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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The polynomial x^6 - 64y^6son factorization gives |
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| 2. |
The Polynomial, p^2 + 2q- q^22+ r^21- 2pr on factorization gives |
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Answer» thanks |
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| 3. |
\begin{array} { l } { \text { Draw the graph of the polynomial } P ( x ) = x ^ { 2 } - x - 12 \text { and find the } } \\ { \text { zeroes? } } \end{array} |
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| 4. |
\left. \begin{array} { l } { \text { The value of the polynomial } x ^ { 2 } + 4 \sqrt { 2 } x + 6 , \text { when } x = \frac { 1 } { \sqrt { 2 } } , \text { i } } \\ { ( A ) \frac { 21 } { 2 } } \\ { \frac { 11 } { \sqrt { 2 } } } \end{array} \right. |
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| 5. |
(i) 4, 8, 16,虱河灭 |
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Answer» 4,8,16........a1=4=2^2a2=8=2^3a3=16=2^4....an=2^(n+1)an=2^(9+1)an=2^10=1024 |
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| 6. |
\begin { equation } \begin{array}{l}{\text { Find all the zeroes of the polynomial }} \\ {x^{3}+3 x^{2}-2 x-6, \text { if two of its zeroes are } \sqrt{2} \text { and }} \\ {-\sqrt{2} .}\end{array} \end { equation } |
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| 7. |
\left. \begin{array} { l } { \text { The polynomial, } x ^ { 2 } + y ^ { 2 } - z ^ { 2 } - 2 x y \text { on factorization gives } } \\ { ( A ) ( x - y - z ) ( x - y + z ) } \\ { ( C ) ( x + y + z ) ( x - y - z ) } \end{array} \right. |
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| 8. |
Find the value of polynomialp(x)=x^{3}-2 x^{2}-2 x+6 \text { at } x=\sqrt{2} |
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| 9. |
\left. \begin{array} { l } { \operatorname { cos } 45 ^ { \circ } } \\ { \operatorname { sec } 30 ^ { \circ } + \operatorname { cosec } 30 ^ { \circ } } \\ { \frac { 5 \operatorname { cos } ^ { 2 } 60 ^ { \circ } + 4 \operatorname { sec } ^ { 2 } 30 ^ { \circ } - \operatorname { tan } ^ { 2 } 45 ^ { \circ } } { \operatorname { sin } ^ { 2 } 30 ^ { \circ } + \operatorname { cos } ^ { 2 } 30 ^ { \circ } } } \end{array} \right. |
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Answer» the answer is wrong |
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| 10. |
6-3(4>(8-:-16)A1 B4C)0.5 D)2 |
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Answer» 6÷3(4×8÷16)=6÷3(32÷16)=6÷3(2)=2(2)=2×2=4 |
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| 11. |
calculated graphically with the help ofcumulative frequency curve?(A) Mean(B) Median(C) Mode.(D) of these3. Which measurement of a central tendencyis known as center of gravity?(A) Mean(B) Median-(C) Mode(D) of these4. The value which is repeated highest time inthe series is known as?(A) Mean(B) Median(C) Mode(D) of these |
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Answer» 3) b is the right answer b is the right answer.... B is the right answer 🙂🙂 b is the correct answer B is the correct answer |
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| 12. |
There is a grass bed of width 1.5 m along the four sides of a square pond of side 27 mn. Find the areaof the grass bed. |
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Answer» (28.5) ^2 - (27)^2= 1.5× 55.5 |
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| 13. |
\left. \begin array l \text polynomial 2 x ^ 4 - 11 x ^ 3 %2B 7 x ^ 2 %2B 13 x - \\ ( 3 %2B \sqrt 2 ) \text and ( 3 - \sqrt 2 ) \end array \right. |
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Answer» If α and β are zeroes of the polynomial 2x⁴ - 11x³ + 7x² + 13x - 7 Then, x² - (α + β)x + αβ . Here, α = (3 + √2) andβ = (3 - √2). So, α+ β = 6 and αβ = 7. Thus, x²- 6x + 7 is a factor of 2x⁴ - 11x³ + 7x²+ 13x-7 Now, Given polynomial 2x⁴ - 11x³ + 7x²+ 13x-7, So, x² - 6x + 7 ) 2x⁴ - 11x³ + 7x² + 13x - 7 ( 2x² + x - 1 2x⁴ - 12x³ + 14x² (substract) ----------------------------- x³- 7x²+ 13x x³ - 6x²+ 7x (substract) -------------------------------------- - x²+ 6x - 7 - x²+ 6x - 7 (substract) ----------------------------- 0 ----------------------------- We have,the Quotient as 2x²+ x -1 = 2x²+ 2x - x -1 = 2x(x + 1) - 1(x + 1) = (2x - 1)(x + 1) ∴ x = 1/2 , -1 are the other zeros of the polynomial . Like my answer if you find it useful! |
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| 14. |
