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 |
|---|---|---|---|---|---|
a ) 1000 , b ) 2500 , c ) 1200 , d ) 5000 , e ) 1600 | b | multiply(multiply(5, const_1000), 5) | find the product of the localvalue and absolutevalue of 5 in 20568 ? | "local value of 5 = 5 x 100 = 500 place value of 5 = 5 there fore = 5 x 500 = 2500 b" | a = 5 * 1000
b = a * 5
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a ) 19.81 % , b ) 20 % , c ) 37.5 % , d ) 25 % , e ) 37.5 % | d | multiply(subtract(add(const_100, 10), add(9.09, subtract(const_100, 20))), const_2) | a man cheats while buying as well as while selling . while buying he takes 10 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 9.09 % below the cost price of the claimed weight . | "( 1 + m 1 % ) ( 1 + m 2 % ) ( 1 - d % ) = ( 1 + p % ) 11 / 10 * 5 / 4 * 10 / 11 = ( 1 + p % ) profit % = 25 % answer : d" | a = 100 + 10
b = 100 - 20
c = 9 + 9
d = a - c
e = d * 2
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a ) rs . 1178.55 , b ) rs . 1978.25 , c ) rs . 1332.5 , d ) rs . 1678 , e ) rs . 1675.55 | c | subtract(2665, divide(multiply(multiply(3, 5), 2665), add(multiply(3, 5), multiply(5, 3)))) | a sum of rs . 2665 is lent into two parts so that the interest on the first part for 5 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 5 * 3 ) / 100 = ( ( 2665 - x ) * 3 * 5 ) / 100 15 x / 100 = 39975 / 100 - 15 x / 100 30 x = 39975 = > x = 1332.5 second sum = 2665 β 1025 = 1332.5 answer : c" | a = 3 * 5
b = a * 2665
c = 3 * 5
d = 5 * 3
e = c + d
f = b / e
g = 2665 - f
|
a ) 42 , b ) 38 , c ) 28 , d ) 29 , e ) 11 | a | subtract(negate(15), multiply(subtract(3, 6), divide(subtract(3, 6), subtract(2, 3)))) | 2 , 3 , 6 , 15 , _ , 123 ? | answer : a | a = negate - (
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a ) 33 , b ) 38 , c ) 32 , d ) 28 , e ) 19 | a | subtract(47, 15) | nitin ranks 15 th in a class of 47 students . what is rank from the last ? | "explanation : number students behind the nitin in rank = ( 47 - 15 ) = 32 nitin is 33 rd from the last answer : a ) 33" | a = 47 - 15
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a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | c | multiply(divide(multiply(105, 2), add(70, 105)), const_60) | cole drove from home to work at an average speed of 70 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 70 time from work to home = x / 105 total time = 2 hrs ( x / 70 ) + ( x / 105 ) = 2 solving for x , we get x = 84 time from home to work in minutes = ( 84 ) * 60 / 70 = 72 minutes ans = c" | a = 105 * 2
b = 70 + 105
c = a / b
d = c * const_60
|
a ) 304 , b ) 14440 , c ) 760 , d ) 28880 , e ) 1520 | c | multiply(add(152, const_1), const_2) | what is the least common multiple of 152 and 190 ? | "yes there is a shorter way 152 = 2 * 2 * 2 * 19 190 = 2 * 5 * 19 i think everyone knows how to do this . then choose 2 * 2 * 2 and choose 5 and choose 19 2 * 2 * 2 * 5 * 19 = 760 answer is c" | a = 152 + 1
b = a * 2
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a ) 1220 , b ) 1650 , c ) 1920 , d ) 2440 , e ) 2860 | c | multiply(subtract(power(3, 5), 3), multiply(4, 4)) | in how many ways can an answer key for a quiz be written if the quiz contains 5 true - false questions followed by 3 multiple - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | "there are 2 ^ 5 = 32 possibilities for the true - false answers . however we need to remove two cases for ttttt and fffff . there are 4 * 4 * 4 = 64 possibilities for the multiple choice questions . the total number of possibilities is 30 * 64 = 1920 . the answer is c ." | a = 3 ** 5
b = a - 3
c = 4 * 4
d = b * c
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a ) 0 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 4 | c | subtract(divide(1, 3), multiply(divide(1, 2), divide(1, 2))) | canister c is 1 / 2 full of water and canister d , which has twice the capacity of canister c , is 1 / 3 full of water . if the water in canister d is poured in canister c until canister c is completely full of water , canister d will still contain what fraction of its capacity of water ? | say canister c has a capacity of 6 liters . it ' s half full , thus there can be poured 3 liters of water . canister b is 12 liters and there are 4 liters of water . we can pour 3 liters from d to c and 1 liter will still be left in d , which is 1 / 12 of its total capacity . answer : c . | a = 1 / 3
b = 1 / 2
c = 1 / 2
d = b * c
e = a - d
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a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 100 | e | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 6000 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 6000 x = 100 the numbers are 3 * 100 , 4 * 100 , 5 * 100 hence required h . c . f . = 100 answer is e" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
|
a ) 1 hour , b ) 45 minutes , c ) 20 minutes , d ) 5 minutes , e ) 1 hour and 10 minutes | c | divide(160, 8) | jose is uploading a file to the internet for a college project . the file weighs 160 megabytes . if jose ' s internet speed for uploading tops 8 megabytes per minute , how long will it take until the upload is completed ? | answer is ( c ) . jose calculates that if the files weighs 160 megabytes , and his internet uploads at 8 megabytes per minute , it would take 20 minutes to upload as 160 divided by 8 megabytes per minute equals 20 . | a = 160 / 8
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a ) 3700 , b ) 3890 , c ) 88798 , d ) 2789 , e ) 2891 | a | subtract(multiply(add(1600, 100), add(20, const_1)), multiply(1600, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1600 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1700 * 21 - 1600 * 20 ) = rs . 3700 . answer : a ) 3700" | a = 1600 + 100
b = 20 + 1
c = a * b
d = 1600 * 20
e = c - d
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a ) $ 10900 , b ) $ 19000 , c ) $ 190000 , d ) $ 1900 , e ) none | c | divide(19000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 19000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 19000 , = [ 1 - 9 / 10 ] * total salary = $ 19000 , total salary = $ 19000 * 10 = $ 190000 , correct answer ( c )" | a = 1 / 5
