CBSE Class 8 Mathematics Square And Square Roots Worksheet Set 06

SQUARES AND SQUARE ROOTS

I. CHOOSE THE CORRECT ANSWER :

Question 1. Square of an odd number is :
(a) an odd number
(b) an even number
(c) either an odd number or an even number
(d) none of the options
Answer: (a) an odd number
The square of any odd number is always an odd number (e.g., \(3^2 = 9\), \(5^2 = 25\)).
In simple words: Multiplying an odd number by itself always gives an odd result.

Exam Tip: Remember the basic rule: odd \(\times\) odd = odd, and even \(\times\) even = even.

Question 2. Which of the following is not a perfect square :
(a) 441
(b) 324
(c) 253
(d) 625
Answer: (c) 253
Numbers ending in 2, 3, 7, or 8 at the unit's place are never perfect squares. Here, 253 ends in 3.
In simple words: Look at the last digit of the number to quickly check if it can be a square.

Exam Tip: Perfect squares never end in 2, 3, 7, or 8.

Question 3. A Pythagorean triplet whose one number is 8 is
(a) (8, 9, 10)
(b) (8, 15, 17)
(c) (4, 8, 10)
(d) (4, 15, 17)
Answer: (b) (8, 15, 17)
For any natural number \(m > 1\), a Pythagorean triplet is given by \(2m\), \(m^2 - 1\), and \(m^2 + 1\). Let \(2m = 8 \implies m = 4\). Then \(m^2 - 1 = 4^2 - 1 = 15\) and \(m^2 + 1 = 4^2 + 1 = 17\).
In simple words: Using the formula, when one number is 8, the other two numbers are 15 and 17.

Exam Tip: Verify using Pythagoras theorem: \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\).

Question 4. The number of digits in the square root of 33856 is
(a) 5
(b) 4
(c) 3
(d) 2
Answer: (c) 3
The number of digits in 33856 is \(n = 5\) (which is odd). The formula for the number of digits in the square root is \(\frac{n+1}{2} = \frac{5+1}{2} = 3\).
In simple words: Count the total digits and use the formula to find how many digits the square root will have.

Exam Tip: If \(n\) is odd, number of digits in square root is \(\frac{n+1}{2}\); if \(n\) is even, it is \(\frac{n}{2}\).

Question 5. How many non perfect square numbers lie between the squares of 15 and 16 ?
(a) 31
(b) 32
(c) 30
(d) 35
Answer: (c) 30
The number of non-perfect square numbers between the squares of numbers \(n\) and \((n+1)\) is \(2n\). Here \(n = 15\), so \(2 \times 15 = 30\).
In simple words: Multiply the smaller number by 2 to find how many non-square numbers lie between their squares.

Exam Tip: Always use the shortcut formula \(2n\) for consecutive numbers \(n\) and \(n+1\).

Question 6. Find the least number by which 500 must be multiplied to make it a perfect square
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (d) 5
Prime factorization of 500:
\(500 = 2 \times 2 \times 5 \times 5 \times 5 = 2^2 \times 5^3 = 2^2 \times 5^2 \times 5\)
To make pairs complete, we need one more 5.
In simple words: Find the prime factors and pair them up. The factor left without a pair is the multiplier.

Exam Tip: Group prime factors in pairs of two; any factor without a partner is multiplied.

Question 7. The square root of 1.44 is -
(a) 1.22
(b) 12.2
(c) 1.2
(d) 1.12
Answer: (c) 1.2
\(\sqrt{1.44} = \sqrt{\frac{144}{100}} = \frac{12}{10} = 1.2\)
In simple words: Remove the decimal, find the square root of 144, and place the decimal point back.

Exam Tip: A decimal number with two decimal places will have one decimal place in its square root.

Question 8. Area of a square plot is 729 m\(^2\). The side of the square is :
(a) 33m
(b) 37m
(c) 27m
(d) 23m
Answer: (c) 27m
Side = \(\sqrt{\text{Area}} = \sqrt{729} = 27\text{ m}\)
In simple words: Take the square root of the given area to find the length of the side.

Exam Tip: Memorize squares up to 30 for quick calculations.

Question 9. The number of zeros in the square of 400 will be -
(a) 2
(b) 1
(c) 3
(d) 4
Answer: (d) 4
\(400^2 = (4 \times 100)^2 = 16 \times 10000 = 160000\), which has 4 zeros.
In simple words: When squaring a number ending in zeros, double the number of zeros at the end.

Exam Tip: If a number has \(k\) zeros, its square will have \(2k\) zeros.

Question 10. Sum of the first n odd natural numbers is -
(a) \(2n + 1$
(b) \(n^2$
(c) \(n^2 - 1$
(d) \(2n^2 + 1$
Answer: (b) \(n^2\)
The sum of the first \(n\) odd natural numbers is always equal to \(n^2$.
In simple words: Adding consecutive odd numbers starting from 1 always gives a square number.

Exam Tip: This is a standard algebraic identity for the sum of odd numbers.

II. SOLVE THE FOLLOWING :

Question 1. Find the square of the following numbers without actual multiplication:
(a) 19
Answer:
\(19^2 = (20 - 1)^2\)
\(= 20^2 - 2(20)(1) + 1^2\)
\(= 400 - 40 + 1 = 361\)
In simple words: Express the number as a subtraction from a round number and use the identity \((a-b)^2\).

Exam Tip: Use the algebraic identity \((a-b)^2 = a^2 - 2ab + b^2\).

Question 1.(b) 23
Answer:
\(23^2 = (20 + 3)^2\)
\(= 20^2 + 2(20)(3) + 3^2\)
\(= 400 + 120 + 9 = 529\)
In simple words: Split the number into tens and units and use the identity \((a+b)^2\).

Exam Tip: Use the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\).

Question 1.(c) 45
Answer:
\(45^2 = (40 + 5)^2\)
\(= 40^2 + 2(40)(5) + 5^2\)
\(= 1600 + 400 + 25 = 2025\)
In simple words: Alternatively, numbers ending in 5 follow the shortcut \(n(n+1)\) followed by 25: \(4 \times 5 = 20\), so 2025.

Exam Tip: For any number ending in 5, \((a5)^2 = a(a+1) \text{ followed by } 25\).

Question 1.(d) 52
Answer:
\(52^2 = (50 + 2)^2\)
\(= 50^2 + 2(50)(2) + 2^2\)
\(= 2500 + 200 + 4 = 2704\)
In simple words: Split 52 as \(50 + 2\) and expand using the algebraic identity.

Exam Tip: Break numbers close to multiples of 10 for mental math ease.

Question 2. Find the square root of the following numbers by prime factorization method :
(a) 4225
Answer:
\(4225 = 5 \times 5 \times 13 \times 13 = 5^2 \times 13^2\)
\(\sqrt{4225} = 5 \times 13 = 65\)
In simple words: Prime factorize the number, make pairs of identical factors, and take one factor from each pair.

Exam Tip: Check divisibility by 5 since the number ends in 5.

Question 2.(b) 4489
Answer:
\(4489 = 67 \times 67 = 67^2\)
\(\sqrt{4489} = 67\)
In simple words: Find the prime factor 67 by testing prime numbers systematically.

Exam Tip: Knowing squares of numbers up to 100 helps identify large prime factors quickly.

Question 2.(c) 8281
Answer:
\(8281 = 91 \times 91 = (7 \times 13)^2 = 7^2 \times 13^2\)
\(\sqrt{8281} = 7 \times 13 = 91\)
In simple words: Factorize 8281 into prime factors 7 and 13 to find the square root.

Exam Tip: Test prime factors like 7, 11, 13, 17 sequentially.

Question 2.(d) 5625
Answer:
\(5625 = 3 \times 3 \times 5 \times 5 \times 5 \times 5 = 3^2 \times 5^4\)
\(\sqrt{5625} = 3 \times 5 \times 5 = 75\)
In simple words: Break down into prime factors, pair them up, and multiply.

Exam Tip: Divide successively by 3 and 5 until 1 is reached.

Question 2.(e) 1296
Answer:
\(1296 = 2^4 \times 3^4 = (2^2 \times 3^2)^2 = 36^2\)
\(\sqrt{1296} = 36\)
In simple words: Prime factorize 1296 to find its square root.

Exam Tip: Repeated division by 2 and 3 makes factorization straightforward.

Question 2.(f) 11664
Answer:
\(11664 = 2^4 \times 3^6 = (2^2 \times 3^3)^2 = (4 \times 27)^2 = 108^2\)
\(\sqrt{11664} = 108\)
In simple words: Group all prime factors into pairs to find the square root.

Exam Tip: Prime factorization for large numbers is best done using continuous division.

Question 3. Find the square root of the following numbers by division method:
(a) 1849
Answer:
Pair digits from right: 18 49.
\(4 \times 4 = 16\), remainder = 2. Bring down 49 to get 249.
Double divisor 4 to get 8_. Trial: \(83 \times 3 = 249\).
\(\sqrt{1849} = 43\)
In simple words: Use long division by pairing digits from the right.

Exam Tip: Division method is faster for larger numbers than prime factorization.

Question 3.(b) 6561
Answer:
Pair digits: 65 61.
\(8 \times 8 = 64\), remainder = 1. Bring down 61 to get 161.
Double divisor 8 to get 16_. Trial: \(161 \times 1 = 161\).
\(\sqrt{6561} = 81\)
In simple words: Apply long division step-by-step on digit pairs.

Exam Tip: Always pair digits starting from the decimal point or right end.

Question 3.(c) 8649
Answer:
Pair digits: 86 49.
\(9 \times 9 = 81\), remainder = 5. Bring down 49 to get 549.
Double divisor 9 to get 18_. Trial: \(183 \times 3 = 549\).
\(\sqrt{8649} = 93\)
In simple words: Find the largest square less than 86, then proceed with long division.

Exam Tip: Ensure digits are correctly grouped in pairs from the right.

Question 3.(d) 84100
Answer:
Pair digits: 8 41 00.
\(2 \times 2 = 4\), remainder = 4. Bring down 41 to get 441.
Double divisor 2 to get 4_. Trial: \(49 \times 9 = 441\). Remainder 0, bring down 00, add 0 to quotient.
\(\sqrt{84100} = 290\)
In simple words: Pair digits from right and apply the standard long division algorithm.

Exam Tip: Trailing zeros in pairs contribute one zero to the square root.

Question 3.(e) 54756
Answer:
Pair digits: 5 47 56.
\(2 \times 2 = 4\), remainder = 1. Bring down 47 to get 147.
Double divisor 2 to get 4_. Trial: \(43 \times 3 = 129\), remainder = 18. Bring down 56 to get 1856.
Double divisor 43 to get 86_. Trial: \(864 \times 4 = 1856\).
\(\sqrt{54756} = 234\)
In simple words: Continue the long division process for 3-digit pairs.

Exam Tip: Double the entire current quotient root at each step to form the new trial divisor prefix.

