Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 42 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 42 + q = > q = 12 answer : c" | a ) 22 , b ) 18 , c ) 12 , d ) 88 , e ) 66 | c | subtract(42, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0)| | general |
what is the sum of the squares of the first 20 natural numbers ( 1 to 20 ) ? | n ( n + 1 ) ( 2 n + 1 ) / 6 20 ( 21 ) ( 21 ) / 6 = 2870 answer : a | ['a ) 2870', 'b ) 2000', 'c ) 5650', 'd ) 6650', 'e ) 7650'] | a | divide(multiply(multiply(20, add(20, 1)), add(multiply(const_2, 20), 1)), multiply(const_2, const_3)) | add(n0,n1)|multiply(n0,const_2)|multiply(const_2,const_3)|add(n1,#1)|multiply(n0,#0)|multiply(#3,#4)|divide(#5,#2) | geometry |
the average of 9 observations was 7 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 7 = 63 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 β 63 = 27 option a" | a ) 27 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(7, 5)), multiply(7, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
the guests at a football banquet consumed a total of 319 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 159 * 2 ( max possible amount of food a guest could consume ) = 318 pounds , so there must be more than 159 guests , next integer is 160 . answer : a ." | a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | a | add(floor(divide(319, 2)), const_1) | divide(n0,n1)|floor(#0)|add(#1,const_1)| | general |
8 persons can build a wall 140 m long in 42 days . in how many days can 30 persons complete a similar wall 100 m long ? | explanation : more persons , less days ( indirect proportion ) more length of the wall , more days ( direct proportion ) β 8 Γ 100 Γ 42 = 30 Γ 140 Γ x β x = ( 8 Γ 100 Γ 42 ) / ( 30 Γ 140 ) = ( 8 Γ 100 Γ 14 ) / ( 10 Γ 140 ) = ( 8 Γ 100 ) / ( 10 Γ 10 ) = 8 . answer : option c | a ) 12 , b ) 10 , c ) 8 , d ) 6 , e ) 5 | c | divide(100, multiply(30, divide(140, multiply(8, 42)))) | multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2) | physics |
what is the prime factors β number of 48 ? | prime factors β number , as i assume , for a number x = a ^ n * b ^ m * c ^ o * d ^ p . . . is = n + m + o + p . . . so , 24 = 2 ^ 4 * 3 ^ 1 prime factors β number will be 4 + 1 = 5 . hence , answer is d . | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(add(add(const_1, add(const_1, const_1)), const_1), const_1) | add(const_1,const_1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1) | other |
in a group of cows and hens , the number of legs are 20 more than twice the number of heads . the number of cows is | "explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 20 4 x + 2 y = 2 x + 2 y + 20 2 x = 20 x = 10 answer : c" | a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | c | subtract(20, const_4) | subtract(n0,const_4)| | general |
the compound interest on a sum of money for 2 years is rs . 832 and the simple interest on the same sum for the same period is rs . 800 . the difference between the compound interest and simple interest for 3 years | explanation : difference in c . i and s . i in 2 years = rs . 32 s . i for 1 year = rs . 400 s . i for rs . 400 for one year = rs . 32 rate = [ 100 * 32 ) / ( 400 * 1 ) % = 8 % difference between in c . i and s . i for 3 rd year = s . i on rs . 832 = rs . ( 832 * 8 * 1 ) / 100 = rs . 66.56 answer : c ) rs . 98.56 | a ) 98.5 , b ) 08.56 , c ) 98.56 , d ) 98.86 , e ) 98.46 | c | add(divide(multiply(832, divide(multiply(subtract(subtract(832, divide(800, 2)), divide(800, 2)), const_100), divide(800, 2))), const_100), subtract(subtract(832, divide(800, 2)), divide(800, 2))) | divide(n2,n0)|subtract(n1,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)|multiply(n1,#4)|divide(#5,const_100)|add(#6,#2) | general |
if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n . what is the value of n ? | "if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n what is the value of n 4 ^ 11 = ( 2 ^ 2 ) ^ 11 = 2 ^ 22 10 ^ n = ( 5 x 2 ) ^ n = 5 ^ n x 2 ^ n 5 ^ 21 x 2 ^ 22 = 2 x 2 ^ n x 5 ^ n = 2 ^ n + 1 x 5 ^ n = 2 ^ 22 x 5 ^ 21 n = 21 option b" | a ) 11 , b ) 21 , c ) 22 , d ) 23 , e ) 32 | b | divide(log(multiply(power(4, 11), power(5, 21))), log(2)) | log(n4)|power(n2,n3)|power(n0,n1)|multiply(#1,#2)|log(#3)|divide(#4,#0)| | general |
from below option 51 is divisible by which one ? | "51 / 3 = 17 d" | a ) a ) 4 , b ) b ) 5 , c ) c ) 6 , d ) d ) 3 , e ) e ) 7 | d | sqrt(51) | sqrt(n0)| | general |
each child has 2 pencils and 13 skittles . if there are 11 children , how many pencils are there in total ? | 2 * 11 = 22 . answer is d . | a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | d | multiply(2, 11) | multiply(n0,n2)| | general |
. a car covers a distance of 390 km in 4 hours . find its speed ? | 390 / 4 = 98 kmph answer : d | a ) 104 , b ) 55 , c ) 66 , d ) 98 , e ) 100 | d | divide(390, 4) | divide(n0,n1) | physics |
find the sum of all 3 digit natural numbers , which are divisible by 8 | "the three digit natural numbers divisible by 8 are 104 , 112 , 120 , β¦ . 992 . let sndenote their sum . that is , sn = 104 112 120 128 , 992 g + + + + + . now , the sequence 104 , 112 , 120 , g , 992 forms an a . p . a = 104 , d = 8 , l = 992 n = l - a / d n = 112 s 112 = n / 2 ( a + l ) = 61376 answer a 61376" | a ) 61376 , b ) 54411 , c ) 612314 , d ) 64170 , e ) 64171 | a | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 8)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 8), add(const_100, const_4)), 8), const_1), const_2)) | add(const_100,const_4)|multiply(const_10,const_100)|subtract(#1,n1)|add(#0,#2)|subtract(#2,#0)|divide(#4,n1)|add(#5,const_1)|divide(#6,const_2)|multiply(#3,#7)| | general |
