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 swimmer can swim in still water at 4 km / h . if the speed of the water current is 2 km / h , how many hours will the swimmer take to swim against the current for 7 km ?
|
"the swimmer can swim against the current at a speed of 4 - 2 = 2 km / h . the time it will take is 7 / 2 = 3.5 hours . the answer is c ."
|
a ) 3.0 , b ) 3.25 , c ) 3.5 , d ) 3.75 , e ) 4.0
|
c
|
divide(7, subtract(4, 2))
|
subtract(n0,n1)|divide(n2,#0)|
|
physics
|
C
|
the radius of a circle is 4 / π cm then its perimeter is ?
|
perimeter = 2 π r 2 π * 4 / π = 8 cm answer : c
|
['a ) 10', 'b ) 9', 'c ) 8', 'd ) 12', 'e ) 15']
|
c
|
multiply(const_2, multiply(const_pi, divide(4, const_pi)))
|
divide(n0,const_pi)|multiply(#0,const_pi)|multiply(#1,const_2)
|
physics
|
C
|
the price of a certain painting increased by 20 % during the first year and decreased by 25 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ?
|
"easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 20 = 120 price at the end of year 2 = 120 - 120 * 0.25 = 120 * 0.75 = 90 hence required answer = ( 90 / 100 ) * 100 % = 90 % answer is b ."
|
a ) 102 % , b ) 90 % , c ) 120 % , d ) 85 % , e ) 95 %
|
b
|
add(subtract(subtract(20, 25), divide(multiply(25, 20), const_100)), const_100)
|
multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|add(#3,const_100)|
|
general
|
B
|
in a tree , 3 / 5 of the birds are parrots while the rest are toucans . if 1 / 3 of the parrots are female and 3 / 4 of the toucans are female , what fraction of the birds in the tree are male ?
|
the fraction of birds that are male parrots is ( 2 / 3 ) ( 3 / 5 ) = 2 / 5 . the fraction of birds that are male toucans is ( 1 / 4 ) ( 2 / 5 ) = 1 / 10 . the total fraction of male birds is 2 / 5 + 1 / 10 = 1 / 2 . the answer is a .
|
a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 5 , d ) 3 / 10 , e ) 7 / 10
|
a
|
add(multiply(divide(3, 5), subtract(const_1, divide(1, 3))), multiply(divide(const_2, 5), subtract(const_1, divide(3, 4))))
|
divide(n0,n1)|divide(n2,n0)|divide(const_2,n1)|divide(n0,n5)|subtract(const_1,#1)|subtract(const_1,#3)|multiply(#0,#4)|multiply(#2,#5)|add(#6,#7)
|
general
|
A
|
a cube of edge 15 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 14 cm , find the rise in water level ?
|
"increase in volume = volume of the cube = 15 * 15 * 15 cm ^ 3 rise in water level = volume / area = 15 * 15 * 15 / 20 * 14 = 12.05 cm answer is b"
|
a ) 6 cm , b ) 12.05 cm , c ) 11.25 cm , d ) 15.12 cm , e ) 20.62 cm
|
b
|
divide(volume_cube(15), multiply(20, 14))
|
multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)|
|
geometry
|
B
|
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 130000 , find the cost of a camel ?
|
"cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 130000 p = rs . [ ( 24 * 4 * 4 * 130000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 49920000 / 9600 ) = > p = rs . 5200 answer : d"
|
a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 5200 , e ) s . 6880
|
d
|
divide(multiply(multiply(multiply(24, 4), 4), 130000), multiply(multiply(multiply(10, 16), 6), 10))
|
multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)|
|
general
|
D
|
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 40 percent on clothing , 30 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ?
