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 ) 26 days , b ) 31 days , c ) 50 days , d ) 65 days , e ) 36 days | c | multiply(const_3, 15) | working together , jose and jane can complete an assigned task in 15 days . however , if jose worked alone and complete half the work and then jane takes over the task and completes the second half of the task , the task will be completed in 45 days . how long will jose take to complete the task if he worked alone ? assume that jane is more efficient than jose | "assume : jose does 1 job in x days , so jose does 1 / x job in a day jane does 1 job in y days , so jane does 1 / y job in a day together , they does ( x + y ) / xy job in a day . this is equals to 1 / 20 . so ( x + y ) / xy = 1 / 15 15 ( x + y ) = xy next , we ' re told 1 job takes 45 days to complete if jose and jane each does half the work . so since jose does 1 job in x days , he wil need x / 2 days to do half the job . jane similarly will need y / 2 days to do the other half . x / 2 + y / 2 = 45 x + y = 90 so xy = 1350 the answer choices are : 25 days 30 days 60 days 65 days 36 days from the answer choices , so i ' ll go for 50 days for jose and 27 days for jane . answer : c" | a = 3 * 15
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a ) 5 / 12 , b ) 7 / 12 , c ) 12 / 5 , d ) 7 / 5 , e ) 11 / 5 | c | divide(multiply(6, 4), add(4, 6)) | in an electric circuit , two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is 4 ohms and y is 6 ohms ? | "1 / r = 1 / x + 1 / y 1 / r = 1 / 4 + 1 / 6 = 5 / 12 r = 12 / 5 the answer is c ." | a = 6 * 4
b = 4 + 6
c = a / b
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a ) 87 , b ) 27 , c ) 25 , d ) 26 , e ) 24 | c | divide(subtract(multiply(90, 5), multiply(5, 40)), subtract(90, 80)) | the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 40 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 80 x - ( 5 * 40 ) = 90 ( x - 5 ) 250 = 10 x = > x = 25 answer : c" | a = 90 * 5
b = 5 * 40
c = a - b
d = 90 - 80
e = c / d
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a ) - 4 , b ) - 2 , c ) 11 , d ) 13 , e ) 22 | d | divide(add(36, 16), add(3, const_1)) | when x is multiplied by 3 , the result is 16 more than the result of subtracting x from 36 . what is the value of x ? | "the equation that can be formed is : 3 x - 16 = 36 - x or , 4 x = 52 or , x = 13 . answer : d" | a = 36 + 16
b = 3 + 1
c = a / b
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a ) 36 kmph , b ) 40 kmph , c ) 30 kmph , d ) 42 kmph , e ) none of these | a | divide(multiply(4, 90), add(6, 4)) | a goods train leaves a station at a certain time and at a fixed speed . after 6 hours , an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph . this train catches up the goods train in 4 hours . find the speed of the goods train . | let the speed of the goods train be x kmph . distance covered by goods train in 10 hours = distance covered by express train in 4 hours . ∴ 10 x = 4 × 90 or x = 36 . so , speed of goods train = 36 kmph . answer a | a = 4 * 90
b = 6 + 4
c = a / b
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 7 | c | multiply(3, divide(4, 4)) | if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 4 people to do 4 times of that particular work ? | "3 people can do the work one time in one day . 1 person can do 1 / 3 of the work in one day . 4 people can do 4 / 3 of the work in one day . 4 people can do 4 times the work in 3 days . the answer is c ." | a = 4 / 4
b = 3 * a
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a ) 36 , b ) 46 , c ) 56 , d ) 59 , e ) 60 | e | divide(factorial(5), factorial(subtract(5, 3))) | 5 ! / ( 5 - 3 ) ! = ? | 5 ! / ( 5 - 3 ) ! = 5 ! / 2 ! = 5 * 4 * 3 = 60 . hence , the correct answer is e . | a = math.factorial(5)
b = 5 - 3
c = math.factorial(b)
d = a / c
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a ) $ 0.94 , b ) $ 0.96 , c ) $ 0.99 , d ) $ 1.00 , e ) $ 1.20 | c | multiply(add(const_1, divide(12, const_100)), divide(0.8, divide(subtract(const_100, 10), const_100))) | the manager of a produce market purchased a quantity of tomatoes for $ 0.80 per pound . due to improper handling , 10 percent of the tomatoes , by weight , were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal to 12 percent of the cost of the tomatoes . | assume the manager bought 100 tomatoes . cost price = 80 given : 10 % are damaged - - > available tomatoes to sell = 90 90 * x - 80 = 0.12 * 80 90 x - 80 = 9.6 90 x = 89.6 x = 89.6 / 90 = 0.995 x is slightly under 0.9955 = 0.99 answer : c | a = 12 / 100
b = 1 + a
c = 100 - 10
d = c / 100
e = 0 / 8
f = b * e
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(divide(5, const_2), multiply(4, 4)) | what is the remainder when 4 ^ 8707 is divided by 5 ? | "4 ^ 0 = 1 4 ^ 1 = 4 4 ^ 2 = 16 4 ^ 3 = 64 4 ^ 4 = 256 4 ^ 5 = . . . . . 4 4 ^ 6 = . . . . . . . 6 and so on in a distinct last - digit pattern of 4,6 , 4,6 . . . after ignoring the first last - digit of 1 8707 / 2 = 4353 ( with a remainder of 1 ) so the last number is a 4 . alternatively , 8707 is an odd number and hence the last digit is a 4 when 4 ^ 8707 is divided by 5 the remainder is 4 , ans = e" | a = 5 / 2
b = 4 * 4
c = a - b
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a ) 127 , b ) 1200 , c ) 1600 , d ) 1400 , e ) 1202 | b | divide(subtract(220, 160), divide(5, const_100)) | if 5 % more is gained by selling an article for rs . 220 than by selling it for rs . 160 , the cost of the article is ? | "let c . p . be rs . x . then , 5 % of x = 220 - 160 = 60 x / 20 = 60 = > x = 1200 answer : b" | a = 220 - 160
b = 5 / 100
c = a / b
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a ) 36 , b ) 38 , c ) 40 , d ) 42 , e ) 44 | b | subtract(170, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 100 and 170 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 170 , inclusive , is 71 . 3 * 34 = 102 and 3 * 56 = 168 so the number of multiples of 3 is 23 . 5 * 20 = 100 and 5 * 34 = 170 so the number of multiples of 5 is 15 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 11 = 165 so the number of multiples of 15 is 5 . thus the total number is 71 - 23 - 15 + 5 = 38 . the answer is b ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 170 - e
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a ) 1 / 8 , b ) 1 / 3 , c ) 3 / 6 , d ) 3 / 8 , e ) 5 / 3 | d | multiply(5, add(divide(const_1, 20), divide(const_1, 40))) | two persons a and b can complete a piece of work in 20 days and 40 days respectively . if they work together , what part of the work will be completed in 5 days ? | "a ' s one day ' s work = 1 / 20 b ' s one day ' s work = 1 / 40 ( a + b ) ' s one day ' s work = 1 / 20 + 1 / 40 = 3 / 40 the part of the work completed in 5 days = 5 ( 3 / 40 ) = 3 / 8 . answer : d" | a = 1 / 20
