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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. |
Three Fair Coins Are Tossed. What Is The Probability Of Getting At Most Two Heads ? |
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Answer» Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = occasion of GETTING at most two heads. At that point E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P (E) = N (E)/n(S) = 7/8 Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = occasion of getting at most two heads. At that point E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P (E) = n (E)/n(S) = 7/8 |
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| 2. |
Tickets Numbered 1 To 20 Are Stirred Up And After That, A Ticket Is Drawn Indiscriminately. What Is The Probability Getting Ticket Drawn Has A Quantity Which Is A Difference Of 3 Or 5 ? |
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Answer» Let, S = {1, 2, 3, 4, 19, 20}. Let E = event of getting a few of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}. P (E) = N (E)/n(S) = 9/20 Let, S = {1, 2, 3, 4, 19, 20}. Let E = event of getting a few of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}. P (E) = n (E)/n(S) = 9/20 |
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| 3. |
A Man Has Rs. 480 In The Sections Of One-rupee Notes, Five-rupee Notes, And Ten-rupee Notes. The Quantity Of Notes In Every Section Is Equivalent. What Is The Aggregate Number Of Notes That He Has? |
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Answer» Give the number of notes of every group a CHANCE to be x. At that point x + 5x + 10x = 480 16x = 480 x = 30. CONSEQUENTLY, add up to number of notes = 3X = 90. Give the number of notes of every group a chance to be x. At that point x + 5x + 10x = 480 16x = 480 x = 30. Consequently, add up to number of notes = 3x = 90. |
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| 4. |
A Man Took Credit From A Bank At The Rate Of 12% P.a. Straightforward Intrigue. Following 3 Years He Needed To Pay Rs. 5400 Intrigue Just For The Period. The Essential Sum Acquired By Him Was |
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Answer» PRINCIPAL = RS. 100 X 5400/1200x3 = Rs. 15000. Principal = Rs. 100 x 5400/1200x3 = Rs. 15000. |
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| 5. |
An Entirety Of Cash At Basic Premium Adds Up To Rs. 815 Out Of 3 Years And To Rs. 854 Out Of 4 Years. The Aggregate Is |
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Answer» S.I. for 1 YEAR = RS. (854 - 815) = Rs. 39. S.I. for a LONG time = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698. S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for a long time = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698. |
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| 6. |
Merchandise Prepares Keeps Running At The Speed Of 72 Km Ph And Crosses A 250 M Long Stage In 26 Seconds. What Is The Length Of The Products Prepared ? |
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Answer»
Time = 26 sec. Then, x + 250 /26= 20 x + 250 = 520 x = 270. Speed = 20 m/sec. Time = 26 sec. Then, x + 250 /26= 20 x + 250 = 520 x = 270. |
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| 7. |
270 Meters In Length Prepare Running At The Speed Of 120 Km Ph Crosses Another Prepare Running Inverse Way At The Speed Of 80 Km Ph In 9 Seconds. What Is The Length Of The Other Prepare ? |
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Answer» Relative SPEED = (120 + 80) km/hr = 500/9 m/sec. Relative speed = (120 + 80) km/hr = 200 x 5/18 m/sec = 500/9 m/sec. |
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| 8. |
A Man Has Rs. 480 In The Divisions Of One-rupee Notes, Five-rupee Notes And Ten-rupee Notes. The Quantity Of Notes In Every Section Is Equivalent. What Is The Aggregate Number Of Notes That He Has ? |
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Answer» GIVE the number of notes of every division a chance to be X. At that point x + 5x + 10X = 480 16x = 480 x = 30. Consequently, add up to number of notes = 3X = 90 Give the number of notes of every division a chance to be x. At that point x + 5x + 10x = 480 16x = 480 x = 30. Consequently, add up to number of notes = 3x = 90 |
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| 9. |
The Speed Of A Boat In Standing Water Is 9 Km Ph And The Speed Of The Stream Is 1.5 Km Ph. A Man Lines To A Place At A Separation Of 105 Km, And Returns To The Beginning Stage. The Aggregate Time Taken By Him Is |
