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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i bought two books ; for rs . 480 . i sold one at a loss of 15 % and other at a loss of 25 % and then i found each book was sold at the same price . find the cost of the book sold at a lower loss ? | "x * ( 85 / 100 ) = ( 480 - x ) 75 / 100 x = 225 answer : c" | a ) a ) 657 , b ) b ) 280 , c ) c ) 225 , d ) d ) 289 , e ) e ) 427 | c | divide(multiply(480, add(const_100, 25)), add(subtract(const_100, 15), add(const_100, 25))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain |
what is the tens digit of 7 ^ 1213 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 1213 = 4 * 304 + 1 the ten ' s digit will be 0 . answer a" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | floor(divide(reminder(power(7, reminder(1213, add(const_4, const_1))), const_100), const_10)) | add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)| | general |
if 5 workers collect 48 kg of cotton in 4 days , how many kg of cotton will 9 workers collect in 2 days ? | "( 3 * 5 ) / 48 = ( 9 * 2 ) / x x = 57.6 kg answer : e" | a ) 47.6 kg , b ) 52.6 kg , c ) 59.6 kg , d ) 56.6 kg , e ) 57.6 kg | e | multiply(multiply(divide(divide(48, 5), 4), 9), 2) | divide(n1,n0)|divide(#0,n2)|multiply(n3,#1)|multiply(n4,#2)| | physics |
in the third grade of windblown school there are 78 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | total = 78 failed in math = 78 / 3 = 26 failed in literature = 78 / 6 = 13 the least failed in both can be 0 while max can be 13 answer a | a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(78, const_3), divide(78, 6)) | divide(n0,const_3)|divide(n0,n2)|min(#0,#1) | other |
david has $ 2,400 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | "suppose total money spent = x not spend ( money he still has ) = x - 800 x + x - 800 = 2400 x = 1600 money not spend = 1600 - 800 = 800 answer : d" | a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | d | subtract(divide(add(add(multiply(const_100, const_10), 800), 800), const_2), 800) | multiply(const_10,const_100)|add(n1,#0)|add(n1,#1)|divide(#2,const_2)|subtract(#3,n1)| | general |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 5 % his yearly income diminishes by rs . 61.50 . his capital is | "explanation : capital = rs . x , then 2 / 5 x = 153.75 answer : a ) rs . 153.75" | a ) 153.75 , b ) 37.78 , c ) 246.0 , d ) 27.88 , e ) 127.71 | a | divide(61.50, divide(const_4, 5)) | divide(const_4,n3)|divide(n4,#0)| | gain |
if 2994 ã · 14.5 = 173 , then 29.94 ã · 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 173 in the place of 2994 / 14.5 ] = 173 / 10 = 17.3 answer is b ." | a ) 17.1 , b ) 17.3 , c ) 17.5 , d ) 17.7 , e ) 17.2 | b | divide(173, divide(14.5, 1.45)) | divide(n1,n4)|divide(n2,#0)| | general |
the ages of 2 persons differ by 16 years . if 6 years ago the elder one be 3 times as old as the younger one , find the present age of younger person . | "age of the younger person = x age of the elder person = x + 16 3 ( x - 6 ) = x + 16 - 6 x = 14 answer is b" | a ) 10 , b ) 14 , c ) 20 , d ) 15 , e ) 32 | b | subtract(add(divide(multiply(2, 16), subtract(6, const_1)), 6), 2) | multiply(n0,n1)|subtract(n2,const_1)|divide(#0,#1)|add(n2,#2)|subtract(#3,n0)| | general |
three numbers are in the ratio 4 : 5 : 6 and their average is 39 . the largest number is : | "explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 39 15 x = 117 x = 7.8 largest number = 6 x = 46.8 . answer e" | a ) 28 , b ) 32 , c ) 36 , d ) 42 , e ) 46.8 | e | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 262 runs ? | "required run rate = [ 262 - ( 3.2 * 10 ) ] / 40 = 230 / 40 = 5.75 answer : d" | a ) 6.25 , b ) 6.22 , c ) 6.29 , d ) 5.75 , e ) 6.13 | d | divide(subtract(262, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
two pipes can fill a tank in 18 minutes and 60 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 18 + 1 / 60 - 1 / 45 = 1 / 20 so , the tank becomes full in 20 minutes . answer : c" | a ) 1 / 15 , b ) 1 / 16 , c ) 1 / 20 , d ) 1 / 10 , e ) 1 / 12 | c | subtract(add(divide(const_1, 18), divide(const_1, 60)), divide(const_1, 45)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 400 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "number of newborns that can die in first month = 1 / 10 * 400 = 40 survived = 360 number of newborns that can die in second month = 1 / 10 * 360 = 36 survived = 324 number of newborns that can die in third month = 1 / 10 * 324 = 32 survived = 292 answer : d" | a ) 240 , b ) 246 , c ) 252 , d ) 292 , e ) 298 | d | multiply(multiply(multiply(400, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10))) | divide(n1,n2)|subtract(n1,#0)|multiply(n3,#1)|multiply(#2,#1)|multiply(#3,#1)| | probability |
the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 50 of the area of the field . what is the length of the field ? | "explanation : a / 50 = 8 * 8 = > a = 8 * 8 * 50 x * 2 x = 8 * 8 * 50 x = 40 = > 2 x = 80 answer : option e" | a ) 73 , b ) 32 , c ) 34 , d ) 43 , e ) 80 | e | sqrt(divide(multiply(square_area(8), 50), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)| | geometry |
a fill pipe can fill 2 / 3 of cistern in 40 minutes in how many minutes , it can fill full of the cistern ? | 2 / 3 of the cistern can fill in 40 min 3 / 3 of the cistern can fill in = 40 * 3 / 2 * 3 / 3 = 60 min answer is a | a ) 60 min , b ) 36 min , c ) 25 min , d ) 30 min , e ) 50 min | a | divide(40, divide(2, 3)) | divide(n0,n1)|divide(n2,#0) | physics |
