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 ) 24 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | e | add(add(18, multiply(add(3, const_1), subtract(divide(18, 3), const_1))), multiply(divide(18, 3), add(3, const_1))) | a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day , but at the same speed as on the second day . how many miles in total did she walk ? | first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 4 miles per hour and 6 hours - 24 miles total 18 + 20 + 24 - 62 answer : option e . | a = 3 + 1
b = 18 / 3
c = b - 1
d = a * c
e = 18 + d
f = 18 / 3
g = 3 + 1
h = f * g
i = e + h
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a ) 14 , 16 , b ) 16 , 14 , c ) 18 , 12 , d ) 12 , 18 , e ) 17 , 13 | b | subtract(30, divide(subtract(30, divide(10, const_2)), const_2)) | sum of two numbers is 30 . two times of the first exceeds by 10 from the three times of the other . then the numbers will be ? | "explanation : x + y = 30 2 x ã ¢ â ‚ ¬ â € œ 3 y = 10 x = 16 y = 14 answer : b" | a = 10 / 2
b = 30 - a
c = b / 2
d = 30 - c
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(40, divide(divide(40, const_2), 4)) | a runner runs the 40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot , and continues to run at half her previous speed . if the second half takes her 4 hours longer than the first half , how many hours did it take the runner to run the second half ? | "the runner runs the first 20 miles at speed v and the second 20 miles at speed v / 2 . the time t 2 to run the second half must be twice the time t 1 to run the first half . t 2 = 2 * t 1 = t 1 + 4 t 1 = 4 and so t 2 = 8 . the answer is c ." | a = 40 / 2
b = a / 4
c = 40 / b
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a ) 6.4 days , b ) 4.4 days , c ) 5.4 days , d ) 8.4 days , e ) 2.4 days | a | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 15), divide(const_1, 20)))) | a man can do a job in 15 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | a 6.4 days 1 day work of the three persons = ( 1 / 15 + 1 / 20 + 1 / 25 ) = 47 / 300 so , all three together will complete the work in 300 / 47 = 6.4 days . - | a = 1 / 25
b = 1 / 15
c = 1 / 20
d = b + c
e = a + d
f = 1 / e
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a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 8 | a | divide(subtract(272, multiply(3.2, 10)), 40) | cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 272 runs ? | required run rate = 272 - ( 3.2 x 10 ) = 240 = 6 40 40 a | a = 3 * 2
b = 272 - a
c = b / 40
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a ) 8 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 22 % | e | subtract(multiply(divide(subtract(37, 34), subtract(const_100, 34)), const_100), const_10) | two numbers are less than third number by 34 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 34 ) % of x = 66 % of x = 66 x / 100 second number is ( 63 x / 100 ) difference = 66 x / 100 - 63 x / 100 = 3 x / 100 so required percentage is , difference is what percent of first number ( 3 x / 100 * 100 / 63 x * 100 ) % = 22 % answer : e" | a = 37 - 34
b = 100 - 34
c = a / b
d = c * 100
e = d - 10
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a ) 30.5 , b ) 49 , c ) 100 , d ) 105 , e ) 140 | a | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 30), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 30 % iodine solution ? | "solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 30 / 100 40 x + 700 = 30 x + 1050 10 x = 305 x = 30.5 lts correct option : a" | a = 35 * 20
b = a / 100
c = b * 30
d = 35 * 20
e = d / 100
f = e * 20
g = c - f
h = 40 / 100
i = 20 * h
j = 35 * 20
k = j / 100
l = i - k
m = g / l
n = m + 35
|
a ) 5.5 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 % | a | multiply(divide(add(multiply(50, divide(5, const_100)), multiply(30, divide(10, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if jill paid a 5 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2.5 tax on others = 3 percentage = 5.5 / 100 = 5.5 % answer : a" | a = 5 / 100
b = 50 * a
c = 10 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 8 , b ) 64 , c ) 120 , d ) 102 , e ) 144 | c | subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 70), const_100), 48) | in a certain warehouse , 70 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ? | if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 70 % of the packages weigh less than 75 pounds so , 70 % oft = number of packages that weigh less than 75 pounds 70 % of 240 = 168 , so 168 packages weigh less than 75 pounds of those 168 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 168 - 48 = 120 = c | a = 100 - 80
b = 48 / a
c = b * 100
d = c * 70
e = d / 100
f = e - 48
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a ) 15 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | b | multiply(multiply(multiply(divide(1, 12), 3), const_2), multiply(30, 2)) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is 3 times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 2 ounces of corn syrup , how many ounces of water does it contain ? | standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 2 . so , x = 32.5 ounces . water = ( 60 / 65 ) * ( 65 / 2 ) = = 30 ounces . ans b | a = 1 / 12
b = a * 3
c = b * 2
d = 30 * 2
e = c * d
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a ) 63 , b ) 65 , c ) 62 , d ) 60 , e ) 45 | a | multiply(divide(add(multiply(1, const_60), 45), 30), 18) | a machine , working at a constant rate , manufactures 18 pens in 30 minutes . how many pens does it make in 1 hr 45 min ? | change 1 hr 45 min to 105 min . for this , we need to set up a simple proportion of pens per time 18 / 30 = s / 105 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead , cancel before you multiply . for what we can see this post . we can cancel 18 / 30 as 6 / 10 6 / 10 = s / 105 we can cross - multiply . 10 * s = 6 * 105 = 630 if we divide both sides by 10 , s = 63 the machine would be 63 pens in 1 hr 45 min . answer : a | a = 1 * const_60
b = a + 45
c = b / 30
d = c * 18
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a ) 220 , b ) 200 , c ) 210 , d ) 225 , e ) 230 | b | divide(1, divide(add(multiply(const_3600, divide(1, 225)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 225 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "b 225 * t = 1 km = > t = 1 / 225 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 225 + 2 / 3600 ) = 1 = > v = 200 km / h" | a = 1 / 225
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
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a ) 5 , b ) - 4 , c ) 1 , d ) 2 , e ) 3 | c | divide(12, 7) | if ( - 7 ) ^ ( 8 x ) = 7 ^ ( 12 - 5 x ) and x is an integer , what is the value of x ? | "since x is an integer , ( - 7 ) ^ ( 8 x ) is always positive . so , 7 ^ 8 x = 7 ^ ( 13 - 5 x ) 8 x = 12 - 5 x 12 x = 12 x = 1 answer : c" | a = 12 / 7
