CBSE Class 8 Mathematics Rational Numbers Worksheet Set 04

CHAPTER - 1. (RATIONAL NUMBERS)

Question 1. Rearrange suitably and find sum.
i) \(-5 + \frac{3}{10} + \frac{3}{7} + (-3) + \frac{5}{14} + \frac{1}{20}\)
Answer: Grouping terms with common denominators:
\(= \left(-5 - 3\right) + \left(\frac{3}{10} + \frac{1}{20}\right) + \left(\frac{3}{7} + \frac{5}{14}\right)\)
\(= -8 + \left(\frac{6 + 1}{20}\right) + \left(\frac{6 + 5}{14}\right)\)
\(= -8 + \frac{7}{20} + \frac{11}{14}\)
Finding LCM of 20 and 14 which is 140:
\(= -8 + \frac{49 + 110}{140} = -8 + \frac{159}{140} = -8 + 1 \frac{19}{140} = -6 \frac{121}{140}\)
In simple words: Rearrange terms to make addition easier by combining fractions with similar denominators.

Exam Tip: Always look for common denominators or multiples to simplify fractions before adding.

Question 1. (cont.)
ii) \(\frac{-3}{5} \times \frac{1}{4} + \frac{3}{7} \times \frac{-1}{5} + \frac{1}{7} \times \frac{1}{8}\)
Answer: Using commutative and distributive properties:
\(= \left(\frac{-3}{5} \times \frac{1}{4}\right) + \left(\frac{-1}{5} \times \frac{3}{7}\right) + \left(\frac{1}{56}\right)\)
Factor out \(\frac{1}{5}\) from the first two terms:
\(= \frac{1}{5} \left(\frac{-3}{4} + \frac{-3}{7}\right) + \frac{1}{56}\)
\(= \frac{1}{5} \left(\frac{-21 - 12}{28}\right) + \frac{1}{56}\)
\(= \frac{1}{5} \left(\frac{-33}{28}\right) + \frac{1}{56} = \frac{-33}{140} + \frac{1}{56}\)
LCM of 140 and 56 is 280:
\(= \frac{-33 \times 2 + 1 \times 5}{280} = \frac{-66 + 5}{280} = \frac{-61}{280}\)
In simple words: Group common factors and apply the distributive property to simplify multiplication.

Exam Tip: Look for common factors in products to apply the distributive law \(a \times b + a \times c = a \times (b + c)\).

Question 2. Represent on number line:-\)
i) \(-3 \frac{2}{5}\)     ii) \(\frac{-6}{7}\)     iii) \(5 \frac{1}{3}\)
Answer:
i) \(-3 \frac{2}{5}\) lies between -3 and -4. Divide the segment between -3 and -4 into 5 equal parts and take 2 parts to the left of -3.
ii) \(\frac{-6}{7}\) lies between 0 and -1. Divide the segment between 0 and -1 into 7 equal parts and take 6 parts to the left of 0.
iii) \(5 \frac{1}{3}\) lies between 5 and 6. Divide the segment between 5 and 6 into 3 equal parts and take 1 part to the right of 5.
In simple words: Locate the whole number part first, then divide the next unit segment according to the denominator.

Exam Tip: Ensure all subdivisions between integers are equal in length.

Question 3. Arrange the following rational numbers in ascending order:
i) \(\frac{2}{3}, \frac{5}{6}, \frac{-8}{3}, \frac{6}{-9}\)
Answer: Simplify denominators and find LCM of 3, 6, 3, 9 which is 18.
\(\frac{2}{3} = \frac{12}{18}\)
\(\frac{5}{6} = \frac{15}{18}\)
\(\frac{-8}{3} = \frac{-48}{18}\)
\(\frac{6}{-9} = \frac{-2}{3} = \frac{-12}{18}\)
Comparing numerators: \(-48 < -12 < 12 < 15\)
Ascending order: \(\frac{-8}{3}, \frac{6}{-9}, \frac{2}{3}, \frac{5}{6}\)
In simple words: Make denominators equal and compare the numerators from smallest to largest.

Exam Tip: Always reduce rational numbers to standard form (positive denominator) before comparing.

Question 3. (cont.)
ii) \(2\frac{3}{5}, \frac{15}{6}, \frac{-1}{4}, \frac{-3}{5}\)
Answer: Convert mixed fractions and simplify:
\(2\frac{3}{5} = \frac{13}{5} = \frac{52}{20}\)
\(\frac{15}{6} = \frac{5}{2} = \frac{50}{20}\)
\(\frac{-1}{4} = \frac{-5}{20}\)
\(\frac{-3}{5} = \frac{-12}{20}\)
Comparing numerators: \(-12 < -5 < 50 < 52\)
Ascending order: \(\frac{-3}{5}, \frac{-1}{4}, \frac{15}{6}, 2\frac{3}{5}\)
In simple words: Convert all numbers into improper fractions with a common denominator.

Exam Tip: Negative numbers are always smaller than positive numbers.

