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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.
| 651. |
Solve the quadratic equation in variable x : abx² = (a+b)²(x-1) |
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Answer» I hope you understand Ans:abx²=(a+b)²(x-1)abx²=(a+b)²x-(a+b)²abx²-(a+b)²x+(a+b)²=0here,a=ab , b=-(a+b)² , c=(a+b)²By quadratic formula x=-b +/- _/b²-4ac /2a = -[-(a+b)²]+/- _/ [-(a+b)²]²-4(ab)(a+b)²] / 2ab =(a+b)²+/- _/(a+b)⁴-4ab(a+b)² / 2ab = (a+b)²+/- _/(a+b)²[(a+b)²-4ab] / 2ab = (a+b)²+/- (a+b)_/a²+b²+2ab-4ab /2ab = (a+b)²+/-(a+b)_/a²+b²-2ab / 2ab = (a+b)²+/-(a+b)_/(a-b)² / 2ab =(a+b)²+/-(a+b)(a-b) / 2ab = a²+b²+2ab +/- a2-b² /2abSo, x=a²+b²+2ab+a²-b² / 2ab = 2a²+2ab /2ab = 2a(a+b) /2ab = a+b/b OR x=a²+b²+2ab-(a²-b²) / 2ab =a²+b²+2ab-a²+b² / 2ab = 2b²+2ab / 2ab =2b(b+a) /2ab = a+b/aThus, the value of x= a+b/b , a+b/a Boubht baar try kara meine Nhi hua yr |
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| 652. |
Math ch.12 , exer. 12.2 question 3 |
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Answer» Yr woh toh mention kar diya ho tha tune Which class |
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| 653. |
If cos(a+b)=0 show that sin(a-b)=? |
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Answer» Cos(a+b)=0We know that, Cos 90 = 0 So, Cos(a+b) = Cos 90=> a+b = 90=> a = 90 - b Now, Sin(a-b) = Sin(90-b-b) = Sin(90-2b) = Cos 2b [ as Sin(90-x) = Cos x ]Thus, Sin(a-b) = Cos 2b cos (a+b) =0=> cos (a+b) = cos 90=> a+b = 90=> a = 90- bNowsin(a+b)= sin(90-b-b)= sin(90-2b)= cos 2b [as sin (90 - x) = cos x] cos (a+b) =0=> cos (a+b) = cos 90=> a+b = 90=> a = 90- bNowsin(a+b)= sin(90-b-b)= sin(90-2b)= cos 2b [as sin (90 - x) = cos x] |
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| 654. |
C.S.A of cube |
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Answer» C. S. A of a cube is4(s)² or(4×side×side) It means we take the area of a cube except the top and bottom side. In cube there are 6 squares in which we only take 4 of them. Hope it is helpful ✌ 4(s)*2 4[ side] 2 4(s)² The curved surface area or the lateral surface area of a cube is 4(side)².#Hope it is helpful.. |
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| 655. |
Find the integers of the word problem started above |
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| 656. |
Find the value of k if x-2 is a factor of x2-kx+10 |
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Answer» X=2X²-kx+10=02²-k(2)+10=04-2k+10=0-2k+4+10=0-2k+14=0-2k=-14K=-14/-2K=7 Given x-2=0.....so, x=22²-2k+10=0-2k=-10-42k=14 (cancelling negative sign)k=14/2k=7#Hope it is helpful... Put x = 2x2-kx+10 = 0\xa0(2)2 - k(2) + 10 = 04 - 2k + 10 = 0-2x +14 = 0-2x = -14x = 14/2 = 7 |
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| 657. |
Find ap of multiples of multiple of 2 between 100 to100 and from 100 to 1000for the same |
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| 658. |
What is the derivation of the Section formula? (Not the definition.ok? ) |
| Answer» Let A (x1, y1) and B (x2, y2) be the endpoints of the given line segment AB and C(x, y) be the point which divides AB in the ratio m : n. We want to find the coordinates (x, y) of C. Now draw perpendiculars of A, C, B parallel to Y coordinate joining at P, Q, and R on X-axis. | |
| 659. |
Find the sum of 32 terms of an A.P. whose third term is 1 and the 6th term is -11. |
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Answer» Third term =a+2d=1---(¡) and sixth term = a+5d=-11 ---(¡¡) subtract equation (¡)from(¡¡) we get d=-4 then put the value of d in equation I we get a = 9 then sum = n÷2×[2a+(n-1)×d] we get sum is -1696 Answer= -1696 |
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| 660. |
Please answer the question from triangle chapter example 7 . Please explain. ??? |
