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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how many boxes do you need if you have to pack 240 pairs ornamental bangles into boxes that each hold 2 dozens of bangles ? | "c 20 240 pairs of bangles = 480 bangles = 40 dozens . 40 ÷ 2 = 20 boxes ." | a ) 40 , b ) 35 , c ) 20 , d ) 25 , e ) 30 | c | divide(240, const_12) | divide(n0,const_12)| | geometry |
in an election , candidate a got 85 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate ? | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 85 % therefore , the number of valid votes polled in favour of candidate a = 85 % of 476000 = 85 / 100 × 476000 = 40460000... | a ) 355600 , b ) 355800 , c ) 404600 , d ) 356800 , e ) 357000 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(85, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
jerry bought a bottle of perfume for a gift for his wife . the perfume cost $ 92 before tax . if the total price including tax was $ 98.90 , find the tax rate | total price including tax is $ 98.90 perfume cost before tax is = 92 ie 92 * 7.5 % + 98.90 answer is 7.5 % | a ) 7.5 % , b ) 5 % , c ) 12 % , d ) 8 % , e ) 10 % | a | multiply(subtract(divide(98.9, 92), const_1), const_100) | divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100) | general |
tea worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , what is the price of the third variety per kg ? | "tea worth rs . 126 ratio 1 : 1 average price = ( 126 + 135 ) / 2 = 130.5 mean price = ( x - 153 ) : 22.50 = > x - 153 = 22.50 x = 175.5 answer a" | a ) 175.5 , b ) 182.5 , c ) 170.0 , d ) 180.0 , e ) 190.0 | a | add(add(add(add(153, const_10), const_10), 2), add(const_0_25, const_0_25)) | add(n5,const_10)|add(const_0_25,const_0_25)|add(#0,const_10)|add(n4,#2)|add(#3,#1)| | other |
the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 3.6 miles in the seventh 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | "( 3.6 miles / 8 minutes ) * 60 minutes / hour = 27 mph let x be the original speed . x - 6 ( 3 ) = 27 x = 45 mph the answer is e ." | a ) 33 , b ) 36 , c ) 39 , d ) 42 , e ) 45 | e | add(add(add(add(divide(3.6, divide(8, const_60)), 3), 3), 3), 3) | divide(n1,const_60)|divide(n2,#0)|add(n0,#1)|add(n0,#2)|add(n0,#3)|add(n0,#4)| | physics |
a 7 - digit number comprises of only 2 ' s and 3 ' s . how many of these are multiples of 12 ? | detailed solution number should be a multiple of 3 and 4 . so , the sum of the digits should be a multiple of 3 . we can either have all seven digits as 3 , or have three 2 ' s and four 3 ' s , or six 2 ' s and a 3 . ( the number of 2 ' s should be a multiple of 3 ) . for the number to be a multiple of 4 , the last 2 d... | a ) 11 , b ) 12 , c ) 10 , d ) 22 , e ) 44 | a | add(divide(factorial(add(3, 2)), multiply(factorial(3), factorial(2))), const_1) | add(n1,n2)|factorial(n2)|factorial(n1)|factorial(#0)|multiply(#1,#2)|divide(#3,#4)|add(#5,const_1) | general |
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8110 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8110 = p + ( 54 p / 100 ) 8110 = ( 1... | a ) rs . 4,900 , b ) rs . 5,000 , c ) rs . 5,100 , d ) rs . 5266 , e ) none of these | d | divide(8110, add(const_1, divide(multiply(6, 9), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)| | gain |
a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 10 percent of the drive from the center , what percent of the round - trip has the technician completed ? | "assuming that the technician makes a round - trip of 40 miles ( each way is 20 miles ) , then the technician would have completed 20 miles + 2 miles ( 10 % of the remaining 20 miles ) . therefore , the total is 22 miles . 22 miles / 40 miles is 55 % of the entire trip . answer : e" | a ) 5 % , b ) 10 % , c ) 25 % , d ) 40 % , e ) 55 % | e | add(divide(const_100, const_2), divide(multiply(10, divide(const_100, const_2)), const_100)) | divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)| | gain |
the banker â € ™ s discount of a certain sum of money is rs . 60 and the true discount on the same sum for the same time is rs . 54 . the sum due is | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 60 * 54 / 60 - 54 ] = rs . [ 72 * 60 / 6 ] = rs . 720 answer d" | a ) 210 , b ) 280 , c ) 360 , d ) 720 , e ) none | d | divide(multiply(60, 54), subtract(60, 54)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | gain |
a certain number of workers can do a work in 25 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | number of workers = 10 * ( 25 - 10 ) / 10 = 15 answer is a | a ) 15 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(25, 10), 10), subtract(25, subtract(25, 10))) | subtract(n0,n1)|multiply(n1,#0)|subtract(n0,#0)|divide(#1,#2) | physics |
a big container is 30 % full with water . if 9 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 9 liters of water is added , the container becomes 75 % full . hence these 9 liters account for 45 % of the container , which means that the capacity of it is 9 / 0.45 = 20 liters . or : if the capacity of the container is x liters then : 0.3 x + 9 = 0.75 x - - > x = 2... | a ) 20 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 60 liters | a | divide(9, subtract(divide(3, 4), divide(30, const_100))) | divide(n2,n3)|divide(n0,const_100)|subtract(#0,#1)|divide(n1,#2)| | general |
a man cheats while buying as well as while selling . while buying he takes 11 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 9.09 % below the cost price of the claimed weight . | there is a one step calculation method too . it requires more thought but is faster . the man takes 11 % more than what he pays for . so if he claims to take 100 pounds , he pays $ 100 but he actually takes 111 pounds for which he will take from the customer $ 111 . hence , in effect , there is a 11 % mark up . while s... | a ) 19.81 % , b ) 20 % , c ) 37.5 % , d ) 25 % , e ) 42.86 % | e | multiply(subtract(add(const_100, 11), add(9.09, subtract(const_100, 20))), const_2) | add(n0,const_100)|subtract(const_100,n1)|add(n2,#1)|subtract(#0,#2)|multiply(#3,const_2) | gain |
