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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. |
\frac { 5 ( x - 2 ) } { 3 } \geq \frac { 3 ( 2 - x ) } { 5 } - \frac { 34 } { 15 } |
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Answer» thanks |
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| 2. |
A cubical block of side 7 cm is surmounted by a hemi-sphere. What is the greatest diameter the hemispherecan have? Find the surface area of solid. |
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Answer» side of cube = 7cmdiameter of hemisphere=7radius=7/2 cmsurface area of solid= (TSA of cube - 1 surface area )+ SA of hemisphere={ 6(7)^2-(7)^2} + 2×22/7×7/2×7/2= 5×49+ 11×7= 405+77= 482 cm^2 the greatest diameter that the hemisphere can have is 7cm and surface area of solid will be =(tsa of cube - πr^2)+CSA of hemisphere edge of the cube=7cmlargest diameter=7cmr=d/2r=3.5cmCSA of solid=5×(side)²+CSA of hemisphere +(area of base of square - area of base of hemisphere)CSA=5 × 49 + 2×22/7×(3.5)²+(7×7- 22/7×(3.5)²CSA= The cube has side = 7cm. So, the hemisphere has greatest diameter = 7cmr = 7cm/2 CSA of hemisphere = 2πr²= 2 * 22/7 * 7cm/2 * 7cm/2= 77cm² TSA of cube= 6*(side)²= 6 * (7cm)²= 6 * 49cm²= 294cm² The hemisphere enclosed a circular top part of the circle, when surmounted on it. Area of Enclosed region= πr²= 22/7 * 7cm/2 * 7cm/2= 38.5cm² TSA of solid = (CSA of hemisphere) + (TSA of cube) - (Area of enclosed region) = 77cm² + 294cm² - 38.5cm²= 332.5cm² 482cm^2 is correct answer edge of the cube =7cmlargest diameter =7cm r=d/2r=3.5cmcsa of solid answer of your questions 482 cm^2 iska answer 7000 hoga side of cube is 7 cm greatest diameter that hemisphere can have is 7 cmradius of hemisphere is 7/2 cmsurface area of soild is ( TSA of cube-pie r^2) + CSAof hemisphere the greatest diameter is 7 greatest diameter of hemisphere is 7cm.surface area of solid is 332.5 side of cube = 7cm diameter of hemisphere=7radius=7/2 cmsurface area of solid=(TSA of cube - 1surface area )+ SA of hemisphere ={ 6 (7)^2-(7^2} + 2×22/7×7/2×7/2=5×49+ 11×7=405+77=482 cm^2 The cube has side=7 cm So, the hemisphere has a greater diameter=7cmr=7cm/2 CSA of hemisphere=2πr²=2*22/7*7cm/2*7cm/2=77cm² TSA of cube=6*(side)I=6*(7cm)²=6*49cm=294cm² The hemisphere enclosed a circular top part of the circle, when surmounted on it.Area of enclosed region= πr²= 22/7*7cm/2*7cm/2=38.5cm²TSA of solid(CSA of hemisphere)+(TSA of cube)-(Area of enclosed region)=77cm² +294cm² -38.5cm²=332.5cm² greatest diameter will be 7and total area will be 482 cm squares okkk 482 cm is the answer i think side of cube 7cm diameter of hemisphere=7 radius=7/2 cm surface area of solid= (tsa of cube 1 surface area )+ sa of hemisphere ={6(7)^2-(7)*2}+ 2x22/7x7/2x7/2 5x49+11x7 405+77 482 cm^2 Actual area of the pot=2πr2+5l2=2×22/7×3.5×3.5+5×7×7=44/7×3.5×3.5+5×49=44×0.5×3.5+245=44×1.75+245=77+245=322cm2 322+10.5=332.5cm2area of square=l2 =7×7=49area of circle=πr2 =22/7×3.5×3.5=22/7×0.5×3.5=22×1.75=38.5=49-38.5=10.5cm2 |
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| 3. |
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatestdiameter the hemisphere can have? Find the surface area of the solid |
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| 4. |
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatestdiameter the hemisphere can have? Find the surface area of the solid. |
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| 5. |
Find the angle between the pair of lines\begin{array}{l}{\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}} \\ {\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}}\end{array} |
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| 6. |
Length of a rectangle is four less than the thrice of its breadth. The perimeter of the rectangleis 110. |
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Answer» Let breadth be x Length of recatngle = 3x - 4 Perimeter = 110 2 × ( L + B) = 110 L + B = 55 x + 3x - 4 = 55 4x = 59 x = 59/4 = 14.75 So, Length is 14.75 units Breadth = 3 × 14.75 - 4 = 40.25 units |
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| 7. |
Find the perimeter and area (in hectares) of a field whose length is 240 m and breadthis 110 dm.. |
