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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. |
The Difference Between A Number And Its Two-fifth Is 510. What Is 10% Of That Number? |
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Answer»
x - 2/5 x = 510 x = (510 * 5)/3 = 850 10% of 850 = 85. Let the number be x. Then, x - 2/5 x = 510 x = (510 * 5)/3 = 850 10% of 850 = 85. |
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| 2. |
Two-fifth Of One-third Of Three-seventh Of A Number Is 15. What Is 40% Of That Number? |
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Answer»
2/5 of 1/3 of 3/7 of x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2 40% of 525/2 = (40/100 * 525/2) = 105. Let the number be x. Then, 2/5 of 1/3 of 3/7 of x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2 40% of 525/2 = (40/100 * 525/2) = 105. |
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| 3. |
If 20% Of A Number, Then 120% Of That Number Will Be? |
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Answer» Let the NUMBER x. Then, 20% of x = 120 x = (120 * 100)/20 = 600 120% of x = (120/100 * 600) = 720. Let the number x. Then, 20% of x = 120 x = (120 * 100)/20 = 600 120% of x = (120/100 * 600) = 720. |
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| 4. |
The Income Of A Broker Remains Uncharged Though The Rate Of Commission Is Increased From 4 % To 5 % The Percentage Of A Slump In Business Is? |
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Answer» Let the business value changes from X to y. Then 4 % of x = 5 % of y of 4/100 × x = 5/100 × y or y = 4/5 x. Change in business = (x - 4/5 x) = 4/5 x. PERCENTAGE SLUMP in business = (1/5 x ×1/x × 100)% = 20%. Let the business value changes from x to y. Then 4 % of x = 5 % of y of 4/100 × x = 5/100 × y or y = 4/5 x. Change in business = (x - 4/5 x) = 4/5 x. Percentage slump in business = (1/5 x ×1/x × 100)% = 20%. |
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| 5. |
The Price Of An Article Is Cut By 10%. To Restore It To The Former Value, The New Price Must Be Increased By? |
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Answer» Let the PRICE of the article = Rs.100 Therefore the new price must be INCREASED by(100−90)×100/90=100/9%=11 1/9%. Let the price of the article = Rs.100 New price = 100 - 10 = 90 Therefore the new price must be increased by(100−90)×100/90=100/9%=11 1/9%. |
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| 6. |
If 35% Of A Number Is 12 Less Than 50% Of That Number, Then The Number Is? |
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Answer» Let the number be x. Then, 50% of x - 35% of x = 12 50/100 x - 35/100 x = 12 x = (12 * 100)/15 = 80. Let the number be x. Then, 50% of x - 35% of x = 12 50/100 x - 35/100 x = 12 x = (12 * 100)/15 = 80. |
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| 7. |
A Man’s Basic Pay For A 40 Hour Week Is Rs. 20 Overtime Is Paid For At 25 % Above The Basic Rate, In A Certain Week He Worked Overtime And His Total Wage Was Rs. 25. He Therefore Worked For A Total Of? |
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Answer» BASIC rate PER hour = Re. (20/40) = Re. 1/2. Overtime per hour = 125 % of Re. 1/2 = Re.5/8. He worked x hours overtime. Then 20 + 5/8 x = 25 (or) = 5/8 x = 5 X = 40/5 = 8 hours => he worked in all for (40 + 8) = 48 hours. Basic rate per hour = Re. (20/40) = Re. 1/2. Overtime per hour = 125 % of Re. 1/2 = Re.5/8. He worked x hours overtime. Then 20 + 5/8 x = 25 (or) = 5/8 x = 5 X = 40/5 = 8 hours => he worked in all for (40 + 8) = 48 hours. |
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| 8. |
On Decreasing The Price Of T.v. Sets By 30 % Its Sale Is Increased By 20 %. What Is The Effect On The Revenue Received By The Shopkeeper? |
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Answer» Let price = Rs. 100, SALE = 100 Then Sale Value = Rs. (100 × 100) = Rs. 10000 NEW Sale Value = Rs. (70 × 120) = Rs. 8400 DECREASE % = (1000/10000 x 100) % = 16 %. Let price = Rs. 100, Sale = 100 Then Sale Value = Rs. (100 × 100) = Rs. 10000 New Sale Value = Rs. (70 × 120) = Rs. 8400 Decrease % = (1000/10000 x 100) % = 16 %. |
