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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a person crosses a 1080 m long street in 14 minutes . what is his speed in km per hour ? | "speed = 1080 / ( 12 x 60 ) m / sec = 1.3 m / sec . converting m / sec to km / hr = 1.3 x ( 18 / 5 ) km / hr = 4.6 km / hr . answer : c" | a ) 4.1 , b ) 4.5 , c ) 4.6 , d ) 5.4 , e ) 5.5 | c | divide(divide(1080, const_1000), divide(multiply(14, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics |
if x is a prime number , and x - 1 is the median of the set { x - 1 , 3 x + 3 , 2 x - 4 } , then what is the average ( arithmetic mean ) of the set ? | if x is a prime number , and x - 1 is the median of the set { x - 1 , 3 x + 3 , 2 x - 4 } , then what is the average ( arithmetic mean ) of the set ? a . 2 b . 5 / 3 c . 3 d . 10 / 3 e . 14 / 3 solution : x - 1 is the median of the set , implies if we arrange the set in ascending order , the set would be [ 2 x - 4 , x ... | a ) 2 , b ) 5 / 3 , c ) 3 , d ) 10 / 3 , e ) 14 / 3 | d | divide(add(add(multiply(3, 2), 3), subtract(2, 1)), 3) | multiply(n2,n4)|subtract(n4,n0)|add(n2,#0)|add(#2,#1)|divide(#3,n2) | general |
the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 – 2552 = 208 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 208 2 x = 358 x = 179 runs answer ... | a ) 179 , b ) 208 , c ) 210 , d ) 223 , e ) 229 | a | divide(add(150, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2) | multiply(n0,n1)|subtract(n1,const_2)|multiply(n3,#1)|subtract(#0,#2)|add(n2,#3)|divide(#4,const_2)| | general |
the sector of a circle has radius of 21 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 49.5 + 42 = 91.5 cm answer : a" | a ) 91.5 cm , b ) 92.2 cm , c ) 28.9 cm , d ) 29.2 cm , e ) 98.2 cm | 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 |
8.008 / 1.001 | "answer is 8 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 8008 / 1001 = 8 answer b" | a ) 0.008 , b ) 8 , c ) 0.8 , d ) 80 , e ) 800 | b | multiply(divide(8.008, 1.001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the radius of a semicircle is 7 . what is the approximate perimeter of the semicircle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 7 + 2 * 7 = 35.98 approximately 36 answer a | ['a ) 36', 'b ) 25', 'c ) 15', 'd ) 28', 'e ) 37'] | a | add(divide(circumface(7), const_2), add(7, 7)) | add(n0,n0)|circumface(n0)|divide(#1,const_2)|add(#0,#2) | geometry |
a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 18 seconds . due to some problem , one boggies were detached . the train now crosses a telegraph post in | length of train = 12 ã — 15 = 180 m . then , speed of train = 180 â „ 18 = 10 m / s now , length of train = 11 ã — 15 = 165 m â ˆ ´ required time = 165 â „ 10 = 16.5 sec . answer a | a ) 16.5 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) none of these | a | divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 18)) | multiply(n0,n1)|divide(#0,n2)|subtract(#0,n1)|divide(#2,#1) | physics |
if a sum of money doubles itself in 8 years at simple interest , the ratepercent per annum is | explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 8 ) = 12.5 option b | a ) 12 , b ) 12.5 , c ) 13 , d ) 13.5 , e ) 14 | b | divide(divide(const_2, divide(8, const_100)), const_2) | divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2) | gain |
natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 2 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 2 km / hour x / 3 = 2 x = 6 km the average speed while climbing = distance / time ... | a ) 1.5 , b ) 2.5 , c ) 3.75 , d ) 5 , e ) 7.5 | a | divide(divide(multiply(add(4, 2), 2), 2), 4) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics |
the average ( arithmetic mean ) of 4 positive integers is 30 . if the average of 2 of these integers is 15 , what is the greatest possible value that one of the other 2 integers can have ? | "a + b + c + d = 120 a + b = 30 c + d = 90 greatest possible = 89 ( just less than 1 ) answer = b" | a ) 55 , b ) 89 , c ) 100 , d ) 109 , e ) 115 | b | subtract(multiply(30, 4), multiply(15, 2)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
the length of the bridge , which a train 135 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 135 + x ) / 30 = 25 / 2 = > 2 ( 135 + x ) = 750 = > x = 240 m . answer : c" | a ) 76 m , b ) 178 m , c ) 240 m , d ) 187 m , e ) 176 m | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 135) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
carol spends 1 / 4 of her savings on a stereo and 1 / 3 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? | "total savings = s amount spent on stereo = ( 1 / 4 ) s amount spent on television = ( 1 - 1 / 3 ) ( 1 / 4 ) s = ( 2 / 3 ) * ( 1 / 4 ) * s = ( 1 / 6 ) s ( stereo + tv ) / total savings = s ( 1 / 4 + 1 / 6 ) / s = 5 / 12 answer : c" | a ) 1 / 4 , b ) 2 / 7 , c ) 5 / 12 , d ) 1 / 2 , e ) 7 / 12 | c | divide(1, 4) | divide(n0,n1)| | general |
jancy had 100 currency notes in all , some of which are of rs 70 denomination and the remaining of rs 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount ( in rs ) did she have in the denomination of rs 50 ? | let the number of 50 - rupee notes = x then , the number of 70 - rupee notes = ( 100 – x ) 50 x + 70 ( 100 – x ) = 5000 : x = 100 answer : c | a ) 103 , b ) 102 , c ) 100 , d ) 105 , e ) 106 | c | divide(5000, 50) | divide(n3,n2) | general |
a line that passes through ( – 1 , – 4 ) and ( 4 , k ) has a slope = k . what is the value of k ? | "slope = ( y 2 - y 1 ) / ( x 2 - x 1 ) = > k = ( k + 4 ) / ( 4 + 1 ) = > 5 k = k + 4 = > k = 1 ans b it is !" | a ) 3 / 4 , b ) 1 , c ) 4 / 3 , d ) 2 , e ) 7 / 2 | b | divide(4, 4) | divide(n1,n2)| | general |
a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 25 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he ... | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.25 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 25 % income the savings increased by 100 % . or s = . 25 i or s =... | a ) 20 % , b ) 25 % , c ) 28.5 % , d ) 30 % , e ) 32 % | b | multiply(divide(subtract(add(add(100, 25), 100), multiply(2, 100)), 100), 100) | add(n0,n1)|multiply(n1,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,n1)|multiply(#4,n1)| | general |
