Dataset Preview Go to dataset viewer
Problem (string)Rationale (string)options (string)correct (string)annotated_formula (string)linear_formula (string)category (string)
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 31.1 % . what would have been the percentage of profit earned if no discount had been offered ?
"giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 ã — ( 100 + 31.1 ) / ( 100 â ˆ ’ 5 ) = 138 hence , required % profit = 138 â € “ 100 = 38 % answer a"
a ) 38 , b ) 27.675 , c ) 30 , d ) data inadequate , e ) none of these
a
subtract(divide(multiply(add(const_100, 31.1), const_100), subtract(const_100, 5)), const_100)
add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)|
gain
what will be the difference between simple and compound interest at 14 % per annum on a sum of rs . 1000 after 4 years ?
"s . i . = ( 1000 * 14 * 4 ) / 100 = rs . 560 c . i . = [ 1000 * ( 1 + 14 / 100 ) 4 - 1000 ] = rs . 689 difference = ( 689 - 560 ) = rs . 129 answer : a"
a ) 129 , b ) 130 , c ) 124 , d ) 133 , e ) 145
a
subtract(subtract(multiply(1000, power(add(divide(14, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(14, const_100)), 4))
divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)|
gain
there are 28 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ?
"the total number of stations = 30 from 30 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 3 ⁰ p ₂ ways . 30 p ₂ = 30 * 29 = 870 . answer : c"
a ) 156 , b ) 167 , c ) 870 , d ) 352 , e ) 380
c
multiply(add(28, const_1), add(add(28, const_1), const_1))
add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)|
physics
the present population of a town is 3888 . population increase rate is 20 % p . a . find the population of town before 2 years ?
"p = 3888 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3888 / ( 1 + 20 / 100 ) ^ 2 = 3888 / ( 6 / 5 ) ^ 2 = 2700 ( approximately ) answer is e"
a ) 2500 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2700
e
add(3888, divide(multiply(3888, 20), const_100))
multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|
gain
the triplicate ratio of 1 : 9 is ?
"13 : 93 = 1 : 729 answer : e"
a ) 1 : 0 , b ) 1 : 8 , c ) 1 : 7 , d ) 1 : 2 , e ) 1 : 729
e
divide(power(const_2.0, 9), power(const_3.0, 9))
power(const_2.0,n1)|power(const_3.0,n1)|divide(#0,#1)|
other
the sum of all the integers s such that - 26 < s < 24 is
"easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 s = - 49 . d is the answer ."
a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 51
d
add(add(negate(26), const_1), add(add(negate(26), const_1), const_1))
negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)|
general
a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 6 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ?
"the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 36 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 360 / 10,000 = 36 / 1000 = 0.036 ft the answer is a ."
a ) 0.036 , b ) 0.36 , c ) 0.6 , d ) 6 , e ) 3.6
a
divide(volume_cylinder(6, 10), circle_area(100))
circle_area(n0)|volume_cylinder(n2,n3)|divide(#1,#0)|
geometry
each week a restaurant serving mexican food uses the same volume of chili paste , which comes in either 35 - ounce cans or 25 - ounce cans of chili paste . if the restaurant must order 20 more of the smaller cans than the larger cans to fulfill its weekly needs , then how manysmallercans are required to fulfill its weekly needs ?
"let x be the number of 35 ounce cans . therefore ( x + 20 ) is the number of 25 ounce cans . total volume is same , therefore 35 x = 25 ( x + 20 ) 10 x = 500 x = 50 therefore , number of 15 ounce cans = 50 + 20 = 70 ans - b"
a ) 60 , b ) 70 , c ) 80 , d ) 100 , e ) 120
b
add(25, 20)
add(n1,n2)|
general
if n is an integer and 101 n ^ 2 is less than or equal to 10000 , what is the greatest possible value of n ?
"101 * n ^ 2 < = 10000 n ^ 2 < = 10000 / 101 which will be less than 100 since 10000 / 100 = 100 which is the square of 9 next closest value of n where n ^ 2 < = 100 is 9 ans c"
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
c
floor(sqrt(divide(10000, 101)))
divide(n2,n0)|sqrt(#0)|floor(#1)|
general
a constructor estimates that 10 people can paint mr khans house in 4 days . if he uses 5 people instead of 10 , how long will they take to complete the job ?
"explanation : use formula for a work members ã — days = constant 10 ã — 4 = 5 ã — a a = 8 so answer is 8 days answer : d"
a ) 10 , b ) 4 , c ) 5 , d ) 8 , e ) 6
d
divide(const_1, multiply(divide(const_1, multiply(const_4.0, 10)), 4))
multiply(n0,n1)|divide(const_1,#0)|multiply(n2,#1)|divide(const_1,#2)|
physics
the population of a town is 8000 . it decreases annually at the rate of 20 % p . a . what will be its population after 3 years ?
"formula : ( after = 100 denominator ago = 100 numerator ) 8000 ã — 80 / 100 ã — 80 / 100 x 80 / 100 = 4096 answer : b"
a ) 5100 , b ) 4096 , c ) 5200 , d ) 5400 , e ) 5500
b
subtract(subtract(8000, multiply(8000, divide(20, const_100))), multiply(subtract(8000, multiply(8000, divide(20, const_100))), divide(20, const_100)))
divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)|
gain
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 40 % profit ?
"let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 140 % of rs . 1600 = 140 / 100 * 1600 = rs . 2240 . answer : e"
a ) 2000 , b ) 2778 , c ) 2299 , d ) 2778 , e ) 2240
e
multiply(divide(add(const_100, 40), const_100), divide(add(1920, 1280), const_2))
add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)|
gain
running at the same constant rate , 6 identical machines can produce a total of 360 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ?
"let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 360 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 360 x = ( 10 x 4 x 360 ) / ( 6 ) x = 2400 . answer : c"
a ) 648 , b ) 1800 , c ) 2400 , d ) 10800 , e ) 10900
c
multiply(multiply(divide(360, 6), 4), 10)
divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)|
gain
there are 1000 buildings in a street . a sign - maker is contracted to number the houses from 1 to 1000 . how many zeroes will he need ?
divide as ( 1 - 100 ) ( 100 - 200 ) . . . . ( 900 - 1000 ) total 192 answer : c
a ) 190 , b ) 191 , c ) 192 , d ) 193 , e ) 194
c
add(add(divide(1000, const_10), multiply(subtract(const_10, 1), const_10)), const_2)
divide(n0,const_10)|subtract(const_10,n1)|multiply(#1,const_10)|add(#0,#2)|add(#3,const_2)
general
a man bought 20 shares of rs . 50 at 5 discount , the rate of dividend being 13 . the rate of interest obtained is :
"investment = rs . [ 20 x ( 50 - 5 ) ] = rs . 900 . face value = rs . ( 50 x 20 ) = rs . 1000 . dividend = rs . 27 x 1000 = rs . 135 . 2 100 interest obtained = 135 x 100 % = 15 % 900 view answer discuss in forum answer : c"
a ) 27 % , b ) 87 % , c ) 15 % , d ) 66 % , e ) 88 %
c
divide(multiply(multiply(20, 50), divide(13, const_100)), multiply(20, subtract(50, 5)))
