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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 pair of linear equations which has a unique solution x = 2 and y = -3 isA. x + y = 1 and 2x - 3y = -5B. 2x + 5y = -11 and 4x + 10y = -22C. 2x - y =1 and 3x + 2y = 0D. x - 4y - 14 = 0 and 5x - y - 13 = 0 |
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Answer» Correct Answer - B If x = 2, y = -3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations. From option (b), LHS = 2x + 5y = 2 (2) + 5(-3) = 4 - 15 = -11 = RHS and LHS = 4x + 10y = 4(2) + 10(-3) = 8 - 30 = -22 = RHS |
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| 2. |
Solve the following system of equations .`x/p+y/q=1`, `p(x-p)-q(y+q)=2p^2 + q^2` |
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Answer» `x/p+y/q = 1` `=>qx+py = pq->(1)` `p(x-p)-q(y+q) = 2p^2+q^2` `=>px-p^2-qy-q^2 = 2p^2+q^2` `=>px-qy = 3p^2+2q^2->(2)` Now, multiplying (1) by `q` and multiplying (2) by `p` and adding them, `=>q^2x+pqy+p^2x-pqy = pq^2+3p^3+2pq^2` `=>(p^2+q^2)x = 3p(p^2+q^2)` `=> x = 3p` Putting values of `x` in (1),`(3p)q +py = pq` `py = -2pq=> y = -2q` |
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| 3. |
One equation of a pair of dependent linear equations is `-5x + 7y -2 = 0`. The second equation can beA. `10x + 14 y + 4 = 0`B. `-10x - 14y +4 =0`C. `-10 +14y +4 =0`D. `10x - 14y + 4 = 0` |
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Answer» Correct Answer - D Condition for dependent linear equations `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = c_(1)/(c_(2)) = (1)/(k)` Give equation of line is, `-5x + 7y - 2 = 0` Here, `" " a_(1) = -5, b_(1) = 7, c_(1) = -2` From Eq. (i), `" " -(5)/(a_(2)) = (7)/(b_(2)) = - (2)/(c_(2)) = (1)/(k) " " ` [say] `rArr " " a_(2) = -5k, b_(2) = 7k, c_(2) = -2k` where, k is any arbitrary constant. Putting k = 2, then `" " a_(2) = -10 , b_(2) = 14` and `" " c_(2) = -4` `:.` The required equation of line becomes `" " a_(2) x + b_(2) y + c_(2) = 0` `rArr " " -10x + 14y - 4 = 0` `rArr " " 10x - 14y + 4 = 0` |
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| 4. |
Vijay had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of RS. 2 for 3 bananas and the second lot at the rate of Rs 1 per banana and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 1 per banana and the second lot at the rate of Rs.4 for 5 bananas , his total collection would have been Rs 460. Find the total number of bananas he had. |
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Answer» Let the number of bananas in lots A and B be x and y, respectively Case I Cost of the first lot at the rate of Rs. 2 for 3 bananas + Cost of the second lot at the rate of Rs. 1 per banana = Amount received `rArr " " (2)/(3)x+y=400` `rArr " " 2x+3y=1200 " " ...(i)` Case II Cost of the first lot at the rate of RS. 1 per banana+ Cost of the second lot at the rate of Rs. 4 for 5 bananas = Amount received `rArr " " x(4)/(5)y=460` `rArr " " 5x+4y=2300 " " ...(ii)` On multiplying in Eq. (i) by 4 and Eq. (ii) by 3 and then subtracting them, we get `{:(8x+12y=4800),(ul(15x+12y=6900)),(" "-7x=-2100):}` `rArr " " x=300` Now, put the value of x in Eq. (i), we get `2xx300+3y=1200` `rArr" " 600+3y=1200` `rArr " " 3y=1200-600` `rArr " " 3y=600` `rArr " " y=200` `:. ` Total number of bananas=Number of bananas in lot A+ Number of bananas in lot B = x+y =300+200=500 Hence, he had 500 bananas. |
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| 5. |
`(3x)/2-(5y)/3=-2`, `x/3+y/2=13/6` |
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Answer» `(3x)/2 - (5y)/3 = -2 ->(1)` `x/3+y/2 = 13/6->(2)` Multiplying (1) with `12` and (2) with `54` and then subtracing (2) from (1), `18x-20y-18x-27y = -24-117` `=>-47y = -141 => y = 3` Putting value of `y` in (1), `(3x)/2 - 5 = -2 => 3x = 6 => x = 2` `:. x = 2, y = 3`. |
