Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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the present population of a town is 4032 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 4032 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 4032 / ( 1 + 20 / 100 ) ^ 2 = 4032 / ( 6 / 5 ) ^ 2 = 2800 ( approximately ) answer is b" | a ) 2500 , b ) 2800 , c ) 3500 , d ) 3600 , e ) 2050 | b | add(4032, divide(multiply(4032, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 2 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 2 = 240 answer : b" | a ) 480 , b ) 240 , c ) 289 , d ) 270 , e ) 927 | b | multiply(24, 20) | multiply(n0,n2)| | physics |
the present population of a town is 10000 . population increase rate is 10 % p . a . find the population of town after 2 years ? | "p = 10000 r = 10 % required population of town = p ( 1 + r / 100 ) ^ t = 10000 ( 1 + 10 / 100 ) ^ 2 = 10000 ( 11 / 10 ) ^ 2 = 12100 answer is d" | a ) 10000 , b ) 11100 , c ) 15000 , d ) 12100 , e ) 14520 | d | add(10000, divide(multiply(10000, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
john has taken 4 ( 4 ) tests that have an average of 82 . in order to bring his course grade up to a ‘ b ’ , he will need to have a final average of 86 . what will he need to average on his final two tests to achieve this grade ? | traditional method : total scored till now 82 * 4 = 328 total score to avg 86 in 6 tests = 86 * 6 = 516 total to be scored on 2 tests = 516 - 328 = 188 avg on 2 tests = 188 / 2 = 94 answer : d | a ) 87 , b ) 90 , c ) 92 , d ) 94 , e ) 97 | d | subtract(multiply(86, const_2), 82) | multiply(n3,const_2)|subtract(#0,n2) | general |
find the surface area of a 8 cm x 4 cm x 2 cm brick . | "sol . surface area = [ 2 ( 8 x 4 + 4 x 2 + 8 x 2 ) ] = ( 2 x 56 ) = 112 cu cm answer b" | a ) 136 cu cm , b ) 112 cu cm , c ) 164 cu cm , d ) 180 cu cm , e ) none | b | surface_rectangular_prism(2, 4, 8) | surface_rectangular_prism(n0,n1,n2)| | geometry |
a car started running at a speed of 38 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | a 47 km the total distance covered by the car in the first 10 hours = 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 = sum of 10 terms in ap whose first term is 30 and last term is 56 = 10 / 2 [ 38 + 56 ] = 470 km . | a ) 47 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(38, multiply(2, 10)), 38), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1) | physics |
3 containers a , b and c are having mixtures of milk and water in the ratio of 1 : 5 and 3 : 5 and 5 : 7 respectively . if the capacities of the containers are in the ratio of all the 3 containers are in the ratio 5 : 4 : 5 , find the ratio of milk to water , if the mixtures of all the 3 containers are mixed together . | assume that there are 500400 and 500 liters respectively in the 3 containers . then , we have , 83.33 , 150 and 208.33 liters of milk in each of the three containers . thus , the total milk is 441.66 liters . hence , the amount of water in the mixture is 1400 - 441.66 = 958.33 liters . hence , the ratio of milk to wate... | a ) 53 : 115 , b ) 55 : 115 , c ) 115 : 55 , d ) 5 : 115 , e ) 115 : 53 | a | divide(add(add(multiply(divide(1, add(1, 5)), 5), multiply(divide(3, add(3, 5)), 4)), multiply(divide(5, add(5, 7)), 5)), subtract(add(add(5, 4), 5), add(add(multiply(divide(1, add(1, 5)), 5), multiply(divide(3, add(3, 5)), 4)), multiply(divide(5, add(5, 7)), 5)))) | add(n1,n2)|add(n0,n2)|add(n2,n6)|add(n2,n9)|add(n2,#3)|divide(n1,#0)|divide(n0,#1)|divide(n2,#2)|multiply(n2,#5)|multiply(n9,#6)|multiply(n2,#7)|add(#8,#9)|add(#11,#10)|subtract(#4,#12)|divide(#12,#13) | other |
the length of a rectangle is reduced by 22 % . by what % would the width have to be increased to maintainthe original area ? | "sol . required change = ( 22 * 100 ) / ( 100 - 22 ) = 28 % c" | a ) 15 % , b ) 20 % , c ) 28 % , d ) 35 % , e ) 40 % | c | multiply(divide(subtract(const_1, divide(subtract(const_100, 22), const_100)), divide(subtract(const_100, 22), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry |
if 5 x = 7 y and xy ≠ 0 , what is the ratio of 1 / 5 * x to 1 / 6 * y ? | "5 x = 7 y = > x / y = 7 / 5 1 / 5 * x to 1 / 6 * y = x / y * 6 / 5 = ( 7 / 5 ) * ( 6 / 5 ) = 42 / 25 ans : b" | a ) 25 / 6 , b ) 42 / 25 , c ) 6 / 5 , d ) 5 / 6 , e ) 25 / 36 | b | multiply(divide(divide(1, const_3.0), divide(1, 5)), divide(7, 5)) | divide(n3,const_3.0)|divide(n3,n0)|divide(n1,n0)|divide(#0,#1)|multiply(#3,#2)| | general |
if 10 litres of an oil of rs . 50 per litres be mixed with 5 litres of another oil of rs . 67 per litre then what is the rate of mixed oil per litre ? | "50 * 10 = 500 67 * 5 = 335 835 / 15 = 55.66 answer : d" | a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 54.17 , d ) rs . 55.66 , e ) none of the above | d | divide(add(multiply(10, 50), multiply(5, 67)), add(10, 5)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | gain |
the sum of first 45 natural numbers is : | "let sn = ( 1 + 2 + 3 + . . . + 45 ) . this is an a . p . in which a = 1 , d = 1 , n = 45 . sn = n [ 2 a + ( n - 1 ) d ] = 45 x [ 2 x 1 + ( 45 - 1 ) x 1 ] = 45 x 46 = ( 45 x 23 ) 2 2 2 = 45 x ( 20 + 3 ) = 45 x 20 + 45 x 3 = 900 + 135 = 1035 . shorcut method : sn = n ( n + 1 ) = 45 ( 45 + 1 ) = 1035 . 2 2 a" | a ) 1035 , b ) 1040 , c ) 1042 , d ) 1045 , e ) 1050 | a | divide(multiply(45, add(45, const_1)), const_2) | add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 226 m , what is its area ? | "length = breadth + 23 . therefore , 4 × breadth + 2 × 23 = 226 m ⇒ breadth = 45 m length = 45 + 23 = 68 m area = 68 × 45 = 3060 m 2 answer is c ." | a ) 2510 , b ) 2535 , c ) 3060 , d ) 2515 , e ) 2520 | c | rectangle_area(add(divide(subtract(226, multiply(const_2, 23)), const_4), 23), divide(subtract(226, multiply(const_2, 23)), const_4)) | multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)| | geometry |
