One of the most important ways to prepare for the Board exams is by practicing the previous year papers. At our website, students can easily access the ICSE Class 10 Maths Previous Years Questions with solutions from 2010 to 2023. Scoring well in ICSE class 10 Math exam requires students to have a good understanding of the concepts and formulas. By solving these PYQs, students can get familiarized with the exam pattern and the types of questions asked in the exam. This can help them to approach the exam with confidence and reduce exam pressure.
ICSE Class 10 Maths Previous Years Questions with Solutions Chapter Quadratic Equations 2010 to 2023
The past year questions from Quadratic Equations available on our website are from the years 2023, 2022, 2020, 2019, 2018, 2017, 2016, 2015, 2014, 2013, 2012, 2011 and 2010. Each sum is solved in detail that helps students understand the marking pattern.
Moreover, practicing the ICSE Class 10 Maths Previous Years Questions Chapter- Quadratic Equations can be particularly beneficial for students. Quadratic Equations is an important topic in Maths and requires a good understanding of formulas and concepts. By practicing the ICSE Class 10 Maths previous years questions related to Quadratic Equations students can get a good grasp of the topic and be better prepared for the exam.
ICSE Class 10 Maths Previous Years Questions Chapter-Quadratic Equations
Q1. Solve the following quadratic equation: x2 + 4x – 8 = 0
Give your answer correct to one decimal place.
(Use mathematical tables if necessary.) [2023]
Answer: x= 1.5, -5.5
Step-by-step explanation:
\[x^2+4x-8=0\\comparing\;the\;above\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=4,\;c=-8\\By\;Quadratic\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-4\pm\sqrt{4^2-4\times1\times(-8)}}{2\times1}\\x=\frac{-4\pm\sqrt{16+32}}2\\x=\frac{-4\pm\sqrt{48}}2\\x=\frac{-4\pm6.928}2\\x=\frac{-4+6.928}2,\;\frac{-4-6.928}2\\x=\frac{2.928}2\;,\;\frac{-10.928}2\\x=\;1.464\;,\;-5.464\\x=\;1.5\;,\;-5.5\\\\\\\]
Q2. If 3 is a root of the quadratic equation x2 – px + 3 = 0 then p is equal to:
(a) 4
(b) 3
(c) 5
(d) 2 [2023]
Answer: (a) 4
Step-by-step explanation:
\[x^2-px+3=0\\Given,\;3\;is\;a\;root\;of\;the\;above\;equation.\\Therefore,\;\\{(3)}^2-p\times3+3=0\\9-3p+3=0\\-3p+12=0\\-3p=-12\\p=4\\\\\\\\\]
Q3. One of the roots of the quadratic equation x2 – 8x + 5 = 0 is 7.3166. The root of the equation correct to 4 significant figures is: [1]
(a) 7.3166
(b) 7.317
(c) 7.316
(d) 7.32 [2021 Semester-1]
Answer: (b) 7.317
Step-by-step explanation:
\[x^2-8x+5=0\\Given,\;7.3166\;is\;a\;root\;of\;the\;above\;equation.\\Therefore,\;\\The\;root\;of\;the\;equation\;correct\;to\;4\;significant\;figures\;is\\7.317\\\\\\\\\\\]
Q4. Which of the following quadratic equations has 2 and 3 as its roots? [1]
(a) x2 – 5x + 6 =0
(b) x2 + 5x + 6 = 0
(c) x2 – 5x – 6 =0
(d) x2 + 5x -6 =0 [2021 Semester-1]
Answer: (a) x2-5x+6=0
Step-by-step Explanation:
\[x^2-5x+6=0\\x^2-3x-2x+6=0\\x(x-3)-2(x-3)=0\\(x-2)(x-3)=0\\Either\;x-3=0\;OR\;x-2=0\\Therefore,\;x=3\;OR\;x=2\\\\\\\\\\\\\]
Q5. Solve the following Quadratic Equation:
𝑥2 − 7𝑥 + 3 = 0
Give your answer correct to two decimal places. [2020]
Answer: x = 6.54 , 0.46
Step-by-step Explanation:
\[x^2-7x+3=0\\Comparing\;the\;given\;equation\;with\;ax^2+bx+c=0\;we\;have,\\a=1,\;b=-7,\;c=3\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-7)\pm\sqrt{{(-7)}^2-4\times1\times3}}{2\times1}\\x=\frac{7\pm\sqrt{49-12}}2\\x=\frac{7\pm\sqrt{37}}2\\x=\frac{7\pm6.083}2\\x=\frac{7+6.083}2\;,\;\frac{7-6.083}2\\x=\frac{13.083}2\;,\;\frac{0.917}2\\x=6.5416\;,\;0.4585\\x=6.54\;,\;0.46\\\\\\\\\\\\\\\\\]
Q6. Solve for x the quadratic equation x2 – 4x – 8 = 0
Give your answer correct to three significant figures. [2019]
Answer: x= 5.46 , -1.46
Step-by-step Explanation:
\[x^2-4x-8=0\\Comparing\;the\;given\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=-4,\;c=-8\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-4)\pm\sqrt{{(-4)}^2-4\times1\times(-8)}}{2\times1}\\x=\frac{4\pm\sqrt{16+32}}2\\x=\frac{4\pm\sqrt{48}}2\\x=\frac{4\pm6.928}2\\x=\frac{4+6.928}2,\;\frac{4-6.928}2\\x=\frac{10.928}2\;,\;\frac{-2.928}2\\x=5.464\;,\;-1.464\\x=5.46\;,\;-1.46\]
You can see video solution of these questions here.
