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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. |
A student Gaurav of weight 50 kg secured 150 marks. Another student Saurav of weight 40 kg proportion to their marks, how much did Saurav score the exam ? |
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Answer» Correct Answer - 120 marks 50 :40 : : 150 : x ` 50/40 = 150/x Rightaarrow 5/4 = 150/x` ` x = 150 xx 4/5 Rightarrow x= 120` Saurav got 120 marks. |
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| 2. |
find the greatest ratio of the following 1 : 6 , 3: 8 (ii) 9: 16: 5 : 12 |
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Answer» Correct Answer - (i) 3:8 (ii) 9 : 16 ` 1/6 " 3/8 ` LCM of 6 and 8 is 24 ` 1/6 xx 24 : 3/8 xx 24 ` ` 4: 9 = 1/6 lt 3/8` ` 3 : 8` is greater. (ii) `9/16 : 5/12` LCM of 26, 12 is 48 ` 9/ 16 xx 48 : 5/12 xx 48 = 27 : 20` ` 9 /16 lt 5 /12 ` 9 : 16 is greater. ] |
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| 3. |
Mr Mallanna wants to distribute watermelon to his three daugters in the ratio 2:3:4. such that the eldest daughter gets the highest share . If the eldest daughter gets least share. If the eldest daughter gets 250 g more than that of the youngest then find the total weight of the watermelon distributed. |
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Answer» Correct Answer - 1125 g ` (4 / (2 + 3+4) - 2 / (2+3+4))` of total weight = 250 gm ` ( 4/9 - 2/9)` of total weight = 250 ` ( 2/9) ` of total weight = 250 total weight = `(259xx9) / 2 = 1125` gm |
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| 4. |
Shantan got 68 marks in English and 85 marks in Mathematics. Find the ratio of the marks scored in English to the Matematics. |
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Answer» Correct Answer - `4:5` 68 : 85 ` 17 xx 4 : 17 xx 5 = 4:5 ` |
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| 5. |
Mr Arunachalam earcned ₹ 48,800 in a month. He spent ₹24,000. find the ratio of (i) the earnings to expenditure and (ii) expenditure to savings. |
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Answer» Correct Answer - (i) 61 : 30 (ii) 30 : 31 Earnings of Aruncachalam = ₹ 48800 Expenditure = ₹ 24000 (i) Ratio of earnings to expenditure 48800 : 24000 488 : 240 ` 8 xx 61 : 8 xx 30` = 61 : 30 (ii) Savings = Income - Expenditure ₹ 48800 - ₹ 24000= ₹ 24800 Expenditure : savings 24000 : 24800 = 240 : 248 = 30:31 |
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| 6. |
A sweet maker mixes sugar, milk and other ingredients in the ratio 4: 9:3. in 2 Kg of such sweet. Find the quantity of sugar. |
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Answer» Correct Answer - 500g 2 kg sweet : sugar : milk : others = 4:9:3 weight of sugar in 2 kg of sweet ` = 4/(4+9+3) xx 2 = 4/16 xx 2 = 1/2 kg = 500 g` |
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| 7. |
If ` x = (8ab)/(a+b) `, then find the value of `((x+4a)/(x - 4a) + (x+4b)/(x-4b))`. |
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Answer» Given that ` x = (8ab)/(a + b) " or, " x/(4a) = (2b)/(a +b)` or, `(x+4a)/(x - 4a) = (2b + a + b)/(2b - a - b) = (a + 3b)/(b - a) ` ......................(1) Again, ` x = (8ab)/(a + b) " or, " x/(4b) = (2a)/(a + b)` or, ` (x + 4b)/(x - 4b) = (2a + a +b)/(2a - a - b) = (3a +b)/(a-b) ` ..................(2) ` :. ` adding (1) and (2) we get, `(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b) = (a+3b)/(b-a) + (3a +b)/(a-b)` ` = (-a - 3b + 3a + b)/(a -b) = (2a - 2b)/(a -b) = (2 (a -b))/((a - b)) = 2`. ` :. (x + 4a)/(x - 4a) + (x + 4b)/(x - 4b) = 2`. |
