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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 Man Has Rs. 480 In The Denominations Of One-rupee Notes, Five-rupee Notes And Ten-rupee Notes. The Number Of Notes Of Each Denomination Is Equal. What Is The Total Number Of Notes That He Has ? |
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Answer» Let NUMBER of notes of each denomination be X. Then x + 5x + 10x = 480 16x = 480 x = 30. Hence, TOTAL number of notes = 3X = 90. Let number of notes of each denomination be x. Then x + 5x + 10x = 480 16x = 480 x = 30. Hence, total number of notes = 3x = 90. |
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| 2. |
What Is The Overall Percentage Of Tarun? |
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Answer» AGGREGATE marks OBTAINED by Tarun = [ (65% of 150) + (35% of 130) + (50% of 120) + ((77% of 100) + (80% of 60) + (80% of 40) ] = [ 97.5 + 45.5 + 60 + 77 + 48 + 32 ] = 360. The maximum marks (of all the six subjects) = (150 + 130 + 120 + 100 + 60 + 40) = 600. Therefore Overall percentage of Tarun =(360/600x 100)% = 60%. Aggregate marks obtained by Tarun = [ (65% of 150) + (35% of 130) + (50% of 120) + ((77% of 100) + (80% of 60) + (80% of 40) ] = [ 97.5 + 45.5 + 60 + 77 + 48 + 32 ] = 360. The maximum marks (of all the six subjects) = (150 + 130 + 120 + 100 + 60 + 40) = 600. Therefore Overall percentage of Tarun =(360/600x 100)% = 60%. |
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| 3. |
A Card Is Drawn From A Pack Of 52 Cards. The Probability Of Getting A Queen Of Club Or A King Of Heart Is? |
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Answer» <P>Here, n(S) = 52. Let E = EVENT of getting a queen of CLUB or a king of heart. Then, n(E) = 2. P(E) =n(E)/ n(S)=2/52=1/26. Here, n(S) = 52. Let E = event of getting a queen of club or a king of heart. Then, n(E) = 2. P(E) =n(E)/ n(S)=2/52=1/26. |
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| 4. |
Two Dice Are Thrown Simultaneously. What Is The Probability Of Getting Two Numbers Whose Product Is Even? |
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Answer» <P>In a simultaneous throw of two dice, we have n(S) = (6 X 6) = 36. Then, E = {(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)} n(E) = 27. P(E) = n(E)/ n(S)=27/36=3/4. In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36. Then, E = {(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)} n(E) = 27. P(E) = n(E)/ n(S)=27/36=3/4. |
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| 5. |
A Rectangular Tract Of Land Measures 860 Feet By 560 Feet. Approximately How Many Acres Is This? (one Acre = 43,560 Square Feet)? |
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Answer» AREA of the rectangular TRACT = Length X Breadth = 860 x 560 Sq.ft. = {860x560}/43560 acres = 11.06 acres. Area of the rectangular tract = Length x Breadth = 860 x 560 Sq.ft. = {860x560}/43560 acres = 11.06 acres. |
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| 6. |
A Is Two Years Older Than B Who Is Twice As Old As C. If The Total Of The Ages Of A, B And C Be 27, The How Old Is B? |
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Answer» LET C's age be X years. Then, B's age = 2x years. A's age = (2x + 2) years. (2x + 2) + 2x + x = 27 5x = 25 x = 5. Hence, B's age = 2x = 10 years. Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years. (2x + 2) + 2x + x = 27 5x = 25 x = 5. Hence, B's age = 2x = 10 years. |
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| 7. |
A Father Said To His Son, "i Was As Old As You Are At The Present At The Time Of Your Birth". If The Father's Age Is 38 Years Now, The Son's Age Five Years Back Was? |
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Answer» LET the son's present AGE be X years. Then, (38 - x) = x 2x = 38. x = 19. Son's age 5 years back (19 - 5) = 14 years. Let the son's present age be x years. Then, (38 - x) = x 2x = 38. x = 19. Son's age 5 years back (19 - 5) = 14 years. |
