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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The weekly expenses of a hostel are partly constant and partly vary directly as the number of inmates . If the expenses are rs 2000 when the number of inmates is 120 and rs 1700 when the number is 100 . Find the number of inmates when the expenses are rs 1880 . |
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Answer» Let the expenses be rs y and the number of inmates = x. As per question , y =`k_1 +k_2x……..(1)` [where `k_1 =constant and k_2 ne 0= ` variation constant ] As per question, if x =120 , then y 2000. `therefore` from (1) we get , 2000 =`k_1 +k_2xx120` or, `k_1+120k_2=2000......(2)` Again , if x =100,then y =1700. `therefore` from (1) we get ,1700=`k_1+k_2xx100` or, `k_1+100k_2=1700........(3)` Now , subtracting (3) from (2) we get , `20k_2=300 or, k_2=15` . `therefore` from (2) we get ,` k_1 +120 xx15 =2000` `rArrk_1=2000-1800` `rArr k_1 =200` `therefore` from (1) we get ,y 200 +15x........(4) so, putting y = 1880 in (4) we get , 1880 =200 +15x or, 15x =1680 or, x=`(1680)/(15)=112` . Hence the required number of inmates =112. |
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| 2. |
Complete the following table considering that the cost of apples and their number are in direct variation.Number of apples (x)14 -12 -Cost of apples (y)83256 -160 |
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Answer» The cost of apples (y) and their number (x) are in direct variation. ∴y \(\propto\) x ∴y = kx …(i) where k is the constant of variation i. When, x = 1, y = 8 ∴ Substituting, x = 1 and y = 8 in (i), we get y = kx ∴ 8 = k × 1 ∴ k = 8 Substituting k = 8 in (i), we get y = kx ∴ y = 8x …(ii) This the equation of variation ii. When, y = 56, x = ? ∴ Substituting y = 56 in (ii), we get y = 8x ∴ 56 = 8x ∴ x = 56/8 ∴ x = 7 iii. When, x = 12, y = ? ∴ Substituting x = 12 in (ii), we get y = 8x ∴ y = 8 × 12 ∴ y = 96 iv. When, y = 160, x = ? ∴ Substituting y = 160 in (ii), we get y = 8x ∴ 160 = 8x ∴ x = 160/8 ∴ x = 20
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| 3. |
If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14. |
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Answer» Given that, m ∝ n ∴ m = kn …(i) where k is constant of variation. When m = 154, n = 7 ∴ Substituting m = 154 and n = 7 in (i), we get m = kn ∴ 154 = k × 7 ∴ k = 154/7 ∴ k = 22 Substituting k = 22 in (i), we get m = kn ∴ m = 22n …(ii) This is the equation of variation. When n = 14, m = ? ∴ Substituting n = 14 in (ii), we get m = 22n ∴ m = 22 × 14 ∴ m = 308 |
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| 4. |
x varies inversely as y, when x = 15 then y = 10, if x = 20, then y = ? |
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Answer» Given that x ∝ (1/√y) x = k × (1/√y) where, k is the constant of variation. ∴ x × y = k …(i) When x = 15, y = 10 ∴ Substituting, x = 15 and y = 10 in (i), we get x × y = k ∴ 15 × 10 = k ∴ k = 150 Substituting, k = 150 in (i), we get x × y = k ∴ x × y = 150 …(ii) This is the equation of variation. When x = 20, y = ? ∴ substituting x = 20 in (ii), we get x × y = 150 ∴ 20 × y = 150 ∴ y = 150/20 ∴ y = 7.5 |
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| 5. |
The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed? |
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Answer» Let x represent the number of apples in each box and y represent the total number of boxes required. The number of apples in each box are varying inversely with the total number of boxes. ∴ x ∝ (1/y) ∴ x = k x (1/y) where, k is the constant of variation, ∴ x × y = k …(i) If 24 apples are put in a box then 27 boxes are needed. i.e., when x = 24, y = 27 ∴ Substituting x = 24 and y = 27 in (i), we get x × y = k ∴ 24 × 27 = k ∴ k = 648 Substituting k = 648 in (i), we get x × y = k ∴ x × y = 648 …(ii) This is the equation of variation. Now, we have to find number of boxes needed when, 36 apples are filled in each box. i.e., when x = 36,y = ? ∴ Substituting x = 36 in (ii), we get x × y = 648 ∴ 36 × y = 648 ∴ y = 648/36 ∴ y = 18 ∴ If 36 apples are filled in a box then 18 boxes are required. |
