Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
the average of first three prime numbers greater than 20 is ? | "23 + 29 + 31 = 83 / 3 = 27.7 answer : d" | a ) 10 , b ) 20 , c ) 30 , d ) 27.7 , e ) 50 | d | add(20, const_1) | add(n0,const_1)| | general |
if greg buys 3 shirts , 4 trousers and 2 ties , the total cost is $ 40 . if greg buys 7 shirts , 2 trousers and 2 ties , the total cost is $ 60 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ? | "solution : 3 x + 4 y + 2 z = 40 7 x + 2 y + 2 z = 60 adding both the equations = 10 x + 6 y + 4 z = 100 5 x + 3 y + 2 z = 50 ans a" | a ) $ 50 , b ) $ 64 , c ) $ 75 , d ) $ 96 , e ) can not be determined | a | divide(add(60, 40), 2) | add(n3,n7)|divide(#0,n2)| | general |
the heights of 3 individuals are in the ratio 4 : 5 : 6 . if the sum of the heights of the heaviest and the lightest boy is 150 cm more than the height of the third boy , what is the weight of the lightest boy ? | let the heights of the three boys be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 150 = > 5 k = 150 = > k = 30 therefore the height of the lightest boy = 4 k = 4 ( 30 ) = 120 cm . answer : a | a ) 120 cm , b ) 150 cm , c ) 160 cm , d ) 190 cm , e ) of these | a | multiply(divide(150, subtract(add(4, 6), 5)), 4) | add(n1,n3)|subtract(#0,n2)|divide(n4,#1)|multiply(n1,#2) | general |
according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 29 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 29 = 61 - - > c = 5 / 9 * 29 = 16.11 . increase = 16.11 degrees . answer : b ." | a ) 9 , b ) 15 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(29, 5), 9) | multiply(n1,n3)|divide(#0,n0)| | general |
a man engaged a servant on the condition that he would pay him rs . 900 and auniform after 1 year service . he served only for 9 months and received uniform and rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a ) rs . 80 , b ) rs . 100 , c ) rs . 120 , d ) rs . 145 , e ) rs . 156 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general |
the sale price of a trolley bag including the sale tax is rs . 1190 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is : | "explanation : 112 % of s . p . = 1190 s . p . = rs . ( 1190 x 100 / 112 ) = rs . 1062.50 . c . p . = rs ( 100 / 125 x 1062.50 ) = rs 850 answer : d" | a ) rs 800 , b ) rs 820 , c ) rs 860 , d ) rs 850 , e ) none of these | d | divide(subtract(1190, multiply(1190, divide(12, const_100))), add(divide(25, const_100), const_1)) | divide(n1,const_100)|divide(n2,const_100)|add(#1,const_1)|multiply(n0,#0)|subtract(n0,#3)|divide(#4,#2)| | gain |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 6 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 6 years , their ages will be 46 and 26 . sum of their ages = 46 + 26 = 72 . answer : e" | a ) 22 , b ) 77 , c ) 70 , d ) 98 , e ) 72 | e | add(add(multiply(divide(60, 6), const_2), 6), add(divide(60, 6), 6)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
the cross - section of a cannel is a trapezium in shape . if the cannel is 5 m wide at the top and 3 m wide at the bottom and the area of cross - section is 3800 sq m , the depth of cannel is ? | "1 / 2 * d ( 5 + 3 ) = 3800 d = 950 answer : d" | a ) 920 , b ) 930 , c ) 940 , d ) 950 , e ) 960 | d | divide(divide(divide(3800, divide(add(5, 3), const_2)), 3), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 74 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he me... | "total distance = 74 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 70 kms relative speed ( opposite side ) ( as they are moving towards each other speed w... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | divide(add(74, 6), add(4, 6)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | physics |
the two sides of a triangle are 32 and 68 . the area is 960 sq . cm . find the third side of triangle ? | we see 68 ^ 2 - 32 ^ 2 = ( 68 + 32 ) * ( 68 - 32 ) = 100 * 36 = 60 ^ 2 now ( 1 / 2 ) * 60 * 32 = 960 ( match with given options ) ( i . e area of a right angled triangle whose sides are 32 , 60,68 ) third side = 60 answer : c | ['a ) 45', 'b ) 50', 'c ) 60', 'd ) 70', 'e ) 80'] | c | subtract(multiply(divide(add(sqrt(subtract(power(add(68, 32), const_2), multiply(divide(add(sqrt(subtract(power(multiply(68, 32), const_2), multiply(power(960, const_2), const_4))), multiply(68, 32)), const_2), const_4))), add(68, 32)), const_2), const_2), add(68, 32)) | add(n0,n1)|multiply(n0,n1)|power(n2,const_2)|multiply(#2,const_4)|power(#0,const_2)|power(#1,const_2)|subtract(#5,#3)|sqrt(#6)|add(#1,#7)|divide(#8,const_2)|multiply(#9,const_4)|subtract(#4,#10)|sqrt(#11)|add(#0,#12)|divide(#13,const_2)|multiply(#14,const_2)|subtract(#15,#0) | geometry |
from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 40 percent . what percent of the stock â € ™ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 40 % = 90 * 1.2 = 126 thus ratio = 126 / 100 = 1.26 ( in % terms = 126 % ) . thus b is the cor... | a ) 80 , b ) 126 , c ) 95 , d ) 100 , e ) 108 | b | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 40), const_100)) | add(n5,const_100)|add(n1,const_100)|subtract(const_100,n3)|divide(#0,const_100)|divide(#2,const_100)|divide(#1,const_100)|multiply(#5,const_100)|multiply(#4,#6)|multiply(#3,#7)| | gain |
if 9 a - b = 10 b + 80 = - 12 b - 2 a , what is the value of 9 a - 11 b ? | "this implies 9 a - b = 10 b + 80 , 9 a - b = - 12 b - 2 a , 10 b + 80 = - 12 b - 2 a manipulating the second equation gives us 9 a - b = 10 b + 80 = = > 9 a - 11 b = 80 answer is e" | a ) - 4 , b ) - 2 , c ) 0 , d ) 2 , e ) 80 | e | multiply(negate(multiply(divide(80, 2), 2)), 9) | divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)| | general |
