InterviewSolution
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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 Bag Contains A Total Of 93 Coins In The Form Of One Rupee And 50 Paise Coins. If The Total Value Of Coins In The Bag Is Rs.56, Find The Number Of 50 Paise Coins In The Bag ? |
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Answer» LET the NUMBER of one rupee coins in the BAG be x. Number of 50 PAISE coins in the bag is 93 - x. Total VALUE of coins [100x + 50(93 - x)]paise = 5600 paise => x = 74 Let the number of one rupee coins in the bag be x. Number of 50 paise coins in the bag is 93 - x. Total value of coins [100x + 50(93 - x)]paise = 5600 paise => x = 74 |
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| 2. |
The Cost Of 16 Pens And 8 Pencils Is Rs.352 And The Cost Of 4 Pens And 4 Pencils Is Rs.96. Find The Cost Of Each Pen ? |
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Answer» Let the COST of each pen and PENCIL be 'p' and 'q' respectively. 16p + 8q = 352 --- (1) 4p + 4q = 96 8p + 8q = 192 --- (2) (1) - (2) => 8p = 160 => p = 20 Let the cost of each pen and pencil be 'p' and 'q' respectively. 16p + 8q = 352 --- (1) 4p + 4q = 96 8p + 8q = 192 --- (2) (1) - (2) => 8p = 160 => p = 20 |
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| 3. |
The Present Average Age Of A Couple And Their Daughter Is 35 Years. Fifteen Years From Now, The Age Of The Mother Will Be Equal To The Sum Of Present Ages Of The Father And The Daughter. Find The Present Age Of Mother ? |
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Answer»
m + 15 = f + d Substituting f + d as m + 15 in (1), we GET 2m + 15 = 105 2m = 90 => m = 45 years. (f + m + d)/3 = 35 => f + m + d = 105 --- (1) m + 15 = f + d Substituting f + d as m + 15 in (1), we get 2m + 15 = 105 2m = 90 => m = 45 years. |
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| 4. |
Ten Years Ago, The Age Of Anand Was One-third The Age Of Bala At That Time. The Present Age Of Bala Is 12 Years More Than The Present Age Of Anand. Find The Present Age Of Anand ? |
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Answer» Let the present ages of Anand and Bala be 'a' and 'b' respectively. b = a + 12 Substituting b = a + 12 in first equation, a - 10 = 1/3 (a + 2) => 3a - 30 = a + 2 => 2a = 32 => a = 16. Let the present ages of Anand and Bala be 'a' and 'b' respectively. a - 10 = 1/3 (b - 10) --- (1) b = a + 12 Substituting b = a + 12 in first equation, a - 10 = 1/3 (a + 2) => 3a - 30 = a + 2 => 2a = 32 => a = 16. |
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| 5. |
Mudit's Age 18 Years Hence Will Be Thrice His Age Four Years Ago. Find Mudit's Present Age ? |
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Answer» LET Mudit's PRESENT age be 'm' YEARS. m + 18 = 3(m - 4) => 2m = 30 => m = 15 years. Let Mudit's present age be 'm' years. m + 18 = 3(m - 4) => 2m = 30 => m = 15 years. |
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| 6. |
When 2 Is Added To Half Of One-third Of One-fifth Of A Number, The Result Is One-fifteenth Of The Number. Find The Number ? |
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Answer»
2 + 1/2[1/3(a/5)] = a/15 Let the number be 2 + 1/2[1/3(a/5)] = a/15 => 2 = a/30 => a = 60 |
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| 7. |
In A Three Digit Number, The Hundred Digit Is 2 More Than The Tens Digit And The Units Digit Is 2 Less Than The Tens Digit. If The Sum Of The Digits Is 18, Find The Number ? |
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Answer» Let the three DIGIT numbers be 100a + 10b + c a = b + 2 c = b - 2 a + b + c = 3b = 18 => b = 6 So a = 8 and b = 4 Hence the three digit NUMBER is: 864 Let the three digit numbers be 100a + 10b + c a = b + 2 c = b - 2 a + b + c = 3b = 18 => b = 6 So a = 8 and b = 4 Hence the three digit number is: 864 |
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| 8. |
The Number Obtained By Interchanging The Two Digits Of A Two-digit Number Is Less Than The Original Number By 45. If The Sum Of The Two Digits Of The Number So Obtained Is 13, Then What Is The Original Number ? |
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Answer» Let the number be in the form of 10a + B Number formed by interchanging a and b = 10b + a. a + b = 13 --- (1) 10b + a = 10a + b - 45 45 = 9a - 9B => a - b = 5 --- (2) Adding (1) and (2), we get 2a = 18 => a = 9 and b = 4 The number is: 94. Let the number be in the form of 10a + b Number formed by interchanging a and b = 10b + a. a + b = 13 --- (1) 10b + a = 10a + b - 45 45 = 9a - 9b => a - b = 5 --- (2) Adding (1) and (2), we get 2a = 18 => a = 9 and b = 4 The number is: 94. |
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| 9. |
The Sum Of Five Consecutive Even Numbers Of Set X Is 440. Find The Sum Of A Different Set Of Five Consecutive Integers Whose Second Least Number Is 121 Less Than Double The Least Number Of Set X ? |
