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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.
| 9001. |
3×25 kya hota ha |
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Answer» 75 25x3=75 35x25=75 75 75 |
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| 9002. |
Please koi muje bta ye ki chapter 6 triangle main kon se exercise cut ho gaye h |
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Answer» 4 and 5 Ex 4 and 5 dono cut he Thanks Ex 6.4 cancel ha kyoki theorem 6.9and6.6 cancel ha na isliya yh exercise us pr based ha yh nhi krni |
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| 9003. |
3x2 - kx +6 sum of zeroes is 3 find the value of l |
| Answer» The value of the k is 9 | |
| 9004. |
If tan theta=√3, find the value of sintheta costheta |
| Answer» Hyeee | |
| 9005. |
Simple form quadratic EQ.is |
| Answer» ax*2+bx+c=0 | |
| 9006. |
Product of zeros polynomial ax square+bx+c is |
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Answer» Product of the zeroes of polynomial ax^2+bx+c is c/a Product of zeroes of polynomial ax^2+bx+c =b/a. |
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| 9007. |
√1-cos theta/1+cos theta =cosec theta -cot theta |
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Answer» Same here Yah chapter muze samaz hi nhi aaya payal ?? |
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| 9008. |
Draw a double bar graph comparing rivers in Kerala and himachal Pradesh |
| Answer» | |
| 9009. |
If alpha and bita are the zeros of the quadratic polynomial ax^2+bx+ c |
| Answer» Sum of zeros = -b/aProduct of zeros = c/a | |
| 9010. |
What is value of root 3 |
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Answer» 1.73 The\xa0square root of 3 is represented using the square root or the radical symbol “√”,\xa0and it is\xa0written as\xa0√3.\xa0The value of\xa0√3\xa0is approximately equal to\xa01.732. This value is widely used in mathematics. Since root 3 is an\xa0irrational number, which cannot be represented in the form of a fraction. It means that it has an infinite number of decimals. Here, we are going to have a look at the value of root and the long division method to find the value of root 3. |
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| 9011. |
Prove that √2 is irrational also prove of √3, √5, √7 |
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Answer» Ans: √2 Let us assume that √2 is a rational number Take a and b are co prime numbers where b is not equal to 0 Such that √2 =a/b √2b= a take it as eqn 1 Squaring both side (√2b) 2 = a2 (2b) 2 = a2 take it as eqn two b2= a2/2 2 divides a2 Also 2 divides a ( if a prime number p divides a2 then divides a also) Let a = 2c Substitute in 2 (2b)2 =( 2c) 2 2b2 = 4c2 2b2 / 4= c2 b2 /2 = c2 2 divides b2 Also 2 divides b Therefore, 2 divides both a and b also a and b are co prime number. Our assumption that √2 is a rational number is wrong. √2 is a irrational number. 1.\xa0Let\xa0√2 be a rational number\xa0Therefore,\xa0√2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q\xa0≠ 0On squaring both sides, we get\xa0 p²= 2q² ...(1)Clearly, 2 is a factor of 2q²⇒ 2 is a factor of p² [since, 2q²=p²]⇒ 2 is a factor of p\xa0Let p =2 m for all m ( where m is a positive integer)Squaring both sides, we get\xa0 p²= 4 m² ...(2)From (1) and (2), we get\xa0 2q² = 4m² ⇒ q²= 2m²Clearly, 2 is a factor of 2m²⇒ 2 is a factor of q² [since, q² = 2m²]⇒ 2 is a factor of q\xa0Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1 Therefore, Our supposition is wrongHence\xa0√2 is not a rational number i.e., irrational number. |
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| 9012. |
Prove that the parallelogram Cicumscribing a circle is a rhombus |
| Answer» Given: ABCD be a parallelogram circumscribing a circle with centre O.To prove: ABCD is a rhombus.We know that the tangents drawn to a circle from an exterior point are equal in length.Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.Adding the above equations,AP + BP + CR + DR = AS + BQ + CQ + DS(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)AB + CD = AD + BC2AB = 2BC(Since, ABCD is a parallelogram so AB = DC and AD = BC)AB = BCTherefore, AB = BC = DC = AD.Hence, ABCD is a rhombus. | |