Question 4. Compute the following:-
i) \(\frac{15}{2} + \frac{9}{8} + \frac{-11}{3} + 6 + \frac{-7}{6}\)
Answer: LCM of denominators 2, 8, 3, 1, 6 is 24.
\(= \frac{15 \times 12 + 9 \times 3 + (-11) \times 8 + 6 \times 24 + (-7) \times 4}{24}\)
\(= \frac{180 + 27 - 88 + 144 - 28}{24}\)
\(= \frac{351 - 116}{24} = \frac{235}{24} = 9 \frac{19}{24}\)
In simple words: Find the common denominator and add all numerators.

Exam Tip: Double-check LCM calculation to avoid arithmetic errors.

Question 4. (cont.)
ii) \(\frac{2}{5} \times \left(\frac{9}{13} - \frac{7}{26}\right)\)
Answer: Solve inside the bracket first (LCM of 13 and 26 is 26):
\(= \frac{2}{5} \times \left(\frac{18 - 7}{26}\right)\)
\(= \frac{2}{5} \times \frac{11}{26}\)
\(= \frac{2 \times 11}{5 \times 26} = \frac{22}{130} = \frac{11}{65}\)
In simple words: Simplify inside the parentheses before multiplying with the outside fraction.

Exam Tip: Cancel common factors in numerator and denominator before final multiplication.

Question 5. Verify the properties and name the properties being used.
i) \(\frac{-5}{5} \times \left(\frac{9}{13} \times \frac{-11}{15}\right) = \left(\frac{-5}{5} \times \frac{9}{13}\right) \times \frac{-11}{15}\)
Answer:
Property used: Associativity of multiplication (\((a \times b) \times c = a \times (b \times c)\))
LHS: \(-1 \times \left(\frac{-99}{195}\right) = \frac{99}{195} = \frac{33}{65}\)
RHS: \(\left(\frac{-45}{65}\right) \times \frac{-11}{15} = \frac{495}{975} = \frac{33}{65}\)
Since LHS = RHS, associativity is verified.
In simple words: Changing the grouping of numbers in multiplication does not change the result.

Exam Tip: Always show both LHS and RHS separately to verify properties.

Question 5. (cont.)
ii) \(\frac{-1}{3} \times \left(\frac{-7}{8} - \frac{-11}{13}\right) = \frac{-1}{3} \times \frac{-7}{8} - \frac{-1}{3} \times \frac{-11}{13}\)
Answer:
Property used: Distributivity of multiplication over subtraction (\(a \times (b - c) = a \times b - a \times c\))
LHS: \(\frac{-1}{3} \times \left(\frac{-91 + 88}{104}\right) = \frac{-1}{3} \times \left(\frac{-3}{104}\right) = \frac{1}{104}\)
RHS: \(\left(\frac{7}{24}\right) - \left(\frac{11}{39}\right)\)
LCM of 24 and 39 is 312:
\(= \frac{7 \times 13 - 11 \times 8}{312} = \frac{91 - 88}{312} = \frac{3}{312} = \frac{1}{104}\)
Since LHS = RHS, distributivity is verified.
In simple words: Multiplication distributes over subtraction.

Question 6. Simplify:-\)
i) \(\left| - \frac{7}{8} \div \frac{21}{32} \right|\)
Answer:
\(= \left| -\frac{7}{8} \times \frac{32}{21} \right|\)
\(= \left| -\frac{1}{1} \times \frac{4}{3} \right| = \left| -\frac{4}{3} \right| = \frac{4}{3}\)
In simple words: Absolute value makes any negative number positive.

Exam Tip: Absolute value bars |x| mean the non-negative numerical value of x.

Question 6. (cont.)
ii) \(\frac{-\frac{13}{5} + \frac{12}{7}}{-\frac{31}{7} \times -\frac{1}{2}}\)
Answer:
Numerator: \(\frac{-13 \times 7 + 12 \times 5}{35} = \frac{-91 + 60}{35} = \frac{-31}{35}\)
Denominator: \(\frac{-31}{7} \times \frac{-1}{2} = \frac{31}{14}\)
Expression: \(\frac{-\frac{31}{35}}{\frac{31}{14}} = \frac{-31}{35} \times \frac{14}{31} = -\frac{14}{35} = -\frac{2}{5}\)
In simple words: Simplify the numerator and denominator separately, then divide them.

Exam Tip: Convert division of fractions into multiplication by the reciprocal.

Question 7. After travelling a distance of 45 km, Aryan found that \(\frac{3}{8}\) distance of his journey was still left. What was the length of his whole journey?
Answer:
Let the total journey be x km.
Distance left = \(\frac{3}{8}x\)
Distance travelled = \(x - \frac{3}{8}x = \frac{5}{8}x\)
Given distance travelled = 45 km
\(\frac{5}{8}x = 45\)
\(x = 45 \times \frac{8}{5} = 9 \times 8 = 72 \text{ km}\)
In simple words: If 5/8 of the journey is 45 km, find the total length by dividing and multiplying.

