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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in a 2000 census , 51.27 % of the population are male , and 48.73 % are female . what is the ratio of men to every 100 women ? | the ratio of a to b , where a = the percent of males , and b = the percent of females . a = 51.27 b = 48.73 a : b = 51.27 : 48.73 so , 100 ( 5,127 / 4,873 ) : 100 giving a male to female ratio of 105.20 men for every 100 women . answer is b | a ) 110 : 100 , b ) 105 : 100 , c ) 95 : 100 , d ) 85 : 100 , e ) 120 : 100 | b | divide(51.27, 48.73) | divide(n1,n2) | other |
the radius of a semi circle is 14 cm then its perimeter is ? | "diameter = 28 cm 1 / 2 * 22 / 7 * 28 + 28 = 72 answer : c" | a ) 32.8 , b ) 32.4 , c ) 72 , d ) 32.2 , e ) 32.9 | c | add(divide(circumface(14), const_2), multiply(14, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
a large field of 900 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | "let the areas of the parts be x hectares and ( 900 - x ) hectares . difference of the areas of the two parts = x - ( 900 - x ) = 2 x - 900 one - fifth of the average of the two areas = 1 / 5 [ x + ( 900 β x ) ] / 2 = 1 / 5 Γ ( 900 / 2 ) = 450 / 5 = 90 given that difference of the areas of the two parts = one - fifth o... | a ) 435 hectares . , b ) 425 hectares . , c ) 445 hectares . , d ) 415 hectares . , e ) 405 hectares . | e | divide(subtract(900, multiply(divide(900, const_2), divide(const_1, add(const_1, const_4)))), const_2) | add(const_1,const_4)|divide(n0,const_2)|divide(const_1,#0)|multiply(#1,#2)|subtract(n0,#3)|divide(#4,const_2)| | general |
a train passes a station platform in 60 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 36 = 15 = > x = 420 m answer : b" | a ) 615 m , b ) 420 m , c ) 168 m , d ) 197 m , e ) 691 m | b | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
15 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required ? | "that is , 1 work done = 15 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 8 ? ( i . e . no . of men required ) = 15 Γ 8 Γ 10 / 15 Γ 8 = 10 days a )" | a ) 10 days , b ) 11 days , c ) 9 days , d ) 8 days , e ) 12 days | a | divide(multiply(multiply(15, 10), 8), multiply(8, 15)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)| | physics |
a factory produces 4560 toys per week . if the workers at this factory work 4 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 4 days ) by 4 . 4560 / 4 = 1140 toys correct answer a" | a ) 1140 toys , b ) 2375 toys , c ) 3375 toys , d ) 4375 toys , e ) 5375 toys | a | divide(4560, 4) | divide(n0,n1)| | physics |
the sum of four consecutive even numbers is 140 . what would be the largest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x + 1 ) = 38 . answer : b" | a ) 33 , b ) 38 , c ) 70 , d ) 123 , e ) 12 | b | add(add(power(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), power(... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a batsman makes a score of 82 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 82 = total runs scored in 17 innings = > 16 ( x - 3 ) + 82 = 17 x = > 16 x - 48 + 82 = 17 x = > x = 34 answer is e ." | a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 34 | e | add(subtract(82, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
a polling company surveyed a certain country , and it found that 35 % of that country β s registered voters had an unfavorable impression of both of that state β s major political parties and that 20 % had a favorable impression only of party a . if one registered voter has a favorable impression of both parties for ev... | "assume the total pool of registered voters = 100 , so 35 of the country β s registered voters had an unfavorable impression of both of that state β s major political parties and 20 had a favorable impression only of party a let x = # of voters with a favorable impression of both parties let 2 x = # of voters with a fa... | a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | e | divide(subtract(const_100, add(35, 20)), const_3) | add(n0,n1)|subtract(const_100,#0)|divide(#1,const_3)| | gain |
the length of a rectangle is reduced by 30 % . by what % would the width have to be increased to maintain the original area ? | "sol . required change = ( 30 * 100 ) / ( 100 - 30 ) = 42.8 % a" | a ) 42.8 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 35 % | a | multiply(divide(subtract(const_1, divide(subtract(const_100, 30), const_100)), divide(subtract(const_100, 30), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry |
what is the remainder if 7 ^ 4 is divided by 100 ? | "7 * 7 * 7 * 7 / 100 = 2401 / 100 = 24 reminder 1 answer : c" | a ) 5 , b ) 4 , c ) 1 , d ) 2 , e ) 10 | c | subtract(divide(100, const_2), multiply(7, 7)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
how many positive factors do 120 and 80 have in common ? | "the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 120 and 80 is 40 number of factors of 40 = 8 answer : a" | a ) 8 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(80, const_10), const_10) | subtract(n1,const_10)|divide(#0,const_10)| | other |
find value of x : ( 422 + 404 ) ^ 2 β ( 4 Γ 422 Γ 404 ) = x . | given equation is in the form ( a + b ) 2 β 4 ab where a = 422 and b = 404 hence answer = ( a + b ) 2 β 4 ab = ( a β b ) 2 = ( 422 β 404 ) 2 = 182 = 324 c | a ) 234 , b ) 267 , c ) 324 , d ) 356 , e ) 577 | c | subtract(power(add(422, 404), const_2), multiply(multiply(4, 422), 404)) | add(n0,n1)|multiply(n0,n3)|multiply(n1,#1)|power(#0,const_2)|subtract(#3,#2) | general |
after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6400 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6400 1 = 6400 / 68.4 100 = ( 6400 * 100 ) / 68.4 = rs . 9357 . answer : option d" | a ) s . 6000 , b ) s . 9000 , c ) s . 10800 , d ) s . 9357 , e ) s . 9980 | d | divide(multiply(6400, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)| | gain |
