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 ) - 0.1 , b ) 0 , c ) 0.05 , d ) 0.2 , e ) 0.3 | d | subtract(add(add(divide(add(floor(multiply(5.45, const_10)), const_1), const_10), divide(add(const_1, floor(multiply(2.95, const_10))), const_10)), divide(add(floor(multiply(3.74, const_10)), const_1), const_10)), divide(floor(multiply(add(add(5.45, 2.95), 3.74), const_10)), const_10)) | tough and tricky questions : decimals . the value of x is derived by summing a , b , and c and then rounding the result to the tenths place . the value of y is derived by first rounding a , b , and c to the tenths place and then summing the resulting values . if a = 5.45 , b = 2.95 , and c = 3.74 , what is y β x ? | to find x , we first sum a , b , and c , then round to the tenths place . 5.45 + 2.95 + 3.74 = 12.14 , which rounds to 12.2 . to find y , we first round a , b , and c to the tenths place and them sum them . so 5.4 + 2.9 + 3.7 = 12.0 . we are looking for y - x , which gives us 12.2 - 12.0 = 0.2 , or answer choice d . | a = 5 * 45
b = math.floor(a)
c = b + 1
d = c / 10
e = 2 * 95
f = math.floor(e)
g = 1 + f
h = g / 10
i = d + h
j = 3 * 74
k = math.floor(j)
l = k + 1
m = l / 10
n = i + m
o = 5 + 45
p = o + 3
q = p * 10
r = math.floor(q)
s = r / 10
t = n - s
|
a ) 298 m , b ) 125 m , c ) 120 m , d ) 217 m , e ) 200 m | e | divide(20, subtract(divide(20, 30), 18)) | a train covers a distance of 20 km in 30 min . if it takes 18 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 20 / 30 * 60 ) km / hr = ( 40 * 5 / 18 ) m / sec = 100 / 9 m / sec . length of the train = 100 / 9 * 18 = 200 m . answer : e" | a = 20 / 30
b = a - 18
c = 20 / b
|
a ) 794 m , b ) 704 m , c ) 454 m , d ) 186 m , e ) 176 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 500), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 22.4 = 70400 cm = 704 m answer : b" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 500
i = h / 100
|
a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 12 cm | c | multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(16, const_2), 16), const_pi), const_3), const_4), 12), const_4), const_4), const_2) | 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 16 cm height . what is the diameter of each sphere ? | "volume of cylinder = pi * r ^ 2 * h volume of a sphere = 4 * pi * r ^ 3 / 3 12 * 4 * pi * r ^ 3 / 3 = pi * r ^ 2 * h r ^ 3 = r ^ 2 * h / 16 = 64 cm ^ 3 r = 4 cm d = 8 cm the answer is c ." | a = 16 / 2
b = volume_cylinder / (
c = b * math.pi
d = c / 3
e = d / 4
f = e / 12
g = f / 4
h = g * 4
|
a ) 28 : 23 , b ) 6 : 11 , c ) 5 : 4 , d ) 3 : 5 , e ) 7 : 11 | a | divide(add(multiply(4, divide(35, add(4, 3))), 8), add(multiply(3, divide(35, add(4, 3))), 8)) | the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 35 years . the ratio of their ages after 8 years will be | let mini β s age = 4 x and minakshi β s age = 3 x then 4 x + 3 x = 35 x = 5 mini β s age = 20 years and minakshi β s age = 15 years ratio of their ages after 8 years = ( 20 + 8 ) : ( 15 + 8 ) = 28 : 23 answer : a | a = 4 + 3
b = 35 / a
c = 4 * b
d = c + 8
e = 4 + 3
f = 35 / e
g = 3 * f
h = g + 8
i = d / h
|
a ) 24 % , b ) 34 % , c ) 22 % , d ) 19 % , e ) 8.5 % | d | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(15, const_100), const_4), divide(15, const_100))), const_100) | one fourth of a solution that was 15 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.15 + 25 * x = 100 * 0.16 - - > x = 0.19 . answer : d ." | a = 16 / 100
b = a * 4
c = 15 / 100
d = c * 4
e = 15 / 100
f = d - e
g = b - f
h = g * 100
|
a ) 7 / 30 , b ) 11 / 30 , c ) 13 / 30 , d ) 17 / 30 , e ) none | a | divide(add(3, 4), multiply(add(2, 1), const_10)) | [ 1 / ( 1 . 2.3 ) ] + [ 1 / ( 2 . 3.4 ) ] + [ 1 / ( 3 . 4.5 ) ] + [ 1 / ( 4 . 5.6 ) ] is equal to | answer given exp . = [ ( 4 x 5 x 6 ) + ( 5 x 6 ) + ( 2 x 6 ) + ( 2 x 3 ) ] / [ ( 2 x 3 x 4 x 5 x 6 ) ] = 168 / ( 24 x 30 ) = 7 / 30 correct option : a | a = 3 + 4
b = 2 + 1
c = b * 10
d = a / c
|
a ) 2 / 21 , b ) 2 / 19 , c ) 1 / 11 , d ) 1 / 7 , e ) 1 / 5 | a | divide(subtract(21, 19), 21) | a β s speed is 21 / 19 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | let d be the full distance . let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 21 v / 19 ) = xd / v ( 19 / 21 ) * d / v = x * d / v x = 19 / 21 b should have a head start of 2 / 21 of the full distance . the answer is a . | a = 21 - 19
b = a / 21
|
a ) 0.201 , b ) 0.000201 , c ) 2.01 , d ) 0.0201 , e ) 0.00201 | e | multiply(divide(2.01, 0.001), const_100) | 2.01 Γ£ β 0.001 = ? | "2.01 Γ£ β 0.001 = ? or , ? = 0.00201 answer e" | a = 2 / 1
b = a * 100
|
a ) 3 β 20 , b ) 11 β 60 , c ) 9 β 40 , d ) 1 β 5 , e ) 2 β 15 | e | divide(subtract(divide(subtract(const_100, 40), const_2), const_4), multiply(40, const_2)) | if there are half as many women as men in a group and an equal number of men and women do not own cars - a group that is 40 % of the total . what fraction of the total is women who own cars ? | "consider a group of 200 men and 100 women , a total of 300 people . 40 % of them , which is 120 , form a group of people who do n ' t own a car . half of them are men , and the other half are women , more precisely 60 . it means that there are 100 - 60 = 40 women who own a car , and this represents 40 / 300 = 2 / 15 of the total . answer e" | a = 100 - 40
b = a / 2
c = b - 4
d = 40 * 2
e = c / d
|
a ) 8500 , b ) 3500 , c ) 4500 , d ) 5000 , e ) 1500 | d | divide(5000, const_1) | the total population of city a and city b is 5000 more than the total population of city b and city c . how much lesser is the population in city c as compared to city a ? | ( a + b ) - ( b + c ) = 5000 a - c = 5000 answer : d | a = 5000 / 1
|
a ) 15 cm , b ) 18 cm , c ) 21 cm , d ) 24 cm , e ) 27 cm | c | divide(294, const_10) | the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 294 sq . cm , what is the length of the rectangle ? | "2 l + 2 w = 5 w l = 3 w / 2 w * l = 294 3 w ^ 2 / 2 = 294 w ^ 2 = 196 w = 14 l = 3 ( 14 ) / 2 = 21 the answer is c ." | a = 294 / 10
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | d | divide(23, add(50, 14)) | some of the 50 % solution of acid was removed and this was replaced with an equal amount of 14 % solution of acid . as a result , a 23 % solution of acid was obtained . what fraction of the original solution was replaced ? | "let x be the fraction of the original solution that was replaced . 0.5 * ( 1 - x ) + 0.14 ( x ) = 0.23 0.36 x = 0.27 x = 3 / 4 the answer is d ." | a = 50 + 14
b = 23 / a
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | a | multiply(1, 7) | if 7 spiders make 7 webs in 7 days , then 1 spider will make 1 web in how many days ? | let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 7 webs 7 : 1 1 x 7 x x = 7 x 1 x 7 = > x = 7 answer is 7 | a = 1 * 7
|
