Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
the amount of time that three people worked on a special project was in the ratio of 4 to 3 to 5 . if the project took 120 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 4 * 120 / 12 = 40 hours b = 3 * 120 / 12 = 30 hours c = 5 * 120 / 12 = 50 hours c is the hardest worker and b worked for the least number of hours . so the difference is 50 - 30 = 20 hours . answer : c" | a ) 25 hours , b ) 30 hours , c ) 20 hours , d ) 26 hours , e ) 32 hours | c | subtract(multiply(divide(120, add(add(4, 3), 5)), 5), multiply(divide(120, add(add(4, 3), 5)), 4)) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|multiply(n0,#2)|subtract(#3,#4)| | physics |
a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 11 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | "explanation : l * 30 = 1200 Γ¨ l = 40 40 + 30 + 50 = 120 120 * 11 = 1320 answer : option e" | a ) 1276 , b ) 1200 , c ) 2832 , d ) 1299 , e ) 1320 | e | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 11) | divide(n1,n3)|power(n3,n2)|add(n3,#0)|power(#0,n2)|add(#1,#3)|sqrt(#4)|add(#2,#5)|multiply(n0,#6)| | geometry |
a call center has two teams . each member of team a was able to process 3 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | "let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 3 calls , so total calls for team a = 15 fraction for team b = 40 / ( 40 + 15 ) = 8 / 11 = answer = c" | a ) 3 / 2 , b ) 3 / 4 , c ) 8 / 11 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 3))) | multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|divide(#0,#2)| | general |
jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much r does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a ) $ 10079.44 , b ) r = $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_4)|multiply(n0,#3)| | gain |
cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 71.4 after it went on sale for 15 % off , what is the difference between the original price and... | "0.85 * { original price } = $ 71.4 - - > { original price } = $ 84 . { final price } = $ 71.4 * 1.25 = $ 89.25 . the difference = $ 89.25 - $ 84 = $ 5.25 answer : d ." | a ) $ 5.625 , b ) $ 1.00 , c ) $ 3.40 , d ) $ 5.25 , e ) $ 6.80 | d | subtract(multiply(71.4, divide(add(const_100, 25), const_100)), divide(71.4, divide(subtract(const_100, 15), const_100))) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#3)|multiply(n2,#2)|subtract(#5,#4)| | general |
if ( n + 2 ) ! / n ! = 110 , n = ? | "( n + 2 ) ! / n ! = 110 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 110 from here , we might just test the answer choices . since ( 11 ) ( 10 ) = 110 , we can see that n... | a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1)| | general |
( 1 / 5 ) ^ e * ( 1 / 4 ) ^ 18 = 1 / ( 2 ( 10 ) ^ 35 ) . what is e ? | we need to arrive at a common base . - - > ( 5 ) ^ ( - e ) * 2 ^ ( - 36 ) = 2 ^ ( - 36 ) * 5 ^ ( - 35 ) 5 ^ ( - e ) = 5 ^ ( - 35 ) - e = - 35 e = 35 = d | a ) 17 , b ) 18 , c ) 34 , d ) 35 , e ) 36 | d | divide(35, 1) | divide(n8,n0) | general |
i bought two books ; for rs . 360 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 360 - x ) 119 / 100 x = 210 answer : b" | a ) 197 , b ) 210 , c ) 189 , d ) 278 , e ) 268 | b | divide(multiply(360, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain |
if n is a natural number , then ( 6 n 2 + 6 n ) is always divisible by : | "( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . answer c ) 6 and 12 both ." | a ) 6 only , b ) 12 only , c ) 6 and 12 both , d ) by 18 only , e ) 15 only | c | add(multiply(6, const_100), multiply(2, 6)) | multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)| | general |
| 13 - 3 | - | 4 - 10 | = ? | "| 13 - 3 | - | 4 - 10 | = | 10 | - | - 6 | = 10 - 6 = 4 correct answer e" | a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | e | subtract(subtract(13, 3), subtract(10, 4)) | subtract(n0,n1)|subtract(n3,n2)|subtract(#0,#1)| | general |
a fill pipe can fill 1 / 6 of cistern in 90 minutes in how many minutes , it can fill 5 / 6 of the cistern ? | "1 / 6 of the cistern can fill in 90 min 5 / 6 of the cistern can fill in = 90 * 6 * 5 / 6 = 450 min answer is a" | a ) 450 min , b ) 360 min , c ) 240 min , d ) 306 min , e ) 500 min | a | divide(90, 1) | divide(n2,n0)| | physics |
increasing the original price of an article by 6 percent and then increasing the new price by 6 percent is equivalent to increasing the original price by | 1.06 * 1.06 * x = 1.1236 * x the answer is b . | a ) 12.1 % , b ) 12.36 % , c ) 12.94 % , d ) 13.65 % , e ) 14.56 % | b | multiply(const_100, subtract(multiply(add(const_1, divide(6, const_100)), add(const_1, divide(6, const_100))), const_1)) | divide(n0,const_100)|add(#0,const_1)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100) | gain |
the monthly salaries of 6 employees in a company are $ 1000 , $ 2500 , $ 3100 , $ 3650 , $ 1500 , $ 2000 . what is the mean of the salaries of 6 employees . | mean of the salaries = ( $ 1000 + $ 2500 + $ 3100 + $ 3650 + $ 1500 + $ 2000 ) / 6 = 13750 / 6 = $ 2291.7 answer is a | a ) $ 2291.7 , b ) $ 1546.8 , c ) $ 2546.5 , d ) $ 1879.3 , e ) $ 3125.3 | a | divide(add(add(add(add(add(1000, 2500), 3100), 3650), 1500), 2000), 6) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|add(n6,#3)|divide(#4,n0) | general |
ravi and sunil are partners in a business . ravi invests rs . 18,000 for 8 months and sunil invested rs . 10000 for 10 months then after one year ratio of their profits will be | "= ( 18000 * 8 ) : ( 10000 * 10 ) = 144000 : 100000 = 36 : 25 answer : b" | a ) 1 : 3 , b ) 36 : 25 , c ) 3 : 2 , d ) 4 : 5 , e ) 20 : 45 | b | divide(multiply(multiply(multiply(multiply(8, const_3), multiply(10, const_2)), const_100), 8), multiply(10000, 10)) | multiply(const_3,n1)|multiply(n3,const_2)|multiply(n2,n3)|multiply(#0,#1)|multiply(#3,const_100)|multiply(n1,#4)|divide(#5,#2)| | gain |
