InterviewSolution
Saved Bookmarks
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. |
20% Of 40 + ? % Of 80 = 120? |
|
Answer»
=> 800/100 + ?/100 * 80 = 120 => ? = 140. 20% of 40 + ? % of 80 = 120 => 800/100 + ?/100 * 80 = 120 => ? = 140. |
|
| 2. |
In How Many Different Ways Can Four Books A, B, C And D Be Arranged One Above Another In A Vertical Order Such That The Books A And B Are Never In Continuous Position? |
|
Answer» The NUMBER of arrangement in which A and B are not TOGETHER = Total number of ARRANGEMENTS = Number of arrangements in which A and B are together =4!-3!x2! = 24-12 =12. The number of arrangement in which A and B are not together = Total number of arrangements = Number of arrangements in which A and B are together =4!-3!x2! = 24-12 =12. |
|
| 3. |
There Are Two Identical Red, Two Identical Black, And Two Identical White Balls.in How Many Different Ways Can The Balls Be Placed In The Cells (each Cell To Contain One Ball) Shown Above Such That Balls Of The Same Colour Do Not Occupy Any Two Consecutive Cells? |
|
Answer» Case I : 2 BALLS of the same COLOUR and two balls are a different colour are ARRANGED. Two balls of the same colour and two balls of different COLOURS can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways Therefore, Total number of arrangements = 6x3 =18 ways Case II : Two colours out of 3 can be selected in = 3C1 = 3ways Now 2 balls of each colour can be arranged alternatively in 2 ways Thus 4 balls can be arranged(two of each colours) = 3x2 = 6ways Hence total number of arrangements = 18+6 =24 ways. Case I : 2 balls of the same colour and two balls are a different colour are arranged. Two balls of the same colour and two balls of different colours can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways Therefore, Total number of arrangements = 6x3 =18 ways Case II : Two colours out of 3 can be selected in = 3C1 = 3ways Now 2 balls of each colour can be arranged alternatively in 2 ways Thus 4 balls can be arranged(two of each colours) = 3x2 = 6ways Hence total number of arrangements = 18+6 =24 ways. |
|
| 4. |
If X = V196 + V200, Then X/2 + 2/x Is? |
|
Answer»
2/x = (7 - 5v2) / (7 + 5v2)(7 - 5v2) = 5v2 - 7 x/2 + 2/x = 10v2 = 2v50. x = 14 + 10v2 x/2 = 7 + 5v2 2/x = (7 - 5v2) / (7 + 5v2)(7 - 5v2) = 5v2 - 7 x/2 + 2/x = 10v2 = 2v50. |
|
| 5. |
The Bus Fare For Two Persons For Travelling Between Agra And Aligarh Id Four-thirds The Train Fare Between The Same Places For One Person. The Total Fare Paid By 6 Persons Travelling By Bus And 8 Persons Travelling By Train Between The Two Places Is Rs.1512. Find The Train Fare Between The Two Places For One Person? |
|
Answer» Let the train fare between the two places for ONE PERSON be Rs.t BUS fare between the two places for two persons Rs.4/3 t => 6/2 (4/3 t) + 8(t) = 1512 => 12t = 1512 => t = 126. Let the train fare between the two places for one person be Rs.t Bus fare between the two places for two persons Rs.4/3 t => 6/2 (4/3 t) + 8(t) = 1512 => 12t = 1512 => t = 126. |
|
| 6. |
