Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
|---|---|---|---|---|---|---|---|
150 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 150 = 90 liters water = 50 liters 90 : ( 50 + p ) = 3 : 4 150 + 3 p = 360 = > p = 70 50 liters of water are to be added for the ratio become 3 : 4 . answer : c" | a ) 90 liters , b ) 20 liters , c ) 50 liters , d ) 20 liters , e ) 70 liters | c | multiply(divide(150, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | general | C |
seller selling an apple for rs . 16 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 16 loss = cp 17 loss = cp − sp = cp − 16 ⇒ cp 17 = cp − 16 ⇒ 16 cp 17 = 16 ⇒ cp 17 = 1 ⇒ cp = 17 c" | a ) 10 , b ) 12 , c ) 17 , d ) 18 , e ) 20 | c | add(16, 1) | add(n0,n1)| | general | C |
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 108 as interest at the end of the loan period , what was the rate of interest ? | let rate = r % and time = r years . then , 1200 x r x r / 100 = 108 12 r 2 = 108 r 2 = 9 r = 3 . answer : d | a ) 3.6 , b ) 6 , c ) 18 , d ) 3 , e ) none of these | d | sqrt(divide(multiply(108, const_100), 1200)) | multiply(n1,const_100)|divide(#0,n0)|sqrt(#1) | gain | D |
70 is subtracted from 30 % of a number , the result is 20 . find the number ? | ( 30 / 100 ) * x – 70 = 20 3 x = 900 x = 300 answer : d | a ) 150 , b ) 997 , c ) 266 , d ) 300 , e ) 271 | d | divide(add(70, 20), divide(30, const_100)) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1) | gain | D |
the ratio between the number of sheep and the number of horses at the stewart farm is 1 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | "let the number of sheeps and horses be 1 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 1 x = 1 * 8 = 8 . hence e" | a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 8 | e | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 1) | multiply(const_2,const_4)|multiply(const_1000,const_12)|multiply(#0,const_10)|multiply(#0,const_100)|add(#2,#3)|add(#4,#1)|divide(#5,n2)|divide(#6,n1)|multiply(n0,#7)| | other | E |
a mixture contains alcohol and water in the ratio 2 : 5 . if 10 liters of water is added to the mixture , the ratio becomes 2 : 7 . find the quality of alcohol in the given mixture . | let the quantity of alcohol and water be 2 x and 5 x 2 x / ( 5 x + 10 ) = 2 / 7 20 x = 4 ( 3 x + 5 ) x = 5 quantity of alcohol = 2 * 5 = 10 liters . answer is b | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | multiply(10, const_1) | multiply(n2,const_1) | general | B |
x and y invested in a business . they earned some profit which they divided in the ratio of 1 : 3 . if x invested rs . 10,000 . the amount invested by y is | "solution suppose y invested rs . y then , 10000 / y = 1 / 3 â € ¹ = â € º y = ( 10000 ã — 3 / 1 ) . â € ¹ = â € º y = 30000 . answer a" | a ) rs . 30,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | a | divide(multiply(multiply(add(const_1, const_4), const_1000), 1), 3) | add(const_1,const_4)|multiply(#0,const_1000)|multiply(n0,#1)|divide(#2,n1)| | gain | A |
when a student joe , weighing 44 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? | "after two persons leave the group the average remains the same . that means the weight of the two persons = 44 + 30 = 74 so , the average the two persons = 37 that gives the answer 44 - 37 = 7 answer e" | a ) 5.5 kg , b ) 11 kg , c ) 30 kg , d ) 36.5 kg , e ) 7 kg | e | subtract(44, divide(subtract(add(multiply(30, subtract(44, add(30, 1))), 44), multiply(subtract(subtract(44, add(30, 1)), 1), 30)), const_2)) | add(n1,n2)|subtract(n0,#0)|multiply(n1,#1)|subtract(#1,n2)|add(n0,#2)|multiply(n1,#3)|subtract(#4,#5)|divide(#6,const_2)|subtract(n0,#7)| | general | E |
from an island , it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for 2 hours at 8 kilometers per hour , while ferry q takes a route that is three times longer . if ferry p is slower than ferry q by 4 kilometers per hour , how many hours longer is the journey of ferry q compared with the journey of ferry p ? | the distance traveled by ferry p is 16 km . then the distance traveled by ferry q is 48 km . ferry q travels at a speed of 12 kph . the time of the journey for ferry q is 48 / 12 = 4 hours , which is 2 hours more than ferry p . the answer is b . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(multiply(const_3, multiply(8, 2)), add(8, 4)), 2) | add(n1,n2)|multiply(n0,n1)|multiply(#1,const_3)|divide(#2,#0)|subtract(#3,n0) | physics | B |
p alone can complete a job in 6 days . the work done by q alone in one day is equal to one - fifth of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 6 q ' s rate is 1 / 30 the combined rate is 1 / 6 + 1 / 30 = 1 / 5 if they work together , the job will take 5 days . the answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(const_1, add(divide(const_1, 6), divide(divide(const_1, 6), const_3))) | divide(const_1,n0)|divide(#0,const_3)|add(#0,#1)|divide(const_1,#2)| | physics | E |
the diameter of the wheel of a car is 120 m . how many revolution / min must the wheel makein order to keep a speed of 60 km / hour approximately ? | distance to be covered in 1 min . = ( 60 x 1000 ) / ( 60 ) m = 1000 m . circumference of the wheel = ( 2 x ( 22 / 7 ) x 0.60 ) m = 3.77 m . number of revolutions per min . = ( 1000 / 3.77 ) = 265 e | a ) 210 , b ) 245 , c ) 230 , d ) 254 , e ) 265 | e | divide(divide(multiply(60, const_1000), const_60), multiply(divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3)), divide(120, const_100))) | add(const_3,const_4)|divide(n0,const_100)|multiply(n1,const_1000)|multiply(const_10,const_2)|add(#3,const_2)|divide(#2,const_60)|divide(#4,#0)|multiply(#6,#1)|divide(#5,#7) | physics | E |