a ) 75.56 % , b ) 66.66 % , c ) 70.05 % , d ) 80 % , e ) 85 % | b | multiply(const_100, divide(40, subtract(const_100, 40))) | if p ' s height is 40 % less than that of q , how much percent q ' s height is more than that of p ? | excess of q ' s height over p = ( 40 / ( 100 - 40 ) ) * 100 = = > 66.66 % answer b | a = 100 - 40
b = 40 / a
c = 100 * b
|
a ) 6 days , b ) 4 days , c ) 5 days , d ) 3 days , e ) 7 days | c | divide(subtract(const_1, add(multiply(divide(const_1, 12), const_2), multiply(divide(const_1, 20), const_2))), divide(const_1, 20)) | a can finish a piece of work in 12 days . b can do it in 20 days . they work together for 9 days and then a goes away . in how many days will b finish the work ? | "9 / 12 + ( 9 + x ) / 20 = 1 = > x = 5 days answer : c" | a = 1 / 12
b = a * 2
c = 1 / 20
d = c * 2
e = b + d
f = 1 - e
g = 1 / 20
h = f / g
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | b | divide(add(sqrt(add(multiply(multiply(120, const_2), const_4), const_1)), const_1), const_2) | if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 120 games will be played during the tournament . find the number of participants . | "let p be the number of participants . pc 2 = 120 ( p ) ( p - 1 ) = 240 = 16 * 15 p = 16 the answer is b ." | a = 120 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
a ) $ 140 , b ) $ 100 , c ) $ 220 , d ) $ 240 , e ) $ 260 | b | subtract(multiply(add(multiply(const_10, add(const_3, const_2)), const_2), 10), multiply(35, const_12)) | a parking garage rents parking spaces for $ 10 per week or $ 35 per month . how much does a person save in a year by renting by the month rather than by the week ? | 10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 35 = 420 $ 520 - 420 = 100 ans b | a = 3 + 2
b = 10 * a
c = b + 2
d = c * 10
e = 35 * 12
f = d - e
|
a ) rs . 962 , b ) rs . 672 , c ) rs . 546 , d ) rs . 10800 , e ) none of these | d | multiply(divide(surface_cube(30), 20), 40) | the cost of the paint is rs . 40 per kg . if 1 kg of paint covers 20 sq . ft , how much will it cost to paint outside of a cube having 30 feet each side | "explanation : surface area of a cube = 6 x 30 ^ 2 = 5400 sq . ft quantity of paint required = ( 5400 / 20 ) = 270 kg cost of painting = 40 x 270 = rs . 10800 answer : d" | a = surface_cube / (
b = a * 20
|
['a ) 378', 'b ) 278', 'c ) 258', 'd ) 287', 'e ) 271'] | c | divide(multiply(multiply(subtract(add(55, 35), 4), const_4), 75), const_100) | a rectangular lawn 55 m by 35 m has two roads each 4 m wide running in the middle of it . one parallel to the length and the other parallel to breadth . the cost of graveling the roads at 75 paise per sq meter is | explanation : area of cross roads = 55 * 4 + 35 * 4 - 4 * 4 = 344 sq m cost of graveling = 344 * ( 75 / 100 ) = rs . 258 answer : c ) rs . 258 | a = 55 + 35
b = a - 4
c = b * 4
d = c * 75
e = d / 100
|
a ) 14 , b ) 10 , c ) 12 , d ) 16 , e ) 20 | a | add(add(power(add(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2))) | the sum of four consecutive even numbers is 52 , find the third number ? | "let the first number be n let the second number be n + 2 let the third number be n + 4 let the third number be n + 6 n + n + 2 + n + 4 + n + 6 = 52 combine like terms : 4 n + 12 = 52 4 n + 12 = 52 4 n = 52 - 12 4 n = 40 n = 10 n + 2 = 12 n + 4 = 14 n + 6 = 16 the third number is 14 answer : a" | a = 52 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 52 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 52 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 52 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 1 minute , b ) 5 minutes , c ) 7 minutes , d ) 3 minutes , e ) 4 minutes | a | divide(60, add(36, divide(36, divide(const_3, const_2)))) | earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 60 envelopes | "earl takes 1 min . for 36 envelopes . ellen takes 3 / 2 mins for the same . so ellen can stuff ( ( 36 ) / ( 3 / 2 ) ) in 1 min . i . e . , 24 envlpes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 60 envelopes they will take 60 / 60 mins . i . e . , 1 mins . answer : a" | a = 3 / 2
b = 36 / a
c = 36 + b
d = 60 / c
|
a ) 17 , b ) 13 , c ) 15 , d ) 14 , e ) 16 | b | divide(multiply(1300, 4.15), 5395) | if 5395 / 4.15 = 1300 , then 539.5 / 41.5 is equal to ? | "answer given expression 539.5 / 41.5 = 5395 / 415 = 5395 / ( 415 x 100 ) = 1300 / 100 = 13 correct option : b" | a = 1300 * 4
b = a / 5395
|
a ) 8 % , b ) 12 % , c ) 16 % , d ) 20 % , e ) 24 % | b | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 80), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 80 % of calculus students are enrolled in the beginner course . if 4 / 5 of students are in the beginner courses , and one student is selected at random , what is the probability that an advanced trigonometry student is selected ? | "let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.8 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.8 x = 1.2 x the number of students in advanced trigonometry is 0.3 x the percentage in advanced trigonometry is 0.3 x / 2.5 x = 12 % the answer is b ." | a = 50 * 2
b = a + 50
c = 50 * 2
d = b + c
e = 4 / 5
f = d * e
g = f - 80
h = 50 * 2
i = h + 50
j = 50 * 2
k = i + j
l = g / k
m = 50 * 2
n = l * m
|
a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 24 | e | multiply(subtract(divide(30, 15), const_1), 24) | pipe a can fill a tank in 15 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | "let the pipe b be closed after x minutes . 30 / 15 - x / 24 = 1 = > x / 24 = 30 / 15 - 1 = 1 = > x = 1 * 24 = 24 . answer : e" | a = 30 / 15
b = a - 1
c = b * 24
|
a ) 292 , b ) 270 , c ) 192 , d ) 144 , e ) 168 | b | multiply(multiply(3, const_2), divide(subtract(242, add(divide(280, const_2), 12)), const_2)) | a train starts full of passengers at the first station it drops 1 / 3 of the passengers and takes 280 more at the second station it drops one half the new total and takes 12 more . on arriving at the third station it is found to have 242 passengers . find the no of passengers in the beginning ? | let no of passengers in the beginning be x after first station no passengers = ( x - x / 3 ) + 280 = 2 x / 3 + 280 after second station no passengers = 1 / 2 ( 2 x / 3 + 280 ) + 12 1 / 2 ( 2 x / 3 + 280 ) + 12 = 242 2 x / 3 + 280 = 2 * 230 = 460 - 280 = 180 2 x / 3 = 180 x = 270 . so the answer is option b ) 270 . | a = 3 * 2
b = 280 / 2
c = b + 12
d = 242 - c
e = d / 2
f = a * e
|
a ) 144 , b ) 255 , c ) 148 , d ) 265 , e ) 320 | a | divide(multiply(multiply(120, 6), 4), multiply(5, 4)) | a farming field can be ploughed by 6 tractors in 4 days . when 6 tractors work together , each of them ploughs 120 hectares a day . if two of the tractors were moved to another field , then the remaining 4 tractors could plough the same field in 5 days . how many hectares a day would one tractor plough then ? | if each of 66 tractors ploughed 120120 hectares a day and they finished the work in 44 days , then the whole field is : 120 β
