Sol : Two numbers differ by 3 and their product is 504. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : Find two consecutive positive integers, the sum of whose squares is 365.<\/strong><\/p>\n\n\n\n Sol : The difference of two numbers is 4. If the difference of their reciprocals is <\/strong>$\\frac{4}{21}$<\/strong>, find the two numbers.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula The sum of two numbers is 18 and the sum of their reciprocals is <\/strong>$\\frac{1}{4}$. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The sum of the squares of three consecutive positive integers is 50. Find the integers.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula Find three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula A two\u2013digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.<\/strong><\/p>\n\n\n\n Sol : 10X+Y+45=10Y+X<\/p>\n\n\n\n On simplifying further, Putting the value of $\\mathrm{X}=\\frac{14}{\\mathrm{Y}}$ from equation \u2013\u2013\u2013\u2013\u2013\u2013 (i) On applying Sreedhracharya formula $\\frac{5 \\pm 9}{2}$<\/p>\n\n\n\n Y = \u2013 2 or 7<\/p>\n\n\n\n \u2234 Y = 7 (Only Positive values) The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula The difference of two numbers is 5 and the difference of their reciprocals is $\\frac{1}{10}$Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, The sum of a number and its reciprocal is <\/strong>$\\frac{10}{3}$<\/strong>. Find the number.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula Divide 12 into two parts such that the sum of their squares is 74.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula The sum of the squares of two consecutive natural numbers is 421. Find the numbers.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula A two\u2013digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula A two\u2013digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.<\/strong><\/p>\n\n\n\n Sol : Putting the value of $X=\\frac{4 Y}{5}$ from equation \u2013\u2013\u2013\u2013\u2013\u2013 (i) On applying Sreedhracharya formula The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2 \\frac{19}{15}$, find the fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The numerator of a fraction is one more than its denominator. If its reciprocal is subtracted from it, the difference is $\\frac{11}{30}$. Find the fraction.<\/strong><\/p>\n\n\n\n Sol : Taking L.C.M we get, Cross-multiplying we get, So, 11x + 5 = 0 or x \u2013 5 = 0<\/p>\n\n\n\n $x=-\\frac{5}{11}$ or $x=5$<\/p>\n\n\n\n $x+1=-\\frac{5}{11}+1=\\frac{6}{11}$ or <\/p>\n\n\n\n x+1=5+1=6<\/p>\n\n\n\n So the possible fractions are:<\/p>\n\n\n\n $\\frac{x+1}{x}=\\frac{\\frac{6}{11}}{-\\frac{5}{11}}=-\\frac{6}{5}$<\/p>\n\n\n\n Or The numerator of a fraction is one more than its denominator. If its reciprocal is added to it the sum is $\\frac{61}{30}$. Find the fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The numerator of a fraction is 3 more than its denominator. If its reciprocal is subtracted from it, the difference is $\\frac{33}{28}$. Find the fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The denominator of a fraction exceeds its numerator by 3. If one is added to both numerator and denominator, the difference between the new and the original fractions 1 becomes $\\frac{1}{24}$. Find the original fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The denominator of a fraction exceeds its numerator by 3. If 3 is added to both numerator and denominator, the difference between the new and the original fraction is $\\frac{9}{88}$. Find the original fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The numerator of a fraction is 3 less than denominator. If 2 is added to both 29 numerator as well as denominator, then sum of the new and original fraction is $\\frac{19}{15}$ . Find the fraction.<\/strong><\/p>\n\n\n\n Sol : On simplifying further, On applying Sreedhracharya formula The numerator of a fraction is 2 less than the denominator. If 1 is added to both numerator and denominator the sum of the new and original fraction is $\\frac{19}{15}$. Find the original fraction.<\/strong><\/p>\n\n\n\n Sol : $=\\frac{x-2+1}{x+1}=\\frac{x-1}{x+1}$<\/p>\n\n\n\n Sum of the fractions $=\\frac{x-2}{x}+\\frac{x-1}{x+1}$ Taking L.C.M we get, Cross-multiplying we get, So, 11x + 5 = 0 or x \u2013 5 = 0<\/p>\n\n\n\n $x=-\\frac{5}{11}$ or x=5 x=5-2=3<\/p>\n\n\n\n Putting these values in fraction $=\\frac{x-2}{x}$ The hypotenuse of a right\u2013angled triangle is 6 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.<\/strong><\/p>\n\n\n\n Sol : On applying Sreedhracharya formula The sum of the areas of two squares is 640 m2<\/sup>. If the difference in their perimeters be 64 m, find the sides of the two squares.<\/strong><\/p>\n\n\n\n Sol : Putting the value of X+Y=16 in equation (i) On applying Sreedhracharya formula The hypotenuse of a right triangle is 3\u221a5cm. If the smaller side is tripled and the longer side doubled, new hypotenuse will be 9\u221a5 cm. How long are the sides of the triangle?<\/strong><\/p>\n\n\n\n Sol : A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that lie was short of 25 students. Find the number of students.<\/strong><\/p>\n\n\n\n Sol : The area of a triangle is 30 sq cm. Find the base if the altitude exceeds the base by 7 cm.<\/strong><\/p>\n\n\n\n Sol : Is it possible to design a rectangular mango grove whose length is twice its breadth, and area is 800 m2<\/sup>? If so, find its length and breadth.<\/strong><\/p>\n\n\n\n Sol :
Let the first number be \u2018X\u2019, so the other number will be \u2019(12\u2013X)\u2019.
