CBSE Class 6 Mathematics Whole Numbers Worksheet Set 03

Whole Numbers

FILL IN THE BLANKS:

Question 1. The whole number which is not a natural number is _________________ .
Answer: 0
In simple words: The set of whole numbers includes zero, whereas natural numbers start from one.

Exam Tip: Remember that natural numbers are counting numbers starting from 1, while whole numbers include 0 along with all natural numbers.

Question 2. The natural number whose predecessor does not exist is ______________ .
Answer: 1
In simple words: 1 is the first natural number, so it has no natural number before it.

Exam Tip: Predecessor means 1 less than a given number. Since natural numbers start at 1, 1 has no natural predecessor.

Question 3. There are __________ whole numbers up to 75.
Answer: 76
In simple words: Whole numbers up to 75 include 0 and all numbers from 1 to 75, making a total of 76 numbers.

Exam Tip: When counting whole numbers up to \(n\), always include 0, so the count is \(n + 1\).

Question 4. Predecessor of 2,90,099 is ____________ .
Answer: 2,90,098
In simple words: Subtract 1 from 2,90,099 to get its predecessor.

Exam Tip: To find the predecessor, always subtract 1 from the given number.

Question 5. There are ____________ natural numbers up to 80.
Answer: 80
In simple words: Natural numbers start from 1, so there are 80 natural numbers up to 80.

Exam Tip: Natural numbers start from 1, so the count of natural numbers up to \(n\) is simply \(n\).

Question 6. The additive identity for whole numbers is __________ .
Answer: 0
In simple words: Adding 0 to any whole number leaves the number unchanged.

Exam Tip: The additive identity is always 0 for both whole numbers and integers.

Question 7. The statement (3 + 5) + 6 = 3 + (5 + 6) shows that addition of whole numbers is _____ .
Answer: associative
In simple words: Changing the grouping of numbers during addition does not change the result.

Question 8. ___________ is the multiplicative identity of whole numbers.
Answer: 1
In simple words: Multiplying any whole number by 1 gives the same number back.

Exam Tip: Multiplicative identity is 1, whereas additive identity is 0.

Question 9. 6 x (7 + 3) = (6 x 7) + (6 x 3). This statement shows that multiplication of whole numbers is _____________ over addition.
Answer: distributive
In simple words: Multiplication distributes over numbers being added inside a bracket.

Exam Tip: This is the distributive property of multiplication over addition, expressed as \(a \times (b + c) = (a \times b) + (a \times c)\).

Question 10. (3 x 5) x 9 = 3 x (5 x 9). This statement shows that multiplication of whole number is ____________ .
Answer: associative
In simple words: Changing the grouping of factors in multiplication does not affect the product.

Exam Tip: Look at the parentheses to identify the associative property of multiplication.

Question 11. ____________ is the only whole number which when divided by itself gives a quotient equal to itself.
Answer: 1
In simple words: When 1 is divided by 1, the result is 1.

Exam Tip: Note that division by zero is not defined, and 0 divided by any non-zero number is 0.

DO THE FOLLOWING:

Question 1. How many times does the digit 5 occur in tens place in the natural numbers from 100 to 1000?
Answer:
We need to count how many times the digit 5 appears in the tens place for numbers between 100 and 1000.
For each hundreds block (like 100 to 199, 200 to 299, etc.), the tens digit can be 5 in numbers from 50 to 59.
There are 10 such numbers in each hundred (e.g., 150, 151, ..., 159).
Since there are 9 such hundreds blocks from 100 to 1000 (100s, 200s, 300s, 400s, 500s, 600s, 700s, 800s, 900s), the total count is:
\(9 \times 10 = 90\)
In simple words: The digit 5 appears in the tens place 10 times in every hundred numbers, giving 90 times across 900 numbers.

Exam Tip: Break range-counting problems down by place value and intervals of hundreds or thousands to avoid counting errors.

Question 2. How many times does the digit 9 occur in the units place in all the natural numbers up to 100?
Answer:
The numbers up to 100 with 9 in the units place are:
9, 19, 29, 39, 49, 59, 69, 79, 89, 99.
Counting these gives a total of 10 times.
In simple words: Every decade from 1 to 100 contains exactly one number ending in 9, making 10 numbers in total.

Exam Tip: For any single digit from 1 to 9, it appears in the units place once every 10 numbers.

Question 3. Find the sum by suitable rearrangement.
a) 534 + 100 + 466
b) 365 + 700 + 635
Answer:
a) Group numbers that add up to a multiple of 100:
\(534 + 466 + 100\)
\(= (534 + 466) + 100\)
\(= 1000 + 100 = 1100\)

b) Group numbers that add up to a multiple of 1000:
\(365 + 635 + 700\)
\(= (365 + 635) + 700\)
\(= 1000 + 700 = 1700\)
In simple words: Rearrange the addends so that units digits sum to 10, making mental addition easier.

Exam Tip: Look for pairs of numbers whose unit digits sum to 10 to simplify addition.

Question 4. Add each of the following and check by reversing the order of the addends.
(a) 3025 + 1825
(b) 2094 + 2942
Answer:
(a) First addition: \(3025 + 1825 = 4850\)
Reversed order: \(1825 + 3025 = 4850\)
Since both results are equal, the addition is verified.

(b) First addition: \(2094 + 2942 = 5036\)
Reversed order: \(2942 + 2094 = 5036\)
Since both results are equal, the addition is verified.
In simple words: Addition follows the commutative property, so changing the order of numbers does not change the sum.

Exam Tip: Use the commutative property of addition (\(a + b = b + a\)) to verify your calculations.

Question 5. Find the product using properties of whole numbers.
(a) 155 x 601
(b) 1800 x 499
(c) 71 x 380
Answer:
(a) \(155 \times 601 = 155 \times (600 + 1)\)
\(= (155 \times 600) + (155 \times 1)\)
\(= 93000 + 155 = 93155\)

(b) \(1800 \times 499 = 1800 \times (500 - 1)\)
\(= (1800 \times 500) - (1800 \times 1)\)
\(= 900000 - 1800 = 898200\)

(c) \(71 \times 380 = (70 + 1) \times 380\)
\(= (70 \times 380) + (1 \times 380)\)
\(= 26600 + 380 = 26980\)
In simple words: Split one number into a round number plus or minus a small number, then apply distributive property.

Exam Tip: Splitting numbers near multiples of 100 or 1000 makes multiplication much faster and less prone to errors.

Question 6. Find the value of the following expressions using properties of whole numbers.
(a) 4,380 x 85 - 4,380 x 75
(b) 270 x 8 + 270 x 9 + 270
(c) 45 x 20 x 5
Answer:
(a) Using distributive property in reverse: \(4380 \times (85 - 75)\)
\(= 4380 \times 10 = 43800\)

(b) Rewrite as \(270 \times 8 + 270 \times 9 + 270 \times 1\):
\(= 270 \times (8 + 9 + 1)\)
\(= 270 \times 18 = 4860\)

(c) Using associative property: \(45 \times (20 \times 5)\)
\(= 45 \times 100 = 4500\)
In simple words: Factor out common terms or regroup numbers to create round numbers before calculating.

Exam Tip: Always look for common multipliers to apply the distributive property backwards.

Question 7. Find the quotient and remainder in each of the following divisions. Check the answer by division algorithm.
(a) 4,567 \(\div\) 234
(b) 3,645 \(\div\) 15
(c) 7,654 \(\div\) 19
Answer:
(a) \(4567 \div 234\):
Quotient = 19, Remainder = 121
Check: \(234 \times 19 + 121 = 4446 + 121 = 4567\). (Verified)

(b) \(3645 \div 15\):
Quotient = 243, Remainder = 0
Check: \(15 \times 243 + 0 = 3645\). (Verified)

(c) \(7654 \div 19\):
Quotient = 402, Remainder = 16
Check: \(19 \times 402 + 16 = 7638 + 16 = 7654\). (Verified)
In simple words: Verify division using the rule: Dividend = (Divisor \(\times\) Quotient) + Remainder.

Exam Tip: Always check your division results using the division algorithm formula.

Question 8. The product of two numbers is 3,75,200. If one of the numbers is 160. Find the other?
Answer:
Product of two numbers = 3,75,200
One number = 160
Other number = \(375200 \div 160\)
\(= 37520 \div 16 = 2345\)
In simple words: Divide the given product by the known number to find the missing number.

Exam Tip: When the product of two numbers and one factor are given, find the other factor by division.

Question 9. On dividing 4,56,789 by a certain number the quotient is 1,952 and a remainder 21 is left. Find the number.
Answer:
Using division algorithm: Dividend = (Divisor \(\times\) Quotient) + Remainder
\(456789 = (\text{Divisor} \times 1952) + 21\)
\(\text{Divisor} \times 1952 = 456789 - 21 = 456768\)
\(\text{Divisor} = 456768 \div 1952 = 234\)
In simple words: Subtract the remainder from the dividend, then divide the result by the quotient to find the divisor.

Exam Tip: Rearrange the division formula: \(\text{Divisor} = (\text{Dividend} - \text{Remainder}) \div \text{Quotient}\).

Question 10. Divide the largest 4-digit number by the largest 2-digit number and write down the quotient and the remainder.
Answer:
Largest 4-digit number = 9999
Largest 2-digit number = 99
Division: \(9999 \div 99\)
Quotient = 101, Remainder = 0
In simple words: Divide 9999 by 99 to get 101 with no remainder.

