Datasets:
ericsun153
/
mathqa_question_token_drop_test

Dataset card Data Studio Files Community
Problem
stringlengths
3
32
Rationale
stringlengths
1
2.74k
options
stringlengths
37
137
correct
stringclasses
5 values
annotated_formula
stringlengths
6
848
linear_formula
stringlengths
7
357
category
stringclasses
6 values
reen years interest?
"let rate = r % and time = r years . then , 1200 x r x r / 100 = 300 12 r 2 = 300 r 2 = 25 r = 5 . answer : a"
a ) 5 , b ) 6 , c ) 18 , d ) can not be determined , e ) none of these
a
sqrt(divide(multiply(300, const_100), 1200))
multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)|
gain
if the x?
"multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 872 , 152,24 x sum of digits at odd places = 8 + 2 + 5 + 2 + x = 17 + x ( i ) sum of digits at even places = 7 + 1 + 2 + 4 = 14 ( ii ) ( i ) - ( ii ) = 17 + x - 14 = x - 3 hence x should be = 3 to make this a multiple of 11 ( 0 ) option c"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
c
multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11)
multiply(const_10,const_3)|add(#0,const_1)|multiply(#1,const_2)|multiply(#2,const_4)|multiply(n2,#3)|
general
d walking x?
"the time it takes darcy to walk to work is ( 1.5 / 3 ) * 60 = 30 minutes the time it takes darcy to take the train is ( 1.5 / 20 ) * 60 + x = 4.5 + x minutes it takes 15 minutes longer to walk , so 30 = 4.5 + x + 20 x = 5.5 minutes answer : a"
a ) 5.5 , b ) 15 , c ) 25.5 , d ) 30 , e ) 60
a
subtract(subtract(divide(const_60, const_2), 20), divide(const_60, divide(20, 1.5)))
divide(const_60,const_2)|divide(n2,n0)|divide(const_60,#1)|subtract(#0,n3)|subtract(#3,#2)|
physics
one is height?
"when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = h ( 1 + 1 + 0.5 ) = 150 2.5 h = 150 h = 60 . a is the answer ."
a ) 60 cm , b ) 90 cm , c ) 100 cm , d ) 120 cm , e ) 130 cm
a
divide(150, add(const_2, divide(50, const_100)))
divide(n0,const_100)|add(#0,const_2)|divide(n1,#1)|
gain
a second schedule?
' approximately ' could actually make such a question ambiguous . not this one though but a similar question with the answer as 9.2 days . you round off 8.89 days as 9 days and everything is fine in this question . what do you do when you get 9.2 days ? do you need 9 days or 10 days ? can you round off 9.2 as 9 even though that is what you do with numbers ? no , because in 9 days your work is not over . you do need 10 days . to finish a work machine w say you need to work full 9 days and a part of the 10 th day . if i ask you how many days do you need to complete the work , will you say 9 or 10 ? you will say 10 even if you do n ' t use the 10 th day fully = d
a ) 14 , b ) 13 , c ) 11 , d ) 9 , e ) 7
d
inverse(add(inverse(divide(multiply(12, const_2), add(const_1, divide(const_1, const_2)))), inverse(divide(multiply(15, const_2), add(const_1, divide(const_1, const_2))))))
divide(const_1,const_2)|multiply(n0,const_2)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#3)|divide(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|inverse(#8)
physics
what 18?
"when n is even , ( x ^ n - a ^ n ) is completely divisible by ( x - a ) . ( 17 ^ 200 - 1 ^ 200 ) is completely divisible by ( 17 + 1 ) , ( 17 ^ 200 - 1 ) is completely divisible by 18 . on dividing 17 ^ 200 by 18 , we get 1 as remainder . answer is d"
a ) 3 , b ) 8 , c ) 5 , d ) 1 , e ) 7
d
power(17, 17)
power(n0,n0)|
general
calculate469 =?
"answer 469200 x 9999 = 469200 x ( 10000 - 1 ) = 4692000000 - 469200 = 4691100843 . option : e"
a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) 4691530800
e
multiply(469200, 9999)
multiply(n0,n1)|
general
how130ph?
"explanation : d = 100 + 130 = 230 s = 36 * 5 / 18 = 10 mps t = 230 / 10 = 23 sec answer : option d"
a ) 26 , b ) 72 , c ) 25 , d ) 23 , e ) 27
d
divide(add(130, 100), multiply(36, const_0_2778))
add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|
physics
the9090 is
"explanation : first six prime numbers which are between 60 and 90 = 61 , 67 , 71 , 73 , 79 , 83 average = ( 61 + 67 + 71 + 73 + 79 + 83 ) / 6 = 72.33 answer : e"
a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 72.33
e
add(60, const_1)
add(n0,const_1)|
general
a years b?
"c age x , then b age is 2 x so a age is 2 x + 2 . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 x = 5 so b is 2 x = 2 ( 5 ) 2 x 5 = 10 answer : b"
a ) 12 years , b ) 10 years , c ) 8 years , d ) 14 years , e ) 16 years
b
divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1))
add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)|
general
a22 year?
total number of installments = 52 payment per installment for the first 22 installments = 410 payment per installment for the remaining 30 installments = 410 + 65 = 475 average = ( 22 * 410 + 30 * 475 ) / 52 = 447.50 answer a
a ) 447.5 , b ) 450 , c ) 465 , d ) 468 , e ) 475
a
divide(add(multiply(22, 410), multiply(add(410, 65), subtract(52, 22))), 52)
add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0)
general
the m is :
"let the numbers be 5 x and 6 x . then their h . c . f = x . so , x = 4 . so , the numbers are 20 and 24 . l . c . m of 20 and 24 = 120 . answer : e"
a ) 12 , b ) 16 , c ) 24 , d ) 48 , e ) 120
e
lcm(multiply(5, 4), multiply(6, 4))
multiply(n0,n2)|multiply(n1,n2)|lcm(#0,#1)|
other
a a divisor?
