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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 ladder of length 20 feet rests against a smooth vertical wall. The lowerend, which is on a smooth horizontal floor is moving away from the wall at the rate of 4 feet/sec. If the lower end is 12 feet away from the wall, then the rate at which the upper end move, isA. `(16)/(3)" ft/sec"`B. `(-16)/(3)" ft/sec"`C. 3 ft/secD. `"-3" ft/sec"` |
| Answer» Correct Answer - D | |
| 2. |
The total cost `C(x)` of producing `x` items in a firm is given by `C(x)=0.0005x^3-0.002x^2+30x+6000` Find the marginal cost when `4` units are produced |
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Answer» Given: `C(x) = 0.005 x^(3) - 0.02 x^(2) + 30x + 6000` `rArr MC = (dC)/(dx)` `= (d)/(dx) (0.005 x^(3) - 0.02x^(2) + 30x + 6000)` `= {(0.005 xx 3x^(2)) - (0.02 xx 2x) + 30}` `rArr [MC]_(x = 4) = {(0.005 xx 3 xx 4^(2)) - (0.02 xx 2 xx 4) + 30}` `= (0.24 - 0.16 + 30) = 30.08` Hence, the required marginal cost is Rs 30.08 |
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| 3. |
Oil is leaking at the rate of 16 mL/s from a vertically kept cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm, find the rate at which the level of the oil is changing when the oil level is 18 cm |
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Answer» Correct Answer - `(16)/(49pi) cm//s` `V = (pi xx 7 xx 7 xx h) rArr (dV)/(dt) = 49pi.(dh)/(dt) rArr 49pi.(dh)/(dt) = 16 rArr (dh)/(dt) = (16)/(49pi) cm//s` |
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| 4. |
A ladder is inclined to a wall making an angle of 30° with it. A man is ascending the ladder at the rate of 3 m/sec. How fast is he approaching the wall ? |
| Answer» Correct Answer - `1.5 m//sec` | |
| 5. |
A square plate is contracting at the uniform rate of `2 cm^(2)//sec`. If side fo the square is 16 cm long, then the rate of decrease of its perimeter isA. `(1)/(2) cm//sec.`B. `(-1)/(2) cm//sec.`C. `(1)/(4) cm//sec`D. `(-1)/(4) cm//sec` |
| Answer» Correct Answer - C | |
| 6. |
A man of height 160 cm walks at a rate of 1.1` m/sec from a lamp post which 6m high. Find the rate at which the length of his shadow increases when he is 1m away from the lamp post. |
| Answer» Correct Answer - `0.4 m//sec` | |
| 7. |
The side of a square in increasing at the rate of 0.2 cm/sec. Find therate of increase of the perimeter of the square. |
| Answer» Correct Answer - `(dP)/(dt) = 0.8 cm//s` | |
| 8. |
If the radius of a circle is 2 cm and is increasing at the rate of 0.5 cm/sec., then the rate of increase of its area isA. `pi cm^(2)//sec`B. `2pi cm^(2)//sec`C. `3pi cm^(2)//sec`D. `4pi cm^(2)//sec` |
| Answer» Correct Answer - B | |
| 9. |
A bullet is shot horizontally and its distance s cms at time `t` sec is given by `s=1200t-15t^(2)` then the distance covered with which the bullet is shot when it comes to the rest isA. 48000 cms.B. 24000 cmsC. 4800 cmsD. 2400 cms |
| Answer» Correct Answer - B | |
| 10. |
A particle moves along the curve `6y = x^3 + 2`.Find the points on the curve at whichy-co-ordinate is changing 8 times as fast as thex-co-ordinate. |
| Answer» Correct Answer - `(4,11),(-4,(-31)/3)` | |
| 11. |
A particle moves along the curve `6y=x^3+2`. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. |
