Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 2 w widgets . | "rx * ( t + 2 ) = w ry * ( t ) = w . also , w / ( t + 2 ) + w / t = ( 5 w / 4 ) * ( 1 / 3 ) or 1 / t + 1 / ( t + 2 ) = 5 / 12 - - > ( 3 + 2 ) / 12 = 1 / 4 + 1 / 6 thus , t = 4 . replacing , rx = w / 6 , w / 6 * t = 2 w or t = 12 days . answer : e" | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(multiply(2, 2), 3) | multiply(n0,n4)|multiply(n3,#0)| | general |
how long will a boy take to run round a square field of side 60 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 60 * 4 = 240 m time taken = 240 * 2 / 5 = 96 sec answer is a" | a ) 96 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(60, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
a 60 cm long wire is to be cut into two pieces so that one piece will be 2 / 4 th of the other , how many centimeters will the shorter piece be ? | "explanation : 1 : 2 / 4 = 4 : 2 2 / 6 * 60 = 20 answer : option b" | a ) a ) 73 , b ) b ) 20 , c ) c ) 83 , d ) d ) 21 , e ) e ) 52 | b | subtract(60, divide(60, add(divide(2, 4), const_1))) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | physics |
a boat running up stram takes 6 hours to cover a certain distance , while it takes 8 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? | "explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 8 ( x - y ) 6 x + 6 y = 8 x - 8 y 14 y = 2 x 7 y = x x / y = 7 / 1 7 : 1 answer : option d" | a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 7 : 1 , e ) 8 : 1 | d | subtract(8, 6) | subtract(n1,n0)| | physics |
if a book is sold at 9 % profit instead of 9 % loss , it would have brought rs 9 more . find out the cost price of the book | let c . p . of the book be rs . ’ x ’ given , 1.09 x - 0.91 x = 9 = > 0.18 x = 9 = 9 / 0.18 = rs 50 answer : c | a ) 75 , b ) 72 , c ) 50 , d ) 70 , e ) 80 | c | divide(multiply(const_100, divide(9, const_2)), 9) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0) | gain |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m cube ) is : | "explanation : l = ( 48 - 16 ) m = 32 m , [ because 8 + 8 = 16 ] b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 32 x 20 x 8 ) m cube = 5120 m cube . option d" | a ) 4120 m cube , b ) 4140 m cube , c ) 5140 m cube , d ) 5120 m cube , e ) none of these | d | volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
aarti can do a piece of work in 6 days . in how many days will she complete three time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 6 days . three times of work of same type can be done in 6 x 3 = 18 days answer b" | a ) 6 days , b ) 18 days , c ) 21 days , d ) 3 days , e ) 13 days | b | multiply(const_3, 6) | multiply(n0,const_3)| | physics |
a man bought an article and sold it at a gain of 5 % . if he had bought it at 5 % less and sold it for re 3 less , he would have made a profit of 10 % . the c . p . of the article was | "explanation : let original cost price is x its selling price = ( 105 / 100 ) * x = 21 x / 20 new cost price = ( 95 / 100 ) * x = 19 x / 20 new selling price = ( 110 / 100 ) * ( 19 x / 20 ) = 209 x / 200 [ ( 21 x / 20 ) - ( 209 x / 200 ) ] = 3 = > x = 600 answer : d ) rs 600" | a ) 344 , b ) 218 , c ) 200 , d ) 600 , e ) 211 | d | multiply(multiply(multiply(5, const_4.0), 10), 3) | multiply(n0,const_4.0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
the average ( arithmetic mean ) of 4 positive integers is 45 . if the average of 2 of these integers is 45 , what is the greatest possible value that one of the other 2 integers can have ? | "a + b + c + d = 180 a + b = 90 c + d = 90 greatest possible = 89 ( just less than 1 ) answer = d" | a ) 55 , b ) 65 , c ) 100 , d ) 89 , e ) 115 | d | subtract(multiply(45, 4), multiply(45, 2)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 20 , what is the value of r - p ? | arithmetic mean expression for p and q : ( p + q ) / 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) / 2 = 20 ; q + r = 40 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 20 hence , the correct answer is a | a ) 20 , b ) 10 , c ) 30 , d ) 40 , e ) 5 | a | subtract(multiply(20, const_2), multiply(10, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1) | general |
a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,052 . how much did she invest at 5 % ? | "et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1052 / 1000 ) - 1 = 0.052 = 5.2 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of ... | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 950 | d | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | multiply(n1,const_2)|multiply(#0,const_100)|multiply(n1,#0)|add(#2,const_4)|multiply(n1,#1)|add(#3,#1)|subtract(#5,#1)|multiply(#6,const_100)|subtract(#7,#4)|subtract(#1,#8)| | gain |
one ball will drop from a certain height . the height it will reach after rebounding from the floor is 50 percent of the previous height . the total travel is 260 cm when it touches the floor on third time . what is the value of the original height ? | when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = h... | a ) 80 cm , b ) 90 cm , c ) 104 cm , d ) 120 cm , e ) 130 cm | c | divide(260, add(const_2, divide(50, const_100))) | divide(n0,const_100)|add(#0,const_2)|divide(n1,#1) | gain |
16.02 ã — 0.00001 = ? | "16.02 ã — 0.00001 = 0.0001602 the answer is e ." | a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | e | multiply(divide(16.02, 0.00001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a distributor sells a product through an online store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 17 per item . what is the price that the buyer observers online if the distributor wants to maintain a 20 % profit on the co... | "let x be the price that buyers see online . the distributor wants to receive 1.2 ( original price ) which should be 80 % of x . 1.2 ( 17 ) = 0.8 x x = 1.2 ( 17 ) / 0.8 = 1.5 ( 17 ) = $ 25.50 the answer is d ." | a ) $ 22.50 , b ) $ 23.50 , c ) $ 24.50 , d ) $ 25.50 , e ) $ 26.50 | d | divide(add(multiply(divide(20, const_100), 17), 17), divide(subtract(const_100, 20), const_100)) | divide(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(n1,#0)|add(n1,#3)|divide(#4,#2)| | gain |