Exam Tip: The largest \(n\)-digit number is formed by writing the digit 9 \(n\) times.

Negative Numbers And Integers

I. ANSWER THE FOLLOWING:

Question 1. Which is the smallest positive integer?
Answer: 1
In simple words: Positive integers start from 1, making it the smallest one.

Exam Tip: Zero is neither positive nor negative, so the smallest positive integer is 1.

Question 2. Which is the greatest negative integer?
Answer: -1
In simple words: On the number line, -1 is closest to zero among all negative numbers.

Exam Tip: As negative numbers increase in magnitude, their value decreases. Thus, -1 is the largest negative integer.

Question 3. Write absolute value of a) | -13 | b) - | -30 |
Answer:
a) \(|-13| = 13\)
b) \(-|-30| = -(30) = -30\)
In simple words: Absolute value measures the distance from zero, so it is always non-negative unless a negative sign is outside the bars.

Exam Tip: Absolute value bars make any inner number positive, but outside signs still apply.

Question 4. Which integer is neither positive nor negative?
Answer: 0
In simple words: Zero acts as the neutral dividing line between positive and negative integers.

Exam Tip: Remember that 0 has no sign.

Question 5. Write the opposite of -875.
Answer: 875
In simple words: The opposite of a negative number is its positive counterpart.

Exam Tip: Changing the sign gives the opposite integer.

Question 6. Which is greater: -41 or -14?
Answer: -14
In simple words: -14 is closer to zero, so it is greater than -41.

Exam Tip: For negative integers, the number with the smaller absolute value is greater.

Question 7. Write the integer which is 5 more than -5.
Answer: 0
In simple words: Moving 5 units to the right from -5 on the number line lands on 0.

Exam Tip: "More than" means addition: \(-5 + 5 = 0\).

Question 8. Find (-19) + 9.
Answer: -10
In simple words: Subtract 9 from 19 and keep the negative sign because 19 is larger.

Exam Tip: When adding integers with different signs, find the difference of their absolute values and use the sign of the larger number.

Question 9. Replace * by < or >
i) -3 * 3
ii) 0 * -125
Answer:
i) -3 < 3
ii) 0 > -125
In simple words: Any positive number is greater than any negative number, and zero is greater than any negative number.

Exam Tip: Negative numbers are always less than zero and positive numbers.

Question 10. Subtract -1140 from -780.
Answer:
\(-780 - (-1140)\)
\(= -780 + 1140\)
\(= 360\)
In simple words: Subtracting a negative is the same as adding its positive value.

Exam Tip: Change subtraction of a negative integer to addition: \(a - (-b) = a + b\).

II. ANSWER THE FOLLOWING:

Question 1. Write all integers between -35 and -31.
Answer: -34, -33, -32
In simple words: The integers strictly between -35 and -31 are -34, -33, and -32.

Exam Tip: Do not include the boundary numbers unless specified as "inclusive".

Question 2. Sum of two integers is 1890. If one of them is -412. Find the other integer.
Answer:
Let the other integer be \(x\).
\(x + (-412) = 1890\)
\(x = 1890 - (-412)\)
\(x = 1890 + 412 = 2302\)
In simple words: Subtract the known integer from the sum to find the other integer.

Exam Tip: Moving a negative term across the equals sign changes it to positive.

Question 3. Fill the boxes with < or >
i) (-9) + (-6) [ ] (-6) - (-9)
Answer:
Left side: \((-9) + (-6) = -15\)
Right side: \((-6) - (-9) = -6 + 9 = 3\)
Comparison: \(-15 < 3$
In simple words: Simplify both sides first: -15 is less than 3.

Exam Tip: Always compute both sides of the inequality completely before comparing.

Question 4. Add:
i) -2050 + 687
ii) 396 - 159
iii) -493 + 108
Answer:
i) \(-2050 + 687 = -(2050 - 687) = -1363\)
ii) \(396 - 159 = 237\)
iii) \(-493 + 108 = -(493 - 108) = -385\)
In simple words: Combine positive and integers by finding the difference and keeping the sign of the larger absolute value.

Question 5. Simplify
i) -37 + 28 - 6 + 23
ii) 1 - 3 + 5 - 7 + 9 - 11 + 13 - 17
iii) 0 + 2 - 4 + 6 - 8 + 10 - 12 + 14
Answer:
i) \((-37 - 6) + (28 + 23) = -43 + 51 = 8\)
ii) Grouping positives and negatives:
Positives: \(1 + 5 + 9 + 13 = 28\)
Negatives: \(-3 - 7 - 11 - 17 = -38\)
Result: \(28 - 38 = -10\)
iii) Grouping positives and negatives:
Positives: \(2 + 6 + 10 + 14 = 32\)
Negatives: \(-4 - 8 - 12 = -24\)
Result: \(32 - 24 = 8\)
In simple words: Group all positive numbers together and all negative numbers together before combining them.

Exam Tip: Combining like signs first simplifies long arithmetic expressions.

Question 6. Represent all integers between -5 and +3 on the number line.
Answer:
The integers between -5 and +3 are: -4, -3, -2, -1, 0, 1, 2.
In simple words: List all integers strictly located between -5 and 3 on a number line.

Exam Tip: Exclude the endpoints (-5 and 3) when asked for numbers "between" them.

Question 7. Write the integer which is:
i) 3 more than -2
ii) 4 less than -1
iii) -2 more than -5
iv) -3 less than -4
Show answers on number line.
Answer:
i) \(-2 + 3 = 1\)
ii) \(-1 - 4 = -5\)
iii) \(-5 + (-2) = -7\)
iv) \(-4 - (-3) = -4 + 3 = -1\)
In simple words: Translate phrases like "more than" into addition and "less than" into subtraction.

Exam Tip: Be careful with phrasing like "-3 less than -4", which translates to \(-4 - (-3)\).

Question 8. Arrange in increasing and decreasing order.
i) -12, 0, 7, -9, 8, -8, -14
ii) -3, 2, -5, 10, 11, -15, 6, 25
iii) -6, 0, -5, -7, 3, 4, -16, 8
Answer:
i) Increasing: -14, -12, -9, -8, 0, 7, 8
Decreasing: 8, 7, 0, -8, -9, -12, -14

ii) Increasing: -15, -5, -3, 2, 6, 10, 11, 25
Decreasing: 25, 11, 10, 6, 2, -3, -5, -15

iii) Increasing: -16, -7, -6, -5, 0, 3, 4, 8
Decreasing: 8, 4, 3, 0, -5, -6, -7, -16
In simple words: Increasing order goes from smallest negative to largest positive, while decreasing order goes from largest positive to smallest negative.

Exam Tip: Remember that larger negative numbers have smaller values.

Question 9. Find the value of:
i) 9 - |-7|
ii) -|-15|
iii) 26 - |-10| + |-13|
iv) 4 + |-4| + 4 + |-4|
Answer:
i) \(9 - 7 = 2\)
ii) \(-(15) = -15\)
iii) \(26 - 10 + 13 = 16 + 13 = 29\)
iv) \(4 + 4 + 4 + 4 = 16\)
In simple words: Evaluate absolute values first, then perform the addition and subtraction.

Exam Tip: Treat absolute value bars like grouping symbols.

Question 10. Draw a number line and answer the following:
a) Which number will we reach if we move 3 numbers to the left of -5.
b) If we are at +3 in which direction should we move to reach -10.
Answer:
a) Moving 3 units to the left of -5: \(-5 - 3 = -8\)
b) To move from +3 to -10, we must move to the left by 13 units.
In simple words: Left movement represents subtraction, and right movement represents addition.

Exam Tip: Always associate "left" with negative/subtraction and "right" with positive/addition.

Playing With Numbers

I) Choose the correct answers:

Question 1. HCF of two consecutive even numbers
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
Consecutive even numbers are of the form \(2n\) and \(2n+2\). Their greatest common factor is always 2.
In simple words: Any two consecutive even numbers are both divisible by 2.

Exam Tip: Consecutive even numbers differ by 2 and share 2 as their HCF.

Question 2. LCM of two Prime numbers is
(a) their sum
(b) their difference
(c) their product
(d) none of the options
Answer: (c) their product
Since prime numbers share no common factors other than 1, their LCM is always their product.
In simple words: Co-prime numbers and prime numbers have an LCM equal to their product.

Exam Tip: For any two prime numbers \(p\) and \(q\), \(\text{LCM} = p \times q\).

Question 3. The Prime factorization of 126 is
(a) 2 \(\times\) 2 \(\times\) 3 \(\times\) 6
(b) 2 \(\times\) 3 \(\times\) 3 \(\times\) 7
(c) 2 \(\times\) 3 \(\times\) 3 \(\times\) 4
(d) none of the options
Answer: (b) 2 \(\times\) 3 \(\times\) 3 \(\times\) 7
Prime factorization of 126: \(126 = 2 \times 63 = 2 \times 3 \times 21 = 2 \times 3 \times 3 \times 7\).
In simple words: Breaking 126 down into prime numbers gives \(2 \times 3 \times 3 \times 7\).

Exam Tip: Check that all factors in the correct option are prime numbers.

