**Collins Maths Solutions Class 7 Chapter 4 Exponents**

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 4, Exponents. Here students can easily find chapter 4 solutions with exercise wise explanation. Students will find proper solutions for Exercise 4.1 and 4.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.

**Exponents Exercise 4.1 Solution :**

**Question no – (1) **

**Solution : **

**(a) (5) ^{0}**

= 1

**(b) 5 ^{12} × 5^{9}**

= 5 ^{(12 + 9)}

= 5^{21}

**(c) (9) ^{5}**

= 59049

**(d) (2) ^{8} ÷ (2)^{5}**

= 2^{8}/2^{5}

= 2^{8} ^{– 5}

= 2^{3}

= 8

**(e) (5 ^{3})^{9}**

= 5^{27}

**(f) (6 ^{5})^{2}**

= 6^{10}

**Question no – (2) **

**Solution : **

**(a) 3 ^{0} × 3^{1}**

= 1 × 3

= 3

**(b) 15 ^{5} ÷ 5^{3}**

= (5 × 3)^{5} /5^{3}

= 5^{2} × 3^{5}

= 25 × 243

= 6075

**(c) (-2) ^{12} × (-2^{2})^{5} × (-2^{3})^{9}**

= (-2) ^{12 + 10 27}

= (-2) ^{49}

**(d) (6/3) ^{3} × (9/4)^{3}**

= (6/3 × 9/4)^{3}

= (9/2)^{3}

= 729/8

**(e) 42 ^{3} × 30^{3}/64 × 729 × 35^{3}**

= (42 × 30)^{3}/64 × 729 35^{3}

= (42 × 30)^{3}/(4 × 9)^{3} × 35^{3}

= (9/5)^{3}

**Exponents Exercise 4.2 Solution : **

**Question no – (1) **

**Solution : **

Total capacity of boilers,

= (9^{3} × 9^{5})

= 9^{(3 +5)
}

= 9^{8
}

Therefore, total capacity of the boilers will be 9^{8} tones

**Question no – (2) **

**Solution :**

Volume,

= (2^{6} × 3^{4} × 2^{3})

= 2^{6+3} × 3^{4}

= (2^{9} × 3^{4})

Therefore, the volume will be (2^{9} × 3^{4})

**Question no – (3) **

**Solution :**

Total number of words,

= (10^{3} × 10^{4}

= 10^{(3+4)
}

= 10^{7}

Therefore, Mohit can type 107 words in 104 hours.

**Next Chapter Solution : **

👉 Chapter 5 👈