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Answer» If you solve this question by not saying unwanted advice I will obliged Thanku for ur advice The answer is very simple my child. First of all read the question twice and is your own brainbecause I am a very busy personGOOD LUCK BOY/GIRL .YOU HAVE MY LEGENDARY BLESSINGS. ?THANK YOU? |
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| 661. |
Coffeceant and zeros |
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| 662. |
39/+0(13+13) |
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Answer» I think Answer will be \'not defined\' 39 /+0 (13+13)=39/+0 (26)=39 /+0*26=39/0=0ans 0 is the correct answer 39/+0(13+13)= 39/+0(26)= 39/0= 0 |
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| 663. |
(4x+9x)² |
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Answer» (4x+9x)²We use (a+b)²=a²+b²+2ab=4x²+9x²+2*4x*9x=16x²+81x²+72=97x²+72x²=169x² 169x2 (4x+9x)²= (13x)2= 169x2 |
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| 664. |
Solve the pare of linear equations 141x+93 y =189 and 93 x + 141 y=45 |
| Answer» 141x+93 y =189 .........(i)93 x + 141 y=45.........(ii)Add (i) and (ii), we get141x+93 y =189\xa093 x + 141 y=45_______________234x + 234 y= 234x + y=1........ (iii)Subtract\xa0(i) and (ii), we get141x+93 y =189\xa093 x + 141 y=45_______________48 x - 48 y=144 x - y = 3........ (iv)Adding (iii) and (iv), we get x + y=1+ x - y = 3____________2x = 4x = 2Put x = 2 in (iii)2 + y = 1y = 1 - 2 = -1 | |
| 665. |
Formulae to find product of zero |
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Answer» For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.For a quadratic equation ax2+bx+c = 0,the sum of its roots = –b/a and the product of its roots = c/a. Chapter 15 |
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| 666. |
The quadratic polynomial whose sum of zeros is 3 and product of zeros is -2 is |
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Answer» X^2 -3x-2 x*2-3x-2 x²-3x-2 x*2 - 3x+2 |
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| 667. |
X^2-5x=14 find value of x ? |
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Answer» X²-5X=14 X²-5X-14=0X²-7X+2X-14=0X(X-7)+2(X-7)=0(X-7)(X+2)=0(X-7)=0 (X+2)=0X=7 and X=-2 it is done by this this x2-5x-14=0 x2+2x-7x-14=0. x(x+2)-7(x+2)=0 (x+2) (x-7) =0 if x+2=0. x=-2 Not Valid x-7=0 x=7 7,-2 x*2-5x-14=0=x*2-7x -2x-14=0=x(x-7)-2(x-7)=0 =(x-7)(x-2)so x=7,2 |
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| 668. |
Find the area of a triangle with A(1,4)and mid-pionts of sides through A being (2,-1)and(0,-1) |
| Answer» Q u e s t i o n : Find the area of a triangle ABC with A(1, - 4) and mid-points of sides through A being (2, - 1) and (0, - 1)\xa0A n s w e r :Let E be the mid point of AB | |
| 669. |
Is (√3+5)² + (5-√3)² a rational number justify |
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Answer» You have to solve both the brackets Separately By (a²+b²+2ab) & (a²+b²-2ab) =3+25+10root3+25+3-10root3=3+25+3+25=5656 is a terminating no so it is a rational no 4 |
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| 670. |
How many natural numbers are there between 1 to 1000 divisible by 5 |
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Answer» a=5,d=5,an=1000We have to use the formula an=a+(n-1)d1000=5+(n-1)51000-5=(n-1)5995=(n-1)5995/5=n-1199=n-1199+1=nn=200 Step - by - step explanation :a =5, d =5, tn =995995 = 5 + (n-1) × 5995 = 5 + 5n - 5995 = 5 nn = 995 ÷ 5 = 199 200 |
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| 671. |
Varg antral 130-140 ka work gyat kijiye |
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| 672. |
Varg antral130-140 ka varg gyat kijiye |
| Answer» Please write in english and clearly... | |
| 673. |
I want to ask that the chapters which are deleted surely will not come? |
| Answer» HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.Deleted syllabus of CBSE Class 10 Mathematics\xa0 | |
| 674. |
If tan theta =4/3 find sin theta+ cos theta/sin theta - cos theta |
