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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Which Of The Following Numbers Are Completely Divisible By 11? |
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Answer» If the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then CLEARLY the number is divisible by 11. [(Sum of digits at odd places) - (Sum of digits at even places)]/11 => [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/11 = 0/11 = 0 similarly others number are divisible by 11. If the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11. [(Sum of digits at odd places) - (Sum of digits at even places)]/11 => [(3+4+6+2) - (2+5+8)]/11 => (15 - 15)/11 = 0/11 = 0 similarly others number are divisible by 11. |
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| 2. |
Exact Time Of A Clock Was Set Right At 5:00 Am Then It Loses 16 Min. Every Day, On The 4th Day When It Shows 10:00 Am, What Is The Exact Time? |
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Answer» Total late in three day = 16*3 = 48 MIN. Total late in one day =16 min. so late in 1 min. = 16/(24*60) = 0.01111; HENCE in 5 hour (5:00 AM to 10:00 AM on 4th day) =.0111*5*60 = 3.333; So, Exact time will be = 10:52 AM. Total late in three day = 16*3 = 48 min. Total late in one day =16 min. so late in 1 min. = 16/(24*60) = 0.01111; Hence in 5 hour (5:00 AM to 10:00 AM on 4th day) =.0111*5*60 = 3.333; So, Exact time will be = 10:52 AM. |
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| 3. |
In A Stream Running At 2 Kmph,a Motar Boat Goes 6 Km Upstream And Back Again To The Starting Point In 33 Minutes. Find The Speed Of The Motarboat In Still Water? |
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Answer» Lets assume speed of boat in still water = x km/hr So boat downstream speed = (x+2) km/hr. boat upstream speed = (x-2) km/hr. 6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20 => 120*[(x-2) + (x+2)] = 11*x^2 -44 => 240 x = 11*x^2-44 => 11* x^2 -240x -44 = 0 ---------(1) By solving EQ. (1) x = 22 km/hr. Lets assume speed of boat in still water = x km/hr So boat downstream speed = (x+2) km/hr. boat upstream speed = (x-2) km/hr. 6/(x-2)+6/(x+2)=33/60 => 6/(x-2)+6/(x+2)=11/20 => 120*[(x-2) + (x+2)] = 11*x^2 -44 => 240 x = 11*x^2-44 => 11* x^2 -240x -44 = 0 ---------(1) By solving eq. (1) x = 22 km/hr. |
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| 4. |
If The Sales Tax Reduced From 3 1/2% To 3 1/3%, Then What Difference Does It Make To A Person Who Purchases An Article With Market Price Of Rs. 8400 ? |
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Answer» TAX reduced from 3 1/2 or 7/2% to 3 1/3% or 10/3%. So, the difference in tax = (7/2 - 10/3)% = 1/6%. => 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14. Tax reduced from 3 1/2 or 7/2% to 3 1/3% or 10/3%. So, the difference in tax = (7/2 - 10/3)% = 1/6%. => 1/6 % of Rs. 8400 = 8400*1/6% = 8400* (1/6 * 1/100) = Rs. 14. |
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| 5. |
A Vessel Is Filled To Its Capacity With Pure Milk. Ten Litres Are Withdrawn From It And Replaced By Water. This Procedure Is Repeated Again. The Vessel Now Has 32 Litres Of Milk. Find The Capacity Of The Vessel (in Litres)? |
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Answer» Lets assume the capacity of the vessel = x litres. x (1- 10/x)^2 = 32 x^2 +100-20x = 32x x^2 +100 - 52x= 0 x^2 - 50X - 2x + 100 = 0 => (x-2) (x-50) = 0 => x=2 & x=50 As 10 litres is withdrawn so vessel capacity will be 50 litres. Lets assume the capacity of the vessel = x litres. x (1- 10/x)^2 = 32 x^2 +100-20x = 32x x^2 +100 - 52x= 0 x^2 - 50x - 2x + 100 = 0 => (x-2) (x-50) = 0 => x=2 & x=50 As 10 litres is withdrawn so vessel capacity will be 50 litres. |
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| 6. |
Two Circles Touch Each Other Externally. The Distance Between Their Centres Is 14 Cm And The Sum Of Their Areas Is 130 Cm^2. Find Their Radii? |
