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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16 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 16 * 10 / 40 = 4 answer is c" | a ) 10 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | c | divide(multiply(16, 10), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 15000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 15000 + 7000 + 9000 / 5 = rs . 8800 correct option : c" | a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8800 , d ) rs . 9000 , e ) none | c | divide(add(add(add(add(8000, 5000), 15000), 7000), 9000), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
if n is a prime number greater than 19 , what is the remainder when n ^ 2 is divided by 12 ? | "there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 19 , square it and see what would be the remainder upon division of it by 12 . n = 23 - - > n ^ 2 = 529 - - > remainder upon division 529 by 12 is 1 . a... | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(power(add(19, 2), 2), multiply(12, const_4)) | add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)| | general |
total number of 4 digit number do not having the digit 3 or 6 . | answer : d | a ) 22 , b ) 28 , c ) 27 , d ) 19 , e ) 11 | d | add(add(add(add(6, 3), 4), 3), 3) | add(n1,n2)|add(n0,#0)|add(n1,#1)|add(n1,#2) | general |
the average weight of a , b and c is 43 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 129 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 37 . b ' s wei... | a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 37 kg | e | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(43, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
the length of rectangle is thrice its breadth and its perimeter is 80 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 80 l = 36 b = 10 lb = 36 * 10 = 360 a" | a ) 360 , b ) 376 , c ) 299 , d ) 276 , e ) 111 | a | multiply(multiply(divide(80, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(80, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)| | geometry |
a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability r that both selected dogs are not littermates ? | "we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one ... | a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9 | c | divide(const_5, 6) | divide(const_5,n1)| | other |
a train running at the speed of 60 km / hr crosses a pole in 21 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 21 ) m = 350 m . answer : a" | a ) 350 m , b ) 278 m , c ) 876 m , d ) 150 m , e ) 267 m | a | multiply(divide(multiply(60, const_1000), const_3600), 21) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a number when divided by 214 gives a remainder 35 , what remainder will be obtained by dividing the same number 14 ? | "explanation : 214 + 35 = 249 / 14 = 11 ( remainder ) answer : c" | a ) 7 , b ) 10 , c ) 11 , d ) 2 , e ) 3 | c | subtract(35, multiply(14, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
a number whose fifth part increased by 3 is equal to its fourth part diminished by 3 is ? | "answer let the number be n . then , ( n / 5 ) + 3 = ( n / 4 ) - 3 â ‡ ’ ( n / 4 ) - ( n / 5 ) = 6 â ‡ ’ ( 5 n - 4 n ) / 20 = 6 â ˆ ´ n = 120 option : a" | a ) 120 , b ) 180 , c ) 200 , d ) 220 , e ) none | a | divide(add(3, 3), subtract(divide(const_1, 3), divide(const_1, add(const_1, 3)))) | add(n0,n1)|add(const_1,n0)|divide(const_1,n0)|divide(const_1,#1)|subtract(#2,#3)|divide(#0,#4)| | general |
a military camp has a food reserve for 250 personnel for 40 days . if after 15 days 50 more personnel are added to the camp , find the number of days the reserve will last for ? | explanation : as the camp has a reserve for 250 personnel that can last for 40 days , after 10 days the reserve left for 250 personnel is for 30 days . now 50 more personnel are added in the camp . hence , the food reserve for 300 personnel will last for : 250 : 300 : : x : 30 … … . . ( it is an indirect proportion as ... | a ) 25 , b ) 67 , c ) 26 , d ) 29 , e ) 18 | a | add(divide(multiply(250, subtract(40, 15)), add(250, 50)), const_3) | add(n0,n3)|subtract(n1,n2)|multiply(n0,#1)|divide(#2,#0)|add(#3,const_3) | general |
what is x if x + 2 y = 12 and y = 3 ? | "x = 12 - 2 y x = 12 - 6 . x = 6 answer : c" | a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | c | subtract(12, multiply(2, 3)) | multiply(n0,n2)|subtract(n1,#0)| | general |
in how many years rs 200 will produce the same interest at 10 % as rs . 1000 produce in 2 years at 12 % | explanation : clue : firstly we need to calculate the si with prinical 1000 , time 2 years and rate 12 % , it will be rs . 240 then we can get the time as time = ( 100 * 240 ) / ( 200 * 10 ) = 12 option d | a ) 13 , b ) 9 , c ) 11 , d ) 12 , e ) 10 | d | divide(multiply(divide(multiply(1000, 12), const_100), 2), multiply(divide(10, const_100), 200)) | divide(n1,const_100)|multiply(n2,n4)|divide(#1,const_100)|multiply(n0,#0)|multiply(n3,#2)|divide(#4,#3) | gain |
a store ’ s selling price of $ 2240 for a certain computer would yield a profit of 40 percent of the store ’ s cost for the computer . what selling price would yield a profit of 60 percent of the computer ’ s cost ? | "1.4 x = 2240 x = 2240 / 1.4 so , 1.6 x = 2240 * 1.6 / 1.4 = 2560 answer : - c" | a ) $ 2400 , b ) $ 2464 , c ) $ 2560 , d ) $ 2732 , e ) $ 2800 | c | multiply(2240, divide(add(const_100, 60), add(const_100, 40))) | add(n2,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(n0,#2)| | gain |
