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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.
| 7401. |
4-1+1+4+6 |
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Answer» 8 14 |
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| 7402. |
What is euclid geometry |
| Answer» Geometry dealing with lines | |
| 7403. |
56+65 |
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Answer» 121 121 |
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| 7404. |
tanA/1-cotA+cotA/1-tanA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 7405. |
If the numbers (x-2), (4x-1), (5x-2) are in Ap fine the vqlu3 of x |
| Answer» Let a=x-2 , b=4x-1, c=5x-2a+c/2 =b (x-2)+(5x-2)/2 =4x-16x-4=2(4x-1)6x-1=8x-28x-6x= -1+2 2x=1x=1/2 | |
| 7406. |
ashwani |
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Answer» Illiana S. M. |
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| 7407. |
trigomontry |
| Answer» Circle | |
| 7408. |
The difference between two numbers is 9 and they are in the ratio 4:8. Find the numbers. |
| Answer» | |
| 7409. |
The first three terms of an AP are respectively (3y-1),(3y+5) and (5y+1) ,find the value of y |
| Answer» Y=5 | |
| 7410. |
456= 10\\4 |
| Answer» | |
| 7411. |
100x2-81 write zeroes |
| Answer» | |
| 7412. |
(a+b)x(a-b)=? |
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Answer» asquare- bsquare (a2-b2) A square + b square |
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| 7413. |
If one zero of polynomial 2x3+x2-7x-6is2 find all zeroes |
| Answer» | |
| 7414. |
If 24 is 2/3 of 3/4 of a number then what will be 1/4 of that number |
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Answer» let the number = x24=2/3 of 3/4 x24=1/2 xx=48now 1/4 of 48=1/4x48=12 12 |
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| 7415. |
View of exam |
| Answer» | |
| 7416. |
Find the LCM of small prime number |
| Answer» | |
| 7417. |
Write central tendencies |
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| 7418. |
Sum of three numbers is 12 sum of their cubes is 288 find the numbers |
| Answer» Let the three numbers in A.P. be {tex}a - d, a, a + d{/tex}.{tex}3a = 12 {/tex}or, {tex}a = 4{/tex}.Also, (4 - d)3 + 43 + (4 + d)3\xa0= 288or, 64 - 48d + 12d2- d3 + 64 + 64 + 48d + 12d2 + d3 = 288or, 24d2 + 192 = 288or, 24d2 = 288 - 192or, 24d2 = 96or, d2 = 96/24or, d2 = 4d ={tex}\\pm{/tex}2The numbers are 2,4, 6, or 6,4,2. | |
| 7419. |
The value of (sin 30+cos30)-(sin60+cos60)is |
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Answer» ० If cos A = 4/5, then tan A = ? The value of tan 60°/cot 30° is equal to:2 points = Sin30 + sin (90-30) - sin60 - sin (90 - 60 )= sin 30 + sin 60 - sin 60 - sin 30 =0 |
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| 7420. |
2x*34 |
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Answer» Class x me रत्वतचकिरचततसलवसहकतकok hi 68x |
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| 7421. |
What is frustum ? Give its formula.. |
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Answer» FORMULA:TSA of a frustum of a cone = πl(r1 + r2) + πr12 + πr22Volume of the frustum of a cone = (1/3) πh (r12 + r22 + r1 r2) Cut a right circular cone with a plane parallel to the base of the cone, then the solid shape between the plane and the base of the cone is called the frustum of a cone. |
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| 7422. |
if √3sinθ find the value ofsinθ.tanθ./sinθ+cosθ |
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| 7423. |
Pls everyone tell me about ur daily routine |
| Answer» | |
| 7424. |
Find value of k 3x+4y+2=0,9x+12y+k=0 represents co-incident lines |
| Answer» | |
| 7425. |
what is formula for hemispheres |
| Answer» TSA =3πrsq. CSA=2πrsq. Volume=2/3πrsq. | |
| 7426. |
Can (x-7) be the remainder on division of a polynomial p(x) by (7x + 2)? Justify |
| Answer» No | |
| 7427. |
ABCD is a trapezium in which AB||DC and AB||EF,then prove that AE/AD=BF/BC.please give answer fast |
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Answer» Give proper answer By BPT |
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| 7428. |
x²+x–² |
| Answer» | |
| 7429. |
Find the value of a if the point, (3,5)and (7,1)are equidistant from the point (a, 0)? |
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| 7430. |
Find the centre of the circle passing through the points (6,-6);(3,-7); and (3,3) |
