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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the ratio 4 : 20 expressed as percent equals to | explanation : actually it means 4 is what percent of 20 , which can be calculated as , ( 4 / 20 ) * 100 = 4 * 5 = 20 answer : option a | a ) 20 % , b ) 25 % , c ) 55 % , d ) 65 % , e ) none of above | a | multiply(divide(4, 20), const_100) | divide(n0,n1)|multiply(#0,const_100) | general |
jack takes a loan of $ 120000 with 12 % annual interest : the interest is paid once , at the end of the year . jill takes a loan of $ 120000 with 12 % annual interest , compounding monthly at the end of each month . at the end of one full year , compared to jack ' s loan interest , approximately how much more does jill... | jack ' s interest = $ 120,000 * 0.12 = $ 14400 or $ 1,200 each month . jills β s interest , 12 % / 12 = 1 % each month : for the 1 st month = $ 120,000 * 0.01 = $ 1,200 ; for the 2 nd month = $ 1,200 + 1 % of 1,200 = $ 1,212 , so we would have interest earned on interest ( very small amount ) ; for the 3 rd month = $ 1... | a ) zero , b ) $ 81.90 , c ) $ 8190.03 , d ) $ 819.00 , e ) $ 8.19 | d | subtract(subtract(multiply(120000, power(add(const_1, divide(12, multiply(12, const_100))), 12)), 120000), divide(multiply(120000, 12), const_100)) | multiply(n1,const_100)|multiply(n0,n1)|divide(n1,#0)|divide(#1,const_100)|add(#2,const_1)|power(#4,n1)|multiply(n0,#5)|subtract(#6,n0)|subtract(#7,#3) | general |
the area of one square is x ^ 2 + 8 x + 16 and the area of another square is 4 x ^ 2 β 12 x + 9 . if the sum of the perimeters of both squares is 32 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 4 ) ^ 2 a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 l 2 = 2 x - 3 p 1 = 4 ( x + 4 ) p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 32 4 ( x + 4 ) + 4 ( 2 x - 3 ) = 32 . . . . . . . . . . . . . . > x = 7 / 3 answer : d" | a ) 0 , b ) 2 , c ) 2.5 , d ) 7 / 3 , e ) 10 | d | divide(subtract(32, subtract(multiply(4, divide(8, 2)), 12)), 12) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n5)|subtract(n7,#2)|divide(#3,n5)| | general |
evaluate 45 / . 05 | "explanation : 45 / . 05 = 4500 / 5 = 900 option b" | a ) 700 , b ) 900 , c ) 705 , d ) none of these , e ) 506 | b | divide(const_100.0, divide(05, 45)) | divide(n1,const_100)|divide(n0,#0)| | general |
how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 1 ? | 510 , 512 , 514 , 516 , and 518 , so total 5 . hence option a . | a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 1), 1) | add(n0,n2)|add(n2,#0) | general |
brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 150 meters . they next meet after sally has run 200 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ? | nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 100 meters , thus second time she runs 2 * 150 = 300 meters . since second time ( when they run full circumference ) brenda runs 300 meters and sally runs 200 meters , thus... | a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 500 | e | add(multiply(const_2, 150), 200) | multiply(n0,const_2)|add(n1,#0) | physics |
a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 4... | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 400 bolts 400 * 6 / 25 = 96 mins ans c" | a ) 36 , b ) 72 , c ) 96 , d ) 144 , e ) 180 | c | divide(400, add(divide(100, const_60), divide(150, const_60))) | divide(n0,const_60)|divide(n1,const_60)|add(#0,#1)|divide(n2,#2)| | physics |
gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 40 and the turban . find the price of the turban . | "let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 40 . thus , 3434 ( 90 + x ) = 40 + x or 270 + 3 x = 160 + 4 x or x = 110 answer : e" | a ) 27 , b ) 36 , c ) 29 , d ) 10 , e ) 110 | e | subtract(multiply(divide(subtract(90, 40), subtract(const_12, 9)), const_12), 90) | subtract(n0,n2)|subtract(const_12,n1)|divide(#0,#1)|multiply(#2,const_12)|subtract(#3,n0)| | general |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.6 km ? | "a 600 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1600 ) = 600 m" | a ) 600 m , b ) 300 m , c ) 400 m , d ) 700 m , e ) 500 m | a | divide(multiply(18, multiply(1.6, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics |
a train passes a station platform in 50 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 ) / 50 = 15 = > x = 300 m answer : d | a ) 615 m , b ) 240 m , c ) 168 m , d ) 300 m , e ) 691 m | d | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0) | physics |
the lowest number which should be added to 8247 so that the sum is exactly divisible by 5 , 6 , 7 , 8 and 9 is : | "l . c . m . of 5 , 6 , 7 , 8 and 9 = 2520 . on dividing 8247 by 2520 , the remainder is 687 . number to be added = ( 2520 - 687 ) = 1833 . answer : option ' a '" | a ) 1833 , b ) 1853 , c ) 1733 , d ) 1233 , e ) 1832 | a | divide(8247, 6) | divide(n0,n2)| | general |
two spherical balls lie on the ground touching . if one of the balls has a radius of 3 cm , and the point of contact is 5 cm above the ground , what is the radius of the other ball ? | "similar triangle properties . . 2 / r + 3 = 3 / r - 3 giving r = 15 . answer : d" | a ) 3 cm , b ) 9 cm , c ) 12 cm , d ) 15 cm , e ) none of the these | d | add(add(3, 5), 3) | add(n0,n1)|add(#0,n0)| | physics |
in a class there are a total of 8 rows of desks and each desk can seat one student . there are 10 desks in the first row . in each subsequent row there are 2 more desks than in the previous row . find the maximum number of students seated in the class ? | the 8 rows form an arithmetic progression with first term 10 and last 24 . avg is 17 . so 17 * 8 = 136 answer is c . | a ) 112 , b ) 144 , c ) 136 , d ) 132 , e ) 118 | c | multiply(divide(add(multiply(8, 2), add(8, 10)), 2), 8) | add(n0,n1)|multiply(n0,n2)|add(#0,#1)|divide(#2,n2)|multiply(n0,#3) | general |
