InterviewSolution
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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.
| 1951. |
3x square -2_/¯6x+2 by completing square method |
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| 1952. |
Practicals of maths have to be done |
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| 1953. |
If sin(A+B)=1 and cos(A-B)=1 , find the values of A and B.Given that 0=< 90° and A>=B |
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Answer» Sin(A+B)=1, cos( A-B)=1A+B=90. A-B=0 A=BThen ,A=B=45 Both are 45 |
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| 1954. |
Can anyone tell me the datesheet of board exam? |
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| 1955. |
Define sector? |
| Answer» Covered by two radius in circle | |
| 1956. |
What is segment? |
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| 1957. |
If the sum of n terms of an ap is given by Sn=3n^2+4n. Determine the ap and the nth term |
| Answer» Sn = 3n2 + 4n.a1\xa0= S1 = 3(1)2\xa0+ 4(1) = 7a1\xa0+ a2\xa0= S2\xa0= 3(2)2 + 4(2)= 12 + 8= 20a2 = S2- S1\xa0= 20 - 7 = 13or, {tex}a + d = 13{/tex}or, {tex}7 + d = 13{/tex}{tex}\\therefore{/tex}\xa0{tex}d = 13 - 7 = 6{/tex}{tex}\\therefore{/tex}\xa0A.P. becomes 7,13,19,......\xa0Now, {tex}a_n =a + (n - 1 )d{/tex}{tex}= 7 + (n - 1)(6){/tex}{tex}= 7 + 6n - 6{/tex}{tex}= 6n + 1{/tex}or, an = 6n + 1 | |
| 1958. |
If the point on the y axis which is equidistance from the point (5,-2)and (-3,2) |
| Answer» Let the given points be A(5, -2 ) and B(-3, 2) and let the required point be P(x, 0). Then,PA = PB\xa0{tex}\\Rightarrow{/tex}\xa0PA2\xa0= PB2{tex}\\Rightarrow{/tex}\xa0(x - 5)2\xa0+ (0 + 2)2= (x + 3)2\xa0+ (0 - 2)2{tex}\\Rightarrow{/tex}\xa0(x - 5)2\xa0+ 4 = (x + 3)2 + 4{tex}\\Rightarrow{/tex}\xa0(x - 5)2 - (x + 3)2 =0{tex}\\Rightarrow{/tex}\xa0(x 2 -10x + 25) - (x2 + 6x + 9)=0{tex}\\Rightarrow{/tex}\xa0-16x + 16 =0{tex}\\Rightarrow{/tex}\xa016x = 16\xa0{tex}\\Rightarrow{/tex}\xa0x = 1.Hence, the required point is P(1, 0). | |
| 1959. |
Find the eleventh term from the last term of the AP,27\'23,19,...,65 |
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Answer» -209 is this correct 105 is it right |
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| 1960. |
What is G. P. |
| Answer» Geometrical Progressions | |
| 1961. |
If alpha and beta are the zeroes of the equation 6x^2+x-2, find alpha/beta + beta/alpha |
| Answer» f(x) = 6x2\xa0+ x - 2a = 6, b = 1, c = -2Let zeroes be {tex}\\alpha{/tex}\xa0and β.ThenSum of zeroes=\xa0{tex}\\alpha{/tex} +\xa0β {tex}=\\;-\\frac ba\\;=-\\frac16{/tex}Product of zeroes {tex}\\alpha{/tex}× β {tex}=\\;\\;\\frac ca\\;=\\;\\frac{-2}6\\;=\\;-\\frac13{/tex}{tex}\\frac { \\alpha } { \\beta } + \\frac { \\beta } { \\alpha } = \\frac { \\alpha ^ { 2 } + \\beta ^ { 2 } } { \\alpha \\beta }{/tex}{tex}= \\frac { ( \\alpha + \\beta ) ^ { 2 } - 2 \\alpha \\beta } { \\alpha \\beta } \\left[ \\because ( \\alpha + \\beta ) ^ { 2 } = \\alpha ^ { 2 } + \\beta ^ { 2 } + 2 \\alpha \\beta \\right]{/tex}{tex}= \\frac { \\left[- \\frac { 1 } { 6 } \\right] ^ { 2 } - 2 \\left[ - \\frac { 1 } { 3 } \\right] } { \\left[ - \\frac { 1 } { 3 } \\right] }{/tex}{tex}= \\frac { \\frac { 1 } { 36 } + \\frac { 2 } { 3 } } { - \\frac { 1 } { 3 } }{/tex}{tex}= \\frac { \\frac { 1 + 24 } { 36 } } { - \\frac { 1 } { 3 } }{/tex}{tex}= \\frac { 25 } { 36 } \\times \\frac { - 3 } { 1 }{/tex}{tex}= \\frac { - 25 } { 12 }{/tex} | |
| 1962. |
Find the rational numbers between underoot 2and underoot3 |
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| 1963. |
Find the value of p in equation px²-5x+p and both the roots are same |
| Answer» B²-4ac= 0, (-5)²-4*p*p= 0, 25-4p² = 0, p= 5/2 | |
| 1964. |
Cos45÷sec30+cos30 |
| Answer» 0 | |
| 1965. |
If the sum of the zeros of the quadratic polynomial kx2-3x+5is 1 then find k |
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Answer» K=-3 is the correct answer K = -2 |
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| 1966. |
SecA(1_sinA)(secA+tanA)=1 |
| Answer» LHS\xa0{tex}= \\sec A(1 - \\sin A)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\sec A \\times \\sin A)(\\sec A + \\tan A){/tex}{tex} = \\left( {\\sec A - \\frac{1}{{\\cos A}} \\times \\sin A} \\right)\\left( {\\sec A + \\tan A} \\right){/tex}\xa0{tex}\\left[ {\\because \\sec A = \\frac{1}{{\\cos A}}} \\right]{/tex}{tex} = \\left( {\\sec A - \\frac{{\\sin A}}{{\\cos A}}} \\right)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\tan A)(\\sec A + \\tan A){/tex}\xa0{tex}\\left[ {\\because \\frac{{\\sin A}}{{\\cos A}} = \\tan A} \\right]{/tex}Using identity (a - b)(a + b) = a2 - b2= sec2 - tan2A= 1\xa0{tex}\\left[ {\\because {{\\sec }^2} - {{\\tan }^2}A = 1} \\right]{/tex}= RHSHence proved | |
| 1967. |
If cos A=2/3 find the value of 4+4 tan A |
