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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. |
3.Find five rational numbers betweenand55 |
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| 2. |
892 - Sab + 36, gab - 6627 302p2 tab - 6add the algebraic expressi |
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Answer» 8a^2-5a b +3b^2 , 2ab- 6b^2+3a^2, b^2+ab+6a^2; 6a^2 + ab + b^2 3a^2 +2ab -6b^2 8a^2 -5ab + 3b^2 _______________________ 17a^2-2ab -2b^2 |
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| 3. |
25. Two cubes each of side 4 cm are joinedend to end. Find the surface area of theresulting cuboid. 4 60 [CBSE 2011[CBSE 2011]3 |
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| 4. |
- are in A.P., finedx+5andthe value of x.[CBSE 2011] |
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| 5. |
EXAMPLE 6.15. In Fig. 6.49 given below, PQ | RS,find x.[CBSE 2011]20120°ox140° |
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| 6. |
26. Dividex'-3+4x + 5 byr-x + 1,find quotientand remainder.[CBSE 2011 | |
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| 7. |
Wuyuu Nou Wuy!10. Reduce each of the following fractions to the lowest terms:161296(ii)(i) 207481 |
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| 8. |
56 and 38220 |
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Answer» the image is too much blurred plz ask the question again 196 and 38320; 196x195=38220 |
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| 9. |
196 और 38220 |
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Answer» this is the number ok 195,will be your answer. |
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| 10. |
1963*(38220*pi) |
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Answer» 38416 is the correct answer of the given question the answer is 196. |
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| 11. |
| Tab+4a में ८ + 8094 को न |
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Answer» 7ab+4a+a+8ba= 15ab+ 5athanks 7ab+8ba+4a+a15ab+5a |
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| 12. |
5a° â Tab + 5b* from 25. 4 |
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Answer» 3ab - 2a² - 2b² - ( 5a² - 7ab + 5b²)3ab - 2a² -2b² - 5a² + 7ab -5b²-7a² -7b² +10ab thank you all 😚😚😚 |
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| 13. |
HDth+c =10O TAB-C =40 |
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| 14. |
Find the meon of the foloaing tab le |
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Answer» Explanation : Mean = sum of f*x / sum of f = (30+78+36+105+207) / (70) = 456 / 70 = 6.514 Solution : 6.514 If you find this answer helpful then like it. |
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| 15. |
is (Q+b+c) Factor of e2+62-c?tab? |
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Answer» a^2+b^2+c^2+2abc( a+b+c) a^2+b^2+c^2+2abc(a+b+c) |
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| 16. |
The cangent of angle between the lines whose intercepts on the axes are a, -band A - a, respectively, is2%. |
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| 17. |
(3)33 5-57-56x42-56x4 का मान होगा-(1) 15625 (2) 78125•(नेतरहाट ५२५,0 (4) 0-625तो होता है? |
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Answer» 5^8 - 5^7 - 5^6*4^2 - 5^6*4 = 5^6(5^2 - 5 - 4^2 - 4) = 5^6(25 - 5 - 16 - 4) = 5^6(25 - 25) = 5^6*0 = 0 (3) is correct option 100डिग्री का संख्यात्म मान क्या होगा 100डिग्री का संख्यात्म मान क्या होगा |
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| 18. |
2 निम्नलिखित संख्याओं का 107 ज्ञात करने के लिए यूक्लिंड विभाजन एल्गोरिथ्म का प्रयोगकीजिए < Xiy 1353k22s (if) 196 और 38220 (00 867 और 255 |
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Answer» (i) 135 और 225 225> 135 से, हम प्राप्त करने के लिए 225 और 135 पर विभाजन लेम्मा लागू करते हैं 225 = 135 × 1 + 90 शेष 90 remainder 0 के बाद से, हम प्राप्त करने के लिए विभाजन लेम्मा को 135 और 90 पर लागू करते हैं 135 = 90 × 1 + 45 हम नए डिविज़र 90 और नए शेष 45 पर विचार करते हैं, और प्राप्त करने के लिए विभाजन लेम्मा लागू करते हैं 90 = 2 × 45 + 0 चूंकि शेष शून्य है, इसलिए प्रक्रिया बंद हो जाती है। चूंकि इस स्तर पर विभाजक 45 है, इसलिए, 135 और 225 का एचसीएफ 45 है। (ii) 196 और 38220 38220> 196 के बाद से, हम प्राप्त करने के लिए 38220 और 196 पर विभाजन लेम्मा लागू करते हैं 38220 = 196 × 195 + 0 चूंकि शेष शून्य है, इसलिए प्रक्रिया बंद हो जाती है। चूँकि इस स्तर पर विभाजक 196 है, इसलिए, 196 और 38220 का एचसीएफ 196 है। (iii) 867 और 255 867> 255 के बाद से, हम प्राप्त करने के लिए विभाजन लेम्मा को 867 और 255 पर लागू करते हैं 867 = 255 × 3 + 102 शेष em 102 remainder 0 से, हम प्राप्त करने के लिए 255 और 102 में विभाजन लेम्मा लागू करते हैं 255 = 102 × 2 + 51 हम नए विभाजक 102 और नए शेष 51 पर विचार करते हैं, और प्राप्त करने के लिए विभाजन लेम्मा लागू करते हैं 102 = 51 × 2 + 0 चूंकि शेष शून्य है, इसलिए प्रक्रिया बंद हो जाती है। चूंकि इस स्तर पर विभाजक 51 है, इसलिए, 867 का HCF और 255 का 51 है। |
