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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.
| 22101. |
Hcf 210,55 |
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Answer» Let 4-√3 be rational So, 4-√3 = p/q where p and q are co-prime integers and q not equal to 0. √3= 4-p/q √3=4q-p/q Since 4 ,p and q are integers. So,4q-p/p is rational (integer/integer).Where as √3 is irrational. It contradicts our assumption.so,4-√3 is irrational. 210=2×3×5×755=5×11So,HCF=5 Let us possible 4root 3 is a rational number and it\'s simplest form be a/b now, 4root 3 =a/b so, root 3 = a/4b for any positive value of a and b gives a /4b is a rational number. But it contradict the fact that root 3 is irrational. The contradiction arises by assuming the fact that 4 root 3 is irrational. Hence, 4 root 3 is an irrational number. Prove that 4-√3 is ir-rational |
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| 22102. |
Today is my maths paper what can I do now for good marks |
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Answer» Brother if u want to get Good marks in maths ,just solve all ncert problems along with examples.Only ncert study is enough to get good marks. you can do good study.Focus more on Ap trignometry and last 2 chapters |
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| 22103. |
Cot A+cosecA=5 Find cosA |
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Answer» Q.Cot A+cosecA=5 Find cosA Here is the required solution: CosA =4.... |
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| 22104. |
(23) -5436.657+66666+(-33) wHat his answee |
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| 22105. |
how to important question |
| Answer» Means | |
| 22106. |
About maths lab manual |
| Answer» What??? | |
| 22107. |
2√45 + 3√20 / 2√5 |
| Answer» Given ,\xa0{tex}{\\frac{2\\sqrt{45}+3\\sqrt{20}}{2\\sqrt{5}}} \\\\\\\\={\\frac{2\\sqrt{3\\times3\\times5}+3\\sqrt{2\\times2\\times5}}{2\\sqrt{5}}}\\\\\\\\={\\frac{2.3\\sqrt{5}+3.2\\sqrt{5}}{2\\sqrt{5}}}\\\\\\\\={\\frac{12\\sqrt{5}}{2\\sqrt{5}}}=6{/tex}Here you can see after simplifying we get 6 , according to definition of rational number [ a number , is in the form of P/Q, Q≠ 0 and P , Q ∈ I ] given number is rational number . | |
| 22108. |
Area of hexazon |
| Answer» {tex}A = \\frac{{3\\sqrt 3 }}{2}{a^2}{/tex} | |
| 22109. |
What is evaluation ?please tell me evidence of revaluation |
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| 22110. |
given a=7 a13=35 find d and s13 |
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Answer» Sorry sum=273 D=7/3 and sum=286 |
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| 22111. |
Find the sum of the following AP 2,7,12,...........upto 10 terms. |
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Answer» 245 is the answer of your question. Who are u sourav i mean frm which state 245 245 245 |
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| 22112. |
Relationship between mean median and mode |
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Answer» 3median =mode + 2mean. Mode = 3Median-2Mean Mode=3Median-2Mean 3 median is equal to Mode plus 2 mean.. 3 median= mode + 2 mean. |
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| 22113. |
The sum of first, third and seventeenth term of an ap is 216.find the first 13th term |
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Answer» 936 nth term of an A.P whose starting term is "a" and common difference is "d"is an = a+(n-1)d1st term = a 3rd term =a+2d 17th term = a+16dAccording to question\xa0a+a+2d+a+16d=2163a+18d=2163(a+6d) = 216a+6d= 216/3a+6d= 72Sum of "n" terms in an AP = n(2a+(n-1)d)/2Sum of thirteen terms = 13(2a+12d)/2 =13(a+6d) = 13*72 = 936 936 |
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| 22114. |
If sin a is equal to Cos A then tan A + sin square A + 1 is equal to |
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Answer» ya its cos square A Cos square A |
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| 22115. |
If sin a is equal to cos tan tan A + sin square A + 1 is equal to |
