Sunday, 4 March 2012

CBSE Class 6th ( VI) Mathematics Chapter 2. Whole Numbers : Exercise 2.2

Question 1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647

Answer:
(a) 837+208+363
  = (837+363)+208
= 1200+208
= 1408



(b) 1962 + 453 + 1538 + 647
 = 1962 + 647+ 453 + 1538
= (1962 +1538) + (647+453)
= (3500)+(1100)
= 4400

Question 2. Find the product by suitable rearrangement:
(a) 2 ×1768× 50 (b) 4 × 166 × 25 (c) 8 × 291 × 125
(d) 625 ×279 × 16 (e) 285 × 5 ×60 (f) 125 × 40 × 8 × 25

Answer:
(a)  2× 1768 ×50
 = 2 × 50 ×1768
= 100 ×1768
= 176800

(b)  4 × 166 ×25
 = 4 × 25 ×166
= 100 × 166
= 16600

(c) 8 × 291 ×125
 = 8 × 125 ×291
= 1000 × 291
= 291000

(d) 625 × 279×16
 = 625 ×16 ×279
= 10000 ×279
= 2790000

(e) 285 ×5 ×60
 = 285 ×300
= 285×100×3
= 285×3×100
= 855×100
= 85500

(f) 125 ×40 ×8 × 25
= 125 × 8 × 40 × 25
= 1000 ×1000
= 1000000


Question 3. Find the value of the following:
(a) 297 × 17 + 297 × 3 (b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 ×69 (d) 3845 × 5 × 782 + 769 × 25 ×218

Answer:

(a) 297 ×17 + 297 ×3
 = 297 × ( 17 +3 )
= 297×(20)
= 297×10×2
= 2970×2
= 5940


(b)  54279 × 92 + 8 × 54279
 = 54279 × (92+8)
= 54279 × (100)
= 54279 ×100
= 5427900


(c)  81265 × 169 – 81265 ×69
 = 81265 ×( 169 – 69)
= 81265 ×(100)
=81265 ×100
=8126500

(d)  3845 × 5 × 782 + 769 × 25 ×218
= 769 ×5×5 × 782 + 769 × 25 ×218
= 769 ×(5×5 ×782+25 ×218)
= 769 ×(25× 782+25 ×218)
= 769 ×25(782+218)
= 769 ×25(1000)
= 769 ×25 ×1000
= 19225000

Question 4. Find the product using suitable properties.
(a) 738 ×103 (b) 854 × 102 (c) 258 × 1008 (d) 1005 × 168

Answer:

(a)  738 ×103
 = 738×(100+3)
= 738×100 + 738×3
= 73800 + 2214
= 76014


(b)  854 × 102
 = 854 ×(100+2)
= 854×100 + 854×2
= 85400 + 1708
= 87108


(c)  258 × 1008
 = 258×(1000+8)
= 258×1000 + 258×8
= 258000 + 2064
= 260064


(d) 1005 × 168
 = (1000+5)×168
= 1000 ×168+168×5
= 168000 + 840
= 168840


Question 5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday.
The next day,he filled the tank with 50 litres of petrol. If the petrol costs
Rs 44 per litre, how much did he spend in all on petrol?

Answer:
A taxidriver filled his car petrol tank on Monday= 40 litres
A taxi driver filled his car petrol tank on next day= 50 litres
The totle litres of petrol he filled his car petrol tank= 40 + 50 = 90 litres
The price of one liter of petrol=Rs 44
There for the price of 90 lirtes of petrol=Rs 90 × 44=Rs 3960

Question 6. A vendor supplies 32 litres of milk to a hotel in the morning and 68litres of milk in the evening. If the milk

costs
Rs 15 per litre, howmuch money is due to the vendor per day?

Question 7. Match the following:
(i) 425 ×136 = 425 × (6 + 30 +100)(a)Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49(b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005(c) Distributivity of multiplication over addition.
Answer:
(i) 425 ×136 = 425 × (6 + 30 +100)(c) Distributivity of multiplication over addition.
(ii) 2 × 49 × 50 = 2 × 50 × 49(a) Commutativity under multiplication.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005(b) Commutativity under addition.

25 comments:

  1. very helpful for parents thank you

    ReplyDelete
    Replies
    1. Yes very helpful thank you

      Delete
    2. I have some problems in maths can you please help me

      Delete
    3. the sum by suitable rearrangement: - 248 + 165 + 352 + 135

      Delete
  2. Thanks for the help
    Yashita Thakur

    ReplyDelete
  3. The correct answer for question 1 part b is 4600

    ReplyDelete
  4. (b) 1962 + 453 + 1538 + 647
    = 1962 + 647+ 453 + 1538
    = (1962 +1538) + (647+453)
    = (3500)+(1100)
    = 4400 - pls make it 4600

    ReplyDelete
  5. I like it thanks for this because I can understand this sum and now I understand it breafle.

    Thanks

    ReplyDelete
  6. Can you give me the and of this sum 81265×169-81265×69

    ReplyDelete
  7. Very very helpful to parents thank you so much

    ReplyDelete
  8. 3845×5×782+3845×5×218 sir i am not understand solve this problem.

    ReplyDelete
    Replies
    1. 3845 × 5 × ( 782 + 218)
      19225 × 1000
      = 19225000

      Delete
  9. 493×99 find the suitable properties

    ReplyDelete
  10. 4x65x25 in suitable rearrangement

    ReplyDelete
  11. where is answer of the question number 6???

    ReplyDelete
  12. Find the sum by suitable rearrangement:I
    In b part your answer is 4400 and my class teacher gave the answer is 4600.

    ReplyDelete
  13. Very nice it helped me a lot

    ReplyDelete
  14. can u pls tell the anwser of Determine the given product by suitable rearrangement

    155 x 102

    ReplyDelete