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| 6. |
x ∝ (1/√y) and when x = 40 then y = 16. If x = 10, find y. |
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Answer» x ∝ (1/√y) ∴ x = k x (1/√y) where, k is the constant of variation. ∴ x × √y = k …(i) When x = 40, y = 16 ∴ Substituting x = 40 andy = 16 in (i), we get x × √y = k ∴ 40 × √16 = k ∴ k = 40 × 4 ∴ k = 160 Substituting k = 160 in (i), we get x × √y = k ∴ x × √y = 160 …(ii) This is the equation of variation. When x = 10,y = ? ∴ Substituting, x = 10 in (ii), we get x × √y = 160 ∴ 10 × √y = 160 ∴ √y = 160/10 ∴ √y = 16 ∴ y = 256 … [Squaring both sides] |
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| 7. |
The total remuneration paid to laborers, employed to harvest soybean is in direct variation with the number of laborers. If remuneration of 4 laborers is Rs 1000, find the remuneration of 17 laborers. |
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Answer» Let, m represent total remuneration paid to laborers and n represent number of laborers employed to harvest soybean. Since, the total remuneration paid to laborers, is in direct variation with the number of laborers. ∴ m ∝ n ∴ m = kn …(i) where, k = constant of variation Remuneration of 4 laborers is Rs 1000. i. e., when n = 4, m = Rs 1000 ∴ Substituting, n = 4 and m = 1000 in (i), we get m = kn ∴ 1000 = k × 4 ∴ k = 1000/4 ∴ k = 250 Substituting, k = 250 in (i), we get m = kn ∴ m = 250 n …(ii) This is the equation of variation Now, we have to find remuneration of 17 laborers. i. e., when n = 17, m = ? ∴ Substituting n = 17 in (ii), we get m = 250 n ∴ m = 250 × 17 ∴ m = 4250 ∴ The remuneration of 17 laborers is Rs 4250. |
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| 8. |
Write the following statements using the symbol of variation. 1. Circumference (c) of a circle is directly proportional to its radius (r). 2. Consumption of petrol (l) in a car and distance traveled by that car (d) are in direct variation |
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Answer» 1. c ∝ r 2. l ∝ d |
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| 9. |
m ∝ n, n = 15 when m = 25. Hence i. Find m when n = 87, ii. Find n when m = 155. |
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Answer» Given that, m ∝ n ∴ m = kn …(i) where, k is the constant of variation. When m = 25, n = 15 ∴ Substituting, m = 25 and n = 15 in (i), we get m = kn ∴ 25 = k × 15 ∴ k = 25/15 ∴ k = 5/3 Substituting k = 5/3 in (i), we get m = kn ∴ m = (5/3)n ... (ii) i. When n = 87, m = ? Substituting n = 87 in (ii), we get m = (5/3)n m = (5/3) x 87 m = 5 × 29 m = 145 ii. When m = 155, n = ? ∴ Substituting m = 155 in (ii), we get m = (5/3)n ∴ 155 = (5/3)n ∴ (155 x 3)/5 = n ∴ n = 31 × 3 ∴ n = 93 |
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| 10. |
y varies inversely with x. If y = 30 when x = 12, find i. y when x = 15, ii. x when y = 18. |
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Answer» Given that, y ∝ (1/x) ∴ y = k x (1/x) where, k is the constant of variation. ∴ y × x = k …(i) When x = 12, y = 30 ∴ Substituting, x = 12 and y = 30 in (i), we get y × x = k ∴ 30 × 12 = k ∴ k = 360 Substituting, k = 360 in (i), we get y × x = k ∴ y × x = 360 ….(ii) i. When x = 15,y = ? ∴ Substituting x = 15 in (ii), we get y × x = 360 ∴ y × 15 = 360 y = 360/15 y = 24 ii. When y = 18, x = ? ∴ Substituting y = 18 in (ii), we get y × x = 360 ∴ 18 × x = 360 x = 360/18 x = 20 |
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| 11. |
Write the following statements using symbol of variation. 1. The wavelength of sound (l) and its frequency (f) are in inverse variation. 2. The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp |
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Answer» 1. l ∝ (1/f) 2. I ∝ (1/d2) |
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| 12. |
