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
values |
|---|---|---|---|---|---|---|
the distance between delhi and mathura is 110 kms . a starts from delhi with a speed of 20 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 25 kmph at 8 p . m . from delhi . when will they meet ? | "d 10 a . m . d = 110 β 20 = 90 rs = 20 + 25 = 45 t = 90 / 45 = 2 hours 8 a . m . + 2 = 10 a . m ." | a ) 08 a . m . , b ) 09 a . m . , c ) 12 a . m . , d ) 10 a . m . , e ) 07 a . m . | d | add(8, divide(110, add(20, 25))) | add(n1,n3)|divide(n0,#0)|add(n4,#1)| | physics |
what is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33 ? | "explanatory answer useful result pertaining to remainders you can solve this problem if you know this rule about remainders . let a number x divide the product of a and b . the remainder will be the product of the remainders when x divides a and when x divides b . using this rule , the remainder when 33 divides 1044 i... | a ) 3 , b ) 27 , c ) 30 , d ) 21 , e ) 18 | c | reminder(multiply(1047, 1044), 1050) | multiply(n0,n1)|reminder(#0,n2)| | general |
how many positive integers less than 100 are neither multiples of 3 or 4 . | "to answer this q we require to know 1 ) multiples of 3 till 100 = 100 / 3 = 33 2 ) multiples of 4 till 100 = 100 / 4 = 25 add the two 25 + 33 = 58 ; subtract common terms that are multiple of both 3 and 4 . . lcm of 4 and 3 = 12 multiples of 6 till 100 = 100 / 12 = 8.3 = 8 so total multiples of 3 and 4 = 58 - 8 = 50 a... | a ) a ) 50 , b ) b ) 31 , c ) c ) 32 , d ) d ) 33 , e ) e ) 34 | a | divide(factorial(subtract(add(const_4, 3), const_1)), multiply(factorial(3), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
shopkeeper rise price by 31 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ? | "let d initial price be 100 31 % rise now price = 131 / 100 * 100 = 131 10 % discount then price = 131 * 90 / 100 = 117.9 15 % discount then price = 117.9 * 85 / 100 = 100.215 so gain = 100.215 - 100 = 0.215 gain % = gain * 100 / cp = = > 0.215 * 100 / 100 = 0.215 % answer : a" | a ) 0.215 % , b ) 1.215 % , c ) 3.215 % , d ) 6.215 % , e ) 7.215 % | a | subtract(multiply(multiply(add(const_100, 31), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
the probability that a computer company will get a computer hardware contract is 4 / 5 and the probability that it will not get a software contract is 3 / 5 . if the probability of getting at least one contract is 9 / 10 , what is the probability that it will get both the contracts ? | "let , a β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ event of getting both hardware and software contract . p ( a ) = 4 / 5 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 3 / 5 ) = 2 / 5 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b )... | a ) 1 / 8 , b ) 2 / 9 , c ) 3 / 10 , d ) 4 / 11 , e ) 5 / 12 | c | subtract(add(divide(const_3.0, 5), subtract(const_1, divide(4, 5))), divide(5, 10)) | divide(n2,n1)|divide(n0,n3)|divide(n3,n5)|subtract(const_1,#1)|add(#0,#3)|subtract(#4,#2)| | other |
in right triangle abc , ac is the hypotenuse . if ac is 20 and ab + bc = 50 , what is the area of the triangle abc ? | "square ab + bc = 50 : ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 2500 . since ( ac ) ^ 2 = ( ab ) ^ 2 + ( bc ) ^ 2 = 20 ^ 2 = 400 , then ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 400 + 2 * ab * bc = 2500 . 400 + 2 * ab * bc = 2500 . ab * bc = 1050 . the area = 1 / 2 * ab * bc = 525 . answer : b ." | a ) 225 , b ) 525 , c ) 25 β 2 , d ) 200 , e ) 200 β 2 | b | triangle_area_three_edges(20, multiply(const_3, const_10), multiply(const_4, const_10)) | multiply(const_10,const_3)|multiply(const_10,const_4)|triangle_area_three_edges(n0,#0,#1)| | geometry |
a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 500 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for s... | a ) 12,000 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(500, add(const_2, const_3)), const_3), multiply(divide(500, add(const_2, const_3)), const_4)) | add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)| | geometry |
if k and w are the dimensions of a rectangle that has area 21 , and if k and w are integers such that k > w , what is the total number of possible values of k ? | "kw = 21 = 21 * 1 = 7 * 3 - - > k can take 2 values , answer : a" | a ) two , b ) three , c ) four , d ) five , e ) six | a | multiply(const_3, const_1) | multiply(const_1,const_3)| | geometry |
how many digits are in the number 50 ^ 8 Γ 8 ^ 3 Γ 11 ^ 2 x 10 ^ 6 ? | "whenever you are asked to find the number of digits , try to bring the number in multiples of 10 . this way , we can wasily calculate the umber of 0 ' s through the powers of 10 50 ^ 8 Γ 8 ^ 3 Γ 11 ^ 2 x 10 ^ 6 = ( 5 ^ 2 * 2 ) ^ 8 * 2 ^ 9 * 11 ^ 2 x 10 ^ 6 = 5 ^ 16 * 2 ^ 17 * 11 ^ 2 x 10 ^ 6 = 2 * 11 ^ 2 * 10 ^ 16 x 1... | a ) 22 , b ) 20 , c ) 19 , d ) 18 , e ) 17 | e | subtract(multiply(10, 2), 2) | multiply(n5,n6)|subtract(#0,n5)| | general |
the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 4 years ? | "explanation : 40,000 * ( 3 / 4 ) ^ 4 answer : c" | a ) 3277 , b ) 2977 , c ) 12656 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 4)) | divide(n0,n1)|multiply(const_100,const_100)|multiply(n1,#1)|power(#0,n3)|multiply(#2,#3)| | general |
the simple interest on a sum of money will be rs . 400 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 400 p - - - 5 - - - - - 200 3 p - - - 5 - - - - - 600 - - - - - - = > 800 answer : a" | a ) 800 , b ) 3799 , c ) 1200 , d ) 2693 , e ) 1732 | a | add(multiply(multiply(divide(400, 10), 5), const_3), multiply(divide(400, 10), 5)) | divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)| | general |