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| 22116. |
if 3x/2-y/3=2 and x/2+y/3=13/6 so what is the value of x and y |
| Answer» The given system of equations may be written as{tex}9x -10y + 12 = 0{/tex} ...(i){tex}2x + 3y - 13 = 0{/tex}.... (ii)From (ii), we get\xa0{tex} y = \\frac { 13 - 2 x } { 3 }{/tex}Substituting\xa0{tex} y = \\frac { 13 - 2 x } { 3 }{/tex}\xa0in (i), we get{tex} 9 x - \\frac { 10 ( 13 - 2 x ) } { 3 } + 12 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}27x - 10(13 - 2x) + 36 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}27x -130 + 20x + 36 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}47x - 94 = 0{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}47x = 94{/tex}{tex} \\Rightarrow x = \\frac { 94 } { 47 } = 2{/tex}Substituting {tex}x = 2{/tex} in (i), we get9\xa0{tex} \\times{/tex} 2 - 10y + 12 = 0\xa0{tex} \\Rightarrow 10 y = 30 {/tex}{tex}\\Rightarrow y = \\frac { 30 } { 10 } = 3{/tex}Hence, x = 2 and y = 3 is the required solution. | |
| 22117. |
1/a+b+x = 1/a + 1/b + 1/x. Solve for x |
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Answer» 1-2b/b=x x=-a and x=-b |
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| 22118. |
Check whether the following are quadratic equation |
| Answer» When the power is the it is qadratic eqation . | |
| 22119. |
Important formula of maths chapter 13 |
| Answer» Frastum ka h | |
| 22120. |
Prove thatSec theta+tan theta-1 / tan theta-sec theta+1=cos theta/1-sin theta |
| Answer» According to the question,L.H.S. =\xa0{tex}\\frac { \\sec \\theta + \\tan \\theta - 1 } { \\tan \\theta - \\sec \\theta + 1 }{/tex}{tex}= \\frac { ( \\sec \\theta + \\tan \\theta ) - \\left( \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta \\right) } { ( \\tan \\theta - \\sec \\theta + 1 ) }{/tex}\xa0[{tex}\\because{/tex} sec2{tex}\\theta{/tex} - tan2{tex}\\theta{/tex} = 1]{tex}= \\frac { ( \\sec \\theta + \\tan \\theta ) [ 1 - ( \\sec \\theta - \\tan \\theta ) ] } { ( \\tan \\theta - \\sec \\theta + 1 ) }{/tex}{tex}= \\frac { ( \\sec \\theta + \\tan \\theta ) ( \\tan \\theta - \\sec \\theta + 1 ) } { ( \\tan \\theta - \\sec \\theta + 1 ) } = ( \\sec \\theta + \\tan \\theta ){/tex}{tex}= \\left( \\frac { 1 } { \\cos \\theta } + \\frac { \\sin \\theta } { \\cos \\theta } \\right) = \\frac { ( 1 + \\sin \\theta ) } { \\cos \\theta } = \\frac { ( 1 + \\sin \\theta ) } { \\cos \\theta } \\times \\frac { ( 1 - \\sin \\theta ) } { ( 1 - \\sin \\theta ) }{/tex}{tex}= \\frac { \\left( 1 - \\sin ^ { 2 } \\theta \\right) } { \\cos \\theta ( 1 - \\sin \\theta ) } = \\frac { \\cos ^ { 2 } \\theta } { \\cos \\theta ( 1 - \\sin \\theta ) } = \\frac { \\cos \\theta } { ( 1 - \\sin \\theta ) }{/tex}\xa0= R.H.S.{tex}\\therefore{/tex}\xa0{tex}\\frac { \\sec \\theta + \\tan \\theta - 1 } { \\tan \\theta - \\sec \\theta + 1 } = \\frac { \\cos \\theta } { ( 1 - \\sin \\theta ) }{/tex} | |
| 22121. |
The perimeter of right angle triangle is 60 CM its hypotenuse is 25 cm find the area of the triangle |
| Answer» Here,\xa0a + b + c = 60, c = 25a + b = 6 0 - ca + b = 6 0 - 2 5 = 35Using Pythagoras theorem|{tex}a ^ { 2 } + b ^ { 2 } = 625{/tex}\xa0Using identity\xa0{tex}( a + b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } + 2 a b{/tex}\xa0{tex}35 ^ { 2 } = 625 + 2 a b{/tex}\xa01225 - 625 = 2abor, ab = 300Hence, Area of\xa0{tex}\\triangle A B C = \\frac { 1 } { 2 } a b = 150 \\mathrm { cm } ^ { 2 }{/tex} | |
| 22122. |
Find the area of the shaded region in where ABCD is a square of side 14 cm |
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Answer» Type karne se pehle ques. dekh le bhai. Where is the shaded regions???? |
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| 22123. |
Surface area and volume ke formulas class 10th |
| Answer» Check revision notes for formulae :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 22124. |
What is the value of sin30 |
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Answer» 1/2 Value of sin 30- 1/2. 1/2 U wil get it in Ncert text book |
