Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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the product of the squares of two positive integers is 9 . how many pairs of positive integers satisfy this condition ? | "ans : b - 1 pairs ( x Λ 2 ) ( y Λ 2 ) = 9 [ square root both sides ] xy = 3 3 = 1 x 3 , 3 x 1 cancel the repeats this leaves us with exactly 1 options . hence , b" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry | B |
the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n ^ 3 / 3 + c * n ^ 2 + n / 6 , where c is a constant . what is the sum of the first 16 positive perfect squares ? | "first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 16 and c = 1 / 2 into the equation and simplify to get the answer 1496 option e ." | a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 1,496 | e | divide(divide(multiply(multiply(16, add(16, const_1)), add(multiply(16, 2), const_1)), 6), divide(multiply(multiply(16, add(16, const_1)), add(multiply(16, 2), const_1)), 6)) | add(n4,const_1)|multiply(n4,n2)|add(#1,const_1)|multiply(n4,#0)|multiply(#2,#3)|divide(#4,n3)|divide(#5,#5)| | general | E |
the reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 so , hcf = 16 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 16 * 312 / 24 = 208 answer : d" | a ) 126 , b ) 136 , c ) 146 , d ) 208 , e ) 266 | d | divide(multiply(16, 312), 24) | multiply(n1,n3)|divide(#0,n4)| | physics | D |
of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 25 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 25 grades are d ' s . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a β s + # b β s + # c β s + # d β s = total number of students in the class 1 / 5 ( t ) + ΒΌ ( t ) + Β½ ( t ) + 25 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 500 = 20 t 19 t + 500 = 20 t 500 = t there are a total of 500 students in the math class . answer is d ." | a ) 80 , b ) 110 , c ) 160 , d ) 500 , e ) 400 | d | divide(25, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general | D |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 80 feet wide by 150 feet long by 10 feet deep and is currently at 100 % capacity , how long will it take to drain the pool ? | "volume of pool = 80 * 150 * 10 cu . ft , 100 % full = 60 * 100 * 10 * 1 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 80 * 150 * 10 * 1 / 60 min = 2000 min e" | a ) 1400 , b ) 1600 , c ) 1800 , d ) 1500 , e ) 2000 | e | divide(multiply(divide(100, const_100), multiply(multiply(80, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0)| | gain | E |
if 3 : 7 : : x : 21 , then find the value of x | explanation : treat 3 : 7 as 3 / 7 and x : 21 as x / 21 , treat : : as = so we get 3 / 7 = x / 21 = > 7 x = 63 = > x = 9 option e | a ) 7 , b ) 8 , c ) 11 , d ) 6 , e ) 9 | e | divide(add(multiply(7, 3), 7), 21) | multiply(n0,n1)|add(n1,#0)|divide(#1,n2)| | general | E |
a salesman sold twice as much pears in the afternoon than in the morning . if he sold 360 kilograms of pears that day , how many kilograms did he sell in the morning and how many in the afternoon ? | "let x be the number of kilograms he sold in the morning . then in the afternoon he sold 2 xkilograms . so , the total is x + 2 x = 3 x this must be equal to 360 . 3 x = 360 x = 360 / 3 x = 120 therefore , the salesman sold 120 kg in the morning and 2 β
120 = 240 kg in the afternoon ." | a ) 560 , b ) 240 , c ) 329 , d ) 443 , e ) 544 | b | multiply(divide(360, const_3), const_2) | divide(n0,const_3)|multiply(#0,const_2)| | other | B |
if 14 men do a work in 80 days , in how many days will 20 men do it ? | "14 * 80 = 20 * x x = 56 days answer : e" | a ) 18 days , b ) 38 days , c ) 42 days , d ) 48 days , e ) 56 days | e | divide(multiply(14, 80), 20) | multiply(n0,n1)|divide(#0,n2)| | physics | E |
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,700 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.7 so , answer is d . | a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,700 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | multiply(const_100,const_100)|multiply(#0,const_4) | general | D |
a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 sec . the speed of the train is ? | "speed of the train relative to man = 125 / 10 = 25 / 2 m / sec . = 25 / 2 * 18 / 5 = 45 km / hr let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 8 = 45 = > x = 53 km / hr . answer : b" | a ) 37 km / hr , b ) 53 km / hr , c ) 36 km / hr , d ) 26 km / hr , e ) 87 km / hr | b | divide(divide(subtract(125, multiply(multiply(8, const_0_2778), 8)), 8), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics | B |
each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.7 % , what fraction of the products was checked by worker y ? | x : 0.5 % is 0.2 % - points from 0.7 % . y : 0.8 % is 0.1 % - points from 0.7 % . therefore the ratio of products checked by y : x is 2 : 1 . thus , worker y checked 2 / 3 of the products . the answer is a . | a ) 2 / 3 , b ) 5 / 6 , c ) 7 / 8 , d ) 4 / 5 , e ) 5 / 8 | a | divide(subtract(0.7, 0.5), subtract(0.8, 0.5)) | subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1) | general | A |
two employees x and y are paid a total of rs . 638 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 638 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 638 β y [ 12 / 10 + 1 ] = 638 β 22 y / 10 = 638 β 22 y = 6380 β y = 6380 / 22 = 580 / 2 = rs . 290 c" | a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 300 | c | divide(multiply(638, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general | C |
