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Maths MCQ

Interest · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 18 questions.

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18 questions · auto-graded multiple-choice test.

MCQ 11 Mark
Mark $(\checkmark)$ against the correct answer in the following:
A borrows Rs. 8000 at $12 \%$ per annum simple interest and B borrows Rs. 9100 at $10 \%$ per annum simple interest. In how many years will their amounts be equal?
  • A
    18 years
  • B
    20 years
  • ✓
    22 years
  • D
    24 years
Answer
Correct option: C.
22 years
22 years

$R _1=12 \%$

$R _2=10 \%$

$P _1=$ Rs. 8000

$P _2=$ Rs. 9100

Let their amounts be equal in $T$ years.

$\text { Amount }_1=\text { S.I. }_1+P_1$

$=\frac{P_1 \times r_1 \times T}{100}+P_1$

$=960 T+8000$

$\text { Amount }_2=\text { S.I. }_2+P_2$

$=\frac{P_2 \times r_2 \times T}{100}+P_2$

$=\frac{9100 \times 10 \times T}{100}+9100$

$=910 T+9100$

$\text { Amount } t_1=\text { Amount }$

$\Rightarrow 960 T+8000=910 T+9100$

$\Rightarrow 960 T-910 T=9100-8000$

$\Rightarrow 50 T=1100$

$\Rightarrow T=22$

Hence, after 22 years their amounts will be equal.
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MCQ 21 Mark
Mark $(\checkmark)$ against the correct answer in the following: A sum amounts to $Rs.3605$ in $219$ days at $5\%$ per annum. The sum is:
  • A
    $Rs. 3250$
  • ✓
    $Rs. 3500$
  • C
    $Rs. 3400$
  • D
    $Rs. 3550$
Answer
Correct option: B.
$Rs. 3500$

Amount $= Rs. 3605$
Time $=\frac{219}{365}$ days $=\frac{219}{365}$ days
Rate $= 5\%$ per annum
Amount $=\text{Sum}+\frac{\text{Sum}\times\text{Rate}\times\text{Time}}{100}$
Amount $=\text{Sum}\Big(1+\frac{\text{Rate}\times\text{Time}}{100}\Big)$
$\text{Sum}=\frac{3605}{1+\frac{5}{100}\times\frac{219}{365}}$
$=\frac{3605\times36500}{37595}$
$\text{Sum}=\text{Rs. }3500$

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MCQ 31 Mark
Mark $(\checkmark)$ against the correct answer in the following: In what time will $Rs. 6000$ amount to $Rs. 6360$ at $8\%$ interest per annum simple interest?
  • ✓
    $9\text{ months}$
  • B
    $\text{8 months}$
  • C
    $1\frac14\text{ years}$
  • D
    $1\frac12\text{ years}$
Answer
Correct option: A.
$9\text{ months}$

$P = Rs. 6000$
$A = Rs. 6360$
$S.I. = A - P = 6300 - 6000$
$= Rs. 360$
$\text{T}=\frac{\text{S.I.}\times100}{\text{P}\times\text{T}}=\frac{360\times100}{6000\times8}$
$=\frac68\text{ years}$
$=\frac67\times12\text{ months}$
$=\frac{72}{8}$
$=9\text{ months}$

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MCQ 41 Mark
Mark $(\checkmark)$ against the correct answer in the following: The simple interest on $Rs. 6250$ at $4\%$ per annum for $6$ months is:
  • ✓
    $Rs. 125$
  • B
    $Rs. 150$
  • C
    $Rs. 175$
  • D
    $Rs. 135$
Answer
Correct option: A.
$Rs. 125$

$Rs. 125$
Princepal $= Rs. 6250$
Simple interest $= 4\%$ per annum
Time $= 6$ months $=\frac12$ years
Simple interest $=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
Simple interest $=\frac{6250\times4\times1}{100\times2}$
Simple interest $=\frac{250}{2}=\text{Rs. }125$

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MCQ 51 Mark
Mark $(\checkmark)$ against the correct answer in the following: The simple interest on a sum for $5$ years is $\frac25$ of the sum. The rate percent per annum is:
  • A
    $10\%$
  • ✓
    $8\%$
  • C
    $6\%$
  • D
    $12\frac12\%$
Answer
Correct option: B.
$8\%$

