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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. |
`{:((x)/(3) + (y)/(15) = 4),((x)/(3) - (y)/(12) = (19)/(4)):}` |
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Answer» Correct Answer - ` x = 18, y = 15 ` The given equations are ` 5x + 2y - 120 = 0 and 4x - y - 57 = 0 ` |
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| 2. |
Solve: `x/a+y/b=a+b``x/(a^2)+y/(b^2)=2` |
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Answer» Correct Answer - ` x = a ^(2), y = b ^(2)` ` b x + ay = a ^(2) b + ab^(2)" " `… (i) and ` b^(2) x + a ^(2) y = 2 a ^(2) b ^(2)" " `… (ii) |
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| 3. |
The students of a class are made to stand in rows. If 3 students areextra in a row, there would be 1 row less. If 3 students are less in a rowthere would be 2 rows more. Find the number of students in the class. |
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Answer» Let the number of rows = x and the number of students in each row = y `therefore` Total number of students in the class = xy According to the I condition, (y + 3) (x - 1) = xy (no. of students in each row `xx` no. of rows = total students) implies xy - y + 3x - 3 = xy implies 3x - y = 3 ...(1) According to the II condition, (y - 3) (x + 2) = xy implies xy + 2y - 3x - 6 = xy implies 3x - 2y = - 6 Subtracting equation (2) from equation (1), we have y = 9 Put y = 9 in equation (1), we get 3x - 9 = 3 implies 3x = 12 implies x = 4 `therefore` Total number of students in the class = xy = 4 `xx` 9 = 36. |
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| 4. |
Students ofa class are preparing for a drill and are made to stand in rows. If 4students are extra in a row, then there would be 2 rows less. But there wouldbe 4 more rows if 4 students are less in a row. The number of students in theclass is(a) 56 (b) 65 (c) 69 (d) 96 |
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Answer» Let the number of rows be x and the number of students in each row be y. Then, the total number of students = xy . Case I When there are 4 more students in each row = ( y + 4) And, number of rows = (x - 2 ) Total number of students = (x - 2) ( y + 4) `therefore (x - 2) ( y + 4) = xy rArr 4x - 2y = 8 ` `" " rArr 2 x - y = 4 " " `... (i) Case II When 4 students are removed from each row. Then, number of students in each row = ( y - 4) And, number of rows = ( x+ 4) Total number of stuents = ( x+ 4 ) ( y - 4) ` therefore (x + 4 ) ( y - 4) = xy rArr 4 y - 4x = 16` `" " rArr 4 ( y - x) = 16 ` ` " " rArr ( y - x ) = 4 " " `... (ii) Adding (i) and (ii), we get x = 8 Putting ` x = 8` in (ii), we get ` y- 8 = 4 rArr y = 12 ` Thus, x = 8 and y = 12 This shows that there are 8 rows and there are 12 students in each row. Hence, the number of students in the class = ` xy = 8 xx 12 = 96`. |
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| 5. |
The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7:5. If each saves ₹ 9000 per month, find the monthly income of each. |
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Answer» Correct Answer - ₹ 30,000, ₹ 24,000 `5x - 7y = 9000 and 4x - 5y = 9000`. Find x and y . |
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| 6. |
A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028 then find the cost of the saree and the list price (price before discount) of the sweater. |