when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in a hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of a ? | "rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * a = 1 = > x = 1 / a when they both work together then their rates get added up combined rate = ( w + x ) r * t = ... | a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2 | b | add(subtract(4, 2), divide(const_1, add(2, 3))) | add(n0,n1)|subtract(n2,n0)|divide(const_1,#0)|add(#2,#1)| | physics |
the total age of a and b is 10 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 10 â € ¹ = â € º a - c = 10 . answer b" | a ) 12 , b ) 10 , c ) c is elder than a , d ) data inadequate , e ) none | b | multiply(10, const_1) | multiply(n0,const_1)| | general |
a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 5 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the p... | "i think b - 43 / 60 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 5 x so old = x , new = 4 x now 1 / 3 of x are occupied and 1 / 5 of 4 x are occupied which simplifies to ( 4 / 5 ) x we are shifting 1 / 3 of x to the new ward so ... | a ) 11 / 30 , b ) 43 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15 | b | multiply(add(multiply(5, 3), 1), multiply(divide(1, 3), divide(1, 5))) | divide(n1,n2)|divide(n1,n4)|multiply(n0,n2)|add(n1,#2)|multiply(#0,#1)|multiply(#3,#4)| | general |
what is the probability of having 53 fridays in an ordinary year | "1 / 7 because 53 friday in ordinary year then 53 * 7 = 371 ( 53 / 371 = 1 / 7 ) answer : c" | a ) 1 / 5 , b ) 1 / 6 , c ) 1 / 7 , d ) 1 / 8 , e ) 1 / 9 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
how many integerskgreater than 100 and less than 900 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be seven pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = ... | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | divide(n0,const_10)|multiply(const_2,const_4)|multiply(#0,#1)|add(#0,#2)| | general |
how many halves are there in 9 - fourth ? | divide 9 / 4 by 1 / 2 = 9 / 4 ÷ 1 / 2 = 9 / 4 * 2 / 1 = 18 / 4 = 4.5 . answer is b . | a ) 8 , b ) 4.5 , c ) 3.2 , d ) 7 , e ) 3 | b | divide(divide(9, const_4), divide(const_1, const_2)) | divide(n0,const_4)|divide(const_1,const_2)|divide(#0,#1) | general |
what is the population standard deviation for the numbers : 75 , 83 , 96 , 100 , 121 and 125 ? | 1 . firstly find the mean : mean = ( 75 + 83 + 96 + 100 + 121 + 125 ) ÷ 6 = 600 ÷ 6 = 100 2 . next find the variance . to calculate the variance , take each difference , square it , and then average the result : ( 75 - 100 ) 2 + ( 83 - 100 ) 2 + ( 96 - 100 ) 2 + ( 100 - 100 ) 2 + ( 121 - 100 ) 2 + ( 125 - 100 ) 2 = ( -... | a ) 18.2 , b ) 20 , c ) 25 , d ) 26 , e ) 28 | a | subtract(subtract(subtract(subtract(add(121, 125), 100), 96), divide(83, const_3)), const_4) | add(n4,n5)|divide(n1,const_3)|subtract(#0,n3)|subtract(#2,n2)|subtract(#3,#1)|subtract(#4,const_4) | general |
63 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 ? | "63 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 - 63 = 153 hence , e is the answer ." | a ) 64 , b ) 128 , c ) 152 , d ) 216 , e ) 153 | e | subtract(volume_cube(add(cube_edge_by_volume(63), const_2)), 63) | cube_edge_by_volume(n0)|add(#0,const_2)|volume_cube(#1)|subtract(#2,n0)| | geometry |
there are two tanks a & b with same capacity of 20 litres . if the water inflow rate ( speed ) in tank a is 2 litres per hour & in tank b is 4 litres per hour . how much longer tank a takes to fill than tank b ? | time taken for tank a to fill = 20 / 2 = 10 hours time taken for tank b to fill = 20 / 4 = 5 hours so tank a takes 5 more hours to fill answer : b | a ) 4 hours , b ) 5 hours , c ) 3 hours , d ) 2 hours , e ) 1 hour | b | subtract(divide(20, 2), divide(20, 4)) | divide(n0,n1)|divide(n0,n2)|subtract(#0,#1) | physics |
while calculating the edge of a square , a worker makes an error of 2 % in excess . what % error does he make in calculating area ? ( % ) | explanation : given error = 2 % while measuring the side of a square . if the correct value of the side of square is 100 , the measured value : = > 100 + 2 % * 100 = 100 + 2 = 102 the area of square with edge 100 = side * side = > 100 * 100 = > 10000 the area of square with side 102 = 102 * 102 = 10404 error in area ca... | ['a ) 4.09', 'b ) 4.0', 'c ) 4.04', 'd ) 4.02', 'e ) 4.01'] | c | divide(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|divide(#3,const_100) | geometry |
how many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 ? | "total multiples of 13 : 15 ( first multiple : 13 , last multiple : 195 ) total multiples of 12 : 16 ( first multiple : 12 , last multiple : 192 ) multiple of 12 and 13 : 1 ( 156 ) 15 + 16 - 1 = 30 answer : c" | a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | c | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
if a / b = 1 / 2 , b / c = 5 , c / d = 2 / 3 , d / e = 1 / 6 and e / f = 2 / 3 , then what is the value of abc / def ? | "say a = 2 . then : a / b = 1 / 2 - - > b = 4 ; b / c = 5 - - > c = 4 / 5 ; c / d = 2 / 3 - - > d = 6 / 5 ; d / e = 1 / 6 - - > e = 36 / 5 ; e / f = 2 / 3 - - > f = 54 / 2 . abc / def = ( 2 * 4 * 4 / 5 ) / ( 6 / 5 * 36 / 5 * 54 / 5 ) = 50 / 729 . answer : a ." | a ) 50 / 729 , b ) 20 / 456 , c ) 45 / 752 , d ) 13 / 452 , e ) 45 / 741 | a | divide(multiply(2, 2), multiply(5, multiply(2, 2))) | multiply(n1,n1)|multiply(n1,n7)|multiply(n2,#1)|divide(#0,#2)| | general |