Times $= 5$ years
Simple interest $=\frac25\text{P}$
$\Rightarrow\frac{\text{P}\times\text{Rate}\times\text{time}}{100}=\frac{2}{5}\text{P}$
$\Rightarrow\frac{\text{Rate}\times5}{100}=\frac25$
$\Rightarrow\text{Rate}=\frac{2\times100}{5\times5}$
$\Rightarrow\text{Rate}=8\%$

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MCQ 61 Mark
Mark $(\checkmark)$ against the correct answer in the following: A sum of $Rs. 600$ amounts to $Rs. 720$ in $4$ years. What will it amount to if the rate of interest is increased by $2\%?$
  • A
    $Rs. 724$
  • B
    $Rs. 648$
  • ✓
    $Rs. 768$
  • D
    $Rs. 792$
Answer
Correct option: C.
$Rs. 768$

Let the rate be $R \%$
$S.I. = A - P$
$= 720 - 600$
$= Rs. 120$
Time $= 4$ years
$\text{R}=\frac{10\times\text{S.I.}}{\text{P}\times\text{T}}$
$\text{R}=\frac{100\times120}{600\times4}$
$=5$
Rate of interest $= 5\%$
Now,
$\text{R}=(5+2)\%=7\%$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{600\times7\times4}{100}$
$\text{Rs. }168$
Amount $= S.I. + P$
$= 600 + 168$
$= Rs. 768$

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MCQ 71 Mark
Mark $(\checkmark)$ against the correct answer in the following: The simple interest at $x %$ per annum for $x$ years will be $Rs. x$ on a sum of:
  • A
    $\text{Rs. }\text{x}$
  • B
    $\text{Rs. 100x}$
  • ✓
    $\text{Rs. }\Big(\frac{100}{\text{x}}\Big)$
  • D
    $\text{Rs. }\Big(\frac{100}{\text{x}^2}\Big)$
Answer
Correct option: C.
$\text{Rs. }\Big(\frac{100}{\text{x}}\Big)$

Simple interest $= Rs. x$
Rate $= x \%$ per annum
Time $= x$ years
Simple interest $=\frac{\text{Princepal}\times\text{Rate}\times\text{Time}}{100}$
$\Rightarrow\text{x}=\frac{\text{Princepal}\times\text{x}\times\text{x}}{100}$
$\Rightarrow\text{Principal}=\text{Rs. }\frac{100}{\text{x}}$

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MCQ 81 Mark
Mark $(\checkmark)$ against the correct answer in the following: At simple interest a sum becomes $\frac65$ of itself in $2\frac12$ years. The rate of interest per annum is:
  • A
    $6\%$
  • B
    $7\frac12\%$
  • ✓
    $8\%$
  • D
    $9\%$
Answer
Correct option: C.
$8\%$

Let the sum be $Rs. x$
Rate of interest $= r%$
Time $=2\frac12$ years $=\frac52$ years
Amount $=\frac65\times\text{Sum}$
Rate $= ?$
Amount $=\frac65\times\text{Sum}$
Principal $+ S.I. =$ Amount
$\text{Principal}+\frac{\text{Principal}\times\text{Rate}\times\text{Time}}{100}=\frac65\times\text{Principal}$
$\Rightarrow\text{x}+\frac{\text{x}\times\text{r}\times5}{100\times2}=\frac65\text{x}$
$\Rightarrow\text{x}\Big(1+\frac{5\text{r}}{100\times2}\Big)=\frac65\text{x}$
$\Rightarrow1+\frac{\text{r}}{40}=\frac65$
$\Rightarrow\text{r}=40\times\frac15$
$\Rightarrow\text{r}=8$
So, the rate of interest is $8\%$

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MCQ 91 Mark
Mark $(\checkmark)$ against the correct answer in the following:A sum amounts to $Rs. 3626$ in $219$ days at $6\%$ per annum simple interest. The sum is:
  • A
    $Rs. 3000$
  • B
    $Rs. 3200$
  • ✓
    $Rs. 3500$
  • D
    $Rs. 3600$
Answer
Correct option: C.
$Rs. 3500$