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a ) 2.5 % , b ) 17.67 % , c ) 28.8 % , d ) 45.2 % , e ) 73.6 % | c | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 3))), subtract(1, divide(1, 4))), const_100) | a library branch originally contained 18360 volumes , 30 % of which were fiction novels . 1 / 4 of the volumes were transferred to another location and 1 / 3 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ? | "fiction novels = 5,508 transferred to another location = 4590 transferred fiction novels = 1530 non transferred fiction novels = 3,978 percent of the remaining collection was fiction novels = 3978 / ( 18360 - 4590 ) * 100 = > 28.888 . . . % hence answer will be ( c )" | a = 30 / 100
b = 1 / 3
c = 1 - b
d = a * c
e = 1 / 4
f = 1 - e
g = d / f
h = g * 100
|
a ) 2525 , b ) 2975 , c ) 3225 , d ) 3775 , e ) 3885 | d | subtract(divide(multiply(100, add(100, const_1)), const_2), divide(multiply(subtract(51, const_1), 51), const_2)) | ( 51 + 52 + 53 + … … … + 100 ) is equal to : | "( 51 + 52 + 53 + … … … + 100 ) = ( 1 + 2 + 3 + … … . + 100 ) - ( 1 + 2 + 3 + 4 + … … + 50 ) = ( 100 * 101 ) / 2 - ( 50 * 51 ) / 2 = ( 5050 - 1275 ) = 3775 . answer : d" | a = 100 + 1
b = 100 * a
c = b / 2
d = 51 - 1
e = d * 51
f = e / 2
g = c - f
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a ) 35.2 , b ) 36.1 , c ) 30.5 , d ) 39.1 , e ) none | c | divide(add(multiply(30, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 30 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 30 × 50 + 48 – 23 ) = 1525 . therefore correct mean = 1525 / 50 = 30.5 . answer c" | a = 30 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 120 metres , b ) 700 metres , c ) 324 metres , d ) 828 metres , e ) 600 metres | b | multiply(divide(multiply(60, const_1000), const_3600), 42) | a train running at the speed of 60 km / hr crosses a pole in 42 seconds . what is the length of the train ? | "speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 42 m = 150 m . option b" | a = 60 * 1000
b = a / 3600
c = b * 42
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a ) 60 kg , b ) 75 kg , c ) 68 kg , d ) 85 kg , e ) 90 kg | c | add(multiply(6, 3.5), 47) | the average weight of 6 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 47 kg . what might be the weight of the new person ? | "total weight increased = ( 6 x 3.5 ) kg = 21 kg . weight of new person = ( 47 + 21 ) kg = 68 kg option c" | a = 6 * 3
b = a + 47
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a ) 5 hrs , b ) 11.6 hrs , c ) 15.6 hrs , d ) 20.1 hrs , e ) 30 hrs | b | divide(70, 6) | ajay can walk 6 km in 1 hour . in how many hours he can walk 70 km ? | "1 hour he walk 6 km he walk 70 km in = 70 / 6 * 1 = 11.6 hours answer is b" | a = 70 / 6
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a ) 11.4 km , b ) 10.9 km , c ) 10.4 km , d ) 6.06 km , e ) 12.56 km | d | multiply(add(18, 6), divide(14, const_60)) | the speed of a boat in still water in 18 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 14 minutes is : | "explanation : speed downstream = ( 18 + 6 ) = 24 kmph time = 14 minutes = 14 / 60 hour = 7 / 30 hour distance travelled = time × speed = ( 7 / 30 ) × 26 = 6.06 km answer : option d" | a = 18 + 6
b = 14 / const_60
c = a * b
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a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 155.50 , e ) none | d | divide(subtract(multiply(133, add(add(1, 1), 2)), add(126, 126)), 2) | tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 133 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 133 / 22.50 = 1 = â € º x - 133 = 22.50 = â € º x = 155.50 . hence , price of the third variety = rs . 155.50 per kg . answer d" | a = 1 + 1
b = a + 2
c = 133 * b
d = 126 + 126
e = c - d
f = e / 2
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a ) 7500 , b ) 6500 , c ) 8875 , d ) 9500 , e ) none of them | c | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 71,000 at 16 2 / 3 % per year for 9 months . | "p = rs . 71000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . simple interest = ( p * r * t ) / 100 = rs . ( 71,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8875 answer is c ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
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a ) 16 % , b ) 10.76 % , c ) 10.69 % , d ) 10.98 % , e ) 10 % | b | multiply(divide(subtract(72, 65), 65), const_100) | john makes $ 65 a week from his job . he earns a raise andnow makes $ 72 a week . what is the % increase ? | "increase = ( 7 / 65 ) * 100 = ( 7 / 65 ) * 100 = 10.76 % . b" | a = 72 - 65
b = a / 65
c = b * 100
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(multiply(20, 15), 20) | if 20 men take 15 days to to complete a job , in how many days can 20 men finish that work ? | ans . 15 days | a = 20 * 15
b = a / 20
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a ) 15 days , b ) 16 days , c ) 19 days , d ) 17 days , e ) 24 days | e | divide(const_1, add(divide(const_1, 40), divide(const_1, 60))) | a can do a piece of work 40 days . b can do work in 60 days . in how many days they will complete the work together ? | "lcm = 120 , ratio = 40 : 60 = 2 : 3 no of days = 120 / ( 2 + 3 ) = 90 / 5 = 24 days answer : e" | a = 1 / 40
b = 1 / 60
c = a + b
d = 1 / c
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a ) s . 260 , b ) s . 160 , c ) s . 360 , d ) s . 50 , e ) s . 90 | b | subtract(divide(divide(800, add(divide(1, 2), divide(1, 3))), 2), divide(divide(800, add(divide(1, 2), divide(1, 3))), 3)) | a profit of rs . 800 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 800 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 480 and 320 . difference in profit share = 480 - 320 = 160 answer : b" | a = 1 / 2
b = 1 / 3
c = a + b
d = 800 / c
e = d / 2
f = 1 / 2
g = 1 / 3
h = f + g
i = 800 / h
j = i / 3
k = e - j
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a ) 48 , b ) 9 , c ) 7 , d ) 60 , e ) 15 | d | subtract(divide(multiply(1.50, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.0, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.50 km and 1.0 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | ": relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.50 + 1.0 = 2.5 km = 2500 m . required time = 2500 * 3 / 125 = 60 sec . answer : d" | a = 1 * 50
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 0
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) a ) 70 , b ) b ) 45 , c ) c ) 200 , d ) d ) 280 , e ) e ) 140 | b | add(const_3, const_4) | what is the smallest integer that is multiple of 3 , 5,9 | "it is the lcm of 3 , 5 and 9 which is 45 . the answer is b ." | a = 3 + 4
