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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.
| 21851. |
Can i use the formula to find the area of segment although it is not mentioned in ncert |
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Answer» Yup u can use ...but dont do so in the exam yes u can use the formula if u wish. i guess so... What is the formula please tell |
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| 21852. |
Find the coordinates of point on y axis which is nearest to the point (-2,5 ). |
| Answer» The nearest point from any point is the point where the perpendicular from the point intersect the line.as shown in the figure this point is B(0,5) | |
| 21853. |
1 mahine me ncert khatam Ho skti hai |
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Answer» Yes......... I think yes |
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| 21854. |
The boat question of quadratic equations... ritu questionn |
| Answer» Upstream downstream wala | |
| 21855. |
Complete square method |
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Answer» Example : Solve 5x2 – 4x – 2 = 0Step 1 Divide all terms by 5x2 – 0.8x – 0.4 = 0Step 2 Move the number term to the right side of the equation:x2 – 0.8x = 0.4Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:(b/2)2 = (0.8/2)2 = 0.42 = 0.16x2 – 0.8x + 0.16 = 0.4 + 0.16(x – 0.4)2 = 0.56Step 4 Take the square root on both sides of the equation:x – 0.4 = ±√0.56 = ±0.748 (to 3 decimals)Step 5 Subtract (-0.4) from both sides (in other words, add 0.4):x = ±0.748 + 0.4 = -0.348 or 1.148 If it is easy so do it instead of saying no ok jatin . rohini i will give u example wait ? To solve ax2+bx+c=0 by completing the square: 1. Transform the equation so that the constant term, c , is alone on the right side. 2. If a , the leading coefficient (the coefficient of the x2 term), is not equal to 1 , divide both sides by a . 3. Add the square of half the coefficient of the x -term, (b2a)2 to both sides of the equation. 4. Factor the left side as the square of a binomial. 5. Take the square root of both sides. (Remember: (x+q)2=r is equivalent to x+q=±r√ .) 6. Solve for x . It is a very siple question so i will not do it |
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| 21856. |
1/2 - 1/5 |
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Answer» 3/10 or 0.3 3/10 |
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| 21857. |
Explain fleming left hand rule |
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Answer» If we strech three fingers in the way that they are mutually perpendicular to each other i.e thumb, middle finger and forefinger. If the forefinger represents the direction of magnetic field, the mid finger represents the direction of current then the thumb represents the direction of motion or force Thumb indicates the direction of force and fore finger indicates direction of magnetic field and mid finger indicates the direction of current |
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| 21858. |
The volume of hemisphere is 2425 1/2cubes. Find its curved surface area |
| Answer» Volume of hemisphere = 2425 1/2 cu cm or 2425.5 cu cm.Volume of hemisphere = 2/3πr³2425.5 = 2/3× 22/7× r³r³ = (2425.5×21)/44r³ = 50935.5/44r³ = 1157.625r = 10.5 cmNow,Curved Surface Area of hemisphere = 2πr²= 2 × 22/7 × 10.5 × 10.5= 693 sq cmCSA of hemisphere is 693 sq cm. | |
| 21859. |
Formula of Diagonal of square? |
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Answer» a root 2 √2a where a is side of a square |
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| 21860. |
State and prove "Basic proportionality theorem" |
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Answer» Basic Proportionality TheoremRefer the below attachment. Please refer NCERT |
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| 21861. |
133 |
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| 21862. |
If (a- b ) , a and ( a+b) are the zeroes of polynomial 2x-6x+5x-7 write the value of a . |