b = 1 / 10
c = a + b
d = 3 / 5
e = c + d
f = 1 - e
g = 19000 / f
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a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 69.10 % , e ) none of these | d | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(30, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 30 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | "girls = 900 , boys = 1100 passed = ( 30 % of 1100 ) + ( 32 % of 900 ) = 330 + 288 = 618 failed = 2000 - 618 = 1382 failed % = [ ( 1382 / 2000 ) x 100 ] % = 69.1 % . answer : d" | a = 100 * 10
b = 2 * a
c = 30 / 100
d = 100 * 10
e = 2 * d
f = e - 900
g = c * f
h = b - g
i = 32 / 100
j = i * 900
k = h - j
l = 100 * 10
m = 2 * l
n = k / m
o = n * 100
|
a ) 1 , b ) 4 / 3 , c ) 17 / 9 , d ) 18 / 5 , e ) 4 | c | divide(add(divide(subtract(multiply(7, 2), 10), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 10), subtract(multiply(2, 2), const_1))))), 3) | if 2 x + y = 7 and x + 2 y = 10 , then ( x + y ) / 3 = | "we have two equations : 2 x + y = 7 x + 2 y = 10 notice that something nice happens when we add them . we get : 3 x + 3 y = 17 divide both sides by 3 to get : x + y = 17 / 3 so , ( x + y ) / 3 = 17 / 9 answer : c" | a = 7 * 2
b = a - 10
c = 2 * 2
d = c - 1
e = b / d
f = 7 * 2
g = f - 10
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 7 - k
m = e + l
n = m / 3
|
a ) 54 , b ) 110 , c ) 120 , d ) 16 , e ) 180 | c | multiply(40, divide(divide(36, 2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 40 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "c 120 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 40 mph = 120 miles" | a = 36 / 2
b = a / 6
c = 40 * b
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a ) 13 , b ) 17 , c ) 18 , d ) 11 , e ) 15 | e | multiply(15, const_1) | puzzle ! ! ! π if 111 = 09 444 = 12 777 = 15 then 888 = ? ? ? | e 15 one + one + one ( 3 + 3 + 3 ) = 09 four + four + four ( 4 + 4 + 4 ) = 12 seven + seven + seven ( 5 + 5 + 5 ) = 15 therefore eight + eight + eight ( 5 + 5 + 5 ) = 15 | a = 15 * 1
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | divide(multiply(multiply(const_2, const_3), 5), const_2) | town m town n town o town p town q town r ; town a town b town c town d town e town f in the table above , what is the least number of table entries that are needed to show the mileage between each town and each of the other 5 towns ? | easy way to go about this problem is we have 6 * 6 = 36 enteries in table the least number of enteries would be ( 36 - 6 ) / 2 since 6 enteries represent the distances between same points . alternatively this can be solved as combination problem . correct answer e | a = 2 * 3
b = a * 5
c = b / 2
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(divide(subtract(33, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 33 ? | "33 / 3 = 11 the three numbers are 10 , 11 , and 12 . the answer is c ." | a = 33 - 3
b = a / 3
c = b + 2
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a ) 875 , b ) 945 , c ) 1425 , d ) 2025 , e ) 2500 | b | divide(multiply(subtract(45, const_3), 45), const_2) | a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 45 - sided polygon have ? | "there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] / 2 , n = number of sides of the regular polygon . here , n = 45 . plugging it in , we get 945 diagonals ! answer ( b ) ." | a = 45 - 3
b = a * 45
c = b / 2
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a ) rs . 800 , b ) rs . 400 , c ) rs . 600 , d ) rs . 500 , e ) rs . 900 | b | divide(1600, const_3) | divide rs . 1600 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 1600 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1600 = > 400 answer : b" | a = 1600 / 3
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a ) 66 , b ) 26 , c ) 49 , d ) 27 , e ) 11 | c | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 112), const_2) | 112 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 112 = 49 answer : c" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 112
f = e / 2
|
a ) 30,000 , b ) 5,000 , c ) 80,000 , d ) 120,000 , e ) none | b | multiply(add(add(multiply(multiply(add(const_2, const_3), multiply(add(const_2, const_3), const_2)), const_100), multiply(multiply(multiply(add(const_2, const_3), multiply(add(const_2, const_3), const_2)), const_100), const_3)), multiply(multiply(multiply(multiply(add(const_2, const_3), multiply(add(const_2, const_3), const_2)), const_100), const_3), const_2)), divide(3000, multiply(multiply(multiply(multiply(add(const_2, const_3), multiply(add(const_2, const_3), const_2)), const_100), const_3), const_2))) | a , b and c started a shop by investing rs . 5,000 , rs . 15,000 and rs . 30,000 respectively . at the end of the year , the profits were distributed among them . if c Γ’ β¬ β’ s share of profit be rs . 3000 , then the total profit was : | sol . a : b : c = 5000 : 15000 : 30000 = 1 : 3 : 6 . so , c Γ’ β¬ β’ s share : total profit = 6 : 10 let the total profit be rs . x . then , 6 / 10 = 3000 / x or x = 3000 * 10 / 6 = 5000 . answer b | a = 2 + 3
b = 2 + 3
c = b * 2
d = a * c
e = d * 100
f = 2 + 3
g = 2 + 3
h = g * 2
i = f * h
j = i * 100
k = j * 3
l = e + k
m = 2 + 3
n = 2 + 3
o = n * 2
p = m * o
q = p * 100
r = q * 3
s = r * 2
t = l + s
u = 2 + 3
v = 2 + 3
w = v * 2
x = u * w
y = x * 100
z = y * 3
A = z * 2
B = 3000 / A
C = t * B
|
a ) 22.5 , b ) 20.0 , c ) 18.75 , d ) 12.5 , e ) 10.0 | c | subtract(25, divide(multiply(25, 25), const_100)) | christine selects an item at a 25 % off ticket price sale . the item is ticket priced at $ 25 . how much should christine expect to pay at the register ? | r . p . = $ 25.00 sale % = 25 s . p . = r . p . * ( 1 - ( sale % / 100 ) ) = 25 * ( 1 - ( 25 / 100 ) ) = 18.75 answer : c | a = 25 * 25
b = a / 100
c = 25 - b
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a ) s . 227 , b ) s . 287 , c ) s . 297 , d ) s . 300 , e ) s . 380 | a | divide(multiply(500, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 500 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 500 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 500 β y [ 12 / 10 + 1 ] = 500 β 22 y / 10 = 500 β 22 y = 5000 β y = 5000 / 22 = rs . 227.77 a )" | a = 1 + 4