6 β
4 = 720 β
4 = 2880120 β
6 β
4 = 720 β
4 = 2880 hectares . let ' s suppose that each of the four tractors ploughed xx hectares a day . therefore in 5 days they ploughed 5 β
4 β
x = 20 β
x 5 β
4 β
x = 20 β
x hectares , which equals the area of the whole field , 2880 hectares . so , we get 20 x = 288020 x = 2880 x = 288020 = 144 x = 288020 = 144 . hence , each of the four tractors would plough 144 hectares a day . correct answer is a ) 144 | a = 120 * 6
b = a * 4
c = 5 * 4
d = b / c
|
a ) 10 gallons , b ) 8.5 gallons , c ) 8 gallons , d ) 8.33 gallons , e ) 9.33 gallons | d | divide(subtract(multiply(divide(multiply(divide(20, const_100), 125), const_100), 125), 25), subtract(const_1, divide(multiply(divide(20, const_100), 125), const_100))) | a mixture of 125 gallons of wine and water contains 20 % of water . how much water must be added to the mixture in order to increase the percentage of water to 25 % of the new mixture ? | in 125 gallons of the solution there are 0.2 β 125 = 25 gallons of water . we want to add w gallons of water to 125 gallons of solution so that 25 + w gallons of water to be 25 % of new solution : 25 + w = 0.25 ( 125 + w ) - - > w = 253 β 8.33 . answer : d | a = 20 / 100
b = a * 125
c = b / 100
d = c * 125
e = d - 25
f = 20 / 100
g = f * 125
h = g / 100
i = 1 - h
j = e / i
|
a ) 1200 , b ) 3000 , c ) 1000 , d ) 3600 , e ) 1500 | e | divide(subtract(multiply(divide(6, const_100), 3000), multiply(3000, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | barbata invests $ 3000 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 ; 3000 + 0.05 * 3000 + x + 0.08 x = 3000 + x + 0.06 ( 3000 + x ) 0.05 * 3000 + 0.08 x = 0.06 x + 0.06 * 3000 0.02 x = 3000 ( 0.06 - 0.05 ) x = 3000 * 0.01 / 0.02 = 1500 ans : ` ` e ' '" | a = 6 / 100
b = a * 3000
c = 5 / 100
d = 3000 * c
e = b - d
f = 8 / 100
g = 6 / 100
h = f - g
i = e / h
|
a ) 60 , b ) 65 , c ) 69 , d ) 68.25 , e ) 67.5 | d | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 15), multiply(subtract(const_100, 15), divide(30, const_100)))), subtract(subtract(const_100, 15), multiply(subtract(const_100, 15), divide(30, const_100))))) | the price of a jacket is reduced by 15 % . during a special sale the price of the jacket is reduced another 30 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | 1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 15 % , therefore bringing down the price to $ 85 . 3 ) again it is further discounted by 30 % , therefore bringing down the price to $ 59.5 4 ) now 59.5 has to be added byx % in order to equal the original price . 59.5 + ( x % ) 59.5 = 100 . solving this eq for x , we get x = 68.25 ans is d | a = 100 - 15
b = 100 - 15
c = 30 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 15
h = 100 - 15
i = 30 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 910 , b ) 920 , c ) 930 , d ) 940 , e ) 950 | e | divide(add(800, multiply(700, 7)), const_3) | village p β s population is 800 greater than village q ' s population . if village q β s population were reduced by 700 people , then village p β s population would be 7 times as large as village q ' s population . what is village q ' s current population ? | "p = q + 800 . p = 7 ( q - 700 ) . 7 ( q - 700 ) = q + 800 . 6 q = 5700 . q = 950 . the answer is e ." | a = 700 * 7
b = 800 + a
c = b / 3
|
a ) 13 , b ) 12 , c ) 9 , d ) 8 , e ) 17 | e | divide(subtract(subtract(divide(56, const_2), 5), subtract(56, add(divide(56, const_2), divide(56, 8)))), const_2) | half of the workers in palabras bookstore have read the latest book by j . saramago , and 5 / 8 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are 56 workers in the palabras bookstore , how many of them have read both books ? | "there are total 56 workers . half of the workers in palabras bookstore have read the latest book by j . saramago , so 28 have read saramago . 5 / 8 of the workers have read the latest book by h . kureishi . so ( 5 / 8 ) * 56 = 35 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books , 20 - b have read saramago but not kureishi . so , ( 28 - b - 1 ) have read neither . total = n ( a ) + n ( b ) - both + neither 56 = 28 + 35 - b + ( 28 - b - 1 ) b = 17 answer ( e )" | a = 56 / 2
b = a - 5
c = 56 / 2
d = 56 / 8
e = c + d
f = 56 - e
g = b - f
h = g / 2
|
a ) 4 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | a | multiply(4, const_1) | for integers x and y , if ( x + 2 ) ( y + 3 ) is even then 4 xy must be divisible by : | just looking at the question , ( x + 2 ) ( y + 3 ) could be even * even , odd * even , or even * odd in order for the whole expression to be even . however , just looking at 4 xy , since x * y will always be an integer , it will always be divisible by 4 . 4 divided by 4 is 1 , hence preserving x * y . answer : a . 4 | a = 4 * 1
|
a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | e | multiply(divide(multiply(1, 10), subtract(3, 1)), 3) | two integers are in the ratio of 1 to 3 . if 10 is added to the smaller number , the ratio becomes 1 to 1 . find the larger integer . | "one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 3 , then 3 x = y , where x is the smaller integer . if adding 10 to the smaller integer makes the ratio 1 to 1 , then x + 10 = y . substituting y = 3 x into the second equation yields x + 10 = 3 x . so , x = 5 ( smaller integer ) and , y = 3 x = 15 ( larger integer ) so e is the correct answer . another option is to test the answer choices . a ) the larger number is 6 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 2 if 10 is added to the smaller number , the ratio becomes 1 to 1 if we add 10 to the smaller number ( 2 ) , we get 12 . so , the new ratio is 12 to 6 which is the same as 2 to 1 . no good . we want a resulting ratio of 1 to 1 eliminate a b ) the larger number is 9 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 3 if 10 is added to the smaller number , the ratio becomes 1 to 1 if we add 10 to the smaller number ( 3 ) , we get 13 so , the new ratio is 13 to 9 . no good . we want a resulting ratio of 1 to 1 eliminate b c ) the larger number is 10 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 3.33 ( which is not an integer ) since both numbers are integers , the larger number can not be 10 . eliminate c d ) the larger number is 12 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 4 if 10 is added to the smaller number , the ratio becomes 1 to 1 if we add 10 to the smaller number ( 4 ) , we get 14 so , the new ratio is 14 to 12 which is 7 to 6 . no good . we want a resulting ratio of 1 to 1 eliminate d e ) the larger number is 15 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 5 if 10 is added to the smaller number , the ratio becomes 1 to 1 if we add 10 to the smaller number ( 5 ) , we get 15 so , the new ratio is 15 to 15 . this is the same as the ratio of 1 to 1 therefore e is the only correct choice ." | a = 1 * 10