there is a 30 % chance sandy will visit china this year , while there is a 60 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | p ( china and not malaysia ) = 0.3 * 0.4 = 0.12 p ( malaysia and not china ) = 0.6 * 0.7 = 0.42 total probability = 0.12 + 0.42 = 0.54 = 54 % the answer is c . | a ) 42 % , b ) 48 % , c ) 54 % , d ) 60 % , e ) 66 % | c | divide(add(30, 60), multiply(multiply(const_5, const_5), const_4)) | add(n0,n1)|multiply(const_5,const_5)|multiply(#1,const_4)|divide(#0,#2) | probability |
mike earns $ 14 per hour and phil earns $ 10.5 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "let it be x % less , then = 14 ( 1 - x / 100 ) = 10.5 1 - x / 100 = 10.5 / 14 x = 350 / 14 x = 25 % answer : a" | a ) 25 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | a | multiply(divide(10.5, 14), const_100) | divide(n1,n0)|multiply(#0,const_100)| | general |
the total price of a basic computer and printer are $ 2500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 6 of that total . what was the price of the basic computer ? | let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 6 of that new total price : p = 1 / 6 * $... | a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | e | subtract(2500, divide(add(2500, 500), 6)) | add(n0,n1)|divide(#0,n3)|subtract(n0,#1) | general |
find the average of first 4 multiples of 9 ? | "average = ( 9 + 18 + 27 + 36 ) / 4 = 22.5 answer is c" | a ) 10 , b ) 12.6 , c ) 22.5 , d ) 31.3 , e ) 40.8 | c | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(9, 4), subtract(9, const_2))), 4) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general |
company a ' s workforce consists of 10 percent managers and 90 percent software engineers . company b ' s workforce consists of 30 percent managers , 10 percent software engineers , and 60 percent support staff . the two companies merge , every employee stays with the resulting company , and no new employees are added ... | let say company a has x employes and b has y employees . now they merge and total no of employees = x + y employees . per the question company a ' s workforce consists of 10 percent managers and 90 percent software engineers . company b ' s workforce consists of 30 percent managers , 10 percent software engineers , and... | a ) 10 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 75 % | c | subtract(subtract(add(divide(multiply(divide(60, 30), 25), 10), 90), 60), 10) | divide(n4,n2)|multiply(n5,#0)|divide(#1,n0)|add(n1,#2)|subtract(#3,n4)|subtract(#4,n0) | general |
how many positive integers less than 100 are neither multiples of 5 or 6 . | "to answer this q we require to know 1 ) multiples of 5 till 100 = 100 / 5 = 20 2 ) multiples of 6 till 100 = 100 / 6 = 16.66 = 16 add the two 20 + 16 = 36 ; subtract common terms that are multiple of both 5 and 6 . . lcm of 5 and 6 = 30 multiples of 30 till 100 = 100 / 30 = 3.3 = 3 so total multiples of 2 and 3 = 36 -... | a ) a ) 30 , b ) b ) 31 , c ) c ) 32 , d ) d ) 33 , e ) e ) 67 | e | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), 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 |
a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 12 : 00 p . m . ? | "sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour . hence , they cross each other 20 times before 12 : 00 p . m . answer e" | a ) 28 , b ) 27 , c ) 26 , d ) 25 , e ) 20 | e | add(add(2, 3), add(2, 3)) | add(n1,n2)|add(#0,#0)| | physics |
if n is a positive integer and n ^ 3 is a multiple of 550 , what is the least possible value of n ? | "cube of a number entails having 3 copies of the original prime factors . since 550 is a factor of n ^ 3 the least possible value of n is 5 ( 11 ) 2 = 110 the answer is d ." | a ) 8 , b ) 27 , c ) 81 , d ) 110 , e ) 125 | d | multiply(sqrt(divide(550, 3)), 3) | divide(n1,n0)|sqrt(#0)|multiply(n0,#1)| | general |
two goods trains each 375 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 375 + 375 = 750 m . required time = 750 * 6 / 125 = 36 sec . answer : option e" | a ) 40 , b ) 45 , c ) 48 , d ) 51 , e ) 36 | e | add(45, 30) | add(n1,n2)| | physics |
if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 300 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 307.50 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 15,000 . now , put the value of x = 3000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 3000 + 5 % of 3000 = 3150 . 2 nd year , amount = 3150 + 5 % of 3150 = 3307.50 . we can see after 2 years interest = 330... | a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | e | divide(power(divide(300, const_2), const_2), subtract(307.50, 300)) | divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)| | gain |
a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 25 % . staff are allowed a further 20 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 25 - 20 + ( - 25 ) ( - 20 ) / 100 = - 40 sale price = d * ( 1 - 40 / 100 ) sale price = . 6 * d answer ( a )" | a ) 0.6 d , b ) 0.7 d , c ) 0.5 d , d ) 0.8 d , e ) 0.9 d | a | subtract(divide(subtract(const_100, 25), const_100), multiply(divide(subtract(const_100, 25), const_100), divide(20, const_100))) | divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)| | gain |