Complete the following table considering that the cost of apples and their number are in direct variation.Number of apples (x)14...12...Cost of apples (y)83256...160 |
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Answer» The cost of apples (y) and their number (x) are in direct variation. ∴ y ∝ x ∴ y = kx …(i) where k is the constant of variation i. When, x = 1, y = 8 ∴ Substituting, x = 1 and y = 8 in (i), we get y = kx ∴ 8 = k × 1 ∴ k = 8 Substituting k = 8 in (i), we get y = kx ∴ y = 8x …(ii) This the equation of variation ii. When,y = 56, x = ? ∴ Substituting y = 56 in (ii), we get y = 8x ∴ 56 = 8x ∴ x = 56/8 x = 7 iii. When, x = 12, y = ? ∴ Substituting x = 12 in (ii), we get y = 8x ∴ y = 8 × 12 ∴ y = 96 iv. When, y = 160, x = ? ∴ Substituting y = 160 in (ii), we get y = 8x ∴ 160 = 8x ∴ x = 160/8 ∴ x = 20
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| 13. |
If n varies directly as m, complete the following table.m356.5...1.25n1220...28... |
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Answer» Given, n varies directly as m ∴ n ∝ m ∴ n = km …(i) where, k is the constant of variation i. When m = 3, n = 12 ∴ Substituting m = 3 and n = 12 in (i), we get n = km ∴ 12 = k × 3 ∴ k = 12/3 ∴ k = 4 Substituting, k = 4 in (i), we get n = km ∴ n = 4m …(ii) This is the equation of variation. ii. When m = 6.5, n = ? ∴ Substituting, m = 6.5 in (ii), we get n = 4m ∴ n = 4 × 6.5 ∴ n = 26 iii. When n = 28, m = ? ∴ Substituting, n = 28 in (ii), we get n = 4m ∴ 28 = 4m ∴ m = 28/4 m = 7 iv. When m = 1.25, n = ? ∴ Substituting m = 1.25 in (ii), we get n = 4m ∴ n = 4 × 1.25 ∴ n = 5
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| 14. |
Identify the variation and solve. It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel? |
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Answer» Let, v represent the speed of the bus and t represent the time required to travel from one town to the other. The speed of the bus varies inversely with the time required to travel from one town to the other. ∴ v ∝ 1/t ∴ v = k x (1/t) where, k is the constant of variation. ∴ v × t = k …(i) It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. i.e., when v = 48, t = 5 ∴ Substituting v = 48 and t = 5 in (i), we get v × t = k ∴ 48 × 5 = k ∴ k = 240 Substituting k = 240 in (i), we get v × t = k ∴ v × t = 240 …(ii) Since, the speed of the bus is reduced by 8 km/hr, ∴ Speed of the bus in second case (v) = 48 – 8 = 40 km/hr ∴ When v = 40, t = ? ∴ Substituting v = 40 in (ii), we get v × t = 240 ∴ 40 × t = 240 ∴ t = 240/40 t = 6 ∴ The problem is of inverse variation and the bus would take 6 hours to travel the distance if its speed is reduced by 8 km/hr. |
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| 15. |
A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in 7(1/2) hours? |
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Answer» Let v represent the speed of car in km/hr and t represent the time required. Since, speed of a car varies inversely as the time required to cover a distance. ∴ v ∝ (1/t) ∴ v = k x (1/t) where, k is the constant of variation. ∴ v × t = k …(i) Since, a car with speed 60 km/hr takes 8 hours to travel some distance. i.e., when v = 60, t = 8 ∴ Substituting v = 60 and t = 8 in (i), we get v × t = k ∴ 60 × 8 = t ∴ k = 480 Substituting k = 480 in (i), we get v × t = k ∴ v × t = 480 …(ii) This is the equation of variation. Now, we have to find speed of car if the same distance is to be covered in 7(1/2) hours. i.e., when t = 7(1/2) = 7.5, v =? ∴ Substituting, t = 7.5 in (ii), we get v × t = 480 ∴ v × 7.5 = 480 v = 480/7.5 = 4800/75 ∴ v = 64 The speed of vehicle should be 64 km/hr to cover the same distance in 7.5 hours. ∴ The increase in speed = 64 – 60 = 4km/hr ∴ The increase in speed of the car is 4 km/hr. |
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| 16. |
The information about number of workers and number of days to complete a work is given in the following table. Complete the table.Number of workers3020...10...Days6912...36 |