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Answer» There is a mistake in ques. You havnt wrote powers.Answering as per write question. f(x) = 2x3\xa0- 6x2\xa0+ 5x - 7Let α, β and γ be the zeroes.Then, α = a - b, β = a and γ = a + bWe know that if α, β and γ are the zeroes of a cubic polynomial,α + β + γ = -b/a(a - b) + a + (a + b) = -(-6)/23a = 3a = 1So, the value of a = 1 |
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| 21863. |
If the point P(x,y) is equidistant from the points A(a+b,b-a) and B(a-b,a+b), prove that bx=ay |
| Answer» |PQ| = |PR{tex}\\begin{aligned} \\sqrt { [ x - ( a + b ) ] ^ { 2 } + [ y - ( b - a ) ] ^ { 2 } } = \\sqrt { [ x - ( a - b ) ] ^ { 2 } + [ y - ( b + a ) ] ^ { 2 } } \\end{aligned}{/tex}Squaring, we get[x - (a + b)]2 + [y - (b\xa0- a)]2\xa0= [x - (a - b)]2 + [y - (a + b)]2or, [x - (a + b)]2 - [x - a + b]2\xa0= (y - a - b)2 - (y - b + a)2or, (x - a - b + x - a + b) ( x - a - b - x + a - b)= (y - a - b + y - b + a)(y - a - b - y + b - a)or, (2x - 2a) (- 2b) = (2y - 2b) (- 2a)or, (x - a)b = (y - b)aor, bx = ay.Hence Proved. | |
| 21864. |
If sec thera +tan theta =p,then find the value of cosec theta |
| Answer» First convest seca into cos and tan into sin upon cos solve it then equation will be 1+ sin=pcos then square to both sides and the solve it uu will get equation (1+p square)sin square +2sin +(1-p square)=0 then solve it by using qudratic formula | |
| 21865. |
RD Sharma chapter 9 question 58 of exercise 9.13 |
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| 21866. |
Make graph of equation x²+5x-1 . According to chapter 2 |
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Answer» http://www.paraspunj.com/?p=1763 How to post graph here |
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| 21867. |
Distance formulea |
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Answer» √(x2-x1)2 + (y2-y1)2 √(x2-x1)^2+(y2-y1)^2 AB=√[(x2-x1)²+(y2-y1)²] |
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| 21868. |
If k-1, k+3 and 3k-1are in AP ;then find the value of k |
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Answer» a3 -a2 = a2 -a1 3k-1 -(k+3) = k+3 -(K-1)3k-1 - k - 3 = k+3-k+12k -4 = 4 2k = 4+4 2k = 8 K = 8/2 K = 4 Send solution If it is in ap then d=(k+3)-(k-1)=> k+3-k+1=> 4(3k-1)-(k+3) should be 4=> 3k-1-k-3=4=> 2k-4=4=> 2k=8=> k = 4 Subtract 2nd term from 3rd term and also 1st term from 2nd term both will be equal as they arr commn diff. of an A.P. 4 |
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| 21869. |
2x+3y=8X-2y+3=0 |
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Answer» By subsitution method 2x= 8-3yx =8-3y/2Put the value of x in second eqn.x -2y+3....8-3y/2 -2y+3=08-3y-4y+6/2=08-7y+6=0-7y=-14Y =2Put he value of y in first eqn.2x+3y=8...2x+3(2)=82x+6=82x=8-62x=2X=1 Y=2, X=1 X=1 and y=2 |
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| 21870. |
What are the important theorem for pre board |
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Answer» Book ki sare theorem important he Pythagorus,theles theorem Pythagorus,theles and theorem 10.2 Pythagoras theorem and Thales theorem.... Pythagoras theorem |
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| 21871. |
Lab activities |
| Answer» It must be clear for examination because it was of 8 Marks in exam | |
| 21872. |
2x+3y=84x+6y=16 |
| Answer» By multiplying equation number 1 by 2 Then we will get equation number 2 | |
| 21873. |
If one zero of a polynomial is given then how we an find others three |
| Answer» Divide the given polynomial by taking one zero | |
| 21874. |
find the condition that zeroes of polynomial p(x)=ax2+bx +c are reciprocal of each other |
| Answer» Take one zero as alpha..and other as one upon alpha | |
| 21875. |
Solve for x and y : 2x+ 3y =7 ,, (a+b+1)x +(a+2b+2)y =4 (a+b)+1 |