| Answer» Let\xa0A → (6, –6), B\xa0→ (3, –7) and C\xa0→ (3, 3).Let the centre of the circle be I(x, y)Then, IA = IB = IC [By definition of a circle]{tex}\\Rightarrow{/tex} IA2 = IB2 = IC2{tex}\\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2Taking first two, we get(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 6x + 2y = 14{tex}\\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]Taking last two, we get(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2{tex}\\Rightarrow{/tex} (y + 7)2 = (y - 3)2{tex}\\Rightarrow{/tex} (y + 7) = {tex}\\pm{/tex}(y-3)taking +e sign, we gety + 7 = y - 3{tex}\\Rightarrow{/tex} 7 = -3which is impossibleTaking -ve sign, we gety + 7 = -(y - 3){tex}\\Rightarrow{/tex} y + 7 = -y + 3{tex}\\Rightarrow{/tex} 2y = -4{tex}\\Rightarrow y = \\frac{{ - 4}}{2} = - 2{/tex}Putting y = -2 in equation (1), we get{tex}\\Rightarrow{/tex} 3x - 2 = 7{tex}\\Rightarrow{/tex} 3x = 9{tex}\\Rightarrow{/tex} x = 3Thus, I {tex}\\rightarrow{/tex} (3, -2)Hence, the centre of the circle is (3, -2). | |
| 7431. |
How many 0zero in million |
| Answer» 6 | |
| 7432. |
Find mean of following data: 10,15,20,16,12,17,30,10,11,13 |
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| 7433. |
Artaamatic progression |
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| 7434. |
What is trigonometry actually means |
| Answer» Tri- three; gono-angle; metry- measurement, trigonometry simply means three angle measurement. | |
| 7435. |
Show that only one of the numbers n,n+2,n+4 is divisible by 3 |
| Answer» Let the number be (3q + r){tex}n = 3 q + r \\quad 0 \\leq r < 3{/tex}{tex}\\text { or } 3 q , 3 q + 1,3 q + 2{/tex}{tex}\\text { If } n = 3 q \\text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}{tex}3 q \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 1 \\text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}{tex}( 3 q + 3 ) \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 2 \\text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}{tex}( 3 q + 6 ) \\text { is divisible by } 3{/tex}.{tex}\\therefore \\text { out of } n , ( n + 2 ) \\text { and } ( n + 4 ) \\text { only one is divisible by } 3{/tex}. | |
| 7436. |
Find the next term of AP : root 7, root 28 ,root 63 ,.......... |
| Answer» Here,{tex}a = \\sqrt { 7 } , a + d = \\sqrt { 28 }{/tex}{tex}\\therefore \\quad d = \\sqrt { 28 } - \\sqrt { 7 } {/tex}{tex}= 2 \\sqrt { 7 } - \\sqrt { 7 }{/tex}{tex}= \\sqrt { 7 }{/tex}or, Next term =\xa0{tex}\\sqrt { 63 } + \\sqrt { 7 }{/tex}or, = {tex}\\sqrt { 9\\times7 } + \\sqrt { 7 }{/tex}or,\xa0{tex}= 3 \\sqrt { 7 } + \\sqrt { 7 } {/tex}or, {tex}= 4 \\sqrt { 7 }{/tex}or,\xa0{tex}= \\sqrt { 7 \\times 16 }{/tex}{tex}= \\sqrt { 112 }{/tex}So, next term is {tex}\\sqrt { 112 }{/tex}. | |
| 7437. |
How to understand the concepts of Algebra? |
| Answer» | |
| 7438. |
speed questions |
| Answer» | |
| 7439. |
In ? PQR Q=90degree,qs perpendicular to or prove that pq square/pr square=rs/ps? |
| Answer» | |
| 7440. |
Find the greatest number of 5 digit that is 24 15 and 36 |
| Answer» The greater number of 6 digits is 999999.LCM of 24, 15, and 36 is 360.{tex}999999 = 360 \\times 2777 + 279{/tex}Required number is = 999999 - 279 = 999720\xa0 | |
| 7441. |
98------------2×2•7×7Is it is terminating ir non terminating? |
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Answer» Yes of course 2.592592592 is terminating |
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| 7442. |
Find the middle term of the AP 6, 13,20,.....216. |
| Answer» Here,a=6 & d= 7 by nth term=a+(n-1)d we know that 216 is 31th term therefore mid term will be 16 now we can easily calculate 16th term. 16th term=6+(16-1)716 term=111 | |
| 7443. |
express trigonometric ratio sinA in term of cotA |
| Answer» For sin A,By using identity {tex}cosec ^ { 2 } A - \\cot ^ { 2 } A = 1 \\Rightarrow \\cos e c ^ { 2 } A = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\frac { 1 } { \\sin ^ { 2 } A } = 1 + \\cot ^ { 2 } A{/tex}{tex}\\Rightarrow \\sin A = \\frac { 1 } { \\sqrt { 1 + \\cot ^ { 2 } A } }{/tex} | |
| 7444. |
Sec theta minus one upon sec theta + 1 is equals to sin theta upon 1 + cos theta whole square |
| Answer» | |
| 7445. |
The sum of two rationals is always rational |
| Answer» | |
| 7446. |
If tan2 A is equal to 3 find A. Plz... |
| Answer» | |
| 7447. |
Xsquare(root2+1)x+rootSolve by completing the square method |
| Answer» | |
| 7448. |
Find the value of y for which the distance between the point p(2,-3) and q(10,y) is 10 units |
| Answer» y=-9,3 | |
| 7449. |
5 multiple 8 |
| Answer» | |
| 7450. |
Mensurationof |
| Answer» | |