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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.
| 42501. |
Find all the zeros of f(x)=2x^-8 |
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Answer» If x=x+j X=4,x=-4 |
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| 42502. |
Write the quadratic polynomial whose zeroes is 2 and sum is -8 |
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| 42503. |
Choose the correct number in place of"?\'.?\' के स्थान पर सही संख्या का चयन कीजिए5476283114169144361 |
| Answer» 3 | |
| 42504. |
Find the least number which when divided by 35 56 and 91 leaving the same remainder 7 in each case |
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Answer» The smallest number which when divided by 35, 56 and 91 =\xa0LCM(35,56,91)35=5×756=23×791=7×13LCM=23×5×7×13=3640The\xa0smallest\xa0number\xa0that\xa0when\xa0divided\xa0by\xa035,\xa056,\xa091\xa0leaves\xa0a\xa0remainder\xa07\xa0in\xa0each\xa0case\xa0=\xa03640\xa0+\xa07\xa0=\xa03647. Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91. Prime factorization of 35, 56 and 91 is: 35 = 5 × 7\xa0& 56 = 23 × 7\xa0& 91 = 7 × 13 LCM =23 × 5 × 7 × 13 = 3640 Least number which can be divided by 35, 56 and 91 is 3640. Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647. |
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| 42505. |
Using the Pythagoras answer the following questions:a. What is the length of the broken part? |
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Answer» 1) c.25m2) a.40m3) c.15m4) a.100m25) a.60m Suresh is having a garden near Delhi. In the garden, there are different types of trees and flower plants. One day due to heavy rain and storm one of the trees got broken as shown in the figure.The height of the unbroken part is 15 m and the broken part of the tree has fallen at 20 m away from\xa0the base of the tree.Using the Pythagoras answer the following questions:1) What is the length of the broken part?A) 15 mB) 20 mC) 25 mD) 30 m2) What was the height of the full tree?A) 40 mB) 50 mC) 35 mD) 30 m3) In the formed right-angle triangle what is the length of the hypotenuse?A) 15 mB) 20 mC) 25 mD) 30m4) What is the area of the formed right angle triangle?A) 100 m2B) 200 m2C) 60 m2D) 150 m25 )What is the perimeter of the formed triangle?A) 60 mB) 50 mC) 45 mD) 100 m\xa0Answer:1)C2)A3)C4)D5)A |
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| 42506. |
Give me plz the easiest way of solving the problem of irrational or rational sums ? |
| Answer» https://rational.irrational.def/..Examples./vj/72411575258?pwd=VTdIWlJRRDFsaGorZFIvanQxVGNFZkhli | |
| 42507. |
Find the area of triangle whose vertices are : (1). (2,3), (-1,0), (2,-4) |
| Answer» Let\xa0A(2,3),\xa0B(−1,0)\xa0and\xa0C(2,−4)\xa0are the vertices of\xa0△ABCArea of triangle= 1/2 \u200b[x1\u200b(y2\u200b−y3\u200b)+x2\u200b(y3\u200b−y1\u200b)+x3\u200b(y1\u200b−y2\u200b)]Here\xa0(x1\u200b,y1\u200b)=(2,3)(x2\u200b,y2\u200b)=(−1,0)(x3\u200b,y3\u200b)=(2,−4)= 1/2 \u200b[2(0+4)−1(−4−3)+2(3−0)]= 1/2 \u200b[8+7+6]= 21/2 \u200b\xa0sq.unit | |
| 42508. |
A milkman |
| Answer» | |
| 42509. |
What is nature of roots |
| Answer» Nature of RootsA quadratic equation ax2\xa0+ bx + c = 0 has(i) two distinct real roots, if b2\xa0– 4ac > 0,(ii) two equal real roots, if b2\xa0– 4ac = 0,(iii) no real roots, if b2\xa0– 4ac < 0.Since\xa0b2\xa0– 4ac\xa0determines whether the quadratic equation\xa0ax2 +\xa0bx\xa0+\xa0c\xa0= 0 has real roots or not,\xa0b2\xa0– 4ac\xa0is called the\xa0discriminant\xa0of this quadratic equation | |
| 42510. |
If the line given by 2x+5y+a=0 and 3x+2ky=b are coincident, then the value of k is |
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| 42511. |
If the quadratic equation px^2-2√5px+15=0 has two equal roots,then find the value of p. |
| Answer» Given quadratic equation is, px² -2√5 px + 15 = 0Compare px² -2√5 px + 15 = 0 with ax² + bx + c = 0a = p , b = -2√5p , c = 15We know that , If the roots of the quadratic equation are equal , then it\'s discriminant (D) equals to zero.discriminant = 0b² - 4ac = 0( -2√5 p )² - 4× p × 15 = 020p² - 60p = 020p ( p - 3 ) = 0p - 3 = 0p = 3\xa0\xa0\xa0\xa0\xa0 | |
| 42512. |
