Square Roots & Cube Roots Questions and Answers

Answer: C

The value of \((4x + 7)\) is 23.

The largest perfect square not exceeding 2605 is \(51^2 = 2601\), since \(52^2 = 2704\). So the smallest number to subtract is

\[ \begin{aligned} x &= 2605 - 2601 = 4 \\ 4x + 7 &= 16 + 7 = 23 \end{aligned} \]

31 would need x = 6, 27 would need x = 5, and 11 would need x = 1, none of which leaves a perfect square.

Answer: B

The value is 10.

Work from the innermost root outwards:

\[ \begin{aligned} \sqrt{121} &= 11 \\ \sqrt{70 + 11} &= \sqrt{81} = 9 \\ \sqrt{91 + 9} &= \sqrt{100} = 10 \end{aligned} \]

9 and 11 are the values of the inner roots, picked up by stopping too early, and 12 does not arise at any step.

Answer: B

There are 28 plants in each row.

If the number of plants in a row equals the number of rows, the plants form a square, so the number in each row is the square root of the total:

\[ \sqrt{784} = 28 \quad (28 \times 28 = 784) \]

A quick check from the last digit: 784 ends in 4, so its root ends in 2 or 8, and 784 lies between 20² = 400 and 30² = 900, so the root is 28. 18² = 324, 38² = 1444 and 48² = 2304 are all wrong.

Answer: D

The value is \(\frac{205}{84}\).

Combine under one root and factorise: \(5043 = 3 \times 41^2\), \(1323 = 3^3 \times 7^2\), \(50 = 2 \times 5^2\) and \(32 = 2^5\).

\[ \begin{aligned} \sqrt{\frac{5043 \times 50}{32 \times 1323}} &= \sqrt{\frac{41^2 \times 5^2}{2^4 \times 3^2 \times 7^2}} \\ &= \frac{41 \times 5}{4 \times 3 \times 7} \\ &= \frac{205}{84} \end{aligned} \]

The key step is \(\sqrt{5043} = 41\sqrt{3}\); taking 5043 as \(43^2 \times 3\) leads to 215, and \(\frac{205}{42}\) is exactly twice the true value. Check: \(\sqrt{32} \times \sqrt{1323} = 4\sqrt{2} \times 21\sqrt{3}\) and \(\sqrt{5043} \times \sqrt{50} = 41\sqrt{3} \times 5\sqrt{2}\), so the ratio is \(\frac{205}{84}\).

Answer: A

The value of \(3x + 2\) is 20.

\[ \begin{aligned} 30^2 &= 900, \quad 31^2 = 961 \\ x &= 961 - 955 = 6 \\ 3x + 2 &= 18 + 2 = 20 \end{aligned} \]

The next perfect square above 955 is 961, so the least number to add is 6. 14, 11 and 5 come from wrong values of x.

Answer: C

The value is 1.2.

Work from the inside: √0.04 = 0.2, √0.0009 = 0.03, √1.44 = 1.2 and √0.0001 = 0.01. Their sum is 1.44.

\[ \sqrt{0.2+0.03+1.2+0.01}=\sqrt{1.44}=1.2 \]

1, 1.1 and 1.3 do not equal the square root of 1.44.

The inside roots are all perfect squares of decimals, 0.2 × 0.2 = 0.04, 0.03 × 0.03 = 0.0009, 1.2 × 1.2 = 1.44 and 0.1 × 0.1 = 0.01 (so the root of 0.0001 is 0.01). Their sum is 1.44, and 1.2 × 1.2 is 1.44, which gives the answer.

Answer: D

The value of (3x - 5) is 16.

Find the largest perfect square not exceeding 7751. Since 88² = 7744 and 89² = 7921, the nearest square below is 7744.

\[ \begin{aligned} x &= 7751 - 7744 = 7 \\ 3x - 5 &= 21 - 5 = 16 \end{aligned} \]

Subtracting any smaller number leaves a number between 7744 and 7751, which is not a perfect square, so 7 is the smallest. The value 10 comes from using x = 5, and 13 from x = 6, neither of which works.

Answer: B

\(\sqrt[3]{0.000216}\\ = \left ( \dfrac{216}{10^6} \right )^{\dfrac{1}{3}}\\ = \left ( \dfrac {6 \times 6 \times 6}{10^2 \times 10^2 \times 10^2} \right )^{\dfrac{1}{3}}\\ = \dfrac{6}{10^2}\\ = \dfrac{6}{100}\\ = 0.06\)

Answer: C

Work outward from the inner root.

\[ \sqrt{64} = 8 \quad\Longrightarrow\quad \sqrt{248 + 8} = \sqrt{256} \]

\[ = 16 \]

Answer: C

Each student gives as many paise as there are students, so the total is the square of the number of students.

\[ \text{Rs. } 59.29 = 5929 \text{ paise} \]

\[ \sqrt{5929} = 77 \]