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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.
| 11401. |
find the some of two consecutive postive integers sum of whose squares is 365\xa0 |
| Answer» Let one of the consecutive number be x.Therefore other consecutive number be (x + 1)According to question,(x)2\xa0+ (x + 1)2\xa0= 365x2\xa0+ x2\xa0+ 2x + 1 = 3652x2\xa0+ 2x - 364 = 0x2\xa0+ x - 182 = 0x2\xa0+ 14x - 13x - 182 = 0x(x + 14) - 13(x + 14) = 0(x + 14)(x - 13) = 0x = -14 (neglected), x = 13Therefore required two positive integers are 13 and 14. | |
| 11402. |
Are the points (-3,-3),(-3,2)and(-3,5)collinear?give reason. |
| Answer» For collinear,Area of triangle = 01/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2) = 0(-3)(2-5) - 3(5+3) - 3(-3-2) = 09 - 24 + 15 = 024 - 24 = 00 = 0Since Area of traingle = 0therefore, points are collinear. | |
| 11403. |
Find the locus of centres of circles which touch two intersecting lines. |
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Answer» let l1 and l2 be two intersecting lines which intersect at point P.\xa0Let O be the centre of circle which touches both l1 and l2 .In triangles OAP and OBP, we haveOA=OB (each equal to radius)\xa0PA=PB(tangents to a circle are equal)\xa0OP=OPso, by sss congruence criteria, we haveΔOAP congruent to ΔBPO | |
| 11404. |
What can be said about the roots of quadratic equation ax*2 +bx +c =0, where a>0,b=0,c>0? |
| Answer» Here Discriminant is b2-4ac which is -4ac( Since b = 0)Now Since a and c are positive which implies that the discriminant is negative, Hence this equation has imaginary roots. | |
| 11405. |
If P(2,4) is equidistant from Q(7,0) and R (X,9), find the value of X. Also find the distance PQ. |
| Answer» Given: P(2,4) is equidistnat from Q(7,0) and R(X,9).then PQ = PRUsing Distance Formula :\xa0=\xa0(x2-x1)2\xa0+\xa0(y2-\xa0y1)2PQ =\xa025\xa0+\xa016\xa0\xa0=\xa041PR\xa0=\xa0(x-2)2\xa0+\xa0(5)2PQ = PR\xa0Sqauring Both Side, We get\xa0x2\xa0+ 4\xa0-4x + 25\xa0= 41=> x2\xa0- 4x -12 = 0=>\xa0x2\xa0- 6x +2x -12 = 0=> x(x-6) +2(x-6) = 0=> (x-6) (x+2) = 0=> x = -2 , 6\xa0 | |
| 11406. |
Solve quadratic equation by factorization\xa0x-a/ x-b +x -b/ x-a = a/b+ b/a |
| Answer» x-ax-b+x-bx-a=ab+balet\xa0x-ax-b=y\xa0⇒x-bx-a=1y⇒y+1y=a2+b2ab⇒y2+1y=a2+b2ab⇒aby2-a2y-b2y-ab=0⇒ayby-a-bby-a=0⇒by-aay-b=0⇒y=ab\xa0or\xa0y=bacase-1if\xa0y=ab⇒x-ax-b=ab⇒bx-ba=ax-ba⇒bx=axbx-ax=0⇒x(b-a)=0⇒x=0Case-2if\xa0y=ba⇒x-ax-b=ba⇒ax-a2=bx-b2⇒ax-bx=a2-b2⇒x(a-b)=(a-b)(a+b)x=a+bHence\xa0x=0\xa0or\xa0a+b | |
| 11407. |
1/a+b+x=1/a+1/b+1/x .........find x |
| Answer» 1/a+b+x\xa0=1/a+1/b+1/x1/a+b+x-1/x = 1/a+1/bx-(a+b+x)/(a+b+x)x = a+b/abx-a-b-x/(a+b+x)x= a+b/ab-(a+b)/(a+b+x)x = a+b/ab\xa0(when we cross multiply).\xa0-ab(a+b) = a+b(a+b+x)x-ab(a+b)/a+b = ax+bx+x2\xa0-ab = ax+bx+x2ax+bx+x2+ab (on transposing left to right).\xa0ax+x2+bx+abx(a+x) +b(x+a)(x+a)(x+b)therefor x=-a x =-b----------------------------- | |
| 11408. |