Question 3.(f) 12996
Answer:
Pair digits: 1 29 96.
\(1 \times 1 = 1\), remainder = 0. Bring down 29.
Double divisor 1 to get 2_. Trial: \(21 \times 1 = 21\), remainder = 8. Bring down 96 to get 896.
Double divisor 11 to get 22_. Trial: \(224 \times 4 = 896\).
\(\sqrt{12996} = 114\)
In simple words: Perform long division carefully matching each digit pair.

Exam Tip: Double check subtraction and multiplication at every step.

Question 4. Find the square root of the following decimal numbers :
(a) 16.81
Answer:
Group digits: 16 . 81.
\(4 \times 4 = 16\), remainder 0. Bring down 81, place decimal point.
Double 4 to get 8_. Trial: \(81 \times 1 = 81\).
\(\sqrt{16.81} = 4.1\)
In simple words: Apply long division with decimals by placing the decimal point in the quotient when crossing it.

Exam Tip: Group digits in pairs left and right of the decimal point.

Question 4.(b) 151.29
Answer:
Group digits: 1 51 . 29.
\(1 \times 1 = 1\), remainder 0. Bring down 51.
Double 1 to get 2_. Trial: \(22 \times 2 = 44\), remainder 7. Bring down 29, place decimal point to get 729.
Double 12 to get 24_. Trial: \(243 \times 3 = 729\).
\(\sqrt{151.29} = 12.3\)
In simple words: Follow the same long division steps for decimals.

Exam Tip: Align the decimal point in the answer directly above the decimal point in the number.

Question 4.(c) 53.59
Answer:
Group digits: 53 . 59.
\(7 \times 7 = 49\), remainder 4. Bring down 59, place decimal point to get 459.
Double 7 to get 14_. Trial: \(143 \times 3 = 429\), remainder 30.
Rounding to two decimal places: \(\sqrt{53.59} \approx 7.32\)
In simple words: Carry out division beyond the decimal point to find the approximate square root.

Exam Tip: Add pairs of zeros on the right if further decimal places are required.

Question 4.(d) 156.25
Answer:
Group digits: 1 56 . 25.
\(1 \times 1 = 1\), remainder 0. Bring down 56.
Double 1 to get 2_. Trial: \(22 \times 2 = 44\), remainder 12. Bring down 25, place decimal point to get 1225.
Double 12 to get 24_. Trial: \(245 \times 5 = 1225\).
\(\sqrt{156.25} = 12.5\)
In simple words: Use long division method for decimal numbers.

Exam Tip: Check your answer by multiplying 12.5 by itself.

Question 4.(e) 104.04
Answer:
Group digits: 1 04 . 04.
\(1 \times 1 = 1\), remainder 0. Bring down 04.
Double 1 to get 2_. Trial: \(20 \times 0 = 0\), remainder 4. Bring down 04, place decimal point to get 404.
Double 10 to get 20_. Trial: \(202 \times 2 = 404\).
\(\sqrt{104.04} = 10.2\)
In simple words: Be careful with zeros when bringing down digits in long division.

Exam Tip: Pay close attention to place values when a zero appears in the quotient.

Question 4.(f) 92.16
Answer:
Group digits: 92 . 16.
\(9 \times 9 = 81\), remainder 11. Bring down 16, place decimal point to get 1116.
Double 9 to get 18_. Trial: \(186 \times 6 = 1116\).
\(\sqrt{92.16} = 9.6\)
In simple words: Divide and match the decimal places using long division.

Exam Tip: Verify the result: \(9.6 \times 9.6 = 92.16\).

Question 5. Find the least perfect square number which is divisible by each of the numbers 4, 5 and 10 .
Answer:
First, find the LCM of 4, 5, and 10.
Prime factorizations: \(4 = 2^2\), \(5 = 5^1\), \(10 = 2 \times 5\).
LCM \(= 2^2 \times 5 = 4 \times 5 = 20\).
To make 20 a perfect square, we need to multiply by 5 (since 5 lacks a pair: \(2^2 \times 5^1\)).
Required perfect square \(= 20 \times 5 = 100\).
In simple words: Find the LCM first, then multiply by the missing prime factors to make it a square.

Exam Tip: The required number must be a multiple of the LCM and also a perfect square.

Question 6. Find the least perfect square number which is divisible by each of the numbers 8, 12, 15 and 20 .
Answer:
Find the LCM of 8, 12, 15, and 20:
\(8 = 2^3\)
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(20 = 2^2 \times 5\)
LCM \(= 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120\).
Prime factorization of 120: \(120 = 2^3 \times 3 \times 5 = 2^2 \times 2 \times 3 \times 5\).
To make it a perfect square, multiply by \(2 \times 3 \times 5 = 30\).
Required number \(= 120 \times 30 = 3600\).
In simple words: Find the LCM, look at unpaired prime factors, and multiply the LCM by them.

Exam Tip: Always prime factorize the LCM to find which factors need pairing.

Question 7. The sides of a rectangular field are 80m and 18m respectively. Find the length of the diagonal.
Answer:
Length \(l = 80\text{ m}\), Breadth \(b = 18\text{ m}\).
Diagonal \(d = \sqrt{l^2 + b^2} = \sqrt{80^2 + 18^2}\)
\(d = \sqrt{6400 + 324} = \sqrt{6724}\)
Using long division for \(\sqrt{6724}\): \(d = 82\text{ m}\).
In simple words: Use Pythagoras theorem where the diagonal is the hypotenuse of a right-angled triangle.

Exam Tip: Formula for diagonal of a rectangle is \(\sqrt{l^2 + b^2}\).

Question 8. Find the smallest number by which 2475 must be multiplied to make it a perfect square.
Answer:
Prime factorization of 2475:
\(2475 = 3 \times 825 = 3 \times 3 \times 275 = 3^2 \times 5 \times 55 = 3^2 \times 5^2 \times 11\).
The prime factor 11 does not have a pair.
Therefore, 2475 must be multiplied by 11 to make it a perfect square.
In simple words: Find prime factors and multiply by the factor that remains unpaired.

Exam Tip: The number which lacks a pair in prime factorization is the required multiplier.

Question 9. Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient?
Answer:
Prime factorization of 9408:
\(9408 = 2 \times 4704 = 2^2 \times 2352 = \dots = 2^6 \times 3 \times 7^2 = (2^3 \times 7)^2 \times 3\).
The factor 3 is left unpaired.
Smallest number to divide by = 3.
Quotient \(= \frac{9408}{3} = 3136\).
Square root of quotient \(= \sqrt{3136} = 56\).
In simple words: Divide by the unpaired prime factor to get a perfect square, then find its square root.

Exam Tip: Division removes the extra unpaired factor from the prime factorization.

Question 10. Find the smallest number by which 1250 must be multiplied to make it a perfect square. Also find the square root of the number so obtained.
Answer:
Prime factorization of 1250:
\(1250 = 2 \times 625 = 2 \times 5^4 = 2 \times (5^2)^2 = 5^4 \times 2\).
The factor 2 is unpaired. Multiply by 2.
New number \(= 1250 \times 2 = 2500\).
Square root \(= \sqrt{2500} = 50\).
In simple words: Multiply by 2 to make all prime factors paired, then take the square root.

Exam Tip: Multiplying adds one more of the unpaired prime factor to complete the pair.

Question 11. Is 2352 a perfect square? If not, find the smallest number by which 2352 must be multiplied so that the product is a perfect square. Find the square root of the quotient.
Answer:
Prime factorization of 2352:
\(2352 = 2 \times 1176 = 2^4 \times 3 \times 7^2 = (2^2 \times 7)^2 \times 3\).
Since 3 is unpaired, 2352 is not a perfect square.
Multiply by 3 to make it a perfect square.
Product \(= 2352 \times 3 = 7056\).
Square root of product \(= \sqrt{7056} = 84\).
In simple words: Check prime factors; multiply by the leftover factor to form a square.

Exam Tip: Ensure every prime factor appears with an even exponent in a perfect square.

Question 12. Find the least number which must be subtracted from 984 to make it a perfect square.
Answer:
Using long division for \(\sqrt{984}\):
Pair digits: 9 84.
\(3 \times 3 = 9\), remainder = 0. Bring down 84.
Double 3 to get 6_. Trial: \(61 \times 1 = 61\), remainder = \(84 - 61 = 23\).
Since the remainder is 23, subtracting 23 from 984 gives a perfect square (\(31^2 = 961\)).
Least number to be subtracted = 23.
In simple words: Find the remainder using long division; that remainder is the number to subtract.

Exam Tip: The remainder in long division directly gives the number to subtract.

Question 13. Find the least number which must be subtracted from 7230 to make it a perfect square. Find the square root of the number so obtained.
Answer:
Using long division for \(\sqrt{7230}\):
Pair digits: 72 30.
\(8 \times 8 = 64\), remainder = 8. Bring down 30 to get 830.
Double 8 to get 16_. Trial: \(164 \times 4 = 656\), remainder = \(830 - 656 = 174\).
Least number to subtract = 174.
New perfect square \(= 7230 - 174 = 7056\).
Square root \(= \sqrt{7056} = 84\).
In simple words: Subtract the long division remainder to get the square, then find its root.

Exam Tip: Subtraction problems rely entirely on the leftover remainder.

Question 14. Find the least number of four digit which is a perfect square.
Answer:
The smallest 4-digit number is 1000.
Using long division for \(\sqrt{1000}\):
\(31^2 = 961\) (3-digit), and \(32^2 = 1024\).
Therefore, the least 4-digit perfect square is 1024.
In simple words: Check squares of numbers around \(\sqrt{1000}\) to find the first 4-digit square.

Exam Tip: Smallest 4-digit number is 1000, and \(31^2 < 1000 < 32^2\), so take \(32^2\).

Question 15. Find the least number which must be added to 9213 to make it a perfect square. Find the square root of the number so obtained.
Answer:
Using long division for \(\sqrt{9213}\):
\(95^2 = 9025\) and \(96^2 = 9216\).
Since \(95^2 < 9213 < 96^2\), the next square is \(96^2 = 9216\).
Number to be added \(= 9216 - 9213 = 3\).
Square root of the new number \(= \sqrt{9216} = 96\).
In simple words: Find the next integer's square and subtract the given number from it.

Exam Tip: For addition problems, square the next number after the root and find the difference.

Question 16. Find the greatest number of five digit, which is a perfect square.
Answer:
The greatest 5-digit number is 99999.
Using long division for \(\sqrt{99999}\), we get quotient 316 and remainder 143.
Greatest 5-digit square \(= 99999 - 143 = 99856\) (which is \(316^2\)).
In simple words: Subtract the long division remainder from the greatest 5-digit number.

Exam Tip: Subtracting the remainder from the number always yields the largest square below it.

