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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. |
Subtract 2/9 from 3/10 |
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Answer» direct how..we want method tq |
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| 2. |
the previous class, we have learnt to add and subtract integers. Using those methods.in the blanks below.5 + 7 =(2) 10+(-5)=L (3) - 4+3 =(-7) + (-2) = O (5) (+8) - (+ 3) = (6)(+8) - (-3) = 0te a number in each bracket to obtain the answer '3' in each operation. |
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Answer» 5+7=12 -7+-2=-910+-5=5 +8-+3=5 -4+3=-1 +8-(-3)=11 |
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| 3. |
57. In the given figure, AABC is right-angled at A. Find the areashaded region if AB- 6 cm, BC 10 cm and O is the centre ofincircle of △ABC. [Take π = 3.141CBSE 2009 |
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| 4. |
7. In the given figure, AABC is right-angled at A. Find the area of theshaded region if AB= 6 cm, BC = 10 cm and O is the centre of theCBSE 2009incircle of △ABC. [Take π = 3.14.] |
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| 5. |
cm.2 ĂD is the diameter of a circle of radius 6 cm and ABBC- CD. Semi-circles are drawn with AB and BD as diametersAas shown in the figure. Find the perimeter and area of theshaded region. |
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Answer» Area of shaded region=Area of smaller semicircle with Diameter AB +area of semicircle with Diameter AD - area of semicircle with Diameter BD AD=6, AB=BC=CD=2cmArea=π(1/2)+π(9/2)-π(4/2)=π/2+9π/2-4π/2=3π=66/7 cm^2 Perimeter=π(1)+π(3)-π(2)=2πPerimeter of shaded region=44/7cm |
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| 6. |
AD is the diameter of a circle of radius 6 cm and AB BCCD-Semi-circles are drawn with AB and BD as diameters Aas shown in the figure. Find the perimeter and area of theshaded region. |
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| 7. |
The dimensions of a metal block are 2.25 m1.5 m. by 27cm. It is melted and recast into cubeach of side 45 cm. How many cubes are form |
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Answer» thanks |
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| 8. |
4:Subtract 72- (90) |
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Answer» 72-(90)=72-90subtract 72 from 90We get, 18Now 90>72 so answer will have minus sign-18 is answer |
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| 9. |
3104Subtract from |
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| 10. |
4 Subtract.a 3/4 from 7/8 |
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| 11. |
r is increased by 25%. By what per cent should the newThe salary of an officedecreased to restore the original salary? |
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| 12. |
centre of the iree, 7t.1LA). In Fig. 12.117, ABC is a triangle right angled at A. Find the area of the shadedAB6 cm, BC 10 cm and I is the centre of incirele of AABC |
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Answer» where is the fig?....plz post an appropriate question this is the one in the qstn it is given that AB=6... but in yhe fig it is AC=6....😒?? hey, if the question was AB=6cm ,how will you answer in the qstn it is AB=6...but in the fig it is different...either the data provided by the qstn is wrong or the fig is wrong... hey my brother answer me when AB =6cm ok wait... don't worry... kk |
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| 13. |
9. Find f (2,2 ex dx1+ |
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| 14. |
d"yIf y = ex + 2x + log x, then find-2.dx |
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Answer» Thanx... More ques are thr... Plzz reply |
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| 15. |
valuate(x2 +3x + ex) dx,as the limit of the sum |
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| 16. |
73. एक समलम्ब 48020 जिसमें 48 | 20 है, के विकर्ण परस्पर बिन्दु 0) पर प्रतिच्छेद करते हैं। यदि 48 न 20105तो न्रिधुजों 4002 और (00) के क्षेत्रफलों का अनुपात ज्ञात कीजिए, CER |
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| 17. |
