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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. |
A Train Crosses A Post In 1515 Seconds And A Platform 100100 Metre Long In 2525 Seconds. Its Length Is? |
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Answer» The TRAIN can cover a distance equal to its length in 1515 SECONDS. The train can cover (a distance equal to its length + 100100 METRE) in 2525 seconds. Therefore, it can be concluded that the train can travel 100100 metre in (25−15)=10 seconds and therefore its SPEED is 100/10=10 m/s Hence, length of the train =10×15=150 metre. The train can cover a distance equal to its length in 1515 seconds. The train can cover (a distance equal to its length + 100100 metre) in 2525 seconds. Therefore, it can be concluded that the train can travel 100100 metre in (25−15)=10 seconds and therefore its speed is 100/10=10 m/s Hence, length of the train =10×15=150 metre. |
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| 2. |
Two Stations P And Q Are 110110 Km Apart On A Straight Track. One Train Starts From P At 77a.m. And Travels Towards Q At 2020 Kmph. Another Train Starts From Q At 88 A.m. And Travels Towards P At A Speed Of 2525 Kmph. At What Time Will They Meet? |
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Answer» Assume both trains meet xx HOURS after 77 a.m. DISTANCE covered by train starting from P in xx hours =20X km=20x km Distance covered by train starting from Q in (x−1)(x−1) hours =25(x−1) km=25(x−1) km TOTAL distance =110 km=110 km ⇒20x+25(x−1)=110 ⇒45x=135 ⇒x=3 Hence, they meet 33 hours after 77 a.m. i.e., they meet at 10 a.m. Assume both trains meet xx hours after 77 a.m. Distance covered by train starting from P in xx hours =20x km=20x km Distance covered by train starting from Q in (x−1)(x−1) hours =25(x−1) km=25(x−1) km Total distance =110 km=110 km ⇒20x+25(x−1)=110 ⇒45x=135 ⇒x=3 Hence, they meet 33 hours after 77 a.m. i.e., they meet at 10 a.m. |
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| 3. |
P Works Twice As Fast As Q. If Q Alone Can Complete A Work In 12 Days, P And Q Can Finish The Work In --- Days? |
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Answer» Work done by Q in 1 DAY = 1/12 Work done by P in 1 day = 2 × (1/12) = 1/6 Work done by P and Q in 1 day = 1/12 + 1/6 =1/4 => P and Q can FINISH the work in 4 days. Work done by Q in 1 day = 1/12 Work done by P in 1 day = 2 × (1/12) = 1/6 Work done by P and Q in 1 day = 1/12 + 1/6 =1/4 => P and Q can finish the work in 4 days. |
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| 4. |
P Takes Twice As Much Time As Q Or Thrice As Much Time As R To Finish A Piece Of Work. They Can Finish The Work In 2 Days If Work Together. How Much Time Will Q Take To Do The Work Alone? |
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Answer» Let P TAKES X days to complete the WORK Then Q takes x/2 days and R takes x/3 days to finish the work Amount of work P does in 1 day = 1/x Amount of work Q does in 1 day = 2/x Amount of work R does in 1 day = 3/x Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x 6/x = 2 => x = 12 => Q takes 12/2 days = 6 days to complete the work. Let P takes x days to complete the work Then Q takes x/2 days and R takes x/3 days to finish the work Amount of work P does in 1 day = 1/x Amount of work Q does in 1 day = 2/x Amount of work R does in 1 day = 3/x Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x 6/x = 2 => x = 12 => Q takes 12/2 days = 6 days to complete the work. |
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| 5. |
A Train Travelled At An Average Speed Of 100100 Km/hr, Stopping For 33 Minutes After Every 7575 Km. How Long Did It Take To Reach Its Destination 600600 Km From The Starting Point? |
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Answer» TIME needed to travel 600 km =600/100=6 hour Now we need to find out the number of stops in the 600600 km JOURNEY. GIVEN that the train stops after every 7575 km. 600/75=8 It means, the train stops 7 times before 600 km and 11 time just after 600 km. Hence we need to take only 7 stops into consideration for the 600 km journey. Hence, TOTAL stopping time in the 600 km journey =7×3=21minutes Total time needed to reach the destination =6 hours +21 minutes =6 hours 21 minutes. Time needed to travel 600 km =600/100=6 hour Now we need to find out the number of stops in the 600600 km journey. Given that the train stops after every 7575 km. 600/75=8 It means, the train stops 7 times before 600 km and 11 time just after 600 km. Hence we need to take only 7 stops into consideration for the 600 km journey. Hence, total stopping time in the 600 km journey =7×3=21minutes Total time needed to reach the destination =6 hours +21 minutes =6 hours 21 minutes. |