the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 54 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 54 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 54 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 54 r 2 = ( 54 * 2 ) / 3 = 36 = > r = 6 rate = 6 % answer : option e" | a ) 8 , b ) 2 , c ) 9 , d ) 4 , e ) 6 | e | sqrt(54) | sqrt(n2)| | gain | E |
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 20 terms of the arithmetic progression ? | "n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 20 . . . we get 20 / 2 [ 2 a + 14 d ] = 20 [ a + 7 d ] = 20 * 10 = 200 . . . answer is e . . ." | a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 200 | e | divide(multiply(20, 20), const_2) | multiply(n0,n1)|divide(#0,const_2)| | general | E |
if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 24 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 24 ) / 10 = ( 54 - 24 ) / 10 = 3 answer : option e" | a ) 4 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | e | divide(subtract(add(multiply(7, 7), 5), 24), 10) | multiply(n1,n1)|add(n0,#0)|subtract(#1,n3)|divide(#2,n4)| | general | E |
two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 50 hr . if all pipes are opened simultaneously , then the cistern will be filled in | solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 − 1 / 50 = 8 / 49 hence , tank will be filled in 49 / 8 = 6.12 hour option ( a ) | a ) 6.12 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 50))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4) | physics | A |
express 5 mps in kmph ? | "5 * 18 / 5 = 18 kmph answer : d" | a ) 22 , b ) 88 , c ) 90 , d ) 18 , e ) 24 | d | multiply(divide(5, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600)| | physics | D |
jacob is 39 years old . he is 3 times as old as his brother . how old will jacob be when he is twice as old ? | "j = 39 ; j = 3 b ; b = 39 / 3 = 13 ; twice as old so b = 13 ( now ) + ( 13 ) = 26 ; jacob is 39 + 26 = 65 answer : c" | a ) 63 , b ) 64 , c ) 65 , d ) 76 , e ) 67 | c | multiply(39, 3) | multiply(n0,n1)| | general | C |
a car traveled 448 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 448 / h = > 336 / c = 448 / ( c + 6 ) = > 336 c + 336 * 6 = 448 c = > c = 336 * 6 / 112 = 18 answer b ." | a ) 14 , b ) 18 , c ) 21 , d ) 22 , e ) 27 | b | divide(336, divide(subtract(448, 336), 6)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics | B |
what is the maximum number of pieces of birthday cake of size 4 ” by 4 ” that can be cut from a cake 16 ” by 16 ” ? | "the prompt is essentially asking for the maximum number of 4 x 4 squares that can be cut from a larger 16 by 16 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 4 ' pieces ' each , we have ( 4 ) ( 4 ) = 16 total smaller squares ( at maximum ) . c" | a ) 5 , b ) 4 , c ) 16 , d ) 20 , e ) 25 | c | divide(multiply(16, 16), multiply(4, 4)) | multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)| | physics | C |
3 partners a , b , c starts a business . twice a ' s capital is equal to thrice b ' s capital and b ' s capital is 4 times c ' s capital . out of a total profit of rs . 16500 at the end of the year , b ' s share is | solution let c = x . then , b = 4 x and 2 a = 3 x 4 x = 12 x or a = 6 x . ∴ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so b ' s capital = rs ( 16500 x 4 / 11 ) = rs . 6000 . answer b | a ) rs . 4000 , b ) rs . 6000 , c ) rs . 7500 , d ) rs . 6600 , e ) none | b | multiply(divide(4, add(add(3, 4), 4)), 16500) | add(n0,n1)|add(n1,#0)|divide(n1,#1)|multiply(n2,#2) | general | B |
if xy = 1 , x / y = 36 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 1 - - - > x = 1 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 1 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 1 / 36 - - - > y = 1 / 6 or - 1 / 6 the question states that x and y are positive integers . therefore , y = 1 / 6 is the answer . answer e ." | a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | e | sqrt(divide(1, 36)) | divide(n0,n1)|sqrt(#0)| | general | E |
the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 9 percent , what was the annual interest rate last year ? | "9 = 1.1 * x x = 8.18 % answer c )" | a ) 1 % , b ) 1.1 % , c ) 8.18 % , d ) 10 % , e ) 10.8 % | c | divide(multiply(9, const_100), add(9, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain | C |
a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 12 km , if the person completes half of the distance by walking and the other half by running ? | "time = 6 / 4 + 6 / 8 = 18 / 8 = 2.25 hours the answer is c ." | a ) 1.25 , b ) 1.75 , c ) 2.25 , d ) 2.75 , e ) 3.25 | c | add(divide(divide(12, const_2), 4), divide(divide(12, const_2), 8)) | divide(n2,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)| | physics | C |
if albert ’ s monthly earnings rise by 20 % , he would earn $ 560 . if , instead , his earnings rise by only 21 % , how much ( in $ ) would he earn this month ? | "= 560 / 1.2 ∗ 1.21 = 564 = 564 answer is d" | a ) 643 , b ) 652 , c ) 660 , d ) 564 , e ) 693 | d | multiply(divide(560, add(const_1, divide(20, const_100))), add(const_1, divide(21, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain | D |
an alloy of zinc and copper contains the metals in the ratio 5 : 3 . the quantity of zinc to be added to 6 kg of the alloy so that the ratio of the metal may be 3 : 1 is : | explanation : in sixteen kg of alloy 10 kg of zinc and 6 kg copper is present to make the ratio 3 : 1 we must add 8 kg of zinc to make it 18 kg of zinc and 6 kg copper so answer is 8 . answer : c | a ) 2 , b ) 9 , c ) 8 , d ) 7 , e ) 4 | c | subtract(multiply(multiply(divide(3, add(5, 3)), add(const_10, 6)), 3), multiply(divide(5, add(5, 3)), add(const_10, 6))) | add(n2,const_10)|add(n0,n1)|divide(n1,#1)|divide(n0,#1)|multiply(#0,#2)|multiply(#0,#3)|multiply(n1,#4)|subtract(#6,#5) | other | C |
