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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. |
(01) s* 95)ع19ع |
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| 2. |
The sides of a triangle are 25 cm, 39 cm and 56 cm. Find (i) area, (ii) the length of the perpendicular drawn from the opposite vertex to the side of 25 cm. |
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| 3. |
7-67-) -(4+0 |
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| 4. |
1568hn (821 |
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| 5. |
12 128 keRES के छा ‘dore LypkL kb 0% 12 Ja phap 0३kb (8 [0० कण (3५५९ ७8मिड परी ॥2७ , "है छह 8. AR2kk bk ‘Db b LB ®bR e kb e lhe 12Il RlIab | 8°L Dblie Sjiis28 छह 1 ले सह 3४LeShy ki db हि 1किफ (के S |
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Answer» From the given graph it can be observed that the coordinate of PointAAis(3,4)(3,4), coordinate of PointBBis(6,7)(6,7), coordinate of PointCCis(9,4)(9,4)and coordinate of PointDDis(6,1)(6,1). The distance between the two pointsP(x1,y1)P(x1,y1)andQ(x2,y2)Q(x2,y2)is given by, PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√PQ=(x2−x1)2+(y2−y1)2 The distance between the points A(3,4)(3,4)and B(6,7)(6,7)is, AB=(6−3)2+(7−4)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitAB=(6−3)2+(7−4)2=9+9=32unit The distance between the points B(6,7)(6,7)and C(9,4)(9,4)is, BC=(9−6)2+(4−7)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitBC=(9−6)2+(4−7)2=9+9=32unit The distance between the points C(9,4)(9,4)and D(6,1)(6,1)is, CD=(6−9)2+(1−4)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitCD=(6−9)2+(1−4)2=9+9=32unit The distance between the points A(3,4)(3,4)and D(6,1)(6,1)is, AD=(3−6)2+(4−1)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitAD=(3−6)2+(4−1)2=9+9=32unit Therefore, the distance between AB, BC, CD and AD is same. All the distances are equal. So it may be a square or a rhombus. We need to find the distance between the diagonals. Now, the distance between the points A(3,4)(3,4)and C(9,4)(9,4)is, AC=(9−3)2+(4−4)2−−−−−−−−−−−−−−−√=36+0−−−−−√=6unitAC=(9−3)2+(4−4)2=36+0=6unit The distance between the points B(6,7)(6,7)and D(6,1)(6,1)is, BD=(6−6)2+(1−7)2−−−−−−−−−−−−−−−√=0+36−−−−−√=6unitBD=(6−6)2+(1−7)2=0+36=6unit The length of the diagonals is equal. So, the quadrilateral is a square. Thus, ABCD is a square and Champa is correct. |
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| 6. |
p %2B 4=15 |
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Answer» p + 4 = 15 p = 15 - 4 p = 11 Therefore,Value of p = 11 p=11 is the correct answer of the given question P + 4 = 15P = 15-4 =11 p + 4 = 15 p = 15 -4 p = 11 If,p + 4 = 15,then,p = 11. P+4 = 15p = 15-4p = 11verification 11 + 4 = 15 p=15-4=11p=11 p+4=15p=15-4p=1111 is the right answer p=11 is right answer... p=11 is right answer....... p+4=15. p=15-4. p=11 |
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| 7. |
1. Draw the gra |
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| 8. |
41 5 2 3210 2 6 5 10 |
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Answer» 4 1/10-[2 1/2-{5/6-(4/10+3/10))}41/10-[5/2-{5/6-7/10}41/10-[5/2-(50-42/60)]41/10-[5/2-8/60]41/10-[300-16/120]41/10-[288/120]41/10-288/1201/10[492-288/12]1/10[204/12]1/10[17)17/10 |
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| 9. |
number.The present ages of Rohit and Mayank are in the ratio 11 :8. 8 years later the sum oftheir ages will be 54 years. What are their present ages?f his twn sons, Five years later he will be |
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Answer» let the present age of Rohit be x and mayank be yx/y=11/8(x+8)+(y+8)=54x+y+16=54x+y=3811y/8 +y =3819y/8=38y=2×8=16yrx=22yr |
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| 10. |
-9 - 12 %2B 3 %2B 15 %2B 15 |
