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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.
| 17251. |
-81 is the ___term of the AP : 21,18,15 |
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Answer» An=-81. 21,18,15......a=21.d=a2-a1=18-21=-3.An=a+(n-1)d.-81=21+(n-1)(-3).-81=21-3n-3.-81=18-3n.-81-18=-3n.-99=-3n.n=-99/-3.n=33.-81 is 33rd term. a²-a¹=............ a³-a²=............ a⁴-a³=............ a=? And d=? a=21, d=a2-a1 =18-21 =-3an=-81an=a+(n-1) d-81=21+(n-1) -3-81=21-3n+3-81=21+3-3n-81=24-3n-81-24=3n-105=3n-105/3=n-35=n |
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| 17252. |
Which digit would occupy the units place of 6¹⁰⁰. |
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Answer» 604,661, 760 6³=216 Hi can u tell me ir number sandhya Thanks for answering all my questions yogita ingle This is related to concept of numbers in the unit digits place of the power of natural number.The power of 6 any index repetation 6i.e\xa0(6)(anypower................) the last digit will be 6 only.Example1) 61=62) 62 = 363) 63\xa0= 216So, the last digit in the expansion of\xa06100\xa0is\xa06. |
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| 17253. |
Solve x and y : 2x – y – 3 =0 , 4x + y – 3 = |
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Answer» hope this will help u ? 2x−y=3\xa0------- (1)4x+y=3\xa0------ (2)From equation 1 :y=2x−3Substitute the value of\xa0y\xa0in equation 2 :\xa04x+y=34x+2x−3=36x=3+3X=66\u200b=1Now, Substitute\xa0x=1\xa0in equation 1 :\xa02x−y=32(1)−y=32−3=yY=−1Therefore the solution is:\xa0x=1\xa0and\xa0y=−1 |
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| 17254. |
If sin theta - x and sec theta -y, then find the value of cot theta |
| Answer» XY | |
| 17255. |
Find phone numbers in AP whose sum is 20 and the sum of whose square is 180 |
| Answer» Let the four numbers in A.P. be.Given,Given,When\xa0a\xa0= 5 and\xa0d\xa0= 2, we getWhen\xa0a\xa0= 5 and\xa0d\xa0= –2, we getThus, four numbers in A.P. are –1, 3, 7 and 11. | |
| 17256. |
Express 156 as the product of prime? |
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Answer» ANSWER:-=2²*3*13 2×2×3×13 |
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| 17257. |
What is substitution method ? |
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Answer» In the substitution method you solve for one variable, and then substitute that expression into the other equation. The important thing here is that you are always substituting values that are equivalent. For example: Sean is 5 years older than four times his daughter\'s age. A way to solve a linear system algebraically is to use the substitutions method. |
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| 17258. |
What is impossible event |
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Answer» Impossible event= 1-possible event The probability of an event which is impossible to occur is 0 . Such an event is called impossible event |
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| 17259. |
Check whether polynomial x-1 is a factor of polynomial x²+³ 8x²+²+19x-12 |
| Answer» No | |
| 17260. |
The probability that will rain tomorrow is 0.85.what is the probability that will not rain tomorrow |
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Answer» ANSWER:-P(E) = 0.85P( not E) = 1- P(E)P( not E) = 1 - 0.85P( not E) =0.15Therefore, probability that it will not rain tomorrow is 0.15.... 0.15 Let E be the event of happening of rainP(E) is given as 0.85E¯\xa0⟶ not happening of rainP(E¯) = 1 – P(E) = 1 – 0.85 = 0.15∴ P(not happening of rain) = 0.15 |
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| 17261. |
Check whether (1 + cot A + tan A) (sin A -Cos A )=sin A tan A - cot A cos A |
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| 17262. |
Can anyone tell me which maths chapter is going on in your school |
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Answer» Syllabus complete and now revision Syllabus completed Areas related to circle syllabus completed... Trigonometry |
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| 17263. |
All questions from height and distance |
| Answer» Yes | |
| 17264. |
How will you show that (17×11×2)+(17×11×5) is a composite number? Explain. |