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Answer» Let the cost price of a saree = Rs. X and the list price of sweater = Rs. Y Case I : (S.P. of a saree at 8% profit) + (S.P. of a sweater at 10% discount) - Rs. 1008 `implies ((100+8))/(100)x + ((100-10))/(100)y = 1008` implies 108x + 90y = 100800 implies 6x + 5y = 5600 ...(1) Case II : (S.P. of a saree at 10% profit) + (S.P. of a sweater at 8% discount) = Rs. 1028 `implies ((100+10))/(100) x + ((100 - 8))/(100) y = 1028` implies 110x + 92y = 102800 ...(2) Dividing equation (2) by 2, we have 55x + 46y = 51400 ...(3) Again, multiplying equation (3) by 5, we get 275x + 230y = 257000 ...(4) Multiplying equation (1) by 46, we get 276x + 230y = 257600 ...(5) Subtracting equation (5) from (4), we get `{:(275x + 230y = 257000),(276x + 230y = 257600),(ul("- - -")),(-x = 257000 - 257600):}` implies -x = - 600 implies x = Rs. 600 Now, 6x + 5y = 5600 [from (1)] implies `6 xx 600` + 5y = 5600 (`because` x = 600) implies 5y = 5600 - 3600 implies `y = (2000)/(5)` implies y = 400 Hence, the C.P. of a saree and C.P. of a sweater are Rs. 600, Rs. 400 respectively. |
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| 7. |
The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number. |
| Answer» Correct Answer - 57 | |
| 8. |
A jeweller has bars of 18 carat gold and 12 carat gold. How much of each must be melted together to obtain a bar of 16 carat gold, weighing 120g (Given pure gold is of 24 carat.)? |
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Answer» Correct Answer - 80 g, 40 g Let x g of 18- carat gld be mixed with y g of 12-carat gold to get 120 g of 16- carat gold. Then, ` x + y = 120 " "` … (i) Gold % in 18 -carat gold = `((18)/(24) xx 100) % = 75% ` Gold % in 12-carat gold = ` ((12)/(24) xx 100 )% = 50 % ` Gold % in 16 - carat gold = ` ((16)/(24) xx 100)% = (200)/(3)%` ` therefore 75% of x + 50 % of y = (200)/(3) % of 120 ` ` rArr (75x )/(100) + ( 50 y )/(100) = (200xx 120)/(3 xx 100) rArr 3x + 2y = 320 " " `... (ii) Now, solve (i) and (ii). |
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| 9. |
90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acids to be mixed to from the mixture. |
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Answer» Correct Answer - 6 litres, 15 litres Let x litres of 90% pure solution be mixed with y litres of 97% pure solution to get 21 litres of 95% pure solution. Then, ` x + y = 21 " " `… (i) and ` (90x)/(100) + (97y)/(100) = ( 95 xx 21)/(100)rArr 90x + 97 y = 1995" "`... (ii) Now, solve (i) and (ii). |
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| 10. |
A borrowed Rupees `2500` from B at `12%` per annum compound interest. After 2 years, A gave Rupees `2936` and a watch to B to clear the account. Find the cost of the watch. |
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Answer» `A` borrowed Rs `2500` from `B` at `12%` per annum compound interest. So, final amount after `2` years `= 2500(1+12/100)^2` `= 2500**112/100**112/100 = 3136` Now, `A` gave `2936` Rs and a watch to clear the account. Let cost of the watch is `x` Rs. `:. x+2936 = 3136` `=> x= 200` `:.` Cost of the watch is `200` Rs. |
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| 11. |
