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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
root 12^{3 m-2}=12^{2}, m =? |
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Answer» 12^(3m~2)=12^4then3m~2=4m=2 |
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| 2. |
Solve: 스 + y = a2 + b2 and __ += a + ba ba b2 |
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| 3. |
2-t 2-12-2 |
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Answer» 2+2=44/2=2 answers |
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| 4. |
Find the mean from the following table:410 12 |
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| 5. |
33. यदिx:y=3:5 हो, तो (10x+3y) : (5x+2y) = ?(a) 9:4 |
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Answer» 9:5 is the right answer x/y=3:5now, (10×3+3×5):(5×3+2×5)45:25=9:5 9:5 is correct answer ha bhai 9:5 is the right answer of this question 9:5 is the correct answer (10x+3y):(5x+2y)=(10×3+3×5):(5×3+2×5)=(30+15):(15+10)=45:25=9:5 |
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| 6. |
Solve for x and y(a-b)x +(a +b)y a2a +b)(x +y)9."-2ab-b2(a2+b |
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Answer» Like if you find it useful |
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| 7. |
काHion % eही &M : "orm0t ks3Kfio lowst न |
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Answer» 3+√2/4√2*4√2/4√2=12√2+4*2/16*2=12√2+8/32=3/8√2+1/4 |
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| 8. |
Problem 1.Draw an ellipse when the distance of its focus from its directrix is equal to 50 mm and2eccentricity is .Also draw a tangent and a normal to the ellipse at a point 60 mm from the directrix. |
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| 9. |
"८ So heoky O&L—M; n Indlo Clorare ped !—, GV/(’/O/(O / e 0T W 9065 S |
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Answer» Farmer suicides have been taking place across India for years now, and studies of rural distress reveal the deeply-rooted, tenacious causes, such as lack of irrigation, fragmentation of land, unsuitability of seeds and inadequate sources of credit. Despite the democratically-elected governments that claim to represent a country where over half the population is dependent on farming, agriculture has been consistently ignored at a steep cost to farmers' lives. |
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| 10. |
Prove that: (sin A + cosec A)+ (cos A + sec A)7+ tan'A + cot A29. A person standing on the bankofa river obseryes that the angleofelevationn |
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| 11. |
7. Evaluate 3+3+{)+ (%) after rearrangement7.Evaluate-+ -+-after rearrangement. |
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Answer» 28 /15 is the correct answer 28/15 is Right answer of this question. |
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| 12. |
ng iaentuty and ind thé vaiue oi me1992^2 |
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Answer» 199^2=(200-1)^2=200^2-2*200+(1)^2 =40000-400+1=39601 |
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| 13. |
The angle between A and B is θ. The value of the tri-ple product A . B x Ais(a) AB(c) A2B sin θ(b) zero(d) A2B cos θ. |
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Answer» zero id the correct answer option b |
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| 14. |
बिक 2 b ps& PRl 12 LR b I %0T & Mdk1ed Ehth k2 122k %0T L klsh LR pib pepe 00८1 ४ |
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Answer» 1000 but i'm no sure |
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| 15. |
(ii) a2b® +3ab - 961 + ् दर भ्ठ ik -5_4 & |
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Answer» (1-9)a^2b^2 +(3+5/2)ab + (3/5-4/5)ab^2= - 8a^2b^2 + 11/2ab - 1/5ab^2 |
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| 16. |
a2 b2Solve for x and y:= 0a2b b2a |
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Answer» a^2/x - b^2/y = 0...................(1)a^2b/x + b^2a/y = a + b....... (2) Multiply eq(1) by aa^3/x - b^2a/y = 0.................(3) Add eq(2) and eq(3)a^3/x = a + bx = a^3/(a + b) Put value of x in eq(1), we geta^2 /[a^3/(a + b)] - b^2/y = 0(a + b)/a = b^2/yy = ab^2/(a + b) |
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| 17. |
रपट 0५ -0t Ay |
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Answer» 3y² - 20 = 160 - 2y² 5y² = 180 y² = 180/5 y² = 36 y = √(36) y = -6 and 6 |
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| 18. |
0T W) + 408 WAZ = (0 सैर शक |
