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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. |
The GCD of two numbers is 4 and their LCM is400. How many pairs of values can the numbersassume? |
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Answer» Since the GCD is 4, the two numbers are of the form 4x and 4y, where x and y are coprime. Since x and y are coprime, their LCM is 4xy. Therefore, we need to find all pairs of coprime numbers that satisfy xy=100. Going over all options (not that many), we get that x and y can be either (1,100) or (4,25). The original numbers are therefore either (4,400) or (16,100). |
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| 2. |
7. The GCD of two numbers is 4 and their LCM is400. How many pairs of values can the numbers 60.assume? |
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Answer» Since the GCD is 4, the two numbers are of the form 4x and 4y, where x and y are coprime. Since x and y are coprime, their LCM is 4xy. Therefore, we need to find all pairs of coprime numbers that satisfy xy=100. Going over all options (not that many), we get that x and y can be either (1,100) or (4,25). The original numbers are therefore either (4,400) or (16,100). |
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| 3. |
3. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the nextrow, 18 in the row next to it and so on. In how may rows are the 200 logs placed andhow many logs are in the top row? |
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| 4. |
t o otl2Describe the flow of blood through different chambers of heart ofman with neat labelled diagram.34.3+2-5 |
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Answer» Theheartconsists of fourchambersin whichblood flows.Bloodenters the right atrium and passesthroughthe right ventricle. ... The left ventricle pumps theblood tothe aorta which will distribute the oxygenatedblood toall parts of the body. |
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| 5. |
The average weight of 150 students in a certain classis 60 kg. The mean weight of the boys from theclass is 70 kg, while that of the girl is 55 kg. Whatis the number of girls in the class?(1) 105(3) 95(2) 100(4) 60 |
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Answer» first of all the total numeber of student is 150certain class 60 kg boys weight 70 kg girl is 55 kgfind girls in class lets starttotal number of student - girl in 55 kg150 -5595 ans is 95 Total students in class = 150mean weight = 60 kgtotal weight = 150×60 = 9000 kg Let the total number of boys = xmean weight of boys = 70kgtotal weight of boys = 70×x = 70x kg total number of girls = total students - no. of boys =150 - xmean weight of girls= 55kgtotal weight of girls = 55×(150-x) = 55×150 - 55x = (8250 - 55x) kg Total weight = weight of boys + weight of girls⇒ 9000 = 70x + (8250 - 55x)⇒ 9000 = 70x + 8250 - 55x⇒ 9000 - 8250 = 70x - 55x⇒ 750 = 15x⇒ x = 750/15⇒ x = 50 So number of boys = 50number of girls = 150 - 50 = 100 |
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| 6. |
1. The extension in an elastic suring varies directly as the weight hung on it. If a weight of 150 gproduces an attension of 2.8 cm, what weight would produce an extension of 19.6 cm? |
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| 7. |
The students of a class are made to stand in rows. If three students are extra in a row, there would be one row less. If three students are less in a row, there would be 2 more rows. Find the number of students in the class. |
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Answer» Let the number of rows be y and number of students in a row be X. Total students of the class= Number of rows × Number of students in a row Total students of the class= xy Condition I: Total number of students = (y -1) (x +3)xy = (y-1) (x+ 3) xy= xy - x + 3y-33y − x-3 = 03y − x= 3 -x+3y=3…………,...(1) Condition II:Total number of students = (y + 2) (x− 3)xy = xy + 2x-3y− 62x− 3y = 6………….. (2) On multiplying equation (1) by (2)2(-x+3y=3) -2x+6y=6……………….(3) On adding equation 2 and 3 2x− 3y = 6 -2x+6y=6 --------------------3y=12 y=12/3 y=4 On putting y= 4 in eq 1. -x+3y=3 -x+3(4)=3 -x+12=3 -x=3-12 -x=-9 x=9 number of students in class= xy=9×4=36. Hence, the number of students in class=36 |
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| 8. |
030. Students of 9 class are made to sta id in rows. If one students is extra in a row, there would be 2 rowless. if one students is less in a row ther vould be 3 rows more. Find the number of students in the class |
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| 9. |
4Prove the following:a. |
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Answer» L.H.S.=(cos(2pie+x)cosec(2pie+x)tan(pie/2+x))/(sec(pie/2+x)cosecxcot(pie+x)) =(cosx*cosecx*(-cotx))/((-cosecx)*cosx**cotx) =1 =R.H.S.Hence Proved. |
