Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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dick and jane each saved $ 4,000 in 1989 . in 1990 dick saved 10 percent more than in 1989 , and together he and jane saved a total of $ 6,500 . approximately what percent less did jane save in 1990 than in 1989 ? | "1990 dick saved = $ 4400 jane saved = $ 2100 ( jane saved $ 2800 less than she did the prior year ) jane saved approximately $ 2800 / 4000 $ ( 70 % ) less in 1990 answer : b" | a ) 20 % , b ) 70 % , c ) 50 % , d ) 30 % , e ) 40 % | b | multiply(subtract(const_1, subtract(const_2, add(divide(10, const_100), const_1))), const_100) | divide(n3,const_100)|add(#0,const_1)|subtract(const_2,#1)|subtract(const_1,#2)|multiply(#3,const_100)| | general |
the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is ? | explanation : ( 5 * 3.5 ) / 2 = 8.75 answer is c | ['a ) 0.35 cm 2', 'b ) 17.5 cm 2', 'c ) 8.75 cm 2', 'd ) 55 cm 2', 'e ) 50 cm 2'] | c | multiply(divide(const_1, const_2), multiply(5, 3.5)) | divide(const_1,const_2)|multiply(n0,n1)|multiply(#0,#1) | physics |
if one positive integer is greater than another positive integer by 3 , and the difference of their cubes is 189 , what is their sum ? | "1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 the two numbers are 3 and 6 . the answer is d ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(189, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), subtract(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(189, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), 3)) | power(n0,n0)|power(n0,const_2)|multiply(#1,const_2)|power(#0,const_2)|subtract(n1,#0)|power(#4,const_2)|add(#3,#5)|sqrt(#6)|add(#0,#7)|divide(#8,#2)|subtract(#9,n0)|add(#9,#10)| | general |
there are 16 bees in the hive , then 5 more fly . how many bees are there in all ? | 16 + 5 = 20 . answer is a . | a ) 20 , b ) 33 , c ) 12 , d ) 17 , e ) 25 | a | add(16, 5) | add(n0,n1)| | general |
a special municipal payroll tax charges not tax on a payroll less than $ 250,000 and only 0.1 % on a company β s payroll above $ 250,000 . if belfried industries paid $ 200 in this special municipal payroll tax , then they must have had a payroll of | "answer : a , ( with different approach ) : the 200 paid is 0.1 % of the additional amount above 250,000 . let it be x now 0.1 % of x = 200 therefore x = 200,000 total = 250,000 + x = 450,000" | a ) $ 450,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 400,000 , e ) $ 2 , 200,000 | a | multiply(multiply(200, const_100), const_10) | multiply(n3,const_100)|multiply(#0,const_10)| | general |
for what value of x , is | x β 3 | + | x + 1 | + | x | = 11 ? | for what value of x , is | x β 3 | + | x + 1 | + | x | = 11 ? it ' s easiest just to plug in answer choices : ( c ) : - 3 | x β 3 | + | x + 1 | + | x | = 11 ? | - 3 - 3 | + | - 3 + 1 | + | - 3 | = 11 ? | 6 | + | 2 | + | 3 | = 11 ( c ) | a ) 0 , b ) 3 , c ) - 3 , d ) 4 , e ) - 2 | c | negate(3) | negate(n0) | general |
an auction house charges a commission of 16 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.16 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 210,000 ( 16 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c ." | a ) $ 115,000 , b ) $ 160,000 , c ) $ 210,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(16, 10), 10) | multiply(n0,n2)|add(n2,#0)| | general |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 13 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 13 + 21 = 34 t = 12 d = 34 * 12 = 408 answer : b" | a ) 11 , b ) 408 , c ) 2881 , d ) 287 , e ) 221 | b | add(multiply(divide(60, subtract(21, 13)), 13), multiply(divide(60, subtract(21, 13)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
what should be the least number to be added to the 1202 number to make it divisible by 4 ? | answer : 2 option : e | a ) 12 , b ) 17 , c ) 18 , d ) 77 , e ) 2 | e | subtract(4, reminder(1202, 4)) | reminder(n0,n1)|subtract(n1,#0) | general |
a sum of money deposited at c . i . amounts to rs . 3250 in 2 years and to rs . 3830 in 3 years . find the rate percent ? | "3250 - - - 580 100 - - - ? = > 17 % answer : d" | a ) 30 % , b ) 10 % , c ) 20 % , d ) 17 % , e ) 50 % | d | multiply(divide(subtract(3830, 3250), 3250), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if | 20 x - 10 | = 150 , then find the product of the values of x ? | "| 20 x - 10 | = 150 20 x - 10 = 150 or 20 x - 10 = - 150 20 x = 160 or 20 x = - 140 x = 8 or x = - 7 product = - 7 * 8 = - 56 answer is b" | a ) - 45 , b ) 56 , c ) - 62 , d ) 35 , e ) - 30 | b | subtract(subtract(subtract(150, 10), add(150, 10)), 10) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general |
a certain car uses 12 gallons of gasoline in traveling 180 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car β s gas mileage be increased ? | "180 / 10 = 18 . the difference is 18 - 15 = 3 . answer b" | a ) 2 , b ) 3 , c ) 6 , d ) 8 , e ) 10 | b | subtract(divide(180, 10), divide(180, 12)) | divide(n1,n2)|divide(n1,n0)|subtract(#0,#1)| | physics |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . what is the present age of his son ? | "let present age of the son = x years then present age the man = ( x + 24 ) years given that in 2 years man ' s age will be twice the age of his son x = 22 answer b" | a ) 23 years , b ) 22 years , c ) 21 years , d ) 20 years , e ) 30 years | b | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
in the hillside summer camp there are 60 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 60 students , 90 % of 60 = 54 boys and remaining 6 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 5 % . . if we add 60 to existing 54 then the count will be 114 and the girls number will be 6 as it ... | a ) 50 . , b ) 45 . , c ) 40 . , d ) 60 . , e ) 25 . | d | add(multiply(divide(subtract(const_100, 5), const_100), 60), multiply(divide(5, const_100), 60)) | divide(n2,const_100)|subtract(const_100,n2)|divide(#1,const_100)|multiply(n0,#0)|multiply(n0,#2)|add(#4,#3)| | general |