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Answer» l=240mb=110dm=11mso perimeter=2(240+11)=502marea=240×11=2640m^2 |
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| 8. |
Find the perimeter and area (in hectares) of a field whose length is 240 m and breadthis 110 dm.4. |
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Answer» l=240mb=110dm=11mso perimeter=2(240+11)=502marea=240×11=2640m^2 |
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| 9. |
28cmdiameter |
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Answer» d = 2π=2×28= 56cm Answer d=2π=2×28=56 cm is correct answer. |
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| 10. |
34. Find the quantity of air present inside the ball of radius 28cm. |
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Answer» Given,Radius of ball r = 28 cm Volume of ball= 4/3 * pi * r^3= 4/3 * 22/7 * 28 * 28 * 28= 4/3 * 22 * 4 * 28 * 28= 91989. 3 cm^3 Therefore,Quantity of air present inside ball = 91989.3 cm^3 |
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| 11. |
Jan B+ (0= P1 (ogecr âAm*H || |
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Answer» hit like if you find it useful |
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| 12. |
3 x + 2 y \leq 12 , x \geq 1 , y \geq 2 |
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| 13. |
A = \left[ \begin{array} { l } { 3 - 4 } \\ { 1 - 1 } \end{array} \right] \text { then for any integer } n \geq 1 \text { show that } A ^ { n } = \left[ \begin{array} { c c } { 1 + 2 n } & { - 4 n } \\ { n } & { 1 - 2 n } \end{array} \right] |
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| 14. |
Jan 12.05(2014) ]tan |
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| 15. |
a _ { 1 } = 1 , a _ { n } = a _ { n - 1 } + 2 , n \geq 2 |
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Answer» a1 = 1a2 = a1+2 = 1+2 = 3a3 = a2 +2 = 3+2 = 5a4 = a3 +2 = 5+2 = 7 a5 = a5 +2 = 7+2 = 9 |
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| 16. |
o) यदि \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi. तो सिद्ध कीजिए किx^{2}+y^{2}+z^{2}+2 x y z=1 |
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| 17. |
The sides of a triangle are 20 cm, 99 cm and101 cm. Find (a) its area (b) altitude on thegreatest side. |
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| 18. |
Sin(x)+Sin(y)+Sin(z)=3Find:Cos(x)+Cos(y)+Cos(z) |
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Answer» The only way to have if sin x + sin y + sin z = 3 is if sin x = sin y = sin z = 1 Since cos²θ = 1 - sin²θ, then cos x = cos y = cos z = 0 So cos x + cos y + cos z = 0 |
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| 19. |
\operatorname { cos } x + \operatorname { cos } y + \operatorname { cos } z = \operatorname { sin } x + \operatorname { sin } y + \operatorname { sin } z = 0 , \text { show that } \operatorname { cos } \frac { x - y } { 2 } |
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Answer» cosx + cosy = -cosz , cos^2x + cos^2y + 2cosx.cosy = cos^2z , in a similar way : sin^2x+sin^2y+2sinx.siny = sin^2z , add eqns & simplify : 1+1+2(cosx.cosy+sinx.siny) = 1 , 2cos(x-y)=1-2=-1 , cos(x-y)=-1/2 |
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| 20. |
apes.Lesson 3: Area & Permeter: Comic17. Find the missing measure in each given planeshape.A. Rectangle with length = 10 cm, ared = 110cm?, perimeter = ? |
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Answer» breadth= 110/10=11cm perimeter = 2(length+breadth) =2(11+10) =2(21)=42 Ans 2(21)=42 is the best answer Area of rectangle = l × b110 = 10 × bTherefore, b = 110 ÷ 10Therefore, b = 11 Perimeter of rectangle = 2(l + b)Perimeter of rectangle = 2(10 + 11)Perimeter of rectangle = 2(21)Perimeter of rectangle = 42 cm L=10,area=110 L.b=area b=area/length 110/10=11 Peimeter=2l+2b=2.10+2.11=20+22=42 4 2is the correct answer of the given question |
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| 21. |
If z( cos θ, sin θ ) , find | z-z |
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Answer» thanks |
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| 22. |
23. Find the area of a square with perimeter 28cm |
