Sol :
Let the two numbers be x and y.
According to the question,
x + y = 18 \u2026(i)
$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{4}$ \u2026(ii)
<\/p>\n\n\n\n
Eq. (ii) can be re – written as
$\\frac{y+x}{x y}=\\frac{1}{4}$ \u2026(iii)
<\/p>\n\n\n\n
On putting the value of x + y = 18 in Eq. (iii), we get
$\\frac{18}{x y}=\\frac{1}{4}$
<\/p>\n\n\n\n
\u21d2 xy = 72<\/p>\n\n\n\n
$\\Rightarrow x=\\frac{72}{y}$
<\/p>\n\n\n\n
On putting the value of $x=\\frac{72}{y}$ in Eq. (i), we get If y=6, then $x=\\frac{72}{6}=12$<\/p>\n\n\n\n If y=12, then $x=\\frac{72}{12}=6$<\/p>\n\n\n\n Hence, the two numbers are 6 and 12.<\/p>\n\n\n\n The sum of two numbers is 15 and sum of their reciprocals is <\/strong>$\\frac{3}{10}$. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Eq. (ii) can be re – written as $\\frac{y+x}{x y}=\\frac{3}{10}$ \u2026(iii)<\/p>\n\n\n\n On putting the value of x+y=18 in Eq. (iii), we get On putting the value of $x=\\frac{50}{y}$ in Eq. (i), we get \u21d2y=5 and 10<\/p>\n\n\n\n If y=5, then $x=\\frac{50}{5}=10$<\/p>\n\n\n\n If y=10, then $x=\\frac{50}{10}=5$<\/p>\n\n\n\n Hence, the two numbers are 5 and 10.<\/p>\n\n\n\n Two numbers are in the ratio of 5 : 6. If 8 is subtracted from each of the numb, they become in the ratio of 4 : 5. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : According to the question,<\/p>\n\n\n\n $\\frac{x}{y}=\\frac{5}{6}$<\/p>\n\n\n\n $\\Rightarrow \\mathrm{y}=\\frac{6 \\mathrm{x}}{5}$ \u2026(i) Also, $\\frac{x-8}{y-8}=\\frac{4}{5}$ On putting the value of $y=\\frac{6 x}{5}$ in Eq. (ii), we get On putting the value of x = 40 in Eq. (i), we get Hence, the two numbers are 40 and 48.<\/p>\n\n\n\n The sum of two numbers is 16 and the sum of their reciprocals is 1\/3. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Eq. (ii) can be re – written as<\/p>\n\n\n\n $\\frac{y+x}{x y}=\\frac{1}{3}$ \u2026(iii) On putting the value of x + y = 16 in Eq. (iii), we get $\\frac{16}{x y}=\\frac{1}{3}$ $\\Rightarrow x=\\frac{48}{y}$<\/p>\n\n\n\n On putting the value of $x=\\frac{48}{y}$ in Eq. (i), we get If y=4 then $x=\\frac{48}{4}=12$<\/p>\n\n\n\n If y=12 then $x=\\frac{48}{12}=4$<\/p>\n\n\n\n Hence, the two numbers are 4 and 12.<\/p>\n\n\n\n Two positive numbers differ by 3 and their product is 54. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On putting the value of $x=\\frac{54}{y}$ in Eq. (i), we get \u21d2 y =6, then $x=\\frac{54}{6}=9$<\/p>\n\n\n\n Hence, the two numbers are 9 and 6.<\/p>\n\n\n\n Two numbers are in the ratio of 3 : 5. If 5 is subtracted from each of the number they become in the ratio of 1 : 2. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : According to the question, On putting the value of x = 15 in Eq. (i), we get Hence, the two numbers are 15 and 25.<\/p>\n\n\n\n Two numbers are in the ratio of 3 : 4. If 8 is added to each number, they become in the ratio of 4 : 5. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Also, $\\frac{x+8}{y+8}=\\frac{4}{5}$ On putting the value of $y=\\frac{6 x}{5}$ in Eq. (ii), we get On putting the value of x = 24 in Eq. (i), we get Hence, the two numbers are 24 and 32.<\/p>\n\n\n\n Two numbers differ by 2 and their product is 360. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On putting the value of $x=\\frac{360}{y}$ in Eq. (i), we get \u21d2 y = 18, then $x=\\frac{360}{18}=20$<\/p>\n\n\n\n Hence, the two numbers are 20 and 18.<\/p>\n\n\n\n Two numbers differ by 4 and their product is 192. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Also, x\u00d7y = 192 On putting the value of $x=\\frac{192}{y}$ in Eq. (i), we get But y = \u2013 16 can\u2019t be the one number as it is given that the numbers are positive.<\/p>\n\n\n\n \u21d2 y = 12, then $x=\\frac{192}{12}=16$<\/p>\n\n\n\n Hence, the two numbers are 16 and 12.<\/p>\n\n\n\n Two numbers differ by 4 and their product is 96. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Also, x\u00d7y = 96 On putting the value of $x=\\frac{96}{y}$ in Eq. (i), we get But y = \u2013 8 can\u2019t be the one number as it is given that the numbers are positive.<\/p>\n\n\n\n \u21d2 y = 12,then $x=\\frac{96}{12}=8$ Hence, the two numbers are 8 and 12.<\/p>\n\n\n\n The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly expenditures are in the ratio of 7 : 5. If each saves Rs. 3000 per month, find the monthly income of each.<\/strong><\/p>\n\n\n\n Sol : Scooter charges consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a person travels 12 km, he pays Rs. 45 and for travelling 20 km, he pays Rs. 73. Express the above statements in the form of simultaneous equations and hence, find the fixed charges and the rate per km.<\/strong><\/p>\n\n\n\n Sol : On putting the value of y = 3.5 in Eq. (i), we get A part of monthly hostel charges in a college is fixed and the remaining depend on the number of days one has taken food in the mess. When a student A, takes food for 22 days, he has to pay Rs. 1380 as hostel charges, whereas a student B, who takes food for 28 days, pays Rs. 1680 as hostel charges. Find the fixed charge and the cost of food per day.<\/strong><\/p>\n\n\n\n Sol : Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels 110 km, he pays Rs. 690, and for travelling 200 km, he pays Rs. 1050. Find the fixed charges per day and the rate per km.<\/strong><\/p>\n\n\n\n Sol : A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs. 1750 as hostel charges whereas a student a d who takes food for 28 days, pays Rs. 1900 as hostel charges. Find the fixed charges and the cost of the food per day.<\/strong><\/p>\n\n\n\n Sol : The total expenditure per month of a household consists of a fixed rent of the house and the mess charges, depending upon the number of people sharing the house. The total monthly expenditure is Rs. 3,900 for 2 people and Rs. 7,500 for 5 people. Find the rent of the house and the mess charges per head per month.<\/strong><\/p>\n\n\n\n Sol : The car rental charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 13 km, the charge paid is Rs. 96 and for a journey of 18 km, the charge paid is Rs. 131. What will a person have to pay for travelling a distance of 25 km?<\/strong><\/p>\n\n\n\n Sol : The sum of a two – digit number and the number formed by interchanging the digits is 132. If 12 is added to the number, the new number becomes 5 times the sum of the digits. Find the number.<\/strong><\/p>\n\n\n\n Sol : A two – digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.<\/strong><\/p>\n\n\n\n Sol : A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.<\/strong><\/p>\n\n\n\n Sol : According to the question, On substituting the value of x = y + 1 in Eq. (i), we get The sum of the digits of a two – digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.