NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.3

Exercise 5.3 of Chapter 5 is about solving real-life word problems based on squares and square roots. These types of questions will allow you to take everything you've learned from the previous exercises and apply them in real-life situations. This exercise adhere to the current NCERT syllabus for Class 8 within the official textbook supplied by NCERT. These questions are beneficial in school as well as improving your problem-solving capabilities for other competitive exams like the Math Olympiads.

The NCERT Solutions by ALLEN  detail every question in an effective and simple manner so that you can understand the solutions better. By working through and practicing these questions regularly, you can gain confidence and enhance your problem-solving skills.

1.0Download NCERT Solutions Class 8 Maths Chapter 5 Squares and Square Root Exercise 5.3: Free PDF

The NCERT Solutions for Class 8 Maths Chapter 5 contains problems and their detailed solutions on square roots in a step by step manner. Download the Frre PDF from below:

2.0Key Concepts in Exercise 5.3 of Class 8 Maths Chapter 5

This exercise includes real-life situations where square and square root concepts are applied. Key concepts include:

• Word problems involving square numbers
• Real-life applications of square roots
• Using known methods to find unknown values
• Interpreting and solving practical questions using logic
• Applying mathematical thinking to daily life scenarios

3.0NCERT Class 8 Maths Chapter 5: Other Exercises

4.0NCERT Class 8 Maths Chapter 5 Exercise 5.3: Detailed Solutions

• What could be the possible one's digits of the square root of each of the following numbers? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 Sol. The possible one's digits of the square root of the numbers:

• Without doing any calculation, find the numbers which are surely not perfect squares: (i) 153 (ii) 257 (iii) 408 (iv) 441 Sol. We know that a number ending in 2,3,7 or 8 is never a perfect square. Therefore, (i) 153 is not a perfect square. (ii) 257 is not a perfect square. (iii) 408 is not a perfect square (iv) 441 may be a perfect square. So, 153, 257 and 408 are surely not perfect squares.
• Find the square roots of 100 and 169 by the method of repeated subtraction. Sol. From 100, we subtract successive odd numbers starting from 1 as under. 100-1 = 99 $99-3=96$ $96-5=91$ $91-7=84$ $84-9=75$ $75-11=64$ $64-13=51$ $51-15=36$ $36-17=19$ 19-19 = 0 and obtain 0 at $10^{\text{th }}$ step. $\therefore \sqrt{100}=10$ From 169, we subtract successive odd numbers starting from 1 as under. $\begin{array}{rl}& 169-1=168\\ & 168-3=165\\ & 165-5=160\\ & 160-7=153\\ & 153-9=144\\ & 144-11=133\\ & 133-13=120\\ & 120-15=105\\ & 105-17=88\\ & 88-19=69\\ & 69-21=48\\ & 48-23=25\\ & 25-25=0\end{array}$ and obtain 0 at $13^{\text{th }}$ step. $\therefore \sqrt{169}=13$
• Find the square roots of the following numbers by the Prime Factorisation Method: (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100 Sol. (i) By prime factorisation, we get
  
  $729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ $\therefore \sqrt{729}=3\times 3\times 3=27$ (ii) By prime factorisation, we get
  
  $400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$ $\therefore \sqrt{400}=2\times 2\times 5=20$ (iii) By prime factorisation, we get
  
  $1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{1764}=2\times 3\times 7=42$ (iv) By prime factorisation, we get
  
