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. |
tper meter squareCarved surface area of a hollow cylinder is 4224 cm' A rectangutar sheethaving width 33 cm is formed cutting it along its height. Find perimeter ofshoet |
| Answer» | |
| 2. |
14. In Figure 14, ABIICD. Transversal t cuts ABat E and CD at F. If EG and FH are thebisectors of alternate angles ZBEF andZEFC, respectively, prove that EG| HF.A.|| CD. |
| Answer» | |
| 3. |
. Divide 32 into four parts, which are in A.P. such that the ratio pf produčt of extremes tothe product of means is 7: 15. |
|
Answer» Let the 4 numbers are: a-3d, a-d, a+d, a+3d where sum of 4 terms = 32, hence 4a = 32, implies a = 8; according to question: (8-3d)(8+3d)/(8-d)(8+d) = 7/15 solving we get answer: 2,6,10,14 can you explain me last 2 line with the answer |
|
| 4. |
Divide 56 in four parts in A.P. such that the ratio of the product of theirproduct of their means is 5:6. |
|
Answer» 4 no =a-3d,a-d,a+d,a+3dsum , 4a=56a=14(a-3d)(a+3d)/(a-d)(a+d)=5:66(a²-9d²)=5(a²-d²)a²=54d² - 5d²14² =49d²d²=14² /7²d=±14/7=±2for d=2, a-3d=14-6=8a-d=14-2=12a+d=14+2=16a+3d=14+6=20Numbers are 8,12,16,20 |
|
| 5. |
। ®)£ w1o |
|
Answer» ( 3y + 1/4y)³ (3y)³ + (1/4y)³ + 3 × 3y × 1/4y ( 3y + 1/4y) 27y³ + 1/64y³+ 9/4 ( 3y + 1/4y) 27y³ + 1/64y³+ 27y/4 + 9/16y |
|
| 6. |
14. In Figure 14, AB|CD. Transversal t cuts ABat E and CD at F. If EG and FH are thebisectors of alternate angles ZBEF andZEFC, respectively, prove that EG HF. |
| Answer» | |
| 7. |
जा 8° 4 sin? 72° W1 WA ¥W \ |
|
Answer» sin^2 18° + sin^2 72°=sin^2 18° + sin^2 (90-18)=sin^2 18° + cos^2 18°=1 |
|
| 8. |
Find five rational numbers betweenUWWA |
|
Answer» multiply 3/5 (both numerator and denominator) with 6similarly multiply 4/5 with 6so, 3×6/5×6=18/30 and 4×6/5×6=24/30 so, 19/30,20/30,21/30,22/30,23/30 are the five rational numbers between 3/5 and 4/5 u just multiply the no with 10 then found 3/5*10=30/50 and 4/5*10=40/50Then the answer is 31/50,32/50,33/50,34/50 and ,35/50 1225this is answer of the your question25 |
|
| 9. |
3.Find five rational numbers betweenuwand |
|
Answer» the given fractions have same denominator so,we want 5 rational no,so 5+1=6 always add one more no than the given rational no. we multiply it and divide it by 63/5*6=18/304/5*6=24/30hence the five rational no are 19/30 ,2 |
|
| 10. |
नलि - PP+ |
|
Answer» x² × ( x/y - y/x)( y/x + x/y) y² (xy)² ( x² - y²)/(xy) ×( x² + y²)/xy (x² - y²) ( x² + y²) x⁴ - y⁴ |
|
| 11. |
e)4Simplify the following using p -(pp-q):(a) 382 -122 (b) 65 x55 (c) 43-17 m |
|
Answer» solve question b |
|
| 12. |
-25. Subtract the sum- andfrom the sum ofUw |
|
Answer» 3/5+7/3=44/15and -11/5+(-2/3)=-43/15let subtract 44/15-(-43/15)=44-(-43)/15=44+43/15=87/15 87/15 in the right way and red 87/15 is the correct answer of the given question |
|
| 13. |
เคนเคฟSTLo PP Y |
|
Answer» Joining of two things is called additionthanksplease like the solution 👍 ✔️ addition of two signed or unsigned numbers called addition😀 addition is thing from which we can add somethingfor eg.12 12 24 |
|
| 14. |