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| 6. |
The Distance Between Two Cities A And B Is 330330 Km. A Train Starts From A At 88 A.m. And Travels Towards B At 6060 Km/hr. Another Train Starts From B At 99 A.m. And Travels Towards A At 7575 Km/hr. At What Time Will They Meet? |
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Answer» Assume that they MEET xx hours after 88 a.m. Then, train 1,1, starting from A, travels xx hours TILL the trains meet. DISTANCE travelled by train 11 in xx hours =60x km=60x km Train 2,2, starting from B, travels (x−1)(x−1) hours till the trains meet. Distance travelled by train 22 in (x−1)(x−1) hours =75(x−1) km=75(x−1) km Total distance travelled = Distance travelled by train 11 + Distance travelled by train 22 ⇒330=60x+75(x−1) ⇒12x+15(x−1)=66 ⇒12x+15x−15=66 ⇒27x=66+15=81 ⇒3x=9 ⇒x=3. Assume that they meet xx hours after 88 a.m. Then, train 1,1, starting from A, travels xx hours till the trains meet. Distance travelled by train 11 in xx hours =60x km=60x km Train 2,2, starting from B, travels (x−1)(x−1) hours till the trains meet. Distance travelled by train 22 in (x−1)(x−1) hours =75(x−1) km=75(x−1) km Total distance travelled = Distance travelled by train 11 + Distance travelled by train 22 ⇒330=60x+75(x−1) ⇒12x+15(x−1)=66 ⇒12x+15x−15=66 ⇒27x=66+15=81 ⇒3x=9 ⇒x=3. |
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| 7. |
Arun Invested Rs. 333000 In 5½ % Stocks At 110 .if Brokerage Is Rs.1, What Is His Annual Income From His Investment? |
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Answer» Investment = Rs.333000 Since face value is not GIVEN, we can take it as Rs.100 and dividend per SHARE = Rs.11/2 MARKET Value = 110 + 1 = 111 Number of shares purchased = 333000/111 = 3000 TOTAL income = 3000×11/2 = Rs.16500. Investment = Rs.333000 Since face value is not given, we can take it as Rs.100 and dividend per share = Rs.11/2 Market Value = 110 + 1 = 111 Number of shares purchased = 333000/111 = 3000 Total income = 3000×11/2 = Rs.16500. |
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| 8. |
A Company Declared A Semi-annual Dividend Of 12%. Find The Annual Dividend Of Sam Owing 2000 Shares Of The Company Having A Par Value Of Rs. 10 Each? |
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Answer» semi-ANNUAL DIVIDEND = 10×12/100 = Rs.1.2 TOTAL semi-annual dividend = 2000 × 1.2 = Rs.2400 Total annual dividend = 2 × Rs.2400 = Rs.4800. semi-annual dividend = 10×12/100 = Rs.1.2 Total semi-annual dividend = 2000 × 1.2 = Rs.2400 Total annual dividend = 2 × Rs.2400 = Rs.4800. |
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| 9. |
A Man Sells 4000 Common Shares Of A Company X (each Of Par Value Rs. 10), Which Pays A Dividend Of 40% At Rs. 30 Per Share. He Invests The Sale Proceeds In Ordinary Shares Of Company Y (each Of Par Value Rs. 25) That Pays A Dividend Of 15%. If The Market Value Of Company Y Is Rs. 15, Find The Number Of Shares Of Company Y Purchased By The Man? |
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Answer» MARKET Value of COMPANY X (his selling price) = Rs.30 Total shares sold = 4000 Amount he gets = Rs.(4000 × 30) He invests this amount in ordinary shares of Company Y Market Value of Company Y(His purchasing price) = 15 Number of shares of company Y which he purchases = 4000×30/15=8000. market Value of Company X (his selling price) = Rs.30 Total shares sold = 4000 Amount he gets = Rs.(4000 × 30) He invests this amount in ordinary shares of Company Y Market Value of Company Y(His purchasing price) = 15 Number of shares of company Y which he purchases = 4000×30/15=8000. |
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| 10. |
To Produce An Annual Income Of Rs. 800 From A 8% Stock At 90, The Amount Of Stock Needed Is? |
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Answer» SINCE face value is not GIVEN, TAKE it as Rs.100. As it is an 8% stock, income (dividend) per stock = Rs.8 IE, For an income of Rs.8,amount of stock NEEDED = Rs.100 For an income of Rs.800, amount of stock needed = 100×800/8=10000. Since face value is not given, take it as Rs.100. As it is an 8% stock, income (dividend) per stock = Rs.8 ie, For an income of Rs.8,amount of stock needed = Rs.100 For an income of Rs.800, amount of stock needed = 100×800/8=10000. |
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| 11. |
