Mathematics · Quantitative Aptitude

Number Operations and Properties

896 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 hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The G.C.D. of two whole numbers is $5$ and their L.C.M. is $60$. If one of the numbers is $20$, then the other number would be

  1. $23$
  2. $13$
  3. $16$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If we are given two numbers $N _1$ and $N _2$ and their $G.C.D$ and $L.C.M.$.

then by property of numbers $N _1$$\times$$N _2=G.C.D$ $\times$ $L.C.M.$

Here Given:
$N _1=20$
$G.C.D.=5$
and $L.C.M=60$
Let, $N _2=x$

then from  above relation
$20$$\times$$x=5$$\times$$60$

$=>x=\dfrac{300}{20}$

$=>x=15$

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The solution of: $8\mod x\equiv 6\mod 14$ is,

  1. ${8, 6}$
  2. ${6, 14}$
  3. ${6, 13}$
  4. ${8, 14}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Solution:-
$8x \equiv 6 \left( mod \ 14 \right)$

$\because \; gcd \left( 8, 14 \right) = 2 \text{ divides } 6$

To find solutions, we first solve

$8x − 14y = 6$

By trial and error method, we find a solution

$\left( x, y \right) = \left( 6, 3 \right)$

This means that $x \equiv 6 \left( mod \ 14 \right)$ is a solution

To the congruence $8x \equiv 6 \left( mod \ 14  \right)$

$\therefore$ Incongruent solutions are,

$x = 6 +\left ( k \times \dfrac{14}{2} \right );  k = 0, 1$

$\therefore \; x = 6, 13$

Hence option $C$ is the answer.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

What is the least number by which $825$ must be multiplied in order to produce a multiple of $715 ?$

  1. $13$
  2. $15$
  3. $11$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$825=3\times5\times5\times 11$

$715=5\times11\times13$
In the factor of both numbers, $13$ is not common. 
Hence, the least number by which $825$ must be multiplied in order to produce a multiple of $715=13.$

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

Find the G.C.F of $2541$ and $3102$ in the scale of seven.

  1. $87$
  2. $54$
  3. $33$
  4. $85$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Prime factors of $2541$ are $3, 7, 11, 11$. Prime factorization of $3102$ in exponential form is

$2541=3^1\times 7^1\times 11^2$


Prime factors of $3102$ are $2, 3, 11, 47$. Prime factorization of $3102$ in exponential form is

$3102=2^1\times 3^1\times 11^1\times 47^1$

We found the factors and prime factorization of $2541$ and $3102$. The biggest common factor number is the $GCF$ number.


So, the greatest common factor $2541$ and $3102$ is $3\times 11=33$.

Hence, this is the answer.

Multiple choice roman numbers and numbers upto hundred roman numerals knowing our numbers maths

Write $39$ in roman numbers

  1. XXXIX

  2. XXXXI

  3. XXXX

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To convert $39$ to Roman Numerals we need to split it up into place values (ones, tens, hundreds, etc.), like this:

Place Value Number Roman Numeral
Tens            $30$ $XXX$
Ones               $9$ $IX$


Please note, we skipped place values that equal 0.

You then combine them all together (starting from the top) to get $XXXIX.$
So option A is the correct answer.

Multiple choice roman numbers and numbers upto hundred roman numerals knowing our numbers maths

The product obtained when $13$ is multiplied by $7$(in roman number ) is 

  1. XXCI

  2. LXI

  3. XCI

  4. CI
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
 $13\times 7=91$
Break the number $91$ (decompose it) into place value subgroups: 
$91 = 90 + 1;$

Convert each subgroup: $90 = 100 - 10 => C - X = XC; 1 = I;$

Wrap up the Roman numeral: 
$91 = 90 + 1 = XC + I = XCI;$
$ XCI$ is a group of numerals in additive and subtractive notation which will be $91.$

So option C is the correct answer.
Multiple choice roman numbers and numbers upto hundred roman numerals knowing our numbers maths

IX + XV + XX = ______

  1. 45

  2. 35

  3. 44

  4. 40
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Roman numerals are represented by the following letters:

I = 1

V = 5

X = 10

L = 50

C = 100

D = 500

M = 1000

If a numeral is followed by another numeral of lower denomination, the two are added together;

if it is preceded by one of lower denomination, the smaller numeral is subtracted from the greater.

 

9 + 15 + 20 = 44

Multiple choice roman numbers and numbers upto hundred roman numerals knowing our numbers maths

Numeral for $CCCXLVII$

  1. $357$
  2. $347$
  3. $367$
  4. $387$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Roman numerals are represented by the following letters:

$I = 1$

$V = 5$

$X = 10$

$L = 50$

$C = 100$

$D = 500$

$M = 1000$

If a numeral is followed by another numeral of lower denomination, the two are added together;

if it is preceded by one of lower denomination, the smaller numeral is subtracted from the greater.

CCCXLVII
$100 + 100 + 100 + 40 + 7 = 347$

Multiple choice roman numbers and numbers upto hundred roman numerals knowing our numbers maths

The product obtained when 13 is multiplied by 7 is

  1. XXCI

  2. LXI

  3. XCI

  4. CI
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Roman numerals are represented by the following letters:

I = 1

V = 5

X = 10

L = 50

C = 100

D = 500

M = 1000

If a numeral is followed by another numeral of lower denomination, the two are added together;

if it is preceded by one of lower denomination, the smaller numeral is subtracted from the greater.

 

13 $\displaystyle \times $ 7 = 91
91 $\displaystyle \rightarrow $ XCI