Saved Bookmarks
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. |
नमक ;ज्ञात कीजिए :sin 42° cos 48° + sin 48° cos 42° |
|
Answer» cos 48 = sin(90-48) = sin 42. sin 48 = cos (90-48) = cos 42Therefore, sin42 cos 48 + sin 48 cos 42 = (sin42)^2 + (cos42)^2= 1. |
|
| 2. |
\9:25.ज्ञात कीजिए :Sin 42° cos 48° + sin 4%° cos 42° |
| Answer» | |
| 3. |
11.Four points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for(A) k 0(B) k = 1(C) k-5/13(D) k 5 |
| Answer» | |
| 4. |
1 1+x 111 1+ ythe value of:(B) 0(C)(D) xy(A)1χof x is-[2011A1 |
| Answer» | |
| 5. |
- हेUV e ey 1 |
|
Answer» cos(90-A)=cos90cosA+sin90sinA=0+sinA=sinA |
|
| 6. |
के 1l L न :L_ :\LS_s°) uv—-—SOJ uE—SOD uZ——SOJ = 5032 उश्पा ग048 i |
|
Answer» Like my answer if you find it useful! |
|
| 7. |
UV).11. If a = -9 and b = -6, show that (a - b) = (b - a). |
| Answer» | |
| 8. |
\lim _ { x \rightarrow - a } \frac { x ^ { 9 } + a ^ { 9 } } { x + a } = 9 , \text { find } a |
| Answer» | |
| 9. |
EXERCISE 13.3LEVEL-1B is a chord of a circle with centre O and radius4 cm. AB is of length 4 cm and dividee circle into two segments. Find the area of the minor segment. |
| Answer» | |
| 10. |
10/A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulledalong the ground, away from the wall, at the rate of 2cm/s. How fast is its heighton the wall decreasing when the foot of the ladder is 4 m away from the wall |
|
Answer» length of ladder is l=5mwhen distance(x )of ladder from wall is 4m its height(h) on wall=(l^2 -x^2)^0.5h=(25-16)^2=3mgiven dx/dt =-2cm/s=-0.02m/snow;h^2 =l^2 -x^2differentiating w.r.t. to time t2h dh/dt =0-2x dx/dtputting values 2×3 dh/dt = -2×4×(-0.02)6 dh/dt = 0.16dh/dt =0.08/3 cm/s=8/3 m/s |
|
| 11. |
o eDivide Q_'x".,.âs-,ÂŤ.-fl by x + 2" |
|
Answer» Thank you |
|
| 12. |
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulledalong the ground, away from the wall, at the rate of 2cm/s. How fast is its heighton the wall decreasing when the foot of the ladder is 4 m away from the wall ? |
| Answer» | |
| 13. |
which its volume Is increasing1o A ladder 5 m long is leaning against a wall The bottom of the ladder is pulledalong the ground, away from the wall, at the rate of 2cm/s. How fast is its heightthe wall decreasing when the foot of the ladder is 4 m away from the wall'? |
|
Answer» length of ladder,l=5mdx/dt=-2cm/s=-0.02cm/s... (x is distance of ladder from ground)-ve sign is used because height is decreasingheight, h^2 =l^2 - x^2...... (1)from here we get h=(5^2 - 4^2)^0.5=3mdifferentiating eq. (1)2h dh/dt =0-2x dx/dt .... (length of ladder is constant so its differentiation is 0)2h dh/dt = -2x dx/dtputting values2×3 dh/dt = -2×4×(-0. 02)dh/dt=0.08/3 cm/s=8/3 m/s |
|
| 14. |
Draw a line segment of lenth 7.6 cm and divide it in the ratio 5: 8. Measureparts.e ratio 8. |
| Answer» | |
| 15. |
(x-9)(x+9) |
| Answer» | |
| 16. |
13. Divide the sum of 13 and 12 by theproduct of 31 and |
| Answer» | |
| 17. |
12by the13. Divide the sum of 13 andproduct of 31 and |
| Answer» | |
| 18. |
A godown measures 40 m x 25 m × 15 m. Find the maximum number ofwooden crateseach measuring 1 .5 m × 1.25 m × 0.5 m that can be stored in the godown. |
| Answer» | |
| 19. |
godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crateach measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. |
| Answer» | |
| 20. |