"easy solution : n = dq 1 + 241 2 n = 2 dq 1 + 482 - ( 1 ) 2 n = dq 2 + 102 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 380 d * some integer = 380 checking all options only ( c ) syncs with it . answer c"
a ) 370 , b ) 365 , c ) 380 , d ) 456 , e ) 460
c
subtract(multiply(241, const_2), 102)
multiply(n0,const_2)|subtract(#0,n1)|
general
the90 due is
"sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 90 * 60 / 90 - 60 ] = rs . [ 90 * 60 / 30 ] = rs . 180 answer b"
a ) 210 , b ) 180 , c ) 360 , d ) 450 , e ) none
b
divide(multiply(90, 60), subtract(90, 60))
multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
gain
s s10?
it is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer . to help see this , consider the following examples : 4 / 10 is 0 with a remainder of 4 14 / 10 is 1 with a remainder of 4 5 / 10 is 0 with a remainder of 5 105 / 10 is 10 with a remainder of 5 it is also essential to remember that the s is a positive integer and multiple of 2 . any integer that is a multiple of 2 is an even number . so , s must be a positive even integer . with these two observations , the question can be simplified to : what is the units digit of 4 raised to an even positive integer ? the units digit of 4 raised to an integer follows a specific repeating pattern : 4 ^ 1 = 4 4 ^ 2 = 16 4 ^ 3 = 64 4 ^ 4 = 256 4 ^ ( odd number ) - - > units digit of 4 4 ^ ( even number ) - - > units digit of 6 there is a clear pattern regarding the units digit . 4 raised to any odd integer has a units digit of 4 while 4 raised to any even integer has a units digit of 6 . since s must be an even integer , the units digit of p = 4 ^ s will always be 6 . consequently , the remainder when p = 4 ^ s is divided by 10 will always be 6 . in case this is too theoretical , consider the following examples : s = 2 - - > p = 4 ^ z = 16 - - > s / 10 = 1 with a remainder of 6 s = 4 - - > p = 4 ^ z = 256 - - > s / 10 = 25 with a remainder of 6 s = 6 - - > p = 4 ^ z = 4096 - - > s / 10 = 409 with a remainder of 6 s = 8 - - > p = 4 ^ z = 65536 - - > s / 10 = 6553 with a remainder of 6 answer : b .
a ) 10 , b ) 6 , c ) 4 , d ) 0 , e ) it can not be determined
b
reminder(power(4, 2), 10)
power(n1,n0)|reminder(#0,n2)
general
if is numbers is
"sol . let the numbers be x and y . then , ( x + y ) = 22 and x 2 + y 2 = 386 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 22 ) 2 - 386 = 484 - 386 = 98 xy = 49 . answer b"
a ) 40 , b ) 49 , c ) 80 , d ) 88 , e ) 90
b
divide(subtract(power(22, const_2), 386), const_2)
power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|
general
the divisible is?
"largest 4 - digit number = 9999 88 ) 9999 ( 113 88 - - - - 119 88 - - - - 319 264 - - - 55 - - - required number = ( 9999 - 55 ) = 9944 . e )"
a ) 9934 , b ) 9938 , c ) 9940 , d ) 9942 , e ) 9944
e
square_area(const_pi)
square_area(const_pi)|
general
a service completed?
"round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 80 % of 50 % i . e 40 % . so he completed 50 + 40 = 90 % of total trip . e"
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90
e
add(divide(const_100, const_2), divide(multiply(80, divide(const_100, const_2)), const_100))
divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)|
gain
if =?
"( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 15 - m ) = 4 ^ ( ( 15 - m ) / 2 ) therefore , m = ( 15 - m ) / 2 2 m = 15 - m m = 5 answer d"
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6
d
divide(15, add(2, const_1))
add(n0,const_1)|divide(n3,#0)|
general
the then is?
"perimeter = distance covered in 10 min . = ( 15000 x 8 ) / 60 m = 2000 m . let length = 3 x metres and breadth = 4 x metres . then , 2 ( 3 x + 4 x ) = 2000 or x = 142.86 . length = 428.58 m and breadth = 571.44 m . area = ( 428.58 x 571.44 ) m 2 = 244907.76 m a"
a ) 244907.04 m , b ) 244000 m , c ) 244900 m , d ) 400000 m , e ) 154300 m
a
rectangle_area(divide(divide(multiply(multiply(divide(15, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 4)), const_2), multiply(divide(divide(multiply(multiply(divide(15, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 4)), const_2), 4))
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
physics
$ as share?
"let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 378 x = 54 4 x = 216 the answer is c ."
a ) $ 200 , b ) $ 208 , c ) $ 216 , d ) $ 224 , e ) $ 232
c
multiply(divide(378, add(add(divide(const_1, const_2), const_1), const_2)), const_2)
divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2)|
general
h perkeeper?
"cost of 8 kg grapes = 75 Γ— 8 = 600 . cost of 9 kg of mangoes = 55 Γ— 9 = 495 . total cost he has to pay = 600 + 495 = 1095 . e )"
a ) a ) 1000 , b ) b ) 1055 , c ) c ) 1065 , d ) d ) 1075 , e ) e ) 1095
e
add(multiply(8, 75), multiply(9, 55))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
gain
if square a?
"let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( sqrt ( 5 ) x ) ^ 2 = 5 x ^ 2 . the answer is e ."
a ) 1.5 , b ) 2 , c ) 3 , d ) 4 , e ) 5
e
power(sqrt(5), const_2)
sqrt(n0)|power(#0,const_2)|
geometry
a35 replaced?
"let ' s say that the total original mixture a is 100 ml the original mixture a thus has 50 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 50 ml and 25 ml is 25 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 25 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 15 % . therefore , 15 out of 25 is 3 / 5 and e is the answer ."
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5
e
inverse(add(divide(subtract(35, 25), subtract(50, 35)), const_1))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|add(#2,const_1)|inverse(#3)|
other
a purchase?
