Squares and Square Roots: MCQ test, Set 10

Which of these is equal to $78 \times 78$ ?
A $2 \times 78$ B $78^2$ C $2^{78}$ D $78^3$

Answer and solution

Answer: (B) $78^2$

$78$ is multiplied by itself $2$ times.

$\therefore 78 \times 78 = 78^2$
The square of $21$ is
A $400$ B $42$ C $441$ D $484$

Answer and solution

Answer: (C) $441$

$21^2 = 21 \times 21 = 441$
The next number in $676, 729, 784, \ldots$ is
A $840$ B $900$ C $839$ D $841$

Answer and solution

Answer: (D) $841$

These are $26^2, 27^2, 28^2$ .

Next is $29^2 = 841$
$100^2 =$
A $100000$ B $1000$ C $200$ D $10000$

Answer and solution

Answer: (D) $10000$

$100^2 = 100 \times 100$

$10 \times 10 = 100$ , then write two zeros: $10000$
$115^2 = \square 25$ . The number in the box is
A $144$ B $121$ C $132$ D $23$

Answer and solution

Answer: (C) $132$

Rule: multiply the number before $5$ by the next number, then write $25$ .

$11 \times 12 = 132$ , so $115^2 = 13225$
If $189^2 = 35721$ , then $\sqrt{35721} =$
A $17860.5$ B $378$ C $189$ D $\sqrt{189}$

Answer and solution

Answer: (C) $189$

$a^2 = b \Rightarrow \sqrt{b} = a$

$189^2 = 35721 \Rightarrow \sqrt{35721} = 189$

(The square root is not half the number.)
$\sqrt{121} =$
A $11$ B $10$ C $60.5$ D $12$

Answer and solution

Answer: (A) $11$

$11^2 = 11 \times 11 = 121$

$\therefore \sqrt{121} = 11$
$\sqrt{\square} = 16$ . The number in the box is
A $256$ B $\sqrt{16}$ C $32$ D $8$

Answer and solution

Answer: (A) $256$

$\sqrt{b} = a \Rightarrow b = a^2$

$16^2 = 16 \times 16 = 256$
$4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ . So $\sqrt{4096} =$
A $128$ B $64$ C $2048$ D $2$

Answer and solution

Answer: (B) $64$

Take one factor from each pair: $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$
$\sqrt{5 \times 5 \times 13 \times 13} =$
A $65$ B $18$ C $4225$ D $13$

Answer and solution

Answer: (A) $65$

$\sqrt{5 \times 5 \times 13 \times 13} = \sqrt{(5 \times 5) \times (13 \times 13)}$

$= 5 \times 13 = 65$
By the long division method, $\sqrt{1681} =$
A $41$ B $49$ C $51$ D $31$

Answer and solution

Answer: (A) $41$

Pairs: $\overline{16}\;\overline{81}$ .

$4 \times 4 = 16 \le 16$ : first digit $4$ , and $16 - 16 = 0$

Bring down $81$ : we get $81$ . New divisor: $4 \times 2 = 8$ ; $81 \times 1 = 81$ ( $82 \times 2 = 164$ is too big); $81 - 81 = 0$

Remainder $0$ . $\therefore \sqrt{1681} = 41$
While finding $\sqrt{7569}$ by long division, the first digit of the root is $8$ . The next divisor begins with
A $8$ B $9$ C $16$ D $64$

Answer and solution

Answer: (C) $16$

Double the root found so far: $8 \times 2 = 16$

Then $167 \times 7 = 1169$ ( $168 \times 8 = 1344$ is too big), so $\sqrt{7569} = 87$
In finding $\sqrt{29584}$ by the long division method, the first digit of the root is
A $2$ B $1$ C $5$ D $3$

Answer and solution

Answer: (B) $1$

Pair from the right: $\overline{2}\;\overline{95}\;\overline{84}$ ; the first group is $2$ .