\operatorname { tan } ( 45 ^ { \circ } + \theta ) - \operatorname { tan } ( 45 ^ { \circ } - \theta ) = 2 \operatorname { tan } 2 \theta |
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Answer» tan (45°+x)=(1+tanx)/(1-tanx)........ (1)tan (45°-x)=(1-tanx)/(1+tanx)........ (2)subtracting eq.2 from1tan (45°+x)-tan (45°-x)=(1+tanx)/(1-tanx) - (1-tanx)/(1+tanx) =[(1+tanx)(1+tanx)-(1-tanx)(1-tanx)]/(1+tanx)(1-tanx) =[(1+tan^2 x +2tanx)-(1+tan^2 x -2tanx)]/(1-tan^2 x) =(2tanx + 2tanx)/(1-tan^2 x) =4tanx/(1-tan^2 x) =2×(2tanx/(1-tan^2 x) =2×tan2x =2tanx here i have used x for theta |
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| 15. |
\frac 3 \operatorname tan ^ 2 30 ^ \circ %2B \operatorname tan ^ 2 60 ^ \circ %2B \operatorname cosec 30 ^ \circ - \operatorname tan 45 ^ \circ \operatorname cot ^ 2 45 ^ \circ |
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Answer» tan30=1/√3 tan60=√3 cosec30=2 Tan45=1cot 45=13*1/3+3+2-1/1=1+3+1/1=5 |
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| 16. |
\frac { \operatorname { sin } 30 + \operatorname { tan } 45 - \operatorname { cosec } 60 } { \operatorname { sec } 30 + \operatorname { cos } 60 + \operatorname { cot } 45 } |
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| 17. |
\left. \begin{array} { l } { 2 \operatorname { tan } ^ { 2 } 45 ^ { \circ } + \operatorname { cos } ^ { 2 } 30 ^ { \circ } - \operatorname { sin } ^ { 2 } 60 ^ { \circ } } \\ { \frac { \operatorname { sin } 30 ^ { \circ } + \operatorname { tan } 45 ^ { \circ } - \operatorname { cosec } 60 ^ { \circ } } { \operatorname { sec } 30 ^ { \circ } + \operatorname { cos } 60 ^ { \circ } + \operatorname { cot } 45 ^ { \circ } } } \end{array} \right. |
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Answer» 1) 2(1)^2+(root(3)/2)^2-(root(3)/2)^2=2+3/4-3/4=22)(1/2+1-2/root(3))/(2/root(3)+1/2+1)=(root(3)+2root(3)-4)/(4+root(3)+2root(3))=(3root(3)-4)/(3root(3)+4)=(3root(3)-4)^2/(27-16)=(27+16-24root(3))/11=(43-24root(3))/11 |
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| 18. |
Z if a dog eats gth of a packet of dog food, then how many dogscan be fed with 15 packets of dog food? |
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| 19. |
:Find the time taken by a person in walkingalong the boundary of a square field of area40,000 m2 at a speed of 4 km/h |
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| 20. |
12 show that there is no value of n for which (2" × 5") ends in 5. |
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Answer» 2^n * 5^n = 10^n^2 Any number multipliedby 10 always ends in 0. (The basic test of divisibility rule to check if anumber is divisible by 10 is whether the final digit of the number is 0). Thus, for any value of n, n^2 multipliedby 10 will also end in 0. As a result, the value of 10n2 will neverend with 5. |
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| 21. |
1 2 3 4 isn. The H.C.F. of 3 À 5 is |
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Answer» HCF of { 1/2 , 2/3 , 3/4 , 4/5 } = HCF of { 1 , 2, 3 , 4 }/ LCM of { 2 , 3 , 4 , 5 } HCF of { 1, 2, 3 , 4 , } = common factor of { 1, 2, 3 , 4 } = 1 LCM of { 2 , 3 , 4 , 5} = 4 × 5 × 3 = 60 so, HCF = 1/60 ( answer ) |
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| 22. |
3. A gulab jamun, contains sugars30% offs volume. Fmdapproximatelynosyrup.would be foundn 5 gulab minsshapedljske a cylinder with tiwo hemispuericat endswith length 5 om and damteroi 5 |
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Answer» solution:-Please give this solution a "like" Total height = 5cmtherefore height of cylinder = 5-2.8=2.2volume of gulab jamun= volume of cylinder + volume of 2 hemispheres.volume= 3πr^2h + 4/3πr^3=22/7x1.4x1.4(3x2.2+4x1.4)/3=6.16(12.2)/3=75.152/3=25.05 cm cube.jamuns = 45therefore 45 gulab jamuns volume= 45x25.05=1127.2830% syrup.therefore volume=1127.28x30/100=338.184cm cube. |
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| 23. |
(c) AB = 5.6cm,t possible to have a triangle, if its angles are given à(a) 90°, 70°, 20°2. Is i(b) 110°, 50°, 50° |
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Answer» The sum of angles of a traingle is 180. So, the first case is possible and the second case is not possible because it's sum exceeds 180. |
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| 24. |
4.The distance between the points (acos8+bsin0,0) and (0,a sino - bcost) is1. à +622. a + b3. 2-6 |
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Answer» Answer:4)√(a^2+b^2)Explanation: |