"let x be the amount of money paid for the first purchase . the second time , the customer paid 0.8 ( 1.2 x ) = 0.96 x . the difference is 4 % . the answer is d ."
a ) 10 % , b ) 8 % , c ) 6 % , d ) 4 % , e ) 2 %
d
multiply(subtract(const_1, multiply(add(divide(20, const_100), const_1), divide(80, const_100))), const_100)
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)|
general
there store store?
amount left = 0.5 x βˆ’ 4 for fifth store this is zero . so x = 8 . that means he entered fifth store with 8 . now for fourth store , amount left = 8 so 0.5 x βˆ’ 4 = 8 β‡’ x = 24 for third store , amount left = 24 so 12 x βˆ’ 4 = 24 β‡’ x = 56 for second store , amount left = 56 so 0.5 x βˆ’ 4 = 56 β‡’ x = 120 for first store , amount left = 120 so 0.5 x βˆ’ 4 = 120 β‡’ x = 248 so he entered first store with 248 . answer : a
a ) 248 , b ) 120 , c ) 252 , d ) 250 , e ) 350
a
multiply(add(multiply(add(multiply(add(multiply(add(multiply(4, const_2), 4), const_2), 4), const_2), 4), const_2), 4), const_2)
multiply(n1,const_2)|add(n1,#0)|multiply(#1,const_2)|add(n1,#2)|multiply(#3,const_2)|add(n1,#4)|multiply(#5,const_2)|add(n1,#6)|multiply(#7,const_2)
general
an square feet?
"area of the poster is 2.4 x 2.4 = 5.76 1 / 2 the area = 2.88 pi * r ^ 2 = 2.88 r ^ 2 = 2.88 / pi r = sqrt ( 2.88 / pi ) answer ( b )"
a ) 1 / pi , b ) sqrt ( 2.88 / pi ) , c ) 1 , d ) 2 / sqrt ( pi ) , e ) pi / 2
b
sqrt(divide(divide(square_area(2.4), 2), const_pi))
square_area(n0)|divide(#0,n2)|divide(#1,const_pi)|sqrt(#2)|
geometry
53551 =?
"5359 x 51 = 5359 x ( 50 + 1 ) = 5359 x 50 + 5359 x 1 = 267950 + 5359 = 273309 e )"
a ) 273232 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273309
e
multiply(divide(5359, 51), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
3, by :
original share of 1 person = 1 / 3 new share of 1 person = 1 / 2 increase = ( 1 / 2 - 1 / 3 = 1 / 6 ) therefore , required fraction = ( 1 / 6 ) / ( 1 / 3 ) = ( 1 / 6 ) x ( 3 / 1 ) = 1 / 2 answer is a .
a ) 1 / 2 , b ) 2 / 7 , c ) 3 / 2 , d ) 4 / 7 , e ) none of them
a
divide(subtract(divide(const_1, const_2), divide(const_1, 3)), divide(const_1, 3))
divide(const_1,const_2)|divide(const_1,n0)|subtract(#0,#1)|divide(#2,#1)
general
two. direction?
solution relative speed = ( 45 - 40 ) kmph = 5 kmph = ( 5 x 5 / 18 ) m / sec = ( 25 / 18 ) m / sec time taken = ( 350 x 18 / 25 ) sec = 252 sec . answer d
a ) 72 sec , b ) 132 sec , c ) 192 sec , d ) 252 sec , e ) none
d
multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(45, 40)))
add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)
physics
a for percentage?
"l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 175 ( 35 * 5 ) profit percentage = ( 175 - 100 ) / 100 * 100 = 75 % answer : c"
a ) 80 % , b ) 50 % , c ) 75 % , d ) 40 % , e ) 53 %
c
subtract(multiply(35, add(const_4, const_1)), multiply(25, const_4))
add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)|
gain
how /2?
"required number = ( 75 / 2 ) / ( 1 / 8 ) = ( 75 / 2 x 8 / 1 ) = 300 . answer : a"
a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700
a
divide(add(37, divide(1, 2)), divide(1, 8))
divide(n0,n4)|divide(n0,n1)|add(n2,#0)|divide(#2,#1)|
general
pipe pipe tank?
"suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 2 / 4 ) rq ] = 4 / 2 also , rp = rr / ( 2 / 3 ) = > 2 = rr / ( 2 / 3 ) = > rr = 4 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 2 + 4 / 3 = 13 / 3 = 1 / ( 3 / 13 ) thus running simultaneously , pipe q will drain 3 / 13 of the liquid . thus answer = a ."
a ) 3 / 13 , b ) 8 / 23 , c ) 3 / 8 , d ) 17 / 29 , e ) 3 / 4
a
divide(multiply(2, 2), add(multiply(multiply(4, 2), const_2), multiply(2, 2)))
multiply(n0,n0)|multiply(n0,n1)|multiply(#1,const_2)|add(#2,#0)|divide(#0,#3)|
physics
in line xz
"look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 5 = 8 : 10 - - > qy = 4.0 . answer : c ."
a ) 3.6 cm , b ) 2.4 cm , c ) 4.0 cm , d ) 2.16 cm , e ) 3.2 cm
c
divide(multiply(5, 8), const_10)
multiply(n1,n2)|divide(#0,const_10)|
geometry
the runs4?
"explanation : average after 11 innings = 24 required number of runs = ( 24 * 11 ) – ( 20 * 10 ) = 264 – 200 = 64 answer : c"
a ) 22 , b ) 77 , c ) 64 , d ) 19 , e ) 17
c
subtract(multiply(add(10, const_1), add(20, 4)), multiply(20, 10))
add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
general
a the b.
