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a towel , when bleached , lost 20 % of its length and 10 % of its breadth . what is the percentage decrease in area ?
"formula for percentage change in area is : = ( βˆ’ x βˆ’ y + ( xy ) / 100 ) % = ( βˆ’ 20 βˆ’ 10 + ( 20 Γ— 10 ) / 100 ) % = βˆ’ 28 % i . e . , area is decreased by 28 % answer is b ."
a ) 26 , b ) 28 , c ) 24 , d ) 20 , e ) 22
b
divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 20), subtract(const_100, 10))), const_100)
multiply(const_100,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#1,#2)|subtract(#0,#3)|divide(#4,const_100)|
gain
two numbers are respectively 35 % and 50 % more than a third number . the percentage that is first of the second is ?
"i ii iii 135 150 100 150 - - - - - - - - - - 135 100 - - - - - - - - - - - ? = > 90 % answer : a"
a ) 90 % , b ) 47 % , c ) 38 % , d ) 52 % , e ) 83 %
a
subtract(const_100, multiply(divide(add(35, const_100), add(50, const_100)), const_100))
add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|
gain
a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 272 less than the sum lent . what was the sum lent ?
"p - 272 = ( p * 4 * 8 ) / 100 p = 400 answer : a"
a ) 400 , b ) 278 , c ) 289 , d ) 500 , e ) 821
a
divide(272, subtract(const_1, divide(multiply(4, 8), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
gain
a marketing firm determined that , of 200 households surveyed , 80 used neither brand r nor brand b soap , 60 used only brand r soap , and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 households surveyed used both brands of soap ?
solution for soap r and soap b ( d ) 40
a ) 15 , b ) 20 , c ) 30 , d ) 40 , e ) 45
d
add(add(divide(subtract(200, add(80, 60)), add(3, const_1)), divide(subtract(200, add(80, 60)), add(3, const_1))), const_10)
add(n1,n2)|add(n3,const_1)|subtract(n0,#0)|divide(#2,#1)|add(#3,#3)|add(#4,const_10)
other
the length of the bridge , which a train 150 meters long and travelling at 45 km / hr can cross in 30 seconds , is ?
"speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 150 + x ) / 30 = 25 / 2 = = > 2 ( 150 + x ) = 750 = = > x = 225 m . answer : c"
a ) 388 , b ) 267 , c ) 225 , d ) 288 , e ) 261
c
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 150)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
if x and y are numbers such that ( x + 4 ) ( y - 4 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2
"from ( x + 4 ) ( y - 4 ) = 0 it follows that either x = - 4 or y = 4 . thus either x ^ 2 = 16 or y ^ 2 = 16 . now , if x ^ 2 = 16 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 16 + 0 = 16 . similarly if y ^ 2 = 16 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 + 16 = 16 . answer : d ."
a ) 0 , b ) 9 , c ) 12 , d ) 16 , e ) 18
d
power(4, 2)
power(n0,n3)|
general
in the set of positive integers from 1 to 90 , what is the sum of all the odd multiples of 5 ?
"reduce 1 - 90 5 - 15 - 25 - 35 - 45 - - 55 - - 65 - - 75 - - 85 are valid multiples . add them - - > 405 d"
a ) 180 , b ) 245 , c ) 320 , d ) 405 , e ) 450
d
divide(multiply(90, multiply(const_3, 5)), 5)
multiply(n2,const_3)|multiply(n1,#0)|divide(#1,n2)|
general
1000 men have provisions for 18 days . if 400 more men join them , for how many days will the provisions last now ?
1000 * 18 = 1400 * x x = 12.8 answer : b
a ) 12.9 , b ) 12.8 , c ) 12.6 , d ) 12.2 , e ) 12.1
b
divide(multiply(18, 1000), add(1000, 400))
add(n0,n2)|multiply(n0,n1)|divide(#1,#0)
physics
if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2014 ) = ?
"f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 β‡’ β‡’ f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 β‡’ β‡’ f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 β‡’ β‡’ f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2014 ) = 2015 ans : a"
a ) 2015 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781
a
add(1, 2014)
add(n3,n4)|
general
a , b and c can do a piece of work in 30 days , 30 days and 40 days respectively . they began the work together but c left 4 days before the completion of the work . in how many days was the work completed ?
"one day work of a , b and c = 1 / 30 + 1 / 30 + 1 / 40 = 1 / 11 work done by a and b together in the last 4 days = 4 * ( 1 / 30 + 1 / 30 ) = 4 / 15 remaining work = 11 / 15 the number of days required for this initial work = 8 days . the total number of days required = 4 + 8 = 12 days . answer : a"
a ) 12 days , b ) 16 days , c ) 18 days , d ) 11 days , e ) 38 days
a
add(divide(subtract(const_1, multiply(add(divide(const_1, 30), divide(const_1, 30)), 4)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 4))), add(divide(const_1, 30), divide(const_1, 30)))), 4)
add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,n3)|add(#1,#2)|multiply(#0,#3)|divide(const_1,#5)|multiply(n3,#4)|add(#4,#6)|subtract(const_1,#7)|divide(#9,#8)|add(n3,#10)|
physics
a man buys an article for $ 100 . and sells it for $ 110 . find the gain percent ?
"c . p . = $ 100 s . p . = $ 110 gain = $ 10 gain % = 10 / 100 * 100 = 10 % answer is a"
a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 %
a
subtract(divide(110, divide(100, const_100)), const_100)
divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)|
gain
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 854 in 4 years . the sum is :
"s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 3 years = rs . ( 39 x 3 ) = rs . 117 . principal = rs . ( 815 - 117 ) = rs . 698 . answer : option c"
a ) rs . 650 , b ) rs . 690 , c ) rs . 698 , d ) rs . 700 , e ) rs . 740
c
subtract(815, divide(multiply(subtract(854, 815), 3), 4))
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
gain
if the average ( arithmetic mean ) of x + 1 , x + 5 , and x + 9 is 0 , then x =
"( x + 1 + x + 5 + x + 9 ) / 3 = 0 = > 3 x + 15 = 0 = > x = - 5 answer a"
a ) – 5 , b ) – 4 , c ) – 2 , d ) – 1 , e ) 0
a
subtract(multiply(5, const_2), multiply(1, const_2))
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
general
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 14 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 14,000 imported car and ended up paying $ 1440 in taxes , what is the first tier ' s price level ?
"let t be the tier price , p be total price = 14000 per the given conditions : 0.14 t + 0.08 ( p - t ) = 1440 - - - - > t = 5333 . b is the correct answer ."
a ) $ 1600 , b ) $ 5333 , c ) $ 6050 , d ) $ 7050 , e ) $ 8000
b
divide(subtract(1440, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(8, const_100))), subtract(divide(14, const_100), divide(8, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)|
general
what should be added to 4750 so that it may become a perfect square ?
"69 x 69 = 4761 4761 - 4750 = 11 if added to 11 get perfect square answer = a"
a ) 11 , b ) 15 , c ) 16 , d ) 17 , e ) 18
a
subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4750)
add(const_10,const_10)|add(const_3,const_4)|add(#0,#0)|add(#2,#0)|add(#3,#1)|multiply(#4,#4)|subtract(#5,n0)|
geometry
john makes $ 60 a week from his job . he earns a raise andnow makes $ 72 a week . what is the % increase ?
