CBSE Class 8 Mathematics Cubes And Cube Roots Worksheet Set 04

Question. Which of the following numbers is a perfect cube ?
(A) 108
(B) 343
(C) 243
(D) 5324
Answer : B

Question. Find the value of √³√0.000064 .
(A) 0.2
(C) 0.002
(B) 0.02
(D) 0.04 
Answer : A

Question. Evaluate 3√343 x 125/0.064 .
(A) 875
(B) 8.75
(C) 807.5
(D) 87.5
Answer : D

Question. What is the volume of a cube of edge 15 cm?
(A) 15625 cm³
(C) 3375cm³
(B) 3.375 cm³
(D) 225 cm³
Answer : C

Question. Find the value of ³√0.000216 .
(A) 0.006
(C) 0.36
(B) 0.6
(D) 0.06
Answer : D

Question. Find the cube root of 0.001728.
(A) 0.12
(B) 1.2
(C) 12
(D) 2.6
Answer : A

Question. The quotient of the cube root of a number and 25 is 5. Identify the number.
(A) 53
(B) 25³
(C) 125²
(D) 125³
Answer : D

Question. The volume of liquid in a measuringjar is 200 cm³. A cube of side 7 em is immersed in the liquid. What is the reading on the measuring jar?
(A) 800 cm³
(C) 643 cm³
(B) 543 cm³
(D) 843 cm³
Answer : B

Question. A metallic cuboid with dimensions 16 em x 8 em x 4 em was melted and recast into a cube. What is the length of the edge of the cube?
(A) 8 cm
(B) 2 cm
(C) 4 cm
(D) 16 cm
Answer : A

Question. A cube is made of 24389 cm³ of wood. What is the measure of its edge ?
(A) 39 cm
(B) 29 cm
(C) 2.9 cm
(D) 19 cm
Answer : B

Question. If P = ³√³√a⁹ what is the value of P?
(A) ( a³)³
(B) a³
(C) a⁹
(D) a
Answer : D

Question. Evaluate ³√2744 ÷ ³√0.008.
(A) 70
(B) - 70
(C)14
(D) - 14
Answer : B

Question. Identify the digit in the units place of ³√21952.
(A) 8
(B) 4
(C) 6
(D) 7
Answer : A

Question. By which number should 5184 be divided to make it a perfect cube ?
(A) 2
(B) 4
(C) 3
(D) 6
Answer : C

Question. The volume of material used to make a cube is 4913 cm³. What is the length of the edge of the cube ?
(A) 16 cm
(B) 27 cm
(C) 18 cm
(D) 17 cm
Answer : D

Question. A tank holds 27000 litres of water. How many litres of water would a similar pond hold, if its dimensions were double the first one?
(A) 17286 litres
(C) 15625 litres
(B) 216000 litres
(D) 432000 litres
Answer : B

Question 1. Find the cube of 18.
Answer: Cube of 18 = \(18^3 = 18 \times 18 \times 18\)
\(18 \times 18 = 324\)
\(324 \times 18 = 5832\)
In simple words: Multiply 18 by itself three times to get its cube.

Exam Tip: Memorize cubes of numbers from 1 to 20 for faster calculations in exams.

Question 2. Express \(6^3\) as the sum of odd numbers using the pattern given below.
\(1 = 1 = 1^3\)
\(3 + 5 = 8 = 2^3\)
\(7 + 9 + 11 = 27 = 3^3\)
\(13 + 15 + 17 + 19 = 64 = 4^3\)
\(21 + 23 + 25 + 27 + 29 = 125 = 5^3\)
Answer: Following the consecutive odd number pattern for each cube:
For \(1^3\): 1 term starting from 1
For \(2^3\): 2 terms starting from 3
For \(3^3\): 3 terms starting from 7
For \(4^3\): 4 terms starting from 13
For \(5^3\): 5 terms starting from 21
Therefore, for \(6^3\), we need 6 consecutive odd numbers starting after 29:
\(6^3 = 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3\)
In simple words: The cube of any number n can be written as the sum of n consecutive odd numbers.

Exam Tip: Notice that the starting number of the sequence for \(n^3\) is given by \(n(n-1) + 1\).

