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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a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is ? | "by the rule of alligation : profit of first part profit of second part so , ratio of 1 st and 2 nd parts = 4 : 6 = 2 : 3 . quantity of 2 nd kind = ( 3 / 5 x 1000 ) kg = 600 kg answer : c" | a ) 700 kg , b ) 500 kg , c ) 600 kg , d ) 800 kg , e ) 900 kg | c | multiply(multiply(add(const_2, const_3), const_2), multiply(subtract(add(divide(14, const_100), const_1), add(const_1, divide(8, const_100))), 1000)) | add(const_2,const_3)|divide(n3,const_100)|divide(n1,const_100)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|subtract(#3,#4)|multiply(n0,#6)|multiply(#5,#7)| | gain |
a rower whose speed is 8 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 4 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? | "time upstream = d / 4 time downstream = d / 12 total time = d / 4 + d / 12 = d / 3 average speed = 2 d / ( d / 3 ) = 6 km / hr the answer is b ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(add(subtract(8, const_0.5), add(8, 4)), const_2) | add(n0,const_0.5)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
how many pounds of salt at 50 cents / lb must be mixed with 40 lbs of salt that costs 25 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 25 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt ... | a ) 20 , b ) 15 , c ) 40 , d ) 60 , e ) 25 | d | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(25, 40))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | add(n3,const_100)|multiply(n1,n4)|multiply(n1,n2)|divide(#0,const_100)|multiply(#3,#2)|multiply(n0,#3)|subtract(#1,#4)|subtract(#5,n4)|divide(#6,#7)| | gain |
if 10 gallons of grape juice are added to 40 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ? | "mixture contains 4 gallons of grape juice out of 40 gallons after addition : 14 gallons of grape juice out of 50 gallons = = > 28 % answer c" | a ) 14 % , b ) 25 % , c ) 28 % , d ) 34 % , e ) 50 % | c | multiply(divide(add(multiply(divide(10, const_100), 40), 10), add(40, 10)), const_100) | add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|add(n0,#2)|divide(#3,#0)|multiply(#4,const_100)| | general |
18 men can complete a piece of work in 30 days . in how many days can 15 men complete that piece of work ? | "b 36 days 18 * 30 = 15 * x = > x = 36 days" | a ) 23 days , b ) 36 days , c ) 22 days , d ) 29 days , e ) 20 days | b | divide(multiply(30, 18), 15) | multiply(n0,n1)|divide(#0,n2)| | physics |
if the sales tax be reduced from 4 ( 1 / 4 ) % to 1 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 4500 ? | explanation : required difference = ( 4 ( 1 / 4 ) of rs . 4500 ) - ( 1 ( 1 / 2 ) of rs . 4500 ) = ( 17 / 4 – 3 / 2 ) % of rs . 4500 = ( 11 / 4 ) x ( 1 / 100 ) x 4500 = rs . 123.75 answer c | a ) rs . 156.66 , b ) rs . 150.23 , c ) rs . 123.75 , d ) rs . 135.06 , e ) none of these | c | subtract(multiply(add(divide(add(4, divide(1, 4)), const_100), 1), divide(4500, add(divide(add(divide(1, 2), 1), const_100), 1))), 4500) | divide(n1,n0)|divide(n1,n5)|add(n0,#0)|add(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|add(n1,#4)|add(n1,#5)|divide(n6,#7)|multiply(#6,#8)|subtract(#9,n6) | general |
two , trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is : | let the slower train have speed of 1 unit and faster train a speed of r units of speed . let t = time to when the trains meet . rt 1 t howrah | - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - | patna meet they meet rt units from howrah and 1 t units from patna for the second part of the journey the ... | a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) 9 : 14 | b | sqrt(divide(16, 9)) | divide(n1,n0)|sqrt(#0) | physics |
in a colony of 70 resident s , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of education to uneducated persons is 8 : 27 , then find the ratio of the number of educated and uneducated men in the colony ? | number of men in the colony = 4 / 7 ( 70 ) = 40 number of women in the colony = 3 / 7 ( 70 ) = 30 number of educated women in the colony = 1 / 5 ( 30 ) = 6 number of uneducated women in the colony = 4 / 5 ( 30 ) = 24 number of educated persons in the colony = 8 / 35 ( 70 ) = 16 as 6 females are educated , remaining 10 ... | a ) 1 : 9 , b ) 1 : 6 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 ratio 3 | e | divide(subtract(multiply(divide(8, add(8, 27)), 70), multiply(divide(1, add(1, 4)), multiply(divide(3, add(4, 3)), 70))), subtract(multiply(divide(4, add(4, 3)), 70), subtract(multiply(divide(8, add(8, 27)), 70), multiply(divide(1, add(1, 4)), multiply(divide(3, add(4, 3)), 70))))) | add(n5,n6)|add(n1,n3)|add(n1,n2)|divide(n5,#0)|divide(n3,#1)|divide(n2,#2)|divide(n1,#2)|multiply(n0,#3)|multiply(n0,#5)|multiply(n0,#6)|multiply(#4,#8)|subtract(#7,#10)|subtract(#9,#11)|divide(#11,#12) | other |
what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 300 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 300 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 30 ( 10 ) = 300 so we know that there are 30 total nu... | a ) 155 , b ) 195 , c ) 200 , d ) 205 , e ) 210 | a | divide(divide(multiply(add(10, 300), add(divide(subtract(300, 10), 10), const_1)), const_2), add(divide(subtract(300, 10), 10), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
if 0.75 : x : : 5 : 8 , then x is equal to | "sol . ( x × 5 ) = ( 0.75 × 8 ) ⇒ x = 6 / 5 = 1.20 . answer c" | a ) 1.12 , b ) 1.25 , c ) 1.20 , d ) 1.3 , e ) none | c | divide(multiply(0.75, 8), 5) | multiply(n0,n2)|divide(#0,n1)| | general |
