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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. |
A Purse Contains 342 Coins Consisting Of One Rupees, 50 Paise And 25 Paise Coins. If Their Values Are In The Ratio Of 11 : 9 : 5 Then Find The Number Of 50 Paise Coins? |
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Answer» Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5X respectively. No. of 1 rupee coins = (11x / 1) =11x No. of 50 paise coins = (9x / 0.5) = 18X No. of 25 paise coins = (5x / 0.25) = 20x 11x + 18x + 9x = 342 38x = 342 x = 9 Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins No. of 50 paise coins = 18 x 9 = 162 coins No. of 25 paise coins = 20 x 9 = 180 coins. Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively. No. of 1 rupee coins = (11x / 1) =11x No. of 50 paise coins = (9x / 0.5) = 18x No. of 25 paise coins = (5x / 0.25) = 20x 11x + 18x + 9x = 342 38x = 342 x = 9 Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins No. of 50 paise coins = 18 x 9 = 162 coins No. of 25 paise coins = 20 x 9 = 180 coins. |
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| 2. |
A Bag Contains Equal Number Of 25 Paise, 50 Paise And One Rupee Coins Respectively. If The Total Value Is Rs 105, How Many Types Of Each Type Are Present? |
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Answer» Bag consists of 25 paise, 50 paise and 1 RUPEE (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4 Total value of 25 paise coins =(1 / 7 ) X 105 = 15 Total value of 50 paise coins = (2 / 7) x 105 = 30 Total value of 100 paise coins = (4 / 7) x 105 = 60 No. of 25 paise coins = 15 x 4 = 60 coins No. of 50 paise coins = 30 x 2 = 60 coins No. of 1 rupee coins = 60 x 1 = 60 coins. Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4 Total value of 25 paise coins =(1 / 7 ) x 105 = 15 Total value of 50 paise coins = (2 / 7) x 105 = 30 Total value of 100 paise coins = (4 / 7) x 105 = 60 No. of 25 paise coins = 15 x 4 = 60 coins No. of 50 paise coins = 30 x 2 = 60 coins No. of 1 rupee coins = 60 x 1 = 60 coins. |
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| 3. |
Find The Lowest Common Multiple Of 24, 36 And 40? |
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Answer» To find the LCM of 24, 36 and 40 36 = 2 x 2 x 3 x 3 40 = 2 x 2 x 2 x 5 Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5 = 8 x 9 x 5 = 72 x 5 = 360. To find the LCM of 24, 36 and 40 24 = 2 x 2 x 2 x 3 36 = 2 x 2 x 3 x 3 40 = 2 x 2 x 2 x 5 Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5 = 8 x 9 x 5 = 72 x 5 = 360. |
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| 4. |
Three Numbers Are In The Ratio Of 3:4:5 And Their L.c.m Is 3600.their Hcf Is? |
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Answer» Let the numbers be 3x, 4x, 5x. Then, their L.C.M = 60X. So, 60x=3600 or x=60. THEREFORE, The numbers are (3 x 60), (4 x 60), (5 x 60). Hence,required H.C.F=60. Let the numbers be 3x, 4x, 5x. Then, their L.C.M = 60x. So, 60x=3600 or x=60. Therefore, The numbers are (3 x 60), (4 x 60), (5 x 60). Hence,required H.C.F=60. |
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| 5. |
If N Is The Greatest Number That Will Divide 1305, 4665 And 6905, Leaving The Same Remainder In Each Case. What Is The Sum Of The Digits Of N ? |
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Answer» N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. SUM of digits in N = ( 1 + 1 + 2 + 0 ) = 4. N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4. |
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| 6. |
If The Sum Of Two Numbers Is 55 And The H.c.f. And L.c.m. Of These Numbers Are 5 And 120 Respectively, Then The Sum Of The Reciprocals Of The Numbers Is Equal To ? |
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Answer» Let the NUMBERS be a and b. We KNOW that product of TWO numbers = Product of their HCF and LCM Then, a + b = 55 and ab = 5 x 120 = 600. => The required sum = (1/a) + (1/b) = (a+b)/ab =55/600 = 11/120. Let the numbers be a and b. We know that product of two numbers = Product of their HCF and LCM Then, a + b = 55 and ab = 5 x 120 = 600. => The required sum = (1/a) + (1/b) = (a+b)/ab =55/600 = 11/120. |