A Man Traveled A Total Distance Of 1800 Km. He Traveled One-third Of The Whole Trip By Plane And The Distance Traveled By Train Is Three-fifth Of The Distance Traveled By Bus. If He Traveled By Train, Plane And Bus, Then Find The Distance Traveled By Bus? |
|
Answer» Total DISTANCE traveled = 1800 KM. Distance traveled by PLANE = 600 km. Distance traveled by BUS = x Distance traveled by train = 3x/5 => x + 3x/5 + 600 = 1800 => 8x/5 = 1200 => x = 750 km. Total distance traveled = 1800 km. Distance traveled by plane = 600 km. Distance traveled by bus = x Distance traveled by train = 3x/5 => x + 3x/5 + 600 = 1800 => 8x/5 = 1200 => x = 750 km. |
|
| 7. |
Ajay And Vijay Have Some Marbles With Them. Ajay Told Vijay "if You Give Me 'x' Marbles, Both Of Us Will Have Equal Number Of Marbles". Vijay Then Told Ajay "if You Give Me Twice As Many Marbles, I Will Have 30 More Marbles Than You Would". Find 'x'? |
|
Answer» If Vijay gives 'x' marbles to AJAY then Vijay and Ajay would have V - x and A + x marbles. V - x = A + x --- (1) If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A - 2x and V + 2x marbles. V + 2x - (A - 2x) = 30 => V - A + 4X = 30 --- (2) From (1) we have V - A = 2x Substituting V - A = 2x in (2) 6x = 30 => x = 5. If Vijay gives 'x' marbles to Ajay then Vijay and Ajay would have V - x and A + x marbles. V - x = A + x --- (1) If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A - 2x and V + 2x marbles. V + 2x - (A - 2x) = 30 => V - A + 4x = 30 --- (2) From (1) we have V - A = 2x Substituting V - A = 2x in (2) 6x = 30 => x = 5. |
|
| 8. |
A Vessel Of Capacity 90 Litres Is Fully Filled With Pure Milk. Nine Litres Of Milk Is Removed From The Vessel And Replaced With Water. Nine Litres Of The Solution Thus Formed Is Removed And Replaced With Water. Find The Quantity Of Pure Milk In The Final Milk Solution? |
|
Answer» Let the INITIAL quantity of MILK in vessel be T litres. Let US say y litres of the mixture is taken out and replaced by water for n times, alternatively. Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T For the given problem, T = 90, y = 9 and n = 2. HENCE, quantity of milk finally in the vessel = [(90 - 9)/90]2 (90) = 72.9 litres. Let the initial quantity of milk in vessel be T litres. Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively. Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T For the given problem, T = 90, y = 9 and n = 2. Hence, quantity of milk finally in the vessel = [(90 - 9)/90]2 (90) = 72.9 litres. |
|
| 9. |
Two Vessels P And Q Contain 62.5% And 87.5% Of Alcohol Respectively. If 2 Litres From Vessel P Is Mixed With 4 Litres From Vessel Q, The Ratio Of Alcohol And Water In The Resulting Mixture Is? |
|
Answer» Quantity of ALCOHOL in VESSEL P = 62.5/100 * 2 = 5/4 litres Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5. Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5. |
|
| 10. |
A Cloth Merchant Has Announced 25% Rebate In Prices. If One Needs To Have A Rebate Of Rs. 40. Then How Many Shirts Each Costing Rs.32 He Should Purchase? |
|
Answer» The number of SHIRTS = X Rebate = (25/100 × 32 X) = 8X 8X = 40 or X = 5. The number of shirts = X Rebate = (25/100 × 32 X) = 8X 8X = 40 or X = 5. |
|
| 11. |