| 9013. |
Proof of BPT theorem Chapter 6 similiar triangle |
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Answer» Thanks BASIC PROPORTIONALITY THEOREM PROOFIf a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.\xa0Given :\xa0In a triangle ABC, a straight line\xa0l\xa0parallel to BC, intersects AB at D and AC at E.\xa0\xa0To prove :\xa0AD/DB = AE/ECConstruction :Join BE, CD.\xa0Draw EF\xa0⊥ AB and DG\xa0⊥ CAProof :Step 1 :Because\xa0EF\xa0⊥ AB, EF is the height of the triangles ADE and DBE.\xa0Area (ΔADE) = 1/2\xa0⋅ base\xa0⋅ height = 1/2\xa0⋅ AD\xa0⋅ EFArea (ΔDBE) = 1/2\xa0⋅ base\xa0⋅ height = 1/2 ⋅ DB ⋅ EFTherefore,\xa0Area (ΔADE) /\xa0Area (ΔDBE) := (1/2\xa0⋅ AD\xa0⋅ EF) / (1/2\xa0⋅ DB\xa0⋅ EF)Area (ΔADE) /\xa0Area (ΔDBE) = AD / DB -----(1)Step 2 :\xa0Similarly, we getArea (ΔADE) /\xa0Area (ΔDCE) := (1/2\xa0⋅ AE\xa0⋅ DG) / (1/2\xa0⋅ EC\xa0⋅ DG)Area (ΔADE) /\xa0Area (ΔDCE) = AE / EC -----(2)Step 3 :\xa0But\xa0ΔDBE and\xa0ΔDCE are on the same base DE and between the same parallel straight lines BC and DE.\xa0Therefore,\xa0Area (ΔDBE) = Area (ΔDCE) -----(3)Step 4 :\xa0From (1), (2) and (3), we can obtainAD / DB = AE / ECHence, the theorem is proved.\xa0 |
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| 9014. |
Calculate the median if the mean=30 and the mode= 40 |
| Answer» 3Median =Mode + 2Mean\xa0Given, Mean = 30Mode = 40\xa0Now, 3Median = Mode + 2Mean=> 3Median = 40 + 2(30)=> 3Median = 40 + 60=> 3Median = 100=> Median = 100/3=> Median = 33.3Hence,Median=33.3 | |
| 9015. |
Find two consecutive positive intiger whose sum is 21 and products is 110 |
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Answer» Formula Let\xa0first number = xsince this no are consecutive sosecond number = x+1\xa0now according to question\xa0x (x+1) = 110\xa0x² + x = 110\xa0x² + x - 110 = 0\xa0by using factorization method\xa0x² + 11x - 10x -110 = 0x(x+11) -10(x+10) = 0(x+11) (x-10) = 0∵ x = -11 and x = 10The two numbers would be 10 and 11.10 x 11 = 11010 + 11 = 21 |
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| 9016. |
Suresh has a garden in delhi |
| Answer» a) the length of the broken part can be figured out by the pythagoras theoram-so, suppose the length of the broken part is x15 square +20 square = x square= 225 + 400 = x squarethus, x square = 625hence, x= sqrt of 625therefore x= 25m\xa0\xa0b) the heinght of the full tree is 15+ 25= 40mc) the length of the hypotenuse is 25md) the area of the triangle= (15*20)/4 = 300/4 = 75m sq. | |
| 9017. |
Find the quadratic polynomial whose sum of zeroes is 3 and product of zeroes is 2. |
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Answer» X² - 3x+2 Given:-a+b=3 and ab=2Required quadratic polynomial =x2-(a+b)x+(ab) =x2-3x+2 Let\xa0α\xa0and\xa0β\xa0are the roots of given Quadratic polynomial.Given:α+β=3,αβ=2Quadratic polynomial:x2−(α+β)x+(αβ)∴\xa0Polynomial is\xa0x2−3x+2\xa0 |
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| 9018. |
If sinA=3/5 then value of 2+2ten square A |
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Answer» Given,sinA= 3/5We know that ,sinA = perpendicular \\hypotenuseSo, P=3 and H=5 By Pythagoras therorm: H^2=P^2+B^2 B=√H2-P2 =√25-9 =√16 =4 A/Q =2+2tan^2A =2+2(P/B)^2 =2+2×(3/4) ^2 =2+2×9/16 =2+9/8 =25/8 25/8 |
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| 9019. |
find the zeroes of 2x^2+root 5 |
| Answer» This polynomial has no zeroes. If you see graphically, the parabola for this equation does not intersect the x-axis at any point and algebraically, since there is no negative number, how can you get 0? | |
| 9020. |
If Sin =a^2-b^2\\a^2+b^2 , find the value of other five trigonometric ratios |
