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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.
| 26551. |
A 6 cm cube is cut into 1cm cubes . Calculate the total surface area of all small cube\'s |
| Answer» TSA =6*1*1=6cm squareBy anurag patil 9901881381-ph no | |
| 26552. |
Is practical is needed for class 10th |
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Answer» Yes Yes |
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| 26553. |
Time table kab aayega |
| Answer» CBSE has not conformed . For my guess Cbse released date sheet tommorow 8th january | |
| 26554. |
Please tell me theorems of triangles all wit their complete solution??? |
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| 26555. |
If tan^2 A =1+2tan^B, prove that 2sin^2 A=1+sin^B. |
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| 26556. |
Given a, b, c, l, m, are in AP then find the value of a-4b+6c-4l+m. |
| Answer» 2 | |
| 26557. |
2+2/2 =???? |
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Answer» 2.5 or 5/2 3.. So difficult.. Almost sweating 2 |
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| 26558. |
Blue print of maths |
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| 26559. |
Activity :to obtain the formula for area of pie r square by paper cutting and pasting method |
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| 26560. |
Ap is 12,24,36 |
| Answer» d=24-12=12 | |
| 26561. |
Find the curved surface area of cone whose radius is 7cm and slant height is 15cm |
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| 26562. |
If AP= 2/3,K.,K5/8 Then find thee value of K |
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| 26563. |
Probilitu |
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| 26564. |
2-2, |
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Answer» Error 0 |
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| 26565. |
Which term of the a. P 3,7,11,15,........,is 91 |
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Answer» 23rd Tumhe formula put karke karna padega a+(n-1)d a=3,d=7-3=4,an=91 |
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| 26566. |
What is the value of 2sintheta.. |
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Answer» 30 degree Only theta |
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| 26567. |
ImMature love says that I love u bcoz I need u but mature love says that I need u BC oz I love u |
| Answer» | |
| 26568. |
If HCF (a,*b)-12 and ab-1800 ,then find LCM (a,b) |
| Answer» HCF(a, b)\xa0{tex}\\times{/tex}\xa0LCM(a, b) = a\xa0{tex}\\times{/tex}\xa0b{tex}\\Rightarrow{/tex}12\xa0{tex}\\times{/tex}\xa0LCM(a, b) = 1,800{tex}\\Rightarrow{/tex}\xa0LCM(a, b) =\xa0{tex}\\frac{1,800}{12}{/tex}\xa0= 150 | |
| 26569. |
(Sin A + secA)^2 + (cosA + cosecA)^2 = ( 1+ secA×cosecA )^2 |
| Answer» | |
| 26570. |
The perimeter of a triangle is 60cm. It\'s hypotenuse is 25cm. Find tha area of tha triangle |
| Answer» 150m2 | |
| 26571. |
Value tan 60 |
| Answer» Root 3 | |
| 26572. |
tan5×tan30×tan85 |
| Answer» | |
| 26573. |
2tangent segment PA are drawn to a circlevwith center at O angle APB=120 prove that OP=2AP |
| Answer» A circle C(O, r). PA and PB are tangents to the circle from point P, outside the circle such that {tex}\\angle{/tex}APB = 120°.Construction: Join OA and OB.\xa0Proof. Consider {tex}\\triangle{/tex}{tex}PAO \\ and\\ \\triangle PBO\xa0{/tex}{tex}PA = PB{/tex}\xa0[Tangents drawn from a point to circle are equal]{tex} OP = OP{/tex}[Common]{tex}\\angle{/tex}OAP = {tex}\\angle{/tex}OBP = 90°{tex}\\therefore{/tex}{tex}\\triangle{/tex}OAP\xa0{tex}\\cong{/tex}{tex}\\triangle{/tex}OBP [by SAS cong.]{tex}\\therefore{/tex}{tex}\\angle{/tex}OPA =\xa0{tex}\\angle{/tex}OPB = {tex}\\frac { 1 } { 2 } \\angle \\mathrm { APB } = \\frac { 1 } { 2 } \\times 120 ^ { \\circ } = 60 ^ { \\circ }{/tex}In right angled {tex}\\triangle{/tex}OAP, {tex}\\frac { \\mathrm { AP } } { \\mathrm { OP } }{/tex}= cos 60° = {tex}\\frac {1}{2}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}OP = 2AP.{/tex} | |
| 26574. |
How 1 degree is equal to 60 min |
| Answer» In latitude longitude scale as it is the time taken to go from 1 degree to other | |
| 26575. |
Is it enough to solve only ncert question in trigno or should do extra q for exam |
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Answer» If I solve some step of trigno q correctly. Could I get some marks? You have to practice more and more and have to do extra also Keep Concept & u can solve ane qstn |
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| 26576. |
find area of tringle by（1，-4），（3. -2），（-3，16） |
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Answer» 24 Esy yr ..use formulae and directly get ans |
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| 26577. |
Which term of the following A. P is - 1 14,11,18,,,,, |
| Answer» Not AP | |
| 26578. |
Can two number have 15 as their HAD and 175 as their LCM ? Give reason. |
| Answer» {tex}\\frac{175}{15}=11.667{/tex}Hence 175 is not divisible by 15But LCM of two numbers should be divisible by their HCF.{tex}\\therefore{/tex}\xa0Two numbers cannot have their HCF as 15 and LCM as 175. | |
| 26579. |
Find the coordinates of the point on y axis which is nearest to the point (-2,5). |
| Answer» The point on y-axis that is nearest to the point(-2,5) is (0,5). | |
| 26580. |
tanA/1-cotA + cotA/1-tanA = 1+secA.cosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 26581. |
Three equal circles, each of radius 6 cm, touch one another.find the area enclosed between them. |
| Answer» Let A, B, C, be the centres of these circles. Joint AB, BC, CA. Radius of each circle = 6cm . AB = BC = CA = 12 cm .Then,ABC is an equilateral triangle.Then, Area of ∆ABC,{tex}=\\frac{\\sqrt3}{4}×(side)^2\\\\ =\\frac{\\sqrt 3}{4}×(12)^2\\\\ =\\frac{\\sqrt 3}{4}×(12)×(12)\\\\ =36×1.73=62.28{/tex}Area of sector of angle 60° and radius 6 cm{tex}=\\frac{\\theta}{360}×πr^2\\\\{/tex}{tex} =\\frac{60}{360}×3.14×6×6\\ cm^2=18.84\\ cm^2{/tex}Required area = Area of ∆ABC- 3 ×area of one sector {tex}{/tex}\xa0{tex}{/tex}{tex}{/tex}=62.28-3× 18.84{tex}{/tex} =(62.28- 56.52) cm{tex}^2{/tex}{tex}{/tex}=5.76 cm{tex}^2{/tex}{tex}{/tex}Area of shaded region = 5.76 cm2 | |
| 26582. |
Solve for x:9x²-9(a+b)x+(2a²+5ab+2b²)=0 |
| Answer» x = 2a+b/3 & a+2b/3 | |
| 26583. |
Find the value of sin30.cos60 |
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Answer» 1/4 1/2 |
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| 26584. |
X-3÷x+3-x+3÷x-3=48÷7 |
| Answer» {tex}\\Rightarrow{/tex}{tex}(x - 3)(x - 3) - (x + 3)(x + 3) = \\frac{48}{7}(x + 3)(x - 3){/tex} [Multiplying both sides by (x + 3)(x -3)]{tex}\\Rightarrow{/tex}{tex}x^2 + 9 - 6x - [x^2 + 9 + 6x] ={/tex}{tex}\\frac{{48}}{7}{/tex}{tex}[(x^2 - 9)]{/tex}{tex}\\Rightarrow{/tex}{tex}x^2 + 9 - 6x - x^2 - 9 - 6x ={/tex}{tex}\\frac{{48}}{7}{/tex}{tex}(x^2 - 9){/tex}{tex}\\Rightarrow - 12x = \\frac{{48}}{7}({x^2} - 9){/tex}{tex} \\Rightarrow - 12x \\times 7 = 48({x^2} - 9){/tex} [Multiplying both sides by 7]{tex}\\Rightarrow - 7x = \\frac{{48}}{{12}}({x^2} - 9){/tex} [Dividing both sides by 12]{tex}\\Rightarrow{/tex}{tex}-7x = 4(x^2 - 9){/tex}{tex}\\Rightarrow{/tex}{tex} -7x = 4x^2 - 36{/tex}{tex}\\Rightarrow{/tex}{tex}4x^2 + 7x - 36 = 0{/tex}In order to factorize 4x2 + 7x - 36, we have to find two numbers \'a\' and \'b\' such that.a + b = 7 and a {tex}\\times {/tex} b = 4 {tex}\\times{/tex} (-36) = -144Clearly, 16 + (-9) = 7 and (16) {tex}\\times{/tex} (-9) = -144{tex}\\therefore{/tex} a = 16 and b = -9Now,4x2 + 7x - 36 = 0{tex}\\Rightarrow{/tex} 4x2 + 16x - 9x - 36 = 0{tex}\\Rightarrow{/tex} 4x(x + 4) - 9(x + 4) = 0{tex}\\Rightarrow{/tex} (x + 4)(4x - 9) = 0{tex}\\Rightarrow{/tex} x + 4 = 0 or 4x - 9 = 0{tex}\\Rightarrow{/tex} x = -4 or {tex}x = \\frac{9}{4}{/tex}Hence, the required roots of given equation are -4 and\xa0{tex}\\frac{{9}}{4}{/tex} | |
| 26585. |
How the statistics formula make of mode |
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| 26586. |
2x/3+4-6=7 |
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| 26587. |
When will 2018 exam datesheet for class 10 release ?? |
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| 26588. |
Find root of equation 16x-10/x=27 |
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| 26589. |
Board exam question will came from this?? |
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| 26590. |
agar hama kisi die koa 2 baar phekte hi toa total outcomes Ketna hoga |
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Answer» 36 12/2 36 |
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| 26591. |
Show that exactly one of the numbers n,n+2,n+4 is divided by 3 |
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| 26592. |
if two lines intersects at right angles then the product of their slope is... |
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| 26593. |
Show thatany positive odd integer is of the form of 6p+1 or 6p+5 where p is some integer |
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| 26594. |
If sinA +sin2A=1,then evaluate cos2A+cos4A |
| Answer» | |
| 26595. |
Find the middle term of A.P 6, 13,20,........216???? |
| Answer» Here,{tex} a = 6, l =216, d = 13 - 6 = 7{/tex}Let the number of terms be n{tex}l = a + ( n - 1) d{/tex}216 = {tex}6 + (n -1)(7){/tex}216 -6 = {tex}7(n - 1){/tex}{tex}7(n - 1){/tex}= 210{tex}n - 1{/tex}=\xa0{tex}\\frac { 210 } { 7 }{/tex}= 30{tex}n{/tex} = 30 + 1 = 31The middle term will be =\xa0{tex}\\frac { 31 + 1 } { 2 }{/tex}= 16th term{tex}\\therefore{/tex}a16\xa0= 6 +(16 - 1)(7){tex}= 6 + 15 \\times 7{/tex}= 6 + 105= 111Middle term will be 111. | |
| 26596. |
What is nth term |
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Answer» Sry yar i m very confused in Sorry aksh it last term of ap given |
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| 26597. |
Solve 2 (ax-by) + a + 4b=02 (ax-ay) + b- 4a = 0 |
| Answer» The given equations may be written as2(ax - by) + (a + 4b) = 02ax - 2by + (a + 4b) = 02ax - 2by = - a - 4 b ... (i)and 2(bx + ay) + (b - 4a) = 0(2bx + 2ay) + (b - 4a) = 02bx + 2ay = 4a - b. ... (ii)Multiplying (i) by a and (ii) by b and adding, we get(2a2+ 2b2) x = (-a2\xa0- b2){tex} \\Rightarrow 2 \\left( a ^ { 2 } + b ^ { 2 } \\right) x = - \\left( a ^ { 2 } + b ^ { 2 } \\right) {/tex}{tex}\\Rightarrow x = \\frac { - 1 } { 2 }{/tex}Putting x = -{tex}\\frac 12{/tex}\xa0in (i), we get{tex}2 a \\times \\left( \\frac { - 1 } { 2 } \\right) - 2 b y = - a - 4 b{/tex}{tex}\\Rightarrow \\quad - a - 2 b y = - a - 4 b{/tex}{tex}\\Rightarrow \\quad 2 b y = 4 b {/tex}{tex}\\Rightarrow y = \\frac { 4 b } { 2 b } = 2{/tex}Hence, x =\xa0{tex}\\frac{-1}{2}{/tex} and y = 2 | |
| 26598. |
If A=15\' ; verify that 4Sin2A .cos6A=1. |
| Answer» | |
| 26599. |
abcd is a rectangle in which diagonal bd bisect |
| Answer» Given: A rectangle ABCD in which diagonal BD bisects {tex}\\angle B{/tex}.\xa0To prove: ABCD is a square.Proof: DC||AB [{tex}\\because{/tex} Opposite sides of a rectangle are parallel]{tex}\\Rightarrow \\;\\angle 4 = \\angle 1{/tex}\xa0..(1) [Alternate interior angles]Similarly,\xa0{tex}\\angle 3 = \\angle 2{/tex} …(2) [Alternate interior angles]And {tex}\\angle 1 = \\angle 2{/tex}…(3) [Given]From equation (1), (2) and (3), we get{tex}\\angle 3 = \\angle 4{/tex}In\xa0{tex}\\Delta BDA{/tex} and\xa0{tex}\\Delta BDC{/tex}, we have{tex}\\angle 1 = \\angle 2{/tex}\xa0[Given]BD = BD [Common side]{tex}\\angle 3 = \\angle 4{/tex}\xa0[proved above]So, By ASA criterion of congruence, we have{tex}\\Delta BDA \\cong \\Delta BDC{/tex}\xa0{tex}\\therefore{/tex} AB = BC [CPCT]So, ABCD is a square.Hence, proved. | |
| 26600. |
Weightage of each chapter for exam 2018 |
| Answer» Check marking scheme in the syllabus :\xa0https://mycbseguide.com/cbse-syllabus.html | |