<\/strong><\/p>\n\n\n\n Sol : A two – digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.<\/strong><\/p>\n\n\n\n Sol : A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.<\/strong><\/p>\n\n\n\n Sol : The sum of the digits of a two – digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.<\/strong><\/p>\n\n\n\n Sol : The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.<\/strong><\/p>\n\n\n\n Sol :<\/p>\n\n\n\n Let the numerator = x So, the fraction $=\\frac{x}{y}$ Condition I: Condition II: Hence, the fraction is $\\frac{4}{7}$<\/p>\n\n\n\n The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, then are in the ratio 2 : 3. Determine the fraction.<\/strong><\/p>\n\n\n\n Sol : So, the fraction $=\\frac{x}{y}$ According to the question, Condition I: Condition II: On putting the value of y in Eq.(ii) , we get On putting the value of x in Eq. (i), we get The sum of the numerator and denominator of a fraction is 8. If 3 is added to both 3 the numerator and the denominator, the fraction becomes 3\/4. Find the fraction.<\/strong><\/p>\n\n\n\n Sol : According to the question, So, the numerator is 3 and the denominator is 5 Hence, the fraction is $\\frac{3}{5}$<\/p>\n\n\n\n The numerator of a fraction is one less than its denominator. If 3 is added to each of the numerator and denominator, the fraction is increased by 3\/28, find the fraction. 28<\/strong><\/p>\n\n\n\n Sol : So, the fraction $=\\frac{x-1}{x}$ According to the question, The age of the father is 3 years more than 3 times the son’s age. 3 years here, the age of the father will be 10 years more than twice the age of the son. Find their present ages.<\/strong><\/p>\n\n\n\n Sol : Two years ago, a man was five times as old as his son. Two years later his age will be 8 more than three times the age of the son. Find the present ages of man and his son.<\/strong><\/p>\n\n\n\n Sol : Father’s age is three times the sum of ages of his two children. After 5 years, his age will be twice the sum of ages of two children. Find the age of father.<\/strong><\/p>\n\n\n\n Sol : Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?<\/strong><\/p>\n\n\n\n Sol : Ten years hence, a man’s age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages.<\/strong><\/p>\n\n\n\n Sol : Find a cyclic quadrilateral ABCD, <\/strong>\u2220A = (2x + 4)\u00b0, \u2220B = (y + 3)\u00b0, \u2220C = (2y + 10) and \u2220D = (4x-5)\u00b0. Find the four angles.<\/strong><\/p>\n\n\n\n Sol :<\/p>\n\n\n\n \u2234\u2220A + \u2220C = 180\u00b0 and \u2220B + \u2220D = 180\u00b0 So, we get pair of linear equation i.e. On subtracting Eq.(i) from (ii), we get On putting the value of x = 33 in Eq. (i) we get, On putting the values of x and y, we calculate the angles as Hence, the angles are \u2220A = 63\u00b0, \u2220B = 57\u00b0, \u2220C = 117\u00b0, \u2220D = 123\u00b0<\/p>\n\n\n\n Find the four angles of a cyclic quadrilateral ABCD in which \u2220A = (2x \u2014 3)\u00b0,<\/strong><\/p>\n\n\n\n \u2220B = (y + 7)\u00b0, \u2220C = (2y + 17)\u00b0 and \u2220D = (4x \u2014 9)\u00b0.<\/strong> \u2234\u2220A + \u2220C = 180\u00b0 and \u2220B + \u2220D = 180\u00b0 On subtracting Eq.(i) from (ii), we get On putting the value of x = 33 in Eq. (i) we get, On putting the values of x and y, we calculate the angles as In a \u0394ABC, \u2220C = 3 \u2220B = 2 (\u2220A + \u2220B). Find the three angles.<\/strong><\/p>\n\n\n\n \u2220A = 20\u00b0, \u2220B = 40\u00b0, \u2220C = 120\u00b0.<\/strong> We know that, in a triangle , the sum of three angles is 180\u00b0 On taking II and III, we get Now, on taking I and II, we get \u21d2 \u2220C = 3(2 \u2220A) (from eq. (i)) On substituting the value of \u2220B and \u2220C in Eq. (a), we get In a \u0394ABC, \u2220A = x\u00b0, \u2220B = (3x)\u00b0 and \u2220C = y\u00b0.<\/strong><\/p>\n\n\n\n If 3y \u2014 5x = 30, show that the triangle is right \u2013 angled.<\/strong><\/p>\n\n\n\n Sol : The area of a rectangle gets reduced by 8 m2<\/sup>, when its length is reduced by 5m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m2<\/sup>. Find the length and the breadth of the rectangle.<\/strong><\/p>\n\n\n\n Sol : The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.<\/strong><\/p>\n\n\n\n Sol : Two places A and B are 120 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 6 hours, and if they move in opposite directions, they meet in 1 hour 12 minutes. Find the speed of each car.<\/strong><\/p>\n\n\n\n Sol : 1hr 12min $=1+\\frac{12}{60}=\\frac{6}{5} \\mathrm{hr}$ Since they are travelling in opposite directions, sign should be positive. On adding (i) and (ii) , we get A train travels a distance of 300 km at a constant speed. If the speed of SE the 2O train is increased by 5 km an hour, the journey would have taken 2 hours less. Find the original speed of the train.<\/strong><\/p>\n\n\n\n Sol :<\/p>\n\n\n\n Total distance travelled = 300km We know that, Hence, time taken by train $=\\frac{300}{x}$ According to the question, Time taken to cover 300km $=\\frac{300}{x+5}$ Given that time taken is 2hrs less from the previous time A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km\/hr from the usual speed. Find its usual speed.<\/strong><\/p>\n\n\n\n Sol :<\/p>\n\n\n\n Let the usual time taken by the aeroplane = x km\/hr Hence, speed $=\\frac{1500}{\\mathrm{x}} \\mathrm{hrs}$ According to the question, 30min $=\\frac{30}{60}=\\frac{1}{2} \\mathrm{hr}$ Time taken by the aeroplane $=x-\\frac{1}{2} h r s$ \u2234 the speed of the plane $=\\frac{1500}{x-\\frac{1}{2}}$ Given that speed has to increase by 250 km\/hr \u2234 $x=\\frac{3}{2}$ or x = 2 Since, time can\u2019t be negative. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.<\/strong><\/p>\n\n\n\n Sol : Time take = 6hrs 30min <\/p>\n\n\n\n $=6+\\frac{30}{60}=6.5 \\mathrm{hrs}$ If he travels 200km by train and rest by car i.e. (600 \u2013 200) = 400km We know that, Let take $\\frac{1}{\\mathrm{x}}=\\mathrm{u}$ and $\\frac{1}{y}=v$<\/p>\n\n\n\n 400u + 200v = 6.5 \u2026(iii) On multiplying Eq. (iii) by 2 and Eq. (iv) by 4, we get On putting the value of v in Eq. (iv), we get So, we get $\\mathrm{u}=\\frac{1}{100}$ and $v=\\frac{1}{80}$ Hence, the speed of the train is 100km\/hr and the speed of the car is 80km\/hr.<\/p>\n\n\n\n Places A and B are 80 km apart from each other on a highway. One car starts from A and another from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find speed of the cars.<\/strong><\/p>\n\n\n\n Sol : Now, Let two cars meet after 1hr 20 min Since they are travelling in opposite directions, sign should be positive.