a person buys 18 local tickets for rs 110 . each first class ticket costs rs 10 and each second class ticket costs rs 3 . what will another lot of 18 tickets in which the numbers of first class and second class tickets are interchanged cost ? | explanation : let , there are x first class ticket and ( 18 - x ) second class tickets . then , 110 = 10 x + 3 ( 18 − x ) . = > 110 = 10 x + 54 − 3 x . = > 7 x = 56 . = > x = 8 . if the number of the first class and second class tickets are interchanged , then the total cost would be 10 × 10 + 3 × 8 = 124 . answer : d | a ) 112 , b ) 118 , c ) 121 , d ) 124 , e ) none of these | d | add(multiply(divide(subtract(110, multiply(18, 3)), subtract(10, 3)), 3), multiply(10, 10)) | multiply(n0,n3)|multiply(n2,n2)|subtract(n2,n3)|subtract(n1,#0)|divide(#3,#2)|multiply(n3,#4)|add(#5,#1) | physics |
for all numbers a and b , the operationis defined by ab = ( a + 2 ) ( b – 3 ) . if 3 y = – 30 , then y = | ( 3 + 2 ) ( y - 3 ) = - 30 . . x - 3 = - 6 . . x = - 3 c | a ) – 15 , b ) – 6 , c ) - 3 , d ) 6 , e ) 15 | c | negate(divide(30, 3)) | divide(n3,n1)|negate(#0) | general |
a train 250 m long running at 72 kmph crosses a platform in 40 sec . what is the length of the platform ? | "d = 72 * 5 / 18 = 40 = 800 â € “ 250 = 450 m answer : a" | a ) 450 m , b ) 200 m , c ) 250 m , d ) 270 m , e ) 300 m | a | subtract(multiply(40, multiply(72, const_0_2778)), 250) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
out of 3 given numbers , the first one is twice the second and 3 times the third . if the average of these numbers is 88 , then the difference between first and third is . | sum of three number is = 88 * 3 = 264 let three numbers are a , b , c and a is the highest and c is the lowest then , 2 b = a so b = a / 2 and 3 c = a so c = a / 3 we can write , a + b + c = 264 a + a / 2 + a / 3 = 264 11 a / 6 = 264 a = 144 so , c = 144 / 3 = 48 so there difference is = 144 - 48 = 96 answer d | a ) 92 , b ) 39 , c ) 87 , d ) 96 , e ) none | d | add(88, add(const_4, const_4)) | add(const_4,const_4)|add(n2,#0) | general |
an item is being sold for $ 10 each . however , if a customer will “ buy at least 3 ” they have a promo discount of 32 % . also , if a customer will “ buy at least 10 ” items they will deduct an additional 8 % to their “ buy at least 3 ” promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 40 % , thus he must pay slightly more than $ 60 . only answer choice e fits . answer : e ." | a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 70.00 | e | multiply(subtract(10, divide(multiply(32, 8), const_100)), 10) | multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,n0)| | gain |
r is the set of positive odd integers less than 200 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive odd integers less than 200 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 1,3 , 5,7 , 9,11 , 13,15 . . . s = 1 , 9,25 , 49,81 . . . numbers : 1 , 9 , 25 , 49 , 81 , 121 , and 169 are odd integers ( less than 200 ) that ... | a ) none , b ) two , c ) four , d ) five , e ) seven | e | subtract(subtract(200, const_4), const_4) | subtract(n0,const_4)|subtract(#0,const_4)| | physics |
the cost to park a car in a certain parking garage is $ 9.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 9 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 9 + 7 * 1.75 = 21.25 thus the average parking price = 21.25 / 9 = 2.36 $ d is the correct answer ." | a ) $ 1.09 , b ) $ 1.67 , c ) $ 2.25 , d ) $ 2.36 , e ) $ 2.50 | d | divide(add(9.00, multiply(1.75, subtract(9, 2))), 9) | subtract(n4,n1)|multiply(n2,#0)|add(n0,#1)|divide(#2,n4)| | general |
find the compound interest on $ 30000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 30000 * ( 1 + 2 / 100 ) ^ 4 = 30000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 32472.96 c . i . = 32472.96 - 10000 = $ 2472.96 answer is d" | a ) $ 645.56 , b ) $ 824.32 , c ) $ 954.26 , d ) $ 2472.96 , e ) $ 1020.45 | d | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 30000), 30000) | divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)| | gain |
the average waight of a , b , c is 45 kg . the avg wgt of a & b be 40 kg & that of b , c be 43 kg . find the wgt of b . | ". let a , b , c represent their individual wgts . then , a + b + c = ( 45 * 3 ) kg = 135 kg a + b = ( 40 * 2 ) kg = 80 kg & b + c = ( 43 * 2 ) kg = 86 kg b = ( a + b ) + ( b + c ) - ( a + b + c ) = ( 80 + 86 - 135 ) kg = 31 kg . answer is e ." | a ) 34 kg , b ) 40 kg , c ) 42 kg , d ) 41 kg , e ) 31 kg | e | subtract(multiply(40, const_2), subtract(multiply(45, const_3), multiply(43, const_2))) | multiply(n1,const_2)|multiply(n0,const_3)|multiply(n2,const_2)|subtract(#1,#2)|subtract(#0,#3)| | general |
of the 600 residents of clermontville , 35 % watch the television show island survival , 40 % watch lovelost lawyers and 50 % watch medical emergency . if all residents watch at least one of these 3 shows and 18 % watch exactly 2 of these shows , then how many clermontville residents z watch all of the shows ? | oa is d . 100 = a + b + c - ab - ac - bc + abc , which is the same as the following formula 100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - 2 abc . the term between parantheses value 18 % so the equation to resolve is 100 = 35 + 40 + 50 - 18 - 2 abc therefore the value of abc is z = 3.5 % of 600 , is 21 . d is... | a ) 150 , b ) 108 , c ) 42 , d ) 21 , e ) - 21 | d | divide(subtract(subtract(add(add(multiply(600, divide(35, const_100)), multiply(600, divide(40, const_100))), multiply(600, divide(50, const_100))), 600), multiply(divide(18, const_100), 600)), 2) | divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|divide(n5,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|add(#4,#5)|add(#8,#6)|subtract(#9,n0)|subtract(#10,#7)|divide(#11,n6) | gain |