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | multiply(divide(subtract(78, 60), 60), const_100) | a book is bought for $ 60 and sold for $ 78 . what is the profit in percentage ? | "78 / 60 = 1.3 the answer is c ." | a = 78 - 60
b = a / 60
c = b * 100
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a ) 50 / 49 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 7), const_2)) | product of two natural numbers is 7 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 7 , as 17 is a prime number , so a = 1 , b = 7 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 7 ( 2 ) = 50 / 49 option a" | a = 1 / 1
b = a ** 2
c = 1 / 7
d = c ** 2
e = b + d
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a ) 37.5 $ , b ) 75 $ , c ) 87.5 $ , d ) 94.5 $ , e ) 98.4 $ | b | add(multiply(multiply(add(32, divide(multiply(32, 75), const_100)), 2), 0.4), multiply(multiply(32, 2), 0.4)) | in a fuel station the service costs $ 1.15 per car , every liter of fuel costs 0.4 $ . assuming that you own 2 sports cars and 2 executive cars and all fuel tanks are empty . how much will it cost to fuel all cars together if a sports car tank is 32 liters and an executive car tank is 75 % bigger ? | "total cars = 4 1.15 * 4 = 4.6 - > service cost fuel cost in sports car = 2 * 32 * 0.4 = 25.6 fuel cost in executive car = 25.6 * 7 / 4 = 44.8 total fuel cost = 25.6 + 44.8 = 70.4 cost to fuel car = 70.4 + 4.6 = 75 answer : b" | a = 32 * 75
b = a / 100
c = 32 + b
d = c * 2
e = d * 0
f = 32 * 2
g = f * 0
h = e + g
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a ) 7.1 days , b ) 8.1 days , c ) 9.1 days , d ) 10 days , e ) 11 days | b | divide(67, divide(add(add(divide(67, 11), divide(67, 20)), add(divide(67, 11), divide(67, 67))), const_2)) | a , band c can do a piece of work in 11 days , 20 days and 67 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | "( a + b ) ' s 1 day ' s work = 1 / 11 + 1 / 20 = 31 / 220 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 67 = 78 / 737 work done in 2 day ' s = 31 / 220 + 78 / 737 = 19 / 77 19 / 77 th work done in 2 days work done = 77 / 19 * 2 = 8.1 days answer : b" | a = 67 / 11
b = 67 / 20
c = a + b
d = 67 / 11
e = 67 / 67
f = d + e
g = c + f
h = g / 2
i = 67 / h
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a ) 20 , b ) 22 , c ) 26 , d ) 24 , e ) 30 | c | subtract(power(3, 3), const_1) | a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 6 , 4 and 3 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | "total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes only the product of 333 and will be odd , the remaining 26 will have either 2 or 4 in them , which ensures that their product will be even . therefore the number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 1 = 26 . answer : c" | a = 3 ** 3
b = a - 1
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a ) 21 . , b ) 28 . , c ) 35 . , d ) 12 . , e ) 52 . | d | add(const_3, const_2) | in a rectangular axis system , what is the area of a parallelogram with the coordinates : ( 3,5 ) , ( 9,5 ) , ( 2,7 ) , ( 8,7 ) ? | "delta x will give us the dimension of one side of the parallelogram = 9 - 3 = 6 unit delta y will give us the dimension of the other side of parallelogram = 7 - 5 = 2 unit area of parallelogram = 6 * 2 = 12 answer is d" | a = 3 + 2
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a ) 10290 , b ) 5290 , c ) 1289 , d ) 1290 , e ) 2290 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | find the c . i . on a sum of rs . 10000 for 6 months at 25 % per annum , interest being compounded quarterly ? | c . i . = 2000 ( 21 / 20 ) ^ 2 - 1800 = 1289 answer : c | a = const_0_25 / 4
b = 1 + a
c = const_0_25 / 4
d = 1 + c
e = b * d
f = 10000 * e
g = f - 10000
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a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | b | subtract(multiply(add(20, 5), const_2), 10) | you collect pens . suppose you start out with 5 . mike gives you another 20 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 10 pens . how many pens do you have at the end ? | "solution start with 5 pens . mike gives you 20 pens : 5 + 20 = 25 pens . cindy doubles the number of pens you have : 25 ã — 2 = 50 pens . sharon takes 10 pens from you : 50 - 10 = 40 pens . so you have 40 at the end . correct answer : b" | a = 20 + 5
b = a * 2
c = b - 10
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a ) rs . 1400 , b ) rs . 1450 , c ) rs . 1500 , d ) rs . 1550 , e ) rs . 1525 | c | subtract(1560, divide(multiply(subtract(1590, 1560), 4), 6)) | a sum of money at simple interest amounts to rs . 1560 in 4 years and to rs . 1590 in 6 years . the sum is : | "s . i . for 2 years = rs . ( 1590 - 1560 ) = rs . 30 . s . i . for 4 years = rs . ( 30 x 2 ) = rs . 60 . principal = rs . ( 1560 - 60 ) = rs . 1500 . answer : option c" | a = 1590 - 1560
b = a * 4
c = b / 6
d = 1560 - c
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a ) 60 , b ) 65 , c ) 70 , d ) 75 , e ) 80 | a | add(add(25, 32), const_2) | a lady has fine gloves and hats in her closet - 18 blue , 32 red , and 25 yellow . the lights are out and it is totally dark . in spite of darkness , she can make out the difference between a hat and a glove . she takes out an item out of the closet only if she is sure that it is a glove . how many gloves must she take out to make sure she has a pair of each color ? | "32 r + 24 y + 1 y + 1 b + 2 b = 60 answer : a" | a = 25 + 32
b = a + 2
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a ) 2.8 % , b ) 3.6 % , c ) 4 % , d ) 5.2 % , e ) 6.0 % | c | multiply(divide(add(multiply(40, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 40 percent on clothing , 30 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "let amount spent by jill = 100 clothing = 40 , food = 30 , others = 30 tax on clothing = 1.6 tax on others = 2.4 percentage = 4 / 100 = 4 % answer : c" | a = 4 / 100
b = 40 * a
c = 8 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
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a ) 48 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 5) | a student chose a number , multiplied it by 5 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 5 ⋅ x − 138 = 102 5 x = 240 x = 48 answer a" | a = 102 + 138
b = a / 5
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a ) 10 , b ) 15 , c ) 4 , d ) 12 , e ) indeterminate | e | subtract(30, 20) | in a class , 30 students pass in english and 20 students in maths , while some students among these pass in both . how many students do only english as compared to those doing only maths ? | "for doing union or intersection we would need three values . in this question the missing information in this question is total no . of students in the class . answer : e" | a = 30 - 20
