Datasets:
pgilliar
/
MNLP_M2_rag_dataset

Dataset card Data Studio Files
xet
Community
Problem
stringlengths
5
967
Rationale
stringlengths
1
2.74k
options
stringlengths
37
164
correct
stringclasses
5 values
annotated_formula
stringlengths
7
1.65k
linear_formula
stringlengths
8
925
category
stringclasses
6 values
answer
stringclasses
5 values
a certain machine produces 1,100 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 6 days ?
since 6 days consist of 24 * 6 hours the total is 144 hours . since every hour the machine produces 1100 units of product p the total product during 144 hours is 144 * 1100 = 158,400 . correct option : e
a ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 100,000 , e ) 158,400
e
multiply(const_4, const_10)
multiply(const_10,const_4)
physics
E
a fair 2 sided coin is flipped 3 times . what is the probability that tails will be the result at least twice , but not more than 3 times ?
"at least twice , but not more than 3 timesmeans exactly 2 times , 3 times the probability of getting exactly k results out of n flips is nck / 2 ^ n 3 c 2 / 2 ^ 3 = 3 / 8 option : c"
a ) 5 / 8 , b ) 3 / 4 , c ) 3 / 8 , d ) 7 / 4 , e ) 5 / 6
c
subtract(const_1, add(multiply(inverse(power(2, 3)), 3), add(inverse(power(2, 3)), inverse(power(2, 3)))))
power(n0,n1)|inverse(#0)|add(#1,#1)|multiply(n1,#1)|add(#2,#3)|subtract(const_1,#4)|
general
C
yesterday ' s closing prices of 2,200 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ?
"lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2200 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 1200 . c is my answer ."
a ) 484 , b ) 726 , c ) 1,200 , d ) 1,320 , e ) 1,694
c
multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100)))
divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6)|
gain
C
a train passes a station platform in 62 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?
"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 62 = 15 = > x = 444 m answer : d"
a ) 615 m , b ) 240 m , c ) 168 m , d ) 444 m , e ) 691 m
d
multiply(20, multiply(54, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics
D
what quantity of water should be added to reduce 5 litres of 45 % acidic liquid to 25 % acidic liquid ?
"quantity of acid in 5 litres = 45 * 5 / 100 = 2.25 litres let x litres of water is added to the solution , then there is 2.25 litres of acid in ( 5 + x ) litres of liquid . 25 % of ( 5 + x ) = 2.25 , 25 / 100 * ( 5 + x ) = 2.25 x = 225 - 125 / 25 = 4 answer : d"
a ) 3 litres , b ) 2 litres , c ) 4 litres , d ) 4.5 litres , e ) 5 litres
d
subtract(divide(multiply(multiply(5, divide(45, const_100)), const_100), 25), 5)
divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_100)|divide(#2,n2)|subtract(#3,n0)|
gain
D
each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.6 % , what fraction of the products was checked by worker y ?
"x : 0.5 % is 0.1 % - points from 0.6 % . y : 0.8 % is 0.2 % - points from 0.6 % . therefore the ratio of products checked by y : x is 1 : 2 . thus , worker y checked 1 / 3 of the products . the answer is c ."
a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 3 / 7
c
divide(subtract(0.6, 0.5), subtract(0.8, 0.5))
subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)|
general
C
in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 12 or larger is drawn is 1 / 3 , what is the probability that a number less than or equal to 20 is drawn ?
"you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 1 / 3 + p ( set 2 ) - 1 / 6 = 1 p ( set 2 ) = 5 / 6 answer ( e )"
a ) 1 / 18 , b ) 1 / 6 , c ) 1 / 3 , d ) 1 / 2 , e ) 5 / 6
e
subtract(add(12, divide(12, 6)), divide(1, 6))
divide(n0,n3)|divide(n5,n3)|add(n0,#0)|subtract(#2,#1)|
general
E
a fill pipe can fill 1 / 2 of cistern in 25 minutes . in how many minutes , it can fill 1 / 2 of the cistern ?
"required time = 25 * 2 * 1 / 2 = 25 minutes answer is d"
a ) 10 min , b ) 15 min , c ) 20 min , d ) 25 min , e ) 30 min
d
divide(25, 1)
divide(n2,n0)|
physics
D
a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 5 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the median of set r ?
the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ) , the median is given to be 3 / 4 * b : ( a + b ) / 2 = ( 3 / 4 ) * b a = b / 2 for set q , consisting of numbers ( b , b + 1 , . . . , c ) , the median is given to be 5 / 8 * c : ( b + c ) / 2 = ( 5 / 8 ) * c b = ( 1 / 4 ) * c for set r , consisting of numbers ( a , a + 1 , . . . c ) , the median needs to be found : a = b / 2 = ( 1 / 4 * c ) / 2 = ( 1 / 8 ) * c median = ( a + c ) / 2 = ( 1 / 8 * c + c ) / 2 = ( 9 / 8 ) * c / 2 = ( 9 / 16 ) * c ( answer c )
a ) 3 / 8 , b ) 1 / 2 , c ) 9 / 16 , d ) 5 / 7 , e ) 3 / 4
c
divide(add(const_1, divide(subtract(multiply(divide(5, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2)
divide(n2,n3)|divide(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,const_1)|subtract(#3,const_1)|divide(const_1,#5)|divide(#4,#6)|add(#7,const_1)|divide(#8,const_2)
general
C
when a laptop is sold for rs . 27000 , the owner loses 10 % . at what price must that laptop be sold in order to gain 10 % ?
"90 : 27000 = 110 : x x = ( 27000 x 110 ) / 90 = 33000 . hence , s . p . = rs . 33,000 . answer : option a"
a ) 33,000 , b ) 34,000 , c ) 35,000 , d ) 36,000 , e ) 37,000
a
multiply(divide(multiply(multiply(multiply(add(const_3, const_4), add(const_3, const_4)), const_100), multiply(add(const_3, const_2), const_2)), subtract(const_100, 10)), add(const_100, 10))
add(n1,const_100)|add(const_3,const_4)|add(const_2,const_3)|subtract(const_100,n1)|multiply(#1,#1)|multiply(#2,const_2)|multiply(#4,const_100)|multiply(#6,#5)|divide(#7,#3)|multiply(#0,#8)|
gain
A
a vending machine randomly dispenses 4 different types of fruit candy . there are twice as many apple candies as orange candies , twice as many strawberry candies as grape candies , and twice as many apple candies as strawberry candies . if each candy cost $ 0.1 , and there are exactly 90 candies , what is the minimum amount of money required to guarantee that you would buy at least 3 of each type of candy ?
