Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 2 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 2 w = 120 = > w = 60 answer d" | a ) 65 , b ) 63 , c ) 52 , d ) 60 , e ) 68 | d | divide(rectangle_area(5, 24), 2) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry |
a man took loan from a bank at the rate of 12 % p . a . s . i . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was ? | principal = ( 100 * 5400 ) / ( 12 * 3 ) = rs . 15000 . answer : c | a ) 15002 , b ) 15001 , c ) 15000 , d ) 15007 , e ) 15009 | c | divide(divide(multiply(5400, const_100), 3), 12) | multiply(n2,const_100)|divide(#0,n1)|divide(#1,n0)| | gain |
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 8 per item . if the manufacturer ’ s selling price per item is $ 12 , how many items the manufacturer produce and sell to earn an annual profit of $ 140,000 ? | "let the items manufactured or sold bex 130000 + 8 x = 12 x - 140000 4 x = 270000 x = 67500 ans : e" | a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 67,500 | e | divide(multiply(divide(divide(multiply(add(add(const_3, 8), 12), power(const_100, const_2)), subtract(12, 8)), multiply(const_4, const_2)), const_3), const_1000) | add(n1,const_3)|multiply(const_2,const_4)|power(const_100,const_2)|subtract(n2,n1)|add(n2,#0)|multiply(#4,#2)|divide(#5,#3)|divide(#6,#1)|multiply(#7,const_3)|divide(#8,const_1000)| | general |
how many numbers between 100 and 714 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 714 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 714 : 714 / 42 = 17 hence , 42 * 14 = 714 is the last number divisible by 42 within 714 h... | a ) 112 , b ) 77 , c ) 267 , d ) 15 , e ) 99 | d | subtract(divide(714, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)| | general |
how many paying stones , each measuring 3 m * 2 m are required to pave a rectangular court yard 60 m long and 14 m board ? | "60 * 14 = 3 * 2 * x = > x = 140 answer : b" | a ) 99 , b ) 140 , c ) 26 , d ) 17 , e ) 12 | b | divide(rectangle_area(60, 14), rectangle_area(3, 2)) | rectangle_area(n2,n3)|rectangle_area(n0,n1)|divide(#0,#1)| | physics |
if a light flashes every 6 seconds , how many times will it flash in 1 / 2 of an hour ? | "in 1 / 2 of an hour there are 30 * 60 = 1800 seconds the number of 6 - second intervals = 1800 / 6 = 300 after the first flash , there will be 300 more flashes for a total of 301 . the answer is a ." | a ) 301 times , b ) 351 times , c ) 401 times , d ) 451 times , e ) 481 times | a | divide(multiply(divide(const_3600, const_4), const_3), 6) | divide(const_3600,const_4)|multiply(#0,const_3)|divide(#1,n0)| | physics |
the average runs scored by a batsman in 30 matches is 40 . in the next 10 matches the batsman scored an average of 13 runs . find his average in all the 30 matches ? | "total score of the batsman in 30 matches = 600 . total score of the batsman in the next 10 matches = 130 . total score of the batsman in the 30 matches = 730 . average score of the batsman = 730 / 30 = 24.3 . answer : a" | a ) 24.3 , b ) 25.3 , c ) 26.3 , d ) 28.3 , e ) 34.5 | a | divide(add(multiply(40, 30), multiply(13, 10)), add(30, 10)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 36 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | let x be the wholesale price . then 1.8 x = 36 and x = 36 / 1.8 = 20 . to achieve a 100 % markup , the price needs to be $ 40 . the retailer needs to increase the price by $ 4 more . the answer is b . | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(multiply(divide(36, add(const_1, divide(80, const_100))), const_2), 36) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|multiply(#2,const_2)|subtract(#3,n1) | general |
in the city of san durango , 60 people own cats , dogs , or rabbits . if 30 people owned cats , 40 owned dogs , 16 owned rabbits , and 12 owned exactly two of the 3 types of pet , how many people owned all 3 ? | you are considering a case when cat dogs and rabbit are not exactly equal to 12 . the solution shall be 60 = 30 + 40 + 16 - 12 - 2 x and hence x shall be 7 . answer d | a ) 2 , b ) 4 , c ) 8 , d ) 7 , e ) 32 | d | divide(subtract(subtract(add(add(30, 40), 16), 12), 60), const_2) | add(n1,n2)|add(n3,#0)|subtract(#1,n4)|subtract(#2,n0)|divide(#3,const_2) | general |
a certain company ’ s profit in 1996 was 17 percent greater than its profit in 1995 , and its profit in 1997 was 25 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ? | "profit in 1995 - 100 profit in 1996 - 117 % increae profit in 1997 in comparison to 1995 = 17 + 117 * 25 % = 46.25 correct option : e" | a ) 5 % , b ) 18 % , c ) 33 % , d ) 35 % , e ) 46.25 % | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(const_1, divide(17, const_100))), const_1), const_100) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
a can give b 80 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ? | a runs 1000 m while b runs 920 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) / 920 = 869.56 m b can give c = 1000 - 869.56 = 130.43 m . answer : b | a ) 130.12 , b ) 130.43 , c ) 130.22 , d ) 111.0 , e ) 111.12 | b | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 80))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4) | physics |
what is the smallest no . that should be added to 45454 to make it exactly divisible by 9 ? | "if a number is divisible by 9 , the sum of its digits must be a multiple of 9 . here , 4 + 5 + 4 + 5 + 4 = 22 , the next multiple of 9 is 27 . 5 must be added to 45454 to make it divisible by 9 c" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | c | add(subtract(9, reminder(45454, 9)), const_2) | reminder(n0,n1)|subtract(n1,#0)|add(#1,const_2)| | general |
30 square stone slabs of equal size were needed to cover a floor area of 58.8 sq . m . find the length of each stone slab ? | "area of each slab = 58.8 / 30 m 2 = 1.96 m 2 length of each slab √ 1.96 = 1.4 m = 140 cm" | a ) 140 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(58.8, 30)), const_100) | divide(n1,n0)|sqrt(#0)|multiply(#1,const_100)| | geometry |
nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days they completed 1 / 3 part of work . than how many more emploies should nhai hire to finish it in time working 10 hours a day | "( m 1 * d 1 * h 1 ) / w 1 = ( m 2 * d 2 * h 2 ) / w 2 ( 100 * 25 * 8 ) / ( 1 / 3 ) = ( ( 100 + a ) * 25 * 10 ) ) / ( 2 / 3 ) 160 = 100 + a a = 60 answer : a" | a ) 60 , b ) 70 , c ) 80 , d ) 50 , e ) 40 | a | subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100) | multiply(n0,n4)|multiply(n4,n7)|multiply(n3,#0)|multiply(n1,#2)|divide(#3,#1)|subtract(#4,n0)| | physics |
the average of first four prime numbers greater than 20 is ? | "23 + 29 + 31 + 37 = 120 / 4 = 30 answer : c" | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | add(20, const_1) | add(n0,const_1)| | general |
jar x is 1 / 4 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | "let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 4 + p / 4 = p / 2 the answer is d ." | a ) 2 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 / 2 , e ) 2 / 3 | d | add(divide(1, const_2.0), multiply(divide(1, 4), divide(1, 4))) | divide(n0,n3)|multiply(#0,#0)|add(#0,#1)| | general |
the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 40 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 40 + 42 + 44 + . . . . . upto 12 terms ) this is an a . p with first term , a = 40 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 40 + { 12 - 1 ) x 2 ] = 6 ( 102 ) = 612 kms . answer : c" | a ) 252 kms , b ) 152 kms , c ) 612 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 40), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics |
in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 60 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 60 * 950 = rs . 57000 c" | a ) rs . 50000 , b ) rs . 52000 , c ) rs . 57000 , d ) rs . 60000 , e ) rs . 67000 | c | add(add(multiply(multiply(multiply(20, 15), const_2), 60), multiply(multiply(multiply(20, 5), const_2), 60)), multiply(multiply(multiply(15, 5), const_2), 60)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#8)| | general |
in the coordinate plane , a circle centered on point ( - 3 , - 3 ) passes through point ( 1 , 1 ) . what is the area of the circle ? | "r ^ 2 = ( - 3 - 1 ) ^ 2 + ( - 3 - 1 ) ^ 2 = 16 + 16 = 32 area of circle = π r ^ 2 = 32 π answer : b" | a ) 9 π , b ) 32 π , c ) 25 π , d ) 37 π , e ) 41 π | b | circle_area(sqrt(add(power(subtract(3, 1), const_2), power(add(1, 3), const_2)))) | add(n1,n2)|subtract(n0,n2)|power(#1,const_2)|power(#0,const_2)|add(#2,#3)|sqrt(#4)|circle_area(#5)| | geometry |
a small island country of 5000 people has an emigration and immigration statistic per 500 persons of 50.4 and 15.4 respectively . how long will it take for the population to be half of its current number . | annual decrease is ( 500 + 15.4 - 50.4 ) * 10 = 4650 hence every year there is a decrease of 350 for the population to become half 2500 must emigrate therefore 2500 / 350 = 133.3 correct option is e ) 7.14 | a ) 6.8 , b ) 8.5 , c ) 6.1 , d ) 9.1 , e ) 7.14 | e | divide(divide(5000, const_2), multiply(subtract(50.4, 15.4), divide(5000, 500))) | divide(n0,const_2)|divide(n0,n1)|subtract(n2,n3)|multiply(#1,#2)|divide(#0,#3) | other |
the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 66 kg . what is the correct weight ? | "actual total weight is ( 20 x 58.4 - 56 + 66 ) = 1178 kgs actual average weight is 1178 / 20 = 58.9 kgs c" | a ) 58 kgs , b ) 58.85 kgs , c ) 58.9 kgs , d ) 59 kgs , e ) 59.85 kgs | c | divide(add(multiply(58.4, 20), subtract(66, 56)), 20) | multiply(n0,n1)|subtract(n3,n2)|add(#0,#1)|divide(#2,n0)| | general |
in a certain store , the profit is 320 % of the cost . if the cost increases by 40 % but the selling price remains constant , approximately what percentage of the selling price is the profit | explanation : let c . p . = rs . 100 . then , profit = rs . 320 , s . p . = rs . 420 . new c . p . = 140 % of rs . 100 = rs . 140 new s . p . = rs . 420 . profit = rs . ( 420 - 140 ) = rs . 280 required percentage = ( 280 / 420 ) * 100 = 66.67 % ( approx ) = 67 % answer : a | a ) 67 % , b ) 80 % , c ) 90 % , d ) 100 % , e ) none of above | a | multiply(divide(subtract(add(const_100, 320), add(const_100, 40)), add(const_100, 320)), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100) | gain |
if the perimeter of a rectangular garden is 480 m , its length when its breadth is 100 m is ? | "2 ( l + 100 ) = 480 = > l = 140 m answer : b" | a ) 299 m , b ) 140 m , c ) 200 m , d ) 167 m , e ) 128 m | b | subtract(divide(480, const_2), 100) | divide(n0,const_2)|subtract(#0,n1)| | physics |
a family pays $ 800 per year for an insurance plan that pays 60 percent of the first $ 1000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ? | assuming the medical expenses are $ 1000 or more , the family pays $ 800 + $ 400 = $ 1200 . the total amount paid by insurance plan for the first $ 1000 of expenses is $ 600 . the insurance will pay another $ 600 when the medical expenses are $ 1600 . the answer is d . | a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,400 , d ) $ 1,600 , e ) $ 2,000 | d | subtract(1000, 800) | subtract(n2,n0) | general |
the h . c . f . of two numbers is 30 and the other two factors of their l . c . m . are 10 and 15 . the larger of the two numbers is : | "explanation : clearly , the numbers are ( 30 x 10 ) and ( 30 x 15 ) . { \ color { blue } \ therefore } larger number = ( 30 x 15 ) = 450 . answer : c ) 450" | a ) 350 , b ) 250 , c ) 450 , d ) 500 , e ) 620 | c | multiply(30, 15) | multiply(n0,n2)| | other |
