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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.
| 17001. |
How and Where we can use the trigonometric ratios??? Plzzzz.... help |
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Answer» In right angle triangles @Hrishikesh.S?? @Hrishiikesh.S tell me the answer not joke In exams ? |
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| 17002. |
Agar app jante ho to please boldo chapter 8 shortcut main |
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Answer» Agar table or formulas ate hai to apko ch agayega And table There are some formulas Learn equation and learn values of theta |
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| 17003. |
p²-54p-225 |
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| 17004. |
Why a+b^2 is ........ |
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| 17005. |
If the product of zeros of quadratic polynomial ax2-6x-6 is 4 ,find the value of a |
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Answer» Alpha×Beta=c/a 4 = -6/a a=-3/2 Given polynomial:\xa0\xa0Product of zeroes of this polynomial = 4.\xa0A polynomial is an expression of more than one or two algebraic term, may be a sum of those terms also.\xa0We know, that the product of zeroes of a polynomial\xa0\xa0can be given by =c/a\xa0Hence, for the polynomial:\xa0, a = a, b = -6 and c = -6.\xa0Product of zeroes of this polynomial = (-6)/a\xa04 = -6/a\xa0a = -6/4= -3/2 Herex is equal to 4 .Put the value 4 in the place of X.a(4×4)-6(4)-6=0.16a-24-6=0.16a = 30 .a= 15/8.I think it is clear to u. -3/2 |
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| 17006. |
The decimal representation of 11/2³×5 will |
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Answer» Terminate after 3 decimal places 11/2³×5=11×5²/2³×5×5²=275/1000=0.275 1.375 It is of one marks |
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| 17007. |
Find the value of k for which the line (K +1)x + 3ky + 15 =0 and 5x + ky + 5 coincident |
| Answer» The given equations are:\xa0(k+1)x+3ky+15=0, and\xa05x+ky+5=0\xa0Since, the given equations are coincident, therefore{tex}\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\frac{k+1}{5}=\\frac{3k}{k}=\\frac{15}{5}{/tex}Taking the first two terms, we have\xa0k=14\xa0OrThe given equations are:, andSince, the given equations are coincident, thereforeTaking the first two terms, we have | |
| 17008. |
3+5= 7 prove that |
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Answer» What is the question? The question is wrong 3+5=77=7 Hence proved!! |
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| 17009. |
What is the sum of 2 + 3 = …......... |
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Answer» 5 5 5 5 5 |
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| 17010. |
For the AP: 3/2,1/2,-1/2,-3/2,....., write the first term a and the common difference d .? |
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Answer» a = 3/2 and d = - 1 -1 Good a = 3/2d = - 1 |
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| 17011. |
if we don\'t study rd sharma we get low marks?? |
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Answer» No Haha, there is nothing like this...Just get perfection in NCERT for perfection you can do NCERT EXEMPLER.For more questions practice you can choose RS AGGARWAL. Ya concept is clear and NCERT solutions are not clear Always read ncert Noo |
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| 17012. |
If125 par power x.25/5par power x find x |
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Answer» 3 I don\'t know bro.. |
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| 17013. |
If i am 14 years what is my brorher age uf he was born on 1800 |
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Answer» What is the age of your parents...??? He is 220 years old ? |
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| 17014. |
Case study question of maths |
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| 17015. |
2 cube each of volume 64cm3 are joined end to end. Find the surface area of the resulting cuboid. |
