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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. |
If Two Dice Are Thrown Together, The Probability Of Getting An Even Number On One Die And An Odd Number On The Other Is -. |
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Answer» The number of exhaustive OUTCOMES is 36. Let E be the event of getting an EVEN number on one die and an ODD number on the other. let the event of getting either both even or both odd then = 18/36 = 1/2 P(E) = 1 - 1/2 = 1/2. The number of exhaustive outcomes is 36. Let E be the event of getting an even number on one die and an odd number on the other. let the event of getting either both even or both odd then = 18/36 = 1/2 P(E) = 1 - 1/2 = 1/2. |
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| 2. |
The Dimensions Of A Room Are 25 Feet * 15 Feet * 12 Feet. What Is The Cost Of White Washing The Four Walls Of The Room At Rs. 5 Per Square Feet If There Is One Door Of Dimensions 6 Feet * 3 Feet And Three Windows Of Dimensions 4 Feet * 3 Feet Each? |
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Answer» Area of the four walls = 2h(L + b) Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft. Total cost = 906 * 5 = Rs. 4530. Area of the four walls = 2h(l + b) Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft. Total cost = 906 * 5 = Rs. 4530. |
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| 3. |
A Wire In The Form Of A Circle Of Radius 3.5 M Is Bent In The Form Of A Rectangule, Whose Length And Breadth Are In The Ratio Of 6 : 5. What Is The Area Of The Rectangle? |
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Answer» The circumference of the CIRCLE is equal to the permeter of the RECTANGLE. Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5 => x = 1 Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 CM2. The circumference of the circle is equal to the permeter of the rectangle. Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5 => x = 1 Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm2. |
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| 4. |
If The Sides Of A Triangle Are 26 Cm, 24 Cm And 10 Cm, What Is Its Area? |
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Answer» The triangle with SIDES 26 cm, 24 cm and 10 cm is right ANGLED, where the hypotenuse is 26 cm. Area of the triangle = 1/2 * 24 * 10 = 120 cm2. The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm. Area of the triangle = 1/2 * 24 * 10 = 120 cm2. |
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| 5. |
Find The Roots Of The Quadratic Equation: 2x2 + 3x - 9 = 0? |
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Answer» 2X2 + 6x - 3x - 9 = 0 (x + 3)(2x - 3) = 0 => x = -3 or x = 3/2. 2x2 + 6x - 3x - 9 = 0 2x(x + 3) - 3(x + 3) = 0 (x + 3)(2x - 3) = 0 => x = -3 or x = 3/2. |
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| 6. |
3/20 Is What Percent Of 12/25? |
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Answer» Let the REQUIRED percentage be X%. 3/20 = x% of 12/25 => 3/20 = x/100 * 12/25 => 12X = (300 * 25)/20 => x = (25 * 25)/20 => x = 31.25%. Let the required percentage be x%. 3/20 = x% of 12/25 => 3/20 = x/100 * 12/25 => 12x = (300 * 25)/20 => x = (25 * 25)/20 => x = 31.25%. |
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| 7. |
64 Is What Percent Of 80? |
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Answer»
80 * x/100 = 64 => x = (64 * 100)/80 => x = 80. 80% of 80 is 64. Let x percent of 80 be 64. 80 * x/100 = 64 => x = (64 * 100)/80 => x = 80. 80% of 80 is 64. |
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| 8. |
Find The 37.5% Of 976 = |
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Answer» 37.5 % of 976 = 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976 37.5 % of 976 = 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976 = 3 * 122 = 366. |
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| 9. |
There Were Two Candidates In An Election. Winner Candidate Received 62% Of Votes And Won The Election By 288 Votes. Find The Number Of Votes Casted To The Winning Candidate? |
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Answer»
62% - 38% = 24% 24% -------- 288 62% -------- ? => 744. W = 62% L = 38% 62% - 38% = 24% 24% -------- 288 62% -------- ? => 744. |
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| 10. |
