MATHS M.C.Q. [1 Marks Each]

Addition is associative for whole numbers. i.e. $a + (b + c) = (a + b) + c$
$\therefore  52 + (53 + 55) = (52 + 53) + 55$
$\therefore$ The missing terms are $52, 53.$

$ 1, 2, 3, 4, 5, .....$ are called "natural numbers" and are also known as counting numbers.

By regrouping, we get
$4 \times 378 \times 25 = 4 \times 25 \times 378$
$= 100 \times 378$
$= 37800$
Hence, the correct option is $(a).$

If two numbers are equal, then their $LCM$ is equal to their $HCF.$

 Successor of $999 = 999 + 1 = 1000$
Predecessor of $1001 = 1001 - 1 = 1000$
Now,
Product$ = ($Successor of $999) × ($Predecessor of $1001)$
$= 1000 × 1000$
$= 1000000$
$=$ One million
Hence, the correct option is $ (c).$

The sum of any two whole numbers and the product of the same are not necessarily equal.
Eg. $2 + 2 = 4 = 2 \times 2,$ but $3 + 3 = 6$ and $3 \times 3 = 9$ which are not equal.
Rest all options depicts commutative property which whole numbers satisfy.

Since, factors of,
$91 = 1 \times 7 \times 13$
$81 = 1 \times 3 \times 3 \times 3 \times 3$
$87 = 1 \times 3 \times 29$
$97 = 1 \times 97$
Thus, $81, 87$ and $91$ all are not prime numbers.

Product of two numbers $= HCF$ of two numbers $\times \ LCM$ of two numbers.
$\Rightarrow 1530 = 15 \times LCM$ of two numbers
$\Rightarrow LCM$ of two numbers $= 153015 = 102$
$\Rightarrow 1530 = 15 \times LCM$ of two numbers
$\Rightarrow LCM$ of two numbers $= 153015 = 102$

Here, $n + 35 = 101.$
Adding $-\ 35$ on both sides, we get
$n + 35 + (-35) = 101 + (-35)$
$n + 0 = 66$
$n = 66$
Hence, the correct option is $(d).$

Natural numbers start at $1$ and as we add $1$ to a natural number, we get the next natural number.
So if we keep adding $1,$ we keep getting next natural numbers, so the natural numbers never ends.

Successor of $1001 = 1001 + 1 = 1002$

Given expression:
$= 3640 - 14 ÷ 7 \times 2$
$= 3640 - 2 \times 2 [$performing division$]$
$= 3640 - 4 [$performing multiplication$]$
$= 3636 [$performing subtraction$]$

We have:
$15 + 5 ÷ 5 \times 2$
$= 15 + 1 \times 2 ($Performing division$)$
$= 15 + 2 ($Performing multiplication$)$
$= 17$

$1$ is the only number which when multiplied to any number $82,$ gives the same number.

Associative property for addition states that for any three numbers,
it does not matter which two are added first and then their sum is added to the third integer the final result will be same$\therefore a + (b + c) = (a + b) + c$

Here, we have to tell that what least number must be added to $10056$ to get a number exactly divisible by $23.$
So, first we will divide $10056$ by $23.$

Remainder $= 5$
Required number $= 23 - 5 = 18$
So, $18$ must be added to $10056$ to get a number exactly divisible by $23.$
i.e., $10056 + 18 = 10074$​​​​​​​

Hence, $10074$ is exactly divisible by $23.$

$LHS = 0 + (8 + 5) = 0+ 13 = 13$
$RHS = (0 + 8) + 5 = (0 + 8) + 5 = 13$
$LHS = RHS$