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| 8. |
In a basket, the number of blue, red and green balls are in the ratio 4 : 7:9. if there are a total of 960 balls, then find the number of green balls. |
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Answer» Correct Answer - 432 The ratio of the number of blue balls, red balls and green balls 4:7:9 The total number of balls is 960. number of green balls = `9/(4+7+9) xx 960` ` 9/20 xx 960 = 9 xx 48 = 432` |
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| 9. |
There are 20 girls and 15 boys in a class.(a) What is the ratio of number of girls to the number of boys?(b) What is the ratio of number of girls to the total number of students in the class? |
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Answer» girls=20 boys=15 total=35 a)ratio=girls/boys=20/15=4:3 b)ratio=girls/total=20/35=4:7. |
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| 10. |
Find the ratio of the following:(a) 30 minutes to 1.5 hours (b) 40 cm to 1.5 m(c) 55 paise to Re 1 (d) 500 ml to 2 litres |
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Answer» a) 1 hr=60 min 1.5 hrs=1.5*60=90 min ratio=30/90=1:3 b)1 m=100 cm 1.5 m=150 cm ratio=40/150=4:15 c) 1 rs = 100 paiseratio=55/100=11:20 d)2 liters = 2000 ml ratio=500/2000=1:4. |
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| 11. |
Ratio of distance of the school from Mary’s home to the distanceof the school from John’s home is 2 : 1.(a) Who lives nearer to the school?(b) Complete the following table which shows some possible distances that Mary and John could live from the schoolIf the ratio of distance of Mary’s home to the distance of Kalam’s home from school is 1 : 2, then who lives nearer to the school? |
| Answer» a)Johnc)Mary | |
| 12. |
Consider the ratio 2:3 . Ther consequent and antecendent are doubled and tripled respectively. Ther resultant ratio is ______ |
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Answer» Correct Answer - `1:1` The required ratio is ` 3xx 2 : 2 xx 3 = 6 : 6 =1 :1 ` |
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| 13. |
Find the mixed ratios of the ratios `a/b : c , b/c : a , c/a : b`. |
| Answer» The mixed ratio of ` a/b : c, b/c : a " and " c/a : b " is " (1/b xx b/c xx c/a) : ( c xx a xx b) = 1 : abc `. | |
| 14. |
Express the following quantities as a ratio . Also write which of them are ratios of equality, which are ratios of greater inequalities and which are ratios of less inequalities : (a) 4 months and 1 year 6 months , (b) 75 paise and 1 rupes 25 paise. |
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Answer» (a) 1 year 6 months ` = ( 1 xx 12 + 6)` = 18 months. ` :. ` 4 moths : 1 year 6 months = 4 months : 18 months ` = "4 months"/"18 months" = 2/9 = 2 :9` Now, since ` 2 lt 9`, the ratio is a ratio of less inequality . (b) 1 rupee 25 paise = ` (1 xx 100 + 25)` paise = 125 paise. ` :. ` 75 paise : 1 rupes 25 paise = 75 paise : 125 paise ` = "75 paise"/(125 "paise" = 3/5 = 3 : 5` Now , since ` 3 lt 5` , antecedent ` lt` consequent. ` :.` the ratio is a less inequality . |
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| 15. |