Question 4. The fundamental theorem of Arithmetic states that:
(a) every number greater than 1 has many factors.
(b) every number greater than 1 has exactly one Prime factorization.
(c) every number greater than 1 has many multiples.
(d) none of the options
Answer: (b) every number greater than 1 has exactly one Prime factorization.
The fundamental theorem of arithmetic states that every composite integer greater than 1 can be factored into prime numbers in a unique way.
In simple words: Every number greater than 1 is either prime or can be written as a unique product of primes.

Exam Tip: Memorize the fundamental theorem of arithmetic for definitions.

Question 5. The number which is divisible by 9 is
(a) 7802
(b) 8100
(c) 3271
(d) none of the options
Answer: (b) 8100
Sum of digits of 8100 is \(8 + 1 + 0 + 0 = 9\), which is divisible by 9.
In simple words: A number is divisible by 9 if the sum of its digits is divisible by 9.

Exam Tip: Use the sum of digits test for divisibility by 9.

II) Find the value of
a) 4 + 3 \(\times\) (5 + 3)
b) 6 \(\times\) 8 - 25 \(\div\) 5 + 3
c) 12 \(\div\) (4 + 2) - 3 \(\times\) 5 + 7 \(\times\) (3 + 1)

Question II(a). Find the value of 4 + 3 \(\times\) (5 + 3)
Answer:
\(= 4 + 3 \times 8\)
\(= 4 + 24 = 28\)
In simple words: Solve inside the parentheses first, then multiply, and finally add.

Exam Tip: Follow the BODMAS rule (Brackets, Orders, Division/Multiplication, Addition/Subtraction).

Question II(b). Find the value of 6 \(\times\) 8 - 25 \(\div\) 5 + 3
Answer:
\(= 48 - 5 + 3\)
\(= 43 + 3 = 46\)
In simple words: Perform multiplication and division first from left to right, then addition and subtraction.

Question II(c). Find the value of 12 \(\div\) (4 + 2) - 3 \(\times\) 5 + 7 \(\times\) (3 + 1)
Answer:
\(= 12 \div 6 - 3 \times 5 + 7 \times 4\)
\(= 2 - 15 + 28\)
\(= -13 + 28 = 15\)
In simple words: Evaluate parentheses first, then do multiplication and division, and finish with addition and subtraction.

Exam Tip: Take each step carefully when multiple operations are present.

III) Simplify :
a) 125 - [18 - \{28 \(\div\) 4 - (18 - 30 \(\div\) 5) \(\div\) 2\}]
b) 5[18 + \{3 + 4(5 - 2)\}]
c) 130 - [100 \(\div\) (2 of 5) - \{8 - 4 - 2\}]

Question III(a). Simplify 125 - [18 - \{28 \(\div\) 4 - (18 - 30 \(\div\) 5) \(\div\) 2\}]
Answer:
Inside innermost parentheses: \(30 \div 5 = 6\), so \(18 - 6 = 12\)
Next: \(12 \div 2 = 6\)
Inside braces: \(28 \div 4 - 6 = 7 - 6 = 1\)
Inside brackets: \(18 - 1 = 17\)
Final subtraction: \(125 - 17 = 108\)
In simple words: Work from the innermost brackets outward: parentheses, braces, then square brackets.

Exam Tip: Maintain proper order of brackets: \(( ), \{ \}, [ ]\).

Question III(b). Simplify 5[18 + \{3 + 4(5 - 2)\}]
Answer:
\(= 5[18 + \{3 + 4(3)\}]\)
\(= 5[18 + \{3 + 12\}]\)
\(= 5[18 + 15]\)
\(= 5[33] = 165\)
In simple words: Evaluate inside parentheses, multiply by 4, simplify braces, and then multiply by 5.

Exam Tip: Clear parentheses before handling braces and square brackets.

Question III(c). Simplify 130 - [100 \(\div\) (2 of 5) - \{8 - 4 - 2\}]
Answer:
Note: "of" means multiplication.
\(2 \text{ of } 5 = 2 \times 5 = 10\)
\(100 \div 10 = 10\)
Inside braces: \(8 - 4 - 2 = 4 - 2 = 2\)
Inside brackets: \(10 - 2 = 8\)
Final subtraction: \(130 - 8 = 122\)
In simple words: Evaluate the word "of" as multiplication before division, then simplify brackets.

Exam Tip: "Of" takes precedence over standard division and multiplication.

Question IV. Using divisibility tests, determine which of the following numbers are divisible by 2, 3, 4, 5, 6, 8 and 9.
a) 14560
b) 726352
c) 12159
d) 12150
e) 639210
Answer:
a) 14560: Divisible by 2 (ends in 0), 4 (60 is div by 4), 5 (ends in 0), 10 (not asked, but valid). Sum of digits = 16 (not div by 3, 9). Not div by 6, 8.
b) 726352: Divisible by 2 (ends in 2), 4 (52 is div by 4), 8 (352 is div by 8). Sum of digits = 25 (not div by 3, 9).
c) 12159: Sum of digits = 18 (divisible by 3 and 9). Ends in 9 (not div by 2, 4, 5, 8).
d) 12150: Divisible by 2, 3, 5, 6, 9 (Sum of digits = 9, ends in 0).
e) 639210: Divisible by 2, 3, 5, 6 (Sum of digits = 21, ends in 0).
In simple words: Apply individual divisibility rules for each prime factor and composite number.

Exam Tip: A number is divisible by 6 if it is divisible by both 2 and 3.

Question V. Determine HCF of numbers given below by Prime factorization method.
a) 36, 84
b) 34, 102
c) 27, 63
Answer:
a) \(36 = 2^2 \times 3^2\), \(84 = 2^2 \times 3 \times 7\)
HCF = \(2^2 \times 3 = 12\)
b) \(34 = 2 \times 17\), \(102 = 2 \times 3 \times 17\)
HCF = \(2 \times 17 = 34\)
c) \(27 = 3^3\), \(63 = 3^2 \times 7\)
HCF = \(3^2 = 9\)
In simple words: Find the prime factors of each number and take the product of the lowest powers of common prime factors.

Exam Tip: HCF uses the lowest exponent of common prime factors.

Question VI. Determine HCF of numbers given below by continued division method.
a) 28, 36
b) 880, 1375
c) 255, 1620
Answer:
a) \(36 \div 28\) gives quotient 1, remainder 8. Then \(28 \div 8\) gives quotient 3, remainder 4. Then \(8 \div 4\) gives remainder 0. HCF = 4.
b) \(1375 \div 880\) gives remainder 495. Then \(880 \div 495\) gives remainder 385. Then \(495 \div 385\) gives remainder 110. Then \(385 \div 110\) gives remainder 55. Then \(110 \div 55\) gives remainder 0. HCF = 55.
c) \(1620 \div 255\) gives remainder 90. Then \(255 \div 90\) gives remainder 75. Then \(90 \div 75\) gives remainder 15. Then \(75 \div 15\) gives remainder 0. HCF = 15.
In simple words: Continuously divide until the remainder is zero; the last non-zero divisor is the HCF.

Exam Tip: Continued division is very efficient for finding the HCF of large numbers.

Question VII. Determine LCM of
a) 16, 24, 40
b) 18, 24, 32
c) 32, 56, 46
Answer:
a) Prime factors: \(16 = 2^4\), \(24 = 2^3 \times 3\), \(40 = 2^3 \times 5\)
LCM = \(2^4 \times 3 \times 5 = 16 \times 15 = 240\)
b) Prime factors: \(18 = 2 \times 3^2\), \(24 = 2^3 \times 3\), \(32 = 2^5\)
LCM = \(2^5 \times 3^2 = 32 \times 9 = 288\)
c) Prime factors: \(32 = 2^5\), \(56 = 2^3 \times 7\), \(46 = 2 \times 23\)
LCM = \(2^5 \times 7 \times 23 = 32 \times 161 = 5152\)
In simple words: Take the highest power of every prime factor present in any of the numbers to find the LCM.

Exam Tip: LCM uses the highest exponent of all prime factors.

Question VIII. Write the numerical expression for each of the following using brackets:
a) Nine subtracted from the product eight and seven.
b) Seventy two divided by the sum of six and three.
c) Four multiplied by the sum of two and five.
Answer:
a) \( (8 \times 7) - 9$
b) \( 72 \div (6 + 3) $
c) \( 4 \times (2 + 5) $
In simple words: Translate word statements into mathematical expressions using brackets where needed.

Exam Tip: Words like "sum" or "product" indicate terms that should be grouped in brackets.

Question IX. 1) Express 71 as the sum of three odd primes.
2) Express 24 as the sum of twin Primes.
3) Find all Prime numbers between 70 and 90.
Answer:
1) \(71 = 3 + 17 + 51\) (Wait, 51 is not prime. Let's use: \(3 + 19 + 49\) - no. Correct: \(3 + 13 + 55\) - no. Correct: \( 31 + 11 + 29 = 71$ ).
2) Twin primes that sum to 24: \(11 + 13 = 24\).
3) Prime numbers between 70 and 90: 71, 73, 79, 83, 89.
In simple words: Find prime numbers that satisfy the given conditions by checking odd numbers.

Exam Tip: Memorize prime numbers up to 100 for quick reference.