john ' s marks wrongly entered as 82 instead of 62 . due to that the average marks for the class got increased by half ( 1 / 2 ) . the number of john in the class is ? | otal increase in marks = x x 1 = x 2 2 x / 2 = ( 82 - 62 ) x / 2 = 40 x = 80 . c | a ) 70 , b ) 78 , c ) 80 , d ) 84 , e ) 90 | c | multiply(divide(subtract(82, 62), divide(1, 2)), 2) | divide(n2,n3)|subtract(n0,n1)|divide(#1,#0)|multiply(n3,#2) | general |
a train running at the speed of 90 km / hr crosses a pole in 10 seconds . find the length of the train . | "speed = 90 * ( 5 / 18 ) m / sec = 25 m / sec length of train ( distance ) = speed * time 25 * 10 = 250 meter answer : c" | a ) 150 , b ) 180 , c ) 250 , d ) 200 , e ) 225 | c | multiply(divide(multiply(90, const_1000), const_3600), 10) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
milk is poured from a full rectangular container with dimensions 4 inches by 9 inches by 10 inches into a cylindrical container with a diameter of 6 inches . if the milk does not overflow , how many inches high will the milk reach ? | let the height of level of milk in the cylinder = h since , volume of milk is constant . therefore 4 * 9 * 10 = π π * 3 ^ 2 * h = > π * h = 40 = > h = 40 / ( π ) answer c | ['a ) 60 / π', 'b ) 24', 'c ) 40 / π', 'd ) 10', 'e ) 3 π'] | c | divide(divide(volume_rectangular_prism(4, 9, 10), 9), const_pi) | volume_rectangular_prism(n0,n1,n2)|divide(#0,n1)|divide(#1,const_pi) | geometry |
if 65 % of a number is greater than 5 % of 60 by 23 , what is the number ? | "explanation : 65 / 100 * x - 5 / 100 * 60 = 23 65 / 100 * x - 3 = 23 65 / 100 * x = 26 x = 26 * 100 / 65 x = 40 answer : option b" | a ) 65 , b ) 40 , c ) 55 , d ) 30 , e ) 60 | b | multiply(const_100, divide(add(23, multiply(divide(60, const_100), 5)), 65)) | divide(n2,const_100)|multiply(n1,#0)|add(n3,#1)|divide(#2,n0)|multiply(#3,const_100)| | gain |
8 men and 2 boys working together can do 4 times as much work as a man and a boy . working capacity of man and boy is in the ratio | explanation : let 1 man 1 day work = x 1 boy 1 day work = y then 8 x + 2 y = 4 ( x + y ) = > 4 x = 2 y = > x / y = 2 / 4 = > x : y = 1 : 2 option a | a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 1 , d ) 2 : 3 , e ) none of these | a | divide(subtract(4, 2), subtract(8, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | other |
the difference between the place value and the face value of 7 in the numeral 856973 is | ( place value of 7 ) - ( face value of 7 ) = ( 70 - 7 ) = 63 answer : option a | a ) 63 , b ) 6973 , c ) 5994 , d ) 6084 , e ) none of these | a | subtract(multiply(const_10, 7), 7) | multiply(n0,const_10)|subtract(#0,n0) | general |
if a person walks at 15 km / hr instead of 5 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 5 = ( x + 20 ) / 15 10 x - 100 = > x = 10 km . answer : d" | a ) 50 km , b ) 76 km , c ) 18 km , d ) 10 km , e ) 97 km | d | multiply(5, divide(20, subtract(15, 5))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
the diameter of a cylindrical tin is 4 cm and height is 5 cm . find the volume of the cylinder ? | "r = 2 h = 5 π * 2 * 2 * 5 = 20 π cc answer : c" | a ) 33 , b ) 45 , c ) 20 , d ) 77 , e ) 21 | c | divide(volume_cylinder(divide(4, const_2), 5), const_pi) | divide(n0,const_2)|volume_cylinder(#0,n1)|divide(#1,const_pi)| | geometry |
the sum of 3 consecutive numbers is definitely | "if 1 st term is x : x + ( x + 1 ) + ( x + 2 ) = 3 x + 3 - - - > always divisible by 3 if 2 nd term is x : ( x - 1 ) + x + ( x + 1 ) = 3 x - - - > always divisible by 3 if 3 rd term is x : ( x - 2 ) + ( x - 1 ) + x = 3 x - 3 - - - > always divisible by 3 answer : d" | a ) positive . , b ) divisible by 2 . , c ) divisible by 4 . , d ) divisible by 3 . , e ) divisible by 5 . | d | add(divide(subtract(3, const_1), multiply(const_2, const_1)), const_1) | multiply(const_1,const_2)|subtract(n0,const_1)|divide(#1,#0)|add(#2,const_1)| | physics |
for all even integers n , h ( n ) is defined to be the sum of the even integers between 6 and n , inclusive . what is the value of h ( 18 ) / h ( 10 ) ? | "concept : when terms are in arithmetic progression ( a . p . ) i . e . terms are equally spaced then mean = median = ( first + last ) / 2 and sum = mean * number of terms h ( 18 ) = [ ( 6 + 18 ) / 2 ] * 7 = 84 h ( 10 ) = ( 6 + 10 ) / 2 ] * 3 = 24 h ( 18 ) / h ( 10 ) = ( 84 ) / ( 24 ) ~ 4 answer : b" | a ) 1.8 , b ) 4 , c ) 6 , d ) 18 , e ) 60 | b | divide(divide(multiply(add(18, 6), add(divide(subtract(18, 6), const_2), const_1)), const_2), divide(multiply(add(divide(subtract(10, 6), const_2), const_1), add(6, 10)), const_2)) | add(n0,n1)|add(n0,n2)|subtract(n1,n0)|subtract(n2,n0)|divide(#2,const_2)|divide(#3,const_2)|add(#4,const_1)|add(#5,const_1)|multiply(#0,#6)|multiply(#7,#1)|divide(#8,const_2)|divide(#9,const_2)|divide(#10,#11)| | general |
20 % of a 6 litre solution and 60 % of 4 litre solution are mixed . what percentage of the mixture of solution | 20 % of 6 litre is ( 6 * 20 / 100 ) = 1.2 litre 60 % of 4 litre is ( 60 * 4 / 100 ) = 2.4 litre the mixture is 3.6 litre so the percentage is ( 3.6 * 100 / 10 ) = 36 % answer : a | a ) 36 % , b ) 35 % , c ) 34 % , d ) 33 % , e ) 32 % | a | multiply(divide(add(divide(multiply(60, 4), const_100), divide(multiply(20, 6), const_100)), add(6, 4)), const_100) | add(n1,n3)|multiply(n2,n3)|multiply(n0,n1)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100) | gain |
