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InterviewSolution
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.
| 41601. |
Is x =2 y= 3 a solution of the linear equation 2x+3y-13 =0 |
| Answer» Yes | |
| 41602. |
KirtyChow YouTube channel kon-kon dekhta h..... |
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| 41603. |
(Sina/1+cosa+1+cosa/sina)(cosa/1+sina+1+sina/cosa)=4seca×coseca |
| Answer» LHS=SinA/1+CosA+1+CosA/SinA(CosA/1+SinA+1+SinA/CosA)=Sin2+(1+CosA)2/SinA(1+CosA){Cos2A+(1+SinA)2/CosA(1+SinA)}=Sin2A(1+Cos2A+2CosA)/SinA+SinA×CosA(Cos2A+1+Sin2A+2SinA/CosA+SinA×CosA)=1+Sin2A+Cos2A+2CosA/SinA+SinA×CosA(1+Sin2A+Cos2A+2SinA/CosA+SinA×CosA)=2+2CosA/SinA(1+CosA){2+2SinA/CosA(1+SinA)} (Formula used : Sin2A+Cos2A = 1)=2(1+CosA)/SinA(1+CosA){2(1+SinA)/CosA(1+SinA}=2×1/SinA×1(2×1/CosA×1)=2/SinA(2/CosA)=2CosecA×2SecA (Formula used : 1- 1/SinA = CosecA 2- 1/CosA = SecA)=4CosecA×SecA So, L.H.S = R.H.S | |
| 41604. |
The sum of n terms of an AP is 4n2+5n. Find the AP. |
| Answer» 9,17,25....... | |
| 41605. |
The sum of the first 20 terms of the sequence 0.7+0.77+0.777+........ |
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Answer» Sorry it is of objective (c)7/81(179+10 to the power -21) Kya answer 20.3 hai...... |
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| 41606. |
If the area of circle 308cm square find the perimeter is |
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| 41607. |
Prove thale\'s theorum |
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Answer» Bpt Refer ncert |
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| 41608. |
Sum of third and seventh terms of an AP is 6 and their product is 8.find the sum of first 16th terms |
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Answer» 76 76 |
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| 41609. |
Can two no. Have 18 as their hcf and 380 as their lcm |
| Answer» No. Becoz the HCF should be the factor of LCM of any given number. | |
| 41610. |
What is the hcf of smallest prime number and the smallest composit number |
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Answer» 2 2 = 2 * 14 = 2 *2Hcf = 2 2 2 |
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| 41611. |
Trigonometry samjha do |
| Answer» It is logical chapter .sin cos....are just names of different creatures . Practice chapter 8(8.4) and u will understand | |
| 41612. |
State and prove thales theorem. |
| Answer» Thales theorem is BPT..and....it was given in textbook i.e N.C.E.R.T on pg no.124 | |
| 41613. |
Roots of fraction alpha beeta product |
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| 41614. |
सिद्ध कीजिए √5 एक अपरिमेय संख्या हैं।यहाँँ से दशाईए कि 3√5-8 भी एक अपरिमेय संख्या है। |
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| 41615. |
The sum of square of two consecutive multiples of 7 is 637. Find the multiple. |
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Answer» (7x) square +(7(x+1)) square= 63749x^2 + 49(x+1)^2 =637Taking 49 common49( x^2 + x^2 +2x +1) = 6372x^2 + 2x +1 = 637 / 49 = 132x^2 + 2x = 13 - 1 = 12x^2 + x = 6x (x + 1) =62 *3 6So multiple s are 7x ie 7*2=14And 21 we have given thatsum of the squares of two consecutive multiples of 7 is 637. we have to findthe numbers are ??solution:-let the number multiple of. 7 = 7x thenconsecutive multiples of 7 = 7(x+1) now according to question(7x)² + {7 (x+1)}² = 637= 49 x² + 49 (x+1)² = 637= 49 x² + 49( x² +1+2x) = 98x² +98 x +49 = 637= 98x² +98x = 637-49 = 588=> x² +x= 6. or = x² + x -6 =0=> x² +3x -2 x -6 = 0=> x(x+3) -2 (x+3) = 0=> x-2 =0. and x+3 = 0=> x= 2 and -3 answerhence (1) if x = 2the number multiple of 7 is 7×2 =14 it\'s consecutive number multiple of 7 is = 7×(2+1) = 7×3 = 21(2) if x =-3the number multiple of 7 is 7×(-3) = -21it\'s consecutive number multiple of 7 is = 7×(-3 +1) = 7× (-2) = -14Thank you?? |
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| 41616. |
Prove areas theorem |
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| 41617. |
In a square abcd 2 quadrants are drawn taking bc as radius find the areaof shaded region |
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| 41618. |
Maximum ek dim me kitne hr padhna chahiye abhi?science ar maths dono |
| Answer» 15 hour | |
| 41619. |
Bpt theorem from chapter 6 |
