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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. |
Kabir Buys An Article With 25% Discount On Its Marked Price. He Makes A Profit Of 10 % By Selling It At Rs. 660. The Marked Price Is? |
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Answer»
C.P = (X – 25% of X) = 3X/4 S.P = (3X/4 + 10% of 3X/4) = 33X/40= 660 => X = 800. Original price be Rs. X C.P = (X – 25% of X) = 3X/4 S.P = (3X/4 + 10% of 3X/4) = 33X/40= 660 => X = 800. |
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| 2. |
A Discount Series Of 10%, 20% And 40% Is Equal To A Single Discount Of? |
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Answer» Original price = Rs. 100 Price after first DISCOUNT = Rs. 90 Price after second discount = Rs. (80/100 × 90) = Rs. 72 Price after third discount = Rs. (60/100 × 72) = Rs. 43.20 Single discount = (100 – 43.20) = 56.8%. Original price = Rs. 100 Price after first discount = Rs. 90 Price after second discount = Rs. (80/100 × 90) = Rs. 72 Price after third discount = Rs. (60/100 × 72) = Rs. 43.20 Single discount = (100 – 43.20) = 56.8%. |
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| 3. |
The Ratio Of The Prices Of Three Different Types Of Cars Is 4:5:7. If The Difference Between The Costliest And The Cheapest Cars Is Rs. 60000 The Price Of The Car Of Modest Price Is? |
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Answer» Let the PRICES be 4X, 5X and 7 X rupees 7X – 4X = 60000 => X = 20000 REQUIRED PRICE = 5X = Rs. 100000. Let the prices be 4X, 5X and 7 X rupees 7X – 4X = 60000 => X = 20000 Required price = 5X = Rs. 100000. |
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| 4. |
Subhash Purchased A Tape Recorder At 9/10th Of Its Selling Price And Sold It At 8% More Than Its S.p. His Gain Is? |
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Answer» S.P be RS. X C.P paid by Subhash = Rs. 9X/10 S.P RECEIVED by Subhash = Rs. (108% of Rs. X) = Rs. 27X/25 GAIN = Rs. (27X/25 – 9X/10) = Rs. 9X/50 Gain = (9X/50 × 10/9X × 100)% = 20%. S.P be Rs. X C.P paid by Subhash = Rs. 9X/10 S.P received by Subhash = Rs. (108% of Rs. X) = Rs. 27X/25 Gain = Rs. (27X/25 – 9X/10) = Rs. 9X/50 Gain = (9X/50 × 10/9X × 100)% = 20%. |
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| 5. |
If The Rate Of Interest Be 4% Per Annum For First Year 5% Per Annum For The Second Year And 6% Per Annum From The Third Year Then The Compound Interest Of Rs 10000 For 3 Years Will Be? |
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Answer» AMOUNT = RS[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)] = Rs(1000 × 26/25 × 21/20 × 53/50) = Rs (57876/5) = Rs 11575.20 C.I = Rs (11575.20 - 10000) = Rs 1575.20. Amount = Rs[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)] = Rs(1000 × 26/25 × 21/20 × 53/50) = Rs (57876/5) = Rs 11575.20 C.I = Rs (11575.20 - 10000) = Rs 1575.20. |
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| 6. |
If The Interest Is Payable Annually Than The Principal On Which The Compound Interest For 3 Years At 10% P.a Is Rs 33/- Is Given By? |
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Answer» Let the principal be Rs X. Then X × (1 + 10/100)3 – x = 331 => (x × 11/10 × 11/10 × 11/10 - x) = 331 => ((1331x-1000x)/1000) = 331 => 331x = 331000 => X = 1000 HENCE the principal is Rs 1000. Let the principal be Rs x. Then X × (1 + 10/100)3 – x = 331 => (x × 11/10 × 11/10 × 11/10 - x) = 331 => ((1331x-1000x)/1000) = 331 => 331x = 331000 => X = 1000 Hence the principal is Rs 1000. |
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| 7. |
The Compound Interest On Rs 16000 For 9 Months At 20% P.a Compounded Quarterly Is? |