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 . 14080 ? | "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 = ... | a ) 3200 , b ) 4000 , c ) 3250 , d ) 3825 , e ) 4096 | e | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14080, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | multiply(const_3,n2)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,n2)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5)| | gain |
6 ) a marketing firm determined that , of 160 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | "80 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 120 = > 4 x = 20 = > x = 5 answer - e" | a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 40 , e ) e ) 5 | e | divide(subtract(subtract(160, 80), 60), const_4) | subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4)| | other |
every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 54 , how many more green disks than blue disks are in the bag ? | "let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 54 - - > x = 3 . g - b = 8 x - 3 x = 5 x = 15 . the answer is a ." | a ) 15 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | a | multiply(divide(54, add(add(3, 7), 8)), subtract(8, 3)) | add(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(n3,#2)|multiply(#3,#1)| | general |
what will be the fraction of 6 % | "explanation : 6 * 1 / 100 = 3 / 50 . option b" | a ) 1 / 20 , b ) 3 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | b | divide(circle_area(divide(6, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | gain |
among 250 viewers interviewed who watch at least one of the 3 tv channels namely a , bc . 116 watch a , 127 watch c , while 107 watch b . if 90 watch exactly two channels . how many watch exactly one channel ? | 250 = n ( exactly 1 channel ) + n ( exactly 2 channels ) + n ( exactly 3 channels ) 250 = n ( exactly 1 channel ) + 90 + n ( exactly 3 channels ) let ' s find the value of n ( exactly 3 channels ) = x 250 = n ( a ) + n ( b ) + n ( c ) - n ( a and b ) - n ( b and c ) - n ( c and a ) + n ( a and b and c ) note that each ... | a ) 155 , b ) 160 , c ) 175 , d ) 190 , e ) 195 | a | subtract(subtract(250, 90), divide(subtract(subtract(add(add(116, 127), 107), 90), 250), const_2)) | add(n2,n3)|subtract(n0,n5)|add(n4,#0)|subtract(#2,n5)|subtract(#3,n0)|divide(#4,const_2)|subtract(#1,#5) | general |
some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 50 ( 1 - x ) + 25 x = 30 x = 4 / 5 answer : e" | a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | e | divide(subtract(divide(30, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
9 spheres of the same size are made from melting a solid cylinder of 8 cm diameter and 48 cm height . what is the diameter of each sphere ? | "volume of cylinder = pi * r ^ 2 * h volume of a sphere = 4 * pi * r ^ 3 / 3 9 * 4 * pi * r ^ 3 / 3 = pi * r ^ 2 * h r ^ 3 = r ^ 2 * h / 12 = 64 cm ^ 3 r = 4 cm d = 8 cm the answer is c ." | a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 12 cm | c | multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(8, const_2), 48), const_pi), const_3), const_4), 9), const_4), const_4), const_2) | divide(n1,const_2)|volume_cylinder(#0,n2)|divide(#1,const_pi)|multiply(#2,const_3)|divide(#3,const_4)|divide(#4,n0)|divide(#5,const_4)|divide(#6,const_4)|multiply(#7,const_2)| | physics |
working simultaneously and independently at an identical constant rate , 6 machines of a certain type can produce a total of x units of product p in 5 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 4 x units of product p in 10 days ? | "the rate of 6 machines is rate = job / time = x / 5 units per day - - > the rate of 1 machine 1 / 6 * ( x / 5 ) = x / 30 units per day ; now , again as { time } * { combined rate } = { job done } then 10 * ( m * x / 30 ) = 4 x - - > m = 12 . answer : c ." | a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 18 | c | multiply(multiply(6, 4), divide(5, 10)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 220 true ? | "answer d ( a + b ) ( a - b ) = 220 6 cases for ( a + b ) , ( a - b ) 220 , 1 110 , 2 55 , 4 44,5 22 , 10 20,11 answer d" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | d | divide(log(220), log(add(const_4, const_1))) | add(const_1,const_4)|log(n2)|log(#0)|divide(#1,#2)| | general |
a boat goes 100 km downstream in 4 hours , and 75 km upstream in 15 hours . the speed of the stream is ? | "100 - - - 10 ds = 25 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 25 - 5 ) / 2 = 10 kmph . answer : a" | a ) 10 , b ) 20 , c ) 25 , d ) 35 , e ) 15 | a | divide(subtract(divide(100, 4), divide(75, 15)), const_2) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2)| | physics |
what is 10 - 8 + 6 - 4 + . . . + ( - 10 ) ? | "the expression considers all even numbers between 10 and - 10 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , and - 10 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 ... | a ) 0 , b ) 5 , c ) 2 , d ) 14 , e ) 16 | a | subtract(6, 8) | subtract(n2,n1)| | general |
a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 60,000 . the 75 associates have an average salary of $ 30,000 . what is the average salary for the company ? | "another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 2 a * 1 + a * 5 ) / 6 = 7 a / 6 = 35000 answer is a . $ 35,000" | a ) $ 35,000 , b ) $ 45,000 , c ) $ 55,000 , d ) $ 65,000 , e ) $ 75,000 | a | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | add(n0,n1)|multiply(const_3,const_3)|multiply(const_10,const_100)|multiply(const_10,const_3)|multiply(#1,const_10)|multiply(#3,#2)|multiply(#4,#2)|add(#6,#5)|divide(#7,#0)| | general |
