Free Practice Questions for Number-system in Maths

We know that, to divide a number by 5, last digit of number should be 0 or 5.
And to divide by 8, last three digits of number should be divisible by 8.
Therefore, minimum possible value for dividing 40 is 57968320.
Required, value of $P=2$ and $q=0$

Let the number $n$ is $6 x+5$
Therefore, required remainder of $\frac{9 \times(6 x+5)}{6}$
Remainder of $\frac{9 \times(6 x)}{6}+\frac{9 \times 5}{6}$
$
=0+3
$
$
=3
$

ATQ,
Let first number $=x$
Second number $=\mathrm{y}$
$\Rightarrow x+y=9$
$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
$\Rightarrow \frac{x+y}{x y}=\frac{1}{2}$
$\Rightarrow x y=18$
Put, $x=6, y=3$
$\Rightarrow x+y=6+3=9$ verify

Slope $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$=\frac{2-5}{5-2}$
$=-1$
$\Rightarrow(\mathrm{y}-5)=-1(x-2)$
$\Rightarrow \mathrm{y}-5=-x+2$
$\Rightarrow \mathrm{y}+x=7 \ldots \ldots$ (i)
$3 x+2 y=17 \ldots \ldots$ (ii)
$\mathrm{y}+x=7 \ldots \ldots$ (i) $\times 3$
$3 y+3 x=21$
$2 y+3 x=17$
$-{ }^{3}+3=4$
Putting the value of ' $y$ ' in equation (i)
$
\begin{aligned}
&\Rightarrow x+y=7 \\
&\Rightarrow x+4=7 \\
&x=3 \\
&\Rightarrow \frac{3}{1}=\frac{5 \lambda+2}{\lambda+1} \\
&\Rightarrow 3 \lambda+3=5 \lambda+2
\end{aligned}
$
$
\begin{aligned}
&\Rightarrow 2 \lambda=1 \\
&\lambda=\frac{1}{2}
\end{aligned}
$
Ratio 1:2

ATQ,
Let number $=x$
$
\begin{aligned}
&x \times \frac{35}{100}=91 \\
&x=260
\end{aligned}
$
Hence, the number is $=260$

ATQ,
4 divisibility Rule - last two digit divisible is
4 then the number is divisible by 4
Option (i)
6542176
last two digit $=\frac{76}{4}=$ Remainder $=0$

ATQ,
$
\begin{array}{r}
0.020 \\
\Rightarrow \frac{-0.002}{0.018} \\
\hline
\end{array}
$

The number be $x=\operatorname{LCM}$ of $(8,12,15,24,25,40)+7$
$
=600+7
$
$
=607
$
Therefore, required remainder $=\frac{607}{29}=21$
Hence, option B is correct.

Given:

247X8 is divisible by 44

Concept used:

If the number formed by the last two digits in a number is divisible by 4, the original number is divisible by 4

If the difference between the sum of the digits at the odd and even places equals 0 or divisible by 11, then the number is divisible by 11

Calculation:

Possible values of X = 0, 2, 4, 6, 8

For 0

(2 + 7 + 8) - (4 + 0) = 13 So, not divisible by 11

For 2

(2 + 7 + 8) - (4 + 2) = 11 So, divisible by 11

For 4

(2 + 7 + 8) - (4 + 4) = 9 So, not divisible by 11

For 6

(2 + 7 + 8) - (4 + 6) = 7 So, not divisible by 11

For 8

(2 + 7 + 8) - (4 + 8) = 5 So, not divisible by 11

∴ Required answer is 2

ATQ,
largest four digit number
= 9999
$\Rightarrow \frac{9999}{49}=204$
$\Rightarrow$ 49 × 204 = 9996
Hence, 9996 is divisible by 49.

ATQ,
Let the first fraction = x
2nd fraction = y

$\Rightarrow x + y = \frac{7}{4}$

$\Rightarrow \frac{5}{3}+y=\frac{7}{4}$

$\Rightarrow y=\frac{7}{4}-\frac{5}{3}$

$y=\frac{1}{12}$

Hence, second fraction $=\frac{1}{12}$

ATQ,
$x^{2}+5 k x+k^{2}+5$ is divisible by (x + 2)
$\Rightarrow \quad x+2=0$
$x=-2$
$\Rightarrow \quad x^{2}+5 k x+k^{2}+5=0$
$\Rightarrow \quad(-2)^{2}+-10 k+k^{2}+5=0$
$\Rightarrow 4-10 k+k^{2}+5=0$
$\Rightarrow \quad k^{2}-9 k-k+9=0$
$\Rightarrow \quad k(k-9)-1(k-9)=0$
k = (9 , 1 )

Again, $\Rightarrow \quad x^{2}+5 k x+k^{2}+5$
Put x + 3 = 0
$x=-3$ 

and k = 1
$9-15+6=0$
$\Rightarrow \quad x^{2}+5 k x+k^{2}+5$

Put k = 9 & $x=-3$
$\Rightarrow$$9-45 \times 3+86 \neq 0$
Hence, value of k = 9

ATQ,
$\Rightarrow \quad 1296$
$=\quad(36)^{2}$
$\Rightarrow \quad 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
$\Rightarrow \quad 2^{4} \times 3^{4}$
Perfect square

ATQ,
Greatest five digit number
= 99999
Divided by 324
$=\frac{99999}{324}$
Remainder = 207
Greatestfive digit number is completely divisible by 324
= 99999 – 207
= 99792

ATQ,
We know that
Product of two number = first number × second number
1728 = 6.4 × second number
Second number = $\frac{1.728}{6 \cdot 4}$
= 0.27

ATQ,
Irrational number $\rightarrow$ Irrational number are the real numbers that cannot be represented as a simple fraction. It cannot be expressed in the form of a ratio, such as $\frac{p}{q}$, when p and q are integers , q $\neq$ 0 $\sqrt{5428}$ is not perfect square of a number.
Hence, this $\sqrt{5428}$ is not a irrational numbers.

Let when $\mathrm{N}$ is divisible by 624 , the remainder will be 53 .
Using remainder theorem:
$N=624 q+53$
Now, divide N by $16 $.

Since 16 is the factor of 624 

Hence to get the required remainder of the N, we just need to divide the given remainder by 16. 

$\operatorname{Remainder}\left(\frac{53}{16}\right)=5$

ATQ,
Number = x 4738
Divisibility rule 9
If the sum of digits of any number is divisible by 9 then the number is also divisible by 9
$\Rightarrow x+4+7+3+8$
$\Rightarrow 22 + x, x=5$
$\Rightarrow$ 27 divisible by 9
Hence, value of $x$ = 5

ATQ,
Let fraction = $x$
$\Rightarrow x+\frac{5}{16}=1$
$\Rightarrow x=1-\frac{5}{16}=\frac{11}{16}$
$\Rightarrow \frac{11}{16}=\frac{22}{32}$
Hence
Option (3) is correct

ATQ,
Square root of
$\Rightarrow 108900$
$\Rightarrow 2 \times 2 \times 5 \times 5 \times 3 \times 3 \times 11 \times 11$
$\Rightarrow 2^{2} \times 3^{2} \times 5^{2} \times 11^{2}$
Square = 2 × 3 × 5 × 11
= 330