Question X. 1) Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
2) Find the least number which divided by 12, 16, 24 and 36 leave remainder 7 in each case.
3) Find the greatest number which divides 1,628, 1,736 and 268 leaving remainders 8, 11 and 13 respectively.
4) Find the largest number which divides 146, 254 and 272 leaving a remainder 2 in each case.
5) Find the smallest 4 digit number which is divisible by 18, 24 and 32.
6) Write the smallest 5-digit number and find its factors.
7) For each of the following pair of numbers, verify that LCM \(\times\) HCF = Product of the numbers.
(i) 21, 28
(ii) 231 and 273
(iii) 36, 90
Answer:
1) LCM of 6, 15, 18 = 90. Required number = \(90 + 5 = 95\).
2) LCM of 12, 16, 24, 36 = 144. Required number = \(144 + 7 = 151\).
3) Subtract remainders first: \(1628 - 8 = 1620\), \(1736 - 11 = 1725\), \(268 - 13 = 255\). HCF of 1620, 1725, and 255 is 15.
4) Subtract remainder 2 from each: 144, 252, 270. HCF of 144, 252, 270 = 18.
5) LCM of 18, 24, 32 = 288. Smallest 4-digit multiple of 288 is \(288 \times 4 = 1152\).
6) Smallest 5-digit number = 10000. Prime factorization: \(10000 = 2^4 \times 5^4\).
7) (i) HCF = 7, LCM = 84. Product = \(21 \times 28 = 588\), \(7 \times 84 = 588\). (Verified)
(ii) HCF = 21, LCM = 3003. Product = \(231 \times 273 = 63063\), \(21 \times 3003 = 63063\). (Verified)
(iii) HCF = 18, LCM = 180. Product = \(36 \times 90 = 3240\), \(18 \times 180 = 3240\). (Verified)
In simple words: Use LCM for "least number divisible by" problems and HCF for "greatest number dividing" problems.

Exam Tip: Always add the common remainder to the LCM, but subtract individual remainders before calculating the HCF.

Decimals

Question 1) Express as rupees using decimals.
a) 640p
b) 29p
c) 4p
d) Rs 25, 50 paise
e) Rs 1, 50p
Answer:
a) \(640 \text{p} = \text{Rs } \frac{640}{100} = \text{Rs } 6.40\)
b) \(29 \text{p} = \text{Rs } \frac{29}{100} = \text{Rs } 0.29\)
c) \(4 \text{p} = \text{Rs } \frac{4}{100} = \text{Rs } 0.04\)
d) \(\text{Rs } 25 \text{ and } 50\text{p} = \text{Rs } 25.50\)
e) \(\text{Rs } 1 \text{ and } 50\text{p} = \text{Rs } 1.50\)
In simple words: Divide paise by 100 to convert into rupees.

Exam Tip: 1 Rupee = 100 Paise.

Question 2) Express as Kg using decimals.
a) 40g
b) 2kg 400g
c) 5g
d) 4312g
e) 804kg 60g
Answer:
a) \(40\text{g} = \frac{40}{1000}\text{kg} = 0.040\text{kg}\)
b) \(2\text{kg } 400\text{g} = 2 + \frac{400}{1000} = 2.400\text{kg}\)
c) \(5\text{g} = \frac{5}{1000}\text{kg} = 0.005\text{kg}\)
d) \(4312\text{g} = \frac{4312}{1000}\text{kg} = 4.312\text{kg}\)
e) \(804\text{kg } 60\text{g} = 804 + \frac{60}{1000} = 804.060\text{kg}\)
In simple words: Divide grams by 1000 to convert into kilograms.

Exam Tip: 1 Kilogram = 1000 Grams.

Question 3) Express as meters using decimals:
a) 40cm
b) 200mm
c) 240cm
d) 3m 34cm
Answer:
a) \(40\text{cm} = \frac{40}{100}\text{m} = 0.4\text{m}\)
b) \(200\text{mm} = \frac{200}{1000}\text{m} = 0.2\text{m}\)
c) \(240\text{cm} = \frac{240}{100}\text{m} = 2.4\text{m}\)
d) \(3\text{m } 34\text{cm} = 3 + \frac{34}{100} = 3.34\text{m}\)
In simple words: Divide centimeters by 100 and millimeters by 1000 to convert to meters.

Exam Tip: 1m = 100cm and 1m = 1000mm.

Question 4) Express as km using decimals
a) 240km 120m
b) 4000m
c) 9999m
d) 4km 20m
e) 22km 8m
Answer:
a) \(240\text{km } 120\text{m} = 240.120\text{km}\)
b) \(4000\text{m} = \frac{4000}{1000}\text{km} = 4\text{km}\)
c) \(9999\text{m} = \frac{9999}{1000}\text{km} = 9.999\text{km}\)
d) \(4\text{km } 20\text{m} = 4.020\text{km}\)
e) \(22\text{km } 8\text{m} = 22.008\text{km}\)
In simple words: Divide meters by 1000 to convert to kilometers.

Exam Tip: 1 km = 1000 m, so meters must occupy three decimal places.

Question 5) Find
a) 420.20 + 21.02
b) 41 + 8.881
c) 0.2 + 2.1 + 2.01
d) 0.001 + 2 + 4.2
e) 12 + 24.2 + 3.24
f) 8.06 - 4.06
g) 0.04 - 0.008
h) 44.444 - 3.33
i) 12.001 - 9
j) 42 - 38.2
Answer:
a) 441.22
b) 49.881
c) 4.31
d) 6.201
e) 39.44
f) 4.00
g) 0.032
h) 41.114
i) 3.001
j) 3.8
In simple words: Line up the decimal points and add or subtract as whole numbers.

Exam Tip: Pad numbers with trailing zeros so all decimals have the same number of decimal places.

Question 6) Reema bought 5.5kg of wheat and 10.75kg of sugar and 15.25kg of rice. Find the total weight of the items she purchased?
Answer:
Total weight = \(5.5 + 10.75 + 15.25\)
\(= 5.50 + 10.75 + 15.25 = 31.50\text{ kg}\)
In simple words: Add the weights of all three items together.

Exam Tip: Align decimal points carefully when adding weights.

Question 7) Hyzam had Rs 20. He bought one pen for Rs 9.75 and one pencil for Rs 2.50. How much Rs does he have now?
Answer:
Total cost = \(9.75 + 2.50 = \text{Rs } 12.25\)
Amount left = \(20.00 - 12.25 = \text{Rs } 17.75\)
In simple words: Add the cost of the pen and pencil, then subtract from the total money he started with.

Exam Tip: Write whole rupees with decimal zeros (\(20.00\)) before subtracting.

Question 8) Ayesha bought 8m 20 cm cloth. She cut 4m 20 cm length from it for dress. How much cloth does she have now?
Answer:
Remaining cloth = \(8\text{m } 20\text{cm} - 4\text{m } 20\text{cm} = 4\text{m } 00\text{cm}\)
In simple words: Subtract the length of cloth used from the total length bought.

Exam Tip: Convert mixed units into decimals or subtract meters and centimeters separately.

Question 9) Munna's school is at a distance of 5 km 350 m from her house. She travels 1 km 70 m on foot and the rest by bus. How much distance does she travel by bus?
Answer:
Total distance = \(5.350\text{ km}\)
Distance on foot = \(1.070\text{ km}\)
Distance by bus = \(5.350 - 1.070 = 4.280\text{ km}\)
In simple words: Subtract the distance walked from the total distance to find the bus distance.

Exam Tip: Note that 70m is written as \(0.070\text{ km}\).

Question 10) Rahul bought 4 kg 90 g of apples, 2 kg 60 g of grapes and 5 kg 300 g of mangoes. Find the total weight of all the fruits he bought?
Answer:
Apples = \(4.090\text{ kg}\)
Grapes = \(2.060\text{ kg}\)
Mangoes = \(5.300\text{ kg}\)
Total weight = \(4.090 + 2.060 + 5.300 = 11.450\text{ kg}\)
In simple words: Convert all weights to kilograms and add them up.

Question 11) Subtract:
a) Rs 18.25 from Rs 20.75
b) 202.54 m from 250 m
c) 2.051 km from 5.20 km
d) 0.314 kg from 2.107 kg
Answer:
a) \(20.75 - 18.25 = \text{Rs } 2.50\)
b) \(250.00 - 202.54 = 47.46\text{ m}\)
c) \(5.200 - 2.051 = 3.149\text{ km}\)
d) \(2.107 - 0.314 = 1.793\text{ kg}\)
In simple words: Subtract the first number from the second number.

Exam Tip: "Subtract A from B" means \(B - A\).

Algebra

I) Fill in the blanks :

Question 1. Letters used to represent numbers are called _________________
Answer: variables (or literals)
In simple words: Letters like \(x\) or \(y\) represent unknown numbers in algebra.

Exam Tip: Variables can take different numerical values.

Question 2. The literals whose value vary from problem to problem are called _____________
Answer: variables
In simple words: Quantities that change or vary are called variables.

Question 3. Numerals whose value is fixed are called _____________________
Answer: constants
In simple words: Fixed numbers like 5, -3, or \(\frac{1}{2}\) are constants.

Exam Tip: Real numbers are constants in algebraic expressions.

Question 4. The statement of equality which involves literals is called an ___________
Answer: equation
In simple words: An equation has an equals sign connecting two expressions.

Exam Tip: An expression has no equals sign, whereas an equation does.