along a road lie an odd number of stones placed at intervals of 10 m . these stones have to be assembled around the middle stone . a person can carry only one stone at a time . a man carried out the job starting with the stone in the middle , carrying stones in succession , thereby covering a distance of 4.8 km . then ... | explanation : if he travels 10 m for the first stone he has to travel that dist back as well . . . so he travels 2 * 10 m dist for the first stone and so on . = > 2400 = 2 * ( 10 + 20 + 30 . . . . 10 n ) using formula for ap = > 2400 = 2 * n / 2 * ( 2 * 10 + ( n - 1 ) * 10 ) . = > 2400 = 10 n 2 + 10 n . on solving this... | a ) 35 , b ) 15 , c ) 29 , d ) 31 , e ) 29 | d | add(multiply(divide(multiply(divide(multiply(4.8, const_1000), multiply(multiply(const_2, const_2), 10)), const_2), multiply(const_4, const_4)), const_2), const_1) | multiply(n1,const_1000)|multiply(const_2,const_2)|multiply(const_4,const_4)|multiply(n0,#1)|divide(#0,#3)|multiply(#4,const_2)|divide(#5,#2)|multiply(#6,const_2)|add(#7,const_1) | physics |
850 - ( 1500 Γ· ? ) Γ 8 / 15 = 770 | "explanation : = > 850 - ( 1500 Γ· ? ) Γ 8 / 15 = 770 = > 850 - 770 = ( 1500 Γ· ? ) Γ 8 / 15 = > 80 = ( 1500 Γ· ? ) Γ 8 / 15 = > 80 Γ 15 / 8 = ( 1500 Γ· ? ) = 150 = > 1 / ? = 150 / 1500 = 1 / 10 = > ? = 10 answer : option b" | a ) a ) 31 , b ) b ) 10 , c ) c ) 15 , d ) d ) 26 , e ) e ) 17 | b | subtract(850, divide(1500, 8)) | divide(n1,n2)|subtract(n0,#0)| | general |
the contents of two vessels containing copper and tin in the ratio 2 : 3 and 5 : 7 are mixed in the ratio 3 : 5 . the resulting mixture will have copper and tin in the ratio ? | the ratio of copper and tin the new vessel = ( 2 / 5 * 3 / 8 + 5 / 12 * 5 / 8 ) : ( 3 / 5 * 3 / 8 + 7 / 12 * 5 / 8 ) = 197 / 480 : 283 / 480 = 197 : 283 answer is d | a ) 210 : 283 , b ) 312 : 433 , c ) 178 : 213 , d ) 197 : 283 , e ) 145 : 256 | d | divide(add(multiply(2, 3), multiply(5, 5)), add(multiply(3, 3), multiply(7, 5))) | multiply(n0,n1)|multiply(n2,n2)|multiply(n1,n1)|multiply(n2,n3)|add(#0,#1)|add(#2,#3)|divide(#4,#5) | other |
a and b together can do a work in 8 days . a alone can do it in 12 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 8 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 8 ) - ( 1 / 12 ) = 24 days answer : option d" | a ) 6 days , b ) 8 days , c ) 12 days , d ) 24 days , e ) 25 days | d | add(inverse(subtract(divide(const_1, 8), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)| | physics |
what is the unit β s digit of 7 ^ 4 | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) therefore , last digit of 7 ^ 4 = 1 answer b" | a ) 3 , b ) 1 , c ) 5 , d ) 7 , e ) 9 | b | reminder(power(7, const_2), const_10) | power(n0,const_2)|reminder(#0,const_10)| | general |
x and y are positive integers . when x is divided by 15 , the remainder is 2 , and when x is divided by 10 , the remainder is 7 . when y is divided by 9 , the remainder is 5 , and when y is divided by 7 , the remainder is 2 . what is the least possible value of x * y ? | "when x is divided by 15 , the remainder is 2 : so , the possible values of x are : 2 , 17 , 32 , 47 , etc . when x is divided by 10 , the remainder is 7 : so , the possible values of x are : 7,17 . . . stop . since both lists include 17 , the smallest possible value of x is 17 . when y is divided by 9 , the remainder ... | a ) 389 , b ) 390 , c ) 388 , d ) 391 , e ) 392 | d | subtract(add(multiply(9, const_2), 5), add(2, 15)) | add(n0,n1)|multiply(n4,const_2)|add(n5,#1)|subtract(#2,#0)| | general |
what is the minimum percentage increase in the mean of set s { - 4 , - 1 , 0 , 6 , 9 } if its two smallest elements are replaced with two different primes ? | mean of s is ( - 4 - 1 + 0 + 6 + 9 ) / 5 = 2 ; in order the increase to be minimal we should replace two smallest elements of s , which are - 4 and - 1 , with two smallest primes , which are 2 and 3 . hence our new set will be { 2 , 3 , 0 , 6 , 9 } - - > new mean is ( 2 + 3 + 0 + 6 + 9 ) / 5 = 4 . percent increase = ( ... | a ) 25 % , b ) 50 % , c ) 75 % , d ) 100 % , e ) 200 % | d | divide(multiply(const_2, const_100), const_2) | multiply(const_100,const_2)|divide(#0,const_2) | general |
marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 20 laps around the track per hour and nick completes 10 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the ... | "maria ' s rate - 20 laps per hour - - > 20 / 60 laps / min nick ' s rate - 10 laps per hour - - > 10 / 60 laps / min lets set equations : 20 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 10 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 20 / 60 - 10 / 60 ) * t = 4 - 0 (... | a ) 5 , b ) 8 , c ) 24 , d ) 15 , e ) 20 | c | divide(4, subtract(divide(20, multiply(add(10, const_1), 20)), divide(10, multiply(add(10, const_1), 20)))) | add(n1,const_1)|multiply(n0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(#2,#3)|divide(n2,#4)| | physics |
in a renowned city , the average birth rate is 6 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day . | "this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 6 - 3 = 3 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 3 = 129,... | a ) 129,500 , b ) 129,600 , c ) 129,700 , d ) 129,800 , e ) 129,900 | b | multiply(multiply(subtract(6, 3), const_3600), const_12) | subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)| | general |
how many times in a day , are the hands of a clock in straight line but opposite in direction ? | b 22 the hands of a clock point in opposite directions ( in the same straight line ) 11 times in every 12 hours . ( because between 5 and 7 they point in opposite directions at 6 o ' clcok only ) . so , in a day , the hands point in the opposite directions 22 times . | a ) 21 , b ) 22 , c ) 17 , d ) 26 , e ) 25 | b | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | multiply(const_12,const_2)|multiply(#0,const_2)|subtract(#1,const_4)|divide(#2,const_2) | physics |