\u2235 X (12 \u2013 X) = 32
12X \u2013 X2<\/sup> = 32
X2<\/sup> \u2013 12X + 32 = 0
On factorising further,
X2<\/sup> \u2013 4X \u2013 8X + 32 = 0
X(X \u2013 4) \u2013 8(X \u2013 4) = 0
(X \u2013 4)(X \u2013 8) = 0
So, X = 4 or 8
\u2234 The numbers are 4 and 8.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(X+3)\u2019.
\u2235 X(X + 3) = 504
X2<\/sup> + 3X\u2013 504=0
On applying Sreedhracharya formula
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-3 \\pm \\sqrt{9-4(1)(-504)}}{2(1)}$
$\\frac{-3 \\pm \\sqrt{9+2016}}{2(1)}$
$\\frac{-3 \\pm \\sqrt{2025}}{2}$
$\\frac{-3 \\pm 45}{2}$
X = \u201324 or 21.
\u2234 if the first number is \u2013 24 , then the other number is \u2013 21.
\u2234 if the first number is 21 , then the other number is 24.
\u2234 The numbers are not (21,24) & (\u2013 21, \u2013 24).<\/p>\n\n\n\n
\n\n\n\nQuestion 3 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(X+1)\u2019.
\u2235 X2<\/sup>+(X+1)2<\/sup> = 365
X2<\/sup> + X2<\/sup> + 1 + 2X \u2013 365 = 0
2X2<\/sup> + 2X \u2013 364 = 0
X2<\/sup> + X \u2013 182 = 0
On applying Sreedhracharya formula
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-1 \\pm \\sqrt{1-4(1)(-182)}}{2(1)}$
$\\frac{-1 \\pm \\sqrt{1+728}}{2(1)}$
$\\frac{-1 \\pm \\sqrt{729}}{2}$
$\\frac{-1 \\pm 27}{2}$
\u2234 X = \u201314 or X = 13.
\u2234 The numbers are 13 & 14.<\/p>\n\n\n\n
\n\n\n\nQuestion 4 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(X+4)\u2019.
$\\because \\frac{1}{X}-\\frac{1}{X+4}=\\frac{4}{21}$
$\\frac{X+4-X}{X^{2}+4X}=\\frac{4}{21}$
On simplifying further,
X2<\/sup> + 4X \u2013 21 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-4 \\pm \\sqrt{16-4(1)(-21)}}{2(1)}$
$\\frac{-4 \\pm \\sqrt{16+84}}{2(1)}$
$\\frac{-4 \\pm \\sqrt{100}}{2}$
$\\frac{-4 \\pm 10}{2}$
\u2234 X = \u20137 or X = 3.
\u2234 The numbers are \u20137 & \u20133 or 3 & 7.<\/p>\n\n\n\n
\n\n\n\nQuestion 5 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(18\u2013X)\u2019.
$\\because \\frac{1}{X}+\\frac{1}{18-X}=\\frac{1}{4}$
$\\frac{18-X+X}{-X^{2}+18 X}=\\frac{1}{4}$<\/p>\n\n\n\n
\u2013X2<\/sup> + 18X \u2013 72 = 0
X2<\/sup> \u2013 18X + 72 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-18) \\pm \\sqrt{(-18)^{2}-4(1)(72)}}{2(1)}$
$\\frac{18 \\pm \\sqrt{324-288}}{2(1)}$
$\\frac{18 \\pm \\sqrt{36}}{2}$
$\\frac{18 \\pm 6}{2}$
X = 6 or X = 12
\u2234 The numbers are 6 & 12.<\/p>\n\n\n\n
\n\n\n\nQuestion 6 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other numbers will be \u2019(X+1)\u2019 & \u2018(X+2).