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Answer» Length of cube\xa0=aa3=64⇒a=4Length of cuboid =\xa08=lbreadth of cuboid =\xa04=bheight of cuboid =\xa04=hSurface area\xa0=2(lb+bh+lh)=2((8×4)+(4×4)+(8×4))=2(32+16+32)=160\xa0cm2 Given,Volume of each cube = 64cm^3length of the cuboid =4+4= 8cmBreadth of the cuboid =4 cmHeight of the cuboid = 4 cmT.S.A of cuboid =2(lb+bh+hl)=2(8×4+4×4+4×8)=2(32+16+32)=2(80) |
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| 17016. |
What is the deleted question of chapter 4 maths class 10 |
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Answer» Thanks only word problem will not be asked( word problem nhi aayega bs ??) HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.Deleted syllabus of CBSE Class 10 Mathematics\xa0\xa0 |
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| 17017. |
I want maths subject 3rd chapter |
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Answer» Polynomial Please explain for me in english |
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| 17018. |
If roots are equal of quadratic equation p(q-r)x^2+q(r-p)x+r(p-q)=0 Prove that: 1/p+1/r= 2/q |
| Answer» Given:The roots of the quadratic equation :p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.To prove :Solution:Compare given Quadratic equation with ax²+bx+c=0, we geta = p(q-r), b = q(r-p), c = r(p-q)Discreminant (D) = 0/* roots are equal given */=> b²-4ac=0=>[q(r-p)]²-4×p(q-r)×r(p-q)=0=>(qr-pq)²-4pr(q-r)(p-q)=0\xa0=> (qr)²+(pq)²-4(qr)(pq)-4pr(pq-q²-pr+qr)=0=> (qr)²+(pq)²-4pq²r-4p²qr+4prq²+4p²r²-4pqr²=0=> (qr)²+(pq)²+(-2pr)²+2pq²r-4p²qr-4pqr²=0=> (qr)²+(pq)²+(-2pr)²+2(qr)(pq)+2(pq)(-2pr)+2(-2pr)(qr)=0/* we know the algebraic identity*//*a²+b²+c²+2ab+2bc+2ca=(a+b+c)² */=> (qr+pq-2pr)² = 0=> qr+pq-2pr = 0Divide each term by pqr , we getTherefore, | |
| 17019. |
State and prove area of similar triangle theorem |
| Answer» Given\xa0:∆ABC ~ ∆PQRRTP:Construction:Draw AM perpendicular to BC and PN perpendicular to QR.Proof:∆ABM~∆PQN ( By AA similarity )Also ∆ABC ~ ∆PQR (given)\xa0\\* From (1),(2) and (3)Now, by using (3) , we get | |
| 17020. |
1. In the figure ,quad. ABCD Circumscribes the circle.Find the length of the side CD. |
| Answer» Tangents drawn from am external point to a circle\xa0are equal in lengthso from the diagram we getAH\u200b= AE\u200b; DH = DG; CG\u200b=CF\u200b4−x=x y=2y−32x=4 y=3\u200bx=2\u200bAs\xa0x=2 DH = 5 − AH = 5−(4−x) =3as\xa0DH=DG DG=3So Length of\xa0CD=CG+GD=3+3=6 | |
| 17021. |
What do you mean by probability of any event? |
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Answer» 1 Thank you A probability event can be defined as a set of outcomes of an experiment. In other words, an event in probability is the subset of the respective sample space. Probability of an event can also be said as the Emperical Probability, P(E). It is the ratio of no : of favourable outcomes / total no: of outcomes. |
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| 17022. |
2cos67/sin23+tan40/cot50+cos0 |
| Answer» These type of questions are cutted | |
| 17023. |
How to find prime factor for 5005 |
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Answer» 5¹x7¹x11¹x13¹Exponential form A n s w e rThe prime factors are: 5 x 7 x 11 x 13or also written as { 5, 7, 11, 13 }Written in exponential form: 51 x 71 x 111 x 131 |
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| 17024. |
What is upstream and downstream.explain it with diagram? |
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Answer» There are a variety of subconcepts which are related to answering questions based on boat and streams. Given below are the four terms which are important for a candidate to know to understand the concept of streams.\tStream –\xa0The moving water in a river is called stream.\xa0\tUpstream –\xa0If the boat is flowing in the opposite direction to the stream, it is called upstream. In this case, the net speed of the boat is called the upstream speed\tDownstream –\xa0If the boat is flowing along the direction of the stream, it is called downstream. In this case, the net speed of the boat is called downstream speed\tStill Water –\xa0Under this circumstance the water is considered to be stationary and the speed of the water is zero The terms\xa0upstream and downstream\xa0oil and gas production refer to an oil or gas company\'s location in the supply chain. Upstream\xa0oil and gas production is conducted by companies who identify, extract, or produce raw materials.\xa0Downstream\xa0oil and gas production companies are closer to the end user or consumer.When a boat travels in the same direction as the current, we say that it is traveling\xa0downstream. Thus if b is the speed of the boat in still water, and c is the speed of the current, then its total speed is.\xa0Downstream\xa0speed = b + c. When a boat travels against the current, it travels\xa0upstream. |