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Answer» Speed upstream = 7.5 km PH. Speed downstream = 10.5 km ph. ADD up to time taken = 105/7.5 + 105/10.5 hours = 24 hours. Speed upstream = 7.5 km ph. Speed downstream = 10.5 km ph. Add up to time taken = 105/7.5 + 105/10.5 hours = 24 hours. |
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| 10. |
Three Pipes A, B And C Can Fill A Tank From Void To Full In 30 Minutes, 20 Minutes, And 10 Minutes Individually. At The Point When The Tank Is Unfilled, All The Three Funnels Are Opened. A, B And C Release Compound Arrangements P, Q And R Separately. What Is The Extent Of The Arrangement R In The Fluid In The Tank Following 3 Minutes ? |
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Answer» Part filled by (A + B + C) in 3 minutes = 3 (1/30+1/20+1/10)= 3*11/60=11/20 Required PROPORTION = 3/10 X 20/11 = 6 /11 Part filled by (A + B + C) in 3 minutes = 3 (1/30+1/20+1/10)= 3*11/60=11/20 Required proportion = 3/10 x 20/11 = 6 /11 |
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| 11. |
An Organic Product Vendor Had A Few Apples. He Offers 40% Apples And Still Has 420 Apples. Initially, He Had |
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Answer» ASSUME initially he had x APPLES. At that point, (100 - 40)% of x = 420. 60/100 x x = 420 x = 420 x 100 = 700. Assume initially he had x apples. At that point, (100 - 40)% of x = 420. 60/100 x x = 420 x = 420 x 100 = 700. |
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| 12. |
16, 33, 65, 131, 261, (....) |
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Answer» Each number is DOUBLE the former one with 1 included or SUBTRACTED then again. Along these lines, the following number is (2 x 261 + 1) = 523. Each number is double the former one with 1 included or subtracted then again. Along these lines, the following number is (2 x 261 + 1) = 523. |
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| 13. |
3, 5, 11, 14, 17, 21 |
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Answer» Every one of the numbers ASIDE from 14 is an ODD NUMBER. The number '14' is the MAIN EVEN number. Every one of the numbers aside from 14 is an odd number. The number '14' is the main EVEN number. |
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| 14. |
A Boat Running Upstream Takes 8 Hours 48 Minutes To Cover A Specific Separation, While It Takes 4 Hours To Cover A Similar Separation Running Downstream. What Is The Proportion Between The Speed Of The Boat And Speed Of The Water Current And Flow Individually ? |
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Answer» At that point, remove canvassed upstream in 8 hrs. 48 min = DISTANCE covered downstream in 4 hrs. X* 8 4 /5= (y x 4) 44/5 x =4y y = 11/5 x. Required proportion = y + x/2: y - x /2 = 16/5 x 1/2: 6/5 x 1/2 = 8/5:3/5 = 8: 3. At that point, remove canvassed upstream in 8 hrs. 48 min = Distance covered downstream in 4 hrs. X* 8 4 /5= (y x 4) 44/5 x =4y y = 11/5 x. Required proportion = y + x/2: y - x /2 = 16/5 x 1/2: 6/5 x 1/2 = 8/5:3/5 = 8: 3. |
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| 15. |
A Brick Measures 20 Cm * 10 Cm * 7.5 Cm How Many Bricks Will Be Required For A Wall 25 M * 2 M * 0.75 M? |
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Answer» 25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * X 25 = 1/100 * x => x = 25000 25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x 25 = 1/100 * x => x = 25000 |
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| 16. |
Five Men And Nine Women Can Do A Piece Of Work In 10 Days. Six Men And Twelve Women Can Do The Same Work In 8 Days. In How Many Days Can Three Men And Three Women Do The Work ? |
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Answer» (5m + 9w)10 = (6m + 12w)8 => 50M + 90W = 48w + 96 W => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m 8 men can do the work in 10 DAYS. 3m +3w = 3m + 1W = 4m So, 4 men can do the work in (10 * 8)/4 = 20 days. (5m + 9w)10 = (6m + 12w)8 => 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m 8 men can do the work in 10 days. 3m +3w = 3m + 1w = 4m So, 4 men can do the work in (10 * 8)/4 = 20 days. |
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| 17. |
What Was Be The Day Of The Week On Fifteenth August 2010 ? |