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| 10. |
The mass of an atom is 1.025 × 10-17 g.What would the mass of 8000 such atomsbe? |
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Answer» Mass of 8000 atoms = 8×10³ × 1.025×10^-17 = 8.2×10-¹⁴ g |
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| 11. |
Prove that 2 C, 44" |
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| 12. |
e the equation.6y + 84(2y +1) |
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Answer» - 6y + 8 = - 4(2y + 1)- 6y + 8y = - 4 - 8 2y = - 12 y = - 12/2 = - 6 Value of y is - 6 |
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| 13. |
f-17 bekar weight 85 kg how much would 28 Bakers weight |
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Answer» 17 beakers weight = 85 kgSo 1 beaker weight =85/17 kgThen 28 beakers weight = (85/17)×28 = 140 kg 140 kilograms ️⚖️ is your answer |
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| 14. |
17) The students of a class are made to stand in rows. If 3 studentsare extra in a row, there would be 1 row less. If 3 students areless in a row there would be 2 rows more. Find the number ofstudents in the class. |
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Answer» Just same question ; I hope u solve Same question but digit changed |
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| 15. |
0 By selling 17 chairs, Rana lost the S.P. of 3 chairs. Find his loss per cent. Had he purchasesthem for3400, what would have been the S.P of one chair ? |
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Answer» selling price of one chair will be 170 ruppes |
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| 16. |
Find each of the following ratios in the simplest form:(i) 24 to 56(ii) 84 paise to R 3 |
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Answer» yeh wrong hai yeh upon me hoga |
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| 17. |
Reduce the following fractions to simplest form.\frac { 48 } { 60 } \quad ( b ) \frac { 150 } { 60 } \quad ( \text { c) } \frac { 84 } { 98 } |
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| 18. |
7. Find the LCM of each of the following by prime factorization method.a. 63.91b. 154, 196c. 63, 84, 98 |
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Answer» correct answer is (a) 819 BY FACTORIZATION WE GET:- |63,91|9,13 SO,9×13×7=819. THEREFORE LCM IS 819. answer is option number a and like my answer Correct answer is a 819 option A is the correct answer 819 is the right answer this question answer is 819 option A is the correct answer 819 is the right answer of given question. |
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| 19. |
log b 10gC, Prove thatlog a logH-4. If y-z -Ë--, Prove that |
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Answer» Like if you find it useful |
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| 20. |
Ive the following: (3 marks each)Which term of the arithemetic prograssion 3, 10, 17... would be 84 more than its 13th(12)term ? |
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Answer» a = 3 d = 7 , 13th term will be , a₁₃ = a + 12d a₁₃ = 3 + 12×7 a₁₃= 3+ 84 = 87 Since , 84 more than its 13th term ,so , value of term will be = 87 + 84 = 171 So value of term is 171 Now using , aₙ = a+(n-1)d 171 = 3 + 7n - 7 171 + 4 = 7n 175/7 = n n = 25 , So 25th term will be 84 more than its 13th term , Verification :- a₂₅ = a + 24 d a₂₅ = 3 + 24×7 a₂₅ = 3 + 168 = 171 |
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| 21. |
189=vii) 2x + 5 = 4 होने से सिद्ध करें कि 27x3 + 83 = 189|37 |
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| 22. |
?18951. What should replace both the question marks in the following equation?-? / 84 = 189 / ? |
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| 23. |
If 2x + 1/3x=4 prove it 27x cube = 189 |
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| 24. |
Prove that cotg - tane2co50-1sinecos |
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| 25. |
Findl aI7c.tq30° |
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Answer» angle l of triangle lmc, = 100 angle x = 180 -(180-100-30)= 130 once accha se please |
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| 26. |
18, How many pieces of plywood, each 0.35 cm thick, are required to make a pile 1.89 m highHint. 1.89 m (1.89x100) cm 189 cm. |
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| 27. |
64--(25x-рел |
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Answer» Roopam ji ka sahi hai |
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| 28. |
Findl |
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| 29. |
e Re S 4‘% Ca लिपि पट कान 1o eYL I el el i |
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| 30. |
Sinecos then findl theCOSCO |