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Answer» Thanks a lot Tan theta=4/3tan theta=p/bp=4,and b=3by paithagoras theorem,H=root under P^2+B^2H=root under 4^2+3^2H=root under 16+9H=root under 25H=5sin theta=p/hsin theta=4/5cos theta=b/hcos theta=3/5sin theta+cos theta/sin theta-cos theta=4/5+3/5 by 4/5-3/5=4+3/5by4-3/5=7/5by1/5=7/5×5/1=7/1=7 |
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| 675. |
For any positive integer n, prove that (n3 -3) is divisible by 6. |
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| 676. |
Show that 3/5 root 2 is irrational,given that root 2 is irrational number |
| Answer» Please anyone give me answer. | |
| 677. |
Use Euclid\'s algorithms to find the HCF of 4052 and 12576 |
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Answer» The HCF of 12576 and 4052 is 4 |
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| 678. |
Find hcf of 420 and 272 |
| Answer» Applying Euclid \'s division algorithm,H.C.F of 420 and 272\xa0Now, The remainder becomes 0.So, HCF of 420 and 272 is 4.Applying fundamental theorem of arithmetic for verifying,The prime factors are\xa0Therefore, HCF of 420 and 272 is 4.\xa0 | |
| 679. |
10/x+y +2/x-y , 15/x+y - 5/x-y solve the pair of equation by them to a pair of linear equation |
| Answer» Given equations are ,\xa010/( x + y ) + 2/( x - y ) = 4 ---( 1 )\xa015/( x + y ) - 5/( x - y ) = -2 --( 2 )\xa0Let ,\xa01/(x+y) = a , 1/(x-y) = b\xa010a + 2b = 4\xa0Divide each term with 2 , we get\xa05a + b = 2\xa0=> b = 2 - 5a ----( 3 )\xa015a - 5b = -2 -----( 4 )\xa0Substitute b = 2 - 5a in equation\xa0( 4 ) , we get\xa015a - 5( 2 - 5a ) = -2\xa0=> 15a - 10 + 25a = -2\xa0=> 40a = -2 + 10\xa0=> 40a = 8\xa0=> a = 8/40\xa0=> a = 1/5\xa0Put a = 1/5 in equation ( 3 ) , we\xa0get\xa0b = 2 - 5 × 1/5\xa0=> b = 2 - 1\xa0b = 1\xa0Therefore ,\xa01/( x + y ) = 1/5 => x + y = 5 --( 5 )\xa01/(x-y) = 1/1 => x - y = 1 ---( 6 )\xa0Add equations ( 5 ) and ( 6 ) ,\xa0We get\xa02x = 6\xa0=> x = 6/2 = 3 ,\xa0Put x = 3 in equation ( 5 ) , we get\xa03 + y = 5\xa0=> y = 5 - 3 = 2\xa0Therefore ,\xa0x = 3 , y = 2 | |
| 680. |
The angles of a quadrilateral are in AP .The greatest angle is double the least .find all the angles |
| Answer» The angle of a quadrilateral are in Arithmetic Progression, now as there are four angles, the greatest of all the angles is double to that of smallest angles. Then to find the 4 angles of the quadrilateral, the angles can be written as (x-3d)(x-d)(x+d)(x+3d)\xa0The common difference in A.P. = d\xa0Therefore, the sum of the four angles should be equal to that of 360°\xa0Hence, (x-3d)+(x-d)+(x+d)+(x+3d)=360°\xa04x=360°\xa0x=90°\xa0Let (x+3d) be more than (x-3d)\xa0Putting x = 90 degree in the equation x+3d = 2 (x-3d)\xa0Then, 90+3d = 2 (90-3d)\xa09d=90;d=10\xa0Hence, the common difference is 10\xa0Therefore putting the value of x=90° and d=10 in (x-3d)+(x-d)+(x+d)+(x+3d)=360°\xa0The angles are 60, 80, 100, and 120 of the quadrilateral. which are in AP. | |
| 681. |
What is thales theorm |
| Answer» If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio. Given : In ∆ABC , DE || BC and intersects AB in D and AC in E. ... between the same || lines.\xa0The\xa0intercept theorem, also known as Thales\' theorem (not to be confused with another theorem with the same name) or\xa0basic proportionality theorem, is about ratios of various\xa0line segments\xa0that are created if two intersecting lines are intercepted by a pair of parallels. | |
| 682. |
X3-3x2 |
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| 683. |
Evaluate :Sin² 63° + Sin² 27°____________________Cos ²17° + cos 73° |
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Answer» (1 answer ( sin ² 63° + sin ² 27° ) / ( cos ² 17° + cos ² 73° )= { sin ² 63° + sin ² ( 90 - 63 ) ° } / { cos ² 17° + cos ² ( 90 - 17 ) ° }= ( sin ² 63° + cos ² 63° ) / ( sin ² 17° + cos ² 17° )= 1 / 1= 1 |
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| 684. |
If sec θ+ tan θ = x, then sec θ = |