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Answer» Let the required point be (x, y) Given: `(dy)/(dt) = 8 (dx)/(dt)`...(i) Given curve is `6y = x^(3) + 2` ...(ii) On differentiating both sides of (ii) w.r.t. t, we get `6(dy)/(dt) = 3x^(2) (dx)/(dt)` `rArr 6(8 (dx)/(dt)) = 3x^(2) (dx)/(dt)` [using (i)] `rArr 3x^(2) = 48 rArr x^(2) = 16 rArr x = +-4` Putting x = 4 in (ii), we get y = 11 Putting `x = -4` in(ii), we get `y = (-62)/(6) = (-31)/(3)` Hence, the required points are (4,11) and `(-4, (-31)/(3))` |
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| 12. |
The radius of a circle is increasing at the rateof 0.7 cm/sec. What is the rate of increase of its circumference? |
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Answer» Let r be the radius and C be the circumference of circle. Given `(dr)/(dt) = 0.7 cm//sec` Now ` C = 2 pi r` `rArr (dC)/(dt) = 2pi (dr)/(dt)` `-2pi xx 0.7 = 1.4 pi cm//sec` |
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| 13. |
A stone is dropped into a quiet lake and waves move in circles at a speed of 3.5 cm per second. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing ? (Take `pi = 22//7`) |
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Answer» Correct Answer - `165 cm^(2)//s` `A = pi r^(2) rArr (dA)/(dt) = 2pir. (dr)/(dt) = (2pi r xx 3.5) = 7 pi r` `rArr [(dA)/(dt)]_(r = 7.5) = (7pi xx 7.5) cm^(2)//s = 165 cm^(2)//s` |
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| 14. |
A stone isdropped into a quiet lake and waves move in circles at the speed of 5 cm/s.At the instant when the radius of the circular wave is 8 cm, how fast is theenclosed area increasing? |
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Answer» Let r be the radius and A be the area of the wave. Given `(dr)/(dt) = 5 cm//sec` Now, `A= pi r^(2)` `rArr (dA)/(dt) = 2 pi r (dr)/(dt)` at r = 8 ` (dA)/(dt)= 2pi xx 8xx5 = 80 pi cm^(2)//sec` |
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| 15. |
A particle moves along the curve `6y = x^3 + 2`.Find the points on the curve at whichy-co-ordinate is changing 8 times as fast as thex-co-ordinate.A. (4,11)B. (4,-11)C. (-4,11)D. (-4,-11) |
| Answer» Correct Answer - A | |
| 16. |
An edge of a variable cube is increasing at therate of 3 cm per second. How fast is the volume of the cube increasing whenthe edge is 10 cm long? |
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Answer» Let x be the edge of cube and V be the volume. Given ` (dx)/(dt) = 3 cm//sec` Now ,` V=x^(3) rArr (dV)/(dt) = 3x^(2)(dx)/(dt)` at x = 10 cm ` (dV)/(dt) = 3 xx 10^(2) xx 3 = 900 cm^(3)//sec` |
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| 17. |
The edge of a cube is decreasing at the rate of 0.04 cm/sec. If the edge of cube is 10 cm, then the rate of decrease of its surface area isA. `0.6 cm^(2)//sec`B. `1.2 cm^(2)//sec`C. `2.4 cm^(2)//sec`D. `4.8 cm^(2)//sec` |
| Answer» Correct Answer - D | |
| 18. |
A spherical snow ball is melting so that its volume is decreasing at the rate of 8 cc/sec. Find the rate at which its radius is decreasing when it is 2 cm.A. `(1)/(pi)` cm/secB. `(1)/(2pi` cm/secC. `pi` cm/secD. `2pi` cm/sec |
| Answer» Correct Answer - B | |
| 19. |
The volume of spherical ball is increasing at the rate of `4pi "cc/sec"`. If its volume is `288 pi" cc",` then the rate of change of its radius isA. `(pi)/(36)` cm/secB. `(4pi)/(6)` cm/secC. `(1)/(36)` cm/secD. `(1)/(6)` cm/sec |
| Answer» Correct Answer - C | |
| 20. |
The equation of tangent to the curve `x=a sec theta, y =a tan theta" at "theta=(pi)/(6)` isA. 2x-y=3aB. 2x+y=3aC. `2x-y=sqrt(3)a`D. `2x+y=sqrt(3)a` |
