Class 7 Ratio and Proportion Exercise 9.1

\begin{array}{l} \text {Q1. If } x: y=3: 5 \text {, find the ratio } 3 x+4 y: 8 x+5 y \\ \end{array}\begin{array}{l} \text {Sol. } \\ x: y=3: 5 \\ \Rightarrow \quad \frac{x}{y}=\frac{3}{5} \\ \Rightarrow \quad 5 x=3 y \\ \Rightarrow \quad x=\frac{3 y}{5} \\ \text {By substituting the value of x in } 3x + 4y: 8x + 5y \text { we get,} \\ 3 x+4 y: 8 x+5 y \\ =\frac{3 \times 3 y}{5}+4 y: \frac{8 \times 3 y}{5}+5 y \\ =\frac{9 y+20 y}{5}: \frac{24 y+25 y}{5}\\ =\frac{29 y}{5}: \frac{49 y}{5}\\ =29 y: 49 y\\ =29: 49 \\ \end{array}\begin{array}{l} \text {Q2. If } x: y=8: 9 \text {, find the ratio } (7 x-4 y): 3 x+2 y \end{array}\begin{array}{l} \text {Sol. } \\ x: y=8: 9 \\ \Rightarrow \quad \frac{x}{y}=\frac{8}{9} \\ \Rightarrow \quad 9 x=8 y \\ \Rightarrow \quad x=\frac{8 y}{9} \\ \text {By substituting the value of x in } (7 x-4 y): 3 x+2 y \text { we get,} \\ 7 x-4 y: 3 x+2 y \\ =\frac{7 \times 8 y}{9}-4 y: \frac{3 \times 8 y}{9}+2 y \\ =\frac{56 y-36 y}{9}: \frac{42 y}{9} \\ =20: 42 \\ =10: 21 \\ \end{array}\begin{array}{l} \text {Q3. If two numbers are in the ratio 6: 13 and their lcm is 312, find the numbers. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets the required numbers be 6x and 13x. } \\ \text {L.C.M of } 6x \text { and } 13x = 78 x \\ \Rightarrow \quad 78 x=312 \\ \Rightarrow \quad x=\frac{312}{78} \\ \Rightarrow \quad x=4 \\ \text {Hence the required numbers are } 6x=6 \times 4=24 \text { and } 13x=13 \times 4=52\end{array}\begin{array}{l} \text {Q4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that the required numbers be 3 x and 5 x } \\ \text {If 8 is added to each number then numbers are 3 x+8 and 5 x+8 respectiviely } \\ \text {According to question, we have } \\ 3 x+8 : 5 x+8=2: 3 \\ \Rightarrow \quad \frac{3 x+8}{5 x+8}=\frac{2}{3} \\ \Rightarrow \quad 3(3 x+8)=2(5 x+8) \\ \Rightarrow \quad 9 x+24=10 x+16 \\ \Rightarrow \quad 10 x-9 x=24-16 \\ \Rightarrow \quad x=8 \\ \text {Hence, the required numbers are } 3x=3 \times 8=24 \text { and } 5 x=5 \times 8=40 \end{array}\begin{array}{l} \text {Q5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let’s assume that the required number to be added be x } \\ \text {According to question, we get } \\ \frac{7+x}{13+x}=\frac{2}{3} \Rightarrow \quad (7+x) \times 3=2 \times (13+x) \\ \Rightarrow \quad 3 x-2 x=26-21 \Rightarrow \quad x=5 \\ \text {Hence, the required number is 5.} \end{array}\begin{array}{l} \text {Q6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that the required numbers be 2x, 3x and 5x } \\ \text {Sum of these numbers } =800 \\ \text {According to question, we get } \\ 2x +3x +5x = 800 \\ \Rightarrow \quad 10x = 800 \\ \Rightarrow \quad x = 80 \\\therefore \text {First number } =2 \times 80=160 \\ \text {Second number } =3 \times 80=240 \\ \text {Third number } =5 \times 80=400 \\\end{array}\begin{array}{l} \text {Q7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let’s assume that the required ages be 5 x and 7 x.} \\ \therefore \text {18 years ago their ages are }(5 x-18) \text { and } (7 x-18) \text {respectively.