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Answer» In ΔABC, Let ∠ABC be θSin θ = (a2 - b2) / (a2 + b2)⇒ AB = (a2 - b2)⇒ AC = (a2 + b2)⇒ BC = √[(a2 + b2)2 - (a2 - b2)2] [According to Pythagoras theorem]⇒ BC = √(4a2b2)⇒ BC = 2abCos θ = 2ab / (a2 + b2)Tan θ = (a2 - b2) / 2ab.Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively. In ΔABC, Let ∠ABC be\xa0θSin θ = (a2\xa0- b2) / (a2\xa0+ b2)⇒ AB = (a2\xa0- b2)⇒ AC = (a2\xa0+ b2)⇒ BC = √[(a2\xa0+ b2)2\xa0- (a2\xa0- b2)2] [According to Pythagoras theorem]⇒ BC = √(4a2b2)⇒ BC = 2abCos θ = 2ab / (a2\xa0+ b2)Tan θ = (a2\xa0- b2) / 2ab.Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively. |
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| 9021. |
12.1 ka question no 3 |
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Answer» . Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.Solution:The radius of 1st\xa0circle, r1\xa0= 21/2 cm (as diameter D is given as 21 cm)So, area of gold region = π r12\xa0= π(10.5)2\xa0= 346.5 cm2Now, it is given that each of the other bands is 10.5 cm wide,So, the radius of 2nd\xa0circle, r2\xa0= 10.5cm+10.5cm = 21 cmThus,∴ Area of red region = Area of 2nd\xa0circle − Area of gold region = (πr22−346.5) cm2= (π(21)2\xa0− 346.5) cm2= 1386 − 346.5= 1039.5 cm2Similarly,The radius of 3rd\xa0circle, r3\xa0= 21 cm+10.5 cm = 31.5 cmThe radius of 4th\xa0circle, r4\xa0= 31.5 cm+10.5 cm = 42 cmThe Radius of 5th\xa0circle, r5\xa0= 42 cm+10.5 cm = 52.5 cmFor the area of nth\xa0region,A = Area of circle n – Area of circle (n-1)∴ Area of blue region (n=3) = Area of third circle – Area of second circle= π(31.5)2\xa0– 1386 cm2= 3118.5 – 1386 cm2= 1732.5 cm2∴ Area of black region (n=4) = Area of fourth circle – Area of third circle= π(42)2\xa0– 1386 cm2= 5544 – 3118.5 cm2= 2425.5 cm2∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle= π(52.5)2\xa0– 5544 cm2= 8662.5 – 5544 cm2= 3118.5 cm2 Ok |
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| 9022. |
Sin2theta + cos2theta = |
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Answer» Ans I 1 As sin ^2 theta + cos ^2 theta = 1 1 1 1 It is ncert identity and it\'s answer is 1 |
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| 9023. |
1/coseca-cota+1/cosecb-cotb+1/cosecc-cotc |
| Answer» | |
| 9024. |
-x²÷x |
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Answer» -x -× |
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| 9025. |
Prove -:(1+cot A- cosec A)(1+tan A+ sec A)=2. |
| Answer» (1+cos¢\\sin¢-1/sin¢) (1+sin¢/cos¢+1/cos¢) =sin²¢+cos²+2sincos¢-1÷cos¢sin¢ =1+2sincos¢-1÷sin¢cos¢ =2 | |
| 9026. |
What is a formula of chapter 2 |
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Answer» Alpha + beta = - b/a Alpha × beta = c/a Alpha +beta +gama = -b/aAlpha ×beta× gama = c/a Alpha ×beta + beta × gama + gama × alpha = - d / a a+b = -b/aa×b = c/aa+b+y = -b/aab+by+ya = c/aaby = -d/aa= alphab= betay= gamma a+b=-b/aab=c/a |
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| 9027. |
Assignment on topic mensuration about 15to 20 pages... |
| Answer» | |
| 9028. |
A man traveling 370 km |
| Answer» | |
| 9029. |
Exercise 3.4 ncert |
| Answer» Yes | |
| 9030. |
Prove that √5 is rational no. |
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Answer» √5 is an irrational number Let us assume that contrary that √5 is rational number√5=a by b√5b=asquare the both sides 5bsquare= asquareASquare is divided by 5 I think it should be irrational instead of rational |
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| 9031. |
Write the common difference of an arithmetic progression 8, 5, 2 - 1 |
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Answer» -3 is the right answer of your question 3 3 is the common difference i.e D Common difference = a2-a1=5-8=-3 (Ans) -3 |
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| 9032. |
If 1 is the root of equations ax 2 + ax + 3 =0 and x 2 + x + b = 0 then find ab . |
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Answer» No solution because lines are parallelA1/A2=B1/B2≠C1/C2 No solution |