Question 17. The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs.1156, find the strength of the class.
Answer:
Let the number of students be \(x\).
Contribution per student = Rs. \(x\).
Total contribution \(= x \times x = x^2 = 1156\).
\(x = \sqrt{1156} = 34\).
Strength of the class = 34 students.
In simple words: Take the square root of the total amount collected to find the number of students.

Exam Tip: Whenever contribution equals total members, the answer is the square root of the total sum.

Question 18. 8649 students are sitting in a lecture room in such a manner that there are many students in a row as there are rows in the lecture room. How many students are there in each row?
Answer:
Let number of rows = \(x\). Number of students in each row = \(x\).
Total students \(= x^2 = 8649\).
\(x = \sqrt{8649} = 93\).
There are 93 students in each row.
In simple words: Square root of total students gives the number of rows and students per row.

Exam Tip: Equal rows and columns problems always require calculating a square root.

Question 19. There are certain number of rows of trees in a garden. The number of trees in each row is twice the number of rows. If the number of trees in the garden is 1250, find the number of rows in the garden.
Answer:
Let number of rows = \(x\).
Number of trees in each row = \(2x\).
Total trees \(= x \times 2x = 2x^2 = 1250\).
\(x^2 = \frac{1250}{2} = 625\).
\(x = \sqrt{625} = 25\).
Number of rows = 25.
In simple words: Set up the algebraic equation, simplify, and take the square root.

Exam Tip: Formulate the product equation carefully based on the given ratio.

Question 20. A gardener has 1500 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Answer:
Find the square root of 1500 using long division:
\(38^2 = 1444\) and \(39^2 = 1521\).
Since \(1444 < 1500 < 1521\), the next square is \(39^2 = 1521\).
Additional plants needed \(= 1521 - 1500 = 21\).
In simple words: Find the next larger square number and find the difference.

Exam Tip: This is a classic "number to be added" problem disguised in a word context.

Question 21. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so 50 soldiers are left out. If the total total number of soldiers be 4811, find the total number of soldiers in each row?
Answer:
Soldiers in the square arrangement \(= 4811 - 50 = 4761\).
Let number of soldiers in each row be \(x\).
Total soldiers in square \(= x^2 = 4761\).
\(x = \sqrt{4761} = 69\).
There are 69 soldiers in each row.
In simple words: Subtract the leftover soldiers first, then take the square root of the remaining number.

Exam Tip: Always subtract leftovers before calculating the square root for the square formation.

Question 22. A yoga instructor wants to arrange maximum possible number of 6000 students in a ground so that the number of rows is same as the number of columns. How many rows will be there if 71 students were left out after the arrangement ?
Answer:
Students in square arrangement \(= 6000 - 71 = 5929\).
Let number of rows = \(x\).
\(x^2 = 5929\).
\(x = \sqrt{5929} = 77\).
There will be 77 rows.
In simple words: Subtract the left out students to get the exact square number, then find its square root.

Exam Tip: Square root of the adjusted student count gives the number of rows or columns.

CUBE AND CUBEROOTS

Question 1. The unit digit of the cube of the number 1127 is
(a) 7
(b) 4
(c) 2
(d) 3
Answer: (d) 3
The unit digit of 1127 is 7. The cube of 7 is \(7^3 = 343\), which ends in 3.
In simple words: Cube only the unit digit of the number to find the last digit of its cube.

Exam Tip: Focus solely on the last digit of the base number for unit digit questions.

Question 2. Volume of a cube with side 2m is
(a) 9m\(^3$
(b) 6m\(^3$
(c) 8m\(^3$
(d) 12m\(^3$
Answer: (c) 8m\(^3\)
Volume of cube \(= (\text{side})^3 = 2^3 = 8\text{ m}^3\).
In simple words: Cube the side length to get the volume.

Exam Tip: Volume formula for a cube is \(s^3\).

Question 3. Number of digits in the cube root of 140608
(a) 3
(b) 2
(c) 4
(d) 6
Answer: (a) 3
Grouping digits from right in blocks of 3: (140) (608). There are 2 blocks, so the cube root has 2 digits? Wait, counting digits of 140608: 6 digits. Cube root of a 6-digit number has 2 digits (\(6/3 = 2\)). Let's check: \(52^3 = 140608\). So 2 digits.
Correction for option: (b) 2.
In simple words: Divide the total number of digits by 3 to find the number of digits in the cube root.

Exam Tip: For cube roots, group digits in threes from right to left.

Question 4. A perfect cube number does not end with _____ zeros.
(a) 2
(b) 6
(c) 3
(d) 9
Answer: (a) 2
A perfect cube always ends in zeros that are multiples of 3 (like 3, 6, 9). It cannot end in 2 zeros.
In simple words: Zeros at the end of a cube always come in groups of three.

Exam Tip: Perfect cubes end in 0, 3, 6, or 9 zeros.

Question 5. The value of \(\sqrt[3]{27} + 3^3\)
(a) 12
(b) 27
(c) 10
(d) 30
Answer: (d) 30
\(\sqrt[3]{27} = 3\), and \(3^3 = 27\).
Sum \(= 3 + 27 = 30\).
In simple words: Evaluate the cube root and cube separately, then add them together.

Exam Tip: Calculate each term independently following PEMDAS/BODMAS order.

Question 6. Find the digit in the ones place of the cubes of the following numbers.
(a) 79 (b) 81 (c) 94 (d) 106 (e) 37 (f) 62 (g) 45 (h) 53
Answer:
(a) 79: Unit digit 9, \(9^3 = 729\), ones place = 9.
(b) 81: Unit digit 1, \(1^3 = 1\), ones place = 1.
(c) 94: Unit digit 4, \(4^3 = 64\), ones place = 4.
(d) 106: Unit digit 6, \(6^3 = 216\), ones place = 6.
(e) 37: Unit digit 7, \(7^3 = 343\), ones place = 3.
(f) 62: Unit digit 2, \(2^3 = 8\), ones place = 8.
(g) 45: Unit digit 5, \(5^3 = 125\), ones place = 5.
(h) 53: Unit digit 3, \(3^3 = 27\), ones place = 7.
In simple words: Cube only the last digit of each number to find its ones place.

Exam Tip: Memorize the cube of digits 0 through 9 for instant answers.

Question 7. Find the cube root of the following numbers by prime factorization method:
(a) 74088 (b) 4096 (c) 13824 (d) 5832 (e) 35937 (f) 42875 (g) 15625 (h) 46656 (i) 175616
Answer:
(a) \(74088 = 2^3 \times 3^3 \times 7^3 \implies \sqrt[3]{74088} = 2 \times 3 \times 7 = 42\).
(b) \(4096 = 2^{12} \implies \sqrt[3]{4096} = 2^4 = 16\).
(c) \(13824 = 2^9 \times 3^3 \implies \sqrt[3]{13824} = 2^3 \times 3 = 24\).
(d) \(5832 = 2^3 \times 3^6 \implies \sqrt[3]{5832} = 2 \times 3^2 = 18\).
(e) \(35937 = 3^9 \implies \sqrt[3]{35937} = 3^3 = 33\).
(f) \(42875 = 5^3 \times 7^3 \implies \sqrt[3]{42875} = 5 \times 7 = 35\).
(g) \(15625 = 5^6 \implies \sqrt[3]{15625} = 5^2 = 25\).
(h) \(46656 = 2^6 \times 3^6 \implies \sqrt[3]{46656} = 2^2 \times 3^2 = 36\).
(i) \(175616 = 2^9 \times 7^3 \implies \sqrt[3]{175616} = 2^3 \times 7 = 56\).
In simple words: Prime factorize and group factors in triplets of three to find the cube root.

Exam Tip: Take one factor from each triplet group of identical prime factors.

Question 8. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. Also find the cube root of the number so obtained.
(a) 1024 (b) 6561 (c) 1125 (d) 3456 (e) 16384 (f) 675 (g) 9000
Answer:
(a) \(1024 = 2^{10} = 2^9 \times 2\). Needs \(2^2 = 4\). Cube root of new number = 16.
(b) \(6561 = 3^8 = 3^6 \times 3^2\). Needs 3. Cube root of new number = 27.
(c) \(1125 = 3^2 \times 5^3\). Needs 3. Cube root of new number = 15.
(d) \(3456 = 2^7 \times 3^3 = 2^6 \times 3^3 \times 2\). Needs \(2^2 = 4\). Cube root of new number = 24.
(e) \(16384 = 2^{14} = 2^{12} \times 2^2\). Needs 2. Cube root of new number = 32.
(f) \(675 = 3^3 \times 5^2\). Needs 5. Cube root of new number = 15.
(g) \(9000 = 2^3 \times 3^2 \times 5^3\). Needs \(3\). Cube root of new number = 30.
In simple words: Make triplets of prime factors; multiply by the factors needed to complete each triplet.

Exam Tip: Cube root prime factors must always be grouped in triplets of three.

Question 9. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. Also find the cube root of the quotient so obtained.
(a) 16875 (b) 1188 (c) 3087 (d) 3327 (e) 5632 (f) 1600 (g) 5103
Answer:
(a) \(16875 = 3^3 \times 5^4 = 3^3 \times 5^3 \times 5\). Divide by 5. Cube root = 15.
(b) \(1188 = 2^2 \times 3^3 \times 11\). Divide by \(2^2 \times 11 = 44\). Cube root = 3.
(c) \(3087 = 3^2 \times 7^3\). Divide by \(3^2 = 9\). Cube root = 7.
(d) \(3327 = 3 \times 31^2\)... Wait, factorize: \(3327 = 3 \times 1109\) (approx). Let's use standard prime factorization: \(3327 = 3 \times 1109\). Divide by 3 to get 1109? Actually \(3327 = 3 \times 31 \times 31 = 3 \times 31^2\). Divide by 3. Cube root = 31.
(e) \(5632 = 2^9 \times 11\). Divide by 11. Cube root = 16.
(f) \(1600 = 2^6 \times 5^2\). Divide by \(5^2 = 25\). Cube root = 4.
(g) \(5103 = 3^5 \times 7 = 3^3 \times 3^2 \times 7\). Divide by \(9 \times 7 = 63\). Cube root = 3.
In simple words: Divide by the extra prime factors that do not form complete triplets.

Exam Tip: Division removes the incomplete extra factors from the prime factorization.

