Class 8 Squares and Square Roots Exercise 3.4-2

\begin{array}{l} \text {Q9. A society collected Rs 92.16. Each member collected } \\ \text {as many paise as there were members. How many members } \\ \text {were there and how much did each contribute? }\\ \\\text {sol. Let’s assume that there are x no. of members in the society. } \\ \Rightarrow \quad \text {Each member contributed x paise as per given in question.} \\ \Rightarrow \quad x^{2}=9216 \\ \Rightarrow \quad x=\sqrt{9216} \\ \Rightarrow \quad x=\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3} \\ \Rightarrow \quad x=2 \times 2 \times 2 \times 2 \times 2 \times 3 \\ \Rightarrow \quad x=96 \\ \therefore \text {Each member contributed 96 paise or Rs. 0.96 and there are 96 members.} \\ \\\text {Q10. A society collected Rs 2304 as fees from its students. } \\ \text {If each student paid as many paise as there were students in the } \\ \text {school, how many students were there in the school? }\\ \\\text {sol. Let’s assume that there are x no. of students in the school. } \\ \Rightarrow \quad \text {Each student contributed x paisa as per given in question.} \\ \text {Total collection = Rs. 2304 } = 2304 \times 100 \text { paise} \\\Rightarrow \quad x^{2}= 2304 \times 100 \\ \Rightarrow \quad x=\sqrt{2304 \times 100} \\ \Rightarrow \quad x=\sqrt{ 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 10 \times 10} \\\Rightarrow \quad x= 2 \times 2 \times 2 \times 2 \times 3 \times 10 \\ \Rightarrow \quad x=480 \\ \therefore \text {School is having 480 students.}\\\\\text {Q11. The area of a square field is 5184 }m^{2}. \text {A rectangular field, whose length } \\ \text {is twice its breadth has its perimeter equal to the perimeter } \\ \text {of the square field. } \\ \text {Find the area of the rectangular field. }\\\\\text {Sol. Let’s assume that the side of square field be x meters.} \\ \text {Area of square field } = x^{2} = 5184 m^{2} \\ \Rightarrow \quad x= \sqrt{5184} \\ \Rightarrow \quad x=2 \times 2 \times 2 \times 9 \\ \Rightarrow \quad x = 72 \text { meters} \\ \text {We know, Perimeter of square } =4 \times \text { Side of Square}\\ =4 \times x =4 \times 72 = 288 \\ \\\text {We know that perimeter of rectangle } =2(\text {length}+ \text {breadth}) \\ \text {As given in the question, } \\ \text {Perimeter of rectangle } =2(\text {length}+ \text {breadth}) \\ = \text {Perimeter of the square field } =288 \\ \text {Also given that lenght of rectangular field is twice its breadth. } \\ \text {Lets assume b is breadth of rectangular field. Hence} \\ 2(2b+b)=288 \\ \Rightarrow \quad 6b=288 \\ \Rightarrow \quad b=48 \\ \text {Area of rectangular field } = l \times b = (2b) \times b \\ =2 \times 48 \times 48 = 4608 \quad m^{2} \\ \\\text {Sol. (i) L.C.M of 6,9,15,20 = 180} \\ \text {The prime factors of 180 }=2^{2} \times 3^{2} \times 5 \\ \text {To make 180 a perfect square, we have to multiply the number with 5} \\ \Rightarrow \quad 180 \times 5=2^{2} \times 3^{2} \times 5^{2} = 900\\ \therefore \text {900 is the least square number divisible by 6,9,15 and 20} \\ \\\text {(i) L.C.M of 8,2,15 and 20 = 360} \\ \text {The prime factors of 360 }=2^{2} \times 3^{2} \times 2 \times 5 \\ \text {To make 360 a perfect square, we have to multiply the number with } 2 \times 5=10 \\ \Rightarrow \quad 360 \times 10=2^{2} \times 3^{2} \times 2^{2} \times 5^{5} = 3600\\ \therefore \text {3600 is the least square number divisible by 8,2,15 and 20 } \\ \\ \\\text {Q13. Find the square roots of 