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| 17025. |
All circles are |
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Answer» Similar Probability means the chance that something will happen. How likely it is that some event will occur. All circles are similar What is probability |
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| 17026. |
In what ratio does the X - axis divide the line segment joining the points (-4, -6) and |
| Answer» ಲೋ ನಂಗ್ ನಿಜವಾಗ್ಲೂ ಗೊತ್ತಿಲ್ಲ ಬ್ರೊ??????? | |
| 17027. |
Explain why 7x13x11+11 |
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Answer» This can also be written as 7*13*11+11=11(7*13+1)=11×(7*13+1)It has more than two factors so it is composite number 1011 |
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| 17028. |
Ex-7. 4, question no. 5 |
| Answer» Solution:(i) Taking A as origin, coordinates of the vertices P, Q and R are,From figure: P = (4, 6), Q = (3, 2), R (6, 5)Here AD is the x-axis and AB is the y-axis.(ii) Taking C as origin,Coordinates of vertices P, Q and R are ( 12, 2) , (13, 6) and (10, 3) respectively.Here CB is the x-axis and CD is the y-axis.Find the area of triangles:Area of triangle PQR in case of origin A:Using formula: Area of a triangle == ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]= ½ ( – 12 – 3 + 24 )= 9/2 sq unit(ii) Area of triangle PQR in case of origin C:Area of a triangle == ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]= ½ ( 36 + 13 – 40)= 9/2 sq unitThis implies, Area of triangle PQR at origin A = Area of triangle PQR at origin CArea is same in both case because triangle remains the same no matter which point is considered as origin. | |
| 17029. |
If 8cotA=7,evaluate (1+sinA)(1-sinA)/(1+cosA)(1-cosA) |
| Answer» Given:\xa0=\xa0We have to find, the value of\xa0\xa0is:Solution:∴\xa0Using the algebraic identity:(a + b)(a - b) =\xa0Using the trigonometric identity:\xa0+\xa0\xa0= 1⇒\xa0\xa0= 1 -\xa0\xa0and\xa0\xa0= 1 -\xa0Using the trigonometric identity:\xa0=\xa0=\xa0Put\xa0\xa0=\xa0,\xa0we get=\xa0=\xa0∴\xa0\xa0=\xa0Thus, the value of\xa0\xa0is "equal to\xa0". | |
| 17030. |
sin theta - cos theta + 1 / cos theta + sin theta -1 = 1/ sec theta - tan theta |
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| 17031. |
If x and y are complementary angles then |
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Answer» Sec x = cosec y X + Y = 90° Then x+y =90° Secx and cosec y |
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| 17032. |
(x+2)³=2x(x²-1) Give me answer |
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Answer» We know that the general form of quadratic equation is\xa0ax2+bx+c=0.The given equation is\xa0(x+2)3=2x(x2−1)\xa0can be simplified as follows:\xa0(x+2)3\xa0= 2x(x2−1)⇒x3+23+(3×x×2)( x+2)=2x3−2x (∵(a+b)3=a3+b3+3ab(a+b))⇒x3+8+6x(x+2)=2x3−2x⇒x3+8+6x2+12x=2x3−2x⇒x3+8+6x2+12x−2x3+2x=0⇒−x3+6x2+14x+8=0Since the variable\xa0x\xa0in the equation\xa0−x3+6x2+14x+8=0\xa0has degree\xa03, therefore, it is not of the form\xa0ax2+bx+c=0.Hence, the equation\xa0(x+2)3=2x(x2−1)\xa0is not a quadratic equation. thank u |
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| 17033. |
What will be the value of 2 tan30° upon 1+tan square 30° |
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Answer» A = 30°= = tan60°= Therefore,The value of = $\\sqrt{3}$ 2tan30°/ (1 + tan²30°)(2/√3)/(1+ 1/3)(2/√3)/ (4/3)(2/√3) × (3/4)3/2√3After rationalisation,√3/2 A\xa0=\xa030°=\xa0=\xa0tan60°=\xa0Therefore,The\xa0value\xa0of\xa0= $\\sqrt{3}$ |
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| 17034. |
2x²-x+⅛=0 Give me answer |
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Answer» 2x²-x+1/8=0Multiplying the eq. by 816x²-8x+1=016x²-4x-4x+1=04x(4x-1) -1(4x-1) =0X=1/4, 1/4 thank u Hence the solution is\xa0 |
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| 17035. |
If 9th term of an ap is zero prove that 29 term is twice of the 19th term |