b = a * 2
c = 500 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
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['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(rectangle_area(2, 3), square_area(divide(const_10, const_4))) | if the perimeter of square region e and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of e | we know perimeter of a square ( pe ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration 2 k : 3 k = 5 k ; we can assume such a number ) therefore , pe = pr = 40 area of the square = 100 sq . units we know 2 ( length + breadth ) = 40 i . e . length + breadth = 20 ( or 5 k = 20 given that l : b ( or b : l ) = 2 : 3 ) therefore length = 8 , breath = 12 area of the rectangle = 8 * 12 = 96 sq . units question asked = area of the rectangle : area of the square = 96 : 100 = = > 24 : 25 = b | a = rectangle_area / (
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a ) 10 , b ) 22 , c ) 25 , d ) 27 , e ) 30 | e | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 30)), const_2), divide(const_1, 30))) | p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 30 days to complete the same work . then q alone can do it in | work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 30 work done by p , q and r in 1 day = 1 / 10 + 1 / 30 = 4 / 30 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day Γ£ β 2 = 4 / 30 = > work done by p in 1 day = 4 / 60 = > work done by q and r in 1 day = 4 / 60 hence work done by q in 1 day = 4 / 60 Γ’ β¬ β 1 / 30 = 2 / 60 = 1 / 30 so q alone can do the work in 30 days answer is e . | a = 1 / 10
b = 1 / 30
c = a + b
d = c / 2
e = 1 / 30
f = d - e
g = 1 / f
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a ) 11 , b ) 10 , c ) 18 , d ) 19 , e ) 21 | d | multiply(divide(add(add(floor(divide(11, 2)), const_1), floor(divide(27, 2))), const_2), 2) | calculate the average of all the numbers between 11 and 27 which are divisible by 2 . | "explanation : numbers divisible by 2 are 12,14 , 16,18 , 20,22 , 24,26 , average = ( 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 , ) / 8 = 152 / 8 = 19 answer : d" | a = 11 / 2
b = math.floor(a)
c = b + 1
d = 27 / 2
e = math.floor(d)
f = c + e
g = f / 2
h = g * 2
|
a ) $ 4600 , b ) $ 4800 , c ) $ 5000 , d ) $ 5200 , e ) $ 5400 | c | divide(3500, subtract(const_1, divide(30, const_100))) | we had $ 3500 left after spending 30 % of the money that we took for shopping . how much money did we start with ? | "let x be the amount of money we started with . 0.7 x = 3500 x = 5000 the answer is c ." | a = 30 / 100
b = 1 - a
c = 3500 / b
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a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | divide(divide(40, const_2), divide(80, const_100)) | at the end of the month , a certain ocean desalination plant β s reservoir contained 40 million gallons of water . this amount is one fifth of the normal level . if this amount represents 80 % of the reservoir β s total capacity , how many million gallons short of total capacity is the normal level ? | "the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 40 mg = 80 % of total . . total = 40 / . 8 = 50 mg . . normal level = 1 / 5 of 50 = 10 mg . . shortage of normal level = 50 - 10 = 40 mg . . c" | a = 40 / 2
b = 80 / 100
c = a / b
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | b | subtract(subtract(multiply(3, divide(add(multiply(15, 5), multiply(15, 3)), subtract(multiply(3, 3), 5))), 15), add(divide(add(multiply(15, 5), multiply(15, 3)), subtract(multiply(3, 3), 5)), 15)) | on a certain farm the ratio of horses to cows is 3 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 5 : 3 . after the transaction , how many more horses than cows would the farm own ? | "originally , there were 3 k horses and k cows . 3 ( 3 k - 15 ) = 5 ( k + 15 ) 9 k - 5 k = 75 + 45 4 k = 120 k = 30 the difference between horses and cows is ( 3 k - 15 ) - ( k + 15 ) = 2 k - 30 = 30 the answer is b ." | a = 15 * 5
b = 15 * 3
c = a + b
d = 3 * 3
e = d - 5
f = c / e
g = 3 * f
h = g - 15
i = 15 * 5
j = 15 * 3
k = i + j
l = 3 * 3
m = l - 5
n = k / m
o = n + 15
p = h - o
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 11 | b | add(divide(subtract(38, 2), 4), const_1) | what is the total number of integers between 2 and 38 that are divisible by 4 ? | "4 , 6 , 8 , . . . , 32,36 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 36 - 4 ) / ( 4 ) + 1 = 32 / 4 + 1 = 8 + 1 = 9 answer is b" | a = 38 - 2
b = a / 4
c = b + 1
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a ) 22 , b ) 84 , c ) 28 , d ) 76 , e ) 21 | b | divide(multiply(multiply(4, 6), 7), const_2) | a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 7 rows or 6 rows or 4 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 7 , 6,4 ) = 84 . answer : b" | a = 4 * 6
b = a * 7
c = b / 2
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a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) all the above value | e | subtract(5, 4) | the number n is 5 , h 64 , where h represents the 100 ' s digit . if n is divisible by 4 , what is the value of h ? | if the number is divisible by 4 , the last two digits must be divisible by 4 . all the values yields such a number . answer : e | a = 5 - 4
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a ) 23 , b ) 20 , c ) 11 , d ) 5 , e ) 2 | a | add(divide(add(22, 20), 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 { 22 } + { 20 } ? | "soln : { 22 } + { 20 } = 22 * { 20 } + { 20 } = 23 * { 20 } answer : a" | a = 22 + 20
b = a / 2
c = 1 * 2
d = b + c
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a ) rs . 300 , b ) rs . 200 , c ) rs . 240 , d ) rs . 400 , e ) none of these | d | divide(1480, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 1480 will be : | "z share = z , y = 1.2 z x = 1.25 Γ£ β 1.2 z , x + y + z = 1480 ( 1.25 Γ£ β 1.2 + 1.2 + 1 ) z = 1480 3.7 z = 1480 , z = 400 answer : . d" | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b * d
f = 20 / 100
g = 1 + f
h = e + g
i = h + 1
j = 1480 / i
|