the reciprocal of the hcf and lcm of two are 1 / 13 and 1 / 312 . if one of the number is 24 then other no . is | reciprocal of the hcf and lcm of two are 1 / 13 and 1 / 312 so , hcf = 13 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 13 * 312 / 24 = 169 answer : e | a ) 126 , b ) 136 , c ) 146 , d ) 156 , e ) 169 | e | divide(multiply(13, 312), 24) | multiply(n1,n3)|divide(#0,n4) | physics |
the ratio of two no . addition and subtraction be 4 : 3 . the what is the ratio of numbers ? | ( x + y ) / ( x - y ) = 4 / 3 dividing numerator and denominator by y ( x / y ) + 1 / ( x / y ) - 1 = 4 / 3 let x / y be z z + 1 / z - 1 = 4 / 3 3 z + 3 = 4 z - 4 z = 7 x / y = 7 answer e | a ) 3 : 2 , b ) 4 : 3 , c ) 5 : 1 , d ) 6 : 5 , e ) 7 : 1 | e | add(4, 3) | add(n0,n1) | general |
among 300 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 90 using your analysis : we see that the total number of students who study either maths or sociology = 132 + 168 - 90 = 210 so , in the image we know that the number of students in the ... | a ) 30 , b ) 90 , c ) 120 , d ) 172 , e ) 188 | b | subtract(300, subtract(add(multiply(divide(300, const_100), 56), multiply(divide(300, const_100), 44)), multiply(divide(300, const_100), 30))) | divide(n0,const_100)|multiply(n1,#0)|multiply(n2,#0)|multiply(n4,#0)|add(#1,#2)|subtract(#4,#3)|subtract(n0,#5)| | other |
6 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 6 / 36 = . 16666 answer ( b )" | a ) 0.004 , b ) 0.16666 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(6, 0.03), divide(6, 0.37))) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|inverse(#2)| | general |
a marketing survey of anytown found that the ratio of trucks to sedans to motorcycles was 3 : 7 : 2 , respectively . given that there are 9,800 sedans in anytown , how many motorcycles are there ? | "let the total number of trucks = 3 x total number of sedans = 7 x total number of motorcycles = 2 x total number of sedans = 9800 = > 7 x = 9800 = > x = 1400 total number of motorcycles = 2 x = 2 * 1400 = 2800 answer a" | a ) 2800 , b ) 2100 , c ) 3600 , d ) 4200 , e ) 5200 | a | multiply(divide(add(multiply(multiply(3, 3), const_1000), const_100), 7), 2) | multiply(n0,n0)|multiply(#0,const_1000)|add(#1,const_100)|divide(#2,n1)|multiply(n2,#3)| | other |
in a can , there is a mixture of milk and water in the ratio 1 : 5 . if it is filled with an additional 2 litres of milk the can would be full and ratio of milk and water would become 2.00001 : 5.00001 . find the capacity of the can ? | let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 1 / 6 ( t - 2 ) after adding milk , quantity of milk in the mixture = 2 / 7 t . 2 t / 7 - 2 = 1 / 6 ( t - 2 ) 5 t = 84 - 14 = > t = 14 . answer : a | a ) 14 , b ) 44 , c ) 48 , d ) 50 , e ) 56 | a | add(add(multiply(5, divide(2, subtract(multiply(divide(2.00001, 5.00001), 5), 1))), divide(2, subtract(multiply(divide(2.00001, 5.00001), 5), 1))), 2) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5) | general |
the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 55 kms . what was the total distance travelled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 55 + 57 + 59 + . . . . . upto 12 terms ) this is an a . p with first term , a = 55 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 55 + { 12 - 1 ) x 2 ] = 6 ( 132 ) = 792 kms . answer : d" | a ) 252 kms , b ) 152 kms , c ) 552 kms , d ) 792 kms , e ) 152 kms | d | multiply(add(multiply(2, 55), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics |
45 x ? = 35 % of 900 | "answer let 45 x a = ( 35 x 900 ) / 100 ∴ a = ( 35 x 9 ) / 45 = 7 correct option : b" | a ) 16.2 , b ) 7 , c ) 5 , d ) 500 , e ) none | b | divide(multiply(divide(35, const_100), 900), 45) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)| | general |
the ratio of radius of a circle and the side of a square is 2 : 13 . find the ratio of their areas : | radius / side = 2 / 13 â ‡ ’ area of circle / area of square = 4 / 169 answer : d | ['a ) 2 : 1', 'b ) 4 : 7', 'c ) 8 : 77', 'd ) 4 : 169', 'e ) none'] | d | power(divide(2, 13), const_2) | divide(n0,n1)|power(#0,const_2) | geometry |
if the average of 54 , 55 , 57 , 58 , 59 , 62 , 62 , 63 , 65 and x is 60 , what is the value of x ? | "sum of the deviations of the numbers in the set from the mean is always zero 54 , 55 , 57 , 58 , 59 , 62 , 62 , 63 , 65 mean is 60 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence t... | a ) 60 , b ) 62 , c ) 64 , d ) 65 , e ) 66 | d | subtract(multiply(60, const_10), add(add(add(add(add(add(add(add(54, 55), 57), 58), 59), 62), 62), 63), 65)) | add(n0,n1)|multiply(n9,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n5,#5)|add(n7,#6)|add(n8,#7)|subtract(#1,#8)| | general |
a ranch has both horses and ponies . exactly 5 / 6 of the ponies have horseshoes , and exactly 2 / 3 of the ponies with horseshoes are from iceland . if there are 4 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ? | "5 / 6 * p have horseshoes , so p is a multiple of 6 . 2 / 3 * 5 / 6 * p = 5 / 9 * p are icelandic ponies with horseshoes , so p is a multiple of 9 . the minimum value of p is 18 . then h = p + 4 = 22 . the minimum number of horses and ponies is 40 . the answer is d ." | a ) 18 , b ) 21 , c ) 38 , d ) 40 , e ) 57 | d | add(add(lcm(6, multiply(divide(6, 2), 3)), 4), lcm(6, multiply(divide(6, 2), 3))) | divide(n1,n2)|multiply(n3,#0)|lcm(n1,#1)|add(n4,#2)|add(#3,#2)| | general |
