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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. |
differentiate between rational numbers and irrational numbers with exampals |
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Answer» Difference between rational number and irrational number areRational Number:-A number which is written in the form p/q where p and q are integers and q is not equal to zero.Irrational Numbers:- A number which do not written in the form P/q where p and q are interger and Q is equal to Zero. |
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| 2. |
1छु2 1७. 532 110 00509 kb5 lefeie विद, 00005 करियर के अप डु नकल % 01 कब * |
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| 3. |
司!and 3Obtain all zeroes of1513+25 30, if two of its zeroes are 3 |
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Answer» Like if you find it useful no answe |
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| 4. |
5. Find five numbers in A.P. whose sum is 12and the ratio of the first to the last terms is2:3 |
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Answer» Sum of an AP is given by n/2 (2a+(n-1)d)Siince sum of first 5 terms is given∴ 5/2 (2a+4d)=12= a+2d=12/5 ___(1)and a/d=2/3 __(2)∴ put value of a from (2) in (1) we get 2d/3 +2d=12/5∴d=9/10 and a=2d/3=3/5=6/10therefore first five numbers are a ,a+d,a+2d,a+3d,a+4d6/10,15/10,24/10,33/10,42/10 |
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| 5. |
7y+4 â4s i 3 |
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| 6. |
|2x-5|≤4 |
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| 7. |
E Find all the zeroes of the polynomial 2x-3-5x2 +9x-3. it is being given that wo ofits zeroes are 3, 3 |
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| 8. |
Example 3 Find the zeroes of the polynomialbetween the zeroes and the coefficients.3 and verify the relationship |
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| 9. |
(1) a - (6? – -4) |
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| 10. |
g T \â4-3\ -g= & = |
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| 11. |
4. Arrange the following in ascending order52 |
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| 12. |
\frac{\sec \theta}{\cos e c}+\frac{\cos e c \theta}{\sec \theta}=\sec \theta \cdot \cos e c d |
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| 13. |
Prove that (\sec \theta+\cos e c \theta)(\sin \theta+\cos \theta)=\sec \theta \cos e c \theta+2 |
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| 14. |
उदाहरण /. !सड करार डकProwg का 9 _ #ल<8+1T il 2secO+1 tan® sa— |
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Answer» LHS= tanA/secA - 1 + tanA/secA + 1 = sinA/cosA / 1/cosA - 1 + sinA/cosA / 1/cosA + 1 [because tan A = sin A / cos A and sec A = 1 / cos A] = sinA/cosA / (1 - cosA) / cosA + sinA/cosA / (1 + cosA) / cos A = sinA / 1 - cosA + sinA / 1 + cos A = [sinA (1 + cosA) + sinA (1 - cosA)] / (1 - cosA) (1 + cosA) = [sinA + sinAcosA + sinA - sinAcosA] /1-cos^2A = 2sinA /sin^2A[From identity:sin^2A+cos^2A= 1] = 2 / sin A = 2 x 1 / sinA = 2 cosec A [because 1 / sin A = cosec A] =RHS 5_-7×-9_-2=-9_2×5_-7 |
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| 15. |
\frac{\cos \theta}{\cos e c \theta+1}+\frac{\cos \theta}{\cos e c \theta^{-1}}=2 \tan \theta |
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| 16. |
1. Define an irrational number.2. Explain, how irrational numbers differ from rational numbers?3. Examine, whether the following numbers are rational or irrational:(1) J7(ii) 4(iii) 2 + 3(iv) B +52(1) 3 +55 (vi) (J2-23 |
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| 17. |
Exercise 11.5(a) 17 m+ 12 m 6 cm |
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Answer» we have17m 12m 6cm1m = 100cmso 1cm = 1/100 m6cm = 6/100 m = 0.06m17m + 12.06 m = 29.06m |
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| 18. |
Calculate the mean, the median and the mode of the numbers3, 2, 6, 3, 3, 1, 1,2 |
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| 19. |
Isaned 532 Add |
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Answer» 1.56+5.32 = 6.88 The solution is 6.88 |
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| 20. |
The average of 7 numbers is 30.If the average offirst 3 numbers is 32 and the average of last 3numbers is 28, then find the 4th result? |
