options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 5 / 2 , b ) 10 / 3 , c ) 15 / 4 , d ) 25 / 3 , e ) 25 / 6 | e | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 5))), const_2)) | if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 5 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 5 = 2 d / 25 the average of november and january is ( n + j ) / 2 = 12 d / 25 / 2 = 6 d / 25 d is 25 / 6 times the average of november and january . the answer is e ." | a = 2 / 5
b = 2 / 5
c = 1 / 5
d = b * c
e = a + d
f = e / 2
g = 1 / f
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | multiply(6, divide(3, 2)) | john was thrice as old as tom 6 years ago . john will be 3 / 2 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 3 / 2 * ( t + 6 ) 2 j + 12 = 3 t + 18 2 ( 3 t - 12 ) + 12 = 3 t + 18 3 t = 30 t = 10 the answer is c ." | a = 3 / 2
b = 6 * a
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a ) 92 , b ) 85 , c ) 136.67 , d ) 140.67 , e ) 150 | c | divide(divide(multiply(160, const_100), 60), const_2) | a rainstorm increased the amount of water stored in state j reservoirs from 130 billion gallons to 160 billion gallons . if the storm increased the amount of water in the reservoirs to 60 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ? | "let total capacity be x we know 160 = 0.60 x x = 160 / 0.60 = 266.67 prior to storm , we had 130 bn gallons 266.67 - 130 = 136.67 answer : c" | a = 160 * 100
b = a / 60
c = b / 2
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a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | b | divide(73, 11) | what is the 8 th digit to the right of the decimal point in the decimal equivalent of 73 / 11 ? | "73 / 11 = 6.6363 . . . . 63 is non - terminating repeating decimal . the 8 th digit to the right of decimal point will be 3 . answer b" | a = 73 / 11
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a ) rs . 300 , b ) rs . 360 , c ) rs . 389 , d ) rs . 368 , e ) rs . 323 | b | divide(510, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | if rs . 510 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 510 * 2 / 17 = rs . 60 b ' s share = 510 * 3 / 17 = rs . 90 c ' s share = 510 * 12 / 17 = rs . 360 . answer : b" | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 510 / f
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a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | if p # q denotes the least common multiple of p and q , then p = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , p = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c ." | a = 18 * 24
b = 12 * 16
c = a - b
d = c / 2
e = d + 24
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a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 3 | d | multiply(const_3.0, 1) | if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 6 = d" | a = 3 * 0
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a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 64 | d | divide(multiply(8, subtract(8, const_1)), const_2) | there are 8 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 7 teams . . . so total no of matches = 8 x 7 = 56 . now , each match is played twice = > 56 x 2 but 2 teams play a match = > 56 x 2 / 2 = 56 . ans : d" | a = 8 - 1
b = 8 * a
c = b / 2
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a ) 13.2 % , b ) 15.2 % , c ) 14.28 % , d ) 32 1 / 3 % , e ) 33 1 / 2 % | c | divide(multiply(3.5, const_100), 4) | a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3.5 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3.5 he selling at 3.5 a rupee , his sp = 4 profit = sp - cp = 4 - 3.5 = 0.5 profit percent = 0.5 / 3.5 * 100 = 14.28 % answer : c" | a = 3 * 5
b = a / 4
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a ) 32.5 % , b ) 40 % , c ) 60 % , d ) 65 % , e ) can not be determined | c | multiply(add(multiply(divide(21, add(21, 4)), divide(5, add(5, 6))), multiply(divide(2, add(2, 3)), divide(6, add(5, 6)))), const_100) | solution a is made up of alcohol and water mixed in the ratio of 21 : 4 by volume ; solution b is made up of alcohol and water mixed in the ratio of 2 : 3 by volume . if solution a and solution b are mixed in the ratio of 5 : 6 by volume , what percent of the resultant mixture is alcohol ? | let mixture after mixing a and b is 110 ml ( number assumed for calculation because of 5 : 6 ) now solution a is 50 ml and solu b is 60 ml further in a 100 ml a contains 84 ml alch so 50 ml contain 42 ml in mix b 100 ml solution contain 40 ml alc , 60 ml of b , it will be 24 ml alcohol . so 42 + 24 = 66 in 110 ml solution which is 60 % answer : c | a = 21 + 4
b = 21 / a
c = 5 + 6
d = 5 / c
e = b * d
f = 2 + 3
g = 2 / f
h = 5 + 6
i = 6 / h
j = g * i
k = e + j
l = k * 100
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a ) 95 , b ) 20 , c ) 85 , d ) 75 , e ) 35 | c | multiply(40, const_1) | at veridux corporation , there are 180 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ? | "well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 180 â € “ 90 = 90 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 180 â € “ 40 = 140 must be associates . now , in the â € œ associate â € row , 135 + e = 140 , which means e = 5 â € ” the other 5 associates must be female . now , to find b , which is what the question is asking , we need only look at the sum in the â € œ female â € column : b + 5 = 90 , which means b = 85 . there are fifteen female managers in this company . thus , the answer = ( c ) ." | a = 40 * 1
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a ) 270 % , b ) 310 % , c ) 350 % , d ) 390 % , e ) 430 % | d | multiply(divide(divide(subtract(multiply(55, const_100), multiply(20, const_100)), subtract(const_100, 55)), 20), const_100) | keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 55 % of the total number of books in the library . if prior to the purchase , only 20 % of the books in keats library were biographies , by what percent has the number of biographies in the library increased ? | "let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.55 ( 5 b + x ) = b + x 1.75 b = 0.45 x x = 3.9 b the answer is d ." | a = 55 * 100
