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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.
| 24051. |
How can I download CBSE latest sample paper |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 24052. |
Using Euclid division algorithm , find the hcf of 56,96 and404 |
| Answer» | |
| 24053. |
Cot=3/5 find the value of other trigonometric ratios |
| Answer» SinA=5/√34 : CosA=3/√34 : TanA=5/3 : SecA=√34/3 : and CosecA=√34/5 ??????? | |
| 24054. |
What is discriminant? |
| Answer» The discriminant of a polynomial is a polynomial function of its coefficients, which allows deducing some properties of the roots without computing them. | |
| 24055. |
Show that there is no positive integer n for which √n-1 +√n+1 is rational |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares.Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 24056. |
Find the area of ring whose outer and inner radio are 23 and 12 respectively |
| Answer» 22/7×23+12×23-12 | |
| 24057. |
Write thé areas of thé first qudrant of a circule of radius R. |
| Answer» 1/4 πr^2 | |
| 24058. |
Sidesof two triangle similrar triangle are n thé ratio 4:9.what are thé ratio of areas of triangles? |
| Answer» 16:81 | |
| 24059. |
Root of equation x^2+x-p(p+1)=0 where p is constant |
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| 24060. |
Draw a line segment of length 7cm and divide it in the ratio of 2:5measure it\'s two part length? |
| Answer» 2 5 | |
| 24061. |
SinA-cosA+1/sinA+cosA-1=1/secA-tanA prove LHS=RHS |
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| 24062. |
Example 4 of chapter 8 |
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| 24063. |
3 digit no which is divisible by 9 how many term in this ap |
| Answer» 108 is smallest and 999 is largest three digit number which is divisible by 9 ,so AP is108,117,...….....,999a=108, d=9, a\u200b\u200b\u200b\u200b\u200b\u200bn=999a\u200b\u200b\u200b\u200b\u200b\u200bn=a+(n-1)d999=108+(n-1)9 9( n-1 )=999-108n-1=891/9n-1=99n =99+1n=100There are \'100 \' terms | |
| 24064. |
If one is zero of 3x^2-4x+p is resiprocal to the others then find the value of p |
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| 24065. |
What is arc |
| Answer» hj | |
| 24066. |
If p-cosec=cot,,then show that p-1\\p+1= cos |
| Answer» Given, cosec θ + cot θ = p...(i)We know that, {tex}cosec^2\\theta-cot^2\\theta=1{/tex}{tex}\\Rightarrow (cosec\\theta+cot\\theta)(cosec\\theta-cot\\theta)=1{/tex}{tex}\\Rightarrow p(cosec\\theta-cot\\theta)=1{/tex}{tex}\\Rightarrow cosec\\theta-cot\\theta=\\frac 1p{/tex}\xa0....(ii)Adding i and ii, we get{tex}2cosec\\theta=p+ \\frac 1p{/tex}{tex}cosec\\theta=\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow sin\\theta= \\frac{1}{cosec\\theta}=\\frac{2p}{p^2+1}{/tex}We know that,{tex}cos\\theta=\\sqrt{1-sin^2\\theta}=\\sqrt{1- \\frac{4p^2}{(p^2+1)^2}}=\\sqrt{\\frac{p^4+1-2p^2}{(p^2+1)^2}}{/tex}{tex}cos\\theta=\\sqrt{\\frac{(p^2-1)^2}{(p^2+1)^2}}=\\frac{p^2-1}{p^2+1} {/tex} | |
| 24067. |
If sin+cos=√3,then prove that tan+cot=1 |
| Answer» | |
| 24068. |
If the pth term of an AP is q and the qth term is p, show that rth term is p+q-r |
| Answer» Its very lengthy | |
| 24069. |
In AN ap: Sum of firme 10 terme is -150 and sum of its next 10 termin is - 550. Fino AP. |
| Answer» According to the question,the sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550Let a be the first term and d be the common difference of the given AP.Then, we haveS10=-150{tex}\\Rightarrow \\frac { 10 } { 2 } [ 2 a + 9 d ] = - 150{/tex}{tex}\\Rightarrow{/tex}5[2a+9d]=-150{tex}\\Rightarrow{/tex}2a+9d=-30...(i)Clearly, the sum of first 20 terms =-150+(-550)=-700{tex}\\therefore{/tex}S20=-700{tex}\\Rightarrow \\frac { 20 } { 2 } [ 2 a + 19 d ] = - 700{/tex}{tex}\\Rightarrow{/tex}10[2a+19d]=-700{tex}\\Rightarrow{/tex}2a+19d=-70...(iii)Subtracting (i) from (ii), we get10d=-40{tex}\\Rightarrow{/tex}d=-4{tex}\\Rightarrow{/tex}2a=-30-9(-4)=-30+36=6{tex}\\Rightarrow{/tex}a=3Thus, we have\xa0First term=a=3Second term= a+d=3+2(-4)=-1Third term=a+2d=3+2(-4)=3-8=-5Fourth term=a+3d=3+3(-4)=3-12=-9Thus, the given AP is 3,-1,-5,-9,.... | |
| 24070. |
If sina+sinb+sinc=3and0< a, b, c |
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| 24071. |
Appropriate method |
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| 24072. |
If P(x,y)is equidistant from the points A(7,-2) and B(3,1), express y in terms of x. |
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| 24073. |
If i fail in one subject than i pramote or not???? |
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| 24074. |
If sin a+cos =root2 then find the value of tan a+cot a |
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| 24075. |
(42}74 |
| Answer» 3108 is the answer. | |
| 24076. |
9^x+2-9^x=240 |
| Answer» 9(x+2)-9x=2409x (92-1)=2409x (81-1)=2409x (80)=2409x =332x=32x=1x=0.5\xa0 | |
| 24077. |
H.c.f. |
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Answer» Highest common factor is the ans. Highest common factor |
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| 24078. |
If sin theta =8 cos theta , then find the value of 1+sin theta /1-cos theta ×cot theta |
