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
two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer e between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ?
"since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 11,12 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11,13 ) , ( 7,17 ) , ( 5 , 19 ) 2 , 3 , 5 , 7 , 11 , 13 , 15,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 13 , 17 ) , ( 11 , 19 ) , ( 7 , 23 ) 2 , 3 , 5 , 7 , 11 , 13,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11 , 23 ) , ( 5 , 29 ) , ( 3 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 18,19 , 23 , 29 , 31 , 37 - four pairs ( 17 , 19 ) , ( 13 , 23 ) , ( 7 , 29 ) , ( 5 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 20,23 , 29 , 31 , 37 - definitely can not be more than 4 since there are only 4 primes more than 20 . so must be less than 4 pairs . ignore . answer ( d ) ."
a ) 12 , b ) 15 , c ) 17 , d ) e = 18 , e ) 20
d
multiply(multiply(3, const_2), 3)
multiply(const_2,n0)|multiply(n0,#0)|
physics
D
when asked what the time is , a person answered that the amount of time left is 3 / 5 of the time already completed . what is the time .
"a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 βˆ’ x = 3 / 5 x β‡’ x = 15 time is 3 pm answer : c"
a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 8 pm , e ) 6 pm
c
subtract(multiply(3, const_12), divide(multiply(3, const_12), add(divide(3, 5), const_1)))
divide(n0,n1)|multiply(const_12,n0)|add(#0,const_1)|divide(#1,#2)|subtract(#1,#3)|
general
C
a man buys a cycle for rs . 1750 and sells it at a loss of 8 % . what is the selling price of the cycle ?
"s . p . = 92 % of rs . 1750 = rs . 92 x 1750 / 100 = rs . 1610 answer : option b"
a ) s . 1090 , b ) s . 1610 , c ) s . 1190 , d ) s . 1202 , e ) s . 1092
b
divide(multiply(subtract(const_100, 8), 1750), const_100)
subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|
gain
B
a man can swim in still water at 15 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?
"m = 15 s = x ds = 15 + x us = 15 - x 15 + x = ( 15 - x ) 2 15 + x = 30 - 2 x 3 x = 15 x = 5 answer : b"
a ) 3.9 , b ) 5 , c ) 5.3 , d ) 1.5 , e ) 5.2
b
divide(15, const_3)
divide(n0,const_3)|
general
B
in a school of 600 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ?
"45 + 23 + 15 = 83 % 100 – 83 = 17 % 600 * 17 / 100 = 102 the answer is a ."
a ) 102 , b ) 112 , c ) 122 , d ) 132 , e ) 142
a
subtract(600, add(add(multiply(divide(45, const_100), 600), multiply(divide(23, const_100), 600)), multiply(divide(15, const_100), 600)))
divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)|
gain
A
find number which is 60 % less than 80 .
"explanation : 60 % less is 40 % of the given number therefore , 40 % of 80 is 32 . answer : b"
a ) 18 , b ) 32 , c ) 28 , d ) 26 , e ) 98
b
divide(multiply(80, 60), const_100)
multiply(n0,n1)|divide(#0,const_100)|
gain
B
a 60 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ?
"explanation : 1 : 2 / 5 = 5 : 2 2 / 7 * 60 = 20 answer : option d"
a ) a ) 73 , b ) b ) 20 , c ) c ) 83 , d ) d ) 17.1 , e ) e ) 52
d
subtract(60, divide(60, add(divide(2, 5), const_1)))
divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|
physics
D
a jar contains only red , yellow , and orange marbles . if there are 1 red , 5 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ?
"i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 1 / 45 . this helped me realise that for the 3 possible orderings the probabbility is the same . so , it should be ( 1 / 45 ) * ( 1 ) , which indeed is 1 / 45 . b"
a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22
b
divide(multiply(choose(5, 2), choose(1, 1)), choose(add(add(5, 4), 1), 1))
add(n1,n2)|choose(n1,n4)|choose(n0,n5)|add(n0,#0)|multiply(#1,#2)|choose(#3,n0)|divide(#4,#5)|
probability
B
in a certain state , the ratio of registered republicans to registered democrats is 3 to 2 , and every registered voter is either a republican or a democrat . if 70 percent of the republicans and 25 percent of the democrats are expected to vote for candidate x , and everyone else is expected to vote for candidate y , by what percent is candidate x expected to win the election ?
since we were expected to find a percentage figure - it thought that it might be easier to pick a ' smart number ' to represent the total number of voters ( republicans and democrats ) . therefore , i picked 100 ( as the total number of voters ) and thus 60 : 40 represents the number ratio of republicans : democrats . if 70 % of republicans ( which is ( 60 * 0.7 ) = 42 ) and 25 % of democrats ( 40 * 0.25 = 10 ) voted for candidate x , means that out of total of 100 voters ; 52 ( 42 + 10 ) voters voted for candidate x and 48 voted for candidate y . thus we can infer that candidate x is expected to win the election by 4 ( 52 - 48 ) votes . therefore candidate x is expected to win the election by ( 4 / 100 ) votes which is equivalent to 4 % . i think the answer is b .
a ) 2 % , b ) 4 % , c ) 8 % , d ) 10 % , e ) 15 %
b
multiply(divide(subtract(add(multiply(divide(25, const_100), 2), multiply(divide(70, const_100), 3)), add(subtract(3, multiply(divide(70, const_100), 3)), subtract(2, multiply(divide(25, const_100), 2)))), add(3, 2)), const_100)
add(n0,n1)|divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#1)|multiply(n0,#2)|add(#3,#4)|subtract(n0,#4)|subtract(n1,#3)|add(#6,#7)|subtract(#5,#8)|divide(#9,#0)|multiply(#10,const_100)
other
B
the market value of a certain machine decreased by 30 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 10,000 , what was its market value two years later ?
"b . market value in 1982 = $ 10000 market value in 1983 = $ 10000 - ( $ 10000 x 30 / 100 ) = 10000 - 3000 = $ 7000 market value in 1984 = market value in 1983 - ( 30 % of $ 10000 ) = 7000 - 3000 = $ 4000"
a ) $ 8,000 , b ) $ 4,000 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 800
b
subtract(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(divide(30, const_100), const_2)))
divide(n0,const_100)|multiply(n0,const_100)|multiply(const_2,const_4)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#3,#4)|subtract(#3,#5)|
gain
B
a sum of money at simple interest amount to rs 720 after 2 years and to rs 1020 after a further period of 5 years . the sum is :
simple interest for 5 year = rs . ( 1020 - 720 ) = rs . 300 . simple interest for 2 year = rs . ( 300 / 5 Γ— 2 ) = rs . 120 . principal = rs . ( 720 - 120 ) = rs . 600 . answer : a
a ) 600 , b ) 450 , c ) 650 , d ) 500 , e ) 550
a
subtract(720, divide(multiply(subtract(1020, 720), 2), 5))
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)
gain
A
how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 45 percent butterfat to obtain milk that is 20 percent butterfat ?
equate the fat : 0.1 x + 0.45 * 8 = 0.2 ( x + 8 ) - - > x = 20 . answer : e .
a ) 6 , b ) 12 , c ) 14 , d ) 16 , e ) 20
e
divide(multiply(subtract(45, 20), 8), 10)
subtract(n2,n3)|multiply(n1,#0)|divide(#1,n0)
general
E
lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 16 more photos than claire . how many photos has claire taken ?