a man buys 25 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 25 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 25 lts is 5 . so water = 25 - 5 = 20 lts . 30 % of 25 lts = 7.5 . so water in 2 nd mixture = 25 - 7.5 = 17.5 lts . now total quantity = 25 + 25 = 50 lts . total water in it will be 20 + 17.5 = 37.5 lts . % of water = ( 100 * 37.5 ) / 50 = 75 . answer : d" | a ) 55 , b ) 82 , c ) 73 , d ) 75 , e ) 85 | d | divide(add(multiply(25, subtract(25, 20)), multiply(25, subtract(25, 30))), add(25, 25)) | add(n0,n2)|subtract(n0,n1)|subtract(n0,n3)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)| | gain |
find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm . | "given : length = 17 cm , breadth = 13 cm perimeter of rectangle = 2 ( length + breadth ) = 2 ( 17 + 13 ) cm = 2 Γ 30 cm = 60 cm we know that the area of rectangle = length Γ breadth = ( 17 Γ 13 ) cm 22 = 221 cm 2 answer : a" | a ) 221 cm 2 , b ) 211 cm 2 , c ) 231 cm 2 , d ) 236 cm 2 , e ) 241 cm 2 | a | square_area(17) | square_area(n0)| | geometry |
john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 45 * 2 + 3 * 55 = 255 miles . answer c" | a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | c | add(multiply(45, 2), multiply(3, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | physics |
a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 555 in all . for how many days was he absent ? | "30 * 25 = 750 455 - - - - - - - - - - - 195 25 + 7.50 = 32.5 195 / 32.5 = 6 d" | a ) 8 , b ) 10 , c ) 15 , d ) 6 , e ) 19 | d | subtract(30, divide(add(multiply(7.50, 30), 555), add(7.50, 25))) | add(n1,n2)|multiply(n0,n2)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3)| | physics |
a does 4 / 5 th of a work in 20 days . he then calls in b and they together finish the remaining work in 3 days . how long b alone would take to do the whole work ? | explanation : a can finish the whole work in 20 Γ 5 / 4 days = 25 days a and b together finish the whole work in 5 Γ 3 days = 15 days therefore , b can finish the whole work in 25 b / 25 + b = 15 25 b = 15 ( 25 + b ) = 375 + 15 b 10 b = 375 and b = 375 / 10 = 37 Β½ days . answer : option c | a ) 23 days , b ) 37 days , c ) 37 Β½ days , d ) 40 days , e ) 41 days | c | inverse(subtract(inverse(multiply(3, 5)), inverse(multiply(divide(5, 4), 20)))) | divide(n1,n0)|multiply(n1,n3)|inverse(#1)|multiply(n2,#0)|inverse(#3)|subtract(#2,#4)|inverse(#5) | physics |
a and b can do a work in 12 days and 36 days respectively . if they work on alternate days beginning with b , in how many days will the work be completed ? | the work done in the first two days = 1 / 12 + 1 / 36 = 1 / 9 so , 9 such two days are required to finish the work . i . e . , 18 days are required to finish the work . answer : e | a ) 11 , b ) 17 , c ) 10 , d ) 17 , e ) 18 | e | subtract(add(inverse(add(inverse(36), inverse(12))), 12), const_3) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|add(n0,#3)|subtract(#4,const_3) | physics |
a train , 140 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 140 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 140 + x ) / 30 m / s = > 45 Γ£ β ( 10 / 36 ) = ( 140 + x ) / 30 = > 45 Γ£ β 10 Γ£ β 30 / 36 = 140 + x = > 45 Γ£ β 10 Γ£ β 10 / 12 = 140 + x = > 15 Γ£ β ... | a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | c | subtract(multiply(multiply(45, const_0_2778), 30), 140) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
if 12 : 18 : : x : 9 , then find the value of x | explanation : treat 12 : 18 as 12 / 18 and x : 9 as x / 9 , treat : : as = so we get 12 / 18 = x / 9 = > 18 x = 108 = > x = 6 option c | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | divide(add(multiply(18, 12), 18), 9) | multiply(n0,n1)|add(n1,#0)|divide(#1,n2)| | general |
an error 2 % in excess is made while measuring the side of asquare . the % of error in the calculated area of the square is ? | "100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 102 ) 2 - ( 100 ) 2 ] = ( 102 + 100 ) x ( 102 - 100 ) = 404 cm 2 . percentage error = 404 x 100 % = 4.04 % 100 x 100 b" | a ) 4.00 % , b ) 4.04 % , c ) 4.26 % , d ) 4.56 % , e ) 4.67 % | b | divide(multiply(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | geometry |
a can do a piece of work in 30 days ; b can do the same in 30 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 / 30 + 10 / 30 + 10 / x = 1 x = 30 days answer : a | a ) 30 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days | a | divide(10, subtract(const_1, divide(add(10, 10), 30))) | add(n2,n2)|divide(#0,n0)|subtract(const_1,#1)|divide(n2,#2) | physics |
the distance between towns a and b is 300 km . one train departs from town a and another train departs from town b , both leaving at the same moment of time and heading towards each other . we know that one of them is 10 km / hr faster than the other . find the speeds of both trains if 2 hours after their departure the... | let the speed of the slower train be xx km / hr . then the speed of the faster train is ( x + 10 ) ( x + 10 ) km / hr . in 2 hours they cover 2 x 2 x km and 2 ( x + 10 ) 2 ( x + 10 ) km , respectively . therefore if they did n ' t meet yet , the whole distance from a to b is 2 x + 2 ( x + 10 ) + 40 = 4 x + 602 x + 2 ( ... | a ) 22 , b ) 27 , c ) 236 , d ) 90 , e ) 81 | d | add(subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10), subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10)) | add(n2,n2)|subtract(n0,n3)|divide(#1,#0)|subtract(#2,n1)|subtract(#3,n1)|add(#4,#4) | physics |
tom read a book containing 480 pages by reading the same number of pages each day . if he would have finished the book 2 days earlier by reading 16 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p + 16 ) ( d - 2 ) = 480 = > pd - 2 p + 16 d - 32 = 480 as the 1 ) stated u can put 1 ) into 2 ) = > 480 - 2 p + 16 d - 32 = 480 = > 16 d - 2 p = 32 put the bold one ... | a ) 12 , b ) 15 , c ) 10 , d ) 20 , e ) 14 | a | divide(subtract(sqrt(add(multiply(multiply(16, 2), const_4), power(16, const_2))), 16), const_2) | multiply(n1,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)| | general |