- बहुपद ' (k - 81 +4 कोकीजिए।2 और + 1 से भाग देने पर प्राप्त शेषफलों का योग शून्य हैं। ५ का गान ज्ञात(NCERT Exemplar) |
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| 18. |
10 gu iree times round the field, if he walks at rate of 1.5 m per second?I0. Two plots of land have the same perimeter. One is a square of side 60 m, while the other is a rectanglewhose breadth is 1.5 dam. Which plot has the greater area and by how much? |
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Answer» perimeter of rectangle = perimeter of square2×(l+b)=4×side2×(?+1.5)=60m1dam=10m1.5 dam=1.5x10=152×(?×15)=60?=60÷2÷15=2solveanswer is 2 |
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| 19. |
(vi)pº +p3-p? + 1 by p - 1 |
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Answer» -p(p^2+p-1) is the correct answer -p(p^2+p-1) must be correct answer p^4+p^3-p^2+1÷p-1= -p(p^2+p-1) p cube+2p square+p-1 is the answer P3+2P2+P-1 is your answer p4 is the correct answer of this question -p(p^2+p-1) is tha correct answar -p(p^2+p-1)is always right answer Answer of this question is -p(p^2+p-1) |
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| 20. |
A. pProve that Sp= p3 |
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| 21. |
\int \cos ^{3} x \cdot e^{\log \sin x} d x |
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| 22. |
At a ticket counter 35 people are standing in a row. Mohit is 15th from ticketcounter. What is his position from the end?(b) 20th(a) 19th(c) 21st(d) 22nd |
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Answer» Answer is :-(b) 20th answer is (b) 20th... answer is B) 20 and please also like my answer He will be at 20 position in the row. ANSWER IS :- 20th. 21st is the correct answer to this question. answer is (b)20th... answer is ( B) 20th and position in row 20th no.35-15=20hence the position of Mohit will be 20th. |
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| 23. |
If $ y=e^{7 x} \cdot \cos 7 x \operatorname{fin} d \cdot \frac{d y}{d x} $ |
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Answer» tysm👍 |
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| 24. |
15. Write any two rational numbers eachgreater thanwhose sum is 0. |
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Answer» I think 0 +0 greater than _3/4 |
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| 25. |
\int x^{2} \cdot e^{x^{3}} d x= |
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| 26. |
2 o e ___W_MMSQ.SOCW)3 x \sin 60^{\circ} \cdot \cos ^{2} 30^{\circ}=\frac{\tan 45^{\circ} \sec 60^{\circ}}{\csc 30^{\circ} \cdot \sin 45^{\circ}}जब जन निर्णमि A |
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| 27. |
ical balloon of radius r subtends an angle 9 at the eye of an observer. If then of its centre is Ń, find the height of the centre of the balloon. [NCERTI |
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Answer» like if you find it useful |
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| 28. |
मिस|, % 8 ABC ~ A POR &, 1/ इनके क्षेत्रफल क्रमश: 64 सेत्रफल क्रमश: 64 सेमी० तथा 121 सेमी।o2 तथा (0# > 15न 154 सेमी०हो, तो 20 न? |
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| 29. |
ULOAn electrons moves such that its velocity is always perpendicular to its radius vector. Show that iis a circle.La vastare cer ) |
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Answer» Since the magnetic force is always perpendicular to the velocity of a charged particle, the particle will undergo circular motion. it is related to the any topic of mathematics i want mathematical solution |
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| 30. |
\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3} |
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| 31. |
Y A29. The line through P(5, 3) intersects y-axis at QオP(5,3)(i) Write the slope of the line.(i) Write the equation of the line.(ii) Find the coordinates of C45°X'(2012)Y' |
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| 32. |
8. If the mean of the following distributions is 2.6, then find the value of K.24f, K 5 81 2 |
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| 33. |
