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 ) 10.5 , b ) 15 , c ) 25.5 , d ) 30 , e ) 23.5 | e | subtract(subtract(divide(const_60, const_2), 2), divide(const_60, divide(20, 1.5))) | darcy lives 1.5 miles from work . she can walk to work at a constant rate of 3 miles per hour , or she can ride the train to work at a constant rate of 20 miles per hour . if she rides the train , there is an additional x minutes spent walking to the nearest train station , waiting for the train , and walking from the final train station to her work . if it takes darcy a total of 2 more minutes to commute to work by walking than it takes her to commute to work by riding the train , what is the value of x ? | "the time it takes darcy to walk to work is ( 1.5 / 3 ) * 60 = 30 minutes the time it takes darcy to take the train is ( 1.5 / 20 ) * 60 + x = 4.5 + x minutes it takes 15 minutes longer to walk , so 30 = 4.5 + x + 2 x = 23.5 minutes answer : e" | a = const_60 / 2
b = a - 2
c = 20 / 1
d = const_60 / c
e = b - d
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a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40 | d | divide(multiply(subtract(1, divide(1, 4)), 12), subtract(1, subtract(1, divide(1, 4)))) | a driver would have reduced the time it took to drive from home to the store by 1 / 4 if the average speed had been increased by 12 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | "since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 4 would give us : d = ( r ) ( t ) = ( 3 t / 4 ) ( x * r ) x = 4 / 3 since ( 3 t / 4 ) ( 4 r / 3 ) = ( r ) ( t ) = d 4 r / 3 = r + 12 r / 3 = 12 r = 36 the answer is d ." | a = 1 / 4
b = 1 - a
c = b * 12
d = 1 / 4
e = 1 - d
f = 1 - e
g = c / f
|
a ) $ 92 , b ) $ 96 , c ) $ 156 , d ) $ 120 , e ) $ 240 | c | multiply(add(divide(0.80, const_2), multiply(multiply(subtract(1.10, 0.80), const_10), 0.04)), 300) | a certain company expects quarterly earnings of $ 0.80 per share of stock , half of which will be distributed as dividends to shareholders while the rest will be used for research and development . if earnings are greater than expected , shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings . if quarterly earnings are $ 1.10 per share , what will be the dividend paid to a person who owns 300 shares of the company ' s stock ? | "eps actual > eps expected . each gets and additional . 12 per share . thus . 52 * 300 - - > $ 156 answer is c" | a = 0 / 80
b = 1 - 10
c = b * 10
d = c * 0
e = a + d
f = e * 300
|
a ) 1006 , b ) 1230 , c ) 1010 , d ) 1200 , e ) 1300 | a | power(subtract(6, 1), const_4) | can you find the smallest non fractional number such that if the number gets divided by 6 , we get the remainder of 1 ; if the number gets divided by 7 , we get the remainder of 2 if the number gets divided by 8 , we get the remainder of 3 ; if the number gets divided by 9 , we get the remainder of 4 . | a . 1006 as we know lcm of the numbers 6 7 8 & 9 is 1008 so if i divide 1008 by any of the four number , it gives remainder as 0 . now to get reminder as desired , is should be 2 short of 1006 . ( 3 - 1,4 - 2,5 - 3,6 - 4 ) ans : option a | a = 6 - 1
b = a ** 4
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a ) 3 , b ) 0.3 , c ) 33 , d ) 3 / 1 , e ) 13 | b | divide(divide(divide(divide(divide(divide(3, const_4), const_3), const_4), const_3), const_2), const_2) | which is less than 3 ? | "0.3 less than 3 . answer : b" | a = 3 / 4
b = a / 3
c = b / 4
d = c / 3
e = d / 2
f = e / 2
|
a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) 269 | e | divide(add(1360, 15), subtract(6, const_1)) | the difference of two numbers is 1360 . 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 + 1360 ) . x + 1360 = 6 x + 15 5 x = 1345 x = 269 smaller number = 270 . answer e" | a = 1360 + 15
b = 6 - 1
c = a / b
|
a ) 36 , b ) 25 , c ) 72 , d ) 29 , e ) 22 | b | multiply(23, 2) | the average of 10 numbers is 23 . if each number is increased by 2 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 2 , the total increase = 2 * 10 = 20 the new sum = 230 + 20 = 250 the new average = 250 / 10 = 25 . answer : b" | a = 23 * 2
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a ) $ 0.15 , b ) $ 0.20 , c ) $ 0.33 , d ) $ 0.40 , e ) $ 0.45 | c | divide(multiply(add(const_1, const_1), 0.50), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.50 per glass on the first day , what was the price per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.5 = 3 * x - - > x = $ 0.33 . answer : c ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
|
a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | a | divide(subtract(const_100, add(35, 20)), const_3) | a polling company surveyed a certain country , and it found that 35 % of that country β s registered voters had an unfavorable impression of both of that state β s major political parties and that 20 % had a favorable impression only of party t . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country β s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "s = 100 not ( t and b ) = 35 only t = 20 ( t and b ) / b = 1 / 2 let ( t and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans" | a = 35 + 20
b = 100 - a
c = b / 3
|
a ) 399 , b ) 700 , c ) 288 , d ) 500 , e ) 211 | d | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 85 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 85 5 x / 100 + 90 β 6 x / 100 = 85 x / 100 = 5 = > x = 500 answer : d" | a = 5 + 6
b = a * 100
|
a ) 16 , b ) 17 , c ) 12 , d ) 15 , e ) 18 | e | subtract(const_60, multiply(const_60, divide(21, 30))) | excluding stoppages , the speed of a train is 30 kmph and including stoppages it is 21 kmph . of how many minutes does the train stop per hour ? | "t = 9 / 30 * 60 = 18 answer : e" | a = 21 / 30
b = const_60 * a
c = const_60 - b
|
a ) 1750 , b ) 4000 , c ) 3500 , d ) 2000 , e ) 1000 | d | divide(divide(multiply(1300, const_100), 13), 5) | a man took loan from a bank at the rate of 5 % p . a . s . i . after 13 years he had to pay rs . 1300 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 1300 ) / ( 5 * 13 ) = rs . 2000 answer : d" | a = 1300 * 100
b = a / 13
c = b / 5
|