john makes $ 50 a week from his job . he earns a raise and now makes $ 70 a week . what is the % increase ? | "increase = ( 20 / 50 ) * 100 = 40 % . e" | a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 40 % | e | multiply(divide(subtract(70, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the perimeter of one face of a cube is 48 cm . its volume will be : | "explanation : edge of cude = 48 / 4 = 12 cm volume = a * a * a = 12 * 12 * 12 = 1728 cm cube option e" | a ) 125 cm 3 , b ) 400 cm 3 , c ) 250 cm 3 , d ) 625 cm 3 , e ) none of these | e | volume_cube(square_edge_by_perimeter(48)) | square_edge_by_perimeter(n0)|volume_cube(#0)| | geometry |
the vertex of a rectangle are ( 1 , 0 ) , ( 5 , 0 ) , ( 1 , 1 ) and ( 5 , 1 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | "if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 3 , 0.5 ) . the slope of a line passing through ( 0,0 ) and ( 3 , 0.5 ) is 0.5 / 3 = 1 / 6 . the answer is c ." | a ) 1 / 2 , b ) 2 , c ) 1 / 6 , d ) 3 , e ) 1 / 4 | c | divide(1, divide(add(subtract(5, 1), 1), 1)) | subtract(n2,n0)|add(#0,n5)|divide(#1,n5)|divide(n0,#2)| | general |
on a game show , there are 3 tables . each table has 3 boxes ( one zonk ! , one cash prize , and one grand prize ) . the contestant can choose one box from each table for a total of 3 boxes . if any of the boxes are a zonk ! , the contestant loses everything . what is the probability of getting no zonk ! from any of th... | no zonk ! : 1 st box = no zonk ! = 2 / 3 2 nd box = no zonk ! = 2 / 3 3 rd box = no zonk ! = 2 / 3 ( 2 / 3 ) ( 2 / 3 ) ( 2 / 3 ) = 8 / 27 8 / 27 is the probability of no zonk ! , so . . . answer : c | a ) 1 / 5 , b ) 1 / 3 , c ) 8 / 27 , d ) 1 / 27 , e ) 7 / 8 | c | divide(power(const_2, 3), power(3, const_3)) | power(const_2,n0)|power(n0,const_3)|divide(#0,#1) | general |
the average weight of a , b and c is 42 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 42 x 3 ) = 126 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 40 . b ' s wei... | a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 40 kg , e ) 37 kg | d | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(42, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
dividing by 3 ⁄ 8 and then multiplying by 5 ⁄ 9 is the same as dividing by what number ? | say x / 3 / 8 * 5 / 9 = x * 8 / 3 * 5 / 9 = x * 40 / 27 d | a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 40 / 27 , e ) 5 ⁄ 16 | d | multiply(divide(8, 3), divide(5, 9)) | divide(n1,n0)|divide(n2,n3)|multiply(#0,#1) | general |
how many integers n greater than and less than 100 are there such that , if the digits of n are reversed , the resulting integer is n + 9 ? | ( 10 x + y ) = ( 10 y + x ) + 9 = > 9 x - 9 y = 9 = > x - y = 1 and only 8 numbers satisfy this condition and the numbers are 21,32 , 43,54 , 65,76 , 87,98 answer : d | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | subtract(divide(100, const_10), const_2) | divide(n0,const_10)|subtract(#0,const_2) | general |
a runs twice as fast as b and gives b a start of 50 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 50 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 50 m is 50 / 1 = 50 m so the total time taken by a and b to reach = 2 * 50 = 100 m answer : d" | a ) 75 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | d | multiply(50, const_2) | multiply(n0,const_2)| | physics |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 12 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 88 2 / 3 sp 1 = 88 sp 1 = 132 100 - - - 32 = > 32 % answer : d" | a ) 20 % , b ) 25 % , c ) 13 1 / 30 % , d ) 32 % , e ) 13 % | d | subtract(divide(subtract(const_100, 12), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
reeya obtained 65 , 67 , 76 , 80 and 95 out of 100 in different subjects , what will be the average | explanation : ( 65 + 67 + 76 + 80 + 95 / 5 ) = 76.6 option a | a ) 76.6 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | divide(add(add(add(add(65, 67), 76), 80), 95), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general |
an ant walks an average of 600 meters in 12 minutes . a beetle walks 15 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 600 meters in 12 minutes 600 meters in 1 / 5 hours the beetle walks 15 % less distance = 600 - 90 = 510 meters in 12 minutes 0.510 km in 12 / 60 = 1 / 5 hours speed = 0.510 * 5 = 2.55 km / h correct answer b = 2.55" | a ) 2.215 . , b ) 2.55 , c ) 2.775 . , d ) 3.2 . , e ) 3.5 . | b | multiply(divide(divide(600, const_1000), divide(12, const_60)), subtract(const_1, divide(15, const_100))) | divide(n0,const_1000)|divide(n1,const_60)|divide(n2,const_100)|divide(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)| | general |
an aeroplane covers a certain distance of 450 kmph in 4 hours . to cover the same distance in 3 2 / 3 hours , it must travel at a speed of | "speed of aeroplane = 450 kmph distance travelled in 4 hours = 450 * 4 = 1800 km speed of aeroplane to acver 1800 km in 11 / 3 = 1800 * 3 / 11 = 490 km answer b ." | a ) 440 , b ) 490 , c ) 640 , d ) 740 , e ) 250 | b | divide(multiply(450, 4), divide(add(multiply(3, 3), 2), 3)) | multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3)| | physics |