"ratio of the capitals of a , b and c = 20000 Γ£ β€” 10 + 15000 Γ£ β€” 2 : 20000 Γ£ β€” 10 + 16000 Γ£ β€” 2 : 20000 Γ£ β€” 10 + 26000 Γ£ β€” 2 = 230000 : 232000 : 252000 = 230 : 232 : 252 . b Γ’ € β„’ s share = ( 71400 Γ£ β€” 232 Γ’  β€ž 714 ) = 23200 ; answer c"
a ) 20,000 , b ) 21,200 , c ) 23,200 , d ) 20,500 , e ) none of these
c
divide(add(multiply(subtract(const_12, 10), 4000), multiply(multiply(const_2, multiply(const_100, const_100)), 10)), multiply(const_100, const_10))
multiply(const_100,const_100)|multiply(const_10,const_100)|subtract(const_12,n1)|multiply(n3,#2)|multiply(#0,const_2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#1)|
gain
the80 k =
"the number of terms in this set would be : n = ( k - 1 ) / 2 ( as k is odd ) last term : k - 1 average would be first term + last term / 2 = ( 2 + k - 1 ) / 2 = ( k + 1 ) / 2 also average : sum / number of terms = 79 * 80 / ( ( k - 1 ) / 2 ) = 158 * 80 / ( k - 1 ) ( k + 1 ) / 2 = 158 * 80 / ( k - 1 ) - - > ( k - 1 ) ( k + 1 ) = 158 * 160 - - > k = 159 answer e ."
a ) 79 , b ) 80 , c ) 81 , d ) 157 , e ) 159
e
add(multiply(79, const_2), 1)
multiply(n1,const_2)|add(#0,n0)|
general
how length?
"speed = 78 * 5 / 18 = 21.7 m / sec total distance covered = 125 + 125 = 250 m . required time = 250 / 21.7 ' = 11.5 sec . answer : c"
a ) 12.7 sec , b ) 10.1 sec , c ) 11.5 sec , d ) 12.1 sec , e ) 11.7 sec
c
divide(add(125, 125), multiply(78, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
ele the today?
ele , the circus elephant , is currently three times older than lyn , the circus lion . ele = 3 * lyn usually , ages are integers so there is a good possibility that the age of ele is 45 ( the only option that is a multiple of 3 ) . then age of lyn would be 15 . in 15 yrs , ele would be 60 and lyn would be 30 - so lyn would be half as old as ele . answer ( d )
a ) 40 , b ) 48 , c ) 43 , d ) 45 , e ) 41
d
multiply(subtract(multiply(const_2, 15), 15), 3)
multiply(n1,const_2)|subtract(#0,n1)|multiply(n0,#1)
general
a age son is
"explanation : let ' s son age is x , then father age is x + 30 . = > 2 ( x + 2 ) = ( x + 30 + 2 ) = > 2 x + 4 = x + 32 = > x = 28 years option a"
a ) 28 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 26 years
a
divide(subtract(30, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))
multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|
general
l farmer is :
"total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 394000 x \ therefore required percentage = \ inline \ frac { 800 x } { 394000 } \ times 100 = 0.20304 e"
a ) 0.20833 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.20304
e
multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100)
divide(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_4)|divide(n3,#0)|multiply(#1,const_1000)|multiply(#3,const_1000)|add(#5,#6)|divide(#4,#7)|multiply(#8,const_100)|
general
the the is :
"speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 110 + x ) / 30 = 25 / 2 = = > 2 ( 110 + x ) = 750 = = > x = 265 m . answer : option a"
a ) 265 , b ) 244 , c ) 245 , d ) 238 , e ) 236
a
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 110)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
cole drove work?
"let the distance one way be x time from home to work = x / 60 time from work to home = x / 100 total time = 2 hrs ( x / 60 ) + ( x / 100 ) = 2 solving for x , we get x = 75 time from home to work in minutes = ( 75 ) * 60 / 60 = 75 minutes ans = d"
a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78
d
multiply(divide(multiply(100, 2), add(60, 100)), const_60)
add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)|
physics
if decreased area?
"area of square = pi * radius ^ 2 new radius = 0.5 * old radius so new area = ( 0.5 ) ^ 2 old area = > 0.25 of old area = > 25 % old area ans : c"
a ) 10 % decrease , b ) 20 % decrease , c ) 75 % decrease , d ) 40 % decrease , e ) 50 % decrease
c
subtract(const_100, multiply(power(divide(50, const_100), const_2), const_100))
divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)|
geometry
sak the work :
"solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 30 : x β‡’ x = ( 30 x 4 / 5 ) β‡’ x = 24 days . hence , tanya takes 16 days is complete the work . answer c"
a ) 15 , b ) 16 , c ) 14 , d ) 25 , e ) 10
c
divide(30, add(const_1, divide(25, const_100)))
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|
physics
what 1?
"3 ^ 8 - 1 = ( 3 ^ 4 - 1 ) ( 3 ^ 4 + 1 ) = 80 * 82 = 8 * 10 * 2 * 41 the answer is d ."
a ) 29 , b ) 31 , c ) 37 , d ) 41 , e ) 43
d
floor(divide(3, divide(8, const_2)))
divide(n1,const_2)|divide(n0,#0)|floor(#1)|
general
calculate of y?
"explanation : 90 x = 60 y x : y = 90 : 60 = 3 : 2 answer : c"
a ) 1 : 2 , b ) 3 : 5 , c ) 3 : 2 , d ) 3 : 4 , e ) 5 : 2
c
divide(90, 60)
divide(n0,n1)|
general
find off cm?
"1 / 2 * 50 ( 15 + 5 ) = 500 cm 2 answer : e"
a ) 100 cm 2 , b ) 150 cm 2 , c ) 200 cm 2 , d ) 250 cm 2 , e ) 500 cm 2
e
multiply(multiply(divide(const_1, const_2), add(5, 15)), 50)
add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)|
geometry
a the is?
"speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 1200 + x ) / 60 = 65 / 3 x = 100 m . answer : d"
a ) 2898 , b ) 277 , c ) 500 , d ) 100 , e ) 435
d
divide(1200, multiply(subtract(78, 1), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
the is?