the cricket team of 11 members is 24 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 24 + 27 ) = 9 ( x - 1 ) = > 11 x - 9 x = 42 = > 2 x = 42 = > x = 21 . so , average age of the team is 21 years . a" | a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | a | divide(subtract(add(add(24, 3), 24), subtract(11, 2)), subtract(11, subtract(11, 2))) | add(n1,n2)|subtract(n0,n3)|add(n1,#0)|subtract(n0,#1)|subtract(#2,#1)|divide(#4,#3)| | general | A |
what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 50 % ? | "the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 50 % increase a surface = 6 * ( ( 1.5 a ) ^ 2 ) = 6 * 2.25 * a ^ 2 the increase in the surface area = ( 6 * 2.25 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 2.25 - 1 ) ) / ( 6 * a ^ 2 ) = 2.25 - 1 = 1.25 = 125 % answer : a" | a ) 125 % , b ) 150 % , c ) 100 % , d ) 90 % , e ) 85 % | a | multiply(subtract(power(add(const_1, divide(50, const_100)), const_2), const_1), const_100) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)| | geometry | A |
a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 70 km downstream . | "speed of boat in still water = 13 km / hr speed of the stream = 5 km / hr speed downstream = ( 13 + 5 ) = 18 km / hr time taken to travel 70 km downstream = 70 Γ’ Β β 18 = 3.8 hours answer is b" | a ) 4.25 hr , b ) 3.8 hr , c ) 8.25 hr , d ) 2.25 hr , e ) 2.50 hr | b | divide(70, add(13, 5)) | add(n0,n1)|divide(n2,#0)| | physics | B |
one train is travelling 45 kmph and other is at 10 meters a second . ratio of the speed of the two trains is ? | c 5 : 4 45 * 5 / 18 = 10 25 : 20 = > 5 : 4 | a ) 3 : 3 , b ) 5 : 7 , c ) 5 : 4 , d ) 7 : 2 , e ) 8 : 5 | c | divide(multiply(45, const_0_2778), 10) | multiply(n0,const_0_2778)|divide(#0,n1) | physics | C |
if n is a positive integer , what is the remainder when ( 7 ^ ( 4 n + 2 ) ) ( 6 ^ n ) is divided by 10 ? | this one took me bout 3 1 / 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * 7 = 9 7 : 7 * 9 = 3 pattern repeats every @ 5 . notice every ^ 4 or multiple of 4 is always going to be 1 . this is just for future notice for similar problems . so 7 ^ 4 n + 3 - - - > if n = 1 then its ( ( 7 ^ 7 ) * 6 ) ) / 10 which can say is going to be 3 * 8 - - > 18 / 10 - - > r = 8 now from here if id double check just to make sure . 7 ^ 4 ( 2 ) + 3 * 6 ^ 2 - - - > 7 ^ 11 * 36 or we can just say again 7 ^ 11 * 6 ( b / c we are only interested in the units digit ) . since ^ 12 is going to be 1 that means ^ 11 is going to be 3 ( as taken from our pattern ) so again 3 * 6 = 18 / 10 - - - > r = 4 . c or j in this problem . | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | c | reminder(multiply(multiply(const_3, const_3), 6), 10) | multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4) | general | C |
how many even number in the range between 8 to 100 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 8,10 , 12,14 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 47 . for 6 , 12,18 , . . . 96 using ap formula , we can say 96 = 12 + ( n - 1 ) * 6 or n = 15 . hence , only divisible by 2 but not 3 = 47 - 15 = 32 . hence , answer c" | a ) 15 , b ) 30 , c ) 32 , d ) 33 , e ) 46 | c | subtract(divide(subtract(subtract(100, 8), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)| | general | C |
jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent r ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . r = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % . c" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics | C |
a and b can finish a work in 10 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 10 β 1 / 20 = 1 / 20 = > 20 days answer : a" | a ) 20 days , b ) 48 days , c ) 98 days , d ) 31 days , e ) 22 days | a | inverse(subtract(inverse(10), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics | A |
a pipe can fill a cistern in 55 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 55 minutes 1 / 11 part filled in = 55 * 1 / 11 = 5 minutes answer is a" | a ) 5 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min | a | multiply(55, divide(1, 11)) | divide(n1,n2)|multiply(n0,#0)| | physics | A |
in a certain village , 200 litres of water are required per household per month . at this rate , if there are 10 households in the village , how long ( in months ) will 2000 litres of water last ? | "i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 10 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 10 = 2000 ( m * n ) water available is 2000 ( p ) it will last 1 months ( p / m * n ) answer ( a )" | a ) 1 , b ) 20 , c ) 2 , d ) 200 , e ) 2.5 | a | divide(2000, multiply(200, 10)) | multiply(n0,n1)|divide(n2,#0)| | gain | A |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 120 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 120 sq . ft . so , b = 6 ft . length of fencing = ( l + 2 b ) = ( 20 + 12 ) ft = 32 ft . answer : e" | a ) 34 , b ) 40 , c ) 68 , d ) 88 , e ) 32 | e | add(multiply(divide(120, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry | E |
the ages of two persons differ by 16 years . 6 years ago , the elder one was 3 times as old as the younger one . what are their present ages of the elder person | "let ' s take the present age of the elder person = x and the present age of the younger person = x β 16 ( x β 6 ) = 3 ( x - 16 - 6 ) = > x β 6 = 3 x β 66 = > 2 x = 60 = > x = 60 / 2 = 30 answer : d" | a ) 11 , b ) 66 , c ) 28 , d ) 30 , e ) 99 | d | add(divide(add(multiply(3, 6), subtract(16, 6)), const_2), 16) | multiply(n1,n2)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)|add(n0,#3)| | general | D |
a searchlight on top of the watchtower makes 4 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | the searchlight completes one revolution every 15 seconds . the probability that the man ' s area will be lit up is 5 / 15 = 1 / 3 . the probability that he will stay in the dark is 1 - 1 / 3 = 2 / 3 the answer is d . | a ) 5 / 6 , b ) 4 / 5 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 2 | d | subtract(const_1, divide(5, multiply(5, const_3))) | multiply(n1,const_3)|divide(n1,#0)|subtract(const_1,#1) | physics | D |