Question 3. Show that 189 is not a perfect cube.
Answer: Finding the prime factorization of 189:
\(189 = 3 \times 63 = 3 \times 3 \times 21 = 3 \times 3 \times 3 \times 7 = 3^3 \times 7\)
Since the prime factor 7 does not appear in a group of three (its exponent is 1, not a multiple of 3), 189 is not a perfect cube.
In simple words: A number is a perfect cube only if all its prime factors can be grouped in triplets.

Exam Tip: Always prime factorize the number and check if every prime factor's power is a multiple of 3.

Question 4. Find the number whose cube is 9261.
Answer: Performing prime factorization of 9261:
\(9261 \div 3 = 3087\)
\(3087 \div 3 = 1029\)
\(1029 \div 3 = 343\)
\(343 \div 7 = 49\)
\(49 \div 7 = 7\)
\(7 \div 7 = 1\)
So, \(9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 = 3^3 \times 7^3 = (3 \times 7)^3 = 21^3\)
Therefore, the number whose cube is 9261 is 21.
In simple words: Take one prime factor from each triplet of identical prime factors and multiply them.

Exam Tip: Group prime factors into triplets and take one factor from each triplet.

Question 5. Find the cube root of 512.
Answer: Finding the prime factorization of 512:
\(512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9\)
Cube root of 512 = \(\sqrt[3]{512} = \sqrt[3]{2^9} = 2^{9/3} = 2^3 = 8\)
In simple words: Group the prime factors of 512 into triplets of 2s, giving \(2 \times 2 \times 2 = 8\).

Exam Tip: Memorize powers of 2 up to \(2^{10} = 1024\) for quick solving.

Question 6. Evaluate: \((0.8)^3\)
Answer: \((0.8)^3 = 0.8 \times 0.8 \times 0.8\)
\(0.8 \times 0.8 = 0.64\)
\(0.64 \times 0.8 = 0.512\)
In simple words: Multiply decimal 0.8 by itself three times, keeping three decimal places in the final result.

Exam Tip: Count the total number of decimal places in the factors to place the decimal point correctly.

Question 7. Find the cube of \(1\frac{2}{3}\).
Answer: Convert mixed fraction into improper fraction:
\(1\frac{2}{3} = \frac{5}{3}\)
Cube of \(\frac{5}{3} = \left(\frac{5}{3}\right)^3 = \frac{5^3}{3^3} = \frac{125}{27}\)
In simple words: Cubing a fraction means cubing both its numerator and denominator separately.

Exam Tip: Always convert mixed fractions to improper fractions before performing operations like squaring or cubing.

Question 8. Evaluate: \(\sqrt[3]{4^3 \times 6^3}\)
Answer: Using properties of exponents and radicals:
\(\sqrt[3]{4^3 \times 6^3} = (4^3 \times 6^3)^{\frac{1}{3}} = (4^3)^{\frac{1}{3}} \times (6^3)^{\frac{1}{3}} = 4 \times 6 = 24\)
Alternatively:
\(\sqrt[3]{(4 \times 6)^3} = \sqrt[3]{24^3} = 24\)
In simple words: The cube root of a product of cubes is simply the product of the numbers themselves.

Exam Tip: Apply the rule \(\sqrt[3]{a^3 \cdot b^3} = a \cdot b\) to save calculation time.

Question 9. Find the one's digit of the cube of each of the following:
(a) 1024
(b) 71
Answer:
(a) The one's digit of 1024 is 4.
The one's digit of the cube of 1024 is the unit digit of \(4^3\), which is 6 (since \(4 \times 4 \times 4 = 64\)).
(b) The one's digit of 71 is 1.
The one's digit of the cube of 71 is the unit digit of \(1^3\), which is 1 (since \(1 \times 1 \times 1 = 1\)).
In simple words: The unit digit of a cube depends entirely on the unit digit of the original number.

Exam Tip: You only need to cube the last digit of the number to find the unit digit of its cube.

Question 10. Find the smallest number by which 12500 must be multiplied so that the product is a perfect cube.
Answer: Finding the prime factorization of 12500:
\(12500 = 125 \times 100 = (5^3) \times (10^2) = 5^3 \times (2 \times 5)^2 = 5^3 \times 2^2 \times 5^2 = 2^2 \times 5^5\)
To make the exponents multiples of 3:
For \(2^2\), we need one more 2 (exponent becomes \(2 + 1 = 3\)).
For \(5^5\), we need one more 5 (exponent becomes \(5 + 1 = 6\)).
Smallest multiplier = \(2^1 \times 5^1 = 2 \times 5 = 10\)
In simple words: Multiply by the missing prime factors needed to complete triplets for every prime factor.

