Free Practice Questions for Number-system in Maths

$520=2 \times 2 \times 2 \times 5 \times 13$
$936=2 \times 2 \times 2 \times 3 \times 3 \times 13$
Lcm of 520 and $936=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 13=4680$
Required number $=4680-17=4663$

$\mathrm{Lcm}$ of 10,15 and 18 is 90
We know the largest 4 digit no. is 9999
If we divide 9999 by 90 we get 9 as a remiander.
so, the required number $=9999-9=9990$

Lcm of 3,4,5 and 6 is 60
When we divide 2438 by 60 we get remainder as 23 .
Required number $=60-23=37$

$95 \times 103=10000+10 a-15$
$(100-5)(100+3)=10000+10 a-15$
$10000-500+300-15=10000+10 a-15$
$10000-200-15=10000+10 a-15$
$10 a=-200$
$a=\frac{-200}{10}=-20$

Hence, If we add 11 in 1510 , It will be a perfect square.

Alternative method,

$
\begin{aligned}
&35^2<1510<40^2 \\
&40^2=1600 \\
&39^2=1521
\end{aligned}
$
Hence, If we add 11 in 1510 , It will be a perfect square.

$
24=3 \times 8
$
Divisibility of 8:  If the last 3 digits of any number is divisible by 8 , the number will also be divisible by 8 .
So, $34 y$ is divisible by 8 , if $y=4$
Divisibility of 3: If the sum of the digits of a number is divisible by 3 , the number will also be divisible by 3 .
So, $9+x+3+4+4=20+x$
$20+x=21,24$ or 27
Since we need the largest value, So
$
20+x=27
$
$
x=7
$
Hence

 $(x+y)=4+7=11$

$\mathrm{Lcm}$ of $10,15,24$ and 32 is 480
The number between 1420 and $1780=480 \times 3=1440$
Number when remainder in each case is $7=1440+7=1447$
Required sum $=1+4+4+7=16$

According to Question

$\begin{aligned}
&m+n=84 \\
&m-n=6
\end{aligned}
$
By substitution method we get
$
2 \mathrm{~m}=90
$
$
\mathrm{m}=45
$
Putting the value of $\mathrm{m}$ in equation 1
$
\begin{aligned}
&45+\mathrm{n}=84 \\
&\mathrm{n}=84-45=39
\end{aligned}
$
Ratio of the required number $=\frac{45}{39}=15: 13$

$
88=8 \times 11
$
Any number is divisible by 8 when the last 3 digits of that number are divisible by 8.
$5 \mathrm{y} 2$ is divisible by 8, so $\mathrm{y}=9$ (for the largest value as 592 is divisible by 8 )
Now, the number is $643x114592$.
Any number is divisible by 11, when the difference between the sum of the digits at even places and that of those at odd places is either 0 or multiple of 11.
$
(6+3+1+4+9)-(4+x+1+5+2)=23-(12+x)
$
So, $\mathrm{x}=0$
$
∴(2 x-3 y)=(0-3 \times 9)=-27
$

A number is divisible by 99 when it is divisible by 9 and 11 .
Any number is divisible by 9 when the sum of digits of that number are divisible by 9
$\Rightarrow(6+5+2+3+6+7+8+p+q)$ should be divisible by 9
$\Rightarrow(37+p+q)$ should be divisible by 9
$\Rightarrow \mathrm{p}+\mathrm{q}=8 \ldots(1)$
Any number is divisible by 11 when the difference between the sum of the digits at even places and that of those at odd places is either 0 or multiple of 11.
$
\Rightarrow(6+2+6+8+q)-(5+3+7+p)=11
$
$
\begin{aligned}
&\Rightarrow 22+q-15-p=11 \\
&\Rightarrow q-p=4 \ldots .(2)
\end{aligned}
$
Solving 1 and 2 we get
$
p=2, q=6
$

$A B \times 3=C A B$

${50}\times{3} =150$
$\mathrm{B}$ can only be 0 , So
$A \times 3=C A$
$5\times3=15$
By elimination method, only option (4) satisfies the equation.
Hence, option (4) is the correct answer.

Divisibility of 11 : If the difference between the sum of digits at odd places that of those at even places is either 0 or multiple of 11, then the number is divisible by 11.

Option (1), $(2+2+2)-(2+2+2)=6-6=0$
Option (2), $(3+3+3)-(3+3+3)=9-9=0$
Option (3), $(0+1+0+3)-(1+1+1+1)=4-4=0$
Option (4), $(0+1+0+3)-(1+0+1+1)=4-3=1$

Hence, 10011013 is not divisible by 11.

Hence, If 6 is subtracted from 9222 , the will be a perfect square.

Alternative method, 

$
\begin{aligned}
&95^2<9222<100^2 \\
&95^2=9025 \\
&96^2=9216
\end{aligned}
$
Hence, If 6 is subtracted from 9222 , the result will be a perfect square.

Hence, If 6 is subtracted from 5935 , it will be a perfect square.

Alternative method :

$\begin{aligned}
&75^2<5935<80^2 \\
&76^2=5776 \\
&77^2=5929 \\
&5935-5929=6
\end{aligned}
$
Hence, If 6 is subtracted from 5935 , it will be a perfect square.

To find unit digit, we need only the digit at ones place, So
$\left(3^3+7^3\right)=(27+343)=370$
Hence, Unit digit is 0 .

Trick : 2, 3, 7 and 8 can never be the unit digit of a perfect square. Hence, 3158 is not a perfect square.
Basic Method :
$
\begin{aligned}
\sqrt{2116} &=46 \\
\sqrt{2916} &=54 \\
\sqrt{3136} &=56 \\
\sqrt{3158} & \approx 56.2
\end{aligned}
$
Hence, 3158 is not a perfect square.

Greatest three digit number = 652

Smallest three digit number = 205

We cannot place 0 at the beginning of any number as it will not be counted as a digit of that number.

Hence, Difference = 652−205=447

$3426 k$ is divisible by $6$ , So the sum of its digits must be divisible by $3$ and the

unit digit can be $0,2,4,6,8$.
Since it is not divisible by $5$, the unit digit cannot be $0$ and $5$ .

Sum of digits =  $3+4+2+6+k=15+k$
⇒$15+k=18,21 \\\ \& \\\ 24$
⇒$k=3, 6 \\\ \& \\\ 9$ 

Since the unit's digit should be $2,4,6,8$.
So, $k=6$

After 10 days, the mess has provisions for 15 days for 350 boys.
Let $x$ boys were shifted to another hostel, then
$
\begin{aligned}
&15 \times 350=21 \times(350-x) \\
&350-x=250 \\
&x=100
\end{aligned}
$
Hence, 100 boys were shifted to another hostel.

$
72=8 \times 9
$
Any number is divisible by 8 when the last 3 digits of that number are divisible by 8.
$87 \mathrm{y}$ is divisible by 8 so $\mathrm{y}=2$ (As 872 is divisible by 8 )
Now the number is $89563x872$
Any number is divisible by 9 when the sum of the digits is divisible by 9.
$
8+9+5+6+3+x+8+7+2=48+x
$
So, $x=6$
Now, $\sqrt{7 x-3 y}=\sqrt{7 \times 6-3 \times 2}$
$
\begin{aligned}
&=\sqrt{42-6} \\
&=\sqrt{36}=6
\end{aligned}
$