MCQ

MCQ 11 Mark

The time in which $₹.6000$ amounts to $₹.7986$ at $10\%\ p.a.$ compounded annually is

A
$2$ years
✓
$3$ years
C
$4$ years
D
$5$ years

Answer

Correct option: B.

$3$ years

Amount $(A) = ₹ .7986$
Principal $(P) = ₹ .6000$
Rate $(R) = 10\%\ p.a$.
$\therefore \frac{ A }{ P }=\left(1+\frac{ R }{100}\right)^n$
$\Rightarrow \frac{7986}{6000}=\left(1+\frac{10}{100}\right)^n$
$\Rightarrow \frac{7986}{6000}=\left(\frac{11}{10}\right)^n$
$\Rightarrow\left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^n$
$\therefore n =3$
$\therefore$ Time $=3$ years

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MCQ 21 Mark

If the number of conversion periods $\geq 2$, then compound interest is

A
less than or equal to simple interest
B
greater than or equal to simple interest
C
less than simple interest
✓
greater than simple interest

Answer

Correct option: D.

greater than simple interest

Number of conversion period $\geq 2$
The $\text {C.I.}$ is greater than simple interest $(\text {S.I}.) $

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MCQ 31 Mark

If $₹.12000 $ taken for $2$ years at $4\%$ per annum compounded quarterly, then time period and rate is

A
$n = 2, R = 16\%$
B
$n = 4, R = 1 \%$
✓
$n = 8, R = 1\%$
D
$n = 8, R = 16\%$

Answer

Correct option: C.

$n = 8, R = 1\%$

Principal $(P) = ₹. 12000$
Rate $(R) = 4\%$ p.a. or $1\%$ quarterly
Time $(n) = 2$ years or $8$ quarter

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MCQ 41 Mark

The time periods and rate for a sum taken at $8 \%$ p.a. for $1 \frac{1}{2}$ years compounded half yearly are

✓
$n=3, R=4 \%$
B
$n=6, R=2 \%$
C
$n=3, R=2 \%$
D
$n=6, R=4 \%$

Answer

Correct option: A.

$n=3, R=4 \%$

Rate $(R)=8 \%$ p.a. $=4 \%$ half$-$yearly
Time $(n)=1 \frac{1}{2}$ years $=3$ half$-$year

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MCQ 51 Mark

The compound interest on $₹.10000$ at $8\%$ per annum for $6$ months compounded quarterly is

A
$₹.408$
✓
$₹.10404$
C
$₹.404$
D
$₹.400$

Answer

Correct option: B.

$₹.10404$

Principal $(P) = ₹. 10000$
Rate $(R) = 8\%$ p.a. or $2\%$ quarterly
Period $(n) = 6$ months $= 2 $ quarters
$\therefore A = R \left(1+\frac{ R }{100}\right)^n=10000 \times\left(1+\frac{2}{100}\right)^2$
=₹. $10000 \times \frac{51}{50} \times \frac{51}{50}=₹. 10404$
$\therefore \text {C.I. = A – P} = ₹.10404 – ₹.10000 = ₹.404$

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MCQ 61 Mark

The compound interest on $₹ .5000$ at $20 \%$ per annum for $1 \frac{1}{2}$ years compounded half yearly is

A
$₹.6655$
✓
$₹.1655$
C
$₹.50$
D
$₹.1000$

Answer

Correct option: B.

$₹.1655$

Principal $(P) = ₹.5000$
Rate $(R) = 20\%$ p.a. or $10\%$ half$-$yearly
Period $(n)=1 \frac{1}{2}$ years or $3$ half$-$years
$\therefore A = P \left(1+\frac{ R }{100}\right)^n=$ ₹. $5000\left(1+\frac{10}{100}\right)^3$
$=5000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}= ₹.6655$
$\therefore \text {C.I. = A – P} = ₹.6655 – ₹.5000 = ₹.1655$

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MCQ 71 Mark

The compound interest on $₹.1000$ at $10\%$ p.a. for $2$ years is

A
$₹.190$
✓
$₹.210$
C
$₹.1210$
D
$₹.200$

Answer

Correct option: B.

$₹.210$

Principal $(P) = ₹.1000$
Rate $(R) = 10\%$ p.a.
Period $(n) = 2$ years
$A=P\left(1+\frac{R}{100}\right)^2= ₹. 1000\left(1+\frac{10}{100}\right)^2$
$= ₹. 1000 \times \frac{11}{10} \times \frac{11}{10}= ₹. 1210$
and $\text{C.I. = A – P}$
$= ₹.1210 – ₹.1000 = ₹.210$

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