"l = r l = 3 c r = c + 16 3 c = c + 16 c = 8 the answer is b ."
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
b
divide(16, subtract(3, const_1))
subtract(n0,const_1)|divide(n1,#0)|
general
B
the price of lunch for 10 people was $ 207 including a 15 % gratuity for service . what was the average price per person , excluding the gratuity ?
clearly e is the answer i used poe here lets consider option ( e ) 10 * 18 = 180 now 180 ( 115 / 100 ) = 207 = > possible answer imo e
a ) 11.73 , b ) 12 , c ) 13.8 , d ) 14 , e ) 18
e
divide(divide(multiply(207, subtract(const_100, 15)), const_100), const_10)
subtract(const_100,n2)|multiply(n1,#0)|divide(#1,const_100)|divide(#2,const_10)
general
E
martin buys a pencil and a notebook for 80 cents . at the same store , gloria buys a notebook and an eraser for 85 cents , and zachary buys a pencil and an eraser for 45 cents . how many cents would it cost to buy 3 pencils , 3 notebooks , and 3 erasers ? ( assume that there is no volume discount . )
pencil + notebook = 80 notebook + eraser = 85 pencil + eraser = 45 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 210 cents the cost to buy 3 of each would be ( 3 / 2 ) ( 210 ) = 315 the answer is b .
a ) 300 , b ) 315 , c ) 330 , d ) 345 , e ) 360
b
add(add(multiply(subtract(45, subtract(80, divide(add(85, subtract(80, 45)), const_2))), 3), multiply(subtract(80, divide(add(85, subtract(80, 45)), const_2)), 3)), multiply(divide(add(85, subtract(80, 45)), const_2), 3))
subtract(n0,n2)|add(n1,#0)|divide(#1,const_2)|multiply(n3,#2)|subtract(n0,#2)|multiply(n3,#4)|subtract(n2,#4)|multiply(n3,#6)|add(#7,#5)|add(#8,#3)
gain
B
there are 100 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of manager to 98 % ?
"we have 99 managers and 1 director . that 1 director to compose 2 % of the total number of people , there must be 50 people in the room , hence 50 managers must leave . answer : c ."
a ) 1 , b ) 2 , c ) 50 , d ) 49 , e ) 97
c
divide(subtract(multiply(100, divide(99, const_100)), multiply(100, divide(98, const_100))), subtract(const_1, divide(98, const_100)))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)|
gain
C
two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 60 kmph , find the rate at which the train b runs .
"if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in β€˜ a ’ and β€˜ b ’ hours respectively ( i . e . a takes β€˜ a hrs ’ to travel from the meeting point to his destination and b takes β€˜ b hrs ’ to travel from the meeting point to his destination ) , then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( 4 / 9 ) = 2 / 3 this gives us that the ratio of the speed of a : speed of b as 2 : 3 . since speed of a is 60 kmph , speed of b must be 80 * ( 3 / 2 ) = 90 kmph d"
a ) 40 , b ) 60 , c ) 120 , d ) 90 , e ) 100
d
multiply(60, sqrt(divide(9, 4)))
divide(n0,n1)|sqrt(#0)|multiply(n2,#1)|
physics
D
ana climbs up stairs 20 floors building and back . she takes the exact same route both ways . on the trip up she climbs an average speed of 2 steps per second . on the trip back she climbs down at an average speed of 4 steps per second . what is her approximate average speed for the round trip in miles per hour ?
average speed = total distance / total time here number of steps to 20 th floor and back is same as she takes the same route . d = 2 t 1 and d = 4 t 2 therefore , t 1 = d / 2 and t 2 = d / 4 t 1 + t 2 = 3 d / 4 therefore , average speed = 2 d / 3 d / 4 = 2 d * 4 / 3 d answer is 1.3 since , the options are far away we need not solve till the decimal points so correct answer is option c
a ) 1 , b ) 1.25 , c ) 1.33 , d ) 1.5 , e ) 1.75
c
divide(20, add(divide(20, 2), divide(20, 4)))
divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(n0,#2)
physics
C
a and b start a business , with a investing the total capital of rs . 50000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 1500 per month from the total profit and any profit remaining is equally shared by both of them . at the end of the year , it was found that the income of a is twice that of b . find the total profit for the year ?
"interest received by a from b = 10 % of half of rs . 50000 = 10 % * 25000 = 2500 . amount received by a per annum for being a working partner = 1500 * 12 = rs . 18000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 2500 + 18000 + p ) total income of b = only his share from the remaining profit = ' p ' , as a and b share the remaining profit equally . income of a = twice the income of b ( 2500 + 18000 + p ) = 2 ( p ) p = 20500 total profit = 2 p + 18000 = 2 * 20500 + 18000 = 59000 . answer : d"
a ) 59028 , b ) 27777 , c ) 29999 , d ) 59000 , e ) 27772
d
add(multiply(multiply(divide(divide(50000, const_2), 10), const_3), const_4), 1500)
divide(n0,const_2)|divide(#0,n1)|multiply(#1,const_3)|multiply(#2,const_4)|add(n2,#3)|
gain
D
jaclyn buys $ 40 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter
"explanation : i = ( p x r x t ) / 100 = 40000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 950 answer : a"
a ) 950 , b ) 1234 , c ) 1289 , d ) 1345 , e ) none of these
a
divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4)
multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_4)|
gain
A
bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 20 seconds and his sister ’ s time is currently 9 minutes 36 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ?
"bob ' s time = 620 secs . his sis ' time = 576 secs . percent increase needed = ( 620 - 576 / 620 ) * 100 = 44 / 640 * 100 = 7 % . ans ( c ) ."
a ) 3 % , b ) 5 % , c ) 7 % , d ) 10 % , e ) 12 %
c
multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(9, const_60), 36), add(multiply(10, const_60), 20))))
multiply(n0,n0)|multiply(n2,const_60)|multiply(n0,const_60)|add(n3,#1)|add(n1,#2)|divide(#3,#4)|subtract(const_1,#5)|multiply(#0,#6)|
physics
C
in a company , 48 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ?