John Buys 100 Shares Of Par Value Rs. 5 Each, Of A Company, Which Pays An Annual Dividend Of 12% At Such A Price That He Gets 10% On His Investment. Find The Market Value Of A Share? |
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Answer» Face value of each SHARE = Rs.5 Total dividend RECEIVED by John = 100×5×12/100 = Rs.60 Let market value of 100 SHARES = Rs.x x×10/100=60 x = 600 ie, Market value of 100 shares = Rs.600 HENCE, Market value of each share = Rs.6 Face value of each share = Rs.5 Total dividend received by John = 100×5×12/100 = Rs.60 Let market value of 100 shares = Rs.x x×10/100=60 x = 600 ie, Market value of 100 shares = Rs.600 Hence, Market value of each share = Rs.6 |
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| 12. |
A Room Has Equal Number Of Men And Women. Eight Women Left The Room, Leaving Twice As Many Men As Women In The Room. What Was The Total Number Of Men And Women Present In The Room Initially? |
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Answer» Let INITIAL number of men = initial number of WOMEN =X 2(x−8)=x ⇒2x−16=x ⇒x=16 Total number of men and women =2x=2×16=32. Let initial number of men = initial number of women =x 2(x−8)=x ⇒2x−16=x ⇒x=16 Total number of men and women =2x=2×16=32. |
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| 13. |
In A Group Of Ducks And Cows, The Total Number Of Legs Is 2828 More Than Twice The Number Of Heads. Find The Total Number Of Cows? |
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Answer»
Number of COWS =c=c Then, total number of legs =2d+4c=2d+4c Total number of heads =d+c=d+c GIVEN that total number of legs is 2828 more than twice the number of heads ⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14 i.e., total number of cows =14. Let number of ducks =d=d Number of cows =c=c Then, total number of legs =2d+4c=2d+4c Total number of heads =d+c=d+c Given that total number of legs is 2828 more than twice the number of heads ⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14 i.e., total number of cows =14. |
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| 14. |
Andrews Earns An Interest Of Rs. 1596 For The Third Year And Rs. 1400 For The Second Year On The Same Sum. Find The Rate Of Interest If It Is Lent At Compound Interest? |
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Answer» Interest earned in 3rd year = Rs. 1596 Interest earned in 2ND year = Rs. 1400 i.e, in 3rd year, Andrews gets ADDITIONAL interest of (Rs. 1596 - Rs. 1400) = Rs.196 This MEANS, Rs.196 is the interest obtained for Rs.1400 for 1 year R=100×SI/PT=100×196/1400×1=196/14=14%. Interest earned in 3rd year = Rs. 1596 Interest earned in 2nd year = Rs. 1400 i.e, in 3rd year, Andrews gets additional interest of (Rs. 1596 - Rs. 1400) = Rs.196 This means, Rs.196 is the interest obtained for Rs.1400 for 1 year R=100×SI/PT=100×196/1400×1=196/14=14%. |
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| 15. |
Find The Odd Man Out. 3576, 1784, 888, 440, 216, 105, 48? |
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Answer» 3576 (3576-8)/2 = 1784 (1784-8)/2 = 888 (888-8)/2 = 440 (440-8)/2 = 216 (216-8)/2 = 104 (104-8)/2 = 48 Hence, 105 is wrong. 104 should have COME in place of 105. 3576 (3576-8)/2 = 1784 (1784-8)/2 = 888 (888-8)/2 = 440 (440-8)/2 = 216 (216-8)/2 = 104 (104-8)/2 = 48 Hence, 105 is wrong. 104 should have come in place of 105. |
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| 16. |
In One Km Race A Beats B By 4 Seconds Or 40 Metres. How Long Does B Take To Run The Kilometer? |
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Answer» This means, B takes 4 SECONDS to RUN 40 metres => B takes 4/40=1/10 seconds to run 1 metre => B takes 1/10×1000=100 seconds to run 1000 metre. This means, B takes 4 seconds to run 40 metres => B takes 4/40=1/10 seconds to run 1 metre => B takes 1/10×1000=100 seconds to run 1000 metre. |
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| 17. |
A Can Run 220 Metres In 41 Seconds And B In 44 Seconds. By How Many Seconds Will B Win If He Has 30 Metres Start? |
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Answer» B has a start of 30 metre => A has to run 220 metre and B has to run (220-30)=190 metre Given that A TAKES 41 SECONDS to cover this 220 metre B takes 44 seconds to cover 220 metre => B takes 44220 seconds to cover 1 metre => B takes 44/220×190=420×190 = 38 seconds to cover 190 metre => B takes 44/220 seconds to cover 1 metre => B takes 44/220×190=4/20×190 = 38 seconds to cover 190 metre i.e., A beats B by (41-38)= 3 seconds. B has a start of 30 metre => A has to run 220 metre and B has to run (220-30)=190 metre Given that A takes 41 seconds to cover this 220 metre B takes 44 seconds to cover 220 metre => B takes 44220 seconds to cover 1 metre => B takes 44/220×190=420×190 = 38 seconds to cover 190 metre => B takes 44/220 seconds to cover 1 metre => B takes 44/220×190=4/20×190 = 38 seconds to cover 190 metre i.e., A beats B by (41-38)= 3 seconds. |