Find 15 term of A.P with second term 11 and common difference\xa0 |
| Answer» | |
| 11409. |
For what value of K are the roots of the quadratic equation kx(x-2)+6=0\xa0 |
| Answer» For Equal roots D = 0kx(x-2) + 6 = 0kx2 -2kx + 6 = 0comparing with ax2+ bx + c = 0, we get\xa0a = k, b = -2k , c = 6D = b2 - 4acPut value of a b c(-2k)2 - 4*6*k = 04k2 - 24k = 04k(k-6) =0\xa04k = 0 or k-6=0k = 6 | |
| 11410. |
Why volume of cone is1/3 of cylinder volume |
| Answer» | |
| 11411. |
Find the probability that a non leap year chosen at random has\xa0\t52 Sundays\t53 Sundays |
| Answer» 2) A non-leap year has 365 daysA year has 52 weeks. Hence there will be 52 Sundays for sure.52 weeks = 52 x 7 = 364 days .365– 364 = 1day extra.In a non-leap year there will be 52 Sundays and 1day will be left.This 1 day can be\xa0Sunday, Monday,\xa0Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.Of these total 7 outcomes, the favourable outcomes are 1.Hence the probability of getting 53 sundays = 1 / 7.1)∴ probability of getting 52 sundays = 1 - 1/ 7 = 6 / 7. | |
| 11412. |
\xa0the sum of intiger and their reciprocal is 145/12. Find the intiger ? |
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Answer» Solution:Let the integer = xThen reciprocal = 1/xAccording to Questionx + 1/x = 145/12=> (x2+1)/x = 145/12=> 12x2 + 12 = 145x=> 12x2\xa0- 145x + 12 = 0=> 12x2- 144x - x + 12 = 0=> 12x(x - 12) - 1(x - 12) = 0=> (x - 12)(12x - 1) = 0=> x = 12 and x = 1/12as X is integer x can\'t be 1/12Therefore the required integer is 12 Let the integer be x.x + 1/x = 145/1212x2\xa0- 145x + 12 = 012x2\xa0- 144x - x + 12 = 012x(x - 12) - 1(x - 12) = 0(x - 12)(12x - 1) = 0x = 12 and x = 1/12Therefore the required integer is 12. |
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| 11413. |
Which term of AP 121,117,113.................is its first negative term |
| Answer» a = 121d = -4let an= 0thenan\xa0= a + (n-1)d=> 0 = 121 + (n-1)(-4)=> 0 = 121 - 4n + 4=> 4n = 125=> n = 125/4=> n = 31.25as n must be natural number then n= 3232nd\xa0term is First negative number.a32\xa0= 121 + (31)-4= 121 - 124 = -3 | |
| 11414. |
an+1+bn+1/an+bn\xa0is the A.M between a and b, then find the value of n.\xa0 |
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Answer» Ans.\xa0As we knowA.M. of two numbers a and b = (a+b)/2So(an+1\xa0+ bn+1)/(an+bn) = (a+b)/2=> 2(an+1+bn+1) = (a+b)(an+bn)=> 2an+1\xa0+ 2bn+1\xa0= an+1\xa0+ abn\xa0+ ban\xa0+ bn+1=> an+1\xa0- ban\xa0+ bn+1\xa0- abn\xa0= 0=> an(a-b) + bn( b-a) = 0=> an(a-b) -\xa0bn(a-b) = 0=> an(a-b) = bn(a-b)=> an/bn\xa0= 1=> (a/b)n\xa0= (a/b)0=> n =0value of n = 0 According to question,an+1+bn+1/an+bn\xa0= (a + b) / 2=> 2(an+1 + bn+1) = (a + b)(an + bn)=> 2an+1 + 2bn+1 =\xa0an+1 + abn + anb + bn+1=> an+1\xa0-\xa0anb +\xa0bn+1 - abn = 0=> an\xa0(a -\xa0b) - bn (a - b)= 0=> (an\xa0-\xa0bn)(a- b)= 0=> (an\xa0-\xa0bn) = 0=> an\xa0= bn=> (an/bn) = 1=> (a/b)n = (a/b)0=> n = 0\xa0\xa0\xa0 |