a train traveled the first d miles of its journey it an average speed of 60 miles per hour , the next d miles of its journey at an average speed of y miles per hour , and the final d miles of its journey at an average speed of 160 miles per hour . if the train ’ s average speed over the total distance was 90 miles per ... | "average speed = total distance traveled / total time taken 3 d / d / 60 + d / y + d / 160 = 90 solving for d and y , 15 y = 11 y + 480 4 y = 440 y = 110 answer d" | a ) 68 , b ) 84 , c ) 90 , d ) 110 , e ) 135 | d | divide(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), subtract(divide(multiply(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), const_3), 90), add(multiply(const_4, const_2), const_3))) | multiply(const_2,const_4)|multiply(const_100,const_4)|add(#0,const_3)|multiply(#0,const_10)|add(#3,#1)|multiply(#4,const_3)|divide(#5,n2)|subtract(#6,#2)|divide(#4,#7)| | physics |
a number is doubled and 15 is added . if resultant is trebled , it becomes 75 . what is that number | "explanation : = > 3 ( 2 x + 15 ) = 75 = > 2 x + 15 = 25 = > x = 5 option c" | a ) 8 , b ) 10 , c ) 5 , d ) 14 , e ) 7 | c | divide(subtract(75, multiply(const_3, 15)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
divide rs . 2800 among a , b and c so that a receives 3 / 4 as much as b and c together and b receives 1 / 4 as a and c together . a ' s share is ? | "a + b + c = 2800 a = 3 / 4 ( b + c ) ; b = 1 / 4 ( a + c ) a / ( b + c ) = 3 / 4 a = 1 / 7 * 8400 = > 1200 answer : b" | a ) 1300 , b ) 1200 , c ) 1375 , d ) 1400 , e ) 8400 | b | divide(2800, const_3) | divide(n0,const_3)| | general |
if 1 + 2 + 3 + . . . + n = n ( n + 1 ) , then 3 ( 1 + 3 + 5 + . . . . + 69 ) = ? | explanation : to solve this use the formula of ap , sn = ( n / 2 ) ( a + l ) . . . . . . . . . . . . . . . . ( 1 ) to find n , use = > tn = a + ( n - 1 ) d = > 69 = 1 + ( n - 1 ) 2 = > n = 35 use value of n in ( 1 ) then , sn = ( 35 / 2 ) ( 1 + 69 ) = 1225 ans : - 3 ( sn ) = 3675 answer : a | a ) 3675 , b ) 3575 , c ) 3475 , d ) 3375 , e ) 3275 | a | multiply(subtract(divide(multiply(69, add(69, const_1)), const_2), multiply(divide(subtract(69, const_1), const_2), add(divide(subtract(69, const_1), const_2), const_1))), const_3) | add(n8,const_1)|subtract(n8,const_1)|divide(#1,const_2)|multiply(n8,#0)|add(#2,const_1)|divide(#3,const_2)|multiply(#4,#2)|subtract(#5,#6)|multiply(#7,const_3) | general |
if the average ( arithmetic mean ) of 8 consecutive odd integers is 414 , then the least of these integers is | a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 414 . there are 8 numbers in the set , and in a set with an even number o... | a ) a ) 407 , b ) b ) 518 , c ) c ) 519 , d ) d ) 521 , e ) e ) 525 | a | add(subtract(414, 8), const_1) | subtract(n1,n0)|add(#0,const_1) | general |
a shipment of 250 smartphones contains 84 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ? a . b . c . d . e . | probability of choosing one defective phone from a lot of 250 which ontains 84 defective phones is = ( 84 / 250 ) probability of choosing one defective phone from a lot of 249 ( we already picked one ) which ontains 83 ( we already picked one ) defective phones is = ( 83 / 249 ) combined probability of series of events... | a ) 1 / 250 , b ) 1 / 84 , c ) 1 / 11 , d ) 1 / 9 , e ) 1 / 3 | d | multiply(divide(84, 250), divide(subtract(84, const_1), add(250, const_1))) | add(n0,const_1)|divide(n1,n0)|subtract(n1,const_1)|divide(#2,#0)|multiply(#1,#3) | other |
what is the positive difference between the sum of the squares of the first 7 positive integers and the sum of the prime numbers between the first square and fourth square ? | "forget conventional ways of solving math questions . in ps , ivy approach is the easiest and quickest way to find the answer . the sum of the squares of the first 4 positive integers = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 7 ^ 2 = 140 the sum of the prime numbers between the first square ( = 1 ) and fourth square ( = 16 ) =... | a ) 11 , b ) 52 , c ) 83 , d ) 94 , e ) 99 | e | subtract(add(add(add(add(add(const_1, power(const_2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_4, const_1), const_2)), power(7, const_2)), add(add(add(const_4, const_3), 7), add(add(add(add(const_2, const_3), add(const_4, const_1)), add(const_4, const_3)), add(add(const_2, const_3),... | add(const_1,const_4)|add(const_3,const_4)|add(const_2,const_3)|power(const_2,const_2)|power(const_3,const_2)|power(const_4,const_2)|power(n0,const_2)|add(#3,const_1)|add(n0,#1)|add(#2,#0)|add(n0,#2)|power(#0,const_2)|add(#7,#4)|add(#9,#1)|add(#12,#5)|add(#13,#10)|add(#14,#11)|add(#8,#15)|add(#16,#6)|subtract(#18,#17)| | general |
517 x 517 + 483 x 483 = ? | "= ( 517 ) ^ 2 + ( 483 ) ^ 2 = ( 500 + 17 ) ^ 2 + ( 500 - 17 ) ^ 2 = 2 [ ( 500 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 250000 + 289 ] = 2 x 250289 = 500578 answer is c" | a ) 79698 , b ) 80578 , c ) 500578 , d ) 81268 , e ) none of them | c | multiply(517, power(517, 483)) | power(n1,n2)|multiply(n0,#0)| | general |
on dividing a number by 5 , we get 3 as remainder . what will be the remainder when the square of this number is divided by 5 ? | let the number be x and on dividing x by 5 , we get k as quotient and 3 as remainder . x = 5 k + 3 x ^ 2 = ( 5 k + 3 ) ^ 2 = ( 25 k ^ 2 + 30 k + 9 ) = 5 ( 5 k ^ 2 + 6 k + 1 ) + 4 on dividing x ^ 2 by 5 , we get 4 as remainder . answer is d | ['a ) 0', 'b ) 1', 'c ) 3', 'd ) 4', 'e ) 2'] | d | multiply(subtract(divide(power(3, const_2), 5), floor(divide(power(3, const_2), 5))), 5) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3) | geometry |