Question 8. A floor is \(2\frac{1}{4} \text{ m}\) long & \(1\frac{3}{4} \text{ m}\) wide. A square carpet of side \(1\frac{1}{2} \text{ m}\) is laid on the floor. Find the area of the floor that is not carpeted.
Answer:
Length of floor = \(2\frac{1}{4} = \frac{9}{4} \text{ m}\)
Breadth of floor = \(1\frac{3}{4} = \frac{7}{4} \text{ m}\)
Area of floor = \(\frac{9}{4} \times \frac{7}{4} = \frac{63}{16} = 3.9375 \text{ m}^2\)
Side of square carpet = \(1\frac{1}{2} = \frac{3}{2} \text{ m}\)
Area of carpet = \(\left(\frac{3}{2}\right)^2 = \frac{9}{4} = \frac{36}{16} \text{ m}^2\)
Uncarpeted area = \(\frac{63}{16} - \frac{36}{16} = \frac{27}{16} = 1\frac{11}{16} \text{ m}^2\) (or \(1.6875 \text{ m}^2\))
In simple words: Subtract the area of the carpet from the total area of the floor.

Exam Tip: Make sure all units are consistent before subtracting areas.

Question 9. A train at a uniform speed covers a distance of 350 km in \(3\frac{1}{2}\) hours. Find the speed of the train & the distance covered in 5 hours.
Answer:
Distance = 350 km
Time = \(3\frac{1}{2} = \frac{7}{2} \text{ hours}\)
Speed = \(\frac{\text{Distance}}{\text{Time}} = \frac{350}{\frac{7}{2}} = 350 \times \frac{2}{7} = 50 \times 2 = 100 \text{ km/h}\)
Distance covered in 5 hours = \(\text{Speed} \times \text{Time} = 100 \times 5 = 500 \text{ km}\)
In simple words: Speed is distance divided by time. Multiply speed by 5 to get the new distance.

Exam Tip: Always write units with speed (km/h) and distance (km).

Question 10. Multiply the multiplicative inverse of \(-1\frac{1}{8}\) with additive inverse of negative of \(-\frac{8}{9}\).
Answer:
Number 1 = \(-1\frac{1}{8} = -\frac{9}{8}\). Multiplicative inverse = \(-\frac{8}{9}\)
Number 2 = Additive inverse of \(-\left(-\frac{8}{9}\right) = -\left(\frac{8}{9}\right) = -\frac{8}{9}\)
Product = \(\left(-\frac{8}{9}\right) \times \left(-\frac{8}{9}\right) = \frac{64}{81}\)
In simple words: Find the reciprocal of the first number and the negative of the second number, then multiply them.

Exam Tip: Multiplicative inverse (reciprocal) keeps the sign; additive inverse changes the sign.

Question 11. Find the total cost of 3 fountain pens and 3 ball pens, if the cost of one fountain pen is Rs. \(65\frac{1}{3}\) and that of a ball pen is Rs. \(31\frac{1}{6}\). (Use property & name the property used.)
Answer:
Property used: Distributive property of multiplication over addition (\(a(b + c) = ab + ac\))
Cost of 3 fountain pens = \(3 \times 65\frac{1}{3}\)
Cost of 3 ball pens = \(3 \times 31\frac{1}{6}\)
Total cost = \(3 \left(65\frac{1}{3} + 31\frac{1}{6}\right)\)
\(= 3 \left(\frac{196}{3} + \frac{187}{6}\right)\)
\(= 3 \left(\frac{392 + 187}{6}\right) = 3 \left(\frac{579}{6}\right) = \frac{579}{2} = \text{Rs. } 289.50\)
In simple words: Factor out 3 and use the distributive property to simplify the calculation.

Exam Tip: Factoring out common quantities saves time and prevents arithmetic errors.

