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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.
| 8051. |
If m time the mth term Of an AP is n times its nth term Show that (m+n)th term is 0 |
| Answer» Let a be the first term and d be the common difference of the given AP. Then, in general the mth and nth terms can be written as\xa0Tm = a + (m - 1) d and Tn = a + (n - 1)d respectively.According to the question,we are given that,\xa0(m.Tm) = (n.Tn){tex}\\Rightarrow{/tex}\xa0m.{a + (m - 1)d} = n.{a + (n - 1)d}{tex}\\Rightarrow{/tex}\xa0a.(m - n) + {(m2 - n2) - (m - n)} . d = 0{tex}\\Rightarrow{/tex}\xa0(m - n).{a + (m + n - 1)}d.{tex}\\Rightarrow{/tex}\xa0(m - n).Tm+n = 0{tex}\\Rightarrow{/tex}Tm+n = 0 [{tex}\\because{/tex}\xa0(m-n){tex}\\neq{/tex}0].Hence, the (m + n)th term is zero. | |
| 8052. |
Which kind of paper comes in cbse 10 ?? |
| Answer» | |
| 8053. |
Do you know that all the heights measure throgh trignometru rule |
| Answer» Many rules | |
| 8054. |
If sin theta and cos theta are the roots of the equation. ax²+bx+c =0 prove that a²-b²+2ac =o |
| Answer» The given equation is {tex}ax^2 + bx + c = 0{/tex}sin\xa0{tex}\\alpha{/tex}\xa0and cos\xa0{tex}\\alpha{/tex}\xa0are roots of the given equation.{tex}\\therefore{/tex}\xa0sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex}\xa0=\xa0-{tex}\\frac{b}{a}{/tex}\xa0...(i)and {tex}sin{/tex}\xa0{tex}\\alpha{/tex}{tex}.cos{/tex}\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{c}{a}{/tex}...(ii)Squaring both sides of equation (i),\xa0{tex}\\Rightarrow {/tex}(sin\xa0{tex}\\alpha{/tex}\xa0+ cos\xa0{tex}\\alpha{/tex})2 =\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}sin^2{/tex}\xa0{tex}\\alpha{/tex}\xa0{tex}+ cos^2{/tex}\xa0{tex}\\alpha{/tex}\xa0+\xa02 sin\xa0{tex}\\alpha{/tex}cos\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}1 + 2{tex}\\frac{c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{a + 2c}{a}{/tex}\xa0=\xa0{tex}\\frac { b ^ { 2 } } { a ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}{tex}a^2 + 2ac = b^2{/tex} | |
| 8055. |
Do you know the difference between median and bisector |
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Answer» A median devides the line in 2 equal parts perpendicularly while a bisector devides the line segment perpendicularly when it is a perpendicular bisector Median divides a triangle in two equal area but bisector bisect the line |
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| 8056. |
Where the π comes from?? |
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Answer» Aryabhatt gave the value of Π . Greek 22/7 is comes from divide the circumferences of circle and it\'s diameter Whooo knowsss |
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| 8057. |
When will be first paper of class 10th in 2017_18 |
| Answer» | |
| 8058. |
What is locus ?? |
| Answer» Location of a certain gene on chromosomes | |
| 8059. |
Please give us cbse pre board exam date sheet |
| Answer» | |
| 8060. |
If a coin toss 2 times what will the probability of getting even? |
| Answer» 1 | |
| 8061. |
The sum of first 7 terms is 49 and that of first 17 terms is 289 . Find the sum of first n terms . |
| Answer» Sn= {tex}\\frac n2{/tex}[2a + (n - 1)d]S7\xa0=\xa0{tex}\\frac 72{/tex}(2a + 6d) = 49or, {tex}a + 3d = 7{/tex}...........(i)S17\xa0=\xa0{tex}\\frac{17}{2}{/tex}(2a + 16d) = 289or, {tex}a + 8d = 17{/tex}..........(ii)On subtracting (i) from (ii), we getor, 5d = 10 or, d = 2Put d = 2 in (i)a + 3d = 7a + 2(3) = 7a + 6 = 7and a = 1{tex}S _ { n } = \\frac { n } { 2 } [ 2 \\times 1 + ( n - 1 ) 2 ]{/tex}{tex}= \\frac { n } { 2 } [ 2 + 2 n - 2 ] {/tex}{tex}= \\frac { n } { 2 } [ 2 n ] {/tex}{tex}= n ^ { 2 }{/tex}Hence, sum of n terms = n2 | |
| 8062. |
WHICH CHAPTERS CONTAIN MORE MARKS IN BOARD EXAM |
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Answer» 20 marks Chapter of algebra unit.☺? |
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| 8063. |
A hemisphere has it\'s volume and surface area numerically equal . What would be it\'s diameter ? |
| Answer» Let radius of the\xa0hemisphere\xa0be r.According to the question,Volume of hemisphere\xa0= Surface area\xa0of hemisphere{tex}\\frac { 2 } { 3 } \\pi r ^ { 3 } = 3 \\pi r ^ { 2 }{/tex}{tex}r = \\frac { 9 } { 2 }{/tex}unitsDiameter of the hemisphere = 2( radius of the hemisphere){tex}\\therefore \\quad \\text { diameter } = \\frac { 9 } { 2 } \\times 2 = 9{/tex} | |
| 8064. |
Hcf of 135&225 by eucild division algorithm |
| Answer» 225=135×1+90135=90 ×1+4590 =45 ×2+00 | |
| 8065. |
2x-2y-2=0 |
| Answer» 2(x-y-1)=0X-y-1=0\\2X-y=1 | |
| 8066. |
Furmula of area of triangle |
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Answer» 1/2×base ×height Half ×base×height |
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| 8067. |
Can we have blueprint for 10th board exam 2017-18 ? |
| Answer» Yes sure | |
| 8068. |
prove pythagores theorem |