a certain car uses one gallon of gasoline every 38 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | "4 miles on the highway = 4 / 38 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 38 + 4 / 20 = 29 / 95 gallons . 8 miles on the highway = 8 / 38 gallons . the % change = ( 29 / 95 - 8 / 38 ) / ( 8 / 38 ) = 0.45 . answer : d ." | a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 45 % , e ) 50 % | d | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 38))), multiply(8, divide(const_1, 38))), multiply(8, divide(const_1, 38))), const_100) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n4,#1)|add(#2,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)| | general |
divide $ 300 among a , b in the ratio 1 : 2 . how many $ that a get ? | "sum of ratio terms = 1 + 2 = 3 a = 300 * 1 / 3 = $ 100 answer is e" | a ) $ 50 , b ) $ 500 , c ) $ 150 , d ) $ 250 , e ) $ 100 | e | divide(300, 1) | divide(n0,n1)| | other |
a man can row upstream at 45 kmph and downstream at 55 kmph , and then find the speed of the man in still water ? | "us = 45 ds = 55 m = ( 45 + 55 ) / 2 = 50 answer : d" | a ) 32 kmph , b ) 34 kmph , c ) 30 kmph , d ) 50 kmph , e ) 65 kmph | d | divide(add(45, 55), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 8 cm each . the radius of the sphere is | solution 4 î r 2 = 2 î 4 x 8 â ‡ ’ r 2 = ( 4 x 8 / 2 ) â ‡ ’ 16 â ‡ ’ r = 4 cm . answer b | ['a ) 3 cm', 'b ) 4 cm', 'c ) 6 cm', 'd ) 8 cm', 'e ) none'] | b | sqrt(divide(multiply(multiply(const_pi, multiply(8, divide(8, const_2))), const_2), multiply(const_pi, const_4))) | divide(n0,const_2)|multiply(const_4,const_pi)|multiply(n0,#0)|multiply(#2,const_pi)|multiply(#3,const_2)|divide(#4,#1)|sqrt(#5) | geometry |
the sum of the non - prime numbers between 30 and 40 , non - inclusive , is | "sum of consecutive integers from 31 to 39 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 31 + 39 ) / 2 * 9 = 35 * 9 = 315 sum of non - prime numbers b / w 30 and 40 , non inclusive = = = > 315 - 68 ( i . e . , 31 + 37 , being the prime # s in the range ) = 247 answer : d" | a ) 202 , b ) 217 , c ) 232 , d ) 247 , e ) 262 | d | add(add(add(add(add(add(add(30, const_1), add(add(30, const_1), const_1)), add(add(add(30, const_1), const_1), const_2)), add(add(add(add(30, const_1), const_1), const_2), const_1)), add(add(add(add(add(30, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(30, const_1), const_1), const_2), cons... | add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)| | general |
find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 9 x + 4 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 9 / 1 = - 9 ab = 4 / 1 = 4 hence a / b + b / a = [ ( - 9 ) 2 - 2 ( 4 ) ] / 4 = 73 / 4 = 18.25 . b )" | a ) 17 , b ) 88 , c ) 14 , d ) 65 , e ) 89 | b | subtract(divide(power(negate(9), 2), 4), 2) | negate(n1)|power(#0,n0)|divide(#1,n2)|subtract(#2,n0)| | general |
6 wires are by average 80 cm long each . if the average length of one third of the wires is 70 cm , what is the average of the other wires ? | edit : given ( x 1 + x 2 . . . + x 6 ) / 6 = 80 ( x 1 + x 2 . . . + x 6 ) = 480 - - > eq 1 . now given avg length of one third wires is 70 . that means out 6 / 3 = 2 wires . let the avg length of two wires be ( x 1 + x 2 ) / 2 = 70 . ( x 1 + x 2 ) = 140 . - - > eq 2 . now we are asked to find the average of the remaini... | a ) 75 . , b ) 85 . , c ) 90 . , d ) 94 . , e ) 100 . | b | divide(subtract(multiply(6, 80), multiply(const_2, 70)), const_4) | multiply(n0,n1)|multiply(n2,const_2)|subtract(#0,#1)|divide(#2,const_4) | general |
when 15 is divided by integer u , the remainder is 4 . for how many values of u is this be true ? | when 15 is divided by u , the remainder is 4 i . e . 4 mangoes left over after grouping , so u must be greater than 4 . it also means that 11 is completely divisible by u . factors of 11 are 1 and 11 . out of these , u could be 11 . answer ( a ) | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(add(subtract(15, 4), 4), 15) | subtract(n0,n1)|add(n1,#0)|divide(#1,n0) | general |
a dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare . compare their speeds ? | "let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare be y then 3 x = 4 y x = 4 / 3 y 4 x = 16 / 3 y ratio of speeds of dog and hare = ratio of distances covered by them in the same time = 4 x : 5 y = 16 / 3 y : 5 = 16 : 15 answer is b" | a ) 4 : 5 , b ) 16 : 15 , c ) 9 : 13 , d ) 5 : 15 , e ) 9 : 17 | b | divide(multiply(4, 3), 5) | multiply(n0,n2)|divide(#0,n1)| | physics |
the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds t... | "given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 4 ) = 4 ^ 2 + 2 * 4 = 24 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 24 = 39 km / hr option c" | a ) 4 , b ) 9 , c ) 39 , d ) 48 , e ) 63 | c | subtract(add(power(7, 2), multiply(7, 2)), add(power(4, 2), multiply(4, 2))) | multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,n0)|add(#0,#2)|add(#1,#3)|subtract(#4,#5)| | physics |
if p # q denotes the least common multiple of p and q , then w = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , ... | a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2)|add(n3,#3)| | general |
a car travels from point a to point b . the average speed of the car is 60 km / hr and it travels the first half of the trip at a speed of 75 km / hr . what is the speed of the car in the second half of the trip ? | let d be the distance and let v be the speed in the second half . the total time = t 1 + t 2 d / 60 = d / 150 + ( d / 2 ) / v d / 100 = d / 2 v and so v = 50 km / hr the answer is d . | a ) 40 , b ) 45 , c ) 48 , d ) 50 , e ) 55 | d | divide(multiply(60, divide(multiply(75, 60), multiply(subtract(75, 60), 60))), divide(60, const_10)) | divide(n0,const_10)|multiply(n0,n1)|subtract(n1,n0)|multiply(n0,#2)|divide(#1,#3)|multiply(n0,#4)|divide(#5,#0) | general |