the contents of a certain box consist of 24 apples and 30 kiwis . how many kiwis must be added to the box so that exactly 30 % of the pieces of fruit in the box will be apples ? | "apples = ( apples + kiwi + x ) * 0.3 24 = ( 30 + 24 + x ) * 0.3 x = 26 . answer : b ." | a ) 24 , b ) 26 , c ) 30 , d ) 46 , e ) 50 | b | subtract(add(24, 30), divide(24, divide(30, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general |
p and q can complete a work in 80 days and 48 days respectively . p alone started the work and q joined him after 16 days till the completion of the work . how long did the work last ? | explanation : work done by p in 1 day = 1 / 80 work done by q in 1 day = 1 / 48 work done by p in 16 days = 16 ã — ( 1 / 80 ) = 1 / 5 remaining work = 1 â € “ 1 / 5 = 4 / 5 work done by p and q in 1 day = 1 / 80 + 1 / 48 = 1 / 30 number of days p and q take to complete the remaining work = ( 4 / 5 ) / ( 1 / 30 ) = 24 t... | a ) 5 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 40 days | e | add(16, divide(subtract(const_1, divide(16, 80)), add(inverse(80), inverse(48)))) | divide(n2,n0)|inverse(n0)|inverse(n1)|add(#1,#2)|subtract(const_1,#0)|divide(#4,#3)|add(n2,#5) | physics |
what is the square root of 83,521 ? | "1 ) checking the last digits in the answer options a , c , e are out since last digit of the square is known to be 1 . 2 ) c = 489 ^ 2 and e = 511 ^ 2 c = ( 300 - 11 ) ^ 2 and e = ( 300 + 11 ) ^ 2 since we need the answer less than 90000 = > e is out . thus we have c as the ans ." | a ) 476 , b ) 489 , c ) 289 , d ) 511 , e ) 311 | c | circle_area(divide(83,521, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 16 : 9 = 4 : 3 . answer : b" | a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 4 : 5 | b | divide(sqrt(16), sqrt(9)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 2 / 3 of the juniors is equal to 1 / 2 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 2 / 3 ) j = ( 1 / 2 ) s = > s = 4 / 3 j total number of students = j + s = ( 7 / 3 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 7 / 3 ) j ] = 3 / 7 answer c" | a ) 3 / 20 , b ) 1 / 3 , c ) 3 / 7 , d ) 2 / 7 , e ) 7 / 2 | c | divide(multiply(3, 3), add(2, multiply(3, 3))) | multiply(n1,n1)|add(n3,#0)|divide(#0,#1)| | general |
a meal cost $ 34.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be : | 10 % ( 34.5 ) = 3.45 15 % ( 34.5 ) = 5.175 total amount could have been 34.5 + 3.45 and 34.5 + 5.175 = > could have been between 37.95 and 39.675 = > approximately between 38 and 40 answer is c . | a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | c | add(multiply(34.5, divide(15, const_100)), 34.5) | divide(n2,const_100)|multiply(n0,#0)|add(n0,#1) | general |
in an examination , questions were asked in 5 sections . out of the total students , 5 % candidates cleared the cut - off in all the sections and 5 % cleared none . of the rest , 25 % cleared only one section and 20 % cleared 4 sections . if 24.5 % of the entire candidates cleared two sections and 300 candidates cleare... | solution : passed in none = 5 % passed in all = 5 % passed in four = 20 % 0 f 90 % = 18 % passed in three = 24.5 % passed in three = ( 100 - 5 - 5 - 22.5 - 24.5 - 18 ) = 25 % . but given 300 students passed in three . hence , 25 % = 300 . so , 100 % = 1200 . 1200 students must have appeared . answer : option b | a ) 1000 , b ) 1200 , c ) 1500 , d ) 2000 , e ) 1800 | b | multiply(divide(300, subtract(const_100, add(add(add(add(5, 5), divide(multiply(subtract(const_100, add(5, 5)), 25), const_100)), divide(multiply(subtract(const_100, add(5, 5)), 20), const_100)), 24.5))), const_100) | add(n0,n0)|subtract(const_100,#0)|multiply(n3,#1)|multiply(n4,#1)|divide(#2,const_100)|divide(#3,const_100)|add(#0,#4)|add(#6,#5)|add(n6,#7)|subtract(const_100,#8)|divide(n7,#9)|multiply(#10,const_100) | general |
a carpenter worked alone for 1 day on a job that would take him 4 more days to finish . he and another carpenter completed the job in 2 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | "a carpenter worked only 1 day on something that takes him 4 more days . means ; carpenter finishes his work in 5 days . let his buddy finish the same task in x days . respective rates per day : 1 / 5 and 1 / x to complete 1 work : first guy worked for 3 days @ rate = 1 / 5 per day . second one worked for 2 days @ rate... | a ) 5 , b ) 7 , c ) 9 , d ) 14 , e ) 24 | a | divide(multiply(2, add(1, 4)), const_2) | add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)| | physics |
g ( x ) is defined as the product of all even integers k such that 0 < k ≤ x . for example , g ( 14 ) = 2 × 4 × 6 × 8 × 10 × 12 × 14 . if g ( z ) is divisible by 4 ^ 11 , what is the smallest possible value for z ? | g ( z ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b | a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44 | b | multiply(2, 12) | multiply(n2,n7) | general |
if 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a 7 - percent vinegar solution , what was the concentration of the original solution ? | let x be the quantity of non - vinegar in the strong vinegar solution thus vinegar quantity will be 12 - x when 50 ounces of water were added the percentage of vinegar becomes 7 % , thus ( 12 - x ) / 62 = 7 / 100 from this equation x = 7.66 answer ( 12 - 7.66 ) / 12 = 36.17 % answer : c | a ) 19.3 % , b ) 17 % , c ) 36.17 % , d ) 15.5 % , e ) 12.5 % | c | divide(multiply(add(50, 12), 7), 12) | add(n0,n1)|multiply(n2,#0)|divide(#1,n0) | other |
if ( a + b ) = 5 , ( b + c ) = 6 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 5 b + c = 6 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 5 - 6 + 3 = 2 . option d . . ." | a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | d | subtract(add(5, 3), 6) | add(n0,n2)|subtract(#0,n1)| | general |