the banker Γ’ β¬ β’ s gain on a sum due 3 years hence at 10 % per annum is rs . 120 . the banker Γ’ β¬ β’ s discount is | "solution t . d = ( b . g x 100 / r x t ) = rs . ( 120 x 100 / 10 x 3 ) = rs . 400 . b . d = rs ( 400 + 120 ) = rs . 520 . answer c" | a ) rs . 960 , b ) rs . 840 , c ) rs . 520 , d ) rs . 760 , e ) none | c | add(divide(multiply(120, const_100), multiply(3, 10)), 120) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.90 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum... | "total no . of boxes = 3060000 / ( 20 Γ 20 Γ 15 ) = 510 total cost = 510 Γ $ 0.9 = $ 459 answer c" | a ) $ 255 , b ) $ 275 , c ) $ 459 , d ) $ 1,250 , e ) $ 2,550 | c | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 0.90) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)| | 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 5.0 β 5.0 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 5.0 * = 4 ( the greatest positive even integer less than or equal to 5.0 is 4 ) . hence , 5.0 β 5.0 * = 5.0 - 4 = 1.0 answer : e ." | a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 1.0 | e | subtract(5.0, subtract(floor(5.0), const_1)) | floor(n0)|subtract(#0,const_1)|subtract(n0,#1)| | general |
the salary of a person was reduced by 10 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 90 increase on 90 = 10 increase on 100 = 10 / 90 * 100 = 11 1 / 9 % answer is b" | a ) 10 % , b ) 11 1 / 9 % , c ) 13 2 / 7 % , d ) 15 3 / 11 % , e ) 12 1 / 3 % | b | multiply(divide(multiply(const_100, divide(10, const_100)), subtract(const_100, multiply(const_100, divide(10, const_100)))), const_100) | divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)| | gain |
if the average ( arithmetic mean ) of 102 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 102 numbers in the set , and in a set with an even numbe... | a ) a ) 160 , b ) b ) 289 , c ) c ) 250 , d ) d ) 313 , e ) e ) 525 | d | add(subtract(414, 102), const_1) | subtract(n1,n0)|add(#0,const_1)| | general |
10 play kabadi , 40 play kho kho only , 5 play both gmaes . then how many in total ? | "10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 40 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 40 + 5 = 45 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 45 - 5 = 50 answer : e" | a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 50 | e | subtract(add(10, add(40, 5)), 5) | add(n1,n2)|add(n0,#0)|subtract(#1,n2)| | general |
set s consists of integers { 6 , 8 , 10 , 12 , 14 } . if integer n is included in the set , the average ( arithmetic mean ) of set s will increase by 30 % . what is the value of integer n ? | "the average of the numbers in set s is 10 . if we increase the mean by 30 % , the new mean is 13 . thus , on average , 5 numbers increase by 3 . therefore n = 13 + 15 = 28 the answer is e ." | a ) 10 , b ) 12 , c ) 16 , d ) 22 , e ) 28 | e | add(add(10, 12), 12) | add(n2,n3)|add(n3,#0)| | general |
1 + 2 = 10 2 + 3 = 26 3 + 4 = 50 then 4 + 5 = ? | 1 + 2 = 10 . . . . . ( 1 + 2 ) ^ 2 + 1 = 10 2 + 3 = 26 . . . . . . . ( 2 + 3 ) ^ 2 + 1 = 26 3 + 4 = 50 then . . . ( 3 + 4 ) ^ 2 + 1 = 50 4 + 5 = 82 . . . . . . . . ( 4 + 5 ) ^ 2 + 1 = 82 answer : c | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | c | add(power(add(4, 5), const_2), const_1) | add(n7,n10)|power(#0,const_2)|add(#1,const_1) | general |
n is a positive integer . when n + 1 is divided by 4 , the remainder is 3 . what is the remainder when n is divided by 3 ? | "n + 1 = 4 a + 3 i . e . n + 1 = 7 , 10 , 13 , 16 , . . . etc . i . e . n = 6 , 9 , 12 , 15 , . . . etc . when n is divided by 3 remainder is always 0 answer : d" | a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 4 | d | divide(4, add(3, 3)) | add(n2,n3)|divide(n1,#0)| | general |
in the list 3 , 4 , 5 , 5 , 5 , 5 , 7 , 11 , 21 , what fraction of the data is less than the mode ? | highest frequent term in the set is 5 ( with frequency 4 ) hence mode = 5 two terms ( 3 , 4 ) out of a total of 9 terms are less than mode of the set . fraction of set that are less than mode of set = 2 / 9 answer : option a | a ) 2 / 9 , b ) 1 / 3 , c ) 2 / 5 , d ) 2 / 3 , e ) 7 / 9 | a | divide(const_2, add(5, 4)) | add(n1,n2)|divide(const_2,#0) | general |
how many multiples of 8 are less than 4500 , and also multiples of 12 ? | "lcm of 8 & 12 = 24 tried dividing 4500 by 24 got quotient 187.5 ' so b is answer" | a ) 104.6 , b ) 187.5 , c ) 625.6 , d ) 450.5 , e ) 467.4 | b | add(divide(subtract(12, 4500), 8), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
john makes $ 60 a week from his job . he earns a raise andnow makes $ 84 a week . what is the % increase ? | "increase = ( 24 / 60 ) * 100 = ( 4 / 10 ) * 100 = 40 % . d" | a ) 16.12 % , b ) 16.66 % , c ) 16.56 % , d ) 40 % , e ) 18.1 % | d | multiply(divide(subtract(84, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
56 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | explanation : let the two parts be ( 56 - x ) and x . then , 10 ( 56 - x ) + 22 x = 780 = > 12 x = 220 = > x = 20 . bigger part = ( 56 - x ) = 37.66 . answer : b ) 37.66 | a ) 33.0 , b ) 37.66 , c ) 32.1 , d ) 31.23 , e ) 19 | b | subtract(56, divide(subtract(780, multiply(56, 10)), subtract(22, 10))) | multiply(n0,n1)|subtract(n2,n1)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3) | general |