"7 + 11 + 13 + 17 + 19 = 67 / 5 = 13.40 answer : a"
a ) 13.4 , b ) 12.26 , c ) 12.3 , d ) 32.8 , e ) 32.4
a
add(5, const_1)
add(n0,const_1)|
general
a of cone?
edge of the cube = 3 √ 334 = 7 cm ∴ radius of cone = 3.5 cm height = 7 cm volume of cone = 1 ⁄ 3 Ο€ r 2 h 1 ⁄ 3 Ο€ r 2 h = 1 ⁄ 3 Γ— 22 ⁄ 7 Γ— ( 3.5 ) 2 Γ— 7 = 1 ⁄ 3 Γ— 22 Γ— 12.25 β‰ˆ 90 sec answer b
['a ) 80 cc', 'b ) 90 cc', 'c ) 110 cc', 'd ) 105 cc', 'e ) 100 cc']
b
volume_cone(divide(cube_edge_by_volume(343), const_2), cube_edge_by_volume(343))
cube_edge_by_volume(n0)|divide(#0,const_2)|volume_cone(#1,#0)
geometry
if what second?
"here , l = 7 and m = 28 therefore , first number = l / m x 100 % of second number = 7 / 28 x 100 % of second number = 25 % of second number answer : b"
a ) 20 % , b ) 25 % , c ) 18 % , d ) 30 % , e ) none of these
b
multiply(divide(divide(7, const_100), divide(28, const_100)), const_100)
divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|
gain
a53 is?
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 83 - 53 ) = > x / 2 = 30 = > x = 60 . answer : c"
a ) 18 , b ) 82 , c ) 60 , d ) 27 , e ) 29
c
multiply(subtract(83, 53), const_2)
subtract(n0,n1)|multiply(#0,const_2)|
general
the. diagonal?
area of rectangle = 20 x 10 = 200 cm Γ’ Β² let ' l ' the length of other diagonal = 0.5 x 32 xl = 200 which gives x = 12.5 cm answer : b
['a ) 10', 'b ) 12.5', 'c ) 15', 'd ) 16', 'e ) 17.5']
b
divide(multiply(rectangle_area(20, 10), const_2), 32)
rectangle_area(n0,n1)|multiply(#0,const_2)|divide(#1,n2)
geometry
if. principal?
p - 2160 = ( p * 5 * 4 ) / 100 p = 2700 answer : c
a ) 1500 , b ) 2500 , c ) 2700 , d ) 3200 , e ) 11500
c
divide(2160, subtract(const_1, divide(multiply(4, 5), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)
gain
a number week?
"for 40 hrs = 40 * 16 = 640 excess = 948 - 640 = 308 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 308 / 28 = 77 / 7 = 11 total hrs = 40 + 11 = 51 answer e 51"
a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 51
e
add(40, divide(subtract(948, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100)))
add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)|
general
a x axis?
"assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point w of intersection is ( 40,0 ) . a is the correct answer ."
a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 )
a
multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10)
divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10)|
general
a the sales?
let his total sales be x total sales - commission = $ 31100 x - [ ( 5 % of 10000 ) + 4 % of ( x - 10000 ) ] = 31100 x - 500 - ( x - 10000 ) / 25 = 31100 x = 32500 answer is b
a ) $ 30000 , b ) $ 32500 , c ) $ 35120 , d ) $ 41520 , e ) $ 25460
b
divide(add(31100, subtract(divide(multiply(5, 10000), const_100), divide(multiply(4, 10000), const_100))), subtract(const_1, divide(4, const_100)))
divide(n2,const_100)|multiply(n0,n1)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,const_100)|subtract(const_1,#0)|subtract(#3,#4)|add(n3,#6)|divide(#7,#5)
general
how of30?
( 60 / 100 ) * 50 – ( 34 / 100 ) * 30 30 - 10.2 = 19.8 answer : b
a ) 18 , b ) 19.8 , c ) 11 , d ) 17 , e ) 12
b
subtract(divide(multiply(60, 50), const_100), divide(multiply(34, 30), const_100))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)
gain
a still man?
"m = 5 s = 2 ds = 7 us = 3 x / 7 + x / 3 = 1 x = 2.1 d = 2.1 * 2 = 4.2 answer : d"
a ) 5.75 , b ) 5.7 , c ) 5.76 , d ) 4.2 , e ) 5.71
d
multiply(divide(multiply(add(5, 2), subtract(5, 2)), add(add(5, 2), subtract(5, 2))), const_2)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|
physics
how60 length?
d = 200 + 180 = 380 m s = 60 * 5 / 18 = 50 / 3 t = 380 * 3 / 50 = 22.8 sec answer : c
a ) 36.7 , b ) 26.8 , c ) 22.8 , d ) 21.1 , e ) 16.2
c
divide(add(200, 180), multiply(60, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)
physics
difference m is?
solving the two equations , we get : l = 30 and b = 40 . area = ( l x b ) = ( 30 x 40 ) m 2 = 1200 m ^ 2 d
['a ) 2400 m ^ 2', 'b ) 1500 m ^ 2', 'c ) 2520 m ^ 2', 'd ) 1200 m ^ 2', 'e ) 2580 m ^ 2']
d
multiply(multiply(10, const_4), multiply(10, const_3))
multiply(n0,const_4)|multiply(n0,const_3)|multiply(#0,#1)
geometry
137 6430
"1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 answer : a ."
a ) 1070 , b ) 6530 , c ) 6630 , d ) 6730 , e ) 6830
a
subtract(negate(6430), multiply(subtract(1320, 180), divide(subtract(1320, 180), subtract(1370, 1320))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
machine machine x produce
"machine y produces 20 percent more widgets in an hour than machine x does in an hour . so if machine x produces 100 widgets , then machine y produces 120 widgets . ratio of 120 / 100 = 6 / 5 . this is their speed of work ( y : x ) . i . e . speed of their work ( x : y ) = 5 / 6 now , time is inversely proportional to speed . hence the ratio of the time spent ( x : y ) = 6 / 5 let us assume that they spend 6 x and 5 x hours . given that 6 x - 5 x = 50 so , x = 50 . hence 6 x = 6 * 50 = 300 hours . hence x takes 120 hours to produce 1080 widgets . so , in 1 hour , it can produce ( 1 * 1080 ) / 300 = 3.6 hence option ( b ) ."
a ) 100 , b ) 3.6 , c ) 25 , d ) 11 , e ) 9
b
divide(1080, multiply(divide(50, const_10), 50))
divide(n0,const_10)|multiply(n0,#0)|divide(n1,#1)|
general
for greatest0?