a rectangular table , kept against a wall has a three sides free and the wall along the fourth side . the side opposite the wall is twice the length of each of the other two free sides . if the area of the rectangular table is 128 square feet , what is the total length of the table free sides , in feet ? | two sides each = x the third = 2 x and the wall side is thus 2 x too x * 2 x = 2 x ^ 2 = 128 ie x ^ 2 = 64 ie x = 8 l = 16 w = 8 total lenght of table ' s free sides = 2 * 8 + 16 = 32 my answer is d | ['a ) 4', 'b ) 8', 'c ) 16', 'd ) 32', 'e ) 64'] | d | add(add(sqrt(divide(128, const_2)), multiply(sqrt(divide(128, const_2)), const_2)), sqrt(divide(128, const_2))) | divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2)|add(#2,#1)|add(#3,#1) | geometry |
a boat having a length 8 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "volume of water displaced = ( 8 x 3 x 0.01 ) m 3 = 0.24 m 3 . mass of man = volume of water displaced x density of water = ( 0.24 x 1000 ) kg = 240 kg . answer : e" | a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 80 kg , e ) 240 kg | e | multiply(multiply(multiply(8, 3), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
excluding stoppages , the speed of a bus is 86 kmph and including stoppages , it is 76 kmph . for how many minutes does the bus stop per hour ? | "d 7 min due to stoppages , it covers 10 km less . time taken to cover 10 km = ( 10 / 86 x 60 ) min = 7 min" | a ) 7.5 min , b ) 16 min , c ) 20 min , d ) 7 min , e ) 40 min | d | multiply(const_60, divide(subtract(86, 76), 86)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
the perimeter of a square is equal to the perimeter of a rectangle of length 15 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | let the side of the square be a cm . perimeter of the rectangle = 2 ( 15 + 14 ) = 58 cm perimeter of the square = 58 cm i . e . 4 a = 58 a = 14.5 diameter of the semicircle = 14.5 cm circumference of the semicircle = 1 / 2 ( ∏ ) ( 14.5 ) = 1 / 2 ( 22 / 7 ) ( 14.5 ) = 22.78 cm to two decimal places answer : a | ['a ) 22.78', 'b ) 23.54', 'c ) 24.5', 'd ) 25.55', 'e ) 23.51'] | a | floor(divide(circumface(divide(divide(multiply(const_2, add(14, 15)), const_4), const_2)), const_2)) | add(n0,n1)|multiply(#0,const_2)|divide(#1,const_4)|divide(#2,const_2)|circumface(#3)|divide(#4,const_2)|floor(#5) | geometry |
the ratio of numbers is 3 : 4 and their h . c . f is 4 . their l . c . m is ? | "let the numbers be 3 x and 4 x . then their h . c . f = x . so , x = 4 . so , the numbers are 12 and 16 . l . c . m of 12 and 16 = 48 . answer : d" | a ) 23 , b ) 77 , c ) 88 , d ) 48 , e ) 11 | d | lcm(multiply(3, 4), multiply(4, 4)) | multiply(n0,n2)|multiply(n1,n2)|lcm(#0,#1)| | other |
john makes $ 60 a week from his job . he earns a raise and now makes $ 120 a week . what is the % increase ? | "increase = ( 60 / 60 ) * 100 = ( 6 / 6 ) * 100 = 100 % . b" | a ) 16 % , b ) 100 % , c ) 17 % , d ) 17.61 % , e ) 17.56 % | b | multiply(divide(subtract(120, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a taxi leaves point a 5 hours after a bus left the same spot . the bus is traveling 20 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in three hours . | "let the speed of bus be v - 20 , speed of taxi be v the bus travelled a total of 8 hrs and taxi a total of 3 hrs . hence 8 * ( v - 20 ) = 3 v 8 v - 160 = 3 v 5 v = 160 v = 32 mph b" | a ) 34 , b ) 32 , c ) 36 , d ) 38 , e ) 40 | b | divide(add(multiply(5, 20), multiply(5, 20)), subtract(add(5, 5), 5)) | add(n0,n0)|multiply(n0,n1)|add(#1,#1)|subtract(#0,n0)|divide(#2,#3)| | physics |
a can finish a work in 12 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 12 * 1 / 3 ) = 4 days . correct option : a" | a ) 4 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these | a | divide(multiply(multiply(divide(const_1, 15), 10), 12), const_2) | divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)| | physics |
two trains are moving in opposite directions with speed of 90 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | "explanation : relative speed = 90 + 90 = 180 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 180 hr = 1 / 90 hr = 3600 / 90 seconds = 40 seconds answer : option b" | a ) 56 , b ) 40 , c ) 47 , d ) 26 , e ) 25 | b | multiply(divide(add(1.10, 0.9), add(90, 90)), const_3600) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600)| | physics |
a can do a piece of work in 40 days ; b can do the same in 20 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 40 + 10 / 20 + 10 / x = 1 x = 40 days answer : d" | a ) 24 days , b ) 65 days , c ) 86 days , d ) 40 days , e ) 17 days | d | divide(10, subtract(const_1, divide(add(10, 10), 40))) | add(n2,n2)|divide(#0,n0)|subtract(const_1,#1)|divide(n2,#2)| | physics |
for any number s , s * is defined as the greatest positive even integer less than or equal to s . what is the value of 5.2 – 5.2 * ? | since s * is defined as the greatest positive even integer less than or equal to s , then 5.2 * = 4 ( the greatest positive even integer less than or equal to 5.2 is 4 ) . hence , 5.2 – 5.2 * = 5.2 - 4 = 1.2 answer : b . | a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | b | subtract(5.2, const_4) | subtract(n0,const_4) | general |
the present ages of a , b & c are in the ratio of 5 : 7 : 8 . 7 years ago , the sum oftheir ages was 59 . what is the present age of eldest one ? | let their present ages be 5 x , 7 x and 8 x . 7 years ago , sum of their ages was 59 . sum of their present ages = 59 + ( 3 x 7 ) = 80 sum of their present ages given is 5 x + 7 x + 8 x = 20 x 20 x = 80 = > x = 80 20 = 4 age of the eldest one = 8 x = 8 x 4 = 32 years . c | a ) 30 , b ) 31 , c ) 32 , d ) 35 , e ) 36 | c | multiply(divide(add(multiply(7, const_3), 59), add(add(5, 7), 8)), 8) | add(n0,n1)|multiply(n1,const_3)|add(n4,#1)|add(n2,#0)|divide(#2,#3)|multiply(n2,#4) | general |
? % of 360 = 129.6 | "? % of 360 = 129.6 or , ? = 129.6 × 100 / 360 = 36 answer b" | a ) 277 , b ) 36 , c ) 64 , d ) 72 , e ) none of these | b | divide(multiply(129.6, const_100), 360) | multiply(n1,const_100)|divide(#0,n0)| | gain |