Q7. Solve x2 + 7x = 7 and give your answer correct to two decimal places. [4] [2018]
Answer: 0.89 , -7.89
Step-by-step Explanation:
\[x^2+7x=7\\x^2+7x-7=0\\Comparing\;the\;given\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=7,\;c=-7\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-7\pm\sqrt{{(7)}^2-4\times1\times(-7)}}{2\times1}\\x=\frac{-7\pm\sqrt{49+28}}2\\x=\frac{-7\pm\sqrt{77}}2\\x=\frac{-7\pm8.775}2\\x=\frac{-7+8.775}2\;,\;\frac{-7-8.775}2\\x=\frac{1.775}2\;,\;\frac{-15.775}2\\x=0.8875\;,\;-7.8875\\x=0.89\;,\;-7.89\]
Q8. Find the value of k for which the following equation has equal roots. [3]
x2 + 4kx + (k2 – k + 2) =0 [2018]
Answer: k= -1 or 2/3
Step-by-step Explanation:
\[x^2\;+\;4kx\;+\;(k^2\;–\;k\;+\;2)\;=0\;\;\\Comparing\;the\;equation\;with\;ax^2+bx+c=0\;we\;have,\\a=1,\;b=4k,\;c=\;(k^2-k+2)\\Given,\;b^2-4ac=0\\Therefore,\;{(4k)}^2-4\times1\times(k^2-k+2)=0\\16k^2-4k^2+4k-8=0\\12k^2+4k-8=0\\4(3k^2+k-2)=0\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(k+1)(3k-2)=0\\Either\;k+1=0\;\;or\;\;3k-2=0\\Either\;k=-1\;or\;\frac23\]
Q9. Solve the equation 4x2 – 5x – 3 = 0 and give your answer correct to two decimal places. [4] [2017]
Answer: x = 1.69 , -0.44
Step-by-step Explanation:
\[4x^2-5x-3=0\;\;\\Comparing\;the\;equation\;with\;ax^2+bx+c=0\;we\;have,\\a=4,\;b=-5,\;c=-3\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-5)\pm\sqrt{{(-5)}^2-4\times4\times(-3)}}{2\times4}\\x=\frac{5\pm\sqrt{25+48}}8\\x=\frac{5\pm\sqrt{73}}8\\x=\frac{5\pm8.544}8\\x=\frac{5+8.544}8\;,\;\frac{5-8.544}8\\x=\frac{13.544}8\;,\;\frac{-3.544}8\\x=1.693\;,\;-0.443\\x=\;1.69\;,\;-0.44\]
Q10. Solve the quadratic equation x2 – 3(x + 3) = 0; Give your answer correct to two significant figures. [3] [2016]
Answer: x = 5.9 , -0.85
Step-by-step Explanation:
\[x^2-3(x+3)=0\;\;\\x^2-3x-9=0\\Comparing\;the\;equation\;with\;ax^2+bx+c=0\;we\;have,\\a=1,\;b=-3,\;c=-9\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-3)\pm\sqrt{{(-3)}^2-4\times1\times(-9)}}{2\times1}\\x=\frac{3\pm\sqrt{9+36}}2\\x=\frac{3\pm\sqrt{45}}2\\x=\frac{3\pm6.708}2\\x=\frac{3+6.708}2\;,\;\frac{3-6.708}2\\x=\frac{9.708}2\;,\;\frac{-3.708}2\\x=4.854\;,\;-1.854\\x=\;4.9\;,\;-1.9\]
You can find ICSE Class 10 Maths Previous Years Questions with solution of each Chapter here.