when the number 72 y 6139 is exactly divisible by 11 , then the smallest whole number that can replace y is ? | "the given number = 72 y 6139 sum of the odd places = 9 + 1 + y + 7 sum of the even places = 3 + 6 + 2 = 11 ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) ( 16 + y ) - ( 11 ) = divisible by 11 y � 5 = divisible by 11 . y must be 6 , to make given number divisible by 11 . d" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | d | subtract(72, 72) | subtract(n0,n0)| | general |
a 12 month project had a total budget of $ 42000 . after 8 months , the project had spent $ 23700 . at this point , how much was the project under budget ? | each month , the project should spend $ 42,000 / 12 = $ 3500 . in 8 months , the project should spend 8 * $ 3500 = $ 28,000 . the project is under budget by $ 28,000 - $ 23,700 = $ 4300 . the answer is b . | a ) $ 4100 , b ) $ 4300 , c ) $ 4500 , d ) $ 4700 , e ) $ 4900 | b | multiply(divide(subtract(42000, multiply(divide(23700, 8), 12)), 12), 8) | divide(n3,n2)|multiply(n0,#0)|subtract(n1,#1)|divide(#2,n0)|multiply(n2,#3) | general |
a is the average ( arithmetic mean ) of the first 7 positive multiples of 4 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 – b ^ 2 is zero , what is the value of n ? | "if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 16 , which also equals b . b is the second positive multiple of n , thus n = 16 / 2 = 8 . the answer is d ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(multiply(4, const_4), 2) | multiply(n1,const_4)|divide(#0,n3)| | general |
a cistern 8 m long and 6 m wide contains water up to a breadth of 1 m 85 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 1.85 + 8 * 1.85 ) ] + 8 * 6 = 100 m square option b" | a ) 102 m sqaure , b ) 100 m sqaure , c ) 152 m sqaure , d ) 164 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(85, const_100), 1), 6), multiply(add(divide(85, const_100), 1), 8))), multiply(6, 8)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
the ratio of buses to cars on river road is 1 to 13 . if there are 60 fewer buses than cars on river road , how many cars are on river road ? | "b / c = 1 / 13 c - b = 60 . . . . . . . . . > b = c - 60 ( c - 60 ) / c = 1 / 13 testing answers . clearly eliminate bcde put c = 65 . . . . . . . . . > ( 65 - 60 ) / 65 = 5 / 65 = 1 / 13 answer : a" | a ) 65 , b ) 60 , c ) 55 , d ) 40 , e ) 45 | a | multiply(divide(60, subtract(13, 1)), 13) | subtract(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | other |
if a takes x days to do a work then b takes 2 x days to do the same work then with in how many day ' s a will alone complete this work ? | 1 / x + 1 / 2 x = 1 / 18 = > 3 / 2 x = 1 / 18 = > x = 27 days . hence , a alone can finish the work in 27 days . answer : a | a ) 27 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | a | add(subtract(multiply(multiply(const_3, const_3), multiply(2, const_2)), add(multiply(const_3, const_3), multiply(2, const_2))), multiply(2, const_2)) | multiply(n0,const_2)|multiply(const_3,const_3)|add(#1,#0)|multiply(#1,#0)|subtract(#3,#2)|add(#0,#4) | physics |
62 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "62 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 62 = 154 hence , d is the answer ." | a ) 64 , b ) 128 , c ) 152 , d ) 154 , e ) 256 | d | subtract(volume_cube(add(cube_edge_by_volume(62), const_2)), 62) | cube_edge_by_volume(n0)|add(#0,const_2)|volume_cube(#1)|subtract(#2,n0)| | geometry |
if you throw two dice at a same time . can you find the probability of getting sum as 10 of the two numbers shown ? | c 1 / 12 all possible cases can be 36 ( 6 * 6 ) case we need : [ ( 4,6 ) , ( 5,5 ) , ( 6,4 ) ] = 3 probability = > 3 / 36 = 1 / 12 | a ) 1 / 16 , b ) 2 / 14 , c ) 1 / 12 , d ) 1 / 19 , e ) 1 / 17 | c | add(add(divide(const_1, power(const_6, const_2)), divide(const_1, power(const_6, const_2))), divide(const_1, power(const_6, const_2))) | power(const_6,const_2)|divide(const_1,#0)|add(#1,#1)|add(#2,#1) | probability |
in an office , 30 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ? | "( 10 / 100 ) workers = 16 = > number of workers = 160 ( 30 / 100 ) * workers = x + 16 = > x = 48 answer b" | a ) 480 , b ) 48 , c ) 50 , d ) 144 , e ) 160 | b | divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4)) | divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)| | gain |
the least number which must be subtracted from 820 to make it exactly divisible by 9 is : | "on dividing 820 by 9 , we get remainder = 1 therefore , required number to be subtracted = 1 answer : c" | a ) a ) 2 , b ) b ) 3 , c ) c ) 1 , d ) d ) 5 , e ) e ) 6 | c | subtract(820, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)| | general |
can n and can в are both right circular cylinders . the radius of can n is twice the radius of can b , while the height of can n is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can n with the same brand of gasoline ? | let x be the radius of b and 2 h be the height of b . therefore , radius of n = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e | ['a ) $ 1', 'b ) $ 2', 'c ) $ 4', 'd ) $ 8', 'e ) $ 16'] | e | multiply(multiply(4, const_2), divide(power(const_2, const_2), const_2)) | multiply(n0,const_2)|power(const_2,const_2)|divide(#1,const_2)|multiply(#2,#0) | geometry |
a certain car can travel 40 minutes on a gallon of gasoline at 60 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 120 miles at 60 mph ? | "total time for travelling 120 miles @ 60 mph = 120 / 60 = 2 hour = 120 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 60 mph . thus in 120 minutes it will use = 3 gallons . thus , full tank = 3 + 8 = 11 gallons - - - > 3 / 11 = 27 % of the fuel used . d is the correct answer ." | a ) 15 % , b ) 20 % , c ) 25 % , d ) 27 % , e ) 40 % | d | divide(divide(multiply(40, 60), multiply(120, 60)), add(divide(multiply(40, 60), multiply(120, 60)), 8)) | multiply(n0,n1)|multiply(n3,n4)|divide(#0,#1)|add(n2,#2)|divide(#2,#3)| | physics |