population of a city decreases by 10 % at the end of first year and increases by 10 % at the end of second year and again decreases by 10 % at the end of third year . if the population of the city at the end of third year is 4455 , then what was the population of the city at the beginning of the first year ? | m . f = 90 / 100 * 110 * 100 * 90 / 100 = 81 * 11 / 1000 population before 3 yrs = i . q / m . f = 4455 * 1000 / 81 * 11 = 5000 answer : a | a ) 5000 , b ) 4500 , c ) 4950 , d ) 1000 , e ) 2000 | a | divide(4455, multiply(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100)))) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|multiply(#3,#2)|divide(n3,#4) | gain |
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 . 425 in all . for how many days was he absent ? | 30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = 10 b | a ) 8 , b ) 10 , c ) 15 , d ) 17 , e ) 19 | b | subtract(30, divide(add(multiply(7.5, 30), 425), add(7.5, 25))) | add(n1,n2)|multiply(n0,n2)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3) | physics |
stalin and heather are 20 miles apart and walk towards each other along the same route . stalin walks at constant rate that is 1 mile per hour faster than heather ' s constant rate of 5 miles / hour . if heather starts her journey 20 minutes after stalin , how far from the original destination has heather walked when t... | original distance between s and h = 20 miles . speed of s = 5 + 1 = 6 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 20 / 60 = t + 2 / 6 hours . now , the total distances traveled by s and h = 20 miles - - - > 6 * ( t + 2 / 6 ) + 5 * t = 20 - - - > t = 8 / 11 hours . thus h has... | a ) 4 miles , b ) 6 miles , c ) 9 miles , d ) 10 miles , e ) 12 mile | a | divide(multiply(divide(subtract(20, multiply(divide(add(1, 5), const_60), 20)), add(5, add(1, 5))), 5), const_2) | add(n1,n2)|add(n2,#0)|divide(#0,const_60)|multiply(n0,#2)|subtract(n0,#3)|divide(#4,#1)|multiply(n2,#5)|divide(#6,const_2) | physics |
what is the square root of 16 ? | "4 x 4 = 16 answer b" | a ) 8 , b ) 4 , c ) 35 , d ) 42 , e ) 86 | b | circle_area(divide(16, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
a bag marked at $ 240 is sold for $ 120 . the rate of discount is ? | "rate of discount = 120 / 240 * 100 = 50 % answer is d" | a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 45 % | d | multiply(divide(subtract(240, 120), 240), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the average weight of 8 person ' s increases by 2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 2 ) kg = 16 kg . weight of new person = ( 65 + 16 ) kg = 81 kg . c )" | a ) 70 kg , b ) 80 kg , c ) 81 kg , d ) 90 kg , e ) 91 kg | c | add(multiply(8, 2), 65) | multiply(n0,n1)|add(n2,#0)| | general |
the mean proportional between 4 and 9 is ? | "7 / 20 * 100 = 35 answer : b" | a ) 33 , b ) 77 , c ) 35 , d ) 88 , e ) 29 | b | sqrt(multiply(4, 9)) | multiply(n0,n1)|sqrt(#0)| | general |
a man buy a book in rs 50 & sale it rs 100 . what is the rate of profit ? ? ? | "cp = 50 sp = 100 profit = 100 - 50 = 50 % = 50 / 50 * 100 = 100 % answer : b" | a ) 10 % , b ) 100 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(100, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 6 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 6 = 720 answer : c" | a ) 480 , b ) 287 , c ) 720 , d ) 270 , e ) 927 | c | multiply(24, 20) | multiply(n0,n2)| | physics |
a group of students was interviewed for that if it was asked whether or not they speak french and / or english . among those who speak french , 20 speak english well , while 60 of them do not speak english . if 60 % of students do not speak french , how many students were surveyed ? | "number of students who speak french are 60 + 20 = 80 of total students , the percentage of students who do not speak french was 60 % - - > percentage of who do is 40 % 80 - - - - - - - 40 % x - - - - - - - 100 % x = 80 * 100 / 40 = 200 = number of all students answer is e" | a ) 250 , b ) 225 , c ) 175 , d ) 195 , e ) 200 | e | divide(add(20, 60), divide(subtract(const_100, 60), const_100)) | add(n0,n1)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,#2)| | gain |
the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.2 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.2 weight of the new student = ( 30 * 27.2 – 29 * 28 ) = 816 - 812 = 4 answer : a" | a ) 4 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 21 kg , e ) none of these | a | subtract(multiply(add(29, const_1), 27.2), multiply(29, 28)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
the average expenditure of a labourer for 6 months was 80 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 × 80 ) – debt = 480 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 750 average monthly income = 750 ÷ 10 = 75 answer c" | a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | c | divide(add(add(multiply(80, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)| | general |
the length of a rectangle is halved , whileits breadth is tripled . wat is the % change in area ? | "let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c" | a ) 40 % , b ) 45 % , c ) 50 % , d ) 65 % , e ) 70 % | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | divide(const_1,const_2)|multiply(#0,const_3)|subtract(#1,const_1)|divide(#2,const_1)|multiply(#3,const_100)| | geometry |
a man saves 25 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 175 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 75 savings = rs . 25 present expenditure 75 + 75 * ( 10 / 100 ) = rs . 82.5 present savings = 100 – 82.50 = rs . 17.50 if savings is rs . 17.50 , salary = rs . 100 if savings is rs . 175 , salary = 100 / 17.5 * 175 = 1000 answer : a" | a ) rs . 1000 , b ) rs . 2000 , c ) rs . 1500 , d ) rs . 3000 , e ) rs . 3100 | a | divide(multiply(175, const_100), subtract(const_100, add(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(10, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)| | general |