a salesperson received a commission of 3 percent of the sale price for each of the first 100 machines that she sold and 4 percent of the sale price for each machine that she sold after the first 100 . if the sale price of each machine was $ 10,000 and the salesperson received a $ 40,000 commission , how many machines d... | "first 100 machines = 3 % commission = 0.03 * 100 * 10000 = 30000 commission from sale of next machines = 40000 - 30000 = 10000 so 25 more machines . . total = 125 machines imo e . . ." | a ) 90 , b ) 103 , c ) 105 , d ) 115 , e ) 125 | e | add(100, divide(subtract(multiply(multiply(multiply(add(4, 3), multiply(3, const_2)), 100), multiply(add(4, const_1), const_2)), multiply(multiply(multiply(100, 100), divide(3, 100)), 100)), multiply(multiply(100, 100), divide(4, 100)))) | add(n0,n2)|add(const_1,n2)|divide(n0,n1)|divide(n2,n1)|multiply(const_2,n0)|multiply(n1,n1)|multiply(#0,#4)|multiply(#1,const_2)|multiply(#2,#5)|multiply(#3,#5)|multiply(#6,n1)|multiply(n1,#8)|multiply(#10,#7)|subtract(#12,#11)|divide(#13,#9)|add(n1,#14)| | gain |
find the average of all the numbers between 6 and 30 which are divisible by 4 . | "sol . average = ( 8 + 12 + 16 + 20 + 24 + 28 / 6 ) = 108 / 6 = 18 answer b" | a ) 15 , b ) 18 , c ) 20 , d ) 22 , e ) none | b | divide(add(add(6, const_4), subtract(30, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
in a renowned city , the average birth rate is 8 people every two seconds and the death rate is 6 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 2 persons are added ( 8 - 6 ) . every second 1 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 1 = 86400 option d" | a ) 32,300 , b ) 172,800 , c ) 468,830 , d ) 86,400 , e ) 259,200 | d | multiply(multiply(subtract(8, 6), const_3600), const_12) | subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)| | general |
in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 24,947 . approximately what was the percent increase ? | "the percent increase is ( 24947 - 8902 ) / 8902 = 16045 / 8902 = 1.80 so the approximate answer is e" | a ) 45 % , b ) 125 % , c ) 145 % , d ) 150 % , e ) 180 % | e | divide(subtract(24,947, 8,902), 8,902) | subtract(n3,n1)|divide(#0,n1)| | general |
in may , the groundskeeper at spring lake golf club built a circular green with an area of 70 π square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | "area = π r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 70 π ) = 280 π the answer is c ." | a ) 1000 π , b ) 400 π , c ) 280 π , d ) 200 π , e ) 20 π | c | multiply(power(multiply(sqrt(70), const_2), const_2), const_pi) | sqrt(n0)|multiply(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi)| | geometry |
find the cost of fencing around a circular field of diameter 30 m at the rate of rs . 5 a meter ? | "2 * 22 / 7 * 15 = 94.25 94.25 * 5 = rs . 471.25 answer : e" | a ) 288 , b ) 132 , c ) 772 , d ) 592 , e ) 471.25 | e | multiply(circumface(divide(30, const_2)), 5) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
if x # y is defined to equal x ^ 2 / y for all x and y , then ( - 1 # 2 ) # - 3 = | ( - 1 ) ^ 2 / 2 = 1 / 2 ( - 1 / 2 ) ^ 2 / - 3 = - 1 / 12 so d is my answer | a ) 1 / 12 , b ) 1 / 3 , c ) 1 / 2 , d ) - 1 / 12 , e ) - 4 / 3 | d | divide(power(divide(power(negate(const_1), 2), 2), 2), negate(3)) | negate(const_1)|negate(n3)|power(#0,n0)|divide(#2,n0)|power(#3,n0)|divide(#4,#1) | general |
39 persons can repair a road in 12 days , working 5 hours a day . in how many days will 15 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 15 : 39 : : 12 : x working hours / day 6 : 5 15 x 6 x x = 39 x 5 x 12 x = ( 39 x 5 x 12 ) / ( 15 x 6 ) x = 26 . answer : d" | a ) 10 , b ) 13 , c ) 14 , d ) 26 , e ) 16 | d | divide(multiply(multiply(39, 12), 5), multiply(15, 6)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics |
let f ( x ) = x ^ 2 + bx + c . if f ( 1 ) = 0 and f ( - 5 ) = 0 , then f ( x ) crosses the y - axis at what y - coordinate ? | "when x = 1 and when x = - 5 , the expression f ( x ) = x ² + bx + c equals 0 . then f ( x ) = ( x - 1 ) ( x + 5 ) f ( 0 ) = - 5 the answer is a ." | a ) - 5 , b ) - 1 , c ) 0 , d ) 1 , e ) 5 | a | negate(divide(subtract(power(5, 2), 1), add(5, 1))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general |
the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 40 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 * 40 . ( formula for ... | a ) 20 ft , b ) 25 ft , c ) 600 ft , d ) 900 ft , e ) 1000 ft | c | 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 |
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 4.5 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4.5 = 6480 . answer : d" | a ) 5729 , b ) 5760 , c ) 2889 , d ) 6480 , e ) 2799 | d | divide(multiply(4.5, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)| | physics |
a and b invests rs . 10000 each , a investing for 8 months and b investing for all the 12 months in the year . if the total profit at the end of the year is rs . 5000 , find their shares ? | "the ratio of their profits a : b = 8 : 12 = 2 : 3 share of a in the total profit = 2 / 5 * 5000 = rs . 2000 share of b in the total profit = 3 / 5 * 5000 = rs . 3000 answer : c" | a ) 2000,8000 , b ) 2000,4000 , c ) 2000,3000 , d ) 1000,3000 , e ) 4000,3000 | c | divide(multiply(10000, const_1), const_3) | multiply(n0,const_1)|divide(#0,const_3)| | gain |
a circular path of 12 m radius has marginal walk 2 m wide all round it . find the cost of leveling the walk at 50 p per m 2 ? | "explanation : π ( 14 ^ 2 - 12 ^ 2 ) = 22 / 7 * 52 = 163.43 163.43 * 1 / 2 = rs . 81.72 answer : option c" | a ) rs . 49 , b ) rs . 40 , c ) rs . 81.72 , d ) rs . 42 , e ) rs . 43 | c | multiply(50, subtract(circle_area(add(12, 2)), circle_area(12))) | add(n0,n1)|circle_area(n0)|circle_area(#0)|subtract(#2,#1)|multiply(n2,#3)| | physics |
last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 8 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | "let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.08 * 1.1 x = 0.088 x 0.088 x / 0.06 x = 88 / 60 = 1.47 = 147 % the answer is b ." | a ) 132 % , b ) 147 % , c ) 158 % , d ) 176 % , e ) 188 % | b | multiply(divide(multiply(divide(8, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100) | add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)| | gain |