in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 10 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 10 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 10 = 4 / 9 ( t - 10 ) 10 t = 990 - 440 = > t = 55 . answer : d" | a ) 40 , b ) 44 , c ) 48 , d ) 55 , e ) none of these | d | add(add(multiply(5, divide(10, subtract(multiply(divide(6, 5), 5), 4))), divide(10, subtract(multiply(divide(6, 5), 5), 4))), 10) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5)| | general |
in a particular state , 70 % of the counties received some rain on monday , and 50 % of the counties received some rain on tuesday . no rain fell either day in 20 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ? | "70 + 50 + 20 = 140 % the number is 40 % above 100 % because 40 % of the counties were counted twice . the answer is b ." | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | b | subtract(add(70, 50), subtract(const_100, 20)) | add(n0,n1)|subtract(const_100,n2)|subtract(#0,#1)| | gain |
a train passes a platform in 16 seconds . the same train passes a man standing on the platform in 10 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "speed of the train = 54 km / hr = ( 54 × 10 ) / 36 m / s = 15 m / s length of the train = speed × time taken to cross the man = 15 × 10 = 150 m let the length of the platform = l time taken to cross the platform = ( 150 + l ) / 15 = > ( 150 + l ) / 15 = 16 = > 150 + l = 15 × 16 = 240 = > l = 240 - 150 = 90 meter answe... | a ) 40 , b ) 50 , c ) 90 , d ) 30 , e ) 20 | c | multiply(multiply(const_0_2778, 54), subtract(16, 10)) | multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)| | physics |
a certain university will select 1 of 4 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidat... | "1 of 7 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more thn one deparments . we can choose any of the 4 candidates for the math dep . , which gives as 4 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections a... | a ) 42 , b ) 70 , c ) 140 , d ) 180 , e ) 315 | d | multiply(multiply(10, 3), 4) | multiply(n3,n5)|multiply(n1,#0)| | other |
sum of two numbers is 25 . first exceeds by 5 from the other . then the numbers will be ? | "explanation : x + y = 25 x - y = 5 x = 15 y = 10 answer : c" | a ) 20 , 5 , b ) 10 , 15 , c ) 15 , 10 , d ) 5 , 20 , e ) 12 , 13 | c | subtract(25, divide(subtract(25, divide(5, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
if 35 % of the 840 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "we know 35 % people study biology , therefore the no of people not studying = 100 - 35 = 65 % > therefore the people not studying biology out of a total 840 people are = 65 % of 840 > ( 65 / 100 ) * 840 = 546 people a" | a ) 546 , b ) 550 , c ) 555 , d ) 532 , e ) 515 | a | multiply(divide(840, const_100), subtract(const_100, 35)) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)| | other |
a train is 410 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 410 + 140 = 550 meter time = distance / speed = 550 * ( 2 / 25 ) = 44 seconds answer : e" | a ) 76 seconds , b ) 18 seconds , c ) 40 seconds , d ) 19 seconds , e ) 44 seconds | e | divide(add(410, 140), divide(multiply(45, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
if a is thrice as fast as b and together can do a work in 30 days . in how many days a alone can do the work ? | a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 30 = 3 + 1 / 3 x = 4 / 3 x = 1 / 30 x = 30 * 4 / 3 = 40 answer : e | a ) 36 , b ) 42 , c ) 28 , d ) 54 , e ) 40 | e | inverse(divide(const_3, multiply(30, const_4))) | multiply(n0,const_4)|divide(const_3,#0)|inverse(#1) | physics |
angelina walked 200 meters from her home to the grocery at a constant speed . she then walked 300 meters to the gym at double the speed . she spent 50 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to t... | "let the speed be x . . . so time taken from home to grocery = 200 / x . . the speed to gym = 2 x . . so time taken = 300 / 2 x = 150 / x . . its given 200 / x - 150 / x = 50 . . 50 / x = 50 . . x = 1 m / secs . . so grocery to gym = 2 * 1 = 2 m / s . . . answer : a" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12 | a | multiply(divide(subtract(200, divide(300, const_2)), 50), const_2) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,n2)|multiply(#2,const_2)| | physics |
the sum of the ages of 5 children born at the intervals of 3 year each is 40 year . what is the age of the eldest child ? | "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 40 â ‡ ” 5 x = 10 â ‡ ” x = 2 . â ˆ ´ age of the youngest child = x + 12 = 2 + 12 = 14 years . answer a" | a ) 14 year , b ) 8 year , c ) 10 year , d ) none of these , e ) 9 year | a | divide(subtract(divide(40, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general |
how many positive integers less than 5,000 are evenly divisible by neither 17 nor 21 ? | "integers less than 5000 divisible by 17 5000 / 17 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 17 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 17 and 21 = 333 + 238 - 47 = 524 now subtract tha... | a ) 4,514 , b ) 4,480 , c ) 4,521 , d ) 4,428 , e ) 4,349 | b | divide(factorial(subtract(add(const_4, 17), const_1)), multiply(factorial(17), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 2 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 2 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 2 * ( 2 / 3 ) ) days = > 90 days answer : e" | a ) 30 , b ) 20 , c ) 40 , d ) 60 , e ) 90 | e | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 2), divide(2, 3))) | divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4)| | physics |
if x and y are positive integers and 42 x = 56 y what is the least possible value of xy ? | "42 x = 56 y = > x / y = 4 / 3 = > 3 x = 4 y 3 ( 14 ) = 4 ( 14 ) = > x * y = 196 e" | a ) 14 , b ) 28 , c ) 63 , d ) 196 , e ) 252 | e | add(56, const_1) | add(n1,const_1)| | general |