a ) 70 , b ) 52 , c ) 55 , d ) 49 , e ) 50 | c | add(add(15, 12), add(13, 15)) | in a games hour 4 different types of players came to the ground ? cricket 15 , hokey 12 , football 13 , softball 15 . how many players are present in the ground ? | "total number of players = 15 + 12 + 13 + 15 = 55 answer is c" | a = 15 + 12
b = 13 + 15
c = a + b
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | e | divide(25, add(add(2, 2), 1)) | if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 25 ) which of these integers could be the value of z ? | solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( β which of these integers β¦ β ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 25 incorrect ( division by zero ) b . 2 / 1 = 2 / 2 + 2 / 26 incorrect c . 2 / 2 = 2 / 3 + 2 / 27 incorrect d . 2 / 3 = 2 / 4 + 2 / 28 incorrect e . 2 / 4 = 2 / 5 + 2 / 30 correct the correct answer is e , because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case | a = 2 + 2
b = a + 1
c = 25 / b
|
a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | e | subtract(multiply(45, 14.66), subtract(multiply(45, 14), 15)) | the average student age of a certain class which has 45 students is 14 . if one student aged 15 go to another class and the age of the class teacher is included the average changes to 14.66 . what is the age of class teacher ? | sum of ages of class before replacement = 45 x 14 = 630 sum of ages of class without teacher = 630 - 15 = 615 sum of ages of class after replacement = 45 x 14.66 = 660 age of teacher = 660 - 615 = 45 answer : e | a = 45 * 14
b = 45 * 14
c = b - 15
d = a - c
|
a ) 797 , b ) 697 , c ) 687 , d ) 699 , e ) 799 | b | divide(multiply(41, 801), 150.17) | 41 % of 801 β 150.17 = ? β 57 % of 910 | "explanation : 328 β 150 = 178 x β 519 = 178 x = 697 answer : option b" | a = 41 * 801
b = a / 150
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a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | e | multiply(2, 33) | the area of a square garden is q square feet and the perimeter is p feet . if q = 2 p + 33 , what is the perimeter of the garden in feet ? | "let x be the length of one side of the square garden . x ^ 2 = 8 x + 33 x ^ 2 - 8 x - 33 = 0 ( x - 11 ) ( x + 3 ) = 0 x = 11 , - 3 p = 4 ( 11 ) = 44 the answer is e ." | a = 2 * 33
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a ) 35.2 , b ) 36.1 , c ) 36.5 , d ) 41.5 , e ) none | d | divide(add(multiply(41, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 41 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 41 Γ 50 + 48 β 23 ) = 2075 . therefore correct mean = 2075 / 50 = 41.5 . answer d" | a = 41 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 15 , b ) 88 , c ) 44 , d ) 54 , e ) 23 | b | add(subtract(87, multiply(17, 3)), 3) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning ? | "let the average after 7 th inning = x then average after 16 th inning = x - 3 16 ( x - 3 ) + 87 = 17 x x = 87 - 48 = 39 answer = b" | a = 17 * 3
b = 87 - a
c = b + 3
|
a ) 5 seconds , b ) 12 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these | b | divide(300, multiply(90, const_0_2778)) | in what time will a train 300 meters long cross an electric pole , if its speed is 90 km / hr | "explanation : first convert speed into m / sec speed = 90 * ( 5 / 18 ) = 25 m / sec time = distance / speed = 300 / 25 = 12 seconds answer : b" | a = 90 * const_0_2778
b = 300 / a
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a ) $ 900 , b ) $ 891 , c ) $ 915 , d ) $ 715 , e ) $ 795 | b | multiply(1100, power(subtract(const_1, divide(10, const_100)), 2)) | a present value of a machine is $ 1100 . its value depletion rate is 10 % per annum then find the machine value after 2 years ? | p = $ 1100 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 1100 * 9 / 10 * 9 / 10 = $ 891 answer is b | a = 10 / 100
b = 1 - a
c = b ** 2
d = 1100 * c
|
a ) 9 , b ) 4 , c ) 5 , d ) 2 , e ) 3 | c | subtract(17, 12) | vanessa , mike and sarah have missed a total of 17 days of school this year . vanessa and mike have missed 14 days total , and mike and sarah have missed 12 days total . how many days of school did vanessa miss alone ? | vanessa + mike + sarah = 17 v + m = 14 m + s = 12 < < use this formula & plug into the original formula , to get vanessa ' s total v + ( m + s ) = 17 v + 12 = 17 subtract 12 from both sides v = 5 answer : c | a = 17 - 12
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a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | c | multiply(divide(7, subtract(9, 7)), 16) | sandy is younger than molly by 16 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | "let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 16 x = 8 sandy is 56 years old . the answer is c ." | a = 9 - 7
b = 7 / a
c = b * 16
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a ) 99 , b ) 147 , c ) 363 , d ) 1200 , e ) 1625 | b | add(add(add(add(add(multiply(multiply(7, 2), 2), multiply(multiply(7, 2), 2)), multiply(multiply(7, 2), 2)), 2), const_4), const_4) | if both 7 ^ 2 and 3 ^ 3 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 * 7 ^ 2 = 147 the answer is b ." | a = 7 * 2
b = a * 2
c = 7 * 2
d = c * 2
e = b + d
f = 7 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 221 / 250 , b ) 199 / 250 , c ) 33 / 50 , d ) 8 / 25 , e ) 209 / 260 | e | divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(850, subtract(const_1, divide(40, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25))) | in a village of 2,600 people , 800 people are over 70 years old and 850 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a male or younger than 70 years old ? | the number of people younger than 70 years old is 2600 - 800 = 1800 the number of females older than 70 years old is 0.6 * 850 = 510 then the number of males older than 70 years old is 800 - 510 = 290 . the number of people who are either male or younger than 70 is 1800 + 290 = 2090 . p ( a person is younger than 70 or male ) = 2090 / 2600 = 209 / 260 the answer is e . | a = 5 * 5
b = a / const_0_25
c = const_26 * b
d = 5 * 5
e = d / const_0_25
f = 40 / e
g = 1 - f
h = 850 * g
i = c - h
j = 5 * 5
k = j / const_0_25
l = const_26 * k
m = i / l
