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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.
| 11751. |
AD/DB=AE/EC |
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| 11752. |
The cost of 5chair and 6cost of table is 245and 567 find the cost of one chairs and table |
| Answer» Hiii, let the cost of one chair be x and let the cost of one table be y then according to question 5x=245 and so on | |
| 11753. |
4x÷4 |
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Answer» X X=1 X |
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| 11754. |
What is Airthmatic progression? |
| Answer» An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common difference of 2. | |
| 11755. |
Given a=5, d=3, an=50, find n and SN |
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Answer» Given a=5,d=3an=50We knowa\u200b\u200b\u200b\u200b\u200b\u200bn\xa0= a+(n-1)d=> 50 = 5+(n-1)3=> 45 = (n-1)3=> 15 = n-1=> n = 16{tex}=> S_ {16}= {16\\over 2}[2\\times 5+(16-1)3]\\\\=> S_{16}=8[10+45]\\\\=> S_{16}=440{/tex} a = 5 d = 3 an = 50 n = ? Sn = ?an = a + (n -1)d50 = 5 + (n - 1)345/3 = n - 115 + 1 = nn = 16Sn = n/2(2a + (n - 1)d S16= 16/2(10 + 15x3) = 8 x 55 = 440\xa0 N=10 and SN=0.1 |
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| 11756. |
If the difference between 10 term of two A.P. is 10 the find the difference between their 20 terms |
| Answer» Let first term and answer difference of 1st ap = a and d ,And first term and difference of 2nd ap =A and D , Now , a10 - A10 = 10 a + 9d -(A + 9 D ) =10 a+9d-A-9D =10 a-A= 10.......(1) Now, a20-A20 a+19d-(A+19D) a+19d-A-19D a-A And we know , a-A=10 ( find in equation (1)) So, a - A = 10 is answer 10 is the difference between their 20th term | |
| 11757. |
What is the role of acid in our body. |
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Answer» It help in the process of digestion The acid creates an acidic medium in the stomach. This acid helps to digestvthe food. It neutralises excess base and helps in digestion of food particles |
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| 11758. |
2x +18=12 |
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Answer» 2x + 18 = 122x = 12 - 182x = - 6x = - 6/2x = - 3 Sorry, x=-3 X=-3 X=-3 X=-6 |
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| 11759. |
Completing the square method |
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| 11760. |
find the value of sin^2 5° + sin^2 10° + sin^2 15° + sin^2 20°+...+sin^2 90° |
| Answer» LHS = {tex}sin^2 5° + sin^2 10° + ....+ sin^2 85° + sin^2 90°{/tex}= (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + ....+ (sin2 40° + sin2 50°) + (sin2 45° + sin2 90°)= {sin2 5° + cos2 (90°-85°)} + {sin210° + cos2 (90° - 80°)} + ....+ {sin240° + cos2(90° - 50°) +\xa0{tex}\\left\\{ \\left( \\frac { 1 } { \\sqrt { 2 } } \\right) ^ { 2 } + ( 1 ) ^ { 2 } \\right\\}{/tex}= {sin2 5° + cos2 5°} + {sin210° + cos2 10°} + ....+ {sin240° + cos240°} +\xa0{tex}\\left\\{ \\left( \\frac { 1 } { \\sqrt { 2 } } \\right) ^ { 2 } + ( 1 ) ^ { 2 } \\right\\}{/tex}= (1) + (1) + 1... 8 times +\xa0{tex}\\frac { 1 } { 2 }{/tex}\xa0+ 1= 8 +\xa0{tex}\\frac { 1 } { 2 }{/tex}\xa0+ 1 =\xa0{tex}9 \\frac { 1 } { 2 }{/tex}\xa0RHS | |
| 11761. |
(1+tan^2A)/(1-tan^2A)=((1-tanA)/(1+tanA))^2=tan^2A |
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| 11762. |
Prove that 1+cosa/Sina=Sina/1-cosa |
| Answer» {tex}\\frac{{1+cosA}}{sinA}{/tex}={tex}\\frac{{(1+cosA)(1-cosA)}}{sinA(1-cosA)}{/tex}={tex}\\frac{{1-cos²A}}{sinA(1-cosA)}{/tex}={tex}\\frac{{sin²A}}{sinA(1-cosA)}{/tex}={tex}\\frac{{sinA}}{(1-cosA)}{/tex} | |
| 11763. |
Use Euclids division algorithm to find the hcf of 240 and 272 |
| Answer» 272=240×1+32240=32×7+1632=16×2+0 therefore HCF=16 | |
| 11764. |
If 5costheta+12sintheta=13,find the value of tantheta |
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Answer» 17 17 |
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| 11765. |