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Answer» Lets assume the radius of 1st circle = x CM. so, the radius of 2ND circle = (14-x)cm. (pi*x*x)+(pi*(14-x)*(14-x))=130 By SOLVING above eq. x = 11 or 3. Lets assume the radius of 1st circle = x cm. so, the radius of 2nd circle = (14-x)cm. (pi*x*x)+(pi*(14-x)*(14-x))=130 By solving above eq. x = 11 or 3. |
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| 7. |
A Constructor Estimates That 3 People Can Paint Mr. Khans House In 4 Days. If He Uses 4 People Instead Of 3,how Long Will They Take To Complete The Job? |
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Answer» Lets assume day REQUIRED to COMPLETE the JOB = x day Men Day --------------------- 3(u) 4(d) 4 x x/4 = 3/4 Lets assume day required to complete the job = x day Men Day --------------------- 3(u) 4(d) 4 x x/4 = 3/4 => x = 3 days. |
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| 8. |
Find The Smallest Number Which Leaves 22,35, 48 And 61 As Remainders When Divided By 26, 39, 52 And 65 Respectively? |
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Answer» 26-22=4 39-35=4 52-48=4 65-61=4 So SMALLEST NUMBER = (780-4) = 776. 26-22=4 39-35=4 52-48=4 65-61=4 LCM of (26,39,52,65)=780 So smallest number = (780-4) = 776. |
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| 9. |
The Average Number Of Visitors Of A Library In The First 4 Days Of A Week Was 58. The Average For The 2nd, 3rd, 4th And 5th Days Was 60.if The Number Of Visitors On The 1st And 5th Days Were In The Ratio 7:8 Then What Is The Number Of Visitors On The 5th Day Of The Library? |
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Answer» If number of VISITORS on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & E RESPECTIVELY, then a+b+c+d=58*4=232 ----(i) b+c+d+e=60*4=240 ----(ii) Subtracting EQ. (i) from (ii) e-a=8 ---(iii) Given, a/e=7/8 ---(iv) So from eq. (iii) & (iv) a=56, e=64. If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then a+b+c+d=58*4=232 ----(i) b+c+d+e=60*4=240 ----(ii) Subtracting eq. (i) from (ii) e-a=8 ---(iii) Given, a/e=7/8 ---(iv) So from eq. (iii) & (iv) a=56, e=64. |
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| 10. |
In An Exam, Ajith, Sachu, Karna, Saheep And Ramesh Scored An Average Of 39 Marks. Saheep Scored 7 Marks More Than Ramesh. Ramesh Scored 9 Fewer Than Ajith. Sachu Scored As Many As Saheep And Ramesh Scored.sachu And Karna Scored 110 Marks Them. If Ajith Scores 32 Marks Then How Many Marks Did Karna Score? |
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Answer» LETS assume marks of Ajith, Sachu, Karna, Saheep and RAMESH RESPECTIVELY u, v, w, X, y So, as per the QUESTION: z+7=x ---(i) u- 9=z ---(ii) x+ z=v ---(iii) v+w=110 ---(iv) Given, u=32 ---- (v) By solving eq. (i), (ii), (iii), (iv) and (v) w=57. Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y So, as per the question: z+7=x ---(i) u- 9=z ---(ii) x+ z=v ---(iii) v+w=110 ---(iv) Given, u=32 ---- (v) By solving eq. (i), (ii), (iii), (iv) and (v) w=57. |
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| 11. |
Two Urns Contain Respectively 2 Red 3 White And 3 Red,5 White Balls.one Ball Is Drawn At Random From The First Urn And Transferred Into The Second.a Ball Is Now Drawn From The Second Urn And It Turns Out To Be Red. What Is The Probability That The Transferred Ball Was White? |
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Answer» <P>According to baye's theorem: p(no of way to DRAWN a BALL from 2nd urns is red and ball transferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/45 Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/45 so probability=(9/45)/(17/45)=9/17. According to baye's theorem: p(no of way to drawn a ball from 2nd urns is red and ball transferred from 1 urns)=(3c1/5c1)*(3c1/9c1)=9/45 Total=(2c1/5c1)*(4c1/9c1)+(3c1/5c1)*(3c1/9c1)=17/45 so probability=(9/45)/(17/45)=9/17. |