| Answer» Correct Answer - C | |
| 21. |
Find the maximum area of an isosceles triangle inscribed in the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`with its vertex at one end of the major axis. |
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Answer» Let any point on the ellipse `x^(2)/a^(a)+y^(2)/b^(2) = 1" be "P (a cos theta, b sin theta)`. ` (##NTN_MATH_XII_C06_E04_270_S01.png" width="80%"> Draw `PM bot OX` from P and produce it to meet the ellipse at Q, then `Delta APQ` is an isosceles triangle, let its area be S, then `S = 2 xx 1/2 xx AM xx MP` ` = (OA - OM) xx MP` ` = (a - a cos theta)* b sin theta` ` rArr S = ab (sin theta-1/2 sin 2 theta)` ` = ab ( sin theta - 1/2 sin 2 theta)` ` rArr (dS)/(d theta) = ab (cos theta- cos 2 theta)` Again differentiate w.r.t.` theta`, `(d^(2)S)/(d theta^(2)) = ab ( - sin theta + 2 sin 2 theta)` For maxima/minima ` (dS)/(d theta) = 0 ` ` rArr cos theta = cos 2 theta rArr 2 theta = 2 pi - 0 rArr theta =(2pi)/3 at theta=(2pi)/3`, `((d^(2)S)/(d theta^(2)))_(theta=(2pi)/3)= ab [-sin(2pi)/2 + 2 sin (2 xx(2pi)/3)]` ` = ab (-sqrt3/2 - (2sqrt3)/2)=ab ((-3sqrt3)/2)` ` = (-3sqrt3 ab)/2 lt 0 ` ` :." S is maximum at "theta = (2pi)/3`, and maximum value of S `= ab (sin (2pi)/3 - 1/2*2 sin.(2pi)/3cos. (2pi)/3)` ` = ab (sqrt3/2 + sqrt3/2 xx 1/2)` `= ab ((2sqrt3+sqrt3)/4)= (3sqrt3)/4` ab sq. units . |
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| 22. |
If `y=sinx`and `x`changes from `pi//2`to `22//14`, what is theapproximate change in `y`? |
| Answer» Correct Answer - No change | |
| 23. |
The equation of normal to the curve `y=sqrt(2) sin (2x+(pi)/(4))" at "x=(pi)/(4)` isA. `x+2y+2-(pi)/(4)=0`B. `x+2y-2+(pi)/(4)=0`C. `x-2y+2-(pi)/(4)=0`D. `x-2y-2+(pi)/(4)=0` |
| Answer» Correct Answer - C | |
| 24. |
The equation of normal to the curve `x= a cos^(3) theta, y=a sin^(3) theta" at "theta=(pi)/(4)` isA. x-y=0B. x+y=0C. `x-y=asqrt(2)`D. `x+y=asqrt(2)` |
| Answer» Correct Answer - A | |
| 25. |
The equation of tangent to the curve `y=sqrt(2) sin (2x+(pi)/(4))" at "x=(pi)/(4)` isA. `2x-y-(pi)/(2)-1=0`B. `2x+y-(pi)/(2)-1=0`C. `2x+y+(pi)/(2)+1=0`D. `2x-y+(pi)/(2)+1=0` |
| Answer» Correct Answer - B | |
| 26. |
The co-ordinates of the point of the curve `y=x-(4)/(x)`, where the tangent is parallel to the line y=2x isA. (2,2)B. (-2,-2)C. `(pm 2,0)`D. (0,2) |
| Answer» Correct Answer - C | |
| 27. |
The equation of tangent to the curve `x=a cos^(3) theta, y= a sin^(3) theta" at "theta=(pi)/(4)` isA. `sqrt(2)x+sqrt(2)y+a=0`B. `sqrt(2)x-sqrt(2)y-a=0`C. `sqrt(2)x-sqrt(2)y+a=0`D. `sqrt(2)x+sqrt(2)y-a=0` |
| Answer» Correct Answer - D | |
| 28. |
The equation of tangent to the curve `y=4 xe^(x)` at `(-1,(-4)/(e ))` isA. x=0B. y=0C. `x=(-4)/(e )`D. `y=(-4)/(e )` |
| Answer» Correct Answer - D | |
| 29. |
Find the equation of tangent of the curve `y^(2) = 4x+5` which is parallel to the line ` 2x-y=5`. |
| Answer» Correct Answer - y = 2x + 3 | |
| 30. |
Find the equation of the normal to the curve `x^2=4y`which passes through the point (1, 2). |
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Answer» Equation of given curve is : `y^(2) = 4x` ` rArr 2y(dy)/(dx) = 4 rArr (dy)/(dx) = 4/(2y) = 2/y` `:." slope of tangent at point "(1, 2) is ((dy)/(dx))_(1, 2) = 2/2 = 1` Slope of normal at point `(1, 2), = - 1/1 =- 1` ` :. ` equation of normal at point (1, 2), ` y-2 =- 1(x-1)` ` rArr y-2=-x+1 rArr x+y-3=0` |
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| 31. |