if x and y are integers such that x ^ 2 = 3 y and xy = 36 , then x β y = ? | here x and y are integers . x ^ 2 = 3 y , xy = 36 . substitute ( x ^ 2 ) / 3 = y in xy = > x ^ 3 = 36 * 3 = > x ^ 3 = 108 . here x ^ 3 is positive , x is also positive . x = 6 then y = 6 . x - y = 0 so option e is correct | a ) - 30 , b ) - 20 , c ) - 5 , d ) 5 , e ) 0 | e | subtract(power(multiply(36, 3), const_0_33), divide(36, power(multiply(36, 3), const_0_33))) | multiply(n1,n2)|power(#0,const_0_33)|divide(n2,#1)|subtract(#1,#2) | general |
49 pumps can empty a reservoir in 13 / 2 days , working 8 hours a day . if 196 pumps are used for 5 hours each day , then the same work will be completed in : | explanation : let the required number of days be x . then , more pumps , less days ( indirect proportion ) less working hrs / day , more days ( indirect proportion ) pumps 196 : 49 working hrs / day 5 : 8 : : 13 / 2 : x 96 x 5 x x = 49 x 8 x 13 / 2 x = 49 x 8 x 13 / 2 x 1 / ( 196 x 5 ) x = 13 / 5 answer c | a ) 2 days , b ) 5 / 2 days , c ) 13 / 5 days , d ) 3 days , e ) none of these | c | multiply(multiply(divide(49, 196), divide(8, 5)), divide(13, 2)) | divide(n1,n2)|divide(n0,n4)|divide(n3,n5)|multiply(#1,#2)|multiply(#0,#3) | physics |
a couple decides to have 3 children . if they succeed in having 4 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 2 girls and 1 boy ? | "sample space = 2 ^ 3 = 8 . favourable events = { bgg } , { bgb } , { bbb } , { ggg } , { gbg } probability = 5 / 8 = 5 / 8 . ans ( d ) ." | a ) 1 , b ) 2 , c ) 3 , d ) 5 / 8 , e ) 4 | d | subtract(2, multiply(divide(factorial(3), factorial(1)), power(divide(2, 1), 3))) | divide(n2,n3)|factorial(n0)|factorial(n3)|divide(#1,#2)|power(#0,n0)|multiply(#3,#4)|subtract(n2,#5)| | general |
working together at their respective constant rates , machine a and machine b can produce 600 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 600 unit total would it produce ? | "1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 600 units in 24 hrs , so it can produce 600 * 8 / 24 = 200 units is 8 hrs . 200 is 30 % of 600 . d is the answer" | a ) 25 , b ) 37 , c ) 50 , d ) 30 , e ) 75 | d | multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100) | multiply(const_4,const_4)|multiply(n1,n2)|divide(#1,const_100)|multiply(#0,const_100)|add(n1,#2)|divide(#3,n1)|divide(#3,#4)|subtract(#5,#6)|multiply(n1,#7)|divide(#8,#3)|multiply(#9,const_100)| | gain |
some of the 50 % solution of acid was removed and this was replaced with an equal amount of 30 % solution of acid . as a result , a 40 % 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 ) + 0.3 ( x ) = 0.4 0.2 x = 0.1 x = 1 / 2 the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 2 / 5 | a | divide(40, add(50, 30)) | add(n0,n1)|divide(n2,#0)| | gain |
in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 5 rows in all and a total of 200 bricks in the wall , how many bricks does the bottom row contain ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 = 200 5 x - 10 = 200 5 x = 190 x = 38 answer : e" | a ) 42 , b ) 35 , c ) 40 , d ) 33 , e ) 38 | e | divide(subtract(subtract(subtract(subtract(200, const_1), const_2), const_3), const_4), 5) | subtract(n1,const_1)|subtract(#0,const_2)|subtract(#1,const_3)|subtract(#2,const_4)|divide(#3,n0)| | general |
a certain business school has 500 students , and the law school at the same university has 800 students . among these students , there are 30 sibling pairs consisting of 1 business student and 1 law student . if 1 student is selected at random from both schools , what is the probability that a sibling pair is selected ... | probability of selecting 1 student from harvard ' s business school - - - 1 / 500 probability of selecting 1 student from harvard ' s law school - - - - - - - - - 1 / 800 probability that these two students are siblings - - - - ( 1 / 500 * 1 / 800 ) since there are 30 siblings , hence ( 1 / 500 * 1 / 800 ) * 30 . 3 / 4... | a ) 3 / 40000 , b ) 3 / 20000 , c ) 3 / 4000 , d ) 9 / 400 , e ) 6 / 130 | a | multiply(multiply(divide(1, 500), divide(1, 800)), 30) | divide(n3,n0)|divide(n3,n1)|multiply(#0,#1)|multiply(n2,#2) | other |
a number increased by 40 % gives 700 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 40 % = 140 % 140 % - - - - - - - > 700 ( 140 Γ 5 = 700 ) 100 % - - - - - - - > 700 ( 100 Γ 5 = 500 ) b )" | a ) 250 , b ) 500 , c ) 450 , d ) 500 , e ) 520 | b | divide(700, add(const_1, divide(40, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
when positive integer k is divided by 5 , the remainder is 2 . when k is divided by 6 , the remainder is 5 . if k is less than 41 , what is the remainder when k is divided by 7 ? | "cant think of a straight approach but here is how i solved it : k is divided by 5 and remainder is 2 . this means k = 5 n + 2 ( n is an integer ) so the possible values of k = { 2 , 7 , 12 , 17 , 22 , 27 , 32 , 37 } ( less than 41 ) secondly , if k is divided by 6 , the remainder is 5 = > k = 6 m + 5 so the possible v... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | reminder(add(const_12, 5), 7) | add(n0,const_12)|reminder(#0,n5)| | general |
the profits of qrs company rose 40 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ? | "assume 100 in march , then 140 in april as 40 % increase , then 112 in may as 20 % decrease from april , and then 168 in june which is 150 % of 112 . so overall increase is from 100 to 168 is 68 % answer d" | a ) 15 % , b ) 32 % , c ) 40 % , d ) 68 % , e ) 80 % | d | multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100)))), const_1)) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)| | gain |