what is the smallest positive integer nn such that 6,480 β n β 6,480 β n is a perfect cube ? | "sol : let ' s factorize 6480 and we get 6480 = 3 ^ 4 * 2 ^ 4 * 5 now we need to see for what minimum value of n β n * 6480 = a ^ 3 where a is an integer so from 6480 we already have 2 ^ 4 * 3 ^ 4 * 5 * n β n = ( 2 ^ 2 ) ^ 3 * ( 3 ^ 2 ) ^ 3 * ( 5 ) ^ 3 why cause a is an integer which will need to be have the same facto... | a ) 5 , b ) 5 ^ 2 , c ) 30 , d ) 30 ^ 2 , e ) 30 ^ 4 | e | add(const_3, const_4) | add(const_3,const_4)| | geometry |
a candidate who gets 20 % of the marks fails by 40 marks . but another candidate who gets 30 % marks gets 20 marks more than necessary for passing . find the number of marks for passing ? | "20 % - - - - - - - - - - - - 40 30 % - - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - - - 10 % - - - - - - - - - - - - - 60 20 % - - - - - - - - - - - - - - 120 120 + 40 = 160 marks answer : d" | a ) 100 marks , b ) 200 marks , c ) 280 marks , d ) 160 marks , e ) 827 marks | d | add(multiply(divide(add(40, 20), subtract(divide(30, const_100), divide(20, const_100))), divide(20, const_100)), 40) | add(n1,n3)|divide(n2,const_100)|divide(n0,const_100)|subtract(#1,#2)|divide(#0,#3)|multiply(#4,#2)|add(n1,#5)| | gain |
10 women can complete a work in 8 days and 10 children take 12 days to complete the work . how many days will 6 women and 3 children together take to complete the work ? | "explanation : 1 women β s 1 day β s work = 1 / 8 / 10 = 1 / 80 1 child β s 1 day β s work = 1 / 12 / 10 = 1 / 120 6 women β s 1 day β s work = 1 / 80 Γ 6 = 3 / 40 3 children β s 1 day β s work = 1 / 120 Γ 3 = 1 / 40 6 women β s + 3 children β s 1 day β s work = 3 / 40 + 1 / 40 = 1 / 10 therefore , they will finish the... | a ) 7 , b ) 10 , c ) 8 , d ) 9 , e ) 11 | b | inverse(add(divide(6, multiply(10, 8)), divide(10, multiply(3, 12)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n5,#1)|add(#2,#3)|inverse(#4)| | physics |
solve for x and check : - 200 x = 1600 | "solution : dividing each side by - 200 , we obtain ( - 200 x / - 200 ) = ( 1600 / - 200 ) therefore : x = - 8 check : - 200 x = 1600 ( - 200 * - 8 ) = 1600 1600 = 1600 answer : c" | a ) 2000 , b ) 2573 , c ) 1600 , d ) 2950 , e ) none of these | c | multiply(200, divide(1600, 200)) | divide(n1,n0)|multiply(n0,#0)| | general |
if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 140 units ? | "here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 ... | a ) 50 hours , b ) 40 hours , c ) 17.5 hours , d ) 12 hours , e ) 8 hours | c | divide(140, multiply(divide(divide(20, 10), 5), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 1 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.10 i ( last year ) = 1.1 i ( last year ) 1 = 1.1 x i ( last year ) i ( last year ) = 1 / 1.1 = 10 / 11 = 0.90909 . . . or i ( last year ) = 0.91 % answer : c" | a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 10 % , e ) 10.8 % | c | divide(multiply(1, const_100), add(1, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
there are 450 birds at a small zoo . the number of birds is 5 times the number of all the other animals combined . how many more birds are there than non bird animals at the zoo ? | total birds = 450 = 5 * other animals ( x ) or x = 90 . so , difference in birds and x = 450 - 90 = 360 . answer is b . | a ) 400 , b ) 360 , c ) 270 , d ) 180 , e ) 90 | b | subtract(450, divide(450, 5)) | divide(n0,n1)|subtract(n0,#0) | general |
a man can row downstream at 20 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 20 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 30 = > x = 15 , y = 5 . answer : d" | a ) 15,2 , b ) 15 , 4 , c ) 15 , 8 , d ) 15 , 5 , e ) 15,7 | d | divide(divide(add(20, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
an seller earns an income of re 1 on the first day of his business . on every subsequent day , he earns an income which is just thrice of that made on the previous day . on the 20 th day of business , he earns an income of : | "2 nd day he earns = 3 ( 2 β 1 ) 3 rd day he earns = 3 ( 3 β 1 ) on 20 th day he earns 3 ( 20 - 1 ) = 57 rupees answer : b" | a ) 29 , b ) 57 , c ) 35 , d ) 20 , e ) 30 | b | multiply(subtract(20, 1), 1) | subtract(n1,n0)|multiply(n0,#0)| | physics |
if 16 ^ y = 4 ^ 16 , what is y ? | "16 ^ y = 4 ^ 2 y = 4 ^ 16 2 y = 16 y = 8 the answer is c ." | a ) 2 , b ) 4 , c ) 8 , d ) 12 , e ) 16 | c | divide(16, const_2) | divide(n2,const_2)| | general |
the sum of the ages of 4 children born at the intervals of 3 years each is 38 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) = 38 4 x = 20 = > x = 5 . age of youngest child = x = 5 years . answer : d" | a ) 2 years , b ) 3 years , c ) 4 years , d ) 5 years , e ) 6 years | d | divide(subtract(divide(38, divide(4, 3)), multiply(subtract(4, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general |
if the selling price of 8 articles is same as the cost price of 10 articles . find the gain or loss percentage ? | "let the c . p of each article be re 1 . then , s . p of 8 articles = c . p of 10 articles = rs . 10 now , c . p of 8 articles = rs . 8 , s . p of 8 articles = rs 10 gain = rs ( 10 - 8 ) = rs 2 . gain % = ( 2 / 8 Γ 100 ) % = 25 % answer : a" | a ) 25 % , b ) 45 % , c ) 35 % , d ) 65 % , e ) 55 % | a | subtract(8, 10) | subtract(n0,n1)| | gain |
a cricket player whose bowling average was 24.85 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.85 . what will be the number of wickets taken by him till the last match ? | "average = total runs / total wickets total runs after last match = 24.85 w + 52 total wickets after last match = w + 5 ( 24.85 w + 52 ) / ( w + 5 ) = 24.85 - 0.85 = 24 w = 80 so total wickets aftr last match = w + 5 = 85 answer : c" | a ) 64 , b ) 72 , c ) 85 , d ) 96 , e ) 108 | c | divide(subtract(multiply(subtract(24.85, 0.85), 5), 52), 0.85) | subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)| | general |