Exam Tip: Identify deficient prime factors whose powers are not multiples of 3 and multiply by their required deficits.

Question 11. Find the smallest number by which 704 must be divided to obtain a perfect cube.
Answer: Finding the prime factorization of 704:
\(704 = 2 \times 352 = 2 \times 2 \times 176 = 2^3 \times 88 = 2^3 \times 2^3 \times 11 = 2^6 \times 11^1\)
To make the quotient a perfect cube, we must eliminate factors that do not form a complete triplet.
Here, \(2^6\) is already a cube, but \(11^1\) has an exponent of 1.
Therefore, 704 must be divided by 11.
In simple words: Divide the number by any leftover prime factors that cannot form a triplet.

Exam Tip: For division questions, divide out the prime factors that lack a triplet companion.

Question 12. Find the cube root of \(140 \times 2450\).
Answer: Let \(N = 140 \times 2450\). Breaking into prime factors:
\(140 = 14 \times 10 = 2 \times 7 \times 2 \times 5 = 2^2 \times 5 \times 7\)
\(2450 = 245 \times 10 = 49 \times 5 \times 2 \times 5 = 2 \times 5^2 \times 7^2\)
Combining both:
\(N = (2^2 \times 5 \times 7) \times (2 \times 5^2 \times 7^2) = 2^3 \times 5^3 \times 7^3\)
Cube root = \(\sqrt[3]{2^3 \times 5^3 \times 7^3} = 2 \times 5 \times 7 = 70\)
In simple words: Factorize both numbers, combine their prime factors, and take one factor from each triplet.

Exam Tip: Combine numbers under a single radical before prime factorization to simplify grouping.

Question 13. Find the cube root of \(\frac{125}{512}\).
Answer: \(\sqrt[3]{\frac{125}{512}} = \frac{\sqrt[3]{125}}{\sqrt[3]{512}}\)
Since \(125 = 5^3\) and \(512 = 8^3\):
\(\frac{\sqrt[3]{5^3}}{\sqrt[3]{8^3}} = \frac{5}{8}\)
In simple words: Take the cube root of the numerator and denominator separately.

Exam Tip: Remember standard cubes like \(5^3 = 125\) and \(8^3 = 512\) to solve instantly.

Question 14. Find the cube root of 10648 by prime factorisation method.
Answer: Finding the prime factorization of 10648:
\(10648 \div 2 = 5324\)
\(5324 \div 2 = 2662\)
\(2662 \div 2 = 1331\)
\(1331 \div 11 = 121\)
\(121 \div 11 = 11\)
\(11 \div 11 = 1\)
So, \(10648 = 2^3 \times 11^3\)
Cube root = \(\sqrt[3]{2^3 \times 11^3} = 2 \times 11 = 22\)
In simple words: Factorize 10648 into primes, group them in triplets, and multiply one from each group.

Exam Tip: 1331 is a well-known cube of 11; recognizing it speeds up factorization.

Question 15. Find the cube of \(5\frac{2}{7}\).
Answer: Convert mixed fraction to improper fraction:
\(5\frac{2}{7} = \frac{5 \times 7 + 2}{7} = \frac{37}{7}\)
Cube of \(\frac{37}{7} = \frac{37^3}{7^3}\)
\(37^3 = 37 \times 37 \times 37 = 1369 \times 37 = 50653\)
\(7^3 = 343\)
Result = \(\frac{50653}{343}\)
In simple words: Convert to an improper fraction and cube both numerator and denominator.

Question 16. Find the cube root of 1.331.
Answer: Convert decimal to fraction:
\(1.331 = \frac{1331}{1000}\)
Cube root = \(\sqrt[3]{\frac{1331}{1000}} = \frac{\sqrt[3]{1331}}{\sqrt[3]{1000}} = \frac{11}{10} = 1.1\)
In simple words: Express the decimal as a fraction with a denominator that is a cube of 10, then take cube roots.

Exam Tip: Convert decimals to fractions over powers of 10 to simplify root calculations.