b = 1 / 40
c = a + b
d = 5 * c
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a ) 32 , b ) 33 , c ) 34 , d ) 45 , e ) 44 | c | subtract(multiply(const_12, const_3), 2) | there are x persons and y chocolates . if each person is given 1 chocolate then 1 chocolate is left , if each person is given 2 chocolates then 1 person will be left . so how many persons and chocolates are there ? ? | x person , y chocolates 1 st equation = = > ( y - 1 ) / x = 1 2 nd equation = = > y / ( x - 1 ) = 2 solving we get x = 3 , y = 4 answer : c | a = 12 * 3
b = a - 2
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a ) 8 , b ) 6 , c ) 4 , d ) 5 , e ) 9 / 2 | e | inverse(add(inverse(subtract(9, const_1)), inverse(add(9, const_2.0)))) | working alone at their respective constant rates , a can complete a task in ‘ a ’ days and b in ‘ b ’ days . they take turns in doing the task with each working 3 days at a time . if a starts they finish the task in exactly 9 days . if b starts , they take a day more . how long does it take to complete the task if they both work together ? | "work done by ab in a day = xy respectively . when a starts : no . of days when a works = 6 no . of days when b works = 3 → 6 x + 3 y = 1 when b starts : no . of days when a works = 4 no . of days when a works = 5 → 5 x + 4 y = 1 solving the above two equations for xy x = 1 / 9 y = 1 / 9 → total work done by ab in a day = 1 / 9 + 1 / 9 = 2 / 9 → no . of days to complete the work when both work together = 9 / 2 ; answer : e" | a = 9 - 1
b = 1/(a)
c = 9 + 2
d = 1/(c)
e = b + d
f = 1/(e)
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a ) 16 , b ) 49 , c ) 19 , d ) 26 , e ) 17 | b | power(divide(13, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | find out the square of a number which when doubled exceeds its one seventh by 13 ? | "a let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 7 ) + 13 = > 14 p = p + 91 = > p = 7 p ^ 2 = 7 ^ 2 = 49 . answer : b" | a = 4 + 1
b = 1 / a
c = 2 - b
d = 13 / c
e = d ** 2
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a ) 190 , b ) 284.6 , c ) 300 , d ) 244 , e ) 312 | d | multiply(12.2, 20) | a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 20 gallons of gasoline ? | "so 12.2 * 20 = 244 . . imo option d is correct answer . ." | a = 12 * 2
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a ) 337 , b ) 400 , c ) 298 , d ) 2798 , e ) 271 | b | divide(multiply(add(130, 14), const_100), 36) | an engineering student has to secure 36 % marks to pass . he gets 130 marks and fails by 14 marks . the maximum number of marks obtained by him is ? | 130 14 - - - - - - - 361 - - - - - - 144 100 % - - - - - - ? = > 400 answer : b | a = 130 + 14
b = a * 100
c = b / 36
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a ) 2 minutes , b ) 10 minutes , c ) 20 minutes , d ) 40 minutes , e ) 60 minutes | e | divide(subtract(12, 9), subtract(0.25, 0.2)) | united telephone charges a base rate of $ 9.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ? | lets take number of minutesx . given that , 9 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 60 minutes ans e | a = 12 - 9
b = 0 - 25
c = a / b
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a ) 7 m , b ) 7.5 m , c ) 5.5 m , d ) 6.5 m , e ) 8.5 m | b | divide(multiply(divide(81, 45), divide(75, const_100)), 18) | the cost of carpeting a room 18 m long with a carpet 75 cm wide at 45 paise per meter is rs . 81 . the breadth of the room is : | "length of the carpet = total cost / rate / m = 8100 / 45 = 180 m area of the carpet = 180 * 75 / 100 = 135 m 2 breadth of the room = ( area / length ) = 135 / 18 = 7.5 m answer : b" | a = 81 / 45
b = 75 / 100
c = a * b
d = c / 18
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a ) 3.0 , b ) 3.25 , c ) 3.5 , d ) 3.75 , e ) 4.0 | c | divide(7, subtract(4, 2)) | a swimmer can swim in still water at 4 km / h . if the speed of the water current is 2 km / h , how many hours will the swimmer take to swim against the current for 7 km ? | "the swimmer can swim against the current at a speed of 4 - 2 = 2 km / h . the time it will take is 7 / 2 = 3.5 hours . the answer is c ." | a = 4 - 2
b = 7 / a
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 2 | c | multiply(const_3.0, divide(7, add(2, 5))) | ratio between 2 numbers is 5 : 7 and their product is 560 . what is the difference between 2 numbers ? | "explanation : x / y = 5 / 7 x × y = 560 ⇒ x = 560 / y substituting this value in first equation , we get 560 / yy = 57560 / yy = 57 ⇒ 560 y 2 = 57560 y 2 = 57 ⇒ y = 28 x = 20 so difference between the numbers could be x – y = – 8 y – x = 8 answer : c" | a = 2 + 5
b = 7 / a
c = 3 * 0
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a ) 3.3 , b ) 3.5 , c ) 3.7 , d ) 3.9 , e ) 4.1 | a | power(add(power(3, 2), add(1, power(2, 2))), const_0_33) | the edges of three metal cubes are 1 cm , 2 cm , and 3 cm respectively . a new cube is made by melting these three cubes together . what is the edge of the new cube ( in centimeters ) ? | "the total volume is 1 ^ 3 + 2 ^ 3 + 3 ^ 3 = 36 the edge of the new cube is the cube root of 36 which is about 3.3 cm . the answer is a ." | a = 3 ** 2
b = 2 ** 2
c = 1 + b
d = a + c
e = d ** const_0_33
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a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 40 inches | e | divide(add(multiply(6, const_12), 8), 2) | a scale 6 ft . 8 inches long is divided into 2 equal parts . find the length of each part . | explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 2 parts = 80 / 2 = 40 inches answer : e | a = 6 * 12
b = a + 8
c = b / 2
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a ) $ 110 , b ) $ 120 , c ) $ 183.3 , d ) $ 220 , e ) $ 260 | c | divide(multiply(22, const_100), subtract(multiply(add(11, 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 / 11 of the sum of their two accounts . if al ’ s account were to increase by 10 % and eliot ’ 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 / 11 * ( a + e ) = = > 10 a = 12 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 a = 220 + 10 a = = > a = 220 ; e = 183.3 c" | a = 22 * 100
b = 11 + 1
c = b * 10
d = 100 + 20
e = c - d
f = a / e