a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 00000 and the monthly manufacturing expenses is rs 1 , 50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over... | explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 69 / 70 ) Γ’ Λ β 1500... | a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | d | divide(subtract(75000, 50000), add(const_2, const_3)) | add(const_2,const_3)|subtract(n5,n4)|divide(#1,#0) | general |
a fruit seller sells mangoes at the rate of rs . 14 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 15 % ? | "solution 85 : 14 = 115 : x x = ( 14 Γ£ β 115 / 85 ) = rs . 18.94 hence , s . p per kg = rs . 18.94 answer c" | a ) rs . 11.81 , b ) rs . 12 , c ) rs . 18.94 , d ) rs . 12.31 , e ) none | c | divide(multiply(14, add(const_100, 15)), subtract(const_100, 15)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)| | gain |
for a 3 - digit number xyz , where x , y , and z are the digits of the number , f ( xyz ) = 5 ^ x 2 ^ y 3 ^ z . if f ( abc ) = 3 * f ( def ) , what is the value of abc - def ? | since f ( abc ) = 3 * f ( def ) , i would assume that f = c - 1 from the function above . the answer should be ( a ) | a ) 1 , b ) 2 , c ) 3 , d ) 9 , e ) 27 | a | subtract(3, 2) | subtract(n0,n2) | general |
how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length | "t = ( 660 + 165 ) / 54 * 18 / 5 t = 55 answer : c" | a ) 33 , b ) 44 , c ) 55 , d ) 77 , e ) 22 | c | divide(add(165, 660), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
a truck covers a distance of 392 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 70 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 392 / 8 = 49 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 48 + 18 ) = 66 kmph distance travelled by car = 392 + 70 = 462 km time taken by car = distance / speed = 462 / 66 = 7 hours . answer β c" | a ) 6 hours , b ) 5 hours , c ) 7 hours , d ) 8 hours , e ) none | c | divide(add(392, 70), add(divide(392, 8), 18)) | add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2)| | physics |
the area of a triangle is with base 4 m and height 8 m ? | "1 / 2 * 4 * 8 = 16 m 2 answer : d" | a ) 11 , b ) 10 , c ) 787 , d ) 16 , e ) 12 | d | triangle_area(4, 8) | triangle_area(n0,n1)| | geometry |
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 70 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 70 = 220000 + 70 = 220070 e" | a ) 145646 , b ) 236578 , c ) 645353 , d ) 456546 , e ) 220070 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 70) | add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)| | general |
the average of 45 results is 10 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 10 * 45 - 15 * 22 + 20 * 22 = 450 - 330 + 440 = 560 answer is e" | a ) 600 , b ) 480 , c ) 750 , d ) 650 , e ) 560 | e | subtract(subtract(multiply(45, 10), multiply(22, 20)), multiply(22, 15)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general |
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 140 , how much more remains to be paid ? | "10 / 100 p = 140 > > p = 140 * 100 / 10 = 1400 1400 - 140 = 1260 answer : e" | a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,260 | e | subtract(multiply(140, divide(const_100, 10)), 140) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
a professional janitor can clean a certain high school in ( 4 + 4 ) hours , working at a constant rate . a student sentenced to detention can clean that same high school in 20 hours , also working at a constant rate . if the student is paid $ 7 total per hour and the janitor is paid $ 21 per hour , how much more would ... | a professional janitor can clean a certain high school in ( 4 + 4 ) or 8 hours so ( applying rule # 1 ) , the janitor can clean 1 / 8 of the school in one hour a student sentenced to detention can clean that same high school in 20 hours so ( applying rule # 1 ) , the student can clean 1 / 20 of the school in one hour s... | a ) - $ 56 , b ) - $ 6 , c ) $ 0 , d ) $ 6 , e ) $ 8 | e | subtract(multiply(add(4, 4), 21), multiply(add(7, 21), divide(const_1, add(divide(const_1, add(4, 4)), divide(const_1, 20))))) | add(n0,n0)|add(n3,n4)|divide(const_1,n2)|divide(const_1,#0)|multiply(n4,#0)|add(#3,#2)|divide(const_1,#5)|multiply(#1,#6)|subtract(#4,#7) | general |
3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 2 men and 3 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 2 m + 3 w = 7 w answer is 7 / 14 = 1 / 2 answer : a" | a ) 1 / 2 , b ) 1 / 10 , c ) 1 / 18 , d ) 1 / 16 , e ) 1 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 2), 3), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | subtract(n1,n3)|subtract(n2,n0)|divide(#0,#1)|multiply(n4,#2)|multiply(n2,#2)|add(n5,#3)|add(n3,#4)|divide(#5,#6)| | physics |
a light has a rating of 60 watts , it is replaced with a new light that has 12 % higher wattage . how many watts does the new light have ? | "final number = initial number + 12 % ( original number ) = 60 + 12 % ( 60 ) = 60 + 7 = 67 answer b" | a ) 105 , b ) 67 , c ) 80 , d ) 60 , e ) 100 | b | multiply(60, add(const_1, divide(12, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)| | gain |
a sum of money becomes 7 / 6 of itself in 6 years at a certain rate of simple interest . the rate per annum is ? | "let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 6 years . rate = ( 100 * x ) / ( x * 6 * 6 ) = 25 / 9 % . answer : d" | a ) 50 / 9 , b ) 50 / 13 , c ) 50 / 17 , d ) 25 / 9 , e ) 5 5 / 1 | d | multiply(divide(subtract(divide(7, 6), const_1), 6), const_100) | divide(n0,n1)|subtract(#0,const_1)|divide(#1,n2)|multiply(#2,const_100)| | gain |