a 15 % stock yielding 12 % is quoted at : | "income of rs 12 on investment of rs 100 income of rs 15 on investment of ? = ( 15 * 100 ) / 12 = 125 answer : d" | a ) s . 83.33 , b ) s . 110 , c ) s . 112 , d ) s . 125 , e ) s . 140 | d | multiply(divide(const_100, 12), 15) | divide(const_100,n1)|multiply(n0,#0)| | gain |
on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 2 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on september 1 than were in stock on january 1 ? | "jan 1 = c feb 1 = c - 2 march 1 = c - 4 april 1 = c - 8 may 1 = c - 10 june 1 = c - 12 july 1 = c - 14 aug 1 = c - 16 sept 1 = c - 18 difference between stock on september 1 than were in stock on january 1 will be - c - ( c - 18 ) = 18 hence answer will be ( a )" | a ) 18 , b ) 12 , c ) 20 , d ) 32 , e ) 36 | a | multiply(subtract(const_10, 1), 2) | subtract(const_10,n0)|multiply(n1,#0)| | gain |
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 $ 0.92 at the candy store . what is john ’ s weekly allowance ? | "x = 3 x / 5 + 1 / 3 * 2 x / 5 + 92 4 x / 15 = 92 x = 345 = $ 3.45 the answer is d ." | a ) $ 2.55 , b ) $ 2.85 , c ) $ 3.15 , d ) $ 3.45 , e ) $ 3.75 | d | divide(0.92, 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 |
a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1300 plus 2.5 % commission on all sales exceeding rs . 4000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | [ 1300 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 18000 answer : c | a ) 12028 , b ) 12000 , c ) 12019 , d ) 12197 , e ) 18000 | c | divide(subtract(subtract(multiply(const_100, const_10), multiply(4000, divide(2.5, const_100))), 600), divide(2.5, const_100)) | divide(n2,const_100)|multiply(const_10,const_100)|multiply(n3,#0)|subtract(#1,#2)|subtract(#3,n4)|divide(#4,#0) | general |
a sum fetched a total simple interest of 4020.75 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4020.75 ) / ( 9 x 5 ) = 402075 / 45 = 8935 . answer e" | a ) 5768 , b ) 8925 , c ) 2345 , d ) 6474 , e ) 8935 | e | divide(divide(multiply(4020.75, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 28 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 28 + 21 = 49 t = 12 d = 49 * 12 = 588 answer : d" | a ) 457 km , b ) 444 km , c ) 547 km , d ) 588 km , e ) 653 km | d | add(multiply(divide(60, subtract(21, 28)), 28), multiply(divide(60, subtract(21, 28)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 11 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ? | the total number of the people in the room must be a multiple of both 7 and 11 ( in order 3 / 7 and 5 / 11 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 11 , which is 77 . since , the total number of the people in the room is greater than 50 and less than 100 ... | a ) 21 , b ) 35 , c ) 33 , d ) 60 , e ) 65 | c | divide(multiply(multiply(7, 11), 3), 7) | multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1) | general |
one night 15 percent of the female officers on a police force were on duty . if 300 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 300 number of female officers on duty on that night = 300 / 2 = 150 ( 15 / 100 ) * f = 150 = > f = 1000 answer e" | a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | e | divide(divide(300, const_2), divide(15, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
with # andeach representing different digits in the problem below , the difference between # and # # is 801 . what is the value of ? # - # # ____ 812 | 100 x - ( 10 x + x ) - - - - - - - - - - - 89 x = 801 x = 9 c | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | subtract(subtract(812, 801), const_2) | subtract(n1,n0)|subtract(#0,const_2) | general |
a cat leaps 5 leaps for every 4 leaps of a dog , but 3 leaps of the dog are equal to 4 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ? | "solution : given ; 3 dog = 4 cat ; or , dog / cat = 4 / 3 ; let cat ' s 1 leap = 3 meter and dogs 1 leap = 4 meter . then , ratio of speed of cat and dog = 3 * 5 / 4 * 4 = 15 : 16 . ' ' answer : option d" | a ) 11 : 15 , b ) 15 : 11 , c ) 16 : 15 , d ) 15 : 16 , e ) 11 : 16 | d | divide(multiply(divide(3, 4), 5), 4) | divide(n2,n3)|multiply(n0,#0)|divide(#1,n1)| | other |
( √ 97 + √ 486 ) / √ 54 = ? | "( √ 96 + √ 486 ) / √ 54 = ( 4 √ 6 + 9 √ 6 ) / 3 √ 6 = 13 √ 6 / 3 √ 6 = 13 / 3 hence , the correct answer is e ." | a ) 2 √ 2 , b ) 2 √ 3 , c ) 3 √ 2 , d ) 3 √ 3 , e ) 13 / 3 | e | divide(add(sqrt(97), sqrt(486)), sqrt(54)) | sqrt(n0)|sqrt(n1)|sqrt(n2)|add(#0,#1)|divide(#3,#2)| | general |
a secret can be told only 2 persons in 5 minutes . the same person tells to 2 more persons and so on . how long will take to tell it to 768 persons ? | at start one person will tell to 2 persons , it will take 5 min , now that 1 + 2 = 3 persons will tell this to next 6 persons , then 1 + 2 + 6 = 9 persons will tell to next 18 persons , then 1 + 2 + 6 + 18 = 27 persons to 54 similarly 1 + 2 + 6 + 18 + 54 = 81 persons will tell this to 162 persons similarly 1 + 2 + 6 + ... | a ) 25 min , b ) 32 min , c ) 33 min , d ) 34 min , e ) 35 min | e | multiply(add(2, 5), 5) | add(n0,n1)|multiply(n1,#0) | physics |
in what time will a train 110 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 110 / 40 = 2.75 sec . answer : d" | a ) 2.35 sec , b ) 2.85 sec , c ) 7.5 sec , d ) 2.75 sec , e ) 1.5 sec | d | divide(110, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 6 miles per hour while the second runner jogs due east at a constant rate that is 2 miles per hour faster than the first runner ' s rate . how far apart , to the neares... | "if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 6 m / h - - > 3 m in 1 / 2 hr side 2 = 8 m / h - - > 4 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 3 ^ 2 d ^ 2 = 25 = 5 answer option c" | a ) 6 , b ) 7 , c ) 5 , d ) 12 , e ) 14 | c | sqrt(add(power(multiply(6, divide(1, 2)), 2), power(multiply(subtract(6, 2), divide(1, 2)), 2))) | divide(n3,n2)|subtract(n1,n2)|multiply(n1,#0)|multiply(#0,#1)|power(#2,n2)|power(#3,n2)|add(#4,#5)|sqrt(#6)| | physics |