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| 9. |
Water Tax Is Increased By 20% But Its Consumption Is Decreased By 20 %. Then, The Increase Or Decrease In The Expenditure Of The Money Is? |
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Answer» Let tax = Rs. 100 and consumption = 100 UNITS ORIGINAL Expenditure = Rs. (100 × 100) = Rs. 10000 New Expenditure = Rs. (120 × 80) = Rs. 9600 Decrease in expenditure = (400/10000 X 100) % = 4 %. Let tax = Rs. 100 and consumption = 100 units Original Expenditure = Rs. (100 × 100) = Rs. 10000 New Expenditure = Rs. (120 × 80) = Rs. 9600 Decrease in expenditure = (400/10000 x 100) % = 4 %. |
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| 10. |
The Length Of A Rectangle Is Increased By 60 %. By What Percent Would The Width Have To Be Decreased To Maintain The Same Area? |
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Answer» Let length = 100 m. BREATH = 100 m. New length = 160 m. New breath = x METERS Then, = 160 × x = 100 × 100 (or) X = (100 ×100)/160 × 125/2 Decrease in breadth = (100- 125/2) % = 37 ½ %. Let length = 100 m. Breath = 100 m. New length = 160 m. New breath = x meters Then, = 160 × x = 100 × 100 (or) X = (100 ×100)/160 × 125/2 Decrease in breadth = (100- 125/2) % = 37 ½ %. |
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| 11. |
The Price Of An Article Has Been Reduced By 25 %. In Order To Restore The Original Price, The New Price Must Be Increased By? |
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Answer»
Reduced Price = Rs. 75. INCREASE on Rs. 75 = Rs. 25 Increase on 100 = (25/75 x 100) % = 33 1/3 %. Let original price = Rs. 100. Reduced Price = Rs. 75. Increase on Rs. 75 = Rs. 25 Increase on 100 = (25/75 x 100) % = 33 1/3 %. |
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| 12. |
Sameer Spends 40 % Of His Salary On Food Article, And 1/3 Rd Of The Remaining On Transport. If He Saves Rs. 450 Per Month Which Is Half Of The Balance After Spending On Food Items And Transport. What Is This Monthly Salary? |
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Answer» Suppose, salary = RS. 100 EXPENDITURE on food = Rs. 40 Balance = Rs. 60 Expenditure on transport = 1/3 × Rs. 60 = Rs. 20 Now, balance = Rs. 40 Saving = Rs. 20 If saving is = Rs. 20 Salary = Rs. 100 If saving is Rs. 450, salary = Rs (100/20 X 450) = Rs. 2250. Suppose, salary = Rs. 100 Expenditure on food = Rs. 40 Balance = Rs. 60 Expenditure on transport = 1/3 × Rs. 60 = Rs. 20 Now, balance = Rs. 40 Saving = Rs. 20 If saving is = Rs. 20 Salary = Rs. 100 If saving is Rs. 450, salary = Rs (100/20 x 450) = Rs. 2250. |
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| 13. |
Pushpa Is Twice As Old As Rita Was Two Years Ago If The Difference Between Their Ages Is 2 Years, How Old Is Pushpa Today? |
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Answer» Rita’s age 2 years ago be x years PUSHPA’s PRESENT age =(2x) years 2x –(x +2) =2 => x =4 Therefore pushpa’s present age = 8 years. Rita’s age 2 years ago be x years Pushpa’s present age =(2x) years 2x –(x +2) =2 => x =4 Therefore pushpa’s present age = 8 years. |
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| 14. |
The Sum Of The Ages Of A Son And Father Is 56 Years After Four Years The Age Of The Father Will Be Three Times That Of The Son. Their Ages Respectively Are? |
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Answer» Present ages of SON and FATHER be X years (56 -x)years (56- x +4) = 3(x + 4) or 4X =48 or x = 12 Ages are 12 years, 44 years. Present ages of son and father be x years (56 -x)years (56- x +4) = 3(x + 4) or 4x =48 or x = 12 Ages are 12 years, 44 years. |
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| 15. |