b = 20 * 100
c = a - b
d = 100 - 55
e = c / d
f = e / 20
g = f * 100
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a ) 1667 , b ) 6789 , c ) 1200 , d ) 1100 , e ) 1421 | d | divide(836, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 836 . find the actual cost of an article ? | cp * ( 76 / 100 ) = 836 cp = 11 * 100 = > cp = 1100 answer : d | a = 24 / 100
b = 1 - a
c = 836 / b
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a ) 4 / 6 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 2 / 3 | a | divide(sqrt(16), sqrt(36)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 36 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 16 / 36 - - > height / height = 4 / 6 . answer : a ." | a = math.sqrt(16)
b = math.sqrt(36)
c = a / b
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a ) 225 , b ) 275 , c ) 325 , d ) 350 , e ) 600 | e | divide(add(1500, 300), const_3) | in a weight - lifting competition , the total weight of joe ' s two lifts was 1500 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | "this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 1500 pounds . we sum the two variables to obtain : f + s = 1500 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f – 300 = s we can now plug in ( 2 f – 300 ) for s into the first equation , so we have : f + 2 f – 300 = 1500 3 f = 1,800 f = 600 answer is e ." | a = 1500 + 300
b = a / 3
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a ) 123 , b ) 109 , c ) 100 , d ) 156 , e ) 240 | b | subtract(109.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 109.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 109.25 - 0.25 ) = rs . 109 . answer : b" | a = 1 / 4
b = 109 - 25
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a ) 110 , b ) 330 , c ) 550 , d ) 462 , e ) 880 | d | multiply(divide(880, const_100), subtract(const_100, 47.5)) | if 47.5 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "students enrolled in biology are 47.5 % and therefore not enrolled are 52.5 % . so of 880 is 880 * . 525 = 462 answer is d 462" | a = 880 / 100
b = 100 - 47
c = a * b
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a ) 173 m , b ) 200 m , c ) 273 m , d ) 300 m , e ) 373 m | c | multiply(add(sqrt(const_3), const_1), 100) | two ships are sailing in the sea on the two sides of a lighthouse . the angle of elevation of the top of the lighthouse is observed from the ships are 30 ° and 45 ° respectively . if the lighthouse is 100 m high , the distance between the two ships is : | let ab be the lighthouse and c and d be the positions of the ships . then , ab = 100 m , acb = 30 ° and adb = 45 ° . ab = tan 30 ° = 1 ac = ab x 3 = 1003 m . ac 3 ab = tan 45 ° = 1 ad = ab = 100 m . ad cd = ( ac + ad ) = ( 1003 + 100 ) m = 100 ( 3 + 1 ) = ( 100 x 2.73 ) m = 273 m . answer = c | a = math.sqrt(3)
b = a + 1
c = b * 100
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a ) 42 , b ) 45 , c ) 48 , d ) 50 , e ) 55 | c | divide(multiply(divide(multiply(10, 60), subtract(80, 60)), 80), add(20, divide(multiply(10, 60), subtract(80, 60)))) | a tank can supply water to a village for 80 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 60 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 10 liters per day results in a loss of 600 liters in 60 days . so , those 600 liters were for 20 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 80 = 2400 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last 2400 / 50 = 48 days . the answer is c ." | a = 10 * 60
b = 80 - 60
c = a / b
d = c * 80
e = 10 * 60
f = 80 - 60
g = e / f
h = 20 + g
i = d / h
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a ) 20 days , b ) 10 days , c ) 6 8 / 7 days , d ) 8 days , e ) 7 days | c | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 25))) | a can do a work in 25 days and b can do it in 10 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 25 b ' s 1 day ' s work = 1 / 10 they work together = 1 / 25 + 1 / 10 = 7 / 50 = 50 / 7 = 6 8 / 7 days answer : option c" | a = 1 / 10
b = 1 / 25
c = a - b
d = 1 / c
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a ) 28 , b ) 29 , c ) 34 , d ) 38 , e ) 36 | d | subtract(add(20, 20), add(1, 1)) | a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the difference between the maximum and minimum possible values of ' a - b ' ? | possible values of a and b can be 54 ; 45 and 120 ; 201 maximum possible value for a - b is 20 - 1 = 19 minimum possible value for a - b is 1 - 20 = - 19 19 - ( - 19 ) = 38 ans is d | a = 20 + 20
b = 1 + 1
c = a - b
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['a ) 2', 'b ) 4', 'c ) 34', 'd ) 38', 'e ) 40'] | c | divide(subtract(power(18, const_2), power(16, const_2)), const_2) | the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 18 - inch screen would be how many square inches greater than the area of a square 16 - inch screen ? | pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 18 ^ 2 / 2 - 16 ^ 2 / 2 = 68 / 2 = 34 ans c . | a = 18 ** 2
b = 16 ** 2
c = a - b
d = c / 2
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a ) 6 , b ) 9 , c ) 12 , d ) 24 , e ) 48 | c | divide(multiply(divide(80, const_100), 24), divide(70, const_100)) | an alloy weighing 24 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 80 percent gold ? | "in 24 ounces , gold is 24 * ( 70 / 100 ) = 16.8 ounces . now we add x ounces of pure gold to make it 90 % gold . so 16.8 + x = ( 24 + x ) * 80 / 100 = > x = 12 . answer is c ." | a = 80 / 100
b = a * 24
c = 70 / 100
d = b / c
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a ) 1000 , b ) 2217 , c ) 2889 , d ) 2777 , e ) 1300 | e | divide(1456, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1456 ? | "1456 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1300 answer : e" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1456 / f