| Answer» Given:sec θ + tan θ = xTo find:Sec θ = ?Solution:sec θ + tan θ = x --- eq 1sec θ = x - tan θSquaring both sides - sec θ = x² - 2x² tan θ + tan²θsec² θ - tan² θ = x - 2xtan θ1 = x² - 2xtan θtan θ = x² - 1/2xSubstituting value in eq 1 sec θ + x² -1/2x = xsec θ = x² - x-1/2x = 2x² - x² +1/2xsec θ = x² + 1/2xAnswer : The value of sec θ = x² + 1/2x | |
| 685. |
Please tell me question no. And examples from maths textbook which are deleted of Ch 4 |
| Answer» HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.Deleted syllabus of CBSE Class 10 Mathematics\xa0\xa0 | |
| 686. |
X2+20x+100 |
| Answer» X2+20x+100=\xa0X2\xa0+ 10x + 10x +100= x ( x+ 10) + 10 (x + 10)= (x +10) (x +10)= ( x + 10)2 | |
| 687. |
The discriminant of quadratic equation 2x²-11 . Anybody please tell |
| Answer» Comparing with standard quadratic equation ax2\xa0+ bx + c = 0a = 2, b = 0, c = -11Discriminant D = b2\xa0– 4ac= (0)2\xa0– 4.2.(-11)= 0 + 88= 88 | |
| 688. |
Prime factor |
| Answer» That number which is divisible by 1 or itself is said to be prime number E.g_ 1,2 ,3,5,7,11......? | |
| 689. |
pg no. 214 probability question no. 43 |
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Answer» Contact me on whats app for any type of solution of phy,che,bio and math Post the question |
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| 690. |
Check whether the following quadratic equation x2-3x+5=(x+5)2 |
| Answer» x2-3x+5=(x+5)2x2\xa0- 3x+5 = x2\xa0\xa0+ 10x + 25x2 - x2 -3x - 10x + 5 - 25 = 0- 13x - 20 = 0It\'s not a quadratic equation. | |
| 691. |
Sin-2sin^3/2cos^3-cos=tanProve it |
| Answer» \xa0LHS = = RHS\xa0ORLHS=sin a -2sin ^3 a/ 2cos^3 a + cos a=sin a (1-2sin^2a)/ cos a ( 2cos2a+1)=sin a/cos a × 1-2sin2a/2cos2a+1=tan a × 2(1-sin2a)/2(1-sin2 a)+1=tana ×1/1=tan a=RHS | |
| 692. |
In the triangle with sides 25cm,5cm and 24 cm a right triangle? Give reasons for your answer. |
| Answer» No according to converse of Pythagoras theoremIf H2 = p2 + b2 then the triangle is right angle.....So, in this triangle square of hypotenuse is not equal to sum of the squares of perpendicular and base.....? That\'s allI hope it is helpful | |
| 693. |
Ganesh |
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| 694. |
Question of ch5 , Q If 8n=4n-n2 find an |
| Answer» We have given thatSum of the first n terms = 4n – n2⇒ Sn\xa0= 4n – n2Sn –1\xa0= 4(n – 1) – (n – 1)2= (4n – 4) – (n2\xa0+ 1 – 2n)= 4n – 4 – n2\xa0– 1 + 2n= 6n – n2\xa0– 5∴ nth term = Sn\xa0– Sn – 1= (4n – n2) – (6n – n2\xa0– 5)= 4n – n2\xa0– 6n + n2\xa0+ 5= 5 –2n. | |
| 695. |
Pythagores theorem |
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Answer» The sum of hypotenuse is equal to the sum of other two side of triangle is called pythagoras theorem. AC²=AB²+BC² \xa0Pythagoras TheoremStatement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.\xa0Given: A right triangle ABC right angled at B.To prove: AC2\xa0= AB2\xa0+ BC2Construction: Draw BD\xa0ACProof :We know that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.ADB\xa0ABCSo,\xa0(Sides are proportional)Or, AD.AC = AB2\xa0... (1)Also,\xa0BDC\xa0ABCSo,\xa0Or, CD. AC = BC2\xa0... (2)Adding (1) and (2),AD. AC + CD. AC = AB2\xa0+ BC2AC (AD + CD) = AB2\xa0+ BC2AC.AC = AB2\xa0+ BC2AC2\xa0= AB2\xa0+ BC2Hence Proved. Hypotenuse^2 = base ^ 2 + altitude ^ 2 |
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| 696. |
Pic ? |
| Answer» What? | |
| 697. |
Can i send puc |
| Answer» Why? | |
| 698. |
Only if i take a pic and send u will be able to answer the math doupt |
| Answer» But u can\'t share a pic here.. | |
| 699. |
the values after using trigonometry indenties can be with minus sign? |
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Answer» No, the trigonometric values are never calculated as minus, see the length will always be positive _+ root 1-sin² A .......why it can\'t be in minus |
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| 700. |
Prove bpt thereom |
| Answer» Please write clearly... | |