0

Sum of numbers - class-VI

Attempted 0/35 Correct 0 Score 0

Sum of an even number and an odd number is always an odd number.

  1. True

  2. False


Correct Option: A
Explanation:
$2+3=5$

$16+5=21$

$Even+Odd=Odd$

Hence, it is true.

Multiplicative inverse of $\dfrac{1}{-5}$ is 

  1. $\dfrac{1}{-5}$

  2. $-5$

  3. $\dfrac{5}{1}$

  4. $Not\ defined$


Correct Option: B
Explanation:

As we know that multiplicative inverse of $\cfrac{a}{b}$ will be $\cfrac{b}{a}$.

Therefore,
Multiplicative inverse of $\cfrac{1}{-5} = \cfrac{-5}{1} = -5$

Given that a, b are odd and c, d are even. Then,

  1. $\displaystyle a^{2}-b^{2}+c^{2}-d^{2}$ is always divisible by 4

  2. $abc + bcd + cda + dac$ is always divisible by 4

  3. $\displaystyle a^{4}+b^{4}+c^{3}+d^{3}+c^{2}b+a^{2}b$ is always odd

  4. $a + 2b + 3c + 4d$ is odd


Correct Option: D
Explanation:

Let $a=1, b= 3, c= 2, d = 4$
Option D
$a+2b+3c+4d$
1+6+6+16 =29 which is odd number
In other 3 option always not correct for different values for a, b , c, d

A book has pages numbered 1 to 192 (totally 96 sheets). Some 25 sheets are pulled out of it at random. Then, the sum of these 50 numbers cannot be

  1. $1001$

  2. $1567$

  3. $2008$

  4. $3003$


Correct Option: C
Explanation:

Each of the pulled out $25$ sheets will have an odd number and an even number, back to back.
Therefore, total of numbers on each sheet is odd.
Hence, total of numbers on $25$ sheets is also odd as
when odd number is multiplied to odd number always gives odd number.
Therefore, the total cannot be $2008$, which is even.

The value of $\displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1+1/2}}}$ on simplification is

  1. 5/8

  2. 6/7

  3. 7/8

  4. 8/6


Correct Option: A
Explanation:
$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2+1}{2}}}}=$
=$\frac{1}{1+\frac{1}{1+\frac{2}{3}}}= \frac{1}{1+\frac{1}{\frac{3+2}{3}}}$
=$\frac{1}{1+\frac{3}{5}}=\frac{1}{\frac{8}{5}}$
=$\frac{5}{8}$

Successor of every even number is

  1. even

  2. prime

  3. odd

  4. none of these


Correct Option: C
Explanation:

The successor of an even number is always an odd number. For example, after 26, 27 will come.

So option C is the correct answer.

a, b, c are even numbers and x, y, z are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a\times b}{c} = x\times y$ (b) $\dfrac{a\times b}{x}=yz$ (c) $\dfrac{xy}{z} = ab$

  1. Only B

  2. Only C

  3. All the three

  4. Only B and C


Correct Option: D
Explanation:

In option B a×b is always even & xyz is always odd therefore equality not holds.

In option C ab is always even therefore abz is also even & xy is always odd hence equality not holds.
In option A xy is always odd but (a×b)/c can be odd or even therefore equality can hold in this case.

If $a$ and $b$ are odd numbers, then which of the following is even?

  1. $a+b$

  2. $a+b+1$

  3. $ab$

  4. $ab+2$

  5. None of these


Correct Option: A
Explanation:
We know the following rule :

odd + odd = even,

even + even = even,

odd + even = odd,

even + odd = odd,

odd × odd = odd.

(A) The given expression is

a + b = odd + odd = even.

(B) The given expression is

a + b + 1 = odd + odd + odd = even + odd = odd.

(C) The given expression is

ab = odd × odd = odd.

(D) The given expression is

ab + 2 = odd × odd + 2= odd + even = odd.

Thus, the correct option is (A) a+b.