b = 3 - 1
c = a / b
d = c * 3
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | e | add(divide(100, 9), const_1) | how many integers are divisible by 9 between 10 ! and 10 ! + 100 inclusive ? | "e - 7 10 ! is divisible by 9 there are 11 numbers between 10 ! and 10 ! + 100 that are divisible by 9 . hence 10" | a = 100 / 9
b = a + 1
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a ) 13 , b ) 12 , 45 , c ) 1 , 25 , d ) 12 , 34 , e ) all of these | d | min(1, 1) | if x and y are multiples of 2 then which are all the multiples of 2 1 ) x * y 2 ) x ^ y 3 ) x + y 4 ) x - y 5 ) x / y | x * y , x ^ y , x + y , x - y are even numbers answer : d | a = min(1)
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a ) 150 , b ) 350 , c ) 300 , d ) 370 , e ) 280 | b | multiply(divide(490, const_3), const_2.0) | the ratio of number of boys and girls in a school is 2 : 5 . if there are 490 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 2 x and 5 x total students = 490 number of girls in the school = 5 * 490 / 7 = 350 answer is b" | a = 490 / 3
b = a * 2
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a ) 150 cms , b ) 151 cms , c ) 152 cms , d ) 154 cms , e ) 156 cms | d | divide(add(multiply(152, 15), multiply(156, const_10)), 30) | the average height of 15 girls out of a class of 30 is 152 cm . and that of the remaining girls is 156 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 15 Γ 152 + 15 Γ 156 / 30 ) = 154 cms answer d" | a = 152 * 15
b = 156 * 10
c = a + b
d = c / 30
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a ) 25 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 100 % | d | multiply(divide(divide(3, 5), divide(6, 7)), const_100) | a cylinder of height h is 6 / 7 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 6 / 7 new cylinder 3 / 5 so 3 / 5 / 6 / 7 = 3 / 5 * 7 / 6 = 0.70 or 70 % . answer d" | a = 3 / 5
b = 6 / 7
c = a / b
d = c * 100
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a ) s . 500 , b ) s . 600 , c ) s . 700 , d ) s . 800 , e ) s . 900 | a | multiply(divide(divide(30, 1), 6), const_100) | a sum was put a simple interest at a certain rate for 1 years . had it been put at 6 % higher rate , it would have fetched rs . 30 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x Γ£ β ( r + 6 ) Γ£ β 1 / 100 Γ’ Λ β x Γ£ β r Γ£ β 1 / 100 = 30 Γ’ β‘ β x Γ£ β 6 Γ£ β 1 / 100 = 30 Γ’ β‘ β x Γ£ β 1 / 100 = 5 Γ’ β‘ β x = 500 answer : option a" | a = 30 / 1
b = a / 6
c = b * 100
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a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 % | c | multiply(multiply(divide(divide(divide(2, 7), 2), add(divide(divide(2, 7), 2), subtract(const_1, divide(2, 7)))), const_100), const_3) | of the 19,210 employees of the anvil factory , 2 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ? | "the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 / 7 portion represents ( 14210 / 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 / 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 / 7 of the total workforce . if 1 / 7 of the workforce is journeymen , and 1 / 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 / 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 / 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 / 7 of it because 1 / 7 was eliminated . the remaining workers now account for 1 / 7 out of 6 / 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 33 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 / 12,180 is exactly 1 / 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you β re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . c" | a = 2 / 7
b = a / 2
c = 2 / 7
d = c / 2
e = 2 / 7
f = 1 - e
g = d + f
h = b / g
i = h * 100
j = i * 3
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a ) 25 sec , b ) 30 sec , c ) 48 sec , d ) 36 sec , e ) 40 sec | c | divide(add(1.10, 0.9), multiply(add(const_100.0, 90), const_0_2778)) | two trains are moving in opposite directions @ 60 kmph and 90 kmph . their length are 1.10 km and 0.9 km . the time taken by the slower train to cross faster train in second is ? | "relative speed = 60 + 90 = 150 * 5 / 18 = 125 / 3 m / s distance covered = 1.10 + 0.9 = 2 km = 2000 m required time = 2000 * 3 / 125 = 48 sec answer is c" | a = 1 + 10
b = 100 + 0
c = b * const_0_2778
d = a / c
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a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 11 | a | add(subtract(divide(67, 5), 9), const_1) | how many integers are between 9 and 67 / 5 , inclusive ? | "67 / 5 = 13 . xx we are not concerned about the exact value of 67 / 5 as we just need the integers . since the values are small , we can write down the integers . the different integers between 9 and 67 / 5 would be 9 , 10 , 11 , 12,13 total number of integers = 5 option a" | a = 67 / 5
b = a - 9
c = b + 1
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a ) 35 , b ) 36 , c ) 21 , d ) 32 , e ) 26 | a | add(36, const_1) | total 36 matches are conducted in knockout match type . how many players will be participated in that tournament ? | "35 players answer : a" | a = 36 + 1
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a ) 1 / 2 , b ) 3 / 7 , c ) 4 / 9 , d ) 3 / 8 , e ) 4 / 7 | a | divide(multiply(divide(3, 4), 8), 12) | a pipe can empty 3 / 4 th of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | 3 / 4 - - - - 12 ? - - - - - 8 = = > 1 / 2 a | a = 3 / 4
b = a * 8
c = b / 12
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a ) 8 , b ) 30 , c ) 3 , d ) 7 , e ) 10 | b | divide(divide(360, 2), 6) | benny goes to the market for buying some apples to be distributed between her 9 kids equally . she takes 360 dollars with her . the cost of each apple is 2 dollars . how many apples does she buy to share them equally between her 6 kids ? | cost of each apple = 2 dollars apples that benny can buy with the amount she has = 360 / 2 = 180 . apples that each kid gets evenly = 180 / 6 = 30 apples . so the answer is b = 30 | a = 360 / 2
b = a / 6
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a ) 7787 , b ) 8000 , c ) 15000 , d ) 1277 , e ) 2081 | c | divide(divide(14784, subtract(const_1, divide(12, const_100))), add(const_1, divide(12, const_100))) | in one year , the population , of a village increased by 12 % and in the next year , it decreased by 12 % . if at the end of 2 nd year , the population was 14784 , what was it in the beginning ? | "x * 112 / 100 * 88 / 100 = 14784 x * 0.9856 = 14784 x = 14784 / 0.9856 = > 15000 answer : c" | a = 12 / 100
b = 1 - a
c = 14784 / b
d = 12 / 100
e = 1 + d
f = c / e