a ) 70 , b ) 25 , c ) 60 , d ) 80 , e ) 55 | a | add(add(multiply(divide(60, 5), const_2), 5), add(divide(60, 5), 5)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 5 years , their ages will be 45 and 25 . sum of their ages = 45 + 25 = 70 . answer a" | a = 60 / 5
b = a * 2
c = b + 5
d = 60 / 5
e = d + 5
f = c + e
|
a ) 2 : 3 , b ) 5 : 3 , c ) 5 : 6 , d ) 1 : 4 , e ) 3 : 4 | c | divide(divide(5, 2), 3) | if the ratio of apples to bananas is 5 to 2 and the ratio of bananas to cucumbers is 1 to 3 , what is the ratio of apples to cucumbers ? | "the ratio of bananas to cucumbers is 1 to 3 which equals 2 to 6 . the ratio of apples to bananas to cucumbers is 5 to 2 to 6 . the ratio of apples to cucumbers is 5 to 6 . the answer is c ." | a = 5 / 2
b = a / 3
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a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 4 / 5 | d | divide(const_4, add(0,0, const_10)) | in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,6 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | "the area of the right triangle is ( 1 / 2 ) * 4 * 6 = 12 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 12 = 2 / 3 the answer is d ." | a = 0 + 0
b = 4 / a
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a ) 3 : 9 , b ) 3 : 0 , c ) 3 : 5 , d ) 3 : 2 , e ) 3 : 1 | c | divide(add(5, subtract(multiply(5, 3), 5)), add(5, multiply(2, subtract(multiply(5, 3), 5)))) | the ratio of the present age of two brothers is 1 : 2 and 5 years back , the ratio was 1 : 3 . what will be the ratio of their ages after 5 years ? | let the present ages of the two brothers be x and 2 x years respectively . then , ( x - 5 ) / ( 2 x - 5 ) = 1 / 3 3 ( x - 5 ) = ( 2 x - 5 ) = > x = 10 required ratio = ( x + 5 ) : ( 2 x + 5 ) = 15 : 25 = 3 : 5 answer : c | a = 5 * 3
b = a - 5
c = 5 + b
d = 5 * 3
e = d - 5
f = 2 * e
g = 5 + f
h = c / g
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | subtract(724946, multiply(floor(divide(724946, 10)), 10)) | find the least number must be subtracted from 724946 so that remaining no . is divisible by 10 ? | "on dividing 724946 by 10 we get the remainder 6 , so 6 should be subtracted d" | a = 724946 / 10
b = math.floor(a)
c = b * 10
d = 724946 - c
|
a ) 500 % , b ) 250 % , c ) 500 / 3 % , d ) 84 % , e ) 60 % | d | multiply(add(const_1, divide(20, const_100)), subtract(const_100, 30)) | if x is 20 percent more than y and y is 30 percent less than z , then x is what percent of z ? | "z = 100 ; y = 70 so x = 84 x as % of z = 84 / 100 * 100 = > 84 % answer will be ( d )" | a = 20 / 100
b = 1 + a
c = 100 - 30
d = b * c
|
a ) 6 hours , b ) 5 hours , c ) 7 hours , d ) 8 hours , e ) none | b | divide(add(240, 0), add(divide(240, 8), 18)) | a truck covers a distance of 240 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 0 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 240 / 8 = 30 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 30 + 18 ) = 48 kmph distance travelled by car = 240 + 0 = 240 km time taken by car = distance / speed = 240 / 48 = 5 hours . answer β b" | a = 240 + 0
b = 240 / 8
c = b + 18
d = a / c
|
a ) 42 , b ) 49 , c ) 62 , d ) 72 , e ) 82 | b | subtract(57, multiply(multiply(12, const_2.0), 2)) | evaluate : 57 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 57 - 12 * 3 * 2 = 57 - 8 = 49 correct answer b" | a = 12 * 2
b = a * 2
c = 57 - b
|
a ) 4 liters , b ) 8 liters , c ) 10 liters , d ) 12 liters , e ) 12.5 liters | d | divide(subtract(multiply(divide(25, const_100), 180), multiply(divide(20, const_100), 180)), subtract(const_1, divide(25, const_100))) | a mixture of 180 liters of wine and water contains 20 % water . how much more water should be added so that water becomes 25 % of the new mixture ? | "number of liters of water in 180 liters of the mixture = 20 % of 120 = 20 / 100 * 180 = 36 liters . p liters of water added to the mixture to make water 25 % of the new mixture . total amount of water becomes ( 36 + p ) and total volume of mixture is ( 180 + p ) . ( 36 + p ) = 25 / 100 * ( 180 + p ) 96 + 4 p = 180 + p p = 12 liters . answer : d" | a = 25 / 100
b = a * 180
c = 20 / 100
d = c * 180
e = b - d
f = 25 / 100
g = 1 - f
h = e / g
|
a ) 0 , b ) - 43 , c ) - 25 , d ) - 49 , e ) - 51 | b | add(add(negate(23), const_1), add(add(negate(23), const_1), const_1)) | the sum of all the integers k such that β 23 < k < 24 is | "- 22 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 22 - 21 sum of which is - 43 . hence option d . b" | a = negate + (
b = a + 1
|
a ) 10 , b ) 12 , c ) 24 , d ) 60 , e ) 100 | a | multiply(factorial(3), factorial(2)) | there are 3 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | "there are 3 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? 10 12 24 60 100 soln : total number of patterns is 5 ! since 3 red chips are identical and 2 blue ones are identical thus we have = 5 ! / ( 2 ! * 3 ! ) = 10 such different patterns a" | a = math.factorial(3)
b = math.factorial(2)
c = a * b
|
a ) 30 , b ) 40 , c ) 50 , d ) 55 , e ) 70 | b | multiply(subtract(const_100, 80), subtract(divide(const_100, 45), divide(80, 80))) | john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 45 miles per hour , what is v in miles per hour ? | "hibunuel the question seems incorrect . it should not be 80 % at the speed of 80 . however if it ' s 20 % at the speed of 80 , answer comes out 55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it , which comes as : 0.8 d / 80 + 0.2 d / v = d / 40 = > v = 40 ( option b ) ." | a = 100 - 80
b = 100 / 45
c = 80 / 80
d = b - c
e = a * d
|
a ) 21 , b ) 7000.0707 , c ) 7777 , d ) 7014 , e ) 7000.0077 | b | add(multiply(7, const_10), multiply(7, const_1000)) | what is the sum between the place values of three 7 ' s in the numeral 87953.0727 | "required sum = 7000 + 0.0700 + 0.0007 = 7000.0707 answer is b" | a = 7 * 10
b = 7 * 1000
c = a + b
|
a ) a ) 23 , b ) b ) 21 , c ) c ) 18 , d ) d ) 56 , e ) e ) 12 | c | add(6, divide(multiply(6, subtract(12000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 6 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | "let the total number of workers be x . then , 8000 x = ( 12000 * 6 ) + 6000 ( x - 6 ) = > 2000 x = 36000 = x = 18 . answer : c" | a = 12000 - 8000
b = 6 * a
c = 8000 - 6000
d = b / c
