**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Savings $=$ Rs $7,50,250$ Cash withdrawn $=$ Rs $5,34,500$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) $x$