The Difference Between A Discount Of 40% On Rs. 500 And Two Successive Discount Of 36% And 4% On The Same Amount Is? |
|
Answer» Sale after 40% DISCOUNT = 60% of Rs. 500 = Rs. 300 Price after 36% discount = 64% of Rs. 500 = Rs. 320 Price after next 4% discount = 96% of Rs. 320 = Rs. 307.20 Difference in TWO prices = Rs. 7.20. Sale after 40% discount = 60% of Rs. 500 = Rs. 300 Price after 36% discount = 64% of Rs. 500 = Rs. 320 Price after next 4% discount = 96% of Rs. 320 = Rs. 307.20 Difference in two prices = Rs. 7.20. |
|
| 12. |
Tarun Bought A T.v With 20% Discount On The Labelled Price. Had He Bought It With 25% Discount? He Would Have Saved Rs. 500. At What Price Did He Buy The T.v? |
|
Answer»
S.P in 1st case = Rs. 80 S.P in 2nd case = Rs. 75 Saving is Rs. 5 labelled price = Rs. 100 Saving is Rs. 500 labelled price = Rs. (100/5 × 500) = Rs. 10000. Labelled price be Rs. 100 S.P in 1st case = Rs. 80 S.P in 2nd case = Rs. 75 Saving is Rs. 5 labelled price = Rs. 100 Saving is Rs. 500 labelled price = Rs. (100/5 × 500) = Rs. 10000. |
|
| 13. |
A Dealer Marks His Goods 20% Above Cost Price. He Then Allows Some Discount On It And Makes A Profit Of 8%. The Rate Of Discount Is? |
|
Answer»
Marked Price = Rs. 120, S.P = Rs. 108 Discount = (12/120 × 100)% = 10%. C.P = Rs. 100 Marked Price = Rs. 120, S.P = Rs. 108 Discount = (12/120 × 100)% = 10%. |
|
| 14. |
The Effective Annual Rate Of Interest Corresponding To A Nominal Rate Of 6% Per Annum Payable Half Yearly Is? |
|
Answer» Let the sum be Rs 100. Then P = Rs 100, R = 3 % per half – year, t = 2 half – YEARS AMOUNT = Rs [100 × (1 + 3/100)2] = Rs (100 × 103/100 × 103/100) = Rs 10609/100 = Rs 106.09 Effective Annual RATE = 6.09%. Let the sum be Rs 100. Then P = Rs 100, R = 3 % per half – year, t = 2 half – years Amount = Rs [100 × (1 + 3/100)2] = Rs (100 × 103/100 × 103/100) = Rs 10609/100 = Rs 106.09 Effective Annual Rate = 6.09%. |
|
| 15. |
The Difference Between Simple And Compound Interest (compounded-annually) On A Sum Of Money For 2 Years At 10% Per Annum Is Rs 65. The Sum Is? |
|
Answer» LET the sum be Rs x. Then [x × (1 +10/100)2 - x] – (x × 10/100 × 2) = 65 => (x × 11/10 × 11/10 - x) – x/5 = 65 => (121x/100 - x) – x/5 = 65 => (21x/100 – x/5) = 65 => (21x – 20X) = 6500 => X = 6500. Let the sum be Rs x. Then [x × (1 +10/100)2 - x] – (x × 10/100 × 2) = 65 => (x × 11/10 × 11/10 - x) – x/5 = 65 => (121x/100 - x) – x/5 = 65 => (21x/100 – x/5) = 65 => (21x – 20x) = 6500 => X = 6500. |
|
| 16. |
A Top Can Fill A Cistern In 8 Hours And Another Can Empty It In 16 Hours. If Both The Taps Are Opened Simultaneously The Time (in Hours) To Fill The Tank Is? |
|
Answer» PART filled by insert in 1 HOUR = 1/8 Part emptied by outlet in 1 hour = 1/16 Net filling in 1 hour = (1/8 - 1/16) = 1/16 Time taken to fill the tank = 1/16 hour= 16 hours. Part filled by insert in 1 hour = 1/8 Part emptied by outlet in 1 hour = 1/16 Net filling in 1 hour = (1/8 - 1/16) = 1/16 Time taken to fill the tank = 1/16 hour= 16 hours. |
|
| 17. |
A Thief Steals A Ca R At 2.30 Pm And Drives It At 60 Kmph.the Theft Is Discovered At 3 Pm And The Owner Sets Off In Another Car At 75 Kmph When Will He Overtake The Thief? |