find the missing figures : 0.3 % of ? = 0.15 | let 0.3 % of x = 0.15 . then , 0.30 * x / 100 = 0.15 x = [ ( 0.15 * 100 ) / 0.3 ] = 50 . answer is a . | a ) 50 , b ) 45 , c ) 150 , d ) 450 , e ) 500 | a | divide(0.15, divide(0.3, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 265 , then the mean profit for the last 15 days would be | average would be : 350 = ( 265 + x ) / 2 on solving , x = 435 . answer : a | a ) rs . 435 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | a | divide(subtract(multiply(30, 350), multiply(15, 265)), 15) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n3) | gain |
the ratio of the number of females to males at a party was 1 : 2 but when 5 females and 5 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 5 ) = 2 x - 5 x = 10 there were 3 x = 30 people at the party originally . the answer is c ." | a ) 26 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | c | add(divide(subtract(multiply(multiply(5, 2), 3), multiply(5, 2)), 2), subtract(multiply(multiply(5, 2), 3), multiply(5, 2))) | multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,n1)|add(#3,#2)| | other |
the least number which when increased by 8 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 8 = 864 - 8 = 856 . answer d" | a ) 427 , b ) 859 , c ) 869 , d ) 856 , e ) none of these | d | subtract(lcm(lcm(lcm(24, 32), 36), 54), 8) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general |
find the length of the wire required to go 10 times round a square field containing 53824 m 2 . | a 2 = 53824 = > a = 232 4 a = 928 928 * 10 = 9280 answer : b | ['a ) 15840', 'b ) 9280', 'c ) 2667', 'd ) 8766', 'e ) 66711'] | b | multiply(square_perimeter(square_edge_by_area(53824)), 10) | square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1) | physics |
a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 4 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 4 / 100 x 12000 ) = rs . 480 . answer e" | a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | e | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(4, 100)) | divide(n3,n1)|divide(n2,n1)|multiply(const_10,const_1000)|multiply(const_1000,const_4)|multiply(n1,const_4)|add(#2,#3)|multiply(n1,#1)|add(#5,#4)|add(n1,#6)|divide(#7,#8)|multiply(n1,#9)|multiply(#0,#10)| | gain |
what should be added to 4329 so that it may become a perfect square ? | "66 x 66 = 4356 4356 - 4329 = 27 if added to 27 get perfect square answer = d" | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | d | subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4329) | add(const_10,const_10)|add(const_3,const_4)|add(#0,#0)|add(#2,#0)|add(#3,#1)|multiply(#4,#4)|subtract(#5,n0)| | geometry |
the number of the members of a club is more than 20 and less than 50 . when 4 people sit at a table , other people exactly divided to 6 - people groups ( 6 people sit at a table ) or 7 - people groups ( 7 people sit at a table ) . if the members are divided to 8 - people groups , how many people will be left ? | "the number of members is 7 k + 4 = 6 j + 4 the only number in this range which satisfies this is 46 . 46 / 8 = 5 ( 8 ) + 6 the answer is d ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | reminder(add(lcm(6, 7), 4), 8) | lcm(n3,n5)|add(n2,#0)|reminder(#1,n7)| | general |
the income of a company increases 20 % per annum . if its income is 2664000 in the year 1999 what was its income in the year 1997 ? | let income in 1997 = x according to the question , income in 1998 = x + x β 5 = 6 x β 5 income in 1999 = 6 x β 5 + 6 x β 25 = 36 x β 25 but given , income in 1999 = 2664000 β΄ 36 x β 25 = 2664000 β x = 1850000 answer e | a ) 2220000 , b ) 2850000 , c ) 2121000 , d ) 1855000 , e ) none of these | e | divide(2664000, multiply(divide(add(20, const_100), const_100), divide(add(20, const_100), const_100))) | add(n0,const_100)|divide(#0,const_100)|multiply(#1,#1)|divide(n1,#2) | gain |
when 242 is divided by a certain divisor the remainder obtained is 11 . when 698 is divided by the same divisor the remainder obtained is 18 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 9 . what is the value of the divisor ? | "let that divisor be x since remainder is 11 or 18 it means divisor is greater than 18 . now 242 - 11 = 231 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 18 = 680 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 231 + 680 ) + 11 + 18 = x ( k + l ) + 29 when we divide th... | a ) 11 , b ) 17 , c ) 13 , d ) 20 , e ) none of these | d | subtract(add(11, 18), 9) | add(n1,n3)|subtract(#0,n6)| | general |
anil brought a scooter for a certain sum of money . he spent 10 % of the cost on repairs and sold the scooter for a profit of rs . 1100 . how much did he spend on repairs if he made a profit of 20 % ? | e c . p . be rs . x . then , 20 % of x = 1100 20 / 100 * x = 1100 = > x = 5500 c . p . = rs . 5500 , expenditure on repairs = 10 % actual price = rs . ( 100 * 5500 ) / 110 = rs . 5000 expenditures on repairs = ( 5500 - 5000 ) = rs . 500 . | a ) 220 , b ) 420 , c ) 250 , d ) 700 , e ) 500 | e | divide(multiply(divide(multiply(1100, divide(10, divide(20, 10))), add(const_1, divide(const_1, 10))), 10), multiply(10, 10)) | divide(n2,n0)|divide(const_1,n0)|multiply(n0,n0)|add(#1,const_1)|divide(n0,#0)|multiply(n1,#4)|divide(#5,#3)|multiply(n0,#6)|divide(#7,#2) | gain |
if 7 persons can do 7 times of a particular work in 7 days , then , 9 persons can do 9 times of that work in ? | that is , 1 person can do one time of the work in 7 days . therefore , 9 persons can do 9 times work in the same 7 days itself . option ' e ' | a ) 3 days , b ) 8 days , c ) 9 days , d ) 11 days , e ) 7 days | e | divide(divide(multiply(7, 7), divide(7, 9)), 9) | divide(n0,n3)|multiply(n0,n0)|divide(#1,#0)|divide(#2,n3) | physics |