let number of apple , orange , strawberry and grape candies be a , o , s and g respectively . a = 2 o s = 2 g a = 2 s a = 4 g a + o + s + g = 90 = > a + a / 2 + a / 2 + a / 4 = 90 = > 9 a / 4 = 90 = > a = 40 o = 20 s = 20 g = 10 cost of each candy = . 1 $ mininum amount of money required to guarantee that you would buy at least three of each type of candy we can buy 40 apple candies , 20 orange candies , 20 strawberry candies and 3 grape candies to ensure atleast 3 of each type of candies . total = 83 candies amount required = 83 * . 1 = 8.30 $ answer a
a ) $ 8.30 , b ) $ 20.75 , c ) $ 22.50 , d ) $ 42.75 , e ) $ 45.00
a
multiply(0.1, add(add(add(multiply(divide(90, add(4, divide(const_1, const_2))), const_2), divide(90, add(4, divide(const_1, const_2)))), divide(90, add(4, divide(const_1, const_2)))), 3))
divide(const_1,const_2)|add(n0,#0)|divide(n2,#1)|multiply(#2,const_2)|add(#2,#3)|add(#4,#2)|add(n3,#5)|multiply(n1,#6)
general
A
convert 600 miles into meters ?
"1 mile = 1609.34 meters 600 mile = 600 * 1609.34 = 965604 meters answer is c"
a ) 784596 , b ) 845796 , c ) 965604 , d ) 784596 , e ) 864520
c
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 600), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
C
a bag contains 3 red , 6 yellow and 8 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ?
"total number of balls = 3 + 6 + 8 = 17 n ( s ) = 17 c 3 = 680 n ( e ) = 3 c 1 * 6 c 1 * 8 c 1 = 144 probability = 144 / 680 = 18 / 85 answer is d"
a ) 2 / 7 , b ) 3 / 5 , c ) 3 / 11 , d ) 18 / 85 , e ) 7 / 16
d
divide(multiply(choose(const_4.0, const_2), choose(add(const_3.0, 6), const_1)), choose(add(add(3, 6), 8), 3))
add(n0,n1)|choose(const_4.0,const_2)|add(n2,#0)|choose(#0,const_1)|choose(#2,n0)|multiply(#1,#3)|divide(#5,#4)|
probability
D
john purchased a grinder and a mobile for rs . 15000 & rs . 10000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 10 % . overall how much he make a profit .
"let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 10000 ( 1 + 10 / 100 ) = 10000 + 1000 total sp - total cp = r + m - ( 15000 + 10000 ) = - 600 + 1000 = rs . 400 as this is positive , an overall profit of rs . 400 was made . a"
a ) s . 400 , b ) s . 120 , c ) s . 420 , d ) s . 450 , e ) s . 290
a
subtract(multiply(10, divide(10000, const_100)), multiply(4, divide(15000, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)|
gain
A
subtracting 2 % of a from a is equivalent to multiplying a by how much ?
answer let a - 2 % of a = ab . ⇒ ( 98 x a ) / 100 = ab ∴ b = 0.98 correct option : a
a ) 0.98 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none
a
divide(subtract(const_100, 2), const_100)
subtract(const_100,n0)|divide(#0,const_100)
general
A
the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 2500 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ?
"x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 2500 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 2500 ) * t = = = > 3 d = ts + 2500 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 2500 t - 2 d = - 2 st + 2500 t 2 d = 2 st - 2500 t d = st - 1250 t x 13 : d = s * t st - 1250 t = s * t s - 1250 = s - 625 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 2500 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 2500 2 = 1250 d = s * t d = 1250 * ( 1 / 2 ) d = 625 answer : b"
a ) 500 , b ) 625 , c ) 1,000 , d ) 1,500 , e ) 2,500
b
multiply(divide(2500, 2), divide(const_1, 2))
divide(n8,n3)|divide(const_1,n3)|multiply(#0,#1)|
physics
B
the average mark of the students of a class in a particular exam is 72 . if 5 students whose average mark in that exam is 40 are excluded , the average mark of the remaining will be 92 . find the number of students who wrote the exam ?
"let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 92 ( x - 5 ) 72 x - ( 5 * 40 ) = 92 ( x - 5 ) 260 = 20 x = > x = 13 answer : b"
a ) 12 , b ) 13 , c ) 14 , d ) 16 , e ) 15
b
divide(subtract(multiply(92, 5), multiply(5, 40)), subtract(92, 72))
multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|
general
B
three machines , each working at the same constant rate , together can complete a certain job in 18 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 9 days ?
"rate of one machine = 1 job / ( 18 * 3 ) days let x = number of machines needed to complete the job in 9 days 1 / ( 3 * 18 ) * 9 * x = 1 job x = 6 6 - 3 = 3 answer : a"
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 8
a
subtract(divide(multiply(18, add(const_4, const_1)), 9), add(const_4, const_1))
add(const_1,const_4)|multiply(n0,#0)|divide(#1,n1)|subtract(#2,#0)|
physics
A
one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , and 5 / 8 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are 72 workers in the palabras bookstore , how many of them have read both books ?
"there are total 72 workers . one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , so 18 have read saramago . 5 / 8 of the workers have read the latest book by h . kureishi . so ( 5 / 8 ) * 72 = 45 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books , 18 - b have read saramago but not kureishi . so , ( 18 - b - 1 ) have read neither . total = n ( a ) + n ( b ) - both + neither 72 = 18 + 45 - b + ( 18 - b - 1 ) b = 4 answer ( e )"
a ) 13 , b ) 12 , c ) 9 , d ) 8 , e ) 4
e
divide(subtract(subtract(add(add(divide(72, const_4), divide(72, const_4)), multiply(divide(72, 8), 5)), 72), const_1), const_2)
divide(n2,const_4)|divide(n2,n1)|add(#0,#0)|multiply(n0,#1)|add(#2,#3)|subtract(#4,n2)|subtract(#5,const_1)|divide(#6,const_2)|
general
E
the sum of the ages of 4 children born at the intervals of 3 years each is 36 years . what is the age of the youngest child ?
