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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.
| 7551. |
Find value of k for the following quadratc equation, so that they have two equal roots. 3x²-k√3x+4 |
| Answer» | |
| 7552. |
Find 3 numbers in AP, whose sum is 15 and whose product is 105. |
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Answer» And is 3,5,7 a-d. , a, a+d. are the three numbers |
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| 7553. |
For final board the text book questions only will come? |
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Answer» ? No man, only few questions will appear from your book in board exams No |
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| 7554. |
Determine if the points (1,5), (2,3) and (-2,-11) are collinear |
| Answer» If area =0 | |
| 7555. |
Is the paper going to be ncert based |
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Answer» No it is not complsery that paper from ncert but exam will not hard according to anouncement by cbse not totoly but about 65% from ncert and rest from RS agarwal |
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| 7556. |
A card has 3,4,5........50 numbers .find the probability of perfect square number. |
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Answer» Sorry it\'s not 3/50 it\'s 1/16 Probability is 3/50 |
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| 7557. |
State and prove converse of Pythagoras theorem |
| Answer» It is very simple go on youtube and find aise likhne se samaj nhi aayega | |
| 7558. |
The hcf of two numbers is 145 and their lcm is 2175 if one number is 725 then find the other number |
| Answer» Hcf x lcm = a x b145 x 2175 = 725 x bb = 435 | |
| 7559. |
Distance between the centre of the circle of radius 3 cm and 8 cm |
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| 7560. |
Solve eqation ax+by=b-a |
| Answer» x\\y = a\\b | |
| 7561. |
Number of question come on each chapter |
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| 7562. |
bhh |
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| 7563. |
IncontinenceHello |
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| 7564. |
What is hcf |
| Answer» Highest Common Factor(HCF) of two or more numbers is the greatest number which divides each of them exactly. Greatest Common Measure(GCM) and Greatest Common Divisor(GCD) are the other terms used to refer HCF. | |
| 7565. |
Ajagsjhsishig |
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| 7566. |
ST line segment joining the points (4,2)&(-6,4)&(-10,5)&(8,1) bisect each other... |
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| 7567. |
If roots of qdrtc eqn (a-b)+(b-c)+(c-a)=0 are equal, PT 2a=b+c.. |
| Answer» We have (a-b)x2 + (b-c)x + (c-a) = 0Here A\xa0= (a-b), B\xa0= (b-c), C= (c-a){tex}\\therefore D = B^2 - 4AC = (b-c)^2 - 4(a-b) (c-a){/tex}For equal roots, D = 0{tex}\\implies{/tex}(b-c)2 - 4 (a-b) (c-a) = 0b2+c2-2bc -4(ac-a2-bc+ab) =0b2+c2-2bc -4ac+4a2+4bc-4ab=04a2+b2+c2+2bc-4ab-4ac=0(2a-b-c)2=0i.e. 2a-b-c =0Hence, b + c = 2a or 2a = b + c | |
| 7568. |
2x+4x |
| Answer» 2x+4x=6x | |
| 7569. |
Kya kar rahe ho |
| Answer» Kuch ni | |
| 7570. |
3square |
| Answer» 9 | |
| 7571. |
Can we answer from last question in board |
| Answer» Yes | |
| 7572. |
What is circle |
| Answer» A closed figure formed by infinite points | |
| 7573. |
(A^2 + b^2 )^2 |
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| 7574. |
5x-4y+8=0 , 7x+6y-9 |
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| 7575. |
CH No 6 Triangle 10 class |
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| 7576. |
Greatest integer |
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| 7577. |
Problem |
| Answer» Solution | |
| 7578. |
Is upgradation for all subjects? |
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| 7579. |
How to form the quadratic eqn by seeing the question |
