Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
which number need to add to 8679921 to get a number exactly divisible by 330 ? | 8679921 / 330 = 26302 and reminder = 261 . 330 - 261 = 69 so , the next number divisible by 339 is 69 places in front of 8679921 which means 69 + 8679921 = 8679990 69 should be added to 8679921 a | a ) 69 , b ) 62 , c ) 61 , d ) 63 , e ) 64 | a | multiply(330, subtract(add(floor(divide(8679921, 330)), const_1), divide(8679921, 330))) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3) | general |
at a certain college , 60 percent of the total number of students are freshmen . if 40 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 20 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal ar... | "let ' s say there is a total of 100 students at this college . 60 percent of the total number of students are freshmen . # of freshmen = 60 % of 100 = 60 40 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 40 % of 60 = 24 . . . and , of these , 20 percent ar... | a ) 4 % , b ) 8 % , c ) 4.2 % , d ) 4.8 % , e ) 5 % | d | divide(multiply(divide(multiply(60, 40), const_100), 20), const_100) | multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)| | general |
what annual installment will discharge a debt of rs . 1008 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 = 1008 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1008 ï ƒ › ( 28 x + 31 x + 25 x ) = ( 1008 * 25 ) x = ( 1008 * 25 ) / 84 = rs . 300 . therefore , each installment = rs . 300 . answer is a ." | a ) 300 , b ) 345 , c ) 325 , d ) 335 , e ) none of them | a | multiply(multiply(const_100.0, divide(12, 1008)), 3) | divide(n2,const_100)|multiply(n0,#0)|multiply(n1,#1)| | gain |
what is the smallest integer k for which 64 ^ k > 4 ^ 16 ? | "64 ^ k > 4 ^ 16 4 ^ ( 3 k ) > 4 ^ 16 3 k > 16 k = 6 the answer is c ." | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | add(floor(divide(16, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 1000 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2200 m t = d / s t = 2200 / 10 t = 220 sec answer : a" | a ) 220 sec , b ) 190 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | a | divide(add(1200, 1000), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics |
if t = 5 / 9 * ( k - 32 ) , and if t = 50 , then what is the value of k ? | "k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 50 ) / 5 + 32 = 122 the answer is c ." | a ) 116 , b ) 119 , c ) 122 , d ) 125 , e ) 128 | c | add(divide(multiply(50, 9), 5), 32) | multiply(n1,n3)|divide(#0,n0)|add(n2,#1)| | general |
let q represent a set of 7 distinct prime numbers . if the sum of the numbers in q is even and x is a member of q , then what is the least possible value that x can be ? | q = p 1 + p 2 + p 3 + p 4 + p 5 + p 6 + p 7 = even ( and all primes are distinct ) if the least prime is 2 then we have sum of q = even . ans . c . 3 | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | c | add(divide(7, 7), const_2) | divide(n0,n0)|add(#0,const_2) | general |
how many digits are in the number 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 4 ? | whenever you are asked to find the number of digits , try to bring the number in multiples of 10 . this way , we can wasily calculate the umber of 0 ' s through the powers of 10 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 4 = ( 5 ^ 2 * 2 ) ^ 8 * 2 ^ 9 * 11 ^ 2 * 10 ^ 4 = 5 ^ 16 * 2 ^ 17 * 11 ^ 2 * 10 ^ 4 = 2 * 11 ^ 2 * 10 ^ 16 * 10... | a ) 22 , b ) 20 , c ) 19 , d ) 18 , e ) 17 | d | subtract(multiply(10, 2), const_2) | multiply(n5,n6)|subtract(#0,const_2) | general |
sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 432 per week . how much does she earn in dollars per hour ? | "explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 432 . hourly wage = 432 / 36 = 12 answer : e" | a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 12 | e | divide(432, add(multiply(8, const_3), multiply(6, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)| | physics |
x , y , and z are consecutive numbers and x > y > z . also , 2 x + 3 y + 3 z = 5 y + 11 . what is the value of z ? | "if x , y , and z are consecutive numbers and x > y > z , then y = z + 1 and x = z + 2 . 2 x + 3 y + 3 z = 5 y + 11 2 z + 4 + 3 z + 3 + 3 z = 5 z + 5 + 11 3 z = 9 z = 3 the answer is b ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(subtract(11, subtract(multiply(const_4, 2), multiply(2, 3))), 3) | multiply(n0,const_4)|multiply(n0,n1)|subtract(#0,#1)|subtract(n4,#2)|divide(#3,n1)| | general |
a and b are two circles . the radius of a is 5 times as large as the radius of b . what is the ratio between the areas of the circles ? | given : the radius of a is 5 times as large as the radius of b . = > r ( a ) = 5 r ( b ) . the radius are in ratio of 1 : 5 thus the area will be in the ratio of square of radius . 1 : 25 . hence d . | ['a ) 1 : 8 .', 'b ) 1 : 2 .', 'c ) 1 : 4 .', 'd ) 1 : 25 .', 'e ) 1 : 6 .'] | d | divide(multiply(const_pi, const_1), multiply(const_pi, power(5, const_2))) | multiply(const_1,const_pi)|power(n0,const_2)|multiply(#1,const_pi)|divide(#0,#2) | geometry |
[ ( 3.242 x 10 ) / 100 ] = ? | "answer multiplying 3.242 x 10 = 3.242 now divide 3.242 by 100 so , 3.242 ÷ 100 = 0.03242 ∴ shift the decimal two places to the left as 100 correct option : c" | a ) 0.045388 , b ) 4.5388 , c ) 0.03242 , d ) 473.88 , e ) none of these | c | divide(divide(multiply(3.242, 10), 100), const_10) | multiply(n0,n1)|divide(#0,n2)|divide(#1,const_10)| | general |
pipe a can fill a pool in 12 hours at a constant rate . if pipe a does the job for 8 hours and pipe b does the rest of the job , which works at 1 / 3 constant rate of pipe a . how long will it take for pipe b alone to do the rest of the job ? | pipe a will do 8 / 12 in 8 hrs , so a does 2 / 3 of the work . . therefore , b will do the remaining 1 / 3 rd work alone . . as the speed of b is 1 / 3 rate of a , b will do the 1 / 3 rd work in same time that a takes to complete full job . . . ans 12 a | a ) 12 hrs , b ) 6 hrs , c ) 8 hrs , d ) 10 hrs , e ) 24 hrs | a | divide(subtract(12, 8), inverse(3)) | inverse(n3)|subtract(n0,n1)|divide(#1,#0) | physics |