Find the compound ratio of the inverse ratios of the ratios ` x^(2) : (yz)/x, y^(2) : (zx)/y " and " z^(2) : (xy)/z`. |
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Answer» The inverse ratios of the ratios ` x^(2) : (yz)/x, y^(2) : (zx)/y " and " z^(2) : (xy)/x : x^(2) , (zx)/y : y^(2) " and " (xy)/z : z^(2) ` respectively. ` :. ` their compound ratio ` = ((yz)/z xx (zx)/y xx (xy)/z) : ( x^(2) xx y^(2) xx z^(2))` ` = xyz : x^(2) y^(2)z^(2)` ` = (xyz)/(x^(2)y^(2)z^(2)) = 1 : xyz`. Hence the required compound ratio = `1 : xyz`. |
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| 16. |
Calculate what ratio and `x^(2) : yz` will form the mixed ratio ` xy : z^(2)`. |
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Answer» Let ` a : b " and " x^(2) : yz ` will form the mixed ratio ` xy : z^(2)`. `:. a xx x^(2) : b xx yz = xy : z^(2)` or, `(ax^(2))/(byz) = (xy)/z^(2) " or, " a/b = (xy xx yz)/(x^(2) xx z^(2)) = y^(2)/(xz)` ` :. a : b = y^(2) : xz `. Hence the required ratio is `y^(2) : xz `. |
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| 17. |
If `x/a = y/b = z/c `, then show that ` (x^(2)-yz)/(a^(2)-bc) = (y^(2)-zx)/(b^(2) - ca) = (z^(2)-xy)/(c^(2)-ab)` |
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Answer» ` x/a = y/b = z/c = k " (let)" [k ne 0]` ` :. X = ak , y = bk " and " z = ck`. Now, `(x^(2)-yz)/(a^(2)-bc) = ((ak)^(2)-(bk)(ck))/(a^(2)-bc)` ` = (a^(2)k^(2)-bck^(2))/(a^(2) - bc) = (k^(2)(a^(2)-bc))/((a^(2)-bc)) = k^(2) ` ............(1) `(y^(2) - zx)/(b^(2) - ca) = ((bk)^(2)-ck.ak)/(b^(2) - ca) = (b^(2)k^(2)-k^(2)ca)/(b^(2)-ca) =(k^(2)(b^(2)-ca))/(b^(2)-ca) =k^(2) `...............(2) `(z^(2)-xy)/(c^(2)-ab) = ((ck)^(2)-(ak)(bk))/(c^(2)-ab)= (c^(2)k^(2)-abk^(2))/(c^(2)-ab) = (k^(2)(c^(2)-ab))/((c^(2) - ab)) = k^(2)` .....................(3) ` :. ` We get from (1) , (2) and (3) , `(x^(2)-yz)/(a^(2)-bc) = (y^(2)-zx)/(b^(2)-ca) = (z^(2)-xy)/(c^(2)-ab)`. |
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| 18. |
The sub-duplicate ratio of `(7+4sqrt3): (7-4sqrt3)` isA. `(7 + 4 sqrt3) : 1`B. `(7 - 4 sqrt3) : 1`C. `4 sqrt3 : 1 `D. None of these. |
| Answer» Correct Answer - A | |
| 19. |
The mean - proportional of `x^(2) y " and " yz^(2)` isA. xyzB. `-xyz `C. ` pm xyz`D. `pm x^(2)y^(2)z^(2)` |
| Answer» Correct Answer - C | |
| 20. |
The sub-triplicate ratio of the ratio `(7 + 5 sqrt2) : (7 - 5 sqrt2)` isA. `-(3 - 2 sqrt2) : 1`B. `(3 + 2 sqrt2) : 1 `C. `-(3 + 2 sqrt2) : 1`D. None of these. |
| Answer» Correct Answer - C | |
| 21. |
Fill in the blanks : (a) The triplicate ratio of `2^(1/3) : 3^(1/3) ` is ________ (b) The sub - triplicate ratio of ` 8 : 27` is ____________ |
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Answer» (a) ` 2 : 3`, since the triplicate ratio of `(2^(1/3))^(3) : (3^(1/3))^(3) = 2 : 3` (b) ` 2 : 3` since the sub-triplicate ratio of ` 8 : 27` is `root(3)(27) , root(3)(27) = root(3)(3^(3)) = 2 : 3` |
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| 22. |
The triplicate ratio of the ratio `x^(1/3)y^(1/3): y^(1/3) z^(1/3)` isA. ` x : z `B. ` x : y`C. ` y : z `D. ` z : x ` |
| Answer» Correct Answer - A | |
| 23. |
The fourth proportional of ` x - y , x^(2) - y^(2) " and " x^(2) - xy + y^(2)` isA. ` x + y`B. ` x - y`C. ` x^(3) + y^(3)`D. ` x^(3) - y^(3)`. |
| Answer» Correct Answer - C | |
| 24. |
The fourth proportional of 4, 6, 10 isA. 10B. 15C. 20D. 22 |
| Answer» Correct Answer - B | |
| 25. |
Do the ratios 15 cm to 2 m and 10 sec to 3 minutes form a proportion? |