For what value of K the roots of the equation x^2+4x+K=0 |
| Answer» Answer:The value of k must be less than or equal to 4.Step-by-step explanation:The given quadratic equation isA quadratic equation\xa0\xa0has real roots ifWe have,Divide both sides by 4.Therefore the value of k must be less than or equal to 4. | |
| 42513. |
find the sum of 1/15,1/12,1/10,......to 11 terms |
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Answer» The given AP is 1/15, 1/12, 1/10, ......... The given AP is 1/15, 1/12, 1/10, ......... |
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| 42514. |
if d=HCF (48,72) find the value of d is |
| Answer» If d is the hcf of 48 and 72 so the value of d will be the hcf of laterfactor of 48 is2,3,4,6,8,12,24 and 48factor of 72 is2,3,4,6,8,9,12,24,36 and 72so the hcf is 24 and d= 24ord = HCF ( 48, 72 )48 = 2 * 2 * 2 * 2 * 372 = 2 * 2 * 2 *3 * 3so HCF = 2 * 2 * 2 * 3➡ 24Answer :- d = HCF ( 48, 72 ) = 24 | |
| 42515. |
Prove root3+2 is irrational |
| Answer» Let 2+√3 is a rational number.A rational number can be written in the form of p/q.2+√3=p/q√3=p/q-2√3=(p-2q)/qp,q are integers then (p-2q)/q is a rational number.But this contradicts the fact that √3 is an irrational number.So,our supposition is false.Therefore,2+√3 is an irrational number.Hence proved. | |
| 42516. |
Sum of n terms |
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Answer» If 4is zero of polynomial of Sn=n/2{2a+(n-1)d} S=n/2[2a+(n-1)d] |
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| 42517. |
find the sum of 5+(-41)+9+(-39)+13+(-37)+17+.....+(-5)+81+(-3) |
| Answer» The series can be written as(5 + 9 + 13 + .... + 81) + [(- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)]For the series (5 + 9 + 13 + ... + 81)a = 8d = 9 - 5 = 5and a(n) = 81Then, a(n) = 5 + (n - 1)4 = 81or, (n - 1)4 = 76⇒\xa0n = 20S(n) = 20/2(5 + 81)=\xa0860For Series (- 41) + (- 39) + (- 37) + (- 35) ... (- 5) + (- 3)a(n) = - 3a = - 41d = 2Then, a(n) = - 41 + (n - 1)(2)⇒\xa0n = 20Now, S(n) = 20/2[- 41 + (- 3)]= - 440Sum of the series = 860 - 440=\xa0420Hence, the sum of the following series is 420. | |
| 42518. |
find the sum of all multiples of 9 lying between 300 and 700 |
| Answer» The multiples of 9 lying between 300 and 700 are 306, 315,………, 693.This is an AP with a = 306, d = 9 and l = 693.Suppose these are n terms in the AP. Then, | |
| 42519. |
Triangle ABC is an equilateral triangle of side 2a units. Find each of its altitudes |
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Answer» Mast questions Anjali Altitude be √3a Answer:Each altitude of the equilateral triangle is equal to\xa0Thus,To find:Each of its altitudes = ?Solution: Given : ABC is an equilateral triangle of\xa0side 2a.In equilateral triangle, △ADB = △ADC& ∠ADB = ∠ADC (Both 90 degree as AD ⊥ BC)Thus, △ ADB ≅ △ ADC (By R. H. S. Congruency)As per CPCT i.e. they are corresponding parts of congruent triangle,BD = DC Using Pythagoras theorem,Hence,Thus, the\xa0altitudes\xa0of the given equilateral triangle is equal to\xa0 |
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| 42520. |
Which term of ap : 3,8,13,18,...is18 |
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Answer» Given a=3,d=8-3=5,an=18An=a+(n-1)×d18=3+(n-1)×518-3/5=n-13=n-1N=3+1=4Ans:4th term of ap is 18 4th term of AP 4 th The given sequence is 3, 8, 13, 18, ....78.\'a\' = 3, \'d\' = 5, last term = 78, \'n\' = ?tn\u200b=a+(n−1)d⟹78=3+(n−1)5⟹78=3+5n−5⟹78=5n−2⟹5n=80orn=16.The 16th term is 78. |
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| 42521. |
Q.no-1 if tan2A=cot(A-18),where 2A is an acute angle find the value of A |
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Answer» tan2A = cot(A-18)tan2A = tan722A= 72A= 72/2A= 36 Value of A is 36 A=36° A ki value 36 Answer is 36 |
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| 42522. |
Please help me.Its urgent. |
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Answer» Kya hua Yes tell me what happen Tell what happened Question? Tell what happened |
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| 42523. |
If tan theta =5/6. Then, (1-cos theta)(1+cos theta)/(1-sin theta)(1+sin theta)=? |