divide(n3,const_100)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#1)|multiply(n0,#2)|divide(#3,#4)|
gain
? % of 360 = 108
"? % of 360 = 108 or , ? = 108 × 100 / 360 = 30 answer a"
a ) 30 , b ) 36 , c ) 64 , d ) 72 , e ) none of these
a
divide(multiply(108, const_100), 360)
multiply(n1,const_100)|divide(#0,n0)|
gain
a train 300 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 150 m long ?
"speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 300 + 150 = 450 m required time = 450 * 2 / 25 = 36 sec answer : option b"
a ) 40 , b ) 36 , c ) 41 , d ) 42 , e ) 34
b
divide(300, multiply(subtract(45, 150), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
what is the sum of all digits for the number 10 ^ 29 - 41 ?
"10 ^ 29 is a 30 - digit number : 1 followed by 29 zeros . 10 ^ 29 - 41 is a 29 - digit number : 27 9 ' s and 59 at the end . the sum of the digits is 27 * 9 + 5 + 9 = 257 . the answer is a ."
a ) 257 , b ) 242 , c ) 231 , d ) 202 , e ) 187
a
multiply(add(divide(subtract(subtract(29, 10), const_2), const_2), 10), divide(add(subtract(29, 10), const_2), const_2))
subtract(n1,n0)|add(#0,const_2)|subtract(#0,const_2)|divide(#2,const_2)|divide(#1,const_2)|add(n0,#3)|multiply(#5,#4)|
general
a train running at the speed of 120 km / hr crosses a pole in 18 seconds . what is the length of the train ?
"speed = ( 120 x ( 5 / 18 ) m / sec = ( 100 / 3 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 100 / 3 ) x 18 ) m = 600 m e"
a ) 560 , b ) 570 , c ) 580 , d ) 590 , e ) 600
e
multiply(divide(multiply(120, const_1000), const_3600), 18)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
a train 100 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ?
"s = ( 100 + 300 ) / 45 = 400 / 45 * 18 / 5 = 32 answer : a"
a ) 32 kmph , b ) 76 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph
a
divide(divide(add(100, 300), const_1000), divide(45, const_3600))
add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)|
physics
each month a retailer sells 100 identical items . on each item he makes a profit of $ 40 that constitutes 10 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ?
"for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 10 % of the cost : sell price = cost + $ 40 sell price = $ 400 + $ 40 thus , the sell price is $ 440 for each item . selling all 100 items gives the retailer . . . 100 ( $ 40 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 440 ) = $ 22 discount $ 418 = $ 400 + $ 18 now , the retailer makes a profit of just $ 18 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 18 ( x ) = $ 2,000 x = 2,000 / 18 x = 222.222222 items you ' ll notice that this is not among the answer choices . . . . 221 and 223 are . selling 221 items would get us 9 ( 221 ) = $ 1989 which is not enough money . to get back to at least $ 2,000 , we need to sell 223 items . final answer : d"
a ) 191 , b ) 213 , c ) 221 , d ) 223 , e ) 226
d
divide(multiply(100, 40), subtract(40, divide(multiply(add(divide(multiply(100, 40), 10), 40), 5), 100)))
multiply(n0,n1)|divide(#0,n2)|add(n1,#1)|multiply(n3,#2)|divide(#3,n0)|subtract(n1,#4)|divide(#0,#5)|
gain
the average age of 15 students of a class is 14 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ?
"age of the 15 th student = [ 15 * 14 - ( 14 * 5 + 16 * 9 ) ] = ( 210 - 214 ) = 4 years . answer : b"
a ) 3 years , b ) 4 years , c ) 5 years , d ) 6 years , e ) 7 years
b
subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
bookman purchased 55 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 460 , how many paperback copies were sold ?
"the bookman had 10 hardback ad 55 - 10 = 45 paperback copies ; 14 copies were sold , hence 55 - 14 = 41 copies were left . let # of paperback copies left be p then 10 p + 20 ( 41 - p ) = 460 - - > 10 p = 360 - - > p = 36 # of paperback copies sold is 45 - 36 = 9 answer : e"
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 9
e
divide(subtract(subtract(add(multiply(subtract(55, 10), 10), multiply(10, 20)), 460), multiply(gcd(55, 10), 20)), 10)
gcd(n0,n1)|multiply(n1,n2)|subtract(n0,n1)|multiply(n1,#2)|multiply(n2,#0)|add(#3,#1)|subtract(#5,n5)|subtract(#6,#4)|divide(#7,n1)|
general
diana is painting statues . she has 1 / 2 of a gallon of paint remaining . each statue requires 1 / 16 gallon of paint . how many statues can she paint ?
"number of statues = all the paint ÷ amount used per statue = 1 / 2 ÷ 1 / 16 = 8 / 16 * 16 / 1 = 8 / 1 = 8 answer is a ."
a ) 8 , b ) 20 , c ) 28 , d ) 14 , e ) 19
a
divide(divide(1, 2), divide(1, 16))
divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|
general
if the price of gasoline increases by 25 % and a driver intends to spend only 20 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ?
"let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.25 * p ) y = 1.2 ( p * x ) y = ( 1.2 / 1.25 ) x = 0.96 x which is a reduction of 4 % . the answer is c ."
a ) 2 % , b ) 3 % , c ) 4 % , d ) 5 % , e ) 6 %
c
multiply(divide(subtract(add(25, const_100), add(20, const_100)), add(25, const_100)), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)|
general
an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 1000 sculptures are not on display , how many pieces of art does the gallery have ?
too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 / 3 of the pieces not on display are sculptures . 1000 sculptures are not on display , so according to ( iii ) 2 / 3 * { not on display } = 1000 - - > { not on display } = 1500 . according to ( i ) 2 / 3 * { total } = 1500 - - > { total } = 2250 . answer : b .
a ) 360 , b ) 2250 , c ) 540 , d ) 640 , e ) 720
b
divide(divide(1000, subtract(const_1, divide(1, 3))), subtract(const_1, divide(1, 3)))
divide(n0,n1)|subtract(const_1,#0)|divide(n6,#1)|divide(#2,#1)
general
john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 60000 and ingrid makes $ 72000 , what is their combined tax rate ?
"( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 72000 / 132000 = 6 / 11 . 6 / 11 is a little above 6 / 12 = 1 / 2 . thus , our answer is just a little above 35 . answer : d"
a ) 32 % , b ) 34.4 % , c ) 35 % , d ) 35.6 % , e ) 36.4 %
d
multiply(divide(add(multiply(divide(30, const_100), 60000), multiply(divide(40, const_100), 72000)), add(72000, 60000)), const_100)
add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|
gain
the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 900 per sq . metre .
"area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 900 ã — 22 ) = rs . 19800 answer : option a"
a ) s . 19,800 , b ) s . 15,600 , c ) s . 16,500 , d ) s . 17,600 , e ) s . 17,900
a
multiply(900, multiply(5.5, 4))
multiply(n0,n1)|multiply(n2,#0)|
physics
a factory that employs 1000 assembly line workers pays each of these workers $ 5 per hour for the first 40 hours worked during a week and 1 ½ times that rate for hours worked in excess of 40 . what was the total payroll for the assembly - line workers for a week in which 30 percent of them worked 15 hours , 50 percent worked 40 hours , and the rest worked 50 hours ?
