Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
two numbers n and 12 have lcm = 48 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 12 × n = 48 × 8 n = 48 × 8 / 12 = 32 correct answer e" | a ) 24 , b ) 34 , c ) 44 , d ) 54 , e ) 32 | e | divide(multiply(48, 8), 12) | multiply(n1,n2)|divide(#0,n0)| | physics |
in a hostel , the number of students decreased by 9 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previ... | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 8 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.91 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.91 * 1.2 ) = > f ( new ) = 0.915 f therefore the ... | a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 8.4 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 9), add(const_100, 20)))), const_100) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general |
a company produces on an average 4000 items per month for the first 3 months . how many items it must produce on an average per month over the next 9 months , to average of 4375 items per month over the whole ? | sol . required average = ( 4375 x 12 ) - ( 4000 x 3 ) / 9 ⇔ 52500 - 12000 / 9 ⇔ 40500 / 9 = 4500 . answer a | a ) 4500 , b ) 4600 , c ) 4680 , d ) 4710 , e ) none | a | divide(subtract(multiply(4375, add(9, 3)), multiply(4000, 3)), 9) | add(n1,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2) | general |
pat , kate and mark charged a total of 144 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate . | "let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 144 - total hours charged for , x = 16 . mat charged 6 x - x or 5 x for more hours than kate , or for 80 hours . c is correct" | a ) 18 , b ) 36 , c ) 80 , d ) 90 , e ) 108 | c | multiply(divide(144, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1)) | add(n1,const_2)|multiply(n2,const_2)|add(#0,#1)|subtract(#1,n1)|divide(n0,#2)|multiply(#4,#3)| | general |
69 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,075 bacteria each . how many bacteria are there approximately ? | "69 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,075 bacteria each . how many bacteria are there approximately ? a . 10 ^ 7 b . 10 ^ 8 c . 10 ^ 9 d . 10 ^ 10 e . 10 ^ 11 - > due to approximately , 69 = 70 , 48 = 50 , 25,075 = 25,000 are derived , which makes (... | a ) 10 ^ 7 , b ) 10 ^ 8 , c ) 10 ^ 9 , d ) 10 ^ 10 , e ) 10 ^ 11 | a | multiply(multiply(69, 48), add(add(multiply(multiply(const_0_25, const_1000), const_100), multiply(add(const_3, const_4), const_10)), const_3)) | add(const_3,const_4)|multiply(const_0_25,const_1000)|multiply(n0,n1)|multiply(#1,const_100)|multiply(#0,const_10)|add(#3,#4)|add(#5,const_3)|multiply(#6,#2)| | general |
divide $ 5000 among x , y in the ratio 2 : 8 . how many $ that x get ? | sum of ratio terms = 2 + 8 = 10 x = 5000 * 2 / 10 = $ 1000 answer is a | a ) $ 1000 , b ) $ 1050 , c ) $ 1100 , d ) $ 1150 , e ) $ 1200 | a | subtract(5000, multiply(const_4, const_1000)) | multiply(const_1000,const_4)|subtract(n0,#0) | general |
a teacher gave the same test to 3 history classes : u , b , and c . the average ( arithmetic mean ) scores for the 3 classes were 65 , 80 , and 77 , respectively . the ratio of the numbers of students in each class who took the test was 4 to 6 to 5 , respectively . what was the average score for the 3 classes combined ... | ans : b ( 75 ) lets say class u indeed has 4 children , b has 6 children and c has 5 children . now , if the average of class u is 65 , hence total marks awarded in the class = 65 * 4 = 260 similarly , class b = 80 * 6 = 480 class c = 77 * 5 = 385 total marks provided = u + b + c = 260 + 480 + 385 = 1125 avg . marks = ... | a ) 74 , b ) 75 , c ) 76 , d ) 77 , e ) 78 | b | divide(add(add(multiply(65, 4), multiply(80, 6)), multiply(77, 5)), add(add(4, 6), 5)) | add(n5,n6)|multiply(n2,n5)|multiply(n3,n6)|multiply(n4,n7)|add(#1,#2)|add(n7,#0)|add(#4,#3)|divide(#6,#5) | general |
a cube of edge 15 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in water level ? | "increase in volume = volume of the cube = 15 * 15 * 15 cm ^ 3 rise in water level = volume / area = 15 * 15 * 15 / 20 * 15 = 11.25 cm answer is c" | a ) 6 cm , b ) 8.25 cm , c ) 11.25 cm , d ) 15.12 cm , e ) 20.62 cm | c | divide(volume_cube(15), multiply(20, 15)) | multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)| | geometry |
how many positive integers less than 28 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 3 odd multiples of 5 : { 5 , 15 , 25 } 11 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 } notice , that 5 is in two sets , thus total # of integers satis... | a ) 27 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | d | subtract(subtract(subtract(28, 2), const_1), const_1) | subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
what are the last two digits of 63 * 35 * 37 * 82 * 71 * 41 * 53 ? | we know that to find the last two digits , we need to find the remainder we get when we divide the product by 100 . remainder of ( 63 * 35 * 37 * 82 * 71 * 41 ) / 100 note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator . but before we do that , here is an important no... | a ) 10 , b ) 30 , c ) 40 , d ) 70 , e ) 80 | e | subtract(82, const_2) | subtract(n3,const_2) | general |
in n is a positive integer less than 200 , and 27 n / 60 is an integer , then n has how many different positive prime factors ? | "( a ) . 27 n / 60 must be an integer . = > 9 n / 20 must be an integer . hence n must be a multiple of 2 * 5 . = > n has 2 different prime integers ." | a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | a | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)| | general |
the ratio of the adjacent angles of a parallelogram is 5 : 10 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 5 x + 10 x = 180 ° or , 15 x = 180 ° or , x = 12 ° hence the angles of the parallelogram are 60 ° and 120 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 ... | a ) 168 ° , b ) 228 ° , c ) 144 ° , d ) 224 ° , e ) none of these | c | multiply(divide(multiply(add(multiply(multiply(5, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 10) | add(n2,n3)|multiply(const_2,n0)|add(n4,#0)|multiply(#1,const_10)|add(#3,const_100)|add(n5,#2)|multiply(#4,const_2)|divide(#6,#5)|multiply(n1,#7)| | other |