what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 3 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 3 = 1.66 so biggest is 5 / 3 and smallest is 2 / 3 their difference is 5 / 3 - 2 / 3 = 3 / 3 = 1 option d" | a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 , e ) none of these | d | subtract(divide(4, 5), divide(2, 3)) | divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)| | general |
10 ^ 222 Γ· 10 ^ 220 = ? | 10 ^ 222 Γ· 10 ^ 220 = 10 ^ ( 222 - 220 ) = 10 ^ 2 = 100 answer is c | a ) 10 , b ) 1000 , c ) 100 , d ) 10000 , e ) 100000 | c | divide(power(10, 222), power(10, 220)) | power(n0,n1)|power(n0,n3)|divide(#0,#1) | general |
a library has an average of 510 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is | "explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 510 * 5 + 240 * 25 ) / 30 = 8550 / 30 = 285 . answer : a" | a ) 285 , b ) 337 , c ) 878 , d ) 227 , e ) 291 | a | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 Γ£ β 2.5 = 20 if x is the weight of the new person , total increase in weight = x Γ’ Λ β 50 = > 20 = x - 50 = > x = 20 + 50 = 70 answer : option c" | a ) 75 kg , b ) 50 kg , c ) 70 kg , d ) 80 kg , e ) 60 kg | c | add(multiply(8, 2.5), 50) | multiply(n0,n1)|add(n2,#0)| | general |
the average of 10 numbers is calculated as 15 . it is discovered later on that while calculating the average one number , namely 36 was wrongly read as 26 . the correct average is : | explanation : sum of numbers = ( 10 Γ Γ 15 - 26 + 36 ) = 160 correct average = 160 / 10 = 16 correct option : c | a ) 12.4 , b ) 14 , c ) 16 , d ) 18.6 , e ) none of these | c | add(15, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1) | general |
if z is not equal to zero , and z = 6 zs β 9 s 2 β β β β β β β β β z = 6 zs β 9 s 2 , then z equals : | squaring on both the sides gives z ^ 2 = 6 zs - 9 s ^ 2 = z ^ 2 - 6 zs + 9 s ^ 2 = z ( z - 3 s ) - 3 s ( z - 3 s ) = ( z - 3 s ) ( z - 3 s ) = z = 3 s answer = b | a ) s , b ) 3 s , c ) 4 s , d ) - 3 s , e ) - 4 s | b | divide(add(divide(9, gcd(6, 9)), gcd(6, 9)), divide(6, gcd(6, 9))) | gcd(n0,n1)|divide(n1,#0)|divide(n0,#0)|add(#1,#0)|divide(#3,#2) | general |
sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 115 . if the ratio of his marks in english and maths is 5 : 1 , find his marks in science ? | "s : e = 1 : 3 e : m = 5 : 1 - - - - - - - - - - - - s : e : m = 5 : 15 : 3 5 / 23 * 115 = 25 answer : c" | a ) 18 , b ) 77 , c ) 25 , d ) 55 , e ) 31 | c | add(add(add(multiply(5, const_3), 5), multiply(multiply(5, const_3), add(multiply(5, const_3), 5))), 5) | multiply(n1,const_3)|add(n1,#0)|multiply(#1,#0)|add(#1,#2)|add(n1,#3)| | general |
two boats are heading towards each other at constant speeds of 5 miles / hr and 23 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 5 + 23 = 25 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 28 * 1 / 60 = 7 / 15 miles apart . answer : b ." | a ) 1 / 12 , b ) 7 / 12 , c ) 1 / 6 , d ) 1 / 3 , e ) 1 / 5 | b | divide(add(23, 5), const_60) | add(n0,n1)|divide(#0,const_60)| | physics |
when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum e of all possible 2 - digit remainders for x / y ? | "ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . e = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b" | a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784 | b | divide(59.32, subtract(2, floor(2))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
there are cats got together and decided to kill the mice of 999936 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999936 can be written as 1000000 Γ’ β¬ β 64 = 10002 Γ’ β¬ β 82 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 8 ) * ( 1000 - 8 ) = ( 1008 ) * ( 992 ) given that number of cats is less than number if mice . so number of cats is 992 and number of mice were 1008 answer d" | a ) 941,1009 , b ) 991,1001 , c ) 991,1009 , d ) 992,1008 , e ) 931,1009 | d | divide(999936, add(multiply(const_100, const_10), add(const_3, const_2))) | add(const_2,const_3)|multiply(const_10,const_100)|add(#0,#1)|divide(n0,#2)| | general |
a man can row 7 Γ’ Β½ kmph in still water . if in a river running at 1.5 km / hr an hour , it takes him 50 minutes to row to a place and back , how far off is the place ? | speed downstream = ( 7.5 + 1.5 ) km / hr = 9 km / hr ; speed upstream = ( 7.5 - 1.5 ) kmph = 6 kmph . let the required distance be x km . then , x / 9 + x / 6 = 50 / 60 . 2 x + 3 x = ( 5 / 6 * 18 ) 5 x = 15 x = 3 . hence , the required distance is 3 km . answer d | a ) 4 km , b ) 2 km , c ) 5 km , d ) 3 km , e ) 1 km | d | subtract(add(add(7, divide(const_1, const_2)), 1.5), subtract(add(7, divide(const_1, const_2)), 1.5)) | divide(const_1,const_2)|add(n0,#0)|add(n1,#1)|subtract(#1,n1)|subtract(#2,#3) | physics |
a chemical supply company has 60 liters of a 30 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | 60 liters of a 30 % hno 3 solution means hno 3 = 18 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 18 + x = 50 % of ( 60 + x ) or x = 24 . hence , d | a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | d | multiply(subtract(divide(60, const_2), divide(multiply(60, 30), const_100)), const_2) | divide(n0,const_2)|multiply(n0,n1)|divide(#1,const_100)|subtract(#0,#2)|multiply(#3,const_2) | gain |
what is the sum of all digits for the number 10 ^ 28 - 44 ? | "10 ^ 28 is a 29 - digit number : 1 followed by 28 zeros . 10 ^ 28 - 44 is a 28 - digit number : 26 9 ' s and 56 at the end . the sum of the digits is 26 * 9 + 5 + 6 = 245 . the answer is b ." | a ) 233 , b ) 245 , c ) 257 , d ) 270 , e ) 285 | b | multiply(add(divide(subtract(subtract(28, 10), const_2), const_2), 10), divide(add(subtract(28, 10), const_2), const_2)) | subtract(n1,n0)|add(#0,const_2)|subtract(#0,const_2)|divide(#2,const_2)|divide(#1,const_2)|add(n0,#3)|multiply(#5,#4)| | general |
