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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. |
G 3 ot At 908 0% doe : |
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| 2. |
Find the area of a semicircular park whose perimeter is 360 m. (Use Ďěť |
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Answer» answer is incorrect correct answer is 7700 m^2 |
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| 3. |
The perimeter of a semicircular protractor is 144cm, find the diameter of the protractorTake Ď= |
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Answer» Perimeter of semicircle = (circumference of circle)/2 + Diameter of circle => P = 2πr/2 + 2r => 144 = (π + 2)r => 144 = 36/7 r => r = 144 x 7/36 = 28 cm Now Area of semicircle = πr^2/2 = 22/7 x 28 x 28 x 1/2 = 11 x 4 x 28 = 1232 sqcm |
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| 4. |
18. In the given figure, find the measures of LAOC, COF,DOE and LBOF. |
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Answer» angle AOC = 35°Angle FOB = 40°angle DOE + 40 + 35 = 180°angle DOE = 180° - 75° = 105° |
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| 5. |
18. In the given figure, find the measures of LAOC, 2COF, <DOE and 40F.c |
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| 6. |
(1) The perimeter of a semicircular protractor is 108 cm. Find the diameter of theprotractor. |
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| 7. |
(i) (5-1 -41)-1 + (2-1- 3-1)-1 |
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Answer» (1/5-1/4)^-1+(1/2+1/3)^-120/9-6/5= 100-54/45= 46/45 |
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| 8. |
41-m(A) 0(C) 1(B)-1 |
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| 9. |
19 El-—41 =p 3 |
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Answer» gdhj go hõõ hhōõ ydykjœò hòœö 9-3×3+1= 9- 9+1= 1 1 is the right answer 9-3÷1/3+19-3×3/1+19-9+110-9=1 1 is correct answer of the following one is correct answer 1 is correct answer of the 1 is the right answer. 9-3÷1/3+1=9-3×3/1+1=9-3×3+1=9-9+1=0+1=1 is the answer the correct answer is 1 of the following question 9-9+11is correct answer 9-3÷1/3+19-3×3+19-9+1=0+1 = 1 Ans. |
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| 10. |
Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cmare drawn in the given figure. Find the area of the shaded region. |
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| 11. |
13. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radiare drawn in the given figure. Find the area of the shaded region.us 45 |
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Answer» Area of semi-circle PQR =½π(9/2)2 = 81π/8 cm2 Area of region circle, A =π(9/4)2 = 81π/16 cm2 Area of region (B + C) =π(3/2)2 = 9π/4 cm2 Area of region D =½π(3/2)2 = 9π/8 cm2 Area of shaded region =Area of semicircle - Area of circle -Area of region (B + C) +Area of region D =81π/8 -81π/16 -9π/4 +9π/8 = 63π/16 cm2 = 99/8 cm2 |
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| 12. |
13. Three semicircles each of diameter 3 cm, a circle of diameter 4-5 cm and a semicircle of radius4-5 cm are drawnin the given figure. Find the area ofthé shaded region.3 cm |
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| 13. |
Three semicircles each of diameter 3 cm, a circle of diameter 4- cmsemicircle of radius 4-5 cm are drawn in the given figure. Find the area othe shaded region.3 cmcm3 cm Âť |
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| 14. |
Three semicircles each of diameter 3 cm,a circle of diameter 4.5 cm and asemicircle of radius 4.5 cm are drawn inthe given figure. Find the area of theshaded region.(CBSE 2017) |
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Answer» Area of shaded region = Area of larger semicircle - Area of circle - 2(Area of small semicircle) + Area of small semicircle= Area of larger semicircle - Area of circle - Area of small semicircle= [π(4.5 cm)²/2] - π(4.5/2 cm)² - [π(3/2 cm)²/2]= π[(4.5)²/2 - (4.5/2)² - (3/2)²/2] cm²= 12.37 cm² |
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| 15. |
X'ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD assemicircles are drawn. Find the area of shaded region.Y'diameter,threeăm14 cm |
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| 16. |
Three semicircles each of diameter 3 cm, a circle of diameter 4-5 em and asemicircle of radius 4-5 cm are drawn in the given figure. Find thethe shaded region. no 29area of3 cm3 cm - |
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Answer» Area of shaded region = Area of larger semicircle - Area of circle - 2(Area of small semicircle) + Area of small semicircle= Area of larger semicircle - Area of circle - Area of small semicircle= [π(4.5 cm)²/2] - π(4.5/2 cm)² - [π(3/2 cm)²/2]= π[(4.5)²/2 - (4.5/2)² - (3/2)²/2] cm²= 12.37 cm² Like my answer if you find it useful! |
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| 17. |
170) 41Albre 1 |
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| 18. |
4.00Solve the equatien by uinthe bross muuti5,0016.0023 Sunday |
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| 19. |