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Answer» FIFTEENTH August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010) ODD days in 1600 years = 0 Odd days in 400 years = 0 9 years = (2 jump years + 7 CUSTOMARY years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days. Jan. Feb. March APRIL May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days 227 days = (32 weeks + 3 days) 3 odd days. Add up to number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days. Given day is Sunday. Fifteenth August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010) Odd days in 1600 years = 0 Odd days in 400 years = 0 9 years = (2 jump years + 7 customary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days. Jan. Feb. March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days 227 days = (32 weeks + 3 days) 3 odd days. Add up to number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days. Given day is Sunday. |
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| 18. |
In A 1000 M Race, A Beats B By 200 Meters Or 25 Seconds. Find The Speed Of B ? |
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Answer» Since A beats B by 200 m or 25 seconds, it implies that B COVERS 200 m in 25 seconds. HENCE speed of B = 200/25 = 8 m/s. Since A beats B by 200 m or 25 seconds, it implies that B covers 200 m in 25 seconds. Hence speed of B = 200/25 = 8 m/s. |
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| 19. |
What Was The Day Of The Week On 28th May 2006 ? |
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Answer» 28 May, 2006 = (2005 years + PERIOD from 1.1.2006 to 28.5.2006) Odd days in 400 years = 0 5 years = (4 conventional years + 1 jump year) = (4 x 1 + 1 x 2) 6 odd days Jan. Feb. March April May (31 + 28 + 31 + 30 + 28) = 148 days 148 days = (21 weeks + 1 day) 1 odd day. Add up to number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day. Given day is SUNDAY. 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006) Odd days in 1600 years = 0 Odd days in 400 years = 0 5 years = (4 conventional years + 1 jump year) = (4 x 1 + 1 x 2) 6 odd days Jan. Feb. March April May (31 + 28 + 31 + 30 + 28) = 148 days 148 days = (21 weeks + 1 day) 1 odd day. Add up to number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day. Given day is Sunday. |
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| 20. |
What Amount Does Kiran Get If He Invests Rs. 18000 At 15% P.a. Simple Interest For Four Years ? |
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Answer» <P>Simple interest = (18000 * 4 * 15)/100 = Rs. 10800 AMOUNT = P + I = 18000 + 10800 = Rs. 28800 Simple interest = (18000 * 4 * 15)/100 = Rs. 10800 Amount = P + I = 18000 + 10800 = Rs. 28800 |
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| 21. |
The Total Ages Of 5 Youngsters Conceived At The Interval Of 3 Years Each Is 50 Years. What Is The Age Of The Youngest Child ? |
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Answer» GIVE the times of youngsters a chance to be X, (x + 3), (x + 6), (x + 9) and (x + 12) years. At that point, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 5x = 20 x = 4. Give the times of youngsters a chance to be x, (x + 3), (x + 6), (x + 9) and (x + 12) years. At that point, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 5x = 20 x = 4. |
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| 22. |
A Man Purchased 15 Pens, 12 Books, 10 Pencils And 5 Erasers. The Cost Of Each Pen Is Rs.36, Each Book Is Rs.45, Each Pencil Is Rs.8, And The Cost Of Each Eraser Is Rs.40 Less Than The Combined Costs Of Pen And Pencil. Find The Total Amount Spent? |
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Answer» Cost of each ERASER = (36 + 8 -40) = Rs.4 Required AMOUNT = 15 * 36 + 12 * 45 + 10 * 8 + 5 * 4 540 + 540 + 80 + 20 = Rs.1180 Cost of each eraser = (36 + 8 -40) = Rs.4 Required amount = 15 * 36 + 12 * 45 + 10 * 8 + 5 * 4 540 + 540 + 80 + 20 = Rs.1180 |
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| 23. |
In A Race Of 600 Meters, A Can Beat B By 60 Meters And In A Race Of 500 Meters; B Can Beat C By 50 Meters. By What Number Of Meters Will A Beat C In A Race Of 400 Meters ? |
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Answer» A RUNS B runs C runs 600 METERS race 600 m 540 m 500 meters race 500 m 450 m COMING ratio A runs B runs C runs 300 meters-2700 meters-2430 meters Unitary A runs B runs C runs Methods 400m-360m-324m A beat C by 400- 324=76 meters. A runs B runs C runs 600 meters race 600 m 540 m 500 meters race 500 m 450 m Coming ratio A runs B runs C runs 300 meters-2700 meters-2430 meters Unitary A runs B runs C runs Methods 400m-360m-324m A beat C by 400- 324=76 meters. |
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| 24. |
A Shopkeeper Sells 20% Of His Stock At 10% Profit Ans Sells The Remaining At A Loss Of 5%. He Incurred An Overall Loss Of Rs. 400. Find The Total Worth Of The Stock ? |