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| 6. |
`2x + y = (7xy)/3, x+3y = (11xy)/3` |
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Answer» `2x+y=(7xy)/3-(1)` `x+3y=(11xy)/3-(2)` from equation 1 and 2 `-5y=(7xy)/3-(-22xy)/3` `x=1` Putting in equation 1 `2+y=(7y)/3` `y=3/2`. |
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| 7. |
If a pair of linearequations in two variables is consistent, then the lines represented by twoequations areA. parallelB. always coincidentC. intersecting or coincidentD. always intersecting |
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Answer» Correct Answer - C Condition for a consistent pair of linear equations `" " (a_(1))/(a_(2)) != (b_(1))/(b_(2)) " " ` [intersecting lines having unique solution] and `" " (a_(1))/(a_(2))=(b_(1))/(b_(2)) = (c_(1))/(c_(2)) " "` [coincident or dependent] |
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| 8. |
There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B but, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B, then find the number of students in the both halls. |
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Answer» Let the number of students in halls A and B are x and y, respectively. Now, by given condition, `" " x-10=y+10` `rArr " " x-y=20 " " ...(i)` and `" " (x+20)=2(y-20)` `rArr " " x-2y=-60 " " ...(ii)` On subtracting Eq. (ii) from Eq. (i) , we get `(x-y)-(x-2y)=20+60` `x-y-x+2y=80 rArr y = 80` On putting y = 80 in Eq. (i), we get x-80=20`rArr`x=100 and `" "`y=80 Hence, 100 students are in hall A and 80 students are in hall B. |
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| 9. |
The sum of a two–digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? |
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Answer» let tens digite of number is`x` let unit digit be`y` number= `(10x+y)` if no is reversed then unit digit `x` & tens digit `y` reversed number= `(10y+x)` acc to question `(10x+y) + (10y+x)= 66` `11x+11y=66` so, `x+y=6` now `x-y=2` or `y-x=2` case-1 `x-y=2 & x+y=6` by adding both, we get`2x= 8` `x=4` `y=2` so number will be `4*10=2 = 42` case 2 : `y-x=2 & x+y=6` by adding both, we get `2y= 8` `y=4` `x=2` now number will be `24` answer is 24 and 42 |
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| 10. |
A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number. |
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Answer» Let the two-digit number = 10x + y Case I Multiplying the sum of the digits by 8 and then subtracting 5 = two-digit number `rArr " " 8xx(x+y)-5=10x+y` `rArr " " 8x+8y-5=10x+y` `rArr " " 2x-7y=-5 " " ...(i)` Case II Multiplying the difference of the digits by 16 and then adding 3 = two-digit number `rArr " " 16xx(x-y)+3=10x+y` `rArr " " 16x-16y+3=10x+y` `rArr " " 6x-17y=-3 " " ...(ii)` Now, multiplying in Eq. (i) by 3 and then subtracting from Eq. (ii), we get `{:(6x-17y=-3),(ul(underset(-)6x+underset(+)21y=underset(+)-15)),(4y=12rArry=3):}` Now, put the value of y in Eq. (i), we get `2x-7xx3=-5` `rArr 2x=21-5=16 rArr x=8` Hence, the required two-digit number `=10x+y` `=10xx8+380+3=83` |
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| 11. |
Ankita travels 14km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus. On the other hand, if she travel 4 km by rickshaw and the remaining distance by bus, she takes 9 minute longer. Find the speed of the rickshaw and of the bus. |
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Answer» Total distance=14 Km Let x be the speed of bus Let y be the speed of Rickshaw `12/x+2/y=1/2-(1)` `10/x+4/y=39/60-(2)` subtracting equation 1 from 2 `10/x-24/x=-21/60` `14/x=21/60` `x=40 km/h` Putting this value in equation 1 `12/40+2/y=1/2` `2/y=2/10` `y=10 km/h` Speed of bus=x=40 km/h Speed of rickshaw=y=10 km/h`. |