20 , 22 , 26 , 34 , 41 , 46 , 56 , 67 , __ ? | "next no . = previous + ( sum of digits ) 20 + ( 2 + 0 ) = 22 22 + ( 2 + 2 ) = 26 26 + ( 2 + 6 ) = 34 - - - - - - - - 56 + ( 5 + 6 ) = 67 67 + ( 6 + 7 ) = 80 answer : b" | a ) 72 , b ) 80 , c ) 88 , d ) 68 , e ) 78 | b | subtract(negate(34), multiply(subtract(22, 26), divide(subtract(22, 26), subtract(20, 22)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
the sale price sarees listed for rs . 278 after successive discount is 18 % and 19 % is ? | "explanation : 278 * ( 88 / 100 ) * ( 81 / 100 ) = 198 answer : e" | a ) 321 , b ) 276 , c ) 342 , d ) 265 , e ) 198 | e | subtract(subtract(278, divide(multiply(278, 18), const_100)), divide(multiply(subtract(278, divide(multiply(278, 18), const_100)), 19), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
in a group of 90 students , 36 are taking history , and 30 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ? | "number of students taking history = h = 36 number of students taking statistics = s = 30 total number of students = t = 90 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 90 - 59 = 31 letxbe the number of students taking both history and s... | a ) 9 , b ) 19 , c ) 23 , d ) 27 , e ) 29 | e | subtract(36, subtract(add(36, 30), 59)) | add(n1,n2)|subtract(#0,n3)|subtract(n1,#1)| | other |
what least number must be subtracted from 427398 so that remaining no . is divisible by 14 | "explanation : on dividing 427398 by 14 we get the remainder 6 , so 6 should be subtracted answer : option c" | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(427398, multiply(floor(divide(427398, 14)), 14)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
if the average ( arithmetic mean ) of a and b is 20 , and c – a = 30 , what is the average of b and c ? | "a + b / 2 = 20 = > a + b = 40 a = c - 30 . . . sub this value c - 30 + b = 40 = > c + b = 70 = > c + b / 2 = 35 answer : d" | a ) 25 , b ) 30 , c ) 40 , d ) 35 , e ) 45 | d | subtract(multiply(30, const_2), multiply(20, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
three unbiased coins are tossed . what is the probability of getting 1 heads and 1 tail ? | "let , h - - > head , t - - > tail here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting 3 heads then e = { tht , hth } p ( e ) = n ( e ) / n ( s ) = 2 / 8 = 1 / 4 answer is b" | a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 8 , d ) 7 / 8 , e ) 1 / 8 | b | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
the average monthly salary of laborers and supervisors in a factory is rs . 1250 per month ; where as the average monthly salary of 6 supervisors is rs . 2450 . if the average monthly salary of the laborers is rs . 950 find the number of laborers ? | 5 x 6 x 2 x 50 25 100 250 x + 150 x + 200 x = 4200 600 x = 4200 x = 7 = > 6 x = 42 answer : b | a ) 87 , b ) 42 , c ) 78 , d ) 76 , e ) 26 | b | subtract(subtract(multiply(divide(subtract(multiply(2450, 6), multiply(1250, 6)), subtract(1250, 950)), const_2), const_4), const_2) | multiply(n1,n2)|multiply(n0,n1)|subtract(n0,n3)|subtract(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|subtract(#5,const_4)|subtract(#6,const_2) | general |
the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 40 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 40 % then output equals 154 to restore new holidy demand output to original 100 . final - initial / final * 100 = 54 / 154 * 100 = 35 % approxiamately . option d is correct ." | a ) 20 % , b ) 24 % , c ) 30 % , d ) 35 % , e ) 79 % | d | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 40), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 40), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|divide(#5,#4)|multiply(#6,const_100)| | general |
balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 424 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the t... | as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 424 ... | a ) 1176 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(424, multiply(const_4, const_100)), add(multiply(subtract(424, multiply(const_4, const_100)), 2), const_1)) | multiply(const_100,const_4)|subtract(n1,#0)|multiply(n0,#1)|add(#2,const_1)|multiply(#3,#1) | general |
a wholesaler wishes to sell 100 pounds of mixed nuts at $ 2.50 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ? | "cashews / peanuts = ( mean price - peanuts price ) / ( cashew price - mean price ) = ( 2.5 - 1.5 ) / ( 4 - 2.5 ) = 2 / 3 cashew = ( 2 / 5 ) * 100 = 40 kg answer is a ." | a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | divide(multiply(subtract(4.00, 2.50), 100), 2.50) | subtract(n3,n1)|multiply(n0,#0)|divide(#1,n1)| | general |
each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 44 hours . if harry and anni... | "annie earned 40 x + 4 ( 2 x ) = 48 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 48 x ( 1.5 x ) ( h ) = 63 x h = 42 hours the answer is b ." | a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | b | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
two pipes a and b can fill a cistern in 9 and 18 minutes respectively , and a third pipe c can empty it in 24 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? | "1 / 9 + 1 / 18 - 1 / 24 = 1 / 8 8 / 1 = 8 answer : c" | a ) 6 min , b ) 7 min , c ) 8 min , d ) 10 min , e ) 17 min | c | add(multiply(9, subtract(const_1, multiply(add(inverse(9), inverse(18)), 24))), 24) | inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n0,#4)|add(n2,#5)| | physics |