Q11. Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation,
(K + 2) x2 – Kx + 6 = 0.Thus find the other root of the equation. [2015]
Answer: k= -4, other root = -1
Step-by-step Explanation:
\[(K\;+\;2)\;x^2\;–\;Kx\;+\;6\;=\;0\\x=3\;is\;a\;solution\;of\;the\;above\;equation.\\Therefore,\;\\(k+2){(3)}^2-k\times3+6=0\\9k+18-3k+6=0\\6k+24=0\\6k=-24\\k=-4\\putting\;k=-4\;in\;the\;given\;equation,\\-2x^2+4x+6=0\\-2(x^2-2x-3)=0\\x^2-2x-3=0\\x^2-3x+x-3=0\\x(x-3)+1(x-3)=0\\(x-3)(x+1)=0\\Either\;x+1=0\;or\;x-3=0\\x=-1\;or\;\;3\\So,\;other\;root\;of\;the\;equation\;is\;-1.\\\]
Q12. Solve for x using the quadratic formula. Write your answer correct to two significant figures,
(x – 1)2 – 3x + 4 = 0. [3] [2014]
Answer: 3.6 , 1.4
Step-by-step Explanation:
\[{(x\;–\;1)}^2\;–\;3x\;+\;4\;=\;0\\\Rightarrow x^2-2x+1-3x+4=0\\\Rightarrow x^2-5x+5=0\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=-5,\;c=5\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-5)\pm\sqrt{{(-5)}^2-4\times1\times5}}{2\times1}\\x=\frac{5\pm\sqrt{25-20}}2\\x=\frac{5\pm\sqrt5}2\\x=\frac{5\pm2.24}2\\x=\frac{5+2.24}2\;or\;\frac{5-2.24}2\\x=\frac{7.24}2\;or\;\frac{2.76}2\\x=3.62\;or\;1.38\\x=3.6\;,\;1.4\]
Q13. Solve the following equation and calculate the answer correct to two decimal places:
x2 – 5x – 10 = 0 [3] [2013]
Answer: 6.53 , -1.53
Step-by-step Explanation:
\[x^2-5x-10=0\;\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=-5,\;c=-10\\By\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-5)\pm\sqrt{{(-5)}^2-4\times1\times(-10)}}{2\times1}\\x=\frac{5\pm\sqrt{25+40}}2\\x=\frac{5\pm\sqrt{65}}2\\x=\frac{5\pm8.062}2\\x=\frac{5+8.062}2\;or\;\frac{5-8.062}2\\x=\frac{13.062}2\;or\;\frac{-3.062}2\\x=6.531\;or\;-1.531\\x=6.53\;,\;-1.53\]
Q14. Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots: x2 + (p – 3)x + p = 0 [2013]
Answer: p= 1 or 9
Step-by-step Explanation:
\[x^2+(p\;–\;3)x+p=0\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=1,\;b=(p-3),\;c=p\\Given,\;b^2-4ac=0\\\Rightarrow{(p-3)}^2-4\times1\times p=0\\\Rightarrow p^2-6p+9-4p=0\\\Rightarrow p^2-10p+9=0\\\Rightarrow p^2-9p-p+9=0\\\Rightarrow p(p-9)-1(p-9)=0\\\Rightarrow(p-9)(p-1)=0\\Either\;p-9\;=\;0\;OR\;p-1\;=0\\p=9\;or\;1\\\\\]
Q15. Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.
