Exercise 1.3 - Chapter 1 Numbers Term 1 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 1.3
Question $1 .$
Fill in the blanks.
1. If Arulmozhi saves $₹ 12$ per day then she saves $₹$ in 30 days.
2. If a person 'A' earns ₹ 1800 in 12 days, then he earns ₹ in a day.
3. $45 \div(7+8)-2=$
Solution:
1. $12 \times 30=₹ 360$
2. $\frac{1800}{12}=150$
3. $45 \div 15-2=3-2=1$
 

Question $2 .$
Say True or False.
1. $3+9 \times 8=96$
2. $7 \times 20-4=136$
3. $40+(56-6) \div 2=45$
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:
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Savings $=$ Rs $7,50,250$ Cash withdrawn $=$ Rs $5,34,500$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 $=1560 \times 25=39000$
$\quad 1560$
$\times \quad 25$
$\quad 7800$
$\quad 3120$
$\underline{39,000}$

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 $62500 \div 25$
$=\operatorname{RS} \frac{62500}{25}$
$=$ Rs 2500

Question $7 .$
Simplify the following numerical expression:
(i) $(10+17) \div 3$
(ii) $12-[3-\{6-(5-1)\}]$
(iii) $100+8 \div 2+\{(3 \times 2)-6 \div 2\}$
Solution:
(i) $(10+17) \div 3$ (Given)
$=27 \div 3$ (Bracket completed first)
$=9(\div$ completed $)$
$\therefore(10+17) \div 3=9$

(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 $)$
$\therefore 12-[3-\{6-(5-1)\}]=11$
(iii) $100+8 \div 2+\{(3 \times 2)-6 \div 2\}$ (Given)
$=100+8 \div 2+\{6-6 \div 2\}$ (Innermost bracket completed first)
$=100+8 \div 2+\{6-3\}$ (To remove the next bracket $\div$ within the bar completed second)
$=100+8 \div 2+3$ (bar completed third)
$=100+4+3(\div$ completed fourth $)$
$=107(+$ completed $)$
$\therefore 100+8 \div 2+\{(3 \times 2)-6 \div 2\}=107$

Objective Type Questions
Question $8 .$
The value of $3+5-7 \times 1$ is $\ldots \ldots .$
(a) 5
(b) 7
(c) 8
(d) 1
Solution:
(d) 1

Question $9 .$
The value of $24 \div\{8-(3 \times 2)\}$ is
(a) 0
(b) 12
(c) 3
(d) 4
Answer:
(b) 12
$24 \div\{8-3 \times 2\}=24 \div\{8-6\}=24 \div 2=12$
 

Question 10 .
Use BIDMAS and put the correct operator in the box. $2 ₹ 6-12 \div(4+2)=10$
(a) +
(b) -
(c) $x$
(d) $\div$
Solution:
(c) $x$