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a ) 1 , b ) 2 , c ) 6 , d ) 7 , e ) 9 | c | add(const_2, const_3) | if 31 / 198 = 0.1565 , what is the 97 nd digit to the right of the decimal point of the fraction ? | "we are not concerned what 31 / 198 means . . we have to look at the decimal . . 0.1565 means 0.1565656 . . . . so leaving girst and second digit to the right of decimal , all odd numbered are 6 and all even numbered are 5 . . here 97 is odd , so ans is 6 c" | a = 2 + 3
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a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 24 , d ) 19 / 30 , e ) 11 / 15 | c | subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(8))) | sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 8 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card , what portion of her limit on the platinum card will remain unspent ? | let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 8 x = 7 / 8 x so if gold card balance is transferred then the rest unspent will be 7 / 8 x - x / 6 = 17 / 24 x so the ans is c | a = 1/(3)
b = 1/(2)
c = a * b
d = 1/(8)
e = c + d
f = 1 - e
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a ) $ 150 , b ) $ 210 , c ) $ 250 , d ) $ 160 , e ) $ 200 | c | add(125, 125) | robert spent $ 100 in buying raw materials , $ 125 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | let the total amount be x then , ( 100 - 10 ) % of x = 100 + 125 90 % of x = 225 90 x / 100 = 225 x = $ 250 answer is c | a = 125 + 125
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a ) rs . 72 , b ) rs . 150 , c ) rs . 112.50 , d ) rs . 116.50 , e ) none of these | b | multiply(divide(const_100, 8), 12) | a 12 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 12 , investment = rs . ( 100 / 8 x 12 ) = rs . 150 ∴ market value of rs . 100 stock = rs . 150 answer b" | a = 100 / 8
b = a * 12
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a ) 13.9 sec , b ) 13.1 sec , c ) 17.9 sec , d ) 61.9 sec , e ) 47.98 sec | b | divide(add(130, 160), multiply(98, const_0_2778)) | how long does a train 130 m long running at the speed of 98 km / hr takes to cross a bridge 160 m length ? | "speed = 98 * 5 / 18 = 27 m / sec total distance covered = 130 + 160 = 290 m . required time = 290 / 22 = 13.1 sec . answer : b" | a = 130 + 160
b = 98 * const_0_2778
c = a / b
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a ) 1000 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | a | multiply(5500, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 5500 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 5500 x ( 2 / 11 ) ) = rs . 1000 . answer is a" | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 5500 * f
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a ) 23 , b ) 20 , c ) 15 , d ) 5 , e ) 2 | c | add(divide(add(14, 12), const_2), multiply(1, const_2)) | if { x } is the product of all even integers from 1 to x inclusive , what is the greatest prime factor of { 14 } + { 12 } ? | "soln : { 14 } + { 12 } = 14 * { 12 } + { 12 } = 15 * { 12 } answer : c" | a = 14 + 12
b = a / 2
c = 1 * 2
d = b + c
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a ) w = 21 , b ) w = 22 , c ) w = 23 , d ) w = 24 , e ) 27 | c | subtract(power(5, 2), 2) | if x + ( 1 / x ) = 5 , what is the value of w = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a = 5 ** 2
b = a - 2
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a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 700 | a | multiply(divide(351, divide(add(const_100, 30), const_100)), 2) | if the price of a certain computer increased 30 percent from d dollars to 351 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 351 ( given ) i . e . d = 351 / 1.3 = $ 270 i . e . 2 d = 2 * 270 = 540 answer : option a" | a = 100 + 30
b = a / 100
c = 351 / b
d = c * 2
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a ) 50 , b ) 48 , c ) 70 , d ) 40 , e ) 50 | b | divide(multiply(add(54, 6), 6), 6) | a certain number of men can do a work in 54 days . if there were 6 men more it could be finished in 6 days less . how many men were there in the beginning ? | "explanation : m ( 54 ) = ( m + 6 ) ( 48 ) 9 m - 8 m = 48 m = 48 answer : option b" | a = 54 + 6
b = a * 6
c = b / 6
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a ) 20 , b ) 15 , c ) 40 , d ) 10 , e ) 12 | a | multiply(4, inverse(subtract(1, add(divide(1, 2), divide(3, 5))))) | in traveling from a dormitory to a certain city , a student went 1 / 2 of the way by foot , 3 / 5 of the way by bus , and the remaining 4 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 2 x + 3 / 5 x + 4 x - 1 / 2 x - 3 / 5 x = 4 x = 20 km option : a" | a = 1 / 2
b = 3 / 5
c = a + b
d = 1 - c
e = 1/(d)
f = 4 * e
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a ) 1000 , b ) 1250 , c ) 1300 , d ) 1600 , e ) 1875 | e | divide(add(divide(multiply(500, const_100), 30), 500), divide(80, const_100)) | a small pool filled only with water will require an additional 500 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 500 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "since pumping in additional 500 gallons of water will increase the amount of water in the pool by 30 % , then initially the pool is filled with 1,000 gallons of water . so , we have that 1,000 + 500 = 0.8 * { total } - - > { total } = 1,875 answer : e ." | a = 500 * 100
b = a / 30
c = b + 500
d = 80 / 100
e = c / d
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a ) 0.09 , b ) 0.15 , c ) 0.54 , d ) 0.85 , e ) 0.95 | e | divide(add(add(22, 18), 17), 60) | a certain bag contains 60 balls — 22 white , 18 green , 17 yellow , 3 red , and 1 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 18 + 17 ) / 60 = 57 / 60 = 0.95 . answer : e ." | a = 22 + 18
b = a + 17
c = b / 60
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a ) 1.8 , b ) 2.0 , c ) 1.52 , d ) 0.52 , e ) 1.3 | d | multiply(divide(40, multiply(multiply(const_4, const_5), const_5)), divide(20, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 40 % of the times , b speaks the truth 20 % of the times . what is the probability that at least one will tell the truth | "probability of a speaks truth p ( a ) = 4 / 10 ; false = 6 / 10 probability of b speaks truth p ( b ) = 2 / 10 ; false = 8 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 6 / 10 ) * ( 8 / 10 ) ] = 1 - 48 / 100 = 1 - 0.48 = 0.52 answer : d" | a = 4 * 5
b = a * 5
c = 40 / b
d = 4 * 5
e = d * 5
f = 20 / e
g = c * f