a “ sophie germain ” prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 4 is | "in that case , the sophie prime numbers greater than 5 are 7 , 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be c ." | a ) 3 , b ) 7 , c ) 189 , d ) 227 , e ) 234 | c | multiply(multiply(multiply(subtract(add(add(subtract(4, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(4, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(4, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | add(n0,n1)|multiply(n0,const_10)|subtract(n2,n1)|add(#0,#2)|add(#3,#0)|multiply(n0,#3)|add(n1,#5)|add(#4,#0)|subtract(#4,const_10)|multiply(n0,#7)|subtract(#6,const_10)|add(n1,#9)|multiply(#8,#10)|multiply(#0,#12)|subtract(#11,#1)|multiply(#13,#14)| | general |
x starts a business with rs . 45000 . y joins in the business after 8 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 — 4 = 45 ã — 12 : 30 ã — 4 = 3 ã — 12 : 2 ã — 4 = 9 : 2 answer : option b" | a ) 1 : 2 , b ) 9 : 2 , c ) 1 : 3 , 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 |
what number has a 5 : 1 ratio to the number 12 ? | "5 : 1 = x : 12 x = 60 answer : e" | a ) 22 , b ) 50 , c ) 88 , d ) 52 , e ) 60 | e | multiply(12, 5) | multiply(n0,n2)| | other |
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 50000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 50000 share of b = 6 / 13 ( 50000 ) = rs . 23076.92 ~ 23077 answer : a" | a ) 23077 , b ) 24000 , c ) 24005 , d ) 24009 , e ) 24002 | a | subtract(50000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
if a * b * c = 130 , b * c * d = 65 , c * d * e = 500 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a â ˆ — b â ˆ — c / b â ˆ — c â ˆ — d = 130 / 65 = > a / d = 2 d â ˆ — e â ˆ — f / c â ˆ — d â ˆ — e = 250 / 500 = > f / c = 1 / 2 a / d * f / c = 2 * 1 / 2 = 1 answer : e" | a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 | e | divide(multiply(130, 250), multiply(500, 65)) | multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)| | general |
how many internal diagonals does a heptagon ( seven sided polygon ) have ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 7 no . of diagonals = 7 ( 7 - 3 ) / 2 = 14 ans c" | a ) 7 , b ) 9 , c ) 14 , d ) 20 , e ) 35 | c | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | multiply(const_2,const_4)|divide(#0,const_2)|subtract(#0,const_3)|multiply(#1,#2)| | geometry |
in the coordinate plane , points ( x , 5 ) and ( 7 , y ) are on line k . if line k passes through the origin and has slope 5 / 7 , then x + y = | "line k passes through the origin and has slope 5 / 7 means that its equation is y = 5 / 7 * x . thus : ( x , 5 ) = ( 7,5 ) and ( 7 , y ) = ( 7,5 ) - - > x + y = 7 + 5 = 12 . answer : e" | a ) 13 , b ) 14 , c ) 9 , d ) 10 , e ) 12 | e | multiply(multiply(7, 7), divide(5, 7)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general |
a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 2.5 grams per cubic centimeter and mineral y weighs 2 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 2 gms in 600,000 cm ^ 3 , y = 2 * 600,000 = 1 ,... | a ) 5 , 500,000 , b ) 2 , 200,000 , c ) 55,000 , d ) 28,000 , e ) 280 | b | subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 2)), 2)), volume_cube(100)) | multiply(const_2,n3)|volume_cube(n5)|divide(#1,const_10)|multiply(#2,n2)|multiply(#2,#0)|multiply(#3,n2)|multiply(#4,n3)|add(#5,#6)|subtract(#7,#1)| | geometry |
if the wheel is 56 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 56 * x = 1056 = > x = 3 answer : c" | a ) 10 , b ) 12 , c ) 3 , d ) 11 , e ) 9 | c | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 56)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 12 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | "explanation : l * 30 = 1200 è l = 40 40 + 30 + 50 = 120 120 * 12 = 1440 answer : option a" | a ) 1440 , b ) 1200 , c ) 2832 , d ) 1299 , e ) 1236 | a | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 12) | divide(n1,n3)|power(n3,n2)|add(n3,#0)|power(#0,n2)|add(#1,#3)|sqrt(#4)|add(#2,#5)|multiply(n0,#6)| | geometry |
a boat moves upstream at the rate of 1 km in 40 minutes and down stream 1 km in 12 minutes . then the speed of the current is : | "rate upstream = ( 1 / 40 * 60 ) = 1.5 kmph rate dowm stream = 1 / 12 * 60 = 5 kmph rate of the current = ½ ( 5 - 1.5 ) = 1.75 kmph answer : c" | a ) 1 kmph , b ) 2 kmph , c ) 1.75 kmph , d ) 2.5 kmph , e ) 3.5 kmph | c | divide(subtract(multiply(divide(1, 12), const_60), multiply(divide(1, 40), const_60)), const_2) | divide(n0,n3)|divide(n0,n1)|multiply(#0,const_60)|multiply(#1,const_60)|subtract(#2,#3)|divide(#4,const_2)| | physics |
if 10 bulls can plough 20 identical fields in 3 days working 10 hours a day , then in how many days can 30 bulls plough 32 same identical fields working 8 hours a day ? | explanation : m 1 * d 1 * w 2 = m 2 * d 2 * w 1 10 * 3 * 10 * 32 = 30 * d * 8 * 20 d = 2 days answer : d | a ) 8 , b ) 7 , c ) 5 , d ) 2.0 , e ) 1 | d | divide(multiply(multiply(multiply(10, 10), 3), 32), multiply(multiply(30, 8), 20)) | multiply(n0,n0)|multiply(n4,n6)|multiply(n2,#0)|multiply(n1,#1)|multiply(n5,#2)|divide(#4,#3) | physics |