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a ) 4016 , b ) 4076 , c ) 4816 , d ) 4916 , e ) 5916 | d | subtract(multiply(add(5, const_1), 5750), add(add(add(add(5266, 5744), 5864), 6122), 6588)) | a grocer has a sale of rs . 5266 , rs . 5744 , rs . 5864 , rs . 6122 and rs . 6588 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5750 ? | "total sale for 5 months = rs . ( 5266 + 5744 + 5864 + 6122 + 6588 ) = rs . 29584 . required sale = rs . [ ( 5750 x 6 ) - 34500 ] = rs . ( 34500 - 29584 ) = rs . 4916 . answer : d" | a = 5 + 1
b = a * 5750
c = 5266 + 5744
d = c + 5864
e = d + 6122
f = e + 6588
g = b - f
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a ) s . 43000 , b ) s . 19000 , c ) s . 57000 , d ) s . 60000 , e ) s . 62000 | b | add(add(multiply(multiply(multiply(20, 15), const_2), 20), multiply(multiply(multiply(20, 5), const_2), 20)), multiply(multiply(multiply(15, 5), const_2), 20)) | in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 20 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 20 * 950 = rs . 19000 b" | a = 20 * 15
b = a * 2
c = b * 20
d = 20 * 5
e = d * 2
f = e * 20
g = c + f
h = 15 * 5
i = h * 2
j = i * 20
k = g + j
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a ) 1 / 250 , b ) 1 / 84 , c ) 11 / 76 , d ) 1 / 9 , e ) 1 / 3 | c | divide(multiply(84, subtract(84, const_1)), multiply(220, subtract(220, const_1))) | a shipment of 220 smartphones contains 84 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ? | "probability of chosing one defective phone from a lot of 220 which ontains 84 defective phones is = ( 84 / 220 ) probability of chosing one defective phone from a lot of 219 ( we already picked one ) which ontains 83 ( we already picked one ) defective phones is = ( 83 / 219 ) combined probability of series of events = product of the probabilities = ( 84 / 220 ) * ( 83 / 219 ) 84 / 220 is close to ( 21 / 55 ) and ( 83 / 219 ) = ( 36 / 95 ) so answer is ( 21 / 55 ) * ( 36 / 95 ) = ( 11 / 76 ) so , answer will be c" | a = 84 - 1
b = 84 * a
c = 220 - 1
d = 220 * c
e = b / d
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a ) 775 , b ) 620 , c ) 53 , d ) 530 , e ) 220 | a | divide(600, subtract(const_1, divide(20, const_100))) | after decreasing 20 % in the price of an article costs rs . 600 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 600 cp = 7.75 * 100 = > cp = 775 answer : a" | a = 20 / 100
b = 1 - a
c = 600 / b
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a ) 25 , b ) 27 , c ) 30 , d ) 35 , e ) 42 | c | divide(add(multiply(2, 27), multiply(3, 32)), 5) | the average score of a cricketer in 2 matches is 27 and in other 3 matches is 32 . then find the average score in all the 5 matches ? | "average in 5 matches = ( 2 * 27 + 3 * 32 ) / 2 + 3 = 54 + 96 / 5 = 30 answer is c" | a = 2 * 27
b = 3 * 32
c = a + b
d = c / 5
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a ) 1 : 3 , b ) 4 : 5 , c ) 1 : 5 , d ) 2 : 5 , e ) 3 : 5 | e | divide(subtract(const_100, 6), add(const_100, 18)) | the number of singles that a baseball player hit increased 18 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 6 percent . if the total number of singles and doubles that she hit increased 3 percent from her first season to her second season , what is the ratio of the number of singles she hit in her first season to the number of doubles she hit in her first season ? | "soln : - 1.18 s + 0.94 d = 1.03 [ s + d ] 0.15 s = 0.09 d s / d = 3 / 5 answer : e" | a = 100 - 6
b = 100 + 18
c = a / b
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a ) - 5 , b ) - 6 , c ) - 7 , d ) - 8 , e ) - 10 | d | floor(divide(19, 3)) | if a and d are both integers , a > d , and - 3 c > 19 , then the largest value of d would be ? | no , your thinking is incorrect . when we know that a > d and a < - 6.33 , the largest value of a can be - 7 while if a = - 7 , then largest value of d < - 7 will be - 8 . for negative numbers , - 7 > - 8 and - 8 > - 10 . you are right in saying that d can take any value less than - 7 - - - > d could be - 8 , - 9 , - 10 . . . . and out of all these values , - 8 is the greatest . look at the numbers on the number line . for any 2 numbers , the ones on the right are greater than the ones on the left : . . . . . . . - 11 - 10 - 9 - 8 - 7 - 6 - 5 . . . . . 0 1 2 3 4 5 6 . . . ( - 11 < - 10 , - 10 < - 8 , 4 < 5 etc ) . so , as per the question if d < a and a = - 7 , then d ' s largest ' possible ' value has to be - 8 . - 10 is smaller than - 8 = d | a = 19 / 3
b = math.floor(a)
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a ) 9 square inches , b ) 10 square inches , c ) 15 square inches , d ) 25 square inches , e ) 350 square inches | a | subtract(multiply(power(9, const_2), const_3), add(add(power(6, const_2), power(9, const_2)), add(power(6, const_2), power(9, const_2)))) | huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 6 inches on a side and costs $ 10 , and a large pizza that measures 9 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ? | "in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 6 * 6 = 36 ) for the small pizza and ( 9 * 9 = 81 ) for the large pizza . in total sq inches that would be ( 36 + 81 ) * 2 = 234 sq inches . in the second case if they pool their money together they can buy 3 large pizzas . in terms of square inches that would be 3 * 81 = 243 sq inches . hence , the difference is 9 square inches more ( 243 - 234 ) . the correct answer is a" | a = 9 ** 2
b = a * 3
c = 6 ** 2
d = 9 ** 2
e = c + d
f = 6 ** 2
g = 9 ** 2
h = f + g
i = e + h
j = b - i
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a ) 12.6 , b ) 1.21 , c ) 1.25 , d ) 12.4 , e ) 12.7 | c | multiply(divide(divide(multiply(divide(20, const_100), 100), 20), multiply(divide(20, const_100), 100)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 20 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 20 ? - - - 1 = > rs . 1 100 - - - 80 ? - - - 1 = > rs . 1.25 answer : c" | a = 20 / 100
b = a * 100
c = b / 20
d = 20 / 100
e = d * 100
f = c / e
g = f * 100
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a ) 52 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 12) | john found that the average of 15 numbers is 40 . if 12 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 52 option a" | a = 40 + 12
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a ) 0.35 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | b | inverse(subtract(inverse(2), inverse(22))) | a and b together can do a piece of work in 2 days . if a alone can do the same work in 22 days , then b alone can do the same work in ? | "b = 1 / 2 β 1 / 22 = 0.45 days answer : b" | a = 1/(2)
b = 1/(22)
c = a - b
d = 1/(c)
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a ) 2 , b ) 3 , c ) - 1 , d ) - 2 , e ) - 3 | e | subtract(add(1, 1), 1) | if 1 / ( x + 3 ) + 1 / ( x - 3 ) = 1 / ( x + 3 ) , what is the value of x ? | "if we solve the question , we get x = - 3 . option : e" | a = 1 + 1