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a ) 11 % , b ) 20 % , c ) 22 % , d ) 21 % , e ) 23 % | d | multiply(divide(subtract(divide(11, 10), divide(10, 11)), divide(10, 11)), const_100) | if 11 oranges bought 10 r and 10 oranges sold 11 r . what is the profit | suppose he bought 110 apples . cp = rs 100 sp = rs 121 % profit = 100 * ( 121 - 100 ) / 100 = 21 % answer : d | a = 11 / 10
b = 10 / 11
c = a - b
d = 10 / 11
e = c / d
f = e * 100
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a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 28 | b | divide(110, multiply(add(15, const_3.0), const_0_2778)) | the speed at which a man can row a boat in still water is 15 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 110 meters ? | "the speed of the boat downstream = 15 + 3 = 18 km / hr 18 km / hr * 5 / 18 = 5 m / s the time taken to cover 110 meters = 110 / 5 = 22 seconds . the answer is b ." | a = 15 + 3
b = a * const_0_2778
c = 110 / b
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a ) s . 357.21 , b ) s . 322.12 , c ) s . 400 , d ) s . 278.9 , e ) s . 300 | a | multiply(567, divide(7, const_100)) | find the simple interest on rs . 567 for 7 months at 9 paisa per month ? | "explanation : i = ( 567 * 7 * 9 ) / 100 = 357.21 answer : option a" | a = 7 / 100
b = 567 * a
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a ) 340 , b ) 342 , c ) 523 , d ) 363 , e ) none of these | b | multiply(lcm(lcm(lcm(4, 6), 7), 8), const_2) | the least number which when divided by 4 , 6 , 7 and 8 leaves a remainder 6 , but when divided by 9 leaves no remainder , is | "explanation : l . c . m of 4 , 6 , 7 , 8 = 168 therefore required number is of the form 168 k + 6 . least value of k for which ( 168 k + 6 ) is divisible by 9 is k = 2 therefore required number = ( 168 x 2 + 6 ) = 342 . answer : b" | a = math.lcm(4, 6)
b = math.lcm(a, 7)
c = math.lcm(b, 8)
d = c * 2
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a ) 56 , b ) 5.6 , c ) 0.56 , d ) 0.056 , e ) none of these | b | divide(add(multiply(1.4, const_3.0), 1.4), 2) | if 0.6 : 1.4 : : 2 : 4 : x , then x = ? | "explanation : 0.6 × x = 2.4 × 1.4 x = ( 2.4 × 1.4 ) / 0.6 = 24 / 10 × 17 / 10 × 1 / ( 6 / 10 ) 24 / 10 × 14 / 10 × 10 / 6 2 / 5 × 14 = 28 / 5 = 5.6 answer : option b" | a = 1 * 4
b = a + 1
c = b / 2
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a ) 1 % decrease , b ) 2 % increase , c ) 3 % decrease , d ) 4 % decrease , e ) none of these | a | subtract(const_100, multiply(multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | the tax on a commodity is diminished by 10 % and its consumption increases by 10 % . find the effects on revenue ? | "since tax consumption = revenue . therefore , net % change in revenue = ( x + y + xy / 100 ) % = [ - 10 + 10 + ( - 10 x 10 ) / 100 ] % or - 1 % . therefore , revenue decreased by 1 % . answer : a" | a = 10 / 100
b = 1 + a
c = 10 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 7 , b ) 14 , c ) 25 , d ) 27 , e ) 30 | d | add(subtract(add(multiply(floor(divide(44, 03)), const_3.0), 20), multiply(floor(divide(44, 8)), 8)), 03) | at 15 : 00 there were 20 students in the computer lab . at 15 : 03 and every three minutes after that , 4 students entered the lab . if at 15 : 10 and every ten minutes after that 8 students left the lab , how many students were in the computer lab at 15 : 44 ? | "initial no of students + 3 * ( 1 + no of possible 3 minute intervals between 15 : 03 and 15 : 44 ) - 8 * ( 1 + no of possible 10 minute intervals between 15 : 10 and 15 : 44 ) 20 + 3 * 14 - 8 * 4 = 27 d" | a = 44 / 3
b = math.floor(a)
c = b * 3
d = c + 20
e = 44 / 8
f = math.floor(e)
g = f * 8
h = d - g
i = h + 3
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a ) 0 , b ) 12 , c ) 13 , d ) 9 , e ) 11 | a | subtract(add(divide(20, const_4), const_4), add(divide(20, const_4), const_4)) | ' a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the difference between the maximum and minimum possible values of ' a + b ' ? | a = 4 , b = 5 as they have to co - primes ( hcf being 1 and lcm = 20 ) so a + b = 9 is their max and min value hence difference = 0 answer - a | a = 20 / 4
b = a + 4
c = 20 / 4
d = c + 4
e = b - d
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a ) 1 , b ) 7 , c ) 8 , d ) 9 , e ) 36 | e | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 6)), divide(const_1, 18))) | if a , b and c together can finish a piece of work in 4 days . a alone in 6 days and b in 18 days , then c alone can do it in ? | "c = 1 / 4 - 1 / 6 – 1 / 18 = 1 / 36 = > 36 days ' answer : e" | a = 1 / 6
b = const_0_25 - a
c = 1 / 18
d = b - c
e = 1 / d
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a ) 227 , b ) 444 , c ) 277 , d ) 298 , e ) 212 | b | multiply(divide(60, subtract(21, 16)), add(16, 21)) | wo passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | 1 h - - - - - 5 ? - - - - - - 60 12 h rs = 16 + 21 = 37 t = 12 d = 37 * 12 = 444 answer : b | a = 21 - 16
b = 60 / a
c = 16 + 21
d = b * c
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a ) 2277 , b ) 2999 , c ) 1000 , d ) 3000 , e ) 1971 | d | divide(3135, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 3135 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 3135 x * ( 11 / 10 ) * ( 1 / 100 ) = 33 x = 3000 answer : d" | a = 10 / 100
b = 1 + a
c = 5 / 100
d = 1 - c
e = b * d
f = 3135 / e
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a ) 195 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | a | multiply(130, subtract(const_2, const_1)) | a train speeds past a pole in 15 sec and a platform 130 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 130 ) / 25 = x / 15 = > x = 195 m . answer : option a" | a = 2 - 1
b = 130 * a
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a ) 5 : 00 , b ) 5 : 30 , c ) 6 : 00 , d ) 6 : 35 , e ) 7 : 00 | d | divide(add(4, const_2), 35) | city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 4 : 35 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? | "train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 4 : 35 pm + 2 hrs = 6 : 35 pm answer : d . 6 : 35" | a = 4 + 2
b = a / 35
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a ) 88 ab , b ) - 89 ab , c ) 89 ab , d ) - 88 a , e ) - 88 b | c | add(multiply(11, 8), const_1) | if - 11 a and - 8 b are negative integers , then ( - 11 a ) * ( - 8 b ) + ab is | answer : c | a = 11 * 8
b = a + 1
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a ) 0 , b ) 1 , c ) 3 , d ) 5 , e ) 8 | e | floor(multiply(const_100, divide(43, 5000))) | what is the thousandths digit in the decimal equivalent of 43 / 5000 ? | "43 / 5000 = 43 / ( 5 * 10 ^ 3 ) = ( 43 / 5 ) * 10 ^ - 3 = 8.6 * 10 ^ - 3 = . 0086 thousandths digit = 8 answer e" | a = 43 / 5000
