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

## 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.

1. 3 + 9 × 8 = 96
2. 7 × 20 – 4 = 136
3. 40 + (56 – 6) ÷ 2 = 45

Solution:

1. False
2. True
3. 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Savings = Rs 7,50,250 Cash withdrawn = Rs 5,34,500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?
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) ×