e = 6 + d
|
a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) none | c | multiply(multiply(500, inverse(multiply(add(45, 30), const_0_2778))), const_2) | two good train each 500 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 500 + 500 ) m = 1000 m . required time = ( 1000 x 6 / 125 ) sec = 48 sec . answer c | a = 45 + 30
b = a * const_0_2778
c = 1/(b)
d = 500 * c
e = d * 2
|
a ) 3 / 5 , b ) 4 / 15 , c ) 7 / 15 , d ) 8 / 15 , e ) 17 / 30 | c | add(multiply(divide(2, 5), divide(2, 3)), multiply(divide(const_3, 5), divide(1, 3))) | in a tree , 2 / 5 of the birds are robins while the rest are bluejays . if 1 / 3 of the robins are female and 2 / 3 of the bluejays are female , what fraction of the birds in the tree are male ? | the fraction of birds that are male robins is ( 2 / 3 ) ( 2 / 5 ) = 4 / 15 . the fraction of birds that are male bluejays is ( 1 / 3 ) ( 3 / 5 ) = 1 / 5 . the total fraction of male birds is 4 / 15 + 1 / 5 = 7 / 15 . the answer is c . | a = 2 / 5
b = 2 / 3
c = a * b
d = 3 / 5
e = 1 / 3
f = d * e
g = c + f
|
a ) 500 , b ) 600 , c ) 289 , d ) 276 , e ) 207 | a | multiply(multiply(90, const_0_2778), 20) | what distance will be covered by a bus moving at 90 kmph in 20 seconds ? | "90 kmph = 90 * 5 / 18 = 25 mps d = speed * time = 25 * 20 = 500 m . answer : a" | a = 90 * const_0_2778
b = a * 20
|
a ) 35 , b ) 24 , c ) 14 , d ) 24 , e ) 12 | d | multiply(6, inverse(subtract(1, add(divide(1, 2), divide(1, 4))))) | in traveling from a dormitory to a certain city , a student went 1 / 2 of the way by foot , 1 / 4 of the way by bus , and the remaining 6 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 + 1 / 4 x + 6 x - 1 / 2 x - 1 / 4 x = 6 x = 24 km option : d" | a = 1 / 2
b = 1 / 4
c = a + b
d = 1 - c
e = 1/(d)
f = 6 * e
|
a ) 76 kmph , b ) 70 kmph , c ) 87 kmph , d ) 56 kmph , e ) 86 kmph | b | divide(add(80, 60), const_2) | the speed of a car is 80 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 80 + 60 ) / 2 = 70 kmph answer : b" | a = 80 + 60
b = a / 2
|
a ) a . 16000 , b ) b . 17000 , c ) c . 18000 , d ) d . 19000 , e ) e . 20000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | if a town of 15,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a = 3 + 4
b = 5 * a
c = b * 100
d = 3 + 4
e = d + 3
f = c * e
|
['a ) 60', 'b ) 50', 'c ) 40', 'd ) 80', 'e ) 70'] | a | multiply(300, divide(divide(40, 100), const_2)) | the ratio , by volume , of bleach ( b ) to detergent ( d ) to water in a certain solution is 2 : 40 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 300 liters of water , how many liters of detergent will it contain ? | b : d : w = 2 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 40 ) = ( 1 / 60 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 300 dnew = wnew / 5 = 300 / 5 = 60 so , answer will be a | a = 40 / 100
b = a / 2
c = 300 * b
|
a ) 10 , b ) 8 , c ) 12 , d ) 14 , e ) 16 | e | divide(400, subtract(26, const_1)) | in a garden , 26 trees are planted at equal distances along a yard 400 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | 26 trees have 25 gaps between them . length of each gap = 400 / 25 = 16 i . e . , distance between two consecutive trees = 16 answer is e . | a = 26 - 1
b = 400 / a
|
a ) 24 , b ) 77 , c ) 6 , d ) 29 , e ) 21 | c | divide(multiply(18, 36), 108) | 36 men can complete a piece of work in 18 days . in how many days will 108 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 108 : 36 : : 18 : x x = 6 answer : c ) 6 days" | a = 18 * 36
b = a / 108
|
a ) 2261 , b ) 2888 , c ) 1400 , d ) 2699 , e ) 2771 | c | add(multiply(multiply(divide(700, 10), 5), const_3), multiply(divide(700, 10), 5)) | the simple interest on a sum of money will be rs . 700 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 700 p - - - 5 - - - - - 350 3 p - - - 5 - - - - - 1050 - - - - - - = > 1400 answer : c" | a = 700 / 10
b = a * 5
c = b * 3
d = 700 / 10
e = d * 5
f = c + e
|
a ) 1200 , b ) 3000 , c ) 1000 , d ) 3600 , e ) 2400 | c | divide(subtract(multiply(divide(6, const_100), 2000), multiply(2000, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | barbata invests $ 2000 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | "let the additional invested amount for 8 % interest be x ; equation will be ; 2000 + 0.05 * 2000 + x + 0.08 x = 2000 + x + 0.06 ( 2000 + x ) 0.05 * 2000 + 0.08 x = 0.06 x + 0.06 * 2000 0.02 x = 2000 ( 0.06 - 0.05 ) x = 2000 * 0.01 / 0.02 = 1000 ans : ` ` c ' '" | a = 6 / 100
b = a * 2000
c = 5 / 100
d = 2000 * c
e = b - d
f = 8 / 100
g = 6 / 100
h = f - g
i = e / h
|
a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | e | multiply(subtract(125, 10), 10) | what is the greatest positive integer x such that 5 ^ x is a factor of 125 ^ 10 ? | "125 ^ 10 = ( 5 ^ 3 ) ^ 10 = 5 ^ 30 answer : e" | a = 125 - 10
b = a * 10
|
a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | a | divide(const_1, divide(20, const_2)) | a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 35 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 40 in what ratio should he mix the water and milk ? | "water = w ( liter ) milk = m ( liter ) = = > cost = price x quantity = 0.35 m = = > revenue = price x quantity = 0.40 ( m + w ) = = > profit = 0.40 ( m + w ) - 0.35 m = 0.2 * ( 0.35 m ) [ 20 % of cost ] = = > 0.40 m + 0.40 w - 0.35 m = 0.07 m = = > 0.02 m = 0.40 w = = > m / w = 0.40 / 0.02 = 20 - - or - - w / m = 1 / 20 a is correct ." | a = 20 / 2
b = 1 / a
|
a ) 68 , b ) 70 , c ) 72 , d ) 74 , e ) 76 | b | multiply(divide(150, 4), divide(150, add(divide(150, 4), divide(150, 3.5)))) | two trains start simultaneously from opposite ends of a 150 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 150 - km trip in 4 hours . train y , travelling at a constant rate , completes the 150 - km trip in 3.5 hours . how many kilometers had train x traveled when it met train y ? | if the two trains cover a total distance d , then train x travels ( 7 / 15 ) * d while train y travels ( 8 / 15 ) * d . if the trains travel 150 km to the meeting point , then train x travels ( 7 / 15 ) * 150 = 70 km . the answer is b . | a = 150 / 4
b = 150 / 4
c = 150 / 3
d = b + c
e = 150 / d
f = a * e
|