Question 5. The value of the variable which satisfies the equation is called the ___________ to the equation.
Answer: solution (or root)
In simple words: The solution makes both sides of the equation equal.

Exam Tip: Always verify your solution by substituting it back into the equation.

II) Write the expression:

Question 1. m increased by 8.
Answer: \(m + 8\)

Exam Tip: "Increased by" means addition.

Question 2. b decreased by 31.
Answer: \(b - 31\)

Exam Tip: "Decreased by" means subtraction.

Question 3. Five times x increased by 5 gives 55.
Answer: \(5x + 5 = 55\)

Exam Tip: "Times" indicates multiplication.

Question 4. Half of p subtracted from the sum of a and b.
Answer: \((a + b) - \frac{p}{2}\)

Exam Tip: Pay attention to the order of subtraction.

Question 5. Product of 7 and x divided by the difference of 7 and x.
Answer: \(\frac{7x}{7 - x}\)

Exam Tip: Form the numerator and denominator separately before dividing.

Question 6. Sum of p, q and r divided by the product of a and 5.
Answer: \(\frac{p + q + r}{5a}\)

Exam Tip: Product of \(a\) and 5 is written as \(5a\).

Question 7. One fourth of the difference of x and y.
Answer: \(\frac{1}{4}(x - y)\)

Exam Tip: Difference is \((x - y)\), and one-fourth of it multiplies the difference by \(\frac{1}{4}\).

Question 8. 4 subtracted from n, gives 36.
Answer: \(n - 4 = 36\)

Exam Tip: "Subtracted from \(n\)" means \(n\) comes first.

Question 9. z minus twice x.
Answer: \(z - 2x\)

Exam Tip: Twice \(x\) is \(2x\).

Question 10. m taken away from 50, gives 15.
Answer: \(50 - m = 15\)

Exam Tip: "Taken away from 50" means start with 50 and subtract \(m\).

III) Identify the solution and show that other values do not satisfy

Question 1. X + 8 = 24 (x = 3, x = 15, x = 16, x = 0)
Answer:
Substitute each value into \(x + 8\):
For \(x = 3\): \( 3 + 8 = 11 \neq 24$
For \(x = 15\): \( 15 + 8 = 23 \neq 24$
For \(x = 16\): \(16 + 8 = 24\) (Satisfies)
For \(x = 0\): \( 0 + 8 = 8 \neq 24$
Solution is \(x = 16\).
In simple words: Test each given value by substitution; only \(x = 16\) makes the equation true.

Exam Tip: Show substitution steps for every test value.

Question 2. 4a = 32 (a = 8, a = 4, a = 6, a = 7)
Answer:
For \(a = 8\): \(4(8) = 32\) (Satisfies)
For \(a = 4\): \( 4(4) = 16 \neq 32$
For \(a = 6\): \( 4(6) = 24 \neq 32$
For \(a = 7\): \( 4(7) = 28 \neq 32$
Solution is \(a = 8\).
In simple words: Multiply each value by 4; only 8 gives 32.

Exam Tip: Check each option systematically.

Question 3. y - 9 = 29 (y = 20, y = 38, y = 37, y = 39)
Answer:
For \(y = 20\): \( 20 - 9 = 11 \neq 29$
For \(y = 38\): \(38 - 9 = 29\) (Satisfies)
For \(y = 37\): \( 37 - 9 = 28 \neq 29$
For \(y = 39\): \( 39 - 9 = 30 \neq 29$
Solution is \(y = 38\).
In simple words: Subtract 9 from each test value; only 38 yields 29.

Exam Tip: Substitution is a reliable way to verify solutions.

Question 4. \(\frac{p}{5} = 16\) (p = 75, p = 85, p = 90, p = 80)
Answer:
For \(p = 75\): \( 75 \div 5 = 15 \neq 16$
For \(p = 85\): \( 85 \div 5 = 17 \neq 16$
For \(p = 90\): \( 90 \div 5 = 18 \neq 16$
For \(p = 80\): \(80 \div 5 = 16\) (Satisfies)
Solution is \(p = 80\).
In simple words: Divide each number by 5; only 80 results in 16.

Exam Tip: Test division by dividing each candidate number.

IV) Solve :

Question 1. X + 11 = 30
Answer:
\( X = 30 - 11 = 19 $
In simple words: Subtract 11 from both sides to find \(X\).

Exam Tip: Transpose terms across the equals sign by changing their sign.

Question 2. 6a = 24
Answer:
\( a = \frac{24}{6} = 4 $
In simple words: Divide both sides by 6.

Exam Tip: The coefficient of the variable divides the other side.

Question 3. m - 5 = 2
Answer:
\( m = 2 + 5 = 7 $
In simple words: Add 5 to both sides.

Exam Tip: Negative terms become positive when transposed.

Question 4. \(\frac{x}{4} = 2\)
Answer:
\( x = 2 \times 4 = 8 $
In simple words: Multiply both sides by 4.

Exam Tip: Division transforms into multiplication when transposed.

Question 5. \(\frac{x}{7} = 5\)
Answer:
\( x = 5 \times 7 = 35 $
In simple words: Multiply 5 by 7.

Question 6. \(\frac{a}{9} = 4\)
Answer:
\( a = 4 \times 9 = 36 $
In simple words: Multiply 4 by 9.

Question 7. 10n = 70
Answer:
\( n = \frac{70}{10} = 7 $
In simple words: Divide 70 by 10.

Question 8. 16 = y + 10
Answer:
\( y = 16 - 10 = 6 $
In simple words: Subtract 10 from 16.

Question 9. t - 14 = 0
Answer:
\( t = 14 $
In simple words: Add 14 to 0.

Question 10. 4x = 32
Answer:
\( x = \frac{32}{4} = 8 $
In simple words: Divide 32 by 4.

Question 11. 17 - p = 15
Answer:
\(17 - 15 = p\)
\( p = 2 $
In simple words: Rearrange to find \(p = 17 - 15\).

Question 12. 23 - y = 0
Answer:
\( y = 23 $
In simple words: \(y\) must equal 23 for the difference to be zero.

Basic Geometrical Ideas

I) Fill in the blanks :

Question 1. Circles having different radii but the same centre are called __________________
Answer: concentric circles
In simple words: Circles with a shared center are concentric.

Exam Tip: Concentric circles never intersect each other.

Question 2. Radius is ______________ of the diameter.
Answer: half (\(\frac{1}{2}\))
In simple words: The radius is half the length of the diameter.

Exam Tip: \(\text{Radius} = \frac{\text{Diameter}}{2}\).

Question 3. The diameter of a circle is the ______________ chord of the circle.
Answer: longest
In simple words: The longest chord passing through the center is the diameter.

Exam Tip: The diameter is twice the radius.

Question 4. The Perimeter of a circle is called the _____________________.
Answer: circumference
In simple words: The boundary length of a circle is its circumference.

Exam Tip: Circumference formula is \(2\pi r\).

Question 5. A quadrilateral has ____________ diagonals.
Answer: 2
In simple words: Connecting opposite vertices gives 2 diagonals.

Exam Tip: A polygon with \(n\) sides has \(\frac{n(n-3)}{2}\) diagonals.

Question 6. A triangle has __________________ elements.
Answer: 6 (3 sides and 3 angles)
In simple words: Triangles consist of 3 sides and 3 angles.

Question 7. The interior of an angle together with the angle (boundary) itself is called the _______________
Answer: triangular region / angle region
In simple words: The inside plus the boundary form the region of the angle.

Question 8. In \(\Delta PQR\), the side opposite to \(\angle P\) is _______________
Answer: QR
In simple words: The side not touching vertex P is QR.

Question 9. Length of a diameter is _________________ the radius of a circle.
Answer: twice (or two times)
In simple words: Diameter equals \(2 \times \text{radius}\).

Question 10. All the radii of a circle are ____________________
Answer: equal in length
In simple words: Every radius from the center to the boundary is equal.

II) Choose the correct answer :

Question 1. The centre of a circle
(i) Lies in its interior
(ii) lies in its exterior
(iii) Lies on the circle
(iv) none of the options
Answer: (i) Lies in its interior
In simple words: The center is always strictly inside the circle.

Question 2. In quadrilateral PQRS, the two diagonals are
(i) PQ and RS
(ii) PR and RS
(iii) PS and PR
(iv) PR and QS
Answer: (iv) PR and QS
In simple words: Diagonals connect opposite vertices P to R and Q to S.

Question 3. A line segment joining any two points on the circle is called a ________________
(i) radius
(ii) diameter
(iii) chord
(iv) secant
Answer: (iii) chord
In simple words: Any line segment connecting two points on a circle's boundary is a chord.

Question 4. A closed figure bounded by three or more segments is called a ____________
(i) Curve
(ii) Polygon
(iii) Circle
(iv) None of the options
Answer: (ii) Polygon
In simple words: Polygons are closed figures made of straight line segments.

Question 5. A line intersecting a circle at two different points is called a __________________ of a circle.
(i) diameter
(ii) Radius
(iii) secant
(iv) centre
Answer: (iii) secant
In simple words: A secant line passes through a circle, intersecting it at two points.

Question 6. A closed figure formed by joining three non-collinear points is called
(i) a triangle
(ii) an angle
(iii) a curve
(iv) none of the options
Answer: (i) a triangle
In simple words: Three non-collinear points form a triangle when connected.

