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 between the number of sheep and the number of horses at the stewar farm is 4 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 4 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 4 * 8 ) = 32 answer : c" | a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 58 | c | multiply(divide(divide(12880, 230), 7), 4) | divide(n3,n2)|divide(#0,n1)|multiply(n0,#1)| | other |
a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 154 . find the unknown rate of two blankets ? | "10 * 154 = 1540 3 * 100 + 5 * 150 = 1050 1540 – 1050 = 490 answer : c" | a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | c | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 15 % greater than any other employee , what is the lowest possible salary that any one employee earned ? | "employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.15 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.15 x * 20 ) + x = 1 , 050,000 solving for x = $ 43,750 answer d" | a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 43,750 , e ) $ 60,000 | d | add(divide(divide(divide(multiply(add(divide(15, const_100), 1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(21, 1), add(divide(15, const_100), 1)), 1)), add(divide(15, const_100), 1)), const_100), add(multiply(const_100, const_2), const_3)) | add(const_1000,const_60)|divide(n3,const_100)|multiply(const_100,const_2)|subtract(n2,n0)|add(#2,const_3)|add(#1,n0)|subtract(#0,const_10)|multiply(#6,const_1000)|multiply(#5,#3)|add(n0,#8)|multiply(#5,#7)|divide(#10,#9)|divide(#11,#5)|divide(#12,const_100)|add(#4,#13)| | general |
in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is married ? | "let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 30 is the total number of students who are married and out of that 20 are males then t... | a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | b | divide(const_10, 30) | divide(const_10,n1)| | gain |
the area of a rectangle is 63 sq m . the width is two meters shorter than the length . what is the width ? | a = l x w w = l - 2 l = w + 2 a = ( w + 2 ) x w a = w ^ 2 + 2 x w 63 = w ^ 2 + 2 w 0 = w ^ 2 + 2 w - 63 0 = ( w + 9 ) ( w - 7 ) w = - 9 and w = 7 , width can not be negative so w = 7 answer is b | ['a ) 9', 'b ) 7', 'c ) - 9', 'd ) 11', 'e ) 6'] | b | divide(subtract(sqrt(add(multiply(63, const_4), power(const_2, const_2))), const_2), const_2) | multiply(n0,const_4)|power(const_2,const_2)|add(#0,#1)|sqrt(#2)|subtract(#3,const_2)|divide(#4,const_2) | geometry |
when the positive integer x is divided by 9 , the remainder is 5 . what is the remainder when 8 x is divided by 9 ? | "i tried plugging in numbers x = 9 q + 5 x = 14 8 x = 112 8 x / 9 = 9 * 12 + 4 remainder is 4 . answer is d ." | a ) 0 , b ) 1 , c ) 3 , d ) 4 , e ) 6 | d | reminder(multiply(5, 8), 9) | multiply(n1,n2)|reminder(#0,n0)| | general |
a man can row upstream at 25 kmph and downstream at 65 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 65 m = ( 65 + 25 ) / 2 = 45 answer : a" | a ) 45 , b ) 86 , c ) 30 , d ) 78 , e ) 38 | a | divide(add(25, 65), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a train crosses a bridge of length 150 m in 7.5 seconds and a lamp post on the bridge in 2.5 seconds . what is the length of the train in metres ? | let length of train = l case - 1 : distance = 150 + l ( while crossing the bridge ) time = 7.5 seconds i . e . speed = distance / time = ( 150 + l ) / 7.5 case - 2 : distance = l ( while passing the lamp post ) time = 2.5 seconds i . e . speed = distance / time = ( l ) / 2.5 but since speed has to be same in both cases... | a ) 37.5 m , b ) 75 m , c ) 25 m , d ) 80 m , e ) 30 m | b | multiply(divide(150, subtract(7.5, 2.5)), 2.5) | subtract(n1,n2)|divide(n0,#0)|multiply(n2,#1) | physics |
at a special sale , 12 tickets can be purchased for the price of 3 tickets . if 12 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 12 tickets ? | "let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 12 tickets cost 1200 12 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 900 , % of 5 tickets = ( 900 / 1200 ) * 100 = 75 % answer : e" | a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 50 % , e ) 75 % | e | divide(multiply(subtract(multiply(12, 12), multiply(3, 12)), const_100), multiply(12, 12)) | multiply(n0,n0)|multiply(n0,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#0)| | gain |
if x > 0 , x / 40 + x / 20 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 40 + 100 / 20 = 2.5 + 5 = 7.5 7.5 is 75 % of 100 = e" | a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | e | multiply(add(divide(const_1, 40), divide(const_1, 20)), const_100) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|multiply(#2,const_100)| | general |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.6 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is : | "let the ratio be k : 1 . then , k * 16.6 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.6 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 1 / 2 required ratio = 1 / 1 : 1 = 1 : 2 . answer : a" | a ) 1 : 2 , b ) 2 : 3 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 9 | a | divide(subtract(15.8, 15.4), subtract(16.6, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
solution a is 20 % salt and solution b is 60 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 60 - 50 ) / ( 50 -... | a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 1 : 3 | e | divide(divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))), subtract(50, divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))))) | divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)| | other |