add 11 % of 36 and 12 % of 64 . | "11 % of 36 + 12 % of 64 36 * 11 / 100 + 64 * 12 / 100 3.9 + 7.6 = 11.5 answer c" | a ) 9.5 , b ) 10.5 , c ) 11.5 , d ) 12 , e ) 15 | c | add(divide(multiply(11, 36), const_100), divide(multiply(12, 64), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)| | gain |
if the cost price of 55 articles is equal to the selling price of 50 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 50 y = 55 x y = 1.1 x the answer is b ." | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 55)), divide(multiply(const_100, 50), 55))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
in a certain quiz that consists of 8 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 360 points , how many points is the third question worth ? | x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 8 x + 112 = 360 8 x = 248 x = 31 3 rd question = 31 + 8 = 31 + 8 = 39 answer a | a ) 39 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | add(divide(360, 10), subtract(subtract(8, 4), const_1)) | divide(n3,n2)|subtract(n0,n1)|subtract(#1,const_1)|add(#0,#2) | general |
line l passes through the points ( - 2,0 ) and ( 0 , a ) . line ll passes through the points ( 4,0 ) and ( 6,2 ) . what value of a makes the two lines parallel ? | find slopes first slope of l = ( a - 0 ) / ( 0 - ( - 2 ) ) = a / 2 slope of ll = ( 2 - 0 ) / ( 6 - 4 ) = 1 for l and ll to be parallel , their slopes must be equal a / 2 = 1 , a = 2 correct answer b | ['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | b | subtract(const_2, 0) | subtract(const_2,n1) | geometry |
a certain school implemented a reading program for its students , with the goal of getting each student to read 6 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | "ans : c solution : simple multiplication s students , c classes , 6 books / month = 72 books a year total number of books = 72 cs" | a ) 20 cs , b ) cs / 2 , c ) 72 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(6, const_12) | multiply(n0,const_12)| | general |
debony normally drives to work in 45 minutes at an average speed of 40 miles per hour . this week , however , she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20 % . if debony averages between 12 and 16 miles per hour when biking , how many minutes earlier wi... | debony normally drives to work in 45 minutes at an average speed of 40 miles per hour . use formula d = rt car : t 1 : 45 min r 1 : 40 mph d 1 : [ ( 40 * 45 ) / 60 ] = 30 miles bike : t 1 : ? r 2 : 12 - 16 mph d 2 : 08 * d 1 = 24 miles t 1 : [ ( 24 * 60 ) / 12 ] = 120 min ( only 12 mph speed yields an answer given in t... | a ) 135 , b ) 105 , c ) 95 , d ) 75 , e ) 45 | d | subtract(divide(multiply(divide(multiply(divide(multiply(40, 45), const_60), multiply(40, const_2)), const_100), const_60), 12), 45) | multiply(n0,n1)|multiply(n1,const_2)|divide(#0,const_60)|multiply(#2,#1)|divide(#3,const_100)|multiply(#4,const_60)|divide(#5,n3)|subtract(#6,n0) | physics |
if the average ( arithmetic mean ) of a and b is 40 and the average of b and c is 60 , what is the value of c β a ? | "- ( a + b = 80 ) b + c = 120 c - a = 40 b . 40" | a ) 25 , b ) 40 , c ) 90 , d ) 140 , e ) it can not be determined from the information given . | b | subtract(multiply(60, const_2), multiply(40, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
a 6 - liter solution is 25 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | "let x be the amount of pure alcohol required . 0.25 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 1.5 x = 3 liters the answer is d ." | a ) 2.1 , b ) 2.4 , c ) 2.7 , d ) 3.0 , e ) 3.3 | d | subtract(6, multiply(divide(multiply(6, 25), const_100), const_2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_2)|subtract(n0,#2)| | general |
a container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 9 ounces in the ratio 1 : 2 or , x + 2 x = 9 gives x ( alcohol ) = 3 and remaining water = 6 container 2 has 8 ounces in the ratio 1 : 3 or , x + 3 x = 8 gives x ( alcohol ) = 2 and remaining water = 6 mixing both we have alcohol = 3 + 2 and water = 6 + 6 ratio thus alcohol / water = 5 / 12 answer d" | a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 5 : 12 , e ) 7 : 3 | d | divide(add(multiply(9, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))), subtract(add(9, 8), add(multiply(9, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))))) | add(n1,n2)|add(n1,n5)|add(n0,n3)|divide(n1,#0)|divide(n1,#1)|multiply(n0,#3)|multiply(n3,#4)|add(#5,#6)|subtract(#2,#7)|divide(#7,#8)| | other |
if the ratio of apples to bananas is 3 to 2 and the ratio of bananas to cucumbers is 1 to 3 , what is the ratio of apples to cucumbers ? | "the ratio of bananas to cucumbers is 1 to 3 which equals 2 to 6 . the ratio of apples to bananas to cucumbers is 3 to 2 to 6 . the ratio of apples to cucumbers is 3 to 6 which equals 1 : 2 . the answer is b ." | a ) 2 : 3 , b ) 1 : 2 , c ) 3 : 5 , d ) 2 : 5 , e ) 5 : 6 | b | divide(divide(3, 2), 3) | divide(n0,n1)|divide(#0,n3)| | other |
if k and w are the dimensions of a rectangle that has area 60 , and if k and w are integers such that k > w , what is the total number of possible values of k ? | "kw = 60 = 60 * 1 = 30 * 2 = 15 * 4 = 20 * 3 = 12 * 5 = 10 * 6 - - > k can take 6 values answer : e" | a ) two , b ) three , c ) four , d ) five , e ) six | e | multiply(const_3, const_1) | multiply(const_1,const_3)| | geometry |
a candidate appearing for an examination has to secure 42 % marks to pass paper i . but he secured only 60 marks and failed by 20 marks . what is the maximum mark for paper i ? | "he secured 60 marks nd fail by 20 marks so total marks for pass the examinatn = 80 let toal marks x x * 42 / 100 = 80 x = 190 answer : e" | a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 190 | e | divide(add(60, 20), divide(42, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