$A = Rs. 3626$
$R = 6\%$
$T = 219$ days $=\frac{219}{365}\text{ years}$
Let the required sum be $Rs. x$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times6\times219}{100\times365}$
$=\frac{1314\text{x}}{36500}$
$\text{A}=\text{P}+\text{S.I.}$
$=\text{x}+\frac{1314\text{x}}{36500}$
$=\frac{36500\text{x}+1314\text{x}}{36500}$
$=\frac{1314\text{x}}{36500}$
But the amount is $Rs. 3626$
$\therefore\frac{37814\text{x}}{36500}=3626$
$\text{x}=\frac{3626\times36500}{37814}=3500$
The required sum is $Rs. 3500$

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MCQ 101 Mark
Mark $(\checkmark)$ against the correct answer in the following: At which sum the simple interest at the rate of $3\frac34\%$ per annum will be $Rs. 210$ in $2\frac13$ years?
  • A
    $Rs. 1580$
  • ✓
    $Rs. 2400$
  • C
    $Rs. 2800$
  • D
    None of these
Answer
Correct option: B.
$Rs. 2400$
Rate $=3\frac34\%$ per annum
$=\frac{15}{4}\%$ per annum
Time $=2\frac13$ years
$=\frac73$ years
$\text{S.I.}=\frac{\text{P}\times\frac{15}{4}\times\frac{7}{3}}{100}$
$\Rightarrow\text{P}=\frac{210\times100}{\Big(\frac{15}{4}\times\frac{7}{3}\Big)}$
$\Rightarrow\text{P}=600\times4$
$\Rightarrow\text{P}=\text{Rs. }2400$
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MCQ 111 Mark
Mark $(\checkmark)$ against the correct answer in the following: The simple interest on a sum for $5$ years is $\frac{3}{3}$ of the sum. The rate percent per annum is:
  • A
    $8\%$
  • B
    $10\%$
  • ✓
    $12\%$
  • D
    $12\frac12\%$
Answer
Correct option: C.
$12\%$
Let the sum be $Rs. x$
Let the rate be $R \%$ per annum
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times\text{R}\times5}{100}$
$=\frac{\text{Rx}}{20}$
$\therefore\frac{3\text{x}}{5}=\frac{\text{Rx}}{20}$
$\Rightarrow\text{R}=\frac{3\text{x}\times20}{5\text{x}}=12\%$
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MCQ 121 Mark
Mark $(\checkmark)$ against the correct answer in the following: In what time will $Rs. 8000$ amount to $Rs. 8360$ at $6\%$ per annum simple interest?
  • A
    $8\text{ months}$
  • ✓
    $9\text{ months}$
  • C
    $1\frac14\text{ years}$
  • D
    $1\frac12\text{ years}$
Answer
Correct option: B.
$9\text{ months}$

Let the time be t years.
Principal $= Rs. 8000$
Amount $= Rs. 8360$
Rate $= 6\%$ per annum
Amount $=\text{Princepal}\Big(1+\frac{\text{Rate}\times\text{Times}}{100}\Big)$
$\frac{8360}{8000}=1+\frac{6\times\text{t}}{100}$
$\Rightarrow\frac{8360}{8000}-1=\frac{\text{6t}}{100}$
$\Rightarrow\text{t}=\Big(\frac{8360-8000}{8000}\Big)\times\frac{100}{6}$
$=\frac{360}{8000}\times\frac{100}{6}$
$=\frac68\times12\text{ months}$
$=9\text{ months}$