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Answer» First ratio, `r_1 = (15cm)/(2m) = (15cm)/(200cm) = 3/40` Second ratio, `r_2 = (10sec)/(3minutes) = (10sec)/(180sec) = 1/18` As, `r_1 != r_2`, So, these two ratios do not form a proportion. |
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| 26. |
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.(a) 25 cm : 1 m and Rs 40 : Rs 160 (b) 39 litres : 65 litres and 6 bottles : 10 bottles(c) 2 kg : 80 kg and 25 g : 625 g (d) 200 ml : 2.5 litre and Rs 4 : Rs 50 |
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Answer» a) 25cm/1m=25cm/100cm=1/4 `40Rs/160 Rs=1/4 They are proportionate middle terms are 25 cm and 40 Rs Extreme terms are 1 m and 160 Rs b)39l/65l 39/65 6/10 bottles=3/5 They are not Proportionate. c)2kg/80kg=1/40 25g/625g=1/25 They are not proportionate d)200/2500=2/25 4/50=2/25 They are proportionate middle terms are 200ml and 4Rs Extreme numbers are 2.5litre and 50Rs. |
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| 27. |
Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of(a) Number of students who opted basketball to the number of students who opted table tennis.(b) Number of students who opted cricket to the number of students optingbasketball.(c) Number of students who opted basketball to the total number of students. |
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Answer» Total number of students `= 1800` Number of students who opted basketball` = 750` Number of students who opted cricket ` = 800` Number of students who opted table tennis ` = 1800-750-800 = 250` (a) Ratio of number of students who opted basketball to the number of students who opted table tennis ` = 750/250 = 3/1 = 3:1` (b) Ratio of number of students who opted cricket to the number of students opting basketball `= 800/750 =16/15 = 16:15` (c) Ratio of number of students who opted basketball to the total number of students `=750/1800 = 15/36 = 5/12` |
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| 28. |
Incomesof A, B and C are in the ratio 7:9:12 and their respective expenditures arein the ratio 8:9:5.If A saves `1/4`ofhis income then the ratio of their saving sis`56 : 99 : 69`b. `15 : 28 : 27`c. `33 : 19 : 23`d.`56 : 69 : 99` |
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Answer» Savings=income-expenditure Let the incomes of A,B and C be 7x,9x and 12x. their respective expenditures be 8z,9z and 5z. A saves `(1/4)^th` of his income=>`x/4=7x-8z` =>`z=(21x)/32` =>ratio of savings=>`x/4:9(x-z):(69x)/32` =>`56:99:69` |
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| 29. |
If `x : y = 3 : 4`, then find ` (3y - x) : (2 x + y)`. |
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Answer» We have, ` x : y = 3 : 4` or, ` x/y = 3/4 rArr x/y = (3k)/(4k) [k ne 0]` ` :. ` Let x = 3k and y = 4k Now, `(3y - x) : (2 x + y) = (3y -x)/(2x + y) = (3 xx 4k - 3k)/(2 xx 3k + 4k) ` ` = (12k - 3k)/(6k+4k) = (9k)/(10k) = 9/10 = 9 : 10`. Hence ` (3y - x) : (2x + y) = 9 : 10`. |
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| 30. |
Find the value of x in each of the following (i) 2 : 5: : x : 20 (ii) x : 10 : : 15 : 30 |
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Answer» Correct Answer - (i) 8 (ii) 5 9i) 2:5 : : x:20 ` Rightarrow 2/5 = x/20 Rightarrow x= 2/5 xx 20 = 8 ` (b) x : 10 :: 15 : 30 ` x/10 = 15/30` ` x/10 = 1/2 Rightarrow x=5` |