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| 12. |
Draw the graphs of the equations `5x y=5`and `3x y=3`. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis. |
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Answer» 5x-y=5 let x=0,then y=-5 letx=1,then y=0 ploting this point on graph 3x-y=3 let x=0, then y=3 let x=1, then y=-3 ploting these points on graph let this triangle be `/_ABC` we get co-ordinates A(0,-3),B(0,-5),c(1,0) |
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| 13. |
Draw the graphs of the equations `x y+1=0` and `3x+2y 12=0`. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. |
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Answer» `x-y+1 =0->Eq(1)` If we put,` x=0`, then ` y = 1` If we put `y=0` then `x=-1` So, we can draw aline with points (0,1) and (-1,0). Please refer to video for graph. `3x+2y-12=0->Eq(2)` If we put,` x=0`, then ` y = 6` If we put `y=0` then `x=4` So, we can draw aline with points (0,6) and (4,0). Please refer to video for graph. Now, from the graph, w can see the coodinates of the vertices of triangles are, `(-1,0),(2,3),(4,0)`. |
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| 14. |
Find the values of x and y in the following rectangle |
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Answer» By property of rectangle), Lengths are equal, i.e., `" " CD=AB` `rArr " " x+3y=13 " " ...(i)` Breadth are equal, i.e., `" " AD=BC ` `rArr " " 3x+y=7 " " ...(ii)` On multiplying Eq. (ii) by 3 and then subtracting Eq. (i) , we get `{:(9x + 3y = 21),( ul(underset(-)x+underset(-)3y=1underset(-)3)),(8x " "=8):}` `" " x=1` On putting x =1 in Eq. (i), we get `3y=12 rArr y=4` Hence, the required values of x and y are 1 and 4, respectively. |
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| 15. |
Find the solution of the pair of equations `(x)/(10) + (y)/(5) - 1 = 0` and `(x)/(8) + (y)/(6) = 15` and find `lambda`, if `y = lambda x + 5`. |
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Answer» Given pair of equations is `(x)/(10)+(y)/(5)-1=0 " " ...(i)` and `" " (x)/(8)+(y)/(6)=15 " " ...(ii)` Now, multiplying both sides of Eq. (i) by LCM (10,5)=10, we get `x+2y-10=0` `rArr " " x+2y=10 " " ...(iii)` Again, multiplying both sides of Eq. (iv) by LCM (8,6)=24, we get `3x+4y=360 " " ...(iv)` On, multiplying Eq. (iii) by 2 and then subtracting from Eq. (iv), we get `{:(3x + 4y = 360),( ul(underset(-)2x+underset(-)4y=underset(-)2 0)),(" "x=340):}` Put the value of x in Eq. (iii), we get 340+2y=10 `rArr " " 2y=10-340=-330` `rArr " " y=-165` Given that, the linear relation between x, y and `lambda` is `y=lambda x + 5` Now, put the values of x and y in above relation, we get `-165=lambda (340)+5` `rArr " " 340 lambda =-170` `rArr " " lambda=-(1)/(2)` Hence, the solution of the pair of equations is x=340, y=-165 and the required value of `lambda` is `-(1)/(2)`. |
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| 16. |
For which value (s) of `lambda`, do the pair of linear equations `lambdax + y = lambda^(2)` and `x + lambda y = 1 ` have (i) no solution ? (ii) infinitely many solutions ? (iii) a unique solution ? |
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Answer» The given pair of linear equations is `lambdax + y = lambda^(2)` and `x + lambda y = 1` Here, `" " a_(1) = lambda, b_(1) = 1, c_(1) = - lambda^(2)` `a_(2) = 1, b_(2) = lambda, c_(2) = -1` (i) For no solution, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) != (c_(1))/(c_(2))` `rArr " " (lambda)/(1) = (1)/(lambda) != (-lambda^(2))/(-1)` `rArr " " lambda^(2) - 1 = 0` `rArr " " (lambda - 1) (lambda + 1) = 0` `rArr " " lambda = 1, -1` Here, we take only `lambda = -1` because at `lambda = 1` the system of linear equations has infinitely many solutions. (ii) For infinitely many solutions, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` `rArr " " (lambda)/(1) = (1)/(lambda) = (lambda^(2))/(1)` `rArr " " (lambda)/(1) = (lambda^(2))/(1)` `rArr " " lambda (lambda - 1) = 0` when `lambda != 0, ` then ` lambda = 1` For a unique solution, ` (a_(1))/(a_(2))!=(b_(1))/(b_(2)) rArr (lambda)/(1) != (1)/(lambda)` `rArr " " lambda^(2) != 1 rArr lambda != +- 1` So, all real values of `lambda ` except `+-1`. |
