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 ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 28 | b | multiply(2, add(divide(subtract(48, multiply(2, 2)), add(6, 2)), 2)) | lionel left his house and walked towards walt ' s house , 48 miles away . four hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 2 miles per hour and walt ' s 6 miles per hour , how many miles had lionel walked when he met walt ? | "in the first 4 hours lionel at the rate of 2 miles per hour covered distance = rate * time = 4 * 2 = 8 miles . so , the distance between him and walt was 48 - 8 = 40 miles when walt left his house . now , their combined rate to cover this distance was 2 + 6 = 8 miles per hour , hence they will meet ( they will cover that distance ) in time = distance / rate = 40 / 8 = 5 hours . total time that lionel was walking is 4 + 5 = 9 hours , which means that he covered in that time interval distance = rate * time = 2 * 9 = 18 miles . answer : b ." | a = 2 * 2
b = 48 - a
c = 6 + 2
d = b / c
e = d + 2
f = 2 * e
|
a ) 4032 , b ) 4000 , c ) 3250 , d ) 3825 , e ) 3985 | a | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(13860, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 13860 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 13860 ã — 16 / 55 = rs . 4032 . share of z in the profit is rs . 4032 . option a" | a = 3 * 4
b = a - 4
c = 48000 * b
d = 3 * 4
e = 36000 * d
f = 3 * 4
g = 42000 * f
h = e + g
i = 3 * 4
j = i - 4
k = 48000 * j
l = h + k
m = 13860 / l
n = c * m
|
a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | b | divide(subtract(280, multiply(divide(subtract(12, 3), const_2), 10)), add(divide(subtract(12, 3), const_2), add(divide(subtract(12, 3), const_2), 3))) | a cyclist traveled for two days . on the second day the cyclist traveled 3 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | "solution : d = 280 mi t = 12 hrs day 1 time = t 1 day 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs day 1 rate = r 1 day 2 rate = r 2 r 1 - r 2 = 10 mph i . e . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 i . e . 280 = 8 r 2 + 4 ( 10 + r 2 ) i . e . r 2 = 20 mph answer : b" | a = 12 - 3
b = a / 2
c = b * 10
d = 280 - c
e = 12 - 3
f = e / 2
g = 12 - 3
h = g / 2
i = h + 3
j = f + i
k = d / j
|
a ) 36 , b ) 60 , c ) 84 , d ) 72 , e ) 86 | b | add(multiply(factorial(subtract(5, const_2)), multiply(const_2, const_3)), multiply(const_4, factorial(subtract(5, const_2)))) | from the digits 2 , 3 , 4 , 5 , 6 and 7 , how many 5 - digit numbers can be formed that have distinct digits and are multiples of 12 ? | detailed solution any multiple of 12 should be a multiple of 4 and 3 . first , let us look at the constraint for a number being a multiple of 3 . sum of the digits should be a multiple of 3 . sum of all numbers from 2 to 7 is 27 . so , if we have to drop a digit and still retain a multiple of 3 , we should drop either 3 or 6 . so , the possible 5 digits are 2 , 4 , 5 , 6 , 7 or 2 , 3 , 4 , 5 , 7 . when the digits are 2 , 4 , 5 , 6 , 7 . the last two digits possible for the number to be a multiple of 4 are 24 , 64 , 52 , 72 , 56 , 76 . for each of these combinations , there are 6 different numbers possible . so , with this set of 5 digits we can have 36 different numbers . when the digits are 2 , 3 , 4 , 5 , 7 . the last two digits possible for the number to be a multiple of 4 are 32 , 52 , 72 , 24 . for each of these combinations , there are 6 different numbers possible . so , with this set of 5 digits we can have 24 different numbers . overall , there are 60 different 5 - digit numbers possible . correct answer : b | a = 5 - 2
b = math.factorial(a)
c = 2 * 3
d = b * c
e = 5 - 2
f = math.factorial(e)
g = 4 * f
h = d + g
|
a ) 672 , b ) 682 , c ) 328 , d ) 964 , e ) 285 | a | multiply(add(8,12, const_1), const_2) | what is the least common multiple of 8,12 , 32,14 | "the given numbers are 8,12 , 32,14 thus the l . c . m is 2 * 2 * 2 * 1 * 3 * 4 * 7 = 672 the answer is a" | a = 8 + 12
b = a * 2
|
a ) 6992 , b ) 6982 , c ) 7982 , d ) 7972 , e ) 7992 | e | subtract(multiply(8, const_1000), 8) | the difference between the local value and the face value of 8 in the numeral 96348621 is | "explanation : ( local value of 8 ) - ( face value of 8 ) = ( 8000 - 8 ) = 7992 e )" | a = 8 * 1000
b = a - 8
|
a ) 5 : 9 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) none of these | a | divide(subtract(6.30, 5.80), subtract(7.20, 6.30)) | find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.80 a kg to produce a mixture worth rs . 6.30 a kg ? | "solution required ratio = 50 : 90 = 5 : 9 answer a" | a = 6 - 30
b = 7 - 20
c = a / b
|
a ) 2 hours , b ) 12 hours , c ) 3 hours , d ) 5 hours , e ) 6 hours | b | add(add(4, 4), 4) | three pipes , a , b , & c are attached to a tank . a & b can fill it in 20 & 30 minutes respectively while c can empty it in 15 minutes . if a , b & c are kept open successively for 4 minute each , how soon will the tank be filled ? | in three minute 1 / 20 + 1 / 30 - 1 / 15 = 1 / 60 part is filled 12 min - - - - - - - - 1 / 60 parts x min - - - - - - - - - 1 part ( full ) x = 720 min = 12 hours answer : b | a = 4 + 4
b = a + 4
|
a ) - 1 , b ) 1 , c ) - 2 , d ) 2 , e ) - 3 | c | subtract(multiply(6, divide(subtract(multiply(2, 2), 4), subtract(multiply(7, 2), multiply(2, 8)))), add(multiply(8, divide(subtract(multiply(2, 2), 4), subtract(multiply(7, 2), multiply(2, 8)))), 2)) | when positive integer x is divided by 8 , the quotient is y and the remainder is 2 . when 2 x is divided by 7 , the quotient is 2 y and the remainder is 4 . what is the value of 6 y – x ? | "( 1 ) x = 8 y + 2 ( 2 ) 2 x = 14 y + 4 ( 2 ) - ( 1 ) : x = 6 y + 2 6 y - x = - 2 the answer is c ." | a = 2 * 2
b = a - 4
c = 7 * 2
d = 2 * 8
e = c - d
f = b / e
g = 6 * f
h = 2 * 2
i = h - 4
j = 7 * 2
k = 2 * 8
l = j - k
m = i / l
n = 8 * m
o = n + 2
p = g - o
|
a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 100 / 3 , e ) 80 / 3 | d | divide(log(32), log(power(2, 0.15))) | if n = 2 ^ 0.15 and n ^ b = 32 , b must equal | "15 / 100 = 3 / 20 n = 2 ^ 3 / 20 n ^ b = 2 ^ 5 ( 2 ^ 3 / 20 ) ^ b = 2 ^ 5 b = 100 / 3 answer : d" | a = math.log(32)
b = 2 ** 0
c = math.log(b)
d = a / c
|