how many numbers from 2 to 7 are exactly divisible by 2 ? | "2 / 2 = 1 and 7 / 2 = 3 3 - 1 = 2 2 + 1 = 3 numbers . b )" | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | add(divide(subtract(multiply(floor(divide(7, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
the product of two numbers is 120 . the sum of their squares is 289 . the sum of the two numbers is : | "explanation : let the number be x and y . we know that , ( x + y ) 2 = ( x 2 + y 2 ) + 2 xy = 289 + 2 × 120 = 289 + 240 = 529 ⇒ x + y = √ 529 = 23 correct option : b" | a ) 20 , b ) 23 , c ) 169 , d ) 172 , e ) 198 | b | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
the least number , which when divided by 8 , 15 , 20 and 54 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 8 , 15 , 20 , 54 ) + 8 = 1080 + 8 = 1088 . answer : e" | a ) 448 , b ) 488 , c ) 542 , d ) 548 , e ) 1088 | e | multiply(54, const_10) | multiply(n3,const_10)| | general |
there are 10 slate rocks , 15 pumice rocks , and 6 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "10 / 31 * 9 / 30 = 3 / 31 the answer is b ." | a ) 2 / 19 , b ) 3 / 31 , c ) 4 / 37 , d ) 5 / 41 , e ) 6 / 53 | b | multiply(divide(10, add(add(10, 15), 6)), divide(subtract(10, const_1), subtract(add(add(10, 15), 6), const_1))) | add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)| | other |
the average of 6 no . ' s is 2.80 . the average of 2 of them is 2.4 , while the average of the other 2 is 2.3 . what is the average of the remaining 2 no ' s ? | sum of the remaining two numbers = ( 2.80 * 6 ) - [ ( 2.4 * 2 ) + ( 2.3 * 2 ) ] = 16.80 - ( 4.8 + 4.6 ) = 16.80 - 9.40 = 7.40 required average = ( 7.4 / 2 ) = 3.7 answer : d | a ) 2.3 , b ) 2.6 , c ) 3.6 , d ) 3.7 , e ) 4.6 | d | divide(subtract(multiply(6, 2.8), add(multiply(2, 2.4), multiply(2, 2.3))), 2) | multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2) | general |
if x and y are integers and | x - y | = 12 , what is the minimum possible value of xy ? | "we are given | x - y | = 12 , minimum possible value of xy would be maximum numeric value with a - ive sign . . so one of x or y will be negative and other negative . . | x - y | = 12 in this case means that the numeric sum of x and y is 12 . . various combinations could be - 1 and 11 , - 2 and 10 , - 6 and 6 , - 11 a... | a ) - 12 , b ) - 18 , c ) - 24 , d ) - 36 , e ) - 48 | d | multiply(negate(add(const_2, const_3)), subtract(12, add(const_2, const_3))) | add(const_2,const_3)|negate(#0)|subtract(n0,#0)|multiply(#1,#2)| | general |
what is the smallest positive perfect square that is divisible by 4 , 10 , and 18 ? | the number needs to be divisible by 2 ^ 2 , 2 * 5 , and 2 * 3 ^ 2 . the smallest such perfect square is 2 ^ 2 * 3 ^ 2 * 5 ^ 2 = 900 the answer is a . | ['a ) 900', 'b ) 1,600', 'c ) 2,500', 'd ) 3,600', 'e ) 4,900'] | a | add(multiply(multiply(multiply(4, power(const_3, const_2)), 10), const_2), multiply(10, 18)) | multiply(n1,n2)|power(const_3,const_2)|multiply(n0,#1)|multiply(n1,#2)|multiply(#3,const_2)|add(#4,#0) | geometry |
if grapes are 91 % water and raisins are 19 % water , then how many kilograms did a quantity of raisins , which currently weighs 9 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transform... | "let x be the original weight . the weight of the grape pulp was 0.09 x . since the grape pulp is 81 % of the raisins , 0.09 x = 0.81 ( 9 kg ) . then x = 9 * 9 = 81 kg . the answer is c ." | a ) 63 , b ) 72 , c ) 81 , d ) 90 , e ) 99 | c | divide(multiply(divide(subtract(const_100, 19), const_100), 9), divide(subtract(const_100, 91), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)| | general |
a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 210 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 210 ⇒ 1.75 x = 210 ⇒ x = 120 coins answer d" | a ) 20 coins , b ) 50 coins , c ) 100 coins , d ) 120 coins , e ) none of these | d | divide(210, add(add(inverse(const_4), inverse(const_2)), const_1)) | inverse(const_4)|inverse(const_2)|add(#0,#1)|add(#2,const_1)|divide(n2,#3)| | general |
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 4 cm . what is the length of the wire ? | "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 4 ) ^ 2 * h = ( 4 / 3 ) π ( 12 ) ^ 3 = > h = 144 cm answer : b" | a ) 116 cm , b ) 144 cm , c ) 168 cm , d ) 173 cm , e ) 189 cm | b | divide(multiply(const_4, divide(power(12, const_3), power(4, const_2))), const_3) | power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)| | physics |
a train crosses a platform of 150 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be ‘ x ’ x + 150 / 15 = x + 250 / 20 4 x + 600 = 3 x + 750 x = 150 m answer : a" | a ) 150 m , b ) 188 m , c ) 267 m , d ) 268 m , e ) 287 m | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(150, divide(20, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
find √ ? / 19 = 4 ? | "answer let √ n / 19 = 4 then √ n = 19 x 4 = 76 ∴ n = 76 x 76 = 5776 . correct option : b" | a ) 76 , b ) 5776 , c ) 304 , d ) 1296 , e ) none | b | power(multiply(4, 19), const_2) | multiply(n0,n1)|power(#0,const_2)| | general |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint , in gallons , that must be purchased in order to measure o... | "ratio ; white : black : gray = 3 : 5 : ( 3 + 5 ) = 3 : 5 : 8 if 2 gallons of the mixture is needed < = = > need 2 gallons of gray paint . then , 3 : 5 : 8 = x : y : 2 3 / 8 = x / 2 ; x = 3 / 4 5 / 8 = y / 2 ; y = 5 / 4 you need 3 / 4 gallon of white . you have to buy 1 gallons . you need 5 / 4 gallons of black . you h... | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | multiply(const_0_25,n2)|multiply(n1,n2)|add(#0,#1)| | general |