<\/p>\n\n\n\n $\\frac{4}{3} x+\\frac{4}{3} y=80$ A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream in the same time. Find the speed of the boat in still water and that of the stream.<\/strong><\/p>\n\n\n\n Sol : According to the question<\/p>\n\n\n\n ConditionI:<\/p>\n\n\n\n When boat goes 16 km upstream, let the time taken be t1<\/sub>. $\\mathrm{t}_{1}=\\frac{16}{\\mathrm{x}-\\mathrm{y}} \\mathrm{h}$ $\\left[\\because\\right.$ time $\\left.=\\frac{\\text { distance }}{\\text { speed }}\\right]$<\/p>\n\n\n\n When boat goes 24 km downstream, let the time taken be t2<\/sub>. Then, But total time taken (t1<\/sub> + t2<\/sub>) = 6 hours<\/p>\n\n\n\n $\\therefore \\frac{16}{x-y}+\\frac{24}{x+y}=6$ \u2026(a)<\/p>\n\n\n\n Condition II:<\/p>\n\n\n\n When boat goes 12 km upstream, let the time taken be T1<\/sub>. $\\mathrm{T}_{1}=\\frac{12}{\\mathrm{x}-\\mathrm{y}} \\mathrm{h}$ $\\left[\\because\\right.$ time $\\left.=\\frac{\\text { distance }}{\\text { speed }}\\right]$ When boat goes 36 km downstream, let the time taken be T2<\/sub>. Then, But total time taken (T1<\/sub> + T2<\/sub>) = 6 hours Now, we solve tis pair of linear equations by elimination method On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients equal of first term, we get the equation as On substracting Eq. (iii) from Eq. (iv), we get On putting the value of x \u2013 y = 12 in Eq. (i), we get Adding Eq. (a) and (b), we get On putting value of x = 8 in eq. (a), we get Hence, x = 8 and y = 4 , which is the required solution. Hence, the speed of the boat in still water is 8km\/hr and speed of the stream is 4km\/hr<\/p>\n\n\n\n A man travels 370 km, partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.<\/strong><\/p>\n\n\n\n Sol : According to the question, If he covers 250km by train and rest by car i.e. (370 \u2013 250) = 120km If he travels 130km by train and rest by car i.e. (370 \u2013 130) = 240km He takes 18min longer i.e. $4+\\frac{18}{60}=4.3$hours So, total time = train time + car time We know that, $\\Rightarrow \\frac{250}{x}+\\frac{120}{y}=4$ \u2026(i)<\/p>\n\n\n\n $\\Rightarrow \\frac{130}{\\mathrm{x}}+\\frac{240}{\\mathrm{y}}=4.3$ \u2026(ii) Let take $\\frac{1}{x}=u$ and $\\frac{1}{\\mathrm{y}}=\\mathrm{v}$ 250u + 120v = 4 \u2026(iii) On multiplying Eq. (iii) by 2 On subtracting Eq. (iv) from Eq. (v), we get On putting the value of v in Eq. (iv), we get $\\Rightarrow \\mathrm{v}=\\frac{1}{80}$ So, we get $u=\\frac{1}{100}$ and $v=\\frac{1}{80}$ \u21d2 x = 100 and y = 80 Hence, the speed of the train is 100km\/hr and the speed of the car is 80km\/hr.<\/p>\n\n\n\n
$\\frac{72}{y}+y=18$
\u21d2 72 + y2<\/sup> = 18y
\u21d2 y2<\/sup> \u2013 18y + 72 = 0
\u21d2 y2<\/sup> \u2013 12y \u2013 6y + 72 = 0
\u21d2 y(y \u2013 12) \u2013 6(y \u2013 12) = 0
\u21d2 (y \u2013 6)(y \u2013 12) = 0
\u21d2 y = 6 and 12<\/p>\n\n\n\n
\n\n\n\nQuestion 2 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x + y = 15 \u2026(i)
$\\frac{1}{x}+\\frac{1}{y}=\\frac{3}{10}$ \u2026(ii)
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\frac{15}{x y}=\\frac{3}{10}$
\u21d2 xy = 50
$\\Rightarrow x=\\frac{50}{y}$
<\/p>\n\n\n\n
$\\frac{50}{y}+y=18$
\u21d2 50 + y2<\/sup> = 15y
\u21d2 y2<\/sup> \u2013 15y + 50 = 0
\u21d2 y2<\/sup> \u2013 10y \u2013 5y + 50 = 0
\u21d2 y(y \u2013 10) \u2013 5(y \u2013 10) = 0
\u21d2 (y \u2013 5)(y \u2013 10) = 0<\/p>\n\n\n\n
\n\n\n\nQuestion 3 <\/h4>\n\n\n\n
Let the two numbers be x and y.
<\/p>\n\n\n\n
<\/p>\n\n\n\n
\u21d2 5(x \u2013 8) = 4(y \u2013 8)
\u21d2 5x \u2013 40 = 4y \u2013 32
\u21d2 5x \u2013 4y = 8 \u2026(ii)
<\/p>\n\n\n\n
$5 x-4\\left(\\frac{6 x}{5}\\right)=8$
$\\Rightarrow \\frac{25 x-24 x}{5}=8$
\u21d2 x = 40
<\/p>\n\n\n\n
$y=\\frac{6 \\times 40}{5}=48$
<\/p>\n\n\n\n
\n\n\n\nQuestion 4 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x + y = 16 \u2026(i)
$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{3}$ \u2026(ii)
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
\u21d2xy = 48<\/p>\n\n\n\n
$\\frac{48}{y}+y=16$
\u21d2 48 + y2<\/sup> = 16y
\u21d2 y2<\/sup> \u2013 16y + 48 = 0
\u21d2 y2<\/sup> \u2013 12y \u2013 4y + 48 = 0
\u21d2 y(y \u2013 12) \u2013 4(y \u2013 12) = 0
\u21d2 (y \u2013 4)(y \u2013 12) = 0
\u21d2 y = 4 and 12<\/p>\n\n\n\n
\n\n\n\nQuestion 5 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x \u2013 y = 3 \u2026(i)
Also, x\u00d7y = 54
$\\Rightarrow \\mathrm{x}=\\frac{54}{\\mathrm{y}}$ \u2026(ii)
<\/p>\n\n\n\n
$\\frac{54}{y}-y=3$
\u21d2 54 \u2013 y2<\/sup> = 3y
\u21d2 y2<\/sup> + 3y \u2013 54 = 0
\u21d2 y2<\/sup> + 9y \u2013 6y \u2013 54 = 0
\u21d2 y(y + 9) \u2013 6(y + 9) = 0
\u21d2 (y \u2013 6)(y + 9) = 0
\u21d2 y = \u2013 9 and 6
But y = \u2013 9 can\u2019t be the one number as it is given that the numbers are positive.<\/p>\n\n\n\n
\n\n\n\nQuestion 6 <\/h4>\n\n\n\n
Let the two numbers be x and y.
<\/p>\n\n\n\n
$\\frac{x}{y}=\\frac{3}{5}$
$\\Rightarrow y=\\frac{5 x}{3}$ \u2026(i)
Also, $\\frac{x-5}{y-5}=\\frac{1}{2}$
\u21d2 2(x \u2013 5) = (y \u2013 5)
\u21d2 2x \u2013 10 = y \u2013 5
\u21d2 2x \u2013 y = 5 \u2026(ii)
On putting the value of $y=\\frac{5 x}{3}$ in Eq. (ii), we get
$2 x-\\left(\\frac{5 x}{3}\\right)=5$
$\\Rightarrow \\frac{6 x-5 x}{3}=5$
\u21d2 x = 15<\/p>\n\n\n\n
$y=\\frac{5 \\times 15}{3}=25$<\/p>\n\n\n\n
\n\n\n\nQuestion 7 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
$\\frac{x}{y}=\\frac{3}{4}$
$\\Rightarrow \\mathrm{y}=\\frac{4 \\mathrm{x}}{3}$ \u2026(i)
<\/p>\n\n\n\n
\u21d2 5(x + 8) = 4(y + 8)
\u21d2 5x + 40 = 4y + 32
\u21d2 5x \u2013 4y = \u2013 8 \u2026(ii)
<\/p>\n\n\n\n
$5 x-4\\left(\\frac{4 x}{3}\\right)=-8$
$\\Rightarrow \\frac{15 x-16 x}{3}=-8$
\u21d2 x = 24
<\/p>\n\n\n\n
$y=\\frac{4 \\times 24}{3}=32$
<\/p>\n\n\n\n
\n\n\n\nQuestion 8 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x \u2013 y = 2 \u2026(i)
Also, x\u00d7y = 360
$\\Rightarrow x=\\frac{360}{y}$ \u2026(ii)
<\/p>\n\n\n\n
$\\frac{360}{y}-y=2$
\u21d2 360 \u2013 y2<\/sup> = 2y
\u21d2 y2<\/sup> + 2y \u2013 360 = 0
\u21d2 y2<\/sup> + 20y \u2013 18y \u2013 360 = 0
\u21d2 y(y + 20) \u2013 18(y + 20) = 0
\u21d2 (y \u2013 18)(y + 20) = 0
\u21d2 y = \u2013 20 and 18
But y = \u2013 20 can\u2019t be the one number as it is given that the numbers are positive.<\/p>\n\n\n\n
\n\n\n\nQuestion 9 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x \u2013 y = 4 \u2026(i)
<\/p>\n\n\n\n
$\\Rightarrow x=\\frac{192}{y}$ \u2026(ii)
<\/p>\n\n\n\n
$\\frac{192}{\\mathrm{y}}-\\mathrm{y}=4$
\u21d2 192 \u2013 y2<\/sup> = 4y
\u21d2 y2<\/sup> + 4y \u2013 192 = 0
\u21d2 y2<\/sup> + 16y \u2013 12y \u2013 192 = 0
\u21d2 y(y + 16) \u2013 12(y + 16) = 0
\u21d2 (y \u2013 12)(y + 16) = 0
\u21d2 y = \u2013 16 and 12
<\/p>\n\n\n\n
\n\n\n\nQuestion 10 <\/h4>\n\n\n\n
Let the two numbers be x and y.