"30 % of 1000 = 300 worked for 15 hours payment @ 5 / hr total payment = 300 * 15 * 5 = 22500 50 % of 1000 = 500 worked for 40 hours payment @ 5 / hr total payment = 500 * 40 * 5 = 100000 remaining 200 worked for 50 hours payment for first 40 hours @ 5 / hr payment = 200 * 40 * 5 = 40000 payment for next 10 hr @ 7.5 / hr payment = 200 * 10 * 7.5 = 15000 total payment = 22500 + 100000 + 40000 + 15000 = 1775000 hence , answer is d"
a ) $ 180,000 , b ) $ 185,000 , c ) $ 190,000 , d ) $ 177,500 , e ) $ 205,000
d
multiply(add(divide(1, 15), 1), divide(multiply(1000, divide(add(add(multiply(add(multiply(divide(const_3, const_2), multiply(5, 15)), multiply(40, 5)), subtract(15, add(const_3, 5))), multiply(multiply(40, 5), 5)), multiply(multiply(5, 15), const_3)), 15)), 1000))
add(n1,const_3)|divide(n3,n6)|divide(const_3,const_2)|multiply(n1,n6)|multiply(n1,n2)|add(n3,#1)|multiply(#2,#3)|multiply(n1,#4)|multiply(#3,const_3)|subtract(n6,#0)|add(#6,#4)|multiply(#10,#9)|add(#11,#7)|add(#12,#8)|divide(#13,n6)|multiply(n0,#14)|divide(#15,n0)|multiply(#5,#16)|
general
a corporation double its annual bonus to 100 of its employees . what percent of the employees ’ new bonus is the increase ?
let the annual bonus be x . a corporation double its annual bonus . so new bonus = 2 x . increase = 2 x - x = x the increase is what percent of the employees ’ new bonus = ( x / 2 x ) * 100 = 50 % hence a .
a ) 50 % , b ) 12 % , c ) 8 % , d ) 6 % , e ) 5 %
a
multiply(divide(subtract(const_2, const_1), const_2), 100)
subtract(const_2,const_1)|divide(#0,const_2)|multiply(n0,#1)
general
a and b together do a work in 20 days . b and c together in 15 days and c and a in 12 days . then b alone can finish same work in how many days ?
"( a + b ) work in 1 day = 1 / 20 , ( b + c ) work in 1 days = 1 / 15 . , ( c + a ) work in 1 days = 1 / 12 ( 1 ) adding = 2 [ a + b + c ] in 1 day work = [ 1 / 20 + 1 / 15 + 1 / 12 ] = 1 / 5 ( a + b + c ) work in 1 day = 1 / 10 b work in 1 days = [ a + b + c ] work in 1 days - work of ( a + c ) in 1 days = [ 1 / 10 - 1 / 12 ] = 1 / 60 b alone finish work in 60 days answer b"
a ) 50 , b ) 60 , c ) 45 , d ) 35 , e ) 48
b
inverse(divide(add(inverse(12), add(inverse(20), inverse(15))), const_2))
inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|inverse(#5)|
physics
two trains of equal length , running with the speeds of 60 and 16 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ?
"rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 50 * 100 / 18 = 2500 / 9 rs = 60 + 16 = 76 * 5 / 18 t = 2500 / 9 * 18 / 380 = 13.15 sec . answer : d"
a ) 10.11 , b ) 8.11 , c ) 77.2 , d ) 13.15 , e ) 22.22
d
multiply(multiply(multiply(const_0_2778, subtract(60, 16)), 50), inverse(multiply(const_0_2778, add(60, 16))))
add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)|
physics
calculate the ratio between x and y if 30 % of x equal to 50 % of y ?
"explanation : 30 x = 50 y x : y = 30 : 50 = 3 : 5 answer : b"
a ) 4 : 5 , b ) 3 : 5 , c ) 3 : 7 , d ) 3 : 2 , e ) 4 : 5
b
divide(30, 50)
divide(n0,n1)|
general
three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 34 square meters . what is the area that is covered with three layers of wallpaper ?
"300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 34 = 86 sq m of the wallpaper overlaps in three layers . 86 sq m makes two extra layers hence the area over which it makes two extra layers is 43 sq m . answer ( a ) ."
a ) 43 square meters , b ) 36 square meters , c ) 42 square meters , d ) 83.3 square meters , e ) 120 square meters
a
divide(subtract(subtract(300, 180), 34), const_2)
subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|
geometry
a meeting has to be conducted with 6 managers . find the number of ways in which the managers may be selected from among 9 managers , if there are 2 managers who refuse to attend the meeting together .
"the total number of ways to choose 6 managers is 9 c 6 = 84 we need to subtract the number of groups which include the two managers , which is 7 c 4 = 35 . 84 - 35 = 49 the answer is e ."
a ) 36 , b ) 40 , c ) 42 , d ) 45 , e ) 49
e
subtract(choose(9, 6), choose(subtract(9, 2), 2))
choose(n1,n0)|subtract(n1,n2)|choose(#1,n2)|subtract(#0,#2)|
probability
a trader bought a car at 20 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ?
"original price = 100 cp = 80 s = 80 * ( 140 / 100 ) = 112 100 - 112 = 12 % answer : c"
a ) 82 % , b ) 52 % , c ) 12 % , d ) 19 % , e ) 22 %
c
multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 40)), const_100), const_100), const_1), const_100)
add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
what is the unit digit in the product ( 3 ^ 65 x 6 ^ 59 x 7 ^ 71 ) ?
explanation : unit digit in 3 ^ 4 = 1 unit digit in ( 3 ^ 4 ) 16 = 1 unit digit in 3 ^ 65 = unit digit in [ ( 3 ^ 4 ) 16 x 3 ] = ( 1 x 3 ) = 3 unit digit in 6 ^ 59 = 6 unit digit in 7 ^ 4 unit digit in ( 7 ^ 4 ) 17 is 1 . unit digit in 7 ^ 71 = unit digit in [ ( 7 ^ 4 ) 17 x 73 ] = ( 1 x 3 ) = 3 required digit = unit digit in ( 3 x 6 x 3 ) = 4 e
a ) 18 , b ) 12 , c ) 69 , d ) 32 , e ) 4
e
subtract(multiply(multiply(3, 6), 3), subtract(multiply(multiply(3, 6), 3), const_4))
multiply(n0,n2)|multiply(n0,#0)|subtract(#1,const_4)|subtract(#1,#2)
general
a start walking from a place at a uniform speed of 6 kmph in a particular direction . after half an hour , b starts from the same place and walks in the same direction as a at a uniform speed and overtakes a after 1 hour 48 minutes . find the speed of b .
"distance covered by a in 30 min = 1 km b covers extra 1 km in 1 hour 48 minutes ( 9 / 5 hr ) i . e . relative speed of b over a = 1 / ( 9 / 5 ) = 5 / 9 so the speed of b = speed of a + 5 / 9 = 6 + 5 / 9 = 6.55 answer b"
a ) 4.7 kmph , b ) 6.6 kmph , c ) 4 kmph , d ) 7 kmph , e ) 5.3 kmph
b
add(divide(1, divide(add(const_60, 48), const_60)), 6)
add(n2,const_60)|divide(#0,const_60)|divide(n1,#1)|add(n0,#2)|
physics
oak street begins at pine street and runs directly east for 2 kilometers until it ends when it meets maple street . oak street is intersected every 400 meters by a perpendicular street , and each of those streets other than pine street and maple street is given a number beginning at 1 st street ( one block east of pine street ) and continuing consecutively ( 2 nd street , 3 rd street , etc . . . ) until the highest - numbered street one block west of maple street . what is the highest - numbered street that intersects oak street ?
2 km / 400 m = 5 . however , the street at the 2 - km mark is not 5 th street ; it is maple street . therefore , the highest numbered street is 4 th street . the answer is a .
a ) 4 th , b ) 5 th , c ) 6 th , d ) 7 th , e ) 8 th
a
subtract(divide(multiply(2, const_1000), 400), const_1)
multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1)
physics