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Answer»
PER quarter, t = 3 quarters C.I = Rs [16000 × (1+ 5/100)3-16000) = Rs [16000 × 21/20 ×21/20× 21/20- 16000) =Rs (18522-16000)=Rs 2522. P = Rs 16000, R = (10/2)% per quarter, t = 3 quarters C.I = Rs [16000 × (1+ 5/100)3-16000) = Rs [16000 × 21/20 ×21/20× 21/20- 16000) =Rs (18522-16000)=Rs 2522. |
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| 8. |
Two Pipes A And B Together Can Fill A Cistern In 4 Hours. Had They Been Opened Separately, Then B Would Have Taken 6 Hours More Than A To Fill Cistern. How Much Time Will Be Taken By A To Fill The Cistern Separately? |
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Answer» Let the CISTERN be filled by PIPE A alone in x hours. Then, pipe B will FILL it in (x + 6) hours. 1/x + 1/(x + 6) = 1/4 x2 - 2x - 24 = 0 (x - 6)(x + 4) = 0 => x = 6. Let the cistern be filled by pipe A alone in x hours. Then, pipe B will fill it in (x + 6) hours. 1/x + 1/(x + 6) = 1/4 x2 - 2x - 24 = 0 (x - 6)(x + 4) = 0 => x = 6. |
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| 9. |
Two Pipes A And B Can Fill A Tank In 15 Min And 20 Min Respectively. Both The Pipes Are Opened Together But After 4 Min, Pipe A Is Turned Off. What Is The Total Time Required To Fill The Tank? |
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Answer» Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15 REMAINING part = 1 - 7/15 = 8/15 Part filled by B in 1 minute = 1/20 1/20 : 8/15 :: 1 ; X x = 8/15 * 1 * 20 = 10 2/3 min = 10 min 40 SEC. The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec. Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15 Remaining part = 1 - 7/15 = 8/15 Part filled by B in 1 minute = 1/20 1/20 : 8/15 :: 1 ; x x = 8/15 * 1 * 20 = 10 2/3 min = 10 min 40 sec. The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec. |
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| 10. |
A Large Tanker Can Be Filled By Two Pipes A And B In 60 And 40 Minutes Respectively. How Many Minutes Will It Take To Fill The Tanker From Empty State If B Is Used For Half The Time And A And B Fill It Together For The Other Half? |
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Answer» Part FILLED by (A + B) in 1 minute = (1/60 + 1/40) = 1/24 SUPPOSE the TANK is filled in x minutes. Then, x/2(1/24 + 1/40) = 1 x/2 * 1/15 = 1 => x = 30 min. Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24 Suppose the tank is filled in x minutes. Then, x/2(1/24 + 1/40) = 1 x/2 * 1/15 = 1 => x = 30 min. |
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| 11. |
A Tank Is Filled By Three Pipes With Uniform Flow. The First Two Pipes Operating Simultaneously Fill The Tank In The Same During Which The Tank Is Filled By The Third Pipe Alone. The Second Pipe Fills The Tank 5 Hours Faster Than The First Pipe And 4 Hours Slower Than The Third Pipe. The Time Required By The First Pipe Is? |
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Answer» Suppose, first pipe alone takes x hours to fill the tank. Then, SECOND and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank. 1/x + 1/(x - 5) = 1/(x - 9) (2x - 5)(x - 9) = x(x - 5) (x- 15)(x - 3) = 0 => x = 15. Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank. 1/x + 1/(x - 5) = 1/(x - 9) (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0 (x- 15)(x - 3) = 0 => x = 15. |
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| 12. |
A Tank Is Filled In 5 Hours By Three Pipes A, B And C. The Pipe C Is Twice As Fast As B And B Is Twice As Fast As A. How Much Time Will Pipe A Alone Take To Fill The Tank? |
