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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.
| 751. |
The root of equation 2x²-6x+7=0 are |
| Answer» Ge wanna | |
| 752. |
If H.C.F (25,15)= 4X+1 then x= |
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Answer» 25 = 5\xa0× 515 = 5\xa0× 3H.C.F (25,15) = 5H.C.F (25,15)= 4X+14x + 1 = 54x = 5 - 1\xa04x = 4x = 4/4x = 1 Find hcf and substitute |
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| 753. |
Chapter 4 ke important questions |
| Answer» What of which subject baby | |
| 754. |
Find the zeroes of the polynomial 5 ?2– 4 – 8x |
| Answer» Dear please write the question properly ???? | |
| 755. |
Prove that sin theta -2sin^3theta /2cos^3 theta - cos theta = tan theta |
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| 756. |
Show that 3√2 is not a irrational |
| Answer» Let us assume, to the contrary, that\xa03\xa0√\xa02\u200b\xa0is\xa0rational. Then, there exist co-prime positive integers\xa0a\xa0and\xa0b\xa0such that3\xa0√\xa02\u200b= a/b⇒ √\xa02 \u200b= a/3b\xa0⇒ √\xa02\u200b\xa0is rational ...[∵3,a\xa0and\xa0b\xa0are integers∴ 1/3b \u200bis a rational number]This contradicts the fact that √\xa02\u200b\xa0is irrational.\xa0So, our assumption is not correct.Hence,\xa03\xa0√\xa02\u200b\xa0is an irrational number. | |
| 757. |
Find a point on y axis which is equidistant from the point A (6,5 )and b (- 4,3) |
| Answer» The point P(0,9) is equidistant from the points A(6,5) and B(-4,3).Step-by-step explanation:Let the point on y axis is P(0,y) which is equidistant from the points A(6,5) and B(-4,3). The distance between PA is equal to PB such that,On squaring,So, the point P(0,9) is equidistant from the points A(6,5) and B(-4,3). | |
| 758. |
6/2 (1+2) |
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Answer» 6/2×3=3×3=9 =6/2 × 3=18/2=9 |
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| 759. |
6devide2(1+2 |
| Answer» Same question ???? | |
| 760. |
State AAA similarty creteripn |
| Answer» AAA similarity theorem or criterion:If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and the triangles are similarIn ΔABC and ΔPQR,\xa0∠A = ∠P , ∠B = ∠Q , and\xa0∠C =\xa0∠R then AB PQ\xa0= BC QR\xa0= ACPRand ΔABC ∼ ΔPQR.Given: In\xa0ΔABC and\xa0ΔPQR,\xa0∠A =\xa0∠P,\xa0∠B =\xa0∠Q,\xa0∠C =\xa0∠R.To prove: AB PQ\xa0= BC QR\xa0= ACPRConstruction : Draw LM such that PL AB\xa0=\xa0PM AC\xa0.Proof: In\xa0ΔABC and\xa0ΔPLM,AB = PL and AC = PM (By Contruction)∠BAC =\xa0∠LPM (Given)∴\xa0ΔABC\xa0≅\xa0ΔPLM (SAS congruence rule)∠B =\xa0∠L (Corresponding angles of congruent triangles)Hence\xa0∠B =\xa0∠Q (Given)∴\xa0∠L = ∠Q\xa0LQ is a transversal to LM and QR.Hence ∠L = ∠Q\xa0(Proved)∴ LM\xa0∥ QR PL LQ\xa0= PM MR\xa0 LQ PL\xa0= MR PM (Taking reciprocals) LQ PL\xa0+ 1 = MR PM\xa0+ 1 (Adding 1 to both sides) LQ+PL PL\xa0= MR+PM PM\xa0 PQ PL\xa0= PR PM PQ AB\xa0= PR AC (AB = PL and AC =PM) AB PQ\xa0= AC PR (Taking Reciprocals) ............... (1) AB PQ\xa0= BC QR AB PQ\xa0= AC PR\xa0= BC QR\xa0∴\xa0ΔABC\xa0~\xa0ΔPQR | |
| 761. |
10th class NCERT test book exercise 1.2 5th sum |
| Answer» Check whether 6n\xa0can end with the digit 0 for any natural number\xa0n.Ans.\xa0If any number ends with the digit 0, it should be divisible by 10.In other words, it will also be divisible by 2 and 5 as\xa0Prime factorisation of\xa0It can be observed that 5 is not in the prime factorisation of.Hence, for any value of\xa0n,\xa0will not be divisible by 5.Therefore,\xa0cannot end with the digit 0 for any natural number\xa0n. | |