"let x = the youngest child . each of the other four children will then be x + 3 , x + 6 , x + 9 we know that the sum of their ages is 36 so , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) = 36 therefore the youngest child is 4.5 years old answer : c"
a ) 22 , b ) 18 , c ) 4.5 , d ) 99 , e ) 38
c
divide(subtract(divide(36, divide(4, 3)), multiply(subtract(4, const_1), 3)), 3)
divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)|
general
C
i remember during the school days , the teacher asked the class ` ` can you tell me the sum of the first 50 odd numbers ? ` ` . i ran quickly to the teacher and told her ` ` the answer is 2500 ' ' . the teacher replied ` ` lucky guess ' ' . she then asked me ` ` can you tell me the sum of first 75 odd numbers ? ` ` . i wait for approx 10 seconds and replied with the correct answer . how can i answer so quickly and whats the correct answer ?
solution : 5625 n ^ 1 75 * 75 = 5625 ( sum of first 75 odd numbers ) . 50 * 50 = 2500 ( sum of first 50 odd numbers ) . answer b
a ) 5624 , b ) 5625 , c ) 5626 , d ) 5627 , e ) none
b
power(75, const_2)
power(n2,const_2)
physics
B
a certain fruit stand sold apples for $ 0.80 each and bananas for $ 0.70 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.50 , what total number of apples and bananas did the customer purchase ?
let ' s start with 1 apple for $ 0.80 . let ' s subtract $ 0.80 from $ 6.50 until we get a multiple of $ 0.70 . $ 6.50 , $ 5.70 , $ 4.90 = 7 * $ 0.70 the customer purchased 7 bananas and 2 apples . the answer is b .
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
b
add(divide(subtract(6.5, multiply(0.8, const_2)), 0.7), const_2)
multiply(n0,const_2)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_2)
other
B
how many bricks , each measuring 80 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ?
"number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 80 x 11.25 x 6 ) = 2000 answer : b"
a ) 6400 , b ) 2000 , c ) 5500 , d ) 7400 , e ) 3000
b
divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(80, 11.25), 6))
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|
physics
B
what is the sum of mean and median of 17 , 45 , 59 , 113 , 76 , 95 , 59 , 35 , 1236 ?
mean = ( 17 + 45 + 59 + 113 + 76 + 95 + 59 + 35 + 12 + 36 ) / 10 = 547 / 10 = 54.7 answer is c
a ) 50.2 , b ) 45.8 , c ) 54.7 , d ) 61.8 , e ) 10.5
c
subtract(113, 59)
subtract(n3,n2)
general
C
the salary of a , b , c , d , e is rs . 9000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is
"answer average salary = 9000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8200 correct option : b"
a ) rs . 7000 , b ) rs . 8200 , c ) rs . 8500 , d ) rs . 9000 , e ) none
b
divide(add(add(add(add(9000, 5000), 11000), 7000), 9000), 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
B
1 * 2 + 2 * 2 ^ 2 + 3 * 2 ^ 3 + 4 * 2 ^ 4 … … … … … … … … + 2012 * 2 ^ 2012
take 2 outside as common we get 2 ( 1 + 2 * 1 ^ 2 + 3 * 1 ^ 3 + . . . . . . . . . 2012 * 1 ^ 2012 ) then 2 ( 1 + 2 + . . . . . 2012 ) we get 8100312 answer : a
a ) 8100312 , b ) 8100313 , c ) 8100412 , d ) 8200312 , e ) 8300312
a
multiply(multiply(2, add(2012, const_1)), 2012)
add(n11,const_1)|multiply(n1,#0)|multiply(n11,#1)
general
A
a , b and c can do a work in 90 , 45 and 6 days respectively . if they work together , in how many days will they complete the work ?
one days ' s work of a , b and c = 1 / 90 + 1 / 45 + 1 / 6 = ( 1 + 2 + 15 ) / 90 = 1 / 5 a , b and c together can do the work in 5 days . answer : a
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 11
a
inverse(add(divide(const_1, 6), add(divide(const_1, 90), divide(const_1, 45))))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)
physics
A
how long will it take a train travelling at 68 kmph to pass an electric pole if the train is 170 m long
"sol . speed = [ 68 x 5 / 18 ] m / sec = 18.9 m / sec . time taken = ( 170 / 18.9 ) sec = 8.99 sec . answer c"
a ) 6.99 , b ) 5.99 , c ) 8.99 , d ) 4.99 , e ) 9.99
c
multiply(divide(divide(170, const_1000), 68), const_3600)
divide(n1,const_1000)|divide(#0,n0)|multiply(#1,const_3600)|
physics
C
a and b invests rs . 5000 and rs . 6000 in a business . after 4 months , a withdraws half of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ?
"a : b ( 5000 * 4 ) + ( 2500 * 8 ) : ( 6000 * 6 ) + ( 4000 * 6 ) 40000 : 60000 2 : 3 answer : a"
a ) 2 : 3 , b ) 82 : 31 , c ) 32 : 45 , d ) 34 : 89 , e ) 35 : 21
a
divide(add(multiply(5000, 4), multiply(divide(6000, const_3), multiply(2, 4))), add(multiply(6000, multiply(2, const_3)), multiply(subtract(6000, divide(6000, const_3)), multiply(2, const_3))))
divide(n1,const_3)|multiply(n0,n2)|multiply(n3,n2)|multiply(n3,const_3)|multiply(#0,#2)|multiply(n1,#3)|subtract(n1,#0)|add(#1,#4)|multiply(#3,#6)|add(#5,#8)|divide(#7,#9)|
gain
A
to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ?
"let the capacity of 1 bucket = x . then , the capacity of tank = 25 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 25 x ) / ( 2 x / 5 ) = ( 25 x ) x 5 / 2 x = 125 / 2 = 62.5 answer is a ."
a ) 62.5 , b ) 65 , c ) 61 , d ) 64 , e ) 60
a
divide(multiply(25, add(const_4, const_1)), const_2)
add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)|
physics
A
70 % of the employees of a company are men . 50 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ?
"no of employees = 100 ( say ) men = 70 women = 30 men speaking french = 0.5 * 70 = 35 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 35 = 5 and women not speaking french = 30 - 5 = 25 % of women not speaking french = 25 / 30 * 100 = 83.33 % answer c"
a ) 4 % , b ) 10 % , c ) 83.33 % , d ) 90.33 % , e ) 20 %
c
multiply(divide(subtract(divide(subtract(const_100, 70), const_100), subtract(divide(40, const_100), multiply(divide(70, const_100), divide(50, const_100)))), divide(subtract(const_100, 70), const_100)), const_100)
divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)|
gain
C
the average of 11 results is 60 . if the average of first 6 results is 58 and that of the last 6 is 63 , find the sixth result ?
sixth result = 58 * 6 + 63 * 6 - 60 * 11 = 66 answer is e
a ) 50 , b ) 52 , c ) 65 , d ) 42 , e ) 66
e
subtract(add(multiply(6, 58), multiply(6, 63)), multiply(11, 60))
multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)
general
E
after decreasing 24 % in the price of an article costs rs . 988 . find the actual cost of an article ?