Find the maximum andminimum values of `f(x)=sinx+1/2cos2x`in `[0, pi//2]`. |
| Answer» Correct Answer - max. value is `(3)/(4) " at " x = (pi)/(6)` and min. value is `(1)/(2) " at " x = (pi)/(2)` | |
| 32. |
The equation of normal to the curve `sqrt(x)-sqrt(y)=1` at (9,4) isA. 3x-2y+35=0B. 3x+2y-35=0C. 3x-2y-35=0D. 3x+2y+35=0 |
| Answer» Correct Answer - B | |
| 33. |
The equation of tangent to the curve `sqrt(x)-sqrt(y)=1` at (9,4) isA. 2x+3y=6B. 2x+3y+6=0C. 2x-3y+6=0D. 2x-3y=6 |
| Answer» Correct Answer - D | |
| 34. |
Find the equation of the tangent to the curve `x^(2) + 3y = 3`, which is parallel to the line `y - 4x + 5 = 0` |
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Answer» Correct Answer - `4x - y + 13 = 0` `x^(2) + 3y = 3 rArr 2x + 3 (dy)/(dx) = 0 rArr (dy)/(dx) = (-2x)/(3)` `y - 4x + 5 = 0 rArr y = 4x + 5 rArr` its slope = 4 `:.` the slope of the tangent = 4 So, `(-2x)/(3) = 4 rArr x = -6` `x^(2) + 3y = 3, x = -6 rArr y = -11` The equation of the tangent at `(-6, -11)` with slope = 4 is `y + 11 = 4 (x +6)` |
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| 35. |
Consider `f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot`A normal to `y=f(x)`at `x=pi/6`also passes through the point:(1) (0, 0)(2) `(0,(2pi)/3)`(3) `(pi/6,0)`(4) `(pi/4,0)`A. (0,0)B. `(0,(2pi)/(3))`C. `((pi)/(6),0)`D. `((pi)/(4),0)` |
| Answer» Correct Answer - B | |
| 36. |
The equation of the normal to the curve `y=sqrt(x-3)`, which is parallel to the curve `6x+3y-4=0` areA. 2x+y+9=0, 2x+y+7=0B. 2x-y+9=0, 2x+y+7=0C. 2x+y-9=0, 2x+y-7=0D. 2x-y-9=0, 2x+y-7=0 |
| Answer» Correct Answer - C | |
| 37. |
The equation of normal to the curve `2x^(2)+3y^(2)-5=0` at (1,1) isA. 3x-2y-1=0B. 3x-2y+1=0C. 3x+2y-1=0D. 3x+2y+1=0 |
| Answer» Correct Answer - A | |
| 38. |
The points on the curve `y=sqrt(x-3)`, where the tangent is perpendicular to the line 6x+3y-5=0 areA. `(4, pm 1)`B. `(4, pm 2)`C. `(5, sqrt(2))`D. (7,2) |
| Answer» Correct Answer - A | |
| 39. |
The equation of tangent to the curve `2x^(2)+3y^(2)-5=0` at (1,1) isA. 2x+3y+5=0B. 2x-3y-5=0C. 2x-3y+5=0D. 2x+3y-5=0 |
| Answer» Correct Answer - D | |
| 40. |
Find the equation of the tangents to the curve `2x^(2) + 3y^(2) = 14`, parallel to the line `x + 3y = 4` |
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Answer» Correct Answer - `x + 3y = 7, x + 3y = -7` The given lies is `x + 3y = 4 rArr y = - (1)/(3) x + (4)/(3)` Slope of tangent = slope of given lines `= (-1)/(3)` Let the point of contact be `(x_(1), y_(1))` Now, `2x^(2) + 3y^(2) = 14 rArr 4x + 6y (dy)/(dx) = 0 rArr (dy)/(dx) = (-4x)/(6y) = (-2x)/(3y) rArr ((dy)/(dx))_((x_(1)"," y_(1)) = (-2x)/(3y_(1))` `:. (-2x_(1))/(3y_(1)) = (-1)/(3) rArr 2x_(1) = y_(1)` Also, `(x_(1), y_(1))` lies on the given curve `:. 2x_(1)^(2) + 3y_(1)^(2) = 14 rArr 2x_(1)^(2) + 3 xx 4x_(1)^(2) = 14 rArr x_(1)^(2) = 1 rArr x_(1) = +- 1` `(x_(1) = 1 rArr y_(1) = 2) and (x_(1) = -1 rArr y_(1) = -2)` So, the points of contact are (1, 2) and `(-1, -2)` The respective tangents are `(y -2)/(x -1) = (-1)/(3) adn (y +2)/(x +1) = (-1)/(3)` `rArr 3y - 6 = -x + 1 and 3y + 6 = -x - 1 rArr 3y + x = 7 and 3y + x = -7` |
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| 41. |
The normal to the curve `x^(2)+2xy-3y^(2)=0,` at (1,1)A. meets the curve again the third quadrantB. meets the curve again in the fourth quadrantC. does not meet the curve againD. meets the curve again in the second quadrant |
