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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.
| 7301. |
(1/tana+cota)(1/1+sina)+(1/1-sina) =2sec2a |
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| 7302. |
what is the HCF of two consecutive even numbers |
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Answer» HCF of two consecutve even nos is always 2for exampleHCF of 2 and 4 is = 2HCF of 22 and 24 is = 2 and so on\xa0 2 |
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| 7303. |
Solve for x if (under root 2x+9)+x=13 |
| Answer» 8 and 20 | |
| 7304. |
Find four no. In AP whose sum is 28 and sum of square is 216 ? With solution |
| Answer» Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)According to the question, sum of the numbers=28{tex}\\therefore{/tex}4a = 28\xa0{tex}\\Rightarrow{/tex}\xa0a = 7Sum of the squares of these numbers=(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=4(a2+5d{tex}^2{/tex})Now, sum of the squares of numbers=216{tex}\\therefore{/tex}4(a2+5d2)=216{tex}\\Rightarrow{/tex}a2+5d2=54 [{tex}\\because {/tex}a=7]{tex}\\Rightarrow{/tex}5d2= 54-49{tex}\\Rightarrow{/tex}5d2=5{tex}\\Rightarrow{/tex}d2=1{tex}\\Rightarrow{/tex}d={tex} \\pm {/tex}1Hence, the required numbers (4, 6, 8, 10). | |
| 7305. |
The value of 2x if the mean of x,x+2,x+7, x+11 and 2x is 16 |
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Answer» 10 value of 2x=20 10 |
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| 7306. |
In triangle ABC AB=√3 AC = 12 BC =6 find angle b |
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| 7307. |
Find zeroes of v²+4√3v-15Also verify relationship between the zeroes and the coefficient |
| Answer» The given quadratic polynomial is: v2\xa0+ 4{tex}\\sqrt 3{/tex}v - 15By factorizing it we have v2\xa0+ 4{tex}\\sqrt 3{/tex}v - 15 = v2\xa0+ 5{tex}\\sqrt 3{/tex}v -\xa0{tex}\\sqrt 3{/tex}v - 15=\xa0v(v\xa0+ 5{tex}\\sqrt 3{/tex})\xa0-\xa0{tex}\\sqrt 3{/tex}(v +\xa05{tex}\\sqrt 3{/tex})\xa0= (v -\xa0{tex}\\sqrt 3{/tex})(v +\xa05{tex}\\sqrt 3{/tex})For zeroes, put the factors equal to zero i.e.,\xa0(v -\xa0{tex}\\sqrt 3{/tex})(v +\xa05{tex}\\sqrt 3{/tex})\xa0= 0{tex}\\Rightarrow v = \\sqrt { 3 } , - 5 \\sqrt { 3 }{/tex}\xa0are zeroes of the polynomial.Verification:\xa0In the given polynomial a = 1, b = 4{tex}\\sqrt 3{/tex}\xa0and c = - 15Now Sum of the zeroes =\xa0{tex}\\sqrt { 3 } + ( - 5 \\sqrt { 3 } ) = - 4 \\sqrt { 3 }{/tex}Also sum of zeroes =\xa0{tex}\\frac { - b } { a }{/tex},\xa0{tex}\\frac { - b } { a } = \\frac { - 4 \\sqrt { 3 } } { a } = - 4 \\sqrt { 3 }{/tex}And product of zeroes =\xa0{tex}\\sqrt { 3 } \\times - 5 \\sqrt { 3 }{/tex}= -15Also,\xa0product of zeroes =\xa0{tex}\\frac { c } { a } = \\frac { - 15 } { 1 } = - 15{/tex} | |
| 7308. |
Sam |
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| 7309. |
How to learn trignometry values |
| Answer» Sorry bcoz it will take time ???? | |
| 7310. |
(X+2)(3x_5)= |
| Answer» (x+2)(3x-5)=3x^2-5x+6x-10=3x^2-x-10=3x^2-6x+5x-10=3x(x-2)+5(x-2)=(3x+5)(x-2). ==x=-5/3 or 2 | |
| 7311. |
Findthe value of at x+y,if 3x-xy=5 and 3y-2x |
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| 7312. |
Are two triangles with equal corresponding sides always similar? |
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Answer» yestwo triangles are similar if they have same shape but not necessary same size, one may be bigger than the other but of same shapetriangles are similar if\xa0AAA (Angle , Angle , Aangle) If all three pairs of corresponding angles are the sameSSS (Side , Side , Side) if all three pairs of corresponding sides are in the same proportionSAS (Side , Angle, Side) if two pairs of sides are in the same proportion and included angle equal Ya |
