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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. |
Solve: 1 + 6 + 11 + 16 ++ x = 148 |
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Answer» 1 + 6 + 11 + 16 + ... + x = 148 a = 1 d = 5 an = a + (n-1)d x = 1 + (n-1)5 x = 5n - 4 148 = n/2 (1 + 5n - 4) 296 = 5n*n - 3n 5n*n - 3n - 296 = 0 5n*n - 40n + 37n - 296 = 0 5n(n-8) + 37(n-8) = 0 (5n+37)(n-8) = 0 n = 8 x = 5*8 - 4 = 40 - 4 x = 36 If you find this answer helpful then like it. |
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| 2. |
3.-19 /]+[9X(-3)]Express the following in kilograms(1) 190 grams (ii) 247 grams |
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Answer» 1 kg=1000gm190gm=190/1000=0.190 kg same 0.247kg 190 grams=0.19 kilogram 247 grams=0.247 kilogram |
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| 3. |
(b) ABCD is a quadrilateral in which the diagonals AC and BD intersect at 0. Prove thatAB+BC + CD + AD < 2(AC + BD).hc |
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Answer» Who was the wrote biology book ABCD is a quadrilateral and AC, and BD are the diagonals. Sum of the two sides of a triangle is greater than the third side. So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > AC ..........(1)CD + AD > AC .........(2)AB + AD > BD .........(3)BC + CD > BD .........(4) Adding all the above equations, 2(AB + BC + CA + AD) > 2(AC + BD)⇒ 2(AB + BC + CA + AD) > 2(AC + BD)⇒ (AB + BC + CA + AD) > (AC + BD)⇒ (AC + BD) < (AB + BC + CA + AD) |
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| 4. |
andShort Answer Type Questions Stab-1)Show that x = 3, y = 1 is a solution of the equation 3x- 247.() |
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| 5. |
(9 91-рео3 ┬в |
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| 6. |
The product of the digits of a 2-digit number is 20. When 9 is added to the number,the digits interchange their places. Determine the number |
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| 7. |
23. Two different dice are thrown together. Find the probability that the numbers obtained have(i) even sum, and(ii) even product. |
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| 8. |
16. g&#72-91?-9-? |
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Answer» 7957/1020 is a best answer 8557/1020 is the correct answer |
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| 9. |
y cpbóards can 14 men make in 14 days?2. In a hostel it costs 1800 to keep 50 students for 8 weeks. For what length of time did the cost of keepi90 students amount to39 nersons can rannir21,060? |
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Answer» The key here is to first determine the cost of 1 child for 1 week.Set x= cost of one childw= number of weeksw*x = 18008(50)x = 1800400x = 1800Divide each side by 400x = 1800/400 = 4.5So 1 child per week = 4.5To solve for number of weeks Rs.21060 pays for,Set y = number of weeksy*4.5 = 21060Divide each side by 4.5y = 21060/4.5 = 4680 |
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| 10. |
49*9+37-91= |
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Answer» 49*9+37-91441+37-91478-91387 |
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| 11. |
4.A verandah 2 m wide is constructed all around a room of dimensions 8 m x 5 m.Farea of the verandah |
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Answer» Area of the room= length×breadth =8×5=40 sq.mLength of the room including verandah=8+2+2=12mBreadth of the room including verandah=5+2+2=9mHence,Area of the room including verandah=12×9=108 sq.m∴,the area of the verandah is=(area of the room including verandah-area of the room)=108-40 =68sq.m |
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| 12. |
3. A and B .con dos o pieceof amork an 28 days. B and Ccan do it in 24 days whilecand a can finish it in 36days. In how many days.car? A, B and Coginish it, itthey call work together. InLour many dous quill eachze of them Einish it workingzione? 0 |
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Answer» wrong |
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| 13. |
C. A varandah 2m wide is constructed all around outside a room of size 8m by 5m. Find the area of theverandah. |
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Answer» Area of the room= length×breadth =8×5=40 sq.mLength of the room including verandah=8+2+2=12mBreadth of the room including verandah=5+2+2=9mHence,Area of the room including verandah=12×9=108 sq.m∴,the area of the verandah is=(area of the room including verandah-area of the room)=108-40 =68sq.m what is Pythagoras property |
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| 14. |
2 /Write three divisions of integers such that the fractional form of each will be-5Write three divisions of integers such that the fractional form of each will be |
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Answer» 2. 48/10 , 72/15 , 96/20 |
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| 15. |
7. ABCD is a parallelogram in which diagonals AC and BD intersect atO. If the area of ABCD is 40 cm2, then what is the area of AAOB? |
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Answer» 10cm^2 is correct answer |
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| 16. |