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| 19. |
1.16071353h225)G() 19638220Gjรก) 1963 38220(iii) . 867255 |
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| 20. |
Using euclid's division algorithm to find the HCF of1) 135 and 2252) 196 and 382203) 867 and 255 |
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Answer» i) 135 = 3×3×3×5 225 = 3×3×5×5 HCF = 3×3×5 = 45 ii) 196 = 2×2×7×7 38220 = 2×2×3×5×7×7×13 HCF = 2×2×7×7 = 196 iii) 867 = 3×17×17 255 = 3×5×17 HCF = 3×17 = 51 |
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| 21. |
. Use Euclid's division algorithm to find the HCF of( 135 and 225(ii)196 and 38220(ii) 867 and 255 |
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Answer» (i) 135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain38220 = 196 × 195 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196. (iii) 867 and 255 Since 867 > 255, we apply the division lemma to 867 and 255 to obtain867 = 255 × 3 + 102Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain102 = 51 × 2 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51. Like my answer if you find it useful! (1) 135 and 225 225=135×1+90135=90×1+4590=45×2+0HCF=45 (2) 196 and 38220 38220=196×195+0HCF=196 |
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| 22. |
Euclid's division algorithm to find the HCF of225(irl 95and 38220tiff )867 and 255 |
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| 23. |
√3 cosec 20°-sec 20°=4 |
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Answer» √3 cosec 20° - sec 20° = √3/ sin 20° - 1/ cos 20°= √3cos 20° - sin 20°/ sin 20° cos 20° = 4 [√3/2 cos 20° - 1/2 sin 20° / 2 sin 20° cos 20°]= 4 [ sin 60° cos 20° - cos 60° sin 20°/ sin 40°] -- ( 2 sin A cos A = sin 2A) Using identity sin ( A - B ) = sin A cos B - cos A sin B= 4 sin (60° - 20°) / sin 40°= 4 sin 40°/ sin 40°= 4 Required Answer is 4 hit like if you find it useful |
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| 24. |
तो सिद्ध कीजिए (4 - 20(4 - 30 न 0. | |e |
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| 25. |
EXERCISE 641. Le J ABC-A D andhemecvaly2. Diagonals of a treperium AbCD with Ab DC intr in he p64 i and 12isAcn tnBcTAB 2CD, fied the ratio of the areos of triangles A0B asd CoD |
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Answer» 1) 2) |
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| 26. |
(8) Find the equations of lines which pass through thepoint (3, 4) and the sum of whose intercepts on theaxes is 14. |
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| 27. |
Find equation of the line passing through the point (2, 2) and cuting off interceptson the axes whose sum is 9 |
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| 28. |
3. ยก529(a) 13(c) 33x. Find the value of x.(b) 23(d) 29 |
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Answer» option c) 23 is the answer |
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| 29. |
find the value of √ 1032 |
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Answer» 32•124 is the right answer. plz accept as best 32.124 is the correct answer the correct answer is 32.124 of the following question 32.124 is correct answer 32.124 is the right answer.... |
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| 30. |
11. Find the value of x:3659(335/ |
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Answer» thanks Thanks |
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| 31. |
In Fig 8.31, OA and OB are opposite rays:0 If x 25°, what is the value of y? (ii) If y 35, what is the value of x?(2y +5)Fig. 8.31Fig. 8.32 |
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| 32. |
If tan (α + θ)-n tan (a-0), show that (n + 1) sin 29-(n-1) sin Dr.Sin 2๕ |
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| 33. |
1535. In the Fig., ABCD is a parallelogram in which:BAO 35°, ZDAO 40° and ZCOD- 105°. Find:(i) <ABO (ii) <ODC (iii) LACB (iv) <CBD105°40°35° |