mr and mrs a are opposite to each other . the distance between mr a and mrs a are 120 km . the speed of mr a and mrs a are 30 kmph , 10 kmph respectively . one bee is running between mr a nose to mrs a nose and returning back to mr a nose . the speed of bee is 60 kmph . then how far bee traveled ? | 30 x + 10 x = 120 x = 3 hrs speed of bee = 60 kmph distance traveled by bee = speed * time taken = 60 * 3 = 180 km answer : c | a ) 160 km , b ) 170 km , c ) 180 km , d ) 190 km , e ) 130 km | c | multiply(divide(120, add(30, 10)), 60) | add(n1,n2)|divide(n0,#0)|multiply(n3,#1) | physics |
each machine of type a has 2 steel parts and 3 chrome parts . each machine of type b has 3 steel parts and 5 chrome parts . if a certain group of type a and type b machines has a total of 40 steel parts and 32 chrome parts , how many machines are in the group | look at the below representation of the problem : steel chrome total a 2 3 40 > > no . of type a machines = 40 / 5 = 8 b 3 5 32 > > no . of type b machines = 32 / 8 = 4 so the answer is 12 i . e a . hope its clear . | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | add(divide(32, add(5, 3)), divide(40, add(2, 3))) | add(n1,n3)|add(n0,n1)|divide(n5,#0)|divide(n4,#1)|add(#2,#3) | general |
because he β s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ? | "let x be mork ' s income , then mindy ' s income is 4 x . the total tax paid is 0.4 x + 1.2 x = 1.6 x 1.6 x / 5 x = 0.32 the answer is a ." | a ) 32 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 38 % | a | multiply(const_100, divide(add(divide(40, const_100), multiply(4, divide(30, const_100))), add(const_1, 4))) | add(n2,const_1)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|multiply(#5,const_100)| | gain |
find the average of all the numbers between 6 and 34 which are divisible by 5 ? | "average = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 100 / 5 = 20 . answer : b" | a ) 32 , b ) 20 , c ) 28 , d ) 11 , e ) 18 | b | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
set # 1 = { a , b , c , d , e } set # 2 = { k , l , m , n , o , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? | so not a vowel in set - 1 : 3 / 5 and not a vowel in ser - 2 : 5 / 6 now , 3 / 5 β 5 / 6 = 12 this is for not a vowel . then for at least one vowel will be = 1 β 1 / 2 = 1 / 2 answer will be c . | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | c | multiply(divide(const_3, add(const_3, 2)), divide(add(const_3, 2), add(add(const_3, 2), 1))) | add(n1,const_3)|add(n0,#0)|divide(const_3,#0)|divide(#0,#1)|multiply(#2,#3) | general |
in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 15 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 15 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 5 ) = 1 / 15 therefore only video cassette recorder = 1 / 15 - 1 / 15 = 0 total = 1 / 5 + 0 + neither cable tv nor videocassette recorders 1 = 1 /... | a ) 23 / 30 , b ) 11 / 15 , c ) 7 / 10 , d ) 4 / 5 , e ) 2 / 15 | d | divide(add(15, 5), multiply(15, const_2)) | add(n1,n3)|multiply(n3,const_2)|divide(#0,#1)| | other |
a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in metres ) is | "explanation : speed = ( 5 Γ 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 Γ 15 Γ 60 ) m = 1250 m . answer : d" | a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these | d | multiply(divide(multiply(5, const_1000), const_60), 15) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | physics |
a student is ranked 17 th from right and 5 th from left . how many students are there in totality ? | "from right 17 , from left 5 total = 17 + 5 - 1 = 21 answer : d" | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | d | subtract(add(17, 5), const_1) | add(n0,n1)|subtract(#0,const_1)| | general |
a reduction of 46 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "explanation : 40 * ( 46 / 100 ) = 18.4 - - - 64 ? - - - 12 = > rs . 3.45 answer : d" | a ) 2.45 , b ) 8.45 , c ) 7.45 , d ) 3.45 , e ) 1.45 | d | multiply(const_12, divide(multiply(46, divide(46, const_100)), 64)) | divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)| | gain |
what should be the least number to be added to the 11002 number to make it divisible by 11 ? | "answer : 9 option : b" | a ) 12 , b ) 9 , c ) 18 , d ) 77 , e ) 26 | b | subtract(11, reminder(11002, 11)) | reminder(n0,n1)|subtract(n1,#0)| | general |
in right triangle abc , ac is the hypotenuse . if ac is 50 and ab + bc = 70 , what is the area of the triangle abc ? | square ab + bc = 70 : ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 4900 . since ( ac ) ^ 2 = ( ab ) ^ 2 + ( bc ) ^ 2 = 50 ^ 2 = 2500 , then ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 2500 + 2 * ab * bc = 4900 . 2500 + 2 * ab * bc = 4900 . ab * bc = 1200 . the area = 1 / 2 * ab * bc = 600 . answer : e . | ['a ) 225', 'b ) 450', 'c ) 25 β 2', 'd ) 200', 'e ) 600'] | e | triangle_area_three_edges(50, multiply(const_3, const_10), multiply(const_4, const_10)) | multiply(const_10,const_3)|multiply(const_10,const_4)|triangle_area_three_edges(n0,#0,#1) | geometry |
the ratio between x and y is 8 / 5 ; x is increased by 10 and y is multiplied by 10 , what is the ratio between the new values of x and y ? | "ratio = 8 k / 5 k = 8 / 5 , 16 / 10 , etc . x and y are decreased by 5 - - > ( 8 k + 10 ) / ( 5 k * 10 ) new ratio can be 18 / 50 , 26 / 100 , etc . answer : e" | a ) 8 / 5 , b ) 5 / 8 , c ) 1 , d ) 10 , e ) it can not be determined | e | divide(multiply(8, 5), multiply(5, 8)) | multiply(n0,n1)|divide(#0,#0)| | general |
the sum of three consecutive multiples of 3 is 72 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 72 9 x = 63 x = 7 largest number = 3 x + 6 = 27 . answer : c" | a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) 39 | c | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(di... | 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)| | general |
cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 22 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 27 shows a cube of a number could be right . here 64 + 22 = 86 72 + 22 = 94 86 + 22 = 108 98 + 22 = 120 103 + 22 = 125 - - - ( 5 * 5 * 5 ) so we have 103 as the answer ! e" | a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 103 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 22) | add(const_1,const_3)|add(#0,const_1)|power(#1,const_3)|subtract(#2,n0)| | geometry |
the average age of an adult class is 40 years . 12 new students with an avg age of 34 years join the class . therefore decreasing the average by 4 years . find what was the original average age of the class ? | "let original strength = y then , 40 y + 12 x 34 = ( y + 12 ) x 36 Γ’ β‘ β 40 y + 408 = 36 y + 432 Γ’ β‘ β 4 y = 48 Γ’ Λ Β΄ y = 6 a" | a ) 6 , b ) 12 , c ) 16 , d ) 20 , e ) 22 | a | divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 34)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general |
the ratio of the volumes of two cubes is 2197 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = Β³ β 2197 : Β³ β 1331 = 13 : 11 ratio of surface areas = 13 ^ 2 : 11 ^ 2 = 169 : 121 answer : option a" | a ) 169 : 121 , b ) 169 : 127 , c ) 169 : 191 , d ) 121 : 169 , e ) 121 : 182 | a | power(divide(2197, 1331), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
a train 150 m long crosses a platform 120 m long in 10 sec ; find the speed of the train ? | "d = 150 + 120 = 270 t = 10 s = 270 / 10 * 18 / 5 = 97 kmph answer : c" | a ) 87 kmph , b ) 65 kmph , c ) 97 kmph , d ) 16 kmph , e ) 18 kmph | c | subtract(multiply(10, multiply(120, const_0_2778)), 150) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
there were totally 100 men . 85 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 85 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 15 - 25 - 15 - 30 = 100 - 85 = 15 answer : e" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | subtract(100, add(add(add(subtract(100, 85), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)| | general |
two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 10 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 5 ... | a ) 1 : 5 , b ) 9 : 25 , c ) 5 : 1 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(10, const_2))) | divide(n1,const_2)|divide(n0,const_2)|circumface(#0)|circumface(#1)|divide(#2,#3)| | physics |
a trader sells 45 meters of cloth for rs . 4500 at the profit of rs . 14 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4500 / 45 = rs . 100 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 14 = rs . 86 . answer : b" | a ) rs . 80 , b ) rs . 86 , c ) rs . 90 , d ) rs . 95 , e ) none of these | b | subtract(divide(4500, 45), 14) | divide(n1,n0)|subtract(#0,n2)| | physics |
a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 20 km , if the person completes half of the distance by walking and the other half by running ? | "time = 10 / 4 + 10 / 8 = 30 / 8 = 3.75 hours the answer is c ." | a ) 2.75 , b ) 3.25 , c ) 3.75 , d ) 4.25 , e ) 4.75 | c | add(divide(divide(20, const_2), 4), divide(divide(20, const_2), 8)) | divide(n2,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)| | physics |
12 times a number gives 156 . the number is | "explanation : let the number be ' n ' 12 Γ n = 156 β n = 13 correct option : c" | a ) 11 , b ) 12 , c ) 13 , d ) none , e ) can not be determined | c | divide(156, divide(156, 12)) | divide(n1,n0)|divide(n1,#0)| | general |
if the sides of a triangle are 30 cm , 28 cm and 10 cm , what is its area ? | "the triangle with sides 30 cm , 28 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 28 * 10 = 140 cm 2 answer : option d" | a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150 | d | divide(multiply(28, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
if the side length of square b is sqrt ( 3 ) times that of square a , the area of square b is how many times the area of square a ? | let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( sqrt ( 3 ) x ) ^ 2 = 3 x ^ 2 . the answer is d . | ['a ) 9', 'b ) 6', 'c ) 4', 'd ) 3', 'e ) 2'] | d | power(sqrt(3), const_2) | sqrt(n0)|power(#0,const_2) | geometry |
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 15 percent , and on day 3 , it is discounted an additional 20 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 15 % discount = 765 price in day 3 after 20 % discount = 612 so , price in day 3 as percentage of the sale price on day 1 will be = 612 / 900 * 100 = > 68 % answer will definitely be ( c )" | a ) 28 % , b ) 40 % , c ) 68 % , d ) 70 % , e ) 72 % | c | add(multiply(divide(divide(20, const_100), subtract(1, divide(1, 10))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
an author received $ 0.80 in royalties for each of the first 100000 copies of her book sold , and $ 0.30 in royalties for each additional copy sold . if she received a total of $ 260000 in royalties , how many copies of her book were sold ? | total royalties for first 100.000 books = . 8 * 100,000 = 80,000 total royalties for the rest of the books = 260,000 - 80,000 = 180,000 remaining books = 180,000 / 0.3 = 600,000 total books = 600,000 + 100,000 = 700,000 answer a | a ) 700,000 , b ) 300,000 , c ) 380,000 , d ) 400,000 , e ) 420,000 | a | divide(add(divide(subtract(260000, multiply(0.8, 100000)), 0.3), 100000), const_1000) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|add(n1,#2)|divide(#3,const_1000) | general |
in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 / 6 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 / 6 of the garden is green , so , since all the flowers must be either green or yellow , we know that 5 / 6 are yellow . we ' re also told there is an equal p... | a ) 1 / 7 , b ) 1 / 8 , c ) 1 / 4 , d ) 3 / 7 , e ) 7 / 8 | d | multiply(subtract(1, divide(1, 6)), divide(1, 2)) | divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)| | probability |