} \\ \text {According to question, we get } \\ \frac{5 x-18}{7 x-18}=\frac{8}{13} \\ \Rightarrow \quad 65 x-13 \times 18=8 \times 7 x-8 \times 18 \\ \Rightarrow \quad 65 x-234=56 x-144 \\ \Rightarrow \quad 65 x-56 x=234-144 \\ \Rightarrow \quad 9 x=90 \\ \Rightarrow \quad x=10 \\ \text {Hence, the ages are } 5x=5 \times 10=50 \text { years and } 7x=7 \times 10=70 \text { years.} \end{array}\begin{array}{l} \text {Q8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, } \\ \text { the ratio becomes 2: 3. Find the numbers } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let’s assume that the required numbers be 7 x and 11 x.} \\ \text {If 7 is added to each of the numbers it becomes } \\ \frac{7 x+7}{11 x+7}=\frac{2}{3} \\ \Rightarrow \quad 21 x+21=22 x+14 \\ \Rightarrow \quad x=21-14=7 \\ \text {Hence, the numbers are } 7 x=7 \times 7=49 \text { and } 11 x=11 \times 7=77\end{array}\begin{array}{l} \text {Q9. Two numbers are in the ratio 2 : 7. If the sum of the numbers is 810, find the numbers.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that the required numbers be 2x and 7x } \\ \text {Sum of these numbers } =810 \\ \text {According to question, we get } \\ 2x +7x = 810 \\ \Rightarrow \quad 9x = 810 \\ \Rightarrow \quad x = 90 \\\therefore \text {First number } =2 \times 90=180 \\ \text {Second number } =7 \times 90=630 \\\end{array}\begin{array}{l} \text {Q10. Divide Rs 1350 between Ravish and Shikha in the ratio 2 : 3 } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that sum of the terms of the ratio } =2+3=5 \\ \text {Ravish money } =\frac{2}{5} \times 1350 =2 \times 270 \\ = \text {Rs. } 540 \\ \text {Shikha money }=\frac{3}{5} \times 1350 =3 \times 270 \\ = \text {Rs. } 810 \end{array}\begin{array}{l} \text {Q11. Divide Rs 2000 among P, Q, R in the ratio 2 : 3: 5 } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that sum of the terms of the ratio } =2+3+5=10 \\ \text {P’s share }=\frac{2}{10} \times 2000 \\ =2 \times 200 \\ = \text {Rs. } 400 \\ \text {Q’s share }=\frac{3}{10} \times 2000 \\ =3 \times 200 \\ = \text {Rs. } 600 \\ \text {R’s share }=\frac{5}{10} \times 2000 \\ =5 \times 200 \\ = \text {Rs. } 1000 \end{array}\begin{array}{l} \text {Q12. The boys and the girls in a school are in the ratio 7: 4 . If total } \\ \text {strength of the school be 550, find the number of boys and girls. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Givent that he boys and girls is in the ratio } =7: 4 \\ \text {Sum of the terms in the ratio }=7+4=11 \\ \text {Total strength } =550 \\ \therefore \text {Boys strength }=\frac{7}{11} \times 550 \\ =7 \times 50 \\ =350 \\ \text {Girls strength }=\frac{4}{11} \times 550 \\ =4 \times 50 \\ =200 \end{array}\begin{array}{l} \text {Q13. The ratio of monthly income to the savings of a family is 7: 2 . If the savings } \\ \text { be of Rs. 500, find the income and expenditure.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that income and savings be 7x and 2x.