b = 100 * a
c = math.floor(b)
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a ) 6084 , b ) 3788 , c ) 4086 , d ) 2721 , e ) 1812 | c | subtract(add(add(add(multiply(multiply(5, const_100), const_10), multiply(4, const_100)), multiply(3, const_10)), 1), add(add(add(const_1000, multiply(3, const_100)), multiply(4, const_10)), 5)) | what is the difference between the largest number and the least number written with the digits 5 , 3 , 1 , 4 ? | explanation : 1345 5431 - - - - - - - - - - - - 4086 answer : c | a = 5 * 100
b = a * 10
c = 4 * 100
d = b + c
e = 3 * 10
f = d + e
g = f + 1
h = 3 * 100
i = 1000 + h
j = 4 * 10
k = i + j
l = k + 5
m = g - l
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a ) 6 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(9, 3) | if ( a – b ) is 9 more than ( c + d ) and ( a + b ) is 3 less than ( c – d ) , then ( a – c ) is : | ( a – b ) – ( c + d ) = 9 and ( c – d ) – ( a + b ) = 3 = > ( a – c ) – ( b + d ) = 9 and ( c – a ) – ( b + d ) = 3 = > ( b + d ) = ( a – c ) – 9 and ( b + d ) = ( c – a ) – 3 = > ( a – c ) – 9 = ( c – a ) – 3 = > 2 ( a – c ) = 6 = > ( a – c ) = 3 answer : c | a = 9 / 3
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a ) 2 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | a | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 75,000 to 2 , 10,000 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 210000 - 175000 ) = 35000 . increase % = ( 35000 / 175000 ã — 100 ) % = 20 % . required average = ( 20 / 10 ) % = 2 % . answer a" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
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a ) 27 , b ) 32 , c ) 35 , d ) 30 , e ) 45 | a | divide(add(408, 240), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | there are 408 boys and 240 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | "explanation : hcf ( 408 , 240 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 240 / 24 = 17 + 10 = 27 answer : a" | a = 408 + 240
b = 2 * 2
c = b * 2
d = c * 3
e = a / d
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a ) 6 , b ) 9 , c ) 12 , d ) 76 , e ) 48 | d | divide(multiply(divide(90, const_100), 20), divide(70, const_100)) | an alloy weighing 20 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 90 percent gold ? | "in 24 ounces , gold is 20 * ( 70 / 100 ) = 14 ounces . now we add x ounces of pure gold to make it 90 % gold . so 14 + x = ( 24 + x ) * 90 / 100 = > x = 76 . answer is d ." | a = 90 / 100
b = a * 20
c = 70 / 100
d = b / c
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['a ) rs . 200', 'b ) rs . 672', 'c ) rs . 546', 'd ) rs . 876', 'e ) none of these'] | a | multiply(divide(surface_cube(5), 15), 20) | the cost of the paint is rs . 20 per kg . if 1 kg of paint covers 15 sq . ft , how much will it cost to paint outside of a cube having 5 feet each side | explanation : surface area of a cube = 6 x 5 ^ 2 = 150 sq . ft quantity of paint required = ( 150 / 15 ) = 10 kg cost of painting = 20 x 10 = rs . 200 answer : a | a = surface_cube / (
b = a * 15
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a ) 6084 , b ) 3788 , c ) 2077 , d ) 2721 , e ) 6174 | e | subtract(add(add(add(multiply(multiply(9, const_100), const_10), multiply(7, const_100)), multiply(3, const_10)), 5), add(add(add(const_1000, multiply(3, const_100)), multiply(7, const_10)), 9)) | what is the difference between the largest number and the least number written with the digits 9 , 3 , 5 , 7 ? | "explanation : 3579 9753 - - - - - - - - - - - - 6174 answer : e" | a = 9 * 100
b = a * 10
c = 7 * 100
d = b + c
e = 3 * 10
f = d + e
g = f + 5
h = 3 * 100
i = 1000 + h
j = 7 * 10
k = i + j
l = k + 9
m = g - l
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a ) 1 / 13 , b ) 2 / 23 , c ) 1 / 6 , d ) 4 / 27 , e ) 3 / 23 | c | divide(choose(4, 2), choose(add(add(4, 3), 2), 2)) | a bag contains 4 red , 3 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 4 c 29 c 2 = 4 c 29 c 2 = 6 / 36 = 1 / 6 c" | a = math.comb(4, 2)
b = 4 + 3
c = b + 2
d = math.comb(c, 2)
e = a / d
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a ) 100 , b ) 300 , c ) 150 , d ) 200 , e ) 250 | d | subtract(multiply(divide(divide(subtract(power(8, 1), power(3, 600)), const_1000), const_1000), 8), divide(divide(subtract(power(8, 1), power(3, 600)), const_1000), const_1000)) | what is the value of 8 1 / 3 % of 600 + 37 ½ of 400 | "25 * 600 / 300 + 75 * 400 / 200 = 50 + 150 = 200 answer : d" | a = 8 ** 1
b = 3 ** 600
c = a - b
d = c / 1000
e = d / 1000
f = e * 8
g = 8 ** 1
h = 3 ** 600
i = g - h
j = i / 1000
k = j / 1000
l = f - k
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a ) 25.0 % , b ) 37.5 % , c ) 50.0 % , d ) 62.5 % , e ) 75 % | c | multiply(add(multiply(divide(subtract(power(multiply(const_2, const_5), const_2), 30), power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2))), multiply(divide(30, power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2)))), power(multiply(const_2, const_5), const_2)) | there is a 30 % chance jen will visit chile this year , while there is a 50 % chance that she will visit madagascar this year . what is the probability that jen will visit either chile or madagascar this year , but not both ? | p ( chile and not madagascar ) = 0.3 * 0.5 = 0.15 p ( madagascar and not chile ) = 0.5 * 0.7 = 0.35 total probability = 0.15 + 0.35 = 0.5 = 50 % the answer is c . | a = 2 * 5
b = a ** 2
c = b - 30
d = 2 * 5
e = d ** 2
f = c / e
g = 2 * 5
h = g ** 2
i = 50 / h
j = f * i
k = 2 * 5
l = k ** 2
m = 30 / l
n = 2 * 5
o = n ** 2
p = 50 / o
q = m * p
r = j + q
s = 2 * 5
t = s ** 2
u = r * t
|
a ) 18 , b ) 13 , c ) 15 , d ) 17 , e ) 71 | b | subtract(divide(multiply(45, 60), const_100), divide(multiply(35, 40), const_100)) | how much 45 % of 60 is greater than 35 % of 40 ? | "( 45 / 100 ) * 60 – ( 35 / 100 ) * 40 27 - 14 = 13 answer : b" | a = 45 * 60
b = a / 100
c = 35 * 40
d = c / 100
e = b - d
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a ) 15 , b ) 20 , c ) 40 , d ) 60 , e ) 70 | a | multiply(2, 4) | three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 180 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 180 x = 15 hcf of 2 x , 3 x and 4 x = x = 15 the answer is a ." | a = 2 * 4
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a ) 15.27 , b ) 16.27 , c ) 17.27 , d ) 7.27 , e ) 7.63 | e | multiply(divide(add(multiply(const_2, 5), const_2), add(const_3, const_4)), multiply(multiply(5.4, 5.4), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | the length of minute hand of a clock is 5.4 cm . what is the area covered by this in 5 minutes | "area of circle is pi * r ^ 2 but in 5 minutes area covered is ( 5 / 60 ) * 360 = 30 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.4 ^ 2 ) * ( 30 / 360 ) = 7.63 cm ^ 2 answer : e" | a = 2 * 5