| Answer» Let the first term of an A.P. =\xa0aAnd common difference =\xa0dGiven: 9th\xa0term of an A.P. is 0. Therefore,We have to prove that\xa0Thus,AndFrom equation (2) and (3),Hence provedc | |
| 17036. |
Sin A=3/4,calculate cosec A |
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Answer» CosecA = 4\\3 Given, sin A=3/4.We know, Cosec A =1/sin A Then, cosec A=1/3/4 =4/3 sinA = 3/4. cosecA = 1/sinA. cosecA = 4/3 Sin A=3/4Since, cosec A=1/sin ATherefore, cosec A=1/3/4=> cosec A =4/3 4/3 |
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| 17037. |
Where can I get solutions for sample papers |
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Answer» U can search in google Type class 10 cbse solved sample papers In Google |
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| 17038. |
Will 2 months be enough for final preparation for boards ? |
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Answer» Yàaa Yes, you need to be thorough with the concepts by then. And last 2 months you can utilise your time in revision |
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| 17039. |
3x2 +8=0 |
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Answer» 14 14 |
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| 17040. |
If a sin theta + b cos theta = c. Prove that a cos theta - b sin theta = root a² + b² - c² |
| Answer» asinθ + bcosθ = ctaking square both sides,(asinθ + bcosθ)² = c²⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)Let acosθ - bsinθ = xSquaring both sides(acosθ - bsinθ)² = x²⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)Add equation (1) and (2),a²sin²θ + b²cos²θ +2abinθ.cosθ + a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x²⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x²⇒(a² + b²)[sin²θ + cos²θ ] = c² + x²⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ]⇒(a² + b² - c²) = x²Take square root both sides,Hence, acosθ - bsinθ =\xa0 | |
| 17041. |
Case study problems |
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| 17042. |
P(-2, 5) and Q(3, 2) are two points. Find the co-ordinates of the point R on PQ such that PR=2QR |
| Answer» Q u e s t i o n :P (-2, 5) and Q (3, 2) are two points. Find the co-ordinates of the point R on PQ such that PR = 2QR.A n s w e r:PR:QR = 2:1R(4 / 3, 3) | |
| 17043. |
Show that any positive odd integer is of the form 6q+1, 6q+3, 6q+5 , where q is some integer |
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Answer» Thank you Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder.according to Euclid’s division lemmaa=bq+ra=6q+rwhere , a=0,1,2,3,4,5then,a=6qora=6q+1ora=6q+2ora=6q+3ora=6q+4ora=6q+5but here,a=6q+1 & a=6q+3 & a=6q+5 are odd. |
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| 17044. |
If-1 is the zero of polynomial 2xsquare-x square -5x-2. Find the other zeros |
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| 17045. |
Total numbers of a factor of a prime numbers is |
| Answer» A\xa0prime number\xa0has exactly two\xa0factors, 1 and the\xa0number\xa0itself. So,\xa0Total numbers of a factor of a prime numbers is 2. | |
| 17046. |
Write all trigonometric ratios of angle A in terms of sec A |
| Answer» We know that,sec A = 1/cos A⇒\xa0cos A = 1/sec Acos²A\xa0+ sin²A = 1⇒ sin²A = 1 – cos²A⇒ sin²A = 1 – (1/sec²A) (cosA= 1/secA)⇒ sin²A = (sec²A-1)/sec²A⇒sinA =\xa0√((sec²A-1)/sec²A)⇒sinA =\xa0√(sec²A-1) ÷ (secA)............................(i)sin A = 1/cosec A⇒ cosec A = 1/sin A⇒cosecA= secA\xa0÷√sec²A-1\xa0 (from eq i)Now,sec²A – tan²A = 1⇒ tan²A = sec²A\xa0+ 1⇒tanA = √sec²A\xa0+\xa01.....................................(ii)tan A = 1/cot A⇒ cot A = 1/tan A⇒cotA = 1/√sec²A\xa0+\xa01 (from eq ii) | |
| 17047. |
F(x)=x2-2√2x+6 |
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Answer» It has no root x=-root2 and x=3root2 |
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| 17048. |
three farmers |
| Answer» So what next?? | |
| 17049. |
. Prove that: cot (90° – 0), cosec (90° - ) sin 0 tan a tan (90° - 0 |
| Answer» Hlo | |
| 17050. |
Math syllabus reduce chapter |
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Answer» In Ch 3 cross multiplication method was deleted Topics are reduced Step deviationCross multiplicationCubic polynomials No any chapter is reduced only the sub topics of some chapters is reduced.. |
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