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Answer» Suppose PIPE A alone takes X hours to fill the tank. Then, pipes B and C will TAKE x/2 and x/4 hours respectively to fill the tank. 1/x + 2/x + 4/x = 1/5 Suppose pipe A alone takes x hours to fill the tank. Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank. 1/x + 2/x + 4/x = 1/5 7/x = 1/5 => x = 35 hrs. |
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| 13. |
A Man Covers A Distance On Scooter .had He Moved 3 Kmph Faster He Would Have Taken 40 Min Less. If He Had Moved 2kmph Slower He Would Have Taken 40 Min More. The Distance Is? |
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Answer» LET distance = x m Usual rate = y kmph x/y x/y+3 = 40/60 hr 2y(y+3) = 9x 1 x/y-2 x/y = 40/60 hr y(y-2) = 3x 2 divide 1 & 2 EQUATIONS by SOLVING we get x = 40. Let distance = x m Usual rate = y kmph x/y x/y+3 = 40/60 hr 2y(y+3) = 9x 1 x/y-2 x/y = 40/60 hr y(y-2) = 3x 2 divide 1 & 2 equations by solving we get x = 40. |
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| 14. |
The Average Weight Of A, B And C Is 45 Kg. If The Average Weight Of A And B Be 40 Kg And That Of B And C Be 43 Kg, Then The Weight Of B Is? |
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Answer» Let A, B, C represent their RESPECTIVE weights. Then, we have: A + B + C = (45 * 3) = 135 --- (i) A + B = (40 * 2) = 80 --- (ii) B + C = (43 * 2) = 86 --- (iii) ADDING (ii) and (iii), we get: A + 2B + C = 166 --- (iv) Subtracting (i) from (iv), we get: B = 31 B's weight = 31 kg. Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 * 3) = 135 --- (i) A + B = (40 * 2) = 80 --- (ii) B + C = (43 * 2) = 86 --- (iii) Adding (ii) and (iii), we get: A + 2B + C = 166 --- (iv) Subtracting (i) from (iv), we get: B = 31 B's weight = 31 kg. |
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| 15. |
The Average Monthly Income Of P And Q Is Rs. 5050. The Average Monthly Income Of Q And R Is 6250 And The Average Monthly Income Of P And R Is Rs. 5200. The Monthly Income Of P Is? |
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Answer» Let P, Q and R represent their respective monthly INCOMES. Then, we have: P + Q = (5050 * 2) = 10100 --- (i) Q + R = (6250 * 2) = 12500 --- (ii) P + R = (5200 * 2) = 10400 --- (iii) ADDING (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 = P + Q + R = 16500 --- (iv) Subtracting (ii) from (iv), we get, P = 4000. P's monthly income = RS. 4000. Let P, Q and R represent their respective monthly incomes. Then, we have: P + Q = (5050 * 2) = 10100 --- (i) Q + R = (6250 * 2) = 12500 --- (ii) P + R = (5200 * 2) = 10400 --- (iii) Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 = P + Q + R = 16500 --- (iv) Subtracting (ii) from (iv), we get, P = 4000. P's monthly income = Rs. 4000. |
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| 16. |
A Cricketer Whose Bowling Average Is 12.4 Runs Per Wicket Takes 5 Wickets For 26 Runs And There By Decreases His Average By 0.4. The Number Age Of The Family Now Is? |
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Answer» Let the number of wickets taken TILL the last MATCH be x. Then, (12.4x + 26)/(x + 5) = 12 = 12.4x + 26 = 12x + 60 = 0.4x = 34 = x = 340/4 = 85. Let the number of wickets taken till the last match be x. Then, (12.4x + 26)/(x + 5) = 12 = 12.4x + 26 = 12x + 60 = 0.4x = 34 = x = 340/4 = 85. |
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| 17. |
A Cricketer Has A Certain Average For 10 Innings. In The Eleventh Inning, He Scored 108 Runs, There By Increasing His Average By 6 Runs. His New Average Is? |
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Answer» Let average for 10 INNINGS be x. Then, = 11x + 66 = 10x + 108 = x = 42. New average = (x + 6) = 48 runs. Let average for 10 innings be x. Then, (10x + 108)/11 = x + 6 = 11x + 66 = 10x + 108 = x = 42. New average = (x + 6) = 48 runs. |