the population of a city increases by 5 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 4 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 4 / 100 ) ^ 3 = 112.4864 ∴ required percentage = 12.48 % answer : d" | a ) 9 % , b ) 9.27 % , c ) 27 % , d ) 12.48 % , e ) none of these | d | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(5, 1)), subtract(5, 1)), const_100), subtract(5, 1))), subtract(5, 1)), const_100), add(divide(multiply(add(const_100, subtract(5, 1)), subtract(5, 1)), const_100), subtract(5, 1))) | subtract(n0,n1)|add(#0,const_100)|multiply(#1,#0)|divide(#2,const_100)|add(#3,#0)|add(#4,const_100)|multiply(#5,#0)|divide(#6,const_100)|add(#4,#7)| | general |
the sum of the present age of henry and jill is 33 . what is their present ages if 6 years ago henry was twice the age of jill ? | "let the age of jill 6 years ago be x , age of henry be 2 x x + 6 + 2 x + 6 = 33 x = 7 present ages will be 13 and 20 answer : a" | a ) and 20 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | a | subtract(33, divide(add(33, 6), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
in a division sum , the remainder is 5 and the divisor is 3 times the quotient and is obtained by adding 3 to the thrice of the remainder . the dividend is : | "diver = ( 5 * 3 ) + 3 = 18 3 * quotient = 18 quotient = 6 dividend = ( divisor * quotient ) + remainder dividend = ( 18 * 6 ) + 5 = 113 d" | a ) 72 , b ) 76 , c ) 100 , d ) 113 , e ) 112 | d | add(multiply(add(multiply(5, const_3), 3), divide(add(multiply(5, const_3), 3), 3)), 5) | multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3)| | general |
what is the sum of the multiples of 4 between 34 and 135 inclusive ? | the fastest way in an ap is to find the average and multiply with total integers . . between 34 and 135 , the smallest multiple of 4 is 36 and largest = 132 . . average = ( 36 + 132 ) / 2 = 84 . . total numbers = ( 132 - 36 ) / 4 + 1 = 24 + 1 = 25 . . sum = 25 * 84 = 2100 ans b | a ) 2000 , b ) 2100 , c ) 2150 , d ) 2200 , e ) 2500 | b | multiply(divide(add(subtract(135, const_3), add(34, const_2)), const_2), add(divide(subtract(subtract(135, const_3), add(34, const_2)), 4), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5) | general |
a train sets off at 9 : 00 am at the speed of 70 km / h . another train starts at 9 : 30 am in the same direction at the rate of 80 km / h . at what time will the second train catch the first train ? | "in thirty minutes the first train travels 35 km . the second train catches the first train at a rate of 80 km / h - 70 km / h = 10 km / h . the second train will catch the first train in 35 / 10 = 3.5 hours , so at 1 : 00 pm . the answer is b ." | a ) 12 : 30 , b ) 1 : 00 , c ) 1 : 30 , d ) 2 : 00 , e ) 2 : 30 | b | divide(add(70, multiply(70, divide(const_1, const_2))), subtract(80, 70)) | divide(const_1,const_2)|subtract(n5,n2)|multiply(n2,#0)|add(n2,#2)|divide(#3,#1)| | physics |
the area of a rectangular plot is 21 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | "l × b = 21 × b ∴ l = 21 m and l – b = 10 ∴ b = 21 – 10 = 11 m answer c" | a ) 10 metres , b ) 5 metres , c ) 11 metres , d ) data inadequate , e ) none of these | c | subtract(21, 10) | subtract(n0,n1)| | geometry |
a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 3600 interest . the principal amount borrowed by him was . | "explanation : s . i . = p â ˆ — r â ˆ — t / 100 = > p = s . i . â ˆ — 100 / r â ˆ — t = > p = 3600 â ˆ — 100 / 12 â ˆ — 3 = rs 10000 option a" | a ) rs 10000 , b ) rs 15000 , c ) rs 16000 , d ) rs 17000 , e ) none of these | a | divide(multiply(3600, const_100), multiply(12, 3)) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)| | gain |
a , b and c started a business with a total investment of rs . 72000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find c ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 72000 = > x + ( x - 3000 ) + ( x + 3000 ) = 72000 = > 3 x = 72000 = > x = 24000 hence , a ' s investment = rs . 27000 b ' s investme... | a ) rs . 3240 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | c | multiply(8640, divide(add(divide(subtract(72000, add(6000, 3000)), const_3), 6000), 72000)) | add(n1,n2)|subtract(n0,#0)|divide(#1,const_3)|add(n1,#2)|divide(#3,n0)|multiply(n3,#4)| | general |
( 3 x + 2 ) ( 2 x - 3 ) = ax ^ 2 + kx + n . what is the value of a - n + k ? | "expanding we have 6 x ^ 2 - 9 x + 4 x - 6 6 x ^ 2 - 5 x - 6 taking coefficients , a = 6 , k = - 5 , n = - 6 therefore a - n + k = 6 - ( - 6 ) - 5 = 12 - 5 = 7 the answer is b ." | a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 11 | b | add(add(multiply(3, 2), multiply(3, 2)), subtract(multiply(2, 2), multiply(3, 3))) | multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5)| | general |
a vendor sells 50 percent of the apples he had and throws away 20 percent of the remainder . the next day , the vendor sells 50 percent of the remaining apples and throws away the rest . in total , what percent of his apples does the vendor throw away ? | "let x be the original number of apples . on day one , the vendor throws away ( 0.2 ) ( 0.5 ) x = 0.1 x . the remaining apples are ( 0.8 ) ( 0.5 ) x = 0.4 x . on day two , the vendor throws away ( 0.5 ) ( 0.4 ) x = 0.2 x . the vendor throws away a total of 0.1 x + 0.2 x = 0.3 x . the vendor throws away 30 percent of th... | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | a | add(divide(subtract(subtract(const_100, 50), multiply(divide(20, const_100), subtract(const_100, 50))), const_2), multiply(divide(20, const_100), subtract(const_100, 50))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|divide(#3,const_2)|add(#4,#2)| | general |