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| 25. |
UP!If the mean of the following distribution is 6, then find the value of p.* 12 14 16 10 (p+5y 3 2 3 1À5964 |
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Answer» 27.5 sigma xy=6+8+18+10+2p+10=2p+52sigma y = 3+2+3+1+2=11mean=sigma xy/sigma y6=2p+52/112p+52=6×112p+52=662p=66-522p=14p=7Abs)The value of p is 7. sigma xy=6+8+18+10+2p+10=2p+52sigma y = 3+2+3+1+2=11mean=sigma xy/sigma y6=2p+52/112p+52=6×112p+52=662p=66-522p=14p=7Abs)The value of p is 7. |
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| 26. |
solve by using formula :2 x^2 - 3x + 1 =0 |
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| 27. |
Find the median of first to prime numbers?Efi ; = 1860 and Its=20 then find mean?I write the formula to find the mean of grouped datausing assumed mean method and explain each termsWickets taken by a bowler in 10 matches are 2,6,41,3,2,3 what is the mode of the data?write the formula of modedata andfor a groupedeach term in it?. Find the average of first 100 natural numbers.1.25 4.32 7.63 8.4 0.8 |
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Answer» ffxfnjkkjg jgffghjjj jhjjjjkkkkkngf. bfdhjjjgt jhg hrrujjbvhjjiu khggjjkknff iffhjkm bgcgv FX is equal to 7 cm is equal to 9 x 9 x is equal to 8.4 first prime numbers=2,3,5,7,11,13,17,19,23,29median=(11+13)/2=24/2=12mean=1860/30=62 |
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| 28. |
53. A square field ABCD of side 90 m is so locatedthat its diagonal AC is from north to south and thecorner B is to the west of D. Rohan and Rahul startwalking along the sides from B and C respectivelyin the clockwise and anti-clockwise directions withspeeds of 8 km/hr and 10 km/hr. Where shall theycross each other the second time?(a) On AD at a distance of 30 m from A(6) On BC at a distance of 10 m from B(c) On AD at a distance of 30 m from D(d) On BC at a distance of 10 m from C(M.A.T. 2005) |
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Answer» Option B is correct answer b.is correct answer........ option B is correct answer it is correct answer d) on BC at A distance of 10m from C d) on BC at a distance of 10m from C the option d is the correct answer option D is correct answer d option is correct 😀😀😀 Option d is correct answer c is the correct answer bro check ✔it A is the right✔ anser option d is the correct answer (D) option is right answer option b= right bc at a distance 30 m from b the option D is the correct answer to this question (D) is correct answer option no d is correct who gave correct answer like here and Mark my answer as best d is coerrect answer option d is the correct answer on Bc at a distance of 10m from C D option is right answer option d is correct answer d) On BC at a distance of 10. from c the correct answer is C option d is the right answer option d hope my answer helps you also some of the students are answering wrong don't believe that d is the correct answer on BC at a distance of 10m from B b is the right answer d)on BC at a distance of 10m from c. option d is the correct answer... option (d) is the right answer. plz like my answer answer hai d because it ln triangle ABC 2b=a+c option d) is correct answer option a us the correct answer |
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| 29. |
Numberof policy holdersAge (in years)Bdow20Bdow 25Bdow30Bdow 35Bdow40Bdow45Bdow 50Bdow 55Bdow60100 |
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Answer» question is incomplete plz click full question |
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| 30. |
1. Define mode? |
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Answer» The mode is the set of data values is the value that appears most often. The mode of a set of data values is the value that appears most often. If X is a discrete random variable, the mode is the value x at which the probability mass function takes its maximum value. In other words, it is the value that is most likely to be sampled. |
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| 31. |
11 If'n'is a natural number, then 4-30 ends with a digit x The numberof(A) 3(C) 5n ends with a digit x, The number of possible values of x is(B) 8(D) 6 |
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Answer» but answer is given 5 |
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| 32. |
If A-B-0 then write relation between A and B |