what is the normal price of an article sold at $ 108 after two successive discounts of 10 % and 20 % ? | "0.8 * 0.9 * cost price = $ 108 cost price = $ 150 the answer is c ." | a ) $ 130 , b ) $ 140 , c ) $ 150 , d ) $ 160 , e ) $ 170 | c | divide(108, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))) | subtract(const_100,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n0,#4)| | gain |
kim has 8 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 16 shoes , what is the probability that she will select 2 shoes of the same color ? | "total pairs = 16 c 2 = 120 ; same color pairs = 8 c 1 * 1 c 1 = 8 ; prob = 1 / 15 ans e" | a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 15 | e | divide(8, choose(16, 2)) | choose(n2,n1)|divide(n0,#0)| | probability |
how many factors does 16 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 2 * 2 = 4 answer c ." | a ) 2 , b ) 8 , c ) 4 , d ) 25 , e ) 26 | c | divide(16, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(n0,#0)| | other |
there are 14 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "14 c 2 = 91 the answer is a ." | a ) 91 , b ) 98 , c ) 105 , d ) 112 , e ) 119 | a | divide(multiply(14, subtract(14, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
a technology company made a $ 5 million profit on its first $ 15 million in sales and a $ 12 million profit on its next $ 30 million in sales . by what percent did the ratio of profit to sales increase from the first $ 15 million in sales to the next $ 30 million in sales ? | solution : this is a percent increase problem . we will use the formula : percent change = ( new – old ) / old x 100 to calculate the final answer . we first set up the ratios of profits to sales . the first ratio will be for the first 15 million in sales , and the second ratio will be for the next 30 million in sales ... | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | multiply(divide(subtract(divide(12, 30), divide(5, 15)), divide(5, 15)), const_100) | divide(n2,n3)|divide(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100) | other |
a number is doubled and 9 is added . if the resultant is trebled , it becomes 63 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 63 2 x = 12 = > x = 6 answer : b" | a ) 3.5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | b | divide(subtract(63, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a bag holds 4 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "given : 4 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 4 red and 3 green marbles after one red has been picked , we need to pick 1 red from a total of 3 red and 3 green marbles . p ( both red ) = ( 4 / 7 ) * ( 3 / 6 )... | a ) 4 / 14 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | a | divide(choose(4, 4), choose(add(const_2.0, 3), const_2)) | add(n0,n1)|choose(const_2.0,n0)|choose(#0,const_2)|divide(#1,#2)| | other |
a train speeds past a pole in 15 seconds and speeds past a platform 100 meters long in 25 seconds . its length in meters is : | "let the length of the train be x metres and its speed be y metres / sec . then x / y = 15 y = x / 15 x + 100 / 25 now , ( x + 100 ) / 25 = x / 15 x = 150 m answer : e" | a ) 60 m , b ) 70 m , c ) 100 m , d ) 120 m , e ) 150 m | e | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
a cab driver 5 days income was $ 300 , $ 150 , $ 750 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 300 + 150 + 750 + 400 + 500 ) / 5 = 420 answer is c" | a ) $ 500 , b ) $ 650 , c ) $ 420 , d ) $ 375 , e ) $ 625 | c | divide(add(add(add(add(300, 150), 750), 400), 500), 5) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)| | general |
if 12401 is divided by any no . then quotient is 76 and remainder is 13 . what is divisor ? | divisor = ( dividend - remainder ) / quotient = ( 12401 - 13 ) / 76 = 12388 / 76 = 163 divisor = 163 answer d | a ) 154 , b ) 124 , c ) 153 , d ) 163 , e ) 183 | d | divide(subtract(12401, 13), 76) | subtract(n0,n2)|divide(#0,n1) | general |
in a games hour 4 different types of players came to the ground ? cricket 12 , hokey 17 , football 11 , softball 10 . how many players are present in the ground ? | "total number of players = 12 + 17 + 11 + 10 = 50 answer is b" | a ) 70 , b ) 50 , c ) 62 , d ) 49 , e ) 50 | b | add(add(12, 17), add(11, 10)) | add(n1,n2)|add(n3,n4)|add(#0,#1)| | physics |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 55 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 55 = rs . 150 cp on 40 metres = 150 x 40 = rs . 6000 profit earned on 40 metres cloth = rs . 8200 – rs . 6000 = rs . 2200 . answer : option d | a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 2200 , e ) none of these | d | multiply(55, 40) | multiply(n0,n2) | gain |
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 20 households had a both a car and a bike and 44 had a car , how many had bike only ? | "{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 20 + 11 - - > { bike } = 55 - - > # those who have bike only is { bike } - { both } = 55 - 20 = 35 . answer : b ." | a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10 | b | subtract(subtract(add(subtract(90, 11), 20), 44), 20) | subtract(n0,n1)|add(n2,#0)|subtract(#1,n3)|subtract(#2,n2)| | other |
what is the units digit of 32 ! + 50 ! + 2 ! + 4 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 2 + 4 = 6 the units digit is 6 . the answer is d ." | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | other |