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| 11415. |
If the sum of first p terms of an ap ap2 + bq find its common difference |
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Answer» Given: Sp\xa0= ap2\xa0+ bqS1\xa0= a = a(1)2\xa0+ bq = a + bqS2\xa0= a + a2\xa0= a(2)2\xa0+ bq = 4a + bqThen,,. a2\xa0= 4a + bq - a - bq = 3aTherefore, common difference = a2 - a = 3a - a - bq = 2a - bq Ans. Sp\xa0= ap2+ bqPutting p= 1, we \'ll get some of first one term or first term itselfa1 = a(1)2+ bq\xa0=> a1\xa0= a + bqPutting p=2 we \'ll get sum of first two terms I.e a1+a2S2 = a(2)2 + bq\xa0= 4a+bqa2\xa0= S2-a2=> a2 = 4a + bq - a - bq = 3aCommon difference = a2-a1 = 3a-a-bq = 2a-bq Given, ap\xa0= ap2\xa0+ bqThen,a = a(1)2\xa0+ bqa2\xa0= a(2)2\xa0+ bq = 4a + bqTherefore,\xa0Common difference = 4a + bq - a - bq = 3a |
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| 11416. |
Find the sum of first n terms of an AP whose n th term is 5n-1.Hence find the sum of first 20 terms. |
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Answer» Ans.\xa0Let\xa0first\xa0term\xa0of\xa0A.P\xa0is\xa0a\xa0and\xa0common\xa0difference\xa0is\xa0d.Given\xa0:\xa0an\xa0=\xa05n-1So,\xa0a1\xa0=\xa05×1-1\xa0=4a2\xa0=\xa05×2\xa0-1\xa0=\xa09Common Difference (d) = 9-4 = 5Sum\xa0of\xa0first\xa0n\xa0terms\xa0Sn\xa0=\xa0n22×4\xa0+\xa0(n-1)\xa0×5=>\xa0Sn\xa0=\xa0n2(8\xa0+5n\xa0-\xa05)=>\xa0Sn\xa0=\xa0n23+5nSum\xa0of\xa0First\xa020\xa0Terms\xa0S20\xa0=\xa0202(3+5×20)\xa0=\xa010×\xa0103\xa0\xa0=1030 Given: an = 5n-1a = 5 x 1 - 1 = 5 - 1 = 4a2 = 5 x 2 - 1 = 10 - 1 = 9a3 = 5 x 3 - 1 = 15 - 1 = 14Therefore, AP is 4, 9, 14, .......Here, a = 4 and d = 5S20 = (20/2)[2 x 4 + (20 - 1) x 5] = 10 [8 + 95] = 10 x 103 = 1030Therefore, sum of 20 terms of given AP is 1030 |
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| 11417. |
solve for x by factorisation method. 3x2+√21x-14=0 |
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Answer» Ans.\xa03x2+\xa021x\xa0-14\xa0=\xa00=>\xa03x2\xa0+\u2009221x\xa0-\xa021x\xa0-14\xa0=\xa00=>\xa03x(3x\xa0-27\xa0)\xa0-7(3x\xa0-\xa027)\xa0=\xa0\xa00\xa0=>\xa0(3x\xa0-7)(3x\xa0-\xa027)\xa0=\xa00Either\xa03x\xa0-\xa07\xa0=\xa00\xa0\xa0\xa0or\xa0\xa03x\xa0-\xa027\xa0=\xa00\xa0x\xa0=\xa073\xa0\xa0or\xa0\xa0273 3x2+21x-14=03x2+221x-21x-14=03x(3x+27)-7(3x+27)=0(3x+27)(3x-7)=03x+27=0\xa0and\xa03x-7=0x=-273\xa0and\xa0x=73 3x2 + 221x - 21x - 14 = 03x(3x+27)\xa0-\xa07(3x+27)(3x+27)\xa0(3x-7)3x+27\xa0=\xa00,\xa03x-7\xa0=\xa00x\xa0=\xa0-273=-273,\xa0x\xa0=\xa073=73 |
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| 11418. |
In construction , even if they have not mentioned , whether we have to write steps of construction?? |
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Answer» Yes, mention steps of construction Ans. Yes, Its a good habbit to Write steps of construction even if it has been asked or not. |
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| 11419. |
