Question 1

Find the cube root of 0.000216.

Accepted Answer

Problem restatement Evaluate √√√(0.000216). Concept/Approach Use the property √√√(a × 10k) = √√√(a) × 10k/3. Recognize 0.000216 = 216 × 10−6. Step-by-step calculation √√√(0.000216) = √√√(216) × √√√(10−6)= 6 × 10−2 = 0.06 Verification/Alternative (0.06)3 = 0.000216, confirming the result. Common pitfalls Taking cube root of the exponent incorrectly (e.g., using 10−3 instead of 10−2). Final Answer 0.06

Question 2

A group collects as many paise from each member as the number of members. If the total collection is ₹59.29, how many members are there in the group?

Accepted Answer

Problem restatement A group has n members. Each member contributes n paise (i.e., n/100 rupees). The total collected is ₹59.29. Find n. Given data Total collection = ₹59.29 = 5929 paise. Each member pays = n paise; number of members = n. Concept/Approach Total in paise = n (members) × n (paise per member) = n2 paise. Equate to 5929. Step-by-step calculation n2 = 5929 n = √5929 = 77 Verification/Alternative 77 members paying 77 paise each gives 77 × 77 = 5929 paise = ₹59.29. Common pitfalls Treating ₹59.29 directly as 59.29 units instead of converting to paise to match the per-member amount. Final Answer 77

Question 3

If x = √3 + 1 and y = √3 − 1, then what is x² + y²?

Accepted Answer

Problem restatement Evaluate x2 + y2 for x = √3 + 1 and y = √3 − 1. Given data x = √3 + 1 y = √3 − 1 Concept/Approach Square each expression and add; the radical cross-terms cancel. Step-by-step calculation x2 = (√3 + 1)2 = 3 + 2√3 + 1 = 4 + 2√3 y2 = (√3 − 1)2 = 3 − 2√3 + 1 = 4 − 2√3 x2 + y2 = (4 + 2√3) + (4 − 2√3) = 8 Verification/Alternative Use identity: (a + b)2 + (a − b)2 = 2(a2 + b2) with a = √3, b = 1 ⇒ 2(3 + 1) = 8. Final Answer 8

Question 4

How many two-digit numbers have the unit digit of their square equal to 8?

Accepted Answer

Problem restatement Count two-digit integers n such that the last digit of n2 is 8. Concept/Approach Squares modulo 10 only take certain last digits: for units digit d ∈ {0,…,9}, d2 ends with {0,1,4,9,6,5,6,9,4,1} respectively. Step-by-step reasoning No square ends with 8 for any digit 0–9. Final Answer 0

Question 5

Evaluate and simplify the expression √5 − 10 + √125 (use √5 = 2.236 if needed).

Accepted Answer

6√5 − 10

Question 6

Compute √625 × 14 × 11.

Accepted Answer

3850

Question 7

Find x if √0.0169 × x = 1.3.

Accepted Answer

Problem restatement Solve for x in √0.0169 × x = 1.3. Given data √0.0169 = 0.13 (since 0.13 × 0.13 = 0.0169). Concept/Approach Isolate x by dividing both sides by 0.13. Step-by-step calculation 0.13 × x = 1.3 x = 1.3 ÷ 0.13 = 10 Verification/Alternative Check: √0.0169 × 10 = 0.13 × 10 = 1.3 (satisfies). Final Answer 10

Question 8

What is √64009 ?

Accepted Answer

Problem restatement Find the integer square root of 64009. Concept/Approach Test nearby squares around 2502 = 62500. Step-by-step calculation (250 + 3)2 = 2502 + 2×250×3 + 32 = 62500 + 1500 + 9 = 64009 Final Answer 253

Question 9

Evaluate √[(7 + 3√5)(7 − 3√5)].

Accepted Answer

Problem restatement Compute the principal square root of the product (7 + 3√5)(7 − 3√5). Concept/Approach Use the identity (a + b)(a − b) = a2 − b2. Step-by-step calculation (7 + 3√5)(7 − 3√5) = 72 − (3√5)2 = 49 − 9×5 = 49 − 45 = 4 √4 (principal) = 2 Common pitfalls Reporting ±2; by convention, “square root” means the non-negative principal root. Final Answer 2

Question 10

Simplify (√3 − 1)².

Accepted Answer

Problem restatement Expand and simplify the binomial square (√3 − 1)2. Concept/Approach Use (a − b)2 = a2 − 2ab + b2. Step-by-step calculation (√3 − 1)2 = (√3)2 − 2(√3)(1) + 12 = 3 − 2√3 + 1 = 4 − 2√3 Verification/Alternative Numeric check: √3 ≈ 1.732 ⇒ (1.732 − 1)2 ≈ 0.7322 ≈ 0.536; 4 − 2√3 ≈ 4 − 3.464 ≈ 0.536 (matches). Final Answer 4 − 2√3

Question 11

Find the least perfect square that is divisible by each of 21, 36, and 66.

