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ericsun153
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mathqa_question_token_drop_test

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some work in?
32 / ( 2 * 2 ) = 8 days answer : e
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8
e
multiply(multiply(32, divide(const_1, const_2)), divide(const_1, const_2))
divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1)
physics
what digit2?
"the units digit of 33 ^ 2 is the units digit of 3 * 3 = 9 which is 9 . the units digit of 17 ^ 3 is the units digit of 7 * 7 * 7 = 343 which is 3 . the units digit of 39 ^ 2 is the units digit of 9 * 9 = 81 which is 1 . the units digit of 9 * 3 * 1 = 27 is 7 . the answer is d ."
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
d
divide(add(multiply(factorial(33), factorial(2)), multiply(factorial(33), factorial(3))), 33)
factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)|
general
a conference send?
they can make a team of 3 men and 2 women . the number of ways to do this is 6 c 3 * 4 c 2 = 20 * 6 = 120 they can make a team of 2 men and 3 women . the number of ways to do this is 6 c 2 * 4 c 3 = 15 * 4 = 60 the total number of distinct groups is 180 . the answer is c .
a ) 120 , b ) 150 , c ) 180 , d ) 210 , e ) 240
c
add(add(multiply(multiply(6, 5), 4), multiply(6, 5)), multiply(6, 5))
multiply(n1,n3)|multiply(n2,#0)|add(#1,#0)|add(#2,#0)
general
an while square?
"percentage error in calculated area = ( 3 + 3 + ( 3 Γ£ β€” 3 ) / 100 ) % = 9.09 % answer : a"
a ) 9.09 % , b ) 4.02 % , c ) 4 % , d ) 3 % , e ) 2 %
a
divide(multiply(subtract(square_area(add(const_100, 3)), square_area(const_100)), const_100), square_area(const_100))
add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|
gain
unc dough chocolate?
"first , you must find the total weight of the mixture given that 80 % of it will be dough . 75 % * total = 40 = > ( 75 / 100 ) total = 40 = > total = 4000 / 75 = > total = 53.33 oz , from there , you must find 25 % of the total 53.33 oz of the mixture . 25 % * total = > ( 25 / 100 ) ( 53.33 ) = 13.33 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 20 - 13.3 = 6.7 oz chocolate left over . answer : d"
a ) 8.5 , b ) 9.5 , c ) 10.5 , d ) 6.7 , e ) 6.5
d
multiply(divide(25, const_100), 25)
divide(n2,const_100)|multiply(n2,#0)|
gain
two are tank?
"1 / 10 + 1 / 15 - 1 / x = 1 / 15 x = 10 10 * 20 = 200 answer : c"
a ) 150 liters , b ) 540 liters , c ) 200 liters , d ) 560 liters , e ) 580 liters
c
multiply(multiply(inverse(subtract(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 20)), divide(const_1, 15))), const_3), 15)
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#4,#2)|subtract(#5,#3)|inverse(#6)|multiply(#7,const_3)|multiply(n1,#8)|
physics
a000 respectively?
"a : b = 90000 : 25000 = 90 : 25 = 18 : 5 answer : d"
a ) 9 : 2 , b ) 3 : 2 , c ) 18 : 20 , d ) 18 : 5 , e ) 17 : 4
d
divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1)))
multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)|
gain
we $ %?
"let x be the money invested at 3 % . 1.03 x + 1.04 ( 1000 - x ) = 1036.70 . 0.01 x = 1040 - 1036.70 . 0.01 x = 3.30 . x = 330 . the answer is b ."
a ) $ 300 , b ) $ 330 , c ) $ 360 , d ) $ 390 , e ) $ 420
b
divide(subtract(multiply(multiply(const_100, multiply(add(const_2, const_3), const_2)), add(divide(4, const_100), const_1)), add(add(multiply(const_100, multiply(add(const_2, const_3), const_2)), multiply(multiply(add(const_2, const_3), const_2), 3)), multiply(const_2, const_3))), subtract(add(divide(4, const_100), const_1), add(divide(3, const_100), const_1)))
add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(const_2,const_3)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|multiply(#6,const_100)|multiply(#6,n1)|subtract(#4,#5)|add(#7,#8)|multiply(#4,#7)|add(#10,#3)|subtract(#11,#12)|divide(#13,#9)|
gain
how digits?
"notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 3 5 5 number of possibilities = 4 * 4 * 4 = 64 . then , just add up the number of 1 digits numbers = 4 , so total is 64 + 4 = 68 . answer : e"
a ) 48 , b ) 52 , c ) 66 , d ) 84 , e ) 68
e
divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
probability
fork knives is :
explanation : let forks = 4 x , spoons = 4 x & knives = 3 x . now , 4 x = 16 hence x = 4 . number of knives = 3 x = 12 . answer : c
a ) 8 , b ) 4 , c ) 12 , d ) 16 , e ) 14
c
multiply(divide(16, 4), 3)
divide(n3,n0)|multiply(n2,#0)
other
what such integer?
"450 = 2 x 3 ^ 2 x 5 ^ 2 now we need two 2 s , one 3 and one 5 to make it perfect cube . so x = 2 ^ 2 x 3 x 5 = 60 . answer is c ."
a ) 2 , b ) 15 , c ) 30 , d ) 60 , e ) 120
c
add(const_3, const_4)
add(const_3,const_4)|
geometry
in. was :
"c number of valid votes = 80 % of 6000 = 4800 . valid votes polled by other candidate = 45 % of 4800 = ( 45 / 100 x 4800 ) = 2160 ."
a ) 2800 , b ) 2700 , c ) 2160 , d ) 2200 , e ) 2300
c
multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6000)
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|
gain
10 seconds6 minutes
"10 * 6 = 4 * x x = 15 answer : b"
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30
b
multiply(6, divide(10, 4))
divide(n0,n1)|multiply(n2,#0)|
physics
a platform platform is
"speed of the train = 54 km / hr = ( 54 Γ— 10 ) / 36 m / s = 15 m / s length of the train = speed Γ— time taken to cross the man = 15 Γ— 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 36 = > 300 + l = 15 Γ— 36 = 540 = > l = 540 - 300 = 240 meter answer is a ."
a ) 240 , b ) 250 , c ) 260 , d ) 230 , e ) 220
a
multiply(multiply(const_0_2778, 54), subtract(36, 20))
multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)|
physics
rs amounts interest?