Question 17. Evaluate: \(\sqrt[3]{8 \times 17 \times 17 \times 17}\)
Answer: \(\sqrt[3]{8 \times 17 \times 17 \times 17} = \sqrt[3]{2^3 \times 17^3}\)
Using radical properties:
\(= \sqrt[3]{2^3} \times \sqrt[3]{17^3} = 2 \times 17 = 34\)
In simple words: Group identical factors into triplets and take one factor out of the cube root.

Exam Tip: Look for ready triplets like \(17 \times 17 \times 17\) to avoid unnecessary multiplication.

Question 18. Find the cube root of \(27 \times 1728\).
Answer: Taking the cube root of each factor separately:
\(\sqrt[3]{27 \times 1728} = \sqrt[3]{27} \times \sqrt[3]{1728}\)
Since \(27 = 3^3\) and \(1728 = 12^3\):
\(= 3 \times 12 = 36\)
In simple words: Find the cube root of each number individually and multiply them.

Exam Tip: Recognizing that \(1728 = 12^3\) directly avoids lengthy prime factorization.

Question 19. Find the cube root of 5832.
Answer: Finding prime factorization of 5832:
\(5832 \div 2 = 2916\)
\(2916 \div 2 = 1458\)
\(1458 \div 2 = 729\)
\(729 = 9^3 = 3^6\)
So, \(5832 = 2^3 \times 3^6 = (2 \times 3^2)^3 = (2 \times 9)^3 = 18^3\)
Cube root = 18
In simple words: Prime factorize 5832 into \(2^3 \times 3^6\) and take one-third of each exponent.

Exam Tip: Since 729 is \(3^6\), you can quickly combine it with \(2^3\) to find the base.

Question 20. Find the cube root of 91125.
Answer: Finding prime factorization of 91125:
\(91125 \div 5 = 18225\)
\(18225 \div 5 = 3645\)
\(3645 \div 5 = 729\)
\(729 = 3^6\)
So, \(91125 = 5^3 \times 3^6 = (5 \times 3^2)^3 = (5 \times 9)^3 = 45^3\)
Cube root = 45
In simple words: Factorize the number into triplets of prime factors to find its cube root.

Exam Tip: Numbers ending in 5 are always divisible by 5; start dividing by 5 successively.

Question 21. Evaluate: \(\sqrt[3]{125 \times 27}\)
Answer: \(\sqrt[3]{125 \times 27} = \sqrt[3]{125} \times \sqrt[3]{27}\)
Since \(125 = 5^3\) and \(27 = 3^3\):
\(= 5 \times 3 = 15\)
In simple words: Take the cube root of 125 and 27 separately and multiply the results.

Exam Tip: Use the property \(\sqrt[3]{a \cdot b} = \sqrt[3]{a} \cdot \sqrt[3]{b}\) for separated terms.

Question 22. Evaluate: \(\sqrt[3]{700 \times 2 \times 49 \times 5}\)
Answer: Let the expression be \(N\):
\(N = \sqrt[3]{(7 \times 100) \times 2 \times 49 \times 5} = \sqrt[3]{(7 \times 10^2) \times 2 \times 7^2 \times 5}\)
Combine numbers:
\(700 = 7 \times 2^2 \times 5^2\)
\(2 = 2^1\)
\(49 = 7^2\)
\(5 = 5^1\)
Total product under root = \((2^2 \times 5^2 \times 7) \times 2 \times 7^2 \times 5 = 2^3 \times 5^3 \times 7^3\)
Cube root = \(2 \times 5 \times 7 = 70\)
In simple words: Group all factors to form cubes of prime numbers, then take their cube root.

Exam Tip: Convert all numbers into prime factorization form inside the cube root for easy combination.

Question 23. Find the cube root of 32768 through estimation.
Answer: Step 1: Form two groups from the right: 32 and 768.
Step 2: Consider the first group (768). Its unit digit is 8. The cube root of a number ending in 8 has a unit digit of 2 (since \(2^3 = 8\)). So, the unit digit of the result is 2.
Step 3: Consider the second group (32). We know that \(3^3 = 27\) and \(4^3 = 64\). Since 27 < 32 < 64, the ten's digit is 3.
Combining them, the cube root of 32768 is 32.
In simple words: Split the number into two parts, estimate the unit digit from the last part, and the ten's digit from the first part.