b = a - 1
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a ) 12 h , b ) 10 h , c ) 8 h , d ) 6 h , e ) none of these | b | divide(multiply(const_4, sqrt(1600)), 16) | the area of a square field is 1600 km 2 . how long will it take for a horse to run around at the speed of 16 km / h ? | explanation area of field = 1600 km 2 . then , each side of field = β 1600 = 40 km distance covered by the horse = perimeter of square field = 40 Γ 4 = 160 km β΄ time taken by horse = distances / peed = 160 / 16 = 10 h answer b | a = math.sqrt(1600)
b = 4 * a
c = b / 16
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a ) 400 , b ) 1728 , c ) 108 , d ) 216 , e ) 432 | d | multiply(divide(multiply(9, subtract(9, 1)), const_2), divide(multiply(4, const_3), const_2)) | how many 4 - digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ? . | the number of ways u can chose two digits from 9 digits is 9 c 2 = ( 9 * 8 ) / 2 . and the no of ways u can arrange two digits like asked is 4 c 2 = ( 4 * 3 ) / 2 . total ways is 36 * 6 = 216 . answer : d | a = 9 - 1
b = 9 * a
c = b / 2
d = 4 * 3
e = d / 2
f = c * e
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a ) 45 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | d | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 50), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 50), subtract(const_100, 10)), const_100)), const_100) | the charge for a single room at hotel p is 50 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | p = 0.5 r = 0.9 g r = 0.9 g / 0.5 = 1.8 * g thus r is 80 % greater than g . the answer is d . | a = 100 - 50
b = 100 - 10
c = a / b
d = c * 100
e = 100 - d
f = 100 - 50
g = 100 - 10
h = f / g
i = h * 100
j = e / i
k = j * 100
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a ) 0.0003 , b ) 0.03 , c ) 0.3 , d ) 3 , e ) 30 | c | divide(0.009, 0.03) | 0.009 / x = 0.03 . find the value of x | "x = 0.009 / 0.03 = 0.3 answer : c" | a = 0 / 9
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a ) 76 kmph , b ) 6 kmph , c ) 3.5 kmph , d ) 4.5 kmph , e ) 4 kmph | c | divide(subtract(divide(90, 3), divide(70, 3)), const_2) | a man rows his boat 90 km downstream and 70 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 70 / ( 3 ) = 23 kmph the speed of the stream = ( 30 - 23 ) / 2 = 3.5 kmph answer : c" | a = 90 / 3
b = 70 / 3
c = a - b
d = c / 2
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a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | d | divide(divide(180, const_2), divide(18, const_100)) | one night 18 percent of the female officers on a police force were on duty . if 180 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 180 number of female officers on duty on that night = 180 / 2 = 90 ( 18 / 100 ) * f = 90 = > f = 500 answer d" | a = 180 / 2
b = 18 / 100
c = a / b
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(multiply(3, 24), 24), subtract(multiply(3, 3), 1)) | there are 24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted 3 roses , every three boys planted 1 birch . by the end of the day they planted 24 plants . how many birches and roses were planted ? | "let x be the number of roses . then the number of birches is 24 β x , and the number of boys is 3 Γ ( 24 β x ) . if each girl planted 3 roses , there are x 3 girls in the class . we know that there are 24 students in the class . therefore x 3 + 3 ( 24 β x ) = 24 x + 9 ( 24 β x ) = 3 β
24 x + 216 β 9 x = 72 216 β 72 = 8 x 1448 = x x = 18 so , students planted 18 roses and 24 - x = 24 - 18 = 6 birches . so answer is d ." | a = 3 * 24
b = a - 24
c = 3 * 3
d = c - 1
e = b / d
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a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,900 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,900 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | "median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.9 so , answer is d ." | a = 100 * 100
b = a * 4
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a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 2 | d | divide(400, 1200) | if shares of two persons in profits are rs . 400 and rs . 1200 then ratio of their capitals is | profit = time * capital so 400 : 1200 = 1 : 3 answer : d | a = 400 / 1200
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a ) 50 , b ) 54 , c ) 47 , d ) 40 , e ) 45 | d | divide(multiply(560, 240), multiply(60, 56)) | rectangular tile each of size 60 cm by 56 cm must be laid horizontally on a rectangular floor of size 560 cm by 240 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 60 * 56 = 3360 area of floor = 560 * 240 = 134400 no of tiles = 134400 / 3360 = 40 so , the no of tile = 40 answer : d" | a = 560 * 240
b = 60 * 56
c = a / b
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a ) 10 , b ) 46 , c ) 70 , d ) 90 , e ) none | a | divide(multiply(5, 495), 10) | the l . c . m of two numbers is 495 and their h . c . f is 5 . if the sum of the numbers is 10 , then their difference is | "solution let the numbers be x and ( 100 - x ) . then , x ( 100 - x ) = 5 Γ 495 βΉ = βΊ x 2 - 100 x + 2475 = 0 . βΉ = βΊ ( x - 55 ) ( x - 45 ) = 0 . x = 55 or x = 45 . therefore , the numbers are 45 and 55 . required difference = ( 55 - 45 ) = 10 . answer a" | a = 5 * 495
b = a / 10
|
a ) 20 , b ) 150 , c ) 225 , d ) 304 , e ) 320 | d | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), const_4)) | how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 9 can be used only once ? | 4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with two 9 - s , 9 x 9 x 4 * 1 * 4 * 1 = 16 . thus there are 320 - 16 = 304 such numbers . answer : d . | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 * 4
e = c * d
f = 4 + 1
g = 4 + 1
h = f * g
i = h * 4
j = e - i
|
a ) 25 , b ) 66 , c ) 77 , d ) 88 , e ) 75 | e | divide(multiply(45, 5), const_4) | there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 45 , then find the smaller number ? | "let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 45 3 x = 45 = > x = 15 = > smaller number = 5 x = 75 . answer : e" | a = 45 * 5
b = a / 4
|
a ) 8 , b ) 9 , c ) 10 , d ) 12 , e ) 17 | d | divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 36), 4) | the sum of ages of 4 children born 2 years different each is 36 years . what is the age of the elder child ? | "let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) = 36 4 x = 24 x = 6 . x + 6 = 6 + 6 = 12 answer : d" | a = 2 + 4
b = 2 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 36
g = f / 4
|
a ) 6 days , b ) 9 days , c ) 2 days , d ) 7 days , e ) 1 days | a | inverse(add(inverse(8), inverse(24))) | if ajay completes a certain work in 8 days and vijay can complete the same work in 24 days , find the time in which the work will be completed if both ajay and vijay work together . | explanation : on working together ajay and vijay can complete the work in ( 8 * 24 ) / ( 8 + 24 ) = 192 / 32 = 6 days answer : a | a = 1/(8)