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| 8. |
A Sum Of Money At Simple Interest Amounts To Rs. 815 In 3 Years And To Rs. 854 In 4 Years. The Sum Is? |
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Answer» S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = 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 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698. |
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| 9. |
The Largest 4 Digit Number Exactly Divisible By 88 Is? |
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Answer» Divide largest four DIGIT number 9999 by 88. You GET 113.625. Obviously 113 would be EXACTLY divisible so we want to know what that number is. We get this by multiplying 113 with 88 = 9944 Divide largest four digit number 9999 by 88. You get 113.625. Obviously 113 would be exactly divisible so we want to know what that number is. We get this by multiplying 113 with 88 = 9944 |
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| 10. |
What Least Number Must Be Added To 1056, So That The Sum Is Completely Divisible By 23 ? |
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Answer» If the number 1056 is completely divisible by 23 means, REMAINDER should come zero. But if we divide 1056 by 23, the remainder is 2. So if 2 is added to the 1056, we get remainder 0. Therefore solution is 2. If the number 1056 is completely divisible by 23 means, remainder should come zero. But if we divide 1056 by 23, the remainder is 2. So if 2 is added to the 1056, we get remainder 0. Therefore solution is 2. |
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| 11. |
On 8th Dec, 2007 Saturday Falls. What Day Of The Week Was It On 8th Dec, 2006? |
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Answer» The year 2006 is an ordinary year. So, it has 1 ODD DAY. So, the day on 8th DEC, 2007 will be 1 day beyond the day on 8th Dec, 2006. But, 8th Dec, 2007 is SATURDAY. 8th Dec, 2006 is Friday. The year 2006 is an ordinary year. So, it has 1 odd day. So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006. But, 8th Dec, 2007 is Saturday. 8th Dec, 2006 is Friday. |
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| 12. |
Today Is Monday. After 61 Days, It Will Be? |
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Answer» Each DAY of the week is REPEATED after 7 DAYS. So, after 63 days, it will be Monday. After 61 days, it will be Saturday. Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday. After 61 days, it will be Saturday. |
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| 13. |
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 1600 years = 0 Odd days in 400 years = 0 5 years = (4 ordinary years + 1 LEAP 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. TOTAL 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 ordinary years + 1 leap 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. Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day. Given day is Sunday. |
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| 14. |
It Was Sunday On Jan 1, 2006. What Was The Day Of The Week Jan 1, 2010? |
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Answer» On 31st December, 2005 it was Saturday. Number of ODD days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 31st December 2009, it was THURSDAY. Thus, on 1ST Jan, 2010 it is Friday. On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 31st December 2009, it was Thursday. Thus, on 1st Jan, 2010 it is Friday. |
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| 15. |
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 RS. 5x, Rs. 2x, Rs. 4x and Rs. 3X respectively. Then, 4x - 3x = 1000 X = 1000. B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000. Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively. Then, 4x - 3x = 1000 x = 1000. B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000. |
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| 16. |