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(const_2, const_2) | find the number of zeroes in 58 ! ( 58 factorial ) | "no of zeroes is 58 / 5 = 11 11 / 5 = 2 11 + 2 = 13 answer : b" | a = 2 + 2
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a ) 438 , b ) 395 , c ) 378 , d ) 279 , e ) 222 | b | multiply(circumface(divide(36, const_2)), 3.50) | find the cost of fencing around a circular field of diameter 36 m at the rate of rs . 3.50 a meter ? | "2 * 22 / 7 * 18 = 113 113 * 3 1 / 2 = rs . 395 answer : b" | a = 36 / 2
b = circumface * (
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a ) 7 , b ) 8 , c ) 9 , d ) 21 , e ) 14 | c | subtract(47, multiply(19, const_2)) | a number when divided by 342 gives a remainder 47 . when the same number ift divided by 19 , what would be the remainder ? | "sol . on dividing the given number by 342 , let k be the quotient and 47 as remainder . then , number – 342 k + 47 = ( 19 x 18 k + 19 x 2 + 9 ) = 19 ( 18 k + 2 ) + 9 . the given number when divided by 19 , gives ( 18 k + 2 ) as quotient and 9 as remainder . option c" | a = 19 * 2
b = 47 - a
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a ) 58 , b ) 68 , c ) 78 , d ) 88 , e ) 90 | d | subtract(1100, add(4, multiply(gcd(1100, lcm(lcm(7, 12), 16)), lcm(lcm(7, 12), 16)))) | which is the least number that must be subtracted from 1100 so that the remainder when divided by 7 , 12 , 16 is 4 ? | "first we need to figure out what numbers are exactly divisible by 7 , 12,16 . this will be the set { lcm , lcmx 2 , lcmx 3 , . . . } lcm ( 7 , 12,16 ) = 48 * 7 = 336 the numbers which will leave remainder 4 will be { 336 + 4 , 336 x 2 + 4 , 336 x 3 + 4 , . . . } the largest such number less than or equal to 1100 is 336 x 3 + 4 or 1012 to obtain this you need to subtract 88 d" | a = math.lcm(7, 12)
b = math.lcm(a, 16)
c = math.gcd(1100, b)
d = math.lcm(7, 12)
e = math.lcm(d, 16)
f = c * e
g = 4 + f
h = 1100 - g
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a ) 8 , b ) 16 , c ) 32 , d ) 48 , e ) 54 | b | multiply(divide(multiply(1, 12), subtract(4, 1)), 4) | two integers are in the ratio of 1 to 4 . if 12 is added to the smaller number , the ratio becomes 1 to 1 . find the larger integer . | "one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 4 , then 4 x = y , where x is the smaller integer . if adding 12 to the smaller integer makes the ratio 1 to 1 , then x + 12 = y . substituting y = 4 x into the second equation yields x + 12 = 4 x . so , x = 4 ( smaller integer ) and , y = 4 x = 16 ( larger integer ) so b is the correct answer . another option is to test the answer choices . a ) the larger number is 8 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 2 if 12 is added to the smaller number , the ratio becomes 1 to 1 if we add 12 to the smaller number ( 2 ) , we get 14 . so , the new ratio is 14 to 8 which is the same as 7 to 4 . no good . we want a resulting ratio of 1 to 1 eliminate a b ) the larger number is 16 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 4 if 12 is added to the smaller number , the ratio becomes 1 to 1 if we add 12 to the smaller number ( 4 ) , we get 16 so , the new ratio is 16 to 16 which is 1 to 1 . therefore ( b ) is correct . c ) the larger number is 32 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 8 if 12 is added to the smaller number , the ratio becomes 1 to 1 if we add 12 to the smaller number ( 8 ) , we get 20 so , the new ratio is 20 to 32 which is 5 to 8 . no good . we want a resulting ratio of 1 to 1 eliminate c d ) the larger number is 48 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 12 if 12 is added to the smaller number , the ratio becomes 1 to 1 if we add 12 to the smaller number ( 12 ) , we get 24 so , the new ratio is 24 to 48 . this is the same as the ratio of 1 to 2 no good . we want a resulting ratio of 1 to 1 eliminate d e ) the larger number is 54 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 13.5 ( which is not an integer ) since both numbers are integers , the larger number can not be 54 . eliminate e therefore b is the only correct choice ." | a = 1 * 12
b = 4 - 1
c = a / b
d = c * 4
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a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 3000 , e ) $ 1200 | d | subtract(4000, divide(4000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | the salaries of a and b together amount to $ 4000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | let a ' s salary is x b ' s salary = 4000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 4000 - x ) x = $ 3000 answer is d | a = 100 - 85
b = 100 - 95
c = a / b
d = c + 1
e = 4000 / d
f = 4000 - e
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a ) 29 / 16 , b ) 19 / 16 , c ) 15 / 16 , d ) 17 / 4 , e ) 0 | d | subtract(add(add(multiply(divide(2, 3), divide(3, 8)), 4), divide(1, 2)), divide(8, 16)) | 1 / 2 + [ ( 2 / 3 * 3 / 8 ) + 4 ] - 8 / 16 = | 1 / 2 + [ ( 2 / 3 * 3 / 8 ) + 4 ] - 8 / 16 = 1 / 2 + [ ( 1 / 4 ) + 4 ] - 8 / 16 = 1 / 2 + [ 17 / 4 ] - 9 / 16 = 8 / 16 + 68 / 16 - 8 / 16 = 68 / 16 = 17 / 4 d | a = 2 / 3
b = 3 / 8
c = a * b
d = c + 4
e = 1 / 2
f = d + e
g = 8 / 16
h = f - g
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a ) 117 % , b ) 120 % , c ) 121 % , d ) 135 % , e ) 140 % | a | multiply(power(add(1, divide(divide(6, const_100), const_2)), add(const_2, const_3)), const_100) | michelle deposited a certain sum of money in a savings account on july 1 st , 2007 . she earns an 6 % interest compounded semiannually . the sum of money in the account on december 31 st , 2009 is approximately what percent of the initial deposit ? | "since michelle earns 6 % interest compounded semiannually , then she earns 3 % interest every 6 months . now , the simple interest earned in 5 periods ( 30 months = 5 * 6 months ) would be 3 % * 5 = 15 % . but , since the interest iscompoundedevery 6 months , then there would be interest earned on interest ( very small amount ) thus the actual interest earned would be a little bit more than 15 % , only answer choice a fits . answer : a" | a = 6 / 100
b = a / 2
c = 1 + b
d = 2 + 3
e = c ** d
f = e * 100
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a ) 12.00 % , b ) 16.67 % , c ) 18.33 % , d ) 20.00 % , e ) 21.50 % | b | multiply(subtract(const_1, divide(multiply(multiply(5, 10), 20), multiply(multiply(6, 10), 20))), const_100) | a company producing fruit juice changed its packaging from boxes measuring 5 x 10 x 20 centimeters to boxes measuring 6 x 10 x 20 centimeters . if the price of a box did not change and all boxes are full of juice , by approximately what percent did the price of the juice decrease ? | suppose when v = 5 x 10 x 20 = 1000 cm , the price is $ 1200 per cm price = $ 1.2 the price is $ 1200 if v = 6 x 10 x 20 = 1000 cm . per cm price = $ 1.0 so the price is decreased by $ 0.2 so the % price decreased by $ 0.2 / 1.20 = 16.67 % answer : b | a = 5 * 10
