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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.
| 24101. |
Write the fourmlas of chapter area related to circle |
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| 24102. |
If sin +cos=√3 then prove tan +cot=1 |
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| 24103. |
√3 |
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| 24104. |
How to prove 1=2 |
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| 24105. |
What is the coordinate of centroid |
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| 24106. |
How can we make time table for study |
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| 24107. |
Math me best sample paper kiska acha hota hai |
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Answer» Rachna sagar &Oswal also Together or arihant |
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| 24108. |
How we can make time table for study, please tell..for all subjects. |
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| 24109. |
When 10 boards date of exams |
| Answer» The board exams are expected to start from first of March 2018. CBSE is working on proposal to start them from mid-February so that results can be declared a bit earlier but not final decision is taken yet. We expect final guidelines by the end of December 2017.\xa0 | |
| 24110. |
Can 5n ends with digit 0 |
| Answer» Yas | |
| 24111. |
What is Unitary Method? |
| Answer» The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.\xa0 | |
| 24112. |
How to construct tangent of circle ? |
| Answer» Take a point outside the circle and then join it with the centre..Then draw the perpendicular bisector of the line..then u will get the mid point from where u will draw a circle.. then where the two circles will meet will be the tangent drawn for the first circle....May be u understand...☺☺ | |
| 24113. |
Which term of the AP :3,8,13,18......is, 78? |
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Answer» 16 th term 16th term 16 |
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| 24114. |
In an AP what is the meaning of double answer |
| Answer» it means that the second terms sum is zero | |
| 24115. |
What is Composite Numbur Defination in Hindi |
| Answer» Jan koi no. ka 2 se Zara factors hota hai toh usko composite number bolte hai...For ex... 4 - 1, 2 and 4 | |
| 24116. |
2 cubes each of volume 64cm are joined end to end .find the surface area of the resulting cuboid |
| Answer» Two cubes each of volume 64 cm3 are joined end to end. We have to find the surface area and volume of the resulting cuboid.Let the length of each edge of the cube of volume 64 cm{tex}^3{/tex} be x cm. Then,Volume = 64 cm3\xa0{tex}\\Rightarrow{/tex}\xa0x3 = 64{tex}\\Rightarrow{/tex}\xa0x3 = 4\u200b\u200b\u200b\u200b\u200b\u200b3{tex}\\Rightarrow{/tex}\xa0x = 4 cmThe dimensions of the cuboid so formed are:L = Length = (4 + 4) cm = 8 cm, b = Breadth = 4 cm and, h = Height = 4 cmSurface area of the cuboid = 2 (lb + bh + Ih)= 2\xa0{tex}( 8 \\times 4 + 4 \\times 4 + 8 \\times 4 ) \\mathrm { cm } ^ { 2 }{/tex}= 160 cm2Volume of the cuboid = Ibh ={tex}8 \\times 4 \\times 4 \\mathrm { cm } ^ { 3 }{/tex}\xa0= 128 cm3 | |
| 24117. |
Will the scholarship given to the students having 10 cgpa |
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| 24118. |
6-8 |
| Answer» -2 | |
| 24119. |
Matrix |
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| 24120. |
What is concentric circles |
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Answer» i.e. the circles are drawn from a common center and both the circle have different radius ? The circles having same center but different radius . |