if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 8 m of the same rod ? | explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 8 : : 42.75 : x = > 11.25 x x = 8 x 42.75 = > x = ( 8 x 42.75 ) / 11.25 = > x = 30.4 answer : d | a ) 22.8 kg , b ) 25.6 kg , c ) 28 kg , d ) 30.4 kg , e ) none of these | d | divide(multiply(8, 42.75), 11.25) | multiply(n1,n2)|divide(#0,n0) | physics |
the radius of the wheel of a bus is 250 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is | "radius of the wheel of bus = 250 cm . then , circumference of wheel = 2 ï € r = 500 ï € = 1571.43440 cm distance covered by bus in 1 minute = 66 â „ 60 ã — 1000 ã — 100 cms distance covered by one revolution of wheel = circumference of wheel = 1571.45 cm â ˆ ´ revolutions per minute = 6600000 / 60 ã — 1571.43 = 70 a... | a ) 70 , b ) 125 , c ) 300 , d ) 500 , e ) none of these | a | divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 250), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2))))) | multiply(n1,const_1000)|multiply(const_1,const_60)|multiply(const_3,const_4)|multiply(const_2,const_4)|multiply(n0,const_2)|add(#2,const_2)|add(#3,const_2)|multiply(#0,const_100)|divide(#7,#1)|power(#6,const_2)|divide(#5,#9)|add(#10,const_3)|multiply(#11,#4)|divide(#8,#12)| | physics |
1000 men have provisions for 21 days . if 800 more men join them , for how many days will the provisions last now ? | "1000 * 21 = 1800 * x x = 11.6 answer : d" | a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 11.6 , e ) 12.1 | d | divide(multiply(21, 1000), add(1000, 800)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 40 square feet of the wall , and no tiles ov... | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 40 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ... | a ) 160 , b ) 220 , c ) 360 , d ) 440 , e ) 560 | b | add(40, multiply(divide(multiply(40, 3), const_2), 3)) | multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)| | geometry |
how many positive integers less than 10,000 are such that the product of their digits is 210 ? | "210 = 2 x 5 x 3 x 7 = 5 x 6 x 7 x 1 = 5 x 6 x 7 those are the only sets of digits we can use to for the numbers ( any other combination of factors will have two digit factors ) . numbers using 2,5 , 3,7 = 4 ! numbers using 5,6 , 7,1 = 4 ! numbers using 5 , 6,7 ( 3 - digit numbers ) = 3 ! answer = 24 + 24 + 6 = 54 answ... | a ) 24 , b ) 30 , c ) 48 , d ) 54 , e ) 72 | d | divide(factorial(subtract(add(const_4, 210), const_1)), multiply(factorial(210), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
find the last term of a g . p whose first term is 9 and common ratio is ( 1 / 3 ) if the sum of the terms of the g . p is ( 40 / 3 ) | sum of the g . p . = ( first term - r * last term ) / 1 – r 40 / 3 = 9 – 1 / 3 ( last term ) / 2 / 3 last term = ( - 40 / 3 * 2 / 3 + 9 ) * 3 = - 80 / 3 + 27 = 1 / 3 answer : a | a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 2 / 3 , e ) 4 / 5 | a | divide(1, 3) | divide(n1,n2) | general |
the average of marks obtained by 120 boys was 36 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 36 24 x = 4320 - 1800 = > x = 2520 / 24 x = 105 . hence , the number of boys passed = 105 . answer : a" | a ) 105 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | a | divide(subtract(multiply(36, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
shekar scored 76 , 65 , 82 , 62 and 85 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 62 + 85 ) / 5 = 370 / 5 = 74 hence average = 74 answer : a" | a ) 74 , b ) 69 , c ) 75 , d ) 85 , e ) 90 | a | divide(add(add(add(add(76, 65), 82), 62), 85), 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 |
a rectangular plot measuring 10 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | "perimeter of the plot = 2 ( 10 + 50 ) = 120 m no of poles = 120 / 5 = 24 m answer : d" | a ) 46 m , b ) 66 m , c ) 26 m , d ) 24 m , e ) 25 m | d | divide(multiply(add(10, 50), const_2), 5) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)| | physics |
x can finish a work in 21 days . y can finish the same work in 15 days . y worked for 5 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 21 work done by y in 1 day = 1 / 15 work done by y in 5 days = 5 / 15 = 1 / 3 remaining work = 1 – 1 / 3 = 2 / 3 number of days in which x can finish the remaining work = ( 2 / 3 ) / ( 1 / 21 ) = 14 d" | a ) 12 , b ) 13 , c ) 16 , d ) 14 , e ) 18 | d | divide(subtract(const_1, multiply(5, divide(const_1, 15))), divide(const_1, 21)) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)| | physics |
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 16 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 14,000 imported car and ... | "let t be the tier price , p be total price = 14000 per the given conditions : 0.16 t + 0.08 ( p - t ) = 1440 - - - - > t = 8000 . e is the correct answer ." | a ) $ 1600 , b ) $ 6000 , c ) $ 6050 , d ) $ 7050 , e ) $ 4000 | e | divide(subtract(1440, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(8, const_100))), subtract(divide(16, const_100), divide(8, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)| | general |
what is the remainder when the number r = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus r = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | divide(n4,n4)| | general |