According to the question,
x \u2013 y = 4 \u2026(i)
<\/p>\n\n\n\n
$\\Rightarrow \\mathrm{x}=\\frac{96}{\\mathrm{y}}$ \u2026(ii)
<\/p>\n\n\n\n
$\\frac{96}{y}-y=4$
\u21d2 96 \u2013 y2<\/sup> = 4y
\u21d2 y2<\/sup> + 4y \u2013 96 = 0
\u21d2 y2<\/sup> + 12y \u2013 8y \u2013 96 = 0
\u21d2 y(y + 12) \u2013 8(y + 12) = 0
\u21d2 (y \u2013 8)(y + 12) = 0
\u21d2 y = \u2013 8 and 12
<\/p>\n\n\n\n
<\/p>\n\n\n\n
\n\n\n\nQuestion 11 <\/h4>\n\n\n\n
Given ratio of incomes = 5:4
And the ratio of their expenditures = 7:5
Saving of each person = Rs. 3000
Let incomes of two persons = 5x and 4x
And their expenditures = 7y and 5y
According to the question,
5x \u2013 7y = 3000 \u2026(i)
4x \u2013 5y = 3000 \u2026(ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 5 to make the coefficients of x equal, we get
20x \u2013 28y = 12000 \u2026(iii)
20x \u2013 25y = 15000 \u2026(iv)
On subtracting Eq. (iii) from (iv), we get
20x \u2013 25y \u2013 20x + 28y = 15000 \u2013 12000
\u21d2 3y = 3000
\u21d2 y = 1000
On putting the y = 1000 in Eq. (i), we get
5x \u2013 7y = 3000
\u21d2 5x \u2013 7(1000) = 3000
\u21d2 5x = 10000
\u21d2 x = 2000
Thus, monthly income of both the persons are 5(2000) and 4(2000), i.e. Rs. 10000 and Rs. 8000<\/p>\n\n\n\n
\n\n\n\nQuestion 12 <\/h4>\n\n\n\n
Let fixed charge = Rs x
and charge per kilometer = Rs y
According to the question,
x + 12y = 45 \u2026(i)
and x + 20y = 73 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 20y \u2013 x \u2013 12y = 73 \u2013 45
\u21d2 8y = 28
$\\Rightarrow \\mathrm{y}=\\frac{28}{8}=3.5$
<\/p>\n\n\n\n
x + 12(3.5) = 45
\u21d2 x + 42 = 45
\u21d2 x = 45 \u2013 42 = 3
Hence, monthly fixed charges is Rs. 3 and charge per kilometer is Rs. 3.5<\/p>\n\n\n\n
\n\n\n\nQuestion 13 <\/h4>\n\n\n\n
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 22y = 1380 \u2026(i)
In case of student B,
x + 28y = 1680 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y \u2013 x \u2013 22y = 1680 \u2013 1380
\u21d2 6y = 300
\u21d2 y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 22(50) = 1380
\u21d2 x + 1100 = 1380
\u21d2 x = 1380 \u2013 1100 = 280
Hence, monthly fixed charges is Rs. 280 and cost of food per day is Rs. 50<\/p>\n\n\n\n
\n\n\n\nQuestion 14 <\/h4>\n\n\n\n
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 110y = 690 \u2026(i)
and x + 200y = 1050 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 200y \u2013 x \u2013 110y = 1050 \u2013 690
\u21d2 90y = 360
\u21d2 y = 40
On putting the value of y = 40 in Eq. (i), we get
x + 110(40) = 690
\u21d2 x + 440 = 690
\u21d2 x = 690 \u2013 440 = 250
Hence, monthly fixed charges is Rs. 250 and charge per kilometer is Rs. 40<\/p>\n\n\n\n
\n\n\n\nQuestion 15 <\/h4>\n\n\n\n
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 25y = 1750 \u2026(i)
In case of student B,
x + 28y = 1900 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y \u2013 x \u2013 25y = 1900 \u2013 1750
\u21d2 3y = 150
\u21d2 y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 25(50) = 1750
\u21d2 x + 1250 = 1750
\u21d2 x = 1750 \u2013 1250 = 500
Hence, monthly fixed charges is Rs. 500 and cost of food per day is Rs. 50<\/p>\n\n\n\n
\n\n\n\nQuestion 16 <\/h4>\n\n\n\n
Let fixed rent of the house = Rs. x
And the mess charges per hesd per month = Rs. y
According to the question,
x + 2y = 3900 \u2026(i)
and x + 5y = 7500 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 5y \u2013 x \u2013 2y = 7500 \u2013 3900
\u21d2 3y = 3600
\u21d2 y = 1200
On putting the value of y = 1200 in Eq. (i), we get
x + 2 (1200) = 3900
\u21d2 x + 2400 = 3900
\u21d2 x = 1500
Hence, fixed rent of the house is Rs. 1500 and the mess charges per head per month is Rs. 1200.<\/p>\n\n\n\n
\n\n\n\nQuestion 17 <\/h4>\n\n\n\n
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 13y = 96 \u2026(i)
and x + 18y = 131 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 18y \u2013 x \u2013 13y = 131 \u2013 96
\u21d2 5y = 35
\u21d2 y = 7
On putting the value of y = 7 in Eq. (i), we get
x + 13 (7) = 96
\u21d2 x + 91 = 96
\u21d2 x = 5
Hence, monthly fixed charges is Rs. 5 and charge per kilometer is Rs. 7
Now, amount to be paid for travelling 25 km
= Fixed charge + Rs 7 \u00d725
= 5 + 175
= Rs. 180
Hence, the amount paid by a person for travelling 25km is Rs. 180<\/p>\n\n\n\n
\n\n\n\nQuestion 18 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
After interchanging the digits, New number = x + 10y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) + (x + 10y) = 132
\u21d2 11x + 11y = 132
\u21d2 11(x + y) = 132
\u21d2 x + y = 12 \u2026(i)
and 10x + y + 12 = 5(x + y)
\u21d2 10x + y + 12 = 5x + 5y
\u21d2 10x \u2013 5x + y \u2013 5y = \u2013 12
\u21d2 5x \u2013 4y = \u2013 12 \u2026(ii)
From Eq. (i), we get
x = 12 \u2013 y \u2026(iii)
On substituting the value of x = 12 \u2013 y in Eq. (ii), we get
5(12 \u2013 y) \u2013 4y = \u2013 12
\u21d2 60 \u2013 5y \u2013 4y = \u2013 12
\u21d2 \u2013 9y = \u2013 12 \u2013 60
\u21d2 \u2013 9y = \u2013 72
\u21d2 y = 8
On putting the value of y = 8 in Eq. (iii), we get
x = 12 \u2013 8 = 4
So, the Original number = 10x + y
= 10\u00d74 + 8
= 48
Hence, the two digit number is 48.<\/p>\n\n\n\n
\n\n\n\nQuestion 19 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the two digit number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) = 4(x + y)
\u21d2 10x + y = 4x + 4y
\u21d2 10x \u2013 4x + y \u2013 4y = 0
\u21d2 6x \u2013 3y = 0
\u21d2 2x \u2013 y = 0
\u21d2 y = 2x \u2026(i)
After interchanging the digits, New number = x + 10y
and 10x + y + 18 = x + 10y
\u21d2 10x + y + 18 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = \u2013 18
\u21d2 9x \u2013 9y = \u2013 18
\u21d2 x \u2013 y = \u2013 2 \u2026(ii)
On substituting the value of y = 2x in Eq. (ii), we get