if the sides of a triangle are 39 cm , 36 cm and 15 cm , what is its area ? | "the triangle with sides 39 cm , 36 cm and 15 cm is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 36 * 15 = 270 cm 2 answer : d" | a ) 120 cm 2 , b ) 765 cm 2 , c ) 216 cm 2 , d ) 270 cm 2 , e ) 275 cm 2 | d | divide(multiply(36, 15), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
what is the dividend . divisor 19 , the quotient is 7 and the remainder is 6 | "d = d * q + r d = 19 * 7 + 6 d = 133 + 6 d = 139" | a ) 136 , b ) 137 , c ) 138 , d ) 139 , e ) 140 | d | add(multiply(19, 7), 6) | multiply(n0,n1)|add(n2,#0)| | general |
obra drove 120 π meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 π square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 π square meters so , π ( r ^ 2 ) = 57,600 π - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 π r = 480 π therefore , part of circumference covered = 120 π / 480 π = 25 % hence , answer is d ." | a ) 6.67 % , b ) 12.5 % , c ) 18.75 % , d ) 25 % , e ) 33.3 % | d | subtract(subtract(multiply(multiply(divide(circumface(120), 120), const_3), const_2), const_4), const_2) | circumface(n0)|divide(#0,n0)|multiply(#1,const_3)|multiply(#2,const_2)|subtract(#3,const_4)|subtract(#4,const_2)| | geometry |
some persons can do a piece of work in 20 days . two times the number of these people will do half of that work in ? | "20 / ( 2 * 2 ) = 5 days answer : c" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | c | multiply(multiply(20, divide(const_1, const_2)), divide(const_1, const_2)) | divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1)| | physics |
if paint costs $ 3.20 per quart , and a quart covers 1200 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 1200 = 0.5 cost will be 0.5 * 3.2 = $ 1.6 ans : a | ['a ) $ 1.60', 'b ) $ 16.00', 'c ) $ 96.00', 'd ) $ 108.00', 'e ) $ 196.00'] | a | multiply(divide(3.2, 1200), surface_cube(10)) | divide(n0,n1)|surface_cube(n2)|multiply(#0,#1) | geometry |
the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 33 ft long . what is the distance traveled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 33 . ( formula for c... | ['a ) 20 ft', 'b ) 25 ft', 'c ) 750 ft', 'd ) 900 ft', 'e ) 1650 ft'] | e | multiply(30, add(divide(multiply(30, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | divide(const_10,const_2)|multiply(n0,#0)|divide(#1,const_3)|add(#2,#0)|multiply(n0,#3) | physics |
the two lines y = x and x = - 4 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 w 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 ( - 4 , - 4 ) 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 ) w = 16 , b ) w = 8 √ 2 , c ) w = 8 , d ) w = 2 √ 2 , e ) ( √ 2 ) / 3 | d | sqrt(divide(multiply(4, 4), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1)| | general |
if 50 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ? | "50 % of ( x - y ) = 20 % of ( x + y ) 50 / 100 ( x - y ) = 20 / 100 ( x + y ) 3 x = 7 y required percentage = y / x * 100 = 3 y / 7 y * 100 = 42.85 % answer is b" | a ) 50.5 % , b ) 42.8 % , c ) 22.2 % , d ) 33.3 % , e ) 25 % | b | multiply(divide(subtract(50, 20), add(50, 20)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 34 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "34 % is 9 % - points above 25 % and 6 % - points below 40 % . thus the ratio of mixture y to mixture x is 2 : 3 . the percent of mixture x is 3 / 5 = 60 % . the answer is d ." | a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | d | divide(subtract(34, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | add(n2,const_1)|divide(n0,const_100)|divide(#0,const_100)|subtract(n4,#0)|subtract(#1,#2)|divide(#3,#4)| | gain |
in a triangle abc , ab = 6 , bc = 8 and ac = 10 . a perpendicular dropped from b , meets the side ac at d . a circle of radius bd ( with centre b ) is drawn . if the circle cuts ab and bc at p and q respectively , then ap : qc is equal to | explanation : triangle abc and triangle adb are similar , ac / ab = bc / bd 10 / 6 = 8 / r r = 24 / 5 bp = bq = bd = r = 24 / 5 ap = ab - r = 6 - 24 / 5 = 6 / 5 cq = bc - r = 8 - 24 / 5 = 16 / 5 ap : cq = 6 / 16 = 3 : 8 answer : d | ['a ) 1 : 1', 'b ) 3 : 2', 'c ) 4 : 1', 'd ) 3 : 8', 'e ) 2 : 5'] | d | divide(subtract(6, divide(multiply(8, 6), 10)), subtract(8, divide(multiply(8, 6), 10))) | multiply(n0,n1)|divide(#0,n2)|subtract(n0,#1)|subtract(n1,#1)|divide(#2,#3) | geometry |
find the mean proportional between 49 & 81 ? | formula = √ a × b a = 49 and b = 81 √ 49 × 81 = 7 × 9 = 63 c | a ) 59 , b ) 61 , c ) 63 , d ) 65 , e ) 67 | c | divide(add(49, 81), const_2) | add(n0,n1)|divide(#0,const_2) | general |
the average monthly income of p and q is rs . 6050 . the average monthly income of q and r is rs . 7050 and the average monthly income of p and r is rs . 8000 . the monthly income of p + q + r is : | "explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 6050 x 2 ) = 12100 . . . . ( i ) q + r = ( 7050 x 2 ) = 14100 . . . . ( ii ) p + r = ( 8000 x 2 ) = 16000 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 42200 or p + q + r = 21100 . . .... | a ) 41100 , b ) 42000 , c ) 21100 , d ) 42200 , e ) 21000 | c | subtract(add(6050, 8000), 7050) | add(n0,n2)|subtract(#0,n1)| | general |