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| 7. |
Which Of The Following Has Most Number Of Divisors? |
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Answer»
101= 1 x 101 176= 1 x 2X 2 x 2 x 2 x 11 182= 1 x 2 x 7 x 13 So, divisors of 99 are 1, 3, 9, 11, 33, 99 divisors of 101 are 1,101 divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176 divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182 Hence , 176 hasthe most number of divisors. 99 = 1 x 3 x 3 x 11 101= 1 x 101 176= 1 x 2x 2 x 2 x 2 x 11 182= 1 x 2 x 7 x 13 So, divisors of 99 are 1, 3, 9, 11, 33, 99 divisors of 101 are 1,101 divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176 divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182 Hence , 176 hasthe most number of divisors. |
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| 8. |
A Drink Vendor Has 368 Liters Of Maaza, 80 Liters Of Pepsi And 144 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 = 368/16 = 23 Number of cans of Pepsi = 80/16 = 5 Number of cans of Sprite = 144/16 = 9 The total number of cans required = 23 + 5 + 9 = 37 cans. The number of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of cans of Maaza = 368/16 = 23 Number of cans of Pepsi = 80/16 = 5 Number of cans of Sprite = 144/16 = 9 The total number of cans required = 23 + 5 + 9 = 37 cans. |
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| 9. |
Two Stations A And B Are 200 Km Apart On A Straight Track. One Train Starts From A At 7 A.m. And Travels Towards B At 20 Kmph. Another Train Starts From B At 8 A.m. And Travels Towards A At A Speed Of 25 Kmph. At What Time Will They Meet? |
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Answer» <P>Assume both trains meet after 'p' hours after 7 a.m. Distance covered by train starting from A in 'p' hours = 20p km Distance covered by train starting from B in (p-1) hours = 25(p-1) Total distance = 200 => 20X + 25(x-1) = 200 => 45X = 225 => p= 5 Means, they meet after 5 hours after 7 am, IE, they meet at 12 p.m. Assume both trains meet after 'p' hours after 7 a.m. Distance covered by train starting from A in 'p' hours = 20p km Distance covered by train starting from B in (p-1) hours = 25(p-1) Total distance = 200 => 20x + 25(x-1) = 200 => 45x = 225 => p= 5 Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m. |
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| 10. |
Running 3/4th Of His Usual Rate, A Man Is 15min Late. Find His Usual Time In Hours ? |
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Answer» WALKING at 3/4th of usual RATE implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time so 1/3rd of the usual time = 15min or usual time = 3 x 15 = 45MIN = 45/60 hrs = 3/4 hrs. Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time so 1/3rd of the usual time = 15min or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs. |
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| 11. |
In A Daily Morning Walk Three Persons Step Off Together. Their Steps Measure 75 Cm, 80 Cm And 85 Cm Respectively. What Is The Minimum Distance Each Should Walk So That Thay Can Cover The Distance In Complete Steps ? |
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Answer» To find the minimum distance, we have to get the LCM of 75, 80, 85 Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400 Hence, the minimum distance each should walk so that THAY can cover the distance in complete STEPS = 20400 cms = 20400/100 = 204 mts. To find the minimum distance, we have to get the LCM of 75, 80, 85 Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400 Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts. |
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| 12. |
Two Trains Start From Same Place At Same Time At Right Angles To Each Other. Their Speeds Are 36km/hr And 48km/hr Respectively. After 30 Seconds The Distance Between Them Will Be ? |
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Answer» Using pythagarous THEOREM, DISTANCE TRAVELLED by first train = 36x5/18x30 = 300m distance travelled by second train = 48x5/18x30 = 400M so distance between them =v( 90000 + 160000) = v250000 = 500mts. Using pythagarous theorem, distance travelled by first train = 36x5/18x30 = 300m distance travelled by second train = 48x5/18x30 = 400m so distance between them =v( 90000 + 160000) = v250000 = 500mts. |