The Tax On A Commodity Is Diminished By 20% But Its Consumption Is Increased By 10%. Find The Decrease Percent In The Revenue Derived From It? |
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Answer»
80 * 110 = 8800 10000------- 1200 100 ------- ? = 12%. 100 * 100 = 10000 80 * 110 = 8800 10000------- 1200 100 ------- ? = 12%. |
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| 11. |
The Sum Of The Present Ages Of Two Persons A And B Is 60. If The Age Of A Is Twice That Of B, Find The Sum Of Their Ages 5 Years Hence? |
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Answer»
2B + B = 60 => B = 20 then A = 40. 5 years, their ages will be 45 and 25. Sum of their ages = 45 + 25 = 70. A + B = 60, A = 2B 2B + B = 60 => B = 20 then A = 40. 5 years, their ages will be 45 and 25. Sum of their ages = 45 + 25 = 70. |
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| 12. |
The Sum Of The Two Digits Of A Number Is 10. If The Number Is Subtracted From The Number Obtained By Reversing Its Digits, The Result Is 54. Find The Number? |
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Answer» Any TWO digit number can be written as (10P + Q), where P is the digit in the TENS place and Q is the digit in the units place. P + Q = 10 ----- (1) (10Q + P) - (10P + Q) = 54 9(Q - P) = 54 (Q - P) = 6 ----- (2) SOLVE (1) and (2) P = 2 and Q = 8 The required number is = 28 Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place. P + Q = 10 ----- (1) (10Q + P) - (10P + Q) = 54 9(Q - P) = 54 (Q - P) = 6 ----- (2) Solve (1) and (2) P = 2 and Q = 8 The required number is = 28 |
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| 13. |
The Sum Of Three Consecutive Integers Is 102. Find The Lowest Of The Three? |
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Answer» THREE consecutive NUMBERS can be taken as (P - 1), P, (P + 1). So, (P - 1) + P + (P + 1) = 102 The lowest of the three = (P - 1) = 34 - 1 = 33. Three consecutive numbers can be taken as (P - 1), P, (P + 1). So, (P - 1) + P + (P + 1) = 102 3P = 102 => P = 34. The lowest of the three = (P - 1) = 34 - 1 = 33. |
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| 14. |
The Radius Of A Cylindrical Vessel Is 7cm And Height Is 3cm. Find The Whole Surface Of The Cylinder? |
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Answer» r = 7 h = 3 2πr(h + r) = 2 * 22/7 * 7(10) = 440. r = 7 h = 3 2πr(h + r) = 2 * 22/7 * 7(10) = 440. |
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| 15. |
The Sum Of Five Consecutive Odd Numbers Of Set P Is 435. What Is The Sum Of Five Consecutive Numbers Of Another Set Q. Whose Largest Number Is 45 More Than The Largest Number Of Set P? |
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Answer» Let the five consecutive ODD numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5. Sum of these five numbers = 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5 = 10n + 5 = 435 => n = 43 Largest number of set p = 2(43) + 5 = 91 The largest number of set Q = 91 + 45 = 136 => The five numbers of set q are 132, 133, 134, 135, 136. Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670. Let the five consecutive odd numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5. Sum of these five numbers = 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5 = 10n + 5 = 435 => n = 43 Largest number of set p = 2(43) + 5 = 91 The largest number of set q = 91 + 45 = 136 => The five numbers of set q are 132, 133, 134, 135, 136. Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670. |
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| 16. |
A Laborer Is Engaged For 30 Days On The Condition That He Receives Rs.25 For Each Day He Works And Is Fined Rs.7.50 For Each Day Is Absent. He Gets Rs.425 In All. For How Many Days Was He Absent? |
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Answer» 30 * 25 = 750 ----------- 325 25 + 7.50 = 32.5 325/32.5 = 10. 30 * 25 = 750 425 ----------- 325 25 + 7.50 = 32.5 325/32.5 = 10. |
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| 17. |
How Many Figures Are Required To Number The Pages The Pages Of A Book Containing 365 Pages? |
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Answer» 1 to 9 = 9 * 1 = 9 100 to 365 = 266 * 3 = 798 ----------- 987 1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 365 = 266 * 3 = 798 ----------- 987 |