Exam Tip: Estimation is much faster than prime factorization for 5-digit perfect cubes in exams.

Question 24. Find the smallest number by which 1600 must be divided so that the quotient is a perfect cube, further find its cube root.
Answer: Prime factorization of 1600:
\(1600 = 16 \times 100 = 2^4 \times (2 \times 5)^2 = 2^4 \times 2^2 \times 5^2 = 2^6 \times 5^2\)
To make the quotient a perfect cube, we must remove the factors that do not form a triplet.
Here, \(2^6\) is a perfect cube, but \(5^2\) lacks one factor of 5.
Thus, 1600 must be divided by \(5^2 = 25\).
Quotient = \(\frac{1600}{25} = 64\)
Cube root of the quotient = \(\sqrt[3]{64} = 4\)
In simple words: Divide by the leftover non-triplet prime factor and then find the cube root of the resulting quotient.

Exam Tip: Always state both the smallest divisor and the final cube root clearly.

Question 25. Sheetal makes a cuboid of sides 5 cm, 2 cm and 5 cm. How many such cuboids will she need to form a cube?
Answer: Volume of one cuboid = \(\text{Length} \times \text{Breadth} \times \text{Height} = 5 \times 2 \times 5 = 5^2 \times 2^1\) cm\(^3\).
To form a cube, the dimensions in prime factorization must have exponents that are multiples of 3.
We have \(5^2\) (needs one more 5) and \(2^1\) (needs two more 2s).
Number of cuboids needed = \(5^1 \times 2^2 = 5 \times 4 = 20\).
In simple words: Multiply the missing prime factors needed to complete triplets to find the number of cuboids.

Exam Tip: This is a practical application of finding the smallest multiplier to make a volume a perfect cube.

Question 26. Find the cube root of the following by prime factorisation:
(i) 8000
(ii) 13824
Answer:
(i) For 8000:
\(8000 = 8 \times 1000 = 2^3 \times 10^3 = (2 \times 10)^3 = 20^3\)
Cube root = 20.
(ii) For 13824:
\(13824 \div 2 = 6912\)
\(6912 \div 2 = 3456\)\(\rightarrow \dots \rightarrow 13824 = 2^9 \times 3^3\)
Cube root = \(\sqrt[3]{2^9 \times 3^3} = 2^3 \times 3 = 8 \times 3 = 24\).
In simple words: Prime factorize each number and take one factor from each triplet of identical primes.

Exam Tip: Separate zeros in powers of 10 to simplify factorization of large numbers.

Question 27. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further find the cube root of the product.
Answer: Prime factorization of 3600:
\(3600 = 36 \times 100 = (2^2 \times 3^2) \times (2^2 \times 5^2) = 2^4 \times 3^2 \times 5^2\)
To make powers multiples of 3:
For \(2^4\), we need \(2^2\) (exponent becomes 6).
For \(3^2\), we need \(3^1\) (exponent becomes 3).
For \(5^2\), we need \(5^1\) (exponent becomes 3).
Smallest multiplier = \(2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60\).
Product = \(3600 \times 60 = 216000\).
Cube root of product = \(\sqrt[3]{216000} = \sqrt[3]{216 \times 1000} = 6 \times 10 = 60\).
In simple words: Multiply by the required deficits to complete triplets, then compute the cube root of the new product.

Exam Tip: Double-check that all prime exponents in the new product are divisible by 3.

Question 28. The three numbers are in the ratio 2 : 3 : 4. The sum of their cubes is 33957. Find the numbers.
Answer: Let the numbers be \(2x\), \(3x\), and \(4x\).
Sum of their cubes = \((2x)^3 + (3x)^3 + (4x)^3 = 33957\)
\(8x^3 + 27x^3 + 64x^3 = 33957\)
\(99x^3 = 33957\)
\(x^3 = \frac{33957}{99} = 343\)
\(x = \sqrt[3]{343} = 7\)
Therefore, the numbers are:
First number = \(2 \times 7 = 14\)
Second number = \(3 \times 7 = 21\)
Third number = \(4 \times 7 = 28\)
In simple words: Assume the numbers with a common variable x, cube them, set up the equation, solve for x, and find the numbers.

Exam Tip: Whenever quantities are given in a ratio, introduce a variable x and write them as \(ax, bx, cx\).

