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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. |
Write like terms from the following algebraic expressions5x,-7y,-13x, 9y*. |
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Answer» like terms are 5x and -13x -5x Ku aayega check it again. i have edited it now. ooo gud |
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| 2. |
Add the following algebraic expressions.a. 4pqr - 3qr + Opr, -7pqr - Oqr + Opr, 3pqr + 2qr |
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| 3. |
11. A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to issides. The width of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find thearea of the roads and the area of the remaining portion of the field. |
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| 4. |
Express the following angles into radians() 50° 37 30" (i)10° 40 30"(1)in degrees |
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Answer» Our angle is = 5° 37' 30" For 30'' We know 1' = 60'' So, 30'' = 1/60 * 30 = 0.5' So our angle will be = 5° 37.5' Now we know, 1° = 60' So, 37.5' = 1/60 * 37.5 = 0.625° So our angle is = 5.625° To convert the angle into radians, we know 1° = π / 180 radians So, 5.625° = 5.625 x π/180 Taking π as 3.14, we get 5.625 x 3.14 / 180 = 0.098125 radiansPlease like the solution 👍 ✔️👍 |
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| 5. |
13. A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel toits sides. The width of the longer and the shorter roads are 2 m and 2.5-m-respectivelyFind the area of the roads and the area of the remaining portion of the field. |
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Answer» please draw the figure of question |
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| 6. |
a rectangular field is 15 by 40 metre it has two roads through its Centre running parallel to its side the width of the longer and the salter roads are 2 metre and 2.5 metre respectively find the area of the roads and the area of the remaining portion of the field |
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| 7. |
13. A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel toits sides. The width of the longer and the shorter roads are 2 m and 2.5-m-respectivelyFind the area of the roads and the area of the remaining portion of the field |
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| 8. |
A rectangular field is 50 m by 40 m. It has two roads thts sides. The width of the longer and the shorterFind the area of the roads and the area of the remaining portion of the fieldrough its centre, running parallel toroads are 2 m and 2.5-m-respectively |
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| 9. |
1. An electric pole is 10 metres high. If its shadow is 10/3 metres in length, find theelevation of the sun. |
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| 10. |
Girija has a gold coin of 150 gms. She distributed- of the coin to herdaughters. Now how much gold is left with Girija?1 3 16 5'5 |
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Answer» total gold = 150first daughter get = 150×1/6= 25gsecond get = 150× 3/5=90gthird will get = 150× 1/5= 30gGold left = 150-(25+90+30)= 5g thanks.. |
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| 11. |
62.एक दुकानदार कुछ चीजें क्रय मूल्य के अनुसार बेच रहा था परन्तु 1 किलो के बदले वका बाट लगा रहा था। उसका लाभ प्रतिशत ज्ञात करो:A. 9%B. 10%C. 11%D. 11-%A shopkeeper is selling his goods at cost price, but he uses 900 gms1kg. His gain percentage is :A. 9%B. 10%C.11%D. 11-%दो संख्याओं में अनपान 3.5 है। यदि दोनो संख्याओं को 10 बढ़ा दिया जाए तो आहो जाता है । संख्या ज्ञाते करीत |
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Answer» SolutionLet the CP of 1000 gm (1 kg) be Rs. 1000Then CP of 900 gm = Rs 90pSP of 900gm = Rs 900 Profit = (1000-900)*100/900 = (100/900)*100 = 100/9 let cp of 1000gm material = 100he sells 900 gm material in ₹ 100profit = 100gm on 1000gmcost of 1 gm material = 100/1000= .1₹sp of 1000gm material = 100×1000/900= 111.11₹so profit = 11.11% 62.. Option D is the right answer |
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| 12. |
(B)2(C)3A, A are the tuo A Ms between two numbern a and b and G, G, be tuo GMs between same tuonumbers, thenG, G |
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Answer» A1 , A2 are inserted bw a,b ... total terms now are 4 ... let common difference is d then a+(n-1)d = b a+(4-1)d = b 3d = b-a .................1 second term is A1 = a+d third term is A2 = a+2d A1+A2 = 2a+3d = 2a+b-a (by using 1) =a+b now if A1,A2 are Gm's bw a,b then arn-1= b ar3= b r3= b/a ............2 2nd term is A1 = ar third term is A2 = ar2 A1A2 = a2r3= ab (by using 2) now A1+A2/A1A2 = a+b/ab |
