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InterviewSolution
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.
| 40651. |
I don\'t like to do maths. Wht we do ? |
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Answer» m also there, can u suggest some link My friend, Try to connect the equations to our daily life. like going to market etc. Try watching some videos which are avail able in Youtube. The pictorical representation will help you catch the topic well. |
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| 40652. |
If theta =30 find the value of sin theta +cos theta |
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Answer» First of all put the value of theta in sin theta and cos theta That\'s, sin 30+cos301/2+root3/21+root3/2 1+√3______ 2 |
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| 40653. |
Find the area of the major segment of a circle whose radius is 14cm and its central angle is 60°. |
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Answer» First of all find the area of minor segment 523.05 |
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| 40654. |
a² +b²/ab = a+c/b |
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| 40655. |
What formula segment |
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Answer» R^2(π x theta/360° - sin theta/2 x cos theta/2). What segment ??? |
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| 40656. |
HCF of two numbers is 113, their LCM is 56952. If one number is 904, then other number |
| Answer» Given ,HCF = 113LCM = 56952a = 904b = ?\xa0HCF ×LCM = a × b (Product of two numbers)113 × 56952 = 904 × b6435576 = 904 ×bTherefore, b = 6435576 / 904b = 7119 | |
| 40657. |
Prove that if x and y are odd positive integer then x^2+y^2 is even which is not divisible by 4 |
| Answer» Let the two odd positive numbers be x = 2k + 1 a nd y = 2p + 1Hence x2\xa0+ y2\xa0= (2k + 1)2\xa0+ (2p + 1)2 = 4k2\xa0+ 4k + 1 + 4p2\xa0+ 4p + 1 = 4k2\xa0+ 4p2\xa0+ 4k + 4p + 2 = 4(k2\xa0+ p2\xa0+ k + p) + 2Clearly notice that the sum of square is even the number is not divisible by 4\xa0 | |
| 40658. |
What is the probabilty of number between 1 to 100 divisible bye 8 |
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Answer» But if we take it as between 1 and 100 then 1 and 100 will be removed from it ,it will remain total outcomes 98 If I say from 1 to 100, we know that 1 and 100 (and everything in between) are possible.That’s 100 numbers, of which 12 (8, 16, …, 96) are divisible by 8.So the probability that a number divisible by 8 = 12/100 = 3/25 12/100 |
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| 40659. |
Prove that Sin A minus 2 Sin cube A upon 2 cos cube A minus Cos A is equal to 10 |
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| 40660. |
Tan360° value |
| Answer» tan 360 = sin 360 / cos 360 = 0/1= 0tan 360 = 0 | |
| 40661. |
sina-2sin3a/2cos3a-cosa=tana |
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Answer» c LHS=sinA(1-2sin2A) / cos A (2cos2A-1)=\xa0sinA(sin2A + cos2A-2sin2A) / cos A (2cos2A-(sin2A + cos2A))=sinA( cos2A-sin2A) / cosA ( cos2A-sin2A)=sinA / cosA = tanA=RHS |
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| 40662. |
If 5tan =4 then find 5sin-3cos/5sin+2cos |
| Answer» say 5tanX=4 then tan X=4/5now(5 sinX-3cosX) / 5 sinX - 2cosXdevide up and diwn with cos x= 5 sinX/cos X-3/5 sinX/cosX-2=(5tanX-3 ) / ( 5tanX-2)=(4-3) / (4-2)=1 / 2 | |
| 40663. |
3×5== |
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Answer» 94 15 |
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| 40664. |
Proove that (6+√5) is irrational |
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Answer» Let 6+√5 is a rational number So 6+√5= p÷ q √5=p÷q-6 √5=p-6q÷ q We know that √5 is an irrational number so p-6q÷q but we know that p-6q÷q is an rational number So that it contracdict our form that 6+√5 is an irrational number This shows that 6+√5 is irrational Let 6+√5 be rational6+√5=a/b (a & b are co-primes)√5=a/b -6LCM√5=a-6b/bIrrational not = rational |
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| 40665. |
Find the value of cos 1.cos2.cos3.......cos180 |
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| 40666. |
In ∆ABC, right angled at B, AB=24cm, BC=7cm. DetermineSinA, cosA |
| Answer» In right angle traingle ABCAngle B= 90°By PGT(paytha goras theorem) in ∆ABCAB=24cm , BC=7cm So, AC×AC=BC×BC+AB×AB= Ac×Ac= 24×24+7×7=Ac×Ac=576+49=Ac×Ac=625=Ac=√625=Ac=25 Now, sinA= p/h=7/25CosA= B/H=4/25 | |
| 40667. |
Give me the all chapters formula |