Question 10. Find the cube root of the following numbers by estimation method.
(a) 4913 (b) 12167 (c) 54872 (d) 68921 (e) 592704 (f) 512000 (g) 857375
Answer:
(a) 4913: Group into (4) (913). Unit digit 3 corresponds to 7. Thousands digit 4 lies between \(1^3\) and \(2^3\), so tens digit is 1. Result = 17.
(b) 12167: Groups (12) (167). Unit digit 7 \(\to\) 3. 12 lies between \(2^3\) (8) and \(3^3\) (27) \(\to\) 2. Result = 23.
(c) 54872: Groups (54) (872). Unit digit 2 \(\to\) 8. 54 lies between \(3^3\) (27) and \(4^3\) (64) \(\to\) 3. Result = 38.
(d) 68921: Groups (68) (921). Unit digit 1 \(\to\) 1. 68 lies between \(4^3\) (64) and \(5^3\) (125) \(\to\) 4. Result = 41.
(e) 592704: Groups (592) (704). Unit digit 4 \(\to\) 4. 592 lies between \(8^3\) (512) and \(9^3\) (729) \(\to\) 8. Result = 84.
(f) 512000: Groups (512) (000). Unit digit 0 \(\to\) 0. 512 is \(8^3\) \(\to\) 8. Result = 80.
(g) 857375: Groups (857) (375). Unit digit 5 \(\to\) 5. 857 lies between \(9^3\) (729) and \(10^3\) (1000) \(\to\) 9. Result = 95.
In simple words: Split into groups of three digits from the right and estimate using unit digits and cube ranges.

Exam Tip: Estimation method is very fast for large numbers if you know cubes from 1 to 10.

Question 11. Raj made a cuboid of plasticine of 15cm, 30cm, 15cm. How many such cuboids will he need to form a cube?
Answer:
Dimensions of cuboid \(= 15 \times 30 \times 15\).
Prime factorization: \(15 = 3 \times 5\), \(30 = 2 \times 3 \times 5\), \(15 = 3 \times 5\).
Combined prime factors \(= 2^1 \times 3^3 \times 5^3\).
To form a cube, each prime factor must have an exponent that is a multiple of 3. The factor 2 has an exponent of 1, so it needs two more 2s (\(2^2 = 4\)).
Number of cuboids needed = 4.
In simple words: Find prime factors of dimensions and multiply by missing factors to complete triplets.

Exam Tip: Volume packing problems are solved by analyzing prime factor triplets.

Question 12. Volume of a cube is 3375m\(^3\). Find the side of the cube.
Answer:
Volume \(V = s^3 = 3375\).
Side \(s = \sqrt[3]{3375}\).
Prime factorization of 3375: \(3375 = 3^3 \times 5^3 = (3 \times 5)^3 = 15^3\).
Side \(s = 15\text{ m}\).
In simple words: Take the cube root of the given volume to find the side length.

Exam Tip: Formula for side from volume is \(s = \sqrt[3]{V}\).

Question 13. Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube?
Answer:
Prime factorization of 53240:
\(53240 = 2^3 \times 11^3 \times 5\).
Since the factor 5 is not in a triplet, 53240 is not a perfect cube.
Dividing by 5 makes the quotient a perfect cube.
Smallest number = 5.
In simple words: Check prime factor triplets; the leftover factor is the number to divide by.

Exam Tip: Division removes any leftover prime factor that ruins the cube property.

Question 14. Shyam has colourful cuboidal blocks of sides 3cm, 5cm, 3cm. How many such blocks will be needed to form a perfect cube?
Answer:
Dimensions = \(3 \times 5 \times 3\).
Prime factors = \(3^2 \times 5^1\).
To complete triplets, we need one 3 and two 5s (\(5^2 = 25\)).
Total blocks needed \(= 3 \times 25 = 75\).
In simple words: Find what is missing to make every prime factor appear in groups of three.

Exam Tip: Multiply the missing prime factors together to find the total blocks required.

Question 15. The side of a cube is 17cm. Find its volume.
Answer:
Side \(s = 17\text{ cm}\).
Volume \(= s^3 = 17^3 = 17 \times 17 \times 17 = 4913\text{ cm}^3\).
In simple words: Cube the given side length to calculate volume.

Exam Tip: Volume = side \(\times\) side \(\times\) side.

Question 16. A box with volume 4096 cm\(^3\) is filled with cubes of edges 4 cm. How many cubes are there in the box?
Answer:
Volume of the box \(= 4096\text{ cm}^3\).
Volume of one small cube \(= 4^3 = 64\text{ cm}^3\).
Number of cubes \(= \frac{\text{Volume of box}}{\text{Volume of one cube}} = \frac{4096}{64} = 64\).
In simple words: Divide the total volume of the box by the volume of one small cube.

Exam Tip: Number of packed items = Total volume \(\div\) Volume of one item.

UNDERSTANDING QUADRILATERALS

Question 1. In fig. \(AB \parallel CD\), If \(\angle A = 110^{\circ}\) and \(\angle C = 50^{\circ}\), name the quadrilateral and find \(\angle B\) and \(\angle D\).
Answer:
Since \(AB \parallel CD\), the quadrilateral \(ABCD\) is a trapezium.
Adjacent angles on the same side of transversal sum to \(180^{\circ}\):
\(\angle A + \angle D = 180^{\circ} \implies 110^{\circ} + \angle D = 180^{\circ} \implies \angle D = 70^{\circ}\).
Similarly, \(\angle B + \angle C = 180^{\circ} \implies \angle B + 50^{\circ} = 180^{\circ} \implies \angle B = 130^{\circ}\).
In simple words: Use interior consecutive angles property of parallel lines to find the missing angles.

Exam Tip: Consecutive interior angles between parallel lines always add up to \(180^{\circ}\).

Question 2. Three angles of a quadrilateral are equal and fourth angle is \(130^{\circ}\). What is the measure of each of the equal angles?
Answer:
Let each of the three equal angles be \(x\).
Sum of angles in a quadrilateral \(= 360^{\circ}\).
\(x + x + x + 130^{\circ} = 360^{\circ}\)
\(3x = 360^{\circ} - 130^{\circ} = 230^{\circ}\)
\(x = \frac{230^{\circ}}{3} = 76.67^{\circ}\) (or \(76^{\circ} 40'\)).
In simple words: Set up an equation with three equal variables and subtract the given fourth angle from \(360^{\circ}\).

Exam Tip: Angle sum property of any quadrilateral is always \(360^{\circ}\).

Question 3. The angles of a quadrilateral are in the ratio 3:4:5:6. Find the measure of each angle.
Answer:
Let the angles be \(3x, 4x, 5x, 6x\).
\(3x + 4x + 5x + 6x = 360^{\circ}\)
\(18x = 360^{\circ} \implies x = 20^{\circ}\).
Angles are:
\(3(20^{\circ}) = 60^{\circ}\)
\(4(20^{\circ}) = 80^{\circ}\)
\(5(20^{\circ}) = 100^{\circ}\)
\(6(20^{\circ}) = 120^{\circ}\)
In simple words: Assign a variable \(x\) to the ratio parts, sum them to \(360^{\circ}\), solve for \(x\), and multiply.

Exam Tip: Always verify that the sum of calculated angles equals \(360^{\circ}\).

Question 4. Two adjacent sides of a parallelogram are in the ratio 2:7 and its perimeter is 63cm. Find all the sides of the parallelogram.
Answer:
Let adjacent sides be \(2x\) and \(7x\).
Perimeter \(= 2(\text{side}_1 + \text{side}_2) = 2(2x + 7x) = 63\)
\(2(9x) = 63 \implies 18x = 63 \implies x = \frac{63}{18} = 3.5\text{ cm}\).
Sides are \(\text{side}_1 = 2(3.5) = 7\text{ cm}\), \(\text{side}_2 = 7(3.5) = 24.5\text{ cm}\).
Opposite sides of a parallelogram are equal, so all sides are 7 cm, 24.5 cm, 7 cm, 24.5 cm.
In simple words: Use the perimeter formula for a parallelogram to find \(x\), then calculate all four sides.

Exam Tip: Perimeter of a parallelogram = \(2(l + b)\).

Question 5. Two adjacent angles of a parallelogram are in the ratio 3:2. Find all the angles of the parallelogram.
Answer:
Adjacent angles of a parallelogram are supplementary (sum to \(180^{\circ}\)).
Let angles be \(3x\) and \(2x\).
\(3x + 2x = 180^{\circ} \implies 5x = 180^{\circ} \implies x = 36^{\circ}\).
Angles are \(3(36^{\circ}) = 108^{\circ}\) and \(2(36^{\circ}) = 72^{\circ}\).
Since opposite angles are equal, all four angles are \(108^{\circ}, 72^{\circ}, 108^{\circ}, 72^{\circ}\).
In simple words: Adjacent angles add up to \(180^{\circ}\) in a parallelogram.

Exam Tip: Do not confuse adjacent angles with opposite angles; adjacent sum to \(180^{\circ}\).

Question 6. Find the measure of each angle of a parallelogram PQRS, if
(a) \(\angle P = (2x + 10)^{\circ}\) and \(\angle R = (3x - 20)^{\circ}\)
(b) \(\angle Q = -5y^{\circ}\) and \(\angle S = (2y + 19)^{\circ}\) (Assuming positive magnitudes or standard opposite angle equality):
Answer:
(a) Opposite angles of a parallelogram are equal: \(\angle P = \angle R\).
\(2x + 10 = 3x - 20 \implies x = 30\).
\(\angle P = 2(30) + 10 = 70^{\circ}\). Thus \(\angle R = 70^{\circ}\).
Adjacent angles sum to \(180^{\circ}\), so \(\angle Q = 180^{\circ} - 70^{\circ} = 110^{\circ}\).
All angles: \(70^{\circ}, 110^{\circ}, 70^{\circ}, 110^{\circ}\).
(b) \(\angle Q = \angle S \implies -5y = 2y + 19\) (assuming typo in sign, let's treat as \(5y = 2y + 19 \implies 3y = 19\)). Alternatively, if \(\angle Q = 5y\) and \(\angle S = 2y + 19\):
\(5y = 2y + 19 \implies 3y = 19 \implies y = \frac{19}{3}\).
In simple words: Opposite angles in a parallelogram are equal, and adjacent angles add up to \(180^{\circ}\).

Exam Tip: Set opposite angles equal to solve for the variable first.

Question 7. In rhombus ABCD, \(AC = 8\text{ cm}\) and \(BD = 6\text{ cm}\). Find the length of AB.
Answer:
Diagonals of a rhombus bisect each other at right angles (\(90^{\circ}\)).
Let diagonals intersect at \(O\).
\(OA = \frac{AC}{2} = \frac{8}{2} = 4\text{ cm}\).
\(OB = \frac{BD}{2} = \frac{6}{2} = 3\text{ cm}\).
In right triangle \(AOB\), \(AB^2 = OA^2 + OB^2 = 4^2 + 3^2 = 16 + 9 = 25\).
\(AB = \sqrt{25} = 5\text{ cm}\).
In simple words: Diagonals bisect at \(90^{\circ}\), forming right-angled triangles with the side as hypotenuse.

Exam Tip: Remember that rhombus diagonals bisect perpendicularly.

Question 8. The side of a rhombus is 13 cm and one of its diagonals is 24 cm. Find the length of the other diagonal and its area.
Answer:
Side \(s = 13\text{ cm}\), Diagonal \(d_1 = 24\text{ cm}\).
Half of diagonal \(d_1\) is \(\frac{24}{2} = 12\text{ cm}\).
Using right triangle relation with side as hypotenuse: \(\text{half of } d_2 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}\).
Other diagonal \(d_2 = 2 \times 5 = 10\text{ cm}\).
Area of rhombus \(= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 24 \times 10 = 120\text{ cm}^2\).
In simple words: Use Pythagoras theorem to find half of the second diagonal, then calculate area.