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a ) 18 Β° , b ) 36 Β° , c ) 54 Β° , d ) 72 Β° , e ) 90 Β° | d | divide(multiply(const_360, subtract(const_100, add(add(add(add(60, 9), 5), 4), 2))), const_100) | a circle graph shows how the budget of a certain company was spent : 60 percent for salaries , 9 percent for research and development , 5 percent for utilities , 4 percent for equipment , 2 percent for supplies , and the remainder for transportation . if the area of each sector of the graph is proportional to the percent of the budget it represents , how many degrees of the circle are used to represent transportation ? | "the percent of the budget for transportation is 100 - ( 60 + 9 + 5 + 4 + 2 ) = 20 % 100 % of the circle is 360 degrees . then ( 20 % / 100 % ) * 360 = 72 degrees the answer is d ." | a = 60 + 9
b = a + 5
c = b + 4
d = c + 2
e = 100 - d
f = 360 * e
g = f / 100
|
a ) 7 , b ) 8.2 , c ) 2 , d ) 10.8 , e ) 11.2 | d | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | two trains 140 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "answer : option d relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 160 = 300 m . required time = 300 * 9 / 250 = 54 / 5 = 10.8 sec" | a = 140 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | inverse(subtract(inverse(4), inverse(6))) | working alone , a can complete a certain kind of job in 6 hours . a and d , working together at their respective rates , can complete one of these jobs in 4 hours . in how many hours can d , working alone , complete one of these jobs ? | "let total time taken by d to complete the job = d total time taken by a to complete the job = 6 work done by a in an hour 1 / a = 1 / 6 working together a and d can complete the job in 4 hours 1 / a + 1 / d = 1 / 4 = > 1 / d = 1 / 4 - 1 / a = 1 / 4 - 1 / 6 = 1 / 12 = > d = 12 hours answer c" | a = 1/(4)
b = 1/(6)
c = a - b
d = 1/(c)
|
a ) 50 , b ) 74 , c ) 78 , d ) 80 , e ) none of these | a | subtract(multiply(70, 6), multiply(74, 5)) | ashok secured average of 70 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 70 therefore total marks in 6 subjects = 70 * 6 = 420 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 420 β 370 = 50 answer a" | a = 70 * 6
b = 74 * 5
c = a - b
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a ) 26630 , b ) 26640 , c ) 36623 , d ) 30613 , e ) 26820 | d | multiply(23000, add(const_1, divide(multiply(3, 10), const_100))) | population is 23000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 23000 * 10 / 100 = 2300 = = = > 23000 + 2000 = 25300 population after 2 nd year = 25300 * 10 / 100 = 2530 = = = > 25300 + 2530 = 27830 population after 3 rd year = 27830 * 10 / 100 = 2783 = = = > 27830 + 2783 = 30613 answer : d" | a = 3 * 10
b = a / 100
c = 1 + b
d = 23000 * c
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a ) 150 meter , b ) 299 meter , c ) 666 meter , d ) 155 meter , e ) 180 meter | e | multiply(divide(multiply(72, const_1000), const_3600), 9) | a train running at the speed of 72 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 72 * ( 5 / 18 ) m / sec = 20 m / sec length of train ( distance ) = speed * time ( 20 ) * 9 = 180 meter answer : e" | a = 72 * 1000
b = a / 3600
c = b * 9
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a ) 45 , 63 , b ) 50 , 70 , c ) 55 , 77 , d ) 60 , 84 , e ) 65 , 91 | c | subtract(multiply(multiply(subtract(multiply(2, const_4.0), multiply(1, 33)), const_2.0), divide(7, 5)), multiply(subtract(multiply(2, 33), multiply(1, 33)), 5)) | two numbers are in the ratio of 5 : 7 . if 33 is subtracted from each number , they are in the ratio of 1 : 2 . what are the two numbers ? | "( 5 x - 33 ) : ( 7 x - 33 ) = 1 y : 2 y 5 x - 33 = y 7 x - 33 = 2 y 231 - 165 = 10 y - 7 y 66 = 3 y y = 22 5 x - 33 = 22 x = 11 the two numbers are 5 x and 7 x which are 55 and 77 . the answer is c ." | a = 2 * 4
b = 1 * 33
c = a - b
d = c * 2
e = 7 / 5
f = d * e
g = 2 * 33
h = 1 * 33
i = g - h
j = i * 5
k = f - j
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a ) 24 , b ) 12 , c ) 64 , d ) 48 , e ) 25 | d | subtract(add(multiply(14, const_2), 5), add(5, 12)) | x and y are positive integers . when x is divided by 12 , the remainder is 5 , and when x is divided by 13 , the remainder is 3 . when y is divided by 14 , the remainder is 5 , and when y is divided by 7 , the remainder is 5 . what is the least possible value of x + y ? | "when x is divided by 12 , the remainder is 5 : so , the possible values of x are : 5 , 17 , 29 , 41 , etc . when x is divided by 13 , the remainder is 3 : so , the possible values of x are : 3 , 16,29 . . . stop . since both lists include 29 , the smallest possible value of x is 29 . when y is divided by 14 , the remainder is 5 : so , the possible values of y are : 5 , 19 , 33,47 etc . when y is divided by 7 , the remainder is 5 : so , the possible values of y are : 5 , 19 . . . stop . since both lists include 19 , the smallest possible value of y is 19 since the smallest possible values of x and y are 29 and 19 respectively , the smallest possible value of x + y is 48 . so , d is the correct answer to the original question ." | a = 14 * 2
b = a + 5
c = 5 + 12
d = b - c
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a ) 530 , b ) 305 , c ) 503 , d ) 430 , e ) 350 | e | multiply(divide(multiply(180, const_1000), const_3600), 7) | a train running at the speed of 180 km / hr crosses a pole in 7 seconds . what is the length of the train ? | "speed = ( 180 * 5 / 18 ) m / sec = ( 50 ) m / sec length of the train = ( speed x time ) = ( 50 * 7 ) m = 350 m . answer : e" | a = 180 * 1000
b = a / 3600
c = b * 7
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a ) 12 , b ) 17 , c ) 18 , d ) 19 , e ) 14 | a | sqrt(divide(432, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 432 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 432 3 b 2 = 432 b 2 = 144 b = 12 m . answer : option a" | a = 432 / 3
b = math.sqrt(a)
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a ) 10 and 3 , b ) 9 and 6 , c ) 10 and 7 , d ) 3 and 10 , e ) 10 and 10 | b | subtract(add(divide(16, 2), 1), 3) | one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 16 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x β 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x β 3 + 1 ) = ( x β 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) β