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| 31. |
In a class, there are 70 students. If the ratio boys to girls is 5:2, then find the number of girls in the class. |
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Answer» Correct Answer - 28 Total student in the class = 70 Let x be the number of grils in the class. ` 70 : x=5:2` ` 70/x = 5/2 Rightarrow 70 xx 2/5 = x ` x= 28 Number of grils in the class = 28 |
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| 32. |
the amount with A is ₹250 and his age is25 years. Amount with B is ₹ 400. if the ratio of the amounts of A and B are in proportion to the ratio of their ages, then find the age of B. |
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Answer» Correct Answer - 40 Let the age of B be x. 25 : x = 250 : 400 `25/x = 250/400 = 5/8 ` ` x= 25 xx 8/5 = 40` The age of B is 40 years. |
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| 33. |
In a proportion the product of extremes is 144. if one of the means is 9 , then find the other. |
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Answer» Correct Answer - 16 Product of extremes = Product of Means 144= `9 xx x` ` x= 144/9 = 16` |
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| 34. |
Find the value of x in each of the followings proportions. (i) 17: 68 :: 7 :x (ii( 14 : 42 :: x: 27 |
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Answer» Correct Answer - (i) 28 (ii) 9 17 : 68 : : 7 :x `Rightarrow 17/68 = 7/x` x = 28 (ii) 14 : 42: : x : 27 ` 14/42 = x/ 27` `1/3 = x/27` x=9 |
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| 35. |
The fourth proportional of 3, 4 and 6 isA. 8B. 10C. 12D. 24 |
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Answer» Correct Answer - A `:. 3/4 = 6/x " or, " 3x = 24 " or," x = 24/3 = 8`. ` :. ` (a) is correct. |
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| 36. |
Third proportional of 8 and 12 isA. 12B. 16C. 18D. 20 |
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Answer» Correct Answer - C Let the third proportional = x. ` :. 8, 12` and x are in continued proportion. ` :. 8/12 = 12/x " or," 8x = 12 xx 12 " or, " x = (12 xx 12)/8 " or, " x = 18`. `:. `(c ) is correct. |
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| 37. |
The mean - proportional of 16 and 25 isA. 400B. 100C. 20D. 40 |
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Answer» Correct Answer - C Let the mean - proportional = x. ` :. ` 16, x and 25 are in continued proportion. ` :. 16/x = x/(25) " or, " x^(2) = 400 " o, " x = sqrt(400) = 20`. `:. ` (c ) is correct. |
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| 38. |
Find the mean - proportional of `(a + b)^(2)" and " (a - b)^(2)`. |
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Answer» Let the mean - proportional of `(a+b)^(2) " and " (a - b)^(2) ` be x. ` :. ((a +b)^(2))/x = x/((a-b)^(2))` or, ` x^(2) = ( a + b)^(2) xx (a - b)^(2)` or, ` x^(2) = {(a + b)(a - b)^(2) " or, " x^(2) = (a^(2) - b^(2))^(2)`. or, ` x = pm(a^(2) - b^(2))` Hence the required mean - proportional `= pm (a^(2) - b^(2))`. |
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| 39. |
If ` a : 2 = b : 5 = c : 8 ` , then 50% of a = 20% of b = ________ % of c . |
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Answer» Given that ` a : 2 = b : 5 = c : 8`. or, ` a/2 = b/5 = c/8` or, `a/2 xx 100% = b/2 xx 100 % = c/8 xx 100%` or, ` 50% " of " a = 20% " of " b = 25/2 % " of " c `. or, ` 50% " of " a = 20% " of " b = underline(12 1/2 %)` of c . |