if the average ( arithmetic mean ) of 16 consecutive odd integers is 414 , then the least of these integers is | "a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 414 . there are 16 numbers in the set , and in a set with an even number... | a ) a ) 399 , b ) b ) 418 , c ) c ) 519 , d ) d ) 521 , e ) e ) 525 | a | add(subtract(414, 16), const_1) | subtract(n1,n0)|add(#0,const_1)| | general |
a car gets 30 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 200 kilometers ? | "each 30 kilometers , 1 gallon is needed . we need to know how many 30 kilometers are there in 200 kilometers ? 200 ã · 30 = 6.7 ã — 1 gallon = 6.7 gallons correct answer is c ) 6.7 gallons" | a ) 3.5 gallons , b ) 2.7 gallons , c ) 6.7 gallons , d ) 4.5 gallons , e ) 7.5 gallons | c | divide(200, 30) | divide(n1,n0)| | physics |
maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft co... | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c =... | a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | add(n2,const_2)|add(#0,const_2)|add(n3,#1)|divide(const_2,#2)|divide(#1,#2)|add(#3,#4)|multiply(#3,#4)|subtract(#5,#6)| | general |
the number 219 can be written as sum of the squares of 3 different positive integers . what is the difference of these 2 different larger integers ? | "sum of the squares of 3 different positive integers = 219 13 ^ 2 + 7 ^ 2 + 1 ^ 2 = 219 now , difference of these 2 different larger integers = 13 - 7 = 6 ans - a" | a ) 6 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | a | multiply(const_2, sqrt(divide(219, const_2))) | divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2)| | general |
this year , mbb consulting fired 10 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool al... | "100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 10 % reduction in employees lead to 90 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 90 employees is 90 * 1100 = 99000 now the new salary is more than pre... | a ) 99 % , b ) 100.0 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | a | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 10), const_4)), multiply(const_100, 10)) | multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)| | general |
what is the greatest positive integer n such that 3 ^ n is a factor of 36 ^ 150 ? | "36 = 3 ^ 2 * 2 ^ 2 . 36 ^ 150 = 3 ^ 300 * 2 ^ 300 the answer is c ." | a ) 100 , b ) 200 , c ) 300 , d ) 600 , e ) 900 | c | multiply(subtract(36, 150), 150) | subtract(n1,n2)|multiply(n2,#0)| | other |
real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 15 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 40 percent discount .... | let the retail price be = x selling price of z = 0.7 x selling price of x = 0.85 * 0.7 x = 0.60 x selling price of y = ( ( 0.7 x + 0.6 x ) / 2 ) * 0.60 = 0.65 x * 0.60 = 0.39 x 0.39 x = k * 0.60 x k = 0.39 / 0.6 = 39 / 6 answer : a | a ) 39 / 6 , b ) 62 / 11 , c ) 11 / 61 , d ) 21 / 61 , e ) 20 / 61 | a | multiply(divide(divide(multiply(divide(add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_2), subtract(const_100, 40)), const_100), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_10) | subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10) | general |
in a certain alphabet , 11 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | "we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 11 + 24 = 35 ; next , we find the total # of letters without line : 40 - 35 = 5 ; final... | a ) 5 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | a | subtract(40, add(11, 24)) | add(n0,n1)|subtract(n2,#0)| | other |
an alloy of copper and zinc contains copper and zinc in the ratio 5 : 3 . another alloy of copper and zinc contains copper and zinc in the ratio 1 : 7 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "alloy - 1 : copper : zinc = 5 : 3 , let quantity = x i . e . copper in alloy - 1 = [ 5 / ( 5 + 3 ) ] * x = ( 5 / 8 ) x and zinc in alloy - 1 = [ 3 / ( 5 + 3 ) ] * x = ( 3 / 8 ) x alloy - 2 : copper : zinc = 1 : 7 , let quantity = y i . e . copper in alloy - 2 = [ 1 / ( 1 + 7 ) ] * y = ( 1 / 8 ) y and zinc in alloy - 1... | a ) 1 : 5 , b ) 7 : 3 , c ) 5 : 3 , d ) 3 : 1 , e ) 4 : 3 | d | divide(subtract(divide(5, add(1, 7)), divide(1, 7)), subtract(divide(1, 7), divide(1, add(1, 7)))) | add(n2,n3)|divide(n2,n3)|divide(n0,#0)|divide(n2,#0)|subtract(#2,#1)|subtract(#1,#3)|divide(#4,#5)| | general |
the average weight of 30 boys sitting in a bus had some value . a new person added to them whose weight was 40 kg only . due to his arrival , the average weight of all the boys decreased by 2 kg . find the average weight of first 30 boys ? | 30 x + 40 = 31 ( x â € “ 2 ) x = 102 e | a ) 150 , b ) 122 , c ) 30 , d ) 120 , e ) 102 | e | add(add(40, add(30, 30)), 2) | add(n0,n0)|add(n1,#0)|add(n2,#1) | general |
the original price of a suit is $ 200 . the price increased 25 % , and after this increase , the store published a 25 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 25 % off the increased price , how much did these consumers pay for the suit ? | "0.75 * ( 1.25 * 200 ) = $ 187.50 the answer is d ." | a ) $ 178.50 , b ) $ 182.50 , c ) $ 185.50 , d ) $ 187.50 , e ) $ 200 | d | subtract(add(200, divide(multiply(200, 25), const_100)), divide(multiply(add(200, divide(multiply(200, 25), const_100)), 25), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|subtract(#2,#4)| | general |
what is the maximum number r of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ? | 27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can ... | ['a ) 36', 'b ) 32', 'c ) 24', 'd ) 21', 'e ) 15'] | c | multiply(multiply(divide(9, const_3), divide(12, cube_edge_by_volume(27))), floor(divide(8, cube_edge_by_volume(27)))) | cube_edge_by_volume(n0)|divide(n2,const_3)|divide(n1,#0)|divide(n3,#0)|floor(#2)|multiply(#1,#3)|multiply(#4,#5) | geometry |