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a ) 1123 , b ) 1234 , c ) 1349 , d ) 1455 , e ) 1567 | c | add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(add(4, 3), add(4, 3))) | what is the 4 digit no . in which the 1 st digit is 1 / 3 of the second , the 3 rd is the sum of the 1 st and 2 nd , and the last is 3 times the second ? | first digit is 1 / 3 second digit = > the numbers can be 1 & 3 , 2 & 6 , 3 & 9 . first + second = third = > we can eliminate 3 & 9 since 3 + 9 = 12 . last is 3 times the second = > we can eliminate option 2 & 6 since 3 * 6 = 18 . hence the number is 1349 c | a = 100 * 10
b = 3 * 100
c = a + b
d = 4 + 3
e = 4 + 3
f = d * e
g = c + f
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a ) 101 , b ) 225 , c ) 304 , d ) 324 , e ) 336 | a | divide(76, subtract(const_1, divide(const_1, const_4))) | what number is 76 more than one - fourth of itself ? | "1 / 4 x + 76 = x that means 76 = 3 / 4 x x = ( 76 * 4 ) / 3 = 304 / 3 = 101 a is the answer" | a = 1 / 4
b = 1 - a
c = 76 / b
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a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint w , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , lets further reduce 0.75 : 1.25 gives 1 + 1.5 = 2.5 gallons of mixture . this looks ok , but lets reduce further just to be sure 0.375 : 0.625 gives 0.5 + 1 = 1.5 gallons of mixture , thats less than 2 gallons of mixture , so not acceptable . so correct ans is 2.5 gallons . b" | a = const_0_25 * 2
b = 2 * 5
c = a + b
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a ) 600 ft , b ) 660 ft , c ) 670 ft , d ) 680 ft , e ) 700 ft | b | multiply(30, divide(multiply(15, 5280), const_3600)) | someone on a skateboard is traveling 15 miles per hour . how many feet does she travel in 30 seconds ? ( 1 mile = 5280 feet ) | "per second = > 15 * 5280 ft / 60 * 60 = 22 ft 30 seconds = > 22 * 30 = 660 ft answer : b" | a = 15 * 5280
b = a / 3600
c = 30 * b
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a ) 2000 , b ) 4500 , c ) 5000 , d ) 7200 , e ) 9000 | d | divide(72000, 10) | a company produces 72000 bottles of water everyday . if a case can hold 10 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 10 number of cases required to hold 72000 bottles = 72000 / 10 = 7200 cases . so the answer is d = 7200" | a = 72000 / 10
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a ) 100 , b ) 120 , c ) 149 , d ) 150 , e ) 151 | d | add(add(power(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2))) | the sum of 99 consecutive integers is 9999 . what is the greatest integer in the set ? | "sum of n numbers = ( n ( 2 a + ( n - 1 ) d ) ) / 2 where n = no of terms a = 1 st term , d - difference between two terms ( ie d = a 2 - a 1 ) here sum = 9999 , n = 99 d = 1 ( since consecutive numbers ) 9999 = ( 99 / 2 ) * ( 2 a + ( 99 - 1 ) 1 ) from this ' a ' ( ie the 1 st term ) = 52 nth term in a sequence : nth term = a + ( n - 1 ) d here a = 52 , n = 99 , d = 1 so nth term = 52 + ( 99 - 1 ) * 1 therefore nth term = 150 answer : d" | a = 99 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 99 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 99 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 99 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 4 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | add(divide(multiply(add(24, const_1), 60), const_1000), 3.5) | a train of 24 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 3.5 km long ? | "d = 25 * 60 + 3500 = 5000 m t = 5000 / 60 * 18 / 5 = 300 sec = 5 mins answer : c" | a = 24 + 1
b = a * 60
c = b / 1000
d = c + 3
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a ) 9 % decrease , b ) 8 % decrease , c ) 6 % decrease , d ) 1 % decrease , e ) 2 % decrease | b | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | the tax on a commodity is diminished by 20 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 80 * 115 = 9200 - - - - - - - - - - - 10000 - - - - - - - - - - - 800 100 - - - - - - - - - - - ? = > 8 % decrease answer : b" | a = 15 / 100
b = 1 + a
c = 20 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | multiply(sqrt(divide(divide(484, 3), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 484 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 484 / 3 = 161.33 sq m l b = 161.33 i . e . , l * l / 3 = 161.33 l ^ 2 = 484 = > l = 22 . d" | a = 484 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
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a ) 125 , b ) 272 , c ) 278 , d ) 277 , e ) 112 | a | multiply(divide(multiply(50, const_1000), const_3600), 9) | a train running at the speed of 50 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 9 = 125 meter answer : a" | a = 50 * 1000
b = a / 3600
c = b * 9
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 33 ? | "first , let ' s translate the expression 21 # 33 , using the definition given : 21 # 33 = 21 × 24 × 27 × 30 × 33 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 33 = 3 ^ 6 ( 7 × 8 × 9 × 10 × 11 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 33 = 3 ^ 5 ( 7 × 2 ^ 3 × 3 ^ 2 × ( 2 × 5 ) × 11 ) let ' s group the prime bases : 21 # 33 = 2 ^ 4 × 3 ^ 7 × 5 × 7 × 11 the sum of the exponents is 4 + 7 + 1 + 1 + 1 = 14 the answer is c ." | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
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a ) 775 , b ) 620 , c ) 750 , d ) 800 , e ) 720 | a | divide(620, subtract(const_1, divide(20, const_100))) | after decreasing 20 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 620 cp = 7.75 * 100 = > cp = 775 answer : a" | a = 20 / 100
b = 1 - a
c = 620 / b
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a ) 32256 , b ) 24000 , c ) 24936 , d ) 25600 , e ) none | d | multiply(divide(const_100, 90), 23040) | 90 % of the population of a village is 23040 . the total population of the village is ? | "answer ∵ 90 % of p = 23040 ∴ p = ( 23040 x 100 ) / 90 = 25600 correct option : d" | a = 100 / 90
b = a * 23040