Let S be a set of all even integers. If the operations:
1. addition 2. subtraction 3. multiplication 4. division
are applied to any pair of numbers from S, then for which operations is the resulting number is S?

  1. $1, 2, 3$ and $4$

  2. $1, 2$ and $3$ only

  3. $1$ and $3$ only

  4. $2$ and $4$ only


Correct Option: B
Explanation:

Addition of two even numbers; subtraction of two even numbers and product of two even numbers is an even number.

Multiplication of one odd and one even integer is always :

  1. Even

  2. Odd

  3. Can't be determined

  4. None of the above


Correct Option: A
Explanation:

Even integer is an integer having unit digit as a multiple of $2$

So on multiplication with odd integer, unit digit will still remain a multiple of $2$, hence, multiplication of odd and even integer gives even integer.

$a, b, c$ are even numbers and $x, y, z$ are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a \times b}{c} = x \times y$  (b)  $\dfrac{a \times b}{x} = yz$  (c) $\dfrac{xy}{z} = ab$

  1. Only a

  2. Only c

  3. All the three

  4. Only b and c


Correct Option: D
Explanation:
Use the rules that 
(a) product of two odd or even numbers are odd and even respectively.
(b) the quotient of the odd number $ \div $ even number or vice-versa may or may not be strictly odd or even.

a. $ ab $ must be a even number and $ \dfrac{ab}{c} $ may be even number or a fraction or odd number and $ xy $ must be a odd number
therefore it can be justified in some case.

b. $ \dfrac { ab }{ x } $ is always a fraction or even number and $ yz $ is always a odd number
therefore it cannot be justified in any case

c.$ \dfrac { xy }{ z } $ maybe odd number or fraction but $ ab $ will always be even.
therefore it cannot be justified.

Relationships given in b and c can't be justified at any cost.

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

  1. an even integer

  2. an odd integer always disable by $3$

  3. an odd integer not divisible by $3$

  4. an odd integer may or not be divisible by $3$


Correct Option: D
Explanation:

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.

If $P$ is an integer between $0$ and $9,R-P=16229$ and $R$ divisible by $11$, then find the value of $\dfrac {P+R-1}{3}$

  1. $5014$

  2. $4514$

  3. $5414$

  4. $5114$


Correct Option: C
Explanation:
$ R-P = 16229 $
P be b/w $0\& 9 $
and R is divisible by 11
So, $ R = \dfrac{16229+P}{11} $
(and Reminder = 0)
So, $ \Rightarrow (\dfrac{11+P}{11}) $ so $ P = 7 $
and $ R = 16229+7 $
$ = 16236 $
So $ \dfrac{P+R+1}{3} $
$ \Rightarrow \dfrac{16236+7-1}{3} $
$ \Rightarrow \dfrac{16242}{3} = 5414 $ 
Option C is correct 

Difference of squares of two odd integers is always divisible by ?

  1. 3

  2. 5

  3. 16

  4. 8


Correct Option: D
Explanation:
Let two consecutive odd integers be $2k +1$ and $2k+3$, where $k$ is any integer
Difference$={\left(2k+3\right)}^{2}-{\left(2k+1\right)}^{2}$
$=4{k}^{2}+9+12k-4{k}^{2}-1-4k=8k$
Hence, the difference is always divisible by $8$.

Consider $n={21}^{52}$, then

  1. number of even divisors of $n$ is $704$

  2. number of odd divisors of $n$ is $2809$

  3. last two digits of $n$ is $41$

  4. number of even divisors of $n$ which are multiple of $9$ is $2705$


Correct Option: B,C
Explanation:

Let,we have


$n = {21^{52}}$

can be written as $n = {\left( {7 \times 3} \right)^{52}}$

$n = {7^{52}}{.3^{52}}$

We know, no. of total divisors of any number

$k = {p^m}.{q^n}$

Total divisors$=(m+1)\,(n+1)$

so, similary here

odd divisors$=(52+1)(52+1)=2809$

Hence the option $(B)$ is correct

But again 

For last two digit

$n = {21^{52}} = {\left( {20 + 1} \right)^{52}}$

${\left( {20 + 1} \right)^{52}}{ = ^{52}}{C _1}{\left( {20} \right)^{52}} + .....{ + ^{52}}{C _{51}}{\left( {20} \right)^1}{ + ^{52}}{C _{52}}{\left( {20} \right)^0}$

For last two digit we notice last two terms 

$=^{52}{C _{51}}\left( {20} \right) + 1$

$ = 52 \times 20 \times 1$

$=1041$

$1041$ has last two digit is $41$

so, option $(C)$ is also correct

Hence both the option $(B)$ and $(C)$ are correct.

The smallest odd number formed by using the digits $1,0,3,4$ and $5$ is

  1. $10345$

  2. $10453$

  3. $10543$

  4. $10534$


Correct Option: A
Explanation:

The smallest odd number using digits $1,0,3,4,5$


$\rightarrow $ We have five digits and we have to make smallest five digit odd numbers.


$\rightarrow$ So, the number cannot start with $0$

$\rightarrow$ For the smallest it should be start with $1$

$\rightarrow$ and second space should be $0$

    $1\\ \overline { 1st } $  $0\\ \overline { 2nd }$  $\;\\ \overline { 3rd } $  $\;\\ \overline { 4rt } $  $\;\\ \overline { 5th } $

$\rightarrow$ Now two space are filled and $3$ are left.

$\rightarrow$ For smallest third place for should be $3$ 

          $\underline { 1 } \underline { 0 } \underline { 3 } \underline {  } \underline {  } $

$\rightarrow $ Now two places are left for and no. should be odd so, last digit should be $5$

So, the number $=10345.$

The integer just below $(\sqrt{53}+7)^{11}-2\times 7^{11}$ is 

  1. Divisible by exactly $4$ primes factors

  2. Divisible by exactly $3$ primes factors

  3. is divisible by $7$

  4. has $53$ as its only two digit prime factor


Correct Option: A

Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

  1. $192$

  2. $375$

  3. $400$

  4. $720$


Correct Option: D
Explanation:

we have the number $0,1,2,3,5,7$

Now the digit should be odd and hence last digit should be 
filled with an odd number 
$ \Rightarrow $ Number of way to filled last number $= 4$ $({\text{i}}{\text{.e}}{\text{. }}1,3,5,7)$
$ \Rightarrow $ Number of way to filled third digit  $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled second digit $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled first digit $= 5$ $({\text{i}}{\text{.e}.}1,2,3,5,7)$
$ \Rightarrow $ Total 4digit number $=4\times6\times6\times5$
$= 720$
hence,
Opton $D$ is correct answer.

The number of even proper divisor of 1008 is

  1. 18

  2. 17

  3. 23

  4. 9


Correct Option: C
Explanation:

We have,

$ 1008=2\times 2\times 2\times 2\times 3\times 3\times 7 $

$ ={{2}^{4}}\times {{3}^{2}}\times {{7}^{1}} $

$ =2\left( {{2}^{3}}\times {{3}^{2}}\times {{7}^{1}} \right) $

Then, the number of even divisors

$ =\left( 3+1 \right)\left( 2+1 \right)\left( 1+1 \right) $

$ =24 $

But the above divisors also contain the 1008 which is not a proper divisors

Then number of proper divisors$=23$

Hence, this is the answer.

The product of two odd numbers is

  1. An even numbers

  2. An odd number

  3. Cannot be determined

  4. None of these


Correct Option: B
Explanation:

An odd number is an integer which is not a multiple of two. For example $1,3,5,7,9,11......$


Let us take two odd numbers $a=3$ and $b=5$ and the product of $a$ and $b$ is as follows:

$a\times b=3\times 5=15$

Since $15$ is not a multiple of two, therefore, $15$ is also an odd number.

