Class 8 Direct and Inverse variations 10.2

\begin{array}{l} \text {Q1. } \quad \text {In which of the following tables x and y vary inversely: } \\ \end{array}\begin{array}{l} \text {(i) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 4 & 3 & 12 & 1 \\ \hline y & 6 & 8 & 2 & 24 \\ \hline \end{array}\begin{array}{l} \text {(ii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 5 & 20 & 10 & 4 \\ \hline y & 20 & 5 & 10 & 25 \\ \hline \end{array}\begin{array}{l} \text {(iii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 4 & 3 & 6 & 1 \\ \hline y & 9 & 12 & 8 & 36 \\ \hline \end{array}\begin{array}{l} \text {(iv) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 9 & 24 & 15 & 3 \\ \hline y & 8 & 3 & 4 & 25 \\ \hline \end{array}\begin{array}{l} \text {Sol. } \\ \text {We know that if a and b vary inversely, then the product xy remains same for all values of x and y.} \\ \end{array}\begin{array}{l} \text {(i) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 4 & 3 & 12 & 1 \\ \hline y & 6 & 8 & 2 & 24 \\ \hline \end{array}\begin{array}{l} \text{Here, }\\4 \times 6= 24, 3 \times 8 = 24, 12 \times 2 =24, 1 \times 24 = 24 \\ \text {As product of all values of x and corresponding values of y is equal to 24, therefore x and y vary inversely.} \end{array}\begin{array}{l} \text {(ii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 5 & 20 & 10 & 4 \\ \hline y & 20 & 5 & 10 & 25 \\ \hline \end{array}\begin{array}{l} \text{Here, }\\5 \times 20 = 100, 20 \times 5 = 100, 10 \times 10 = 100, 4 \times 25 = 100 \\ \text {As product of all values of x and corresponding values of y is equal to 100, therefore x and y vary inversely.} \end{array}\begin{array}{l} \text {(iii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 4 & 3 & 6 & 1 \\ \hline y & 9 & 12 & 8 & 36 \\ \hline \end{array}\begin{array}{l} \text{Here, }\\4 \times 9 = 36, 3 \times 12 = 36, 6 \times 8 = 48, 1 \times 36 = 36 \\ \text {As product of all values of x and corresponding values of y is not equal, therefore x and y doesn’t vary inversely.} \end{array}\begin{array}{l} \text {(iv) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 9 & 24 & 15 & 3 \\ \hline y & 8 & 3 & 4 & 25 \\ \hline \end{array}\begin{array}{l} \text{Here, }\\9 \times 8 = 72, 24 \times 3 = 72, 15 \times 4 = 60, 3 \times 25 = 75 \\ \text {As product of all values of x and corresponding values of y is not equal, therefore x and y doesn’t vary inversely.} \end{array}\begin{array}{l} \text {Q2. } \quad \text {If x and y vary inversely, fill in the following blanks: } \end{array} \begin{array}{l} \text {(i) } \\ \end{array} \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 16 & … & 8 & … \\ \hline y & … & 6 & 4 & … & 0.25 \\ \hline \end{array}\begin{array}{l} \text {(ii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 16 & 32 & 8 & 128 \\ \hline y & 4 & … & … & 0.25 \\ \hline \end{array}\begin{array}{l} \text {(iii) } \\ \end{array} \begin{array}{|l|l|l|l|l|} \hline x & 9 & … & 81 & 243 \\ \hline y & 27 & 9 & … & 1 \\ \hline \end{array}\begin{array}{l} \text {Sol. If x and y vary inversely then xy = k (constant) } \\ \text {(i) Here, } x=16 \text { and } y=6 \\ \Rightarrow \quad 16 \times 6=k \\ \Rightarrow \quad k=96 \\ \Rightarrow \quad 12 \times y = 96, x \times 4 = 96, 8 \times y = 96, x \times 0.25 = 96 \\ \Rightarrow \quad y = 8, x = 24, y = 12, x = 384 \\\text {(ii) Here, } x=16 \text { and } y=4 \\ \Rightarrow \quad 16 \times 4=k \\ \Rightarrow \quad k=64 \\ \Rightarrow \quad 32 \times y = 64, 8 \times y = 64 \\ \Rightarrow \quad y = 2, y = 8 \\\text {(iii) Here, } x=9 \text { and } y=27 \\ \Rightarrow \quad 9 \times 27=k \\ \Rightarrow \quad k=243 \\ \Rightarrow \quad x \times 9 = 243, 81 \times y = 243 \\ \Rightarrow \quad x = 27, y = 3 \\ \end{array}\begin{array}{l} \text {Q3. } \quad \text {3. Which of the following quantities vary inversely as each other? } \\ \text {(i) } \quad \text {The number of x men hired to construct a wall and the time y taken to finish the job. } \\\text {(ii) } \quad \text {The length x of a journey by bus and price y of the ticket.} \\\text {(iii) } \quad \text {Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.} \\ \end{array}\begin{array}{l} \text {(i) With increase in number of men for construction, total time taken to complete the job will reduce and vice versa. } \\ \text {Therefore no. of men and time are inversely proprotional.