a ) 21 , b ) 22 , c ) 20 , d ) 18 , e ) 23 | e | divide(subtract(25, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 25 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is | "explanation : let the son ' s present age be x years . then , man ' s present age = ( x + 25 ) years = > ( x + 25 ) + 2 = 2 ( x + 2 ) = > x + 27 = 2 x + 4 so , x = 23 . answer : e" | a = 2 * 2
b = a - 2
c = 25 - b
d = 2 - 1
e = c / d
|
a ) 80 % , b ) 107 % , c ) 116.7 % , d ) 120 % , e ) 140 % | b | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(10, subtract(10, const_1)), subtract(const_100, 10)), const_100), 5), 25)), const_100), multiply(subtract(const_1, divide(add(5, divide(multiply(add(10, const_4), subtract(const_100, 10)), const_100)), 20)), const_100)), const_100) | during a special promotion , a certain filling station is offering a 10 percent discount on gas purchased after the first 5 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella ‟ s total per - gallon discount is what percent of kim ‟ s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 5 + 0.9 * 15 = 18.5 gallons , so the overall discount she got was 1.5 / 20 = 7.5 % . isabella purchased 25 gallons of gas . she paid for 5 + 0.9 * 20 = 23 gallons , so the overall discount she got was . 5 / 25 = 8 % . 8 is 8 / 7.5 * 100 = 107 % of 7.5 . answer : b ." | a = 10 - 1
b = 10 + a
c = 100 - 10
d = b * c
e = d / 100
f = e + 5
g = f / 25
h = 1 - g
i = h * 100
j = 10 + 4
k = 100 - 10
l = j * k
m = l / 100
n = 5 + m
o = n / 20
p = 1 - o
q = p * 100
r = i / q
s = r * 100
|
a ) 10,000 , b ) 11,600 , c ) 12,000 , d ) 14,000 , e ) 16,400 | c | add(5, 6) | jerome anticipated that the webweb . com stock price would fall and sold all his webweb . com stocks for $ 5 per stock . he paid $ 10,000 tax on the revenue . a week later , jerome became convinced that the webweb . com stock price would rise , and he used the money that he had gotten from selling the webweb . com stocks to purchase them again , this time at $ 6 per stock . if jerome ended up having 2000 webweb . com stocks fewer than the number he had before selling them , how many webweb . com stocks did he have originally ? | let the number of shares be x . 5 * x - 10000 ( money paid in taxes ) = 6 ( x - 2000 ) solving for x , we get the shares as 12000 . ans : ( option c ) | a = 5 + 6
|
a ) 3 , b ) 6 , c ) 18 , d ) 12 , e ) 16 | c | multiply(const_3, divide(divide(54, const_3), const_3)) | if m is a positive integer and m ^ 2 is divisible by 54 , then the largest positive integer that must divide m is ? | m ^ 2 is divisible by 48 so m ^ 2 must be multiple of 48 . if the value of m is multiples of 18 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is c but if m = 54 then answer should be 16 . is the question right ? or am i missing some thing ? c | a = 54 / 3
b = a / 3
c = 3 * b
|
a ) 276 , b ) 350 , c ) 425 , d ) 512 , e ) 648 | a | divide(factorial(24), multiply(factorial(subtract(24, const_2)), factorial(const_2))) | if 24 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 24 lines is 24 c 2 = 276 the answer is a ." | a = math.factorial(24)
b = 24 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) p = 8 | e | divide(240, 30) | a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations p are possible ? | "the combinations could be p { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pairs , and we are left only with . { ( 8,30 , ( 10,24 ) , ( 12,20 ) , ( 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) } hence 8 . e" | a = 240 / 30
|
a ) 2 / 5 , b ) 3 / 5 , c ) 3 / 7 , d ) 4 / 7 , e ) 2 / 9 | e | divide(divide(multiply(4, 4), const_2), multiply(4, add(const_1, 4))) | in the x - y plane , there are 4 points ( 0,0 ) , ( 0,4 ) , ( 9,4 ) , and ( 9,0 ) . if these 4 points makes a rectangle , what is the probability that x + y < 4 ? | "the line y = - x + 4 intersects the rectangle and these three points of intersection ( 0,0 ) , ( 0,4 ) and ( 4,0 ) form a triangle . the points below the line y = - x + 4 satisfy x + y < 4 . the area of this triangle is ( 1 / 2 ) ( 4 ) ( 4 ) = 8 the area of the rectangle is 36 . p ( x + y < 4 ) = 8 / 36 = 2 / 9 the answer is e ." | a = 4 * 4
b = a / 2
c = 1 + 4
d = 4 * c
e = b / d
|
a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 . | d | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.2) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.2 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.2 = 6 * 0 + 2 * 7.2 + 8 - 6.2 = 0 + 14.4 + 1.8 16.2 answer d" | a = 2 * 7
b = 4 / 10
c = 8 - 4
d = a + c
e = d - 6
|
a ) 24.65 , b ) 22.27 , c ) 2.273 , d ) 11.62 , e ) 21.13 | c | inverse(subtract(divide(subtract(const_1, multiply(inverse(100), const_2)), subtract(subtract(multiply(6, const_2), 6), const_2)), inverse(100))) | machine a takes 100 hours to complete a certain job and starts that job at 6 am . after ten hour of working alone , machine a is joined by machine b and together they complete the job at 6 pm . how long would it have taken machine b to complete the job if it had worked alone for the entire job ? | "let us assume total job = 100 units a finishes 100 units in 100 hrs ( given ) hence a ( working rate ) = 1 units / hr now given that a works for 10 hr ( so 10 units done ) then a and b finish total work in 12 hours . hence a and b finish 90 units in 2 hours . of these 1 x 2 = 2 units were done by a . hence b did 88 units in 2 hours . hence b ( working rate ) = 44 units / hr hence b takes 100 / 44 = 2.273 hours to complete the job . answer c ." | a = 1/(100)
b = a * 2
c = 1 - b
d = 6 * 2
e = d - 6
f = e - 2
g = c / f
h = 1/(100)
i = g - h
j = 1/(i)
|
a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | d | add(sqrt(divide(subtract(power(12, const_2), power(const_2, const_2)), const_2)), const_2) | efrida and frazer who live 12 miles apart , meet at a restaurant that is directly north of efrida ' s home and directly east of frazer ' s home . if the restaurant is two miles closer to efrida ' s home , than to frazer ' s home , how many miles is the restaurant from frazer ' s home ? | "it ' s a simple geometry problem . look at the diagram below : answer : d ." | a = 12 ** 2
b = 2 ** 2
c = a - b
d = c / 2
e = math.sqrt(d)
f = e + 2
|