"the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 52 % of all employees are women , so non - union women are 52 % - 18 % = 34 % 40 % of all employees are non - union . the percent of non - union employees who are women is 34 % / 40 % = 85 % the answer is a ."
a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 %
a
multiply(const_100, divide(subtract(subtract(const_100, 48), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60)))
divide(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(n1,#0)|subtract(n1,#3)|subtract(#1,#4)|divide(#5,#2)|multiply(#6,const_100)|
gain
A
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10092 sq . m , the cost of fencing the field @ 25 paise per metre is
"solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x Γ— 4 x = 10092 ⇔ 12 x 2 = 10092 ⇔ x 2 = 841 ⇔ x = 29 . so , length = 87 m and breadth = 116 m . perimeter = [ 2 ( 87 + 116 ) ] m = 406 m . ∴ cost of fencing = rs . ( 0.25 Γ— 406 ) = rs . 101.50 . answer d"
a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 101.50 , e ) none of these
d
divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(10092, multiply(3, 4)))), multiply(4, sqrt(divide(10092, multiply(3, 4))))), 25), const_100)
multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4)|multiply(n3,#5)|divide(#6,const_100)|
physics
D
7 people average age is 30 . youngest person age is 7 . find average of the people when youngest was born .
"average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 49 = 161 so average age would be 161 / 7 = 23 answer : a"
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
a
divide(subtract(multiply(30, 7), multiply(7, 7)), 7)
multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n0)|
general
A
it is currently 7 : 16 pm . at what time in the morning was it exactly 19,443 minutes earlier ?
converting 19,443 minutes to hours , we get 19,443 / 60 = 324 r 3 ; that is 324 hours and 3 minutes . all of the answers are during the same hour of the morning , thus the hours can be assumed to bring us into the 7 am hour evenly from 7 : 16 pm . thus 324 hours ago was 7 : 16 am . take an extra 3 minutes off , and it was 7 : 13 am . b
a ) 7 : 11 , b ) 7 : 13 , c ) 7 : 17 , d ) 7 : 19 , e ) 7 : 21
b
divide(7, subtract(16, const_3))
subtract(n1,const_3)|divide(n0,#0)
physics
B
in a renowned city , the average birth rate is 6 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day .
"this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 6 - 3 = 3 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 3 = 129,600 . hence b ."
a ) 129,500 , b ) 129,600 , c ) 129,700 , d ) 129,800 , e ) 129,900
b
multiply(multiply(subtract(6, 3), const_3600), const_12)
subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)|
general
B
the ratio of the present age of two brothers is 1 : 2 and 5 years back , the ratio was 1 : 3 . what will be the ratio of their ages after 5 years ?
let the present ages of the two brothers be x and 2 x years respectively . then , ( x - 5 ) / ( 2 x - 5 ) = 1 / 3 3 ( x - 5 ) = ( 2 x - 5 ) = > x = 10 required ratio = ( x + 5 ) : ( 2 x + 5 ) = 15 : 25 = 3 : 5 answer : c
a ) 3 : 9 , b ) 3 : 0 , c ) 3 : 5 , d ) 3 : 2 , e ) 3 : 1
c
divide(add(5, subtract(multiply(5, 3), 5)), add(5, multiply(2, subtract(multiply(5, 3), 5))))
multiply(n2,n4)|subtract(#0,n2)|add(n2,#1)|multiply(n1,#1)|add(n2,#3)|divide(#2,#4)
other
C
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 30 % , then find the population of city atthe end of the year 2007
"required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 30 / 100 ) = 971750 b"
a ) 976374 , b ) 971750 , c ) 980241 , d ) 2356677 , e ) 1083875
b
multiply(1000000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100))))
divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)|
gain
B
the speed of a railway engine is 84 km per hour when no compartment is attached , and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached , the maximum number of compartments that can be carried by the engine is :
"the reduction in speed is directly proportional to the square root of the number of compartments attached doesreductionmean amount subtracted ? or percentage decrease ? there are at least two interpretations , and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words , if s = speed , and n = number of compartments , then s = 84 - k * sqrt ( n ) wherekis a constant of the proportionality . in general , if a is directly proportional to b , we can write a = k * b and solve for k . if n = 9 , then s = 24 24 = 84 - k * sqrt ( 9 ) = 84 - 3 k k = 20 now , we need to know : what value of n makes s go to zero ? 0 = 84 - 20 * sqrt ( n ) 20 * sqrt ( n ) = 84 sqrt ( n ) = 4.2 n = 4.2 ^ 2 > 17 with 18 compartments , the train does not budge . therefore , it would budge if there were one fewer cars . thus , 17 is the maximum number of cars the engine can pull and still move . d"
a ) 19 , b ) 18 , c ) 16 , d ) 17 , e ) 14
d
power(divide(84, divide(subtract(84, 24), sqrt(9))), const_2)
sqrt(n2)|subtract(n0,n1)|divide(#1,#0)|divide(n0,#2)|power(#3,const_2)|
physics
D
a train running at the speed of 54 km / hr crosses a pole in 9 sec . what is the length of the train ?
"speed = 54 * 5 / 18 = 15 m / sec length of the train = speed * time = 15 * 9 = 135 m answer : c"
a ) 288 , b ) 279 , c ) 135 , d ) 272 , e ) 150
c
multiply(divide(multiply(54, const_1000), const_3600), 9)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
C
a and b invests rs . 4000 and rs . 6000 in a business . after 4 months , a withdraws three - fourth of his capital and 4 months later , b withdraws half of his capital . in what ratio should they share the profits at the end of the year ?
"a : b ( 4000 * 4 ) + ( 1000 * 8 ) : ( 6000 * 8 ) + ( 3000 * 4 ) 24000 : 60000 2 : 5 answer : a"
a ) 2 : 5 , b ) 3 : 5 , c ) 4 : 5 , d ) 3 : 7 , e ) 5 : 7
a
divide(add(multiply(4000, 4), multiply(divide(6000, const_3), multiply(4, 4))), add(multiply(6000, multiply(4, const_3)), multiply(subtract(6000, divide(6000, const_3)), multiply(4, 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
if x is an integer such that 1 < x < 9 , 2 < x < 15 , 7 > x > – 1 , 4 > x > 0 , and x + 1 < 5 , then x is
"1 < x < 9 , 2 < x < 15 , - 1 < x < 7 0 < x < 4 x + 1 < 5 from above : 2 < x < 4 - - > x = 3 . answer : a ."
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
subtract(subtract(5, 1), 1)
subtract(n9,n8)|subtract(#0,n8)|
general
A
the radius of a wheel is 25.4 cm . what is the distance covered by the wheel in making 500 resolutions ?