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| 18. |
If Selling Price Of An Article Is Rs. 250,rs. 250, Profit Percentage Is 25%.25%. Find The Ratio Of The Cost Price And The Selling Price? |
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Answer» SELLING price =250 profit =25% cost price =250×100125=200 Required RATIO =200:250=4:5. selling price =250 profit =25% cost price =250×100125=200 Required ratio =200:250=4:5. |
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| 19. |
Arun Bought A Computer With 15%15% Discount On The Labelled Price. He Sold The Computer For Rs. 2880rs. 2880 With 20%20% Profit On The Labelled Price. At What Price Did He Buy The Computer? |
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Answer» selling price =2880 LABELLED price =2880×100/120=2400 price at which he BOUGHT the COMPUTER =2400−2400×15/100=2040. selling price =2880 labelled price =2880×100/120=2400 price at which he bought the computer =2400−2400×15/100=2040. |
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| 20. |
Two Dice Are Rolled Together. What Is The Probability Of Getting Two Numbers Whose Product Is Even? |
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Answer» Total number of outcomes possible when a die is ROLLED = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Let E = the event of getting two numbers WHOSE product is even = {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Hence, n(E) = 27 P(E) = n(E)/n(S)=27/36=3/4. Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Let E = the event of getting two numbers whose product is even = {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Hence, n(E) = 27 P(E) = n(E)/n(S)=27/36=3/4. |
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| 21. |
One Ball Is Picked Up Randomly From A Bag Containing 8 Yellow, 7 Blue And 6 Black Balls. What Is The Probability That It Is Neither Yellow Nor Black? |
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Answer» TOTAL number of balls, n(S) = 8 + 7 + 6 = 21 n(E) = Number of ways in which a ball can be SELECTED which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black) P(E) = n(E)/n(S)=7/21=1/3. Total number of balls, n(S) = 8 + 7 + 6 = 21 n(E) = Number of ways in which a ball can be selected which is neither yellow nor black = 7 (∵ there are only 7 balls which are neither yellow nor black) P(E) = n(E)/n(S)=7/21=1/3. |
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| 22. |
Two Pipes A And B Can Fill A Tank In 2 And 6 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long Will It Take To Fill The Tank? |
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Answer» Part filled by FIRST pipe in 11 minute =1/2 Part filled by second pipe in 11 minute =1/6 Net part filled by pipe A and B in 11 minute =1/2+1/6=2/3 i.e, pipe A and B TOGETHER can FILL the TANK in 3/2 minutes =1.5minutes. Part filled by first pipe in 11 minute =1/2 Part filled by second pipe in 11 minute =1/6 Net part filled by pipe A and B in 11 minute =1/2+1/6=2/3 i.e, pipe A and B together can fill the tank in 3/2 minutes =1.5minutes. |
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| 23. |
In How Many Different Ways Can 5 Girls And 5 Boys Form A Circle Such That The Boys And The Girls Alternate? |
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Answer» Around a circle, 5 boys can be ARRANGED in 4! ways. Given that the boys and the GIRLS alternate. Hence there are 5 places for the girls. THEREFORE the girls can be arranged in 5! ways. TOTAL number of ways =4!×5!=24×120=2880. Around a circle, 5 boys can be arranged in 4! ways. Given that the boys and the girls alternate. Hence there are 5 places for the girls. Therefore the girls can be arranged in 5! ways. Total number of ways =4!×5!=24×120=2880. |
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| 24. |
25 Buses Are Running Between Two Places P And Q. In How Many Ways Can A Person Go From P To Q And Return By A Different Bus? |
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Answer» He can go in any of the 25 buses (25 ways). Since he cannot come back in the same BUS, he can RETURN in 24 ways. TOTAL number of ways =25×24=600. He can go in any of the 25 buses (25 ways). Since he cannot come back in the same bus, he can return in 24 ways. Total number of ways =25×24=600. |