121 and 169 by the method of repeated subtraction. }\\ \\\text {Sol. To find sq root of 121, we have } \\ \quad 121-1=120 \\ \quad 120-3=117 \\ \quad 117-5=112 \\ \quad 112-7=105 \\ \quad 105-9=96 \\ \quad 96-11=85 \\ \quad 85-13=72 \\ \quad 72-15=57 \\ \quad 57-17=40 \\ \quad 40-19=21 \\ \quad 21-21=0 \\ \text {In total we did subtraction 11 times from 121. } \\ \therefore \quad \sqrt{121} = 11 \\ \\\text {To find sq root of 169, we have } \\ \quad 169-1=168 \\ \quad 168-3=165 \\ \quad 165-5=160 \\ \quad 160-7=153 \\ \quad 153-9=144 \\ \quad 144-11=133 \\ \quad 133-13=120 \\ \quad 120-15=105 \\ \quad 105-17=88 \\ \quad 88-19=69 \\ \quad 69-21=48 \\ \quad 48-23=25 \\ \quad 25-25=0$ \\ \text {In total we did subtraction 13 times from 169. } \\ \therefore \quad \sqrt{169} = 13\\\\\text {Q14. Write the prime factorization of the following numbers } \\ \text {and hence find their square roots. } \\ (i) \quad 7744 \quad (ii) \quad 9604 \quad (iii) \quad 5929 \quad (iv) \quad 7056\\\text {Sol. (i) The prime factors for 7744 }=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11 \\ \text {By Grouping them into pair of equal factors, we get } \\ 7744=(2 \times 2) \times(2 \times 2) \times(2 \times 2) \times(11 \times 11) \\ \Rightarrow \quad \sqrt{7744} =2 \times 2 \times 2 \times 11=88 \\ \\\text {(ii) The prime factors for 9604 }=2 \times 2 \times 7 \times 7 \times 7 \times 7 \\ \text {By Grouping them into pair of equal factors, we get} \\ 9604=(2 \times 2) \times(7 \times 7) \times(7 \times 7)$ \Rightarrow \quad \sqrt{9604}=2 \times 7 \times 7=98 \\ \\\text {(iii) The prime factors for 5929 }=7 \times 7 \times 11 \times 11 \\ \text {By grouping them into pair of equal factors, we get} \\ 5929=(7 \times 7) \times(11 \times 11) \\ \Rightarrow \quad \sqrt{5929}=7 \times 11=77 \\ \\\text {(iv) The prime factors for 7056}=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7 \\ \text {By grouping them into pair of equal factors, we get } \\ 7056=(2 \times 2) \times(2 \times 2) \times(3 \times 3) \times(7 \times 7) \\ \Rightarrow \quad \sqrt{7056}=2 \times 2 \times 3 \times 7=84 \\\\\text {Q15. The students of class VIII of a school donated Rs. 2401 for PM’s National } \\ \text {Relief Fund. Each student donated as many rupees as the number of students } \\ \text {in the class, find the number of students in the class.}\\ \\\text {Sol. Let’s assume that there are x no. of students in the class } \\ \Rightarrow \quad \text {Each student contributed Rs. x as per given in question.} \\ \Rightarrow \quad x^{2}=2401 \\ \Rightarrow \quad x=\sqrt{2401} \\ \Rightarrow \quad x=\sqrt{7 \times 7 \times 7 \times 7} \\ \Rightarrow \quad x=7 \times 7 \\ \Rightarrow \quad x=49 \\ \therefore \text {There are 49 students in the class.} \\ \\\text {Q16. A PT teacher wants to arrange maximum possible number of 6000 students } \\ \text {in a field such that the number of rows is equal to the number of columns. } \\ \text {Find the number of rows if 71 were left out after arrangement.}\\\text {sol. Let’s assume that there are no.of rows are x. } \\ \text {So, no. of columns are also x.} \\ \Rightarrow \quad \text {According to given question, we have} \\ \Rightarrow \quad x^{2} + 71 =6000 \\ \Rightarrow \quad x^{2} =6000 -71 \\ \Rightarrow \quad x^{2} =5929 \\ \Rightarrow \quad x=\sqrt{5929} \\ \Rightarrow \quad x=\sqrt{7 \times 7 \times 11 \times 11} \\ \Rightarrow \quad x=7 \times 11 \\ \Rightarrow \quad x=77 \\ \therefore \text {There are a total of 77 rows.} \end{array}