"increase = ( 12 / 60 ) * 100 = ( 1 / 5 ) * 100 = 20 % . a"
a ) 20 % , b ) 16.66 % , c ) 16.56 % , d ) 17.66 % , e ) 18.1 %
a
multiply(divide(subtract(72, 60), 60), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
50 + 5 * 12 / ( 180 / 3 ) = ?
"explanation : 50 + 5 * 12 / ( 180 / 3 ) = 50 + 5 * 12 / ( 60 ) = 50 + ( 5 * 12 ) / 60 = 50 + 1 = 51 . answer : d"
a ) 23 , b ) 78 , c ) 27 , d ) 51 , e ) 81
d
add(50, divide(multiply(5, 12), divide(180, 3)))
divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)|
general
find number which is 50 % less than 60 .
explanation : 50 % less is 50 % of the given number therefore , 50 % of 60 is 30 . answer : b
a ) 18 , b ) 30 , c ) 32 , d ) 26 , e ) 98
b
divide(multiply(60, 50), const_100)
multiply(n0,n1)|divide(#0,const_100)
gain
mark bought a set of 6 flower pots of different sizes at a total cost of $ 7.80 . each pot cost 0.25 more than the next one below it in size . what was the cost , in dollars , of the largest pot ?
this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 7.80 ( a 25 - cent increment ) and the the difference in sequential prices of the pots is 25 cents , the largest pot probably has a price that is a 25 - cent increment . from the answer choices , i would then test answer c first ( since answers b and d are not in 25 - cent increments ) . if . . . . the largest pot = $ 1.75 the other 5 pots are . . . . 1.75 1.50 1.25 1.00 0.75 total = $ 7.80 so this must be the answer . b
a ) $ 1.75 , b ) $ 1.93 , c ) $ 2.00 , d ) $ 2.15 , e ) $ 2.30
b
add(divide(subtract(7.8, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.25)), 6), multiply(subtract(6, const_1), 0.25))
subtract(n0,const_1)|multiply(n0,#0)|multiply(n2,#0)|divide(#1,const_2)|multiply(n2,#3)|subtract(n1,#4)|divide(#5,n0)|add(#6,#2)
general
a certain basketball team that has played 2 / 3 of its games has a record of 18 wins and 2 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ?
"18 wins , 2 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 2 games so it can lose another 5 at most . answer ( c )"
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3
c
subtract(multiply(divide(3, 2), add(18, 2)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(18, 2)))), const_1))
add(n2,n3)|divide(n1,n0)|divide(n1,n5)|multiply(#0,#1)|multiply(#2,#3)|floor(#4)|add(#5,const_1)|subtract(#3,#6)|
general
a circular well with a diameter of 2 metres , is dug to a depth of 10 metres . what is the volume of the earth dug out ?
"solution volume = Ο€ r 2 h β€Ή = β€Ί ( 22 / 7 Γ— 1 Γ— 1 Γ— 10 ) m 3 β€Ή = β€Ί 31.4 m 3 . answer b"
a ) 32 m 3 , b ) 31.4 m 3 , c ) 40 m 3 , d ) 44 m 3 , e ) none
b
volume_cylinder(divide(2, 2), 10)
divide(n0,n0)|volume_cylinder(#0,n1)|
geometry
a sum fetched a total simple interest of $ 4241.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?
"e 8930 principal = $ 100 x 4241.25 / 9 x 5 = $ 424125 / 45 = $ 8930 ."
a ) $ 8829 , b ) $ 2840 , c ) $ 6578 , d ) $ 7782 , e ) $ 8930
e
divide(divide(multiply(4241.25, const_100), 9), 5)
multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|
gain
the unit digit in the product ( 635 * 767 * 984 * 489 ) is :
"explanation : unit digit in the given product = unit digit in ( 5 * 7 * 4 * 9 ) = 0 answer : a"
a ) 0 , b ) 8 , c ) 3 , d ) 2 , e ) 1
a
subtract(multiply(multiply(multiply(635, 767), 984), 489), subtract(multiply(multiply(multiply(635, 767), 984), 489), add(const_4, const_4)))
add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)|
general
in a survey of political preferences , 78 % of those asked were in favour of at least one of the proposals : i , ii and iii . 50 % of those asked favoured proposal i , 30 % favoured proposal ii , and 20 % favoured proposal iii . if 5 % of those asked favoured all 3 of the proposals , what x percentage of those asked favoured more than one of the 3 proposals .
bunuel , my answer for exactly 2 people was 17 and this was my approach : 100 % = ( a + b + c ) - ( anb + anc + bnc ) - 5 % + 22 % which leads me to x = 100 % = ( 50 + 30 + 20 ) - ( at least 2 people ) - 5 % + 22 % . c
a ) 10 , b ) 12 , c ) 17 , d ) 22 , e ) 30
c
add(subtract(subtract(add(subtract(add(50, 30), 5), 20), 5), 78), 5)
add(n1,n2)|subtract(#0,n4)|add(n3,#1)|subtract(#2,n4)|subtract(#3,n0)|add(n4,#4)
general
a certain bag contains 60 balls β€” 22 white , 18 green , 2 yellow , 15 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ?
"according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 18 + 2 ) / 60 = 42 / 60 = 0.7 . answer : a ."
a ) 0.7 , b ) 0.15 , c ) 0.54 , d ) 0.85 , e ) 0.91
a
divide(add(add(22, 18), 2), 60)
add(n1,n2)|add(n3,#0)|divide(#1,n0)|
other
if the personal income tax rate is lowered from 38 % to 30 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 36500 ?
"saving = ( 38 - 33 ) % of 36500 = 1825 . answer : a"
a ) $ 1825 , b ) $ 5000 , c ) $ 3150 , d ) $ 7000 , e ) $ 10000
a
multiply(divide(36500, const_100), subtract(38, 30))
divide(n2,const_100)|subtract(n0,n1)|multiply(#0,#1)|
gain
elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ 48 and elvin ' s total telephone bill for february was 90 $ . if elvin ' s charge for the calls he made in february was twice the charge for the calls he made in january , what is elvin ' s fixed monthly charge for internet service ?
"bill = fixed charge + charge of calls made in jan , bill = fixed charge ( let , y ) + charge of calls made in jan ( let , x ) = $ 48 in feb , bill = fixed charge ( let , y ) + charge of calls made in feb ( then , 2 x ) = $ 90 i . e . x + y = 48 and 2 x + y = 90 take the difference if two equations i . e . ( 2 x + y ) - ( x + y ) = 90 - 48 i . e . x = 42 i . e . fixed monthly charge , y = 6 answer : option d"
a ) $ 5 , b ) $ 10 , c ) $ 14 , d ) $ 6 , e ) $ 28
d
subtract(multiply(48, const_2), 90)
multiply(n0,const_2)|subtract(#0,n1)|
general
in the recent , climate conference in new york , out of 500 men , 300 women , 500 children present inside the building premises , 10 % of the men , 60 % of the women and 70 % of the children were indians . find the percentage of people who were not indian ?