A medicine capsule is in the shape ofacylinder with two hemispheres stuck to eachof its ends (see Fig. 13.10). The length ofthe entire capsule is 14 mm and the diameterof the capsule is 5 mm. Find its surface area. |
| Answer» | |
| 15. |
A medicine capsule is in the shape of acylinder with two hemispheres stuck to eaclhof its ends (see Fig. 13.10). The length ofthe entire capsule is 14 mm and the diameterof the capsule is 5 mm. Find its surface area. |
| Answer» | |
| 16. |
A medicine capsule is in the shape of acylinder with two hemispheres stuck to eachof its ends (see Fig. 13.10). The length ofthe entire capsule is 14 mm and the diameterf the capsule is 5 mm. Find its surface area. |
| Answer» | |
| 17. |
6. A medicine capsule is in the shape of acylinder with two hemispheres stuck to eachof its ends (see Fig. 13.10). The length ofthe entire capsule is 14 mm and the diameterof the capsule is 5 mm. Find its surface arca. |
| Answer» | |
| 18. |
-1.. 62 0) त्रिज्या वाले एक वृत्त के एक बिन्वखंड का क्षेत्रफल ज्ञात कीजिए, जिसका कोण 607 है।(oo e D el L L LR |
| Answer» | |
| 19. |
gt ki एच e दि कर पुलिस-jee |
|
Answer» गणितमें किसीवास्तविक संख्याकानिरपेक्ष मानया 'निरपेक्ष मूल्य' (absolute value) या 'मापांक' (modulus) |a| उस संख्या के चिह्न के बिना उसके आंकिक मान के बराबर होता है। उदाहरण के लिये 3 का निरपेक्ष मान 3 है, तथा -3 का भी निरपेक्ष मान भी 3 ही है। किसी संख्या के निरपेक्ष मान को उस संख्या की शून्य से दूरी के बराबर समझा जा सकता है। |
|
| 20. |
ISR T~ =e A - -(दाbz7=9 |
|
Answer» -2xy =20-xy= 20/2-xy= 10 |
|
| 21. |
6. A medicine capsule is in the shape of acylinder with two hemispheres stuck to eachof its ends (see Fig. 13.10). The length ofthe entire capsule is 14 mm and the diameterof the capsule is 5 mm. Find its surface area. |
| Answer» | |
| 22. |
7—4J_l47 +8—3\/—_l92 —Sl,/‘75 |
|
Answer» 4/7V147 + 3/8 V192 - 1/5V75 = 4/7 V3x49 + 3/8 V3x 64 - 1/5V3x25 = 4/7 (7)V3 + 3/8 (8)V3- 5V3/5 = 4V3 + 3V3 -V3==7V3-V3=6V3 |
|
| 23. |
“Tr | d Jdo), sl£ ent) रत 3लाफट गन 2an+27 |
| Answer» | |
| 24. |
ŕ¤ŕ¤Śqbz % b LE 2:05 |
| Answer» | |
| 25. |
2 [ 22 EEBl LI B bz eधवन 5 “5:2८ w77, _[}-_2‘ # 4 |
|
Answer» hit like if you find it useful |
|
| 26. |
s bz b i el L 1l OLk |
|
Answer» 7 km= 7000mअनुपात= 140/7000= 14/700= 2/100= 1/50 |
|
| 27. |
6. A medicine capsule is in the shape of acylinder with two hemispheres stuck to eachof its ends (see Fig. 13.10), The length ofthe entire capsule is 14 mm and the diameterof the capsule is 5 mm. Find its surface area. |
| Answer» | |
| 28. |
120201o (b +d) ) iy SRy 0 ‘ d b e bageह 20 (2 मे 1५० छ 20IBE W) Db 0T k&K - |
| Answer» | |
| 29. |
35. Find the condition that r + (p + qx + a is divisibleby (x + p + 9) |
|
Answer» Let f(x) = x^3 + (p+q) x + a. We use Remainder theorem to derive our result. Now for (x + p + q) to be a factor of f(x), f(-[p+q]) should be equal to zero. => -(p+q)^3 + (p+q)[-(p+q)] + a = 0. => a = (p+q)^2 + (p+q)^3 => a = (p+q+1)(p+q)^2 |
|
| 30. |
नमक v w : ‘ aw ‘ Oyl AR 0-2५ 020९ दाग °Q 8P ,,‘awy ‘9 128 k¥ whg करू छा लाल f)e‘ - O Wlebkie (p) |
|
Answer» k + g = 48 2k + 4g = 140 2k + 4g - 2k - 2g = 140 - 96 2g = 44 g = 22 k = 26 |
|
| 31. |
whose perimeter is 320 m'indthealeadlasquaréparkFind the breadth of a rectangular plot of land, if its area is 440 m2 and the length is22 m. Also find its perimeter. |