x \u2013 y = \u2013 18
\u21d2 x \u2013 2x = \u2013 2
\u21d2 \u2013 x = \u2013 2
\u21d2 x = 2
On putting the value of x = 2 in Eq. (i), we get
y = 2\u00d72 = 4
So, the Original number = 10x + y
= 10\u00d72 + 4
= 20 + 4
= 24
Hence, the two digit number is 24.<\/p>\n\n\n\n
\n\n\n\nQuestion 20 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
<\/p>\n\n\n\n
$\\frac{(10 x+y)}{x+y}=6$
\u21d2 10x + y = 6(x + y)
\u21d2 10x + y = 6x + 6y
\u21d2 10x + y \u2013 6x \u2013 6y
\u21d2 4x \u2013 5y = 0 \u2026(i)
The reverse of the number = x + 10y
and 10x + y \u2013 9 = x + 10y
\u21d2 10x + y \u2013 9 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = 9
\u21d2 9x \u2013 9y = 9
\u21d2 x \u2013 y = 1
\u21d2 x = y + 1 \u2026(ii)
<\/p>\n\n\n\n
4x \u2013 5y = 0
\u21d2 4(y + 1) \u2013 5y = 0
\u21d2 4y + 4 \u2013 5y = 0
\u21d2 4 \u2013 y = 0
\u21d2 y = 4
On substituting the value of y = 4 in Eq. (ii), we get
x = y + 1
\u21d2 x = 4 + 1
\u21d2 x = 5
So, the Original number = 10x + y
= 10\u00d75 + 4
= 50 + 4
= 54
Hence, the two digit number is 54.<\/p>\n\n\n\n
\n\n\n\nQuestion 21 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 12 \u2026(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 18 = x + 10y
\u21d2 10x + y + 18 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = \u2013 18
\u21d2 9x \u2013 9y = \u2013 18
\u21d2 x \u2013 y = \u2013 2 \u2026(ii)
On adding Eq. (i) and (ii) , we get
x + y + x \u2013 y = 12 \u2013 2
\u21d2 2x = 10
\u21d2 x = 5
On substituting the value of x = 5 in Eq. (i), we get
x + y = 12
\u21d2 5 + y = 12
\u21d2 y = 7
So, the Original number = 10x + y
= 10\u00d75 + 7
= 50 + 7
= 57
Hence, the two digit number is 57.<\/p>\n\n\n\n
\n\n\n\nQuestion 22 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 3 + 4(x + y)
\u21d2 10x + y = 3 + 4x + 4y
\u21d2 10x + y \u2013 4x \u2013 4y = 3
\u21d2 6x \u2013 3y = 3
\u21d2 2x \u2013 y = 1 \u2026(i)
The reverse number = x + 10y
and 10x + y + 18 = x + 10y
\u21d2 10x + y + 18 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = \u2013 18
\u21d2 9x \u2013 9y = \u2013 18
\u21d2 x \u2013 y = \u2013 2 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii) , we get
x \u2013 y \u2013 2x + y = \u2013 2 \u2013 1
\u21d2 \u2013 x = \u2013 3
\u21d2 x = 3
On substituting the value of x = 3 in Eq. (i), we get
2(3) \u2013 y = 1
\u21d2 6 \u2013 y = 1
\u21d2 \u2013 y = 1 \u2013 6
\u21d2 \u2013 y = \u2013 5
\u21d2 y = 5
So, the Original number = 10x + y
= 10\u00d73 + 5
= 30 + 5
= 35
Hence, the two digit number is 35.<\/p>\n\n\n\n
\n\n\n\nQuestion 23 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 7(x + y)
\u21d2 10x + y = 7x + 7y
\u21d2 10x + y \u2013 7x \u2013 7y = 0
\u21d2 3x \u2013 6y = 0
\u21d2 x \u2013 2y = 0
\u21d2 x = 2y \u2026(i)
The reverse number = x + 10y
and 10x + y \u2013 27 = x + 10y
\u21d2 10x + y \u2013 27 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = 27
\u21d2 9x \u2013 9y = 27
\u21d2 x \u2013 y = 3 \u2026(ii)
On substituting the value of x = 2y in Eq. (ii), we get
x \u2013 y = 3
\u21d2 2y \u2013 y = 3
\u21d2 y = 3
On putting the value of y = 3 in Eq. (i), we get
x = 2(3) = 6
So, the Original number = 10x + y
= 10\u00d76 + 3
= 60 + 3
= 63
Hence, the two digit number is 63.<\/p>\n\n\n\n
\n\n\n\nQuestion 24 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 15 \u2026(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 9 = x + 10y
\u21d2 10x + y + 9 = x + 10y
\u21d2 10x \u2013 x + y \u2013 10y = \u2013 9
\u21d2 9x \u2013 9y = \u2013 9
\u21d2 x \u2013 y = \u2013 1 \u2026(ii)
On adding Eq. (i) and (ii) , we get
x + y + x \u2013 y = 15 \u2013 1
\u21d2 2x = 14
\u21d2 x = 7
On substituting the value of x = 5 in Eq. (i), we get
x + y = 15
\u21d2 7 + y = 15
\u21d2 y = 8
So, the Original number = 10x + y
= 10\u00d77 + 8
= 70 + 8
= 78
Hence, the two digit number is 78.<\/p>\n\n\n\n
\n\n\n\nQuestion 25 <\/h4>\n\n\n\n
and the denominator = y
<\/p>\n\n\n\n
According to the question,
<\/p>\n\n\n\n
x + y = 2y \u2013 3
\u21d2 x + y \u2013 2y = \u2013 3
\u21d2 x \u2013 y = \u2013 3 \u2026(i)
<\/p>\n\n\n\n
$\\frac{x-1}{y-1}=\\frac{1}{2}$
\u21d2 2(x \u2013 1) = y \u2013 1
\u21d2 2x \u2013 2 = y \u2013 1
\u21d2 2x \u2013 y = 1 \u2026(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2x \u2013 y \u2013 x + y = 1 + 3
\u21d2 x = 4
On putting the value of x in Eq. (i), we get
4 \u2013 y = \u2013 3
\u21d2 y = 7
So, the numerator is 4 and the denominator is 7
<\/p>\n\n\n\n
\n\n\n\nQuestion 26 <\/h4>\n\n\n\n
Let the numerator = x
and the denominator = y
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
x + y = 2x + 4
\u21d2 x + y \u2013 2x = 4
\u21d2 \u2013 x + y = 4
\u21d2 y = 4 + x \u2026(i)
<\/p>\n\n\n\n
$\\frac{x+3}{y+3}=\\frac{2}{3}$
\u21d2 3(x + 3) = 2( y + 3)
\u21d2 3x + 9 = 2y + 6
\u21d2 3x \u2013 2y = \u2013 3 \u2026(ii)
<\/p>\n\n\n\n
3x \u2013 2(4 + x) = \u2013 3
\u21d2 3x \u2013 8 \u2013 2x = \u2013 3
\u21d2 x = 5
<\/p>\n\n\n\n
y = 4 + 5
\u21d2 y = 9
So, the numerator is 5 and the denominator is 9
Hence, the fraction is $\\frac{5}{9}$<\/p>\n\n\n\n
\n\n\n\nQuestion 27 <\/h4>\n\n\n\n
Let the numerator = x
and the denominator = y
So, the fraction $=\\frac{x}{y}$
<\/p>\n\n\n\n
Condition I:
x + y = 8
\u21d2 y = 8 \u2013 x \u2026(i)
Condition II:
$\\frac{x+3}{y+3}=\\frac{3}{4}$
\u21d2 4(x + 3) = 3( y + 3)
\u21d2 4x + 12 = 3y + 9
\u21d2 4x \u2013 3y = \u2013 3 \u2026(ii)
On putting the value of y in Eq.(ii) , we get
4x \u2013 3(8 \u2013 x) = \u2013 3
\u21d2 4x \u2013 24 + 3x = \u2013 3
\u21d2 7x = 21
\u21d2 x = 3
On putting the value of x in Eq. (i), we get
y = 8 \u2013 3
\u21d2 y = 5
<\/p>\n\n\n\n
<\/p>\n\n\n\n
\n\n\n\nQuestion 28 <\/h4>\n\n\n\n
Let the denominator = x
Given that numerator is one less than the denominator