"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 25.4 = 79829 cm = 798.3 m answer : b"
a ) 724 m , b ) 798.3 m , c ) 287 m , d ) 278 m , e ) 927 m
b
divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 25.4), const_2), 500), const_100)
add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)|
physics
B
a can do a piece of work 30 days . b can do work in 55 days . in how many days they will complete the work together ?
"lcm = 330 , ratio = 30 : 45 = 6 : 11 no of days = 330 / ( 6 + 11 ) = 330 / 17 = 19.4 days answer : c"
a ) 15 days , b ) 16 days , c ) 19.4 days , d ) 17.2 days , e ) 18 days
c
divide(const_1, add(divide(const_1, 30), divide(const_1, 55)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|
physics
C
a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 36 . by how many more dollars does he need to increase the price to achieve a 100 % markup ?
let x be the wholesale price . then 1.8 x = 36 and x = 36 / 1.8 = 20 . to achieve a 100 % markup , the price needs to be $ 40 . the retailer needs to increase the price by $ 4 more . the answer is b .
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
b
subtract(multiply(divide(36, add(const_1, divide(80, const_100))), const_2), 36)
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|multiply(#2,const_2)|subtract(#3,n1)
general
B
what is the remainder when 46 * 49 is divided by 8 ?
"we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 46 * 49 ) / 8 } } = remainder of ( 46 / 8 ) * remainder of ( 49 / 8 ) = 6 * 1 = 6 answer : d"
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7
d
subtract(divide(8, const_2), multiply(46, 46))
divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)|
general
D
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40,000 , the amount invested by y is ?
"suppose y invested rs . y . then 40000 / y = 2 / 3 or y = 60000 . answer : c"
a ) 60229 , b ) 62887 , c ) 60000 , d ) 61298 , e ) 61291
c
divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3)
add(const_1,const_4)|multiply(#0,const_1000)|multiply(n0,#1)|divide(#2,n1)|
gain
C
what is the next number : 3 , 18 , 258 , __
"4 ^ 0 + 2 = 3 4 ^ 2 + 2 = 18 4 ^ 4 + 2 = 258 4 ^ 6 + 2 = 4098 the answer is c ."
a ) 2928 , b ) 3638 , c ) 4098 , d ) 4418 , e ) 5378
c
subtract(subtract(subtract(multiply(258, const_10), const_100), 3), 3)
multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)|
general
C
how many 4 - digit positive integers are there , where each digit is positive , and no 3 adjacent digits are same ?
first digit . . 9 posibilities second digit , 8 possibilities third digit , 7 possibilities fourth digit , 7 possibilities . 9 * 8 * 7 * 7 = 3528 . b
a ) 1236 , b ) 3528 , c ) 4096 , d ) 4608 , e ) 6561
b
multiply(multiply(add(const_4, const_3), add(const_4, const_3)), multiply(add(4, 4), multiply(3, 3)))
add(const_3,const_4)|add(n0,n0)|multiply(n1,n1)|multiply(#0,#0)|multiply(#1,#2)|multiply(#3,#4)
general
B
if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 13 is divided by 41 ?
"x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 13 = 82 k + 5 + 13 = 82 k + 18 or x + 13 = 18,100 , 182 etc . when divided by 41 , we will get the remainder 18 . b"
a ) 3 , b ) 18 , c ) 6 , d ) 16 , e ) 5
b
add(5, 13)
add(n1,n2)|
general
B
what is the smallest positive perfect square that is divisible by 9 , 15 , and 25 ?
"the number needs to be divisible by 3 ^ 2 , 3 * 5 , and 5 ^ 2 . the smallest such perfect square is 3 ^ 2 * 5 ^ 2 = 225 the answer is b ."
a ) 100 , b ) 225 , c ) 900 , d ) 1,600 , e ) 4,900
b
add(multiply(multiply(multiply(9, power(const_3, const_2)), 15), const_2), multiply(15, 25))
multiply(n1,n2)|power(const_3,const_2)|multiply(n0,#1)|multiply(n1,#2)|multiply(#3,const_2)|add(#4,#0)|
geometry
B
ashis ' s height is 25 % more than babji , by how much percent babji ' s height is less than ashis .
explanation : solution : babji is less than ashis by ( 25 / ( 100 + 25 ) * 100 ) % = 20 % answer : d
a ) 30 % , b ) 25 % , c ) 75 % , d ) 20 % , e ) none of these
d
multiply(divide(25, add(const_100, 25)), const_100)
add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)
general
D
how many integers from 10 to 180 , inclusive , are divisible by 3 but not divisible by 7 ?
"we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 10 to 180 , inclusive is ( 168 - 21 ) / 21 + 1 = 8 ; 57 - 8 = 49 . answer : b ."
a ) 45 , b ) 49 , c ) 50 , d ) 52 , e ) 56
b
divide(180, const_10)
divide(n1,const_10)|
general
B
a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 3600 interest . the principal amount borrowed by him was .
"explanation : s . i . = p Γ’ Λ† β€” r Γ’ Λ† β€” t / 100 = > p = s . i . Γ’ Λ† β€” 100 / r Γ’ Λ† β€” t = > p = 3600 Γ’ Λ† β€” 100 / 12 Γ’ Λ† β€” 3 = rs 10000 option a"
a ) rs 10000 , b ) rs 15000 , c ) rs 16000 , d ) rs 17000 , e ) none of these
a
divide(multiply(3600, const_100), multiply(12, 3))
multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|
gain
A
if n is a natural number , then ( 6 n ^ 2 + 6 n ) is always divisible by :
explanation : ( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . e
a ) 6 , b ) 12 , c ) 24 , d ) 6 only , e ) 6 and 12 both
e
add(multiply(6, const_100), multiply(2, 6))
multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)
general
E
find the least number that must be subtracted from 1387 so that the remaining number is divisible by 15 .
on dividing 1387 by 15 we get the remainder 7 , so 7 should be subtracted . the answer is c .
a ) 1 , b ) 5 , c ) 7 , d ) 9 , e ) 13
c
subtract(1387, multiply(subtract(const_100, multiply(const_2, const_4)), 15))
multiply(const_2,const_4)|subtract(const_100,#0)|multiply(n1,#1)|subtract(n0,#2)
general
C
a person can swim in still water at 20 km / h . if the speed of water 12 km / h , how many hours will the man take to swim back against the current for 40 km ?