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Answer» Let the five consecutive even numbers be 2(x - 2), 2(x - 1), 2X, 2(x + 1) and 2(x + 2) Their sum = 10X = 440 x = 44 => 2(x - 2) = 84 Second least number of the other set = 2(84) - 121 = 47 This set has its least number as 46. Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50 = 48 - 2 + 48 - 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240 Let the five consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) and 2(x + 2) Their sum = 10x = 440 x = 44 => 2(x - 2) = 84 Second least number of the other set = 2(84) - 121 = 47 This set has its least number as 46. Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50 = 48 - 2 + 48 - 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240 |
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| 10. |
In A Series Of Six Consecutive Even Numbers, The Sum Of The Second And Sixth Numbers Is 24. What Is The Fourth Number ? |
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Answer» Let the NUMBERS be x, x + 2, x + 4, x + 6, x + 8 and x + 10. Given (x + 2) + (x + 10) = 24 => 2X + 12 = 24 => x = 6 The fourth number = x + 6 = 6 + 6 = 12. Let the numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10. Given (x + 2) + (x + 10) = 24 => 2x + 12 = 24 => x = 6 The fourth number = x + 6 = 6 + 6 = 12. |
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| 11. |
The Sum Of The Digits Of A Two-digit Number Is 12. The Difference Of The Digits Is 6. Find The Number? |
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Answer» Let the two-digit number be 10a + b If a>b, a - b = 6 If b>a, b - a = 6 If a - b = 6, ADDING it to equation (1), we get 2a = 18 => a =9 so b = 12 - a = 3 Number would be 93. if b - a = 6, adding it to the equation (1), we get 2b = 18 => b = 9 a = 12 - b = 3. Number would be 39. THEREFORE, Number would be 39 or 93. Let the two-digit number be 10a + b a + b = 12 --- (1) If a>b, a - b = 6 If b>a, b - a = 6 If a - b = 6, adding it to equation (1), we get 2a = 18 => a =9 so b = 12 - a = 3 Number would be 93. if b - a = 6, adding it to the equation (1), we get 2b = 18 => b = 9 a = 12 - b = 3. Number would be 39. Therefore, Number would be 39 or 93. |
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| 12. |
The Tens Digit Of A Two-digit Number Is Two More Than Its Unit Digit. The Two-digit Number Is 7 Times The Sum Of The Digits. Find The Units Digits ? |
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Answer» Let the two-digit number be 10a + b a = b + 2 --- (1) Substituting a = 2b in EQUATION (1), we get 2b = b + 2 => b = 2 HENCE the units digit is: 2. Let the two-digit number be 10a + b a = b + 2 --- (1) 10a + b = 7(a + b) => a = 2b Substituting a = 2b in equation (1), we get 2b = b + 2 => b = 2 Hence the units digit is: 2. |
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| 13. |
The Cost Of 10 Kg Of Apples Is Equal To The Cost Of 24 Kg Of Rice. The Cost Of 6 Kg Of Flour Equals The Cost Of 2 Kg Of Rice. The Cost Of Each Kg Of Flour Is Rs.20.50. Find The Total Cost Of 4 Kg Of Apples, 3 Kg Of Rice And 5 Kg Of Flour? |
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Answer» Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively. 10a = 24r and 6 * 20.50 = 2r a = 12/5 r and r = 61.5 a = 147.6 REQUIRED total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5 = 590.4 + 184.5 + 102.5 = Rs.877.40 Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively. 10a = 24r and 6 * 20.50 = 2r a = 12/5 r and r = 61.5 a = 147.6 Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5 = 590.4 + 184.5 + 102.5 = Rs.877.40 |
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| 14. |
Solve The Equation For X : 19(x + Y) + 17 = 19(-x + Y) – 21. |
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Answer» 19X + 19y + 17 = -19x + 19y - 21 19x + 19y + 17 = -19x + 19y - 21 38x = -38 => x = -1 |
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| 15. |
Solve The Equation For X : 6x - 27 + 3x = 4 + 9 – X |
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Answer»
9 x + x = 13 + 27 10 x = 40 => x = 4 |
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| 16. |
If The Numerator Of A Fraction Is Increased By 20% And Its Denominator Is Diminished By 25% Value Of The Fraction Is 2/15. Find The Original Fraction ? |
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Answer» X * (120/100) --- = 2/15 Y * (75/100) X/Y = 1/12 X * (120/100) --- = 2/15 Y * (75/100) X/Y = 1/12 |
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| 17. |
Subtracting 10% From X Is The Same As Multiplying X By What Number ? |
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Answer» X - (10/100) X = X * ? ? = 90% X - (10/100) X = X * ? ? = 90% |
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| 18. |
If The Price Of Gold Increases By 50%, Find By How Much The Quantity Of Ornaments Must Be Reduced, So That The Expenditure May Remain The Same As Before? |
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Answer»
150-------50 100-------? => 331/3% 100 150 150-------50 100-------? => 331/3% |
