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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on dividing 14698 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 14698 - 14 ) / 89 = 165 answer ( e )" | a ) 743 , b ) 154 , c ) 852 , d ) 741 , e ) 165 | e | divide(subtract(14698, 14), 89) | subtract(n0,n2)|divide(#0,n1)| | general |
how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 190 m in length ? | "d = 110 + 190 = 300 m s = 60 * 5 / 18 = 50 / 3 t = 300 * 3 / 50 = 18.0 sec answer : d" | a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 18.0 sec , e ) 72.0 sec | d | divide(add(110, 190), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
paulson spends 75 % of his income . his income is increased by 20 % and he increased his expenditure by 10 % . find the percentage increase in his savings ? | "let original income = $ 100 expenditure = $ 75 savings = $ 25 new income = $ 120 new expenditure = $ 110 / 100 * 75 = $ 165 / 2 new savings = 120 - 165 / 2 = $ 75 / 2 increase in savings = 75 / 2 - 25 = $ 25 / 2 increase % = 25 / 2 * 1 / 25 * 100 = 50 % answer is c" | a ) 20 % , b ) 25 % , c ) 50 % , d ) 75 % , e ) 60 % | c | multiply(divide(subtract(subtract(add(const_1, divide(20, const_100)), multiply(add(const_1, divide(10, const_100)), divide(75, const_100))), subtract(const_1, divide(75, const_100))), subtract(const_1, divide(75, const_100))), const_100) | divide(n1,const_100)|divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#2)|subtract(#3,#6)|subtract(#7,#5)|divide(#8,#5)|multiply(#9,const_100)| | general |
rice weighing 33 / 4 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces ) | "33 / 4 ÷ 4 = 33 / 16 pounds in each container 33 / 16 pounds * 16 ounces / pound = 33 ounces in each container the answer is c ." | a ) 27 , b ) 30 , c ) 33 , d ) 36 , e ) 39 | c | divide(multiply(divide(33, 4), 16), 4) | divide(n0,n1)|multiply(n4,#0)|divide(#1,n2)| | general |
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 22 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 22 = > x = 4 so deepak age is = 3 ( 4 ) = 12 answer : e" | a ) 22 , b ) 15 , c ) 77 , d ) 266 , e ) 12 | e | divide(multiply(subtract(22, 6), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 112 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 112 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 112 = 16 / 30 x x = 210 210 - 112 = 98 answer b" | a ) 80 , b ) 98 , c ) 105 , d ) 120 , e ) 210 | b | multiply(divide(112, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), mul... | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain |
if an object travels 70 feet in 2 seconds , what is the object ’ s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "70 feet / 2 seconds = 35 feet / second ( 35 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 23.86 miles / hour ( approximately ) the answer is e ." | a ) 15.95 , b ) 17.28 , c ) 19.54 , d ) 21.73 , e ) 23.86 | e | divide(divide(70, 5280), multiply(2, divide(1, const_3600))) | divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)| | physics |
85 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 85 = 113.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capa... | a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | d | multiply(1200, multiply(800, divide(85, multiply(800, 600)))) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n4,#2)| | physics |
evaluate 28 % of 650 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 650 + ( 45 / 100 ) * 280 = 182 + 126 = 308 answer : option b" | a ) 232 , b ) 308 , c ) 252 , d ) 262 , e ) 282 | b | divide(28, divide(650, 28)) | divide(n1,n0)|divide(n0,#0)| | gain |
a can give b 300 meters start and c 600 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 700 m and c runs 400 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 400 ) / 700 = 571.43 m . b can give c = 1000 - 571.43 = 428.57 m . answer : c" | a ) 111.12 , b ) 111.67 , c ) 428.57 , d ) 111.11 , e ) 101.12 | c | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 600)), subtract(multiply(const_100, const_10), 300))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)| | physics |
there are 2000 students in a school and among them 10 % of them attends chess class . 50 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "10 % of 2000 gives 200 . so 200 attends chess and 50 % of 200 gives 100 so 100 enrolled for swimming answer : c" | a ) 1 , b ) 10 , c ) 100 , d ) 50 , e ) 20 | c | divide(multiply(divide(multiply(10, 2000), const_100), 50), const_100) | multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)| | gain |
a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 6 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the p... | "i think e - 11 / 15 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 6 x so old = x , new = 5 x now 1 / 3 of x are occupied and 1 / 5 of 5 x are occupied which simplifies to ( 5 / 5 ) x we are shifting 1 / 3 of x to the new ward so ... | a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15 | e | multiply(add(multiply(6, 3), 1), multiply(divide(1, 3), divide(1, 5))) | divide(n1,n2)|divide(n1,n4)|multiply(n0,n2)|add(n1,#2)|multiply(#0,#1)|multiply(#3,#4)| | general |