find the smallest number which when divided by 13 and 15 leaves respective remainders of 2 and 4 | let ' n ' is the smallest number which divided by 13 and 15 leaves respective remainders of 2 and 4 . required number = ( lcm of 13 and 15 ) - ( common difference of divisors and remainders ) = ( 195 ) - ( 11 ) = 184 . answer : c | a ) 187 , b ) 197 , c ) 184 , d ) 219 , e ) 227 | c | subtract(multiply(13, 15), add(const_10, const_1)) | add(const_1,const_10)|multiply(n0,n1)|subtract(#1,#0) | general |
if sharon ' s weekly salary increased by 16 percent , she would earn $ 406 per week . if instead , her weekly salary were to increase by 20 percent , how much would she earn per week ? | "( 406 / 116 ) 120 = 420 in this case long division does not take much time . ( 406 / 116 ) = 3.5 35 * 12 = 420 ( 350 + 70 ) answer a" | a ) $ 420 , b ) $ 382 , c ) $ 385 , d ) $ 392 , e ) $ 399 | a | add(divide(406, add(const_1, divide(16, const_100))), multiply(divide(20, const_100), divide(406, add(const_1, divide(16, const_100))))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4)| | general |
two tains are running in opposite directions with the same speed . if the length of each train is 120 metres and they cross each other in 12 seconds , then the speed of each train ( in km / hr ) is : | sol . let the speed of each train be x m / sec . then , relative speed of the two trains = 2 x m / sec . so , 2 x = ( 120 + 120 ) / 12 ⇔ 2 x = 20 ⇔ x = 10 . ∴ speed of each train = 10 m / sec = [ 10 * 18 / 5 ] km / hr = 36 km / hr . answer c | a ) 12 , b ) 24 , c ) 36 , d ) 48 , e ) 38 | c | multiply(divide(divide(add(120, 120), 12), const_2), const_3_6) | add(n0,n0)|divide(#0,n1)|divide(#1,const_2)|multiply(#2,const_3_6) | physics |
a taxi company charges $ 2.5 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ? | "if we start out with $ 4.90 and have to spend $ 2.5 for the first quarter - mile , we will have $ 2.40 left to spend on quarter - mile intervals . since $ 2.40 / $ 0.15 = 16 , we can buy 16 more quarter - miles , and will travel 17 quarter miles in all : 17 × 1 / 4 = 4 1 / 4 miles . the correct answer is choice ( b ) ... | a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 2 miles , e ) 6 1 / 4 miles | b | divide(divide(multiply(subtract(4.90, 2.5), const_100), const_3), const_4) | subtract(n1,n0)|multiply(#0,const_100)|divide(#1,const_3)|divide(#2,const_4)| | physics |
a student has to obtain 60 % of the total marks to pass . he got 160 marks and failed by 20 marks . the maximum marks are ? | "let the maximum marks be x then , 60 % of x = 160 + 20 60 x / 100 = 180 60 x = 180 * 100 = 18000 x = 300 answer is c" | a ) 210 , b ) 280 , c ) 300 , d ) 320 , e ) 340 | c | divide(add(160, 20), divide(60, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 7 - 2 = 5 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 5 = 216,0... | a ) 215,000 , b ) 216,000 , c ) 217,000 , d ) 218,000 , e ) 219,000 | b | multiply(multiply(subtract(7, const_2), const_3600), const_12) | subtract(n0,const_2)|multiply(#0,const_3600)|multiply(#1,const_12) | general |
two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 36.00001 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 36 = 25 / 9 = > x = 50 . answer : a" | a ) 50 , b ) 99 , c ) 77 , d ) 26 , e ) 23 | a | divide(multiply(36.00001, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
two numbers n and 14 have lcm = 56 and gcf = 10 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 14 × n = 56 × 10 n = 56 × 10 / 14 = 40 correct answer b" | a ) 24 , b ) 40 , c ) 44 , d ) 54 , e ) 64 | b | divide(multiply(56, 10), 14) | multiply(n1,n2)|divide(#0,n0)| | physics |
how many liters of water must be evaporated from 50 liters of a 2 percent sugar solution to get a 4 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.02 ( 50 ) = 0.04 ( 50 - x ) 0.04 x = 2 - 1 x = 1 / 0.04 = 25 liters the answer is b ." | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | subtract(50, multiply(divide(50, const_100), 4)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
a batsman scored 60 runs which included 2 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 60 - ( 2 x 4 + 2 x 6 ) = 60 - ( 20 ) = 20 now , we need to calculate 20 is what percent of 60 . = > 20 / 60 * 100 = 33.33 % option b" | a ) 30 % , b ) 33.33 % , c ) 40 % , d ) 60 % , e ) 50 % | b | multiply(divide(subtract(60, add(multiply(2, 2), multiply(2, 2))), 60), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 5 people to do 5 times of that particular work ? | 3 people can do the work one time in one day . 1 person can do 1 / 3 of the work in one day . 5 people can do 5 / 3 of the work in one day . 5 people can do 5 times the work in 3 days . the answer is c . | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | c | multiply(3, divide(5, 5)) | divide(n3,n3)|multiply(n0,#0) | physics |
what is the value of â ˆ š 64 % ? | "explanation : br > â ˆ š 64 % = â ˆ š 64 / â ˆ š 100 = 8 / 10 = 80 / 100 = 80 % correct answer is a ) 80 %" | a ) 80 % , b ) 20 % , c ) 40 % , d ) 90 % , e ) 26 % | a | circle_area(divide(64, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | gain |
