Free Practice Questions for Number-system in Maths

ATQ,
We know that
$\Rightarrow$ Product of two number $=\mathrm{I}\times \mathrm{II}$
$\Rightarrow 0.646=1.7\times \mathrm{II}$
$\Rightarrow \mathrm{II}=\frac{0.646}{1.7}$
$\Rightarrow \mathrm{II}=0.38$
Other number $=0.38$

ATQ,
$\Rightarrow x^2-2x-1=0$
$\Rightarrow \alpha +\beta =2$
$\alpha \times \beta =-1$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2=(2{)}^2-2\times (+1)$
$=6$
${\alpha }^2\cdot {\beta }^2=(-1{)}^2=1$
Equation
$\Rightarrow x^2-(\alpha +\beta )x+\alpha \cdot \beta =0$
$x^2-6x+1=0$

ATQ,
Square root
$=\sqrt{4624}$
$=\sqrt{2\times 2\times 2\times 2\times 17\times 17}$
$=2\times 2\times 17$
$=68$

ATQ,
Let fraction $=x$
$\Rightarrow \frac{3}{4}-x=\frac{5}{12}$
$\Rightarrow \frac{3\times 3}{4\times 3}-x=\frac{5}{12}$
$\Rightarrow x=\frac{1}{3}$

ATQ,

Red and yellow color bought a customers $=$ 20
Yellow and maroon colors bought a customers $=10$
Red and maroon colors bought a customers $=15$
All three bought color $=19$
Only one color of 2 meters of cloth bought
$=(20-19)+(19-10)+(19-15)$
$=1+9+4$
$=14$

ATQ,
Total number of students were interviewed $=75$
Number of students interviewed in Math = 39
Number of students interviewed in Biology $=32$
Neither Math $+$Nor Biology $=12$

$n(MUB)=75-12=63$

Number of students who took interviewed in both math and biology $=39+32-63=8$

ATQ,
Square root of
60025
$60025\Rightarrow 5\times 5\times 7\times 7\times 7\times 7$
Square root $=5\times 7\times 7$ $=245$

ATQ,

The weight of $\frac{7}{16}$ part of brick is $\frac{21}{8}\mathrm{kg}$.

$\frac{5}{12}$ part of the brick $=\frac{5}{12}\times 6$
$=\frac{5}{2}\mathrm{kg}$

Rules of divisibility of 72 - number should be divisible by 9 and 8
Rules of divisibility of 9 - sum of digits of number should be divisible by 9
Rules of divisibility of 8 - last three digits of number should be divisible by 8
Therefore, possible number is $=65784432672$
Value of $x=4,y=2$
Required value of $\sqrt{x+6y}=\sqrt{4+6\times 2}=\sqrt{4+12}=\sqrt{16}=4$

ADDITIONAL INFORMATION:

$n=475AB$ is a positive integer whose tens and units digits are $A$ and $B$, respectively.

Divisibility by 5 : A number is divisible by 5 if its unit digit is either 0 or 5.
$\Rightarrow B=0\text{ or }5$

Case 1 - 
$B=0$
$n=475A0$

Divisibility by 8 : A number is divisible by 8 if number formed by last three digits of the given number is divisible by 8.
Therefore, $475\mathrm{A}0$ will be divisible by 8 if $5\mathrm{AO}$ is divisible by 8 .
As 520 is divisibly by 8 so $A=2$.
Resultant number $=47520$
Now check whether 47520 is divisible by 9 or not.

Divisibility by 9 : A number is divisible by 9 if sum of digits of that number is divisible by 9.
Sum of digits of $47520=4+7+5+2+0=18$
As 18 is divisible by 9 therefore 47520 is divisible by 9.