the sum of two numbers is 40 and their difference is 4 . the ratio of the numbers is | sol . let the numbers be x and y . then , x + y / x - y = 40 / 4 = 10 ⇔ ( x + y ) = 10 ( x - y ) ⇔ 9 x = 11 y ⇔ x / y = 11 / 9 . answer a | a ) 11 : 9 , b ) 11 : 18 , c ) 21 : 19 , d ) 21 : 19 , e ) none | a | divide(divide(subtract(40, subtract(divide(40, const_2), const_2)), const_2), divide(subtract(divide(40, const_2), const_2), const_2)) | divide(n0,const_2)|subtract(#0,const_2)|divide(#1,const_2)|subtract(n0,#1)|divide(#3,const_2)|divide(#4,#2) | general |
on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number q of races we need to conduct to get the top 3 fastest horses ? | q = 7 is the correct answer . good solution buneul . b | a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | divide(n1,n0)|add(#0,const_2) | general |
the ages of 2 persons differ by 21 years . if 15 years ago the elder one be 8 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 21 8 ( x - 15 ) = x + 21 - 15 x = 18 age of elder person = 18 + 21 = 39 answer is e" | a ) 30 , b ) 48.5 , c ) 50.4 , d ) 62.6 , e ) 47.9 | e | subtract(add(divide(multiply(2, 21), subtract(21, const_1)), 21), 2) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
in a 100 m race , a covers the distance in 30 seconds and b in 45 second . in this race a beats b by : | solution distance covered by b in 15 sec . = ( 100 / 45 x 15 ) m = 33.33 m . ∴ a beats b by 33.33 metres . answer a | a ) 33.33 m , b ) 25 m , c ) 45 m , d ) 30 m , e ) none of these | a | multiply(divide(100, 45), subtract(45, 30)) | divide(n0,n2)|subtract(n2,n1)|multiply(#0,#1) | physics |
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 12 ) , what is the value of x ? | i am guessing the question is : ( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 12 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 12 hence x = 14 . answer is e | a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 14 | e | add(12, 2) | add(n0,n5) | general |
a , b , c can do a work in 15 , 20,45 days respectively . they get rs 500 for their work . what is the share of a ? | lcm = 180 share of a = ( lcm / a x total amount ) / lcm / a + lcm / b + lcm / c = ( 180 / 15 ) / ( 180 / 15 + 180 / 20 + 180 / 45 ) = ( 12 / 25 ) * 500 = rs . 240 answer : a | a ) rs . 240 , b ) rs . 340 , c ) rs . 260 , d ) rs . 280 , e ) rs . 440 | a | multiply(500, divide(inverse(15), add(add(inverse(multiply(const_2, const_10)), inverse(add(multiply(const_2, const_10), multiply(const_100, const_0_25)))), inverse(15)))) | inverse(n0)|multiply(const_10,const_2)|multiply(const_0_25,const_100)|add(#1,#2)|inverse(#1)|inverse(#3)|add(#4,#5)|add(#6,#0)|divide(#0,#7)|multiply(n2,#8) | physics |
express a speed of 252 kmph in meters per second ? | "e 70 mps 252 * 5 / 18 = 70 mps" | a ) 10 mps , b ) 05 mps , c ) 09 mps , d ) 12 mps , e ) 70 mps | e | multiply(const_0_2778, 252) | multiply(n0,const_0_2778)| | physics |
the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 4... | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant “ a ” . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 40 minutes after it was poured is : f = 120 ( 2 ^ - 40 / 10 ) + 60 f = 120 * 1 / 16 + 60 f = 15 / 2 + 60 f = 135 ... | a ) 67.5 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general |
a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 3 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 3 ( pir ^ 2 ) pir ^ 2 = 6 pir r = 6 answer : a" | a ) 6 , b ) 2 , c ) 4 , d ) 8 , e ) 16 | a | multiply(const_2, sqrt(power(3, const_2))) | power(n1,const_2)|sqrt(#0)|multiply(#1,const_2)| | geometry |
jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 25 / 49 , what... | "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 / 49 = ( 5 / 7 ) * ( 5 / 7 ) therefore probability that the next one she draws is pink = 2 / 7 option ( b )" | a ) 1 / 49 , b ) 2 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | b | subtract(const_1, sqrt(divide(25, 49))) | divide(n0,n1)|sqrt(#0)|subtract(const_1,#1)| | general |
the areas of the two spheres are in the ratio 1 : 4 . the ratio of their volume is ? | 4 π r 12 : 4 π r 22 = 1 : 4 r 1 : r 2 = 1 : 2 4 / 3 π r 13 : 4 / 3 π r 23 r 13 : r 23 = 1 : 8 answer : c | ['a ) 1 : 4', 'b ) 1 : 16', 'c ) 1 : 8', 'd ) 1 : 64', 'e ) 2 : 3'] | c | divide(1, multiply(4, const_2)) | multiply(n1,const_2)|divide(n0,#0) | geometry |
a hostel had provisions for 250 men for 40 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 40 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 40 250 × 40 = 200 x 5 × 40 = 4... | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | c | divide(multiply(250, 40), subtract(250, 50)) | multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)| | gain |
the distance between parallel sides of the hexagon is 9 . determine the length of the hexagon | length of each side of regular hexagon will be 3 * sqrt 3 . total length ( perimeter ) of regular hexagon will be 18 * sqrt 3 answer : a | ['a ) 18 * sqrt 3', 'b ) 18 * sqrt 2', 'c ) 18 * sqrt 4', 'd ) 19 * sqrt 3', 'e ) 17 * sqrt 3'] | a | multiply(multiply(const_2, const_3), divide(9, sqrt(const_3))) | multiply(const_2,const_3)|sqrt(const_3)|divide(n0,#1)|multiply(#2,#0) | physics |
a train covers a distance of 12 km in 10 min . if it takes 3 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 3 = 60 m . answer : c" | a ) 70 m , b ) 16 m , c ) 60 m , d ) 80 m , e ) 88 m | c | divide(12, subtract(divide(12, 10), 3)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