| Answer» Correct Answer - B | |
| 42. |
Find the equation of normal to the curve `y(x-2)(x-3)-x+7=0` at that point at which the curve meets X-axis. |
| Answer» Correct Answer - `20x+y - 140= 0` | |
| 43. |
Find the equation of tangent of tangent of the curve y = `b * e^(-x//a)` at that point at which the curve meets the Y-axis. |
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Answer» `y=b*e^(-x//a) …(1)` At Y-axis, x=0 `:. Y=be^(0)= b` So, we will find the equation of tantion of tangent at point (0, b). Now differentiare eq. (1) with respect to x. `(dy)/(dx) =-b/a * e^(-x//a)` Slope of tnagent at point (0, b) is `m =-b/a * e^(0)=-b/a` and equation of tangent ` y-b=-b/a*(x-0)` `rArr ay - ab=-bx` `rArr bx + ay = ab`. |
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| 44. |
If the tangent at `(1,1)`on `y^2=x(2-x)^2`meets the curve again at `P ,`then find coordinates of `Pdot`A. (4, 4)B. (2, 0)C. (9/4, 3/8)D. `(3,3^(1//2))` |
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Answer» Correct Answer - C `2y(dy)/(dx)=(2-x)^(2)-2(2-x)` `"So "(dy)/(dx):|_("(1,1)")``=-(1)/(2)` Therefore, the equation of tangent at (1,1) is `y-1=-(1)/(2)(x-1)` `rArr" "y=(-x+3)/(2)` The intersection of the tangent and the curve is given by `(1//4)(-x+3)^(2)=x(4+x^(2)-4x)` `rArr" "4x^(3)-17x^(2)+22x-9=0` `rArr" "(x-1)(4x^(2)-13x+9)=0` `rArr" "(x-1)(4x^(2)-13x+9)=0` `rArr" "(x-1)^(2)(4x-9)=0` Since x = 1 is already the point of tangency, `x=9//4` and `y^(2)=(9)/(2)(2-(9)/(4))^(2)=(9)/(64)`. Thus, the required point is `(9//4, 3//8)`. |
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| 45. |
The tangent at any point on the curve `x = at^3. y = at^4` divides the abscissa of the point of contact in the ratio m:n, then `|n + m|` is equal to (m and n are co-prime)A. `1//4`B. `3//4`C. `3//2`D. `2//5` |
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Answer» Correct Answer - B `(dy)/(dx)=(4t)/(3)` Tangent is `y-at^(4)=(4t)/(3)(x-at^(3))` x-intercept`=(at^(3))/(4)` y-intercept `=(at^(4))/(3)` the point of intersection of tangent with the axes are `((at^(3))/(4),0) and (0,-(at^(4))/(3))` `therefore" "(m)/(n)=-(3)/(4)` |
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| 46. |
A spherical balloon is filled with 4500p cubic meters ofhelium gas. If a leak in the balloon causes the gas to escape at the rate of `72pi`cubic meters per minute, then the rate (in metersper minute) at which the radius of the balloon decreases 49 minutes after theleakage began is(1) `9/7`(2) `7/9`(3) `2/9`(4) `9/2`A. `(2)/(9)` meters/minB. `(-2)/(9)` meters/minC. `(4)/(9)` meters/minD. `(-4)/(9)` meters/min |
| Answer» Correct Answer - A | |
| 47. |
Find the maximum and minimum values of `2x^3-24x+107` on the interval `[-3, 3]` |
| Answer» Correct Answer - max. value is 139 at x `= -2` and min. value is 89 at x = 3 | |
| 48. |
Find the points oflocal maxima or minima and corresponding local maximum and minimum values of `f(x)=x^3-6x^2+9x+15`. Also, find the pointsof inflection, if any: |
| Answer» Correct Answer - local max. value is 19 at x = 1 and local min. value is 15 at x = 3 | |
| 49. |
Divide 15 into two parts such that product of square of one part and cube of other is maximum |
| Answer» Correct Answer - 6,9 | |
| 50. |
Divide a into two parts such that the product of the pth power of one part and the qth power of the second part may be maximum |
| Answer» Correct Answer - `(ap)/(p + q), (aq)/(p + q)` | |