\u2235X2<\/sup> + (X + 1)2<\/sup> + (X + 2)2<\/sup> = 50
X2<\/sup> + X2<\/sup> + 1 + 2X + X2<\/sup> + 4 + 4X = 50<\/p>\n\n\n\n
3X2<\/sup> + 6X\u2013 45 = 0
X2<\/sup> + 2X \u2013 15 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-2 \\pm \\sqrt{4-4(1)(-15)}}{2(1)}$
$\\frac{-2 \\pm \\sqrt{4+60}}{2(1)}$
$\\frac{-2 \\pm \\sqrt{64}}{2}$
$\\frac{-2 \\pm 8}{2}$
X = \u2013 5 or 3
\u2234 X = 3 (Only Positive values)
\u2234 X + 1 = 4
\u2234 X +2 = 5
\u2234 The numbers are 3, 4 & 5.<\/p>\n\n\n\n
\n\n\n\nQuestion 7 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other numbers will be \u2019(X+1)\u2019 & \u2018(X+2).
\u2235 X2<\/sup> + (X + 1)(X + 2) = 154
X2<\/sup> + X2<\/sup> + 2X + X + 2 = 154
On simplifying further,
2X2<\/sup> + 3X \u2013 152 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-3 \\pm \\sqrt{3^{2}-4(2)(-152)}}{2(2)}$
$\\frac{-3 \\pm \\sqrt{9+1216}}{4}$
$\\frac{-3 \\pm \\sqrt{1225}}{4}$
$\\frac{-3 \\pm 35}{4}$
X = \u2013 9.5 or 8
\u2234 X = 8 (Only whole values)
\u2234 X +1 = 9
\u2234 X + 2 = 10
\u2234 The numbers are 8, 9 & 10.<\/p>\n\n\n\n
\n\n\n\nQuestion 8 <\/h4>\n\n\n\n
Let the units digit be \u2018Y\u2019, so the tens digit will be X.
\u2234 the number is 10X + Y![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990186867771.png\")
X . Y =14 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
9X \u2013 9Y + 45 = 0
X \u2013 Y + 5 = 0<\/p>\n\n\n\n
$\\frac{14}{\\mathrm{Y}}-\\mathrm{Y}+5=0$
14 \u2013 Y2<\/sup> + 5Y = 0
Y2<\/sup> \u2013 5Y \u2013 14 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-5) \\pm \\sqrt{(-5)^{2}-4(1)(-14)}}{2(1)}$
$\\frac{5 \\pm \\sqrt{25+56}}{2}$
$\\frac{5 \\pm \\sqrt{81}}{2}$<\/p>\n\n\n\n
$\\therefore \\mathrm{X}=\\frac{14}{7}$
\u2234 X = 2
\u2234 The number is 27 {2 (10) + 7}.<\/p>\n\n\n\n
\n\n\n\nQuestion 9 <\/h4>\n\n\n\n
Let the larger number be \u2018X\u2019, so the smaller number will be \u2018Y\u2019.
\u2234 X2<\/sup> \u2013 Y2<\/sup> = 45 \u2013\u2013\u2013\u2013\u2013 (i)
\u22354X = Y2<\/sup> \u2013\u2013\u2013\u2013\u2013\u2013 (ii)
On simplifying further,
Putting the value of Y2<\/sup> = 4X from equation \u2013\u2013\u2013\u2013\u2013\u2013 (ii)
X2<\/sup> \u2013 4X = 45
X2<\/sup> \u2013 4X \u2013 45 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-4) \\pm \\sqrt{(-4)^{2}-4(1)(-45)}}{2(1)}$
$\\frac{4 \\pm \\sqrt{16+180}}{2}$
$\\frac{4 \\pm \\sqrt{196}}{2}$
$\\frac{4 \\pm 14}{2}$
X = \u2013 5 or 9
\u2234 X = 9 (Only natural number, as given in the question)
\u2234 Y=\u221a(4*9)
\u2234 Y = 6
\u2234 The numbers are 9 & 6.<\/p>\n\n\n\n
\n\n\n\nQuestion 10 <\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(X+5)\u2019.
$\\because \\frac{1}{X}-\\frac{1}{X+5}=\\frac{1}{10}$
$\\frac{X+5-X}{X^{2}+5 X}=\\frac{1}{10}$<\/p>\n\n\n\n
X2<\/sup> + 5X \u2013 50 = 0
On applying Sreedhracharya formula
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-5 \\pm \\sqrt{5^{2}-4(1)(-50)}}{2(1)}$
$\\frac{-5 \\pm \\sqrt{25+200}}{2(1)}$
$\\frac{-5 \\pm \\sqrt{225}}{2}$
$\\frac{-5 \\pm 15}{2}$
\u2234 X = \u201310 or X = 5.
Then other numbers will be (X+5) = {\u201310+5} & {5+5} i.e., \u20135 or 10.
\u2234 The numbers are \u201310 & \u20135 or 5 & 10.<\/p>\n\n\n\n
\n\n\n\nQuestion 11<\/h4>\n\n\n\n
Let the number be \u2018X\u2019, so the reciprocal will be \u2019$\\left(\\frac{1}{x}\\right)$\u2019.