"cp * ( 76 / 100 ) = 988 cp = 13 * 100 = > cp = 1300 answer : c"
a ) 1667 , b ) 6789 , c ) 1300 , d ) 6151 , e ) 1421
c
divide(988, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
C
a train 220 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is :
"speed of the train relative to man = 220 / 20 = 11 m / sec . = 11 x 18 / 5 km / hr = 198 / 5 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 198 / 540.6 km / hr . answer : b"
a ) 41 , b ) 40.6 , c ) 40.8 , d ) 42 , e ) 42.6
b
divide(divide(subtract(220, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778)
multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
B
a person want to give his money of $ 12000 to his 5 children a , b , c , d , e in the ratio 2 : 4 : 3 : 1 : 5 . what is the d + e share ?
"d ' s share = 12000 * 1 / 15 = $ 800 d ' s share = 12000 * 5 / 15 = $ 4000 a + d = $ 4800 answer is e"
a ) $ 1800 , b ) $ 2800 , c ) $ 3800 , d ) $ 5800 , e ) $ 4800
e
multiply(divide(5, add(add(const_2.0, const_3.0), 3)), 12000)
add(n1,n2)|add(n4,#0)|divide(n3,#1)|multiply(n0,#2)|
other
E
it takes 40 identical printing presses 15 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ?
40 printing presses can do 1 / 15 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 15 = 1 / 20 of the job each hour . the answer is c .
a ) 16 , b ) 18 , c ) 20 , d ) 21 , e ) 24
c
divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 15), 30))
divide(const_1000,const_2)|multiply(#0,const_1000)|divide(#1,n0)|divide(#2,n1)|multiply(n3,#3)|divide(#1,#4)
physics
C
the ratio of the ages of 3 people is 4 : 6 : 7 . 8 years ago , the sum of their ages was 44 . find their present ages .
let the present ages be 4 x , 6 x , 7 x . ( 4 x - 8 ) + ( 6 x - 8 ) + ( 7 x - 8 ) = 44 17 x = 68 x = 4 their present ages are 16 , 24 , and 28 . the answer is d .
a ) 28 , 42,49 , b ) 24 , 36,42 , c ) 20 , 30,35 , d ) 16 , 24,28 , e ) 12 , 18,21
d
multiply(6, divide(add(multiply(8, 3), 44), add(add(4, 6), 7)))
add(n1,n2)|multiply(n0,n4)|add(n5,#1)|add(n3,#0)|divide(#2,#3)|multiply(n2,#4)
general
D
find the cost of fencing around a circular field of diameter 18 m at the rate of rs . 2.50 a meter ?
"2 * 22 / 7 * 9 = 56 56 * 2 1 / 2 = rs . 140 answer : d"
a ) 138 , b ) 132 , c ) 150 , d ) 140 , e ) 222
d
multiply(circumface(divide(18, const_2)), 2.50)
divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|
physics
D
bhanu spends 30 % of his income on petrol on scooter 10 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ?
"given 30 % ( income ) = 300 ⇒ ⇒ income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 10 % ( 700 ) = rs . 70 answer : b"
a ) 2287 , b ) 70 , c ) 128 , d ) 797 , e ) 120
b
multiply(subtract(divide(300, divide(30, const_100)), 300), divide(10, const_100))
divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)|
gain
B
how many two - element subsets of { 6,7 , 8,9 } are there that do not contain the pair of elements 8 and 9 ?
required subsets are = { 6,7 } , { 6,8 } , { 6,9 } , { 7,8 } , { 7,9 } = 5 answer : a
a ) 5 , b ) 4 , c ) 3 , d ) 6 , e ) 7
a
subtract(9, const_4)
subtract(n3,const_4)
general
A
find the compound ratio of ( 2 : 5 ) , ( 6 : 11 ) and ( 11 : 2 ) is
"required ratio = 2 / 5 * 6 / 11 * 11 / 2 = 2 / 1 = 6 : 5 answer is d"
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 6 : 5 , e ) 2 : 3
d
multiply(divide(2, 5), multiply(divide(2, 5), divide(6, 5)))
divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)|
other
D
q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 60 , what is the greatest possible value of q minus r ?
"a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 60 . the greatest multiple of 9 which is less than 60 is 54 . the answer is b ."
a ) 52 , b ) 54 , c ) 55 , d ) 56 , e ) 58
b
multiply(reminder(60, subtract(const_10, const_1)), subtract(const_10, const_1))
subtract(const_10,const_1)|reminder(n0,#0)|multiply(#1,#0)|
general
B
a no . when divided by 221 gives a remainder 43 , what remainder will be obtainedby dividingthe same no . 17 ?
"221 + 43 = 264 / 17 = 9 ( remainder ) c"
a ) 3 , b ) 6 , c ) 9 , d ) 11 , e ) 14
c
divide(add(221, 43), 17)
add(n0,n1)|divide(#0,n2)|
general
C
how many 2 digit number contain number 3 ?
"total 2 digit no . = 9 * 10 * = 90 not containing 3 = 8 * 9 = 72 total 2 digit number contain 3 = 90 - 72 = 18 answer : a"
a ) 18 , b ) 28 , c ) 38 , d ) 68 , e ) 58
a
add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(2, 2)), multiply(const_4, const_2))), const_10)
multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)|
general
A
a dishonest dealer professes to sell goods at the cost price but uses a weight of 750 grams per kg , what is his percent ?
"750 - - - 250 100 - - - ? = > 33.33 % answer : b"
a ) 25 % , b ) 33 % , c ) 29 % , d ) 55 % , e ) 45 %
b
subtract(multiply(divide(const_100, 750), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)
add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|
gain
B
a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 360 . how much does c have ?
"a + b + c = 500 a + c = 200 b + c = 360 - - - - - - - - - - - - - - a + b + 2 c = 560 a + b + c = 500 - - - - - - - - - - - - - - - - c = 60 answer : a"
a ) 60 , b ) 78 , c ) 267 , d ) 29 , e ) 27
a
subtract(add(200, 360), 500)
add(n1,n2)|subtract(#0,n0)|
general
A
lilly has 10 fish and rosy has 9 fish . in total , how many fish do they have in all ?
"10 + 9 = 19 the answer is d ."
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20
d
add(10, 9)
add(n0,n1)|
general
D
the manufacturing cost of a shoe is rs . 180 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains
explanation : total cost of a watch = 180 + ( 500 / 100 ) = 185 . gain = 20 % = > sp = 1.2 cp = 1.2 x 185 = 222 answer : a
a ) rs 222 , b ) rs 216 , c ) rs 220 , d ) rs 210 , e ) rs 217
a
add(add(180, divide(500, 100)), multiply(divide(20, const_100), add(180, divide(500, 100))))
divide(n1,n2)|divide(n3,const_100)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)
gain
A
if rs . 595 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ?
"( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 595 * 2 / 17 = rs . 70 b ' s share = 595 * 3 / 17 = rs . 105 c ' s share = 595 * 12 / 17 = rs . 420 . answer : a"
a ) s . 420 , b ) s . 360 , c ) s . 389 , d ) s . 368 , e ) s . 323
a
divide(595, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1))
divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)|
general
A
3 investors , a , b , and c , divide the profits from a business enterprise in the ratio of 5 : 7 : 8 , respectively . if investor a earned $ 2500 , how much money did investors b and c earn in total ?
let profit for a = 5 x profit for b = 7 x profit for c = 8 x investor a earned = 2500 5 x = 2500 = > x = 500 total profit for b and c = 7 x + 8 x = 15 x therefore , 15 x = 15 * 500 = 7,500 answer c
a ) $ 4,000 , b ) $ 4,900 , c ) $ 7,500 , d ) $ 9,500 , e ) $ 10,500
c
add(divide(divide(add(multiply(divide(7, add(add(5, 7), 8)), divide(2500, divide(5, add(add(5, 7), 8)))), multiply(divide(8, add(add(5, 7), 8)), divide(2500, divide(5, add(add(5, 7), 8))))), const_1000), const_2), const_0_25)
add(n1,n2)|add(n3,#0)|divide(n2,#1)|divide(n1,#1)|divide(n3,#1)|divide(n4,#3)|multiply(#2,#5)|multiply(#4,#5)|add(#6,#7)|divide(#8,const_1000)|divide(#9,const_2)|add(#10,const_0_25)
general
C
a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of 42 . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with is ?
sum of length and breadth will be 21 units . 42 / 2 = 21 area will be max when lxb = 11 x 10 = 110 sq units area will be min when lxb = 20 x 1 = 20 sq units . . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with = 110 - 20 = 90 sq units answer : e
['a ) 50 sq units', 'b ) 60 sq units', 'c ) 70 sq units', 'd ) 80 sq units', 'e ) 90 sq units']
e
subtract(multiply(subtract(divide(42, const_2), const_10), const_10), multiply(subtract(divide(42, const_2), const_1), const_1))
divide(n0,const_2)|subtract(#0,const_10)|subtract(#0,const_1)|multiply(#1,const_10)|multiply(#2,const_1)|subtract(#3,#4)
geometry
E
in a garden , 26 trees are planted at equal distances along a yard 500 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ?
"26 trees have 25 gaps between them . length of each gap = 500 / 25 = 20 i . e . , distance between two consecutive trees = 20 answer is a ."
a ) 20 , b ) 8 , c ) 12 , d ) 14 , e ) 16
a
divide(500, subtract(26, const_1))
subtract(n0,const_1)|divide(n1,#0)|
physics
A
a courtyard is 18 meter long and 12 meter board is to be paved with bricks of dimensions 15 cm by 13 cm . the total number of bricks required is :
"explanation : number of bricks = courtyard area / 1 brick area = ( 1800 ã — 1200 / 15 ã — 13 ) = 11076 option c"
a ) 16000 , b ) 18078 , c ) 11076 , d ) 11456 , e ) none of these
c
divide(multiply(multiply(18, const_100), multiply(12, const_100)), multiply(15, 13))
multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)|
physics
C
two goods trains each 125 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ?
"relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 125 + 125 = 250 m . required time = 250 * 6 / 125 = 12 sec . answer : option b"
a ) 40 , b ) 12 , c ) 48 , d ) 51 , e ) 44
b
add(45, 30)
add(n1,n2)|
physics
B
in a group of ducks and cows , the total number of legs are 24 more than twice the number of heads . find the total number of cows .
"let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 24 more than twice the number of heads = > 2 ( d + 2 c ) = 24 + 2 ( c + d ) = > d + 2 c = 12 + c + d = > 2 c = 12 + c = > c = 12 i . e . , total number of cows = 12 answer is a ."
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20
a
divide(24, const_2)
divide(n0,const_2)|
general
A
there are 1200 jelly beans divided between two jars , jar x and jar y . if there are 400 fewer jelly beans in jar x than 3 times the number of beans in jar y , how many beans are in jar x ?
x + y = 1200 so y = 1200 - x x = 3 y - 400 x = 3 ( 1200 - x ) - 400 4 x = 3200 x = 800 the answer is d .
a ) 650 , b ) 700 , c ) 750 , d ) 800 , e ) 850
d
subtract(1200, divide(add(1200, 400), const_4))
add(n0,n1)|divide(#0,const_4)|subtract(n0,#1)
general
D
how long does a train 100 m long running at the speed of 65 km / hr takes to cross a bridge 145 m length ?
"speed = 65 * 5 / 18 = 18 m / sec total distance covered = 100 + 145 = 245 m . required time = 245 / 18 = 13.6 sec . answer : a"
a ) 13.6 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 61.9 sec , e ) 47.98 sec
a
divide(add(100, 145), multiply(65, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
A
an outlet pipe empties a tank which is full in 10 hours . if the inlet pipe is kept open , which lets liquid in at the rate of 8 litres / min then outlet pipe would take 6 hours longer . find the capacity of the tank .
"let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 10 hours = ( x - 480 ) * ( 10 + 6 ) hours 10 x = ( x - 480 ) * 16 - - > x = 1280 - - > capacity = 10 x = 12800 liters . answer : b"
a ) 8600 litres , b ) 12800 litres , c ) 12100 litres , d ) 11200 litres , e ) 13200 litres
b
multiply(divide(multiply(multiply(8, const_60), add(10, 6)), 6), 10)
add(n0,n2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,n2)|multiply(n0,#3)|
physics
B
a man can row downstream at 18 kmph and upstream at 6 kmph . find the speed of the man in still water and the speed of stream respectively ?
let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 6 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 24 = > x = 12 , y = 6 . answer : c
a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9
c
divide(divide(add(18, 6), const_2), const_2)
add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)
physics
C
two trains of length 160 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?
"relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 160 + 280 = 440 m . the time required = d / s = 440 / 20 = 22 sec . answer : e"
a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 22
e
divide(add(160, 280), multiply(add(42, 30), const_0_2778))
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
E
the product of two numbers is 120 and the sum of their squares is 289 . the sum of the number is :
"let the numbers be x and y . then , xy = 120 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 120 ) = 529 x + y = 529 = 23 . answer a"
a ) 23 , b ) 20 , c ) 12 , d ) 128 , e ) 171
a
sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120)))
multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)|
general
A
thirty percent of the women in a college class are science majors , and the non - science majors make up 60 % of the class . what percentage of the men are science majors if 40 % of the class are men ?
science majors make up 0.4 of the class . 60 % of the class are women and 0.3 * 0.6 = 0.18 of the class are female science majors . then 0.22 of the class are male science majors . 0.4 x = 0.22 x = 0.55 = 55 % the answer is c .
a ) 35 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 %
c
multiply(divide(subtract(subtract(const_100, 60), multiply(divide(multiply(const_10, const_3), const_100), subtract(const_100, 40))), 40), const_100)
multiply(const_10,const_3)|subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|multiply(#3,#2)|subtract(#1,#4)|divide(#5,n1)|multiply(#6,const_100)
gain
C
a certain telescope increases the visual range at a particular location from 100 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ?