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| 25. |
An Event Manager Has Ten Patterns Of Chairs And Eight Patterns Of Tables. In How Many Ways Can He Make A Pair Of Table And Chair? |
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Answer» He has 10 PATTERNS of chairs and 8 patterns of TABLES A chair can be SELECTED in 10 WAYS. A table can be selected in 8 ways. Hence one chair and one table can be selected in 10×810×8 ways=80 ways. He has 10 patterns of chairs and 8 patterns of tables A chair can be selected in 10 ways. A table can be selected in 8 ways. Hence one chair and one table can be selected in 10×810×8 ways=80 ways. |
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| 26. |
30% Of The Men Are More Than 25 Years Old And 80% Of The Men Are Less Than Or Equal To 50 Years Old. 20% Of All Men Play Football. If 20% Of The Men Above The Age Of 50 Play Football, What Percentage Of The Football Players Are Less Than Or Equal To 50 Years? |
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Answer» Let total number of men = 100 Then, 20 men play football. 80 men are LESS than or equal to 50 years OLD. Remaining 20 men are above 50 years old. Number of football players above 50 years old =20×20/100=4 Number of football players less than or equal to 50 years old =20−4=16 Required percentage =16/20×100=80%. Let total number of men = 100 Then, 20 men play football. 80 men are less than or equal to 50 years old. Remaining 20 men are above 50 years old. Number of football players above 50 years old =20×20/100=4 Number of football players less than or equal to 50 years old =20−4=16 Required percentage =16/20×100=80%. |
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| 27. |
P, Q, R Enter Into A Partnership. P Initially Invests 25 Lakh And Adds Another 10 Lakh After One Year. Q Initially Invests 35 Lakh And Withdraws 10 Lakh After 2 Years. R's Investment Is Rs 30 Lakh. In What Ratio Should The Profit Be Divided At The End Of 3 Years? |
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Answer»
=(25×1+35×2):(35×2+25×1)=(25×1+35×2):(35×2+25×1) :(30×3):(30×3) =95:95:90=19:19:18 P : Q : R =(25×1+35×2):(35×2+25×1)=(25×1+35×2):(35×2+25×1) :(30×3):(30×3) =95:95:90=19:19:18 |
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| 28. |
A And B Starts A Business Investing Rs.85000 And Rs.15000 Respectively. Find Out The Ratio In Which The Profits Should Be Shared? |
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Answer» Here A's and B's capitals are there for equal time.Hence A : B = 85000 : 15000 = 85 : 15 Here A's and B's capitals are there for equal time.Hence A : B = 85000 : 15000 = 85 : 15 = 17 : 3. |
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| 29. |
What Is The Smallest 6 Digit Number Exactly Divisible By 111? |
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Answer» Smallest 6 digit number = 100000 100000÷111 = 900, remainder = 100. Hence 11 should be added to 100000 to get the smallest 6 digit number exactly divisible by 111 THEREFORE, smallest 6 digit number exactly divisible by 111= 100000 + 11 = 100011 Smallest 6 digit number = 100000 100000÷111 = 900, remainder = 100. Hence 11 should be added to 100000 to get the smallest 6 digit number exactly divisible by 111 Therefore, smallest 6 digit number exactly divisible by 111= 100000 + 11 = 100011 |
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| 30. |
? + 3699 + 1985 - 2047 = 31111? |
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Answer» LET x+3699+1985−2047=31111x+3699+1985−2047=31111 x=31111−3699−1985+2047=27474. Let x+3699+1985−2047=31111x+3699+1985−2047=31111 x=31111−3699−1985+2047=27474. |
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| 31. |
3 Litre Of Water Is Added To 11 Litre Of A Solution Containing 42% Of Alcohol In The Water. The Percentage Of Alcohol In The New Mixture Is? |
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Answer» We have a 11 litre solution containing 42% of ALCOHOL in the WATER. => Quantity of alcohol in the solution =11×42/100 Now 3 litre of water is ADDED to the solution. => Total quantity of the NEW solution = 11 + 3 = 14 Percentage of alcohol in the new solution =11×42/100/14×100 =11×3/100=33%. We have a 11 litre solution containing 42% of alcohol in the water. => Quantity of alcohol in the solution =11×42/100 Now 3 litre of water is added to the solution. => Total quantity of the new solution = 11 + 3 = 14 Percentage of alcohol in the new solution =11×42/100/14×100 =11×3/100=33%. |