a can do a piece of work in 12 days and b alone can do it in 14 days . how much time will both take to finish the work ? | this question can be solved by different methods . we need to conserve time in exams so solving this problem using equations is the good idea . time taken to finish the job = xy / ( x + y ) = 12 x 14 / ( 12 + 14 ) = 150 / 25 = 6.46 days answer : c | a ) a ) 3.2222 , b ) b ) 5 , c ) c ) 6.46 , d ) d ) 8.3333 , e ) e ) 9 | c | divide(const_1, add(divide(const_1, 12), divide(const_1, 14))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 26 % . what quantity of whisky is replaced ? | let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 -... | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | b | divide(subtract(40, 26), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1) | gain |
two numbers are in the ratio 3 : 4 . if their l . c . m . is 96 . what is sum of the numbers ? | "explanation : let the numbers be 3 x and 4 x lcm of 3 x and 4 x = 12 x ( since lcm of 3 and 4 is 12 . hence lcm of 3 x and 4 x is 12 x ) given that lcm of 3 x and 4 x is 96 = > 12 x = 96 = > x = 96 / 12 = 8 sum of the numbers = 3 x + 4 x = 7 x = 7 x 8 = 56 answer : option b" | a ) 48 , b ) 56 , c ) 76 , d ) 84 , e ) 85 | b | add(multiply(multiply(3, const_4.0), const_100), multiply(4, 96)) | multiply(n0,const_4.0)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
x ^ y + y ^ x = 2408 find the values of x ? | 2407 ^ 1 + 1 ^ 2407 = 2408 answer : a | a ) 2407 , b ) 2408 , c ) 2409 , d ) 2405 , e ) 32 | a | subtract(2408, const_1) | subtract(n0,const_1) | general |
the speed of a boat in still water in 20 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 12 minutes is : | "speed downstream = ( 20 + 6 ) = 26 kmph time = 24 minutes = 12 / 60 hour = 1 / 5 hour distance travelled = time × speed = 1 / 5 × 26 = 5.20 km answer is b ." | a ) 10.6 , b ) 5.2 , c ) 10.8 , d ) 5.4 , e ) 5.0 | b | multiply(add(20, 6), divide(12, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
15 litres of mixture contains 20 % alcohol and the rest water . if 5 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 15 litres of mix . = 20 % of 15 litres = ( 20 * 15 / 100 ) = 3 litres water in it = 15 - 3 = 12 litres new quantity of mix . = 15 + 5 = 20 litres quantity of alcohol in it = 3 litres percentage of alcohol in new mix . = 3 * 100 / 20 = 15 % answer is c" | a ) 16.67 % , b ) 23 % , c ) 15 % , d ) 19.75 % , e ) 21.23 % | c | multiply(divide(subtract(add(15, 5), add(multiply(divide(subtract(const_100, 20), const_100), 15), 5)), add(15, 5)), const_100) | add(n0,n2)|subtract(const_100,n1)|divide(#1,const_100)|multiply(n0,#2)|add(n2,#3)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1111 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1111 96 + 225 + 490 + 6 x = 1111 6 x = 300 = > x = 50 . answer : d" | a ) 25 , b ) 66 , c ) 77 , d ) 50 , e ) 91 | d | divide(subtract(subtract(subtract(1111, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general |
the average of 3 numbers is 48 . the average of two of these numbers is 56 % . what is the third number ? | b 32 the total of three numbers must be 48 x 3 = 144 . the total of two numbers must be 56 x 2 = 112 . therefore , 144 - 112 = 32 . | a ) 64 , b ) 32 , c ) 46 , d ) 75 , e ) 47 | b | subtract(multiply(3, 48), multiply(const_2, 56)) | multiply(n0,n1)|multiply(n2,const_2)|subtract(#0,#1) | general |
by selling an article for $ 180 , a person gains $ 30 . what is the gain % ? | "s . p . = $ 180 gain = $ 30 c . p . = 180 - 30 = 150 gain % = 30 / 150 * 100 = 20 % answer is d" | a ) 25 % , b ) 30 % , c ) 50 % , d ) 20 % , e ) 10 % | d | divide(multiply(30, const_100), subtract(180, 30)) | multiply(n1,const_100)|subtract(n0,n1)|divide(#0,#1)| | gain |
two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 2 litres from vessel p is mixed with 6 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | quantity of alcohol in vessel p = 62.5 / 100 * 2 = 5 / 4 litres quantity of alcohol in vessel q = 87.5 / 100 * 6 = 21 / 4 litres quantity of alcohol in the mixture formed = 5 / 4 + 21 / 4 = 13 / 2 = 6.5 litres as 8 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 6.5 : 1.5 = 13 : 3 . ans... | a ) 13 : 1 , b ) 13 : 3 , c ) 13 : 8 , d ) 13 : 5 , e ) 13 : 2 | b | divide(add(divide(multiply(62.5, 2), const_100), divide(multiply(87.5, 6), const_100)), add(subtract(2, divide(multiply(62.5, 2), const_100)), subtract(6, divide(multiply(87.5, 6), const_100)))) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n2,#2)|subtract(n3,#3)|add(#5,#6)|divide(#4,#7) | other |
what is 2 / 5 of 5 / 9 of 1 / 2 ? | "2 / 5 * 5 / 9 * 1 / 2 = 1 / 9 answer : b" | a ) 1 / 4 , b ) 1 / 9 , c ) 9 / 16 , d ) 5 / 8 , e ) 16 / 9 | b | multiply(divide(2, 5), multiply(divide(5, 9), divide(1, 2))) | divide(n0,n1)|divide(n1,n3)|divide(n4,n0)|multiply(#1,#2)|multiply(#0,#3)| | general |
john completes a piece of work in 10 days , rose completes the same work in 40 days . if both of them work together , then the number of days required to complete the work is ? | "if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 10 × 40 / 50 = 8 days b )" | a ) 5 days , b ) 8 days , c ) 10 days , d ) 12 days , e ) 14 days | b | divide(const_1, add(divide(const_1, 10), divide(const_1, 40))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
at a time a , do 20 % less work than b . if a do any work in 15 / 2 hour , then in how many hour b will finish work ? | let b will finish a work in x hour . then , in x hour a , 80 / 100 = 4 / 5 work do ratio of work done by a and b = inverse ratio of time taken 4 / 5 : 1 = 2 / 15 : 1 / x 4 / 5 * 1 / x = 1 * 2 / 15 = 4 / 5 x = 2 / 15 ; x = 6 . time = 6 hour answer c | a ) 5 hour , b ) 11 / 2 hour , c ) 6 hour , d ) 13 / 2 hour , e ) 8 hour | c | multiply(divide(15, 2), subtract(const_1, divide(20, const_100))) | divide(n1,n2)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2) | physics |
worker a takes 8 hours to do a job . worker b takes 10 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "one day work of a = 1 / 8 one day work of b = 1 / 10 so one day work of a and b together = 1 / 8 + 1 / 10 = 9 / 40 so total days required = 40 / 9 answer : b" | a ) 20 / 9 , b ) 40 / 9 , c ) 50 / 9 , d ) 60 / 9 , e ) 80 / 9 | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