rs . 1210 were divided among a , b , c so that a : b = 5 : 4 and b : c = 9 : 10 . then , c gets : | "solution a : b = 5 : 4 , b : c = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 90 = 4 : 40 / 9 ∴ a : b : c = 5 : 4 : 40 / 9 = 45 : 36 : 40 . sum of ratio terms = ( 45 + 36 + 40 ) = 121 ∴ c ’ s share = rs . ( 1210 x 40 / 121 ) = rs . 400 . answer b" | a ) rs . 340 , b ) rs . 400 , c ) rs . 450 , d ) rs . 475 , e ) none | b | add(add(divide(subtract(1210, add(add(5, 5), 4)), const_3), 4), 5) | add(n1,n1)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|add(n2,#3)|add(n1,#4)| | general |
two sacks together contained 40 kilograms of sugar . if after 1 kilogram of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 60 % the weight of the sugar in the second , what was the original difference in the weights of the sacks ? | say the weight of the second sack after change is x kilograms , then the weight of the first sack after change would be 0.6 x . since the total weight of sugar in both sacks remained the same then x + 0.6 x = 40 , which gives x = 25 . now , if the weight of the second sack after change is 25 kilograms then initially it... | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | subtract(subtract(divide(40, add(const_1, divide(60, const_100))), 1), subtract(40, subtract(divide(40, add(const_1, divide(60, const_100))), 1))) | divide(n2,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(#2,n1)|subtract(n0,#3)|subtract(#3,#4) | other |
find a positive number , which when increased by 8 is equal to 128 times reciprocal of the number . | explanation : let the number be x . then , x + 8 = 128 x ( 1 / x ) = > x < sup > 2 < / sup > + 8 x - 128 = 0 = > ( x + 16 ) ( x - 8 ) = 0 = > x = 8 answer : option c | a ) 5 , b ) 6 , c ) 8 , d ) 7 , e ) 9 | c | divide(subtract(sqrt(add(multiply(128, const_4), power(8, const_2))), 8), const_2) | multiply(n1,const_4)|power(n0,const_2)|add(#0,#1)|sqrt(#2)|subtract(#3,n0)|divide(#4,const_2) | general |
find the value of x : 2 a ^ 2 - 7 a + 10 = 14 | "using the elimination method substitute options for x and find the correct option . answer : d" | a ) 7 , b ) 5 , c ) 6 , d ) 4 , e ) 2 | d | divide(add(7, 2), power(2, 2)) | add(n0,n2)|power(n0,n1)|divide(#0,#1)| | general |
sheela deposits rs . 3400 in bank savings account . if this is 15 % of her monthly income . what is her monthly income in ? | "explanation : 15 % of income = rs . 3400 100 % of income = 3400 x 100 / 15 = rs . 22666 answer : a" | a ) 22666 , b ) 20000 , c ) 25000 , d ) 22235 , e ) none of these | a | divide(multiply(3400, const_100), 15) | multiply(n0,const_100)|divide(#0,n1)| | gain |
when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of r ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a pers... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,n3)| | general |
a number when multiplied by 3 is odd and divisible by 9 . this number when multiplied by 4 is 108 . what is the original number ? | working backward , 108 divided by 4 is 27 . this number fits the rules of three and nine , where the digits add up to a sum divisible by nine . answer is e . | a ) 10 , b ) 18 , c ) 13 , d ) 21 , e ) 27 | e | divide(108, 4) | divide(n3,n2) | general |
if there is one larger cube with surface area 600 and no . of smaller cubes with surface area 24 . then how many small cubes are required to make a larger cube with surface area of 600 ? | the ratio between larger and smaller is 6 a 12 : 6 a 22 = 600 : 24 a 12 : a 22 = 100 : 4 a 1 : a 2 = 10 : 2 then length of cube is 10 and 2 so make a cube with length 10 , we have to 5 small cubes each in length , breadth and height . so total 125 cubes are required . then the answer is b | ['a ) 5', 'b ) 125', 'c ) 4', 'd ) 100', 'e ) 24'] | b | volume_cube(divide(sqrt(divide(600, divide(600, const_100))), sqrt(divide(24, divide(600, const_100))))) | divide(n0,const_100)|divide(n0,#0)|divide(n1,#0)|sqrt(#1)|sqrt(#2)|divide(#3,#4)|volume_cube(#5) | geometry |
in the coordinate plane , line a has a slope of - 1 and an x - intercept of 1 . line b has a slope of 5 and a y - intercept of - 5 . if the two lines intersect at the point ( a , b ) , what is the sum a + b ? | "the equation of line a is y = - x + 1 the equation of line b is y = 5 x - 5 5 x - 5 = - x + 1 x = 1 y = 0 the point of intersection is ( 1,0 ) and then a + b = 1 . the answer is b ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(subtract(subtract(5, 5), 1), 1) | subtract(n3,n2)|subtract(#0,n0)|divide(#1,n1)| | general |
the interest on a certain deposit at 4.5 % p . a . is rs . 405 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a . ? | s . i . = rs . 405 r = 4.5 % t = 1 year principal = ( 100 * 405 ) / ( 4.5 * 1 ) = rs . 9000 s . i at 5 % interest = ( 9000 * 5 * 1 ) / 100 = rs . 450 difference in interest = 450 – 405 = rs . 45 answer : b | a ) rs . 50 , b ) rs . 45 , c ) rs . 40.5 , d ) rs . 48.5 , e ) rs . 49.5 | b | subtract(divide(multiply(divide(405, divide(4.5, const_100)), 5), const_100), 405) | divide(n0,const_100)|divide(n1,#0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1) | gain |
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 16 cm . what is the length of the wire ? | volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 16 ) 2 * h = ( 4 / 3 ) π ( 12 ) 3 = > h = 9 cm answer : e | a ) 45 cm , b ) 18 cm , c ) 90 cm , d ) 180 cm , e ) none of these | e | divide(multiply(const_4, divide(power(12, const_3), power(16, const_2))), const_3) | power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3) | physics |
approximately how many cubic feet of water are needed to fill a circular swimming pool that is 100 feet across and 4 feet deep ? | "answer should be b . v = \ pir ^ 2 h = \ pi * 50 ^ 2 * 4 = approximately 30000" | a ) 15000 , b ) 30000 , c ) 50000 , d ) 70000 , e ) 90000 | b | volume_cylinder(divide(100, const_2), 4) | divide(n0,const_2)|volume_cylinder(#0,n1)| | geometry |
in a box of 12 pens , a total of 4 are defective . if a customer buys 3 pens selected at random from the box , what is the probability that neither pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 4 defective in a lot of 12 pens i . e . probability of first pen not being defective = ( 8 / 12 ) i . e . probability of second pen not being defective = ( 7 / 11 ) i . e . probability of third pen not being defective = ( 6 / 10 ) probability of both pen being non - defect... | a ) 1 / 6 , b ) 2 / 55 , c ) 14 / 55 , d ) 12 / 55 , e ) 3 / 4 | c | multiply(divide(subtract(12, 4), 12), divide(subtract(subtract(12, 4), const_1), subtract(12, const_1))) | subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)| | general |
