Class 8 Linear Equations in One Variable 9.4

\begin{array}{l} \text {Q1. } \quad \text {Four-fifth of a number is more than three-fourth of the number by 4. Find the number. } \\ \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the number be x.} \\ \text {Three-fourth of the number x is } \frac {3}{4} x \\ \text {Fourth-fifth of x is } \frac {4}{5} x \\ \text{According to question }\frac {4}{5} x -\frac {3}{4} x =4 \\ \Rightarrow \quad \frac {16 x-15 x}{20}=4 \\ \Rightarrow \quad 16 x-15 x=4(20) \\ \Rightarrow \quad x=80 \\\therefore \text {The required number is 80.} \\ \end{array}\begin{array}{l} \text {Q2. } \quad \text {The difference between the squares of two consecutive numbers is 31. Find the numbers. } \\ \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the two consecutive numbers be x and (x+1).} \\ \text{According to question } \\ (x+1)^{2} – x^{2} =31 \Rightarrow \quad x^{2}+ 2x +1 -x^{2}=31 \quad \quad [\text {Using :} (a+b )^{2}=a^{2}+b^{2} + 2ab]\\ \Rightarrow \quad 2 x+1=31 \Rightarrow \quad 2 x=31-1 \\ \Rightarrow \quad 2 x=30 \\ \Rightarrow \quad x= \frac {30}{2} \Rightarrow \quad x= 15 \\\therefore \text {Requied two consecutive numbers are 15 and 16.} \\ \end{array}\begin{array}{l} \text {Q3. } \quad \text {Find a number whose double is 45 greater than its half.} \\ \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the number be x.} \\ \text{According to question } \\ 2 x – \frac {x}{2} =45 \\ \Rightarrow \quad \frac {4x-x}{2}=45 \\ \Rightarrow \quad 4x-x=45 \times 2 \\ \Rightarrow \quad 3x=90 \\\Rightarrow \quad x = \frac {90}{3}=30 \\\therefore \text {The required number is 30.} \\ \end{array}\begin{array}{l} \text {Q4. } \quad \text {Find a number such that when 5 is subtracted from 5 times that number, } \\ \text {the result is 4 more than twice the number. } \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the number be x.} \\ \text{According to question } \\ 5x – 5 = 2x+4 \\ \Rightarrow \quad 5 x-2 x=5+4 \\ \Rightarrow \quad 3 x=9 \\ \Rightarrow \quad x=\frac{9}{3} \\ \Rightarrow \quad x=3 \\\therefore \text {The required number is 3.} \\ \end{array}\begin{array}{l} \text {Q5. } \quad \text {A number whose fifth part increased by 5 is equal to its fourth part diminished by 5.} \\ \text {Find the number.} \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the number be x.} \\ \therefore \text {The fourth part of x is } \frac {x}{4} \\ \text {The fifth part of x is } \frac {x}{5} \\ \text{According to question } \\ \frac{x}{5} + 5 =\frac {x}{4} – 5 \\\Rightarrow \quad \frac{x+25}{5} =\frac {x-20}{4} \\ \Rightarrow \quad 4(x+25) = (x-20)5 \\ \Rightarrow \quad 4x + 100 = 5x -100 \\ \Rightarrow \quad 5x-4x=100+100 \\ \Rightarrow \quad x=200 \\\therefore \text {The required number is 200.} \\ \end{array}\begin{array}{l} \text {Q6. } \quad \text {A number consists of two digits whose sum is 9. If 27 is subtracted from the number} \\ \text {the digits are reversed. Find the number. } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume that x be the digit at unit place.} \\ \therefore \text {The other digit is 9-x as its given that sum of two digits is 9.} \text {So the two digit number is } 10(9-x)+x. \text {And, the number obtained after interchanging the digits is } 10x+(9-x) \\\Rightarrow \quad 10(9-x)+x-27=10 x+(9-x) \\ \Rightarrow \quad 90-10x + x – 27=10x + 9 – x \\ \Rightarrow \quad -9x-9x=9 – 90 + 27 \\ \Rightarrow \quad -18x=-54 \\ \Rightarrow \quad x= \frac {54}{18} \\ \Rightarrow \quad x= \frac {9}{3} \\ \Rightarrow \quad x=3 \\\text {Hence the two digit number is } 10(9-3) +3 = 63 \\ \end{array}\begin{array}{l} \text {Q7. } \quad \text {Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8 } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume first part of 184 be x.