if the radius of a circle that centers at the origin is 5 , how many j points on the circle have integer coordinates ? | i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : j = c | ['a ) 4', 'b ) 8', 'c ) 12', 'd ) 15', 'e ) 20'] | c | subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2)) | add(n0,const_4)|multiply(n0,n0)|power(const_2,const_2)|subtract(#1,#0)|subtract(#3,#2) | geometry |
an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 50 packages of gaskets . if 20 percent of the gaskets went to company x , 15 perc... | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 / 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 20 % of the gaskets to x = 10 gaskets = 25 * 10 = 250 2 . he sold 30 % of the gaskets to y = 15 gaskets = 25 * 10 + 20 * 5 = 350 3 . ... | a ) 1105 , b ) 1245 , c ) 1275 , d ) 1215 , e ) 1150 | e | add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(50, multiply(50, divide(20, const_100))), multiply(50, divide(20, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(50, divide(20, const_100))), multiply(25, multiply(50, divide(20, const_100))))) | divide(n2,n3)|divide(n6,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n5,#1)|multiply(n0,#4)|subtract(n5,#4)|add(#5,#5)|subtract(#6,#4)|subtract(#8,n1)|multiply(#3,#9)|add(#10,#2)|add(#11,#7)| | general |
ravi can do a piece of work in 15 days while prakash can do it in 30 days . in how many days will they finish it together ? | "1 / 15 + 1 / 30 = 3 / 30 10 / 1 = 10 days answer : e" | a ) 11 days , b ) 9 days , c ) 8 days , d ) 12 days , e ) 10 days | e | divide(const_1, add(inverse(15), inverse(30))) | inverse(n0)|inverse(n1)|add(#0,#1)|divide(const_1,#2)| | physics |
( 1000 ) 7 ÷ ( 10 ) 16 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 16 = ( 10 ) 21 / ( 10 ) 16 = 10 ( 5 ) = 100000 option d" | a ) 10 , b ) 100 , c ) 1000 , d ) 100000 , e ) none of these | d | multiply(1000, 10) | multiply(n0,n2)| | general |
what is remainder of the division ( 1425 * 1427 * 1429 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : e" | a ) 5 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | e | subtract(multiply(multiply(1425, 1427), 1429), subtract(multiply(multiply(1425, 1427), 1429), const_3)) | multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)| | general |
two men a and b start from place x walking at 4 ½ kmph and 5 ¾ kmph respectively . how many km apart they are at the end of 5 ½ hours if they are walking in the same direction ? | "rs = 5 ¾ - 4 ½ = 1 ¼ t = 3 ½ h . d = 5 / 4 * 11 / 2 = 55 / 8 = 6 7 / 8 km answer : c" | a ) 7 km , b ) 6 km , c ) 6 7 / 8 km , d ) 9 km , e ) 5 km | c | add(multiply(add(5, divide(const_1, const_2)), subtract(add(5, divide(const_3, 4)), add(4, divide(const_1, const_2)))), const_2) | divide(const_1,const_2)|divide(const_3,n0)|add(n2,#0)|add(n1,#1)|add(n0,#0)|subtract(#3,#4)|multiply(#2,#5)|add(#6,const_2)| | physics |
a man walking at a rate of 10 km / hr crosses a bridge in 10 minutes . the length of the bridge is ? | "speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 10 minutes = 50 / 18 * 10 * 60 = 1667 m answer is c" | a ) 1521 , b ) 1492 , c ) 1667 , d ) 1254 , e ) 1112 | c | multiply(divide(multiply(10, const_1000), const_60), 10) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | gain |
the distance from steve ' s house to work is 40 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 40 km = > speed = 20 kmph e" | a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | e | multiply(divide(add(40, divide(40, const_2)), 6), const_2) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|multiply(#2,const_2)| | physics |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 67 kg . what is the weight of the new person ? | "total increase in weight = 8 × 2.5 = 20 if x is the weight of the new person , total increase in weight = x − 67 = > 20 = x - 67 = > x = 20 + 67 = 87 answer is c ." | a ) 75 , b ) 65 , c ) 87 , d ) 95 , e ) 80 | c | add(multiply(8, 2.5), 67) | multiply(n0,n1)|add(n2,#0)| | general |
if a # b = ab – b + b ^ 2 , then 3 # 8 = | "solution - simply substitute 3 and 8 in equation in the place of a and b respectively . 3 # 8 = 3 * 8 - 8 + 8 ^ 2 = 24 - 8 + 64 = 80 . ans a" | a ) 80 , b ) 8 , c ) 15 , d ) 21 , e ) 35 | a | add(subtract(multiply(3, 8), 8), power(8, 2)) | multiply(n1,n2)|power(n2,n0)|subtract(#0,n2)|add(#1,#2)| | general |
a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations z are possible ? | "the combinations could be { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pa... | a ) 3 , b ) z = 4 , c ) z = 5 , d ) z = 6 , e ) z = 8 | e | divide(240, 30) | divide(n0,n2)| | general |
find the sum of first 10 odd numbers | "explanation : n 2 = 102 = 100 answer : option d" | a ) 10 , b ) 40 , c ) 200 , d ) 100 , e ) 540 | d | add(divide(divide(10, divide(divide(divide(divide(divide(10, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(10, divide(divide(divide(divide(divide(10, const_2), const_2), const_2), const_2), const_2)), const_2)))) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)| | general |