What is date sheet of non board of class 10\xa0 |
| Answer» Every school has different datesheet for Class 10 Non-boards. Most probably the exams will start either in first or second week of March. The school will provide you with the datesheet in the next few days.All the best ! | |
| 11420. |
In what ratio does the line x-y-2=0 divide the line joining the points A(3,-1) and B(8,9)? |
| Answer» Let the ratio be k : 1.The given points are A(3, -1) and B(8, 9).Then the coordinates of point of intersection are: m1x2+m2x1m1+m2,m1y2+m2y1m1+m2 = k×8+1×3k+1,\xa0k×9+1×-1k+1 = 8k+3k+1,\xa09k-1k+1Now, these coordinates is on the line x - y - 2 = 0Then putting the coordinates of intersection point in this line, we get8k+3k+1-9k-1k+1-2=0=> 8k+3-9k+1-2k-2=0=> -3k+2=0=> k=23Therefore, the required ratio is 2 : 3. | |
| 11421. |
Solve the equation. 22x+3 is equal to 65(2x_2) + 122 |
| Answer» 22x+3=65(2x-2)+1228.22x=65.22x-130+1228.22x=65.22x-88D2-65D+8=0(take D=2x)(D-8)(8D-1)=0(BY MID TERM FACTOR)THEN,D-8=0 OR 8D-1=0D=8 OR\xa0182x=8 OR\xa018x=3 or 1/3\xa0 | |
| 11422. |
A circle is inscribed in a triangleABC. If AB=12cm,AC=10cm,BC=8cm Find AD,BE and CF |
| Answer» Ans. Given : A Circle inscribed in A\xa0triangle ABCAB = 12cm, AC = 10cm, BC = 8cmWe Know that Tangent drawn from an external point are equal in length.\xa0Let AD = x\xa0Then AD = AE = x [Tangent from Same point]Similarly\xa0BD = BF = y\xa0CE = CF = z\xa0AB = AD + BD => x + y =\xa012 (1)AC = AE + EC\xa0=> x + z = 10 (2)BC = BF + FC=> y + z = 8 (3)Adding all three equations, we get2(x+y+z) = 30=> x + y + z = 15 (4)from (1) x + y = 12then z = 3 cmSimilarly y = 5 cm and z = 7cm | |
| 11423. |
A circle is inscribed in a triangleABC If AB=12cm,AC=10cm,BC=8cm Find AD,BE and CF |
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Answer» Let AD be x cm, BE = y cm, CF = z cmThen, AD = AF = x cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal] BD = BE = y cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal] CE = CF = z cm [Since tangents from\xa0an external point to\xa0the circle\xa0are equal]According to question,AD + BD = 12 cm => x\xa0+ y = 12 ..........(i)BE + EC =\xa08 cm => y\xa0+\xa0z =\xa08 ..........(ii)CF + AC =\xa010 cm => x\xa0+\xa0z =\xa010 ..........(iii)Adding eq.(i), (ii) and (iii), we get2(x + y + z) = 30 => x + y + z = 15 ..........(iv)Solving eq. (iv) and (ii), we get x = 7 cm => AD = 7 cmSolving eq. (iv) and (iii), we get\xa0y =\xa05 cm => BE =\xa05 cmSolving eq. (iv) and (i), we get\xa0z =\xa03 cm => CF =\xa03 cm Ans. Given : A Circle inscribed in A\xa0triangle ABCAB = 12cm, AC = 10cm, BC = 8cmWe Know that Tangent drawn from an external point are equal in length.\xa0Let AD = x\xa0Then AD = AE = x [Tangent from Same point]Similarly\xa0BD = BF = y\xa0CE = CF = z\xa0AB = AD + BD => x + y =\xa012 (1)AC = AE + EC\xa0=> x + z = 10 (2)BC = BF + FC=> y + z = 8 (3)Adding all three equations, we get2(x+y+z) = 30=> x + y + z = 15 (4)from (1) x + y = 12then z = 3 cmSimilarly y = 5 cm and z = 7cm |
