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 ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | c | add(4, 2) | find the value of log y ( x 4 ) if logx ( y 3 ) = 2 | logx ( y 3 ) = 2 : given x 2 = y 3 : rewrite in exponential form x 4 = y 6 : square both sides x 4 = y 6 : rewrite the above using the log base y logy ( x 4 ) = logy ( y 6 ) = 6 correct answer c | a = 4 + 2
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a ) 16 days . , b ) 17 days . , c ) 18 days . , d ) 19 days . , e ) 42 days . | e | subtract(70, multiply(divide(70, 10), 4)) | arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 70 days . | "they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 70 * 6 / 10 = 42 days . answer : e" | a = 70 / 10
b = a * 4
c = 70 - b
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a ) 37.5 litres , b ) 36.67 litres , c ) 37.67 litres , d ) 36.5 litres , e ) 33.33 litres | e | multiply(100, subtract(const_1, sqrt(divide(16, 36)))) | a 100 - litre mixture of milk and water contains 36 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 16 % . what is the value of x ? | "working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 36 % ( 1 - x / 100 ) ^ 2 = 16 % * 100 % ( 1 - x / 100 ) ^ 2 = 16 / 36 - - - - - - > ( 1 - x / 100 ) ^ 2 = ( 4 / 6 ) ^ 2 100 - x = 400 / 6 x = 33.33 . . . ans e" | a = 16 / 36
b = math.sqrt(a)
c = 1 - b
d = 100 * c
|
a ) 40 , b ) 25 , c ) 38 , d ) 50 , e ) 39 | b | divide(subtract(5, multiply(5, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 5 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 5 ] = ( 1 / 2 ) ( a + 5 ) 2 ( 2 a + 25 ) = 5 ( a + 8 ) a = 25 b" | a = 1 / 2
b = 5 * a
c = 5 - b
d = 1 / 2
e = 2 / 5
f = d - e
g = c / f
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a ) 2.5 , b ) 3.0 , c ) 4.0 , d ) 6.5 , e ) 12.0 | e | divide(divide(0.75, const_0_25), const_0_25) | after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.75 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | "let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.75 x = 12 alternate solution : try to backsolve . initial volume = 12 after one hour - - > ( 1 / 4 ) 12 = 3 after two hours - - > ( 1 / 4 ) 3 = 0.75 answer : e" | a = 0 / 75
b = a / const_0_25
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a ) 146.2 cms , b ) 146.5 cms , c ) 146.9 cms , d ) 142.2 cms , e ) 136.2 cms | a | divide(add(multiply(142, 20), multiply(149, const_10)), 50) | the average height of 20 girls out of a class of 50 is 142 cm . and that of the remaining girls is 149 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 20 × 142 + 30 × 149 / 50 ) = 146.2 cms answer a" | a = 142 * 20
b = 149 * 10
c = a + b
d = c / 50
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a ) 92.5 % , b ) 89.8 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(3, const_100)))), add(600, 400)), const_100) | a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 3 % of bananas were rotten . find the percentage of fruits in good condition ? | total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 × 600 = 9000 / 100 = 90 number of rotten bananas = 3 % of 400 = 12 therefore , total number of rotten fruits = 90 + 12 = 102 therefore number of fruits in good condition = 1000 - 102 = 898 therefore percentage of fruits in good condition = ( 898 / 1000 × 100 ) % = ( 89800 / 1000 ) % = 89.8 % answer : b | a = 600 + 400
b = 15 / 100
c = 600 * b
d = 3 / 100
e = 400 * d
f = c + e
g = a - f
h = 600 + 400
i = g / h
j = i * 100
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a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.20 , e ) $ 6.60 | e | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.40)) | if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.40 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.4 ) = $ 5.6 ( as the charge is 0.4 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 5.6 = 6.6 answer is e ." | a = 1 / 5
b = 1 / 0
c = b - 1
d = c * 0
e = 1 + 0
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a ) a ) 1040 , b ) b ) 1050 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1125 | e | add(multiply(9, 70), multiply(9, 55)) | bruce purchased 9 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 9 kg grapes = 70 × 9 = 630 . cost of 9 kg of mangoes = 55 × 9 = 495 total cost he has to pay = 630 + 495 = 1125 e" | a = 9 * 70
b = 9 * 55
c = a + b
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a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 22 , e ) e ) 15 | d | subtract(35, divide(subtract(35, divide(5, const_2)), const_2)) | sum of two numbers is 35 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 35 2 x – 3 y = 5 x = 22 y = 13 d )" | a = 5 / 2
b = 35 - a
c = b / 2
d = 35 - c
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a ) 179 , b ) 194 , c ) 269 , d ) 177 , e ) 191 | b | divide(add(180, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 180 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 â € “ 2552 = 208 runs . let the highest score be x , hence the lowest score = x â € “ 180 x + ( x - 180 ) = 208 2 x = 388 x = 194 runs answer : b" | a = 60 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 180 + d
f = e / 2
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a ) 10.6 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | c | subtract(multiply(add(19, const_1), 14.8), multiply(19, 15)) | the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.8 kg . the weight of the new student is ? | answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.8 - 19 x 15 ) kg = 11 kg . correct option : c | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
|
a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | b | subtract(add(multiply(const_100, const_10), 800), divide(add(800, add(multiply(const_100, const_10), 800)), const_2)) | john has $ 1,600 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | "let the money spent be x money he is left with after spending = x - 800 total money - - > x + ( x - 800 ) = 1600 solving for x will give x = 1200 , therefore the money he is left with = x - 800 = 1200 - 800 = 400 answer : b" | a = 100 * 10
b = a + 800
c = 100 * 10
d = c + 800
e = 800 + d
f = e / 2
g = b - f
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a ) 352 , b ) 268 , c ) 236 , d ) 252 , e ) 354 | d | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(3, 3)), multiply(const_4, const_2))), const_10) | how many 3 digit number contain number 4 ? | "total 3 digit no . = 9 * 10 * 10 = 900 not containing 4 = 8 * 9 * 9 = 648 total 3 digit number contain 4 = 900 - 648 = 252 answer : d" | a = 1000 - 10
b = 3 * 3