| Answer» You can use phythogorus theorem to sin and cos value than pit it in the equation then simplify it... you will get answers. | |
| 42524. |
Find the maximum volume of a cone that can be curved out of a solid hemisphere of a radius r |
| Answer» Radius of the base of cone= Radius of the hemisphere =rHeight of cone= Radius of the hemisphere Volume of the cone1/3pie*(r)²*r=1/3pie*(r)³ | |
| 42525. |
If 3x-4y+14=0 and 5x-3y+5=0 have a common solution (a, b) then the value of a is ____ |
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| 42526. |
Proof of SSS |
| Answer» Show the proof of SSS | |
| 42527. |
PQ is a tangent at point C at circle with center o.if AB is a diameter and ACB-30 find PCA |
| Answer» In the given figure,In Δ ACO,OA=OC (Radii of the same circle)Therefore,ΔACO is an isosceles triangle.∠CAB = 30°\xa0(Given)∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)∠PCO = 90° …(radius is drawn at the point ofcontact is perpendicular to the tangent)Now ∠PCA = ∠PCO – ∠CAOTherefore,∠PCA = 90° – 30° = 60° | |
| 42528. |
Find the sum 2,7,12,......,to 100 terms |
| Answer» Here\xa0a=2,n=100And\xa0d=7−2=5Sn\u200b= n/2 \u200b[2a+(n−1)d]S100 \u200b= 100/2 \u200b[2(2)+(100−1)5]\xa0= 50[4 + 495 ]= 50(499)=24950 | |
| 42529. |
Find the quadratic polynomial whose the zeroes are -2√3 and √3÷2 |
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Answer» 2.59807621135 -2.59807621135 |
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| 42530. |
Find the distance of the point (-3,-4) from the x-axis ( in units) |
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Answer» 4 unit When we are asked to calculate the distance between a point and any of the axes, we calculate the perpendicular distance between them.Therefore, the point (-3,-4) is 3 units away from the y-axis and 4 units from the x-axis. This is because in case of either axes the co-ordinate of the other axis is zero.Therefore the distance between the point P(-3,-4) and x-axis is 4 units. For x axis our point x2=-3 & y2=0 D=√[-3-(-3)]^2. + [0-(-4)]^2 =√(0)^2. + (4)^2 =√16 D=4 unit For x axis our point x2=-3 & y2=0D=√[-3-(-3)]^2. + [0-(-4)]^2 =√(0)^2. + (4)^2=√16D=4 unit 4 |
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| 42531. |
Sin squared theta divided by cos squared theta |
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Answer» [email\xa0protected]/[email\xa0protected][email\xa0protected] sec squared theta minus cos square theta whole divided by sin square theta minus cos square theta into cos square theta minus sin square theta divided by sin square theta minus cos square theta |
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| 42532. |
Find the solutionof the equation x2_6x+5=0 |
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Answer» X^2 - 6x +5 =0X^2 - 5x - x +5=0X(x -5 ) -1(x-5)=0(X-1)(x-5)=0X=1 or x=5 X2_5x_x+5X (x_5)_1 (x_5)(X_5) (x_1)X=5 or x=1 X = 5;1 X²+6x+5X²+5x+x+5X(x+5) +1(x+5) (x+1) (x+5) takex+1=0 and x+5=0 Hence x=-1, -5 given,\xa0x2−6x+5=0x2−5x−x+5=0x(x−5)−1(x−5)=0(x−5)(x−1)=0x=5,1 |
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| 42533. |
Proof That 2 root 5 irrational |
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Answer» Let as assume that to the contrary ,that 2√5 is rational. 2√5 =a/b (a and b is coprime and integers and q is not equal to 0 ) √5=a/2b a and b are integers , a/2b is rational , and so √5 is rational .But this contradicts the fact that √5 is irrational .This contradiction has arisen becz of our assumption was wrong that 2√5 is rational.Hence 2√5 is irrational number. Let 2√5 be rational number.it can be written in the form of p/q.2√5=p/q√5=p/q. 1/2We know that √5 is irrational number.It has arisen our assumtion is worng that 2√5 is rational number.Hence 2√5 is irrational number. |
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| 42534. |
If alpha and beta be the zeores and polynomial 2x²-5x+8 evaluate alpha/beta+ beta/alpha |
| Answer» α +β= (-b/a),α+β=5/2,αβ=c/a,αβ=8/2=4, (α+β)^2 = α^2+β^2+2αβ,(5/2)^2=α^2+β^2+825/4-8=α^2+β^2,α^2+β^2=(-7/4)Then, α/β+β/α =α^2+β^2/αβ=(-7/4)/4=(-7/16) | |
| 42535. |
cot θ(sec θ−1)2(1+sin θ)sec θ(sin θ−1)2(1+sec θ) |
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| 42536. |