if one root of the equation 2 x ^ 2 + 3 x β k = 0 is 7 , what is the value of k ? | we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 7 ^ 2 + 3 * 7 - k = 0 k = 98 + 21 = 119 the answer is c | a ) 100 , b ) 110 , c ) 119 , d ) 120 , e ) 112 | c | add(multiply(2, power(7, const_2)), multiply(3, 7)) | multiply(n2,n4)|power(n4,const_2)|multiply(n0,#1)|add(#2,#0) | general | C |
a straight line in the xy - plane has y - intercept of 79 and slope of 6 . on this line the y - coordinate of the point is 14 , then what is the x - coordinate of the point ? | "eq of line = y = mx + c c = 79 m = 6 y = 14 substitute given : x = ( y - c ) / m = ( 14 - 79 ) / 6 = - 65 / 6 = - 10.83 correct option is b" | a ) 11 , b ) - 10.83 , c ) 71 , d ) 20.83 , e ) - 12 | b | divide(subtract(14, 79), 6) | subtract(n2,n0)|divide(#0,n1)| | general | B |
walking 7 / 6 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 7 / 6 = 6 : 7 time ratio = 7 : 6 1 - - - - - - - - 7 4 - - - - - - - - - ? Γ¨ 28 m answer : d" | a ) 11 , b ) 15 , c ) 16 , d ) 28 , e ) 19 | d | multiply(4, 7) | multiply(n0,n2)| | gain | D |
a man swims downstream 28 km and upstream 12 km taking 2 hours each time , what is the speed of the man in still water ? | "28 - - - 2 ds = 14 ? - - - - 1 12 - - - - 2 us = 6 ? - - - - 1 m = ? m = ( 14 + 6 ) / 2 = 10 answer : a" | a ) 10 , b ) 8 , c ) 5 , d ) 2 , e ) 4 | a | divide(add(divide(12, 2), divide(28, 2)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics | A |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 18 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 18 5 x = 1347 x = 269.4 number = 269.4 + 1365 = 1634.4 d" | a ) 1235 , b ) 1346 , c ) 1378 , d ) 1634.4 , e ) 1489 | d | multiply(divide(subtract(1365, 18), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general | D |
a person saved $ 10 in buying an item on sale . if he spent $ 400 for the item , approximately how much percent he saved in the transaction ? | "actual price = 400 + 10 = $ 410 saving = 10 / 410 * 100 = 100 / 41 = 2.4 % approximately answer is d" | a ) 8 % , b ) 10 % , c ) 2 % , d ) 2.4 % , e ) 6 % | d | add(floor(multiply(divide(10, add(10, 400)), const_100)), const_1) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|floor(#2)|add(#3,const_1)| | general | D |
what will be the fraction of 5 % | "explanation : 5 * 1 / 100 = 1 / 20 . option a" | a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | a | divide(circle_area(divide(5, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | gain | A |
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 27000 . a ' s share is ? | x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 27000 = 9000 answer : a | a ) 9000 , b ) 2778 , c ) 6000 , d ) 2889 , e ) 6612 | a | multiply(multiply(const_1, const_12), divide(27000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0) | gain | A |
find the simple interest on $ 500 for 3 years at 10 % per annum ? | "si = ptr / 100 = 500 * 3 * 10 / 100 = $ 150 answer is d" | a ) $ 250 , b ) $ 300 , c ) $ 500 , d ) $ 150 , e ) $ 100 | d | multiply(500, divide(3, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain | D |
the sub - duplicate ratio of 1 : 9 is | "root ( 1 ) : root ( 9 ) = 1 : 3 answer : c" | a ) 4 : 3 , b ) 1 : 2 , c ) 1 : 3 , d ) 1 : 4 , e ) 2 : 3 | c | divide(sqrt(1), sqrt(9)) | sqrt(n0)|sqrt(n1)|divide(#0,#1)| | other | C |
if y is 80 % greater than x , than x is what % less than y ? | "let x = 100 so y = 180 we want percentage change . so , ( 100 - 180 ) / 180 = - 44.44 % = 44 4 / 9 % lesser than y . hence option ( d ) ." | a ) 20 , b ) 25 , c ) 33 1 / 3 , d ) 44 4 / 9 , e ) 80 | d | multiply(divide(80, add(80, const_100)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)| | general | D |
the average of 25 results is 45 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ? | "solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 45 or , x = 708 . answer : option e" | a ) 740 , b ) 750 , c ) 690 , d ) 780 , e ) 708 | e | subtract(subtract(multiply(25, 45), multiply(12, 17)), multiply(12, 14)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general | E |
if x + y = - 10 , and x = 25 / y , what is the value of x ^ 2 + y ^ 2 ? | "x ^ 2 + y ^ 2 should make you think of these formulas : ( x + y ) ( x + y ) = x ^ 2 + y ^ 2 + 2 xy we already know ( x + y ) = - 10 and x * y = 25 ( x + y ) ( x + y ) = ( - 10 ) ( - 10 ) = x ^ 2 + y ^ 2 + 2 * ( 25 ) x ^ 2 + y ^ 2 = 100 - 50 = 50 answer : c" | a ) 55 , b ) 65 , c ) 50 , d ) 75 , e ) 85 | c | subtract(power(10, 2), multiply(2, 25)) | multiply(n1,n2)|power(n0,n2)|subtract(#1,#0)| | general | C |
how many different positive integers exist between 10 ^ 4 and 10 ^ 5 , the sum of whose digits is equal to 2 ? | "10001 10010 10100 11000 20000 total no . = 5 c" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(add(const_4, const_3), const_1) | add(const_3,const_4)|add(#0,const_1)| | general | C |
a drink vendor has 50 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | "the number of liters in each can = hcf of 50 , 144 and 368 = 2 liters . number of cans of maaza = 50 / 2 = 25 number of cans of pepsi = 144 / 2 = 72 number of cans of sprite = 368 / 2 = 184 the total number of cans required = 25 + 72 + 184 = 281 cans . answer : c" | a ) 135 , b ) 237 , c ) 281 , d ) 300 , e ) 380 | c | add(divide(368, gcd(gcd(50, 144), 368)), add(divide(50, gcd(gcd(50, 144), 368)), divide(144, gcd(gcd(50, 144), 368)))) | gcd(n0,n1)|gcd(n2,#0)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)| | general | C |