a ) 0.07 , b ) 0.049 , c ) 0.7 , d ) 0.49 , e ) 4.9 | b | divide(volume_cylinder(7, 10), circle_area(100)) | a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 7 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ? | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 49 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 490 / 10,000 = 49 / 1000 = 0.049 ft the answer is b ." | a = volume_cylinder / (
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 33 % , e ) 44 % | e | multiply(subtract(divide(add(const_100, 80), add(const_100, 25)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 25 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 80 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | "another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 80 = 25 + b + 25 b / 100 b = 44 answer ( e )" | a = 100 + 80
b = 100 + 25
c = a / b
d = c - 1
e = d * 100
|
a ) 8.33 % , b ) 9.33 % , c ) 10.00 % , d ) 11.11 % , e ) 7.56 % | a | subtract(const_100, divide(multiply(1300, const_100), 1200)) | an article is bought for rs . 1200 and sold for rs . 1300 , find the loss percent ? | "1200 - - - - 100 100 - - - - ? = > 16 2 / 3 % answer : a" | a = 1300 * 100
b = a / 1200
c = 100 - b
|
a ) 25 % , b ) 30 % , c ) 50 % , d ) 20 % , e ) 10 % | a | divide(multiply(20, const_100), subtract(100, 20)) | by selling an article for $ 100 a man gains $ 20 then his gain % ? | "s . p . = $ 100 gain = $ 20 c . p . = 100 - 20 = 80 gain % = 20 / 80 * 100 = 25 % answer is a" | a = 20 * 100
b = 100 - 20
c = a / b
|
a ) 20 , b ) 40 , c ) 50 , d ) 55 , e ) 70 | a | multiply(subtract(const_100, 80), subtract(divide(const_100, 50), divide(80, 80))) | john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 50 miles per hour , what is v in miles per hour ? | hibunuel the question seems incorrect . it should not be 80 % at the speed of 80 . however if it ' s 20 % at the speed of 80 , answer comes out 55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it , which comes as : 0.8 d / 80 + 0.2 d / v = d / 50 = > v = 20 ( option a ) . | a = 100 - 80
b = 100 / 50
c = 80 / 80
d = b - c
e = a * d
|
a ) s . 90 , b ) s . 120 , c ) s . 200 , d ) s . 50 , e ) s . 290 | d | subtract(multiply(10, divide(8000, const_100)), multiply(5, divide(15000, const_100))) | john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 5 % and the mobile phone at a profit of 10 % . overall how much he make a profit . | let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 5 / 100 ) = 15000 - 750 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 750 + 800 = rs . 50 as this is positive , an overall profit of rs . 50 was made . d | a = 8000 / 100
b = 10 * a
c = 15000 / 100
d = 5 * c
e = b - d
|
a ) s 282 , b ) s 216 , c ) s 220 , d ) s 210 , e ) s 217 | a | add(add(230, divide(500, 100)), multiply(divide(20, 100), add(230, divide(500, 100)))) | the manufacturing cost of a shoe is rs . 230 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 230 + ( 500 / 100 ) = 235 . gain = 20 % = > sp = 1.2 cp = 1.2 x 235 = 282 answer : a" | a = 500 / 100
b = 230 + a
c = 20 / 100
d = 500 / 100
e = 230 + d
f = c * e
g = b + f
|
a ) 28 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | e | divide(subtract(87, 15), const_2) | we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 15 , and the difference between the total number of strawberries and the total number of blueberries would increase by 87 . each blue box contains how many blueberries ? | "let x be the number of blueberries in each blue box . then there are x + 15 strawberries in each red box . x + ( x + 15 ) = 87 x = 36 the answer is e ." | a = 87 - 15
b = a / 2
|
a ) 1220 , b ) 1250 , c ) 1280 , d ) 1310 , e ) 1340 | b | divide(subtract(700, multiply(12.50, 40)), 0.16) | suppose you work for a manufacturing plant that pays you $ 12.50 an hour plus $ 0.16 for each widget you make . how many widgets must you produce in a 40 hour week to earn $ 700 ( before payroll deductions ) ? | "total pay = 40 * $ 12.50 + $ 0.16 * x = 700 x = 200 / 0.16 = 1250 the answer is b ." | a = 12 * 50
b = 700 - a
c = b / 0
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a ) 48 , b ) 52 , c ) 66 , d ) 68 , e ) 84 | d | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | how many positive integers less than 500 can be formed using the numbers 1 , 2 , 3 and 5 for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 number of possibilities = 4 * 4 * 4 = 64 . then , just add up the number of 1 digits numbers = 4 , so total is 64 + 4 = 68 . answer : d" | a = 4 + 1
b = a - 1
c = math.factorial(b)
d = math.factorial(1)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 1400 , b ) 141.71 , c ) 1210 , d ) 1511.71 , e ) 1500 | d | multiply(subtract(power(25, const_2), power(const_10, const_2)), divide(add(multiply(12, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 12 m to 25 m , how much additional grassy ground shall it graze ? | "π ( 252 – 122 ) = 1511.71 answer : d" | a = 25 ** 2
b = 10 ** 2
c = a - b
d = 12 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
a ) 10 , b ) 20 , c ) 30 , d ) 24 , e ) 25 | b | add(multiply(30, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 30), divide(const_1, 60)), const_2)), 30), const_60)) | two pipes a and b can fill a tank in 30 and 60 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 1 / 30 part filled by b in 1 min . = 1 / 60 part filled by ( a + b ) in 1 min . = 1 / 30 + 1 / 60 = 1 / 20 . both the pipes can fill the tank in 20 minutes . answer : b" | a = 30 * 100
b = 1 / 30
c = 1 / 60
d = b + c
e = d * 2
f = 1 - e
g = f * 30
h = g * const_60
i = a + h
|
a ) 8.56 , b ) 9.35 , c ) 9.43 , d ) 6.56 , e ) 5.5 | c | multiply(divide(const_1, multiply(add(const_100, 40), divide(const_1, subtract(const_100, 40)))), 22) | by selling 22 pencils for a rupee a man loses 40 % . how many for a rupee should he sell in order to gain 40 % ? | "60 % - - - 22 140 % - - - ? 60 / 140 * 22 = 9.43 answer : c" | a = 100 + 40
b = 100 - 40
c = 1 / b
d = a * c
e = 1 / d
f = e * 22
|
a ) 29997 , b ) 299997 , c ) 297 , d ) 0 , e ) none of them | b | divide(add(40378273, const_1), const_2) | the difference between the place values of two threes in the numerical 40378273 is | "required difference = ( 300000 - 3 ) = 299997 . answer is b" | a = 40378273 + 1
b = a / 2
|