sum of the digits of a two-digit number is 8. If the digits are reversed, the new number increases by 18.TheFind the number. |
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Answer» x + y = 8 {equation 1} The value of xy is 10x + y The value of yx is 10y + x 10y + x = 10x + y + 18 9y - 9x = 18 9(y - x) = 18 y - x = 2 From equation 1: y = 8-x 8 - x - x = 2 8 - 2x = 2 -2x = -6 x = 3 y - x = 2 y - 3 = 2 y = 5 Originalnumber xy = 35 |
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| 17. |
1- How many combinations of two digit numbersaving & can be made from the follewingnumbers?10 |
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Answer» One position will be filled with 8 and other will be filled with one of the five other numbers. There are 5 such combinations with numbers starting with 8, i.e. 85, 82, and so on. There 5 such combinations with 8 in the units place i.e. 58, 28, and so on. Since there is not mention of repetition not allowed, we will assume it is allowed. Hence, 88 is also an option.Total numbers possible are 5+5+1 = 11 Like my answer if you find it useful! |
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| 18. |
10, In a 3-digit number, unit's digit is one more than the hundred's digit and ten's digit isone less than the hundred's digit. If the sum of the original 3-digit number and numbersobtained by changing the order of digits cyclically is 2664, find the number. |
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Answer» where is the solution |
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| 19. |
5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9which are divisible by 10 and no digit is repeated 2 |
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Answer» 120 6 digits numbers can be formed. |
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| 20. |
ete the horse can graze.of one circular field is 5 m and that of other is 3 m. Find the radius of the circular field whoseis the difference of the areas of the first and second field. |
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Answer» radius 5m the other is 3m it's not a13m |
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| 21. |
10. How many two digit numbers are divisible by 13? |
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| 22. |
A circular field has an area of 628 m2. A horse is tied at the centre offield with the help of a rope 8 m long. What percent of the area offield will the horse be able to graze?[Use a = 3.14] |
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Answer» The horse is able to graze an area of (2π8^2)=2×3.14×64=401.92 m^2the percent of the area is (401.92÷628)×100=64% |
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| 23. |
cirele7 The diameter of a circular field is 98 m. If a child walks at the rate of 14 m per minute, how long will ittake to walk round the field once? |
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Answer» Circumference of the field = π * Diameter= 308m. Time = Distance/Speed= 308/14= 22 minutes. Please hit the like button if this helped you |
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| 24. |
wo numbors aPe TH hOTiTwo numbers are in the ratio 3:4. If their ICF is A, sind the numbers.Which ratio is greater?12((iii) (2:3) or (9:7) |
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Answer» Explanation: Let the numbers be 3x and 4x . Then their H.C.F = x. So, x=4 Therefore, The numbers are 12 and 16 Thanks |
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| 25. |
The area of a rectangular piece of cardboard is 54 sq cm and its length is 9 cm. What ihe cardboard?s the width |
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Answer» Area of a rectangular piece of cardboard = 54 cm² Length of the cardboard = 9 cm Then,---------- Area = Length × Width => 54 = 9 × Width => Width = 54/9 => Width = 6 cm |
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| 26. |
The area of a rectangular piece of cardboard is 54 sq cm and its length is 9 cm. What is the width ofthe cardboard?There are 36 nersons working in an office. If the number of females is 20 and remaining are males, find |
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| 27. |
000.Leer days on which(8) Mrs. Saxena's income is 23% more than Mrs. Sharma's. What per cent is Mrs. Sharma'sinemenMrs. Saxena's? |
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Answer» 77% less is less than Mrs.Saxena's of Mrs.Sharma's income |
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| 28. |
Find the smallest number which when divided by35, 56 and 91 leaves remainder 7 in each case. |
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| 29. |
10. Find the least number which when divided by 35, 56 and 91 leaves thesame remainder 7 in each case |
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| 30. |
Find the least number which when divided by 35, 56 and 91 leaves thesame remainder 7 in each case. |
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| 31. |
Find the least number which whendivided by 12,16,24 and 36 leavesas remainder 7 in each case |
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| 32. |
95. Ifp :qr s t u 2: 3, then (mp+nr+ot):(mo+ns+ou) is equal to |
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| 33. |