virginia , adrienne , and dennis have taught history for a combined total of 102 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 102 = > 3 d = 102 + 27 = 129 = > d = 43 answer d" | a ) 23 , b ) 32 , c ) 35 , d ) 43 , e ) 44 | d | add(divide(subtract(102, add(add(9, 9), 9)), const_3), add(9, 9)) | add(n1,n2)|add(n1,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)| | general |
the age of father 10 years ago was thirce the age of his son . 10 years hence , father β s age will be twice that of his son . the ration of their present ages is : | solution let the ages of father and son 10 year ago be 3 x and x years respectively . then , ( 3 x + 10 ) + 10 = 2 [ ( x + 10 ) + 10 β 3 x + 20 = 2 x + 40 β x = 20 . β΄ required ratio = ( 3 x + 10 ) : ( x + 10 ) = 70 : 30 : 7 : 3 . answer b | a ) 5 : 2 , b ) 7 : 3 , c ) 9 : 2 , d ) 13 : 4 , e ) none of these | b | divide(add(multiply(subtract(add(multiply(10, const_2), multiply(10, const_2)), multiply(10, const_2)), const_3), 10), add(subtract(add(multiply(10, const_2), multiply(10, const_2)), multiply(10, const_2)), 10)) | multiply(n0,const_2)|add(#0,#0)|subtract(#1,#0)|add(n0,#2)|multiply(#2,const_3)|add(n0,#4)|divide(#5,#3) | general |
a one - foot stick is marked in 1 / 2 and 1 / 4 portion . how many total markings will there be , including the end points ? | "lcm of 8 = 4 1 / 2 marking are ( table of 2 ) 0 . . . . . . 2 . . . . . . . . . . . 4 ( total = 3 ) 1 / 4 marking are ( table of 1 ) 0 . . . . . . . 1 . . . . . . 2 . . . . . . 3 . . . . . . . . 4 ( total = 5 ) overlapping markings are 0 . . . . . . . . 2 . . . . . . . . . 4 ( total = 3 ) total markings = 3 + 5 - 3 = ... | a ) 8 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(2, 4) | add(n1,n3)| | general |
if integer k is equal to the sum of all even multiples of 25 between 200 and 600 , what is the greatest prime factor of k ? | "if we break down what the stem is asking what is the sum of all mult of 50 between 200 and 600 . using arithmetic progression to find n : 600 = 200 + ( n - 1 ) 50 400 + 50 = 50 n 450 = 50 n = > n = 9 the sum would be : 9 * mean mean = [ 600 + 200 ] / 2 = 400 9 * 400 = 3600 d" | a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | d | divide(add(divide(subtract(multiply(floor(divide(600, 25)), 25), multiply(floor(divide(200, 25)), 25)), 25), const_1), const_2) | divide(n2,n0)|divide(n1,n0)|floor(#0)|floor(#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,#5)|divide(#6,n0)|add(#7,const_1)|divide(#8,const_2)| | general |
how many integers between 324,700 and 448,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 448,513 the number of integers is 4486 - 3247 = 1239 the answer is e ." | a ) 10,300 , b ) 8,030 , c ) 1,253 , d ) 1,252 , e ) 1,239 | e | subtract(448,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
g ( x ) is defined as the product of all even integers k such that 0 < k β€ x . for example , g ( 14 ) = 2 Γ 4 Γ 6 Γ 8 Γ 10 Γ 12 Γ 14 . if g ( r ) is divisible by 4 ^ 11 , what is the smallest possible value for r ? | "g ( r ) = 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 ... | a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44 | b | multiply(2, 12) | multiply(n2,n7)| | general |
if 1 / 12 of the passengers on a ship are from north america , 1 / 8 are europeans , 1 / 3 are from africa , 1 / 6 are from asia and the remaining 35 people are citizens of other continents , then how many passengers are on board the ship ? | 1 / 12 + 1 / 8 + 1 / 3 + 1 / 6 = ( 2 + 3 + 8 + 4 ) / 24 = 17 / 24 let x be the number of passengers on the ship . 35 = ( 7 / 24 ) x x = 120 the answer is c . | a ) 110 , b ) 115 , c ) 120 , d ) 125 , e ) 130 | c | divide(35, subtract(const_1, add(add(add(divide(1, 12), divide(1, 8)), divide(1, 3)), divide(1, 6)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|divide(n0,n7)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(const_1,#6)|divide(n8,#7) | general |
54671 - 14456 - 33466 = ? | "c if we calculate we will get 6749" | a ) 2449 , b ) 5449 , c ) 6749 , d ) 6449 , e ) 6468 | c | subtract(multiply(divide(54671, const_100), 14456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 14456))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
one side of a rectangular field is 15 m and one of its diagonal is 17 m . find the area of the field . | solution other side = β ( 17 ) 2 - ( 15 ) 2 = β 289 - 225 = β 64 = 8 m . β΄ area = ( 15 x 8 ) m 2 = 120 m 2 . answer b | ['a ) 100', 'b ) 120', 'c ) 150', 'd ) 180', 'e ) none'] | b | rectangle_area(15, sqrt(subtract(power(17, const_2), power(15, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3) | geometry |
a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 13 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 3 more or 3 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could ha... | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that... | a ) $ 2,040 , b ) $ 2,120 , c ) $ 2,240 , d ) $ 1,920 , e ) $ 1,400 | c | subtract(multiply(const_3, const_1000), add(multiply(divide(add(13, 3), 3), 20), multiply(subtract(divide(add(13, const_2.0), 3), 3), 100))) | add(n3,n5)|multiply(const_1000,const_3)|divide(#0,const_2.0)|multiply(n1,#2)|subtract(#2,const_2.0)|multiply(n2,#4)|add(#3,#5)|subtract(#1,#6)| | general |
the ages of two person differ by 20 years . if 7 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 7 ) = ( x + 20 - 7 ) = > 4 x = 48 = > x = 12 their present ages are 32 years and 12 year . answer : b" | a ) 30 , 10 , b ) 32 , 12 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | b | add(divide(add(multiply(5, 7), subtract(20, 7)), subtract(5, const_1)), 20) | multiply(n1,n2)|subtract(n0,n1)|subtract(n2,const_1)|add(#0,#1)|divide(#3,#2)|add(n0,#4)| | general |