b = a * 20
c = 6 * 10
d = c * 20
e = b / d
f = 1 - e
g = f * 100
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a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 950 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 19,000 is invested at the same interest rate , what is the interest from the investment ? | "- > 250 / 5,000 = 5 % and 19,000 * 5 % = 950 . thus , d is the answer ." | a = 2 * 100
b = a * 100
c = 250 * b
d = 2 * 100
e = d * 100
f = e / 4
g = c / f
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a ) 1.9 , b ) 1.7 , c ) 1.4 , d ) 1.5 , e ) 1.1 | c | divide(subtract(multiply(4.2, const_2), 4.2), const_3) | a man can row 4.2 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | "speed of boat in still water ( b ) = 4.2 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.4 km / hr . answer : c" | a = 4 * 2
b = a - 4
c = b / 3
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a ) 29491 , b ) 28491 , c ) 39491 , d ) 49491 , e ) 59491 | a | multiply(116000, divide(multiply(3, 5), add(add(multiply(3, 5), multiply(4, 5)), multiply(4, 6)))) | divide rs . 116000 among 3 persons a , b and c such that the ratio of the shares of a and b is 3 : 4 and that of b : c is 5 : 6 . find the share of a ? | compound ratio of a : b : c a : b = 3 : 4 b : c = 5 : 6 - - - - - - - - - - a : b : c = 15 : 20 : 24 we can divide rs . 116000 in this ratio . share of a = 15 / 59 * 116000 = 29491 answer : a | a = 3 * 5
b = 3 * 5
c = 4 * 5
d = b + c
e = 4 * 6
f = d + e
g = a / f
h = 116000 * g
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 7 / 10 , e ) 11 / 15 | c | add(divide(multiply(subtract(const_1, divide(40, multiply(40, 40))), 150), 450), divide(add(multiply(divide(40, multiply(40, 40)), 300), multiply(divide(40, multiply(40, 40)), 150)), 450)) | in a certain corporation , there are 300 male employees and 150 female employees . it is known that 40 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | "p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.4 * 300 / 450 = 120 / 450 = 4 / 15 the sum of the probabilities is 9 / 15 = 3 / 5 the answer is c ." | a = 40 * 40
b = 40 / a
c = 1 - b
d = c * 150
e = d / 450
f = 40 * 40
g = 40 / f
h = g * 300
i = 40 * 40
j = 40 / i
k = j * 150
l = h + k
m = l / 450
n = e + m
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a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | d | multiply(divide(add(multiply(20, divide(20, const_100)), multiply(80, divide(15, const_100))), add(20, 80)), const_100) | in a park there are two ponds with both brown ducks and green ducks . in the smaller pond there are 20 ducks and in the larger pond there are 80 ducks . if 20 % of the ducks in the smaller pond are green and 15 % of the ducks in the larger pond are green , then what percentage of ducks are green ? | "number of ducks in small pond = 20 green ducks in small pond = 20 % of 20 = 4 ducks number of ducks in large pond = 80 green ducks in large pond = 15 % of 80 = 12 ducks total number of ducks = 20 + 80 = 100 total number of green ducks = 4 + 12 = 16 ducks percentage of green ducks = 16 / 100 * 100 = 16 % answer : d" | a = 20 / 100
b = 20 * a
c = 15 / 100
d = 80 * c
e = b + d
f = 20 + 80
g = e / f
h = g * 100
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a ) a ) 7.63 , b ) b ) 2.6 , c ) c ) 8.83 , d ) d ) 6.69 , e ) e ) 3 | c | divide(power(5, 2), 2) | the current of a stream runs at the rate of 5 kmph . a boat goes 6 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 5 m = x ds = x + 5 us = x - 5 6 / ( x + 5 ) + 6 / ( x - 5 ) = 2 x = 8.83 answer : c" | a = 5 ** 2
b = a / 2
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a ) 80 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | c | multiply(divide(1386, add(445, 545)), const_60) | if two projectiles are launched at the same moment from 1386 km apart and travel directly towards each other at 445 km per hour and 545 km per hour respectively , how many minutes will it take for them to meet ? | "the projectiles travel a total of 990 km per hour . the time to meet is 1386 / 990 = 1.4 hours = 84 minutes the answer is c ." | a = 445 + 545
b = 1386 / a
c = b * const_60
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a ) 15 , b ) 18 , c ) 16 , d ) 22 , e ) none | c | divide(add(add(11, const_4), subtract(21, const_4)), const_2) | find the average of all the numbers between 11 and 21 which are divisible by 2 . | "sol . average = ( 12 + 14 + 16 + 18 + 20 / 5 ) = 80 / 5 = 16 . answer c" | a = 11 + 4
b = 21 - 4
c = a + b
d = c / 2
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a ) 238 , b ) 150 , c ) 988 , d ) 210 , e ) 171 | d | multiply(140, subtract(const_2, const_1)) | a train speeds past a pole in 15 sec and a platform 140 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 140 ) / 25 = x / 15 = > x = 210 m . answer : d" | a = 2 - 1
b = 140 * a
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a ) rs . 25000 , b ) rs . 20000 , c ) rs . 15000 , d ) rs . 22000 , e ) none of these | b | divide(400, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 400 . find the total worth of the stock ? | let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 400 = > x = 20000 answer : b | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 400 / h
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a ) 21 / 26 , b ) 25 / 26 , c ) 11 / 26 , d ) 6 / 26 , e ) 22 / 33 | d | multiply(divide(add(const_2, const_1), add(multiply(12, const_2), const_2)), const_2) | the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 12 times the probability that both events a and b occur . what is the probability that event a occurs ? | "let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 12 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 12 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 12 * 2 b * b 3 b - 2 b ^ 2 = 24 b ^ 2 3 b = 26 b ^ 2 b = 3 / 26 = 3 / 26 so , a = 2 b = 6 / 26 the correct answer is d ." | a = 2 + 1
b = 12 * 2
c = b + 2
d = a / c
e = d * 2
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a ) 212 , b ) 200 , c ) 220 , d ) 230 , e ) 240 | a | subtract(320, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(320, 15.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | a snooker tournament charges $ 40.00 for vip seats and $ 15.00 for general admission ( “ regular ” seats ) . on a certain night , a total of 320 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | "let no of sits in vip enclosure is x then x * 40 + 15 ( 320 - x ) = 7500 or 25 x = 7500 - 4800 , x = 2700 / 25 = 108 vip = 108 general 212 a" | a = 3 + 4
b = a * 1000
c = 2 + 3
d = c * 100
