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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.
| 7901. |
Draw a line segment Ab=4.8cm and find a point o on ab such that Ap =1/4ab. |
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| 7902. |
X-2y=5 and 2x- 3y = -4 |
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| 7903. |
x/a-y/b=a-b and ax+by=a cube +b cube |
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| 7904. |
Where I come to know about blueprint for 10 board exam |
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| 7905. |
Sec20° tan45°? |
| Answer» .342 | |
| 7906. |
If circumference of circle and area of circle is equal then value of radius |
| Answer» R=2 unitd | |
| 7907. |
Prove that line from to center to the point of touching tangent perpendicular to tangent |
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| 7908. |
Find the zeroes of p(x) = 4x²+5x√2-3 |
| Answer» I got its solution4x²+6x√2-x√2-3 =(2x√2-1)(x√2+3) | |
| 7909. |
Exercise 2a question no. 4 |
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| 7910. |
Is sample papers are available |
| Answer» Yess | |
| 7911. |
15+15 |
| Answer» 30 | |
| 7912. |
Composite number |
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| 7913. |
Examples for composite number |
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| 7914. |
If mean, median and mode of a grouped data is x, x², 1 respectively.find thevalue of mean and median |
| Answer» Use empirical formula. | |
| 7915. |
Find the value of k to 2k+1 3k+3 5k-1 |
| Answer» If {tex}2k + 1, 3k + 3, 5k -1{/tex} are in A.P.then {tex} (5k - 1) - (3k+ 3)= (3k + 3) - (2k + 1){/tex}or, {tex}5k- 1 - 3k- 3 = 3k + 3 - 2k -1{/tex}or, {tex}2k - 4 = k + 2{/tex}or, {tex}2k - k = 4 + 2{/tex}or {tex}k = 6{/tex} | |
| 7916. |
(sec square thita -1) ( 1 - cosec square thita) = -1 |
| Answer» (Sec square thita- 1)(1-cosec square thita) =-1LHS:-(sec square thita- 1)(1- cosec square thita ) = (tan square thita) (-cot Square thita) = -1 proved | |
| 7917. |
theorem of centroid |
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| 7918. |
What\'s thyroxin? |
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Answer» The hormone secreted by thyroid gland which is present below the neck It is a hormone which is released by thyroid gland. |
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| 7919. |
In a ABCD square which side is 10.5 cm . |
| Answer» All side are equal | |
| 7920. |
CosA=2/5 prove 4-4(tanA)w |
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| 7921. |
12,16,20........,248 |
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Answer» A=12,d=16-12=4,an=248n=?An=a+(n-1)4248=12(n-1)4248-12=(n-1)4236=(n-1)4236÷4=(n-1)59+1=n60=n 9000 |
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| 7922. |
What is area of cylinder |
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| 7923. |
If the points (k,2k), (3k,3k) and (3,1) are collinear then k is equal to |
| Answer» -1/3 | |
| 7924. |
Has any friend buyed sample papers from this app..If someone has bought plse reply me? |
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| 7925. |
Show sin tita-cos tita+1÷sin tita + cos tita -1=1÷sec tita -tan tita |
| Answer» sintita-cos Tita+1/sintita+costita-1=1/sectita-tantita | |
| 7926. |
Determine if the points (1,5)(3,3)and(-2,-11) collinear |
| Answer» Let A = (1, 5), B = (2, 3) and C = (-2, -11){tex}AB = \\sqrt {{{(2 - 1)}^2} + {{(3 - 5)}^2}} = \\sqrt 5 {/tex}{tex} BC = \\sqrt {{{( - 11 - 3)}^2} + {{( - 2 - 2)}^2}} = \\sqrt {212} {/tex}{tex}AC = \\sqrt {{{( - 2 - 1)}^2} + {{( - 11 - 5)}^2}} = \\sqrt {265} {/tex}AB + BC\xa0{tex} \\ne {/tex}\xa0ACHence, A, B and C are not collinear | |
| 7927. |
find the sum of first 15 terms of an A.P. if an =3+4n |
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| 7928. |
Find sum of all two digit whole numbers divisible by 2 |
| Answer» Two digit no. divisible by 2=10,12,14......98a=10,d=2,l=98l=a+(n-1)d98=10+(n-1)298=10+2n-298=8+2n98-8=2n2n=90n=45Sn=n/2{2a+(n-1)d}Sn=45/2{2×10+(45-1)2}Sn=45/2{20+44×2}Sn=45/2{20+88}Sn=45/2×108Sn=45×54Sn=2436 | |
| 7929. |
SinA-cosA+1/sinA+cosA-1 |
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| 7930. |
Evaluate sin 18÷cos 72 |
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Answer» 1 Sin 4 |
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| 7931. |
Prove that. cosA-sinA+1/cosA+sinA-1 =cosecA +cotA |
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| 7932. |
Solve 2x+8=14 |
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Answer» 2x=14-82x=6x=3???? 2x+8=142x=14-82x=6x=6/2X=3 This is very easy question. Answer is 3 3 |
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| 7933. |
If cos a is reverse of sin a then why tan a is not reverse of sec a |
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| 7934. |
What is angles in alternate segment?Explain using theoram. |
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| 7935. |
What is electric susceptibility and polarisation density explanation |
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| 7936. |
What do you mean by similar triangle? |
