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. |
4.f(x)(A) lim fx) -(C) xlim-f(x) =(D) lim f(x)-LLLLL-xxâ2(E) ConclusionsI-12, 12] by I-10, 10]2x23x 5f(x) =-x-4 |
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Answer» 1) limx→-∞ , value becomes 2x²/x² = 2 2) limx→∞ , value becomes 2x²/x² = 23) limx→ 2- , value becomes = -∞4) lim x→ 2 , value doesn't exist |
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| 2. |
20 gram is what percent of 1 kg? |
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| 3. |
41. 20 gram and 745 milligram is equal to |
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Answer» 745 milligram=0.745 gmso total=20+0.745=20.745gm1 gm=1/1000 kg20.745 gm=0.020745 kg |
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| 4. |
-15/8 %2B 8/2 |
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| 5. |
(7/2)*(3*(3/4) %2B 8/2) |
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| 6. |
2 \varepsilon \wedge S %2B 8 / 2 \cdot |
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| 7. |
Which term of the A.P. 10+ 8 + 6 +_____ is zero? |
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Answer» here first term is 10 comman difference = 2 so 6th term will be 010,8,6,4,2,0 |
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| 8. |
(5 %2B 8)^2=p |
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Answer» a Formula: (a+b)²=a²+b²+2ab a^2+2ab+b^2 is answer |
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| 9. |
find the zero of polynomial 6 |
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| 10. |
(x %2B 8)^2 - 7=0 |
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Answer» this question the answer is 1x |
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| 11. |
¥ . PAbunf o vVid-stR |
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Answer» √(1 + sin∅)/√(1 - sin∅) = √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅) = √(1 + sin∅)²/√(1 -sin²∅) = (1 + sin∅)/√cos²∅ = (1 + sin∅)/cos∅ = 1/cos∅ + sin∅/cos∅ = sec∅ + tan∅ |
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| 12. |
1/15 %2B 8*(2/5) |
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| 13. |
(A) 2, 3(C) 6,-1(B) 2, 3(D) 6, 1STR (A) |
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| 14. |
Find the ratio of:(a) 5 to 50 paise |
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Answer» rs 5 => 5×100 paisa => 500 paisa 500 paisa : 50 paisa => 500/50 => 10/1 => 10 paisa : 1 paisa |
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| 15. |
L Find the ratio of(a)5 to 50 paise |
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Answer» We know that, 1 rupee = 100 paise. 5 rupee = 500 paise. Given, 5 : 50 = 500/50 = 10/1 Ratio of 5:50 = 10:1 5:50 Right click on the link below thk u ₹5=500p500/50=100/10=10/1=10:1 |
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| 16. |
IL Find the ratio of(a) Rs. 5 to 50 Paise |
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| 17. |
MooeL. II6. Find the area of a quadrilateral PORS shown in the alongside figure.7. The area of a quadrilateral is 300 cm2. The perpendiculars from twovertices to the diagonal are 15 cm and 15 cm. What is the length ofits diagonal?2.55.5 cm1.5 cm i |
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| 18. |
find l and b if x + 1 and X+ 2 are factors of x3+3x2-2ax+b |
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| 19. |
ax +b, if x <11, ifx=12ax-b, If x > 1f(x) =is continuous at x = 1, then find a and b. |
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| 20. |
The abscissa of two points A and B are the roots of the equation x+ 2ax-b 0 and their ordinates are theroots of the equation x2+2px-q 0. Find the equation and the radius of the circle with AB as diameter |
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| 21. |
x2 - 2ax – (462 - a?) = 0 |
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Answer» -b+-Vb^2-4ac/2a=2a+-V(2a)^2-4(2a)4(a^2-b^2)/2=V4a^a-16a^3/2=V4(a^2-4a^3/2=Va^2-4a^3 |
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| 22. |