in a 200 m race , if a gives b a start of 25 metres , then a wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres the race ends in a dead heat . how long does a take to run 200 m ? | explanatory answer a gives b a start of 25 metres and still wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres , then the race ends in a dead heat . therefore , the additional 20 metres start given to b compensates for the 10 seconds . i . e . , b runs 20 metres in 10 seconds . hence , b will take 100 seconds to run 200 metres . we know that a gives b a start of 45 metres . b will take 22.5 seconds to run this 45 metres as b runs 20 metres in 10 seconds or at the speed of 2 m / s . hence , a will take 22.5 seconds lesser than b or 100 - 22.5 = 77.5 seconds to complete the race . answer c | a ) 100 seconds , b ) 112.5 seconds , c ) 77.5 seconds , d ) 87.5 seconds , e ) none | c | divide(subtract(200, 45), const_2) | subtract(n0,n3)|divide(#0,const_2) | physics | C |
the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 40 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 40 kg . as the price falls by 20 % , new price for 40 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 40 = rs . 2 per kg . rice can bought now at = 100 / 2 = 50 kg . answer : option d" | a ) 5 kg , b ) 15 kg , c ) 25 kg , d ) 50 kg , e ) none | d | divide(const_100, divide(subtract(const_100, 20), 40)) | subtract(const_100,n0)|divide(#0,n1)|divide(const_100,#1)| | gain | D |
if 1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter 1000 microns = 1 , 000,000 angstroms 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : b" | a ) 10 , b ) 1000 , c ) 100 , d ) 0.01 , e ) 0.0001 | b | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)| | general | B |
the sum of the first 50 positive even integers is 2550 . what is the sum r of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : r = 2550 + ( 100 * 50 ) = 7550 . b" | a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)| | general | B |
the perimeter of a semi circle is 140 cm then the radius is ? | "36 / 7 r = 140 = > r = 27 answer : a" | a ) 27 , b ) 28 , c ) 19 , d ) 11 , e ) 12 | a | divide(140, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics | A |
what percent is 300 gm of 1 kg ? | "1 kg = 1000 gm 300 / 1000 ã — 100 = 30000 / 1000 = 30 % answer is c" | a ) 25 % , b ) 40 % , c ) 30 % , d ) 8 % , e ) 12 % | c | multiply(divide(300, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain | C |
the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability q that the next couple the maitre ' d seats will not order dessert ? | "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . q = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %" | a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | multiply(const_2,const_5)|multiply(#0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(const_1,#3)|divide(#2,#4)|subtract(const_1,#5)|multiply(#1,#6)| | probability | B |
what will be the vulgar fraction of 0.70 | "explanation : 0.70 = 70 / 100 = 7 / 10 option b" | a ) 3 / 5 , b ) 7 / 10 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.70, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | general | B |
on a certain transatlantic crossing , 20 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 50 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ? | "0.20 p = rt + c 0.5 ( rt ) = no c = > 0.50 ( rt ) had c 0.20 p = 0.50 ( rt ) rt / p = 40 % answer - b" | a ) 33.3 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 66.6 % | b | divide(20, divide(subtract(const_100, 50), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain | B |
how many digits are in ( 8 × 10 ^ 8 ) ( 10 × 10 ^ 10 ) ? | "the question simplfies to ( 8 × 10 ^ 8 ) ( 10 ^ 11 ) = > 8 * 10 ^ 19 = > will contain 19 zeros + 1 digit 8 = > 20 ans a" | a ) 20 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | add(10, add(const_3, const_4)) | add(const_3,const_4)|add(#0,n1)| | general | A |
train e and f , 455 miles apart , are traveling toward each other at constant rates and in the same time zone . if train e left at 4 pm traveling at a speed of 60 miles per hour , and train f left at 5 : 45 pm and traveling at 45 miles per hour , then at what time would they pass each other ? | first , since e has a headstart then in that 1 hr 45 min or 1.75 hrs he travels 105 miles then remaining distance to be traveled will be 455 - 105 = 350 miles now , using relative rates ( 105 ) ( t ) = 350 this gives 10 / 3 hours now 5.45 pm + 10 / 3 hours gives us 9.05 pm hence answer is b | a ) 9.03 pm , b ) 9.05 pm , c ) 9.07 pm , d ) 9.08 pm , e ) 9.10 pm | b | divide(add(divide(455, 60), divide(455, 45)), const_2) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(#2,const_2) | physics | B |
on a certain road 12 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "answer is c . this question is in the og and thus well explained by ets . those who exceed : x so x = 12 % + 0,2 x id est x = 15 %" | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 30 % | c | divide(const_100, multiply(multiply(divide(12, const_100), divide(20, const_100)), const_100)) | divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)| | gain | C |
the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number q ? | ( 1 + 1 / 8 ) x - ( 1 - 1 / 4 ) x = 30 ( 9 / 8 ) x - ( 3 / 4 ) x = 30 x = 80 = q answer : b | a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#2,#3)|divide(n2,#4) | general | B |