a ) 8 , b ) 10 , c ) 9 , d ) 14 , e ) 16 | c | add(floor(divide(22, 3)), floor(divide(22, power(3, const_2)))) | if m = 3 ^ n , what is the greatest value of n for which m is a factor of 22 ! | "solution - consider multiples of 25 ! = > 3 , 6,9 , 12,15 , 18,21 count no . of 3 in each multiple . 3 = 3 x 1 - > 1 6 = 3 x 2 - > 1 9 = 3 x 3 - > 2 12 = 3 x 4 - > 1 15 = 3 x 5 - > 1 18 = 3 x 3 x 2 - > 2 21 = 3 x 7 - > 1 - - - - count 3 ' s = 9 so answer is 9 answer : c" | a = 22 / 3
b = math.floor(a)
c = 3 ** 2
d = 22 / c
e = math.floor(d)
f = b + e
|
a ) 140 , b ) 120 , c ) 130 , d ) 110 , e ) 150 | b | multiply(const_100, divide(add(const_100, 14), subtract(const_100, 5))) | a shopkeeper sold an book offering a discount of 5 % and earned a profit of 14 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be $ 100 . then , s . p . = $ 114 let marked price be $ x . then , 95 / 100 x = 114 x = 114 / 95 = $ 120 now , s . p . = $ 120 , c . p . = $ 100 profit % = 20 % . b" | a = 100 + 14
b = 100 - 5
c = a / b
d = 100 * c
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a ) 18 square inches , b ) 20 square inches , c ) 24 square inches , d ) 28 square inches , e ) 30 square inches | c | divide(subtract(subtract(212, 24), multiply(175, divide(80, const_100))), const_2) | three table runners have a combined area of 212 square inches . by overlapping the runners to cover 80 % of a table of area 175 square inches , the area that is covered by exactly two layers of runner is 24 square inches . what is the area of the table that is covered with three layers of runner ? | "total = a + b + c - ( sum of exactly 2 - group overlaps ) - 2 * ( all three ) + neither 80 % * 175 = 212 - 24 - 2 * ( all three ) + 0 2 * ( all three ) = 212 - 24 - 140 all three = 24 answer : c" | a = 212 - 24
b = 80 / 100
c = 175 * b
d = a - c
e = d / 2
|
a ) 48 , b ) 30 , c ) 26 , d ) 23 , e ) 12 | b | add(45, 75) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 75 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 75 = 120 km / hr . 120 * 5 / 18 = 100 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 100 = 30 sec . answer : b" | a = 45 + 75
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a ) 1 / 5 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 1 / 3 | e | divide(subtract(6, 5), subtract(6, 3)) | patty ran a marathon at an average speed of 3 mph for some time and then at an average speed of 6 mph for the rest of the marathon . if she made no stops during the run and her average speed for the entire marathon was 5 mph , for what fraction of the total time did she run at 3 mph ? | we do n ' t need to get into calculations for solving this question . we can use the concept of weighted averages . we know that if the average speed for the entire marathon was 5 mph , she ran at 6 mph for a longer duration . 2 1 3 - - 5 - - - - 6 this shows that you can divide the entire journey in 3 equal parts . thus , 2 / 3 part she ran at 6 mph and 1 / 3 part she ran at 3 mph . answer : e | a = 6 - 5
b = 6 - 3
c = a / b
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a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) none | b | divide(add(add(add(add(66, 65), 77), 62), 75), add(const_1, const_4)) | kamal obtained 66 , 65 , 77 , 62 and 75 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | sol . average = 66 + 65 + 77 + 62 + 75 / 5 ) = ( 345 / 5 ) = 69 . answer b | a = 66 + 65
b = a + 77
c = b + 62
d = c + 75
e = 1 + 4
f = d / e
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a ) 00 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 50 % | a | subtract(multiply(multiply(subtract(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))), const_100), const_100) | a furniture store owner decided to drop the price of her recliners by 20 % to spur business . by the end of the week she had sold 25 % more recliners . what is the percentage increase of the gross ? | "say a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 % , she will earn $ 80 per recliner and she sells 25 % more ie . , 125 recliners hence her sales tields 125 * 80 = $ 10000 increase in sales = 10000 - 10000 = $ 0 so % increase = 0 * 100 / 10000 = 0 % a is the answer" | a = 20 / 100
b = 1 - a
c = 25 / 100
d = 1 + c
e = b * d
f = e * 100
g = f - 100
|
a ) s . 4,000 , b ) s . 6,000 , c ) s . 30,000 , d ) s . 40,000 , e ) s . 50,000 | b | subtract(multiply(5, const_4), const_12) | a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 750 more than that by the previous schema , his sales were worth ? | "[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 750 x = 6000 answer : b" | a = 5 * 4
b = a - 12
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a ) 10 : 7 , b ) 9 : 8 , c ) 1 : 2 , d ) 13 : 11 , e ) 14 : 8 | c | divide(subtract(25, 22), subtract(22, 16)) | what ratio must a shopkeepermix peas and soybean of rs . 16 and rs . 25 / kg , as to obtain a mixture of rs . 22 ? | "correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 3 : 6 = 1 : 2" | a = 25 - 22
b = 22 - 16
c = a / b
|
a ) 6.67 , b ) 1.67 , c ) 2.67 , d ) 3.67 , e ) 4.67 | d | divide(7, subtract(4.4, 2.5)) | calculate how long it will take a swimmer to swim a distance of 7 km against the current of a river which flows at 2.5 km / hr , given that he can swim in still water at 4.4 km / h | "swim in still water at = 4.4 speed of river = 2.5 us = 4.4 - 2.5 = 1.9 distance = 7 t = 7 / 1.9 = 3.67 answer : d" | a = 4 - 4
b = 7 / a
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['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | c | divide(210, add(subtract(220, multiply(6, divide(subtract(220, divide(200, const_2)), 4))), divide(subtract(220, divide(200, const_2)), 4))) | a sporting good store sells one type of baseball bat and one type of baseball . the cost for 2 bats and 4 balls is $ 200 . the cost for 1 bat and 6 balls is $ 220 , as well . if someone were to buy an equal number of bats and balls , at most how many bats can he purchase if he has a budget of $ 210 for the purchase ? | imo it should be c that is 3 reason : formed an equation . . . bat = b ball = c 2 b + 4 c = 200 1 b + 6 c = 220 solving both we get b that is bat = 40 and c that is ball = 30 new equation 210 to be divided in equal 3 b + 3 c = 210 3 * 40 + 3 * 30 = 210 120 + 90 = 210 | a = 200 / 2
b = 220 - a
c = b / 4
d = 6 * c
e = 220 - d