a ) 17 , b ) 18 , c ) 34 , d ) 35 , e ) 36 | c | subtract(subtract(multiply(18, 2), const_1), 1) | ( ( 1 ^ ( m + 1 ) ) / ( 5 ^ ( m + 1 ) ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) what is m ? | ( ( 1 ^ ( m + 1 ) ) / ( 5 ^ ( m + 1 ) ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) ( ( 1 / 5 ) ^ ( m + 1 ) ) * ( ( 1 / 2 ) ^ 36 ) = 1 / ( 2 * ( 2 * 5 ) ^ 35 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ ( m + 1 ) = ( 1 / 5 ) ^ 35 ( ( 1 / 5 ) ^ ( m + 1 ) ) * ( ( 1 / 2 ) ^ 36 ) = 1 / [ ( 2 ^ 36 ) * ( 5 ^ 35 ) ] so , m = 34 answer c | a = 18 * 2
b = a - 1
c = b - 1
|
a ) 123 , b ) 267 , c ) 277 , d ) 122 , e ) 120 | d | divide(add(106, 138), 2) | a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 106 . what was the number he chose ? | "let xx be the number he chose , then 2 Γ’ βΉ β¦ x Γ’ Λ β 138 = 106 x = 122 answer : d" | a = 106 + 138
b = a / 2
|
a ) 1.4 , b ) 1.5 , c ) 1.6 , d ) 1.8 , e ) 1.0 | d | divide(add(multiply(6, 3), 18), 10) | a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 18 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ? | "let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 18 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 1.8 m / s d" | a = 6 * 3
b = a + 18
c = b / 10
|
a ) 4.07 % , b ) 4 % , c ) 2.7 % , d ) 2.04 % , e ) 2.08 % | d | multiply(divide(subtract(const_100, 98), 98), const_100) | if the cost price is 98 % of sp then what is the profit % | "sol . sp = rs 100 : then cp = rs 98 : profit = rs 2 . profit = { ( 2 / 98 ) * 100 } % = 2.04 % answer is d ." | a = 100 - 98
b = a / 98
c = b * 100
|
a ) 1.5 , b ) 1.3 , c ) 1.25 , d ) 1.6 , e ) 2 | c | multiply(const_60, divide(multiply(50, divide(15, const_60)), subtract(60, 50))) | mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least 3 hours , how long after he passed the gas station did paul catch up with mary ? | "d = rt m : r = 50 mph , t = t + 1 / 4 hr d = 50 ( t + 1 / 4 ) p : r = 60 , t = t d = 60 t since they went the same distance : 50 t + 50 / 4 = 60 t 10 t = 50 / 4 t = 1.25 or 1 hr , 15 min c" | a = 15 / const_60
b = 50 * a
c = 60 - 50
d = b / c
e = const_60 * d
|
a ) 429 , b ) 280 , c ) 360 , d ) 450 , e ) none | a | divide(multiply(78, 66), subtract(78, 66)) | the banker Γ’ β¬ β’ s discount of a certain sum of money is rs . 78 and the true discount on the same sum for the same time is rs . 66 . the sum due is | sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 78 * 66 / 78 - 66 ] = rs . [ 78 * 66 / 12 ] = rs . 429 answer a | a = 78 * 66
b = 78 - 66
c = a / b
|
a ) $ 110 , b ) $ 152.3 , c ) $ 180 , d ) $ 220 , e ) $ 260 | b | divide(multiply(22, const_100), subtract(multiply(add(10, 1), 10), add(const_100, 20))) | right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 10 of the sum of their two accounts . if al β s account were to increase by 10 % and eliot β s account were to increase by 20 % , then al would have exactly $ 22 more than eliot in his account . how much money does eliot have in his account right now ? | "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 10 * ( a + e ) = = > 9 a = 11 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 22 + 1.2 e = = > 11 a = 220 + 12 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 11 e / 9 * 11 = 220 + 12 e or ( 121 / 9 - 12 ) e = 220 or 13 / 9 e = 220 e = 220 * 9 / 13 = 152.3 b" | a = 22 * 100
b = 10 + 1
c = b * 10
d = 100 + 20
e = c - d
f = a / e
|
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number q of tickets sold by the male members of the association | "given that , total average q sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ." | a = 1 + 2
b = 66 * a
c = 70 * 2
d = b - c
|
a ) 196 , b ) 1996 , c ) 2996 , d ) 3996 , e ) 6000 | d | multiply(multiply(4, const_100), const_10) | subtract the absolute value from the local value of 4 in 564823 | explanation : place value = local value face value = absolute value the place value of 4 in 564823 is 4 x 1000 = 4000 the face value of 4 in 564823 is nothing but 4 . = > 4000 - 4 = 3996 answer : option d | a = 4 * 100
b = a * 10
|
a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(660, divide(multiply(multiply(add(divide(10, const_100), const_1), 660), 10), subtract(multiply(add(divide(10, const_100), const_1), 660), 660))) | machine a and machine g are each used to manufacture 660 sprockets . it takes machine a 10 hours longer to produce 660 sprockets than machine g . machine g produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | [ reveal ] spoiler : timeg : 660 / x timea : [ 660 / x + 10 ] 660 / x = [ 660 / x + 10 ] * 110 / 100 660 / x = 66 * 11 / x + 10 660 x + 10 = 66 * 11 * x 660 x + 6600 = 66 * 11 * x x = 100 plug in back to timea 660 / 100 + 10 = > 660 / 110 = 6 | a = 10 / 100
b = a + 1
c = b * 660
d = c * 10
e = 10 / 100
f = e + 1
g = f * 660
h = g - 660
i = d / h
j = 660 / i
|
a ) [ 28 ] , b ) [ 84 ] , c ) [ 42 ] , d ) [ 48 ] , e ) [ 14 ] | b | divide(multiply(divide(4, 2), multiply(3, 7)), 3) | for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 7 ] * [ 4 ] equivalent to ? | "[ 7 ] * [ 4 ] = 21 * 2 = 42 = ( 1 / 2 ) ( 84 ) = [ 84 ] the answer is b ." | a = 4 / 2
b = 3 * 7
c = a * b
d = c / 3
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | floor(divide(reminder(power(6, reminder(21, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 6 ^ 21 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . . . 56 is 5 . the tens digit of 6 ^ 7 = . . . 36 is 3 again . etc . . . 21 has the form 5 n + 1 , so the tens digit of 6 ^ 21 is 5 . the answer is c ." | a = 4 + 1
b = 6 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(660, multiply(add(const_1, divide(10, const_100)), const_100)) | machine x and machine b are each used to manufacture 660 sprockets . it takes machine x 10 hours longer to produce 660 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine x . how many sprockets per hour does machine x produces ? | i think the correct answer is a . machine x produces at a speed of x sp / hour and b at a speed of b sp / hour . so , 660 / x = ( 660 / b ) + 10 and x = 1,1 x - - - > 1,1 * 660 = 660 + 11 x - - - > a = 6 , so answer a is correct | a = 10 / 100
b = 1 + a
c = b * 100
d = 660 / c
|