50 percent of the members of a study group are women , and 30 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ?
"say there are 100 people in that group , then there would be 0.5 * 0.30 * 100 = 15 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable / total = 15 / 100 . answer : e"
a ) 0.16 , b ) 0.25 , c ) 0.45 , d ) 0.35 , e ) 0.15
e
multiply(divide(50, multiply(multiply(const_5, const_5), const_4)), divide(30, multiply(multiply(const_5, const_5), const_4)))
multiply(const_5,const_5)|multiply(#0,const_4)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)|
gain
the cross - section of a cannel is a trapezium in shape . if the cannel is 14 m wide at the top and 8 m wide at the bottom and the area of cross - section is 550 sq m , the depth of cannel is ?
"1 / 2 * d ( 14 + 8 ) = 550 d = 50 answer : c"
a ) 76 , b ) 28 , c ) 50 , d ) 80 , e ) 25
c
divide(divide(divide(550, divide(add(14, 8), const_2)), 8), const_2)
add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)|
physics
kim finds a 4 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ?
"3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is c"
a ) 2 / 5 , b ) 7 / 5 , c ) 3 / 5 , d ) 8 / 15 , e ) 1 / 2
c
subtract(const_1, add(add(divide(4, multiply(add(const_2, 4), 4)), divide(const_2, multiply(add(const_2, 4), 4))), divide(const_1, multiply(add(const_2, 4), 4))))
add(const_2,n0)|multiply(n0,#0)|divide(n0,#1)|divide(const_2,#1)|divide(const_1,#1)|add(#2,#3)|add(#5,#4)|subtract(const_1,#6)|
physics
$ 350 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ?
"let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 350 x = 50 4 x = 200 the answer is c ."
a ) $ 200 , b ) $ 225 , c ) $ 250 , d ) $ 275 , e ) $ 300
c
multiply(divide(350, add(add(divide(const_1, const_2), const_1), const_2)), const_2)
divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2)|
general
in an it company , there are a total of 90 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ?
"you could pick 40 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 45 . b"
a ) 10 , b ) 45 , c ) 55 , d ) 35 , e ) 65
b
add(subtract(80, 90), subtract(50, 35))
subtract(n2,n0)|subtract(n1,n3)|add(#0,#1)|
general
the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if arthur and david have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ?
"if arthur and david submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combined no discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) ."
a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54
b
divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2)
multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2)|
gain
( 3 x + 1 ) ( 2 x - 5 ) = ax ^ 2 + kx + n . what is the value of a - n + k ?
"expanding we have 6 x ^ 2 - 15 x + 2 x - 5 6 x ^ 2 - 13 x - 5 taking coefficients , a = 6 , k = - 13 , n = - 5 therefore a - n + k = 6 - ( - 13 ) - 5 = 19 - 5 = 14 the answer is d ."
a ) 5 , b ) 8 , c ) 9 , d ) 14 , e ) 11
d
add(add(multiply(3, 1), multiply(5, 1)), subtract(multiply(1, 1), multiply(5, 3)))
multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5)|
general
if 6 men and 8 women can do a piece of work in 10 days while 26 men and 48 women can do the same in 2 days , the time taken by 15 men and 20 women in doing the same type of work will be ?
let 1 man ' s 1 day ' s work = x and 1 women ' s 1 day ' s work = y . then , 6 x + 8 y = 1 and 26 x + 48 y = 1 . 10 2 solving these two equations , we get : x = 1 and y = 1 . 100 200 ( 15 men + 20 women ) ' s 1 day ' s work = 15 + 20 = 1 . 100 200 4 15 men and 20 women can do the work in 4 days . hence answer will be b
a ) 5 , b ) 4 , c ) 6 , d ) 7 , e ) 8
b
divide(multiply(add(divide(8, divide(subtract(multiply(48, 2), multiply(8, 10)), subtract(multiply(6, 10), multiply(26, 2)))), 6), 10), add(divide(20, divide(subtract(multiply(48, 2), multiply(8, 10)), subtract(multiply(6, 10), multiply(26, 2)))), 15))
multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|divide(n1,#6)|divide(n7,#6)|add(n0,#7)|add(n6,#8)|multiply(n2,#9)|divide(#11,#10)
physics
the maximum number of students among them 1345 pens and 775 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :
"explanation : required number of students = h . c . f of 1345 and 775 = 5 . answer : d"
a ) 91 , b ) 10 , c ) 6 , d ) 5 , e ) none of these
d
gcd(1345, 775)
gcd(n0,n1)|
general
a sum of rs . 1360 has been divided among a , b and c such that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets . b ' s share is :
"let c ' s share = rs . x then , b ' s share = rs . x / 4 ; a ' s share = rs . 2 / 3 * x / 4 = rs . x / 6 therefore x / 6 + x / 4 + x = 1360 17 x / 12 = 1360 x = 1360 * 12 / 17 = rs . 960 hence , b ' s share = rs . 960 / 4 = rs . 240 answer : c"
a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 300 , e ) rs . 500
c
subtract(subtract(multiply(divide(1360, const_10), const_2), const_12), const_12)
divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)|
general
how many cuboids of length 5 m , width 3 m and height 2 m can be farmed from a cuboid of 18 m length , 15 m width and 2 m height .
"( 18 × 15 × 12 ) / ( 5 × 3 × 2 ) = 108 answer is c ."
a ) 106 , b ) 109 , c ) 108 , d ) 101 , e ) 104
c
divide(multiply(multiply(18, 15), 2), multiply(multiply(5, 3), 2))
multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)|
physics
two - third of a positive number and 16 / 216 of its reciprocal are equal . find the positive number .
"explanation : let the positive number be x . then , 2 / 3 x = 16 / 216 * 1 / x x 2 = 16 / 216 * 3 / 2 = 16 / 144 x = √ 16 / 144 = 4 / 12 . answer : a"
a ) 4 / 12 , b ) 4 / 17 , c ) 4 / 15 , d ) 4 / 11 , e ) 4 / 03
a
sqrt(divide(multiply(16, const_3), multiply(216, const_2)))
multiply(n0,const_3)|multiply(n1,const_2)|divide(#0,#1)|sqrt(#2)|
general
find large number from below question the difference of two numbers is 1365 . 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 + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 d"
a ) 1235 , b ) 1346 , c ) 1378 , d ) 1635 , e ) 1489
d
multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6)
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|
general
the average of first three prime numbers greater than 5 is ?
"7 + 11 + 13 = 31 / 3 = 10.3 answer : d"
a ) 12.6 , b ) 12.9 , c ) 22.3 , d ) 10.3 , e ) 12.7
d
add(5, const_1)
add(n0,const_1)|
general
in a recent election , james received 0.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 50 percent of the vote . how many additional votes would james have needed to win the election ?
james = ( 0.5 / 100 ) * 2000 = 10 votes to win = ( 50 / 100 ) * total votes + 1 = ( 50 / 100 ) * 2000 + 1 = 1001 remaining voted needed to win election = 1001 - 10 = 991 answer : option d
a ) 901 , b ) 989 , c ) 990 , d ) 991 , e ) 1,001
d
subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 0.5))
add(const_1000,const_1000)|multiply(n0,#0)|subtract(#0,#1)
general
if 12 : 8 : : x : 16 , then find the value of x