e = b + d
f = 320 * 15
g = e - f
h = 2 + 3
i = 2 + 3
j = h * i
k = g / j
l = 320 - k
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a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52 | b | divide(multiply(32, const_2), const_3) | a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money r did bob receive ? | "notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ r = 20 answer : b" | a = 32 * 2
b = a / 3
|
a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 6 / 20 , e ) 3 / 10 | d | divide(subtract(20, 12), 20) | a ’ s speed is 20 / 12 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 12 meters in that time . so if the race is 20 meters long , when a reaches the finish line , b would be 8 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 8 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 20 meters ) would be the same as the time required by b ( to cover 12 meters ) to reach the finish line . so b should get a head start of 8 / 20 th of the race . answer ( d )" | a = 20 - 12
b = a / 20
|
a ) 175 , b ) 185 , c ) 195 , d ) 205 , e ) 215 | c | multiply(130, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 130 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 130 = 25 * ( x / 15 ) 10 x = 1950 x = 195 meters the answer is c ." | a = 2 - 1
b = 130 * a
|
a ) 59 hours , b ) 54 hours , c ) 59 hours , d ) 56 hours , e ) 26 hours | d | inverse(subtract(divide(const_1, 7), divide(const_1, 8))) | calculate the time it will take for a full tank to become completely empty due to a leak given that the tank could be filled in 7 hours , but due to the leak in its bottom it takes 8 hours to be filled ? | part filled without leak in 1 hour = 1 / 7 part filled with leak in 1 hour = 1 / 8 work done by leak in 1 hour = 1 / 7 â ˆ ’ 1 / 8 = 56 hours answer : d | a = 1 / 7
b = 1 / 8
c = a - b
d = 1/(c)
|
a ) 23 , b ) 13 , c ) 21 , d ) 24 , e ) 25 | b | divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 2))) | a is 2 times as fast as b . a alone can do the work in 20 days . if a and b working together in how many days will the work be completed ? | "a can finish 1 work in 20 days b can finish 1 / 2 work in 20 days - since a is 2 faster than b this means b can finish 1 work in 20 * 2 days = 40 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 20 * 40 / ( 20 + 40 ) = 13 days so answer is b" | a = 1 / 20
b = 1 / 20
c = b / 2
d = a + c
e = 1 / d
|
a ) 3 / 8 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | a | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 1.2 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | "the easiest way for me is to plug in numbers . let the number of basic stereos produced be 40 , and number of delux stereos produced be 20 . total of 60 stereos . if it takes an hour to produce a basic stereo then it will take 1.2 hours to produce a deluxe stereo . 40 basic stereos = 40 hours . 20 delux stereos = 24 hours . total hours = 64 . then the fraction would be 24 / 64 = 3 / 8 . therefore answer a ." | a = 3 + 2
b = a / 10
|
a ) 27 , b ) 279 , c ) 60 , d ) 278 , e ) 379 | c | multiply(divide(multiply(30, const_1), subtract(60, 30)), 60) | a train leaves mumabai at 9 am at a speed of 30 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 60 kmph . when and at what distance from mumbai do the two trains meet ? | when the second train leaves mumbai the first train covers 30 * 1 = 30 km so , the distance between first train and second train is 30 km at 10.00 am time taken by the trains to meet = distance / relative speed = 30 / ( 60 - 30 ) = 1 hours so , the two trains meet at 11 a . m . the two trains meet 1 * 60 = 60 km away from mumbai . answer : c | a = 30 * 1
b = 60 - 30
c = a / b
d = c * 60
|
a ) none , b ) one , c ) two , d ) three , e ) four | c | subtract(15, multiply(3, const_4)) | a = 4 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itself . on the other hand the stem says that x could get how many different number not must get different number ( this is very important issue ) as our friends say , if we simplify numerator more we can obtain : 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 ( 124 ) = 5 ^ 12 ( 31 * 2 * 2 ) divided by x and we are told that this fraction is an integer . so , x could be ( not must be ) 5 , 31 , or 2 ! ! ! so , x could get 2 different values and answer is c . . . ." | a = 3 * 4
b = 15 - a
|
a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.9 kg , e ) 7.8 kg | b | divide(multiply(8.5, 800), const_1000) | a envelop weight 8.5 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ? | "800 * 8.5 6800.0 gm 6.8 kg answer : b" | a = 8 * 5
b = a / 1000
|
a ) 324138 , b ) 248888 , c ) 378908 , d ) 277880 , e ) 379010 | a | add(add(divide(multiply(48, subtract(728400, divide(multiply(728400, 35), const_100))), const_100), divide(multiply(30, divide(multiply(divide(multiply(728400, 35), const_100), 20), const_100)), const_100)), divide(multiply(40, subtract(divide(multiply(728400, 35), const_100), divide(multiply(divide(multiply(728400, 35), const_100), 20), const_100))), const_100)) | in a city , 35 % of the population is composed of migrants , 20 % of whom are from rural areas . of the local population , 48 % is female while this figure for rural and urban migrants is 30 % and 40 % respectively . if the total population of the city is 728400 , what is its female population ? | explanation : total population = 728400 migrants = 35 % of 728400 = 254940 local population = ( 728400 - 254940 ) = 473460 . rural migrants = 20 % of 254940 = 50988 urban migrants = ( 254940 - 50988 ) = 203952 female population = 48 % of 473460 + 30 % of 50988 + 40 % of 203952 = 324138 answer : a | a = 728400 * 35
b = a / 100
c = 728400 - b
d = 48 * c
e = d / 100
f = 728400 * 35
g = f / 100
h = g * 20
i = h / 100
j = 30 * i
k = j / 100
l = e + k
m = 728400 * 35
n = m / 100
o = 728400 * 35
p = o / 100
q = p * 20
r = q / 100
s = n - r
t = 40 * s
u = t / 100
v = l + u
|
a ) 40 % , b ) 50 % , c ) 65 % , d ) 68 % , e ) 70 % | b | multiply(divide(add(multiply(3, const_4), multiply(add(3, 3), 8)), 120), const_100) | a batsman scored 120 runs whichincluded 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets ? | "number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b" | a = 3 * 4
b = 3 + 3
c = b * 8
d = a + c
e = d / 120
f = e * 100
|
a ) 234341 , b ) 235633 , c ) 234677 , d ) 315656 , e ) 251501 | e | subtract(add(multiply(397, 397), multiply(104, 104)), multiply(multiply(2, 397), 104)) | 397 x 397 + 104 x 104 + 2 x 400 x 104 = x ? | "given exp . = ( 397 ) 2 + ( 104 ) 2 + 2 x 397 x 104 = ( 397 + 104 ) 2 = ( 501 ) 2 = ( 500 + 1 ) 2 = ( 5002 ) + ( 1 ) 2 + ( 3 x 500 x 1 ) = 250000 + 1 + 1500 = 251501 e" | a = 397 * 397
b = 104 * 104