the cost per pound of green tea and coffee were the same in june . in july , the price of coffee shot up by 100 % and that of green tea dropped by 90 % . if in july , a mixture containing equal quantities of green tea and coffee costs $ 3.15 for 3 lbs , how much did a pound of green tea cost in july ? | "lets assume price of coffee in june = 100 x price of green tea in june = 100 x price of coffee in july = 200 x ( because of 100 % increase in price ) price of green tea in july = 10 x ( because of 90 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of green tea in july will be = 300 x + 15 x = 315 x as per... | a ) $ 4 , b ) $ 0.1 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | multiply(divide(add(multiply(divide(add(100, 100), 100), divide(3, const_2)), multiply(divide(3, const_2), divide(subtract(100, 90), 100))), 3.15), divide(subtract(100, 90), 100)) | add(n0,n0)|divide(n3,const_2)|subtract(n0,n1)|divide(#0,n0)|divide(#2,n0)|multiply(#3,#1)|multiply(#1,#4)|add(#5,#6)|divide(#7,n2)|multiply(#8,#4)| | general |
if xy > 0 , 1 / x + 1 / y = 4 , and 1 / xy = 6 , then ( x + y ) / 2 = ? | "( 1 / x + 1 / y ) = 4 canbe solved as { ( x + y ) / xy } = 6 . substituting for 1 / xy = 6 , we get x + y = 4 / 6 = = > ( x + y ) / 2 = 4 / ( 6 * 2 ) = 1 / 3 . d" | a ) 1 / 25 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) 6 | d | divide(divide(4, 6), 6) | divide(n3,n5)|divide(#0,n5)| | general |
what is 0.01 percent of 12,356 ? | "soln : - 0.01 % of 12,356 = 0 . 011000.01100 x 12,356 = 1100 β 1001100 β 100 x 12,356 = 12,356100 β 10012,356100 β 100 = 1.2356 answer : b" | a ) 0.12356 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356 | b | divide(multiply(0.01, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
indu gave bindu rs . 2500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 2 % per annum simple interest ? | "2500 = d ( 100 / 2 ) 2 d = 1 answer : a" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(subtract(multiply(2500, power(add(const_1, divide(4, const_100)), 2)), 2500), multiply(multiply(2500, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent e of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is e = . 25 / 1 * 100 = 25 % d is the answer ." | a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
for what values of k will the pair of equations 3 ( 3 x + 4 y ) = 36 and kx + 12 y = 30 does not have a unique solution ? | "we have 2 equations 1 . 3 ( 3 x + 4 y ) = 36 - - > 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b" | a ) 12 , b ) 9 , c ) 3 , d ) 7.5 , e ) 2.5 | b | divide(multiply(12, 3), const_4.0) | multiply(n1,n4)|divide(#0,n2)| | general |
working at constant rate , pump x pumped out half of the water in a flooded basement in 5 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constan... | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 26 hours d" | a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 26 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 5)), const_1), const_1), const_1), const_1), 5), const_1), const_1), const_1), const_1), const_1), const_1) | add(n0,n1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,n0)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1)| | physics |
there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total r of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) r 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is ... | a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | add(n0,n5)| | other |
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 5 cups of tea . last week mo drank a total of 36 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how ma... | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 36 t - c = 14 - > t = 25 . c = 11 . mo drinks 5 cups of tea a day then number of days that are not rainy = 25 / 5 = 5 so number of rainy days = 7 - 5 = 2 d is the answer ." | a ) 5 , b ) 3 , c ) 4 , d ) 2 , e ) 6 | d | subtract(add(const_4, const_3), divide(divide(add(36, 14), const_2), 5)) | add(const_3,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)| | general |
there are 2 sections a and b in a class , consisting of 16 and 14 students respectively . if the average weight of section a is 20 kg and that of section b is 25 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 16 * 20 + 14 * 25 = 670 average weight of the class is = 670 / 20 = 33.5 kg answer is a" | a ) 33.5 kg , b ) 37.25 kg , c ) 42.45 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(16, 20), multiply(14, 25)), add(16, 14)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general |
a car traveled 75 % of the way from town a to town b at an average speed of 60 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 40 miles per hour . what is s ? | total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip β 75 / 60 = 1.25 2 nd part of the trip β 25 / s = t total trip β ( 75 + 25 ) / 40 = 1.25 + t Β» 100 / 40 = 1.25 + t Β» 2.5 = 1.25 + t Β» t = 1.25 back to 2 nd part of the trip formula ... | a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 37.5 | b | divide(subtract(add(40, 60), 75), subtract(divide(add(40, 60), 40), divide(75, 60))) | add(n1,n2)|divide(n0,n1)|divide(#0,n2)|subtract(#0,n0)|subtract(#2,#1)|divide(#3,#4) | physics |
the price of a jacket is reduced by 35 % . during a special sale the price of the jacket is reduced another 10 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 35 % , therefore bringing down the price to $ 65 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 58.5 4 ) now 58.5 has to be added byx % in order to equal the original price . 58.5 + ( x % ) 58.5 = 100 . ... | a ) 70.9 , b ) 75 , c ) 48 , d ) 65 , e ) 67.5 | a | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100)))), subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