laura took out a charge account at the general store and agreed to pay 5 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 5 % interest = ( 5 / 100 ) * 35 = 1.75 $ total amount that laura owes a year later = 35 + 1.75 = 36.75 $ answer b" | a ) $ 2.10 , b ) $ 36.75 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80 | b | add(multiply(divide(5, const_100), 35), 35) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in ... | "first $ 20 million : royalties / sales ratio = 3 / 20 = 36 / 240 next $ 108 million : royalties / sales ratio = 9 / 108 = 1 / 12 = 20 / 240 answer : c" | a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | c | multiply(divide(subtract(multiply(divide(3, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(3, 20), const_100)), const_100) | divide(n0,n1)|divide(n2,n3)|multiply(#0,const_100)|multiply(#1,const_100)|subtract(#2,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
what is the sum of the integers from - 194 to 195 inclusive ? | "sum / n = average . sum = ( average ) ( n ) average = a + b / 2 = 194 + 195 / 2 = 0.5 number of items ( n ) = b - a + 1 = 195 - ( - 194 ) + 1 = 195 + 195 = 390 . sum = average * n = 0.5 * 390 = 195 . answer is c" | a ) 0 , b ) 5 , c ) 195 , d ) 875 , e ) 965 | c | divide(multiply(194, 195), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
if x and y are integers such that | y + 3 | ≤ 3 and 2 y – 3 x + 6 = 0 , what is the least possible value q of the product xy ? | "how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + 3 | ≤ 3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + 3 ≤ 3 y ≤ 0 so y is 0 or negative second scenario - remove the... | a ) - 12 , b ) - 3 , c ) 0 , d ) 2 , e ) none of the above | c | multiply(divide(add(6, multiply(negate(add(3, 3)), 2)), 3), negate(add(3, 3))) | add(n0,n0)|negate(#0)|multiply(n2,#1)|add(n4,#2)|divide(#3,n0)|multiply(#4,#1)| | general |
the average of first seven prime numbers which are between 30 and 70 is | "explanation : first seven prime numbers which are between 30 and 70 = 31 , 37 , 41 , 43 , 47 , 53 , 59 average = ( 31 + 37 + 41 + 43 + 47 + 53 + 59 ) / 7 = 44.4 answer : c" | a ) 35.4 , b ) 42 , c ) 44.4 , d ) 57 , e ) 67 | c | add(30, const_1) | add(n0,const_1)| | general |
a batsman makes a score of 90 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 90 = total runs scored in 17 innings = > 16 ( x - 3 ) + 90 = 17 x = > 16 x - 48 + 90 = 17 x = > x = 42 answer is e ." | a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 42 | e | add(subtract(90, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
if x / y = 3 / z , then 4 x ^ 2 = | "this question is most easily solved by isolating y in the equation and substituting into the expression 4 x ² : x / y = 3 / z x = 3 y / z if we substitute 3 y / z into the expression for x , we get : 4 ( 3 y / z ) ² = 4 ( 9 y ² / z ² ) = 36 y ² / z ² . the correct answer is choice ( d ) ." | a ) y / z , b ) xy , c ) y ² / z ² , d ) 36 y ² / z ² , e ) 15 y ² / z ² | d | divide(add(3, 4), subtract(3, 4)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
the captain of a cricket team of 11 members is 25 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x â € “ ( 25 + 28 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 44 2 x = 44 x = 22 . so , average age of the team is 22 years . answer c" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 24 years | c | divide(subtract(add(25, add(25, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orang... | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is ... | a ) $ 0.15 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.6 | e | divide(multiply(add(const_1, const_1), 0.90), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
if a person walks at 25 km / hr instead of 10 km / hr , he would have walked 31 km more . the actual distance traveled by him is ? | "let the actual distance traveled is ' x ' km then by given conditions , we have x / 10 = ( x + 31 ) / 25 25 x = 10 x + 310 x = 20.67 km ans - c" | a ) 30 , b ) 25 , c ) 20.67 , d ) 12.33 , e ) 19.48 | c | multiply(10, divide(31, subtract(25, 10))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
the speed of a boat in still water is 8 kmph . if it can travel 1 km upstream in 1 hr , what time it would take to travel the same distance downstream ? | "speed of the boat in still water = 8 km / hr speed upstream = 1 ⁄ 1 = 1 km / hr speed of the stream = 8 - 1 = 7 km / hr speed downstream = ( 8 + 7 ) = 15 km / hr answer is b time taken to travel 1 km downstream = 1 / 15 hr = ( 1 × 60 ) / 15 = 4 minutes" | a ) 14 minutes , b ) 4 minutes , c ) 24 minutes , d ) 28 minutes , e ) 23 minutes | b | subtract(8, 1) | subtract(n0,n1)| | physics |
# p is defined as 2 p - 20 for any number p . what is p , if # ( # ( # p ) ) = 12 ? | # p = 2 p - 20 - - - > # ( # p ) = 2 ( 2 p - 20 ) - 20 = 4 p - 60 and thus # ( 4 p - 60 ) = 2 ( 4 p - 60 ) - 20 = 8 p - 140 = 12 - - - > 8 p = 152 - - - > p = 19 , b is the correct answer . | a ) – 108 , b ) 19 , c ) 10 , d ) 16 , e ) 18 | b | divide(add(add(20, multiply(20, 12)), 12), multiply(2, const_4)) | multiply(n1,n2)|multiply(n0,const_4)|add(n1,#0)|add(n2,#2)|divide(#3,#1)| | general |
if 90 percent of 600 is 50 percent of x , then x = ? | "0.9 * 600 = 0.5 * x x = 9 / 5 * 600 = 1080" | a ) 100 , b ) 1000 , c ) 1080 , d ) 1020 , e ) 1200 | c | divide(multiply(90, 600), 50) | multiply(n0,n1)|divide(#0,n2)| | general |
a number increased by 20 % gives 1080 . the number is ? | "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 1080 ( 120 ã — 9 = 1080 ) 100 % - - - - - - - > 900 ( 100 ã — 9 = 900 ) option ' d '" | a ) 800 , b ) 700 , c ) 500 , d ) 900 , e ) 600 | d | divide(1080, add(const_1, divide(20, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