the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 60,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42... | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 60,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 4... | a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 480,000 , e ) $ 540,000 | d | divide(const_3600, const_10) | divide(const_3600,const_10)| | general |
if a and b are multiples of 3 then which are all the multiples of 3 1 ) a / b 2 ) a ^ b 3 ) a + b 4 ) a - b 5 ) a * b | a * b a ^ b a + b a - b are multiples of 3 except a / b answer : e | a ) 13 , b ) 12 , 45 , c ) 1 , 25 , d ) all of these , e ) 12 , 35 | e | multiply(multiply(3, 4), add(5, multiply(3, const_10))) | multiply(n0,const_10)|multiply(n0,n5)|add(n6,#0)|multiply(#2,#1) | general |
75 persons major in physics , 83 major in chemistry , 10 not at major in these subjects u want to find number of students majoring in both subjects | consider the number of total students = n ( t ) = 100 number of persons major in physics = n ( p ) = 75 number of persons major in chemistry = n ( c ) = 83 according to the question ; 10 not at major in these subjects = n ( p ' ∩ c ' ) = 10 n ( p ' ∩ c ' ) = n ( p u c ) ' = 10 n ( p u c ) ' = n ( t ) - n ( p u c ) 10 =... | a ) 65 , b ) 66 , c ) 67 , d ) 68 , e ) 69 | d | subtract(add(75, 83), subtract(const_100, 10)) | add(n0,n1)|subtract(const_100,n2)|subtract(#0,#1) | other |
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 320 square feet . how many feet offence does frank use ? | "area = length x breadth 320 = 40 x breadth so , breadth = 8 units fencing required is - breadth + breadth + length 8 + 8 + 40 = > 56 feet answer must be ( d ) 56" | a ) 14 , b ) 47 , c ) 54 , d ) 56 , e ) 240 | d | add(add(divide(320, 40), divide(320, 40)), 40) | divide(n1,n0)|add(#0,#0)|add(n0,#1)| | geometry |
if a is a positive integer , and if the units digit of a ^ 2 is 4 and the units digit of ( a + 1 ) ^ 2 is 9 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 4 , then the units digit of a is either 2 or 8 . if the units digit of ( a + 1 ) ^ 2 is 9 , then the units digit of a + 1 is either 3 or 7 . to satisfy both conditions , the units digit of a must be 2 . then a + 2 has the units digit of 4 , thus the units digit of ( a + 2 ) ^ 2 will be 6... | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | power(add(multiply(4, 2), 2), 2) | multiply(n0,n1)|add(n0,#0)|power(#1,n0)| | general |
lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 54,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 354000 x \ therefore required percentage = \ inline \ frac { 800 x } { 354000 } \ times 100 = 0.22598 a" | a ) 0.22598 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.63435 | a | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | divide(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_4)|divide(n3,#0)|multiply(#1,const_1000)|multiply(#3,const_1000)|add(#5,#6)|divide(#4,#7)|multiply(#8,const_100)| | general |
if 15 students in a class average 70 % on an exam and 10 students average 90 % on the same exam , what is the average in percent for all 25 students ? | ( 15 * 70 + 10 * 90 ) / 25 = 78 % the answer is a . | a ) 78 % , b ) 79 % , c ) 80 % , d ) 81 % , e ) 82 % | a | divide(add(multiply(15, 70), multiply(10, 90)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4) | general |
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 87 , and the average of the female students ’ grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 87 thus m = 696 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 696 + 92 f ) / ( 8 + f ) = ... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | divide(subtract(multiply(90, 8), multiply(87, 8)), subtract(92, 90)) | multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
ajith and rana walk around a circular course 115 km in circumference , starting together from the same point . if they walk at speed of 4 and 5 kmph respectively , in the same direction , when will they meet ? | rana is the faster person . he gains 1 km in 1 hour . so rana will gain one complete round over ajith in 115 hours . i . e . they will meet after 115 hours . answer : b | a ) after 20 hours , b ) after 115 hours , c ) after 115 minutes , d ) after 20 minutes , e ) after 30 minutes | b | multiply(115, subtract(5, 4)) | subtract(n2,n1)|multiply(n0,#0) | physics |
two pipes a and b fill at a certain rate . b is filled at 1020 , 4080 ( 10 in 1 hour , 20 in 2 hours , 40 in 3 hrs and so on ) . if 1 / 16 of b is filled in 17 hrs , what time it will take to get completely filled ? | 1 / 16 in 17 hrs so 1 / 8 in 18 hrs 1 / 4 in 19 hrs 1 / 2 in 20 hrs and complete tank in 21 hrs answer : b | a ) 20 hrs , b ) 21 hrs , c ) 22 hrs , d ) 23 hrs , e ) 24 hrs | b | add(17, log(16)) | log(n9)|add(n10,#0) | gain |
two numbers are respectively 20 % and 50 % more than a third number . the ratio of the two numbers is : | "third no is x first no - 120 % of x = 6 x / 5 second no = 150 x / 100 = 3 x / 2 ration of first nos = 6 x / 5 : 3 x / 2 = 12 x : 15 x = 4 : 5 answer b" | a ) 3 : 5 , b ) 4 : 5 , c ) 5 : 5 , d ) 4 : 3 , e ) 2 : 1 | b | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | other |
a car is purchased on hire - purchase . the cash price is $ 26 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ? | "explanation : cash price = $ 26 000 deposit = 10 % ã — $ 26 000 = $ 2600 loan amount = $ 26000 â ˆ ’ $ 2600 number of payments = 60 = $ 23400 i = p * r * t / 100 i = 14040 total amount = 23400 + 14040 = $ 37440 regular payment = total amount / number of payments = 624 answer : b" | a ) $ 603 , b ) $ 624 , c ) $ 625 , d ) $ 626 , e ) $ 627 | b | add(divide(multiply(multiply(26, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(26, const_1000), subtract(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)| | gain |