how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 2 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with two 2 - s , 2 x 2 ... | a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 320 | d | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics |
in one hour , a boat goes 9 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 9 + 5 ) kmph = 7 kmph . answer c" | a ) 2 , b ) 4 , c ) 7 , d ) 12 , e ) 15 | c | divide(add(9, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a 70 cm long wire is to be cut into two pieces so that one piece will be 2 / 3 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 3 = 3 : 2 2 / 5 * 70 = 28 answer : c" | a ) 35 , b ) 20 , c ) 28 , d ) 36 , e ) 30 | c | subtract(70, divide(70, add(divide(2, 3), const_1))) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | physics |
165 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 * 165 = 99 liters water = 66 liters 99 : ( 66 + p ) = 3 : 4 198 + 3 p = 396 = > p = 66 66 liters of water are to be added for the ratio become 3 : 4 . answer : a" | a ) 66 liters , b ) 32 liters , c ) 41 liters , d ) 50 liters , e ) 34 liters | a | multiply(divide(165, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | general |
raman mixed 54 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 54 ( 150 ) + 36 ( 125 ) ] / ( 54 + 36 ) = rs . 140 sp = cp [ ( 100 + profit % ) / 100 ] = 140 * [ ( 100 + 40 ) / 100 ] = rs . 196 . answer : c" | a ) 129 , b ) 287 , c ) 196 , d ) 188 , e ) 112 | c | add(divide(add(multiply(54, 150), multiply(36, 125)), add(36, 54)), multiply(divide(add(multiply(54, 150), multiply(36, 125)), add(36, 54)), divide(40, const_100))) | add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6)| | gain |
average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 25 % the new mean is $ 850 . what was original rent of friend whose rent is increased ? | "0.25 x = 4 ( 850 - 800 ) 0.25 x = 200 x = 800 answer a" | a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | a | divide(multiply(subtract(850, 800), 4), divide(25, const_100)) | divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0)| | general |
a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 5 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 4 × 2 / 100 = 400 he also lends it at 5 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 5 × 2 / 100 = 500 his overall gain in 2 years = rs . 500 - rs . 4... | a ) 50 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | a | divide(subtract(divide(multiply(multiply(5000, 5), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain |
a chemist mixes one liter of pure water with x liters of a 30 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 30 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 30 - 15 ) / ( 15 - 0 ) = 1 / 1 1 / x = 1 / 1 x = 1 answer : d" | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | d | divide(15, subtract(30, 15)) | subtract(n0,n1)|divide(n1,#0)| | gain |
the surface of a cube is 96 sq cm . find its volume ? | 6 a 2 = 96 = 6 * 16 a = 4 = > a 3 = 64 cc answer : e | ['a ) 8 cc', 'b ) 9 cc', 'c ) 2 cc', 'd ) 4 cc', 'e ) 64 cc'] | e | volume_cube(sqrt(divide(96, add(const_2, const_4)))) | add(const_2,const_4)|divide(n0,#0)|sqrt(#1)|volume_cube(#2) | geometry |
the largest four digit number which is a perfect cube , is : | "explanation : 21 * 21 * 21 = 9261 option c" | a ) 7000 , b ) 8000 , c ) 9261 , d ) 9999 , e ) none of these | c | square_area(const_pi) | square_area(const_pi)| | geometry |
the angle between two hands at 3.45 is | "theta degree = 11 / 2 m - 30 h = 11 / 2 ( 45 ) - 30 ( 3 ) = 247.5 - 90 = 157.5 or 157 1 / 2 degree answer : e" | a ) 110 degree , b ) 115 degree , c ) 112 1 / 2 degree , d ) 117 degree , e ) 157 1 / 2 degree | e | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | geometry |
pipe a and pipe b fill water into a tank of capacity 2000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated unti... | "tank capacity : 2000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 2000 l tank : 2000 / 250 = 8 time taken to fill : 8 * total time = 8 ... | a ) 14 min , b ) 18 min , c ) 25 min , d ) 32 min , e ) 40 min | e | multiply(add(add(1, 2), 2), divide(2000, subtract(add(200, multiply(50, 2)), multiply(25, 2)))) | add(n4,n5)|multiply(n2,n5)|multiply(n3,n5)|add(n5,#0)|add(n1,#1)|subtract(#4,#2)|divide(n0,#5)|multiply(#3,#6)| | physics |
a wooden cube whose edge length is 9 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least ... | "there are a total of 9 * 9 * 9 = 729 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 7 * 7 * 7 = 343 cubes . the number of cubes with at least one side painted red is 729 - 343 = 386 cubes the probability that a cube has at least one side painted red is 386 / 729 whic... | a ) 52.9 % , b ) 55.4 % , c ) 58.3 % , d ) 61.7 % , e ) 64.5 % | a | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(9)))) | multiply(const_1,const_4)|volume_cube(n0)|volume_cube(#0)|divide(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)| | geometry |
the sector of a circle has radius of 21 cm and central angle 110 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 110 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 40.3 + 42 = 82.3 cm answer : d" | a ) 91.5 , b ) 91.4 , c ) 91.7 , d ) 82.3 , e ) 91.1 | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
