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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.
| 24801. |
A biggest sphere is made by a cube of 14 cm find the volume |
| Answer» The base of cone is the largest circle that can be inscribed in the face cube and the height will be equal to edge of the cube.Radius of cone =\xa0{tex}\\frac{14}{2}{/tex}\xa0= 7 cmHeight of cone = 14 cm{tex}\\therefore{/tex}\xa0Volume of cone =\xa0{tex}\\frac{1}{3}{/tex}{tex}\\pi{/tex}r2h=\xa0{tex}\\frac{1}{3}{/tex}\xa0{tex}\\times{/tex}\xa0{tex}\\frac{22}{7}{/tex}\xa0{tex}\\times{/tex}\xa07\xa0{tex}\\times{/tex}\xa07\xa0{tex}\\times{/tex}\xa014=\xa0{tex}\\frac{2156}{3}{/tex}\xa0{tex}\\approx{/tex} 718.67. | |
| 24802. |
If x=√2 is a solution of kx+√2-4=0 |
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Answer» 2√2-1/2 Bhai ye ho sakta Hai kay2✓2-1 |
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| 24803. |
If p and q are the roots of equation x^2+px-q=0 then find the value of p and q. |
| Answer» {tex}x^2\xa0+\xa0px -\xa0q = 0{/tex}and roots are given as p and qsum of the roots\xa0= {tex}\\frac{ - coefficient \\: of \\: x}{coefficient \\: of \\: {x}^{2} }{/tex}So,{tex}\\Rightarrow{/tex}\xa0{tex}p + q ={/tex}\xa0{tex}\\frac{-p}{1}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}p + q = -p{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}q = -p -p{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}q = -2p{/tex} ..(1)Now, product\xa0of roots\xa0= {tex}\\frac{constant \\: term \\: }{coefficient\\ of\\: {x}^{2} } {/tex}So,pq = {tex}\\frac{-q}{1}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0pq = -q\xa0{tex}\\Rightarrow{/tex}\xa0p =\xa0{tex}\\frac{-q}{q}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0p = -1 ..(2)Put the value of (2) in (1), we get{tex}\\Rightarrow{/tex}\xa0{tex}q = -2(-1) = 2{/tex}. | |
| 24804. |
Why cosec A =1/sin A |
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Answer» Because 1/sinA is cosecA. It is considered. Because the product is 1 See in trigonometric table |
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| 24805. |
What is section formula? |
| Answer» (m1(x2)+m2(x1)/m1 + m2 , m1(y2)+m2(y1)/m1+m2) | |
| 24806. |
(sina+cosa)^2=1+2sina×cosa |
| Answer» By using identity (a+b)^2= a^2 + b^2 + 2ab Therefore,LHS, = (SinA +CosA)^2= Sin^2A + Cos^2A + 2SinACosA= 1 + 2SinACosA [ Sin^2 + Cos^2A =1]PROVED | |
| 24807. |
How to prepare trigonometric identities |
| Answer» | |
| 24808. |
The nth term of Ap is 6n+2.find it\'s common difference |
| Answer» an = 6n + 2{tex}\\Rightarrow{/tex}\xa0a1\xa0= 6\xa0{tex}\\times{/tex} 1 + 2 = 8a2 = 6\xa0{tex}\\times{/tex}\xa02\xa0+ 2 = 14{tex}\\therefore{/tex}\xa0common difference = a2\xa0- a1\xa0= 14 - 8 = 6 | |
| 24809. |
what is arithmetic progression mean |
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| 24810. |
What is the role of appendix in our body ? How does the disease occurs from this |
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Answer» When the food particles enters in it, when we eat or drink in stand position then it cause intense problem which could only be cured after operation. Nothing |
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| 24811. |
Which is know as stem of brain? |
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| 24812. |
How did the scientists find the distance b/w sun to planets & one planet to other? |
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Answer» By trigonometry I think they use trigonometry |
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| 24813. |
Find 31 st term of an AP whose 11 th term is 38 and 16 th term is 73 |