if g and f are both odd prime numbers andg < f , then how many different positive integer factors does 2 gfhave ? | g and f are both odd prime numbers - it means either g or f is not 2 and since prime numbers have only two factors - 1 and the number itself g and f each will have ( 1 + 1 ) = 2 factors hence 2 gf will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(2, multiply(2, 2)) | multiply(n0,n0)|multiply(n0,#0) | other |
a and b go around a circular track of length 600 m on a cycle at speeds of 30 kmph and 60 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 30 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 72 sec . answer : d" | a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 72 sec , e ) 189 sec | d | divide(600, subtract(multiply(60, const_0_2778), multiply(30, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 18,000 is invested at the same interest rate , what is the interest from the investment ? | "- > 250 / 5,000 = 5 % and 18,000 * 5 % = 900 . thus , e is the answer ." | a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 850 , e ) $ 900 | e | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | multiply(const_100,const_2)|multiply(#0,const_100)|divide(#1,const_4)|multiply(n1,#1)|divide(#3,#2)| | gain |
m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 18 ) both lie on the line defined by the equation x = ( y / 6 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 6 ) - ( 2 / 5 ) , and so y = 6 x + 12 / 5 . the slope is 6 . ( n + 18 - n ) / ( m + p - m ) = 6 p = 3 the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(18, 6) | divide(n0,n1)| | general |
if w is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of w ? | w = 30 ! 8 w = 30 x 29 x 28 x 27 x 26 x 25 x 24 x 24 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 out of these 30 , 27 , 24 , 21 , 18 , 15 , 12 , 09 , 06 , 3 are factors of 3 3 x 10 , 3 x 3 x 3 , 3 x 8 , 3 x 3 x 2 , 3 x 5 , 3 x 4 , 3 x 3 x 3 , 3 x 2 , 3 so we hav... | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | divide(subtract(30, const_2), const_2) | subtract(n1,const_2)|divide(#0,const_2) | general |
a and b started a business in partnership investing rs . 20000 and rs . 15000 respectively . after 6 months , c joined them with rs . 20000 . what will be b ' s share in the total profit of rs . 25000 earned at the end of 2 years from the starting of the business ? | solution a : b : c = ( 20000 × 24 ) : ( 15000 × 24 ) : ( 20000 × 18 ) = 4 : 3 : 3 b ' s share = rs . ( 25000 × 3 / 10 ) = rs . 7500 . answer a | a ) rs . 7500 , b ) rs . 9000 , c ) rs . 9500 , d ) rs . 10,000 , e ) none | a | multiply(divide(multiply(15000, multiply(const_12, const_2)), add(add(multiply(20000, multiply(const_12, const_2)), multiply(15000, multiply(const_12, const_2))), multiply(20000, subtract(multiply(const_12, const_2), 6)))), 25000) | multiply(const_12,const_2)|multiply(n1,#0)|multiply(n0,#0)|subtract(#0,n2)|add(#2,#1)|multiply(n0,#3)|add(#4,#5)|divide(#1,#6)|multiply(n4,#7) | gain |
length of a rectangular plot is 14 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 14 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 14 ) + x ] = 200 2 x + 14 = 100 2 x = 86 x = 43 . hence , length = x + 14 = 57 m b" | a ) 50 m , b ) 57 m , c ) 80 m , d ) 82 m , e ) 84 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 14)), const_4) | divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)| | physics |
9 log 9 ( 7 ) = ? | "exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 7 ) = 7 correct answer c" | a ) 1 , b ) 2 , c ) 7 , d ) 4 , e ) 5 | c | divide(log(multiply(9, 9)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)| | other |
if ( a + b ) = 4 , ( b + c ) = 5 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 4 b + c = 5 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 4 - 5 + 3 = 2 . option d . . ." | a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | d | subtract(add(4, 3), 5) | add(n0,n2)|subtract(#0,n1)| | general |
indu gave bindu rs . 6500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "6500 = d ( 100 / 4 ) 2 d = 10.4 answer : c" | a ) 12.8 , b ) 20 , c ) 10.4 , d ) 11.6 , e ) 15 | c | subtract(subtract(multiply(6500, power(add(const_1, divide(4, const_100)), 2)), 6500), multiply(multiply(6500, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
if the product of two numbers is 84942 and their h . c . f . is 33 , find their l . c . m . | "explanation : hcf * lcm = 84942 , because we know product of two numbers = product of hcf and lcm lcm = 84942 / 33 = 2574 option a" | a ) 2574 , b ) 2500 , c ) 1365 , d ) 1574 , e ) none of these | a | divide(84942, 33) | divide(n0,n1)| | physics |
if n is a natural number , then ( 6 n ^ 2 + 6 n ) is always divisible by | "( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is divisible by both 6 and 12 , since n ( n + 1 ) is always even . answer b ." | a ) 6 only , b ) 6 and 12 both , c ) 12 only , d ) 18 only , e ) 12 and 18 both | b | add(multiply(6, const_100), multiply(2, 6)) | multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)| | general |
express the ratio 3 2 / 3 : 7 1 / 3 in its simplest form . | solution we first convert the mixed numbers 3 2 / 3 and 7 1 / 3 into fractions 3 2 / 3 = 3 * 3 / 3 + 2 / 3 = 11 / 3 7 1 / 3 = 7 * 3 / 3 + 1 / 3 = 22 / 3 the ratio 3 2 / 3 : 7 1 / 3 can be expressed as 11 / 3 ÷ 22 / 3 = 11 / 3 × 3 / 22 simplify = 11 / 22 = 1 / 2 the ratio is 1 / 2 or 1 : 2 answer is d | a ) 1 : 21 , b ) 1 : 1 , c ) 2 : 1 , d ) 1 : 2 , e ) 2 : 11 | d | divide(divide(add(multiply(3, 3), 2), 3), divide(add(multiply(3, 7), 1), 3)) | multiply(n0,n0)|multiply(n0,n3)|add(n1,#0)|add(n4,#1)|divide(#2,n0)|divide(#3,n0)|divide(#4,#5) | general |