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Answer» Perimeter of square = 4×side28 = 4×sideside = 28/4 = 7Area of square = side × side = 7×7 = 49 Area of square = 1/2×(diagonal)²49 = 1/2×(diagonal)²Diagonal ² = 49×2 = 98Diagonal = ✓98 = 9.8cm |
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| 23. |
हे» sin? (2xyf1—x?) =2 cos" % (z—l"< g |
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Answer» Let x= cos theta. Then cos^-1x = theta. sin^-1(2x(sqrt(1-x^2)))= sin^-1(2cos theta(sqrt(1-cos^2theta))) = sin^-1(2cos theta sin theta) = sin^-1(sin 2theta) = 2theta = 2cos^-1x |
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| 24. |
A1.2-m-tallgirlspotsa balloon moving with the wind in aLE 28horizontal line at a height of 88.2 metres from the ground. The angleof elevation of the balloon from the eyes of the girl at any instant is60°.After some time, the angle of elevation reduces to 30°. Find thedistance travelled by the balloon during the interval |
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Answer» in the light of tha importance of mathematice difine mathematice |
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| 25. |
29. A 1.2. m tall girl spots a balloon movingwith the wind in a horizontal line at aheight of 88.2 m from the ground. Theangle of elevation of the balloon from theeyes of the girl at any instant is 60°. Aftersometime, the angle of elevation'reducesto 30 (see fig-). Find the distance travelledby the balloon during the interval.60°Use 3 1.73]1.2 m |
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| 26. |
%Wt(०5५३) |
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Answer» (5x/y + y/5x)^3 =(5x/y)^3+(y/5x)^3+3(5x/y)(y/5 )(5x/y + y/5x) =(125/y^3)+(y^3/125)+3(0)(5x/y+y/5x)= =(125/y^3 + y^3/125) |
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| 27. |
Find the coordinates of the foci, thevertices, the length of major axis, the minoraxis, the eccentricity and the length of thelatus rectum of the ellipse |
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| 28. |
(c)07.Draw an ellipse having major axis. 120 mm and minor axis 80 mmby using concentric circle method.52580140 ONSDV86.101SS10602991SP99 ASISSASEBS3891980N2889 4 OROS |
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Answer» for eg: |
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| 29. |
14. A1.2 m tall girl spots a balloon movingwith the wind in a horizontal line at aheight of 88.2 m from the ground. Theangle of elevation of the balloon fromthe eyes of the girl at any instant is60°. After some time, the angle ofelevation reduces to 30° (see Fig. 9.13)Find the distance travelled by theballoon during the interval. |
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| 30. |
\frac { 1 } { 2 } ( \frac { 3 x } { 5 } + 4 ) \geq \frac { 1 } { 3 } ( x - 6 ) |
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Answer» thanks |
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| 31. |
14. A1.2 m tall girl spots a balloon movingwith the wind in a horizontal line at aheight of 88.2 m from the ground. Theangle of elevation of the balloon fromthe eyes of the girl at any instant is60°. After some time, the angle of 6elevation reduces to 30° (see Fig. 9.13). 130°Find the distance travelled by the88.2m0,-Af |
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| 32. |
EXERCISE 23ره سه1. Find a b whenA وA Aa=-24k and h (ن)2 - 4 - 3 = |
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| 33. |
1. निम्न द्विघात बहुपदों के शून्यक ज्ञात कीजिए और शून्यकों तथा गुणांकोंके बीच के सम्बन्ध की सत्यता की जाँच कीजिए :| (i) x2 - 23 - 8[NCERT EXERCISE]FAVARAA |
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Answer» x^2-2x-8=x^2+4x-2x-8=x(x+4)-4(x-8)=(x-2)(x-4); x=2, x=-4 x=-2 &x=4is right answer |
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| 34. |
Lesson 23 Perimeter & AreaExercise 23 aand the - - -7th = 15 cm, breadth = 8cm |
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Answer» what is the question 46 is right answer , 46 is the correct answer |
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| 35. |
A 1.2 m tall girl spots a balloon movingwith the wind in a horizontal line at aheight of 88.2 m from the ground. Theangle of elevation of the balloon fromthe eyes of the girl at any instant is60°. After sometime, the angle ofelevation reduces to 30°. Find thedistance travelled by the balloonduring the interval.1.AS |
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| 36. |
EXERCISE 23Find the perimeter and area of a rectangle having:(i) Length = 16 cm, Breadth = 12 cm |
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Answer» Perimeter=2(l+b)=2(16+12) =2(28) =56 cm Area=lb =16×12 = 192 cm sq. |
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| 37. |
A 1.2 m tall girl spots a balloon moving withthe wind in a horizontal line at a height of88.2 m from the ground. The angle of eleva-tion of the balloon from the eyes of the girlat any instant is 60. After some time, theangle of elevation reduces to 30. Find thedistance travelled by the balloon during theinterval. |
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| 38. |
5. If sin x+sin y+sinz-3, then cos x + cos y + cos z- |
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Answer» sinx+siny+sinz=3.....(given)it is possible only if sinx=siny=sinz=1then we know that,cos^2x=1-sin^2xand sinx=1hence cosx=0.....(1)similarly cosy=cosz=0hence cosx+cosy+cosz=0 |
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| 39. |
क43, 35,27, 19; या क्रमिकेची पुढील दोन पदे लिहा. |
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| 40. |
\int_{-\infty}^{\infty} \frac{d x}{x^{2}+2 x+2} |
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| 41. |
7. Find the equation of the ellipse in the standard form whose minor axis isequal tothe distance between foci and whose latusrectum is 10. |
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| 42. |
\left. \begin{array} { l } { 1.8 + 4 \sqrt { 2 } + 4 + 2 \sqrt { 2 } + \ldots \infty } \\ { 3 . \sqrt { 2 } - \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { 2 \sqrt { 2 } } - \frac { 1 } { 4 \sqrt { 2 } } + \ldots \infty } \end{array} \right. |
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Answer» a /1-r = 8/1-1/√2 = 8√2/√2-1 |
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| 43. |
\left. \begin{array} { l } { \text { Solution of } | 3 x - 2 | \geq 1 \text { is } } \\ { ( a ) [ 1 / 3,1 ] } \\ { ( c ) \{ 1 / 3,1 \} } & { ( d ) ( - \infty , 1 / 3 ] \cup [ 1 , \infty ) } \end{array} \right. |
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| 44. |
If a^{2}+b^{2}+c^{2}=2 then the range ofab + bc + ca is\begin{array}{ll}{\text { 1) }[-1 / 2,1]} & {\text { 2) }[-1 / 2, \infty)} \\ {\text { 3) }[-1,2]} & {\text { 4) }[1, \infty)}\end{array} |
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Answer» thank u very much teacher |
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| 45. |
Find the smallest number that has to be subtracted from 256, so that it becomes exactlydivisible by 10?9. |
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Answer» 1 divide 256 by 10 2 the remainder will be=6 ans = 6(it is the smallest no which should be subtracted from 256 so that it is exactly divisible by 10) When we divide 256 by 10 , the quotient is 25 and the remainder is 6 256= 25×10 + 6 Hence 6 is the smallest no. to be subtracted to 256 so that it is divisible by 10 = 256-6=250 |
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| 46. |
(-9) x (-18)9-(-18) m |
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| 47. |
ids. TMdtch the word. altophobia |
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Answer» Altophobiais the fear of heights. |
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| 48. |
resourcesoControl ids(MP0.4, Describe the reasons of population explosion.Or |
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Answer» The most importantcauseof famine is therefore not thepopulation explosion. ... As mentioned, poverty is also an underlyingcauseof rapidpopulationgrowth. Social factors are at the base of both poverty andpopulationgrowth |
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| 49. |
tan+ xtan(4+x)-(1-tanx)26. Prove that4 |
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| 50. |
EXERCISE 231. Divide the polynomial p(x) by the polynomialgt) and find the quotient and remainderin each of the following:p(x) - 3 +5x-3, 8(x)= -2plx) *-31-4x+5. (x)=r+1- |
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