p is able to do a piece of work in 20 days and q can do the same work in 10 days . if they can work together for 2 days , what is the fraction of work left ? | "explanation : amount of work p can do in 1 day = 1 / 20 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 20 + 1 / 10 = 3 / 20 amount of work p and q can together do in 2 days = 2 × ( 3 / 20 ) = 3 / 10 fraction of work left = 1 – 3 / 10 = 7 / 10 answer : option c" | a ) 5 / 10 , b ) 9 / 10 , c ) 7 / 10 , d ) 6 / 10 , e ) 4 / 10 | c | subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 20)), 2)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
carina has 70 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | "lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 70 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 70 - > y = 60 / 15 . = 4 c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(subtract(70, multiply(5, 2)), add(10, 5)) | add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0)| | general |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 42 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 42 x 2 ) = 84 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 170 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 35 b ' s weigh... | a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 35 kg | e | subtract(add(multiply(42, const_2), multiply(43, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
john found that the average of 15 numbers is 40 . if 11 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 51 option a" | a ) 51 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 11) | add(n1,n2)| | general |
a boat goes 100 km downstream in 8 hours , and 75 km upstream in 15 hours . the speed of the stream is ? | "100 - - - 10 ds = 12.5 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 12.5 - 5 ) / 2 = 3.75 kmph . answer : e" | a ) 3 , b ) 6.5 , c ) 5.5 , d ) 4 , e ) 3.75 | e | divide(subtract(divide(100, 8), divide(75, 15)), const_2) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2)| | physics |
what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 30 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here s... | a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | a | subtract(1000, subtract(add(divide(1000, 11), divide(1000, 30)), divide(1000, multiply(11, 30)))) | divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)| | other |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 26 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 26 + 21 = 47 t = 12 d = 47 * 12 = 564 answer : b" | a ) 288 , b ) 564 , c ) 877 , d ) 278 , e ) 178 | b | add(multiply(divide(60, subtract(21, 26)), 26), multiply(divide(60, subtract(21, 26)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
what is the smallest no . which must be added to 25268 so as to obtain a sum which is divisible by 11 ? | "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 25268 , difference = ( 2 + 2 + 8 ) - ( 5 + 6 ) = 12 - 11 = 1 . the units digit is at odd place . so we add 10 to the number = > 25268 + 10 = 25278 now , ( 2 + 2 + 8 ) - ( 5 + 7 ) = 12 - 12 = 0 ... | a ) 5 , b ) 10 , c ) 11 , d ) 20 , e ) 30 | b | divide(multiply(25268, 11), 25268) | multiply(n0,n1)|divide(#0,n0)| | general |
a train 455 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 455 * 2 / 35 = 39 sec answer : d" | a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 39 sec , e ) 12 sec | d | multiply(divide(455, multiply(63, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics |
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 15 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 15 % , price = 81 - 12.15 = 68.85 which is 68.85 / 90 * 100 of the sale price on day 1 = ~ 76.5 % answer d" | a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 76.5 % , e ) 72 % | d | add(multiply(divide(divide(15, 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 aeroplane covers a certain distance at a speed of 120 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 240 x 5 ) = 480 km . speed = distance / time speed = 480 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 480 x 3 / 5 ) km / hr = 288 km / hr answer b ) 288 km / hr" | a ) 520 , b ) 288 , c ) 820 , d ) 740 , e ) 720 | b | divide(divide(multiply(120, 4), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)| | physics |
the present age of a father is 3 years more than 3 times the age of his son . 5 years hence , father ' s age will be 10 years more than twice the age of the son . find the present age of the father . | if the present age be x years . father ' s will be ( 3 x + 3 ) years . . so , ( 3 x + 3 + 5 ) = 2 ( x + 3 ) + 10 or , x = 8 so the fathers present age = ( 3 x + 3 ) = ( 3 * 8 + 3 ) years = 27 years . . answer : option c | a ) 33 , b ) 38 , c ) 27 , d ) 40 , e ) 48 | c | subtract(subtract(add(3, multiply(3, subtract(subtract(add(multiply(const_2, 5), 10), 5), 3))), const_10), const_1) | multiply(n2,const_2)|add(n3,#0)|subtract(#1,n2)|subtract(#2,n0)|multiply(n0,#3)|add(n0,#4)|subtract(#5,const_10)|subtract(#6,const_1) | general |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 20 , find the other ? | "other number = 12 * 600 / 20 = 360 answer is b" | a ) 100 , b ) 360 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 20) | multiply(n0,n2)| | physics |
what is the next no . 4 12 84 | "3 ^ 0 + 3 = 4 3 ^ 2 + 3 = 12 3 ^ 4 + 3 = 84 3 ^ 6 + 3 = 732 answer : b" | a ) 632 , b ) 732 , c ) 832 , d ) 850 , e ) 902 | b | add(4, reminder(4, 12)) | reminder(n0,n1)|add(n0,#0)| | general |