two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 44 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "solution relative speed = ( 44 - 40 ) kmph = 4 kmph = ( 4 x 5 / 18 ) m / sec = ( 20 / 18 ) m / sec time taken = ( 350 x 18 / 20 ) sec = 140 sec . answer a" | a ) 140 sec , b ) 132 sec , c ) 192 sec , d ) 252 sec , e ) none | a | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(44, 40))) | add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)| | physics |
tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 37 pages per hour . if tom starts reading a novel at 4 : 00 , and jan begins reading an identical copy of the same book at 4 : 56 , at what time will they be reading the same page ? | "since tom reads an average of 1 page every 2 minutes , tom will read 28 pages in the first 56 minutes . jan can catch tom at a rate of 7 pages per hour , so it will take 4 hours to catch tom . the answer is d ." | a ) 7 : 26 , b ) 7 : 56 , c ) 8 : 26 , d ) 8 : 56 , e ) 9 : 26 | d | divide(add(4, divide(multiply(add(subtract(4, 4), divide(subtract(56, 00), const_60)), 30), subtract(37, 30))), 56) | subtract(n5,n3)|subtract(n4,n2)|subtract(n1,n0)|divide(#0,const_60)|add(#3,#1)|multiply(n0,#4)|divide(#5,#2)|add(n4,#6)|divide(#7,n5)| | general |
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 42 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 42 = > x = 9 so deepak age is = 3 ( 9 ) = 27 option e" | a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 27 | e | divide(multiply(subtract(42, 6), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
an outlet pipe can empty 2 / 3 of a cistern in 20 minutes . in 8 minutes , what part of the cistern will be emptied ? | "8 / 20 * 2 / 3 = 4 / 15 the answer is d ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 9 , d ) 4 / 15 , e ) 9 / 20 | d | divide(multiply(8, divide(2, 3)), 20) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics |
the number of diagonals of a polygon of n sides is given by the formula f = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | "f = n ( n - 3 ) f = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b" | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | add(n0,n1)| | general |
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 16 sec . what is the length of the platform ? | "speed = 300 / 16 = 75 / 4 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 75 / 4 = > x = 731.25 m answer : b ( 431.25 )" | a ) 287 m , b ) 431.25 m , c ) 267 m , d ) 287.25 m , e ) 656 m | b | subtract(multiply(speed(300, 16), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
two poles of height 6 meters and 11 meters stand on a plane ground . if the distance between their feet is 12 meters then find the difference in the distance between their tops : | distance between their tops = sqrt ( 12 ^ 2 + ( 11 - 6 ) ^ 2 ) = sqrt 169 = 13 m answer : c | a ) 12 m , b ) 5 m , c ) 13 m , d ) 11 m , e ) 12 m | c | sqrt(add(power(12, const_2), power(subtract(11, 6), const_2))) | power(n2,const_2)|subtract(n1,n0)|power(#1,const_2)|add(#0,#2)|sqrt(#3) | physics |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind loe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of loe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . loe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena ca... | a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
a bag contains 4 red , 5 blue and 3 green balls . if 3 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 4 c 3 / 12 c 3 = 1 / 55 e" | a ) 1 / 33 , b ) 1 / 22 , c ) 1 / 12 , d ) 1 / 44 , e ) 1 / 55 | e | divide(choose(4, 3), choose(add(add(4, 5), 3), 3)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 507 sq meters , then what is the width ( in meters ) of the rectangular plot ? | "area = l * w = 3 w ^ 2 = 507 w ^ 2 = 169 w = 13 the answer is c ." | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | c | divide(divide(divide(507, const_3), const_3), const_4) | divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)| | geometry |
the amount of time that three people worked on a special project was in the ratio of 3 to 5 to 6 . if the project took 140 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 3 * 140 / 14 = 30 hours b = 5 * 140 / 14 = 50 hours c = 6 * 140 / 14 = 60 hours c is the hardest worker and a worked for the least number of hours . so the difference is 60 - 30 = 30 hours . answer : d" | a ) 32 hours , b ) 40 hours , c ) 35 hours , d ) 30 hours , e ) 36 hours | d | subtract(multiply(divide(140, add(add(3, 5), 6)), 6), multiply(divide(140, add(add(3, 5), 6)), 3)) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|multiply(n0,#2)|subtract(#3,#4)| | physics |
a student gets 60 % in one subject , 80 % in the other . to get an overall of 75 % how much should get in third subject . | let the 3 rd subject % = x 60 + 80 + x = 3 * 75 140 + x = 225 x = 225 - 140 = 85 answer : c | a ) 75 % , b ) 25 % , c ) 85 % , d ) 55 % , e ) 65 % | c | subtract(multiply(75, const_3), add(60, 80)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0) | gain |
according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 20 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 20 oz can = 300 oz / 20 oz = 15 answer a , 15 cans" | a ) 15 , b ) 34 , c ) 50 , d ) 67 , e ) 100 | a | divide(multiply(6, 200), divide(200, const_10)) | divide(n2,const_10)|multiply(n2,n3)|divide(#1,#0)| | general |