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 11 , b ) 16 , c ) 13 , d ) 14 , e ) 15 | b | multiply(16, const_1) | the total age of a and b is 16 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 16 + b + c = > a ? c = 16 + b ? b = 16 = > c is younger than a by 16 years answer : option b" | a = 16 * 1
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a ) 14 , b ) 13 , c ) 12 , d ) 11 , e ) 8 | e | subtract(subtract(subtract(14, 2), const_4), const_1) | how many positive integers less than 14 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | "the number = 2 a + 3 b < 20 when a = 1 , b = 1 , 2 , 3 , 4 , 5 - > 2 a = 2 ; 3 b = 3 , 6 , 9 - > the number = 5 , 8 , 11 - - > 3 numbers when a = 2 , b = 1 , 2,3 - > . . . . - - > 3 numbers when a = 3 , b = 1,2 , 3,4 - - > . . . . - - > 2 numbers total number is already 8 . look at the answer there is no number greater than 8 - - > we dont need to try any more answer must be e" | a = 14 - 2
b = a - 4
c = b - 1
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a ) 4 days , b ) 5 days , c ) 6 days , d ) 8 days , e ) 2 days | a | divide(multiply(add(6, const_4), 10), add(15, 10)) | if 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days , the time taken by 15 men and 20 boys in doing the same type of work will be : | let 1 man ' s 1 day ' s work = x and 1 boy ' s 1 day ' s work = y . then , 6 x + 8 y = 1 / 10 and 26 x + 48 y = 1 / 2 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 ( 15 men + 20 boy ) ' s 1 day ' s work = ( 15 / 100 + 20 / 200 ) = 1 / 4 . 15 men and 20 boys can do the work in 4 days . answer : a | a = 6 + 4
b = a * 10
c = 15 + 10
d = b / c
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a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 35 | d | add(subtract(87, multiply(17, 3)), 3) | a batsman makes a score of 87 runs in the 17 th match and thus increases his average by 3 . find his average after 17 th match | "explanation : let the average after 17 th match is x then the average before 17 th match is x - 3 so 16 ( x - 3 ) + 87 = 17 x = > x = 87 - 48 = 39 option d" | a = 17 * 3
b = 87 - a
c = b + 3
|
a ) $ 3.50 , b ) $ 4.00 , c ) $ 4.25 , d ) $ 4.50 , e ) $ 5.00 | d | divide(multiply(multiply(3, 5), 1.2), const_4) | having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.20 at the candy store . what is this student ’ s weekly allowance ? | let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 120 cents ( 4 / 15 ) x = 120 x = $ 4.50 the answer is d . | a = 3 * 5
b = a * 1
c = b / 4
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a ) 76 kmph , b ) 6 kmph , c ) 14 kmph , d ) 7 kmph , e ) 4 kmph | d | divide(subtract(divide(78, 2), divide(50, 2)), const_2) | a man rows his boat 78 km downstream and 50 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 78 / ( 2 ) = 39 kmph speed upstream = d / t = 50 / ( 2 ) = 25 kmph the speed of the stream = ( 39 - 25 ) / 2 = 7 kmph answer : d" | a = 78 / 2
b = 50 / 2
c = a - b
d = c / 2
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a ) a ) 270 , b ) b ) 280 , c ) c ) 290 , d ) d ) 300 , e ) e ) 310 | c | divide(add(1465, 15), subtract(6, const_1)) | the difference of two numbers is 1465 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1465 ) . x + 1465 = 6 x + 15 5 x = 1450 x = 290 smaller number = 290 . c )" | a = 1465 + 15
b = 6 - 1
c = a / b
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a ) 1 ⁄ 3 , b ) 4 ⁄ 3 , c ) 80 , d ) 200 , e ) 180 | d | divide(multiply(2, const_60), divide(3, 5)) | if it takes a machine 3 ⁄ 5 minute to produce one item , how many items will it produce in 2 hours ? | "1 item takes 3 / 5 min so it takes 120 min to produce x 3 x / 5 = 120 the x = 200 answer : d" | a = 2 * const_60
b = 3 / 5
c = a / b
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a ) $ 200 , b ) $ 204 , c ) $ 208 , d ) $ 212 , e ) $ 216 | c | multiply(divide(364, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | $ 364 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | "let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 364 x = 52 4 x = 208 the answer is c ." | a = 1 / 2
b = a + 1
c = b + 2
d = 364 / c
e = d * 2
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a ) 32.5 , b ) 35 , c ) 48 , d ) 65 , e ) 66.67 | e | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(20, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(20, const_100))))) | the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 20 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 20 % , therefore bringing down the price to $ 60 . 4 ) now 60 has to be added byx % in order to equal the original price . 60 + ( x % ) 60 = 100 . solving this eq for x , we get x = 66.67 ans is e ." | a = 100 - 25
b = 100 - 25
c = 20 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 25
h = 100 - 25
i = 20 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8 | a | add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12)))) | two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many e hours will it take if jim types the report and mary edits it immediately after he is done ? | "break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many e hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a ." | a = 1 / 12
b = 1 / 30
c = a - b
d = 1/(c)
e = 1 / 7
f = 1 / 12
g = e - f
h = 1/(g)
i = d + h
|
a ) rs . 4500 , b ) rs . 4000 , c ) rs . 4167 , d ) rs . 4200 , e ) rs . 3000 | c | divide(multiply(500, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(10, const_100))))) | a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 500 per month . what is his monthly salary ? | income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 + 80 * ( 10 / 100 ) = rs . 88 present savings = 100 – 88 = rs . 12 if savings is rs . 12 , salary = rs . 100 if savings is rs . 500 , salary = 100 / 12 * 500 = 4167 answer : c | a = 500 * 100
b = 100 - 20
c = 100 - 20
d = 10 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) s . 42028 , b ) s . 42000 , c ) s . 42003 , d ) s . 42029 , e ) s . 24029 | b | subtract(78000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 78000 , then the share of a is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 78000 share of b = 7 / 13 ( 78000 ) = rs . 42000 answer : b" | a = const_60 * 100
b = 78000 - a
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a ) - 3 , b ) 3 , c ) 5 , d ) - 12 , e ) 12 | e | add(7, 5) | what is the minimum value of | x - 4 | + | x + 7 | + | x - 5 | ? | "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 7 | > = 0 - - > minimum occurs at x = - 7 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 7 - - > result = 11 + 0 + 12 = 23 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 11 + 1 = 12 x = 5 - - > result = 1 + 12 + 0 = 13 x = 7 - - > result = 3 + 14 + 2 = 19 so minimum value of the expression occurs at x = 4 and the resultant value = 12 answer : e" | a = 7 + 5