|
"let amount spent by jill = 100 clothing = 40 , food = 30 , others = 30 tax on clothing = 1.6 tax on others = 2.4 percentage = 4 / 100 = 4 % answer : c"
|
a ) 2.8 % , b ) 3.6 % , c ) 4 % , d ) 5.2 % , e ) 6.0 %
|
c
|
multiply(divide(add(multiply(40, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100)
|
divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)|
|
general
|
C
|
a certain car uses 12 gallons of gasoline in traveling 360 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car ’ s gas mileage be increased ?
|
"360 / 10 = 36 . the difference is 36 - 30 = 6 . answer c"
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
|
c
|
subtract(divide(360, 10), divide(360, 12))
|
divide(n1,n2)|divide(n1,n0)|subtract(#0,#1)|
|
physics
|
C
|
payal finished 2 / 3 of a book . she calculated that she finished 20 more pages than she has yet to read . how long is her book ?
|
let x be the total number of pages in the book , then she finished 2 / 3 * x pages . then she has x − 2 / 3 * x = 1 / 3 * x pages left . 2 / 3 * x − 1 / 3 * x = 20 1 / 3 * x = 20 x = 60 so the book is 270 pages long . answer is a .
|
a ) a ) 60 , b ) b ) 150 , c ) c ) 130 , d ) d ) 90 , e ) e ) 210
|
a
|
divide(20, divide(divide(2, 3), const_2))
|
divide(n0,n1)|divide(#0,const_2)|divide(n2,#1)
|
general
|
A
|
the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 20 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 ) 80 x - ( 5 * 20 ) = 92 ( x - 5 ) 360 = 12 x = > x = 30 answer : c"
|
a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 55
|
c
|
divide(subtract(multiply(92, 5), multiply(5, 20)), subtract(92, 80))
|
multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|
|
general
|
C
|
nick took a bus from home to market , that travels at 80 kmph . while walking back at 8 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 60 kmph . find the average speed .
|
let the distance be 2 x ( one way ) time taken by bus = 2 x / 80 , by walking = x / 8 , by cycling = x / 60 hours : . average speed = total distance / total time = 8 x / x / 40 + x / 8 + x / 60 = 8 * 60 / 1.5 + 7.5 + 1 = 12.0 answer : d
|
a ) 49.5 kmph , b ) 49.0 kmph , c ) 48.5 kmph , d ) 48.0 kmph , e ) none of these
|
d
|
inverse(add(divide(inverse(60), const_2), inverse(80)))
|
inverse(n2)|inverse(n0)|divide(#0,const_2)|add(#2,#1)|inverse(#3)
|
physics
|
D
|
find the sum of all odd number upto 250 .
|
"explanation : number of odd numbers upto 250 = n / 2 = 250 / 2 = 125 sum of first 125 odd numbers = ( 125 ) 2 = 15625 answer : option c"
|
a ) 15345 , b ) 15645 , c ) 15625 , d ) 15342 , e ) 15683
|
c
|
add(divide(divide(250, divide(divide(divide(divide(divide(250, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(250, divide(divide(divide(divide(divide(250, const_2), const_2), const_2), const_2), const_2)), const_2))))
|
divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)|
|
general
|
C
|
two good train each 250 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 .
|
"sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 250 + 250 ) m = 1000 m . required time = ( 500 x 6 / 125 ) sec = 24 sec . answer b"
|
a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) none
|
b
|
multiply(multiply(250, inverse(multiply(add(45, 30), const_0_2778))), const_2)
|
add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2)|
|
physics
|
B
|
addison high school ’ s senior class has 200 boys and 160 girls . if 60 % of the boys and 75 % of the girls attended college , what percentage of the total class that did not attend college ?
|
"number of boys attending the college : 0.60 * 200 = 120 number of girls attending the college : 0.75 * 160 = 120 total attending = 120 + 120 = 240 total people = 200 + 160 = 360 % of people attending = 240 / 360 = 66.67 % . d is the correct answer"
|
a ) 33.33 , b ) 60 , c ) 80 , d ) 66.67 , e ) 64
|
d
|
multiply(divide(add(multiply(divide(60, const_100), 200), multiply(subtract(const_1, divide(75, const_100)), 160)), add(200, 160)), const_100)
|
add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|subtract(const_1,#2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#0)|multiply(#7,const_100)|
|
general
|
D
|
sarah is driving to the airport . after driving at 40 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 50 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did she drive in all ?