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| 32. |
Six Bells Start Ringing Together And Ring At Intervals Of 4, 8, 10, 12, 15 And 20 Seconds Respectively. How Many Times Will They Ring Together In 60 Minutes ? |
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Answer» LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes Hence all the SIX bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes =1+60/2=31. LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes Hence all the six bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes =1+60/2=31. |
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| 33. |
Three Numbers Which Are Co-prime To Each Other Are Such That The Product Of The First Two Is 119 And That Of The Last Two Is 391. What Is The Sum Of The Three Numbers? |
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Answer» Since the numbers are co-prime, their HCF = 1 Product of first TWO numbers = 119 Product of last two numbers = 391 The middle number is common in both of these products. Hence, if we take HCF of 119 and 391, we GET the common middle number. HCF of 119 and 391 = 17 => Middle Number = 17 First Number =119/17=7 Last Number =391/17=23 Sum of the three numbers = 7 + 17 + 23 = 47. Since the numbers are co-prime, their HCF = 1 Product of first two numbers = 119 Product of last two numbers = 391 The middle number is common in both of these products. Hence, if we take HCF of 119 and 391, we get the common middle number. HCF of 119 and 391 = 17 => Middle Number = 17 First Number =119/17=7 Last Number =391/17=23 Sum of the three numbers = 7 + 17 + 23 = 47. |
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| 34. |
What Is The Least Multiple Of 7 Which Leaves A Remainder Of 4 When Divided By 6, 9, 15 And 18 ? |
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Answer»
REQUIRED Number = (90k + 4) which is a multiple of 7 PUT K = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7 Put k = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7 Put k = 3. We get number as (90 × 3) + 4 = 274. But this is not a multiple of 7 Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7 Hence 364 is the answer. LCM of 6, 9, 15 and 18 = 90 Required Number = (90k + 4) which is a multiple of 7 Put k = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7 Put k = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7 Put k = 3. We get number as (90 × 3) + 4 = 274. But this is not a multiple of 7 Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7 Hence 364 is the answer. |
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| 35. |
A Hostel Had Provisions For 250 Men For 40 Days. If 50 Men Left The Hostel, How Long Will The Food Last At The Same Rate? |
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Answer» A hostel had provisions for 250 men for 40 days If 50 men leaves the hostel, remaining men = 250 - 50 = 200 We need to FIND out how LONG the food will last for these 200 men. Let the required NUMBER of days = XX days More men, Less days (Indirect Proportion) (men) 250 : 200 :: xx : 40 ⇒250×40=200x ⇒5×40=4x ⇒x=5×10=50. A hostel had provisions for 250 men for 40 days If 50 men leaves the hostel, remaining men = 250 - 50 = 200 We need to find out how long the food will last for these 200 men. Let the required number of days = xx days More men, Less days (Indirect Proportion) (men) 250 : 200 :: xx : 40 ⇒250×40=200x ⇒5×40=4x ⇒x=5×10=50. |
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| 36. |
A Garrison Of 500 Persons Had Provisions For 27 Days. After 3 Days A Reinforcement Of 300 Persons Arrived. For How Many More Days Will The Remaining Food Last Now? |
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Answer» GIVEN that fort had provision for 500 persons for 27 days HENCE, after 3 days, the remaining food is sufficient for 500 persons for 24 days Remaining persons after 3 days = 500 + 300 = 800 Assume that after 3 days,the remaining food is sufficient for 800 persons for XX days More MEN, Less days (Indirect Proportion) (men) 500 : 800 :: xx : 24 ⇒500×24=800x ⇒5×24=8x ⇒x=5×3=15. Given that fort had provision for 500 persons for 27 days Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days Remaining persons after 3 days = 500 + 300 = 800 Assume that after 3 days,the remaining food is sufficient for 800 persons for xx days More men, Less days (Indirect Proportion) (men) 500 : 800 :: xx : 24 ⇒500×24=800x ⇒5×24=8x ⇒x=5×3=15. |
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| 37. |
