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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.
| 24551. |
What is cos a |
| Answer» Cos a is equal to base upon hypotenuse | |
| 24552. |
In the following APs find the missing term in the box 2 , , 26 |
| Answer» 14 | |
| 24553. |
A die is thrown.Probability of prime number |
| Answer» 3/6 | |
| 24554. |
Value of root 3 |
| Answer» 1.73 | |
| 24555. |
The difference between the circumference and diameter of circle is 135cm find radius |
| Answer» 31.5 | |
| 24556. |
Free sample paper |
| Answer» Check Sample Papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 24557. |
How many terms of the A.P 9,17,25,......must be tak3n to give a sum of 636 |
| Answer» a=9 d=4 Sn=636 n=?Sn=n/2(2a+(n-1)d)636=n/2(18+(n-1)4)636*2=n(18+4n-4)1272=n(14+4n)1272=14n+4n*n4n square+14n-1272=0Now solve this by splitting the middle term | |
| 24558. |
hubham |
| Answer» | |
| 24559. |
Is it true that some questions will come in options in exam of class 10 |
| Answer» Yes for sure..?? | |
| 24560. |
Proving of a^3+b^3+c^-3abc |
| Answer» | |
| 24561. |
Find volume of cylinder radius is 3cm n height is 4cm |
| Answer» Radius=3cmHeight=4cmVolume of cylinder=pie.r^2.hVolume=3.14×3×3×4Volume=113.04cm^3 | |
| 24562. |
(X-1)(X-2)=(X-3)(x+1) |
| Answer» | |
| 24563. |
Please give me 10th cbse board exam 2017 18 datesheet |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 24564. |
What is degree of polynomials? |
| Answer» The highest power of. Variable is known as degree of polynomial | |
| 24565. |
CBSE syllabus and marks |
| Answer» Full syllabus except SST because in SST there is choice between chapters and it is of total 80 marks | |
| 24566. |
prove that the diagonal of a rectangle bisect each other are equal. |
| Answer» Let ABCD be a rectangle. Taking A as origin, the vertices of a rectangle are A(0, 0), B(a, 0), C(a, b) and D(0, b).\xa0Diagonal {tex}AC = \\sqrt {{{(a - 0)}^2} + {{(b - 0)}^2}} = \\sqrt {{a^2} + {b^2}} {/tex}Diagonal {tex}DB = \\sqrt {{{(a - 0)}^2} + {{(b - 0)}^2}} = \\sqrt {{a^2} + {b^2}} {/tex}{tex}\\therefore \\;AC = DB\\;\\left[ {\\because \\;{\\text{Each}}\\,\\sqrt {{a^2} + {b^2}} } \\right]{/tex}Mid-point of {tex}AC = \\left( {\\frac{{0 + a}}{2},\\frac{{0 + b}}{2}} \\right) = \\left( {\\frac{a}{2},\\frac{b}{2}} \\right){/tex}Mid-point of {tex}BD = \\left( {\\frac{{a + 0}}{2},\\frac{{0 + b}}{2}} \\right) = \\left( {\\frac{a}{2},\\frac{b}{2}} \\right){/tex}{tex}\\Rightarrow{/tex}\xa0Mid-point of AC = Mid-point of BD{tex}\\Rightarrow{/tex}\xa0Mid-point of AC and BD is same{tex}\\Rightarrow{/tex}\xa0AC and BD bisect each other.Hence the diagonals of a rectangles bisect each other and are equal proved. | |
| 24567. |
If n\'th of an ap i n and m\'th of an ap is n prove that (m+n)=0 |
| Answer» mam = nanm[a + (m - 1)d] = n [a + (n - 1)d]{tex} \\Rightarrow {/tex}\xa0ma + m2d - md = na + n2d - nd{tex} \\Rightarrow {/tex}\xa0a(m - n) + (m2 - n2)d - md + nd = 0{tex} \\Rightarrow {/tex}\xa0a(m - n) + (m - n) (m + n)d - (m - n)d = 0{tex} \\Rightarrow {/tex}\xa0(m - n) [a + (m + n - 1)d] = 0{tex} \\Rightarrow {/tex}\xa0a + (m + n - 1)d = 0{tex} \\Rightarrow {/tex}\xa0am+n = 0Hence proved. | |