a train 150 m long is running with a speed of 98 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going | "explanation : speed of the train relative to man = ( 98 - 8 ) kmph = ( 90 × 5 / 18 ) m / sec = 25 m / sec time taken by the train to cross the man i = time taken by it to cover 150 m at 25 m / sec = 150 × 1 / 25 sec = 6 sec answer : option b" | a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec | b | divide(150, multiply(add(98, 8), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
what is the sum of the greatest common factor and the lowest common multiple of 36 and 56 ? | "prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 56 = 2 x 2 x 2 x 7 gcf = 4 lcm = 504 sum = 508 ans : b" | a ) 512 , b ) 508 , c ) 451 , d ) 516 , e ) 451 | b | divide(multiply(36, 56), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 7225 , how much amount did raghu invest ? | "let money invested by raghu = rs . x money invested by trishul = 9 / 10 x = 0.9 x money invested by vishal = 9 / 10 x * 110 / 100 = 0.99 x also , x + 0.9 x + 0.99 x = 7225 = x = 7225 / 2.89 = 2500 therefore , amount invested by raghu is rs . 2500 . answer : c" | a ) 1287 , b ) 2887 , c ) 2500 , d ) 1129 , e ) 1192 | c | divide(7225, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1)) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|add(#3,#2)|add(#4,const_1)|divide(n2,#5)| | general |
the difference between two parallel sides of a trapezium is 4 cm . perpendicular distance between them is 10 cm . if the area of the trapezium is 250 cm 2 find the lengths of the parallel side | explanation : let the two parallel sides of the trapezium be a cm and b cm . then , a - b = 4 - - - - - - ( 1 ) and , ( 1 / 2 ) x ( a + b ) x 10 = 475 = > ( a + b ) = ( ( 250 x 2 ) / 10 ) = > a + b = 50 - - - - - - - ( 2 ) solving 1 and 2 , we get : a = 27 , b = 23 so , the two parallel sides are 27 cm and 23 cm . answ... | ['a ) 27 , 23', 'b ) 25 , 25', 'c ) 24 , 26', 'd ) 23 , 29', 'e ) 24 , 45'] | a | add(multiply(add(divide(subtract(divide(multiply(250, const_2), 10), 4), const_2), const_4), const_100), divide(subtract(divide(multiply(250, const_2), 10), 4), const_2)) | multiply(n2,const_2)|divide(#0,n1)|subtract(#1,n0)|divide(#2,const_2)|add(#3,const_4)|multiply(#4,const_100)|add(#3,#5) | physics |
in a box of 8 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 6 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 8 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 6 c 3 / 8 c 3 = 20 / 56 = 5 / 14 answer is b" | a ) 1 / 12 , b ) 5 / 14 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | b | divide(divide(factorial(subtract(8, 2)), multiply(factorial(subtract(subtract(8, 2), 3)), factorial(3))), divide(factorial(8), multiply(factorial(subtract(8, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9)| | general |
what is the area of square field whose side of length 17 m ? | "17 * 17 = 289 sq m answer : d" | a ) 225 , b ) 777 , c ) 266 , d ) 289 , e ) 261 | d | square_area(17) | square_area(n0)| | geometry |
a sum is divided between a and b in the ratio of 1 : 2 . a purchased a car from his part , which depreciates 14 2 / 7 % per annum and b deposited his amount in a bank , which pays him 20 % interest per annum compounded annually . by what percentage will the total sum of money increase after two years due to this invest... | let the total money = 300 a ' s part = 100 b ' s part = 200 after 2 years a ' s part = 73.46 ( as it depreciates by 14.29 % per annum ) after 2 years b ' s part = 288 ( using the compound interest formula , a = p ( 1 + r / n ) ^ nt ) , where p = 200 , r = 0.2 , n = 1 , t = 2 ) so , after 2 years total amount = 361.46 t... | a ) 22.48 % , b ) 20.48 % , c ) 24.48 % , d ) 25.48 % , e ) 26.48 % | b | add(20, divide(add(add(add(20, 20), 7), 1), const_100)) | add(n5,n5)|add(n4,#0)|add(n0,#1)|divide(#2,const_100)|add(n5,#3) | general |
how many seconds will a train 100 meters long take to cross a bridge 250 meters long if the speed of the train is 36 kmph ? | "d = 100 + 250 = 350 s = 36 * 5 / 18 = 10 mps t = 350 / 10 = 35 sec answer : b" | a ) 54 sec , b ) 35 sec , c ) 25 sec , d ) 45 sec , e ) 24 sec | b | divide(add(250, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
two trains of length 100 m and 160 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 160 = 260 m . the time required = d / s = 260 / 20 = 13 sec . answer : c" | a ) 18 sec , b ) 70 sec , c ) 13 sec , d ) 20 sec , e ) 19 sec | c | divide(add(100, 160), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
pipe a fills a tank in 30 minutes . pipe b can fill the same tank 5 times as fast as pipe a . if both the pipes were kept open when the tank is empty , how much time will it take for the tank to overflow ? | "let the total capacity of tank be 90 liters . capacity of tank filled in 1 minute by a = 3 liters . capacity of tank filled in 1 minute by b = 15 liters . therefore , capacity of the tank filled by both a and b in 1 minute = 18 liters . hence , time taken by both the pipes to overflow the tank = 90 / 18 = 5 minutes . ... | a ) 4 mins , b ) 6 mins , c ) 5 mins , d ) 3 mins , e ) 7 mins | c | inverse(add(divide(const_1, 30), divide(5, 30))) | divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2)| | physics |
the average age of a class of 15 students is 10 years . if the age of the teacher be included , then the average increased by 1 year . find the age of the teacher ? | "total age of 15 persons = 15 * 10 = 150 years average age of 16 persons = 11 years total age of 16 persons = 16 * 11 = 176 years age of the teacher = 176 - 150 = 26 years answer is c" | a ) 22 , b ) 23 , c ) 26 , d ) 29 , e ) 31 | c | subtract(multiply(add(15, 1), add(10, 1)), multiply(15, 10)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the sum of digits of a two digit number is 12 , the difference between the digits is 4 . find the number | "description : = > x + y = 12 , x - y = 4 adding these 2 x = 16 = > x = 8 , y = 4 . thus the number is 84 answer a" | a ) 84 , b ) 94 , c ) 83 , d ) 72 , e ) none | a | add(multiply(divide(add(12, 4), const_2), 12), subtract(12, divide(add(12, 4), const_2))) | add(n0,n1)|divide(#0,const_2)|multiply(n0,#1)|subtract(n0,#1)|add(#2,#3)| | general |