find the value of 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 4 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 4 x 4 x 4 x 50 = 3200 correct option : b" | a ) 800 , b ) 3200 , c ) 900 , d ) 1600 , e ) none | b | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 4) | multiply(n1,n2)|multiply(n4,n5)|multiply(n3,#0)|multiply(n6,#1)|divide(#2,#3)|multiply(n0,#4)| | general |
an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 1... | "fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage o... | a ) 4 , b ) 9 , c ) 6 , d ) 8 , e ) 10 | b | add(add(const_2, add(const_1, const_4)), const_2) | add(const_1,const_4)|add(#0,const_2)|add(#1,const_2)| | general |
from the sale of sleeping bags , a retailer made a gross profit of 12 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | cost price * 1.12 = selling price - - > cost price * 1.12 = $ 28 - - > cost price = $ 25 . answer : d . actually even without any math only c and d make any sense , but since 24.64 * 1.12 wo n ' t be an integer ( $ 28 ) then only answer choice d remains . | a ) 3.0 , b ) 3.36 , c ) 24.64 , d ) 25.0 , e ) 31.36 | d | divide(multiply(28, const_100), add(const_100, 12)) | add(n0,const_100)|multiply(n1,const_100)|divide(#1,#0) | gain |
there is enough provisions for 700 men in an army camp for 25 days . if there were 300 men less , how long will the provision last ? | "exp : we have , m 1 d 1 = m 2 d 2 700 * 25 = 300 * d 2 d 2 = 700 * 25 / 300 = 59 days . answer : d" | a ) 30 days , b ) 40 days , c ) 50 days , d ) 59 days , e ) 65 days | d | divide(multiply(700, 25), 300) | multiply(n0,n1)|divide(#0,n2)| | physics |
the distance between delhi and mathura is 140 kms . a starts from delhi with a speed of 24 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 46 kmph at 8 a . m . from delhi . when will they meet ? | "d = 140 β 24 = 116 rs = 46 + 24 = 70 t = 116 / 70 = 1.6 hours 8 a . m . + 1.6 = 9 1 / 2 a . m . . answer : d" | a ) 11 , b ) 23 3 / 4 , c ) 16 1 / 2 , d ) 9 1 / 2 , e ) 98 | d | add(8, divide(140, add(24, 46))) | add(n1,n3)|divide(n0,#0)|add(n4,#1)| | physics |
3 distinct single - digit numbers . a , b , c are in gp . if abs ( x ) for real x is the absolute value of x ( x if x is + ve or 0 and - x if x is - ve ) , then the number of different possible values of abs ( a + b + c ) is | a , b , c may be ( 1 , 2,4 ) & ( 4 , 2,1 ) ( 1 , 3,9 ) & ( 9 , 3,1 ) ( 2 , 4,8 ) & ( 8 , 4,2 ) ( 4 , 6,9 ) & ( 9 , 6,4 ) find abs ( a + b + c ) for these 8 gps 7,13 , 15,19 so , we get 4 different values answer : d | a ) 5 , b ) 6 , c ) 3 , d ) 4 , e ) 7 | d | add(3, const_1) | add(n0,const_1) | general |
the charge for a single room at hotel p is 25 percent less than the charge for a single room at hotel r and 20 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 ? | "let rate in r = 100 x then p = 75 x g = 100 y p = 80 y thus 75 x = 80 y or x = 1.06 y ans r = 106 y so increase = 6 % answer : e" | a ) 15 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 6 % | e | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 25), subtract(const_100, 20)), const_100)), multiply(divide(subtract(const_100, 25), subtract(const_100, 20)), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 53 , what is the value of x ? | "am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 53 x = 51 answer : e" | a ) 52 , b ) 53 , c ) 54 , d ) 55 , e ) 51 | e | subtract(multiply(4, const_2), multiply(2, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
a courtyard is 18 meter long and 12 meter board is to be paved with bricks of dimensions 12 cm by 6 cm . the total number of bricks required is : | "explanation : number of bricks = courtyard area / 1 brick area = ( 1800 Γ£ β 1200 / 12 Γ£ β 6 ) = 30000 option d" | a ) 16000 , b ) 18000 , c ) 20000 , d ) 30000 , e ) none of these | d | divide(multiply(multiply(18, const_100), multiply(12, const_100)), multiply(12, 6)) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)| | physics |
a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 52 minutes . the total distance is ? | "c 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 52 / 60 47 x / 60 = 52 / 60 = > x = 1.1 total distance = 3 * 1.1 = 3.3 km ." | a ) 1 km , b ) 2 km , c ) 3.3 km , d ) 4 km , e ) 5 km | c | multiply(multiply(divide(divide(52, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics |
a batch of cookies was divided among 3 tins : 2 / 3 of all the cookies were placed in either the blue tin or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction w of the cookies that were placed in the other tins were placed in the green tin ... | blue tin or red tin : 2 / 3 ( n ) red tin : ( 1 / 3 ) n blue tin : ( 1 / 4 ) n what the last statment meant , is it wants this fraction : ( # of cookies in green tin ) / ( # of cookies in red and green tin ) # of cookies in green tin = 2 n / 3 - n / 4 = 8 n - 3 n / 12 = 5 n / 12 # of cookies in red and green tin = n / ... | a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 9 / 7 | c | divide(subtract(divide(2, 3), divide(1, 4)), add(subtract(divide(2, 3), divide(1, 4)), divide(1, 3))) | divide(n1,n0)|divide(n3,n4)|divide(n3,n0)|subtract(#0,#1)|add(#2,#3)|divide(#3,#4) | general |
two trains of equal lengths take 12 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 15 = 8 m / sec . relative speed = 10 + 8 = 18 m / sec . required time = ( 120 + 120 ) / 18 = 13 sec . answer : b" | a ) 22 , b ) 13 , c ) 77 , d ) 99 , e ) 21 | b | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 12))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
the difference of two numbers is 11 and one - fifth of their sum is 9 . find the numbers . | "let the number be x and y . then , x β y = 11 - - - - ( i ) and 1 / 5 ( x + y ) = 9 = > x + y = 45 - - - - ( ii ) adding ( i ) and ( ii ) , we get : 2 x = 56 or x = 28 . putting x = 28 in ( i ) , we get : y = 17 . hence , the numbers are 28 and 17 . answer is a ." | a ) 28 and 17 , b ) 27 and 18 , c ) 25 and 15 , d ) 27 and 15 , e ) 25 and 18 | a | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | divide(const_10,const_2)|multiply(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#1,#3)|add(#0,#4)| | general |
