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. |
show that the lines x-1\2=y-2\3=z-3\4and x-4\5=y-1\2=z-0\1 intersect |
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| 2. |
41. 115 + 57 ~ 1 3 + 1का मान क्या है?4=(2) 139784ॐ| (4) 1189420 |
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Answer» Take l.c.m of 5,7,3,4=420So,after addition answer =1189/420(4) is the right answer 1189/420 is right answer 1189/420 is the right answer |
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| 3. |
3. Convert each of the following into a decimal:...2516(115)(1) 156liv) 3524100100(iv) 1000(vi) 17(vii) 25into libre derimals: |
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| 4. |
i. Let Î ABC . A DEF and their areas be, respectively, 64cm2 and 121 cm2. If15.4 cm, find BC.2. Diagonals of a trapezium ABCD with AB lI DC intersect each other at the poIfAB-2 CD, find the ratio of the areas of triangles AOB and COD. |
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| 5. |
The straight lines x +y-0, 3x + y-4-0, x+3y-4-0from aAJísosceles(D) none of these(B) equilateral(C) right angledints then locus of the po |
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Answer» Since none of the lines are perpendicular therefore it is not a right angled triangle. now find the intersection points of x+y=0, 3x+y-4=0, x+3y-4=0, we get(2,-2),(-2,2) & (1,1) let A(2,-2) B(-2,2), C(1,1) By distance formula we get AC=AB but not equal to AB Therefore it is a isoscels triangle. |
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| 6. |
4,-Find the angle between the lines xa and by+(10. |
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Answer» The line x = a is a straight vertical line, and the line by + c = 0, if rearranged, is y = -c/b. Since there is no x variable in y = -c/b, it is also a straight line, but a horizontal straight line. If one line is going straight up and down, and the other left and right, and the angle between them is 90° |
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| 7. |
In the given figure, Q> R. PS is the bisector of ZOPR, PM L QR. Find CMPS80°Q M s |
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Answer» 15° will be tha answer right not sure what time you going to the gym today and tomorrow is a good time to go back to the gym now I'm going 🔙 to you later on tonight at work today and it is not a problem |
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| 8. |
4.Find the angle between the lines x,, a and by + c-0 |
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| 9. |
The diagonals of a parallelogram PQRS are along the lines x +3y 4 and 6x -2y 7. ThenPQRS is |
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| 10. |
X+2x+4 |
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Answer» 3x+4 is the answer to this and if you equate the equation to zero, you get the value of x as -4/3 |
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| 11. |
Solve:x-3=27 + 2x4,) 9x + 5 = 4(x-2) + 8 |
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Answer» 1) 8x + 3 = 27 +2x6x = 24x = 42)9x + 5 = 4(x-2)+85x = -5x= -1 |
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| 12. |
In the given figure, PQ is a diameter of a circlewith centre O. If 4PQR 65, SPR 40and ZPQM 50° find ZOPR, Z0PM andPRS.40°659O 50 |
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| 13. |
Fi 10 37,4 POR 100,whre P.sd are fpoints on a circle with centre O. Find zOPR |
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| 14. |
11. In the given figure. LPQR =120°, where P,Q and R are points on a circle with centreO. Then ZOPR is(A) 20°(B) 10°(C) 30°(D) 40°P120 |
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Answer» thank |
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| 15. |
ejWhich angles are adjacent angles?ZNMK and LOPMNMK and LLMPZNMK and ZOPRZNMK and ZLMK |
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Answer» angle NMK and angle LMK are adjacent angles (4) option is correct |
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| 16. |
हर्टc DX 8 C |
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Answer» What is called Japan in Japanese language |
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| 17. |
1. Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 121 cm2.15.4 cm, find BC. |
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| 18. |
Let Δ ABC ~ Δ DEF and their areas be respectively 64 cm2and 121 cm2. If EF 15.4 cm. Find BC. |
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| 19. |
Ans. ZIMC 1151? in the given Fig. 47, ZOPR LPOR and M and N are points respectively on sides QR and PR of APOR such thatOM- PN. Prove that OP 0Q. where O is the point of intersection of PM and ON |
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| 20. |
ADEF and their areas be respectively 64 cm2 and 121 cm2. If EF 15.4 cm.213.Letfind BC.AABC |
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| 21. |
2. In Δ ABC, DEIBC., If DE3 BC and area of Δ ABC-81 cm2 , find the area of Δ ADE. |
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Answer» In triangle ADE and triangle ABC angle A = angle A ( common ) angle ADE = angle ABC ( corresponding angles, DE is parallel to BC) therefore, triangle ADE is similar to triangle ABC ( by AA corollary of AAA) area of triangleADE / area of triangle ABC = (DE)^2 / (BC)^2 area of triangle ADE / 81 = (2/3 BC)^2 / (BC)^2 area of trangle ADE / 81 = 4BC^2 / 9 / BC^2 area of triangle ADE = 4 / 9 * 81 area of triangle ADE= 36 cm^2 it's not answer |
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| 22. |
tan(32 )14. If x-200s 0-cos 2 e and y-2sin e-sin 2 θ, prove that aye-sin 26,provethatartan |30tCBSE2013] |
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| 23. |
26. Using integration, find the area of theregion in the first quadrant enclosed by theX-axis, the line y = x and the circlex" + y" = 32 |
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| 24. |
1. Find the area of the region bounded by the curve y- z and the linesx 4 and the x-axis in the first quadrant. |
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| 25. |
Starting from ()*5, write various producteshowing some pattern to show(-1)x(1)-1 |
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| 26. |
ली1 है? क्यों अथवा क्यों नहीं?8- का प्रतिलोम 0.3 हैक्या 8: का IR |
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| 27. |
PInFig. 10.37, < PQR =100, where P, Q and R arepoints on a circle with centre O. Find ZOPR. |
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| 28. |
2. Find the area of the region bounded by y9x, x 2x 4 and the x-axis inthefirst quadrant |
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| 29. |
Find the slope of a line (i) which bisects the first quadrant angle (iikie direction of /-axis measured anticl |
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| 30. |
LAB C ~ Δ PQR M is the mid-point of BC and B is the mid-point of QR. The area ofAABC-t 100 sq. cm and that of ΔPOR . 144 sq. cm. If AM cm, then find PN. |
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Answer» Like if you find it useful |
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| 31. |
chord of a circle is equal to the radius of thecircle-. on the minor arc and also at a point on theFi. 10.37, PQR- 1009, where P. Q and R areP185angle subtended by the chord atmajor ar100ircle with centre O. FindOPRpoints on aFig. 10.37ăŁ. 10.38,L ABC = 69 , < ACB = 31°, findn FigZ BDCFig. 10.38 . |
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| 32. |
Find the area of the region bounded by xfirst quadrant.4y, y2, y4 and the y-axis in the |
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| 33. |
ar (PQR).ABCD is a rectangle. If AB 20 cm and ar (ABCD) 100 cm2find BC.7. 1AnsmFind the length |
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Answer» area of rectangle ABCD = base × height =AB ×BC100 cm^2= 20 cm × BCBC= 5cm hit like if you find it useful |
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| 34. |
2' 2(C)し22)2The equation of the circle which touches both the axes and the line+1 and lies in the first3 4quadrant is (x-c)2 + (y-c)2弌2 where c is(A) 1(B) 2(C) 4(D) 6 |
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| 35. |
^ { 41 } C _ { 40 } ^ { 8 } C _ { 5 } , 11 |
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Answer» 1 2 3 |
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| 36. |