if a is a positive integer , and if the units digit of a ^ 2 is 9 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units z digit of ( a + 2 ) ^ 2 ? | "i also got a . by punching in numers : z . . . 7 ^ 2 = . . . 9 . . . 8 ^ 2 = . . . 4 . . . 9 ^ 2 = . . . 1 . a" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) c . 14 | a | power(add(multiply(9, 2), 2), 2) | multiply(n0,n1)|add(n0,#0)|power(#1,n0)| | general |
106 ã — 106 + 94 ã — 94 = ? | "explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1062 + 942 = ( 100 + 6 ) 2 + ( 100 â ˆ ’ 6 ) 2 = 2 ( 1002 + 62 ) = 2 ( 10000 + 36 ) = 20072 . answer : option a" | a ) 20072 , b ) 20062 , c ) 10072 , d ) 20172 , e ) 10272 | a | add(subtract(power(106, const_2), 106), subtract(power(94, const_2), 94)) | power(n0,const_2)|power(n2,const_2)|subtract(#0,n0)|subtract(#1,n2)|add(#2,#3)| | general |
in an office , totally there are 6400 employees and 65 % of the total employees are males . 25 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ? | "number of male employees = 6400 * 65 / 100 = 4160 required number of male employees who are less than 50 years old = 4160 * ( 100 - 25 ) % = 4160 * 75 / 100 = 3120 . answer : c" | a ) 1040 , b ) 2080 , c ) 3120 , d ) 4160 , e ) none of these | c | multiply(divide(multiply(6400, 65), const_100), divide(subtract(const_100, 25), const_100)) | multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)| | general |
the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 75 percent of a , and m equals b decreased by 80 percent of b , what is the value of m / x ? | "a / b = 4 / 5 m / x = ( 1 / 5 ) * 5 / ( 7 / 4 ) * 4 = 1 / 7 the answer is a ." | a ) 1 / 7 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 4 , e ) 3 / 2 | a | multiply(divide(subtract(const_100, 80), add(const_100, 75)), divide(5, 4)) | add(n2,const_100)|divide(n1,n0)|subtract(const_100,n3)|divide(#2,#0)|multiply(#3,#1)| | general |
9 . the least number which should be added to 28523 so that the sum is exactly divisible by 3 , 5 , 7 and 8 is | lcm of 3 , 5 , 7 and 8 = 840 28523 ÷ 840 = 33 remainder = 803 hence the least number which should be added = 840 - 803 = 37 answer : option d | a ) 41 , b ) 42 , c ) 32 , d ) 37 , e ) 39 | d | subtract(lcm(lcm(lcm(3, 5), 7), 8), reminder(28523, lcm(lcm(lcm(3, 5), 7), 8))) | lcm(n2,n3)|lcm(n4,#0)|lcm(n5,#1)|reminder(n1,#2)|subtract(#2,#3) | general |
find the value of ( 75983 * 75983 - 45983 * 45983 / 30000 ) | "given expression = ( 75983 ) 2 - ( 45983 ) 2 / ( 75983 - 45983 ) = ( a - b ) 2 / ( a - b ) = ( a + b ) ( a - b ) / ( a - b ) = ( a + b ) = 75983 + 45983 = 121966 answer is d ." | a ) 129166 , b ) 121696 , c ) 126196 , d ) 121966 , e ) none of them | d | multiply(add(divide(75983, 45983), 75983), 45983) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
what is the value of 10 ^ 6 - 6 ^ 4 ? | "as 10 ^ n will always have last digit as 0 and 6 ^ n will always as last digit 6 . . hence difference of such sum should always be ending with 4 and there is only on option . . answer c" | a ) 9 , 97,111 , b ) 9 , 97,322 , c ) 9 , 98,704 , d ) 9 , 98,851 , e ) 9 , 98,900 | c | divide(multiply(add(add(const_100, const_60), const_1), 10), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 1250 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 1250 resolutions . = 1250 * 2 * 22 / 7 * 22.4 = 176000 cm = 1760 m answer : b" | a ) 1277 m , b ) 1760 m , c ) 1278 m , d ) 1288 m , e ) 1378 v | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 1250), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
a man gains 20 % by selling an article for a certain price . if the sells it at double the price , the percentage of profit will be | "explanation : let c . p . = rs . x . then , s . p . = rs . ( 12 % of x ) = rs . 6 x / 5 new s . p . = 2 * 6 x / 5 = rs . 12 x / 5 profit = 12 x / 5 - x = rs . 7 x / 5 profit = 7 x / 5 * 1 / x * 100 = 140 % . answer : d" | a ) 740 % , b ) 540 % , c ) 640 % , d ) 140 % , e ) 240 % | d | add(multiply(subtract(multiply(add(const_1, divide(20, const_100)), const_2), const_1), const_100), const_100) | divide(n0,const_100)|add(#0,const_1)|multiply(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|add(#4,const_100)| | gain |
the salary of a worker is first increased by 10 % and afterwards reduced by 10 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.9 ( 1.1 x ) = 0.99 x the answer is b ." | a ) 1 % increase , b ) 1 % decrease , c ) 2 % increase , d ) 2 % decrease , e ) no change | b | subtract(const_100, subtract(add(10, const_100), divide(multiply(add(10, const_100), 10), const_100))) | add(n0,const_100)|multiply(n0,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)| | gain |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1051 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1051 96 + 225 + 490 + 6 x = 1051 6 x = 240 = > x = 40 . answer : b" | a ) 25 , b ) 40 , c ) 77 , d ) 99 , e ) 91 | b | divide(subtract(subtract(subtract(1051, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general |
the cost of painting the 4 walls of a room is rs . 350 . the cost of painting a room 3 times in length , breadth and height will be : | explanation : area of 4 walls of the room = [ 2 ( l + b ) × h ] m 2 area of 4 walls of new room = [ 2 ( 3 l + 3 b ) × 3 h ] m 2 = 9 [ 2 ( l + b ) × h ] m 2 cost of painting the 4 walls of new room = rs . ( 9 × 350 ) = rs . 3150 correct option : c | a ) rs . 1050 , b ) rs . 1400 , c ) rs . 3150 , d ) rs . 4200 , e ) none | c | multiply(350, power(3, const_2)) | power(n2,const_2)|multiply(n1,#0) | general |
a number is doubled and 5 is added . if the resultant is trebled , it becomes 123 . what is that number ? | "explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 123 6 x + 15 = 123 6 x = 108 x = 18 answer : d" | a ) 12 , b ) 29 , c ) 27 , d ) 18 , e ) 99 | d | divide(subtract(123, multiply(const_3, 5)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a certain number when divided by 899 leaves the remainder 63 . find the remainder when the same number is divided by 29 . | "sol . ( a ) number = 899 q + 63 , where q is quotient = 31 × 29 q + ( 58 + 5 ) = 29 [ 31 q + 2 ] + 5 ∴ remainder = 5 answer a" | a ) 5 , b ) 4 , c ) 1 , d ) 0 , e ) can not be determined | a | reminder(63, 29) | reminder(n1,n2)| | general |
