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. |
trecofA contractor plans to install two slides for the chaildren to play in a park For the childremhelow the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, ande A contractor plans to install f |
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Answer» Sir iska video hai kya |
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| 2. |
The ratio between the lengths of the edges oftwo cubes are in the ratio 3:2. Find the ratiobetween their:(0 total surface area(i) volume. |
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| 3. |
3The Cure of a douane is 59536 Square miltufind the lingth of its side |
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| 4. |
5. Do not sit watMathematicsI. Divide:2 4 8 4 |
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Answer» as 2 * 242 is 484 the quotient should be 242 242 |
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| 5. |
Q.5. Why does the poet call man's last age as "second childishnessand mere oblivion" ? |
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Answer» In the monologue of Jaques from Shakespeare's "As You Like It," theborther to Orlandocapitalizes on the words of Duke Senior who states that "we are not all alone unhappy...in this wide and universal theatre." Jaques calls the world a stage upon which men and women are players, making their entrances and exits. In the first stage, the infant is "mewling and puking in his nurse's arms." Finally, in the seventh stage, man has returned to the infantile stage as he exists in "mere oblivion" and is "sans"/without teeth, without sight, without taste, without everything. Like babies who have yet to reach the development of their senses, the aged have their senses mitigated to a similar point, but it is the loss rather than the nascence as in infants. |
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| 6. |
(5) 2x3 + 5x2-7 |
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Answer» kya krna hai iska😅 |
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| 7. |
Prove that any two opposite edges of a regulartetrahedron are perpendicular |
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| 8. |
रर यदि ८0 ८70, 070 न 4 सेमी तथा 12२ >5 सभा, तने 1 ९की लंबाई जात कीजिए। ‘ (NCERT Exemplar) |
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Answer» Here angle R= angle P they are corresponding anglesSo QP=QRHence QR=QP=4cmTherefore length of QP=4cm hindi me bata dijiega |
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| 9. |
व कक है. ** e d,’ है. _ ८.2 + 4 - 070. 3548. AL |
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| 10. |
17, 18, 24, 25, 35, 36, 46 का माध्यक होगा हे |
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Answer» Thanks bhai |
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| 11. |
(b) At present, the sum of the ages of Ruby and her daughter is 49. Seven years ago Ruby was sixtimes as old as her daughter. Findtheir present ages. |
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Answer» let age of ruby=xage of her daughter=ySEVEN YEARS AGOruby=x-7her daughter's age= y-7ruby age= 6(y-7)x-7=6(y-7) SUM OF THEIR AGES=49 ruby age + daughter age=496(y-7)+ (y-7)=496y - 42 + y -7=498y - 49-7=498y = 2×49y=98 / 8y=12 years 3 months |
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| 12. |
2. The ratio of surface area of a sphere and curvedarea of a hemisphere is 9:2. Find ratio of their volumes. |
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Answer» Let radius of sphere be r1Let radius of hemisphere be r2 Surface area of Sphere A1 = 4*pi*r1^2 Curved surface area of Hemisphere A2 = 2*pi*r2^2 As per given conditionA1/A2 = 9/22*r1^2/ = 9/2r2^2 r1/r2 = 3/2 Ratio of volume of sphere V1 to volume of hemisphere V2 V1/V2 = 4/3*pi*r1^3/2/3*pi*r2^3 = 2*(r1/r2)^3 = 2(27/8) = 27/4 wrong answer |
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| 13. |
2. The ratio of surface area of a sphere and curved surfacearea of a hemisphere is 9:2. Find ratio of their volumes. |
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Answer» CSA of hemisphere = 2πr² [r= radius of hemisphere] CSA of sphere = 4πR² [R=radius of sphere] thus ratio of their CSA= 2:9 ∴2/9 = 2πr²/4πR² 2/9= r²/2R² 2/9×2R²=r² 2×2R²=9×r² 4R²=9r² r²/R²=4/9 r/R=√4/√9 ∴r/R=2/3 ratio of the radii of the hemisphere and sphere r:R=2:3 |
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| 14. |
The ratio of surface area of a sphere and curved surfacearea of a hemisphere is 9:2. Find ratio of their volumes. |
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| 15. |
80: 384(v) 186ach of the following ratios in the simplest form:oo F16 80(i11) 3 m 5 c(ii) 3 weeks: 30 days (i) 3 m 5 c070 ml (vi) 4 kg: 2 |
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| 16. |
(v) am?ÂŽ+ bm* + bn* + anÂŽg AT - 0 g |