if | x | = 5 x - 3 , then x = ? | "answer : a approach : substituted option a i . e x = 1 . inequality satisfied . c" | a ) 1 , b ) 1 / 2 , c ) 2 , d ) - 1 / 2 , e ) - 1 | c | divide(3, subtract(5, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
the area of a square is 4489 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4489 a = 67 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 43 : 134 answer : e" | a ) 5 : 28 , b ) 5 : 19 , c ) 5 : 12 , d ) 5 : 13 , e ) 43 : 134 | e | divide(subtract(sqrt(4489), 24), multiply(sqrt(4489), const_2)) | sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)| | geometry |
a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 12 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 12 = > m = 36 answer e" | a ) 39 , b ) 33 , c ) 26 , d ) 21 , e ) 36 | e | subtract(divide(12, divide(2, 3)), 12) | divide(n0,n1)|divide(n2,#0)|subtract(#1,n2)| | general |
a rectangular cube has sides measuring 3 inches long by 2 inches wide by 0.5 inches high . if the surface area of the rectangle is the same as a cube , what do the sides / walls of the cube measure ? round to the nearest whole number . | first calculate the surface area of the rectangle by multiplying the length and width of the rectangle together , then multiply by 2 to get both sides of the rectangle . this calculates to 12 inches . find the surface area of the sides of the rectangular cube . multiply the height by the length of the rectangle . multi... | ['a ) 4', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 2'] | e | floor(divide(surface_rectangular_prism(3, 2, 0.5), multiply(const_2, const_3))) | multiply(const_2,const_3)|surface_rectangular_prism(n0,n1,n2)|divide(#1,#0)|floor(#2) | geometry |
company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 7.00 each , what was company c β s gross profit ? | "cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 7 = $ 3500 profit : 3500 - 1900 = $ 1600 option c is correct" | a ) $ 2,250 , b ) $ 2,500 , c ) $ 1,600 , d ) $ 1,250 , e ) $ 1,500 | c | floor(divide(subtract(subtract(multiply(500, 7.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000)) | multiply(n3,n4)|multiply(n0,n1)|subtract(n3,n1)|multiply(n2,#2)|subtract(#0,#1)|subtract(#4,#3)|divide(#5,const_1000)|floor(#6)| | gain |
a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 100 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ? | "let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.15 * 1.07 x = 100 x = 81.27 hence , the correct option is d" | a ) 79.55 $ , b ) 80.63 $ , c ) 81.63 $ , d ) 80.27 $ , e ) 83.15 $ | d | divide(100, add(divide(add(7, 15), const_100), const_1)) | add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2)| | general |
a garrison of 2000 men has provisions for 54 days . at the end of 21 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ? | "2000 - - - - 54 2000 - - - - 33 x - - - - - 20 x * 20 = 2000 * 33 x = 3300 2000 - - - - - - - 1300 answer : e" | a ) 1977 , b ) 1893 , c ) 1979 , d ) 1900 , e ) 1300 | e | subtract(divide(subtract(multiply(2000, 54), multiply(2000, 21)), 20), 2000) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
a person buys an article at $ 200 . at what price should he sell the article so as to make a profit of 10 % ? | "c 220 cost price = $ 200 profit = 10 % of 200 = $ 20 selling price = cost price + profit = 200 + 20 = 220" | a ) 240 , b ) 260 , c ) 220 , d ) 300 , e ) 360 | c | add(200, multiply(200, divide(10, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
200 liters of a mixture contains milk and water in the ratio 4 : 1 . if 20 liters of this mixture be replaced by 20 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 200 liters if mix = 200 * 4 / 5 = 160 liters quantity of milk in 210 liters of new mix = 160 + 20 = 180 liters quantity of water in it = 210 - 180 = 30 liters ratio of milk and water in new mix = 180 : 30 = 6 : 1 answer is a" | a ) 6 : 1 , b ) 1 : 6 , c ) 2 : 3 , d ) 3 : 2 , e ) 4 : 1 | a | divide(multiply(subtract(20, const_1), 1), multiply(4, 1)) | multiply(n2,n1)|subtract(n3,const_1)|multiply(#1,n2)|divide(#2,#0)| | other |
if the speed of a man is 66 km per hour , then what is the distance traveled by him in 30 seconds ? | "the distance traveled in 30 sec = 66 * ( 5 / 18 ) * 30 = 550 m answer : a" | a ) 550 m , b ) 500 m , c ) 375 m , d ) 420 m , e ) 440 m | a | multiply(multiply(66, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
when positive integer x is divided by positive integer y , the remainder is 6 . if x / y = 6.12 , what is the value of y ? | "guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 6.12 and remainder is 6 so . 12 x y = 6 hence y = 50 ans c" | a ) 96 , b ) 75 , c ) 50 , d ) 25 , e ) 12 | c | divide(6, subtract(6.12, floor(6.12))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
two identical trains aa and bb running in opposite direction at same speed tale 2 min to cross each other completely . the number of bogies of aa are increased from 12 to 16 . how much more time would they now require to cross each other ? | total initial bogies is 12 + 12 = 2412 + 12 = 24 additional bogies = 16 Γ’ Λ β 12 = 4 24 bogies is 12 + 12 = 24 additional bogies = 16 - 12 = 4 24 bogies take 2 min 2 * 60 / 24 * 4 = 20 sec answer d | a ) 40 sec , b ) 50 sec , c ) 60 sec , d ) 20 sec , e ) 25 sec | d | multiply(divide(multiply(2, const_60), add(12, 12)), subtract(16, 12)) | add(n1,n1)|multiply(n0,const_60)|subtract(n2,n1)|divide(#1,#0)|multiply(#3,#2) | general |
express 5 mps in kmph ? | "5 * 18 / 5 = 18 kmph answer : d" | a ) 22 , b ) 88 , c ) 90 , d ) 18 , e ) 24 | d | multiply(divide(5, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600)| | physics |