Case - 2 -

$B=5$
$n=475A5$
Divisibility by 8 : A number is divisible by 8 if number formed by last three digits of the given number is divisible by 8.
Therefore, $475\mathrm{A}5$ will be divisible by 8 if $5\mathrm{A}5$ is divisible by 8 .
There is no such value of A for which $5\mathrm{A}5$ is divisible by $8.$

Hence, $B=0$ and $A=2$
Now, $10A+B=10(2)+0=20$

We know that number should be divisible by 11 and 8 to be divisible by 88 .

Divisibility rule of 11:  The difference of the sum of alternatives digits of a number is divisible by 11

Option $1\to 2+6+4+6-(7+7+1)=18-15=3$ it is not divisible by 11

Option $2\to 2+6+4+0-(7+7+4)=12-18=-6$ it is not divisible by 11

Option $3\to 2+7+4+8-(7+6+0)=21-13=8$ it is not divisible by 11

Option $4\to 2+7+4+0-(7+6+0)=13-13=0$ it is divisible by 11

Divisibility rule of 8: Last three digits of number should be divisible 8.

400 is divisible by 8.

Hence, only option 4satisfy the rule of divisibility of 11 and 8.

Given:
a is a prime,
$b$ is a composite number
$a+b=240$
LCM of $a$ and $b$ is $4199.$
Formula Used:
$\mathrm{HCF}\times \mathrm{LCM}=$ Product of two numbers
Solution :
If a is a prime number.
The factors are 1 , a
If $b$ is a composite number.
The factors are $1,a,b,c\dots ,n$
The HCF of $a$ and $b$ is 1 .
The LCM of $a$ and $b$ is 4199 .
HCF $\times$ LCM $=$ Product of two numbers
$\begin{array}{rl}& a\times b=1\times 4199\\ & \Rightarrow ab=4199\\ & \Rightarrow a=4199/b\cdots (1)\end{array}$
Here $a+b=240$
$\cdots -(2)$
Solving (1) and (2),
$\begin{array}{rl}& \Rightarrow 4199/b+b=240\\ & \Rightarrow 4199+b^2=240b\\ & \Rightarrow b^2-240b=4199\\ & \Rightarrow b^2-240b-4199=0\\ & \Rightarrow b^2-221b-19b-4199=0\\ & \Rightarrow b(b-221)-19(b-221)=0\\ & \Rightarrow b=221,19\end{array}$
Since b is a composite number we take
$\Rightarrow \mathrm{b}=221$
Substitute in (2),
$\begin{array}{rl}& a+221=240\\ & a=240-221\Rightarrow 19\end{array}$
$\therefore$ The value of $a$ and $b$ is $19\mathrm{\&}221$
Note:
The question was incorrect in the official paper. However, we have changed it and update it accordingly.

ATQ,
quiet decimal - A decimal that is divisible by a whole number is called a Quiet Decimal. 

option(3),

 $\frac{57}{120}$ is a quiet decimal

ATQ,
Number of entries expected $=300$
Entry fees $=200$ Rs. per head
$\Rightarrow$ Expected amount $=$ Rs. $(300\times 200)=$ Rs. 60000
Number of Actual entries $=200$
$\Rightarrow$ Amount Actual received $=$ Rs. $(200\times 200)$
$=$ Rs. 40000
$\Rightarrow$ Less money received $=$ Rs. $(60000-40000)$
$=20000\mathrm{Rs}$.

पाँच अंकों की सबसे छोटी संख्या $=10000$

पाचँ अंकों की संख्या छोटी संख्या, जोकि 41 से भाज्य है।
$10000+41-37=10004$

$=(122{)}^{173}$ में इकाई का अंक
$=(2{)}^{4\times 43+1}$ में इकाई का अंक
$={\left(2^4\right)}^{43}\times 2$ में इकाई का अंक
$=16\times 2$ में इकाई का अंक
$=16\times 2$ में इकाई का अंक $=2$

तीन अंकों की बड़ी से बड़ी संख्या $=999$

अत: तीन अंकों की सबसे बड़ी अभीष्ट संख्या $=999-19$
$=980$