a positive number x is multiplied by 10 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "sq rt ( 10 x / 3 ) = x = > 10 x / 3 = x ^ 2 = > x = 10 / 3 ans - e" | a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 10 / 3 | e | divide(10, 3) | divide(n0,n1)| | general | E |
p and q started a business investing rs . 45,000 and rs . 10,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 45000 : 10000 = 9 : 2 answer : a" | a ) 9 : 2 , b ) 4 : 5 , c ) 9 : 3 , d ) 17 : 9 , e ) 17 : 4 | a | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain | A |
what is the remainder when 1003 * 1005 * 1007 is divided by 15 | "1003 / 15 = = > remainder = 13 1005 / 15 = = > remainder = 0 1007 / 15 = = > remainder = 2 13 * 0 * 2 = 0 / 15 = = > remainder = 0 answer : d" | a ) 3 , b ) 17 , c ) 14 , d ) 0 , e ) 7 | d | reminder(multiply(1005, 1003), 1007) | multiply(n0,n1)|reminder(#0,n2)| | general | D |
a man can row his boat with the stream at 24 km / h and against the stream in 10 km / h . the man ' s rate is ? | "explanation : ds = 24 us = 10 s = ? s = ( 24 - 10 ) / 2 = 7 kmph answer : c" | a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 4 kmph , e ) 9 kmph | c | divide(subtract(24, 10), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain | C |
5 n + 2 > 12 and 7 n - 19 < 44 ; n must be between which numbers ? | 5 n + 2 > 12 5 n > 10 n > 2 7 n - 19 < 44 7 n < 63 n < 9 so n must be between 2 and 9 2 < n < 9 correct answer c | a ) 1 and 8 , b ) 2 and 6 , c ) 2 and 9 , d ) 2 and 7 , e ) 2 and 9 | c | add(multiply(const_2, const_10), divide(add(44, 19), 7)) | add(n4,n5)|multiply(const_10,const_2)|divide(#0,n3)|add(#2,#1) | general | C |
a company that ships boxes to a total of 15 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter ) | "back - solving is the best way to solve this problem . you basically need 12 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - d" | a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 9 | d | subtract(divide(factorial(subtract(divide(15, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(15, const_2), const_1)) | divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)| | general | D |
at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 10 on candy , how much did john spend at the supermarket ? | "let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t . we are also given that he spent the remaining $ 10 on candy . since we know where all his money was allocated , we can sum these values together and set the sum to t . so we have : ( 1 / 2 ) t + ( 1 / 3 ) t + ( 1 / 10 ) t + 10 = t to get rid of the fractions we can multiply the entire equation by 30 , and we obtain : 15 t + 10 t + 3 t + 300 = 30 t 28 t + 300 = 30 t 300 = 2 t t = 150 john spent $ 90 at the supermarket . answer : e" | a ) $ 60 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 150 | e | divide(10, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2)))) | divide(n0,n5)|divide(n0,n3)|divide(n0,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general | E |
in one alloy there is 12 % chromium while in another alloy it is 8 % . 10 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 10 + 30 = 40 kg alloy is 0.12 * 10 + 0.08 * 30 = 3.6 kg , so the percentage is 3.6 / 40 * 100 = 9 % . answer : c" | a ) 9.4 % , b ) 9.6 % , c ) 9 % , d ) 9.8 % , e ) 10 % | c | multiply(divide(add(divide(multiply(12, 10), const_100), divide(multiply(8, 30), const_100)), add(10, 30)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain | C |
in a 400 m race , a covers the distance in 60 seconds and b in 100 second . in this race a beats b by : | "solution distance covered by b in 40 sec . = ( 400 / 100 x 40 ) m = 160 m . β΄ a beats b by 160 metres . answer d" | a ) 20 m , b ) 125 m , c ) 120 m , d ) 160 m , e ) none of these | d | multiply(divide(400, 100), subtract(100, 60)) | divide(n0,n2)|subtract(n2,n1)|multiply(#0,#1)| | physics | D |
a boy is travelling from his home to school at 12 km / hr and reached 7 min late . next day he traveled at 16 km / hr and reached 8 min early . distance between home and school ? | "let the distance be x t 1 = x / 12 hr t 2 = x / 16 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 12 - x / 16 = 1 / 4 x / 48 = 1 / 4 x = 12 km answer is a" | a ) 12 km , b ) 13 km , c ) 14 km , d ) 15 km , e ) 16 km | a | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 16)) | divide(n1,const_60)|divide(n3,const_60)|divide(const_1,n2)|add(#0,#1)|divide(#3,#2)| | physics | A |
find the highest value of ' a ' so that 365 a 16 is divisible by 8 . | explanation : given , number is divisible by 8 only if ' a 16 ' is divisible by 8 . . : highest value of a is ' 8 ' . answer : option b | a ) 9 , b ) 8 , c ) 0 , d ) 2 , e ) 1 | b | divide(multiply(lcm(const_100, 8), gcd(const_100, 8)), const_100) | gcd(n2,const_100)|lcm(n2,const_100)|multiply(#0,#1)|divide(#2,const_100) | general | B |
how long does a train 165 meters long running at the rate of 36 kmph take to cross a bridge 660 meters in length ? | "explanation : t = ( 660 + 165 ) / 36 * 18 / 5 t = 82.5 answer : option d" | a ) 33 , b ) 72 , c ) 55 , d ) 82.5 , e ) 62 | d | divide(add(165, 660), multiply(36, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | D |