"let the amount lent at 2 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 2 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 3 x / 100 = > 600 - 3 / 100 x = 450 = > x = 1875 amount lent at 10 % = 4125 required ratio = 5 : 11 answer : a"
a ) 5 : 11 , b ) 5 : 6 , c ) 5 : 2 , d ) 5 : 8 , e ) 5 : 2
a
divide(divide(subtract(multiply(450, const_100), multiply(6000, 2)), subtract(10, 2)), divide(subtract(multiply(450, const_100), multiply(6000, 2)), subtract(10, 2)))
multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)|
gain
if the dice is
"the number of exhaustive outcomes is 36 . let e be the event of getting doublet on the dies is 6 / 36 = 1 / 6 p ( e ) = 1 / 6 . a )"
a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2
a
divide(const_6, multiply(const_6, const_6))
multiply(const_6,const_6)|divide(const_6,#0)|
probability
how are64?
"8 Γ— 8 = 4 ^ 2 Γ— 2 ^ 2 so total factors = ( 3 + 1 ) ( 3 + 1 ) = 16 answer : d"
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
d
add(power(const_2, const_2), const_2)
power(const_2,const_2)|add(#0,const_2)|
other
the512 areas?
"ratio of the sides = Γ’ Β³ Γ’ Λ† Ε‘ 343 : Γ’ Β³ Γ’ Λ† Ε‘ 512 = 7 : 8 ratio of surface areas = 49 : 64 answer : e"
a ) 12 : 24 , b ) 8 : 16 , c ) 45 : 25 , d ) 18 : 56 , e ) 49 : 64
e
power(divide(343, 512), divide(const_1, const_3))
divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)|
geometry
a is :
"explanation : for an income of rs . 9 , investment = rs . 100 . for an income of rs 6 , investment = rs . 100 / 9 x 6 = rs 66.66 market value of rs . 100 stock = rs . 66.66 answer is e"
a ) rs 66.55 , b ) rs 68.55 , c ) rs 69.55 , d ) rs 65.55 , e ) rs 66.66
e
multiply(divide(const_100, 9), 6)
divide(const_100,n1)|multiply(n0,#0)|
gain
jeall pink?
"the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 / 36 = ( 5 / 6 ) * ( 5 / 6 ) therefore probability that the next one she draws is pink = 1 / 6 option ( a )"
a ) 1 / 6 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49
a
subtract(const_1, sqrt(divide(25, 36)))
divide(n0,n1)|sqrt(#0)|subtract(const_1,#1)|
general
if during participants?
"let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 253 n * ( n - 1 ) = 506 = 23 * 22 ( trial and error ) the answer is c ."
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
c
divide(add(sqrt(add(multiply(multiply(253, const_2), const_4), const_1)), const_1), const_2)
multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|
general
it15 days?
more days , more length ( direct ) less breadth , more length ( indirect ) more depth , less length ( indirect days 10 : 30 ; breadth 25 : 50 ; : : 100 : x depth 15 : 10 ; : . 10 * 25 * 15 * x = 30 * 50 * 10 * 100 x = ( 30 * 50 * 10 * 100 ) / 10 * 25 * 15 = 400 so the required length = 400 m answer : a
a ) 400 m , b ) 200 m , c ) 100 m , d ) 89 m , e ) 79 m
a
divide(multiply(multiply(multiply(30, 50), 10), 100), multiply(15, multiply(10, 25)))
multiply(n2,n6)|multiply(n0,n4)|multiply(n0,#0)|multiply(n5,#1)|multiply(n1,#2)|divide(#4,#3)
physics
if can days?
"explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 3 : 3 : : 56 : x therefore ( 20 x 3 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 60 = 98 hence , the required length is 98 m . answer : e"
a ) 40 m , b ) 64 m , c ) 77 m , d ) 89 m , e ) 98 m
e
multiply(divide(56, multiply(20, 3)), multiply(35, 3))
multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)|
physics
p much have?
p = ( 2 / 9 ) * p + 63 ( 7 / 9 ) * p = 63 p = 81 the answer is e .
a ) $ 69 , b ) $ 72 , c ) $ 75 , d ) $ 78 , e ) $ 81
e
divide(63, subtract(const_1, multiply(divide(1, 9), const_2)))
divide(n1,n2)|multiply(#0,const_2)|subtract(const_1,#1)|divide(n0,#2)
general
a the train?
"speed = 110 * 5 / 18 = 275 / 9 m / sec length of the train = speed * time = 275 / 9 * 9 = 275 m answer : b"
a ) 298 m , b ) 275 m , c ) 208 m , d ) 988 m , e ) 299 m
b
multiply(divide(multiply(110, const_1000), const_3600), 9)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
if y?