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| 17. |
10, There are 1,000 sealed envelopes in a box, 10 of them contains in cash prizeof R 100 each, 100 of them contains a cash prize of 50 each and 200 of themcontains a cash prize of?10 each and rest do not contains any cash prize. If theyare well-shuffled and an envelope is picked up out, what is the probability thatit contains no cash prize.Sol. See Q .No. 16 on page 198 |
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Answer» Hindi mewat mewat aao |
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| 18. |
T V4 T25608 —+4sec’ ——tan’ —0 : जोन निवशिकाओ 6 4i Alsin” 30°+cos —6 |
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| 19. |
12:3525 18 2TIIn a bag containing red, green and pink tokens, the ratio of red and green tokens was 12:35 while the ratio of pink and red tokens was 25 : 18. What was the ratio of green and pinktokens? |
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Answer» R, G, P tokens R:G = 12:35 R:P = 18:25 G:P = ? R:G = 12:35 x 3 = 36:105 R:P = 18:25 x 2 = 36:50 we have made ratio of red in both the cases equal so we can say G:P = 105:50 = 21:10 |
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| 20. |
(1) 4 (p + c) (3a - b) + 6 (p+al (2-3)g. Factorise:ab-bc - ab + caxy + bcxy-az-bcz |
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Answer» (i)ab²-bc²-ab+c²=ab²-ab-bc²+c²=ab(b-1)-c²(b-1)=(b-1)(ab-c²) |
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| 21. |
the20. Io an equilateral triangle, the length of a methan is 3 3. Then, the length of cachside of the triangle isA, 6 B, 2-3 C.2alD.3having rentre al the origin 1s |
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| 22. |
(al 6. 92 Find the LCM by listing the multiples:(d)18 and 2(c) 6 and8(b) 2 and 6liating the multiples:ai 3 and 4d and 5(d) 7 10 |
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Answer» Please hit the like button |
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| 23. |
SOLO29. An electronic device makes a beep after every 15 minutes. Another device makafter every 20 minutes. They beeped together at 6 a.m. At what time will theya.m. At what time will they next beeptogether? |
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| 24. |
Find the area of shaded region30 m50 m |
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Answer» thank you |
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| 25. |
2. Find the area of shaded region :30 m50 m |
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Answer» 750 sq.m is the area of this right angle triangle. formula-- 1/2 bh as area= 30*50= 1500now half of 1500 = 750 it is the ans The correct answer for this question is 750 area of rec=30×501500cm sq.area of triangle = 1/2*1500 = 750cm sq. 750 sq. m is the area of this angle triangle. formula -1/2 bh 750 sq.m is the answer by applying the formula 1/2 base * height |
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| 26. |
21. In given figure, PQ and AB are respectively thearea of two concentric circles of radii 7 cm and3.5 cm and centre O. If LPOQ 30, then findthe area of the shaded region.30°8 |
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| 27. |
14 AB and CD are respectively arcs oftwo concentric circles of radi 21 cm and 7 cm andcentre O (see Fig. 12.32). If ZAOB-30°, find the area of the shaded region. |
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| 28. |
14 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm andCentre O (see Fig. 12.32). IfZAOB-30°, find the area of the shaded region. |
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| 29. |
Fig. 12.314. AB and.CD are respectively arcs of two concentric circles of radi 21 cm and 7 cm andcentre O (see Fig. 12.32). If ZAOB 30°, find the area of the shaded region. |
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| 30. |
Blow en is dissible by 6 for every positius interesa40427-22MA |
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| 31. |
Find the area of the shaded region in figure where ABCD is a square of side 14 cmthe area of the shaded region in ligure, |
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Answer» Shaded area= area of square - (area of 4 circle14*14-(4*22/7*(3.5)^2as 3.5 will be radius of circle 196-(154) 42cm^2 |
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| 32. |
16. What should be added toto get |
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| 33. |
The difference between two numbers is-8a2-5a + 3. If the bigger number is 10a? + 6 -6,then what is the smallar number? |
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Answer» Given:Difference between two numbers is 8a²-5a+3Bigger number is 10a²+6-6Let x be smaller number Now,8a²-5a+3=10a²+6-6-x8a²-5a+3=10a²-xx=10a²-8a²+5a-3x=2a²+5a-3Thus the smaller number is 2a²+5a-3 difference between two numbers is 8a^2-5a+3, bigger number is 10a^2+6-6, x small number. 8a^2-5a+3=10a^2+6-6; 8a^2-5a+3=10a^2;-X, X=10a^2-8a^2+5a-3=2a^2+5a-3=0, smaller number is 2a^2+5a-3 |