| Answer» If infinite number of solutions,{tex}\\frac{{{a_1}}}{{{a_2}}} = \\frac{{{b_1}}}{{{b_2}}} = \\frac{{{c_1}}}{{{c_2}}}{/tex}or {tex}\\frac{2}{{a + b + 1}} = \\frac{3}{{a + 2b + 2}} = \\frac{7}{{4a + 4b + 1}}{/tex}If\xa0{tex}\\frac{2}{{a + b + 1}} = \\frac{3}{{a + 2b + 2}}{/tex}{tex}\\Rightarrow{/tex} a - b = 1and if {tex}\\frac{3}{{a + 2b + 2}} = \\frac{7}{{4a + 4b + 1}}{/tex}{tex}\\Rightarrow{/tex} 5a - 2b = 11On solving we get,a = 3 and b = 2 | |
| 21876. |
Can we have any natural number \'n\' for which 6^n ends with the digit \'0\'?justify you answer. |
| Answer» Let 6^n ends with 0 so its prime factor is 2^n×5^mBut as v know Prime factor of 6^n is 2^n ×3^n Not 2^n×5^m so No any natural number for which 6^n ends with 0 | |
| 21877. |
,hindi me maths |
| Answer» गणित | |
| 21878. |
Find thw value of k so that the following system of equation has no sollution:3x-y-5=0; 6x-2y-k=0 |
| Answer» 3x-y-5=06x-2y-k=0a1/a2=b1/b2 is not equal c1/c2 (no solution)3=1 is not equal 56 2 kK is not equal to 10 ( cross multiply) | |
| 21879. |
What is the relation between mass and radius of sphere? |
| Answer» density = mass / volumeAnd then find the radius from the volumevolume = 4πr3/3 | |
| 21880. |
In triangle ABC right angle at A and un triangle PQR right angle at P and cosB=cos theta |
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| 21881. |
Fundamental theorum of arithmetic |
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Answer» theorem, states that every integer greater than either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors.[For example,1200 = 24 × 31 × 52 = 2 x 2 x 2 x 2 x 3 x 5 x 5 = 5 × 2 × 5 × 2 × 3 × 2 × 2 = ...HOPE IT WILL HELP YoU Fundamental Theorem of Arithmetic states that every composite number greater than 1 can be expressed or factorised as a unique product of prime numbers except in the order of the prime factors.The HCF of two numbers is equal to the product of the terms containing the least powers of common prime factors of the two numbers.The LCM of two numbers is equal to the product of the terms containing the greatest powers of all prime factors of the two numbers.The product of the given numbers is equal to the product of their HCF and LCM. This result is true for all positive integers and is often used to find the HCF of two given numbers if their LCM is given and vice versa. For any two positive integers a and b, HCF(a , b) x LCM(a , b) = a x b |
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| 21882. |
Tringle theroms |
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Answer» Take help of reference for triangles theromsIt may be help u alot Its given in NCERT |
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| 21883. |
√6+√6+√6 |
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Answer» {tex}\\begin{array}{l}\\sqrt6+\\sqrt6+\\sqrt6\\\\=3\\sqrt6\\;=3\\times2.45=7.35\\end{array}{/tex} Root 12+root 6 |
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| 21884. |
Show that the square of any positive integer is of the formof 3m or 3m+1 for some integer m. |
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Answer» As per Euclid\'s Division LemmaIf a & b are 2 positive integers, thena = bq + r {tex}0\\;\\leq b |
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| 21885. |
Can two numbers have 16 as their HCF and 380 as their LCM ? Give reason |
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Answer» \xa0HCF of two numbers is a factor of the LCM of those numbers.Thus, we cannot have two numbers whose HCF is 16 and LCM is 380.Because, when we divide 380 by 16, we get a remainder of 12.Thus, 16 is not a factor of 380.Hence, we cannot have two numbers whose HCF is 16 and LCM is 380. HCF of two numbers should be the factor of LCM of those numbers. But, 16 is not a Factor of 380. Hence, two numbers can not have their HCF and LCM as 16 and 380 |
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| 21886. |
What is substitution reaction Describe with example |
