WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Example Problems

Chapter 1 Simplification Solved Example Problems

Question 1: Simplify each of the cases (1), (2), (3), (4) Verify whether the results of the cases are equal or not.

1.10 + 8 ÷ (5 – 2)

Solution:

Given

10 + 8 ÷ (5 – 2)

= 10 + 8 ÷ 3

= 10 + \(\frac{8}{3}\)

= \(\frac{10+3}{3}\)

= \(\frac{38}{3}\)

10 + 8 ÷ (5 – 2)  = \(\frac{38}{3}\)

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

WBBSE Class 6 Simplification Example Problems

2. (10 + 8) ÷ (5 – 2)

Solution:

Given

(10 + 8) ÷ (5 – 2)

= 18 ÷ 3

= 6.

(10 + 8) ÷ (5 – 2) = 6.

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Example Problems

3. (10 – 8)(5 – 2)

Solution:

Given

(10 – 8)(5 – 2)

= (2)(3)

= 6

(10 – 8)(5 – 2) = 6

Short Questions on Simplification Problems

4. 10 – 8 (5 – 2).

Solution:

Given

10 – 8 (5 – 2)

= 10 – 8 (3)

= 10 – 24

= – 14.

10 – 8 (5 – 2) = – 14.

Question 2.

Simplify :

1. (12 – 2) ÷ 2,

2. {90 – (48 – 21)} ÷ 7

Solution: 

Given

1. (12 – 2) ÷ 2

= 10 ÷ 2

= 5.

(12 – 2) ÷ 2 = 5.

2. {90 – (48 – 21)} ÷ 7

= {90 – 27} ÷ 7

= 63 ÷ 7

= 9.

{90 – (48 – 21)} ÷ 7= 9.

Common Simplification Problems and Solutions

Question 3

Evaluate:

1. (72 ÷ 8 x 9) – (72 ÷ 8 of 9)

2. {25 x 16 ÷ (60 ÷ 15) – 4 x (77 – 62)} ÷ (20 x 6 + 3)

3. 200 ÷ [88 – {(12 x 13) – 3 x (40 – 9)}]

Solution:

Given

1. (72 ÷ 8 x 9)- (72 ÷ 8 of 9)

= (9 x 9) – (72 ÷ 72)

= 81 – 1

= 80.

(72 ÷ 8 x 9)- (72 ÷ 8 of 9) = 80.

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2. {25 x 16 ÷ (60 + 15) – 4 x (77 – 62)} ÷ (20 x 6 ÷ 3)

= {25 x 16 + 4 – 4 xl5} ÷ (20 x 2)

= {25 x 4 – 4 x 15} ÷ 40

= {100 – 60} ÷ 40

= 40 v 40

= 1

{25 x 16 ÷ (60 + 15) – 4 x (77 – 62)} ÷ (20 x 6 ÷ 3) = 1

Practice Problems on Simplification with Solutions

3. 200 ÷ [88 – {(12 x 13) – 3 x (40 – 9)}]

= 200 ÷ [88 – {156 – 3 x 31}] – 200 + [88 – {156 – 93}]

= 200 ÷ [88 – 63]

= 200 ÷ 25

= 8.

200 ÷ [88 – {(12 x 13) – 3 x (40 – 9)}] = 8.

Question 4

1. 256 ÷ \(\overline{16 \div 2}\) ÷ \(\overline{18 \div 9}\)  x 2

Solution :

256 + 16 + 2 + 18 + 9 x 2

= 256 + 8 – 2 x2

= 32 + 2 x 2

= 16 x 2

= 32.

256 + 16 + 2 + 18 + 9 x 2 = 32.