two friends plan to walk along a 33 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 20 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ? | "if q complete x kilometers , then p completes 1.2 x kilometers . x + 1.2 x = 33 2.2 x = 33 x = 15 then p will have have walked 1.2 * 15 = 18 km . the answer is d ." | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | multiply(divide(33, add(add(divide(20, const_100), const_1), const_1)), add(divide(20, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n0,#2)|multiply(#1,#3)| | gain |
the number of degrees that the hour hand of a clock moves through between noon and 2.30 in the afternoon of the same day is | "explanation : the hour hand moves from pointing to 12 to pointing to half way between 2 and 3 . the angle covered between each hour marking on the clock is 360 / 12 = 30 . since the hand has covered 2.5 of these divisions the angle moved through is 75 . answer : c" | a ) 720 , b ) 180 , c ) 75 , d ) 65 , e ) none of these | c | multiply(multiply(add(floor(2.30), multiply(divide(subtract(2.30, floor(2.30)), const_60), const_100)), const_3), const_10) | floor(n0)|subtract(n0,#0)|divide(#1,const_60)|multiply(#2,const_100)|add(#0,#3)|multiply(#4,const_3)|multiply(#5,const_10)| | physics |
a train running at the speed of 90 km / hr crosses a pole in 15 seconds . what is the length of the train ? | "speed = ( 90 x ( 5 / 18 ) m / sec = ( 25 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 25 ) x 15 ) m = 375 m c" | a ) 325 , b ) 350 , c ) 375 , d ) 400 , e ) 425 | c | multiply(divide(multiply(90, const_1000), const_3600), 15) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
two cars a and b start from boston and new york respectively simultaneously and travel towards each other at constant speeds along the same route . after meeting at a point between boston and new york the two cars a and b proceed to their respective destinations of new york and boston . car a reaches new york 60 minute... | "both cars leave at the same time both cars travel at constant speed stealing a useful piece of information from paragkan : if two objects a and b start from opposite points and , after having met en route , reach their respective destinations in a and b mins ( or any other measure of time ) respectively , then the rat... | a ) 1 hour , b ) 1 hour 10 minutes , c ) 2 hours 30 minutes , d ) 2 hours , e ) 2 hours 10 minutes | d | add(multiply(sqrt(divide(60, 90)), 90), 60) | divide(n0,n1)|sqrt(#0)|multiply(n1,#1)|add(n0,#2)| | physics |
four numbers are in the ratio 2 : 3 : 4 : 5 add up to give a sum of 1344 . find the biggest number . | answer : a | a ) 480 , b ) 239 , c ) 270 , d ) 282 , e ) 112 | a | add(multiply(multiply(2, 4), const_100), multiply(3, 4)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
five machines , each working at the same constant rate , together can complete a certain job in 20 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 10 days ? | rate of one machine = 1 job / ( 20 * 5 ) days let x = number of machines needed to complete the job in 10 days 1 / ( 5 * 20 ) * 10 * x = 1 job x = 10 10 - 5 = 5 answer : d | a ) 4 , b ) 6 , c ) 8 , d ) 5 , e ) 10 | d | subtract(divide(multiply(20, add(const_4, const_1)), 10), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,n1)|subtract(#2,#0) | physics |
find ( 7 x + 4 y ) / ( x - 2 y ) if x / 2 y = 5 / 2 ? | "x / 2 y = 3 / 2 = > x = 10 y / 2 = 5 y = > ( 7 x + 4 y ) / ( x - 2 y ) = ( ( 7 * ( 5 y ) ) + 4 y ) / ( 5 y - 2 y ) = > 39 y / 3 y = 13 answer : a" | a ) 13 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | a | divide(add(multiply(multiply(divide(5, 2), 2), 7), 4), subtract(multiply(divide(5, 2), 2), 2)) | divide(n4,n2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#1,n2)|add(n1,#2)|divide(#4,#3)| | general |
there is 60 % increase in an amount in 6 years at si . what will be the ci of rs . 11,000 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 . 11000 . t = 3 years and r = 10 % p . a . c . i . = rs . 11000 x 1 + 10 3 - 1 100 = rs . 11000 x 331 1000 = 3641 . a" | a ) 3641 , b ) 2572 , c ) 4542 , d ) 2343 , e ) 3972 | a | 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 |
a train travels from albany to syracuse , a distance of 120 miles , at the average rate of 40 miles per hour . the train then travels back to albany from syracuse . the total travelling time of the train is 5 hours and 24 minutes . what was the average rate of speed of the train on the return trip to albany ? | "40 * t = 120 therefore t = 120 / 40 = 3 on return speed * ( 5.4 - 3 ) = 120 therefore t = 120 / 2.4 = 50 e" | a ) 45 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | e | speed(120, subtract(add(5, divide(24, const_60)), divide(120, 40))) | divide(n3,const_60)|divide(n0,n1)|add(n2,#0)|subtract(#2,#1)|speed(n0,#3)| | physics |
the ratio between the present ages of a and b is 4 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ? | let the present ages of a and b be 4 x and 3 x years respectively . then , ( 4 x - 4 ) / ( 3 x + 4 ) = 1 / 1 1 x = 8 = > x = 8 required ratio = ( 4 x + 4 ) : ( 3 x - 4 ) = 36 : 20 = 9 : 5 . answer : e | a ) 3 : 4 , b ) 3 : 0 , c ) 3 : 1 , d ) 3 : 2 , e ) 9 : 5 | e | divide(add(multiply(4, divide(add(4, 4), subtract(4, 3))), 4), subtract(multiply(3, divide(add(4, 4), subtract(4, 3))), 4)) | add(n0,n0)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n0,#3)|subtract(#4,n0)|divide(#5,#6)| | other |
find value of x ( 11 ^ 2 + 12 ^ 2 + 13 ^ 2 + . . . + 20 ^ 2 ) = x | ( 11 ^ 2 + 12 ^ 2 + 13 ^ 2 + . . . + 20 ^ 2 ) = ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 20 ^ 2 ) - ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 10 ^ 2 ) 6 = 20 x 21 x 41 - 10 x 11 x 21 6 6 = ( 2870 - 385 ) = 2485 e | a ) 1200 , b ) 1340 , c ) 1234 , d ) 1203 , e ) 2485 | e | subtract(divide(multiply(multiply(20, add(20, const_1)), add(multiply(2, 20), const_1)), multiply(const_2, const_3)), divide(multiply(multiply(const_10, add(const_10, const_1)), add(multiply(2, const_10), const_1)), multiply(const_2, const_3))) | add(n6,const_1)|add(const_1,const_10)|multiply(n1,n6)|multiply(const_2,const_3)|multiply(n1,const_10)|add(#2,const_1)|add(#4,const_1)|multiply(n6,#0)|multiply(#1,const_10)|multiply(#5,#7)|multiply(#6,#8)|divide(#9,#3)|divide(#10,#3)|subtract(#11,#12) | general |