how many positive integers less than 25 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 2 odd multiples of 5 : { 5 , 15 } 10 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 } notice , that 5 is in two sets , thus total # of integers satisfying the ... | a ) 27 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | e | subtract(subtract(subtract(25, 2), const_1), const_1) | subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = [ 282 - ( 3.2 * 10 ) ] / 40 = 250 / 40 = 6.25 ' answer : a" | a ) 6.25 , b ) 6.27 , c ) 6.23 , d ) 6.29 , e ) 6.39 | a | divide(subtract(282, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
company m produces two kinds of stereos : basic and deluxe . of the stereos produced by company m last month , 2 / 3 were basic and the rest were deluxe . if it takes 7 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last... | # of basic stereos was 3 / 4 of total and # of deluxe stereos was 1 / 4 of total , let ' s assume total = 16 , then basic = 12 and deluxe = 4 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 7 / 5 units . total time for basic would be 12 * 1 = 12 and to... | a ) 5 / 22 , b ) 4 / 22 , c ) 3 / 22 , d ) 9 / 22 , e ) 7 / 22 | e | divide(7, add(add(divide(multiply(multiply(3, 5), 2), 3), 7), 5)) | multiply(n1,n3)|multiply(n0,#0)|divide(#1,n1)|add(n2,#2)|add(n3,#3)|divide(n2,#4) | general |
rectangular tile each of size 80 cm by 40 cm must be laid horizontally on a rectangular floor of size 130 cm by 230 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 80 * 40 = 3200 area of floor = 130 * 230 = 29900 no of tiles = 29900 / 3200 = 9.34 so , the no of tile = 9 answer : d" | a ) 6 , b ) 2 , c ) 8 , d ) 9 , e ) 7 | d | divide(multiply(130, 230), multiply(80, 40)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
find large number from below question the difference of two numbers is 1395 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1395 ) . x + 1395 = 6 x + 15 5 x = 1380 x = 276 large number = 276 + 1395 = 1671 e" | a ) 1235 , b ) 1345 , c ) 1678 , d ) 1767 , e ) 1671 | e | multiply(divide(subtract(1395, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 1740 ? | "explanation : 1740 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 1500 answer : option d" | a ) rs . 1000 , b ) rs . 1550 , c ) rs . 1510 , d ) rs . 1500 , e ) none of these | d | divide(1740, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general |
the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 45 , what is the possible maximum range of the temperatures ? | "average = 50 , sum of temperatures = 50 * 5 = 250 as the min temperature is 45 , max would be 250 - 4 * 45 = 70 - - > the range = 70 ( max ) - 45 ( min ) = 25 answer : b ." | a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 75 | b | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(45, const_4)), 45) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1)| | general |
a grocer has a sale of rs . 800 , rs . 900 , rs . 1000 , rs . 700 and rs . 800 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 850 ? | "total sale for 5 months = rs . ( 800 + 900 + 1000 + 700 + 800 ) = rs . 4200 required sale = rs . [ ( 850 x 6 ) - 4200 ] = rs . ( 5100 - 4200 ) = rs . 900 . option d" | a ) s . 440 , b ) s . 850 , c ) s . 450 , d ) s . 900 , e ) s . 950 | d | subtract(multiply(add(5, const_1), 850), add(add(add(add(800, 900), 1000), 700), 800)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 18 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 2 ? - - - - - - 60 12 h rs = 16 + 18 = 34 t = 12 d = 34 * 12 = 408 answer : e" | a ) 565 , b ) 444 , c ) 676 , d ) 767 , e ) 408 | e | add(multiply(divide(60, subtract(18, 16)), 16), multiply(divide(60, subtract(18, 16)), 18)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
a can finish a piece of work in 5 days . b can do it in 16 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 16 = 1 = > x = 6 days answer : c" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 16), const_2))), divide(const_1, 16)) | divide(const_1,const_4.0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics |
what is the greatest positive integer n such that 3 ^ n is a factor of 54 ^ 100 ? | "54 = 3 ^ 3 * 2 . 54 ^ 100 = 3 ^ 300 * 2 ^ 100 the answer is c ." | a ) 100 , b ) 200 , c ) 300 , d ) 600 , e ) 900 | c | multiply(subtract(54, 100), 100) | subtract(n1,n2)|multiply(n2,#0)| | other |
when a random experiment is conducted , the probability that event a occurs is 1 / 5 . if the random experiment is conducted 5 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 5 * 1 / 5 * 4 / 5 * 4 / 5 * 4 / 5 = 64 / 3125 the total number of possible cases is 5 c 2 = 10 p ( event a occurs exactly twice ) = 10 * ( 64 / 3125 ) = 128 / 625 the answer is e ." | a ) 36 / 625 , b ) 48 / 625 , c ) 64 / 625 , d ) 98 / 625 , e ) 128 / 625 | e | subtract(1, divide(const_2, 5)) | divide(const_2,n1)|subtract(n0,#0)| | general |
in a lake , there is a patch of lily pads . every day , the patch doubles in size . if it takes 48 days for the patch to cover the entire lake , how long would it take for the patch to cover half of the lake ? | "explanation : with all the talk of doubling and halves , your brain jumps to the conclusion that to solve the problem of when the lily patch covers half the lake , all you have to do is divide the number of days it took to fill the lake ( 48 ) in half . it ' s understandable but wrong . the problem says that the patch... | a ) 45 , b ) 56 , c ) 47 , d ) 57 , e ) 49 | c | subtract(48, const_1) | subtract(n0,const_1)| | physics |