|
Answer» Let the thief is OVERTAKEN x hrs after 2.30 PM DISTANCE covered by the thief in x hrs = distance covered by the owner in x-1/2 hr 60x = 75 ( x- ½) x= 5/2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm. Let the thief is overtaken x hrs after 2.30 pm distance covered by the thief in x hrs = distance covered by the owner in x-1/2 hr 60x = 75 ( x- ½) x= 5/2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm. |
|
| 18. |
Three Taps A, B And C Can Fill A Tank In 12, 15 And 20 Hours Respectively. If A Is Open All The Time And B And C Are Open For One Hour Each Alternately, The Tank Will Be Full Is? |
|
Answer» (A + B)'s 1 HOUR work = (1/12 + 1/15) = 3/20 (A + C)'s 1 hour work = (1/12 + 1/20) = 2/15 Part filled in 2 hrs = (3/20 + 2/15) = 17/60 Part filled in 6 hrs = 3 * 17/60 = 17/20 Remaining part = 1 - 17/20 = 3/20 Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour. Total TIME taken to FILL the tank = (6 + 1) = 7 hrs. (A + B)'s 1 hour work = (1/12 + 1/15) = 3/20 (A + C)'s 1 hour work = (1/12 + 1/20) = 2/15 Part filled in 2 hrs = (3/20 + 2/15) = 17/60 Part filled in 6 hrs = 3 * 17/60 = 17/20 Remaining part = 1 - 17/20 = 3/20 Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour. Total time taken to fill the tank = (6 + 1) = 7 hrs. |
|
| 19. |
The Average Of 18 Observations Was Calculated And It Was 124. Later On It Was Discovered That Two Observations 46 And 82 Were Incorrect. The Correct Values Are 64 And 28. The Correct Average Of 18 Observations Is? |
|
Answer» Sum of 18 OBSERVATIONS = 18x124 =2232 CORRECT sum of 18 observations = 2232-46-82+64+28 = 2196 Therefore, Correct average =2196/18 = 122. Sum of 18 observations = 18x124 =2232 Correct sum of 18 observations = 2232-46-82+64+28 = 2196 Therefore, Correct average =2196/18 = 122. |
|
| 20. |
A Person Travels From X To Y At A Speed Of 40km/h And Returns By Increasing His Speed 50%. What Is His Average Speed For Both The Trips? |
|
Answer» Speed of PERSON from x to y =40 km/h Speed of person from y to x =(40x150)/100 = 60km/h Since the distance travelled is same THEREFORE, Average Speed = (2x40x60)/40+60 = 48km/h. Speed of person from x to y =40 km/h Speed of person from y to x =(40x150)/100 = 60km/h Since the distance travelled is same Therefore, Average Speed = (2x40x60)/40+60 = 48km/h. |
|
| 21. |
The Average Of Women And Child Workers In A Factory Was 15%yr. The Average Age Of All The 16 Children Was 8yr And Average Age Of Women Workers Was 22 Yrs If Ten Women Workers Were Married Then The Number Of Unmarried Women Workers Were? |
|
Answer» LET the number of WOMEN workers be x According to the question 22*x+16*8 = 15(16+x) => 22x-15x = 240 -128 => 7x=112 Therefore, x=112/7 =16 Therefore, Unmarried women workers =(16-10) =6. Let the number of women workers be x According to the question 22*x+16*8 = 15(16+x) => 22x+128 =240 +15x => 22x-15x = 240 -128 => 7x=112 Therefore, x=112/7 =16 Therefore, Unmarried women workers =(16-10) =6. |
|
| 22. |
Rs.1000 Is Invested At 5 % P.a Simple Interest If The Interest Is Added To The Principle After Every 10 Years. The Amount Will Become Rs. 2000 After? |