| Answer» ( 17 x 11 x 2 ) + ( 17 x 11 x 5 )By taking ( 17 x 11 ) common ,= ( 17 x 11 ) ( 2 + 5 )= ( 17 x 11 ) ( 7 )= 17 x 11 x 7.If any number can be represented in the form of product of prime numbers then that number is composite number.Hence , ( 17 x 11 x 2 ) + ( 17 x 11 x 5 ) is a composite number. | |
| 17265. |
Which of the following holds true for all value of x? (2) |
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| 17266. |
Solve the equation X + y equal to 14 |
| Answer» 7+7=14 | |
| 17267. |
How to find five digit largest perfect cube |
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Answer» 99999 316 We know that the largest 5 digit number is 99999. So here we can see that 99999 is not a perfect square as it leaves 143 as the remainder while finding it s square root.So for finding 5 digit perfect square we will have to subtract 143 from 99999=> 99999-143 = 99856Hence\xa099856 is a perfect square whose square root is\xa0√99856 = 316. |
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| 17268. |
Example 9 from coordinate geometry |
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| 17269. |
A quadratic polynomial, whose zeroes are -3 and 4 is |
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Answer» (Ans) :- X square + 3x + 4 x^2-x-12=0 |
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| 17270. |
What is Pythagoras theorom. |
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Answer» the square of the hypotenuse side is equal to the sum of squares of the other two sides is know as phythagoras theorem Formula : Hypotenuse2\xa0= Perpendicular2\xa0+ Base2\xa0c2\xa0= a2\xa0+ b2 The square of two the hypotenuse is the sum of the other two sides is known as pythagoras theoremFORMULA:H2 =B2+P2H Square =B Square +P Square The square of two the hypotenuse is the sum of other two sides is known as pythagoras therom The square of the hypotenuse is equal to the sum of squares of other two sides is known as Pythagoras theorem |
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| 17271. |
gud afternoon everyday . have a nyc day ..?????? |
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Answer» Good afternoon Good afternoon. Good afternoon |
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| 17272. |
if sin A = 3/4 calculate sec A |
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Answer» SecA= 4√7/7 sin A = 3/4 P = 3H= 4B=? H² = P² + B²B²= 16 - 9B = √7 Sec A = H/B = 4/√74/√7 × √7/√7 = 4√7/7 So, Sec A = 4√7/7 Sin A= 3/2=p/hb²=h²-p²=b²=4-9=-5b=√-5SecA=h/b=2/-√5 Sec A = h / b = 4/√7 secA = 4/√7 |
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| 17273. |
Find the. Lcm of 6and 20 |
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Answer» LCM(6,20)= 60 60 6 = 21\xa0× 3120 = 22\xa0× 51LCM(6,20) = 22\xa0× 31\xa0× 51LCM(6,20) = 60 |
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| 17274. |
The comon point of a tangent to a circle and the circle is called__________ . |
| Answer» The common point of a tangent to a circle and the circle is called___Point of contact | |
| 17275. |
If α and β are the zeroes of the polynomial ax2+ bx + c, find the value of α2+ β2 |
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Answer» Value of α2+ β2. Value of\xa0α2+ β2.\xa0 |
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| 17276. |
Solve the following system of linear equations by substitution method: 2x – y = 2 x + 3y =15 |
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Answer» 2x-y=22x-2=y ....(1)x+3y=15 ....(2)By substituting the value from 1 and 2 eq.x+3(2x-2)=15x+6x-6=157x-6=157x = 21x= 21÷7x=3From 1 eq we can find out the value of y .....2x-2=y2(3)-2=y6-2 =yx=3 and y=4 2x - y = 2y = 2x - 2\xa0...(1)x + 3y = 15\xa0...(2)Substituting the value of y from (1) in (2), we get,x + 6x - 6 = 157x = 21x = 3From (1),y = 2\xa0\xa03 - 2 = 4 |
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| 17277. |
Use the Quadratic formula and find the roots of equation , 5 7 6 |
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| 17278. |
Determine whether the given quadratic equation has real roots or not. x 2- 2x + 1 = 0. |