explanation : treat 12 : 8 as 12 / 8 and x : 16 as x / 16 , treat : : as = so we get 12 / 8 = x / 16 = > 8 x = 192 = > x = 24 option b
a ) 18 , b ) 24 , c ) 28 , d ) 16 , e ) 20
b
divide(add(multiply(8, const_3.0), 8), 16)
multiply(const_3.0,n1)|add(n1,#0)|divide(#1,n2)|
general
what least number must be subtracted from 3832 so that the remaining number is divisible by 5 ?
"on dividing 3832 by 5 , we get remainder = 2 . required number be subtracted = 2 answer : b"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
b
subtract(3832, multiply(floor(divide(3832, 5)), 5))
divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|
general
find the length of the wire required to go 14 times round a square field containing 5625 m 2 .
"a 2 = 5625 = > a = 75 4 a = 300 300 * 14 = 4200 answer : c"
a ) 15840 , b ) 3388 , c ) 4200 , d ) 8766 , e ) 66711
c
multiply(square_perimeter(square_edge_by_area(5625)), 14)
square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)|
physics
how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 1 m x 2 m x 20 cm ?
"number of bricks = volume of the wall / volume of 1 brick = ( 100 x 200 x 20 ) / ( 25 x 11.25 x 6 ) = 237 . answer : option c"
a ) 5600 , b ) 6000 , c ) 237 , d ) 7200 , e ) 8600
c
divide(multiply(multiply(multiply(1, const_100), multiply(2, const_100)), 20), multiply(multiply(25, 11.25), 6))
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|
physics
spanish language broadcast records last 90 min on each of two sides . if it takes 3 hours to translate one hour of broadcast , how long will it take to translate 16 full records ?
records last 90 min on each of 2 sides , = = > record last 90 * 2 = 180 min = 3 hours 16 full records - - > 16 * 3 = 48 hour broadcast given , 3 hours to translate 1 hour of broadcast let x be the time required to translate 48 hour broadcast ( 16 full records ) x = 48 * 3 = 144 hours answer : a
a ) 144 hours , b ) 124 hours , c ) 134 hours , d ) 154 hours , e ) 164 hours
a
multiply(multiply(divide(multiply(90, const_2), const_60), 16), 3)
multiply(n0,const_2)|divide(#0,const_60)|multiply(n2,#1)|multiply(n1,#2)
physics
a began business with rs . 45000 and was joined afterwards by b with rs . 5400 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ?
"45 * 12 : 54 * x = 2 : 1 x = 5 12 - 5 = 7 . answer : a"
a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5
a
subtract(multiply(const_4, const_3), divide(divide(multiply(45000, multiply(const_4, const_3)), 5400), 2))
multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3)|
other
a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and c together can do it in 2 hours . how long will b alone take to do it ?
"a ' s 1 hour work = 1 / 4 ; ( b + c ) ' s 1 hour work = 1 / 3 ; ( a + c ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 3 ) = 7 / 12 b ' s 1 hour work = ( 7 / 12 + 1 / 2 ) = 1 / 12 b alone will take 12 hours to do the work . answer : c"
a ) 15 hours , b ) 14 hours , c ) 12 hours , d ) 74 hours , e ) 79 hours
c
divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 4))))
divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)|
physics
how many seconds will a 650 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ?
"explanation : here distance d = 650 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 39 sec . answer : d"
a ) 48 , b ) 36 , c ) 26 , d ) 39 , e ) 18
d
divide(650, multiply(subtract(63, 3), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
today is thursday . i came home from a trip 3 days before the day after last monday . how many days have i been home ?
d 6 days the day after last monday was tuesday . if i came home 3 days before that , i came home on saturday , sunday , monday , tuesday , wednesday , and thursday = 6 days .
a ) 1 day , b ) 2 days , c ) 7 days , d ) 6 days , e ) 10 days
d
add(add(3, const_1), const_2)
add(n0,const_1)|add(#0,const_2)
physics
an article is bought for rs . 675 and sold for rs . 1100 , find the gain percent ?
"675 - - - - 425 100 - - - - ? = > = 63 % answer : c"
a ) 65 % , b ) 64 % , c ) 63 % , d ) 62 % , e ) 61 %
c
subtract(const_100, divide(multiply(1100, const_100), 675))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|
gain
in one hour , a boat goes 19 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 ( 19 + 5 ) kmph = 12 kmph . answer d"
a ) 2 , b ) 4 , c ) 7 , d ) 12 , e ) 15
d
divide(add(19, 5), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
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 $ 45,000 commission , how many machines did she sell ?
"first 100 machines = 3 % commission = 0.03 * 100 * 10000 = 30000 commission from sale of next machines = 46000 - 30000 = 16000 so 40 more machines . . total = 140 machines imo a . ."
a ) 140 , b ) 103 , c ) 105 , d ) 115 , e ) 120
a
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
solve the equation for x : 6 x - 27 + 3 x = 4 + 9 - x
"d 4 9 x + x = 13 + 27 10 x = 40 = > x = 4"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
d
divide(add(27, 4), add(6, 6))
add(n1,n3)|add(n0,n0)|divide(#0,#1)|
general
a profit of rs . 900 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ?
a profit of rs . 900 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 540 and 360 . difference in profit share = 540 - 360 = 180 answer : b
a ) s . 280 , b ) s . 180 , c ) s . 380 , d ) s . 50 , e ) s . 90
b
subtract(divide(divide(900, add(divide(1, 2), divide(1, 3))), 2), divide(divide(900, add(divide(1, 2), divide(1, 3))), 3))
divide(n1,n2)|divide(n1,n4)|add(#0,#1)|divide(n0,#2)|divide(#3,n2)|divide(#3,n4)|subtract(#4,#5)
general
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , the how old is b ?
"explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 ⇒ 5 x = 25 ⇒ x = 5 . hence , b ' s age = 2 x = 10 years . answer : e"
a ) 7 , b ) 9 , c ) 8 , d ) 11 , e ) 10
e
divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1))
add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)|
general
calculate the share of y , if rs . 2880 is divided among x , y and z in the ratio 3 : 5 : 8 ?
3 + 5 + 8 = 16 2880 / 16 = 180 so y ' s share = 3 * 180 = 540 answer : a
a ) 540 , b ) 520 , c ) 140 , d ) 560 , e ) 240
a
multiply(divide(2880, add(add(3, 5), 8)), 3)
add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n1,#2)
general
3 years ago , paula was 3 times as old as karl . in 9 years , paula will be twice as old as karl . what is the sum of their ages now ?
"p - 3 = 3 ( k - 3 ) and so p = 3 k - 6 p + 9 = 2 ( k + 9 ) ( 3 k - 6 ) + 9 = 2 k + 18 k = 15 p = 39 p + k = 54 the answer is d ."
a ) 42 , b ) 46 , c ) 50 , d ) 54 , e ) 58
d
add(subtract(multiply(add(negate(subtract(9, multiply(9, const_2))), subtract(multiply(3, 3), 3)), 3), subtract(multiply(3, 3), 3)), add(negate(subtract(9, multiply(9, const_2))), subtract(multiply(3, 3), 3)))
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 train running at the speed of 50 km / hr crosses a post in 4 seconds . what is the length of the train ?