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| 24121. |
1/a+b+x=1/a+1/b+1/x find the value of x in the term of a and b |
| Answer» x=-a and x=-b | |
| 24122. |
Find the HCF and lCM of 222,664 |
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Answer» Factors of 222 = 2*3*37Factors of 664= 2*2*2*83The only common factor is 2. So HCF is 2.We know that HCF * LCM = 222*6642*lcm = 222*664LCM = 222*664/2LCM= 73704\xa0 HCF 2 and LCM 12 |
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| 24123. |
What is Euclid\'s division lemma? |
| Answer» a = bq+r | |
| 24124. |
If cosec A +cot A=P prove that cosA=p×p-1÷p×p+1 |
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| 24125. |
X^2+2x+5 |
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| 24126. |
Find value of sin60 numerically |
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Answer» {tex}√3/2{/tex} √3÷2 |
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| 24127. |
If in Traingle ABC, AD is median and AM perpendicular to BC, then prove that AB2+AC2= 2AD2+1/2BC2 |
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| 24128. |
How many multiple of 9 lie between 10 and 300 |
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Answer» 32 First multiple of 9 is 18 and last multiple of 9 is 297 between 10 and 300.So, First term a = 18, Common difference d= 9an\xa0= a+(n-1)d297 = 18+(n-1)9297-18= (n-1)9279 = (n-1)9(n-1) = 279/9n-1= 31n = 31+1n=32So, there are 32 multiples of 9 between 10 and 300.\xa0 |
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| 24129. |
find the mean of First 10 Natural number |
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Answer» 5.5 First 10 natural numbers are 1,2,3,4,5,6,7,8,9 and 10.Mean = Sum of numbers / total numbers\xa0Mean = 1+2+3+4+5+6+7+8+9+10/10mean = 55/10Mean = 5.5Another Method\xa0When difference between all numbers is constant then,\xa0Mean = First term + last term /2mean = 1+10/2Mean= 11/2Mean= 5.5\xa0 |
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| 24130. |
Abcg=5 |
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| 24131. |
2+2÷2= |
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Answer» 3 3 3 |
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| 24132. |
Fourth term of an ap is 0 prove 25th term is 3time it 11th term |
| Answer» We have,a4\xa0= 0{tex}a + 3d = 0{/tex}{tex}3d = -a{/tex}or {tex}-3d = a{/tex}..........(i)Now,a25\xa0= {tex}a + 24d{/tex}= {tex}-3d + 24d{/tex} [Putting value of a from eq(i)]= {tex}21d{/tex}...........(ii)a11\xa0= {tex}a + 10d{/tex}= {tex}-3d + 10d{/tex}= {tex}7d{/tex}.........(iii)From eq(ii) and (iii), we geta25 = 21 da25 = 3(7d)a25\xa0= 3a11Hence Proved | |
| 24133. |
7 sinA+3cosA =1 upon root 3 tan |
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| 24134. |
What is the approx date for maths board examination |
| Answer» 10 March | |
| 24135. |
Find the coordinate of the point on y-axis which is nearest to tge point (-2,5) |
| Answer» /29 | |
| 24136. |
16x²—10/x=27 |
| Answer» We have, {tex}16x - \\frac{{10}}{x} = 27{/tex}{tex}\\Rightarrow \\frac{{16{x^2} - 10}}{x} = 27{/tex}{tex}\\Rightarrow{/tex} 16x2 - 10 = 27x{tex}\\Rightarrow{/tex} 16x2 - 27x - 10 = 0{tex}\\Rightarrow{/tex} 16x2 - 32x + 5x - 10 = 0{tex}\\Rightarrow{/tex} 16x(x - 2) + 5(x - 2) = 0{tex}\\Rightarrow{/tex} (16x + 5)(x - 2) = 0{tex}\\Rightarrow{/tex} (16x + 5) = 0 or (x - 2) = 0{tex}\\Rightarrow x = - \\frac{{ 5}}{{16}}{/tex} or, x = 2Hence, the roots of given quadratic equation are 2 and\xa0{tex}-\\frac{{5}}{16}{/tex} | |