on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 25 inch from the base of the mountain . how many km is he from the base of the mountain ? | "explanation : since 312 inch = 136 km so 1 inch = 136 / 312 km so 25 inch = ( 136 ã — 25 ) / 312 = 10.89 km answer : a" | a ) 10.89 , b ) 14.81 , c ) 14.8 , d ) 14.82 , e ) 14.12 | a | divide(multiply(25, 136), 312) | multiply(n1,n2)|divide(#0,n0)| | physics |
a garrison of 400 men had a provision for 31 days . after 27 days 200 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | 400 - - - 31 400 - - - 4 200 - - - ? 400 * 4 = 200 * x = > x = 8 days . answer : c | a ) 65 days , b ) 45 days , c ) 8 days , d ) 16 days , e ) 18 days | c | divide(multiply(subtract(31, 27), 400), 200) | subtract(n1,n2)|multiply(n0,#0)|divide(#1,n3) | other |
a salesman commission is 10 % on all sales upto $ 5000 and 5 % on all sales exceeding this . he remits $ 15000 to his parent company after deducting his commission . find the total sales ? | "let his total sales be x total sales - commission = $ 15000 x - [ ( 10 % of 5000 ) + 5 % of ( x - 5000 ) ] = 15000 95 x / 100 = 15000 x = 15789 approximately answer is c" | a ) 15456 , b ) 14758 , c ) 15789 , d ) 13250 , e ) 12450 | c | divide(add(15000, subtract(divide(multiply(10, 5000), const_100), divide(multiply(const_4.0, 5000), const_100))), subtract(const_1, divide(5, const_100))) | divide(const_4.0,const_100)|multiply(n0,n1)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,const_100)|subtract(const_1,#0)|subtract(#3,#4)|add(n3,#6)|divide(#7,#5)| | general |
( 7.5 × 7.5 + 3.75 + 2.5 × 2.5 ) is equal to | "solution given expression = ( 7.5 x 7.5 + 2 × 7.5 × 2.5 + 2.5 × 2.5 ) ² = ( a ² + 2 ab + b ² ) = ( a + b ) ² = ( 7.5 + 2.5 ) ² = 10 ² = 100 . answer d" | a ) 30 , b ) 60 , c ) 80 , d ) 100 , e ) 90 | d | add(add(multiply(7.5, 7.5), multiply(2.5, 2.5)), 3.75) | multiply(n0,n0)|multiply(n3,n3)|add(#0,#1)|add(n2,#2)| | general |
a merchant has selected two items to be placed on sale , one of which currently sells for 40 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 60 ; we must increase $ 60 to $ 100 , so by $ 40 , which is approximately 60 % increase : ( 100 - 60 ) / 60 = 2 / 3 = ~ 0.66 . answer : c ." | a ) 40 % , b ) 50 % , c ) 60 % , d ) 30 % , e ) 20 % | c | multiply(divide(divide(40, const_100), subtract(const_1, divide(40, const_100))), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(#0,#1)|multiply(#2,const_100)| | general |
a owes b rs . 50 . he agrees to pay b over a number of consecutive days on a monday , paying single note or rs . 10 or rs . 20 on each day . in how many different ways can a repay b . | he can pay by all 10 rupee notes = 1 way 3 ten rupee + 1 twenty rupee = 4 ! 3 ! × 1 ! 4 ! 3 ! × 1 ! = 4 ways 1 ten rupee + 2 twenty rupee notes = 3 ! 2 ! × 1 ! 3 ! 2 ! × 1 ! = 3 ways total ways = 1 + 4 + 3 = 8 answer : b | a ) 2 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(add(divide(factorial(const_3), factorial(const_2)), divide(factorial(const_4), factorial(const_3))), divide(factorial(divide(50, 10)), factorial(divide(50, 10)))) | divide(n0,n1)|factorial(const_3)|factorial(const_2)|factorial(const_4)|divide(#1,#2)|divide(#3,#1)|factorial(#0)|add(#4,#5)|divide(#6,#6)|add(#7,#8) | physics |
if two positive numbers are in the ratio 1 / 12 : 1 / 9 , then by what percent is the second number more than the first ? | "given ratio = 1 / 12 : 1 / 9 = 9 : 12 let first number be 9 x and the second number be 12 x . the second number is more than first number by 3 x . required percentage = 3 x / 9 x * 100 = 33.33 % . answer : c" | a ) 70 % , b ) 90 % , c ) 33.33 % , d ) 50 % , e ) 65 % | c | multiply(divide(1, 9), const_100) | divide(n0,n3)|multiply(#0,const_100)| | general |
a and b invests rs . 10000 each , a investing for 6 months and b investing for all the 12 months in the year . if the total profit at the end of the year is rs . 4500 , find their shares ? | "the ratio of their profits a : b = 6 : 12 = 1 : 2 share of a in the total profit = 1 / 3 * 4500 = rs . 1500 share of b in the total profit = 2 / 3 * 4500 = rs . 3000 answer : d" | a ) 1500,6000 , b ) 4500,3000 , c ) 3500,3000 , d ) 1500,3000 , e ) 2500,3000 | d | divide(multiply(10000, const_1), const_3) | multiply(n0,const_1)|divide(#0,const_3)| | gain |
a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 32.49 , the number of the member is the group is : | "explanation : money collected = ( 32.49 x 100 ) paise = 3249 paise . ∴ number of members = √ ( 3249 ) = 57 . answer : a" | a ) 57 , b ) 67 , c ) 77 , d ) 87 , e ) 97 | a | sqrt(multiply(32.49, const_100)) | multiply(n0,const_100)|sqrt(#0)| | general |
q ' = 3 q - 3 , what is the value of ( 6 ' ) ' ? | ( 6 ' ) ' = ( 3 * 6 - 3 ) ' = 15 ' = 15 * 15 - 15 = 210 answer d | a ) 120 , b ) 150 , c ) 180 , d ) 210 , e ) 240 | d | multiply(subtract(multiply(subtract(multiply(3, 6), 3), 3), 3), subtract(6, const_1)) | multiply(n0,n2)|subtract(n2,const_1)|subtract(#0,n0)|multiply(n0,#2)|subtract(#3,n0)|multiply(#4,#1) | general |
in 1979 approximately 1 / 3 of the 38.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 2 the number that used kennedy airport and 4 times the number that used logan airport , approximately how many millions of the... | number of passengers using kennedy airport = 38 / 3 = ~ 12.67 passengers using miami airport = 12.67 / 2 = ~ 6.3 passengers using logan airport = 6.3 / 4 = ~ 1.58 so e | a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 3.1 , e ) 1.6 | e | divide(divide(38.3, 3), multiply(4, 2)) | divide(n3,n2)|multiply(n5,n6)|divide(#0,#1) | general |
when n is divided by 24 , the remainder is 3 . find thee difference between previous remainder and the remainder when 9 n is divided by 7 ? | "let n = 3 ( leaves a remainder of 3 when divided by 24 ) 9 n = 9 ( 3 ) = 27 , which leaves a remainder of 6 when divided by 7 . difference = 6 - 3 = 3 . answer d" | a ) 2 , b ) 7 , c ) 5 , d ) 3 , e ) 6 | d | subtract(3, reminder(9, 7)) | reminder(n2,n3)|subtract(n1,#0)| | general |