2 ^ y = 4 ^ ( 5 x + 3 ) 2 ^ y = 2 ^ 2 ( 5 x + 3 ) y = 10 x + 6 . . . . . . . . . . . 1 3 ^ ( x - 7 ) = 9 ^ y 3 ^ ( x - 7 ) = 3 ^ 2 y x - 7 = 2 y . . . . . . . . . . . . . 2 put value of y = 10 x + 6 in eq 2 x - 7 = 2 ( 10 x + 6 ) x - 7 = 20 x + 12 19 x = - 19 x = - 1 therefore , y = - 10 + 6 y = - 4 x + y = - 1 - 4 = - 5 answer : b
a ) - 10 , b ) - 5 , c ) - 4 , d ) 3 , e ) 7
b
add(divide(add(divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1)), negate(7)), 2), divide(add(negate(7), negate(multiply(multiply(2, 3), 2))), subtract(multiply(2, const_10), const_1)))
multiply(n0,n3)|multiply(n0,const_10)|negate(n5)|multiply(n0,#0)|subtract(#1,const_1)|negate(#3)|add(#2,#5)|divide(#6,#4)|add(#7,#2)|divide(#8,n0)|add(#9,#7)
general
a /ger?
"speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 140 = 380 m . time taken = 380 / 10 = 38 sec . answer : c"
a ) 28 sec , b ) 16 sec , c ) 38 sec , d ) 18 sec , e ) 17 sec
c
divide(add(240, 140), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))
add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|
general
of - pool?
"neither car nor garage = total - garage - ( swim - common ) = 75 - 50 - ( 40 - 35 ) = 75 - 55 = 20 answer c"
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30
c
subtract(75, add(add(subtract(50, 35), subtract(40, 35)), 35))
subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)|
other
if price a =
"before price increase price = a after 30 % price increase price = a + ( 30 / 100 ) * a = 1.3 a = 351 ( given ) i . e . a = 351 / 1.3 = $ 270 i . e . 2 a = 2 * 270 = 540 answer : option a"
a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 700
a
multiply(divide(351, divide(add(const_100, 30), const_100)), 2)
add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)|
general
in at degree?
"in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability w of cgd / ( cg ) - > 3 / 19 answer d"
a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43
d
divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1))
divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|add(#2,n0)|divide(#2,#3)|
other
what % %?
"sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 75 2 / 3 sp 1 = 75 sp 1 = 112.50 100 - - - 12.5 = > 12.5 % answer : e"
a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 12.5 %
e
subtract(divide(subtract(const_100, 25), divide(2, 3)), const_100)
divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|
gain
how the number.
n = ( 4 * 5 - 1 ) * 5 * 5 * 5 = 2375 where 4 cases of first digit { 2,4 , 6,8 } 5 cases of second digit { 0 , 2,4 , 6,8 } 1 case of 22 for two leftmost digit 5 cases of third digit { 1 , 3,5 , 7,9 } 5 cases of fourth digit { 1 , 3,5 , 7,9 } 5 cases of fifth digit { 1 , 3,5 , 7,9 } c
a ) 2200 , b ) 2295 , c ) 2375 , d ) 2380 , e ) 2385
c
multiply(add(multiply(const_3, const_4), add(const_3, const_4)), power(5, const_3))
add(const_3,const_4)|multiply(const_3,const_4)|power(n0,const_3)|add(#0,#1)|multiply(#3,#2)
general
a eats nuts?
"in one hour , the crow eats 1 / 24 of the nuts . ( 1 / 4 ) / ( 1 / 24 ) = 6 hours the answer is a ."
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
a
divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 4))
add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,#0)|divide(#2,n0)|divide(#1,#3)|
general
al rs cup?
"let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1081 96 + 225 + 490 + 6 x = 1081 6 x = 270 = > x = 45 . answer : e"
a ) 25 , b ) 66 , c ) 77 , d ) 99 , e ) 45
e
divide(subtract(subtract(subtract(1081, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6)
multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)|
general
selling clearance is?
"explanation : marked price = rs . 30 c . p . = 100 / 140 * 30 = rs . 21.42 sale price = 90 % of rs . 30 = rs . 27 required gain % = 5.57 / 21.42 * 100 = 26 % . answer : e"
a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 26 %
e
multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(40, const_100))), divide(multiply(30, const_100), add(40, const_100))), const_100)
add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)|
gain
how numbers1?
"prime factors of 30 are 2 ^ 1,3 ^ 1,5 ^ 1 total divisors = ( power if a prime factor + 1 ) total no . of odd factors ( 3,5 , ) = ( 1 + 1 ) ( 1 + 1 ) = 4 since we need odd divisors other than 1 = > 4 - 1 = 3 odd divisors b is the answer"
a ) 3 , b ) 3 , c ) 5 , d ) 6 , e ) 7
b
divide(30, multiply(const_10, const_2))
multiply(const_10,const_2)|divide(n0,#0)|
other
at serves x?
"withoutusing the formula , we can see that today the restaurant served 40 customers above the average . the total amount above the average must equal total amount below the average . this additional 40 customers must offset the β€œ deficit ” below the average of 80 created on the x days the restaurant served only 75 customers per day . 40 / 5 = 8 days . choice ( a ) . withthe formula , we can set up the following : 80 = ( 75 x + 120 ) / ( x + 1 ) 80 x + 80 = 75 x + 120 5 x = 40 x = 8 answer choice ( d )"
a ) 2 , b ) 5 , c ) 7 , d ) 8 , e ) 20
d
subtract(divide(subtract(120, 80), subtract(80, 75)), divide(subtract(120, const_100), const_100))
subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)|
general
div number?
"say x / 3 / 6 * 5 / 6 = x * 6 / 3 * 5 / 6 = x * 5 / 3 e"
a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 9 ⁄ 20 , e ) 5 ⁄ 3
e
multiply(divide(6, 3), divide(5, 6))
divide(n1,n0)|divide(n2,n3)|multiply(#0,#1)|
general
an the15?