| Answer» | |
| 32. |
W PUDLve:20-KX+424-5-+3-0 |
|
Answer» jzjshsjskskskkskskaoa |
|
| 33. |
The three sides of a triangle are 6.2 cm, 4.2 cm and 2.1 cm long. Find its perimeter. |
|
Answer» 13.5 cm is the correct answer of the given question peri. of triangle= S+S+S = 6.2+4.2+2.1 = 12.5cm the correct answer is 12.5cm 12.5cm is the correct answer perimeter of triangle equals to sum of all side therefore add 6.2 + 4.2 + 2.1 CM equals to 12.5 CM |
|
| 34. |
aleaaldutotalSond its curvedA road roller is 2m long and its diameter is 112cm. How many revolutions will itmake to level a road with area 7040m2?surlace |
|
Answer» PLEASE LIKE IT, IF YOU FIND THE ANSWER HELPFUL. |
|
| 35. |
tan(900-6)cosec%, tan θcos2 θ |
|
Answer» tan(90-∅) = cot∅ cot∅/cosec²∅.tan∅= cot²∅/cosec²∅= cos²∅/sin²∅*cosec²∅= cos²∅ |
|
| 36. |
हे... 25 0515... 2. ८५ shomb aLy cm, 5706... (5 नगर i”!‘zs,_cm.‘_"“’ and |
| Answer» | |
| 37. |
Dsp a+b = 90, tan a= 2, then Rind Cot Bz.atb- 90 |
| Answer» | |
| 38. |
12. Find the area of a square inscribed in a circle ofradius x cm. |
| Answer» | |
| 39. |
दि « psrpTpu) Mo T पथदः। फै opvg U S P TP Y, /9o कप ए हा ImPW 0% o "mpraeso) il st ) e W हाकै - bus cag i |
| Answer» | |
| 40. |
PJea356. Find ZP, ZQ, ZR and ZS in the adjoining figure.REA21bi379R |
|
Answer» ∠R + ∠RQS + ∠RSQ = 180 (Angle sum property) => ∠R + 35 + 37 = 180=> ∠R = 180 - 72 = 108∠R = ∠P = 108 (∵ Opposite angles of a parallelogram are equal) ∠R + ∠S = 180 (∵ Adjacent angles of a parallelogram are supplementary) ∠S = 72 ∠Q = 72 |
|
| 41. |
a)Find Linn | |
| Answer» | |
| 42. |
Evaluate: linn,-0 |
| Answer» | |
| 43. |
की स्व राiy B s> e [हाफSt (i €F i e Mo 2 T bl e il कप |
| Answer» | |
| 44. |
12 cmcm4 In the adjoining figure, BD and CE intersect eachother at the point P. Is ΔΡBC ~ ΔΡDE? Give reasonsfor your answer6 cm10 cm |
|
Answer» thanks a lot |
|
| 45. |
A rectangular sheet of paper is 12Find its perimetercm long and 10cm wide |
|
Answer» 139/3 |
|
| 46. |
9. A rectangular sheet of paper is 15 cm long and 12 cm wide. Find its perimeter. |
| Answer» | |
| 47. |
a of quadrilateral PQRS as shown in the adjoining fFind the areadjoining figure12 cm8 cmv |
|
Answer» Area of quadrilateral =Area of PQR + Area of PSR= 0.5*12*8 + 0.5*12*4= 48+24.= 72 cm² Please hit the like button if this helped you thanks |
|
| 48. |
लिन... Rtpe ¢f I BYSTRE 07 Mo V7 OV 1 0-0.. dek 2 O¥a7 = ovgy e (W)% DR ७ न o3 GraY = Opgy e 0deb 2 0907 न 09827 e (M)2कि o (8 20aV = H0gy ke (Wदे ८ 30-28 e Oवह ८ (0 न 80 pe e | gg St थार ३ gy b B 2ean s 0 2 |’7 &0 ogy डा पेकि¥ |
|
Answer» आमने सामने की भुजाएं बराबर और समान्तर होती हैं। विकर्ण एक दूसरे को समद्विभाजित करते हैं। आमने सामने के कोण बराबर होते हैं। विकर्ण आमने सामने के कोण को समद्विभाजित करते हैं wrong पूरा प्रश्न हल करो |
|
| 49. |
Mathematd) In the adjoining figure, AB isdiameter of the circle, AC = 6 cmand BC-8 cm. Find the area of Athe shaded portion. (UseT 314)4 |
| Answer» | |
| 50. |
From a circle ofradius 15 cm. a sectorwith angie 216% cut out and its boundingare bent so as to form a cone. Find its volume.radii |
| Answer» | |