john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.36 more than jane , what was the original price of the dish ? | "the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.36 - - > 15 p - 13.5 p = 36 - - > p = 24 . answer : a ." | a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84 | a | divide(0.36, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#0,#3)|divide(n2,#4)| | gain |
a , b and c invested rs . 8000 , rs . 4000 and rs . 8000 respectively in a business . a left after six months . if after eight months , there was a gain of rs . 4005 , then what will be the share of b ? | "explanation : a : b : c = ( 8000 * 6 ) : ( 4000 * 8 ) : ( 8000 * 8 ) = 48 : 32 : 64 = 3 : 2 : 4 so b share = ( 2 / 9 ) * 4005 = rs 890 option c" | a ) rs 690 , b ) rs 790 , c ) rs 890 , d ) rs 990 , e ) none of these | c | multiply(divide(8000, add(add(8000, 4000), 8000)), 4005) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
two pumps are connected to an empty tank . pump x fills the tank with water at a constant rate , while pump y drains water out of the tank at a constant rate . the two pumps finish filling the tank in 6 times the duration it would take pump x alone to fill the tank . if pump y alone can empty a whole tank in 48 minutes... | let v be the volume of the tank . let r be the rate per minute that pump x fills the tank . let t be the time it takes pump x to fill the tank . the rate at which pump y empties the tank is v / 48 per minute . ( r - v / 48 ) * 6 t = v = rt . ( r - v / 48 ) * 6 = r 5 r = v / 8 . r = v / 40 . it takes pump x 40 minutes t... | a ) 32 , b ) 34 , c ) 36 , d ) 38 , e ) 40 | e | subtract(48, divide(48, 6)) | divide(n1,n0)|subtract(n1,#0) | physics |
a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 16.80 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 16.80 when c . p = ( 100 * 16.80 ) / 21 = 50 answer : e" | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 80 | e | divide(multiply(16.80, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | add(n0,const_100)|add(n3,const_100)|multiply(n2,const_100)|subtract(const_100,n1)|multiply(#1,#3)|divide(#4,const_100)|subtract(#0,#5)|divide(#2,#6)| | gain |
given a two - digit number , the unit ' s digit exceeds its 10 ' s digit by 5 and the product of the given number and the sum of its digits is equal to 90 , which of the options is the number ? | using the elimination method the option that fits this description is 27 7 - 2 = 5 ( unit ' s digit that exceeds its ten ' s digit by 3 ) 27 * 9 = 243 ( the product of the given number and the sum of its digits is equal to 175 ) answer : a | a ) 27 , b ) 16 , c ) 38 , d ) 61 , e ) 83 | a | add(multiply(const_2, 10), add(divide(subtract(sqrt(subtract(multiply(add(multiply(5, add(10, const_1)), 10), add(multiply(5, add(10, const_1)), 10)), multiply(multiply(subtract(multiply(5, 5), add(multiply(const_60, const_4), const_3)), multiply(add(10, const_1), const_2)), const_4))), add(multiply(5, add(10, const_1)... | add(n0,const_1)|multiply(n1,n1)|multiply(const_4,const_60)|multiply(n0,const_2)|add(#2,const_3)|multiply(n1,#0)|multiply(#0,const_2)|add(n0,#5)|multiply(#6,const_2)|subtract(#1,#4)|multiply(#7,#7)|multiply(#6,#9)|multiply(#11,const_4)|subtract(#10,#12)|sqrt(#13)|subtract(#14,#7)|divide(#15,#8)|add(n1,#16)|add(#17,#3) | general |
if â € œ * â € is called â € œ + â € , â € œ / â € is called â € œ * â € , â € œ - â € is called â € œ / â € , â € œ + â € is called â € œ - â € . 240 - 80 / 60 - 40 * 10 = ? | explanation : given : 240 - 80 / 60 - 40 * 10 = ? substituting the coded symbols for mathematical operations , we get , 240 / 80 * 60 / 40 + 10 = ? 3 * 1.5 + 10 = 14.5 answer : a | a ) 14.5 , b ) 13.5 , c ) 12.5 , d ) 17.5 , e ) 11.5 | a | add(multiply(divide(60, 40), divide(240, 80)), 10) | divide(n2,n3)|divide(n0,n1)|multiply(#0,#1)|add(n4,#2) | general |
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 5.5 hours , then the distance covered at 40 kmph is | "let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 5.5 or , x / 40 + ( 250 - x ) / 60 = 5.5 or , ( 3 x + 500 - 2 x ) / / 120 = 5.5 or 500 + x = 660 x = 160 answer : b" | a ) 130 km , b ) 160 km , c ) 100 km , d ) 140 km , e ) none of these | b | multiply(divide(subtract(multiply(60, 5.5), 250), negate(subtract(40, 60))), 40) | multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)| | physics |
the least number which when increased by 3 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 3 = 864 - 3 = 861 . answer d" | a ) 427 , b ) 859 , c ) 869 , d ) 861 , e ) none of these | d | subtract(lcm(lcm(lcm(24, 32), 36), 54), 3) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.15 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.15 = 1.25 km = 1250 m . required time = 1250 * 3 / 125 = 30 sec . answer : b" | a ) 65 sec , b ) 30 sec , c ) 48 sec , d ) 33 sec , e ) 12 sec | b | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.15, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics |
what is the compound interest on rs : 80,000 for 4 months at the rate of 5 % per annum | "it is monthly compound rate = 5 / 12 % per month 80000 * ( 1 + 5 / 1200 ) ^ 4 - 80000 = 1341.69 answer : a" | a ) 1341.69 , b ) 1342.69 , c ) 1343.69 , d ) 1344.69 , e ) 1345.69 | a | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(4, const_3)))), const_100) | multiply(const_100,const_3)|multiply(const_3,n1)|divide(n1,#1)|multiply(#0,const_100)|multiply(n2,#2)|multiply(#3,#4)|divide(#5,const_100)| | gain |
tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ? | "explanation : here , s = { 1 , 2 , 3 , 4 , . . . . , 19 , 20 } . let e = event of getting a multiple of 3 or 5 = { 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 20 } . p ( e ) = n ( e ) / n ( s ) = 9 / 20 answer : d" | a ) 1 / 2 , b ) 1 / 2 , c ) 8 / 15 , d ) 9 / 20 , e ) 9 / 24 | d | divide(3, const_10) | divide(n2,const_10)| | general |
the cross - section of a water channel is a trapezium in shape . if the channel is 14 meters wide at the top and 8 meters wide at the bottom and the area of cross - section is 770 square meters , what is the depth of the channel ( in meters ) ? | "1 / 2 * d * ( 14 + 8 ) = 770 d = 70 the answer is c ." | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(770, divide(add(14, 8), const_2)) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)| | physics |
what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 5 , 12 and 13 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 13 ... | a ) 2.5 , b ) 6.5 , c ) 5 , d ) 6.0 , e ) 7 | b | divide(triangle_area_three_edges(5, 12, 13), divide(triangle_perimeter(5, 12, 13), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry |
the cost price of 20 articles is the same as the selling price of x articles . if the profit is 25 % then determine the value of x . | "explanation : let the cost price 1 article = re 1 cost price of x articles = x s . p of x articles = 20 gain = 20 - x = > 25 = ( 20 − x / x ∗ 100 ) = > 2000 − 100 x = 25 x = > x = 16 option c" | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | divide(multiply(20, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain |
how many positive integers less than 260 are multiple of 4 but not multiples of 6 ? | "260 / 4 = 65 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 65 - 20 = 45 answer : c" | a ) 20 , b ) 31 , c ) 45 , d ) 53 , e ) 64 | c | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
the area of sector of a circle whose radius is 10 metro and whose angle at the center is 42 â ° is ? | "42 / 360 * 22 / 7 * 10 * 10 = 36.7 m 2 answer : a" | a ) 36.7 , b ) 52.9 , c ) 52.8 , d ) 52.1 , e ) 52.2 | a | multiply(multiply(power(10, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry |
120 is what percent of 80 ? | "120 = x * 80 / 100 x = 150 % ans ; d" | a ) 0.2 % , b ) 2 % , c ) 5 % , d ) 150 % , e ) 500 % | d | multiply(divide(120, 80), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a rectangular room has the rectangular shaped rug shown as above figure such that the rug ’ s area is 140 square feet and its length is 4 feet longer than its width . if the uniform width between the rug and room is 4 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | "rug ' s area = 140 which is ( x ) x ( 4 + x ) = 140 so x = 10 rug maintains a uniform distance of 4 feet so room has dimension 10 + 8 and 14 + 8 i . e . 18 and 22 area of room 18 x 22 = 396 area covered is 140 so uncovered area is 396 - 140 = 256 ( answer e )" | a ) 32 , b ) 36 , c ) 40 , d ) 46 , e ) 256 | e | subtract(multiply(add(subtract(divide(140, const_10), const_2), const_10), add(divide(140, const_10), subtract(divide(140, const_10), const_2))), 140) | divide(n0,const_10)|subtract(#0,const_2)|add(#1,const_10)|add(#0,#1)|multiply(#2,#3)|subtract(#4,n0)| | geometry |
at a meeting , 25 attendants used a pencil and 15 attendants used a pen . if exactly 20 attendants used only one of the two types of writing tools , how many attendants wrote with both types of writing tools ? | say x attendants wrote with both writing tools . ( 25 - x ) + ( 15 - x ) = 20 - - > x = 10 . answer : a . | a ) 10 , b ) 6 , c ) 7 , d ) 4 , e ) 14 | a | divide(subtract(add(25, 15), 20), const_2) | add(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | other |
in the xy - plane the point ( - 2 - 3 ) is the centre of a circle , the point ( - 2 , 2 ) lies inside the circle and the point ( 5 , - 3 ) lies outside the circle . if the radius r of the circle r is an integer then r = | "can be solved without much calculations . you are given that ( - 2 , - 3 ) is the center of the circle . point ( 5 , - 3 ) lies inside the circle - - - > the radius is lesser than distance of ( - 2 , - 3 ) from ( 5 , - 3 ) - - - > lesser than 7 units but the radius will also be greater than the distance of ( - 2 , - 3... | a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | divide(add(subtract(5, negate(2)), 5), 2) | negate(n0)|subtract(n4,#0)|add(n4,#1)|divide(#2,n0)| | general |
working together , tim and tom can type 15 pages in one hour . if they would be able to type 18 pages in one hour if tom increases his typing speed by 60 % , what is the ratio of tom ' s normal typing speed to that of tim ? | "lets say tim types x pages an hour and tom types y pages an hour . we know that x + y = 15 tom increase speed by 60 % means he will type 1.6 y pages an hour . so we get x + 1.6 y = 18 we need to know the ratio of tom ' s speed to tim ' s speed . this is going to be proportional to the number of pages each can type in ... | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(divide(subtract(18, 15), divide(60, multiply(const_100, const_1))), subtract(15, divide(subtract(18, 15), divide(60, multiply(const_100, const_1))))) | multiply(const_1,const_100)|subtract(n1,n0)|divide(n2,#0)|divide(#1,#2)|subtract(n0,#3)|divide(#3,#4)| | physics |