a person starting with rs . 64 and making 6 bets , wins 3 times and loses 3 times , the wins and loses occurring in random order . the chance for a win is equal to the chance for a loss . if each wager is for 1 / 2 the money remaining at the time of the bet , then the final result is ? | as the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5 , we will multiply initial amount by 1.5 thrice and by 0.5 thrice ( in any order ) . the overall resultant will remain same . so final amount with the person will be ( in all cases ) : = 64 ( 1.5 ) ( 1.5 ) ( 1.5 ) ( 0.5 ) ... | a ) rs . 23 , b ) rs . 37 , c ) rs . 45 , d ) rs . 67 , e ) rs . 77 | b | subtract(64, multiply(multiply(64, power(add(const_1, divide(1, 2)), const_3)), power(divide(1, 2), const_3))) | divide(n4,n5)|add(#0,const_1)|power(#0,const_3)|power(#1,const_3)|multiply(n0,#3)|multiply(#4,#2)|subtract(n0,#5) | general |
a boat can travel with a speed of 25 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 90 km downstream . | "speed downstream = ( 25 + 5 ) km / hr = 30 km / hr . time taken to travel 90 km downstream = 90 / 30 hrs = 3 hrs . answer : c" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | c | divide(90, add(25, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
how many ounces of a 60 % salt solution must be added to 60 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ? | "let x = ounces of 60 % salt solution to be added . 2 * 60 + . 6 x = . 4 ( 60 + x ) x = 60 answer d" | a ) 16.67 , b ) 30 , c ) 50 , d ) 60 , e ) 70 | d | divide(subtract(multiply(divide(40, const_100), 60), multiply(divide(20, const_100), 60)), subtract(divide(40, const_100), divide(20, const_100))) | divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#0,#1)|subtract(#2,#3)|divide(#5,#4)| | gain |
in a college , 50 % of total 400 arts students are locals . 25 % of students from 100 science students are locals and 85 % of total 120 commerce students are locals . what is the total percentage of locals from arts , science and commerce . | locals from arts = 50 % of 400 = 200 locals from science = 25 % of 100 = 25 locals from commerce = 85 % of 120 = 102 total locals = 200 + 25 + 102 = 327 d | a ) 322 , b ) 340 , c ) 333 , d ) 327 , e ) 500 | d | add(add(multiply(400, divide(50, const_100)), multiply(divide(25, const_100), 100)), multiply(divide(85, const_100), 120)) | divide(n0,const_100)|divide(n2,const_100)|divide(n4,const_100)|multiply(n1,#0)|multiply(n3,#1)|multiply(n5,#2)|add(#3,#4)|add(#6,#5) | gain |
12 : 6 seconds : : ? : 8 minutes | "12 * 8 = 6 * x x = 16 answer : c" | a ) 10 , b ) 15 , c ) 16 , d ) 25 , e ) 30 | c | multiply(8, divide(12, 6)) | divide(n0,n1)|multiply(n2,#0)| | physics |
a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also ... | "simply look at the question from the other side . what is the probability that you can β t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability of something happening is 1 β the probability that it doesn β t happen , you can use the following equation to reach the right answer y :... | a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25 | e | add(multiply(divide(3, 5), divide(3, 5)), add(multiply(divide(3, 5), divide(const_2, 5)), multiply(divide(3, 5), divide(const_2, 5)))) | divide(n0,n1)|divide(const_2,n1)|multiply(#0,#1)|multiply(#0,#0)|add(#2,#2)|add(#4,#3)| | general |
if jake loses 15 pounds , he will weigh twice as much as his sister . together they now weigh 132 pounds . what is jake β s present weight , in pounds ? | "lets say j is the weight of jack and s is the wt of his sister . if he loses 15 pounds , he s twice as heavy as his sister . j - 15 = 2 * s also , together they weight 132 pounds j + s = 132 solvong the 2 equation , we get j = 93 pounds ! b" | a ) 63 , b ) 93 , c ) 139 , d ) 147 , e ) 188 | b | add(multiply(divide(subtract(132, 15), const_3), const_2), 15) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
a and b complete a work in 7 days . a alone can do it in 14 days . if both together can do the work in how many days ? | "1 / 7 + 1 / 14 = 3 / 14 14 / 3 = 4.6 days answer : a" | a ) 4.6 days , b ) 4.78 days , c ) 5.65 days , d ) 3.77 days , e ) 5.75 days | a | inverse(add(inverse(7), inverse(14))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
a car traveled the first quarter of a certain distance at 4 times the speed it traveled the remaining distance . what proportion of the total time traveled , was the time taken to travel the first quarter of the distance ? | these problems can be solved through algebra or sly number picking . being a big fan of solving problems with numbers , let ' s pick a total distance divisible by 4 ( say 40 ) so we can break it up into quarters , and a speed that can easily be one fourth , say 10 . each quarter is thus 10 kilometers ( or miles or feet... | a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) 1 / 13 | e | inverse(add(multiply(4, subtract(4, const_1)), const_1)) | subtract(n0,const_1)|multiply(n0,#0)|add(#1,const_1)|inverse(#2) | physics |
one fourth of one third of two fifth of a number is 17 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 17 then x = 17 * 30 = 510 40 % of 510 = 204 answer : option b" | a ) 140 , b ) 204 , c ) 180 , d ) 200 , e ) 250 | b | divide(multiply(divide(17, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain |
what is the decimal equivalent of ( 1 / 5 ) ^ 2 ? | "( 1 / 5 ) Β² = ( 1 / 5 ) ( 1 / 5 ) = 1 / 25 approach # 1 : use long division to divide 25 into 1 to get 1 / 25 = 0.04 b" | a ) 0.0016 , b ) 0.04 , c ) 0.16 , d ) 0.25 , e ) 0.5 | b | power(divide(1, 5), 2) | divide(n0,n1)|power(#0,n2)| | general |
