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Question 12 Marks
Find the remainder when $2^{100}$ is divided by 11 .
Answer
We know that if $\mathrm{a} \equiv \mathrm{b}(\bmod \mathrm{m})$ and $0 \leq \mathrm{b} \leq \mathrm{m}$, then b is the remainder when a is divided by m . Therefore, to find the remainder when $2^{100}$ is divided by 11 , its is sufficient to find an integer $b$ such that $2^{100} \equiv b(\bmod 11)$, where $0 \leq b \leq 11$
Now,
$2^{1} \equiv 2(\bmod 11)$
$\Rightarrow 2^{2} \equiv 2 \times 2=4(\bmod 11)$
$\Rightarrow 2^{3} \equiv 2 \times 4=8(\bmod 11)$
$\Rightarrow 2^{4} \equiv 2 \times 8 \equiv 5(\bmod 11)\left[\because 2^{4} \equiv 16(\bmod 11)\right.$ and $\left.16 \equiv 5(\bmod 11) \therefore 2^{4} \equiv 5(\bmod 11)\right]$
$\Rightarrow 2^{5} \equiv 2 \times 5 \equiv 10(\bmod 11)$
$\Rightarrow 2^{5} \equiv-1(\bmod 11)[\because 10 \equiv-1(\bmod 11)]$
$\Rightarrow\left(2^{5}\right)^{20} \equiv(-1)^{20}(\bmod 11)$
$\Rightarrow 2^{100} \equiv 1(\bmod 11)$
Hence, 1 is the remainder when $2^{100}$ is divided by 11 .
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Question 22 Marks
At what rate per cent, per annum compounded annually, will the sum of money become 4 times of itself in 2 years?
Answer
Interest for 1 year $=$₹$(4320-4000)=$₹ $320$
Let rate of interest be r\%
$\therefore \frac{4000 \times r \times 1}{100}=320 \Rightarrow r=8$
$\therefore$ Rate of interest $=8 \%$
$\therefore$ Amount after 3 years $=4000\left(1+\frac{8}{100}\right)^{3}=4000(1.08)^{3}=4000 \times 1.259=$₹ $5036$
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Question 32 Marks
Mrs. Dubey borrowed ₹ $ 500000$ from a bank to purchase a car and decided to repay by monthly installments in 5 years. The bank charges interest at $8 \%$ p.a. compounded monthly. Calculate the EMI. $\left(\right.$ Given $(1.0067)^{60}=$ 1.4928)
Answer
Given $\mathrm{P}=$₹ $500000, \mathrm{n}=12 \times 5=60$ months, $\mathrm{i}=\frac{8}{1200}=0.0067$
$\therefore \mathrm{EMI}=\frac{\mathrm{P} \times i(1+i)^{n}}{(1+i)^{n}-1}=\frac{500000 \times 0.0067 \times(1.0067)^{60}}{(1.0067)^{60}-1}$
$=\frac{500000 \times 0.0067 \times 1.4928}{0.4928}=$₹ $10147.89$
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Question 42 Marks
Evaluate: $\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x$
Answer
Put $9 x^{2}-1=t \Rightarrow 18 \mathrm{xdx}=\mathrm{dt} \Rightarrow 3 \mathrm{xdx}=\frac{1}{6} \mathrm{dt}$.
When $\mathrm{x}=1, \mathrm{t}=9.1^{2}-1=8$ and when $\mathrm{x}=2, \mathrm{t}=9.2^{2}-1=35$.
$\therefore \mathrm{I}=\frac{1}{6} \int_{8}^{35} \frac{1}{t} d t=\frac{1}{6}[\log |t|]_{8}^{35}=\frac{1}{6}(\log 35-\log 8)=\frac{1}{6} \log \frac{35}{8}$.
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Question 52 Marks
A company ABC Ltd has raised funds in the form of 1,000 zero-coupon bonds worth $₹ 1,000$ each. The company wants to set up a sinking fund for repayment of the bonds, which will be after 10 years. Determine the amount of the periodic contribution if the annualized rate of interest is $5 \%$, and the contribution will be done half-yearly. Given that $(1.025)^{20}=1.6386$.
Answer
Sinking Fund, $A=$₹ $1,000 \times 1000=$₹ $1,000,000, r=5 \%$ or 0.05 , No. of years, $n=10$ years and No. of payments per year, $m=2$ (Half Yearly)
Periodic Contribution, $\mathrm{P}=\frac{A \times\left(\frac{r}{m}\right)}{\left[\left(1+\left(\frac{r}{m}\right)\right)^{n \times m}\right]-1}$
$P=\frac{1,000,000 \times\left(\frac{0.05}{2}\right)}{\left[\left(1+\left(\frac{0.05}{2}\right)\right)^{10 \times 2}\right]-1}$
$=\frac{1,000,000 \times 0.025}{1.6386-1}$
$=\frac{25,000}{0.6386}$
= ₹39,148.136~₹39,148
Therefore, the company will be required to contribute a sum of ₹ 39,148 half-yearly in order to build the sinking fund to retire the zero-coupon bonds after 10 years.
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Question 62 Marks
Find the compound interest on ₹ 7000 at $6 \%$ p.a for 18 months compounded quarterly. [Use $(1.015)^{6}=1.093$ ]
Answer
$\mathrm{P}=$₹ $7000, \mathrm{r}=6 \%$ p.a. $=1.5 \%$ quarterly $\mathrm{n}=18$ months $=6$ quarters
$\therefore$ C.I. $=7000\left[\left(1+\frac{1.5}{100}\right)^{6}-1\right]$
$=7000\left[(1.015)^{6}-1\right]$
$=7000(1.093-1)$
$=7000 \times 0.093$
= ₹ 651
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Question 72 Marks
The Production of cement by a firm in year 1 to 9 is given below:
Year123456789
Production in (Tonnes)4556789810
Calculate the trend values for the above series by the 3-yearly moving average method.
Answer
YearProduction (in Tonnes)Three yearly moving totalsThree yearly moving averages
14--
25144.67
35165.33
46186
57217
68248
79258.33
88279
910--
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2 Marks Questions - Applied Maths STD 12 Science Questions - Vidyadip