the sides of a cube measures 8 cm . if the surface area of the cube is the same as a ball , what is the radius of the ball ? round to the nearest whole number . | first calculate the surface area of the cube . the cube is made of 6 squares with the sides measuring 8 cm each . the surface area of 1 side of the cube is 64 cm . multiply the surface area of the 1 side by 6 will give you the cube ' s total surface area . the cubes total surface area is 384 cm . now we know that the s... | a ) 3 , b ) 8 , c ) 7 , d ) 6 , e ) 5 | d | floor(sqrt(divide(divide(surface_cube(8), const_4), const_pi))) | surface_cube(n0)|divide(#0,const_4)|divide(#1,const_pi)|sqrt(#2)|floor(#3)| | geometry |
find the value for x from below equation : x / 2 = - 2 ? | "1 . multiply both sides by 2 : x * 2 / 2 = - 2 / 2 simplify both sides : x = - 4 e" | a ) - 6 , b ) 1 , c ) - 2 , d ) - 3 , e ) - 4 | e | multiply(subtract(2, const_4), 2) | subtract(n1,const_4)|multiply(#0,n0)| | general |
a man sold 18 toys for rs . 18900 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 18900 . gain = sp Γ’ β¬ β cp 3 x = 18900 Γ’ β¬ β 18 x 21 x = 18900 x = rs . 900 . answer : option d" | a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 990 | d | divide(18900, add(18, 3)) | add(n0,n2)|divide(n1,#0)| | gain |
how many of the positive factors of 75 are not factors of 95 ? | "factors of 75 - 1 , 3 , 5 , 15 , 25 , 75 , factors of 95 - 1 , 5 , 19 , 95 , comparing both , we have three factors of 75 which are not factors of 95 - 3 , 15 , 25 , 75 , the answer is e" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | divide(95, 75) | divide(n1,n0)| | other |
q is as much younger than r as he is older than t . if the sum of the ages of r and t is 50 years , what is definitely the difference between r and q β s age ? | "explanation given that : 1 . the difference of age b / w r and q = the difference of age b / w q and t . 2 . sum of age of r and t is 50 i . e . ( r + t ) = 50 . question : r β q = ? . explanation : r β q = q β t ( r + t ) = 2 q now given that , ( r + t ) = 50 so , 50 = 2 q and therefore q = 25 . question is ( r β q )... | a ) 1 year , b ) 2 years , c ) 25 years , d ) data inadequate , e ) none of these | d | divide(50, const_2) | divide(n0,const_2)| | general |
if it is assumed that 62 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 62 % = 300 / 0.62 = 483.8 option d" | a ) 400 , b ) 420 , c ) 480 , d ) 483.8 , e ) 600 | d | divide(300, divide(62, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
what is the lowest positive integer that is divisible by 1 through 2 , inclusive ? | the integer should be divisible by : 1 and 2 . the least common multiple of these integers is lcm = 1 * 2 = 2 answer : a | a ) 2 , b ) 4 , c ) 8 , d ) 6 , e ) 1 | a | lcm(1, 2) | lcm(n0,n1) | general |
the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 12 % . | "sol . let marked price = rs . 100 . then , c . p . = rs . 64 . s . p = rs . 88 . β΄ gain % = [ 24 / 64 * 100 ] % = 37.5 % . answer a" | a ) 37.5 % , b ) 38.5 % , c ) 42 % , d ) 27.5 % , e ) none | a | multiply(subtract(divide(subtract(const_100, 12), 64), const_1), const_100) | subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)| | gain |
two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 1 : 2 and 3 : 2 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | vessel a = 300 gallons - - > milk = 100 , water = 200 ; vessel b = 500 gallons - - > milk = 300 , water = 200 ; vessel a + b = 800 gallons - - > milk = 400 , water 400 . the ratio = 400 / 400 - - > 1 : 1 answer : b | ['a ) 99 : 61', 'b ) 1 : 1', 'c ) 61 : 160', 'd ) 61 : 99', 'e ) 99 : 160'] | b | power(1, 3) | power(n2,n0) | other |
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 30 % profit ? | "c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 130 % of rs . 640 = 130 / 100 * 640 = rs . 832 . d" | a ) 220 , b ) 230 , c ) 960 , d ) 832 , e ) 400 | d | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(30, const_100))) | divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)| | gain |
each year for 4 years , a farmer increased the number of trees in a certain orchard by 1 / 4 of the number of trees in the orchard of the preceding year . if all of the trees thrived and there were 1250 trees in the orchard at the end of 4 year period , how many trees were in the orchard at the beginning of the 4 year ... | trees increase by 1 / 4 the number of trees in preceding year . hence , correct answer must be divisible by 4 . based on divisibility rules , if last 2 digits are divisible by 4 then the number is divisible by 4 . thus , we can eliminate a , c , d , e the answer to be b again , trees increase by 1 / 4 the number of tre... | a ) 513 , b ) 512 , c ) 517 , d ) 519 , e ) 521 | b | divide(1250, power(add(divide(1, 4), 1), 4)) | divide(n1,n0)|add(n1,#0)|power(#1,n0)|divide(n3,#2) | general |
a river boat leaves silver town and travels upstream to gold town at an average speed of 6 kilometers per hour . it returns by the same route at an average speed of 5 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? | "pick a number which is lcm of 5 and 6 = 30 . upstream time = 30 / 6 = 5 hrs downstream time = 30 / 5 = 6 hrs total time = 11 hrs total distance = 60 average speed = 60 / 11 = 5.5 km / hr" | a ) 5.5 , b ) 7.1 , c ) 7.2 , d ) 7.0 , e ) 8.0 | a | divide(multiply(multiply(6, 5), const_2), add(6, 5)) | add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|divide(#2,#0)| | physics |
a cistern is normally filled in 10 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 10 - 1 / x = 1 / 12 x = 60 answer : c" | a ) 33 , b ) 88 , c ) 60 , d ) 99 , e ) 11 | c | inverse(subtract(divide(const_1, 10), divide(const_1, add(10, const_2)))) | add(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)| | physics |
three numbers are in the ratio 4 : 5 : 6 and their average is 42 . the largest number is : | "explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 42 15 x = 126 x = 8.4 largest number = 6 x = 50.4 . answer d" | a ) 28 , b ) 32 , c ) 36 , d ) 50.4 , e ) 45 | d | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