at an international conference , β red β world countries and β blue β world countries are the only participants . the ratio of β red β world participants to β blue β world participants is 10 : 5 . if one - third of β red β world participants are left - handed and two - thirds of β blue β world participants are left - h... | red : blue = 10 : 5 let red = 10 x and blue = 5 x 1 / 3 of red are left handed = > 1 / 3 * 10 x = 10 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 5 x = 10 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left hande... | a ) 4 / 10 , b ) 4 / 9 , c ) 6 / 5 , d ) 9 / 2 , e ) 6 / 4 | b | divide(add(multiply(10, const_0_33), multiply(5, divide(const_2, const_3))), add(10, 5)) | add(n0,n1)|divide(const_2,const_3)|multiply(n0,const_0_33)|multiply(n1,#1)|add(#2,#3)|divide(#4,#0) | other |
the average salary / head of allthe workers in a workshop is rs . 750 , if the average salary / head of 5 technician is rs . 900 and the average salary / head of the rest is rs . 700 , the total no . of workers in the work - shop is ? | let the total number of workers be y . so sum of salary for all workers = sum of salary of 5 technician + sum of salary for other y - 5 workers . 5 x 900 + 700 ( y - 5 ) = 750 y β 4500 + 700 y - 3500 = 750 y β 50 y = 1000 β΄ y = 20 so total number of workers = 20 b | a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | b | divide(subtract(multiply(5, 900), multiply(5, 700)), subtract(750, 700)) | multiply(n1,n2)|multiply(n1,n3)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
the average age of 10 men increases by 6 years when two women are included in place of two men of ages 18 and 22 years . find the average age of the women ? | "explanation : 18 + 22 + 10 * 6 = 100 / 2 = 50 answer : d" | a ) 37 , b ) 26 , c ) 30 , d ) 50 , e ) 11 | d | divide(add(add(18, 22), multiply(10, 6)), const_2) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
a started a business with an investment of rs . 70000 and after 3 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 43000 , then the share of b is ? | ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 9 ) = 7 : 36 total profit = rs . 43000 share of b = 36 / 43 ( 43000 ) = rs . 36000 answer : b | a ) 33888 , b ) 36000 , c ) 27778 , d ) 27772 , e ) 81122 | b | subtract(43000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0) | gain |
working together , printer a and printer b would finish the task in 10 minutes . printer a alone would finish the task in 30 minutes . how many pages does the task contain if printer b prints 5 pages a minute more than printer a ? | "10 * a + 10 * b = x pages in 10 mins printer a will print = 10 / 30 * x pages = 1 / 3 * x pages thus in 10 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 5 more pages per min that printer a . in 10 mins printer b will print 50 more pages than printer a thus 2 /... | a ) 125 , b ) 150 , c ) 175 , d ) 200 , e ) 225 | b | multiply(divide(5, subtract(divide(30, 10), const_1)), 30) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)| | physics |
the ratio of boys to girl in a school is 5 : 4 . if there are 1500 boys in the school , how many girls are there ? | think of the ratio as 5 parts : 4 parts . divide 1500 by 5 to find 1 ` ` part ' ' of the ratio . 1500 / 5 = 300 multiply this by 4 to get 4 ` ` parts ' ' of the ratio . 300 * 4 = 1200 there are 1200 girls in the school . the answer is b | a ) 300 , b ) 1200 , c ) 1500 , d ) 1800 , e ) 2000 | b | multiply(divide(1500, 5), 4) | divide(n2,n0)|multiply(n1,#0) | other |
a tank can be filled by two pipes a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tank from empty state if b is used for the first half time and then a and b fill it together for the other half | "explanation : let the total time be x mins . part filled in first half means in x / 2 = 1 / 40 part filled in second half means in x / 2 = 1 / 60 + 1 / 40 = 1 / 24 total = x / 2 β 1 / 40 + x / 2 β 1 / 24 = 1 = > x / 2 ( 1 / 40 + 1 / 24 ) = 1 = > x / 2 β 1 / 15 = 1 = > x = 30 mins option d" | a ) 15 mins , b ) 20 mins , c ) 25 mins , d ) 30 mins , e ) none of these | d | divide(multiply(multiply(40, const_3), const_2), add(const_4, const_4)) | add(const_4,const_4)|multiply(n1,const_3)|multiply(#1,const_2)|divide(#2,#0)| | physics |
the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 4 revolutions ? ( use Γ― β¬ = 22 Γ’ Β β 7 ) | "required area covered in 5 revolutions = 4 Γ£ β 2 Γ― β¬ rh = 4 Γ£ β 2 Γ£ β 22 Γ’ Β β 7 Γ£ β 0.7 Γ£ β 2 = 35.2 m 2 answer a" | a ) 35.2 m 2 , b ) 44 m 2 , c ) 48 m 2 , d ) 36 m 2 , e ) none of these | a | multiply(4, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | divide(n0,n1)|divide(n3,n4)|multiply(#1,n1)|multiply(#0,#2)|multiply(n1,#3)|multiply(n2,#4)| | physics |
in a group of ducks and cows , the total number of legs are 48 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 48 = > 2 x = 48 = > x = 24 c" | a ) 28 , b ) 12 , c ) 24 , d ) 16 , e ) 19 | c | divide(48, const_2) | divide(n0,const_2)| | general |
two - third of a positive number and 64 / 216 of its reciprocal are equal . the number is : | "let the number be x . then , 2 / 3 x = 64 / 216 * 1 / x x 2 = 64 / 216 * 3 / 2 = 64 / 144 = 4 / 9 x = 2 / 3 answer : d" | a ) 5 / 12 , b ) 12 / 5 , c ) 25 / 144 , d ) 2 / 3 , e ) 146 / 25 | d | sqrt(divide(multiply(64, const_3), multiply(216, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|divide(#0,#1)|sqrt(#2)| | general |
when positive integer m is divided by positive integer n , the remainder is 12 . if m / n = 24.2 , what is the value of n ? | n = decimal part of 24.2 * remainder i . e . ( 12 ) hence 0.2 * n = 12 n = 60 . b | a ) 120 , b ) 60 , c ) 30 , d ) 24 , e ) 12 | b | divide(12, reminder(24.2, const_1)) | reminder(n1,const_1)|divide(n0,#0) | general |
if x and y are positive integers and 12 x = 21 y what is the least possible value of xy ? | "12 x = 21 y = > x / y = 7 / 4 = > 4 x = 7 y 4 ( 3 ) = 7 ( 3 ) = > x * y = 9 a" | a ) 9 , b ) 28 , c ) 63 , d ) 84 , e ) 252 | a | add(21, const_1) | add(n1,const_1)| | general |
a train crosses a platform of 350 m in 15 sec , same train crosses another platform of length 500 m in 20 sec . then find the length of the train ? | "length of the train be β x β x + 350 / 15 = x + 500 / 20 4 x + 1400 = 3 x + 1500 x = 100 m answer : a" | a ) 100 m , b ) 180 m , c ) 159 m , d ) 250 m , e ) 152 m | a | subtract(multiply(500, divide(15, divide(15, const_3))), multiply(350, divide(20, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