5. A godown measures 40 m × 25 m × 15 m. Find the maximum numberof wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can bestored in the godown. |
| Answer» | |
| 21. |
(1) x3 + x2 + x + 1 |
|
Answer» (x+1) (x+1) Hey please like my answer 🙏 and select the best 😃 x³+x²+x+1=x²(x+1)+1(x+1)=(x²+1)(x+1) |
|
| 22. |
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crateseach measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.7. |
| Answer» | |
| 23. |
An iron tank 20 m long, 10 m wide and 8 m deep is to be made from iron sheet 4 m wide. Iittank is open from the top, find the cost of the iron sheet at the rate or 5 per metre. |
|
Answer» Given that length of theopen iron tank = 20 m. Breadth of the closed iron tank = 10 m The height of the closed iron tank = 8 m. Surface Area of the tank = 2[ bh + hl]= 2[10×8+8×20]= 2×240=480m^2 Area of the rectangular iron sheet = 480m^2so that length × breadth = 480m^2 But breadth of sheet = 4 m (given) ∴ length = 480 / 4 = 120 m . Given that cost of iron, the sheet used at the rate of Rs 5 per metre.The total cost of the sheet = Rs. (120 × 5) = Rs. 600. |
|
| 24. |
2 32.32 a t3 |
| Answer» | |
| 25. |
A godown measures 40 m × 25 m × 1 5 m. Find the maximum number of wooden crateseach measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.7. |
| Answer» | |
| 26. |
(1) x3 + x2 + x +1 |
|
Answer» (x+1) x(x+1)is the best answer |
|
| 27. |
A godown measures 40 m × 25 m × 1 5 m. Find the maximum number ofwooden crateseach measuring 1.5 m × 1 .25 m × 0.5 m that can be stored in the godown.7. |
| Answer» | |
| 28. |
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crateseach measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. |
| Answer» | |
| 29. |
Divide.13. -- 16 |
|
Answer» 4/15÷16 4/15×1/16=1/60 1/60 is the correct answer 60 is right answer for question |
|
| 30. |
Divide E by 13 |
|
Answer» like if this is right please 🙏 the. correct answer is. 48/49 48/49 is right answer of this question. please like my answer. 48/49 is the answer of the question (12/7)÷(7/4)(12/7)×(4/7)48/49is correct answer 48/49 is the right to 48/49 is the answer.... 48/49 is the right answer 48/49 is the correct answer of the given question 48/49 is the right answer 48/49 is the correct answer (12/7) ÷ (7/4)(12/7) ÷ (4/7)48/49 is the correct answer |
|
| 31. |
x3 + 8x2 + 19x + 12 |
| Answer» | |
| 32. |
1. x38x2 19x 12 |
| Answer» | |
| 33. |
1. Construct△XYZ in which X、4.5 cm. YZ5 cm and ZX-6 cm. |
| Answer» | |
| 34. |
Simplity(al2014 -5) + 2 and find its values for 0) a23xia x+1)-1 and find its values for () x1 (i)xm(m - 3m -2) +5 and find its values for (i) m 2 () m2() as. Add:(a) 2m(2 -m) and 4m 8md 4pg (p - q) and Sp'qią 2q)(b) Sabla - b) and 2ab(2a 4b)(d) 7xylx+ 2y) and Sy(we 2xy)6. Subtract:(a) 5xtx-3y+ 22) from 3x(4z -2y + 5x)(b) 310- 4m + 2n) from 2/(3n - 4m +5)c 3abla+ b+ d -2ab (a - b+ 2c) from Sabl- a+b+ 3c |
| Answer» | |
| 35. |
Construct&XYZ in which XY = 4.5 cm, YZ5 cm and ZX = 6 cm. |
| Answer» | |
| 36. |
Solve for x:x2 + 5x-(a2 + a-6) = 0 |
|
Answer» x^2 + 5x - (a^2 + a - 6) = 0 x^2 + 5x - (a^2 + 3a - 2a - 6) = 0 x^2 + 5x - [a(a + 3) - 2(a + 3)] = 0 x^2 + 5x + (a + 3)(a - 2) = 0 x^2 + (a+3)x - (a-2)x + (a+3)(a-2) = 0 x(x + a + 3) - (a - 2)(x + a + 3) = 0 (x - a + 2)(x + a + 3) = 0 x = (a - 2), - (a + 3) Sorry I can't di these |
|
| 37. |
The maximum volume (in cu. m) of the rightcircular cone having slant height 3m is:(2) 6 π(3) 213 π(4)-π |