| 762. |
D=3+2√2-3+√2=? |
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Answer» 3root2 3+2√2-3+√2= 2√2+√2= 3√2 3+2√2-3+√22√2+√22+1(√2) 3√2 3+2√2-3+√2=3-3+2√2+√2=2√2+√2= (2+1) √2= (2+1) √2 |
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| 763. |
The sum of three terms of an Ap is -3 & there product is 8. find Ap. |
| Answer» Let the terms be (a-d), a, (a+d) Sum=3a=-3 So a=-1Now, product = (a+d)a(a-d) = 8 (d-1) (-1) (-1-d) =8 (d-1) (-1-d) =-8 -d^2 +d -d +1=-8 d^2= 9So d= -3 or 3So AP will be -1, -4, -7\'......... ORAP will be -1, 2,5 | |
| 764. |
The sum of three no of an AP is27 and their product is 405 . Find the number |
| Answer» Let the numbers be a - d, a, a + d which are in AP.(a - d) + (a) + (a + d) = 27 .........(1)(a - d)(a)(a + d) = 405 ..........(2)(1) => 3a = 27a = 9(2) => (a2 - d2)(a) = 405[(9)2 - d2] (9) = 405(81 - d2) (9) = 405729 - 9d2 = 4059d2 = 729 - 4059d2 = 324d2 = 36d = 6Now, substitute the values in the 1st equation.(a-d) = 9 - 6 = 3a = 9(a+d) = 9 + 6 = 15Hence the numbers are 3, 9 and 15. | |
| 765. |
3x+2y=12x+3y=4 |
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Answer» Write the value of k for which the system of eq x+ y -4=0,2x+ky-3=0 has no sok 3x+2y=112x+3y=4by adding them5x+5y=15x+y=3x=3-yput the value of x3x+2y=113(3-y)+2y=119-3y+2y=119-y=11-y=11-9-y=2y=-2put the value of yx=3-yx=3-(-2)x=3+2x=5hence x=5 and y=-2 |
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| 766. |
Formulas for all the chapters |
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| 767. |
Solve:x+y=a+bax-by=a²-b² |
| Answer» x + y = a + b …(i)ax – by = a2\xa0– b2…(ii)To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.Lets multiply equation (i) by b, so that variable y in both the equations have same coefficient.Recalling equations (i) & (ii),x + y = a + b\xa0[×bax – by = a2\xa0– b2⇒\xa0bx + ax = ab + a2⇒\xa0(b + a)x = a(b + a)⇒\xa0x = aSubstitute x = a in equations (i)/(ii), as per convenience of solving.Thus, substituting in equation (i), we geta + y = a + b⇒\xa0y = bHence, we have x = a and y = b. | |
| 768. |
Last year paper real |
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| 769. |
The larger of two supplementary angles exceed the smaller by 18 degrees find them |
| Answer» Let the angle be x\xa0other angle (it\'s supplement) =( 180-x)x+18= (180-x)2x = 162x = 162/2 = 81\xa0other angle(its supplement) = 180- x => 180-81 => 99 | |
| 770. |
Which term of the ap 121 |
| Answer» Suzmvggm | |
| 771. |
The first four terms of the sequence an = 2n +3 are |
| Answer» Solution:an = 2n + 3n = 1an = 2n + 3a1 = 2(1) + 3a1 = 2 + 3a1 = 5n = 2an = 2n + 3a2 = 2(2) + 3a2 = 4 + 3a2 = 7n = 3an = 2n + 3a3 = 2(3) + 3a3 = 6 + 3a3 = 9n = 4an = 2n + 3a4 = 2(4) + 3a4 = 8 + 3a4 = 11 | |
| 772. |
The roots of the equation a²b²x²-(4b⁴-3a⁴)x-12a²b²=0 |
| Answer» A²b²x² - (4b⁴ - 3a⁴)x - 12a²b² = 0⇒a²b²x² - 4b⁴x + 3a⁴x - 12a²b² = 0⇒b²x(a²x - 4b²) + 3a²(a²x - 4b²) = 0⇒(b²x + 3a²)(a²x - 4b²) = 0⇒x = 4b²/a², -3a²/b²now, using quadratic formula ,x = {(4b⁴ - 3b⁴) ± √{(4b⁴ -3a⁴)²+48(a²b²)²}}/2a²b²= {(4b⁴ - 3a⁴) ± √{(4b⁴ + 3a⁴)²}}/2a²b² [ use formula, (x-y)²+4xy=(x+y)²]= {4b⁴ - 3a⁴ ± (4b⁴ + 3a⁴)}/2a²b²= 4b²/a², -3a²/b²LE BHAI | |
| 773. |
Lcm of 12576 |
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Answer» lcm\xa0(12,576; 4,052) = 12,739,488:\xa0least common multiple, calculated. The numbers have common prime factors. \t125762628823144215722786239331311311\tLCM\xa0= 25.\xa031.\xa01311\xa0=\xa012576 |
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| 774. |