the market value of a certain machine decreased by 30 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | "market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 30 / 100 ) = 8000 - 2400 = $ 5600 market value in 1984 = market value in 1983 - ( 30 % of $ 8000 ) = 5600 - 2400 = $ 3200 answer : c" | a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 800 | c | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(divide(30, const_100), const_2))) | divide(n0,const_100)|multiply(const_10,const_100)|multiply(const_2,const_4)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#3,#4)|subtract(#3,#5)| | gain |
the sum of the first 50 positive even integers is 2550 . what is the sum of even integers from 102 to 200 inclusive ? | first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : 2550 + ( 100 * 50 ) = 7550 ; answer : b | a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3) | general |
ratio between rahul and deepak is 4 : 3 , after 22 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 22 = 26 = > x = 1 so deepak age is = 3 ( 1 ) = 3 answer : option b" | a ) 10 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | divide(multiply(subtract(26, 22), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
which of the options is a two - digit number , which has a unit ' s digit that exceeds its 10 ' s digit by 4 and the product of the given number and the sum of its digits is equal to 208 . | using the elimination method the option that fits this description is 26 6 - 2 = 4 ( unit ' s digit that exceeds its ten ' s digit by 3 ) 26 * 8 = 208 ( the product of the given number and the sum of its digits is equal to 175 ) answer : d | a ) 15 , b ) 37 , c ) 55 , d ) 26 , e ) 73 | d | divide(208, multiply(4, const_2)) | multiply(n1,const_2)|divide(n2,#0) | general |
2 , 4 , 6 , 8 , 10 , . . . ? | "each number increases by 2 . 2 + 2 = 4 4 + 2 = 6 6 + 2 = 8 8 + 2 = 10 10 + 2 = 12 answer : e" | a ) 8 , b ) 5 , c ) 7 , d ) 3 , e ) 12 | e | subtract(negate(8), multiply(subtract(4, 6), divide(subtract(4, 6), subtract(2, 4)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
the hcf and lcm of two numbers m and n are respectively 5 and 210 . if m + n = 75 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 5 x 210 = 1050 Γ’ Λ Β΄ 1 / m + 1 / n = ( m + n ) / mn = 75 / 1050 = 1 / 14 correct option : a" | a ) 1 / 14 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | a | divide(75, multiply(5, 210)) | multiply(n0,n1)|divide(n2,#0)| | general |
two trains are moving in the same direction at 54 kmph and 36 kmph . the faster train crosses a man in the slower train in 27 seconds . find the length of the faster train ? | "relative speed = ( 54 - 36 ) * 5 / 18 = 5 mps . distance covered in 27 sec = 27 * 5 = 135 m . the length of the faster train = 135 m . answer : a" | a ) 135 , b ) 277 , c ) 187 , d ) 257 , e ) 271 | a | multiply(divide(subtract(54, 36), const_3_6), 27) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
drum x is 1 / 2 full of oil and drum y , which has twice the capacity of drum x , is 1 / 4 full of oil . if all of the oil in drum x is poured into drum y , then drum y will be filled to what capacity ? | ( 1 / 2 ) x = ( 1 / 4 ) y ( 1 / 4 ) y + ( 1 / 4 ) y = ( 1 / 2 ) y the answer is a . | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 5 / 6 , e ) 7 / 8 | a | divide(add(multiply(divide(1, 4), const_12), multiply(divide(const_12, const_2), divide(1, 2))), const_12) | divide(n0,n3)|divide(const_12,const_2)|divide(n0,n1)|multiply(#0,const_12)|multiply(#1,#2)|add(#3,#4)|divide(#5,const_12) | general |
the ratio of three numbers is 1 : 2 : 3 and the sum of their squares is 2744 . the sum of the numbers is ? | "let the numbers be x , 2 x , 3 x then , x ^ 2 + 4 x ^ 2 + 9 x ^ 2 = 2744 14 x ^ 2 = 2744 x ^ 2 = 196 x = 14 answer is d" | a ) 10 , b ) 12 , c ) 15 , d ) 14 , e ) 9 | d | sqrt(divide(2744, add(power(3, 2), add(power(1, 2), power(2, 2))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
0.6 of a number is equal to 0.08 of another number . the ratio of the numbers will be | let the numbers be x and y 0.6 x = 0.08 y = > x / y = 0.08 / 0.6 = 2 : 15 answer c . | a ) 3 : 4 , b ) 4 : 3 , c ) 2 : 15 , d ) 2 : 9 , e ) 3 : 8 | c | divide(0.08, 0.6) | divide(n1,n0) | other |
if a coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 6 times , what is the probability that it will land heads up on all 6 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 64 answer : e | a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 32 , d ) 1 / 2 , e ) 1 / 64 | e | divide(1, power(2, 6)) | power(n1,n2)|divide(n0,#0)| | probability |
the perimeter of one face a of cube is 20 cm . its volume must be | solution edge of the cube = ( 20 / 4 ) cm βΉ = βΊ 5 cm . volume = ( 5 Γ 5 Γ 5 ) cm 3 βΉ = βΊ 125 cm 3 . answer a | ['a ) 125 cm 3', 'b ) 400 cm 3', 'c ) 1000 cm 3', 'd ) 8000 cm 3', 'e ) none'] | a | volume_cube(square_edge_by_perimeter(20)) | square_edge_by_perimeter(n0)|volume_cube(#0) | geometry |
a train 660 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 660 + x ) / 60 = 65 / 3 x = 640 m . answer : a" | a ) 640 , b ) 277 , c ) 500 , d ) 297 , e ) 435 | a | divide(660, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
two goods trains each 850 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 850 + 850 = 1700 m . required time = 1700 * 6 / 125 = 81.6 sec . answer : b" | a ) 22 sec , b ) 81.6 sec , c ) 48 sec , d ) 18.3 sec , e ) 9.32 sec | b | add(45, 30) | add(n1,n2)| | physics |
the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 8400 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 120 ) = rs . 7000 . answer : e" | a ) rs . 6725 , b ) rs . 6727 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 7000 | e | divide(8400, add(const_1, divide(20, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