a , b and c rents a pasture for rs . 638 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 638 = 198 answer : b" | a ) 270 , b ) 198 , c ) 676 , d ) 156 , e ) 122 | b | multiply(divide(638, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
if x / 4 - x - 3 / 6 = 1 , then find the value of x . | ( x / 4 ) - ( ( x - 3 ) / 6 ) = 1 = > ( 3 x - 2 ( x - 3 ) ) / 12 = 1 = > 3 x - 2 x + 6 = 12 = > x = 6 . answer is a | a ) 6 , b ) 4 , c ) 5 , d ) 1 , e ) 2 | a | multiply(add(divide(3, 6), 1), 4) | divide(n1,n2)|add(n3,#0)|multiply(n0,#1) | general |
bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 42 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 42 0.07 = 42 p = 6... | a ) $ 480 , b ) $ 570 , c ) $ 660 , d ) $ 720 , e ) $ 850 | c | multiply(divide(42, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|subtract(#5,#2)|divide(n3,#6)|multiply(#2,#7)| | general |
x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 14300 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = ... | a ) 3200 , b ) 4000 , c ) 3250 , d ) 4160 , e ) 3985 | d | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14300, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | multiply(const_3,n2)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,n2)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5)| | gain |
in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 10 , 0 ) , ( 0 , - 3.5 ) , ( - 10 , 0 ) ? | "area of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 10 + 10 = 20 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 20 * 7 = 70 c is the answer" | a ) 56 , b ) 88 , c ) 70 , d ) 116 , e ) 120 | c | rhombus_area(multiply(10, const_2), multiply(3.5, const_2)) | multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)| | geometry |
hcf of two numbers is 15 and their lcm is 180 . if their sum is 105 , then the numbers are : | "explanation : let the numbers be 15 a and 15 b . then , 15 a + 15 b = 105 or a + b = 7 . . ( i ) lcm = 15 ab = 180 ab = 12 . . ( ii ) solving equations ( i ) and ( ii ) , we get a = 4 , b = 3 so , the numbers are 15 × 4 and 15 × 3 , i . e . , 60 and 45 answer : d" | a ) 30 and 75 , b ) 35 and 70 , c ) 40 and 65 , d ) 45 and 60 , e ) 55 and 70 | d | divide(180, 15) | divide(n1,n0)| | physics |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 10 overs to reach the target of 282 runs ? | "explanation : runs scored in the first 10 overs = 10 × 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 10 run rate needed = 250 / 10 = 25 answer : option c" | a ) 6.25 , b ) 5.5 , c ) 25 , d ) 15 , e ) 6 | c | divide(subtract(282, multiply(10, 3.2)), 10) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
if the cost price of 50 articles is equal to the selling price of 25 articles , then the gain or loss percent is ? | "percentage of profit = 25 / 25 * 100 = 100 % answer : e" | a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 100 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 50)), divide(multiply(const_100, 25), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
jake can dig a well in 16 days . paul can dig the same well in 24 days . jake , paul and hari together dig the well in 8 days . hari alone can dig the well in | jake 1 day work = 1 / 16 paul 1 day work = 1 / 24 j + p + h 1 ady work = 1 / 8 1 / 16 + 1 / 24 + 1 / x = 1 / 8 1 / x = 1 / 48 x = 48 so , hari alone can dig the well in 48 days answer : b | a ) 96 days , b ) 48 days , c ) 32 days , d ) 24 days , e ) 28 days | b | inverse(subtract(divide(const_1, 8), add(divide(const_1, 16), divide(const_1, 24)))) | divide(const_1,n2)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|subtract(#0,#3)|inverse(#4) | physics |
3 numbers are in the ratio 4 : 5 : 6 and their average is 20 . the largest number is : | explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 20 15 x = 60 x = 4 largest number = 6 x = 24 . answer a | a ) 24 , b ) 32 , c ) 36 , d ) 42 , e ) 45 | a | multiply(divide(20, divide(add(add(4, 5), 6), 3)), 6) | add(n1,n2)|add(n3,#0)|divide(#1,n0)|divide(n4,#2)|multiply(n3,#3) | general |
a father told his son ` ` i was as old as you are at present , at the time of your birth ' ' . if the father is 38 years old now , then what was the son ' s age 5 years ago in years ? | et son ' s present age is = x then 38 - x = x x = 19 son ' s age 5 years back is 19 - 5 = 14 years . answer : a | a ) 14 , b ) 19 , c ) 38 , d ) 33 , e ) 35 | a | subtract(divide(38, const_2), 5) | divide(n0,const_2)|subtract(#0,n1) | general |
a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 120 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 120 total = 120 + 120 = 240 answer ( e )" | a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 240 | e | divide(subtract(multiply(120, divide(60, const_100)), multiply(120, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n3,const_100)|divide(n0,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#1,#2)|subtract(#3,#4)|divide(#6,#5)| | general |
the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 5600 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 5600 ( 100 / 125 ) = rs . 4480 . answer : a" | a ) rs . 4480 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6725 , e ) none of these | a | divide(5600, add(const_1, divide(25, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio o | "4 : 5 3 : 2 - - - - - - - 12 : 15 : 10 12 : 10 = = > 6.5 answer a" | a ) 6 : 5 , b ) 4 : 7 , c ) 3 : 5 , d ) 5 : 7 , e ) 8 : 9 | a | divide(multiply(4, 3), multiply(5, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
the sector of a circle has radius of 18 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 18 ) + 2 ( 18 ) = 42.4 + 36 = 78.4 cm answer : a" | a ) 78.4 cm , b ) 11.5 cm , c ) 91.8 cm , d ) 92.5 cm , e ) 99.5 cm | a | multiply(multiply(const_2, divide(multiply(subtract(18, const_3), const_2), add(const_4, const_3))), 18) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
