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 $=60-23=37$

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

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

Alternative method,

$\begin{array}{rl}& {35}^2<1510<{40}^2\\ & {40}^2=1600\\ & {39}^2=1521\end{array}$
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, $34y$ 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

$\begin{array}{rl}& m+n=84\\ & m-n=6\end{array}$
By substitution method we get
$2\mathrm{m}=90$
$\mathrm{m}=45$
Putting the value of $\mathrm{m}$ in equation 1
$\begin{array}{rl}& 45+\mathrm{n}=84\\ & \mathrm{n}=84-45=39\end{array}$
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, $\mathrm{x}=0$
$\therefore (2x-3y)=(0-3\times 9)=-27$

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
$\Rightarrow \mathrm{p}+\mathrm{q}=8\dots (1)$
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.
$\Rightarrow (6+2+6+8+q)-(5+3+7+p)=11$
$\begin{array}{rl}& \Rightarrow 22+q-15-p=11\\ & \Rightarrow q-p=4\dots .(2)\end{array}$
Solving 1 and 2 we get
$p=2,q=6$

$AB\times 3=CAB$

$50\times 3=150$
$\mathrm{B}$ can only be 0 , So
$A\times 3=CA$
$5\times 3=15$
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, 

$\begin{array}{rl}& {95}^2<9222<{100}^2\\ & {95}^2=9025\\ & {96}^2=9216\end{array}$
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 :

$\begin{array}{rl}& {75}^2<5935<{80}^2\\ & {76}^2=5776\\ & {77}^2=5929\\ & 5935-5929=6\end{array}$
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
$(3^3+7^3)=(27+343)=370$
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 :
$\begin{array}{rl}\sqrt{2116}& =46\\ \sqrt{2916}& =54\\ \sqrt{3136}& =56\\ \sqrt{3158}& \approx 56.2\end{array}$
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

$3426k$ 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 =  $3+4+2+6+k=15+k$
⇒$$\begin{gathered} 15+k=18,21 \\ \mathrm{\&} \\ 24 \end{gathered}$$
⇒$$\begin{gathered} k=3,6 \\ \mathrm{\&} \\ 9 \end{gathered}$$ 

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
$\begin{array}{rl}& 15\times 350=21\times (350-x)\\ & 350-x=250\\ & x=100\end{array}$
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, $\sqrt{7x-3y}=\sqrt{7\times 6-3\times 2}$
$\begin{array}{rl}& =\sqrt{42-6}\\ & =\sqrt{36}=6\end{array}$