} \\ \therefore \text {The second part is } 184-x. \\ \Rightarrow \text {One third of x is } \frac {x}{3} \\ \Rightarrow \text {One seventh of 184-x is } \frac {184-x}{7} \\ \text{According to question } \\ \frac {x}{3} – \frac {184-x}{7}=8 \\ \Rightarrow \quad \frac {7x-3(184-x)}{21} = 8 \\ \Rightarrow \quad 7 x-552+3 x = 8(21)\Rightarrow \quad 10 x-552=168 \\ \Rightarrow \quad 10 x=168+552 \\ \Rightarrow \quad 10 x=720 \\ \Rightarrow \quad x= \frac {720}{10} \\ \Rightarrow \quad x=72 \\\therefore \text {Reqired two parts of 184 are 72 and } (184-72) = 112 \\ \end{array}\begin{array}{l} \text {Q8. } \quad \text {The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator,} \\ \text {the fraction is equal to } \frac {2}{3} \text {What is the original fraction equal to? } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume the denominator be x. } \\ \therefore \text {The numerator is } (x-6) \\ \text{According to question } \\ \frac{(x-6)+3}{x}=\frac{2}{3} \\ \Rightarrow \quad \frac{(x-3}{x}=\frac{2}{3} \\ \Rightarrow \quad 3(x-3) = 2x \\ \Rightarrow \quad 3x-9 = 2x \\ \Rightarrow \quad 3x – 2x = 9 \\ \Rightarrow \quad x=9 \\\therefore \text {The denominator is } =9 \text {and numerator is } (x-6)=(9-6)=3 \\ \text {Hence, the required fraction } =\frac {x-6}{x} =\frac {3}{9}=\frac {1}{3} \\ \end{array}\begin{array}{l} \text {Q9. } \quad \text {A sum of Rs. 800 is in the form of denominations of Rs. 10 and Rs. 20. If the total number of notes be 50.} \\ \text {Find the number of notes of each type.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume number of Rs. 10 notes be x.} \\ \therefore \text {Number of Rs. 20 notes be } 50-x. \\\Rightarrow \quad \text {Value of Rs. 10 notes } = 10x \\ \Rightarrow \quad \text {And Value of Rs. 20 notes } = 20(50-x) \\ \text {According to question } \\ 10x+ 20(50-x) = 800 \\ \Rightarrow \quad 10x + 1000 – 20x = 800 \\ \Rightarrow \quad -10x=800-1000 \\ \Rightarrow \quad -10x=-200 \\ \Rightarrow \quad x=\frac {200}{10} \\ \Rightarrow \quad x=20 \\\therefore \text {The number of Rs. 10 notes are 20. And Number of Rs. 20 notes are } 50-20 =30 \\ \end{array}\begin{array}{l} \text {Q10. } \quad \text {Seeta Devi has Rs. 9 in fifty-paise and twenty five-paise coins. } \\ \text {She has twice as many twenty-five paise coins as she has fifty- paise coins. How many coins of each kind does she have? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that number of 50 paise coins be x. } \\ \therefore \text {Number of 25 paise coins be 2x.} \\ \Rightarrow \quad \text {Value of 50 paise coins } = (50 paise) \times x = Rs. 0.5 x \\ \Rightarrow \quad \text {Value of 25 paise coins} = (25 paise) \times 2x = Rs. \frac {1}{4} \times 2x= Rs. 0.5 x \\ \text{According to question } \\ 0.5x + 0.5x = 9 \Rightarrow \quad x = 9 \\\therefore \text {Number of 50 paise coins } =x =9. \text {Number of 25 five paise coins } =2x =2 \times 9=18 \\ \end{array}\begin{array}{l} \text {Q11. } \quad \text {Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age,} \\ \text { then Sunita will be four times Ashima’s age. How old were they two years ago? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let’s assume present age of Ashima be x years.