( x β 2 ) cm 2 , which is 16 cm 2 more than the first area . therefore a 1 + 16 = a 2 x ( x β 3 ) + 16 = ( x + 1 ) ( x β 2 ) x 2 β 3 x + 16 = x 2 + x β 2 x β 2 2 x = 18 x = 9 . so , the sides of the rectangle are 9 cm and ( 9 β 3 ) = 6 cm long . so answer is b ." | a = 16 / 2
b = a + 1
c = b - 3
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a ) rs . 2.50 , b ) rs . 3 , c ) rs . 3.75 , d ) rs . 4 , e ) rs . 5 | b | subtract(subtract(multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 1200), 1200), divide(multiply(1200, 10), const_100)) | the difference between simple interest and compound interest on rs . 1200 for one year at 10 % per annum reckoned half - yearly is : | s . i = ( 1000 * 10 * 4 ) / 100 = rs . 400 c . i = 1200 * 1 + 5 / 100 ) 2 β 1200 = 123 . difference = rs . ( 123 - 120 ) = rs . 3 answer : b | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 1200
f = e - 1200
g = 1200 * 10
h = g / 100
i = f - h
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['a ) $ 980', 'b ) $ 1060', 'c ) $ 1220', 'd ) $ 1440', 'e ) $ 1650'] | d | multiply(surface_rectangular_prism(3, 6, 2), 20) | a rectangular tank needs to be coated with insulation . the tank has dimensions of 3 feet , 6 feet , and 2 feet . each square foot of insulation costs $ 20 . how much will it cost to cover the surface of the tank with insulation ? | the total surface area is 2 ( 2 * 3 + 3 * 6 + 2 * 6 ) = 72 square feet the total cost is 72 * $ 20 = $ 1440 the answer is d . | a = surface_rectangular_prism * (
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a ) 1.69 % , b ) 1.74 % , c ) 1.84 % , d ) 1.47 % , e ) 1.24 % | a | divide(multiply(13, 13), const_100) | if a trader sold two cars each at rs . 325475 and gains 13 % on the first and loses 13 % on the second , then his profit or loss percent on the whole is ? | "sp of each car is rs . 325475 , he gains 13 % on first car and losses 13 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 13 ) ( 13 ) / 100 % = 1.69 % answer : a" | a = 13 * 13
b = a / 100
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a ) rs . 3000 , b ) rs . 3600 , c ) rs . 2000 , d ) rs . 4000 , e ) none of these | c | multiply(divide(5000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | p , q and r have rs . 5000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 5000 - r ) = > 3 r = 10000 - 2 r = > 5 r = 10000 = > r = 2000 . answer : c" | a = 2 / 3
b = 1 + a
c = 5000 / b
d = 2 / 3
e = c * d
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a ) $ 440 , b ) $ 450 , c ) $ 460 , d ) $ 470 , e ) $ 480 | c | divide(add(add(add(add(250, 400), 750), 400), 500), 5) | a cab driver 5 days income was $ 250 , $ 400 , $ 750 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 250 + 400 + 750 + 400 + 500 ) / 5 = 460 answer is c" | a = 250 + 400
b = a + 750
c = b + 400
d = c + 500
e = d / 5
|
a ) a ) 3623216 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3621409 , e ) e ) 3625216 | d | multiply(divide(1903, 1903), const_100) | 1903 x 1903 = ? | "1903 x 1903 = ( 1903 ) 2 = ( 1900 + 3 ) 2 = ( 1900 ) 2 + ( 3 ) 2 + ( 2 x 1900 x 3 ) = 3610000 + 9 + 11400 . = 3621409 . d )" | a = 1903 / 1903
b = a * 100
|
a ) 12 days , b ) 15 days , c ) 16 days , d ) 24 days , e ) 18 days | e | inverse(subtract(inverse(3), add(inverse(6), inverse(9)))) | a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 9 days , how long will a woman take to complete the work ? | explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 3 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 9 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 9 = 5 / 18 therefore , 1 woman β s 1 day β s work = 1 / 3 β 5 / 18 = 1 / 18 therefore , the woman alone can finish the work in 18 days . answer : option e | a = 1/(3)
b = 1/(6)
c = 1/(9)
d = b + c
e = a - d
f = 1/(e)
|
a ) s . 4991 , b ) s . 5991 , c ) s . 6001 , d ) s . 6191 , e ) s . 7991 | d | subtract(multiply(add(5, const_1), 6700), add(add(add(add(6435, 6927), 6855), 7230), 6562)) | a grocer has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6700 ? | "total fr 5 mnths = ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs 34009 . reqd . sale = rs . [ ( 6700 * 6 ) - 34009 ] = rs . ( 40200 - 34009 ) = rs . 6191 . answer : d" | a = 5 + 1
b = a * 6700
c = 6435 + 6927
d = c + 6855
e = d + 7230
f = e + 6562
g = b - f
|
a ) 25 , b ) 30 , c ) 32 , d ) 35 , e ) 40 | c | multiply(add(3, const_1), add(7, const_1)) | if p and q are prime numbers , how many divisors does the product p ^ 3 * q ^ 7 have ? | "when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is c . 4 * 8 = 32" | a = 3 + 1
b = 7 + 1
c = a * b
|
a ) 4000 , b ) 24000 , c ) 8000 , d ) 10000 , e ) none of these | b | divide(multiply(1200, const_100), 15) | a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 15 % on conveyance . if his savings at the end of the month is rs . 1200 , then his salary per month in rupees is : | "total expenditure = 40 + 25 + 15 + 15 = 95 % saving = ( 100 - 95 ) = 5 % 5 / 100 Γ salary = 1200 , salary = 24000 rs . answer : b" | a = 1200 * 100
b = a / 15
|
a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | e | subtract(multiply(multiply(45, const_0_2778), 30), 135) | a train , 135 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 135 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 135 + x ) / 30 m / s = > 45 Γ£ β ( 10 / 36 ) = ( 135 + x ) / 30 = > 45 Γ£ β 10 Γ£ β 30 / 36 = 135 + x = > 45 Γ£ β 10 Γ£ β 10 / 12 = 135 + x = > 15 Γ£ β 10 Γ£ β 10 / 4 = 135 + x = > 15 Γ£ β 25 = 135 + x = 375 = > x = 375 - 135 = 240 answer : option e" | a = 45 * const_0_2778
b = a * 30
c = b - 135
|
a ) 9 : 16 , b ) 5 : 9 , c ) 12 : 17 , d ) 13 : 24 , e ) none of these | a | divide(add(multiply(1, const_60), 30), add(multiply(2, const_60), 40)) | find the ratio between 1 hr 30 minutes and 2 hrs 40 minutes . | solution : the ratio = 1 hr 30 min : 2 hr 40 min . 90 min : 160 min = 9 : 16 . answer is a | a = 1 * const_60
b = a + 30
c = 2 * const_60
d = c + 40
e = b / d
|