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Answer» Let, n represent the number of workers and d represent the number of days required to complete a work. Since, number of workers and number of days to complete a work are in inverse poportion. ∴ n ∝ (1/d) ∴ n = k x (1/d) where k is the constant of variation. ∴ n × d = k …(i) i. When n = 30, d = 6 ∴ Substituting n = 30 and d = 6 in (i), we get n × d = k ∴ 30 × 6 = k ∴ k = 180 Substituting k = 180 in (i), we get ∴ n × d = k ∴ n × d = 180 …(ii) This is the equation of variation ii. When d = 12, n = 7 ∴ Substituting d = 12 in (ii), we get n × d = 180 ∴ n × 12 = 180 ∴ n = 180/12 ∴ n = 15 iii. When n = 10, d = ? ∴ Substituting n = 10 in (ii), we get n × d = 180 10 × d = 180 ∴ d = 180/10 ∴ d = 18 iv. When d = 36, n = ? ∴ Substituting d = 36 in (ii), we get n × d = 180 ∴ n × 36 = 180 ∴ n = 180/36 ∴ n = 5
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| 17. |
If x `prop 1/y ` and y `prop 1/z ` , then x `prop ` |
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Answer» x `prop 1/y rArr x =k_(1)1/y` (`k_(1) ne `0= variation constant . ) `rArr y =(k_(1))/(x) ……..(1)` Again , y `prop 1/z rArr =k_(2) 1/z` (`k_(2) ne`0= variation constant .) `rArr (k_(1))/(x) =k_(2)1/z` [from (1) ] `rArr x =(k_(1))/(k_(2))z ` `rArr x=k.Z(when (k_(1))/(k_(2))kne0="variation constant")` `rArr x prop z (because k ne0="variation constant")` `therefore x prop z` |
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| 18. |
If x `prop ` y then x^(n) `prop` |
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Answer» `x propy rArr x =ky` ( k =non zero variation constant ) `rArr (x)^(n) =(ky)^(n) ` (taking n-th power ) `rArr x^(n) =k^(n).y^(n). ` `rArrn^(n)=my^(n)` (when m =`k^(n) ne0=`variation constant ) `rArr x^(n)prop y^(n)` (`because` m =non -zero variation constant ). `therefore x^(n) propy^(n) .` |
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| 19. |
If y `prop 1/x ` , then `y/x` = non -zero constant , |
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Answer» `yprop1/x rArr y=k.1/x` (where k = non -zero variation constant ) `rArr xy =k ` `rArr` xy = non -zero variation constant . `therefore ` The statement is false. |
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| 20. |
y varies directly as square root of x. When x = 16, y = 24. Find the constant of variation and equation of variation. |
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Answer» Given, y varies directly as square root of x. ∴ y ∝ √4x ∴ y = k √x …(i) where, k is the constant of variation. When x = 16 ,y = 24. ∴ Substituting, x = 16 and y = 24 in (i), we get y = k√x ∴24 = k√16 ∴24 = 4k \(\therefore\,k = \cfrac{24}{4}\) ∴ k = 6 Substituting k = 6 in (i), we get y = k√x ∴ y = 6√x This is the equation of variation ∴ The constant of variation is 6 and the equation of variation is y = 6√x . |
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| 21. |
If `x prop y^(2)` and y =4 when x =8 , if x =32 , then the positive value of y isA. 4B. 8C. 16D. 32 |
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Answer» `x prop y^(2), therefore x=k.y^(2)` (where k=non -zero constant )…….(i) As per question , y =4 when x =8 `therefore` from (1) we get , 8 =k.`4^(2) ` or , 8 =16k or k=`(8)/(16)=1/2` `therefore` x =`1/2y^(2)` [from (1)]…….(2) Now ,putting x =32 in (2) we get , 32 =`1/2y^(2)` or , `Y^(2)`=64 or , y=`sqrt(64)=pm8`. `therefore` y =8 (taking positive value ) `therefore` (b) is correct. |
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| 22. |
If x `prop y and x prop z,` then (y+z) `prop` |
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Answer» `x prop y rArr k_(1) y ( k_(1)` = non -zero variation constant )…..(1) x `prop z rArr x = k_(2)z (k_(2)`= non -zero variation constant ) ……..(2) From (1) we get , y `(x) /(k_(1))`…….(3) From (2) we get ,z = `(x)/(k_(2)) `………(4) Now , adding (3) and (4) we get , `y+z =(x)/(k_(1))+(x)/(k_(2)) ` ` or ,y+z =((x)/(k_(1))+(x)/(k_(2)))x ` ` or ,y+z =k.x[when(x)/(k_(1))+(x)/(k_(2))=k] ` `therefore y +zpropx [becausekne0="variation constant"]` `thereforey +z propx. ` |