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a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 8,820 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 5 * 8820 = 44 , 100 , which is a square of 210 b" | a = 2 + 3
b = a * 2
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a ) 36.21 , b ) 77.11 , c ) 54.12 , d ) 33.0 , e ) 20.25 | e | subtract(multiply(40, multiply(const_60.0, const_0_2778)), 100) | a train 100 m long crosses a platform 125 m long in 40 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 40 s = 225 / 40 * 18 / 5 = 20.25 kmph answer : e" | a = const_60 * 0
b = 40 * a
c = b - 100
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a ) 15 , b ) 48 , c ) 54 , d ) 33 , e ) 45 | b | divide(multiply(8, 18), subtract(18, 15)) | a group of men decided to do a work in 15 days , but 8 of them became absent . if the rest of the group did the work in 18 days , find the original number of men ? | "original number of men = 8 * 18 / ( 18 - 15 ) = 48 answer is b" | a = 8 * 18
b = 18 - 15
c = a / b
|
a ) 1 / 5 , b ) 3 / 20 , c ) 1 / 7 , d ) 2 / 15 , e ) 3 / 28 | e | multiply(divide(const_2, 5), divide(const_1, const_4)) | x , y , and z are all unique numbers . if x is chosen randomly from the set { 5 , 6 , 7 , 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "p ( x is prime ) = 3 / 7 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 3 / 7 * 1 / 4 = 3 / 28 the answer is e ." | a = 2 / 5
b = 1 / 4
c = a * b
|
a ) 14 days , b ) 20 days , c ) 22 days , d ) 24 days , e ) 26 days | d | divide(multiply(2, const_3), subtract(divide(add(divide(multiply(2, const_3), 4), add(divide(multiply(2, const_3), 2), divide(multiply(2, const_3), 3))), const_2), divide(multiply(2, const_3), 2))) | a and b can do a piece of work in 2 days , b and c in 3 days , c and a in 4 days . how long will c take to do it ? | "2 c = 1 / 3 + 1 / 4 – 1 / 2 = 1 / 12 c = 1 / 24 = > 24 days answer : d" | a = 2 * 3
b = 2 * 3
c = b / 4
d = 2 * 3
e = d / 2
f = 2 * 3
g = f / 3
h = e + g
i = c + h
j = i / 2
k = 2 * 3
l = k / 2
m = j - l
n = a / m
|
a ) $ 54.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 134.40 | e | divide(multiply(subtract(1720, 600), 12), const_100) | a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 12 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : e the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.12 ( $ 1120 ) = $ 134.40 . the correct answer is e ." | a = 1720 - 600
b = a * 12
c = b / 100
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | a | divide(subtract(36, subtract(multiply(4, divide(10, 2)), 10)), 10) | the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 36 , what is the value of x ? | "the areas are ( x + 5 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 5 and 2 x - 5 . if we add the two perimeters : 4 ( x + 5 ) + 4 ( 2 x - 5 ) = 36 12 x = 36 x = 3 the answer is a ." | a = 10 / 2
b = 4 * a
c = b - 10
d = 36 - c
e = d / 10
|
a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0003 , e ) 0.0002 | d | divide(0.3, 1000) | when magnified 1000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 0.3 centimeter . the actual diameter of the tissue , in centimeters , is | it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 0.3 then x = 0.3 / 1000 = 0.0003 . the answer is d | a = 0 / 3
|
a ) 66 , b ) 46 , c ) 49 , d ) 53 , e ) 86 | a | add(add(multiply(5, const_2), 3), add(multiply(5, multiply(const_2, 3)), 3)) | a group of n students can be divided into equal groups of 4 with 3 student left over or equal groups of 5 with 3 students left over . what is the sum of the two smallest possible values of n ? | "4 x + 3 = 5 y + 3 . . . . . . . . . . . ie : 4 x - 5 y = 0 x , y must be > 1 and y is even ie ( 2 , 4,6 , . . etc ) if y = 2 thus x = fraction ( not possible ) if y = 4 thus x = 5 n = 23 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 43 23 + 43 = 66 . . . . . a" | a = 5 * 2
b = a + 3
c = 2 * 3
d = 5 * c
e = d + 3
f = b + e
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(add(add(add(add(1, 1), 1), 3), 4), 1) | q - 1 : how many two digit numbers of distinct digits can be formed by using digits 1 , 2 , 3 , 4 , 5 , 6 and 7 such that the numbers are divisible by 5 ? | "6 two digit numbers divisible by 5 must end in either 0 or 5 . since there is no 0 , all correct answers are a combination of the other digits and 5 . there are six other numbers plus the 5 so there are 6 correct answers . correct answer is d ." | a = 1 + 1
b = a + 1
c = b + 3
d = c + 4
e = d + 1
|
a ) 125 miles , b ) 225 miles , c ) 585 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 9) | a car travels at a speed of 65 miles per hour . how far will it travel in 9 hours ? | "during each hour , the car travels 65 miles . for 9 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 = 9 × 65 = 585 miles correct answer is c ) 585 miles" | a = 65 * 9
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a ) 21 , b ) 30 , c ) 11 , d ) 10 , e ) 9 | b | divide(subtract(105, 45), const_2) | how many of the integers between 45 and 105 are even ? | "number start between 45 to 105 is 60 numbers half of them is even . . which is 30 answer : b" | a = 105 - 45
b = a / 2
|
a ) 654 , b ) 655 , c ) 656 , d ) 657 , e ) 470 | e | multiply(divide(subtract(const_100, 50), const_100), 940.00) | yearly subscription to professional magazines cost a company $ 940.00 . to make a 50 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 50 / 100 = 470 so the cut in amount is 470 the less amount to be spend is 940 - 470 = 470 answer : e" | a = 100 - 50
b = a / 100
c = b * 940
|
a ) 32.25 % , b ) 23.34 % , c ) 36 % , d ) 39 % , e ) 29 % | a | multiply(subtract(multiply(divide(add(const_100, 15), const_100), divide(add(const_100, 15), const_100)), const_1), const_100) | price of a book increases 15 % successively ( 2 times ) what is the new price of the book more compared to that of the old price : | new price is 1.15 * 1.15 * old price = 1.3225 * old price increase in price = 0.3225 * old price % increase = 100 * 0.3225 * old price / old price = 32.25 % answer : a | a = 100 + 15
b = a / 100
c = 100 + 15