a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 3 | e | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 2)))) | a can do a piece of work in 2 hours ; b and c together can do it in 3 hours , while a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour ' s work = 1 / 2 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 2 + 1 / 3 ) = 5 / 6 . b ' s 1 hour ' s work = ( 5 / 6 - 1 / 2 ) = 1 / 3 . therefore a alone will take 3 hours to do the work . e" | a = 1 / 3
b = 1 / 2
c = 1 / 2
d = b - c
e = a - d
f = 1 / e
|
a ) 11 , b ) 66 , c ) 28 , d ) 30 , e ) 99 | d | add(divide(add(multiply(3, 6), subtract(16, 6)), const_2), 16) | the ages of two persons differ by 16 years . 6 years ago , the elder one was 3 times as old as the younger one . what are their present ages of the elder person | "let ' s take the present age of the elder person = x and the present age of the younger person = x β 16 ( x β 6 ) = 3 ( x - 16 - 6 ) = > x β 6 = 3 x β 66 = > 2 x = 60 = > x = 60 / 2 = 30 answer : d" | a = 3 * 6
b = 16 - 6
c = a + b
d = c / 2
e = d + 16
|
a ) 9.4 % , b ) 9.6 % , c ) 9 % , d ) 9.8 % , e ) 10 % | c | multiply(divide(add(divide(multiply(12, 10), const_100), divide(multiply(8, 30), const_100)), add(10, 30)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 8 % . 10 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 10 + 30 = 40 kg alloy is 0.12 * 10 + 0.08 * 30 = 3.6 kg , so the percentage is 3.6 / 40 * 100 = 9 % . answer : c" | a = 12 * 10
b = a / 100
c = 8 * 30
d = c / 100
e = b + d
f = 10 + 30
g = e / f
h = g * 100
|
a ) 100 , b ) 120 , c ) 180 , d ) 200 , e ) 240 | c | multiply(divide(37, add(divide(const_1, 6), add(divide(const_1, 4), divide(const_1, 5)))), const_3) | an automotive test consisted of driving a car the same distance 3 separate times , first at an average rate of 4 miles per hour , then 5 miles per hour , then 6 miles per hour . if the test took 37 hours to complete , how many miles was the car driven during the entire test ? | x is the segment to be traveled x / 4 + x / 5 + x / 6 = 37 or 74 x / 120 = 37 x = 60 miles . total distance = 3 * 60 = 180 miles c | a = 1 / 6
b = 1 / 4
c = 1 / 5
d = b + c
e = a + d
f = 37 / e
g = f * 3
|
a ) 3 , b ) 5 , c ) 8 , d ) 2 , e ) 6 | c | divide(multiply(12, 24), add(12, 24)) | a can do a work in 12 days . b can do the same work in 24 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 12 b rate = 1 / 24 ( a + b ) rate = ( 1 / 12 ) + ( 1 / 24 ) = 1 / 8 a & b finish the work in 8 days correct option is c" | a = 12 * 24
b = 12 + 24
c = a / b
|
a ) 460 , b ) 289 , c ) 220 , d ) 400 , e ) 640 | e | multiply(divide(2000, add(2000, add(multiply(3000, 2), multiply(4000, 2)))), 2000) | a , b and c invests rs . 2000 , rs . 3000 and rs . 4000 in a business . after 2 year a removed his money ; b and c continued the business for two more year . if the net profit after 3 years be rs . 4000 , then a ' s share in the profit is ? | "2 * 24 : 3 * 36 : 4 * 36 4 : 9 : 12 4 / 25 * 4000 = 640 answer : e" | a = 3000 * 2
b = 4000 * 2
c = a + b
d = 2000 + c
e = 2000 / d
f = e * 2000
|
a ) 465 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(19, multiply(const_4, const_100)), add(multiply(subtract(19, multiply(const_4, const_100)), 2), const_1)) | balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 19 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did . how many balls made up the triangle ? | "as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 19 ) ( x - 8 ) ^ 2 = t + 19 - - - - - ( b ) now the hardest part of the question will be to solve these 2 equations and this looks like time consuming but the easy way will be plug and play . also , we ' ve to find a value of t ( from 5 optiosn given below ) which can make a square of a a number . one we know this , it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - 465 + 19 = 484 = 22 ^ 2 = ( x - 8 ) ^ 2 hence , x = 30 cross check by putting in eq ( a ) = x ( x + 1 ) / 2 = t = > 30 * 31 / 2 = 465 hence , answer is a ." | a = 4 * 100
b = 19 - a
c = 4 * 100
d = 19 - c
e = d * 2
f = e + 1
g = b * f
|
a ) 19 , b ) 14 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 10), multiply(3, 4)), 2) | the average of 5 quantities is 10 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | answer : a ( 5 x 10 - 3 x 4 ) / 2 = 19 | a = 5 * 10
b = 3 * 4
c = a - b
d = c / 2
|
a ) rs . 258.80 , b ) rs . 358.80 , c ) rs . 458.80 , d ) rs . 520.80 , e ) none of these | d | multiply(divide(70, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 70 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) Γ£ β h ] + ( l Γ£ β b ) = [ 2 ( 25 + 12 ) Γ£ β 6 ] + ( 25 Γ£ β 12 ) = 744 sq m cost of plastering = 744 Γ£ β ( 70 / 100 ) = rs . 520.80 answer : d" | a = 70 / 100
b = 25 * 12
c = 25 * 6
d = 2 * c
e = 12 * 6
f = e * 2
g = d + f
h = b + g
i = a * h
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | b | add(add(add(const_4, const_2), const_1), const_1) | how many odd factors does 520 have ? | "start with the prime factorization : 520 = 2 * 5 * 13 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 520 . answer : b ." | a = 4 + 2
b = a + 1
c = b + 1
|
a ) 40 , b ) 44 , c ) 12 , d ) 88 , e ) 48 | c | divide(subtract(power(22, const_2), 460), const_2) | if the sum of two numbers is 22 and the sum of their squares is 460 , then the product of the numbers is | "according to the given conditions x + y = 22 and x ^ 2 + y ^ 2 = 460 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 22 ^ 2 = 460 + 2 xy so xy = 24 / 2 = 12 answer : c" | a = 22 ** 2
b = a - 460
c = b / 2
|
a ) 47 , b ) 25 , c ) 37 , d ) 33 , e ) 29 | b | add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1) | what is the value of n if the sum of the consecutive odd integers from 1 to n equals 169 ? | # of terms = ( n - 1 / 2 ) + 1 { ( last term - first term ) / 2 + 1 | sum = ( 1 + n ) / 2 * # of terms = ( n + 1 ) ^ 2 / 4 = 169 n + 1 = 13 * 2 n + 1 = 26 n = 25 . answer : b | a = math.sqrt(169)
b = a * 2
c = 2 * 1
d = b - c
e = d + 1
|