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| 23. |
If `xprop y` and y =8 when x =2, if y=16 , then the value of x isA. 2B. 4C. 6D. 8 |
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Answer» `x prop y rArr x=ky ` (where k= non zero constant )…..(i) Given that y=8 when x =2 . `therefore 2 =k xx8rArr2/8=1/4` `therefore` form (1) we get , ` x =1/4 y …….(2)` putting y =16 in (2) we get , x `1/4 xx16or , x=4` `therefore` (b) is correct. |
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| 24. |
If ` x prop y, y prop z ` and ` z prop x, ` then find the product of three non-zero variation constant. |
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Answer» `x prop y rArr x = k_1y(K_1 =` non- zero variation constant ) `rArr k_1 = x/y ….. (1)` `y prop z rArr k_2 z (k_2= ` non-zero variation constant ) `rArr k_1 =y/z …….. (2)` `z prop x rArr z = k_2 x (k_3 ` non - zero variation constant ) `rArr k_3 = z/x ....... (3)` now multiplying (1),(2) and (3) we get `k_1k_2 k_3 = x/y xx y/z xx z/x ` or d`k_1 k_2 k_3= 1 ` Hence the produc of non- zero variation constant is 1. |
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| 25. |
If `xprop y^(2) and y= 2a when x=a,` then find the relation between x and y . |
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Answer» `xprop y^(2) rArr x =k.y^(2) .....(1) (when k ne0="variation constant")` If x =a , y=2a , `therefore ` from (1) we get , a =k =`(2a)^(2)` `rArr a=l.4a^(2)rArr k=(a)/(4a)^(2)=(1)/(4a)` `therefore` from (1) we get , x `1/4ay^(2) rArr y^(2) `=4ax. Hence the required relation between x and y is `y^(2) ` = 4ax. |
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| 26. |
If `(y-z)prop1/x, (z-x)prop and (x-y)prop1/z,` then find the sum of three variation constants . |
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Answer» `(y-z) prop1/x rArry-z =k_(1). 1/x(where k_(1)ne0="variation constant".)` `rArr k_(1) =x(y-z)…..(1) ` `(z-x) prop1/y , therefore(z-x)=k_(2). 1/y(where k_(2)ne0="variation constant" )` `therefore k_(2) =y (z-x) …….(2) ` Also , `(x-y) prop1/z , therefore x-y =k_(3) x 1/z(" where " k_(3) ne0= "variation constant")` `therefore k_(3) =z(x-y)......(3)` Now , adding (1) +(2) +(3) we get , `k_(1)+k_(2)+k_(3) =x(y-z)+y(z-x)+z(x-y)` `=xy-xz+yz-xy+zx-yz=0` `therefore k_(1) +k_(2)+ k_(3) =0.` Hence the sum of three variation constant =0. |
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| 27. |
If x `propsqrt(y)` and y=16 when x =3 . If y =64 ,then the positive value of x will beA. 4B. 6C. 8D. 10 |
| Answer» Correct Answer - B | |
| 28. |
If a `prop1/b`, thenA. `a=1/b`B. `b=1/a`C. `ab=1`D. ab=non-zero constant. |
| Answer» Correct Answer - D | |
| 29. |
If a `prop ` b and `prop ` c , then show that `(a^3b^3+b^3c^3+c^3a^3) prop abc (a^3+b^3+c^3)`. |
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Answer» `a prop b rArr a =k_(1)b (k_(1)` =non -zero variation constant ) `b prop c rArr b= k_(2)c (k_(2)`=non -zero variation constant ) `therefore a= k_(1) k_(2) c ` . Now , `(a^3b^3+b^3c^3+c^3a^3)/(abc(a^3+b^3+c^3))=((k_(1)k_2c)^3+(k_2c)^3+(k_2c)^3+(c)^3+c^3.(k_1k_2c)^3)/(k_1k_2c.k_2c.c{(k_1k_2c)^3(k_2c)^3+c^3}` `(k_1^3k_2^6c^6+k_2^3c^6+k_1^3k_2^3c^6)/(k_1k_2^2c^3(k_1^3k_2^3c^3+k_2^3c^3+c^3))` `=(k_2^3c^6(k_1^3k_2^3+1+k_1^3))/(k_1k_2^2c^6(k_1^3k_2^3+k_2^3+1))` `=(k_2(k_1^3k_2^3+k_1^3+1))/(k_1(k_1^3k_2^3+k_2^3+1))` =Non -zero constant `[because k-1 and k_2 ne0]` =k (let) `therefore a^3b^3+b^3c^3+c^3a^3=k{abc(a^3+b^3+c^3)}` `therefore a^3b^3+b^3c^3+c^3a^3 prop abc (a^3+b^3+c^3)` [`because k ne0=` variation constant. ] (proved ) |
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| 30. |
If ` xprop1/y` thenA. ` x=1/y`B. ` y=1/x`C. ` xy=1`D. xy=non -zero constant |
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Answer» `xprop1/yrArrk.1/y` (where k =non -zero constant ) `thereforexy=krArrxy= "non zero constant"` `therefore `(d) id correct. |
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| 31. |
