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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. |
Find the slope of a line 2x-3y+6=0? |
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Answer» Given :2x-3y+6=03y=2x+6Dividing throughout by 3y=2/3x+2it is of form y=mx+chere m = slope= 2/3 2x-3y-6=02x-y =6x-y= -2/6 |
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| 2. |
Find the slope of the line 2x + 3y-4 0 |
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| 3. |
astne-beose3. Find the slope of line 2x+3y 8. |
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| 4. |
Determine the ratio in which the line 2x+y-4 0 divides the line segment joining thepoints A(2-2) and B (3,7). |
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| 5. |
Q24. Determine the ratio in which the line 2x + y-4segment joining the points A(2,-2) and B(3,7).divides the line |
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| 6. |
9-3x=3 |
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Answer» Given:9-3x=39-3=3x6=3xx=6/3x=2 X=2 is the correct answer of the given question x=2 is the answer X = 2 |
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| 7. |
CONSULULLLLLS6. A flat costs 36% more today than when it was built. If the originaln when it was built the original cost of the law19,00,000, find its price today.come |
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| 8. |
a ball of length 10 m was to be built across an open ground the height of the wall is to be built up with bricks whose dimensions are 24cm by 12cm by 8 cm how many bricks would be required ? |
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| 9. |
I. Used to repair body cells |
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Answer» Protein. Your body uses the amino acids in dietary protein to form other proteins, build new body cells and generate energy for cellular functions. ... Fats. ... Vitamin C. |
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| 10. |
(3 Ex—zx धन टब्नध |
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Answer» (3x^-3)^2 * (2x^3)^4 ÷ (12x^4)^2= 9x^[-6] * 8x^[12] ÷ 144x^[8]= (72/144)*x^[-6+12-8]= 1/2*x^-2= 1/2x^2 |
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| 11. |
5 [__I‘__j\zx+I e é |
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Answer» y= tan^-1[{√(1+x^2)-1}/x]. Putting x=cot(a) y= tan^-1[{√(1+cot^2a)-1}/cota}]. y= tan^-1[{ 1/sina -1}×sina/cosa}]. y = tan^-1[(1-sina)/cosa ] y=tan^-1[(cos^2a/2+sin^2a/2–2.sina/2.cosa/2)/(cos^2a/2-sin^a/2)] y= tan^-1[(cosa/2-sina/2)^2/(cosa/2+sina/2)(cosa/2-sina/2)] y= tan^-1[(cosa/2-sina/2)/(cosa/2+sina/2)] Dividing in Nr and Dr by cosa/2 y= tan^-1[(1-tana/2)/(1+tana/2)] y= tan^-1[(tanπ/4-tana/2)/(1+tanπ/4.tanA/2)] y=tan^-1 tan(π/4-a/2) y = π/4 -a/2 y= π/4 -1/2.cot^-1(x) dy/dx = 0 -1/2.[-1/(1+x^2)]. dy/dx = 1/{2(1+x^2)}. Answer. |
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| 12. |
X0« Xकान s —/\Zx +71 it |
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| 13. |
Simplify 108 of 5 +Zx 87 |
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| 14. |
A shopkeeper sold two electronic gadgets for 44000 each. The shopkeeper made aoss of 12% on one and a profit of 10% on the other. Find his overall gain or loss(Hint. Find C.P. of each) |
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Answer» here is a solution to a similar question. A shopkeeper sold two fans for rs 990 each. On one he gains 10% and on the other he loses 10%. Calculate his gain or loss per cent in the whole transaction C.P of first fan =( 990×100)÷110= rs900. C.P of second fan =(990×100)÷90=rs1100 Total C.P. = 1100+900 =2000. Total S.P = 990×2=1980 Loss = 2000-1980 =rs20 Loss%= (20×100)÷2000= 1 % |
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| 15. |
A ietal parallelopiped of measures 16 cm x 1lem x 10 cm was melted tomoke coins. How many coins were mode if the thickness and diameter of eachcoin was 2 mm and 2 cm respectively ?s4 cm resnectively. To level the |
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Answer» Given : A metal parallelepiped of measures 16 cm ×11 cm ×10cm was melted to make coins. To Find : How many coins were made if the thickness and diameter of each coin was 2mm and 2 cm respectively Solution : Volume of metal piped = 16 cm ×11 cm ×10cm =1760 cubic cm Now Thicknes of coin = 2mm 1 mm = 0.1 cm 2 mm = 0.2 cm Thus the thickness(h) of coin is 0.2 cm Radius of coin(r) = diameter /2 =2/2 =1 cm Since coin is in cylindrical shape . So, volume of coin = πr²h use π =3.14 = 3.14*1²*0.2 =0.628 cubic cm Thus no. of coins were made = volume of pipe/volume of 1 coin =1760/0.628 =2802.54 Thus no. of coins were made is 2802 like my answer if you find it useful! |
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| 16. |