A man invested an amount at 12 % per annum simple interest and another amount at 10 % per annum simpel interest. He received an annual interest fo ₹ 2600. But, if he had intercharged the amounts invested, he would havr received ₹ 140 less. What amounts did he invest at the different rates ? |
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Answer» Let the amound invested at 12 % by ₹ x and that invested at 10 % be ₹ y . Then , total annual interest `" " = ₹ ((x xx 12 xx 1 )/( 100) + (y xx 10 xx 1)/(100)) = ₹ ((6x + 5y)/( 50))` ` therefore ( 6x + 5y )/( 50) = 2600 rArr 6x + 5y = 130000 " " `... (i) Again, the amount invested at 12 % is ₹ y and the invested at 10 % is ₹ x. Total annual interest at the new rates `" " = ₹(( yxx 12 xx 1)/( 100 ) + (x xx 10 xx 1 )/( 100)) = ₹ ((6y + 5x )/( 50))` But, interest received at the new rates = ₹ (2600- 140 ) = ₹ 2460. ` therefore ( 6y + 5x)/( 50 ) = 2460 rArr 5 x + 6y = 123000" " `... (ii) Adding (i) and (ii), we get ` 11 x + 11y = 253000 rArr 11 (x + y ) = 253000 ` ` rArr x + y = 23000" " `... (iii) Subtracting (ii) from (i), we get ` x - y = 7000 " " `...(iv) Adding (iii) and (iv), we get ` 2x = 30000 rArr x = 15000 ` Putting ` x = 15000 ` in (i) , we get `15000 + y = 23000 rArr y =23000 - 15000 = 8000` ` therefore x = 15000 and y = 8000` Hence, the amount at 12 % is ₹ 15000 and that at 10 % is ₹ 8000. |
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| 12. |
A man purchased 47 stamps of 20 p and 25 p for ₹ 10. Find the number of each type of stamps. |
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Answer» Let x stamps of 20 p and y stamps of 25 p be purchased. Then, ` x + y = 47" " `… (i) and ` 20 x + 25 y = 1000 rArr 4x + 5y = 200" " `… (ii) Solve (i) and (ii). |
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| 13. |
If ` 2 x + 3y = 12 and 3 x - 2y = 5 ` thenA. ` x =2, y = 3 `B. `x =2, y = - 3 `C. `x = 3, y = 2 `D. `x= 3, y = - 2 ` |
| Answer» Correct Answer - C | |
| 14. |
Solve the following pairs of equation for `x and y`, `15/(x-y)+22/(x+y)=5` and `40/(x-y) + 55/(x+y)=13` |
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Answer» Putting ` (1)/(x - y ) = uand (1)/( x + y ) = v `, the given equations become ` 15u + 22v = 5 " " `… (i) ` 40 u + 55v = 13" " `… (ii) Multiplying (ii)by 2 and (i) by 5 and subtracting the results, we get `80 u - 75 u = 26 - 25` `rArr 5u = 1 ` ` rArr u = (1)/(5)` Putting ` u = (1)/(5)` in (i), we get ` ( 15 xx (1)/(5) ) + 22 v = 5` ` rArr 3 + 22 v = 5 ` ` rArr 22 v = 2 rArr v = (2)/(22) = (1)/(11)` Now, ` u = (1)/(5) rArr (1)/(x- y ) = (1)/(5) rArr x - y = 5 " " `... (iii) And, ` v = (1)/(11) rArr (1)/( x+ y ) = (1)/(11) rArr x +y = 11 " " `... (iv) On adding (iii) and (iv), we get ` 2x= 16 rArr x = 8`. On subtracting (iii) from (iv), we get ` 2y = 6 rArr y = 3.` Hence, ` x = 8 and y = 3`. |
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| 15. |
`3x-2y+3=0;4x+3y-47=0` |
| Answer» Correct Answer - ` x = 5, y = 9 ` | |
| 16. |
` 6x - 5y - 16 = 0`, ` 7x - 13 y + 10 = 0 `. |
| Answer» Correct Answer - ` x = 6, y = 4`. | |
| 17. |
`(44)/(x+y)+(30)/(x-y)=10``(55)/(x+y)+(40)/(x-y)=13` |
| Answer» Correct Answer - ` x = 8, y = 3 ` | |
| 18. |
If the sum of two numbers is divided by 13, the quotient is 4 and the remainder is 8. If the difference of the same two numbers is divided by 4, the quotient is 3 and the remainder 2. Find the numbers |