Ayesha's Father Was 38 Years Of Age When She Was Born While Her Mother Was 36 Years Old When Her Brother Four Years Younger To Her Was Born. What Is The Difference Between The Ages Of Her Parents? |
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Answer» Mother's AGE when AYESHA's brother was born = 36 years. Father's age when Ayesha's brother was born = (38 + 4) = 42 years. Required difference = (42 - 36) = 6 years. Mother's age when Ayesha's brother was born = 36 years. Father's age when Ayesha's brother was born = (38 + 4) = 42 years. Required difference = (42 - 36) = 6 years. |
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| 16. |
The Sum Of The Ages Of 5 Children Born At The Intervals Of 3 Years Each Is 50 Years. What Is The Age Of The Youngest Child? |
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Answer» Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) YEARS. Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 Age of youngest child = x = 4 years. Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) years. Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 5x = 20 => x = 4. Age of youngest child = x = 4 years. |
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| 17. |
Rajan Got Married 8 Years Ago. His Present Age Is 6/5 Times His Age At The Time Of His Marriage. Rajan's Sister Was 10 Years Younger To Him At The Time Of His Marriage. The Age Of Rajan's Sister Is? |
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Answer» Let Rajan's present age be x years. Then, his age at the time of MARRIAGE = (x - 8) years. x = 6/5 (x - 8) 5X = 6x - 48 => x = 48 Rajan's SISTER's age at the time of his marriage = (x - 8) - 10 = 30 years. Rajan's sister's present age = (30 + 8) = 38 years. Let Rajan's present age be x years. Then, his age at the time of marriage = (x - 8) years. x = 6/5 (x - 8) 5x = 6x - 48 => x = 48 Rajan's sister's age at the time of his marriage = (x - 8) - 10 = 30 years. Rajan's sister's present age = (30 + 8) = 38 years. |
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| 18. |
If The Length, Breadth And The Height Of A Cuboid Are In The Ratio 6: 5: 4 And If The Total Surface Area Is 33300 Cm2, Then Length Breadth And Height In Cms, Are Respectively? |
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Answer» Length = 6X Breadth = 5x Height = 4x in CM Therefore, 2(6x × 5x + 5x × 4x + 6x × 4x) = 33300 148x2 = 33300 => x2 = 33300/148 = 225 => X = 15. Therefore, Length = 90cm, Breadth = 75CM, Height = 60CM 90, 75 , 60 cm. Length = 6x Breadth = 5x Height = 4x in cm Therefore, 2(6x × 5x + 5x × 4x + 6x × 4x) = 33300 148x2 = 33300 => x2 = 33300/148 = 225 => x = 15. Therefore, Length = 90cm, Breadth = 75cm, Height = 60cm 90, 75 , 60 cm. |
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| 19. |
The Surface Area Of A Cube Is 726m2, Its Volume Is? |
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Answer» 6a2= 726 => A2 = 121 => a = 11 CM = (11 × 11 × 11) cm3 = 1331 cm3. 6a2= 726 => a2 = 121 => a = 11 Cm Therefore, Volume of the cube = (11 × 11 × 11) cm3 = 1331 cm3. |
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| 20. |
9 3/4 + 7 2/17 - 9 1/15 = ? |
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Answer» Given sum = 9 + 3/4 + 7 + 2/17 - (9 + 1/15) = (9 + 7 - 9) + (3/4 + 2/17 - 1/15) = 7 + (765 + 120 - 68)/1020 = 7 817/1020. Given sum = 9 + 3/4 + 7 + 2/17 - (9 + 1/15) = (9 + 7 - 9) + (3/4 + 2/17 - 1/15) = 7 + (765 + 120 - 68)/1020 = 7 817/1020. |
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| 21. |
Which One Of The Following Numbers Is Exactly Divisible By 11? |
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Answer» (4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11. (2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11. (4 + 6 + 1) - (2 + 5 + 3) = 1, not divisible by 11. (4 + 6 + 1) - (2 + 5 + 4) = 0. So, 415624 is divisible by 11. (4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11. (2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11. (4 + 6 + 1) - (2 + 5 + 3) = 1, not divisible by 11. (4 + 6 + 1) - (2 + 5 + 4) = 0. So, 415624 is divisible by 11. |