$\\because \\frac{1}{X}+X=\\frac{10}{3}$
$\\frac{1+X^{2}}{X}=\\frac{10}{3}$<\/p>\n\n\n\n
3 + 3X2<\/sup> = 10X
3X2<\/sup> \u2013 10X + 3 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-10) \\pm \\sqrt{(-10)^{2}-4(3)(3)}}{2(3)}$
$\\frac{10 \\pm \\sqrt{100-36}}{6}$
$\\frac{10 \\pm \\sqrt{64}}{6}$
$\\frac{10 \\pm 8}{6}$
\u2234 X = 3 or $X=\\frac{1}{3}$
Then other numbers will be $\\frac{1}{X}=\\frac{1}{3}$& 3![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990212719241.png\")
\u2234 The numbers are 3 or $\\frac{1}{3}$<\/p>\n\n\n\n
\n\n\n\nQuestion 12<\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(12\u2013X)\u2019.
\u2235 X2<\/sup> + (12 \u2013 X)2<\/sup> = 74
X2<\/sup> + 144 + X2<\/sup> \u2013 24X = 74
On simplifying further,
2X2<\/sup> \u2013 24X + 70 = 0
X2<\/sup> \u2013 12X + 35 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-12) \\pm \\sqrt{(-12)^{2}-4(1)(35)}}{2(1)}$
$\\frac{12 \\pm \\sqrt{144-140}}{2(1)}$
$\\frac{12 \\pm \\sqrt{4}}{2}$
$\\frac{12 \\pm 2}{2}$
\u2234 X = 5 or X = 7.
Then other numbers will be (12\u2013X) = {12\u20135} & {12\u20137} i.e., 7 or 5.
\u2234 The number 12 is divided into two parts namely 5 & 7.<\/p>\n\n\n\n
\n\n\n\nQuestion 13<\/h4>\n\n\n\n
Let the first number be \u2018X\u2019, so the other number will be \u2019(X+1)\u2019.
\u2235 X2<\/sup> + (X + 1)2<\/sup> = 421
X2<\/sup> + X2<\/sup> + 1 + 2X = 421
On simplifying further,
2X2<\/sup>+ 2X \u2013 420 = 0
X2<\/sup> + X \u2013 210 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(1) \\pm \\sqrt{(1)^{2}-4(1)(-210)}}{2(1)}$
$\\frac{-1 \\pm \\sqrt{1+840}}{2(1)}$
$\\frac{-1 \\pm \\sqrt{841}}{2}$
$\\frac{-1 \\pm 29}{2}$
\u2234 X = \u201315 or X = 14.
\u2234 X = 14 (Only natural number, as given in the question)
\u2234 Then other numbers will be 14 & 15.<\/p>\n\n\n\n
\n\n\n\nQuestion 14<\/h4>\n\n\n\n
Let the units digit be \u2018Y\u2019, so the tens digit will be X.
\u2234 the number is 10X + Y
\u2235 X . Y =18 \u2013\u2013\u2013\u2013\u2013\u2013 (i)
10X+Y\u201363=10Y+X
On simplifying further,
9X \u2013 9Y \u2013 63 = 0
X \u2013 Y \u2013 7 = 0
Putting the value of $X=\\frac{18}{Y}$ from equation \u2013\u2013\u2013\u2013\u2013\u2013 (i)
$\\frac{18}{\\mathrm{Y}}-\\mathrm{Y}-7=0$
18 \u2013 Y2<\/sup> \u2013 7Y = 0
Y2<\/sup> + 7Y \u2013 18 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-7 \\pm \\sqrt{(7)^{2}-4(1)(-18)}}{2(1)}$
$\\frac{-7 \\pm \\sqrt{49+72}}{2}$
$\\frac{-7 \\pm \\sqrt{121}}{2}$
$\\frac{-7 \\pm 11}{2}$
Y = \u2013 9 or 2![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990227016543.png\")
Y = 2 (Only Positive values)
$\\therefore \\mathrm{X}=\\frac{18}{2}$
\u2234 X = 9
\u2234 The number is 92 {9(10)+2}.<\/p>\n\n\n\n
\n\n\n\nQuestion 15<\/h4>\n\n\n\n
Let the units digit be \u2018Y\u2019, so the tens digit will be X.![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990228504398.png\")
the number is 10X + Y![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990229269744.png\")
10X + Y = 5 (X + Y)
On simplifying further,
10X \u2013 5X \u2013 5Y + Y = 0
5X \u2013 4Y = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990230011270.png\")
10X + Y = 2XY + 5
10X\u2013 2XY + Y \u2013 5 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (ii)<\/p>\n\n\n\n