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a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | d | subtract(subtract(multiply(3, 16), add(subtract(13, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 13 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 13 a – d = 3 a + d = 13 2 d = 10 d = 5" | a = 3 * 16
b = 13 - 16
c = b + 3
d = a - c
e = d - 16
|
a ) 18 % , b ) 26.57 % , c ) 20 % , d ) 19 % , e ) none of these | b | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 456), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 456 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 456 , discount = 621 - 456 = 165 discount % = 165 / 621 * 100 = 26.57 % answer : b" | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 456
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
|
a ) 99 , b ) 97 , c ) 95 , d ) 91 , e ) 96 | a | multiply(11, 77) | the h . c . f . of two numbers is 11 and their l . c . m . is 693 . if one of the numbers is 77 , find the other . | "other number = 11 x 693 / 77 = 99 answer is a ." | a = 11 * 77
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a ) 5 / 16 , b ) 1 / 2 , c ) 12 / 30 , d ) 15 / 32 , e ) 3 / 8 | d | divide(add(divide(factorial(5), multiply(factorial(const_4), factorial(const_1))), divide(factorial(5), multiply(factorial(const_3), factorial(const_2)))), power(const_2, 5)) | kate and danny each have $ 10 . together , they flip a fair coin 5 times . every time the coin lands on heads , kate gives danny $ 1 . every time the coin lands on tails , danny gives kate $ 1 . after the 5 coin flips , what is the probability that kate has more than $ 10 but less than $ 15 ? | the probability of the coin landing tails up either 3 or 4 times = p ( 3 t ) + p ( 4 t ) binomial distribution formula : nck p ^ k ( 1 - p ) ^ ( n - k ) p ( 3 t ) = 5 c 3 ( 1 / 2 ) ^ 3 ( 1 / 2 ) ^ 2 = 10 ( 1 / 2 ) ^ 5 p ( 4 t ) = 5 c 4 ( 1 / 2 ) ^ 4 ( 1 / 2 ) ^ 1 = 5 ( 1 / 2 ) ^ 5 = > p ( 3 t ) + p ( 4 t ) = 15 / 32 answer : d | a = math.factorial(5)
b = math.factorial(4)
c = math.factorial(1)
d = b * c
e = a / d
f = math.factorial(5)
g = math.factorial(3)
h = math.factorial(2)
i = g * h
j = f / i
k = e + j
l = 2 ** 5
m = k / l
|
a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | c | multiply(divide(add(divide(multiply(30, 20), const_100), divide(multiply(50, 12), const_100)), add(30, 50)), const_100) | there are two groups of students in the sixth grade . there are 30 students in group a , and 50 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 12 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their homework ? | total students = 30 + 50 = 80 20 % of 30 = 6 12 & of 50 = 6 total students who forget homework = 6 + 6 = 12 percentage = 1280 ∗ 100 = 15 = 12 / 80 ∗ 100 = 15 ; answer = c = 15 % | a = 30 * 20
b = a / 100
c = 50 * 12
d = c / 100
e = b + d
f = 30 + 50
g = e / f
h = g * 100
|
a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 75 % , e ) none of these | b | multiply(add(const_1, divide(divide(multiply(100, 40), add(100, divide(multiply(multiply(const_3, const_10), 100), const_100))), 40)), const_10) | a rectangle having length 100 cm and width 40 cm . if the length of the rectangle is increased by fifty percent then how much percent the breadth should be decreased so as to maintain the same area . | "explanation : solution : ( 50 / ( 100 + 50 ) * 100 ) % = 33.33 % answer : b" | a = 100 * 40
b = 3 * 10
c = b * 100
d = c / 100
e = 100 + d
f = a / e
g = f / 40
h = 1 + g
i = h * 10
|
a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(13, const_2), choose(13, const_3)), choose(13, const_4)), choose(13, 13)), power(const_2, 13)) | a fair coin is tossed 13 times . what is the probability of getting more heads than tails in 13 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ." | a = math.comb(13, 2)
b = math.comb(13, 3)
c = a + b
d = math.comb(13, 4)
e = c + d
f = math.comb(13, 13)
g = e + f
h = 2 ** 13
i = g / h
|
a ) 129 , b ) 287 , c ) 192 , d ) 188 , e ) 112 | c | add(divide(add(multiply(34, 150), multiply(36, 125)), add(36, 34)), multiply(divide(add(multiply(34, 150), multiply(36, 125)), add(36, 34)), divide(40, const_100))) | raman mixed 34 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 34 ( 150 ) + 36 ( 125 ) ] / ( 34 + 36 ) = rs . 137.14 sp = cp [ ( 100 + profit % ) / 100 ] = 137.14 * [ ( 100 + 40 ) / 100 ] = rs . 192 . answer : c" | a = 34 * 150
b = 36 * 125
c = a + b
d = 36 + 34
e = c / d
f = 34 * 150
g = 36 * 125
h = f + g
i = 36 + 34
j = h / i
k = 40 / 100
l = j * k
m = e + l
|
a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 3 / 8 | e | divide(subtract(5, multiply(const_2, const_3)), 5) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 5 | "total number of cases = 4 * 4 = 16 favourable cases = [ ( 2,4 ) , ( 3,3 ) , ( 3,4 ) , ( 4,2 ) , ( 4,3 ) , ( 4,4 ) ] = 6 so probability = 6 / 16 = 3 / 8 answer is e" | a = 2 * 3
b = 5 - a
c = b / 5
|
a ) 1 / 2 , b ) 7 / 18 , c ) 5 / 13 , d ) 5 / 12 , e ) 6 / 17 | b | divide(subtract(9, multiply(const_2, const_3)), 9) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 9 | "total number of cases = 8 * 8 = 64 favourable cases = [ ( 2,8 ) , ( 3,6 ) , ( 3,7 ) , ( 4,6 ) , ( 4,7 ) , ( 4,8 ) , ( 5,5 ) , ( 5,6 ) , ( 5,7 ) , ( 5,8 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) , ( 6,7 ) , ( 6,8 ) , ( 7,3 ) , ( 7,4 ) , ( 7,5 ) , ( 7,6 ) , ( 7,7 ) , ( 7,8 ) , ( 8,2 ) , ( 8,3 ) , ( 8,4 ) , ( 8,5 ) , ( 8,6 ) , ( 8,7 ) , ( 8,8 ) ] = 28 so probability = 28 / 64 = 7 / 18 answer is b" | a = 2 * 3
b = 9 - a
c = b / 9
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 3 / 4 | a | divide(2, divide(32, const_10)) | a cubical tank is filled with water to a level of 2 feet . if the water in the tank occupies 32 cubic feet , to what fraction of its capacity is the tank filled with water ? | "the volume of water in the tank is h * l * b = 32 cubic feet . since h = 2 , then l * b = 16 and l = b = 4 . since the tank is cubical , the capacity of the tank is 4 * 4 * 4 = 64 . the ratio of the water in the tank to the capacity is 32 / 64 = 1 / 2 the answer is a ." | a = 32 / 10
b = 2 / a
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a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 45 % | d | multiply(divide(2200, 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 . 2200 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 = 2200 100 r + 120 r = 2200 r = 10 % answer : d" | a = 5000 * 2
b = 3000 * 4
c = a + b
d = 2200 / c
e = d * 100