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| 13. |
Kamal Consistently Runs 240 Meters A Day And On Saturday He Runs For 400 Meters. How Many Kilometers Will He Have To Run In Four Weeks ? |
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Answer» Total running distance in FOUR weeks = (24 X 240) + (4 x 400) = 5760 + 1600 = 7360 meters = 7360/1000 => 7.36 kms. Total running distance in four weeks = (24 x 240) + (4 x 400) = 5760 + 1600 = 7360 meters = 7360/1000 => 7.36 kms. |
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| 14. |
If Hema Walks At 12 Km/hr Instead Of 8 Km/hr, She Would Have Walked 20 Km More. The Actual Distance Travelled By Hema Is ? |
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Answer» Let the actual DISTANCE travelled be x km. Then x/8=(x+20)/12 => 12x = 8x + 160 => 4x = 160 => x = 40 km. Let the actual distance travelled be x km. Then x/8=(x+20)/12 => 12x = 8x + 160 => 4x = 160 => x = 40 km. |
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| 15. |
K And L Starts Walking Towards Each Other At 4 Pm At Speed Of 3 Km/hr And 4 Km/hr Respectively. They Were Initially 17.5 Km Apart. At What Time Do They Meet ? |
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Answer» Suppose they meet after 'h' hours Then 7H = 17.5 h = 2.5 hours So they meet at => 4 + 2.5 = 6:30 pm. Suppose they meet after 'h' hours Then 3h + 4h = 17.5 7h = 17.5 h = 2.5 hours So they meet at => 4 + 2.5 = 6:30 pm. |
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| 16. |
In What Time A 360 M. Long Train Moving At The Speed Of 44 Km/hr Will Cross A 140 M. Long Bridge ? |
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Answer» SPEED = 44 kmph x 5/18 = 110/9 m/s We KNOW that, Time = distance/speed Time = (360 + 140) / (110/9) = 500 x 9/110 = 41 SEC. Speed = 44 kmph x 5/18 = 110/9 m/s We know that, Time = distance/speed Time = (360 + 140) / (110/9) = 500 x 9/110 = 41 sec. |
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| 17. |
Robert Is Traveling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Km/hr; He Will Reach There At 12 Noon If He Travels At 15 Km/hr. At What Speed Must He Travel To Reach A At 1 P.m. ? |
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Answer» Let the distance traveled be x km. Then, x/10 - x/15 = 2 3x - 2x = 60 => x = 60 km. Time taken to travel 60 km at 10 km/hr = 60/10 = 6 HRS. So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m. REQUIRED speed = 60/5 = 12 kmph. Let the distance traveled be x km. Then, x/10 - x/15 = 2 3x - 2x = 60 => x = 60 km. Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs. So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m. Required speed = 60/5 = 12 kmph. |
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| 18. |
A Bus Covers Its Journey At The Speed Of 80km/hr In 10hours. If The Same Distance Is To Be Covered In 4 Hours, By How Much The Speed Of Bus Will Have To Increase ? |
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Answer»
TOTAL distance = 80 x 10 = 800KM New speed = 800/4 =200km/hr Increase in speed = 200 - 80 = 120km/hr. Initial speed = 80km/hr Total distance = 80 x 10 = 800km New speed = 800/4 =200km/hr Increase in speed = 200 - 80 = 120km/hr. |
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| 19. |
The Numbers X, Y, Z Are Such That Xy = 96050 And Xz = 95625 And Y Is Greater Than Z By One. Find Out The Number Z ? |
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Answer» XY = 96050 ...(i) and XZ = 95625 ...(ii) and y - z = 1 ... (iii) Dividing (i) by (ii) we get y/z = 96050 / 95625 = 3842 / 3825 = 226 / 225 ... (IV) COMBINING (iii) and (iv) we get z = 225. xy = 96050 ...(i) and xz = 95625 ...(ii) and y - z = 1 ... (iii) Dividing (i) by (ii) we get y/z = 96050 / 95625 = 3842 / 3825 = 226 / 225 ... (iv) Combining (iii) and (iv) we get z = 225. |
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| 20. |
If A Number Is Decreased By 4 And Divided By 6 The Result Is 9. What Would Be The Result If 3 Is Subtracted From The Number And Then It Is Divided By 5 ? |