Exam Tip: Area formula using diagonals is \(\frac{1}{2} d_1 d_2\).

Question 9. One of the angles of a parallelogram is \(85^{\circ}\), find the other angles.
Answer:
Opposite angle = \(85^{\circ}\).
Adjacent angle = \(180^{\circ} - 85^{\circ} = 95^{\circ}\).
All angles are \(85^{\circ}, 95^{\circ}, 85^{\circ}, 95^{\circ}\).
In simple words: Opposite angles are equal, and adjacent angles are supplementary.

Exam Tip: Quickly find remaining angles using angle properties of parallelograms.

Question 10. The sum of the opposite angles of parallelogram is \(130^{\circ}\). Find the measures of the angles of the parallelogram.
Answer:
Opposite angles in a parallelogram are equal. Let each opposite angle be \(x\).
\(x + x = 130^{\circ} \implies 2x = 130^{\circ} \implies x = 65^{\circ}\).
The other two opposite angles sum to \(360^{\circ} - 130^{\circ} = 230^{\circ}\), so each is \(115^{\circ}\).
Angles are \(65^{\circ}, 115^{\circ}, 65^{\circ}, 115^{\circ}\).
In simple words: Equal opposite angles sum to \(130^{\circ}\), so each is \(65^{\circ}\).

Exam Tip: Sum of all four angles in any parallelogram is always \(360^{\circ}\).

Question 11. In parallelogram ABCD, the length of the sides are \(AD = (6y - 3)\text{cm}\) and \(BC = (3y + 3)\text{cm}\). Find the values of x and y.
Answer:
Opposite sides of a parallelogram are equal, so \(AD = BC\).
\(6y - 3 = 3y + 3\)
\(6y - 3y = 3 + 3 \implies 3y = 6 \implies y = 2\).
In simple words: Equate the lengths of opposite sides to solve for \(y\).

Exam Tip: Opposite sides of a parallelogram are always equal in length.

Question 12. Find the length of the diagonals of a rectangle whose sides are 12cm and 5cm.
Answer:
Diagonal \(d = \sqrt{l^2 + b^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\).
Both diagonals of a rectangle are equal in length, so each diagonal is 13 cm.
In simple words: Use Pythagoras theorem where the sides form the legs and the diagonal is the hypotenuse.

Exam Tip: Diagonals of a rectangle are equal and bisect each other.

Question 13. One of the diagonals of a rhombus is equal to one of its sides. Find the angles of the rhombus.
Answer:
Let side of rhombus be \(s\). One diagonal equals \(s\).
This divides the rhombus into two equilateral triangles with side length \(s\).
Angles of equilateral triangle are \(60^{\circ}\).
Therefore, the angles of the rhombus are \(60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}\).
In simple words: A diagonal equal to the side creates equilateral triangles inside, giving angles of \(60^{\circ}\) and \(120^{\circ}\).

Exam Tip: Equilateral triangles have all angles equal to \(60^{\circ}\).

Question 14. How many sides does a regular polygon has if each of its interior angles is
a. \(135^{\circ}\)
b. \(108^{\circ}\)
Answer:
a. Interior angle \(= 135^{\circ}\). Exterior angle \(= 180^{\circ} - 135^{\circ} = 45^{\circ}\).
Number of sides \(n = \frac{360^{\circ}}{45^{\circ}} = 8\).
b. Interior angle \(= 108^{\circ}\). Exterior angle \(= 180^{\circ} - 108^{\circ} = 72^{\circ}\).
Number of sides \(n = \frac{360^{\circ}}{72^{\circ}} = 5\).
In simple words: Find the exterior angle first, then divide \(360^{\circ}\) by the exterior angle to get the number of sides.

Exam Tip: Formula for sides: \(n = \frac{360^{\circ}}{\text{Exterior Angle}}\).

Question 15. How many sides does a regular polygon has if each of its exterior angles is
a. \(15^{\circ}\)
b. \(45\)
Answer:
a. \(n = \frac{360^{\circ}}{15^{\circ}} = 24\).
b. \(n = \frac{360^{\circ}}{45^{\circ}} = 8\).
In simple words: Divide \(360^{\circ}\) by the given exterior angle.

Exam Tip: Sum of all exterior angles in any convex polygon is always \(360^{\circ}\).

Question 16. Opposite sides of a parallelogram are _____ and _____
Answer: equal, parallel
Opposite sides in a parallelogram are always equal in length and parallel to each other.
In simple words: Opposite sides match in length and never intersect.

Exam Tip: Remember both properties: equality of length and parallelism.

Question 17. A quadrilateral having only one pair of opposite sides parallel is called a_____
Answer: trapezium
A trapezium is defined as a quadrilateral with exactly one pair of parallel opposite sides.
In simple words: A four-sided figure with one pair of parallel sides is a trapezium.

Exam Tip: Distinguish between trapezium (one pair parallel) and parallelogram (two pairs parallel).

Question 18. A parallelogram having all sides equal is called a_____
Answer: rhombus
A parallelogram with equal sides is a rhombus.
In simple words: A tilted square with all equal sides is a rhombus.

Exam Tip: Rhombus has all sides equal, unlike a generic rectangle or parallelogram.

PRACTICAL GEOMETRY

Question 1. Construct a quadrilateral ABCD in which AB = 4.4cm, BC = 4cm, CD = 6.4cm, DA = 2.8 cm and BD = 6.6 cm. Measure the length of AC.
Answer:
Steps of construction:
1. Draw diagonal \(BD = 6.6\text{ cm}\).
2. With \(B\) as center and radius \(4.0\text{ cm}\), draw an arc.
3. With \(D\) as center and radius \(2.8\text{ cm}\), draw an arc cutting the previous arc at point \(A\). Join \(AB\) and \(AD\).
4. With \(B\) as center and radius \(4.4\text{ cm}\), draw an arc.
5. With \(D\) as center and radius \(6.4\text{ cm}\), draw an arc cutting at point \(C\). Join \(BC\) and \(DC\).
6. Measure the diagonal \(AC\) with a ruler (approx \(7.2\text{ cm}\)).
In simple words: Construct two triangles sharing the diagonal BD, then connect points to form the quadrilateral.

Exam Tip: Always construct the base diagonal first when given four sides and one diagonal.

Question 2. Construct a quadrilateral PQRS where PQ = 3 cm, QR = 5 cm, QS = 5 cm, PS = 4 cm and SR = 7 = 4 cm (SR = 4 cm).
Answer:
Steps of construction:
1. Draw diagonal \(QS = 5\text{ cm}\).
2. With \(Q\) as center and radius \(3\text{ cm}\), and \(S\) as center and radius \(4\text{ cm}\), draw arcs intersecting at \(P\). Join \(PQ\) and \(PS\).
3. With \(Q\) as center and radius \(5\text{ cm}\), and \(S\) as center and radius \(4\text{ cm}\), draw arcs intersecting at \(R\). Join \(QR\) and \(SR\).
In simple words: Form two triangles using the diagonal QS as the common side.

Exam Tip: Label all vertices correctly according to the problem statement.

Question 3. Construct a parallelogram "MATH" where MA = 3.6 cm, AT = 4.2 cm and MT = 6.5 cm.
Answer:
In parallelogram MATH, opposite sides are equal: \(HT = MA = 3.6\text{ cm}\) and \(MH = AT = 4.2\text{ cm}\).
Steps of construction:
1. Draw diagonal \(MT = 6.5\text{ cm}\).
2. With \(M\) as center and radius \(3.6\text{ cm}\), and \(T\) as center and radius \(4.2\text{ cm}\), draw arcs intersecting at \(A\). Join \(MA\) and \(TA\).
3. With \(M\) as center and radius \(4.2\text{ cm}\), and \(T\) as center and radius \(3.6\text{ cm}\), draw arcs intersecting at \(H\). Join \(MH\) and \(TH\).
In simple words: Use opposite side properties to find missing dimensions, then construct as a general quadrilateral using a diagonal.

Exam Tip: Parallelogram opposite sides are always equal.

Question 4. Construct a quadrilateral ABCD where AB = 5 cm, BC = 4 cm, AD = 3 cm, CD = 6 cm and BD = 5 cm.
Answer:
Steps of construction:
1. Draw diagonal \(BD = 5\text{ cm}\).
2. With \(B\) as center and radius \(5\text{ cm}\), and \(D\) as center and radius \(3\text{ cm}\), draw arcs intersecting at \(A\). Join \(AB\) and \(AD\).
3. With \(B\) as center and radius \(4\text{ cm}\), and \(D\) as center and radius \(6\text{ cm}\), draw arcs intersecting at \(C\). Join \(BC\) and \(DC\).
In simple words: Draw the diagonal first, then construct triangles ABD and CBD using compass arcs.

Exam Tip: Check compass openings carefully against given measurements.

Question 5. Construct a quadrilateral "KIND" where KI = 3.8 cm, IN = 3 cm, KD = 2.3 cm, KN = 4.5 cm and ID = 3.8 cm.
Answer:
Steps of construction:
1. Draw base \(KI = 3.8\text{ cm}\).
2. From \(K\), draw an arc of radius \(4.5\text{ cm}\) for point \(N\), and from \(I\) draw an arc of radius \(3\text{ cm}\) to locate \(N\). Join \(KN\) and \(IN\).
3. From \(K\), draw an arc of radius \(2.3\text{ cm}\) for point \(D\), and from \(I\) draw an arc of radius \(3.8\text{ cm}\) to locate \(D\). Join \(KD\) and \(ID\).
In simple words: Construct triangles KIN and KID sharing the base KI.

Exam Tip: Ensure arcs intersect cleanly at the required vertex points.

Question 6. Construct a rhombus whose diagonals are of lengths 6.8 cm and 8.4 cm.
Answer:
Steps of construction:
1. Draw diagonal \(AC = 8.4\text{ cm}\).
2. Draw the perpendicular bisector of \(AC\) intersecting it at \(O\).
3. From \(O\), cut off arcs of length \(\frac{6.8}{2} = 3.4\text{ cm}\) on both sides along the perpendicular bisector to mark points \(B\) and \(D\).
4. Join \(AB, BC, CD\), and \(DA\).
In simple words: Draw one diagonal, construct its perpendicular bisector, and mark half lengths of the second diagonal.

Exam Tip: Diagonals of a rhombus bisect each other at right angles.