in a mixture 60 litres , the ratio of milk and water 2 : 1 . if the this ratio is to be 1 : 2 , then the quanity of water to be further added is | "explanation : quantity of milk = 60 * ( 2 / 3 ) = 40 liters quantity of water = 60 - 40 = 20 liters as per question we need to add water to get quantity 2 : 1 = > 40 / ( 20 + x ) = 1 / 2 = > 20 + x = 80 = > x = 60 liters option d" | a ) 20 liters , b ) 30 liters , c ) 50 liters , d ) 60 liters , e ) none of these | d | subtract(multiply(subtract(60, divide(60, add(2, 1))), 2), divide(60, add(2, 1))) | add(n1,n2)|divide(n0,#0)|subtract(n0,#1)|multiply(n1,#2)|subtract(#3,#1)| | general |
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,600 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | "median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.6 so , answer is d ." | a ) 38,000 , b ) 40,000 , c ) 42,600 , d ) 45,600 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | multiply(const_100,const_100)|multiply(#0,const_4)| | general |
if m and n are positive integers and m ^ 2 + n ^ 2 = 58 , what is the value of m ^ 3 + n ^ 3 ? | "you need to integers which squared are equal 58 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 4 ^ 2 = 16 5 ^ 2 = 25 6 ^ 2 = 36 7 ^ 2 = 49 stop . the integers ca n ' t be greater than 7 or we will score above 58 . the second integer need to be picked up the same way . 1 ^ 2 ... | a ) 72 , b ) 224 , c ) 370 , d ) 512 , e ) 1,600 | c | gcd(58, const_4) | gcd(n2,const_4)| | general |
a circular ground whose radius is 15 metres , has a 1.2 metre - broad garden around it . what is the area of the garden in square metres ? | req . area = ï € [ ( 16.2 ) 2 â € “ ( 15 ) 2 ] = 22 â „ 7 ã — ( 31.2 ã — 1.2 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 22 ã — 31.2 ã — 1.2 ) / 7 = 117.67 sq m answer c | ['a ) 118.67', 'b ) 116.67', 'c ) 117.67', 'd ) 120.56', 'e ) 120.57'] | c | subtract(circle_area(add(15, 1.2)), circle_area(15)) | add(n0,n1)|circle_area(n0)|circle_area(#0)|subtract(#2,#1) | geometry |
rahim bought 65 books for rs . 1160 from one shop and 50 books for rs . 920 from another . what is the average price he paid per book ? | average price per book = ( 1160 + 920 ) / ( 65 + 50 ) = 2080 / 115 = rs . 18.08 answer : c | a ) rs . 18.28 , b ) rs . 20 , c ) rs . 18.08 , d ) rs . 28.08 , e ) rs . 38.08 | c | divide(add(1160, 920), add(65, 50)) | add(n1,n3)|add(n0,n2)|divide(#0,#1) | general |
the ratio of the volumes of two cubes is 512 : 729 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 512 : ³ √ 729 = 8 : 9 ratio of surface areas = 82 : 92 = 41 : 46 answer : c" | a ) 1 : 22 , b ) 81 : 12 , c ) 41 : 46 , d ) 41 : 19 , e ) 84 : 12 | c | power(divide(512, 729), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
x starts a business with rs . 45000 . y joins in the business after 7 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 ã — 12 : 30000 ã — 5 = 45 ã — 12 : 30 ã — 5 = 3 ã — 12 : 2 ã — 5 = 18 : 5 answer : option b" | a ) 1 : 2 , b ) 18 : 5 , c ) 1 : 5 , d ) 3 : 1 , e ) 1 : 1 | b | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
if rs . 782 be divided into 3 parts , proportional to 1 / 2 : 1 / 3 : 3 / 4 , then the first part is ? | given ratio = 1 / 2 : 2 / 3 : 3 / 4 = 6 : 4 : 9 1 st part = 782 * 6 / 19 = rs . 247 answer : c | a ) a ) 426 , b ) b ) 526 , c ) c ) 247 , d ) d ) 204 , e ) e ) 297 | c | multiply(divide(divide(1, 2), add(add(divide(1, 2), divide(1, 3)), divide(3, 4))), 782) | divide(n2,n3)|divide(n2,n1)|divide(n1,n7)|add(#0,#1)|add(#3,#2)|divide(#0,#4)|multiply(n0,#5) | general |
a palindrome is a word or a number that reads the same forward and backward . for example , 2442 and 111 are palindromes . if 5 - digit palindromes are formed using one or more of the digits 1 , 2 , 3 , 4 , 5 , 6 , and 7 , how many palindromes are possible ? | "there are 7 choices for each of the first three digits . the number of possible palindromes is 7 ^ 3 = 343 . the answer is e ." | a ) 35 , b ) 49 , c ) 105 , d ) 255 , e ) 343 | e | power(5, 3) | power(n2,n5)| | general |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 63.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 63.50 32 x - 31 x = 6350 * 4 x = 25,400 . answer : e" | a ) 24,602 , b ) 28,888 , c ) 24,600 , d ) 24,628 , e ) 24,400 | e | divide(63.50, divide(const_4, 4)) | divide(const_4,n3)|divide(n4,#0)| | gain |
in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 5 nches . find the % correction made by the candidate from his stated height to his actual height ? | "his height was = 5 feet 5 inch = 5 + 60 = 65 inch . required % correction = 65 * ( 1.25 - 1 ) * 100 = 20 b" | a ) 10 , b ) 20 , c ) 30 , d ) 35 , e ) 43 | b | multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100) | add(const_0_25,const_0_25)|add(n0,#0)|subtract(#1,n1)|add(#2,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 6 t seconds , then what is the length of the platform ? | "the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 6 t ( l + p ) / ( l / t ) = 6 t p = 5 l the answer is c ." | a ) l , b ) 3 l , c ) 5 l , d ) 7 l , e ) 9 l | c | subtract(6, const_1) | subtract(n0,const_1)| | physics |
a bus 15 m long is running with a speed of 40 km / hr . in what time will it pass a man who is skateboarding at 8 km / hr in the direction opposite to that in which the car is going ? | speed of bus relative to man = 40 + 8 = 48 km / hr . = 48 * 5 / 18 = 40 / 3 m / sec . time taken to pass the man = 5 * 3 / 40 = 1.125 sec . answer : a | a ) 1.125 , b ) 1.5 , c ) 1.75 , d ) 1.25 , e ) 1 | a | divide(divide(multiply(15, const_3600), add(40, 8)), const_1000) | add(n1,n2)|multiply(n0,const_3600)|divide(#1,#0)|divide(#2,const_1000) | physics |
the end of a blade on an airplane propeller is 10 feet from the center . if the propeller spins at the rate of 660 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | distance traveled in 1 revolution = 2 π r = 2 π 10 / 5280 revolutions in one second = 660 revolutions in 60 seconds ( one minute ) = 660 * 60 total distance traveled = total revolutions * distance traveled in one revolution 660 * 60 * 2 π 10 / 5280 = 150 π a is the answer | a ) 150 π , b ) 240 π , c ) 300 π , d ) 480 π , e ) 1,200 π | a | multiply(multiply(multiply(multiply(divide(10, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 660), const_60) | add(const_3,const_4)|add(const_2,const_3)|multiply(const_10,const_2)|multiply(const_100,const_2)|add(#2,const_2)|multiply(#1,const_1000)|add(#3,#5)|divide(#4,#0)|divide(n0,#6)|multiply(#8,const_2)|multiply(#7,#9)|multiply(n1,#10)|multiply(#11,const_60) | physics |