"factorization of 45 = 3 * 3 * 5 factorization of 80 = 5 * 2 ^ 4 means the smallest value of k is \ sqrt { 3 ^ 2 * 5 * 2 ^ 4 } = 3 * 5 * 2 ^ 2 k ^ 3 = 3 ^ 3 * 5 ^ 3 * 2 ^ 6 * x = 3 ^ 3 * 2 * 4000 * x where x can be any integer k ^ 3 / 4000 = 3 ^ 3 * 2 * x = 54 x which obviously has different units digit depending on x answer : d"
a ) 0 , b ) 1 , c ) 27 , d ) 54 , e ) can not be determined
d
divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4)
multiply(n6,n9)|multiply(#0,#0)|multiply(#1,#0)|divide(#2,const_4)|
general
p s income?
make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get s = 7.6 and q 10.8 so 10.8 / 7.6 = 2.7 / 1.9 ans : e
a ) 11 / 12 , b ) 13 / 17 , c ) 13 / 19 , d ) 12 / 19 , e ) 27 / 19
e
divide(add(multiply(60, const_100), multiply(60, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 60)))
add(n0,n1)|multiply(n4,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|multiply(n4,#3)|multiply(n4,#0)|add(#1,#4)|add(#2,#5)|divide(#6,#7)
general
if / =?
"let p = 3 , q = 7 then 2 * 3 + 7 = 13 so 2 p + q = 13 . answer : c"
a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 16
c
add(multiply(3, 2), 7)
multiply(n0,n2)|add(n1,#0)|
general
in, defective?
"# defective pens = 2 # good pens = 7 probability of the 1 st pen being good = 7 / 9 probability of the 2 nd pen being good = 6 / 8 total probability = 7 / 9 * 6 / 8 = 7 / 12 the answer is d ."
a ) 4 / 7 , b ) 5 / 9 , c ) 6 / 11 , d ) 7 / 12 , e ) 8 / 15
d
multiply(divide(subtract(9, 2), 9), divide(subtract(subtract(9, 2), const_1), subtract(9, const_1)))
subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)|
general
sum m m is
"explanation : let the numbers be x and y according to the problem x + y = 63 x - y = 1 / 8 ( x + y ) x - y = 1 / 8 * 63 , x - y = 9 2 x = 72 , x = 36 and y = 27 l . c . m of 36 and 27 is 351 answer : option c"
a ) 231 , b ) 153 , c ) 351 , d ) 345 , e ) 355
c
subtract(63, divide(subtract(63, divide(1, const_2)), const_2))
divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)|
general
if in it?
"explanation : let the required no of hours be x . then less men , more hours ( indirct proportion ) \ inline \ fn _ jvn \ therefore 15 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 15 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 15 } = 60 hence , 15 men can do it in 60 hours . answer : c ) 60"
a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 72
c
divide(multiply(36, 25), 15)
multiply(n0,n1)|divide(#0,n2)|
physics
the62 is :
"d 5 % increase in 10 years = ( 262500 - 175000 ) = 87500 . increase % = ( 87500 / 175000 x 100 ) % = 50 % . required average = ( 50 / 10 ) % = 5 % ."
a ) 1 % , b ) 2 % , c ) 3 % , d ) 5 % , e ) 4 %
d
add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4)
add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)|
general
last range year?
let the lowest price be x . therefore , highest price is x + 100 . now price of each variety is increased by 10 % . therefore the price will remain arranged in the same order as before . or lowest price = 1.1 x and highest = 1.1 * ( x + 100 ) or range = highest - lowest = 1.1 * ( x + 100 ) - 1.1 x = 110 , hence , c
a ) $ 50 , b ) $ 100 , c ) $ 110 , d ) $ 600 , e ) $ 300
c
multiply(100, divide(add(10, const_100), const_100))
add(n3,const_100)|divide(#0,const_100)|multiply(n0,#1)
general
the by profit?
"explanation : let the c . p . of the article be rs . x given that % profit earned by selling article at rs . 1920 = % loss incurred by selling article at rs . 1280 ( 1920 βˆ’ x / x ) βˆ— 100 = ( x βˆ’ 1280 / x ) βˆ— 100 = > 1920 - x = x - 1280 = > 2 x = 3200 = > x = 1600 s . p . for 25 % profit = rs . 1600 + 25 % of rs . 1600 = rs . 1600 * ( 125 / 100 ) = rs . 2000 answer : a"
a ) rs . 2000 , b ) rs . 2200 , c ) rs . 2400 , d ) data inadequate , e ) can not be determined
a
multiply(divide(add(const_100, 25), 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
the fabric cut?
the tailors would need to cut the fabric 49 times thus the total amount spent would be 245 minutes . the answer is d
a ) 150 , b ) 200 , c ) 188 , d ) 245 , e ) 123
d
multiply(subtract(divide(multiply(10, const_1000), 200), const_1), 5)
multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1)|multiply(n3,#2)
physics
find42 4
"answer = 60 / 42 * 4 = 60 / 168 = 0.3571 option d is correct"
a ) 14.7653 , b ) 0.5897 , c ) 3.6974 , d ) 0.3571 , e ) 1.2587
d
divide(60, 42)
divide(n0,n1)|
general
find average50?
"there are five prime numbers between 30 and 50 . they are 31,37 , 41,43 and 47 . therefore the required average = ( 31 + 37 + 41 + 43 + 47 ) / 5  199 / 5  39.8 answer c 39.8"
a ) 18.9 , b ) 19.8 , c ) 39.8 , d ) 29.8 , e ) 20.8
c
divide(add(add(add(30, const_1), add(add(30, const_1), const_2)), add(subtract(50, 30), subtract(50, const_2))), 30)
add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)|
general
in get?