if a square mirror has a 20 - inch diagonal , what is the approximate perimeter q of the mirror , in inches ? | "if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle , addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than 20 and perimeter q must be less than 80 , so we can eliminate answer choice c , d and e . so side 1 + side 2 > 20 , that means side 1 or side 2 must be > 10 . so we can eliminate the answer choice a . now we are left with is b" | a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | b | square_perimeter(divide(20, power(add(const_1, const_1), inverse(const_2)))) | add(const_1,const_1)|inverse(const_2)|power(#0,#1)|divide(n0,#2)|square_perimeter(#3)| | geometry | B |
if 0.5 % of a = 80 paise , then the value of a is ? | "answer ∵ 0.5 / 100 of a = 80 / 100 ∴ a = rs . ( 80 / 0.5 ) = rs . 160 correct option : b" | a ) rs . 170 , b ) rs . 160 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | b | divide(80, 0.5) | divide(n1,n0)| | gain | B |
at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 45 seconds . how long does it tick at 12 ′ o clock | "explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 45 / 5 = 9 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 9 = 99 secs . answer : e" | a ) 47 , b ) 76 , c ) 28 , d ) 66 , e ) 99 | e | multiply(divide(45, subtract(6, const_1)), subtract(12, const_1)) | subtract(n0,const_1)|subtract(n3,const_1)|divide(n2,#0)|multiply(#2,#1)| | physics | E |
a train speeds past a pole in 15 seconds and a platform 100 m long in 20 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 20 = x / 15 x = 300 m . answer : d" | a ) 188 m , b ) 876 m , c ) 251 m , d ) 300 m , e ) 145 m | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics | D |
a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . if 25 cubic centimeters of the solution were poured into an empty container , how many grams of salt would be in the container ? | we are given that a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of salt is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of salt would be in the container when we have 25 cubic centimeters of solution , we can set up a proportion . we can say : “ 15 grams of salt is to 1000 cubic centimeters of solution as x grams of salt is to 25 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 15 / 1000 = x / 25 when we cross multiply we obtain : ( 15 ) ( 25 ) = 1000 x 375 = 1000 x 0.375 = x there are 0.375 grams of salt in the solution in the container . the answer is e . | ['a ) 0.550', 'b ) 0.350', 'c ) 0.450', 'd ) 0.700', 'e ) 0.375'] | e | divide(multiply(15, 25), 1000) | multiply(n0,n2)|divide(#0,n1) | physics | E |
a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . metre | "area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 80 = 68000 p in rs = 68000 / 100 = rs 680 answer : a" | a ) rs 680 , b ) rs 780 , c ) rs 880 , d ) rs 480 , e ) rs 980 | a | divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 80), const_100) | multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics | A |
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 | D |
9 men went to a theater . 8 of them spent rs . 3 each over their tickets and the ninth spent rs . 2 more than the average expenditure of all the 9 . determine the total money spent by them ? | average of 9 = x 9 x = 8 * 3 + x * 2 x = 3.25 total = 9 * 3.25 = 29.25 b | a ) 29 , b ) 29.25 , c ) 31 , d ) 31.23 , e ) 32 | b | multiply(divide(add(multiply(3, 8), 2), 8), 9) | multiply(n1,n2)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2) | general | B |
one hour after matthew started waking from q to y , a distance of 45 km , johnny started walking along the same road from y to q . matthew ' s walking rate was 3 km per hour and johnny ' s was 4 km per hour , how many km had johnny walked when they met ? | just an alternative method . . . after the first hour the distance q is 42 km ( 45 - 3 ) . now the problem can be treated as if bothof them started at the same time . since the speeds are in the ratio 3 : 4 , the distances will also be in the same ratio . splitting 42 in that ratio we get 18 : 24 . so answer is 24 . hence a . | a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 19.5 | a | multiply(divide(subtract(45, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2) | physics | A |
a drink vendor has 60 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | "the number of liters in each can = hcf of 60 , 144 and 368 = 4 liters . number of cans of maaza = 60 / 4 = 15 number of cans of pepsi = 144 / 4 = 36 number of cans of sprite = 368 / 4 = 92 the total number of cans required = 15 + 36 + 92 = 143 cans . answer : c" | a ) 135 , b ) 137 , c ) 143 , d ) 310 , e ) 380 | c | add(divide(368, gcd(gcd(60, 144), 368)), add(divide(60, gcd(gcd(60, 144), 368)), divide(144, gcd(gcd(60, 144), 368)))) | gcd(n0,n1)|gcd(n2,#0)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)| | general | C |
a driver goes on a trip of 60 kilometers , the first 30 kilometers at 60 kilometers per hour and the remaining distance at 30 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | "the time for the first part of the trip was 30 / 60 = 1 / 2 hours . the time for the second part of the trip was 30 / 30 = 1 hour . the total time for the trip was 3 / 2 hours . the average speed for the trip was 60 / ( 3 / 2 ) = 40 kph the answer is c ." | a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 45 | c | divide(60, add(divide(30, 60), divide(subtract(60, 30), 30))) | divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#0,#2)|divide(n0,#3)| | physics | C |
according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 10 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounces of water , that would mean we would need 150 packets of smoothie mix ( along with 12 ( 150 ) ounces of water for a total of 150 packets . however , we know , after a more careful reading of the stimulus , that the 3 ounces are not included in the 12 ounces . as such , the answer has to be less than 150 packets , since 150 would be too much powder considering you already have 150 ( 12 ) ounces of water and need less packets than water to make a smoothie . as such , the only answer less than 150 is 120 , a . does this make sense ? or am i way off base ? b" | a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | b | add(150, multiply(3, const_10)) | multiply(n1,const_10)|add(n4,#0)| | general | B |