"m = 20 s = 12 us = 20 - 12 = 8 d = 40 t = 40 / 8 = 5 answer : c"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
c
divide(40, subtract(20, 12))
subtract(n0,n1)|divide(n2,#0)|
physics
C
out of 3 consecutive odd numbers 8 times the first number is equal to addition of thrice the third number and adding 5 to twice the second . what is the first number ?
description : = > 8 x = 2 ( x + 2 ) + 5 + 3 ( x + 4 ) = > 3 x = 21 , x = 7 = > x + 4 = 11 x = 11 - 4 = 7 answer c
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
c
divide(add(add(multiply(const_2, const_2), multiply(3, const_4)), 5), subtract(8, add(3, const_2)))
add(n0,const_2)|multiply(const_2,const_2)|multiply(n0,const_4)|add(#1,#2)|subtract(n1,#0)|add(n2,#3)|divide(#5,#4)
general
C
find 12 Γ— Γ— 19
"mentally imagine this number as ( 10 + 2 ) Γ— Γ— 19 = 190 + 38 = 228 . answer : b"
a ) 238 , b ) 228 , c ) 208 , d ) 277 , e ) 101
b
divide(12, 19)
divide(n0,n1)|
general
B
a train is moving at a speed of 100 km / hour . the length of the train is 100 metre . in how much time will it cross a pole ?
in order to cross the pole , the train will need to cover its own length of 100 meter . to cover 100 x 1000 meter , train takes 60 minutes or 60 x 60 = 3600 second . so , to cover 100 meter , train would take 3600 x 100 / 100 x 1000 = 3.6 second . answer : d .
a ) 1 second , b ) 2.5 second , c ) 4 second , d ) 3.6 second , e ) 4.5 second
d
divide(100, multiply(100, const_0_2778))
multiply(n0,const_0_2778)|divide(n0,#0)
physics
D
the guests at a football banquet consumed a total of 411 pounds of food . if no individual guest consumed more than 2.5 pounds of food , what is the minimum number of guests that could have attended the banquet ?
"to minimize one quantity maximize other . 164 * 2.5 ( max possible amount of food a guest could consume ) = 410 pounds , so there must be more than 164 guests , next integer is 165 . answer : a ."
a ) 165 , b ) 161 , c ) 162 , d ) 163 , e ) 164
a
add(floor(divide(411, 2.5)), const_1)
divide(n0,n1)|floor(#0)|add(#1,const_1)|
general
A
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 854 in 4 years . the sum is :
"s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 3 years = rs . ( 39 x 3 ) = rs . 117 . principal = rs . ( 815 - 117 ) = rs . 698 . answer : option c"
a ) rs . 650 , b ) rs . 690 , c ) rs . 698 , d ) rs . 700 , e ) rs . 740
c
subtract(815, divide(multiply(subtract(854, 815), 3), 4))
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
gain
C
city x has a population 7 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ?
x = 7 y , y = 2 * z x : y , y : z 7 : 1 , 2 : 1 14 : 2 , 2 : 1 so , x : z = 14 : 1 ( d )
a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 14 : 1 , e ) 18 : 1
d
multiply(7, const_2)
multiply(n0,const_2)
general
D
there are 810 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ?
"let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 810 . hence , we have m + f = 810 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f / 2 + m / 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 810 / 3 = 270 solving the three equations yields the solution f = 270 , m = 540 , and d = 270 . the number of female democratic participants equals half the female participants equals 270 / 2 = 135 . answer : d"
a ) 75 , b ) 100 , c ) 125 , d ) 135 , e ) 225
d
divide(subtract(multiply(divide(810, const_3), const_4), 810), const_2)
divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)|
general
D
if y is 75 % greater than x , than x is what % less than y ?
"y = 1.75 x x = y / 1.75 = 100 y / 175 = 4 y / 7 x is 3 / 7 less which is about 42.9 % less than y . the answer is d ."
a ) 30.6 % , b ) 34.4 % , c ) 38.7 % , d ) 42.9 % , e ) 46.5 %
d
multiply(divide(75, add(75, const_100)), const_100)
add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)|
general
D
at what rate percent on simple interest will rs . 750 amount to rs . 975 in 5 years ?
"225 = ( 750 * 5 * r ) / 100 r = 6 % . answer : e"
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
e
multiply(divide(divide(subtract(975, 750), 750), 5), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
E
if rs . 10 be allowed as true discount on a bill of rs . 110 at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is ?
explanation : present worth = amount - truediscount = 110 - 10 = rs . 100 si on rs . 100 for a certain time = rs . 10 si on rs . 100 for doube the time = rs . 20 truediscount on rs . 120 = 120 - 100 = rs . 20 truediscount on rs . 110 = = rs . 18.33 answer : b
a ) 68.33 , b ) 18.33 , c ) 28.33 , d ) 48.33 , e ) 98.33
b
multiply(110, divide(subtract(add(10, 110), const_100), add(10, 110)))
add(n0,n1)|subtract(#0,const_100)|divide(#1,#0)|multiply(n1,#2)
general
B
the average of first five multiples of 4 is
"solution average = 4 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 60 / 5 . = 12 answer d"
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15
d
add(4, const_1)
add(n0,const_1)|
general
D
the overall age of x and y is 10 year greater than the overall age of y and z . z is how many decades younger that x ?
"b 10 ( x + y ) Γ’ € β€œ ( y + z ) = 10 x Γ’ € β€œ z = 10"
a ) 11 , b ) 10 , c ) 12 , d ) 17 , e ) 19
b
divide(10, const_1)
divide(n0,const_1)|
general
B
a man can row upstream at 27 kmph and downstream at 35 kmph , and then find the speed of the man in still water ?
"us = 27 ds = 35 m = ( 35 + 27 ) / 2 = 31 answer : a"
a ) 31 , b ) 92 , c ) 30 , d ) 32 , e ) 23
a
divide(add(27, 35), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
A
on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 30 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ?
"0.1 m = 0.70 e = > e / m = 1 / 7 * 100 = 14.28 % so answer is e . m - # of motorists e - # of motorists exceeding speed"
a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 14.28 %
e
divide(const_100, multiply(multiply(divide(10, const_100), divide(30, const_100)), const_100))
divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)|
gain
E
how many bricks , each measuring 40 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 ) / ( 40 x 11.25 x 6 ) = 4000 answer : c"
a ) 1400 , b ) 2400 , c ) 4000 , d ) 7000 , e ) 3400
c
divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(40, 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
C
a 40 kg metal bar made of alloy of tin and silver lost 4 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ?
you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 4 kg when 40 kg alloy is immersed . this is a loss of ( 4 / 40 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e
a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3
e
divide(divide(subtract(4, multiply(divide(0.375, 5), 40)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(40, divide(subtract(4, multiply(divide(0.375, 5), 40)), subtract(divide(1.375, 10), divide(0.375, 5)))))
divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)
other
E
the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 60 kmph its consumption is 1 litre per hour . if each litre costs $ 60 and other expenses per hous is $ 60 , then what would be the minimum expenditure required to cover a distance of 600 km ?