"then , 15 % of x = 15 ( 15 / 100 ) x = 15 x = ( 15 * 100 * ) / 15 = 100 answer is a"
a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160
a
divide(multiply(15, const_100), 15)
multiply(n1,const_100)|divide(#0,n0)|
gain
pipe minute full?
"in one cycle they fill 40 + 30 - 20 = 50 liters 700 = 50 * n = > n = 14 here n = number of cycles . total time = 14 * 3 = 42 as in one cycle there are 3 minutes . thus 42 minutes answer : a"
a ) 42 minutes , b ) 14 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these
a
multiply(divide(700, subtract(add(40, 30), 20)), const_3)
add(n1,n2)|subtract(#0,n3)|divide(n0,#1)|multiply(#2,const_3)|
physics
the is is?
"2 x * x = 72 = > x = 6 answer : a"
a ) 6 , b ) 16 , c ) 8 , d ) 36 , e ) none
a
sqrt(divide(72, const_2))
divide(n0,const_2)|sqrt(#0)|
geometry
a downstream water?
"us = 20 ds = 30 m = ( 30 + 20 ) / 2 = 25 answer : c"
a ) 11 , b ) 77 , c ) 25 , d ) 88 , e ) 34
c
divide(add(10, 20), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
a a work?
here lcm of 12 and 16 is taken as total work . ( becomes easy to solve ) assume total work = 48 units then workdone by ( a + b ) in one day = 48 / 12 = 4 units similarly , by ( b + c ) in one day = 48 / 16 = 3 units now according to question , a works 5 days , b for 7 days and c for 13 days to complete total work so , 5 a + 7 b + 13 c = 48 units 5 ( a + b ) + 2 ( b + c ) + 11 c = 48 units 5 * 4 + 2 * 3 + 11 c = 48 units 11 c = 22 units c = 2 units ( c does 2 units of work daily ) therefore , 48 / 2 = 24 days c requires 24 days to complete the work alone . answer d
a ) 22 days , b ) 21 days , c ) 25 days , d ) 24 days , e ) 23 days
d
divide(const_1, divide(subtract(const_1, add(divide(5, 12), divide(const_2, 16))), subtract(add(5, 13), 7)))
add(n2,n4)|divide(n2,n0)|divide(const_2,n1)|add(#1,#2)|subtract(#0,n3)|subtract(const_1,#3)|divide(#5,#4)|divide(const_1,#6)
physics
the20 are respectively
let their ages be x and ( x + 20 ) years ( x - 5 ) * 5 = ( x + 20 - 5 ) after solving this we get x = 10 years the age of elder one = 10 + 20 = 30 years so the present ages are 30 and 10 years answer : a
a ) 30 , 10 , b ) 2010 , c ) 3515 , d ) 5117 , e ) 20,17
a
add(20, const_10)
add(n0,const_10)
general
the last stock?
( 100 % + 8 % ) * ( 100 % - 6 % ) = 1.08 * 0.94 = 1.0152 = 101.52 % . the net percentage change in the price of the stock is ( + ) 1.52 % the answer is d
a ) 0.2 % , b ) 0.8 % , c ) 1.2 % , d ) 1.52 % , e ) 2 %
d
subtract(multiply(multiply(divide(add(const_100, 8), const_100), divide(subtract(const_100, 6), const_100)), const_100), const_100)
add(n0,const_100)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100)
general
the imeters )
"400 kg - 1 cubic meter ; 400,000 g - 1 cubic meter ; 400,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 400,000 = 10 / 4 = 2.5 cubic centimeters . answer : b ."
a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 4.5 , e ) 5.5
b
divide(multiply(1,000, 1,000), multiply(400, 1,000))
multiply(n4,n4)|multiply(n1,n4)|divide(#0,#1)|
geometry
for the roots?
"for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 5 k ) ^ 2 - 4 * 2 k * 1 = 0 by solving this equation we get k = 0 , k = 8 / 25 answer : c"
a ) 2 / 7 , b ) 9 / 4 , c ) 8 / 25 , d ) 7 / 1 , e ) 7 / 2
c
divide(power(2, add(2, 2)), power(5, 2))
add(n0,n0)|power(n2,n0)|power(n0,#0)|divide(#2,#1)|
general
in any borrowed?
"the average number of books per student was 2 means that total of 2 * 25 = 50 books were borrowed ; 2 + 10 + 8 = 20 students borrowed total of 2 * 0 + 10 * 1 + 8 * 2 = 26 books ; so 50 - 26 = 24 books are left to distribute among 25 - 20 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed . minimum these 4 students could have borrowed is 4 books per student , so total number of books they could have borrowed is 4 * 3 = 12 books . so the 5 th student could have borrowed is 24 - 12 = 12 books . answer : c ."
a ) 8 , b ) 10 , c ) 12 , d ) 13 , e ) 15
c
add(subtract(multiply(25, 2), add(add(10, multiply(8, 2)), multiply(2, 3))), 3)
multiply(n0,n5)|multiply(n4,n5)|multiply(n1,n6)|add(n2,#1)|add(#3,#2)|subtract(#0,#4)|add(n6,#5)|
general
a whose wire?
"volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . Ο€ ( 24 ) ^ 2 * h = ( 4 / 3 ) Ο€ ( 12 ) ^ 3 = > h = 4 cm answer : b"
a ) 6 cm , b ) 4 cm , c ) 8 cm , d ) 3 cm , e ) 9 cm
b
divide(multiply(const_4, divide(power(12, const_3), power(24, const_2))), const_3)
power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)|
physics
boys the write?
there are twice as many girls as boys in the class - - > g = 2 b . each girl writes 3 more letters than each boy - - > boys write x letters , girls write x + 3 letters . boys write 24 letters - - > bx = 24 . girls write 90 - 24 = 66 letters - - > ( 2 b ) ( x + 3 ) = 66 - - > 2 bx + 6 b = 66 - - > 2 * 24 + 6 b = 66 - - > b = 3 . bx = 24 - - > 3 x = 24 - - > x = 8 . answer : d .
a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12
d
divide(24, divide(subtract(90, multiply(3, 24)), multiply(3, const_2)))
multiply(n0,n1)|multiply(n0,const_2)|subtract(n2,#0)|divide(#2,#1)|divide(n1,#3)
general
a / is :
"speed = ( 5 * 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 * 15 * 60 ) m = 1250 m . correct option : d"
a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these
d
multiply(divide(multiply(5, const_1000), const_60), 15)
multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|
physics
a the rose?
