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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. |
This is a picture of a triangularpyramid This shape hasi.facesedges-vertices |
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Answer» Number of faces:4 Number of edges:6 Number of vertices:4 |
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| 2. |
20. Mean value of the weekly income of 40 families is 265. But in the calculation, income of onefamily was read as? 150 instead ofAns, 264.125]115. Find the 'Corrected' mean. |
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Answer» Wrongly stipulated sum was 265 x 40= 10600 Subtract the wrong value and add the correct income value to get the correct sumCorrect sum= 10600- 150+ 115 = 10565 Correct mean= 10565/40= 264.125 |
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| 3. |
Il104 X 96 |
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Answer» What is this question may turn out the correct answer is 9984 the answer of your question is 9984 9984 is multification 9984 is the correct answer 9984this is answer plzzz like 9984“ψ(｀∇´)ψ (^～^)(^～^)(^～^) 9,984 this is answer 9984 is correct answer 😀😃😃 9984 is right answer 😃 104×96(100+4)(100-4)Use- (a+b)(a-b)=a^2-b^2100^2-4^210000-16 104×96(100+4)(100-4)Use (a+b)(a-b)=a^2-b^2100^2-4^210000-169984 104×96(100+4)(100-4)10000-169984 is your answer. 9984 is correct answer |
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| 4. |
The sum of the length and the breadth of a rectangle is 240 cm. If the length isdecreased by 20% and the breadth is increased by 10% the perimeter remains thesame. Find the dimensions of the rectangle. |
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Answer» Let length be 10 And breadth be 10 Area 100 Now length is increased by 20% New length 12 And breadth is decreased by 10% New breadth is 9 New area is 12*9=108 Change in area 108-100= 8 So overall areaincreases by 8% |
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| 5. |
8A horse is tied to a peg at one corner of a squareshaped grass field of side 15 m by means of a 5 mlong rope (see Fig. 12.11). Firn(0) the area of that part of the field in which thehorse can graze.(ii) the increase in the grazing area if the rope wereFig. 12.1110 m long instead of 5 m. (Use π= 314) |
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| 6. |
horse is tied to a peg at one corner of a squareshaped grass field of side 15 m by means of a 5 rmlong rope (see Fig. 12.11). Find0) the area of that part of the field in which thehorse can graze.(i) the increase in the grazing area if the rope were10 m long instead of 5 m. (Use3.14) |
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| 7. |
1. Fill in the blanks:(expressed as a rational number with denominater 24 -7-8 |
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Answer» 7/-8*(-3/3)=21/24 denominater 24 |
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| 8. |
8. A horse is tied to a peg at one corner of a squareshaped grass field of side 15 m by means of a 5 mlong rope (see Fig. 5.11). Find(0 the area of that part of the field in which thehorse can graze() the increase in the grazing area if the rope were10m long instead of 5 m. (Use Tt -3.14)Fig. |
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| 9. |
8. A horse is tied to a peg at one corner of a squareshaped grass field of side 15 m by means of a 5 mlong rope (see Fig. 12.11). Find(0 the area of that part of the field in which thehorse can graze.(ii) the increase in the grazing area if the rope were10 m long instead of 5 m. (Use Tu 3.14) |
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| 10. |
2. Evaluate the following products without multiplying directly:(ii) 95 x 96(iii) 104 x 96103 21076) 95496104x96 |
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| 11. |
Evaluate the following products without multiplying directly:(i) 103 x 107(ii) 95×96 |
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| 12. |
End the value s--n-Lon-is-a-pas allelo g ramSo° |
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| 13. |
Abba is tiling her kitchen. Each tile is 20 cm by 20 cm. The kitchen is 2200 cm wide. How many tiles will be required to cover the wall.en is 240 cmlon240cm20 om |
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Answer» Number of tiles = Area of kitchen / Area of a tile= (240*200)/(20*20)= 12*10= 120 tiles. Please hit the like button if this helped you |
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| 14. |
65. Find the prime factorization of the denominator of the rational number expressed as 6.12CBSE 2015, 16]543in simplest form. |
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Answer» x = 6.121212121212… Multiply 100 on both side 100 x = 612.12121212121….. 100x = 606 + 6.12121212… 100x = 606 + x 100x – x = 606 99x = 606 X = 606/99 = 203/33 The rational number = 101 * 2 / 3 * 11 The denominator is 3 * 11 = 33 |
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| 15. |