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Answer» LET the total worth of the stock be Rs. x. The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50 The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50 Total SP = 11x/50 + 38x/50 = 49x/50 Overall LOSS = x – 49x/50 = x/50 Let the total worth of the stock be Rs. x. The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50 The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50 Total SP = 11x/50 + 38x/50 = 49x/50 Overall loss = x – 49x/50 = x/50 x/50 = 400 => x = 20000 |
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| 25. |
If 20 Men Can Construct A Divider 112 Meters In Length In 6 Days, What Length Of A Comparable Divider Can Be Worked By 25 Men In 3 Days ? |
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Answer» 20 men in 6 days can build 112 METERS 25 men in 30 days can build=112*(25/20)*(3/6) = 70 meters 20 men in 6 days can build 112 meters 25 men in 30 days can build=112*(25/20)*(3/6) = 70 meters |
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| 26. |
Find The Least Number Which When Divided By 35 And 11 Leaves A Remainder Of 1 In Each Case. |
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Answer» The LEAST number which when divided by different divisors leaving the same remainder in each case = LCM(different divisors) + remainder left in each case. HENCE the required least number = LCM(35, 11) + 1 = 386. The least number which when divided by different divisors leaving the same remainder in each case = LCM(different divisors) + remainder left in each case. Hence the required least number = LCM(35, 11) + 1 = 386. |
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| 27. |
A, B And C Mutually Thought Of Connecting With Themselves In A Business Wander. It Concurred That A Would Contribute Rs. 6500 For A Half Year, B, Rs. 8400 For 5 Months And C, Rs. 10,000 For 3 Months. A Needs To Be The Working Part For Which, He Was To Get 5% Of The Benefits. The Benefit Earned Was Rs. 7400. Calculate The Share Of B In The Profit. |
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Answer» For DEALING with, A got = 5% of RS. 7400 = Rs. 370. Equalization = Rs. (7400 - 370) = Rs. 7030. PROPORTION of their INVESTMENT = (6500 x 6): (8400 x 5): (10000 x 3) = 39000: 42000: 30000 = 13: 14: 10 B's offer = Rs. 7030 x14/37= Rs. 2660. For dealing with, A got = 5% of Rs. 7400 = Rs. 370. Equalization = Rs. (7400 - 370) = Rs. 7030. Proportion of their investment = (6500 x 6): (8400 x 5): (10000 x 3) = 39000: 42000: 30000 = 13: 14: 10 B's offer = Rs. 7030 x14/37= Rs. 2660. |
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| 28. |
A Heap Of Stones Can Be Made Up Into Groups Of 21. When Made Up Into Groups Of 16, 20, 25 And 45 There Are 3 Stones Left In Each Case. How Many Stones At Least Can There Be In The Heap? |
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Answer»
1 * 3600 + 3 = 3603 not divisible by 21 2 * 3600 + 3 = 7203 is divisible by 21 LCM of 16, 20, 25, 45 = 3600 1 * 3600 + 3 = 3603 not divisible by 21 2 * 3600 + 3 = 7203 is divisible by 21 |
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| 29. |
If A Card Is Drawn From A Well Shuffled Pack Of Cards, The Probability Of Drawing A Spade Or A King Is |
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Answer» <P>P(SNK) = P(S) + P(K) – P(SnK), where S denotes SPADE and K denotes king. P(SnK) = 13/52 + 4/52 – 1/52 = 4/13 P(SnK) = P(S) + P(K) – P(SnK), where S denotes spade and K denotes king. P(SnK) = 13/52 + 4/52 – 1/52 = 4/13 |
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| 30. |
There Are Three Numbers. 5/7th Of The First Number Is Equal To 48% Of The Second Number. The Second Number Is 1/9th Of The Third Number. If The Third Number Is 1125, Then Find 25% Of The First Number? |
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Answer» Let the first NUMBER and the second number be F and S RESPECTIVELY. 5/2 F = 48/100 S —-> (1) S = 1/9 * 1125 = 125 (1) => 5/7 F = 48/100 * 125 => F = 84 25% of F = 1/4 * 84 = 21. Let the first number and the second number be F and S respectively. 5/2 F = 48/100 S —-> (1) S = 1/9 * 1125 = 125 (1) => 5/7 F = 48/100 * 125 => F = 84 25% of F = 1/4 * 84 = 21. |
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| 31. |
A Man Purchases A Book For Rs.29.50 And Offers It For Rs. 31.10. Discover His Gain Percent |
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Answer» So we have C.P. = 29.50 S.P. = 31.10 GAIN = 31.10 - 29.50 = RS. 1.6 Gain %=( Gain/Cost*100)% = (1.6/29.50?100)%=5.4% So we have C.P. = 29.50 S.P. = 31.10 Gain = 31.10 - 29.50 = Rs. 1.6 Gain %=( Gain/Cost*100)% = (1.6/29.50?100)%=5.4% |