6 persons standing in queue with different age group , after two years their average age will be 43 and seventh person joined with them . hence the current average age has become 45 . find the age of seventh person ? | explanation : let the sum of current ages of 6 persons = x given average age of 6 person after 2 years = 43 = > x + 6 ( 2 ) / 6 = 43 = > x + 12 = 258 = x = 246 let the seventh ' s person age will be y given current average age of 7 persons = 45 [ sum of current 6 person ' s age ( x ) + seventh person ' s age ( y ) ] / 7 = 45 = > 246 + y = 45 ( 7 ) = > y = 315 - 246 = > y = 69 hence seventh person ' s age = 69 hence ( d ) is the correct answer . answer : d | a ) 65 , b ) 67 , c ) 68 , d ) 69 , e ) 50 | d | subtract(multiply(45, add(6, const_1)), multiply(subtract(43, const_2), 6)) | add(n0,const_1)|subtract(n1,const_2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#2,#3) | general | D |
a river 5 m deep and 35 m wide is flowing at the rate of 2 kmph , calculate the amount of water that runs into the sea per minute ? | "rate of water flow - 2 kmph - - 2000 / 60 - - 33.33 m / min depth of river - - 5 m width of river - - 35 m vol of water per min - - 33.33 * 5 * 35 - - - 5832.75 answer a" | a ) 5832.75 , b ) 5839.75 , c ) 5837.75 , d ) 5222.75 , e ) 5835.75 | a | divide(multiply(multiply(5, 35), multiply(2, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics | A |
what is the sum of the 1 st 6 prime numbers greater than 10 ? | required numbers are = ( 11 + 13 + 17 + 19 + 23 + 29 ) = 112 note : 1 is not a prime number answer a | a ) 112 , b ) 111 , c ) 212 , d ) 115 , e ) 113 | a | add(add(add(add(add(add(1, 10), add(add(1, 10), const_2)), add(add(add(1, 10), const_2), const_4)), add(add(add(add(1, 10), const_2), const_4), const_2)), add(add(add(add(add(1, 10), const_2), const_4), const_2), const_4)), add(add(add(add(add(add(1, 10), const_2), const_4), const_2), const_4), 6)) | add(n0,n2)|add(#0,const_2)|add(#0,#1)|add(#1,const_4)|add(#2,#3)|add(#3,const_2)|add(#4,#5)|add(#5,const_4)|add(#6,#7)|add(n1,#7)|add(#8,#9) | general | A |
the ratio of length to width of a rectangular showroom window is 3.3 to 2 . if the width of the window is 8 feet , what is the approximate length of the display in feet ? | explanation : letting l be the length of the window , the proportion for the ratio of the length to the width can be expressed in the following equation : 3.3 / 2 = l / 8 26.4 = 2 l 13.2 = l answer : option c | ['a ) 7', 'b ) 11', 'c ) 13', 'd ) 16', 'e ) 26'] | c | multiply(3.3, divide(8, 2)) | divide(n2,n1)|multiply(n0,#0) | other | C |
there are 4 more women than there are men on a local co - ed softball team . if there are a total of 18 players on the team , what is the ratio of men to women ? | "w = m + 4 w + m = 18 m + 4 + m = 18 2 m = 14 m = 7 w = 11 ratio : 7 : 11 ans : c" | a ) 10 / 16 , b ) 6 / 16 , c ) 7 / 11 , d ) 6 / 10 , e ) 4 / 10 | c | divide(divide(subtract(18, 4), add(const_1, const_1)), add(divide(subtract(18, 4), add(const_1, const_1)), 4)) | add(const_1,const_1)|subtract(n1,n0)|divide(#1,#0)|add(n0,#2)|divide(#2,#3)| | general | C |
an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 8 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | "explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 8 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 2400 . hence , ( r - p ) = ( 2400 - 900 ) = 1500 answer : b" | a ) s 1200 , b ) s 1500 , c ) s 1600 , d ) s 1900 , e ) s 1700 | b | multiply(divide(1500, 5), subtract(8, 3)) | divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)| | other | B |
how many hours are there in 660 minutes ? | "solution : number of minutes in 1 hour = 60 minutes so just divide 660 by 60 660 Γ£ Β· 60 = 11 , so there are 11 hours in 660 minutes option d" | a ) 60 hours , b ) 6 hours , c ) 30 hours , d ) 11 hours , e ) 10 hours | d | divide(660, const_60) | divide(n0,const_60)| | physics | D |
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 35 % profit ? | let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 130 % of rs . 1600 = 130 / 100 * 1600 = rs . 2080 . answer : a | a ) 2080 , b ) 2778 , c ) 2299 , d ) 2778 , e ) 2771 | a | multiply(divide(add(const_100, 35), const_100), divide(add(1920, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3) | gain | A |
12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 4 days , working 15 hours a day , the number of men required ? | that is , 1 work done = 12 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 6 ? ( i . e . no . of men required ) = 12 Γ 8 Γ 10 / 15 Γ 6 = 16 days c ) | a ) 15 days , b ) 12 days , c ) 16 days , d ) 18 days , e ) 20 days | c | divide(multiply(multiply(12, 10), 8), multiply(4, 15)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1) | physics | C |
in a class of students , 2 / 3 of the number of girls is equal to 1 / 4 of the total number of students . what is the ratio of boys to girls in the class ? | "( 2 / 3 ) g = ( 1 / 4 ) ( b + g ) 8 g = 3 b + 3 g 5 g = 3 b b / g = 5 / 3 . the answer is d ." | a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | d | divide(subtract(divide(2, 3), divide(1, 4)), divide(1, 4)) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#1)| | general | D |
a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many q apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons q = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | add(const_10,const_10)|add(#0,const_2)|divide(n0,#1)|add(n1,#2)|divide(n0,#3)|add(n2,#4)| | general | C |
5 n + 2 > 12 and 7 n - 12 < 44 ; n must be between which numbers ? | "5 n + 2 > 12 5 n > 10 n > 2 7 n - 12 < 44 7 n < 56 n < 8 so n must be between 2 and 8 2 < n < 8 correct answer a" | a ) 2 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 2 and 7 , e ) 2 and 9 | a | add(multiply(2, const_10), divide(add(44, 12), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general | A |
carrie likes to buy t - shirts at the local clothing store . they cost $ 9.15 each . one day , she bought 22 t - shirts . how much money did she spend ? | $ 9.15 * 22 = $ 201.3 . answer is d . | a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 201.3 , e ) $ 190 | d | floor(multiply(22, 9.15)) | multiply(n0,n1)|floor(#0)| | general | D |
