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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. |
uc9. The product of two nu . If one ofthe numbers is — find the other number. |
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| 2. |
14 Find the HCF of each of the following by prime factorization method.a. 65, 91b. 120, 40c. 700, 900d. 154,242. |
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| 3. |
CUI Duys.11: A sum of 7000 is divided among A, B, C in such a way that the shares of A and B are in the ratio 2:3and those of B and C are in the ratio 4:5. Find B's share. |
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Answer» the answer of b share is 2400 A:B= 2:3, B:C = 4:5(A:B)×4 and (B:C)×3then. A:B:C = 8:12:15SO A+B+C = 70008x+12x+15x = 7000 35x = 7000 x = 7000/35 x = 200so B's share = 12×200 = 2400 A:B=2.3. B:C=4:5(A;B)×4. (B:C)×3 SO. A:B:C = 8:12:15 A+B+C. =70008X+12X+15X = 700036X = 7000X=35/7000X=200B IS EQUAL TO 12×200=2,400 |
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| 4. |
2 opandy ulinclos 4ind the ctthe walls |
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Answer» Length of a door = 2.5m Breadth of the door = 1.2m area of the door = l * b = 2.5m * 1.2m = 3m^2 ∴ area of 2 doors = 2 * 3 = 6m^2 Length of the window = 1.5m Breadth of the window = 1m Area of the window = l * b = 1.5 * 1 = 1.5m^2 ∴ area of4 windows= 4* 1.5 = 6m^2 Area of the doors + Area of the windows = 6m^2+ 6m^2= 12m^2 Length of the wall = 12.5m Breadth of the wall = 9m Height of the wall = 7m Area of the wall = LSA of a cuboid = 2h(l+b) =2 * 7 (12.5 + 9) = 14 * 21.5 = 301m^2 Required area of the wall to be painted = Area of the wall - Area of the doors and windows i.e., area of the wall to be painted = 301m^2- 12m^2 area of the wall to be painted = 289m^2 Cost of painting 1m^2of the wall = Rs. 3.50 Cost of painting 289m^2of the wall = Rs. (3.50 * 289) Cost of painting the wall = Rs. 1011.5 Like my answer if you find it useful! |
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| 5. |
58. OP,OQ, OR&are 4'rays prove that <POQ + <QOR + |
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| 6. |
17. Find the area of the shaded region in Fig. 12.22, if ABCD is a square ofside 14 cm and APD and BPC are semicircles.NCERT |
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Answer» Where is the figure? |
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| 7. |
8.In figure, area of shaded region is:2 |
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Answer» what was the reason for putting barriers to foreign trade and investment by the Indian government? why did it wish to remove these barriers? |
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| 8. |
R=A¢—x {91=AG — x¢ |
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| 9. |
3. In Fig. 10.33, find x. Further find 4BOC, LCOD and ZAODl pairs11. In Fig. 10.40, ALDCA and 412. Given ZPORline. (Fig. 10.(x + 10)r+20°t 20。Fig. 10.33Fiq. 10.34 |
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Answer» sum of angles AOD+DOC+COB = 180°... (linear pair) => x+10+x+x+20 = 180°=> 3x+30 = 180°=> 3x = 180-30 = 150°=> x = 150/3 = 50° now, angle AOD = x+10 = 60° angle DOC = 50°angle BOC = x+20 = 70° |
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| 10. |
(NOD) 2 22 145 91 02 टो शुरू ६.090 0£( ()ophiterz R 1k OF IR YT TT TB i afie 2 28- |
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Answer» 12= 2*2*324= 2*2*2*330= 2*2*5LCM= 2*2*2*2*3*5= 240 |
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| 11. |
TRY THESEcind the area of the following trapemFeilFig 11% |
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| 12. |
८0८० छ (व) 1-94] ४8 2४ 2 है 10० 1 | हु ४ 91 151 डक ०i i S X R N |
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Answer» hit like if you find it useful |
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| 13. |
छ्यि 6.33 उऊ, ?0 जाक 1२४ मात्नाग फूयनकA FAGAIF 331 (2R | AB Sieifers Ry0 माटनानव 9 विन्यूज शविद खाक धंजिकनिऊ 2BC #1223 91 RS it C R sificg | 2वश्धि 'याएकी (1) निटिगएव विधवीऊकमऊ थडिकनिऊटड्टड। घमाग कदा (¥ AB || CDi |
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Answer» Apologies but we are currently providing answers to questions in English only. We are going to include other languages very soon. So keep checking this app in the coming months. |