a train 500 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 500 / 20 s = 25 m / sec speed = 25 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 90 kmph answer : d" | a ) 87 kmph , b ) 97 kmph , c ) 72 kmph , d ) 90 kmph , e ) 19 kmph | d | divide(divide(500, const_1000), divide(20, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
if 5 < x < 12 and y = x + 3 , what is the greatest possible integer value of x + y ? | "x + y = x + x + 3 = 2 x + 3 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 11.5 then x + y = 11.5 + 14.5 = 26 . the answer is d ." | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | d | add(add(3, const_10), const_10) | add(n2,const_10)|add(#0,const_10)| | general |
on dividing a number by 357 , we get 42 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 42 = ( 17 * 21 * y ) + ( 17 * 2 ) + 8 = 17 * ( 21 y + 2 ) + 8 . required number = 8 . answer is c" | a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | c | multiply(subtract(divide(power(const_3.0, const_2), 357), floor(divide(power(42, const_2), 357))), 357) | power(const_3.0,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
evaluate : 20 - 16 ÷ 4 × 3 = | "according to order of operations , 16 ÷ 4 × 3 ( division and multiplication ) is done first from left to right 16 ÷ 4 × 3 = 4 × 3 = 12 hence 20 - 16 ÷ 4 × 3 = 20 - 12 = 8 correct answer d ) 8" | a ) 16 , b ) 10 , c ) 4 , d ) 8 , e ) 6 | d | subtract(20, multiply(multiply(16, 4), 3)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a train 1020 m long running at 102 kmph crosses a platform in 50 sec . what is the length of the platform ? | "d = 102 * 5 / 18 = 50 = 1416 â € “ 1020 = 396 answer : c" | a ) 287 , b ) 298 , c ) 396 , d ) 726 , e ) 267 | c | subtract(multiply(50, multiply(102, const_0_2778)), 1020) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a seller of used cars has 15 cars to sell and each of his clients selected 3 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ? | "ifno caris selected more than once then the number of clients = 15 / 3 = 5 but since every car is being selected three times so no . of clients must be thrice as well = 5 * 3 = 15 answer : option e" | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | e | multiply(divide(15, 3), const_3) | divide(n0,n1)|multiply(#0,const_3)| | general |
if a rectangular room measures 11 meters by 6 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "e . 264 , 000,000 11 * 100 * 6 * 100 * 4 * 100 = 264 , 000,000" | a ) 24,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 24 , 000,000 , e ) 264 , 000,000 | e | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(6, 100)), multiply(11, 100)) | divide(n3,const_10)|multiply(n2,n4)|multiply(n1,n4)|multiply(n0,n4)|multiply(#0,#1)|multiply(#4,#2)|multiply(#5,#3)| | geometry |
a rock is dropped into a well and the distance traveled is 16 t 2 feet , where t is the time . if the water splash is heard 3 seconds after the rock was dropped , and that the speed of sound is 1100 ft / sec , approximate the height of the well | let t 1 be the time it takes the rock to reach the bottom of the well . if h is the height of the well , we can write h = 16 t 1 2 let t 2 be the time it takes sound wave to reach the top of the well . we can write h = 1100 t 2 the relationship between t 1 and t 2 is t 1 + t 2 = 3 eliminate h and combine the equations ... | a ) 134.8 feet , b ) 136.2 feet , c ) 132.7 feet , d ) 132.2 feet , e ) 115.3 feet | c | multiply(16, multiply(divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)), divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)))) | multiply(n3,n3)|multiply(n2,n3)|multiply(n0,const_4)|multiply(n0,const_2)|multiply(#1,#2)|add(#0,#4)|sqrt(#5)|subtract(#6,n3)|divide(#7,#3)|multiply(#8,#8)|multiply(n0,#9) | physics |
a man can row his boat with the stream at 18 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 18 us = 4 s = ? s = ( 18 - 4 ) / 2 = 7 kmph answer : d" | a ) 1 kmph , b ) 3 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | d | divide(subtract(18, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
how is 2 % expressed as a decimal fraction ? | 2 / 100 = 0.02 answer : b | a ) 0.2 , b ) 0.02 , c ) 0.002 , d ) 0.0002 , e ) 2 | b | divide(2, const_100) | divide(n0,const_100) | gain |
for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers v are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | add(n0,n0)| | general |
25.25 / 3000 is equal to : | "25.25 / 3000 = 2525 / 300000 = 0.008416667 answer : a" | a ) 0.008416667 , b ) 0.110773333 , c ) 0.12526234 , d ) 0.01072333 , e ) 0.12725002 | a | divide(25.25, 3000) | divide(n0,n1)| | general |
a man can row downstream at 28 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 28 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 38 = > x = 19 , y = 9 . answer : d" | a ) 2 , 9 , b ) 4 , 9 , c ) 8 , 9 , d ) 19 , 9 , e ) 7 , 9 | d | divide(divide(add(28, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
two bullet trains of equal lengths take 8 seconds and 15 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | speed of the first bullet train = 120 / 8 m / sec = 15 m / sec . speed of the second bullet train = 120 / 15 m / sec = 8 m / sec . relative speed = ( 15 + 8 ) = 23 m / sec . required time = ( 120 + 120 ) / 23 sec = 10.4 sec . b | a ) 13 sec . , b ) 10.4 sec . , c ) 12 sec . , d ) 17 sec . , e ) 19 sec . | b | divide(add(120, 120), add(speed(120, 8), speed(120, 15))) | add(n2,n2)|speed(n2,n0)|speed(n2,n1)|add(#1,#2)|divide(#0,#3) | physics |