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Answer» Let two numbers be x and y `x+y=13*4+8` `x+y=60-(1)` `x-y=4*3+2` `x-y=14-(2)` from equation 1 and 2 `2x=74` `x=37` pitting this value in equation 2 `x-y=14` `y=23`. |
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| 19. |
If three times the larger of the two numbers isdivided by the smaller one, we get 4 as quotient and 3 as remainder. Also ifseven times the smaller number is divided by the larger one, we get 5 asquotient and 1 as remainder . Find the numbers. |
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Answer» Let the larger number be x and the smaller one by y . We know that dividend = ( divisor ` xx` quotient ) + remainder Using the above result and the given conditions, we have ` 3 x = 4y + 3 rArr 3 x - 4y = 3 " " `... (i) and ` 7y = 5 x + 1 rArr 5 x - 7y = - 1 " " `... (ii) Multiplying (i) by 5, (ii) by 3 and subtracting, we get ` y = 18` Putting `y = 18 ` in (i), we get ` 3 x - ( 4 xx 18) = 3 rArr 3x - 72 = 3 ` `rArr 3x = 75 rArr x = 25 ` Hence, the required numbers are 25 and 18. |
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| 20. |
A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of the journey. |
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Answer» Correct Answer - 1080 km Let the original speed of the train be x kmph and let the time taken to complete the journey be y hours. Then, length of the journey = xy km. Case I Speed = `( x + 5)` kmph and time taken = ` ( y - 3) ` hours. ` therefore xy = (x + 5) ( y - 3 ) rArr 5y - 3x = 15" "`... (i) Case II Speed ` = ( x - 4) `kmph and time taken ` = ( y + 3) ` hours ` therefore xy = (x - 4) ( y + 3) rArr 3x - 4y = 12 " " `... (ii) From (i) and (ii), we get ` x = 40 and y = 27` . ` therefore ` length of the journey = ` x y = 40 xx 27 ` km = 1080 km. |
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| 21. |
5. After covering a distance of 30 km with a uniform speed there is some defect in train engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, train reaches its destination late by 45 minutes. Had it happened after covering 18 km more, reached 9 minutes earlier. Find the speed of the train and the distance of covered. |
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Answer» Let the initial speed of train = x km/hr and the distance of journey = y km `therefore` Usual time = `(y)/(x)` hr Case I : When defect in engine, after 30 km `therefore` Time taken in initial speed + time taken in reduced speed = `(y)/(x) + (45)/(60)` implies `(30)/(x) + (y - 30)/((4)/(5)x) = (y)/(x) + (3)/(4)` implies `(120 + 5(y - 30))/(4x) = (4y + 3x)/(4x)` ` implies 120 + 5y - 150 = 4y + 3x` implies 3x - y = - 30 implies 12x - 4y = - 120 ...(1) Case II : When defect in engine, after 48 km `therefore (48)/(x) + (y - 48)/((4)/(5)x) = (y)/(x) + (36)/(60)` (now he is 9 minutes earlier means still he is late by 36 minutes) `implies (192 + 5(y - 48))/(4x) = (60y + 36x)/(60x)` `implies 15(192 + 5y - 240) = 60y + 36x` implies 5(5y - 48) = 20y + 12x implies 25y - 20y - 12x = 240 implies 5y - 12x = 240 ...(2) Adding equations (1) and (2), we get y = 120 Putting y = 120 in equation (1), we get 3x = 120 - 30 implies x = 30 Hence, initial speed of train = 30 km/hr and the distance travelled = 120 km. |
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| 22. |