Question 7. Construct a rectangle of adjacent sides 4.2 cm and 2.5 cm.
Answer:
Steps of construction:
1. Draw side \(AB = 4.2\text{ cm}\).
2. At points \(A\) and \(B\), construct \(90^{\circ}\) angles using a protractor or compass.
3. Cut off arcs of length \(2.5\text{ cm}\) along these perpendicular rays to mark points \(D\) and \(C\).
4. Join \(DC\).
In simple words: Draw the base, erect \(90^{\circ}\) angles at both ends, cut off the breadth, and connect the top.

Exam Tip: All four corner angles of a rectangle are exactly \(90^{\circ}\).

Question 8. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, \(\angle A = 90^{\circ}\) and \(\angle B = 105^{\circ}\).
Answer:
Steps of construction:
1. Draw base \(AB = 3\text{ cm}\).
2. At \(A\), construct an angle of \(90^{\circ}\) and cut off \(AD = 5\text{ cm}\).
3. At \(B\), construct an angle of \(105^{\circ}\) and cut off \(BC = 3\text{ cm}\).
4. Join points \(D\) and \(C\).
In simple words: Draw base, construct given angles at endpoints, mark side lengths, and close the figure.

Exam Tip: Construct \(105^{\circ}\) by bisecting the angle between \(90^{\circ}\) and \(120^{\circ}\).

Question 9. Construct a quadrilateral "RENT" in which RE = 5 cm, EN = 6 cm, RT = 5.5 cm, \(\angle R = 60^{\circ}\), \(\angle E = 105^{\circ}\).
Answer:
Steps of construction:
1. Draw base \(RE = 5\text{ cm}\).
2. At \(R\), construct \(\angle R = 60^{\circ}\) and cut off \(RT = 5.5\text{ cm}\).
3. At \(E\), construct \(\angle E = 105^{\circ}\) and cut off \(EN = 6\text{ cm}\).
4. Join points \(T\) and \(N\).
In simple words: Start with base RE, make angles at both ends, mark side lengths, and connect the remaining vertices.

Exam Tip: Use a protractor or compass for precise angle construction.

Question 10. Construct a quadrilateral "FAST" in which FA = 5.5 cm, AS = 4 cm, \(\angle F = 85^{\circ}\), \(\angle A = 100^{\circ}\) and \(\angle S = 90^{\circ}\).
Answer:
Steps of construction:
1. Draw base \(FA = 5.5\text{ cm}\).
2. At \(A\), construct \(\angle A = 100^{\circ}\) and cut off \(AS = 4\text{ cm}\).
3. At \(S\), construct \(\angle S = 90^{\circ}\) and draw a ray.
4. At \(F\), construct \(\angle F = 85^{\circ}\) and draw a ray intersecting the ray from \(S\) at point \(T\).
In simple words: Draw base, construct given angles, and extend lines until they intersect.

Exam Tip: When three angles are given, build the known sides first and find the remaining vertex by intersection.

Question 11. Construct a rhombus whose diagonals are of lengths 8 cm and 7 cm.
Answer:
Steps of construction:
1. Draw diagonal \(AC = 8\text{ cm}\).
2. Construct perpendicular bisector of \(AC\).
3. From the midpoint, cut arcs of length \(\frac{7}{2} = 3.5\text{ cm}\) above and below to mark \(B\) and \(D\).
4. Join \(AB, BC, CD, DA\).
In simple words: Draw one diagonal, find its midpoint, and mark half lengths of the second diagonal perpendicularly.

Exam Tip: Perpendicular bisector ensures diagonals cross at \(90^{\circ}\).

Question 12. Construct a parallelogram LIFE, given that LI = 6 cm, IF = 5 cm and \(\angle L = 60^{\circ}\).
Answer:
Steps of construction:
1. Draw base \(LI = 6\text{ cm}\).
2. At \(L\), construct \(\angle L = 60^{\circ}\) and cut off \(LE = 5\text{ cm}\) (since opposite sides are equal, \(FE = 6\text{ cm}\) and \(IE = 5\text{ cm}\)).
3. With \(I\) as center and radius \(5\text{ cm}\), and \(E\) as center and radius \(6\text{ cm}\), draw arcs intersecting at \(F\).
4. Join \(IF\) and \(EF\).
In simple words: Use parallel property to find adjacent side and angle, then complete the parallelogram.

Exam Tip: Adjacent angles of a parallelogram are supplementary.

Question 13. Construct a square of side 5.4 cm.
Answer:
Steps of construction:
1. Draw base \(AB = 5.4\text{ cm}\).
2. Construct \(90^{\circ}\) angles at \(A\) and \(B\).
3. Cut off arcs of length \(5.4\text{ cm}\) from \(A\) and \(B\) to mark \(D\) and \(C\).
4. Join \(DC\).
In simple words: Draw the base, erect perpendiculars at both ends, and mark equal side lengths.

Exam Tip: All four sides of a square are equal and meet at \(90^{\circ}\).

Question 14. Construct a square of side 4.8 cm.
Answer:
Steps of construction:
1. Draw base \(PQ = 4.8\text{ cm}\).
2. Construct \(90^{\circ}\) at \(P\) and \(Q\).
3. Cut arcs of \(4.8\text{ cm}\) along perpendiculars to get \(S\) and \(R\).
4. Join \(SR\).
In simple words: Same as constructing a square with the given side length.

Exam Tip: Verify diagonals are equal to confirm accuracy.

Question 15. Construct a quadrilateral "RICE" in which RI = 5cm, IC = 6.5cm, \(\angle R = 60^{\circ}\), \(\angle I = 100^{\circ}\), \(\angle E = 95^{\circ}\).
Answer:
Find fourth angle: \(\angle C = 360^{\circ} - (60^{\circ} + 100^{\circ} + 95^{\circ}) = 360^{\circ} - 255^{\circ} = 105^{\circ}\).
Steps of construction:
1. Draw base \(RI = 5\text{ cm}\).
2. At \(I\), construct \(\angle I = 100^{\circ}\) and cut off \(IC = 6.5\text{ cm}\).
3. At \(C\), construct \(\angle C = 105^{\circ}\).
4. At \(R\), construct \(\angle R = 60^{\circ}\) and extend the ray to intersect the ray from \(C\) at point \(E\).
In simple words: Calculate the remaining angle first using angle sum property, then draw rays from endpoints.

Exam Tip: Sum of interior angles of a quadrilateral is always \(360^{\circ}\).

RATIONAL NUMBERS

Question 1. How many rational nos. are there in between 0 and 1
(1, 100, 0, infinite)
Answer: (infinite)
There are infinitely many rational numbers between any two distinct rational numbers.
In simple words: You can always find more fractions between 0 and 1, so there is no end.

Exam Tip: Between any two numbers, there exist infinitely many rational numbers.

Question 2. Multiplicative inverse of -2/7 is _____
Answer: \(- \frac{7}{2}\)
The multiplicative inverse (reciprocal) of a number \(\frac{a}{b}\) is \(\frac{b}{a}\).
In simple words: Flip the fraction upside down while keeping the same sign.

Exam Tip: Product of a number and its multiplicative inverse is always 1.

Question 3. The standard form of 8/(-64) is _____
Answer: \(-\frac{1}{8}\)
Divide numerator and denominator by their HCF, which is 8: \(\frac{8 \div 8}{-64 \div 8} = -\frac{1}{8}\).
In simple words: Simplify the fraction by dividing top and bottom by their common factor.

Exam Tip: In standard form, the denominator must be positive and the fraction in lowest terms.

Question 4. Write any five rational nos. in between (i) -1 and 1 (ii) -3/4 and -4/3 (iii) 1/5 and 1/6
Answer:
(i) Between -1 and 1: \(-\frac{1}{2}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}\).
(ii) Between \(-\frac{3}{4}\) (or -0.75) and \(-\frac{4}{3}\) (or -1.33): \(-0.8, -0.9, -1.0, -1.1, -1.2\).
(iii) Between \(\frac{1}{5} = 0.2\) and \(\frac{1}{6} \approx 0.167\): convert with common denominator 30: \(\frac{6}{30}\) and \(\frac{5}{30}\). Expand to denominator 60: \(\frac{11}{60}, \frac{12}{60}, \frac{13}{60}, \frac{14}{60}, \frac{15}{60}\).
In simple words: Make denominators equal or convert to decimals to easily insert numbers in between.

Exam Tip: Use equivalent fractions with larger denominators to find numbers between fractions.

Question 5. Subtract (i) 3/7 from 2/3 (ii) 6 from 3/5 (iii) -4/7 from -3/11
Answer:
(i) \(\frac{2}{3} - \frac{3}{7} = \frac{14 - 9}{21} = \frac{5}{21}\).
(ii) \(\frac{3}{5} - 6 = \frac{3 - 30}{5} = -\frac{27}{5}\).
(iii) \(-\frac{3}{11} - \left(-\frac{4}{7}\right) = -\frac{3}{11} + \frac{4}{7} = \frac{-21 + 44}{77} = \frac{23}{77}\).
In simple words: Remember "subtract A from B" means \(B - A\). Find common denominators to combine.

Exam Tip: Watch out for double negative signs when subtracting negative numbers.

Question 6. What should be added to 3/4 to get -1/4
Answer:
Let the number be \(x\).
\(\frac{3}{4} + x = -\frac{1}{4}\)
\(x = -\frac{1}{4} - \frac{3}{4} = \frac{-1 - 3}{4} = -\frac{4}{4} = -1\).
In simple words: Form an equation and isolate \(x\) by subtraction.

Exam Tip: "What should be added" equations always translate to \(\text{Target} - \text{Given}\).

Question 7. Represent the following no.s on a number line (i) 1/4, 1/2, 3/8 (ii) -3/4, 3/4, 2/4
Answer:
(i) Convert to common denominator 8: \(\frac{2}{8}, \frac{4}{8}, \frac{3}{8}\). Mark equal divisions between 0 and 1 on the number line and plot \(\frac{2}{8}, \frac{3}{8}, \frac{4}{8}\).
(ii) \(-\frac{3}{4}, \frac{2}{4} \left(=\frac{1}{2}\right), \frac{3}{4}\). Plot points on negative and positive sides of 0 divided into quarters.
In simple words: Convert fractions to a common denominator to space them evenly on a number line.

Exam Tip: Equal spacing between tick marks is essential for number line graphs.

Question 8. Arrange the following rational numbers in descending order
(i) 4/5, -2/3, -1/2, -4/7
(ii) 4/9, -5/12, 7/-18, -2/3
Answer:
(i) LCM of denominators 5, 3, 2, 7 = 210.
\(\frac{4}{5} = \frac{168}{210}\), \(-\frac{2}{3} = \frac{-140}{210}\), \(-\frac{1}{2} = \frac{-105}{210}\), \(-\frac{4}{7} = \frac{-120}{210}\).
Descending order (largest to smallest): \(\frac{4}{5}, -\frac{1}{2}, -\frac{4}{7}, -\frac{2}{3}\).
(ii) LCM of 9, 12, 18, 3 = 36.
\(\frac{4}{9} = \frac{16}{36}\), \(-\frac{5}{12} = \frac{-15}{36}\), \(-\frac{7}{18} = \frac{-14}{36}\), \(-\frac{2}{3} = \frac{-24}{36}\).
Descending order: \(\frac{4}{9}, -\frac{7}{18}, -\frac{5}{12}, -\frac{2}{3}\).
In simple words: Convert all fractions to a common denominator and compare their numerators.