if 90 % of a = 30 % of b and b = c % of a , then the value of c is ? | "answer ∵ 90 a / 100 = 30 b / 100 = ( 30 / 100 ) x ac / 100 ∴ c = 100 x ( 100 / 30 ) x ( 90 / 100 ) = 300 correct option : d" | a ) 900 , b ) 800 , c ) 600 , d ) 300 , e ) none | d | divide(30, 90) | divide(n1,n0)| | gain | D |
the average of marks obtained by 120 boys was 36 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 36 24 x = 4320 - 1800 = > x = 2520 / 24 x = 105 . hence , the number of boys passed = 105 . answer : a" | a ) 105 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | a | divide(subtract(multiply(36, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)| | general | A |
a woman complete a journey in 30 hours . she travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km . | "0.5 x / 21 + 0.5 x / 24 = 30 - - > x / 21 + x / 24 = 60 - - > x = 672 km . e" | a ) 334 km . , b ) 216 km . , c ) 314 km . , d ) 224 km . , e ) 672 km . | e | multiply(multiply(divide(multiply(30, 24), add(24, 21)), 21), const_2) | add(n1,n2)|multiply(n0,n2)|divide(#1,#0)|multiply(n1,#2)|multiply(#3,const_2)| | physics | E |
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 272 runs ? | "required run rate = [ 272 - ( 3.2 * 10 ) ] / 40 = 240 / 40 = 6.00 answer : e" | a ) 6.25 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.0 | e | divide(subtract(272, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain | E |
the list price of an article is rs . 70 . a customer pays rs . 61.11 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "explanation : 70 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 61.11 x = 3 % d" | a ) 8 % , b ) 7 % , c ) 10 % , d ) 3 % , e ) 4 % | d | multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 61.11), subtract(70, multiply(70, divide(10, const_100)))), const_100) | divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)| | gain | D |
each of the three people individually can complete a certain job in 4 , 6 , and 8 hours , respectively . what is the lowest fraction of the job that can be done in 1 hour by 2 of the people working together at their respective rates ? | "the two slowest people work at rates of 1 / 6 and 1 / 8 of the job per hour . the sum of these rates is 1 / 6 + 1 / 8 = 7 / 24 of the job per hour . the answer is c ." | a ) 3 / 8 , b ) 5 / 12 , c ) 7 / 24 , d ) 11 / 36 , e ) 13 / 48 | c | add(divide(1, 6), divide(1, 8)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)| | physics | C |
to fill a tank , 100 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to four - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 100 x . new capacity of bucket = 4 / 5 x therefore , required number of buckets = ( 100 x ) / ( 4 x / 5 ) = ( 100 x ) x 5 / 4 x = 500 / 4 = 125 answer is b ." | a ) 100 , b ) 125 , c ) 150 , d ) 175 , e ) 200 | b | divide(multiply(100, add(const_4, const_1)), const_2) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)| | physics | B |
the salaries of a and b together amount to $ 7000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 7000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 7000 - x ) x = $ 5250 answer is b" | a ) $ 1000 , b ) $ 5250 , c ) $ 2500 , d ) $ 4500 , e ) $ 1200 | b | subtract(7000, divide(7000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4)| | gain | B |
a grocery shop has a sale of rs . 6535 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6535 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6535 + 6927 + 6855 + 7230 + 6562 + x = 6 × 6500 = > 34109 + x = 39000 = > x = 39000 − 34109 = 4891 answer : d" | a ) 4857 , b ) 4184 , c ) 4012 , d ) 4891 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6535, 6927)), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | D |
a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 , c receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d | a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800 | d | divide(subtract(divide(multiply(divide(divide(subtract(50000, add(add(4000, 5000), 5000)), const_3), 50000), 35000), const_4), const_100), const_10) | add(n1,n2)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|divide(#3,n0)|multiply(n3,#4)|divide(#5,const_4)|subtract(#6,const_100)|divide(#7,const_10) | general | D |
jim is able to sell a hand - carved statue for $ 750 which was a 35 % profit over his cost . how much did the statue originally cost him ? | "750 = 1.35 * x x = 750 / 1.35 = 555.5555556 . . . which rounds to $ 555.56 , which is ( c ) ." | a ) $ 496.30 , b ) $ 512.40 , c ) $ 555.56 , d ) $ 574.90 , e ) $ 588.20 | c | divide(750, add(divide(35, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | gain | C |
if ( m - 8 ) is a factor of m ^ 2 - pm - 24 , then p = | "( m - 8 ) ( m - a ) = m ^ 2 - pm - 24 a = - 3 p = 8 + a = 5 = b" | a ) 3 , b ) 5 , c ) 6 , d ) 11 , e ) 16 | b | subtract(8, divide(24, 8)) | divide(n2,n0)|subtract(n0,#0)| | general | B |
a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 2 cm x 7 cm x 3 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 2 * 7 * 3 = 8000000 answer : a | a ) 8000000 , b ) 1000000 , c ) 7500000 , d ) 1200000 , e ) none of these | a | divide(multiply(multiply(multiply(6, const_100), multiply(7, const_100)), multiply(8, const_100)), multiply(multiply(8, 7), 7)) | multiply(n3,const_100)|multiply(const_2.0,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n1,#2)|multiply(#3,#0)|divide(#5,#4)| | physics | A |