"let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 14 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 2 m / s a"
a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10
a
divide(add(multiply(6, 3), 14), 10)
multiply(n1,n3)|add(n2,#0)|divide(#1,n0)|
physics
mr amount borrowed.
"explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8010 = p + ( 54 p / 100 ) 8010 = ( 100 p + 54 p ) / 100 8010 = 154 p / 100 p = ( 8010 * 100 ) / 154 = rs . 5201.298 answer : d"
a ) s . 5266 , b ) s . 5269 , c ) s . 5228 , d ) s . 5201 , e ) s . 52192
d
divide(8010, add(const_1, divide(multiply(6, 9), const_100)))
multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)|
gain
in votes was :
"d number of valid votes = 80 % of 6500 = 5200 . valid votes polled by other candidate = 45 % of 5200 = ( 45 / 100 x 5200 ) = 2340 ."
a ) 2800 , b ) 2700 , c ) 2900 , d ) 2340 , e ) 2300
d
multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 6500)
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|
gain
if x x?
9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 4 is allowed . imo b .
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
b
add(divide(9, 3), divide(3, divide(9, 3)))
divide(n0,n1)|divide(n1,#0)|add(#0,#1)
general
if is x?
"21 - 7 * 3 18 - 3 * 3 * 2 15 - 5 * 3 12 - 4 * 3 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 9 is allowed . imo d ."
a ) 3 , b ) 4 , c ) 5 , d ) d , e ) 7
d
add(divide(21, 3), divide(3, divide(21, 3)))
divide(n0,n1)|divide(n1,#0)|add(#0,#1)|
general
the that result?
sum of all the 13 results = 13 * 60 = 780 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 780 = 46 . c
a ) 35 , b ) 37 , c ) 46 , d ) 48 , e ) 50
c
subtract(add(multiply(7, 57), multiply(7, 61)), multiply(13, 60))
multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)
general
23,. )
"explanation : all are prime numbers in their order , starting from 23 hence , next number is 67 answer : b"
a ) 53 , b ) 67 , c ) 48 , d ) 59 , e ) 45
b
multiply(43, 23)
multiply(const_3.0,n0)|
general
if a x?
"x / ( 11 p ) = even prime number x = even prime number * 11 p least value of x = lowest even prime number * 11 * lowest value of p = 2 * 11 * 2 = 44 answer c"
a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99
c
multiply(multiply(9, const_2), const_2)
multiply(n0,const_2)|multiply(#0,const_2)|
general
after150 article?
"cp * ( 75 / 100 ) = 1500 cp = 20 * 100 = > cp = 2000 answer : d"
a ) 1400 , b ) 1300 , c ) 1200 , d ) 2000 , e ) 1500
d
divide(1500, subtract(const_1, divide(25, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
if is area?
"the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 / 2 * 48 * 20 = 480 cm 2 answer : a"
a ) 480 cm 2 , b ) 765 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2
a
divide(multiply(48, 20), const_2)
multiply(n1,n2)|divide(#0,const_2)|
geometry
5 how beginning?
"x * ( 95 / 100 ) * ( 85 / 100 ) = 3443 x = 4264 answer : a"
a ) 4264 , b ) 2776 , c ) 4400 , d ) 2871 , e ) 881
a
floor(divide(3443, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100))))
subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|
general
adam of borrow?
let the sum borrowed be x . then , ( x Γ— 6 Γ— 21 / 00 ) + ( x Γ— 9 Γ— 3 / 100 ) + ( x Γ— 14 Γ— 4 / 100 ) = 11900 β‡’ ( 3 ⁄ 25 x + 27 ⁄ 100 x + 14 ⁄ 25 x ) = 11400 β‡’ 95 ⁄ 100 x = 11900 β‡’ x = ( 11900 Γ— 100 / 95 ) = 12526 hence , sum borrowed = 12,526 answer b
a ) 10,526 , b ) 12,526 , c ) 14,000 , d ) 16,000 , e ) 16,536
b
subtract(divide(11900, add(add(divide(multiply(6, const_2), const_100), divide(multiply(9, 3), const_100)), divide(multiply(14, 4), const_100))), multiply(const_12, const_1000))
multiply(n0,const_2)|multiply(n1,n2)|multiply(n3,n4)|multiply(const_1000,const_12)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|add(#4,#5)|add(#7,#6)|divide(n5,#8)|subtract(#9,#3)
gain
8 work?
"explanation : 1 man ’ s 1 day ’ s work = 1 / 8 Γ— 12 = 1 / 96 6 men ’ s 1 day ’ s work = 1 Γ— 6 / 96 = 1 / 16 1 woman ’ s 1 day ’ s work = 1 / 192 4 women ’ s 1 day ’ s work = 1 / 192 Γ— 4 = 1 / 48 1 child ’ s 1 day ’ s work = 1 / 240 10 children ’ s 1 day ’ s work = 1 / 24 therefore , ( 6 men + 4 women + 10 children ) ’ s 1 day ’ s work = 1 / 16 + 1 / 48 + 1 / 24 = 1 / 8 the required no . of days = 8 days answer : option d"
a ) 5 days , b ) 15 days , c ) 28 days , d ) 8 days , e ) 7 days
d
inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 6), multiply(inverse(multiply(48, 4)), 4))))
multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n6,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10)|
physics
if6 full?
