Distributivity and Algebra | IT

Question 17

Use this to find (i) 89×10189 \times 101, (ii) 949×101949 \times 101, (iii) 265831×1001265831 \times 1001, (iv) 1111×10011111 \times 1001, (v) 9734×999734 \times 99 and (vi) 23478×99923478 \times 999.

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Solution

We will use the distributive property to simplify these multiplication problems. This property helps us break down complex multiplications into simpler ones.

Step 1 — Multiplying by 101 (Part i)

We want to calculate 89×10189 \times 101. We can write 101101 as 100+1100 + 1. The distributive property is a×(b+c)=a×b+a×ca \times (b+c) = a \times b + a \times c. Let a=89a = 89, b=100b = 100, and c=1c = 1. So, we multiply 8989 by 100100 and then by 11. Then we add the results together.

89×101=89×(100+1)89 \times 101 = 89 \times (100 + 1)

=(89×100)+(89×1)= (89 \times 100) + (89 \times 1)

=8900+89= 8900 + 89

8989\boxed{8989}

Diagram 1

Step 2 — Multiplying by 101 (Part ii)

We need to find the product of 949949 and 101101. Again, we write 101101 as 100+1100 + 1. We use the distributive property. We multiply 949949 by 100100 and then by 11. Then we add these products.

949×101=949×(100+1)949 \times 101 = 949 \times (100 + 1)

=(949×100)+(949×1)= (949 \times 100) + (949 \times 1)

=94900+949= 94900 + 949

95849\boxed{95849}

Step 3 — Multiplying by 1001 (Part iii)

We want to calculate 265831×1001265831 \times 1001. We can write 10011001 as 1000+11000 + 1. We apply the distributive property. We multiply 265831265831 by 10001000 and then by 11. Then we add the results.

265831×1001=265831×(1000+1)265831 \times 1001 = 265831 \times (1000 + 1)

=(265831×1000)+(265831×1)= (265831 \times 1000) + (265831 \times 1)

=265831000+265831= 265831000 + 265831

266096831\boxed{266096831}

Diagram 2

Step 4 — Multiplying by 1001 (Part iv)

Let us find the product of 11111111 and 10011001. We express 10011001 as 1000+11000 + 1. Using the distributive property, we multiply 11111111 by 10001000 and by 11. Then we add these two products.

1111×1001=1111×(1000+1)1111 \times 1001 = 1111 \times (1000 + 1)

=(1111×1000)+(1111×1)= (1111 \times 1000) + (1111 \times 1)

=1111000+1111= 1111000 + 1111

1112111\boxed{1112111}

Step 5 — Multiplying by 99 (Part v)

We need to calculate 9734×999734 \times 99. We can write 9999 as 1001100 - 1. The distributive property also works for subtraction: a×(bc)=a×ba×ca \times (b-c) = a \times b - a \times c. Let a=9734a = 9734, b=100b = 100, and c=1c = 1. So, we multiply 97349734 by 100100 and then by 11. Then we subtract the second result from the first.

9734×99=9734×(1001)9734 \times 99 = 9734 \times (100 - 1)

=(9734×100)(9734×1)= (9734 \times 100) - (9734 \times 1)

=9734009734= 973400 - 9734

963666\boxed{963666}

Diagram 3

Step 6 — Multiplying by 999 (Part vi)

Let us find the product of 2347823478 and 999999. We write 999999 as 100011000 - 1. We use the distributive property for subtraction. We multiply 2347823478 by 10001000 and by 11. Then we subtract the second product from the first.

23478×999=23478×(10001)23478 \times 999 = 23478 \times (1000 - 1)

=(23478×1000)(23478×1)= (23478 \times 1000) - (23478 \times 1)

=2347800023478= 23478000 - 23478

23454522\boxed{23454522}

Answer

(i) 8989 (ii) 95849 (iii) 266096831 (iv) 1112111 (v) 963666 (vi) 23454522

More questions in IT

Q1

Context: Consider the multiplication of two numbers, say, 23 × 27.

Q. By how much does the product increase if the first number (23) is increased by 1?

Q2

Context: Consider the multiplication of two numbers, say, 23 × 27.

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Q3

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Q4

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Q. Do you see a pattern that could help generalise our observations to the product of any two numbers?

Q5

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Q6

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Q7

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Check by substituting different values for aa and bb in each of the above cases. For example, a=5,b=8;a=4,b=5a = -5, b = 8; a = -4, b = -5; etc.

Q8

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Q9

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Q11

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Q12

Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.

Evaluate: (i) 94×1194 \times 11 (ii) 495×11495 \times 11 (iii) 3279×113279 \times 11 (iv) 4791256×114791256 \times 11

Q13

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Q14

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Q15

Use this to multiply 3874×1013874 \times 101 in one line.

Q16

What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, ...

Q17

Use this to find (i) 89×10189 \times 101, (ii) 949×101949 \times 101, (iii) 265831×1001265831 \times 1001, (iv) 1111×10011111 \times 1001, (v) 9734×999734 \times 99 and (vi) 23478×99923478 \times 999.

Q18

The area of a square of sidelength 60 units is 3600 sq. units (60260^2) and that of a square of sidelength 5 units is 25 sq. units (525^2). Can we use this to find the area of a square of sidelength 65 units?

Q19

What if we write 65265^2 as (30+35)2(30 + 35)^2 or (52+13)2(52 + 13)^2? Draw the figures and check the area that you get.

Q20

If aa and bb are any two integers, is (a+b)2(a + b)^2 always greater than a2+b2a^2 + b^2? If not, when is it greater?

Q21

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Q22

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(i) (m+3)2(m + 3)^2

(ii) (6+p)2(6 + p)^2

Q23

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Q24

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Q25

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Q26

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Q27

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Q28

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Q29

Pattern 2

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Q30

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Q31

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Q32

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Q33

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Q36

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Q37

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Q38

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Q39

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Q41

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Q42

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