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 50 % profit ? | "let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 150 % of rs . 640 = 150 / 100 * 640 = rs . 960 . answer : b" | a ) 277 , b ) 960 , c ) 277 , d ) 266 , e ) 121 | b | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(50, const_100))) | divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)| | gain | B |
when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 96.25 , what is the value of y ? | "guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 96.25 and remainder is 9 so . 25 x y = 9 hence y = 36 ans c" | a ) 96 , b ) 75 , c ) 36 , d ) 25 , e ) 12 | c | divide(9, subtract(96.25, floor(96.25))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general | C |
in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 65 , what is the greatest possible value of b x a ? | "i also was confused when i was looking forabove number : d as far as i understood , 65 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 65 x 1 ( a = 6 , b = 5 respectively we have = 30 ) . this is the greatest possible value of b x a . imo e ." | a ) 0 , b ) 10 , c ) 15 , d ) 20 , e ) 30 | e | multiply(add(const_3, const_4), add(const_2, const_3)) | add(const_3,const_4)|add(const_2,const_3)|multiply(#0,#1)| | general | E |
last year , for every 100 million vehicles that traveled on a certain highway , 100 vehicles were involved in accidents . if 2 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 ) | "to solve we will set up a proportion . we know that β 100 million vehicles is to 100 accidents as 2 billion vehicles is to x accidents β . to express everything in terms of β millions β , we can use 2,000 million rather than 2 billion . creating a proportion we have : 100 / 100 = 2,000 / x cross multiplying gives us : 100 x = 2,000 * 100 x = 20 * 100 = 2000 answer : d" | a ) 500 , b ) 1500 , c ) 2500 , d ) 2000 , e ) 1000 | d | multiply(100, multiply(2, const_10)) | multiply(n2,const_10)|multiply(n1,#0)| | general | D |
a certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360 . what amount of total annual earnings would result in a 65 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) | for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960 ans : d | a ) $ 15,360 , b ) $ 17,360 , c ) $ 18,000 , d ) $ 21,960 , e ) $ 27,360 | d | add(multiply(const_100, const_3), const_60) | multiply(const_100,const_3)|add(#0,const_60) | general | D |
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 23.5 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 Γ ( 100 + 23.5 ) / ( 100 β 5 ) = 100 Γ 247 / 190 = 130 hence , required % profit = 130 β 100 = 30 % answer c" | a ) 28.5 , b ) 27.675 , c ) 30 , d ) data inadequate , e ) none of these | c | subtract(divide(multiply(add(const_100, 23.5), const_100), subtract(const_100, 5)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain | C |
nitin ranks 23 th in a class of 60 students . what is rank from the last ? | explanation : number students behind the nitin in rank = ( 60 - 23 ) = 37 nitin is 38 nd from the last answer : b ) 38 | a ) 33 , b ) 38 , c ) 32 , d ) 28 , e ) 19 | b | subtract(60, 23) | subtract(n1,n0) | other | B |
there are 2 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | "using anagram method : 4 _ 3 _ 2 _ 1 r _ r _ b _ b so . . 4 ! / number of repeated letters ( 2 ! ) ( 2 ! ) = 6 ans : d" | a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 60 , e ) e ) 100 | d | multiply(factorial(2), factorial(2)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | general | D |
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 950 gm . for a kg . his gain is β¦ % . | he sells 950 grams of pulses and gains 50 grams . if he sells 100 grams of pulses then he will gain ( 50 / 950 ) * 100 = 5.26 4 . a software engineer has the capability of thinking 100 lines of code in five minutes and can type 100 lines of code in 10 minutes . he takes a break for five minutes after every ten minutes . how many lines of codes will he complete typing after an hour ? answer : a | a ) 5.3 % , b ) 8.2 % , c ) 4.3 % , d ) 8.3 % , e ) 8.0 % | a | multiply(subtract(inverse(divide(950, 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 | A |
a sum of money is sufficient to pay p ' s wages for 24 days and q ' s wages for 40 days . the same money is sufficient to pay the wages of both for ? | let the total money be $ x p ' s 1 day work = $ x / 24 q ' s 1 day work = $ x / 40 p + q 1 day work = $ x / 15 money is sufficient to pay the wages of both for 15 days answer is d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(const_1, add(divide(const_1, 24), divide(const_1, 40))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | general | D |
how long will a boy take to run round a square field of side 50 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 50 * 4 = 200 m time taken = 200 * 2 / 5 = 80 sec answer is a" | a ) 80 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(50, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain | A |
a car covers a distance of 624 km in 6 Γ’ Β½ hours . find its speed ? | explanation : 624 / 6 = 104 kmph answer : c | a ) 104 , b ) 190 , c ) 109 , d ) 278 , e ) 211 | c | divide(624, 6) | divide(n0,n1) | physics | C |
if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 210 responses are needed , what is the minimum number of questionnaires that should be mailed ? | let x be the minimum number of questionnaires to be mailed . 0.6 x = 210 x = 350 the answer is c . | a ) 290 , b ) 320 , c ) 350 , d ) 380 , e ) 410 | c | divide(210, divide(60, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain | C |