a ) 39 , b ) 35 , c ) 42 , d ) 40.5 , e ) 41 | a | add(subtract(87, multiply(17, 3)), 3) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "explanation : let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 87 = total runs scored in 17 innings = > 16 ( x - 3 ) + 87 = 17 x = > 16 x - 48 + 87 = 17 x = > x = 39 answer : option a" | a = 17 * 3
b = 87 - a
c = b + 3
|
a ) 24 , b ) 27 , c ) 29 , d ) 31 , e ) 33 | c | subtract(add(add(30, 29), 10), 40) | mrs . evans gave a test to her freshmen economics class , which has 40 students enrolled and 30 of them answered question 1 correctly . if 29 answered question 2 correctly and 10 did not take the test then how many answered both questions correctly ? | "total number of enrolled students = 40 number of students who did not take test = 10 hence , number of students who took test = 40 - 10 = 30 number of students who answered q 2 correctly = 29 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 29 . number of students who answered both q 1 & q 2 correctly = 29 . answer : c" | a = 30 + 29
b = a + 10
c = b - 40
|
a ) $ 16.7 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 40 | a | multiply(divide(multiply(1.25, 8), subtract(8, 5)), 5) | nina has exactly enough money to purchase 5 widgets . if the cost of each widget were reduced by $ 1.25 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "a its is . let price = x ( x - 1.25 ) 8 = 5 x x = 10 / 3 hence total money = 5 * 10 / 3 = 16.7" | a = 1 * 25
b = 8 - 5
c = a / b
d = c * 5
|
a ) 11 , b ) 13 , c ) 15 , d ) 18 , e ) 10 | c | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 9000), const_100) | if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of rs . 9000 ? | "required difference = [ 3 Β½ % of rs . 9000 ] β [ 3 1 / 3 % of rs . 9000 ] = [ ( 7 / 20 - ( 10 / 3 ) ] % of rs . 9000 = 1 / 6 % of rs . 9000 = rs . [ ( 1 / 6 ) 8 ( 1 / 100 ) * 9000 ] = rs . 15 . answer is c ." | a = 1 / 2
b = a + 3
c = 1 / 3
d = c + 3
e = b - d
f = e * 9000
g = f / 100
|
a ) 277 sec , b ) 285 sec , c ) 267 sec , d ) 167 sec , e ) 276 sec | b | subtract(divide(multiply(const_1, const_1000), divide(40, 15)), 15) | in a kilometer race , a beats b by 40 meters or 15 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 15 ) / 50 = 300 sec . time taken by a = 300 - 15 = 285 sec . answer : b" | a = 1 * 1000
b = 40 / 15
c = a / b
d = c - 15
|
a ) 1.86 , b ) 2.86 , c ) 4.86 , d ) 3.86 , e ) 5.86 | b | multiply(const_12, divide(multiply(40, divide(40, const_100)), 67)) | a reduction of 40 % in the price of bananas would enable a man to obtain 67 more for rs . 40 , what is reduced price per dozen ? | "40 * ( 40 / 100 ) = 16 - - - 67 ? - - - 12 = > rs . 2.86 answer : b" | a = 40 / 100
b = 40 * a
c = b / 67
d = 12 * c
|
a ) 1 / 9 , b ) 1 / 6 , c ) 2 / 3 , d ) 4 / 9 , e ) 7 / 11 | e | divide(7, add(7, 4)) | a waitress ' s income consists of her salary and tips . during one week , her tips were 7 / 4 of her salary . what fraction of her income for the week came from tips ? | "her tips were 7 / 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 7 / 4 ) ( $ 4 ) = $ 7 so , her total income = $ 4 + $ 7 = $ 11 what fraction of her income for the week came from tips $ 7 / $ 11 = 7 / 11 = e" | a = 7 + 4
b = 7 / a
|
a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | a | divide(multiply(divide(1.5, 1.5), 3), 3) | if one cat eats 1.5 rats in 1.5 days , then in howmany days 3 cats will eat 3 rats ? ? | 1 cat - - - - - - 3 / 2 rat - - - - - - - - - 3 / 2 days 1 cat - - - - - - ( 3 / 2 ) * 2 rat - - - - - ( 3 / 2 ) * 2 days ( rat doubled , days doubled ) 1 * 3 cat - - - - - 3 rat - - - - - - - - - - - 3 / 3 = 1 day ( cat * 3 , same rat , days / 3 ) answer : a | a = 1 / 5
b = a * 3
c = b / 3
|
a ) 09 , b ) 07 , c ) 12 , d ) 05 , e ) 03 | c | add(divide(add(465, 75), add(60, 75)), 8) | the distance between two cities a and b is 465 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 465 therefore , 60 x + 75 ( x - 1 ) = 465 = > x = 3 . so , they meet at ( 8 + 4 ) i . e , 12 noon answer : c )" | a = 465 + 75
b = 60 + 75
c = a / b
d = c + 8
|
a ) 81 , b ) 77 , c ) 54 , d ) 33 , e ) 99 | a | subtract(multiply(10, multiply(125, const_0_2778)), 100) | a train 100 m long crosses a platform 125 m long in 10 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 10 s = 225 / 10 * 18 / 5 = 81 kmph answer : a" | a = 125 * const_0_2778
b = 10 * a
c = b - 100
|
a ) 911 , b ) 969 , c ) 979 , d ) 977 , e ) 971 | c | multiply(add(const_100, const_2), 89) | calculate the largest 3 digit number which is exactly divisible by 89 ? | "largest 4 digit number is 999 after doing 999 Γ· 89 we get remainder 20 hence largest 3 digit number exactly divisible by 89 = 999 - 20 = 979 c" | a = 100 + 2
b = a * 89
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | floor(divide(83, 9)) | on dividing 83 by a number , the quotient is 9 and the remainder is 2 . find the divisor ? | "d = ( d - r ) / q = ( 83 - 2 ) / 9 = 81 / 9 = 9 c )" | a = 83 / 9
b = math.floor(a)
|
a ) s . 1300 , b ) s . 1005 , c ) s . 1031 , d ) s 1331 , e ) s . 1120 | d | multiply(1000, power(add(const_1, divide(10, const_100)), 3)) | albert invested rs . 1000 in a scheme for 3 years at compound interest rate 10 % p . a . how much amount will albert get on maturity of the fixed deposit ? | "amount = [ 1000 * ( 1 + 10 / 100 ) 3 ] = 1000 * 11 / 10 * 11 / 10 * 11 / 10 = rs . 1331 answer : d" | a = 10 / 100
b = 1 + a
c = b ** 3
d = 1000 * c
|
a ) 66.66 % , b ) 66.68 % , c ) 53.84 % , d ) 86.66 % , e ) 66.65 % | c | multiply(divide(35, subtract(const_100, 35)), const_100) | if a ' s height is 35 % less than that of b , how much percent b ' s height is more than that of a ? | "excess of b ' s height over a ' s = [ ( 35 / ( 100 - 35 ) ] x 100 % = 53.84 % answer : c )" | a = 100 - 35
b = 35 / a
c = b * 100
|
a ) 62.5 % , b ) 63 % , c ) 63.5 % , d ) 64.5 % , e ) 65 % | a | divide(multiply(5, 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 5 / 8 as a percent | "5 / 8 as a percent = ( 5 / 8 ) x 100 = 0.625 x 100 = 62.5 % answer : a i . e . 62.5 %" | 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 = 5 * n
p = o / 100
|