"solution : number of indians men present there = 500 * 10 / 100 = 50 indian women = 300 * 60 / 100 = 180 indian children = 500 * 70 / 100 = 350 . total member present in climate conference = 500 + 300 + 500 = 1300 total indian = 50 + 180 + 350 = 580 hence , % of indian present there = 580 * 100 / 1300 = 44.61 % % of people who were not indian = 100 - 44.61 = 55.39 % . answer c"
a ) 59.54 % , b ) 66.25 % , c ) 55.39 % , d ) 63.21 % , e ) 64.22 %
c
multiply(divide(add(add(multiply(500, subtract(const_1, divide(10, const_100))), multiply(300, subtract(const_1, divide(60, const_100)))), multiply(500, subtract(const_1, divide(70, const_100)))), add(add(500, 300), 500)), const_100)
add(n0,n1)|divide(n3,const_100)|divide(n4,const_100)|divide(n5,const_100)|add(n2,#0)|subtract(const_1,#1)|subtract(const_1,#2)|subtract(const_1,#3)|multiply(n0,#5)|multiply(n1,#6)|multiply(n2,#7)|add(#8,#9)|add(#11,#10)|divide(#12,#4)|multiply(#13,const_100)|
gain
two numbers are in the ratio of 5 : 8 . if 30 be subtracted from each , they are in the ratio of 5 : 9 . find the numbers ?
"( 5 x - 30 ) : ( 8 x - 30 ) = 5 : 9 x = 24 = > 120,192 answer : c"
a ) 122,195 , b ) 121,199 , c ) 120,192 , d ) 102,192 , e ) 126,109
c
multiply(5, 30)
multiply(n0,n2)|
other
find k if 56 / k = 4 .
"since 56 / k = 4 and 56 / 14 = 4 , then k = 14 correct answer d"
a ) 16 , b ) 17 , c ) 18 , d ) 14 , e ) 13
d
divide(56, 4)
divide(n0,n1)|
general
a car traveled from san diego to san francisco at an average speed of 45 miles per hour . if the journey back took twice as long , what was the average speed of the trip ?
"let the time taken be = x one way distance = 45 x total distance traveled = 2 * 45 x = 90 x total time taken = x + 2 x = 3 x average speed = 90 x / 3 x = 30 answer : a"
a ) 30 . , b ) 32 . , c ) 36 . , d ) 42 . , e ) 44 .
a
inverse(add(inverse(45), divide(inverse(45), const_2)))
inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)|
physics
a 400 meter long train crosses a man standing on the platform in 10 sec . what is the speed of the train ?
"s = 400 / 10 * 18 / 5 = 144 kmph answer : a"
a ) 144 , b ) 108 , c ) 278 , d ) 126 , e ) 112
a
multiply(divide(400, 10), const_3_6)
divide(n0,n1)|multiply(#0,const_3_6)|
physics
a man took a loan from a bank at the rate of 9 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was :
"solution principal = rs . ( 100 x 5400 / 9 x 3 ) = rs . 20000 . answer d"
a ) rs . 2000 , b ) rs . 10,000 , c ) rs . 15,000 , d ) rs . 20,000 , e ) none
d
divide(divide(multiply(5400, const_100), multiply(3, 9)), multiply(divide(const_60, const_4), const_100))
divide(const_60,const_4)|multiply(n2,const_100)|multiply(n0,n1)|divide(#1,#2)|multiply(#0,const_100)|divide(#3,#4)|
gain
what is the value of ( 44444445 * 88888885 * 44444442 + 44444438 / 44444444 ^ 2 )
"let x = 44444444 , then 44444445 = x + 1 88888885 = 2 x - 3 44444442 = x - 2 44444438 = x - 6 44444444 ^ 2 = x ^ 2 then substitute it in equation ( x + 1 ) ( 2 x - 3 ) ( x - 2 ) + ( x - 6 ) / x ^ 2 ans is 2 x - 5 i . e 88888883 answer : a"
a ) 88888883 , b ) 88888884 , c ) 88888885 , d ) 88888886 , e ) 88888873
a
power(44444445, negate(88888885))
negate(n1)|power(n0,#0)|
general
jack spent 1 / 6 of his lifetime in adolescence . his facial hair started growing after 1 / 12 more of his life . he successfully married his girlfriend diana after 1 / 7 more of his life . their son was born after 5 years from then . the son lived just 1 / 2 of what jack lived . jack dies 4 years after his son died . how long did jack live ?
e 84 explanation : let us assume that jack lived for p years . p / 6 + p / 12 + p / 7 + 5 + p / 2 + 4 = x = > p = 84 . therefore , jack lived for 84 years .
a ) 75 , b ) 90 , c ) 45 , d ) 80 , e ) 84
e
divide(add(5, 4), subtract(const_1, add(add(add(divide(1, 6), divide(1, 12)), divide(1, 7)), divide(1, 2))))
add(n6,n9)|divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|divide(n0,n8)|add(#1,#2)|add(#5,#3)|add(#6,#4)|subtract(const_1,#7)|divide(#0,#8)
general
if 7 a - 3 b = 10 b + 40 = - 12 b - 2 a , what is the value of 9 a + 9 b ?
( i ) 7 a - 13 b = 40 ( ii ) 2 a + 22 b = - 40 adding ( i ) and ( ii ) : 9 a + 9 b = 0 the answer is c .
a ) - 5 , b ) - 3 , c ) 0 , d ) 3 , e ) 5
c
divide(const_0_33, const_1000)
divide(const_0_33,const_1000)
general
the price of a book is increased from $ 300 to $ 420 . what is the % of increase in its price ?
"explanation : change in the price = rs 420 Γ’ € β€œ rs 300 = rs 120 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 120 300 ) * 100 = 40 % c"
a ) 10 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 60 %
c
multiply(divide(subtract(420, 300), 300), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
a train 210 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ?
"speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 210 + 140 = 350 m required time = 350 * 2 / 25 = 28 sec answer : e"
a ) 40 sec , b ) 50 sec , c ) 44 sec , d ) 49 sec , e ) 28 sec
e
divide(210, multiply(subtract(45, 140), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 8 of that total . what was the price of the basic computer ?
"let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 8 of that new total price : p = 1 / 8 * $ 3,000 = $ 375 . plug this value in the first equation : c + 375 = $ 2,500 - - > c = $ 2,125 . answer : e ."
a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2125
e
subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 8))
multiply(const_0_25,const_1000)|multiply(#0,const_10)|add(n1,#1)|divide(#2,n3)|subtract(#1,#3)|
general
39 ! is divided by 41 then what is the remainder ?
"substituting p = 41 in the wilson ' s theorem , we get 40 ! + 141 = 0 40 Γ— 39 ! + 141 = 0 βˆ’ 1 Γ— 39 ! 41 = βˆ’ 1 cancelling - 1 on both sides , 39 ! 41 = 1 a"
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 7
a
reminder(factorial(39), 41)
factorial(n0)|reminder(#0,n1)|
other
if x is equal to the sum of the integers from 10 to 30 , inclusive , and y is the number of even integers from 10 to 30 , inclusive , what is the value of x + y ?