6 % of customers that enter a store on any given day must not pay tax . if 1000 people shop every day , how many people pay taxes in the store every week | 6 % of customers do not pay taxes . 6 / 100 * 1000 customers do not pay taxes . 6 / 100 * 1000 = 60 therefore 1000 - 60 customers do pay taxes . 1000 - 60 = 940 940 * 7 days per week = 6580 b ) | a ) 6500 , b ) 6580 , c ) 7200 , d ) 7120 , e ) 6800 | b | multiply(add(6, const_1), subtract(1000, multiply(1000, divide(6, const_100)))) | add(n0,const_1)|divide(n0,const_100)|multiply(n1,#1)|subtract(n1,#2)|multiply(#0,#3) | gain |
find the compound ratio of ( 2 : 3 ) , ( 5 : 3 ) and ( 2 : 5 ) is | "required ratio = 2 / 3 * 5 / 3 * 2 / 5 = 4 / 9 = 4 : 9 answer is d" | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 9 , e ) 3 : 2 | d | multiply(divide(2, 3), multiply(divide(2, 3), divide(5, 3))) | divide(n0,n5)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other |
1 = 5,2 = 10,3 = 253,4 = 254,5 = 255,6 = 256,7 = 257,8 = 258 then 10 = ? | "1 = 5,2 = 10,3 = 253,4 = 254,5 = 255,6 = 256,7 = 257,8 = 258 then 10 = ? 10 = 2 check the first eqn . answer : a" | a ) 2 , b ) 255 , c ) 345 , d ) 445 , e ) 235 | a | divide(subtract(subtract(255,6, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2) | add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)| | general |
p and q started a business investing rs . 85,000 and rs . 30,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 30000 = 17 : 6 . answer : a" | a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 2 , e ) 17 : 3 | 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 raz ' s company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio of revenue from car sales in... | a raz ' s company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . 1996 - x - - - - - y - - - - - - - x + y 1997 - x 1 - - - - y 1 - - x 1 + y 1 x 1 = a * x . . . . . ( p ) y 1 = 1.07 * y . . . ( q ) ... | a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(divide(add(const_100, 7), const_100), divide(add(const_100, 1), const_100)), subtract(divide(add(const_100, 1), const_100), divide(subtract(const_100, 11), const_100))) | add(n3,const_100)|add(n6,const_100)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|subtract(#3,#4)|subtract(#4,#5)|divide(#6,#7) | other |
what is the units digit of 23 ^ 3 * 17 ^ 2 * 39 ^ 2 ? | "the units digit of 23 ^ 3 is the units digit of 3 * 3 * 3 = 27 which is 7 . the units digit of 17 ^ 2 is the units digit of 7 * 7 = 49 which is 9 . the units digit of 39 ^ 2 is the units digit of 9 * 9 = 81 which is 1 . the units digit of 7 * 9 * 1 = 63 is 3 . the answer is d ." | a ) 9 , b ) 7 , c ) 5 , d ) 3 , e ) 1 | d | divide(add(multiply(factorial(23), factorial(3)), multiply(factorial(23), factorial(2))), 23) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
three 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 | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
a customer went to a shop and paid a total of $ 25 , out of which 90 cents was for sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "the total cost was $ 25 . the tax was $ 0.90 let the original price of the taxable items = x given that tax rate = 6 % 0.06 x = 0.90 x = $ 15 the cost of the tax free items was $ 25 - $ 15 - $ 0.90 = $ 9.10 the answer is c ." | a ) $ 8.50 , b ) $ 8.70 , c ) $ 9.10 , d ) $ 10.00 , e ) $ 10.40 | c | subtract(subtract(25, 90), multiply(6, const_2)) | multiply(n2,const_2)|subtract(n0,n1)|subtract(#1,#0)| | gain |
45 x ? = 60 % of 900 | "answer let 45 x a = ( 60 x 900 ) / 100 β΄ a = ( 60 x 9 ) / 45 = 12 correct option : b" | a ) 16.2 , b ) 12 , c ) 5 , d ) 500 , e ) none | b | divide(multiply(divide(60, const_100), 900), 45) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)| | general |
set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 6 , while all other numbers remain the same , what is the new average of set s ? | "old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 6 then total increased to 62 + 6 = 68 new avg - 68 / 10 = 6.8 . hence answer is c ." | a ) 6.6 , b ) 6.7 , c ) 6.8 , d ) 6.85 , e ) 6.9 | c | divide(add(multiply(10, 6.2), 6), 10) | multiply(n0,n1)|add(n2,#0)|divide(#1,n0)| | general |
a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 3 days . for how many number of days should a and c should work together to complete the remaining work ? | "b work 1 / 8 * 3 = 3 / 8 remaining work = 1 - 3 / 8 = 5 / 8 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 remaining work is done in = 8 / 5 * 1 / 4 = 2 / 5 = 0.4 answer : d" | a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | d | add(divide(3, 6), divide(3, 8)) | divide(n3,n0)|divide(n3,n1)|add(#0,#1)| | physics |
what is the greatest 6 - digit number when divided by 8 , 9 , and 10 leaves a remainder of 6 , 7 , and 8 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 8 when we divide by 10 , so b is the only possible answer . | a ) 456780 , b ) 678918 , c ) 997479 , d ) 997916 , e ) 997920 | b | subtract(subtract(multiply(const_1000, const_1000), const_1), multiply(multiply(lcm(lcm(8, 9), lcm(10, 9)), lcm(10, 9)), 10)) | lcm(n1,n2)|lcm(n2,n3)|multiply(const_1000,const_1000)|lcm(#0,#1)|subtract(#2,const_1)|multiply(#3,#1)|multiply(n3,#5)|subtract(#4,#6) | general |
a thief is spotted by a policeman from a distance of 160 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 160 m / 1000 ) x 1 / 2 hr = 2 / 25 hr . in 2 / 25 hr / s , the thief covers a distance of 8 x 2 / 25 km = 16 / 25 km = 640 m answer is d ." | a ) 350 m , b ) 240 m , c ) 440 m , d ) 640 m , e ) none of them | d | divide(multiply(160, 8), subtract(10, 8)) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)| | physics |