"speed = ( 54 x 5 / 18 ) = 15 m / sec . length of the train = ( speed x time ) . length of the train = 15 x 4 m = 60 m . answer : c"
a ) 90 , b ) 120 , c ) 60 , d ) 95 , e ) 75
c
multiply(divide(multiply(50, const_1000), const_3600), 4)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
if soundharya rows 49 km upstream and 77 km down steam taking 7 hours each , then the speed of the stream
speed upstream = 49 / 7 = 7 kmph speed down stream = 77 / 7 = 11 kmph speed of stream = ½ ( 11 - 7 ) = 2 kmph answer : c
a ) 6 kmph , b ) 5 kmph , c ) 2 kmph , d ) 3 kmph , e ) 4 kmph
c
divide(subtract(77, 49), multiply(7, const_2))
multiply(n2,const_2)|subtract(n1,n0)|divide(#1,#0)
physics
the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 140 inclusive is the digital sum a multiple of 7 ?
is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 95 106 115 124 133 15 ways . . d
a ) 7 , b ) 8 , c ) 14 , d ) 15 , e ) 20
d
subtract(subtract(24, 7), const_2)
subtract(n0,n2)|subtract(#0,const_2)
general
the ratio of 2 numbers is 2 : 5 and their h . c . f . is 6 . their l . c . m . is ?
"let the numbers be 2 x and 5 x their h . c . f . = 6 so the numbers are 2 * 6 , 5 * 6 = 12,30 l . c . m . = 60 answer is e"
a ) 20 , b ) 24 , c ) 52 , d ) 36 , e ) 60
e
sqrt(divide(6, add(power(5, 2), add(power(2, 2), power(2, 2)))))
power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)|
other
20 beavers , working together in a constant pace , can build a dam in 6 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ?
"total work = 20 * 6 = 120 beaver hours 12 beaver * x = 120 beaver hours x = 120 / 12 = 10 answer : a"
a ) 10 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 8 .
a
divide(multiply(6, 20), 12)
multiply(n0,n1)|divide(#0,n2)|
physics
if a student loses 6 kilograms , he will weigh twice as much as his sister . together they now weigh 126 kilograms . what is the student ' s present weight in kilograms ?
"let x be the weight of the sister . then the student ' s weight is 2 x + 6 . x + ( 2 x + 6 ) = 126 3 x = 120 x = 40 kg then the student ' s weight is 86 kg . the answer is c ."
a ) 82 , b ) 84 , c ) 86 , d ) 88 , e ) 90
c
subtract(126, divide(subtract(126, 6), const_3))
subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)|
other
a person buys an article at rs . 575 . at what price should he sell the article so as to make a profit of 15 % ?
"cost price = rs . 575 profit = 15 % of 575 = rs . 86.25 selling price = cost price + profit = 575 + 86.25 = 661.25 answer : d"
a ) 600 , b ) 277 , c ) 269 , d ) 661.25 , e ) 281
d
add(575, multiply(575, divide(15, const_100)))
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many t hours will it take if jim types the report and mary edits it immediately after he is done ?
"break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many t hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a ."
a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8
a
add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12))))
divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|
physics
an amount of rs . 1638 was divided among a , b and c , in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of a ?
let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 / 12 : 4 / 12 : 3 / 12 = 6 : 4 : 3 . share of a = 6 / 13 * 1638 = rs . 756 answer : c
a ) 656 , b ) 456 , c ) 756 , d ) 745 , e ) 720
c
multiply(divide(1638, add(add(2, 3), 4)), 4)
add(n2,n4)|add(n6,#0)|divide(n0,#1)|multiply(n6,#2)
general
what least value should be replaced by * in 2551112 * so the number become divisible by 6
"explanation : trick : number is divisible by 6 , if sum of all digits is divisible by 3 and 2 , so ( 2 + 5 + 5 + 1 + 1 + 1 + 2 + * ) = 17 + * should be divisible by , 17 + 1 will be divisible by 3 , but we ca n ' t take this number because 1 is not dividable by 2 ( 2 only dividable by those numbers who contain even number at last position ) so that least number is 4 . answer : option b"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
b
subtract(6, subtract(6, 6))
subtract(n1,n1)|subtract(n1,#0)|
general
what number is obtained by adding the units digits of 734 ^ 100 and 347 ^ 83 ?
"the units digit of 734 ^ 100 is 6 because 4 raised to the power of an even integer ends in 6 . the units digit of 347 ^ 83 is 3 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 83 is in the form 4 n + 3 , the units digit is 3 . then 6 + 3 = 9 . the answer is c ."
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
c
subtract(subtract(100, 83), divide(100, add(const_1, const_10)))
add(const_1,const_10)|subtract(n1,n3)|divide(n1,#0)|subtract(#1,#2)|
general
in two triangles , the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4 . find the ratio of their bases .
sol . let the bases of the two triangles be x and y and their heights be 3 h and 4 h respectively . then , ( ( 1 / 2 ) x xx 3 h ) / ( 1 / 2 ) x y x 4 h ) = 4 / 3  x / y = ( 4 / 3 x 4 / 3 ) = 16 / 9 required ratio = 16 : 9 . ans : c
['a ) 2 : 3', 'b ) 4 : 5', 'c ) 16 : 9', 'd ) 7 : 9', 'e ) 8 : 5']
c
multiply(divide(4, 3), inverse(divide(3, 4)))
divide(n0,n1)|divide(n1,n0)|inverse(#1)|multiply(#0,#2)
geometry
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 15 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ?
"explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 15 = rs . 190 cp on 40 metres = 190 x 40 = rs . 7600 profit earned on 40 metres cloth = rs . 8200 – rs . 7600 = rs . 600 . answer : option c"
a ) rs . 950 , b ) rs . 1500 , c ) rs . 600 , d ) rs . 1200 , e ) none of these
c
multiply(15, 40)
multiply(n0,n2)|
gain
what is the probability of drawing a queen from a deck of 52 cards ?
"total number of cards , n ( s ) = 52 total number of queen cards , n ( e ) = 4 p ( e ) = n ( e ) / n ( s ) = 4 / 52 = 1 / 13 option b is answer"
a ) 4 / 13 , b ) 1 / 13 , c ) 4 , d ) 1 , e ) 2 / 13
b
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
a thief goes away with a santro car at a speed of 50 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 60 kmph when will the owner over take the thief from the start ?
"explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 60 50 d = 20 rs = 60 – 50 = 10 t = 20 / 10 = 2 hours answer : option a"
a ) a ) 2 , b ) b ) 5 , c ) c ) 7 , d ) d ) 5 , e ) e ) 8
a
subtract(divide(multiply(divide(const_1, const_2), 50), subtract(60, 50)), divide(const_1, const_2))
divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)|
physics
the average weight of 18 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class .
"explanation : average weight of 18 boys = 50.25 total weight of 18 boys = 50.25 × 18 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 18 ) + ( 45.15 × 8 ) total boys = 18 + 8 = 26 average weight of all the boys = ( 50.25 × 18 ) + ( 45.15 × 8 ) / 26 = 48.68077 answer : option a"
a ) 48.68077 , b ) 42.25983 , c ) 50 , d ) 51.25388 , e ) 52.25
a