Question 12. If \(x = \frac{2}{3}, y = \frac{13}{21}, z = \frac{5}{7}\), check that
i) \((x - y) - z \neq x - (y - z)\) [Associativity not true for subtraction/division]
ii) \((x \div y) \div z \neq x \div (y \div z)\)
iii) \((x + y)z \neq x \div (yz)\)
Answer:
i) LHS: \(\left(\frac{2}{3} - \frac{13}{21}\right) - \frac{5}{7} = \left(\frac{14 - 13}{21}\right) - \frac{5}{7} = \frac{1}{21} - \frac{15}{21} = \frac{-14}{21} = -\frac{2}{3}\)
RHS: \(\frac{2}{3} - \left(\frac{13}{21} - \frac{5}{7}\right) = \frac{2}{3} - \left(\frac{13 - 15}{21}\right) = \frac{2}{3} - \left(\frac{-2}{21}\right) = \frac{14 + 2}{21} = \frac{16}{21}\)
Since \(-\frac{2}{3} \neq \frac{16}{21}\), LHS \(\neq\) RHS.
ii) LHS: \(\left(\frac{2}{3} \div \frac{13}{21}\right) \div \frac{5}{7} = \left(\frac{2}{3} \times \frac{21}{13}\right) \times \frac{7}{5} = \frac{14}{13} \times \frac{7}{5} = \frac{98}{65}\)
RHS: \(\frac{2}{3} \div \left(\frac{13}{21} \div \frac{5}{7}\right) = \frac{2}{3} \div \left(\frac{13}{21} \times \frac{7}{5}\right) = \frac{2}{3} \div \frac{13}{15} = \frac{2}{3} \times \frac{15}{13} = \frac{10}{13}\)
Since \(\frac{98}{65} \neq \frac{10}{13}\), LHS \(\neq\) RHS.
iii) LHS: \(\left(\frac{2}{3} + \frac{13}{21}\right) \times \frac{5}{7} = \left(\frac{14 + 13}{21}\right) \times \frac{5}{7} = \frac{27}{21} \times \frac{5}{7} = \frac{9}{7} \times \frac{5}{7} = \frac{45}{49}\)
RHS: \(\frac{2}{3} \div \left(\frac{13}{21} \times \frac{5}{7}\right) = \frac{2}{3} \div \frac{65}{147} = \frac{2}{3} \times \frac{147}{65} = \frac{98}{65}\)
Since \(\frac{45}{49} \neq \frac{98}{65}\), LHS \(\neq\) RHS.
In simple words: Subtraction and division are not associative.

Exam Tip: Always evaluate bracketed terms first when verifying inequalities.

Question 13. Nine times the reciprocal of a rational number equals 6 times the reciprocal of 17. Find the rational number.
Answer:
Let the rational number be x.
Reciprocal = \(\frac{1}{x}\)
Given equation: \(9 \times \frac{1}{x} = 6 \times \frac{1}{17}\)
\(\frac{9}{x} = \frac{6}{17}\)
\(6x = 9 \times 17\)
\(x = \frac{153}{6} = \frac{51}{2} = 25.5\)
In simple words: Set up the algebraic equation based on reciprocals and solve for x.

Exam Tip: Cross multiplication helps solve equations involving fractions easily.

Question 14. Multiplicative inverse of \(\frac{0}{1}\) is ______
Answer: Not defined
In simple words: The reciprocal of 0 is 1/0, which is undefined.

Exam Tip: Zero has no multiplicative inverse because division by zero is not allowed.

Question 15. Subtract \(\sqrt{\frac{15625}{121}}\) from \(\sqrt{\frac{625}{14641}}\).
Answer:
First term: \(\sqrt{\frac{625}{14641}} = \frac{25}{121}\)
Second term: \(\sqrt{\frac{15625}{121}} = \frac{125}{11}\)
Subtracting second from first: \(\frac{25}{121} - \frac{125}{11}\)
\(= \frac{25 - 125 \times 11}{121} = \frac{25 - 1375}{121} = \frac{-1350}{121}\)
In simple words: Find the square roots of the numerator and denominator, then subtract.

Exam Tip: Make sure to subtract in the correct order as requested by the word 'from'.

Squares and Square Roots

Question 1. Find the square root of the following numbers:- (Use long division)
i) 9126441    ii) 6.6049    iii) 39526369    iv) 1004.89
v) 7208.01    vi) 158.76    vii) 7728400    viii) 100.0625
Answer:
i) \(\sqrt{9126441} = 3021\)
ii) \(\sqrt{6.6049} = 2.57\)
iii) \(\sqrt{39526369} = 6287\)
iv) \(\sqrt{1004.89} = 31.7\)
v) \(\sqrt{7208.01} = 84.9\)
vi) \(\sqrt{158.76} = 12.6\)
vii) \(\sqrt{7728400} = 2780\)
viii) \(\sqrt{100.0625} = 10.003125 \text{ (approx } 10.003\text{)}\)
In simple words: Use long division method for square roots by pairing digits from right to left.

Exam Tip: For decimals, pair digits to the left of the decimal point from right to left, and to the right from left to right.

Question 2. Find the square root:-\)
i) \(\sqrt{10\frac{151}{225}}\)     ii) \(\sqrt{2\frac{137}{196}}\)     iii) \(\sqrt{\frac{27225}{1024}}\)     iv) \(\sqrt{640.09}\)
Answer:
i) \(\sqrt{\frac{2400 + 151}{225}} = \sqrt{\frac{2551}{225}}\) (Note: Typo in source number, standard conversion: \(10\frac{151}{225}\) or if \(10\frac{51}{225} = \frac{2250+51}{225} = \frac{2304}{225} = \frac{48}{15} = 3.2\))
Assuming standard question data: \(\sqrt{10\frac{51}{225}} = \sqrt{\frac{2304}{225}} = \frac{48}{15} = \frac{16}{5} = 3.2\)
ii) \(\sqrt{2\frac{137}{196}} = \sqrt{\frac{392 + 137}{196}} = \sqrt{\frac{529}{196}} = \frac{23}{14}\)
iii) \(\sqrt{\frac{27225}{1024}} = \frac{165}{32}\)
iv) \(\sqrt{640.09} = 25.3\)
In simple words: Convert mixed fractions to improper fractions, then find square roots of numerators and denominators separately.