Exam Tip: Descending order means arranging from largest (positive) to smallest (negative).

Question 9. Fill in the blanks
(i) -14/63 = ----/9 = -18/----
(ii) ----/60 = -5/-12 = -35/----
Answer:
(i) Simplify \(-\frac{14}{63} = -\frac{2}{9}\).
First blank: \(\frac{x}{9} = -\frac{2}{9} \implies x = -2\).
Second blank: \(-\frac{2}{9} = \frac{-18}{y} \implies -2y = -162 \implies y = 81\).
Answers: -2, 81.
(ii) Simplify \(-\frac{-5}{-12} = \frac{5}{12}\).
First blank: \(\frac{x}{60} = \frac{5}{12} \implies 12x = 300 \implies x = 25\).
Second blank: \(\frac{5}{12} = \frac{-35}{y} \implies 5y = -420 \implies y = -84\).
Answers: 25, -84.
In simple words: Use equivalent fraction ratios by cross-multiplying or simplifying.

Exam Tip: Equivalent fractions maintain the same ratio when multiplied or divided by the same factor.

Question 10. Using suitable rearrangement find the sum of the following
(i) 4/3 + 3/5 + -2/3 + -11/5
(ii) -13/20 + 11/14 + -5/7 + 1
Answer:
(i) Group terms with same denominators:
\(\left(\frac{4}{3} - \frac{2}{3}\right) + \left(\frac{3}{5} - \frac{11}{5}\right) = \frac{2}{3} + \frac{-8}{5} = \frac{10 - 24}{15} = -\frac{14}{15}\).
(ii) Rearrange and combine fractions:
\(\frac{11}{14} - \frac{5}{7} = \frac{11 - 10}{14} = \frac{1}{14}\).
Now add \(-\frac{13}{20} + \frac{1}{14} + 1\)... LCM of 20, 14, 1 = 140.
\(= \frac{-91 + 10 + 140}{140} = \frac{59}{140}\).
In simple words: Group fractions with common denominators together to simplify addition.

Exam Tip: Commutative and associative properties allow rearranging terms for easier calculation.

Question 11. Using suitable properties find
(i) \(2/5 \times -3/7 - 1/14 - 3/7 \times 3/5$
(ii) \(-5/7 \times 2/3 + 7/9 - 5/7 \times 1/3$
(iii) \(-3/2 \times 4/5 + 9/5 \times -10/3 - 1/2 \times 3/4$
(iv) \(-4/9 \times 2/3 - 1/3 \times 4/9 + 1/2 \times 1/6$
(v) \(9/16 \times 4/12 + 9/16 \times -5/15$
Answer:
(i) Rearrange using distributive property: \(\frac{2}{5} \times -\frac{3}{7} - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14}\)
\(= \frac{3}{5} \left(-\frac{2}{7} - \frac{3}{7}\right) - \frac{1}{14} = \frac{3}{5}(-1) - \frac{1}{14} = -\frac{3}{5} - \frac{1}{14} = -\frac{42+5}{70} = -\frac{47}{70}\).
(ii) Use distributive property with \(-\frac{5}{7}\):
\(-\frac{5}{7} \left(\frac{2}{3} + \frac{1}{3}\right) + \frac{7}{9} = -\frac{5}{7}(1) + \frac{7}{9} = -\frac{5}{7} + \frac{7}{9} = \frac{-45+49}{63} = \frac{4}{63}\).
(iii) Simplify each term: \(-\frac{12}{10} - \frac{30}{15} - \frac{3}{8} = -\frac{6}{5} - 2 - \frac{3}{8} = -\frac{16}{5} - \frac{3}{8} = \frac{-128-15}{40} = -\frac{143}{40}\).
(iv) Factor out \(\frac{4}{9}\): \(\frac{4}{9} \left(-\frac{2}{3} - \frac{1}{3}\right) + \frac{1}{12} = \frac{4}{9}(-1) + \frac{1}{12} = -\frac{4}{9} + \frac{1}{12} = \frac{-16+3}{36} = -\frac{13}{36}\).
(v) Simplify fractions: \(\frac{4}{12} = \frac{1}{3}\), \(-\frac{5}{15} = -\frac{1}{3}\).
\(\frac{9}{16} \times \frac{1}{3} + \frac{9}{16} \times -\frac{1}{3} = \frac{3}{16} - \frac{3}{16} = 0\).
In simple words: Apply distributive and associative properties to factor out common terms and simplify.

Exam Tip: Distributive property \(a(b+c) = ab + ac\) is extremely useful for simplifying rational number expressions.

Question 12. Verify the following . Also mention the property used
(i) \(2 \times -3/5 = -3/5 \times 2$
(ii) \(-12/5 \times (4/15 \times 25/-16) = (-12/5 \times 4/15) \times 25/-16$
(iii) \(-5/2 \times (-2 + 11/3) = (-5/2 \times -2) + (-5/2 \times 11/3)$
Answer:
(i) LHS \(= -\frac{6}{5}\), RHS \(= -\frac{6}{5}\). Property: Commutative property of multiplication (\(a \times b = b \times a\)).
(ii) Both sides evaluate to \(\frac{5}{4}\). Property: Associative property of multiplication (\(a \times (b \times c) = (a \times b) \times c\)).
(iii) Both sides evaluate to \(\frac{5}{6}\). Property: Distributive property of multiplication over addition (\(a \times (b + c) = ab + ac\)).
In simple words: Compute both sides separately to verify equality and name the algebraic property applied.

Exam Tip: Learn standard mathematical property names (commutative, associative, distributive).

Question 13. Divide the product of 5/9 and -6/5 by their difference
Answer:
Product \(= \frac{5}{9} \times -\frac{6}{5} = -\frac{6}{9} = -\frac{2}{3}\).
Difference \(= \frac{5}{9} - \left(-\frac{6}{5}\right) = \frac{5}{9} + \frac{6}{5} = \frac{25 + 54}{45} = \frac{79}{45}\).
Division of product by difference \(= \frac{-2/3}{79/45} = -\frac{2}{3} \times \frac{45}{79} = -\frac{30}{79}\).
In simple words: Find the product first, then the difference, and divide the two results.

Exam Tip: Dividing by a fraction is the same as multiplying by its reciprocal.

Question 14. By what rational number should we multiply -3/7 to get 4/49 ?
Answer:
Let the required number be \(x\).
\(-\frac{3}{7} \times x = \frac{4}{49}\)
\(x = \frac{4}{49} \div \left(-\frac{3}{7}\right) = \frac{4}{49} \times \left(-\frac{7}{3}\right) = -\frac{4}{21}\).
In simple words: Divide the target product by the given number to find the multiplier.

Exam Tip: Set up a linear equation with \(x\) and solve using division.

EXPONENTS AND POWERS

Question 1. Square of \(\left(\frac{-2}{3}\right)\)
(a) \(\left(\frac{-2}{3}\right)\)
(b) \(\left(\frac{2}{3}\right)\)
(c) \(\left(\frac{-2}{3}\right)^2$
(d) \(\left(\frac{4}{9}\right)\)
Answer: (d) \(\left(\frac{4}{9}\right)\)
\(\left(\frac{-2}{3}\right)^2 = \frac{(-2)^2}{3^2} = \frac{4}{9}\).
In simple words: Square both the numerator and denominator.

Exam Tip: Square of a negative number is always positive.

Question 2. Cube of \(\left(\frac{-1}{2}\right)\)
(a) \(\left(\frac{1}{8}\right)\)
(b) \(\left(\frac{1}{16}\right)\)
(c) \(\left(\frac{-1}{8}\right)\)
(d) \(\left(\frac{1}{16}\right)\)
Answer: (c) \(\left(\frac{-1}{8}\right)\)
\(\left(\frac{-1}{2}\right)^3 = \frac{(-1)^3}{2^3} = \frac{-1}{8}\).
In simple words: Cube the top and bottom numbers while preserving the negative sign.

Exam Tip: Cube of a negative number remains negative.

Question 3. \(\left(\left(\frac{1}{3}\right)^2\right)^4\)
(a) \(\left(\frac{1}{3}\right)^6$
(b) \(\left(\frac{1}{3}\right)^8$
(c) \(\left(\frac{1}{3}\right)^{24}$
(d) \(\left(\frac{1}{3}\right)^{16}$
Answer: (b) \(\left(\frac{1}{3}\right)^8\)
Using power of a power rule \((a^m)^n = a^{m \times n}\): \((2 \times 4) = 8\).
In simple words: Multiply the exponents when raising a power to a power.

Exam Tip: \((a^m)^n = a^{mn}\).

Question 4. \(\left(\frac{-1}{5}\right)^3 \div \left(\frac{-1}{5}\right)^8\) is equal to
(a) \(\left(\frac{-1}{5}\right)^5$
(b) \(\left(\frac{-1}{5}\right)^{11}$
(c) \(\left(\frac{-1}{5}\right)^5$
(d) \((-5)^5$
Answer: (a) \(\left(\frac{-1}{5}\right)^{-5}\) or check options: wait, quotient rule \(a^m \div a^n = a^{m-n}\): \(3 - 8 = -5\).
Since \(\left(\frac{-1}{5}\right)^{-5} = (-5)^5\), option (d) is correct!
In simple words: Subtract exponents when dividing same bases, then simplify negative exponents.

Exam Tip: \(a^m \div a^n = a^{m-n}\).

Question 5. Simplify the following: \(\left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2}\)
Answer:
\(\left(\frac{1}{4}\right)^{-2} = 4^2 = 16\)
\(\left(\frac{1}{2}\right)^{-2} = 2^2 = 4\)
\(\left(\frac{1}{3}\right)^{-2} = 3^2 = 9\)
Sum \(= 16 + 4 + 9 = 29\).
In simple words: Flip fractions to make negative exponents positive, then evaluate.

Exam Tip: \((a/b)^{-n} = (b/a)^n\).

Question 6. \(\{6^{-1} + \left(\frac{3}{2}\right)^{-1}\}^{-1}\)
Answer:
\(6^{-1} = \frac{1}{6}\)
\(\left(\frac{3}{2}\right)^{-1} = \frac{2}{3}\)
Sum inside braces \(= \frac{1}{6} + \frac{2}{3} = \frac{1 + 4}{6} = \frac{5}{6}\).
Raise to \(-1\): \(\left(\frac{5}{6}\right)^{-1} = \frac{6}{5}\).
In simple words: Simplify inside the brackets first, then apply the outer negative exponent.

Exam Tip: Follow BODMAS/PEMDAS order of operations carefully.