b = 1/(24)
c = a + b
d = 1/(c)
|
a ) 6 cm , b ) 12.05 cm , c ) 11.25 cm , d ) 15.12 cm , e ) 20.62 cm | b | divide(volume_cube(15), multiply(20, 14)) | a cube of edge 15 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 14 cm , find the rise in water level ? | "increase in volume = volume of the cube = 15 * 15 * 15 cm ^ 3 rise in water level = volume / area = 15 * 15 * 15 / 20 * 14 = 12.05 cm answer is b" | a = volume_cube / (
|
a ) 18 , b ) 144 , c ) 104 , d ) 216 , e ) 250 | c | add(multiply(divide(360, 9), 1), divide(subtract(360, multiply(divide(360, 9), 1)), 5)) | joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 9 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 9 ) * 360 = 40 gallons . remaning paint = 320 gallons paint used in the second week = ( 1 / 5 ) * 320 = 64 gallons total paint used = 104 gallons . option c" | a = 360 / 9
b = a * 1
c = 360 / 9
d = c * 1
e = 360 - d
f = e / 5
g = b + f
|
a ) - 29 , b ) 29 , c ) - 30 , d ) 0 , e ) 60 | d | add(29, const_1) | what is the sum of 59 consecutive integers from - 29 inclusive , in a increasing order ? | "from - 29 to - 1 - - > 29 nos . zero - - > 1 number from + 1 to + 29 - - > 29 nos . when we add up nos . from - 29 to + 29 sum will be zero for total 59 nos . sum of these 59 nos . = 0 . d is the answer ." | a = 29 + 1
|
a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | a | divide(const_2, 8) | if an integer n is to be selected at random from 1 to 400 , inclusive , what is probability n ( n + 1 ) will be divisible by 8 ? | "for n ( n + 1 ) to be a multiple of 8 , either n or n + 1 has to be a multiple of 8 . thus n must be of the form 8 k or 8 k - 1 . the probability is 2 / 8 = 1 / 4 . the answer is a ." | a = 2 / 8
|
a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 2 : 1 , e ) 3 : 1 | d | divide(subtract(180, 170), subtract(185, 180)) | students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 185 cms . what was the ratio of men to women ? | "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 185 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 185 m ) / ( w + m ) = 180 170 w + 185 m = 180 w + 180 m 5 m = 10 w m = 2 w m / w = 2 / 1 the ratio of men to women is 2 to 1 . d" | a = 180 - 170
b = 185 - 180
c = a / b
|
a ) 8 pm , b ) 6 pm , c ) 7 pm , d ) 10 pm , e ) 5 pm | a | add(6, divide(18, add(4, 6))) | a and b start walking towards each other at 6 pm at speed of 4 kmph and 6 kmph . they were initially 18 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 18 / 6 + 3 = 18 / 9 = 2 hrs after 6 pm = 8 pm answer is a" | a = 4 + 6
b = 18 / a
c = 6 + b
|
a ) 560 , b ) 364 , c ) 320 , d ) 330 , e ) 360 | e | multiply(multiply(5, 4), multiply(9, 8)) | calculate the l . c . m of 5 / 6 , 4 / 7 , 9 / 10 , 8 / 9 is : | "required l . c . m = l . c . m . of 5 , 4 , 9 , 8 / h . c . f . of 6 , 7 , 10 , 9 = 360 / 1 = 360 answer is e" | a = 5 * 4
b = 9 * 8
c = a * b
|
a ) 560 , b ) 380 , c ) 900 , d ) 2560 , e ) 2520 | e | multiply(multiply(multiply(power(const_2, const_3), power(const_3, const_2)), 5), add(const_3, const_4)) | find the least positive integer divisible by each of the integers between 5 through 10 ( including 10 ) . | the integer should be divisible by : 5 , 6 , 7 , 89 , and 10 . the least common multiple of these integers is lcm = 420 . answer : 5 * 3 ^ 2 * 7 * 2 ^ 3 the answer is e ) | a = 2 ** 3
b = 3 ** 2
c = a * b
d = c * 5
e = 3 + 4
f = d * e
|
a ) 12 , b ) 13 , c ) 35 , d ) 38 , e ) 59 | d | add(25, 11) | a number when divided by a divisor leaves a remainder of 25 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 25 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 38 as d is also integer and 38 is a prime number , the d should be 38 to satisfy the above equation . hence answer is ' d '" | a = 25 + 11
|
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 22 | c | subtract(divide(1000, 50), const_1) | dexter and prexter are competing with each other in a friendly community competition in a pool of 50 m length and the race is for 1000 m . dexter crosses 50 m in 2 min and prexter in 3 min 15 sec . each time they meet / cross each other , they do handshake ' s . how many such handshake ' s will happen if they start from the same end at the same time ? | explanation : when dexter completes second round , they do handshake once . now for every round which dexter completes , there will be one hand shake as the ratio of speed is 13 : 8 . d and p will meet at the pool end only after d completes 26 rounds . in the 20 th round , d finish the race and the total handshake ' s will be : 20 β 1 = 19 . answer : c | a = 1000 / 50
b = a - 1
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | b | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 6 and 34 which are divisible by 5 | "explanation : average = ( 10 + 15 + 20 + 25 + 305 ) = 1005 = 20 answer : option b" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
a ) 8 % , b ) 10 % , c ) 33 % , d ) 41 % , e ) 42 % | c | subtract(multiply(divide(subtract(37, 35), subtract(const_100, 35)), const_100), const_10) | two numbers are less than third number by 35 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 35 ) % of x = 65 % of x = 65 x / 100 second number is ( 63 x / 100 ) difference = 65 x / 100 - 63 x / 100 = x / 50 so required percentage is , difference is what percent of first number ( x / 50 * 100 / 65 x * 100 ) % = 33 % answer : c" | a = 37 - 35
b = 100 - 35
c = a / b
d = c * 100
e = d - 10
|
a ) $ 70 , b ) $ 72 , c ) $ 73 , d ) $ 74 , e ) $ 75 | e | add(multiply(add(multiply(add(divide(1, 2), add(divide(1, 4), divide(1, 3))), 15), 15), 2), const_10) | sam β s car was fined when he gave joe and peter a ride , so they decided to help sam pay the fine . joe paid $ 8 more than 1 / 4 of the fine and peter paid $ 8 less than 1 / 3 of the fine , leaving pay $ 15 less than 1 / 2 the fine to complete the payment . what fraction of the fine did sam pay ? | "call the fine f . joe paid ( 1 / 4 ) f + 4 and peter paid ( 1 / 3 ) f β 4 , leaving ( 1 / 2 ) f β 5 left . if we add those three up , they should add up to f . f = [ ( 1 / 4 ) f + 8 ] + [ ( 1 / 3 ) f β 8 ] + [ ( 1 / 2 ) f β 15 ] f = ( 1 / 4 ) f + ( 1 / 3 ) f + ( 1 / 2 ) f β 15 multiply all terms by 12 to clear the fractions . 12 f = 3 f + 4 f + 6 f β 180 12 f = 13 f β 180 β f = β 180 f = 180 well , if the fine cost $ 180 , then sam paid the part not covered by joe or peter . half the fine is $ 90 , and sam paid $ 15 less than this : $ 75 answer = e" | a = 1 / 2
b = 1 / 4
c = 1 / 3
d = b + c
e = a + d
f = e * 15
g = f + 15
h = g * 2
i = h + 10
|