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| 18. |
The Captain Of A Cricket Team Of 11 Members Is 26 Years Old And The Wicket Keeper Is 3 Years Older. If The Ages Of These Two Are Excluded, The Average Age Of The Remaining Players Is One Year Less Than The Average Age Of The Whole Team. What Is The Average Of The Team? |
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Answer» Let the AVERAGE of the whole TEAM be x YEARS. 11x - (26 + 29) = 9(x - 1) = 11x - 9X = 46 = 2x = 46 => x = 23 So, average age of the team is 23 years. Let the average of the whole team be x years. 11x - (26 + 29) = 9(x - 1) = 11x - 9x = 46 = 2x = 46 => x = 23 So, average age of the team is 23 years. |
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| 19. |
Kruti Took A Loan At Simple Interest At 6 % In The First Year With An Increase Of 0.5 % In Each Subsequent Year. She Paid Rs. 3375 As Interest After 4 Years. How Much Loan Did She Take? |
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Answer» LET the loan taken be Rs. X , then X × 6/100 × 1 + X × 6.5/100 × 1 + X × 7/100 × 1 + X × 7.5/100 × 1 = 3375 => (6+ 6.5 +7 +7.5) × X/100 = 3375 => X = (3375 × 100/27) = 12500. Let the loan taken be Rs. X , then X × 6/100 × 1 + X × 6.5/100 × 1 + X × 7/100 × 1 + X × 7.5/100 × 1 = 3375 => (6+ 6.5 +7 +7.5) × X/100 = 3375 => X = (3375 × 100/27) = 12500. |
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| 20. |
Rs.6000 Becomes Rs.7200 In 4 Years At A Certain Rate Of Interest. If The Rate Becomes 1.5 Times Of Itself. The Amount Of The Same Principle In 5 Years Will Be? |
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Answer» S.I on Rs. 600 for 4 YEARS = Rs (7200 - 600) = Rs 1200 Therefore RATE = (120000/24000) % p.a = 5 % p.a New Rate = (5 x 3/2) % = 15/2 % p.a Required Amount = [6000 + (6000 x 5/100 x 15/2)] = Rs(6000 + 2250) = Rs 8250. S.I on Rs. 600 for 4 Years = Rs (7200 - 600) = Rs 1200 Therefore Rate = (120000/24000) % p.a = 5 % p.a New Rate = (5 x 3/2) % = 15/2 % p.a Required Amount = [6000 + (6000 x 5/100 x 15/2)] = Rs(6000 + 2250) = Rs 8250. |
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| 21. |
A Borrows Rs.800 At The Rate Of 12 % Per Annum. Simple Interest And B Borrows Rs.910 At The Rate Of 10 % Per Annum Simple Interest. In How Many Years Will Their Amounts At Debts Be Equal? |
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Answer» Let the REQUIRED TIME be x YEARS. Then 800 + 800 x 12/100 × x = 910 + 910 x 10/100 × x => (96X -91 x) = 110 => 5x =110 => x =22. Let the required time be x years. Then 800 + 800 x 12/100 × x = 910 + 910 x 10/100 × x => (96x -91 x) = 110 => 5x =110 => x =22. |
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| 22. |
Simple Interest On Rs.500 For 4 Years At 6.25 % Per Annum Is Equal To The Simple Interest On Rs. 400 At 5 % Per Annum For A Certain Period Of Time. The Period Of Time Is? |
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Answer» LET the required period of time be x YEARS. Then 500 x 4 x 6.25/100 = 400 x 5/100 × x => 20 x 6.25 = 20 × x => X = 6.25 = 625/100 = 25/4 = 6 ¼ years. Let the required period of time be x years. Then 500 x 4 x 6.25/100 = 400 x 5/100 × x => 20 x 6.25 = 20 × x => X = 6.25 = 625/100 = 25/4 = 6 ¼ years. |
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| 23. |
A Person Invests Money In Three Different Schemes For 6 Years, 10 Years And 12 Years At 10 %, 12 %, And 15 %. Simple Interest Respectively. At The Completion Of Each Scheme, He Gets The Same Interest. The Ratio Of His Investments Is? |