a ) 57 , b ) 67 , c ) 77 , d ) 88 , e ) 97 | d | sqrt(multiply(77.44, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 77.44 , the number of the member is the group is : | "money collected = ( 77.44 x 100 ) paise = 7744 paise . number of members = square root of 7744 = 88 . answer : option d" | a = 77 * 44
b = math.sqrt(a)
|
a ) 288 , b ) 278 , c ) 267 , d ) 285 , e ) 231 | d | divide(add(multiply(add(const_4, const_1), 510), multiply(multiply(add(const_4, const_1), add(const_4, const_1)), 240)), 30) | a library has an average of 510 visitors on sunday and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "since the month begins with a sunday , so there will be five sundays in the month . required average = [ ( 510 * 5 ) + ( 240 * 25 ) ] / 30 = 8550 / 30 = 285 . answer : d" | a = 4 + 1
b = a * 510
c = 4 + 1
d = 4 + 1
e = c * d
f = e * 240
g = b + f
h = g / 30
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 3 / 5 , e ) 4 / 5 | c | divide(20, add(50, 10)) | some of the 50 % solution of acid was removed and this was replaced with an equal amount of 10 % solution of acid . as a result , a 20 % solution of acid was obtained . what fraction of the original solution was replaced ? | "let x be the fraction of the original solution that was replaced . 0.5 * ( 1 - x ) + 0.1 ( x ) = 0.2 0.4 x = 0.3 x = 3 / 4 the answer is c ." | a = 50 + 10
b = 20 / a
|
a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | c | multiply(divide(add(multiply(30, divide(20, const_100)), multiply(50, divide(12, const_100))), add(30, 50)), const_100) | in a park there are two ponds with both brown ducks and green ducks . in the smaller pond there are 30 ducks and in the larger pond there are 50 ducks . if 20 % of the ducks in the smaller pond are green and 12 % of the ducks in the larger pond are green , then what percentage of ducks are green ? | number of ducks in small pond = 30 green ducks in small pond = 20 % of 30 = 6 ducks number of ducks in large pond = 50 green ducks in large pond = 12 % of 50 = 6 ducks total number of ducks = 30 + 50 = 80 total number of green ducks = 6 + 6 = 12 ducks percentage of green ducks = 12 / 80 * 100 = 15 % answer : c | a = 20 / 100
b = 30 * a
c = 12 / 100
d = 50 * c
e = b + d
f = 30 + 50
g = e / f
h = g * 100
|
a ) 302 , b ) 304 , c ) 306 , d ) 308 , e ) 310 | a | multiply(17, 289) | the h . c . f of two numbers is 17 and their l . c . m is 5134 . if one of the numbers is 289 , then the other is ? | "other number = ( 17 * 5134 ) / 289 = 302 . answer : a" | a = 17 * 289
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a ) 120 , b ) 140 , c ) 160 , d ) 180 , e ) 200 | b | add(add(35, divide(multiply(35, 15), subtract(const_100, add(60, 15)))), divide(multiply(35, 60), subtract(const_100, add(60, 15)))) | a fruit drink is made of orange , watermelon , and grape juice , where 15 percent of the drink is orange juice and 60 percent is watermelon juice . if the drink is made with 35 ounces of grape juice , how many ounces is the drink total ? | "let the total number of ounces in the drink be x . % of orange = 15 % % of watermelon = 60 % % of grape = 100 % - 75 % = 25 % 0.25 x = 35 x = 140 therefore there are a total of 140 ounces in the drink . the answer is b ." | a = 35 * 15
b = 60 + 15
c = 100 - b
d = a / c
e = 35 + d
f = 35 * 60
g = 60 + 15
h = 100 - g
i = f / h
j = e + i
|
['a ) s . 4529', 'b ) s . 4586', 'c ) s . 7248', 'd ) s . 4530', 'e ) s . 4528'] | c | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, const_3)), multiply(3, multiply(4, 3))), 8) | the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 8 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 8 = rs . 7248 answer : c | a = 2 * 12
b = 15 + 25
c = a * b
d = 6 * 3
e = c - d
f = 4 * 3
g = 3 * f
h = e - g
i = h * 8
|
a ) 24,607 , b ) 24,609 , c ) 24,600 , d ) 24,602 , e ) 24,604 | c | divide(61.50, divide(const_4, 4)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 61.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 61.50 32 x - 31 x = 6150 * 4 x = 24,600 . answer : c" | a = 4 / 4
b = 61 / 50
|
a ) 290 , b ) 280 , c ) 270 , d ) 260 , e ) 250 | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 210) | a 210 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 210 ) / 9 = 500 / 9 = > x = 290 . answer : option a" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 210
|
a ) 16 min , b ) 26 min , c ) 34 min , d ) 24 min , e ) 12 min | d | multiply(divide(6, divide(5, 4)), 5) | walking with 4 / 5 of my usual speed , i miss the bus by 6 minutes . what is my usual time ? | "speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 5 1 - - - - - - - - 6 4 - - - - - - - - - ? Γ¨ 24 answer : d" | a = 5 / 4
b = 6 / a
c = b * 5
|
a ) 47 : 11 , b ) 37 : 21 , c ) 27 : 21 , d ) 47 : 21 , e ) 47 : 31 | d | divide(add(multiply(subtract(multiply(const_2, add(8, 5)), add(8, 5)), const_3), 8), add(subtract(multiply(const_2, add(8, 5)), add(8, 5)), 8)) | the age of father 8 years ago was thrice the age of his son . 5 years hence , father ' s age will be twice that of his son . the ratio of their present ages is : | let the ages of father and son 8 years ago be 3 x and x years respectively . then , ( 3 x + 8 ) + 5 = 2 [ ( x + 8 ) + 5 ] 3 x + 13 = 2 x + 26 x = 13 . required ratio = ( 3 x + 8 ) : ( x + 8 ) = 47 : 21 . answer : option d | a = 8 + 5
b = 2 * a
c = 8 + 5
d = b - c
e = d * 3
f = e + 8
g = 8 + 5
h = 2 * g
i = 8 + 5
j = h - i
k = j + 8
l = f / k
|
a ) 26 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | a | divide(subtract(multiply(69, 2), subtract(16, 8)), add(2, 3)) | if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 69 , what is the value of a ? | ( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 69 a = 26 the answer is a . | a = 69 * 2
b = 16 - 8
c = a - b
d = 2 + 3
e = c / d
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(43, 43)) | what is the remainder when 43 ^ 89 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 89 has the form 4 k + 1 so the units digit of 43 ^ 89 is 3 . the remainder when dividing by 5 is 3 . the answer is d ." | a = 5 / 2
b = 43 * 43
c = a - b
|
a ) 9989 , b ) 9896 , c ) 9892 , d ) 9809 , e ) 9810 | c | multiply(power(99, divide(91, 99)), 99) | the value of 99 ^ ( 91 / 99 ) x 99 is : | ( 100 Γ’ β¬ β 8 / 99 ) x 99 = 9900 Γ’ β¬ β 8 = 9892 . answer : c | a = 91 / 99
b = 99 ** a
c = b * 99
