Ncert Solutions For Class 6 Maths Ex 2.2

Ncert Solutions For Class 6 Maths Chapter 2 Ex 2.2

Question 1:

Frame the numbers in proper order and find the sum

a) 820 + 200 +325

b) 1960 + 448 + 1520 + 638

Answer:

a) (800 + 325) + 200 = 1325

b) (1960 +1520) + (448 + 638) = 3480 + 1086 = 4566

 

Question 2:

Find the product by framing it in suitable order

a) 2\(\times\) 1750\(\times\) 45

b) 3\(\times\) 165\(\times\)25

c) 6\(\times\)290\(\times\)120

d) 600\(\times\)286\(\times\)18

e) 286\(\times\)8\(\times\)70

f) 130\(\times\)45\(\times\)6\(\times\)30

Answer:

a) 2\(\times\)1750\(\times\)45

= (2\(\times\)45) \(\times\)1750 = 90\(\times\)1750 = 157,500

b) 3\(\times\)165\(\times\)25

= (3\(\times\)25) \(\times\)165 = 75\(\times\)165 = 12,375

c) 6\(\times\)290\(\times\)120

= (6\(\times\)120) \(\times\)290 = 720\(\times\)290 = 208,800

d) 600\(\times\)286\(\times\)18

= (600\(\times\)18) \(\times\)286 = 10800\(\times\)286 = 3,088,800

e) 286\(\times\)8\(\times\)70

= 286\(\times\) (8\(\times\)70) = 286\(\times\)560 = 160,160

f) 130\(\times\)45\(\times\)6\(\times\)30

= (130\(\times\)6) \(\times\) (45\(\times\)30) = 780\(\times\)1350 = 1,053,000

 

Question 3:

Find the solution of the following

a) 290\(\times\)15 + 290\(\times\)5

Answer:

Since 290 is repeated, we can take it as a common term

290\(\times\) (15+5) = 290\(\times\)20 = 5800

 

b) 54270\(\times\)95 + 6\(\times\)54270

Answer:

Since 54270 is repeated, we can take it as a common term

54270\(\times\) (95+6) = 54270\(\times\)101 = 5,481,270

 

c) 8250\(\times\)62 – 8250\(\times\)50

Answer:

Since 8250 is repeated, we can take it as a common term

8250\(\times\) (62-50) = 8250\(\times\)12 = 99,000

 

d) 3845\(\times\)5\(\times\)720 + 769\(\times\)25\(\times\)220

Answer:

= 3845\(\times\)5\(\times\)720 + 769\(\times\)5\(\times\)5\(\times\)220

= 3845\(\times\)5\(\times\)720 + (769\(\times\)5)5\(\times\)220

= 3845\(\times\)5\(\times\)720 + 3845\(\times\)5\(\times\)220

Since 3845 and 5 is repeated, we can take it as a common term

= 3845\(\times\)5\(\times\) (720 +220)

= 19,225\(\times\)940

= 18,071,500

 

Question 4:

Use suitable properties to find out the product

a) 642\(\times\)105

Answer:

= 642\(\times\) (100+5)

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 642\(\times\)100 + 642\(\times\)5

= 64200 + 3210

= 67,410

 

b) 850\(\times\)103

Answer:

= 850\(\times\) (100+3)

By distributive law,

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 850\(\times\)100 + 850\(\times\)3

= 85000 + 2550

= 87,550

 

c) 260\(\times\)1006

Answer:

= 260\(\times\) (1000+6)

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 260\(\times\)1000 + 260\(\times\)6

= 260000 + 1560

= 261,560

 

d) 1008\(\times\)168

Answer:

= (1000\(\times\)8) \(\times\)168

By distributive law

A\(\times\) (B+C) = A\(\times\)B + A\(\times\)C

= 1000\(\times\)168 + 8\(\times\)168

= 168000 + 1344

= 169,344

 

Question 5:

A cab driver filled his vehicle tank with 35litres of diesel on Tuesday. The next day, he filled the tank with 45litres of diesel. If the diesel costs Rs.32 per litre. How much did he spend on diesel?

Answer:

The quantity of diesel filled on Tuesday = 35L

The quantity of diesel filled on Wednesday = 45L

The total quantity filled = (35+45)L

Cost of diesel (per L) = Rs.32

The total money spent on diesel = 32\(\times\) (35+45)

= 32\(\times\)80

= Rs.2560

The total amount spent on diesel = Rs.2560

Question 6:

A milkman supplies 30 liters of milk to a teashop in the morning and 60 liters of milk in the evening. If the cost of milk is 12 per liter, how much does the milkman earn a day?

Answer:

The quantity of milk supplied in the morning = 30L

The quantity of milk supplied in the evening = 60L

The total quantity filled = (30+60)L

Cost of milk (per L) = Rs.12

(L represents litres)

The total money spent on diesel = 12\(\times\) (30+60)

= 12\(\times\)90

= Rs.1080

The total amount spent on diesel = Rs.1080

Question 7:

Write the answers using correct property

a) 424\(\times\)128

Answer:

424\(\times\)128 = 424\(\times\) (100+20+8)

This is distributive property of multiplication over addition

 

b) 2\(\times\)49\(\times\)50

Answer:

2\(\times\)48\(\times\)51 = 2\(\times\)51\(\times\)48

This is commutative property under multiplication

 

c) 82+2000+20

Answer:

82+2000+20 = 82+20+2000

This is commutative property under addition

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