x and y are positive integers . when x is divided by 9 , the remainder is 2 , and when x is divided by 7 , the remainder is 4 . when y is divided by 11 , the remainder is 3 , and when y is divided by 13 , the remainder is 12 . what is the least possible value of y - x ? | "when x is divided by 9 , the remainder is 2 : so , the possible values of x are : 2 , 11 , 21 , 29 , etc . when x is divided by 7 , the remainder is 4 : so , the possible values of x are : 4 , 11,18 , . . . stop . since both lists include 11 , the smallest possible value of x is 11 . when y is divided by 11 , the rema... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | subtract(add(multiply(11, const_2), 3), add(2, 9)) | add(n0,n1)|multiply(n4,const_2)|add(n5,#1)|subtract(#2,#0)| | general |
at joes steakhouse the hourly wage for a chef is 22 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 8.50 per hour , how much less than a manager does a chef earn each hour ? | "manager wages per hour = $ 8.50 dishwasher wages per hour = half of manager ' s wages . = 1 / 2 ( $ 8.50 ) = = > $ 4.25 chef wages per hour = 22 % greater than dishwasher wages - - > 22 % of $ 4.25 = ( 22 * ( $ 4.25 ) ) / 100 - - > ( $ 93.5 ) / 100 - - > $ 0.935 therefore , chef wages per hour = $ 4.25 + $ 0.935 = = >... | a ) $ 1.40 , b ) $ 2.40 , c ) $ 3.315 , d ) $ 4.40 , e ) $ 5.40 | c | multiply(subtract(const_1, multiply(divide(add(const_100, 22), const_100), divide(const_1, const_2))), 8.50) | add(n0,const_100)|divide(const_1,const_2)|divide(#0,const_100)|multiply(#2,#1)|subtract(const_1,#3)|multiply(n1,#4)| | general |
if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 15000 sheets of xerox paper ? | 30 sec approach - solve it using approximation 15000 sheet at full price , 5 cent = 750 15000 sheet at max discount price , 4 cent = 600 your ans got to be between these two . ans b it is . | a ) $ 1250 , b ) $ 700 , c ) $ 1350 , d ) $ 900 , e ) $ 1000 | b | multiply(subtract(10000, 2000), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | divide(const_1,n1)|subtract(n4,n2)|subtract(const_1,#0)|multiply(#0,#2)|multiply(#3,#1) | gain |
if the sum of two positive integers is 18 and the difference of their squares is 36 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 18 - - 1 x ^ 2 - y ^ 2 = 36 = > ( x + y ) ( x - y ) = 36 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 10 y = 8 product = 10 * 8 = 80 answer b" | a ) 108 , b ) 80 , c ) 128 , d ) 135 , e ) 143 | b | multiply(divide(subtract(18, divide(36, 18)), divide(36, 18)), add(divide(subtract(18, divide(36, 18)), divide(36, 18)), divide(36, 18))) | divide(n1,n0)|subtract(n0,#0)|divide(#1,#0)|add(#2,#0)|multiply(#3,#2)| | general |
a certain number when divided by 80 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 80 + 25 = 105 / 15 = 7 ( remainder ) d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | d | reminder(25, 15) | reminder(n1,n2)| | general |
a rectangular circuit board is designed to have a width of w inches , a length of l inches , a perimeter of p inches , and an area of c square inches . which of the following equations must be true ? | p = 2 ( l + w ) - - - - - - - - - - - - - - - - - 1 ) c = lw - - - - - - - - - - - - - - - - - - - - - - - - 2 ) option a is not possible why ? ? because all the terms are positive . lets try option b , put value of p and a from 1 and 2 we have , 2 w ^ 2 - 2 ( l + w ) w + 2 ( lw ) 2 w ^ 2 - 2 lw - 2 w ^ 2 + 2 lw = 0 . ... | ['a ) 2 w ^ 2 + pw + 2 c = 0', 'b ) 2 w ^ 2 − pw + 2 c = 0', 'c ) 2 w ^ 2 − pw − 2 c = 0', 'd ) w ^ 2 + pw + c = 0', 'e ) w ^ 2 − pw + 2 c = 0'] | b | rhombus_perimeter(const_4) | rhombus_perimeter(const_4) | geometry |
a candidate appearing for an examination has to secure 35 % marks to pass paper i . but he secured only 42 marks and failed by 23 marks . what is the maximum mark for paper i ? | "he secured 42 marks nd fail by 23 marks so total marks for pass the examinatn = 65 let toal marks x x * 35 / 100 = 65 x = 186 answer : c" | a ) 110 , b ) 120 , c ) 186 , d ) 140 , e ) 150 | c | divide(add(42, 23), divide(35, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
how many odd integers from 1 to 50 ( both inclusive ) have odd number of factors ? | "integers having odd number of factors will be perfect squares . odd numbers will have odd perfect squares . thus , the possible values for the perfect squares are : 1,9 , 25,49 and the corresponding integers are 1,3 , 5,7 ( more than 3 ) . thus b is the correct answer ." | a ) 13 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(subtract(divide(divide(50, const_2), const_2), const_10), const_10) | divide(n1,const_2)|divide(#0,const_2)|subtract(#1,const_10)|subtract(#2,const_10)| | other |
a rectangular plot measuring 90 metres by 40 metres is to be enclosed by wire fencing . if the poles of the fence are kept 5 metres apart , how many poles will be needed ? | "solution perimeter of the plot = 2 ( 90 + 40 ) = 260 m . ∴ number of poles = [ 260 / 5 ] = 52 m answer d" | a ) 55 , b ) 56 , c ) 57 , d ) 52 , e ) none of these | d | divide(rectangle_perimeter(90, 40), 5) | rectangle_perimeter(n0,n1)|divide(#0,n2)| | physics |
find the value for x from below equation : x / 3 = - 2 ? | 1 . multiply both sides by 3 : x * 3 / 3 = - 2 / 3 2 . simplify both sides : x = - 6 a | a ) - 6 , b ) 1 , c ) - 2 , d ) - 3 , e ) 4 | a | multiply(subtract(2, const_4), const_3) | subtract(n1,const_4)|multiply(#0,const_3) | general |
dan can do a job alone in 12 hours . annie , working alone , can do the same job in just 9 hours . if dan works alone for 4 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 12 of the job per hour . in 4 hours , dan completes 4 ( 1 / 12 ) = 1 / 3 of the job . annie can complete 1 / 9 of the job per hour . to complete the job , annie will take 2 / 3 / 1 / 9 = 6 hours . the answer is c ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | multiply(subtract(const_1, divide(4, 12)), 9) | divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)| | physics |