the length of a rectangle is twice its breadth . if its length is decreased by 10 cm and breadth is increased by 10 cm , the area of the rectangle is increased by 75 sq . cm . what is the length of the rectangle ? | "let breadth = x cm then , length = 2 x cm area = x × 2 x = 2 x ^ 2 sq . cm . new length = ( 2 x − 10 ) cm new breadth = ( x + 10 ) cm new area = ( 2 x − 10 ) ( x + 10 ) sq . cm . given that , new area = initial area + 75 sq . cm . ⇒ ( 2 x − 10 ) ( x + 10 ) = 2 x ^ 2 + 75 ⇒ 2 x ^ 2 + 20 x − 10 x − 100 = 2 x ^ 2 + 75 ⇒ ... | a ) 25 cm , b ) 35 cm , c ) 30 cm , d ) 45 cm , e ) 55 cm | b | divide(const_100, const_3) | divide(const_100,const_3)| | geometry |
factory x ' s bulbs work for over 5000 hours in 78 % of cases , whereas factory y ' s bulbs work for over 5000 hours in 65 % of cases . it is known that factory x supplies 90 % of the total bulbs available . what is the chance that a purchased bulb will work for longer than 5000 hours ? | "for x , 90 % of 78 % will work . for y , 10 % of 65 % will work . * 10 % is the rest of the bulb supply in the market . so , the probability that a purchased bulb will work is : 0.90 ( 0.78 ) = . 702 0.10 ( 0.65 ) = 0.065 the combined probability then is 70.2 + 6.5 = 76.7 % ans b" | a ) 76.4 % , b ) 76.7 % , c ) 77.4 % , d ) 75 % , e ) 73.9 % | b | divide(add(78, 65), const_2) | add(n1,n3)|divide(#0,const_2)| | physics |
7 years ago , paula was 3 times as old as karl . in 2 years , paula will be twice as old as karl . what is the sum of their ages now ? | "p - 7 = 3 ( k - 7 ) and so p = 3 k - 14 p + 2 = 2 ( k + 2 ) ( 3 k - 14 ) + 2 = 2 k + 4 k = 16 p = 34 p + k = 50 the answer is e ." | a ) 26 , b ) 32 , c ) 38 , d ) 44 , e ) 50 | e | add(subtract(multiply(add(negate(subtract(2, multiply(2, const_2))), subtract(multiply(7, 3), 7)), 3), subtract(multiply(7, 3), 7)), add(negate(subtract(2, multiply(2, const_2))), subtract(multiply(7, 3), 7))) | multiply(n2,const_2)|multiply(n0,n1)|subtract(n2,#0)|subtract(#1,n0)|negate(#2)|add(#4,#3)|multiply(#5,n1)|subtract(#6,#3)|add(#5,#7)| | general |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 5 : 4 . the ratio of economy cars to sport utility vehicles is 3 : 2 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 5 : 4 - - > e : l = 5 : 4 = 15 : 12 . the ratio of economy cars to sport utility vehicles is 3 : 2 - - > e : s = 3 : 2 = 15 : 10 . thus , l : s = 12 : 10 = 6 : 5 . answer : c ." | a ) 9 : 8 , b ) 8 : 9 , c ) 6 : 5 , d ) 2 : 3 , e ) 1 : 2 | c | divide(divide(multiply(const_4, 2), multiply(2, 2)), divide(multiply(2, const_4), multiply(4, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other |
a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 65 % . staff are allowed a further 60 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 65 - 60 + ( - 65 ) ( - 60 ) / 100 = - 86 sale price = d * ( 1 - 86 / 100 ) sale price = . 14 * d answer ( d )" | a ) 0.15 d , b ) 0.16 d , c ) 0.65 d , d ) 0.14 d , e ) 0.05 d | d | subtract(divide(subtract(const_100, 65), const_100), multiply(divide(subtract(const_100, 65), const_100), divide(60, const_100))) | divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)| | gain |
the difference of two numbers is 11 . one third of their sum is 7 . what are the two numbers ? | "let the two numbers be x and y . equation ( i ) : x - y = 11 equation ( ii ) : ( x + y ) / 3 = 7 solve system of equations : x + y = 21 x - y = 11 ( add equations together ) - - - 2 x = 32 - - - x = 16 16 - y = 11 - - - y = 5 since x = 16 and y = 5 , answer b ( 16,5 ) is correct ." | a ) 21,10 , b ) 16,5 , c ) 3,14 , d ) 11,10 , e ) 7,14 | b | add(subtract(multiply(divide(const_10, const_2), 7), divide(add(11, multiply(divide(const_10, const_2), 7)), const_2)), divide(const_10, const_2)) | divide(const_10,const_2)|multiply(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#1,#3)|add(#0,#4)| | general |
75 % of 480 = ( ? ) x 15 ? | "answer let 75 % of 480 = a x 15 . ⇒ ( 75 x 480 ) / 100 = 15 a ∴ a = ( 75 x 480 ) / ( 100 x 15 ) = 24 correct option : d" | a ) 32 , b ) 18 , c ) 360 , d ) none of these , e ) can not be determined | d | divide(multiply(divide(multiply(15, 480), const_100), 75), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | general |
if a truck is traveling at a constant rate of 54 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 54 km / hr = > 54,000 m / hr in one minute = > 54000 / 60 = 900 meters in one sec = > 900 / 60 = 15 meters time = total distance need to be covered / avg . speed = > 600 / 15 = 40 and hence the answer : e" | a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 40 | e | multiply(divide(divide(600, 1000), 54), const_3600) | divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)| | physics |
what number times ( 1 ⁄ 7 ) ^ 2 will give the value of 7 ^ 3 ? | "x * ( 1 / 7 ) ^ 2 = 7 ^ 3 x = 7 ^ 2 * 7 ^ 3 = 7 ^ 5 = 16,807 the answer is e ." | a ) 7 , b ) 49 , c ) 343 , d ) 2401 , e ) 16,807 | e | multiply(power(7, 7), power(7, 3)) | power(n1,n2)|power(n1,n4)|multiply(#0,#1)| | general |
the ratio 2 : 10 expressed as percent equals to | "explanation : actually it means 2 is what percent of 10 , which can be calculated as , ( 2 / 10 ) * 100 = 2 * 10 = 20 answer : option d" | a ) 50 % , b ) 30 % , c ) 25 % , d ) 20 % , e ) 90 % | d | multiply(divide(2, 10), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a watch was sold at a loss of 21 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "79 % 104 % - - - - - - - - 25 % - - - - 140 100 % - - - - ? = > rs . 560 answer : a" | a ) 560 , b ) 882 , c ) 799 , d ) 778 , e ) 901 | a | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 21))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 8 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 1 to 4 . how many cooks does the restaurant have ? | originally there were 3 k cooks and 8 k waiters . the new ratio is 1 : 4 which equals 3 : 12 . 12 k = 8 k + 12 k = 3 there are 9 cooks . the answer is c . | a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | multiply(divide(12, 4), divide(12, 4)) | divide(n2,n4)|multiply(#0,#0) | other |