a ) a ) 41 , b ) b ) 38 , c ) c ) 35 , d ) d ) 29 , e ) e ) 28 | c | subtract(subtract(subtract(50, const_4), add(const_4, const_1)), 6) | the total number of plums that grew during each year on a certain plum tree was equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded to the nearest lower integer . ) during its 3 rd year , this plum tree grew 50 plums . how many plums did it grow during its 6 th year ? | the answer shud be 38 . ( 50 - 3 - 4 - 5 = 38 ) yes , if you go the 50 - 3 way , then it will be 38 . if instead you go the 50 - 4 way ( which they have suggested ) , the answer will be 35 . as i said , there is ambiguity here . answer is c | a = 50 - 4
b = 4 + 1
c = a - b
d = c - 6
|
a ) 3 / 5 , b ) 3 / 6 , c ) 3 / 7 , d ) 3 / 8 , e ) none of these | d | add(divide(const_1, 8), multiply(divide(const_1, 8), const_2)) | a can finish a work in 8 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ? | "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 8 b ' s 1 day work = 1 / 4 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 / 8 + 1 / 4 ) = 3 / 8 so in one day 3 / 8 work will be done answer : d" | a = 1 / 8
b = 1 / 8
c = b * 2
d = a + c
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(divide(500, subtract(divide(500, 40), divide(500, 45))), 40) | donovan and michael are racing around a circular 500 - meter track . if donovan runs each lap in 45 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 500 / 45 = > 100 / 9 2 . speed / rate of michael = distance / time = > 500 / 40 = > 50 / 4 relative speed between them = 50 / 4 - 100 / 9 = > 25 / 36 ( we subtract the rates if moving in the same direction and add the rates if moving in the opposite direction ) in order to pass donovan - distance to be covered = 500 , relative rate = 50 / 36 total time taken by micheal to surpass donovan = distance / rate = > 500 * 36 / 50 = > 3600 / 10 = > 360 no . of laps taken by michael = total time / michael ' s rate = > 360 / 40 = > 9 hence correct answer is 9 laps . a" | a = 500 / 40
b = 500 / 45
c = a - b
d = 500 / c
e = d / 40
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | e | subtract(add(6, 2), 6) | if m = | | n β 3 | β 2 | , for how many values of n is m = 6 ? | "m = | | n β 3 | β 2 | can be 4 only and only when n - 3 = + / - 8 . so there are 2 values of n answer : e" | a = 6 + 2
b = a - 6
|
a ) 3488 , b ) 6240 , c ) 2776 , d ) 2889 , e ) 7721 | b | multiply(divide(96, 10), 650) | in order to obtain an income of rs . 650 from 10 % stock at rs . 96 , one must make an investment of : | "explanation : to obtain rs . 10 , investment = rs . 96 . to obtain rs . 650 , investment = = rs . 6240 answer : b ) 6240" | a = 96 / 10
b = a * 650
|
a ) 13 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | e | subtract(subtract(add(add(add(20, const_1), add(20, 6)), add(add(20, const_10), 4)), 68), multiply(const_2, 3)) | in a group of 68 students , each student is registered for at least one of 3 classes β history , math and english . 20 - one students are registered for history , 20 - 6 students are registered for math , and thirty - 4 students are registered for english . if only 3 students are registered for all 3 classes , how many students are registered for exactly two classes ? | a u b u c = a + b + c - ab - bc - ac + abc 68 = 21 + 26 + 34 - ab - bc - ac + 3 = > ab + bc + ac = 16 exactly two classes = ab + bc + ac - 3 abc = 16 - 3 * 3 = 7 hence e | a = 20 + 1
b = 20 + 6
c = a + b
d = 20 + 10
e = d + 4
f = c + e
g = f - 68
h = 2 * 3
i = g - h
|
a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | b | subtract(210, add(add(60, 90), divide(60, const_3))) | 210 reputed college students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | we are told that 60 students claimed that they preferred mac to windows , which means that 60 preferred mac but not windows , so # of students who preferred mac ( p ( a ) as you wrote ) , does not equal to 60 , it equals to 60 + 20 ( 20 is # of students who equally preferred both brands ) . also we are asked to find # of the students who preferred windows to mac , so if you denote x as those who prefer windows then you should calculate x - 20 . so , if we use your formula it should be : 210 = { mac } + { windows } - { both } + { neither } = ( 60 + 20 ) + x - 20 + 90 - - > x = 60 ( # of student who prefer windows ) - - > # of the students who preferred windows to mac is x - 20 = 40 . | a = 60 + 90
b = 60 / 3
c = a + b
d = 210 - c
|
a ) 8 kmph , b ) 9 kmph , c ) 7 kmph , d ) 6 kmph , e ) 5 kmph | d | subtract(18, divide(multiply(18, 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 18 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 = 18 / 3 = 6 kmph . answer : d" | a = 18 * 2
b = a / 3
c = 18 - b
|
a ) 63 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) can not be determined | a | divide(subtract(multiply(35, 62), subtract(85, 50)), 35) | a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 62 . if the marks secured by reema was written as 50 instead of 85 then find the correct average marks up to two decimal places . | "total marks = 35 x 62 = 2170 corrected total marks = 2170 - 50 + 85 = 2205 correct average = 2205 / 35 = 63 answer : a" | a = 35 * 62
b = 85 - 50
c = a - b
d = c / 35
|
['a ) ( 8 , - 3 )', 'b ) ( 4 , 5 )', 'c ) ( - 5 , 7 )', 'd ) ( - 2 , - 20 )', 'e ) ( - 15 , - 3 )'] | d | divide(20, const_2) | on the xy - coordinate plane , point a lies on the y - axis and point b lies on the x - axis . points a , b , and c form a right triangle with a 90 - degree angle at point c and the triangle has an area of 20 . if ac is parallel to the x - axis , and bc is parallel to the y - axis , which of the following could be the coordinates of point c ? | the area of the triangle formed will be : 1 / 2 * base * height = 1 / 2 * | ( x - coordinate of point c ) | * | ( y - coordinate of point c ) | = 20 thus | product of coordinates of point c | = 40 the answer is d . | a = 20 / 2
|
a ) a ) 87 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 97 | b | subtract(divide(multiply(divide(multiply(63, 8), 30), 50), 6), 63) | 63 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 63 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 140 140 β 63 = 77 answer : b" | a = 63 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 63
|
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