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| 34. |
16. By what number should (-5) be multiplied to get |
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Answer» (-5)^3 × (1/5)^3(1/-15)×(1/15)225 |
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| 35. |
H-12.65Area ofshaded region |
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Answer» Area = length of base x height = 6 x 12.65 cm2 area = 6×12.65= 75.9cm² don't forget to loke the answer of the question is 75.90 |
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| 36. |
Mensuration59ea of the shaded region in figure itfuare of side 14 cm and APD andBPC are semi-circles.14 cmDr14 cmthe shaded region = |
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| 37. |
Find the area of shaded region |
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Answer» shaded area is right angled triangleso area=bh/2=1/2*50*30=750m*m |
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| 38. |
(iii): 4. Find the area of an isosceles right triangle if one of the right sides is 20 cm long.Hint In the isos A hase altitude 20 cml |
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| 39. |
In trapezium ABCD, AB DC and diagonals AC andintersect at O. Prove that ar (AOD) ar (BOC). |
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| 40. |
By what number 6be multiplied to get 16 |
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Answer» (32/5)*x=16X=16*5/32X=5/2=2 1/2 =2.5 suppose the required number be X=6 &2/5 × X=16=6×5+2/5 ×X=16=32/5 × X=16= X=16 ÷32/5= X=16×5/32= X=5/2so the required number is 5/2. 5/2 is the correct answer of the given question 5/2 is the right answer for given question. |
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| 41. |
What should be added to a to get 16? |
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| 42. |
13. Mala has 36 toffees. Sheare left with her?hem to her friend. How manyIf we |
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Answer» Given to friend =4/9*36=4*4=16Toffeees left with her =36-16=20Hope I helped you 3.5 20 toffees left with her. 20 toffees left her. 20 toffee left with her 20 toffees left with her. 20 toffees left with her. 20 toffees are left with mala. 20 toffees left with her |
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| 43. |
Qs PR |
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Answer» Yes it is the correct solution for the problem |
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| 44. |
ए-0) आर11-21) कशून्यका का गुणपणा होगा।3.In the adjoining fiqure e || m ||n.If y:z=3:7. Find the value of x.संलग्न चित्र में 2 ||m ||n है,यदि y:z=3:7 तो का मान क्या होगा?m |
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Answer» 3x + 7x = 180° 10x = 180° X = 18° here x = z and X + y or X + z is equal to 180°therefore z = x = 7x X = 7*18 X = 126° Ans. 3x+7x=18010x=180x=18here x=z and x+y or x+z =180therefore z=x=7xx=7*18x=126 hughjhbghggvgfybcchcc |
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| 45. |
Rakesh's income is 25%more than of Rohan. What percent is Rohan's income less than Rakesh's income? |
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Answer» Let us assume B's Salary is Rs. 100. Then,A's Salary = (100 + 25% of 100) = Rs. 125.Difference between A's & B's Salary = 125 - 100 = Rs. 25.Hence, % difference (lower) = (25/125)*100 = 20% |
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| 46. |
se 14.2-. In the given fiqure, find the length of unknownside x.13 cm90°12 cm |
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Answer» 5 is the correct answer of the given question 5 is the correct answer you should use Pythagoras theorem13^2=12^2+×^2 169=144+×^2 x^2=169-144 x^2=25 x=5. the answer will be 16 |
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| 47. |
In a figure a tpwer AB is 20m BC its shadow on the ground is 20 m long Find the suns altitude |
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| 48. |
In a figure A tower AB is 20 m high BC its shadow on the ground is 20 root2 long find the suns altitude |
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| 49. |
ORIn figure, a tower AB is 20 m high and BC,its shadow on the ground, is 205 m long. Find the Sunsaltitude20 m820-3 |
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Answer» Givenh=20m=20m x=20√3 tanθ=h/x=20/20√3 =>θ=30∘θ=30∘ Answer :30∘ |
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| 50. |
82 12 पुरुष अथवा 15 औरतें एक कार्य को 21 दिनों मेंपूरा कर पाते हैं। उसी कार्य को 6 पुरुष तथा 10 औरतेंकितने दिनों में पूरा करेंगे?(1) 15 ) 18 (3) 21 (4) 24 |
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Answer» 1 people jaldi Nahi kar sakta |
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