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Answer» average of 7 numbers is 30 so total.of 7 numbers 7×30 = 210 average of first 3 numbers is 32 so total of first 3 numbers 32×3 = 96 average of last 3 numbers is 28 so total of last 3 numbers 28×3 = 84 so fourth number = 210 - 96 - 84 = 210 - 180 = 30 |
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| 21. |
nrnduct of two numbers is 972 and their quotient isFind the numbers.3 |
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Answer» let the 1st number be x the 2nd number will be 972/x. Here, x/972/x = 4/3 = x^2/972 = 4/3 by cross multiplication,we get, = 3x^2 = 3888 = x^2 = 3888/3 = x^2 = 1296 x = 36 Hence, 1st number (x) = 36 2nd number (972/x) = 27 |
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| 22. |
allArrangeascending order3 7 5 11a)12'6'122. Add the following. |
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| 23. |
.Write the following in ascending order: |
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Answer» a) Order of ascending .1 , .5, . 7, .9 |
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| 24. |
(c) Compare2. (a)Arrange the following ratios in ascending order:(i) 1:4,2:3, 3:9,4:5 (ii) 2:7,9:12,3:6, 8:9.in descending order: |
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Answer» asscending order1:4 ,3:9 ,2:3,4:5 2:7, 3:6,9:12,8:9 |
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| 25. |
The value of 932 is |
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Answer» 3/2 can be written as 3*(1/2) Therefore, (√9)³ = 3³ = 27 |
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| 26. |
rrange the following ratios in ascending order5:4, 7: 6, 3: 2, 5:8 |
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Answer» 5:4 = 5/4 = 1.257:6 = 7/6 = 1.163:2 = 3/2 = 1.55:8 = 5/8 = 0.625 so, 5:8 < 7:6 < 5:4 < 3:2 |
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| 27. |
" - v F Aडे (6 39503- 8109 + ) (932 +8 ig फल >“ ले =0 |
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| 28. |
The value of 932 |
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Answer» Value of (9)^3/2= [(3)^2]^3/2= (3)^3= 3*3*3= 27 |
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| 29. |
The value of 932 is1S |
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Answer» 9^(3/2) (3²)^(3/2) =3^(2*3/2) =3^3 =27 |
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| 30. |
The value of 932 isas(A) 8I(EC) 18(D) 24 |
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Answer» (9)^3/2=(3^2)^3/2=3^(2×3/2)=3^3=27 |
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| 31. |
1.The value of 932 is(A) 81(B) 27C) 18(D) 24 |
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| 32. |
duestion il2 1 1wt(a) How many terms of 932 will make sum 72 4 |
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Answer» Given,a=2/9, r= (-1/3×9/2)= -3/2, Sn= 55/72, n=? Sn=a(1-r^n)/(1-r) 55/72 = (2/9)[1-(-3/2)^n]/(1+3/2) (55/72)×(9/2)×(5/2)=1-(-3/2)^n 275/32=1-(-3/2)^n (-3/2)^n = 1–275/32 = (32–275)/32 = (-243)/32 or (-3/2)^n=(-3/2)^5 or n = 5 terms |
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| 33. |
write the following in the ascending order if thier magnitude 4^√3 , 3^√2 , 3^√4 |
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Answer» 3^sqrt(2), 3^sqrt(4), 4^sqrt(3) fuck off rfhsxvnhdzcvhfsdhjfssswfjbvfddhjjbcd |
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| 34. |
Arrange in ascending order:2/7 , -2/5, -3/10 |
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| 35. |
1.EXERCISE 6.2In Fig. 6.28, tind the values of x and y and thenshow that ABCDFig. 6.28 |
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| 36. |
nves are ia and 3A2 , Eeo) cea, b) and DI. 2) are the vertices of a parallelogram ABCD, tind the values of a anedHence find the lengths of its sides.OR |
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| 37. |
Eg). There is an Ensenute collection of white reklackballs and any hall is equally likely to meCohite or black, 10 balts are drawn fromcollection and are put in a bag of aball as asawra from bag es white what is thepob it is the only white ball intebag |
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| 38. |
I box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cmA cubicawide and 8 cm high.Which box has the greater lateral surface area and by how much?Which box has the smaller total surface area and by how much? |
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| 39. |
Mohit painted the outside of a box of length 2.5 m, breadth 2 m and height 1 m. How much surface areadid he cover, if he painted all except the bottom of the box? |