c = a + b
d = 2 * 397
e = d * 104
f = c - e
|
a ) 44 / 7 , b ) 9 / 8 , c ) 25 / 8 , d ) 3 / 4 , e ) 7 / 8 | d | multiply(divide(add(const_2, const_1), add(multiply(3, const_2), const_2)), const_2) | the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 3 times the probability that both events a and b occur . what is the probability that event a occurs ? | "let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 3 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 3 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 3 * 2 b * b 3 b - 2 b ^ 2 = 6 b ^ 2 3 b = 8 b ^ 2 b = 3 / 8 = 3 / 8 so , a = 2 b = 3 / 4 the correct answer is d ." | a = 2 + 1
b = 3 * 2
c = b + 2
d = a / c
e = d * 2
|
a ) 18 km , b ) 20 km , c ) 22 km , d ) 24 km , e ) 30 km | d | multiply(40, divide(multiply(30, divide(add(8, 4), const_60)), subtract(40, 30))) | john goes to his office by car at a speed of 40 kmph and reaches 8 minutes earlier . if he goes at a speed of 30 kmph , he reaches 4 mins late . what is the distance from his house to office ? | let his office time be 10 a . m . he reaches office at 8 minutes earlier when he travels at 40 kmph = > he reaches at 9 : 52 a . m he reaches his office 4 minutes late when travels at 30 kmph speed = > he reaches at 10.04 a . m the time difference is ( 10.04 – 9.52 ) = 12 minutes let the distance be d and time = distance speed t 1 – t 2 = 12 minutes = 12 / 60 hours = 1 / 5 hours d / 30 – d / 40 = 1 / 5 ( 4 d - 3 d ) / 120 = 1 / 5 d = 24 km | a = 8 + 4
b = a / const_60
c = 30 * b
d = 40 - 30
e = c / d
f = 40 * e
|
a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 3 / 5 , e ) 1 / 6 | a | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(10, 5), 11), 5), const_2), 5), const_3)) | a certain list consists of 11 different numbers . if n is in the list and n is 5 times the average ( arithmetic mean ) of the other 10 numbers in the list , then n is what fraction of the sum of the 11 numbers in the list ? | "series : a 1 , a 2 . . . . a 10 , n sum of a 1 + a 2 + . . . + a 10 = 10 * x ( x = average ) so , n = 5 * x hence , a 1 + a 2 + . . + a 10 + n = 15 x so , the fraction asked = 5 x / 15 x = 1 / 3 answer is a" | a = 1 * 1
b = 10 / 5
c = b + 11
d = c / 5
e = d * 2
f = e - 5
g = f - 3
h = a / g
|
a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 630 , e ) none of the above | d | subtract(multiply(4000, power(add(const_1, divide(divide(10, const_2), const_100)), multiply(add(1, divide(1, 2)), const_2))), 4000) | compound interest of rs . 4000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | 10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 4000 [ 1 + ( 5 / 100 ) ] ^ 3 - 4000 = 4000 [ ( 21 / 20 ) ^ 3 - 1 ] = 4000 ( 9261 - 8000 ) / 8000 = 4 * 1261 / 8 = 630 answer : d | a = 10 / 2
b = a / 100
c = 1 + b
d = 1 / 2
e = 1 + d
f = e * 2
g = c ** f
h = 4000 * g
i = h - 4000
|
a ) 31 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | d | sqrt(add(power(16, const_2), add(power(24, const_2), power(18, const_2)))) | find the length of the longest pole that can be placed in an indoor stadium 24 m long , 18 m wide and 16 m high | "sqrt ( 1156 ) = 34 becoz max length is cuboid diagonal d = sqrt ( l ^ 2 + b ^ 2 + h ^ 2 ) where l lenght b breadth & h height answer : d" | a = 16 ** 2
b = 24 ** 2
c = 18 ** 2
d = b + c
e = a + d
f = math.sqrt(e)
|
a ) 6 miles , b ) 8,4 miles , c ) 9 miles , d ) 9,6 miles , e ) 12 miles | c | multiply(divide(13, add(3, 2)), 2) | one hour before john started walking from p to q , a distance of 13 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | "ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 13 - 3 = 10 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 10 miles so it will take them 10 / 5 = 2 hours to meet . in 2 hrs , ann would have covered 2 hrs * 3 miles per hour = 6 miles . adding this to the 3 miles she covered before john , ann covered a total of 3 + 6 = 9 miles . answer ( c )" | a = 3 + 2
b = 13 / a
c = b * 2
|
a ) 22000 , b ) 20000 , c ) 25123 , d ) 17272 , e ) none of these | d | divide(multiply(3800, const_100), 22) | sheela deposits rs . 3800 in bank savings account . if this is 22 % of her monthly income . what is her monthly income in ? | "explanation : 22 % of income = rs . 3800 100 % of income = 3800 x 100 / 22 = rs . 17272 answer : d" | a = 3800 * 100
b = a / 22
|
a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 9 , e ) 10 | e | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.5))), 1.5) | one pipe can fill a pool 1.5 times faster than a second pipe . when both pipes are opened , they fill the pool in four hours . how long would it take to fill the pool if only the slower pipe is used ? | "say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.5 r = 3 r / 2 . since when both pipes are opened , they fill the pool in four hours , then their combined rate is 1 / 4 pool / hour . thus we have that r + 3 r / 2 = 1 / 4 - - > r = 1 / 10 pool / hour - - > time is reciprocal of rate thus it ' s 10 / 1 = 10 hours . answer : e ." | a = 3 + 2
b = 1/(a)
c = 1 + 1
d = b / c
e = 1/(d)
f = e / 1
|
a ) 40 , b ) 50 , c ) 60 , d ) 110 , e ) 80 | d | subtract(multiply(60, const_2), 10) | 3 photographers , lisa , mike and norm , take photos of a wedding . the total of lisa and mikes photos is 60 less than the sum of mike ' s and norms . if norms photos number 10 more than twice the number of lisa ' s photos , then how many photos did norm take ? | l + m = m + n - 60 / n = 2 l + 10 60 = m + n - l - m 60 = n - l 60 = 2 l + 10 - l 50 = l 2 ( 50 ) + 10 = 110 d | a = 60 * 2
b = a - 10
|
a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph | b | subtract(39, divide(multiply(39, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 39 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 39 / 3 = 13 kmph answer : b" | a = 39 * 2
b = a / 3
c = 39 - b
|
a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.58 % | e | subtract(subtract(7, 6), divide(multiply(7, 6), const_100)) | in measuring the sides of a rectangle , one side is taken 7 % in excess , and the other 6 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 61 / 57 ) x ( 47 / 50 ) y = ( 867 / 862 ) ( xy ) error in measurement = ( 867 / 862 ) xy - xy = ( 5 / 862 ) xy error percentage = [ ( 5 / 862 ) xy ( 1 / xy ) 100 ] % = ( 29 / 50 ) % = 0.58 % . answer is e ." | a = 7 - 6
b = 7 * 6
c = b / 100
d = a - c
|
['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | a | subtract(floor(sqrt(2300)), floor(sqrt(2011))) | how many squares are there between 2011 to 2300 ? ? ? ? | nos . are 2025 , 2116 , 2209 answers is 3 answer : a | a = math.sqrt(2300)
b = math.floor(a)
c = math.sqrt(2011)
d = math.floor(c)
e = b - d
|
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