|
a ) 5 , b ) 5 / 4 , c ) 4 / 5 , d ) 1 / 4 , e ) 1 / 5 | b | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 3 * / 5 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . based on this definition . . . . if . . . . k = 2 k * = ( 1 / 1 ) ( 1 / 2 ) = 1 / 2 if . . . . k = 3 k * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) = 1 / 6 we ' re asked to find the value of 5 * / 4 * now that we know how the symbol ' works ' , solving this problem should n ' t be too difficult . you can actually choose to do the math in a couple of different ways . . . . 5 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) do n ' t calculate this just yet though . . . . since we ' re dividing by 4 * , many of those fractions will ' cancel out . ' 4 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) we ' re looking for the value of : ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) / ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) since the first four fraction in the numerator and denominator cancel out , we ' re left with just one fraction : 5 / 3 b" | a = 1 / 3
b = a / 3
c = 1 + 4
d = b / c
|
a ) 701 , b ) 120 , c ) 42 , d ) 868 , e ) 1 | c | subtract(multiply(223, add(const_2, const_4)), add(684, add(2, add(507, add(55, 48))))) | if the average of 55 , 48 , 507 , 2 , 684 and y is 223 , then what is y ? | y = 42 , because : ( 55 + 48 + 507 + 2 + 684 + y ) / 6 = 223 ( 55 + 48 + 507 + 2 + 684 + y ) / 6 * 6 = 223 * 6 ( 55 + 48 + 507 + 2 + 684 + y ) - 1296 = 1338 - 1296 y = 42 so the correct answer is c , 42 . | a = 2 + 4
b = 223 * a
c = 55 + 48
d = 507 + c
e = 2 + d
f = 684 + e
g = b - f
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | d | divide(subtract(add(5, 5), 5), 2) | in the xy - coordinate system , if ( m , n ) and ( m + 5 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 5 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k - - - > 5 = 2 k - - - > k = 5 / 2 . the answer is , therefore , ( d ) ." | a = 5 + 5
b = a - 5
c = b / 2
|
a ) 6 / Ο , b ) 9 / Ο , c ) 6 , d ) 9 , e ) 12 | d | add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi))) | the surface area of a sphere is 4 Ο r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area r of that hemisphere ? | "given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 / pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 / pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 / 2 = 6 ( curved part , not including the flat rounded base ) . but the total surface area = 6 + area of the base of a hemisphere . = 6 + 3 = 9 . answer is d ! !" | a = 4 * math.pi
b = 3 / math.pi
c = a * b
d = c / 2
e = 3 / math.pi
f = math.pi * e
g = d + f
|
a ) 4500 , b ) 4600 , c ) 4300 , d ) 4900 , e ) 5000 | d | add(4970, const_1) | the sum of the first 70 positive even integers is 4970 . what is the sum of the first 70 odd integers ? | sum of first n even numbers = n ( n + 1 ) = 4970 sum of first n odd numbers = n ^ 2 = 70 * 70 = 4900 ( here n = 70 ) answer : d | a = 4970 + 1
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a ) 150 m , b ) 180 m , c ) 159 m , d ) 250 m , e ) 50 m | e | subtract(multiply(250, divide(10, divide(10, const_3))), multiply(150, divide(15, divide(10, const_3)))) | a train crosses a platform of 150 m in 10 sec , same train crosses another platform of length 250 m in 15 sec . then find the length of the train ? | "length of the train be β x β x + 150 / 10 = x + 250 / 15 3 x + 450 = 2 x + 500 x = 50 m answer : e" | a = 10 / 3
b = 10 / a
c = 250 * b
d = 10 / 3
e = 15 / d
f = 150 * e
g = c - f
|
a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 40 | d | add(divide(add(subtract(subtract(5, const_1), const_1), add(50, subtract(5, const_1))), const_3), const_3) | the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded down to the nearest integer ) . during its 5 th year , the plum tree grew 50 plums . if this trend continues , how many plums will it grow during its 7 th year ? | "1 st year : 0 - 1 ( age ) , we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear : 1 - 2 ( age ) , we take age = 1 3 rd year : 2 - 3 ( age ) , we take age = 2 4 th year : 3 - 4 ( age ) , we take age = 3 5 th year : 4 - 5 ( age ) , we take age = 4 6 th year : 5 - 6 ( age ) , we take age = 5 7 th year : 6 - 7 ( age ) , we take age = 6 thus for the 5 th year = 50 , 6 th year = 50 - 5 = 45 7 th year = 45 - 6 = 39 the correct answer is d ." | a = 5 - 1
b = a - 1
c = 5 - 1
d = 50 + c
e = b + d
f = e / 3
g = f + 3
|
a ) 10.6 sec , b ) 18.8 sec , c ) 14.8 sec , d ) 10.88 sec , e ) 11.88 sec | e | divide(add(170, 160), multiply(add(60, 40), const_0_2778)) | two trains 170 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 170 + 160 = 330 m . required time = 330 * 9 / 250 = 297 / 25 = 11.88 sec . answer : e" | a = 170 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 30 % | c | divide(const_100, multiply(multiply(divide(12, const_100), divide(20, const_100)), const_100)) | on a certain road 12 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "answer is c . this question is in the og and thus well explained by ets . those who exceed : x so x = 12 % + 0,2 x id est x = 15 %" | a = 12 / 100
b = 20 / 100
c = a * b
d = c * 100
e = 100 / d
|
a ) 9 % , b ) 12 % , c ) 15 % , d ) 18 % , e ) 21 % | a | multiply(divide(add(multiply(40, divide(5, const_100)), 2.5), add(40, add(2.5, 7.5))), const_100) | a 40 - liter solution of alcohol and water is 5 percent alcohol . if 2.5 liters of alcohol and 7.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 2.5 ) / 50 = 4.5 / 50 = 9 % the answer is a ." | a = 5 / 100
b = 40 * a
c = b + 2
d = 2 + 5
e = 40 + d
f = c / e
g = f * 100
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | c | subtract(subtract(add(add(50, 30), 20), 80), 5) | in a consumer survey , 80 % of those surveyed liked at least one of 3 products : 1 , 2 , and 3 . 50 % of those asked liked product 1 , 30 % liked product 2 , and 20 % liked product 3 . if 5 % of the people in the survey liked all 3 of the products , what percentage of the survey participants liked more than one of the 3 products ? | pick 100 and x = all set with exactly two items 80 = 30 + 50 + 20 - ( x ) - 10 x = 10 so the answer is 10 + 5 / 100 = 15 % answer c | a = 50 + 30