kelsey travelled for 10 hours . he covered the first half of the distance at 25 kmph and remaining half of the distance at 40 kmph . find the distance travelled by kelsey ? | let the distance travelled be x km . total time = ( x / 2 ) / 25 + ( x / 2 ) / 40 = 10 = > x / 50 + x / 80 = 10 = > ( 8 x + 5 x ) / 400 = 10 = > x = 400 km answer : b | a ) 583 km , b ) 400 km , c ) 670 km , d ) 360 km , e ) 234 km | b | multiply(divide(25, add(divide(const_1, 40), divide(const_1, 10))), const_2) | divide(const_1,n2)|divide(const_1,n0)|add(#0,#1)|divide(n1,#2)|multiply(#3,const_2) | physics |
a gambler has won 40 % of his 30 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 80 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ? | "let x be the number of additional games the gambler needs to play . 0.4 ( 30 ) + 0.8 x = 0.6 ( x + 30 ) 0.2 x = 6 x = 30 the answer is c ." | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | c | divide(subtract(multiply(30, divide(60, const_100)), multiply(30, divide(40, const_100))), subtract(divide(80, const_100), divide(60, const_100))) | divide(n3,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)| | gain |
lola has $ 240.00 in her checking account . she spent $ 180.00 . what percentage does she have left in her account ? | explanation : amount left : 240 - 180 = 60 amount left percentage = ( 60 / 240 x 100 ) % = 25 % answer : e | a ) 75 % , b ) 50 % , c ) 40 % , d ) 55 % , e ) none of these | e | multiply(divide(subtract(240, 180), 240), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100) | gain |
the length of the rectangular field is double its width . inside the field there is square shaped pond 5 m long . if the area of the pond is 1 / 8 of the area of the field . what is the length of the field ? | a / 8 = 5 * 5 = > a = 5 * 5 * 8 x * 2 x = 5 * 5 * 8 x = 10 = > 2 x = 20 answer : d | ['a ) 54', 'b ) 32', 'c ) 75', 'd ) 20', 'e ) 11'] | d | sqrt(divide(multiply(square_area(5), 8), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3) | geometry |
if the length of a rectangle is halved and its breadth is tripled , what is the percentage change in its area ? | length is halved . i . e . , length is decreased by 50 % breadth is tripled i . e . , breadth is increased by 200 % formula for change in area is : = ( - x + y - xy / 100 ) % = ( − 50 + 200 − 50 × 200 / 100 ) % = 50 % i . e . , area is increased by 50 % answer is c . | ['a ) 30', 'b ) 40', 'c ) 50', 'd ) 60', 'e ) 70'] | c | multiply(subtract(divide(const_3, const_2), const_1), const_100) | divide(const_3,const_2)|subtract(#0,const_1)|multiply(#1,const_100) | geometry |
the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8587 for a computer table , then what was the cost price of the computer table ? | ": cp = sp * ( 100 / ( 100 + profit % ) ) = 8587 ( 100 / 124 ) = rs . 6925 . answer : a" | a ) 6925 , b ) 6887 , c ) 6728 , d ) 6725 , e ) 2871 | a | divide(8587, add(const_1, divide(24, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
a train covers a distance of 100 km in 1 hour . if its speed is decreased by 25 km / hr , the time taken by the car to cover the same distance will be ? | "speed = 100 / 1 = 100 km / hr new speed = 100 - 25 = 75 km / hr time taken = 100 / 75 = 1 hr 20 min answer is b" | a ) 1 hr , b ) 1 hr 20 min , c ) 50 min , d ) 1 hr 30 min , e ) 1 hr 45 min | b | divide(100, subtract(divide(100, 1), 25)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
how many positive integers less than 600 can be formed using the numbers 1 , 2 , 3 and 5 for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 5 5 5 number of possibilities = 5 * 4 * 4 = 80 . then , just add up the number of 1 digits numbers = 4 , so total is 80 + 4 = 84 . answer : e" | a ) 48 , b ) 52 , c ) 66 , d ) 68 , e ) 84 | e | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | probability |
in a certain state , the ratio of registered republicans to registered democrats is 3 to 2 , and every registered voter is either a republican or a democrat . if 90 percent of the republicans and 15 percent of the democrats are expected to vote for candidate x , and everyone else is expected to vote for candidate y , b... | "since we were expected to find a percentage figure - it thought that it might be easier to pick a ' smart number ' to represent the total number of voters ( republicans and democrats ) . therefore , i picked 100 ( as the total number of voters ) and thus 30 : 20 represents the number ratio of republicans : democrats .... | a ) 2 % , b ) 5 % , c ) 8 % , d ) 10 % , e ) 20 % | e | multiply(divide(subtract(add(multiply(divide(15, const_100), 2), multiply(divide(90, const_100), 3)), add(subtract(3, multiply(divide(90, const_100), 3)), subtract(2, multiply(divide(15, const_100), 2)))), add(3, 2)), const_100) | add(n0,n1)|divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#1)|multiply(n0,#2)|add(#3,#4)|subtract(n0,#4)|subtract(n1,#3)|add(#6,#7)|subtract(#5,#8)|divide(#9,#0)|multiply(#10,const_100)| | other |
a man can row downstream at 22 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 32 = > x = 16 , y = 6 . answer : d" | a ) 16 , 2 , b ) 16 , 4 , c ) 16 , 8 , d ) 16 , 6 , e ) 16 , 7 | d | divide(divide(add(22, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
if | 7 x + 2 | = 16 , then find the product of the values of x ? | "| 7 x + 2 | = 16 7 x + 2 = 16 or 7 x + 2 = - 16 7 x = 14 or 7 x = - 18 x = 2 or x = - 2.57 product = - 2.57 * 2 = - 5.14 answer is a" | a ) - 5.14 , b ) 6.19 , c ) - 7.18 , d ) - 8.62 , e ) 5.69 | a | divide(subtract(16, 2), 7) | subtract(n2,n1)|divide(#0,n0)| | general |
the cost of an article is decreased by 15 % . if the original cost is $ 80 , find the decrease cost . | "original cost = $ 80 decrease in it = 15 % of $ 80 = 15 / 100 × 80 = 1200 / 100 = $ 12 therefore , decrease cost = $ 80 - $ 12 = $ 68 answer : c" | a ) 33 , b ) 11 , c ) 68 , d ) 36 , e ) 91 | c | divide(multiply(80, 15), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