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Answer» A substitution reaction is a reaction in which an atom or a group of atom substitutes another atom in an organic compound. For example, chlorination of alkanes occurs by substitution of H-atom in alkane by Cl-atom.The reaction of magnesium with copper chlorideMg + CuCl2\xa0→ MgCl2\xa0+ Cu A substitution reaction is a type of chemical reaction where an atom or functional group of a molecule is replaced by another atom or functional group. Examples: CH3Cl reacted with a hydroxy ion (OH-) will produce CH3OH and chlorine |
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| 21887. |
28.5✓3 - 9.5✓3 = ?? |
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Answer» Anshika,The answer is{tex}\\begin{array}{l}28.5\\sqrt3-9.5\\sqrt3\\\\=9.5\\sqrt3(3-1)\\\\=2\\times9.5\\sqrt3\\\\=19\\sqrt3\\end{array}{/tex}Here no need to cut {tex} \\sqrt 3{/tex} ,You can leave the answer as it. Root 3( 28.5 - 9.5) = 19 root 3.........??? Par kese puja ✓3 - ✓3 cut nhi hoga kya bolo 19 root 3.... Are koi to bata do answer 19 hoga ki 19✓3 ??????? please batado |
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| 21888. |
In ∆ABC, angle abc = 45°.prove that AC² = AB² + BC² - 4ar(∆abc). |
| Answer» Angle ABC measures 45° (given). Hence angle OAB will also be 45°As a result x = hNow the area of triangle ABC = 1/2(BC)(h) = 1/2(BC)(x) ; since h = x,\xa0Or we can write BC = 2(Area of Triangle ABC)/x ----------------(i)We can also write AC² = h² + y²But y is nothing but (BC - x)So we can write AC² = h² + (BC - x)²Simplifying this we get,AC² = h² + BC² + x² - 2.x.BCNow x² + h² = AB² (by Pythagoras theorem for triangle AOB)Hence AC² = AB² + BC² -2(x)(BC)Substituting the value of BC obtained from (i), we getAC² = AB² + BC² - 2(x)(2Area of Triangle ABC)/xSimplifying this we get:AC² = AB² + BC² -4.Area of Triangle ABC | |
| 21889. |
If s1=n ,s2=2n , and s3=3n than prove that s3=3 (s2-s1) |
| Answer» Let ‘a’ be the first term of the AP and ‘d’ be the common difference S1 = (n/2)[2a + (n – 1)d] --- (1) S2 = (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2) S3 = (3n/2)[2a + (3n – 1)d] --- (3) Consider the RHS: 3(S2 – S1) = S3 = L.H.S ∴ S3 = 3(S2 - S1) . | |
| 21890. |
can two numbers have18 as their HCF and 380 as their LCM ? Give resion. |
| Answer» Noooo, kyunki H.C.F ko L.C.M ka factor hona chahiye bt here 18 is nt a factor of 380........ | |
| 21891. |
Compartment question paper 2018 |
| Answer» Check on the internet | |
| 21892. |
SecA+tanA=x then prove sinA =x sqare -1divided by x sqare+1 |
| Answer» SecA+tanA=X; secA-tanA=1/x adding we get 2secA=x+1/x>>>>secA=x square+1 divide 2x that is cosA=2x/x square+1 i.e sinA=√1-cosA^ | |
| 21893. |
Solve the following quadratic equation in variable x. abx^=(a+b)^(x-1) |
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| 21894. |
How to prove 1+2cos^2x=4sinxcosx |
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| 21895. |
Find the real number whose product with irrational number,4-/3 is 39 |
| Answer» Let the no be xThen x(4-√3)=39 X=39/4-√3 rationalise it you will get X=3(4+√3) | |
| 21896. |
If sin (A-B) =1/2,cos(A+B)=1/2 then find the value of A and B |
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Answer» Thx a lot kunal bro Sin (A-B) = sin 30 [ Here we cancel sin as they are on the both side]and cos ( A+B)= cos60 { similarly we also cancel cos} Now, A-B =30 A+B= 60 --------------- 2A= 90 ------------- A= 45 and B=15 |
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| 21897. |
Which chapter have more marks give me chapter mark in decreasing order |
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| 21898. |
CosA-sinA+1______________=cosecA+cotACosA+sinA-1 |
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| 21899. |
Use e.d.l methid to find following pair of number is coprime |
| Answer» Which are the pair of no.s?? | |
| 21900. |
How many terms of the AP 9\' 17\' 25\'•••••••••••••••must be taken to give a sum of 616 |
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Answer» Sn=n/2 (2(9)+(n-1)8)616=n/2(18+8n-8)616= n/2( 10+8n)1272= 10n+8n2Whole divided by 24n2+5n-636=0By solving n= 12,-53/4Therefore n=12 153 |
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