the measurement of a rectangular box with lid is 25 cmx 18 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 18 , r = 9 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 9 * 9 * 9 = 972 π cm 3 answer : d" | a ) 288 , b ) 48 , c ) 72 , d ) 972 , e ) 964 | d | multiply(divide(const_4, 3), power(3, 3)) | divide(const_4,n3)|power(n3,n3)|multiply(#0,#1)| | geometry |
the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 5 / 9 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ? | "let , a ≡ event of getting hardware contract b ≡ event of getting software contract ab ≡ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 5 / 9 ) = 4 / 9 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b )... | a ) 11 / 30 , b ) 31 / 60 , c ) 41 / 80 , d ) 51 / 120 , e ) 71 / 180 | e | subtract(add(divide(5, 4), subtract(const_1, divide(const_3.0, 4))), divide(9, 5)) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|subtract(const_1,#1)|add(#0,#3)|subtract(#4,#2)| | other |
working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 3 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 3 x boxes ? | "rate = work / time given rate of machine a = x / 10 min machine b produces 3 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 6 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combin... | a ) 3 minutes , b ) 4 minutes , c ) 4.2 minutes , d ) 6 minutes , e ) 12 minutes | c | divide(multiply(3, 10), add(speed(10, 10), speed(multiply(3, 10), 5))) | multiply(n0,n3)|multiply(n0,n1)|speed(n0,n0)|speed(#1,n2)|add(#2,#3)|divide(#0,#4)| | physics |
a student chose a number , multiplied it by 4 , then subtracted 138 from the result and got 102 . what was the number he chose ? | solution : let xx be the number he chose , then 4 ⋅ x − 138 = 102 4 x = 240 x = 60 answer a | a ) 60 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 4) | add(n1,n2)|divide(#0,n0) | general |
what is the speed of the stream if a canoe rows upstream at 4 km / hr and downstream at 12 km / hr | "sol . speed of stream = 1 / 2 ( 12 - 4 ) kmph = 4 kmph . answer b" | a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph | b | divide(subtract(12, 4), const_2) | subtract(n1,n0)|divide(#0,const_2)| | physics |
when asked what the time is , a person answered that the amount of time left is 3 / 5 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 3 / 5 x ⇒ x = 15 time is 3 pm answer : c" | a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 8 pm , e ) 6 pm | c | subtract(multiply(3, const_12), divide(multiply(3, const_12), add(divide(3, 5), const_1))) | divide(n0,n1)|multiply(const_12,n0)|add(#0,const_1)|divide(#1,#2)|subtract(#1,#3)| | general |
the side of a cube is 15 m , find it ' s surface area ? | "surface area = 6 a ( power ) 2 sq . units 6 a ( power ) 2 = 6 × 225 = 1350 m ( power ) 2 answer is a ." | a ) 1350 , b ) 1750 , c ) 1150 , d ) 1450 , e ) 1570 | a | rectangle_area(add(multiply(const_4, const_10), const_3), 15) | multiply(const_10,const_4)|add(#0,const_3)|rectangle_area(n0,#1)| | geometry |
a water tank is three - fifths full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how long will it take to empty or fill the tank completely ? | "the combined rate of filling / emptying the tank = 1 / 10 - 1 / 6 = - 1 / 15 since the rate is negative , the tank will be emptied . a full tank would take 15 minutes to empty . since the tank is only three - fifths full , the time is ( 3 / 5 ) * 15 = 9 minutes the answer is d ." | a ) 6 min , b ) 8 min , c ) 7 min , d ) 9 min , e ) 1 min | d | divide(divide(const_4, add(const_2, const_3)), subtract(inverse(6), inverse(10))) | add(const_2,const_3)|inverse(n1)|inverse(n0)|divide(const_4,#0)|subtract(#1,#2)|divide(#3,#4)| | physics |
two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 4 litres from vessel p is mixed with 8 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 4 = 5 / 2 litres quantity of alcohol in vessel q = 87.5 / 100 * 8 = 7 / 1 litres quantity of alcohol in the mixture formed = 5 / 2 + 7 / 1 = 19 / 2 = 9.50 litres as 12 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 9.50 : 2.50 = 38 : 10 .... | a ) 19 : 1 , b ) 19 : 4 , c ) 19 : 8 , d ) 38 : 10 , e ) 38 : 2 | d | divide(add(divide(multiply(62.5, 4), const_100), divide(multiply(87.5, 8), const_100)), add(subtract(4, divide(multiply(62.5, 4), const_100)), subtract(8, divide(multiply(87.5, 8), const_100)))) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n2,#2)|subtract(n3,#3)|add(#5,#6)|divide(#4,#7)| | other |
( 228 % of 1265 ) ÷ 6 = ? | "explanation : ? = ( 228 x 1265 / 100 ) ÷ 6 = 288420 / 600 = 481 answer : option c" | a ) a ) 125 , b ) b ) 175 , c ) c ) 481 , d ) d ) 375 , e ) e ) 524 | c | divide(multiply(divide(228, const_100), 1265), 6) | divide(n0,const_100)|multiply(n1,#0)|divide(#1,n2)| | general |
the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked from one cow ? | "let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 − x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % 15 % or 10 % so 8100 + 15100 ⋅ x + 10100 ⋅ ( 8100 − x ) = 9100... | a ) 2178 lt , b ) 3697 lt , c ) 6583 lt , d ) 4370 lt , e ) 5548 lt | d | subtract(8100, divide(subtract(9100, multiply(add(const_1, divide(10, const_100)), 8100)), subtract(add(const_1, divide(15, const_100)), add(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n0,#2)|subtract(#3,#2)|subtract(n3,#4)|divide(#6,#5)|subtract(n0,#7)| | general |