find number which is 80 % less than 80 . | "explanation : 80 % less is 20 % of the given number therefore , 20 % of 80 is 16 . answer : e" | a ) 18 , b ) 22 , c ) 28 , d ) 26 , e ) 16 | e | divide(multiply(80, 80), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
there are some birds in cage , 1 / 3 rd fly away . then 2 / 5 th of the remaining left the cage , then 2 / 3 rd of the remaining left the cage . at last 8 were remaining . how many birds were in total ? | sol : if x birds were there in total , then x × ( 1 – 1 / 3 ) × ( 1 – 2 / 5 ) × ( 1 – 2 / 3 ) = 8 x × 2 / 3 × 3 / 5 × 1 / 3 = 8 x = 60 answer : a | a ) 60 , b ) 27 , c ) 28 , d ) 26 , e ) 91 | a | divide(8, multiply(divide(3, 5), multiply(divide(2, 3), divide(1, 3)))) | divide(n1,n3)|divide(n2,n1)|divide(n0,n1)|multiply(#1,#2)|multiply(#0,#3)|divide(n6,#4) | general |
x is 3 times as fast as y and working together , they can complete a work in 20 days . in how many days can y alone complete the work ? | x = 3 y x + y = 3 y + y = 4 y these 4 y people together can do the work in 20 days , which means y can alone do the work in 4 * 20 = 80 days . answer : b | a ) 11 days , b ) 80 days , c ) 13 days , d ) 15 days , e ) 19 days | b | multiply(add(3, const_1), 20) | add(n0,const_1)|multiply(n1,#0) | physics |
the cross - section of a cannel is a trapezium in shape . if the cannel is 9 m wide at the top and 7 m wide at the bottom and the area of cross - section is 560 sq m , the depth of cannel is ? | "1 / 2 * d ( 9 + 7 ) = 560 d = 70 answer : c" | a ) 39 , b ) 28 , c ) 70 , d ) 80 , e ) 71 | c | divide(divide(divide(560, divide(add(9, 7), const_2)), 7), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
on dividing 199 by a number , the quotient is 11 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 199 - 1 ) / 11 = 198 / 11 = 18 d )" | a ) a ) 12 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | d | floor(divide(199, 11)) | divide(n0,n1)|floor(#0)| | general |
of the diplomats who attended a summit conference : 20 spoke french , 32 did not speak hindi and 20 % of the diplomats spoke neither french nor hindi . if 10 % of the diplomats spoke both french and hindi , then how many diplomats attended the conference ? | "2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer" | a ) 70 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | d | divide(subtract(32, 20), divide(10, const_100)) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,#0)| | other |
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 59 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 * 59 ) = 481.20 hence answer will be ( e )" | a ) 320.20 $ , b ) 380.20 $ , c ) 421.20 $ , d ) 451.20 $ , e ) 481.20 $ | e | multiply(multiply(0.65, 59), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
what is the nearest no . to 2304 which is divisible by 21 ? | remainder when 2304 is divided by 21 = 15 so , no . should added to make 15 to 21 = 6 no . = 2304 + 6 = 2310 answer b | a ) 2289 , b ) 2310 , c ) 2333 , d ) 2456 , e ) 2309 | b | add(2304, reminder(2304, 21)) | reminder(n0,n1)|add(n0,#0) | general |
the population of a town is 10000 . it increases annually at the rate of 25 % p . a . what will be its population after 2 years ? | formula : 10000 × 125 / 100 × 125 / 100 = 15625 answer : d | a ) 14000 , b ) 14400 , c ) 14500 , d ) 15625 , e ) 14700 | d | add(10000, multiply(divide(multiply(10000, 25), const_100), const_2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_2)|add(n0,#2) | gain |
when asked what the time is , a person answered that the amount of time left is 4 / 8 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 4 / 8 x ⇒ x = 20 time is 4 pm answer : c" | a ) 2 pm , b ) 9 pm , c ) 4 pm , d ) 8 pm , e ) 6 pm | c | subtract(multiply(const_2, const_12), divide(multiply(const_2, const_12), add(divide(4, 8), const_1))) | divide(const_2.0,n1)|multiply(const_12,const_2)|add(#0,const_1)|divide(#1,#2)|subtract(#1,#3)| | general |
if x and y are integers , what is the least positive number of 24 x + 20 y ? | 24 x + 20 y = 4 ( 6 x + 5 y ) which will be a minimum positive number when 6 x + 5 y = 1 . 6 ( 1 ) + 5 ( - 1 ) = 1 then 4 ( 6 x + 5 y ) can have a minimum positive value of 4 . the answer is c . | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | subtract(24, 20) | subtract(n0,n1) | general |
of 25 applicants for a job , 14 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 25 - 3 = 22 aub = a + b - aib aib = 18 + 14 - 22 = 10 answer b" | a ) 14 , b ) 10 , c ) 9 , d ) 7 , e ) 5 | b | add(subtract(add(14, 18), subtract(25, 3)), subtract(18, 14)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general |
the sector of a circle has radius of 28 cm and central angle 90 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 90 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 28 ) = 44 + 56 = 100 cm answer : a" | a ) 100 cm , b ) 85 cm , c ) 75 cm , d ) 95 cm , e ) 110 cm | a | multiply(multiply(const_2, divide(multiply(subtract(28, const_3), const_2), add(const_4, const_3))), 28) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . [ ( 2 - 1 ) / 1 ] * 100 % = 100 % answer : c" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
3 * 11 + 3 * 12 + 3 * 15 + 11 = ? | "we have : 3 * 11 = 33 3 * 12 = 36 3 * 15 = 45 all of this is : ( 33 + 36 + 45 ) + 11 = 114 + 11 = 125 correct answer a" | a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 151 | a | add(add(add(multiply(3, 11), multiply(3, 12)), multiply(3, 15)), 11) | multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|add(#0,#1)|add(#3,#2)|add(n6,#4)| | general |