| Answer» | |
| 38. |
tent is made of 4.5 m wide tarpaulin. Vertical height of the conical tent is 4 m and baselength of the tarpaulin used, assuming that 10% of extra material is required for25/ Aine margins and wastage in cuttingTake T3.14 |
|
Answer» Applying Pythagors' theorem,slant height, l² = r² +h²⇒ l² = 3² +4²⇒l² =25⇒slant height = 5mnow, curved surface area =π× r×l = 3.14×3×5 = 47.1 m²now, 10% extra material is required for stitching and wastage = 0.1×47.1=4.71 m²total area = 47.1 +4.71 =51.81 m²now, L×B = 51.81⇒Length ,L = 51.81/4.5 = 11.51 m |
|
| 39. |
puleentage increase in thesurlaca1. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made Deteminecost of iron sheet used at the rate of 2 5 per metre sheet, sheet being 2 m wideby making, Ravish wanted to make a temporary shelter for his carstructure with tarpaulin that covers all the four sid |
|
Answer» Given that length of theclosed iron tank = 12 m. Breadth of the closed iron tank = 9 m The height of the closed iron tank = 4 m. Total Surface Area of the tank = 2[lb + bh + hl] = 2[12×9+9×4+4×12] = 384 Area of the rectangular iron sheet = 384So that length × breadth = 384 But the breadth of sheet = 2 m (given) ∴ length = 384 / 2 = 192 m . Given that cost of iron, the sheet used at the rate of Rs 5 per metre. The total cost of the sheet = Rs. (192 × 5) = Rs. 960. |
|
| 40. |
1./X3 + 8x2 + 19x + 12 |
| Answer» | |
| 41. |
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crateseach measuring 1.5 mx 1.25 m x 0.5 m that can be stored in the godown. |
| Answer» | |
| 42. |
x3-19x+30x3x- 40 |
|
Answer» (x*x*x-19x+30)/(x*x-3x-40) = (x*x*x-2x*x+2x*x-4x-15x+30)/(x*x-8x+5x-40) = ((x-2)(x*x+2x-15))/((x-8)(x+5)) = ((x-2)(x+5)(x-3))/((x-8)(x+5)) = ((x-2)(x-3))/(x-8) If you find this answer helpful then like it. |
|
| 43. |
A godown measures 40 m * 25 m x 15 m. Find the maximum number of wooden crates eachmeasuring 1.5 mx 1.25 m 0.5 m that can be stored in the godown. |
| Answer» | |
| 44. |
7मा(s)*2 से3 सेमी1331 सेमी1 सेमी2 ममीआकृति से घिरे क्षेत्र का क्षेत्रफल क्या होगा?(A) 10 वर्ग सेमी(B) 13 वर्ग सेमी(C) 14 वर्ग सेमी(D) 20 वर्ग सेमी |
|
Answer» the answer is that 20 varg cm |
|
| 45. |
A godown measures 40 m × 25 m × 15 m. Find the maximum number of woodeneach measuring 1.5 m x 1.25 mx 0.5 m that can be stored in the godown.7.ce |
| Answer» | |
| 46. |
13. (a + b)3-a-b |
| Answer» | |
| 47. |
A godown measures 40 m x 25 m × 15 m. Find the maximum number of wooden craeach measuring 1.5 mx 1.25 m x 0.5 m that can be stored in the godown.7. |
| Answer» | |
| 48. |
0=6+ + a2 + x91 - + zx pue 6= zk + xx |
|
Answer» x^2 + y^2=9; c1=0(0,0), r1=3; x^2+y^2-16x+2y+49=0=( x-8)^2+(y+1)^2= c2=(8, -1), r2=4; c1c2=V65> r1+r2; p=(-8a, a), q=(8b, -b), b>0, c1=c1c2x r2/r1=V65x3/(4-3)=3V65=a=3, p(-24, 3); pp1 perpencular p1c1=(y1-3)/( x1+24)×y1/ x1=-1=x1^2+ y^2+24x1-3y1=0; 8x1-y11+3 (or)y1=8×1+3=65x1^2+48x1=0, p1=(0,3) p2=(-48/65, -189/65), t1:y=3, t2=16x+63y=195 |
|
| 49. |
)A2 = {x : x2-4x-5-0) |
|
Answer» A = { x : x² - 4x - 5 = 0 } x² - 4x - 5 = 0 x² + x - 5x - 5 = 0 x ( x + 1) -5 ( x +1) = 0 (x+1) ( x - 5) = 0 So in roaster form A = { -1, 5} |
|
| 50. |
x² + 5x-(a2+a-6) = 0 |
|
Answer» ⇒x2+5x−(a+3)(a−2)=0 ⇒x2+[(a+3)−(a−2)]x−(a+3)(a−2)=0 ⇒x2+(a+3)x−(a−2)x−(a+3)(a−2)=0 ⇒x[x+(a+3)]−(a−2)[x+(a+3)]=0 ⇒[x+(a+3)][x−(a−2)]=0 ⇒[x+(a+3)]=0 or [x−(a−2)]=0 ⇒x=−(a+3) or x=(a−2) |
|