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| 14. |
r(彚) is the midpoint of theine segmentjoining the points (2,0)and(0,) then show391that the line 5x + 3y2point (-1, 3p.0 passes through the |
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Answer» Let ( x₁,y₁ = A( 2 , 0 ) , ( x₂ , y₂ ) = B ( 0 , 2/9 ) ; midpoint of joining of A and B = ( x₁+ x₂/2, y₁+ y₂ /2 ) ( 1 , p/ 3 ) = ( 0 + 2 /2 , 0 + 2/9 / 2 ) = ( 1 , 1/9 ) p/3 = 1/9 [∵ If ( a , b ) = ( c , d ) then a = c and b = d ] p = 1/3 --- ( 1 ) according to the problem given , put ( -1 , 3p ) in the equation 5x + 3y + 2 =0 5 ( -1 ) + 3× ( 3p ) + 2 = 0 -5 + 3× 3 ( 1/3 ) + 2 =0 [ from ( 1 ) ] -5 + 3 + 2 =0 0 = 0 [ true ] Therefore , 5x + 3y + 2 =0 line passes through the point ( -1 , 3p ) . |
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| 15. |
1. Find the value of m and s in the given table if r and y vary directly.12 15a. x 565 91 156 |
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Answer» a. m=7 n=195b. m=30 n=2 |
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| 16. |
Dund+)R-25.500-25 6121 The diameter of the wheel of a bus ts 91 cm. How many kilometres will it travel in500 roladons of the wheel?buta wulan 9 duys. How many days will 12 luhourer's take to build the same |
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| 17. |
In the figure LABC = 80°. Find 2DGC andDEF given ABDE and BCIEF |
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| 18. |
80. Find the area of the shaded region givNCERT Exemplarfigure. |
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| 19. |
te A wire can be bent in the form of a civele of radius 35 cm. If it is hent in the frm f swhat will be its areal |
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Answer» Circumference of circle=2π35 cm circumference=70*22/7=220 cm circumference of circle=perimeter of square 220cm=perimeter of square 220cm=4*side 55cm=side area =55*55cm^2 area of square=3025 cm^2 |
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| 20. |
R. A shopkeeper bought wheat for 35000. Due to leakage in the godown of the total wheat was spoiledsold the good wheat at again of 10% and the spoiled wheat at a loss of 25%. Find his total gain or loss perce |
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Answer» Find amount remaining after leakage: leakage = 1/7 x 35000 = 5000 Remaining = 35000 - 5000 = 30000 . Amount earned from the good wheat: 10% Gain = 10% x 30000 = 0.1 x 30000 = 3000 Total gain = 30000 + 3000 = 33 000 . Amount gain from the spoilt wheat: 25% loss = 25% x 5000 = 0.25 x 5000 = 1250 Total wheat sold = 5000 - 1240 = 3750 . Find total amount : Total amount = 33 000 + 3750 = 36750 . Find gain percentage: Gain = 36750 - 35000 = 1750 Gain percentage = 1750/35000 x 100 = 5% Answer: The gain was 5% |
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| 21. |
11.31,a) Which square has the larger periWhich is larger, perimeter of smaller square or thecircumference of the circle?gat2Fig 11.31 |
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| 22. |
A shopkeeper bought wheat forHe sold the good wheat at a gain of 10% and the spoiled wheat at a loss of 25%. Find his total gain or losspercent.35,000. Due to leakage in the godown 2 of the total wheat was spoiled6. |
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| 23. |
Cost of 5 kg of wheat is R 91.50.(a) What will be the cost of 8 kg of wheat?(b) What quantity of wheat can be purchased in? 183?4. |
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Answer» My question is another |
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| 24. |
7. A chord of a circle of radius 12 cm subtends anangle of 120o at the centre. Find the area of thecorresponding segment of the circle.(Use π= 3.14 and、/3 = 1 .73) |
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a of themajor segment APB, in the figure, of a circle of radius 35 cm and ZAOB-90.segmenta athe area of the8Or |
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Answer» wrong answer is710 2/7 |
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| 26. |
4. What is the angle in major segment? |