sandy bought 65 books for $ 1280 from one shop and 55 books for $ 880 from another shop . what is the average price that sandy paid per book ? | "average price per book = ( 1280 + 880 ) / ( 65 + 55 ) = 2160 / 120 = $ 18 the answer is d ." | a ) $ 12 , b ) $ 14 , c ) $ 16 , d ) $ 18 , e ) $ 20 | d | divide(add(1280, 880), add(65, 55)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)| | general |
a fruit seller had some apples . he sells 35 % and still has 6500 apples . originally , he had ? | "answer â ˆ µ 65 % of n = 800 â ˆ ´ n = ( 650 x 100 ) / 65 = 1000 correct option : e" | a ) 650 apples , b ) 600 apples , c ) 772 apples , d ) 700 apples , e ) none | e | original_price_before_loss(35, 6500) | original_price_before_loss(n0,n1)| | gain |
each of the integers from 0 to 8 , inclusive , is written on a separate slip of blank paper and the 10 slips are dropped into hat . if the slips are then drawn one at a time without replacement , how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10 ? | you should consider the worst case scenario : if you pick numbers 0 , 1 , 2 , 3 , 4 and 5 then no two numbers out of these 6 add up to 10 . now , the next , 7 th number whatever it ' ll be ( 6,7 , or 8 ) will guarantee that two number will add up to 10 . so , 7 slips must be drawn to ensure that the numbers on two of t... | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | c | add(const_1, subtract(8, const_2)) | subtract(n1,const_2)|add(#0,const_1) | gain |
at 15 : 00 there were 20 students in the computer lab . at 15 : 03 and every three minutes after that , 3 students entered the lab . if at 15 : 10 and every ten minutes after that 7 students left the lab , how many students were in the computer lab at 15 : 44 ? | "initial no of students + 3 * ( 1 + no of possible 3 minute intervals between 15 : 03 and 15 : 44 ) - 8 * ( 1 + no of possible 10 minute intervals between 15 : 10 and 15 : 44 ) 20 + 3 * 14 - 8 * 4 = 25 c" | a ) 7 , b ) 14 , c ) 25 , d ) 27 , e ) 30 | c | add(subtract(add(multiply(floor(divide(44, 03)), 03), 20), multiply(floor(divide(44, 7)), 7)), 03) | divide(n10,n4)|divide(n10,n8)|floor(#0)|floor(#1)|multiply(n4,#2)|multiply(n8,#3)|add(n2,#4)|subtract(#6,#5)|add(n4,#7)| | physics |
an ore contains 25 % of an alloy that has 90 % iron . other than this , in the remaining 75 % of the ore , there is no iron . how many kilograms of the ore are needed to obtain 60 kg of pure iron ? | solution : let there is 100 kg of ore . 25 % ore contains 90 % off iron that means 25 kg contains ; 25 * 90 / 100 = 22.5 kg iron . 22.5 kg iron contains 100 kg of ore . then , 1 kg of iron contains = 25 / 100 kg ore ; hence , 60 kg iron contains = 100 * 60 / 22.5 = 266.66 kg ore . answer : option d | a ) 250 kg , b ) 275 kg , c ) 300 kg , d ) 266.66 kg , e ) none | d | divide(multiply(const_100, 60), divide(multiply(25, 90), const_100)) | multiply(n3,const_100)|multiply(n0,n1)|divide(#1,const_100)|divide(#0,#2) | other |
how many integers are between 5 and 74 / 5 , inclusive ? | "74 / 5 = 14 . xx we are not concerned about the exact value of 74 / 5 as we just need the integers . the different integers between 5 and 74 / 5 would be 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13,14 . total number of integers = 10 option e" | a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 10 | e | add(subtract(divide(74, 5), 5), const_1) | divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)| | general |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 8 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 8 x + 15 7 x = 1350 x = 192.9 large number = 192.9 + 1365 = 1557.9 b" | a ) 1235 , b ) 1557.9 , c ) 1378 , d ) 1635 , e ) 1489 | b | multiply(divide(subtract(1365, 15), subtract(8, const_1)), 8) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
a is the average ( arithmetic mean ) of the first 7 positive multiples of 6 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 – b ^ 2 is zero , what is the value of n ? | if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 24 , which also equals b . b is the second positive multiple of n , thus n = 24 / 2 = 12 . the answer is c . | a ) 8 , b ) 10 , c ) 12 , d ) 16 , e ) 24 | c | divide(multiply(6, const_4), const_2) | multiply(n1,const_4)|divide(#0,const_2) | general |
cereal a is 11 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 3 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "2 % is 1 % - points below 3 % and 11 % is 8 % - points above 3 % . the ratio of a : b should be 1 : 8 . the answer is d ." | a ) 2 : 9 , b ) 3 : 7 , c ) 2 : 5 , d ) 1 : 8 , e ) 1 : 5 | d | divide(subtract(3, 2), subtract(11, 3)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
in a class of 20 students , 2 students did not borrow any books from the library , 10 students each borrowed 1 book , 5 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ... | "the total number of books the students borrowed is 20 * 2 = 40 . the students who borrowed zero , one , or two books borrowed 10 * 1 + 5 * 2 = 20 books . the 3 students who borrowed at least three books borrowed 40 - 20 = 20 books . if 2 of these students borrowed exactly 3 books , then the maximum that one student co... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | subtract(multiply(20, 2), add(multiply(subtract(subtract(20, add(add(multiply(10, 1), 5), 2)), 1), 3), add(multiply(10, 1), multiply(5, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)| | general |
a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations g are possible ? | "the combinations could be { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pa... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) g = 8 | e | divide(240, 30) | divide(n0,n2)| | general |
the reciprocal of the hcf and lcm of two are 1 / 15 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 15 and 1 / 312 so , hcf = 15 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 15 * 312 / 24 = 195 answer : e" | a ) 126 , b ) 136 , c ) 146 , d ) 156 , e ) 195 | e | divide(multiply(15, 312), 24) | multiply(n1,n3)|divide(#0,n4)| | physics |