Class 6 Simplify Questions

2. 76 – 4 – [6 + {19 – (48 – 57-17)}] \(\overline{57 – 17}\)

Solution:

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 4 Q 2

 

3. [16 + {42 – \(\overline{38 + 2}\) }]12 + (24 + 6) x 2 + 4.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 4 Q 3

Examples of Real-Life Applications of Simplification

Question 5. Evaluate:

1. 4 x [24 – {(110 – \(\overline{11 + 3}\) x 4) + 9}] ÷ 2 of 9

Solution:

4 x [24 – {(110 – \(\overline{11 + 3}\) x 4) + 9}] ÷ 2 of 9

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 1

 

2. (987 – \(\overline{43 + 25}\)) – 10 [5 + {(999 ÷ \(\overline{9 * 3}\)) + (\(\overline{8 * 9}\) ÷ 6) 4 }].

Solution:

(987 – \(\overline{43 + 25}\)) – 10 [5 + {(999 ÷ \(\overline{9 * 3}\)) + (\(\overline{8 * 9}\) ÷ 6) 4 }].

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 2

Conceptual Questions on BODMAS and Simplification

Example 6. Evaluate:

[latex100-[60 \div \overline{3+2}\} \div\{(5 \text { of } 3) \div \overline{1+4}\}] \text { of } \overline{12+13}[/latex]

Solution:

\(100-[60 \div \overline{3+2}\} \div\{(5 \text { of } 3) \div \overline{1+4}\}] \text { of } \overline{12+13}\)

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6

 

Example 7. Simplify:

\(4-[4+\{4-(\overline{4-4}) \text { of } 4\} \div 4]\)

Solution:

4-[4+\{4-(\overline{4-4}) \text { of } 4\} \div 4]

Simplification Questions For Class 6

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7

Example 8. Simplify:

\(a-[a+\{a-(a+\overline{a-a}) \div a\}-a]\)

Solution:

\(a-[a+\{a-(a+\overline{a-a}) \div a\}-a]\)

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 8

Real-Life Scenarios Involving Mathematical Problem Solving

Question 9. Express the following in Mathematical Language and then solve it:

Father of Subhas plucked 125 guavas from their guava orchard and sold them at the rate of 2 each. Then he purchased 2 pens at the cost of 5 each and 2 exercise books at the cost of 20 each from the money he got by selling the navas. Then the remaining money was distributed between Subhas and his sister equally for eating sweets. How much money did Subhas receive?

Solution:

Expressing the given information in the language of mathematics, we get

Subhas received = [125 x 2 -((5 x 2) + (20 × 2)}] ÷ 2

Solution is Subhas received = [125 x 2 -((5 x 2) + (20 × 2)}] ÷ 2

= [250 (10+40)] ++ 2

= [125 x 2 -((5 x 2) + (20 x 2)}] ÷ 2 [25050] ÷ 2 

= [200] ÷ 2 

= 100.

∴ Subhas received 100.

 

Example 10. Express the following statement in Mathematical Language and then solve it:

Debapriyo, a student of class Six, received 10,000 as the first prize in the talent search competition in the village. Then he went to his house and he gave to his mother half the prize money and to his elder sister studying Physics Honours 1 half of the remaining money. After this, he purchased a watch by \(\frac{1}{20}\)th of the 20 prize money and purchased 10 pens at a cost of 5 each, 10 exercise books at a cost of 20 each and a book at a cost of 150 for his youngest brother. After these expenses, he kept the remaining money with his father for savings. What was the savings of Debapriyo?

Solution:

Expressing the given information in the mathematical language, we get

The savings of Debapriyo

= [{10,000 – (10,000 ÷ 2)) (10,000 (10,000 ÷ 2)} ÷ 2] – [(10,000 ÷ 20) + (5 x 10+ 20 x 10 + 150)]

= [{10,000 – 5000) – (10,000 – 5000) ÷ 2] – [500+ (50+ 200+ 150)]

= [5000 – 5000 ÷ 2] – [500 +400]

= [ 5000 – 2500] – 900 

= 2500 – 900

= 1600

The savings of Debapriyo was 1600.

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