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 3600 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 3600 ) = rs . 450 answer = b | a ) s . 375 , b ) s . 450 , c ) s . 600 , d ) s . 800 , e ) s . 850 | b | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 3600), 3) | inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5) | physics | B |
the l . c . m of two numbers is 48 . the numbers are in the ratio 4 : 3 . the sum of numbers is : | "let the numbers be 4 x and 3 x . then , their l . c . m = 12 x . so , 12 x = 48 or x = 4 . the numbers are 16 and 12 . hence , required sum = ( 16 + 12 ) = 28 . answer : a" | a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | divide(multiply(4, 48), 3) | multiply(n0,n1)|divide(#0,n2)| | other | A |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 72 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 72 5 x = 70 = > x = 14 hence , b ' s age = 2 x = 28 years . answer : a" | a ) 28 years , b ) 19 years , c ) 29 years , d ) 10 years , e ) 12 years | a | divide(multiply(subtract(72, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general | A |
two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 40 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 30 miles ahead of the black car , how many hours will it take the black car to overtake the red car ? | option e 20 + 40 t = 50 t t = 3 | a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.2 , e ) 3 | e | divide(30, subtract(50, 40)) | subtract(n1,n0)|divide(n2,#0) | physics | E |
a train 140 m long running at 84 kmph crosses a platform in 16 sec . what is the length of the platform ? | "d = 84 * 5 / 18 = 16 = 373 â € “ 140 = 233 answer : c" | a ) 287 , b ) 298 , c ) 233 , d ) 726 , e ) 267 | c | subtract(multiply(16, multiply(84, const_0_2778)), 140) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics | C |
the cost of the paint is rs . 36.50 per kg . if 1 kg of paint covers 16 square feet , how much will it cost to paint outside of a cube having 8 feet each side . | "explanation : we will first calculate the surface area of cube , then we will calculate the quantity of paint required to get answer . here we go , surface area = 6 a 2 = 6 ∗ 8 ( 2 ) = 384 sq feet quantity required = 38416 = 24 kg cost of painting = 36.50 ∗ 24 = rs . 876 option c" | a ) rs . 850 , b ) rs . 860 , c ) rs . 876 , d ) rs . 886 , e ) none of these | c | multiply(divide(surface_cube(8), 16), 36.50) | surface_cube(n3)|divide(#0,n2)|multiply(n0,#1)| | geometry | C |
find the least number of five digits which is exactly divisible by 12 and 18 ? | "the smallest five digit numbers are 10025 , 10080,11080 10025 is not divisible by 12 10080 is divisible by both 12 and 18 answer : b" | a ) a ) 1080 , b ) b ) 10080 , c ) c ) 10025 , d ) d ) 11080 , e ) e ) 12080 | b | add(divide(divide(multiply(floor(add(divide(subtract(multiply(const_10, const_1000), const_1), 18), const_1)), 18), const_4), const_100), const_2) | multiply(const_10,const_1000)|subtract(#0,const_1)|divide(#1,n1)|add(#2,const_1)|floor(#3)|multiply(n1,#4)|divide(#5,const_4)|divide(#6,const_100)|add(#7,const_2)| | general | B |
the area of a square is equal to three times the area of a rectangle of dimensions 25 cm * 27 cm . what is the perimeter of the square ? | "area of the square = s * s = 3 ( 25 * 27 ) = > s = 3 * 5 * 3 = 45 cm perimeter of the square = 4 * 45 = 180 cm . answer : a" | a ) 180 cm , b ) 190 cm , c ) 170 cm , d ) 150 cm , e ) 160 cm | a | multiply(sqrt(multiply(rectangle_area(25, 27), divide(27, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry | A |
the total cost of a vacation was divided among 4 people . if the total cost of the vacation had been divided equally among 5 people , the cost per person would have been $ 50 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 4 * p c = 5 * p - 250 substituting the value of p from the first equation onto the second we get p = 250 . plugging in the value of p in the first equation , we get c = 1000 . which leads us to answer choice e" | a ) $ 200 , b ) $ 300 , c ) $ 400 , d ) $ 500 , e ) $ 1000 | e | multiply(multiply(5, 4), divide(50, subtract(5, 4))) | multiply(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#2,#0)| | general | E |
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 5000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 5000 ) = rs . 625 answer = c" | a ) s . 375 , b ) s . 425 , c ) s . 625 , d ) s . 800 , e ) s . 850 | c | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 5000), 3) | inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5)| | physics | C |
in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 58 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "so 57 days answer a = 57" | a ) 57 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | a | subtract(58, const_1) | subtract(n0,const_1)| | physics | A |
how many times digit 6 is used while writing numbers from 100 to 1100 ? | "there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' so far we have calculated 190 . the total now comes to 280 . the nearest to which is 320 . hence b ." | a ) 648 , b ) 320 , c ) 252 , d ) 225 , e ) 26 | b | add(add(divide(subtract(1100, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | add(const_1,const_10)|multiply(n0,const_2)|subtract(n2,n1)|divide(#2,const_10)|multiply(#0,#0)|add(#3,#4)|add(#5,#1)| | general | B |