"original visual range = 100 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 100 ) / 100 * 100 % = 1 / 2 * 100 % = 50 % answer e"
a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 50 %
e
multiply(divide(subtract(150, 100), 100), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
general
E
calculate the dividend from caleb ' s stock , if caleb he invested rs . 2500 at 79 to obtain an income of rs . 200 .
by investing rs . 2500 , income = rs . 200 by investing rs . 79 , income = 200 ã — 79 / 2500 = 6.32 ie , dividend = 6.32 % answer is a .
a ) 6.32 , b ) 5.32 , c ) 7.32 , d ) 2.32 , e ) 1.32
a
divide(200, divide(2500, 79))
divide(n0,n1)|divide(n2,#0)
gain
A
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 70 / 10 sq . metre , whose base & perpendicular distance from the other side being 84 m & 24 m respectively ?
"area of the parallelogram = length of the base * perpendicular height = 84 * 24 = 2016 m . total cost of levelling = rs . 2016 b"
a ) s . 2400 , b ) s . 2016 , c ) s . 1400 , d ) s . 3480 , e ) s . 2000
b
multiply(multiply(84, 24), divide(70, 10))
divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1)|
physics
B
there has been successive increases of 25 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ?
"let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.25 * p = x * p y = x / ( 1.1 * 1.25 ) which is about 0.73 x which is a decrease of about 27 % . the answer is b ."
a ) 23 % , b ) 27 % , c ) 31 % , d ) 35 % , e ) 39 %
b
multiply(subtract(const_1, divide(const_100, add(add(const_100, 25), divide(multiply(add(const_100, 25), 10), const_100)))), const_100)
add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100)|
general
B
what is the sum of all the multiples of 10 between 0 and 100 ?
"the multiples of 10 between 0 and 100 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 and 100 . if these are all added together , the result is 550 . final answer : c"
a ) 500 , b ) 620 , c ) 550 , d ) 340 , e ) 440
c
add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4))
add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)|
general
C
it costs $ 4 for the first 1 / 4 hour to use the laundry machine at the laundromat . after the first ¼ hour it costs $ 12 per hour . if a certain customer uses the laundry machine for 3 hours and 25 minutes , how much will it cost him ?
"3 hrs 25 min = 205 min first 15 min - - - - - - > $ 4 time left is 190 min . . . now , 60 min costs $ 12 1 min costs $ 12 / 60 190 min costs $ 12 / 60 * 190 = > $ 38 so , total cost will be $ 38 + $ 4 = > $ 42 hene answer will be ( e ) $ 42"
a ) $ 25 , b ) $ 32 , c ) $ 36 , d ) $ 40 . , e ) $ 42 .
e
add(add(multiply(12, 3), 4), multiply(divide(12, const_60), divide(25, 4)))
divide(n3,const_60)|divide(n5,n2)|multiply(n3,n4)|add(n0,#2)|multiply(#0,#1)|add(#3,#4)|
physics
E
the volume of a cube is 1000 cc . find its surface .
"a 3 = 1000 = > a = 10 6 a 2 = 6 * 10 * 10 = 600 answer : d"
a ) 864 , b ) 556 , c ) 255 , d ) 600 , e ) 267
d
surface_cube(cube_edge_by_volume(1000))
cube_edge_by_volume(n0)|surface_cube(#0)|
geometry
D
xavier starts from p towards q at a speed of 60 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 60 km , then how much time does he take to cover the distance ?
"first 12 min = 60 * 12 / 60 = 12 km 2 nd 12 min = 70 * 12 / 60 = 14 km 3 rd 12 min = 80 * 12 / 60 = 16 km 4 th 12 min = 90 * 12 / 60 = 18 km total time 12.4 = 48 min a"
a ) 48 , b ) 59 , c ) 60 , d ) 56 , e ) 50
a
add(add(add(12, 12), 12), 12)
add(n1,n1)|add(n1,#0)|add(n1,#1)|
physics
A
x starts a business with rs . 45000 . y joins in the business after 7 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ?
"explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 ã — 12 : 30000 ã — 5 = 45 ã — 12 : 30 ã — 5 = 3 ã — 12 : 2 ã — 5 = 18 : 5 answer : option b"
a ) 1 : 2 , b ) 18 : 5 , c ) 1 : 5 , d ) 3 : 1 , e ) 1 : 1
b
divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3)))
add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)|
other
B
a 15 by 20 rectangle is inscribed in circle . what is the circumference of the circle ?
"the diagonal of the rectangle will be the diameter of the circle . and perimeter = 2 * pi * r ans : e"
a ) 5 π , b ) 10 π , c ) 15 π , d ) 20 π , e ) 25 π
e
circumface(divide(sqrt(add(power(15, const_2), power(20, const_2.0))), 20))
power(n0,const_2)|power(n1,const_2.0)|add(#0,#1)|sqrt(#2)|divide(#3,n1)|circumface(#4)|
geometry
E
if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 3 ø 1 ) ø 1 = ?
"3 ø 1 = 2 ^ 3 / 2 ^ 1 = 4 4 ø 1 = 2 ^ 4 / 2 = 8 the answer is c ."
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32
c
divide(power(2, divide(power(2, 3), power(2, 2))), power(2, 1))
power(n0,n2)|power(n0,n0)|power(n0,n4)|divide(#0,#1)|power(n0,#3)|divide(#4,#2)|
general
C
tough and tricky questions : word problems . ak car company wants to manufacture a new car known as model x , and it is trying to determine how many cars it needs to sell in order to make an annual profit of $ 30 , 500000 . the annual fixed costs for model x total $ 50 , 200000 . in addition , each model x car has an average cost of $ 5000 per vehicle to manufacture . if the company forecasts it will sell 20000 model x cars this year , at what price must the company sell the car to achieve the desired annual profit ?
the ak company plans to make a profit of $ 30 , 500000 and it needs to earn the fixed costs of $ 50 , 200000 , so it needs to earn $ 80 , 700000 on top of the costs for the production of the 20000 cars . therefore , price of each car needs to be the unit costs plus $ 80 , 700000 / 20000 . after removing the 0 ' s , we end with the unit costs plus $ 8070 / 2 , which is $ 5000 plus $ 4035 . therefore , answer c is correct .
a ) $ 4035 , b ) $ 4036 , c ) $ 9035 , d ) $ 16140 , e ) $ 36140
c
divide(add(add(add(multiply(50, multiply(const_1000, const_1000)), 200000), multiply(20000, 5000)), add(multiply(30, multiply(const_1000, const_1000)), 500000)), 20000)
multiply(const_1000,const_1000)|multiply(n4,n5)|multiply(n2,#0)|multiply(n0,#0)|add(n3,#2)|add(n1,#3)|add(#4,#1)|add(#6,#5)|divide(#7,n5)
general
C
a man can do a job in 10 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ?