Question 7. An Angle has
(i) One vertex and one arm
(ii) One vertex and two arms
(iii) Two vertex and two arms
(iv) None of the options
Answer: (ii) One vertex and two arms
In simple words: An angle is formed by two rays meeting at a common vertex.

Question 8. A circle is
(i) a polygon
(ii) an open curve
(iii) a closed curve
(iv) none of the options
Answer: (iii) a closed curve
In simple words: A circle forms a continuous closed curve.

Question 9. A Point equidistant from all the Points on a circle is called the ______________ of the circle.
(i) Centre
(ii) Radius
(iii) diameter
(iv) None of the options
Answer: (i) Centre
In simple words: The center is the fixed point equidistant from all points on the boundary.

Perimeter And Area

I) Fill in the blanks :

Question 1. _______________________ is the distance around the figure.
Answer: Perimeter
In simple words: Perimeter measures the total boundary distance of a 2D shape.

Question 2. Perimeter of a square ___________________________.
Answer: \(4 \times \text{side}\)
In simple words: Multiply the length of one side by 4.

Question 3. Perimeter \(\div\) 4 = a ______________ of a square.
Answer: side
In simple words: Dividing a square's perimeter by 4 gives its side length.

Question 4. Perimeter of a regular pentagon is 125cm. What is the length of the side?
Answer:
Side = \(125 \div 5 = 25\text{ cm}\)
In simple words: Divide the perimeter of a 5-sided regular polygon by 5.

Question 5. Length of the rectangle = _________________ \(\div\) breadth.
Answer: Area
In simple words: Area divided by breadth gives length.

II) Do as directed :

Question 1. Find the perimeter of the following:
a. Rectangle whose l=5cm and b=4cm.
b. Square whose side = 15cm
c. A regular hexagon, side = 7.2cm
d. Equilateral triangle, side = 5cm
Answer:
a) Perimeter of rectangle = \(2(l + b) = 2(5 + 4) = 18\text{ cm}\)
b) Perimeter of square = \(4 \times 15 = 60\text{ cm}\)
c) Perimeter of regular hexagon = \(6 \times 7.2 = 43.2\text{ cm}\)
d) Perimeter of equilateral triangle = \(3 \times 5 = 15\text{ cm}\)
In simple words: Add up all side lengths or use standard formulas for regular shapes.

Question 2. Find the side of the square whose perimeter is:
a) 22cm
b) 16cm
Answer:
a) Side = \(22 \div 4 = 5.5\text{ cm}\)
b) Side = \(16 \div 4 = 4\text{ cm}\)
In simple words: Divide the perimeter by 4.

Question 3. Shika runs around a square of side 75m. Priya runs around a rectangle with length 60m and breadth 45m. Who covers the smaller distance?
Answer:
Shika's distance (Perimeter of square) = \(4 \times 75 = 300\text{ m}\)
Priya's distance (Perimeter of rectangle) = \(2(60 + 45) = 2(105) = 210\text{ m}\)
Priya covers the smaller distance.
In simple words: Compare the perimeters of both running paths.

Question 4. The length of a rectangular field is 100m. If its perimeter is 300m. What is its breadth?
Answer:
Perimeter = \(2(\text{length} + \text{breadth})\)
\(300 = 2(100 + \text{breadth})\)
\(150 = 100 + \text{breadth}\)
Breadth = \(50\text{ m}\)
In simple words: Half the perimeter minus the length gives the breadth.

Question 5. Find the cost of fencing a rectangular park of length 175m and breadth 125m at the rate of Rs 12/m.
Answer:
Perimeter = \(2(175 + 125) = 2(300) = 600\text{ m}\)
Cost = \(600 \times 12 = \text{Rs } 7200\)
In simple words: Multiply the perimeter of the park by the cost per meter.

Question 6. The area of a rectangle is 240 sq cm. If its length is 20cm. Find its breadth.
Answer:
Breadth = \(\text{Area} \div \text{length} = 240 \div 20 = 12\text{ cm}\)
In simple words: Divide the total area by the given length.

Question 7. A marble tile measures 25cm by 20cm. How many tiles will be required to cover a floor of size 4m by 3m?
Answer:
Floor dimensions in cm: \(400\text{ cm} \times 300\text{ cm}\)
Area of floor = \(400 \times 300 = 120000\text{ sq cm}\)
Area of one tile = \(25 \times 20 = 500\text{ sq cm}\)
Number of tiles = \(\frac{120000}{500} = 240 \text{ tiles}\)
In simple words: Divide the total floor area by the area of a single tile after making units consistent.

Question 8. Find the perimeter of a rectangle whose area is 650 sq cm and its breadth is 13cm.
Answer:
Length = \(650 \div 13 = 50\text{ cm}\)
Perimeter = \(2(50 + 13) = 2(63) = 126\text{ cm}\)
In simple words: Find the length first by dividing area by breadth, then calculate the perimeter.

Question 9. The side of a square is 70cm. Find its area and perimeter.
Answer:
Area = \(\text{side} \times \text{side} = 70 \times 70 = 4900\text{ sq cm}\)
Perimeter = \(4 \times 70 = 280\text{ cm}\)
In simple words: Square the side length for area and multiply by 4 for perimeter.

Ratio, Proportion And Unitary Method

Question 1. Write the ratios in the simplest form:
i) 15 minutes to one hour
ii) 600g to 1 kg
iii) 50p to Rs. 2.50
iv) 2l to 800 ml
v) 80cm to 4 m
vi) 3 km to 600m
vii) 3 dozen to 3 scores
viii) 2cm to 4 cm
ix) 4 days to 2 weeks
x) 3 yrs. to 6 months
xi) 45 sec to 3 min
xii) 45cm to 3m
xiii) 30 min to 4 hours
xiv) Rs. 1500 to Rs. 18000
xv) 750g to 3kg
xvi) 8days to 6 weeks
xvii) 450ml to 3l
xviii) 8mm to 1cm
xix) Rs. 1.50 to 25p
xx) 4 months to 4yrs
xxi) 294 : 343
Answer:
i) \(15 : 60 = 1 : 4\)
ii) \(600 : 1000 = 3 : 5\)
iii) \(50 : 250 = 1 : 5\)
iv) \(2000 : 800 = 5 : 2\)
v) \(80 : 400 = 1 : 5\)
vi) \(3000 : 600 = 5 : 1\)
vii) \(36 : 60 = 3 : 5\)
viii) \(2 : 4 = 1 : 2\)
ix) \(4 : 14 = 2 : 7\)
x) \(36 : 6 = 6 : 1\)
xi) \(45 : 180 = 1 : 4\)
xii) \(45 : 300 = 3 : 20\)
xiii) \(30 : 240 = 1 : 8\)
xiv) \(1500 : 18000 = 1 : 12\)
xv) \(750 : 3000 = 1 : 4\)
xvi) \(8 : 42 = 4 : 21\)
xvii) \(450 : 3000 = 3 : 20\)
xviii) \(8 : 10 = 4 : 5\)
xix) \(150 : 25 = 6 : 1\)
xx) \(4 : 48 = 1 : 12\)
xxi) \(294 : 343 = 6 : 7\)
In simple words: Convert both quantities to the same unit, then simplify the fraction to lowest terms.

Question 2. Divide Rs. 1450 between A and B in the ratio 2 : 3.
Answer:
Sum of ratio terms = \(2 + 3 = 5\)
A's share = \(\frac{2}{5} \times 1450 = 2 \times 290 = \text{Rs } 580\)
B's share = \(\frac{3}{5} \times 1450 = 3 \times 290 = \text{Rs } 870\)
In simple words: Divide the total amount by the sum of parts, then multiply by each person's ratio share.

Question 3. Dr. Suma earns Rs. 50,000 p.m. She spends Rs. 42,000 of it and saves the rest. Find the ratio of:
i) Her expenditure to her income
ii) Her saving to her income
iii) Her expenditure to her savings
Answer:
Income = Rs. 50,000, Expenditure = Rs. 42,000, Savings = \(50000 - 42000 = \text{Rs } 8,000\)
i) Expenditure to income = \(42000 : 50000 = 21 : 25\)
ii) Savings to income = \(8000 : 50000 = 4 : 25\)
iii) Expenditure to savings = \(42000 : 8000 = 21 : 4\)
In simple words: Find savings first, then write the required ratios in simplest form.

Question 4. Fill in:
i) 3 : 5 = ____ : 15
ii) 6 : ____ = 12 : 30
iii) ____ : 11 = 12 : 22
iv) 9 : 13 = 27 : ____
v) 6 : 7 = ____ : 49
vi) ____ : 5 = 20 : 25
vii) 3 : ____ = 5 : 10
viii) 45 : 15 = 3 : ____
ix) 5 : 30 = ____ : 6
x) 7 : ____ = 49 : 63
xi) 1 : 11 = 9 : ____
xii) ____ : 13 = 18 : 117
Answer:
i) 9
ii) 15
iii) 6
iv) 39
v) 42
vi) 4
vii) 6
viii) 1
ix) 1
x) 9
xi) 99
xii) 2
In simple words: Use cross-multiplication or equivalent fractions to find the missing term.