the initial price of an article is rs . 6000 which increases 40 % increse in its price in the first year , a 20 % decrease in the second year and a 20 % increase in the next year . what is the final price of the article ? | "the initial price of the article is rs . 6000 . in the 1 st year , price of the article = 6000 + 2400 = rs . 8400 . in the 2 nd year , price = 8400 - 20 % of 8400 = 8400 - 1680 = rs . 6720 . in the 3 rd year , price = 6720 + 10 % of 6720 = 6720 + 672 = rs . 7392 required price = = rs . 7392 . answer : d" | a ) rs . 5392 , b ) rs . 6392 , c ) rs . 8392 , d ) rs . 7392 , e ) rs . 9392 | d | add(subtract(6000, divide(multiply(6000, 20), const_100)), divide(multiply(subtract(6000, divide(multiply(6000, 20), const_100)), 20), const_100)) | multiply(n0,n2)|divide(#0,const_100)|subtract(n0,#1)|multiply(n3,#2)|divide(#3,const_100)|add(#4,#2)| | general |
a and b can finish a work in 16 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 16 – 1 / 20 = 1 / 80 = > 80 days answer : c" | a ) 76 days , b ) 48 days , c ) 80 days , d ) 31 days , e ) 22 days | c | inverse(subtract(inverse(16), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
one side of a rectangular field is 13 m and one of its diagonal is 17 m . find the area of the field . | "solution other side = √ ( 17 ) 2 - ( 13 ) 2 = √ 289 - 169 = √ 120 = 10.9 m . ∴ area = ( 13 x 10.9 ) m 2 = 141.7 m 2 . answer b" | a ) 100 , b ) 141.7 , c ) 150 , d ) 180 , e ) none | b | rectangle_area(13, sqrt(subtract(power(17, const_2), power(13, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry |
a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 16000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 16000 , = [ 1 - 9 / 10 ] * total salary = $ 16000 , total salary = $ 16000 * 10 = $ 160000 , correct answer ( c )" | a ) $ 10800 , b ) $ 18000 , c ) $ 160000 , d ) $ 1800 , e ) none | c | divide(16000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n4,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
a has 10000 chocolates . 20 % of the chocolates are eaten . what percentage of the uneaten chocolates must be eaten so that 3 / 2 of all chocolates are eaten ? | number of chocolates = 10,000 currently eaten = 10,000 * 20 / 100 = 2000 uneaten chocolates = 10000 - 2000 = 8000 2 / 3 of all chocolates = 5000 the number of chocolates to be eaten to make 2 / 3 of all chocolates eaten = 5000 - 2000 = 3000 so now the question remains - 3000 is how much percentage of uneaten chocolates... | a ) 39.25 , b ) 37.25 , c ) 37.26 , d ) 38.21 , e ) 38.32 | b | divide(multiply(subtract(divide(10000, 2), divide(multiply(20, 10000), const_100)), const_100), subtract(10000, divide(multiply(20, 10000), const_100))) | divide(n0,n3)|multiply(n0,n1)|divide(#1,const_100)|subtract(#0,#2)|subtract(n0,#2)|multiply(#3,const_100)|divide(#5,#4) | gain |
a school has 4 section of chemistry in class x having 60 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | "required average marks = 60 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 60 + 35 + 45 + 42 = 3000 + 2100 + 2475 + 1890 / 182 = 9465 â „ 182 = 52 answer b" | a ) 50 , b ) 52 , c ) 51 , d ) 53 , e ) none of the above | b | divide(add(add(add(multiply(60, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(60, 35), 45), 42)) | add(n1,n2)|multiply(n1,n5)|multiply(n2,n6)|multiply(n3,n7)|multiply(n3,n4)|add(#1,#2)|add(n3,#0)|add(#5,#3)|add(n4,#6)|add(#7,#4)|divide(#9,#8)| | general |
the number of employees in obelix menhir co . is a prime number and is less than 300 . the ratio of the number of employees who are graduates and above , to that of employees who are not , can possibly be : - | explanation : going through the given options , option 1 : - 101 : 88 = > 189 . [ divisible by 3 ] option 2 : - 87 : 100 = > 187 . [ divisible by 11 ] option 3 : - 85 : 98 = > 183 . [ divisible by 3 ] option 4 : - 97 : 84 = > 181 . [ prime number ] hence , only 181 is the prime number i . e the total number of employee... | a ) 101 : 88 , b ) 87 : 100 , c ) 110 : 111 , d ) 85 : 98 , e ) 97 : 84 | e | divide(subtract(const_100, divide(300, const_100)), subtract(subtract(const_100, divide(300, const_100)), add(const_10, divide(300, const_100)))) | divide(n0,const_100)|add(#0,const_10)|subtract(const_100,#0)|subtract(#2,#1)|divide(#2,#3) | general |
rs . 825 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 825 become in 3 years ? | "solution s . i . = rs . ( 956 - 825 ) = rs . 131 rate = ( 100 x 131 / 825 x 3 ) = 524 / 99 % new rate = ( 524 / 99 + 4 ) % = 920 / 99 % new s . i . = rs . ( 825 x 920 / 99 x 3 / 100 ) rs . 230 . ∴ new amount = rs . ( 825 + 230 ) = rs . 1055 . answer c" | a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1055 , d ) data inadequate , e ) none of these | c | add(825, divide(multiply(multiply(825, add(divide(multiply(subtract(956, 825), const_100), multiply(825, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)| | gain |
mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 496.80 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 496.80 , discount = 621 - 496.80 = 124.20 discount % = 124.2 / 621 * 100 = 20 % . answer : c" | a ) 22 , b ) 27 , c ) 20 , d ) 21 , e ) 12 | c | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 496.80), const_100), multiply(540, divide(add(const_100, 15), const_100))) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|subtract(#2,n2)|multiply(#3,const_100)|divide(#4,#2)| | gain |
a 45 ° - 45 ° - 90 ° right triangle has hypotenuse of length h . what is the area of the triangle t in terms of h ? | if . . . each of the two shorter sides = 3 , then the hypotenuse = h = 3 ( root 2 ) . the area t = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 3 ) ( 3 ) = 9 / 2 . so we ' re looking for an answer that = 9 / 2 when h = 3 ( root 2 ) . there ' s only one answer that matches . . . e | ['a ) h / √ 2', 'b ) h / 2', 'c ) h / 4', 'd ) ( h ) ^ 2', 'e ) ( h ) ^ 2 / 4'] | e | power(divide(45, 90), const_2) | divide(n0,n2)|power(#0,const_2) | geometry |