"1 day work of the three persons = ( 1 / 10 + 1 / 20 + 1 / 25 ) = 19 / 100 so , all three together will complete the work in 100 / 19 = 5.3 days . answer : c"
a ) 6.3 , b ) 6.9 , c ) 5.3 , d ) 6.1 , e ) 6.2
c
divide(const_1, add(divide(const_1, 25), add(divide(const_1, 10), divide(const_1, 20))))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4)|
physics
C
m is 30 % of q . q is 20 % of p . n is 50 % of p . find m : n ratio
let p be 100 n = 50 % of 100 ( p = 100 ) = 50 q = 20 % of 100 ( p = 100 ) = 20 m = 30 % of 20 ( q = 20 ) = 6 m : n = 6 : 50 m : n = 3 : 25 answer : c
a ) 1 : 25 , b ) 2 : 25 , c ) 3 : 25 , d ) 4 : 25 , e ) 5 : 25
c
divide(multiply(divide(30, const_100), divide(20, const_100)), divide(50, const_100))
divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|divide(#3,#2)
other
C
a man can row 10 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?
m = 10 s = 1.2 ds = 11.2 us = 8.8 x / 11.2 + x / 8.8 = 1 x = 4.93 d = 4.93 * 2 = 9.86 answer : c
a ) 6.24 km , b ) 6 km , c ) 9.86 km , d ) 5.66 km , e ) 10 km
c
multiply(divide(multiply(add(10, 1.2), subtract(10, 1.2)), add(add(10, 1.2), subtract(10, 1.2))), const_2)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)
physics
C
if 5 a = 3125 , then the value of 5 ( a - 3 ) is
"sol . 5 a = 3125 = 5 a = 55 a = 5 . 5 ( 1 - 3 ) = 5 ( 5 - 3 ) = 5 ( 2 ) = 25 . answer a"
a ) 25 , b ) 50 , c ) 250 , d ) 300 , e ) none
a
multiply(3, multiply(divide(5, 5), divide(5, 3125)))
divide(n2,n0)|divide(n2,n1)|multiply(#0,#1)|multiply(n3,#2)|
general
A
the ratio of the number of females to males at a party was 1 : 2 but when 4 females and 4 males left , the ratio became 1 : 3 . how many people were at the party originally ?
"the total number of people are x females + 2 x males . 3 * ( x - 4 ) = 2 x - 4 x = 8 there were 3 x = 24 people at the party originally . the answer is c ."
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28
c
add(divide(subtract(multiply(multiply(4, 2), 3), multiply(4, 2)), 2), subtract(multiply(multiply(4, 2), 3), multiply(4, 2)))
multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,n1)|add(#3,#2)|
other
C
162 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ?
"let b be the number of boys in the school . 162 = xb / 100 b = 0.5 x 16200 = 0.5 x ^ 2 x ^ 2 = 32400 x = 180 the answer is c ."
a ) 120 , b ) 150 , c ) 180 , d ) 250 , e ) 440
c
sqrt(divide(multiply(162, const_100), divide(50, const_100)))
divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)|
gain
C
how many seconds will a 770 m long train take to cross a man walking with a speed of 2.5 km / hr in the direction of the moving train if the speed of the train is 46.5 km / hr ?
"speed of train relative to man = 46.5 - 2.5 = 44 km / hr . = 44 * 5 / 18 = 110 / 9 m / sec . time taken to pass the man = 770 * 9 / 110 = 63 sec . answer : a"
a ) 63 , b ) 88 , c ) 66 , d ) 44 , e ) 35
a
divide(770, multiply(subtract(46.5, 2.5), const_0_2778))
subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
A
stacy has a 63 page history paper due in 3 days . how many pages per day would she have to write to finish on time ?
"63 / 3 = 21 answer : e"
a ) 9 , b ) 8 , c ) 10 , d ) 8.5 , e ) 21
e
divide(63, 3)
divide(n0,n1)|
physics
E
find the 21 th term of an arithmetic progression whose first term is 3 and the common difference is 5 .
"n th term of a . p = a + ( n - 1 ) * d = 3 + ( 21 - 1 ) * 5 , = 3 + 100 = 103 . answer : c"
a ) 100 , b ) 102 , c ) 103 , d ) 150 , e ) 46
c
add(multiply(subtract(21, const_1), 5), 3)
subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1)|
general
C
a work can be finished in 13 days by ten women . the same work can be finished in fifteen days by ten men . the ratio between the capacity of a man and a woman is
work done by 10 women in 1 day = 1 / 13 work done by 1 woman in 1 day = 1 / ( 13 ã — 10 ) work done by 16 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 ã — 10 ) ratio of the capacity of a man and woman = 1 / ( 15 ã — 10 ) : 1 / ( 13 ã — 10 ) = 1 / 15 : 1 / 13 = 1 / 15 : 1 / 13 = 13 : 15 answer is b .
a ) 14 : 13 , b ) 13 : 15 , c ) 12 : 13 , d ) 13 : 12 , e ) 14 : 15
b
divide(divide(const_1, multiply(add(13, const_2), const_10)), divide(const_1, multiply(13, const_10)))
add(n0,const_2)|multiply(n0,const_10)|divide(const_1,#1)|multiply(#0,const_10)|divide(const_1,#3)|divide(#4,#2)
physics
B
how many numbers from 2 to 11 are exactly divisible by 2 ?