victor gets 92 % marks in examinations . if these are 368 marks , find the maximum marks . | "let the maximum marks be m then 92 % of m = 368 ⇒ 92 / 100 × m = 368 ⇒ m = ( 368 × 100 ) / 92 ⇒ m = 36800 / 92 ⇒ m = 400 therefore , maximum marks in the examinations are 400 . answer : c" | a ) 334 , b ) 500 , c ) 400 , d ) 288 , e ) 271 | c | divide(368, divide(92, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
a train running at the speed of 300 km / hr crosses a pole in 15 seconds . what is the length of the train ? | "speed = ( 300 * 5 / 18 ) m / sec = ( 250 / 3 ) m / sec length of the train = ( speed x time ) = ( 250 / 3 * 15 ) m = 1250 m . answer : b" | a ) 1200 m , b ) 1250 m , c ) 1300 m , d ) 1350 m , e ) 1450 m | b | multiply(divide(multiply(300, const_1000), const_3600), 15) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a starts business with rs . 3500 and after 6 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b â € ™ s contribution in the capital ? | "explanation : a invested rs . 3500 for 12 months . let b joined with investment x . and he invested for 12 - 6 = 6 months . so there profit ratio = ( 3500 ã — 12 ) : ( 6 x ) = 2 : 3 â ‡ ’ x = 10500 answer : d" | a ) 24887 , b ) 20778 , c ) 23788 , d ) 10500 , e ) 2811 | d | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 6), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)| | other |
the population of a village is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 × 120 / 100 × 120 / 100 = 14400 e" | a ) 10000 , b ) 12000 , c ) 13000 , d ) 14000 , e ) 14400 | e | multiply(10000, power(add(const_1, divide(20, const_100)), 2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain |
in a class of 30 students , 5 students did not borrow any books from the library , 12 students each borrowed 1 book , 8 students each borrowed 2 books , and the rest of the students each borrowed at least 3 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum numbe... | the average number of books per student was 2 means that total of 2 * 30 = 60 books were borrowed ; 5 + 12 + 8 = 25 students borrowed total of 5 * 0 + 12 * 1 + 8 * 2 = 28 books ; so 60 - 28 = 32 books are left to distribute among 30 - 25 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe n... | a ) 5 , b ) 8 , c ) 15 , d ) 18 , e ) 20 | e | add(subtract(multiply(30, 2), add(add(12, multiply(8, 2)), multiply(5, 3))), 3) | multiply(n0,n5)|multiply(n4,n5)|multiply(n1,n6)|add(n2,#1)|add(#3,#2)|subtract(#0,#4)|add(n6,#5) | general |
two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 4 carpenters build in 3 hours ? | "w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 × r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 4 carpenters in 3 hrs = 4 × rate of each carpenter x time = 4 × 2 / 5 × 3 = 4.8 desks c is the correct answer ." | a ) 2.4 . , b ) 3.6 . , c ) 4.8 . , d ) 5.5 . , e ) 6.4 | c | multiply(multiply(divide(divide(2, divide(3, 2)), 2), 4), 3) | divide(n2,n0)|divide(n0,#0)|divide(#1,n0)|multiply(n1,#2)|multiply(n2,#3)| | physics |
mr . kutty has only hens and sheep . if the total number of their heads is 38 and the total number of legs is 100 then what is the ratio between the numbers of hens and sheep ? | let the total number of hens and sheep be x and y respectively . i . e . , x + y = 38 and 2 x + 4 y = 100 ∴ ratio = 13 : 6 answer d | a ) 2 : 1 , b ) 1 : 2 , c ) 6 : 13 , d ) 13 : 6 , e ) none of these | d | divide(subtract(38, divide(subtract(100, multiply(38, const_2)), const_2)), divide(subtract(100, multiply(38, const_2)), const_2)) | multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_2)|subtract(n0,#2)|divide(#3,#2) | other |
a boat covers a certain distance downstream in 1 hour , while it comes back in 1.30 hours . if the speed of the stream be 3 kmph , what is the speed of the boat in still water ? | "let the speed of the water in still water = x given that speed of the stream = 3 kmph speed downstream = ( x + 3 ) kmph speed upstream = ( x − 3 ) kmph he travels a certain distance downstream in 1 hour and come back in 1 1 ⁄ 2 hour . i . e . , distance travelled downstream in 1 hour = distance travelled upstream in 1... | a ) 10 , b ) 13 , c ) 11 , d ) 08 , e ) 15 | e | divide(add(multiply(divide(3, const_2), 3), 3), subtract(divide(3, const_2), 1)) | divide(n2,const_2)|multiply(n2,#0)|subtract(#0,n0)|add(n2,#1)|divide(#3,#2)| | physics |
if x is a positive number and 1 / 4 the square root of x is equal to 3 x , then x = | "1 / 4 of sqrt ( x ) = 3 x , which means that sqrt ( x ) = 12 x or x = 144 x ^ 2 - > divide by x 1 = 144 x x = 1 / 144 e ." | a ) 1 / 16 , b ) 12 , c ) 1 / 12 , d ) 144 , e ) 1 / 144 | e | power(4, multiply(4, const_3)) | multiply(n1,const_3)|power(n1,#0)| | general |
a library has an average of 510 visitors on sundays and 240 on other day . the average number of visitors in a month of 15 days starting with sunday is | "explanation : as the month begin with sunday , so there will be five sundays in the month . so result will be : = ( 510 × 5 + 240 × 25 / 15 ) = ( 8550 / 15 ) = 570 answer : option e" | a ) 280 , b ) 285 , c ) 290 , d ) 295 , e ) 570 | e | divide(add(multiply(add(floor(divide(15, add(const_3, const_4))), const_1), 510), multiply(subtract(15, add(floor(divide(15, add(const_3, const_4))), const_1)), 240)), 15) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
in how many ways can 11 books on english and 9 books on french be placed in a row on a shelf so that two books on french may not be together ? | we have 11 english and 9 french books , no french books should be adjacent . imagine 11 english books in a row and empty slots like below : * e * e * e * e * e * e * e * e * e * e * e * now if 9 french books would be placed in 12 empty slots , all french books will be separated by english books . so we can ` ` choose '... | a ) 220 . , b ) 240 , c ) 260 , d ) 280 , e ) 300 | a | add(add(choose(subtract(11, const_1), const_3), permutation(subtract(11, const_1), const_2)), choose(subtract(11, const_1), const_1)) | subtract(n0,const_1)|choose(#0,const_3)|choose(#0,const_1)|permutation(#0,const_2)|add(#1,#3)|add(#4,#2) | probability |