Question 5. Meena got Rs. 600 as her share from Rs. 2000. Tina got Rs. 750 as her share from Rs. 3000. Express their shares as ratios in the lower terms. Compare and find who got more.
Answer:
Meena's ratio = \(600 : 2000 = 6 : 20 = 3 : 10\) (or \(\frac{3}{10} = 30\%\))
Tina's ratio = \(750 : 3000 = 75 : 300 = 1 : 4\) (or \(\frac{1}{4} = 25\%\))
Comparing fractions: \(\frac{3}{10} > \frac{1}{4}\), so Meena got a greater share.
In simple words: Convert each share to a fraction or percentage to compare them accurately.

Question 6. Fill in the blanks with >, = or <
i) 7:10 ____ 5:10
ii) 6:7 ____ 6:11
iii) 4:5 ____ 8:10
iv) 9:13 ____ 9:15
v) 11:20 ____ 9:20
vi) 4:9 ____ 12:27
Answer:
i) >
ii) >
iii) =
iv) >
v) >
vi) =
In simple words: Convert ratios to fractions with common denominators or cross-multiply to compare.

Question 7. The ratio of the length of a rectangle to its breadth is 3:2. If the perimeter of the rectangle is 20 cm find the length, breadth and area.
Answer:
Let length = \(3x\) and breadth = \(2x\).
Perimeter = \(2(3x + 2x) = 20\)
\( 10x = 20 \implies x = 2$
Length = \(3(2) = 6\text{ cm}\), Breadth = \(2(2) = 4\text{ cm}\)
Area = \(6 \times 4 = 24\text{ sq cm}\)
In simple words: Use variables based on the ratio to find dimensions from the perimeter.

Question 8. Fill in:
i) 2:4 :: ____:10
ii) ____:8 :: 1:4
iii) 3:8 :: 15: ____
iv) 1:____ :: 3:15
v) 9:____ :: 90:100
Answer:
i) 5
ii) 2
iii) 40
iv) 5
v) 10
In simple words: In proportions, product of extremes equals product of means.

Question 9. The cost of 4 pens is Rs. 40. The cost of 11 pens is Rs. ____________.
Answer:
Cost of 1 pen = \(40 \div 4 = \text{Rs } 10\)
Cost of 11 pens = \(11 \times 10 = \text{Rs } 110\)
In simple words: Find the unit cost first, then multiply by the required quantity.

Question 10. The weight of 15 boxes is 60 kg. The weight of 12 boxes is ___________.
Answer:
Weight of 1 box = \(60 \div 15 = 4\text{ kg}\)
Weight of 12 boxes = \(12 \times 4 = 48\text{ kg}\)
In simple words: Find the weight of one box, then multiply by 12.

Question 11. Maya can walk 6km in 2 hours. In 3 hours she can walk ___________.
Answer:
Speed = \(6 \div 2 = 3\text{ km/h}\)
Distance in 3 hours = \(3 \times 3 = 9\text{ km}\)
In simple words: Calculate her walking speed per hour and multiply by 3.

Question 12. Dinner at Marhaba costs SR 1050 for 10 people. How many people can have dinner for SR 1995?
Answer:
Cost per person = \(1050 \div 10 = \text{SR } 105\)
Number of people = \(1995 \div 105 = 19 \text{ people}\)
In simple words: Divide total money by cost per person.

Question 13. A car travels 95 km in 5 litres of petrol. How far can it go in 11 litres of petrol?
Answer:
Distance per litre = \(95 \div 5 = 19\text{ km/l}\)
Distance in 11 litres = \(19 \times 11 = 209\text{ km}\)
In simple words: Multiply mileage per liter by 11 liters.

Elementary Shapes

Question 1. Given below are the lengths of the sides of triangles. Classify them as equilateral, isosceles or scalene:
i) 6cm, 2.4cm, 6cm
ii) 7cm, 9cm, 5.5cm
iii) 5.4cm, 7cm, 6.1cm
iv) 7.2cm, 7.2cm, 7.2cm
v) 10.1cm, 8.6cm, 8.6cm
vi) 3.5cm, 4.5cm, 5.1cm
vii) 4.8cm, 4.8cm, 4.8cm
viii) 6.8cm, 6.8cm, 8.6cm
ix) 3cm, 4cm, 5cm
Answer:
i) Isosceles
ii) Scalene
iii) Scalene
iv) Equilateral
v) Isosceles
vi) Scalene
vii) Equilateral
viii) Isosceles
ix) Scalene
In simple words: Equilateral has 3 equal sides, Isosceles has 2 equal sides, and Scalene has all different sides.

Question 2. Draw a triangle ABC using a protractor, measure \(\angle A\), \(\angle B\) and \(\angle C\). Find their sum. What do you notice. Also measure the sides AB, BC, AC. What kind of a triangle is this?
Answer:
The sum of all three interior angles in any triangle is always \(180^\circ\).
In simple words: The angle sum property states that \(\angle A + \angle B + \angle C = 180^\circ\).

Question 3. Given below are the measures of the angles of some triangles. Classify them as acute-angled, obtuse-angled or right-angled:
i) \(60^\circ, 90^\circ, 30^\circ\( ii) \(40^\circ, 100^\circ, 40^\circ\)
iii) \(60^\circ, 60^\circ, 60^\circ\( iv) \(20^\circ, 40^\circ, 120^\circ\)
v) \(50^\circ, 60^\circ, 70^\circ\( vi) \(45^\circ, 45^\circ, 90^\circ\)
Answer:
i) Right-angled
ii) Obtuse-angled
iii) Acute-angled
iv) Obtuse-angled
v) Acute-angled
vi) Right-angled
In simple words: Classified by the largest angle: acute (< \(90^\circ\)), right (= \(90^\circ\)), or obtuse (> \(90^\circ\)).

Question 4. Draw a triangle ABC. Measure the sides AB, BC, AC. Verify i) AB + BC > AC ii) BC + AC > AB iii) AB + AC > BC
Answer:
The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. This is known as the triangle inequality theorem.
In simple words: Any two sides added together must exceed the remaining side.

Question 5. Write the number of sides of the following Polygons: Triangles, Pentagon, Quadrilateral, Heptagon, Hexagon, Nonagon, Octagon, Decagon.
Answer:
Triangle: 3, Pentagon: 5, Quadrilateral: 4, Heptagon: 7, Hexagon: 6, Nonagon: 9, Octagon: 8, Decagon: 10.
In simple words: Match each polygon name with its respective number of sides.

Question 6. Fill in the table for 3D shapes (Faces, Edges, Vertices):
- Cube
- Cuboid
- Square Pyramid
- Triangular Pyramid
- Triangular Prism
Answer:
- Cube: Faces = 6, Edges = 12, Vertices = 8
- Cuboid: Faces = 6, Edges = 12, Vertices = 8
- Square Pyramid: Faces = 5, Edges = 8, Vertices = 5
- Triangular Pyramid (Tetrahedron): Faces = 4, Edges = 6, Vertices = 4
- Triangular Prism: Faces = 5, Edges = 9, Vertices = 6
In simple words: Count faces, edges, and vertices according to Euler's formula (\(F + V - E = 2\)).

Question 7. Write Yes or No for properties of quadrilaterals (Trapezium, Parallelogram, Rhombus, Rectangle, Square).
Answer:
- Parallelogram: Opposite sides parallel & equal (Yes), all sides equal (No), opposite angles equal (Yes), diagonals bisect each other (Yes).
- Rectangle: Opposite sides parallel & equal (Yes), all sides equal (No), all angles equal (Yes), diagonals equal and bisect each other (Yes).
- Square: All properties are "Yes".
In simple words: Review properties of standard quadrilaterals regarding parallelism, equal sides, and diagonals.

Symmetry

Question 1. Look around and list any five things from your surroundings that are symmetrical.
Answer:
1. Blackboard
2. Notebook
3. Glass window
4. Ceiling fan blade
5. Carrom board
In simple words: Objects that can be divided into identical halves are symmetrical.

Question 2. Write any 3 English alphabets, which are symmetrical and have:
i) One axis of symmetry
ii) Two axes of symmetry
iii) More than two axes of symmetry
Answer:
i) One axis: A, M, T
ii) Two axes: H, I, X
iii) More than two: None in English alphabet (O has 2 principal axes, but some consider ellipse-like, though O has infinitely many lines of symmetry if circular).
In simple words: Count how many fold-lines divide each capital letter into matching halves.

Question 3. Write the number of axes of symmetry of:
i) Square
ii) Rectangle
iii) Circle
iv) Scalene Triangle
v) Equilateral Triangle
vi) Rhombus
Answer:
i) Square = 4
ii) Rectangle = 2
iii) Circle = Infinite (\(\infty\))
iv) Scalene Triangle = 0
v) Equilateral Triangle = 3
vi) Rhombus = 2
In simple words: Count how many lines can fold each shape perfectly in half.

Geometrical Constructions

Question 1. Construct the following angles using a ruler and compasses:
i) \(60^\circ\( ii) \(120^\circ\)
iii) \(30^\circ\( iv) \(90^\circ\)
v) \(45^\circ$
Answer:
Steps for \(60^\circ\): Draw a ray, draw an arc from the endpoint, and cut the arc with the same radius to form an angle of \(60^\circ\). Bisecting \(60^\circ\) gives \(30^\circ\). Constructing a perpendicular at \(90^\circ\) and bisecting it gives \(45^\circ$.
In simple words: Use standard compass arcs to construct fundamental angles like \(60^\circ\) and \(90^\circ\), then bisect them.