how many digits are in ( 12 × 10 ^ 14 ) ( 10 × 10 ^ 15 ) ? | "the question simplfies to ( 12 × 10 ^ 14 ) ( 10 ^ 16 ) = > 12 * 10 ^ 30 = > will contain 30 zeros + 2 digit 12 = > 32 ans e" | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 32 | e | add(10, add(const_3, const_4)) | add(const_3,const_4)|add(#0,n1)| | general |
20 men can complete a piece of work in 30 days . in how many days can 25 men complete that piece of work ? | "20 * 30 = 25 * x = > x = 24 days answer : c" | a ) 23 , b ) 27 , c ) 24 , d ) 27 , e ) 11 | c | divide(multiply(30, 20), 25) | multiply(n0,n1)|divide(#0,n2)| | physics |
a lamp is put on one corner of a square plot of side 50 m . it ' s light reaches 21 m . find the area of that plot that is lit by that lamp ? | area covered by lamp = pi * r ^ 2 / 4 ( here we divide by 4 because lamp is put in the corner of the plot and only 1 / 4 part lit is of the plot ) where r = 21 m = length of part lit so area = ( 22 / 7 ) * 21 * 21 / 4 = 346.5 sq m answer : c | ['a ) 248', 'b ) 312.5', 'c ) 346.5', 'd ) 392.5', 'e ) 424'] | c | divide(circle_area(21), const_4) | circle_area(n1)|divide(#0,const_4) | geometry |
what is 12.5 % of 4 / 12 of 600 ? | "12.5 % = 12.5 / 100 = 1 / 8 of 4 / 12 = 1 / 8 * 4 / 12 = 1 / 24 of 600 = 1 / 24 * 600 = 25 ans - a" | a ) 25 , b ) 24 , c ) 30 , d ) 27 , e ) 28 | a | divide(multiply(12.5, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
what is 985 * 985 ? | "if you take a base of 1000 then 985 is 15 less than 1000 to get the product of 985 x 985 write like this 985 - 15 ( as 15 less than base 1000 ) 985 - 15 now 15 x 15 = 225 and 985 - 15 = 970 so 985 x 985 = 970225 . . . ( bingo the answer is e . you can even have a shortcut . . . . . . 15 x 15 = 225 . . . only answer ch... | a ) 954,169 , b ) 964,219 , c ) 964,549 , d ) 965,019 , e ) 970,225 | e | circle_area(divide(985, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | general |
find the slope of the line perpendicular to the line y = ( 1 / 6 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 6 . if m is the slope of the line perpendicular to the given line , then m × ( 1 / 6 ) = - 1 solve for m m = - 6 correct answer c ) - 6" | a ) 1 , b ) 2 , c ) - 6 , d ) 4 , e ) 5 | c | divide(1, 6) | divide(n0,n1)| | general |
find the mean proportional between 36 & 49 ? | "formula = √ a × b a = 36 and b = 49 √ 36 × 49 = 6 × 7 = 42 a" | a ) a ) 42 , b ) b ) 61 , c ) c ) 63 , d ) d ) 65 , e ) e ) 67 | a | divide(add(36, 49), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 1 : 3 and b : c = 1 : 5 . if the total runs scored by all of them are 95 , the runs scored by c are ? | a : b = 1 : 3 b : c = 1 : 5 a : b : c = 1 : 3 : 15 15 / 19 * 95 = 75 answer : c | a ) 20.23 , b ) 20.13 , c ) 75 , d ) 20.93 , e ) 10.93 | c | multiply(divide(95, add(add(divide(1, 3), divide(5, 1)), const_1)), 5) | divide(n0,n1)|divide(n3,n0)|add(#0,#1)|add(#2,const_1)|divide(n4,#3)|multiply(n3,#4) | general |
dacid obtained 61 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 61 + 65 + 82 + 67 + 85 ) / 5 = 72 answer : b" | a ) 29 , b ) 72 , c ) 39 , d ) 37 , e ) 75 | b | divide(add(add(add(add(61, 65), 82), 67), 85), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
7500 + ( 1250 / 50 ) | explanation : as per bodmas rule first we will solve the terms in the bracket then other . = 7500 + ( 25 ) = 7525 option b | a ) 7500 , b ) 7525 , c ) 7550 , d ) 8000 , e ) none of these | b | add(7500, divide(1250, 50)) | divide(n1,n2)|add(n0,#0) | general |
the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 50 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 500 gm ) of tea and 1 / 2 kg ( 500 gm ) of coffee ( equal quantities ) = 50 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 50 = > x = 50 thus proved . . . option a ." | a ) 50 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | a | divide(50, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | general |
in a certain town , the ratio of ny yankees fans to ny mets fans is 2 : 1 , and the ratio of ny mets fans to boston red sox fans is 3 : 7 . if there are 320 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ? | "the ratio of yankees : mets : red sox = 6 : 3 : 7 the mets fans are 3 / 16 of the population . ( 3 / 16 ) * 320 = 60 the answer is d ." | a ) 48 , b ) 52 , c ) 56 , d ) 60 , e ) 64 | d | multiply(divide(320, add(add(multiply(2, 1), 3), 7)), 3) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)| | other |
solve below question 2 x + 1 = - 11 | "2 x + 1 = - 11 x = - 6 b" | a ) - 8 , b ) - 6 , c ) 9 , d ) 8 , e ) - 7 | b | divide(negate(add(11, 1)), 2) | add(n1,n2)|negate(#0)|divide(#1,n0)| | general |
if 7 / w + 7 / x = 7 / y and wx = y , then the average ( arithmetic mean ) of w and x is | given : 7 / w + 7 / x = 7 / ywx = y find : ( w + x ) / 2 = ? 7 ( 1 / w + 1 / x ) = 7 ( 1 / y ) - divide both sides by 7 ( 1 / w + 1 / x ) = 1 / y ( x + w ) / wx = 1 / wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) / 2 = 1 / 2 ans : a | a ) 1 / 2 , b ) 1 , c ) 2 , d ) 4 , e ) 8 | a | divide(7, add(7, 7)) | add(n0,n0)|divide(n0,#0) | general |
a man can row 10 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 10 s = 1.2 ds = 10 + 1.2 = 11.2 us = 10 - 1.2 = 8.8 x / 11.2 + x / 8.8 = 1 x = 4.92 . answer : b" | a ) 5.02 , b ) 4.92 , c ) 4.82 , d ) 4.72 , e ) 4.61 | b | multiply(divide(multiply(add(10, 1.2), subtract(10, 1.2)), add(add(10, 1.2), subtract(10, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
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