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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. |
6, Find the length and surface area of a wooden plank of width 3m, thickness 75 cm andvolume 33.75m3. |
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| 2. |
6. A well of diameter 140cm has a stone parapet around it. If the length of the outer edgeof theparapet is 616 cm, find the width of the parapet. |
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| 3. |
Ex. 6. A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of theparapet is 660 cm, then find the width of the parapet. |
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| 4. |
13. If the zeroes of the polynomial + p + q aredouble in value to the zeroes of 2r - 51 - 3. findthe value of p and aFCBSE 20121 |
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| 5. |
11.12.(ü) 2.7 X 1.5 X 2.1The product of two rational number isFind the other.From a rope of the length 40m, a marHow piany pieces can be cut if each p!For x ,y and z =13Find the value of the expression:(x-y) -- z and x-(y-z). Are they equal.*********** |
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Answer» x=1/8, y=-8/6, z= 7/20; ( x-y)- z = (1/8-(-8/5))-7/20= (1/8+8/5)-7/20=5-7/20=100-8/20=92/8 = 11. 5, x-(y-z)=1/8-(--8/6-7/20)= 1/8+8/6+7/20=1/6+7/20= 20+42/20=62/20=3.1 |
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| 6. |
6. Find the cost of carpeting a room 13 m by 9 m with a carpet of width 75 cm at the rate of7 105 per metre. |
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Answer» nice |
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| 7. |
sphereThe external and internal diameters of a hollow hemispherical bowl are 12 cm and 10respectively. Find the cost of polishing its total surface area at the rate of 10 per cm38.cm |
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Answer» wrong answer |
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| 8. |
. The internal and external diameters of a hollow hemispherical vessel are 42 cm and 45.5 cm, respectivelyFind its capacity and also its outer curved surface area. |
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Answer» Inner diameter = 42 cminner radius ,r= 21 cm outer diameter =45.5 cm outer radius ,R= 22.75 cm Capacity = Volume = 2 / 3π (R³ - r³ ) = 2 / 3× 22 / 7× [ (22.75)³ - (21 )³] = 2/3× 22 / 7× [11774.5469 - 9261] = 2 / 3× 22 / 7×2513.5469 =5266.47922 cm³ Outer C.S.A = 2πR² = 2× 22 / 7×22.75×22.75 =3253.25 cm² |
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| 9. |
3(5a - 7) + 2(9a - 11) = 4(8a - 7) - 111 |
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Answer» 3(5a-7)+2(9a-11)=4(8a-7)-111 15a-21+18a-22=32a-28-11 33a-43=32a-39a=43-39a=4. |
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| 10. |
find the cube root of each of the following numbers by the prime factorisation method. 91125 |
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| 11. |
Solve the following equation:\frac{8 x-3}{3 x}=2 |
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Answer» 8x-3=6x8x-6x=2x=3 so x=3/2 15y+20=-4+12y3y=-24so y=-8 |
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| 12. |
nemarie9. Write the natural numbers from 102 to 113. What fraction of them anprime numbers?Vinthem |
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Answer» same as above it is right 1/3 is the correct answer of the given question 1/3 is the correct answer |
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| 13. |
Solve the following equation:č13 + 3x = 5(x-3) |
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Answer» (x +3)/2 + 3x = 5/2(x - 3) x/2 + 3/2 + 3x = 5x/2 - 15/2x/2 + 3x - 5x/2 = - 3/2 - 15/2 Take LCM(1 + 6 - 5)x/2 = - 18/22x/2 = - 9x = - 9 Value of x = - 9 |
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| 14. |
प्रश्न 2. & &छ80 में ८ & समकोण है । यदि &छ - 20,AC = 21, BC = 29, & L6sin B, cos C 3R tan Bका माप लिखिए | |- |
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Answer» sinB 20/21cosC 29/20tanB. 20/29agar answers galt hai to photo pura bhejo |
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| 15. |
8. Find the roots of the following quadratiequation.3)(c-i)" (x-)ICE SE 201213 xICBSE 2012 |
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Answer» (x+3)(x-1) = 3(x-1/3) x*x + 3x - x - 3 = 3x - 1 x*x - x - 2 = 0 (x-2)(x+1) = 0 x = -1,2 If you find this answer helpful then like it. |