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| 12. |
The Age Of The Grand Father Is The Sum Of His Three Grandsons.the Second Is 2 Year Younger Than First One And The Third One Is 2 Year Younger Than The Second One. Then What Will Be The Age Of The Grandfather? |
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Answer» Lets assume second GRANDSON = X year. So, FIRST grandson AGE = (x+2) year , third grandson = (x-2) year So, x+2+x+x-2=3x i.e grandfather age is 3 times as OLDER as his second grandson. Lets assume second grandson = x year. So, first grandson age = (x+2) year , third grandson = (x-2) year So, x+2+x+x-2=3x i.e grandfather age is 3 times as older as his second grandson. |
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| 13. |
Nitish Sold His Watch And Sunglasses At A Loss Of 4% And Gain Of 4% Respectively For 2600 To Kamal. Kamal Sold The Same Sun Glasses And Watch At A Loss Of 4% And Gain Of 4% Respectively For 2700. The Price Of Watch And Sun Glasses To Nitish Were? |
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Answer» LETS assume the CP of watch = Rs X and SUNGLASSES = Rs y. 2600=96x/100 + 104y/100 2700= 104x/100 + 96y/100 By solving, y=700 x=1960. Lets assume the CP of watch = Rs x and sunglasses = Rs y. 2600=96x/100 + 104y/100 2700= 104x/100 + 96y/100 By solving, y=700 x=1960. |
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| 14. |
If 20 Men Or 24 Women Or 40 Boys Can Do A Job In 12 Days Working For 8 Hours A Day, How Many Men Working With 6 Women And 2 Boys Take To Do A Job Four Times As Big Working For 5 Hours A Day For 12 Days? |
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Answer» AMOUNT of work done by 20 men = 24 WOMEN = 40 boys i.e 5man = 6 woman = 10 boys. -----------------(i) According to the 1st condition, 5 men can do a job in 12 days working for 8 hours a day. Required :- how many more men are required to work with 6 women and 2 boys. Let the required NUMBER of men be M. According to the given information, =>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/4. ( 6 women= 5men & 2 boys =1 men) =>32= M + 6. =>M=26. Hence, 26 men are working with 6 women and 2 boys. Amount of work done by 20 men = 24 women = 40 boys i.e 5man = 6 woman = 10 boys. -----------------(i) According to the 1st condition, 5 men can do a job in 12 days working for 8 hours a day. Required :- how many more men are required to work with 6 women and 2 boys. Let the required number of men be M. According to the given information, =>(5 x 12 x 8 )/1=[(M + 5 + 1) x 5 x 12 ]/4. ( 6 women= 5men & 2 boys =1 men) =>32= M + 6. =>M=26. Hence, 26 men are working with 6 women and 2 boys. |
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| 15. |
A Certain Quantity Of 40% Solution Is Replaced With 25% Solution Such That The New Concentration Is 35%. What Is The Fraction Of The Solution That Was Replaced? |
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Answer» Consider a solution = 1 LTR X is the certain QUANTITY which has to be REPLACED Now, 40%of (1-x)+(25%of x)=35/100 => 40/100 * (1-x) + 25x/100 = 35/100 => 40/100 - 40x/100 + 25x/100 = 35/100 => 15x/100 = 5/100 => x = 1/3. Consider a solution = 1 ltr X is the certain quantity which has to be replaced Now, 40%of (1-x)+(25%of x)=35/100 => 40/100 * (1-x) + 25x/100 = 35/100 => 40/100 - 40x/100 + 25x/100 = 35/100 => 15x/100 = 5/100 => x = 1/3. |
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| 16. |
The Ratio Of Two Numbers Is 3:4 And Their Hcf Is 4.their Lcm Is? |
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Answer» GIVEN TWO No, ratio = 3:4 and their HCF = 4 So, No. = 3*4 =12 and 4*4=16 Given Two No, ratio = 3:4 and their HCF = 4 So, No. = 3*4 =12 and 4*4=16 LCM of 12,16 = 48. |
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| 17. |
A Gets Rs.33 When A Sum Of Money Was Distributed Among A, B And C In The Ratio 3:2:5, What Will Be The Sum Of Money? |