tom opened a shop investing rs . 500 . jose joined him 2 months later , investing rs . 350 . they earned a profit of rs . 1720 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 1720 = 80 a" | a ) 80 , b ) 20 , c ) 30 , d ) 41 , e ) 15 | a | multiply(1720, subtract(const_1, divide(multiply(500, multiply(2, multiply(2, const_3))), add(multiply(350, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(500, multiply(2, multiply(2, const_3))))))) | multiply(n1,const_3)|multiply(#0,n1)|multiply(n0,#1)|subtract(#1,n1)|multiply(n2,#3)|add(#4,#2)|divide(#2,#5)|subtract(const_1,#6)|multiply(n3,#7)| | gain |
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 8748 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 8748 ⇔ 12 x 2 = 8748 ⇔ x 2 = 729 ⇔ x = 27 . so , length = 81 m and breadth = 108 m . perimeter = [ 2 ( 81 + 108 ) ] m = 378 m . ∴ cost of fencing = rs . ( 0.25 × 378 ) = rs . 94.50 . answer d" | a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 94.50 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(8748, multiply(3, 4)))), multiply(4, sqrt(divide(8748, multiply(3, 4))))), 25), const_100) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics |
the angle between the minute hand and the hour hand of a clock when the time is 4.20 , is : | "angle traced hr 13 / 3 = ( 360 / 12 * 13 / 3 ) = 130 traced by min hand 20 min = ( 360 / 60 * 20 ) = 120 req = ( 130 - 120 ) = 10 d answer a" | a ) 10 d , b ) 0 d , c ) 15 d , d ) 5 d , e ) 20 d | a | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | physics |
the positive numbers w , x , y , and z are such that x is 40 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 23.2 - answer d" | a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | d | multiply(const_100, subtract(multiply(multiply(divide(add(40, const_100), const_100), divide(add(40, const_100), const_100)), divide(subtract(const_100, 40), const_100)), const_1)) | add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#3,#3)|multiply(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general |
if 10 men do a work in 80 days , in how many days will 20 men do it ? | "10 * 80 = 20 * x x = 40 days answer : c" | a ) 18 days , b ) 38 days , c ) 40 days , d ) 48 days , e ) 50 days | c | divide(multiply(10, 80), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
a trader sells 60 meters of cloth for rs . 8400 at the profit of rs . 12 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 8400 / 60 = rs . 140 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 140 - rs . 12 = rs . 128 . answer : a" | a ) 128 , b ) 140 , c ) 123 , d ) 110 , e ) 150 | a | subtract(divide(8400, 60), 12) | divide(n1,n0)|subtract(#0,n2)| | physics |
48 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 48 = 21 answer : e" | a ) 66 , b ) 26 , c ) 42 , d ) 27 , e ) 21 | e | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 48), const_2) | add(const_3,const_4)|add(#0,const_2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_2)| | general |
a car covers a distance of 624 km in 6 â ½ hours . find its speed ? | explanation : 624 / 6 = 104 kmph answer : c | a ) 104 , b ) 190 , c ) 109 , d ) 278 , e ) 211 | c | divide(624, 6) | divide(n0,n1) | physics |
when average age of 23 members are 0 , how many members greater than 0 ? | "average of 23 numbers = 0 . sum of 23 numbers ( 0 x 23 ) = 0 . it is quite possible that 22 of these numbers may be positive and if their sum is a then 23 rd number is ( - a ) answer is 22 ( b )" | a ) 17 , b ) 22 , c ) 21 , d ) 24 , e ) 25 | b | subtract(23, const_1) | subtract(n0,const_1)| | general |
if 2 : 9 : : x : 18 , then find the value of x | "explanation : treat 2 : 9 as 2 / 9 and x : 18 as x / 18 , treat : : as = so we get 2 / 9 = x / 18 = > 9 x = 36 = > x = 4 option c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(add(multiply(9, 2), 9), 18) | multiply(n0,n1)|add(n1,#0)|divide(#1,n2)| | general |
during the first two weeks of january , the total rainfall in springdale was 35 inches . if the rainfall during the second week was 1.5 times the rainfall during the first week , what was the rainfall during the second week of january ? | total rainfall in 2 weeks = 35 inches . assume the rainfall in second week = 1 . x rainfall in first week = x total rainfall = 2.5 x = 35 inches x = 14 and 1.5 x = 21 rainfall during second week = 21 inches option e | a ) 5 inches , b ) 6 inches , c ) 9 inches , d ) 10 inches , e ) 21 inches | e | multiply(35, divide(1.5, add(const_1, 1.5))) | add(n1,const_1)|divide(n1,#0)|multiply(n0,#1) | general |
if you roll one fair 6 - sided die , what is the probability that the number is even or less than 3 ? | the numbers which satisfy the conditions are 1 , 2 , 4 , and 6 . the probability is 4 / 6 = 2 / 3 the answer is a . | a ) 2 / 3 , b ) 1 / 2 , c ) 3 / 4 , d ) 5 / 6 , e ) 1 / 3 | a | add(divide(const_3, 6), divide(const_1, 6)) | divide(const_3,n0)|divide(const_1,n0)|add(#0,#1) | probability |
a fifth of arun â € ™ s marks in mathematics exceed a third of his marks in english by 20 . if he got 260 marks in two subjects together how many marks did he got in english ? | let arun â € ™ s marks in mathematics and english be x and y then ( 1 / 5 ) x - ( 1 / 3 ) y = 20 3 x - 5 y = 300 â € ¦ â € ¦ > ( 1 ) x + y = 260 â € ¦ â € ¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 200 and y = 60 answer is c . | a ) 12080 , b ) 18060 , c ) 20060 , d ) 20040 , e ) none of them | c | multiply(multiply(multiply(260, 20), const_2), const_2) | multiply(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2) | general |