0.001 Γ£ β 0.001 = ? | "0.001 Γ£ β 0.001 = ? or , ? = 0.000001 answer d" | a ) 0.1 , b ) 0.001 , c ) 0.01 , d ) 1.0 e - 06 , e ) none of these | d | multiply(divide(0.001, 0.001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
if doubling a number and adding 20 to the result gives the same answer as multiplying the number by 8 and taking away 4 from the product , the number is | solution let the number be x . = then , 2 x + 20 = 8 x β 4 = 6 x = 24 βΉ = βΊ x = 4 . answer c | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | c | divide(add(20, 4), subtract(8, const_2)) | add(n0,n2)|subtract(n1,const_2)|divide(#0,#1) | general |
x + ( 1 / x ) = 3.5 find x ^ 2 + ( 1 / x ^ 2 ) | squaring on both sides ( x + 1 / x ) ^ 2 = 3.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 12.25 - 2 x ^ 2 + 1 / x ^ 2 = 10.25 answer : c | a ) 2.25 , b ) 3.25 , c ) 10.25 , d ) 5.25 , e ) 6.25 | c | subtract(power(3.5, 2), const_2) | power(n1,n2)|subtract(#0,const_2) | general |
what is the difference between the largest number and the least number written with the digits 9 , 2 , 1 , 5 ? | "explanation : 1259 9521 - - - - - - - - - - - - 8262 answer : d" | a ) 6084 , b ) 3788 , c ) 2077 , d ) 8262 , e ) 1812 | d | subtract(add(add(add(multiply(multiply(9, const_100), const_10), multiply(5, const_100)), multiply(2, const_10)), 1), add(add(add(const_1000, multiply(2, const_100)), multiply(5, const_10)), 9)) | multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)| | general |
the ratio of flour to water to sugar in a recipe is 7 : 2 : 1 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 2 cups of water , how much sugar is required ? | "the ratio of flour to water is 14 : 2 . the ratio of flour to sugar is 3.5 : 1 = 14 : 4 . the new ratio of flour to water to sugar is 14 : 2 : 4 if we need 2 cups of water , then we need 4 cups of sugar . the answer is c ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | multiply(divide(2, 2), multiply(const_4, 1)) | divide(n3,n1)|multiply(n2,const_4)|multiply(#0,#1)| | other |
a train , 130 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | explanation : assume the length of the bridge = x meter total distance covered = 130 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 130 + x ) / 30 m / s = > 45 Γ ( 10 / 36 ) = ( 130 + x ) / 30 = > 45 Γ 10 Γ 30 / 36 = 130 + x = > 45 Γ 10 Γ 10 / 12 = 130 + x = > 15 Γ 10 Γ 10 / 4 =... | a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | b | subtract(multiply(multiply(45, const_0_2778), 30), 130) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0) | physics |
the unit digit in the product 1258 * 6754 * 4512 * 9783 is ? | "unit digit in the given product = unit digit in 8 * 4 * 2 * 3 = 2 answer is c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(multiply(multiply(multiply(1258, 6754), 4512), 9783), subtract(multiply(multiply(multiply(1258, 6754), 4512), 9783), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 900 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 900 kilometers per hour is 1 / 900 hours = 3,600 / 900 seconds = 4 seconds ; time to cover this distance at regular speed is 900 + 2 = 902 seconds = 902 / 3,600 hours = 1 / 600 hours ; so , we get that to cove... | a ) 671.5 , b ) 600 , c ) 672.5 , d ) 673 , e ) 773.5 | b | divide(1, divide(add(multiply(const_3600, divide(1, 900)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
a school has 4 section of chemistry in class x having 70 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | "required average marks = 70 Γ£ β 50 + 35 Γ£ β 60 + 45 Γ£ β 55 + 42 Γ£ β 45 / 70 + 35 + 45 + 42 = 3500 + 2100 + 2475 + 1890 / 192 = 9965 Γ’ Β β 192 = 51.90 answer c" | a ) 50.9 , b ) 52.9 , c ) 51.9 , d ) 53.25 , e ) none of the above | c | divide(add(add(add(multiply(70, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(70, 35), 45), 42)) | add(n1,n2)|multiply(n1,n5)|multiply(n2,n6)|multiply(n3,n7)|multiply(n3,n4)|add(#1,#2)|add(n3,#0)|add(#5,#3)|add(n4,#6)|add(#7,#4)|divide(#9,#8)| | general |
how much is 40 % of 60 is greater than 4 / 5 of 25 ? | "( 40 / 100 ) * 60 Γ’ β¬ β ( 4 / 5 ) * 25 24 - 20 = 4 answer : e" | a ) 2 , b ) 5 , c ) 12 , d ) 8 , e ) 4 | e | subtract(multiply(60, divide(40, const_100)), multiply(divide(4, 5), 25)) | divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)| | general |
today joelle opened an interest - bearing savings account and deposited $ 5,000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | "interest for 1 st year = 5000 * 5 / 100 = 250 interest for 2 nd year = 5250 * 5 / 100 = 262.50 total = 5000 + 250 + 262.50 = 5512.50 answer : c" | a ) $ 6512.50 , b ) $ 7512.50 , c ) $ 5512.50 , d ) $ 4512.50 , e ) $ 5250.00 | c | add(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 5), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), co... | multiply(n2,const_3)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)| | gain |
if the sides of a triangle are 30 cm , 26 cm and 10 cm , what is its area ? | "the triangle with sides 30 cm , 26 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 26 * 10 = 130 cm 2 answer : option c" | a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 150 | c | divide(multiply(26, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
a store owner estimates that the average price of type a products will increase by 15 % next year and that the price of type b products will increase by 18 % next year . this year , the total amount paid for type a products was $ 4600 and the total price paid for type b products was $ 8400 . according to the store owne... | "cost of type a products next year = 1.15 * 4600 = 5290 cost of type b products next year = 1.18 * 8400 = 9912 total 5290 + 9912 = 15202 answer : a" | a ) $ 15,202 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general |