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| 11424. |
sector of circle explain\xa0 |
| Answer» The region enclosed by an arc of the circle is called sector. | |
| 11425. |
If a2,b2,c2\xa0are in A.P to prove that 1/b+c,1/c+a,1/a+b are in A.P |
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Answer» If a2, b2, c2\xa0are in AP, then2b2\xa0= a2\xa0+ c2\xa0b2\xa0+ b2\xa0= a2\xa0+ c2\xa0b2\xa0- a2\xa0= c2\xa0- b2\xa0(b - a)(b + a) = (c - b)(c + b)(b - a)/(c + b) = (c - b)/(b + a)Dividing both sides by (c + a),(b - a)/{(c + b)(c + a)} = (c - b)/{(b + a)(c + a)}{(b + c) - (c + a)}/{(b + c)(c + a) = {(c + a) - (a + b)}/{(a + b)(c + a)}{1/(c + a)} - {1/(b + c)} = {1/(a + b)} - {1/(c + a)}{2/(c + a)} = {1/(a + b)} - {1/(b + c)}Therefore 1/(a + b), 1/(b + c), 1/(c + a) are in AP. Ans. Given : a2 b2 c2 are in A.P.To Prove :\xa0\\({1 \\over b +c } ,{1 \\over c+a}, {1\\over a+b} \\)\xa0Are in A.P.Proof :\xa02b2 = a2 + c2 [as a2 b2 c2 are in A.P.]=> b2 + b2 = a2 + c2=> b2 - a2 = c2 - b2\xa0=> (b-a) (b+a) = (c-b) (c+b)=>\xa0\\({(b-a) \\over (c+b) } = {(c-b) \\over (b+a)}\\)Divide both side by\xa0\\(1\\over (c+a)\\), We get\xa0=>\xa0\\({(b-a) \\over (c+b) \\times (c+a) } = {(c-b) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({(b-a+c-c) \\over (c+b) \\times (c+a) } = {(c-b+a-a) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({(b+c) - (c+a) \\over (c+b) \\times (c+a) } = {(c+a) -(a+b)) \\over (b+a)\\times(c+a)}\\)=>\xa0\\({1 \\over(c+a)} - {1\\over (c+b) } = {1\\over (a+b)} -{ 1\\over(c+a)}\\)=>\xa0\\({2 \\over(c+a)} = {1\\over (a+b)} + {1\\over (c+b) }\\)Hence by this equation we, can say that\xa0\\({1 \\over b +c } ,{1 \\over c+a}, {1\\over a+b} \\)are in A.P. |
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| 11426. |
For an AP Sm=20 and Sn=10 and n-m=1,then prove that n=10÷a where a=d. |
| Answer» \\(I \\space think \\space in \\space question S_m = 10 \\space and \\space S_n = 20 \\)Ans. Given :\xa0\\(S_m = 10\\)\\(S_n = 20 \\space \\space \\space \\space ..... (1)\\)=> n-m = 1\xa0=> m = n-1\xa0The sum of m terms i.e (n-1) terms\xa0\\(S_{n-1} =S_m = 10 \\space \\space \\space \\space \\space \\space ...... (2)\\)=>\xa0\\(n^{th} term = S_n - S_{n-1}\\)=>\\(n^{th} term = 20 - 10 = 10\\)=>\xa0\\(a + (n-1)d = 10\\)=>\xa0\\(a + (n-1)a = 10 \\space \\space \\space \\space \\space [as \\space a =d ]\\)=>\xa0\\(a+ na -a = 10 \\)=>\xa0\\(na = 10 \\)=>\xa0\\(n = {10\\over a}\\)Hence proved | |
| 11427. |
How to improve my minor mistake in maths.\xa0 |
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Answer» Work without tension. Enjoy doing maths. Never think it as meri attma hamesha satayegee. Best of luck.\xa0 Practice maths for 3 hours daily with full concentration\xa0 |
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| 11428. |
The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm.find the area of a circle. |
