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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
10 Books Are Placed At Random In A Shelf. The Probability That A Pair Of Books Will Always Be Together Is -? |
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Answer» 10 books can be rearranged in 10! ways consider the two books taken as a pair then NUMBER of FAVOURABLE ways of getting these two books together is 9! 2! REQUIRED probability = 1/5. 10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2! Required probability = 1/5. |
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| 2. |
What Is The Probability That A Leap Year Has 53 Sundays And 52 Mondays? |
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Answer» A LEAP year has 52 weeks and two days Number of favourable cases = 1 i.e., {Saturday, SUNDAY} Required Probability = 1/7. A leap year has 52 weeks and two days Total number of cases = 7 Number of favourable cases = 1 i.e., {Saturday, Sunday} Required Probability = 1/7. |
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| 3. |
From A Pack Of Cards Two Cards Are Drawn One After The Other, With Replacement. The Probability That The First Is A Red Card And The Second Is A King Is -? |
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Answer» LET E1 be the event of drawing a red card. Let E2 be the event of drawing a KING . P(E1 n E2) = P(E1) . P(E2) (As E1 and E2 are independent) = 1/2 * 1/13 = 1/26. Let E1 be the event of drawing a red card. Let E2 be the event of drawing a king . P(E1 n E2) = P(E1) . P(E2) (As E1 and E2 are independent) = 1/2 * 1/13 = 1/26. |
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| 4. |
The Probability That A Speaks Truth Is 3/5 And That Of B Speaking Truth Is 4/7. What Is The Probability That They Agree In Stating The Same Fact? |
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Answer» If both AGREE STATING the same fact, either both of them speak TRUTH of both speak false. PROBABILITY = 3/5 * 4/7 + 2/5 * 3/7 = 12/35 + 6/35 = 18/35. If both agree stating the same fact, either both of them speak truth of both speak false. Probability = 3/5 * 4/7 + 2/5 * 3/7 = 12/35 + 6/35 = 18/35. |
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| 5. |
The Sector Of A Circle Has Radius Of 21 Cm And Central Angle 135o. Find Its Perimeter? |
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Answer» PERIMETER of the sector = LENGTH of the arc + 2(RADIUS) = (135/360 * 2 * 22/7 * 21) + 2(21) = 49.5 + 42 = 91.5 cm. Perimeter of the sector = length of the arc + 2(radius) = (135/360 * 2 * 22/7 * 21) + 2(21) = 49.5 + 42 = 91.5 cm. |
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| 6. |
The Length Of A Rectangle Is Two - Fifths Of The Radius Of A Circle. The Radius Of The Circle Is Equal To The Side Of The Square, Whose Area Is 1225 Sq.units. What Is The Area (in Sq.units) Of The Rectangle If The Rectangle If The Breadth Is 10 Units? |
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Answer» Given that the area of the square = 1225 sq.units => Side of square = v1225 = 35 units The radius of the circle = side of the square = 35 units LENGTH of the rectangle = 2/5 * 35 = 14 units Given that breadth = 10 units Area of the rectangle = lb = 14 * 10 = 140 sq.units. Given that the area of the square = 1225 sq.units => Side of square = v1225 = 35 units The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units Given that breadth = 10 units Area of the rectangle = lb = 14 * 10 = 140 sq.units. |
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| 7. |
Find The Roots Of Quadratic Equation: X2+x - 42 = 0? |
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Answer»
X(x + 7) - 6(x + 7) = 0 (x + 7)(x - 6) = 0 => x = -7, 6. x2 + 7x - 6x + 42 = 0 x(x + 7) - 6(x + 7) = 0 (x + 7)(x - 6) = 0 => x = -7, 6. |
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| 8. |
Find The Roots Of Quadratic Equation: 2x2+ 5x + 2 = 0? |