a runs twice as fast as b and gives b a start of 64 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 64 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 64 m is 64 / 1 = 64 m so the total time taken by a and b to reach = 2 * 64 = 128 m answer : b" | a ) 75 m . , b ) 128 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | b | multiply(64, const_2) | multiply(n0,const_2)| | physics |
on a map , 2.5 inches represent 25 miles . how many miles approximately is the distance if you measured 55 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 2.5 inch = 2.54 * 2.5 cm 6.35 cm = 25 miles 55 cms = 25 / 6.35 * 55 = 216.5 miles answer : e" | a ) 212.5 , b ) 213.5 , c ) 214.5 , d ) 215.5 , e ) 216.5 | e | multiply(divide(55, 2.54), divide(25, 2.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics |
a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 920 in total compens... | "for 40 hrs = 40 * 16 = 640 excess = 920 - 640 = 280 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 280 / 28 = 70 / 7 = 10 total hrs = 40 + 10 = 50 answer d 50" | a ) 36 , b ) 40 , c ) 44 , d ) 50 , e ) 52 | d | add(40, divide(subtract(920, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 2 centimeter . the actual diameter of the tissue , in centimeters , is | it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 2 then x = 2 / 1000 = 0.002 . the answer is b | ['a ) 0.005', 'b ) 0.002', 'c ) 0.001', 'd ) 0.0005', 'e ) 0.0002'] | b | divide(2, const_1000) | divide(n1,const_1000) | geometry |
the average age of a class of 20 students is 21 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "total age of all students = 20 ã — 21 total age of all students + age of the teacher = 21 ã — 22 age of the teacher = 21 ã — 22 â ˆ ’ 21 ã — 20 = 21 ( 22 â ˆ ’ 20 ) = 21 ã — 2 = 42 answer is c ." | a ) 40 , b ) 41 , c ) 42 , d ) 45 , e ) 43 | c | subtract(multiply(add(20, 1), add(21, 1)), multiply(20, 21)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
when the positive integer k is divided by the positive integer n , the remainder is 11 . if k / n = 71.2 , what is the value of n ? | "here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . ... | a ) 9 , b ) 20 , c ) 55 , d ) 70 , e ) 80 | e | divide(add(multiply(divide(subtract(multiply(71.2, const_10), const_2), const_10), divide(11, divide(const_2, const_10))), 11), divide(11, divide(const_2, const_10))) | divide(const_2,const_10)|multiply(n1,const_10)|divide(n0,#0)|subtract(#1,const_2)|divide(#3,const_10)|multiply(#4,#2)|add(n0,#5)|divide(#6,#2)| | general |
if a and b are integers and ( a * b ) ^ 5 = 128 y , y could be : | distribute the exponent . a ^ 5 * b ^ 5 = 128 y find the prime factorization of 96 . this is 2 ^ 5 * 2 ^ 2 . we need 2 ^ 3 ( or some other power of 2 that will give us a multiple of 2 ^ 5 as our second term ) . 2 ^ 3 = 8 the answer is d . | a ) 1 , b ) 2 , c ) 4 , d ) 8 , e ) 16 | d | multiply(divide(128, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2) | general |
a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed t of the train . | "let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y... | a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | multiply(n2,n4)|subtract(n3,n5)|add(n0,#1)|divide(#0,n1)|divide(#2,#1)|subtract(#3,n4)|multiply(#4,#5)| | physics |
jacob is now 12 years younger than michael . if 5 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | "jacob = x years , mike = x + 12 years 5 years from now , 2 ( x + 5 ) = x + 17 2 x + 10 = x + 17 x = 7 x + 4 = 11 years answer c" | a ) 3 , b ) 7 , c ) 11 , d ) 21 , e ) 25 | c | add(4, subtract(add(12, 5), multiply(const_2, 5))) | add(n0,n1)|multiply(n1,const_2)|subtract(#0,#1)|add(n2,#2)| | general |
( 1000 ) 7 ÷ ( 10 ) 19 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 19 = ( 10 ) 21 / ( 10 ) 19 = 10 ( 2 ) = 100 option b" | a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these | b | multiply(1000, 10) | multiply(n0,n2)| | general |
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 143 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 143 / 2... | a ) rs . 165.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none | a | divide(subtract(multiply(143, add(add(1, 1), 2)), add(126, 126)), 2) | add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)| | other |
3 numbers are in the ratio 3 : 5 : 7 . the largest number is 56 . what is the difference between smallest and largest number ? | the three numbers are 3 x , 5 x , and 7 x . the largest number is 56 = 7 * 8 , so x = 8 . the smallest number is 3 * 8 = 24 . 56 - 24 = 32 the answer is d . | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | subtract(56, multiply(3, divide(56, 7))) | divide(n4,n3)|multiply(n0,#0)|subtract(n4,#1) | other |
mixture a is 20 % oil and 80 % material b by weight . if 2 more kilograms of oil are added to the 8 kilograms mixture a , how many kilograms of mixture a must be added to make a 70 % material b in the new mixture ? | old mix has 20 % oil . you add 2 kgs of 100 % oil . you get new mix with 30 % oil ( 70 % material b ) . w 1 / w 2 = ( 100 - 30 ) / ( 30 - 20 ) = 70 / 10 = 7 / 1 for every 7 parts of mix a , you have added 1 part of oil . since you have added 2 kgs oil , mix a must be 7 * 2 = 14 kgs . you already had 8 kgs of mix a , so... | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | subtract(multiply(divide(70, subtract(subtract(const_100, 70), 20)), 2), 8) | subtract(const_100,n4)|subtract(#0,n0)|divide(n4,#1)|multiply(n2,#2)|subtract(#3,n3) | general |