| Answer» area of sector = 1/2*l*r = 1/2*6*5.2=3*5.2=15.6cm^2perimeter-radius=length of arc | |
| 11429. |
The sum of the number and its reciprocal is 2 ×1/30.find the number |
| Answer» Let the number be x and its reciprocal be {tex}{1 \\over x}{/tex}According to the question{tex}x + {1 \\over x} = 2{1 \\over 30}{/tex}{tex}\\implies {x^2 +1 \\over x} = {61 \\over 30}{/tex}{tex}\\implies{/tex}30x2 - 61 x+30 = 0{tex}\\implies{/tex}30 x2 - 36x - 25x + 30 =0{tex}\\implies{/tex}6x ( 5x -6) - 5 (5x -6) = 0{tex}\\implies{/tex}(6x - 5) (5x - 6) = 0Either 6x - 5 = 0 or 5x - 6 = 0{tex}\\implies x = {5 \\over 6}, {6 \\over 5}{/tex}Hence, the required number is {tex}{5 \\over 6} \\,or \\,{6 \\over5}{/tex} | |
| 11430. |
√x + y = 8√y + x = 24 |
| Answer» | |
| 11431. |
An unbiased die is thrown find probability of getting number between 1 and 4. |
| Answer» Number of outcomes in throwing a die : 6Number of favourable outcomes : 2 [ i.e. number b/w 1 and 4 are 2 and 3 only]P( getting a no. b/w 1 and 4) = 2/6\xa0= 1/3 | |
| 11432. |
Find the sum of the first 30 terms of an AP\xa0whose nth term is 2-3n?\xa0 |
| Answer» an\xa0= 2-3nTo form an A.P. from it, substitute different values of\xa0na1\xa0= 2-3(1) = 2-3 = -1a2\xa0= 2-3(2) = 2-6 = -4a3\xa0= 2-3(3) = 2-9 = -7a4\xa0= 2-3(4) = 2-12 = -10-1 , -4 , -7 , -10 forms an A.P.The first term a = -1, common difference d = -4-(-1) = -4+1 = -3 , number of terms n =30S30 =\xa0\\({ n \\over 2}\\)\\({[ 2a+ (n-1) d]}\\)= 15 [ -2 + 29 X -3]= 15[-89]S30= -1335 | |
| 11433. |
determine the condition for the roots of the equation ax2 +bx +c= 0 to differ by 2 |
| Answer» Ans. Let one root = xThen other root = x +2As We know,Difference of roots = \\(\\sqrt D \\over a\\)where D = b2-4ac\\(=> 2= {\\sqrt{b^2-4ac} \\over a}\\)\\(=> 2a= \\sqrt{b^2-4ac}\\)Squaring Both Sides We Get\\(=> 4a^2 = b^2 - 4ac\\)This is the required Condition. | |
| 11434. |
in an AP the sum of the first 10 terms is -150. And the sum of next 10 terms is -550. Find the AP. |
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Answer» hence your ap series is -3,1,5,9,13,17............................\xa0 \xa0S10=150 here n=10 ;a=? ;d=?n/2{2a+(n-1)d}150=10/2{2a+9d}30=2a+9deq.1than for next 10th terms we can write S10+S10=150+550S20\xa0=700 than n=20700=20/2{2a+19d}eq.2subtract eq.1 and eq.2 we get\xa040=10dd=4put the value of d in eq.1 we get\xa030=2a+9*4a=-6\xa0\xa0 Ans. Let First Term = a\xa0Common difference = dSum of first 10 Terms S10\xa0= -150\xa0\\(=> S_{10} = {10\\over 2} [2a+ (10-1)d]\\)\\(=> -150 = 5 (2a+9d)\\)=>\xa0\\(2a+9d = -30\\) ........... (1)For Next 10 terms, 11th\xa0ll be first term,\\(=> a_{11} = a + 10d\\)\\(=> S_{10} = {10\\over 2}[2\\times (a+10d) + (10-1)d]\\)\\(=> -550 = 5[2a+20d + 9d]\\)\\(=> 2a+29d = -110 \\) ........ (2)Subtract (1) from (2), we get=> 20d = -80\xa0=> d = -4\xa0Put value of d in(1) we , get\xa0=> a = 3AP : 3 , -1, -5, -9, ........ |