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Answer» LETS assume SUM of money = Rs. x ; A gets Rs. 33 when sum DISTRIBUTED in the ratio of 3:2:5 so, 33=x*3/10 => x=110. Lets assume sum of money = Rs. x ; A gets Rs. 33 when sum distributed in the ratio of 3:2:5 so, 33=x*3/10 => x=110. |
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| 18. |
An Alloy Of Zinc And Copper Contains The Metals In The Ratio 5 : 3. The Quantity Of Zinc To Be Added To 16 Kg Of The Alloy So That The Ratio Of The Metal May Be 3 : 1 Is? |
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Answer» Lets ASSUME quantity of zinc to added = X kg. zinc quantity in alloy = 5/8*16=10 Kg. and copper quantity = 3/8*16=6 kg. Alloy NEW RATIO of zinc and copper = 3:1 Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg. Lets assume quantity of zinc to added = x kg. zinc quantity in alloy = 5/8*16=10 Kg. and copper quantity = 3/8*16=6 kg. Alloy new ratio of zinc and copper = 3:1 Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg. |
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| 19. |
A Cycled From P To Q At 10 Kmph And Returned At The Rate Of 9 Kmph. B Cycled Both Ways At 12 Kmph. In The Whole Journey B Took 10 Minutes Less Than A. Find The Distance Between P And Q? |
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Answer» <P>Lets assume distance between P and Q = d km. Time TAKEN by A in both side = d/10 + d/9 = 19d/90 HRS. Time taken by B in both side = 2d/12 = d/6 hrs. B tool 10 min. or 1/6 hrs. less than A. So, 19d/90 - d/6 = 1/6 => (19d-15d)/90 = 1/6 => 4d/90 = 1/6 => d = 15/4 km = 3.75 km. Lets assume distance between P and Q = d km. Time taken by A in both side = d/10 + d/9 = 19d/90 hrs. Time taken by B in both side = 2d/12 = d/6 hrs. B tool 10 min. or 1/6 hrs. less than A. So, 19d/90 - d/6 = 1/6 => (19d-15d)/90 = 1/6 => 4d/90 = 1/6 => d = 15/4 km = 3.75 km. |
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| 20. |
Find The Approximate Value Of The Following Equation 6.23% Of 258.43 - ? + 3.11% Of 127 = 13.87? |
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Answer» [(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87 16.100189-X+3.9497=13.87 X=6.179889. [(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87 16.100189-X+3.9497=13.87 X=6.179889. |
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| 21. |
At 10:00 Am 2 Trains Started Travelling Towards Each Other From Station 287 Miles Apart They Passed Each Other At 1:30 Pm The Same Dayy .if Average Speed Of The Faster Train Exceeded By 6 Miles /hr What Is Speed Of Faster Train In Miles/hr? |
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Answer» Lets assume the SPEED of slower train = x miles/hrs. So, speed of FASTER train is = (x+6) miles/hrs. Given, passed each other at 1:30 PM i.e after 3 1/2 hrs. Both train travelling towards each other so total RELATIVE speed = x+(x+6) = (2x+6) So, 287/(2x+6) = 7/2 => 574 = 14x + 42 => 14x = 542 => x = ~38 So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs. Lets assume the speed of slower train = x miles/hrs. So, speed of faster train is = (x+6) miles/hrs. Given, passed each other at 1:30 PM i.e after 3 1/2 hrs. Both train travelling towards each other so total relative speed = x+(x+6) = (2x+6) So, 287/(2x+6) = 7/2 => 574 = 14x + 42 => 14x = 542 => x = ~38 So spee of faster train = (x+6) miles/hrs. = 44 miles/hrs. |
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| 22. |
A Ball Dropped From H Height And Moves 80% Of Height Each Time. Total Distance Covered Is? |
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Answer» First time distance= H Second time = 80H/100 = 4H/5 similarly THIRD time 80% of 4H/5 = H(4^2)/(5^2) and so on.. This will lead to infinite terms of geometric progression i.e H+2*4H/5+2*16H/25.. So, Sum = H+ 2*4H/(5(1-4/5)) = 9H. First time distance= H Second time = 80H/100 = 4H/5 similarly third time 80% of 4H/5 = H(4^2)/(5^2) and so on.. This will lead to infinite terms of geometric progression i.e H+2*4H/5+2*16H/25.. So, Sum = H+ 2*4H/(5(1-4/5)) = 9H. |