"x = 10 + 11 + . . . + 30 = 21 ( 20 ) = 420 y = 11 x + y = 431 the answer is d ."
a ) 401 , b ) 411 , c ) 421 , d ) 431 , e ) 441
d
add(multiply(divide(add(10, 30), const_2), add(subtract(30, 10), const_1)), add(divide(subtract(30, 10), const_2), const_1))
add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)|
general
the product of two numbers is 4107 . if the h . c . f . of these numbers is 37 , then the greater number is : explanation :
let the numbers be 37 a and 37 b . then , 37 a x 37 b = 4107 ab = 3 . now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 x 1 , 37 x 3 ) i . e . , ( 37 , 111 ) . greater number = 111 . answer : option c
a ) 101 , b ) 107 , c ) 111 , d ) 117 , e ) 121
c
multiply(divide(4107, power(37, const_2)), 37)
power(n1,const_2)|divide(n0,#0)|multiply(n1,#1)
general
if 42.36 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ?
"42.36 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.36 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.36 = km / 50 . . . . . . 36 / 100 = 3 m / 50 . . . . . . m = 6 k + m = 3 + 6 = 9 answer : d"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
d
add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.36, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.36, subtract(const_0_25, divide(const_1, const_100))), 14))
divide(const_1,const_100)|subtract(const_0_25,#0)|multiply(n2,#1)|subtract(n0,#1)|divide(#3,n1)|divide(#2,#4)|add(#5,#4)|
general
if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 14 integers , y consists of 18 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ?
the number of integers that belong to set x only is 14 - 6 = 8 ; the number of integers that belong to set y only is 18 - 6 = 12 ; the number of integers that belong to set x or set y , but not both is 8 + 12 = 20 . answer : c .
a ) 6 , b ) 16 , c ) 20 , d ) 30 , e ) 174
c
add(subtract(18, 6), subtract(14, 6))
subtract(n1,n2)|subtract(n0,n2)|add(#0,#1)
other
how many seconds will a 700 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ?
"let length of tunnel is x meter distance = 700 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 700 + x = ( 65 / 3 ) * 60 700 + x = 20 * 65 = 1300 x = 1300 - 700 = 600 meters answer : c"
a ) 287 , b ) 288 , c ) 600 , d ) 277 , e ) 121
c
multiply(multiply(subtract(divide(700, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2)
subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2)|
physics
a bakery has 6 employees . it pays annual salaries of $ 16,000 to each of 2 employees , $ 18,000 to 1 employee , and $ 19,000 to each of the remaining 3 employees . which number is closest to the average ( arithmetic mean ) annual salary of these employees ?
"average = 2 ( 16000 ) + ( 18000 ) + 3 ( 19000 ) / 6 = $ 17,833 the answer is c ."
a ) $ 17,200 , b ) $ 17,500 , c ) $ 17,800 , d ) $ 18,000 , e ) $ 18,400
c
divide(add(add(multiply(add(add(add(const_10, const_10), 1), 1), 3), multiply(subtract(add(const_10, const_10), 1), 2)), add(add(const_10, const_10), 1)), 6)
add(const_10,const_10)|add(#0,n4)|subtract(#0,n4)|add(#1,n4)|multiply(#2,n2)|multiply(#3,n6)|add(#5,#4)|add(#6,#1)|divide(#7,n0)|
general
find large no . 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 35 as remainder
"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 35 5 x = 1330 x = 266 large number = 266 + 1365 = 1631 e"
a ) 1235 , b ) 1456 , c ) 1567 , d ) 1678 , e ) 1631
e
add(multiply(divide(subtract(1365, 35), subtract(6, const_1)), 6), 35)
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)|
general
how many odd numbers between 10 and 1400 are the squares of integers ?
"the numbers are the squares of 5 , 7 , 9 , . . . , 37 which includes 17 numbers . the answer is b ."
a ) 14 , b ) 17 , c ) 20 , d ) 23 , e ) 26
b
add(10, const_4)
add(n0,const_4)|
geometry
n is a whole number which when divided by 4 gives 3 as remainder . what will be the remainder when 2 * n is divided by 4 ?
explanation : let n = 4 * q + 3 . then , 2 * n = 8 * q + 6 = 4 ( 2 * q + 1 ) + 2 . thus when 2 * n is divided by 4 , the reminder is 2 . answer : b ) 2
a ) 8 , b ) 2 , c ) 6 , d ) 7 , e ) 1
b
reminder(multiply(3, 2), 4)
multiply(n1,n2)|reminder(#0,n0)
general
a person want to give his money of $ 2700 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ?
"b ' s share = 2700 * 3 / 9 = $ 900 answer is b"
a ) $ 200 , b ) $ 900 , c ) $ 700 , d ) $ 600 , e ) $ 400
b
multiply(divide(3, add(add(2, 3), 4)), 2700)
add(n1,n2)|add(n4,#0)|divide(n1,#1)|multiply(n0,#2)|
other
the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 18 . the larger of the two numbers is :
"clearly , the numbers are ( 23 x 13 ) and ( 23 x 18 ) . larger number = ( 23 x 18 ) = 414 . answer : option b"
a ) 276 , b ) 414 , c ) 322 , d ) 345 , e ) 355
b
multiply(23, 18)
multiply(n0,n2)|
other
if 6 ^ k = 4 , then 6 ^ ( 2 k + 3 ) =
6 ^ k = 4 6 ^ 2 k = 4 ^ 2 6 ^ 2 k = 16 6 ^ ( 2 k + 3 ) = 6 ^ 2 k * 6 ^ 3 = 16 * 216 = 3456 answer : b
a ) 2925 , b ) 3456 , c ) 4581 , d ) 3683 , e ) 1308
b
multiply(power(4, 2), power(6, 3))
power(n1,n3)|power(n0,n4)|multiply(#0,#1)
general
a can finish a work in 18 days and b can do the same work in half the time taken by a . then , working together , what part of the same work they can finish in a day ?
given that b alone can complete the same work in days = half the time taken by a = 9 days a ’ s one day work = 1 / 18 b ’ s one day work = 1 / 9 ( a + b ) ’ s one day work = 1 / 18 + 1 / 9 = 1 / 6 answer : b .
a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 1 , e ) 1 / 4
b
add(divide(const_1, divide(18, const_2)), divide(const_1, 18))
divide(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|add(#2,#1)
physics
an empty fuel tank with a capacity of 212 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ?
"say there are a gallons of fuel a in the tank , then there would be 212 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 212 - a gallons of fuel b is 0.16 ( 212 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 212 - a ) = 30 - - > a = 98 . answer : b ."
a ) 160 , b ) 98 , c ) 100 , d ) 80 , e ) 50
b
divide(subtract(multiply(212, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100)))
divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)|
gain
the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 Ο€ Β· r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams r ?
"weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )"
a ) 16 , b ) r = 32 , c ) r = 64 , d ) 128 , e ) 512
b
multiply(8, 4)
multiply(n0,n3)|
geometry
10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 2 women and 10 children take to complete the work ?
"1 women ' s 1 day work = 1 / 70 1 child ' s 1 day work = 1 / 140 ( 2 women + 10 children ) ' s 1 day work = ( 2 / 70 + 10 / 140 ) = 1 / 10 2 women and 10 children will complete the work in 10 days . e"
a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 10
e
inverse(add(divide(2, multiply(10, 7)), divide(10, multiply(10, 14))))
multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)|
physics
a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 660 - mile journey ? assume that the train makes no intermediate stops .
a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 660 - mile journey ? assume that the train makes no intermediate stops . i think the easiest way to solve this problem would be simply to count the number of miles it travels per hour ( and in total ) hour miles / hour total miles 1 10 10 2 20 30 3 30 60 4 40 100 5 50 150 6 60 210 7 70 280 8 80 360 9 90 450 10 100 550 11 110 660 it takes a total of nine hours to cover the 660 mile distance . answer : c . 11
a ) 8 , b ) 8.5 , c ) 11 , d ) 9.5 , e ) 10
c
divide(660, add(30, add(10, 20)))
add(n0,n1)|add(n2,#0)|divide(n3,#1)
physics
how many integers are divisible by 2 between 10 ! and 10 ! + 20 inclusive ?
"c = 21 10 ! is divisible by 2 there are 20 numbers between 10 ! and 10 ! + 20 that are divisible by 1 . hence 21"
a ) 6 , b ) 7 , c ) 21 , d ) 9 , e ) 10
c
add(divide(20, 2), const_1)
divide(n3,n0)|add(#0,const_1)|
general
ramesh purchased a refrigerator for rs . 14500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 10 % if no discount was offered ?
"price at which the tv set is bought = rs . 14,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 18125 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 18125 + 375 = rs . 18500 the price at which the tv should be sold to get a profit of 10 % if no discount was offered = 18500 * 110 / 100 = rs . 20350 answer : c"
a ) 34778 , b ) 26888 , c ) 20350 , d ) 17600 , e ) 12778
c
divide(multiply(add(const_100, 10), add(divide(multiply(14500, const_100), subtract(const_100, 20)), add(125, 250))), const_100)
add(n4,const_100)|add(n2,n3)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#2,#3)|add(#1,#4)|multiply(#0,#5)|divide(#6,const_100)|
gain
if equation | x / 2 | + | y / 2 | = 4 encloses a certain region on the coordinate plane , what is the area of this region ?
the equation can be reduced to intercept form as | x / 8 | + | y / 8 | = 1 , so these are lines in four quadrants with x and y intercept as 8 , so it is a rhombus with diagonals of 16 each and hence area = 1 / 2 * d 1 * d 2 = 1 / 2 * 16 * 16 = 108 . answer c .
a ) 20 , b ) 50 , c ) 108 , d ) 200 , e ) 400
c
add(power(multiply(2, 2), 2), power(multiply(2, 2), 2))
multiply(n0,n0)|power(#0,n0)|add(#1,#1)|
general
let f ( x ) = x ^ 2 + bx + c . if f ( 6 ) = 0 and f ( - 1 ) = 0 , then b + c =
"f ( x ) = x ^ 2 + bx + c . if f ( 6 ) = 0 and f ( - 1 ) = 0 , then b + c = f ( 6 ) = 0 = 36 + 6 b + c - - - taking 36 to the other side - > 6 b + c = - 36 f ( - 1 ) = 0 = 1 - b + c - - - taking - b + c to the other side - > b - c = 1 when we add these 2 equations , we get 7 b = - 35 - - - > b = - 5 and while substituting b = - 5 we get c = - 6 . b + c = - 11 - - - answer d"
a ) 18 , b ) 15 , c ) - 15 , d ) - 11 , e ) - 24
d
negate(divide(subtract(power(1, 2), 6), add(1, 6)))
add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)|
general
approximately how many revolutions will be made by a car tire with a 13 - inch diameter if the car travels 1 ⁄ 2 mile ?
"diameter of car tire = 13 inches radius of car tire = 6.5 inches = 6.5 * 2.54 cm = 16.51 cm circumference of the car tire = 2 * pi * 16.51 = 2 * 3.14 * 16.51 = 103.68 cm distance traveled by car = . 5 mile = . 5 * 1.6 km = . 8 km = 800 m = 8 * 10 ^ 4 cm number of revolutions = distance traveled / circumference of tire = 8 * 10 ^ 4 / 103.68 = 771.60 since the answer choices are far apart we should use estimation in this question . ideally , in the gmat the conversion between units would be provided . answer c"
a ) 120 , b ) 180 , c ) 772 , d ) 720 , e ) 1,440
c
divide(divide(1, 2), divide(multiply(add(divide(add(divide(add(divide(add(divide(multiply(const_3, 2), const_10), 1), const_10), const_4), const_10), 1), const_10), const_3), 13), multiply(add(add(multiply(add(multiply(add(multiply(multiply(const_3, 2), const_10), const_3), const_10), const_3), const_10), const_3), 2), const_10)))
divide(n1,n2)|multiply(n2,const_3)|divide(#1,const_10)|multiply(#1,const_10)|add(#2,n1)|add(#3,const_3)|divide(#4,const_10)|multiply(#5,const_10)|add(#6,const_4)|add(#7,const_3)|divide(#8,const_10)|multiply(#9,const_10)|add(#10,n1)|add(#11,const_3)|add(#13,n2)|divide(#12,const_10)|add(#15,const_3)|multiply(#14,const_10)|multiply(n0,#16)|divide(#18,#17)|divide(#0,#19)|
general
if x / y = 2 / 5 , then ( x + y ) / ( x - y ) = ?
any x and y satisfying x / y = 2 / 5 should give the same value for ( x + y ) / ( x - y ) . say x = 2 and y = 5 , then ( x + y ) / ( x - y ) = ( 2 + 5 ) / ( 2 - 5 ) = - 7 / 3 . answer : e .
a ) 5 , b ) 1 / 5 , c ) - 1 / 6 , d ) - 1 / 5 , e ) - 7 / 3
e
divide(add(2, 5), subtract(2, 5))
add(n0,n1)|subtract(n0,n1)|divide(#0,#1)
general
if the radius of a circle is increased by 60 % then the area is increased by :
"initially a = pi * r 2 now r 2 = 160 / 100 r then area = pi * ( 160 r / 100 ) 2 area = 256 r / 100 that is area increases from 100 to 256 = increase in area = 156 % answer : c"
a ) 44 % , b ) 120 % , c ) 156 % , d ) 40 % , e ) none of the above
c
multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(60, const_100)), const_2)), const_100)
divide(const_100,const_100)|divide(n0,const_100)|subtract(#0,#1)|power(#2,const_2)|subtract(#0,#3)|multiply(#4,const_100)|
geometry
a company that ships boxes to a total of 12 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter )
"back - solving is the best way to solve this problem . you basically need 12 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - b"
a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) 24
b
subtract(divide(factorial(subtract(divide(12, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(12, const_2), const_1))
divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)|
general
10 binders can bind 1400 books in 21 days . how many binders will be required to bind 1800 books in 20 days ?