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a ) 0 , b ) 50 , c ) 450 , d ) 693 , e ) 500 | d | divide(multiply(10, 7000), const_100) | when 1 / 10 percent of 7000 is subtracted from 1 / 10 of 7000 , the difference is | we can break this problem into two parts : 1 ) what is 1 / 10 percent of 7,000 ? 2 ) what is 1 / 10 of 7,000 ? to calculate 1 / 10 percent of 7,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100 = 1 / 1,000 thus , 1 / 10 percent of 7,000 = 1 / 1,000 x 7,000 = 7 . now let ' s concentrate on part 2 . we need to calculate 1 / 10 of 7,000 . to do this we simply multiply 1 / 10 by 7,000 . 1 / 10 x 7,000 = 700 the answer to part 1 is 7 , and the answer to part 2 is 700 . their difference is 700 – 7 = 693 . answer d . | a = 10 * 7000
b = a / 100
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a ) 288 , b ) 279 , c ) 277 , d ) 272 , e ) 150 | e | multiply(divide(multiply(60, const_1000), const_3600), 9) | a train running at the speed of 60 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 9 = 150 m answer : e" | a = 60 * 1000
b = a / 3600
c = b * 9
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a ) z / 2 , b ) 2 z , c ) z / 3 , d ) 3 z / 5 , e ) 4 z / 9 | e | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | if 3 x = 9 y = z , what is x + y , in terms of z ? | "3 x = 9 y = z x = z / 3 and y = z / 9 x + y = z / 3 + z / 9 = 4 z / 9 answer is e" | a = 3 * 100
b = a / 2
c = b - 2
d = 3 * 100
e = d / 2
f = e + 2
g = c / f
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a ) 2 / 25 , b ) 1 / 50 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3 | e | multiply(multiply(divide(20, add(20, 5)), divide(5, add(20, 5))), const_2) | into a bag there are 20 honey and 5 cherry candies . if a boy pick only two candies simultaneous and randomly , what is the probability that he picks one candy of each flavor ? | we are told that we have 25 candies , 20 honey and 5 cherry candies . the candies are picked simultaneous and randomly , c 1 and c 2 , in different flavors . there are two acceptable outcomes : 1 ) c 1 is honey and c 2 is cherry ; 2 ) c 1 is cherry and c 2 is honey . let ' s go : 1 ) c 1 = ( 20 / 25 ) ( 5 / 24 ) = 1 / 6 chance of this happening . 2 ) c 2 = ( 5 / 25 ) ( 20 / 24 ) = 1 / 6 chance of this happening . then : ( 1 / 6 ) + ( 1 / 6 ) = 1 / 3 , chance of getting the result that you wanted . answer e . | a = 20 + 5
b = 20 / a
c = 20 + 5
d = 5 / c
e = b * d
f = e * 2
|
a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | e | add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4) | if x = 1 + √ 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 2 ? | "answer x = 1 + √ 2 ∴ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 2 = x 2 ( x - 2 ) 2 + 2 = ( 1 + √ 2 ) 2 ( 1 + √ 2 - 2 ) 2 + 2 = ( √ 2 + 1 ) 2 ( √ 2 - 1 ) 2 + 2 = [ ( √ 2 ) 2 - ( 1 ) 2 ] 2 + 2 = ( 2 - 1 ) 2 = 1 + 2 = 3 correct option : e" | a = math.sqrt(2)
b = 1 + a
c = b ** 2
d = math.sqrt(2)
e = 1 + d
f = e - 2
g = f ** 2
h = c * g
i = h + 4
|
a ) ( 400 ) , b ) ( 300 ) , c ) ( 040 ) , d ) ( 4030 ) , e ) ( 030 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | a line has a slope of 3 / 4 and intersects the point ( - 12 , - 39 ) . at which point e does this line intersect the x - axis ? | assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point e of intersection is ( 400 ) . a is the correct answer . | a = negate - (
b = negate * (
c = 3 / 4
d = a / b
e = negate * (
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a ) 50 % , b ) 32 % , c ) 25 % , d ) 54 % , e ) 29 % | d | multiply(divide(multiply(const_100, divide(35, const_100)), subtract(const_100, multiply(const_100, divide(35, const_100)))), const_100) | the salary of a person was reduced by 35 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 65 increase on 65 = 35 increase on 100 = 35 / 65 * 100 = 54 % ( approximately ) answer is d" | a = 35 / 100
b = 100 * a
c = 35 / 100
d = 100 * c
e = 100 - d
f = b / e
g = f * 100
|
a ) - 5 , b ) - 6 , c ) - 7 , d ) - 8 , e ) - 10 | d | divide(19, 3) | if v and d are both integers , v > d , and - 3 v > 19 , then the largest value of d would be ? | no , your thinking is incorrect . when we know that v > d and v < - 6.33 , the largest value of v can be - 7 while if v = - 7 , then largest value of d < - 7 will be - 8 . for negative numbers , - 7 > - 8 and - 8 > - 10 . you are right in saying that d can take any value less than - 7 - - - > d could be - 8 , - 9 , - 10 . . . . and out of all these values , - 8 is the greatest . look at the numbers on the number line . for any 2 numbers , the ones on the right are greater than the ones on the left : . . . . . . . - 11 - 10 - 9 - 8 - 7 - 6 - 5 . . . . . 0 1 2 3 4 5 6 . . . ( - 11 < - 10 , - 10 < - 8 , 4 < 5 etc ) . so , as per the question if d < v and v = - 7 , then d ' s largest ' possible ' value has to be - 8 . - 10 is smaller than - 8 = d | a = 19 / 3
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a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 5 | e | multiply(3, 1) | if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 5 = e" | a = 3 * 1
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a ) 2900 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | a | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(100, subtract(multiply(4, 5), multiply(3, 6)))) | a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 100 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 100 total earning = \ inline \ frac { 58 xy } { 100 } = 2900 answer : a ) rs . 2900" | a = 3 * 6
b = 4 * 5
c = a + b
d = 5 * 4
e = c + d
f = 4 * 5
g = 3 * 6
h = f - g
i = 100 / h
j = e * i
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a ) 9 , b ) 16 , c ) 19 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(35, divide(multiply(3, 2), multiply(4, 3))) | a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 35 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 35 pints will require 9 / 17 * 35 = 19 pints . . c" | a = 3 * 2
b = 4 * 3
c = a / b
d = 35 * c
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a ) 275 days , b ) 178 days , c ) 185 days , d ) 175 days , e ) 675 days | d | multiply(divide(multiply(25, 35), subtract(35, 25)), const_2) | a can build a wall in the same time in which b and c together can do it . if a and b together could do it in 25 days and c alone in 35 days , in what time could b alone do it ? | explanation : no explanation is available for this question ! answer : d | a = 25 * 35
b = 35 - 25
c = a / b
d = c * 2
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | divide(multiply(add(6, const_4), 10), add(15, 10)) | if 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days , the time taken by 15 men and 20 boys in doing the same type of work will be ? | "let 1 men ' s 1 day work = x and 1 boy ' s 1 day work = y . then , 6 x + 8 y = 1 / 10 and 26 x + 48 y = 1 / 2 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 ( 15 men + 20 boys ) ' s 1 day work = ( 15 / 100 + 20 / 200 ) = 1 / 4 15 men and 20 boys can do the work in 4 days . answer : a" | a = 6 + 4
b = a * 10