Question 29. The volume of a cube is 9261000 m\(^3\). Find the side of the cube.
Answer: Volume of cube = \(\text{side}^3\)
\(\text{side}^3 = 9261000\)
\(\text{side} = \sqrt[3]{9261000} = \sqrt[3]{9261 \times 1000}\)
Since \(9261 = 21^3\) and \(1000 = 10^3\):
\(\text{side} = 21 \times 10 = 210 \text{ m}\)
In simple words: The side length is the cube root of the given volume.

Exam Tip: Remember units (meters) in your final answer when dealing with word problems.

Question 30. Find the cube root of the following by prime factorisation:
(i) 8000
(ii) 13824
Answer:
(i) For 8000:
\(8000 = 2^3 \times 10^3\)
Cube root = \(2 \times 10 = 20\).
(ii) For 13824:
\(13824 = 2^9 \times 3^3\)
Cube root = \(2^3 \times 3 = 8 \times 3 = 24\).
In simple words: Factorize into prime triplets and take one factor from each triplet.

Exam Tip: This is a recurring question type; master prime factorization for numbers up to 15000.

Most Important Questions

Question 1. Find the cubes of the following numbers:
(a) 5
(b) 13
(c) 50
(d) 120
Answer:
(a) \(5^3 = 5 \times 5 \times 5 = 125\)
(b) \(13^3 = 13 \times 13 \times 13 = 169 \times 13 = 2197\)
(c) \(50^3 = 5^3 \times 1000 = 125 \times 1000 = 125000\)
(d) \(120^3 = 12^3 \times 1000 = 1728 \times 1000 = 1728000\)
In simple words: Multiply each number by itself three times or use powers of 10 shortcuts.

Exam Tip: For numbers ending in zeros, cube the non-zero digit and add three times the number of zeros.

Question 2. The smallest number by which 120393 must be divided, so that the quotient is a perfect cube:
(a) 7
(b) 12
(c) 13
(d) 3
Answer: (a) 7
Prime factorization of 120393:
\(120393 \div 3 = 40131\)
\(40131 \div 3 = 13377\)
\(13377 \div 3 = 4459\)
\(4459 \div 7 = 637\)
\(637 \div 7 = 91\)
\(91 \div 7 = 13\)
\(13 \div 13 = 1\)
So, \(120393 = 3^3 \times 7^3 \times 13^1\). The leftover prime factor with a non-triplet power is 13? Wait, let us check carefully:
\(3^3 \times 7^3 \times 13^1\)? Let us re-verify division:
\(120393 \div 7 = 17199\)
\(17199 \div 7 = 2457\)
\(2457 \div 7 = 351\)
\(351 \div 3 = 117\)
\(117 \div 3 = 39\)
\(39 \div 3 = 13\)
Thus \(120393 = 7^3 \times 3^3 \times 13^1 \dots\) wait, let's check option (a) 7. If divided by 7, one 7 cancels out, leaving \(7^2\), which is not a cube. The question asks to divide so quotient is a perfect cube. If factors are \(7^3 \times 3^3 \times 13^1\), we must divide by 13. Wait, let's re-read option (a) 7, (c) 13. Let's check 13: if divided by 13, quotient is \(7^3 \times 3^3\), which is a perfect cube! Therefore, the smallest number is 13 (Option c).
In simple words: Divide by the prime factor that does not form a triplet.

Exam Tip: Ensure all prime factors in the quotient have exponents that are multiples of 3.

Question 3. Which of the following is the cube of an odd natural number:
(a) 32678
(b) 4096
(c) 6859
(d) 1728
Answer: (c) 6859
The cube of an odd number is always odd. Out of the options:
(a) 32678 is even.
(b) 4096 is even.
(c) 6859 is odd (\(19^3 = 6859\)).
(d) 1728 is even.
In simple words: Odd numbers always have odd cubes, so eliminate all even numbers first.

Exam Tip: Use parity rules: cube of even is even, cube of odd is odd.

Question 4. Which of the following are the cubes of even natural numbers?
(a) 729
(b) 3375
(c) 1331
(d) 13824
Answer: (d) 13824
An even natural number's cube must be even. Out of the given numbers, only 13824 is even (\(24^3 = 13824\)). 729, 3375, and 1331 are all odd numbers.
In simple words: Only an even number can produce an even cube.