if 35 percent of 400 is 20 percent of x , then x = | "5 / 100 ( 400 ) = 2 / 10 ( x ) x = 700 . . answer : c" | a ) 200 , b ) 350 , c ) 700 , d ) 900 , e ) 1,400 | c | divide(multiply(35, 400), 20) | multiply(n0,n1)|divide(#0,n2)| | general |
the side of a rhombus is 26 m and length of one of its diagonals is 20 m . the area of the rhombus is ? | 262 – 102 = 242 d 1 = 20 d 2 = 48 1 / 2 * 20 * 48 = 480 answer : d | ['a ) 488', 'b ) 429', 'c ) 418', 'd ) 480', 'e ) 410'] | d | divide(multiply(20, sqrt(subtract(power(multiply(const_2, 26), const_2), power(20, const_2)))), const_2) | multiply(n0,const_2)|power(n1,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|multiply(n1,#4)|divide(#5,const_2) | geometry |
80 370 ? 1550 2440 3530 | "10 ^ 2 - 20 = 80 20 ^ 2 - 30 = 370 30 ^ 2 - 40 = 860 40 ^ 2 - 50 = 1550 50 ^ 2 - 60 = 2440 60 ^ 2 - 70 = 3530 . answer : d" | a ) 900 , b ) 840 , c ) 750 , d ) 860 , e ) 960 | d | divide(multiply(add(80, divide(370, 1550)), 2440), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | general |
how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 4 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 4 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 4 % of the solution . set up an inequality and solve for x : 1.5 ... | a ) 35 , b ) 33 1 / 3 , c ) 21 , d ) 16 2 / 3 , e ) 15 | c | subtract(50, multiply(divide(50, const_100), 4)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
three coins are tossed . find the probability of at least 1 head and 1 tail ? | "n ( s ) = 2 ^ 3 = 8 . . . . ( 3 coins tossed ) let e is the event of getting at least 1 head and 1 tail n ( e ) = tth , htt , tht , hht , hth , thh = 6 p ( e ) = n ( e ) / n ( s ) = 6 / 8 = 3 / 4 ans - c" | a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 7 / 8 , e ) 1 / 8 | c | multiply(const_4, multiply(divide(const_1, 1), divide(const_1, 1))) | divide(const_1,n0)|multiply(#0,#0)|multiply(#1,const_4)| | probability |
10 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "though 2 * ( 10 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 9 , so he ' ll play 18 matches . similarly , second guy has to play with the rest of 8 as his 2 games with the ... | a ) 90 , b ) 98 , c ) 35 , d ) 99 , e ) 80 | a | multiply(10, subtract(10, const_1)) | subtract(n0,const_1)|multiply(n0,#0)| | general |
john is building a rectangular fence around a garden . the length will be 5 feet shorter than its width . if the total area of the garden is 176 square feet , what is the length of the diagonal ( in feet ) ? | to solve this , we first need to set up an equation for the area of the garden . if x is the width , then we have x ( x - 5 ) = 176 . by putting the equation in standard form , we get x ^ 2 - 5 x - 176 = 0 . by using the quadratic formula , we get roots of - 11 and 16 . we know that x is the width , and x - 5 is the le... | ['a ) between 18 and 19', 'b ) between 19 and 20', 'c ) between 20 and 21', 'd ) between 21 and 22', 'e ) between 22 and 23'] | b | sqrt(add(power(add(const_10, const_1), const_2), power(add(const_10, add(5, const_1)), const_2))) | add(const_1,const_10)|add(n0,const_1)|add(#1,const_10)|power(#0,const_2)|power(#2,const_2)|add(#3,#4)|sqrt(#5) | geometry |
divide $ 450 among a , b in the ratio 1 : 4 . how many $ that a get ? | "sum of ratio terms = 1 + 4 = 5 a = 450 * 1 / 5 = $ 90 answer is a" | a ) $ 90 , b ) $ 500 , c ) $ 150 , d ) $ 250 , e ) $ 600 | a | divide(450, 1) | divide(n0,n1)| | other |
88 % of 370 + 24 % of 210 - ? = 118 | "let 88 % of 370 + 240 % of 210 - x = 118 then , x = ( 88 / 100 * 370 ) + ( 24 / 100 * 210 ) - 118 - 325.60 + 50.40 - 118 = 376 - 118 = 258 correct option : b" | a ) 256 , b ) 258 , c ) 268 , d ) 358 , e ) none of these | b | divide(multiply(88, 370), 24) | multiply(n0,n1)|divide(#0,n2)| | general |
a train passes a platform in 22 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "speed of the train = 54 km / hr = ( 54 × 10 ) / 36 m / s = 15 m / s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 12 = > 300 + l = 15 × 22 = 330 = > l = 330 - 300 = 30 meter answe... | a ) 40 , b ) 50 , c ) 60 , d ) 30 , e ) 20 | d | multiply(multiply(const_0_2778, 54), subtract(22, 20)) | multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)| | physics |
in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 184 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 184 = > 4 x / 10 = 184 = > 4 x = 1840 = > x = 460 . answer : e" | a ) 430 , b ) 438 , c ) 436 , d ) 434 , e ) 460 | e | divide(184, divide(subtract(70, subtract(const_100, 70)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts four hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 4 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 *... | a ) 3 , b ) 4.5 , c ) 8 , d ) d ) 5.5 , e ) e ) 5 | c | subtract(40, 30) | subtract(n1,n0)| | physics |
a and b together can complete work in 5 days . a alone starts working and leaves it after working for 3 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ? | "explanation : ( a + b ) one day work = 1 / 5 now a does half of the work in 3 days so a can complete the whole work in 6 days a ’ s one day work = 1 / 6 b ’ s one day work = 1 / 5 - 1 / 6 = 1 / 30 b alone can complete the work in 30 days so half of the work in 15 days answer : option c" | a ) 4 , b ) 8 , c ) 15 , d ) 6 , e ) 7 | c | multiply(5, divide(3, 5)) | divide(n1,n0)|multiply(n0,#0)| | physics |
a rectangular plot measuring 15 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | "perimeter of the plot = 2 ( 15 + 50 ) = 130 m no of poles = 130 / 5 = 26 m answer : b" | a ) 46 m , b ) 26 m , c ) 26 m , d ) 56 m , e ) 25 m | b | divide(multiply(add(15, 50), const_2), 5) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)| | physics |
if a train , travelling at a speed of 270 kmph , crosses a pole in 5 sec , then the length of train is ? | "d d = 270 * 5 / 18 * 5 = 375 m" | a ) 281 m , b ) 325 m , c ) 117 m , d ) 375 m , e ) 350 m | d | multiply(multiply(270, const_0_2778), 5) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