c = 15 + 10
d = b / c
|
a ) 1.5 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph | a | divide(subtract(11, 8), const_2) | a man goes downstream at 11 kmph , and upstream 8 kmph . the speed of the stream is | "speed of the stream = 1 / 2 ( 11 - 8 ) kmph = 1.5 kmph . correct option : a" | a = 11 - 8
b = a / 2
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a ) 145 m , b ) 176 m , c ) 168 m , d ) 163 m , e ) 218 m | d | subtract(1000, divide(multiply(subtract(1000, 70), subtract(1000, 100)), 1000)) | in a 1000 m race , a beats b by 70 m and b beats c by 100 m . in the same race , by how many meters does a beat c ? | by the time a covers 1000 m , b covers ( 1000 - 70 ) = 930 m . by the time b covers 1000 m , c covers ( 1000 - 100 ) = 900 m . so , the ratio of speeds of a and c = 1000 / 930 * 1000 / 900 = 1000 / 837 so , by the time a covers 1000 m , c covers 837 m . so in 1000 m race a beats c by 1000 - 837 = 163 m . answer : d | a = 1000 - 70
b = 1000 - 100
c = a * b
d = c / 1000
e = 1000 - d
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a ) 560 , b ) 660 , c ) 800 , d ) 760 , e ) 480 | b | multiply(divide(add(subtract(80, const_3), add(40, const_2)), const_2), add(divide(subtract(subtract(80, const_3), add(40, const_2)), 4), const_1)) | what is the sum of the multiples of 4 from 40 to 80 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 4 and add 1 . ( 80 / 4 ) - ( 40 / 4 ) + 1 multiply these together so what we show average * total numbers = sum ( 80 + 40 ) / 2 * ( 80 / 4 ) - ( 40 / 4 ) + 1 = sum 60 * 11 = 660 b" | a = 80 - 3
b = 40 + 2
c = a + b
d = c / 2
e = 80 - 3
f = 40 + 2
g = e - f
h = g / 4
i = h + 1
j = d * i
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | e | multiply(power(2, 2), power(2, 3)) | what number times ( 1 ⁄ 2 ) ^ 2 will give the value of 2 ^ 3 ? | x * ( 1 / 2 ) ^ 2 = 2 ^ 3 x = 2 ^ 2 * 2 ^ 3 = 2 ^ 5 = 32 the answer is e . | a = 2 ** 2
b = 2 ** 3
c = a * b
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a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 40 % | b | multiply(divide(multiply(50, 0.00008), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.00008 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 00008 * 50 = . 00008 * 100 / 2 = . 008 / 2 = . 004 percent of the original amount of water evaporated during this period = ( . 004 / 10 ) * 100 % = 0.04 % answer b" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 8.9 kmph , b ) 2.96 kmph , c ) 1.09 kmph , d ) 45.9 kmph , e ) 4.8 kmph | c | divide(multiply(divide(12, add(2, divide(add(multiply(14, const_60), 28), const_3600))), 3), 2) | a car travelling with 2 / 3 km of its actual speed covers 12 km in 2 hr 14 min 28 sec find the actual speed of the car ? | "time taken = 2 hr 14 min 28 sec = 823 / 50 hrs let the actual speed be x kmph then 2 / 3 x * 823 / 50 = 12 x = = 1.09 kmph answer ( c )" | a = 14 * const_60
b = a + 28
c = b / 3600
d = 2 + c
e = 12 / d
f = e * 3
g = f / 2
|
a ) 150 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | a | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 30 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 30 miles . teena can cover 7.5 miles in 30 mins teena will cover 30 miles in 120 mins so answer 30 + 120 = 150 mins . ( answer a )" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 427 , b ) 859 , c ) 860 , d ) 4320 , e ) none of these | c | subtract(lcm(lcm(lcm(22, 32), 36), 54), 4) | the least number which when increased by 4 each divisible by each one of 22 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 4 = 864 - 4 = 860 . answer c" | a = math.lcm(22, 32)
b = math.lcm(a, 36)
c = math.lcm(b, 54)
d = c - 4
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a ) rs . 80 , b ) rs . 185 , c ) rs . 140 , d ) rs . 295 , e ) none of these | c | subtract(divide(4500, 30), 10) | a trader sells 30 meters of cloth for rs . 4500 at the profit of rs . 10 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4500 / 30 = rs . 150 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 150 - rs . 10 = rs . 140 . answer : c" | a = 4500 / 30
b = a - 10
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a ) 200 , b ) 490 , c ) 700 , d ) 900 , e ) 1,400 | b | divide(multiply(35, 400), 20) | if 35 percent of 400 is 20 percent of x , then what is 70 percent of x ? | "35 / 100 ( 400 ) = 2 / 10 ( x ) x = 700 . . 70 percent of x = 70 / 100 ( 700 ) = 490 option b ." | a = 35 * 400
b = a / 20
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a ) 3 , b ) 4 , c ) 10 , d ) 12 , e ) 14 | a | divide(300, const_100) | for any even integer p , 300 multiplied by p is square of an integer . what is the least value of p ? | p ∗ 3 ∗ 10 ^ 2 = s ^ 2 so , s = √ p ∗ 3 ∗ 10 ^ 2 so , if p = 3 we get a perfect square number ! ! hence answer will be ( a ) 3 | a = 300 / 100
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a ) 45 % , b ) 10 % , c ) 20 % , d ) 60 % , e ) 56 % | b | subtract(50, 45) | if the selling price of 50 articles is equal to the cost price of 45 articles , then the loss or gain percent is : | "c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 45 . loss % = 5 / 50 * 100 = 10 % answer b" | a = 50 - 45
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a ) 44.28 , b ) 54.28 , c ) 34.28 , d ) 64.28 , e ) 74.28 | b | divide(add(multiply(20, 40), multiply(50, 60)), add(20, 50)) | the average marks of a class of 20 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 20 students = 20 * 40 = 800 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 70 students = 800 + 3000 = 3800 average marks of all the students = 3800 / 70 = 54.28 answer : b" | a = 20 * 40
b = 50 * 60
c = a + b
d = 20 + 50
e = c / d
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a ) 12 , b ) 84 , c ) 108 , d ) 120 , e ) 132 | b | divide(factorial(9), multiply(factorial(divide(const_10, const_2)), factorial(3))) | in a certain circle there are 9 points . what is the number of the triangles connecting 3 points of the 9 points ? | "imo : b here we have to select 3 points out of 9 points . order is not important so the answer will be 9 c 3 = 84 answer b" | a = math.factorial(9)
b = 10 / 2
c = math.factorial(b)
d = math.factorial(3)
e = c * d
f = a / e
|
a ) 2 / 19 , b ) 3 / 29 , c ) 4 / 15 , d ) 1 / 15 , e ) 1 / 2 | d | subtract(divide(4, subtract(10, 4)), divide(subtract(10, 4), 10)) | in shop there are 10 bulbs , a total of 4 are defective . if a customer buys 4 bulbs selected at random from the box , what is the probability that neither bulbs will be defective ? | first , there are 6 c 4 ways you can select 6 good bulbs from 4 good ones . second , there are 10 c 4 ways you select 4 bulbs from 10 ones in the box . then , the probability that neither bulb will be defective is : 6 c 4 / 10 c 4 = 15 / 210 = 1 / 15 answer is d | a = 10 - 4