Ex 6.3 3question |
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| 11766. |
Find the greatest number of 6 digits exactly divisible by 24,15 and 36. |
| Answer» 24= 2×2×2×315= 3×536=2×2×3×3LCM is= 3×2×2×2×3×5 =360Now you have find the multiple of 360 which is belongs to highest 6 digit number also.Now, 999999÷360=2777.775That means, 360×2777=999720\xa0 | |
| 11767. |
Probability |
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| 11768. |
Ax/b -by/a =a+bShort trick to solve this |
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| 11769. |
(A×b)(b×a) |
| Answer» (A×B)^2 | |
| 11770. |
What is the sa1 syllabus |
| Answer» N.c.r.t lesson 1,2,3,4,5,6,8,14,15 | |
| 11771. |
Linear equations |
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| 11772. |
25÷5 |
| Answer» 5 | |
| 11773. |
Chapter 7 NCERT Example no. 1 |
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| 11774. |
2x=54-78y |
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| 11775. |
How to solve trigonometry in easy ways? |
| Answer» You can solve trigonometry in easy way by learn formulas | |
| 11776. |
Solve the following system of linear equation 4/x + 5y =7 and 3/x + 4y= 5 |
| Answer» {tex}4/x + 5y =7 and 3/x + 4y= 5{/tex}\xa0Let, 1/x=m and y=n, then\xa04m+5n=7 ------(i) ×33m+4n=5 -------(ii) ×4Now, solving for \'n\',we write,12m+15n=2112m+16n=20Now, substracting,we get,n=-1i.e. y=-1Putting the value of y into (i), we get,4m+5.(-1)=7or, 4m=7+5=12or, m=3or, x=\xa0{tex}\\frac{{1}}{3}{/tex}So, the solution is,x=\xa0{tex}\\frac{{1}}{3}{/tex}and y=-1 | |
| 11777. |
What is the proof of \'ceva theoram\' |
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| 11778. |
Proof Pythagoras theorm. |
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| 11779. |
Quadratic eqation solve for x 2 |
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| 11780. |
Sa1syllabus |
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| 11781. |
Find the roots of quadratic equation (3x+2)(3x+4)=12. Find solutions for this question. |
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| 11782. |
Tignometry explain |
| Answer» Trignometry is the study of the relationship between sides & angles of a triangle. Tri means three , gon means sides & metron means measures | |
| 11783. |
Cos48_sin42 |
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| 11784. |
Trignometry explain |
| Answer» trigonometric branch of mathematics which deals with the measurements of sides and angles of a triangle | |
| 11785. |
Can you write the criteria of the similarity |
| Answer» Aaa/Sas/aa/Sss | |
| 11786. |
How make trigonometry identies |
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| 11787. |
M=4 ,N=6 m^+n^= |
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| 11788. |
The third term of an AP is 3 and 11th term is - 21. Find its first term and common difference. |
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| 11789. |
How find zero of cubiic polynomial |
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| 11790. |
Tan /1-cot+cot/ 1-tan=tan+cot |
| Answer» {tex}\\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta } = 1 + \\tan \\theta + \\cot \\theta{/tex}{tex}\\text { L.H.S. } = \\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta }{/tex}{tex}= \\frac { \\frac { \\sin \\theta } { \\cos \\theta } } { 1 - \\frac { \\cos \\theta } { \\sin \\theta } } + \\frac { \\frac { \\cos \\theta } { \\sin \\theta } } { 1 - \\frac { \\sin \\theta } { \\cos \\theta } }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } - \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 3 } \\theta - \\cos ^ { 3 } \\theta } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { ( \\sin \\theta - \\cos \\theta ) \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta + \\sin \\theta \\cos \\theta \\right) } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}\\left[ {\\because {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}} \\right){/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\sin \\theta \\cos \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\tan \\theta + \\cot \\theta + 1 = 1 + \\tan \\theta + \\cot \\theta = R . H S \\text { proved }{/tex}Since,\xa0{tex}\\tan A = \\frac{{\\sin A}}{{\\cos A}}{/tex}{tex}\\cot A = \\frac{{\\cos A}}{{\\sin A}}{/tex} | |