$8 Y-\\frac{8 Y^{2}}{5}+Y-5=0$
40Y \u2013 8Y2<\/sup> + 5Y \u2013 25 = 0
8Y2<\/sup> \u2013 45Y + 25 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-45) \\pm \\sqrt{(-45)^{2}-4(8)(25)}}{2(8)}$
$\\frac{45 \\pm \\sqrt{2025-800}}{16}$
$\\frac{45 \\pm \\sqrt{1225}}{16}$
$\\frac{45 \\pm 35}{16}$
Y = $\\frac{5}{8}$ or 5
\u2234 Y = 5
\u2234 $X=\\frac{4 * 5}{5}$ from equation (i) $X=\\frac{4 Y}{5}$
\u2234 X = 4
\u2234 The number is 45 {4(10)+5}.<\/p>\n\n\n\n
\n\n\n\nQuestion 16<\/h4>\n\n\n\n
Let the numerator be \u2018X\u2019, so the denominator will be \u2018(2X+1)\u2019.![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990238917926.png\")
the fraction is $\\frac{X}{2 X+1}$
$\\because \\frac{X}{2 X+1}+\\frac{2 X+1}{X}=\\frac{58}{21}$<\/p>\n\n\n\n
$\\frac{X^{2}+4 X^{2}+4 X+1}{2 X^{2}+X}=\\frac{58}{21}$
21(5X2<\/sup> + 4X + 1) = 58 (2X2<\/sup> + X)
105X2<\/sup> + 84X + 21 = 116X2<\/sup> + 58X
11X2<\/sup> \u2013 26X \u2013 21 = 0\u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-26) \\pm \\sqrt{(-26)^{2}-4(11)(-21)}}{2(11)}$
$\\frac{26 \\pm \\sqrt{676+924}}{22}$
$\\frac{26 \\pm \\sqrt{1600}}{22}$
$\\frac{26 \\pm 40}{22}$
X= $-\\frac{7}{11}$ or 3
\u2234 X = 3 (Only positive values)
\u2234 Numerator is = 3 & denominator is (2X+1) = 7
\u2234 The fraction is $\\frac{3}{7}$.<\/p>\n\n\n\n
\n\n\n\nQuestion 17<\/h4>\n\n\n\n
Given: Numerator of a fraction is one more than its denominator.
To Find: The fraction
Assumption: Let the denominator be x.
Numerator = x + 1
Therefore, the fraction $=\\frac{x+1}{x}$
From the second case, we get,
$\\frac{x+1}{x}-\\frac{x}{x+1}=\\frac{11}{30}$<\/p>\n\n\n\n
$\\frac{(x+1)^{2}-x^{2}}{x(x+1)}=\\frac{11}{30}$
$\\frac{x^{2}+1+2 x-x^{2}}{x(x+1)}=\\frac{11}{30}$
$\\frac{2 x+1}{x^{2}+x}=\\frac{11}{30}$<\/p>\n\n\n\n
30(2x + 1) = 11(x2<\/sup> + x)
60x + 30 = 11x2<\/sup> + 11x
11x2<\/sup> + 11x \u2013 60x \u2013 30 = 0
11x2<\/sup> \u2013 49x \u2013 30 = 0
Now, we need to factorise such that, on multiplication we get 330 and on substraction we get 49.
Therefore, 55 and 6 can be the factors.
So, equation becomes,
11x2<\/sup> \u2013 (55x \u2013 6x) \u2013 30 = 0
11x2<\/sup> \u2013 55x + 6x \u2013 30 = 0
11x(x \u2013 5) + 6(x \u2013 5) = 0
(11x + 6)(x \u2013 5) = 0<\/p>\n\n\n\n
$\\frac{x+1}{x}=\\frac{6}{5}$<\/p>\n\n\n\n
\n\n\n\nQuestion 18<\/h4>\n\n\n\n
Let the denominator be \u2018X\u2019, so the numerator will be \u2018(X+1)\u2019.
\u2234 the fraction is $\\frac{\\mathrm{XX}}{\\mathrm{XX}+1}$
$\\because \\frac{X}{X+1}+\\frac{X+1}{X}=\\frac{61}{30}$<\/p>\n\n\n\n
$\\frac{X^{2}+X^{2}+2 X+1}{X^{2}+X}=\\frac{61}{30}$
30(2X2<\/sup> + 2X + 1) = 61 (X2<\/sup> + X)
60X2<\/sup> + 60X + 30 = 61X2<\/sup>+ 61X
X2<\/sup> + X \u2013 30 = 0\u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-1 \\pm \\sqrt{(1)^{2}-4(1)(-30)}}{2(1)}$
$\\frac{-1 \\pm \\sqrt{1+120}}{2}$
$\\frac{-1 \\pm \\sqrt{121}}{2}$
$\\frac{-1 \\pm 11}{2}$
X= \u2013 6 or 5
\u2234 X = 5 (Only positive values)
\u2234 Denominator is = 5 & numerator is (X+1) = 6
\u2234 The fraction is $\\frac{6}{5}$.<\/p>\n\n\n\n
\n\n\n\nQuestion 19<\/h4>\n\n\n\n
Let the denominator be \u2018X\u2019, so the numerator will be \u2018(X+3)\u2019.