Question 5. Match the items in list A with suitable items in list B.
List A:
A. \(\sqrt[3]{\frac{27}{5832}}\)
B. \(\sqrt[3]{\frac{343}{125}}\)
C. \(\sqrt[3]{\frac{-2197}{-512}}\)
List B:
(i) \(\frac{7}{5}\)
(ii) \(\frac{13}{8}\)
(iii) \(\frac{1}{6}\)
(iv) \(\frac{-13}{8}\)
Code:
(a) A-ii, B-iv, C-i
(b) A-iii, B-i, C-ii
(c) A-iv, B-ii, C-iii
(d) A-i, B-iii, C-iv
Answer: (b) A-iii, B-i, C-ii
Evaluating each item:
A. \(\sqrt[3]{\frac{27}{5832}} = \frac{3}{18} = \frac{1}{6}\) (Matches iii)
B. \(\sqrt[3]{\frac{343}{125}} = \frac{7}{5}\) (Matches i)
C. \(\sqrt[3]{\frac{-2197}{-512}} = \sqrt[3]{\frac{2197}{512}} = \frac{13}{8}\) (Matches ii)
In simple words: Simplify each fraction's cube root individually to find the correct match.

Exam Tip: Negative signs in both numerator and denominator cancel each other out.

Question 6. If \(x = \left(\frac{729}{2197}\right)^{1/3} + \left(\frac{9261}{42875}\right)^{1/3}\), then x =
(a) \(\frac{84}{65}\)
(b) \(\frac{94}{65}\)
(c) \(\frac{104}{65}\)
(d) \(\frac{124}{65}\)
Answer: (a) \(\frac{84}{65}\)
Evaluate each cube root term:
\(\left(\frac{729}{2197}\right)^{1/3} = \frac{\sqrt[3]{729}}{\sqrt[3]{2197}} = \frac{9}{13}\)
\(\left(\frac{9261}{42875}\right)^{1/3} = \frac{\sqrt[3]{9261}}{\sqrt[3]{42875}} = \frac{21}{35} = \frac{3}{5}\)
Now add them:
\(x = \frac{9}{13} + \frac{3}{5} = \frac{9 \times 5 + 3 \times 13}{65} = \frac{45 + 39}{65} = \frac{84}{65}\)
In simple words: Take the cube root of each fraction separately and add them using a common denominator.

Exam Tip: Simplify fractions before adding if possible (e.g. \(\frac{21}{35} = \frac{3}{5}\)).

Question 7. If \(x = \left(\frac{4}{9}\right)^3 + \sqrt[3]{\frac{2744}{729}}\), then x =
(a) \(\frac{1298}{729}\)
(b) \(\frac{1200}{729}\)
(c) \(\frac{1198}{729}\)
(d) \(\frac{1100}{729}\)
Answer: (a) \(\frac{1298}{729}\)
Evaluate each term:
\(\left(\frac{4}{9}\right)^3 = \frac{64}{729}\)
\(\sqrt[3]{\frac{2744}{729}} = \frac{\sqrt[3]{2744}}{\sqrt[3]{729}} = \frac{14}{9}\)
Now add them with a common denominator of 729:
\(\frac{14}{9} = \frac{14 \times 81}{9 \times 81} = \frac{1134}{729}\)
\(x = \frac{64}{729} + \frac{1134}{729} = \frac{1198}{729}\) Wait, let us re-add: \(64 + 1134 = 1198\), which matches option (c). Let us check options: (c) is \(\frac{1198}{729}\).
In simple words: Compute the cube and cube root separately, then convert to a common denominator to add.

Exam Tip: Be careful with fraction addition and ensure denominators match before adding numerators.

Question 8. One's place digit in the cube of 833 is
(a) 7
(b) 3
(c) 9
(d) 1
Answer: (a) 7
The one's digit of 833 is 3.
The cube of 3 is \(3^3 = 27\).
The one's digit of 27 is 7.
In simple words: Cube the last digit of the number to find the unit digit of its cube.

Exam Tip: Only the unit digit of the base number determines the unit digit of any power.

Question 9. What is the cube root of 531441?
Answer: Finding the prime factorization of 531441:
\(531441 \div 9 = 59049 \dots\) Actually, let's use prime factorization by 3:
\(531441 = 3^{12} = (3^4)^3 = 81^3\)
Cube root = 81.
In simple words: Express the number as a cube of 81 using prime factorization.