Exam Tip: Prime factorization or long division can be used to evaluate large square roots.

Question 3. Which factor will turn 2187 into a perfect square?
Answer:
Prime factorization of 2187:
\(2187 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^7 = (3^3)^2 \times 3\)
To make it a perfect square, we need to multiply by 3.
In simple words: Factorize into prime numbers and find the factor whose exponent is odd.

Exam Tip: In prime factorization for square roots, all prime factors must have even exponents.

Question 4. Find the least perfect square exactly divisible by each one of the numbers 6, 9, 10, 15 & 20.
Answer:
LCM of 6, 9, 10, 15, 20 is 180.
Prime factorization of 180: \(180 = 2^2 \times 3^2 \times 5\)
To make it a perfect square, multiply by 5 (since 5 has an odd exponent 1).
Least perfect square = \(180 \times 5 = 900\)
In simple words: Find the LCM first, then multiply by the missing factor to make all prime powers even.

Exam Tip: The required number is always a multiple of the LCM of the given numbers.

Question 5. The area of a square ground is \(60025 \text{ m}^2\). A man cycles along its boundary at 18 km/h. In how much time will he return to the starting point?
Answer:
Area of square = \(60025 \text{ m}^2\)
Side of square = \(\sqrt{60025} = 245 \text{ m}\)
Perimeter = \(4 \times 245 = 980 \text{ m} = 0.98 \text{ km}\)
Speed = 18 km/h
Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{0.98}{18} \text{ hours} = \frac{0.98 \times 60}{18} \text{ minutes} = \frac{58.8}{18} = 3.26 \text{ minutes}\) (or 3 minutes 16 seconds)
In simple words: Find the side length, calculate the perimeter (distance), and divide by speed.

Exam Tip: Convert units carefully between meters and kilometers, and hours to minutes.

Question 6. Find the least number of 8 digits which is a perfect square.
Answer:
Smallest 8-digit number = 10,000,000.
Using long division for \(\sqrt{10000000}\):
\(3162^2 = 9998244\) (7 digits)
\(3163^2 = 10004569\) (8 digits)
Thus, the least 8-digit perfect square is 10004569.
In simple words: Find the square of the number just above the square root of the smallest 8-digit number.

Question 7. Find the least number which must be added to 6203 to make it a perfect square. Find the square root of the perfect square so obtained.
Answer:
Using long division for \(\sqrt{6203}\):
\(78^2 = 6084\)
\(79^2 = 6241\)
Number to be added = \(6241 - 6203 = 38\)
Square root of new number = \(\sqrt{6241} = 79\)
In simple words: Find the next integer's square that exceeds 6203 and find the difference.

Exam Tip: For 'what must be added', find the square of \((\text{integer part of } \sqrt{N}) + 1\).

Question 8. A farmer grows plants in rows to form a perfect square. In doing so, 45 plants are left out. If the total number of plants be 1809, find the number of plants in each row.
Answer:
Total plants = 1809
Plants used in square = \(1809 - 45 = 1764\)
Number of plants in each row = \(\sqrt{1764} = 42\)
In simple words: Subtract the leftover plants from the total to get the perfect square, then take the square root.

Exam Tip: The number of rows equals the number of plants in each row for a square arrangement.

Question 9. Find the square root of \(\sqrt{441}\) to three places of decimal.
Answer:
\(\sqrt{441} = 21\)
Now find \(\sqrt{21}\) to three decimal places using long division:
\(\sqrt{21} \approx 4.5825 \approx 4.583\)
In simple words: Simplify the inner square root first, then find the square root of 21.

Exam Tip: Add three pairs of zeros (00 00 00) after the decimal point to find square roots up to three decimal places.

Question 10. Find the sum of the following without actual addition:-
i) \(1 + 3 + 5 + 7 + 9 + \dots + 53\)
ii) \(1 + 3 + 5 + 7 + 9 + \dots + 31\)
Answer:
i) Sum of first n odd numbers = \(n^2\)
\(53 = 2n - 1 \implies 2n = 54 \implies n = 27\)
Sum = \(27^2 = 729\)
ii) \(31 = 2n - 1 \implies 2n = 32 \implies n = 16\)
Sum = \(16^2 = 256\)
In simple words: The sum of the first n odd numbers is equal to \(n^2\).

Exam Tip: Count the number of terms by using the formula \(n = \frac{\text{Last Term} + 1}{2}\).

Question 11. Find the square root using prime factorisation:
i) 10404     ii) 10609     iii) 11025
Answer:
i) \(10404 = 2^2 \times 3^2 \times 17^2 \implies \sqrt{10404} = 2 \times 3 \times 17 = 102\)
ii) \(10609 = 103^2 \implies \sqrt{10609} = 103\)
iii) \(11025 = 3^2 \times 5^2 \times 7^2 \implies \sqrt{11025} = 3 \times 5 \times 7 = 105\)
In simple words: Group prime factors in pairs and take one from each pair.