The abscissae of two points A and B are the roots of the equation x^2 + 2ax-b^2=0, and their coordinates are the roots of the equation x^2 + 2px q^2=0. Then the radius of the circle with AB as diameter is |
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Answer» Thank you for giving me a solution of given question What are you doing |
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| 23. |
St83πProve that sin41l+ sin+ sin47-2+ sin8 |
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Answer» Solution -sin⁴(π/8) + sin⁴(3π/8) + sin⁴(5π/8) + sin⁴(7π/8) = [sin²(π/8)]² + [sin²(3π/8)]² + [sin²(5π/8)]² + [sin²(7π/8)]² = [{1 - cos(π/4)}/2]² + [{1 - cos(3π/4)}/2]² + [{1 - cos(5π/4)}/2]² + [{1 - cos(7π/4)}/2]² = [{1 - √2/2 }/2]² + [{1 + √2/2 }/2]² + [{1 + √2/2}/2]² + [{1 - √2/2}/2]² = 2[{1 - √2/2 }/2]² + 2[{1 + √2/2 }/2]² = 2[{2 - √2}/4]² + 2[{2 + √2}/4]² = (1/8)(2 - √2)² + (1/8)(2 + √2)² = (1/8)[(4 - 4√2 + 2) + (4 + 4√2 + 2)] = 12/8 = 3/2 |
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| 24. |
θ <stif sin θthen25a) cos θ =--tan θ-5/3 |
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| 25. |
Solve the polynomial inequality 3x2 – 23x > 36.(192) |
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| 26. |
\operatorname st \theta - \operatorname tan \theta = \frac 2 \operatorname cos ^ 2 \theta - 1 \operatorname sin \theta \operatorname cos \theta |
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Answer» Let theta = xLHS:=cot x - tan x=cos x/sin x - sin x/cos x=(cos^2 x - sin^2 x)/sin x.cos x=[cos^2 x - (1 - cos^2 x)] / sin x. cos x= 2cos^2 x - 1/sin x. cos x= RHS Hence proved |
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| 27. |
A rectangle is (8x + 5) cm long and (5x +3) cm broad. Find its area |
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Answer» Area = lb = (8x+5)(5x+3)= 40x² + 49x + 15 Please hit the like button if this helped you. |
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| 28. |
(a) 2 days 18 hours and 3 days 12 hours |
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Answer» (2+3)days+(18+12 )hours5 days+ 30 hour'6 days + 6 hoursas 24 hours= 1 daysplease like the solution 👍 ✔️ |
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| 29. |
4.8m45cm25 cmii) Area of || gmPORS 252 cm2cmRT = ?cm. Find its arel the formula for area of a gm is applicable t |
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| 30. |
1.Express the following ratios in the simplest terms:(ii) 15:25(iii) 2.5:751.75 (v) 15 kg: 1020 gm (vi) 15 km :2500 cm.tn he divided between Amit and Anshu in the ratio 25 Paise : |
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Answer» i) 6:15 = 2:5 ii) 15:25 = 3:5 iii) 2.5:7.5 = 1:3 iv) 15:175 = 3:35 v) 1500:1020 = 150:102 = 75:51 = 25:17 vi) 1500:2500 = 3:5 If you find this answer helpful then like it. |
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| 31. |
Exercise1. Compare 3:5 and 7:9.2. Write the ratio for each of the following:(1) 4 dozen is to 96(ii) 400 cm to 20 km(IT) 25 paise is to 3.75 (iv) 6 kg is to 720 gm:: |
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Answer» divide 4 by 96 an then write the answer like1:3 : stands for is to and real ans is 1: 24 4 dozen is 48 48/96= 1/2 1kg = 1000 gm 6000/720= 25/3 |
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| 32. |
3tFind the value of sin'(sin ST |
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| 33. |
5. If the circumference and area of a circle are numerically equal, then the diameter of the circle is eto(a) 2(b) π(c) 2π(d) 4 |
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Answer» Circumference of circle = 2*pi*rArea of circle = pi*r*rGiven,2*pi*r = pi*r*rr = 2 Diameter of circle = 2r = 2*2 = 4 (d) is correct option |
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| 34. |
(B) Attempt the following:1 mark/If the surface area of a sphere is numerically equalto its volume. Find its radius.Ans. |
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Answer» Volume of a sphere is numerically equal to its surface area, =) 4/3 * pi * r^3 = 4 * pi * r^2 =) r = 3 Hence d = 2r= 2*3= 6. |
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| 35. |