"60 kmph consumption is 1 lt / hr so 600 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 60 so 10 lt costs $ 600 extra expenses for 1 hr - $ 60 10 hrs - $ 600 total expense - $ 600 + $ 600 = $ 1200 answer : c"
a ) 120 , b ) 1250 , c ) 1200 , d ) 1100 , e ) 1150
c
add(multiply(divide(600, 60), 60), multiply(divide(600, 60), 60))
divide(n4,n0)|multiply(n2,#0)|multiply(n3,#0)|add(#1,#2)|
physics
C
9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 2 miles per hour slower than the first runner ' s rate . how far apart , to the nearest mile , will they be after 1 / 2 hour ?
if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 8 m / h - - > 4 m in 1 / 2 hr side 2 = 6 m / h - - > 3 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 3 ^ 2 d ^ 2 = 25 = 5 answer option a
a ) 5 , b ) 7 , c ) 8 , d ) 12 , e ) 14
a
sqrt(add(power(multiply(8, divide(1, 2)), const_2), power(multiply(subtract(8, 2), divide(1, 2)), const_2)))
divide(n3,n2)|subtract(n1,n2)|multiply(n1,#0)|multiply(#0,#1)|power(#2,const_2)|power(#3,const_2)|add(#4,#5)|sqrt(#6)
physics
A
n ^ ( n / 2 ) = 6 is true when n = 6 in the same way what is the value of n if n ^ ( n / 2 ) = 10 ?
n ^ ( n / 2 ) = 10 apply log n / 2 logn = log 10 nlogn = 2 log 10 = log 10 ^ 2 = log 100 logn = log 100 now apply antilog n = 100 / n now n = 10 . answer : c
a ) 8 , b ) 12 , c ) 10 , d ) 16 , e ) 18
c
divide(power(10, 2), 10)
power(n4,n0)|divide(#0,n4)|
general
C
of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 110 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ?
"of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 110 geese survived the first year : 1 / 5 * ( total ) = 110 - - > ( total ) = 550 . answer : d ."
a ) 280 , b ) 400 , c ) 540 , d ) 550 , e ) 840
d
divide(divide(divide(110, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(2, 3))
divide(n1,n5)|divide(n1,n3)|divide(n0,n1)|subtract(const_1,#0)|divide(n6,#3)|divide(#4,#1)|divide(#5,#2)|
general
D
a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 30 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ?
pastry is divided in 30 customers equally . so , total number of pastry must be a multiple of 30 only option a satisfies the condition , and hence is the answer
a ) 450 . , b ) 412 . , c ) 432 . , d ) 502 . , e ) 522 .
a
multiply(divide(multiply(49, 6), subtract(49, 30)), 30)
multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)
general
A
a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is three - fifth of the distance traveled by bus . if he traveled by train , plane and bus , then find the distance traveled by bus ?
"explanation : total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = 3 x / 5 = > x + 3 x / 5 + 600 = 1800 = > 8 x / 5 = 1200 = > x = 750 km . answer : d"
a ) 239 , b ) 247 , c ) 277 , d ) 270 , e ) 898
d
divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3))
add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)|
physics
D
suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 30 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ?
"the part of job that suresh completes in 9 hours = 9 Γ’  β€ž 15 = 3 Γ’  β€ž 5 remaining job = 1 - 3 Γ’  β€ž 5 = 2 Γ’  β€ž 5 remaining job can be done by ashutosh in 2 Γ’  β€ž 5 Γ£ β€” 30 = 12 hours answer d"
a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) none of these
d
multiply(subtract(const_1, divide(9, 15)), 30)
divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)|
physics
D
( 128.5 x 60 ) + ( 13.8 x 65 ) = ? x 25
"explanation : ? = ( 128.5 x 60 ) + ( 13.8 x 65 ) / 25 = 7710 + 897 / 25 = 344.28 answer : option b"
a ) 524.48 , b ) 344.28 , c ) 574.36 , d ) 585.64 , e ) 595.46
b
multiply(128.5, power(add(const_4, const_1), const_4))
add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)|
general
B
a train 150 m long is running with a speed of 98 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going
"explanation : speed of the train relative to man = ( 98 - 8 ) kmph = ( 90 Γ— 5 / 18 ) m / sec = 25 m / sec time taken by the train to cross the man i = time taken by it to cover 150 m at 25 m / sec = 150 Γ— 1 / 25 sec = 6 sec answer : option b"
a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec
b
divide(150, multiply(add(98, 8), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
B
it takes 2 team of farm workers 12 days to completely prepare a piece of land for planting . if both team of workers were to work separately , one of them can complete the work 10 days earlier than the other , how many days will it take each of them to separately complete the work ?
work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here , we ' re told that ( working together ) the two team of workers would complete a job in 12 days . this means that ( individually ) each of team would take more than 12 days to do the job . answers d , a and b are illogical , since the individual times must both be greater than 12 days . so we can test the values for answers c and e . using the values for answers b and d . . . answer b : ( 15 ) ( 25 ) / ( 15 + 25 ) = 375 / 40 = 9.4 this is a match answer e : ( 20 ) ( 30 ) / ( 20 + 30 ) = 12 final answer : e
a ) 12 and 22 , b ) 11 and 21 , c ) 15 and 25 , d ) 9 and 19 , e ) 20 and 30
e
add(subtract(subtract(add(multiply(add(multiply(10, 12), multiply(10, 2)), 10), multiply(multiply(10, 12), 10)), add(multiply(add(const_3, const_4), const_10), const_100)), multiply(add(multiply(10, 12), multiply(10, 2)), 10)), multiply(10, const_100))
add(const_3,const_4)|multiply(n2,const_100)|multiply(n1,n2)|multiply(n0,n2)|add(#2,#3)|multiply(n2,#2)|multiply(#0,const_10)|add(#6,const_100)|multiply(n2,#4)|add(#8,#5)|subtract(#9,#7)|subtract(#10,#8)|add(#1,#11)
physics
E
if 36 men can do a piece of work in 25 hours , in how many hours will 10 men do it ?
explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 10 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 10 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 10 } = 90 hence , 10 men can do it in 90 hours . answer : e ) 90
a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 90
e
divide(multiply(36, 25), 10)
multiply(n0,n1)|divide(#0,n2)
physics
E
3 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ?
"approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 3 / 36 = . 08 something answer ( b )"
a ) 0.004 , b ) 0.08333 , c ) 2.775 , d ) 3.6036 , e ) 36.036
b
inverse(add(divide(3, 0.03), divide(3, 0.37)))
divide(n0,n2)|divide(n0,n4)|add(#0,#1)|inverse(#2)|
general
B
in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 80 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?