kanul spent $ 5000 in buying raw materials , $ 200 in buying machinery and 30 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 20 ) % of x = 5000 + 200 70 % of x = 5200 70 x / 100 = 5200 x = $ 52000 / 7 x = $ 7428.57 answer is c" | a ) $ 7456.00 , b ) $ 7500.55 , c ) $ 7428.57 , d ) $ 7852.56 , e ) $ 7864.00 | c | divide(add(5000, 200), subtract(const_1, divide(30, const_100))) | add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2)| | gain |
there is 60 % increase in an amount in 6 yrs at si . what will be the ci of rs . 12,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 12000 . t = 3 years and r = 10 % p . a . c . i . = rs . 12000 x 1 + 10 3 - 1 100 = rs . 12000 x 331 1000 = 3972 . e" | a ) 2354 , b ) 2450 , c ) 2540 , d ) 2650 , e ) 3972 | e | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multip... | divide(n0,n1)|multiply(const_100,const_100)|divide(#0,const_100)|multiply(#0,const_100)|add(#2,const_1)|multiply(#3,n3)|add(#1,#5)|multiply(#4,#4)|multiply(#4,#7)|multiply(#6,#8)|subtract(#9,#6)| | gain |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.8 km ? | "distance covered by amar = ( 18 / 48 ) * 1800 = 670 m answer : d" | a ) 660 m , b ) 650 m , c ) 570 m , d ) 670 m , e ) 680 m | d | divide(multiply(18, multiply(1.8, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics |
find the value of x from the below equation : x ^ 2 − 7 x + 10 = 0 | "here we need to find out a and b such that a + b = - 7 and ab = + 10 a = - 5 and b = - 2 satisfies the above condition . hence x ^ 2 − 7 x + 10 = ( x − 5 ) ( x − 2 ) x 2 − 7 x + 10 = ( x − 5 ) ( x − 2 ) x ^ 2 − 7 x + 10 = 0 ⇒ ( x − 5 ) ( x − 2 ) = 0 x 2 − 7 x + 10 = 0 ⇒ ( x − 5 ) ( x − 2 ) = 0 step 3 : equate each fac... | a ) 5 , b ) 7 , c ) 6 , d ) 11 , e ) 12 | a | divide(subtract(7, sqrt(subtract(power(7, 2), multiply(10, 10)))), 2) | multiply(n2,n2)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|subtract(n1,#3)|divide(#4,n0)| | general |
find the largest 7 digit number which is exactly divisible by 88 ? | "largest 7 digit number is 9999999 after doing 9999999 ÷ 88 we get remainder 31 hence largest 7 digit number exactly divisible by 88 = 9999999 - 31 = 9999968 e" | a ) 9998765 , b ) 9998907 , c ) 9999944 , d ) 9999954 , e ) 9999968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(7, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7)| | general |
in a theater , the first row has 15 seats and each row has 2 more seats than previous row . if the last row has 53 seats , what is the total number of seats in the theater ? | "the number of seats in the theater is 15 + ( 15 + 2 ) + . . . + ( 15 + 38 ) = 20 ( 15 ) + 2 ( 1 + 2 + . . . + 19 ) = 20 ( 15 ) + 2 ( 19 ) ( 20 ) / 2 = 20 ( 15 + 19 ) = 20 ( 34 ) = 680 the answer is c ." | a ) 600 , b ) 640 , c ) 680 , d ) 720 , e ) 760 | c | multiply(divide(add(15, 53), const_2), divide(add(subtract(53, 15), 2), 2)) | add(n0,n2)|subtract(n2,n0)|add(n1,#1)|divide(#0,const_2)|divide(#2,n1)|multiply(#3,#4)| | general |
evaluate : 20 - 12 ÷ 12 × 2 = | "according to order of operations , 12 ÷ 12 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 12 × 2 = 1 × 2 = 2 hence 20 - 12 ÷ 12 × 2 = 20 - 12 = 8 correct answer is b ) 8" | a ) a ) 4 , b ) b ) 8 , c ) c ) 12 , d ) d ) 16 , e ) e ) 20 | b | subtract(20, multiply(multiply(12, 12), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
shipment - - - no . of defective chips / shipment - - - total chips in shipment s 1 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5,000 s 2 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - ... | for a total of 51000 chips ( adding s 1 , s 2 , s 3 , s 4 ) total number of defective chips is 17 ( ( adding defective chips of s 1 , s 2 , s 3 , s 4 ) so ratio is 12 / 48000 or 1 every 4000 chips . keeping this ratio constant for 60000 chips number of defective chips will be ( 1 / 4000 ) * 60000 = 15 e | a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 15 | e | multiply(add(2, 3), 3) | add(n1,n6)|multiply(n6,#0) | general |
three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 276 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 276 x = 23 hcf of 2 x , 3 x and 4 x = x = 23 the answer is b ." | a ) 18 , b ) 23 , c ) 28 , d ) 32 , e ) 36 | b | multiply(2, 4) | multiply(n0,n2)| | other |
if the length of a rectangle is increased by 10 % and the area is unchanged , then its corresponding breadth must be decreased by ? | suppose length = 100 m and breadth = x m area = 100 m new length = 110 m and breadth = ( x - y % of x ) then , \ inline 110 \ times ( x - \ frac { y } { 100 } x ) = 100 \ times x \ inline \ rightarrow \ inline 110 \ times ( 1 - \ frac { y } { 100 } ) = 100 \ inline \ rightarrow \ inline 1 - \ frac { y } { 100 } = \ fra... | ['a ) 9 1 / 19', 'b ) 9 1 / 15', 'c ) 9 1 / 16', 'd ) 9 1 / 11', 'e ) 9 1 / 10'] | e | subtract(const_100, multiply(divide(const_1, add(const_1, divide(10, const_100))), const_100)) | divide(n0,const_100)|add(#0,const_1)|divide(const_1,#1)|multiply(#2,const_100)|subtract(const_100,#3) | geometry |