A Reduction Of 20% In The Price Of Mangoes Enables A Person To Purchase 12 More For Rs. 15. What Was The Price Of 16 Mangoes Before Reduction Of Price ? |
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Answer» PRICE x Consumption = Expenditure Consumption = Exp. Price (15 / 8x) - (15 / x) = 12 x = (15 x 2) / (12 x 8) For 16 Mangoes = [(15 x 2) / (12 x 8)] x 16 = 5 Price x Consumption = Expenditure Consumption = Exp. Price (15 / 8x) - (15 / x) = 12 x = (15 x 2) / (12 x 8) For 16 Mangoes = [(15 x 2) / (12 x 8)] x 16 = 5 |
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| 17. |
How Many Numbers Between 1 And 100 Are Divisible By 7 ? |
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Answer» No. of DIVISIBLE by 7 7, 14 --------- 98, N = a + (N - 1)d 98 = 7 + (N - 1) 7, 98 = 7 + 7N – 7 98/7= N = 14 No. of divisible by 7 7, 14 --------- 98, n = a + (N - 1)d 98 = 7 + (N - 1) 7, 98 = 7 + 7N – 7 98/7= N = 14 |
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| 18. |
A Person Travels 48 Kms At 12 Kms/hour And Further 48 Kms At 16km.s/hour. His Average Speed For The Whole? |
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Answer» Speed = (Total DISTANCE / Total Time) TD = 48 + 48= 96, T1 = 48/12 = 4hrs T2 = 48/16 = 3HRS T1 + T2 = 4 + 3 = 7 AVG Speed = 96/7 = 13(5/7) km/hr Speed = (Total Distance / Total Time) TD = 48 + 48= 96, T1 = 48/12 = 4hrs T2 = 48/16 = 3hrs T1 + T2 = 4 + 3 = 7 Avg Speed = 96/7 = 13(5/7) km/hr |
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| 19. |
If G (0)=g (1)=1 And G (n)= G (n-1) + G (n -2) Find G (6)? |
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Answer» G(0) = 1 G(1) = 1 G(2) = G(1) + g(0) = 2 G(3) = g(2) + g(1) = 3 G(4) = g(3) + g(2) = 5 G(5) = g(4) + g(3) = 8 G(6) = g(5) +g (4) = 13 G(0) = 1 G(1) = 1 G(2) = G(1) + g(0) = 2 G(3) = g(2) + g(1) = 3 G(4) = g(3) + g(2) = 5 G(5) = g(4) + g(3) = 8 G(6) = g(5) +g (4) = 13 |
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| 20. |
Look At This Series: 2, 1, (1/2), (1/4), ... What Number Should Come Next? |
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Answer» This is a simple division series; each NUMBER is one-half of the previous number. In other terms to say, the number is divided by 2 SUCCESSIVELY to get the next RESULT. 4/2 = 2 2/2 = 1 1/2 = 1/2 (1/2)/2 = 1/4 (1/4)/2 = 1/8 and so on. This is a simple division series; each number is one-half of the previous number. In other terms to say, the number is divided by 2 successively to get the next result. 4/2 = 2 2/2 = 1 1/2 = 1/2 (1/2)/2 = 1/4 (1/4)/2 = 1/8 and so on. |
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| 21. |
Six Bells Commence Tolling Together And Toll At Intervals Of 2, 4, 6, 8 10 And 12 Seconds Respectively. In 30 Minutes, How Many Times Do They Toll Together ? |
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Answer» L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll TOGETHER after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30/2+ 1 = 16 times. L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30/2+ 1 = 16 times. |
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| 22. |
How Many Children Did Not Try Any Of The Rides. ? |
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Answer» 20 kids * 3 rides = Rs. 60 (55-20=)35 kids * 2 rides = Rs. 70 60 + 70 = Rs. 130 So, Rs. (145 – 130 = ) 15 are left for the other (85 – 55 = ) 30 kids . so only 15 of them can take a RIDE and REST 15 will be left out. 145 rides were taken. 20 of them TOOK all three, i.e. Rs. 60 were spent, so 145-60= Rs. 85 are left for the others. Total kids were 85, so rest were 65. out of these 65. 20 kids * 3 rides = Rs. 60 (55-20=)35 kids * 2 rides = Rs. 70 60 + 70 = Rs. 130 So, Rs. (145 – 130 = ) 15 are left for the other (85 – 55 = ) 30 kids . so only 15 of them can take a ride and rest 15 will be left out. 145 rides were taken. 20 of them took all three, i.e. Rs. 60 were spent, so 145-60= Rs. 85 are left for the others. Total kids were 85, so rest were 65. out of these 65. |
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