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Answer» PROOF OF BPTGiven: In ΔABC, DE is parallel to BCLine DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.To Prove: ADBD=AECEConstruction: Join segments DC and BEProof: In ΔADE and ΔBDE,A(ΔADE)A(ΔBDE)=ADBD (triangles with equal heights)In ΔADE and ΔCDE,A(ΔADE)A(ΔCDE)=AECE (triangles with equal heights)Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,A(ΔBDE)=A(ΔCDE)Therefore,A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)Therefore,ADBD=AECEHence Proved. This is base proportionality theorem. ?? If in triangli one line intersect each sides theh this triangle are divide in ratio whic are proptional to eaxh other |
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| 41620. |
Every even number is a whole number |
| Answer» No... For example 3 is a whole number but it is not a even number | |
| 41621. |
Quickly answereded 1/sec-tan-1/cot = |
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| 41622. |
1/sec-tan-1/cot = |
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| 41623. |
The area of square is 64.find the side |
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Answer» The area of square=64 (given)Let the side be xBy using the formula of area of square i.e Square of side=area of squareWe obtainX2 = 64 hence, x=8 8 |
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| 41624. |
When will the NTSE M.P result declared |
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| 41625. |
Find the zeroes of the given polynimial 5x^2_135 |
| Answer» Alpha=+3√3 Beta=-3√3 | |
| 41626. |
Sum of four consecutive terms of an AP is 50 and fourth term is four times the first find the AP |
| Answer» Let the 4 numbers are a , a + d , a + 2d , a + 3d.Sum of 4 numbers AP = 50a + a + d + a + 2d + a + 3d = 50⇒ 4a + 6d = 50⇒ 2a + 3d = 25 --------------(1)Also given the forth term is four times the first4(a) = a + 3d4a - a = 3da = dputting a = d in (1) , we obtain5d = 25d = 5a = 5 but d = a.∴ First four terms are 5 , 10 ,15 , 20.\xa0 | |
| 41627. |
If one zero of the polynomials 5z+13z-p is reciprocal of the other then find p |
| Answer» Does the answer is 5? | |
| 41628. |
solve for x and y:2x+3y=94x+6y=18 |
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Answer» 2x + 3y = 9......... (i)4x + 6y = 18 .... (ii)On multiplying equation (1) with 2, we get2(2x + 3y) = 2(9)⇒ 4x + 6y = 18 .... (iii)Here, both the equation (ii) and (iii) are same and we know that when both equations are exactly same there are infinitely many solutions and the value of x and y can\'t be predicted. Y=5/3; x=6{substition method} |
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| 41629. |
Find the sum of all three digit number which leave the remainder 3 when divided by 5. |
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| 41630. |
ABC is an equilateral triangle of side 2 a. Find each altitudes |
| Answer» Root 3 /2 of side of equilateral triangle is altitude | |
| 41631. |
For what value of p,2p-1,7 and 3p are three consecutive terms of an AP |
| Answer» Formula 2b=a + c | |
| 41632. |
Find the (HCF×LCM)for the numbers 100 and 190 |
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Answer» Hcf=10. Lcm=1900Hcf×lcm=10×1900=19000 Lcm x hcf= 19000 Hcf=10Lcm=18400 HCF*LCM=product of two numbers |
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| 41633. |
If Sin4A/a +cos4A/b =1/a+b,then prove that sin8/Aa³ +cos8A/b³=1/(a+b)³. |
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Answer» Mujhe bhi 10th ki worksheet mein hi Mila hai When ethyne is burnt in air incomplete burning takes place and produce sooty flame due to limited supply oxygen present in the air. ... Hence, This oxy-acetylene flame is used for welding and it is not possible to attain such a high temperature with air. Due this reason a mixture of ethyne and air is not used for welding. Great.But we cant answer bcuz our level is 10th.We cant solve a 12th level question 10th mein 12th ka |
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| 41634. |
A metallic cube of edge 1cm is drawn into a wire of diameter 4mm. Then find the length of the wire |
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| 41635. |
Does steps of construction under chapter constructions to be written as per cbse syllabus 2018-19 |
| Answer» Yes,u can use bpt or pythagoras theorem to justify ur construction | |
| 41636. |
What is the syllabus of 2019 |
| Answer» Full syllabus | |
| 41637. |