40 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 40 % of companies which offer wireless internet also offer snacks . the answer is b ." | a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | b | multiply(40, const_1) | multiply(n0,const_1)| | general |
a boat moves down stream at the rate of 1 km in 6 minutes and upstream at the rate of 1 km in 10 minutes . the speed of current is | if speed ( in kmph ) of the boat = b and current = c , then in downstream time taken ( in hrs . ) = 1 / ( b + c ) = 6 / 60 = 1 / 10 or b + c = 10 - - - ( i ) & in upstream time taken ( in hrs . ) = 1 / ( b - c ) = 10 / 60 = 1 / 6 or b - c = 6 - - - ( ii ) from ( i ) & ( ii ) , b = 8 , c = 2 answer : a | a ) 2 kmph , b ) 3 kmph , c ) 4 kmph , d ) 5 kmph , e ) 6 kmph | a | divide(subtract(divide(1, divide(6, const_60)), divide(1, divide(10, const_60))), const_2) | divide(n1,const_60)|divide(n3,const_60)|divide(n0,#0)|divide(n0,#1)|subtract(#2,#3)|divide(#4,const_2) | physics |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 5 : 2 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 5 : 2 - - > e : l = 5 : 2 = 20 : 8 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 20 : 15 . thus , l : s = 8 : 15 . answer : b ." | a ) 9 : 8 , b ) 8 : 15 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | b | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other |
the sector of a circle has radius of 21 cm and central angle 130 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 130 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 47.7 + 42 = 89.7 cm answer : a" | a ) 89.7 , b ) 91.4 , c ) 91.7 , d ) 91.3 , e ) 91.1 | a | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
the population of a city increases by 8 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 7 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 7 / 100 ) ^ 3 = 122.5043 β΄ required percentage = 22.50 % answer : c" | a ) 9 % , b ) 9.27 % , c ) 22.50 % , d ) 12 % , e ) none of these | c | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(8, 1)), subtract(8, 1)), const_100), subtract(8, 1))), subtract(8, 1)), const_100), add(divide(multiply(add(const_100, subtract(8, 1)), subtract(8, 1)), const_100), subtract(8, 1))) | subtract(n0,n1)|add(#0,const_100)|multiply(#1,#0)|divide(#2,const_100)|add(#3,#0)|add(#4,const_100)|multiply(#5,#0)|divide(#6,const_100)|add(#4,#7)| | general |
pipe a can fill a tank in 3 hours . due to a leak at the bottom , it takes 9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 3 - 1 / x = 1 / 9 = > 1 / x = 1 / 3 - 1 / 9 = ( 3 - 1 ) / 9 = 2 / 9 = > x = 9 / 2 = 4.5 . answer : a" | a ) 4.5 , b ) 17 , c ) 18 , d ) 19 , e ) 12 | a | divide(multiply(9, 3), subtract(9, 3)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
the radius of a semi circle is 4.8 cm then its perimeter is ? | "36 / 7 r = 4.8 = 24.69 answer : e" | a ) 32.51 , b ) 32.4 , c ) 32.1 , d ) 32.92 , e ) 24.69 | e | add(divide(circumface(4.8), const_2), multiply(4.8, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
the number of stamps that p and q had were in the ratio of 9 : 2 respectively . after p gave q 33 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 6 : 5 . as a result of the gift , p had how many more stamps than q ? | "p started with 9 k stamps and q started with 2 k stamps . ( 9 k - 33 ) / ( 2 k + 33 ) = 6 / 5 45 k - 12 k = 178 + 165 33 k = 343 k = 11 p has 9 ( 11 ) - 33 = 66 stamps and q has 2 ( 11 ) + 33 = 55 stamps . the answer is a ." | a ) 11 , b ) 33 , c ) 45 , d ) 67 , e ) 80 | a | divide(add(multiply(33, 6), multiply(33, 5)), add(6, 5)) | add(n3,n4)|multiply(n2,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | other |
a man can row a distance of 5 km in 60 min with the help of the tide . the direction of the tide reverses with the same speed . now he travels a further 40 km in 10 hours . how much time he would have saved if the direction of tide has not changed ? | explanation : he covered 5 km in 1 hour , so he might cover 40 km in 8 hours . but he took 10 hours . he would have saved 10 Γ’ β¬ β 8 = 2 hours . answer : a | a ) 2 , b ) 8 , c ) 1 , d ) 6 , e ) 5 | a | subtract(10, divide(40, 5)) | divide(n2,n0)|subtract(n3,#0) | physics |
how many 3 - digit numerals end with a digit that represents a prime number ? | "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 4 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 10 ways total = 4 * 10 * 10 = 400 ans : e" | a ) 16 , b ) 80 , c ) 160 , d ) 180 , e ) 400 | e | add(multiply(const_100, 3), const_100) | multiply(n0,const_100)|add(#0,const_100)| | general |
machine a and machine b process the same work at different rates . machine c processes work as fast as machines a and b combined . machine d processes work 3 times as fast as machine c ; machine d β s work rate is also exactly 4 times machine b β s rate . assume all 4 machines work at fixed unchanging rates . if machin... | c = a + b d = 3 c = 3 ( a + b ) = 4 b then b = 3 a and c = 4 a the combined rate of the four machines is a + 3 a + 4 a + 12 a = 20 a machine a can complete the work in 380 minutes , so its rate is 1 / 380 of the work per minute . the combined rate is 20 / 380 = 1 / 19 so the work will be completed in 19 minutes . the a... | a ) 15 , b ) 17 , c ) 19 , d ) 21 , e ) 24 | c | divide(add(multiply(6, const_60), 20), add(add(add(const_1, 3), add(3, const_1)), multiply(add(3, const_1), 3))) | add(n0,const_1)|multiply(n3,const_60)|add(n4,#1)|add(#0,#0)|multiply(n0,#0)|add(#3,#4)|divide(#2,#5) | physics |
in a fuel station the service costs $ 1.50 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.50 + 0.65 * 12 * 55 = 447 hence - e | a ) 320 $ , b ) 380 $ , c ) 410 $ , d ) 430 $ , e ) 447 $ | e | multiply(multiply(0.65, 55), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