a bag contains 5 green balls and 8 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 13 c 2 = 78 the number of ways to draw two green balls is 5 c 2 = 10 the number of ways to draw two white balls is 8 c 2 = 28 p ( two balls of the same colour ) = 38 / 78 = 19 / 39 the answer is d ." | a ) 3 / 7 , b ) 6 / 11 , c ) 12 / 21 , d ) 19 / 39 , e ) 29 / 49 | d | add(multiply(divide(8, add(5, 8)), divide(subtract(8, const_1), subtract(add(5, 8), const_1))), multiply(divide(subtract(5, const_1), subtract(add(5, 8), const_1)), divide(5, add(5, 8)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#7,#4)|add(#8,#9)| | other |
if 60 % of 3 / 5 of a number is 36 , then the number is ? | "let the number be x . then 60 % of 3 / 5 of x = 36 60 / 100 * 3 / 5 * x = 36 x = ( 36 * 25 / 9 ) = 100 required number = 100 . correct option : b" | a ) 80 , b ) 100 , c ) 75 , d ) 90 , e ) none of these | b | divide(36, multiply(divide(60, const_100), divide(3, 5))) | divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)| | gain |
a sum of money amounts to rs . 9800 after 5 years and rs . 12005 after 8 years at the same rate of simple interest . the rate of interest per annum is ? | "s . i . for 3 years = ( 12005 - 9800 ) = rs . 2205 s . i . for 5 years = rs . 2205 / 3 * 5 = rs . 3675 . principal = ( 9800 - 3675 ) = rs . 6125 hence , rate = ( 100 * 3675 ) / ( 6125 * 5 ) = 12 % answer : c" | a ) 16 % , b ) 15 % , c ) 12 % , d ) 22 % , e ) 19 % | c | multiply(divide(subtract(12005, 9800), subtract(multiply(9800, 8), multiply(5, 12005))), const_100) | multiply(n0,n3)|multiply(n1,n2)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,const_100)| | gain |
two years ago , ram put $ 1,000 into a savings account . at the end of the first year , his account had accrued $ 200 in interest bringing his total balance to $ 1,200 . the next year , his account balance increased by 20 % . at the end of the two years , by what percent has ram ' s account balance increased from his i... | "investment 1000 dollars 1 st year total gained = 200 total amount end of first year = 1200 second year account increased by 20 % = 1200 * 0.2 = 240 therefore total amount by second year end = 1420 so total percentage increase in money = ( 1420 - 1000 ) * 100 / 1000 = 42 % correct answer c = 42 %" | a ) 19 % , b ) 30 % , c ) 42 % , d ) 45 % , e ) 25 % | c | multiply(divide(add(multiply(divide(20, 200), 1,200), 200), 1,000), 200) | divide(n3,n1)|multiply(n2,#0)|add(n1,#1)|divide(#2,n0)|multiply(#3,n1)| | gain |
in a college , 1 percent of students hate math , 2 percent of students hate english , 1 percent hate french and 4 percent hate german . can you find out the percentage of students who hate all 4 subjects ? | a 8 % of student hate all four subjects . | a ) 8 % , b ) 5 % , c ) 7 % , d ) 12 % , e ) 19 % | a | multiply(4, multiply(1, 2)) | multiply(n0,n1)|multiply(n3,#0) | gain |
walking at 6 / 7 th of his usual speed , a man is 25 mins too late . his usual time is | "as the distance is same s * t = 6 / 7 s * ( t + 25 ) solving this we get t = 150 sec 150 / 60 = 2.5 hrs answer : c" | a ) 7.5 hrs , b ) 1.5 hrs , c ) 2.5 hrs , d ) 1.67 hrs , e ) 2.67 hrs | c | multiply(6, 25) | multiply(n0,n2)| | physics |
amit and ananthu can do a work in 15 days and 30 days respectively . amit started the work and left after 3 days . ananthu took over and completed the work . in how many days was the total work completed ? | "amit ’ s one day ’ s work = 1 / 15 amit ’ s 3 day ’ s work = 1 / 15 * 3 = 1 / 5 work left = 1 - 1 / 5 = 4 / 5 ananthu ’ s one day ’ s work = 1 / 30 ananthu can do work in = 4 / 5 * 30 = 24 days so total days = 24 + 3 = 27 days answer : e" | a ) 28 days , b ) 20 days , c ) 23 days , d ) 25 days , e ) 27 days | e | add(divide(subtract(const_1, multiply(inverse(15), 3)), inverse(30)), 3) | inverse(n0)|inverse(n1)|multiply(#0,n2)|subtract(const_1,#2)|divide(#3,#1)|add(#4,n2)| | physics |
if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 39 , find the value of ab . | "2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 39 - 9 = 30 ab = 15 . answer is a ." | a ) 15 , b ) 8 , c ) 4 , d ) 10 , e ) 3 | a | divide(subtract(39, power(3, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
a cylinder of height h is 2 / 3 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 2 / 3 new cylinder 3 / 5 so 3 / 5 / 2 / 3 = 3 / 5 * 3 / 2 = 0.90 or 90 ... | a ) 25 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | e | multiply(divide(divide(3, 5), divide(2, 3)), const_100) | divide(n3,n4)|divide(n0,n1)|divide(#0,#1)|multiply(#2,const_100)| | gain |
a and b together have $ 1210 . if of a ' s amount is equal to of b ' s amount , what is the ratio between a and b ? | "( 4 / 15 ) a = ( 2 / 5 ) b a = ( ( 2 / 15 ) * ( 15 / 4 ) ) b a = ( 3 / 2 ) b a / b = 3 / 2 a : b = 3 : 2 option b" | a ) 2 : 3 , b ) 3 : 2 , c ) 4 : 5 , d ) 5 : 3 , e ) can not be determined | b | divide(multiply(multiply(1210, const_2), const_2), const_10) | multiply(n0,const_2)|multiply(#0,const_2)|divide(#1,const_10)| | other |
the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 294 sq . cm , what is the width of the rectangle ? | "2 l + 2 w = 5 w l = 3 w / 2 w * l = 294 3 w ^ 2 / 2 = 294 w ^ 2 = 196 w = 14 the answer is c ." | a ) 10 cm , b ) 12 cm , c ) 14 cm , d ) 16 cm , e ) 18 cm | c | divide(294, const_10) | divide(n2,const_10)| | geometry |
express a speed of 108 kmph in meters per second ? | "108 * 5 / 18 = 30 mps answer : a" | a ) 30 mps , b ) 76 mps , c ) 26 mps , d ) 97 mps , e ) 16 mps | a | multiply(const_0_2778, 108) | multiply(n0,const_0_2778)| | physics |
