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

 Given that
we have to find the value of given expression
$​​​​\frac{1}{1+\frac{1}{3}}-\frac{1}{1+\frac{1}{2}}$
$=\frac{1}{\frac{1+3}{3}}-\frac{1}{\frac{1+2}{2}}$
$=\frac{3}{4}-\frac{2}{3} $
$=\frac{3\times3-2\times4}{12}$
$=\frac{1}{12}$

 Decimals having the same number of decimal places are called like decimals i.e.
decimals having the same number of digits on the right of the decimal point are known as like decimals.
For example, $16.37$ and $18.97$ are like decimals as both of these decimal numbers are written up to $2$ places of decimal.
Hence, $16.37$ and $18.97$ are like decimal fractions.

$ 16 = 2 \times 2 \times 2 \times 2$ The factors are $1, 2, 4, 8, 16.$

 Factors of the given numbers are,$4 = 2 \times 2$
$9 = 3 \times 3$
$\therefore LCM$ of $4$ and $9 = 2 \times 2 \times 3 \times 3 = 36$

By getting factors of $24 = 2 \times 2 \times 2 \times 3$
The mean is the average of the numbers. It is easy to calculate: add up all the numbers,
then divide by how many numbers there are.
Mean of the factors will be = $\frac{2 + 2 + 2 + 3}{4} = \frac{9}{4}$
So, the correct answer is option $B.$

 Example:
$HCF$ of $8$ and $21$ is $1.$
$HCF$ of $6$ and $9$ is $3.$
$HCF$ of $9$ and $36$ is $9.$
So there is no fixed number that can be the $HCF$ of an even number and an odd number.

The $LCM$ of two co-prime numbers is their product.

$6th$ multiple of $7 = 42$
$10th$ multiple of $7 = 70$
Now, $x = 2 \times (70 − 42) = 2 \times 28 = 56$

 first five multiple of $6$ are $6 × 1 = 66 × 2 = 126 × 3 = 186 × 4 = 246 × 5 = 30$
Their sum will be $6 + 12 + 18 + 24 + 30 = 90$

Prime numbers out of $17, 8, 21, 13, 41, 2, 27, 31, 51$ are $17, 13, 41, 2, 31.$
Sum of prime numbers $= 17 + 13 + 41 + 2 + 31 = 104.$

$LCM$ of $A$ and $B$ is $B$ it means that $B$ is multiple of $A.\ LCM$ of $B$ and $C$ is $C$ it means $C$ is multiple of $B$ or we can say that $C$ is multiple of $A$ also.
So $LCM$ of $A, B, C$ is $C$

 Factors are $17 = 1 \times 17$
$5 = 1 \times 5$
$\therefore LCM$ of $17$ and $5 = 1 \times 17 \times 5 = 85$

$59 = 1 \times 59$
$1$ and $59$ are the factors.So, options $A$ and $B$ are correct.

 A number is divisible by $5$ if its last digit is either $0$ or $5$ out of which $5$ is maxim
$\therefore b = 5$
A number is divisible by $3$ if the sum of its digits is divisible by $3$
$2 + 3 + 4 + 5 + 6 + 0 + 5 = 25$
So, we can add maximum $8$ to $25$ which will give us $33$ which is divisible by $3$
$\therefore a = 8$
Now, $a + b = 8 + 5 = 13$
Hence, the correct answer is option $(b).$

 The time after which all the three will be together will be $LCM$ of
$15\frac{1}{6}, 16\frac{1}{4}, 18\frac{2}{3}$
$LCM$ of $\frac{91}{6}, \frac{65}{4}, \frac{56}{3}=\frac{\text{LCM of 91,65,56}}{\text{HCF of 6,4,3}}=3640$ second $=1$ hour $40$ minits

 Factor $\times $ factor $=$ product

Prime numbers between $60$ and $75$ are $61, 67, 71,$ and $73.$
Their sum is given by:
$61 + 67 + 71 + 73 = 272$

$0.35=\frac{35}{100}=\frac{7}{20}$

 Every number is a factor and a multiple of itself.
For example, $10$ has a factor $10$ as well as a multiple $10.$