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Answer»
2x(x + 2) + 1(x + 2) = 0 (x + 2)(2x + 1) = 0 => x = -2, -1/2. 2x2+ 4x + x + 2 = 0 2x(x + 2) + 1(x + 2) = 0 (x + 2)(2x + 1) = 0 => x = -2, -1/2. |
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| 9. |
The Price Of A Certain Item Is Increased By 15 %. If A Consumer Wants To Keep His Expenditure On The Item Same As Before How Much Percent Must He Reduce His Consumption Of That Item? |
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Answer» REDUCTION % in CONSUMPTION = {R/(100+r) ×100} % =(15/115 × 100)% = 300/23 % = 13 1/23 %. Reduction % in consumption = {r/(100+r) ×100} % =(15/115 × 100)% = 300/23 % = 13 1/23 %. |
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| 10. |
10 % Of The Electorate Did Not Cast Their Votes In An Election Between Two Candidates. 10 % Of The Votes Polled Were Found Invalid. The Successful Candidate Got 54 % Of The Valid Votes And Won By A Majority Of 1620 Votes. The Number Of Voters Enrolled On The Voter's List Was? |
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Answer» Let the number of VOTES enrolled be X Votes CAST = (90/100 × x) = 9x/10 Valid votes = 90% of 9x/10 = (90/100 × 9x/10) = 81x/100 Votes polled by successful candidate = (54/100 × 81x/100) = 4374x/10000 Votes polled by defeated candidate = (46/100 × 81x/100) = 3726x/10000 => 4374x – 3726x = 16200000 => x = 16200000/648 = 25000. Let the number of votes enrolled be x Votes cast = (90/100 × x) = 9x/10 Valid votes = 90% of 9x/10 = (90/100 × 9x/10) = 81x/100 Votes polled by successful candidate = (54/100 × 81x/100) = 4374x/10000 Votes polled by defeated candidate = (46/100 × 81x/100) = 3726x/10000 => 4374x – 3726x = 16200000 => x = 16200000/648 = 25000. |
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| 11. |
In A Company, 60% Of The Employees Are Men, Of This 40 % Are Drawing More Than Rs. 50000 Per Year. 36% Of The Total Employees Of The Company Draw More Than Rs.50000 Per Year Then What Is The Percentage Of Women Who Are Drawing Less Than Rs. 5000 Per Year? |
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Answer» Total number of employees be 100 Then number of MEN = 6000/100 = 60 Number of women = 4000/100 = 40 Therefore, number of men drawing more than Rs. 50000 = 2400/100 = 24 men Since the number of total employees drawing more than Rs. 50000 = 3600/100 = 36 Number of women who draw more than Rs. 50000 = 36- 24 = 12 Number of women who draw less than Rs. 50000 = 40 -12 = 28 Therefore, Percentage of the women who draw less than Rs. 50000 per year = 28/40 x 100% = 70%. Total number of employees be 100 Then number of men = 6000/100 = 60 Number of women = 4000/100 = 40 Therefore, number of men drawing more than Rs. 50000 = 2400/100 = 24 men Since the number of total employees drawing more than Rs. 50000 = 3600/100 = 36 Number of women who draw more than Rs. 50000 = 36- 24 = 12 Number of women who draw less than Rs. 50000 = 40 -12 = 28 Therefore, Percentage of the women who draw less than Rs. 50000 per year = 28/40 x 100% = 70%. |
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| 12. |
A City Has A Population Of 300000 Out Of Which 180000 Are Males 50% Of The Population Is Illiterate If 70 % Of The Males Are Literate, Then The Number Of Literate Females Is? |
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Answer»
Total NUMBER of males = 180000 Total literates = 5% of total population = 150000 Number of literate males = 70% of males = 126000. Total population = 300000 Total number of males = 180000 Total literates = 5% of total population = 150000 Number of literate males = 70% of males = 126000. |
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| 13. |
In A Test A Candidate Attempted Only 8 Questions And Secured 50% Marks In Each Of The Questions If The Obtained A Total Of 40% In The Test And All Questions In The Test Carried Equal Marks, How Many Questions Were There In The Test? |