the ratio of the two natural numbers is 6 : 5 . if a certain number is subtracted to both the numbers , the ratio becomes 5 : 4 . if the larger number exceeds the smaller number by 5 , find the number subtracted ? | let the two numbers be 6 x and 5 x . let the numbers subtracted to both so that their ratio becomes 5 : 4 be k . ( 6 x - k ) / ( 5 x - k ) = 5 / 4 = > 24 x - 4 k = 25 x - 5 k = > k = x . 6 x - 5 x = 5 = > x = 5 k = x = 5 . answer : b | a ) 8 , b ) 5 , c ) 15 , d ) 22 , e ) 6 | b | subtract(multiply(multiply(5, 5), 5), multiply(add(multiply(5, 5), 5), 4)) | multiply(n1,n1)|add(n1,#0)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3) | other |
200 is what percent of 50 ? | "200 = x * 50 / 100 x = 400 % ans ; e" | a ) 0.2 % , b ) 2 % , c ) 5 % , d ) 20 % , e ) 400 % | e | multiply(divide(200, 50), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a big container is 30 % full with water . if 54 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 54 liters of water is added , the container becomes 75 % full . hence these 54 liters account for 45 % of the container , which means that the capacity of it is 54 / 0.45 = 120 liters . or : if the capacity of the container is x liters then : 0.3 x + 54 = 0.75 x - - > ... | a ) 36 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 120 liters | e | divide(54, subtract(divide(3, 4), divide(30, const_100))) | divide(n2,n3)|divide(n0,const_100)|subtract(#0,#1)|divide(n1,#2)| | general |
at what rate percent per annum will the simple interest on a sum of money be 6 / 5 of the amount in 10 years ? | "let sum = x . then , s . i . = 6 x / 5 , time = 10 years . rate = ( 100 * 6 x ) / ( x * 5 * 10 ) = 12 % answer : e" | a ) 4 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 12 % | e | divide(multiply(divide(6, 5), const_100), 10) | divide(n0,n1)|multiply(#0,const_100)|divide(#1,n2)| | gain |
if x is real , find the maximum value of the expression - 2 x ^ 2 + 5 x + 7 . | this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 5 x + 7 = ( - 2 x + 7 ) ( x + 1 ) the roots are 7 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 1.25 the maximum value is - 2 ( 1.25 ) ^ 2 + 5 ( 1.25 ) + 7 = 10... | a ) 6.125 , b ) 8.125 , c ) 10.125 , d ) 12.125 , e ) 14.125 | c | subtract(add(multiply(5, divide(multiply(2, 2), 5)), 7), multiply(2, power(divide(multiply(2, 2), 5), 2))) | multiply(n0,n0)|divide(#0,n2)|multiply(n2,#1)|power(#1,n0)|add(n3,#2)|multiply(n0,#3)|subtract(#4,#5) | general |
the perimeter of one face of a cube is 40 cm . its volume will be : | "explanation : edge of cude = 40 / 4 = 10 cm volume = a * a * a = 10 * 10 * 10 = 1000 cm cube option d" | a ) 125 cm 3 , b ) 400 cm 3 , c ) 250 cm 3 , d ) 1000 cm 3 , e ) none of these | d | volume_cube(square_edge_by_perimeter(40)) | square_edge_by_perimeter(n0)|volume_cube(#0)| | geometry |
a starts business with rs . 3500 and after 10 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b â € ™ s contribution in the capital ? | "explanation : a invested rs . 3500 for 12 months . let b joined with investment x . and he invested for 12 - 10 = 2 months . so there profit ratio = ( 3500 ã — 12 ) : ( 2 x ) = 2 : 3 â ‡ ’ x = 31500 answer : d" | a ) 24887 , b ) 20778 , c ) 23788 , d ) 31500 , e ) 2811 | d | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 10), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)| | other |
if the personal income tax rate is lowered from 45 % to 30 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 48000 ? | saving = ( 45 - 30 ) % of 48000 = 7200 . answer : d | a ) $ 3500 , b ) $ 5000 , c ) $ 3150 , d ) $ 7200 , e ) $ 10000 | d | multiply(divide(48000, const_100), subtract(45, 30)) | divide(n2,const_100)|subtract(n0,n1)|multiply(#0,#1) | gain |
a trader sells 45 meters of cloth for rs . 4500 at the profit of rs . 12 per metre of cloth . what is the cost price of one metre of cloth ? | sp of 1 m of cloth = 4500 / 45 = rs . 100 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 100 - rs . 12 = rs . 88 . answer : c | a ) rs . 80 , b ) rs . 85 , c ) rs . 88 , d ) rs . 100 , e ) none of these | c | subtract(divide(4500, 45), 12) | divide(n1,n0)|subtract(#0,n2) | physics |
subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 10 pages . | "subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 10 / 2.5 ) = 4 hrs answer : d" | a ) 8 , b ) 10 , c ) 12 , d ) 4 , e ) 16 | d | floor(subtract(divide(300, 40), divide(50, 10))) | divide(n2,n3)|divide(n0,n1)|subtract(#0,#1)|floor(#2)| | general |
two trains are running in opposite directions in the same speed . the length of each train is 120 meter . if they cross each other in 16 seconds , the speed of each train ( in km / hr ) is | "explanation : distance covered = 120 + 120 = 240 m time = 16 s let the speed of each train = v . then relative speed = v + v = 2 v 2 v = distance / time = 240 / 16 = 15 m / s speed of each train = v = 15 / 2 = 7.5 m / s = 7.5 × 36 / 10 km / hr = 27 km / hr answer : option b" | a ) 42 , b ) 27 , c ) 28 , d ) 20 , e ) 24 | b | multiply(const_3_6, divide(divide(add(120, 120), 16), const_2)) | add(n0,n0)|divide(#0,n1)|divide(#1,const_2)|multiply(#2,const_3_6)| | physics |
how much is 80 % of 60 is greater than 4 / 5 of 25 ? | "( 80 / 100 ) * 60 â € “ ( 4 / 5 ) * 25 48 - 20 = 28 answer : b" | a ) 15 , b ) 28 , c ) 16 , d ) 12 , e ) 73 | b | subtract(multiply(60, divide(80, const_100)), multiply(divide(4, 5), 25)) | divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)| | general |