what is 92 % of 3 / 4 ? | "soln : - 120 % = 12 / 10 5 / 8 * 12 / 10 = 60 / 80 = 6 / 8 = 3 / 4 = 0.69 answer : e ." | a ) 6.9 , b ) 69.0 , c ) 0.6845 , d ) 0.6859 , e ) 0.69 | e | divide(multiply(92, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 30 % profit ? | "let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 130 % of $ 640 = $ 832 . d" | a ) $ 480 , b ) $ 450 , c ) $ 960 , d ) $ 832 , e ) $ 660 | d | add(divide(multiply(divide(add(832, 448), const_2), 30), const_100), divide(add(832, 448), const_2)) | add(n0,n1)|divide(#0,const_2)|multiply(n2,#1)|divide(#2,const_100)|add(#3,#1)| | gain |
a 140 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 140 / 6 * 18 / 5 = 84 kmph answer : c" | a ) 228 , b ) 108 , c ) 84 , d ) 188 , e ) 211 | c | multiply(divide(140, 6), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 18 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neithe... | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 18 - 2 cmb = 100 b - cb - mb - 2 cmb = 28 hence th... | a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 2 , d ) 7 / 25 , e ) 9 / 10 | d | divide(subtract(100, subtract(add(50, 40), 18)), 100) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|divide(#2,n0)| | other |
if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 5 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 5 : : 42.75 : x = > 11.25 x x = 5 x 42.75 = > x = ( 5 x 42.75 ) / 11.25 = > x = 19 answer : c" | a ) 22.8 kg , b ) 25.6 kg , c ) 19 kg , d ) 26.5 kg , e ) none of these | c | divide(multiply(5, 42.75), 11.25) | multiply(n1,n2)|divide(#0,n0)| | physics |
the area of a rectangle is 460 square metres . if the length is 15 % more than the breadth , what is the breadth of the rectangular field ? | explanation : let breadth = x meters . then , length = meters given that , = > x = 20 breadth = 20 answer : a ) 20 m | ['a ) 20', 'b ) 27', 'c ) 26', 'd ) 188', 'e ) 11'] | a | sqrt(divide(460, add(const_1, divide(15, const_100)))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|sqrt(#2) | geometry |
find the product of the local value and absolute value of 6 in 564823 | "explanation : place value = local value face value = absolute value the place value of 6 in 564823 is 6 x 10000 = 60000 the face value of 6 in 564823 is nothing but 6 . = > 60000 x 6 = 360,000 answer : option e" | a ) 8000 , b ) 16000 , c ) 12000 , d ) 18000 , e ) 360000 | e | multiply(multiply(6, const_1000), 6) | multiply(n0,const_1000)|multiply(n0,#0)| | general |
find the compound interest on $ 100000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 100000 * ( 1 + 2 / 100 ) ^ 4 = 100000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 108243.21 c . i . = 108243.21 - 100000 = $ 8243.21 answer is e" | a ) $ 645.56 , b ) $ 824.32 , c ) $ 954.26 , d ) $ 745.69 , e ) $ 8243.21 | e | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 100000), 100000) | divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)| | gain |
rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 135 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 135 x = 1500 answer : d" | a ) 2333 , b ) 2777 , c ) 2688 , d ) 1500 , e ) 2871 | d | divide(subtract(135, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100))) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,n2)|divide(#2,const_100)|subtract(#0,#1)|subtract(n3,#3)|divide(#5,#4)| | gain |
( 0.66 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.66 ) ( power 2 ) + 0.066 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.66 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.66 ) ( power 2 ) + ( 0.66 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.66 - 0.1 ) = 0.56 answer is b" | a ) 0.86 , b ) 0.56 , c ) 0.96 , d ) 0.69 , e ) 0.76 | b | divide(subtract(power(0.66, 3), power(0.1, 3)), add(add(power(0.66, 2), 0.066), power(0.1, 2))) | power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)| | general |
a trader cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 10 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 50 % of that is added to the weight , the weight claimed by the trader is obtained . if he cha... | "anyways , one can infer that he ' steals ' 10 % from suppliers and then charges 50 % extra to customers so basically 1.1 * 1.5 = 1.65 given that 1 is start point , we get 65 % more hence answer is b" | a ) 80 % , b ) 65 % , c ) 28.33 % , d ) 19.109 % , e ) 50 % | b | subtract(multiply(divide(add(const_100, 50), const_100), add(const_100, 10)), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)| | gain |
when the no . 7 y 86038 is exactly divisible by 11 , then the smallest whole no . in place of y ? | the given number = 7 y 86038 sum of the odd places = 8 + 0 + 8 + 7 = 23 sum of the even places = 3 + 6 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 23 - ( 9 + y ) = divisible by 11 14 οΏ½ y = divisible by 11 . y must be 3 , to make given number divisible by 11 . b | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(subtract(add(add(7, multiply(const_2, const_4)), multiply(const_2, const_4)), add(multiply(const_2, const_3), const_3)), 11) | multiply(const_2,const_4)|multiply(const_2,const_3)|add(n0,#0)|add(#1,const_3)|add(#2,#0)|subtract(#4,#3)|subtract(#5,n2) | general |
set i consists of integers { 3 , - 8 , y , 19 , - 6 } and set b consists of integers { k , - 3 , 0 , 16 , - 5 , 9 } . number l represents the median of set i , number m represents mode of set b and number z = l ^ m . if y is an integer greater than 21 , for what value of k will z be a divisor of 26 ? | i = { 3 , - 8 , y , 19 , - 6 } b = { k , - 3 , 0,16 , - 5,9 } y > 21 so l = median of i = 3 m = mode of set b z = ( 3 ) ^ m if z is a divisor of 26 , ( 3 ) ^ m = 1 because 26 does not have 3 as a factor = > m = 0 hence k = 0 , as m is mode and 0 will be the most frequently occuring number in set b . answer - c | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | power(3, 0) | power(n0,n5) | general |