a certain car uses one gallon of gasoline every 36 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | "4 miles on the highway = 4 / 36 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 36 + 4 / 20 = 14 / 45 gallons . 8 miles on the highway = 8 / 30 gallons . the % change = ( 14 / 45 - 8 / 36 ) / ( 8 / 36 ) = 0.40 . answer : e ." | a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 25 % , e ) 40 % | e | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 36))), multiply(8, divide(const_1, 36))), multiply(8, divide(const_1, 36))), const_100) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n4,#1)|add(#2,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)| | general |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along theoundary of the park at the speed of 12 km / hr completes one round in 8 min , then the area of the park ( in sq . m ) is ? | "perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1600 or x = 160 . length = 480 m and breadth = 320 m . area = ( 480 x 320 ) m 2 = 153600 m c" | a ) 145010 m , b ) 146400 m , c ) 153600 m , d ) 164500 m , e ) 167550 m | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
the calendar of the year 2060 can be used again in the year ? | "explanation : given year 2060 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2060 + 28 = 2088 answer : c" | a ) 2081 , b ) 2083 , c ) 2088 , d ) 1971 , e ) 1973 | c | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2060) | multiply(const_4,const_4)|subtract(#0,const_2)|multiply(#1,const_2)|add(n0,#2)| | gain |
mixture a is 40 percent alcohol , and mixture b is 80 percent alcohol . if the two are poured together to create a 4 - gallon mixture that contains 50 percent alcohol , approximately how many gallons of mixture a are in the mixture ? | "( 80 - 50 ) / ( 50 - 40 ) = qa / qb 30 / 10 = qa / qb 3 / 1 = qa / qb qa = ( 3 / 5 ) * 4 = 12 / 5 = 2.4 approx answer : e" | a ) 1.5 , b ) 1.7 , c ) 2.3 , d ) 2.5 , e ) 2.4 | e | divide(divide(multiply(40, 80), 80), const_2) | multiply(n0,n1)|divide(#0,n1)|divide(#1,const_2)| | general |
the length of each side of square a is increased by 100 percent to make square b . if the length of the side of square b is increased by 75 percent to make square c , by what percent is the area of square c greater than the sum of the areas of squares a and b ? | "let length of each side of square a be 10 area of a = 10 ^ 2 = 100 since , length of each side of square a is increased by 100 percent to make square b length of each side of square b = 2 * 10 = 20 area of b = 20 ^ 2 = 400 since , length of the side of square b is increased by 75 percent to make square c length of eac... | a ) 75 % , b ) 80 % , c ) 145 % , d ) 150 % , e ) 180 % | c | multiply(divide(subtract(subtract(square_area(add(add(const_1, divide(100, 100)), divide(multiply(add(const_1, divide(100, 100)), 75), 100))), const_1), const_4), add(const_1, square_area(add(const_1, divide(100, 100))))), 100) | divide(n0,n0)|add(#0,const_1)|multiply(n1,#1)|square_area(#1)|add(#3,const_1)|divide(#2,n0)|add(#1,#5)|square_area(#6)|subtract(#7,const_1)|subtract(#8,const_4)|divide(#9,#4)|multiply(n0,#10)| | geometry |
the temperature of a certain cup of coffee 15 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... | "first , we have to find a . we know that after t = 15 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 15 a ) + 60 60 = 120 * ( 2 ^ - 15 a ) 60 / 120 = 2 ^ - 15 a 1 / 2 = 2 ^ - 15 a 2 ^ - 1 = 2 ^ - 15 a - 1 = - 15 a 1 / 15 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 15 *... | a ) 65 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | e | add(multiply(power(2, multiply(divide(60, 15), 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 |
p and q are the only two applicants qualified for a short - term research project that pays 480 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job ... | "let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 480 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 480 / 30 = $ 16 p ' s hourly wage is 480 / ( t - ... | a ) $ 5 , b ) $ 8 , c ) $ 11 , d ) $ 14 , e ) $ 17 | b | subtract(divide(480, divide(10, subtract(divide(const_3, const_2), const_1))), divide(480, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | divide(const_3,const_2)|subtract(#0,const_1)|divide(n2,#1)|add(n2,#2)|divide(n0,#2)|divide(n0,#3)|subtract(#4,#5)| | general |
the area of the largest circle that can be drawn inside a rectangle with sides 18 cm by 14 cm is | explanation : the diameter is equal to the shortest side of the rectangle . so radius = 14 / 2 = 7 cm . answer : b ) 154 | ['a ) 237', 'b ) 154', 'c ) 2376', 'd ) 198', 'e ) 171'] | b | circle_area(divide(14, const_2)) | divide(n1,const_2)|circle_area(#0) | geometry |
what profit percent is made by selling an article at a certain price , if by selling at 1 / 3 rd of that price , there would be a loss of 20 % ? | "sp 2 = 1 / 3 sp 1 cp = 100 sp 2 = 80 1 / 3 sp 1 = 80 sp 1 = 240 100 - - - 240 = > 140 % answer : a" | a ) 140 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 28 % | a | subtract(divide(subtract(const_100, 20), divide(1, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
a certain company had a total annual expenditure of 1.89 ∗ 10 ^ 7 on employee salaries last year . if the company employed 420 people , what was the average employee salary ? | "given : total annual expenditure of 1.89 ∗ 10 ^ 7 on employee salaries total employees = 420 observe that 420 * 4 = 1890 therefore try to bring the numerator in terms of 1890 average salary = ( 1890 * 10 ^ 4 ) / 420 = 4.5 * 10 ^ 4 = 45,000 option e" | a ) $ 20,000 , b ) $ 25,000 , c ) $ 35,000 , d ) $ 40,000 , e ) $ 45,000 | e | multiply(divide(multiply(1.89, multiply(10, 7)), 420), const_100) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n3)|multiply(#2,const_100)| | general |
the timing of a college is from 12 p . m to 4.00 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | "explanation : total time a student spends in college = 4 hours 00 minutes = 240 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 240 â € “ 20 ) / 5 = 44 minutes answer e" | a ) 52 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | e | divide(multiply(4.00, const_60), 5) | multiply(n1,const_60)|divide(#0,n2)| | physics |