x and y start a business with rs . 3000 and rs . 6000 respectively . hoe should they share their profits at the end of one year ? | explanation : they should share the profits in the ratio of their investments . the ratio of the investments made by x and y = 3000 : 6000 = > 1 : 2 answer : c | a ) 3 : 1 , b ) 2 : 1 , c ) 1 : 2 , d ) 1 : 3 , e ) 1 : 4 | c | divide(3000, 6000) | divide(n0,n1) | gain |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 60 liters and they are all empty , how much money total will it cost to fuel all cars ? | total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 60 ) = 489 hence answer will be ( e ) | a ) 320 $ , b ) 389 $ , c ) 420 $ , d ) 450 $ , e ) 489 $ | e | multiply(multiply(0.65, 60), 12) | multiply(n1,n3)|multiply(n2,#0) | general |
set x consists of 10 integers and has median of 10 and a range of 10 . what is the value of the greatest possible integer that can be present in the set ? | "note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between an... | a ) 32 , b ) 37 , c ) c . 40 , d ) 20 , e ) 50 | d | add(10, 10) | add(n1,n2)| | general |
a boy is traveling from his house to school at 3 km / hr and reached school 9 minutes late . next day he traveled 4 km / hr and reached 6 minutes early . then find the distance between house and school ? | "let distance be x s 1 = 3 km / hr s 2 = 4 km / hr t 1 = x / 3 hr t 2 = x / 4 hr difference in time = 9 + 6 = 15 m = 1 / 4 hr ( x / 3 ) - ( x / 4 ) = 1 / 4 x = 3 km answer is a" | a ) 3 km , b ) 4.5 km , c ) 5 km , d ) 2.5 km , e ) 6 km | a | multiply(add(divide(add(multiply(9, 3), 4), subtract(4, 3)), 9), 3) | multiply(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(#2,#1)|add(n1,#3)|multiply(n0,#4)| | physics |
a student chose a number , multiplied it by 2 , then subtracted 140 from the result and got 102 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 140 = 102 2 x = 242 x = 121 correct answer e" | a ) 90 , b ) 100 , c ) 120 , d ) 160 , e ) 121 | e | divide(add(102, 140), 2) | add(n1,n2)|divide(#0,n0)| | general |
a can do a work in 15 days and b in 20 days . if they work on it together for 6 days , then the fraction of the work that is left is | "person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( - ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 / 35 ) days to complete entire work for 1 st 4 days they work 35 * 6 = 210 remaining work is 300 - 210 = 90 remaining fraction of work is = 90 / 300 = 3 / 10 answer c" | a ) 8 / 17 , b ) 7 / 15 , c ) 3 / 10 , d ) 8 / 15 , e ) 1 / 4 | c | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 6)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
a certain quantity is measured on two different scales , the p - scale and the s - scale , that are related linearly . measurements on the p - scale of 6 and 24 correspond to measurements on the s - scale of 30 and 60 , respectively . what measurement on the p - scale corresponds to a measurement of 100 on the s - scal... | first , we have to understand what linearly means . it ' s not a straight ratio ( since 6 : 30 does not equal 24 : 60 ) . we need to look at the increases in each measurement to see what the scalar actually is . from 6 to 24 we have an increase of 18 . from 30 to 60 we have an increase of 30 . therefore , the increase ... | a ) 20 , b ) 36 , c ) 48 , d ) 60 , e ) 84 | c | add(multiply(divide(subtract(100, 30), 30), subtract(24, 6)), 6) | subtract(n4,n2)|subtract(n1,n0)|divide(#0,n2)|multiply(#2,#1)|add(n0,#3) | general |
set a contains all the even numbers between 2 and 50 inclusive . set b contains all the even numbers between 102 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "set a contains 2,4 , 6 . . . 50 set b contains 102 , 104 , 106 . . . 150 number of terms in each set = 25 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 25 = 2500 answer a" | a ) 2500 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(102, 2), add(divide(subtract(50, 2), const_2), const_1)) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)| | general |
a whale goes on a feeding frenzy that lasts for 12 hours . for the first hour he catches and eats 30 kilos of plankton . in every hour after the first , it consumes 2 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 600 ... | therefor in 10 th hour he consume x + 2 * 9 = 30 + 18 = 48 correct option is d | a ) 36 , b ) 45 , c ) 28 , d ) 48 , e ) 52 | d | divide(600, 12) | divide(n3,n0) | general |
the average age of 19 students in a group is 17 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "age of the teacher = ( 20 × 18 – 19 × 17 ) years = 37 years . answer b" | a ) 36 , b ) 37 , c ) 38 , d ) can not be determined , e ) none of these | b | add(19, const_1) | add(n0,const_1)| | general |
machine x takes 10 hours longer than machine y to produce 1080 widgets . machine y produces 20 percent more widgets in an hour than machine x does in an hour . how many widgets per hour does machine x produce | "machine y produces 20 percent more widgets in an hour than machine x does in an hour . so if machine x produces 100 widgets , then machine y produces 120 widgets . ratio of 120 / 100 = 6 / 5 . this is their speed of work ( y : x ) . i . e . speed of their work ( x : y ) = 5 / 6 now , time is inversely proportional to ... | a ) 100 , b ) 65 , c ) 25 , d ) 11 , e ) 18 | e | divide(1080, multiply(divide(const_60.0, const_10), 10)) | divide(const_60.0,const_10)|multiply(n0,#0)|divide(n1,#1)| | general |