x2 + 2 (m – 1) x + (m + 5) = 0 [3] [2012]
Answer: m = -1 , 4
Step-by-step Explanation:
\[x^2\;+\;2\;(m\;–\;1)\;x\;+\;(m\;+\;5)\;=\;0\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0\\a=1,\;b=2(m-1)\;,\;c=(m+5)\\given,\;b^2-4ac=0\\{\lbrack2(m-1)\rbrack}^2-4\times1\times(m+5)=0\\4(m^2-2m+1)-4m-20=0\\4m^2-8m+4-4m-20=0\\4m^2-12m-16=0\\4(m^2-3m-4)=0\\m^2-3m-4=0\\m^2-4m+m-4=0\\m(m-4)+1(m-4)=0\\(m+1)(m-4)=0\\Either\;(m+1)=0\;or\;(m-4)=0\\m=-1\;or\;4\]
Q16. Solve the following equation and give your answer correct to 3 significant figures:
5x2 – 3x – 4 = 0 [3] [2012]
Answer: 1.24 , -0.643
Step-by-step Explanation:
\[5x^2\;–\;3x\;–\;4\;=\;0\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0\\a=5,\;b=-3\;,\;c=-4\\by\;quadratic\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-3)\pm\sqrt{{(-3)}^2-4\times5\times(-4)}}{2\times5}\\x=\frac{3\pm\sqrt{9+80}}{10}\\x=\frac{3\pm\sqrt{89}}{10}\\x=\frac{3\pm9.434}{10}\\x=\frac{3+9.434}{10}\;,\;\frac{3-9.434}{10}\\x=\frac{12.434}{10}\;,\;\frac{-6.434}{10}\\x=1.2434\;,\;-0.6434\\x=1.24\;,\;-0.643\]
Q17. Solve the following equation:
\[x-\frac{18}x=6\]
Give your answer correct to two significant figures. [3] [2011]
Answer: 8.2 , -2.2
Step-by-step Explanation:
\[x-\frac{18}x=6\\\Rightarrow\frac{x^2-18}x=6\\\Rightarrow x^2-18=6x\\\Rightarrow x^2-6x-18=0\\comparing\;the\;given\;equation\;with\;ax^2+bx+c=0\\a=1,\;b=-6\;,\;c=-18\\by\;quadratic\;formula,\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-6)\pm\sqrt{{(-6)}^2-4\times1\times(-18)}}{2\times1}\\x=\frac{6\pm\sqrt{36+72}}2\\x=\frac{6\pm\sqrt{108}}2\\x=\frac{6\pm10.392}2\\x=\frac{6+10.392}2\;,\;\frac{6-10.392}2\\x=\frac{16.392}2\;,\;\frac{-4.392}2\\x=8.196\;,\;-2.196\\x=8.2\;,\;-2.2\]
Q18. Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px2 – 4x + 3 = 0 [3] [2010]
Answer: p = 4/3
Step-by-step Explanation:
\[px^2\;–\;4x\;+\;3\;=\;0\\comparing\;the\;above\;equation\;with\;ax^2+bx+c=0,\;we\;have,\\a=p\;,\;b=-4\;,\;c=3\\given,\;b^2-4ac=0\\{(-4)}^2-4\times p\times3=0\\16-12p=0\\-12p=-16\\p=\frac{16}{12}\\p=\frac43\]
Q19. A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed. [2023]
Answer: 20 km/h
Step-by-step Explanation:
Original speed = x km/hr
Distance = 100 km
Therefore, Time taken = 100/x hr
Now, if speed = (x+5) km/hr
Then, by the problem,
\[\frac{100}{x+5}=\frac{100}x-1\\\Rightarrow\frac{100}{x+5}-\frac{100}x=-1\\\Rightarrow\frac{100x-100(x+5)}{x(x+5)}=-1\\\Rightarrow\frac{100x-100x-500}{x^2+5x}=-1\\\Rightarrow-500=-x^2-5x\\\Rightarrow x^2+5x-500=0\\\Rightarrow x^2+25x-20x-500=0\\\Rightarrow x(x+25)-20(x+25)=0\\\Rightarrow(x+25)(x-20)=0\\either\;(x+25)=0\;or\;(x-20)=0\\x=-25\;or\;20\\As\;speed\;cannot\;be\;negative,\;therefore\\original\;speed\;is\;20\;km/h\\\\\]
Q20. The difference of two natural numbers is 7 and their product is 450. Find the numbers. [2020]
Answer: 18 and 25
Step-by-step Explanation:
Let the two numbers be x and (x-7).
by the problem,
x(x-7) = 450
or, x2 – 7x – 450 = 0
or, x2 – 25x + 18x – 450 = 0
or, x(x – 25) + 18(x – 25) = 0
or, (x – 25)(x + 18) = 0
Either (x – 25) = 0 or (x + 18) = 0
x = 25 or -18
As natural numbers cannot be negative, therefore
One number is 25 and
the other number is (25-7) = 18.