b = a + 20
c = b - 80
d = c - 5
|
a ) $ 32 , b ) $ 50 , c ) $ 480 , d ) $ 400 , e ) $ 500 | c | divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 8), const_100) | in township k each property is taxed at 8 percent of its assessed value . if the assessed value of a property in township k is increased from $ 180,000 to $ 24,000 , by how much will the property tax increase ? | "increase in house value = $ 24,000 - $ 18,000 = $ 8000 so , tax increase = 8 % of $ 6000 = $ 480 answer : c" | a = 3 + 4
b = a * 4
c = b * 1000
d = 4 + 1
e = d * 4
f = e * 1000
g = c - f
h = g * 8
i = h / 100
|
a ) 27 , b ) 29 , c ) 33 , d ) 39 , e ) 41 | b | gcd(subtract(1255, 8), subtract(1490, 11)) | find the greatest number which on dividing 1255 and 1490 , leaves a reminder of 8 and 11 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 1255 - 8 ) and ( 1490 - 11 ) = hcf ( 1247 , 1479 ) = 29 option b" | a = 1255 - 8
b = 1490 - 11
c = math.gcd(a, b)
|
a ) 9800 , b ) 96288 , c ) 26667 , d ) 1662 , e ) 2882 | a | add(multiply(multiply(add(divide(2.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2.5 % more . she would get how much ? | "( 8000 * 3 * 2.5 ) / 100 = 600 9200 - - - - - - - - 9800 answer : a" | a = 2 / 5
b = 9200 - 8000
c = b / 3
d = c / 8000
e = a + d
f = e * 8000
g = f * 3
h = g + 8000
|
a ) 144 , b ) 131 , c ) 115 , d ) 84 , e ) 45 | d | add(divide(multiply(8, subtract(8, const_1)), const_2), multiply(8, 7)) | 8 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 8 business exec and in each handshake 2 business execs are involved . hence 8 c 2 = 28 also , each of 8 exec will shake hand with every 7 other chairmen for total of 56 handshake . total = 28 + 56 = 84 ans : d" | a = 8 - 1
b = 8 * a
c = b / 2
d = 8 * 7
e = c + d
|
a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 3 , d ) 1 / 5 , e ) 1 / 9 | b | divide(2, const_4) | there are 2 positive integers x and y . what is the probability that x + y is odd ? | explanation : the addition of two integer results in an odd number only if any one of the integers is an odd number . the probability of an odd among the two integer is 1 / 2 . hence , the required probability is 1 / 2 . answer : b | a = 2 / 4
|
a ) 40 % , b ) 60 % , c ) 80 % , d ) 100 % , e ) 110 % | c | multiply(divide(divide(multiply(80, const_1000), const_100), const_100), const_10) | solution x contains 20 % of material a and 80 % of material b . solution y contains 30 % of material a and 70 % of material b . a mixture of both these solutions contains 22 % of material a in the final product . how much solution x is present in the mixture ? | we can assume the total weight of the mixture = 100 conc of a in the final mixture = 22 let weight of a in the mixture be x . conc given = 20 % = . 2 therefore , weight of b = 100 - x . conc given = 30 % = . 3 now , acc to the problem , . 2 x + . 3 ( 100 - x ) = 22 solving , we get x = 80 . since we assumed the weight of the mixture = 100 , therefore presence of a in the mixture = 80 % . answer : c | a = 80 * 1000
b = a / 100
c = b / 100
d = c * 10
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(6, 4) | set a consists of the integers from 4 to 17 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 4 , 5,6 , 7,8 , 9,10 , 11,12 , . . . 17 } b = { 6 , 7,8 , 9,10 , 11,12 . . . , 20 } thus we see that there are 12 distinct integers that are common to both . e is the correct answer ." | a = 6 + 4
|
a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | c | multiply(multiply(10, 16), divide(subtract(16, 10), add(2, 2))) | a student travels from his house to school at 10 km / hr and reaches school 2 hours late . the next day he travels 16 km / hr and reaches school 1 hour early . what is the distance between his house and the school ? | "let x be the distance from his house to the school . x / 10 = x / 16 + 3 8 x = 5 x + 240 3 x = 240 x = 80 km the answer is c ." | a = 10 * 16
b = 16 - 10
c = 2 + 2
d = b / c
e = a * d
|
a ) 4 / 7 , b ) 5 / 7 , c ) 8 / 11 , d ) 7 / 12 , e ) 11 / 15 | c | divide(multiply(2, 4), add(multiply(2, 4), multiply(3, 1))) | at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 1 / 2 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( x / 2 ) ( 3 y / 4 ) = 3 xy / 8 the total number of boxes is xy + 3 xy / 8 = 11 xy / 8 the fraction loaded by the day crew is xy / ( 11 xy / 8 ) = 8 / 11 the answer is c ." | a = 2 * 4
b = 2 * 4
c = 3 * 1
d = b + c
e = a / d
|
a ) 20 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | c | subtract(subtract(200, 80), 20) | of the 200 employees at company x , 80 are full - time , and 150 have worked at company x for at least a year . there are 20 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? | "200 employees 80 are full - time 150 have worked at company x for at least a year 20 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? 200 - 80 = 120 employees not full time 120 - 20 = 100 employees not full time who worked over a year 150 employees have worked at company x for at least a year - 100 employees not full time who worked over a year = 50 full - time employees of company x have worked at the company for at least a year c . 50" | a = 200 - 80
b = a - 20
|
a ) 61.5 , b ) 60.5 , c ) 63.5 , d ) 62.5 , e ) 50 | e | divide(multiply(25, add(const_4, const_1)), const_2) | to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to 1 / 2 of its present ? | "let capacity of 1 bucket = x capacity of the tank = 25 x new capacity of the bucket = 1 x / 2 hence , number of buckets needed = 25 x / ( 1 x / 2 ) = ( 25 Γ 2 ) / 1 = 50 answer is e ." | a = 4 + 1
b = 25 * a
c = b / 2
|
a ) 15 , b ) 16 , c ) 19 , d ) 20 , e ) 24 | a | divide(divide(divide(480, 4), const_2), const_3) | how many of the positive divisors of 480 are also multiples of 4 not including 480 ? | "480 = 2 ^ 6 * 3 * 5 = ( 4 ) * 2 ^ 3 * 3 * 5 besides ( 4 ) , the exponents of 2 , 3 , and 5 are 3 , 1 , and 1 . there are ( 3 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ways to make multiples of 4 . we must subtract 1 because one of these multiples is 480 . the answer is a ." | a = 480 / 4
b = a / 2
c = b / 3
|
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