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| 40. |
If ` a : b = 3 : 2 " and " b : c = 3 : 2 `, then find the value of the ratio `(a + b) : (b + c)`. |
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Answer» ` a : b = 3 : 2 rArr a/b = 3/2`. `:. " Let " a = 3k_(1) " and " b = 2k_(1) [k_(1) ne 0]` Also, ` b : c = 3 : 2 rArr b/c = 3/2`. ` :. " Let " b = 3k_(2) " and " c = 2k_(2) [k_(2) ne 0]` Now, ` (a + b) : (b +c) = (a + b)/(b + c) = (3k_(1) + 2k_(1))/(3k_(2) = 2k_(2))` ` = (5k_(1))/(5k_(2)) = k_(1)/k_(2) ` .............. (1) Also, ` a/b = 3/2 rArr (3k_(1))/(3k_(2)) = 3/2 rArr k_(1)/k_(2) = 3/2`. ` :. " from (1) we get , " (a + b)/(b + c) = 3/2` . Hence ` (a +b) : (b + c) = 3 : 2` |
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| 41. |
If ` a : b = 3 : 4 " and " x : y = 5 : 7`, then find the value of ` (3ax - by) : (4by - 7ax)`. |
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Answer» Given that ` a : b = 3 : 4` or, ` a/b = 3/4 " or, " a = (3b)/4 ` .....................(1) Again , `x : y = 5 : 7 " or, " x/y = 5/7 " or, " x = (5y)/7` .....................(2) Now, `(3ax-by) : (4by - 7ax)` ` = (3ax - by)/(4by - 7ax) = (3 xx(3b)/4 xx (5y)/7 - by)/(4by - 7 xx (3b)/4 xx(5y)/7)` ` = (45/28 by -by)/(4by - (15by)/4) = ((45by-28by)/28)/((16by - 15by)/4) = ((17by)/28)/((by)/4)` ` = (17by)/28 xx 4/(by) = 17/7` Hence the required value ` = 17/7`. Aliter : Given that ` a : b = 3 : 4` `:. " Let " a = 3k_(1) , b = 4k_(1)` Again, ` x : y = 5 : 7` ` :. " Let " x = 5k_(2), y = 7k_(2) . [k_(1), k_(2) ne 0]` ` :. ` Given quantity ` = (3ax - by) : (4by - 7ax)` ` = (3ax - by )/(4by - 7ax) = (3 xx 3k_(1) xx 5k_(2) - 4k_(1) xx 7k_(2))/(4 xx 4k_(1) xx 7k_(2) - 7 xx 3k_(1) xx 5k_(2))` ` = (45 k_(1) k_(2) - 28 k_(1) k_(2))/(112 k_(1)k_(2)- 105 k_(1)k_(2)) = (17 k_(1)k_(2))/(7 k_(1) k_(2)) = 17/7`. Hence the required value = `17/7`. |
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| 42. |
If `(3x-5y)/(3x+5y) = 1/2`. Then find the value of `(3x^(2)-5y^(2))/(3x^(2) + 5y^(2))`. |
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Answer» Given that `(3x - 5y)/(3x + 5y) = 1/2` or , ` 6x - 10y = 3x + 5y` or, ` 6x - 3x = 5y + 10y` or, ` 3x = 15y` or, ` x = (15y)/3 " or, " x = 5y` `:. ` Given quantity ` = (3x^(2) - 5y^(2))/(3x^(2) + 5y^(2))` ` = (3 xx (5y)^(2) - 5y^(2))/(3 xx (5y)^(2) + 5y(2))` ` = (75y^(2) - 5y^(2))/(75y^(2) + 5y^(2)) = (70y^(2))/(80y^(2)) = 7/8` Hence the required value = ` 7/8`. |
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| 43. |
If ` x/(y+z) = y/(z+x) = z/(x + y)`, then prove that the value of each ratio is equal to `1/2 " or " (-1)`. |
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Answer» ` x/(y + z) = y/(z +x) = z/(x+y) = k " (let)" [kne 0]` ` :. X = k (y + z), y = k (z + x) , z = k ( x + y)`. Now, ` x + y + z = k (y + z) + k (z + x) + k(x + y)` ` = k (y + z + z + x + x + y)` ` = k (2x + 2y + 2z)` ` = 2k (x + y +z)` or, ` x + y + z - 2k(x + y +z) = 0` or, ` (x + y+z) (1 - 2k) = 0`. ` :. " either" x + y + z = 0 " or, " 1 - 2k = 0` ` rArr 2k = 1 rArr k = 1/2`. ` :. ` each ratio ` = 1/2` Again , ` x + y + z rArr y + z = - x` ` rArr x/(y+z) = x/(-x) = - 1` ltbr. Similarly , ` z + x = - y rArr y/(z + x) = y/(-y) = - 1` and ` x + y = - z rArr z/(x + y) = z/(-z) = - 1 `. ` :. x/(y+z) = y/(z+x) = z/(x+y)` implies that the value of each ratio is equal to ` 1/2 " or " (-1)`. |