how many integers between 1 and 10 ^ 18 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 17 and there are 18 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 17 thus , the total number of integers is 1 + 2 + . . . + 17 + 18 = 18 ... | a ) 161 , b ) 171 , c ) 181 , d ) 191 , e ) 201 | b | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
how many positive integers less than 27 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 3 odd multiples of 5 : { 5 , 15 , 25 } 11 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 } notice , that 5 is in two sets , thus total # of integers satis... | a ) 27 , b ) 25 , c ) 24 , d ) 20 , e ) 22 | e | subtract(subtract(subtract(27, 2), const_1), const_1) | subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
in a corporation , 50 percent of the male employees and 40 percent of the female employees are at least 35 years old . if 42 percent of all the employees are at least 35 years old , what fraction of the employees in the corporation are females ? | you can use the weighted averages formula for a 10 sec solution . no of females / no of males = ( 50 - 42 ) / ( 42 - 40 ) = 4 / 1 no of females as a fraction of total employees = 4 / ( 4 + 1 ) = 4 / 5 ; answer : d | a ) 3 / 5 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | d | divide(subtract(50, 42), subtract(50, 40)) | subtract(n0,n3)|subtract(n0,n1)|divide(#0,#1) | other |
if 4 men working 10 hours a day earn rs . 1400 per week , then 9 men working 6 hours a day will earn how much per week ? | "explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1400 : x hence 4 * 10 * x = 9 * 6 * 1400 or x = 9 * 6 * 1400 / 4 * 10 = 1890 answer : d" | a ) rs 840 , b ) rs 1320 , c ) rs 1620 , d ) rs 1890 , e ) none of these | d | multiply(divide(multiply(9, 6), multiply(4, 10)), 1400) | multiply(n3,n4)|multiply(n0,n1)|divide(#0,#1)|multiply(n2,#2)| | physics |
if a rectangular billboard has an area of 120 square feet and a perimeter of 46 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 120 and a perimeter of 46 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 120 , the shorter side will almost certainly be... | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | c | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 120)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
a car gets 33 miles to the gallon . if it is modified to use a solar panel , it will use only 75 percent as much fuel as it does now . if the fuel tank holds 16 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ? | "originally , the distance the car could go on a full tank was 16 * 33 = 528 miles . after it has been modified , the car can go 33 / 0.75 = 44 miles per gallon . on a full tank , the car can go 16 * 44 = 704 miles , thus 176 miles more . the answer is c ." | a ) 168 , b ) 172 , c ) 176 , d ) 180 , e ) 184 | c | subtract(multiply(multiply(divide(33, 75), const_100), 16), multiply(33, 16)) | divide(n0,n1)|multiply(n0,n2)|multiply(#0,const_100)|multiply(n2,#2)|subtract(#3,#1)| | physics |
for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 2 / 7 , what are the odds of it burning out during ... | "p ( of not burning out in a six mnth period ) = 1 / 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 2 / 7 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 / 7 = 5 / 7 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 5 / 7 = 1 / 3 - - - > p ( of burning ... | a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 2 , d ) 4 / 9 , e ) 2 / 3 | c | multiply(subtract(1, divide(2, 7)), subtract(1, divide(subtract(1, divide(2, 7)), 2))) | divide(n3,n4)|subtract(n6,#0)|divide(#1,n3)|subtract(n6,#2)|multiply(#1,#3)| | general |
shekar scored 76 , 65 , 82 , 67 and 55 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 67 + 55 ) / 5 = 375 / 5 = 69 hence average = 69 answer : b" | a ) 65 , b ) 69 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 55), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 2500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 2500 resolutions . = 2500 * 2 * 22 / 7 * 22.4 = 352000 cm = 3520 m answer : d" | a ) 1187 m , b ) 1704 m , c ) 2179 m , d ) 3520 m , e ) 4297 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 2500), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
the average of 10 numbers is 23 . if each number is increased by 4 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 4 , the total increase = 4 * 10 = 40 the new sum = 230 + 40 = 270 the new average = 270 / 10 = 27 . answer : b" | a ) 20 , b ) 27 , c ) 72 , d ) 28 , e ) 82 | b | multiply(23, 4) | multiply(n1,n2)| | general |
if a person has two rectangular fields . the larger field has thrice the length and 4 times the width of the smaller field . if the smaller field has a length 50 % more than the width . if a person takes 20 minutes to complete one round of a smaller field then what is the time required to complete one round of a larger... | let the width of the smaller rectangle is 4 units then , the length of the smaller rectangle is 6 units ( that is 50 % more than the width ) now the perimeter of the rectangle is 2 ( 6 + 4 ) = 20 units so , 20 units is covered in 20 units , implies covers one unit in one minute so now coming to the larger rectangle the... | a ) 69 minutes , b ) 68 minutes , c ) 58 minutes , d ) 48 minutes , e ) 67 minutes | b | add(add(add(multiply(20, multiply(4, divide(50, const_100))), 20), 4), 4) | divide(n1,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(n2,#2)|add(n0,#3)|add(n0,#4) | general |
the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 7 miles per gallon , will not get him there . by how many miles per gallon must the aircraft β s fuel efficiency be improved to make the flig... | "actual miles / gallon is = 480 / 40 = 12 miles / gallon . current engine miles / gallon is 7 miles / gallon . additional 5 miles / gallon is required to match the actual mileage . imo option c ." | a ) 2 , b ) 4 , c ) 5 , d ) 40 , e ) 160 | c | subtract(divide(480, 40), 7) | divide(n1,n0)|subtract(#0,n2)| | physics |