\u2234 the fraction is $\\frac{X+3}{X}$
$\\because \\frac{X+3}{X}-\\frac{X}{X+3}=\\frac{33}{28}$<\/p>\n\n\n\n
$\\frac{X^{2}+6 X+9-X^{2}}{X^{2}+3 X}=\\frac{33}{28}$
28(6X + 9) = 33(X2<\/sup> + 3X)
168X + 252 = 33X2<\/sup> + 99X
33X2<\/sup> \u2013 69X \u2013 252 = 0
11X2<\/sup> \u2013 13X \u2013 84 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-13) \\pm \\sqrt{(-13)^{2}-4(11)(-84)}}{2(11)}$
$\\frac{13 \\pm \\sqrt{169+3696}}{22}$
$\\frac{13 \\pm \\sqrt{3865}}{22}$![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990269753884.png\")
it does not have any real values.<\/p>\n\n\n\n
\n\n\n\nQuestion 20<\/h4>\n\n\n\n
Let the denominator be \u2018(X+3)\u2019, so the numerator will be \u2018X\u2019.
\u2234 the original fraction is $\\frac{X}{X+3}$
\u2234 the new fraction is $\\frac{X+1}{X+4}$
$\\because \\frac{X+1}{X+4}-\\frac{X}{X+3}=\\frac{1}{24}$<\/p>\n\n\n\n
$\\frac{X^{2}+3 X+X+3-X^{2}-4 X}{X^{2}+3 X+4 X+12}=\\frac{1}{24}$
24(3) = X2<\/sup> + 7X + 12
X2<\/sup> + 7X \u2013 60 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(7) \\pm \\sqrt{(7)^{2}-4(1)(-60)}}{2(1)}$
$\\frac{-7 \\pm \\sqrt{49+240}}{2}$
$\\frac{-7 \\pm \\sqrt{289}}{2}$
$\\frac{-7 \\pm 17}{2}$
\u2234 X = \u201312 or 5.
\u2234 the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\\frac{5}{8}$, as taking (\u201312) will form a fraction i.e., $\\frac{4}{3}$ (not satisfying the conditions) .<\/p>\n\n\n\n
\n\n\n\nQuestion 21<\/h4>\n\n\n\n
Let the denominator be \u2018(X+3)\u2019, so the numerator will be \u2018X\u2019.
\u2234 the original fraction is $\\frac{X}{X+3}$
\u2234 the new fraction is $\\frac{X+3}{X+6}$
$\\because \\frac{X+3}{X+6}-\\frac{X}{X+3}=\\frac{9}{88}$<\/p>\n\n\n\n
$\\frac{X^{2}+6 X+9-X^{2}-6 X}{X^{2}+3 X+6 X+18}=\\frac{9}{88}$
88(9) = 9(X2<\/sup> + 9X + 18)
X2<\/sup> + 9X \u2013 70 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(9) \\pm \\sqrt{(9)^{2}-4(1)(-70)}}{2(1)}$
$\\frac{-9 \\pm \\sqrt{81+280}}{2}$
$\\frac{-9 \\pm \\sqrt{361}}{2}$
$\\frac{-9 \\pm 19}{2}$
\u2234 X = \u201314 or 5.
\u2234 the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\\frac{5}{8}$, as taking (\u201314) will form a fraction i.e., $\\frac{14}{11}$ (not satisfying the conditions) .<\/p>\n\n\n\n
\n\n\n\nQuestion 22<\/h4>\n\n\n\n
Let the denominator be \u2018(X+3)\u2019, so the numerator will be \u2018X\u2019.