} \\ \\\text {(ii) If we need to travel a long distance the we need to pay more fare or cost of ticket and vice-versa. } \\ \text {Hence distance and cost of journey are directly proportional.} \\ \\\text {(iii) More petrol is required to cover more distance and vice-versa. } \\ \text {Hence distance and petrol required to cover that distance are directly proportional.} \\ \\\end{array}\begin{array} \text {Q4. } \quad \text {It is known that for a given mass of gas, the volume v varies inversely as the pressure p.} \\ \text {Fill in the missing entries in the following table:}\end{array}\begin{array}{|l|l|l|l|l|l|l|l|} \hline \text {V in } cm^{3} & … & 48 & 60 & … & 100 & … & 200 \\ \hline \text {P (in atmosphere)} & 2 & … & \frac{3}{2} & 1 & … & \frac {1}{2} & … \\ \hline \end{array}\begin{array}{l} \text {Sol. Given that V and P vary inversely therefore VP = k (constant) } \\ \text {Here, } V=60 \text { and } P =\frac{3}{2} \\ \Rightarrow \quad 60 \times \frac{3}{2} = K \\ \Rightarrow \quad k=90 \\ \Rightarrow \quad V \times 2 = 90, 48 \times P = 90, V \times 1 = 90, 100 \times P = 90,V \times \frac {1}{2} = 90, 200 \times P = 90\\ \Rightarrow \quad V = 45, P = \frac {15}{8}, P = 0.9, V = 180, P = \frac {9}{20} \\ \end{array}\begin{array}{l} \text {Q5. } \quad \text {If 36 men can do a piece of work in 25 days, in how many days will 15 men do it?} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume 15 men will do the work in x days.}\\ \end{array} \begin{array}{|l|l|l|} \hline \text {No. of men} & 36 & 15 \\ \hline \text {Time (days)} & 25 & x \\ \hline \end{array}\begin{array}{l}\text {Time taken to complete the work will go down if we increase the no. of men and vice versa therefore it is a case of inverse variation and hence } \\36 \times 25 = 15 \times x \\\Rightarrow \quad x = \frac {36 \times 25}{15} \\ \Rightarrow \quad x =60 \\ \therefore \text {15 men will take 60 days to complete the work.} \\ \end{array}\begin{array}{l} \text {Q6. } \quad \text {A work force of 50 men with a contractor can finish a piece of work in 5 months. } \\ \text {In how many months the same work can be completed by 125 men?} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume 125 men will complete the work in x months.}\\ \end{array} \begin{array}{|l|l|l|} \hline \text {No. of men} & 50 & 125 \\ \hline \text {Time (months)} & 5 & x \\ \hline \end{array}\begin{array}{l}\text {Time taken to complete the work will go down if we increase the no. of men and vice versa therefore it is a case of inverse variation and hence } \\50 \times 5 = 125 \times x \\\Rightarrow \quad x = \frac { 50 \times 5}{125} \\ \Rightarrow \quad x =2 \\ \therefore \text {125 men will complete the work in 2 months.} \\ \end{array}\begin{array}{l} \text {Q7. } \quad \text {A work-force of 420 men with a contractor can finish a certain piece of work in 9 months. } \\ \text {How many extra men must he employ to complete the job in 7 months? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume x men will complete the work in 7 months.}\\ \end{array} \begin{array}{|l|l|l|} \hline \text {No. of men} & 420 & x \\ \hline \text {Time (months)} & 9 & 7 \\ \hline \end{array}\begin{array}{l}\text {Time taken to complete the work will go down if we increase the no. of men and vice versa therefore it is a case of inverse variation and hence } \\420 \times 9 = x \times 7 \\\Rightarrow \quad x = \frac { 420 \times 9}{7} \\ \Rightarrow \quad x =540 \\ \therefore \text 540-420 = 120 \text { extra men are required to complete the work in 7 months.} \\ \end{array}\begin{array}{l} \text {Q8. } \quad \text {1200 men can finish a stock of food in 35 days. } \\ \text {How many more men should join them so that the same stock may last for 25 days? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume x men will finish the food stock in 25 days.}\\ \end{array} \begin{array}{|l|l|l|} \hline \text {No. of men} & 1200 & x \\ \hline \text {Time (days)} & 35 & 25 \\ \hline \end{array}\begin{array}{l}\text {More men will finish the food in less time and vice versa therefore it is a case of inverse variation and hence } \\1200 \times 35 = x \times 25 \\\Rightarrow \quad x = \frac {1200 \times 35}{25} \\ \Rightarrow \quad x =1680 \\ \text {Extra men required to finish the food in 25 days } = 1680 – 1200 = 480 \therefore \text {480 extra men are required to finish the food in 25 days.