"total fish = x percentage of second catch = ( 2 / 80 ) * 100 = 2.5 % so , x * 2.5 % = 80 x = 3200 ans c ."
a ) 400 , b ) 625 , c ) 3,200 , d ) 4,500 , e ) 10,000
c
divide(80, divide(2, 80))
divide(n2,n1)|divide(n0,#0)|
gain
C
100 liter solution of cool - drink is made from 10 % jasmine water . if 5 liters of jasmine and 10 liters of water were added to the solution , what percent of the solution is jasmine ?
denominator 100 + 5 + 10 115 numerator 100 * 0.1 = 10 - - - > jasmine water 10 + 5 = 15 - - - > new content of jasmine water in new solution ratio 3 / 23 answer is b
a ) 1 / 13 , b ) 2 / 23 , c ) 3 / 31 , d ) 4 / 15 , e ) 2 / 3
b
divide(10, add(5, add(100, 10)))
add(n0,n1)|add(n2,#0)|divide(n1,#1)
gain
B
sahil purchased a machine at rs 9000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it .
"explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 9000 + 5000 + 1000 = 15000 150 % of 15000 = 150 / 100 * 15000 = 22500 option a"
a ) rs . 22500 , b ) rs . 24500 , c ) rs . 26500 , d ) rs . 28500 , e ) none of these
a
add(add(add(9000, 5000), 1000), multiply(divide(add(add(9000, 5000), 1000), const_100), 50))
add(n0,n1)|add(n2,#0)|divide(#1,const_100)|multiply(n3,#2)|add(#1,#3)|
gain
A
shipment - - - no . of defective chips / shipment - - - total chips in shipment s 1 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5,000 s 2 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12,000 s 3 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 15,000 s 4 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 16,000 a computer chip manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ratio for shipments s 1 , s 2 , s 3 , and s 4 combined , as shown in the table above . what ’ s the expected number of defective chips in a shipment of 60,000 chips ?
for a total of 51000 chips ( adding s 1 , s 2 , s 3 , s 4 ) total number of defective chips is 17 ( ( adding defective chips of s 1 , s 2 , s 3 , s 4 ) so ratio is 12 / 48000 or 1 every 4000 chips . keeping this ratio constant for 60000 chips number of defective chips will be ( 1 / 4000 ) * 60000 = 15 e
a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 15
e
multiply(add(2, 3), 3)
add(n1,n6)|multiply(n6,#0)
general
E
if the cost price of 20 articles is equal to the selling price of 15 articles , what is the % profit or loss made by the merchant ?
"let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 15 articles = cost price of 20 articles = $ 15 . now , we know the selling price of 15 articles . let us find the cost price of 15 articles . cost price of 15 articles = 15 * 1 = $ 15 . therefore , profit made on sale of 15 articles = selling price of 15 articles - cost price of 15 articles = 20 - 15 = $ 5 . as the profit is in the positive , the merchant has made a profit of $ 5 . therefore , % loss = loss / cp * 100 % loss = 5 / 15 * 100 = 33.33 % profit . b"
a ) 25 % loss , b ) 33.33 % profit , c ) 20 % loss , d ) 20 % profit , e ) 5 % profit
b
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 15), 20)), divide(multiply(const_100, 15), 20)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
B
how much interest will $ 10,000 earn in 9 months at an annual rate of 3 % ?
"soln : - 9 months = 3 / 4 of year ; 3 % = 3 / 100 = 3 / 100 ; $ 10,000 ( principal ) * 3 / 100 ( interest rate ) * 3 / 4 ( time ) = $ 225 . answer : e"
a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 550 , e ) $ 225
e
multiply(multiply(power(const_100, const_2), divide(3, const_100)), divide(const_3, 3))
divide(const_3,n2)|divide(n2,const_100)|power(const_100,const_2)|multiply(#1,#2)|multiply(#0,#3)|
gain
E
the population of locusts in a certain swarm doubles every two hours . if 4 hours ago there were 1,000 locusts in the swarm , in approximately how many hours will the swarm population exceed 512,000 locusts ?
"- 4 hours : 1,000 - 2 hours : 2,000 now : 4,000 + 2 hours : 8,000 + 4 hours : 16,000 + 6 hours : 32,000 + 8 hours : 64,000 + 10 hours : 128,000 + 12 hours : 256,000 + 14 hours : 512,000 answer : d"
a ) 18 , b ) 15 , c ) 13 , d ) 14 , e ) 12
d
subtract(multiply(log(divide(power(4, 4), const_2)), const_2), 4)
power(n0,n0)|divide(#0,const_2)|log(#1)|multiply(#2,const_2)|subtract(#3,n0)|
general
D
the sum of the ages of 5 children born at the intervals of 2 years each is 50 years . what is the age of the youngest child ?
let x = the youngest child . each of the other four children will then be x + 2 , x + 4 , x + 6 , x + 8 . we know that the sum of their ages is 50 . so , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 50 therefore the youngest child is 6 years old answer : a
a ) 6 , b ) 18 , c ) 10 , d ) 99 , e ) 38
a
divide(subtract(divide(50, divide(5, const_2)), multiply(subtract(5, const_1), 2)), const_2)
divide(n0,const_2)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,const_2)
general
A
jar x is 1 / 4 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ?
"let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 4 + p / 4 = p / 2 the answer is d ."
a ) 2 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 / 2 , e ) 2 / 3
d
add(divide(1, const_2.0), multiply(divide(1, 4), divide(1, 4)))
divide(n0,n3)|multiply(#0,#0)|add(#0,#1)|
general
D
the cost of registration at a professional association meeting was $ 50 per person ; a lunch for registrants only was available for an additional $ 22 per person . if the number of registrants who paid for lunch was 20 more than the number who did not , and if receipts for registration and lunch totaled $ 50,240 , how many people paid just for registration at the meeting ?
hope this might be useful to you . let the number of people who have opted only to register = x now since the registration cost is 50 $ per person , the total amount sums to = 50 x $ as per the information given in the question , the number of registrants who paid for lunch was 20 more than the number who did not . that means , total number of people who registered and paid for lunch = 20 + x . for the people who registered for lunch the cost is 50 $ ( for the event registration ) + 22 $ ( for lunch ) = 72 $ . total amount in this case sums to = 72 ( 20 + x ) = 1440 + 72 x now , total amount received was 50240 . thus , from the above data , 50 x + 1440 + 72 x = 50240 122 x = 50240 - 1440 122 x = 48800 x = 400 . hence the correct ans is e
a ) 700 , b ) 800 , c ) 1,300 , d ) 1,500 , e ) 400
e
multiply(const_1, const_1)
multiply(const_1,const_1)
general
E
there are 10 slate rocks , 11 pumice rocks , and 4 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ?