the least common multiple of positive integer d and 3 - digit integer n is 690 . if n is not divisible by 3 and d is not divisible by 2 , what is the value of n ? | the lcm of n and d is 690 = 2 * 3 * 5 * 23 . d is not divisible by 2 , thus 2 goes to n n is not divisible by 3 , thus 3 goes to d . from above : n must be divisible by 2 and not divisible by 3 : n = 2 * . . . in order n to be a 3 - digit number it must take all other primes too : n = 2 * 5 * 23 = 230 . answer : b . | a ) 115 , b ) 230 , c ) 460 , d ) 575 , e ) 690 | b | multiply(multiply(2, add(const_3, const_2)), divide(divide(divide(690, const_2), const_3), add(const_3, const_2))) | add(const_2,const_3)|divide(n1,const_2)|divide(#1,const_3)|multiply(n3,#0)|divide(#2,#0)|multiply(#4,#3) | general |
in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day . | "this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 7 - 3 = 4 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 4 = 172,... | a ) 172,700 , b ) 172,800 , c ) 172,900 , d ) 173,000 , e ) 173,100 | b | multiply(multiply(subtract(7, 3), const_3600), const_12) | subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)| | general |
two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 54 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 54 = 25 / 9 = > x = 75 . answer : c" | a ) 50 , b ) 26 , c ) 75 , d ) 28 , e ) 21 | c | divide(multiply(54, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
a number is doubled and 8 is added . if the resultant is trebled , it becomes 84 . what is that number ? | "solution let the number be x . then , 3 ( 2 x + 8 ) ‹ = › 84 ‹ = › 2 x + 8 = 28 ‹ = › 2 x = 20 x = 10 . answer a" | a ) 10 , b ) 6 , c ) 8 , d ) none of these , e ) can not be determined | a | divide(subtract(84, multiply(const_3, 8)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a train 100 m in length crosses a telegraph post in 12 seconds . the speed of the train is ? | "s = 100 / 12 * 18 / 5 = 30 kmph answer : b" | a ) 16 kmph , b ) 30 kmph , c ) 54 kmph , d ) 18 kmph , e ) 19 kmph | b | multiply(const_3_6, divide(100, 12)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
by selling a book for 250 , 20 % profit was earned . what is the cost price of the book ? | "sp = 120 % of cp ; : . cp = 250 × 100 / 120 = 208 option ' b '" | a ) a ) 215 , b ) b ) 208 , c ) c ) 230 , d ) d ) 235 , e ) e ) 240 | b | original_price_before_gain(20, 250) | original_price_before_gain(n1,n0)| | gain |
50 liters of a mixture contains milk and water in the ratio 3 : 2 . if 5 liters of this mixture be replaced by 5 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 50 liters if mix = 50 * 3 / 5 = 30 liters quantity of milk in 55 liters of new mix = 30 + 5 = 35 liters quantity of water in it = 55 - 35 = 20 liters ratio of milk and water in new mix = 35 : 20 = 7 : 4 answer is d" | a ) 6 : 5 , b ) 6 : 2 , c ) 6 : 4 , d ) 7 : 4 , e ) 7 : 2 | d | divide(multiply(subtract(5, const_1), 2), multiply(3, 2)) | multiply(n2,n1)|subtract(n3,const_1)|multiply(#1,n2)|divide(#2,#0)| | other |
the area of a square garden is q square feet and the perimeter is p feet . if q = 2 p + 48 , what is the perimeter of the garden in feet ? | "let x be the length of one side of the square garden . x ^ 2 = 8 x + 48 x ^ 2 - 8 x - 48 = 0 ( x - 12 ) ( x + 4 ) = 0 x = 12 , - 4 p = 4 ( 12 ) = 48 the answer is d ." | a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | d | multiply(2, 48) | multiply(n1,n0)| | geometry |
the average of 10 numbers is calculated as 5 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "10 * 5 + 36 – 26 = 60 / 10 = 6 answer : e" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(5, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
if ( 10 ^ 4 * 3.456789 ) ^ 14 is written as a single term , how many digits would be to the right of the decimal place ? | 3.456789 ^ 14 has 6 * 14 = 84 decimal places . 10 ^ 56 moves the decimal place to the right 56 places . ( 10 ^ 4 * 3.456789 ) ^ 14 has 84 - 56 = 28 digits after the decimal point . the answer is d . | a ) 9 , b ) 14 , c ) 21 , d ) 28 , e ) 42 | d | multiply(14, const_2) | multiply(n3,const_2) | general |
find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 5 ) is | "required ratio = 2 / 3 * 6 / 11 * 11 / 5 = 2 / 1 = 4 : 5 answer is d" | a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 2 : 3 | d | multiply(divide(2, 3), multiply(divide(2, 3), divide(6, 3))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other |
how many digits are required to write numbers between 1 to 100 . | "explanation : single digits are from 1 to 9 = 9 digits doubt digits are from 10 to 99 = 90 x 2 = 180 digits 100 needs 3 digits . total 192 digits correct option : c" | a ) 196 , b ) 158 , c ) 192 , d ) 200 , e ) none | c | add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(1, const_100), const_1), const_3)) | subtract(const_10,const_1)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#4,#0)|add(#6,#5)| | general |
a , b , c can complete a piece of work in 16 , 6,12 days . working together , they complete the same work in how many days ? | a + b + c 1 day work = 1 / 16 + 1 / 6 + 1 / 12 = 15 / 48 a , b , c together will complete the job in 48 / 15 days answer is b | a ) 2 , b ) 48 / 15 , c ) 7 / 9 , d ) 10 , e ) 24 / 7 | b | divide(const_1, add(add(divide(const_1, 16), divide(const_1, add(const_4, const_2))), divide(const_1, multiply(const_2, add(const_4, const_2))))) | add(const_2,const_4)|divide(const_1,n0)|divide(const_1,#0)|multiply(#0,const_2)|add(#1,#2)|divide(const_1,#3)|add(#4,#5)|divide(const_1,#6) | physics |