Q21. The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers. [3] [2019]
Answer: 27 and 30
Step-by-step Explanation:
Let the two numbers be 3x and 3(x+1).
by the problem,
\[3x\times3(x+1)=810\\\Rightarrow3x(3x+3)=810\\\Rightarrow9x^2+9x=810\\\Rightarrow9(x^2+x)=810\\\Rightarrow x^2+x=90\\\Rightarrow x^2+x-90=0\\\Rightarrow x^2+10x-9x-90=0\\\Rightarrow x(x+10)-9(x+10)=0\\\Rightarrow(x+10)(x-9)=0\\either\;(x+10)\;=\;0\;or\;(x-9)=0\\x=-10\;or\;9\\As\;natural\;numbers\;cannot\;be\;negative,\;therefore,\\x=9\\Therefore\;the\;two\;numbers\;are\;3x\;=\;27\;and\\3(x+1)\;=30\\\\\]
You can see video solutions to these questions here.
Q22. ₹ 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹ 100 more. Find the original number of children. [2018]
Answer: 50
Step-by-step Explanation:
\[Let\;the\;original\;number\;of\;children\;be\;x.\\Each\;child\;gets\;Rs.\frac{7500}x.\\By\;the\;problem,\\\frac{7500}{x-20}-\frac{7500}x=100\\\Rightarrow\frac{7500x-7500(x-20)}{x(x-20)}=100\\\Rightarrow\frac{7500x-7500x+150000}{x^2-20x}=100\\\Rightarrow100x^2-2000x=150000\\\Rightarrow100x^2-2000x-150000=0\\\Rightarrow100(x^2-20x-1500)=0\\\Rightarrow x^2-20x-1500=0\\\Rightarrow x^2-50x+30x-1500=0\\\Rightarrow x(x-50)+30(x-50)=0\\\Rightarrow(x+30)(x-50)=0\\either\;(x+30)=0\;or\;(x-50)=0\\x=-30\;or\;50\\As\;number\;of\;children\;cannot\;be\;negative,\;therefore,\\x=50.\\Therefore,\;original\;number\;of\;children\;is\;50.\\\\\\\]
Q23. Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then:
i. The number of litres of diesel used by car X is: [1]
\[(a)\;\frac{72}{x-3}litres\\(b)\;\frac{72}{x+3}\;litres\\(c)\;\frac{72}xlitres\\(d)\;\frac{12}xlitres\]
ii. The number of litres of diesel used by car Y is: [1]
\[(a)\;\frac{72}{x-3}litres\\(b)\;\frac{72}{x+3}\;litres\\(c)\;\frac{72}xlitres\\(d)\;\frac{12}{x+3}litres\]
iii. If car X used 4 litres of diesel more than car Y in the journey, then: [1]
\[(a)\;\frac{72}{x-3}-\frac{12}x=4\\(b)\;\frac{72}{x+3}-\frac{72}x=4\\(c)\;\frac{72}x-\frac{72}{x+3}=4\\(d)\;\frac{72}{x-3}-\frac{72}{x+3}=4\]
iv. The amount of diesel used by the car X is: [1]
(a) 6 litres
(b) 12 litres
(c) 18 litres
(d) 24 litres [2021 Semester-1]
Answer: i. (c) , ii. (b) , iii. (c) , iv. (b)
Step-by-step Explanation:
\[By\;the\;problem,\\\frac{72}x-\frac{72}{x+3}=4\\\Rightarrow\frac{72(x+3)-72x}{x(x+3)}=4\\\Rightarrow\frac{72x+216-72x}{x^2+3x}=4\\\Rightarrow4x^2+12x=216\\\Rightarrow4x^2+12x-216=0\\\Rightarrow4(x^2+3x-54)=0\\\Rightarrow x^2+3x-54=0\\\Rightarrow x^2+9x-6x-54=0\\\Rightarrow x(x+9)-6(x+9)=0\\\Rightarrow(x+9)(x-6)=0\\either\;(x+9)=0\;or\;(x-6)=0\\x\;=\;-9\;or\;6\\therefore,\\x=6\;litres.\\Therefore,\;The\;amount\;of\;diesel\;used\;by\;the\;car\;X\;is\\\\\\\\\\\]
\[\frac{72}x\;=\;\frac{72}6\;=\;12\;litres.\\option\;(b).\]
Q24. The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages. [4] [2017]
Answer: 25 years, 22 years
Step-by-step Explanation:
Let the age of Vivek be x years and that of his younger brother be (47-x) years.
