)\times\Big(1+\frac{12}{100}\Big)$
$=\text{Rs}.\ 10000\times\Big(\frac{110}{100}\Big)\times\Big(\frac{112}{100}\Big)$
$=\text{Rs}.\ 10000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{28}{25}\Big)$
$=\text{Rs}.(40\times11\times28)$
$=\text{Rs}.\ 12320$
$\therefore$ CI = A - P = Rs. (12320 - 10000) = Rs. 2320
Hence, the amount is Rs. 12320 and the CI is Rs. 2320

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Question 114 Marks

Ratna obtained a loan of Rs. 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?

Answer

Amount of loan (p) = Rs. 25000

Rate of interest (r) = 8% p.a.

Period (t) = 2 years

Interest for the first year $=\frac{\text{prt}}{100}$

$=\text{Rs.}\ \frac{25000\times8\times1}{100}$

$=\text{Rs.}\ 2000$

$\therefore$ Amount at the end of first year = Rs. 25000 + Rs. 2000 = Rs. 27000

Principal for the second year = Rs. 27000

Interest for the second year $=\text{Rs.}\ \frac{27000\times8\times1}{100}$

$=\text{Rs.}\ 2000$

$\therefore$ Amount at the end of second year = Rs. 27000 + Rs. 2160 = Rs. 29160

$\therefore$ She would repay her debt = Rs. 29160

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Question 124 Marks

Find the amount and the compound interest on Rs. 160000 for 2 years at 10% per annum, compounded half-yearly.

Answer

Principal (P) = Rs. 160000
Rate (R) 10% p.a. or 5% half yearly
Period (n) = 2 year or 4 half years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{R}{100}\Big)^{\text{n}}$
$=\text{Rs. }160000\Big(1+\frac{5}{100}\Big)^4$
$=\text{Rs. }160000\times\Big(\frac{21}{20}\Big)^4$
$=\text{Rs. }160000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}$
$=\text{Rs. }194481$
$\therefore$ C.I. = A - P = Rs. 194481 - Rs. 160000
= Rs. 34481.

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Question 134 Marks

The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs. 90. Find the sum.

Answer

Difference between C.I and S.I. = Rs. 90 Rate (R) = 6% p.a Period (n) = 2 years Let principal (P) = Rs. 100 $\therefore$ Amount on C.I $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=100\Big(1+\frac{6}{100}\Big)^2$ $=\text{Rs. }100\times\Big(\frac{53}{50}\Big)^2$ $=\text{Rs. }100\times\frac{53}{50}\times\frac{53}{50}=\text{Rs. }\frac{2809}{25}$ $\therefore\text{C.I}=\text{A}-\text{P}=\text{Rs. }\frac{2809}{25}-100$ $=\text{Rs. }\frac{2809-2500}{25}=\frac{309}{25}$ and S.I $=\frac{\text{PRT}}{100}=\frac{100\times6\times2}{100}=\text{Rs. }12$ Difference between C.I and S.I. $=\text{Rs. }\frac{309}{25}-12$ $\text{Rs. }\frac{309-300}{25}=\frac{9}{25}$ Now if difference is $\text{Rs. }\frac{9}{25}$ then Principal = Rs. 100 and if difference is Rs. 90, then principal$=\frac{100\times25\times90}{9}=\text{Rs. }25000$

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Question 144 Marks

Find the amount and the compound interest on Rs. 8000 for 1 year at 10% per annum, compounded half-yearly.

Answer

Principal (P) = Rs. 12800
Rate (R) = 10% p.a 5% half yearly
Period (n) = 1 year or 2 half years
$\therefore$ Amount after 1 year (A) $\text{P}= \Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }8000\times\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs. }8000\times\Big(\frac{21}{20}\Big)^2$
$= \text{Rs. }8000\times\frac{21}{20}\times\frac{21}{20}=\text{Rs. 8820}$
and C.I. = A - P
= Rs. 8820 - Rs. 8000
= Rs. 820.

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Question 154 Marks

Arun took a loan of Rs. 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?

Answer

Amount taken from finance company (P) = Rs. 390625Rate of interest (R) = 16% p.a. or 4% quarterly
Period (n) = 1 year or 4 quarterly
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=390625\Big(1+\frac{4}{100}\Big)^4$
$=\text{Rs. }390625\times\Big(\frac{26}{25}\Big)^4$
$=\text{Rs. }390625\times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}$
$=\text{Rs. }456976$

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Question 164 Marks

Harpreet borrowed Rs. 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years.