a certain industrial loom weaves 0.129 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | "let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.129 : 15 : : 1 : x = > 0.129 * x = 15 = > x = 15 / 0.129 = > x = 116 answer : c" | a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | c | divide(15, 0.129) | divide(n1,n0)| | physics |
a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 6 rows or 4 rows or 8 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 6 , 4,8 ) = 24 answer : a" | a ) 24 , b ) 60 , c ) 28 , d ) 76 , e ) 21 | a | divide(multiply(multiply(8, 4), 6), const_2) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_2)| | general |
a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 2 or 19 ? | "number of multiples of 2 from 1 through 30 = 30 / 2 = 15 number of multiples of 19 from 1 through 30 = 30 / 19 = 1 number of multiples of 2 and 19 both from 1 through 30 = number of multiples of 19 * 2 ( = 38 ) = 0 total favourable cases = 15 + 1 - 0 = 16 probability = 16 / 30 = 8 / 15 answer : option d" | a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 8 / 15 , e ) 11 / 30 | d | divide(add(floor(divide(30, 2)), divide(30, 19)), 30) | divide(n0,n2)|divide(n0,n1)|floor(#1)|add(#0,#2)|divide(#3,n0)| | probability |
the cash difference between the selling prices of an article at a profit of 6 % and 8 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 106 % of x ) / ( 108 % of x ) = 106 / 108 = 53 / 54 = 53 : 54 . answer : c" | a ) 51 : 52 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | c | divide(add(const_100, 6), add(const_100, 8)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
working simultaneously and independently at an identical constant rate , 30 machines of a certain type can produce a total of x units of product p in 3 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 5 x units of product p in 10 days ? | "the rate of 30 machines is rate = job / time = x / 3 units per day - - > the rate of 1 machine 1 / 30 * ( x / 3 ) = x / 90 units per day ; now , again as { time } * { combined rate } = { job done } then 10 * ( m * x / 90 ) = 5 x - - > m = 45 . answer : c ." | a ) 41 , b ) 44 , c ) 45 , d ) 47 , e ) 50 | c | multiply(multiply(30, 5), divide(3, 10)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
30 men shake hands with each other . maximum no of handshakes without cyclic handshakes . | "or , if there are n persons then no . of shakehands = nc 2 = 30 c 2 = 435 answer : e" | a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 435 | e | multiply(subtract(30, const_1), divide(30, const_2)) | divide(n0,const_2)|subtract(n0,const_1)|multiply(#0,#1)| | general |
a car traveled 448 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 448 / h = ... | a ) 14 , b ) 18 , c ) 21 , d ) 22 , e ) 27 | b | divide(336, divide(subtract(448, 336), 6)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
at what rate percent on simple interest will rs . 750 amount to rs . 950 in 5 years ? | "200 = ( 750 * 5 * r ) / 100 r = 5.33 % answer : d" | a ) 3.33 % , b ) 5.93 % , c ) 4.33 % , d ) 5.33 % , e ) 6.33 % | d | multiply(divide(divide(subtract(950, 750), 750), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
from the sale of sleeping bags , a retailer made a gross profit of 15 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | "cost price * 1.15 = selling price - - > cost price * 1.15 = $ 28 - - > cost price = $ 24.34 . answer : c ." | a ) 3.0 , b ) 3.36 , c ) 24.34 , d ) 25.0 , e ) 31.36 | c | divide(multiply(28, const_100), add(const_100, 15)) | add(n0,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain |
find 40 % of 240 | "we know that r % of m is equal to r / 100 × m . so , we have 40 % of 240 40 / 100 × 240 = 96 answer : a" | a ) 96 , b ) 94 , c ) 86 , d ) 74 , e ) 110 | a | divide(40, 240) | divide(n0,n1)| | gain |
sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 3.5 per pound , and german chocolate cost $ 2.5 per pound . if the total the she spent on chocolate was $ 70 and both types of chocolate were purchased in whole number of pounds , how many total pounds of ch... | if there were all the expensive ones , 3.5 . . . . there would be 70 / 3.5 or 20 of them but since 2.5 $ ones are also there , answer has to be > 20 . . . . if all were 2.5 $ ones , there will be 70 / 2.5 or 28 . . so only 24 is left ans d . . | a ) 7 , b ) 8 , c ) 10 , d ) 24 , e ) 15 | d | divide(70, divide(add(3.5, 2.5), const_2)) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1) | other |
two men start from opposite banks of a river . they meet 340 meters away from one of the banks on forward journey . after that they meet at 170 meters from the other bank of the river on their backward journey . what will be the width of the river ( in meters ) ? | let p , q are two persons then their speeds be a , b - > m / hr x - - - - 340 - - - - - - - - - - - - - - - - - - - - - - | - - - ( d - 340 ) - - - - - - - - - y in forward journey time taken by p = 340 / a ; time taken by q = d - 340 / b ; in forward journey , time taken by p = time taken by q so 340 / a = d - 340 / b... | a ) 750 , b ) 850 , c ) 900 , d ) 980 , e ) 950 | b | add(add(add(340, 170), 170), 170) | add(n0,n1)|add(n1,#0)|add(n1,#1) | physics |
if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 34 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 34 ) / 10 = ( 54 - 34 ) / 10 = 2 answer : option d" | a ) 4 , b ) 7 , c ) 8 , d ) 2 , e ) 3 | d | divide(subtract(add(multiply(7, 7), 5), 34), 10) | multiply(n1,n1)|add(n0,#0)|subtract(#1,n3)|divide(#2,n4)| | general |