"let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 4720 32.5 k = 32.5 k / 100 k * 4720 = 1534 the number of valid votes polled in favour of b is 1534 . answer : a"
a ) 1534 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2991
a
divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 4720), divide(multiply(15, 4720), const_100)), const_2)
divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(const_1,#0)|multiply(n2,#3)|subtract(#4,#2)|divide(#5,const_2)|
general
if the get?
"worker b will get 11 / 16 = 68.75 % the answer is d ."
a ) 62.45 % , b ) 64.55 % , c ) 66.65 % , d ) 68.75 % , e ) 70.85 %
d
divide(1088, add(5, 11))
add(n1,n2)|divide(n0,#0)|
other
the when women?
"20 + 24 + 6 * 2 = 56 / 2 = 28 answer : e"
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28
e
divide(add(add(20, 24), multiply(6, 2)), const_2)
add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)|
general
operation 100?
official solution : ( b ) we know that a # b = 100 and a # b = 4 a Β² + 4 b Β² + 8 ab . so 4 a Β² + 4 b Β² + 8 ab = 100 we can see that 4 a Β² + 4 b Β² + 8 ab is a well - known formula for ( 2 a + 2 b ) Β² . therefore ( 2 a + 2 b ) Β² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbers . so we can conclude that 2 ( a + b ) = 10 . ( a + b ) + 4 = 10 / 2 + 4 = 9 . the correct answer is b .
a ) 5 , b ) 9 , c ) 10 , d ) 13 , e ) 17
b
add(sqrt(divide(100, 4)), 4)
divide(n6,n0)|sqrt(#0)|add(n0,#1)
general
the m metre.
"area = 5.5 Γ— 3.75 sq . metre . cost for 1 sq . metre . = $ 500 hence , total cost = 5.5 Γ— 3.75 Γ— 500 = $ 10312.50 a"
a ) $ 10312.50 , b ) $ 13512.50 , c ) $ 16512.50 , d ) $ 14512.50 , e ) $ 18512.50
a
multiply(500, multiply(5.5, 3.75))
multiply(n0,n1)|multiply(n2,#0)|
physics
a on required?
"we have : l = 20 ft and lb = 10 sq . ft . so , b = 0.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 1 ) ft = 21 ft . answer : d"
a ) 34 , b ) 40 , c ) 68 , d ) 21 , e ) 78
d
add(multiply(divide(10, 20), const_2), 20)
divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|
geometry
each, investment is
the problem asks for the approximate chance that no more than 1 / 3 of the original investment is lost . we can apply the β€œ 1 – x ” technique : what ’ s the chance that more than 1 / 3 of the original investment is lost ? there are two outcomes we have to separately measure : ( a ) all 3 investments become worthless . ( b ) 2 of the 3 investments become worthless , while 1 doesn ’ t . outcome ( a ) : the probability is ( 0.2 ) ( 0.2 ) ( 0.2 ) = 0.008 , or a little less than 1 % . outcome ( b ) : call the investments x , y , and z . the probability that x retains value , while y and z become worthless , is ( 0.8 ) ( 0.2 ) ( 0.2 ) = 0.032 . now , we have to do the same thing for the specific scenarios in which y retains value ( while x and z don ’ t ) and in which z retains value ( while x and y don ’ t ) . each of those scenarios results in the same math : 0.032 . thus , we can simply multiply 0.032 by 3 to get 0.096 , or a little less than 10 % . the sum of these two probabilities is 0.008 + 0.096 = 0.104 , or a little more than 10 % . finally , subtracting from 100 % and rounding , we find that the probability we were looking for is approximately 90 % . the correct answer is a . this problem illustrates the power of diversification in financial investments . all else being equal , it ’ s less risky to hold a third of your money in three uncorrelated ( independent ) but otherwise equivalent investments than to put all your eggs in one of the baskets . that said , be wary of historical correlations ! housing price changes in different us cities were not so correlated β€” and then they became highly correlated during the recent housing crisis ( they all fell together ) , fatally undermining spreadsheet models that assumed that these price changes were independent .
a ) 90 % , b ) 80 % , c ) 70 % , d ) 60 % , e ) 40 %
a
subtract(add(multiply(20, const_2), multiply(20, 3)), const_10)
multiply(n1,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#2,const_10)
general
how than6?
"multiples of 4 less than 250 = { 4 , 8,12 , . . . . . . 248 } = 62 numbers multiples of 4 which are multiples of 3 too = { 12 , 24,36 . . . . . 240 } = 20 numbers so required number = 62 - 20 = 42 choice c"
a ) 20 , b ) 31 , c ) 42 , d ) 53 , e ) 64
c
divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
general
a the ferry?
to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 50 tons will be 100,000 pounds from the answer choices : let max number of vehicles be 62 total weight will be = 62 * 1600 = 99200 pounds , which is lesser than the maximum weight allowed . ans : d
a ) 23 , b ) 41 , c ) 48 , d ) 62 , e ) 86
d
divide(multiply(multiply(50, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100))
add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)
general
what the other?
"d = 8 d d = d a √ 2 = 8 d a √ 2 = d a = 8 d / √ 2 a = d / √ 2 = > 8 : 1 answer : d"
a ) 3 : 6 , b ) 3 : 3 , c ) 3 : 8 , d ) 8 : 1 , e ) 3 : 2
d
divide(8, divide(8, 8))
divide(n0,n0)|divide(n0,#0)|
geometry
a10 year?
bank balance is doubled with accumulation of interest tp 5080 . . this means interest is 5080 / 2 = 2540 for entire year . . although since interest is 10 % of avg monthly balance , it becomes 25400 . . d
a ) 2840 , b ) 5680 , c ) 6840 , d ) 25400 , e ) 28400
d
multiply(5080, divide(10, const_2))
divide(n0,const_2)|multiply(n1,#0)
general
a bus going?