Answer

In case of Harpreet:

Amount borrowed by Harpreet (P) = Rs. 20000

Rate (r) = 12%

Period (t) = 2 Years

$\therefore\text{S.I.}=\frac{\text{prt}}{100}$

$=\frac{20000\times12\times2}{100}$

$=\text{Rs.}\ 4800$

In case of Alam:

$\therefore$ Interest for the 1st year $=\frac{20000\times12\times1}{100}$

$=\text{Rs.}\ 2400$

Amount at the end of first year = Rs. 20000 + Rs. 2400 = Rs. 22400

Interest for the second year $=\frac{22400\times12\times1}{100}$

$=\text{Rs.}\ 2688$

$\therefore$ C.I. for 2 years = Rs. 2400 + 2688 = Rs. 5088

$\therefore$ Amount of gain = Rs. 5088 - Rs. 4800 = Rs. 288

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Question 174 Marks

Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs. 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.

Answer

Loan recieved for the cost of the house (P) = Rs. 125000Rate of interest (R) = 12% p.a. or 6% half yearly
Period (n) $=1\frac{1}{2}$ year or 3 half years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }125000\Big(1+\frac{6}{100}\Big)^3$
$=\text{Rs. }125000\times\Big(\frac{53}{50}\Big)^3$
$=\text{Rs. }125000\times\frac{53}{50}\times\frac{53}{50}\times\frac{53}{50}$
$=\text{Rs. }148877$
$\therefore$ C.I. paid = A - P
= Rs. 148877- Rs. 125000
= Rs. 23877.

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Question 184 Marks

Find the amount and the compound interest on Rs. 6000 for 1 year at 10% per annum compounded half-yearly.

Answer

Let the principal amount be P = Rs. 6000.
Rate of interest = 10% p.a.= 5% for half yearly.
Time (n) = 1 year = 2 half years
Now, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs}.\ 6000\times\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs}.\ 6000\times\Big(\frac{105}{100}\Big)^2$
$=\text{Rs}.\ 6000\times\Big(\frac{21}{20}\Big)\times\Big(\frac{21}{20}\Big)$
$=\text{Rs}.(15\times21\times21)$
$=\text{Rs}. 6615$
$\therefore$ CI = A - P = Rs. (6615 - 6000) = Rs. 615
Hence, the amount is Rs. 6615 and the CI is Rs. 615

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Question 194 Marks

Sheela deposited Rs. 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?

Answer

Amount deposite in the bank = Rs. 20000Rate of interest (R) = 6% p.a. or 3% half yearly
Period (n) = 1 year or 2 half years
Amount recieved after 1 year
$=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }20000\Big(1+\frac{3}{100}\Big)^2$
$=20000\Big(\frac{103}{100}\Big)^2$
$=\text{Rs. }20000\times\frac{103}{100}\times\frac{103}{100}\times= \text{Rs. }21218$

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Question 204 Marks

Divakaran deposited a sum of Rs. 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets.

Answer

Principal (p) = Rs. 6250

Rate (r) 8% pa or 4% half yearly

Period (t) = 1 year = 2 half years

Interest for the first half year $=\frac{\text{prt}}{100}$

$=\frac{6250\times4\times1}{100}$

$=\text{Rs.}\ 250$

Amount at the end of first half years = Rs. 6250 + 250 = Rs. 6500

Principal for the second half year = Rs. 6500

Interest for the second half year $=\text{Rs.}\ \frac{6500\times4\times1}{100}$

$=\text{Rs.}\ 260$

Compound Interest for one year = Rs. 250 + 260 = Rs. 510

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Question 214 Marks

Sudershan deposited Rs. 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?

Answer

Amount deposit in the bank (P) = Rs. 32000 Rate of interest (R) = 5% p.a. or $\frac{5}{4}\%$ quarterly Period (n) = 6 months or 2 quarters$\therefore$ Amount $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }32000\Big(1+\frac{5}{4\times100}\Big)^2$ $=\text{Rs. }32000\times\Big(\frac{81}{80}\Big)^2$ $=\text{Rs. }32000\times\frac{81}{80}\times\frac{81}{80}=\text{Rs. }32805$

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Question 224 Marks

A sum amounts to Rs. 23762 in 2 years at 9% per annum, compounded annually. Find the sum.

Answer

Amount (A) = Rs. 23762
Rate of interest (R) = 9%
Time (n) = 2 years
Now, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$\Rightarrow\text{Rs}.\ 23762=\text{P}\times\Big(1+\frac{9}{100}\Big)^2$
$\Rightarrow\text{Rs}.\ 23762=\text{P}\times\Big(\frac{109}{100}\Big)\times\Big(\frac{109}{100}\Big)$
$\Rightarrow\text{P}=\text{Rs}. \frac{23762\times100\times100}{100\times100}$
$\Rightarrow\text{P}=\text{Rs.}\ 20000$
$\therefore$ The principal amount is Rs. 20000.

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