Accepted Answer

Problem restatement Determine the smallest square number that is a multiple of 21, 36, and 66. Concept/Approach The least such square is the square of the least common multiple with all prime exponents adjusted to be even. Compute LCM, then multiply by missing prime factors to make exponents even. Step-by-step calculation Prime factors: 21 = 3 × 7; 36 = 22 × 32; 66 = 2 × 3 × 11 LCM = 22 × 32 × 7 × 11 = 4 × 9 × 77 = 2772 To be a perfect square, exponents must be even. Current odd primes: 71, 111. Multiply LCM by 7 × 11 = 77 ⇒ N = 2772 × 77 = 213444 Check: 213444 = (2 × 3 × 7 × 11)2 = 4622 Verification/Alternative Confirm divisibility: 4622 is divisible by 21, 36, and 66 since 462 contains factors 2, 3, 7, 11 with sufficient multiplicity when squared. Common pitfalls Taking just the LCM (2772) instead of making it a perfect square. Forgetting to square the odd-power primes. Final Answer 213444

Question 12

Find x such that both occurrences of x in the equation x = (√162 × √128) / x are equal.

Accepted Answer

Problem restatement Determine the common value of x that satisfies x = (√162 × √128) / x. Given data Equation: x = (√162 × √128) / x Concept/Approach Use √a × √b = √(ab). Multiply both sides by x to obtain a quadratic in x. Step-by-step calculation x = √162 × √128 / x = √(162 × 128) / xx2 = √(20736) = 144x = √144 = 12 (taking the positive principal value) Verification/Alternative Check: √162 × √128 = √20736 = 144; then (144) / 12 = 12, which equals x. Common pitfalls Forgetting to combine radicals or missing that x appears in denominator; also note that in typical aptitude contexts, the positive root is expected. Final Answer 12

Question 13

Given 3√5 + √125 = 17.88, find the value of √80 + 6√5.

Accepted Answer

Problem restatement Use the given radical equation to evaluate √80 + 6√5. Given data 3√5 + √125 = 17.88 Concept/Approach Simplify √125 and √80 in terms of √5; then substitute the derived value of √5. Step-by-step calculation √125 = √(25 × 5) = 5√53√5 + 5√5 = 8√5 = 17.88 ⇒ √5 = 17.88 / 8 = 2.235√80 = √(16 × 5) = 4√5√80 + 6√5 = (4 + 6)√5 = 10√5 = 10 × 2.235 = 22.35 Verification/Alternative Directly, √80 + 6√5 = 4√5 + 6√5 = 10√5, which follows from the given equation. Common pitfalls Not simplifying radicals to a common base (√5), or approximating √5 prematurely. Final Answer 22.35

Question 14

If a = 0.1039, evaluate √(4a² − 4a + 1) + 3a.

Accepted Answer

Problem restatement Compute √(4a2 − 4a + 1) + 3a for a = 0.1039. Concept/Approach Recognize a perfect-square trinomial: 4a2 − 4a + 1 = (2a − 1)2. Then √( (2a − 1)2 ) = |2a − 1|. Step-by-step calculation 2a − 1 = 2(0.1039) − 1 = 0.2078 − 1 = −0.7922|2a − 1| = 0.79223a = 3 × 0.1039 = 0.3117Sum = 0.7922 + 0.3117 = 1.1039 Verification/Alternative Algebraic shortcut: |2a − 1| + 3a = (1 − 2a) + 3a = 1 + a, since 2a − 1 < 0. With a = 0.1039 ⇒ 1 + a = 1.1039. Common pitfalls Dropping the absolute value when taking the square root of a square. Final Answer 1.1039

Question 15

Find √1.5625.

Accepted Answer

Problem restatement Compute the principal square root of 1.5625. Concept/Approach Note that 1.252 = (5/4)2 = 25/16 = 1.5625. Step-by-step calculation √1.5625 = √(25/16) = 5/4 = 1.25 Common pitfalls Incorrect decimal squaring; using 1.24 or 1.26 by rough rounding. Final Answer 1.25

Question 16

Find the cube root of 19683.

Accepted Answer

In 19683, 19 lies between 23 and 33, so left digit is 2 and 683 ends with 3, so right digit is 7. thus 27 is a cube root of 19683.

Question 17

If √ y/169 = 54/39, then y is equal to?

Accepted Answer

√y/169= 54/39 ⇒ y/169 = (54/39) x (54/39) ∴ y= (54/39) x (54/39) x 169 = 324

Question 18

112/√196 x √579/12 x √256/8 =?

Accepted Answer

Given Expression 112/√196 x √579/12 x √256/8 =(112/14) x (24/12) x (16/8) = 32

Question 19

Given that √ 4096 = 64, the value of √ 4096 + √ 40.96 + √ .004096 is?

Accepted Answer

√4096 + √40.96 + √.004096 = √4096 + √4096/100 + √4096/1000000 =√4096 + √4096/10 + √4096/1000 = 64 + 64/10 + 64/1000 = 70.464

Question 20

√ .04

Accepted Answer

√.04 = √4/100 = 2/10 = 0.2

Square Root and Cube Root Questions