a 6 litre sol is 30 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ? | 30 % of 6 = 1.8 50 % of 6 = 3 shortage is 1.2 so we need to have 1.2 / 50 % to get 50 % alcohol content . = 2.4 e | a ) a . 0.6 , b ) b . 1 , c ) c . 2.1 , d ) d . 3 , e ) e . 2.4 | e | subtract(6, multiply(const_2, multiply(divide(30, const_100), 6))) | divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2) | gain |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 470 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 -... | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 470 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 770 a + b + c = 70 per hour so for eight hrs = 70 * 8 = 560 =... | a ) 400 , b ) 475 , c ) 550 , d ) 560 , e ) 700 | d | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0)| | physics |
can you find a 7 digit number which describes itself . the first digit is the number of zeros in the number . the second digit is the number of ones in the number , etc . for example , in the number 21200 , there are 2 zeros , 1 one , 2 twos , 0 threes and 0 fours . | e 3211000 | a ) 4211000 , b ) 1211000 , c ) 5211000 , d ) 2211000 , e ) 3211000 | e | subtract(subtract(subtract(subtract(subtract(subtract(subtract(subtract(add(add(divide(divide(multiply(divide(multiply(multiply(21200, const_10), const_3), 2), const_1000), const_10), const_10), 21200), 21200), const_1000), const_1000), const_3600), const_3600), const_1000), const_1000), const_100), const_100) | multiply(n1,const_10)|multiply(#0,const_3)|divide(#1,n2)|multiply(#2,const_1000)|divide(#3,const_10)|divide(#4,const_10)|add(n1,#5)|add(n1,#6)|subtract(#7,const_1000)|subtract(#8,const_1000)|subtract(#9,const_3600)|subtract(#10,const_3600)|subtract(#11,const_1000)|subtract(#12,const_1000)|subtract(#13,const_100)|subtra... | physics |
what decimal fraction is 50 ml of a litre ? | "answer required fraction = 50 / 1000 = 5 / 100 = . 05 correct option : b" | a ) . 5 , b ) . 05 , c ) . 05 , d ) 0.05 , e ) none of these | b | divide(50, const_1000) | divide(n0,const_1000)| | physics |
gary ’ s gas station serves an average of 13 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary ’ s station serve over the course of a typical week ? | "6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 13 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 13 + 10 ) = 16 * 68 = 1088 answer : a" | a ) 1,088 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(13, 10), multiply(9, add(const_4, const_1)))), const_1000)) | add(n1,n3)|add(n0,n1)|add(const_1,const_4)|multiply(n2,#2)|add(#1,#3)|multiply(#0,#4)|divide(#5,const_1000)|floor(#6)| | physics |
a man ' s speed with the current is 15 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is | sol . man ' s rate in still in water = ( 15 - 2.5 ) km / hr = 12.5 km / hr . man ' s rate against the current = ( 12.5 - 2.5 ) km / hr = 10 km / hr . answer c | a ) 8.5 km / hr , b ) 9 km / hr , c ) 10 km / hr , d ) 12.5 km / hr , e ) none | c | subtract(subtract(15, 2.5), 2.5) | subtract(n0,n1)|subtract(#0,n1) | gain |
in what time will a train 100 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 100 / 40 = 2.5 sec . answer : a" | a ) 2.5 sec , b ) 2.8 sec , c ) 8.5 sec , d ) 2.2 sec , e ) 4.5 sec | a | divide(100, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
if r is a positive integer , and r / 15 = 8.2 , what is the remainder when r is divided by 15 ? | if r were a multiple of 15 , the quotient r / 15 would be an integer . the fact that it ' s a decimal tells us that 15 goes into r eight whole times and some decimal part of a time . this decimal part , 0.2 is the remainder divided by the divisor . let b be the remainder . b / 15 = 0.2 = 1 / 5 b = ( 15 ) * ( 1 / 5 ) = ... | a ) a . 1 , b ) b . 2 , c ) c . 3 , d ) d . 4 , e ) e . 8 | c | reminder(multiply(8.2, 15), 15) | multiply(n0,n1)|reminder(#0,n0) | general |
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 15 , 12 , and 18 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at... | "the time it takes printer x is 15 hours . the combined rate of y and z is 1 / 12 + 1 / 18 = 5 / 36 the time it takes y and z is 36 / 5 the ratio of times is 15 / ( 36 / 5 ) = 5 * 15 / 36 = 25 / 12 the answer is d ." | a ) 4 / 15 , b ) 1 / 2 , c ) 15 / 22 , d ) 25 / 12 , e ) 11 / 4 | d | divide(15, divide(const_1, add(divide(const_1, 12), divide(const_1, 18)))) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)| | general |
what is the 20 - fifth decimal to the right in the fraction 10 / 11 ? | 10 / 11 = 0.9090909090 . . . the odd - numbered decimal places are 9 . the answer is e . | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | floor(multiply(divide(10, 11), const_10)) | divide(n1,n2)|multiply(#0,const_10)|floor(#1) | general |
one cutting machine cuts 100 metal bars in 30 minutes . another machine does the same job in 30 seconds . what time will it take if both machines are put on the same work ? | by guess it is clear that the time taken will be less than 30 seconds and very near to it . therefore , answer 29.5 seconds will be correct . answer - b | a ) 25.0 seconds , b ) 29.5 seconds , c ) 30.0 seconds , d ) 30.5 seconds , e ) 30.7 seconds | b | divide(const_1, add(divide(const_1, multiply(30, const_60)), divide(const_1, 30))) | divide(const_1,n1)|multiply(n1,const_60)|divide(const_1,#1)|add(#2,#0)|divide(const_1,#3) | physics |
what is the average ( arithmetic mean ) of the numbers 16 , 17 , 17 , 18 , 19 , 19 and 20 ? | "{ 16 , 17 , 17 , 18 , 19 , 19 , 20 } = { 18 - 2,18 - 1,18 - 1 , 18 , 18 + 1 , 18 + 1,18 + 2 } - - > the average = 18 . answer : a ." | a ) 18 , b ) 17 , c ) 16 , d ) 15 , e ) 19 | a | divide(divide(multiply(add(16, 17), add(divide(subtract(17, 16), 16), const_1)), const_2), add(divide(subtract(17, 16), 16), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
the sum of 5 th and 21 th term of a . p . is equal to the sum of 8 th , 15 th and 13 th term . find the term which is 0 | "t 5 + t 21 = t 8 + t 15 + t 13 = > a + 4 d + a + 20 d = a + 7 d + a + 14 d + a + 12 d = > a + 9 d = 0 = > t 10 = 0 i . e . 10 th term is zero . answer : b" | a ) 11 , b ) 10 , c ) 12 , d ) 13 , e ) 14 | b | subtract(add(add(8, 15), 13), add(5, 21)) | add(n2,n3)|add(n0,n1)|add(n4,#0)|subtract(#2,#1)| | general |