the average of runs of a cricket player of 20 innings was 36 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average = total runs / no . of innings = 36 so , total = average x no . of innings = 36 * 20 = 720 now increase in avg = 4 runs . so , new avg = 36 + 4 = 40 runs total runs = new avg x new no . of innings = 40 * 21 = 840 runs made in the 11 th inning = 840 - 720 = 120 answer : e" | a ) 96 , b ) 106 , c ) 122 , d ) 116 , e ) 120 | e | subtract(multiply(add(20, const_1), add(4, 36)), multiply(20, 36)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the product of a and b is equal to 14 more than twice the sum of a and b . if b = 8 , what is the value of b - a ? | "ab = 14 + 2 ( a + b ) 8 a = 14 + 2 a + 16 6 a = 30 a = 5 b - a = 8 - 5 = 3 b is the answer" | a ) 2 , b ) 3 , c ) 7 , d ) 24 , e ) 35 | b | subtract(8, divide(add(multiply(8, const_2), 14), subtract(8, const_2))) | multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3)| | general |
excluding stoppages , the speed of a train is 42 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | t = 6 / 42 * 60 = 8.5 answer : b | a ) 7.5 , b ) 8.5 , c ) 9.5 , d ) 4.5 , e ) 6.5 | b | subtract(const_60, multiply(const_60, divide(36, 42))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1) | physics |
a can do a piece of work in 4 hours . a and c together can do it in just 2 hours , while b and c together need 3 hours to finish the same work . in how many hours b can complete the work ? | "explanation : work done by a in 1 hour = 1 / 4 work done by b and c in 1 hour = 1 / 3 work done by a and c in 1 hour = 1 / 2 work done by a , b and c in 1 hour = ( 1 / 4 ) + ( 1 / 3 ) = 7 / 12 work done by b in 1 hour = ( 7 / 12 ) β ( 1 / 2 ) = 1 / 12 = > b alone can complete the work in 12 hour option b" | a ) 10 hours , b ) 12 hours , c ) 16 hours , d ) 18 hours , e ) 20 hours | b | inverse(subtract(inverse(3), subtract(inverse(2), inverse(4)))) | inverse(n2)|inverse(n1)|inverse(n0)|subtract(#1,#2)|subtract(#0,#3)|inverse(#4)| | physics |
a man can row downstream at the rate of 32 kmph and upstream at 17 kmph . find the man β s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 32 + 17 ) = 24.5 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 32 - 17 ) = 1 / 2 ( 15 ) = 7.5 kmph answer is a ." | a ) 7.5 , b ) 8.0 , c ) 9.5 , d ) 9.0 , e ) 8.25 | a | divide(subtract(32, 17), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 15 minutes , a combustion reaction creates 15,000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | "solution : 15,000 carbon dioxide molecules are created over a period of 15 minutes . therefore 15,000 / 15 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 Γ 1,000 = 1,000 carbon a... | a ) 1,000 , b ) 500 , c ) 250 , d ) 50 , e ) 0 | a | divide(divide(multiply(subtract(multiply(15, const_2), 15), const_1000), 15), const_1000) | multiply(n0,const_2)|subtract(#0,n0)|multiply(#1,const_1000)|divide(#2,n0)|divide(#3,const_1000)| | general |
a train speeds past a pole in 15 seconds and a platform 100 m long in 20 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 20 = x / 15 x = 300 m . answer : d" | a ) 188 m , b ) 876 m , c ) 251 m , d ) 300 m , e ) 145 m | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
the smallest number when increased by ` ` 1 ` ` is exactly divisible by 618 , 3648 , 60 is : | lcm = 720 720 - 1 = 719 answer : d | a ) 724 , b ) 721 , c ) 720 , d ) 719 , e ) 700 | d | add(const_100, add(618, 1)) | add(n0,n1)|add(#0,const_100) | general |
a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days g will the work get over ? | "we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 there... | a ) 60 , b ) 70 , c ) g = 75 , d ) g = 80 , e ) 100 | c | divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2)) | multiply(n1,n2)|subtract(const_1,const_0_25)|subtract(n1,n3)|divide(#0,const_0_25)|multiply(#3,#1)|divide(#4,#2)| | physics |
having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.96 at the candy store . what is john β s weekly allowance ? | "x = 3 x / 5 + 1 / 3 * 2 x / 5 + 96 4 x / 15 = 96 x = $ 3.60 = $ 3.60 the answer is d ." | a ) $ 3.00 , b ) $ 3.20 , c ) $ 3.40 , d ) $ 3.60 , e ) $ 3.80 | d | divide(0.96, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5)))))) | divide(n0,n1)|divide(const_1,n0)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(const_1,#4)|divide(n2,#5)| | general |
how many prime numbers are between 78 / 14 and 55 / 4 ? | "78 / 14 = 39 / 7 = 6 - 55 / 4 = 14 - prime numbers between 6 and 14 are 7 , 11 and 13 - sign signifies that the number is marginally less . answer b" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | floor(const_2) | floor(const_2)| | general |
a person can swim in still water at 12 km / h . if the speed of water 10 km / h , how many hours will the man take to swim back against the current for 12 km ? | "m = 12 s = 10 us = 12 - 10 = 2 d = 12 t = 12 / 2 = 6 answer : d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(12, subtract(12, 10)) | subtract(n0,n1)|divide(n2,#0)| | physics |
if 15 % of a class averages 100 % on a test , 50 % of the class averages 78 % on the test , and the remainder of the class averages 63 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) . | "this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 15 % - 50 % = 35 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.15 x 100 + 0.50 x 78 + 0.35 x 63 = 76.05 the class a... | a ) 76 % , b ) 77 % , c ) 78 % , d ) 79 % , e ) 80 % | a | divide(add(add(multiply(15, 100), multiply(50, 78)), multiply(subtract(const_100, add(15, 50)), 63)), const_100) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)| | gain |