\u21d2 numerator = x \u2013 1
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\frac{x-1+3}{x+3}=\\frac{x-1}{x}+\\frac{3}{28}$
$\\Rightarrow \\frac{x+2}{x+3}=\\frac{x-1}{x}+\\frac{3}{28}$
$\\Rightarrow \\frac{x+2}{x+3}-\\frac{x-1}{x}=\\frac{3}{28}$
$\\Rightarrow \\frac{(x+2) x-(x+3)(x-1)}{(x+3)(x)}=\\frac{3}{28}$
\u21d2 28{(x2<\/sup> + 2x) \u2013 (x2<\/sup> \u2013x + 3x \u2013 3)} = 3 (x2<\/sup> + 3x)
\u21d2 28x2<\/sup> + 56x \u2013 28x2<\/sup> \u2013 56x + 84 = 3x2<\/sup> + 9x
\u21d2 3x2<\/sup> + 9x \u2013 84 = 0
\u21d2 x2<\/sup> + 3x \u2013 28 = 0
\u21d2 x2<\/sup> + 7x \u2013 4x \u2013 28 = 0
\u21d2 x(x + 7) \u2013 4(x + 7) = 0
\u21d2 (x \u2013 4) (x + 7) = 0
\u21d2 x = 4 and \u2013 7
But x is a natural number
Hence, x = 4
So, the fraction is $\\frac{x-1}{x}=\\frac{4-1}{4}=\\frac{3}{4}$<\/p>\n\n\n\n
\n\n\n\nQuestion 29 <\/h4>\n\n\n\n
Let the age of father = x years
And the age of his son = y years
According to the question,
x = 3 + 3y …(i)
Three year here,
Father\u2019s age = (x + 3) years
Son\u2019s age = (y + 3) years
According to the question,
(x + 3) = 10 + 2(y + 3)
\u21d2 x + 3 = 10 + 2y + 6
\u21d2 x = 2y + 13 \u2026(ii)
From Eq. (i) and (ii), we get
3 + 3y = 13 + 2y
\u21d2 3y \u2013 2y = 13 \u2013 3
\u21d2 y = 10
On putting the value of y = 7 in Eq. (i), we get
x = 3 + 3(10)
\u21d2 x = 3 + 30
\u21d2 x = 33
Hence, the age of father is 33 years and the age of his son is 10 years.<\/p>\n\n\n\n
\n\n\n\nQuestion 30 <\/h4>\n\n\n\n
Let the age of a man = x years
And the age of his son = y years
Two years ago,
Man\u2019s age = (x \u2013 2) years
Son\u2019s age = (y \u2013 2) years
According to the question,
(x \u2013 2) = 5(y \u2013 2)
\u21d2 x \u2013 2 = 5y \u2013 10
\u21d2 x = 5y \u2013 10 + 2
\u21d2 x = 5y \u2013 8 …(i)
Two years later,
Father\u2019s age = (x + 2) years
Son\u2019s age = (y + 2) years
According to the question,
(x + 2) = 8 + 3(y + 2)
\u21d2 x + 2 = 8 + 3y + 6
\u21d2 x = 3y + 12 \u2026(ii)
From Eq. (i) and (ii), we get
5y \u2013 8 = 3y + 12
\u21d2 5y \u2013 3y = 12 + 8
\u21d2 2y = 20
\u21d2 y = 10
On putting the value of y = 11 in Eq. (i), we get
x = 5(10) \u2013 8
\u21d2 x = 50 \u2013 8
\u21d2 x = 42
Hence, the age of man is 42 years and the age of his son is 10 years.<\/p>\n\n\n\n
\n\n\n\nQuestion 31 <\/h4>\n\n\n\n
Let the age of two children be x and y
So, the father\u2019s present age = 3(x + y)
After five years,
Age of two children = (x + 5) + (y + 5) years
= ( x + y + 10) years
So, the age of father after five years = 3(x + y) + 5
= 3x + 3y + 5
According to the question,
3x + 3y + 5 = 2(x + y + 10)
\u21d2 3x + 3y + 5 = 2x + 2y + 20
\u21d2 3x \u2013 2x + 3y \u2013 2y = 20 \u2013 5
\u21d2 x + y = 15
So, the age of two children = 15 years
And the age of father = 3(15) = 45years
Hence, the age of father is 45 years and the age of his two children is 15 years.<\/p>\n\n\n\n
\n\n\n\nQuestion 32 <\/h4>\n\n\n\n
Let the age of A = x years
And the age of B = y years
Five years ago,
A\u2019s age = (x \u2013 5) years
B\u2019s age = (y \u2013 5) years
According to the question,
(x \u2013 5) = 3(y \u2013 5)
\u21d2 x \u2013 5 = 3y \u2013 15
\u21d2 x = 3y \u2013 10 \u2026(i)
Ten years later,
A\u2019s age = (x + 10)
B\u2019s age = (y + 10)
According to the question,
(x + 10) = 2(y + 10)
\u21d2 x + 10 = 2y + 20
\u21d2 x = 2y + 10 \u2026(ii)
From Eq. (i) and (ii), we get
3y \u2013 10 = 2y + 10
\u21d2 3y \u2013 2y = 10 + 10
\u21d2 y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 3(20) \u2013 10
\u21d2 x = 50
Hence, the age of person A is 50years and Age of B is 20years.<\/p>\n\n\n\n
\n\n\n\nQuestion 33 <\/h4>\n\n\n\n
Let the age of a man = x years
And the age of his son = y years
Ten years hence,
Man\u2019s age = (x + 10) years
Son\u2019s age = (y + 10) years
According to the question,
(x + 10) = 2(y + 10)
\u21d2 x + 10 = 2y + 20
\u21d2 x = 2y + 20 \u2013 10
\u21d2 x = 2y + 10 …(i)
Ten years ago,
Father\u2019s age = (x \u2013 10) years
Son\u2019s age = (y \u2013 10) years
According to the question,
(x \u2013 10) = 4(y \u2013 10)
\u21d2 x \u2013 10 = 4y \u2013 40
\u21d2 x = 4y \u2013 30 \u2026(ii)
From Eq. (i) and (ii), we get
2y + 10 = 4y \u2013 30
\u21d2 2y \u2013 4y = \u2013 30 \u2013 10
\u21d2 \u2013 2y = \u2013 40
\u21d2 y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 2y + 10
\u21d2 x = 2(20) + 10
\u21d2 x = 50
Hence, the age of man is 50 years and the age of his son is 20 years.<\/p>\n\n\n\n
\n\n\n\nQuestion 34 <\/h4>\n\n\n\n
![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1_1544158536206177.png\")
We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180\u00b0
<\/p>\n\n\n\n
\u21d2 2x + 4 + 2y + 10 = 180 and y + 3 + 4x \u2013 5 = 180
\u21d2 2x + 2y = 180 \u2013 14 and 4x + y \u2013 2 = 180
\u21d2 x + y = 83 and 4x + y = 182
<\/p>\n\n\n\n
x + y = 83 \u2026(i)
4x + y = 182 \u2026(ii)
<\/p>\n\n\n\n
4x + y \u2013 x \u2013 y = 182 \u2013 83
\u21d2 3x = 99
\u21d2 x = 33
<\/p>\n\n\n\n
33 + y = 83
\u21d2 y = 83 \u2013 33 = 50
<\/p>\n\n\n\n
\u2220A = (2x + 43)\u00b0 = 2(33) + 4 = 66 + 4 = 70\u00b0
\u2220B = (y + 3)\u00b0 = 50 + 3 = 53\u00b0
\u2220C = (2y + 10)\u00b0 = 2(50) + 10 = 100 + 10 = 110\u00b0
and \u2220D = (4x \u2014 5)\u00b0 = 4(33) \u2013 5 = 132 \u2013 5 = 127\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 35 <\/h4>\n\n\n\n
\u2220A = 63\u00b0, \u2220B = 57\u00b0, \u2220C = 117\u00b0, \u2220D = 123\u00b0.<\/strong>
Sol :
We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180\u00b0
<\/p>\n\n\n\n
\u21d2 2x \u2013 3 + 2y + 17 = 180 and y + 7 + 4x \u2013 9 = 180
\u21d2 2x + 2y + 14 = 180 and 4x + y \u2013 2 = 180
\u21d2 2x + 2y = 180 \u2013 14 and 4x + y = 182
\u21d2 x + y = 83 and 4x + y = 182
So, we get pair of linear equation i.e.