cole drove from home to work at an average speed of 60 kmh . he then returned home at an average speed of 90 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 60 time from work to home = x / 90 total time = 2 hrs ( x / 60 ) + ( x / 90 ) = 2 solving for x , we get x = 72 time from home to work in minutes = ( 72 ) * 60 / 60 = 72 minutes ans = c" | a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | c | multiply(divide(multiply(90, 2), add(60, 90)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
a certain sum of money is divided among a , b and c so that for each rs . a has 80 paisa , b has 65 paisa and c 40 paisa . if c ' s share is rs . 40 , find the sum of money ? | "a : b : c = 80 : 65 : 40 = 16 : 13 : 8 8 - - - - 40 37 - - - - ? = > rs . 185 answer : a" | a ) rs . 185 , b ) rs . 410 , c ) rs . 285 , d ) rs . 385 , e ) rs . 475 | a | multiply(divide(40, 65), add(add(const_100, 80), 65)) | add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)| | general |
from a group of 5 boys and 5 girls , 6 children are to be randomly selected . what is the probability that 3 boys and 3 girls will be selected ? | "the total number of ways to choose 6 children from 10 is 10 c 6 = 210 the number of ways to choose 3 boys and 3 girls is 5 c 3 * 5 c 3 = 10 * 10 = 100 p ( 3 boys and 3 girls ) = 100 / 210 = 10 / 21 the answer is d ." | a ) 4 / 15 , b ) 6 / 17 , c ) 8 / 19 , d ) 10 / 21 , e ) 12 / 23 | d | divide(multiply(choose(5, const_2), choose(5, const_2)), choose(add(5, 5), 6)) | add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)| | probability |
what is the area of square field whose side of length 14 m ? | "14 * 14 = 196 sq m answer : b" | a ) 225 , b ) 196 , c ) 266 , d ) 288 , e ) 261 | b | square_area(14) | square_area(n0)| | geometry |
3 types of tea the a , b , c costs rs . 95 / kg , 100 / kg and 70 / kg respectively . how many kgs of each should be blended to produce 100 kg of mixture worth rs . 90 / kg , given that the quantities of band c are equal | given quantities of b and c are equal . therefore instead of considering them as a different quantities take average of both and consider it as a single entity . so the cost of the mixture ` ` d ' ' ( equal quantities of ` ` b ' ' and ` ` c ' ' ) is 170 / 2 = 85 rs / kg now the tea contains only ` ` a ' ' of 95 rs / pe... | a ) 70 , 1515 , b ) 50 , 2525 , c ) 60 , 2020 , d ) 40 , 3030 , e ) 20 , 3030 | b | multiply(multiply(subtract(100, 70), subtract(100, 95)), const_10) | subtract(n2,n3)|subtract(n2,n1)|multiply(#0,#1)|multiply(#2,const_10) | general |
# p is defined as 2 p - 20 for any number p . what is p , if # ( # ( # p ) ) = 6 ? | # p = 2 p - 20 - - - > # ( # p ) = 2 ( 2 p - 20 ) - 20 = 4 p - 60 and thus # ( 4 p - 60 ) = 2 ( 4 p - 60 ) - 20 = 8 p - 140 = 6 - - - > 8 p = 146 - - - > p = 18.25 , e is the correct answer . | a ) β 108 , b ) β 44 , c ) 10 , d ) 16 , e ) 18.25 | e | divide(add(add(20, multiply(20, 6)), 6), multiply(2, const_4)) | multiply(n1,n2)|multiply(n0,const_4)|add(n1,#0)|add(n2,#2)|divide(#3,#1) | general |
a watch was sold at a loss of 9 % . if it was sold for rs . 220 more , there would have been a gain of 4 % . what is the cost price ? | "91 % 104 % - - - - - - - - 13 % - - - - 220 100 % - - - - ? = > rs . 1692 answer : e" | a ) 1000 , b ) 2876 , c ) 1977 , d ) 2778 , e ) 1692 | e | divide(multiply(220, const_100), subtract(add(const_100, 4), subtract(const_100, 9))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
two persons a and b can complete a piece of work in 30 days and 60 days respectively . if they work together , what part of the work will be completed in 10 days ? | a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 60 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 60 = 1 / 20 the part of the work completed in 10 days = 10 ( 1 / 20 ) = 1 / 2 . answer : d | a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 5 / 3 | d | multiply(10, add(divide(const_1, 30), divide(const_1, 60))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2) | physics |
by travelling at 30 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "let the time taken to reach the destination be 3 x hours . total distance = 30 * 3 x = 90 x km he covered 2 / 3 * 90 x = 60 x km in 1 / 3 * 3 x = x hours so , the remaining 30 x km , he has to cover in 2 x hours . required speed = 30 x / 2 x = 15 kmph . answer : b" | a ) 19 kmph , b ) 15 kmph , c ) 12 kmph , d ) 20 kmph , e ) 23 kmph | b | divide(subtract(multiply(30, const_3), divide(multiply(multiply(30, const_3), const_2), const_3)), subtract(const_3, const_1)) | multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)| | physics |
33 1 / 3 % of 270 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 270 = 90 b )" | a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | b | divide(multiply(add(33, divide(1, 3)), 270), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | gain |
mr . hernandez , who was a resident of state x for only 8 months last year , had a taxable income of $ 22,500 for the year . if the state tax rate were 6 percent of the year β s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez β... | "total tax for the year = 22,500 x 6 % = 1350 as stated annual tax is prorated as per the duration of stay . prorated tax = 1350 ( 8 / 12 ) = 900 answer a" | a ) $ 900 , b ) $ 720 , c ) $ 600 , d ) $ 300 , e ) $ 60 | a | divide(multiply(multiply(divide(add(multiply(multiply(6, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), 6), 8), multiply(const_3, 6)) | multiply(n2,const_100)|multiply(const_0_25,const_100)|multiply(const_3,n2)|multiply(#0,const_100)|multiply(#1,const_100)|add(#3,#4)|divide(#5,const_100)|multiply(#6,n2)|multiply(n0,#7)|divide(#8,#2)| | gain |