} \\\therefore \text {The present age of Sunita is 2x years. } \\ \text {Ashima’s new age when 6 yrs subtracted from her present age is (x-6) yrs }\\ \text {Sunita’s new age when 4 yrs added to her present age (2x+4) years } \\ \text {According to question } \\(2 x+4)=4(x-6) \\ \Rightarrow \quad 2 x+4=4 x-24 \\ \Rightarrow \quad 2 x-4 x=-24-4 \\ \Rightarrow \quad -2 x=-28 \\ \Rightarrow \quad x=\frac {28}{2} \\ \Rightarrow \quad x=14 \\\Rightarrow \quad \text {Present age of Ashima is 14 years and present age of Sunita is 2x =2(14)=28 years. } \\ \therefore \text {Two years ago, age of Ashima is } 14-2=12 \text {years, and age of Sunita is } 28-2 = 26 \text {years.} \\ \end{array}\begin{array}{l} \text {Q12. } \quad \text {The ages of Sonu and Monu are in the ratio 7: 5. Ten years hence, the ratio of their ages will be 9: 7.} \\ \text {Find their present ages. } \end{array}\begin{array} \text {Sol. } \\ \text {It’s given that the ratio of the ages of Sonu and Monu is 7 : 5. } \\ \text {Let’s assume the present age of Sonu and Mony be 7x and 5x years respectively.} \\ \text {Sonu’s age after 10 years } =(7x+10) \text { years } \\ \text {Monu’s age after 10 years } =(5 x+10) \text { years } \\ \text {According to question } \\ \frac {7 x+10}{5 x+10}=\frac {9}{7} \\ \Rightarrow \quad 7(7 x+10)=9(5 x+10) \\ \Rightarrow \quad 49 x+70=45 x+90 \\ \Rightarrow \quad 49 x-45 x=90-70 \\\Rightarrow \quad 4 x=20 \\ \Rightarrow \quad x=\frac {20}{4} \\ \Rightarrow \quad x=5 \\ \therefore \text {The present age of Sonu is } 7x=7(5)=35 \text { years.} \\ \therefore \text {The present age of Monu is } 5x=5(5)=25 \text { years.} \\ \end{array}\begin{array}{l} \text {Q13. } \quad \text {Five years ago a man was seven times as old as his son. Five years hence, } \\ \text { the father will be three times as old as his son. Find their present ages. } \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume the age of son five years ago be x years.} \\ \therefore \text {The age of man five years ago be 7x years} \\\Rightarrow \quad \text {The present age of son is (x+5) years } \\ \text {And present age of father is (7x+5) years} \\\text {According to question } \\ (7x+5) + 5 =3[(x+5)+5] \quad \quad [\text {Using given age relation after 5 yrs}] \\\Rightarrow \quad 7 x+10 =3 x+30 \\ \Rightarrow \quad 7 x-3 x=30-10 \\ \Rightarrow \quad 4 x=20 \\ \Rightarrow \quad x=5 \\\therefore \text {Present age of father is } 7x +5 = 7(5)+5 = 40 \text { years. } \\ \text {Present age of son is } x+5 = 5+5=10 \text { years.}\\ \end{array}\begin{array}{l} \text {Q14. } \quad \text {I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now? } \end{array}\begin{array} \text {Sol. } \\ \text {Let assume the present of son be x years. Therefore, the present age of father be 5x years.} \\\text {After 6 years } \\ \text {Age of son be (x+6) years and age of father be (5x+6) years.} \\ \text {According to question } \\ 5x+6=3(x+6) \\ \Rightarrow \quad 5x + 6=3 x+18 \\ \Rightarrow \quad 5x – 3x=18-6 \\ \Rightarrow \quad 2x=12 \\ \Rightarrow \quad x=\frac {12}{2} \\ \Rightarrow \quad x=6 \\\therefore \text {The present age of son is } x=6 \text { years.} \\ \text {And present age of father is } 5x = 5(6) = 30 \text { years. }\\ \end{array}\begin{array}{l} \text {Q15. } \quad \text {I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that } \\ \text { I have is ten more than the number of five rupee notes, how many notes do I have in each denomination? } \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume that number of five rupee notes be x. } \\ \therefore \text {Number of ten rupee notes be (x+10).