f = 200 / 2
g = 220 - f
h = g / 4
i = e + h
j = 210 / i
|
a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 35 | c | multiply(divide(divide(divide(686, const_3), add(const_3, const_4)), add(const_3, const_4)), const_3) | if the product of the integers from 1 to n is divisible by 686 , what is the least possible value of n ? | "686 = 7 x 7 x 7 x 2 n must include at least 7 , 2 * 7 , and 3 * 7 . the answer is c ." | a = 686 / 3
b = 3 + 4
c = a / b
d = 3 + 4
e = c / d
f = e * 3
|
a ) 200 , b ) 216 , c ) 345 , d ) 244 , e ) 242 | b | multiply(12, 18) | find the area of a parallelogram with base 12 cm and height 18 cm . | "area of a parallelogram = base * height = 12 * 18 = 216 cm 2 answer : option b" | a = 12 * 18
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | subtract(divide(subtract(add(166, 1), 4), 2), subtract(divide(subtract(add(166, 1), 4), 2), 1)) | if a number p is prime , and 2 p + 4 = q , where q is also prime , then the decimal expansion of 1 / q will produce a decimal with q - 1 digits . if this method produces a decimal with 166 digits , what is the units digit of the product of p and q ? | "1 / 6 = 0.1666666 . . . ( a repeating pattern one digit long ) a" | a = 166 + 1
b = a - 4
c = b / 2
d = 166 + 1
e = d - 4
f = e / 2
g = f - 1
h = c - g
|
a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph | b | divide(divide(subtract(110, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 110 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | speed of train relative to man : 110 / 6 * 18 / 5 km / hr = 66 km / hr let speed of train = x therefore x + 6 = 66 x = 66 - 6 x = 60 km / hr answer : b | a = 6 * const_0_2778
b = a * 6
c = 110 - b
d = c / 6
e = d / const_0_2778
|
a ) 10 years , b ) 12 years , c ) 14 years , d ) 15 years , e ) 16 years | b | divide(multiply(subtract(32, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 32 , the how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 32 5 x = 30 x = 6 . hence , b ' s age = 2 x = 12 years . b )" | a = 32 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 150000 , b ) 90000 , c ) 85000 , d ) 190000 , e ) 250000 | b | divide(45000, multiply(divide(2, 3), divide(3, 4))) | a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 45000 rs , what is the value of buzness ? | "if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 45000 - > x = 150000 answer : b" | a = 2 / 3
b = 3 / 4
c = a * b
d = 45000 / c
|
a ) 10 , b ) 50 , c ) 150 , d ) 200 , e ) 300 | c | divide(divide(7500, multiply(const_2, sqrt(25))), sqrt(25)) | in a certain company , the formula for maximizing profits is p = - 25 x ^ 2 + 7500 x , , where p is profit and x is the number of machines the company operates in its factory . what value for x will maximize p ? this one is from integratedlearning . here ' s what i did : the oa states : to find a maximum or minimum value of an equation with an exponent in it , you take the derivative of the equation , set it to zero , and solve . i dont really get what that means . so whoever solves it , could you plz post explanation of what the above sentence mean as well ? | recognize that 25 is a factor of 7500 . if we take this out , we have two parts to the equation : - x ^ 2300 x one part of the equation brings our value down , whereas the other part brings our value up . at this point , we can test the numbers in the answer choice . notice that they are very straight forward to square , and multiplication by 300 is very easy . a ) 10 - 3000 - 100 = 2900 b ) 50 - 15000 - 2500 = 12500 c ) 150 - 45000 - 22500 = 22500 d ) 200 - 60000 - 40000 = 20000 e ) 300 - recognize that this is zero therefore , answer is 150 . hope this helps those - like myself - who have n ' t thought about calculus for over half a decade . c | a = math.sqrt(25)
b = 2 * a
c = 7500 / b
d = math.sqrt(25)
e = c / d
|
a ) 100 , b ) 225 , c ) 169 , d ) 256 , e ) 324 | d | multiply(multiply(5, 3), multiply(5, 3)) | in the coordinate plane , one of the vertices of a square is the point ( - 5 , - 4 ) . if the diagonals of that square intersect at point ( 3 , 2 ) , what is the area of that square ? | "one point ( - 5 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 5 - 3 = - 8 is the midpoint of the square - - > whole side 16 , 16 * 16 = 256 d" | a = 5 * 3
b = 5 * 3
c = a * b
|
a ) rs . 962 , b ) rs . 672 , c ) rs . 540 , d ) rs . 876 , e ) none of these | c | multiply(divide(surface_cube(6), 18), 30) | the cost of the paint is rs . 30 per kg . if 1 kg of paint covers 18 sq . ft , how much will it cost to paint outside of a cube having 6 feet each side | "explanation : surface area of a cube = 6 x 6 ^ 2 = 216 sq . ft quantity of paint required = ( 216 / 18 ) = 12 kg cost of painting = 30 x 18 = rs . 540 answer : c" | a = surface_cube / (
b = a * 18
|
a ) 746 , b ) 750 , c ) 789 , d ) 678 , e ) 897 | a | add(lcm(lcm(8, 14), lcm(26, 28)), 18) | what is the smallest number which when diminished by 18 , is divisible 8 , 14 , 26 and 28 ? | "required number = ( lcm of 8 , 14 , 26 and 28 ) + 18 = 728 + 18 = 746 option a" | a = math.lcm(8, 14)
b = math.lcm(26, 28)
c = math.lcm(a, b)
d = c + 18
|
a ) 2 / 1 , b ) 4 / 1 , c ) 2 / 3 , d ) 9 / 5 , e ) 3 / 2 | d | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 11)) | in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 11 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 11 ) = 36 / 20 = 9 / 5 . answer : d ." | a = 4 + 1
b = 45 / a
c = 45 - b
d = 4 + 1
e = 45 / d
f = e + 11
g = c / f
|
a ) 10 hours , b ) 12 hours , c ) 17 hours , d ) 15 hours , e ) 20 hours | c | divide(subtract(5.5, 5), 8.5) | two men started from the same place walk at the rate of 5 kmph and 5.5 kmph respectively . what time will they take to be 8.5 km apart , if they walk in the same direction ? | "to be 0.5 km apart they take 1 hour to be 8.5 km apart they take = 1 / 0.5 * 8.5 = 17 hours answer is c" | a = 5 - 5
b = a / 8
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a ) 277 , b ) 36 , c ) 45 , d ) 72 , e ) none of these | c | divide(multiply(162, const_100), 360) | ? % of 360 = 162 | "? % of 360 = 162 or , ? = 162 × 100 / 360 = 45 answer c" | a = 162 * 100
b = a / 360
|
a ) 2 , b ) 8 , c ) 24 , d ) 25 , e ) 26 | c | divide(35, multiply(const_10, const_2)) | how many factors does 35 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 6 * 4 = 24 answer c ." | a = 10 * 2