x+2 ) x^{3}+3 n^{2}+4 n+5 |
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| 17. |
\frac { x ^ { 7 + 2 n } \times ( x ^ { 2 } ) ^ { 3 n + 2 } } { ( x ^ { 4 } ) ^ { 2 n + 3 } } |
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Answer» maybe you'll understand |
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| 18. |
x = \frac { x _ { 1 } + x _ { 2 } + x _ { 3 } + \ldots x _ { n } } { N } |
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| 19. |
\operatorname { lim } _ { n \rightarrow 0 } \frac { ( x - 3 ) ^ { n } } { n + 2 } |
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Answer» when n →0 the value will be (-3)^0/(0+2) = 1/2. |
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| 20. |
EXERCISE 4.11. Check whether the following are quadratic equations)r3(2r+1)- x(r+ 5).mx2 + 3x +1=(x-2)2(viii) x3-4E-x + I-(x-2)3(iii)(x-2)(x + I )-(x-1 )(r + 3)2. Represent the following situations in the form of quadratic equations() The area of a rectangular plot is 528 m2. The length of the plot (in metres) is onemore than twice its breadth. We need to find the length and breadth of the plot. |
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| 21. |
Example 2: An electrician has to repair an electricfault on a pole of height 5 m. She needs to reach apoint 1.3m below the top of the pole to undertake therepair work (see Fig. 9.5). What should be the lengthof the ladder that she should use which, when inclinedat an angle of 60° to the horizontal, would enable herto reach the required position? Also, how far fromthe foot of the pole should she place the foot of theladder? (You may take V3 = 1.73) |
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| 22. |
AABCH £ZB=72° £C=64° ८ & को ज्ञात कीजिए। |
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Answer» By angle sum property sum of all 3 angles of triangle should be = 180 Then,. A + B + C = 180A + 72 + 64 = 180A = 180 - 136 = 144 |
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| 23. |
. Identify which of the following aresupplementary, adjacent andlinear pairs of angles.A. ZD and LCB. ZA and ZBC. ZB and ZFD. B and ZC |
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Answer» A . adjacent B . linear pair , adjacent , supplementaryC. vertically opposite anglesD. adjacent |
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| 24. |
7. A field whose shape is triangular. Its sides are 20 m, 51 m and 37 m, then howmany small beds of measure of 2 x 3n? can be plotted in the field? |
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| 25. |
iu urdl 8f the table7. A field whose shape is triangular. Its sides are 20 m, 51 m and 37 m, then howmany small beds of measure of 2 x 3 m2 can be plotted in the field?12.3. Application of Harnn'n |
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| 26. |
d whose shape is triangular. Its sides are 20 m, 51 m and 37 m, then how mbeds of measure of 2 x3 m2 can be plotted in the field?any |
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| 27. |
The perimeter of a triangular field is6p2 - 4p + 9 and two of its sides arep2-2p+ 1 and 3p2-5p+3. Find the thirdside of the field. |
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Answer» sir this is the easiest question is to solve please solve it by yourself if you don't appreciate it should be good |
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| 28. |
Q21. Factorise: 8x3-ัะท-12x2y+6xy2 |
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Answer» PLEASE LIKE THE SOLUTION |
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| 29. |
of the tower.Q21.The angle of elevation of the top of a vertical tower from a point on theground is 600. From another point 10m vertically above the first, its angle ofelevation is 30°. Find the height of the tower. |
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| 30. |
Choose the correct answer in Exercises 21 and 22.Q21.The anti-derivative of\ equalsx+ |
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| 31. |
Q21. In an isosceles triangle ABC with AB=AC. D and E are points on BC such thatBE=CD. Show that AD=AE.22 Throne |
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Answer» InΔabe and inΔacdab =ac ∠B=∠C BE=CDBY SAS RULEΔABE≡ΔACDBY CPCTAE=AD |
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| 32. |
EXERCISE 7.5 (Optional)*AB is a triangle. Locate a point in the interior of Δ ABC which is equidistant from allhe vertices of A ABC.idictant from all the sides of the |
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| 33. |
In a AABC, ZA+ ZB = 65 and 2B + 2C = 140 . Find the angles. |
