Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

The amount on Rs. $20,500$ at $7%$ per annum compunded annually for $2$ years, is:

  1. $22470$
  2. $23470$
  3. $24470$
  4. $25470$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$  $P=Rs.20,500,\,R=7\%$ and $T=2\,years$


$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=20500\times (1+\dfrac{7}{100})^2$

$\Rightarrow$  $A=20500\times (\dfrac{107}{100})^2$

$\Rightarrow$  $A=20500\times (1.07)^2$

$\Rightarrow$  $A=20500\times 1.1449$

$\therefore$    $A=Rs.23470.$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

A sum of Rs $15,000$ is invested for $3$ years at $13$ % per annum compound interest. Calculate the approx interest for the second year.

  1. $2100$
  2. $2200$
  3. $2300$
  4. $2400$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Interst for the first year 

$=Rs \cfrac{15000\times 13\times 1}{100}$
$=Rs1950$
Amount after the first year
$=Rs15000+Rs1950$
$Rs16950$
Interest for the second year
$=Rs\cfrac{16950\times 13\times 1}{100}$
$=Rs2203.5$
$=Rs2200$(approx)

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Vikram borrowed Rs. $20,000$ for $4\dfrac{1}{2}$ years at $10\%$ per annum, compound annually. How much compound interest would he pay at the end of $4\dfrac{1}{2}$ years?

  1. $30711.22$
  2. $20711.22$
  3. $40711.22$
  4. $10711.22$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given:
$P =$ Rs. $20000, r = 10\%, n = 1$ and $t =$ $4\dfrac{1}{2}$ years
$A = 20000\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$A = 20000\times 1.1^{4.5}$
$A = 20000\times 1.535561$
$A =$ Rs. $30711.22$
To find interest we use formula $A = P + I$, since $A = 30711.22$ and $P = 20000$ we have:
$A = P + I$
$30711.22 = 20000 + I$
$I = 30711.22 - 20000 = 10711.22$
Interest, I $=$ Rs. $10711.22$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Joshita is having a bank account whose principal is Rs. $12000$ and her bank compounds the interest thrice a year at an interest rate of $15\%$, how much money did she have in her account at the year's end?

  1. $18250.50$
  2. $28250.50$
  3. $38250.50$
  4. $48250.50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $P = 12000, r = 15\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 12000\left [\left (1+\dfrac{15}{100}\right)^3\right]$
$A =$ Rs. $18250.50$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If $A$ and $B$ are two sets containing four and two elements, respectively. Then the number of subsets of the set $A\times B$ each having at least three elements is

  1. $219$
  2. $256$
  3. $275$
  4. $510$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A=\left{a,b,c,d\right}$

$B=\left{x,y\right}$
$A\times B=2\times 4=8$ Elements.
Total number of subsets of $A\times B$ having $3$ or more elements $=2^8=256$
$\Rightarrow 256-(1$ null set $+$ $8$ singleton set $+$ $^8e _2$having $2$ elements$)$
$=256-1-8-28$
$=219$