72 marks is included, calculate the new mean.The mean of marks obtained by 15 students is 63. If marks of two students were read wrongly 73instead of 78 and 59 instead of 69, find the correct mean.9.or loft the mean was calculated again an |
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Answer» Mean of 15 students in 63.Total marks = 63*15 = 945.Corrected total marks = 945 - 73 - 59 + 78 + 69= 960. Correct Mean = 960 / 15 = 64. PLEASE HIT THE LIKE BUTTON |
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| 16. |
The students of Class VⅢ of a school donated? 2401 in all, for Prime Minister'sNational Relief Fund. Each student donated as many rupees as the number of studentin the class. Find the number of students in the class. |
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| 17. |
P50% instead of DeSuaraNore(a) *182an article is sold at a gain of 5% instead of bem(c) 60What is the cost price of the article?(b) * 40(d) 380 |
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| 18. |
2.The students of Class VIII B of a school donated 2,304 for the Prime MinisterieNational Relief Fund. Each student donated as many rupees as the number mestudents in the class.(a)Find the number of students in VIII B.(b) What quality of the students do you appreciate here? |
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Answer» Given, No. rupee=No.of students=x Donated money 2304 ATQ •°•x×x=2304x^2=2304x=√2304x=48No.of students=48 |
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| 19. |
f sug9) A man donates 2% of his annualearnings to charity. If his monthly salaryis 15,250, find the amount donated. |
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| 20. |
7. The students of Class VII of a school donated 2401 in all, for Prime Minister'sNational Relief Fund. Each student donated as many rupees as the number of studentsin the class. Find the number of students in the class. |
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| 21. |
8. The students of class VII of a school donated6,625 for Drought Relief Fund. Dach student donated asmanyrupees as the number of students in the class. Find the number of students in the class.Hint: Find the square root of 5,625.] |
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Answer» Let there be x students.Each student donates ₹x.Total amount = x² = 5625∴ Number of students in the class = √5625 = 75 nice bhaiya |
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| 22. |
7. The students of Class VIII of a school donatedR 2401 in all, for Prime MinistersNational Relief Fund. Each student donated as many rupees as the number of studenin the class. Find the number of students in the class. |
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| 23. |
AL-%)-18 ° iM |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 24. |
Al हा 11 I 208 =¢9] ‘57 Mli\जरा पा kenki 18 lapalie Khe o3 |
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Answer» Some more examples are as follows:(1) 163 = 1 6 36 216 12 72 ----------------------- 4 0 9 6 ----------------------- |
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| 25. |
A factory produce 10000 apples in 3 hours so how much time will the factory take to produce 2500 apples? |
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Answer» 10000-3 hours2500-xx=2500*3/10000X=3/4 hrs or 45 minutes |
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| 26. |
37. Three ropes of length 60 m, 80 m and 100 ns have to be divided into pieces of same length. The largest possiblelength will be(A)120m (B)600m (C)1.20m (D)20m |
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| 27. |
simplest form of -8over 10 |
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Answer» -8/10 = -4/5 = -0.8 no wrong oh sorry write |
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| 28. |
30. A field is in the shape of a trapeziĂşm having parallel sides 90 m and30 m. These sides meet the third side at right angles. The length of thefourth side is 100 m. If it costs 5 to plough 1 m2 of the field, find thetotal cost of ploughing the field. |
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Answer» we have to multiply it by 5no matter I understood thanks 😊 |
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| 29. |
The simplest form of200 |
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Answer» 5/8 is the right answer. yes 5/8 is correct answer |
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| 30. |
Evaluatethe following limits:\operatorname{lit}_{x \rightarrow 0} \frac{x \tan x}{1-\cos 2 x} |
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Answer» Lim(x→0) {xtanx}/(1 - cos2x ) we know, (1 - cos2x) =2sin²x , use this here, Lim(x→0){xtanx}/(2sin²x) =1/2 × Lim(x→0){ x ×( tanx/x)× x }/(sinx/x)² ×x² we know, Lim(f(x) →0) tanf(x)/f(x) = 1 Lim(f(x)→0) sinf(x)/f(x) = 1 use this concept now, = 1/2 × Lim(x→0) { x²/x²} = 1/2 × 1 = 1/2 ( answer ) |
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| 31. |
Evaluate the following limits:\lim _{x \rightarrow 0} \frac{5^{x}-1}{\sqrt{4+x}-2} |