One specific you is in the shape of a come mounted on a cylinder total height of conicalpartis |
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| 42537. |
find distance of (2,3) from origin |
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Answer» Distance of (2,3)from the origin =root under 2square+3squareRoot under 13 Ans. Under root 13 |
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| 42538. |
Prove that ( 2,2),(-2,1),(5,2) are the vertices of right angle triangle |
| Answer» Distance between (2,2)&(-2,1) is root 17Distance between (-2,1)& (5,2) isRoot 50Distance between (5,2)&(2,2) isRoot 9So as per this Question is mistake | |
| 42539. |
If bcosx = a then prove cosecx + cotx = √ b + a / b - a |
| Answer» Given bcosx=aCosx=a/b. Equation (1)From identity Sinx=√1-cos²xSinx=√1-(a/b)²Sinx=√b²-a²/b. Equation (2)Now Cosecx+cotx=1/sinx. + Cosx/sinx =(1+cosx)/sinxFrom equation (1)&(2)=(1+a/b)÷(√b²-a²)/b=(b+a)÷(√b²-a²)=√(b+a)²÷(√b²-a²)=√(b+a)²/(b²-a²)=√(b+a)(b+a)/(b+a)(b-a)=√(b+a)/(b-a)Hence proved | |
| 42540. |
Find the value of 2cos67°/sin23°+tan 40°/cot 50°+cos0° |
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Answer» 2cos(90_23)/sin23+tan (90_50)/cot50+1 ( cos0=1)2sin23/sin23+cot50/cot50+12×1+1+14 ans. This topic is deleted |
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| 42541. |
Sec theta/ cot theta+tan theta = |
| Answer» [email\xa0protected]/[email\xa0protected][email\xa0protected] = 1/[email\xa0protected]([email\xa0protected]/[email\xa0protected][email\xa0protected]/[email\xa0protected]) = 1/[email\xa0protected]/(sin²@+cos²@)/[email\xa0protected]@ (:- sin²@+cos²@=1) 1/[email\xa0protected]/1/[email\xa0protected]@ = [email\xa0protected]@/[email\xa0protected] = [email\xa0protected] ans | |
| 42542. |
2¾×⅘ |
| Answer» 2⅗ | |
| 42543. |
Prove that√2 +√3 is irrational. |
| Answer» Yaha √2 hi h kya...ya fir 2+√3 h.. | |
| 42544. |
If the zeroes of quadratic polynomial x^2 +(a+1)x+b are 2 and-3 , then find the values of a and b. |
| Answer» 2²+(a+1)2+b=0_________. 4+2a+2+b=0_________. 2a+b+6=0--------1 ________. -3²+(a+1)-3+b=0_________.-3a+b+6=0------------ 2________ :- by using elimination method, 2a +3a-b+b-6+6=0_______. 5a =0 a=0 then b=-6 | |
| 42545. |
Koi hamko btaega ki cbse sample paper kaise download kare |
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Answer» Just go to official website of ncert and get downloaded just search on google u will get . search as cbse 2021 sample papers |
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| 42546. |
The distance between the point (3-2) and (-3,2) |
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Answer» X1= 3 ,X2=( -2) ,y1= (-3) ,y2=2 Distance formula X=√(x2-x1)+(y2+y1) =√ (-2-3)2+{2-(-3)}2=√(-5)2+(2+3)2=√(-5)2+(5)2=√25+25=√50 Sorry by mistake value write wrong ....x1 & x2= 3,-3 and y 1&y2= -2,2 Distance formula = underroot ( x2-x1)+(y2-y1)Here x1=3,x2=-2 and y1=-3 and y2=2 |
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| 42547. |
The area of a square inscribed in a circle of diameter \'d\' is |
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Answer» AB=BC because it is a square. Let ABCD be square............ Then d will be it\'s diagonal ( vertices of square lies on circum.).................. Area of sq. =side² i.e AB² In right triangle ABD, AB²+BC²=AC²............ 2AB²=d²..........AB²=d²/2 units² |
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| 42548. |
The angle of elevation |
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Answer» Angle between observer\'s eye, looking at the top of anything like tower, and horizontal line is called angle of elevation. From ground we can see to top of anything |
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| 42549. |
Draw a 90 angle |
| Answer» ⚔️. 9️⃣0️⃣. ? | |
| 42550. |
If x-2 is a factor of X3+ax2+bx=16and a-b=6then the value of a and b |
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Answer» Thanks ? P(x)=x^3+ax^2+bx=16P(2)=2^3+a×2^2+2×b=16..........8+4a+2b=16...........2a+b=4......(1)a-b =6.............. (2) {given}......(1)+(2):......2a+b=4…....a-b =63a=10, (a=10/3).......(3)Put (3) in (2);.....10/3-b=6......10/3-6=b(B=-8/3) |
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