b = 35 / a
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a ) 3 / 4 , b ) 2 / 5 , c ) 4 / 5 , d ) 2 / 3 , e ) 1 / 4 | e | subtract(const_1, divide(3, 4)) | at an elementary school , 60 % of the faculty members are women and 60 % of the faculty members are married . if 3 / 4 of the men are single , what fraction of the men are married ? | - - - - - - - - - - - - - - - - - - - - m - - - - - - w - - - - - - - - total marrried - - - - - - - - - 10 - - - - - 50 - - - - - - - - - 60 not married - - - - - 30 - - - - - 10 - - - - - - - - - 40 total - - - - - - - - - - - - - 40 - - - - - 60 - - - - - - - - 100 need married man / total man , so 10 / 40 = 1 / 4 e | a = 3 / 4
b = 1 - a
|
a ) 108 , b ) 120 , c ) 134 , d ) 144 , e ) none of these | a | divide(multiply(378, multiply(3, 2)), multiply(add(4, 3), 3)) | p , q , r enter into a partnership & their share are in the ratio 1 / 2 : 1 / 3 : 1 / 4 , after two months , p withdraws half of the capitals & after 10 months , a profit of rs 378 is divided among them . what is r ' s share ? | explanation : the ratio of their initial investment = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 let ' s take the initial investment of p , q and r as 6 x , 4 x and 3 x respectively a : b : c = ( 6 x * 2 + 3 x * 10 ) : 4 x * 12 : 3 x * 12 = ( 12 + 30 ) : 4 * 12 : 3 * 12 = ( 4 + 10 ) : 4 * 4 : 12 = 14 : 16 : 12 = 7 : 8 : 6 r ' s share = 378 * ( 6 / 21 ) = 108 . answer : option a | a = 3 * 2
b = 378 * a
c = 4 + 3
d = c * 3
e = b / d
|
a ) 281 , b ) 277 , c ) 283 , d ) 384 , e ) 385 | b | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 18) | light glows for every 18 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | "the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 18 = 276 . so total 277 times light ll glow answer : b" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 18
|
a ) 67 , b ) 26 , c ) 99 , d ) 26 , e ) 91 | d | add(divide(48, const_2), const_2) | find the smallest number which leaves 2235 , 48 and 61 as remainders when divided by 26 , 39 , 52 and 65 respectively . | answer : d | a = 48 / 2
b = a + 2
|
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | b | divide(divide(168, const_2), const_2) | if k ^ 3 is divisible by 168 , what is the least possible value of integer k ? | k ^ 3 = 168 * x = 2 ^ 3 * 3 * 7 * x the factors of k must at minimum include 2 * 3 * 7 = 42 . the answer is b . | a = 168 / 2
b = a / 2
|
a ) 10 days , b ) 12 days , c ) 16 days , d ) 20 days , e ) 25 days | c | divide(const_1, divide(add(add(inverse(20), inverse(25)), inverse(30)), const_2)) | a and b can do a work in 20 days , b and c in 25 days , c and a in 30 days . if a , b and c work together , they will complete the work in ? | "a + b 1 day work = 1 / 20 b + c 1 day work = 1 / 25 c + a 1 day work = 1 / 30 adding we get 2 ( a + b + c ) = 1 / 20 + 1 / 25 + 1 / 30 = 37 / 300 a + b + c 1 day work = 37 / 600 a , b , c can finish the work in 600 / 37 days = 16 days approximately answer is c" | a = 1/(20)
b = 1/(25)
c = a + b
d = 1/(30)
e = c + d
f = e / 2
g = 1 / f
|
a ) 2.6 sec , b ) 1.9 sec , c ) 8.9 sec , d ) 6.9 sec , e ) 2.9 sec | a | divide(100, multiply(136, const_0_2778)) | in what time will a train 100 m long cross an electric pole , it its speed be 136 km / hr ? | "speed = 136 * 5 / 18 = 37.8 m / sec time taken = 100 / 37.8 = 2.6 sec . answer : a" | a = 136 * const_0_2778
b = 100 / a
|
a ) 34 , b ) 36 , c ) 33 , d ) 42 , e ) 24 | b | divide(5220, 15) | 5220 / ? * 15 = 2175 | "b ( 5220 / ? ) * 15 = 2175 ? = ( 5220 * 15 ) / 2175 = > ? = 36" | a = 5220 / 15
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 70 % | d | multiply(divide(126, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 126 ° of the circle , what percentage of megatek employees are in manufacturing ? | answer : d 126 ° divided by 360 ° equals 0.35 , therefore the sector is equal to 35 % of the total | a = 3600 / 10
b = 126 / a
c = b * 100
|
a ) 2 , b ) 3 , c ) 6 , d ) 12 , e ) 24 | b | divide(divide(divide(divide(divide(divide(864, const_2), const_2), const_2), const_2), const_3), const_3) | if n is the smallest integer such that 864 times n is the square of an integer , what is the value of n ? | "865 can written as = 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 - - > 2 ^ 5 * 3 ^ 3 - - - ( 1 ) so for 864 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 3 to make the power of 3 in ( 1 ) even option b is correct" | a = 864 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 3
f = e / 3
|
a ) 80 , b ) 30 , c ) 40 , d ) 70 , e ) 60 | b | multiply(add(5, const_1), 5) | tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 5 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | "b 30 sum = ( n x n ) + n hence , 5 x 5 = 25 + 5 = 30" | a = 5 + 1
b = a * 5
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a ) 197 , b ) 201 , c ) 205 , d ) 209 , e ) 213 | e | multiply(add(divide(subtract(subtract(25, 10), const_2), const_2), 10), divide(add(subtract(25, 10), const_2), const_2)) | what is the sum of all digits for the number 10 ^ 25 - 58 ? | "10 ^ 25 is a 26 - digit number : 1 followed by 25 zeros . 10 ^ 25 - 58 is a 25 - digit number : 23 9 ' s and 42 at the end . the sum of the digits is 23 * 9 + 4 + 2 = 213 . the answer is e ." | a = 25 - 10
b = a - 2
c = b / 2
d = c + 10
e = 25 - 10
f = e + 2
g = f / 2
h = d * g
|
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | divide(24, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 24 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 24 = 6 k + 4 p = > k = 2 , p = 3 total red ' s in bag a will be 2 k = 4 c" | a = 3 * 2
b = a + 4
c = 24 / b
|
a ) 24 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days | a | divide(10, subtract(const_1, divide(add(10, 10), 40))) | a can do a piece of work in 40 days ; b can do the same in 30 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 40 + 10 / 30 + 10 / x = 1 x = 24 days answer : a" | a = 10 + 10
b = a / 40
c = 1 - b
d = 10 / c
|