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Answer» {tex}\\eqalign{ & {t_{31}} = a + 30d \\cr & = - 32 + 30(7) \\cr & = 178 \\cr} {/tex} {tex}\\eqalign{ & Given\\,\\,that\\,:{t_{11}} = 38 \\cr & {t_{16}} = 73 \\cr & Let\\,first\\,term\\,be\\,a\\,and\\,common\\,difference\\,is\\,d \\cr & a + 10d = 38\\,\\,\\,\\,\\,\\,eq.\\left( 1 \\right) \\cr & a + 15d = 73\\,\\,\\,\\,\\,\\,\\,eq.(2) \\cr & eqs.(1)\\, - (2) \\cr & - 5d = - 35 \\cr & d = 7 \\cr & so\\,from\\,(1) \\cr & a = - 32 \\cr & {t_{38}} = a + 37d \\cr & = - 32 + 37(7) \\cr & = 227 \\cr & \\cr} {/tex} 178 |
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| 24814. |
When class 10 board exam will start?When our datesheet will be released?? |
| Answer» Check date sheets here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 24815. |
Resolve into factors: 6x^8+2x^4+4x6 |
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| 24816. |
Find the value A . If sec 4A =cose A _18 |
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| 24817. |
How can we find tha zero of PX |
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| 24818. |
What number is less than 20 by c? |
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| 24819. |
One upon 10 + cot theta into one upon 1 + sin a + 1 upon 1 minus sin a = 26 query |
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| 24820. |
In an Ap of 50 terms the sum of first 10 terms is 210 and sum of its 15 terms is 2565.find the ap |
| Answer» Let a be the first term and d be the common difference of the given AP.Sum of the first n terms is given bySn = n/2 {2a + (n - 1)d}Putting n = 10, we getS₁₀ = 10/2 {2a + (10 - 1)d}210 = 5 (2a + 9d) 2a + 9d = 210/52a + 9d = 42 ...............(1)Sum of the last 15 terms is 2565⇒ Sum of the first 50 terms - sum of the first 35 terms = 2565S₅₀ - S₃₅ = 2565⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 256525 (2a + 49d) - 35/2 (2a + 34d) = 2565⇒ 5 (2a + 49d) - 7/2 (2a + 34d) = 513⇒ 10a + 245d - 7a + 119d = 513⇒ 3a + 126d = 513 ⇒ a + 42d = 171 ........(2)Multiply the equation (2) with 2, we get2a + 84d = 342 .........(3)Subtracting (1) from (3) 2a + 84d = 342 2a + 9d = 42- - -_______________ 75d = 300_______________ d= 4Now, substituting the value of d in equation (1)2a + 9d = 422a + 9*4 = 422a = 42 - 362a = 6a = 3So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 | |
| 24821. |
Find the non |
| Answer» | |
| 24822. |
What is common difference of ap 2 and 4 |
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Answer» a2 -- a1 4 -- 2 = 2 4 - 2 = 2 Common difference is 2 Common deference is2 The difference is 2. |
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| 24823. |
Sec |
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| 24824. |
By quadratic formula find x,36x²+12ax+(a²-b²) |
| Answer» X=-b+-(b2-4ac)0.5/2a-36+-(144a2-144a2+144b2)/72(-36+144b2)/72-1+4b2/2X=(-1+4b2)/2Or x=-36-(144b2)/2 ( - 1-4b2)/2 | |
| 24825. |
What is the maths |
| Answer» Math is a subject which improve our mental capacity to calculate anything | |
| 24826. |
which is the best mathematics book? |
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Answer» Depend on us how we solve it Together with maths Ncert a RD sharma and NCERT NCERT AND RATNA sagar Cordova-Mathematics In Everyday Life\xa0 Rd sharma & Rs aggrawal NCERT |
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| 24827. |
If the product of zeroes of the polynomial ax2-6x-6 is 4, find the value of a. (2 means square) |
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Answer» C/A=4-6/a=4a=-6/4=-3/2 30/16 a=3/2 |
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| 24828. |
CosA-SinA=sin to show CosA+sinA=2sinA |
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| 24829. |
Why maths is so tough |
