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## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.3

Question 1.

Fill in the blanks.

- If Arulmozhi saves ₹ 12 per day then she saves ₹ ____ in 30 days.
- If a person ‘A’ earns ₹ 1800 in 12 days, then he earns ₹ ____ in a day.
- 45 ÷ (7 + 8) – 2 = _____

Solution:

- 12 × 30 = ₹ 360
- \(\frac{1800}{12}=150\)
- 45 ÷ 15 – 2 = 3 – 2 = 1

Question 2.

Say True or False.

- 3 + 9 × 8 = 96
- 7 × 20 – 4 = 136
- 40 + (56 – 6) ÷ 2 = 45

Solution:

- False
- True
- False

Question 3.

The number of people who visited the Public Library for the past 5 months was 1200, 2000, 2450, 3060 and 3200 respectively. How many people visited the library in the last 5 months.

Solution:

People visited the library for past 5 months = 1200 + 2000 + 2450 + 3060 + 3200 = 11910

Total people visited = 11910

Question 4.

Cheran had a bank savings of Rs 7,50,250. He withdrew Rs 5,34,500 for educational purpose. Find the balance amount in his account.

Solution:

Savings = Rs 7,50,250 Cash withdrawn = Rs 5,34,500

Balance amount = Rs 7,50,250 – Rs 5,34,500 = Rs 2,15,750

Question 5.

In a cycle factory, 1560 bicycles were manufactured every day. Find the number of bicycles manufactured in 25 days.

Solution:

Number of bicycles manufactured in one day = 1560

Number of bicycles manufactured in 25 days = 1560 × 25 = 39000

Question 6.

Rs 62500 was equally distributed as a New Year bonus for 25 employees of a company. How much did each receive?

Ans:

Total amount = Rs 62500

Total number of employees = 25

Bonus amount received by each employee = Rs 62500 ÷ 25

= RS \(\frac{62500}{25}\)

= Rs 2500

Question 7.

Simplify the following numerical expression:

(i) (10 + 17) ÷ 3

(ii) 12 – [3 – {6 – (5 – 1)}]

(iii) 100 + 8 ÷ 2 + {(3 × 2) – 6 ÷ 2}

Solution:

(i) (10 + 17) ÷ 3 (Given)

= 27 ÷ 3 (Bracket completed first)

= 9 (÷ completed)

∴ (10 + 17) ÷ 3 = 9

(ii) 12 – [3 – {6 – (5 – 1)}] (Given)

= 12 – [3 – {6 – 4}] (Innermost bracket completed first)

= 12 – [3 – 2] (Again Inner bracket completed second)

= 12 – 1 (Bracket completed third)

= 11 (- completed)

∴ 12 – [3 – {6 – (5 – 1)}] = 11

(iii) 100 + 8 ÷ 2 + {(3 × 2) – 6 ÷ 2} (Given)

= 100 + 8 ÷ 2 + {6 – 6 ÷ 2} (Innermost bracket completed first)

= 100 + 8 ÷ 2 + {6 – 3} (To remove the next bracket ÷ within the bar completed second)

= 100 + 8 ÷ 2 + 3 (bar completed third)

= 100 + 4 + 3 (÷ completed fourth)

= 107 (+ completed)

∴ 100 + 8 ÷ 2 + {(3 × 2) – 6 ÷ 2} = 107

**Objective Type Questions**

Question 8.

The value of 3 + 5 – 7 × 1 is …….

(a) 5

(b) 7

(c) 8

(d) 1

Solution:

(d) 1

Question 9.

The value of 24 ÷ {8 – (3 × 2)} is ______

(a) 0

(b) 12

(c) 3

(d) 4

Answer:

(b) 12

24 ÷ {8 – 3 × 2} = 24 ÷ {8 – 6} = 24 ÷ 2 = 12

Question 10.

Use BIDMAS and put the correct operator in the box.

2₹6 – 12 ÷ (4 + 2) = 10

(a) +

(b) –

(c) ×

(d) ÷

Solution:

(c) ×