in a mayoral election , candidate x received 1 / 3 more votes than candidate y , and candidate y received 1 / 7 fewer votes than z . if z received 21,000 votes how many votes did candidate x received ? | "z = 21 - - > y received 1 / 7 fewer votes than z - - > y = z - 1 / 7 * z = 18 ; x received 1 / 3 more votes than y - - > x = y + 1 / 3 * y = 24 . answer : c ." | a ) 18000 , b ) 22000 , c ) 24000 , d ) 26000 , e ) 32000 | c | multiply(multiply(subtract(1, divide(3, 7)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(1, 3), 1)) | divide(n0,n1)|divide(n1,n3)|multiply(const_0_25,const_100)|add(n0,#0)|multiply(#2,const_100)|subtract(n0,#1)|multiply(#4,const_10)|multiply(#6,#5)|multiply(#3,#7)| | general |
each factor of 200 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 42 ? | 200 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 1 / 9 . answer : e . | a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 9 | e | divide(divide(choose(42, const_1), 42), power(const_3, const_2)) | choose(n1,const_1)|power(const_3,const_2)|divide(#0,n1)|divide(#2,#1) | other |
a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 18 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 18 ) x solve for r , r = 66 66 - 18 = 48 mpg e" | a ) 14 , b ) 16 , c ) 21 , d ) 32 , e ) 48 | e | divide(336, divide(subtract(462, 336), 18)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
can v and can в are both right circular cylinders . the radius of can v is twice the radius of can b , while the height of can v is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can v with the same brand of gasoline ? | let x be the radius of b and 2 h be the height of b . therefore , radius of v = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e | ['a ) $ 1', 'b ) $ 2', 'c ) $ 4', 'd ) $ 8', 'e ) $ 16'] | e | multiply(4, divide(multiply(multiply(power(const_2, const_4), const_2), const_2), divide(multiply(multiply(const_2, power(const_2, const_2)), const_4), const_2))) | power(const_2,const_4)|power(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_4)|divide(#5,const_2)|divide(#4,#6)|multiply(n0,#7) | geometry |
x can give y 100 meters start and z 200 meters start in x kilometer race . how much start can y give z in x kilometer race ? | c 111.12 m x runs 1000 m while y runs 900 m and z runs 800 m . the number of meters that z runs when y runs 1000 m , = ( 1000 * 800 ) / 900 = 8000 / 9 = 888.88 m . y can give z = 1000 - 888.88 = 111.12 m . | a ) 112.2 m , b ) 216.3 m , c ) 111.12 m , d ) 213.27 m , e ) 222.40 m | c | divide(multiply(const_1000, subtract(200, 100)), subtract(const_1000, 100)) | subtract(n1,n0)|subtract(const_1000,n0)|multiply(#0,const_1000)|divide(#2,#1) | general |
a sum of money deposited at c . i . amounts to rs . 1890 in 2 years and to rs . 2350 in 3 years . find the rate percent ? | "1890 - - - 460 100 - - - ? = > 24 % answer : b" | a ) 30 % , b ) 24 % , c ) 20 % , d ) 19 % , e ) 50 % | b | multiply(divide(subtract(2350, 1890), 1890), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
what is the least number that should be added to 1053 , so the sum of the number is divisible by 23 ? | "( 1053 / 23 ) gives a remainder 18 so we need to add 5 . the answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(subtract(add(const_1, floor(divide(1053, 23))), divide(1053, 23)), 23) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general |
the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 16000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 16000 + 7000 + 9000 / 5 = rs . 9000 correct option : d" | a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none | d | divide(add(add(add(add(8000, 5000), 16000), 7000), 9000), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
find out the c . i on rs . 3000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | "a = 3000 ( 51 / 50 ) 3 = 3183.62 3000 - - - - - - - - - - - 183.62 answer : a" | a ) 183.62 , b ) 106.07 , c ) 106.04 , d ) 106.03 , e ) 186.01 | a | subtract(multiply(3000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 3000) | divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)| | general |
a number when divided by 20 leaves 6 as a remainder . what will be the remainder if the number is divided by 19 ? | let the minimum whole number be 20 + 6 = 26 where 6 is the remainder . 26 when divided by 19 then leaves 7 . correct answer a . | a ) 7 , b ) 6 , c ) 5 , d ) 11 , e ) 13 | a | reminder(add(20, 6), 19) | add(n0,n1)|reminder(#0,n2) | general |
how many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "let length of tunnel is x meter distance = 800 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 800 + x = ( 65 / 3 ) * 60 800 + x = 20 * 65 = 1300 x = 1300 - 800 = 500 meters answer : option c" | a ) 65 , b ) 69 , c ) 30 , d ) 31 , e ) 32 | c | multiply(multiply(subtract(divide(500, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2)| | physics |
the compound interest earned by sunil on a certain amount at the end of two years at the rate of 5 % p . a . was rs . 492 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned . | "let the sum be rs . p p { [ 1 + 5 / 100 ] 2 - 1 } = 492 p ( 5 / 100 ) ( 2 + 5 / 100 ) = 492 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 492 / ( 0.05 ) ( 2.05 ) = 4800 . amount = 4800 + 492 = rs . 5292 answer : e" | a ) rs . 4935 , b ) rs . 4925 , c ) rs . 5390 , d ) rs . 5290 , e ) rs . 5292 | e | add(divide(492, subtract(power(add(const_1, divide(5, const_100)), const_2), const_1)), 492) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|divide(n1,#3)|add(n1,#4)| | gain |