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Answer» LET the REQUIRED ratio be x: 1: y. Then S.I on Rs x for 6 YEARS at 10% p.a = S.I on re.1 10 years at 12 %p.a X × 10/100 × 6 = 1 × 12/100 × 10 => X = 120/60 =2 S.I on Re.1 for 10 years at 12 % p.a = S. I on Rs. Y for 12 years at 15 % p.a Therefore, (1 x 12/100 x 10) = (y x 15/100 x 12) => y = 120/180 = 2/3 Required ratio = 2 : 1 = 2/3 = 6 : 3 : 2. Let the required ratio be x: 1: y. Then S.I on Rs x for 6 years at 10% p.a = S.I on re.1 10 years at 12 %p.a X × 10/100 × 6 = 1 × 12/100 × 10 => X = 120/60 =2 S.I on Re.1 for 10 years at 12 % p.a = S. I on Rs. Y for 12 years at 15 % p.a Therefore, (1 x 12/100 x 10) = (y x 15/100 x 12) => y = 120/180 = 2/3 Required ratio = 2 : 1 = 2/3 = 6 : 3 : 2. |
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| 24. |
Seats For Mathematics, Physics And Biology In A School Are In The Ratio 5:7:8. There Is A Proposal To Increase These Seats By 40%, 50% And 75% Respectively. What Will Be The Ratio Of Increased Seats? |
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Answer» Originally, let the number of seats for Mathematics, Physics and Biology be 5X, 7x and 8x RESPECTIVELY. Number of increased sears are (140% of 5x), (150% of 7x) and (175% of 8x) i.e., (140/100 * 5x), (150/100 * 7x) and (175/100 * 8x) Required ratio = 7x:21x/2:14x = 14x : 21x : 28x = 2:3:4. Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively. Number of increased sears are (140% of 5x), (150% of 7x) and (175% of 8x) i.e., (140/100 * 5x), (150/100 * 7x) and (175/100 * 8x) i.e., 7x, 21x/2 and 14x Required ratio = 7x:21x/2:14x = 14x : 21x : 28x = 2:3:4. |
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| 25. |
In Covering Distance, The Speed Of A & B Are In The Ratio Of 3:4.a Takes 30min More Than B To Reach The Destination. The Time Taken By A To Reach The Destination Is? |
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Answer» Ratio of speed = 3:4 Ratio of time = 4:3 let A takes 4X hrs, B takes 3x hrs then 4x-3x = 30/60 hr X = ½ hr Time taken by A to REACH the destination is 4x = 4 * ½ = 2 hr. Ratio of speed = 3:4 Ratio of time = 4:3 let A takes 4x hrs, B takes 3x hrs then 4x-3x = 30/60 hr x = ½ hr Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr. |
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| 26. |
If Rs. 510 Be Divided Among A, B, C In Such A Way That A Gets 2/3 Of What B Gets And B Gets 1/4 Of What C Gets, Then Their Shares Are Respectively? |
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Answer» (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4 A:B = 2:3 and B:C = 1:4 = 3:12 A:B:C = 2:3:12 A;s share = 510 * 2/17 = Rs. 60 B's share = 510 * 3/17 = Rs. 90 C's share = 510 * 12/17 = Rs. 360. (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4 A:B = 2:3 and B:C = 1:4 = 3:12 A:B:C = 2:3:12 A;s share = 510 * 2/17 = Rs. 60 B's share = 510 * 3/17 = Rs. 90 C's share = 510 * 12/17 = Rs. 360. |
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| 27. |
Salaries Of Ravi And Sumit Are In The Ratio 2:3. If The Salary Of Each Is Increased By Rs. 4000, The New Ratio Becomes 40:57. What Is Sumit's Present Salary? |
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Answer» Let the original salaries of Ravi and SUMIT be RS. 2X and Rs. 3x RESPECTIVELY. Then, (2x + 4000)/(3x + 4000) = 40/57 6x = 68000 => 3x = 34000 Sumit's present SALARY = (3x + 4000) = 34000 + 4000 = Rs. 38,000. Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then, (2x + 4000)/(3x + 4000) = 40/57 6x = 68000 => 3x = 34000 Sumit's present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000. |
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| 28. |