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Answer» option 30 is correct answer correct answer is the 20 12 is the right answer by using sub and add option 12 is right answer 36 is correct answer 12 is the right answer 12 is the correct answer 2 is the correct answer 30 is correct answer 30 answer for example option 30 is correct 15+15 - × -3 + × -2 - 9 than15+15+3-12-933- 2112 ans 12 is the correct answer |
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| 11. |
(4*y)/3 - 8*821^(1/3) %2B 2*15 %2B 15 |
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Answer» Answer of this question is 0 by make it in fraction form u get the answer |
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| 12. |
-49/4 %2B (-3)^3 %2B 15^0*2^5 |
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Answer» 2^5 * 15^0 + (-3)^3 - (2/7)^-2 = 32 * 1 + (-27) - (7/2)^2 = 32 + (-27 - 14/4) = 32 - 27 - 14/4 = 5 - 14/4 = (20 - 14)/4 = 6/4 = 3/2 w |
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| 13. |
-(sqrt(3) %2B sqrt(5))/(sqrt(15) %2B 4) %2B (sqrt(2) %2B sqrt(3))/(-sqrt(2) %2B sqrt(3)) |
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| 14. |
tell me important questions of maths ncert 9 class |
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Answer» Please check the books tab on app or visit our blog for more details regarding study materials. what a great question, this is??? |
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| 15. |
2. Ramesh solved part of an exercise, while Seema solved of it. Who solved lesser part?1.3.abf fruits |
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Answer» 2/7 is the right answer Ramesh Is the answer of the following RAMESH solved lesser part because 2/7(0.28) is lesser than 4/5(0.8) 2/7 is the right answer 2/ 7 is the right answer 2/7 is the right answer 2/7 is the following question answer. |
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| 16. |
10.3x +1)-(2x-x31-2x + 4 |
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Answer» 3*(3x+1)/5 - (2x - (x+1)/3) = 2x/3 + 4x/5 (9x+3)/5 - 2x + (x+1)/3 = (10x+12x)/15 (27x+9-30x+5x+5)/15 = 22x/15 2x + 14 = 22x 20x = 14 x = 14/20 = 7/10 |
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| 17. |
2。76×(a)6 is equal to36(b) 6 x 0c) 5 |
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| 18. |
Which of the following is the highest value?A) 129B) 1011C) 1110D) All are same |
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Answer» 10^11 will be be the highest value |
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| 19. |
Crs(x-8)-2(3x-5)-2-5x |
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Answer» 3(x-8)-2(3x-5)=2-5x 3x-24-6x+10=2-5x -3x+5x=2+24-10 2x=16 x=16/2x=8 |
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| 20. |
12 Rahul, an engineering student, prepared a model shaped like athecylinder with two cones attached at its two ends. The diameter ofmodel is 3 cm and its length is 12 cm. Ifeach conical part has a heightof 2 cm, find the () volume of air contained in Rahul's model(assume the outer and inner dimensions of the model to be nearlytsame), (ii) cost of painting the outer surface of the model atR 12.50per cm3 |
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| 21. |
\begin{array} { l } { \text { Simplify: } } \\ { 48 + [ 25 - \{ 20 - ( 11 - 16 \div 2 \times 4 ) \} ] = ? } \\ { \text { (A) } 40 } \\ { \text { (B) } 32 } \\ { \text { (C) } 62 } \\ { \text { (D) } 58 } \end{array} |
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| 22. |
If a+b 7 and ab12. Then find the value of (a2 + b2)(a+b2) T |
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| 23. |
a2 b2(i)3x+2y = 5 ;2-3x=7 |
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| 24. |
7,x+y=a-band ax-bya2 + b228. |
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Answer» x+y=a-bax-by=a^2+b^2Multiply equation 1 with bbx+by=ab-b^2ax-by=a^2+b^2Add equations(a+b)x=a(a+b)x=aNow, y=a-b-a; y=-b kya ise substitution method se kar sakte ho |