| 24568. |
Find the distance between the points ( a cos 35° ,0) and ( 0,a cos55°) |
| Answer» Distance between the points P(x1, y1) and Q(x2, y2) is given by PQ =\xa0{tex}\\sqrt { \\left( x _ { 2 } - x _ { 1 } \\right) ^ { 2 } + \\left( y _ { 2 } - y _ { 1 } \\right) ^ { 2 } }{/tex}{tex}\\therefore{/tex}\xa0Distance between {tex}\\left( 0 , a \\cos 55 ^ { \\circ } \\right){/tex} and {tex}\\left( a \\cos 35 ^ { \\circ } , 0 \\right){/tex}{tex}= \\sqrt { \\left( a \\cos 35 ^ { \\circ } - 0 \\right) ^ { 2 } + \\left( 0 - a \\cos 55 ^ { \\circ } \\right) ^ { 2 } }{/tex}{tex}= \\sqrt { \\left( a \\cos 35 ^ { \\circ } \\right) ^ { 2 } + \\left( - a \\cos 55 ^ { \\circ } \\right) ^ { 2 } }{/tex}{tex}= \\sqrt { a ^ { 2 } \\cos ^ { 2 } 35 ^ { \\circ } + a ^ { 2 } \\cos ^ { 2 } 55 ^ { \\circ } }{/tex}{tex}= \\sqrt { a ^ { 2 } \\left( \\cos ^ { 2 } 35 ^ { \\circ } + \\cos ^ { 2 } 55 ^ { \\circ } \\right) }{/tex}{tex}= a \\sqrt { \\cos ^ { 2 } \\left( 90 ^ { \\circ } - 55 ^ { \\circ } \\right) + \\cos ^ { 2 } 55 ^ { \\circ } }{/tex}{tex}= a \\sqrt { \\sin ^ { 2 } 55 ^ { \\circ } + \\cos ^ { 2 } 55 ^ { \\circ } }{/tex}{tex}= a \\sqrt { 1 }{/tex}{tex}= a{/tex}\xa0units. | |
| 24569. |
2x+3x-1 |
| Answer» | |
| 24570. |
Two tangent PA and PB are drawn to circle such that angle APB=30Deegre .prove that OP=2AP |
| Answer» In {tex}\\triangle{/tex}AOP and {tex}\\triangle{/tex}BOP, we have,{tex}\\angle OAP = \\angle OBP = 90^\\circ{/tex}OP = OP [Common]PA = PB [{tex}\\because{/tex} Tangents from an external point are equal in length]So, by RHS congruence criterion, we have{tex}\\triangle AOP \\cong \\triangle BOP{/tex}{tex}\\therefore \\angle APO = \\angle BPO{/tex} [By C.P.C.T.]{tex}\\Rightarrow \\angle APO = \\angle BPO = \\frac{1}{2}\\angle APB{/tex}{tex} = \\frac{1}{2} \\times 120^\\circ{/tex}= 60°{tex} \\Rightarrow \\angle APO = \\angle BPO = 60^\\circ {/tex}In {tex}\\triangle{/tex}OAP, we have{tex}\\cos 60^\\circ = \\frac{{AP}}{{OP}}{/tex}{tex} \\Rightarrow \\frac{1}{2} = \\frac{{AP}}{{OP}}{/tex}{tex}\\Rightarrow{/tex} OP = 2APHence proved. | |
| 24571. |
Is there any change in question of board exam 2017-18 because system is changed ?? |
| Answer» | |
| 24572. |
When we download this app so why to buy sample paper to solve |
| Answer» | |
| 24573. |
In a triangle ABC is DE||BC,AE=8cm and BC=6cm, then find DE |
| Answer» | |
| 24574. |
4+4-4×4 |
| Answer» -8 | |
| 24575. |
X square + X + 1 find the root of this equation |
| Answer» No real roots are there in given equation | |
| 24576. |
How I do construction ??? |
| Answer» | |
| 24577. |
Is 10th class datesheet upload? |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 24578. |
αβ |
| Answer» | |
| 24579. |
What is A.P. |
| Answer» Arithmetic progression | |
| 24580. |
In AP chapter formula of Sn= |
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Answer» n/2{2a+(n-1)d} What do you want to convey by writing this? Pvdjjdhhsjehhkshejz |
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| 24581. |
When frequency is not given and mode is given how can we get the median class |
| Answer» | |
| 24582. |
Class interval. |
| Answer» | |