a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | c | add(150, multiply(3, const_10)) | according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 12 ounces of water to make a smoothie . how many 2 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounces of water , that would mean we would need 150 packets of smoothie mix ( along with 12 ( 150 ) ounces of water for a total of 150 packets . however , we know , after a more careful reading of the stimulus , that the 3 ounces are not included in the 12 ounces . as such , the answer has to be less than 150 packets , since 150 would be too much powder considering you already have 150 ( 12 ) ounces of water and need less packets than water to make a smoothie . as such , the only answer less than 150 is 120 , a . does this make sense ? or am i way off base ? c | a = 3 * 10
b = 150 + a
|
a ) 130 % , b ) 320 % , c ) 150 % , d ) 160 % , e ) 170 % | b | add(multiply(subtract(multiply(add(const_1, divide(60, const_100)), const_2), const_1), const_100), const_100) | a man gains 60 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be . | "explanation : let the c . p . = x , then s . p . = ( 160 / 100 ) x = 8 x / 5 new s . p . = 2 ( 8 x / 5 ) = 16 x / 5 profit = 16 x / 5 - x = 16 x / 5 profit % = ( profit / c . p . ) * 100 = > ( 16 x / 5 ) * ( 1 / x ) * 100 = 320 % option b" | a = 60 / 100
b = 1 + a
c = b * 2
d = c - 1
e = d * 100
f = e + 100
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(54, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 54 . what is the difference between the two digits of that number ? | sol . let the ten β s digit be x and unit β s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 54 β 9 ( x - y ) = 54 β x - y = 6 answer c | a = 10 - 1
b = 54 / a
|
a ) 500 , b ) 600 , c ) 650 , d ) 700 , e ) 720 | d | divide(divide(divide(divide(divide(volume_rectangular_prism(100, 70, 40), const_3), const_2), const_4), const_4.0), const_4) | the ratio , by volume , of soap to alcohol to water in a certain solution is 2 : 40 : 70 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water will it contain ? | "soap : alcohol initial ratio soap : alcohol : water - - > 2 : 40 : 70 initial soap : alcohol = 2 / 40 = 2 : 40 after doubled soap : alcohol = 2 * 2 / 40 = 4 : 40 initial soap : water = 2 / 70 = 2 : 70 after halved soap : water : 1 / 2 * 2 / 70 = 1 / 70 = 1 : 70 after soap : alcohol : water - - > 4 : 40 : 280 - - > 1 : 10 : 70 given alcohol 100 cubic centimeter . ratio is 10 : 100 : 700 ( 1 : 10 : 70 ) for 100 cubic centimeter of alcohol - - - 700 cubic cm water is required . answer - d" | a = volume_rectangular_prism / (
b = a / 3
c = b / 2
d = c / 4
e = d / 4
|
a ) 4857 , b ) 4184 , c ) 4012 , d ) 4991 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6435, 6927)), 7230), 6562)) | a grocery shop has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6435 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6435 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ 6500 = > 34009 + x = 39000 = > x = 39000 β 34009 = 4991 answer : d" | a = 5 + 1
b = 6500 * a
c = 6435 + 6927
d = 6855 + c
e = d + 7230
f = e + 6562
g = b - f
|
a ) s . 575 , b ) s . 595 , c ) s . 590 , d ) s . 600 , e ) s . 585 | d | subtract(multiply(subtract(540, 420), 6), subtract(540, 420)) | if rs . 420 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | 80 = ( 420 * 4 * r ) / 100 r = 7.14 % i = ( 420 * 6 * 7.14 ) / 100 = 180 420 + 180 = 600 answer : d | a = 540 - 420
b = a * 6
c = 540 - 420
d = b - c
|
a ) 3 : 7 , b ) 9 : 98 , c ) 3 : 21 , d ) 1 : 7 , e ) 3 : 49 | b | divide(3, const_60) | what is the ratio of 3 / 7 to the product 2 * ( 7 / 3 ) ? | "3 / 7 / 14 / 3 = 9 / 98 . . . imo option b ." | a = 3 / const_60
|
a ) 60 , b ) 50 , c ) 47 , d ) 55 , e ) 80 | a | add(divide(12, divide(subtract(40, 20), const_100)), 20) | if 40 % of a number exceeds 20 % of it by 12 , then find the number ? | "use the elimination method to find the correct option . of all the options only 60 fits 40 % of 60 = 24 20 % of 60 = 12 24 - 12 = 12 required number is 60 . answer : a" | a = 40 - 20
b = a / 100
c = 12 / b
d = c + 20
|
a ) $ 130.00 , b ) $ 160.75 , c ) $ 168.75 , d ) $ 182.00 , e ) $ 168.00 | c | multiply(300, divide(add(25, 25), const_100)) | a discount electronics store normally sells all merchandise at a discount of 5 percent to 25 percent off the suggested retail price . if , during a special sale , an additional 25 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 300 before any discount ? | "original price : 300 $ max first discount = - 25 % thus : 300 Γ’ Λ β ( 1 Γ’ Λ β 25 / 100 ) = 225 second discount on the discounted price = - 25 % thus : 225 Γ’ Λ β ( 1 Γ’ Λ β 25 / 100 ) = 168.75 answer c" | a = 25 + 25
b = a / 100
c = 300 * b
|
a ) 120 m , b ) 140 m , c ) 240 m , d ) 300 m , e ) none | a | divide(const_100.0, subtract(divide(const_100.0, 10), 6)) | a train covers a distance of 12 km in 10 minutes . if it takes 6 seconds to pass a telegraph post , then the length of the train is : | "sol . speed = [ 12 / 10 * 60 ] km / hr = [ 72 * 5 / 18 ] m / sec = 20 m / sec . length of the train = ( speed * time ) = ( 20 * 6 ) m = 120 m . answer a" | a = 100 / 0
b = a - 6
c = 100 / 0
|
a ) 220 , b ) 400 , c ) 500 , d ) 560 , e ) 590 | b | divide(multiply(multiply(subtract(3.20, 3), const_1000), const_100), 50) | workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.20 lacs . how many workers were they ? | "n * 50 = ( 320000 - 300000 ) = 20000 n = 20000 / 50 = 400 b" | a = 3 - 20
b = a * 1000
c = b * 100
d = c / 50
|
a ) 48 , b ) 66 , c ) none of these , d ) 78 , e ) 90 | c | subtract(multiply(multiply(multiply(5, 2), 1), const_2), multiply(multiply(multiply(2, 1), const_2), const_2)) | each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 1 and the 2 can not be adjacent digits in the integer , how many 5 - digit integers are possible ? | "number of arrangements using 5 distinct digits = 5 ! number of arrangements in which 1 and 2 are adjacent - consider 1 and 2 together as one group . now you have 4 numbers / groups to arrange which can be done in 4 ! ways . in each of these arrangements , 1 and 2 can be arranged as 12 or 21 . number of arrangements in which 3 and 4 are not adjacent = 5 ! - 2 * 4 ! = 72 answer : c" | a = 5 * 2
b = a * 1
c = b * 2
d = 2 * 1
e = d * 2
f = e * 2
g = c - f
|
a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 130)))), const_2) | if a rectangular billboard has an area of 130 square feet and a perimeter of 46 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 130 and a perimeter of 46 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 130 , the shorter side will almost certainly be closer to 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 130 , if . . . the shorter side = 10 . . . the area = 130 / 10 = 13 = the longer side perimeter = 10 + 10 + 13 + 13 = 46 d" | a = 46 / 2
b = 46 / 2
c = b ** 2
d = 4 * 130
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
|
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