one pump drains one - half of a pond in 8 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? | first pump drains 1 / 2 of the tank in 8 hours so 16 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 / 16 + 1 / a ) part in 1 hour son in 1 / 2 hour they drain 1 / 2 * ( 1 / 16 + 1 / a ) part of the tank given that in 1 / 2 hour they drain 1 / 2... | a ) 1 hour , b ) 1.1 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours | b | divide(const_1, subtract(const_1, divide(const_1, multiply(8, const_2)))) | multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2) | physics |
there are 2 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | "using anagram method : 4 _ 3 _ 2 _ 1 r _ r _ b _ b so . . 4 ! / number of repeated letters ( 2 ! ) ( 2 ! ) = 6 ans : d" | a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 60 , e ) e ) 100 | d | multiply(factorial(2), factorial(2)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | general |
the speed of a car is 50 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 55 + 60 ) / 2 = 55 kmph answer : c" | a ) 79 kmph , b ) 85 kmph , c ) 55 kmph , d ) 23 kmph , e ) 14 kmph | c | divide(add(50, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
the top and bottom of a tower were seen to be at angles of depression 30 ° and 60 ° from the top of a hill of height 100 m . find the height of the tower | by making a diagram ac represents the hill and de represents the pole given that ac = 100 m anglexad = angleadb = 30 ° ( ∵ ax | | bd ) anglexae = angleaec = 60 ° ( ∵ ax | | ce ) let de = h then , bc = de = h , ab = ( 100 - h ) ( ∵ ac = 100 and bc = h ) , bd = ce tan 60 ° = acce = > 3 √ = 100 ce = > ce = 1003 √ - - - ( ... | a ) 64.67 m , b ) 66.67 m , c ) 68.67 m , d ) 56.67 m , e ) 46.67 m | b | subtract(100, multiply(const_0_33, const_100)) | multiply(const_0_33,const_100)|subtract(n2,#0) | physics |
the age of the daughter is same as the digit in the mother ' s age , but in reverse order . in thirteen years mother will be twice as old as daughter will be then . what is the difference in their current age ? | daughter ' s age = 10 x + y so mother ' s age = 10 y + x . . after 13 years , 10 x + y + 13 = 2 * ( 10 y + x + 13 ) . . . . so 13 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 4 . . . so ages are 41 and 14 and ans as found correctly by u is c . . 27 | a ) 15 , b ) 23 , c ) 27 , d ) 46 , e ) 32 | c | subtract(add(multiply(divide(divide(add(add(const_12, const_1), add(subtract(const_10, const_1), const_10)), const_4), const_2), const_10), const_1), add(const_10, divide(divide(add(add(const_12, const_1), add(subtract(const_10, const_1), const_10)), const_4), const_2))) | add(const_1,const_12)|subtract(const_10,const_1)|add(#1,const_10)|add(#0,#2)|divide(#3,const_4)|divide(#4,const_2)|add(#5,const_10)|multiply(#5,const_10)|add(#7,const_1)|subtract(#8,#6) | general |
working together , wayne and his son can shovel the entire driveway in three hours . if wayne can shovel four times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | "w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 4 and w = 4 s then we have 1 / ( 4 * s ) + 1 / s = 1 / 4 < = > 5 / ( 4 * s ) = 1 / 4 < = > s = 5 ans : c" | a ) 4 , b ) 6 , c ) 5 , d ) 9 , e ) 12 | c | multiply(multiply(add(inverse(multiply(const_2, const_4)), const_1), const_3), multiply(const_2, const_4)) | multiply(const_2,const_4)|inverse(#0)|add(#1,const_1)|multiply(#2,const_3)|multiply(#3,#0)| | physics |
a larger cube has 125 cubic inch as a volume and in the cube there are 125 smaller cubes such that their volume is 1 cubic inch . what is the difference between the surface areas ’ sum of the 125 smaller cubes and the surface area of the larger cube , in square inch ? | "volume of larger cube = 125 = 5 ^ 3 side of larger cube = 5 volume of smaller cube = 1 - - > side of smaller cube = 1 surface area of larger cube = 6 * 5 ^ 2 = 150 surface area of 27 smaller cubes = 125 * 6 * 1 = 750 difference = 750 - 150 = 600 answer : b" | a ) 54 , b ) 600 , c ) 81 , d ) 108 , e ) 120 | b | subtract(multiply(surface_cube(cube_edge_by_volume(1)), 125), surface_cube(cube_edge_by_volume(125))) | cube_edge_by_volume(n2)|cube_edge_by_volume(n0)|surface_cube(#0)|surface_cube(#1)|multiply(n0,#2)|subtract(#4,#3)| | geometry |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 12 , find the other ? | "other number = 12 * 600 / 12 = 600 answer is b" | a ) 100 , b ) 600 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 12) | multiply(n0,n2)| | physics |
if below equation are true 24 $ 6 = 64 35 $ 5 = 87 63 $ 7 = 99 by using same logic what is the value of the below question 88 $ 8 = ? | correct answer : c the number on the right side is formed as ( sum of digits of the 1 st number ) no 1 / no 2 ex : 35 $ 5 = 87 ( 3 + 5 = 8 ) ( 35 / 5 = 7 ) 63 $ 7 = 99 ( 6 + 3 = 9 ) ( 63 / 7 = 9 ) | a ) 1621 , b ) 1711 , c ) 1611 , d ) 1511 , e ) 1311 | c | add(multiply(add(8, 8), const_100), divide(88, 8)) | add(n10,n10)|divide(n9,n10)|multiply(#0,const_100)|add(#1,#2) | other |
the average weight of 10 persons increases by 3.2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "solution total weight increased = ( 10 x 3.2 ) kg = 32 kg . weight of new person = ( 65 + 32 ) kg = 97 kg . answer d" | a ) 76 kg , b ) 80 kg , c ) 85 kg , d ) 97 kg , e ) none | d | add(65, multiply(10, 3.2)) | multiply(n0,n1)|add(n2,#0)| | general |
the average of runs of a cricket player of 10 innings was 37 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 37 so , total = average x no . of innings = 37 x 10 = 370 . now increase in avg = 4 runs . so , new avg = 37 + 4 = 41 runs total runs = new avg x new no . of innings = 41 x 11 = 451 runs made in the 11 th inning = 451 - 370 = 81 answer : b" | a ) 76 , b ) 81 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 37)), multiply(10, 37)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