b = a + 2
c = 3 + 4
d = b / c
e = 5 * 4
f = 1 * const_60
g = 100 * const_3_6
h = f / g
i = e * h
j = d * i
|
a ) 8 , b ) 7 , c ) 9 , d ) 6 , e ) 5 | c | divide(divide(600, subtract(divide(600, 40), divide(600, 45))), 40) | donovan and michael are racing around a circular 600 - 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 = > 600 / 45 = > 40 / 3 2 . speed / rate of michael = distance / time = > 600 / 40 = > 15 relative speed between them = 15 - 40 / 3 = > 5 / 3 ( 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 = 600 , relative rate = 5 / 3 total time taken by micheal to surpass donovan = distance / rate = > 600 * 3 / 5 = > 360 no . of laps taken by michael = total time / michael ' s rate = > 360 / 40 = > 9 hence correct answer is 9 laps . c | a = 600 / 40
b = 600 / 45
c = a - b
d = 600 / c
e = d / 40
|
a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(4, multiply(2, power(const_3, 2))) | if the function q is defined by the formula q = 5 w / ( 4 m ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , m is doubled , and z is tripled ? | we just need to find the factor thats all , w - > quadrupled - > 4 w m - > doubled - > 2 m z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and m * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b | a = 3 ** 2
b = 2 * a
c = 4 / b
|
a ) 1 / 4 , b ) 2 / 7 , c ) 5 / 12 , d ) 1 / 2 , e ) 7 / 12 | a | divide(1, 4) | carol spends 1 / 4 of her savings on a stereo and 2 / 3 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? | total savings = s amount spent on stereo = ( 1 / 4 ) s amount spent on television = ( 1 - 2 / 3 ) ( 1 / 4 ) s = ( 1 / 3 ) * ( 1 / 4 ) * s = ( 1 / 12 ) s ( stereo + tv ) / total savings = s ( 1 / 4 + 1 / 12 ) / s = 1 / 4 answer : a | a = 1 / 4
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a ) 18 , b ) 16 , c ) 12 , d ) 8 , e ) 4 | c | add(12, multiply(18, divide(50, const_100))) | one week , a certain truck rental lot had a total of 18 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 12 trucks on the lot that saturday morning , what is the greatest number of different trucks that could have been rented out during the week ? | "n - not rented trucks ; r - rented trucks n + r = 18 n + r / 2 = 12 r = 12 c" | a = 50 / 100
b = 18 * a
c = 12 + b
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a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 40 % | c | multiply(divide(add(multiply(divide(25, const_100), 400), multiply(100, divide(60, const_100))), add(400, 100)), const_100) | a grocer has 400 pounds of coffee in stock , 25 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer ’ s stock of coffee is decaffeinated ? | 1 . 25 % of 400 = 100 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 160 pounds of decaffeinated out of 500 pounds , that means 160 / 500 * 100 % = 32 % . the correct answer is c . | a = 25 / 100
b = a * 400
c = 60 / 100
d = 100 * c
e = b + d
f = 400 + 100
g = e / f
h = g * 100
|
a ) 2.5 hrs , b ) 2.9 hrs , c ) 4.5 hrs , d ) 3.5 hrs , e ) 1.5 hrs | a | add(divide(reminder(subtract(multiply(11, const_60), multiply(multiply(11, divide(6, 8)), const_60)), const_60), const_100), floor(divide(subtract(multiply(11, const_60), multiply(multiply(11, divide(6, 8)), const_60)), const_60))) | calculate how much time could be saved if the train would run at its usual speed , given that it ran at 6 / 8 of its own speed and got to its destination in 11 hours ? | new speed = 6 / 8 of usual speed new time = 6 / 8 of usual time 6 / 8 of usual time = 10 hrs usual time = 10 * 6 / 8 = 7.5 hrs time saved = 10 - 7.5 = 2.5 hrs answer is a | a = 11 * const_60
b = 6 / 8
c = 11 * b
d = c * const_60
e = a - d
f = reminder / (
g = f + 100
|
a ) 73,075 , b ) 34,075 , c ) 23,075 , d ) 33,075 , e ) 32,075 | d | divide(add(multiply(add(const_4, const_1), const_100), 60), sqrt(const_100)) | a person spent rs . 7,540 from his salary on food and 5,690 on house rent . after that he was left with 60 % of his monthly salary . what is his monthly salary ? | "total money spent on food and house rent = 7,540 + 5,690 = 13,230 which is 100 - 60 = 40 % of his monthly salary ∴ his salary = 13230 x 100 / 40 = 33075 answer : d" | a = 4 + 1
b = a * 100
c = b + 60
d = math.sqrt(100)
e = c / d
|
a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 69 % | e | floor(multiply(subtract(divide(12, 63), divide(9, 80)), const_100)) | a survey was sent to 80 customers , 9 of whom responded . then the survey was redesigned and sent to another 63 customers , 12 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 9 / 80 rate of second survey = 12 / 63 % response rate increase ( 12 / 63 - 9 / 80 ) / ( 9 / 80 ) = 69 % answer is e = 69 %" | a = 12 / 63
b = 9 / 80
c = a - b
d = c * 100
e = math.floor(d)
|
a ) 1.2 km , b ) 1.8 km , c ) 2.4 km , d ) 3.6 km , e ) 1.6 km | d | multiply(divide(12, const_60), add(15, 3)) | the speed of a boat in still water is 15 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 12 minutes is | "b = 15 km / hr , s = 3 km / hr , downstream b + s = 15 + 3 = 18 km / hr down stream time t = 12 min , we have to change it to hours so t = 12 / 60 = 1 / 5 hrs . distance = speed * time = 18 * 1 / 5 = 3.6 km . answer : d" | a = 12 / const_60
b = 15 + 3
c = a * b
|
a ) rs . 195 , b ) rs . 180 , c ) rs . 98 , d ) rs . 120 , e ) rs . 130 | d | multiply(divide(subtract(const_100, 25), const_100), divide(395, add(add(divide(subtract(const_100, 25), const_100), divide(subtract(const_100, 20), const_100)), const_1))) | a sum of rs . 395 was divided among a , b , and c in such a way that b gets 25 % more than a and 20 % more than c . what is the share of a ? | let each one ’ s share is a , b and c respectively , then b = 125 a / 100 = 120 c / 100 a = 100 b / 125 = 4 / 5 b , c = 100 b / 120 = 5 / 6 b 4 / 5 b + b + 5 / 6 b = 395 , 79 b / 30 = 395 , b = 395 * 30 / 79 = 150 a = 4 * 150 / 5 = 120 answer : d | a = 100 - 25
b = a / 100
c = 100 - 25
d = c / 100
e = 100 - 20
f = e / 100
g = d + f
h = g + 1
i = 395 / h
j = b * i
|
a ) 12.5 % , b ) 40 % , c ) 80 % , d ) 125 % , e ) none | d | multiply(divide(5, 4), const_100) | the ratio 5 : 4 expressed as a percent equals | "solution 5 : 4 = 5 / 4 = ( 5 / 4 x 100 ) % . = 125 % . answer d" | a = 5 / 4
b = a * 100