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| 16. |
The construction of a Δ4BC in which BC -6 cm andLB =to50° is not possible when (AB AC) is equal |
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| 17. |
4 252.l6 |
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| 18. |
न 2 कक.s’ 25 7 Saradsad.)4’:.-2‘—2' b) Gy 3T L6 SN35200 6ed)ssn’nm;.)’ |
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Answer» a)22/5×3/8=22×3+3×5/40=66+15/40=81/40 |
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| 19. |
Q.25) In an AP, if Snn4n +1), find the AP. |
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| 20. |
1. Drawa circle of radius 3.2 cm. |
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| 21. |
en figure, AB || CD and CA has been produced to E so4. In the given figure, ARt LBAE = 125°.C=X",LABD =x®.LBDC =y'and L ACD = z.find thevalues of x, y, z.TE |
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| 22. |
हनन हनन| ला हा | 24ocllenf ca|enS | ™’ |
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Answer» (2/3)^2 ÷ (2/3)^5 /(2/3)^3 ÷ (3/2)^4 = (2/3)^[2 - 5] / (2/3)^3 ÷ (2/3)^-4 = (2/3)^[2 - 5]/ (2/3)^[3 - (-4)] = (2/3)^-3/ (2/3)^7 = (2/3)^[-3 - 7] = (2/3)^-10 = (3/2)^10 0.0003048316 is the answer after step |
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| 23. |
2.How much iron is needed to make a rod of length 90 cm and diametar 1.4 Cm |
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Answer» Here , diameter,d= 1.4 cmradius of rod,r=d/2=0.7 cm length of rod, h= 90 cm Amount of iron needed is equal to the volume of the rod. Amount of iron needed= πr²h=22*0.7²*90/7=> Amount of iron needed= 138.6 cm³ |
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| 24. |
x multiplied by itself is 5 more than it. |
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| 25. |
find the cube root of each of the following numbers by the prime factorisation method. 110592 |
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Answer» to questions |
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| 26. |
3(5a- 7)+2 (9a-11)=4(8a-7)-111 |
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Answer» Thanks thanks |
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| 27. |
solve the following equation 3(5t-7)+2(9t-11)=4(8t-7)-111 |
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Answer» 3(5t-7)+2(9t-11) = 4(8t-7)-11115t - 21 + 18t - 22 = 32t-28 - 11133t - 32t = 45 - 139t = - 94 |
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| 28. |
139. The average of the first 9 natural primenumbers 1s(2) 1029(3) 11(4) 1119 |
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| 29. |
Draw seg AB of length 5.7 cm and bisect it. |
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| 30. |
2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same? |
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| 31. |
5. In the adjoining figure, the circles with centres A and Btouch each other at E. Line I is a common tangent whichtouches the circles at C and D respectively. Find thelength of seg CD if the radii of the circles are 4 cm, 6 emConstruction : Draw seg AF I seg BD. |
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Answer» Construct AY || CD => BY = ( 6 - 4 ) = 2cm Apply Pythagoras to get :x²+2²= 10² Hence, CD = 4√6 cm |
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| 32. |
If P is the midpoint of seg AB and AB = 7 cm, find AP. |
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Answer» AB=7cmP is the mid point.AP=(1/2)AB=(1/2)(7)=7/2=3.5cm |
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| 33. |
(1) If P is the midpoint of seg AB and AB 7 cm, find AP. |
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Answer» As P is the midpoint to line segment AB, the midpoint divides the line segment into two equal parts ie AP=BP=1/2AB.AB=AP+BPsince AP=BP thereforeAB=2APor AB=2BPAP=1/2AB = 3.5 cm Thnxc Hasina di |
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| 34. |
Tutcfttuledataabove.9. 100 sunames were tandomly picked up from a local telephone directory and a frequencydistribution of the number of letters in the English alphabet in the surnames was founas follows:Number of lettersNumber of surnames4-6308 - 1212 -20l64() Drawa histogrun to depict the given information. |
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| 35. |
atoto - at totfind the value ofre. |