A plane left 30 minutes late than its scheduled time and in order to reach the destination1500 km away in time, it had to increase the speed by 250 km/h from the usual speed.Find its usual speed. |
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Answer» Let the usual speed = x km/hr and distance travelled = 1500 km `therefore` time taken in usual speed = `(1500)/(x)` hr Increased speed = (x + 250) km/hr `therefore` time taken in increased speed = `(1500)/(x + 250)` hr But difference in both timings = 30 min = `(1)/(2)` hr `implies (1500)/(x) - (1500)/(x + 250) = (1)/(2)` implies` 1500 [(x + 250 - x)/(x(x + 250))] = (1)/(2)` implies x(x + 250) = 500 `xx` 1500 implies `x^(2) + 250x - 750000 = 0` implies (x + 1000)(x - 750) = 0 `therefore x = 750` or x = - 1000 But speed cannot be negative. Hence, the usual speed = 750 km/hr. |
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| 23. |
Points `A`and `B`are 70 km. apart on ahighway. A car starts from `A`and another car startsfrom `B`simultaneously. If theytravel in the same direction, they meet in 7 hours, but if they traveltowards each other, they meet in one hour. Find the speed of the two cars. |
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Answer» Correct Answer - 40 km/hr, 30 km/hr Let the speed of the car from A be x kmph and that of the car from B be y kmph. Then, ` 7x - 7y = 70 rArr x - y = 10 " " `… (i) And, ` x + y = 70." " `… (ii) |
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| 24. |
Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car. |
| Answer» Correct Answer - 50 kmph, 30 kmph | |
| 25. |
For which value (s) of k will the pair of equations `kx + 3y = k - 3`, `12x + ky = k` has no solution ? |
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Answer» The given equations are ` kx + 3y + ( 2- k ) = 0 and 12 x + ky - k = 0 ` These equations are of the form ` a_ 1 x + b_1 y + c_ 1 = 0 and a_ 2 x + b_ 2 y + c_ 2 = 0 `, where ` a_ 1 = k , b_ 1 = 3, c_ 1 = (2 - k ) and a_ 2 = 12, b_ 2 = k, c_ 2 = - k ` ` therefore (a_ 1 )/(a _ 2 ) = (k)/( 12) , (b_ 1 ) /(b _ 2 ) = ( 3)/(k) and (c_ 1 ) /(c_ 2 ) = ((2 - k ))/(- k ) = ((k - 2 ))/( k )` Let the given sytem of equations have no solution. Thenm ` (a_ 1 ) /(a_ 2 ) = (b_ 1 ) /( b _ 2 ) ne (c_ 1 ) /(c_ 2 )` ` rArr ( k ) /(12) = ( 3 ) /(k ) ne (( k - 2 ))/( k ) ` ` rArr (k)/( 12) = (3 ) /(k ) and ( 3 ) /( k ) ne ((k - 2 ))/( k ) ` ` rArr k ^(2) = 36 and k ^(2) - 2k ne 3k ` ` rArr k ^(2) = 36 and k ^(2) - 5k ne 0 ` `rArr ( k = 6 or k = - 6) and k ( k - 5) ne 0 ` `rArr ( k = 6 or k = - 6) and k ( k - 5) ne 0 ` Case 1. When k = 6 In this case, `k ( k - 5)= 6( 6- 5) = 6xx 1 = 6 ne 0 ` . Case 2. When `k = - 6` In this case, `k (k -l 5) = (-6) ( - 6- 5) = (-6) xx (-11) = 66 ne 0` Thus, in each case, the given system has no solution. Hence, `k = 6 or k = -6` |
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| 26. |
Solve 41x + 53y = 135 and 53x + 41y = 147. |
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Answer» Given equations are 41x + 53y = 135 ….(1) and 53x + 41y = 147 ….(2) On adding equations (1) and (2) we get 94x + 94y = 282 implies x + y = 3 ….(3) On subtracting equation (2) from (1), we get -12x + 12y = - 12 implies x = y = 1 ....(4) Adding equations (3) and (4), we get 2x = 4 implies x = 2 Putting x = 2 in equation (3) 2 + y = 3 implies y = 1 Hence, the solution is `{:(x = 2),(y = 1):}}`. |