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1300 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1300 female employees , 60 % of 1300 . = ( 60 * 1300 ) / 100 = 780 . then male employees , = 520 50 % of male are computer literate , = 260 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1300 ) / 100 = 806 computer literate . thus , female computer lite... | a ) 546 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1300) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)| | gain |
each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 5 and the 4 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 4 and 5 are adjacent - consider 4 and 5 together as one group . now you have 4 numbers / groups to arrange which can be done in 4 ! ways . in each of these arrangements , 4 and 5 can be arranged as 45 or 54 . number of arrangements in... | a ) 48 , b ) 66 , c ) 76 , d ) 78 , e ) 72 | e | subtract(multiply(multiply(multiply(5, 4), 5), const_2), multiply(multiply(multiply(4, 5), const_2), const_2)) | multiply(n1,n4)|multiply(n3,n4)|multiply(n3,#0)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|subtract(#4,#5)| | general |
susan drove an average speed of 15 miles per hour for the first 40 miles of a tripthen at a average speed of 60 miles / hr for the remaining 20 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles / hr for the entire trip | avg . speed = total distance / total time total distance = 60 miles total time = 40 / 15 + 20 / 60 = 3 avg . speed = 20 . answer - b | a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55 | b | divide(add(40, 20), add(divide(40, 15), divide(20, 60))) | add(n1,n3)|divide(n1,n0)|divide(n3,n2)|add(#1,#2)|divide(#0,#3) | physics |
the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 480 m , its area is ? | "5 x + 12 x + 13 x = 480 = > x = 16 a = 80 , b = 192 , c = 208 s = ( 80 + 192 + 208 ) / 2 = 240 answer : b" | a ) 150 , b ) 240 , c ) 277 , d ) 261 , e ) 281 | b | multiply(480, divide(480, add(add(5, 12), 13))) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n3,#2)| | geometry |
the perimeter of a semi circle is 113 cm then the radius is ? | "36 / 7 r = 113 = > r = 22 answer : e" | a ) 17 , b ) 28 , c ) 19 , d ) 11 , e ) 22 | e | divide(113, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics |
a boat having a length 4 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "explanation : volume of water displaced = ( 4 x 3 x 0.01 ) m 3 = 0.12 m 3 . β΄ mass of man = volume of water displaced x density of water = ( 0.12 x 1000 ) kg = 120 kg . answer : b" | a ) 12 kg , b ) 120 kg , c ) 72 kg , d ) 96 kg , e ) none of these | b | multiply(multiply(multiply(4, 3), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
a train 450 m long running at 108 kmph crosses a platform in 25 sec . what is the length of the platform ? | "length of the platform = 108 * 5 / 18 * 25 = 750 β 450 = 300 answer : e" | a ) 271 , b ) 266 , c ) 350 , d ) 277 , e ) 300 | e | subtract(multiply(25, multiply(108, const_0_2778)), 450) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
when the price of sugar was increased by 32 % , a family reduced its consumption in such a way that the expenditure on sugar was only 10 % more than before . if 30 kg were consumed per month before , find the new monthly consumption . | since , expenditure = price Γ consumption β΄ 110 % of 30 = 132 β 100 Γ new consumption β 110 β 100 Γ 30 = 132 β 100 Γ new consumption β new consumption = 25 kg answer b | a ) 20 kg , b ) 25 kg , c ) 30 kg , d ) 35 kg , e ) none of these | b | multiply(divide(add(const_100, 10), add(const_100, 32)), 30) | add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|multiply(n2,#2) | general |
find the 25 th term of an arithmetic progression whose first term is 5 and the common difference is 7 . | n th term of a . p = a + ( n - 1 ) * d = 5 + ( 25 - 1 ) * 7 , = 5 + 168 = 173 . answer : c | a ) 145 , b ) 38 , c ) 173 , d ) 156 , e ) 189 | c | add(multiply(subtract(25, const_1), 7), 5) | subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1) | general |
in a graduating class , 40 percent of the students are male . in this class , 50 percent of the male students and 30 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | let x be the total number of students . the number students who are younger than 25 is 0.5 * 0.4 x + 0.7 * 0.6 x = 0.62 x the answer is b . | a ) 0.56 , b ) 0.62 , c ) 0.68 , d ) 0.74 , e ) 0.8 | b | subtract(const_1, add(multiply(divide(40, const_100), divide(50, const_100)), multiply(divide(30, const_100), subtract(const_1, divide(40, const_100))))) | divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|add(#3,#5)|subtract(const_1,#6) | general |
the function g ( a ) is defined for integers a such that if a is even , g ( a ) = a / 2 and if a is odd , g ( a ) = a + 5 . given that g ( g ( g ( g ( g ( a ) ) ) ) ) = 19 , how many possible values for a would satisfy this equation ? | let me define terms : in g ( a ) = r a is argument , r is result , g ( ) is function , in g ( g ( g ( g ( g ( a ) ) ) ) ) , g 1 is inner most , g 5 is outermost for identification . from definition of function g , we can deduce that : if result is even then two possibilities for argument = 1 even 1 odd if result is odd... | a ) 1 , b ) 5 , c ) 7 , d ) 8 , e ) 11 | d | add(add(5, 2), const_1) | add(n0,n1)|add(#0,const_1) | general |
there are 16 bees in the hive , then 9 more fly . how many bees are there in all ? | 16 + 9 = 25 . answer is e . | a ) 7 , b ) 33 , c ) 12 , d ) 17 , e ) 25 | e | add(16, 9) | add(n0,n1) | general |
seller selling an apple for rs . 20 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 20 loss = cp 21 loss = cp β sp = cp β 20 β cp 21 = cp β 20 β 20 cp 21 = 20 β cp 21 = 1 β cp = 21 c" | a ) 10 , b ) 12 , c ) 21 , d ) 18 , e ) 20 | c | add(20, 1) | add(n0,n1)| | general |