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| 18. |
How Many Minimum Number's Of Whole Square Slabs Are Required For Paving The Floor 12.96 Meters Long And 3.84 Meters Side? |
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Answer» HCF of 384, 1296 = 48 48 * 48 * X = 384 * 1296 x = 216. HCF of 384, 1296 = 48 48 * 48 * x = 384 * 1296 x = 216. |
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| 19. |
A Room Is 4 Meters 37 Cm Long And 3 Meters 23cm Broad. It Is Required To Pave The Floor With Minimum Square Slabs. Find The Number Of Slabs Required For This Purpose? |
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Answer» HCF of 323, 437 = 19 323 * 437 = 19 * 19 * X x = 391. HCF of 323, 437 = 19 323 * 437 = 19 * 19 * x x = 391. |
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| 20. |
Sides Of A Rectangular Park Are In The Ratio 3: 2 And Its Area Is 3750 Sq M, The Cost Of Fencing It At 50 Ps Per Meter Is? |
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Answer»
250 * 1/2 = Rs.125. 3x * 2x = 3750 => x = 25 2(75 + 50) = 250 m 250 * 1/2 = Rs.125. |
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| 21. |
A Rectangular Field Has Area Equal To 150 Sq M And Perimeter 50 M. Its Length And Breadth Must Be? |
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Answer»
l – b = 5 l = 15 b = 10. 2(l + b) = 50 => l + b = 25 l – b = 5 l = 15 b = 10. |
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| 22. |
A Man Can Row At 5 Km/hr In Still Water, If The River Is Running At 1 Km/hr It Takes Him 75 Minutes To Row To A Place And Back. How Far Is The Place? |
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Answer» SPEED downstream = (5+1)km/hr = 6 km/hr Speed upstream = (5-1)km/hr = 4 km/hr LET the required distance be x km. Then x/6 + x/4 = 75/60 = 5/4 => (2x + 3x) =15 => x =3. Required distance = 3 km. Speed downstream = (5+1)km/hr = 6 km/hr Speed upstream = (5-1)km/hr = 4 km/hr Let the required distance be x km. Then x/6 + x/4 = 75/60 = 5/4 => (2x + 3x) =15 => x =3. Required distance = 3 km. |
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| 23. |
The Time Taken By A Man To Row His Boat Upstream Is Twice The Time Taken By Him To Row The Same Distance Downstream. If The Speed Of The Boat In Still Water Is 42 Kmph, Find The Speed Of The Stream? |
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Answer» The ratio of the TIMES taken is 2:1. The ratio of the SPEED of the boat in STILL WATER to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1 Speed of the stream = 42/3 = 14 kmph. The ratio of the times taken is 2:1. The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1 Speed of the stream = 42/3 = 14 kmph. |
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| 24. |
A Man Rows His Boat 85 Km Downstream And 45 Km Upstream, Taking 2 1/2 Hours Each Time. Find The Speed Of The Stream? |
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Answer» Speed downstream = d/t = 85/(2 1/2) = 34 kmph Speed upstream = d/t = 45/(2 1/2) = 18 kmph The speed of the stream = (34 - 18)/2 = 8 kmph. Speed downstream = d/t = 85/(2 1/2) = 34 kmph Speed upstream = d/t = 45/(2 1/2) = 18 kmph The speed of the stream = (34 - 18)/2 = 8 kmph. |
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| 25. |
A Man Can Swim In Still Water At 4.5 Km/h, But Takes Twice As Long To Swim Upstream Than Downstream. The Speed Of The Stream Is? |
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Answer» M = 4.5 S = x DS = 4.5 + x US = 4.5 + x 4.5 + x = (4.5 - x)2 4.5 + x = 9 -2x 3X = 4.5 x = 1.5. M = 4.5 S = x DS = 4.5 + x US = 4.5 + x 4.5 + x = (4.5 - x)2 4.5 + x = 9 -2x 3x = 4.5 x = 1.5. |
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| 26. |
Two Persons A And B Take A Field On Rent. A Puts On It 21 Horses For 3 Months And 15 Cows For 2 Months; B Puts 15 Cows For 6months And 40 Sheep For 7 1/2 Months. If One Day, 3 Horses Eat As Much As 5 Cows And 6 Cows As Much As 10 Sheep, What Part Of The Rent Should A Pay? |
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Answer» 3h = 5c A = 21h*3 + 15c*2 = 63h + 30c = 105c + 30c = 135c B = 15c*6 + 40S*7 1/2 = 90c + 300s = 90c + 180c = 270c A:B = 135:270 27:52 A = 27/79 = 1/3. 3h = 5c 6c = 10s A = 21h*3 + 15c*2 = 63h + 30c = 105c + 30c = 135c B = 15c*6 + 40s*7 1/2 = 90c + 300s = 90c + 180c = 270c A:B = 135:270 27:52 A = 27/79 = 1/3. |