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Answer» And their side ratio equal If two triangles are similar then there corresponding angles are equal nd sides are proportional. |
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| 7937. |
6+8 |
| Answer» 14 | |
| 7938. |
Sum to f a number and its reciprocal is 7/36.find number |
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| 7939. |
Solve this equation by middle term splitting 1)x²+14x+24=0 |
| Answer» x=2,x==-12 | |
| 7940. |
Solve the following system of linear equationA)4x-5y-20B)3x+5y-15=0 |
| Answer» On a graph paper, draw a horizontal line X\'OX and a vertical line YOY\' as the x-axis and the y-axis respectively.Graph of {tex}4x - 5y - 20 = 0{/tex}{tex}4x - 5y - 20 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (4x - 20){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 4 x - 20 ) } { 5 }{/tex}.....(i)Table for {tex}4x - 5y - 20 = 0.{/tex}\tx025y-4-2.40\tNow, plot the points {tex}A(0, -4),\\ B(2, -2.4)\\ and\\ C(5, 0){/tex} on the graph paper.Join AB and BC to get the graph line ABC. Extend it on both ways.Thus, the line ABC is the graph of {tex}4x - 5y - 20 = 0.{/tex}{tex}3x\xa0+ 5y -15 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (15 -3x){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 15 - 3 x ) } { 5 }{/tex}.........(ii)Table for {tex}3x\xa0+ 5y -15 = 0.{/tex}\tx-505y630\tOn the same graph paper as above, plot the points P (-5, 6) and Q(0, 3).The third point C(5, 0) has already been plotted.Join PQ and QC to get the graph line PQC. Extend it on both ways.Thus, the line PQC is the graph of 3x + 5y -15 = 0.The two graph lines intersect at the point C(5,0).{tex}\\therefore{/tex}\xa0x = 5, y = 0 is the solution of the given system of equations.Clearly, the given equations are represented by the graph lines ABC and PQC respectively.The vertices of {tex}\\triangle{/tex}AQC formed by these lines and the y-axis are A(0, -4), Q(0,3) and C(5,0). | |
| 7941. |
Solve the following system of equations1.4x-5y-20=02.3x-5y-15=0 |
| Answer» On a graph paper, draw a horizontal line X\'OX and a vertical line YOY\' as the x-axis and the y-axis respectively.Graph of {tex}4x - 5y - 20 = 0{/tex}{tex}4x - 5y - 20 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (4x - 20){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 4 x - 20 ) } { 5 }{/tex}.....(i)Table for {tex}4x - 5y - 20 = 0.{/tex}\tx025y-4-2.40\tNow, plot the points {tex}A(0, -4),\\ B(2, -2.4)\\ and\\ C(5, 0){/tex} on the graph paper.Join AB and BC to get the graph line ABC. Extend it on both ways.Thus, the line ABC is the graph of {tex}4x - 5y - 20 = 0.{/tex}{tex}3x\xa0+ 5y -15 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (15 -3x){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 15 - 3 x ) } { 5 }{/tex}.........(ii)Table for {tex}3x\xa0+ 5y -15 = 0.{/tex}\tx-505y630\tOn the same graph paper as above, plot the points P (-5, 6) and Q(0, 3).The third point C(5, 0) has already been plotted.Join PQ and QC to get the graph line PQC. Extend it on both ways.Thus, the line PQC is the graph of 3x + 5y -15 = 0.The two graph lines intersect at the point C(5,0).{tex}\\therefore{/tex}\xa0x = 5, y = 0 is the solution of the given system of equations.Clearly, the given equations are represented by the graph lines ABC and PQC respectively.The vertices of {tex}\\triangle{/tex}AQC formed by these lines and the y-axis are A(0, -4), Q(0,3) and C(5,0). | |
| 7942. |
square root of 15 |
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Answer» 3.87 Hiiiiiiiiiiiiiiiiiiiiiiiiiiii anjali 225 |
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| 7943. |
What is the value of log |
| Answer» 1/conc of H+ions | |
| 7944. |
X3_4x+5x_2; 2,1,1 |
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| 7945. |
12(12+21) |
| Answer» 396 | |
| 7946. |
44+55 |
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Answer» 99 99 |
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| 7947. |
What is lenght of arc? |
| Answer» Length of Arc of circle =(angle/360°)*2*(22/7)radius of circle | |
| 7948. |
Relation between a mean,median and mode? |
| Answer» I think you talk about empirical relation i.e. mean = 3median -2mode | |
| 7949. |
Alfa,bita n gama r zeroes of polynomial x³+px²+qx+2Such that ab+1=0find 2p+q+5. |
| Answer» P(x) =\xa0x3\xa0+ px2\xa0+ qx + 2Here, a = 1, b = p, c = q, d = 2Now,\xa0{tex}\\alpha + \\beta + \\gamma = \\frac { - b } { a }{/tex}= -p.......(i){tex}\\alpha \\beta + \\beta \\gamma + \\alpha \\gamma = \\frac { c } { a }{/tex}= q{tex}\\Rightarrow \\alpha \\beta + \\gamma ( \\beta + \\alpha ){/tex}\xa0= q..............(ii)and\xa0{tex}\\alpha \\cdot \\beta \\cdot \\gamma = \\frac { - d } { a }{/tex}= -2{tex}\\Rightarrow \\quad \\alpha . \\beta \\cdot \\gamma{/tex}= -2.......................(iii)Also,\xa0{tex}\\alpha \\beta + 1 = 0 \\Rightarrow \\alpha \\beta = - 1{/tex}Therefore, (iii) becomes -1{tex}\\times \\gamma = - 2 \\Rightarrow \\gamma = 2{/tex}Substituting in (i), we get{tex}\\alpha + \\beta + 2 = - p \\Rightarrow \\alpha + \\beta{/tex}\xa0=- p - 2Substituting these value in (ii), we get-1 + 2(-p - 2) = q{tex}\\Rightarrow{/tex}\xa0-1 - 2p - 4 = q{tex}\\Rightarrow{/tex}\xa02p + q + 5 = 0 | |
| 7950. |
The roots of the e |
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