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| 22. |
How Many Of The Following Numbers Are Divisible By 132? |
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Answer» 264, 396, 462, 792, 968, 2178, 5184, 6336 132 = 11 * 3 * 4 CLEARLY, 968 is not divisible by 3 NONE of 462 and 2178 is divisible by 4 And, 5284 is not divisible by 11 Each one of the REMAINING four numbers is divisible by each one of 4, 3 and 11. So, there are 4 such numbers. 264, 396, 462, 792, 968, 2178, 5184, 6336 132 = 11 * 3 * 4 Clearly, 968 is not divisible by 3 None of 462 and 2178 is divisible by 4 And, 5284 is not divisible by 11 Each one of the remaining four numbers is divisible by each one of 4, 3 and 11. So, there are 4 such numbers. |
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| 23. |
The Sum Of The Two Numbers Is 12 And Their Product Is 35. What Is The Sum Of The Reciprocals Of These Numbers? |
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Answer» LET the numbers be a and B. Then, a + b = 12 and ab = 35. (a + b)/ab = 12/35 = (1/b + 1/a) = 12/35 SUM of reciprocals of given numbers = 12/35. Let the numbers be a and b. Then, a + b = 12 and ab = 35. (a + b)/ab = 12/35 = (1/b + 1/a) = 12/35 Sum of reciprocals of given numbers = 12/35. |
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| 24. |
A Drink Vendor Has 80 Liters Of Maaza, 144 Liters Of Pepsi And 368 Liters Of Sprite. He Wants To Pack Them In Cans, So That Each Can Contains The Same Number Of Liters Of A Drink, And Doesn't Want To Mix Any Two Drinks In A Can. What Is The Least Number Of Cans Required? |
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Answer» The NUMBER of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of CANS of Maaza = 80/16 = 5 Number of cans of PEPSI = 144/16 = 9 Number of cans of Sprite = 368/16 = 23 The total number of cans required = 5 + 9 + 23 = 37 cans. The number of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of cans of Maaza = 80/16 = 5 Number of cans of Pepsi = 144/16 = 9 Number of cans of Sprite = 368/16 = 23 The total number of cans required = 5 + 9 + 23 = 37 cans. |
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| 25. |
A Room Is 6 Meters 24 Centimeters In Length And 4 Meters 32 Centimeters In Width. Find The Least Number Of Square Tiles Of Equal Size Required To Cover The Entire Floor Of The Room? |
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Answer» Length = 6 m 24 cm = 624 cm HCF of 624 and 432 = 48 Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117. Length = 6 m 24 cm = 624 cm Width = 4 m 32 cm = 432 cm HCF of 624 and 432 = 48 Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117. |
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| 26. |
Find The Greatest Number That, While Dividing 47, 215 And 365, Gives The Same Remainder In Each Case? |
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Answer» Calculate the differences, taking TWO numbers at a time as follows: (365-215) = 150 (365-47) = 318 HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 GIVES the same. emainder in each cases is 5. Calculate the differences, taking two numbers at a time as follows: (215-47) = 168 (365-215) = 150 (365-47) = 318 HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same. emainder in each cases is 5. |
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| 27. |
Find The Lowest 4-digit Number Which When Divided By 3, 4 Or 5 Leaves A Remainder Of 2 In Each Case? |
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Answer» LOWEST 4-digit number is 1000. LCM of 3, 4 and 5 is 60. Dividing 1000 by 60, we get the REMAINDER 40. Thus, the lowest 4-digit number that exactly DIVISIBLE by 3, 4 and 5 is 1000 + (60 - 40) = 1020. Now, add the remainder 2 that's required. Thus, the answer is 1022. Lowest 4-digit number is 1000. LCM of 3, 4 and 5 is 60. Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020. Now, add the remainder 2 that's required. Thus, the answer is 1022. |