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| 27. |
A, B And C Are Entered Into A Partnership. A Invested Rs.6500 For 6 Months, B Invested Rs.8400 For 5 Months And C Invested For Rs.10000 For 3 Months. A Is A Working Partner And Gets 5% Of The Total Profit For The Same. Find The Share Of C In A Total Profit Of Rs.7400? |
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Answer»
26:28:20 C share = 74000 * 95/100 = 7030 * 20/74 => 1900. 65 * 6 : 84 * 5 : 100 * 3 26:28:20 C share = 74000 * 95/100 = 7030 * 20/74 => 1900. |
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| 28. |
If Rs.3250 Be Divided Among Ram, Shyam And Mohan In The Ratio Of 1/2:1/3:1/4 Then The Share Of Each Are? |
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Answer» 1/2:1/3:1/4 = 6:4:3 Ram = 6/13 * 3250 = 1500 1/2:1/3:1/4 = 6:4:3 Ram = 6/13 * 3250 = 1500 Shyam = 4/13 * 3250 = 1000 Mohan = 3/13 * 3250 = 750. |
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| 29. |
A Train Crosses A Platform Of 120 M In 15 Sec, Same Train Crosses Another Platform Of Length 180 M In 18 Sec. Then Find The Length Of The Train? |
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Answer»
X + 120/15 = X + 180/18 X = 180m . Length of the train be ‘X’ X + 120/15 = X + 180/18 6X + 720 = 5X + 900 X = 180m . |
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| 30. |
A Train Speeds Past A Pole In 15 Seconds And A Platform 100 M Long In 25 Seconds. Its Length Is? |
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Answer» Let the length of the TRAIN be x meters and its speed be y m/sec. They, x / y = 15 => y = x/15 x = 150 m. Let the length of the train be x meters and its speed be y m/sec. They, x / y = 15 => y = x/15 x + 100 / 25 = x / 15 x = 150 m. |
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| 31. |
A Train 110 Meters Long Is Running With A Speed Of 60 Kmph. In What Time Will It Pass A Man Who Is Running At 6 Kmph 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) km/hr = 66 km/hr [66 * 5/18] m/sec = [55/3] m/sec. Time taken to pass the man = [110 * 3/55] sec = 6 sec. Speed of train relative to man = (60 + 6) km/hr = 66 km/hr [66 * 5/18] m/sec = [55/3] m/sec. Time taken to pass the man = [110 * 3/55] sec = 6 sec. |
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| 32. |
A 300 Meter Long Train Crosses A Platform In 39 Seconds While It Crosses A Signal Pole In 18 Seconds. What Is The Length Of The Platform? |
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Answer» SPEED = [300 / 18] m/sec = 50/3 m/sec. LET the length of the platform be X meters. Then, x + 300 / 39 = 50/3 3(x + 300) = 1950 è x = 350M. Speed = [300 / 18] m/sec = 50/3 m/sec. Let the length of the platform be x meters. Then, x + 300 / 39 = 50/3 3(x + 300) = 1950 è x = 350m. |
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| 33. |
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 Seconds. The Speed Of The Train Is? |
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Answer» SPEED of the TRAIN relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 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) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 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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| 34. |
Which Of The Following Groups Of Fractions Is In Descending Order? |
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Answer» The fractions CONSIDERED are 8/15 9/13 6/11 To compare them we make the denominators the same. So the fractions are (8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145 1144/2145, 1485/2145 and 1170/2145 so in descending order the fractions will be 1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15. The fractions considered are 8/15 9/13 6/11 To compare them we make the denominators the same. So the fractions are (8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145 1144/2145, 1485/2145 and 1170/2145 so in descending order the fractions will be 1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15. |
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| 35. |
{20 - [7 - (3 -2)] + 1/3 (4.2)} / 1.4 = ? |
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Answer» {20 - [7 - (3 -2)] + 1/3 (4.2)} / 1.4 => [20 - (7 - 1) + 1.4] / 1.4 => (20 - 6 + 1.4)/1.4 = 10 + 1 = 11. {20 - [7 - (3 -2)] + 1/3 (4.2)} / 1.4 => [20 - (7 - 1) + 1.4] / 1.4 => (20 - 6 + 1.4)/1.4 = 10 + 1 = 11. |