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| 7313. |
The HCF of two number is 23 and their LCM is 1449 . If one of the number is 161 , find the other |
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Answer» Let the two numbers be a and bHCF = 23LCM = 1449a = 161We have to find b.HCF(a,b)xLCM(a,b) = axb∴23x1449 = 161x b∴b = 207Thus, the other number is 207. let the numbers = 161 , aproduct of two nos = hcf x lcm161 x a = hcf x lcm\xa0 a = 23 x 1449 /161 = 1449/7 = 207 |
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| 7314. |
2÷10=2 |
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| 7315. |
Find the number of terms in the following (x+y)+(x-y)+(x-3y)+...........=22(-20) |
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| 7316. |
Is in 2018 all questions comes from ncert |
| Answer» Yes most but not all? | |
| 7317. |
I want circle lesson 10.2 9sum |
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| 7318. |
hallo |
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| 7319. |
If 3 tanx =4 then find the value of sinx +cos x |
| Answer» Tanx=4/3 now hypotenuse becomes 5 therefore sinx4/5 and coax becomes 3/5 adding them we get 7/5 | |
| 7320. |
if the ratio of the sum of n terms of AP\'s is (7n+1):(4n+27),then find the ratio of their mth term |
| Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex} | |
| 7321. |
The radius of a circle is increase twice then its perimeter is increase |
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Answer» Why no Yes perimeter get doubledIf radius is r then perimeter of circle(p) = 2πr And if radius(r) increased twise thenNew radius(R) = 2(r)New perimeter of circle(P)= 2πR = 2π(2r) = 4πr =2(2πr)And the 2πr is equal to the old perimeter of circle(p) So new perimeter of circle = 2pSo new perimeter of circle is double if we increase the radius twise No |
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| 7322. |
19 ,21,23 find commom difference |
| Answer» It is an A.P. so common difference isT2-T1=dd=21-19d=2 | |
| 7323. |
What is the common difference of an AP in which a(18)-a(13)=25? |
| Answer» let first term of AP is a and the common difference is d.\xa0We knowan\xa0= a + (n-1)dSo,\xa0{tex}a_{18} - a_{13} = 25 \\\\=> a+ (18-1)d - a - (13-1)d=25\\\\=> 17d -12d = 25 \\\\=> 5d = 25 \\\\=> d = 25\\div 5 = 5{/tex} | |
| 7324. |
What is the formula to find ui,xi,di,cf |
| Answer» ui=x-A/h , do=xi-A | |
| 7325. |
Can we use set square in construction in board exam |
| Answer» No you can not use\xa0 | |
| 7326. |
Prove pi is an irrational number |
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| 7327. |
The fraction 2(√2+√6)/3(√2+√√3)is equal to |
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| 7328. |
Hcf of 210,270,360 |
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| 7329. |
Hcf of 210,280,360 |
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| 7330. |
Find the sum of the following AP 1/10, 1/2 , 1/10,....,to 11 term |
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| 7331. |
Two cubes of sides 4cm are joined and to ends. Find the surface area of cuboid |
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Answer» 160sq units Cxjc |
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| 7332. |
Introduction of AP |
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| 7333. |
sec6a-tan6a=1+3tan²a *sec²a |
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| 7334. |
cosec alpha =√2. cot beta= 1find sin(alpha+beta) |
