Mathematics · Quantitative Aptitude

Number Sums and Series

268 Questions

Number sums and series questions test the ability to find patterns and calculate the sum of number sequences. These problems are a staple in the quantitative aptitude sections of various competitive exams. Practice this collection to improve speed and accuracy in solving arithmetic and geometric series problems.

Sum of consecutive integersGeometric series sumsRatio and proportion sumsOdd and even number propertiesRoman numeral calculations

Number Sums and Series Questions

Multiple choice maths numbers and place value forming numbers formation of greatest and smallest numbers identifying the largest and smallest numbers with given digits

Find the sum of smallest three digit even number and greatest three-digit odd number using the digits $6, 2, 3$

  1. $859$
  2. $958$
  3. $598$
  4. $985$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The three digit numbers that can be formed using the digits $6,2,3$ are $623,632,263,236,362,326$ in which the odd numbers are $623,263$ and the even numbers are $632,236,362,326$.

The smallest even number is $236$ and the greatest odd number is $623$ and their sum can be determined as:

$236+623=859$

Hence, the sum of smallest three digit even number and greatest three-digit odd number using the digits $6,2,3$ is $859$.
Multiple choice maths tenths and hundredths more money decimal fractions in the units of currency, length, weight, capacity understanding tenth and hundredth part of a number

A certain sum consists of $x$ pounds $y$ shillings, and it is half of $y$ pounds $x$ shillings; find the sum.

  1. Pounds $6$. $13s$.
  2. Pounds $6$. $25s$.
  3. Pounds $7$. $13s$.
  4. Pounds $7$. $25s$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the sum be 20x + y = 1/2 * (20y + x). Solving this: 40x + 2y = 20y + x, which simplifies to 39x = 18y, or 13x = 6y. For integer values, x=6 and y=13. The sum is 6 pounds and 13 shillings.

Multiple choice maths numbers and place value face value of digit large numbers general form of number

What is the sum of all integers between $50$ and $350$ which have $1$ as the units digit?

  1. $5880$
  2. $5985$
  3. $6230$
  4. $6800$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sequence $51+61+71...+341$ is an arithmetic progression.

$\Rightarrow$  To find the sum of n terms of an AP we use the formula.
$\Rightarrow$  Here, $n=30,\,a=51$ and $d=10$.
$\therefore$   $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\therefore$   $S _n=\dfrac{30}{2}[2\times 51+(30-1)10]$

$\therefore$   $S _n=15[102+290]$
$\therefore$   $S _n=15\times 392$
$\therefore$   $S _n=5880$

Multiple choice maths negative numbers and integers odd and even numbers operations on integers different types of numbers

If the number of consecutive odd integers whose sum can be expressed as $50^2 - 13^2$ is k then k, can be 

  1. 33

  2. 35

  3. 37

  4. 39
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sum of odd $n$ consecutive numbers $n^2$

$\therefore (1+3+5\dots\dots (2n-1))=n^2$
where $n$ represents the number of terms.
$\therefore 50^2=1+3+5\dots 99=50\text{ }terms$
$\therefore 13^2=1+3+5\dots 25=13\text{ }terms$
$\therefore 50^2-13^2$$=(1+3+5\dots 99)-(1+3+5\dots 25)\=(27+29\dots 99)\ =37\text{ }terms.$

Multiple choice maths negative numbers and integers odd and even numbers operations on integers different types of numbers

The sum of even numbers between $1$ and $31$ is:

  1. $6$
  2. $28$
  3. $240$
  4. $512$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let ${S} _{n}=(2+4+6+.....+30)$. This is an A.P in which $a=2,d=2$ and $l=30$
Let the number of terms be $n$. Then,
$a+(n-1)d=30$
$\Rightarrow$ $2+(n-1)\times 2=30$
$\Rightarrow$ $n=15$
$\therefore$ ${S} _{n}=\cfrac{n}{2}(a+l)=\cfrac{15}{2}\times (2+30)=(15\times 16)=240$.

Multiple choice maths hcf-lcm introduction to multiples multiples lcm

The sum of the first five multiples of $6$  is

  1. $90$
  2. $60$
  3. $30$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

first five multiple of 6 are 

$6\times 1=6$
$6\times 2=12$
$6\times 3=18$
$6\times 4=24$
$6\times 5=30$
Their sum will be $6+12+18+24+30=90$
So correct answer will be option A

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

If the sum of the first n integers is 15. What is n? (use Gauss method)

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Gauss formula,
$S _n = \dfrac{n}{2} (n + 1)$
$15 = \dfrac{n}{2} (n + 1)$
$30 = n (n + 1)$
$30 = n^2 + n$   ....... (1)
By factorization we can write equation (1) as
$n^2 - 6n + 5n - 30 = 0$
$n(n - 6) + 5 (n - 6) = 0$
$n = 6, - 5$
$\therefore$ The n integers will start from -5 to 6.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If the sum of two integers is $-2$ and their product is $-24$, the numbers are

  1. $6$ and $4$
  2. $-6$ and $4$
  3. $-6$ and $-4$
  4. $6$ and $-4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $p$ and $q$ be the required integers

If $p+q$ and $pq$ are known then quadratic equation corresponding to roots as $p$ and $q$ is given by,
$x^2-(p+q)x+pq=0$
$\Rightarrow x^2+2x-24=0$, substitute the given values
$\Rightarrow x^2+6x-4x-24=0$, split the middle term
$\Rightarrow (x^2+6x)+(-4x-24)=0$, group pair of terms
$\Rightarrow x(x+6)-4(x+6)=0$, factor each binomials 
$\Rightarrow (x+6)(x-4)=0$, factor out common factor 
$\Rightarrow x=-6$ or $x=4$, set each factor to $0$

Hence $p$ and $q$ are $-6$ and $4$
Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

The sum of first 10 natural numbers is____.

  1. 100

  2. 55

  3. 50

  4. 90
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The series of first 10 natural numbers is an arithmetic progressions with first tern as 1 and common difference 1. So the sum of the series will be Sn = n/2 { 2a+ ( n-1 ) d } where n is the number of terms in the series, a is the first term and d is the common difference.

S10= 10/2 { 2(1) + ( 10-1 ) 1 }

     = 5 ( 2+9)

    = 5 ( 11 )

    = 55

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

The sum of two numbers is $4000$. $10\%$ of one number is $\displaystyle 6\frac{2}{3}$ $\%$ of the other The difference of the number is

  1. $600$
  2. $800$
  3. $1025$
  4. $1175$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let one number be $x$. 

Then the other number $= 4000 - x$
Given, $10\%$ of $\displaystyle x=6\frac{2}{3}\%$ of $ (4000-x)$
$\displaystyle \Rightarrow \frac{10}{100}\times x=\frac{20}{3}\times \frac{1}{100}\times (4000-x)$
$\Rightarrow 10x=\dfrac{20}{3}\times 4000-\dfrac{20x}{3}$
$\displaystyle \Rightarrow 10x+\frac{20x}{3}=\frac{20}{3}\times 4000$
$\Rightarrow \dfrac{50x}{3}=\dfrac{20}{3}\times 4000$
$\displaystyle \Rightarrow x=\frac{20\times 4000}{50}=1600$
The two numbers are $1600$ and $2400$. 
$\displaystyle \therefore$ Their difference is $2400 - 1600 = 800$.