CBSE Class 6 Mathematics Whole Numbers Worksheet Set 07

Question 1. Find the sum by suitable rearrangement:
(a) \(538 + 308 + 462\)
Answer: Group the numbers to make tens or hundreds.
\(= (538 + 462) + 308\)
\(= 1000 + 308\)
\(= 1308\)
In simple words: Rearrange the numbers so that their addition results in a multiple of 10 or 100 first.

Exam Tip: Look for unit digits that sum up to 10 (like 8 and 2) to group them together.

Question 1(b). Find the sum by suitable rearrangement:
(b) \(235 + 103 + 765\)
Answer: Group the numbers with matching unit digits.
\(= (235 + 765) + 103\)
\(= 1000 + 103\)
\(= 1103\)
In simple words: Add 235 and 765 first because 5 + 5 makes 10, giving a clean round number.

Exam Tip: Associative and commutative properties of addition allow us to reorder terms freely.

Question 1(c). Find the sum by suitable rearrangement:
(c) \(2896 + 489 + 3204 + 511\)
Answer: Group pairs that add up to round numbers:
\(= (2896 + 3204) + (489 + 511)\)
\(= 6100 + 1000\)
\(= 7100\)
In simple words: Pair 2896 with 3204 and 489 with 511 to make addition effortless.

Exam Tip: Always check numbers that complete thousands or hundreds first.

Question 2. Find the product by suitable rearrangement:
(a) \(5 \times 9685 \times 20\)
Answer: Rearrange to multiply 5 and 20 first:
\(= (5 \times 20) \times 9685\)
\(= 100 \times 9685\)
\(= 968500\)
In simple words: Multiply 5 by 20 to get 100, then just append two zeros to 9685.

Exam Tip: Look for pairs like 2 and 5, or 4 and 25, or 8 and 125 that yield powers of 10.

Question 2(b). Find the product by suitable rearrangement:
(b) \(5 \times 293 \times 20\)
Answer: Rearrange to multiply 5 and 20 first:
\(= (5 \times 20) \times 293\)
\(= 100 \times 293\)
\(= 29300\)
In simple words: Multiplying 5 and 20 gives 100, which makes multiplying by 293 very easy.

Exam Tip: Multiplication is associative, so group numbers that result in 10, 100, or 1000.

Question 2(c). Find the product by suitable rearrangement:
(c) \(396 \times 60 \times 5\)
Answer: Group 60 and 5:
\(= 396 \times (60 \times 5)\)
\(= 396 \times 300\)
\(= 396 \times 3 \times 100\)
\(= 1188 \times 100\)
\(= 118800\)
In simple words: Multiply 60 by 5 to get 300, then multiply 396 by 3 and attach two zeros.

Question 2(d). Find the product by suitable rearrangement:
(d) \(125 \times 8 \times 40 \times 25\)
Answer: Group numbers to form convenient products:
\(= (125 \times 8) \times (40 \times 25)\)
\(= 1000 \times 1000\)
\(= 1000000\)
In simple words: Since 125 × 8 = 1000 and 40 × 25 = 1000, multiply them to get 10,00,000.

Exam Tip: Memorize standard pairs: \(125 \times 8 = 1000\) and \(25 \times 4 = 100\).

Question 3. Find the product using suitable properties:
(a) \(493 \times 105\)
Answer: Use distributive property of multiplication over addition:
\(= 493 \times (100 + 5)\)
\(= (493 \times 100) + (493 \times 5)\)
\(= 49300 + 2465\)
\(= 51765\)
In simple words: Split 105 into 100 + 5 and multiply separately with 493.

Exam Tip: Expand numbers close to multiples of 100 (like 105, 102) to make multiplication mental.

Question 3(b). Find the product using suitable properties:
(b) \(296 \times 102\)
Answer: Use distributive property:
\(= 296 \times (100 + 2)\)
\(= (296 \times 100) + (296 \times 2)\)
\(= 29600 + 592\)
\(= 30192\)
In simple words: Write 102 as \(100 + 2\) and apply distribution.

Exam Tip: Splitting numbers saves time and reduces calculation errors.

Question 3(c). Find the product using suitable properties:
(c) \(245 \times 1006\)
Answer: Use distributive property:
\(= 245 \times (1000 + 6)\)
\(= (245 \times 1000) + (245 \times 6)\)
\(= 245000 + 1470\)
\(= 246470\)
In simple words: Split 1006 into \(1000 + 6\) and distribute.

Exam Tip: For numbers just above a thousand, expand using 1000.