Exam Tip: Recognize powers of 3 like \(3^6 = 729\) and \(3^{12} = 531441\).

Question 10. Find the smallest number by which 243 must be multiplied so that the product is a perfect cube.
Answer: Prime factorization of 243:
\(243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5\)
To make the exponent a multiple of 3, we need two more factors of 3 (since \(5 + 1 = 6\)).
Smallest multiplier = \(3^2 = 9\).
In simple words: Multiply by enough prime factors to complete a triplet.

Exam Tip: Subtract the exponent from the next multiple of 3 to find how many factors are needed.

Question 11. If \(\sqrt[3]{\frac{x}{y}} = \frac{2}{3}\), then the value of \(\frac{x}{y}\) is
(a) \(\frac{2}{3}\)
(b) \(\frac{4}{9}\)
(c) \(\frac{8}{27}\)
(d) None of the options
Answer: (c) \(\frac{8}{27}\)
Cubing both sides of the equation:
\(\left(\sqrt[3]{\frac{x}{y}}\right)^3 = \left(\frac{2}{3}\right)^3\)
\(\frac{x}{y} = \frac{2^3}{3^3} = \frac{8}{27}\)
In simple words: Cube both sides of the equation to eliminate the cube root.

Question 12. If 9A is a perfect cube number what will be the value of A.
Answer: Let \(9A = k^3\).
Prime factorization of 9 is \(3^2\).
For \(3^2 \times A\) to be a cube, A must supply at least one 3 and another 9 to make \(3^3 \times 3^3 = 27^3\) or similar.
Specifically, \(9A = 3^2 \times A\). To make the exponent of 3 a multiple of 3, A must contain \(3^1 \times \text{(some cube)}\).
If \(A = 3 \times 3^2 = 27\), then \(9A = 9 \times 27 = 243\) (not a cube).
Let us test values: if \(A = 243\), then \(9 \times 243 = 2187\) (not a cube).
Wait, let's solve \(9A = k^3 \implies 3^2 \times A = k^3\). The smallest value for A is when \(A = 3\), giving \(9 \times 3 = 27 = 3^3\).
Therefore, the smallest value of A is 3.
In simple words: A must provide the missing prime factors to make the total product a perfect cube.

Exam Tip: Set the prime factorization equal to a cube and solve for the unknown variable.

Question 13. What will be the Unit place digit in cube root of 1331?
Answer: The cube root of 1331 is 11.
The unit place digit of 11 is 1.
In simple words: The cube root of 1331 is 11, so its unit digit is 1.

Exam Tip: A number ending in 1 always has a cube root ending in 1.

Question 14. Find the cube root of \(2300 \times 5290\).
Answer: Let \(N = 2300 \times 5290\).
\(2300 = 23 \times 100 = 23 \times 10^2\)
\(5290 = 529 \times 10 = 23^2 \times 10^1 = 23^2 \times 10\)
Combining them:
\(N = (23 \times 10^2) \times (23^2 \times 10) = 23^3 \times 10^3\)
Cube root = \(\sqrt[3]{23^3 \times 10^3} = 23 \times 10 = 230\)
In simple words: Express both numbers in prime factored form and take one factor from each triplet.

Exam Tip: Notice that \(529 = 23^2\), which makes combining factors straightforward.

Question 15. Find the value of \(\sqrt[3]{392} \times \sqrt[3]{448}\).
Answer: Combine the cube roots under a single radical:
\(\sqrt[3]{392 \times 448}\)
Prime factorization of 392: \(392 = 8 \times 49 = 2^3 \times 7^2\)
Prime factorization of 448: \(448 = 64 \times 7 = 2^6 \times 7^1\)
Product under radical = \((2^3 \times 7^2) \times (2^6 \times 7^1) = 2^9 \times 7^3\)
Cube root = \(\sqrt[3]{2^9 \times 7^3} = 2^3 \times 7 = 8 \times 7 = 56\)
In simple words: Multiply the numbers under a single cube root, then factorize to find the result.

Exam Tip: Use the property \(\sqrt[3]{a} \cdot \sqrt[3]{b} = \sqrt[3]{a \cdot b}\) to simplify complex root multiplications.