a train 220 m long is running with a speed of 90 km / hr . in what time will it pass a bus that is running with a speed of 60 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to bus = 90 + 60 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . time taken to pass the bus = 220 * 3 / 125 = 5.28 sec . answer : b" | a ) 4.37 , b ) 5.28 , c ) 6.75 , d ) 8 , e ) 3.1 | b | divide(220, multiply(add(90, 60), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a number is doubled and 9 is added . if resultant is trebled , it becomes 57 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 57 = > 2 x + 9 = 19 = > x = 5 answer : option a" | a ) 5 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(57, 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 |
tim and é lan are 30 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice a meet the 2 : 1 ( tim : elan = 20 : 10 ) ratio of maintaining total distance traveled 30 miles so correct answer a" | a ) 20 miles . , b ) 35 miles . , c ) 45 miles . , d ) 60 miles . , e ) 65 miles . | a | multiply(divide(10, add(5, 10)), 30) | add(n1,n2)|divide(n1,#0)|multiply(n0,#1)| | physics |
2 pipes can separately fill a tank in 20 hrs and 30 hrs respectively . both the pipes are opened tofill the tank , a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? | "1 / 20 + 1 / 30 = 1 / 12 1 + 1 / 3 = 4 / 3 1 - - - 12 4 / 3 - - - ? 4 / 3 * 12 = 16 hrs c" | a ) 10 hrs , b ) 12 hrs , c ) 16 hrs , d ) 18 hrs , e ) 20 hrs | c | divide(add(const_1, divide(const_1, const_3)), add(inverse(20), inverse(30))) | divide(const_1,const_3)|inverse(n1)|inverse(n2)|add(#0,const_1)|add(#1,#2)|divide(#3,#4)| | general |
there are 99 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | "lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 99 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 99 = 15 + 10 + 5 + 3 + x + y + s we need to know tot... | a ) 2 , b ) 4 , c ) 8 , d ) 69 , e ) 32 | d | add(subtract(99, add(add(add(15, 10), 5), 3)), 3) | add(n1,n2)|add(n3,#0)|add(n4,#1)|subtract(n0,#2)|add(n4,#3)| | general |
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 9 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 candidate... | "ans : 252 7 c 1 * 9 c 2 answer e )" | a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 252 | e | multiply(multiply(9, 3), 7) | multiply(n3,n5)|multiply(n1,#0)| | other |
the simple interest on rs . 8 for 12 months at the rate of 5 paise per rupeeper month is | "sol . s . i . = rs . [ 8 * 5 / 100 * 12 ] = rs . 4.8 answer b" | a ) 1.2 , b ) 4.8 , c ) 4.4 , d ) 3.21 , e ) none | b | divide(multiply(multiply(8, 12), 5), const_100) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)| | gain |
an athlete runs 360 metres race in 24 seconds . what is his speed ? | "speed = distance / time = 360 / 20 = 18 m / s = 18 * 18 / 5 = 64.8 km / hr answer : b" | a ) 23 , b ) 64.8 , c ) 37 , d ) 30 , e ) 28 | b | multiply(divide(360, 24), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 4 hours and 36 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 36 : √ 4 = 6 : 2 = 3 : 1 answer : d" | a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 5 , d ) 3 : 1 , e ) 4 : 2 | d | divide(sqrt(36), sqrt(4)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
a batsman scored 120 runs which included 10 boundaries and 6 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 120 - ( 10 x 4 + 6 x 6 ) = 120 - ( 76 ) = 44 now , we need to calculate 60 is what percent of 120 . = > 44 / 120 * 100 = 36.67 % c" | a ) 40 % , b ) 50 % , c ) 36.67 % , d ) 70 % , e ) 75 % | c | multiply(divide(subtract(120, add(multiply(10, 6), multiply(6, 10))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
the sum of the present ages of two persons a and b is 54 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 54 , a = 2 b 2 b + b = 54 = > b = 18 then a = 36 . 5 years , their ages will be 41 and 23 . sum of their ages = 41 + 23 = 64 . answer : b" | a ) 50 , b ) 64 , c ) 70 , d ) 80 , e ) 90 | b | add(add(multiply(divide(54, 5), const_2), 5), add(divide(54, 5), 5)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
if 12 % of x is equal to 6 % of y , then 18 % of x will be equal to how much % of y ? | we have , 12 % of x = 6 % of y = > 2 % of x = 1 % of y = > ( 2 x 9 ) % of x = ( 1 x 9 ) % of y thus , 18 % of x = 9 % of y . answer : b | a ) 7 % , b ) 9 % , c ) 11 % , d ) 12 % , e ) none of these | b | divide(multiply(6, 18), 12) | multiply(n1,n2)|divide(#0,n0) | general |
heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 80 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 80 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 10 liters . answer : e" | a ) 2.8 liters . , b ) 2.5 liters . , c ) 8.5 liters . , d ) 2.6 liters . , e ) 10 liters . | e | divide(multiply(divide(subtract(const_100, 90), const_100), 80), divide(subtract(const_100, 20), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)| | gain |
a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 30 percent of the drive from the center , what percent of the round - trip has the technician completed ? | round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 30 % of 50 % i . e 15 % . so he completed 50 + 15 = 65 % of total trip . b | a ) 60 , b ) 65 , c ) 55 , d ) 50 , e ) 55 | b | add(divide(const_100, const_2), divide(multiply(30, divide(const_100, const_2)), const_100)) | divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2) | gain |
on dividing 127 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 127 - 1 ) / 9 = 126 / 9 = 14 b )" | a ) a ) 12 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | b | floor(divide(127, 9)) | divide(n0,n1)|floor(#0)| | general |