jolene entered an 18 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 10 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ? | if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 726.24 . answer : b . | a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24 | b | add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2))), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2)))) | divide(n1,const_100)|divide(n3,const_100)|divide(n5,const_100)|power(const_100,const_2)|multiply(#0,#3)|add(#4,#3)|multiply(#5,#1)|add(#5,#6)|multiply(#7,#2)|add(#8,#6) | gain | B |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts five hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 5 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 5 + t ) = 40 * t t = 15 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 15 / 60 = 10 hrs time taken by a = 5 + t = 5 + 15 = 20 hrs time taken by k = 10 hrs so k starts 20 - 10 = 10 hrs after a . ( answer d ) | a ) 3 , b ) 4.5 , c ) 4 , d ) d ) 10 , e ) e ) 5 | d | subtract(40, 30) | subtract(n1,n0) | physics | D |
7386038 is divisible by | "explanation : the given number is 7386038 . the sum of the digits at even places is = 3 + 6 + 3 = 12 . the sum of the digits at odd places is = 7 + 8 + 0 + 8 = 23 . the difference is given by 23 - 12 = 11 answer : d" | a ) 3 , b ) 4 , c ) 9 , d ) 11 , e ) 13 | d | add(floor(divide(7386038, add(const_1, const_4))), floor(divide(7386038, power(add(const_1, const_4), const_2)))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|divide(n0,#2)|floor(#1)|floor(#3)|add(#4,#5)| | general | D |
if the annual increase in the population of a town is 10 % and the present number of people is 14000 , what will the population be in 2 years ? | "the required population is = 14000 ( 1 + 10 / 100 ) ^ 2 = 14000 * 11 / 10 * 11 / 10 = 16940 answer is d" | a ) 12100 , b ) 15240 , c ) 12456 , d ) 16940 , e ) 10002 | d | multiply(multiply(divide(add(10, const_100), const_100), 14000), divide(add(10, const_100), const_100)) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|multiply(#1,#2)| | gain | D |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 34 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 11 / 45 * 60 = 14.6 answer : option d" | a ) e 982 , b ) 27 , c ) 12 , d ) 14.6 , e ) 28 | d | subtract(const_60, multiply(const_60, divide(34, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics | D |
what is the least number that should be added to 1053 , so the sum of the number is divisible by 23 ? | "( 1053 / 23 ) gives a remainder 18 so we need to add 5 . the answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(subtract(add(const_1, floor(divide(1053, 23))), divide(1053, 23)), 23) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general | E |
in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 30 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 130 / 100 = 78 in june total income = 78 + 40 = 118 % of mrs lee ' s income = 72 / 112 = 66.10 ( b )" | a ) 64 % , b ) 66 % , c ) 72 % , d ) 76 % , e ) 80 % | b | multiply(divide(add(const_100, 30), add(add(const_100, 30), const_100)), const_100) | add(n1,const_100)|add(#0,const_100)|divide(#0,#1)|multiply(#2,const_100)| | general | B |
the ratio between the number of sheep and the number of horses at the stewar farm is 2 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 2 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 2 * 8 ) = 16 answer : b" | a ) 18 , b ) 16 , c ) 32 , d ) 56 , e ) 58 | b | multiply(divide(divide(12880, 230), 7), 2) | divide(n3,n2)|divide(#0,n1)|multiply(n0,#1)| | other | B |
a box contains 10 apples , 9 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ? | "when you choose ( and then eat ) the first apple , the probability of that apple being red is 9 / 10 . so if we do the activity 1 times , the probability of it being red is 9 / 10 . for 2 times , it is ( 9 / 10 ) * ( 8 / 9 ) for 3 times , it is ( 9 / 10 ) * ( 8 / 9 ) * ( 7 / 8 ) you can notice that the numerator of the first term cancels with the denominator of the second . so we can see that the probability becomes 0.5 when the last term is 5 / 6 & it becomes less than 0.5 when the last term is 4 / 5 . 9 accounts for n = 1 , so 4 will account for n = 6 , answer . d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(multiply(0.5, const_10), const_1) | multiply(n2,const_10)|add(#0,const_1)| | general | D |
two pipes can fill a tank in 15 minutes and 15 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 / 15 + 1 / 15 - 1 / 45 = 1 / 9 so , the tank becomes full in 9 minutes . answer : d" | a ) 1 / 15 , b ) 1 / 16 , c ) 1 / 11 , d ) 1 / 9 , e ) 1 / 12 | d | subtract(add(divide(const_1, 15), divide(const_1, 15)), divide(const_1, 45)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics | D |
a man walking at a rate of 10 km / hr crosses a bridge in 3 minutes . the length of the bridge is ? | "speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 5 minutes = 50 / 18 * 5 * 60 = 500 m answer is a" | a ) 500 , b ) 650 , c ) 250 , d ) 111 , e ) 236 | a | multiply(divide(multiply(10, const_1000), const_60), 3) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | gain | A |
find compound interest on rs . 7500 at 4 % per annum for 2 years , compounded annually . | "solution amount = rs [ 7500 x ( 1 + 4 / 100 ) ² ] = rs . ( 7500 x 26 / 25 x 26 / 25 ) = rs . 8112 . c . i = rs ( 8112 - 7500 ) = rs . 612 . answer c" | a ) rs . 512 , b ) rs . 552 , c ) rs . 612 , d ) rs . 622 , e ) none | c | subtract(add(add(7500, divide(multiply(7500, 4), const_100)), divide(multiply(add(7500, divide(multiply(7500, 4), const_100)), 4), const_100)), 7500) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)| | gain | C |