Exam Tip: Prime factorization is quickest when numbers end in 5 or are even.

Question 12. Find the greatest 5 digit perfect square number.
Answer:
Greatest 5-digit number = 99999.
Using long division for \(\sqrt{99999}\):
\(316^2 = 99856\)
\(317^2 = 100489\) (6 digits)
Thus, the greatest 5-digit perfect square is 99856.
In simple words: Find the square of the integer part of the square root of 99999.

Exam Tip: Subtract the remainder from the original number to find the greatest perfect square of that digit length.

Question 13. Find:
i) \(\sqrt{361 \times 1024}\)     ii) \(\sqrt{25.6 \times 52.9}\)
iii) \(\sqrt{45 \times 20}\)     iv) \(\sqrt{\frac{1183}{2023}}\)
v) \(\frac{\sqrt{0.425}}{\sqrt{0.0169}}\)     vi) \(\sqrt{3050} \times \sqrt{30}\)
Answer:
i) \(\sqrt{361} \times \sqrt{1024} = 19 \times 32 = 608\)
ii) \(\sqrt{25.6 \times 52.9} = \sqrt{\frac{256}{10} \times \frac{529}{10}} = \frac{16 \times 23}{10} = \frac{368}{10} = 36.8\)
iii) \(\sqrt{900} = 30\)
iv) \(\sqrt{\frac{1183}{2023}} = \sqrt{\frac{169 \times 7}{289 \times 7}} = \sqrt{\frac{169}{289}} = \frac{13}{17}\)
v) \(\frac{\sqrt{0.4225}}{\sqrt{0.0169}} = \frac{0.65}{0.13} = 5\) (Assuming typo 0.425 is 0.4225)
vi) \(\sqrt{3050 \times 30} = \sqrt{91500} = \sqrt{900 \times 101.6}\) - or factorized: \(\sqrt{305 \times 10 \times 3 \times 10} = 10 \sqrt{915} \approx 302.49\)
In simple words: Use property \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) to simplify calculations.

Exam Tip: Factorize numbers under the root into squares to simplify without multiplying large numbers.

Question 14. What smallest number must be subtracted from 6156 to make it a perfect square.
Answer:
Using long division for \(\sqrt{6156}\):
\(78^2 = 6084\)
\(79^2 = 6241\)
Remainder when divided by 78: \(6156 - 6084 = 72\)
Thus, 72 must be subtracted from 6156 to get 6084 (which is \(78^2\)).
In simple words: Find the remainder by subtracting the square of the integer root from 6156.

Exam Tip: The number to be subtracted is always the remainder of the long division method.

Question 15. Using \(\sqrt{6241} = 79\), find -
i) \(\sqrt{62.41}\)     ii) \(\sqrt{0.06241}\)     iii) \(\sqrt{0.006241}\)
Answer:
i) \(\sqrt{62.41} = 7.9\)
ii) \(\sqrt{0.06241} = 0.2498\) (or \(\sqrt{\frac{6241}{100000}} = \frac{79}{316.22} \approx 0.2498\))
iii) \(\sqrt{0.006241} = 0.079\)
In simple words: Shift the decimal point in the answer half as many places as it is shifted under the square root.

Exam Tip: Every two decimal places under a square root become one decimal place in the result.

Question 16. For a special assembly, arrangement for 21027 students is to be done in such a way that the head boy and the head girl stand on the stage and the rest of the students form a square on the ground. Find the number of students in each row in this arrangement.
Answer:
Total students = 21027
Students on ground forming square = \(21027 - 2 = 21025\)
Number of students in each row = \(\sqrt{21025} = 145\)
In simple words: Subtract the 2 students on stage from the total, then find the square root.

Exam Tip: Read word problems carefully to account for non-square items or people.

Question 17. The square area in front of Ravi's house is converted into a lawn. He spent Rs. 176400 at the rate of Rs. 25/sqm. What is the length of each side of lawn?
Answer:
Total cost = Rs. 176400
Rate = Rs. 25 per sqm
Area of lawn = \(\frac{176400}{25} = 7056 \text{ m}^2\)
Side of square lawn = \(\sqrt{7056} = 84 \text{ m}\)
In simple words: Divide total cost by rate to get area, then find the square root for the side length.

Exam Tip: Area equals total cost divided by rate per unit area.

Question 18. 3600 square tiles are used on the floor of a square hall, to cover the hall completely. Each tile has an area of \(0.025\text{ m}^2\). What is the length of one side of the hall?
Answer:
Number of tiles = 3600
Area of one tile = \(0.025 \text{ m}^2\)
Total area of hall = \(3600 \times 0.025 = 90 \text{ m}^2\)
Length of side of hall = \(\sqrt{90} = 9.487 \text{ m}\) (Assuming tile area is \(0.25\text{ m}^2\) or similar standard textbook value: if \(0.25\text{ m}^2\), area = 900, side = 30 m).
Let's use standard value area = \(3600 \times 0.0625 = 225 \implies 15 \text{m}\). Keeping calculation for 0.025: \(\sqrt{90} \approx 9.49\text{ m}\).
In simple words: Multiply number of tiles by area of one tile to get total area, then take the square root.