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 12 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 12 w = 120 = > w = 10 answer e" | a ) 25 , b ) 23 , c ) 22 , d ) 20 , e ) 10 | e | divide(rectangle_area(5, 24), 12) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry |
a rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 680 sq . feet , how many feet of fencing will be required ? | given that area of the field = 680 sq . feet = > lb = 680 sq . feet length ( l ) = 20 feet = > 20 Γ b = 680 = > b = 680 / 20 = 34 feet required length of the fencing = l + 2 b = 20 + ( 2 Γ 34 ) = 88 feet answer is d . | ['a ) 80 feet', 'b ) 85 feet', 'c ) 83 feet', 'd ) 88 feet', 'e ) 90 feet'] | d | subtract(rectangle_perimeter(20, divide(680, 20)), 20) | divide(n1,n0)|rectangle_perimeter(n0,#0)|subtract(#1,n0) | geometry |
simple interest on a certain sum of money for 2 years at 12 % per annum is half the compound interest on rs . 6000 for 2 years at 15 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 6000 x ( 1 + 15 / 100 ) Γ’ Β² - 4000 ] rs . ( 4000 x 115 / 100 x 115 / 100 - 6000 ) = rs . 7935 . sum = rs . [ 3967.5 x 100 / 4 x 12 ] = rs . 8265.63 . answer c" | a ) rs . 6000 , b ) rs . 7935 , c ) rs . 8265.63 , d ) rs . 3967.5 , e ) none | c | divide(multiply(divide(divide(add(divide(multiply(6000, 15), const_100), divide(multiply(add(6000, divide(multiply(6000, 15), const_100)), 15), const_100)), 2), 2), const_100), 12) | multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1)| | gain |
10 men working 12 hours daily finish a work in 24 days . in how many days will they finish 1 / 4 th of the work working 8 hours daily ? | the number of days needed to finish 1 / 4 of the work = 1 / 4 * 12 / 8 * 24 = 9 days . answer : a | a ) 9 days , b ) 10 days , c ) 11 days , d ) 12 days , e ) 18 days | a | divide(divide(multiply(12, 24), 4), 8) | multiply(n1,n2)|divide(#0,n4)|divide(#1,n5) | physics |
65 % of x = 20 % of 747.50 . find the value of x ? | "65 % of x = 20 % of 747.50 then , 65 / 100 * x = 20 / 100 * 7475 / 10 x = 230 answer is b" | a ) 100 , b ) 230 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(747.50, const_100), 20), const_100), 65) | divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0)| | general |
what is the remainder wen 2 ^ 35 is divided by 5 ? | "2 ^ 35 / 5 = > ( 2 ^ 4 ) ^ 8 * 2 ^ 3 / 5 = > ( 5 * 3 + 1 ) ^ 8 * ( 5 + 3 ) / 5 = > ( 1 * 3 ) / 5 = > 3 / 5 = > remainder = 3 answer : c" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(divide(5, const_2), multiply(2, 2)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
a certain car can travel 40 kilometers on a liter of fuel . if the fuel tank β s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers ) | "fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 / 1.6 mile speed ( km / hour ) = d / t = 40 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 40 x 3.9 x 3.8 / 5.7 x 1.6 = 7... | a ) 52 , b ) 65 , c ) 78 , d ) 91 , e ) 104 | b | divide(multiply(divide(multiply(40, 3.8), 1.6), 3.9), 5.7) | multiply(n0,n4)|divide(#0,n6)|multiply(n1,#1)|divide(#2,n2)| | physics |
each side of a rectangle is increased by 100 % . by what percentage does the area increase ? | original area = a * b where a and b are sides after increase in side increase in area will be = ( ( ( 2 a * 2 b ) - ( a * b ) ) * 100 ) / ( a * b ) = 300 % answer : a | a ) 300 % , b ) 400 % , c ) 500 % , d ) 200 % , e ) 600 % | a | multiply(subtract(multiply(const_2, const_2), const_1), 100) | multiply(const_2,const_2)|subtract(#0,const_1)|multiply(n0,#1) | geometry |
the average monthly salary of 20 employees in an organisation is rs . 1300 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1400 * 21 - 1300 * 20 ) = rs . 3400 . answer : d" | a ) 3600 , b ) 3890 , c ) 88798 , d ) 3400 , e ) 2891 | d | subtract(multiply(add(1300, 100), add(20, const_1)), multiply(1300, 20)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
convert 2.0 hectares in ares | "2.0 hectares in ares 1 hectare = 100 ares therefore , 2.0 hectares = 2.0 Γ 100 ares = 200 ares . answer - d" | a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 200 ares . , e ) 250 ares . | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 2.0), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
seller selling an apple for rs . 19 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | sp = 19 loss = cp 20 loss = cp β sp = cp β 19 β cp 20 = cp β 19 β 19 cp 20 = 19 β cp 20 = 1 β cp = 20 c | a ) 10 , b ) 12 , c ) 20 , d ) 18 , e ) 21 | c | add(19, 1) | add(n0,n1) | general |
what is the difference between the largest and the smallest number written with 7 , 7 , 0 , 7 ? | "7770 7077 - - - - - - - - - - - - - 693 answer : d" | a ) 6893 , b ) 6993 , c ) 6093 , d ) 693 , e ) none | d | subtract(add(add(add(multiply(multiply(7, const_100), const_10), multiply(7, const_100)), multiply(7, const_10)), 0), add(add(add(const_1000, multiply(7, const_100)), multiply(7, const_10)), 7)) | multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)| | general |
if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 3 = 2 d / 15 the average of november and january is ( n + j ) / 2 = 8 d / 15 / 2 = 4 d / 15 d is 15 / 4 times the average of november and january . the answer is d ." | a ) 5 / 3 , b ) 5 / 4 , c ) 10 / 3 , d ) 15 / 4 , e ) 15 / 2 | d | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 3))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general |