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Answer» According to the trigonometric identity, tan70 = tan(20+50)tan70 = (tan20+tan50)/ 1-tan20tan50tan70 - tan20tan50tan70 = tan20 + tan50 Also, tan70tan20 = tan70cot70 = 1 Hence, tan70 - tan50 = tan20 + tan50 Therefore, tan70 = tan20 + 2tan50 |
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| 19. |
Find perpendicular distance from the origin ofthe line joining the points (cosa, sin θ)and (cos φ, sin φ). |
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| 20. |
9 500 + 0t usLR |
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Answer» sin30+cos60=1/2+1/2=2/2=1 |
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| 21. |
So,Fig. 6.31Substituting 2 450°That is,1279R DI Recall that youthe earlier clas |
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Answer» y = 127° -50° = 77° In ∆ PQR, PRQ = 180° - 127° = 53°x + y + 53° = 180°x = 180° -53° - 77°x = 50° |
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| 22. |
Solve 5x-3y=1and 2x+5y=19 by substituting method |
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Answer» 5x-3y =1 and 2x - 5y =19 5x = 1+3yx = 1+3y / 5putting the value in 2nd equation2[1+3y / 5} + 5y = 192+ 6y + 25y / 5 = 19 2+ 31y = 9531y = 95 - 231 y = 93y =3 and x = 1+3y/5 = 1 + 3{3} /5 = 1 + 9 /5 = 10/5 =2so x = 2 and y =3 |
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| 23. |
] URltek] B Lo OF 1% *" $ L 0T : AV |
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Answer» a = 10,d = -3, an = a + (n-1) × da10 = 10 + 9×(-3)a10 = 10 + (-18)a10 = -8 Ans. oh Maine 10va pad gyaat Kar Diya |
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| 24. |
कार्तीय तल में स्थित किसी बिन्दु (3, -4) के कोटी का मान है-The value of ordinate of the point (3,-4) in a Cartesian plane is -B) 3|A) |
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Answer» Ordinate is the value of y coordinateOrdinate = -4 |
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| 25. |
OR)The distance of the point (-3, 4) from x-axis is |
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Answer» Distance will be |
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| 26. |
Find the centroid of the triangle whose vertices are given below.(4,7),(8,4),(7,11) |
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Answer» centroid of the triangle is given by(x1+x2+x3/3,y1+y2+y3/3)hence(4+8+7/3,7+4+11/3)(19/3,22/3) |
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| 27. |
Or find the centroid of a whose vertices are (3-5) (-7, 4) and 10,-2) |
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Answer» centriod = (3×-7×-2)+(-5×4×-2)=(21×-2)+(-20×2)=(-42+-42=0 |
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| 28. |
Costprice(in)Selling Profitprice or(in ) LossHowmuch?1.450050002410040903.7007991000920 |
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Answer» 1. Loss : 5002. Loss : 103. Gain : 994. Loss : 80 |
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| 29. |
Find the coordinates of the centroid of a triangle whose v(0,6),(8,12) and(8,0).of 1 2cot2 θ + 2) |
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Answer» X1+x2+x3/3=centroid x of triangle0+8+8/3=16/3x of centroid is16/3 y1+y2+y3/3=centoid y6+12+0/3=18/3centroid y =6 The point = (16/3,6) Like my answer if you find it useful! |
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| 30. |
\left. \begin{array} { c } { b + \frac { 7 + 3 \sqrt { 5 } } { 3 + \sqrt { 5 } } - \frac { 7 - 3 \sqrt { 5 } } { 3 - \sqrt { 5 } } = a + \sqrt { 5 b } } \\ { [ 2 ] } \end{array} \right. |
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Answer» (7+3root(5))/(3+root(5))-(7-3root(5))/(3-root(5))=((21-7root(5)+9root(5)-15)-(21+7root(5)-9root(5)-15))/(9-5)=(6+2root(5)-6+2root(5))/(9-5)=4root(5)/4=root(5) so a=0 b=1 |
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| 31. |
7+3 b^{2} a-4 a^{2} b \text { from } a^{2} b-2 b^{2} a |
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Answer» friend thank you please give me your mobile number |
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| 32. |
\left. \begin{array} { l } { \frac { 3 + \sqrt { 7 } } { 3 - \sqrt { 7 } } = a + b \sqrt { 7 } \text { then } ( a , b ) = } \\ { ( A ) ( 8 , - 3 ) } \end{array} \right. |
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Answer» thank you very much for helping me😄😄😄😄 |
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| 33. |
Find the Co-ordinates of a centroid of a triangle whose vertices are (3, -5), (-7, 4) and (10, -2) |
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| 34. |