a ) 3 , b ) 4 , c ) 5 , d ) 2 , e ) 7 | d | subtract(4, subtract(4, 4)) | what least value should be replaced by * in 2631 * 4 so the number become divisible by 3 | "explanation : trick : number is divisible by 3 , if sum of all digits is divisible by 3 , so ( 2 + 6 + 3 + 1 + * + 4 ) = 16 + * should be divisible by 9 , 16 + 2 will be divisible by 2 , so that least number is 2 . answer : option d" | a = 4 - 4
b = 4 - a
|
a ) s . 4335 , b ) s . 2330 , c ) s . 1190 , d ) s . 1620 , e ) s . 2680 | a | multiply(divide(multiply(divide(735, 7), 17), 7), 17) | the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 735 , krishan has ? | ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 735 : n & there n = 289 x 735 / 49 = rs . 4335 answer : a | a = 735 / 7
b = a * 17
c = b / 7
d = c * 17
|
a ) 6 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 15 sec | a | divide(146, multiply(subtract(98, 12), const_0_2778)) | a train 146 m long is running at a speed of 98 kmph . how long does it take to pass a man who is running at 12 kmph in the same direction as the train ? | "answer : a . speed of the train relative to man = ( 98 - 12 ) kmph = ( 86 * 5 / 18 ) m / sec = ( 215 / 9 ) m / sec time taken by the train to cross the man = time taken by it to cover 146 m at 219 / 9 m / sec = 146 * 9 / 219 sec = 6 sec" | a = 98 - 12
b = a * const_0_2778
c = 146 / b
|
a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 5 / 4 | d | divide(add(multiply(3, const_4), 3), const_10) | nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 5 screws and 1 nails ? | let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 5 s and 1 n = 5 s + 1 * 5 s = 10 s . . ratio = 15 s / 10 s = 15 / 10 = 3 / 2 d | a = 3 * 4
b = a + 3
c = b / 10
|
a ) 9 : 100 , b ) 100 : 9 , c ) 12 : 100 , d ) 12 : 20 , e ) 13 : 100 | a | divide(multiply(multiply(5, 3), 3), multiply(multiply(5, 5), 5)) | a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 5 , and the ratio of soap to salt is four times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is quartered by water evaporation . at that time , what is the ratio of water to salt in the solution ? | "water : soap = 3 : 5 soap : salt = 12 : 20 = > for 12 soap , salt = 20 = > for 5 soap , salt = ( 20 / 12 ) * 5 = 100 / 12 = 25 / 3 so , water : soap : salt = 3 : 5 : 25 / 3 = 9 : 15 : 25 after open container , water : soap : salt = 2.25 : 15 : 25 so , water : salt = 2.25 : 25 = 9 : 100 answer : a" | a = 5 * 3
b = a * 3
c = 5 * 5
d = c * 5
e = b / d
|
a ) $ 83 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 104 | a | subtract(multiply(90, 15), add(multiply(87, 7), multiply(94, 7))) | the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 94 per day . what was his wage on the 8 th day ? | "average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 94 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 94 * 7 ) = 1350 - ( 609 + 658 ) = 1350 - 1267 = 83 answer a" | a = 90 * 15
b = 87 * 7
c = 94 * 7
d = b + c
e = a - d
|
a ) 12 , b ) 3 , c ) 14 , d ) 5 , e ) 6 | a | multiply(divide(divide(subtract(1200, 750), 750), 5), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 1200 in 5 years ? | "450 = ( 750 * 5 * r ) / 100 r = 12 % . answer : a" | a = 1200 - 750
b = a / 750
c = b / 5
d = c * 100
|
a ) rs . 7 , b ) rs . 3 , c ) rs . 8 , d ) rs . 1 , e ) rs . 4 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1200) | the difference between simple and compound interest on rs . 1200 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1200 * 10 * 1 ) / 100 = rs . 120 c . i . = [ 1200 * ( 1 + 5 / 100 ) 2 - 1200 ] = rs . 123 difference = ( 123 - 120 ) = rs . 3 . answer : b" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = 10 / 100
f = e + 1
g = d - f
h = g * 1200
|
a ) 9.0 mile , b ) 13.0 mile , c ) 11.0 mile , d ) 10.2 mile , e ) 12 mile | d | multiply(divide(subtract(25, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | stacy and heather are 25 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ? . | ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 25 - 2.4 = 22.6 time taken = distance / speed = 22.6 / 11 = 2.05 hrs heather will cover = 5 * 2.4 = 10.2 miles . answer d | a = 1 + 5
b = a / const_60
c = 24 * b
d = 25 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
|
a ) 48 , b ) 58 , c ) 68 , d ) 78 , e ) 100 | a | divide(const_1, subtract(divide(const_1, 16), divide(const_1, 24))) | ganesh , ram and sohan together can do a work in 16 days . if ganesh and ram together can do a the same work in 24 days then , how long will take sohan to do the same work ? | solution : work done by three of them together in 1 day = 1 / 16 . . . . . . . . . . . . ( 1 ) work done by ganesh and ram together in 1 day = 1 / 24 . . . . . ( 2 ) so , work done by sohan in one day = ( 1 ) - ( 2 ) = ( 1 / 16 ) - ( 1 / 24 ) = 1 / 48 = > total days taken by sohan to complete the work alone = 48 days . answer a | a = 1 / 16
b = 1 / 24
c = a - b
d = 1 / c
|
a ) a ) 140 , b ) b ) 150 , c ) c ) 180 , d ) d ) 200 , e ) e ) 300 | e | divide(multiply(divide(25, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 25 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 25 then x = 25 * 30 = 750 40 % of 750 = 300 answer : option e" | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 25 / f
h = g * 40
i = h / 100
|
a ) 23.4 % , b ) 44.5 % , c ) 55.3 % , d ) 57.4 % , e ) none | c | multiply(subtract(divide(subtract(const_100, 13), 56), const_1), const_100) | the cost price of an article is 56 % of the marked price . calculate the gain percent after allowing a discount of 13 % . | "sol . let marked price = rs . 100 . then , c . p . = rs . 56 . s . p = rs . 87 . â ˆ ´ gain % = [ 31 / 56 * 100 ] % = 55.3 % . answer c" | a = 100 - 13
b = a / 56
c = b - 1
d = c * 100
|
a ) 15 sec , b ) 19 sec , c ) 12.48 sec , d ) 10 sec , e ) 11 sec | c | divide(260, multiply(80, const_0_2778)) | two trains each 260 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively . in what time will they cross each other completely ? | "explanation : d = 260 m + 260 m = 520 m rs = 80 + 70 = 150 * 5 / 18 = 125 / 3 t = 520 * 3 / 125 = 12.48 sec answer : option c" | a = 80 * const_0_2778
b = 260 / a
|
a ) 13 / 128 , b ) 12 / 128 , c ) 11 / 126 , d ) 11 / 127 , e ) 11 / 128 | e | divide(subtract(1024, add(add(subtract(562, 346), subtract(720, 346)), 346)), 1024) | 1024 people in village 720 speak tamil , 562 speak english while 346 speak both . person who don not speack english or tamil speak hindi . what is probibility that person speak hindi | n ( t ) = 720 n ( e ) = 562 n ( t ^ e ) = 346 n ( tue ) = 720 + 562 - 346 = 936 now , n ( h ) = 1024 - 936 = 88 p ( h ) = 88 / 1024 = 11 / 128 = 0.085 answer : e | a = 562 - 346