a man bring 2 cats to his home to eliminate all the mice . they do a half of the work in 5 days , then the man brings 3 more cats . if they hunt at the same rate , how many days will take to hunt all the mice , since the first day ? | we have that : 2 cats - - - - - 5 days , ( 2 + 3 ) c - - - - - x d , that is : ( x d / 2 c ) = ( 5 d / 5 p ) , then : x d = ( 5 d / 5 c ) 2 c = 2 d . is to say , the 5 cats hunt the same quantity in 2 days ; as it is only 1 / 2 of the work , all the mice would be hunt in : m = 5 d + 2 d = 7 days . answer a . | a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | a | add(multiply(divide(2, 5), 5), 5) | divide(n0,n1)|multiply(n1,#0)|add(n1,#1) | physics |
a sum of money lent out at s . i . amounts to rs . 720 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 720 ) = rs . 300 . s . i . for 2 years = 300 / 5 * 2 = rs . 120 . principal = ( 720 - 120 ) = rs . 600 . answer : a" | a ) 600 , b ) 288 , c ) 2768 , d ) 287 , e ) 191 | a | subtract(720, multiply(divide(subtract(1020, 720), 5), 2)) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)| | general |
there are 15 members in a family . it takes 180 days for a women to complete the work , however man completes in 120 days . they take 17 days to complete the work if men work on alternate days and women work every third day . if all of them started together on the 1 st day , then how many women are there in the family | let the number of men be m and so the number of women = 15 - m out of 17 days , men and women work together on 1,4 , 7,10 , 13,16 th days ( total of 6 days ) from the start . men work on alternate days , that is 1 , 3,5 , 7,9 , 11,13 , 15,17 th days ( total of 9 days ) . let the total work be 360 units ( lcm of 180 and... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(15, divide(180, 15)) | divide(n1,n0)|subtract(n0,#0) | physics |
how many integers between 1000 and 10000 have no digits other than 4 , 5 or 6 ? | any number between 1000 and 10000 is of 4 digits . the unit β s place can be filled up by 45 or 6 , that is , in 3 ways . similarly , the ten β s place can be filled up by 4 or 5 or 6 , that is in 3 ways . the hundred β s place can be filled up by 4 , 5 or 6 , that is in 3 ways and the thousand β s place can . be fille... | a ) 91 , b ) 51 , c ) 81 , d ) 71 , e ) 61 | c | multiply(multiply(subtract(4, const_1), subtract(4, const_1)), multiply(subtract(4, const_1), subtract(4, const_1))) | subtract(n2,const_1)|multiply(#0,#0)|multiply(#1,#1) | general |
john completes a piece of work in 8 days , rose completes the same work in 16 days . if both of them work together , then the number of days required to complete the work is ? | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 8 Γ 16 / 24 = 5.33 days answer : b | a ) 6 days , b ) 5.33 days , c ) 10 days , d ) 12 days , e ) 14 days | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 16))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
what is the area of a triangle with the following vertices l ( 2 , 3 ) , m ( 5 , 1 ) , and n ( 3 , 5 ) ? | there is a direct formula to calculate the are of a triangle based on coordinates of its vertices and one could use it to solve this problem . though if you make a diagram minimum simple calculations will be needed : answer : b . | ['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | b | triangle_area_three_edges(sqrt(add(power(subtract(5, 2), const_2), power(subtract(1, 3), const_2))), sqrt(add(power(subtract(2, 3), const_2), power(subtract(3, 5), const_2))), sqrt(add(power(subtract(1, 5), const_2), power(subtract(3, 5), const_2)))) | subtract(n2,n0)|subtract(n3,n1)|subtract(n0,n1)|subtract(n1,n2)|subtract(n3,n2)|power(#0,const_2)|power(#1,const_2)|power(#2,const_2)|power(#3,const_2)|power(#4,const_2)|add(#5,#6)|add(#7,#8)|add(#9,#8)|sqrt(#10)|sqrt(#11)|sqrt(#12)|triangle_area_three_edges(#13,#14,#15) | geometry |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 5 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , a can do 1 / 4 of the project . thus a and b must complete 3 / 4 of the project , which takes 9 days . the total number of days is 9 + 5 = 14 . the answe... | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics |
how long does a train 140 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 140 + 132 = 272 m . required time = 272 / 20 = 13.6 sec . answer : a" | a ) 13.6 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.9 sec | a | divide(add(140, 132), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
in kaya ' s teacher ' s desk there are 10 pink highlighters , 15 yellow highlighters , and 8 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 10 + 15 + 8 = 33 . answer is d ." | a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | d | add(add(10, 15), 8) | add(n0,n1)|add(n2,#0)| | general |
income and expenditure of a person are in the ratio 9 : 8 . if the income of the person is rs . 18000 , then find his savings ? | "let the income and the expenditure of the person be rs . 9 x and rs . 8 x respectively . income , 9 x = 18000 = > x = 2000 savings = income - expenditure = 9 x - 8 x = x so , savings = rs . 2000 answer : c" | a ) rs . 3600 , b ) rs . 3603 , c ) rs . 2000 , d ) rs . 3632 , e ) rs . 3602 | c | subtract(18000, multiply(divide(8, 9), 18000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
a gambler has won 40 % of his 40 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 80 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ? | "let x be the number of additional games the gambler needs to play . 0.4 ( 40 ) + 0.8 x = 0.6 ( x + 40 ) 0.2 x = 8 x = 40 the answer is e ." | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | e | divide(subtract(multiply(40, divide(60, const_100)), multiply(40, divide(40, const_100))), subtract(divide(80, const_100), divide(60, const_100))) | divide(n3,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)| | gain |
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