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| 32. |
Evaluate the following limits:\lim _{x \rightarrow 0}\left(7 x^{2}-5 x+1\right) |
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Answer» thanks |
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| 33. |
49in its simplest form? |
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Answer» 0.7656 is right answer problem sum 1210 people had 49/64=(7*7)/(8*8)=7^2/8^2=(7/8)^2 right answer is 0.7656 0.7656 is the correct answer |
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| 34. |
simplest form 0.36 whole bar |
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| 35. |
Reduce 42 o simplest formS1 |
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Answer» The reduced form of42/35is6/5 Steps to simplifying fractions Find the GCD (or HCF) of numerator and denominatorGCD of 42 and 35is7 Divide both the numerator and denominator by the GCD42 ÷ 7/35 ÷ 7 Reduced fraction:6/5 You are really cool thanks for helping me 😍😘😘 |
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| 36. |
The simplest form of 0.54 is |
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Answer» 0.54bar=0.54444...0.5444...=x5.4444...=10x_eq154.444...=100x_eq2subtract eq1 from eq2 49=90x49/90=x The bar is on both 5 and 4 |
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| 37. |
A rectangular plot has sides 120m x 90m.Find the length of wire needed to surroundthe plot four times. |
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Answer» Perimeter of rectangular plot=2(l+b)=2(120+90) m=420 m Therefore, length of wire needed to surround the plot 4 times= 4×perimeter of plot=4×420 m=1680 m |
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| 38. |
0. A rectangular plot has sides 120m x 90m.Find the length of wire needed to surroundthe plot four times. |
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Answer» Perimeter of rectangular plot=2(l+b)=2(120+90) m=420 m Therefore, length of wire needed to surround the plot 4 times= 4×perimeter of plot=4×420 m=1680 m |
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| 39. |
Three ropes of iength 60 m, 30 m and 100 m have to be divided into pieces of same length. The largest possiblelength will be(A)120m (B)600m (C)120m (D)20m |
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Answer» 60 can be divided in 20,20,2080 can be divided in 20,20,20,20100 can be divided in 20,20,20,20,20so largest possible length=20m |
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| 40. |
What is the length of wire needed to fence the plot shown, if three strands ofwire are used?11.173 m|-90m |
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Answer» length of wire =perimeter of fig shown=173+90+71+173+90+71=346+180+142=668 |
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| 41. |
e| sin? 0)tan2 6=sinâ 0 |
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| 42. |
CE AREAS ANDVOLUMESEXERCISE 131astic box 1.5 m long, 1.25 m wide and 65 cm deep is to be ntop. Ignoring the thickness of the plastic sheet, determine:C) The area of the sheet required for making the box.(i) The cost of sheet for it, if a sheet measuring im-costs l |
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| 43. |
The area of a triangle is equal to the area of a square of side 30 m. Calculate the base of the triangle if itsaltitude is 90m. |
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| 44. |
1.Write as decimals.100 |
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Answer» 0.09 |
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| 45. |
Show that: cosec20 -tan2(90-sin20+ sin(90° -0) |
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| 46. |
(a) 130(6) 25 |
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Answer» a = 5 - b...... (1)b = 6/a........(2) Put value of n from eq(2) in eq(1)a = 5 - 6/aa^2 - 5a + 6 = 0a^2 - 3a - 2a + 6 = 0a(a - 3) - 2(a - 3) = 0(a - 2)(a - 3) = 0a = 2, 3 for a = 2,b = 6/2 = 3 for a = 3, b = 6/3 = 2 Value of a^3 + b^3 = (2)^3 + (3)^3 = 8 + 27 = 35 |
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| 47. |
Limits x-->0 log tan2 /log tan3x |
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Answer» please check the questionis it log tan2x/ log tan3x |
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| 48. |
(8IN+3A) _ठा के >कापी (b |
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Answer» √8 = √(2*2*2) = 2√2 √18 = √(2*3*3) = 3√2 = √2 (2√2+3√2) = √2*5√2 = 5*2 = 10 Like my answer if you find it useful! |
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| 49. |
3 ....: 88in'5 +sin!17i= cosâ˘! g5 |
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| 50. |
).1. If the perpendicular distance of a point Pfrom x-axis is 5 units and the foot of theperpendicular lies on the negative directionof x-axis then, what will the direction ofPoint P |
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Answer» We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5. |
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