a train is running at a speed of 40 km / hr and it crosses a post in 16.2 seconds . what is the length of the train ? | "speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 16.2 s distance covered , d = vt = ( 400 / 36 ) ã — 16.2 = 160 m distance covered is equal to the length of the train = 160 m correct answer is 160 metres b" | a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 250 metres | b | multiply(16.2, multiply(40, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 30 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ? | find the product of the two increases : ( 14 / 10 ) * ( 13 / 10 ) which is 1.82 and a 82 % increase . d | a ) 65 % , b ) 70 % , c ) 75 % , d ) 82 % , e ) 95 % | d | multiply(subtract(divide(multiply(multiply(const_100, add(const_1, divide(40, const_100))), add(const_1, divide(30, const_100))), const_100), const_1), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,const_100)|multiply(#2,#4)|divide(#5,const_100)|subtract(#6,const_1)|multiply(#7,const_100) | gain |
if 4 % more is gained by selling an article for rs . 350 than by selling it for rs . 340 , the cost of the article is | "explanation : let c . p . be rs . x . then , 4 % of x = 350 - 340 = 10 x / 25 = 10 = > x = 250 answer : a" | a ) 250 , b ) 231 , c ) 200 , d ) 288 , e ) 111 | a | divide(subtract(350, 340), divide(4, const_100)) | divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)| | gain |
on increasing the price of t . v . sets by 50 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ? | "explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 × 100 ) = rs . 10000 new sale value = rs . ( 150 × 80 ) = rs . 12000 increase % = 2000 / 10000 × 100 = 20 % answer : d" | a ) 4 , b ) 5 , c ) 56 , d ) 20 , e ) 7 | d | subtract(add(const_100, 50), add(divide(multiply(add(const_100, 50), 20), const_100), const_100)) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#2,const_100)|subtract(#0,#3)| | gain |
p software has coding line 5 % more than n , n software has coding line 4 / 13 more than m . m software has 78 lines of coding . find p lines . | "m s / w has 78 line of code n s / w has = 78 + 78 * 4 / 13 = 102 line of code p s / w 5 % more n ' code 102 + 5.1 = 107.1 or 107 line of code answer : b" | a ) 106 , b ) 107 , c ) 108 , d ) 109 , e ) 110 | b | add(78, divide(78, 13)) | divide(n3,n2)|add(n3,#0)| | general |
a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 1600 . the difference between the profit shares of a and c is ? | ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 1600 = > 5 parts out of 15 parts is rs . 1600 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1600 ) = rs . 640 answer : e | a ) 288 , b ) 266 , c ) 155 , d ) 600 , e ) 640 | e | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1600) | divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2) | gain |
machine a can finish a job in 4 hours , machine в can finish the job in 12 hours , and machine с can finish the job in 6 hours . how many hours will it take for a , b , and с together to finish the job ? | "the combined rate is 1 / 4 + 1 / 12 + 1 / 6 = 1 / 2 of the job per hour . the time to complete the job is 2 / 1 = 2 hours . the answer is b ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | inverse(add(add(inverse(4), inverse(12)), inverse(6))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 720 sq . feet , how many feet of fencing will be required ? | "explanation we have : l = 20 ft and lb = 720 sq . ft . so , b = 36 ft . length of fencing = ( l + 2 b ) = ( 20 + 72 ) ft = 92 ft . answer b" | a ) 34 , b ) 92 , c ) 68 , d ) 88 , e ) none | b | add(multiply(divide(720, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 4 . what is the value of x ? | "y = 0.5 * 0.5 * x = x / 4 y % * x = 4 ( y / 100 ) * x = 4 ( x / 400 ) * x = 4 x ^ 2 = 4 * 400 x = 40 the answer is a ." | a ) 40 , b ) 100 , c ) 200 , d ) 400 , e ) 600 | a | multiply(multiply(divide(50, 4), divide(50, 4)), const_1000) | divide(n0,n2)|divide(n1,n2)|multiply(#0,#1)|multiply(#2,const_1000)| | general |
walking with 3 / 5 of my usual speed , i miss the bus by 5 minutes . what is my usual time ? | "speed ratio = 1 : 3 / 5 = 5 : 3 time ratio = 3 : 5 1 - - - - - - - - 5 3 - - - - - - - - - ? è 15 answer : e" | a ) 16 min , b ) 26 min , c ) 34 min , d ) 20 min , e ) 15 min | e | multiply(divide(5, divide(5, 3)), 5) | divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | physics |
[ ( 3.241 x 14 ) / 100 ] = ? | "answer multiplying 3.241 x 14 = 4.5374 now divide 4.5374 by 100 so , 4.5374 ÷ 100 = 0.045374 ∴ shift the decimal two places to the left as 100 correct option : a" | a ) 0.045374 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) none of these | a | divide(divide(multiply(3.241, 14), 100), const_10) | multiply(n0,n1)|divide(#0,n2)|divide(#1,const_10)| | general |
a person borrows rs . 4000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 4000 * 6 * 2 ) / 100 ] - [ ( 4000 * 4 * 2 ) / 100 ] 480 - 320 = 160 gain in 1 year = ( 160 / 2 ) = 80 rs answer : b" | a ) 100 rs , b ) 80 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | b | divide(subtract(divide(multiply(multiply(4000, 6), 2), const_100), divide(multiply(multiply(4000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain |
365 : 90 : : 623 : ? | "365 - - - 3 * 6 * 5 . . . . 90 623 - - - 6 * 2 * 3 . . . . 36 answer a" | a ) 36 , b ) 45 , c ) 123 , d ) 63 , e ) none of these | a | multiply(623, divide(365, 90)) | divide(n0,n1)|multiply(n2,#0)| | general |
the volumes of two cones are in the ratio 1 : 20 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ? | "the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 1 ^ 2 . h 1 / r 2 ^ 2 . h 2 = > 1 / 20 = ( 1 ) ^ 2 h 1 / ( 2 ) ^ 2 h 2 = > h 1 / h 2 = 4 / 5 i . e . h 1 : h 2 = 4 : 5 answer : a" | a ) 4 : 5 , b ) 4 : 9 , c ) 4 : 2 , d ) 5 : 2 , e ) 2 : 8 | a | divide(divide(1, 20), power(divide(1, 2), 2)) | divide(n0,n1)|divide(n0,n3)|power(#1,n3)|divide(#0,#2)| | geometry |
let p and q be the root of the quadratic equation x 2 - ( ∝ - 2 ) x - ∝ - 1 = 0 . what is the minimum possible value of p 2 + q 2 ? | explanation : p + q = α – 2 and pq = – α – 1 ( p + q ) 2 = p 2 + q 2 + 2 pq , thus ( α – 2 ) 2 = p 2 + q 2 + 2 ( – α – 1 ) p 2 + q 2 = α 2 – 4 α + 4 + 2 α + 2 p 2 + q 2 = α 2 – 2 α + 6 p 2 + q 2 = α 2 – 2 α + 1 + 5 p 2 + q 2 = ( α – 1 ) 2 + 5 thus , minimum value of p 2 + q 2 is 5 . answer : d | a ) 0 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | subtract(add(add(2, 2), 2), 1) | add(n0,n0)|add(n0,#0)|subtract(#1,n2) | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.