|
Answer» S. I for 10 Years = RS (1000 × 5/100 × 10) = Rs 500 Principle after 10 Years becomes = Rs (1000 + 500) = Rs 1500 S. I on it = Rs (2000 - 1500) = Rs 500 Time = (50000/7500) Years = 6 2/3 Years TOTAL Time = (10 + 6 2/3) Years = 16 2/3 Years. S. I for 10 Years = Rs (1000 × 5/100 × 10) = Rs 500 Principle after 10 Years becomes = Rs (1000 + 500) = Rs 1500 S. I on it = Rs (2000 - 1500) = Rs 500 Time = (50000/7500) Years = 6 2/3 Years Total Time = (10 + 6 2/3) Years = 16 2/3 Years. |
|
| 23. |
On A Certain Sum, The Simple Interest At The End Of 12 ½ Years Becomes ¾ Of The Sum. What Is The Rate Percent Per Annum? |
|
Answer» Let the sum be Rs X. Then S.I = Rs 3x/4, TIME = 25/2 YEARS Therefore, Rate = (100 × 3x/4 × 1/x × 2/25) % p.a = 6 % p.a. Let the sum be Rs x. Then S.I = Rs 3x/4, Time = 25/2 Years Therefore, Rate = (100 × 3x/4 × 1/x × 2/25) % p.a = 6 % p.a. |
|
| 24. |
Ratio Of The Earnings Of A And B Is 4:7. If The Earnings Of A Increases By 50% And Those Of B Decreased By 25%, The New Ratio Of Their Earnings Becomes 8:7. What Are A's Earnings? |
|
Answer» Let the ORIGINAL earnings of A and B be Rs. 4x and Rs. 7x. New earnings of A = 150% 0f Rs. 4x = (150/100 * 4x) = Rs. 6x New earnings of B = 75% of Rs. 7x = (75/100 * 7x) = Rs. 21x/4 6x:21x/4 = 8:7 This does not GIVE X. So, the given data is inadequate. Let the original earnings of A and B be Rs. 4x and Rs. 7x. New earnings of A = 150% 0f Rs. 4x = (150/100 * 4x) = Rs. 6x New earnings of B = 75% of Rs. 7x = (75/100 * 7x) = Rs. 21x/4 6x:21x/4 = 8:7 This does not give x. So, the given data is inadequate. |
|
| 25. |
If 40% Of A Number Is Equal To Two-third Of Another Number, What Is The Ratio Of First Number To The Second Number? |
|
Answer»
40A/100 = 2B/3 => 2A/5 = 2B/3 A/B = (2/3 * 5/2) = 5/3 A:B = 5:3. Let 40% of A = 2/3 B. Then, 40A/100 = 2B/3 => 2A/5 = 2B/3 A/B = (2/3 * 5/2) = 5/3 A:B = 5:3. |
|
| 26. |
The Ratio Of The Number Of Boys And Girls In A College Is 7:8. If The Percentage Increase In The Number Of Boys And Girls Be 20% And 10% Respectively. What Will Be The New Ratio? |
|
Answer» Originally, LET the number of boys and girls in the COLLEGE be 7X and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). i.e., (120/100 * 7x) and (110/100 * 8x) i.e., 42x/5 and 44x/5 Required ratio = 42x/5 : 44x/5 = 21:22. Originally, let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). i.e., (120/100 * 7x) and (110/100 * 8x) i.e., 42x/5 and 44x/5 Required ratio = 42x/5 : 44x/5 = 21:22. |
|
| 27. |
If 10% Of X = 20% Of Y, Then X:y Is Equal To? |
|
Answer»
10x/100 = 20y/100 => x/10 = y/5 x/y = 10/5 = 2/1 x:y = 2:1. 10% of x = 20% of y 10x/100 = 20y/100 => x/10 = y/5 x/y = 10/5 = 2/1 x:y = 2:1. |
|
| 28. |
8 Children And 12 Men Complete A Certain Piece Of Work In 9 Days. If Each Child Takes Twice The Time Taken By Man To Finish The Work. In How Many Days Will 12 Men Finish The Same Work? |
|
Answer» 2 children = 1 man THEREFORE (8 children +12 MEN) = 16 men Now less men more days 12: 16 : : 9 : X Therefore, x = (144/12) = 12 days. 2 children = 1 man Therefore (8 children +12 men) = 16 men Now less men more days 12: 16 : : 9 : x Therefore, x = (144/12) = 12 days. |