if x + | x | + y = 7 and x + | y | - y = 6 , then x + y = | "if x + | x | + y = 7 and x + | y | - y = 6 , then x + y = can be done in 2.3 mins : there are 4 cases to be tested : 1 ) x is - ve and y is - ve substituting in the equation , we get x - x + y = 7 and x - y - y = 6 solve for x and y we get x = 20 and y = 7 , so x + y = 27 reject 2 ) x is + ve and y is + ve substitute ... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 9 | a | multiply(6, const_2) | multiply(n1,const_2)| | general |
a 300 meter long train crosses a platform in 45 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 45 = 50 / 3 3 ( x + 300 ) = 2250 è x = 450 m . answer : b" | a ) 99 , b ) 450 , c ) 350 , d ) 882 , e ) 281 | b | subtract(multiply(divide(300, 18), 45), 300) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
the ratio between the number of sheep and the number of horses at the stewart farm is 3 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | let the number of sheeps and horses be 3 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 3 x = 3 * 8 = 24 . hence a | a ) 24 , b ) 28 , c ) 32 , d ) 56 , e ) 60 | a | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 3) | multiply(const_2,const_4)|multiply(const_1000,const_12)|multiply(#0,const_10)|multiply(#0,const_100)|add(#2,#3)|add(#4,#1)|divide(#5,n2)|divide(#6,n1)|multiply(n0,#7) | other |
one day a car rental agency rented 3 / 4 of its cars , including 4 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 5 ) ( 3 / 5 ) = 3 / 25 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 3 / 25 ) / ( 1 / 4 ) = 12 / 25 the answer is c ." | a ) 23 / 40 , b ) 17 / 30 , c ) 12 / 25 , d ) 7 / 15 , e ) 3 / 10 | c | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), divide(multiply(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), 4), 4)), subtract(multiply(multiply(4, 4), 5), divide(multiply(3, multiply(multiply(4, 4), 5)), 4))), const_100) | multiply(n1,n2)|multiply(n3,#0)|multiply(n1,#1)|multiply(#1,const_2)|divide(#2,n3)|divide(#3,n1)|multiply(n1,#4)|subtract(#1,#5)|divide(#6,n2)|subtract(#4,#8)|divide(#9,#7)|multiply(#10,const_100)| | general |
if a ( k ) = ( k + 1 ) ^ 2 , and k = 1 , what is the value of a ( a ( a ( a ( k ) ) ) ) ? | a ( a ( a ( a ( 1 ) ) ) ) = a ( a ( a ( 4 ) ) ) = a ( a ( 25 ) ) = a ( 676 ) = 677 ^ 2 = 458329 thus , the answer is d . | a ) 328359 , b ) 557321 , c ) 258327 , d ) 458329 , e ) 4422 | d | power(add(power(add(power(add(power(add(1, 1), 2), 1), 2), 1), 2), 1), 2) | add(n0,n0)|power(#0,n1)|add(n0,#1)|power(#2,n1)|add(n0,#3)|power(#4,n1)|add(n0,#5)|power(#6,n1) | general |
from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 20 percent . what percent of the stock ’ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 20 % = 90 * 1.2 = 108 thus ratio = 108 / 100 = 1.08 ( in % terms = 108 % ) . thus e is the cor... | a ) 80 , b ) 90 , c ) 95 , d ) 100 , e ) 108 | e | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 20), const_100)) | add(n5,const_100)|add(n1,const_100)|subtract(const_100,n3)|divide(#0,const_100)|divide(#2,const_100)|divide(#1,const_100)|multiply(#5,const_100)|multiply(#4,#6)|multiply(#3,#7)| | gain |
a 8 % stock yields 20 % . the market value of the stock is : | "explanation : for an income of rs . 20 , investment = rs . 100 . for an income of rs 8 , investment = rs . 100 / 20 x 8 = rs 40 market value of rs . 100 stock = rs 40 answer is c" | a ) rs 48 , b ) rs 45 , c ) rs 40 , d ) rs 50 , e ) rs 55 | c | multiply(divide(const_100, 20), 8) | divide(const_100,n1)|multiply(n0,#0)| | gain |
the difference between a number and its two fifth is 510 . what is 10 % of that number ? | "solution let the number be x . then , x - 2 / 5 x = 510 . ‹ = › 3 / 5 x = 510 ‹ = › x = [ 510 x 5 / 3 ] = 850 . therefore 10 % of 850 = 85 . answer c" | a ) 12.75 , b ) 50 , c ) 85 , d ) 204 , e ) none | c | power(add(510, const_4), const_4) | add(n0,const_4)|power(#0,const_4)| | gain |
in a certain accounting class of 100 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 65 % , and the students on the make - up date had an average score of 95 % , what wa... | "70 % of the class scored 65 % and 30 % of the class scored 95 % . the difference between 65 % and 95 % is 30 % . the average will be 65 % + 0.3 ( 30 % ) = 74 % . the answer is d ." | a ) 68 % , b ) 70 % , c ) 72 % , d ) 74 % , e ) 77 % | d | divide(add(multiply(70, 65), multiply(95, subtract(100, 70))), 100) | multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|add(#0,#2)|divide(#3,n0)| | general |
the maximum number of student amoung them 1020 pens and 860 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "solution required number of student = h . c . f of 1020 and 860 = 20 . answer a" | a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) none of these | a | gcd(1020, 860) | gcd(n0,n1)| | general |
count the numbers between 10 - 99 which yield a remainder of 3 when divided by 9 and also yield a remainder of 2 when divided by 5 ? | answer = a ) two numbers between 10 - 99 giving remainder 3 when divided by 9 = 12 , 21 , 30 , 39 , 48 , 57 , 66 , 75 , 84 , 93 the numbers giving remainder 2 when divided by 5 = 12 , 57 = 2 | a ) two , b ) five , c ) six , d ) four , e ) one | a | multiply(2, const_1) | multiply(n4,const_1) | general |
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