} \\ \text {Value of five rupee notes }=5 \times x=5 x \\ \text {Values of ten rupee notes }=10 \times (x+10) = 10 x+100 \\ \text {As we know that total value Rs.} = 1000 \\ \Rightarrow \quad 5x+10x+100=1000 \\ \Rightarrow \quad 15 x=900 \\ \Rightarrow \quad x=\frac {900}{15} \\ \Rightarrow \quad x=60 \\ \therefore \text {The number of five rupee notes are x } =60.\\ \text {And number of ten rupee notes are x+10} = 60+10=70 \\ \end{array}\begin{array}{l} \text {Q16. } \quad \text {At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, } \\ \text { a third drank squash, two fifths drank fruit juice and just three did not drink anything. How many guests were in all?} \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume that number of guests be x.} \\ \therefore \text {The number of guests who drank colas are } \frac {x}{4} \\ \text {Number of guests who drank squash are } \frac {x}{3} \\ \text {Number of guests who drank fruit juice are } \frac {2}{5}x \\ \text {Number of guests who did not drink anything } = 3 \\ \Rightarrow \quad \frac {x}{4} + \frac {x}{3} + \frac {2x}{5} +3 = x \\ \Rightarrow \quad \frac {15 x+20 x+24 x+ 160}{60}=x \\ \Rightarrow \quad 15 x+20 x+24 x+ 180=60x \\ \Rightarrow \quad 180=60x -15 x-20 x-24 x \\ \Rightarrow \quad x= 180 \\\therefore \text {There are a total of 180 guests.}\\ \end{array}\begin{array}{l} \text {Q17. } \quad \text {There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every } \\ \text { correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers.} \\ \text {If a candidate scored 450 marks in the test, how many questions did he answer correctly? } \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume that number of correct answers be x. } \\ \therefore \text {Number of unattempted or wrongly answered question are (180-x)} \\ \text {Total score for correctly answered questions }=4 \times x \\ \text {Marks deducted when answered wrongly } = 1 \times (180-x) =180-x \\ \Rightarrow \quad 4x- (180-x) = 450 \\ \Rightarrow \quad 4 x-180+x=450 \\ \Rightarrow \quad 5 x=450+180 \\ \Rightarrow \quad 5 x=630 \\ \Rightarrow \quad x= \frac {630}{5} \\ \Rightarrow \quad x=126 \\ \therefore \text {Candidate answered 126 questions correctly.}\\ \end{array}\begin{array}{l} \text {Q18. } \quad \text {A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day,} \\ \text { he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all } \\ \text { for how many days he remained absent? } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume that number of absent days be x.} \\ \therefore \text {Number of present days are (20-x)} \\ \text {Earning for one day work Rs. } =60 \\ \text {Fine for absent day Rs. }=5 \\ \text {According to question } \\ 60(20-x) – 5x=745 \\ \Rightarrow \quad 1200-60 x-5 x=744 \\ \Rightarrow \quad -65 x=744-1200 \\ \Rightarrow \quad -65 x=-456 \\ \Rightarrow \quad x= \frac {456}{65} \\ \Rightarrow \quad x=7 \\ \therefore \text {The labourer was absent for 7 days.}\\ \end{array}\begin{array}{l} \text {Q19. } \quad \text {Ravish has three boxes whose total weight is }60 \frac{1}{2} \text { kg. Box B weighs } 3 \frac{1}{2} \text { kg more than } \\ \text {box A and box C weighs } 5 \frac{1}{3} \text { kg more than box B. Find the weight of box A.