"10 / 25 * 9 / 24 = 3 / 20 the answer is a ."
a ) 3 / 20 , b ) 5 / 26 , c ) 7 / 34 , d ) 9 / 37 , e ) 11 / 49
a
multiply(divide(10, add(add(10, 11), 4)), divide(subtract(10, const_1), subtract(add(add(10, 11), 4), const_1)))
add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)|
other
A
a train passes a station platform in 50 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 ) / 50 = 15 = > x = 300 m answer : d
a ) 615 m , b ) 240 m , c ) 168 m , d ) 300 m , e ) 691 m
d
multiply(20, multiply(54, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)
physics
D
a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 60 , how many horses are there ?
"legs 12 * 4 = 48 now half on their horses so remaining on the walk so 6 men 6 men has 12 legs so , 12 + 48 = 60 legs walking answer : b"
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
b
divide(60, add(const_4, divide(const_2, const_2)))
divide(const_2,const_2)|add(#0,const_4)|divide(n0,#1)|
general
B
if two integers x , y ( x > y ) are selected from - 10 to 5 ( inclusive ) , how many possible cases are there ?
"if two integers x , y ( x > y ) are selected from - 10 to 5 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 16 c 2 = 16 * 15 / 2 = 120 . therefore , the answer is b ."
a ) 105 , b ) 120 , c ) 190 , d ) 210 , e ) 240
b
add(add(add(add(add(add(add(5, 10), add(5, const_2)), add(5, const_1)), 5), 10), const_2), const_1)
add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)|
probability
B
a certain tax rate is $ 82 per $ 100.00 . what is the rate , expressed as a percent ?
"here in question it is asking $ . 82 is what percent of $ 100 . suppose $ . 82 is x % of 100 means 100 * ( x / 100 ) = 82 hence x = 82 % so answer is a"
a ) 82 % , b ) 8.2 % , c ) 0.82 % , d ) 0.082 % , e ) 0.0082 %
a
multiply(divide(82, 100.00), 100.00)
divide(n0,n1)|multiply(#0,n1)|
gain
A
a and b invests rs . 8000 and rs . 9000 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 ( 8000 * 4 ) + ( 4000 * 8 ) : ( 9000 * 6 ) + ( 6000 * 6 ) 64000 : 90000 32 : 45 . answer : a"
a ) 32 : 45 , b ) 32 : 47 , c ) 32 : 45 , d ) 32 : 41 , e ) 32 : 42
a
divide(add(multiply(8000, 4), multiply(divide(9000, const_3), multiply(2, 4))), add(multiply(9000, multiply(2, const_3)), multiply(subtract(9000, divide(9000, 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
5 friends visited a fruit stall , and everyone decided to contribute equally to the total bill of $ 100 . if one of the friends had a coupon for 6 % off the total bill , and if each friend still contributed equally after the coupon was applied to the bill , how much did each friend pay ?
at the non - discounted price , each friend would pay $ 20 , as $ 100 divided by 5 friends is $ 20 per friend . but if the bill is 6 % off , then each friend would pay 6 % less . 6 % of $ 20 is $ 1.2 , so each friend saves $ 1.2 and pays the remaining 18.8 correct option : option b
a ) 18 , b ) 18.8 , c ) 19 , d ) 17.8 , e ) 17.9
b
divide(subtract(100, divide(multiply(6, 100), const_100)), 5)
multiply(n1,n2)|divide(#0,const_100)|subtract(n1,#1)|divide(#2,n0)
general
B
a box contains 9 pairs of shoes ( 18 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ?
"the problem with your solution is that we do n ' t choose 1 shoe from 18 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 17 ( as after taking one at random there are 17 shoes left and only one is the pair of the first one ) = 1 / 17 answer : b ."
a ) 1 / 190 , b ) 1 / 17 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9
b
divide(const_1, subtract(18, const_1))
subtract(n1,const_1)|divide(const_1,#0)|
general
B
a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 54 . find the number of notes of rs . 50 denomination .
"total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 54 - x now , 50 * x + 500 * ( 54 - x ) = 10350 50 x + 27000 - 500 x = 10350 - 450 x = - 16650 x = 37 . no . of 50 rupees note = 37 . answer : option c"
a ) 15 , b ) 21 , c ) 37 , d ) 19 , e ) 21
c
divide(subtract(multiply(500, 54), 10350), subtract(500, 50))
multiply(n2,n3)|subtract(n2,n1)|subtract(#0,n0)|divide(#2,#1)|
general
C
a can do a piece of work in 12 days . he worked for 15 days and then b completed the remaining work in 10 days . both of them together will finish it in .
explanation : 15 / 25 + 10 / x = 1 = > x = 25 1 / 25 + 1 / 25 = 2 / 25 25 / 2 = 12 1 / 2 days answer a
a ) 12 1 / 2 days , b ) 10 1 / 2 days , c ) 12 1 / 3 days , d ) 10 1 / 3 days , e ) 11 1 / 2 days
a
divide(const_1, subtract(divide(const_1, 12), divide(const_1, 15)))
divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)
physics
A
the perimeter of one face of a cube is 48 cm . its volume will be :
"explanation : edge of cude = 48 / 4 = 12 cm volume = a * a * a = 12 * 12 * 12 = 1728 cm cube option e"
a ) 125 cm 3 , b ) 400 cm 3 , c ) 250 cm 3 , d ) 625 cm 3 , e ) none of these
e
volume_cube(square_edge_by_perimeter(48))
square_edge_by_perimeter(n0)|volume_cube(#0)|
geometry
E
the difference between a 7 digit number and the number formed by reversing its digit is not a multiple of
"another approach is to test a number . let ' s say the original number is 1231231 so , the reversed number is 1321321 the difference = 1321321 - 1231231 = 90090 no check the answer choices 90090 is a multiple of 3,9 , 10,11 121 is not a multiple of 90090 answer : a"
a ) 121 , b ) 11 , c ) 9 , d ) 10 , e ) 3
a
power(add(7, const_4), const_4)
add(n0,const_4)|power(#0,const_4)|
general
A
a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part
"explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : b"
a ) 17 inches , b ) 20 inches , c ) 15 inches , d ) 18 inches , e ) 19 inches
b
divide(add(multiply(6, const_12), 8), 4)
multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)|
general
B
the product of a and b is equal to 11 more than twice the sum of a and b . if b = 5 , what is the value of b - a ?
"ab = 11 + 2 ( a + b ) 5 a = 11 + 2 a + 10 3 a = 21 a = 7 b - a = 5 - 7 = - 2 a is the answer"
a ) - 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35
a
subtract(5, divide(add(multiply(5, const_2), 11), subtract(5, const_2)))
multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3)|
general
A
the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is ?
"correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer : c"
a ) 36.6 , b ) 36.1 , c ) 36.5 , d ) 36.2 , e ) 36.9
c
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
C