|
"after driving at 40 miles per hourfor one hour , this distance left to cover is d - 40 . say this distance is x miles . now , we know that the difference in time between covering this distance at 40 miles per hour and 50 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 40 - x / 50 = 3 / 2 - - > 5 x / 200 - 4 x / 200 = 3 / 2 - - > x / 200 = 3 / 2 - - > x = 300 . total distance = x + 40 = 340 miles . answer : b"
|
a ) 140 , b ) 340 , c ) 210 , d ) 245 , e ) 280
|
b
|
subtract(30, subtract(40, 30))
|
subtract(n0,n2)|subtract(n2,#0)|
|
physics
|
B
|
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 75 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ?
|
"the actual answer is obtained by multiplying 175 % by 70 % and subtracting 100 % from this total . that is : 175 % × 70 % = 122.5 % ; 122.5 % − 100 % = 122.5 % . answer : e"
|
a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 22.5 %
|
e
|
multiply(subtract(multiply(add(const_1, divide(75, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100)
|
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|
|
gain
|
E
|
a book is bought for $ 60 and sold for $ 78 . what is the profit in percentage ?
|
"78 / 60 = 1.3 the answer is c ."
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
|
c
|
multiply(divide(subtract(78, 60), 60), const_100)
|
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
|
gain
|
C
|
money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 3500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ?
|
"x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 3500 triples to $ 10500 , in 28 years , it will again triple to $ 10500 * 3 = $ 31,500 answer e"
|
a ) $ 3,750 , b ) $ 5,600 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 31,500
|
e
|
multiply(3500, power(const_3, divide(28, divide(112, 8))))
|
divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2)|
|
gain
|
E
|
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 7 ounces of corn syrup , how many ounces of water does it contain ?
|
"f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 7 * 180 / 12 = 105 ounces of water e"
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 105
|
e
|
multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 7)
|
divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)|
|
other
|
E
|
a , b and c start a business each investing 20000 . after 5 months a withdrew 10000 , b withdrew 15000 and c invests 5000 more . at the end of the year , a total profit of 58000 was recorded . find the share of b .
|
ratio of the capitals of a , b and c = 20000 ã — 5 + 10000 ã — 7 : 20000 ã — 5 + 5000 ã — 7 : 20000 ã — 5 + 25000 ã — 7 = 170000 : 135000 : 275000 = 170 : 135 : 275 . b â € ™ s share = ( 58000 ã — 135 â „ 580 ) = 13500 answer d
|
a ) 20,000 , b ) 21,200 , c ) 28,200 , d ) 13,500 , e ) none of these
|
d
|
divide(multiply(58000, divide(add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5))), add(add(add(multiply(20000, 5), multiply(subtract(20000, 10000), subtract(const_12, 5))), add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5)))), add(multiply(20000, 5), multiply(add(20000, 5000), subtract(const_12, 5)))))), add(const_60, 5))
|
add(n0,n4)|add(n1,const_60)|multiply(n0,n1)|subtract(n0,n3)|subtract(const_12,n1)|subtract(n0,n2)|multiply(#3,#4)|multiply(#5,#4)|multiply(#0,#4)|add(#2,#6)|add(#2,#7)|add(#2,#8)|add(#10,#9)|add(#12,#11)|divide(#9,#13)|multiply(n5,#14)|divide(#15,#1)
|
gain
|
D
|
each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 5 of the slips are the drawn , without replacement , what is the probability that all 5 have a even number written on it ?