when positive integer n is divided by positive integer p , the quotient is 18 , with a remainder of 7 . when n is divided by ( p + 2 ) , the quotient is 10 and the remainder is 3 . what is the value of n ? | "n / p = 18 7 / p = 18 p + 7 n / ( p + 2 ) = 10 1 / ( p + 2 ) = 10 p + 20 + 3 solving these two equations we get p = 2 n = 43 answer is d" | a ) 151 , b ) 331 , c ) 511 , d ) 43 , e ) 87 | d | multiply(18, divide(subtract(add(multiply(10, 2), 3), 7), subtract(18, 10))) | multiply(n2,n3)|subtract(n0,n3)|add(n4,#0)|subtract(#2,n1)|divide(#3,#1)|multiply(n0,#4)| | general |
ashok borrowed some money at the rate of 6 % p . a . for the first two years , at the rate of 9 % p . a . for the next 3 years and at the rate of 14 % p . a . for the period beyond 5 years . if he pays a total interest of 11400 / - at the end of 9 years , how much money did he borrow ? | we have , si = p Γ r Γ t / 100 β΄ 11400 = p Γ 6 Γ 2 / 100 + p Γ 9 Γ 3 / 100 + p Γ 14 Γ 4 / 100 or , 12 p + 27 p + 56 p = 11400 Γ 100 or , 95 p = 11400 Γ 100 β΄ p = 12000 answer d | a ) 16060 , b ) 14000 , c ) 18000 , d ) 12000 , e ) none of these | d | divide(11400, add(divide(multiply(14, subtract(subtract(9, const_3), const_2)), const_100), add(divide(multiply(6, const_2), const_100), divide(multiply(9, 3), const_100)))) | multiply(n0,const_2)|multiply(n1,n2)|subtract(n1,const_3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,const_2)|add(#3,#4)|multiply(n3,#5)|divide(#7,const_100)|add(#6,#8)|divide(n5,#9) | general |
` ` i am 8 times as old as you were when i was as old as you are ` ` , said a man to his son . find out their present ages if the sum of the their ages is 75 years . | present age of boy x and previous age of boy y so present age of dad : 8 y and previous age is x so the gap between duration must be same so 8 y - x = x - y - - > x = 9 y / 2 and x + 8 y = 75 ( given ) - - > 9 y / 2 + y = 75 so y = 6 and and x = 27 therefore fathers age is 8 y = 8 * 6 = 48 and child ' s age = x = 27 an... | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | e | subtract(75, divide(75, add(divide(add(8, const_1), multiply(8, const_2)), const_1))) | add(n0,const_1)|multiply(n0,const_2)|divide(#0,#1)|add(#2,const_1)|divide(n1,#3)|subtract(n1,#4) | general |
ada and paul received their scores on 3 tests . on the first test , ada ' s score was 10 points higher than paul ' s score . on the second test , ada ' s score was 4 points higher than paul ' s score . if paul ' s average ( arithmetic mean ) score on the 3 tests was 4 points higher than ada ' s average score on the 3 t... | my take is option d ( 23 ) i followed a simple approach ( explained below ) : test 1 : ada ' s score = paul ' s score + 10 test 2 : ada ' s score = paul ' s score + 4 avg . of paul ' s score = 4 points higher than avg . of ada ' s score this implies that : sum of paul ' s score [ 3 tests ] = 12 points higher than sum o... | a ) 9 , b ) 14 , c ) 17 , d ) 23 , e ) 26 | e | add(add(multiply(4, 3), 10), 4) | multiply(n0,n2)|add(n1,#0)|add(n2,#1) | general |
the simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on $ 4000 for 2 years at 10 % per annum . the sum placed on simple interest is ? | c . i . = 4000 * ( 1 + 10 / 100 ) ^ 2 - 4000 = 4000 * 11 / 10 * 11 / 10 - 4000 = $ 840 sum = 420 * 100 / 3 * 8 = $ 1750 answer is c | a ) $ 1250 , b ) $ 1300 , c ) $ 1750 , d ) $ 2015 , e ) $ 1452 | c | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), const_2)), 4000), 2), multiply(3, divide(8, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,const_2)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3) | gain |
teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 156 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to fin... | a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 178.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(156, add(add(1, 1), 2)), add(126, 135)), 2) | add(n2,n2)|add(n0,n1)|add(n4,#0)|multiply(n5,#2)|subtract(#3,#1)|divide(#4,n4)| | other |
in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number of employees who take the subway to work ? | "let part time emp = x let full time emp = y then , 48 = x + y . . . . . . . . . ( 1 ) ( 1 / 3 ) x + ( 1 / 4 ) y = no . of ppl taking the subway 4 x + 3 y / 12 = no . of ppl taking the subway using 1 x / 12 + 3 * 48 / 12 = no . of ppl taking the subway so , the minimum value of x has to be 12 . hence to maximize the no... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)| | general |
what is the maximum number of pieces of birthday cake of size 3 β by 3 β that can be cut from a cake 21 β by 21 β ? | "the prompt is essentially asking for the maximum number of 3 x 3 squares that can be cut from a larger 21 by 21 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 7 ' pieces ' each , we have ( 7 ) ( 7 ) = 49 total smaller squares ( at maximum ) . c" | a ) 5 , b ) 7 , c ) 49 , d ) 20 , e ) 25 | c | divide(multiply(21, 21), multiply(3, 3)) | multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)| | physics |
in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 37 . in the company , what is the maximum number of non - managers in a department that has 9 managers ? | "9 / 7 * 37 = 47.6 the answer is d ." | a ) 44 , b ) 45 , c ) 46 , d ) 47 , e ) 48 | d | floor(divide(multiply(37, 9), 7)) | multiply(n1,n2)|divide(#0,n0)|floor(#1)| | general |
a train covers a distance of 12 km in 10 min . if it takes 7 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 7 = 140 m . answer : c" | a ) 177 m , b ) 189 m , c ) 140 m , d ) 178 m , e ) 188 m | c | divide(12, subtract(divide(12, 10), 7)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