  $4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}$ $\therefore \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$ (v) By prime factorisation, we get $\begin{array}{ll}2& 7744\\ 2& 3872\\ 2& 1936\\ 2& 968\\ 2& 484\\ 2& 242\\ 11& 121\\ 11& 11\\ & 1\end{array}$ 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 $\therefore \sqrt{7744}=2\times 2\times 2\times 11=88$ (vi) By prime factorisation, we get $\begin{array}{ll}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$ $9604=\underline{2\times 2\times 7\times 7\times \underline{7\times 7}}$ $\therefore \sqrt{9604}=2\times 7\times 7=98$ (vii) By prime factorisation, we get $\begin{array}{ll}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}$ $5929=\underline{7\times 7}\times \underline{11\times 11}$ $\therefore \sqrt{5929}=7\times 11=77$ (viii) By prime factorisation, we get $\begin{array}{ll}2& 9216\\ 2& 4608\\ 2& 2304\\ 2& 1152\\ 2& 576\\ 2& 288\\ 2& 144\\ 2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}$ $9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$ $\therefore \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3$ $=96$ (ix) By prime factorisation, we get $\begin{array}{ll}23& 529\\ 23& 23\\ & 1\end{array}$ $529=23\times 23$ $\therefore \sqrt{529}=23$ (x) By prime factorisation, we get $\begin{array}{ll}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}$ $8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}$ $\therefore \sqrt{8100}=2\times 3\times 3\times 5=90$
• For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768 Sol. (i) By prime factorisation we get $\begin{array}{ll}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$ $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$ It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. $\therefore 252\times 7=1764$ is a perfect square. Thus, $1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{1764}=2\times 3\times 7=42$ (ii) By prime factorisation, we get $\begin{array}{ll}2& 180\\ 2& 90\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$ $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$ It is clear that in order to get a perfect square, one more 5 is required. So, the given number should be multiplied by 5 to make the product a perfect square. $\therefore 180\times 5=900$ is a perfect square. Thus, $900=\underline{2\times 2}\times \underline{3\times 3}\times \underline{5\times 5}$ $\therefore \sqrt{900}=2\times 3\times 5=30$ (iii) By prime factorisation, we get $\begin{array}{ll}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$ $1008=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times 7$ It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. $\therefore 1008\times 7=7056$ is a perfect square. Thus, $7056=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{7056}=2\times 2\times 3\times 7=84$ (iv) By prime factorisation, we get $\begin{array}{rl}2& 2028\\ 2& 1014\\ 3& 507\\ 3& 169\\ 13& 13\\ & 1\end{array}$ $2028=\underline{2\times 2}\times 3\times \underline{13\times 13}$ It is clear that the given number should be multiplied by 3 to make the product a perfect square $\therefore 2028\times 3=6084$ is a perfect square Thus $6084=\underline{2\times 2}\times \underline{3\times 3}\times \underline{13\times 13}$ $\therefore \sqrt{6084}=2\times 3\times 13=78$ (v) By prime factorisation, we get $\begin{array}{rl}2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$ $1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ It is clear that the given number should be multiplied by 2 to make the product a perfect square. $\therefore 1458\times 2=2916$ is a perfect square. Thus $2916=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ $\therefore \sqrt{2916}=2\times 3\times 3\times 3=54$ (vi) By prime factorisation, we get $\begin{array}{rl}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$ $768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3$ It is clear that the given number should be multiplied by 3 to make the product a perfect square. $\therefore 768\times 3=2304$ $\sqrt{2304}=2\times 2\times 2\times 2\times 3=48$
• For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained. (i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620 Sol. (i) By prime factorisation, we get $\begin{array}{rl}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}$ $252=\underline{2\times 2}\times \underline{3\times 3}\times 7$ Since the prime factor 7 cannot be paired. $\therefore$ The given number should be divided by $\therefore \frac{252}{7}=\frac{2\times 2\times 3\times 3\times 7}{7}=\underline{2\times 2}\times \underline{3\times 3}$ $=36$ is a perfect square and, $\sqrt{36}=\sqrt{\underline{2\times 2\times 3\times 3}}$ $=2\times 3=6$ (ii) By prime factorisation, we get $\begin{array}{rl}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}$ $2925=\underline{3\times 3}\times \underline{5\times 5}\times 13$ Since the prime factor 13 cannot be paired. $\therefore$ The given number should be divided by 13 . $\therefore \frac{2925}{13}=\frac{3\times 3\times 5\times 5\times 13}{13}$ $=\underline{3\times 3}\times \underline{5\times 5}=225$ is a perfect square and, $\sqrt{225}=\sqrt{\underline{3\times 3}\times \underline{5\times 5}}$ $=3\times 5=15$ (iii) By prime factorisation, we get $\begin{array}{rl}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}$ $396=\underline{2\times 2}\times \underline{3\times 3}\times 11$ Since the prime factor 11 cannot be paired so 396 should be divided by 11 . $\frac{396}{11}=36$ is a perfect square. and $\sqrt{36}=\sqrt{2\times 2\times \underline{3\times 3}}=2\times 3=6$ (iv) By prime factorisation, we get $\begin{array}{rl}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}$ $2645=5\times 23\times 23$ Since the prime factor 5 cannot be paired, so 2645 should be divided by 5 . $\frac{2645}{5}=529$ is a perfect square and $\sqrt{529}=\sqrt{23\times 23}=23$ (v) By prime factorisation, we get $\begin{array}{rl}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}$ $2800=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\times 7$ Since the prime factor 7 cannot be paired, so 2800 should be divided by 7 . $=\frac{2800}{7}$ $=400$ is a perfect square and $\sqrt{400}=20$ (vi) By prime factorisation, we get $\begin{array}{rl}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$ $1620=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times 5$ Since the prime factor 5 cannot be paired, so 1620 should be divided by 5 . $\frac{1620}{5}$ $=324$ is a perfect square. and $\sqrt{324}=18$
• The students of class VIII of a school donated ₹2401 in all, for Prime Minister's National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. Sol. Let x be the number of students in the class. Therefore, total money denoted by students $=₹(x\times x)=₹x^2$. But it is given as $₹2401$. $\therefore x^2=2401$ $\Rightarrow \mathrm{x}=\sqrt{2401}$ $\begin{array}{rl}7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}$ $=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}$ $=7\times 7=49$ $\Rightarrow \mathrm{x}=\sqrt{\underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}}$ $\Rightarrow \mathrm{x}=3\times 3\times 5$ $\Rightarrow \mathrm{x}=45$ number of rows as well as number of plants.
• Find the smallest square number that is divisible by each of the numbers 4,9 and 10. Sol. The smallest number divisible by each one of the numbers 4,9 and 10 is their LCM., which is $(2\times 2\times 9\times 5)$, i.e., 180 . Now, $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$ To make it a perfect square, it must be multiplied by 5 . $\therefore$ Required number $=180\times 5=900$
• Find the smallest square number that is divisible by each of the numbers 8,15 and 20. Sol. LCM of 8, 15 and 20
  
  =2x2x2x3x5=120 To make 120 a perfect square, we will multiply this number by $2\times 3\times 5$. So the number will be $2\times 2\times 2\times 2\times 3\times 3\times 5\times 5=3600$ Hence, 3600 is the smallest square number, which is divisible by 8,15 and 20.

5.0Key Features and Benefits of Class 8 Maths Chapter 5 Exercise 5.3

• The exercise contains linking Maths to real-life contexts.
• Fully based on the Class 8 NCERT Maths and CBSE syllabus.
• Helps develop critical thinking and reasoning skills in word problems.
• Increases confidence in application based problems in Olympiads and other competitive tests.
• Improves the ability to break down solutions to the exam-style problems in CBSE.
• Regular practice improves the logical thought process and smart approach to problem-solving.