"binders books days 10 1400 21 x 1600 20 x / 10 = ( 1800 / 1400 ) * ( 21 / 20 ) = > x = 12 answer : e"
a ) 87 , b ) 18 , c ) 17 , d ) 16 , e ) 12
e
divide(divide(1800, 20), divide(1400, multiply(10, 21)))
divide(n3,n4)|multiply(n0,n2)|divide(n1,#1)|divide(#0,#2)|
physics
an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 27 metre of cloth ?
"explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 27 : x = > x = { \ color { blue } \ frac { 27 \ times 1 } { 0.128 } } = > x = 210.94 so time will be approx 211 seconds answer : a"
a ) 211 seconds , b ) 195 seconds , c ) 155 seconds , d ) 115 seconds , e ) 115 seconds
a
divide(27, 0.128)
divide(n1,n0)|
physics
a man cheats while buying as well as while selling . while buying he takes 10 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 11 % below the cost price of the claimed weight .
"there is a one step calculation method too . it requires more thought but is faster . the man takes 10 % more than what he pays for . so if he claims to take 100 pounds , he pays $ 100 but he actually takes 110 pounds for which he will take from the customer $ 110 . hence , in effect , there is a 10 % mark up . while selling , he sells 20 % less . this means , he claims to sell 100 pounds and gets $ 100 but actually sells only 80 pounds and should have got only $ 80 for it . so this is again a mark up of $ 20 on $ 80 which is 25 % . but he also sells at 11 % less ( 1 + m 1 % ) ( 1 + m 2 % ) ( 1 - d % ) = ( 1 + p % ) 11 / 10 * 5 / 4 * 89 / 100 = ( 1 + p % ) profit % = 22.38 % b"
a ) 19.81 % , b ) 22.38 % , c ) 37.5 % , d ) 25 % , e ) 37.5 %
b
multiply(subtract(add(const_100, 10), add(11, subtract(const_100, 20))), const_2)
add(n0,const_100)|subtract(const_100,n1)|add(n2,#1)|subtract(#0,#2)|multiply(#3,const_2)|
gain
the diameter of the driving wheel of a bus in 140 cm . how many revolutions per minute must the wheel make in order to keep a speed of 11 kmph ?
"distance covered in 1 min = ( 66 * 1000 ) / 60 = 183.33 cm circumference of the wheel = ( 2 * ( 22 / 7 ) * . 70 ) = 4.4 m no of revolution per min = 183.33 / 4.4 = 41.67 answer : d"
a ) 21.23 , b ) 52.85 , c ) 23.26 , d ) 41.67 , e ) 25.15
d
divide(divide(multiply(11, const_1000), const_60), multiply(multiply(divide(add(11, const_2), add(const_4, const_3)), const_2), divide(divide(140, const_100), const_2)))
add(n1,const_2)|add(const_3,const_4)|divide(n0,const_100)|multiply(n1,const_1000)|divide(#3,const_60)|divide(#2,const_2)|divide(#0,#1)|multiply(#6,const_2)|multiply(#5,#7)|divide(#4,#8)|
physics
1 * 2 + 2 * 2 ^ 2 + 3 * 2 ^ 3 + 4 * 2 ^ 4 … … … … … … … … + 2012 * 2 ^ 2012
take 2 outside as common we get 2 ( 1 + 2 * 1 ^ 2 + 3 * 1 ^ 3 + . . . . . . . . . 2012 * 1 ^ 2012 ) then 2 ( 1 + 2 + . . . . . 2012 ) we get 8100312 answer : a
a ) 8100312 , b ) 8100313 , c ) 8100412 , d ) 8200312 , e ) 8300312
a
multiply(multiply(2, add(2012, const_1)), 2012)
add(n11,const_1)|multiply(n1,#0)|multiply(n11,#1)
general
stock / number of shares v - - - - - - - - - - - - 68 w - - - - - - - - - - 112 x - - - - - - - - - - - - 56 y - - - - - - - - - - - - 94 z - - - - - - - - - - - - 45 the table shows the number of shares of each of the 5 stocks owned by mr sami . if mr . sami was to sell 20 shares of stock x and buy 23 shares of stock y , what would be the increase in the range of the numbers of shares of the 5 stocks owned by mr . sami ?
based on the original number of shares , the range would be : 112 - 45 = 67 however , the prompt tells us that two of the values in the table are going to change . after the changes , the values will be : v 68 w 112 x 36 y 117 z 45 now the range is 117 - 36 = 81 the increase in the range requires us to compare thenewrange to theoldrange : 81 - 67 = 14 d
a ) 4 , b ) 6 , c ) 9 , d ) 14 , e ) 20
d
subtract(subtract(add(94, 23), subtract(56, 20)), subtract(112, 45))
add(n3,n7)|subtract(n2,n6)|subtract(n1,n4)|subtract(#0,#1)|subtract(#3,#2)
general
if the cost price of 20 articles is equal to the selling price of 25 articles , what is the % profit or loss made by the merchant ?
"let the cost price of 1 article be rs . 1 . therefore , cost price of 20 articles = 20 * 1 = rs . 20 the selling price of 25 articles = cost price of 20 articles = rs . 20 . let us find the cost price of 25 articles . cost price of 25 articles = 25 * 1 = rs . 25 . therefore , profit made on sale of 25 articles = selling price of 25 articles - cost price of 25 articles = 20 - 25 = - rs . 5 . because the profit is in the negative , the merchant has made a loss of rs . 5 . therefore , % loss = loss / cost price βˆ— 100 % loss = 5 / 25 βˆ— 100 = 20 % loss . answer : b"
a ) 25 % loss , b ) 20 % loss , c ) 25 % profit , d ) 20 % profit , e ) 5 % profit
b
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 20)), divide(multiply(const_100, 25), 20)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
the mean of 50 observations was 32 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is
"sol . therefore correct sum = ( 32 Γ— 50 + 48 – 23 ) = 1625 . therefore correct mean = 1625 / 50 = 32.5 . answer b"
a ) 35.2 , b ) 32.5 , c ) 36.5 , d ) 39.1 , e ) none
b
divide(add(multiply(32, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
0.009 / x = 0.05 . find the value of x
"x = 0.009 / 0.18 = 0.05 answer : d"
a ) 0.00018 , b ) 0.09 , c ) 0.018 , d ) 0.18 , e ) 90
d
divide(0.009, 0.05)
divide(n0,n1)|
general
by selling an article at rs . 700 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ?