b = 720 - 346
c = a + b
d = c + 346
e = 1024 - d
f = e / 1024
|
a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 78 kmph | e | divide(divide(subtract(140, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 140 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | "speed of train relative to man : 140 / 6 * 18 / 5 km / hr = 84 km / hr let speed of train = x therefore x + 6 = 84 x = 84 - 6 x = 78 km / hr answer : e" | a = 6 * const_0_2778
b = a * 6
c = 140 - b
d = c / 6
e = d / const_0_2778
|
a ) 15 kmph , b ) 11 kmph , c ) 16 kmph , d ) 18 kmph , e ) 12 kmph | c | multiply(divide(8, const_60), 120) | the speed of a train is 120 kmph . what is the distance covered by it in 8 minutes ? | "120 * 8 / 60 = 16 kmph answer : c" | a = 8 / const_60
b = a * 120
|
a ) 260 m , b ) 270 m , c ) 643 m , d ) 832 m , e ) 270 m | a | multiply(subtract(26, divide(260, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | a goods train runs at the speed of 72 kmph and crosses a 260 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 260 / 26 = 20 x + 260 = 520 x = 260 . answer : a" | a = const_0_2778 * 72
b = 260 / a
c = 26 - b
d = const_0_2778 * 72
e = c * d
|
a ) 643 , b ) 689 , c ) 683 , d ) 690 , e ) 693 | c | multiply(divide(678, add(const_1, divide(14, const_100))), add(const_1, divide(15, const_100))) | if albert ’ s monthly earnings rise by 14 % , he would earn $ 678 . if , instead , his earnings rise by only 15 % , how much ( in $ ) would he earn this month ? | "= 678 / 1.14 ∗ 1.15 = 683 = 683 answer is c" | a = 14 / 100
b = 1 + a
c = 678 / b
d = 15 / 100
e = 1 + d
f = c * e
|
a ) 1 km , b ) 2.87 km , c ) 3 km , d ) 4 km , e ) 5 km | b | multiply(multiply(divide(divide(45, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 45 minutes . the total distance is ? | "b 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 45 / 60 47 x / 60 = 45 / 60 = > x = 0.95 . total distance = 3 * 0.95 = 2.87 km ." | a = 45 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 26.34 litres , b ) 27.36 liters , c ) 28 litres , d ) 29.16 litres , e ) can not be determined | d | multiply(power(subtract(const_1, divide(4, 40)), const_3), 40) | a container contains 40 litres of milk . from this container 4 litres of milk was taken out and replaced by water . this process was repeated further two times . how much milk is now contained by the container . | solution amount of milk left after 3 operations [ 40 ( 1 - 4 / 40 ) 3 ] litres = ( 40 x 9 / 10 x 9 / 10 x 9 / 10 ) = 29.16 litres answer d | a = 4 / 40
b = 1 - a
c = b ** 3
d = c * 40
|
a ) 108.6 , b ) 110.4 , c ) 112.6 , d ) 114.8 , e ) 116.2 | e | divide(multiply(divide(subtract(const_100, 17), const_100), 14), divide(subtract(const_100, 90), const_100)) | if grapes are 90 % water and raisins are 17 % water , then how many kilograms did a quantity of raisins , which currently weighs 14 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 83 % of the raisins , 0.1 x = 0.83 ( 14 kg ) . then x = 8.3 * 14 = 116.2 kg . the answer is e ." | a = 100 - 17
b = a / 100
c = b * 14
d = 100 - 90
e = d / 100
f = c / e
|
a ) 104345 , b ) 107375 , c ) 108385 , d ) 109395 , e ) 105355 | d | add(add(multiply(add(const_3, const_4), const_10), multiply(multiply(add(const_3, const_4), const_100), const_10)), 102325) | on multiplying a number f by 153 , the result obtained was 102325 . however , it is found that both the 2 ' s are wrong . find the correct result . | the only thing you actually know about the correct number f is that it is divisible by 153 and has 5 as a factor . you should immediately try to find the factors of 153 and look for them in the options . 153 = 9 * 17 divisibility by 9 is easy to check . only ( d ) satisfies . | a = 3 + 4
b = a * 10
c = 3 + 4
d = c * 100
e = d * 10
f = b + e
g = f + 102325
|
a ) 11520 , b ) 12600 , c ) 15000 , d ) 13560 , e ) 12560 | b | add(const_100, multiply(divide(5000, subtract(subtract(100, 30), 30)), const_100)) | at an election 2 candidates are participated a candidate got 30 % of votes and defeated by 5000 votes . 100 votes are invalid . find the total polled votes ? | winners votes = 100 - 30 = 70 % invalid votes = 100 polled votes = [ ( 100 * 5000 ) / ( 2 * 70 - 100 ) ] + 100 = 12600 answer is b | a = 100 - 30
b = a - 30
c = 5000 / b
d = c * 100
e = 100 + d
|
a ) 6 : 8 , b ) 6 : 2 , c ) 6 : 5 , d ) 6 : 4 , e ) 6 : 2 | c | divide(multiply(4, 3), multiply(5, 2)) | the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ? | "4 : 5 3 : 2 - - - - - - - 12 : 15 : 10 12 : 10 6 : 5 answer : c" | a = 4 * 3
b = 5 * 2
c = a / b
|
a ) 15 % , b ) 21 % , c ) 25 % , d ) 30 % , e ) 36 % | b | multiply(subtract(power(add(const_1, divide(10, const_100)), const_2), const_1), const_100) | if a large pizza has a radius that is 10 % larger that that of a medium pizza , what is the percent increase in area between a medium and a large pizza ? | "let the radius of medium pizza be r . then the radius of large pizza is 1.1 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.1 * r ) ^ 2 = 1.21 * pi * r ^ 2 , an increase of 21 % . the answer is b ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 3 / 5 , b ) 1 / 32 , c ) 1 / 5 , d ) 1 / 4 , e ) 1 / 2 | b | inverse(power(2, 5)) | if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 2 flips and not on the last 3 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b | a = 2 ** 5
b = 1/(a)
|
a ) 24 kg , b ) 44 kg , c ) 20 kg , d ) 30 kg , e ) 52 kg | b | add(multiply(divide(60, add(3, 2)), 2), multiply(divide(100, add(1, 4)), 1)) | 60 kg of an alloy a is mixed with 100 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 2 and alloy b has tin and copper in the ratio 1 : 4 , then the amount of tin in the new alloy is ? | quantity of tin in 60 kg of a = 60 * 2 / 5 = 24 kg quantity of tin in 100 kg of b = 100 * 1 / 5 = 20 kg quantity of tin in the new alloy = 24 + 20 = 44 kg answer is b | a = 3 + 2
b = 60 / a
c = b * 2
d = 1 + 4
e = 100 / d
f = e * 1
g = c + f
|
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