the average height of 40 girls out of a class of 50 is 169 cm . and that of the remaining girls is 167 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 40 × 169 + 10 × 167 / 50 ) = 168.6 cms answer c" | a ) 138.9 cms , b ) 149.2 cms , c ) 168.6 cms , d ) 159.2 cms , e ) 142.5 cms | c | divide(add(multiply(169, 40), multiply(167, const_10)), 50) | multiply(n0,n2)|multiply(n3,const_10)|add(#0,#1)|divide(#2,n1)| | general |
equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 7 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 7 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 7 + 1 / 7 = 2 / 7 full . answer : d ." | a ) 1 / 7 , b ) 7 / 12 , c ) 1 / 2 , d ) 2 / 7 , e ) 2 / 3 | d | divide(const_2, 7) | divide(const_2,n1)| | general |
given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] – 10 = f ( x – 2 ) ? | "2 ( 3 x - 5 ) - 10 = 3 ( x - 2 ) - 5 3 x = 9 x = 3 the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(add(multiply(2, 5), 10), add(multiply(3, 2), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | multiply(n1,n2)|multiply(n4,n0)|multiply(n0,n2)|multiply(n0,const_1)|add(n3,#0)|add(n1,#1)|subtract(#2,#3)|subtract(#4,#5)|divide(#7,#6)| | general |
the population of a town is 10000 . it increases annually at the rate of 40 % p . a . what will be its population after 2 years ? | "formula : 10000 × 140 / 100 × 140 / 100 = 19600 answer : c" | a ) 14000 , b ) 14400 , c ) 19600 , d ) 14600 , e ) 14700 | c | add(10000, multiply(divide(multiply(10000, 40), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers e are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | add(n0,n0)| | general |
the unit digit in the product ( 784 x 618 x 917 x 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 x 8 x 7 x 3 ) = ( 672 ) = 2 d" | a ) 5 , b ) 9 , c ) 16 , d ) 2 , e ) 42 | d | subtract(multiply(multiply(multiply(784, 618), 917), 463), subtract(multiply(multiply(multiply(784, 618), 917), 463), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
the diameter of a cylindrical tin is 8 cm and height is 5 cm . find the volume of the cylinder ? | "r = 4 h = 5 π * 4 * 4 * 5 = 80 π cc answer : d" | a ) 33 , b ) 45 , c ) 66 , d ) 80 , e ) 21 | d | divide(volume_cylinder(divide(8, const_2), 5), const_pi) | divide(n0,const_2)|volume_cylinder(#0,n1)|divide(#1,const_pi)| | geometry |
whats the reminder when 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 is divided by 2 | "a number ending in a 0 is divisible by 2 . given the obscene number , you should immediately be convinced that you will need to focus on a very small part of it . 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 = 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,540 + 7 the first number is divisible by 16 ... | a ) 1 , b ) 7 . , c ) 2 . , d ) 3 , e ) 9 | a | add(multiply(const_4, const_2), reminder(add(add(multiply(subtract(const_10, const_1), const_100), multiply(multiply(add(const_3, const_2), const_100), const_10)), multiply(add(const_12, add(const_3, const_2)), add(const_3, const_2))), 2)) | add(const_2,const_3)|multiply(const_2,const_4)|subtract(const_10,const_1)|add(#0,const_12)|multiply(#2,const_100)|multiply(#0,const_100)|multiply(#5,const_10)|multiply(#3,#0)|add(#4,#6)|add(#8,#7)|reminder(#9,n6)|add(#1,#10)| | 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 , 26 have long fur , 30 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | "no of dogs = 45 long fur = 26 brown = 30 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 26 + 30 - both both = 19 answer b" | a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6 | b | subtract(add(26, 30), subtract(45, 8)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)| | other |
a person bought 114 glass bowls at a rate of rs . 13 per bowl . he sold 108 of them at rs . 17 and the remaining broke . what is the percentage gain for a ? | "cp = 114 * 13 = 1482 and sp = 108 * 17 = 1836 gain % = 100 * ( 1836 - 1482 ) / 1482 = 5900 / 247 answer : c" | a ) 40 , b ) 3000 / 11 , c ) 5900 / 247 , d ) 2790 / 11 , e ) 2709 / 8 | c | multiply(divide(subtract(multiply(108, 17), multiply(114, 13)), multiply(114, 13)), const_100) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 147 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16... | a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | d | divide(147, add(16, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
the average age of 38 students in a group is 14 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ? | "sol . age of the teacher = ( 39 ã — 15 â € “ 38 ã — 14 ) years = 53 years . answer c" | a ) 31 , b ) 36 , c ) 53 , d ) 58 , e ) none | c | add(38, const_1) | add(n0,const_1)| | general |
the area of a square is 4761 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4761 a = 69 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 45 : 138 = 15 : 46 answer : e" | a ) 5 : 28 , b ) 5 : 19 , c ) 15 : 12 , d ) 5 : 13 , e ) 15 : 46 | e | divide(subtract(sqrt(4761), 24), multiply(sqrt(4761), const_2)) | sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)| | geometry |
a positive integer n is a perfect number provided that the sum of all the positive factors of n , including 1 and n , is equal to 2 n . what is the sum of the reciprocals of all the positive factors of the perfect number 28 ? | soln : 28 = 1 * 28 2 * 14 4 * 7 sum of reciprocals = 1 + 1 / 28 + 1 / 2 + 1 / 14 + 1 / 4 + 1 / 7 = 56 / 28 = 2 answer : c | a ) 1 / 4 , b ) 56 / 27 , c ) 2 , d ) 3 , e ) 4 | c | multiply(multiply(const_2, divide(const_1, 28)), 28) | divide(const_1,n2)|multiply(#0,const_2)|multiply(n2,#1) | general |