divide(add(multiply(18, 50.25), multiply(8, 45.15)), add(18, 8))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|
general
how many words , with or without meaning , can be formed using all letters of the word good using each letter exactly once ?
"the word good has exactly 4 letters which are all different . therefore the number of words that can be formed = number of permutations of 4 letters taken all at a time . = p ( 4 , 4 ) = 4 ! = 4 x 3 x 2 × 1 = 24 answer : e"
a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24
e
factorial(const_3)
factorial(const_3)|
general
in right triangle abc , ac is the hypotenuse . if ac is 40 and ab + bc = 60 , what is the area of the triangle abc ?
"square ab + bc = 60 : ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 3600 . since ( ac ) ^ 2 = ( ab ) ^ 2 + ( bc ) ^ 2 = 40 ^ 2 = 1600 , then ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 1600 + 2 * ab * bc = 3600 . 1600 + 2 * ab * bc = 3600 . ab * bc = 1000 . the area = 1 / 2 * ab * bc = 500 . answer : d ."
a ) 225 , b ) 450 , c ) 25 √ 2 , d ) 500 , e ) 200 √ 2
d
triangle_area_three_edges(40, multiply(const_3, const_10), multiply(const_4, const_10))
multiply(const_10,const_3)|multiply(const_10,const_4)|triangle_area_three_edges(n0,#0,#1)|
geometry
the ratio of the area of a square to that of the square drawn on its diagonal is
answer : a ) 1 : 2
a ) 1 : 2 , b ) 1 : 0 , c ) 1 : 7 , d ) 1 : 5 , e ) 1 : 6
a
power(divide(const_1, sqrt(const_2)), const_2)
sqrt(const_2)|divide(const_1,#0)|power(#1,const_2)|
geometry
a 4 digit number divisible by 7 becomes divisible by 3 when 19 is added to it . the largest such number is :
out of all the 5 options , only 4487 is not divisible by 3 . all others are divisible so answer = d ( no further calculation required ) addition of any two non - divisible numbers by 3 gives the resultant divisible by 3 19 is non - divisible by 3 ; we are adding a number to that so that the resultant becomes divisible by 3 applying the above rule , it means that the number which we are going to add should be non - divisible by 3 so comes the answer = 4487 answer : d
a ) 4461 , b ) 4473 , c ) 4479 , d ) 4487 , e ) 4491
d
add(multiply(multiply(multiply(4, 7), multiply(3, 19)), const_2), subtract(multiply(multiply(4, 7), multiply(3, 19)), multiply(const_2, const_100)))
multiply(n0,n1)|multiply(n2,n3)|multiply(const_100,const_2)|multiply(#0,#1)|multiply(#3,const_2)|subtract(#3,#2)|add(#4,#5)
general
what is the probability for a family with 3 children to have a girl and two boys ( assuming the probability of having a boy or a girl is equal ) ?
one possible case is : girl - boy - boy the probability of this is 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 there are 3 c 2 = 3 such cases so we should multiply by 3 . p ( one girl and two boys ) = 3 / 8 the answer is d .
a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 2 , d ) 3 / 8 , e ) 5 / 8
d
divide(subtract(const_1, multiply(power(divide(const_1, const_2), 3), const_2)), const_2)
divide(const_1,const_2)|power(#0,n0)|multiply(#1,const_2)|subtract(const_1,#2)|divide(#3,const_2)
general
a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 / 5 - inch increase in the length of the shoe . if the largest size of this shoe is 30 % longer than the smallest size , how long , in inches , is the shoe in size 15 ?
"let x be the length of the size 8 shoe . then 0.3 x = 9 / 5 x = 90 / 15 = 6 inches the size 15 shoe has a length of 6 + 7 / 5 = 7.4 inches the answer is b ."
a ) 6.8 , b ) 7.4 , c ) 7.7 , d ) 8.2 , e ) 8.6
b
add(divide(multiply(divide(1, 5), subtract(17, 8)), divide(30, const_100)), multiply(subtract(15, 8), divide(1, 5)))
divide(n2,n3)|divide(n4,const_100)|subtract(n1,n0)|subtract(n5,n0)|multiply(#0,#2)|multiply(#0,#3)|divide(#4,#1)|add(#6,#5)|
general
if x + | x | + y = 7 and x + | y | - y = 5 what is x + y = ?
"if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 5 hence answer should be d"
a ) 1 , b ) - 1 , c ) 3 , d ) 5 , e ) 13
d
multiply(5, const_2)
multiply(n1,const_2)|
general
a 240 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ?
"relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 240 ) / 9 = 500 / 9 = > x = 260 . answer : option a"
a ) 260 , b ) 250 , c ) 240 , d ) 230 , e ) 220
a
subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 240)
add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)|
physics
34 . the side surface of a cylinder is rolled with a rectangular plate . if the perimeter of the circular base is 9 feet , and the diagonal of the rectangular plate was 15 ft . what is height of the of the cylinder ?
think of a pringles can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both , one side and thehypothenuse use pythagoreans theorem or properties of 3 - 4 - 5 triangles to solve for the other side , 12 . correct answer a
['a ) 12', 'b ) 15', 'c ) 10', 'd ) 8', 'e ) 14']
a
sqrt(subtract(power(15, const_2), power(9, const_2)))
power(n2,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)
geometry
what quantity of water should be added to reduce 9 liters of 50 % acidic liquid to 30 % acidic liquid ?
acid in 9 liters = 50 % of 9 = 4.5 liters suppose x liters of water be added . then 4.5 liters of acid in 9 + x liters of diluted solution 30 % of 9 + x = 4.5 27 + 3 x = 45 x = 6 liters answer is a
a ) 6 liters , b ) 8 liters , c ) 10 liters , d ) 12 liters , e ) 15 liters
a
subtract(divide(multiply(multiply(9, divide(50, const_100)), const_100), 30), 9)
divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_100)|divide(#2,n2)|subtract(#3,n0)
gain
a man gains 20 % by selling an article for a certain price . if the sells it at double the price , the percentage of profit will be :
"let c . p . = rs . x . then , s . p . = rs . ( 12 % of x ) = rs . 6 x / 5 new s . p . = 2 * 6 x / 5 = rs . 12 x / 5 profit = 12 x / 5 - x = rs . 7 x / 5 profit = 7 x / 5 * 1 / x * 100 = 140 % . \ answer : d"
a ) 327 , b ) 140 , c ) 277 , d ) 178 , e ) 112
d
add(multiply(subtract(multiply(add(const_1, divide(20, const_100)), const_2), const_1), const_100), const_100)
divide(n0,const_100)|add(#0,const_1)|multiply(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|add(#4,const_100)|
gain
the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 46 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ?
"20 x + 46 = 21 ( x – 5 ) x = 59 answer : e"
a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59
e
subtract(multiply(add(20, const_1), 5), 46)
add(n0,const_1)|multiply(n2,#0)|subtract(#1,n1)|
general
a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 18 days , a finished the remaining work . in how many days a alone can finish the job ?
a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 18 days whole work will be done by a in 18 * 4 / 3 = 24 days answer is b
a ) 10 , b ) 24 , c ) 60 , d ) 30 , e ) 20
b
divide(multiply(18, 40), subtract(40, 10))
multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)
physics
End of preview (truncated to 100 rows)