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| 27. |
Draw the graph of each of the following linear equations in two variables:(i) `x+y=4` (ii) `x-y=2`(iii) `y=3x` (iv) `3=2x+y` |
| Answer» graph is shown in the video. | |
| 28. |
Draw the graphs of thefollowing equations on the same graph paper:`2x+y=2; 2x+y=6`. Find the coordinatesof the vertices of the trapezium formed by these lines. Also, find the areaof the trapezium so formed. |
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Answer» Correct Answer - 8 sq units The lines ` 2 x + y = 2 ` cuts the x-axis at `A(1, 0)` and the y- axis at `B(0, 2 )` The lines ` 2x + y = 6 ` cuts the x-axis at `C(3, 0)` and the y -axis at `D(0, 6)` Area of trap. ABDC = `ar(Delta OCD) - ar(Delta OAB)` ` " " = ((1)/(2)xx 3 xx 6) - ((1)/(2) xx 1 xx 2) = 8` sq units. |
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| 29. |
There are two examination rooms A and B. If 10 candidates are sent A and B, the number of students in each room is same. If20 candidates are sent from B to A, the number of students in A is double thenumber of students in B. Find the number of students in each room. |
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Answer» Correct Answer - 100 in A and 80 in B Let the number of students in room A and room B be x and y respectively. Then, ` x - 10 = y + 10 rArr x - y = 20 " "`… (i) And, `(x + 20 ) = 2 ( y - 20) rArr x- 2y = -60 ` Now, solve (i) and (ii) to get x and y. |
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| 30. |
The difference of two numbers is 4.If the difference of their reciprocals is 4/21, find the numbers |
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Answer» Let the larger number be x and the smaller number be y . Then , ` x- y = 4 " "`… (i) And , ` (1)/( y) - (1)/(x) = ( 4)/( 21)" " [ because x gt y rArr (1)/( y) gt (1)/(x)]` `rArr (x - y )/( xy ) = ( 4)/( 21)` ` rArr ( 4) /( xy ) = ( 4)/( 21) rArr xy = 21 " "`[using (i)] ` therefore ( x+ y ) = sqrt ((x- y ) ^(2) + 4xy) ` ` " " = sqrt( 4^(2) + 4 xx 21 ) = sqrt ( 16 + 84 ) = sqrt (100) = pm 10 ` Thus, we have `{:(x- y = 4,,... (i)),(x+y= 10,,... (ii)):}} or {{:(x- y = 4,,... (iii)),(x+y = - 10,, ... (iv)):}` On solving (i) and (ii), we get x = 7 and y = - 3 On solving (iii) and (iv), we get x = - 3 and y = - 7. Hence, the required numbers are ( 7 and 3) or (-3 and - 7). |
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| 31. |
The denominator of a fraction is greater that its numerator by 11. If 8 is added to both its numerator and denominator, it becomes ` (3)/(4)`. Find the fraction . |
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Answer» Correct Answer - `(25)/( 36)` Let the required fraction be ` (x)/(y)`. Then, ` x + 11 = y rArr x - y =-11`. And, ` (x + 8)/(y + 8) = (3)/(4) rArr 4x + 32 = 3y + 24 rArr 4x - 3y = - 8` |
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| 32. |
A fraction becomes ` (1)/(3)`, if 2 is added to both of its numerator and denominator. If 3 is added to both of its numerator and denominator then it becomes ` ( 2)/(5)`. Find the fraction. |
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Answer» Let the required fraction be `(x)/(y) `. Then, ` (x+ 2 )/( y + 2 ) = (1)/(3) rArr 3x + 6 = y + 2 ` ` " " rArr 3x - y = - 4 " " `… (i) Also, ` ( x + 3 ) /( y + 3 ) = ( 2) /( 5) rArr 5x + 15 = 2y + 6 ` `" " rArr 5x - 2y = - 9 " " `... (ii) Multiplying (i) by 2 and subtracting (ii) from the result, we get ` 6 x - 5 x = - 8 + 9 rArr x = 1 ` Putting x = 1 in (i), we get ` 3- y = - 4 rArr y = 7`. Thus, ` x= 1 and y = 7` Hence, the required fraction is ` (1)/(7)` |