how many even number in the range between 10 to 170 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 170 using ap formula , we can say 170 = 10 + ( n - 1 ) * 2 or n = 81 . for 6 , 12,18 , . . . 168 using ap formula , we can say 168 = 12 + ( n... | a ) 15 , b ) 30 , c ) 31 , d ) 33 , e ) 54 | e | subtract(divide(subtract(subtract(170, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(170, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)| | general |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 600 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 600 = > breadth = 30 feet area to be fenced = 2 b + l = 2 * 30 + 20 = 80 feet answer : option a" | a ) 80 feet , b ) 70 feet , c ) 60 feet , d ) 50 feet , e ) 20 feet | a | add(multiply(divide(600, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
when positive integer x is divided by positive integer y , the remainder is 8 . if x / y = 96.16 , what is the value of y ? | "by the definition of a remainder , the remainder here is equal to 8 / y . the remainder in decimal form is given as . 16 therefore , 8 / y = . 16 solve for y and get 50 . c" | a ) 96 , b ) 75 , c ) 50 , d ) 25 , e ) 12 | c | divide(8, subtract(96.16, floor(96.16))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
the mean of 30 values was 140 . it was detected on rechecking that one value 145 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 140 × 30 − 135 + 145 / 30 = 4200 − 135 + 145 / 30 = 4210 / 30 = 140.33 answer b" | a ) 151 , b ) 140.33 , c ) 152 , d ) 148 , e ) none of the above | b | divide(add(multiply(30, 140), subtract(145, 135)), 30) | multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)| | general |
two trains 110 meters and 180 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 110 + 180 ) / ( 80 + 65 ) * 18 / 5 t = 7.20 answer : a" | a ) 7.2 , b ) 7.85 , c ) 6.85 , d ) 5.85 , e ) 6.15 | a | divide(add(110, 180), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the perimeter of a triangle is 20 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 20 / 2 = 25 cm 2 answer : d" | a ) 28 cm 2 , b ) 27 cm 2 , c ) 29 cm 2 , d ) 25 cm 2 , e ) 35 cm 2 | d | triangle_area(2.5, 20) | triangle_area(n0,n1)| | geometry |
you buy a piece of land with an area of √ 900 , how long is one side of the land plot ? | try filling the numbers into the answer y x y = find the closest to 900 . answer c | ['a ) 28', 'b ) 29', 'c ) 30', 'd ) 31', 'e ) 32'] | c | sqrt(900) | sqrt(n0) | geometry |
a large box contains 21 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 21 * 25 = 525 correct answer e" | a ) 250 , b ) 350 , c ) 450 , d ) 550 , e ) 525 | e | multiply(21, 25) | multiply(n0,n1)| | general |
a typist uses a sheet measuring 20 cm by 30 cm lenghtwise . if a margin of 2 cm is left on each side and a 3 cm margin on the top and bottom , then what is the percentage of page used by the typist ? | area of the sheet = ( 20 * 30 ) cm 2 = 600 cm 2 area used for typing = ( [ 20 - 4 ] * [ 30 - 6 ] ) cm 2 = 384 cm 2 therefore required percentage = ( 384 / 600 ) * 100 = 64 % answer c | a ) 62 % , b ) 63 % , c ) 64 % , d ) 65 % , e ) 66 % | c | multiply(divide(multiply(subtract(20, multiply(2, const_2)), subtract(30, multiply(3, const_2))), multiply(20, 30)), const_100) | multiply(n2,const_2)|multiply(n3,const_2)|multiply(n0,n1)|subtract(n0,#0)|subtract(n1,#1)|multiply(#3,#4)|divide(#5,#2)|multiply(#6,const_100) | gain |
an association of mathematics teachers has 1,260 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 62 percent of the votes cast ? | "total number of members = 1260 number of members that cast votes = 525 since , winning candidate received 62 percent of the votes cast number of votes for winning candidate = ( 62 / 100 ) * 525 = 325.5 percent of total membership that voted for winning candidate = ( 325.5 / 1260 ) * 100 = 25.83 % answer e" | a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 25.83 % | e | multiply(divide(multiply(divide(62, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | divide(n2,const_100)|power(const_4,const_2)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 161 ? | "100 + 102 + . . . + 160 = 100 * 31 + ( 2 + 4 + . . . + 60 ) = 100 * 31 + 2 * ( 1 + 2 + . . . + 30 ) = 100 * 31 + 2 ( 30 ) ( 31 ) / 2 = 100 * 31 + 30 * 31 = 130 ( 31 ) = 4030 the answer is d ." | a ) 2670 , b ) 2980 , c ) 3550 , d ) 4030 , e ) 4540 | d | add(divide(subtract(subtract(161, 1), add(99, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
a clock shows the time as 11 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 6 p . m . ? | "there are 7 hours in between 11 a . m . to 6 p . m . 7 * 5 = 35 minutes . answer : c" | a ) 45 minutes , b ) 55 minutes , c ) 35 minutes , d ) 25 minutes , e ) 40 minutes | c | multiply(add(const_3, const_4), 5) | add(const_3,const_4)|multiply(n1,#0)| | physics |
a man is 26 years older than his son . in two years , his age will be twice the age of his son . what is the present age of his son ? | "let present age of the son = x years then , present age the man = ( x + 26 ) years given that , in 2 years , man ' s age will be twice the age of his son â ‡ ’ ( x + 26 ) + 2 = 2 ( x + 2 ) â ‡ ’ x = 24 answer : c" | a ) 23 years , b ) 22 years , c ) 24 years , d ) 20 years , e ) 19 years | c | divide(subtract(26, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a and b can do a piece of work in 6 days . with the help of c they finish the work in 5 days . c alone can do that piece of work in ? | "c = 1 / 5 – 1 / 6 = 1 / 30 = > 30 days answer : e" | a ) 15.5 days , b ) 19.5 days , c ) 17.5 days , d ) 16.5 days , e ) 30 days | e | inverse(subtract(5, divide(5, 6))) | divide(n1,n0)|subtract(n1,#0)|inverse(#1)| | physics |