Question 3(d). Find the product using suitable properties:
(d) \(1004 \times 138\)
Answer: Use distributive property:
\(= (1000 + 4) \times 138\)
\(= (1000 \times 138) + (4 \times 138)\)
\(= 138000 + 552\)
\(= 138552\)
In simple words: Break 1004 into \(1000 + 4\) and multiply each part by 138.

Exam Tip: The distributive property works whether the expanded number is on the left or right.

Question 3(e). Find the product using suitable properties:
(e) \(135 \times 101\)
Answer: Use distributive property:
\(= 135 \times (100 + 1)\)
\(= (135 \times 100) + (135 \times 1)\)
\(= 13500 + 135\)
\(= 13635\)
In simple words: Expand 101 into \(100 + 1\) for quick calculation.

Exam Tip: Multiplying any number by 101 is just adding the number to 100 times the number.

Question 4(a). Find using distributive property:
(a) \(352 \times 101\)
Answer:
\(= 352 \times (100 + 1)\)
\(= (352 \times 100) + (352 \times 1)\)
\(= 35200 + 352\)
\(= 35552\)
In simple words: Express 101 as \(100 + 1\) and distribute 352 over both terms.

Exam Tip: Always show the expansion step clearly to get full marks for properties.

Question 4(b). Find using distributive property:
(b) \(5723 \times 1001\)
Answer:
\(= 5723 \times (1000 + 1)\)
\(= (5723 \times 1000) + (5723 \times 1)\)
\(= 5723000 + 5723\)
\(= 5728723\)
In simple words: Split 1001 into \(1000 + 1\) and multiply.

Exam Tip: Multiplying by 1001 appends 3 zeros and adds the number itself.

Question 4(c). Find using distributive property:
(c) \(9275 \times 125\)
Answer:
\(= 9275 \times (100 + 25)\)
\(= (9275 \times 100) + (9275 \times 25)\)
\(= 927500 + 231875\)
\(= 1159375\)
In simple words: Split 125 into \(100 + 25\) and distribute.

Exam Tip: Alternatively, rewrite 125 as \(\frac{1000}{8}\) if divisible, but distributive property works universally.

Question 4(d). Find using distributive property:
(d) \(605 \times 45\)
Answer:
\(= 605 \times (40 + 5)\)
\(= (605 \times 40) + (605 \times 5)\)
\(= 24200 + 3025\)
\(= 27225\)
In simple words: Split 45 into \(40 + 5\) and multiply with 605.

Exam Tip: Breaking the smaller number simplifies the multiplication steps.

Question 4(e). Find using distributive property:
(e) \(12 \times 35\)
Answer:
\(= 12 \times (30 + 5)\)
\(= (12 \times 30) + (12 \times 5)\)
\(= 360 + 60\)
\(= 420\)
In simple words: Express 35 as \(30 + 5\) and distribute 12.

Exam Tip: Distributive property is extremely helpful for mental math calculations.

Question 5. Find the product of greatest 4-digit number and least 3-digit number.
Answer:
Greatest 4-digit number = 9999
Least 3-digit number = 100
Product = \(9999 \times 100 = 999900\)
In simple words: The greatest 4-digit number is 9999 and the least 3-digit number is 100. Multiply them to get 9,99,900.

Exam Tip: Greatest n-digit number is written as 'n' nines, and least n-digit number is '1' followed by \(n-1\) zeros.

Question 6. Simplify:
(a) \(126 \times 45 + 126 \times 55\)
Answer: Using distributive property in reverse (taking common factor out):
\(= 126 \times (45 + 55)\)
\(= 126 \times 100\)
\(= 12600\)
In simple words: Factor out 126, add 45 and 55 to get 100, then multiply.

Exam Tip: Look for common terms in addition/subtraction to apply reverse distribution.

Question 6(b). Simplify:
(b) \(92651 \times 129 - 92651 \times 29\)
Answer: Factor out 92651:
\(= 92651 \times (129 - 29)\)
\(= 92651 \times 100\)
\(= 9265100\)
In simple words: Take 92651 common, subtract 29 from 129 to get 100, and multiply.

Question 6(c). Simplify:
(c) \(54689 \times 8 + 54689 \times 92\)
Answer: Factor out 54689:
\(= 54689 \times (8 + 92)\)
\(= 54689 \times 100\)
\(= 5468900\)
In simple words: Factor out 54689, add 8 and 92 to get 100, then multiply.

Exam Tip: Always check if terms repeat across the plus or minus sign.

Question 6(d). Simplify:
(d) \(436 \times 18 + 436 \times 2\)
Answer: Factor out 436:
\(= 436 \times (18 + 2)\)
\(= 436 \times 20\)
\(= 436 \times 2 \times 10\)
\(= 872 \times 10\)
\(= 8720\)
In simple words: Take 436 common, add 18 and 2 to get 20, then multiply.