anil grows tomatoes in his backyard which is in the shape of a square . each tomato takes 1 cm 2 in his backyard . this year , he has been able to grow 131 more tomatoes than last year . the shape of the backyard remained a square . how many tomatoes did anil produce this year ? | detailed solution let the area of backyard be x 2 this year and y 2 last year ∴ x 2 - y 2 = 131 = ) ( x + y ) ( x - y ) = 131 now , 131 is a prime number ( a unique one too . check out its properties on google ) . also , always identify the prime number given in a question . might be helpful in cracking the solution . = ) ( x + y ) ( x - y ) = 131 x 1 = ) x + y = 131 x - y = 1 = ) 2 x = 132 = ) x = 66 and y = 65 ∴ number of tomatoes produced this year = 662 = 4356 correct choice ( c ) | ['a ) 4225', 'b ) 4096', 'c ) 4356', 'd ) insufficient data', 'e ) none of these'] | c | power(divide(add(131, 1), const_2), const_2) | add(n0,n2)|divide(#0,const_2)|power(#1,const_2) | physics | C |
sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 90 ps per meter is ? | "3 x * 2 x = 3750 = > x = 25 2 ( 75 + 50 ) = 250 m 250 * 0.9 = rs . 225 answer : c" | a ) 287 , b ) 369 , c ) 225 , d ) 279 , e ) 361 | c | divide(multiply(90, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics | C |
in a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 48 inches and the width is 72 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | l = 48 / 12 = 4 ft w = 72 / 12 = 6 ft area of paper = 24 area of cube = 4 * side ^ 2 side of cube = 2 v of cube = 8 | ['a ) a 8', 'b ) b 22', 'c ) c 27', 'd ) d 40', 'e ) e 51'] | a | volume_cube(square_edge_by_area(divide(rectangle_area(divide(48, 12), divide(72, 12)), divide(72, 12)))) | divide(n0,n3)|divide(n1,n3)|rectangle_area(#0,#1)|divide(#2,#1)|square_edge_by_area(#3)|volume_cube(#4) | geometry | A |
find the value of 85 p 3 . | "85 p 3 = 85 ! / ( 85 - 3 ) ! = 85 ! / 82 ! = 85 * 84 * 83 * 82 ! / 82 ! = 85 * 84 * 83 = 595650 answer : b" | a ) 565350 , b ) 595650 , c ) 535950 , d ) 565350 , e ) 575350 | b | multiply(subtract(3, const_4), 85) | subtract(n1,const_4)|multiply(#0,n0)| | general | B |
find the value of y from ( 12 ) ^ 2 x 6 ^ 3 ÷ 432 = y ? | "72 e" | a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 72 | e | divide(multiply(power(12, 2), power(6, 3)), 432) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4)| | general | E |
george is 8 years more than christopher and ford is 2 years younger than christopher . the sum of their ages is 60 . find the ages of christopher . | christopher age = x george age , y = x + 8 - - - - - - - - - - > ( 1 ) ford age , z = x - 2 - - - - - - - - - - - - > ( 2 ) sum of their ages , x + y + z = 60 - - - - > ( 3 ) substitute z and y values in equation ( 3 ) therefore , x + ( x + 8 ) + ( x - 2 ) = 60 = > 3 x + 8 - 2 = 60 = > 3 x = 60 - 6 = > 3 x = 54 = > x = 54 / 3 x = 18 answer : a | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | a | divide(subtract(60, subtract(8, 2)), const_3) | subtract(n0,n1)|subtract(n2,#0)|divide(#1,const_3) | general | A |
there are 23 students in a class . in how many different ways can a committee of 3 students be formed ? | "23 c 3 = 23 * 22 * 21 / 6 = 1771 the answer is c ." | a ) 1254 , b ) 1482 , c ) 1771 , d ) 1875 , e ) 1923 | c | multiply(subtract(const_1, divide(3, 23)), 23) | divide(n1,n0)|subtract(const_1,#0)|multiply(#1,n0)| | probability | C |
a bag consists of 20 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | "bag consists of 20 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 20 - 5 - 9 = 6 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 6 / 20 = 15 / 20 = 3 / 4 hence , answer will be a ." | a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | divide(add(subtract(20, add(5, 9)), 9), 20) | add(n1,n2)|subtract(n0,#0)|add(n2,#1)|divide(#2,n0)| | probability | A |
if 25 % of x is 15 less than 15 % of 1500 , then x is ? | "25 % of x = x / 4 ; 15 % of 1500 = 15 / 100 * 1500 = 225 given that , x / 4 = 225 - 15 = > x / 4 = 210 = > x = 840 . answer : d" | a ) 720 , b ) 750 , c ) 820 , d ) 840 , e ) 860 | d | divide(subtract(multiply(1500, divide(15, const_100)), 15), divide(25, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)| | general | D |
a man can do a piece of work in 5 days , but with the help of his son , he can do it in 4 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 4 ) - ( 1 / 5 ) = 1 / 20 the son alone can do the work in 20 days answer is c" | a ) 15 , b ) 17 , c ) 20 , d ) 8 , e ) 9 | c | divide(multiply(5, 4), subtract(5, 4)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics | C |
if a car went the first third of the distance at 80 kmh , the second third at 24 kmh , and the last third at 54 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 2160 ( this number is convenient because it is divisible by 80 , 24 and 54 ) so : 2160 = 80 * t 1 = 27 hrs 2160 = 24 * t 2 = 90 hrs 2160 = 54 * t 3 = 40 hrs t = t 1 + t 2 + t 3 = 157 hrs d = rt ( 240 * 3 ) = r * 157 r = 41.27 answer : c" | a ) 36 kmh , b ) 40 kmh , c ) 41 kmh , d ) 44 kmh , e ) 50 kmh | c | divide(const_3, add(add(divide(const_1, 80), divide(const_1, 24)), divide(const_1, 54))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_3,#4)| | physics | C |
a train 110 m long is running with a speed of 65 km / hr . in what time will it pass a man who is running at 7 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 65 + 7 = 72 km / hr . = 72 * 5 / 18 = 20 m / sec . time taken to pass the men = 110 * 1 / 20 = 5.5 sec . answer : option e" | a ) 5 , b ) 6 , c ) 4 , d ) 8.2 , e ) 5.5 | e | divide(110, multiply(add(65, 7), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | E |
suppose 8 monkeys take 8 minutes to eat 8 banana ' s . how many minutes would it take 3 monkeys to eat 3 banana ' s ? | there are equal no . of monkeys and equal no . of bananas and they take equal time and the time is 8 mint to eat a banana so each monkey take 8 mints to eat a banana ' s so 3 monkeys will take 8 mints to eat 3 banana ' s . answer : d | a ) 5 min , b ) 6 min , c ) 7 min , d ) 8 min , e ) 9 min | d | divide(3, divide(3, 8)) | divide(n3,n0)|divide(n3,#0) | physics | D |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.