an aeroplane covers a certain distance at a speed of 120 kmph in 5 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 120 x 5 ) = 600 km . speed = distance / time speed = 600 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 600 x 3 / 5 ) km / hr = 360 km / hr answer a ) 360 km / hr" | a ) 520 , b ) 620 , c ) 820 , d ) 740 , e ) 720 | a | divide(divide(multiply(120, 5), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)| | physics |
a man swims downstream 18 km and upstream 12 km taking 3 hours each time , what is the speed of the man in still water ? | "18 - - - 3 ds = 6 ? - - - - 1 12 - - - - 3 us = 4 ? - - - - 1 m = ? m = ( 6 + 4 ) / 2 = 5 answer : b" | a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 8 | b | divide(add(divide(12, 3), divide(18, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.3 km ? | "b 325 m distance covered by amar = 18 / 4.8 ( 1.3 km ) = 3 / 8 ( 1300 ) = 325 m answer is b" | a ) 600 m , b ) 325 m , c ) 300 m , d ) 400 m , e ) 100 m | b | divide(multiply(18, multiply(1.3, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics |
a sum of rs . 2678 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2678 - x ) * 3 * 5 ) / 100 24 x / 100 = 40170 / 100 - 15 x / 100 39 x = 40170 = > x = 1030 second sum = 2678 â € “ 1030 = 1648 answer : b" | a ) 1629 , b ) 1648 , c ) 2677 , d ) 2986 , e ) 2679 | b | subtract(2678, divide(multiply(multiply(3, 5), 2678), add(multiply(3, 5), multiply(8, 3)))) | multiply(n2,n4)|multiply(n1,n2)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|subtract(n0,#4)| | gain |
the sum of two numbers is 24 and their product is 148 . find the sum of the squares of that numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 24 ab = 148 so , 24 ^ 2 = a ^ 2 + b ^ 2 + 2 * 148 576 = a ^ 2 + b ^ 2 + 296 a ^ 2 + b ^ 2 = 280 ans b | a ) 250 , b ) 280 , c ) 230 , d ) 290 , e ) 250 | b | divide(subtract(148, power(24, const_2)), const_2) | power(n0,const_2)|subtract(n1,#0)|divide(#1,const_2)| | general |
two numbers n and 12 have lcm = 54 and gcf = 8 . find n . | the product of two integers is equal to the product of their lcm and gcf . hence . 12 × n = 54 × 8 n = 54 × 8 / 12 = 36 correct answer c | a ) 24 , b ) 34 , c ) 36 , d ) 54 , e ) 64 | c | divide(multiply(54, 8), 12) | multiply(n1,n2)|divide(#0,n0) | physics |
the two lines y = x and x = - 6 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 6 , - 6 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length ... | a ) 16 , b ) 3 √ 2 , c ) 8 , d ) 2 √ 2 , e ) ( √ 2 ) / 3 | b | sqrt(divide(multiply(6, 6), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1)| | general |
the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 14 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 14 = 19 years . answer c" | a ) 14 years , b ) 17 years , c ) 19 years , d ) 21 years , e ) none | c | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
in the coordinate plane , one of the vertices of a square is the point ( - 6 , - 4 ) . if the diagonals of that square intersect at point ( 3 , 2 ) , what is the area of that square ? | one point ( - 6 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 6 - 3 = - 9 is the midpoint of the square - - > whole side 18 , 18 * 18 = 324 c | ['a ) 100', 'b ) 169', 'c ) 324', 'd ) 196', 'e ) 225'] | c | multiply(multiply(6, 3), multiply(6, 3)) | multiply(n0,n2)|multiply(#0,#0) | geometry |
mr . jones gave 40 % of the money he had to his wife . he also gave 20 % of the remaining amount to his 3 sons . half of the amount now left was spent on miscellaneous items and the remaining amount of rs . 12000 was deposited in the bank . how much money did mr . jones have initially ? | "let the initial amount with mr . jones be rs . x then , ( 1 / 2 ) [ 100 - ( 3 * 20 ) ] % of x = 12000 ( 1 / 2 ) * ( 40 / 100 ) * ( 60 / 100 ) * x = 12000 x = ( ( 12000 * 25 ) / 3 ) = 100000 answer a" | a ) 100000 , b ) 12000 , c ) 15000 , d ) 13000 , e ) 65000 | a | divide(12000, multiply(divide(divide(const_100, const_2), const_100), multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100))))) | divide(const_100,const_2)|divide(n0,const_100)|divide(n1,const_100)|divide(#0,const_100)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|divide(n3,#7)| | gain |
if n = 3 * 4 * p where p is a prime number greater than 3 , how many different positive non - prime divisors does n have , excluding 1 and n ? | n = 3 ∗ 2 ^ 2 ∗ p number of divisors = 2 * 3 * 2 = 12 the 12 divisors includes 1 , n , 3 , 2 and p number of different non prime divisors excluding 1 and n = 12 - 5 = 7 answer : b | a ) six , b ) seven , c ) eight , d ) nine , e ) ten | b | subtract(multiply(3, 4), add(4, 1)) | add(n1,n3)|multiply(n0,n1)|subtract(#1,#0) | general |
of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 1 / 2 times the fraction of first - year students who have dec... | "tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 1 / 2 ( 1 / 5 ( x / 2 ) ) = 1 / 20 ( x ) - - - > non major = x / 2 - 1 / 20 ( x ) = 9 / 20 ( x ) hence 9 / 20 c" | a ) 1 / 15 , b ) 1 / 5 , c ) 9 / 20 , d ) 1 / 3 , e ) 2 / 5 | c | subtract(divide(1, 2), divide(1, multiply(2, const_10))) | divide(n0,n1)|multiply(n5,const_10)|divide(n0,#1)|subtract(#0,#2)| | general |