a 160 meter long train crosses a man standing on the platform in 9 sec . what is the speed of the train ? | "s = 160 / 9 * 18 / 5 = 64 kmph answer : b" | a ) 96 kmph , b ) 64 kmph , c ) 52 kmph , d ) 86 kmph , e ) 76 kmph | b | multiply(divide(160, 9), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
the difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4 % per annum is re . 1 . the sum ( in rs . ) is ? | "let the sum be rs . x . then , [ x ( 1 + 4 / 100 ) 2 - x ] = ( 676 / 625 x - x ) = 51 / 625 x s . i . = ( x * 4 * 2 ) / 100 = 2 x / 25 51 x / 625 - 2 x / 25 = 1 or x = 625 . answer : a" | a ) 625 , b ) 827 , c ) 657 , d ) 726 , e ) 634 | a | inverse(multiply(power(divide(4, const_100), 2), 1)) | divide(n1,const_100)|power(#0,n0)|multiply(n2,#1)|inverse(#2)| | gain |
how many pieces of 0.85 metres can be cut from a rod 42.5 metres long ? | "number of pieces = 42.5 / 0.85 = 42.50 / 0.85 = 4250 / 85 = 50 . answer : c" | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(42.5, 0.85) | divide(n1,n0)| | physics |
60 % of a number is added to 160 , the result is the same number . find the number ? | ": ( 60 / 100 ) * x + 160 = x 2 x = 800 x = 400 answer : b" | a ) 300 , b ) 400 , c ) 266 , d ) 99 , e ) 121 | b | divide(160, divide(160, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
difference of 2 numbers is 1660 . if 7.5 % of one number is 12.5 % of the other number , find the 2 numbers ? | "let the numbers be x and y 7.5 % of x = 12.5 % of y x = 125 y / 75 = 5 y / 3 x - y = 1660 5 y / 3 - y = 1660 y = 2490 x = 5 y / 3 = 4150 answer is c" | a ) 125,150 , b ) 5124,4515 , c ) 4150,2490 , d ) 3250,4510 , e ) 1254,3210 | c | multiply(divide(1660, 7.5), divide(2, subtract(const_1, divide(1660, 7.5)))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2)| | gain |
an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 2,395 , what was the total number of pairs of contact lenses that she sold ? | "( x + 5 ) * 150 + x * 85 = 2395 = > x = 7 total lens = 7 + ( 7 + 5 ) = 19 answer e" | a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | e | add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5) | add(n0,n1)|multiply(const_100,const_4)|multiply(n2,const_10)|multiply(n0,n2)|add(#1,#2)|add(n2,#4)|add(#5,const_1000)|subtract(#6,#3)|divide(#7,#0)|multiply(#8,const_2)|add(n2,#9)| | general |
cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 1 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 1 hrs therefore , 1 = d / 75 + d / 105 lcm of 75 and 105 = 525 1 = d / 75 + d / 105 = > 1 = 7 d / 525 + 5 d / ... | a ) 35 , b ) 50 , c ) 62 , d ) 75 , e ) 78 | a | multiply(divide(multiply(105, 1), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
sum of the squares of 3 no . ' s is 179 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 179 + 2 * 131 a + b + c = √ 441 = 21 b" | a ) 20 , b ) 21 , c ) 26 , d ) 28 , e ) 30 | b | sqrt(add(multiply(131, const_2), 179)) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1)| | general |
if 20 % of 30 is greater than 25 % of a number by 2 , the number is : | explanation : = > 20 / 100 * 30 - 25 / 100 * x = 2 = > x / 4 = 4 so x = 16 answer : a | a ) 16 , b ) 24 , c ) 20 , d ) 32 , e ) none of these | a | divide(subtract(multiply(divide(20, const_100), 30), 2), divide(25, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1) | gain |
how many positive integers less than 5000 are evenly divisible by neither 15 nor 23 ? | integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 23 5000 / 23 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 23 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 23 = 333 + 238 - 47 = 524 now subtract that... | a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | c | floor(divide(subtract(subtract(5000, const_1), subtract(add(floor(divide(5000, 15)), floor(divide(5000, 23))), floor(divide(5000, lcm(15, 23))))), const_1000)) | divide(n0,n1)|divide(n0,n2)|lcm(n1,n2)|subtract(n0,const_1)|divide(n0,#2)|floor(#0)|floor(#1)|add(#5,#6)|floor(#4)|subtract(#7,#8)|subtract(#3,#9)|divide(#10,const_1000)|floor(#11) | general |
p alone can complete a job in 12 days . the work done by q alone in one day is equal to one - half of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 12 q ' s rate is 1 / 24 the combined rate is 1 / 12 + 1 / 24 = 1 / 8 if they work together , the job will take 8 days . the answer is d ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(const_1, add(divide(const_1, 12), divide(divide(const_1, 12), const_3))) | divide(const_1,n0)|divide(#0,const_3)|add(#0,#1)|divide(const_1,#2)| | physics |
suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 460 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 560 . how much more do pairs of boots cost than pairs of shoes... | let x be pair of shoes and y be pair of boots . 22 x + 16 y = 460 . . . eq 1 8 x + 32 y = 560 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 920 8 x = 560 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 560 . . . then we get 32 y = 480 then y = 25 differenece between x and y is 15 answer... | a ) 15 , b ) 18 , c ) 21 , d ) 23 , e ) 25 | a | divide(subtract(460, multiply(22, divide(subtract(multiply(460, const_2), 560), subtract(multiply(22, const_2), 8)))), 16) | multiply(n2,const_2)|multiply(n0,const_2)|subtract(#0,n5)|subtract(#1,n3)|divide(#2,#3)|multiply(n0,#4)|subtract(n2,#5)|divide(#6,n1) | general |