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| 33. |
The sum of numerator and denominator of a fraction is 3 less than twice thedenominator. If each of the numerator and denominator is decreased by 1,the fraction becomes 1/2 Find the fraction. |
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Answer» Let the required fraction be `(x)/(y)`. Then, ` (x+ y ) = 2y- 3 rArr x - y = - 3 ` And , ` (x - 1 ) /( y - 1 ) = (1)/(2)rArr 2 x - 2 = y - 1 ` `" " rArr 2 x - y = 1 ` On substracting (i) from (ii), we get ` x= 4`. Putting ` x= 4` in (i), we get ` y = x+ 3 = 4 + 3 = 7` ` therefore x = 4 and y = 7` Hence, the required faction is ` ( 4)/(7)` |
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| 34. |
If the numerator of a fraction is multiplied by 2 and the denominatoris reduced by 5 the fraction becomes 6/5. And, if the denominator is doubledand the numerator is increased by 8, the fraction becomes 2/5. Find thefraction. |
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Answer» Let the required fraction be ` (x)/( y)`. Then, `(2x )/( y - 5) = (6)/(5) rArr 10 x = 6 ( y - 5) ` ` " " rArr 10 x - 6y = - 30 ` `" " rArr 5 x - 3y = - 15 " " ` … (i) and ` (x + 8 )/( 2y ) = ( 2 ) /( 5) rArr 5 ( x + 8) = 4 y ` ` " " rArr 5 x - 4y = - 40 " " `... (ii) On subtracting (ii) from (i), we get `y = 25` Putting ` y = 25 ` in (i), we get ` 5x - ( 3 xx 25) = - 15 rArr 5x - 75 = - 15 rArr 5 x = 60 rArr x = 12.` ` therefore x = 12 and y = 25 ` Hence, the required fraction is ` (12)/( 25)` |
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| 35. |
Solve for x and y `sqrt(2)x-sqrt(3)y=0,sqrt5x +sqrt 2y=0` |
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Answer» The given system of equation is ` sqrt 2 x - sqrt3 y = 0 " " `… (i) `sqrt5 x + sqrt 2 y = 0 " " ` …(ii) Multiplying (i) by ` sqrt2 ` and (ii) by `sqrt3 ` and adding, we get ` ( 2 + sqrt(15)) x = 0 rArr x = 0 ` Putting ` x = 0 `in (ii), we get `sqrt 2 y = 0 rArr y = 0 ` Hence, x = 0 and y =0. |
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| 36. |
Solve: `a x+b y=a-b , b x-a y=a+b` |
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Answer» The given equations may be written as ` a x + by = a - b " " `… (i) ` bx - ay = a + b " " `… (ii) Multiplying (i) by a and (ii) by b and adding the results, we get `(a ^(2) + b ^(2)) x = (a ^(2) + b ^(2)) rArr x = 1 ` Putting ` x = 1 ` in (i) , we get ` ( a xx 1 ) + by = a- b ` ` rArr a + by = a - b ` ` rArr by = - b rArr y = (-b)/(b) = - 1 ` Hence, ` x = 1 and y = - 1`. |
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| 37. |
` (bx ) /(a) + ( ay ) /( b) = a ^(2) + b ^(2)`, ` x + y = 2ab ` |
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Answer» Correct Answer - ` x = ab, y = ab ` ` b ^(2)x + a ^(2) y = a ^( 3) b + a b ^(3)" " `… (i) and ` x + y = 2ab" " `… (ii) Multiply (ii) by ` b ^(2)` and subtract the result from (i). |
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| 38. |
Solve the following pair of linear equations 2(ax-by) + (a + 4b) = 0 2(bx + ay) + (b-40 = 0 |