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| 11435. |
how many terms of the AP 9,7,25..... must be taken to give a sum of 636 |
| Answer» it ll be 17 in place of 7.Ans. First Term (a) = 9Common Difference (d) = 17-9 = 8\xa0Sn\xa0= 636We Know\\(=> S_n = {n\\over 2}[2a+(n-1)d]\\)\\(=> 636= {n\\over 2}[2\\times 9 +(n-1)8]\\)\\(=> 636= {n\\over 2}[18 +8n-8]\\)\\(=> 636= {n\\over 2}[10 +8n] => 636=n[5 +4n]\\)=> 4n2 + 5n - 636 = 0=> 4n2 + 53n - 48n - 636 = 0=> n (4n+53) - 12(4n+ 53) = 0=> (4n+53) (n-12) = 0Either 4n + 53 = 0 or n -12 = 0n = (-53\\4) or 12\xa0Number of terms = 12\xa0 | |
| 11436. |
Quardic equation\xa0(x+1)(x+2)(x+3)(x+4)-8=0 |
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| 11437. |
ABCD is a cyclic quadrilateral if |
| Answer» 70o | |
| 11438. |
Find the probability of 53 Sundays in one leap year |
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Answer» the correct answer is 1/7 \xa0 1/7 It\'s 2/7 |
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| 11439. |
If the length of the shadow is 60 m high is 20√2 find sun\'s altitude |
| Answer» | |
| 11440. |
Wt is d probability of getting 5 friday n 5 saturday in month of March\xa0\xa0 |
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| 11441. |
Find the area of the largest triangle that can inscribe in a semicircle of radius 21cm\xa0 |
| Answer» The area of triangle =1/2. ×. base ×. height Where base= 42 diameter of the\xa0 circle And. Height = 21 radius. of the circle perpendicular to diameter\xa0Answer =. 441\xa0 | |
| 11442. |
Which term of the A.P. 20,77/4,37/2,75/4,..........is the first negative term?Find the term also. |
| Answer» The given AP is 20,\xa0{tex}19 \\frac { 1 } { 4 } , 18 \\frac { 1 } { 2 } , 17 \\frac { 3 } { 4 } , \\dots{/tex}common difference ={tex}19 \\frac { 1 } { 4 } - 20 = \\frac { 77 } { 4 } - 20 = \\frac { - 3 } { 4 }{/tex}the general term of an AP is given byan\xa0=a+(n-1)d<0{tex}\\Rightarrow{/tex}\xa0a+(n-1)d<0{tex}\\Rightarrow{/tex}\xa020+(n-1)({tex}\\frac{{ - 3}}{4}{/tex})<0{tex}\\Rightarrow{/tex}\xa020-{tex}\\frac{{3n}}{4}{/tex}+{tex}\\frac{3}{4}{/tex}<0{tex}\\Rightarrow{/tex}\xa0-\u200b\u200b\u200b\u200b\u200b\u200b{tex}\\frac{{3n}}{4}{/tex}+{tex}\\frac{{83}}{4}{/tex}<0{tex}\\Rightarrow{/tex}\xa0-3n+83<0{tex}\\Rightarrow{/tex}\xa0-3n< -83{tex}\\Rightarrow{/tex}\xa0n>{tex}\\frac{{83}}{3}{/tex}{tex}\\Rightarrow{/tex}\xa0n> 27.6So, n=28a28\xa0= a + 27d = 20 + 27\xa0{tex}\\times \\frac{-3}{4}{/tex}= -0.25Hence, the first negative term would be the 28th term.\xa0 | |
| 11443. |
Along a road lie an odd no. Of stones..............rd example 34 .......... |
| Answer» Wrong question\xa0Give a clear data and a clear question | |
| 11444. |
draw a line segment of length 7cm. Find a point P on it which divides it in the ratio 3:5 |