| Answer» We have been given,\xa0x2\xa0- 2x\xa0+ 1 = 0Now we also know that for an equation ax2\xa0+ bx + c = 0, the discriminant is given by the following equation:D = b2\xa0- 4acNow, according to the equation given to us, we have,a = 1, b = -2 and c = 1.Therefore, the discriminant is given as,D = (-2)2\xa0- 4(1)(1)= 4 - 4= 0Since, in order for a quadratic equation to have real roots, D\xa0≥ 0.Here we find that the equation satisfies this condition, hence it has real and equal roots.Now, the roots of an equation is given by the following equation,x=-b± underroot D / 2aTherefore, the roots of the equation are given as follows,x=-(-2)±0 / 2(1)=2 / 2= 1Therefore, the roots of the equation are real and equal and its value is 1. | |
| 17279. |
Sin square + cos square |
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Answer» Sin Square θ + cos square θ= 1. 1 Sin^2A + Cos ^2A =1 1 |
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| 17280. |
A- b 2 |
| Answer» Hdjc ok | |
| 17281. |
If the mean of 5 observations X,X+2,X+4,X+6,X+8 is 11 , find the value of X |
| Answer» (X+X+2+X+4+X+6+X+8)/5=11(5X+20)/5=115X+20=11×55X=55-205X=35X=35/5X=7 | |
| 17282. |
Find HCF of 246 |
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Answer» 246÷2=123÷3=41÷41=1 246÷2=123÷3=41÷41=1 246÷2=123÷3=41÷41=1 |
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| 17283. |
Sin x=cos x find the value of x |
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Answer» 45 degrees Sin x =cos x(as given) =>We can write sin x as cos (90-x) =cosx =>now ,cos from both the sides will b canceled and you would be getting as (90-x) =x=>on solving, 90=x+x90=2x90/2=x45=xHence your answer is x=45 X=45° |
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| 17284. |
9.1 2nd questions |
| Answer» 2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.Solution:Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°BC = 8 mTo Find: Height of the tree, which is ABFrom figure: Total height of the tree is the sum of AB and AC i.e. AB+ACIn right ΔABC,Using Cosine and tangent angles,cos 30° = BC/ACWe know that, cos 30° = √3/2√3/2 = 8/ACAC = 16/√3 …(1)Also,tan 30° = AB/BC1/√3 = AB/8AB = 8/√3 ….(2)Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m. | |
| 17285. |
The decimal representation of 14587/2power1×5power4 will terminate after how many decimal places |
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Answer» After 4decimal places 14587/2×5⁴ =14587×2³/2×5⁴×2³ =116696/10⁴ =116696/10000 =11.6696 (It will terminate after 4 decimal places) =14587/2²×5⁴=14587×2²/2²×5⁴×2²=58348/10⁴=58348/10000=5.8348 (It will terminate after 4 decimal places) |
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| 17286. |
Exercise 3.4 questions 2ka 4 |
| Answer» (iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.Solution:Let the number of Rs.50 notes be A and the number of Rs.100 notes be BAccording to the given information,A + B = 25 ……………………………………………………………………….. (i)50A + 100B = 2000 ………………………………………………………………(ii)When equation (i) is multiplied with (ii) we get,50A + 50B = 1250 …………………………………………………………………..(iii)Subtracting the equation (iii) from the equation (ii) we get,50B = 750B = 15Substituting in the equation (i) we get,A = 10Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100. | |
| 17287. |
Given the HCF (306, 657) = 9, find LCM (306, 657) |
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Answer» LCM×HCF=Product of two numbers LCM×9=306×657 LCM×9=201042LCM=201042÷9 LCM=22338 Let the LCM = x. HCF *LCM =Product of two numbers. 9*x=306*657 x=306*657/9 x=34*657 x=22,338 |
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| 17288. |
Prove that the range ts drawn at the ends of a diameter of a circle are parallel |
| Answer» Let\xa0AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.Radius drawn to these tangents will be perpendicular to the tangents.Thus, OA ⊥ RS and OB ⊥ PQ∠OAR = 90º∠OAS = 90º∠OBP = 90º∠OBQ = 90ºIt can be observed that∠OAR = ∠OBQ\xa0(Alternate interior angles)∠OAS = ∠OBP\xa0(Alternate interior angles)Since\xa0alternate interior angles are equal, lines PQ and RS will be parallel. | |
| 17289. |
Find the coordinate of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3 |