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 90 , b ) 120 , c ) 315 , d ) 180 , e ) 200 | c | divide(multiply(multiply(7, const_4), multiply(6, 7)), power(factorial(2), 2)) | there are 7 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 7 c 2 number of ways of selecting 2 non fiction books = 6 c 2 7 c 2 * 6 c 2 = 21 * 15 = 315 answer : c" | a = 7 * 4
b = 6 * 7
c = a * b
d = math.factorial(2)
e = d ** 2
f = c / e
|
a ) 1000 , b ) 975 , c ) 980 , d ) 1020 , e ) 1080 | b | add(multiply(6, 74), multiply(9, 59)) | andrew purchased 6 kg of grapes at the rate of 74 per kg and 9 kg of mangoes at the rate of 59 per kg . how much amount did he pay to the shopkeeper ? | cost of 6 kg grapes = 74 × 6 = 444 . cost of 9 kg of mangoes = 59 × 9 = 531 . total cost he has to pay = 444 + 531 = 975 b | a = 6 * 74
b = 9 * 59
c = a + b
|
a ) $ 24000 , b ) $ 23000 , c ) $ 22000 , d ) $ 21000 , e ) $ 20000 | e | multiply(4, 5000) | the ratio of the incomes of uma and bala is 4 : 3 and the ratio of their expenditure is 3 : 2 . if at the end of the year , each saves $ 5000 then the income of uma is ? | "let the income of uma and bala be $ 4 x and $ 3 x let their expenditures be $ 3 y and $ 2 y 4 x - 3 y = 5000 - - - - - - - 1 ) 3 x - 2 y = 5000 - - - - - - - 2 ) from 1 ) and 2 ) x = 5000 uma ' s income = 4 x = 4 * 5000 = $ 20000 answer is e" | a = 4 * 5000
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a ) 392 , b ) 229 , c ) 756 , d ) 493 , e ) 540 | c | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5040) | 3 / 4 of 1 / 2 of 2 / 5 of 5040 = ? | c 756 ? = 5040 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 756 | a = 3 / 4
b = 1 / 2
c = a * b
d = 2 / 5
e = c * d
f = e * 5040
|
a ) 156 , b ) 167 , c ) 157 , d ) 342 , e ) 380 | d | multiply(add(17, const_1), add(add(17, const_1), const_1)) | there are 17 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 19 from 19 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 19 p ₂ ways . ² ⁰ p ₂ = 19 * 18 = 342 . answer : d" | a = 17 + 1
b = 17 + 1
c = b + 1
d = a * c
|
a ) 800 , b ) 380 , c ) 360 , d ) 478 , e ) 566 | b | multiply(add(18, const_2), subtract(add(18, const_2), const_1)) | there are 18 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 20 from 20 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 20 p 2 ways . 20 p 2 = 20 * 19 = 380 answer : b" | a = 18 + 2
b = 18 + 2
c = b - 1
d = a * c
|
a ) 2 / 15 , b ) 7 / 15 , c ) 10 / 30 , d ) 7 / 30 , e ) 8 / 30 | b | divide(3, const_10) | tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ? | "here , the sample space s = ( 1 , 2 , 3 , 4 , 5 , . . . , 29 , 30 ) . let e = the event of getting a multiple of 3 or 5 . e = ( 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 20 , 21 , 24 , 25 , 27 , 30 ) p ( e ) = n ( e ) / n ( s ) = 14 / 30 = 7 / 15 answer : b" | a = 3 / 10
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a ) 96 kmph , b ) 94 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 6), const_3_6) | a 160 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | s = 160 / 6 * 18 / 5 = 96 kmph answer : a | a = 160 / 6
b = a * const_3_6
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a ) 12 : 1 , b ) 1 : 2 , c ) 4 : 1 , d ) 2 : 5 , e ) 2 : 6 | b | divide(subtract(divide(6, add(3, 7)), divide(3, 7)), subtract(divide(3, 7), divide(3, add(3, 7)))) | an alloy of copper and zinc contains copper and zinc in the ratio 6 : 4 . another alloy of copper and zinc contains copper and zinc in the ratio 3 : 7 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 6 x / 10 , and fraction of zinc in alloy _ 1 = 4 x / 10 similarly , fraction of copper in alloy _ 2 = 3 y / 10 , and fraction of zinc in alloy _ 2 = 7 y / 10 . mixing them , we get copper = 6 x / 10 + 3 y / 10 ; zinc = 4 x / 10 + 7 y / 10 so , 6 x + 3 y = 4 x + 7 y - > 2 x = 4 y - > x / y = 2 / 4 = 1 / 2 so , they must be mixed in the ratio 1 : 2 answer : b" | a = 3 + 7
b = 6 / a
c = 3 / 7
d = b - c
e = 3 / 7
f = 3 + 7
g = 3 / f
h = e - g
i = d / h
|
a ) 5.18 % , b ) 6.18 % , c ) 7.18 % , d ) 8.18 % , e ) . 7 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 7,500 gallons of a solution that is 6 percent sodium chloride by volume . if 2,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 7,500 gallons of a solution that is 6 % sodium chloride by volume . this means that there are 0.06 x 7,500 = 450 gallons of sodium chloride . when 2,000 gallons of water evaporate we are left with 5,500 gallons of solution . from here we can determine what percent of the 5,500 gallon solution is sodium chloride . ( sodium chloride / total solution ) x 100 = ? ( 450 / 5,500 ) x 100 = ? 0.081 x 100 = ? = 8.18 % answer is d ." | a = 100 * 100
b = 6 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | b | divide(subtract(add(multiply(40.2, 10), add(13, 16)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 16 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 16 = 404 hence , new average = 404 ⁄ 10 = 40.4 answer b" | a = 40 * 2
b = 13 + 16
c = a + b
d = c - 31
e = d / 10
|
a ) 177 , b ) 150 , c ) 180 , d ) 716 , e ) 616 | c | divide(705, add(add(divide(4, 3), divide(4, 5)), const_1)) | rs . 705 is divided amongst a , b , c so that 3 times a ' s share , 5 times b ' s share and 4 times c ' s share are all equal . find b ' s share ? | a + b + c = 705 3 a = 5 b = 4 c = x a : b : c = 1 / 3 : 1 / 5 : 1 / 4 = 20 : 12 : 15 12 / 47 * 705 = rs . 180 answer : c | a = 4 / 3
b = 4 / 5
c = a + b
d = c + 1
e = 705 / d
|
a ) 187 , b ) 133 , c ) 142 , d ) 178 , e ) 175 | b | subtract(subtract(175, divide(multiply(175, 20), const_100)), divide(multiply(subtract(175, divide(multiply(175, 20), const_100)), 5), const_100)) | the sale price sarees listed for rs . 175 after successive discount is 20 % and 5 % is ? | "175 * ( 80 / 100 ) * ( 95 / 100 ) = 133 answer : b" | a = 175 * 20
b = a / 100
c = 175 - b
d = 175 * 20
e = d / 100
f = 175 - e
g = f * 5
h = g / 100
i = c - h