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Answer» (x - 4) / 6 = 9 Multiply both sides by 6: x - 4 = 54 ADD 4 to both sides: x = 58 (58 - 3) / 5 = 55 / 5 = 11. (x - 4) / 6 = 9 Multiply both sides by 6: x - 4 = 54 Add 4 to both sides: x = 58 (58 - 3) / 5 = 55 / 5 = 11. |
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| 21. |
A Two-digit Number Is Seven Times The Sum Of Its Digits, If Each Digit Is Increased By 2, The Number Thus Obtained Is 4 More Than Six Times The Sum Of Its Digits, Find The Number ? |
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Answer» LET the two-digit number be 10x + y 10x + y = 7(X + y) ? x = 2y ...(i) 10(x +2 ) + (y + 2) = 6(x + y + 4) + 4 or 10x + y + 22 = 6X + 6y + 28 ? 4x - 5y = 6 ...(ii) Solving equations (i) and (ii) We GET x = 4 and y = 2. Let the two-digit number be 10x + y 10x + y = 7(x + y) ? x = 2y ...(i) 10(x +2 ) + (y + 2) = 6(x + y + 4) + 4 or 10x + y + 22 = 6x + 6y + 28 ? 4x - 5y = 6 ...(ii) Solving equations (i) and (ii) We get x = 4 and y = 2. |
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| 22. |
The Digit In The Units Place Of A Number Is Equal To The Digit In The Tens Place Of Half Of That Number And The Digit In The Tens Place Of That Number Is Less Than The Digit In Units Place Of Half Of The Number By 1, If The Sum Of The Digits Of The Number Is Seven, Then What Is The Number ? |
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Answer»
and the no. = 10v + w From the given conditions, w= x and V = y-1 Thus the no. = 10 (y-1) + x ? 2(10x + y ) = 10 (y-1) + x ? 8y - 19x = 10 ...(i) v + w = 7 ? y-1 + x = 7 ? x + y = 8 Solving equations (i) and (ii) , we get x = 2 and y = 6. Let 1/2 of the no. = 10x + y and the no. = 10v + w From the given conditions, w= x and v = y-1 Thus the no. = 10 (y-1) + x ? 2(10x + y ) = 10 (y-1) + x ? 8y - 19x = 10 ...(i) v + w = 7 ? y-1 + x = 7 ? x + y = 8 Solving equations (i) and (ii) , we get x = 2 and y = 6. |
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| 23. |
How Many Figures (digits) Are Required To Number A Book Containing 200 Pages ? |
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Answer» Number of one DIGIT pages from 1 to 9 = 9 Number of two digit pages from 10 to 99 = 90 Number of THREE digit pages from 100 to 200 = 101. Number of one digit pages from 1 to 9 = 9 Number of two digit pages from 10 to 99 = 90 Number of three digit pages from 100 to 200 = 101. |
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| 24. |
In A Question Divisor Is 2/3 Of The Dividend And 2 Times The Remainder. If The Remainder Is 5, Find The Dividend? |
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Answer»
and Divisor = 2 x remainder or 2/3 x dividend = 2 x 5 Dividend = 2 x 5 x 3 / 2 = 15. Divisor = 2/3 x dividend and Divisor = 2 x remainder or 2/3 x dividend = 2 x 5 Dividend = 2 x 5 x 3 / 2 = 15. |
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| 25. |
Pumps Working 8 Hours A Day, Can Empty A Tank In 2 Days. How Many Hours A Day Must 4 Pumps Work To Empty The Tank In 1 Day? |
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Answer» Let the REQUIRED no of working hours per day be X. More pumps , LESS working hours per day (Indirect Proportion) Less days, More working hours per day (Indirect Proportion) Pumps4 : 3Days1 : 2}?? 8:x => (4 * 1 * x) = (3 * 2 * 8) => x=12 Let the required no of working hours per day be x. More pumps , Less working hours per day (Indirect Proportion) Less days, More working hours per day (Indirect Proportion) Pumps4 : 3Days1 : 2}?? 8:x => (4 * 1 * x) = (3 * 2 * 8) => x=12 |
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| 26. |
If The Simple Interest On A Sum Of Money At 5% Per Annum For 3 Years Is Rs. 1200, Find The Compound Interest On The Same Sum For The Same Period At The Same Rate? |
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Answer» Clearly, RATE = 5% p.a ., S.I =Rs.1200. So,Principal =Rs.(100 x 1200/3x5) =Rs.8000. Amount =Rs.[8000 x (1+5/100)³] =Rs(8000x21/20x21/20x21/20) = Rs.9261 C.I =Rs.(9261-8000) =Rs.1261. Clearly, Rate = 5% p.a ., Time = 3 years S.I =Rs.1200. So,Principal =Rs.(100 x 1200/3x5) =Rs.8000. Amount =Rs.[8000 x (1+5/100)³] =Rs(8000x21/20x21/20x21/20) = Rs.9261 C.I =Rs.(9261-8000) =Rs.1261. |