what is x if x + 3 y = 10 and y = 3 ? | "substitute y by 3 in x + 3 y = 10 x + 3 ( 3 ) = 10 x + 9 = 10 if we substitute x by 1 in x + 9 = 10 , we have 1 + 9 = 10 . hence x = 1 correct answer a" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(10, multiply(3, 3)) | multiply(n0,n2)|subtract(n1,#0)| | general |
arun borrowed a sum of money from jayant at the rate of 8 % per annum simple interest for the first 4 years , 10 % per annum for the next 6 years and 12 % per annum for the period beyond 10 years . if he pays a total of 12160 as interest only at the end of 15 years , how much money did he borrow ? | let the principal = p then p Γ 8 Γ 4 / 100 + p Γ 10 Γ 6 / 100 + p Γ 12 Γ 5 / 100 = 12160 β 152 p = 12160 Γ 100 or 12160 Γ 100 / 152 = 8000 answer a | a ) 8000 , b ) 10000 , c ) 12000 , d ) 9000 , e ) none of these | a | divide(multiply(12160, const_100), add(add(multiply(8, 4), multiply(10, 6)), multiply(12, subtract(15, 10)))) | multiply(n6,const_100)|multiply(n0,n1)|multiply(n2,n3)|subtract(n7,n2)|add(#1,#2)|multiply(n4,#3)|add(#4,#5)|divide(#0,#6) | general |
what is the largest 4 digit number exactly divisible by 88 ? | "largest 4 digit number = 9999 9999 Γ· 88 = 113 , remainder = 55 hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944 answer is b" | a ) 9999 , b ) 9944 , c ) 9988 , d ) 9900 , e ) 9991 | b | multiply(floor(divide(power(const_10, 4), 88)), 88) | power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)| | general |
a women travels from x to y distance of 1000 miles in 10 hours . she returns to x in 4 hours . find her average speed | speed from x to y = 1000 / 10 = 100 mph speed from y to x = 1000 / 4 = 250 mph average speed = 2 * 100 * 250 / 350 = 142.8 mph answer is c | a ) 300 mph , b ) 150.2 mph , c ) 142.8 mph , d ) 200 mph , e ) 111.01 mph | c | divide(1000, divide(add(10, 4), const_2)) | add(n1,n2)|divide(#0,const_2)|divide(n0,#1) | physics |
in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 45 , what is the greatest possible value of b Γ a ? | "you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 3 = 15 ( next number wo n ' t have 110 as the first 3 digit... | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 25 | d | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 45), const_1), 45), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 45), const_1), 45), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000),... | power(n0,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,n0)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)| | general |
a vendor sells 20 percent of the pears he had and throws away 50 percent of the remainder . the next day , the vendor sells 20 percent of the remaining pears and throws away the rest . in total , what percent of his pears does the vendor throw away ? | let x be the original number of pears . on day one , the vendor throws away ( 0.5 ) ( 0.8 ) x = 0.4 x . the remaining pears are ( 0.5 ) ( 0.8 ) x = 0.4 x . on day two , the vendor throws away ( 0.8 ) ( 0.4 ) x = 0.32 x . the vendor throws away a total of 0.4 x + 0.32 x = 0.72 x . the vendor throws away 72 percent of th... | a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72 | e | multiply(const_100, add(subtract(subtract(subtract(const_1, divide(20, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(20, const_100)))), multiply(subtract(subtract(const_1, divide(20, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(20, const_100)))), divide(20, const_100))),... | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#2,#3)|multiply(#1,#4)|subtract(#4,#5)|add(#3,#6)|multiply(#7,const_100) | general |
john want to buy a $ 100 trouser at the store , but he think it Γ’ β¬ β’ s too expensive . finally , it goes on sale for $ 80 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it Γ’ β¬ β’ s 100 Γ’ β¬ β 80 = 20 . the Γ’ β¬ Ε original Γ’ β¬ Β is our starting point ; in this case , it Γ’ β¬ β’ s 100 . ( 20 / 100 ) * 100 = ( 0.2 ) * 100 = 20 % . a" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | a | subtract(100, 80) | subtract(n0,n1)| | general |
a number is doubled and 9 is added . if the resultant is trebled , it becomes 75 . what is that number ? | "explanation : let the number be v . then , 3 ( 2 x + 9 ) = 75 . 2 x + 9 = 25 = > 2 x = 16 = > x = 8 answer : d ) 8" | a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 9 | d | divide(subtract(75, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
the value of 489.1375 x 0.0483 x 1.956 / 0.0873 x 92.581 x 99.749 is closest to : | = 489.1375 x 0.0483 x 1.956 / 0.0873 x 92.581 x 99.749 = 489 x 0.05 x 2 / 0.09 x 93 x 100 = 489 / 9 x 93 x 10 = 163 / 279 x 1 / 10 = 0.58 / 10 = 0.058 ( or ) 0.06 . answer is d . | a ) 0.48 , b ) 5.8 , c ) 0.58 , d ) 0.058 , e ) 0.0058 | d | divide(multiply(multiply(489.1375, 0.0483), 1.956), multiply(multiply(0.0873, 92.581), 99.749)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3) | general |
an uneducated retailer marks all his goods at 50 % above the cost price and thinking that he will still make 15 % profit , offers a discount of 15 % on the marked price . what is his actual profit on the sales ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 150 . s . p . = 85 % of rs . 150 = rs . 127.5 . β΄ gain % = 27.50 % . answer b" | a ) 12.50 % , b ) 27.50 % , c ) 14 % , d ) 14.50 % , e ) none | b | multiply(subtract(subtract(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(50, const_100)), divide(15, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
on teacher ' s day , sweets were to be equally distributed among 190 children in a school . actually , on the teacher ' s day , 70 children were absent and therefore each child got 14 sweets extra . how many sweets did each child get on that day ? | total children = 190 . let each student gets x chocolates if all were present . so , total chocolates = 190 x . 70 students were absent . so , no of extra chocolates distributed among 120 students , = 14 * 70 = 980 . now , total chocolates = 190 x 120 * ( x + 14 ) = 190 120 x + 1680 = 190 x 70 x = 1680 x = 24 . total n... | a ) 18 , b ) 22 , c ) 28 , d ) 32 , e ) 38 | e | divide(multiply(190, 14), 70) | multiply(n0,n2)|divide(#0,n1) | general |
x , a , z , and b are single digit positive integers . x = ΒΌ a . z = ΒΌ b . ( 10 a + b ) β ( 10 x + z ) could not equal | from the given equations , a = 4 x , b = 4 z ( 10 a + b ) - ( 10 x + z ) = 40 x + 4 z - 10 x - z = 30 x + 3 z = 3 ( 10 x + z ) therefore the resultant should be divisible by 3 which is n ' t possible only in the case of c . hence c is the answer | a ) 33 , b ) 36 , c ) 44 , d ) 63 , e ) 66 | c | add(multiply(const_4, 10), const_4) | multiply(n0,const_4)|add(#0,const_4) | general |
in what time will a train 140 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 140 / 40 = 3.5 sec . answer : a" | a ) 3.5 sec , b ) 4.25 sec , c ) 5 sec , d ) 12.5 sec , e ) 6 sec | a | divide(140, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 1.5 * 3 ) / 100 = > 157.50 answer : c" | a ) 157.20 , b ) 157.29 , c ) 157.50 , d ) 157.30 , e ) 157.23 | c | subtract(divide(multiply(multiply(3500, 11.5), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)| | gain |
if a : b is 2 : 3 and b : c is 3 : 4 then a : c is equal to | "since a : b = 2 : 3 and b : c = 3 : 4 , we can deduce that a : b : c = 2 : 3 : 4 so , a : c = 2 : 4 = 1 : 2 . d is the correct answer ." | a ) 2 : 3 , b ) 6 : 7 , c ) 5 : 6 , d ) 1 : 2 , e ) 3 : 4 | d | multiply(divide(3, 4), divide(2, 3)) | divide(n1,n3)|divide(n0,n1)|multiply(#0,#1)| | general |
carl can wash all the windows of his house in 6 hours . his wife maggie can wash all the windows in 5 hours . how many hours will it take for both of them working together to wash all the windows ? | "work hrs = ab / ( a + b ) = 30 / 11 = 11 8 / 2 answer is c" | a ) 2 , b ) 2 1 / 4 , c ) 11 8 / 2 , d ) 4 1 / 2 , e ) 5 | c | multiply(divide(const_60, add(divide(const_1, 6), divide(const_1, 5))), add(divide(const_1, 6), divide(const_1, 5))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_60,#2)|multiply(#2,#3)| | physics |
on a two - dimensional coordinate plane , the line z = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | apparently it ' s z = x ^ 2 - x ^ 3 instead of z = x ^ 2 - z ^ 3 . in this case : the x - intercept is the value ( s ) of x for z = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c . | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | max(n0,n1)|subtract(#0,const_1) | general |
find the value of 72514 x 9999 = m ? | "72514 x 9999 = 72514 x ( 10000 - 1 ) = 72514 x 10000 - 72514 x 1 = 725140000 - 72514 = 725067486 a" | a ) 725067486 , b ) 436567874 , c ) 653658791 , d ) 725117481 , e ) 357889964 | a | multiply(subtract(9999, const_4), 72514) | subtract(n1,const_4)|multiply(#0,n0)| | general |
a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is 9 feet , and the sum of the lengths of the painted sides is 37 feet , find out the area of the parking space in square feet ? | "length = 9 feet breadth = 37 β 92 = 14 feet area = 9 Γ 14 = 126 square feet answer is e ." | a ) 120 , b ) 122 , c ) 124 , d ) 128 , e ) 126 | e | add(add(add(add(37, 37), 37), const_10), const_4) | add(n1,n1)|add(n1,#0)|add(#1,const_10)|add(#2,const_4)| | geometry |
a man ' s speed with the current is 20 kmph and speed of the current is 1 kmph . the man ' s speed against the current will be | explanation : speed with current is 20 , speed of the man + it is speed of the current speed in s Ι΅ ll water = 20 - 1 = 19 now speed against the current will be speed of the man - speed of the current = 19 - 1 = 18 kmph answer : c | a ) 11 kmph , b ) 12 kmph , c ) 18 kmph , d ) 17 kmph , e ) none of these | c | subtract(subtract(20, 1), 1) | subtract(n0,n1)|subtract(#0,n1) | physics |
if | x - 12 | = 100 , what is the sum of all the possible values of x ? | "there will be two cases x - 12 = 100 or x - 12 = - 100 = > x = 112 or x = - 88 sum of both the values will be - 88 + 112 = 24 answer is c" | a ) - 12 , b ) - 22 , c ) 24 , d ) 36 , e ) 42 | c | subtract(add(100, 12), subtract(100, 12)) | add(n0,n1)|subtract(n1,n0)|subtract(#0,#1)| | general |
calculate the l . c . m of 4 / 9 , 5 / 7 , 9 / 13 , 7 / 15 is : | "required l . c . m = l . c . m . of 4 , 5 , 9 , 7 / h . c . f . of 9 , 7 , 13 , 15 = 1260 / 1 = 1260 answer is c" | a ) 1230 , b ) 1290 , c ) 1260 , d ) 1240 , e ) 2260 | c | multiply(multiply(4, 5), multiply(9, 7)) | multiply(n0,n2)|multiply(n4,n6)|multiply(#0,#1)| | physics |
kanul spent $ 3000 in buying raw materials , $ 2000 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | let the total amount be x then , ( 100 - 10 ) % of x = 3000 + 2000 90 % of x = 5000 90 x / 100 = 5000 x = $ 50000 / 9 x = $ 5555.555 answer is a | a ) $ 5555.555 , b ) $ 5543.22 , c ) $ 5568.000 , d ) $ 5366.521 , e ) $ 5365.66 | a | divide(add(3000, 2000), subtract(const_1, divide(10, const_100))) | add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2) | gain |
the sum of all solutions for x in the equation x ^ 2 β 8 x + 21 = | x β 5 | + 4 is equal to : | x ^ 2 - 8 x + 17 = | x - 5 | rhs can be - ve or + ve x ^ 2 - 9 x + 22 = 0 x ^ 2 - 7 x + 12 = 0 x = 11,4 , 3,2 we test all 3 values in original equation , all ok . thus , sum = 11 + 4 + 3 + 2 = 20 ans ( c ) | a ) β 7 , b ) 7 , c ) 20 , d ) 12 , e ) 14 | c | multiply(4, 5) | multiply(n3,n4) | general |