what are the last two digits of ( 301 * 402 * 503 * 604 * 645 * 547 * 448 * 349 ) ^ 2 | ( ( 301 * 402 * 503 * 604 * 645 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 5,7 , 8,9 ; 6 is missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,45 and final three as 47 * 48 * 49 . solvin... | a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 00 | e | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(const_2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | divide(const_100,const_4)|multiply(const_2,const_3)|subtract(const_1,n8)|multiply(#1,#2)|multiply(#3,#2)|multiply(#4,#2)|add(#0,#5)|multiply(#6,const_4) | general |
nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 1.15 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "b its is . let price = x ( x - 1.15 ) 8 = 6 x x = 4.6 hence total money = 6 * 4.6 = 27.6" | a ) $ 22 , b ) $ 27.6 , c ) $ 30 , d ) $ 36 , e ) $ 40 | b | multiply(divide(multiply(1.15, 8), subtract(8, 6)), 6) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2)| | general |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 60 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease o... | the actual answer is obtained by multiplying 160 % by 70 % and subtracting 100 % from this total . that is : 160 % Γ 70 % = 112 % ; 112 % β 100 % = 12 % . answer : c | a ) β 5 % , b ) 5 % , c ) 12 % , d ) 20 % , e ) 80 % | c | multiply(subtract(multiply(add(const_1, divide(60, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain |
a fair coin is tossed 3 times . what is the probability of getting at least 2 tails ? | "let ' s find the probability of 2 tails , 3 tails p ( ttt ) = ( ( 1 / 2 ) ^ 3 = 1 / 8 . p ( htt ) = ( 3 ! / 2 ! ) * ( 1 / 2 ) ^ 3 = 3 / 8 total probablity = 1 / 8 + 3 / 8 = 1 / 2 answer ( c )" | a ) 3 / 4 , b ) 3 / 2 , c ) 1 / 2 , d ) 3 / 4 , e ) 1 / 4 | c | divide(add(add(add(choose(3, const_2), choose(3, const_3)), choose(3, const_4)), choose(3, 3)), power(const_2, 3)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
a 75 - liter solution of cool - drink is made from 10 % jasmine water . if 3.5 liters of jasmine and 9.5 liters of water were added to the solution , what percent of the solution is jasmine ? | "the percent of jasmine in the resulting solution is : ( amount of jasmine ) / ( total volume ) ( 0.1 ( 75 ) + 3.5 ) / 88 = 11 / 88 = 1 / 8 = 12.5 % the answer is a ." | a ) 12.5 % , b ) 14 % , c ) 15.5 % , d ) 17 % , e ) 18.5 % | a | add(3.5, multiply(divide(10, const_100), 75)) | divide(n1,const_100)|multiply(n0,#0)|add(n2,#1)| | gain |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 1200 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2400 m t = d / s t = 2400 / 10 t = 240 sec answer : b" | a ) 200 sec , b ) 240 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | b | divide(add(1200, 1200), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics |
a 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform . | "explanation : speed = distance / time = 300 / 18 = 50 / 3 m / sec let the length of the platform be x meters then distance = speed β time x + 300 = 50 / 3 β 39 = > 3 ( x + 300 ) = 1950 = > x = 350 meters option d" | a ) 310 meter , b ) 335 meter , c ) 345 meter , d ) 350 meter , e ) none of these | d | subtract(multiply(divide(300, 18), 39), 300) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
a and b go around a circular track of length 100 m on a cycle at speeds of 18 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 100 / ( 18 * 5 / 18 ) , 100 / ( 36 * 5 / 18 ) } = lcm ( 20 , 10 ) = 20 sec . answer : e" | a ) 120 sec , b ) 198 sec , c ) 178 sec , d ) 665 sec , e ) 20 sec | e | divide(100, subtract(multiply(36, const_0_2778), multiply(18, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
zoey won the lottery and got $ 7 , 348340 . she wants to split it among herself and 5 friends evenly . how much money must she add if she wants to split it evenly ? | zoey and her 5 friends is 6 people in all . $ 7 , 348340 must be divisible by 6 if she wants to split it evenly . the money is divisible by 6 if it ' s divisible by 2 and 3 . 7 + 3 + 4 + 8 + 3 + 4 + 0 = 29 . 2 and 3 cant go into 29 . adding 1 dollar makes it 30 . 2 and 3 can go into 30 so 6 can also . the answer is d . | a ) $ 3 , b ) $ 2 , c ) $ 9 , d ) $ 1 , e ) $ 4 | d | subtract(reminder(add(multiply(7, multiply(const_1000, const_1000)), 348340), add(5, const_1)), const_1) | add(n2,const_1)|multiply(const_1000,const_1000)|multiply(n0,#1)|add(n1,#2)|reminder(#3,#0)|subtract(#4,const_1) | general |
in a certain code language , ' book ' is coded as ' pencil ' , ' pencil ' is coded as ' mirror ' , ' mirror ' is coded as ' board ' . then what is useful to write on a paper ? | we use pencil to write on a paper but here pencil is coded as mirror . so , the answer is mirror . answer : a | a ) 2 , b ) 6278 , c ) 277 , d ) 281 , e ) 22 | a | multiply(const_2, const_1) | multiply(const_1,const_2) | general |
what is the angle between the hands of a clock when time is 8 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 8 - 40 * 11 / 2 = 320 - 220 = 100 deg answer : e" | a ) 65 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 100 deg | e | subtract(multiply(40, multiply(const_3, const_2)), 8) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)| | geometry |
how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 6 ? | "the numbers should be of the form 6 c + 3 . the minimum is 3 when c = 0 . the maximum is 45 when c = 7 . there are 8 such numbers . the answer is d ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(50, const_10) | divide(n1,const_10)| | general |
a box contains 4 red balls and 4 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 4 black balls be bbbband 4 red balls be rrrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 8 ! ) / ( 4 ! x 4 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 7 slots and 3 black ( bbb ) and 4 red ( rrrr ) the... | a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 | e | divide(add(4, 4), 4) | add(n0,n1)|divide(#0,n1)| | probability |