$1 \times 1 = 1 \le 2$ , so the first digit is $1$
$\sqrt{6} \times \sqrt{96} =$
A $576$ B $288$ C $\sqrt{102}$ D $24$

Answer and solution

Answer: (D) $24$

$\sqrt{6} \times \sqrt{96} = \sqrt{6 \times 96} = \sqrt{576} = 24$
$(\sqrt{34})^2 =$
A $34$ B $\sqrt{34}$ C $1156$ D $68$

Answer and solution

Answer: (A) $34$

$(\sqrt{a})^2 = \sqrt{a} \times \sqrt{a} = a$

$\therefore (\sqrt{34})^2 = 34$
$\displaystyle \frac{\sqrt{72}}{\sqrt{2}} =$
A $\sqrt{70}$ B $36$ C $6$ D $18$

Answer and solution

Answer: (C) $6$

$\displaystyle \frac{\sqrt{72}}{\sqrt{2}} = \sqrt{\frac{72}{2}} = \sqrt{36} = 6$
$\displaystyle \frac{42}{\sqrt{42}} =$
A $42\sqrt{42}$ B $1$ C $42$ D $\sqrt{42}$

Answer and solution

Answer: (D) $\sqrt{42}$

$\displaystyle \frac{42}{\sqrt{42}} = \frac{\sqrt{42} \times \sqrt{42}}{\sqrt{42}} = \sqrt{42}$
$\sqrt{16} + \sqrt{81} =$
A $\sqrt{97}$ B $13$ C $169$ D $97$

Answer and solution

Answer: (B) $13$

$\sqrt{16} + \sqrt{81} = 4 + 9 = 13$

$\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ , so do not add first.
$\sqrt{900} - \sqrt{576} =$
A $6$ B $324$ C $54$ D $18$

Answer and solution

Answer: (A) $6$

$\sqrt{900} - \sqrt{576} = 30 - 24 = 6$

$\sqrt{a} - \sqrt{b} \ne \sqrt{a - b}$ , so do not subtract first.
$\sqrt{12^2 + 35^2} =$
A $47$ B $1369$ C $37$ D $\sqrt{47}$

Answer and solution

Answer: (C) $37$

$\sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37$

$\sqrt{a^2 + b^2} \ne a + b$
$\sqrt{(23 - 8)^2} =$
A $\sqrt{465}$ B $15$ C $225$ D $465$

Answer and solution

Answer: (B) $15$

$\sqrt{(a - b)^2} = a - b$ when $a - b > 0$

$\sqrt{(23 - 8)^2} = \sqrt{15^2} = 15$
$\sqrt{0.36} =$
A $6$ B $0.18$ C $0.06$ D $0.6$

Answer and solution

Answer: (D) $0.6$

$\displaystyle \sqrt{0.36} = \sqrt{\frac{36}{100}}$

$\displaystyle = \frac{\sqrt{36}}{\sqrt{100}}$

$\displaystyle = \frac{6}{10}$

$= 0.6$
$\sqrt{0.0729}$ written as a fraction is
A $\displaystyle \frac{729}{100}$ B $\displaystyle \frac{27}{10}$ C $\displaystyle \frac{27}{100}$ D $\displaystyle \frac{27}{1000}$

Answer and solution

Answer: (C) $\displaystyle \frac{27}{100}$

$\displaystyle \sqrt{0.0729} = \sqrt{\frac{729}{10000}} = \frac{\sqrt{729}}{\sqrt{10000}} = \frac{27}{100}$
$\sqrt{7.84} =$
A $0.28$ B $3.92$ C $28$ D $2.8$

Answer and solution

Answer: (D) $2.8$

$\displaystyle \sqrt{7.84} = \sqrt{\frac{784}{100}} = \frac{\sqrt{784}}{\sqrt{100}} = \frac{28}{10} = 2.8$
$\sqrt{\square} = 4.5$ . The number in the box is
A $9$ B $20.25$ C $202.5$ D $2.025$

Answer and solution

Answer: (B) $20.25$

$\displaystyle 4.5 = \frac{45}{10}$

$\displaystyle \left(\frac{45}{10}\right)^2 = \frac{2025}{100} = 20.25$

Check: $\displaystyle \sqrt{20.25} = \frac{\sqrt{2025}}{\sqrt{100}} = 4.5$

Menu

MCQ test, Set 10