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| 25. |
e measures of two angles of a triangle are 72 and 58. Find the measure of the ture nf rach of |
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| 26. |
Write the fractions of each of the following fractional numbers:a) one-sixthe) thirteen-nineteenths(b) three-elevenths(f) sixteen-twentieths(c) two-eighths(g) seven-thirteenths(d) five-twelfths(b) six-tenthsinator in rach of the following fractions: |
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Answer» B) 3/11C) 2/8 = 1/4D) 5/12F) 16/20 = 4/5G)7/13H) 6/10 = 3/5 |
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| 27. |
The difference between inside and outside surfaces of a cylindrical tub14 cm long, is 88 cm2. If the volume of the tube is 176 cm3, find the inneand outer radii of the tube. |
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| 28. |
3.32*(text*((2*x^2)*(S*(o*(l*(e*v)))))) - x - 1=0 |
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Answer» (X+1)(2x-1) is the correct answer of the given question 2x^2+2x-x-1=2x(x+1)-1(x+1)=(x+1)(2x-1); x=-1; 2x=1; x=1/2 |
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| 29. |
At T L UL e YOID2; (рео02 5 .v e o |
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Answer» cos^-1[ (3^0.5) / 2] = 30 degrees |
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| 30. |
2^x=100*(text*(2^(x - 2)*(t*(h*(e*(n*(f*(i*(n*(dt*(h*(e*(v*(a*(l*(e*u)))))))))))))))) |
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Answer» 2^x=100now 2^x-2=2^x/2^2100/2^2=100/4=25 |
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| 31. |
Which of theVands e are never repeatedV. X andDel V. L andDay L. R |
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| 32. |
4^x - 4^(x - 1)=24*(text*(x*(f*(i*(n*(dt*(h*(e*(v*(a*(l*(u*(e*(f*o)))))))))))))) |
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Answer» 4^x-4^x-1=244^x(1-1/4)=244^x(3/4)=244^x=24*4/3=8*4=32now4^x=(2^2)^x=2^2xhencenow 32=2^5hence2x=5X=5/2 |
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| 33. |
| gy 1 v : ikCpmethat| a2 B2 7 |=Ben-0)apey 1+a a+p Le |
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Answer» after the transformation of rows into coloum |
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| 34. |
2. A test tubetest tube has diameter 20 mm and height is(HUIS)15 cm. The lower part is a hemisphere. Find thecapacity of the test tube. (r = 3.14). |
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Answer» capacity is πr^2h3.14×0.2cm×0.2cm×15=1.884vm^3is the right answer capacity= ~pie(r^2)h=3.14×(0.2)^2×15= 1.884 1.884 the correct answer of the given question |
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| 35. |
the duadinutIf the non-parallel sides of a trapezium are equal, prove that tis cyelie. |
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| 36. |
Thedifferencebetweeninside and outside surfaces of a cylindrical tube 14 cmLenghis8cmIf the volume of the tube is 176 cm', find the inner and outer radi of the tube |
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| 37. |
2. /The difference between inside and outside surfaces of a cylindrical tube14 cm long, is 88 cm.If the volume of the tube is 176 cm' find the innerand outer radii of the tube |
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| 38. |
1kg vegetable Rs. 40. then what price of 4.15 kg vegetable. |
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Answer» ₹16640*4.15 = 166 43.33333333333333333 cost of 1 kg vegetables = Rs40cost of 4.15 kg vegetables =Rs40×14.5=Rs 166 |
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| 39. |
Mathem126CCE TA. 1. Convert 4into a decimal fraction8 |
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| 40. |
aniciesCCE. Pattern Based QuestionsAnswers(w) F(G) F4.20%6. 500 |