By the problem,
x(47 – x) = 550
or, 47x – x2 =550
or, x2 – 47x + 550 = 0
or, x2 – 25x – 22x + 550 = 0
or, x(x – 25) – 22(x – 25) = 0
or, (x – 25)(x – 22) = 0
either (x – 25) = 0 or (x – 22) = 0
x = 25 or 22
Therefore, Vivek’s age is 25 years and his younger brother’s age is 22 years.
Q25. A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’. [3] [2016]
Answer: 40 km/h
Step-by-step Explanation:
\[Uniform\;speed\;=x\;km/h\;\;\\dis\tan ce\;=240\;km\;\;\\time\;=\;\;\frac{Dis\tan ce}{speed}=\frac{240}x\;hours\;\;\\Due\;to\;heavy\;rain\;speed\;=(x-10)\;km/h\;\;\\Now\;time\;=\;\frac{240}{x-10}hours\\By\;the\;problem,\\\frac{240}{x-10}-\frac{240}x=2\\\Rightarrow\frac{240x-240(x-10)}{x(x-10)}=2\\\Rightarrow\frac{240x-240x+2400}{x^2-10x}=2\\\Rightarrow2x^2-20x=2400\\\Rightarrow2x^2-20x-2400=0\\\Rightarrow2(x^2-10x-1200)=0\\\Rightarrow x^2-10x-1200=0\\\Rightarrow x^2-40x+30x-1200=0\\\Rightarrow x(x-40)+30(x-40)=0\\\Rightarrow(x-40)(x+30)=0\\either\;(x-40)=0\;or\;(x+30)=0\\x\;=\;40\;or\;-30\\As\;speed\;cannot\;be\;negative,\;therefore,\\x\;=\;40\\Therefore,\;speed\;of\;bus\;is\;40\;km/h.\\\\\\\\\\\\\]
Q26. Sum of two natural numbers is 8 and the difference of their reciprocal is 2 / 15. Find the numbers. [3] [2015]
Answer: 3 and 5
Step-by-step Explanation:
Let the two natural numbers be x and (8-x).
Now, by the problem,
\[\frac1x-\frac1{8-x}=\frac2{15}\\\Rightarrow\frac{8-x-x}{x(8-x)}=\frac2{15}\\\Rightarrow\frac{8-2x}{8x-x^2}=\frac2{15}\\\Rightarrow120-30x=16x-2x^2\\\Rightarrow2x^2-46x+120=0\\\Rightarrow2(x^2-23x+60)=0\\\Rightarrow x^2-23x+60=0\\\Rightarrow x^2-3x-20x+60=0\\\Rightarrow x(x-3)-20(x-3)=0\\\Rightarrow(x-3)(x-20)=0\\either\;(x-3)=0\;or\;(x-20)=0\\x=3,20\\20\;is\;not\;possible.\\Therefore,\;the\;two\;numbers\;are\;3\;and\;(8-3)=5.\\\\\\\\\\\\\\\\\\\]
Q27. A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. [4] [2014]
Answer: 23
Step-by-step Explanation:
\[Let\;the\;tens’\;digit\;be\;x\;and\;unit’s\;digit\;be\;\frac6x.\;\;\\Therefore\;the\;two\;digit\;number\;=\;10x\;+\;\frac6x.\\According\;to\;the\;problem,\\10x+\frac6x+9=(10\times\frac6x)+x\\\Rightarrow10x+\frac6x-\frac{60}x-x=-9\\\Rightarrow\frac{10x^2+6-60-x^2}x=-9\\\Rightarrow9x^2-54=-9x\\\Rightarrow9x^2+9x-54=0\\\Rightarrow9(x^2+x-6)=0\\\Rightarrow x^2+x-6=0\\\Rightarrow x^2+3x-2x-6=0\\\Rightarrow x(x+3)-2(x+3)=0\\\Rightarrow(x+3)(x-2)=0\\Either\;(x+3)=0\;or\;(x-2)=0\\x=-3,2\\\sin ce\;number\;is\;positive,\;therefore\\x=2\\Therefore\;the\;two-digit\;number\;is\\10x+\frac6x\\20+\frac62\\23.\\\\\]
Q28. A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was 8 less, the number of books that could be purchased for Rs. 960 would be 4 more. Write an equation, taking the original cost of each book to be Rs. x, and solve it to find the original cost of the books.[4] [2013]