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| 9033. |
Pf of 632 and 123 |
| Answer» | |
| 9034. |
In exercise 10.2 which are the deleted questions? |
| Answer» In 10.2 Exercise of Maths, No questions are deletedAnd in the same, 9-13 are IMP questions | |
| 9035. |
Find the unknown entires a, b, c, d, e and f in the following distribution and hence find their mode |
| Answer» | |
| 9036. |
Why π=180 degree |
| Answer» | |
| 9037. |
Definition polynomial |
| Answer» In which power of variables is whole number.. | |
| 9038. |
If the system of equations 3x+y=1 and 2(k-1)x+(k-1)y=2k+1 is inconsistent then k=? |
| Answer» Adgfxg | |
| 9039. |
A circle has number of tangents |
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Answer» A circle has infinite number of tangents. In the circle there are only one tangent in a point of a circle |
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| 9040. |
Which of the following metals is less reactive than hydrogen |
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Answer» Platinum,gold,silver,Mercury,copper is less reactive than Hydrogen. Platinum,gold,silver,Mercury,copper is less reactive than Hydrogen... Platinum Lead < Iron < Zinc < Aluminium < Magnesium < Calcium < Sodium < Potassium In the reactivity seires hydrogen all metal in the upper |
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| 9041. |
4x²-4x+3 |
| Answer» (4x²-4x+3)(4x²-6x+2x+3)[2x(2x-3)+1(2x-3)][(2x+1)(2x-3)][(X=-1/2)(X=3/2)] | |
| 9042. |
In figure |
| Answer» In which | |
| 9043. |
2 tan30°__________1-tan2 30° |
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Answer» 2tan30°>1-tan²30° Mc |
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| 9044. |
Prove that aSin + bcos=+-√a2+b2- C2 If acos -bsin =c |
| Answer» | |
| 9045. |
Find the area of quadrilateral ABCD whose vertices are A(1,1), B(7,-3), C(12,2) and D(7,21). |
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Answer» The coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21\u200b ∣[x 1\u200b (y 2\u200b −y 3\u200b )+x 2\u200b (y 3\u200b −y 1\u200b )+x 3\u200b (y 1\u200b −y 2\u200b )]∣Area of triangle ABC = 21\u200b ∣[1(3−2)+7(2−1)+12(1−3)]∣= 21\u200b ∣[1+7−24]∣= 21\u200b (16)=8 sq. unitsArea of triangle ACD = 21\u200b ∣[1(2−21)+12(21−1)+7(1−2)]∣= 21\u200b ∣[−19+240−7]∣= 21\u200b (214)=107 sq. unitsHence, area of quadrilateral ABCD = 107+8=115 sq. units Rtfho |
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| 9046. |
sintan |
| Answer» | |
| 9047. |
2x^2+x-6=0 in complete square method |
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Answer» Solution verified Verified by Toppr 2x 2 +x−6 Completing the square, ⇒x 2 + 2 x\u200b−3=0 ⇒x 2 + 2 x\u200b+ 16 1\u200b− 16 1\u200b−3=0 ⇒(x+ 4 1\u200b) 2 − 16 49\u200b=0 ⇒(x+ 4 1\u200b− 4 7\u200b)(x+ 4 1\u200b+ 4 7\u200b)=0 ⇒(x− 2 3\u200b)(x+2)=0 ⇒x= 2 3\u200b,−2 Solve any question of Quadratic Equations with:-. This is wrong. 3/2,-2 this right SolutionverifiedVerified by Toppr2x 2 +x−6Completing the square,⇒x 2 + 2x\u200b −3=0⇒x 2 + 2x\u200b + 161\u200b − 161\u200b −3=0⇒(x+ 41\u200b ) 2 − 1649\u200b =0⇒(x+ 41\u200b − 47\u200b )(x+ 41\u200b + 47\u200b )=0⇒(x− 23\u200b )(x+2)=0⇒x= 23\u200b ,−2Solve any question of Quadratic Equations with:- Completing the square method paper me nahi ayega What is family law |
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| 9048. |
what is the probability of p(E) |
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Answer» Hii Probability of event Probability of event p(E) = No. of outcomes ____________________ Total no. of events |
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| 9049. |
The discriminate of the quadratic equation 7x^2-5+11=0 is |
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Answer» Q 2 ans is parallel D=285 The gemotrical represtation of the pair of linear equations 6x-3y=-10 and 2x-y+9=0 represent lines |
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| 9050. |
Ncjcjfj |
| Answer» Sdfff | |