Dataset Card for "math_qa"

Dataset Summary

Our dataset is gathered by using a new representation language to annotate over the AQuA-RAT dataset. AQuA-RAT has provided the questions, options, rationale, and the correct options.

Supported Tasks and Leaderboards

More Information Needed

Languages

More Information Needed

Dataset Structure

Data Instances

default

  • Size of downloaded dataset files: 6.96 MB
  • Size of the generated dataset: 21.90 MB
  • Total amount of disk used: 28.87 MB

An example of 'train' looks as follows.

{
    "Problem": "a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many r ways can the test be completed if every question is unanswered ?",
    "Rationale": "\"5 choices for each of the 4 questions , thus total r of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c .\"",
    "annotated_formula": "power(5, 4)",
    "category": "general",
    "correct": "c",
    "linear_formula": "power(n1,n0)|",
    "options": "a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024"
}

Data Fields

The data fields are the same among all splits.

default

  • Problem: a string feature.
  • Rationale: a string feature.
  • options: a string feature.
  • correct: a string feature.
  • annotated_formula: a string feature.
  • linear_formula: a string feature.
  • category: a string feature.

Data Splits

name train validation test
default 29837 4475 2985

Dataset Creation

Curation Rationale

More Information Needed

Source Data

Initial Data Collection and Normalization

More Information Needed

Who are the source language producers?

More Information Needed

Annotations

Annotation process

More Information Needed

Who are the annotators?

More Information Needed

Personal and Sensitive Information

More Information Needed

Considerations for Using the Data

Social Impact of Dataset

More Information Needed

Discussion of Biases

More Information Needed

Other Known Limitations

More Information Needed

Additional Information

Dataset Curators

More Information Needed

Licensing Information

More Information Needed

Citation Information


Contributions

Thanks to @thomwolf, @lewtun, @patrickvonplaten for adding this dataset.

Update on GitHub
Papers with Code