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| 36. |
[(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 Of (2 * 38)] = ? |
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Answer» [(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 of (2 * 38)] = [(550 * 152) / 440] / [(110/44) * 38] = (5 * 38)/(5/2 * 38) = 2. [(523 + 27) * (187 - 35) / (424 + 16)] / [110/22 of (2 * 38)] = [(550 * 152) / 440] / [(110/44) * 38] = (5 * 38)/(5/2 * 38) = 2. |
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| 37. |
1860 + 4/7 Of 21.21 - 41.4 = ? |
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Answer» 1860 + 4/7 of 21.21 - 41.4 = 1860 + 4(3.03) - 41.4 = 1860 + 12.12 - 41.4 = 1872.12 - 41.4 = 1830.72. 1860 + 4/7 of 21.21 - 41.4 = 1860 + 4(3.03) - 41.4 = 1860 + 12.12 - 41.4 = 1872.12 - 41.4 = 1830.72. |
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| 38. |
In A Class There Are 20 Boys And 25 Girls. In How Many Ways Can A Boy And A Girl Be Selected? |
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Answer» We can select ONE boy from 20 boys in 20 ways. We select one girl from 25 girls in 25 ways We select a boy and girl in 20 * 25 ways i.e., = 500 ways. We can select one boy from 20 boys in 20 ways. We select one girl from 25 girls in 25 ways We select a boy and girl in 20 * 25 ways i.e., = 500 ways. |
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| 39. |
P, Q And R Have Rs.6000 Among Themselves. R Has Two-thirds Of The Total Amount With P And Q. Find The Amount With R? |
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Answer»
r = 2/3 (total amount with P and Q) r = 2/3(6000 - r) => 3r = 12000 - 2r => 5r = 12000 => r = 2400. Let the amount with R be Rs.r r = 2/3 (total amount with P and Q) r = 2/3(6000 - r) => 3r = 12000 - 2r => 5r = 12000 => r = 2400. |
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| 40. |
The Sum Of Four Consecutive Even Numbers Is 292. What Would Be The Smallest Number? |
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Answer» Let the four CONSECUTIVE even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) Their sum = 8x - 4 = 292 => x = 37 Smallest number is: 2(x - 2) = 70. Let the four consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) Their sum = 8x - 4 = 292 => x = 37 Smallest number is: 2(x - 2) = 70. |
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| 41. |
The Denominator Of A Fraction Is 1 Less Than Twice The Numerator. If The Numerator And Denominator Are Both Increased By 1, The Fraction Becomes 3/5. Find The Fraction? |
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Answer» LET the numerator and denominator of the fraction be 'N' and 'd' respectively. d = 2n - 1 (n + 1)/(d + 1) = 3/5 5n + 5 = 3d + 3 5n + 5 = 3(2n - 1) + 3 => n = 5 d = 2n - 1 => d = 9 Hence the fraction is : 5/9. Let the numerator and denominator of the fraction be 'n' and 'd' respectively. d = 2n - 1 (n + 1)/(d + 1) = 3/5 5n + 5 = 3d + 3 5n + 5 = 3(2n - 1) + 3 => n = 5 d = 2n - 1 => d = 9 Hence the fraction is : 5/9. |
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| 42. |
The Cost Of 2 Chairs And 3 Tables Is Rs.1300. The Cost Of 3 Chairs And 2 Tables Is Rs.1200. The Cost Of Each Table Is More Than That Of Each Chair By? |
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Answer» 2C + 3T = 1300 --- (1) Subtracting 2nd from 1st, we GET -C + T = 100 => T - C = 100. 2C + 3T = 1300 --- (1) 3C + 3T = 1200 --- (2) Subtracting 2nd from 1st, we get -C + T = 100 => T - C = 100. |
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| 43. |
Two Varieties Of Wheat - A And B Costing Rs. 9 Per Kg And Rs. 15 Per Kg Were Mixed In The Ratio 3 : 7. If 5 Kg Of The Mixture Is Sold At 25% Profit, Find The Profit Made? |
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Answer» Let the quantities of A and B mixed be 3X kg and 7x kg. Cost of 3x kg of A = 9(3x) = Rs. 27X Cost of 7x kg of B = 15(7x) = Rs. 105x Cost of 10x kg of the MIXTURE = 27x + 105x = Rs. 132x Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66 Profit made in SELLING 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50. Let the quantities of A and B mixed be 3x kg and 7x kg. Cost of 3x kg of A = 9(3x) = Rs. 27x Cost of 7x kg of B = 15(7x) = Rs. 105x Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66 Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50. |