the length of the rectangular field is double its width . inside the field there is square shaped pond 5 m long . if the area of the pond is 1 / 8 of the area of the field . what is the length of the field ? | a / 8 = 5 * 5 = > a = 5 * 5 * 8 x * 2 x = 5 * 5 * 8 x = 10 = > 2 x = 20 answer : d | ['a ) 54', 'b ) 32', 'c ) 75', 'd ) 20', 'e ) 11'] | d | sqrt(divide(multiply(square_area(5), 8), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3) | geometry | D |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 525 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 525 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 825 a + b + c = 75 per hour so for eight hrs = 75 * 8 = 600 = answer = e" | a ) 400 , b ) 475 , c ) 550 , d ) 625 , e ) 600 | e | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0)| | physics | E |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 64 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 25 : β 64 = 5 : 8 answer : d" | a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 5 , d ) 5 : 8 , e ) 4 : 2 | d | divide(sqrt(25), sqrt(64)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics | D |
find the sum of first 35 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 35 ( 35 + 1 ) / 2 = 35 ( 36 ) / 2 = 630 answer : option c" | a ) 470 , b ) 468 , c ) 630 , d ) 463 , e ) 520 | c | add(divide(divide(35, divide(divide(divide(divide(divide(35, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(35, divide(divide(divide(divide(divide(35, const_2), const_2), const_2), const_2), const_2)), const_2)))) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)| | general | C |
a student was asked to find 4 / 5 of a number . but the student divided the number by 4 / 5 , thus the student got 45 more than the correct answer . find the number . | "let the number be x . ( 5 / 4 ) * x = ( 4 / 5 ) * x + 45 25 x = 16 x + 900 9 x = 900 x = 100 the answer is c ." | a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140 | c | divide(divide(multiply(multiply(45, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5)) | divide(n0,n1)|multiply(n4,#0)|multiply(#0,#0)|multiply(#0,#1)|subtract(const_1,#2)|divide(#3,#4)|divide(#5,#0)| | general | C |
if 0.8 of a number is equal to 0.09 of another number , the ratio of the numbers i | "sol . 0.8 a = 0.09 b Γ’ β‘ β a / b = 0.09 / 0.8 = 9 / 80 = . Γ’ Λ Β΄ a : b = 9 : 80 . answer c" | a ) 2 : 3 , b ) 3 : 4 , c ) 9 : 80 , d ) 20 : 3 , e ) none | c | divide(multiply(0.09, const_100), multiply(0.8, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other | C |
company z has 61 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 61 . now , plug options for x . only c and e give an integer value for y . but only for c x > y as needed . answer : c . | a ) 6 , b ) 7 , c ) 13 , d ) 9 , e ) 12 | c | add(const_4, add(floor(divide(61, add(const_4, const_3))), const_1)) | add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4) | general | C |
karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many w miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance traveled by tom when karen is 4 miles ahead of him w : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) = 24 miles . d is the correct answer ." | a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | divide(n3,n2)|divide(n2,n2)|multiply(n0,#0)|subtract(#1,#0)|add(n0,#2)|divide(#4,#3)|subtract(#5,n0)| | physics | D |
two trains a and b are 200 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross each other ? the length that needs to be crossed = 200 + 150 = 350 m . time taken = 350 / 25 = 14 s . correct answer c ." | a ) 10 s , b ) 6 s , c ) 14 s , d ) 8 s , e ) 12 s | c | divide(add(200, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | add(n0,n1)|multiply(n2,const_1000)|multiply(n3,const_1000)|divide(#1,const_3600)|divide(#2,const_3600)|add(#3,#4)|divide(#0,#5)| | physics | C |
for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fourth of his earnings to charity . if the worker ' s charitable contributions totaled one - eighth of his earnings for the entire year , what is the value of n ? | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 4 = 1 / 8 ( 11 x + nx ) ; 11 / 10 + n / 4 = 1 / 8 ( 11 + n ) = > 44 + 10 n / 40 = 11 + n / 8 352 + 80 n = 440 + 40 n = > 40 n = 88 n = 88 / 40 = 22 / 10 answer : d ." | a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 22 / 10 , e ) 4 | d | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | add(const_1,const_4)|inverse(const_10)|subtract(const_12,const_1)|subtract(const_12,const_3)|inverse(#3)|inverse(#0)|subtract(#4,#1)|subtract(#5,#4)|multiply(#2,#6)|divide(#8,#7)| | general | D |
a person buys an article at rs . 460 . at what price should he sell the article so as to make a profit of 18 % ? | "cost price = rs . 460 profit = 18 % of 460 = rs . 82 selling price = cost price + profit = 460 + 82 = 542 answer : a" | a ) 542 , b ) 882 , c ) 772 , d ) 662 , e ) 521 | a | add(460, multiply(460, divide(18, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | A |
in a city , 35 % of the population is composed of migrants , 20 % of whom are from rural areas . of the local population , 48 % is female while this figure for rural and urban migrants is 30 % and 40 % respectively . if the total population of the city is 728400 , what is its female population ? | explanation : total population = 728400 migrants = 35 % of 728400 = 254940 local population = ( 728400 - 254940 ) = 473460 . rural migrants = 20 % of 254940 = 50988 urban migrants = ( 254940 - 50988 ) = 203952 female population = 48 % of 473460 + 30 % of 50988 + 40 % of 203952 = 324138 answer : a | a ) 324138 , b ) 248888 , c ) 378908 , d ) 277880 , e ) 379010 | a | add(add(divide(multiply(48, subtract(728400, divide(multiply(728400, 35), const_100))), const_100), divide(multiply(30, divide(multiply(divide(multiply(728400, 35), const_100), 20), const_100)), const_100)), divide(multiply(40, subtract(divide(multiply(728400, 35), const_100), divide(multiply(divide(multiply(728400, 35), const_100), 20), const_100))), const_100)) | multiply(n0,n5)|divide(#0,const_100)|multiply(n1,#1)|subtract(n5,#1)|divide(#2,const_100)|multiply(n2,#3)|divide(#5,const_100)|multiply(n3,#4)|subtract(#1,#4)|divide(#7,const_100)|multiply(n4,#8)|add(#6,#9)|divide(#10,const_100)|add(#11,#12) | gain | A |