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a ) 4000 , b ) 2000 , c ) 3000 , d ) 5000 , e ) 1000 | c | multiply(const_100, divide(subtract(52, 16), 1.2)) | the current birth rate per certain number of people is 52 , whereas corresponding death rate is 16 per same number of people . if the net growth rate in terms of population increase is 1.2 percent , find number of persons . ( initally ) | "sol . net growth on x = ( 52 - 16 ) = 36 . net growth on 100 = ( 36 / x ã — 100 ) % = 1.2 % . then x = 3000 answer : c" | a = 52 - 16
b = a / 1
c = 100 * b
|
a ) 10 , b ) 25 , c ) 56 , d ) 21 , e ) 36 | c | subtract(negate(40), multiply(subtract(28, 20), divide(subtract(28, 20), subtract(14, 28)))) | 14 , 28 , 20 , 40 , 32 , 64 , . . . ? | "first , multiply by 2 and then subtract 8 . 14 14 * 2 = 28 28 - 8 = 20 20 * 2 = 40 40 - 8 = 32 32 * 2 = 64 64 - 8 = 56 answer : c" | a = negate - (
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a ) 96 , b ) 106 , c ) 158 , d ) 116 , e ) 122 | c | subtract(multiply(add(20, const_1), add(6, 32)), multiply(20, 32)) | the average of runs of a cricket player of 20 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 6 ? | "average = total runs / no . of innings = 32 so , total = average x no . of innings = 32 * 20 = 640 now increase in avg = 4 runs . so , new avg = 32 + 6 = 38 runs total runs = new avg x new no . of innings = 38 * 21 = 798 runs made in the 11 th inning = 798 - 640 = 158 answer : c" | a = 20 + 1
b = 6 + 32
c = a * b
d = 20 * 32
e = c - d
|
a ) 98 days , b ) 21 days , c ) 17 days , d ) 39 days , e ) 19 days | d | multiply(26, divide(const_3, const_2)) | a is half good a work man as b and together they finish a job in 26 days . in how many days working alone b finish the job ? | "wc = 1 : 2 2 x + x = 1 / 26 = > x = 1 / 78 2 x = 1 / 78 = > 39 days answer : d" | a = 3 / 2
b = 26 * a
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a ) 12 days , b ) 15 days , c ) 16 days , d ) 18 days , e ) 20 days | a | add(add(divide(subtract(subtract(const_1, multiply(const_4, divide(const_1, 20))), multiply(add(const_4, const_4), divide(const_1, 10))), add(divide(const_1, 20), divide(const_1, 10))), add(const_4, const_4)), const_4) | jane and ashley take 20 days and 10 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an eight day leave from work . this led to jane ' s working for four extra days on her own to complete the project . how long did it take to finish the project ? | "let us assume that the work is laying 40 bricks . jane = 2 bricks per day ashley = 4 brick per day together = 6 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 32 last 4 days jane works alone , no . of bricks = 8 remaining bricks = 40 - 40 = 0 so together , they would take 0 / 6 = 0 total no . of days = 8 + 4 + 0 = 12 answer is a" | a = 1 / 20
b = 4 * a
c = 1 - b
d = 4 + 4
e = 1 / 10
f = d * e
g = c - f
h = 1 / 20
i = 1 / 10
j = h + i
k = g / j
l = 4 + 4
m = k + l
n = m + 4
|
a ) 2 , b ) 4 , c ) 5 , d ) 3 , e ) 6 | a | multiply(divide(300, multiply(9, const_1000)), const_60) | if john runs in the speed of 9 km / hr from his house , in what time will he reach the park which is 300 m long from his house ? | speed = 9 * 5 / 18 = 2.5 m / sec time taken = 300 / 2.5 = 120 sec ie . 2 mins . answer : a | a = 9 * 1000
b = 300 / a
c = b * const_60
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a ) 1 : 20 , b ) 1 : 10 , c ) 17 : 8 , d ) 17 : 4 , e ) 17 : 60 | e | divide(subtract(33, divide(36, divide(add(const_100, 40), const_100))), divide(36, divide(add(const_100, 40), const_100))) | a dishonest person wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing 33 $ per litre so as to make a profit of 40 % on cost when he sells the resulting milk and water mixture for 36 $ . in what ratio should he mix the water and milk ? | first of all , let ' s consider 1 liter of the stuff he is going to sell - - - naive customers think it ' s pure milk , but we know it ' s some milk - water mixture . he is going to sell this liter of milk - water for $ 36 . this $ 36 should be a 40 % increase over cost . here , we need to think about percentage increases as multipliers . using multipliers ( cost ) * 1.40 = $ 36 cost = 36 / 1.4 = 360 / 12 = $ 25.71 if he wants a 40 % increase over cost on the sale of one liter of his milk - water , the cost has to be $ 25.71 . well , a liter of milk costs $ 33 , so if he is going to use just $ 30 of milk in his mixture , that ' s 25.71 / 33 = 60 / 77 of a liter . if milk is 60 / 77 of the liter , then water is 17 / 77 of the liter , and the ratio of water to milk is 17 : 60 . answer choice ( e ) | a = 100 + 40
b = a / 100
c = 36 / b
d = 33 - c
e = 100 + 40
f = e / 100
g = 36 / f
h = d / g
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['a ) 90', 'b ) 110', 'c ) 130', 'd ) 140', 'e ) 150'] | a | divide(multiply(18, 10), const_2) | if the sides of a triangle are 26 cm , 18 cm and 10 cm , what is its area ? | the triangle with sides 26 cm , 18 cm and 10 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 18 * 10 = 90 cm 2 answer : option a | a = 18 * 10
b = a / 2
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a ) 0 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | e | multiply(100, const_1) | if x is the sum of first 100 positive even integers and y is the sum of first 100 positive odd integers , what is the value of x - y ? please try to explain your answers | first even minus first odd = 2 - 1 = 1 ; the sum of first 2 even integers minus the sum of first 2 odd integers = ( 2 + 4 ) - ( 1 + 3 ) = 2 ; the sum of first 3 even integers minus the sum of first 3 odd integers = ( 2 + 4 + 6 ) - ( 1 + 3 + 5 ) = 3 ; we can see the patterns here , so the sum of first 100 positive even integers minus the sum of first 100 positive odd integers will be 100 . answer : e . | a = 100 * 1
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a ) $ 120 , b ) $ 150 , c ) $ 180 , d ) $ 210 , e ) $ 240 | c | divide(30, subtract(divide(5, 3), divide(3, 2))) | the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 3 : 2 . if the oil bill for february had been $ 30 more , the corresponding ratio would have been 5 : 3 . how much was the oil bill for january ? | "3 : 2 = 9 : 6 and 5 : 3 = 10 : 6 . an increase in $ 30 increases the ratio by 1 : 6 . therefore , january ' s bill was 6 ( $ 30 ) = $ 180 . the answer is c ." | a = 5 / 3
b = 3 / 2
c = a - b
d = 30 / c
|