A Sum Of Money Is To Be Distributed Among A, B, C, D In The Proportion Of 5:2:4:3. If C Gets Rs. 1000 More Than D, What Is B's Share? |
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Answer» Let the shares of A, B, C and D be 5X, 2X, 4X and 3x Rs. respectively. Then, 4x - 3x = 1000 => x = 1000. B's share = Rs. 2x = 2 * 1000 = Rs. 2000. Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively. Then, 4x - 3x = 1000 => x = 1000. B's share = Rs. 2x = 2 * 1000 = Rs. 2000. |
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| 29. |
Megha Purchased 120 Calendars At A Rate Of Rs 3 Each And Sold 1/3 Of Them At The Rate Of Rs. 4 Each. 1/2 Of Them At The Rate Of Rs 5 Each And Rest At The Cost Price. Her Profit Per Cent Was? |
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Answer» CP of 120 calendars = 120 x 3 = Rs. 360 SP of 40 at Rs. 4 each = Rs. 40 x 4 = Rs.160 SP of 60 at Rs 5 each = Rs 60 x 5 = Rs. 300 SP of remaining 20 calendars = Rs 20 x 3 = 60 TOTAL SP = Rs.( 160 + 300 + 60) = 520 Profit % = 520 – 360 = 160 Profit % = 160/360 x 100 = 400/9. CP of 120 calendars = 120 x 3 = Rs. 360 SP of 40 at Rs. 4 each = Rs. 40 x 4 = Rs.160 SP of 60 at Rs 5 each = Rs 60 x 5 = Rs. 300 SP of remaining 20 calendars = Rs 20 x 3 = 60 Total SP = Rs.( 160 + 300 + 60) = 520 Profit % = 520 – 360 = 160 Profit % = 160/360 x 100 = 400/9. |
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| 30. |
X Varies Inversely As Square Of Y. Given That Y = 2 For X = 1. The Value Of X For Y = 6 Will Be Equal To? |
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Answer» Given X = k/y2, where k is a constant. Now, y = 2 and x = 1 gives k = 4. x = 4/y2 => x = 4/62, when y = 6 => x = 4/36 = 1/9. Given x = k/y2, where k is a constant. Now, y = 2 and x = 1 gives k = 4. x = 4/y2 => x = 4/62, when y = 6 => x = 4/36 = 1/9. |
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| 31. |
A Can Do A Price Of Work In 30days While B Can Do It In 40 Days. In How Many Days Can A And B Working Together Do It? |
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Answer» (A +B)’s 1 day’s WORK = (1/30 + 1/40) = 7/120 Time is TAKEN by both to finish the work = 120/7 DAYS = 17 1/7 days. (A +B)’s 1 day’s work = (1/30 + 1/40) = 7/120 Time is taken by both to finish the work = 120/7 days = 17 1/7 days. |
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| 32. |
A Can Finish A Work In 18 Days B Can Do The Same Work In 15 Days. B Worked For 10 Days And Left The Job. In How Many Days, A Alone Can Finish The Remaining Work? |
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Answer» B's 10 DAY's work = 1/15 * 10 = 2/3 REMAINING work = (1 - 2/3) = 1/3 Now, 1/18 work is done by A in 1 day. 1/3 work is done by A in (18 * 1/3) = 6 days. B's 10 day's work = 1/15 * 10 = 2/3 Remaining work = (1 - 2/3) = 1/3 Now, 1/18 work is done by A in 1 day. 1/3 work is done by A in (18 * 1/3) = 6 days. |
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| 33. |
A Is 30% More Efficient Than B. How Much Time Will They, Working Together, Take To Complete A Job Which A Alone Could Have Done In 23 Days? |
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Answer» Ratio of times taken by A and B = 100:130 = 10:13 Suppose B takes X days to do the WORK. x = (23 * 13)/10 = 299/10 A's 1 day work = 1/23; B's 1 day work = 10/299 (A + B)'s 1 day work = (1/23 + 10/299) = 1/13 A and B together can COMPLETE the job in 13 days. Ratio of times taken by A and B = 100:130 = 10:13 Suppose B takes x days to do the work. x = (23 * 13)/10 = 299/10 A's 1 day work = 1/23; B's 1 day work = 10/299 (A + B)'s 1 day work = (1/23 + 10/299) = 1/13 A and B together can complete the job in 13 days. |
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| 34. |