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Answer» Let the marks of each QUESTION be 10 Total marks got by the candidate = 8 x 5 =40 marks 40% = 40 : 100 = 100 THEREFORE, Total NUMBER of questions = 100/10 =10. Let the marks of each question be 10 Total marks got by the candidate = 8 x 5 =40 marks 40% = 40 : 100 = 100 Therefore, Total number of questions = 100/10 =10. |
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| 14. |
Candidates In A Competitive Examination Consisted Of 60% Men And 40% Women 70% Men And 75% Women Cleared The Qualifying Test And Entered The Final Test Where 80% Men And 70% Women Were Successful? Which Of The Following Statements Is Correct? |
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Answer» LET there are 100 persons in which 60 men and 40 WOMEN NUMBER of men who cleared the qualifying test = 70 x 60/100 = 42 Number of women who cleared the qualifying test = 40 x ¾ = 30 Number of men who get success in final test = 42 x 4/5 = 33.6 Number of women who get success in final test = 30 x 70/100 = 21 Hence more men cleared the examination than a woman. Let there are 100 persons in which 60 men and 40 women number of men who cleared the qualifying test = 70 x 60/100 = 42 Number of women who cleared the qualifying test = 40 x ¾ = 30 Number of men who get success in final test = 42 x 4/5 = 33.6 Number of women who get success in final test = 30 x 70/100 = 21 Hence more men cleared the examination than a woman. |
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| 15. |
A Gardener Increases The Area Of His Rectangular Garden By Increasing Its Length By 40% And Decreasing By 20%. The Area Of The New Garden? |
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Answer» LET the length and breadth of garden be L and b respectively NEW length after 40 % increase = l x 140/100 = 1.4% New breadth after 20% decrease = b x 80/100 =0.8b New area = 1.41 x 0.8b = 1.12lb Therefore, change in area percentage =1.12ib-ib/ib*100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase). Let the length and breadth of garden be l and b respectively Therefore, Area = l x b New length after 40 % increase = l x 140/100 = 1.4% New breadth after 20% decrease = b x 80/100 =0.8b New area = 1.41 x 0.8b = 1.12lb Therefore, change in area percentage =1.12ib-ib/ib*100%= 12% increase (New area – original area = 0.12> 0 i.e.., Increase). |
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| 16. |
In A Survey It Was Found That 80% Of Those Surveyed Owned A Car While 60% Of Those Surveyed Owned A Mobile Phone, If 55% Owned Both A Car And A Mobile Phone, What Percent Of Those Surveyed Owned A Car Or A Mobile Phone Or Both? |
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Answer» GIVEN that percentage of car owners = 80% Percentage of mobile phone owners = 60% Percentage of people having both car and mobile phone = 55% Percentage of people having only car = 80 -55 = 25% Percentage of people having only mobile phone = 60 -55 =5% Percentage of people having car or mobile phone or both = 55% + 25% + 5% = 85%. Given that percentage of car owners = 80% Percentage of mobile phone owners = 60% Percentage of people having both car and mobile phone = 55% Percentage of people having only car = 80 -55 = 25% Percentage of people having only mobile phone = 60 -55 =5% Percentage of people having car or mobile phone or both = 55% + 25% + 5% = 85%. |
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| 17. |
In An Examination, Every Candidate Took Physics Or Mathematics Or Both 65.8% Took Physics And 59.2% Took Mathematics The Total Number Of Candidates Was 2000. How Many Candidates Took Both Physics And Mathematics? |