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| 28. |
The Wheels Revolve Round A Common Horizontal Axis. They Make 15, 20 And 48 Revolutions In A Minute Respectively. Starting With A Certain Point On The Circumference Down Wards. After What Interval Of Time Will They Come Together In The Same Position? |
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Answer» Time for one revolution = 60/15 = 4 60/ 20 = 3 60/48 = 5/4 LCM of 4, 3, 5/4 LCM of Numerators/HCF of DENOMINATORS = 60/1 = 60. Time for one revolution = 60/15 = 4 60/ 20 = 3 60/48 = 5/4 LCM of 4, 3, 5/4 LCM of Numerators/HCF of Denominators = 60/1 = 60. |
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| 29. |
How Many Meters Of Carpet 63cm Will Be Required To Be A Floor Of A Room 14m By 9cm? |
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Answer» WIDTH of the carpet = 63/100 m let it’s length be X METRES Then x × 63/100 = 14 x 9 => x =(14 x 9 x 100/63) = 200m Length = 200m. Width of the carpet = 63/100 m let it’s length be x metres Then x × 63/100 = 14 x 9 => x =(14 x 9 x 100/63) = 200m Length = 200m. |
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| 30. |
The Length Of A Rectangular Increased By 10% And It;s Breadth Is Decreased By 10 %. Then The Area Of The New Rectangle Is? |
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Answer» LENGTH of the be let ‘l’ units and breadth be b units Area =LB sq. Units Now length =(110/100 l) = 11l/10 NEW breadth =(90/100 b) = 96/10 Now area (11l/10 x 96/10)Sq.units = (99/100 lb)sq.unit Area decreased = (lb -99/100 lb)sq.units =lb/100 sq.Units Percentage decreased =(lb/100 x 1/lb x 100)% = 1%. Length of the be let ‘l’ units and breadth be b units Area =lb sq. Units Now length =(110/100 l) = 11l/10 New breadth =(90/100 b) = 96/10 Now area (11l/10 x 96/10)Sq.units = (99/100 lb)sq.unit Area decreased = (lb -99/100 lb)sq.units =lb/100 sq.Units Percentage decreased =(lb/100 x 1/lb x 100)% = 1%. |
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| 31. |
The Area Of A Rectangle Is 12 Sq.metres And It’s Length Is 3 Times That Of It’s Breadth. What Is The Perimeter Of The Rectangle? |
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Answer» LET the breadth be x metres , then it’s LENGTH = 3X metres =>3x × x =12 => x2= 4 => x =2 l=6m, b =2m => PERIMETER = 2(6+2)m = 16m. Let the breadth be x metres , then it’s length = 3x metres =>3x × x =12 => x2= 4 => x =2 l=6m, b =2m => Perimeter = 2(6+2)m = 16m. |
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| 32. |
The Perimeter Of A Rectangle Is 60m. If It’s Length Is Twice Its Breadth Then Its Area Is? |
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Answer» Let the breadth be x metres then length = 2x metres => 6x =60 => x =10 l = 20m and b =10m => AREA =(20 x 10) Sq.m = 200 Sq.m. Let the breadth be x metres then length = 2x metres => 2(2x+x) = 60 => 6x =60 => x =10 l = 20m and b =10m => Area =(20 x 10) Sq.m = 200 Sq.m. |
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| 33. |
If The Length Of A Rectangle Is Increased By 20% And It’s Breadth Is Decreased By 20% Then It Area? |
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Answer» LET length = X METRES and breadth = y metres, then area =(XY) New length =(120/100x)m = 6x/5 m New breadth =(80/100 y)m New area =(6x/5 x 4y/5)m2 = (24/25 xy)m2 Decreased in area =(xy – 24/25 xy)m2= xy/25 m2 Decreased % =(xy/25 x 1/xy x 100)% = 4%. Let length = x metres and breadth = y metres, then area =(xy) New length =(120/100x)m = 6x/5 m New breadth =(80/100 y)m New area =(6x/5 x 4y/5)m2 = (24/25 xy)m2 Decreased in area =(xy – 24/25 xy)m2= xy/25 m2 Decreased % =(xy/25 x 1/xy x 100)% = 4%. |
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| 34. |
The Length Of A Rectangular Plot Is Increased By 25%. To Keep It’s Area Uncharged The Width Of The Plot Should Be? |