Answer: Rs. 48
Step-by-step Explanation:
\[original\;\cos t\;of\;each\;book\;is\;Rs\;x.\;\\Then,the\;no.of\;books\;=\frac{960}x.\;\;\\Also,\;\cos t\;of\;each\;books\;=\;Rs\;(x-8)\;\;\;\\Therefore,\;by\;the\;problem,\\\frac{960}{x-8}-\frac{960}x\;=4\;\;\\\Rightarrow\frac{960x-960(x-8)}{x(x-8)}=4\\\Rightarrow\frac{960x-960x+7680}{x^2-8x}=4\\\Rightarrow4x^2-32x=7680\\\Rightarrow4x^2-32x-7680=0\\\Rightarrow\;x^2-8x-1920=0\;\\\Rightarrow\;x^2+40x-48x-1920=0\;\\\Rightarrow(x-48)(x+40)=0\;\\Either\;x-48=0\;or\;x+40=0\\x\;=\;48\;or\;-\;40\;\;\\\therefore\;\;Original\;\cos t\;of\;book\;=\;Rs\;48.\]
Q29. A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. [4] [2012]
Answer: 48 km/h
Step-by-step Explanation:
\[Let\;the\;original\;speed\;be\;x\;km/h.\;\\Then\;\;Time\;=\;\frac{400}x\;hrs\\If\;speed\;=x+12,\;then\\Time=\frac{400}{x+12}\;\\By\;the\;problem,\\\frac{400}x-\frac{400}{x+12}=1hr\;+\frac{40}{60}\\\Rightarrow\frac{400(x+12)-400x}{x(x+12)}=\frac53\\\Rightarrow\frac{400x+4800-400x}{x^2+12x}=\frac53\\\Rightarrow5x^2+60x=14400\\\Rightarrow5x^2+60x-14400=0\\\Rightarrow x^2+12x-2880=0\\\Rightarrow x^2+60x-48x-2880=0\\\Rightarrow x(x+60)-48(x+60)=0\\\Rightarrow(x+60)(x-48)=0\\Either\;x+60=0\;or\;x-48=0\\x=-60\;or\;48\\Speed\;cannot\;be\;negative.\;therefore,\\speed\;of\;the\;car\;is\;48\;km/h.\]
Q30. Rs.480 is divided equally among ‘x’ children. If the number of children was 20 more, then each would have got Rs. 12 less. Find ‘x’. [3] [2011]
Answer: 20
Step-by-step Explanation:
\[Given,\;Number\;\;of\;children\;=\;x\\Therefore,\;money\;received\;by\;each\;child\;=\;Rs.\frac{480}x\\Now,\;if\;Number\;of\;children\;=\;x+20,\;then,\\by\;the\;problem,\\\frac{480}x-\frac{480}{x+20}=12\\\Rightarrow\frac{480(x+20)-480x}{x(x+20)}=12\\\Rightarrow\frac{480x+9600-480x}{x^2+20x}=12\\\Rightarrow12x^2+240x=9600\\\Rightarrow x^2+20x-800=0\\\Rightarrow x^2+40x-20x-800=0\\\Rightarrow x(x+40)-20(x+40)=0\\\Rightarrow(x+40)(x-20)=0\\Either\;x+40=0\;or\;x-20=0\\x=-40\;or\;20.\\Children\;cannot\;be\;in\;negative.\\Therefore,\;number\;of\;children\;is\;20.\]
Q31. A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number. [4] [2010]
Answer: 6
Step-by-step Explanation:
\[Let\;smaller\;part\;be\;x.\\Then,\;{(larger\;part)}^2=8x\;Therefore,\;larger\;part=\sqrt{8x}\\By\;the\;problem,\\x^2+8x=20\\\Rightarrow x^2+8x-20=0\\\Rightarrow x^2+10x-2x-20=0\\\Rightarrow x(x+10)-2(x+10)=0\\\Rightarrow(x+10)(x-2)=0\\Either\;(x+10)=0\;or\;(x-2)=0\\x=-10\;or\;2.\\As\;the\;number\;is\;positive,\;therefore,\\smaller\;part\;is\;2\;and\\larger\;part\;is\;\sqrt{8\times2}\;=\;4\\Therefore\;the\;number\;is\;4+2\;=\;6.\]
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