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Answer» The angle in the major segment is obtuse |
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| 27. |
16. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is60". Also find the area of the corresponding major segment. [Use Ď-] |
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Answer» Let the center of circle be O.Let the radii of the sector be OA and OB.Given∠AOB = 60°. In theΔAOB, OA=OB, So it is isosceles. => ∠OAB = OBA = [180° -∠AOB ] /2 = 60°Hence,ΔAOB is equilateral. Hence, AB = OA = 14 cm. Area ofΔAOB =√3/4 * R²Area of sector OAB =π R² * (60°/360°) =π R² / 6 Area of minor segment AB : πR²/ 6 -√3/4 R² = 22/7 * 14² /6 - √3 /4 * 14² cm² = 17.75 cm² |
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| 28. |
15. Find the area of the major segment APB in adjoining figure, of a cirele of radius 35 ent and90°"フ |
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Answer» Radius of circle = 35 cm see attachment , you can say that , major segment is the sum of right angled triangle and 3/4 th part of circle { as you know, centre of circle is 360° ,means area of complete circle is πr² , means for 360° , area of circle =πr², so, for 270° , area of circular part = 270°/360° πr² = 3/4 πr²} area of major segment = area of triangle + 3/4 area of circle = 1/2 height × base + 3/4 πr² = 1/2 × r × r + 3/4πr² = r²/4( 2 + 3π)= r²/4 × (2 + 66/7) = 35 × 35/4 × 80/7 = 35 × 5/4 × 80= 35 × 5 × 20 = 3500 cm² Hence, area of major segment is 3500 cm² Like my answer if you find it useful! |
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| 29. |
Q.4.188x18x8-8088(1) 18989?(2) 19022Q.5. 9984 384(1) 636(2) 646Q.6.(1) 17(2) 120.7× 0.25 × 140 = ?+ 5 |
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Answer» ? =(9984 ÷384)^2 =(26)^2 =676 |
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| 30. |
.-t .tR . g. o bसाय खाते ¢ ¥4 & 4 to है १ दर एए है. WA |
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Answer» STEPS OF CONSTRUCTION: 1.Draw a line segment AB = 9 CM2. Draw a circle with Centre O and radius 5 cm and the circle with Centre B and radius 3 cm.3. Now bisect AB. Let O be the midpoint of AB.4. Take O as Centre and AO as radius and draw a dotted circle which intersects the two given circles at N, Q, M and P.5. Join AN, AQ,BM and BP. These are the required tangents to each circle from the centre of the Other circle. hit like if you find it useful |
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| 31. |
In figure, there is achord AB of a circle,with centre O andradius 10 cm, thatsubtends a rightangle at the centre ofthe circle. Find thearea of the minorsegment AQBPHence find the areaof major segment ALBQA. [take π=3.14] |
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| 32. |
1. Construct a quadrilateral ABCD, given that AB = 4.2 cm, BC5 cm, CD = 45 cm, 4-75°, andLC = 100". |
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| 33. |
e adjoining figure, find the area of the major segment APB ofle of radius 35 cm and <AOB = 90°. [Use Ď =(C.B.S.E. 2011) |
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Answer» Radius of circle = 35 cm you can say that , major segment is the sum of right angled triangle and 3/4 th part of circle { as you know, centre of circle is 360° ,means area of complete circle is πr² , means for 360° , area of circle =πr², so, for 270° , area of circular part = 270°/360° πr² = 3/4 πr²} area of major segment = area of triangle + 3/4 area of circle = 1/2 height × base + 3/4 πr² = 1/2 × r × r + 3/4πr² = r²/4( 2 + 3π)= r²/4 × (2 + 66/7) = 35 × 35/4 × 80/7 = 35 × 5/4 × 80= 35 × 5 × 20 = 3500 cm² Hence, area of major segment is 3500 cm² |
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| 34. |
Q.6. Solve the following questions. (Any one)(1) Find the value of k so that PQ will be parallel to RS, where P(2, 4), Q(3, 6), R(8, l), and13 Marks]S(10, k). |
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| 35. |
Q 3: a andare zeroes of polynomial 4x-2x+k+7 Find the value of k. |
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Answer» Since, a and 1/a are the zeroes of the polynomial given, therefore a* 1/a = (k+7)/4. [ Product of zeroes= c/a]Therefore, (k+7)/4=1Ie,. k+7=4So,. k= -3 Like my answer if you find it useful! |