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Answer» Let x% candidates TAKE both the subjects. Therefore, Percentage of candidates who opted physics = 65.8% And percentage of candidates who opted mathematics = 59.2% Therefore, x =(65.8 + 59.2 - 100)% = (125 -100)% = 25% ALSO total NUMBER of candidates = 2000 Therefore, Number of candidates who opted both the subjects = 25/100 x 2000 =500. Let x% candidates take both the subjects. Therefore, Percentage of candidates who opted physics = 65.8% And percentage of candidates who opted mathematics = 59.2% Therefore, x =(65.8 + 59.2 - 100)% = (125 -100)% = 25% Also total number of candidates = 2000 Therefore, Number of candidates who opted both the subjects = 25/100 x 2000 =500. |
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| 18. |
If Mohan While Selling Two Goats At The Same Price Makes A Profit Of 10% On One Goat And Suffers A Loss Of 10% On The Other, He? |
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Answer» LET SP of each goat be Rs. X On one goat profit earned = 10% Therefore CP of FIRST goat = Rs. 10X/11 On other goat loss occurred = 10% Therefore CP of second goat = Rs. 10X/9 Thus, total CP of TWO goats =(10x/11+10x/9) Here CP > SP Therefore loss percentage = 2x/99 x 99/200x x 100% =1% SUFFERS a loss of 1%. Let SP of each goat be Rs. X On one goat profit earned = 10% Therefore CP of first goat = Rs. 10X/11 On other goat loss occurred = 10% Therefore CP of second goat = Rs. 10X/9 Thus, total CP of two goats =(10x/11+10x/9) Here CP > SP Therefore loss percentage = 2x/99 x 99/200x x 100% =1% Suffers a loss of 1%. |
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| 19. |
The Tank Full Of Petrol In Arun’s Motorcycle Lasts For 10 Days, If He Starts Using 25% More Every Day For How Many Days Will The Tank Full Of Petrol Last? |
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Answer» Let us assume that Arun USES X units of petrol everyday. So the amount of petrol in the tank when it is fuel will be 10X. If he started using 25% more petrol every DAY, then the amount of petrol he how uses everyday will be X (1 +25/100) =1.25x Therefore, number of DAYS his petrol will how LAST = Amount of petrol in tank / amount of petrol used everyday = 10x/1.25x = 10/1.25 = 8 Days. Let us assume that Arun uses X units of petrol everyday. So the amount of petrol in the tank when it is fuel will be 10X. If he started using 25% more petrol every day, then the amount of petrol he how uses everyday will be X (1 +25/100) =1.25x Therefore, number of days his petrol will how last = Amount of petrol in tank / amount of petrol used everyday = 10x/1.25x = 10/1.25 = 8 Days. |
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| 20. |
Kamala Got Married 6 Years Ago, Today Her Age Is 1 ¼ Times Her Age At The Time Of Marriage. Her Son’s Age Is (1/10) Times Her Age. Her Son’s Age Is? |
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Answer» Kamala’s AGE 6 YEARS ago be X years Kamala’s present age =(x +6) years Therefore x +6 = 5/4 x or 4x + 24 =5x or x =24 Kamala’s present age = 30 years SON’s present age = (1/10 x 30) = 3 years. Kamala’s age 6 years ago be x years Kamala’s present age =(x +6) years Therefore x +6 = 5/4 x or 4x + 24 =5x or x =24 Kamala’s present age = 30 years Son’s present age = (1/10 x 30) = 3 years. |
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| 21. |
10 Years Ago, Chandravati’s Mother Was 4 Times Older Than Her Daughter, After 10 Years The Mother Will Be Twice Older Than The Daughter. The Present Age Of Chandravati Is? |
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Answer» Chandravati’s age 10 YEARS ago be x years Mother’s age 10 years ago =(4X)years 2(x +20) = (4x +20) => x =10 Present age of Chandravati = (x +10) = 20 years. Chandravati’s age 10 years ago be x years Mother’s age 10 years ago =(4x)years 2(x +20) = (4x +20) => x =10 Present age of Chandravati = (x +10) = 20 years. |
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| 22. |
The Age Of Arvind's Father Is 4 Times His Age If 5 Years Ago Father’s Age Was 7 Times Of The Age Of His Son At The Time. What Is Arvind's Father’s Present Age? |