|
| 29. |
10 Men Can Finish A Piece Of Work In 10 Days. Whereas It Takes 12 Women To Finish It In 10 Days If 15 Men And 6 Women Undertake To Complete The Work. How Many Days Will They Take To Complete It? |
|
Answer» 10 men = 12 women or 1 man = 6/5 women THEREFORE 15 men + 6 women = (15 x 6/5 +6)women i.e,, 24 women Now 12 women can do the WORK in 10 DAYS Therefore, 24 women can do it in = 5 days. 10 men = 12 women or 1 man = 6/5 women Therefore 15 men + 6 women = (15 x 6/5 +6)women i.e,, 24 women Now 12 women can do the work in 10 days Therefore, 24 women can do it in = 5 days. |
|
| 30. |
If 3 Men Or 4 Women Can Construct A Wall In 43 Days, Then The Number Of Days That 7 Men And 5 Women Take To Construct It Is? |
|
Answer» 3 MEN = 4 women or 1 man = 4/3 women THEREFORE 7 men + 5 women = (7 × 4/3 + 5)women i.e.., 43/3 women 4 women can construct the WALL in 43 days Therefore, 43/3 women can construct it in = 12 days. 3 men = 4 women or 1 man = 4/3 women Therefore 7 men + 5 women = (7 × 4/3 + 5)women i.e.., 43/3 women 4 women can construct the wall in 43 days Therefore, 43/3 women can construct it in = 12 days. |
|
| 31. |
12 Men Or 18 Women Can Reap A Field In 14 Days. The Number Of Days That 8 Men And 16 Women Will Take To Reap It? |
|
Answer» 12 men = 18 WOMEN or 1 man = 3/2 women Therefore 8 men + 16 women = (8 × 3/2 + 16) women i.e.., 28 women Now 18 women can reap the field in 14 days Therefore, 28 women can reap it in = 9 days. 12 men = 18 women or 1 man = 3/2 women Therefore 8 men + 16 women = (8 × 3/2 + 16) women i.e.., 28 women Now 18 women can reap the field in 14 days Therefore, 28 women can reap it in = 9 days. |
|
| 32. |
A Takes 2 Hours More Than B To Walk D Km. If A Doubles His Speed Then He Can Make It In 1 Hour Less Than B. How Much Time Does B Require For Walking D Km? |
|
Answer» SUPPOSE B takes x hours to walk d km Then A takes (x+2) hours to walk d km With double of the SPEED. A will TAKE ½(x+2)hours THEREFORE, x -1/2(x+2) = 1 2x –(x+2) X =4 Hence B takes 4 hours to walk d km. Suppose B takes x hours to walk d km Then A takes (x+2) hours to walk d km With double of the speed. A will take ½(x+2)hours Therefore, x -1/2(x+2) = 1 2x –(x+2) X =4 Hence B takes 4 hours to walk d km. |
|
| 33. |
Sunil Covers A Distance By Walking For 6 Hours While Returning His Speed Decreases By 1km/hr And He Takes 9 Hours To Covers The Same Distance What Was His Speed In Turns Journey? |
|
Answer» Let the speed in return JOURNEY be X km/hr, then 6(x +1) =9x => 3x =6 => x =2 Hence the speed in return journey is 2km/hr. Let the speed in return journey be x km/hr, then 6(x +1) =9x => 3x =6 => x =2 Hence the speed in return journey is 2km/hr. |
|
| 34. |
A Boy Is Running At A Speed Of P Km/hr To Cover A Distance Of 1km But Due To The Slippery Ground, His Speed Is Reduced By Q Km/hr(p>q).if The Takes R Hours To Cover The Distance Then? |
|
Answer» Actual SPEED = (P-Q)km/hr , time taken =R HRS Therefore, 1 =(P-Q)R => 1/R =(P-Q). Actual speed = (P-Q)km/hr , time taken =R hrs Distance =(Speed x time) Therefore, 1 =(P-Q)R => 1/R =(P-Q). |
|