A Rope Can Make 70 Rounds Of The Circumference Of A Cylinder Whose Radius Of The Base Is 14cm. How Many Times Can It Go Round A Cylinder Having Radius 20 Cm? |
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Answer» Let the required number of rounds be X More RADIUS, less rounds(Indirect PROPORTION) Hence we can write as (radius) 14 : 20 :: xx : 70 ⇒14×70=20x ⇒14×7=2x ⇒x=7×7=49. Let the required number of rounds be x More radius, less rounds(Indirect proportion) Hence we can write as (radius) 14 : 20 :: xx : 70 ⇒14×70=20x ⇒14×7=2x ⇒x=7×7=49. |
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| 38. |
If The Seventh Day Of A Month Is Three Days Earlier Than Friday, What Day Will It Be On The Nineteenth Day Of The Month? |
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Answer» Given that seventh DAY of a month is THREE days earlier than FRIDAY => Seventh day is Tuesday => 14th is Tuesday => 19th is Sunday. Given that seventh day of a month is three days earlier than Friday => Seventh day is Tuesday => 14th is Tuesday => 19th is Sunday. |
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| 39. |
1.12.91 Is The First Sunday. Which Is The Fourth Tuesday Of December 91? |
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Answer» GIVEN that 1.12.91 is the first SUNDAY Hence we can assume that 3.12.91 is the first Tuesday If we ADD 7 days to 3.12.91, we will get second Tuesday If we add 14 days to 3.12.91, we will get THIRD Tuesday If we add 21 days to 3.12.91, we will get fourth Tuesday => fourth Tuesday = (3.12.91 + 21 days) = 24.12.91. Given that 1.12.91 is the first Sunday Hence we can assume that 3.12.91 is the first Tuesday If we add 7 days to 3.12.91, we will get second Tuesday If we add 14 days to 3.12.91, we will get third Tuesday If we add 21 days to 3.12.91, we will get fourth Tuesday => fourth Tuesday = (3.12.91 + 21 days) = 24.12.91. |
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| 40. |
If 1st October Is Sunday, Then 1st November Will Be? |
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Answer» Given that 1st OCTOBER is Sunday Number of DAYS in October = 31 31 days = 3 odd days (As we can REDUCE multiples of 7 from odd days which will not change anything) Hence 1st NOVEMBER = (Sunday + 3 odd days) = Wednesday. Given that 1st October is Sunday Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Hence 1st November = (Sunday + 3 odd days) = Wednesday. |
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| 41. |
If The First Day Of A Year (other Than Leap Year) Was Friday, Then Which Was The Last Day Of That Year? |
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Answer» Given that FIRST day of a normal YEAR was Friday Odd days of the mentioned year = 1 (Since it is an ORDINARY year) Hence First day of the next year = (Friday + 1 Odd day) = Saturday Therefore, last day of the mentioned year = Friday. Given that first day of a normal year was Friday Odd days of the mentioned year = 1 (Since it is an ordinary year) Hence First day of the next year = (Friday + 1 Odd day) = Saturday Therefore, last day of the mentioned year = Friday. |
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| 42. |
A Boatman Can Row 96 Km Downstream In 8 Hr. If The Speed Of The Current Is 4 Km/hr, Then Find In What Time Will Be Able To Cover 8 Km Upstream? |
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Answer» Speed DOWNSTREAM =96/8 = 12 kmph Speed of current = 4 KM/hr Speed of the boatman in still water = 12-4 = 8 kmph Speed upstream = 8-4 = 4 kmph Time TAKEN to cover 8 km upstream =8/4 = 2 hours. Speed downstream =96/8 = 12 kmph Speed of current = 4 km/hr Speed of the boatman in still water = 12-4 = 8 kmph Speed upstream = 8-4 = 4 kmph Time taken to cover 8 km upstream =8/4 = 2 hours. |
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| 43. |
If A Man Rows At The Rate Of 5 Kmph In Still Water And His Rate Against The Current Is 3 Kmph, Then The Man's Rate Along The Current Is? |
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Answer» LET the RATE ALONG with the CURRENT is xx km/hr x+3/2=5 ⇒x+3=10 ⇒x=7 kmph. Let the rate along with the current is xx km/hr x+3/2=5 ⇒x+3=10 ⇒x=7 kmph. |
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| 44. |
The Average Age Of A Husband And His Wife Was 23 Years At The Time Of Their Marriage. After Five Years They Have A One Year Old Child. What Is The Average Age Of The Family ? |