a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 1419 , what is the value of n ? | factorise 1419 . . 3 * 11 * 43 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 43 = 86 . . ans : a | a ) 86 , b ) 38 , c ) 62 , d ) 122 , e ) 672 | a | multiply(2, divide(divide(1419, add(const_10, const_1)), const_3)) | add(const_1,const_10)|divide(n4,#0)|divide(#1,const_3)|multiply(n0,#2) | general |
what is the dividend ? the divisor is 17 , the quotient is 4 and the remainder is 8 . | "divided = divisor * quotient + remainder ? = 17 * 4 + 8 68 + 8 76 ( e . g . answer : a )" | a ) 76 , b ) 67 , c ) 176 , d ) 671 , e ) 0 | a | subtract(multiply(17, 4), 8) | multiply(n0,n1)|subtract(#0,n2)| | general |
line m lies in the xy - plane . the y - intercept of line m is - 1 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 8 ) . what is the slope of line m ? | ans : e solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 2 ) as we can see that the y axis of intercept point is ( 0 , - 1 ) means line m is parallel to x axis slope m = 0 ans : e | a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | e | divide(add(divide(subtract(4, 8), 2), const_1), divide(add(2, 6), 2)) | add(n1,n3)|subtract(n2,n4)|divide(#1,n1)|divide(#0,n1)|add(#2,const_1)|divide(#4,#3) | general |
the average of first 10 prime numbers is ? | "sum of 10 prime no . = 129 average = 129 / 10 = 12.9 . answer : c" | a ) 12.6 , b ) 12.5 , c ) 12.9 , d ) 12.2 , e ) 12.1 | c | add(10, const_1) | add(n0,const_1)| | general |
if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 50 , what is the value of f ( 5 ) ? | "f ( x ) = 2 x ^ 2 + y f ( 2 ) = 50 = > 2 * ( 2 ) ^ 2 + y = 50 = > 8 + y = 50 = > y = 42 f ( 5 ) = 2 * ( 5 ) ^ 2 + 42 = 92 answer c" | a ) 104 , b ) 60 , c ) 92 , d ) 50 , e ) 25 | c | add(subtract(50, multiply(power(2, 2), 2)), multiply(2, power(5, 2))) | power(n4,n0)|power(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|subtract(n3,#3)|add(#2,#4)| | general |
a man has an investment of $ 3000 which yields a fixed 305 for every $ 500 invested . if the man takes out $ 12.6 for every $ 500 how long will it take for him to double his investment assuming no compounding of interest . | annual increase is ( 500 + 30.5 - 12.6 ) * 6 = 3107.4 hence every year there is an increase of 107.4 for his saving to double he needs additional $ 4000 therefore $ 4000 / 107.4 = 37.3 correct option is e ) 37.3 | a ) 4 , b ) 33.8 , c ) 10 , d ) 30 , e ) 37.3 | e | add(add(add(multiply(divide(3000, 305), const_3), const_3), const_1), const_4) | divide(n0,n1)|multiply(#0,const_3)|add(#1,const_3)|add(#2,const_1)|add(#3,const_4) | general |
each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 47 hours . if harry and anni... | "annie earned 40 x + 7 ( 2 x ) = 54 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 54 x ( 1.5 x ) ( h ) = 69 x h = 46 hours the answer is d ." | a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | d | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
car a runs at the speed of 80 km / hr & reaches its destination in 5 hr . car b runs at the speed of 100 km / h & reaches its destination in 2 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 80 × 5 = 400 km distance travelled by car b = 100 × 2 = 200 km ratio = 400 / 200 = 2 : 1 answer : c" | a ) 11 : 5 , b ) 11 : 8 , c ) 2 : 1 , d ) 15 : 7 , e ) 16 : 9 | c | divide(multiply(80, 5), multiply(100, 2)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 12 x + 9 . if the sum of the perimeters of both squares is 32 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 5 ) ^ 2 & a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 & l 2 = 2 x - 3 p 1 = 4 ( x + 5 ) & p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 32 4 ( x + 5 ) + 4 ( 2 x - 3 ) = 32 . . . . . . . . . . . . . . > x = 2 answer : b" | a ) 0 , b ) 2 , c ) 2.5 , d ) 4.67 , e ) 10 | b | divide(subtract(32, subtract(multiply(4, divide(10, 2)), 10)), 10) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general |
sandy had $ 217 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ? | "let the money sandy took for shopping be x . 0.7 x = 217 x = 310 the answer is c ." | a ) $ 270 , b ) $ 290 , c ) $ 310 , d ) $ 330 , e ) $ 350 | c | divide(217, divide(subtract(const_100, 30), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain |
the cube root of . 000216 is : | answer : b ) . 06 | ['a ) 0.6', 'b ) 0.06', 'c ) 0.9', 'd ) 0.2', 'e ) 0.61'] | b | power(divide(divide(216, const_1000), const_1000), divide(const_1, const_3)) | divide(n0,const_1000)|divide(const_1,const_3)|divide(#0,const_1000)|power(#2,#1) | geometry |
at a special sale , 12 tickets can be purchased for the price of 4 tickets . if 12 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 12 tickets ? | "let the price of a ticket be rs . 100 , so 4 tickets cost 400 & 12 tickets cost 1200 12 tickets purchased at price of 4 tickets ie . , for 400 , so amount saved s rs . 800 , % of 5 tickets = ( 800 / 1200 ) * 100 = 66.6 % answer : e" | a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | e | divide(multiply(subtract(multiply(12, 12), multiply(4, 12)), const_100), multiply(12, 12)) | multiply(n0,n0)|multiply(n0,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#0)| | gain |
a person can swim in still water at 12 km / h . if the speed of water 6 km / h , how many hours will the man take to swim back against the current for 6 km ? | "m = 12 s = 6 us = 12 - 6 = 6 d = 36 t = 36 / 6 = 6 answer : d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(6, subtract(12, 6)) | subtract(n0,n1)|divide(n2,#0)| | physics |
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