Hence, the product of two odd numbers is an odd number.

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.

If n is an integer, which of the following cannot be odd?

  1. $n+3$

  2. $n+1$

  3. $2n$

  4. $3n$


Correct Option: C
Explanation:

We know $2n=2\times n$

So, $2n$ is a multiple of $2$. Hence, it can not be an odd number.
It is always even.

If $|a|$ denotes the absolute value of an integer, then which of the following are correct?
1.$|ab| = |a| |b|$
2. $|a+b| \le |a|+|b|$
3. $|a-b| \ge| |a| -|b||$
Select the correct answer using the code given below.

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3


Correct Option: D
Explanation:

Given $\left| a \right| $ is the absolute value of an integer,

From the definition,
$\left| a \right| =a$ if $ a\ge 0$,
$=-a\quad $ is $a\le0$
$\therefore$ $\left| ab \right| =\left| a \right| \left| b \right| $ where $a,b$ are real numbers.
We know that from the triangle inequality sum of any two sides is always greater than the third side,
i.e.,$\left| a+b \right| \le \left| a \right| +\left| b \right| $,
We can also prove by considering 
Absolute part of the difference between any two sides is always less than the third side,
$\Longrightarrow \left| a-b \right| \ge \left| \left| a \right| -\left| b \right|  \right| $

The difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$. What is the difference between the two digits of number.

  1. $3$

  2. $2$

  3. $1$

  4. Cannot be determined

  5. None of these


Correct Option: C
Explanation:
Let the unit's digit be $y$ and ten's digit be $x$.

Then, the number $= 10x + y$. When we interchange the digits, the number will be $10y + x$.

Now, it is given that the difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$, therefore, we have:

$(10x+y)−(10y+x)=9\\ \Rightarrow 9x-9y=9\\ \Rightarrow 9(x-y)=9\\ \Rightarrow x-y=\frac { 9 }{ 9 } \\ \Rightarrow x-y=1$

Hence, the difference between the two digits of number is $1$.

What will come in place of the question mark $(?)$ in the following question?
$34.667-15.597-8.491-0.548=?$

  1. $14.403$

  2. $10.031$

  3. $18.301$

  4. $21.043$

  5. None of these


Correct Option: B
Explanation:

Let the missing place in the given question be $x$, then we have:


$34.667-15.597-8.491-0.548=x\ \Rightarrow \dfrac { 34667 }{ 1000 } -\dfrac { 15597 }{ 1000 } -\dfrac { 8491 }{ 1000 } -\dfrac { 548 }{ 1000 } =x\quad \quad \quad \quad \quad \left{ \because \quad \dfrac { 1 }{ 10 } =0.1,\dfrac { 1 }{ 100 } =0.01,.... \right} \ \Rightarrow \dfrac { 34667-15597-8491-548 }{ 1000 } =x\ \Rightarrow \dfrac { 34667-(15597+8491+548) }{ 1000 } =x$

$\Rightarrow \dfrac { 34667-24636 }{ 1000 } =x$

$\ \Rightarrow \dfrac { 10031 }{ 1000 } =x\ \Rightarrow x=10.031$

Hence, $34.667-15.597-8.491-0.548=10.031$

Find three consecutive even integers such that the sum of first two integers is same as the sum of third integer and $6$.

  1. $4,6,8$

  2. $6,8,10$

  3. $8,10,12$

  4. $10,12,14$


Correct Option: C
Explanation:

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


Now, it is given that the sum of first two integers is same as the sum of the third integer and $6$ which means:

$x+(x+2)=(x+4)+6\ \Rightarrow 2x+2=x+10\ \Rightarrow 2x-x=10-2\ \Rightarrow x=8$

Therefore, the first even integer is $8$ then the second integer is $x+2=8+2=10$ and the third integer is $x+4=8+4=12$

Hence, the three consecutive even integers are $8,10,12$.

- Hide questions