Free Practice Questions for Number-system in Maths

Let $a$ and $b$ be two numbers
According to question
$
\begin{aligned}
&\frac{a+b}{2}=13 \\
&a+b=26 \ldots . .(1) \\
&\text { Also, } \sqrt{a b}=12 \\
&a b=144
\end{aligned}
$
Now, $(a+b)^2=26^2$
$
\begin{aligned}
&a^2+b^2+2 a b=676 \\
&a^2+b^2+2 a b-4 a b=676-4 a b \\
&a^2+b^2+2 a b-4 a b=676-4 \times 144 \\
&a^2+b^2+2 a b-4 a b=676-576 \\
&(a-b)^2=100 \\
&a-b=10 \ldots .(2)
\end{aligned}
$
Solving (1) and (2)
$
a=18, b=8
$
Required number $=18-8=10$

Numbers after substracting the remainder

$398-7=391$
$437-12=425$
$5425-2=5423$
We can write it
$391=17 \times 23$
$425=5 \times 5 \times 17$
$5423=11 \times 17 \times 29$
Hcf of 391,425 and $5423=17$
So, greatest number is $=17$

Difference $= (10-6)=(15-11)=(20-16)=(24-20)=(30-26)=4$
LCM of $10,15,20,24$ and $30=120$
Number $=120 k-4$, which is closest to the largest 7 digit number 9999999.
$120 k=9999960$
Hence, number $=9999960-4=9999956$

LCM of $12,16$ and $20=240$
Required Number $=240 k+6$, Which is divisible by $9$
$240 k+6=234 k+6 k+6$
$234 k$ is completely divisible by $9$ , so $6 k+6$ too, has to be divisible by $9$.
If $k=2$, then $6 k+6$ is divisible by $9$.
Hence, Required number $=240 \times 2+6=486$
Sum of the digits $=4+8+6=18$

As per the question,
Eight smallest even natural numbers which are NOT divisible by 3 $=2,4,8,10,14,16,20,22$
So, average of these numbers $=\frac{2+4+8+10+14+16+20+22}{8}$
$
=\frac{96}{8}=12
$

Lcm of $3,5,7$ and $9$ is $315$.
four digits least number $=1000$
When we divide $1000$ by $315$ we get $55$ as remainder
$\Rightarrow 315-55=260$
Required number $=1000+260=1260$

LCM of $42,44,26,39$ and $40$ is $120120$
Now greatest $6$ digit number is $999999$
Now the greatest $6$ digit number is $=120120 \times 8=960960$

LCM of $3,4,6,15$ and $20$ is $60$
Greatest $3$ digit number $=999$
when we divide $999$ we get Remainder $39$
$3$ digit greatest number $=999-39=960$
When the number $961$  is divided by $3,4,6,15,20$ the remainder in each case is $1$
So, The number $=960+1=961$
Required sum $=9+6+1=16$

LCM of 28,32 and 56
$
\begin{aligned}
&28=2 \times 2 \times 7 \\
&32=2 \times 2 \times 2 \times 2 \times 2 \\
&56=2 \times 2 \times 2 \times 7
\end{aligned}
$
LCM of 28,32 and $56=2 \times 2 \times 2 \times 2 \times 2 \times 7=224$
To get the least perfect square we need to multiply 224 $\Rightarrow 224 \times 7 \times 2=3136$

Let two digit number be $10 \mathrm{a}+\mathrm{b}$
When the number are interchanged $=10 \mathrm{~b}+\mathrm{a}$
Required number $=10 \mathrm{a}+\mathrm{b}+10 \mathrm{~b}+\mathrm{a}$
$
\begin{aligned}
&=11 \mathrm{a}+11 \mathrm{~b} \\
&=11(\mathrm{a}+\mathrm{b})
\end{aligned}
$
Clearly, 11 is factor of $11(a+b)$.

LCM of $6.9,15,18$ and 21 is 630
The required no. is $630+4=634$

Descending order = 30.5 > 3.50 > 3.055 > 3.05 > 3.005 > 0.355  

So, option (3) is correct.

Let the whole number be x.
According to question,

(50 + x) + (50 - x)

=50 + 50 = 100

Let the number of total students be 'x '
According to question
Students playing Badminton=12x
Students playing Volleyball=14x
Students playing Tennis=18x
Students playing Chess=116x
Given
Number of students playing Volleyball=14x

160 ==14x

 x=160×4

X=640

So, total no of students is640

LCM of 18,36 and 48 is 144.
Greatest 4 digit number $=9999$

when we divide 9999 by 144 we get Remainder 63
4 digit greatest number $=9999-63=9936$
When number is divided by 18,36 and 48 the remainder in each case is 8

So, The number $=9936+8=9944$

Let the two digit number is $10 x+y$
As per the question,
$
\begin{aligned}
&y=x+4 \\
&(10 x+y) \times(x+y)=370
\end{aligned}
$
Putting the value of $y$
$
\begin{aligned}
&(10 x+x+4) \times(x+x+4)=370 \\
&(11 x+4) \times(2 x+4)=370 \\
&11 x^2+26 x-177=0 \\
&11 x^2+(59-33) x-177=0 \\
&(11 x+59)(x-3)=0
\end{aligned}
$
By solving the equation we get $x=3$
Then $y=7$
So the number is 37 .

$
\begin{aligned}
&=\left(3 \frac{1}{7}+4 \frac{3}{7}\right)+\frac{7}{6} \\
&=\left(\frac{22}{7}+\frac{31}{7}\right)+\frac{7}{6} \\
&=\frac{53}{7}+\frac{7}{6} \\
&=\frac{318+49}{42}=\frac{367}{42} \\
&=\frac{318+49}{42}=\frac{367}{42} \\
\end{aligned}
$
$\frac{367}{42} $ is not a whole number .
So if we subtract $\frac{367}{42}$ from $\frac{31}{42}$ it becomes $=\frac{336}{42}=8$ which is a whole number .

We know that
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
According to question
Divisor $=8 \times$ Quotient $=9 \times$ Remainder
So,
Divisor $=9 \times 24=216$
Quotient $=\frac{216}{8}=27$
Hence,
Dividend $=216 \times 27+24=5832+24=5856$

Let the numbers are $x$ and $y$ , then

$x+y=14$ and $x-y=4$

So, $x=\frac{14+4}{2} = 9$

and $y=\frac{14-4}{2} = 5$

So, $x\times y=9\times5 = 45$

Let the larger number be $a$ and the smaller number be $b$, then

$a-b=1208$

$a=b+1208$

and $a=4b+182$ , (As Dividend = Divisor$\times$Quotient + Remainder)  

We can say that, 

$b+1208=4b+182$

$3b=1026$

$b=342$

So larger number $a=342+1208 = 1550$