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Answer» LET the length be X metres and BREADTH be y metres Then it’s area =(xy)m2 New length =(125/100 x)m =(5x/4)m. Let the new breadth be 2 metres Then xy 5x/4 x 2 => 2 = 4/5 y Decreased in WIDTH =(y -4/5 y) = y/5 metres Decreased % in width =(y/5 x 1/y x 100)% = 20%. Let the length be x metres and breadth be y metres Then it’s area =(xy)m2 New length =(125/100 x)m =(5x/4)m. Let the new breadth be 2 metres Then xy 5x/4 x 2 => 2 = 4/5 y Decreased in width =(y -4/5 y) = y/5 metres Decreased % in width =(y/5 x 1/y x 100)% = 20%. |
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| 35. |
The Speed Of A Motor Boat Is That Of The Current Of Water As 36:5. The Boat Goes Along With The Current In 5 Hours 10 Min, It Will Come Back In? |
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Answer» LET the speed of motor boat be 36X KM/hr and that of current of WATER be 5x km/hr Speed downstream = (36x +5x)km/hr = 41x km/hr Speed upstream = (36 -5x)km/hr = 31x km/hr Distance covered downstream = (41x × 31/6)km Distance upstream = [1271x/6 × 1/31x]hrs = 41/6 hrs = 6hrs 50 min. Let the speed of motor boat be 36x km/hr and that of current of water be 5x km/hr Speed downstream = (36x +5x)km/hr = 41x km/hr Speed upstream = (36 -5x)km/hr = 31x km/hr Distance covered downstream = (41x × 31/6)km Distance upstream = [1271x/6 × 1/31x]hrs = 41/6 hrs = 6hrs 50 min. |
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| 36. |
A Boat Covers A Distance Of 30 Km In 2 ½ Hours Running Downstream While Returning It Covers The Same Distance In 3 ¾ Hours. What Is The Speed Of The Boat In Still Water? |
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Answer» SPEED downstream = (30 x 2/5)km/hr = 12km/hr Speed upstream = (30 x 4/15) km/hr = 8 km/hr Speed of BOAT in STILL water = ½ (12 + 8)km/hr = 10 km/hr. Speed downstream = (30 x 2/5)km/hr = 12km/hr Speed upstream = (30 x 4/15) km/hr = 8 km/hr Speed of boat in still water = ½ (12 + 8)km/hr = 10 km/hr. |
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| 37. |
A Steamer Goes Downstream From One Port To Another In 4 Hours, It Covers The Same Distance Upstream In 5 Hours. If The Speed Of The Stream Is 2km/hr,the Distance Between The Two Parts Is? |
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Answer» Let the distance between the two parts be x KM. Then speed DOWNSTREAM = x/4 km/hr. Speed Upstream = x/5 km/hr Speed of the stream = ½ (x/4 –x/5) THEREFORE, ½(x/4 –x/5) = 2. => x/4 –x/5 = 4 => x = 80. Hence, the distance between the two ports is 80km. Let the distance between the two parts be x km. Then speed downstream = x/4 km/hr. Speed Upstream = x/5 km/hr Speed of the stream = ½ (x/4 –x/5) Therefore, ½(x/4 –x/5) = 2. => x/4 –x/5 = 4 => x = 80. Hence, the distance between the two ports is 80km. |
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| 38. |
A Boat Goes 24 Km Downstream In 10 Hours, It Takes 2 Hours More To Cover The Same Distance Against The Stream. What Is The Speed Of The Boat In Still Water? |
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Answer» Speed DOWNSTREAM = 24/10 km/hr = 2.4 km/hr Speed upstream = 24/12 km/hr = 2 km/hr Speed of the BOAT in still WATER = ½ (2.4 +2)km/hr = 2.2 km/hr. Speed downstream = 24/10 km/hr = 2.4 km/hr Speed upstream = 24/12 km/hr = 2 km/hr Speed of the boat in still water = ½ (2.4 +2)km/hr = 2.2 km/hr. |
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| 39. |
A And B Start A Business, With A Investing The Total Capital Of Rs.50000, On The Condition That B Pays A Interest @ 10% Per Annum On His Half Of The Capital. A Is A Working Partner And Receives Rs.1500 Per Month From The Total Profit And Any Profit Remaining Is Equally Shared By Both Of Them. At The End Of The Year, It Was Found That The Income Of A Is Twice That Of B. Find The Total Profit For The Year? |