if ( a + b ) = 4 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 4 = > a = 4 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 4 - b + 3 - c = > 7 - ( b + c ) = > 7 - 9 = - 2 . option e . . ." | a ) 16 . , b ) 8 . , c ) 7 , d ) 2 . , e ) - 2 | e | subtract(add(4, 3), 9) | add(n0,n2)|subtract(#0,n1)| | general |
the difference between a 8 digit number and the number formed by reversing its digit is not a multiple of | "another approach is to test a number . let ' s say the original number is 12000002 so , the reversed number is 20000021 the difference = 20000021 - 12000002 = 8000019 no check the answer choices 8000019 is a multiple of 3,296297 , 888891,2666673 5 is not a multiple of 8000019 answer ; b" | a ) 3 , b ) 5 , c ) 296297 , d ) 888891 , e ) 2666673 | b | power(add(8, const_4), const_4) | add(n0,const_4)|power(#0,const_4)| | general |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 = > x = 5 hence , b ' s age = 2 x = 10 years . answer : b" | a ) 11 years , b ) 10 years , c ) 18 years , d ) 189 years , e ) 28 years | b | divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
15 51 216 1100 ? 46452 | "the format of the series 15 * 3 + 6 = 51 51 * 4 + 12 = 216 216 * 5 + 20 = 1100 1100 * 6 + 30 = 6630 6630 * 7 + 42 = 46452 answer : a" | a ) 6630 , b ) 6650 , c ) 6560 , d ) 6530 , e ) none of these | a | divide(multiply(add(15, divide(51, 216)), 1100), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | general |
if a * b denotes the greatest common divisor of a and b , then ( ( 12 * 15 ) * ( 18 * 24 ) ) = ? | "the greatest common divisor of 12 and 15 is 3 . hence 12 * 15 = 3 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 24 is 6 . hence 18 * 24 = 6 . hence ( ( 12 * 16 ) * ( 18 * 24 ) ) = 3 * 6 . the greatest common divisor of 3 and 6 is 3 . answer ; e ." | a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 3 | e | divide(divide(18, const_3), const_3) | divide(n2,const_3)|divide(#0,const_3)| | general |
what annual installment will discharge a debt of rs . 1260 due in 3 years at 12 % simple interest ? | "let each installment be rs . x then , ( x + ( ( x * 12 * 1 ) / 100 ) ) + ( x + ( ( x * 12 * 2 ) / 100 ) ) + x = 1260 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1260 ï ƒ › ( 28 x + 31 x + 25 x ) = ( 1260 * 25 ) x = ( 1260 * 25 ) / 84 = rs . 375 . therefore , each installment = rs . 375 . answer is d ." | a ) 315 , b ) 345 , c ) 325 , d ) 375 , e ) none of them | d | multiply(multiply(const_100.0, divide(12, 1260)), 3) | divide(n2,const_100)|multiply(n0,#0)|multiply(n1,#1)| | gain |
what is the sum of all even numbers from 1 to 801 ? | "explanation : 800 / 2 = 400 400 * 401 = 160400 answer : a" | a ) 160400 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 128111 | a | divide(multiply(1, 801), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
there are 225 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 225 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 225 = 2 / 5 t , or 2 / 5 ( t - m ) = 225 , from here we get that t - m = 562.5 , that would b... | a ) 300 , b ) 400 , c ) 562.5 , d ) 600 , e ) none of these | c | divide(225, divide(2, 5)) | divide(n1,n2)|divide(n0,#0)| | general |
if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 8 } . what is the probability that the median of the set will increase but the range still remains the same ? | "we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the... | a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6 | d | divide(const_4, const_10) | divide(const_4,const_10)| | general |
a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 300 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to c... | "let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - ... | a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | b | divide(multiply(divide(multiply(divide(40, const_100), 300), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general |
how many prime numbers are between 15 / 7 and 131 / 6 ? | "15 / 7 = 3 - 131 / 6 = 22 - prime numbers between 3 and 22 are 5 , 7 , 11 , 13 , 17 , and 19 - sign signifies that the number is marginally less . answer d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | floor(const_2) | floor(const_2)| | general |
a can do a job in 18 days and b can do it in 30 days . a and b working together will finish twice the amount of work in days ? | "1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 45 / 4 = 11 ¼ * 2 = 22 ½ days answer : b" | a ) 29 ½ days , b ) 22 ½ days , c ) 92 ½ days , d ) 28 ½ days , e ) 92 ½ days | b | add(divide(const_1, 18), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics |
the cost of manufacturing a popular model car is made up of 3 items : cost of raw material , labour and overheads - in a year the cost of 3 items were in the ration of 4 : 3 : 2 . next year the cost of the raw material rose by 10 % , labour cost increased by 8 % but overhead reduced by 5 % . then % increase int the pri... | before increase total cost = 4 + 3 + 2 = 9 , after increasing the cost = 9.54 so increase of . 54 over 9 , equal to 6 % answer : b | a ) 7.67 % , b ) 6 % , c ) 0.54 % , d ) 9.54 % , e ) 8.54 % | b | subtract(add(add(multiply(divide(4, add(add(4, 3), 2)), add(const_100, 10)), multiply(divide(3, add(add(4, 3), 2)), add(const_100, 8))), multiply(divide(2, add(add(4, 3), 2)), subtract(const_100, 5))), const_100) | add(n5,const_100)|add(n0,n2)|add(n6,const_100)|subtract(const_100,n7)|add(n4,#1)|divide(n2,#4)|divide(n0,#4)|divide(n4,#4)|multiply(#0,#5)|multiply(#2,#6)|multiply(#7,#3)|add(#8,#9)|add(#11,#10)|subtract(#12,const_100) | general |