"7 / 8 x - 3 / 4 x = 6 galls 1 / 8 * x = 6 gallons x = 48 gallons answer b"
a ) 25 , b ) 48 , c ) 64 , d ) 80 , e ) 96
b
multiply(6, divide(const_1, subtract(divide(7, 8), divide(3, 4))))
divide(n3,n4)|divide(n1,n2)|subtract(#0,#1)|divide(const_1,#2)|multiply(n0,#3)|
general
in pass way?
"t = 120 / 70 * 18 / 5 = 6 sec answer : d"
a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec
d
divide(120, multiply(70, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
there how %?
"there are 396 managers and 4 others . the 4 others would compose 2 % of the total number of people if there were 200 people in the room . thus 200 managers must leave . the answer is d ."
a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250
d
divide(subtract(multiply(400, divide(99, const_100)), multiply(400, divide(98, const_100))), subtract(const_1, divide(98, const_100)))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)|
gain
each % pounds?
"out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 97 % water and 3 % of non - water , so now 1 pound of non - water composes 3 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.03 = 34 pounds . answer : b ."
a ) 2 , b ) 33 , c ) 92 , d ) 96 , e ) 98
b
multiply(divide(subtract(100, 99), subtract(100, 97)), 100)
subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|
gain
the faster train?
length of the two trains = 600 m + 400 m speed of the first train = x speed of the second train = 48 kmph 1000 / x - 48 = 180 1000 / x - 48 * 5 / 18 = 180 50 = 9 x - 120 x = 68 kmph answer : b
a ) 76 kmph , b ) 68 kmph , c ) 87 kmph , d ) 56 kmph , e ) 10 kmph
b
add(48, multiply(divide(add(400, 600), 180), const_3_6))
add(n0,n1)|divide(#0,n2)|multiply(#1,const_3_6)|add(n3,#2)
physics
a an hour?
"1 flash = 5 sec for 1 min = 12 flashes so for 1 hour = 12 * 60 = 720 flashes . answer : a"
a ) 720 , b ) 600 , c ) 650 , d ) 700 , e ) 750
a
divide(const_3600, 5)
divide(const_3600,n0)|
physics
what units )?
"each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 3333 ) ^ 222 ] we can ' combine ' some of the pieces and rewrite this product as . . . . ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) [ ( 2222 ) ^ 111 ] ( 2222 ) ( 3333 ) = a big number that ends in a 6 taking a number that ends in a 6 and raising it to a power creates a nice pattern : 6 ^ 1 = 6 6 ^ 2 = 36 6 ^ 3 = 216 etc . thus , we know that ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) will be a gigantic number that ends in a 6 . 2 ^ 111 requires us to figure out thecycleof the units digit . . . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 so , every 4 powers , the pattern of the units digits repeats ( 2 , 4 , 8 , 6 . . . . . 2 , 4 , 8 , 6 . . . . ) . 111 = 27 sets of 4 with a remainder of 3 . . . . this means that 2 ^ 111 = a big number that ends in an 8 so we have to multiply a big number that ends in a 6 and a big number that ends in an 8 . ( 6 ) ( 8 ) = 48 , so the final product will be a gigantic number that ends in an 6 . final answer : d"
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
d
add(add(const_4, const_3), const_2)
add(const_3,const_4)|add(#0,const_2)|
general
a how alcohol?
"40 % of 6 = 2.4 50 % of 6 = 3 shortage is 0.6 so we need to have 0.6 / 50 % to get 50 % alcohol content . = 1.2 b"
a ) a . 0.6 , b ) b . 1.2 , c ) c . 2.1 , d ) d . 3 , e ) e . 5.4
b
subtract(6, multiply(const_2, multiply(divide(40, const_100), 6)))
divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2)|
gain
in many integers?
"54 = ( 3 ^ 3 ) * 2 since 54 is not a perfect square , no of ways = 4 answer d"
a ) 10 , b ) 8 , c ) 5 , d ) 4 , e ) 2
d
subtract(divide(divide(54, const_1), const_3), const_3)
divide(n0,const_1)|divide(#0,const_3)|subtract(#1,const_3)|
general
find factors2400
"explanation : l . c . m of 2400 = 3 x 2 x 2 x 2 x 2 x 2 x 5 x 5 3 , 2 , 5 number of different prime factors is 3 . answer : option c"
a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6
c
add(const_2, const_2)
add(const_2,const_2)|
other
in arun?
"let arun ' s weight be x kg . according to arun , 63 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 66 . the values satisfying all the above conditions are 64 and 65 . required average = ( 64 + 65 ) / 2 = 64.5 kg answer : b"
a ) 86.5 kg , b ) 64.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg
b
divide(add(66, add(63, const_1)), const_2)
add(n0,const_1)|add(n4,#0)|divide(#1,const_2)|
general
a but dog?
given ; 2 dog = 3 cat ; or , dog / cat = 3 / 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 5 / 3 * 4 = 5 : 6 . ' ' answer : 5 : 6 ;
a ) 5 : 6 , b ) 3 : 2 , c ) 4 : 8 , d ) 1 : 2 , e ) 7 : 8
a
divide(multiply(divide(2, 3), 5), 4)
divide(n2,n3)|multiply(n0,#0)|divide(#1,n1)
other
the for 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 . = 115 % of rs . 1600 = 115 / 100 * 1600 = rs . 1840 . answer : c"
a ) 2000 , b ) 2778 , c ) 1840 , d ) 2778 , e ) 2771
c
multiply(divide(add(const_100, 15), 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
a but b?
