RD Sharma Solutions Class 7 Integers Exercise 1.3

RD Sharma Solutions Class 7 Chapter 1 Exercise 1.3

RD Sharma Class 7 Solutions Chapter 1 Ex 1.3 PDF Download

Exercise 1.3

Find the value of

Q1) \(36\div 6+3\).

Solution:

\(36\div 6+3=6+3\)

= \(9\)

Q2) \(24+15\div 3\).

Solution:

\(24+15\div 3=24+5\)

= \(29\)

Q3) \(120-20\div 4\).

Solution:

\(120-20\div 4=120-5\)

= 115

Q4) \(32-(3\times 5)+4\).

Solution:

32 – (3 x 5) + 4 = 32 – 15 +4

= 17 + 4

= 21

Q5) \(3-(5-6\div 3)\).

Solution:

\(3-(5-6\div 3)=3-(5-2)\)

= 3 – 3

= 0

Q6) \(21-12\div 3\times 2\).

Solution:

\(21-12\div 3\times 2=21-\frac{12}{3}\times 2\)

= \(21-4\times 2\)

= 21 – 8

= 13

Q7) \(16+8\div 4-2\times 3\).

Solution:

\(16+8\div 4-2\times 3\)

= 16 + 2 – 6

= 18 – 6

= 12

∴ \( 16+8\div 4-2\times 3=12\)

Q8) \(28-5\times 6+2\).

Solution:

\(28-5\times 6+2\) = \(28-(5\times 6)+2\)

= 28 – 30 + 2

= 30 – 30

= 0

Q9) \((-20)\times (-1)+(-28)\div 7\).

Solution:

\((-20)\times (-1)+(-28)\div 7=20+\frac{\mid -28\mid}{\mid 7\mid}\)

= \(20-\frac{28}{7}\)

= 20 – 4

= 16

Q10) \((-2)+(-8)\div (-4)\).

Solution:

\((-2)+(-8)\div (-4)=-2+\frac{\mid -8\mid}{\mid -4\mid}\)

= – 2 + 2

= 0

Q11) \((-15)+4\div (5-3)\).

Solution:

\(-15+4\div (5-3)=-15+4\div 2\)

= – 15 + 2

= – 13

\(-15+4\div (5-3)=-13\)

Q12) \((-40)\times (-1)+(-28)\div 7\).

Solution:

\((-40)\times (-1)+(-28)\div 7=40+(-4)\)

= 40 – 4

= 36

Q13) \((-3)+(-8)\div (-4)-2\times (-2)\).

Solution:

\((-3)+(-8)\div (-4)-2\times (-2)=(-3)+\frac{(-8)}{(-4)}-2\times (-2)\)

= – 3 + 2 + 4

= 6 – 3

= 3

Q14) \((-3)\times (-4)\div (-2)+(-1)\).

Solution:

\((-3)\times (-4)\div (-2)+(-1)=12\div (-2)+(-1)\)

= – 6 – 1

= – 7

∴ \( (-3)\times (-4)\div (-2)+(-1)=-7\)