| 24583. |
if AB is tangent and APQ is secant then prove that AB2=AQ*AP |
| Answer» | |
| 24584. |
In a right circular cone ,the cross section made by the plane parallel to its base is a |
| Answer» The vertex of a\xa0cone\xa0(the point, the apex) is not in the same\xa0plane\xa0as the\xa0base. Allcross sections\xa0of a\xa0cone parallel to the base\xa0will be similar to the\xa0base. ... If the segments joining the center of the circle\xa0base\xa0and vertex point is perpendicular to the\xa0base, the\xa0cone\xa0is a\xa0right circular cone. | |
| 24585. |
MM.graph for the equation x square plus x plus one equal zero |
| Answer» | |
| 24586. |
Vol. of cude formula |
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Answer» lbh L×B×H |
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| 24587. |
Did all the questions in board exam will be from the qiven topics in my cbse quide for class 10 |
| Answer» | |
| 24588. |
How to draw triangle of side 7,6and5 and ratio is 3:5 |
| Answer» | |
| 24589. |
find HCF 196,32880 |
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Answer» Ncert question Silly question |
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| 24590. |
The perimeter of a sector of a circle of radius 5.3cm is 16.4cm . Find the area of the sector |
| Answer» let POQ is the sector, radius of circle =5.3PO+QO+arcQP=16.4\xa05.3+5.3+arc PQ=16.4l=arc pq=5.8area of sector POQ =1/2 X L X r = 1/2 x 5.8 x 5.3 = 15.37 cm2 | |
| 24591. |
Prove that cos\\ 1-sin = sec +tan |
| Answer» | |
| 24592. |
2or 2 |
| Answer» | |
| 24593. |
Prove that the tangents drawn at the ends of a diameter of a circle are parallel |
| Answer» They are parallel because diameter is perpendicular to the tangent as of 90 ° they are parallel H.P | |
| 24594. |
When y= 2 , and y=1/2 ,where y=4power x ,then what is the value of x |
| Answer» | |
| 24595. |
Find the ratio in which the joining of points (-3,10) and (6,-8) is divided by the points (-1,6). |
| Answer» Let A →\xa0(–3, 10), B → (6, –8) and P → (–1, 6)Let P divide AB in the ratio K: 1.{tex}P \\to \\left\\{ {\\frac{{(K)(6) + (1)( - 3)}}{{K + 1}},\\frac{{(K)( - 8) + (1)(10)}}{{K + 1}}} \\right\\}{/tex}or {tex}P \\to \\left( {\\frac{{6K - 3}}{{K + 1}},\\frac{{ - 8K + 10}}{{K + 1}}} \\right){/tex}But P {tex}\\rightarrow{/tex} (-1, 6){tex}\\therefore \\;\\frac{{6K - 3}}{{K + 1}} = - 1{/tex}{tex}\\Rightarrow{/tex} 6K - 3 = -K - 1{tex}\\Rightarrow{/tex} 7K = 2\xa0{tex}\\Rightarrow K = \\frac{2}{7}{/tex}and {tex}\\frac{{ - 8K + 10}}{{K + 1}} = 6{/tex}{tex}\\Rightarrow{/tex} -8k + 10 = 6K + 6{tex}\\Rightarrow{/tex} 14K = 4{tex}\\Rightarrow K = \\frac{4}{{14}} = \\frac{2}{7}{/tex} | |
| 24596. |
Fgg |
| Answer» | |
| 24597. |
Ab - Ab |
| Answer» 0 | |
| 24598. |
5-4 |
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Answer» Great question??? 1? |
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| 24599. |
the perimeter of a rectangle feild is 130m and area is 1000m2 find dimension |
| Answer» on solving its dimension would come 40cm and 25 cm | |
| 24600. |
Prove tan1.tan2.tan3............tan89=1 |
| Answer» tan1.tan(90-1).tan2.tan(90-2).....tan45tan1.cot1.tan2.cot2.tan45= 1 | |