Question 7. \(\left\{\left(\frac{5}{3}\right)^{-1} - \left(\frac{5}{3}\right)^{-1}\right\}^{-1}\)
Answer:
\(\left(\frac{5}{3}\right)^{-1} - \left(\frac{5}{3}\right)^{-1} = 0\).
\(0^{-1}\) is undefined.
In simple words: Any number minus itself is 0, and division by zero is undefined.

Exam Tip: Watch out for terms that cancel out to zero.

Question 8. \((2^{-1} \times 5^{-1})^2 \times \left(\frac{5}{8}\right)^{-1}\)
Answer:
\((2^{-1} \times 5^{-1})^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}\)
\(\left(\frac{5}{8}\right)^{-1} = \frac{8}{5}\)
Product \(= \frac{1}{100} \times \frac{8}{5} = \frac{8}{500} = \frac{2}{125}\).
In simple words: Convert negative exponents to fractions, simplify, and multiply.

Exam Tip: \((ab)^{-n} = a^{-n}b^{-n}\).

Question 9. \((6^{-1} - 8^{-1})^{-1} + (2^{-1} + 3^{-1})^{-1}\)
Answer:
\(6^{-1} - 8^{-1} = \frac{1}{6} - \frac{1}{8} = \frac{4 - 3}{24} = \frac{1}{24}\). Inverse \(= 24\).
\(2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\). Inverse \(= \frac{6}{5}\).
Sum \(= 24 + \frac{6}{5} = \frac{120 + 6}{5} = \frac{126}{5}\).
In simple words: Evaluate each bracket separately and add the results.

Exam Tip: Take reciprocals when dealing with \(-1\) exponents.

Question 10. \((4^{-1} + 8^{-1}) \div \left(\frac{2}{3}\right)^{-1}\)
Answer:
\(4^{-1} + 8^{-1} = \frac{1}{4} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8}\).
\(\left(\frac{2}{3}\right)^{-1} = \frac{3}{2}\).
Division \(= \frac{3}{8} \div \frac{3}{2} = \frac{3}{8} \times \frac{2}{3} = \frac{1}{4}\).
In simple words: Simplify the sum first, then divide by the reciprocal of the divisor.

Exam Tip: Division of fractions becomes multiplication by the reciprocal.

Question 11. \(\left[\left(\frac{1}{3}\right)^{-3} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-3}\)
Answer:
\(\left(\frac{1}{3}\right)^{-3} = 3^3 = 27\)
\(\left(\frac{1}{2}\right)^{-3} = 2^3 = 8\)
Difference \(= 27 - 8 = 19\).
\(\left(\frac{1}{4}\right)^{-3} = 4^3 = 64\).
Result \(= \frac{19}{64}\).
In simple words: Cube the reciprocals inside the brackets, subtract, and divide by 64.

Exam Tip: Negative exponents flip the fraction before applying the positive power.

Question 12. Find 'x' so that \(\left(\frac{7}{3}\right)^{-5} \times \left(\frac{7}{3}\right)^{-11} = \left(\frac{7}{3}\right)^{8x}\)
Answer:
Using product rule on LHS: \(-5 + (-11) = -16\).
\(\left(\frac{7}{3}\right)^{-16} = \left(\frac{7}{3}\right)^{8x}\)
Equating exponents: \(8x = -16 \implies x = -2\).
In simple words: Add exponents on the same base, then set them equal to the right-hand exponent.

Exam Tip: When bases are equal, equate the exponents directly.

Question 13. Find 'm' so that \(\left(\frac{2}{9}\right)^3 \times \left(\frac{2}{9}\right)^{-6} = \left(\frac{2}{9}\right)^{2m-1}\)
Answer:
LHS exponent: \(3 + (-6) = -3\).
\(2m - 1 = -3 \implies 2m = -2 \implies m = -1\).
In simple words: Combine exponents on the left and solve the linear equation for \(m\).

Exam Tip: Set exponents equal when bases match on both sides.

Question 14. Find the value of 'p' for which \((7^5) \div 7^3 = 7^{12}\)... Wait, check equation: \((7^p) \div 7^{-3} = 7^{12}\)
Answer:
\(7^{p - (-3)} = 7^{p+3} = 7^{12}\)
\(p + 3 = 12 \implies p = 9\).
In simple words: Subtract the denominator exponent from the numerator exponent.

Exam Tip: Be careful with subtraction of negative exponents: \(a^m \div a^{-n} = a^{m+n}\).

Question 15. By what number should \(\left(\frac{1}{2}\right)^{-1}\) be multiplied so that the product may be equal to \(\left(\frac{-4}{7}\right)^{-1}\)
Answer:
\(\left(\frac{1}{2}\right)^{-1} = 2\)
\(\left(\frac{-4}{7}\right)^{-1} = -\frac{7}{4}\)
Let multiplier be \(x\): \(2 \times x = -\frac{7}{4} \implies x = -\frac{7}{8}\).
In simple words: Divide the target value by the given number.

Exam Tip: Negative exponent of \(-1\) simply inverts the fraction.

Question 16. By what number should \((-15)^{-1}\) be divided so that the quotient may be equal to \((-5)^{-1}\)
Answer:
\((-15)^{-1} = -\frac{1}{15}\), \((-5)^{-1} = -\frac{1}{5}\).
Let divisor be \(x\): \(-\frac{1}{15} \div x = -\frac{1}{5} \implies x = \frac{-1/15}{-1/5} = \frac{1}{3}\).
In simple words: Set up division equation and solve for the unknown divisor.

Exam Tip: \(a^{-1} = \frac{1}{a}\).

Question 17. By what number should \(\left(\frac{-3}{2}\right)^{-3}\) be divided so that the quotient may be \(\left(\frac{4}{27}\right)^{-2}\)
Answer:
\(\left(\frac{-3}{2}\right)^{-3} = \left(-\frac{2}{3}\right)^3 = -\frac{8}{27}\).
\(\left(\frac{4}{27}\right)^{-2} = \left(\frac{27}{4}\right)^2 = \frac{729}{16}\).
Let divisor be \(x\): \(-\frac{8}{27} \div x = \frac{729}{16} \implies x = -\frac{8}{27} \times \frac{16}{729} = -\frac{128}{19683}\).
In simple words: Simplify both exponential expressions first, then divide.

Exam Tip: Simplify negative exponents by reciprocating the base first.

Question 18. Find 'x' if \(\left(\frac{3}{2}\right)^{-3} \times \left(\frac{3}{2}\right)^5 = \left(\frac{3}{2}\right)^{2x+1}\)
Answer:
LHS exponent: \(-3 + 5 = 2\).
\(2x + 1 = 2 \implies 2x = 1 \implies x = \frac{1}{2}\).
In simple words: Add the exponents on the left and equate to the right side exponent.

Exam Tip: Equate exponents when the base is identical on both sides.

Question 19. Express the following number in usual form
(i) \(3.52 \times 10^5\)
(ii) \(7.54 \times 10^{-4}\)
(iii) \(7.9 \times 10^{-11}\)
(iv) \(9.021 \times 10^{-12}\)
Answer:
(i) \(352,000\)
(ii) \(0.000754\)
(iii) \(0.000000000079\)
(iv) \(0.000000000009021\)
In simple words: Move the decimal point right for positive powers and left for negative powers.

Exam Tip: Positive exponent means large number, negative exponent means decimal number.

Question 20. Express the following number in standard form
(i) 0.0000000942
(ii) 0.000072984
(iii) 0.0000000685
(iv) 87500000000
(v) 954321
Answer:
(i) \(9.42 \times 10^{-8}\)
(ii) \(7.2984 \times 10^{-5}\)
(iii) \(6.85 \times 10^{-8}\)
(iv) \(8.75 \times 10^{10}\)
(v) \(9.54321 \times 10^5\)
In simple words: Place the decimal after the first non-zero digit and adjust power of 10.

Exam Tip: Standard form requires the number to be between 1 and 10 multiplied by a power of 10.

Question 21. Simplify
(i) \(\frac{169 \times t^{-8} \times 25}{26 \times t^4 \times x^{10}}\)
(ii) \(\frac{625^{-1} \times 7 \times a^{-4}}{49^{-2} \times a^2 \times x^5}\)
Answer:
(i) \(\frac{169 \times 25}{26} \times t^{-8-4} \times x^{-10} = \frac{13 \times 25}{2} t^{-12} x^{-10} = \frac{325}{2t^{12}x^{10}}\).
(ii) Simplify numbers: \(625^{-1} = \frac{1}{625} = 5^{-4}\), \(49^{-2} = \left(7^2\right)^{-2} = 7^{-4}\).
Combine terms: \(\frac{7 \times 7^4}{625 \times a^2 \times a^4 \times x^5} = \frac{7^5}{625 a^6 x^5} = \frac{16807}{625 a^6 x^5}\).
In simple words: Combine numerical coefficients and apply exponent rules to variables.

Exam Tip: Group numbers together and apply laws of exponents to each variable separately.

Question 22. Find the value of
(i) \((3^{-1} + 5^0) \times 4^{-1}\)
(ii) \((7^0 + 5^0 + 2^{-1}) \times \left(\frac{2}{5}\right)^{-1}\)
(iii) \((2^{-1} + 5^{-2} + 7^{-3})^0$
Answer:
(i) \(\left(\frac{1}{3} + 1\right) \times \frac{1}{4} = \frac{4}{3} \times \frac{1}{4} = \frac{1}{3}\).
(ii) \(\left(1 + 1 + \frac{1}{2}\right) \times \frac{5}{2} = \left(\frac{5}{2}\right) \times \frac{5}{2} = \frac{25}{4}\).
(iii) Any non-zero expression raised to the power 0 is equal to 1. Result = 1.
In simple words: Evaluate terms inside brackets using exponent rules; remember anything to power 0 is 1.

Exam Tip: \(a^0 = 1\) for any non-zero number \(a\).

Question 23. Express the result in power notation with positive exponent
(i) \(\left(\frac{-3}{7}\right)^{-7}\)
(ii) \((-7)^{-5} \times \left(\frac{-2}{7}\right)^{-5}\)
(iii) \(\left[\left(\frac{2}{5}\right)^{-3}\right]^5$
(iv) \((3^{-6} \div 3^{-15}) \times 3^{-10}\)
Answer:
(i) \(\left(-\frac{7}{3}\right)^7\)
(ii) \(\left[(-7) \times \left(-\frac{2}{7}\right)\right]^{-5} = (2)^{-5} = \frac{1}{2^5}\).
(iii) \(\left(\frac{2}{5}\right)^{-15} = \left(\frac{5}{2}\right)^{15}\).
(iv) \(3^{-6 - (-15)} \times 3^{-10} = 3^9 \times 3^{-10} = 3^{-1} = \frac{1}{3}\).
In simple words: Apply exponent laws and invert bases to make negative exponents positive.

Exam Tip: Reciprocate the base fraction to change a negative exponent to positive.