Tag: maths

Questions Related to maths

The column sum in an incidence matrix for a simple graph is ________________.

maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

  1. depends on number of edges

  2. always greater than 2

  3. equal to 2

  4. equal to the number of edges

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Correct Option: C

Which of the following ways can be used to represent a graph?

maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

  1. Adjacency List and Adjacency Matrix

  2. Incidence Matrix

  3. Adjacency List, Adjacency Matrix as well as Incidence Matrix

  4. None of the mentioned

Show answer
Correct Option: C

In an isosceles $\Delta A B C$ the base $A B$ is produced both the ways to $P$ and $Q$ such that $A P \times BO = A C ^ { 2 }$ then $\Delta A P C \sim \Delta B C Q$

maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

  1. True

  2. False

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Correct Option: A

Mr Bittu deposits a certain sum of money each month in a recurring deposit account. If the rate of interest is $8\%$ per annum and Mr Bittu gets Rs $8,088$ from the bank after $3$ years. Find the value of his monthly instalment.

maturity value of recurring deposits banking maths

  1. $400$

  2. $800$

  3. $200$

  4. $140$

Show answer
Correct Option: C
Explanation:

Let the monthly installment be $P$

Given, maturity value $=8088$, $n=3$ years $=3\times 12=36$ months, $r=8\%$
Interest $=\dfrac {Pn(n+1)r}{2400}$
$=\dfrac {P\times 36\times 37 \times 8}{2400}$
$=\dfrac {111P}{25}$
We know, amount $=Pn+\dfrac {111P}{25}$
$8088=36P+\dfrac {111P}{25}$
$=\dfrac {1011P}{25}$
$\therefore 8088= \dfrac {1011P}{25}$
$\therefore P=200$
Therefore, monthly installment  is Rs. $200$.

Divya has a recurring deposit in a bank at $ 5\%$ per annum simple interest. If she pay monthly installment of Rs. $2000$ for annually. Find her Maturity value.

maturity value of recurring deposits banking maths

  1. $6049.931$

  2. $6249.931$

  3. $6149.931$

  4. $6549.931$

Show answer
Correct Option: A
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\dfrac{-1}{3}}}$ 
R $=$ Monthly Installment $= 2000 $
i $=$ Rate of Interest $5\%$
n $=$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{2000 \times [(1+5/400)^{n} - 1]}{1-(1+5/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{2000\times\dfrac{5}{400}}{1-(\dfrac{405}{400})^{-1/3}}$
$M = 6049.931$

Sheela has a recurring deposit in a bank at $2\%$ per annum simple interest. If she pay monthly installments Rs.$200$ for annually. Find her Maturity value.

maturity value of recurring deposits banking maths

  1. $621.99$

  2. $611.99$

  3. $631.99$

  4. $601.99$

Show answer
Correct Option: D
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 200$ 
$i =$ Rate of Interest/ 400 $= \dfrac {2}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{200 \times [(1+2/400)^{n} - 1]}{1-(1+2/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{200\times\dfrac{2}{400}}{1-(\dfrac{402}{400})^{-1/3}}$
$M = 601.99$

______ type of account is operated by salaried persons and petty traders.

maturity value of recurring deposits banking maths

  1. fixed deposit

  2. save deposit

  3. recurring deposit

  4. current deposit

Show answer
Correct Option: C
Explanation:

Recurring deposit type of account is operated by salaried persons and petty traders.

Rita has a recurring deposit in a bank at $4\%$ per annum simple interest. If she pay monthly installment s Rs.$1000$ for annually. Find her Maturity value.

maturity value of recurring deposits banking maths

  1. $3019.97$

  2. $4019.97$

  3. $2019.97$

  4. $5019.97$

Show answer
Correct Option: A
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R $= Monthly Installment $= 1000 $
$i =$ Rate of Interest/ 400 $= \dfrac {4}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{1000 \times [(1+4/400)^{n} - 1]}{1-(1+4/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{1000\times\dfrac{4}{400}}{1-(\dfrac{404}{400})^{-1/3}}$
$M = 3019.97$

Ahmed has a recurring deposit account in a bank.He deposits Rs $5,500$ per month for 2 years.If he gets Rs $1,46,437.5$ at the time of maturity. Find the interest paid by the bank.

maturity value of recurring deposits banking maths

  1. $10.5$%

  2. $30$%

  3. $35$%

  4. None of the above

Show answer
Correct Option: A
Explanation:

Given, maturity value $=146437.5$, monthly installment $=5,500$, $n=2$ years $=2\times 12=24$ months

Interest $=\dfrac {Pn(n+1)r}{2400}$
$=\dfrac {5500\times 24 \times 25 \times r}{2400}$
$=1375r$
Therefore, amount $=(Pn+1375r)$
$\Rightarrow 146437.5=(5500\times 24+1375r)$
$\Rightarrow 146437.5=132,000+1375r$
$\Rightarrow 14437.5=1375r$
$\Rightarrow  r=10.5$

Nithya deposited Rs. $100$ per month for $12$ months in a bank's recurring deposit account. If the bank pays interest at the rate of $10\%$ per annum, find the amount she gets on maturity.

maturity value of recurring deposits banking maths

  1. $1205$

  2. $1265$

  3. $1345$

  4. $1450$

Show answer
Correct Option: B
Explanation:

Given: $P = 100, T(n) = 12$ months, $R = 10\%$
Equivalent principal for $12$ months $=$ $100\times \dfrac{n(n+1)}{2}$
$=$ $100\times\dfrac{12(13)}{2}$
$= 50 \times  12 \times  13$
Interest $=$ $\dfrac{PRT}{100}$
$=$ $\dfrac{50\times 12 \times 13\times 10\times 1}{100\times 12}$
$=$ $65$
Maturity amount $=$ P$\times$T $+$ Interest
$=$ $100 \times 12 + 65$
$=$ Rs. $1265$