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MCQ 131 Mark
Mark $(\checkmark)$ against the correct answer in the following:
$x, y$ and $z$ are three sums of money such that $y$ is the simple interest on $x$ and $z$ is the simple interest on $y$ for the same time and same rate. Which of the following is correct?
  • A
    $x y z=1$
  • B
    $z^2=x y$
  • C
    $x^2=y z$
  • ✓
    $y^2=z x$
Answer
Correct option: D.
$y^2=z x$
$y = S.I.$ on $x =\frac{ x \times R \times T }{100} \ldots (i)$
$z = S.I.$ on $y =\frac{ y \times R \times T }{100} \ldots (ii)$
Dividing equation $(i)$ by $(ii):$
$\Rightarrow \frac{y}{z}=\left(\frac{x \times R \times T}{100} \times \frac{100}{y \times R \times T}\right)$
$\Rightarrow \frac{y}{z}=\frac{x}{y}$
$\Rightarrow y^2=zX$
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MCQ 141 Mark
Mark $(\checkmark)$ against the correct answer in the following: The simple interest at $x\%$ per annum for $x$ years will be $Rs. x$ on a sum of:
  • A
    $\text{Rs. x}$
  • B
    $\text{Rs. 10x}$
  • C
    $\text{Rs. 100x}$
  • ✓
    $\text{Rs.}\Big(\frac{100}{\text{x}}\Big)$
Answer
Correct option: D.
$\text{Rs.}\Big(\frac{100}{\text{x}}\Big)$
Let the sum be $Rs. P$
Times $= x$ years
Rate of interest
$r = x \%$ per annum
$S.I. = Rs. x$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\text{P}=\frac{\text{x}\times\text{100}}{\text{x}\times\text{x}}$
$\text{P}=\text{Rs. }\frac{100}{\text{x}}$
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MCQ 151 Mark
Mark $(\checkmark)$ against the correct answer in the following: At simple interest a sum becomes $\frac{49}{40}$ of it self in $\frac12$ years. The rate of interest per annum is:
  • A
    $7\%$
  • B
    $8\%$
  • ✓
    $9\%$
  • D
    $12\%$
Answer
Correct option: C.
$9\%$

Let the sum be $Rs x$
$\text{A}=\frac{49\text{x}}{40}$
We know:
$A = P + S.I.$
$S.I. = A - P$
$=\Big(\frac{\text{49x}}{40}-\text{x}\Big)$
$=\frac{49\text{x}-40\text{x}}{40}$
$=\frac{9\text{x}}{40}$
Let the rate be $R\%$ per annum.
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\frac{9\text{x}}{40}=\frac{\text{x}\times\text{R}\times5}{100\times2}$
$\Rightarrow\text{R}=\frac{9\times20\times2}{10}$
$\Rightarrow\text{R}=9$

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MCQ 161 Mark
Mark $(\checkmark)$ against the correct answer in the following:In how much time would the simple interest on a certain sum be $0.125$ times the principal at $10\%$ per annum?
  • ✓
    $1\frac14\text{ years}$
  • B
    $1\frac34\text{ years}$
  • C
    $2\frac14\text{ years}$
  • D
    $2\frac34\text{ years}$
Answer
Correct option: A.
$1\frac14\text{ years}$
Rate $= 10\%$ per annum
Simple interest $= 0.125 \times$ Principal
$\Rightarrow\frac{\text{Princepal}\times\text{Rate}\times\text{Time}}{100}=0.125\times\text{Principal}$
$\Rightarrow\frac{\text{Time}}{10}=0.125$
$\Rightarrow\text{Time}=1.25=1\frac14\text{ years}$
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MCQ 171 Mark
Mark $(\checkmark)$ against the correct answer in the following: At what rate percent per annum simple interest will a sum double it self in $10$ years?
  • A
    $8\%$
  • ✓
    $10\%$
  • C
    $12\%$
  • D
    $12\frac12\%$
Answer
Correct option: B.
$10\%$
Let the sum be $Rs. x$ and the rate be $r\%$
Amount $= 2x$
$\Rightarrow\text{P}+\text{S.I.}=2\text{x}$
$\Rightarrow\text{P}+\frac{\text{P}\times\text{r}\times\text{t}}{100}=\text{2x}$
$\Rightarrow\text{x}\Big(1+\frac{\text{r}\times100}{100}\Big)=\text{2x}$
$\Rightarrow\frac{100+10\text{r}}{100}=2$
$\Rightarrow10\text{r}=200-100$
$\Rightarrow10\text{r}=100$
$\Rightarrow\text{r}=\frac{100}{10}$
$\Rightarrow\text{r}=10$
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MCQ 181 Mark
Mark $(\checkmark)$ against the correct answer in the following: At what rate percent per annum simple interest will a sum double it self in $10$ years?
  • A
    $8\%$
  • ✓
    $10\%$
  • C
    $12\%$
  • D
    $12\frac12\%$
Answer
Correct option: B.
$10\%$
Let the sum be $Rs. x$
Amount $Rs. 2x$
Time $= 10$ years
$S.I. = A - P = 2x - x = Rs. x$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\text{x}=\frac{\text{x}\times\text{R}\times10}{100}$
$\Rightarrow\text{R}=\frac{\text{x}\times100}{\text{x}\times100}$
$\Rightarrow\text{R}=10\%$
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