What type of 4 mark questions are excepted??? From which lesson class 10...which model sums?? |
| Answer» 4 marks ques generally comes frm statistics trigonometry and theorems of triangle chapter | |
| 41638. |
Can anyone tell me is it necessary to practice RD sharma for boards |
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Answer» No it is not necessary to practice rd sharma best book is ncert fist you practice ncert Nooo Firstly solve ncert uska har question ghor k pilo phir rd sirf example lagao First complete ur ncert....and if uu have tym left then definitely u have to go for rd sharma..and other sample papers |
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| 41639. |
Prove that product of three consecutive positive integer is divisible by 6 |
| Answer» Let three consecutive numbers be x, (x + 1) and (x + 2)Let x = 6q + r 0 {tex}\\leq r < 6{/tex}{tex}\\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}{tex}\\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}if x = 6q then which is divisible by 6{tex}\\text { if } x = 6 q + 1{/tex}{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}{tex}= 6 ( 3 q + 1 ) \\cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}which is divisible by 6if x = 6q + 2{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}{tex}= 6 ( 2 q + 1 ) \\cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}Which is divisible by 6{tex}\\text { if } x = 6 q + 3{/tex}{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}which is divisible by 6{tex}\\text { if } x = 6 q + 4{/tex}{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}which is divisible by 6if x = 6q + 5{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}which is divisible by 6{tex}\\therefore {/tex}\xa0the product of any three natural numbers is divisible by 6. | |
| 41640. |
In a quadrilateral ABCD, angleA + angleD =90°.prove that AC^2 +BD^2=AD^2 + BC^2 |
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Answer» Q.In a quadrilateral ΔBCD, ∠A+ ∠D = 90°. Prove that AC2\xa0+ BD2\xa0= AD2\xa0+ BC2.\xa0Given Quadrilateral ΔBCD, in which ∠A+ ∠D = 90°To prove\xa0AC2\xa0+ BD2\xa0= AD2\xa0+ BC2Construct Produce AB and CD to meet at E. \xa0 ^ iska matlab to the power h Iska matlab AC ki power 2 hota h Ismain (^) this means?? Plz post clearly this one... |
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| 41641. |
In a quadrilateral ABCD,angleA + angleD=90° |
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| 41642. |
acos³ø+3acosø.sin²ø=masin³ø+3acos²ø. sinø=nTo proove- (m+n)⅔+(m-n)⅔=2a⅔ |
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| 41643. |
Am going to depression now because my maths was not prepared ??? |
| Answer» Bhai math se jitna daraoge utna wo aapko daragi to dont be scared and do your 100% | |
| 41644. |
NCRT exercise 3.1 , question no.1 |
| Answer» What is your doubt in the question? | |
| 41645. |
If the sum of squares of the zeroes of the polynomial 6xsquare + x +k is 25/36 . Find k |
| Answer» | |
| 41646. |
Prove that a parallelogram circumscribing a circle is a rhombus. |
| Answer» Since ABCD is a parallelogram,AB = CD ---- i)BC = AD ---- ii)\xa0It can be observed thatDR = DS (Tangents on the circle from point D)CR = CQ (Tangents on the circle from point C)BP = BQ (Tangents on the circle from point B)AP = AS (Tangents on the circle from point A)Adding all these equations, we obtainDR + CR + BP + AP = DS + CQ + BQ + AS(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)CD + AB = AD + BCOn putting the values of equations (1) and (2) in this equation, we obtain2AB = 2BCAB = BC …(3)Comparing equations (1), (2), and (3), we obtainAB = BC = CD = DAHence, ABCD is a rhombus. | |
| 41647. |
Guya who do u think is the best teacher of maths on YouTube??? |
| Answer» RB Gautam | |
| 41648. |
What are the properties of a rhombus?? |
| Answer» *Sides :- all 4 sides are congruent *Angles :- diagonals bisect vertex angles*Diagonals :- diagonals are perpendicularArea ............. | |
| 41649. |
(A+b) hole squre |
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Answer» (a+b)² = a²+ b²+2ab (a+b)2=a2 + 2ab + b2 (a + b)2 = a2 + b2 + 2ab |
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| 41650. |
Find the value of x if x=√6√6√6 |
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Answer» X=√6√6√6√6......X=√6+xX^2= 6+xX^2-x-6=0X^2-3x+2x-6=0(X-3)(X+2)=0X=3,-2 6√6 or 14.69 |
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