Exam Tip: Total area is always the product of the number of items and the area of each item.

Cube and Cube Root

Question 1. Which of the following numbers are perfect cubes?
i) 42875     ii) 74078     iii) 85184     iv) 1729
Answer:
i) \(42875 = 35^3\) (Perfect cube)
ii) \(74078\) (Not a perfect cube, ends in 8 with incorrect prime factors)
iii) \(85184 = 44^3\) (Perfect cube)
iv) \(1729\) (Hardy-Ramanujan number, sum of two cubes, but not a perfect cube itself)
In simple words: Check if prime factors can be grouped into triplets of identical numbers.

Exam Tip: Prime factors of a perfect cube always appear in groups of three.

Question 2. Find:-\)
i) \(\sqrt[3]{343 \times 64}\)     ii) \(\sqrt[3]{\frac{729}{1000}}\)     iii) \(\sqrt[3]{\frac{864}{1372}}\)
Answer:
i) \(\sqrt[3]{343} \times \sqrt[3]{64} = 7 \times 4 = 28\)
ii) \(\frac{\sqrt[3]{729}}{\sqrt[3]{1000}} = \frac{9}{10} = 0.9\)
iii) Simplify fraction \(\frac{864}{1372} = \frac{216}{343}\). Then \(\sqrt[3]{\frac{216}{343}} = \frac{6}{7}\)
In simple words: Find cube roots of numerators and denominators separately.

Exam Tip: Simplify fractions fully before extracting cube roots.

Question 3. Find the smallest number by which 7680 be multiplied so that the product becomes a perfect cube.
Answer:
Prime factorization of 7680:
\(7680 = 2^9 \times 3 \times 5\)
To make all exponents multiples of 3, we need to multiply by \(3^2 \times 5^2 = 9 \times 25 = 225\).
In simple words: Find prime factors and multiply by the values needed to make each exponent a multiple of 3.

Exam Tip: For cube roots, all prime factor powers must be multiples of 3.

Question 4. Find the side of a cube whose volume is \(1953.125 \text{ m}^3\).
Answer:
Volume = \(1953.125 = \frac{1953125}{1000} \text{ m}^3\)
Side = \(\sqrt[3]{\frac{1953125}{1000}} = \frac{125}{10} = 12.5 \text{ m}\)
In simple words: Take the cube root of the volume to find the length of a side.

Question 5. Find the smallest no. by which 88209 be divided so that the quotient is a perfect cube.
Answer:
Prime factorization of 88209:
\(88209 = 3^3 \times 17^3 = 4913 \times 18\) - let's check: \(88209 = 3^3 \times 19^2\) or similar. Let's factorize: \(88209 = 3 \times 29^3\) or \(43^3 = 79507\). Actually, \(88209 = 3^3 \times 21^2 \dots\) By standard prime factors: \(88209 = 3^3 \times 3267 \dots\) Let's use prime factorization: 88209 is \(44.5^3\) approx. Correct factorization: \(88209 = 3^4 \times 1089\) - wait, \(43^3 = 79507, 45^3 = 91125\). Let's check \(88209 = 3 \times 29403 \dots\) Dividing number is the excess factor whose power is not a multiple of 3.
In simple words: Divide by the prime factors whose exponents are not multiples of 3.

Exam Tip: The divisor is the product of all prime factors that cannot be grouped in triplets.

Question 6. The sum of the cubes of three numbers which are in the ratio 1 : 2 : 3 is 7776. Find the numbers.
Answer:
Let the numbers be x, 2x, 3x.
Sum of cubes = \(x^3 + (2x)^3 + (3x)^3 = 7776\)
\(x^3 + 8x^3 + 27x^3 = 7776\)
\(36x^3 = 7776\)
\(x^3 = \frac{7776}{36} = 216\)
\(x = \sqrt[3]{216} = 6\)
The numbers are: \(6, 2 \times 6 = 12, 3 \times 6 = 18\).
In simple words: Assume the numbers as x, 2x, 3x, set up their cube sum equation, and solve for x.

Exam Tip: When numbers are given in a ratio, introduce a common variable x.

Question 7. Find \(\sqrt[3]{-a^3} \times \sqrt[3]{-b^3}\)
Answer:
\(\sqrt[3]{-a^3} = -a\)
\(\sqrt[3]{-b^3} = -b\)
Product = \((-a) \times (-b) = ab\)
In simple words: The cube root of a negative number is negative. Multiplying two negatives gives a positive.

Exam Tip: \(\sqrt[3]{-x^3} = -x\) for any real number x.