\u2234 the original fraction is $\\frac{X}{X+3}$
\u2234 the new fraction is $\\frac{X+2}{X+5}$
$\\because \\frac{X+2}{X+5}+\\frac{X}{X+3}=\\frac{19}{15}$<\/p>\n\n\n\n
$\\frac{X^{2}+5 X+6+X^{2}+5 X}{X^{2}+3 X+5 X+15}=\\frac{19}{15}$
15(2X2<\/sup> + 10X + 6) = 19(X2<\/sup> + 8X + 15)
30X2<\/sup> + 150X + 90 = 19X2<\/sup> + 152X + 285
11X2<\/sup> \u2013 2X \u2013 195 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-2) \\pm \\sqrt{(-2)^{2}-4(11)(-195)}}{2(11)}$
$\\frac{2 \\pm \\sqrt{4+8580}}{22}$
$\\frac{2 \\pm \\sqrt{8584}}{22}$
\u2234 it does not have real values.<\/p>\n\n\n\n
\n\n\n\nQuestion 23<\/h4>\n\n\n\n
Given: Numerator of a fraction is 2 less than the denominator
To find: The fraction
Assumption: Let the denominator be x
Numerator = x \u2013 2
Therefore, the fraction $=\\frac{x-2}{x}$
If one is added to the numerator and denominator, fraction becomes <\/p>\n\n\n\n
Sum of the fractions $=\\frac{19}{15}$
Therefore,
$\\frac{x-2}{x}+\\frac{x-1}{x+1}=\\frac{19}{15}$<\/p>\n\n\n\n
$\\frac{(x-2)(x+1)+x(x-1)}{x(x+1)}=\\frac{19}{15}$
$\\frac{x^{2}+x-2 x-2+x^{2}-x}{x^{2}+x}=\\frac{19}{15}$
$\\frac{2 x^{2}-2 x-2}{x^{2}+x}=\\frac{19}{15}$<\/p>\n\n\n\n
30x2<\/sup> \u2013 30x \u2013 30 = 19x2<\/sup> + 19x
30x2<\/sup> \u2013 19x2<\/sup> \u2013 30x \u2013 19x \u2013 30 = 0
11x2<\/sup> \u2013 49x \u2013 30 = 0
Now we need to factorise such that, on multiplication we get 330 and on substraction we get 49.
So, equation becomes,
11x2<\/sup> \u2013 (55x \u2013 6x) \u2013 30 = 0
11x2<\/sup> \u2013 55x + 6x \u2013 30 = 0
11x(x \u2013 5) + 6(x \u2013 5) = 0
(11x + 6)(x \u2013 5) = 0<\/p>\n\n\n\n
$x-2=-\\frac{5}{11}-2=-\\frac{27}{11}$ or <\/p>\n\n\n\n
Hence, the possible fractions are,
$\\frac{x-2}{x}=\\frac{-\\frac{27}{11}}{-\\frac{5}{11}}=\\frac{27}{5}$
$\\frac{x-2}{x}=\\frac{3}{5}$<\/p>\n\n\n\n
\n\n\n\nQuestion 24<\/h4>\n\n\n\n
Let the shortest side(AC) be \u2018(X)\u2019cms, so the hypotenuse (BC) will be \u2018(2X+6)\u2019 cms, as demonstrated in the figure drawn below:![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/15459903117445.png\")
\u2234 AB = 2X + 4 cms
\u2235 (BC)2<\/sup> = (AC)2<\/sup> + (AB)2<\/sup>
\u2234(2X + 6)2<\/sup> = X2<\/sup> +(2X + 4)2<\/sup>
On simplifying further,
4X2<\/sup> + 24X + 36 = X2<\/sup> + 4X2<\/sup> + 16X + 16
Using the identity of a2<\/sup> + b2<\/sup> + 2ab = (a + b)2<\/sup>
X2<\/sup> \u2013 8X \u2013 20 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(-8) \\pm \\sqrt{(-8)^{2}-4(1)(-20)}}{2(1)}$
$\\frac{8 \\pm \\sqrt{64+80}}{2}$
$\\frac{8 \\pm \\sqrt{144}}{2}$
$\\frac{8 \\pm 12}{2}$
\u2234 X = \u20132 or 10.
\u2234 the shortest side (AC) is X = 10 cms (Only positive values), hypotenuse (2X+6) i.e., (BC) is 26 cms and the other side (AB) is (2X+4) i.e., 24 cms .<\/p>\n\n\n\n
\n\n\n\nQuestion 25<\/h4>\n\n\n\n
Let the sides of the squares be \u2018(X)\u2019cms and \u2018(Y) cms.
\u2234 X2<\/sup> + Y2<\/sup> = 640 \u2013\u2013\u2013\u2013\u2013(i)
\u2235 Area = (Side)2<\/sup>
\u2235Perimeter = 4 (Side)
\u22344X \u2013 4Y = 64
On simplifying further,
X \u2013 Y = 16 \u2013\u2013\u2013\u2013\u2013\u2013(ii)
Squaring the above mentioned equation, i.e., equation (ii)
X2<\/sup> + Y2<\/sup> \u2013 2XY = 256
Using the identity of a2<\/sup> + b2<\/sup> \u2013 2ab = (a \u2013 b)2<\/sup>
Putting the value of equation (i) in equation (ii)
\u2234 640 \u2013 2XY = 256
2XY = 384
XY = 192<\/p>\n\n\n\n
(Y+16)2<\/sup> + Y2<\/sup> = 640
Y2<\/sup> + 256 + 32Y + Y2<\/sup> = 640
2Y2<\/sup> + 32Y \u2013 384 = 0
Y2<\/sup> + 16Y \u2013 192 = 0<\/p>\n\n\n\n
$\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}$
$\\frac{-(16) \\pm \\sqrt{(16)^{2}-4(1)(-192)}}{2(1)}$
$\\frac{-16 \\pm \\sqrt{256+768}}{2}$
$\\frac{-16 \\pm \\sqrt{1024}}{2}$
$\\frac{-16 \\pm 32}{2}$
\u2234 X = \u2013 12 or 24.
