NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Class 6 Maths Chapter 2: NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

1. Find the sum by suitable rearrangement:
(A) 837+208+363
(B) 1962+453+1538+647

Solution:
(A) Given, 837+208+363
By using associativity of addition, we can rewrite the question as follows:
=(837+363)+208
=1200+208
=1408
Hence, the required sum is 1408

(B) Given,1962+453+1538+647
By using associativity of addition, we can rewrite the question as follows:
=(1962+1538)+(453+647)
=3500+1100
=4600
Hence, the required sum is 4600


2 Find the product by suitable arrangement:
(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

Solution:
(A) Given, 2×1768×50
By using the associativity of multiplication, the above question can be rewritten as follows: =(2×50)×1768 =100×1768 =176800
Hence, the required product is 176800

(B) Given, 4×166×25
By using the associativity of multiplication, the above question can be rewritten as follows:
=(4×25)×166
=100×166
=16600
Hence, the required product is 16600

(C) Given, 8×291×125
By using the associativity of multiplication, the above question can be rewritten as follows: =(8×125)×291
=1000×291
=291000
Hence, the required product is 291000

(D) Given, 625×279×16
By using the associativity of multiplication, the above question can be rewritten as follows: =(625×16)×279
=10000×279
=2790000
Hence, the required product is 2790000

(E) Given, 285×5×60
By using the associativity of multiplication, the above question can be rewritten as follows:
=284×(5×60) =284×300 =85500
Hence, the required product is 85500

(F) Given, 125×40×8×25
By using the associativity of multiplication, the above question can be rewritten as follows:
=(125×8)×(40×25)
=1000×1000
=1000000
Hence, the required product is 1000000


3. Find the value of the following:
(A) 297×17+297×3
(B) 54279×92+8×54279
(C) 8126×169−81265×69
(D) 3845×5×782+769×25×218

Solution:

(A) Given, 297×17+297×3
By using the principle of distributivity of multiplication over addition, we can rewrite the given question as follows:
=297×(17+3)
=297×20 =5940
Hence, the required value is 5940

(B) Given, 54279×92+8×542379
By using the principle of distributivity of multiplication over addition, we can rewrite the given question as follows:
=54279×(92+8)
=54279×100
=5427900
Hence, the required value is 5427900

(C) Given, 81265×169−81265×69
By using the principle of distributivity of multiplication over subtraction, we can rewrite the given question as follows:
=81265×(169−69)
=81265×100
=8126500
Hence, the required value is 8126500

(D) Given, 3845×5×782+769×25×218
=3845×5×782+769×5×5×218
= 3845×5×782+3845 ×5×218
=3845×5×(782+218)
=3845×5×1000
=19225000
Hence, the required value is 19225000


4. Find the product using suitable properties:
(A) 738×103
(B) 854×102
(C) 258×1008
(D) 1005×168

Solution:

(A) Given, 738×103
=738×(100+3) [Using distributive property]
=738×100+738×3
=73800+2214
=76014
Hence, the required product is 76014

(B) Given, 854×102
=854×(100+2) [Using distributive property]
=854×100+854×2
=85400+1708
=87108
Hence, the required product is 87108

(C) Given, 258×1008
=258×(1000+8) [Using distributive property]
=258×1000+258×8
=258000+2064
=260064
Hence, the required product is 260064

(D) Given, 1005×168
=(1000+5)×168 [Using distributive property]
=1000×168+5×168
=168000+840
=168840
Hence, the required product is 168840


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 ₹ 44 per litre, how much did he spend in all on petrol?

Solution:

Given,
Quantity of petrol filled on Monday=40 litres
Quantity of petrol filled on next day =50 litres
Total petrol filled =40+50=90 litres
Now,
Cost of 1-litre petrol =₹ 44
Cost of 90 litres petrol =44×90
=44×(100−10) [Using distributive property]
=44×100−44×10
=4400−440
=₹ 3960
Therefore, he spent ₹ 3960 on petrol.


6. A vendor supplies 32 litres of milk to a hotel in a morning and 68 litres of milk in the evening. If the milk costs ₹ 15 per litre, how much money is due to the vendor per day?

Solution:

Given,
Milk supplied in the morning =32 litres
Milk supplied in the evening =68 litres
Total supply =32+68=100 litres

Now
Cost of 1 litre of milk =₹ 15
Cost of 100 litres of milk =15×100=₹ 1500
Therefore, ₹ 1500 is due to the vendor per day.


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

Solution:

(i) We know that, Distributivity of multiplication over addition:
a×(b+c)=a×b+a×c
Similarly,
a×(b+c+d)=a×b+a×c+a×d
Therefore, 425×136=425×(6+30+100)
follows distributivity of multiplication over addition property.

(ii) We know that Commutativity under multiplication:
a×b=b×a
Similarly,
a×b×c=a×c×b=b×c×a
Therefore, 2×49×50=2×50×49 follows commutativity under the multiplication property.

(iii) We know that Commutativity under addition:
a+b=b+a
Similarly,
a+b+c=c+b+a=c+a+b
Therefore, 80+2005+20=80+20+2005 follows commutativity under addition property.

(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