a ) 3 , b ) 4 , c ) 5 , d ) 2 , e ) 1 | a | divide(divide(multiply(divide(add(multiply(add(const_1, const_4), add(22, multiply(const_4, 22))), add(22, multiply(const_2, 22))), add(add(45, 2), multiply(add(const_1, const_4), 7))), 7), 22), add(const_1, const_4)) | if x - y - z = 03 x + 4 y + 3 z = 45 x + 2 y + 7 z = 22 , what is the value of z ? | explanation : 4 x + 2 y - 5 z = - 21 â € “ - - - - - - - i 2 x - 2 y + z = 7 â € “ - - - - - - ii 4 x + 3 y - z = - 1 - - - - - - - - iii solve the first equation for x . x - y - z = 0 x = y + z substitute the solution for x into the second and third equations . ii - - - > 3 * ( y + z ) + 4 y + 3 z = 4 3 y + 3 z + 4 y + 3 z = 4 7 y + 6 z = 4 iii - - - - > 5 * ( y + z ) + 2 y + 7 z = 22 5 y + 5 z + 2 y + 7 z = 22 7 y + 12 z = 22 subtract the new second equation from the new third equation and solve for z . 7 y + 12 z = 22 - ( 7 y + 6 z = 4 ) 6 z = 18 z = 3 answer is a | a = 1 + 4
b = 4 * 22
c = 22 + b
d = a * c
e = 2 * 22
f = 22 + e
g = d + f
h = 45 + 2
i = 1 + 4
j = i * 7
k = h + j
l = g / k
m = l * 7
n = m / 22
o = 1 + 4
p = n / o
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a ) 3 / 2 , b ) 3 / 4 , c ) 8 / 9 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 1))) | a call center has two teams . each member of team a was able to process 1 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | "let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 1 calls , so total calls for team a = 5 fraction for team b = 40 / ( 40 + 5 ) = 8 / 9 = answer = c" | a = 8 * 5
b = 8 * 5
c = 5 * 1
d = b + c
e = a / d
|
a ) 112 , b ) 111 , c ) 110 , d ) 109 , e ) 108 | e | multiply(subtract(6, multiply(3, const_2)), multiply(9, 11)) | how much space , in cubic units , is left vacant when maximum number of 3 x 3 x 3 cubes are fitted in a rectangular box measuring 6 x 9 x 11 ? | "no of cubes that can be accommodated in box = ( 6 * 9 * 11 ) / ( 3 * 3 * 3 ) 6 * 9 in numerator can be perfectly divided by 3 * 3 in denominator . side with length 11 ca n ' t be perfectly divided by 3 and hence is the limiting factor . closet multiple of 3 less that 11 is 9 . so vacant area in cube = = 6 * 9 * ( 11 - 9 ) = 6 * 9 * 2 = 108 ans - e" | a = 3 * 2
b = 6 - a
c = 9 * 11
d = b * c
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a ) 50 min , b ) 40 min , c ) 35 min , d ) 37 min , e ) 20 min | d | add(20, 17) | the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km / hr and 17 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 1000 m apart to be 20 + 17 = 37 km apart , they take 1 hour to be 1000 m apart , they take 37 * 1000 / 1000 = 37 min . answer is d" | a = 20 + 17
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a ) 72 , b ) 44 , c ) 80 , d ) 88 , e ) 48 | a | divide(subtract(power(18, const_2), 180), const_2) | if the sum of two numbers is 18 and the sum of their squares is 180 , then the product of the numbers is | "according to the given conditions x + y = 18 and x ^ 2 + y ^ 2 = 180 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 18 ^ 2 = 180 + 2 xy so xy = 144 / 2 = 72 answer : a" | a = 18 ** 2
b = a - 180
c = b / 2
|
a ) rs . 16,500 , b ) rs . 16 , 525.50 , c ) rs . 16 , 537.50 , d ) rs . 18,150 , e ) rs . 19,150 | c | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 15000) | sam invested rs . 15000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 15000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 15000 * ( 1 + 5 / 100 ) 2 ] = ( 15000 * 21 / 20 * 21 / 20 ) = rs . 16537.50 answer : c" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 15000
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a ) 3 , b ) 8 , c ) 12 , d ) 16 , e ) 18 | a | divide(power(3, const_2), 3) | if the area of a square with sides of length 3 centimeters is equal to the area of a rectangle with a width of 3 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 3 ^ 2 = l * 3 = > l = 9 / 3 = 3 answer a" | a = 3 ** 2
b = a / 3
|
a ) 23 , b ) 22 , c ) 21 , d ) 36 , e ) 25 | d | divide(const_1, add(divide(const_1, 30), divide(divide(const_1, 30), 1.5))) | a is 1.5 times as fast as b . a alone can do the work in 30 days . if a and b work on alternative days , beginning with a , in how many days will the work be completed ? | "a can finish 1 work in 30 days b can finish 1 / 1.5 work in 30 days - since a is 1.5 faster than b this means b can finish 1 work in 30 * 1.5 days = 45 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 45 * 30 / ( 45 + 30 ) = 20 * 30 / 50 = 18 days since they alternate on days , it will take them double this time so total time = 18 * 2 = 36 days so answer is d" | a = 1 / 30
b = 1 / 30
c = b / 1
d = a + c
e = 1 / d
|
a ) 273105 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273258 | a | multiply(divide(5355, 51), const_100) | 5355 x 51 = ? | "5355 x 51 = 5355 x ( 50 + 1 ) = 5355 x 50 + 5355 x 1 = 267750 + 5355 = 273105 a" | a = 5355 / 51
b = a * 100
|
a ) 2 : 1 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 5 | a | inverse(subtract(const_1, divide(1, 3))) | he ratio between the sale price and the cost price of an article is 3 : 1 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . x and s . p . = rs . 3 x . then , gain = rs . 2 x required ratio = 2 x : x = 2 : 1 . answer : a" | a = 1 / 3
b = 1 - a
c = 1/(b)
|
a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 9 , e ) 17 : 3 | d | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 85,000 and rs . 45,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 45000 = 17 : 9 . answer : d" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 49 , b ) 53 , c ) 56 , d ) 59 , e ) 62 | e | subtract(100, divide(subtract(376, multiply(3, 100)), const_2)) | there are a total of 100 jars on the shelf . small jars hold 3 liters and large jars hold 5 liters . the jars can hold a total of 376 liters . how many small jars are on the shelf ? | let s be the number of small jars and let l be the number of large jars . s + l = 100 . l = 100 - s . 3 s + 5 l = 376 . 3 s + 5 ( 100 - s ) = 376 . - 2 s + 500 = 376 . 2 s = 124 . s = 62 . the answer is e . | a = 3 * 100
b = 376 - a
c = b / 2
d = 100 - c
|
a ) 34 , b ) 35 , c ) 38 , d ) 40 , e ) ( f ) 42 | b | add(add(add(const_10, add(1, 5)), add(2, 6)), add(7, 5)) | trapezoid jklm in the x - y plane has coordinates j = ( – 2 , – 5 ) , k = ( – 2 , 1 ) , l = ( 6 , 7 ) , and m = ( 6 , – 5 ) . what is its perimeter ? | "jk = 6 lm = 11 kl = using distance formula 10 jm = using distance formula 8 sum of all is 35 b" | a = 1 + 5
b = 10 + a
c = 2 + 6
d = b + c
e = 7 + 5
f = d + e
|
a ) 1235 , b ) 1456 , c ) 1567 , d ) 1636 , e ) 1635 | d | add(multiply(divide(subtract(1365, 10), subtract(6, const_1)), 6), 10) | find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 10 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 10 5 x = 1355 x = 271 large number = 271 + 1365 = 1636 d" | a = 1365 - 10