| 11791. |
159 percent of ............=15900 |
| Answer» {tex}{159\\times x\\over 100} = 15900{/tex}=> x = 10000 | |
| 11792. |
Tan/1-cot+cot/1-tan=tan+cot |
| Answer» {tex}\\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta } = 1 + \\tan \\theta + \\cot \\theta{/tex}{tex}\\text { L.H.S. } = \\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta }{/tex}{tex}= \\frac { \\frac { \\sin \\theta } { \\cos \\theta } } { 1 - \\frac { \\cos \\theta } { \\sin \\theta } } + \\frac { \\frac { \\cos \\theta } { \\sin \\theta } } { 1 - \\frac { \\sin \\theta } { \\cos \\theta } }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } - \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 3 } \\theta - \\cos ^ { 3 } \\theta } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { ( \\sin \\theta - \\cos \\theta ) \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta + \\sin \\theta \\cos \\theta \\right) } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}\\left[ {\\because {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}} \\right){/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\sin \\theta \\cos \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\tan \\theta + \\cot \\theta + 1 = 1 + \\tan \\theta + \\cot \\theta = R . H S \\text { proved }{/tex}Since,\xa0{tex}\\tan A = \\frac{{\\sin A}}{{\\cos A}}{/tex}{tex}\\cot A = \\frac{{\\cos A}}{{\\sin A}}{/tex} | |
| 11793. |
Exam paper for sa12017 |
| Answer» Check last year papers here : https://mycbseguide.com/cbse-question-papers.html | |
| 11794. |
What is alpha |
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| 11795. |
√x+y=11X+√y=11 |
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| 11796. |
Prove that sec^2 A+cos^2 A never less then 0. |
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| 11797. |
25x-30x-1=0 quadratic equation |
| Answer» Give,\xa025x2\xa0- 30x + 11 = 0 ...(i)On comparing Eq. (i) with ax2\xa0+ bx + c = 0, we geta = 25, b = - 30 and c = 11.{tex}\\because{/tex}\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac { - b + \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}and\xa0{tex}\\beta = \\frac { - b - \\sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { ( - 30 ) ^ { 2 } - 4 \\times 25 \\times 11 } } { 2 \\times 25 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { 900 - 1100 } } { 50 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha = \\frac { 30 + \\sqrt { - 200 } } { 50 } \\Rightarrow \\alpha = \\frac { 30 + 10 i \\sqrt { 2 } } { 50 }{/tex}and\xa0{tex}\\beta = \\frac { 30 - \\sqrt { ( - 30 ) ^ { 2 } - 4 \\times 25 \\times 11 } } { 2 \\times 25 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\beta = \\frac { 30 - \\sqrt { 900 - 1100 } } { 50 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\beta = \\frac { 30 - \\sqrt { - 200 } } { 50 } \\Rightarrow \\beta = \\frac { 30 - 10 i \\sqrt { 2 } } { 50 }{/tex}{tex}\\therefore{/tex}\xa0{tex}\\alpha = \\frac { 3 } { 5 } + \\frac { \\sqrt { 2 } } { 5 } i , \\beta = \\frac { 3 } { 5 } - \\frac { \\sqrt { 2 } } { 5 } i{/tex}Also, b2\xa0- 4ac = (- 30)2\xa0- 4\xa0{tex}\\times{/tex}\xa025\xa0{tex}\\times{/tex}\xa011 = 900 - 1100= - 200 < 0Hence, the roots are complex conjugate. | |
| 11798. |
How to prove B.P.T |
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| 11799. |
Find value of x+y if 3x-2y=5and -2x+3y=3. |
| Answer» x+y=4.2+3.8=8 the answer is 8 | |
| 11800. |
Deraviation of area of segment |
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