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| 17. |
Solve by cross multiplication `x+2y+1=0 `, `2x-3y-12=0` |
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Answer» `x+2y+1 = 0` `2x-3y-12 = 0` Solving these two equations with cross-multiplication method, `x/(-24+3) = y/(2-(-12)) = 1/(-3-4)` `=>x/-21 = -1/7=> x = 3` `=>y/14 = -1/7 => y = -2` |
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| 18. |
solve graphically the following system of equations`2x-y-2=0` `4x-6y-10=0` |
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Answer» Equation of line 1, `2x-y =2` When `y = 0`, then `x = 1`. When `x = 0`,then, ` y = -2`. So, we can draw a line joining points `(1,0) and (0,-2)` to draw equation of line 1. Equation of line 2, `4x-6y =10` When `y = 0`, then `x = 5/2`. When `x = 0`,then, ` y = -5/3`. So, we can draw a line joining points `(5/2,0) and (0,-5/3)` to draw equation of line 2. When we draw these lines on the graph, we find the intersection point is `(1/4,-3/2)`, which is the required solution. So, ` x = 1/4 and y = -3/2` |
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| 19. |
Two straight paths are represented by the equations ` x - 3y = 2` and `-2x + 6y = 5`. Check whether the paths cross each other or not. |
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Answer» Given linear equations are `x-3y-2=0 " " ...(i)` and `-2x+6y-5=0" " ...(ii)` On comparing both the equations with `ax+by+c=0`, we get `a_(1)=1,b_(1)=-3` and `c_(1)=-2" " `[from Eq.(i)] `a_(2)=-2,b_(2)=6` and `c_(2)=-5" " `[from Eq.(ii)] Here, `" " (a_(1))/(a_(2))=(1)/(-2)` `(b_(1))/(b_(2))=(-3)/(6)=-(1)/(2)` and `(c_(1))/(c_(2))=(-2)/(-5)=(2)/(5)` i.e., `" " (a_(1))/(a_(2))=(b_(1))/(b_(2))!=(c_(1))/(c_(2))" "` [parallel lines] Hence, two straight paths represented by the given equation never cross each other, because they are parallel to each other. |
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| 20. |
The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. |
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Answer» `x-7= 7(y-7)` `x-7y+42=0` (1) `x+3= 3(y+3)` `x-3y-6=0` (2) in eqn (1) when `y=0, x=-42` when `x=0, y=6` in eqn (2) when `y=0, x=6` when`x=0, y=-2` the intersection of lines is point` (42,12)` |
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| 21. |
If `m_1` and `m_2` are the roots of the equation `x^2+(sqrt(3)+2)x+sqrt(3)-1=0`, then the area of the `Delta` formed by lines `y=m_1x`, `y=m_2x`, `y=c` is:a.`((sqrt(33)+sqrt(11))/4)c^2`b.`((sqrt(32)+sqrt(11))/16)c`c.`((sqrt(33)+sqrt(10))/4)c^2`d.`((sqrt(33)+sqrt(21))/4)c^3` |
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Answer» `(c/m_1,c),(c/m_2,c)` `b=|c/m_1-c/m_2|=c|1/m_11/m_2|=c|(m_1-m_2)/(m_1m_2)|` `=(csqrt(m_1-m_2)^2)/|m_1m_2|` `=c{(m_1+m_2)^2-m_1m_2}/|m_1m_2|` `=(csqrt((sqrt3+2)^2-4(sqrt3-1))/(sqrt3-1))` `=c*sqrt(3^2+2^2+4sqrt3-4sqrt3+4)/(sqrt3-1)` `=c*sqrt11/(sqrt3-1)` h=c Area of triangle=`1/2*b*h=1/2*c*(sqrt33+sqrt11)/2*c` `=(sqrt33+sqrt11)/4*c^2` option a is correct. |
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| 22. |
Solve the following pair of equations by substitution method : `7x 15 y=2` . . . (1)`x+2y=3` . . . (2) |
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Answer» given that`x+2y=3` `x=3-2y` so, putting it in another equation `7(3-2y)-15y=2` `21-14y-15y=2` `21-2=29y` y=19/29`` `x= 3-2y` `x=3-2(19/29)` `=3-38/29` `= (87-38)/29` `=49/29` value of x & y are`49/29,19/29` |