Simplify question is -17/5,7/5,13/4 |
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| 775. |
The value of k for which the numbers x, 2x+k, 3x+6 are in A. P. Is? |
| Answer» It is given that ,x\xa0,2x+k\xa0and\xa03x+6\xa0are\xa0three\xa0consecutive\xa0terms\xa0.we\xa0know\xa0that\xa0,Difference\xa0of\xa0any\xa0two\xa0consecutive\xa0terms\xa0is\xa0equal\xa0in\xa0A.P\xa0Now,(2x+k)-x\xa0=\xa0(3x+6)-(2x+k)=>\xa02x+k-x\xa0=\xa03x+6-2x-k=>\xa0x+k\xa0=\xa0x+6-k=>\xa0k+k\xa0=\xa0x+6-x=>\xa02k\xa0=\xa06=>k\xa0=\xa06/2=>\xa0k\xa0=\xa03Therefore,If\xa0k\xa0=\xa03\xa0then the given three consecutive terms x,2x+k and 3x+6 are in\xa0A.P | |
| 776. |
NCERT Question is exercise 3.5 |
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| 777. |
prove that Under root 2 |
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Answer» Thank\'s gourav seth Kya re kya bol raha raha hai be Thank bhai MC kahi ke lodu Let\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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| 778. |
If costheta - sintheta =x , sintheta + costheta =y prove that x2 + y2 =2 |
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Answer» Hk honey ki behen chod Hey |
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| 779. |
3.9×10 |
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Answer» 39 39 30.9 Hey 39 |
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| 780. |
Ex2.2Q2 |
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Answer» 4th part I can explain Write question also Hello 2 |
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| 781. |
Ex 5.3 me konse question delete hai plz tell me that question number. Immediately. |
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Answer» Question number 15 to 20 delete ho gaya ha 15 to 20 delete hoa hai Hey Do all |
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| 782. |
If the answer is given is 25 and i have got 26 then (surface area chapter) marks would be rewarded? |
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Answer» Hy No the ans is totally wrong but u would get half |
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| 783. |
Q7. If di = xi – A, = 25, A = 250, = 250, then find the value of . |
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| 784. |
in ∆ ABC \' |
| Answer» Three side | |
| 785. |
In a ∆abc; |
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| 786. |
Why the (a+b)squre |
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Answer» (a+b)=a²+2ab+b² Hey, here is the answer(a+b)²= a²+b²+2ab Some more ↓↓(a-b)²=a²+b²-2aba²-b²= (a+b)(a-b)Mark it as expert answer!!;););) (a + b)2 = (a + b) (a + b)= a(a+b) + b(a +b)= a2 + ab + ab + b2= a2 + 2ab + b2 |
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| 787. |
What is credit card. How we can use it. ?? |
| Answer» A\xa0credit card\xa0is issued by a credit card provider, like Capital One, and they are designed to pay for things in shops or online. You can also use credit cards for\xa0balance transfers\xa0and taking out\xa0cash\xa0(also known as cash advance or cash withdrawal) from an ATM. | |
| 788. |
Please tell all the formulas used in Arithmetic progressions |
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Answer» Here is your answer @diya GuptaSeries - a,a+d,a+2d,a+3d,......an= a+(n-1)dSn = n/2 (a+an)Sn= n/2(2a+(n-1)d)Here are all Mark it as expert answer;);) Refer to ncert mathematics book class 10 ch5 |
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| 789. |
