Mathematics · Quantitative Aptitude

Number Operations and Properties

974 Questions

Improve your quantitative aptitude with these number operations and properties questions. The topics range from basic subtraction and division to highest common factor and roman numerals. Regular practice of these fundamentals builds speed for exams.

Basic arithmetic operationsHCF and number propertiesRoman numeral calculationsNumber series evaluation

Number Operations and Properties Questions

Multiple choice maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

$45.678\times \text{ ? }=1187.628$. Find $?$.

  1. $26$
  2. $27$
  3. $28$
  4. $29$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x$ be the missing number.


$45.678\times x=1187.628\ \Rightarrow 45.678x=1187.628\ \Rightarrow \dfrac { 45678x }{ 1000 } =\dfrac { 1187628 }{ 1000 } \quad \quad \quad \quad \quad \left{ \because \quad \dfrac { 1 }{ 10 } =0.1,\dfrac { 1 }{ 100 } =0.01,.... \right} \ \Rightarrow x=\dfrac { 1187628 }{ 1000 } \times \dfrac { 1000 }{ 45678 } \ \Rightarrow x=26$

Hence, the missing number is $26$.

Multiple choice maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

Multiply $43.09$ with $23$.

  1. $99.107$
  2. $991.07$
  3. $9910.7$
  4. $9.9107$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let us first multiply the two given numbers $43.09$ and $23$ without decimal point:
$4309\times 23=99107$
Since, $43.09$ has two decimal places, therefore the answer $99107$ should also have two decimal places that is $991.07$.
Hence, $43.09\times 23=991.07$.
Multiple choice maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

Without actual multiplication, the value $79.01 \times 79.01 + 2 \times 79.01 \times 20.99 + 20.99 \times 20.99$

  1. $10,009$
  2. $1000.06$
  3. $10,000$
  4. $1007$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$79.01\times 79.01+2\times 79.01\times 20.99+20.99\times 20.99....(1)$
$\dfrac {7901}{100}\times \dfrac {7901}{100}+2\times \dfrac {7901}{100}\times  \dfrac {2099}{100}+\dfrac {2099}{100}\times \dfrac {2099}{100}....(2)$

$7901\times 7901=(7901)^2 =(7900 +1)^2$    $\quad [\because \ 79^2=(80-1)^2\\ =80^2+1-2.80\\ =6400+1-160\\ =6241]$
$=(7900)^2 +1^2+2(7900)(1)$
$=62410000+1+15800$
$=62425801$ 

$2099\times 2099 =(2099)^2=(2100-1)^2$ $\quad [\because \ 21^2=(20+1)^2\\ =20^2+1+2.20\\ =400+40+1\\ =441]$
$=(2100)^2+1^2-2(2100)(1)$
$=4410000+1-4200$
$=4405801$

$7901\times 2099=(7900+1) (2100-1)$
$=(7900)(2100)-7900+2100-1$
$=16590000-7900+2100-1$
$=16584299\ =16584199$
from $(2)$
$\dfrac {62425801}{10000}+\dfrac {2\times 16584199}{10000}+\dfrac {4405801}{10000}$
$6242.5801+\dfrac {33168398}{10000}+440.5801$
$6242.5801+3316.8398+440.5801$
$=10,000$
Multiple choice maths introduction to euclid's geometry conditional statements and converse euclid's postulates axioms, postulates and theorems euclid's fifth postulate

By applying Euclid's division lemma $72$ and $28$ can be expressed as

  1. $28 = (72 - 16) \times 2$
  2. $72 = (28 \times 2) + 16$
  3. $72 = (28 \times 2) - 16$
  4. $16 = 72 - (28 + 2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:

According to Euclid's division lemma if $a$ and $b$ are two numbers then they can be expressed as $b=ap+r.$
Therfore,
$72$ and $28$ can be expressed as
$72=(28\times2)+16$
So, $B$ is the correct option. 

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If $2l - 3m = -1$ and $lm = 20$, then the value of $4l^{2} + 9m^{2}$ is ________.

  1. $239$
  2. $240$
  3. $241$
  4. $361$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know the identity $a^{2}+b^{2}-2ab = (a-b)^{2}$

Given, $2l-3m= -1$

Squaring on both sides, we get 

$(2l-3m)^{2}= (-1)^{2}$

$\Rightarrow 4l^{2}+9m^{2}-12lm = 1$     .....Also given that $lm =20 $

$\Rightarrow 4l^{2}+9m^{2}-12 \times 20 = 1$

$\Rightarrow 4l^{2}+9m^{2}-240 = 1$

$\Rightarrow 4l^{2}+9m^{2}= 241$

Hence, option C is correct.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $(8x)^2 + (6x)^2 = d^2$ and $d = 200$, then $8x \times 6x$ is equal to

  1. 18,200

  2. 18,500

  3. 18,900

  4. 19,200
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(8x)^2+(6x)^2=d^2$
or $64x^2+36x^2=d^2$
or $100x^2=d^2$
or $d=\sqrt {100x^2}=10x$
$10x=200$
$\therefore x=\frac {200}{10}=20$
Hence, $8x\times 6x=8\times 20\times 6\times 20=19,200$

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

If $^{ 56 }{ { P } _{ r+6 } }:^{ 54 }{ { P } _{ r+3 }}=30800$, then $r$ is

  1. $39$
  2. $41$
  3. $28$
  4. $43$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$^{ 56 }{ P } _{ r+6 }:^{ 54 }{ P } _{ r+3 }=30800$
$\cfrac { \cfrac { 56! }{ \left( 50-r \right) ! }  }{ \cfrac { 54! }{ \left( 51-r \right) ! }  } =30800$
$\cfrac { 56!\times \left( 51-r \right) ! }{ 54!\left( 50-r \right) ! } =30800$
$56\times 55\times \left( 51-r \right) =30800$
$\left( 51-r \right) =\cfrac { 30800 }{ 56\times 55 }$
$\left( 51-r \right) =10$
$r=41$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

How many $4$-letter words, with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

  1. $5040$
  2. $1000$
  3. $2500$
  4. $2060$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are $10$ letters in the word 'LOGARITHMS'.
So, the number of $4$-letter word$=$Number of arrangements of $10$ letters, taken $4$ at a time
$=$ $^{10}P _4=5040$.