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| 36. |
k4t&81 291 () ६८1 (पी eyl ()८ 8 दा %8 हा L T[t BIE [ 1o दर b Yot k) (07T "B K) IRQEY |
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Answer» difference is increasing by double so answer is 162 |
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| 37. |
If ΔΑΒC-aQRP, ar(AORP. 4, and BC- 15 cm, then find PR. |
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Answer» Like if you find it useful |
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| 38. |
Ar(AABC) q5. If AABC AQRP,-and BC-15 cm, then find PR.Ar(AQRP) 4 |
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Answer» Ar(ABC)/ar(PQR)=K² K IS CONSTANT K=3/2 SINCE TRIANGLES ARE SIMILAR BC/PR=K PR=BC/KPR=15x2/3 PR=10cm |
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| 39. |
If AABC AQRP, arQRP)WAQP 4 and BC 15 cm, then find PR. |
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| 40. |
A chord ofa circle ofradius l 2cm. subtends an angle of I 20° at the centre. Find﹄of the corresponding minor segment of the circle (uset -3.14 and /3-1.732 |
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| 41. |
the difference between circumference and radius of s circle is 37m. find the circumference of the circle |
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Answer» 2πr-r=37r=37/(2π-1)radius=7m Circumference=2π7=14π=44m |
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| 42. |
ill. Hind its circumference.he circumference of a circle exceeds its diameter by 45 cm. Find thercumference of the circle. |
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| 43. |
Selling price of a book is 5/3 of its cost price. Find its loss percent? |
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| 44. |
Achord of a circle of radius 10cm. subtends a right angle at the centre. Find the areathe corresponding: (uset 3.14)ofi. Minor segmentii. Major segmentf 120o at the centre Find the an |
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| 45. |
14. Find the value of x, y. 7if STUV is a square.v-1) cm24ă(x 12) cmS (+ 2) cm T |
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| 46. |
003, tample 6:If the number obtained by interchanging the digits of a 2-digit number is 18 morethan the original number and the sum of the digits is 8, what is the original number?.. hehe 10Then is the tens digit and vis the ones digit |
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| 47. |
IDG=21X6!ISI=zlet!2911 1=214hI-74112122 |
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Answer» 12x 12 =144, 13x12 = 156; 14x12 =168, 15×12 = 180; 16x12 =192; 17x12 =204; 18x12 = 216, 19x12 = 228; 20x12 = 240 12×12=244, 13×12=156; 14×12=168, 15×12=180; 16×12=192; 17×12=204; 18×12=216; 19×12=228;20×12=240 answer 144156168180192204216228240 1. 1442.1563.1684.1805.1926.2047.2168.2289.240 |
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| 48. |
9. If the radius of a circle isius of a circle is increased by 75%, then by what percent will its circumference increase? |
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Answer» Let original radius beRcm. Then, original circumference=2πR=cm. New radius = (175%ofR)cm=(175/100×R)=7R/4cm. New circumference=(2π×7R/4)cm==7πR/2cm Increase in circumference=(7πR/2−2πR)cm=3πR/2cm. Increase %=(3πR/2×1/2πR×100)% =75%. Please like the solution 👍 ✔️👍 |
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| 49. |
BC- AQRP arAR2 and BC- 15 em, then find PR.2 |
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Answer» Like my answer if you find it useful! |
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| 50. |
A sphere of radius 10 cm is melted andmade into a cone of height 10 cm. Findthe diameter of the cone.Q54 |
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Answer» Volume of spher=(4/3)πr³=(4/3)(22/7)(10)³ cm³=4190.48 cm³ Volume of cone=4190.48 cm³(1/3)r²h=4190.48r²=4190.48(3)/(10)r²=1257.14r=35.45 cm Therefore diameter of cone=70.9 cm mam option mey 10 , 20, 40, 80 hai If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k. If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k. |
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