Question 2. Draw a line-segment AB of any length. Draw the perpendicular bisector of it.
Answer:
Place compass pointer at A and open it more than half the length of AB. Draw arcs above and below AB. With the same radius, place pointer at B and cut the previous arcs. Join the intersection points to get the perpendicular bisector.
In simple words: Draw intersecting arcs from both endpoints above and below the line segment to find its midpoint and perpendicular.

Question 3. Draw a circle of radius 4cm. Inscribe a regular hexagon in it.
Answer:
Draw a circle of radius 4 cm. Starting from any point on the circumference, step the radius along the circle 6 times using the compass. Connect consecutive points to form a regular hexagon.
In simple words: The side length of a regular inscribed hexagon equals the radius of the circle.

Question 4. Draw a circle and any two of its diameters. What is the figure obtained? What figure is obtained if the diameters are perpendicular to each other?
Answer:
- Joining the endpoints of any two diameters gives a rectangle.
- Joining the endpoints of two perpendicular diameters gives a square.
In simple words: Connecting the endpoints of two diameters forms a rectangle, which becomes a square when the diameters are perpendicular.

Mensuration

Question 1. The length and breadth of a Rectangle are 11 cm and 9 cm. Find its area and Perimeter.
Answer:
Area = \(11 \times 9 = 99\text{ sq cm}\)
Perimeter = \(2(11 + 9) = 2(20) = 40\text{ cm}\)
In simple words: Multiply length by breadth for area, and take twice their sum for perimeter.

Question 2. The area of a Rectangle is 144 sq cm and its length is 16 cm. Find the breadth of the Rectangle?
Answer:
Breadth = \(\text{Area} \div \text{length} = 144 \div 16 = 9\text{ cm}\)
In simple words: Divide the area by the given length to get the breadth.

Question 3. Find the perimeter of a regular hexagon of side x cm.
Answer:
Perimeter = \(6 \times x = 6x\text{ cm}\)
In simple words: Multiply the side length by 6 for a hexagon.

Question 4. If the perimeter of a regular pentagon is 65 cm. Find its side?
Answer:
Side = \(65 \div 5 = 13\text{ cm}\)
In simple words: Divide the perimeter by 5.

Question 5. The perimeter of a triangle is 42 cm. If two of its sides are 16 cm and 12 cm. Find its third side?
Answer:
Third side = \(42 - (16 + 12) = 42 - 28 = 14\text{ cm}\)
In simple words: Subtract the sum of the two given sides from the total perimeter.

Question 6. A piece of string is 45 cm long. It is bent to form an equilateral triangle. Find the side of triangle.
Answer:
Side = \(45 \div 3 = 15\text{ cm}\)
In simple words: Divide the total length of the string by 3.

Question 7. An athlete takes 5 rounds of a rectangular park 120 m long and 80m wide. Find the total distance covered by him.
Answer:
Perimeter of park = \(2(120 + 80) = 2(200) = 400\text{ m}\)
Total distance = \(5 \times 400 = 2000\text{ m} = 2\text{ km}\)
In simple words: Multiply the perimeter by the number of rounds.

Question 8. Find the area of a square whose perimeter is 260 cm.
Answer:
Side = \(260 \div 4 = 65\text{ cm}\)
Area = \(65 \times 65 = 4225\text{ sq cm}\)
In simple words: Find the side length first by dividing perimeter by 4, then square it to find the area.

Question 9. The total cost of fencing a square park at Rs. 20 per metre is Rs. 2880. Find the side of the square park.
Answer:
Perimeter = \(2880 \div 20 = 144\text{ m}\)
Side = \(144 \div 4 = 36\text{ m}\)
In simple words: Divide total cost by cost per meter to get the perimeter, then divide by 4 to get the side length.

Question 10. The floor of a room with dimensions 5 m and 3m is to be covered with square tiles. If each square tile is of side 25 cm. Find the number of tiles required.
Answer:
Room dimensions in cm: \(500\text{ cm} \times 300\text{ cm}\)
Area of room = \(500 \times 300 = 150000\text{ sq cm}\)
Area of one tile = \(25 \times 25 = 625\text{ sq cm}\)
Number of tiles = \(\frac{150000}{625} = 240 \text{ tiles}\)
In simple words: Divide the total floor area by the area of one tile after converting meters to centimeters.

Fill in the blanks :

Question. 25 8 125 4 = ___________
Answer : 100000

Question. 315 105 = 315 100 + ____________ 5
Answer : 315

Question. Division by zero is _____________
Answer : not defined

Question. The whole number ______________ has no number .
Answer : zero

Question. The smallest natural number is ________________.
Answer : one

Question. The sum of 3 odd numbers is _____________.
Answer : odd

Question. _______________ is the additive identity for the whole numbers.
Answer : zero

Question. ( 7 8) 5 = 7 ( 8 5) This statement shows that multiplication of whole numbers is _____________.
Answer : associate

Question. How many numbers between 102 and 211.
Answer : 108

Question. 3 + 7 = 7 + 3 . This statement shows that addition of whole numbers is __________
Answer : commutative

Question. Determining the product by suitable rearrangements
a) 2 125 50 8
b) 16 279 625
c) 2 1735 50
Answer : a) 100000
b) 2790000
c) 173500

Question. A teacher purchases 42 Mathematics books and 42 English books for his class. If the cost of a Mathematics book is Rs 52 and the cost of an English book is Rs48. Find the total amount paid by the teacher to the shopkeeper.
Answer : 24200

Question. If the cost of a pack of mango drink is Rs.14. Then how many packs of the drink can be purchased for Rs.76 and what is the balance ?
Answer : 5, Balance Rs.6

Knowing our numbers

Question. Estimate using general rule :
(i) 830 + 976
(ii) 496 215
(iii) 13,804 + 3,777
(iv) 61,292 21, 496
Answer : (i) 1,800 (ii) 300 (iii) 18,000 (iv) 40,000

Question. Estimate the products using general rule :
(i) 758 151
(ii) 4391 2300
(iii) 2187 456
(iv) 6978 43
Answer : (i) 16,0000 (ii) 80,00,000 (iii) 10,00,000 (iv) 2,80,000

Question. Write the Roman Numeral for :
(i) 99
(ii) 48
(iii) 67
(iv) 81
(v) 17
(vi) 76
(viii) 54
Answer : (i) XCIX (ii) XLVIII (iii) LXVII (iv) LXXXI (v) XVII (vi) LXXXI (v) XVII (vi) LXXVI (vii) XCIII (viii) LIV

Answer the following :

Question. The town newspaper is published every day. One copy has 15 pages. Everyday 12, 500 copies are printed. How man total pages are printed every day?
Answer : 1,87,500 pages

Question. Apples are packed in boxes, each weighing 5kg 500gm. How many such boxes can be loaded in a van which cannot carry beyond 1000kg?
Answer : 181 boxes

Q1.Estimate the following products
(a) 898 × 795

Q2.Estimate each of the following using general rule:
(a) 730 + 968
(b) 796 – 314

Q3. Estimate the following products using general rule:
(a) 1391 × 592

Q4. Write in Roman numerals.
a) 654 b) 1012

Q5.Write the Roman numerals in number
a) XLVIII
(b). XXIX+LX

Q6.Place commas correctly and write the numerals:
(a) Seven crore fifty two lakh twenty one thousand three hundred two.

Q7.Make the greatest four digit number by using any one digit twice by 3,8,7

Q8. Fill in the blanks to make the statement true
6245+ (631+751) = 631+……….+751

Q9. Find the value of 1507 – (625÷25)

Q10. Find the whole number if n +4 = 9

Q11. The line on which we represent the natural number is known as________

Q12 Write the successor of : (a) 2440701 (b) 100199

Q13. Simplify: 126 × 55 + 126 × 45

Q14. Find the sum by suitable rearrangement:
(a) 1962 + 453 + 1538 + 647

Q15.Find the value of the following:
(a) 297 × 17 + 297 × 5

Q16. Find the product of the largest 2 digit number and the largest 3 digit number,using suitable method

Q17. What is H.C.F OF 4 × 27× 3125

Q18 Write all the factors of the following numbers :
(a) 24
(b) 15

Q19 Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

Q20. Two numbers having only 1 as common factor are called_____

Q21. Common factors of 12 and 24 are

Q22. The product of L.C.M and H.C.F. of two numbers is equal to
(a) Sum of numbers
(b) Difference of numbers

Q23. What is the H.C.F. of 48 and 14

Q24 Find the least number which when divided by 6,15 and 18 leave remainder 5 in each case.

Q25. Find the L.C.M OF 36,75,80

Q26. Reduce 2211/5025 to the lowest terms by dividing the numerator and denominator their by H.C.F

Q27. In one state, the number of bicycles sold in the year 2002-2003 was 8,65,000. In the year 2003-2004, the number of bicycles sold was 12,00,100. In which year were more bicycles sold? And how many more.

Project:
Design your own small book of Maths (using your creativity and knowledge) covering all the topics from integers and playing with numbers explain each subtopic in detail and solve 2 questions each related to every subtopic. Also include the real life situations where we are applying integers and playing with numbers (For eg. +ve Whether, -ve whether and explain the conditions of different whether).correlate it with other subjects (foreg.whether in geography).