} \\ \end{array}\begin{array}{l} \text {Q9. } \quad \text {In a hostel of 50 girls, there are food provisions for 40 days. } \\ \text {If 30 more girls join the hostel, how long will these provisions last?} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume food will last for x days with } (50+30) = 80 \text { girls. Therefore }\\ \end{array} \begin{array}{|l|l|l|} \hline \text {No. of girls} & 50 & 80 \\ \hline \text {Time (days)} & 40 & x \\ \hline \end{array}\begin{array}{l}\text {More girls will finish the food in less time and vice versa therefore it is a case of inverse variation and hence } \\50 \times 40 = 80 \times x \\\Rightarrow \quad x = \frac {50 \times 40}{80} \\ \Rightarrow \quad x =25 \\\therefore \text {Food will be finised in 25 days when 30 more girls join the hostel.} \\ \end{array}\begin{array}{l} \text {Q10. } \quad \text {A car can finish a certain journey in 10 hours at the speed of 48 km/hr.} \\ \text { By how much should its speed be increased so that it may take only 8 hours to cover the same distance? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume the speed required to cover the distance in 8 hrs be x km/hr. Therefore, }\\ \end{array}\begin{array}{|l|l|l|} \hline \text {Speed (km/hr)} & 48 & x \\ \hline \text {Time (hrs)} & 10 & 8 \\ \hline \end{array}\begin{array}{l}\text {Less time is required to cover the distance with increased speed and vice versa therefore it is a case of inverse variation and hence } \\48 \times 10 = x \times 8 \\\Rightarrow \quad x = \frac {48 \times 10}{8} \\ \Rightarrow \quad x =60 \\ \therefore \text {Speed of 60 km/hr is required to cover the distance in 8 hrs.} \\ \text {So we need to increase the speed by } = (60-48) = 12 \text { km/hr} \\\end{array}\begin{array}{l} \text {Q11. } \quad \text {1200 soldiers in a fort had enough food for 28 days. After 4 days, } \\ \text { some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort? } \end{array}\begin{array}{l} \text {Sol. It is given that there is enough food for 28 days but some soldiers left after 4 days} \\ \text { so now food is enough for } 28 – 4 = 24 \text { days } \\ \text {Lets assume that x soldiers were transferred. } \\\Rightarrow \text {No. of soldiers left } = 1200 -x\\ \end{array}\begin{array}{|l|l|l|} \hline \text {No. of Soldiers} & 1200 & 1200-x \\ \hline \text {Time (days)} & 24 & 32 \\ \hline \end{array}\begin{array}{l}\text {With Less soldiers food will last for more days and vice versa therefore it is a case of inverse variation and hence } \\1200 \times 24 = (1200-x) \times 32 \\\Rightarrow \quad 1200-x = \frac {1200 \times 24}{32} \\ \Rightarrow \quad 1200-x =900 \\ \Rightarrow \quad x =1200-900 = 300\\ \therefore \text {300 soldiers left the fort.} \\\end{array}\begin{array}{l} \text {Q12. } \quad \text {Three spraying machines working together can finish painting a house in 60 minutes.} \\ \text {How long will it take for 5 machines of the same capacity to do the same job? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume 5 machines will take x minutes to complete the painting. Therefore, }\\ \end{array}\begin{array}{|l|l|l|} \hline \text {Speed (km/hr)} & 3 & 5 \\ \hline \text {Time (min)} & 60 & x \\ \hline \end{array}\begin{array}{l}\text {More machiens will take less time to paint and vice versa therefore it is a case of inverse variation and hence } \\3 \times 60 = 5 \times x \\\Rightarrow \quad x = \frac { 3 \times 60}{5} \\ \Rightarrow \quad x =36 \\ \therefore \text {5 machines will take 36 minutes to complete the painint work.} \end{array}\begin{array}{l} \text {Q13. } \quad \text {A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends } \\ \text { join this group and they find that the same amount of wheat lasts for 18 days. How many new members are there in this group now? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that x new friends joined the group. So now total friends are x+3 } \\ \text {We are assuming 30 days in a month} \\\end{array}\begin{array}{|l|l|l|} \hline \text {No. of friends} & 3 & x+3 \\ \hline \text {Time (days)} & 30 & 18 \\ \hline \end{array}\begin{array}{l}\text {More people will consume more food so wheat will be finised in less time when more friends joined therefore it is a case of inverse variation and hence } \\3 \times 30 = (x+3) \times 18 \\\Rightarrow \quad x+3 = \frac {3 \times 30 }{18} \\ \Rightarrow \quad x+3 =5 \\ \Rightarrow \quad x =5-3 = 2\\ \therefore \text {2 more friends joined the groups.