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Answer» 1/a+b+x - 1/x= 1/a+1/bx-(a+b+x)/x(x+a+b)= b+a/abx-a-b-x/x^2+ax+bx= (a+b)/ab-(a+b)/x^2+ax+bx= (a+b)/abx^2+ax+bx=-abx^2+ax+bx+ab=0x(x+a)+b(x+a)=0(x+a)(x+b)=0either x+a=0, x= -aor x+b=0, x= -b |
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| 36. |
How much iron is needed to make a rod of length 90 cm and diametar 1.4 cm? |
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Answer» Length of rod , l = 90 cm diameter of rod , d = 1.4 cm so, radius of rod , r = 1.4/2 = 0.7 cm volume of rod = πr²l = 22/7 × 0.7 × 0.7 × 90 = 22 × 0.1 × 0.7 × 90 = 22 × 0.7 × 9= 198 × 0.7 = 138.6 cm³ hence, 138.6 cm³ iron is needed to make a rod of length 90 cm and diameter 1.4 cm. |
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| 37. |
prove that.470casosTot ce en74 ca |
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Answer» cos^2 pi/10 + cos^2 3pi/10 + cos^2 6pi/10 + cos^2 8pi/10 = cos^2 pi/10 + cos^2 3pi/10 + cos^2 3pi/5 + cos^2 84pi/5 = cos^2 pi/10 + cos^2 3pi/10 + sin^2 (pi/2 - 3pi/5) + sin^2(pi/2 - 4pi/5) = cos^2 pi/10 + cos^2 3pi/10 + sin^2 (-pi/10) + sin^2(-3pi/10) = (cos^2 pi/10 + sin^2 pi/10) + cos^2 (3pi/10) + sin^2 (3pi/10) = 1 + 1 = 2 |
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| 38. |
By what number should|be multiplied tot55.) Find the value of x if |
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| 39. |
2.Find LCM of numbers whose prime factorisation are expressible as 3 x 52 and 32 x 72 |
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| 40. |
It is required to make a closed cylindrical tank of height I m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same?. |
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| 41. |
d the mean deviation about the mean TOT4,7, 8, 9, 10, 12, 13, 17 |
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Answer» marks x - mean l x - mean l4 -6 67 -3 38 -2 29 -1 110 0 012 2 213 3 317 7 7____ ____80 24 mean= sum of marks/total no. = 80/8 = 10 MDm = sum of marks - mean / total number = 24/8 = 3 |
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| 42. |
. It is required to make a closed cylindrical tank of height 1 m and basediameter 140 cm from a metal sheet. How many square metres of thesheet are required for the same? |
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| 43. |
2. It is required to make a closed cylindrical tank of leight 1m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same? |
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| 44. |
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cmfrom a metal sheet. How many square metres of the shcet are required for the same? |
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| 45. |
it is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm fromametal sheet. How many square metres of the metal sheet are required for the same?: |
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Answer» thnx |
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| 46. |
2.It is required to make a closed cylindrical tank of height ! m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same? |
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| 47. |
5^(1/3)*5^(1/3) |
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Answer» (25)^1/3 * (5)^1/3 = (5)^2/3 * (5)^1/3 = (5)^[2/3 + 1/3] = (5)^[3/3] = (5)^1 = 5 25)1/3X (5)1/3= (25 X 5)1/3 since(amX bm= (ab)m)= (52X 5)1/3= (53)1/3 (amX an= (a)m+n)= 5 first write in cube root form 25 ,5 and multiple it become cube root 25×5= cube root 125=5answer |
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| 48. |
It is required to make a closed cylindrical tank of height I m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same? |
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Answer» thankvyou do much for helping me |
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| 49. |
A well of diameter 150 cm has a 30 cm wide parapet running around it.find the area of the parapet. |
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Answer» Diameter of well = 150 cmRadius of well = 75 cm Area of well Aw = pi*r*r = 22/7*75*75 Area of outer space Ao = pi*R*R = 22/7*(105)*(105)Area of parapet = Ao - Aw= 22/7*15*15(7*7 - 5*5)= 22/7*15*15(24)= 16971.4 cm^2 |
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| 50. |
A well of diameter 140 cm has a stone parapet around it. If the length of the outer edge of theparapet is 616 cm, find the width of the parapet |
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