last year , for every 100 million vehicles that traveled on a certain highway , 96 vehicles were involved in accidents . if 3 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 “ 100 million vehicles is to 96 accidents as 3 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” , we can use 3,000 million rather than 3 billion . creating a proportion we have : 100 / 96 = 3,000 / x cross multiplying gives us : 10... | a ) 288 , b ) 320 , c ) 2,880 , d ) 3,200 , e ) 28,800 | c | multiply(96, multiply(3, const_10)) | multiply(n2,const_10)|multiply(n1,#0) | general |
for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.2 – 6.2 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 6.2 * = 4 ( the greatest positive even integer less than or equal to 6.2 is 4 ) . hence , 6.2 – 6.2 * = 6.2 - 4 = 2.2 answer : d ." | a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | d | subtract(6.2, subtract(floor(6.2), const_1)) | floor(n0)|subtract(#0,const_1)|subtract(n0,#1)| | general |
108 . triangle a ’ s base is 20 % greater than the base of triangle b , and a ’ s height is 20 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | wish the question specified that we are talking about corresponding height . base of a = 21 / 20 * base of b height of a = 19 / 20 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 21 / 20 * 19 / 20 * area of b = 399 / 400 * area of b area of a is 0.25 % less than the area of b . answer ( a ) | ['a ) 0.25 % less', 'b ) 1 % less', 'c ) equal to each other', 'd ) 1 % more', 'e ) 9 % more'] | a | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 20), subtract(const_100, 20)))) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4) | geometry |
a circular rim 28 inches in diameter rotates the same number of inches per second as a circular rim 35 inches in diameter . if the smaller rim makes x revolutions per second , how many revolutions per second does the larger rim makes in terms of x ? | let ' s try the explanation . we have two wheels . one with 28 pi and the other one with 35 pi . they have the same speed . in the smaller wheel it ' s 28 pi * x , which must be equal to the speed of the bigger one ( 35 pi * a number of revolutions ) . they are asking that number of revolutions ( but in minutes , which... | ['a ) 4 x / 5', 'b ) 75 x', 'c ) 48 x', 'd ) 24 x', 'e ) x / 75'] | a | multiply(multiply(multiply(const_2, const_3), const_10), divide(28, 35)) | divide(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|multiply(#0,#2) | physics |
a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 11 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 11 = > m = 33 answer b" | a ) 39 , b ) 33 , c ) 26 , d ) 21 , e ) 10 | b | subtract(divide(11, divide(2, 3)), 11) | divide(n0,n1)|divide(n2,#0)|subtract(#1,n2)| | general |
walking with 5 / 4 of my usual speed , i miss the bus by 5 minutes . what is my usual time ? | "speed ratio = 1 : 5 / 4 = 4 : 5 time ratio = 5 : 4 1 - - - - - - - - 5 4 - - - - - - - - - ? è 20 answer : c" | a ) 18 , b ) 19 , c ) 20 , d ) 22 , e ) 24 | c | multiply(divide(const_3.0, divide(4, 5)), 4) | divide(n1,const_4.0)|divide(n2,#0)|multiply(n1,#1)| | physics |
if n = 2 ^ 0.25 and n ^ b = 16 , b must equal | "25 / 100 = 1 / 4 n = 2 ^ 1 / 4 n ^ b = 2 ^ 4 ( 2 ^ 1 / 4 ) ^ b = 2 ^ 4 b = 16 answer : d" | a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 16 , e ) 80 / 3 | d | divide(log(16), log(power(2, 0.25))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
in how many q ways can a 4 - letter password be chosen , using the letters a , b , c , d , e , and / or f , such that at least one letter is repeated within the password ? | total number of four letter passwords = 6 * 6 * 6 * 6 = 1296 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 6 c 4 * 4 ! = 15 * 24 = 360 - - - - - - ( 2 ) therefore required value q = ( 1 ) - ( 2 ) = 1296 - 360 = 936 . d | a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 1296 | d | multiply(multiply(divide(divide(factorial(4), factorial(const_2)), factorial(const_2)), divide(divide(factorial(4), factorial(const_2)), factorial(const_2))), multiply(subtract(multiply(const_2, const_3), const_1), subtract(multiply(const_2, const_3), const_1))) | factorial(n0)|factorial(const_2)|multiply(const_2,const_3)|divide(#0,#1)|subtract(#2,const_1)|divide(#3,#1)|multiply(#4,#4)|multiply(#5,#5)|multiply(#7,#6) | general |
the distance between delhi and mathura is 150 kms . a starts from delhi with a speed of 25 kmph at 5 a . m . for mathura and b starts from mathura with a speed of 40 kmph at 6 p . m . from delhi . when will they meet ? | d = 150 – 25 = 125 rs = 40 + 25 = 65 t = 125 / 65 = 2 hours 6 a . m . + 2 = 8 a . m . . answer : c | a ) 11 , b ) 77 , c ) 8 , d ) 10 , e ) 12 | c | add(6, divide(150, add(25, 40))) | add(n1,n3)|divide(n0,#0)|add(n4,#1) | physics |
water boils at 212 ° f or 100 ° c and melts at 32 ° f or 0 ° c . if the temperature of the particular day is 35 ° c , it is equal to | let f and c denotes the temparature in fahrenheit anid celcsius respectively . then , ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) , if c = 35 , then f = 95 . c | a ) 50 ° f , b ) 76 ° f , c ) 95 ° f , d ) 110 ° f , e ) 120 ° f | c | add(multiply(divide(35, 100), subtract(212, 32)), 32) | divide(n4,n1)|subtract(n0,n2)|multiply(#0,#1)|add(n2,#2) | physics |
the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 96 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 96 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 96 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 96 r 2 = ( 96 * 2 ) / 3 = 64 = > r = 8 rate = 8 % answer : a" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | a | sqrt(96) | sqrt(n2)| | gain |
a monkey ascends a greased pole 26 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 24 meters . i mean this will continue for first 24 * 2 = 48 mins . he would have reached 24 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 48 + 1 = 49 mins . so , asnwer will be e" | a ) 50 th minute , b ) 41 st minute , c ) 45 th minute , d ) 42 nd minute , e ) 49 th minute | e | add(multiply(multiply(const_4, 2), 2), 1) | multiply(n1,const_4)|multiply(n1,#0)|add(n2,#1)| | physics |