the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 100,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 100 ( 100,000 / 1,000 = 100 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 100 , where x represents the number of times the population doubles ... | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 2) | add(const_1,const_4)|multiply(#0,n1)|multiply(#1,const_100)|divide(#2,n1)|log(#3)|multiply(n0,#4)| | general |
certain stocks in january were 10 % less than they were in february and 20 % greater than they were in march . what was the percentage decrease in the stocks from february to march ? | let j , f , m be the values of the stock in jan , feb and march . thus , per the question , j = 0.9 f = 1.2 m - - - - > m = 0.75 f thus the % decrease from f to m = ( f - m ) / ( f ) * 100 = ( f - 0.75 f ) / f * 100 = 0.25 * 100 = 25 % , d is the correct answer . | a ) 5 % , b ) 10 % , c ) 20 % , d ) 25 % , e ) 50 % | d | multiply(divide(subtract(divide(const_100, divide(subtract(const_100, 10), const_100)), divide(const_100, divide(add(const_100, 20), const_100))), divide(const_100, divide(subtract(const_100, 10), const_100))), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|divide(const_100,#2)|divide(const_100,#3)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100) | general |
to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 15 hours a day ? | "man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 15 = 300 therefore number of days = 5760 / 300 = 19.2 days ans = e" | a ) 23 , b ) 22 , c ) 21 , d ) 20 , e ) 19.2 | e | divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 15)) | multiply(n1,n2)|multiply(n4,n6)|multiply(n3,#0)|multiply(n5,#2)|divide(#3,#1)| | physics |
a scuba diver descends at a rate of 80 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 4000 feet below the sea level . . how long will he take to reach the ship ? | "time taken to reach = 4000 / 80 = 50 minutes answer : c" | a ) 70 minutes , b ) 72 minutes , c ) 50 minutes , d ) 66 minutes , e ) 67 minutes | c | divide(4000, 80) | divide(n1,n0)| | gain |
the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 784 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 784 x ^ 2 = 196 x = 14 length = 4 * 14 = 56 difference = 56 - 14 = 42 c is the answer" | a ) 38 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 . | c | subtract(multiply(sqrt(divide(784, 4)), 4), sqrt(divide(784, 4))) | divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)| | geometry |
two water pumps , working simultaneously at their respective constant rates , took exactly 5 hours to fill a certain swimming pool . if the constant rate of one pump was 1.5 times the constant rate of the other , how many hours would it have taken the slower pump to fill the pool if it had worked alone at its constant ... | let x be the rate of the slower pump . then 1.5 x is the rate of the faster pump . both pumps together can fill 1 / 5 of the pool each hour . 2.5 x = 1 / 5 x = 1 / 12.5 = 2 / 25 the slower pump could fill the pool in 25 / 2 = 12.5 hours . the answer is c . | a ) 8.5 , b ) 10.5 , c ) 12.5 , d ) 14.5 , e ) 16.5 | c | multiply(add(1.5, const_1), 5) | add(n1,const_1)|multiply(n0,#0) | physics |
a certain box has 8 cards and each card has one of the integers from 1 to 8 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on t... | "the average of the numbers is 4.5 the total number of ways to choose 2 cards from 8 cards is 8 c 2 = 28 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } the probability is 2 / 28 = 1 / 14 the answer is c ." | a ) 1 / 8 , b ) 1 / 12 , c ) 1 / 14 , d ) 1 / 18 , e ) 1 / 28 | c | divide(const_4, divide(factorial(8), multiply(factorial(2), factorial(subtract(8, 2))))) | factorial(n0)|factorial(n3)|subtract(n0,n3)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)|divide(const_4,#5)| | general |
a positive number x is multiplied by 4 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | sq rt ( 4 x / 3 ) = x = > 4 x / 3 = x ^ 2 = > x = 4 / 3 ans - c | a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 1 / 2 | c | divide(4, 3) | divide(n0,n1) | general |
a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 100 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 100 total = 100 + 100 = 200 answer ( d )" | a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | d | divide(subtract(multiply(100, divide(60, const_100)), multiply(100, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n3,const_100)|divide(n0,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#1,#2)|subtract(#3,#4)|divide(#6,#5)| | general |
a 270 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | "relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 270 / 9 = 500 / 9 x = 230 ans is 230 m answer : c" | a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 250 m | c | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 270) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
car z travels 51 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per ho... | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 51 miles / gallon and at 60 miles / hour , that distance decreases by 20 %... | a ) 320 , b ) 375.2 , c ) 400 , d ) 408 , e ) 440 | d | multiply(multiply(subtract(const_1, divide(20, const_100)), 51), 10) | divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)| | gain |
at what rate percent on simple interest will rs . 1800 amount to rs . 2000 in 5 years ? | "explanation : 250 = ( 1800 x 5 xr ) / 100 r = 3.47 % answer : option d" | a ) 4 % , b ) 3 6 / 7 % , c ) 2 6 / 7 % , d ) 3.47 % , e ) 6 % | d | multiply(divide(divide(subtract(2000, 1800), 1800), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
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