two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 46 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "solution relative speed = ( 46 - 40 ) kmph = 6 kmph = ( 6 x 5 / 18 ) m / sec = ( 30 / 18 ) m / sec time taken = ( 350 x 18 / 30 ) sec = 210 sec . answer b" | a ) 72 sec , b ) 210 sec , c ) 192 sec , d ) 252 sec , e ) none | b | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(46, 40))) | add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)| | physics |
an employee ’ s annual salary was increased 50 % . if her old annual salary equals $ 80,000 , what was the new salary ? | "old annual salary = $ 80,000 salary increase = 50 % . original salary = $ 80,000 * 50 / 100 = $ 40,000 new salary = $ 80,000 + $ 40,000 = $ 120,000 hence b ." | a ) $ 128,000 , b ) $ 120,000 , c ) $ 110,000 , d ) $ 139,000 , e ) $ 125,000 | b | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | multiply(const_0_25,const_100)|subtract(const_100,const_10)|multiply(#1,const_1000)|multiply(#0,const_1000)|subtract(#2,#3)|divide(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general |
a can do a job in 15 days and b in 20 days . if they work on it together for 8 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 15 + 1 / 20 = 7 / 60 a + b 8 days work = 7 / 60 * 8 = 14 / 15 remaining work = 1 - 14 / 15 = 1 / 15 answer is e" | a ) 2 / 15 , b ) 8 / 15 , c ) 3 / 11 , d ) 1 / 12 , e ) 1 / 15 | e | subtract(const_1, multiply(8, add(divide(const_1, 15), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 33 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 36 in what ratio should he mix the water and milk ? | cost needed to net a 20 % profit : ( 36 - x ) / x = . 2 x = 30 actual cost : 33 solution ( x = liters of water needed to be added to the 1 liter of milk ) : 33 / ( 1 + x ) = 30 x = 1 / 10 so to get the cost down to 30 milk : water 1 : 1 / 10 or ( 10 / 10 ) : ( 1 / 10 ) answer : b | a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | b | divide(const_1, divide(20, const_2)) | divide(n1,const_2)|divide(const_1,#0) | gain |
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 980 gm . for a kg . his gain is … % . | "his percentage gain is 100 * 20 / 980 as he is gaining 20 units for his purchase of 980 units . so 2.04 % . . answer : a" | a ) 2.04 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(980, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
if 20 % of a number = 400 , then 120 % of that number will be ? | "let the number x . then , 20 % of x = 400 x = ( 400 * 100 ) / 20 = 2000 120 % of x = ( 120 / 100 * 2000 ) = 2400 . answer : d" | a ) 20 , b ) 120 , c ) 360 , d ) 2400 , e ) 2820 | d | divide(multiply(120, 400), 20) | multiply(n1,n2)|divide(#0,n0)| | gain |
at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 17 employees are trained to work in the family buffet , 18 are trained to work in the dinin... | "39 = 17 + 18 + 12 - 4 - 2 x 2 x = 17 + 18 + 12 - 4 - 39 = 43 - 39 = 4 x = 2 a" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(subtract(subtract(add(add(17, 18), 12), 4), 39), 2) | add(n4,n5)|add(n6,#0)|subtract(#1,n7)|subtract(#2,n0)|divide(#3,n8)| | physics |
a cistern is filled by pipe a in 20 hours and the full cistern can be leaked out by an exhaust pipe b in 25 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 20 - 1 / 25 ) hrs = 1 / 100 = 100 hrs answer : e" | a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 100 hrs | e | divide(const_1, subtract(divide(const_1, 20), divide(const_1, 25))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
the population of a town increased from 1 , 75,000 to 2 , 27,500 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 227500 - 175000 ) = 52500 . increase % = ( 52500 / 175000 ã — 100 ) % = 30 % . required average = ( 30 / 10 ) % = 3 % . answer c" | a ) 4.37 % , b ) 5 % , c ) 3 % , d ) 8.75 % , e ) none | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
- 54 x 29 + 100 = ? | given exp . = - 54 x ( 30 - 1 ) + 100 = - ( 54 x 30 ) + 54 + 100 = - 1620 + 154 = - 1466 answer is a | a ) - 1466 , b ) 2801 , c ) - 2801 , d ) - 2071 , e ) none of them | a | multiply(subtract(const_1, const_2), subtract(multiply(54, 29), 100)) | multiply(n0,n1)|subtract(const_1,const_2)|subtract(#0,n2)|multiply(#1,#2) | general |
if 50 % of 100 is greater than 20 % of a number by 47 , what is the number ? | "explanation : 50 / 100 * 100 - 20 / 100 * x = 47 50 - 20 / 100 * x = 47 3 = 20 / 100 * x 3 * 100 / 20 = x 15 = x answer : option c" | a ) 60 , b ) 30 , c ) 15 , d ) 75 , e ) 100 | c | divide(subtract(multiply(divide(50, const_100), 100), 47), divide(20, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)| | gain |
running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 14 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 14 ã — 4 ã — 270 â ‡ ” x = 14 x 4 x 270 / 6 = 2520 . answer a" | a ) 2520 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none of these | a | multiply(multiply(divide(270, 6), 4), 14) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 6 gallons of water and 12 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 12 gallons salt is adde... | a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 90 | e | divide(subtract(multiply(12, const_2), 6), subtract(subtract(subtract(const_1, divide(20, const_100)), multiply(subtract(const_1, divide(20, const_100)), divide(const_1, const_4))), multiply(const_2, divide(20, const_100)))) | divide(n0,const_100)|divide(const_1,const_4)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#2,n1)|subtract(const_1,#0)|multiply(#1,#5)|subtract(#5,#6)|subtract(#7,#3)|divide(#4,#8) | general |