if 50 % of ( x - y ) = 30 % of ( x + y ) , then what percent of x is y ? | "50 % of ( x - y ) = 30 % of ( x + y ) 50 / 100 ( x - y ) = 30 / 100 ( x + y ) x = 4 y required percentage = y / x * 100 = y / 4 y * 100 = 25 % answer is c" | a ) 20 % , b ) 30 % , c ) 25 % , d ) 15 % , e ) 50 % | c | multiply(divide(subtract(50, 30), add(50, 30)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
30 ^ 10 / 180 ^ 5 = ? | "30 ^ 10 / 180 ^ 5 = ? a . 5 ^ 5 b . 5 ^ 6 c . 3 ^ 6 d . 6 ^ 3 e . 15 ^ 3 - > 30 ^ 10 / 180 ^ 5 = ( 30 ^ 10 ) / ( 6 ^ 5 ) ( 30 ^ 5 ) = ( 30 ^ 5 ) / ( 6 ^ 5 ) = ( 6 ^ 5 ) ( 5 ^ 5 ) / ( 6 ^ 5 ) = 5 ^ 5 . thus , a is the answer ." | a ) 5 ^ 5 , b ) 5 ^ 6 , c ) 3 ^ 6 , d ) 6 ^ 3 , e ) 15 ^ 3 | a | divide(power(30, 10), power(30, 5)) | power(n0,n1)|power(n0,n3)|divide(#0,#1)| | general |
an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 1... | fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of... | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | e | add(add(const_2, add(const_1, const_4)), const_2) | add(const_1,const_4)|add(#0,const_2)|add(#1,const_2) | general |
when the price of an article was reduced by 30 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 30 % , the new price is 0.7 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.7 p ) ( 1.8 n ) = 1.26 np , which is 1.26 times the old revenue , so is 26 % greater . answer : a" | a ) 26 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 30)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | gain |
if the simple interest on a certain sum of money for 4 years is one β fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 4 years rate of interest per annum ( r ) = ( 100 Γ si ) / pt = ( 100 Γ ( x / 5 ) / ( x Γ 4 ) = 20 / 4 = 5 % answer : option d" | a ) 4 % , b ) 7 % , c ) 6 % , d ) 5 % , e ) 3 % | d | divide(divide(const_100, add(const_1, const_4)), 4) | add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0)| | gain |
the average of 11 results is 60 . if the average of first 6 results is 58 and that of the last 6 is 63 , find the sixth result ? | sixth result = 58 * 6 + 63 * 6 - 60 * 11 = 66 answer is e | a ) 50 , b ) 52 , c ) 65 , d ) 42 , e ) 66 | e | subtract(add(multiply(6, 58), multiply(6, 63)), multiply(11, 60)) | multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general |
if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 116 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 116 3 x = 111 x = 37 kg then the student ' s weight is 79 kg . the answer is d ." | a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | d | subtract(116, divide(subtract(116, 5), const_3)) | subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)| | other |
a circular path of 13 m radius has marginal walk 2 m wide all round it . find the cost of leveling the walk at 25 p per m 2 ? | "Ο ( 152 - 132 ) = 176 176 * 1 / 4 = rs . 44 answer : c" | a ) rs . 48 , b ) rs . 64 , c ) rs . 44 , d ) rs . 46 , e ) rs . 34 | c | multiply(25, subtract(circle_area(add(13, 2)), circle_area(13))) | add(n0,n1)|circle_area(n0)|circle_area(#0)|subtract(#2,#1)|multiply(n2,#3)| | physics |
tom drives from town q to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between q and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per ... | let ' s assume that it takes 4 hours to go from point q to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b . | a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36 | b | divide(multiply(60, 36), multiply(36, const_3)) | multiply(n0,n1)|multiply(n1,const_3)|divide(#0,#1) | physics |
a watch was sold at a loss of 10 % . if it was sold for rs . 540 more , there would have been a gain of 8 % . what is the cost price ? | "90 % 108 % - - - - - - - - 18 % - - - - 540 100 % - - - - ? = > rs . 3000 answer : c" | a ) 1000 , b ) 2998 , c ) 3000 , d ) 2788 , e ) 2991 | c | divide(multiply(540, const_100), subtract(add(const_100, 8), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
a man took a loan from a bank at the rate of 8 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was : | solution principal = rs . ( 100 x 5400 / 8 x 3 ) = rs . 22,500 . answer d | a ) rs . 2000 , b ) rs . 10,500 , c ) rs . 15,500 , d ) rs . 22,500 , e ) none | d | divide(divide(multiply(5400, const_100), multiply(3, 8)), multiply(divide(const_60, const_4), const_100)) | divide(const_60,const_4)|multiply(n2,const_100)|multiply(n0,n1)|divide(#1,#2)|multiply(#0,const_100)|divide(#3,#4) | gain |
what is the remainder when the number q = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | 14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus q = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a ) | a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | divide(n4,n4) | general |
a jar contains 3 black , 3 white and 1 green balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ? | p ( 1 st black , 2 nd white ) = 3 / 7 * 3 / 6 = 9 / 42 ; p ( 1 st white , 2 nd black ) = 3 / 7 * 3 / 6 = 9 / 42 . p = 9 / 42 + 9 / 42 = 18 / 42 = 3 / 7 . answer : d . | a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 2 | d | multiply(divide(3, subtract(add(add(3, 3), 1), 1)), divide(add(3, 3), add(add(3, 3), 1))) | add(n0,n0)|add(n2,#0)|divide(#0,#1)|subtract(#1,n2)|divide(n0,#3)|multiply(#4,#2) | probability |
of the two square fields , the area of the one is 1 hectare , while anothe one is broader by 1 % . there differences in area is : | area of one square field = 10000 m ( power ) 2 10000 Γ 1 = 10000 side of this field = β 10000 m = 100 m side of another square = 101 m difference of areas = [ 101 ( power ) 2 - 100 ( power ) 2 ] m ( power ) 2 [ 101 + 100 ] [ 101 - 100 ] m ( power ) 2 ( 201 ) ( 1 ) m 2 = 201 m ( power ) 2 answer is a . | ['a ) 201 m ( power ) 2', 'b ) 220 m ( power ) 2', 'c ) 211 m ( power ) 2', 'd ) 219 m ( power ) 2', 'e ) 205 m ( power ) 2'] | a | subtract(square_area(add(1, sqrt(multiply(const_10, const_1000)))), multiply(const_10, const_1000)) | multiply(const_10,const_1000)|sqrt(#0)|add(n0,#1)|square_area(#2)|subtract(#3,#0) | geometry |
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