b = 4 / a
c = 10 - 4
d = c / 10
e = b - d
|
a ) 1 , b ) - 1 , c ) 3 , d ) 5 , e ) 13 | a | multiply(6, const_2) | if x + | x | + y = 2 and x + | y | - y = 6 what is x + y = ? | "if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 1 hence answer should be a" | a = 6 * 2
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | multiply(multiply(const_2, const_3), 2) | employees of a certain company are each to receive a unique 5 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , and 4 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ? | there are 3 ! ways to make codes starting with 12 . there are 3 ! ways to make codes starting with 24 . the number of codes is 2 * 3 ! = 12 . the answer is b . | a = 2 * 3
b = a * 2
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a ) 24 , b ) 17 , c ) 18 , d ) 9 , e ) 11 | d | divide(multiply(18, 36), 72) | 36 men can complete a piece of work in 18 days . in how many days will 72 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 72 : 36 : : 18 : x x = 9 answer : d ) 9 days" | a = 18 * 36
b = a / 72
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a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 32 / 35 | e | add(multiply(divide(4, 7), divide(subtract(7, 3), subtract(7, 1))), multiply(divide(4, 7), divide(subtract(7, 4), subtract(7, 1)))) | in a room filled with 7 people , 4 people have exactly 1 sibling in the room and 3 people have exactly 3 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | "there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssds ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssdssibl ∈ gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and viceversa ) . . . now remaning efg are 3 people who have exactly 3 siblings . . . . ( e has f and g as his / her sibling and so on . . ) there are now 3 different set of siblings ( a and b ) ( c and d ) ; ( efg ) now first selecting 3 people out of 7 is 7 c 3 = 35 first sibling pair - - - - ( a and b ) - - selecting 3 people - - 3 c 3 = 1 second sibling pair ( c and d ) - - selecting 3 people - - 3 c 3 = 1 third sibling pair ( e f g ) - - selecting 3 out of 3 - - 3 c 3 = 1 total = 1 + 1 + 1 = 3 but , a / c to formula p ( success ) - 1 - p ( fail ) here , p ( failure ) is selecting 3 people who are siblings = 3 / 35 ( 35 is 7 c 3 ) = 1 - 3 / 35 = 32 / 35 ans e" | a = 4 / 7
b = 7 - 3
c = 7 - 1
d = b / c
e = a * d
f = 4 / 7
g = 7 - 4
h = 7 - 1
i = g / h
j = f * i
k = e + j
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a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | add(multiply(3, const_4), 3) | if p is the product of the integers from 1 to 34 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of p ? | 34 ! has 3 , 6 , 9 , . . . . 30 , 33 as factors , which are 11 multiples of 3 . we need to add 4 more to these 11 because of 9 , 18 , and 27 . the greatest integer of k is 15 . the answer is c . | a = 3 * 4
b = a + 3
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a ) $ 2100 , b ) $ 2546 , c ) $ 2256 , d ) $ 2451 , e ) $ 2345 | c | multiply(2500, power(subtract(const_1, divide(5, const_100)), 2)) | a present value of a machine is $ 2500 . its value depletiation rate is 5 % per annum then find the machine value after 2 years ? | "p = $ 2500 r = 5 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 2500 * 19 / 20 * 19 / 20 = $ 2256 approximately answer is c" | a = 5 / 100
b = 1 - a
c = b ** 2
d = 2500 * c
|
a ) 2480 , b ) 3490 , c ) 6785 , d ) 8255 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1280) | the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1280 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | "you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 15 + 15 ) ^ 2 now we can expand each square ; they are all in the ( x + y ) ^ 2 = x ^ 2 + 2 xy + y ^ 2 pattern . = ( 15 ^ 2 + 2 * 15 + 1 ^ 2 ) + ( 15 ^ 2 + 4 * 15 + 2 ^ 2 ) + ( 15 ^ 2 + 6 * 15 + 3 ^ 2 ) + . . . + ( 15 ^ 2 + 30 * 15 + 15 ^ 2 ) now we have fifteen 15 ^ 2 terms , so adding these gives 15 * 15 ^ 2 = 15 ^ 3 = 3375 . we also have the entire sum 1 ^ 2 + 2 ^ 2 + . . . + 15 ^ 2 , which we know is equal to 1240 . finally adding the middle terms , we have : 2 * 15 + 4 * 15 + 6 * 15 + . . . + 30 * 15 = 15 ( 2 + 4 + 6 + . . . . + 30 ) = 15 * 2 * ( 1 + 2 + 3 + . . . + 15 ) = 15 * 2 * 8 * 15 = 3600 so the sum must be 3375 + 1240 + 3600 = 8215 alternatively , we can use a different factoring pattern . we want to find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 . well if we subtract 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 from this , the answer will be 1240 less than what we want to find . so if we can find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 - ( 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 ) then we can add 1240 to get the answer . now grouping the terms above to get differences of squares , we have = ( 30 ^ 2 - 15 ^ 2 ) + ( 29 ^ 2 - 14 ^ 2 ) + . . . + ( 16 ^ 2 - 1 ^ 2 ) and factoring each of these using x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) , we have = 45 * 15 + 43 * 15 + 41 * 15 + . . . + 17 * 15 = 15 ( 45 + 43 + 41 + . . . + 17 ) in brackets we have an equally spaced sum with fifteen terms , which we can evaluate using the familiar formula . so the above equals 15 * 15 * 62 / 2 = 6975 and adding back the 1280 , we get the answer of 8255 . ( ans d )" | a = 30 + 1
b = 30 * a
c = 2 * 30
d = c + 1
e = b * d
f = 3 + 3
g = e / f
h = g - 1280
|
a ) 210 , b ) 245 , c ) 230 , d ) 254 , e ) 265 | e | divide(divide(multiply(60, const_1000), const_60), multiply(divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3)), divide(120, const_100))) | the diameter of the wheel of a car is 120 m . how many revolution / min must the wheel makein order to keep a speed of 60 km / hour approximately ? | distance to be covered in 1 min . = ( 60 x 1000 ) / ( 60 ) m = 1000 m . circumference of the wheel = ( 2 x ( 22 / 7 ) x 0.60 ) m = 3.77 m . number of revolutions per min . = ( 1000 / 3.77 ) = 265 e | a = 60 * 1000
b = a / const_60
c = 2 * 10
d = c + 2
e = 4 + 3
f = d / e
g = 120 / 100
h = f * g
i = b / h
|
a ) 4.5 % , b ) 8.6 % , c ) 12.3 % , d ) 16.7 % , e ) 20 % | b | multiply(divide(multiply(subtract(multiply(divide(divide(subtract(multiply(multiply(const_4, const_2), multiply(add(const_4, const_1), const_2)), 70), subtract(70, multiply(const_3, multiply(add(const_4, const_1), const_2)))), add(const_4, const_1)), multiply(add(const_4, const_1), const_2)), divide(multiply(const_3, multiply(add(const_4, const_1), const_2)), const_100)), multiply(const_3, multiply(add(const_4, const_1), const_2))), 70), const_100) | each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , thirty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 70 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on are incandescent ? | let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.3 i + 0.8 f = 0.7 ( i + f ) 0.1 f = 0.4 i f = 4 i this means that for every 1 incandescent bulb , there are 4 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.3 i / ( 0.3 i + 0.8 f ) = 0.3 i / ( 0.3 i + 0.8 * 4 i ) = 0.3 i / 3.5 i = 3 / 35 which is about 8.6 % . the answer is b . | a = 4 * 2