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Answer» Interest received by A from B = 10% of half of Rs.50000 = 10% * 25000 = 2500. Amount received by A per annum for being a working partner = 1500 * 12 = Rs.18000. LET 'P' be the part of the remaining profit that A receives as his SHARE. TOTAL income of A = (2500 + 18000 + P) Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally. Income of A = Twice the income of B (2500 + 18000 + P) = 2(P) P = 20500 Total profit = 2P + 18000 = 2*20500 + 18000 = 59000. Interest received by A from B = 10% of half of Rs.50000 = 10% * 25000 = 2500. Amount received by A per annum for being a working partner = 1500 * 12 = Rs.18000. Let 'P' be the part of the remaining profit that A receives as his share. Total income of A = (2500 + 18000 + P) Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally. Income of A = Twice the income of B (2500 + 18000 + P) = 2(P) P = 20500 Total profit = 2P + 18000 = 2*20500 + 18000 = 59000. |
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| 40. |
A, B And C Started A Business With Capitals Of Rs. 8000, Rs. 10000 And Rs. 12000 Respectively. At The End Of The Year, The Profit Share Of B Is Rs. 1500. The Difference Between The Profit Shares Of A And C Is? |
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Answer» Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6 And also given that, PROFIT share of B is Rs. 1500 => 5 PARTS out of 15 parts is Rs. 1500 Now, required DIFFERENCE is 6 - 4 = 2 parts Required difference = 2/5 (1500) = Rs. 600. Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6 And also given that, profit share of B is Rs. 1500 => 5 parts out of 15 parts is Rs. 1500 Now, required difference is 6 - 4 = 2 parts Required difference = 2/5 (1500) = Rs. 600. |
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| 41. |
A Started A Business With An Investment Of Rs. 70000 And After 6 Months B Joined Him Investing Rs. 120000. If The Profit At The End Of A Year Is Rs. 52000, Then The Share Of B Is? |
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Answer» Ratio of INVESTMENTS of A and B is (70000 * 12) : (120000 * 6) = 7 : 6 SHARE of B = 6/13 (52000) = Rs. 24000. Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6 Total profit = Rs. 52000 Share of B = 6/13 (52000) = Rs. 24000. |
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| 42. |
P And Q Started A Business With Respective Investments Of Rs. 4 Lakhs And Rs. 10 Lakhs. As P Runs The Business, His Salary Is Rs. 5000 Per Month. If They Earned A Profit Of Rs. 2 Lakhs At The End Of The Year, Then Find The Ratio Of Their Earnings? |
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Answer» Ratio of investments of P and Q is 2 : 5 Total salary claimed by P = 12 * 5000 = Rs. 60000 Total profit = Rs. 2 lakhs. Profit is to be shared = Rs. 140000 Share of P = (2/7) * 140000 = Rs. 400000 Share of Q = Rs. 100000 Total earnings of P = (60000 + 40000) = Rs. 100000 Ratio of their earnings = 1 : 1. Ratio of investments of P and Q is 2 : 5 Total salary claimed by P = 12 * 5000 = Rs. 60000 Total profit = Rs. 2 lakhs. Profit is to be shared = Rs. 140000 Share of P = (2/7) * 140000 = Rs. 400000 Share of Q = Rs. 100000 Total earnings of P = (60000 + 40000) = Rs. 100000 Ratio of their earnings = 1 : 1. |
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| 43. |
A Is A Working Partner And B Is A Sleeping Partner In The Business. A Puts In Rs.15000 And B Rs.25000, A Receives 10% Of The Profit For Managing The Business The Rest Being Divided In Proportion Of Their Capitals. Out Of A Total Profit Of Rs.9600, Money Received By A Is? |
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Answer» 15:25 => 3:5 9600*10/100 = 960 9600 - 960 = 8640 8640*3/8 = 3240 + 960 = 4200. 15:25 => 3:5 9600*10/100 = 960 9600 - 960 = 8640 8640*3/8 = 3240 + 960 = 4200. |
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| 44. |
A And B Invests Rs.10000 Each, A Investing For 8 Months And B Investing For All The 12 Months In The Year. If The Total Profit At The End Of The Year Is Rs.25000, Find Their Shares? |