Find the distance of the point (1,-2,3) from the plane x-y+Z= 5 measured parallel to the line \begin { equation } \frac{x}{2}=\frac{y}{3}=\frac{z}{-6} \end { equation } |
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| 35. |
\begin{array} { c } { \text { 8. 49. Find the distance of the point } ( - 2,3 , - 4 ) \text { from the } } \\ { \frac { x + 2 } { 3 } = \frac { 2 y + 3 } { 4 } = \frac { 3 z + 4 } { 5 } } \\ { \text { measured parallel to the plane } 4 x + 12 y - 5 z + 1 = 0 } \end{array} |
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| 36. |
9. The distance ofthe point (3,-4) from the x-axis is |
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Answer» plzz can you say this ans |
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| 37. |
2. Where does the point (-3, 4) lie in the coordinate plane? |
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Answer» It lies in second quadrantplease like the solutionand check previous Answer as well |
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| 38. |
Find the distance of the point (-3, 4) from the x-axis. |
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| 39. |
Q.2) Find the distance of the point (-3, 4) from the origin. |
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Answer» wrong answer |
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| 40. |
e) For what value of x, is Il m25x+350> m450 |
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Answer» 180-45° = 5x+35 ( corresponding angle)135 = 5x+35100 = 5xx = 20 |
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| 41. |
7.Solve the pair of linear equations by the substituting method only4x + 5y = 23 and 2x + 3y = 13. |
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Answer» the linear pair of questions about its answer 4x+5y=235y= 23-4xy= 23-4x/5 2x+3y=132x+3(23-4x)/5=1310x+69-12x/5=13-2x+69= 65-2x= 65-69-2x=-4x=4/2=2y= 23-4x/5 = 23-4×2/5= 23-8/5=15/5=3 589974igfjoiGvkofp00 n=2 is the correct answer 4x + 5y = 23. 23=4x +5y. 23=9xy. 23+9=xy. =32xy 2)2x + 3y=13 13=2x + 3y. 13=5xy 13+5= xy =18xy 4x+5y=23; 2x+3y=13; 2(2x+3y=13)=4x×6y=26; 4x+5y=23/y=3; 4x+5(3)=23; 4x=23-15=8; x=8/4=2 |
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| 42. |
solve x+ y=3 and 4x-3y=26 using substituting method |
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| 43. |
The cost price and selling price are 1given in the following table. Find outwhether there was a profit or a lossand how much it was.Cost Selling Profit HowEx. priceprice or much?(in ) (in ) Loss1.450050002.409041007007999201000 |
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Answer» 1. loss Rs 500. 2 profit Rs 10. 3. loss Rs99 4.profit Rs 80 In 1. he got profit of 500 2. loss of 10 3. profit of 99 4.loss of 80 1. profit of 5002. loss of 103. profit of 994. loss of 80 five hundred nineteen rupees profit five hundred nineteen rupees profit 1loss500. 2loss is 10..3 profit 99..4loss 80 a.) profit 500b.)loss 10c.) loss 99d.) loss 80 1.profit=₹5002.loss=₹103.profit=₹994.loss=₹80 |
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| 44. |
MPLE 1. Divide:ution We have: |
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Answer» 8x^2y^3/-2xy=-4xy^2 |
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| 45. |
21. Find the equations of the altitudes of a AABC, whose vertices areA(2,-2), B(1,1) andE(-1, o, |
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Answer» No |
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| 46. |
17. Find the lengths of the medians of AABC whose vertices are A(7,-3), B(5, 3) andC(3,-1) |
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| 47. |
Median of a triangle divides it to two triangle ofequal area, verify this result for AABC whosevertices are A(4,-6), B (3,-2). C5, 21(b) |
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| 48. |
Find the distance of the line 3x + y + 4= 0 from the point (2,5) measured parallel to the line 3x - 4y8 = 0.Res |
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Answer» let P (2,5) be the point from which we are measuring the distance .the equation of the line passing through the point P and having slope 3/4 is: useful and harmful bacteria |
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| 49. |
27. A(6, y), B- 4, 4) and C(x, -1) are the vertices of AABC whose centroid is the origin. Cthe values of x and y |
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| 50. |
Find equation of the line parallel to the line 3x-4y+2=0and passing through the point (-2, 3). |
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