the number of stamps that p and q had were in the ratio of 7 : 3 respectively . after p gave q 13 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 5 : 4 . as a result of the gift , p had how many more stamps than q ? | "p started with 7 k stamps and q started with 3 k stamps . ( 7 k - 13 ) / ( 3 k + 13 ) = 5 / 4 28 k - 15 k = 117 k = 9 p has 7 ( 9 ) - 13 = 50 stamps and q has 3 ( 9 ) + 13 = 40 stamps . the answer is a ." | a ) 10 , b ) 25 , c ) 40 , d ) 65 , e ) 90 | a | divide(add(multiply(13, 5), multiply(13, 4)), add(5, 4)) | add(n3,n4)|multiply(n2,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | other |
if x and y are positive integers and 25 x = 10 y what is the least possible value of xy ? | "25 x = 10 y = > x / y = 2 / 5 = > 5 x = 2 y 5 ( 3 ) = 2 ( 3 ) = > x * y = 9 but it is not given 5 ( 5 ) = 2 ( 5 ) = > x * y = 25 b" | a ) 14 , b ) 25 , c ) 63 , d ) 84 , e ) 252 | b | add(10, const_1) | add(n1,const_1)| | general |
a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 27 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | "let x be the wholesale price . then 1.8 x = 27 and x = 27 / 1.8 = 15 . to achieve a 100 % markup , the price needs to be $ 30 . the retailer needs to increase the price by $ 3 more . the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(multiply(divide(27, add(const_1, divide(80, 100))), const_2), 27) | divide(n0,n2)|add(#0,const_1)|divide(n1,#1)|multiply(#2,const_2)|subtract(#3,n1)| | general |
the ratio of the volumes of two cubes is 2744 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = Β³ β 2744 : Β³ β 1331 = 14 : 11 ratio of surface areas = 14 ^ 2 : 11 ^ 2 = 196 : 121 answer : option a" | a ) 196 : 121 , b ) 81 : 127 , c ) 181 : 196 , d ) 81 : 161 , e ) 81 : 182 | a | power(divide(2744, 1331), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
evaluate 28 % of 350 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 350 + ( 45 / 100 ) * 280 = 98 + 126 = 224 answer : option a" | a ) 224 , b ) 242 , c ) 252 , d ) 262 , e ) 282 | a | divide(28, divide(350, 28)) | divide(n1,n0)|divide(n0,#0)| | gain |
three unbiased coins are tossed . what is the probability of getting 1 heads , 1 tail ? | "let , h - - > head , t - - > tail here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting 3 heads then e = { tht , hth } p ( e ) = n ( e ) / n ( s ) = 2 / 8 answer is e" | a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 8 , d ) 7 / 8 , e ) 2 / 8 | e | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
divide rs . 32000 in the ratio 3 : 7 ? | 3 / 10 * 32000 = 9600 7 / 8 * 32000 = 22400 answer : b | a ) 12000 , 20000 , b ) 9600 , 22400 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 12000 , 20001 | b | multiply(subtract(7, const_2), divide(32000, add(3, subtract(7, const_2)))) | subtract(n2,const_2)|add(n1,#0)|divide(n0,#1)|multiply(#2,#0) | other |
a train is running with a speed of 60 kmph and its length is 110 metres . calculate the time by which it will pass a man running opposite with speed of 6 kmph | explanation : from the given question , we will first calculate the speed of train relative to man , = > ( 60 + 6 ) = 66 km / hr ( we added 6 because man is running opposite ) convert it in metre / second = 66 Γ£ β 5 / 18 = 55 / 3 m / sec time it will take to pass man = 110 Γ£ β 3 / 55 = 6 seconds answer is c | a ) 2 second , b ) 4 second , c ) 6 second , d ) 8 second , e ) 10 second | c | divide(110, multiply(add(60, 6), const_0_2778)) | add(n0,n2)|multiply(#0,const_0_2778)|divide(n1,#1) | physics |
the total of the ages of amar , akbar and anthony is 66 years . what was the total of their ages 4 years ago ? | explanation : required sum = ( 66 - 3 x 4 ) years = ( 66 - 12 ) years = 54 years . answer : c | a ) 71 , b ) 44 , c ) 54 , d ) 16 , e ) 18 | c | subtract(66, multiply(4, const_3)) | multiply(n1,const_3)|subtract(n0,#0) | general |
kim finds a 3 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? | 3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is b | a ) 2 / 5 , b ) 3 / 5 , c ) 8 / 15 , d ) 1 / 2 , e ) 7 / 5 | b | subtract(const_1, add(add(divide(const_3, multiply(add(const_2, const_3), 3)), divide(const_2, multiply(add(const_2, const_3), 3))), divide(const_1, multiply(add(const_2, const_3), 3)))) | add(const_2,const_3)|multiply(n0,#0)|divide(const_3,#1)|divide(const_2,#1)|divide(const_1,#1)|add(#2,#3)|add(#5,#4)|subtract(const_1,#6) | physics |
a certain country is divided into 8 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 12 the total number of progressives in the entire country , what fraction of t... | "let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 12 the total number of traditionalists is 8 p / 12 = 2 p / 3 . the total population is p + 2 p / 3 = 5 p / 3 p / ( 5 p / 3 ) = 3 / 5 the answer is e ." | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5 | e | subtract(1, divide(divide(8, 12), add(divide(8, 12), 1))) | divide(n0,n2)|add(n1,#0)|divide(#0,#1)|subtract(n1,#2)| | general |
n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 45 and 6 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | you have 6 digits : 12 , 3 , 4 , 5 , 6 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small a... | a ) 59 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | b | subtract(multiply(const_10, 6), const_10) | multiply(n5,const_10)|subtract(#0,const_10) | general |