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Answer» A - B = фSo, it means that from set A all elements are eliminated therefore set A must be equal to set B |
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| 33. |
(7x-1)/4-(2x-(1-x)/2)/3=19/3 |
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Answer» Aapko photo bhej rahe hain. Answer bataiye. tum par kuch bhi nahi ata |
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| 34. |
2. Solve using formula.(1)x2 + 6x + 5 = 0 |
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Answer» Here we will useSri dharacharya formulaa= 1 b= 6 c= 5Discriminant = b^2-4ac36-4*536-20= 16x= -b+-√D/2a-6+√16/2----( taking +ve sign)= -6+4/2=-1-6-√16/2-------(taking -ve sign)= -6-4/2=-5so -1 and -5 are the roots |
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| 35. |
(1) 2x² – 7x +3=0 |
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Answer» x=1\2,x=3 is the right answer x= 1/2, x= 3is the correct answer me x=1/2,x=3 this is right answer |
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| 36. |
द्विघात समीकरणA1निम्नलिखित द्विघात समीकरणों को हल करें।(ii) 2- +2-6=0(ii)3-26y+2=0(iv) 25-y-=0विषद-बोध र न(ii) (4z -1)(x +2)= 2+7xहल करें-| (i) (x + 3)(x -3)=7(iii) (2x-1)* = (x-1)(x -3)+1हल करें |
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| 37. |
1/ x+3 + 1/ 2x-1 =11/7x+9 |
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Answer» thank you so much |
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| 38. |
-3 1 83 1ll)ニ+-+-+3 5 6 15 |
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| 39. |
Solve using formula: 5m2-4m-2 0 |
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Answer» hit like if you find it useful |
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| 40. |
5. What should be addedger |
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Answer» = 4/5 - (-2/7)= 4/5 + 2/7= 38/35 ans |
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| 41. |
10: Find the least number that must be subtracted from 5607 so as to gersquare. Also find the square root of the perfect square.ampleerfect |
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Answer» Step 1- Find the square root of 5607, and round it of to a nearest value, but it need to lower = 74.88 = 74 (lower value) Step 2- find the square of 74 74*74= 5476 Step 3- find the difference between 5607 and 5476 5607-5476= 131 And that is the least value that needs to subtracted from 5607 to get a perfect square (5476) Step 4- You already now the square root of 5476 which is 74 |
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| 42. |
x %2B 1/x=3*(text*(x^2*(f*(i*(d*n))))) %2B 1/(x^2) |
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Answer» Given,x + 1/x = 3 Squaring both sides, we get (x + 1/x)^2 = 9x^2 + 1/x^2 + 2*x*1/x = 9x^2 + 1/x^2 + 2 = 9x^2 + 1/x^2 = 9 - 2x^2 + 1/x^2 = 7 |
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| 43. |
x A die is thrown 10 times. Getting a numbergreater than 4 is considered a success. Findthe mean and variance of the number ofsuccesses. |
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Answer» Let p is probability of successand q is probability of failure Success is when we get number > 4(5 and 6)p = 2/6 = 1/3q = 1- 1/3 = 2/3 Mean of success is given byX = np = 10*2/3 = 20/3 Variance of success is given by= npq= 10*1/3*2/3= 20/9 |
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| 44. |
1/(-b - a %2B x)=1/x - 1/a - 1/b |
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| 45. |
text*(x*(b*y)) %2B 2*x^2 %2B 3*x %2B 2 |
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| 46. |
\left. \begin array l l ( \text A) & x ^ 2 %2B 3 x %2B 2 \\ \text (C) & x ^ 2 - 3 x - 2 \end array \right. |
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Answer» Answer :A) x^2+3x+2Explanation : |
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| 47. |
Solve(7-5x=5-7x) |
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| 48. |
Q Solve 2x^2+7x+3 |
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Answer» PLEASE LIKE THE SOLUTION |
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| 49. |
2x 1 37x-2 5Solve: |
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| 50. |
solve using formula x2 _ 7x + 12 = 0 |
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Answer» According toquadratic formulaifax^2+bx+c=0, thenx=−b±√b^2−4ac/2a Here we havex^2−7x+12i.e.a=1,b=−7andc=12 andx=−(−7)±√(−7)^2−4⋅1⋅12/2⋅1 =7±√49−48/2 =7±√1/2 =7±1/2 i.e.x=4or3 |
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