a man distributed rs . 100 equally among his friends . if there had been 5 more friends , each would have received one rupee less . how many friends had he ? | firstly we have 20 friends and money wiill be distribute each of them 5 rupee then 5 more friends it means total no frnd = 25 each recv 4 rupee which means all recv 1 less rupee answer : b | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | add(divide(100, 5), 5) | divide(n0,n1)|add(n1,#0) | general |
a , b and c can do a work in 6 , 9 and 12 days respectively doing the work together and get a payment of rs . 1800 . what is b ’ s share ? | "wc = 1 / 6 : 1 / 9 : 1 / 12 = > 6 : 4 : 3 4 / 13 * 1800 = 653.8 answer : b" | a ) rs . 245.8 , b ) rs . 653.8 , c ) rs . 300 , d ) rs . 400 , e ) rs . 748.5 | b | multiply(1800, divide(inverse(9), add(inverse(12), add(inverse(6), inverse(9))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 6270 . what was his original salary ? | x * ( 110 / 100 ) * ( 95 / 100 ) = 6270 x * ( 11 / 10 ) * ( 1 / 100 ) = 66 x = 6000 answer : a | a ) 6000 , b ) 2999 , c ) 1000 , d ) 2651 , e ) 1971 | a | divide(6270, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4) | gain |
the perimeter of one face of a cube is 28 cm . its volume will be : | "explanation : edge of cude = 28 / 4 = 7 cm volume = a * a * a = 7 * 7 * 7 = 343 cm cube option b" | a ) 125 cm 3 , b ) 343 cm 3 , c ) 250 cm 3 , d ) 625 cm 3 , e ) none of these | b | volume_cube(square_edge_by_perimeter(28)) | square_edge_by_perimeter(n0)|volume_cube(#0)| | geometry |
the true discount on a bill of rs . 270 is rs . 45 . the banker ' s discount is | solution p . w = rs . ( 270 - 45 ) = rs . 225 s . i on rs . 270 = rs . ( 45 / 225 x 270 ) = rs . 54 . answer a | a ) 54 , b ) 55 , c ) 56 , d ) 57 , e ) none of these | a | multiply(divide(45, subtract(270, 45)), 270) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1) | gain |
a part of certain sum of money is invested at 9 % per annum and the rest at 21 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "21 : 9 = 7 : 3 answer : c" | a ) 4 : 2 , b ) 4 : 8 , c ) 7 : 3 , d ) 4 : 0 , e ) 4 : 9 | c | multiply(divide(21, const_100), 9) | divide(n1,const_100)|multiply(n0,#0)| | gain |
a carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet . if the carpenter made instead a sandbox which was twice as long , twice as wide and twice as high as the original , what would the capacity be of the new larger sandbox ? | explanation : when all the dimensions of a three dimensional object are in fact doubled , then the capacity increases by a factor of 2 x 2 x 2 = 2 ^ 3 = 8 . thus the capacity of the new sandbox is 8 x 10 = 80 cubic feet . answer : option d | ['a ) 20', 'b ) 40', 'c ) 60', 'd ) 80', 'e ) 100'] | d | multiply(power(const_2, const_3), 10) | power(const_2,const_3)|multiply(n0,#0) | geometry |
a pipe can empty 2 / 3 rd of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | "2 / 3 - - - - 12 ? - - - - - 8 = = > 4 / 9 c" | a ) 2 / 3 , b ) 2 / 5 , c ) 4 / 9 , d ) 5 / 7 , e ) 4 / 11 | c | divide(multiply(divide(2, 3), 8), 12) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics |
the toll for crossing a certain bridge is $ 0.75 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 12.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the b... | "option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than... | a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | a | add(multiply(divide(multiply(divide(12.00, multiply(subtract(0.75, 0.30), const_2)), const_2), const_10), const_2), multiply(divide(12.00, multiply(subtract(0.75, 0.30), const_2)), const_2)) | subtract(n0,n2)|multiply(#0,const_2)|divide(n1,#1)|multiply(#2,const_2)|divide(#3,const_10)|multiply(#4,const_2)|add(#5,#3)| | general |
if 20 % of a is the same as 25 % of b , then a : b is : | "expl : 20 % of a i = 25 % of b = 20 a / 100 = 25 b / 100 = 5 / 4 = 5 : 4 answer : c" | a ) 3 : 4 , b ) 4 : 3 , c ) 5 : 4 , d ) 6 : 7 , e ) 5 : 7 | c | divide(divide(25, const_100), divide(20, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain |
4 , 7 , 16 , 43 , 124 , ( . . . ) | "explanation : 4 4 × 3 - 5 = 7 7 × 3 - 5 = 16 16 × 3 - 5 = 43 43 × 3 - 5 = 124 124 × 3 - 5 = 367 answer : option b" | a ) 22 , b ) 367 , c ) 27 , d ) 32 , e ) 25 | b | subtract(negate(43), multiply(subtract(7, 16), divide(subtract(7, 16), subtract(4, 7)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a man buys an article for $ 100 . and sells it for $ 120 . find the gain percent ? | "c . p . = $ 100 s . p . = $ 120 gain = $ 20 gain % = 20 / 100 * 100 = 20 % answer is d" | a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 % | d | subtract(divide(120, divide(100, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)| | gain |
the average weight of a class of 24 students is 35 kg . if the weight of the teacher be included , the average rises by 400 g . the weight of the teacher is | "solution weight of the teacher = ( 35.4 × 25 - 35 × 24 ) kg = 45 kg . answer a" | a ) 45 kg , b ) 50 kg , c ) 53 kg , d ) 55 kg , e ) none of these | a | subtract(multiply(add(35, divide(400, const_1000)), add(24, const_1)), multiply(24, 35)) | add(n0,const_1)|divide(n2,const_1000)|multiply(n0,n1)|add(n1,#1)|multiply(#3,#0)|subtract(#4,#2)| | general |
what percent of 200 is 55 ? | "200 * x / 100 = 55 x = 27.5 ans : d" | a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 27.5 % , e ) 250 % | d | multiply(divide(200, 55), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 25 % ? | "cost price = rs . 500 profit = 25 % of 500 = rs . 125 selling price = cost price + profit = 500 + 125 = 625 answer : c" | a ) 600 , b ) 887 , c ) 625 , d ) 654 , e ) 712 | c | add(500, multiply(500, divide(25, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
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