a , b , and c started a partnership business by investing $ 45,000 , $ 63,000 , and $ 72,000 respectively . at the end of the year , the profit was distributed among them . if c ' s share of profit is $ 24,000 , what is the total profit ? | a : b : c = 45000 : 63000 : 72000 which is a ratio of 5 : 7 : 8 . let the total profit be x . x * 8 / 20 = 24000 x = 24000 * 20 / 8 = 60,000 . the answer is d . | a ) $ 48,000 , b ) $ 52,000 , c ) $ 56,000 , d ) $ 60,000 , e ) $ 64,000 | d | add(add(multiply(multiply(add(multiply(const_2, multiply(add(const_2, const_3), const_2)), const_4), multiply(multiply(add(const_2, const_3), const_2), const_100)), divide(multiply(add(multiply(const_4, multiply(add(const_2, const_3), const_2)), add(const_2, const_3)), multiply(multiply(add(const_2, const_3), const_2),... | add(const_2,const_3)|add(const_3,const_4)|add(const_2,const_4)|multiply(#0,const_2)|multiply(#3,const_4)|multiply(#3,const_100)|multiply(#1,#3)|multiply(#3,const_2)|multiply(#2,#3)|add(#0,#4)|add(#6,const_2)|add(#7,const_4)|add(#8,const_3)|multiply(#9,#5)|multiply(#10,#5)|multiply(#11,#5)|multiply(#12,#5)|divide(#13,#1... | gain |
what will be the vulgar fraction of 0.75 | "explanation : 0.75 = 75 / 100 = 3 / 4 option b" | a ) 3 / 5 , b ) 3 / 4 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.75, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | general |
a company decreased the price of its main product by 75 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | "for the total revenue to remain the same when the price is 3 / 4 , the number of products sold must 3 / 4 times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 75 % = 1.33 answer : a" | a ) 1.33 , b ) 2.33 , c ) 3.33 , d ) 4.33 , e ) 1.45 | a | divide(const_100, 75) | divide(const_100,n0)| | general |
a ' s speed is 15 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 25 % of the length of the race ? | first calculate the distance , b has covered with his speed in the time , in which a reached 80 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 15 / 13 of b , therefore , b ' s speed is 13 / 15 of a distance covered by b ... | a ) 35 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 46 % | a | multiply(subtract(const_1, multiply(divide(subtract(const_100, 25), const_100), divide(13, 15))), const_100) | divide(n1,n0)|subtract(const_100,n2)|divide(#1,const_100)|multiply(#2,#0)|subtract(const_1,#3)|multiply(#4,const_100) | general |
company c sells a line of 20 products with an average retail price of $ 1200 . if none of these products sells for less than $ 400 , and exactly 10 of the products sell for less than $ 1000 , what is the greatest possible selling price of the most expensive product ? | the average price of 20 products is $ 1,200 means that the total price of 20 products is 20 * 1,200 = $ 24,000 . next , since exactly 10 of the products sell for less than $ 1,000 , then let ' s make these 10 items to be at $ 400 each ( min possible ) . now , the remaining 9 items can not be priced less than $ 1,000 , ... | a ) 12000 , b ) 11000 , c ) 10000 , d ) 13000 , e ) 15000 | b | subtract(multiply(20, 1200), add(multiply(subtract(10, const_1), 1000), multiply(400, 10))) | multiply(n0,n1)|multiply(n2,n3)|subtract(n3,const_1)|multiply(n4,#2)|add(#3,#1)|subtract(#0,#4) | general |
in a box of 6 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 4 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 6 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 4 c 3 / 6 c 3 = 4 / 20 = 1 / 5 answer is b" | a ) 1 / 2 , b ) 1 / 5 , c ) 2 / 3 , d ) 2 / 5 , e ) 1 / 7 | b | divide(divide(factorial(subtract(6, 2)), multiply(factorial(subtract(subtract(6, 2), 3)), factorial(3))), divide(factorial(6), multiply(factorial(subtract(6, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9)| | general |
a candidate appearing for an examination has to secure 52 % marks to pass paper i . but he secured only 45 marks and failed by 35 marks . what is the maximum mark for paper i ? | "he secured 45 marks nd fail by 35 marks so total marks for pass the examinatn = 80 let toal marks x x * 52 / 100 = 80 x = 154 answer : b" | a ) 110 , b ) 154 , c ) 130 , d ) 140 , e ) 150 | b | divide(add(45, 35), divide(52, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
jim drove 768 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ? | "the number of miles to drive to finish his journey is given by 1200 - 768 = 432 miles correct answer a" | a ) 432 miles , b ) 651 miles , c ) 387 miles , d ) 954 miles , e ) 556 miles | a | subtract(1200, 768) | subtract(n1,n0)| | physics |
if | 20 x - 10 | = 110 , then find the product of the values of x ? | | 20 x - 10 | = 110 20 x - 10 = 110 or 20 x - 10 = - 110 20 x = 120 or 20 x = - 100 x = 6 or x = - 5 product = - 5 * 6 = - 30 answer is e | a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 30 | e | subtract(subtract(subtract(110, 10), add(110, 10)), 10) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1) | general |
when positive integer n is divided by positive integer j , the remainder is 10 . if n / j = 205.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 205.08 here 205 is the quotient . given that remainder = 10 so , 205.08 = 205 + 10 / j so , j = 125 answer : a" | a ) 125 , b ) 100 , c ) 75 , d ) 150 , e ) 50 | a | divide(10, subtract(205.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
in a certain kindergarten , half of the children are allergic to peanuts , 10 children are not allergic to cashew nuts , 10 children are allergic to both peanuts and cashew nuts , and 18 children are allergic to cashew nuts . how many of the kindergarten ' s children are not allergic to peanuts and are not allergic to ... | we are given children allergic to cashew nuts as 18 and not allergic to cashew nuts as 10 , so total = 10 + 18 = 28 . . . children allergic to both are 10 . . . 18 are allergic to cashew nuts , so and half or 28 / 2 = 14 are allergic to peanuts . . so children allergic to at least one of the two are 18 + 14 - 10 = 22 .... | a ) 14 , b ) 10 , c ) 8 , d ) 6 , e ) 4 | d | subtract(add(18, 10), subtract(add(18, divide(add(18, 10), const_2)), 10)) | add(n0,n2)|divide(#0,const_2)|add(n2,#1)|subtract(#2,n0)|subtract(#0,#3) | other |