} \\ \text {Family monthly savings } =500 \\ \Rightarrow \quad 2x=500 \\ \Rightarrow \quad x=250 \\\therefore \text { Income of family} =7 \times 250 =1750 \\ \text {And expenditure of family =Income – savings }=1750-500 \\ =\text {Rs. } 1250 \end{array}\begin{array}{l} \text {Q14. The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {The sides of the triangle are in the ratio 1: 2: 3 } \\ \text {Sum of the terms in the ratio }=1+2+3=6 \\ \text {Perimeter } =36 \text { cm} \\ \text {First side }=\frac{1}{6} \times 36=6 \text { cm} \\ \text {Second side }=\frac{2}{6} \times 36=12 \text { cm} \\ \text {Third side }=\frac{3}{6} \times 36=18 \text { cm} \\ \end{array}\begin{array}{l} \text {Q15. A sum of Rs 5500 is to be divided between Raman and Aman in the ratio 2: 3.} \\ \text {How much will each get? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that the sum of the terms in the ratio }=2+3=5 \\ \text {Raman share }=\frac{2}{5} \times 5500 \\ =2 \times 1100 \\ =\text {Rs. } 2200 \\ \text {Aman share } =\frac{3}{5} \times 5500 \\ =3 \times 1100 \\ =\text {Rs. } 3300\end{array}\begin{array}{l} \text {Q16. The ratio of zinc and copper in an alloy is 7: 9. If the weight of the copper } \\ \text { in the alloy is 11.7 kg, find the weight of the zinc in the alloy.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that zinc and copper in an alloy be 7x kg and 9x kgs respectively.} \\ \text {Given weight of copper in the alloy} =11.7 \text { kg} \\ \Rightarrow \quad 9 x=11.7 \\ \Rightarrow \quad x=\frac{11.7}{9} \text { kg} \\ \text {Hence, weight of zinc in the alloy} = 7x =1.3 \times 7 =9.10 \text { kg} \\\end{array}\begin{array}{l} \text {Q17. In the ratio 7: 8, if the consequent is 40, what is the antecedent? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let’s assume that antecedent and consequent be 7 x and 8 x } \\ \text {Given that Consequent }=40 \\ \Rightarrow \quad 8 x=40 \\ \Rightarrow \quad x=\frac {40}{8} = 5 \\ \therefore \text {Antecedent } =7 x=7 \times 5=35 \end{array}\begin{array}{l} \text {Q18. Divide Rs 351 into two parts such that one may be to the other as 2: 7. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {The sum of the terms in the ratio }=2+7=9 \\ \text {Total amount } = \text {Rs. } 351 \\ \text {First part } =\frac{2}{9} \times 351 \\ =2 \times 39=\text {Rs. } 78 \\ \text {Second part } =\frac{7}{9} \times 351 \\ =7 \times 39=\text {Rs. } 273 \end{array}\begin{array}{l} \text {Q19. Find the ratio of the price of pencil to that of ball pen, if pencils cost Rs. 16 per score and ball pens cost Rs. 8.40 per dozen.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {One score contains 20 pencils} \\ \text {And cost per score }= \text {Rs. } 16 \\ \therefore \text {Cost of 1 pencil }=\frac {16}{20} \\ =\text {Rs. } 0.80 \\ \text {Cost of one dozen ball pen }=\text {Rs. } 8.40 \\ \text {1 dozen } =12 \\ \therefore \text {Cost of one pen }= \frac {8.40}{12} \\ =\text {Rs. } 0.70 \\ \text {Hence, ratio of the price of pencil to that of ball pen }=\frac{0.80}{0.70} \\ =\frac{8}{7} \\ =8 : 7 \\\end{array}\begin{array}{l} \text {Q20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Given that one out of 6 student fails.} \\ \text {lets assume that total no. of failed students be x} \\ \Rightarrow \quad \text {x out of 42 students fails} \\ \Rightarrow \quad \frac{1}{6}=\frac{x}{42} \\ \Rightarrow \quad x=\frac{42}{6} \Rightarrow \quad x=7 \\ \text {Number of students who fail} =7 \\\ \text {No of students who pass = Total students – Number of students who fail} =42-7=35 \text { students.} \end{array}