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| 44. |
If `a^(2)/(b + c) = b^(2)/(c +a) = c^(2)/(a + b) = 1`, then show that ` 1/(1+a) + 1/(1 + b) + 1/(1+c) = 1`. |
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Answer» `a^(2)/(b+c) = 1 or, " a/(b+c) = 1/a` [by componendo process] ` :. 1/(1 + a) = a/(a + b + c) ` ……………..(1) Similarly , ` 1/(1+b) = b/(a + b + c) ` ………….(2) and , ` 1/(1 +c) = c/(a + b + c) ` ……………(3) Now, adding (1), (2) and (3) we get, `1/(1+a) + 1/(1+b) + 1/(1+c) = a/(a+b+c) + b/(a+b+c) + c/(a + b+c)` ` = (a + b + c)/(a + b + c) = 1`. ` :. 1/(1 + a) + 1/(1 + b) + 1/(1 + c) = 1`. |
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| 45. |
The mean - proportional of `1/2 " and " 1/8` isA. `pm 1/3`B. `pm 1/5`C. `pm 1/4`D. `pm 1/8` |
| Answer» Correct Answer - C | |
| 46. |
If ` a/(y+z) = b/(z + x) = c/(x+y)`, then prove that ` (a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2)-y^(2))`. |
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Answer» `a/(y + z) = b/(z + x) = c/(x+ y) " or . " a/(y + z) = (b-c)/(z + x - x - y)` [by addendo process] ` = (b-c)/(z-y)` ` :. a/(y+z) = (b-c)/(z-y) ` …………..(1) Similarly, it can be proved that ` b/(z + x) = (c-a)/(x - z) ` …………..(2) and, ` c/(x + y) = (a - b)/(y - x) ` ..............(3) Now, expressions on the LHS (1) , (2) and (3) are equal (Given) ` :. ` All the 6 expressions of (1), (2) and (3) of both the sides are equal. `:. a/(y+z) xx ( b -c)/(z - y) = b/(z + x) xx (c -a)/(x+y) xx (a-b)/(y-x)`. or, `(a(b-c))/(-(y+z)(y-z)) = (b(c-a))/(-(z+x)(z-x)) = (c(a-b))/(-(x+y)(x-y))` or, `(a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2)-y^(2))`. ` :. (a(b-c))/(y^(2)-z^(2)) = (b(c-a))/(z^(2)-x^(2)) = (c(a-b))/(x^(2) - y^(2))`. |
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| 47. |
if `1/2, 1/3, 1/4 ` : x then x is _____A. `1/8 `B. `1/6`C. `1/12`D. `1/16` |
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Answer» Correct Answer - B ` 1/2 : 1/3 : : 1/4 : x ` ` 1/2 xx x = 1/3 xx 1/4` ` x 1/3 xx 1/4 xx 2` ` x= 1/6` Hence, the correct option is (b) |
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| 48. |
The ratio of 75 paise to ₹ 10 is _____A. `15 :2`B. `3:40`C. `3 : 4`D. ` 75 : 1000` |
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Answer» Correct Answer - B 75 paise : ₹ 10 = 75 : 1000 = ` 25 xx 3 : 25 xx 40 ` = 3: 40 Hence the correct option is (b). |
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| 49. |
If x : 12 : : 12 : 18. then x isA. 6B. 8C. 10D. 4 |
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Answer» Correct Answer - B x : 12 : : 12 : 18 Product of extremes = Product of means `Rightarrow 18 xx x = 12 xx 12` ` Rightarrow x = 12/18 xx 12 = (12 xx 2) /3` x = 8 Hence, the correct option is (b). |
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| 50. |
Which of the following is true ?A. 2,5,7,10 are in proportion.B. 4:3 : : 8 : 6C. if a : b : c :d , then b : c x isD. None of the above |
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Answer» Correct Answer - B (a) 2, 5 , 7 , 10 ` 2 xx 10 = 20` ` 5 xx 7 = 35` ` 2 xx 10 ne 5xx 7` (b) 4 : 3 : : 8:6 ` Rightarrow 4/3 = 8/6` ( true) (c) a, b,c ,d are in proportion. ` Rightarrow a/b = c/d Rightarrow bc = ad` ` b/c` need not be equal to `d/a` Hence, the correct option is (b) |
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