b = 6 - 1
c = a / b
d = c * 6
e = d + 10
|
a ) 0.0584 , b ) 0.0484 , c ) 0.0434 , d ) 0.0384 , e ) none | d | subtract(negate(0.48), multiply(subtract(0.2, 0.08), divide(subtract(0.2, 0.08), subtract(1, 0.2)))) | 1 , 0.2 , 0.08 , 0.48 , 0.0384 , ____ | "1 , 0.2 , 0.08 , 0.48 , 0.0384 , . . . . . 1 * 0.2 = 0.2 0.2 * 0.4 = 0.08 0.08 * 0.6 = 0.048 0.048 * 0.8 = 0.0384 so 0.0384 * 1 = 0.0384 answer : d" | a = negate - (
|
a ) a - 352000 , b ) b - 356000 , c ) c - 360000 , d ) d - 348000 , e ) e - 340000 | c | multiply(divide(45000, 500), 4000) | a person is saving money to purchase a car . he earns â £ 4000 per month and saves â £ 500 to buy the car . if he needs â £ 45000 to buy his dream car . how much money he will earn , before he saves enough money to buy the car ? | he is saving â £ 500 per month . he needs â £ 45000 and it takes â £ 45000 / â £ 500 = 90 months to save the money . in 90 months , he earns 90 * â £ 4000 = â £ 36000 . answer : c | a = 45000 / 500
b = a * 4000
|
a ) rs . 135 , b ) rs . 96 , c ) rs . 85 , d ) rs . 122 , e ) rs . 142 | a | multiply(multiply(divide(1800, 120), divide(9, const_100)), const_100) | by investing rs . 1800 in 9 % stock , syam earns rs . 120 . the stock is then quoted at | "explanation : assume that face value = rs . 100 . dividend per share = rs . 9 ( as it is a 9 % stock ) by investing rs . 1800 , he earns rs . 120 investment needed to earn rs . 9 = 1800 × 9 / 120 = rs . 135 ie , stock is then quoted ( then market value ) = rs . 135 answer : option a" | a = 1800 / 120
b = 9 / 100
c = a * b
d = c * 100
|
a ) 60 , b ) 70 , c ) 75 , d ) 84 , e ) 90 | a | divide(multiply(36, 5), 3) | the rowing athletes in a rowing conference voted for coach of the year . each rower who voted chose exactly 3 coaches to vote for among the 36 coaches in the conference . if each of the 36 coaches received exactly 5 votes ( a 36 way tie ) , how many rowers voted for coach of the year ? | there were 36 * 5 = 180 total votes made . if each voter picked 3 coaches , there were 180 / 3 = 60 voters . a | a = 36 * 5
b = a / 3
|
a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | a | subtract(2, add(3, 1)) | i . x + 2 y + 3 z = 2 ii . x + y - z = 0 iii . 2 x + 2 y - z = 1 what is the value of y in the system above ? | 1 . from the given system of equations i can make equation ii . to be x + y = z 2 . now put that in iii . : 2 x + 2 y - ( x + y ) = 1 = 2 x + 2 y - x - y = 1 = x + y = 1 that gives usz = 1 ! 3 . put z = 1 in i . and solve : x + 2 y + 3 = 2 ; calculate minus 3 = x + 2 y = - 1 ; calculate minus x = 2 y = - 1 - x ; 4 . now put 2 y = - 1 - x in iii . and solve : 2 x + ( - 1 - x ) - 1 = 1 = x - 2 = 1 ; calculate plus 2 that gives usx = 3 5 . now solve for y with x = 3 and z = 1 : x + y = 1 = 3 + y = 1 ; calculate minus 3 that gives y = - 2 , and hence answer ( a ) - 2 | a = 3 + 1
b = 2 - a
|
a ) 96 , b ) 98 , c ) 100 , d ) 102 , e ) 104 | d | divide(multiply(divide(subtract(const_100, 15), const_100), 12), divide(subtract(const_100, 90), const_100)) | if grapes are 90 % water and raisins are 15 % water , then how many kilograms did a quantity of raisins , which currently weighs 12 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transformation . ) | "let x be the original weight of the grapes . the weight of the grape pulp was 0.1 x . since the grape pulp is 85 % of the raisins , 0.1 x = 0.85 ( 12 kg ) . then x = 8.5 * 12 = 102 kg . the answer is d ." | a = 100 - 15
b = a / 100
c = b * 12
d = 100 - 90
e = d / 100
f = c / e
|
a ) $ 3500 , b ) $ 5500 , c ) $ 7500 , d ) $ 9500 , e ) $ 11,500 | c | divide(add(divide(subtract(210, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(210, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 210 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "210 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 150 x = $ 7,500 the answer is c ." | a = 6 / 100
b = a * 1
c = 210 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 210 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
|
a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 36 kmph | d | divide(add(100, 50), add(divide(100, 30), divide(50, 50))) | a car travels uphill at 30 km / hr and downhill at 50 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | "avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 50 = 5 / 5 ; avg speed = 150 / ( 10 / 3 + 5 / 5 ) = 35 kmph answer : d" | a = 100 + 50
b = 100 / 30
c = 50 / 50
d = b + c
e = a / d
|
a ) 3 , b ) 9 , c ) 15 , d ) 67 , e ) 25 | d | add(add(power(2, 2), multiply(2, 2)), 4) | if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 7 ] ] ? | "these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 7 ] ] = 7 ^ 2 + 2 * 7 + 4 = 67 . option d" | a = 2 ** 2
b = 2 * 2
c = a + b
d = c + 4
|
a ) 700 , b ) 900 , c ) 720 , d ) 730 , e ) 740 | b | add(540, multiply(540, divide(40, const_100))) | a fruit seller had some oranges . he sells 40 % oranges and still has 540 oranges . how many oranges he had originally ? | "60 % of oranges = 540 100 % of oranges = ( 540 × 100 ) / 6 = 900 total oranges = 900 answer : b" | a = 40 / 100
b = 540 * a
c = 540 + b
|
a ) 1 , b ) - 2 , c ) 6 , d ) - 3 , e ) 4 | c | subtract(subtract(subtract(100, 15), add(100, 15)), 15) | if | 5 x - 15 | = 100 , then find the sum of the values of x ? | "| 5 x - 15 | = 100 5 x - 15 = 100 or 5 x - 15 = - 100 5 x = 115 or 5 x = - 85 x = 23 or x = - 17 sum = 23 - 17 = 6 answer is c" | a = 100 - 15
b = 100 + 15
c = a - b
d = c - 15
|
a ) 26 minutes , b ) 23 minutes , c ) 15 minutes , d ) 19 minutes , e ) 28 minutes | c | subtract(25, 10) | the speed of a boat in still water is 25 kmph . if it can travel 10 km upstream in 1 hr , what time it would take to travel the same distance downstream ? | speed of boat in still water = 25 km / hr speed upstream = 10 ⁄ 1 = 10 km / hr speed of the stream = ( 25 - 10 ) = 15 km / hr speed downstream = ( 25 + 15 ) = 40 km / hr time taken to travel 10 km downstream = 10 / 40 hours = ( 10 × 60 ) / 40 = 15 minutes answer is c | a = 25 - 10
|
a ) 36 , b ) 50 , c ) 55 , d ) 26 , e ) 29 | c | divide(divide(subtract(125, multiply(multiply(10, const_0_2778), 10)), 10), const_0_2778) | a train 125 m long passes a man , running at 10 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 10 ) km / hr . x - 10 = 45 = = > x = 55 km / hr answer : c" | a = 10 * const_0_2778
b = a * 10
c = 125 - b
d = c / 10
e = d / const_0_2778
|
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