two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 90 kmph , find the rate at which the train b runs . | "if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) , then the ra... | a ) 40 , b ) 60 , c ) 135 , d ) 80 , e ) 100 | c | multiply(90, sqrt(divide(9, 4))) | divide(n0,n1)|sqrt(#0)|multiply(n2,#1)| | physics |
a vessel of capacity 90 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem , t = 90 , y = 9 and n = 2 . hence , quantity of milk finall... | a ) 73.9 , b ) 72.9 , c ) 73.0 , d ) 73.5 , e ) 92.2 | b | subtract(subtract(90, multiply(const_3, const_3)), multiply(divide(subtract(90, multiply(const_3, const_3)), 90), multiply(const_3, const_3))) | multiply(const_3,const_3)|subtract(n0,#0)|divide(#1,n0)|multiply(#2,#0)|subtract(#1,#3)| | physics |
377 ÷ 13 ÷ 29 × 1 / 4 ÷ 2 = ? | explanation : 377 ÷ 13 ÷ 29 × 1 / 4 ÷ 2 = ? = > ? = ( 337 / ( 13 × 29 × 4 × 2 ) ) = 1 / 8 answer : option b | a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 10 , d ) 1 / 12 , e ) 1 / 15 | b | divide(divide(multiply(divide(divide(377, 13), 29), 1), 4), 2) | divide(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|divide(#3,n5) | general |
if x = 2 ^ 8 and x ^ x = 2 ^ k , what is k ? | "solution : we know that x = 2 ^ 8 which implies x ^ x = ( 2 ^ 8 ) ^ ( 2 ^ 8 ) = 2 ^ ( 8 * 2 ^ 8 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 2 ^ ( 2 ^ 3 * 2 ^ 8 ) = 2 ^ ( 2 ^ ( 3 + 8 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 2 ^ ( 2 ^ 11 ) given that x ^ x = 2 ^ k so 2 ^ ( 2 ^ 11 ) = 2 ^ k si... | a ) 2 ^ 15 , b ) 2 ^ 14 , c ) 2 ^ 13 , d ) 2 ^ 12 , e ) 2 ^ 11 | e | multiply(power(2, 8), 8) | power(n0,n1)|multiply(n1,#0)| | general |
if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 6 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 6 : : 42.75 : x = > 11.25 x x = 6 x 42.75 = > x = ( 6 x 42.75 ) / 11.25 = > x = 22.8 answer : a" | a ) 22.8 kg , b ) 25.6 kg , c ) 28 kg , d ) 26.5 kg , e ) none of these | a | divide(multiply(6, 42.75), 11.25) | multiply(n1,n2)|divide(#0,n0)| | physics |
david obtained 70 , 60 , 78 , 60 and 65 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | "explanation : average = ( 70 + 60 + 78 + 60 + 65 ) / 5 = 333 / 5 = 66.6 . answer : c" | a ) 75 , b ) 27.6 , c ) 66.6 , d ) 16.5 , e ) 11 | c | divide(add(add(add(add(70, 60), 78), 60), 65), add(const_2, const_3)) | add(n0,n1)|add(const_2,const_3)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
a car drives 60 miles on local roads at 30 mph , and 65 miles on the highway at 65 mph , what is the average speed of the entire trip ? | "so the answer is plainly d . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 60 miles , and v = 30 mph so we can obtain time : 30 * t = 60 or t = 60 / 30 = 2 hours the needed time to cover 60 miles in the same way we should divide... | a ) 36 mph , b ) 40 mph , c ) 44 mph , d ) 42 mph , e ) 58 mph | d | divide(add(60, 65), add(divide(60, 30), divide(65, 65))) | add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)| | physics |
set a consists of the integers from 4 to 15 , inclusive , while set b consists of the integers from 7 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 4 , 5,6 , 7,8 , 9,10 , 11,12 } b = { 6 , 7,8 , 9,10 , 11,12 . . . } thus we see that there are 9 distinct integers that are common to both . d is the correct answer ." | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(7, 4) | add(n0,n2)| | other |
the length of the bridge , which a train 120 metres long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 120 + x ) / 30 = 25 / 2 = > 2 ( 120 + x ) = 750 = > x = 255 m . answer : option d" | a ) 230 , b ) 240 , c ) 245 , d ) 255 , e ) 260 | d | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 120) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
the length of a rectangle is halved while its width is also halved . what is the % change in area ? | "the original area is l * w the new area is 0.5 l * 0.5 w = 0.25 * l * w = l * w - 0.75 * l * w the area decreased by 75 % . the answer is e ." | a ) 25 % , b ) 30 % , c ) 50 % , d ) 65 % , e ) 75 % | e | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | divide(const_1,const_2)|multiply(#0,const_3)|subtract(#1,const_1)|divide(#2,const_1)|multiply(#3,const_100)| | geometry |
the floor of a rectangular room is 15 m long and 12 m wide . the room is surrounded by a verandah of width 2 m on all its sides . the area of the verandah is : | area of the outer rectangle = 19 × 16 = 304 m 2 area of the inner rectangle = 15 × 12 = 180 m 2 required area = ( 304 – 180 ) = 124 m 2 answer a | ['a ) 124 m 2', 'b ) 120 m 2', 'c ) 108 m 2', 'd ) 58 m 2', 'e ) none of these'] | a | subtract(rectangle_area(add(multiply(2, const_2), 15), add(12, multiply(2, const_2))), rectangle_area(15, 12)) | multiply(n2,const_2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1) | geometry |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 6600 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 6600 x ( 2 / 11 ) ) = rs . 1200 . answer is d ." | a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | d | multiply(6600, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | divide(const_2,n0)|multiply(const_2,n0)|multiply(#0,n0)|add(#1,#2)|add(#3,n0)|divide(const_2,#4)|multiply(n1,#5)| | gain |
a reduction of 40 % in the price of apples would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "b 40 * ( 40 / 100 ) = 16 - - - 64 ? - - - 12 = > rs . 3" | a ) 1 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | b | multiply(divide(divide(multiply(40, 40), const_100), 64), const_12) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|multiply(#2,const_12)| | gain |
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