ordinary bread 320fruit bread 80cakes and pastries160 Draw a pie chart for thisbiscuits 120others 40Total 720 |
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Answer» what's your question. what u are not understanding we can draw a pie chart |
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| 20. |
-990Gita had 15 slices of bread. She ate 7 of them. How many of themare left ?Slices of bread = 15Slices eaten= - 7Slices left |
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Answer» what is your question 8 is the correct answer 8 bread slices are left for gita 8 is the answer of the following 15 - 7 = 8 . 8 is the correct answer. slices of bread = 15slices eaten. =. 7slices left = 8 8 is the answer.when 15-7 8 is the absolutely right answer bread of 15 slices, ate 7 breads, 15-7/8 breads the correct answer is 8 |
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| 21. |
a grocer buys 480 lovesof bread at ru 216 per dozen . he soldthemat gain of 15 percent find the cp of 15 loves of bread. |
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Answer» Cost .Price. of 480/12 loaves of bread is 40*216= 8640profit=15%selling .price.=(100+gain%)/100 * Cost .Price. =100+15/100*8640 =9936= selling .price. of 480 loaves of bread s.p. of 1 bread =20.7selling .price. of 15 bread =20.7*15= 310.5 |
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| 22. |
10. There are 81 calories in a slice of bread that weighs 30 grams. How many caloriesain a pack of this bread that weighs 600 grams? |
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| 23. |
naparticularthesmartday, the sales (in ?) of different items of a bakers shop are given belowboardinclass to do ExerciseowltemsBread Cake Pastries Biscuits Fruit BreadSale (in)3008016012060Draw a pie-chart. |
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| 24. |
, निम्नलिखित समीकरण के मूल ज्ञात कीजिए:2 o —_.-_‘] o __..l] (X #E— 38x+3 x—-8 33 |
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Answer» 1/x+3 - 1/x-8=11/33=1/3 x-8-(x+3)=(1/3)(x+3)(x-8) 3(x-8-x-3)=x²-5x-24 3(-11)=x²-5x-24-33=x²-5x-24x²-5x-24+33=0x²-5x+9=0 x=5±√25-36/2x=5±√-11/2x=(5±11i)/2 |
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| 25. |
1. यदि 4 वस्तुओं का क्रय मूल्य, 3 वस्तुओं के विक्रय मूल्य केबराबर हो, तो लाभ प्रतिशत होगा--1 133-% 90—(8) 33% ® 97;%(©) '} lg»l% D) 6632—% |
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Answer» answer a is correct |
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| 26. |
Fill in the blank indicated by a star in thenumber 4* 56 so as to make it divisible by 33(a) 3(6) 4(c) 5(d) of these |
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Answer» d is the right answer a) 3 4356/3=132 4456/3=135.03 4556/3=138.06 number (d) is the right answer |
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| 27. |
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed Find the areaof the remaining sheet. (Take π = 3.14) |
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Answer» area of remaining sheet will be π{R^2-r^2}π{4^2-3^2}=π{16-9}7πcm^2 |
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| 28. |
Q. 2. The surface area of a sphere of radius 5 cm is fivetimes the curved surface area of a cone of radius 4cm. Find the height and volume of the cone. |
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| 29. |
(8/33)*(quad*text) |
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Answer» Answer:b)33/8Explanation: |
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| 30. |
x +3 x-8 33 |
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Answer» Given that 1/(x + 3) - 1/(x - 8) = 11/30 Taking L.C.M, we get [(x - 8) - (x + 3)]/(x + 3)(x - 8) = 11/30 -11/(x + 3)(x - 8) = 11/30 On cross multiplying by 30(x + 3)(x - 8)/11 on both sides, we get -30 = (x + 3)(x - 8) x² -8x + 3x - 24 = -30 x² - 5x + 6 = 0 (x - 2)(x - 3) = 0 x = 2 or x = 3 Like my answer if you find it useful! |
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| 31. |
-8 CPX-33)=-56-63847].(-6alt |
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Answer» -8(8x-3x)=-5(-6x+7)-64x+24x=30x-35-64x+24x-30x=-35-70x=-35x=-35/-70=0.5 or 1/2 0.5 my real answer. of this 0.5 is correct answer 0.5 is the correct answer 0.5 is the correct answer |
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| 32. |
ABCD is a rectangle with 4cm and 8cm. Taking 8cm as the diameter, two semicircles are formed. Find the area overlaped by the two semicircle. |
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| 33. |
5. Find the length of a wooden strip required to frame a photograph of breadth 32 cm and length 45 cmAlso find the cost if 1 m frame costs1.20. |
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Answer» The length of the photograph= 45 cmThe breadth of the photograph= 32 cmLength of the wooden strip required to frame the photograph= Perimeter of the photograph= 2 × (length + breadth)=2 × (45 + 32) cm=2 × 77 cm=154 cm=1.54mcost of frame=1.54×1.20=Rs1.848 |
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| 34. |
The side of a square is 2 cm. Semicircles are constructed on two sides of the square, then the area of thewhole fiyure is |