"sp = 700 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 700 * [ 100 / 125 ] = 560 loss = 25 % = 25 % of 560 = rs . 140 sp = cp - loss = 560 - 140 = rs . 420 answer : e"
a ) s . 600 , b ) s . 480 , c ) s . 500 , d ) s . 450 , e ) s . 420
e
subtract(divide(multiply(700, const_100), add(25, const_100)), divide(multiply(divide(multiply(700, const_100), add(25, const_100)), 25), const_100))
add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
in a office work is distribute between p persons . if 1 / 6 members are absent then work increased for each person is ?
let total % of work is 100 % total person = p 1 / 6 person are absent of total person . so absent person is 1 / 6 p ie p / 6 . left person is , p - p / 6 = 5 p / 6 . p person do the work 100 % 1 person do the work 100 * p % 5 p / 6 person do the work ( 100 * p * 6 ) / 5 p % = 120.00 % work increased for each person is = ( 120 - 100 ) % = 20 % answer : e
a ) 10 % , b ) 14.28 % , c ) 15 % , d ) 16 % , e ) 20 %
e
subtract(divide(multiply(const_100, 6), subtract(6, 1)), const_100)
multiply(n1,const_100)|subtract(n1,n0)|divide(#0,#1)|subtract(#2,const_100)
general
a cistern which could be filled in 9 hours takes 1 hour more to be filled owing to a leak in its bottom . if the cistern is full , in what time will the leak empty it ?
"let the leak empty the full cistern in x hours 9 x / x - 9 = 9 + 1 x = 90 hours answer is c"
a ) 50 hours , b ) 62 hours , c ) 90 hours , d ) 75 hours , e ) 100 hours
c
inverse(subtract(divide(const_1, 9), divide(const_1, const_10)))
divide(const_1,n0)|divide(const_1,const_10)|subtract(#0,#1)|inverse(#2)|
physics
the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 10 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .
"let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 10 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 10 ) / 2 = 35 . answer : a"
a ) 35 , b ) 30 , c ) 25 , d ) data inadequate , e ) none of these .
a
divide(add(60, 10), const_2)
add(n0,n1)|divide(#0,const_2)|
general
mrs . evans gave a test to her freshmen economics class , which has 50 students enrolled and 25 of them answered question 1 correctly . if 20 answered question 2 correctly and 20 did not take the test then how many answered both questions correctly ?
"total number of enrolled students = 50 number of students who did not take test = 20 hence , number of students who took test = 50 - 20 = 30 number of students who answered q 2 correctly = 20 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 20 . number of students who answered both q 1 & q 2 correctly = 20 . answer : d"
a ) 23 , b ) 22 , c ) 21 , d ) 20 , e ) 19
d
subtract(add(add(25, 20), 20), 50)
add(n1,n3)|add(n5,#0)|subtract(#1,n0)|
other
carl can wash all the windows of his house in 8 hours . his wife maggie can wash all the windows in 4 hours . how many hours will it take for both of them working together to wash all the windows ?
"work hrs = ab / ( a + b ) = 32 / 12 = 3 2 / 2 answer is c"
a ) 2 , b ) 2 1 / 4 , c ) 3 2 / 2 , d ) 4 1 / 2 , e ) 5
c
multiply(divide(const_60, add(divide(const_1, 8), divide(const_1, 4))), add(divide(const_1, 8), divide(const_1, 4)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_60,#2)|multiply(#2,#3)|
physics
in a certain parking lot , 3 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what % of cars in the parking lot are parked illegally . '
"so total no . of cars parked illegally : 300 total no . of cars : 2000 300 / 2000 * 100 , so 15 % answer : c"
a ) 11 , b ) 13 , c ) 15 , d ) 60 , e ) 75
c
divide(subtract(const_1, divide(80, const_100)), divide(3, const_100))
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|divide(#2,#1)|
gain
if the given two numbers are respectively 5 % and 25 % of a third number , then what percentage is the first of the second ?
"here , l = 5 and m = 25 therefore , first number = l / m x 100 % of second number = 5 / 25 x 100 % of second number = 20 % of second number answer : a"
a ) 20 % , b ) 25 % , c ) 18 % , d ) 30 % , e ) none of these
a
multiply(divide(divide(5, const_100), divide(25, const_100)), const_100)
divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|
gain
if the radius of a cylinder is doubled and height 4 times , what is the new volume of the cylinder divided by the old one ?
"let v and v ' be the original and the changed volume now v = pir ^ 2 h v ' = pi ( 2 r ) ^ 2 ( 4 h ) v ' = 16 v e ) 16"
a ) 8 . , b ) 2 . , c ) 6 . , d ) 4 . , e ) 16 .
e
divide(volume_cylinder(multiply(const_1, const_2), multiply(const_1, const_2)), volume_cylinder(const_1, const_1))
multiply(const_1,const_2)|volume_cylinder(const_1,const_1)|volume_cylinder(#0,#0)|divide(#2,#1)|
general
starting from his house one day , a student walk at a speed of 2 kmph and reaches his school 10 minutes late . next he increased his speed by 1 kmph and reaches the school 10 minutes early . how far is the school from his house ?
let the distance be x km difference in timings = 20 min = 20 / 60 = 1 / 3 hr x / 2 - x / 3 = 1 / 3 x = 2 km answer is b
a ) 1 km , b ) 2 km , c ) 3 km , d ) 4 km , e ) 5 km
b
multiply(multiply(divide(add(10, 10), const_60), 2), add(2, 1))
add(n0,n2)|add(n1,n1)|divide(#1,const_60)|multiply(n0,#2)|multiply(#0,#3)
physics
if 23 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ?
"if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 23 lines is 23 c 2 = 253 the answer is b ."
a ) 176 , b ) 253 , c ) 342 , d ) 458 , e ) 560
b
divide(factorial(23), multiply(factorial(subtract(23, const_2)), factorial(const_2)))
factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)|
physics
what is the compound interest on rs . 6500 at 5.5 % p . a . compounded half - yearly for 1 year .
"compound interest : a = p ( 1 + r / n ) nt a = 6 , 862.42 c . i . > > 6 , 862.42 - 6500 > > rs . 362.42 answer : d"
a ) 369.42 , b ) 762.42 , c ) 162.42 , d ) 362.42 , e ) 333.42
d
divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(1, divide(5.5, multiply(5.5, const_3)))), const_100)
multiply(const_100,const_3)|multiply(const_3,n1)|divide(n1,#1)|multiply(#0,const_100)|multiply(n2,#2)|multiply(#3,#4)|divide(#5,const_100)|
gain
the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 6 units ?
"given that the area of the square = 1225 sq . units = > side of square = √ 1225 = 35 units the radius of the circle = side of the square = 35 units length of the rectangle = 2 / 5 * 35 = 14 units given that breadth = 6 units area of the rectangle = lb = 14 * 6 = 84 sq . units answer : b"
a ) 140 , b ) 84 , c ) 88 , d ) 72 , e ) 12
b
multiply(6, multiply(const_2, divide(sqrt(1225), divide(6, const_2))))
divide(n1,const_2)|sqrt(n0)|divide(#1,#0)|multiply(#2,const_2)|multiply(n1,#3)|
geometry
the sum of the non - prime numbers between 20 and 30 , non - inclusive , is
"sum of consecutive integers from 21 to 29 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 21 + 29 ) / 2 * 9 = 25 * 9 = 225 sum of non - prime numbers b / w 20 and 30 , non inclusive = = = > 225 - 52 ( i . e . , 23 + 29 , being the prime # s in the range ) = 173 answer : a"
a ) 173 , b ) 185 , c ) 197 , d ) 209 , e ) 221
a
add(add(add(add(add(add(add(20, const_1), add(add(20, const_1), const_1)), add(add(add(20, const_1), const_1), const_2)), add(add(add(add(20, const_1), const_1), const_2), const_1)), add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(20, const_1), const_1), const_2), const_1), const_1), const_1), const_1))
add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)|
general
there are 60 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 22 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ?