with both inlets open , a water tank will be filled with water in 36 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 36 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 36 ) - ( 1 / 120 ) = 1 / 51 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 51 ) - ( 1 / 120 ) = 50 cubic meter i . ... | a ) 9,000 , b ) 4,500 , c ) 1,750 , d ) 1,000 , e ) 2,000 | b | divide(divide(50, subtract(divide(subtract(divide(const_60, 36), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | divide(const_60,n0)|inverse(n1)|multiply(n1,const_60)|inverse(#2)|subtract(#0,#1)|divide(#4,const_60)|subtract(#5,#3)|divide(n2,#6)|divide(#7,const_1000)| | physics |
what is the angle between the hands of a clock when time is 16 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 16 - 40 * 11 / 2 = 640 - 220 = 420 deg answer : b" | a ) 625 deg , b ) 420 deg , c ) 145 deg , d ) 150 deg , e ) 300 deg | b | subtract(multiply(40, multiply(const_3, const_2)), 16) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)| | geometry |
solve the equation for x : 8 ( x + y + 2 ) = 8 y - 8 | "a - 3 8 ( x + y + 2 ) = 8 y - 8 8 x + 8 y + 16 = 8 y - 8 8 x + 16 = - 8 8 x = - 24 = > x = - 3" | a ) - 3 , b ) - 2 , c ) - 1 , d ) 1 , e ) 2 | a | divide(add(2, 8), add(8, 8)) | add(n1,n3)|add(n0,n0)|divide(#0,#1)| | general |
during a special promotion , a certain filling station is offering a 8 percent discount on gas purchased after the first 10 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella β s total per - gallon discount is what percent of kim β s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 4 + 0.9 * 16 = 18.4 gallons , so the overall discount she got was 1.6 / 20 = 8 % . isabella purchased 25 gallons of gas . she paid for 4 + 0.9 * 21 = 22.9 gallons , so the overall discount she got was 2.1 / 25 = 8.4 % . 8.4 / 8 * 100 = 105 % . answer : b ." | a ) 80 % , b ) 105 % , c ) 115 % , d ) 120 % , e ) 140 % | b | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(8, subtract(8, const_1)), subtract(const_100, 8)), const_100), 10), 25)), const_100), multiply(subtract(const_1, divide(add(10, divide(multiply(add(8, const_4), subtract(const_100, 8)), const_100)), 20)), const_100)), const_100) | add(n0,const_4)|subtract(n0,const_1)|subtract(const_100,n0)|add(n0,#1)|multiply(#0,#2)|divide(#4,const_100)|multiply(#3,#2)|add(n1,#5)|divide(#6,const_100)|add(n1,#8)|divide(#7,n2)|divide(#9,n3)|subtract(const_1,#10)|multiply(#12,const_100)|subtract(const_1,#11)|multiply(#14,const_100)|divide(#15,#13)|multiply(#16,cons... | gain |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 90 only . men Γ’ β¬ β’ s wages are ? | "answer : option a 5 m = xw = 8 b 5 m + xw + 8 b - - - - - 90 rs . 5 m + 5 m + 5 m - - - - - 90 rs . 15 m - - - - - - 90 rs . = > 1 m = 6 rs ." | a ) rs . 6 , b ) rs . 6.5 , c ) rs . 8 , d ) rs . 5 , e ) rs . 2 | a | divide(90, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
stacy has a 33 page history paper due in 3 days . how many pages per day would she have to write to finish on time ? | "33 / 3 = 11 answer : c" | a ) 9 , b ) 8 , c ) 11 , d ) 8.5 , e ) 6 | c | divide(33, 3) | divide(n0,n1)| | physics |
the smallest value of n , for which 2 n + 1 is not a prime number , is | "solution ( 2 Γ 1 + 1 ) = 3 . ( 2 Γ 2 + 1 ) = 5 . ( 2 Γ 3 + 1 ) = 7 . ( 2 Γ 4 + 1 ) = 9 . which is not prime , n = 4 . answer b" | a ) 3 , b ) 4 , c ) 5 , d ) none , e ) 6 | b | add(2, 2) | add(n0,n0)| | general |
a train running at the speed of 50 km / hr crosses a pole in 18 seconds . find the length of the train . | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 18 = 250 meter answer : c" | a ) 150 , b ) 188 , c ) 250 , d ) 288 , e ) 300 | c | multiply(divide(multiply(50, const_1000), const_3600), 18) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
what sum of money put at c . i amounts in 2 years to rs . 7000 and in 3 years to rs . 9261 ? | "7000 - - - - 2261 100 - - - - ? = > 32.3 % x * 1323 / 100 * 1323 / 100 = 7000 x * 1.75 = 7000 x = 7000 / 1.75 = > 3999.25 answer : a" | a ) 4000 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(7000, power(add(subtract(divide(9261, 7000), const_1), const_1), 2)) | divide(n3,n1)|subtract(#0,const_1)|add(#1,const_1)|power(#2,n0)|divide(n1,#3)| | general |
two employees m and n are paid a total of $ 583 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 583 2.2 n = 583 n = 265 the answer is c ." | a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | c | divide(583, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
if remainder is 1 , quotient is 54 and dividend is 217 then what is divisor ? | "we know dividend = divisor * quotient + remainder = = = > 217 = divisor * 54 + 1 = = = = = > 216 / 54 = divisor = = = > divisor = 4 ans - b" | a ) 2 , b ) 4 , c ) 6 , d ) 14 , e ) 16 | b | divide(subtract(217, 1), 54) | subtract(n2,n0)|divide(#0,n1)| | general |
4 , 7 , 13 , 25 , 49 , ( . . . ) | "explanation : 4 4 Γ 2 - 1 = 7 7 Γ 2 - 1 = 13 13 Γ 2 - 1 = 25 25 Γ 2 - 1 = 49 49 Γ 2 - 1 = 97 answer : option c" | a ) 22 , b ) 35 , c ) 97 , d ) 32 , e ) 25 | c | subtract(negate(25), multiply(subtract(7, 13), divide(subtract(7, 13), subtract(4, 7)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
find the least number that must be subtracted from 1387 so that the remaining number is divisible by 15 . | on dividing 1387 by 15 we get the remainder 7 , so 7 should be subtracted . the answer is c . | a ) 1 , b ) 5 , c ) 7 , d ) 9 , e ) 13 | c | subtract(1387, multiply(subtract(const_100, multiply(const_2, const_4)), 15)) | multiply(const_2,const_4)|subtract(const_100,#0)|multiply(n1,#1)|subtract(n0,#2) | general |