some of 40 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 20 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 40 ( 1 - x ) + 20 x = 20 x = 1 / 1 answer : e" | a ) 2 / 5 , b ) 2 / 3 , c ) 1 / 3 , d ) 1 / 2 , e ) 1 / 1 | e | divide(subtract(divide(20, const_100), divide(40, const_100)), subtract(divide(20, const_100), divide(40, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
there are two concentric circles with radii 10 and 6 . if the radius of the outer circle is increased by 20 % and the radius of the inner circle decreased by 50 % , by what percent does the area between the circles increase ? | "the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 100 pi . the area of the small circle is 36 pi . the area a 1 between the circles is 64 pi . when the big circle ' s radius increases , the new area is 144 pi . when the small circle ' s radius decreases , the new area is 9 pi . th... | a ) 11 , b ) 111 , c ) 211 , d ) 311 , e ) 411 | b | multiply(divide(subtract(subtract(power(multiply(10, divide(add(const_100, 20), const_100)), const_2), power(multiply(6, divide(subtract(const_100, 50), const_100)), const_2)), subtract(power(10, const_2), power(6, const_2))), subtract(power(10, const_2), power(6, const_2))), const_100) | add(n2,const_100)|power(n0,const_2)|power(n1,const_2)|subtract(const_100,n3)|divide(#0,const_100)|divide(#3,const_100)|subtract(#1,#2)|multiply(n0,#4)|multiply(n1,#5)|power(#7,const_2)|power(#8,const_2)|subtract(#9,#10)|subtract(#11,#6)|divide(#12,#6)|multiply(#13,const_100)| | gain |
a man ' s speed with the current is 15 km / hr and the speed of the current is 3.2 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 3.2 km / hr hence , speed of the man = 15 - 3.2 = 11.8 km / hr man ' s speed against the current = speed of the man - speed of the current = 11.8 - 3.2 = 8.6 km / hr answer is d ." | a ) 10 , b ) 20 , c ) 50 , d ) 8.6 , e ) 40 | d | subtract(subtract(15, 3.2), 3.2) | subtract(n0,n1)|subtract(#0,n1)| | gain |
the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 46,875 , how much profit did the business make ? | "let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = 46875 p = $ 75,000 the answer is b ." | a ) $ 55,000 , b ) $ 75,000 , c ) $ 95,000 , d ) $ 115,000 , e ) $ 125,000 | b | divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2))) | add(const_1,n1)|divide(n0,const_100)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,const_100)|multiply(#1,#3)|multiply(#5,const_2)|add(#1,#6)|divide(#4,#7)| | gain |
x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 24 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 24 days whole work will be done by y in 24 * 5 / 4 = 30 days answer is a" | a ) 30 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | a | multiply(24, inverse(subtract(const_1, divide(8, 40)))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1)|multiply(n2,#2)| | physics |
linda bought 3 notebooks at $ 1.20 each ; a box of pencils at $ 1.50 and a box of pens at $ 1.70 . how much did linda spend ? | linda spent 1.20 ? 3 = $ 3.60 on notebooks the total amount of money that linda spent is equal to 3.60 + 1.50 + 1.70 = $ 6.80 correct answer a | a ) $ 6.80 , b ) $ 8.40 , c ) $ 7.70 , d ) $ 4.70 , e ) $ 3.90 | a | add(add(multiply(3, 1.2), 1.5), 1.7) | multiply(n0,n1)|add(n2,#0)|add(n3,#1) | general |
a park square in shape has a 3 metre wide road inside it running along its sides . the area occupied by the road is 1764 square metres . what is the perimeter along the outer edge of the road ? | solution let the length of the outer edges be x metres . then , length of the inner edge = ( x - 6 ) m . β΄ x 2 - ( x - 6 ) 2 = 1764 βΉ = βΊ x 2 - ( x 2 - 12 x + 36 ) = 1764 βΉ = βΊ 12 x = 1800 βΉ = βΊ x = 150 . β΄ required perimeter = ( 4 x ) m βΉ = βΊ ( 4 x 150 ) m = 600 m . answer c | ['a ) 576 metres', 'b ) 589 metres', 'c ) 600 metres', 'd ) 700 metres', 'e ) none'] | c | multiply(divide(add(1764, power(multiply(3, const_2), const_2)), multiply(const_2, multiply(3, const_2))), const_4) | multiply(n0,const_2)|multiply(#0,const_2)|power(#0,const_2)|add(n1,#2)|divide(#3,#1)|multiply(#4,const_4) | geometry |
two numbers are respectively 20 % and 50 % more than a third number . the percentage that is first of the second is ? | "i ii iii 120 150 100 150 - - - - - - - - - - 120 100 - - - - - - - - - - - ? = > 80 % answer : b" | a ) 45 % , b ) 80 % , c ) 78 % , d ) 94 % , e ) 68 % | b | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 22932 / - in 3 years at the same rate of interest . find the rate percentage ? | "explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 22932 / - - 17640 / - = rs . 5292 / - rate of interest = ( 5292 / 22932 ) Γ ( 100 / 1 ) = > 30 % answer : option e" | a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 30 % | e | multiply(divide(subtract(22932, 17640), 17640), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | general |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 60 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 60 4 n < 53 n < 13 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 13 ) combining the two - 1.5 < n < 13.25 n = 1 to 13 ( 48 integers ) and... | a ) 10 , b ) 15 , c ) 8 , d ) 12 , e ) 20 | b | subtract(floor(divide(subtract(60, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5)| | general |
the average of first 25 natural numbers is ? | "sum of 25 natural no . = 650 / 2 = 325 average = 325 / 25 = 13 answer : b" | a ) 14 , b ) 13 , c ) 15 , d ) 18 , e ) 12 | b | add(25, const_1) | add(n0,const_1)| | general |
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