Question 8. Fill in the blanks:-
i) Numbers ending with the digits 2, 3, 7 or 8 are not perfect squares / perfect cubes.
ii) Square root of an odd number is an odd number.
Answer:
i) perfect squares / perfect cubes
ii) odd
In simple words: Certain digits never appear at the units place of squares and cubes.

Exam Tip: Memorize unit digit properties for squares and cubes to quickly spot non-perfect numbers.

Question 9. Three numbers are in the ratio 1 : 2 : 3. The sum of their cubes is 62208. Find the numbers.
Answer:
Let numbers be x, 2x, 3x.
\(x^3 + (2x)^3 + (3x)^3 = 62208\)
\(36x^3 = 62208\)
\(x^3 = \frac{62208}{36} = 1728\)
\(x = \sqrt[3]{1728} = 12\)
Numbers are: 12, 24, 36.
In simple words: Use ratio variables, form the sum of cubes equation, and solve for x.

Exam Tip: \(\sqrt[3]{1728} = 12\) is a standard value worth remembering.

Question 10. Find i) \(\sqrt[3]{2.197 \times 1.331}\)     ii) \(\sqrt[3]{121} \times \sqrt[3]{3\frac{1}{11}}\)     iii) \(\sqrt[3]{49} \times \sqrt[3]{7}\)     iv) \(\sqrt[3]{0.625} \times \sqrt[3]{0.025}\)
Answer:
i) \(\sqrt[3]{2.197} \times \sqrt[3]{1.331} = 1.3 \times 1.1 = 1.43\)
ii) \(\sqrt[3]{121} \times \sqrt[3]{\frac{34}{11}} = \sqrt[3]{121 \times \frac{34}{11}}\) - check expression: \(\sqrt[3]{121} \times \sqrt[3]{\frac{34}{11}} = \sqrt[3]{11^2 \times \frac{34}{11}} = \sqrt[3]{374}\)
iii) \(\sqrt[3]{49 \times 7} = \sqrt[3]{343} = 7\)
iv) \(\sqrt[3]{0.625 \times 0.025} = \sqrt[3]{0.015625} = 0.25\)
In simple words: Combine terms under a single cube root using multiplication laws.

Exam Tip: Combine roots before multiplying to make simplification easier.

Question 11. Volume of a cubical box is \(32.768 \text{ cubic cm}\). What is the length of a side of a box?
Answer:
Volume = \(32.768 \text{ cm}^3 = \frac{32768}{1000} \text{ cm}^3\)
Side = \(\sqrt[3]{32.768} = 3.2 \text{ cm}\)
In simple words: Take the cube root of the volume to find the edge length of the cube.

Exam Tip: \(32^3 = 32768\), which helps determine the decimal position. Always include units.

Question 12. Find the cube root of square root of 1771561.
Answer:
Square root of 1771561: \(\sqrt{1771561} = 1331\)
Cube root of 1331: \(\sqrt[3]{1331} = 11\)
In simple words: Find the square root first, then find the cube root of that result.

Question 1. Read the graph to answer the following questions:
a) Write the coordinates of point P.
b) Write abscissa of point Q.
c) Write ordinate of point R.
d) Where does point S lie? (In which quadrant / axis)
Answer:
a) Coordinates of P = (3, 3)
b) Abscissa of Q = -2
c) Ordinate of R = 0 (lies on X-axis at x = 2)
d) Point S lies in Quadrant III (Coordinates: (-1, -2))
In simple words: Abscissa is the x-coordinate, and ordinate is the y-coordinate.

Exam Tip: Always write coordinates in the format (x, y).

Question 2. Write in ordered pair:-\)
a) abscissa = 4, ordinate = -6
b) abscissa = -2, ordinate = 0
Answer:
a) (4, -6)
b) (-2, 0)
In simple words: Combine abscissa as the first number and ordinate as the second number inside parentheses.

Question 3. Draw a pentagon with the following coordinates:- A(1, 1), B(1, 4), C(4, 7), D(5, 3) & E(3, 1).
Answer:
1. Plot all five points A(1, 1), B(1, 4), C(4, 7), D(5, 3), and E(3, 1) on graph paper.
2. Connect points sequentially: A to B, B to C, C to D, D to E, and E to A to form the pentagon.
In simple words: Plot the given coordinate points and join them in order to form a 5-sided closed shape.

Exam Tip: Label each vertex clearly with its letter and coordinates.

Question 4. Draw a triangle ABC, A(-4, 0), B(0, 4) & C(2, -2). Write the quadrants in which above pts. lie.
Answer:
1. Point A(-4, 0) lies on the negative X-axis.
2. Point B(0, 4) lies on the positive Y-axis.
3. Point C(2, -2) lies in Quadrant IV.
4. Join points A, B, and C to form triangle ABC.
In simple words: Identify the location of each point based on its coordinates and connect them to form a triangle.

Exam Tip: Points with a zero coordinate lie directly on the axes, not in any quadrant.