in a colony of 70 residents , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of education to uneducated persons is 8 : 27 , then find the ratio of the number of educated and uneducated men in the colony ? | "number of men in the colony = 4 / 7 ( 70 ) = 40 number of women in the colony = 3 / 7 ( 70 ) = 30 number of educated women in the colony = 1 / 5 ( 30 ) = 6 number of uneducated women in the colony = 4 / 5 ( 30 ) = 24 number of educated persons in the colony = 8 / 35 ( 70 ) = 16 as 6 females are educated , remaining 10... | a ) 1 : 7 , b ) 1 : 1 , c ) 1 : 2 , d ) 1 : 9 , e ) 1 : 3 | e | divide(subtract(multiply(divide(70, add(8, 27)), 8), divide(multiply(divide(70, add(4, 3)), 3), add(1, 4))), subtract(multiply(divide(70, add(8, 27)), 27), multiply(divide(multiply(divide(70, add(4, 3)), 3), add(1, 4)), 4))) | add(n5,n6)|add(n1,n2)|add(n1,n3)|divide(n0,#0)|divide(n0,#1)|multiply(n5,#3)|multiply(n2,#4)|multiply(n6,#3)|divide(#6,#2)|multiply(n1,#8)|subtract(#5,#8)|subtract(#7,#9)|divide(#10,#11)| | other |
40 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 40 orange + mango + apple = 4 40 + 10 + 4 - 7 = 47 like apple answer : a" | a ) 47 , b ) 46 , c ) 54 , d ) 58 , e ) 62 | a | add(add(subtract(40, const_3), 7), subtract(10, 7)) | subtract(n0,const_3)|subtract(n2,n1)|add(n1,#0)|add(#2,#1)| | general |
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 50,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 150000 - x ) respectively . then , ( 1 / 3 ) x = ( 1 / 2 ) ( 150000 - x ) ( x / 3 ) + ( x / 2 ) = 75000 5 x / 6 = 75000 x = 75000 x 6 / 5 = 90000 savings in public provident fund = rs . ( 150000 - 90000 ) = rs . 60000 answer is b ." | a ) 30000 , b ) 60000 , c ) 50000 , d ) 90000 , e ) 70000 | b | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(n0,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)| | general |
a girl goes to her school from her house at a speed of 6 km / hr and returns at a speed of 4 km / hr . if she takes 10 hours in going and coming back , the distance between her school and house is | let distance be d 10 = d / 4 + d / 6 answer : d | a ) 12 kms , b ) 16 kms , c ) 20 kms , d ) 24 kms , e ) none of above | d | divide(multiply(10, multiply(6, 4)), add(6, 4)) | add(n0,n1)|multiply(n0,n1)|multiply(n2,#1)|divide(#2,#0) | physics |
if x = - 6 and y = - 3 , what is the value of 4 ( x - y ) ^ 2 - xy ? | x = - 6 and y = - 3 x - y = - 6 - ( - 3 ) = - 6 + 3 = - 3 x * y = - 6 * - 3 = 18 now we apply it in the equation 4 ( x - y ) ^ 2 - xy = 4 ( - 3 ) ^ 2 - 18 = = > 4 * 9 - 18 = 36 - 18 = 18 answer : b | a ) 20 , b ) 18 , c ) 17 , d ) 22 , e ) 23 | b | subtract(multiply(power(subtract(negate(6), negate(3)), 2), 4), multiply(negate(6), negate(3))) | negate(n0)|negate(n1)|multiply(#0,#1)|subtract(#0,#1)|power(#3,n3)|multiply(n2,#4)|subtract(#5,#2) | general |
if the sum of the 4 th term and the 12 th term of an arithmetic progression is 12 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 12 i . e . , ( a + 3 d ) + ( a + 11 d ) = 12 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 12 - - - - - - - - - - - - - - - from ( 1 ) = 90 answer : a" | a ) 90 , b ) 80 , c ) 70 , d ) 60 , e ) 50 | a | multiply(divide(15, const_2), 12) | divide(n3,const_2)|multiply(n2,#0)| | general |
( 51 + 52 + 53 + . . . + 100 ) = ? | "explanation : this is an a . p . in which a = 51 , l = 100 and n = 50 . sum = n ( a + l ) = 50 x ( 51 + 100 ) = ( 25 x 151 ) = 3775 . 2 2 answer : d" | a ) 2525 , b ) 2975 , c ) 3225 , d ) 3775 , e ) 2753 | d | divide(add(51, 52), 53) | add(n0,n1)|divide(#0,n2)| | general |
the edges of a cuboid are 2 cm , 5 cm and 8 cm . find the volume of the cuboid ? | "2 * 5 * 8 = 80 answer : b" | a ) 90 , b ) 80 , c ) 40 , d ) 120 , e ) 70 | b | volume_rectangular_prism(2, 5, 8) | volume_rectangular_prism(n0,n1,n2)| | physics |
a car was driving at 50 km / h for 30 minutes , and then at 40 km / h for another 30 minutes . what was its average speed ? | "driving at 50 km / h for 30 minutes , distance covered = 50 * 1 / 2 = 25 km driving at 40 km / h for 30 minutes , distance covered = 40 * 1 / 2 = 20 km average speed = total distance / total time = 45 / 1 = 45 km / h answer : a" | a ) 80 . , b ) 75 . , c ) 70 . , d ) 65 . , e ) 54 . | a | divide(add(multiply(divide(30, add(30, 30)), 50), multiply(divide(30, add(30, 30)), 40)), divide(add(30, 30), const_60)) | add(n1,n3)|divide(n1,#0)|divide(n3,#0)|divide(#0,const_60)|multiply(n0,#1)|multiply(n2,#2)|add(#4,#5)|divide(#6,#3)| | general |
a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 11 : 00 a . m . ? | "sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour . hence , they cross each other 15 times before 11 : 00 a . m . answer d" | a ) 8 , b ) 7 , c ) 6 , d ) 15 , e ) 3 | d | add(add(2, 3), add(2, 3)) | add(n1,n2)|add(#0,#0)| | physics |
a can do a work in 21 days and b alone can do it 28 days , they begin the work together , but a left after some days . b completed the remaining work in 21 days . after how many days did a leave ? | explanation : ( a + b ) 1 day work = { 1 / 21 + 1 / 28 } = 1 / 12 = 12 days is required for a & b b ’ s 1 day work = 1 / 28 b ’ s 21 days work = 21 / 28 1 - 21 / 28 = 7 / 28 - 1 / 4 1 / 4 * 12 = 3 days answer : option c | a ) 10 , b ) 8 , c ) 3 , d ) 15 , e ) 16 | c | divide(subtract(const_1, multiply(speed(const_1, 28), 21)), add(speed(const_1, 21), speed(const_1, 28))) | speed(const_1,n1)|speed(const_1,n0)|add(#1,#0)|multiply(n0,#0)|subtract(const_1,#3)|divide(#4,#2) | physics |
a started a business with an investment of rs . 10000 and after 7 months b joined him investing rs . 12000 . if the profit at the end of a year is rs . 24000 , then the share of b is ? | "ratio of investments of a and b is ( 10000 * 12 ) : ( 12000 * 5 ) = 2 : 1 total profit = rs . 24000 share of b = 1 / 3 ( 24000 ) = rs . 8000 answer : b" | a ) 10000 , b ) 8000 , c ) 12000 , d ) 6000 , e ) 14000 | b | subtract(24000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
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