the perimeter of a triangle is 40 cm and the in radius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 40 / 2 = 50 cm 2 answer : e" | a ) 76 , b ) 88 , c ) 66 , d ) 55 , e ) 50 | e | triangle_area(2.5, 40) | triangle_area(n0,n1)| | geometry |
a man swims downstream 54 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "54 - - - 3 ds = 18 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 18 + 6 ) / 2 = 12 answer : e" | a ) 2 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | e | divide(add(divide(18, 3), divide(54, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
the average of 11 results is 52 , if the average of first 6 results is 49 and that of the last 6 is 52 . find the sixth result ? | 1 to 11 = 11 * 52 = 572 1 to 6 = 6 * 49 = 294 6 to 11 = 6 * 52 = 312 6 th = 294 + 312 – 572 = 34 answer : b | a ) a ) 46 , b ) b ) 34 , c ) c ) 66 , d ) d ) 76 , e ) e ) 74 | b | subtract(add(multiply(52, 6), multiply(49, 6)), multiply(52, 11)) | multiply(n1,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general |
sum of the squares of 3 no . is 138 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 * 131 a + b + c = √ 400 = 20 a" | a ) 20 , b ) 21 , c ) 25 , d ) 26 , e ) 27 | a | sqrt(add(138, multiply(131, const_2))) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1)| | general |
the total of 324 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is : | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 324 - x ) . 0.20 × ( x ) + 0.25 ( 324 - x ) = 71 = > x = 200 . answer : d" | a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | d | divide(subtract(multiply(324, 25), multiply(71, const_100)), subtract(25, 20)) | multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)| | general |
120 liters of a mixture contains milk and water in the ratio 4 : 2 . if 10 liters of this mixture be replaced by 10 liters of milk , the ratio of milk to water in the new mixture would be ? | quantity of milk in 120 liters if mix = 120 * 4 / 6 = 80 liters quantity of milk in 130 liters of new mix = 80 + 10 = 90 liters quantity of water in it = 130 - 90 = 40 liters ratio of milk and water in new mix = 90 : 40 = 9 : 4 answer is c | a ) 8 : 6 , b ) 9 : 3 , c ) 9 : 4 , d ) 9 : 6 , e ) 8 : 2 | c | divide(multiply(subtract(10, const_1), const_2), multiply(const_4, const_2)) | multiply(const_2,const_4)|subtract(n3,const_1)|multiply(#1,const_2)|divide(#2,#0) | other |
170 kg of an alloy a is mixed with 250 kg of alloy b . if alloy a has lead and tin in the ratio 1 : 3 and alloy b has tin and copper in the ratio 3 : 5 , then the amount of tin in the new alloy is ? | "quantity of tin in 170 kg of a = 170 * 3 / 4 = 127.5 kg quantity of tin in 250 kg of b = 250 * 3 / 8 = 93.75 kg quantity of tin in the new alloy = 127.5 + 93.75 = 221.25 kg answer is b" | a ) 210.6 kg , b ) 221.3 kg , c ) 229.5 kg , d ) 110.8 kg , e ) 114 kg | b | add(multiply(divide(170, add(1, 3)), 3), multiply(divide(250, add(3, 5)), 3)) | add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)| | other |
a train 360 m long can cross an electric pole in 30 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 360 / 30 s = 12 m / sec speed = 12 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 43 kmph answer : b" | a ) 88 kmph , b ) 43 kmph , c ) 72 kmph , d ) 16 kmph , e ) 18 kmph | b | divide(divide(360, const_1000), divide(30, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
a can do a piece of work in 4 hours ; b and c together can do it in 2 hours , which a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour work = 1 / 4 ; ( b + c ) ' s 1 hour work = 1 / 2 ; ( a + c ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 2 ) = 3 / 4 b ' s 1 hour work = ( 3 / 4 - 1 / 2 ) = 1 / 4 b alone will take 4 hours to do the work . answer : c" | a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 5 hours , e ) 6 hours | c | divide(const_1, subtract(divide(const_1, 2), subtract(divide(const_1, 2), divide(const_1, 4)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)| | physics |
the maximum number of student amoung them 1204 pens and 840 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "solution required number of student = h . c . f of 1204 and 840 = 28 . answer b" | a ) 91 , b ) 28 , c ) 56 , d ) 89 , e ) none of these | b | gcd(1204, 840) | gcd(n0,n1)| | general |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation .... | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 7 * 180 / 12 = 105 ounces of water e" | a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 105 | e | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 7) | divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)| | other |
the product of three consecutive numbers is 120 . then the sum of the smallest two numbers is ? | "product of three numbers = 120 120 = 2 * 2 * 5 * 6 = 4 * 5 * 6 so , the three numbers are 4,5 and 6 . and sum of smallest of these two = 4 + 5 = 9 . answer : option a" | a ) 9 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | multiply(power(const_2, 120), factorial(120)) | factorial(n0)|power(const_2,n0)|multiply(#0,#1)| | general |
a company pays 12.5 % dividend to its investors . if an investor buys rs . 40 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 12.5 * 40 ) / 100 = rs . 5 rs . 25 is income on an investment of rs . 100 rs . 5 is income on an investment of rs . ( 5 * 100 ) / 25 = rs . 20 answer : a" | a ) 20 , b ) 66 , c ) 18 , d ) 19 , e ) 01 | a | divide(multiply(divide(multiply(12.5, 40), const_100), const_100), 25) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)| | gain |
how many even number in the range between 10 to 140 inclusive are not divisible by 3 | we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 140 using ap formula , we can say 140 = 10 + ( n - 1 ) * 2 or n = 66 . for 6 , 12,18 , . . . 138 using ap formula , we can say 138 = 12 + ( n ... | a ) 15 , b ) 30 , c ) 31 , d ) 33 , e ) 44 | e | subtract(divide(subtract(subtract(140, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(140, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9) | general |