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| 27. |
In How Many Ways Can The Letters Of The Word 'leader' Be Arranged ? |
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Answer» No. of letters in the WORD = 6 No. of 'E' repeated = 2 TOTAL No. of arrangement = 6!/2! = 360. No. of letters in the word = 6 No. of 'E' repeated = 2 Total No. of arrangement = 6!/2! = 360. |
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| 28. |
Find The Compound Interest On Rs.16,000 At 20% Per Annum For 9 Months, Compounded Quartely? |
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Answer» Principal = RS.16,000; =Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20] = Rs.18522. C.I = Rs.(18522 - 16000) = Rs.2522. Principal = Rs.16,000; Time=9 months = 3 quarters; Amount =Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20] = Rs.18522. C.I = Rs.(18522 - 16000) = Rs.2522. |
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| 29. |
The Difference Between Simple Interest And Compound On Rs. 1200 For One Year At 10% Per Annum Reckoned Half-yearly Is? |
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Answer» S.I. = RS.(1200*10*1)/100=rs.120 C.I. =rs[1200*(1+5/100)2-1200]=rs.123 DIFFERENCE = Rs.(123-120) =Rs.3 S.I. = Rs.(1200*10*1)/100=rs.120 C.I. =rs[1200*(1+5/100)2-1200]=rs.123 Difference = Rs.(123-120) =Rs.3 |
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| 30. |
Find The Principal If The Interest Compounded At The Rate Of 10% Per Annum For Two Years Is Rs. 420 ? |
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Answer» Given, Compound rate, R = 10% PER annum Time = 2 years C.I = RS. 420 Let P be the required principal. A = (P+C.I) Amount, A = P(1 + (r/100))N (P+C.I) = P[1 + (10/100)]2 (P+420) = P[11/10][11/10] P-1.21P = -420 0.21P = 420 Hence, P = 420/0.21 = Rs. 2000. Given, Compound rate, R = 10% per annum Time = 2 years C.I = Rs. 420 Let P be the required principal. A = (P+C.I) Amount, A = P(1 + (r/100))n (P+C.I) = P[1 + (10/100)]2 (P+420) = P[11/10][11/10] P-1.21P = -420 0.21P = 420 Hence, P = 420/0.21 = Rs. 2000. |
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| 31. |
What Will Be The Compound Interest On A Sum Of Rs.25,000 After 3 Years At The Rate Of 12 P.c.p.a? |
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Answer»
= RS.(25000x(1+12/100)³ = Rs.(25000x28/25x28/25x28/25) = Rs. 35123.20. C.I = Rs(35123.20 -25000) = Rs.10123.20. Amount = Rs.(25000x(1+12/100)³ = Rs.(25000x28/25x28/25x28/25) = Rs. 35123.20. C.I = Rs(35123.20 -25000) = Rs.10123.20. |
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| 32. |
Simple Interest On A Certain Sum Of Money For 3 Years At 8% Per Annum Is Half The Compound Interest On Rs. 4000 For 2 Years At 10% Per Annum. The Sum Placed On Simple Interest Is? |
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Answer» C.I.= RS.[4000*(1+10/100)^2-4000] =Rs.840 sum=Rs.(420 * 100)/3*8=Rs.1750. C.I.= Rs.[4000*(1+10/100)^2-4000] =Rs.840 sum=Rs.(420 * 100)/3*8=Rs.1750. |
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| 33. |
The Effective Annual Rate Of Interest Corresponding To A Nominal Rate Of 6% Per Annum Payable Half-yearly Is? |
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Answer»
when COMPOUNDED half-yearly = Rs.[100*(1+3/100)^2]=Rs.106.09 EFFECTIVE rate=(106.09-100)%=6.09%. Amount of Rs. 100 for 1 year when compounded half-yearly = Rs.[100*(1+3/100)^2]=Rs.106.09 Effective rate=(106.09-100)%=6.09%. |
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| 34. |
In A Simultaneous Throw Of Pair Of Dice. Find The Probability Of Getting The Total More Than 7? |
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Answer»
Let E = event of getting a total more than 7 = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)} THEREFORE,P(E) = n(E)/n(S) = 15/36 = 5/12. Here n(S) = (6 x 6) = 36 Let E = event of getting a total more than 7 = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)} Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12. |
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| 35. |