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Answer» A=(-1, 7) B=(4, -3) Ratio=2:3 Let the coordinate point P(x, y) P(x, y) =(m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2) P(x, y) ={2×4+3×(-1) /2+3, 2×(-3) +3×7/2+3} P(x, y) =(8-3/5, -6+21/5) P(x, y) =(5/5,15/5) P(x, y) =(1, 3) Hence, coordinate point are P(1, 3) A=(-1, 7) B=(4, -3) Ratio=2:3Let the coordinate point P(x, y) P(x, y) =(m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2) P(x, y) ={2×4+3×(-1) /2+3, 2×(-3) +3×7/2+3}P(x, y) =(8-3/5, -6+21/5) P(x, y) =(5/5,15/5) P(x, y) =(1, 3) Hence, coordinate point are P(1, 3) (1,3) is the dividing point |
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| 17290. |
HCF of 39,91 |
| Answer» What is the formula of class mark | |
| 17291. |
What is the difference between step deviation method and assumed mean method in statistics |
| Answer» Both assumed mean method and step deviation method are ways to find mean without any complications...step deviation method is just an elaboration of assumed mean.1. Assumed Mean method:Step 1- Tabulate the continuous class intervals and frequency alongwith class marks. Class Mark =\xa0Upper Limit + Lower Limit 2Step 2- Assume any of the class marks to be\xa0your mean .Step 3-\xa0Take out the differences of the assumed mean and class marks.{ xi\xa0- a = di}Example if the class marks are 20, 30,40,50,60,70 and so on......and if the a =50 then,di\xa0is -30,-20,-10,0,10,20and so on.....Step 4: Multiply fi\xa0and di\xa0.Step5:\xa0\xa0 d =\xa0sum of fidi sum of\xa0fiStep 6: Mean = a + d | |
| 17292. |
Guys u all know, ab humara 50% syllabus delete hone wala haiiiii |
| Answer» Yes 50%? you are right | |
| 17293. |
1+sin theta /1-sin theta =1+2tan theta/cos theta +2tan^2 theta |
| Answer» | |
| 17294. |
Find discrimant of 3x^2-9x+20 |
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Answer» A quadratic equation is written in the form ax2+bx+c=0 Always change to this form firstThe discriminant is Δ=b2−4ac The solutions to an equation are called the \'roots\' and are referred to as αandβ The value of Δ tells us about the nature of the roots.If Δ>0⇒ the roots are real and unequal (2 distinct roots)If Δ>0 and a prefect square⇒ the roots are real, unequal and ................................................... rationalIf Δ=0⇒ the roots are real and equal (1 root)If Δ<0⇒ the roots are imaginary and unequalNote that if a or b are irrational, the roots will be irrational.−3x2+9x=4 ⇒3x2−9x+4=0 Δ=b2−4ac Δ=(−9)2−4(3)(4)=33 33>0 and is not a perfect squareTherefore, there are two roots which will be real and unequal. -159 by b² - 4ac |
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| 17295. |
For what value of k, the pair of equations 2x+3y+5=0 and kx+4y=10 has a unique solutions |
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Answer» 64 The equation will have unique solution for all number except 8/3 » 2x + 3y + 5 = 0» kx + 4y = 10➡ kx + 4y - 10 = 0we have to find the value of\xa0k\xa0such that the pair of equations have unique solution.we know that, when pair of equations have unique solution,a1/a2 ≠ b1/b2➡ 2/k ≠ 3/5just put any value of\xa0k\xa0so that it doesn\'t get equal to 3/5.you can put 1, 2 etc.for example, k is 2, then a1/a2 = 2/2 = 1LHS ≠ RHS |
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| 17296. |
. Find the roots of the quadratic equation 2x2+x-4=0 by the method of completing thesquare. |
| Answer» 2x^2 + x - 4 = 0 x = -b +\\- √b^2 - 4ac _____________________ 2a x = -1 +\\- √1^2 - 4*2*4 ____________________ 2*2 x = -1 +\\- √1-32 _______________ 4 x = -1 +\\- √31 ___________ 4 Therefore : x = -1 + √31 x= -1 - √31 _________ ________ 4 4 | |
| 17297. |
1+23 |
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Answer» 24 24 24 24 24 |
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| 17298. |
The 10th term of an A.P. is -4 and the 22nd term is -16. Find its 38th term. |
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Answer» -32 -‐32 |
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| 17299. |
The sum of three consecutive numbers in an A.P. is -9 and their product is 21. Find the numbers. |
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| 17300. |
Find the three numbers in A.P., whose sum is 15 and the sum of whose squares is 77. |
| Answer» Let the three numbers in AP be a-d, a, a+dThen,\xa0sum = a-d+a+a+d=153a = 15So, a = 5Also, Sum of squares = 83(a-d)2+a2+(a+d)2=83(5-d)2+25+(5+d)2=8325+d2-10d+25+25+d2+10d=8375+2d2=832d2=8d=+2 or -2When d = 2,The numbers are 3, 5, 7When d = -2,The numbers are 7, 5, 3 | |