} \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume the weight of box A be x kg. } \\ \therefore \text {Weight of box B is } x+3 \frac{1}{2} = x + \frac{7}{2} \text { kg.} \\ \text {And Weight of box C is } x + \frac{7}{2} + 5 \frac{1}{3} = x + \frac{7}{2} + \frac{16}{3} \text { kg.} \\\text {According to question total weight of 3 boxes together are } 60 \frac{1}{2} = \frac{121}{2}\text { kg., therefore}x + ( x + \frac{7}{2}) + (x + \frac{7}{2} + \frac{16}{3}) = \frac{121}{2}\Rightarrow \quad 3x+ \frac{7}{2} + \frac{7}{2} + \frac{16}{3} =\frac{121}{2} \\\Rightarrow \quad \frac {3x \times 6 +21 + 21 + 32}{6} = \frac{121}{2} \\\Rightarrow \quad 3x \times 6 +21 + 21 + 32 = \frac{121}{2} \times 6 \\\Rightarrow \quad 18x +74 = 363 \Rightarrow \quad 18x = 363 – 74 \\ \Rightarrow \quad 18x = 289 \\ \Rightarrow \quad x =\frac {289}{18} \\\therefore \text {The weight of box A is } \frac {289}{18} \text { kg.} \\ \end{array}\begin{array}{l} \text {Q20. } \quad \text {The numerator of a rational number is 3 less than the denominator. If the denominator} \\ \text { is increased by 5 and the numerator by 2, we get the rational number } \frac {1}{2} \text {. Find the rational number. } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume thedenominator of number be x.} \\ \therefore \text {The numerator be } (x-3) \\\text {According to question } \\\frac {(x-3)+2}{x+5}=\frac {1}{2} \\ \Rightarrow \quad \frac {x-1}{x+5}=\frac {1}{2} \\ \Rightarrow \quad 2(x-1)=x+5 \\ \Rightarrow \quad 2 x-2=x+5 \\ \Rightarrow \quad 2 x-x=2+5 \\ \Rightarrow \quad x=7 \\ \therefore \text {The denominator of rational number be } x=7 \\ \text {Numerator is } (x-3) =7-3=4 \\ \text {Hence, the required rational number is } = \frac {4}{7}\\ \end{array}\begin{array}{l} \text {Q21. } \quad \text {In a rational number, twice the numerator is 2 more than the denominator If 3 is added } \\ \text { to each, the numerator and the denominator. The new fraction is } \frac {2 }{3}. \text {Find the original number. } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume the numerator be x} \\ \text {And because its given that double of numerator is 2 more than denominator, therefore} \\ \text {Denominator be (2x-2) } . \\ \Rightarrow \text {The fraction will be } \frac {x}{2x-2} \\ \text {When numerator and denominator are increased by 3 the fraction will be } \frac {x+3}{2x-2+3} \text { and which is equal to } \frac {2}{3} \\ \Rightarrow \quad \frac {x+3}{2x-2+3}=\frac {2}{3} \\ \Rightarrow \quad 3(x+3)=2(2 x+1) \\ \Rightarrow \quad 3 x+9=4 x+2 \\ \Rightarrow \quad 3 x-4 x=2-9 \\ \Rightarrow \quad x=7 \\\therefore \text {The reqired fraction is } \frac {x}{2x-2} = \frac {7}{2(7)-2} = \frac {7}{12} \\ \end{array}\begin{array}{l} \text {Q22. } \quad \text {The distance between two stations is 340 km. Two trains start simultaneously from these } \\ \text { stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other} \\ \text { by 5 km / hr. If the distance between the two trains after 2 hours of their start is 30 km, } \\ \text {find the speed of each train. } \end{array}\begin{array} \text {Sol. } \\ \text {Let assume that speed of first train be x km/hr.} \\ \therefore \text {The Speed of second train be (x+5) km\hr.} \\ \text {Total distance between two stations is 340 Kms.