Three Unbiased Coins Are Tossed.what Is The Probability Of Getting At Least 2 Heads? |
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Answer» Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}. Let E = EVENT of getting at least TWO heads = {THH, HTH, HHT, HHH}. P(E) = n(E) / n(S) = 4/8= 1/2. Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}. Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}. P(E) = n(E) / n(S) = 4/8= 1/2. |
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| 36. |
Three Unbiased Coins Are Tossed. What Is The Probability Of Getting At Most Two Heads? |
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Answer» <P>Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} LET E = event of getting at most two heads. Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P(E) =n(E)/n(S)=7/8. Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event of getting at most two heads. Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}. P(E) =n(E)/n(S)=7/8. |
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| 37. |
Four Dice Are Thrown Simultaneously. Find The Probability That All Of Them Show The Same Face? |
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Answer» The total number of ELEMENTARY EVENTS associated to the random experiments of throwing FOUR dice SIMULTANEOUSLY is: = 6*6*6*6=64 n(S) = 64 Let X be the event that all dice show the same face. X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} n(X) = 6 Hence required probability = n(X)n(S)=6/64=1216. The total number of elementary events associated to the random experiments of throwing four dice simultaneously is: = 6*6*6*6=64 n(S) = 64 Let X be the event that all dice show the same face. X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} n(X) = 6 Hence required probability = n(X)n(S)=6/64=1216. |
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| 38. |
If Two Letters Are Taken At Random From The Word Home, What Is The Probability That None Of The Letters Would Be Vowels? |
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Answer» P(first LETTER is not VOWEL) = 2/4 P(second letter is not vowel) = 1/3 So, probability that none of LETTERS WOULD be VOWELS is = 2/4×1/3=1/6. P(first letter is not vowel) = 2/4 P(second letter is not vowel) = 1/3 So, probability that none of letters would be vowels is = 2/4×1/3=1/6. |
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| 39. |
The Table Is Bought For Rs. 1950 And Sold At Rs. 2340. Find The Profit Percent? |
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Answer» Cost Price = Rs. 1950 SELLING Price = Rs. 2340 PROFIT = S.P – C.P Profit = Rs. 2340 – 1950 = 390 Profit % = (Profit/C.P) X 100 Profit % = (390/1950) x 100 Profit % = 20 %. Cost Price = Rs. 1950 Selling Price = Rs. 2340 Profit = S.P – C.P Profit = Rs. 2340 – 1950 = 390 Profit % = (Profit/C.P) x 100 Profit % = (390/1950) x 100 Profit % = 20 %. |
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| 40. |
Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is? |
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Answer»
Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326. Let E = event of getting 1 spade and 1 heart. n(E)= number of WAYS of CHOOSING 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169. P(E) = n(E)/n(S) = 169/1326 = 13/102. Let S be the sample space. Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326. Let E = event of getting 1 spade and 1 heart. n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169. P(E) = n(E)/n(S) = 169/1326 = 13/102. |
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| 41. |
On Selling 17 Balls At Rs. 720, There Is A Loss Equal To The Cost Price Of 5 Balls. The Cost Price Of A Ball Is ? |
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Answer» LET the COST price of a ball is RS.x Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls The equation is : 17X - 720 = 5x Solving the equation we get x = 60 Therefore, cost price of a ball is Rs. 60. Let the cost price of a ball is Rs.x Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls The equation is : 17x - 720 = 5x Solving the equation we get x = 60 Therefore, cost price of a ball is Rs. 60. |
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| 42. |