| Answer» We have to draw a line segment of length 7 cm.Then,we have to find a point P on it, which divides it in the ratio 3 : 5.Steps of construction:\tDraw a line segment {tex}AB=7{/tex} cm.\tDraw a ray AX, making an acute {tex}\\angle BAX{/tex} with AB.\tMark {tex}3+5=8{/tex} points, i.e, {tex}A_1, A_2, A_3, A_4 ...A_8{/tex} on AX, such that {tex}AA_1 = A_1A_2 = A_2A_3 = A_3A_4 ... = A_7A_8{/tex}\tJoin A{tex}_8{/tex}B\tFrom A3, draw {tex}A_3P || A_8B{/tex} which intersects AB at point P [ by making an angle at A3 equal to {tex}\\angle AA_8B{/tex}\xa0\tThen, P is the point on AB which divides it in the ratio 3:5. So, {tex}AP : PB = 3:5{/tex}Justification: In {tex}\\triangle ABA_8{/tex}, we have {tex}A_3P || A_8B{/tex}\xa0{tex}\\therefore \\frac{AP}{PB}=\\frac{AA_3}{A_3A_8}{/tex} [ by basic proportionality theorem]By construction, {tex}\\frac{AA_3}{A_3A_8}=\\frac{3}{5}{/tex}\xa0Hence, {tex}\\frac{AP}{PB}=\\frac{3}{5}{/tex} | |
| 11445. |
The common tangents AB and CD to two circles O and O\' intersecct at E . Prove that AB =CD.\xa0 |
| Answer» It isEA\xa0and EC are the tangents of circle with centre O soEA=EC (Tangents drawn from the external point of circle are equal)-1NowSimilarly,ED and EB are the tangents of circle with centre O\' soEB=ED (Tangents drawn from external point of circle are equal)-2Adding 1 and 2EA+EB=EC+EDAB=CD | |
| 11446. |
In an ap , the sum of first n terms is 3n2/2 + 13n/2 . Find the 25th\xa0\xa0term. |
| Answer» Sn\xa0= 3n2/2 + 13n/2S1\xa0= a = 3/2 + 13/2 = 8S2\xa0= a + a2\xa0= 12/2 + 26/2 = 19a2\xa0= 19 - 8 = 11Then,. d = 11 -8 = 3Now, a25\xa0= a +(25 - 1)d= 8 + 24 × 3= 80The 25th term is 80 | |
| 11447. |
The mid point of the line segment joining (2a,4) and (-2,2b) is (1,2a+1).find the value of b? |
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Answer» Midpoint of two given points are\xa0{(x1\xa0+ x2)/2, (y1\xa0+ y2)/2}According to question,(2a - 2)/2 = 12a - 2 = 22a = 4a = 2Again, (4 + 2b)/2 = 2a + 14 + 2b = 4a + 22b = 4 × 2 + 2 - 42b = 8 - 2b = 3 Mid point = (x1+x2)/2 ; (y1+y2)/21 = (2a+1)/22 = 2a+12a = 1a =1/22a+ 1 = (4+2b)/24a+2 = 4 +2b4a-2 = 2b2a-1 = b1 - 1 =bb=0 |
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| 11448. |
If one root of the quadratic equation is 3+2root5/4 then what will be the other root |
| Answer» | |
| 11449. |
if one root of the quadratic equation is 3+2root5/4.then find the other root\xa0 |
| Answer» According to the question,\xa0One of the\xa0roots of the quadratic equation =\xa0{tex}\\frac { 3 + 2 \\sqrt { 5 } } { 4 }{/tex}{tex}\\because{/tex}\xa0Coefficient are rational{tex}\\therefore{/tex}\xa0other root of the equation\xa0will be\xa0{tex}\\frac { 3 - 2 \\sqrt { 5 } } { 4 }{/tex}\xa0(If coefficient are rational, then the roots are always conjugate rational) | |
| 11450. |
If 4 times the fourth term og an AP is equal to 18 times its 18th term find its 22 term |
| Answer» Solution: let first term = acommon difference = dATQ.=> 4a4= 18a18=> 4 (a+3d) = 18(a+17d)=> 4a + 12d = 18a + 306d=> 14a - 294d = 0divide by 14=> a+21d = 0=> a+(22-1)d =0=> a22\xa0= 0so 22th term is 0\xa0 | |