a train 1100 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 1100 + x ) / 60 = 65 / 3 x = 200 m . answer : c" | a ) 288 m , b ) 256 m , c ) 200 m , d ) 278 m , e ) 287 m | c | divide(1100, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the sum of number of boys and girls in a school is 300 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 300 = 300 x + x / 100 * 300 = 300 4 * x = 300 x = 75 answer is e" | a ) 50 , b ) 40 , c ) 60 , d ) 100 , e ) 75 | e | divide(300, add(divide(300, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orang... | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is ... | a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.55 | e | divide(multiply(add(const_1, const_1), 0.82), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department take... | answer : d 162 ° divided by 360 ° equals 0.45 , therefore the sector is equal to 45 % of the total | a ) 20 % , b ) 25 % , c ) 30 % , d ) 45 % , e ) 72 % | d | multiply(divide(162, divide(const_3600, const_10)), const_100) | divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)| | physics |
1100 boys and 700 girls are examined in a test ; 42 % of the boys and 30 % of the girls pass . the percentage of the total who failed is ? | total number of students = 1100 + 700 = 1800 number of students passed = ( 42 % of 1100 + 30 % of 700 ) = 462 + 210 = 672 number of failures = 1128 * 100 / 1800 = 190 / 3 % = 63.3 % answer is a | a ) 63.3 % , b ) 52.4 % , c ) 81.2 % , d ) 75.4 % , e ) 69.1 % | a | multiply(const_100, divide(subtract(add(1100, 700), add(multiply(1100, divide(42, const_100)), multiply(700, divide(30, const_100)))), add(1100, 700))) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100) | general |
john makes $ 60 a week from his job . he earns a raise and nowmakes $ 70 a week . what is the % increase ? | "increase = ( 10 / 60 ) * 100 = ( 1 / 6 ) * 100 = 16.66 % . b" | a ) 16 % , b ) 16.66 % , c ) 17 % , d ) 17.2 % , e ) 18 % | b | multiply(divide(subtract(70, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
two heavily loaded sixteen - wheeler transport trucks are 690 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , ... | "i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 690 - 90 = 600 k ratio of speeds 9 : 8 - > 510 / 17 = 30 truck a = 90 + 30 * 9 = 360 truck b = 30 * 8 = 240 d... | a ) 120 , b ) 130 , c ) 150 , d ) 320 , e ) 450 | a | add(90, multiply(divide(subtract(690, 90), add(90, 80)), subtract(90, 80))) | add(n1,n2)|subtract(n0,n1)|subtract(n1,n2)|divide(#1,#0)|multiply(#3,#2)|add(n1,#4)| | general |
on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 42 inch from the base of the mountain . how many km is he from the base of the mountain ? | "explanation : since 312 inch = 136 km so 1 inch = 136 / 312 km so 42 inch = ( 136 ã — 42 ) / 312 = 18.3 km answer : e" | a ) 14.83 , b ) 14.81 , c ) 14.8 , d ) 14.82 , e ) 18.3 | e | divide(multiply(42, 136), 312) | multiply(n1,n2)|divide(#0,n0)| | physics |
on a certain day , tim invested $ 500 at 10 percent annual interest , compounded annually , and lana invested 800 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ s ... | "compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 500 ( 1.10 ) ^... | a ) $ 5 , b ) $ 15 , c ) $ 23 , d ) $ 100 , e ) $ 105 | c | subtract(subtract(multiply(500, power(add(const_1, divide(10, const_100)), 2)), 500), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 800), 800)) | divide(n1,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|power(#2,n4)|power(#3,n4)|multiply(n0,#4)|multiply(n2,#5)|subtract(#6,n0)|subtract(#7,n2)|subtract(#8,#9)| | gain |
the h . c . f of two numbers is 11 and their l . c . m is 7700 . if one of the numbers is 275 , then the other is : | "other number = ( 11 * 7700 ) / 275 = 308 . answer : c" | a ) 279 , b ) 283 , c ) 308 , d ) 318 , e ) 328 | c | multiply(11, 275) | multiply(n0,n2)| | physics |
in measuring the sides of a rectangle , one side is taken 8 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements . | "let the sides of the rectangle be a and b . actual area of the rectangle without error = ab new side a ( 8 % in excess ) = a + 0.08 a = 1.08 a new side b ( 5 % in deficit ) = b - 0.05 b = 0.95 b area of the new rectangle = ( 1.08 a ) ( 0.95 b ) = ( 1.0260 ) ab say a = 1 and b = 1 area of the original rectangle = 1 are... | a ) 3.5 % , b ) 2.4 % , c ) 3 % , d ) 5 % , e ) 2.6 % | e | subtract(subtract(8, 5), divide(multiply(8, 5), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 3200 ? | "3200 * 9 / 8 * 9 / 8 = 4050 answer : d" | a ) 8100 , b ) 3388 , c ) 7767 , d ) 4050 , e ) 2288 | d | add(add(3200, multiply(divide(1, 8), 3200)), multiply(divide(1, 8), add(3200, multiply(divide(1, 8), 3200)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
on a certain test , bill scored 20 more points than john , but half as many points as sue . if the sum of the scores of the 3 students was 160 points , how many points did bill receive ? | consider john ' s points as x so bill ' s points are x + 20 so sue ' s points are 2 x + 40 total of all three students is 4 x + 60 , but this total is given as 160 x = 25 so bill ' s points = 25 + 20 = 45 correct answer option b | a ) 30 , b ) 45 , c ) 50 , d ) 80 , e ) 100 | b | divide(add(160, 20), add(add(const_1, const_1), const_2)) | add(n0,n2)|add(const_1,const_1)|add(#1,const_2)|divide(#0,#2) | general |
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