a ^ { 3 } - b ^ { 3 } + 8 c ^ { 3 } + c |
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Answer» (a)³ +(-b)³ +(2c)³ - 3(a)(-b)(2c) = (a-b+2c)(a²+b²+4c² +ab +2bc -2ca) |
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| 37. |
< ig uRl pue g WIS AR[NOEd * 7 न 6509 भर! |
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Answer» Cos theta = 3/5 means base= 3 and height is 5by pythagoras theoremP^2 + B^2 = H^2P^2 = 25 -9P = Root 16P= 4sin theta = 4/5tan theta = 4/3 |
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| 38. |
1. Write various uses of concave mirror in everyday life. |
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Answer» Large concave mirrors are used to focus sunlight to produce heat in the solar furnace. They are also used in solar ovens to collect a large amount of solarenergyin the focus of the concave mirror for heating,cooking, melting metals, etc. These were some common uses of concave mirror in daily life. |
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| 39. |
3 pusie 1 By be b 8 RET IR By pae) LT |
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Answer» Ans :- बिन्दु की X-अक्ष से दूरी को उस बिन्दु का Y-निर्देशांक याकोटिकहते है। |
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| 40. |
v3+!3 +12 का- हो, तो x2 +43.और yयदि x ==3-13+1मान है- |
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Answer» x=√3+1/√3-1*√3+1/√3+1=3+1+2√3/3-1=2+√3and y=√3-1/√3+1*√3-1/√3-1=3+1-2√3/2=2-√3and x^2+y^2=4+3+4√3+4+3-4√3=7+7=14 |
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| 41. |
θ lies in the first quadrant and 5 tan θ-4, then |
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Answer» angle is in first quadrant5tan=4so tan=4/5that means sin=4/root(41)so (5sin-3)/(5sin+2)=(5*4/root(41)-3)/(5*4/roo(41)+2)=(20-3root(41))/(20+2root(41)) |
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| 42. |
E is the mid-point of side QR of a ÎÎĄ0RD is the mid-point of PE when producedmeets PR in G. Prove that PG =-PR.3 |
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| 43. |
3. Find the area of the region bounded by ' 4y, y 2, y 4 and the >-axisfirst quadrant. |
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Answer» The area can be found out by the integration of x = 2√y , from y = 2 to y = 4 so the integration is 2y^(3/2)/(3/2) = 4/3y^(3/2) now putting the limit we get area = (4/3)[4^(3/2) -2^(3/2)] = (4/3)[8-√8] = (32-8√2)/3 sq.units |
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| 44. |
In figure, the line segment ST is parallel to side PR of ÎPOR and it divides the triangle into two partsPSPQofegual area. Find the ratio- |
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Answer» Like my answer if you find it useful! |
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| 45. |
98037+ 9Sl BUE 5 [ elव.6 500९ न छ पाधटु & |
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Answer» 5 tan theta=4 So tan theta=4/5 therefore, sin theta/cos theta = 4/5 Hence we can say that, sin theta= 4x , cos theta= 5x (5*4x-3*5x)/(4x+2*5x) 5x/14x = 5/14 like my answer if you find it useful! |
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| 46. |
जगावान काना 5 6 (रू) "सन हिरू2,6]3 L o NG . - N कि RS A (Sl S T |
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Answer» Thank you |
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| 47. |
L é-’y\,g-{ e atif o~ O{ o ESZL(Q;‘(J WShose $2deकं श्ग1y तिल - |
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Answer» Area of square = (side)²Area of square = (20 m)² = 400 m² |
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| 48. |
Fig. 5.382. In figure 5.39, âĄPORS and âĄMNRL arerectangles. If point M is the midpoint of side PRthen prove that, (i) SL LR, (ii) LNSQFig. 5.39 |
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| 49. |
20% of a number is 1 more than 79. Findthe number |
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Answer» let number be XhenceX*20/100+1=790.2x+1=790.2x=80x=400 thans |
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| 50. |
~ 4 [Prove dted oppa; E Sl gL omSen 6 coodo | Cebele G R/ *‘a#/(;"’“: at Lo oenh9 el . |
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