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a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 12 | e | add(7, divide(multiply(7, subtract(12000, 9500)), subtract(9500, 6000))) | the average salary of all the workers in a workshop is rs . 9500 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | let the total number of workers be x . then , 9500 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = > 3500 x = 42000 = x = 12 . answer : e | a = 12000 - 9500
b = 7 * a
c = 9500 - 6000
d = b / c
e = 7 + d
|
a ) 1,4 , b ) - 1,4 , c ) 2,3 , d ) 2 , - 3 , e ) 3,4 | b | multiply(4, 2) | solve x 2 – 3 x – 4 = 0 . | "this one factors easily : x 2 – 3 x – 4 = 0 ( x + 1 ) ( x – 4 ) = 0 x + 1 = 0 or x – 4 = 0 x = – 1 or x = 4 the solution is x = – 1 , 4 answer b - 1,4" | a = 4 * 2
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a ) 223.78 , b ) 37.78 , c ) 246.0 , d ) 278.88 , e ) 102.5 | e | divide(61.50, divide(const_4, 5)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 2 / 5 % his yearly income diminishes by rs . 61.50 . his capital is | "explanation : capital = rs . x , then 3 / 5 x = 61.5 x = 102.50 answer : e ) rs . 102.50" | a = 4 / 5
b = 61 / 50
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a ) 200000 , b ) 20000 , c ) 200029 , d ) 288778 , e ) 277789 | b | divide(482, subtract(power(add(divide(divide(20, const_2), const_100), const_1), const_4), power(add(divide(20, const_100), const_1), const_2))) | a sum of money lent at compound interest for 2 years at 20 % per annum would fetch rs . 482 more , if the interest was payable half yearly than if it was payable annually . the sum is | explanation : let sum = rs . x c . i . when compounded half yearly = c . i . when compounded annually = = > x = 20000 answer : b ) 20000 | a = 20 / 2
b = a / 100
c = b + 1
d = c ** 4
e = 20 / 100
f = e + 1
g = f ** 2
h = d - g
i = 482 / h
|
a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | b | subtract(divide(20, subtract(3, 2)), 2) | a monkey start climbing up a tree 20 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 17 hrs it wil be 17 ft . bt since the height of the tree is 20 ft , so if the monkey hops up the tree in the next hr i . e 18 th hr then it reaches at the top of the tree . hence it takes 18 hrs for monkey to reach at the top answer : b | a = 3 - 2
b = 20 / a
c = b - 2
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a ) 25.5 , b ) 16.7 , c ) 25 , d ) . 25 , e ) none of these | b | divide(1, 0.06) | 1 / 0.06 is equal to | "explanation : 1 / 0.06 = ( 1 * 100 ) / 6 = 100 / 6 = 16.7 option b" | a = 1 / 0
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a ) s 4000 , b ) s 2800 , c ) s 1800 , d ) s 3500 , e ) s 3100 | c | divide(subtract(multiply(3600, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3600 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 3600 ) * 100 = 4 ratio = 1 : 1 so , first part = ( 1 / 2 ) * 3600 = rs 1800 . answer : c" | a = 5 / 100
b = 3600 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(const_60, multiply(divide(160, 200), const_60)) | without stoppages , a train travels certain distance with an average speed of 200 km / h , and with stoppages , it covers the same distance with an average speed of 160 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 40 km less . time taken to cover 40 km = 40 â „ 200 h = 1 â „ 5 h = 1 â „ 5 ã — 60 min = 12 min answer e" | a = 160 / 200
b = a * const_60
c = const_60 - b
|
a ) 320.8 $ , b ) 387.8 $ , c ) 420.8 $ , d ) 457.8 $ , e ) 480.8 $ | d | multiply(multiply(0.65, 56), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 56 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 56 ) = 457.8 hence answer will be ( d )" | a = 0 * 65
b = a * 12
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a ) 66 % , b ) 67.5 % , c ) 69 % , d ) 70 % , e ) 72 % | a | add(multiply(divide(divide(25, const_100), subtract(1, divide(1, 2))), const_100), 2) | the price of an item is discounted 2 percent on day 1 of a sale . on day 2 , the item is discounted another 12 percent , and on day 3 , it is discounted an additional 25 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 2 % discount = 980 price in day 2 after 12 % discount = 862.4 price in day 3 after 25 % discount = 646.8 so , price in day 3 as percentage of the sale price on day 1 will be = 646.8 / 980 * 100 = > 66 % answer will definitely be ( a )" | a = 25 / 100
b = 1 / 2
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 1150 , b ) 1160 , c ) 1140 , d ) 1152 , e ) 1145 | b | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 5,7) | find the smallest number of four digits exactly divisible by 4 , 5,7 and 8 . | "smallest number of four digits is 1000 . required number must be divisible by l . c . m . of 4,5 , 7,8 i . e 280 , on dividing 1000 by 280 , we get 120 as remainder . therefore , required number = 1000 + ( 280 â € “ 120 ) = 1160 . answer is b ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 5
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a ) 25000 , b ) 20000 , c ) 18000 , d ) 14000 , e ) 15000 | e | divide(divide(multiply(2700, const_100), 6), 3) | a sum fetched a total simple interest of rs . 2700 at the rate of 6 p . c . p . a . in 3 years . what is the sum ? | "sol . principal = rs . [ 100 * 2700 / 6 * 3 ] = rs . [ 270000 / 18 ] = rs . 15000 . answer e" | a = 2700 * 100
b = a / 6
c = b / 3
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a ) 10 m , b ) 14 m , c ) 17 m , d ) 19 m , e ) 20 m | b | multiply(divide(56, multiply(20, 21)), multiply(35, 3)) | if 20 men can build a water fountain 56 metres long in 21 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 21 : 3 : : 56 : x therefore ( 20 x 21 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 420 = 14 hence , the required length is 14 m . answer : b" | a = 20 * 21
b = 56 / a
c = 35 * 3
d = b * c
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a ) 1 , b ) 2 , c ) 10 , d ) 8 , e ) 16 | c | multiply(const_2, sqrt(power(5, const_2))) | a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 5 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 5 ( pir ^ 2 ) pir ^ 2 = 10 pir r = 10 answer : c" | a = 5 ** 2