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| 24079. |
How u can prove 0=1 |
| Answer» | |
| 24080. |
A copper diameter is 1cm and length is 8cm drawn into a wire 18cm. Find its diameter |
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| 24081. |
Determine the ratio in which the 27x+9y-8=0 divide the line segment into (1,3)and(2,7). |
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Answer» You answer yourself? 3/4 |
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| 24082. |
9/6 |
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Answer» 9/6 = 3/2 = 1.5 3/2 i. e 1.5 |
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| 24083. |
Wright a relation between 2 and 3 |
| Answer» There are many relation such as--1. 2 is the presecessor of 3.2. 3 is the succeasor of 2.3. 2 and 3 are natural, prime, whole and positive integers.4. 2+1=35. 3-1=2Or kisi ko pta h to vo b btade.. | |
| 24084. |
√p is an irrational number |
|
Answer» Yes Yes |
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| 24085. |
What is the greatest number amongst 2 1/2,3 1/3,8 1/8and9 1/9 |
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Answer» 91/9 81/8 |
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| 24086. |
Sum of supplymentry angle |
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Answer» 360 180 90° |
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| 24087. |
I want maths marking scheme of class 10 ? |
| Answer» You can check the marking scheme in the syllabus :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 24088. |
The height of a cone is 10 cm . The cone is divide into 2 parts |
| Answer» Let the radius of original cone be r2Radius of\xa0cut of cone be r1According to the questionHeight of the original\xa0cone = 10 cm (given)The cone is cut off from the midpoint of the height,therefore, height the cone cut off = 5 cm{tex}\\Delta A O C \\sim \\Delta A\' O\' C{/tex}OA = Radius of original cone = r2O\'A\' = Radius of cutoff cone = r1Ratio\xa0of radius of two cones = Ratio of the height of cones{tex}\\therefore \\quad \\frac { A O } { A\' O ^ { \\prime } } = \\frac { r _ { 2 } } { r _ { 1 } } = \\frac { 10 } { 5 }{/tex}{tex}\\Rightarrow \\quad r _ { 2 } = 2 r _ { 1 }{/tex}Radius of original cone = 2 (radius of the cut off cone)Volume of the cut off cone\xa0{tex}= \\frac { 1 } { 3 } \\pi r _ { 1 } ^ { 2 } \\times 5{/tex}{tex}= \\frac { 5 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsVolume of original cone\xa0{tex}= \\frac { 1 } { 3 } \\pi \\left( 2 r _ { 1 } \\right) ^ { 2 } \\times 10{/tex}{tex}= \\frac { 40 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsVolume of frustum = Volume of an original cone - Volume of cut off cone{tex}= \\frac { 40 } { 3 } \\pi r _ { 1 } ^ { 2 } - \\frac { 5 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}{tex}= \\frac { 35 } { 3 } \\pi r _ { 1 } ^ { 2 }{/tex}\xa0cubic\xa0unitsRequired ratio = Volume of frustum: Volume of cut off cone{tex}= \\frac { 35 \\pi r _ { 1 } ^ { 2 } } { 5 \\pi r _ { 1 } ^ { 2 } } = \\frac { 7 } { 1 }{/tex}Therefore, the required ratio\xa0= 7: 1. | |
| 24089. |
Is 68 a term of the ap 7 ,10 ,13 ,....? |
| Answer» No | |
| 24090. |
Trigonometry ratio |
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Answer» sin ,cos, tan ,cosec ,sec ,cot are the trigonometric ratios. Sin, cos, tan, cosec, sec, cot are the trigonometric ratios |
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| 24091. |
A circle touches all four sides of a quadrilateral abcd with ab =6 cm bc=7cm and cd =4cm find ad |
| Answer» Given,\xa0a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm.\xa0If a circle touches all the four sides of quadrilateral ABCD, then\xa0AB + CD = AD + BC{tex}\\therefore{/tex}\xa06 + 8 = AD + 9\xa0{tex}\\Rightarrow{/tex}\xa014 = AD + 9{tex}\\Rightarrow{/tex}\xa014-9 = AD{tex}\\Rightarrow{/tex}\xa0AD = 5 cm | |
| 24092. |
Teroram of triangle |
| Answer» | |
| 24093. |
The number (17)to the power 2001 is divided by 7 than what will be the possible remainder. |
| Answer» | |
| 24094. |
Trigonometry formula |
| Answer» Service enable | |
| 24095. |
Find the value of k if the quadratic equation 3x square -k root 3x +4=0 has equal roots |
| Answer» K = +- 4 | |
| 24096. |
The angle of elevation is 60°on site of tower find the height of tower on the ground level is 10m. |
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Answer» Tan Q=p/b, tan 60=p/10 root 3 =p/10 10×root 3=p p=10 root 3 it\'s a answer 10 underroot 3m Tan30° = √3 Let the height be h so h/10 = √3 & h= 10√3 m |
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| 24097. |
Divide a line segment of length 11 cm in the ratio of 2:5 internally |
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| 24098. |
In a circle the radius is 4 CM. Find circumference. |
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Answer» 2×3.14×4=25.12 cm 25.14cm |
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| 24099. |
When to apply total surface area and lateral surface area? |
| Answer» | |
| 24100. |
How to find the square root of 12.25 |
| Answer» First find the square root of 1225 and the divide it by the square root of 100 (point ke baad do digit h) .The answer will be 35 .Got it! | |