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Answer» Please hit the like button if this helped you |
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| 35. |
9. In parallelogram ABCD, the angles A and C are obtuse. Points X andYare taken on the diagonal BD such that the angles XAD and ZYCB areright angles. Prove that XA YC. |
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| 36. |
rcles are drawn inside a big circle of diameter 24 cm. Theof the diameter of the big11. Two c1 ,2diameter of the two circles areandcircle as shown in the adjoining figure. Find the ratio of the areasof the shaded part to the unshaded part of the circle. |
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| 37. |
19. In the figure, two small circlestouch each other externally at thecentre of a big circle and thesesmall circles are touched internallyby a big circle with radius 4 cmFind the area of the shaded region. |
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| 38. |
Prove that the area of the semicircle drawn on the hypotenuse of a right-angled triangle is equalthe sum of the areas of the semicircles drawn on the other two sides of the triangle. |
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| 39. |
32 m24 m40 + 24 +32en s=48 m13.40 m(s - a) (48-40) 8 m,(s -b) (48-24) 24 m,33 8.5 |
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Answer» you have get the answer plese send to me |
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| 40. |
(A) 24, 48, 72(C) 12, 24, 36(B) 8, 16, 24(D) 96, 192, 288 |
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Answer» 8 और 12 के तिन गुनज 24,48,72 |
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| 41. |
LAENGUL1. In the following pair of angles, say which angle is greater without measuring it. Whichmethod have you applied? |
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Answer» fig 2 is bigger than fig 1 fig. 2 is bigger than fig. 1 |
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| 42. |
H which when multiplied2176.hence find the value of1. Since 3 = (-3) = 9,> 3 and 3 both can be square root of 9. But when theonly positive square root.Thus, V9 = 3 and not -3.2. It is not possible to find the square root of a negative rational number3. The square root of a natural number is never greater than . But if the numay say it is a proper fraction then its square root is always greater then the.g. 49 = 7 and 7 < 49-andExercise 4 (6)1. Find the square root of the following:2. Evaluate1681V1024 (b)63 |
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Answer» 1 (a) 36/49=6/7(b) 121/225=11/15 a) 6/7b) 11/15c) 4/9d) 25/27e) 3&13/36 = 36×3+13/36=121/36= 11/6f) 4&73/324 = 324×4+73/324 = 1369/324 = 37/18g) 3&33/289 = 289×3+33/289 = 900/289 = 30/17h) 80/405 = 16/81 ( Both divided by 5) = 4/9 a) 6/7b) 11/15c) 4/9d) 25/27h) 4/9 a.6/7b.11/15c.4/9d.25/27e.11/6f.37/18g.30/17h.4/9 (a) 6/7(b) 11/15(c) 4/9(d) 25/27(e) a=6/7b=11/15c=4/9 d =25/7e=11/6f=1369/324g=30/17h=4/9 |
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| 43. |
A +B +C =180A, B, C angles of a triangle(a) True(c) can not say(b) false(d) none of these |
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Answer» Option (c) can not say is the correct answer. A,B,C can be angles of a triangle or they can be three angles where their arms don't form a triangle |
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| 44. |
ना 7 नो गा पक =w ATt edgh Sandeep Singh ¥Santosh Tarsem SinSingh Anet Verma _Jaglecp Kamar' Amanbir Singh A |
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Answer» hi Rai shahb kuch baat kr skte hai |
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| 45. |
u u3 Ul tHe above results.Example 3: Show that a median of a triangle divides it into two triangles of equalareas |
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| 46. |
7. Gajendra Singh purchased a house from Avas Parishad on credit. If the cost of the house is64000 and the rate of interest is 5% per annum compounded half yearly. Find the interest paidby Gajendra Singh after one and half years. |
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| 47. |
की,” 1+508 3 |
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| 48. |
508 7800 — .G एछ१ दर + ०09 IS |
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Answer» Sin60° = √3/2= square= 3/4tan45°= 1cos30° = √3/2 square= 3/4 3/4+3/4+13/2+1= 5/2 |
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| 49. |
SANSKRUTI SCHOOLHOLIDAY HOMEWORK (CLASS X)13. Which of the following is not the graph of a quadratic polynomial? |
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Answer» A ISTHE CORRECT ANSWER The answer a is correct option : Abecause did not intersect axis |
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| 50. |
Birla Vidya Mandir, NainitaiHoliday Homework, Class-1oMathematics1. If HCF(26,169)-13, then LCM is (a) 26(b) 52 (c) 338(d) 13 |
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Answer» Product of numbers = HCF*LCM 26*169 = 13*LCM LCM = 26*169/13 LCM = 26*13 LCM = 338 Like my answer if you find it useful! |
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