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

Question 1.

Fill in the Blanks.

(i) If Arulmozhi saves Rs 12 per day, then she saves Rs ………. in 30 days.

(ii) If a person ‘A’ earns Rs 1800 in 12 days, then he earns Rs ……… in a day.

(iii) 45 + (7 + 8) – 2 = ………

Solution:

(i) Rs 150

(ii) Rs. 360

(iii) 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:

Total number of people who visited the public library for the past 5 months

= 1200 + 2000 + 2450 + 3060 + 3200

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Savings = Rs 7,50,250 Cash withdrawn = Rs 5,34,500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

Answer:

Total amount distributed = ₹ 62500

Number of employees received bonus = 25

Amount received by one employee = 62500 ÷ 25 = 2,500.

Each employee received = ₹ 2,500

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

Solution:

(b) 12

Question 10.

Use BODMAS and put the correct operator in the box.

2¤6 – 12 ÷ (4 + 2) = 10

(a) +

(b) –

(c) ×

(d) ÷

Solution:

(c) ×