x + y = 83 \u2026(i)
4x + y = 182 \u2026(ii)
<\/p>\n\n\n\n
4x + y \u2013 x \u2013 y = 182 \u2013 83
\u21d2 3x = 99
\u21d2 x = 33
<\/p>\n\n\n\n
33 + y = 83
\u21d2 y = 83 \u2013 33 = 50
<\/p>\n\n\n\n
\u2220A = (2x \u2014 3)\u00b0 = 2(33) \u2013 3 = 66 \u2013 3 = 63\u00b0
\u2220B = (y + 7)\u00b0 = 50 + 7 = 57\u00b0
\u2220C = (2y + 17)\u00b0 = 2(50) + 17 = 100 + 17 = 117\u00b0
and \u2220D = (4x \u2014 9)\u00b0 = 4(33) \u2013 9 = 132 \u2013 9 = 123\u00b0
Hence, the angles are \u2220A = 63\u00b0, \u2220B = 57\u00b0, \u2220C = 117\u00b0, \u2220D = 123\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 36 <\/h4>\n\n\n\n
Sol :<\/p>\n\n\n\n
\u2234 \u2220A + \u2220B + \u2220C = 180\u00b0 \u2026(a)
According to the question,
$\\begin{matrix}\\angle \\mathrm{C}&=3 \\angle \\mathrm{B}&=2(\\angle \\mathrm{A}+\\angle \\mathrm{B})\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$
<\/p>\n\n\n\n
\u21d2 3\u2220B = 2 (\u2220A + \u2220B)
\u21d2 3\u2220B = 2 \u2220A + 2 \u2220B
\u21d2 \u2220B = 2 \u2220A \u2026(i)
<\/p>\n\n\n\n
\u2220C = 3 \u2220B<\/p>\n\n\n\n
\u21d2 \u2220C = 6 \u2220A \u2026(ii)
<\/p>\n\n\n\n
\u2220A + 2\u2220A + 6\u2220A = 180\u00b0
\u21d2 9\u2220A = 180\u00b0
\u21d2 \u2220A = 20\u00b0
On puuting the value of \u2220A = 20\u00b0 in Eq. (i) and (ii), we get
\u2220B = 2 \u2220A = 2(20) = 40\u00b0
\u2220C = 6 \u2220A = 6(20) = 120\u00b0
Hence, the angles are\u2220A = 20\u00b0, \u2220B = 40\u00b0, \u2220C = 120\u00b0<\/p>\n\n\n\n
\n\n\n\nQuestion 37 <\/h4>\n\n\n\n
We know that, in a triangle , the sum of three angles is 180\u00b0
\u2234 \u2220A + \u2220B + \u2220C = 180\u00b0
According to the question,
x + 3x + y = 180
\u21d2 4x + y = 180
\u21d2 y = 180 \u2013 4x \u2026(i)
Given that 3y \u2014 5x = 30 \u2026(ii)
On substituting the value of y in Eq. (ii), we get
3(180 \u2013 4x) \u2013 5x = 30
\u21d2 540 \u2013 12x \u2013 5x = 30
\u21d2 540 \u2013 17x = 30
\u21d2 \u2013 17x = 30 \u2013 540
\u21d2 \u2013 17x = \u2013 510
\u21d2 x = 30
Now, we substitute the value of x in Eq.(i), we get
\u21d2 y = 180 \u2013 4(30)
\u21d2 y = 60
On putting the value of x and y, we calculate the angles
\u2220A = x\u00b0 = 30\u00b0
\u2220B = (3x)\u00b0 = 3(30) = 90\u00b0
and \u2220C = y\u00b0 = 60\u00b0
Here, we can see that \u2220B = 90\u00b0 , so triangle is a right angled.<\/p>\n\n\n\n
\n\n\n\nQuestion 38 <\/h4>\n\n\n\n
Let the length of a rectangle = x m
and the breadth of a rectangle = y m
Then, Area of rectangle = xy m2<\/sup>
Condition I :
Area is reduced by 8m2<\/sup>, when length = (x \u2013 5) m and breadth = (y + 3) m
Then, area of rectangle = (x \u2013 5)\u00d7(y + 3) m2<\/sup>
According to the question,
xy \u2013 (x \u2013 5)\u00d7(y + 3) = 8
\u21d2 xy \u2013 (xy + 3x \u2013 5y \u2013 15) = 8
\u21d2 xy \u2013 xy \u2013 3x + 5y + 15 = 8
\u21d2 \u2013 3x + 5y = 8 \u2013 15
\u21d2 3x \u2013 5y = 7 \u2026(i)
Condition II:
Area is increased by 74m2<\/sup>, when length = (x + 3) m and breadth = (y + 2) m
Then, area of rectangle = (x + 3)\u00d7(y + 2) m2<\/sup>
According to the question,
(x + 3)\u00d7(y + 2) \u2013 xy = 74
\u21d2 (xy + 3y + 2x + 6) \u2013 xy = 74
\u21d2 xy + 2x + 3y + 6 \u2013 xy = 74
\u21d2 2x + 3y = 74 \u2013 6
\u21d2 2x + 3y = 68 \u2026(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3, we get
6x \u2013 10y = 14 \u2026(iii)
6x + 9y = 204 \u2026(iv)
On subtracting Eq. (i) from Eq. (ii), we get
6x + 9y \u2013 6x + 10y = 204 \u2013 14
\u21d2 19y = 190
\u21d2 y = 10
On putting the value of y = 10 in Eq. (i), we get
3x \u2013 5 (10) = 7
\u21d2 3x \u2013 50 = 7
\u21d2 3x = 57
\u21d2 x = 19
Hence, the length of the rectangle is 19m and the breadth of a rectangle is 10m<\/p>\n\n\n\n
\n\n\n\nQuestion 39 <\/h4>\n\n\n\n
Let the breadth of a room = x m
According to the question,
Length of the room = x + 3
Then, Area of room = (x + 3)\u00d7 (x) m2<\/sup>
= x2<\/sup> + 3x
Condition II:
Area remains same,
when length = (x + 3 + 3) m = (x + 6) m
and breadth = (x \u2013 2) m
According to the question,
x2<\/sup> + 3x = (x + 6)(x \u2013 2)
\u21d2 x2<\/sup> + 3x = x2<\/sup> \u2013 2x + 6x \u2013 12
\u21d2 3x = 4x \u2013 12
\u21d2 3x \u2013 4x = \u2013 12
\u21d2 x = 12
So, length of the room = (x + 3) = 12 + 3 = 15m
Hence, the length of the room is 15m and the breadth of a room is 12m<\/p>\n\n\n\n
\n\n\n\nQuestion 40 <\/h4>\n\n\n\n
Let the speed of car I = x km\/hr
And the speed of car II = y km\/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 6h.
Distance travelled by car I in 6h = 6x km
Distance travelled by car II in 6h = 6y km
Since, they are travelling in same direction, sign should be negative
6x \u2013 6y = 120
\u21d2 x \u2013y = 20 \u2026(i)
Now, Let two cars meet after 1hr 12min
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\frac{6}{5} x+\\frac{6}{5} y=120$
\u21d2 6x + 6y = 120 \u00d7 5
\u21d2 x + y = 100 \u2026(ii)
<\/p>\n\n\n\n
x \u2013 y + x + y = 20 + 100
\u21d2 2x = 120
\u21d2 x = 60
Putting the value of x = 60 in Eq. (i), we get
60 \u2013 y = 20
\u21d2 y = 40
So, the speed of the two cars are 60km\/h and 40 km\/hr respectively.<\/p>\n\n\n\n
\n\n\n\nQuestion 41 <\/h4>\n\n\n\n
Let the speed of train = x km\/hr
<\/p>\n\n\n\n
time $=\\frac{\\text { distance }}{\\text { speed }}$
<\/p>\n\n\n\n
<\/p>\n\n\n\n
Speed of the train is increased by 5km an hour
\u2234 the new speed of the train = (x + 5)km\/hr
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\Rightarrow \\frac{300}{x}-\\frac{300}{x+5}=2$
$\\Rightarrow \\frac{300(\\mathrm{x}+5)-300(\\mathrm{x})}{\\mathrm{x}(\\mathrm{x}+5)}=2$
\u21d2 300x + 1500 \u2013 300x = 2x (x + 5)
\u21d2 1500 = 2x2<\/sup> + 10x
\u21d2 750 = x2<\/sup> + 5x
\u21d2 x2<\/sup> + 5x \u2013 750 = 0
\u21d2 x2<\/sup> + 30x \u2013 25x \u2013 750 = 0
\u21d2 x (x + 30) \u2013 25 (x + 30) = 0
\u21d2 (x \u2013 25) (x + 30) = 0
\u21d2 (x \u2013 25) = 0 or (x + 30) = 0
\u2234 x = 25 or x = \u2013 30
Since, speed can\u2019t be negative.