Twelve Men Can Complete A Work In 8 Days. Three Days After They Started The Work, 3 More Men Joined Them. In How Many Days Will All Of Them Together Complete Remaining Work? |
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Answer» 1 man’s ONE DAY’s work = 1/96 12 men’s 3 day’s work = (3 x 1/8) = 3/8 REMAINING work = (1 – 3/8) = 5/8 15 men’s 1 day’s work = 15/96 Now 15/96 work is done by them in 1day Therefore 5/8 work will be done by them in (96/15 x 5/8) i.e., 4 days. 1 man’s one day’s work = 1/96 12 men’s 3 day’s work = (3 x 1/8) = 3/8 Remaining work = (1 – 3/8) = 5/8 15 men’s 1 day’s work = 15/96 Now 15/96 work is done by them in 1day Therefore 5/8 work will be done by them in (96/15 x 5/8) i.e., 4 days. |
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| 35. |
Mahesh And Umesh Can Complete A Work In 10 Days And 15 Days Respectively. Umesh Starts The Work And After 5 Days Mahesh Also Joins Him In All The Work Would Be Completed In? |
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Answer» Umesh’s 5 DAY’s work = 5 x 1/15 = 1/3 REMAINING work = (1 – 1/3) = 2/3 (1/10 +1/ 15) work is done by both in 1 day Therefore 2/3 work is done by both in (6 x 2/3) = 4days. The work was completed in 9 days. Umesh’s 5 day’s work = 5 x 1/15 = 1/3 Remaining work = (1 – 1/3) = 2/3 (1/10 +1/ 15) work is done by both in 1 day Therefore 2/3 work is done by both in (6 x 2/3) = 4days. The work was completed in 9 days. |
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| 36. |
A Can Do A Piece Of Work In 80 Days. He Works At It For 10 Days And Then B Alone Finishes The Work In 42 Days. The Two Together Could Complete The Work In? |
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Answer» A’s 10 DAYS WORK = (10 x 1/80) = 1/8 Remaining work = (1 -1/8) = 7/8 Therefore 7/8 work is done by A in 42 days. Whole work will be done by A in (42 x 8/7)i.e.., 48 days Therefore, (A+ B)’s 1 DAY work = (1/80 + 1/48) = 8/240 = 1/30. A and B together can finish it in 30days. A’s 10 days work = (10 x 1/80) = 1/8 Remaining work = (1 -1/8) = 7/8 Therefore 7/8 work is done by A in 42 days. Whole work will be done by A in (42 x 8/7)i.e.., 48 days Therefore, (A+ B)’s 1 day work = (1/80 + 1/48) = 8/240 = 1/30. A and B together can finish it in 30days. |
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| 37. |
A, B And C Contract A Work For Rs. 550. Together A And B Are To Do 7/11 Of The Work. The Share Of C Should Be? |
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Answer» Work to be done by C = (1 – 7/11) = 4/11 Therefore, (A +B) = C = 7/11 : 4/11 = 7 : 4 Therefore C’s share = Rs.(550 X 4/11) = Rs. 200. Work to be done by C = (1 – 7/11) = 4/11 Therefore, (A +B) = C = 7/11 : 4/11 = 7 : 4 Therefore C’s share = Rs.(550 x 4/11) = Rs. 200. |
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| 38. |
A Can Do (1/3) Of A Work In 5 Days And B Can Do (2/5) Of The Work In 10 Days. In How Many Days Both A And B Together Can Do The Work? |
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Answer» 1/3 of work is DONE by A in 5 days Therefore, whole work will be done by A in 15 days 2/5 of work is done by B in 10 days Whole work will be done by B in (10 x 5/2) i.e.., 25 days Therefore (A + B)’s 1 DAY’s work = (1/15 + 1/25) = 8/75 So both together can finish 75/8 days i.e.., 9 3/8 days. 1/3 of work is done by A in 5 days Therefore, whole work will be done by A in 15 days 2/5 of work is done by B in 10 days Whole work will be done by B in (10 x 5/2) i.e.., 25 days Therefore (A + B)’s 1 day’s work = (1/15 + 1/25) = 8/75 So both together can finish 75/8 days i.e.., 9 3/8 days. |
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| 39. |