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| 13. |
5-5. लालिग्न चित्र में लम्बाई एशर ज्ञात करो, यदि 0 वृत्त का पर'त्रिज्या 5 सेमी. है, और 0-13 सेमी, ए(0 और PR faसे वृत्त पर स्पर्श रेखाएँ हैं- i |
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| 14. |
Factorise a(a-1)-b(b-1) |
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| 15. |
y3+z'xyzis equal to1#x be the AM and y, z be two GMs between two positive numbers then(A) 1(B) 2(C) 3(D) 4 |
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| 16. |
1. Factorise a-ab6 |
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| 17. |
24. Two windmills of height 50 m and 40 m are on either side of the field. A person observesthe top of the windmills from a point in between the towers. The angle of elevation wasfound to be 45° in the both cases. Find the distance between the windmills. |
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Answer» Like if you find it useful tq |
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| 18. |
Les or 1 hour !45/ How many 1 litre bottles can be filled from a cane containing 91 litres of oil ? |
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Answer» 1 1/2 litres=3/2 litresand 91 1/2 litres=183/2 litreshenceBottles filled will be 183/2÷3/2=31bottles |
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| 19. |
..Utour !45/ How many 1 litre bottles can be filled from a cane containing 91 litres of oil ? |
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| 20. |
144 cartons of coke can and 90 cartons of Pepsi cane are to be stacked icanteen If each stack is of the same height and is to contain cartons of the sdrink What would be the greater number of cartons each stack woul |
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| 21. |
man got 1 0% increase in his salary. n msAn electric pole 14 metres5 mets under similar conditions. dm12.high, casts a shadowof 10 metres. Find the height of a tree that casts a shadow of |
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Answer» The ratio of the real height and the shadow is always equal at the same time.So, the equation made here is,14/10 = x/15x = 7*3x = 21MSo, the height of that tree is 21m. |
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| 22. |
kishore observes a person standing on the ground from a helicopter at an angle of depression 30 degrees. if the helicopter flies at a height of 100meters from the ground. what is the distance of the person from kishore? |
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| 23. |
n tong 30 em wide and 4 m high is made of bricks, each measuringm 12.5 cm x 75 cm. Ifi of the total volume of the wall consists of mortar, how manycss are there in the wall?12 |
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| 24. |
n iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm3 of iron wei7.5 g. |
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| 25. |
2. Factorise the following.a. 4x^2 + 12 |
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Answer» 4x²+12 = 4(x²+3) x²+3 = 0 x² = -3 x = +-√3 |
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| 26. |
Factorise y^2-7y + 12 |
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Answer» Y^2 - 7y + 12= y^2 -(3+4)y + 12= y(y-3)-4(y-3)= (y-3)(y-4) |
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| 27. |
Factorise12 x^{2}-7 x+1 |
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Answer» 12x^2-(4+3)x+1=012x^2-4x-3x+1=04x( 3x-1)-1(3x-1)= 0(4x-1) (3x-1) = 0 |
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| 28. |
Factorise:12 x^{2}-7 x+1 |
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| 29. |
Factorise the binomial 15 x^{2} y^{3}+12 x^{2} y |
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Answer» 15x*x*y*y + 12x*x*y = 3x*x*y(5y+4) |
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| 30. |
Factorise: x^{3}+12 x^{2}+32 x+20 |
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| 31. |
A vertical pole of length 7.5 m casts a shadow 5 m long on the groundand at the same time a tower casts a shadow 24 m long. Find the heightof the tower.3. |
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| 32. |
D440 & ¢ S0 =57 |
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| 33. |
= x (S0 + XU ŕ¤ŕĽ. |
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| 34. |
1 वध DIk Bhin & —— 67 |
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Answer» 29791= 31*31*311000= 10*10*10cube root= 31/10 |
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| 35. |
s0AP in21 |