A Shopkeeper Sold A Mobile Phone For Rs. 12000. Had He Offered Discount Of 10% On The Selling Price, There Would Be A Loss Of 4%. What Is The Cost Price Of That Mobile Phone? |
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Answer» GIVEN that SP = RS. 12000 - 10% = Rs. 10,800 Loss% = 4 We know that, C.P = 100/(100 - Loss%) x 100 => 100/100-4 x 10800 => 1080000/96 C.P = Rs. 11,250. Given that SP = Rs. 12000 - 10% = Rs. 10,800 Loss% = 4 We know that, C.P = 100/(100 - Loss%) x 100 => 100/100-4 x 10800 => 1080000/96 C.P = Rs. 11,250. |
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| 43. |
A Seller Uses 840 Gm In Place Of 1 Kg To Sell His Goods. Find His Actual Profit/loss % When He Sells His Article On 4% Loss On Cost Price ? |
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Answer» Let 1KG of Rs. 100 then 840gm is of Rs. 84. Now (LABEL on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% DISCOUNT [he sells his ARTICLE on 4% loss on cost price.] So now S.P = Rs. 96 But ACTUALLY it contains 840 gm so C.P for shopkeeper = Rs. 84 S.P = Rs. 96 C.P = Rs. 84 Profit% = {(S.P-C.P)/C.P}x100 {(96-84)/84} x 100 = 14.28571429% PROFIT. Let 1kg of Rs. 100 then 840gm is of Rs. 84. Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.] So now S.P = Rs. 96 But actually it contains 840 gm so C.P for shopkeeper = Rs. 84 S.P = Rs. 96 C.P = Rs. 84 Profit% = {(S.P-C.P)/C.P}x100 {(96-84)/84} x 100 = 14.28571429% PROFIT. |
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| 44. |
Raghu Earns 25% On An Investment But Loses 10% On Another Investment. If The Ratio Of The Two Investment Is 3:5. What Is The Gain Or Loss On Two Investments Taken Together ? |
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Answer» Taking the 2 investments to be 3x and 5x respectively Total INCOME of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25 THEREFORE, Gain% = 0.25/8 x 100 = 3.125 %. Taking the 2 investments to be 3x and 5x respectively Total income of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25 Therefore, Gain% = 0.25/8 x 100 = 3.125 %. |
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| 45. |
A Man Saves 20% Of His Monthly Salary. If An Account Of Dearness Of Things He Is To Increase His Monthly Expenses By 15%, He Is Only Able To Save Rs. 400 Per Month. What Is His Monthly Salary? |
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Answer»
Expenditure = Rs. 80 Savings = Rs. 20 Present Expenditure 80x(15/100) = Rs. 12 = 80 + 12 = Rs. 92 Present Savings = 100 – 92 = Rs. 8 100 ------ 8 ? --------- 400 => 5000 His salary = Rs. 5000. Income = Rs. 100 Expenditure = Rs. 80 Savings = Rs. 20 Present Expenditure 80x(15/100) = Rs. 12 = 80 + 12 = Rs. 92 Present Savings = 100 – 92 = Rs. 8 100 ------ 8 ? --------- 400 => 5000 His salary = Rs. 5000. |
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| 46. |
In A Certain Store, The Profit Is 320% Of The Cost. If The Cost Increases By 25% But The Selling Price Remains Constant, Approximately What Percentage Of The Selling Price Is The Profit? |
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Answer»
Then, Profit = Rs. 320, S.P. = Rs. 420. New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. = Rs. 420. Profit = Rs. (420 - 125) = Rs. 295 Required percentage = (295/420) * 100 = 70%(approx). Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420. New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. = Rs. 420. Profit = Rs. (420 - 125) = Rs. 295 Required percentage = (295/420) * 100 = 70%(approx). |
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| 47. |
A Person Sold Two Cows Each For Rs.9900. If He Gained 10% On One And Lost 20% On The Other, Then Which Of The Following Is True? |
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Answer» The CP of profitable COW = 9900/1.1 = 9000 and profit = Rs. 900 The CP of LOSS yielding cow = 9900/0.8 = 12375 and loss = Rs. 2475 so, the NET loss = 2475 - 900 = 1575. The CP of profitable cow = 9900/1.1 = 9000 and profit = Rs. 900 The CP of loss yielding cow = 9900/0.8 = 12375 and loss = Rs. 2475 so, the net loss = 2475 - 900 = 1575. |
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