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Answer» You can get it from your book and make thier short note. You have to work yourself then ,you will get success. Ok search at net u Will get easily. Here no one is there to give u the formula of all ch |
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| 40668. |
Prove that the points (4,5),(7,6),(6,3)and(3,2)are the vertices of a parallelogram.is it a rectangle |
| Answer» Let A(4, 5), B(7, 6), C(6, 3) and D(3, 2) be the given points.And, P the point of intersection of AC and BD.Coordinates of the mid-point of Ac are\xa0{tex}\\left( \\frac { 4 + 6 } { 2 } , \\frac { 5 + 3 } { 2 } \\right) = ( 5,4 ){/tex}Coordinates of the mid-point of BD are\xa0{tex}\\left( \\frac { 7 + 3 } { 2 } , \\frac { 6 + 2 } { 2 } \\right) = ( 5,4 ){/tex}Thus, AC and BD have the same mid-point.Hence, ABCD is a parallelogram.Now we shall see whether ABCD is a rectangle.We have,{tex}A C = \\sqrt { ( 6 - 4 ) ^ { 2 } + ( 3 - 5 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad A C = \\sqrt { 4 + 4 }{/tex}{tex}\\Rightarrow \\quad A C = \\sqrt { 8 }{/tex}And,\xa0{tex}B D = \\sqrt { ( 7 - 3 ) ^ { 2 } + ( 6 - 2 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad B D = \\sqrt { 16 + 16 }{/tex}{tex}\\Rightarrow \\quad B D = \\sqrt { 32 }{/tex}Since, AC\xa0{tex}\\neq{/tex}\xa0BDSo, ABCD is not a rectangle. | |
| 40669. |
Can we draw perpendicular on a line through a point |
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Answer» We draw a straight line and cut of by the perpendicular bisector. Yes Yes |
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| 40670. |
If -5 is a root of the quadratic equation 2x |
| Answer» Given that -5 is the root of\xa0{tex}2 x^{2}+p x-15=0{/tex}Put x = -5 in\xa0{tex}2 x^{2}+p x-15=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2(-5)^{2}+p(-5)-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}50-5 p-15=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}35 - 5p = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5p = 35\xa0{/tex}{tex}\\therefore{/tex}\xa0{tex}p = 7{/tex}Hence the quadratic equation p {tex}(x^2 + x) + k = 0{/tex} becomes,\xa0{tex}7\\left(x^{2}+x\\right)+k=0{/tex}{tex}7 x^{2}+7 x+k=0{/tex}\xa0Here {tex}a = 7,\\ b = 7\\ and\\ c = k{/tex} Given that this quadratic equation has equal roots\xa0{tex}\\therefore{/tex}\xa0{tex}b^{2}-4 a c=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}7^{2}-4(7)(\\mathrm{k})=0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}49 - 28k = 0\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}49 = 28k\xa0{/tex}{tex}\\therefore{/tex}\xa0k = {tex}\\frac{49} {28}{/tex}\xa0=\xa0{tex}\\frac{7} {4}{/tex} | |
| 40671. |
Find the 0s of the polynomial 5√5x^2+30x+8√5 |
| Answer» {tex}5{/tex}{tex}\\sqrt{5}{/tex}{tex}x^2\xa0+ 30x + 8{/tex}{tex}\\sqrt{5}{/tex}\xa0{tex}= 5{/tex}{tex}\\sqrt{5}{/tex}{tex}x^2\xa0+ 20x + 10x + 8{/tex}{tex}\\sqrt{5}{/tex}= {tex}5x ({/tex}{tex}\\sqrt{5} x{/tex}\xa0{tex}+ 4) + 2{/tex}\xa0{tex}\\sqrt{5}{/tex}{tex}\xa0({/tex}{tex}\\sqrt{5}{/tex}{tex}x + 4){/tex}= ({tex}\\sqrt{5} x{/tex}\xa0+ 4)(5x + 2{tex}\\sqrt{5}{/tex})=\xa0{tex}\\sqrt{5}{/tex}\xa0({tex}\\sqrt{5}{/tex}x + 2)({tex}\\sqrt{5}{/tex}x + 4) | |
| 40672. |
1plus tan squaretheta is equal |
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Answer» Sec^2theta Sec square theta |
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| 40673. |
1 cos = |
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| 40674. |
(1+sin/1-sin)^1/2 +(1-sin/1+sin)^1/2 =2sec |
| Answer» (1+sinA)1/2\xa0/ (1-sinA)1/2\xa0\xa0+ (1-sinA)1/2\xa0/ (1+sinA)1/2multiply up and down by (1+sinA)1/2\xa0* (1-sinA)1/2 we get(1 + sinA +1 -sinA)/\xa0(1+sinA)1/2\xa0*(1-sinA)1/2=2 / (1-sin2A)1/2=2/cosA=2secA =Rhs | |
| 40675. |
Using properties find 78×82 |
| Answer» 78 × 82= (80 - 2) (80 + 2)= (80)2 - (2)2 [∵ a2 - b2 = (a - b) (a + b)]= 6400 - 4= 6396 | |
| 40676. |
2018-2019 syllabus of maths |
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Answer» Nooooo rider sunil Exam me full syllabus aaegaMeans ch 1 to 15 But it shows half of book lessons will appear in exam board priya ji Please kindly check the app mycbseapp you are using.. |
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| 40677. |
From where does sums come of cbse maths exam |
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Answer» Lagbhag ncert se hote h From NCERT |
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| 40678. |
Solve this sum 0.686868............ +0.717171.............. |
| Answer» 1.404039i.e,1.404040... | |
