2. Whole Numbers Mathematics Exercise - 2.2 class 6 Maths in English - CBSE Study
NCERT Solutions for Class 6 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Whole Numbers with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.2 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 6 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 6 English Medium Mathematics All Chapters:
2. Whole Numbers
2. Exercise 2.2
Exercise-2.2
Q1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
Solution: 837 + 208 + 363
= (837 + 208) + 363
= 1045 + 363
= 1408
Or Second Method
= 837 + 363 + 208
= (837 + 363) + 208
= 1200 + 208
= 1408
(b) 1962 + 453 + 1538 + 647
Solution: 1962 + 453 + 1538 + 6477
= (1962 + 453) + (1538 + 647)
= 2415 + 2185
= 4600
Or Second Method
= 1962 + 453 + 1538 + 6477
= (1538 + 1962) + (453 + 647)
= 3500 + 1100
= 4600
Q2. 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
Solution:
(a) 2 x 1768 x 50
Solution:
2 x 50 x 1768
= (2 x 50) x 1768
= 100 x 1768
= 176800
(b) 4 x 166 x 25
Solution:
4 x 25 x 166
= (4 x 25) x 166
= 100 x 166
= 16600
(c) 8 x 291 x 125
Solution:
8 x 125 x 291
= 1000 x 291`
= 291000
(d) 125 x 40 x 8 x 25
Solution:
125 x 8 x 40 x 25
= (125 x 8) x (40 x 25)
= 1000 x 1000
= 1000000
(e) 285 × 5 × 60
Solution:
285 × 5 × 60
= 285 × (5 × 60)
= 285 × 300
= 85500
(f) 125 × 40 × 8 × 25
Solution:
125 × 40 × 8 × 25
= (125 × 8) × (40 × 25)
= 1000 × 1000
= 1000000
Q3. Find the value of the following:
(a) 297 × 17 + 297 × 3
Solution:
297 x (17 + 3)
= 297 x 20
= 5940
(b) 54279 × 92 + 8 × 54279
Solution:
54279 x (92 + 8)
= 54279 x 100
= 5427900
(c) 81265 × 169 – 81265 × 69
Solution:
81265 × 169 – 81265 × 69
= 81265 (169 – 69)
= 81265 (100)
= 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
Solution:
3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218
= 3845 × 5 × 782 + 3845 × 5 × 218
= 3845 × 5 (782 + 218)
= 19225 × 1000
= 19225000
Q4. Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
Solution:
(a) 738 × 103
= 738 (100 + 3)
= 738 × 100 + 738 × 3
= 73800 + 2214
= 76014
Solution:
(b) 854 × 102
= 854(100 + 2)
= 854 × 100 + 854 × 2
= 85400 + 1708
= 87108
Solution:
(c) 258 × 1008
= 258(1000 + 8)
= 258 × 1000 + 258 × 8
= 258000 + 2064
= 260064
Solution:
(d) 1005 × 168
= (1000 + 5) x 168
= 1000 x 168 + 5 x 168
= 168000 + 840
= 168840
5. A taxidriver 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?
Solution: Taxi driver filled his car petrol tank on Monday = 40 L
He filled the petrol tank on Tuesday = 50 L
The petrol costs per liter = 44 Rs
He spend in all on petrol = (40 + 50) x 44
= 90 x 44
= 3960 Rs
6. A vendor supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs Rs 15 per liter, how much money is due to the vendor per day?
Solution: A vendor supplies milk to a hotel in the morning = 32 liters
=> A vendor supplies milk to a hotel in the evening = 68 liters
=> The cost of milk per liter = Rs 15
7. Match the following:
Solution:
(i) ----------- (c)
(ii) ---------- (a)
(iii) ---------- (b)
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