b = 4 + 1
c = b * 2
d = a * c
e = d - 70
f = 4 + 1
g = f * 2
h = 3 * g
i = 70 - h
j = e / i
k = 4 + 1
l = j / k
m = 4 + 1
n = m * 2
o = l * n
p = 4 + 1
q = p * 2
r = 3 * q
s = r / 100
t = o - s
u = 4 + 1
v = u * 2
w = 3 * v
x = t * w
y = x / 70
z = y * 100
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | d | divide(20, subtract(40, 20)) | a chemist mixes one liter of pure water with x liters of a 40 % salt solution , and the resulting mixture is a 20 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 40 % concentration of salt in the mixed solution = 20 % the pure solution and the salt solution is mixed in the ratio of - - > ( 40 - 20 ) / ( 20 - 0 ) = 1 / 1 1 / x = 1 / 1 x = 1 answer : d" | a = 40 - 20
b = 20 / a
|
a ) 2 : 6 , b ) 2 : 3 , c ) 2 : 5 , d ) 2 : 1 , e ) 1 : 2 | e | divide(subtract(15.8, 15.5), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.5 years . the ration of the number of boys to the number of girls in the class is ? | "let the ratio be k : 1 . then , k * 16.4 + 1 * 15.5 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.5 ) = k = 0.3 / 0.6 = 1 / 2 required ratio = 1 / 2 : 1 = 1 : 2 . answer : e" | a = 15 - 8
b = 16 - 4
c = a / b
|
['a ) 100 π / 3', 'b ) 80 π / 3', 'c ) 120 π / 7', 'd ) 130 π / 9', 'e ) 110 π / 7'] | a | multiply(const_pi, multiply(power(multiply(divide(divide(10, 4), subtract(divide(10, 4), const_1)), const_2), const_2), const_3)) | consider a right circular cone of base radius 4 and height 10 cm . a cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone . find the largest possible total surface area ( in sq cm ) of the cylinder . | explanation : let the radius of cylinder be r and its height is h . so , the height remained above cylinder is 10 - h also , triangles abc and ade are similar . so ratio of corresponding sides must be equal i . e ab / ad = bc / de . i . e . ( 10 - h ) / 10 = r / 4 . so 40 - 4 h = 10 r ( by cross multiply ) h = ( 40 - 10 r ) / 4 = 10 - 2.5 r now we have total surface area of cyl = 2 * π * r * h + 2 * π * r 2 . putting value of h = 10 - 2.5 r here , we get s = 2 * π * r * ( 10 - 2.5 r ) + 2 * π * r 2 . s = 20 * π * r - 5 * π * r 2 + 2 * π * r 2 . s = 20 * π * r - 3 * π * r . now , we have to maximize s , so differentiate it w . r . t r , we get s = 20 * π - 6 * π * r setting it to zero , we get = > 20 * π - 6 * π * r = 0 = > r = 20 / 6 = 10 / 3 . so , h = 10 - 2.5 r = 10 - 2.5 * 10 / 3 = 5 / 3 so , = > s = 2 * π * r * h + 2 * π * r 2 . = > s = 2 * π * 10 / 3 * 5 / 3 + 2 * π * 100 / 9 . = > s = 2 * π * 50 / 9 + 2 * π * 100 / 9 . = > s = 2 π 150 / 9 . = > s = 100 π / 3 . answer : a | a = 10 / 4
b = 10 / 4
c = b - 1
d = a / c
e = d * 2
f = e ** 2
g = f * 3
h = math.pi * g
|
a ) 400 , b ) 600 , c ) 750 , d ) 2400 , e ) 3200 | a | multiply(3600, divide(1200, 10800)) | a football field is 10800 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 3600 square yards ? | "answer a ) 10800 yards need 1200 lbs 1 yard will need 1200 / 10800 = 1 / 9 lbs 3600 yards will need 1 / 9 * 3600 yards = 400 lbs" | a = 1200 / 10800
b = 3600 * a
|
a ) 9 : 39 , b ) 9 : 31 , c ) 9 : 36 , d ) 9 : 32 , e ) 9 : 34 | b | divide(add(5, 4), add(multiply(const_3, const_10), 1)) | two same glasses are respectively 1 / 4 th 1 / 5 th full of milk . they are then filled with water and the contents mixed in a tumbler . the ratio of milk and water in the tumbler is ? | 1 / 4 : 3 / 4 = ( 1 : 3 ) 5 = 5 : 15 1 / 5 : 4 / 5 = ( 1 : 4 ) 4 = 4 : 16 - - - - - - 9 : 31 answer : b | a = 5 + 4
b = 3 * 10
c = b + 1
d = a / c
|
['a ) 3 and 5', 'b ) 3 and 6', 'c ) 3 and 7', 'd ) 3 and 8', 'e ) 3 and 9'] | a | divide(subtract(15, sqrt(subtract(power(15, const_2), multiply(const_4, 16)))), const_2) | the area of a rectangle is 15 square centimeters and the perimeter is 16 square centimeters . what are the dimensions of the rectangle ? | let x and y be the length and width of the rectangle . using the formulas for the area and the perimeter , we can write two equations . 15 = x y and 16 = 2 x + 2 y solve the second equation for x x = 8 - y substitute x in the equation 15 = x y by 8 - y to rewrite the equation as 15 = ( 8 - y ) y solve for y to find y = 3 and y = 5 use x = 8 - y to find x when y = 3 , x = 5 and when y = 5 , x = 3 . the dimensions of the rectangle are 3 and 5 . as an exercise , check that the perimeter of this rectangle is 16 and its area is 15 . answer a | a = 15 ** 2
b = 4 * 16
c = a - b
d = math.sqrt(c)
e = 15 - d
f = e / 2
|
a ) 32 , b ) 33 , c ) 39 , d ) 40 , e ) 56 | e | subtract(91, divide(91, add(const_1, divide(160, const_100)))) | sales price is $ 91 , gross profit is 160 % of cost , what is the value of gross profit ? | "cost + profit = sales cost + ( 160 / 100 ) cost = 91 cost = 35 profit = 91 - 35 = 56 answer ( e )" | a = 160 / 100
b = 1 + a
c = 91 / b
d = 91 - c
|
a ) 38 , b ) 29 , c ) 27 , d ) 40 , e ) 91 | d | divide(multiply(25, 8), subtract(8, 3)) | a number exceeds by 25 from its 3 / 8 part . then the number is ? | "explanation : x â € “ 3 / 8 x = 25 x = 40 answer : d" | a = 25 * 8
b = 8 - 3
c = a / b
|
a ) 10 , b ) 25 , c ) 55 , d ) 35 , e ) 65 | b | add(subtract(80, 70), subtract(50, 35)) | in an it company , there are a total of 70 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ? | you could pick 70 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 25 . b | a = 80 - 70
b = 50 - 35
c = a + b
|
a ) 6.25 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.0 | e | divide(subtract(272, multiply(10, 3.2)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 272 runs ? | "required run rate = [ 272 - ( 3.2 * 10 ) ] / 40 = 240 / 40 = 6.00 answer : e" | a = 10 * 3
b = 272 - a
c = b / 40
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | subtract(multiply(divide(3, 2), add(16, 4)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(16, 4)))), const_1)) | a certain basketball team that has played 2 / 3 of its games has a record of 16 wins and 4 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ? | "16 wins , 4 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 4 games so it can lose another 3 at most . answer ( e )" | a = 3 / 2
b = 16 + 4
c = a * b
d = 3 / 4
e = 3 / 2
f = 16 + 4
g = e * f
h = d * g
i = math.floor(h)
j = i + 1
k = c - j
|
