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

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 $=12$

$n(M U B)=75-12=63$

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 $\sqrt{x+6 y}=\sqrt{4+6 \times 2}=\sqrt{4+12}=\sqrt{16}=4$

ADDITIONAL INFORMATION:

$n=475 A B$ 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 - 
$B=0$
$n=475 A 0$

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 -

$B=5$
$n=475 A 5$
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, $10 A+B=10(2)+0=20$

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

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

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

Option $4 \rightarrow 2+7+4+0-(7+6+0)=13-13=0$ 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 $1, a, b, c \ldots, n$
The HCF of $a$ and $b$ is 1 .
The LCM of $a$ and $b$ is 4199 .
HCF $\times$ LCM $=$ Product of two numbers
$
\begin{aligned}
&a \times b=1 \times 4199 \\
&\Rightarrow a b=4199 \\
&\Rightarrow a=4199 / b \quad \cdots(1)
\end{aligned}
$
Here $a+b=240$
$\cdots-(2)$
Solving (1) and (2),
$
\begin{aligned}
&\Rightarrow 4199 / b+b=240 \\
&\Rightarrow 4199+b^{2}=240 b \\
&\Rightarrow b^{2}-240 b=4199 \\
&\Rightarrow b^{2}-240 b-4199=0 \\
&\Rightarrow b^{2}-221 b-19 b-4199=0 \\
&\Rightarrow b(b-221)-19(b-221)=0 \\
&\Rightarrow b=221,19
\end{aligned}
$
Since b is a composite number we take
$
\Rightarrow \mathrm{b}=221
$
Substitute in (2),
$
\begin{aligned}
&a+221=240 \\
&a=240-221 \Rightarrow 19
\end{aligned}
$
$\therefore$ The value of $a$ and $b$ is $19 \& 221$
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 $=999-19$
$=980$