the output of a factory was increased by 5 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 20 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 5 % and then 20 % . total % change = a + b + ab / 100 total % change = 5 + 20 + 5 * 20 / 100 = 26 % now , you want to change it to 0 , so , 0 = 26 + x + 26 x / 100 x = - 26 ( 100 ) / 126 = 20 % approximately answer is a" | a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | a | divide(multiply(subtract(add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100))), const_100), const_100), add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100)))) | add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|add(#0,#2)|subtract(#3,const_100)|multiply(#4,const_100)|divide(#5,#3)| | general |
if n is a positive integer and n ^ 2 is divisible by 62 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i ... | a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(62, 2)), 2) | divide(n1,n0)|sqrt(#0)|multiply(n0,#1)| | general |
if a and b get profits of rs . 60,000 and rs . 6,000 respectively at the end of year then ratio of their investments are | ratio = 60000 / 6000 = 10 : 1 answer : b | a ) 4 : 1 , b ) 10 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 2 : 5 | b | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#1,const_100)|multiply(#0,const_2)|multiply(#2,const_100)|divide(#4,#3)|divide(#4,#5) | gain |
what is the 99 th digit after the decimal point in the decimal expansion of 2 / 9 + 3 / 11 ? | 2 / 9 = 0.22222 . . . . = = > 99 th digit is 2 3 / 11 = 0.27272727 . . . . = = > every odd digit is 2 . so , 99 th digit will be 2 . 2 + 2 = 4 answer : c | a ) 1 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | c | floor(multiply(divide(add(multiply(divide(99, 9), 2), multiply(divide(99, 11), 3)), 99), const_10)) | divide(n0,n2)|divide(n0,n4)|multiply(n1,#0)|multiply(n3,#1)|add(#2,#3)|divide(#4,n0)|multiply(#5,const_10)|floor(#6) | general |
the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of bren 10 years from now ? | let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren ' s age 10 years hence = 40 + 10 = 50 answer = e | a ) 17 , b ) 25 , c ) 27 , d ) 35 , e ) 50 | e | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 10) | add(n0,n1)|multiply(n1,const_3)|add(n4,#1)|add(n2,#0)|divide(#2,#3)|multiply(n1,#4)|add(n5,#5) | general |
a trainer is standing in one corner of a square ground of side 25 m . his voice can be heard upto 140 m . find the area of the ground in which his voice can be heard ? | area covered by goat = pi * r ^ 2 / 4 ( here we divide by 4 because the trainer is standing in a corner of the ground and only in 1 / 4 part , the voice can be heard ) where r = 14 m = length reaching the voice so area = ( 22 / 7 ) * 140 * 140 / 4 = 15400 sq m answer : c | ['a ) 12300', 'b ) 14500', 'c ) 15400', 'd ) 16700', 'e ) 18200'] | c | divide(circle_area(140), const_4) | circle_area(n1)|divide(#0,const_4) | geometry |
machine m , n , o working simultaneously machine m can produce x units in 3 / 4 of the time it takes machine n to produce the same amount of units . machine n can produce x units in 2 / 7 the time it takes machine o to produce that amount of units . if all 3 machines are working simultaneously , what fraction of the to... | now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o , say 12 hrs to make x units ( assuming 6 because we can see we will need to divide by 3 and 4 mach o makes x units in 12 hrs so , mach n = 2 / 7 of o = 2 / 7 * 12 = 24 / 7 hrs to make x units and mach m = 3 / 4 of... | a ) 1 / 2 , b ) 7 / 13 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33 | b | divide(7, subtract(multiply(7, 2), const_1)) | multiply(n2,n3)|subtract(#0,const_1)|divide(n3,#1) | general |
a and b can do a work in 2 days , b and c in 6 days and c and a in 8 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 2 one day work of b and c = 1 / 6 one day work of c and a = 1 / 8 2 ( a + b + c ) = 1 / 2 + 1 / 6 + 1 / 8 2 ( a + b + c ) = 19 / 24 ( a + b + c ) = 19 / 48 number of days required = 48 / 19 days . answer : b" | a ) 12 / 28 , b ) 48 / 19 , c ) 16 / 48 , d ) 18 / 48 , e ) 12 / 64 | b | divide(const_1, divide(add(add(inverse(2), inverse(6)), inverse(8)), 2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,n0)|divide(const_1,#5)| | physics |
a man is 18 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 18 . = > 2 ( x + 2 ) = ( x + 18 + 2 ) = > 2 x + 4 = x + 20 = > x = 16 years option e" | a ) 21 years , b ) 22 years , c ) 23 years , d ) 12 years , e ) 16 years | e | divide(subtract(18, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = | "( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = 0.0088 * 450 / 5 * ( 0.1 ) ( 0.008 ) = 0.088 * 90 / 1 * 0.008 = 88 * 90 / 8 = 11 * 90 = 990 answer : e" | a ) 950 , b ) 940 , c ) 980 , d ) 960 , e ) 990 | e | divide(multiply(0.0088, 4.5), multiply(multiply(0.05, 0.1), 0.008)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2)| | general |
a watch was sold at a loss of 10 % . if it was sold for rs . 210 more , there would have been a gain of 4 % . what is the cost price ? | explanation : 90 % 104 % - - - - - - - - 14 % - - - - 210 100 % - - - - ? = > rs . 1500 answer : e | a ) s . 1000 , b ) s . 1009 , c ) s . 1007 , d ) s . 1006 , e ) s . 1500 | e | divide(multiply(210, const_100), subtract(add(const_100, const_4), subtract(const_100, 10))) | add(const_100,const_4)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3) | gain |
( 4300631 ) - ? = 2535618 | "let 4300631 - x = 2535618 then x = 4300631 - 2535618 = 1765013 answer is c" | a ) 1865113 , b ) 1775123 , c ) 1765013 , d ) 1675123 , e ) none of them | c | multiply(4300631, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general |
two trains 250 m and 500 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 250 + 500 = 750 m . required time = 750 * 9 / 250 = 27 sec answer : d" | a ) 10.6 , b ) 10.9 , c ) 10.4 , d ) 27 , e ) 10.1 | d | divide(add(250, 500), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
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