b = math.sqrt(a)
c = 2 * b
|
a ) 4 , b ) 2 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(multiply(3, 6), add(subtract(13, 6), 3)), 6) | the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 3 . if the sum of the first and the last number is 13 . what is the last numbers ? | a + b + c = 18 b + c + d = 9 a + d = 13 a – d = 9 a + d = 13 2 d = 4 d = 2 answer : b | a = 3 * 6
b = 13 - 6
c = b + 3
d = a - c
e = d - 6
|
a ) 16 , b ) 24 , c ) 32 , d ) 48 , e ) 54 | c | multiply(multiply(const_2, 8), const_2) | a classroom has equal number of boys and girls . 8 girls left to play kho - kho , leaving twice as many boys as girls in the classroom . what was the total number of girls and boys present initially ? | after 8 girls left remaining 8 girls now boys 16 are twice as many as remaining girls . initially boys = 16 and girls = 16 . answer : c | a = 2 * 8
b = a * 2
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a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | d | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . what is the present age of the mother ? | "let present age of the mother = 5 x then , present age of the person = 2 x 5 x + 8 = 2 ( 2 x + 8 ) 5 x + 8 = 4 x + 16 x = 8 present age of the mother = 5 x = 40 answer is d ." | a = 2 + 3
b = 8 - a
c = 1 / 2
d = 2 + 3
e = 2 / d
f = c - e
g = b / f
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(add(const_4, const_2), const_1) | a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube ? | "given : positive integer n is a perfect square and a perfect cube - - > n is of a form of n = x 6 for some positive integer x - - > 0 < x ^ 6 < 10 ^ 3 - - > 0 < x ^ 2 < 10 - - > x can be 1 , 2 or 3 hence nn can be 1 ^ 6 , 2 ^ 6 or 3 ^ 6 . answer : b ." | a = 4 + 2
b = a / 1
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a ) 30 kmph , b ) 40 kmph , c ) 78 kmph , d ) 15 kmph , e ) 23 kmph | a | subtract(50, 20) | the speed of a boat in still water is 50 kmph and the speed of the current is 20 kmph . find the speed and upstream ? | speed downstream = 50 + 20 = 70 kmph speed upstream = 50 - 20 = 30 kmph answer : a | a = 50 - 20
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a ) 48 / 25 , b ) 36 / 25 , c ) 6 / 5 , d ) 5 / 6 , e ) 25 / 36 | a | multiply(divide(divide(1, const_3.0), divide(1, 5)), divide(6, 5)) | if 5 x = 6 y and xy ≠ 0 , what is the ratio of 1 / 5 * x to 1 / 8 * y ? | "5 x = 6 y = > x / y = 6 / 5 1 / 5 * x to 1 / 8 * y = x / y * 8 / 5 = ( 6 / 5 ) * ( 8 / 5 ) = 48 / 25 ans : a" | a = 1 / 3
b = 1 / 5
c = a / b
d = 6 / 5
e = c * d
|
a ) 4.5 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 6.5 | b | multiply(divide(10, 12), 6) | when a number is divided by 6 & then multiply by 12 the answer is 10 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 10 = > 2 x = 10 = > x = 5 b" | a = 10 / 12
b = a * 6
|
a ) 6 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | subtract(divide(factorial(subtract(divide(13, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(13, const_2), const_1)) | a company that ships boxes to a total of 13 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter ) | "back - solving is the best way to solve this problem . you basically need 13 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - b" | a = 13 / 2
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = math.factorial(2)
f = d * e
g = c / f
h = 13 / 2
i = h - 1
j = g - i
|
a ) 160 km , b ) 170 km , c ) 180 km , d ) 190 km , e ) 150 km | a | add(multiply(20, subtract(12, 10)), multiply(40, const_3)) | a and b start from house at 10 am . they travel on the mg road at 20 kmph and 40 kmph . there is a junction t on their path . a turns left at t junction at 12 : 00 noon , b reaches t earlier , and turns right . both of them continue to travel till 2 pm . what is the distance between a and b at 2 pm | at 12 a will travel = 20 * 2 = 40 km b will travel this 40 km in 40 / 40 = 1 hr i . e . by 11 am . after t junction for a - distance travelled = 2 * 20 = 40 km for b distance travelled = 40 * 3 = 120 so distance between a & b is = 120 + 40 = 160 km answer : a | a = 12 - 10
b = 20 * a
c = 40 * 3
d = b + c
|
a ) 4 / 7 , b ) 5 / 7 , c ) 3 / 7 , d ) 2 / 7 , e ) 1 / 7 | d | divide(4, add(add(add(subtract(1980, 1972), const_1), 4), const_1)) | the circulation for magazine p in 1971 was 4 times the average ( arithmetic mean ) yearly circulation for magazine p for the years 1972 - 1980 . what is the ratio of the circulation in 1971 to the total circulation during 1971 - 1980 for magazine p ? | there are 9 years from 1972 - 1980 , inclusive . let ' s say the average circulation every year between 1972 - 1980 inclusive is x . so the total circulation is 9 x from 1972 - 1980 , inclusive . in 1971 , the circulation is 4 x . so total circulation for 1971 - 1980 is 4 x + 9 x = 13 x . ratio of circulation in 1971 to total circulation during 1971 - 1980 is 4 x to 14 x or 4 / 14 = 2 / 7 answer : d | a = 1980 - 1972
b = a + 1
c = b + 4
d = c + 1
e = 4 / d
|
a ) $ 16.32 , b ) $ 18.36 , c ) $ 21.60 , d ) $ 24.48 , e ) $ 28.80 | b | multiply(divide(subtract(const_100, 15), const_100), multiply(0.30, 72)) | the regular price per can of a certain brand of soda is $ 0.30 . if the regular price per can is discounted 15 percent when the soda is purchased in 24 - can cases , what is the price of 72 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.85 ) ( $ 0.30 ) , or $ 0.255 . therefore , the price of 72 cans of soda at the discounted price would be ( 72 ) ( $ 0.255 ) = 18.36 answer : b ." | a = 100 - 15
b = a / 100
c = 0 * 30
d = b * c
|