"min value of a / b will be when b is highest and a is lowest - - - > a = 20 and b = 40 so , a / b = 1 / 2 max value of a / b will be when b is lowest and a is highest - - - > a = 30 and b = 30 so , a / b = 1 range is 1 - ( 1 / 2 ) = 1 / 2 . answer should be b"
a ) 1 / 4 , b ) 1 / 2 , c ) 3 / 4 , d ) 1 , e ) 5 / 4
b
subtract(divide(subtract(32, const_1), add(29, const_1)), divide(add(19, const_1), subtract(42, const_1)))
add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)|
general
rs6 %?
"( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 125 x = 2500 answer : e"
a ) 2333 , b ) 2777 , c ) 2688 , d ) 1000 , e ) 2500
e
divide(subtract(125, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100)))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,n2)|divide(#2,const_100)|subtract(#0,#1)|subtract(n3,#3)|divide(#5,#4)|
gain
j average trip?
"total time taken by jerry = ( 8 / 40 ) * 60 minutes + 13 minutes + ( 20 / 60 ) * 60 minutes = 35 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 35 / 60 ) hours = 28 * 60 / 35 = 48 miles per hour answer : option a"
a ) 48 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5
a
divide(add(8, 20), add(add(divide(8, 40), divide(13, 60)), divide(20, 60)))
add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|
physics
the, squares?
first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 18 and c = 1 / 2 into the equation and simplify to get the answer 2109 . option e .
a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 2,109
e
divide(divide(multiply(multiply(18, add(18, const_1)), add(multiply(18, const_2), const_1)), 6), divide(multiply(multiply(18, add(18, const_1)), add(multiply(18, const_2), const_1)), 6))
add(n4,const_1)|multiply(n4,const_2)|add(#1,const_1)|multiply(n4,#0)|multiply(#2,#3)|divide(#4,n3)|divide(#5,#5)
general
surface the volumes?
1 : 8 answer : b
['a ) 1 : 9', 'b ) 1 : 8', 'c ) 1 : 3', 'd ) 1 : 4', 'e ) 1 : 5']
b
power(sqrt(divide(1, 4)), const_3)
divide(n0,n1)|sqrt(#0)|power(#1,const_3)
geometry
if decreases decreases by
"if area of a circle decreased by x % then the radius of a circle decreases by ( 100 βˆ’ 10 √ 100 βˆ’ x ) % = ( 100 βˆ’ 10 √ 100 βˆ’ 30 ) % = ( 100 βˆ’ 10 √ 70 ) % = 100 - 84 = 16 % answer b"
a ) 20 % , b ) 16 % , c ) 36 % , d ) 64 % , e ) none of these
b
multiply(subtract(const_1, sqrt(divide(subtract(const_100, 30), const_100))), const_100)
subtract(const_100,n0)|divide(#0,const_100)|sqrt(#1)|subtract(const_1,#2)|multiply(#3,const_100)|
geometry
two unbiased head?
"s = { hh , tt , ht , th } e = event of getting at most one head . e = { tt , ht , th } . p ( e ) = n ( e ) / n ( s ) = 3 / 4 answer is option c"
a ) 2 / 3 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 1 / 2
c
negate_prob(divide(const_1, power(const_2, const_3)))
power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)|
probability
the6 flour?
"let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 25 = 2 r a = 12 / 5 r and r = 75 a = 180 required total cost = 4 * 180 + 3 * 75 + 5 * 25 = 720 + 225 + 125 = $ 1070 c"
a ) 347 , b ) 987 , c ) 1070 , d ) 1371 , e ) 1667
c
add(add(multiply(4, multiply(divide(24, 10), divide(multiply(25, 6), 2))), multiply(3, divide(multiply(25, 6), 2))), multiply(5, 25))
divide(n1,n0)|multiply(n2,n4)|multiply(n4,n7)|divide(#1,n3)|multiply(#0,#3)|multiply(n6,#3)|multiply(n5,#4)|add(#6,#5)|add(#7,#2)|
general
in students old?
"percent of students who are 25 years old or older is 0.4 * 48 + 0.2 * 52 = ~ 30 , so percent of people who are less than 25 years old is 100 - 30 = 70 . answer : b ."
a ) 0.9 , b ) 0.7 , c ) 0.45 , d ) 0.3 , e ) 0.25
b
subtract(const_1, multiply(divide(40, const_100), divide(subtract(const_100, 20), const_100)))
divide(n2,const_100)|subtract(const_100,n3)|divide(#1,const_100)|multiply(#0,#2)|subtract(const_1,#3)|
general
a pass in?
"speed = 54 * 5 / 18 = 15 m / sec time taken = 300 * 1 / 15 = 20 sec answer : c"
a ) 17 sec , b ) 16 sec , c ) 20 sec , d ) 14 sec , e ) 12 sec
c
multiply(divide(300, multiply(54, const_1000)), const_3600)
multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|
physics
a pay?
"12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 899 = 279 answer : b"
a ) 270 , b ) 279 , c ) 267 , d ) 255 , e ) 552
b
multiply(divide(899, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9))
multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)|
general
a twice correctly?
"explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 27 3 x = 27 x = 9 . answer : d"
a ) 12 , b ) 16 , c ) 18 , d ) 9 , e ) 12
d
divide(27, add(const_1, const_2))
add(const_1,const_2)|divide(n0,#0)|
general
the800 will be
explanation : given that simple interest for 2 years is rs . 800 i . e . , simple interest for 1 st year is rs . 400 and simple interest for 2 nd year is also rs . 400 compound interest for 1 st year will be 400 and compound interest for 2 nd year will be 832 - 400 = 432 you can see that compound interest for 2 nd year is more than simple interest for 2 nd year by 432 - 400 = rs . 32 i . e , rs . 32 is the interest obtained for rs . 400 for 1 year rate , r = 100 Γ— si / pt = ( 100 Γ— 32 ) / ( 400 Γ— 1 ) = 8 % difference between compound and simple interest for the 3 rd year = simple interest obtained for rs . 832 = prt / 100 = ( 832 Γ— 8 Γ— 1 ) / 100 = rs . 66.56 total difference between the compound and simple interest for 3 years = 32 + 66.56 = rs . 98.56 answer : option b
a ) rs . 48 , b ) rs . 98.56 , c ) rs . 66.56 , d ) rs . 66.58 , e ) none of these
b
add(subtract(832, 800), multiply(832, divide(subtract(832, 800), divide(800, 2))))
divide(n2,n0)|subtract(n1,n2)|divide(#1,#0)|multiply(n1,#2)|add(#3,#1)
general
when import item?
let x be the value in excess of $ 1,000 . 0.07 x = 94.5 x = $ 1,350 the total value was $ 1,350 + $ 1,000 = $ 2,350 . the answer is a .
a ) $ 2,350 , b ) $ 2,850 , c ) $ 3,250 , d ) $ 3,400 , e ) $ 3,750
a
floor(divide(add(divide(94.5, divide(7, const_100)), 1000), const_1000))
divide(n0,const_100)|divide(n2,#0)|add(n1,#1)|divide(#2,const_1000)|floor(#3)
general
the than table?
": cp = sp * ( 100 / ( 100 + profit % ) ) = 8463 ( 100 / 124 ) = rs . 6825 . answer : b"
a ) 6727 , b ) 6825 , c ) 6728 , d ) 6725 , e ) 2871
b
divide(8463, add(const_1, divide(24, const_100)))
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
gain
15 a dam?
"total work = 15 * 4 = 60 beaver hours 20 beaver * x = 60 beaver hours x = 60 / 20 = 3 answer : e"
a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 3 .
e
divide(multiply(4, 15), 20)
multiply(n0,n1)|divide(#0,n2)|
physics
find area14 cm
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 14 ) = 266 cm 2 answer : c"
a ) 178 cm 2 , b ) 179 cm 2 , c ) 266 cm 2 , d ) 167 cm 2 , e ) 197 cm 2
c
quadrilateral_area(14, 18, 20)
quadrilateral_area(n2,n1,n0)|
physics
the6070 is
a prime number is a number that has only two factors : 1 and itself . therefore , a prime number is divisible by two numbers only . let ' s list the numbers from 61 to 69 . 61 , 62 , 63 , 64 , 65 , 66 , 67 , 68 , 69 immediately we can eliminate the even numbers because they are divisible by 2 and thus are not prime . we are now left with : 61 , 63 , 65 , 67 , 69 we can next eliminate 65 because 65 is a multiple of 5 . we are now left with 61 , 63 , 67 , 69 . to eliminate any remaining values , we would look at those that are multiples of 3 . if you don ’ t know an easy way to do this , just start with a number that is an obvious multiple of 3 , such as 60 , and then keep adding 3 . we see that 60 , 63 , 66 , 69 are all multiples of 3 and therefore are not prime . thus , we can eliminate 63 and 69 from the list because they are not prime . finally , we are left with 61 and 67 , and we must determine whether they are divisible by 7 . they are not , and therefore they must be both prime . thus , the sum q of 61 and 67 is 128 . answer b .
a ) 67 , b ) 128 , c ) 191 , d ) 197 , e ) 260
b
add(add(60, const_1), subtract(70, const_3))
add(n0,const_1)|subtract(n1,const_3)|add(#0,#1)
general
at with200?
"we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 400 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 400 x 8 = 3200 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3300 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3300 ( answer b )"
a ) 1,108 , b ) 3,300 , c ) 2,108 , d ) 2,124 , e ) 2,256
b
multiply(divide(200, 22.95), 400)
divide(n6,n5)|multiply(n4,#0)|
general
if, n?
"you need to primarily find the primes of 990 : 2 , 3,3 , 5,11 the integer must contain all these primes by knowing 11 is an prime , we know the answer can be b , c , d or e as these all contain 11 . 11 = 1 , 2,3 , 4,5 , 6,7 , 8,9 , 10,11 11 contains the 2 , 3,5 and 11 explicitly and the second 3 comes from the 6 ( 2 * 3 ) , therefore the answer is 11 . this is because we know a number that contains all the primes of 990 will be a multiple of 990 . answer : b"
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
b
divide(divide(divide(divide(990, const_2), const_3), const_4), divide(const_10, const_2))
divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)|
general
the , is
"sol . required probability = pg . ) x p ( b ) = ( 1 β€” d x ( 1 β€” i ) = : x 1 = 1 / 3 ans . ( c )"
a ) 1 / 2 , b ) 1 , c ) 1 / 3 , d ) 3 / 4 , e ) 2
c
multiply(subtract(1, divide(1, 2)), subtract(1, divide(1, 3)))
divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)|
general
if equal to
3 * 5 + 5 * 3 = ( 2 x 3 - 3 x 5 + 3 x 5 ) + ( 2 x 5 - 3 x 3 + 5 x 3 ) = 22 answer a 22
a ) 22 , b ) 25 , c ) 26 , d ) 28 , e ) 23
a
add(multiply(2, 3), multiply(3, 5))
multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)
general
a in of?
answer : d , ( with different approach ) : the 200 paid is 0.2 % of the additional amount above 200,000 . let it be x now 0.2 % of x = 200 therefore x = 100,000 total = 200,000 + x = 300,000
a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 300,000 , e ) $ 2 , 200,000
d
multiply(multiply(200, const_100), const_10)
multiply(n3,const_100)|multiply(#0,const_10)
general