Exam Tip: Factoring out common numbers turns complex arithmetic into simple steps.

Question 6(e). Simplify:
(e) \(3845 \times 5 \times 782 + 769 \times 25 \times 218\)
Answer: Simplify the second term: \(25 = 5 \times 5\) and \(769 \times 5 = 3845\).
\(= 3845 \times 5 \times 782 + (769 \times 5) \times 5 \times 218\)
\(= 3845 \times 5 \times 782 + 3845 \times 5 \times 218\)
Now factor out \(3845 \times 5\):
\(= (3845 \times 5) \times (782 + 218)\)
\(= 19225 \times 1000\)
\(= 19225000\)
In simple words: Adjust the numbers so that a common term like 3845 × 5 appears in both parts, then factor it out.

Exam Tip: Spot hidden common factors by breaking down numbers into prime factors.

Question 6(f). Simplify:
(f) \(44 \times 20 \times 16 + 880 \times 4\)
Answer: Simplify both terms:
First term: \(44 \times 20 \times 16 = 880 \times 16\)
Second term: \(880 \times 4\)
Expression becomes:
\(= 880 \times 16 + 880 \times 4\)
Factor out 880:
\(= 880 \times (16 + 4)\)
\(= 880 \times 20\)
\(= 88 \times 2 \times 100\)
\(= 17600\)
In simple words: Convert \(44 \times 20\) into 880, then factor 880 out from both terms.

Exam Tip: Simplify partial products to discover common factors.

Question 6(h). Simplify:
(h) \(64 \times 75 + 75 \times 36\)
Answer: Factor out 75:
\(= 75 \times (64 + 36)\)
\(= 75 \times 100\)
\(= 7500\)
In simple words: Take 75 common, add 64 and 36 to get 100, then multiply.

Exam Tip: Always look for identical numbers multiplying different addends.

Fill in the blanks:

Question 7(1). Division by zero is _________
Answer: not defined
In simple words: You cannot divide any number by zero.

Exam Tip: Division by zero is mathematically meaningless.

Question 7(2). _________ is the smallest natural number.
Answer: 1
In simple words: Natural numbers start counting from 1.

Exam Tip: Natural numbers are positive counting numbers starting from 1.

Question 7(3). _________ is the smallest whole number.
Answer: 0
In simple words: Whole numbers include zero along with all natural numbers.

Exam Tip: Whole numbers start from 0, unlike natural numbers which start from 1.

Question 7(4). _________ is the identity for addition of whole numbers.
Answer: 0
In simple words: Adding zero to any number leaves the number unchanged.

Exam Tip: 0 is the additive identity because \(a + 0 = a\).

Question 7(5). _________ is the identity for multiplication of whole numbers.
Answer: 1
In simple words: Multiplying any number by 1 gives the same number back.

Exam Tip: 1 is the multiplicative identity because \(a \times 1 = a\).

Question 7(6). The value of \(1 + 0 = \text{_________}\)
Answer: 1
In simple words: Adding zero to 1 gives 1.

Exam Tip: Any number plus zero is the number itself.

Question 7(7). The value of \(\frac{0}{3} = \text{_________}\)
Answer: 0
In simple words: Zero divided by any non-zero number is zero.

Exam Tip: \(0 \div a = 0\) for any \(a \neq 0\).

Question 7(8). The value of \(\frac{4 - 4}{2} = \text{_________}\)
Answer: 0
Calculation: \(\frac{4 - 4}{2} = \frac{0}{2} = 0\)
In simple words: Subtracting 4 from 4 gives 0, and 0 divided by 2 is 0.

Exam Tip: Simplify the numerator completely before dividing.

Question 7(9). The value of \(0 \times 5 = \text{_________}\)
Answer: 0
In simple words: Any number multiplied by zero is zero.

Exam Tip: Multiplication property of zero states that \(a \times 0 = 0\).

Question 7(10). The value of \((2 \times 3) + (2 \times 7)\) is \(\text{_________}\)
Answer: 20
Calculation: \(2 \times (3 + 7) = 2 \times 10 = 20\)
In simple words: Using distributive property, \(2 \times 3 + 2 \times 7 = 2 \times 10 = 20\).

Exam Tip: Look for common factors to speed up evaluation.

Question 7(11). The difference between greatest two digit number and least two digit number is \(\text{_________}\)
Answer: 89
Calculation: Greatest 2-digit number = 99, Least 2-digit number = 10, Difference = \(99 - 10 = 89\).
In simple words: Subtract the smallest 2-digit number (10) from the largest 2-digit number (99) to get 89.

Exam Tip: Read carefully whether the question asks for sum or difference.