"2 / 2 = 1 and 11 / 2 = 5 5 - 1 = 4 4 + 1 = 5 numbers . answer : c"
a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 7 , e ) e ) 8
c
add(divide(subtract(multiply(floor(divide(11, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1)
divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)|
general
C
tom drives from town q to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between q and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per hour ?
let ' s assume that it takes 4 hours to go from point q to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b .
a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36
b
divide(multiply(60, 36), multiply(36, const_3))
multiply(n0,n1)|multiply(n1,const_3)|divide(#0,#1)
physics
B
a man has some hens and cows . if the number of heads be 44 and the number of feet equals 128 , then the number of hens will be
"explanation : let number of hens = h and number of cows = c number of heads = 44 = > h + c = 44 - - - ( equation 1 ) number of feet = 128 = > 2 h + 4 c = 128 = > h + 2 c = 64 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 64 - 44 = > c = 20 substituting the value of c in equation 1 , we get h + 22 = 44 = > h = 44 - 20 = 24 i . e . , number of hens = 24 answer : b"
a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 28
b
divide(subtract(multiply(44, const_4), 128), const_2)
multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|
general
B
what is 1 percent of 12,356 ?
since , percent = 1 / 100 , what = something ( s ) , and is : = . we can write the question as s = 1 ( 1 / 100 ) 12,356 . the answer is 123.56 . hence , the correct answer is a .
a ) 123.56 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356
a
divide(multiply(1, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
A
linda spent 3 / 5 of her savings on furniture and the rest on a tv . if the tv cost her $ 400 , what were her original savings ?
"if linda spent 3 / 5 of her savings on furnitute , the rest 5 / 5 - 3 / 5 = 2 / 5 on a tv but the tv cost her $ 400 . so 2 / 5 of her savings is $ 400 . so her original savings are 5 / 2 times $ 400 = $ 2000 / 2 = $ 1000 correct answer d"
a ) $ 900 , b ) $ 300 , c ) $ 600 , d ) $ 1000 , e ) $ 800
d
divide(400, subtract(const_1, divide(3, 5)))
divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|
general
D
if x is an integer that has exactly 3 positive divisors ( these include 1 and x ) , how many positive divisors does x ^ 3 have ?
since factors of ' 4 ' are { 1 , 2,4 } 4 ^ 3 = 64 number of factors / divisors of 64 = 2 ^ 6 we know that when a number is expressed as a product of the prime factors as below : n = a ^ x * b ^ y * c ^ z then no . of divisors = ( x + 1 ) * ( y + 1 ) * ( z + 1 ) then here ( 6 + 1 ) = 7 answer : d
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
d
add(add(3, 3), 1)
add(n0,n0)|add(n1,#0)
general
D
3 / 4 of 1 / 2 of 2 / 5 of 5060 = ?
"b 759 ? = 5060 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 759"
a ) 392 , b ) 759 , c ) 753 , d ) 493 , e ) 540
b
multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5060)
divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)|
general
B
if the areas of the three adjacent faces of a cuboidal box are 120 cm ² , 72 cm ² and 60 cm ² respectively , then find the volume of the box .
sol . let the length , breadth and height of the box be , l , b , and h respectively , then , volume = lbh = √ ( lbh ) ² = √ lb * bh * lh = √ 120 * 72 * 60 = 720 cm ³ answer d
['a ) 120 m ³', 'b ) 400 m ³', 'c ) 660 m ³', 'd ) 720 cm m ³', 'e ) none']
d
volume_rectangular_prism(sqrt(divide(multiply(120, 60), 72)), divide(120, sqrt(divide(multiply(120, 60), 72))), divide(60, sqrt(divide(multiply(120, 60), 72))))
multiply(n0,n2)|divide(#0,n1)|sqrt(#1)|divide(n0,#2)|divide(n2,#2)|volume_rectangular_prism(#3,#4,#2)
geometry
D
if | 5 x - 30 | = 100 , then find the sum of the values of x ?
"| 5 x - 30 | = 100 5 x - 30 = 100 or 5 x - 30 = - 100 5 x = 130 or 5 x = - 70 x = 26 or x = - 14 sum = 26 - 14 = 12 answer is c"
a ) 1 , b ) - 2 , c ) 12 , d ) - 3 , e ) 4
c
subtract(subtract(subtract(100, 30), add(100, 30)), 30)
add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)|
general
C
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 17 cm
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 17 ) = 323 cm 2 answer : b"
a ) 178 cm 2 , b ) 323 cm 2 , c ) 285 cm 2 , d ) 167 cm 2 , e ) 197 cm 2
b
quadrilateral_area(17, 18, 20)
quadrilateral_area(n2,n1,n0)|
physics
B
a soccer store typically sells replica jerseys at a discount of 30 percent to 70 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the lowest possible sale price ?
"let the list price be 2 x for min sale price , the first discount given should be 70 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , d is the answer"
a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50
d
divide(80, const_2)
divide(n3,const_2)|
general
D
a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 40 % , what should be the sale price of the mixed tea ?
c . p . of mixture = 80 × 15 + 20 × 20 / 80 + 20 = 16 ∴ s . p . = ( 100 + 40 ) / 100 × 16 = 22.4 answer a
a ) 22.4 , b ) 22 , c ) 20 , d ) 19.2 , e ) none of these
a
add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 40), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n2)|add(#1,#2)|divide(#3,#0)|multiply(n4,#4)|divide(#5,const_100)|add(#6,#4)
gain
A
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is b ' s share ?
"let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 b ' s share = 2 x = 2 * $ 500 = $ 1000 the answer is b ."
a ) $ 500 , b ) $ 1000 , c ) $ 1500 , d ) $ 2000 , e ) $ 2500
b
multiply(multiply(subtract(4, 3), 500), 3)
subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)|
general
B
what is the value of n if the sum of the consecutive odd intergers e from 1 to n equals 169 ?
"before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n / 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is ."
a ) 47 , b ) 25 , c ) 37 , d ) 33 , e ) 29
b
add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1)
multiply(n0,const_2)|sqrt(n1)|multiply(#1,const_2)|subtract(#2,#0)|add(n0,#3)|
general
B
how many numbers between 190 and 580 are divisible by 4,5 and 6 ?
every such number must be divisible by l . c . m of 4 , 5,6 i . e , 60 . such numbers are 240,300 , 360,420 , 480,540 . clearly , there are 6 such numbers answer : a
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
a
subtract(floor(divide(580, lcm(lcm(const_4, add(const_4, const_1)), 6))), floor(divide(190, lcm(lcm(const_4, add(const_4, const_1)), 6))))
add(const_1,const_4)|lcm(#0,const_4)|lcm(n3,#1)|divide(n1,#2)|divide(n0,#2)|floor(#3)|floor(#4)|subtract(#5,#6)
general
A
there are 250 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers .
as per question stem 2 / 5 m ( portion of men employees who are managers ) + 250 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 250 = 2 / 5 t , or 2 / 5 ( t - m ) = 250 , from here we get that t - m = 625 , that would be total number of female employees and the answer ( c )
a ) 325 , b ) 425 , c ) 625 , d ) 700 , e ) none of these
c
divide(250, divide(2, 5))
divide(n1,n2)|divide(n0,#0)|
general
C