} \\ \text {As we know Distace = Speed x Time taken} \\ \text {Distance covered by first train in 2 hrs } = 2x \text { km } \\ \text {Distance covered by second train in 2 hrs } = 2(x+5) \text { km } \\ \text {Distance between two trains after 2 hrs is 30 kms} \\ \text {According to question } \\2x+2(x+5)+30=340 \\ \Rightarrow \quad 4 x+40=340 \\ \Rightarrow \quad 4 x=340-40 \\ \Rightarrow \quad 4 x=300 \\ \Rightarrow \quad x=\frac {300}{4} \\ \Rightarrow \quad x=75 \\ \therefore \text {The speed of first train is } x=75 \text { km/hr }\\ \text {Speed of second train is } (x+5) = 75+5 =80 \text { km/hr }\\ \end{array}\begin{array}{l} \text {Q23. } \quad \text {A steamer goes downstream from one point another in 9 hours. It covers the same distance upstream in 10 hours. } \\ \text {If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports. } \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume the speed of steamer in still water be x km/hr.} \\ \text {Speed of stream is given as } =1 \text { km/hr.} \\ \therefore \text {Downstream speed of streamer }= (x+1) \text { km/hr.} \\ \text {Upstream speed of streamer } =(x-1) \text { km/hr.} \\\text {As we know Distace = Speed x Time taken} \\ \text {Downstream distance covered in 9 hrs } = (x+1) \times 9 \text { km } \\ \text {Upstream distance covered in 10 hrs } = (x-1) \times 10 \text { km } \\ \text {But Downstrean and Upstream distance is same, therefore } \\ (x+1) \times 9=(x-1) \times 10 \Rightarrow \quad 9x +9 = 10x -10 \\ \Rightarrow \quad 9 x-10 x=-10-9 \\ \Rightarrow \quad -x=-19 \\ \Rightarrow \quad x=19 \text { km/hr.} \\ \therefore \text {The speed of the steamer in still water is } 19 \text { km/hr.} \\ \text {Distance between the ports is } 9(x+1) =9(19+1)=9(20)=180 \text { km/hr.} \\ \end{array}\begin{array}{l} \text {Q24. } \quad \text {Bhagwanti inherited Rs. 12000.00. She invested part of it as 10% and the rest at 12%. } \\ \text {Her annual income from these investments is Rs 1280.00 How much did she invest at each rate? } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume Bhagwanti invested Rs. x at the rate of 10%.} \\ \therefore \text {Bhagwanti invested Rs. (12000 – x) at the rate of 12%.} \\ \Rightarrow \quad \text {Annual interest earned on Rs. x at the rate of 10% } = 10 \% \text { of } x \\ \text {Annual interest earned on Rs. (12000 – x) at the rate of 12% } = 12 \% \text { of } (12000 – x) \\\text {According to question } \\ 10 \% \text { of } x + 12 \% \text { of } (12000 – x) = 1280 \\ \Rightarrow \quad \frac {10}{100}x + \frac {12}{100}(12000 – x) = 1280 \\ \Rightarrow \quad \frac {10 x+144000-12 x}{100} = 1280 \\ \Rightarrow \quad 144000-2 x=1280 \times 100 \\ \Rightarrow \quad -2 x=128000-144000 \\ \Rightarrow \quad -2 x=-16000 \\ \Rightarrow \quad x= \frac {16000}{2} \\ \Rightarrow \quad x=8000 \\ \therefore \text {Bhagwanti invested Rs. 8000 at the rate of 10%.} \\ \text {And she invested Rs. } (12000 – x) = 12000-8000 = 4000 \text { at the rate of 12%.} \end{array}\begin{array}{l} \text {Q25. } \quad \text {The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm } \\ \text {,the area of the new rectangle will be } 84 cm^{2} \text { more than that of the given rectangle.} \\ \text {Find the length and breadth of the given rectangle. } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume the breadth of the rectangle be x cm.} \\ \therefore \text{The Length of the rectangle be (x+9)cm. } \\ \Rightarrow \text{Area of the rectangle } = x(x+9) cm^{2} \quad \quad [\text {Area of rectangle }=l \times b ]\\\text {And area of rectangle when length and breadth increased by 3 cm } = (x+3)[(x+9)+3] cm^{2} \\\text {According to question } \\ x(x+9)+84 = (x+3)(x+12) \\ \Rightarrow \quad x^{2}+9 x+84 = x^{2}+15 x+36 \\\Rightarrow \quad 15 x-9 x=84-36 \\ \Rightarrow \quad 6 x=48 \\ \Rightarrow \quad x=\frac {48}{6} \\ \Rightarrow \quad x=8 \\\therefore \text {The Length of given rectangle is } (x+9)=(8+9)=17 \text { cm.