120 bags of half litre milk can be filled by a machine within 3 minutes find the time to fill such 1800 bags? |
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Answer» Let b represent the number of bags of half litre milk and t represent the time required to fill the bags. Since, the number of bags and time required to fill the bags varies directly. ∴ b ∝ t ∴ b = kt …(i) where k is the constant of variation. Since, 120 bags can be filled in 3 minutes i.e., when b = 120, t = 3 ∴ Substituting b = 120 and t = 3 in (i), we get b = kt ∴ 120 = k × 3 ∴ k = 120/3 ∴ k = 40 Substituting k = 40 in (i), we get b = kt ∴ b = 40 t …(ii) This is the equation of variation. Now, we have to find time required to fill 1800 bags. ∴ Substituting b = 1800 in (ii), we get b = 40 t ∴ 1800 = 40 t ∴ t = 1800/40 ∴ t = 45 ∴ 1800 bags of half litre milk can be filled by the machine in 45 minutes. |
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| 32. |
Which of the following statements are of inverse variation? i. Number of workers on a job and time taken by them to complete the job. ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank. iii. Petrol filled in the tank of a vehicle and its cost. iv. Area of circle and its radius. |
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Answer» i. Let, x represent number of workers on a job, and y represent time taken by workers to complete the job. As the number of workers increases, the time required to complete the job decreases. ∴ x ∝ 1/y ii. Let, n represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank. As the number of pipes increases, the time required to fill the tank decreases. ∴ n ∝ 1/t iii. Let, p represent the quantity of petrol filled in a tank and c represent the cost of the petrol. As the quantity of petrol in the tank increases, its cost increases. ∴ p ∝ c iv. Let, A represent the area of the circle and r represent its radius. As the area of circle increases, its radius x ∝ 1 y n ∝ 1 t / increases. ∴ A ∝ r ∴ Statements (i) and (ii) are examples of inverse variation. |
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| 33. |
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours? |
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Answer» Let, n represent the number of workers building the wall and t represent the time required. Since, the number of workers varies inversely with the time required to build the wall. ∴ n ∝ 1/t ∴ n = k x (1/t) where k is the constant of variation ∴ n × t = k …(i) 15 workers can build a wall in 48 hours, i.e., when n = 15, t = 48 ∴ Substituting n = 15 and t = 48 in (i), we get n × t = k ∴ 15 × 48 = k ∴ k = 720 Substituting k = 720 in (i), we get n × t = k ∴ n × t = 720 …(ii) This is the equation of variation. Now, we have to find number of workers required to do the same work in 30 hours. i.e., when t = 30, n = ? ∴ Substituting t = 30 in (ii), we get n × t = 720 ∴ n × 30 = 720 ∴ n = 720/30 ∴ n = 24 ∴ 24 workers will be required to build the wall in 30 hours. |
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| 34. |
Which of the following statements are of inverse variation?i. Number of workers on a job and time taken by them to complete the job. ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank. iii. Petrol filled in the tank of a vehicle and its cost. iv. Area of circle and its radius. |
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Answer» i. Let, x represent number of workers on a job, and y represent time taken by workers to complete the job. As the number of workers increases, the time required to complete the job decreases. \(\therefore x\propto \cfrac{1}{y}\) ii. Let, n represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank. As the number of pipes increases, the time required to fill the tank decreases. \(\therefore n\propto \cfrac{1}{t}\) iii. Let, p represent the quantity of petrol filled in a tank and c represent the cost of the petrol. As the quantity of petrol in the tank increases, its cost increases. ∴ p ∝ c iv. Let, A represent the area of the circle and r represent its radius. As the area of circle increases, its radius increases. ∴ A ∝ r ∴ Statements (i) and (ii) are examples of inverse variation. |