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Answer» For an AP nth term isan = a + (n-1)d Where, a = first term and d = common difference Given,a21 - a7 = 84(a + 20d) - (a + 6d) = 8420d - 6d = 8414d = 84d = 84/14 = 6 Therefore, common difference= 6 |
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| 36. |
21.The shadow of a tower standing on a ground level is formed to be 40 m longer when theSun's altitude is 30 than when it is 60%Find the height of the tower.24. Two windmills of height 50 m and 40 m are on either side of the field. A person observesthe top of the windmills from a point in between the towers. The angle of elevation wasfound to be 45 in the both cases. Find the distance between the windmills. |
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| 37. |
002kl 2eeh Š DIk Tlkah (& DAl |
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Answer» भारत में परमाणु कार्यक्रम की नींव रखने का श्रेय डॉ होमी जहांगीर भाभा को जाता है, पर डॉ राजा रामन्ना का योगदान भी इस कार्य में कम नहीं है। 18 मई, 1974 में देश के पहले सफल परमाणु परीक्षण कार्यक्रम में अहम भूमिका निभाने के लिए राजा रामन्ना को याद किया जाता है। रामान्ना उन आरंभिक भारतीय वैज्ञानिकों में से हैं, जिन्होंने ऊर्जा क्षेत्र में आत्मनिर्भरता के लिए नाभिकीय ऊर्जा के उपयोग के लिए पथप्रदर्शक कार्य किए और देश के स्वदेशी परमाणु कार्यक्रम को आगे बढ़ाने के लिए कार्य किया। please like the solution 👍 ✔️ |
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| 38. |
EXERCISE8.3Calculate the amount and compound interest on(a) r10.800 for 3 years at 125%per annumcompounded |
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| 39. |
; ZHXE1#1याक 215 DIK Lb —— kibhs e l’dl," |
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Answer» Domain = x belong to R -{2/3} |
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| 40. |
5. The side of a square lawn is 21 m. Find the length ofthe fence around it. |
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| 41. |
5 g bR PR 220 L) e 2Liy § f DIk b (208 /olae) S det 22k w1यों |
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| 42. |
MB+2423. If ι, β are the zeroes ofx2 +px + q, find the value of |-+ 2O. |
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Answer» As ALpha+ beeta= -b/a= -pand alpha * beeta= qso (alpha+2beeta/beeta)(beeta+2alpha/alpha) No Thank you Hii We're do u live |
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| 43. |
9iuc ol le transversal(iv)the vertically opposite angles.125°3. In the adjoining figure, pllq. Find the unknownangles |
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Answer» e=180-125=55° |
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| 44. |
The sum of two vertically opposite anglesis 176^{\circ}. Find the measure of each of the angles. |
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| 45. |
prove that vertically opposite angles are equal |
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Answer» Given two lines AB and CD intersect each other at the point O.To prove: ∠1 = ∠3 and ∠2 = ∠4Proof:From the figure, ∠1 + ∠2 = 180° [Linear pair] → (1)∠2 + ∠3 = 180° [Linear pair] → (2)From (1) and (2), we get∠1 + ∠2 = ∠2 + ∠3∴ ∠1 = ∠3Similarly, we can prove ∠2 = ∠4 also. |
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| 46. |
Recall, trolI Ulangles are equal. Let us prove this ICproof, and keep those in mind while studying the proUlother, then the vertically o Now,eorem 6.1 : If two lines intersect each other, then tangles are equal.Proof : In the statement above, it is giventhat 'two lines intersect each other'. So, letAB and CD be two lines intersecting at O asshown in Fig. 6.8. They lead to two pairs ofvertically opposite angles, namely(i) L AOC and BOD (ii) L AOD andZ BOCThand刁Examplangle bisSolutionTherefBut,Fig. 6.8: Vertically opposite an TheretSo, |
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| 47. |
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify thatthe bisecting rays are in the same line. |
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| 48. |
A pole 6 m high casts a shadow 24/3 m long on the ground, then find the angle of elevationof the sun.(Ans. 60) |
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| 49. |
. What will be the expenditure of painting the wall up to 5 m height of arectangular hall of length 15 m and breadth 12 m at the rate of 60 per m2? |
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| 50. |
12. A 115-m-long and 64-m-broad lawn has two roads at right angles one 2 m wide, runningparallel to its length, and the other 2.5 m wide, running parallel to its breadth. Fud thecost of gravelling the roads at 60 per m2 |
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Answer» 👍👍👍 |
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