| Answer» Oh.. Prabhas aa gaye | |
| 1968. |
Find coordinates of y axis which is nearest to the point (-2,5) |
| Answer» It is (0,5) | |
| 1969. |
X2-48x+343 =0 find x ...p...t.....y |
| Answer» Ans is 24-√233 & 24+√233. | |
| 1970. |
The area of acircle inscribed in an equilateral triangle is 154cm2 find the perimeter of triangle |
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Answer» Aprrox 73 .. 72.7cm |
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| 1971. |
X²+y²=25What is the value of x+y |
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| 1972. |
X2+y2 = 25 then find x+y =?? Answer quickly .. |
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| 1973. |
5x=secA and 5/x=tanA find the value of 5(x*x-1/x*x) |
| Answer» 1/5 | |
| 1974. |
Exercise 14.3 |
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| 1975. |
Find the number of terms of AP 54,51,48..............so that their sum is 513 |
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Answer» a=54; d=-3; n=?; Sn=513 Sn=n÷2{2a+(n-1)d}513 =n÷2{2(54)+(n-1)(-3)}513×2=n{108-3n+3} 1026=n{111-3n}1026=111n-3n^23n^2-111n+1026=0Divide by 3n^2-37n+342=0n^2-19n-18n+342=0n(n-19)-18(n-19)=0(n-18)(n-19)=0therefore;n=18or19 Answer bhai sn wale formula par rakhoaajayega.pakka No. Of n term is 19.. |
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| 1976. |
Find the root of the equation 16x-10/x=27 |
| Answer» 16x-10/x = 2716 x2\xa0-10 = 27 x16x2\xa0-27x -10 = 0Here a= 16, b= -27 and c= -10x= -b+√ b2\xa0- 4ac/ 2a and x = -b-√b2-4ac/2ax = -(-27)+√(-27)2\xa0-4 *16*(-10)/ 2*16 and x= -(-27) - √(-27)2\xa0-4*16*-10/2*16x= 27 +√729+640/32 and x = 27-√729+640/32x = 27+37/32 and x = 27-37/32x = 64/32 and x = -10/32x = 2 and x = -5/16\xa0\xa0 | |
| 1977. |
Chapter 13 exercise 13.2 |
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| 1978. |
If the areas of 2 similar triangles are equal ,then prove that they are congruent |
| Answer» Prove one angle common and take another two angles equal. | |
| 1979. |
Prove that Sec power 4 theta - secsquare theta = tan power4 +tansquare theta |
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Answer» Sec4A - Sec2A = tan\xa04A + tan\xa02ASec4A - Sec2A = Sec2A( Sec2A-1) = (1+tan2A) tan2A = tan\xa02A+ tan\xa04A Sec⁴A- sec² = sec²( sec² - 1 ) = sec²*tan² = 1/cos²*sin²/cos² = sin²*sec⁴ LHS.. Now, tan⁴+ tan² = tan² (tan²+1) = tan²*sec² = sin²/cos²*1/cos² = sin²*sec⁴ = RHS. Hence LHS=RHS. |
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| 1980. |
Locate some important center of trade and artisanal production in the political map of india |
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| 1981. |
Explain why 7×11×13+13 is a composite number |
| Answer» Because it has more than two factors ... 13{7×11+1}+ 1 | |
| 1982. |
How to study without maths any guide |
| Answer» Without any body help maths is very difficult | |
| 1983. |
If cosec +cotò show that cosecò -cotò=1/q and hence find tge value os secòand sinò |
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| 1984. |
a3+b3=10a2+b2=7a+b=? |
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| 1985. |
Two book rs 500 if 5book rs is find |
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Answer» cost of 2 books = Rs500cost of 1 book = 500/2cost of 5 books = 500 x5/2cost of 5 books = 2500/2= 1250 1250 |
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| 1986. |
How much paper I get from only ncear questions as it is |
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| 1987. |
1+3-4/6+66-45 |
| Answer» 0 | |
| 1988. |
Find the value of p of a e |
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| 1989. |
Find the area (in square units) of the triangle whose vertices are (a, b+c),(a, b-c) &(-a, c) . |
| Answer» 0 | |
| 1990. |
Derivation of mode |
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| 1991. |
What is a number? |
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| 1992. |
Date sheet 10th class |
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| 1993. |
cos/1-tan + sin/1-cot=? |
| Answer» Cos+sin | |
| 1994. |
Detu+1234= |
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| 1995. |
246789×3355 |
| Answer» 827977095 | |
| 1996. |
What is the distance between two parallel tangent of circle of radius 14 cm |
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| 1997. |
Area of eq triangle |
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Answer» A = √3/4 × a2 √3/4 × (side)2 |
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| 1998. |
If m th term of AP is 1/n and n th term is 1/m.find sum of first term |
| Answer» Smn=1/2(1+MN) | |
| 1999. |
Evaluate cosecQ- sec²Q÷cosec²Q+sec²Q |
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| 2000. |
If tan |
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