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| 40. |
322. Arrange the following in ascending order:(0)4873357 |
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Answer»  The given fractions are12,34,56and7812,34,56and78.L.C.M. of 2, 4, 6 and 8 = (2××2××2××3) = 24We convert each of the given fractions into an equivalent fraction with denominator 24.Now, we have:12=1×122×12=1224;34=3×64×6=182456=5×46×4=2024;78=7×38×3=212412=1×122×12=1224;34=3×64×6=182456=5×46×4=2024;78=7×38×3=2124 Clearly,1224<1824<2024<21241224<1824<2024<2124 ∴12<34<56<7812<34<56<78Hence, the given fractions can be arranged in the ascending order as follows:12,34,56, so simple 4/35<2/5<4/7<8/7 |
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| 41. |
© By observing the following figure NitchitaSays "L1 and Lr are adjacent angles. Do you9 agree? Cine reasons, |
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Answer» yes I agree with the answer No ,1 and 2 are not adjacent angles since they have no common arm. no because adjacent angles should have a common arm but here there is no common arm between angles 1and2 |
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| 42. |
13 4Arrange the following fractions in ascending order 575 |
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Answer» 3/7....1/2...7/9...4/5 |
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| 43. |
Arrange Big this in ascending order |
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Answer» 9/35, 13/28, 3/7, 11/14 9/35, 3/7,13/28,11/14 is the correct answer 9/35,13/28,3/7,11/14 is the correct answer of the following question 9/35, 3/7,13/28,11/14 is the correct answer 9/35, 3/7, 13/28, 11/14 is correct answer. please like my answer 9/35,13/28, 3/7 ,11/14 |
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| 44. |
Arrange 7878; 7,77,888;8787; and8,88,888in ascending order |
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Answer» 7,8,77,88,888,7878,8787 |
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| 45. |
- IO 5(D) TRIVANDRUM, August 31,2008 -निम्नलिखित आकृति में त्रिभुजों की “ख़ंख्या :बताइये- z f |
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Answer» 15 tribhuj honge isme |
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| 46. |
The perimeter of a field is 932 metres. If its breadth is 125 metres, what is its len |
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Answer» perimeter=932mbreadth=125mperimeter=2(l+b)932m=2(l+125)m932m=2l+ 250m932m-250m=2l682m=2l682÷2=l341=ll=341ans. |
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| 47. |
Fromthe given figure, calculate: (6) the area of the shaded region and (i) the len ginh of the boundary19n |
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| 48. |
2.If both a and b are rational numbers, then tindvalues of a and b in each case.V3 -1=a+63V3+1 |
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| 49. |
4.Rukhsar painted the outside of the cabinetofmeasure 1 m x 2 m 15m. How muchsurface area did she cover if she painted all except the bonomcol the cine |
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| 50. |
1. Find the volume of cubical box whose edge is 7 cm. |
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Answer» 200000000 +666677 - 5555 Volume of cube = 7cm × 7cm × 7cm Volume of cube = 343cube.cm volume of cubical box = 7cm×7cm×7cm = 343 cubic cm. FORMULA OF VOLUME OF CUBE=a³SO BY APPLYING THE FORMULA, a=7, SO VOLUME OF CUBICAL BOX=7.7.7=343 CUBE CM Volume of a cube = edge × edge × edge =7 cm × 7 cm × 7 cm = 343 cu cm 343 cube. cm is the right answer. 343 cube.cm is a right answer 343 cubic cm is correct answer ...... I am agree with harsh singh 343 is correct and perfect answer volume of cuboid =7cm×7cm volume of cuboid =343 cubic centimetre 343 is the right answer 343 is correct answer for your question volume of a cube = edge * edge* edge =7*7*7=343 cubic cm 343cube.cm is the right ans 343 cube is the right answer Volume of cuboid box=7cm×7cm×7cm=343cm cubic volume of cube=7cm*7cm*7cmvolume of cube=343 cube cm answer Vol. of cube=side×side×side =7×7×7 =343 cubic cm =7×7×7 =343 cubic cm the correct ans. is 1.08 volume of cube=343cube.cm volume is 343 cm³ and the best answer. Volume of cube =edgecube =7 cube =7×7×7=343 to find the volume of a cubical box we have to multiply edge*edge*edge=volumeso 7*7*7=343cmalso we can convert it into metre 3m43 cm. |
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