what is 0.01 percent of 15,356 ? | since , percent = 1 / 100 , what = something ( s ) , and is : = . we can write the question as s = 0.01 ( 1 / 100 ) 15,356 . the answer is 1.5356 . hence , the correct answer is b . | a ) 0.15356 , b ) 1.5356 , c ) 15.356 , d ) 0.015356 , e ) 0.0015356 | b | divide(multiply(0.01, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
evaluate : 12 + sqrt ( - 5 + 7 ã — 8 ã · 4 ) = ? | "according to order of operations , inner brackets first where 7 x 8 ã · 4 is first calculated since it has a multiplication and a division . 7 x 8 ã · 4 = 56 ã · 4 = 14 hence 12 + sqrt ( - 5 + 7 ã — 8 ã · 4 ) = 12 + sqrt ( - 5 + 14 ) = 12 + sqrt ( 9 ) = 12 + 81 = 93 correct answer d ) 93" | a ) 90 , b ) 91 , c ) 92 , d ) 93 , e ) 94 | d | add(12, sqrt(subtract(divide(multiply(7, 5), 4), 5))) | multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3)| | general |
in 5 football games thus far this season , barry has run for 98 , 107 , 85 , 89 , and 91 yards . at a minimum , how many yards n will he need to gain this sunday if he wants to keep his season average above 100 yards ? | sorry the answer n = 131 is d . | a ) 101 , b ) 116 , c ) 130 , d ) n = 131 , e ) n = 141 | d | subtract(multiply(add(5, const_1), 100), add(add(add(add(98, 107), 85), 89), 91)) | add(n0,const_1)|add(n1,n2)|add(n3,#1)|multiply(n6,#0)|add(n4,#2)|add(n5,#4)|subtract(#3,#5) | general |
ab + cd = jjj , where ab and cd are two - digit numbers and jjj is a 3 digit number ; a , b , c , and d are distinct positive integers . in the addition problem above , what is the value of c ? | ab and cd are two digit integers , their sum can give us only one three digit integer of a kind of jjj it ' s 111 . so , a = 1 . 1 b + cd = 111 now , c can not be less than 9 , because no to digit integer with first digit 1 ( mean that it ' s < 20 ) can be added to two digit integer less than 90 to have the sum 111 ( i... | a ) 1 , b ) 3 , c ) 7 , d ) 9 , e ) can not be determined | d | multiply(3, 3) | multiply(n0,n0) | general |
8 friends went to a hotel and decided to pay the bill amount equally . but 7 of them could pay rs . 80 each as a result 8 th has to pay rs . 70 extra than his share . find the amount paid by him . | "explanation : average amount paid by 7 persons = rs . 80 increase in average due to rs . 70 paid extra by the 8 th men = rs . 70 / 7 = rs . 10 therefore , average expenditure of 8 friends = rs . 80 + rs . 10 = rs . 90 therefore , amount paid by the 11 th men = rs . 90 + rs . 70 = rs . 160 correct option : d" | a ) 140 , b ) 121 , c ) 110 , d ) 160 , e ) none | d | add(70, divide(add(multiply(80, 7), 70), 7)) | multiply(n1,n2)|add(n4,#0)|divide(#1,n1)|add(n4,#2)| | general |
if 4 men can colour 48 m long cloth in 2 days , then 8 men can colour 36 m long cloth in | "the length of cloth painted by one man in one day = 48 / 4 × 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 × 8 = 0.75 day . a" | a ) 0.75 day , b ) 2 days , c ) 1.2 days , d ) 4 days , e ) 5 days | a | divide(36, multiply(divide(48, multiply(4, 2)), 8)) | multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)| | physics |
a person lent a certain sum of money at 6 % per annum at simple interest and in 8 years the interest amounted to $ 572 less than the sum lent . what was the sum lent ? | "p - 572 = ( p * 6 * 8 ) / 100 p = 1100 the answer is b ." | a ) 1000 , b ) 1100 , c ) 1200 , d ) 1300 , e ) 1400 | b | divide(572, subtract(const_1, divide(multiply(6, 8), const_100))) | multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)| | gain |
a circular ground whose diameter is 35 metres , has a 1.4 metre - broad garden around inside of it . what is the area of the garden in square metres ? | "req . area = ï € [ ( 35 ) 2 â € “ ( 33.6 ) 2 ] = 22 â „ 7 ã — ( 68.6 ã — 1.4 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] = ( 22 ã — 68.6 ã — 0.2 = 301.84 sq m . answer e" | a ) 301.5 , b ) 484.12 , c ) 401.84 , d ) 301.0 , e ) 301.84 | e | subtract(circle_area(add(divide(35, 1.4), 1.4)), circle_area(divide(35, 1.4))) | divide(n0,n1)|add(n1,#0)|circle_area(#0)|circle_area(#1)|subtract(#3,#2)| | geometry |
the edges of three metal cubes are 4 cm , 5 cm , and 6 cm respectively . a new cube is made by melting these three cubes together . what is the edge of the new cube ( in centimeters ) ? | "the total volume is 4 ^ 3 + 5 ^ 3 + 6 ^ 3 = 405 the edge of the new cube is the cube root of 405 which is about 7.4 cm . the answer is e ." | a ) 6.6 , b ) 6.8 , c ) 7.0 , d ) 7.2 , e ) 7.4 | e | power(add(power(6, 5), add(4, power(5, 5))), const_0_33) | power(n1,n1)|power(n2,n1)|add(n0,#0)|add(#2,#1)|power(#3,const_0_33)| | physics |
mr . hernandez , who was a resident of state x for only 8 months last year , had a taxable income of $ 22,500 for the year . if the state tax rate were 10 percent of the year ’ s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez ... | "total tax for the year = 22,500 x 10 % = 2250 as stated annual tax is prorated as per the duration of stay . prorated tax = 900 ( 8 / 12 ) = 1500 answer a" | a ) $ 1500 , b ) $ 1720 , c ) $ 1600 , d ) $ 1300 , e ) $ 1160 | a | divide(multiply(multiply(divide(add(multiply(multiply(10, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), 10), 8), multiply(const_3, 10)) | multiply(n2,const_100)|multiply(const_0_25,const_100)|multiply(const_3,n2)|multiply(#0,const_100)|multiply(#1,const_100)|add(#3,#4)|divide(#5,const_100)|multiply(#6,n2)|multiply(n0,#7)|divide(#8,#2)| | gain |
walking at 3 / 4 of her normal speed , a worker is 12 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is | let v be her normal speed and let t be her normal time . d = ( 3 / 4 ) v * ( t + 12 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 3 / 4 ) v * ( t + 12 ) t / 4 = 9 t = 36 the answer is a . | a ) 36 , b ) 48 , c ) 60 , d ) 62 , e ) 66 | a | multiply(3, 12) | multiply(n0,n2) | physics |
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