Hence, the speed of the train is 25km\/hr<\/p>\n\n\n\n
\n\n\n\nQuestion 42 <\/h4>\n\n\n\n
Distance to the destination = 1500km
We know that,
speed $=\\frac{\\text { distance }}{\\text { time }}$
<\/p>\n\n\n\n
<\/p>\n\n\n\n
Plane left 30min later than the scheduled time
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\Rightarrow \\frac{1500}{x-\\frac{1}{2}}-\\frac{1500}{x}=250$
$\\Rightarrow \\frac{1500}{\\frac{2 \\mathrm{x}-1}{2}}-\\frac{1500}{\\mathrm{x}}=250$
$\\Rightarrow \\frac{2}{2 x-1}-\\frac{1}{x}=\\frac{250}{1500}$
$\\Rightarrow \\frac{2 x-(2 x-1)}{(2 x-1) x}=\\frac{1}{6}$
\u21d2 6(2x \u2013 2x + 1) = 2x2<\/sup> \u2013 x
\u21d2 6 = 2x2<\/sup> \u2013 x
\u21d2 2x2<\/sup> \u2013x \u2013 6 = 0
\u21d2 2x2<\/sup> \u2013 4x + 3x \u2013 6 = 0
\u21d2 2x (x \u2013 2) + 3 (x \u2013 2) = 0
\u21d2 (2x + 3) (x \u2013 2) = 0
\u21d2 (2x + 3) = 0 or (x \u2013 2) = 0
<\/p>\n\n\n\n
<\/p>\n\n\n\n
Hence, the time taken by the aeroplane is 2hrs and the speed is 750km\/hr<\/p>\n\n\n\n
\n\n\n\nQuestion 43 <\/h4>\n\n\n\n
Let the speed of a train = x km\/hr
And the speed of a car = y km\/hr
Total distance travelled = 600km
According to the question,
If he covers 400km by train and rest by car i.e. (600 \u2013 400) = 200km
<\/p>\n\n\n\n
<\/p>\n\n\n\n
He takes half hour longer i.e. 7 hours
So, total time = train time + car time
<\/p>\n\n\n\n
time $=\\frac{\\text { distance }}{\\text { speed }}$
$\\Rightarrow \\frac{400}{x}+\\frac{200}{y}=6.5$ \u2026(i)
$\\Rightarrow \\frac{200}{x}+\\frac{400}{y}=7$ \u2026(ii)<\/p>\n\n\n\n
and 200u + 400v = 7 \u2026(iv)
<\/p>\n\n\n\n
800u + 400v = 13 \u2026(a)
800u + 1600v = 28 \u2026(b)
On subtracting Eq. (a) from Eq. (b), we get
800u + 1600v \u2013 800u \u2013 400v = 28 \u2013 13
\u21d2 1200v = 15
$\\Rightarrow \\mathrm{v}=\\frac{15}{1200}$
$\\Rightarrow \\mathrm{v}=\\frac{1}{80}$
<\/p>\n\n\n\n
$200 u+400\\left(\\frac{1}{80}\\right)=7$
\u21d2 200u + 5 = 7
\u21d2 200u = 2
$\\Rightarrow \\mathrm{u}=\\frac{1}{100}$
<\/p>\n\n\n\n
\u21d2 x = 100 and y = 80
<\/p>\n\n\n\n
\n\n\n\nQuestion 44 <\/h4>\n\n\n\n
Let the speed of car I = x km\/hr
And the speed of car II = y km\/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 8h.
Distance travelled by car I in 8h = 8x km
Distance travelled by car II in 8h = 8y km
Since, they are travelling in same direction, sign should be negative
8x \u2013 8y = 80
\u21d2 x \u2013y = 10 \u2026(i)<\/p>\n\n\n\n
1hr 20min $=1+\\frac{20}{60}=\\frac{4}{3} \\mathrm{hr}$<\/p>\n\n\n\n
\u21d2 4x + 4y = 240
\u21d2 x + y = 60 \u2026(ii)
On adding (i) and (ii) , we get
x \u2013 y + x + y = 10 + 60
\u21d2 2x = 70
\u21d2 x = 35
Putting the value of x = 25 in Eq. (i), we get
35 \u2013 y = 10
\u21d2 y = 25
So, the speed of the two cars are 35km\/h and 25 km\/hr respectively.<\/p>\n\n\n\n
\n\n\n\nQuestion 45 <\/h4>\n\n\n\n
Let speed of the boat in still water = x km\/hr
and speed of the stream = y km\/hr
Then, the speed of the boat downstream = (x + y)km\/hr
And speed of the boat upstream = (x \u2013 y)km\/hr
<\/p>\n\n\n\n
Then,<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\mathrm{t}_{2}=\\frac{24}{\\mathrm{x}+\\mathrm{y}} \\mathrm{h}$
<\/p>\n\n\n\n
Then,<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
$\\mathrm{T}_{2}=\\frac{36}{\\mathrm{x}+\\mathrm{y}} \\mathrm{h}$
<\/p>\n\n\n\n
$\\therefore \\frac{12}{x-y}+\\frac{36}{x+y}=6$ \u2026(b)
<\/p>\n\n\n\n
$\\frac{16}{x+y}+\\frac{24}{x-y}=6$ \u2026(i)
And $\\frac{12}{x+y}+\\frac{36}{x-y}=6$ \u2026(ii)
<\/p>\n\n\n\n
$\\frac{48}{x+y}+\\frac{72}{x-y}=18$ \u2026(iii)
$\\frac{48}{x+y}+\\frac{144}{x-y}=24$ \u2026(iv)
<\/p>\n\n\n\n
$\\frac{48}{x+y}+\\frac{144}{x-y}-\\frac{48}{x+y}-\\frac{72}{x-y}=24-18$
$\\Rightarrow \\frac{144-72}{x-y}=6$
$\\Rightarrow \\frac{72}{x-y}=6$
\u21d2 x \u2013 y = 12 \u2026(a)
<\/p>\n\n\n\n
$\\Rightarrow \\frac{16}{x+y}+\\frac{24}{12}=6$
$\\Rightarrow \\frac{16}{x+y}=6-2$
$\\Rightarrow \\frac{16}{x+y}=4$
\u21d2 x + y = 4 \u2026(b)
<\/p>\n\n\n\n
\u21d2 2x = 16
\u21d2 x = 8
<\/p>\n\n\n\n
8 \u2013 y = 12
\u21d2 y = \u2013 4 but speed can\u2019t be negative
\u21d2 y = 4
<\/p>\n\n\n\n
<\/p>\n\n\n\n
\n\n\n\nQuestion 46 <\/h4>\n\n\n\n
Let the speed of a train = x km\/hr
And the speed of a car = y km\/hr
Total distance travelled = 370km
<\/p>\n\n\n\n
<\/p>\n\n\n\n
Time take = 4hrs
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
time $=\\frac{\\text { distance }}{\\text { speed }}$<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
and 130u + 240v = 4.3 \u2026(iv)
<\/p>\n\n\n\n
500u + 240v = 8 \u2026(v)
<\/p>\n\n\n\n
500u + 240v \u2013 130u \u2013 240v = 8 \u2013 4.3
\u21d2 370u = 3.7
$\\Rightarrow \\mathrm{u}=\\frac{3.7}{370}$
$\\Rightarrow \\mathrm{u}=\\frac{1}{100}$
<\/p>\n\n\n\n
$130\\left(\\frac{1}{100}\\right)+240 \\mathrm{v}=4.3$
\u21d2 1.3 + 240v = 4.3
\u21d2 240v = 3<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n