"x + ( x - 22 ) = 60 - - > x = 41 . answer : c ."
a ) 20 , b ) 31 , c ) 41 , d ) 53 , e ) 64
c
divide(add(60, 22), const_2)
add(n0,n1)|divide(#0,const_2)|
general
a salt manufacturing company produced a total of 1000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ?
total production of salt by the company in that year = 1000 + 1100 + 1200 + . . . . + 2100 = 18600 . average monthly production of salt for that year = 18600 / 365 Γ’ ‰ Λ† 51 answer : a
a ) 51 , b ) 53 , c ) 55 , d ) 56 , e ) 57
a
divide(divide(multiply(add(multiply(1000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, const_100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3)))
add(const_2,const_3)|multiply(n0,const_2)|multiply(const_100,const_3)|multiply(const_2,const_3)|subtract(const_12,const_1)|multiply(n1,#4)|multiply(#3,const_10)|add(#1,#5)|add(#2,#6)|add(#8,#0)|multiply(#7,const_12)|divide(#10,const_2)|divide(#11,#9)
general
how much water should be added to 15 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ?
"let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 15 + x ) ) ) 2 = 5 % * ( 15 + x ) - - - - - - - - > x = 25 ans a"
a ) 25 liters , b ) 27 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters
a
subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 15)
divide(n2,const_100)|multiply(const_100,const_2)|multiply(n1,#0)|subtract(n1,#2)|divide(#1,#3)|subtract(#4,n0)|
general
the sum of ages of 5 children born 3 years different each is 70 yrs . what is the age of the elder child ?
"let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 70 5 x = 40 x = 8 . x + 12 = 8 + 12 = 20 e"
a ) 17 , b ) 19 , c ) 16 , d ) 18 , e ) 20
e
divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 70), 5)
add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)|
general
in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 27 percent are juniors and 75 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ?
"200 are sophomores . the number of freshmen is 600 - 160 - 0.27 ( 800 ) = 224 the answer is c ."
a ) 12 , b ) 18 , c ) 24 , d ) 30 , e ) 36
c
subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 27), subtract(const_100, 75)), 800), const_100), divide(multiply(subtract(const_100, 75), 800), const_100))
divide(n3,n0)|subtract(const_100,n2)|multiply(#0,const_100)|multiply(n0,#1)|divide(#3,const_100)|subtract(const_100,#2)|subtract(#5,n1)|subtract(#6,#1)|multiply(n0,#7)|divide(#8,const_100)|subtract(#9,#4)|
gain
find the average of all the numbers between 6 and 36 which are divisible by 5 .
"solution average = ( 10 + 15 + 20 + 25 + 30 + 35 ) / 6 ) = 135 / 6 = 22.5 answer e"
a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 22.5
e
divide(add(add(6, const_4), subtract(36, const_4)), const_2)
add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|
general
tom , working alone , can paint a room in 10 hours . peter and john , working independently , can paint the same room in 5 hours and 2 hours , respectively . tom starts painting the room and works on his own for two hour . he is then joined by peter and they work together for two hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?
"let the time when all three were working together be t hours . then : tom worked for t + 4 hour and has done 1 / 10 * ( t + 4 ) part of the job ; peter worked for t + 2 hour and has done 1 / 5 * ( t + 2 ) part of the job ; john worked for t hours and has done 1 / 2 * t part of the job : 1 / 10 * ( t + 4 ) + 1 / 5 * ( t + 2 ) + 1 / 2 * t = 1 - - > multiply by 10 - - > ( t + 4 ) + ( 2 t + 2 ) + 5 t = 10 - - > t = 1 / 2 ; hence peter has done 1 / 5 * ( 1 / 2 + 2 ) = 1 / 5 * 5 / 2 = 1 / 2 answer : b"
a ) 1 / 3 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6
b
divide(const_4, add(multiply(const_4, 2), const_1))
multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)|
physics
a train 440 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ?
"speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 440 * 3 / 55 = 24 sec . answer : c"
a ) 80 , b ) 26 , c ) 24 , d ) 54 , e ) 15
c
divide(440, multiply(add(60, 6), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
a can do a piece of work in 10 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ?
"b 10 2 / 3 2 / 10 + 4 / 15 + x / 20 = 1 x = 32 / 3 = 10 2 / 3"
a ) 09 2 / 3 , b ) 10 2 / 3 , c ) 11 2 / 3 , d ) 12 2 / 3 , e ) 16 2 / 3
b
add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(10), inverse(15))), 2))), inverse(15)), 4)
inverse(n1)|inverse(n2)|inverse(n0)|subtract(n4,n3)|add(#0,#1)|add(#2,#0)|add(#5,#1)|multiply(#4,#3)|multiply(n3,#6)|add(#7,#8)|subtract(const_1,#9)|divide(#10,#0)|add(n4,#11)|
physics
find two integers , neither of which ends in a zero , and whose product is exactly 1 , 00,000
1 , 00,000 = 10 ^ 5 = 10 x 10 x 10 x 10 x 10 = ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) = ( 2 ^ 5 ) x ( 5 ^ 5 ) = 32 x 3125 so the numbers are 32 and 3,125 answer : b
a ) 64 and 15,625 , b ) 32 and 3,125 , c ) 64 and 15,620 , d ) 64 and 15,635 , e ) 64 and 16,625
b
divide(multiply(multiply(const_100, const_100), const_100), divide(multiply(multiply(const_100, const_100), const_100), subtract(multiply(const_3, const_12), const_4)))
multiply(const_100,const_100)|multiply(const_12,const_3)|multiply(#0,const_100)|subtract(#1,const_4)|divide(#2,#3)|divide(#2,#4)
general
a man starts walking at 3 pm . ha walks at a speed of 4 km / hr on level ground and at a speed of 3 km / hr on uphill , 6 km / hr downhill and then 4 km / hr on level ground to reach home at 9 pm . what is the distance covered on one way
avg speed = 2 * 3 * 6 / ( 3 + 6 ) = 4 so speed of man = 4 km / hr distance = 4 * 6 = 24 km since we want one way distance it will be 24 / 2 = 12 km answer : b
a ) 10 km , b ) 12 km , c ) 14 km , d ) 16 km , e ) 18 km
b
multiply(4, divide(6, const_2))
divide(n3,const_2)|multiply(n1,#0)
physics
a train 175 m long running at 36 kmph crosses a platform in 40 sec . what is the length of the platform ?
"length of the platform = 36 * 5 / 18 * 40 = 400 – 175 = 225 answer : c"
a ) 271 , b ) 266 , c ) 225 , d ) 277 , e ) 232
c
subtract(multiply(40, multiply(36, const_0_2778)), 175)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics