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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.
| 21801. |
10.2question 5 |
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Answer» Diagram should be like this :- a circle, AB ia a tangent , OC is perpendicular to AB and O\' is also a point around O(centre) and join O\'C This can be done by contradiction. Therefore let assume that O\'C is perpendicular to Ab so we can say that angle OC\'b is equal to 90°- © we know that angle ACB = 90°- ® From ©and ® O and O\' are same point therefore angle O\'CB = angleOCB . Which chapter?? Dude |
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| 21802. |
In given figure, find the value of X |
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Answer» Where is the fig.???? What\'s the figure |
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| 21803. |
What is the sum of all even numbers |
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Answer» Sum of all even numbers is -1/6 n(n+1)/2 |
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| 21804. |
What is the sum of all odd numbers |
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Answer» Sum of all odd numbers is 1/12 n×n or n square . |
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| 21805. |
What is the sum all natural numbers |
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Answer» Sum of all natural numbers is -1/12 -1/12 Sn =n/2 (n+1) eg:- 100/2 (100+1)= 50×101=5050 |
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| 21806. |
Euclid division lemma (a=bq+r) |
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Answer» I didn\'t understood anything Yes u r right |
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| 21807. |
Who discovers the mathematic |
| Answer» Pythagora | |
| 21808. |
What is the difference between division algorithm and the fundamental theorem of arithmetic??? |
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| 21809. |
If x=a and y=b |
| Answer» | |
| 21810. |
Example 13 to solve |
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Answer» Frm which ch? Which chapter |
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| 21811. |
586% |
| Answer» Plz write full question | |
| 21812. |
Use Euclid algorithm to find hcf of 1190 and 1445 express the hcf in the form of 1190m +1445n |
| Answer» Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.1445 = 1190 ×\xa01 + 2551190 = 255 ×\xa04 + 170255 = 170 ×\xa01 + 85170 = 85 ×\xa02 + 0So, now the remainder is 0, then HCF is 85Now,85 = 255 - 170= (1445 - 1190) - (1190 - 255 ×\xa04)= 1445 - 1190 - 1190 + 255 ×\xa04= 1445 - 1190 ×\xa02 + (1445 - 1190) ×\xa04= 1445 - 1190 ×\xa02 + 1445 × 4 - 1190 ×\xa04= 1445 ×\xa05 - 1190 ×\xa06= 1190 ×\xa0(- 6) + 1445 ×\xa05= 1190m + 1445n , where m = - 6 and n = 5 | |
| 21813. |
If the diametre of a circle is increased by40% what is the increase area of the circle Ans. 96% |
| Answer» The area of a circle is determined by the formula (pi)r2.\xa0The best way for you to approach this problem is by using a sample diameter. Let\'s use 4, since it\'s easy.\xa01: Multiply 4 by 0.40= 1.6. So your new diameter is 5.6.\xa02: Divide the diameter by two to get the radius => 2.8.\xa03: Square the radius => 7.84. Now multiply by pi, using 3.14 ==> 24.62.\xa04: Find the original area of the circle ==> 22*3.14 ==> 12.56.\xa05: 24.62/12.56=1.96 ==> The new is 196% of the original, so the area changes by 96%. | |
| 21814. |
Explain upstream and downstream i have many doubt in this question |
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Answer» When it goes in the direction of flow of river means downstream then we have to add (x+y) and if goes opposite to the direction of flow of river means upstream then we will subtract (x-y) If any boat is moving in direction of river flow is known as DOWNSTREAM andmoving opposite to river flow is UPSTREAM. Downstream=x+y and upstream =x-y It\'s easy yrr.. Just remember that downstream me speed add hoti hai and upstream me subtract Sorry i want to say in this concept |
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| 21815. |
Find the coordinates of the point on y-axis which is nearest to the point(-2, 5) |
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Answer» The nearest point is (0,5)Refer the attactment given below: (0,5) |
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| 21816. |
find alll the zeros 3rd chapter |
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Answer» PLEASE DO NOT CHAT AND DOWNLOAD SOLUTION FROMhttps://www.vedantu.com/ncert-solutions/campve-ncert-solutions-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables Linear me x aur y ka value nikalte h Polynomial me zeros nikalte h Rohan Bro... Equation to batao pahle.. Tabhi to zero find karenge |
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| 21817. |
√3cot²+4 cot +3 then cot²+tan²=? |
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Answer» The answer is 10/3.... ???? Full ques yhi h |
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| 21818. |
Find the coordinate of the point on y - axis which is nearest to the point (- 2,3). |
| Answer» 0,3 | |
| 21819. |
2 cubes 9 each volume 64 cm are joined end to end .Find the surface area of resulting cuboid |
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Answer» And answer is 320 cm2 Sis 9 side h??? |
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| 21820. |
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. |
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Answer» sankey sanket Ayush sanket <“https://images.app.goo.gl/eaJm55Pj3HSHpkne6"> =\'\'https://images.app.goo.gl/eaJm55Pj3HSHpkne6\'\' 28-8=20 & 32-12=20 So the reqd no will be 20 less than the L. C. M. of 28 and 32 . L. C. M. of 28 & 32 is = 224 Reqd smallest no =224-20 = 204. |
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| 21821. |
Circal |
| Answer» It is circle | |
| 21822. |
Tan theta/1-cot theta+cot theta /1-tan theta=1+tan theta +cot thetha |
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Answer» is ur ques right?? Pura solve kar rakha hai https://photos.app.goo.gl/WAQVikzrdniDJXyP8 |
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| 21823. |
Tan theeta +cot theeta =sec theeta+cosec theeta prove it |
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Answer» sec theeta×cosec theeta Ya, u both r ryt... Mera answer sec theta ×Cosec theta aa rha hai yes mera bhi yahi aa raha h Check your question.........i think ye sec theeta × cosec thita hoga....... Bhai sec theta cosec theta to aa raha h but bich me plus nahi multiply h |
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| 21824. |
Marking scheme class 10 maths |
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Answer» 1 or 2 or 3 page of any math\'s reference Google.. |
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| 21825. |
What is curved surface area? |
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Answer» Kisi bhi solid figure ga bahar ka area Area of curved surface of solid. Csa |
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| 21826. |
sin 18°÷cos72° |
| Answer» Answer=1 | |
| 21827. |
If sum of frist n term of an ap is 5n=4nsquare -3n, find nth term and 30th term |
| Answer» nth term 4nsquare-8n=02nsquare-4n=0n(2n-4)=0n=0 OR n=2 | |
| 21828. |
Find the HCF of 92690, 7378,7161 use its Euclid division algorithm |
| Answer» We have to find the HCF, by Euclid\'s division algorithm of the numbers 92690, 7378 and 7161.Here, we will apply Euclid\'s Division Lemma, we have92690 = 7378 ×\xa0{tex}{/tex}12 + 4154Again we apply Euclid\'s Division Lemma of divisor 7,378 and remainder 4154, thus we have7378 = 4154 ×\xa0{tex}{/tex}1 + 3,2244154 = 3224 ×\xa0{tex}{/tex}1 + 9303224 = 930 × 3\xa0{tex}{/tex}\xa0+ 434930 = 434 × 2\xa0{tex}{/tex}+ 62434 = 62 × {tex}{/tex}7 + 0HCF of 92690 and 7378= 62Now, using Euclid\'s Division Lemma on 7161 and 62, we have7161 = 62 ×\xa0{tex}{/tex}115 + 31Again, applying Euclid\'s Division Lemma on divisor 62 and remainder 31, we have62 = 31 ×\xa0{tex}{/tex}2 + 0Clearly, HCF of 7161 and 62 = 31Hence, HCF of 92690, 7378 and 7161 is 31. | |
| 21829. |
If a + b = 90 degree and sec a= 5 / 3 then find the value of cosec b |
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Answer» Sorry sorry i think it is cosec A sec a=5/3a+b =90 so b=90-acosec b =cosec(90-a)=sec a=5/3\xa0 sec a=5/3......h^2=p^2+b^2......5^2=p^2+3^2.....25=p^2+9.......p=4....... cosec=h/p.......so cosec=5/4 How please explain ?? 5/3......... |
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| 21830. |
Find the perpendicular distance of A(0,12) From the y axis. |
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Answer» This question is wrong How please explain The answer is 12 as the y coordinate is 12 Mam ne aaj hi ye question dia its correct question dude Plz check your question............agar 0 hai to perpendicular kaise hoga y axis se |
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| 21831. |
the volume of a cube whose edge is 5a is |
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Answer» Volume of cube = a×a×a =5a×5a×5a= 125a³ 125^3 sorry before answer is wrong 155a^3 625 |
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| 21832. |
2x2+4x +2=0 |
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Answer» 2x2 + 4x + 2 = 02x2 + 2x + 2x + 2 = 02x(x + 1) + 2 (x + 1) = 0(2x + 2 )(x + 1)= 02x + 2 = 0 or x + 1 = 02x = - 2 or x = -1x = -1 -3/2 2(x+1)(x+1)&zero is -1 |
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| 21833. |
Speed |
| Answer» What speed | |
| 21834. |
The hcf and Lcm of two number are 9 and 360 respectively if one number is 45,find the other number. |
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Answer» 72 It is solved byHCF×LCM=product of two no. |
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| 21835. |
What is collinear formula |
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| 21836. |
If tanA+secA=p find cosec A |
| Answer» Given,{tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also, we know that,\xa0{tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1 [{tex}\\because a^2-b^2=(a+b)(a-b){/tex}]{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex})p = 1 [using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(i)+(ii), we get,{tex}sec\\theta + tan\\theta+ sec\\theta - tan\\theta = p+ \\frac{1}{p}{/tex}{tex}\\Rightarrow 2sec\\theta = \\frac{p^2+1}{p}{/tex}{tex}\\Rightarrow sec\\theta = \\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow \\frac{1}{cos\\theta} =\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow cos\\theta =\\frac{2p}{p^2+1}{/tex}------(iii)Now, we know that,{tex}sin\\theta = \\sqrt( 1- cos^2\\theta) {/tex}put the value of\xa0{tex}cos\\theta{/tex}\xa0from eq. (iii), we get,{tex}sin\\theta = \\sqrt(1-(\\frac{2p}{p^2+1})^2){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(1-\\frac{4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1+2p^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1-2p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2-1)^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\frac{p^2-1}{p^2+1}{/tex}{tex}cosec\\theta = \\frac{p^2+1}{p^2-1} [\\because cosec\\theta =\\frac{1}{sin\\theta}]{/tex}hence, {tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |
| 21837. |
Find the sum of 21term of ap whose middle term is 20 |
| Answer» What\'s the common difference?? Or any other particular?? | |
| 21838. |
Class 10 example no 1 chapter 14 |
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| 21839. |
How to solve the problem based on ages |
| Answer» Compare it to the real life examples and form equations and then solve it | |
| 21840. |
2+8+0+56+ |
| Answer» 66+ | |
| 21841. |
Find HCF of the numbers given belowk,2k,3k,4k and 5k,where k is any positive integer |
| Answer» K | |
| 21842. |
If m th term of an AP is 1/n and n th term is 1/m . Then prove that: mn th term is 1. |
| Answer» Let a and d be the first term and common difference respectively of the given A.P. Thenan = a + (n - 1)d{tex}\\frac { 1 } { n } ={/tex}\xa0mth term\xa0{tex}\\Rightarrow \\frac { 1 } { n } {/tex}= a + ( m - 1 ) d\xa0...(i){tex}\\frac { 1 } { m }{/tex}= nth term{tex}\\Rightarrow \\frac { 1 } { m } {/tex}= a + ( n - 1 ) d\xa0...(ii)On subtracting equation (ii) from equation (i), we get{tex}\\frac { 1 } { n } - \\frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]= a + md - d - a - nd + d{tex}= ( m - n ) d{/tex}{tex} \\Rightarrow \\frac { m - n } { m n } = ( m - n ) d {/tex}{tex}\\Rightarrow d = \\frac { 1 } { m n }{/tex}Putting d = {tex}\\frac { 1 } { m n }{/tex}\xa0in equation (i), we get{tex}\\frac { 1 } { n } = a + \\frac { ( m - 1 ) } { m n } {/tex}{tex}\\Rightarrow \\frac { 1 } { n } = a + \\frac { 1 } { n } - \\frac { 1 } { m n } {/tex}{tex}\\Rightarrow a = \\frac { 1 } { m n }{/tex}{tex}\\therefore{/tex}\xa0(mn)th term = a + (mn - 1) d=\xa0{tex}\\frac { 1 } { m n } + ( m n - 1 ) \\frac { 1 } { m n } {/tex}{tex}\\left[ \\because a = \\frac { 1 } { m n } = d \\right]{/tex}= {tex}\\frac { 1 } { m n } + \\frac { mn } { m n } - \\frac { 1 } { m n }{/tex}= 1 | |
| 21843. |
Interval is. 0-10,10-20......... 90-100No. Of Student are 5,3,4,3,3,4,7,9,7,8Find mean |
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| 21844. |
If secA+tanA=x thn find value of secA-tanA in terms of x |
| Answer» 1/x | |
| 21845. |
Why the total outcomes of a dice thrown twice is 36 |
| Answer» Becoz dice have 6 faces and 6 outcomes | |
| 21846. |
If tan(A+B)=√3 and tan (A+B)= 1\\√3;0° |
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Answer» Tan(A+B)=√3. Tan(A-B)=1/√3Tan(A+B)=Tan60°. Tan(A-B)=tan30°A+B=60°. A-B=30° A+B=60° A-B=30° 2A=90° A=45° A+B=60° 45°+B=60° B=60°-45° B=15° tan(A+B)=tan 60\' so A+B=60 similarly Tan(A+B) =tan30\' so A + B =30 now solve these equations you will get answers |
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| 21847. |
Find the distance between the following pairs of points. (a,b) , (_a,_b) |
| Answer» Let a,b be x1, y1 and -a,-b be x2, y2The distance formula is √(x2-x1)²+ (y2-y1)²√(-a-a)²+(-b-b)²√2a²+2b²2√a²+b² | |
| 21848. |
Solve the following equation for x : 9 x square - 9(p+q)x+(2p square+5pq+2q square)=0 |
| Answer» X=2p+q/3 or p+2q/3 | |
| 21849. |
If sum and product of zeroes given 1/4,-1 respectively find its quadratic polynomial how do we find |
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Answer» Sorry the first two lines be like that ?let the two zeroes be in the form of alpha and beeta. Use formula x^2 - (a+b)x + (a×b) =0___X^2 - (1/4)x + (-1)=0 ____ take 1/ 4 comon____ 1/4(4x^2 - x - 4 )=0 ____ 4x^2 - x - 4=0 Let the sum of zeroes be alpha and product of zeroes be beeta.... Then A/Q, alpha+beeta=1/4 and alphaXbeeta=-1... Now put sum and product in these ? x²-(alpha+beeta)x+alpha.beeta......we get,x²-1/4x+(-1)..i,e. the required polynomial is x²-1/4x-1. |
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| 21850. |
If [email\xa0protected] + [email\xa0protected]=p then find the value of [email\xa0protected] |
| Answer» | |