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Answer» Given<A+<B=65°<B+<C=140°We know that <A+<B+<C=180°(angle sum property) (<A+<B)+<C=18065+<C=180<C=180-65=115° <B+<C=140°=<B+115=140=<B=140-115=25° <A+<B=65°=<A+25=65=<A=65-25=40° Thus <A=40°<B=25°<C=115° |
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| 34. |
cach angle of16In a triangle ABC, 2A-ZB = 15°, ZB-ZC-30. Find the measure othe triangle.11. |
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Answer» A-B=15 so A=B+15B-C=30 so C=B-30by adding both A-C=45now its triangleso A+B+C=180so B+15+B-30+B=180so 3B-15=180so 3B=195so B=65so A=80so C=35 |
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| 35. |
EXERCISE 7.4 (Optional)*1. Determine the ratio in which the line 2x ty-4 0 divides the line segment joining thepoints A(2, 2) and B(3, 7) |
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| 36. |
Q4 In triangle ABC, ZA+ZB = 105°, 2B+2C = 120°, find the value of 251 |
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Answer» angle.A+angle.B+angle.C=180°(Angle Sum Property)105°+angle.C=180°Angle .C= 75°Now,Angle.B+ Angle.C=120°(Given)Angle.B+75°=120°Angle.B=45° |
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| 37. |
cma parallelogram ABCD, 2D=140°. Find the measure of A and ZB, |
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Answer» In a Parallelogram if angle D = 140 Then,Angle A = Angle DIn Parallelogram vertically opposite angles are equalSo, Angle A = 140° Angle A + Angle B = 180°(sum of Co interior angles is 180°)Angke B = 180 - 140 = 40° |
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| 38. |
The adjacent sides of rectagle are 9 Cm & 40 Cm, Find the length of its diagonal.In ⳠLSN ,if 4 = 80°,(s = 400 then find out the greatest and smallest side of ALSN. |
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| 39. |
the smallest sideFind the area of a triangular field whose sides are 91 m, 98 m and 105 min length. Find the height corresponding to the longest sidef a trianglse are in the ratio 5: 12: 13 and its perimeter is |
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| 40. |
square/6 %2B 12=13/6 |
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| 41. |
2x+3y=9 3x+4y=5 |
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Answer» Like my answer if you find it useful! |
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| 42. |
ZX + 3y = 9; 3x+4y=5 |
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Answer» 2x + 3y = 9 ......(1) 3x + 4y = 5 .......(2) 3 × (1) - 2 × (2) 6x + 9y = 27 6x + 8y = 10 So, we get y = 17 So, 2x + 51 = 9 2x = 9 - 51 = -42 x = -21 |
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| 43. |
Determine the ratio in which the line 2x +y-4 0 divides the line segment joining the points A(2,-2) and B(3,7). |
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| 44. |
-Solve: 8x - 3=9-2x |
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Answer» 10x=12x=1.2 8x - 3 = 9 - 2x=> 10x = 12=> x = 12/10=> x = 6/5 = 1.2 your answer is =(1.2). 8x+2x = 9+3x=12/10x=1.2 |
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| 45. |
(2007)Find the equation of a line parallel to the line 2x + 5y = 9 and passing through the mid-pointof the line segment joining A(2, 7) and B- 4, 1). |
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Answer» thank u soo much |
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| 46. |
4x/9+1/3+13x/108=8x+9/18 |
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Answer» -18/803 |
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| 47. |
Determine the ratio in which the line 2x +y-4 =0 divides the line segment joining thepoints A(2.-2) and B(3, 7), |
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Answer» LetP(x,y)be the point of intersection of the line2x+y=4and the line joiningA(2,−2)andB(3,7),Assume thatPdivides line segmentABin the ration ofn:1By section formula, |
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| 48. |
8x +5y =93x +2y=4 |
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| 49. |
Figure 6, two circles touch eåch other at A. A common tangent tom at B and C and another common tangent at A meets the prenmon tangentat P. Prove that < BAC = 90°.,Figure 610 |
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Answer» Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively. To find : ∠APB Proof: let ∠CAP = α and ∠CBP = β. CA = CP [lengths of the tangents from an external point C] In a triangle PAC, ∠CAP = ∠APC = α similarly CB = CP and ∠CPB = ∠PBC = β now in the triangle APB, ∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle] α + β + (α + β) = 180° 2α + 2β = 180° α + β = 90° ∴ ∠APB = α + β = 90° |
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| 50. |
Construct a quadrilateral ABCD in which AB-5.6 cm, BCand LD = 8004cm, LA-50°,<B-105° |
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