Questions Related to maths

Multiple choice maths square and square root perfect square or square number squares and triangles powers and roots

If a four-digit perfect square number is such that the number formed by the first two digits and the number formed by the last two digits are also perfect squares, identify the four digit number.

  1. $6416$
  2. $3616$
  3. $1681$
  4. $1664$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Four digit number $accd$

$ab$ is a perfect square
$cd$ is also a perfect square
Consider $6416$
$64$and$16$ are perfect square but $6416$ is not a perfect square.
Consider $3616$
$36$and$16$ are perfect square but $3616$ is not a perfect square.

Consider $1681$
$16$and$81$ are perfect square and $1681$ is a perfect square.

Consider $1664$
$16$and$64$ are perfect square but $1664$ is not a perfect square.
Hence, Option C is correct.


Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

A rectangular veranda is of dimension $18$m $72$cm $\times 13$ m $20$ cm. Square tiles of the same dimensions are used to cover it. Find the least number of such tiles.

  1. $4290$
  2. $4540$
  3. $4620$
  4. $4230$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The edge of rectangular veranda are $18\ m\ 72\ cm=1872\ cm$ and $13\ m\ 20\ cm=1320\ cm$.


On taking $HCF$ of $1872$ and $1320$, we get

$HCF=24$

Therefore,
No. of tiles required $=$ $\dfrac{Area\ of\ Veranda}{Area\ of\ tiles}$

                                  $=\dfrac{1872\times 1320}{24\times 24}$

                                  $=4290$

Hence, this is the answer.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

Determine the HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+12a+35$

  1. (a-5)(a+7)

  2. (a+5)(a-7)

  3. (a-7)

  4. (a+5)

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $a^2 - 25 = (a-5)(a+5) $
$ a^2 -2a -35 = a^2 -7a +5a -35 $
                         $= a(a-7)+5(a-7) $
                         $= (a+5)(a-7) $
and
$a^2+ 12a + 35 =a^2 +7a +5a +35 $
                          $=(a+7)(a+5) $
Clearly HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+ 12a + 35$ i.e $ (a-5)(a+5), (a+5)(a-7)$ and $(a+7)(a+5)$ is $a+5$
Option D is correct.