a ) 48 , b ) 24 , c ) 60 , d ) 72 , e ) 84 | a | divide(12, subtract(const_1, sqrt(divide(9, 16)))) | a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 9 / 16 . how many marbles are in the bag ? | "ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) / x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probability of pulling a non - red marble out is ( x - 12 ) / x probability theorm states that if the probability of event a occuring is m and the probability of event b occuring is n then the probability of both a and b occuring is m * n so therefore the probability of 2 non - red marbles getting pulled out is [ ( x - 12 ) / x ] * [ ( x - 12 ) / x ] this is given as 9 / 16 - - > ( x - 12 ) ^ 2 = 16 / 25 x ^ 2 a square rooting u have x - 12 / x = 3 / 4 or 4 x - 48 = 3 x or x = 48" | a = 9 / 16
b = math.sqrt(a)
c = 1 - b
d = 12 / c
|
a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 53.1 | e | divide(add(multiply(26, 40), multiply(50, 60)), add(26, 50)) | the average marks of a class of 26 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | sum of the marks for the class of 26 students = 26 * 40 = 1040 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 76 students = 1040 + 3000 = 4040 average marks of all the students = 4040 / 76 = 53.15 answer : e | a = 26 * 40
b = 50 * 60
c = a + b
d = 26 + 50
e = c / d
|
a ) 51 , b ) 50 , c ) 88 , d ) 65 , e ) 22 | a | divide(divide(subtract(125, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 125 m long passes a man , running at 6 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "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 - 6 ) km / hr . x - 6 = 45 = = > x = 51 km / hr . answer : a" | a = 6 * const_0_2778
b = a * 6
c = 125 - b
d = c / 6
e = d / const_0_2778
|
a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min | c | multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10) | tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many l minutes it takes john to reach the station ? | since the question states “ after 15 minutes ” , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john ’ s speed is 10 km / hour , which means 1 km / 6 minutes . as he has to travel 5 / 4 km more , it is going to take him 6 ( 5 / 4 ) minutes . hence , l 6 ( 5 / 4 ) = 15 / 2 minutes . the correct answer is c . | a = 15 - 10
b = 15 / const_60
c = a * b
d = c * const_60
e = d / 10
f = e / 2
g = f - const_0_25
h = g * 10
|
a ) 5 , b ) 24 , c ) 78 , d ) 90 , e ) 54 | b | divide(multiply(12, 8), subtract(12, 8)) | pipe a can fill a tank in 8 hours . due to a leak at the bottom , it takes 12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 8 - 1 / x = 1 / 12 = > 1 / x = 1 / 8 - 1 / 12 = ( 3 - 2 ) / 24 = 1 / 24 = > x = 24 . answer : b" | a = 12 * 8
b = 12 - 8
c = a / b
|
a ) 1 / 38 , b ) 1 / 48 , c ) 1 / 56 , d ) 1 / 64 , e ) 1 / 68 | d | multiply(multiply(multiply(divide(const_1, 8), divide(const_1, 8)), divide(const_1, 8)), divide(const_1, 8)) | three 8 faced dice are thrown together . the probability that all the three show the same number on them is ? | "it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 , 666 , 777 and 888 . those are eight in number . further the three dice can fall in 8 * 8 * 8 = 512 ways . hence the probability is 8 / 512 = 1 / 64 answer : d" | a = 1 / 8
b = 1 / 8
c = a * b
d = 1 / 8
e = c * d
f = 1 / 8
g = e * f
|
a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | c | add(50, multiply(50, divide(subtract(90, 50), subtract(90, 50)))) | an escalator is descending at constant speed . a walks down and takes 50 steps to reach the bottom . b runs down and takes 90 steps in the same time as a takes 10 steps . how many steps are visible when the escalator is not operating ? | lets suppose that a walks down 1 step / min and escalator moves n steps / min it is given that a takes 50 steps to reach the bottom in the same time escalator would have covered 50 n steps so total steps on escalator is 50 + 50 n . again it is given that b takes 90 steps to reach the bottom and time taken by him for this is equal to time taken by a to cover 10 steps i . e 10 minutes . so in this 10 min escalator would have covered 10 n steps . so total steps on escalatro is 90 + 10 n again equating 50 + 50 n = 90 + 10 n we get n = 1 hence total no . of steps on escalator is 100 . answer : c | a = 90 - 50
b = 90 - 50
c = a / b
d = 50 * c
e = 50 + d
|
a ) 2.6 kg , b ) 4 kg , c ) 3.4 kg , d ) 1.8 kg , e ) none of these | a | divide(subtract(multiply(30, 10), multiply(multiply(divide(2, const_100), 30), const_100)), multiply(30, const_3)) | how many kg of sugar must be added to 30 kg of 2 % solution of sugar and water to increase it to a 10 % solution ? | explanation : solution : amount of sugar in 30 kg solution = ( 2 / 100 * 30 ) kg = 0.6 kg . let x kg of sugar be added . then , ( 0.6 + x ) / ( 30 + x ) = 10 / 100 60 + 100 x = 300 + 10 x x = 240 / 90 = 8 / 3 = 2.6 answer : a | a = 30 * 10
b = 2 / 100
c = b * 30
d = c * 100
e = a - d
f = 30 * 3
g = e / f
|
a ) 19 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | d | add(divide(18, const_2), 1) | each of the integers from 1 to 18 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even | out of the 18 integers : 9 are odd and 9 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 9 odd numbers first 2 . 10 th number will be the first even number so we need to withdraw at least 10 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 10 . ( d ) | a = 18 / 2
b = a + 1
|
a ) - 1 , b ) - 1 / 2 , c ) - 1 / 4 , d ) 1 / 4 , e ) 1 / 2 | d | divide(1, add(2, 2)) | suppose f ( x ) is an odd function for which f ( x + 2 ) = f ( x ) for all x , and f ( x ) = x 2 for x 2 ( 0 , 1 ) . compute f ( - 3 / 2 ) + f ( 1 ) . | because f is periodic , we know that f ( - 3 / 2 ) = f ( 1 / 2 ) = ( 1 / 2 ) 2 = 1 = 4 . because f is odd , we know that f ( 1 ) = - f ( / 1 ) , but because f is periodic , f ( 1 ) = f ( - 1 ) . therefore , f ( 1 ) = 0 and the answer is 1 = 4 . correct answer d | a = 2 + 2
b = 1 / a
|
a ) 16 % , b ) 12 % , c ) 9 % , d ) 10 % , e ) 45 % | c | multiply(divide(1980, add(multiply(5000, 2), multiply(3000, 4))), const_100) | a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1980 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 1980 100 r + 120 r = 1980 r = 9 % answer : c" | a = 5000 * 2
b = 3000 * 4
c = a + b
d = 1980 / c
e = d * 100
|
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