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| 23. |
Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions:`5x 8y+1=0` . . . (1)`3x-(24)/5y+3/5=0` . . . (2) |
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Answer» `5x-8y+1=0-(1)` `3x-24/5y+3/5=0-(2)` multiplying 5/3 with equation (2) 5x-8y+1=0 So, it is observed that equation of both the lines are same, lines are coincident Both the linnes will have infinite many solution. |
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| 24. |
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water. |
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Answer» let the speed of boat `= x`km/hr speed of stream `=y `km/hr speed of boat downstream =`x+y `km/hr speed of boat upstream `x-y`km/hr time taken `t_1= 30/(x-y)`hrs `t_2 = 44/(x+y)` hrs `t_1 + t_2 = 10`hrs `30/(x-y) + 44/(x+y) = 10` eqn(1) `t_3 = 40/(x-y)`hrs `t_4 = 55/(x+y)` `= 40/(x-y) + 55/(x+y) = 13` eqn(2) let `1/(x-y) = A & 1/(x+y) = B` `30A + 44B = 10` `40A + 55B = 13` `B= (13-40A)/55` `30A + 44((13-40A)/55) = 10` `15A + (22*((13-40A)/55))= 5` `75A + 2(13-40A) = 5` `75A + 26 - 80A = 25` `A=1/5` `B = (13-40*1/5)/55 = (13-8)/55 = 1/11` `1/(x-y) = 1/5` `1/(x+y) = 1/11` `x-y= 5` `x= 5+y` & `x+y =11` `5+y +y =11` `5+2y=11` `y = 6/2=3`km/hr `x = 3+5 = 8`km/hr answer |
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| 25. |
A boat goes 25 km upstream and 35 km downstream in 10 hours. In 15 hours, it can go 40 km upstream and 49 km downstream. Determine the speed of the stream and that of the boat in still water. |
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Answer» Let spped of water is `x` km/h and speed of stream is `y` km/h. Then, speed at upstream `= x-y` km/h Speed at downstream ` = x+y` km/h Case -1 : `25/(x-y)+35/(x+y) = 10` `=>5/(x-y)+7/(x+y) = 2->(1)` Case-2 : `40/(x-y)+49/(x+y) = 15->(2)` Let `1/(x-y) = a, 1/(x+y) = b` So, the equation (1) and (2) becomes, ` 5a+7b = 2->(3)` `40a+49b = 15->(4)` Now, multiplying `(3)` with 7 and subtracting it from (4), `=>40a+49b - 35a -49b = 15-14` `=>5a = 1` `=>a = 1/5` Putting value of `a` in (3), `=>5(1/5) +7b = 2` `=>b = 1/7` `:. x-y = 5 and x+y = 7` `=>x-y+x+y = 5+7` `=>2x = 12 => x = 6` `=> y = 7-6 = 1` `:.` Speed of the stream is `1` km/h and speed of boat is `6` km/h. |
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| 26. |
If `(by+cz)/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)` then prove that each ratio is equal to `x/a = y/b = z/c`. |
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Answer» We are given, `(by+cz)/(b^2+c^2)=(cz+ax)/(c^2+a^2)=(ax+by)/(a^2+b^2)->Eq(1)` Also,`a/b=c/d=e/f`, can be written as `(a+c+e)/(b+d+f)` So, Eq(1) becomes `(by+cz)/(b^2+c^2)=(cz+ax)/(c^2+a^2)=(ax+by)/(a^2+b^2)=(ax+by+cz)/(a^2+b^2+c^2)` Taking, `(by+cz)/(b^2+c^2)=(ax+by+cz)/(a^2+b^2+c^2)` `=>(by+cz)(a^2+b^2+c^2))=(ax+by+cz)(b^2+c^2)` `=>a^2(by+cz)+(b^2+c^2)(by+cz) = ax(b^2+c^2)+(b^2+c^2)(by+cz)` `=>(by+cz)/(b^2+c^2)=x/a-> Eq(2)` Similarly, we can prove that `(cz+ax)/(c^2+a^2) = y/b->Eq(3)` and `(ax+by)/(a^2+b^2) = z/c->Eq(4)` From `Eq(1),Eq(2),Eq(3),and Eq(4)`, we can conclude that, `x/a=y/b=z/c` |
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| 27. |
If the angles of a triangle are x, y and `40^(@)` and the difference between the two angles x and y is `30^(@)`. Then, find the value of x and y. |
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Answer» Given that, x, y and `40^(@)` are the angles of a triangle. `:. " " x+y+40^(@)=180^(@)` `" " `[since, the sum of all the angles of a triangle is `180^(@)`] `rArr " " x+y=140^(@) " " ...(i)` Also, `" " x-y=30^(@) " " ...(ii)` On putting `x=85^(@)` in Eq. (i), we get `" " 85^(@)+y=140^(@)` `rArr " " y=55^(@)` Hence, the required values of x and y are `85^(@)` and `55^(@)`, respectively. |
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