a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 4 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company ch... | "let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 4 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 4 ) = 0.08 t - - > 60 + 6 ( t - 4 ) = 8 t - - > t = ... | a ) 2 , b ) 9 , c ) 15 , d ) 18 , e ) 30 | d | divide(subtract(0.60, multiply(0.06, 4)), subtract(0.08, 0.06)) | multiply(n1,n2)|subtract(n3,n2)|subtract(n0,#0)|divide(#2,#1)| | other |
three solid cubes of sides 1 cm , 6 cm and 8 cm are melted to form a new cube . find the surface area of the cube so formed | explanation : volume of new cube = = edge of new cube = = 9 cm surface area of the new cube = ( 6 x 9 x 9 ) = 486 answer : a ) 486 | ['a ) 486', 'b ) 366', 'c ) 299', 'd ) 278', 'e ) 1888'] | a | multiply(multiply(const_3, power(power(add(add(power(1, const_3), power(6, const_3)), power(8, const_3)), const_0_33), const_2)), const_2) | power(n0,const_3)|power(n1,const_3)|power(n2,const_3)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)|power(#5,const_2)|multiply(#6,const_3)|multiply(#7,const_2) | geometry |
a man cheats while buying as well as while selling . while buying he takes 10 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 12 % below the cost price of the claimed weight . | "there is a one step calculation method too . it requires more thought but is faster . the man takes 10 % more than what he pays for . so if he claims to take 100 pounds , he pays $ 100 but he actually takes 110 pounds for which he will take from the customer $ 110 . hence , in effect , there is a 10 % mark up . while ... | a ) 19.81 % , b ) 21 % , c ) 37.5 % , d ) 25 % , e ) 37.5 % | b | multiply(subtract(add(const_100, 10), add(12, subtract(const_100, 20))), const_2) | add(n0,const_100)|subtract(const_100,n1)|add(n2,#1)|subtract(#0,#2)|multiply(#3,const_2)| | gain |
in a recent head - to - head run - off election , 12,000 absentee ballets were cast . 1 / 2 of the absentee ballets were thrown out and 1 / 2 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | "1 / 2 * 1 / 2 ( total absentee votes ) = 1 / 4 ( total votes ) = 1 / 4 * 12000 = 3000 answer is b" | a ) 2,000 , b ) 3,000 , c ) 6,000 , d ) 8,000 , e ) 9,000 | b | multiply(divide(divide(subtract(subtract(multiply(multiply(const_3, const_4), const_1000), multiply(multiply(multiply(1, const_4), const_1000), divide(1, 2))), multiply(subtract(multiply(multiply(1, const_4), const_1000), multiply(multiply(multiply(1, const_4), const_1000), divide(1, 2))), divide(1, 2))), const_1000), ... | divide(n1,n2)|divide(const_3.0,n4)|multiply(n3,const_4)|multiply(#2,const_1000)|multiply(#0,#3)|subtract(#3,#4)|multiply(#1,#5)|subtract(#5,#6)|divide(#7,const_1000)|divide(#8,const_4)|multiply(#9,n3)| | general |
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 12 % , then the cost price of the article is | "solution 110 % of s . p . = 616 ‹ = › s . p . = ( 616 x 100 / 110 ) . ‹ = › rs . 560 . c . p . ‹ = › rs . ( 100 / 112 x 560 ) ‹ = › rs . 500 . answer a" | a ) rs . 500 , b ) rs . 515 , c ) rs . 550 , d ) rs . 600 , e ) none | a | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 12)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain |
a man swims downstream 45 km and upstream 25 km taking 5 hours each time , what is the speed of the man in still water ? | "45 - - - 5 ds = 9 ? - - - - 1 25 - - - 5 us = 5 ? - - - - 1 m = ? m = ( 9 + 5 ) / 2 = 14 / 2 = 7 answer : c" | a ) 17 , b ) 24 , c ) 7 , d ) 24 , e ) 20 | c | divide(add(divide(25, 5), divide(45, 5)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
the length of a bridge in meters , which a train 90 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | 45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 90 = 375 meters x = 285 meters the answer is d . | a ) 145 , b ) 215 , c ) 265 , d ) 285 , e ) 375 | d | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 90) | multiply(n1,const_1000)|divide(#0,const_3600)|multiply(n2,#1)|subtract(#2,n0) | physics |
if 3 < x < 6 < y < 8 , then what is the greatest possible positive integer difference of x and y ? | 3 < x < 6 < y < 8 ; 3 < x y < 8 3 + y < x + 8 y - x < 5 . positive integer difference is 4 ( for example y = 7.5 and x = 3.5 ) answer : b . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(subtract(8, 3), const_1) | subtract(n2,n0)|subtract(#0,const_1) | general |
a couple decides to have 4 children . if they succeed in having 4 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 2 girls and 2 boys ? | "no of ways of getting p ( ggbb ) is 4 ! / 2 ! * 2 ! ; total no of ways is 2 ^ n = 2 ^ 4 = 16 ; 6 / 16 = 3 / 8 ; we can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads . p ( all out comes ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 ; p ( favorable outco... | a ) 3 / 8 , b ) 1 / 4 , c ) 3 / 16 , d ) 1 / 8 , e ) 1 / 16 | a | subtract(2, multiply(divide(factorial(4), factorial(2)), power(divide(2, 2), 4))) | divide(n2,n3)|factorial(n0)|factorial(n3)|divide(#1,#2)|power(#0,n0)|multiply(#3,#4)|subtract(n2,#5)| | general |
p can lay railway track between two stations in 16 days . q can do the same job in 12 days . with the help of r , they completes the job in 4 days . how much days does it take for r alone to complete the work ? | amount of work p can do in 1 day = 1 / 16 amount of work q can do in 1 day = 1 / 12 amount of work p , q and r can together do in 1 day = 1 / 4 amount of work r can do in 1 day = 1 / 4 - ( 1 / 16 + 1 / 12 ) = 3 / 16 – 1 / 12 = 5 / 48 = > hence r can do the job on 48 / 5 days = 9 ( 3 / 5 ) days b ) | a ) 9 days , b ) 9 ( 3 / 5 ) days , c ) 11 1 / 5 days , d ) 13 days , e ) 13 1 / 7 | b | inverse(subtract(subtract(divide(const_1, 4), divide(const_1, 12)), divide(const_1, 16))) | divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|subtract(#3,#2)|inverse(#4) | physics |
two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 20 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 20 = 10 hours answer is b" | a ) 5 hours , b ) 10 hours , c ) 8 hours , d ) 10 hours , e ) 12 hours | b | divide(subtract(12, 10), 20) | subtract(n1,n0)|divide(#0,n2)| | gain |
in a games hour 4 different types of players came to the ground ? cricket 16 , hokey 12 , football 18 , softball 13 . how many players are present in the ground ? | "total number of players = 16 + 12 + 18 + 13 = 59 answer is d" | a ) 70 , b ) 52 , c ) 62 , d ) 59 , e ) 50 | d | add(add(16, 12), add(18, 13)) | add(n1,n2)|add(n3,n4)|add(#0,#1)| | physics |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.