1.20 can be expressed in terms of percentage as | "explanation : while calculation in terms of percentage we need to multiply by 100 , so 1.20 * 100 = 120 answer : option a" | a ) 120 % , b ) 0.120 % , c ) 1.20 % , d ) 0.209 % , e ) none of these | a | multiply(1.20, const_100) | multiply(n0,const_100)| | general |
if the radius of a cylinder is doubled and height 5 times , what is the new volume of the cylinder divided by the old one ? | "let v and v ' be the original and the changed volume now v = pir ^ 2 h v ' = pi ( 2 r ) ^ 2 ( 5 h ) v ' = 20 v d ) 20" | a ) 8 . , b ) 2 . , c ) 6 . , d ) 20 . , e ) 10 . | d | divide(volume_cylinder(multiply(const_1, const_2), multiply(const_1, const_2)), volume_cylinder(const_1, const_1)) | multiply(const_1,const_2)|volume_cylinder(const_1,const_1)|volume_cylinder(#0,#0)|divide(#2,#1)| | general |
( x ) + 4671 + 6514 - 7687 = 19190 . calculate the value of x | "x + 4671 + 6514 - 7687 = 19190 = x + 4671 + 6514 = 19190 + 7687 = x + 11185 = 26877 = x = 26877 - 11185 = 15692 answer is b" | a ) 15615 , b ) 15692 , c ) 15687 , d ) 15112 , e ) 15690 | b | multiply(subtract(subtract(add(19190, 7687), 6514), 4671), divide(const_60, const_2)) | add(n2,n3)|divide(const_60,const_2)|subtract(#0,n1)|subtract(#2,n0)|multiply(#1,#3)| | general |
in some quantity of ghee , 60 % is pure ghee and 40 % is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was ? | let the original quantity be x then , vanaspati ghee in xkg = 40 x / 100 kg = 2 x / 5 kg ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 2 x / ( 5 x + 50 ) = 1 / 5 x = 10 answer is a | a ) 10 , b ) 15 , c ) 20 , d ) 18 , e ) 22 | a | divide(multiply(20, 10), subtract(40, 20)) | multiply(n2,n3)|subtract(n1,n3)|divide(#0,#1) | gain |
a motorist travels for 6 hours , the first half at 60 kmph and the rest at 48 kmph . find the distance traveled by him . | distance = 3 * 60 + 3 * 48 = 180 + 144 = 324 km answer c . | a ) 234 , b ) 124 , c ) 324 , d ) 452 , e ) 352 | c | add(multiply(60, divide(6, const_2)), multiply(48, divide(6, const_2))) | divide(n0,const_2)|multiply(n1,#0)|multiply(n2,#0)|add(#1,#2) | physics |
in measuring the sides of a rectangle , one side is taken 5 % in excess , and the other 4 % in deficit . find the error percent in the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 105 / 100 * x ) * ( 96 / 100 * y ) = 504 / 500 * xy . error in measurement = 504 / 500 * xy ) - xy = 4 / 500 * xy . error % = [ 4 / 500 * xy * 1 / xy * 100 ] % = 4 / 5 = 0.8 % . answer is a" | a ) 0.8 % , b ) 0.2 % , c ) 0.4 % , d ) 0.9 % , e ) 0.6 % | a | subtract(subtract(5, 4), divide(multiply(5, 4), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
in a group of ducks and cows , the total number of legs are 24 more than twice the number of heads . find the total number of cows . | "let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 24 more than twice the number of heads = > 2 ( d + 2 c ) = 24 + 2 ( c + d ) = > d + 2 c = 12 + c + d = > 2 c = 12 + c = > c = 12 i . e . , total... | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | a | divide(24, const_2) | divide(n0,const_2)| | general |
a can do a piece of work in 6 days . b can do it in 5 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 6 - 1 / 5 = 2 / 15 = > 7.5 days answer : d" | a ) 8.7 days , b ) 2.0 days , c ) 6.6 days , d ) 7.5 days , e ) 4.4 days | d | divide(multiply(6, 5), divide(subtract(multiply(6, 5), multiply(add(divide(multiply(6, 5), 6), divide(multiply(6, 5), 5)), 2)), 2)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics |
if 250 ! / 10 ^ n is an integer , what is the largest possible value of n ? | "the question actually asks the highest power of 10 which divides 250 ! ( for a number to be an integer - without any remainder all the trailing zeroe ' s must be divided by the denominator ) 10 = 2 x 5 250 factorial will have 62 as - 250 / 5 = 50 50 / 5 = 10 10 / 5 = 2 so answer will be ( c ) 62" | a ) 65 , b ) 66 , c ) 62 , d ) 69 , e ) 97 | c | add(divide(divide(250, add(const_4, const_1)), add(const_4, const_1)), divide(250, add(const_4, const_1))) | add(const_1,const_4)|divide(n0,#0)|divide(#1,#0)|add(#2,#1)| | general |
a machine , working at a constant rate , manufactures 72 staplers in 28 minutes . how many staplers does it make in 1 hr 52 min ? | "change 1 hr 52 min to 112 min . for this , we need to set up a simple proportion of staplers per time 72 / 28 = s / 112 . the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of 28 in the two denominat... | a ) 268 , b ) 262 , c ) 300 , d ) 288 , e ) 250 | d | multiply(divide(add(multiply(1, const_60), 52), 28), 72) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)| | physics |
wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 1 / 8 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability... | "p ( 1 and not 2 ) = 1 / 8 * ( 1 - 3 / 5 ) = 1 / 20 . answer : a ." | a ) 1 / 20 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22 | a | multiply(divide(1, 8), subtract(1, divide(3, 1))) | divide(n0,n1)|divide(n3,n0)|subtract(n2,#1)|multiply(#0,#2)| | probability |
when a = x + ( 2 / x ) and b = x - ( 2 / x ) , ( 2 ^ a ^ 2 ) / ( 2 ^ b ^ 2 ) = ? | - - > ( 2 ^ a ^ 2 ) / ( 2 ^ b ^ 2 ) = { ( 2 ) ^ ( a ^ 2 - b ^ 2 ) } = 2 ^ ( a - b ) ( a + b ) . since a - b = 4 / x and a + b = 2 x , 2 ^ ( a - b ) ( a + b ) = 2 ^ ( 4 / x ) ( 2 x ) = 2 ^ 8 = 256 answer is b | a ) 2 , b ) 256 , c ) 8 , d ) 16 , e ) 32 | b | power(2, add(multiply(const_2, 2), multiply(const_2, 2))) | multiply(n0,const_2)|add(#0,#0)|power(n0,#1) | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.