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.

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