"speed of bus relative to woman = 21 + 3 = 24 km / hr . = 24 * 5 / 18 = 20 / 3 m / sec . time taken to pass the woman = 75 * 3 / 20 = 11.25 sec . answer : c"
a ) 5.75 , b ) 7.62 , c ) 11.25 , d ) 4.25 , e ) 3.25
c
divide(divide(multiply(75, const_3600), add(21, 3)), const_1000)
add(n1,n2)|multiply(n0,const_3600)|divide(#1,#0)|divide(#2,const_1000)|
physics
what largest200!
"in real basic terms , we ' re asked to find all of the ' 3 s ' in 200 ! we can figure out that 200 / 3 = 66 , so we know that there are at least 66 ' 3 s ' in 200 ! while that answer is among the 5 choices , it seems a bit too ' easy ' , so let ' s do a bit more work and list out the first few numbers that we know have a ' 3 ' in them : 3 = 3 x 1 6 = 3 x 2 9 = 3 x 3 notice how both 3 and 6 have just one 3 in them , but 9 has two 3 s ( there ' s an ' extra ' 3 that we have to account for ) . this implies that there are probably other numbers that include ' extra 3 s ' that we have to figure out : to find those extra 3 s , we have to look at numbers that contain ' powers of 3 ' . . . 3 ^ 2 = 9 3 ^ 3 = 27 3 ^ 4 = 81 3 ^ 5 = 243 , but that ' s too big ( we ' re only going up to 200 ) . keep in mind that a multiple of 81 is also a multiple of 9 and 27 , so we do n ' t want to count any of those values more than once . 200 / 9 = 22 , so we know that there are at least 22 extra 3 s ( and certainly more because of the 27 and 81 ) . with the 66 3 s that we already have , those 22 extra 3 s increase the total to 88 . with the other extra 3 s , we ' ll end up with more than 88 3 s . there ' s only one answer that fits that logic . . . answer : d"
a ) 88 , b ) 48 , c ) 66 , d ) 97 , e ) 39
d
multiply(floor(divide(power(const_10, 3), 200)), 200)
power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)|
other
ree70 the average
"explanation : ( 40 + 60 + 70 + 80 + 80 / 5 ) = 66 option a"
a ) 66 , b ) 75 , c ) 80 , d ) 85 , e ) 90
a
divide(add(add(add(add(40, 60), 70), 80), 80), add(const_4, const_1))
add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|
general
the leaves bob?
"the time it takes bob to drive to city b is 100 / 40 = 2.5 hours . alice needs to take less than 2 hours for the trip . alice needs to exceed a constant speed of 100 / 2 = 50 miles per hour . the answer is c ."
a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 54
c
divide(100, subtract(divide(100, 40), divide(30, const_60)))
divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2)|
physics
on water day?
"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 3 units of water was used to make 4 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 4 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 4 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 4 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 4 * x - - > x = $ 0.3 . answer : c ."
a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45
c
divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2))
add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|
general
859 -4358
"d 7429 - 4358 = 3071 let 8597 - x = 3071 then , x = 8597 - 3071 = 5526"
a ) 3567 , b ) 6424 , c ) 6835 , d ) 5526 , e ) none of these
d
subtract(multiply(divide(8597, const_100), 7429), multiply(divide(const_1, const_3), multiply(divide(8597, const_100), 7429)))
divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)|
general
when,7?
"cant think of a straight approach but here is how i solved it : k is divided by 5 and remainder is 2 . this means k = 5 n + 2 ( n is an integer ) so the possible values of k = { 2 , 7 , 12 , 17 , 22 } ( less than 24 ) secondly , if k is divided by 6 , the remainder is 5 = > k = 6 m + 5 so the possible value set for k = { 5 , 11 , 17 , 23 } ( less than 24 ) 17 is the only common number in both the sets . hence k = 17 answer : d"
a ) 2 , b ) 5 , c ) 6 , d ) 3 , e ) 8
d
reminder(add(const_12, 5), 7)
add(n0,const_12)|reminder(#0,n5)|
general
is c b?
"let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 32 5 x = 30 = > x = 6 hence , b ' s age = 2 x = 12 years . answer : e"
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12
e
multiply(divide(subtract(32, const_2), add(const_3, const_2)), const_2)
add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|
general
how m length?
"speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 20 = 12.1 sec . answer : b"
a ) 12.6 , b ) 12.0 , c ) 12.1 , d ) 12.3 , e ) 12.2
b
divide(add(110, 132), multiply(72, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
agraph is?
"speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 6 = 120 m . answer : c"
a ) 298 , b ) 288 , c ) 120 , d ) 776 , e ) 991
c
divide(const_100.0, subtract(divide(const_100.0, 10), 6))
divide(const_100.0,n1)|subtract(#0,n2)|divide(const_100.0,#1)|
physics
mary60 income.
"even i got 96 % j = 100 t = 100 * 0.4 = 40 m = 40 * 1.6 = 64 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 64 * 100 / 100 = 64 ans : c"
a ) 124 % , b ) b . 120 % , c ) 64 % , d ) 80 % , e ) 64 %
c
add(subtract(const_100, 60), multiply(subtract(const_100, 60), divide(60, const_100)))
divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)|
general
11 can type?
"no . of ways of picking 2 biology books ( from 11 books ) = 11 c 2 = ( 11 * 10 ) / 2 = 55 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 55 * 28 = 1540 ( option e )"
a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 1540
e
multiply(divide(divide(factorial(11), factorial(subtract(11, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2))
factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9)|
other
pipe minutes together?
pipe a fills 1 / 30 th of the tank in a minute and pipe b empties 1 / 90 th of the tank ( 1 / 30 ) - ( 1 / 90 ) = ( 1 / x ) 2 / 90 = 1 / x = > x = 45 answer : d
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50
d
divide(const_1, subtract(divide(const_1, 30), divide(const_1, 90)))
divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)
physics