the average of 15 result is 55 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th result ? | "sum of all the 13 results = 15 * 55 = 825 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 825 + 399 - 427 = 797 . e" | a ) 760 , b ) 720 , c ) 700 , d ) 786 , e ) 797 | e | subtract(add(multiply(7, 57), multiply(7, 61)), multiply(15, 55)) | multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
a train of length 250 m crosses a bridge of length 150 m in 32 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 150 ) = 32 x speed speed = 400 / 32 = 12.5 m / s = 45 km / h answer = e" | a ) 33 , b ) 27 , c ) 25 , d ) 22 , e ) 45 | e | multiply(const_3_6, divide(add(250, 150), 32)) | add(n0,n1)|divide(#0,n2)|multiply(#1,const_3_6)| | physics |
a cricketer scored 142 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 142 - ( 12 x 4 + 2 x 6 ) = 142 - ( 60 ) = 82 now , we need to calculate 82 is what percent of 142 . = > 82 / 142 * 100 = 57.75 % answer : a" | a ) 57.75 % , b ) 54.54 % , c ) 63 % , d ) 70 % , e ) none of these | a | multiply(divide(subtract(142, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 142), const_100) | multiply(n1,const_4)|multiply(n2,const_3)|multiply(n2,#1)|add(#0,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | general |
a train running at the speed of 54 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 54 * 5 / 18 = 15 m / sec length of the train = speed * time = 15 * 9 = 135 m answer : c" | a ) 288 , b ) 279 , c ) 135 , d ) 272 , e ) 150 | c | multiply(divide(multiply(54, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
one half of a two digit number exceeds its one third by 6 . what is the sum of the digits of the number ? | "explanation : x / 2 – x / 3 = 6 = > x = 6 3 + 6 = 9 b )" | a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 15 | b | add(reminder(multiply(6, const_4), const_10), const_1) | multiply(n0,const_4)|reminder(#0,const_10)|add(#1,const_1)| | general |
a box contains 5 pairs of shoes ( 10 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 20 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 9 ( as after taking one at random there are 9 shoes left and only one is the pair of the first... | a ) 1 / 190 , b ) 1 / 20 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 92 | c | divide(const_1, subtract(10, const_1)) | subtract(n1,const_1)|divide(const_1,#0)| | general |
in a company of 16 employees , 8 employees earn $ 38,000 , 4 employees earn $ 42,000 , and the 4 highest - paid employees earn the same amount . if the average annual salary for the 16 employees is $ 44,000 , what is the annual salary for each of the highest - paid employees ? | "8 * 38,000 + 4 * 42,000 + 4 x = 16 * 44,000 4 x = 704,000 - 304,000 - 168,000 4 x = 232,000 x = 58,000 the answer is e ." | a ) $ 54,000 , b ) $ 55,000 , c ) $ 56,000 , d ) $ 57,000 , e ) $ 58,000 | e | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(4, 4), add(4, const_1)), const_2), divide(const_1, const_2)), 16), multiply(multiply(16, 4), add(4, const_1))), multiply(multiply(multiply(4, 4), add(4, const_1)), 4)), 4), const_1000), const_1000), 8) | add(const_1,n3)|divide(const_1,const_2)|multiply(n5,n5)|multiply(n0,n5)|multiply(#0,#2)|multiply(#0,#3)|add(#4,const_2)|multiply(n3,#4)|add(#6,#1)|multiply(n0,#8)|subtract(#9,#5)|subtract(#10,#7)|divide(#11,n5)|multiply(#12,const_1000)|divide(#13,const_1000)|subtract(#14,n1)| | general |
if the price of petrol increases by 32 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 32 ) i . e 132 . now , his consumption should be reduced to : - = ( 132 − 100 ) / 132 ∗ 100 . hence , the consumption should be reduced to 24.2 % . answer : c | a ) 25 % , b ) 20 % , c ) 24.2 % , d ) 33.33 % , e ) none of these | c | multiply(subtract(const_1, divide(const_100, add(const_100, 32))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100) | general |
108 . triangle a ’ s base is 7 % greater than the base of triangle b , and a ’ s height is 7 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | "wish the question specified that we are talking about corresponding height . base of a = 8 / 7 * base of b height of a = 6 / 7 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 8 / 7 * 6 / 7 * area of b = 48 / 49 * area of b area of a is 2.1 % more than the area of b . answer ( e )" | a ) 9 % less , b ) 1 % less , c ) equal to each other , d ) 1 % more , e ) 2.1 % more | e | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 7), subtract(const_100, 7)))) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4)| | geometry |
if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 64 ) which of these integers could be the value of z ? | "solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 64 incorrect ( division by zero ) b . 2 / 1 = 2 / 2 + 2 / 65 incorrect c . 2 / 2 = 2 / 3 + 2 / 66 incorrect d . 2 / 3 ... | a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 8 | e | divide(64, add(add(2, 2), 1)) | add(n0,n0)|add(n2,#0)|divide(n4,#1)| | general |
praveen starts business with rs . 3360 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ? | let hari ’ s capital be rs . x . then , 3360 * 12 / 7 x = 2 / 3 = > 14 x = 120960 = > x = 8640 . answer : b | a ) s . 7500 , b ) s . 8640 , c ) s . 8500 , d ) s . 9000 , e ) s . 6000 | b | divide(divide(3360, subtract(const_1, divide(5, const_12))), divide(2, 3)) | divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1) | other |
the average ( arithmetic mean ) of 10 , 20 , and 60 is 5 more than the average of 10 , 40 , and what number ? | "a 1 = 90 / 3 = 30 a 2 = a 1 - 5 = 25 sum of second list = 25 * 3 = 75 therefore the number = 75 - 50 = 25 a" | a ) 25 , b ) 60 , c ) 50 , d ) 30 , e ) 40 | a | subtract(add(add(10, 20), 60), add(add(multiply(5, const_3), 10), 40)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general |
jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much does she have in her account after two years ? | solution : first of all , notice the magic word “ approximately ” — the test - writer is letting us know estimation is perfectly fine . furthermore , the answer choices are nicely spread out , which will facilitate estimating . ok , get ready for some fast & furious estimation . the interest rate 3.96 % is an ugly numb... | a ) $ 10079.44 , b ) $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_4)|multiply(n0,#3) | gain |
250 men work for 16 days and finish a work . how many men would do a job 3 times the previous one in 20 days ? | number of men needed to finish the required work ( variation , proportion method ) = 3 * 16 * 250 / 20 = 600 answer : c | a ) 700 , b ) 666 , c ) 600 , d ) 400 , e ) 800 | c | divide(multiply(3, multiply(250, 16)), 20) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3) | physics |
solution for 1.35 + . 123 + . 321 | "1.35 + . 123 + . 321 = 0 0 = 0 - 1.35 - 0.123 - 0.321 0 = - 1.794 answer : b" | a ) 2.91 , b ) 1.794 , c ) 1.338 , d ) 1.986 , e ) 1.999 | b | add(add(1.35, divide(123, const_1000)), divide(321, const_1000)) | divide(n1,const_1000)|divide(n2,const_1000)|add(n0,#0)|add(#2,#1)| | general |
the population of a town increases 25 % and 15 % respectively in two consecutive years . after the growth the present population of the town is 1150 . then what is the population of the town 2 years ago ? | explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1150 * 100 / 125 * 100 / 115 = 800 answer : option e | a ) a ) 400 , b ) b ) 500 , c ) c ) 600 , d ) d ) 700 , e ) e ) 800 | e | divide(1150, multiply(add(const_1, divide(25, const_100)), add(const_1, divide(15, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4) | gain |
if 16 machine can finish a job in 45 days , then how many more machines would be needed to finish the job in one - third less time ? | "you might think of this in a management context - we can use the principle of ' person - hours ' to solve any problem where we have identical workers . so , using simpler numbers , suppose you know that 6 identical employees , working simultaneously , would finish a job in 5 hours . then that job requires 6 * 5 = 30 t... | a ) a . 3 , b ) b . 8 , c ) c . 10 , d ) d . 12 , e ) e . 16 | a | subtract(divide(multiply(16, 45), multiply(divide(const_3, const_4), 45)), 16) | divide(const_3,const_4)|multiply(n0,n1)|multiply(n1,#0)|divide(#1,#2)|subtract(#3,n0)| | general |
two trains start simultaneously from opposite ends of a 160 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 160 - km trip in 5 hours . train y , travelling at a constant rate , completes the 160 - km trip in 3 hours . how many kilometers had train x tr... | if the two trains cover a total distance d , then train x travels ( 3 / 8 ) * d while train y travels ( 5 / 8 ) * d . if the trains travel 160 km to the meeting point , then train x travels ( 3 / 8 ) * 160 = 60 km . the answer is c . | a ) 56 , b ) 58 , c ) 60 , d ) 62 , e ) 64 | c | multiply(divide(160, 5), divide(160, add(divide(160, 5), divide(160, 3)))) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(n0,#2)|multiply(#0,#3)| | physics |
a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the total share of a & b in profit of rs . 12100 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 a ' share = 3 / 10 * 12100 = 3630 b ' s share = 2 / 10 * 12100 = 2420 required total share = 3630 + 2420 = 6050 answer : c" | a ) rs . 6030 , b ) rs . 4050 , c ) rs . 6050 , d ) rs . 7050 , e ) rs . 6080 | c | multiply(divide(6300, add(add(6300, 4200), 10500)), 12100) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
an athlete runs 200 metres race in 40 seconds . what is his speed ? | "speed = distance / time = 200 / 40 = 5 m / s = 5 * 18 / 5 = 3618 km / hr answer : c" | a ) 23 , b ) 38 , c ) 18 , d ) 30 , e ) 28 | c | multiply(divide(200, 40), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
company p had 15 percent more employees in december than it had in january . if company p had 490 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 490 j = 490 / 1.15 j = 49,000 / 115 = 426 thus b is the correct answer ." | a ) 391 , b ) 426 , c ) 410 , d ) 423 , e ) 445 | b | divide(490, add(const_1, divide(15, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
the present worth of rs . 1404 due in two equal half - yearly instalments at 8 % per annum simple interest is : | solution required sum = p . w . of rs . 702 due 6 month hence + p . w . of rs . 702 due 1 year hence = rs . [ ( 100 x 702 / 100 + 8 x 1 / 2 ) + ( 100 x 702 / 100 + ( 8 x 1 ) ) ] = rs . ( 675 + 650 ) = rs . 1325 . answer a | a ) rs . 1325 , b ) rs . 1300 , c ) rs . 1350 , d ) rs . 1500 , e ) none | a | add(divide(divide(1404, const_2), divide(add(const_100, 8), const_100)), divide(divide(1404, const_2), divide(add(const_100, divide(8, const_2)), const_100))) | add(n1,const_100)|divide(n0,const_2)|divide(n1,const_2)|add(#2,const_100)|divide(#0,const_100)|divide(#1,#4)|divide(#3,const_100)|divide(#1,#6)|add(#5,#7) | gain |
the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 3 type b widgets , and the weight of 5 type b widgets is equal to the weight of 7 type c widgets . what is the r... | "5 b = 7 c and so b = 7 c / 5 7 a = 3 b and so a = 3 b / 7 = 3 c / 5 a + b = 3 c / 5 + 7 c / 5 = 10 c / 5 b + c = 7 c / 5 + c = 12 c / 5 the ratio of a + b : b + c = 10 : 12 = 5 : 6 the answer is e ." | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | e | divide(add(1, divide(7, 3)), add(divide(7, 3), divide(5, 3))) | divide(n0,n1)|divide(n2,n1)|add(n4,#0)|add(#0,#1)|divide(#2,#3)| | general |
a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 4 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 24 . if he buys 6 a rupee , his cp = 4 he selling at 4 a rupee , his sp = 6 profit = sp - cp = 6 - 4 = 2 profit percent = 2 / 4 * 100 = 50 % answer : c" | a ) 34 % , b ) 39 % , c ) 50 % , d ) 35 % , e ) 33 % | c | divide(multiply(4, const_100), 6) | multiply(n1,const_100)|divide(#0,n0)| | gain |