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Answer» TOTAL AGE of HUSBAND and wife (at the time of their marriage) = 2 × 23 = 46 Total age of husband and wife after 5 years + Age of the 1 year old child = 46 + 5 + 5 + 1 = 57 AVERAGE age of the FAMILY = 57/3 = 19. Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46 Total age of husband and wife after 5 years + Age of the 1 year old child = 46 + 5 + 5 + 1 = 57 Average age of the family = 57/3 = 19. |
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| 45. |
The Average Of Six Numbers Is X And The Average Of Three Of These Is Y. If The Average Of The Remaining Three Is Z, Then? |
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Answer» Average of 6 numbers = x => SUM of 6 numbers = 6x Average of the 3 numbers = y => Sum of these 3 numbers = 3Y Average of the REMAINING 3 numbers = Z => Sum of the remaining 3 numbers = 3z Now we know that 6x = 3y + 3z => 2x = y + z. Average of 6 numbers = x => Sum of 6 numbers = 6x Average of the 3 numbers = y => Sum of these 3 numbers = 3y Average of the remaining 3 numbers = z => Sum of the remaining 3 numbers = 3z Now we know that 6x = 3y + 3z => 2x = y + z. |
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| 46. |
The Average Of Five Numbers Id 27. If One Number Is Excluded, The Average Becomes 25. What Is The Excluded Number? |
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Answer» SUM of 5 numbers = 5 × 27 Sum of 4 numbers after EXCLUDING ONE number = 4 × 25 EXCLUDED number = 5 × 27 - 4 × 25 = 135 - 100 = 35. Sum of 5 numbers = 5 × 27 Sum of 4 numbers after excluding one number = 4 × 25 Excluded number = 5 × 27 - 4 × 25 = 135 - 100 = 35. |
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| 47. |
The Average Age Of 36 Students In A Group Is 14 Years. When Teacher's Age Is Included To It, The Average Increases By One. Find Out The Teacher's Age In Years? |
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Answer» average age of 36 students in a group is 14 Sum of the ages of 36 students = 36 × 14 When teacher's age is INCLUDED to it, the average INCREASES by one => average = 15 Sum of the ages of 36 students and the teacher = 37 × 15 Hence teachers age = 37 × 15 - 36 × 14 = 37 × 15 - 14(37 - 1) = 37 × 15 - 37 × 14 + 14 = 37(15 - 14) + 14 = 37 + 14 = 51. average age of 36 students in a group is 14 Sum of the ages of 36 students = 36 × 14 When teacher's age is included to it, the average increases by one => average = 15 Sum of the ages of 36 students and the teacher = 37 × 15 Hence teachers age = 37 × 15 - 36 × 14 = 37 × 15 - 14(37 - 1) = 37 × 15 - 37 × 14 + 14 = 37(15 - 14) + 14 = 37 + 14 = 51. |
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| 48. |
Kamal Was 44 Times As Old As His Son 88 Years Ago. After 88 Years, Kamal Will Be Twice As Old As His Son. Find Out The Present Age Of Kamal? |
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Answer» Let age of the SON before 88 years =X Then, age of Kamal before 88 years AGO =4x After 88 years, Kamal will be TWICE as old as his son ⇒4x+16=2(x+16) ⇒x=8 Present age of Kamal =4x+8=4×8+8=40. Let age of the son before 88 years =x Then, age of Kamal before 88 years ago =4x After 88 years, Kamal will be twice as old as his son ⇒4x+16=2(x+16) ⇒x=8 Present age of Kamal =4x+8=4×8+8=40. |
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| 49. |
Present Age Of A Father Is 33 Years More Than Three Times The Age Of His Son. Three Years Hence, Father's Age Will Be 1010 Years More Than Twice The Age Of The Son. What Is Father's Present Age? |
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Answer» Let the present age the SON =x Then, present age of the father =3x+3 Given that, THREE years hence, father's age will be 1010 years more than TWICE the age of the son ⇒(3x+3+3)=2(x+3)+10 ⇒x=10 Father's present age =3x+3=3×10+3=33. Let the present age the son =x Then, present age of the father =3x+3 Given that, three years hence, father's age will be 1010 years more than twice the age of the son ⇒(3x+3+3)=2(x+3)+10 ⇒x=10 Father's present age =3x+3=3×10+3=33. |
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| 50. |
The Ages Of Two Persons Differ By 1616 Years. 66 Years Ago, The Elder One Was 33 Times As Old As The Younger One. What Are Their Present Ages Of The Elder Person? |
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Answer» LET present age of the elder PERSON =X and present age of the YOUNGER person =x−16 (x−6)=3(x−16−6) ⇒x−6=3x−66 ⇒2x=60 ⇒x=60/2=30. Let present age of the elder person =x and present age of the younger person =x−16 (x−6)=3(x−16−6) ⇒x−6=3x−66 ⇒2x=60 ⇒x=60/2=30. |
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