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K.V. I.A.t. girinagar, pune 25

K.V. I.A.t. girinagar, pune 25. Slides prepared By Mrs. Pushpa prakash Class- iv a Subject- mathematics. Unit-8 DIVISION 8.1 Relationship between Multiplication and Division 1. For every multiplication fact with two distinct factors, we get two division facts. Example: 7x8=56 gives

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K.V. I.A.t. girinagar, pune 25

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  1. K.V. I.A.t. girinagar, pune 25 Slides prepared By Mrs. Pushpa prakash Class- iv a Subject- mathematics

  2. Unit-8 DIVISION 8.1 Relationship between Multiplication and Division 1. For every multiplication fact with two distinct factors, we get two division facts. Example: 7x8=56 gives 567=8And, 568=7

  3. 2.2But a multiplication fact with identical factors gives only one division fact: Ex Example: 5x5=25 gives 255=5 And 0x4=0 gives 04=0 But not 00=4 NOTE: Division of a number by 0 is not possible.

  4. Exercise-8.1 QIFor each of the following multiplication fact write both the division facts: (a)3x7=21 (b)7x9=63 (c)8x5=40 QII Find the Quotient: (a)729 (b)488 (c)287

  5. 8.2 Terms Associated with Division In division, the number that we divide is called the DIVIDEND, the number by which we divide is called the DIVISOR, and the answer is the QUOTIENT. If there is a number left over, it is called the REMAINDER. _121__QUOTIENT DIVISOR 2) 243 DIVIDEND - 2 4 - 4 3 - 2 1 REMAINDER

  6. 1.It can be VERIFIED the answer of a division sum. DIVIDEND = QUOTIENT X DIVISIOR + REMAINDER 243= 121 x 2 + 1 = 242 + 1 = 243

  7. EXERCISE- 8.2 1.Copy the sums in your notebook,Work out each sum and verify the Solution. • 9 78 • 5 960 • 5 448 • 2 2153 • 6 5107 • 8 9607

  8. DIVISION BY 10 ,20,30----100 8.4 DIVISION BY 10 AND 100 STUDY THE FOLLOWING EXAMPLE: 5 16 3 10)---- 10)--------- 100)---------- 59 167 317 - 50 -100 -300 ------- --------- ------------- 9 67 17 -60 --------- 7

  9. FROM THE ABOVE FOUR EXAMPLES, WE OBSERVE: WHEN WE DIVIDE NUMBER BY 10,THE QUOTIENT IS OBTAINED BY REMOVING THE ONE'S DIGIT.THE ONE'S DIGIT OF THE DIVIDEND IS THE REMAINDER. BY EXTENDING THE ABOVE FACTS, WE ALSO HAVE A DIVISION BY 100. IT IS STATED AS FOLLOWS: WHEN WE DIVIDE NUMBER 100, THE QUOTIENT IS OBTAINED BY REMOVING THE TEN'S AND ONE'S DIGITS.THE NUMBER FORMED OF THE TEN'S AND ONE'S DIGIT OF THE DIVIDEND IS THE REMAINDER.

  10. 8.5 DIVISION OF A NUMBER BY 10,20,30,40............90 1. CONSIDER 60  20 METHOD 1. SINCE 20= 2 x 10, TO DIVIDE 60 BY 20, WE DIVIDE 60 FIRST BY 2 AND THEN QUOTIENT SO OBTAINED BY 10. 30 3 20) 60 FOLLOWED BY 10)30 - 60 - 30 0 0

  11. 2. CONSIDER 160 ) 40 METHOD- 2 160 40 4 40) 160 - 160 0

  12. EXERCISE: - 8.4 FIND THE QUOTIENT 1. 3010 = 2. 5710= 3. 18030= 4. 80020= 5. 10010=

  13. EXERCISE 8.5 FIND THE ESTIMATED QUITIENTS FOR THE FOLLOWING DIVISION - 1. 71523= 2. 27525= 3. 27931= 4. 63233=

  14. 8.7 DIVISION BY A TWO DIGIT NUMBER 1.LET US DIVIDE 87 BY 13. WE TAKE THE FOLLOWING STEPS. STEP 1: DETERMINING ESTIMATED QUOTIENT 87)13 IS APPROXIMATE 90)10=9)1=9 THUS,THE ESTIMATED QUOTIENT IS 9.

  15. STEP 3: SOLVING THE SUM: 6 13)----- 87 - 78 (13 x 6) ------------------- 9 CHECK: HERE, DIVIDEND=87,DIVISOR=13,QUOTIENT=6 and remainder =9. NOW DIVISOR x QUOTIENT + REMAINDER = 13 x 6 + 9=78 + 9 =87=DIVIDEND

  16. EXERCISE 8.6 1.COPY THE SUM IN YOUR NOTEBOOK.WORK OUT EACH SUM AND VERIFY THE SOLUTION 1. 98 27 = 2. 8924= 3. 82634= 4. 76845= 5.73535=

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