Here $1\%$ can be written as $\frac{1}{100}$
So, $44\%\Rightarrow(44\times1)\%$
$=44\times\frac{1}{100}$
$=\frac{44}{100}$
$=\frac{4\times11}{4\times25}$
$=\frac{11}{25}$
$44\%\Rightarrow\frac{11}{25}$ (Fraction)

There is only one prime number between $90$ and $100,$ i.e. $97.$

Here, $HCF = 12$
Product of two number $= 2160$
We know:
$LCM \times HCF =$ Product of the two numbers
$LCM =\frac{2160}{\text{HCF}}$
$=\frac{2160}{12}$$= 180$
$LCM = 180$

Lets consider prime and composite numbers separately and prove $1$ is a factor in each case.
Prime:
Example: $7$
Factors of $7$ are $1,7.$

Composite:
Example: $10$
Factors of $10$ are $1, 10, 2, 5.$

$6 + 7 + 3 + 0 + 1 + * + 2 = 19 + *$
$8$ is the least number that should be added to $19$ such that number will be divisible by $9.$
Sum of the digits:
$6 + 7 + 3 + 0 + 1 + 8 + 2 = 27$
$27$ is divisible by $9.$

$2\frac{1}{3}=\frac{3\times2+1}{3}=\frac{7}{3}$

Pairs of prime numbers that differ by $2$ are called twin primes.
The difference between $17$ and $23$ is $6.$
Hence, $17$ and $23$ are not twin primes.

$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}17&391\\\hline23&23\\\hline&1\end{array}$
$289 = 17 × 17$
$391 = 17 × 23$
The $HCF$ of $289$ and $391$ is $17.$
Dividing both the numerator and the denominator by $17:$
$\frac{289\div17}{391\div17}=\frac{17}{23}$

Multiple of $6$ like $6, 12, 18, 24, 30$ and they can all be divided evenly by $3.$
So option $C$ is correct.

$p = 4, 4 ÷ 2 = 2$
$p + 2 = 4 + 2, 6 ÷ 2 = 3$
$p + 4 = 4 + 4, 8 ÷ 2 = 4$
Product is divisible by $2.$
Therefore, $C$ is the correct answer.

Sum of the digits at odd places $= 6 + 3 + 4 = 13$
Sum of the digits at even places $= 5 + * + 7 = 12 + *$
Difference $= 13 - [12 + *] = 1 − *$
If $6,53,*47$ is divisible by $11,$ then $1 - *$ must be zero or multiple of $11.$
$1 - * = 0 or 11$
$* = 1$ or $- 10$
But $*$ is a digit, so $*$ must be $1.$

$LCM$ of two co -prime numbers is their product.
Example: Consider $6$ and $7,$
Multiple of $6 = 6, 12, 18, 24, 30, 36, 42, 48$
Multiple of $7 = 7, 14, 21, 28, 35, 42$
$L.C.M$ of $6$ and $7 = 42$
The product of $6$ and $7 = 6 \times 7 = 42$

The $HCF$ of any two consecutive natural numbers is $1$ because two consecutive natural numbers are always co-prime.

$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}+\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$=\frac{\big(\sqrt3+\sqrt2\big)^2+\big(\sqrt3-\sqrt2\big)^2}{3-2}$
$=\frac{3+2+3+2}{1}=10$

$GCD$ of two numbers is $17$
So, the numbers can be $17a$ and $17b.$
Now, $17a \times 17b = 17 \times 765$
$\Rightarrow ab = 45$
So, we can get two pairs
$a = 5$ and $b = 9$ or $a = 9$ and $b = 5$
Thus, the numbers are $17 \times 5 = 85$ and $17 \times 9 = 153.$
Also, we can get the other pair $17 \times 1 = 17$ and $765.$
Hence, the correct answer is option $(b).$

$\frac{7}{4}=1\frac{3}{4}$

Factors of $512 = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512$
Therefore, number of factors of $512 = 10.$

Two fractions are called as unlike fractions, if the denominators of those fractions are different.
For example: Consider $\frac {1}{5}$ and $\frac {3}{6},$ here both the fractions have different denominators, so they are unlike fractions.
Hence, fractions with different denominators are called unlike fractions.