Tag: sum of numbers

Questions Related to sum of numbers

A, Band C are three consecutive even intergers such that three times the first is two more the twice the third one. What is third one?

  1. 11

  2. 12

  3. 14

  4. 10


Correct Option: C
Explanation:

Let the first even consecutive integer be $x$ then the next two integers would be $x+2$ and $x+4$.


It is given that three times the first number is two more than twice the third number, therefore, we have:

$3x=2(x+4)+2\ \Rightarrow 3x=2x+8+2\ \Rightarrow 3x=2x+10\ \Rightarrow 3x-2x=10\ \Rightarrow x=10$

Thus, the third integer will be $x+4=10+4=14$.

Hence, the third integer is $14$.

If $p$ is an integer, then every square integer is of the form

  1. $2p$ or $(4p-1)$

  2. $4p$ or $(4p-1)$

  3. $3p$ or $(3p+1)$

  4. $4p$ or $(4p+1)$


Correct Option: D
Explanation:
Let $x$ be any positive integer.
For example Let $x=2\Rightarrow\,{x}^{2}=4$ is a square number is of the form $4p$ where $p\in Z$
Let $x=3\Rightarrow\,{x}^{2}=9=2\times 4+1$ is a square number is of the form $4p+1$ where $p\in Z$
Let $x=4\Rightarrow\,{x}^{2}=16$ is a square number is of the form $4p$ where $p\in Z$
Let $x=5\Rightarrow\,{x}^{2}=25=4\times 6+1$ is a square number is of the form $4p+1$ where $p\in Z$
Let $x=4p\Rightarrow\,{x}^{2}=16{p}^{2}=4\left(4{p}^{2}\right)=4q$ where $q=4{p}^{2}$ is a square number is of the form $4p$ where $p\in Z$
If $x=4p+1\Rightarrow\,{x}^{2}={\left(4p+1\right)}^{2}=16{p}^{2}+8p+1=4\left(4{p}^{2}+2p\right)+1=4q+1$ where $q=4{p}^{2}+2p$
If $x=4p+3\Rightarrow\,{x}^{2}={\left(4p+3\right)}^{2}=16{p}^{2}+24p+9=4\left(4{p}^{2}+6p+2\right)+1=4q+1$ where $q=4{p}^{2}+6p+2$
$\therefore\,$in each of the above cases,${x}^{2}$ is of the form $4p$ or $4p+1$

Given that the sum of the odd integers from $1$ to $99$ inclusive is $2500$, what is the sum of the even integers from $2$ to $100$ inclusive?

  1. 2450

  2. 2550

  3. 2460

  4. 22500


Correct Option: B
Explanation:

Odd integers $=1,3,5,7,9.....$

Sum of odd integers $=2500$
Even integers $=2,4,6,8,,10......$
In the series of even integers each term is one more than the each term of odd integers.
Hence, there are $50$ terms.
So, the sum of even integers $=$ $2500+50=2550$

The largest odd integer from  $-10$ to $0$ is:

  1. $-9$

  2. $10$

  3. $-10$

  4. $-1$


Correct Option: D
Explanation:
An odd number is an integer which is not divisible by two. If it is divided by two, then the result is a fraction. The set of odd integers is $-11,-9,-7,-5,-3,-1,1,3,5,7,.....$

The following is the set of odd integers between $-10$ and $0$:

$-9,-7,-5,-3,-1$ where $-9<-7<-5<-3<-1$

Therefore, $-1$ is the largest.

Hence, the largest odd integer from $-10$ and $0$ is $-1$.

Addition of odd integers between  $-3 \ and\ 3$ is

  1. $0$

  2. $2$

  3. $-2$

  4. $3$


Correct Option: A
Explanation:
An odd number is an integer which is not divisible by two. If it is divided by two, then the result is a fraction. The set of odd integers is $-5,-3,-1,1,3,5,7,.....$

The following is the set of odd integers between $-3$ and $3$:

$-1,1$

Therefore, the sum is $-1+1=0$

Hence, addition of odd integers between $-3$ and $3$ is $0$.

The 6th consecutive odd integer after $-5$ is

  1. $0$

  2. $-10$

  3. $-13$

  4. $7$


Correct Option: D
Explanation:

$-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9$


$6$th consecutive odd integer after $-5$, as seen from above is $7$

Addition of largest odd number and smallest even number from the integers $-5$ to $5$ is

  1. $9$

  2. $-9$

  3. $1$

  4. $-1$


Correct Option: C
Explanation:

$-5,-4,-3,-2,-1,0,1,2,3,4,5$


Largest odd number $= 5$
Smallest even number $= -4$

$\therefore$ Answer $= 5 + (-4) = 1$

Find three consecutive odd integers such that the sum of first and third integers is same as the second integer when decreased by $9$.

  1. $-9,-7,-5$

  2. $-13,-11,-9$

  3. $-15,-13,-11$

  4. $-11,-9,-7$


Correct Option: D
Explanation:

Let us say the first odd integer be $x$. The second consecutive odd integer would be $x+2$ (zit would not be $x+1$ because that would result in an even integer. The sum of two odd integers is even). The third consecutive odd integer would be $(x+2)+2$ or $x+4$.


Now, it is given that the sum of first and third integers is same as the second integer when decreased by $9$ which means:

$x+(x+4)=(x+2)-9\ \Rightarrow 2x+4=x-7\ \Rightarrow 2x-x=-7-4\ \Rightarrow x=-11$

Therefore, the first odd integer is $-11$ then the second integer is $x+2=-11+2=-9$ and the third integer is $x+4=-11+4=-7$

Hence, the three consecutive odd integers are $-11,-9,-7$.

Sum of one odd and one even integers is :

  1. Even

  2. Odd

  3. Both

  4. Can't be determined


Correct Option: B
Explanation:

An even integer is an integer that is evenly divisible by $2$, that is, division by $2$ results in an integer without any remainder. The set of even integers is:


$.....-8,-6,-4,-2,2,4,6,8,.....$

Whereas, an odd integer is an integer that is not divisible by $2$. The set of odd integers is:

$.....-9,-7,-5,-3,-1,1,3,5,7,9.....$

Now let us take an even integer say, $2$ and an odd integer say $3$, then their sum will be $2+3=5$ which is an odd integer.

Hence, the sum of one odd and one even integer is always odd.

Sum of two even integers is :

  1. Even

  2. Odd

  3. Both

  4. Can't be determined


Correct Option: A
Explanation:

An even integer is an integer that is evenly divisible by $2$, that is, division by $2$ results in an integer without any remainder. The set of even integers is:


$.....-8,-6,-4,-2,2,4,6,8,.....$

Now let us take two even integers say, $2$ and $4$, then their sum will be $2+4=6$ which is also an even integer.

Hence, the sum of even integers is always even.