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| 35. |
Find constant of variation and write equation of variation for every example given below:i. p ∝ (1/q); if p = 15 then q = 4.ii. z ∝ (1/w); when z = 2.5 then w = 24iii. s ∝ (1/t2); if s = 4 then t = 5.iv. x ∝ (1/√y); if x = 15 then y = 9. |
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Answer» i. p ∝ (1/q) … [Given] ∴ p = k × (1/q) where, k is the constant of variation. ∴ p × q = k …(i) When p = 15, q = 4 ∴ Substituting p = 15 and q = 4 in (i), we get p × q = k ∴ 15 × 4 = k ∴ k = 60 Substituting k = 60 in (i), we get p × q = k ∴ p × q = 60 This is the equation of variation. ∴ The constant of variation is 60 and the equation of variation is pq = 60. ii. z ∝ (1/w) …[Given] ∴ z = k × (1/w) where, k is the constant of variation, ∴ z × w = k …(i) When z = 2.5, w = 24 ∴ Substituting z = 2.5 and w = 24 in (i), we get z × w = k ∴ 2.5 × 24 = k ∴ k = 60 Substituting k = 60 in (i), we get z × w = k ∴ z × w = 60 This is the equation of variation. ∴ The constant of variation is 60 and the equation of variation is zw = 60. iii. s ∝ (1/t2) …[Given] ∴ s = k x (1/t2) ∴ where, k is the constant of variation, ∴ s × t2 = k …(i) When s = 4, t = 5 ∴ Substituting, s = 4 and t = 5 in (i), we get s × t2 = k ∴ 4 × (5)2 = k ∴ k = 4 × 25 ∴ k = 100 Substituting k = 100 in (i), we get s × t2 = k ∴ s × t2 = 100 This is the equation of variation. ∴ The constant of variation is 100 and the equation of variation is st2 = 100. iv. x ∝ (1/√y) …[Given] ∴ x = k x (1/√y) where, k is the constant of variation, ∴ x × √y = k …(i) When x = 15, y = 9 ∴ Substituting x = 15 and y = 9 in (i), we get x × √y = k ∴ 15 × √9 = k ∴ k = 15 × 3 ∴ k = 45 Substituting k = 45 in (i), we get x × √y = k ∴ x × √y = 45. This is the equation of variation. ∴ The constant of variation is k = 45 and the equation of variation is x√y = 45. |
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| 36. |
An agricultural co- operative society of village Naraynpur has purchased a tractor . Previously 2400 bighas of land were cultivated by 25 ploughs in 36 days . Now half of the land can be cultivated only by that tractor in 30 days . Calculate by using the theory of variation , the number of ploughs work equaly with one tractor . |
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Answer» Let the quantity of land G bighas . Number of ploughs =P and time of cultivation =D days . We know that if number of ploughs is cosntant , then the quantity of land varies directly with time of cultivation , i.e , G `prop` D when P is constant Again if time of cultivation is cosntant , the quantity of land varies directly with number of ploughs , i.e G `prop ` P , when D is constant. Then by theorem of joint variation , G `prop` PD (where P and D are both variation ) `rArr` G =k . PD (where k `ne` 0= variation constant ) `therefore` G =k.PD .........(1) As per question , if G =2400 and P =25 , then D =36 `therefore` from (1) we get , 2400 =k.`25xx36` or , k =`(2400)/(25xx36)=(96)/(36) =8/3 ` `therefore` (1) becomes G =`8/3` PD ..........(2) Now putting G =`(2400)/2=1200 and D =30` in (2) we get , 1200=`8/3xxPxx30 or, P =(1200xx3)/(8xx30) =15` . Hence the required number of ploughs work equally with one tractor is 15 . |
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| 37. |
If (x+y) `prop`(x-y), then prove that `(x^(2)+y^(2)) prop` xy . |
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Answer» (x+y) `prop` (x-y) `rArr x+y =k (x-y)` ( where k `ne ` 0= variation constant ) `rArr (x+y)^(2) =k^(2) (x-y)^(2) ` (squaring both the sides ) `rArrx^(2) +2xy +y^(2) =k^(2) (x^(2)-2xy+y^(2))` `rArrx^(2) +y^(2) =k^(2) (x^(2)+y^(2)) -2k^(2)xy-2xy` `rArr(x^(2)+y^(2)) -k^(2)(x^(2)+y^(2))=-2k^(2)xy-2xy` `rArrx^(2)+y^(2) -k^(2)(x^(2)+y^(2))=-xy(2k^(2)+2)` `rArr (x^(2) +y^(2))(1-k^(2))=-(2k^(2)+2) xy ` `rArr (x^(2) +Y^(2)) (k^(2)-1)=(2k^(2)+2)xy` `rArr x^(2) + y^(2) =((2k^(2)+2)xy)/(k^(2)-1)` `rArr x^(2) + y^(2) =(2k^(2)+2)/(k^(2)-1).xy` `rArr x^(2)+y^(2) =mxy.(where m=(2k^(2)+2)/(k^(2)-1)ne0)` `rArrx^(2)+y^(2) propxy [because mne0="variation constant"]` `therefore(x^(2)+y^(2)) propxy. (proved )` |