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| 44. |
A Mixture Of 150 Liters Of Wine And Water Contains 20% Water. How Much More Water Should Be Added So That Water Becomes 25% Of The New Mixture? |
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Answer» Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters. P liters of water added to the mixture to make water 25% of the new mixture. Total amount of water becomes (30 + P) and total volume of mixture is (150 + P). (30 + P) = 25/100 * (150 + P) 120 + 4P = 150 + P => P = 10 liters. Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters. P liters of water added to the mixture to make water 25% of the new mixture. Total amount of water becomes (30 + P) and total volume of mixture is (150 + P). (30 + P) = 25/100 * (150 + P) 120 + 4P = 150 + P => P = 10 liters. |
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| 45. |
In A Game Of Billiards, A Can Give B 20 Points In 60 And He Can Give C 30 Points In 60. How Many Points Can B Give C In A Game Of 100? |
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Answer» A SCORES 60 while B SCORE 40 and C scores 30. The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75. In a game of 100 points, B gives (100 - 75) = 25 points to C. A scores 60 while B score 40 and C scores 30. The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75. In a game of 100 points, B gives (100 - 75) = 25 points to C. |
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| 46. |
A Can Run A Kilometer Race In 4 1/2 Min While B Can Run Same Race In 5 Min. How Many Meters Start Can A Give B In A Kilometer Race, So That The Race Mat End In A Dead Heat? |
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Answer» A can GIVE B (5 MIN - 4 1/2 min) = 30 SEC start. The distance covered by B in 5 min = 1000 m. Distance covered in 30 sec = (1000 * 30)/300 = 100 m. A can give B 100m start. A can give B (5 min - 4 1/2 min) = 30 sec start. The distance covered by B in 5 min = 1000 m. Distance covered in 30 sec = (1000 * 30)/300 = 100 m. A can give B 100m start. |
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| 47. |
A Can Give B 100 Meters Start And C 200 Meters Start In A Kilometer Race. How Much Start Can B Give C In A Kilometer Race? |
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Answer» A runs 1000 m while B runs 900 m and C runs 800 m. The NUMBER of meters that C runs when B runs 1000 m, = (1000 * 800)/900 = 8000/9 = 888.88 m. B can GIVE C = 1000 - 888.88 = 111.12 m. A runs 1000 m while B runs 900 m and C runs 800 m. The number of meters that C runs when B runs 1000 m, = (1000 * 800)/900 = 8000/9 = 888.88 m. B can give C = 1000 - 888.88 = 111.12 m. |
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| 48. |
If Goods Be Purchased For Rs.840 And One-fourth Be Sold At A Loss Of 20% At What Gain Percent Should The Remainder Be Sold So As To Gain 20% On The Whole Transaction? |
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Answer» 1/4 CP = 210 SP = 21*(80/100) = 168 1008 - 168 = 840 3/4 SP = 630 Gain = 210 630 --- 210 100 --- ? => 33 1/3%. 1/4 CP = 210 SP = 21*(80/100) = 168 SP = 840*(120/100) = 1008 1008 - 168 = 840 3/4 SP = 630 Gain = 210 630 --- 210 100 --- ? => 33 1/3%. |
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| 49. |
By Selling 50 Meters Of Cloth. I Gain The Selling Price Of 15 Meters. Find The Gain Percent? |
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Answer» SP = CP + g 50 SP = 50 CP + 15 SP 35 SP = 50 CP 35 --- 15 CP gain SP = CP + g 50 SP = 50 CP + 15 SP 35 SP = 50 CP 35 --- 15 CP gain 100 --- ? => 42 6/7%. |
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A Bank Offers 5% C.i. Calculated On Half-yearly Basis . A Customer Deposits Rs. 1600 Each On 1st January And 1st July Of A Year. At The End Of The Year, The Amount He Would Have Gained By Way Of Interest Is? |
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Answer» AMOUNT = [1600 * (1 + 5/(2 * 100)2 + 1600 * (1 + 5/(2 * 100)] = [1600 * 41/40(41/40 + 1) = [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321. C.I. = 3321 - 3200 = Rs. 121. Amount = [1600 * (1 + 5/(2 * 100)2 + 1600 * (1 + 5/(2 * 100)] = [1600 * 41/40(41/40 + 1) = [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321. C.I. = 3321 - 3200 = Rs. 121. |
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