| 40679. |
Prove that the points A(-3,0), B(1,-3) and C(4,1) are the vertices of an isosceles right triangle |
| Answer» Let A (-3, 0), B (1, -3) and C (4, 1) be the given points. Then,AB =\xa0{tex}\\sqrt { ( 1 - ( - 3 ) ) ^ { 2 } + ( - 3 - 0 ) ^ { 2 } } = \\sqrt { 4 ^ { 2 } + ( - 3 ) ^ { 2 } } = \\sqrt { 16 + 9 }{/tex}= 5BC =\xa0{tex}\\sqrt { ( 4 - 1 ) ^ { 2 } + ( 1 + 3 ) ^ { 2 } } = \\sqrt { 9 + 16 }{/tex}= 5 unitsand, CA =\xa0{tex}\\sqrt { ( 4 + 3 ) ^ { 2 } + ( 1 - 0 ) ^ { 2 } } = \\sqrt { 49 + 1 } = 5 \\sqrt { 2 }{/tex}unitsClearly, AB = BC. Therefore, {tex}\\triangle{/tex}ABC is isosceles.Also, AB2 + BC2 = 25 + 25 = (5{tex}\\sqrt2{/tex})2 = CA2{tex}\\therefore{/tex}{tex}\\triangle{/tex}ABC is right-angled at B.Thus, {tex}\\triangle{/tex}ABC is a right-angled isosceles triangle. | |
| 40680. |
What is the formula of curved surface area of cone |
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Answer» πrl , Pie*radius*slant height πr2h Pie × radius ×slant height (l) |
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| 40681. |
(a+b) ^3 |
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Answer» a^3 +b^3 +3ab (a+b) 3a+3b |
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| 40682. |
2+2 2crore rupees price mony |
| Answer» 2+2 = 4 | |
| 40683. |
Chapter triangle exercise 6.3 questions no 14? With solution |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 40684. |
If (1+cosA)(1-cosA)=3÷4 find the value of secA |
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Answer» Or secA=30degree SecA=2÷root3 |
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| 40685. |
Find all the zeroes of 2x⁴-3x³-3x³+6x-2,if you know thar two of it zeroes are √2 and -√2. |
| Answer» It is there in NCERT book ch-2 ex. 9 pg. No. 35 | |
| 40686. |
What is real number define |
| Answer» A collection of rational numbers and irrational numbers make up the set of real number. A real number can be expressed on the number line and has some specific properties. They satisfy:\tThe commutative law of addition. That is, when a and b are two real numbers then a + b = b + a. For example 1 + 3 = 3 + 1 = 4\tThe commutative law of multiplication. That is, when a and b are two real numbers then a x b = b x a. For example 1 x 3 = 3 x 1 = 3\tThe associative law of addition. That is, when a, b and c are three real numbers then a + (b + c) = (a + b) + c. For example 1 + (3 + 4) = (1 + 3) + 4 = 8\tThe associative law of multiplication. That is, when a, b and c are three real numbers then a x (b x c) = (a x b) x c. For example, 1 x (3 x 4) = (1 x 3) x 4 = 12\tThe law of distribution. That is, when a, b, c are three real numbers then a x (b +c) = (a x b) + (a x c). For example 1 x (3 + 4) = (1 x 3) + (1 x 4) = 7 | |
| 40687. |
Xsquare |
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| 40688. |
5 number find to lcm |
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| 40689. |
Is it true that in 2019 examination of math conpect is from NCERT but question are not from NCERT |
| Answer» Yes you must have clear all concept of N.C.E.R.T | |
| 40690. |
Give questions |
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| 40691. |
Find the coordinates of the point on y-axis which is nearest to the point (2,-5) |
| Answer» Plzzz, anyone tell the answer | |
| 40692. |
Draw an angel AOB equal to 60 through point A draw a parallel line to OB |
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| 40693. |
I if sin theta is equals to cos theta then find the value of 2 tan theta + cos square theta |
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| 40694. |
What is blueprint of board of maths |
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| 40695. |
In triangle abc angle b=90° |
| Answer» It is only in right angle triangle | |
| 40696. |
Find the area of the largest triangle that can be inscribed in a semicircle of radius 7cm |
| Answer» It is equal to the square of radius. R^2. Answer=49 | |
| 40697. |
how many irrational numbers are there between 1 and 2 |
| Answer» Infinitely many | |
| 40698. |
Solve this sum step by step 0.686868.............+ 0.717171....…....... |
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Answer» Just solve it as usual by puting bar over the digits Calculator hoga tmhare phone me solve kar po |
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| 40699. |
√√3+√3+√3 |
| Answer» mma ki chu | |
| 40700. |
find the sum of of first 15 multiples of 8 |
| Answer» a=8 ,d=8, l=120 Sn=n/2(a+l)= 15/2(8+120) =15/2×128 15×64 =960 | |