| 24137. |
When 35 is divided by 653768 than what will be thebanswer |
| Answer» 18679.08 | |
| 24138. |
0.2x+0.3y=1.3 0.4x+0.5y=2.3Solve with substituion method |
| Answer» X=2 and y=3 | |
| 24139. |
What is mean of group data |
| Answer» Calculating the sample mean of gruped data | |
| 24140. |
Write the AP |
| Answer» Arithmetic progression.....?? | |
| 24141. |
If m and n are the zeroes of the polynomial p(x) 3x square +11x+4 then find the value of m/n +n/m |
| Answer» According to the question,\xa0m and n are the zeroes of the polynomial 3x2 + 11x - 4,\xa0Let p(x) = 3x2 + 11x - 4 = 0= 3x2 + 12x - x - 4= 3x(x + 4) - 1(x + 4)= (3x - 1) (x + 4)So, zeroes are m =\xa0{tex}\\frac 13{/tex}\xa0and n =-4Now,\xa0{tex}\\frac { m } { n } + \\frac { n } { m } = \\frac { \\left( \\frac { 1 } { 3 } \\right) } { - 4 } + \\frac { - 4 } { \\left( \\frac { 1 } { 3 } \\right) }{/tex}{tex}=\\frac { 1 } { - 12 } - 12{/tex}{tex}= \\frac { - 145 } { 12 }{/tex} | |
| 24142. |
Is ncrt preferred for board exams |
| Answer» Yes, NCERT books are prescribed by CBSE. | |
| 24143. |
Find the zeros of f(x)=x3-kx2+39x-28 if it is given that the zeroes are in a.p |
| Answer» Let\xa0{tex}\\begin{array}{l}\\mathrm\\alpha,\\mathrm\\beta\\;\\mathrm{and}\\;\\mathrm \\gamma\\;\\mathrm{be}\\;\\mathrm{the}\\;\\mathrm{zeroes}\\;\\mathrm{of}\\;\\mathrm f(\\mathrm x)\\\\\\end{array}{/tex}and\xa0the zeroes are in AP.Suppose\xa0{tex} \\alpha = a - d , \\beta = a \\text { and } \\gamma = a + d{/tex}\xa0f(x) =\xa0x3 - 12x2 + 39x - 28{tex} \\therefore \\quad \\alpha + \\beta + \\gamma = - \\left( \\frac { - 12 } { 1 } \\right){/tex}= 12 ---------(1)and,\xa0{tex} \\alpha \\beta \\gamma = \\left( \\frac { - 28 } { 1 } \\right){/tex}= - 28 ---------(2)From equation\xa0(1) , we havea - d + a + a + d = 123a = 12\xa0{tex}\\Rightarrow{/tex}\xa0a = 4.Now from equation (2), we have(a-d)a(a+d) = 28a(a2-d2) = 284(16-d2) = 284d2\xa0= 64 - 28 = 36d2 = 36d =\xa0{tex}\\pm 6{/tex}Case I:\xa0When a = 4 and d = +6 : In this case,{tex}\\alpha{/tex}\xa0= a - d = 4 - 6\xa0= - 2,{tex} \\beta{/tex}\xa0= a = 4 and {tex} \\gamma{/tex}\xa0= a + d = 10CASE II: When a = 4 and d = -6: In this case,{tex}\\alpha{/tex}\xa0= a - d = 4 - (- 6 ) = 10 ,{tex} \\beta{/tex} = a = 4 and {tex} \\gamma{/tex}\xa0= a + d = 4 - 6 = - 2Hence, in either case the zeroes\xa0of the given polynomial are 10,4 and - 2. | |
| 24144. |
Find the largest number which which divides 615 and 963 leaving remainder 6 in each case. |
| Answer» The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.Therefore,The required number = H.C.F. of 609 and 957.By applying Euclid’s division lemma957 = 609 {tex}\\times{/tex}\xa01+ 348609 = 348 {tex}\\times{/tex}\xa01 + 261348 = 216 {tex}\\times{/tex}\xa01 + 87261 = 87 {tex}\\times{/tex}\xa03 + 0.Therefore, H.C.F. of 957 and 609 is = 87.Hence, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87. | |
| 24145. |
660/7+36√3 |
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| 24146. |
Datesheet for 10 class exam |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 24147. |
Sin60 value |
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Answer» √3/2 ?? Root3/2 |
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| 24148. |
Which book is considered best for exam preparation |
| Answer» R D Sharma | |
| 24149. |
Will the paper come easy? |
| Answer» No. ??? | |
| 24150. |
√3(tan10.tan30.tan40.tan50.tan80) |
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Answer» Sonu....kaise hua.... 1/root 3 or0.5777 1.44 (Approx) |
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