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 12375 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 Γ ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) Γ 12375 ] = rs . 3375 answer : option b" | a ) rs . 2250 , b ) rs . 3375 , c ) rs . 6750 , d ) rs . 5625 , e ) none of these | b | multiply(12375, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)| | gain | B |
3 friends a , b , c went for week end party to mcdonald β s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 160 kg then find the average weight of all the 7 rounds ? | average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 160 / 7 = 91.4 kgs answer : a | a ) 91.4 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, const_3), 160), 7) | add(const_1,const_3)|multiply(n2,#0)|divide(#1,n1) | general | A |
if two positive numbers are in the ratio 1 / 8 : 1 / 7 , then by what percent is the second number more than the first ? | given ratio = 1 / 8 : 1 / 7 = 7 : 8 let first number be 7 x and the second number be 8 x . the second number is more than first number by 1 x . required percentage = 1 x / 7 x * 100 = 14.3 % . answer : a | a ) 14.3 % . , b ) 70 % . , c ) 60 % . , d ) 68 % . , e ) 80 % . | a | multiply(divide(1, 7), const_100) | divide(n0,n3)|multiply(#0,const_100) | general | A |
x , y and z , each working alone can complete a job in 2 , 4 and 6 days respectively . if all three of them work together to complete a job and earn $ 2000 , what will be z ' s share of the earnings ? | the dollars earned will be in the same ratio as amount of work done 1 day work of z is 1 / 6 ( or 2 / 12 ) 1 day work of the combined workforce is ( 1 / 2 + 1 / 4 + 1 / 6 ) = 11 / 12 z ' s contribution is 2 / 9 of the combined effort translating effort to $ = 6 / 11 * 2000 = $ 1090.90 hence : e | a ) $ 1080.90 , b ) $ 1000.90 , c ) $ 1070.90 , d ) $ 1050.90 , e ) $ 1090.90 | e | multiply(divide(2000, add(add(inverse(2), inverse(4)), inverse(6))), inverse(2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|multiply(#5,#0) | physics | E |
a certain fruit stand sold apples for $ 0.70 each and cherry for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ? | some multiple of 7 + some multiple of 5 should yield 63 . to get to a some multiple of 5 , we should ensure that a 3 or 8 ( 5 + 3 ) should be a multiple of 7 . 63 is a direct multiple of 7 , however in this case there wo n ' t be any cherry . hence the next option is to look for a multiple of 7 that has 8 as the unit digit . 28 satisfies this hence no . of apples is 4 and no of bananas is 7 b | a ) 12 , b ) 11 , c ) 13 , d ) 14 , e ) 15 | b | add(multiply(0.7, const_10), const_4) | multiply(n0,const_10)|add(#0,const_4) | other | B |
a milk man has 15 liters of milk . if he mixes 5 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is : | "explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 20 liters of milk . therefore , profit percentage = 20 % . answer : a" | a ) 20 % , b ) 25 % , c ) 33.33 % , d ) 18 % , e ) none of these | a | multiply(divide(subtract(multiply(add(add(add(20, 5), const_10), add(divide(5, const_2), const_3)), 18), multiply(15, 18)), multiply(15, 18)), const_100) | add(n1,n2)|divide(n1,const_2)|multiply(n0,n3)|add(#0,const_10)|add(#1,const_3)|add(#3,#4)|multiply(n3,#5)|subtract(#6,#2)|divide(#7,#2)|multiply(#8,const_100)| | gain | A |
q ' = 3 q - 3 , what is the value of ( 5 ' ) ' ? | "( 5 ' ) ' = ( 3 * 5 - 3 ) ' = 12 ' = 12 * 12 - 12 = 132 answer d" | a ) 96 , b ) 108 , c ) 120 , d ) 132 , e ) 144 | d | multiply(subtract(multiply(subtract(multiply(3, 5), 3), 3), 3), subtract(5, const_1)) | multiply(n0,n2)|subtract(n2,const_1)|subtract(#0,n0)|multiply(n0,#2)|subtract(#3,n0)|multiply(#4,#1)| | general | D |
the sum of ages of 5 children born 3 years different each is 75 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 75 5 x = 45 x = 9 . x + 12 = 9 + 12 = 21 e" | a ) 7 , b ) 9 , c ) 16 , d ) 18 , e ) 21 | e | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 75), 5) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general | E |
united telephone charges a base rate of $ 11.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ? | "lets take number of minutesx . given that , 11 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 20 minutes ans c" | a ) 2 minutes , b ) 10 minutes , c ) 20 minutes , d ) 40 minutes , e ) 60 minutes | c | divide(subtract(12.00, 11.00), subtract(0.25, 0.20)) | subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)| | general | C |
a circular ground whose diameter is 34 metres , has a 2 metre - broad garden around it . what is the area of the garden in square metres ? | req . area = Γ― β¬ [ ( 19 ) 2 Γ’ β¬ β ( 17 ) 2 ] = 22 Γ’ Β β 7 Γ£ β ( 36 Γ£ β 2 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 22 Γ£ β 36 Γ£ β 2 ) / 7 = 226.3 sq m answer a | ['a ) 226.3', 'b ) 226.2', 'c ) 228.2', 'd ) 227', 'e ) 226'] | a | subtract(circle_area(add(divide(34, const_2), 2)), circle_area(divide(34, const_2))) | divide(n0,const_2)|add(n1,#0)|circle_area(#0)|circle_area(#1)|subtract(#3,#2) | geometry | A |
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