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| 38. |
If 15 farmers can cultivate 18 bighas of land in 5 days, determine by using theory of variation the number of days required by 10 farmers to cultivate 12 bighas of land . |
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Answer» Let the number of farmers = F , the amount of land =G and the number of days =D , `therefore` if the number of farmers be increased the amount of caltivated land also increases , but the number of days will decreased. `therefore ` F varies directly with G , but inversely with D . ` therefore F prop (G)/(D) rArr F =k .(G) /(D) ` [when k `ne ` 0 = variation constant ] ` therefore F =k.(G)/(D) ............(1) ` As per question , if F = 15, G =18 , then D =5. ` therefore` from (1) we get , 15 =k `(18) /(5) rArr k=(75)/(18)=(25)/(6) ` . `therefore `(1) becomes , F =`(25)/(6).(G) /(D) ...........(2)` Now , putting F =10 , G 12 in (2) we get , 10 =`(25)/(6) .(12)/(D) or , 10 D =50 or , D (50)/(10) =5` Hence the required number of days =5 . |
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| 39. |
50 villages had taken 18 days to dig a pond . Calculate by using theory of variation how many extra persons will be required to dig the pond in 15 days. |
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Answer» Let the number of persons =M and the number of days =D . `because` If the number of persons increases, the number of days will decrease. `therefore` M varies inversely as D . `therefore M proprArr =k . 1/D[where k ne0=v.c]` `rArr M=k/D........(1)` As per question , if M =50 , then D=18. `therefore` from (1) we get , 50 =`(k) /(18)rArr =900`. `therefore M =(900)/(D) ........(2) ["from" (1)]` Now ,putting D =15 we get , M =`(900)/(15)=60. ` The extra persons =60-50 =10. Hence the required number of extra persons = 10. |
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| 40. |
If (x+y) `prop` (x-y) , then show that `(x^(3) +y^(3)) prop (x^(3)-y^(3))`. |
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Answer» (x+y) `prop` (x-y) `rArr (x+y)` =k (x-y) (where k `ne` 0= variation constant ) `rArr (x+y)/(x-y)=k ` `rArr (x+y+x-y)/(x+y-x+y)=(k+1)/(k-1)` [by componendo and dividendo] `rArr (2x)/(2y)=(k+1)/(k-1)rArr(x)/(y)=(k+1)/(k-1)=m("let") [when (k+1)/(k-1)=m]` `rArr (x/y)^(3) =m^(3) ` (cubing both the sides ) `rArr x^(3)/y^(3)=m^(3)rArr (x^(3)+y^(3))/(x^(3)-y^(3))=(m^(3)+1)/(m^(3)-1)` [by componendo and dividendo] `(x^(3)+y^(3))/(x^(3)-y^(3))=n ["whene" n =(m^(3)+1)/(m^(3)-1)ne0]` ` x^(3)+y^(3) =n(x^(3)-y^(3)) and n ne0="variation constant."` `(x^(3)+y^(3))prop(x^(3)-y^(3))(proved)` |
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| 41. |
The value of diamond varies as the square of its weight . A diamond broke into three pieces whose weights are in the ratio `3:4:5` . For this there is a loss rs 9400. Find the value of the original diamond . |
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Answer» Let the weight of the the diamond = n gm and its value =rs A . As per question , A `prop m^2 , therefore A =km^2[k ne 0 `variation constant ] Let the weights of three pieces of diamond be 3n gm , 4n gm and 5n gm . `therefore` The weight of the original diamond =(3n+4n+5n)gm =12n gm. `therefore` The values of the three pieces are respectively . `A_1=rs k(3n)^2=rs9kn^2` `A_2=rsk(4n)^2=rs16kn^2` `A_3 =rsk(5n)^2=rsrs25kn^@(where k ne0` =variation constant) Also , the value of the original diamond =rs k.`(12n)^2` `=rs144kn^2` `therefore` Loss for breacking = rs `{144kn^2-(9kn^2 +25kn^2)}` `=rs94kn^2 ` As per question , `94kn^2 =9400 thereforekn^2=100 ` `therefore` The value of the original diamond =` rs 144kn^2 ` `=rs144xx100 ` `rs14400` . Hence the value of the originla diamond =rs 14400 . |
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