| Answer» Data :Consider a right angled triangle ABC right angled at bTo prove: hypotenuse square=base square + altitude squareConst: draw a perpendicular from b on side ac and name as d Proof:consider ∆abd and ∆abc They are similar by aa criteria AB/AD=AC/AB. (Cpst) Implies AB²=AC×AD...........(1)SIMILARLY consider ∆abc and ∆bdcThey are similar by aa criterionBy cpst AC/BC=BC/DCBC²=AC×DC...............(2)Adding 1 and 2AB²+BC²= AC×BC+AC×ADAB²+BC²=AC(BC+AD)AB²+BC²=AC(AC)AB²+BC²=AC²HENCE PROVED | |
| 8069. |
. Write the formula for the mid-point of a class interval. |
| Answer» Upper limit plus lower limit divided by two | |
| 8070. |
In which month cbse board exam held2018 |
| Answer» March | |
| 8071. |
Find the H.C.F of 81 and 237and express it as a linear combination of 81 and 237. |
| Answer» Since, 237 > 81On applying Euclid\'s division algorithm, we get237 = 81 {tex}\\times{/tex}\xa02 + 75 ..........(i)81 = 75 {tex}\\times{/tex}\xa01 + 6 ..........(ii)75 = 6 {tex}\\times{/tex}\xa012 + 3 .........(iii)6 = 3 {tex}\\times{/tex}\xa02 + 0 .............(iv)Hence, HCF (81,237) = 3.In order to write 3 in the form of 81x + 237y,Now,{tex}\\style{font-family:Arial}{\\begin{array}{l}3=75-6\\times12(\\;from(iii))\\\\=75-(81-75\\times1)\\times12\\\\=75-(81\\times12-75\\times12)\\\\=75-81\\times12+75\\times12\\\\=75(1+12)-81\\times12\\\\=75\\times13-81\\times12\\\\=13\\times(237-81\\times2)-81\\times12\\;(\\;Substuting\\;75\\;from(i))\\\\=13\\times237-13\\times81\\times2-81\\times12\\\\=237\\times13-81\\times(26+12)\\\\=81(-38)+237\\times13\\\\=81x+237y\\end{array}}{/tex}Hence, x = -38 and y = 13\xa0 | |
| 8072. |
Way to study trigonometry ratios |
| Answer» | |
| 8073. |
HCF OF (306,954,1315)=18LCM=? |
| Answer» LCM=306 x 1314/18=17 x 1314 = 22338 | |
| 8074. |
What is a quadratic equation? |
| Answer» | |
| 8075. |
Which of the following numbers have terminating decimal expansion?A.8/225B.5/18C.11/21D.21/150 |
| Answer» 21/150=7/5050=2×5×5=2(1)^×5 (2)^ | |
| 8076. |
SinQ+cotQ=1 |
| Answer» | |
| 8077. |
O is the center of circle area of sector is 5/11 find the angle |
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| 8078. |
How the school will give 20 marks for each subjects on what basis |
| Answer» 5 marks for copy 5 for lab in maths/ science 10 in perodic tests where best 2 are selected | |
| 8079. |
When will be the exams of class 10 session 2017- 2018 |
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Answer» From 15february From march |
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| 8080. |
27-5 |
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Answer» 22 22 |
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| 8081. |
When the time table come |
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| 8082. |
How to prove a parallogram surcumscribing a circle is a rhombus |
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| 8083. |
formula of cube |
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| 8084. |
Are the triangle having same area congruent? |
| Answer» True | |
| 8085. |
Trigonometric |
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| 8086. |
Is maths easy.. how to learn that easily |
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Answer» T k ? Just think it as a great puzzle of life and you will die if you can\'t solve it. |
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| 8087. |
Sin4x/a+cos4x/b=1/a+bProve that Sin8x/a3+cos8x/b3=1\\(a+b)3 |
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| 8088. |
All formula of area related to circle |
| Answer» Area of circle is πr² | |
| 8089. |
Find the values of p for which the quadratic equation 4x2+px+3=0 has equal roots |
| Answer» This equation have equal roots so B square- 4 AC = 0=>p (2)^-4×4×3=0=>p (2)^=4×4×3=>p=root (4×4×3)=4root3 | |
| 8090. |
What do you mean by ap? |
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Answer» Arithmetic progression...??? It is maths |
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| 8091. |
Write 294 as products of its prime number |
| Answer» 2×3×7×7......;)?????? | |
| 8092. |
When 2^256 is divided by 17 the remainder would be (1). 1 (2).16 (3). 14 (4). None of these |
| Answer» (4) none of these Because right answer is 1.65 | |
| 8093. |
Construction last exercise last 3 question plz |
| Answer» | |
| 8094. |
6-2+5-9-1+55*0= |
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Answer» Right answer is -1 According to BODMAS rule. 0 is the answer????????? |
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| 8095. |
Wat is ashock |
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| 8096. |
2+1-3= |
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Answer» 0 We know that from DMAS...so by seeing this question we can say that the answer is 0 because 2 is added to one and then subtracted to 3 According to BODMAS Answer is 0 0 |
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| 8097. |
9+9+9+ |
| Answer» 27 | |
| 8098. |
Cos2a =sin2a |
| Answer» | |
| 8099. |
What do you mean by quadrant |
| Answer» One - fourth of a circular disc is called a quadrant .The central angle of a quadrant is 90 | |
| 8100. |
why is raninbow is not in circular? |
| Answer» | |