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| 22125. |
Check whether the following are quadratic equations (x-2)^2+1=2x-3 |
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| 22126. |
What is the formula of distance formula |
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Answer» √(x2-x1)^2+(y2-y1)^2 Distance = under root (x-y)^2 + (x1-y1)^2 |
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| 22127. |
√5 is irrational .prove (3-√5)is irrational |
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Answer» Let 3-_/5 be rational and co prime 3-_/5=a/b Therefore,3- a/b=_/5 _/5 = 3-a/b = 3b - a/bSince a and b are integers , we get 3 - a/b is rational, and so _/5 is rationalBut this contradicts the fact that _/5 is irrational.This contradiction has arison because of our incorrect assumption that 3 -_/5 is rational.So, we conclude that 3-_/5 is irrational. See in ncert book in chapter 1examples |
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| 22128. |
BPT and converse of BPT theorem also Theorem related to area and PT and converse of PT |
| Answer» It is given in ncert.. | |
| 22129. |
If sec theta + tan theta = p then find cosec theta |
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| 22130. |
If the n^th term of an AP is 2n+1 what is the sum of its first 3 terms. |
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Answer» A= 3, d= 2 hence ap. obtained is 3,5,7 so the sum becomes 15 15 |
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| 22131. |
Use euclid \'s division algorithm to find the HCF of 1) 135 and 225 |
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Answer» 45 Dfj |
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| 22132. |
How to post foto of ques in this app? |
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Answer» Don\'t know.... If anyone know plzz tell how to tke foto of ques and post in this app? Dont know |
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| 22133. |
Markingsheme of maths chapters.... |
| Answer» Number systems. 06 Algebra. 20Coordinate Geometry. 06Geometry. 15Trigonometry. 12Mensuration. 10Statistics and Probability. 11TOTAL = 80 | |
| 22134. |
What is the common difference of an A.P. in which a21-a7=84 |
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Answer» nisha tera galat hai The Common difference of AP in which a21-a7=84 is 6. a+20d-a+6d=84.26d=84d=6 a + 20d - a - 6d =84 14 d = 84 d = 84 / 14 d = 6 a+20d-a+6d=84d=6 d=6 |
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| 22135. |
proof that 1 equal to 2 |
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Answer» This question post by GK This question are proof by sakuntala devi (mathematics expert) right tanmay. 2-1=1+11=2 Is this a question or a joke. |
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| 22136. |
ax/b - by/a = a+bax - 2ab = bySolve the equations |
| Answer» The given equations are{tex}\\frac { a x } { b } - \\frac { b y } { a } - ( a + b ) = 0{/tex}{tex}ax - by - 2ab = 0{/tex}By cross multiplication, we have{tex}\\frac { x } { \\left( - \\frac { b } { a } \\right) \\times ( - 2 a b ) - ( - b ) \\times ( - ( a + b ) ) } ={/tex}{tex}\\frac { y } { - ( a + b ) \\times a - ( - 2 a b ) \\times \\frac { a } { b } }{/tex}{tex}= \\frac { 1 } { \\frac { \\mathrm { a } } { \\mathrm { b } } \\times ( - \\mathrm { b } ) - \\mathrm { a } \\times \\left( - \\frac { \\mathrm { b } } { \\mathrm { a } } \\right) } {/tex}{tex}\\Rightarrow \\frac { x } { 2 b ^ { 2 } - b ( a + b ) } = \\frac { y } { - a ( a + b ) + 2 a ^ { 2 } } = \\frac { 1 } { - a + b }{/tex}or,\xa0{tex}\\frac { x } { 2 b ^ { 2 } - a b - b ^ { 2 } } = \\frac { y } { - a ^ { 2 } - a b + 2 a ^ { 2 } } = \\frac { 1 } { - a + b }{/tex}{tex}\\Rightarrow \\frac { x } { b ^ { 2 } - a b } = \\frac { y } { a ^ { 2 } - a b } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\Rightarrow \\frac { x } { - b ( a - b ) } = \\frac { y } { a ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\therefore \\frac { x } { - b ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}\xa0and\xa0{tex}\\frac { y } { a ( a - b ) } = \\frac { 1 } { - ( a - b ) }{/tex}{tex}\\therefore x = \\frac { - b ( a - b ) } { - ( a - b ) } \\text { and } y = \\frac { a ( a - b ) } { - ( a - b ) }{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}x = b,\\ and\\ y = -a{/tex}{tex}\\therefore{/tex} the solution is {tex}x = b, y = -a{/tex} | |