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Answer» ARVIND’s age be X years Father’s age = 4X years. Therefore (4x -5) =7(x -5) or 3x = 30 => x =10 Arvind’s father’s age is 40 years. Arvind’s age be x years Father’s age = 4x years. Therefore (4x -5) =7(x -5) or 3x = 30 => x =10 Arvind’s father’s age is 40 years. |
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| 23. |
The Sum Of The Ages Of A Father And Son Is 45 Years Five Years Ago The Product Of Their Ages Was 4 Times The Father's Age At That Time. The Present Ages Of The Father And Son Respectively Are? |
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Answer» Present ages of father and SON be x years and (45-x) years (x -5)(45 –x -5) = 4(x -5) Therefore, present ages of father and son are 36 years and 9 years. Present ages of father and son be x years and (45-x) years (x -5)(45 –x -5) = 4(x -5) -x2 +41x -180 =0 or x =36. Therefore, present ages of father and son are 36 years and 9 years. |
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| 24. |
Three Years Ago The Average Age Of A And B Was 18 Years With C Joining Them, Then The Average Becomes 22 Years. How Old Is C Now? |
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Answer» Sum of ages of A and B, 3 Years ago =(18 x 2) = 36 Years Sum of ages of A, B and C, Sum of ages of A and B, Now = (36 +6)Years = 42 Years THEREFORE, C’s age = (66 - 42)Years = 24 Years. Sum of ages of A and B, 3 Years ago =(18 x 2) = 36 Years Sum of ages of A, B and C, Now = (22 x 3) = 66 Years Sum of ages of A and B, Now = (36 +6)Years = 42 Years Therefore, C’s age = (66 - 42)Years = 24 Years. |
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| 25. |
Ratio Of Ashok's Age To Pradeep's Age Is Equal To 4: 3. Ashok Will Be 26 Years Old After 6 Years How Old Is Pradeep Now? |
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Answer» Ashok’s AGE = 4x and pradeep’s age =3X YEARS THEREFORE, 4x +6 = 26 => x = 5 Pradeep age = 3x = 15 years. Ashok’s age = 4x and pradeep’s age =3x years Therefore, 4x +6 = 26 => x = 5 Pradeep age = 3x = 15 years. |
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| 26. |
Jayesh Is As Much Younger To Anil As He Is Older To Prashant If The Sum Of The Ages Of Anil And Prashant Is 48 Years, What Is The Age Of Jayesh? |
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Answer» ANIL’s age = x years Prashant’s age = (48 -x)Years The age of JAYESH be P years P –(48 -x) = x –p => 2P =48 => P= 24 Years. Anil’s age = x years Prashant’s age = (48 -x)Years The age of Jayesh be P years P –(48 -x) = x –p => 2P =48 => P= 24 Years. |
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| 27. |
The Ratio Of Vimal's Age And Aruna's Age Is 3: 5 And Sum Of Their Ages Are 80 Years. The Ratio Of Their Ages After 10 Years Will Be? |
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Answer»
Ratio of their ages after 10 Years = (3x +10 : 5x + 10) = 40 : 60 = 2 : 3. 3x + 5x = 80 => x =10 Ratio of their ages after 10 Years = (3x +10 : 5x + 10) = 40 : 60 = 2 : 3. |
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| 28. |
Mayank Buys Some Candies For Rs 15 A Dozen And An Equal Number Of Different Candies For Rs 12 A Dozen. He Sells All For Rs 16.50 A Dozen And Makes A Profit Of Rs 150. How Many Dozens Of Candies Did He Buy Altogether? |
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Answer» Mayank BUYS some candies for RS 15 a dozen and an equal number of different c et the number of dozens of candies he bought of each variety be x Hence total cost = 12x + 15x = 27x Total selling price = 16.50*2X = 33X Profit = 33x - 27x = 6x Given 6x = 150 => x = 25 Hence he bought 50 dozens of candies in total. Mayank buys some candies for Rs 15 a dozen and an equal number of different c et the number of dozens of candies he bought of each variety be x Hence total cost = 12x + 15x = 27x Total selling price = 16.50*2x = 33x Profit = 33x - 27x = 6x Given 6x = 150 => x = 25 Hence he bought 50 dozens of candies in total. |