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Answer» a sum of money doubles in 6 yearsLet the sum of money that's principal be x. Time = 6years Amount after 6 years becomes double = 2x Therefor, Simple Interest = Amount - Principal = 2x-x = x Rate per annum = SI×100/ P×T = x×100/x×6 = 100/6 = 50/3 = 16.66 % approx |
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| 41. |
37. 10 bags each containing 50 kg of vegetable were bought for10 per kg?3750. What willbe the profit or loss, if the vegetable is sold at |
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Answer» Total weight of vegetables bought=10*50=500KgCost of 500kg=3750Rs1kg=3750/500=7.5RsSo,Cost price=7.5RsSelling price =10RsSelling price of 500kg=500*10=5000RsSP>CP.... ProfitProfit=5000-3750=1250Rs ans in percent |
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| 42. |
The diameters of the front and rear wheels of a tractor are 80 cm and 2 mrespectively. Find the number of revolutions that a rear wheel makes tocover the distance which the front wheel covers in 800 revolutions. |
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| 43. |
जि फेक उ: O %&fl‘ J&\ in dhe I&BNL क्र स्न्ज्लकःparbadan , प०ए दफन छा७- छठ फोर ज3औसरछ1 dhe ethon by लग पछओ Sueh ol o) Ma %A_flll\\ 3 |
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Answer» As we know opposite sides of Parallelogram are equal in length Let length of four sides of Parallelogram are x, x+ 79, x + 79 Hence,Perimeter of Parallelogram =x + (x + 79) + x + (x + 79)= 2x + 2x + 158= 4x + 158 So, 4x + 158 = 4504x = 450-158 = 292x = 292/4 = 73 Therefore, length of all sides of Parallelogram are 73 cm, 152 cm, 73 cm, 152 cm |
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| 44. |
Dale.s tis Is |
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Answer» x+3/3=7/15x+3=7/55x+15=75x=-15+75x= -8 x=-8/5 plz accept as best _8/5 is the correct answer for this question x = -8/5 is the correct answer -8/5is the correct answer |
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| 45. |
ile/yOnlainena at tis |
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Answer» To find the maximum capacity of the container that can measure the diesel of three tankers we have to take their HCF Factors of 403 = 1, 13, 31, 403 Factors of 434 = 1, 2, 7, 14, 31, 62, 217, 434 Factors of 465 = 1, 3, 5, 15, 31, 93, 155, 465 Highest common factor of 403, 434 and 465 is 31. Hence the maximum capacity of the container should be 31 litres. |
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| 46. |
8. How much calories will be required to change 1gm of ice at 0°C to water a0Ca) 60 Caloriesc) 80 Caloriesb) 70 Caloriesd) 90 Calories, |
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Answer» 1 calorie will raise 1 gram of water 1 degree Celsius. If 1 g of water is given2 calories, its temperature will go up 2 degrees. Melting ice at 0 degrees Celsius: It takes80 caloriesto melt 1 gram of ice. |
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| 47. |
23of banana of 1009 gives 90 calories Calculate how much calorieswill be provided by 24 bananas where the weight of each bananais 100g |
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Answer» 2430 calories will be provided by 27 bananas |
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| 48. |
रा. _उ06 वन C . Pag & A"> = -3 =5 =-20°20HC PaT आताo |
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| 49. |
Pag: o) o cas? L0 |
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Answer» = cos^2(45 + theta) + cos^2(45 - theta) = cos^2(45 + theta) + sin^2{90 - (45 -theta)} = cos^2(45 + theta) + sin^2(90 - 45 + theta) = cos^2(45 + theta) + sin^2(45 + theta) = 1 Like my answer if you find it useful! |
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| 50. |
(x2_6xy + 9y2)-25 |
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Answer» (x^2 - 6xy + 9y^2) - 25= (x - 3y)^2 - (5)^2 We know (a^2 - b^2) = (a +b)(a-b) = (x - 3y + 5)(x - 3y - 5) ans ty |
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