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Answer» Maths is not tough your thinking makes it tough .if you want make your mathematics strong then practise is very necessary We take other subjects too lighter. solving maths is a passion.It can enhance our thinking capability. Initially it takes time but once you have started to spend few hours daily in maths , it will become easier automatically. It is due to the presence of nibromium in our mind which makes maths difficult to increase maths power see boring things and boring movies also plays games which you dont like...☺☺☺☺☺☺☺☺☺ It\'s very easy |
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| 24830. |
If the common difference of an AP is 5,then what is a18 _ a13 |
| Answer» The answer is 25. | |
| 24831. |
Find the area of circle whose radius is5cm |
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Answer» 550/7 Use formula. 22/7*rsquare 78.5cm2 The answer is 78.5cm2. 7.86 |
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| 24832. |
What is the standard form of AP |
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Answer» A,a+d,a+2d,a+3d a+d,a+2d,a+3d... |
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| 24833. |
In p,q,r are in A.P then find the value of (p+q-r )(q+r-p) |
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| 24834. |
If 3x=cosec thita and 3/x=cot thita, find the value of 3(x2-1/x2) |
| Answer» We know that{tex}\\cos e c ^ { 2 } \\theta - \\cot ^ { 2 } \\theta = 1{/tex}{tex}\\Rightarrow ( 3 x ) ^ { 2 } - \\left( \\frac { 3 } { x } \\right) ^ { 2 } = 1{/tex}{tex}\\Rightarrow 9 x ^ { 2 } - \\frac { 9 } { x ^ { 2 } } = 1{/tex}{tex}\\Rightarrow 3 \\times 3 \\left( x ^ { 2 } - \\frac { 1 } { x ^ { 2 } } \\right) = 1{/tex}{tex}\\Rightarrow 3 \\left( x ^ { 2 } - \\frac { 1 } { x ^ { 2 } } \\right) = \\frac { 1 } { 3 }{/tex} | |
| 24835. |
What is relation b/w mean mode and median |
| Answer» 3median =2mode+mean | |
| 24836. |
Lab manual is important in class 10th or not |
| Answer» lab manual is of 5 marks | |
| 24837. |
Ex 6.5 |
| Answer» First solve all the questions. Then ask the answers/problems. | |
| 24838. |
What is area of square? |
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Answer» Area of square=side ×side Side square Area of square is side into side |
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| 24839. |
Guy ! Tum logo kidiwali Kaiser rhe |
| Answer» | |
| 24840. |
How to put the formulaes in the questions of triangle |
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| 24841. |
Do we really need to practice rd sharma or rs aggarwal for class 10th mathematics |
| Answer» Yes | |
| 24842. |
what is the altitude of an equilateral triangle ofeach side 6 cm |
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Answer» 3 under root 3 Under root 3 |
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| 24843. |
If one zero of 2x²-3x+k is reciprocal to the other,then fund the value of k? |
| Answer» Zeroes b alpha(a) and 1/alpha(1/a)C/A=k/2a×1/a=1Thus, 1= k/2K= 2 | |
| 24844. |
Sin.sin.cos.cos |
| Answer» Tan tan cos cos | |
| 24845. |
If root 3 =Sine theta find sintheta.tantheta (1+cot theta /sin theta +cot theta |
| Answer» | |
| 24846. |
Can I get sample question papers of social and science |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 24847. |
Pt*pt=pa*pb |
| Answer» | |
| 24848. |
What is similar figures |
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Answer» Have same shape but not necessarily the same size When all angles are equal in two triangle then it is similar. All congruence triangle are similar but all similar triangle are not congruence. |
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| 24849. |
how to find xi |
| Answer» | |
| 24850. |
find the greatest number of 6digit which is exactly divisible by 15,24 and36? |
| 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 | |