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Answer» if sin theta= cos thetatan theta= 1it will be 2+ cos²theta= (2+1/2)= 3/2tan theta= 1 |
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| 31. |
findl osB4 method of Compleh |
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| 32. |
3x*6x*25x |
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Answer» Ans :- 3x × 6x × 25x = 450 x^3 |
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| 33. |
25x(x + 1) =-4 |
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Answer» 25 x (x+1) = -425x² + 25x + 4 = 025x² + 5x + 20x + 4 = 05x ( 5x + 1) + 4 ( 5x + 1) = 0(5x+4) ( 5x+1) = 0x = -4/5 and -1/5 |
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| 34. |
25x(x + 1)-4 |
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| 35. |
12x2-25x + 12 = 0 |
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| 36. |
x+25x+1=0 |
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Answer» x+25x+1=026x+1=026x=-1 x=-1/26 |
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| 37. |
23. x2 -25x-84 |
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| 38. |
A fan is sold for 644. The gain is one-sixth of thecost price of the fan. Find the profit per cent. |
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| 39. |
Ifi-Tio(зі + i) and b-\vec{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k}) \text { and } \vec{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k}), then the value7(2i+3]-6k),then the value2 \vec{a}-\overline{b})[(\vec{a} \times \vec{b}) \times(\vec{a}+2 \vec{b})]1 |
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Answer» thanks |
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| 40. |
The circumference of two circles are inthe raareas ?tio 2:3. What is the ratio of their |
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| 41. |
vessel which supplies blood to the walls of the heart is :The bloodA Pulmonary ArteryC. Pulmonary veinv)B. Carotid ArteryD. Coronary Artery |
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Answer» The heart receives its own supply of blood from thecoronary artery.option d |
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| 42. |
dnical tank of dione tsd hight t0m and thus the lanbpetely iled Findl tha high iof ninuaton on the noof |
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| 43. |
1. Find the area of the triangle whose vertices are2 In each of the following find the value of K', for which the pointsla formed by ioining the mid-points of the sides of s |
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| 44. |
(2) Ratio of corresponding sides of two similar triangles is 2:5. If the area of theslle tiangle is 64 sq cm, then what is the area of the bigger riangle? |
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Answer» It is that ratio of square of corresponding sides of two similar triangles is equal to the ratio of their corresponding areas.Let x be area of bigger triangle. Then, (2*2)/(5*5) = 64/x4/25 = 64/xX = 64/4 * 25X = 16 * 25 = 400 sq. units |
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| 45. |
How is peninsular plateau formed? Write down its featurs. |
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Answer» The Peninsular Plateau.The peninsular plateauisatableland. It is composed oftheoldest rocks because it wasformedfromthedrifted part oftheGondwana land. Broad and shallow valleys and rounded hills arethecharacteristicfeaturesof thisplateau. |
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| 46. |
( 2467 ) ^ { 153 } \times ( 341 ) ^ { 72 } |
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Answer» thanks |
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| 47. |
333+4556 |
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Answer» 4889 is the right answer 333+4556= 4889 is the correct answer . 333+4556 =4889 |
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| 48. |
666666/?=333 |
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Answer» 666666/?=333?=666666/333?=2,002 ans |
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| 49. |
2. Construct a AXYZ in which XY = 4 cm, YZ = 3 cm and XZ = 6 cm. Draw its med |
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Answer» Given : In Δ XYZ with XY = 4 cm, YZ = 3 cm and XZ = 6 cm. Steps for construction : 1.Draw a line segment XY = 4 cm 2. From vertex X draw an arc of radius 6 cm the help of compass. 3.From vertex Y draw an arc of radius 3 cm the help of compass intersecting the previous arc. 4. The point of intersection is Z. 5. Join XZ & YZ ∆XYZ is a required triangle Draw and thanks |
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| 50. |
EXE RCISE 11ind the coordinates of the point whieh divides the join of t 1,7) and ttio2 3iaunent ioining |
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