there are 250 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 250 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 250 = 2 / 5 t , or 2 / 5 ( t - m ) = 250 , from here we get that t - m = 625 , that would be ... | a ) 325 , b ) 425 , c ) 625 , d ) 700 , e ) none of these | c | divide(250, divide(2, 5)) | divide(n1,n2)|divide(n0,#0)| | general |
income and expenditure of a person are in the ratio 15 : 8 . if the income of the person is rs . 15000 , then find his savings ? | "let the income and the expenditure of the person be rs . 15 x and rs . 8 x respectively . income , 15 x = 15000 = > x = 1000 savings = income - expenditure = 15 x - 8 x = 7 x = 7 ( 1000 ) so , savings = rs . 7000 . answer : b" | a ) 6999 , b ) 7000 , c ) 7001 , d ) 7002 , e ) 7003 | b | subtract(15000, multiply(divide(8, 15), 15000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
ashok secured average of 78 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 78 therefore total marks in 6 subjects = 78 * 6 = 468 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 468 – 370 = 98 answer d" | a ) 66 , b ) 74 , c ) 78 , d ) 98 , e ) none of these | d | subtract(multiply(78, 6), multiply(74, 5)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 4 km on foot . his total journey is | "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 4 = x 5 x + 24 = 6 x x = 24 km answer : option d" | a ) 16 km , b ) 10 km , c ) 12 km , d ) 24 km , e ) 25 km | d | multiply(3, 4) | multiply(n3,n4)| | general |
a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company ’ s latest order required a truck with at least 384,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | number of boxes = total volume / volume of one box = 384,000 / ( 8 * 10 * 12 ) = 400 one box costs 7.50 , so 400 box will cost = 400 * 7.5 = 3000 a is the answer | a ) $ 3,000 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250 | a | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, const_10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.5), const_1000) | multiply(const_100,const_3)|multiply(n0,const_10)|multiply(n0,n1)|add(#0,#1)|multiply(n2,#2)|add(#3,const_4)|multiply(#5,const_1000)|divide(#6,#4)|multiply(n3,#7)|divide(#8,const_1000) | general |
if p / q = 3 / 5 , then 2 p + q = ? | "let p = 3 , q = 5 then 2 * 3 + 5 = 11 so 2 p + q = 11 . answer : b" | a ) 12 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | b | add(multiply(3, 2), 5) | multiply(n0,n2)|add(n1,#0)| | general |
fred and sam are standing 100 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 5 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ? | "relative distance = 100 miles relative speed = 5 + 5 = 10 miles per hour time taken = 100 / 10 = 15 hours distance travelled by sam = 15 * 5 = 75 miles = e" | a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 75 | e | multiply(5, divide(100, add(5, 5))) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | physics |
the difference between a two digit number and the number obtained by interchanging the digits is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | sol . since the number is greater than the number obtained on reversing the digits , so the ten ' s is greater than the unit ' s digit . let the ten ' s and units digit be 2 x and x respectively . then , ( 10 × 2 x + x ) - ( 10 x + 2 x ) = 36 ⇔ 9 x = 36 ⇔ x = 4 . ∴ required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x ... | a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 18 | b | multiply(subtract(const_3, 1), divide(36, subtract(add(multiply(const_10, 2), 1), add(const_10, 2)))) | add(n2,const_10)|multiply(n2,const_10)|subtract(const_3,n1)|add(n1,#1)|subtract(#3,#0)|divide(n0,#4)|multiply(#5,#2) | general |
40 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 40 + 5 * 12 / ( 180 / 3 ) = 40 + 5 * 12 / ( 60 ) = 40 + ( 5 * 12 ) / 60 = 40 + 1 = 41 . answer : e" | a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 41 | e | add(40, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general |
on dividing 13787 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 13787 - 14 ) / 89 = 155 answer ( b )" | a ) 743 , b ) 155 , c ) 852 , d ) 741 , e ) 785 | b | divide(subtract(13787, 14), 89) | subtract(n0,n2)|divide(#0,n1)| | general |
in a mixture 60 litres , the ra ɵ o of milk and water 2 : 1 . if the this ra ɵ o is to be 1 : 2 , then the quanity of water to be further added is | explanation : quantity of milk = 60 * ( 2 / 3 ) = 40 liters quantity of water = 60 - 40 = 20 liters answer : d | a ) 20 liters , b ) 30 liters , c ) 50 liters , d ) 60 liters , e ) none of these | d | subtract(multiply(divide(multiply(60, 2), add(2, 1)), 2), subtract(60, divide(multiply(60, 2), add(2, 1)))) | add(n1,n2)|multiply(n0,n1)|divide(#1,#0)|multiply(n1,#2)|subtract(n0,#2)|subtract(#3,#4) | general |
a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : b" | a ) 17 inches , b ) 20 inches , c ) 15 inches , d ) 18 inches , e ) 19 inches | b | divide(add(multiply(6, const_12), 8), 4) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general |
ramu bought an old car for rs . 42000 . he spent rs . 12000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 12000 = rs . 54000 and sp = rs . 64900 profit ( % ) = ( 64900 - 54000 ) / 54000 * 100 = 20.18 % answer : c" | a ) 12 % , b ) 16 % , c ) 20.18 % , d ) 82 % , e ) 23 % | c | multiply(divide(subtract(64900, add(42000, 12000)), add(42000, 12000)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain |
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