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Answer» The RATIO of their PROFITS A:B = 8:12 = 2:3 Share of A in the total PROFIT = 2/5 * 25000 = Rs.10000 Share of A in the total profit = 3/5 * 25000 = Rs.15000. The ratio of their profits A:B = 8:12 = 2:3 Share of A in the total profit = 2/5 * 25000 = Rs.10000 Share of A in the total profit = 3/5 * 25000 = Rs.15000. |
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| 45. |
A And B Starts A Business With Rs.8000 Each, And After 4 Months, B Withdraws Half Of His Capital How Should They Share The Profits At The End Of The 18 Months? |
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Answer» A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for REMAINING 14 months is Rs.4000 only. A : B 8000*18 : (8000*4) + (4000*14) 14400 : 88000 A:B = 18:11. A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only. A : B 8000*18 : (8000*4) + (4000*14) 14400 : 88000 A:B = 18:11. |
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| 46. |
A Train 125 M Long Passes A Man, Running At 5 Km/hr In The Same Direction In Which The Train Is Going, In 10 Sec. The Speed Of The Train Is? |
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Answer» SPEED of the train RELATIVE to man = 125/10 = 25/2 m/sec. = 25/2 * 18/5 = 45 km/hr LET the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 => x = 50 km/hr. Speed of the train relative to man = 125/10 = 25/2 m/sec. = 25/2 * 18/5 = 45 km/hr Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 => x = 50 km/hr. |
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| 47. |
Two Trains Are Moving In Opposite Directions At 60 Km/hr And 90 Km/hr. Their Lengths Are 1.10 Km And 0.9 Km Respectively. The Time Taken By The Slower Train To Cross The Faster Train In Seconds Is? |
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Answer» Relative speed = 60 + 90 = 150 km/hr. = 150 * 5/18 = 125/3 m/sec. Distance COVERED = 1.10 + 0.9 = 2 km = 2000 m. Required TIME = 2000 * 3/125 = 48 sec. Relative speed = 60 + 90 = 150 km/hr. = 150 * 5/18 = 125/3 m/sec. Distance covered = 1.10 + 0.9 = 2 km = 2000 m. Required time = 2000 * 3/125 = 48 sec. |
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| 48. |
Two Trains 140 M And 160 M Long Run At The Speed Of 60 Km/hr And 40 Km/hr Respectively In Opposite Directions On Parallel Tracks. The Time Which They Take To Cross Each Other Is? |
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Answer» RELATIVE speed = 60 + 40 = 100 km/hr. = 100 * 5/18 = 250/9 m/sec. Distance COVERED in crossing each other = 140 + 160 = 300 m. REQUIRED time = 300 * 9/250 = 54/5 = 10.8 sec. Relative speed = 60 + 40 = 100 km/hr. = 100 * 5/18 = 250/9 m/sec. Distance covered in crossing each other = 140 + 160 = 300 m. Required time = 300 * 9/250 = 54/5 = 10.8 sec. |
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| 49. |
A Train 110 M Long Is Running With A Speed Of 60 Km/hr. In What Time Will It Pass A Man Who Is Running At 6 Km/hr In The Direction Opposite To That In Which The Train Is Going? |
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Answer» Speed of TRAIN RELATIVE to man = 60 + 6 = 66 km/hr. = 66 * 5/18 = 55/3 m/SEC. Time taken to pass the men = 110 * 3/55 = 6 sec. Speed of train relative to man = 60 + 6 = 66 km/hr. = 66 * 5/18 = 55/3 m/sec. Time taken to pass the men = 110 * 3/55 = 6 sec. |
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| 50. |
A Jogger Running At 9 Km/hr Along Side A Railway Track Is 240 M Ahead Of The Engine Of A 120 M Long Train Running At 45 Km/hr In The Same Direction. In How Much Time Will The Train Pass The Jogger? |
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Answer» Speed of train RELATIVE to jogger = 45 - 9 = 36 km/hr. = 36 * 5/18 = 10 m/sec. Distance to be COVERED = 240 + 120 = 360 m. Time taken = 360/10 = 36 sec. Speed of train relative to jogger = 45 - 9 = 36 km/hr. = 36 * 5/18 = 10 m/sec. Distance to be covered = 240 + 120 = 360 m. Time taken = 360/10 = 36 sec. |
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