in a class of 78 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "explanation : you could solve this by drawing a venn diagram . a simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking french to find the number taking only french . likewise find those taking only german . then we have : total = only french + only german ... | a ) 26 , b ) 27 , c ) 24 , d ) 18 , e ) 11 | c | subtract(78, subtract(add(41, 22), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
if ( n + 2 ) ! / n ! = 182 , n = ? | "( n + 2 ) ! / n ! = 182 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 132 from here , we might just test the answer choices . since ( 14 ) ( 13 ) = 182 , we can see that n... | a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1)| | general |
in what ratio must tea at rs . 64 per kg be mixed with tea at rs . 74 per kg so that the mixture must be worth rs . 69 per kg ? | "required ratio = 500 : 500 = 1 : 1 answer a" | a ) 1 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 3 , e ) none | a | divide(subtract(74, 69), subtract(69, 64)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | other |
36 . a 240 m long train crosses a platform of equal length in 27 s . what is the speed of the train in km / h ? | by the formula , speed = distance / time = ( 240 + 240 ) / 27 x ( 18 / 5 ) = 480 / 27 x 18 / 5 = 64 km / h answer : d | a ) 48 , b ) 60 , c ) 56 , d ) 64 , e ) 68 | d | multiply(divide(multiply(const_2, 240), 27), divide(const_3600, const_1000)) | divide(const_3600,const_1000)|multiply(n1,const_2)|divide(#1,n2)|multiply(#2,#0) | physics |
the base of a triangular field is three times its altitude . if the cost of cultivating the field at rs . 24.68 per hectare be rs . 333.18 , find its base and height . | sol . area of the field = total cost / rate = ( 333.18 / 25.6 ) hectares = 13.5 hectares ( 13.5 x 10000 ) m ^ 2 = 135000 m ^ 2 . let altitude = x metres and base = 3 x metres . then , ( 1 / 2 ) * 3 x * x = 135000 < = > x ^ 2 = 90000 < = > x = 300 . base = 900 m and altitude = 300 m . ans : e | ['a ) 100 m', 'b ) 125 m', 'c ) 150 m', 'd ) 278 m', 'e ) 300 m'] | e | divide(divide(divide(multiply(divide(333.18, 24.68), const_1000), const_3), const_10), divide(const_3, const_2)) | divide(n1,n0)|divide(const_3,const_2)|multiply(#0,const_1000)|divide(#2,const_3)|divide(#3,const_10)|divide(#4,#1) | geometry |
a , b and c enter into a partnership by investing $ 11000 , $ 15000 and $ 23000 respectively . at the end of 8 months , b receives $ 1540 as his share . find the share of a . | "the ratio of capital of a , b and c = 11000 : 15000 : 23000 = 11 : 15 : 23 a ' s share = ( 11 / 15 ) * 1560 = $ 1144 the answer is d ." | a ) $ 811 , b ) $ 922 , c ) $ 1033 , d ) $ 1144 , e ) $ 1255 | d | divide(multiply(multiply(11000, divide(1540, divide(multiply(15000, 8), const_100))), 8), const_100) | multiply(n1,n3)|divide(#0,const_100)|divide(n4,#1)|multiply(n0,#2)|multiply(n3,#3)|divide(#4,const_100)| | gain |
if each year the population of the country grows by 30 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 130 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 169 . . . . . . . . . . . . . 2 year elapsed year 2004 : population > 200 . . . . . . . . . . . . . . . . . . 3 year elapsed answer : d" | a ) 2 , b ) 4 , c ) 5 , d ) 3 , e ) 7 | d | add(const_4, const_4) | add(const_4,const_4)| | gain |
if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 ? | "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 24 , 26 , 30 , 33 , 34 , 35 , 36 , 42 . . . numbers less than median are 24 , 26 , 30 , 33 . . the median of these numbers = center of 26 and 30 =... | a ) 33 , b ) 28 , c ) 27 , d ) 24 , e ) 23 | b | divide(add(24, 26), const_2) | add(n1,n4)|divide(#0,const_2)| | general |
how many integers are divisible by 3 between 20 ! and 20 ! + 20 inclusive ? | "b - 7 20 ! is divisible by 3 there are 6 numbers between 10 ! and 10 ! + 20 that are divisible by 3 . hence 7" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(divide(20, 3), const_1) | divide(n3,n0)|add(#0,const_1)| | general |
the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 27 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 27 / 60 ) ) km = 11.25 km . answer : e" | a ) 9.25 , b ) 5.25 , c ) 7.25 , d ) 6.25 , e ) 11.25 | e | multiply(divide(27, const_60), add(20, 5)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
if 4 people are selected from a group of 10 married couples , what is the probability that none of them would be married to each other ? | "if we are to select 4 people from 10 couples without any restriction , how many ways can we make the selection ? 20 ! / 4 ! 16 ! = 4845 if we are to select 4 people from 10 couples with restriction that no married couple can both make it to the group , only a representative ? 10 ! / 4 ! 6 ! = 210 but we know that to s... | a ) 1 / 33 , b ) 2 / 63 , c ) 1 / 3 , d ) 43 / 62 , e ) 11 / 12 | d | multiply(multiply(multiply(divide(multiply(10, const_2), multiply(10, const_2)), divide(multiply(4, 4), subtract(multiply(10, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | multiply(n0,n0)|multiply(n1,const_2)|divide(#1,#1)|subtract(#0,const_2)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#0)|divide(#0,#5)|subtract(#3,const_2)|divide(#8,#4)|multiply(#2,#7)|multiply(#6,#10)|multiply(#9,#11)| | probability |
what is the least number of square tiles required to pave the floor of a room 50 m 00 cm long and 11 m 25 cm broad ? | solution length of largest tile = h . c . f . of 5000 cm & 1125 cm = 25 cm . area of each tile = ( 25 x 25 ) cm 2 β΄ required number of tiles = [ 5000 x 1125 / 25 x 25 ] = 9000 . answer d | a ) 724 , b ) 804 , c ) 814 , d ) 9000 , e ) none | d | divide(multiply(add(multiply(11, const_100), 25), multiply(50, const_100)), power(25, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2) | physics |