} \\\end{array}\begin{array}{l} \text {Q14. } \quad \text {55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that x cows can graze the field in 10 days. } \\ \end{array}\begin{array}{|l|l|l|} \hline \text {No. of cows} & 55 & x \\ \hline \text {Time (days)} & 16 & 10 \\ \hline \end{array}\begin{array}{l}\text {More cows can graze the field in less time and vice versa therefore it is a case of inverse variation and hence } \\55 \times 16 = x \times 10 \\\Rightarrow \quad x = \frac { 55 \times 16}{10} \\ \Rightarrow \quad x = 88 \therefore \text {88 cows can graze the field in 10 days.} \\\end{array}\begin{array}{l} \text {Q15. } \quad \text {18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required?} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that x men can reap the field in 15 days. } \\ \end{array}\begin{array}{|l|l|l|} \hline \text {No. of men} & 18 & x \\ \hline \text {Time (days)} & 35 & 15 \\ \hline \end{array}\begin{array}{l}\text {More men can reap the field in less time and vice versa therefore it is a case of inverse variation and hence } \\18 \times 35 = x \times 15 \\\Rightarrow \quad x = \frac { 18 \times 35 }{15} \\ \Rightarrow \quad x = 42 \therefore \text {42 men can reap the field in 15 days.} \\\end{array}\begin{array}{l} \text {Q16. } \quad \text {A person has money to buy 25 cycles worth Rs. 500 each. } \\ \text {How many cycles he will be able to buy if each cycle is costing Rs. 125 more? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that x cycles can be bought with increased cost i.e Rs. } 500 + 125 = 625 \\ \end{array}\begin{array}{|l|l|l|} \hline \text {Cost per cycle} & 500 & 625 \\ \hline \text {No. of Cycles } & 25 & x \\ \hline \end{array}\begin{array}{l}\text {Less cycles can be bought with increased price/cost and vice versa therefore it is a case of inverse variation and hence } \\500 \times 25 = 625 \times x \\\Rightarrow \quad x = \frac {500 \times 25}{625} \\ \Rightarrow \quad x = 20 \therefore \text {20 cycles can be bought with increased price..} \\\end{array}\begin{array}{l} \text {Q17. } \quad \text {Raghu has enough money to buy 75 machines worth Rs. 200 each. } \\ \text {How many machines can he buy if he gets a discount of Rs. 50 on each machine? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Lets assume that x machines can be bought with discounted price i.e Rs. } 200 – 50 = 150 \\ \end{array}\begin{array}{|l|l|l|} \hline \text {Cost per machine} & 200 & 150 \\ \hline \text {No. of machines } & 75 & x \\ \hline \end{array}\begin{array}{l}\text {More machines can be bought with less price/cost and vice versa therefore it is a case of inverse variation and hence } \\200 \times 75 = 150 \times x \\\Rightarrow \quad x = \frac { 200 \times 75 }{150} \\ \Rightarrow \quad x = 100 \therefore \text {100 machines can be bought with discounted price..} \\\end{array}\begin{array}{l} \text {Q18. } \quad \text {If x and y vary inversely as each other and } \\ \text {(i) x = 3 when y = 8, find y when x = 4 } \\ \text {(ii) x = 5, when y = 15, find x when y = 12 } \\ \text {(iii) x = 30, find y when constant of variation = 900 } \\ \text {(iv) y = 35, find x when constant of variation = 7 } \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) x and y vary inversely so xy = k (constant).} \\ \therefore 3 \times 8 = 4 \times y \\ \Rightarrow \quad y = \frac {3 \times 8}{4} = 6 \\\text {(ii) x and y vary inversely so xy = k (constant).} \\ \therefore 5 \times 15 = x \times 12 \\ \Rightarrow \quad y = \frac {5 \times 15}{12} = \frac {25}{4} \\\text {(iii) x and y vary inversely so xy = k (constant).} \\ \text { Its given that constant of variation is 900 i.e k = 900} \\ \therefore 30 \times y = 900 \\ \Rightarrow \quad y = \frac {900}{30} = 30 \\\text {(iv) x and y vary inversely so xy = k (constant).} \\ \text { Its given that constant of variation is 900 i.e k = 7} \\ \therefore x \times 35 = 7 \\ \Rightarrow \quad y = \frac {7}{35} = \frac {1}{5} \\\end{array}