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| 23. |
A Car Averages 55 Mph For The First 4 Hours Of A Trip And Averages 70 Mph For Each Additional Hour. The Average Speed For The Entire Trip Was 60 Mph. How Many Hours Long Is The Trip? |
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Answer» Lets assume additional hours = X hrs. So TOTAL No. hours in journey = (4+x) [(55*4)+(70*x)]/(4+x)=60 => =>x=2 Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs. Lets assume additional hours = x hrs. So total No. hours in journey = (4+x) [(55*4)+(70*x)]/(4+x)=60 => =>x=2 Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs. |
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| 24. |
The Cost Of Two Varieties Of Paint Is Rs.3969 Per 2 Kg And Rs.1369 Per 2 Kg Respectively. After How Many Years, The Value Of Both Paint Will Be The Same, If Variety 1 Appreciates At 26% Per Annum And Variety 2 Depreciates At 26% Per Annum? |
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Answer» Simply appreciates variety 1 by 26% and DEPRECIATES variety 2 by 26% as: 3969(1-(1/26))^n=1369(1+)1/26))^n For n = 2 we GET both values equal. Variety 2 Variety1 3969.00 1369 Initially 2937.06 1724.94 after I year 2173.42 2173.42 after II year So the PRICE become same after 2 years. Simply appreciates variety 1 by 26% and depreciates variety 2 by 26% as: 3969(1-(1/26))^n=1369(1+)1/26))^n For n = 2 we get both values equal. Variety 2 Variety1 3969.00 1369 Initially 2937.06 1724.94 after I year 2173.42 2173.42 after II year So the price become same after 2 years. |
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| 25. |
If 6 Is Subtracted From The Present Age Of Ritu And The Remainder Is Divided By 6, Then The Present Age Of Sheema Is Obtained. If Sheema Is 4 Years Younger To Raju Whose Age Is 12 Years, Then Find The Age Of Ritu? |
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Answer» Lets assume Ritu present age = x YEARS so SHEEMA age = (x-6)/6 years As per question: sheema age [(x-6)/6 + 4] = 12 (x-6)/6 = 12 - 4 => (x-6)/6 = 8 => x - 6 = 48 => x = 54 So Ritu age = 54 years. Lets assume Ritu present age = x years so sheema age = (x-6)/6 years As per question: sheema age [(x-6)/6 + 4] = 12 (x-6)/6 = 12 - 4 => (x-6)/6 = 8 => x - 6 = 48 => x = 54 So Ritu age = 54 years. |
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| 26. |
The Cost Price Of An Article Ls 80% Of Its Marked Price For Sale. How Much Percent Does The Trades-man Gain After Allowing A Discount Of 12%? |
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Answer» C.P. of the article = Rs. 100 So MARKED PRICE = (100*100)/80 = Rs. 125 SP after the discount = Rs.(125*88)/100 = Rs. 110 therefore Gain percent = 10. C.P. of the article = Rs. 100 So Marked price = (100*100)/80 = Rs. 125 SP after the discount = Rs.(125*88)/100 = Rs. 110 therefore Gain percent = 10. |
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| 27. |
A Train Covers A Distance In 50 Min ,if It Runs At A Speed Of 48 Kmph On An Average.the Speed At Which The Train Must Run To Reduce The Time Of Journey To 40 Min Will Be? |
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Answer» Time=50/60 hr=5/6hr SPEED=48mph distance=S*T=48*5/6=40km time=40/60hr=2/3hr NEW speed = 40* 3/2 KMPH= 60KMPH. Time=50/60 hr=5/6hr Speed=48mph distance=S*T=48*5/6=40km time=40/60hr=2/3hr New speed = 40* 3/2 kmph= 60kmph. |
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| 28. |
Find Three Consecutive Odd Integers Whose Sum Is 117? |
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Answer» LET the three integers be 2x + 1, 2x + 3 , 2x + 5. x = 18 The three integers are 37, 39, 41. Let the three integers be 2x + 1, 2x + 3 , 2x + 5. Therefore, 6x + 9 = 117, x = 18 The three integers are 37, 39, 41. |
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