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 65 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 * 65 ) = 21 + 507 = > 528 hence answer will be ( e ) 528" | a ) 320 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 528 $ | e | multiply(multiply(0.65, 65), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
if the average of 37 , 69 , 47 and x is 46.5 , what is x ? | "x = 33 because : ( 37 + 69 + 47 + x ) / 4 = 46.5 ( 37 + 69 + 47 + x ) / 4 * 4 = 46.5 * 4 ( 37 + 69 + 47 + x ) = 186 ( 37 + 69 + 47 + x ) - 153 = 186 - 153 x = 33 therefore , the answer is e , 33 ." | a ) 37 , b ) 41 , c ) 50 , d ) 59 , e ) 33 | e | add(multiply(37, const_2), multiply(subtract(69, multiply(37, const_2)), 37)) | multiply(n0,const_2)|subtract(n1,#0)|multiply(n0,#1)|add(#0,#2)| | general |
joe β s average ( arithmetic mean ) test score across 4 equally weighted tests was 60 . he was allowed to drop his lowest score . after doing so , his average test score improved to 65 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 60 . so what we can assume is that we have 4 test scores , each 60 . he dropped his lowest score and the avg went to 65 . this means that the lowest score was not 60 and other three scores had given the lowest score 5 each to make it up to 60 too . when the lowest sc... | a ) 20 , b ) 45 , c ) 55 , d ) 65 , e ) 80 | b | subtract(multiply(60, 4), multiply(65, const_3)) | multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)| | general |
if the number 992 , 132,24 x is divisible by 11 , what must be the value of x ? | multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 992 , 132,24 x sum of digits at odd places = 9 + 2 + 3 + 2 + x = 16 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 4 = 16 ( ii ) ( i ) - ( ii ) = 16 + x - 16 = x - 0 hence x should b... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 0 | e | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | multiply(const_10,const_3)|add(#0,const_1)|multiply(#1,const_2)|multiply(#2,const_4)|multiply(n2,#3) | general |
in kaya ' s teacher ' s desk there are 9 pink highlighters , 8 yellow highlighters , and 5 blue highlighters . how many highlighters are there in all ? | add the numbers of highlighters . 9 + 8 + 5 = 22 . answer is b . | a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | b | add(add(9, 8), 5) | add(n0,n1)|add(n2,#0) | general |
a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 5... | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 500 bolts 500 * 6 / 25 = 120 mins ans c" | a ) 36 , b ) 72 , c ) 120 , d ) 144 , e ) 180 | c | divide(500, add(divide(100, const_60), divide(150, const_60))) | divide(n0,const_60)|divide(n1,const_60)|add(#0,#1)|divide(n2,#2)| | physics |
the average mark of a class of thirty two students is 75 . if 3 students whose marks are 28 and 34 are removed , then find the approximate average mark of the remaining students of the class . | exp . total mark of 32 students = 75 * 32 = 2400 , total mark after the removal of 2 students = 2400 β ( 28 + 34 ) = 2400 β 62 = 2338 approximate average mark = 2338 / ( 32 - 2 ) = 2338 / 30 = 78 answer : c | a ) 71 , b ) 74 , c ) 78 , d ) 70 , e ) 80 | c | divide(subtract(subtract(multiply(75, subtract(34, const_2)), 28), 34), subtract(subtract(34, const_2), 3)) | subtract(n3,const_2)|multiply(n0,#0)|subtract(#0,n1)|subtract(#1,n2)|subtract(#3,n3)|divide(#4,#2) | general |
a can do a piece of work in 6 days , and b can do it in 5 days . if c , who can do the work in 12 days , joins them , how long will they take to complete the work ? | a , b , and c do the work in = 6 * 5 * 12 / 6 * 5 + 5 * 12 + 6 * 12 = 360 / 162 = 2.2 days answer is c | a ) 2 days , b ) 3.5 days , c ) 2.2 days , d ) 4 days , e ) 5.7 days | c | divide(const_1, add(divide(const_1, 12), add(divide(const_1, 6), divide(const_1, 5)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4) | physics |
the perimeter of a triangle is 39 cm and the inradius of the triangle is 1.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 1.5 * 39 / 2 = 29 cm 2 answer : e" | a ) 22 , b ) 35 , c ) 77 , d ) 54 , e ) 29 | e | triangle_area(1.5, 39) | triangle_area(n0,n1)| | geometry |
the average age of 20 students of a class is 20 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 20 th student is | "explanation : age of the 20 th student = [ 20 * 20 - ( 14 * 5 + 16 * 9 ) ] = ( 400 - 214 ) = 186 years . answer : a" | a ) 186 , b ) 120 , c ) 152 , d ) 220 , e ) 220 | a | subtract(multiply(20, 20), add(multiply(5, 14), multiply(9, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 5 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the ... | "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 21 / 30 ) = 10 / 21 hours peter worked for 31 / 21 hours so he painted 31 / 21 * 1 / 3 = 31 / 63 of t... | a ) 19 / 35 , b ) 23 / 42 , c ) 27 / 55 , d ) 31 / 63 , e ) 35 / 74 | d | divide(const_4, add(multiply(const_4, 5), const_1)) | multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)| | physics |
little roshini had $ 10.50 . he spent some money on sweets and gave to his two friends $ 3.40 each . how much money was spent for sweets ? | roshini spent and gave to his two friends a total of x + 3.40 + 3.40 = $ 3.65 money spent for sweet = 10.50 - 6.80 = $ 3.70 correct answer is e ) $ 3.70 | a ) $ 4.70 , b ) $ 4.80 , c ) $ 3.85 , d ) $ 3.80 , e ) $ 3.70 | e | subtract(10.5, multiply(const_2, 3.4)) | multiply(n1,const_2)|subtract(n0,#0) | general |
a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used th... | a ) 3 , b ) 4 , c ) 18 / 5 , d ) 9 , e ) 12 | c | divide(const_1, multiply(9, add(divide(const_1, 36), divide(const_1, const_2.0)))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(const_2.0,#2)|divide(const_1,#3)| | gain |
a batsman makes a score of 110 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 110 = 11 x x = 110 - 50 = 60 answer is c" | a ) 40 , b ) 50 , c ) 60 , d ) 45 , e ) 56 | c | add(subtract(110, multiply(11, 5)), 5) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.