1. Answer the following:(G) The value of log2 log2 log2 16 is equal to(a) 1(b) 2(c) -I(d) of the above |
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| 34. |
\lim _{x \rightarrow 0} \frac{x\left(e^{x}-1\right)}{1-\cos x} (Ans. 2) S N e x |
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Answer» using L hospitaldifferentiating both numerator and denominatorhencee^x-1+x(e^x)/-sinxagain differentiate both numerator and denominatore^x+e^x+xe^x/cosxput in the limitse^0+e^0+0/11+1/1=2 |
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| 35. |
dy14.If y=e'sin x, show that= 2 e cos x. |
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| 36. |
(छाती ह ५७६८८ है न फ ह (७८ 0(७० कि है! SmA. |
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Answer» thank you |
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| 37. |
1.Write down the following products.(i) ( y + 2)(y - 2)(ii) (p + 1)( p - 1) |
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| 38. |
\int \frac{\sin x-\cos x}{\sqrt{1-\sin 2 x}} e^{\sin x} \cdot \cos x d x |
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| 39. |
Find two rational numbers whose absolute value is3 |
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Answer» Definition: Absolute value of an integer is the numerical value of the integer regardless of its sign. If ' x ' is an integer then its absolute value is denoted by l x l and is -1/3 and -5/15 Write the rational number whose absolute value is 5/6 |
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| 40. |
2 910 /५० हा 6दन 6] नाम ELHEESE & |
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Answer» √(5+√(11+√(19+√(29+√(49)))))=√(5+√(11+√(19+√(29+7))))=√(5+√(11+√(19+√(36))))=√(5+√(11+√(19+6)))=√(5+√(11+√(25)))=√(5+√(11+5))=√(5+√(16))=√(5+4)=√(9)=3 |
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| 41. |
oe(40—io0=8910-y |
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Answer» 1/4 is correct answer. 1/4 is right question.......... this question answer is 0.25 |
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| 42. |
a) Evaluate \log _{3} 81 |
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| 43. |
if (3)x+y=81 and (81)x-y=3, find the value of x and y |
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| 44. |
Evaluate:81 )-2/32163 |
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| 45. |
(I)ëŹnArt §el 3 : log(logy?) _ log(log r) = log2 |
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Answer» 1) LHS= log(logx^2)-log(logx) = log(2logx)-log(logx) =log(2logx/logx) =log2= RHS |
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| 46. |
न नो A 2पुहे न 2 हा ह |
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| 47. |
प्ये :Af‘{/tran 0 + 560 9 ८ द, तो सिद्ध करें कि ४00 न हX |
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Answer» As we know that 1 + tan²A = sec ²A sec²A - tan²A = 1 (secA + tan A) (secA - tanA) = 1 (secA - tanA) = 1/(secA + tanA) secA - tanA = 1/x ( As givensecA + tanA = x) Now adding (secA + tanA ) + ( secA - tanA) = x + 1/x 2secA = (x² + 1) /x 1/cosA = (x²+1)/2x ( As we know thatsecA = 1/cosA) cosA = 2x/(x² + 1) Cos²A = 4x²/(x² + 1)² 1 - cos²A = 1 - 4x²/(x² + 1)² sin²A = (x² + 1 )² - 4x²/(x² + 1)²(sin²A + cos²A = 1) sin²A = [(x²)² + 2x² + 1 -4x²]/(x² + 1) sin²A = [(x²)² - 2x² + 1]/(x² +1)² sinA = √(x² - 1)² / √(x² + 1)² sinA = (x² - 1)/ (x² + 1) Hence proved. |
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| 48. |
. I log2, log (2-1) and log (2+3) are in A.P., write the value of x |
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Answer» you are solving it by taking difference. but if I am solving it by :- let; a , b , c are in AP..taking, 2b=a+b....... getting ans; log4^5 |
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| 49. |
4Q.46 limis equal to(a) 8(c) 2(b) -8(d) of these |
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| 50. |
warte the absolute value asthe 10 |
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Answer» The answer of 5by8. is _40 The absolute value of zero is never zero but any value greater or smaller than zero absolute value of 5/8 is 5/8 |
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