Two Trains Approach Each Other At 30 Km/hr And 27km/hr From Two Places 342km Apart After How Many Hours Will They Meet? |
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Answer» Suppose the two TRAINS meet after x HOURS, then => 30x + 27X = 342 => 57X = 342 => X = 6 So the two trains will meet after 6 hours. Suppose the two trains meet after x hours, then => 30x + 27x = 342 => 57x = 342 => X = 6 So the two trains will meet after 6 hours. |
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| 40. |
The Ratio Between The Speed Of Two Trains Is 7: 8.if The Second Train Runs 400km In 5 Hours. The Speed Of The First Train Is? |
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Answer» Let the SPEED of the first train be 7x km/hr Then the speed of the SECOND train is 8x km/hr But speed of the second train = 400/5 km/hr = 80km/hr => X=10 Hence the speed of first train is (7 x 10)km/hr = 70km/hr. Let the speed of the first train be 7x km/hr Then the speed of the second train is 8x km/hr But speed of the second train = 400/5 km/hr = 80km/hr => 8x =80 => X=10 Hence the speed of first train is (7 x 10)km/hr = 70km/hr. |
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| 41. |
The Speeds Of A And B Are In The Ratio 3: 4. A Takes 20 Minutes More Than B To Reach A Destination. In What Time Does A Reach The Destination?let The Time Taken By A Be X Hours? |
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Answer» The time TAKEN by B =(X-20/60)hrs = (x-1/3)hrs Ratio of speeds = INVERSE ratio of time taken Therefore 3: 4 =(x-1/3) : x =>3x-1/3x=3/4 => 12X -4 =9x => 3x =4 => X=4/3 hrs. X=1 1/3 hours. Required time = 1 1/3 hrs. The time taken by B =(x-20/60)hrs = (x-1/3)hrs Ratio of speeds = Inverse ratio of time taken Therefore 3: 4 =(x-1/3) : x =>3x-1/3x=3/4 => 12x -4 =9x => 3x =4 => X=4/3 hrs. X=1 1/3 hours. Required time = 1 1/3 hrs. |
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| 42. |
A Certain Distance Covered At A Certain Speed. If Half Of This Distance Is Covered In Double The Time The Ratio Of The Two Speeds Is? |
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Answer» Let x KMS be covered in y HOURS, then first speed = x / y km/hr Again x/2 km is covered in 2y hrs Therefore, new speed = (x/2 x 1/2y)km/hr = (x/ 4Y) km/hr Ratio of speeds = x/y : x/4y = 1 : ¼ = 4 : 1. Let x kms be covered in y hours, then first speed = x / y km/hr Again x/2 km is covered in 2y hrs Therefore, new speed = (x/2 x 1/2y)km/hr = (x/ 4y) km/hr Ratio of speeds = x/y : x/4y = 1 : ¼ = 4 : 1. |
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| 43. |
R And S Start Walking Towards Each Other At 10 Am At Speeds Of 3km/hr And 4km/hr Respectively. They Were Initially 17.5km Apart At What Time Do They Meet? |
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Answer» Suppose they meet after x hours , then 3x +4X = 17.5 => 7X = 17.5 => X = 2.5hours So they meet at 12: 30 p.m. Suppose they meet after x hours , then 3x +4x = 17.5 => 7x = 17.5 => X = 2.5hours So they meet at 12: 30 p.m. |
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| 44. |
Bombay Express Left Delhi For Bombay At 14.30 Hours. Travelling At A Speed Of 60kmph And Rajadhani Express Left Delhi For Bombay On The Same Day At 16.30 Hours Travelling Speed Of 80kmph. How Far Away From Delhi Will The Two Trains Meet? |
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Answer» LET the TRAIN meet at a distance of X km from Delhi. Then x/60 –x/80 = 2 => 4x -3x => x =480 Therefore, Required distance = 480km. Let the train meet at a distance of x km from Delhi. Then x/60 –x/80 = 2 => 4x -3x => x =480 Therefore, Required distance = 480km. |
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