} \\ \text {And the breadth of given rectangle is 8 cm.}\\ \end{array}\begin{array}{l} \text {Q26. } \quad \text {The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, } \\ \text {he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, } \\ \text {when Anup is as old as his father. What are their ages now? } \end{array}\begin{array} \text {Sol. } \\ \text {Lets assume the age of Anup be x years.} \\ \therefore \text {Anup’s father will be (100-x) years.} \\ \text {When Anup is as old as his father now, he will be 5 times older then his son Anuj now} \\ \Rightarrow \quad 5 \times (\text {Anuj’s present age}) = \text {Anup’s father present age.} \\ \Rightarrow \quad 5 \times (\text {Anuj’s present age}) = 100 -x \\ \Rightarrow \quad \text {Anuj’s present age} = \frac {100-x}{5} \text { years.} \\\text {Anup will be as old as his father’s present age after }= \text {Anup’s father present age} – \text {Anup’s present age} \\ = (100-x) -x = 100-2x \text { years.} \\\text {After (100-2x) yrs, Anuj’s age will be} = \text {Anuj present age} + 100 -2x \\ = \frac {100-x}{5} + 100 -2x = \frac {600-11x}{5} \text { years.} \\\text {As its given that Anuj is 8 yrs older than his father Anup’s present age after (100-2x) years, therefore } \\ \frac {600-11x}{5} = x +8 \\ \Rightarrow \quad 600-11x = 5 \times (x+8) \\ \Rightarrow \quad 600-11x = 5x+40 \\ \Rightarrow \quad 600-40 = 5x+11x \\ \Rightarrow \quad 16x = 540 \\ \Rightarrow \quad x = \frac {540}{16} \\ \Rightarrow \quad x =35 \\\therefore \text {The present age of Anup is 35 years.} \\ \text {The age of Anup’s father } (100-x)=100-35= 65 \text { years.} \\ \text {The age of Anuj is } \frac {100-x}{5} =\frac {100-35}{5} =\frac {65}{5}=13 \text { years.} \\ \end{array}\begin{array}{l} \text {Q27. } \quad \text {A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a begger} \\ \text { waiting outside the shop. She spent half of what was left on a lunch and followed that up with} \\ \text { a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home,} \\ \text { she found that she had exactly one rupee left. How much money did she start with?} \end{array}\begin{array} \text {Sol. } \\ \text {Let’s assume that the lady had Rs. x in the beginning.} \\ \text {Money spent on buying hankies Rs. } = \frac{x }{2} \\ \text {Money given to begger is Rs. } = 1 \\ \text {Money left with the lady i.e A } = x-(\frac {x}{2}+1)=\frac {x}{2}-1=\frac {x-2}{2} \\\text {Money spent for lunch } = \text {Half of the remaining money i.e A} = \frac {x-2}{2} \times \frac {1}{2} = \frac {x-2}{4} \\ \text {Money given as tip (Rs.)} = 2 \\\text {Money left with the lady after lunch i.e B } = \frac {x-2}{2} – \frac {x-2}{4} – 2 = \frac {2x-4-x+2 -8}{4} = \frac {x-10}{4} \\\text {Money spent for books } =\frac {1}{2} \times \frac {x-10}{4} =\frac {x-10}{8} \\\text {Money spent on Bus (Rs.)} = 3 \\\text {Final amount left after all expenditures } =\frac {x-10}{4} – \frac {x-10}{8} – 3 =\frac {2 x-20-x+10-24}{8}=\frac {x-34}{8} \\\text {Its given that lady is left with Rs. 1, so} \\ \frac {x-34}{8} = 1 \\ \Rightarrow \quad x-34=8 \\ \Rightarrow \quad x=8+34 \\ \Rightarrow \quad x=42 \\ \therefore \text {The lady had 42 rupees in the starting.} \\ \end{array}