\u2234 the side is X = 24 cms (Only positive values), other square\u2019s side is Y=X\u201316 i.e., 8 cms.<\/p>\n\n\n\n
\n\n\n\nQuestion 26<\/h4>\n\n\n\n
Let the shortest side(AC) be \u2018(X)\u2019cms, and the longer side (AB) be \u2018(Y)\u2019 cms, as demonstrated in the figure drawn below:![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990320709339.png\")
\u2234 BC = 3\u221a10 cms
\u2235 (BC)2<\/sup> = (AC)2<\/sup> + (AB)2<\/sup>
\u2234 (3\u221a10)2<\/sup> = X2<\/sup> + (Y)2<\/sup>
On simplifying further,
(Y)2<\/sup> = 90 \u2013 X2<\/sup>
X2<\/sup> + Y2<\/sup> \u2013 90 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (i)
As per the question,
New smaller side = \u2018(3X)\u2019 cms
New longer side = \u2018(2Y)\u2019 cms
\u2234 BC = 9\u221a5 cms
\u2235 (BC)2<\/sup> = (AC)2<\/sup> + (AB)2<\/sup>
\u2234(9\u221a5)2<\/sup> = (3X)2<\/sup> + (2Y)2<\/sup>
On simplifying further,
4Y2<\/sup> + 9X2<\/sup> = 405
4X2<\/sup> + 9Y2<\/sup> \u2013 405 = 0 \u2013\u2013\u2013\u2013\u2013\u2013 (ii)
Putting the value of X2<\/sup>, from equation (i) in equation (ii)
X2<\/sup> = 90 \u2013 Y2<\/sup>
On simplifying further,
(360\u20134Y)2<\/sup> + 9Y2<\/sup> \u2013 405 = 0
5Y2<\/sup> = 45
Y = \u00b1 3 cms(Only positive values),
\u2234 X = \u221a(90\u20139).
\u2234 X = 9 cms
\u2234 the shortest side (AC) is X = 9 cms hypotenuse i.e., (BC) is $3 \\sqrt{10}$ cms and the other side (AB) is 3 cms .<\/p>\n\n\n\n
\n\n\n\nQuestion 27<\/h4>\n\n\n\n
Ok now we know that area of square = side2<\/sup>
So no. of students in the line if side was x (assumption)
No. of students = x2<\/sup>+24 (as there were 24 exrltra students)
No. Of students after increasing 1 student in square = (x+1)2<\/sup>\u201325 (as there were 25 less students)
So x2<\/sup> + 24 = (x + 1)2<\/sup> \u2013 25
x2<\/sup>+ 24 = x2<\/sup> + 1 + 2x \u2013 25
x2<\/sup> \u2013 x2<\/sup> + 24 + 25 \u2013 1 = 2x
48 = 2x
So x = 24
So no. Of students = x2<\/sup> + 24 = 242<\/sup> + 24 = 576 + 24 = 600
\u2234 the no. of students = 600.<\/p>\n\n\n\n
\n\n\n\nQuestion 28<\/h4>\n\n\n\n
![](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545990323691535.png\")
Let the length of base = P cm.
As base exceeds the base by 7cm ,
then , length of altitude = (P + 7)cm
now, area of triangle $=\\frac{1}{2} \\times$ altitude $\\times$ base
given, area of triangle = 30 cm2<\/sup>
so, $30 \\mathrm{cm}^{2}=\\frac{1}{2} \\times(\\mathrm{P}+7) \\times \\mathrm{P}$
\u21d2 30 \u00d7 2 = P2<\/sup> + 7P
\u21d2 60 = P2<\/sup> + 7P
\u21d2 P2<\/sup> + 7P \u2013 60 = 0
\u21d2 P2<\/sup> + 12P \u2013 5P \u2013 60 = 0
\u21d2 P(P + 12) \u2013 5(P + 12) = 0
\u21d2 (P + 12)(P \u20135) = 0
\u21d2 P = 5 , \u201312
but length can’t be negative so, P \u2260 \u201312
hence, P = 5 cm e.g., base = 5cm<\/p>\n\n\n\n
\n\n\n\nQuestion 29<\/h4>\n\n\n\n
Let breadth be X cm
length = 2X cms
area=800
2X (X)=800
X(X)=400
X2<\/sup> = 400