EXERCISE 8.1PS PT1. In ΔPQRST is a line such thatandsQ TRalso ZPST LPRQProve that APQR is an isosceles triangle. |
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| 34. |
11. The radius of a circular field is 20 m. Inside it runs apath 5 m wide all around. Find the area of the path. |
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| 35. |
The area of a rectangular piece of cardboard is 54 sq cm and its length is 9 cm. What is the width ofthe cardboard? |
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Answer» Area of a rectangular piece of cardboard = 54 cm² Length of the cardboard = 9 cm Then,---------- Area = Length × Width => 54 = 9 × Width => Width = 54/9 => Width = 6 cm |
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| 36. |
5. Divide the line segment AB9cm in the ratio 2:3 |
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Answer» Let the ratio be xSo, parts are 2x and 3xAccording to question,2x + 3x = 95x = 9x = 9/5 = 1.8So, 2x = 3.6 cm3x = 5.4 cmSo, Draw one part of length 3.6 cm and other 5.4 cm |
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| 37. |
11. A man spends To th part of his monthly income as house rent and 5part to maintain his family and saves the rest. If his monthly savings is4500, find his monthly income |
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| 38. |
The length and the breadth of a rectangular piece of land are 500 mrespectively. Find(i) its area(ii) the cost of the land, if I m2 of the land costs |
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| 39. |
Sales Price to make a profit of 20% |
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| 40. |
1. /The length and the breadth of a rectangular piece of land are 500 m andrespectively. Find(i) its area(ii) the cost of the land, if 1 m2 of the land costs10 |
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Answer» a) Area =length × breadth = 500 ×300= 150000m* b) cost of 1m* land = Rs 10000×150000=Rs 150,00,00,000 |
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| 41. |
Construct a AXYZ in which ¿Y = 30°, ZZ=90°and XY + YZ + ZX = 11cm. (Also write steps ofconstructions). |
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| 42. |
The length and the breadth of a rectangular piece of land are 500 m and 300mrespectively. Find(i) its are:a1.(i)the cost of the land, if 1 m2 of the land costs? 10,000. |
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| 43. |
1. The lengtit and the breadth of a rectangular piece of land are 500 m and 300 mrespectively. Find0 its area(i)the cost of the land, if 1 m2 of the land costs10,000. |
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| 44. |
Do these constructions,1. Bisect AB |
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Answer» Thank you |
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| 45. |
CONSTRUCTIONSDraw a line segment AB5.4 cm. From this segment cut off AK3.8 cm. |
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Answer» to do this use following steps1. draw a line segment of 5.4cm using ruler, name the end points as A and B.2. Use compass and scale to 3.8 cm, then point the one end in A then cut an arc of 3.8 cm as K in AB.This is our requried construction. |
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| 46. |
CHAPTER 11 CONSTRUCTIONSQ1. Construct an angle whose measure is 75 and then construct its bisectorWritethe measures of the angles so formed.stuet an angle of 52.5 with the helw |
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Answer» the bisector will be 37.5° thanks |
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| 47. |
In an examination out of 100 students, 75 passedin English, 60 passed in Mathematics and 45passed in both English and Mathematics. Whatis the number of students passed in exactly one19of the two subjects? |
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Answer» 45 students passed in exactly one of the subjects. since 75-45 = 30 students passed in english only while 60 - 45 = 15 students passed only in mathematics total = 15+30 |
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| 48. |
Find the quotients without actual divisions.(i)(a3 - 8) = (a -2)(ii)(27+3+8) = (3t+2)(iii)(8a3+27) + (4a2-6a+9)(iv) (64a3-125) + (16a2+20a+25). |
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Answer» 1)a^3-8/(a-2)=(a)^3 -(2)^3/(a-2)=(a-2)(a^2+2a+4)/(a-2)=a^2+2a+42)27t^3+8/(3t+2)(3t)^3+(2)^3/(3t+2)(3t+2)(9t^2-6t+4)/(3t+2)9t^2-6t+4 1) a^2+4+2a2)9t^2+4+6t3)2a+3 4)4a-5 |
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| 49. |
27 - 125 a ^ { 3 } - 135 a + 225 a ^ { 2 } |
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| 50. |
3. Diameter of the circular garden is42 m. There is a 3.5 m wide roadaround the garden. Find the area of |
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Answer» Diameter of the Circular Garden = d d = 42 m Radius ( r )= OB = d/2 = 42/2 = 21 m width of the around the Garden (w)=3.5m Radius of the larger circle ( R ) = OA = r + w R = 21 + 3.5 = 24.5 m Area of the Road ( A ) = π( R+r )( R-r ) A = ( 22/7 ) ( 24.5 + 21 )( 24.5 - 21 ) A = ( 22/7 ) × 45.5 × 3.5 A = 22 × 45.5 × 0.5 A = 500.5 m² Therefore , Area of the Road = A = 500.5 m² please like the solution 👍 ✔️👍 |
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