a ) 16.3 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | a | add(divide(add(800, 50), subtract(500, divide(800, const_2))), 8) | sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 06 : 00 and starts climbing at a constant pace . ken starts climbing only at 08 : 00 , when sari is already 800 meters ahead of him . nevertheless , ken climbs at a constant pace of 500 meters per hour , and reaches the summit before sari . if sari is 50 meters behind ken when he reaches the summit , at what time did ken reach the summit ? | both sari and ken climb in the same direction . speed of sari = 800 / 2 = 400 meters / hr ( since she covers 800 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 800 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / relative speed = ( 800 + 50 ) / ( 500 - 400 ) = 8.5 hrs starting from 8 : 00 , in 8.5 hrs , the time will be 16 : 30 answer ( a ) | a = 800 + 50
b = 800 / 2
c = 500 - b
d = a / c
e = d + 8
|
a ) 1100 , b ) 450 , c ) 550 , d ) 590 , e ) 600 | a | add(divide(800, subtract(const_1, divide(20, const_100))), multiply(divide(800, subtract(const_1, divide(20, const_100))), divide(10, const_100))) | boy sells a book for rs . 800 he gets a loss of 20 % , to gain 10 % , what should be the sp ? | cost price = 800 / 80 x 100 = 1000 to gain 10 % = 1000 x 10 / 100 = 100 sp = cp + gain = 1000 + 100 = 1100 answer : a | a = 20 / 100
b = 1 - a
c = 800 / b
d = 20 / 100
e = 1 - d
f = 800 / e
g = 10 / 100
h = f * g
i = c + h
|
a ) 11 , b ) 18 , c ) 14 , d ) 12 , e ) 10 | b | divide(subtract(multiply(add(const_4, const_1), 9), 9), subtract(add(const_4, const_1), const_3)) | the age of somu is one - third his father ' s . 9 years back he was one - fifth of his father ' s age . what is his persent age ? | explanation : let somu ' s age be x and that of his father be 3 x . so , x - 9 = 3 x - 9 / 5 = x = 18 answer : option b | a = 4 + 1
b = a * 9
c = b - 9
d = 4 + 1
e = d - 3
f = c / e
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | divide(add(16, 11), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 11 pencils and 16 charts have been brought to the meeting , how many people are present ? | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 11 total charts are 11 so a + 2 b = 16 add both : 3 a + 3 b = 27 so a + b = 9 total number of people = 9 d" | a = 16 + 11
b = 2 + 1
c = a / b
|
a ) 65 days , b ) 45 days , c ) 20 days , d ) 16 days , e ) 18 days | c | divide(multiply(subtract(31, 20), 400), 180) | a garrison of 400 men had a provision for 31 days . after 20 days 180 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | "400 - - - 31 400 - - - 11 220 - - - ? 400 * 11 = 220 * x = > x = 20 days . answer : c" | a = 31 - 20
b = a * 400
c = b / 180
|
a ) 15 kmph , b ) 36 kmph , c ) 21 kmph , d ) 17 kmph , e ) 19 kmph | b | divide(multiply(90, 4), add(4, add(const_3, const_3))) | a goods train leaves a station at a certain time and at a fixed speed . after ^ hours , an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph . this train catches up the goods train in 4 hours . find the speed of the goods train . | let the speed of the goods train be x kmph . distance covered by goods train in 10 hours = distance covered by express train in 4 hours 10 x = 4 x 90 or x = 36 . so , speed of goods train = 36 kmph . ans : b | a = 90 * 4
b = 3 + 3
c = 4 + b
d = a / c
|
a ) 7,10 , b ) 8,10 , c ) 9,10 , d ) 10,10 , e ) 10,12 | b | subtract(divide(divide(subtract(2000, 200), divide(1000, 200)), 6), multiply(divide(1000, 200), const_10)) | ram and shakil run a race of 2000 meters . first , ram gives shakil a start of 200 meters and beats him by one minute . if , ram gives shakil a start of 6 minutes ram is beaten by 1000 meters . find the time in minutes in which ram and shakil can run the races separately . | let x and y are the speeds of ram and shakil . . then by problem we got following equation 2000 / x = ( 1800 / y ) - 1 1000 / x = ( 2000 / y ) - 6 solve equation 1 and 2 , we get x = 250 and y = 200 therefore , time taken by ram and shakilk to complete a race of 2000 m is 8 min and 10 min answer : b | a = 2000 - 200
b = 1000 / 200
c = a / b
d = c / 6
e = 1000 / 200
f = e * 10
g = d - f
|
a ) $ 50 , b ) $ 142 , c ) $ 60 , d ) $ 100 , e ) $ 90 | b | multiply(divide(2, add(3, 2)), 355) | rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 355 . what is the share of rahul ? | "rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 355 * 2 / 5 = $ 142 answer is b" | a = 3 + 2
b = 2 / a
c = b * 355
|
a ) 12 , b ) 11 , c ) 13 , d ) 14 , e ) 15 | b | add(multiply(0.7, const_10), const_4) | a certain fruit stand sold apples for $ 0.70 each and cherry for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ? | some multiple of 7 + some multiple of 5 should yield 63 . to get to a some multiple of 5 , we should ensure that a 3 or 8 ( 5 + 3 ) should be a multiple of 7 . 63 is a direct multiple of 7 , however in this case there wo n ' t be any cherry . hence the next option is to look for a multiple of 7 that has 8 as the unit digit . 28 satisfies this hence no . of apples is 4 and no of bananas is 7 b | a = 0 * 7
b = a + 4
|
a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 125 | d | multiply(const_2, floor(divide(multiply(add(100, const_1), divide(100, const_2)), 100))) | the element being searched for is not found in an array of 100 elements . what is the average number of comparisons needed in a sequential search to determine that the element is not there , if the elements are completely unordered ? | let us assume k be the element we need to search then let array be n 1 , n 2 , n 3 . . . . . . . . . . . . . . n 100 now to prove that the element searched is not there in the array we have to compare the element k with every element in the array . so , totally we need 100 comparisons . answer : d | a = 100 + 1
b = 100 / 2
c = a * b
d = c / 100
e = math.floor(d)
f = 2 * e
|
a ) 0.07 , b ) 0.9 , c ) 9 , d ) 90 , e ) none of the above | a | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 15 % of 2 / 3 of 0.7 ? | "the best way to solve these questions is to convert every term into fraction ( 15 / 100 ) * ( 2 / 3 ) * ( 7 / 10 ) = 210 / 3000 = 0.07 option a" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 15 * n
p = o / 100
|
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | d | floor(divide(50, divide(14, const_2))) | what is the greatest integer m for which the number 50 ! / 14 ^ m is an integer ? | "14 ^ m = 2 ^ m * 7 ^ m . let ' s figure out how many 7 ' s are in the prime factorization of 50 ! the multiples of 7 are : 7 , 14 , 21 , 28 , 35 , 42 , 7 * 7 . thus 7 ^ 8 will divide 50 ! but 7 ^ 9 will not . clearly 2 ^ 8 will divide 50 ! so m = 8 is the largest possible integer . the answer is d ." | a = 14 / 2
b = 50 / a
c = math.floor(b)
|
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