| 22137. |
How to prove pythagorous therom |
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Answer» Ncert m diya h This is given in ncert. use AC^2=AB^2+BC^2 Rs agarwal me diya h......?? |
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| 22138. |
(2+5) -(43-31) |
| Answer» (7)-(12)=-5 | |
| 22139. |
Hii.....everyone!!! Can anyone plz tell me the most important chapters in maths?? |
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Answer» Ohh..thnks Ch 1,5,8,14 Trigonometry identies and triangle ????? Only 15 chapters are important in math.... Hi best |
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| 22140. |
Find the area of the shaded region in the following figure |
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Answer» Figureee? Figure ??? Which figure¿ Where is the figure??? |
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| 22141. |
find the distance between ((a+b),(a-b)), ((a-b),(a+b)) |
| Answer» Dude it\'s easy zero is the answer | |
| 22142. |
Example 8 ,chapter 6 |
| Answer» | |
| 22143. |
Show that the square of any positive odd integer is of the form 4m+1 for some integer m. |
| Answer» Let a be any positive integer and b = 4\xa0Then, by Euclid\'\'s algorithm a = 4q + r for some integer q\xa00 and 0\xa0\xa0r < 4\xa0Since, r = 0, 1, 2, 3\xa0Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3\xa0Since, a is an odd integer, o a = 4q + 1 or 4q + 3\xa0Case I:\xa0When a = 4q + 1\xa0Squaring both sides, we have,\xa0a2\xa0= (4q + 1)2\xa0\xa0a2\xa0= 16q2\xa0+ 1 + 8q\xa0 = 4(4q2\xa0+ 2q) + 1\xa0 = 4m + 1 where m = 4q2\xa0+ 2q Case II:\xa0When a = 4q + 3Squaring both sides, we have,a2\xa0= (4q +3)2 = 16q2\xa0+ 9 + 24q = 16 q2\xa0+ 24q + 8 + 1 = 4(4q2\xa0+ 6q + 2) +1 = 4m +1 where m = 4q2\xa0+7q + 2Hence, a is of the form 4m + 1 for some integer m.\xa0 | |
| 22144. |
Sin (1+tan)+ cos(1+cot)=sec+cosec |
| Answer» According to given sum,sin (1+tan)+cos (1+cot)=sec+cosec\xa0=> LHS= sin (1+tan)+cos (1+1/tan)=> sin (1+tan)+cos (1+tan)/tan=> (1+tan)(sin+cos/tan)=> (1+tan)(sin.tan+cos)/tan=> (1+tan)(sin2/cos+cos)/tan=> (1+tan)(sin2+cos2)/tan.cos=>(1+tan)/tan.cos=>(1/tan.cos)+tan/tan.cos=>( cot/cos)+(1/cos)=> cosec +secLHS= RHS. | |
| 22145. |
Show that 12th cannot end with digit 0 or 5 for any natural number \'n\' |
| Answer» If 12n ends with 0 then it must have 5 as a factor.But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.So, 5 is not a factor of 12n.Hence, 12n can never end with the digit 0. | |
| 22146. |
Find the distance between two parallel tangent of a circle of radius 6 CM |
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Answer» sb aap jaise faltu nhi baithe h 12 cm |
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| 22147. |
If p and q are co-prime, then find HCF of (p,q) |
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Answer» Thanks to all of you ???? Answer will be 1 Two integers p and q are said to be coprime if they have no common factor other than 1.∴ HCF of coprime numbers p and q = 1 1 Wlcm Thanks HCF is one ....becoz coprime nos are having 1 as the common factor |
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| 22148. |
What is the relationship between opposite angle of rhombus |
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Answer» Opposite angles r equal in a rhombus Opposite angles of rhombus are equal |
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| 22149. |
Solve for x and y:99x+101y = 499 : 101x + 99y = 501 |
| Answer» y=2,x=3 | |
| 22150. |
What is the schedule of toppers for studies |
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Answer» Ye to kisi topper se hi puchna padega.. Mujhe to koi idea nhi hai Search in YouTube ?✌?✌️ |
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