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| 7580. |
How many numbers lie between 10 and 300 which divided by 4 leave a remainder 3 |
| Answer» The numbers lying between 10 and 300, which when divided by 4 leave a remainder 3 are11, 15, 19............,299This is an A.P. with a = 11, d = 4 and l = 299Let the number of terms be n.then, an = 299{tex}\\Rightarrow{/tex}\xa0a + (n - 1)d = 299{tex}\\Rightarrow{/tex}\xa011 + (n - 1)4 = 299{tex}\\Rightarrow{/tex}\xa0(n - 1)4 = 288{tex}\\Rightarrow{/tex}\xa0n - 1 = 72{tex}\\Rightarrow{/tex}\xa0n = 73Thus, the required number of terms are 73. | |
| 7581. |
□+□+□=30With the number (1,3,5,7,9,11,13,15)you can repeat the numbers |
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Answer» (3+5) +(7+11) +(3+1). Thanks for asking ❎ ? question. Galat hia 11+9+11 |
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| 7582. |
4565+5675×56+56 |
| Answer» 727421 | |
| 7583. |
Pls help me for ogive graph |
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| 7584. |
How to understand the circle chapter example 1 |
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| 7585. |
What is cf |
| Answer» Cumulative frequency | |
| 7586. |
1/(2a+b+2x)=1/2a+1/b+1/2x |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |
| 7587. |
What is the probability of getting t3 Sunday\'s match leap year |
| Answer» If you mean 53rd sunday , then the probability is 2/7 | |
| 7588. |
If the mth term of an AP is 1/n and it\'s nth term be 1/m then show that it\'s mnth 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 | |
| 7589. |
Syllabus for board exams |
| Answer» It\'s clearly given in mycbseguide.So u can refer this. | |
| 7590. |
If x=cot^ + cosec^ find the value of 1+cos^/1-cos^ |
| Answer» And : x | |
| 7591. |
A boiler is in the form of a cylinder 2 m long with hemispherical ends of 2 m find volume???? |
| Answer» According to the question,we are given that,Diameter of common base = 2 mThen, radius of common base =\xa0{tex}\\frac { 2 } { 2 }{/tex}\xa0= 1 mHeight of the cylinder = 2 mVolume of boiler = Volume of cylinder + 2(Volume of hemisphere){tex}= \\pi r ^ { 2 } h + 2 \\times \\frac { 2 } { 3 } \\pi r ^ { 3 }{/tex}{tex}= \\frac { 22 } { 7 } \\times 1 \\times 1 \\times 2 + 2 \\times \\frac { 2 } { 3 } \\times \\frac { 22 } { 7 } \\times 1 \\times 1 \\times 1{/tex}{tex}= \\frac { 44 } { 7 } + \\frac { 88 } { 21 }{/tex}{tex}= \\frac { 132 + 88 } { 21 }{/tex}{tex}= \\frac { 220 } { 21 } \\mathrm { m } ^ { 3 }{/tex} | |
| 7592. |
5+5 |
| Answer» 10 | |
| 7593. |
4 term of an ap is zero prove that the 25th term of the AP 3 times its 11 term |
| Answer» 4th term of AP is 0 This implies that 4th term = a+ (4-1)d. 0=a+3d. a=-3d11th term = a+(11-1)d. 11th =-3d+10d. 11th=7d. equation-----(1). 25th term=a+(25-1)d. 25th= -3d+24d25th =21d. equation--------(2)On comparing equation--(1) & equation--(2) ,we get3 times of 11th term =25th term | |
| 7594. |
ABCD IS A SQUARE OF SIDE14CM AMB &CMD ARE SEMICIRCLES FIND THE PERIMETER OF THE SHADED REGION |
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Answer» 44cm 42cm |
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| 7595. |
What is locus |
| Answer» Locus is\xa0a curve or other figure formed by all the points satisfying a particular equation of the relation between coordinates, or by a point, line, or surface moving according to mathematically defined conditions. | |
| 7596. |
The sum of the interior angle of a pentegon |
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Answer» 540 540 |
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| 7597. |
Xraised to the power y=y raised to the power z=z raised to the power x ,then x.y.z= ??? |
| Answer» | |
| 7598. |
1/secA = |
| Answer» CosA | |
| 7599. |
Last 10 years question papers of class 10 |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 7600. |
Cot^-cos^/cot^+cos^=2-√3/2+√3 |
| Answer» 60° | |