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Answer» The given equations may be written as ` 2ax - 2by = -a - 4b " " `… (i) ` 2bx + 2ay = 4a - b " " `… (ii) Multiplying (i) by a and (ii) by b and adding , we get ` ( 2 a ^(2) + 2 b ^(2)) x = ( - a ^(2) - b ^(2))` ` rArr 2 (a ^(2) + b (2)) x = - (a ^(2) + b ^(2)) rArr x = (-1)/(2)` Putting ` x = (-1)/(2) ` in (i), we get ` 2a xx ((-1)/(2)) - 2 by = - a - 4 b ` ` rArr - a - 2by = - a - 4 b ` ` rArr 2by = 4 b rArr y = ( 4 b )/( 2 b ) = 2 ` Hence, ` x = (-1)/(2) and y = 2 ` |
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| 39. |
` 2x + 3y - 4=0 ` , `3x - y + 5 = 0 ` |
| Answer» Correct Answer - ` x = - 1, y = 2 ` | |
| 40. |
` x + 2y + 2 = 0 `, ` 3x + 2y - 2 = 0 ` |
| Answer» Correct Answer - ` x = 2 , y = - 2 ` | |
| 41. |
` x- y + 3 = 0, 2x + 3y - 4 = 0 `. |
| Answer» Correct Answer - ` (x = - 1, y = 2) ; A(- 1, 2 ), B (- 3, 0), C(2, 0) ; ar(Delta ABC) = 5 ` sq units | |
| 42. |
` 2x - 3y + 4 =0 , x + 2y - 5 =0 ` |
| Answer» Correct Answer - ` ( x =1, y = 2 ) ; A(1, 2 ), B(-2, 0), C (5, 0) ; ar ( Delta ABC)` = 7 sq units | |
| 43. |
`,x-y=-1, 3x-2y=12` |
| Answer» Correct Answer - `(x = 2, y =3 ); A (2, 3 ), B (-1, 0 ), C (4, 0) ; ar( Delta ABC) = 7.5 ` sq units | |
| 44. |
`2x+5y=8/3, 3x-2y=5/6` |
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Answer» Correct Answer - ` x = (1)/(2), y = (1)/(3)` Given equations are `6x+15y=8,18x-12y=5`. |
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| 45. |
` 4x - 3y + 4 = 0, 4x + 3y - 20 = 0 `. |
| Answer» Correct Answer - `(x = 2, y = 4 ) ; A (2, 4), B (-1, 0 ), C(5, 0 ) ; ar (Delta ABC) = 12 ` sq units. | |
| 46. |
` 2x + 3y + 1 = 0 ` , ` ( 7- 4x )/(3) = y` |
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Answer» Correct Answer - ` x = 4, y = - 3` `(7-4x)/(3)=yrArr7-4x=3y rArr 4x+3y=7`. |
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| 47. |
` 0.4 x + 0.3 y = 1.7`, ` 0.7 x - 0.2 y = 0.8` |
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Answer» Correct Answer - ` x = 2, y = 3 `. Multiply each of the given equations throughout by 10. |
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| 48. |
Solve the followingsystem of equations:`7(y+3)-2(x+2)=14 , 4(y-2)+3(x-3)=2` |
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Answer» Correct Answer - ` x = 5, y = 1 ` The given equations are `2x-7y=3,3x+4y=19`. |
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| 49. |
If `0.5x+0.7y=0.74` and `0.3x+0.5y=0.5` then |
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Answer» Correct Answer - `x = 0.5, y = 0. 7 ` Multiply each of the given equations throughout by 10. |
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| 50. |
A man when asked 'how many hens and buffaloes he has ?'. He replied that they have 78 eyes and 110 legs in all. Find the number of hens and buffaloes separately. |
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Answer» Let the number of hens = x and the numbers of buffaloes = y We know that, hen as well as buffalo has 2 eyes while hen has 2 legs and buffalo has 4 legs. So, according to the problem, 2x + 2y = 78 implies x + y = 39 …(1) Also, 2x + 4y = 110 implies x + 2y = 55 ...(2) Subtracting equation (1) from (2), we get y = 16 Putting y = 16 in equation (1), we get x + 16 = 39 implies x = 23 `therefore` Number of hens = 23 and number of buffaloes = 16 |
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