s fick the correet answer in the following and justify your choice: If the perimeter and areaofàcircle are numerically equal, then the radius of the circle is(b) t unitsa) 2 units(c) 4 units(d) 7 units |
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| 36. |
If P, Q and R can complete a job in 24, 16 and 12 days respectively, how long will they take to complete theJob working together?A5 days 12 hoursB15 days |
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Answer» 48/95 real 3/9 and 3/9 is 1/3 then ans will be 5 ..1/3 |
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| 37. |
Sohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice theside along the road. If the area of this field is 10500 square metres and the perpendiculardistance between two parallel sides is 100 m, find the length of the side along the river100 mRiver |
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| 38. |
An alloyof the alloy.is made of 25on and the rest being zinc. Find the mass of zinc in 720 gm13.is made of 25% copper, 30% ir |
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Answer» 25% copper 30% iron so 100-25-30=45% zincnow alloy is of 720 gm so weight of zinc=(720×45)/100=324 gm |
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| 39. |
5. निम्न में से बेमेल को चुनिए :(A) LEAN (B) LIEN(C) LINE(D) NILE |
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Answer» In LIEN, LINE and NILE letters uesd are L, I, N, E But in LEAN letters used are L, E, A, N Therefore,LEAN is different(A) is correct option |
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| 40. |
rectangle.Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find theratio of speed of Hamid to the speed of Akhtar.4. |
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| 41. |
rectangle.4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Findthe ratio of speed of Hamid to the speed of Akhtar. |
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| 42. |
ŃŃpe 2Find the volume of cuboids withSa. 9a(ii) 3,(v) 4y, 2,4(vi) aD Find the products of the follow(ii) a(i) 2a, 5a2, 9a3 |
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Answer» Volume of cuboid= length*breadth*height=(2a)(5a^2)(9a^3)=90a^6 |
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| 43. |
If 12 labourers take 70 days to complete one work, how many days will 21 labourers take tocomplete the same work? |
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| 44. |
If 8 labourers can earn 9000 in 15 days, how many labourers can earn 6300 in7 days? |
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Answer» No. of labourers. days income 8 15 9000 x 7 6300 No. of men=(8*15)/9000=(x*7)/6300x=(8*15*6300)/(9000*7)=12Ans: 12 men |
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| 45. |
The ride BC, CA and AB, of a AABC are producedin order, forming exterior angles Z ACD, LBAEand 2CBF, then(1) 540(3) 180°(2) 360) of these |
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| 46. |
1. A ABC and A DBC are two isosceles triangle on thesame base BC and vertices. A and D are on the sameside of BC (see Fig 7.39). If AD is extended tointersect BC at P, show that(i) A ABD A ACD(ii) Δ ΑΒΡ Δ ACP(ii) AP bisects A as well as Z D(iv) AP is the perpendicular bisector of BC.Fig. 7.39 |
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| 47. |
aCD is a cyclic quadrilateral (oee Fig 3)Find the angles of the cyclic quadrilateral.-Ax3y-5Flg, 3.1 |
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| 48. |
istance between them2+7 |
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Answer» 2+5/√4²+6² = 7/√16+36 = 7/√52 |
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| 49. |
The minute hand of a circular clock is 15 cm.s How far does the tip of the minute hand move in I hour?(Take π = 3.14) |
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Answer» Inone 1 hour, the tip will complete one rotationradius = 15 cmso, distance = 2πr = 2×π×15 =30π cm = 94.2 cm |
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| 50. |
The minute hand of a watch is 1.5 cm long. How far does its tip move in40 minutes? (Use π = 3.14). |
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