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| 7335. |
Prove √3 is irrarional |
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Answer» Let us assume that |
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| 7336. |
Proof completing square method lesson quadratic equation? |
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| 7337. |
If n=2³*3 |
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| 7338. |
Exercise 7.1 ka 1 question |
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| 7339. |
Is there easy methods to solve trigonometric equations? |
| Answer» yes!!u can do by transforming whole equation in sin and cos | |
| 7340. |
Logical questions I |
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| 7341. |
Chapter 2 Ex 2.2 questions 1 |
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| 7342. |
Draw a pair of tangents of radius 5 cm which are inclined to each other at an angle of 45 |
| Answer» Follow the given steps in order to construct the tangents.Step 1 : Draw a circle of radius 5\xa0cm and centre O.Step 2 : Take a point A on the circle. Join OA.Step 3 : Draw a perpendicular to OA at A.Step 4 : Draw a radius OB, making an angle of 135° (180° – 45°) with OA.Step 5 : Draw a perpendicular to OB at point B. Let these perpendiculars intersect at P.Here, PA and PB are the required tangents inclined at angle of 45°. | |
| 7343. |
How to make the equations in word problems in quadratic equations? |
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| 7344. |
Find the area of rhombus if its vertices are (3,0),(4,5),(-1,4)and(-2,-1)taken in order. |
| Answer» Let A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1){tex}AC = \\sqrt {{{( - 1 - 3)}^2} + {{(4 - 0)}^2}} = 4\\sqrt 2 {/tex}{tex}BD = \\sqrt {{{( - 2 - 4)}^2} + {{( - 1 - 5)}^2}} = \\sqrt {36 + 36} = 6\\sqrt 2 {/tex}Area of rhombus = {tex}\\frac{1}{2}{d_1} \\times {d_2}{/tex}{tex} = \\frac{1}{2}AC \\times BD{/tex}{tex} = \\frac{1}{2} \\times 4\\sqrt 2 \\times 6\\sqrt 2 = 24{/tex} Sq. unit. | |
| 7345. |
If root3sintheta-costheta=0 |
| Answer» {tex}\\sqrt 3 \\sin \\theta - \\cos \\theta = 0 \\\\=> \\sqrt 3 \\sin \\theta = \\cos \\theta \\\\ => \\tan \\theta = {1\\over \\sqrt 3} \\\\=> \\tan \\theta = \\tan 30^o \\\\=> \\theta = 30^o{/tex} | |
| 7346. |
√2+√3 is irrational |
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| 7347. |
Sinα-2sin3α\\2cos3α-cosα=tanα |
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| 7348. |
Find the value of K if one of the root of quadratic eqn Kx2-14x+8=0 is six times to the other? |
| Answer» Kx2\xa0- 14x + 8 = 0\xa0On Comparing with standard form of polynomial, we get a = k, b = -14 and c = 8\xa0let one root =\xa0{tex}\\alpha {/tex}then another root =\xa0{tex}6\\alpha {/tex}We know,Sum of roots =\xa0{tex}{-b\\over a}{/tex}=>\xa0{tex}\\alpha + 6\\alpha = {14\\over K} \\\\=> 7\\alpha = {14\\over K}\\\\=> \\alpha = {2\\over K} \\ ... (i){/tex}Also,\xa0Product of roots =\xa0{tex}c\\over a{/tex}{tex}\\alpha \\times 6\\alpha = {8\\over K}\\\\6\\alpha ^2 = {8\\over K}{/tex}{tex}6\\times ({2\\over K})^2= {8\\over K} \\ using \\ (i) \\\\=> {24\\over K^2} = {8\\over K} \\\\=> K = 3 {/tex} | |
| 7349. |
What is formula of Ap |
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Answer» a+(n-1)d a+(n-1)d where a is first term d is common difference |
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| 7350. |
what is corresponding angle |
| Answer» Corresponding angles are\xa0the angles which occupy the same relative position at each intersection where a straight line crosses two others. If the two lines are parallel, the corresponding angles are equal. | |