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Answer» in triangle AOB, angle AOB + angle ABO + Angle BAO=180°ANGLE AOB = ANGLE COD= 105° (vertically opposite angles)=>angle ABO = 180°-35°-105°=40° angle ABD = angle BDC=angle ODC=40°(alternate interior angles where AB||DC) angleACB=angle CAD=40°(alternate interior angles)in triangle CBO, angle BOC=75°(supplementary pair)therefore angle CBO=180°-75°-40°=65° |
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| 34. |
aid pUSITVUe37Ď sin (n log x)5. Evaluatedr |
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Answer» thank u so much |
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| 35. |
i) 6n 38220Wo arnd 2552. Sow thai any pasitive odngs is of the torm 6, or i3,o6q5, whete qsone inreger |
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Answer» Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder. according to Euclid's division lemmaa=bq+r a=6q+r where , a=0,1,2,3,4,5then,a=6qora=6q+1ora=6q+2ora=6q+3ora=6q+4ora=6q+5 but here,a=6q+1 & a=6q+3 & a=6q+5 are odd. |
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| 36. |
Subtract the sum of -1032 and 878 from -34. |
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| 37. |
OMthe same sialaal intrthe natio'3 |
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Answer» Let the angles be 2x and 3xTherefore2x+3x=180°( sum of co interior angles is 180°)5x=180°x=180/5x=36°Therefore larger angle =3×36°=108° |
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| 38. |
5. Subtract the sum of -1032 and 878 from/-34-? |
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Answer» correct answer is 188 |
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| 39. |
11. By taking the values of a, b and c as 20,-4 and 2 respectively verify that:a +(b+c)+(a+b)+(a+c) |
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Answer» a=20 ,b=-4 ,c=2 now, a÷(b+c)=20÷(-4+2) =20÷(-2) =-10 Now,(a÷b)+(a÷c)={20÷(-4)}+(20÷2)=(-5)+(10)=5 -10 is not equal to 5. |
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| 40. |
Asl-::|.fatfe (show thai) 12A1=41ě |
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Answer» |2A|=8-32=-24 4|A|=|A|=2-8=-6×4 |
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| 41. |
4) If A, B, C are the sets of the letters in the wordscollege', 'marriage' and 'luggage' respectivelythen verify that [A-(BUC)][(A-B)n(A-C)ST |
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| 42. |
5. Subtract the sum of -1032 and 878 from -34. |
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Answer» Sum of -1032 and 878 = -1032+878 = -154-34 - ( -154) = -34+154 = 120Hence , the solution is 120. |
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| 43. |
Fig. 343 In Fig 35 4lm and O is any point show such that Z BAO-30and LOCO35: Find LAOC.3035Fig. 35 |
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Answer» thanxxxx bro.. |
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| 44. |
_2=0, 4x-4y-5-05. Half the perimeter of a rectangular garden, whose length B36 m. Find the dimensions of the garden.width, isis 4 m more than itsn variables |
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| 45. |
Alice.. boarded-a-thai ndr 6:ulas she in thethain |
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Answer» From 6:50pm to 10:50pm--4 hoursnow from 10:50pm to 11pm=10minuteshence total time is 4 hours 10minutes. |
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| 46. |
Use Euclid's division lemma to shoio thaiany Positive integes is either o |
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| 47. |
33. If 5-1 (25)2500, then find the value of x. |
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| 48. |
15. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are-3). The origin is the mid-point of the base. Find the coordinates of the points A and B. |
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| 49. |
tower with à unifospeeu. BIA Snto be 60°. Find the time taken by the car to reach the foot of the tower from this poll.s polit.16. The angles of elevation of the top of a tower from two points at a distance of 4 m and9 m from the base of the tower and in the same straight line with it are complementary.Prove that the height of the tower is 6 m. |
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| 50. |
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the pointC are (0, -3). If origin is the mid-point of the base BC, find the coordinates of thepoints A and B.1 |
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