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| 32. |
Four Car Rental Agencies A, B, C And D Rented A Plot For Parking Their Cars During The Night. A Parked 15 Cars For 12 Days, B Parked 12 Cars For 20 Days, C Parked 18 Cars For 18 Days And D Parked 16 Cars For 15 Days. If A Paid Rs. 1125 As Rent For Parking His Cars, What Is The Total Rent Paid By All The Four Agencies? |
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Answer» The ratio in which the four AGENCIES will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15 = 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60 Let us consider the four amounts to be 45k, 60k, 81k and 60k RESPECTIVELY. The total RENT paid by the four agencies = 45k + 60k + 81k + 60k= 246k It is given that A paid Rs. 1125 45k = 1125 => k = 25 246k = 246(25) = Rs. 6150 Thus the total rent paid by all the four agencies is Rs. 6150. The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15 = 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60 Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively. The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k It is given that A paid Rs. 1125 45k = 1125 => k = 25 246k = 246(25) = Rs. 6150 Thus the total rent paid by all the four agencies is Rs. 6150. |
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| 33. |
A Is Twice As Fast As B Is Thrice As Fast As C. The Journey Covered By C In 42 Minutes, What Will Be Covered By A Is |
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Answer»
C COVERED in 42 minutes B covered in 42/3=14 MIN B is thrice as fast as C C covered in 42 minutes B covered in 42/3=14 min |
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| 34. |
The Length Of A Rectangular Floor Is More Than Its Breadth By 200%. If Rs. 324 Is Required To Paint The Floor At The Rate Of Rs. 3 Per Sq M, Then What Would Be The Length Of The Floor? |
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Answer» Let the length and the breadth of the floor be L m and b m respectively. l = b + 200% of b = l + 2b = 3b AREA of the floor = 324/3 = 108 SQ m l b = 108 i.e., l * l/3 = 108 l2 = 324 => l = 18. Let the length and the breadth of the floor be l m and b m respectively. l = b + 200% of b = l + 2b = 3b Area of the floor = 324/3 = 108 sq m l b = 108 i.e., l * l/3 = 108 l2 = 324 => l = 18. |
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| 35. |
A Train 125 Meter Long Is Running At 50 Km/hr. In What Time Will It Pass A Man Running At 5 Km/hr In A Similar Bearing In Which The Train Is Going? |
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Answer» Distance=125 METER speed=50-5=45km/hr=>45*5/18=12.5 m/s Time=125/12.5=10sec Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s Time=125/12.5=10sec |
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| 36. |
What Sum Of Money Put At C.i Amounts In 2 Years To Rs.8820 And In 3 Years To Rs.9261? |
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Answer» 8820 —- 441 X *105/100 * 105/100 = 8820 x*1.1025=8820 x=8820/1.1025 => 8000 8820 —- 441 100 —- ? => 5% x *105/100 * 105/100 = 8820 x*1.1025=8820 x=8820/1.1025 => 8000 |
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| 37. |
A Man Pushes Downstream 30 Km And Upstream 18 Km, Taking 5 Hours Each Time. What Is The Speed Of The Stream (current)? |
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Answer» Let X=speed of BOAT and y=speed of current =30/ (x + y) =18/(x-y) =5 by SOLVING y=1.2 km/hr Let x=speed of boat and y=speed of current =30/ (x + y) =18/(x-y) =5 by solving y=1.2 km/hr |
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| 38. |
In An Exam, Amar Scored 64 Percent, Bhavan Scored 36 Percent And Chetan 44 Percent. The Maximum Score Awarded In The Exam Is 800. Find The Average Mark Scored By All The Three Boys? |
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Answer» Average mark SCORED by all the THREE BOYS = [64/100 (800) + 36/100 (800) + 44/100 (800)] / 3 = 384 Average mark scored by all the three boys = [64/100 (800) + 36/100 (800) + 44/100 (800)] / 3 = 384 |
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| 39. |
A, B And C Can Hire A Taxi For Rs. 2400 For One Day. A, B And C Utilized The Auto For 6 Hours, 8 Hours And 10 Hours Separately. What Amount Did C Pay? |
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Answer»
C share=10/24*2400=1000 Let total fair rs.2400 C share=10/24*2400=1000 |
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