in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 60 percent of the male and 20 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 wi... | "percent of students who are 25 years old or older is 0.6 * 48 + 0.2 * 52 = ~ 39 , so percent of people who are less than 25 years old is 100 - 39 = 61 . answer : b ." | a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(60, const_100), divide(subtract(const_100, 20), const_100))) | divide(n2,const_100)|subtract(const_100,n3)|divide(#1,const_100)|multiply(#0,#2)|subtract(const_1,#3)| | general |
the cross - section of a cannel is a trapezium in shape . if the cannel is 15 m wide at the top and 4 m wide at the bottom and the area of cross - section is 1900 sq m , the depth of cannel is ? | "1 / 2 * d ( 15 + 4 ) = 1900 d = 200 answer : b" | a ) 150 , b ) 200 , c ) 140 , d ) 240 , e ) 145 | b | divide(divide(divide(1900, divide(add(15, 4), const_2)), 4), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
20 percent of andrea ' s living room floor is covered by a carpet that is 4 feet by 9 feet . what is the area of her living room floor ? | "20 % of area of the floor = 4 * 9 square feet = 36 square feet i . e . 100 % area of floor = ( 36 / 20 ) * 100 = 180 square feet answer : option b" | a ) 14.4 , b ) 180 , c ) 50.4 , d ) 60 , e ) 90 | b | divide(multiply(4, 9), divide(20, const_100)) | divide(n0,const_100)|multiply(n1,n2)|divide(#1,#0)| | gain |
a student scored an average of 60 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 60 * 3 = 180 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 180 = 140 answer : d" | a ) 86 , b ) 160 , c ) 76 , d ) 140 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(60, 3)) | multiply(n2,const_2)|multiply(n3,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 21 - m ) then m = ? | "2 m = 21 - m 3 m = 21 m = 7 the answer is c ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(21, add(2, const_1)) | add(n0,const_1)|divide(n3,#0)| | general |
each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 29 have long fur , 17 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | no of dogs = 45 long fur = 29 brown = 17 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 29 + 17 - both both = 9 answer e | a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 9 | e | subtract(add(29, 17), subtract(45, 8)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1) | other |
set a contains all the even numbers between 42 and 90 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 20 more than the corresponding term in set a . the difference of the sums = 25 * 20 = 500 . the answer is e ." | a ) 300 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | e | multiply(subtract(62, 42), add(divide(subtract(90, 42), const_2), const_1)) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)| | general |
of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 6 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 6 is 1200 * 1 / 6 so it reduced 200 people , so there is 1000 clerical people left but since 200 people left , it also reduced from the total of 3600 so there are ... | a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 29.4 % , e ) 11.1 % | d | multiply(divide(multiply(divide(1, 3), subtract(1, divide(1, 6))), add(multiply(divide(1, 3), subtract(1, divide(1, 6))), subtract(1, divide(1, 3)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)| | general |
an artist wishes to paint a circular region on a square poster that is 5 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | "area of the poster is 5 x 5 = 25 1 / 2 the area = 12.5 pi * r ^ 2 = 12.5 r ^ 2 = 12.5 / pi r = sqrt ( 12.5 / pi ) answer ( d )" | a ) 1 / pi , b ) sqrt ( 2 / pi ) , c ) 2 / sqrt ( pi ) , d ) sqrt ( 12.5 / pi ) , e ) pi / 2 | d | sqrt(divide(divide(square_area(5), 2), const_pi)) | square_area(n0)|divide(#0,n2)|divide(#1,const_pi)|sqrt(#2)| | geometry |
the tax on a commodity is diminished by 17 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 83 * 115 = 9545 - - - - - - - - - - - 10000 - - - - - - - - - - - 455 100 - - - - - - - - - - - ? = > 4.55 % decrease answer : c" | a ) 2 % , b ) 8 % , c ) 4.5 % , d ) 6.5 % , e ) 1.5 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(17, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
if x is 20 percent greater than 80 , then x = | "x is 20 % greater than 80 means x is 1.2 times 80 ( in other words 80 + 20 / 100 * 80 = 1.2 * 80 ) therefore , x = 1.2 * 80 = 96.00 answer : c" | a ) 68 , b ) 70.4 , c ) 96 , d ) 105.6 , e ) 108 | c | add(80, multiply(divide(20, const_100), 80)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
a pet groomer has 7 animals to groom for the day ( 2 cats and 5 dogs ) . if she randomly selects 4 animals to groom before lunch , what is the probability she will finish all the cats before lunch ? | combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 4 animals from 7 animals = 7 c 4 = 7 ! / [ 4 ! ( 7 - 4 ) ! ] = 35 . to finish all 2 cats there must be 2 dogs , select 2 dogs from 5 = 5 c 2 = 10 . and , select 2 cats from 2 = 2 c 2 = 1 . 2 cats and 2 dogs = ( 1 ) ( 10 ) = 10 pro... | a ) 3 / 7 , b ) 2 / 7 , c ) 12 / 13 , d ) 5 / 11 , e ) 4 / 7 | b | divide(2, 7) | divide(n1,n0) | physics |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.75 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 250 ? | "to maximize number of hot dogs with 250 $ total number of hot dogs bought in 250 - pack = 22.95 * 10 = 229.5 $ amount remaining = 250 - 229.5 = 20.5 $ total number of hot dogs bought in 20 - pack = 3.05 * 6 = 18.3 $ amount remaining = 20.5 - 18.3 = 2.2 $ total number of hot dogs bought in 8 - pack = 1.55 * 1 = 1.55 $ ... | a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,628 , e ) 2,256 | d | multiply(divide(250, 22.95), 250) | divide(n6,n5)|multiply(n4,#0)| | general |
a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 70 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 70 ⇒ 1.75 x = 70 ⇒ x = 40 coins answer c" | a ) 20 coins , b ) 30 coins , c ) 40 coins , d ) 25 coins , e ) none of these | c | divide(70, add(add(inverse(const_4), inverse(const_2)), const_1)) | inverse(const_4)|inverse(const_2)|add(#0,#1)|add(#2,const_1)|divide(n2,#3)| | general |
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