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| 19. |
A Candidate Got 35% Of The Votes Polled And He Lost To His Rival By 2250 Votes. How Many Votes Were Cast ? |
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Answer» 35%-----------L 30%----------2250 35%-----------L 65%-----------W 30%----------2250 100%---------? => 7500 |
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| 20. |
The Tax On A Commodity Is Diminished By 20% And Its Consumption Increased By 15%. The Effect On Revenue Is ? |
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Answer» 100 * 100 = 10000 10000-----------800 100-----------? => 8% DECREASE 100 * 100 = 10000 80 * 115 = 9200 10000-----------800 100-----------? => 8% decrease |
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| 21. |
The Salary Of A Typist Was At First Raised By 10% And Then The Same Was Reduced By 5%. If He Presently Draws Rs.1045.what Was His Original Salary ? |
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Answer» X * (110/100) * (95/100) = 1045 X * (11/10) * (1/100) = 11 X = 1000 X * (110/100) * (95/100) = 1045 X * (11/10) * (1/100) = 11 X = 1000 |
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| 22. |
A Sells His Goods 50% Cheaper Than B But 50% Dearer Than C. The Cheapest Is ? |
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Answer»
A = 50 C * (150/100) = 50 3C = 100 C = 33.3 then 'C' Cheapest Let B = 100 A = 50 C * (150/100) = 50 3C = 100 C = 33.3 then 'C' Cheapest |
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| 23. |
Two Numbers Are Respectively 20% And 25% More Than A Third Number. The Percentage That Is First Of The Second Is ? |
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Answer» I II III 125----------120 100-----------? => 96% I II III 120 125 100 125----------120 100-----------? => 96% |
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| 24. |
How Much 60% Of 50 Is Greater Than 40% Of 30? |
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Answer»
(60/100) * 50 – (40/100) * 30 30 - 12 = 18 |
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| 25. |
If Y Exceeds X By 20%, Then X Is Less Than Y By? |
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Answer»
120------20 X=100 y=120 120------20 100-------? => 16 2/3% |
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| 27. |
Three Candidates Contested An Election And Received 1136, 7636 And 11628 Votes Respectively. What Percentage Of The Total Votes Did The Winning Candidate Get ? |
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Answer» Total number of votes polled = (1136 + 7636 + 11628) = 20400. REQUIRED PERCENTAGE = (11628 / 20400 x 100) % = 57%. Total number of votes polled = (1136 + 7636 + 11628) = 20400. Required percentage = (11628 / 20400 x 100) % = 57%. |
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| 28. |
A Pupil's Marks Were Wrongly Entered As 83 Instead Of 63. Due To That The Average Marks For The Class Got Increased By Half (1/2). The Number Of Pupils In The Class Is ? |
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Answer» LET there be X pupils in the CLASS. Total increase in marks =(x*1/2)=x/2 Therefore x/2=(83-63) =x/2=20 =x=40 Let there be x pupils in the class. Total increase in marks =(x*1/2)=x/2 Therefore x/2=(83-63) =x/2=20 =x=40 |
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| 29. |
A Man Rows To A Place 48 Km Distant And Come Back In 14 Hours. He Finds That He Can Row 4 Km With The Stream In The Same Time As 3 Km Against The Stream. The Rate Of The Stream Is ? |
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Answer» Suppose he MOVE 4 km downstream in x hours. Then, Speed downstream =(4/x)km/hr Speed up STREAM =(3/x)km/hr 48/4/x+48(3/x)=14 or x=1/2 So,speed downstream = 8 km/hr,speed UPSTREAM =6km/hr Rate of the stream = 1/2(8-6)km/hr =1 km/hr. Suppose he move 4 km downstream in x hours. Then, Speed downstream =(4/x)km/hr Speed up stream =(3/x)km/hr 48/4/x+48(3/x)=14 or x=1/2 So,speed downstream = 8 km/hr,speed upstream =6km/hr Rate of the stream = 1/2(8-6)km/hr =1 km/hr. |
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| 30. |
A Train Overtakes Two Persons Who Are Walking In The Same Direction In Which The Train Is Going, At The Rate Of 2 Kmph And 4 Kmph And Passes Them Completely In 9 And 10 Seconds Respectively. The Length Of The Train Is ? |
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Answer» 2 kmph = ( 2 x5 /18)m/sec = 5 /9m/sec. 4 kmph = ( 4 x 518)m/sec = 10 /9m/sec. Let the length of the train be x metres and its SPEED by y m/sec. Then, (x /y-5/9) = 9 and (x/y-10/9 )= 10. 9y-5=xand 10 (9y-10)=9X 9y-x=5 and 90y-9x=100 on solving we get x=50 Length of the train is 50 m 2 kmph = ( 2 x5 /18)m/sec = 5 /9m/sec. 4 kmph = ( 4 x 518)m/sec = 10 /9m/sec. Let the length of the train be x metres and its speed by y m/sec. Then, (x /y-5/9) = 9 and (x/y-10/9 )= 10. 9y-5=xand 10 (9y-10)=9x 9y-x=5 and 90y-9x=100 on solving we get x=50 Length of the train is 50 m |
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