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| 29. |
If The Volumes Of Two Cubes Are In The Ratio 8: 1, The Ratio Of Their Edges Is? |
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Answer» Volumes be 8x3and X3 2X and x respectively Therefore, Ratio of their EDGES = 2 : 1. Volumes be 8x3and x3 2x and x respectively Therefore, Ratio of their edges = 2 : 1. |
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| 30. |
A Metal Sheet 27 Cm Long 8cm Broad And 1 Cm Thick Is Melted Into A Cube. The Difference Between The Surface Areas Of Two Solids Is? |
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Answer» Volume of sheet = (27 × 8 × 1)cm3 = 216 cm3 Volume of cube FORMED = 216 cm3 Therefore, EDGE of the cube = (6 × 6 × 6)1/3 = 6 cm SURFACE area of original cuboid = 2 (27 × 8 + 8 × 1 + 27 × 1) cm2 = 502 cm2 Surface area of the cube formed = [6 × (6) 2] cm2 = 216 cm2 Therefore, Difference in AREAS = (502 - 216)cm2= 286 cm2. Volume of sheet = (27 × 8 × 1)cm3 = 216 cm3 Volume of cube formed = 216 cm3 Therefore, Edge of the cube = (6 × 6 × 6)1/3 = 6 cm Surface area of original cuboid = 2 (27 × 8 + 8 × 1 + 27 × 1) cm2 = 502 cm2 Surface area of the cube formed = [6 × (6) 2] cm2 = 216 cm2 Therefore, Difference in areas = (502 - 216)cm2= 286 cm2. |
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| 31. |
A Train Travelling At Constant Speed Crosses A 96m Long Platform In 12 Seconds And Another 141m Long Platform In 15 Seconds. The Length Of The Train And It’s Speed Are? |
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Answer» LET the length of the TRAIN be x metres. => 15x(96 +x) = 12 x (141 + x) => 3x = (1692 -1440) = 252 => X = 84m Speed of the train = 15 m/sec = (15 x 18/5)kmph = 54kmph. Let the length of the train be x metres. =>(96+x)/12=(141+x)/15 => 15x(96 +x) = 12 x (141 + x) => 3x = (1692 -1440) = 252 => X = 84m Speed of the train = 15 m/sec = (15 x 18/5)kmph = 54kmph. |
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| 32. |
A Train Crosses A Pole In 15 Seconds While It Crosses A 100m Long Platform In 25 Seconds. The Length Of The Train Is? |
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Answer» Let the LENGTH of the train be x meters. Then x/15 = (100+x)/25 => 10x = 1500 => x =150. Hence the LENGTHS of the trains is 150M. Let the length of the train be x meters. Then x/15 = (100+x)/25 => 25x = 1500 + 15x => 10x = 1500 => x =150. Hence the lengths of the trains is 150m. |
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| 33. |
Two Trains 105m And 90m Long Run At The Speeds Of 45 Kmph And 72 Kmph Respectively In Opposite Direction On Parallel Tracks. The Time Which They Take To Cross Each Other Is? |
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Answer» SUM of the lengths of the trains = (105 + 90)m = 195m Relative speed = (72 +45)KMPH = 117kmph =(117 x 5/18)m/SEC = 585/18 m/sec REQUIRED Time = (195 x 18/585)sec = 6 Sec. Sum of the lengths of the trains = (105 + 90)m = 195m Relative speed = (72 +45)kmph = 117kmph =(117 x 5/18)m/sec = 585/18 m/sec Required Time = (195 x 18/585)sec = 6 Sec. |
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| 34. |
Two Trains A And B Start Running Together From The Same Point In The Same Direction At 60kmph And 72kmph Respectively. If The Length Of Each Train Is 240m. How Long Will It Take For The Train B To Cross Train A? |
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Answer» Relative speed =(72 -60)kmph = 12kmph = (12 X 5/18)m/SEC = 10/3 m/sec Total DISTANCE covered =(240 +240)m =480m Required time =(480 x 3/10)sec = 144sec = 2min 24 sec. Relative speed =(72 -60)kmph = 12kmph = (12 x 5/18)m/sec = 10/3 m/sec Total distance covered =(240 +240)m =480m Required time =(480 x 3/10)sec = 144sec = 2min 24 sec. |
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