Can anybody explain me about trigometrical identities |
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| 790. |
2×2= |
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Answer» 4 4 4 4 4 |
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| 791. |
ax/b -by/a=a+b,ax -by =2ab |
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| 792. |
Show that nsqaur is divisible by 8 if n is odd positive integer |
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Answer» Any odd positive integer\xa0n\xa0can be written in form of 4q\xa0+ 1 or 4q\xa0+ 3.\xa0If\xa0n\xa0= 4q\xa0+ 1, when\xa0n2\xa0- 1 = (4q\xa0+ 1)2\xa0- 1 = 16q2\xa0+ 8q\xa0+ 1 - 1 = 8q(2q\xa0+ 1) which is divisible by 8.If\xa0n\xa0= 4q\xa0+ 3, when\xa0n2\xa0- 1 = (4q\xa0+ 3)2\xa0- 1 = 16q2\xa0+ 24q\xa0+ 9 - 1 = 8(2q2\xa0+ 3q\xa0+ 1) which is divisible by 8.Here is the answerMark it as expert answer;);) Show that n2\xa0- 1 is divisible by 8, if n is an odd positive integer.Any odd positive integer\xa0n\xa0can be written in form of 4q\xa0+ 1 or 4q\xa0+ 3.\xa0If\xa0n\xa0= 4q\xa0+ 1, when\xa0n2\xa0- 1 = (4q\xa0+ 1)2\xa0- 1 = 16q2\xa0+ 8q\xa0+ 1 - 1 = 8q(2q\xa0+ 1) which is divisible by 8.If\xa0n\xa0= 4q\xa0+ 3, when\xa0n2\xa0- 1 = (4q\xa0+ 3)2\xa0- 1 = 16q2\xa0+ 24q\xa0+ 9 - 1 = 8(2q2\xa0+ 3q\xa0+ 1) which is divisible by 8. |
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| 793. |
Find the value of k if p(4 -2)is the midpoint of the line segment joining the points A(5k 3) |
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| 794. |
Find the distance of the point -5,-6 from Y-axis |
| Answer» The point (-5,-6) lies on x\'= 5 unit and y\'= 6 units So,the distance from y axis is 5 unitsMark it as expert answer;);) | |
| 795. |
Explain why 7×13×15×5×6+6 is a composite number |
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Answer» 7×13×15×5×6+6=6(7×13×15×5+1)=6×6826=40956Here is the answer Mark it as expert answer;);)Here is the short trick for mcqIf these type of questions answer is always composite ? Because 6 is a factor solution = 7×13×15×5×6+6 6(7×13×15×1+1) 6(1365+1) 6×1366 hence it is composite number becouse 6 is it factor |
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| 796. |
FOR WHAT VALUE OF K WILL THE EQUATIONS X+2Y+KY+14=0 REPRESENT COINCIDENT LINES ? |
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Answer» Give other equation to solve it incomplete question You have given only one equation where\'s other it\'s so easy So do yourself You are in 10th Do your self...... Hey |
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| 797. |
What is answer of 2^0? |
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Answer» 1 Anything power 0 is 1 1 1 1 1 |
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| 798. |
find the mean of the following distribution: x|10|30|50|70|89f|7|8|10|15|10 |
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Answer» http://cbseacademic.nic.in/SQP_CLASSX_2020-21.html let 5667 fhtuufhujjhgghguvhjjghhop |
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| 799. |
Proof that x°=1 |
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Answer» No need to proof as it is already is a axiom The process which our body parts maintain and necessary for survival is known as life process The process which our body part s What is life process |
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| 800. |
x square+12x+4=0 by completing square method |
| Answer» x=212−160\u200b\u200b=6−210\u200b=−0.325x=212+160\u200b\u200b=6+210\u200b=12.325 | |