|
"key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 p ( + on 5 rd pick ) = 1 / 6 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 a"
|
a ) 1 / 252 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 2 , e ) 5 / 9
|
a
|
multiply(multiply(multiply(divide(add(const_4, const_1), add(9, const_1)), divide(subtract(add(5, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(5, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(5, const_1), const_1), const_1), const_1), subtract(subtract(subtract(add(9, const_1), const_1), const_1), const_1)))
|
add(const_1,const_4)|add(n1,const_1)|divide(#0,#1)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#4)|subtract(#3,const_1)|subtract(#4,const_1)|divide(#6,#7)|multiply(#2,#5)|subtract(#6,const_1)|subtract(#7,const_1)|divide(#10,#11)|multiply(#8,#9)|multiply(#12,#13)|
|
gain
|
A
|
a bag contains 6 black and 9 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ?
|
"let number of balls = ( 6 + 9 ) = 15 . number of white balls = 9 . p ( drawing a white ball ) = 9 / 15 = 3 / 5 hence answer is e"
|
a ) 4 , b ) 4 / 3 , c ) 4 / 5 , d ) 4 / 9 , e ) 3 / 5
|
e
|
divide(add(divide(divide(factorial(9), factorial(subtract(9, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 9)), factorial(subtract(add(6, 9), const_2))), factorial(const_2)))
|
add(n0,n1)|factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|divide(#1,#6)|divide(#3,#7)|factorial(#9)|divide(#10,#2)|divide(#11,#2)|divide(#8,#12)|add(#13,#14)|divide(#15,#2)|divide(#16,#17)|
|
probability
|
E
|
an association of mathematics teachers has 1600 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ?
|
total umber of members = 1600 number of members that cast votes = 525 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315 / 1600 ) * 100 = 19.6 % answer e
|
a ) 75 % , b ) 58 % , c ) 42 % , d ) 14.5 % , e ) 19.6 %
|
e
|
divide(multiply(divide(multiply(525, 60), const_100), const_100), 1600)
|
multiply(n1,n2)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)
|
gain
|
E
|
the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 5 . if the sum of the first and the last number is 17 . what is the last numbers ?
|
"a + b + c = 18 b + c + d = 15 a + d = 17 a – d = 3 a + d = 17 2 d = 14 d = 7 answer : d"
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
|
d
|
subtract(subtract(multiply(3, 6), add(subtract(17, 6), 3)), 6)
|
multiply(n1,n3)|subtract(n6,n3)|add(n1,#1)|subtract(#0,#2)|subtract(#3,n3)|
|
general
|
D
|
7 people average age is 30 . youngest person age is 6 . 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 - 42 = 161 so average age would be 161 / 7 = 24 answer : b"
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
|
b
|
divide(subtract(multiply(30, 7), multiply(6, 7)), 7)
|
multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n0)|
|
general
|
B
|
the sum of 4 consecutive even numbers is 52 , find the third number ?
|
let the first number be n let the second number be n + 2 let the third number be n + 4 let the third number be n + 6 n + n + 2 + n + 4 + n + 6 = 52 combine like terms : 4 n + 12 = 52 4 n + 12 = 52 4 n = 52 - 12 4 n = 40 n = 10 n + 2 = 12 n + 4 = 14 n + 6 = 16 the third number is 14 answer : a
|
a ) 14 , b ) 10 , c ) 12 , d ) 16 , e ) 20
|
a
|
add(divide(52, 4), const_1)
|
divide(n1,n0)|add(#0,const_1)
|
physics
|
A
|
how many numbers between 1000 and 5000 are exactly divisible by 225 ?
|
explanation : first multiple of 225 after 1000 is 1125 ( 225 ã — 5 ) and last multiple of 225 before 5000 is 4950 ( 225 ã — 22 ) total number are l + a / d = 4950 - 1125 / 225 + 1 = 18 answer is b
|
a ) 16 , b ) 18 , c ) 19 , d ) 12 , e ) 21
|
b
|
divide(subtract(5000, 1000), 225)
|
subtract(n1,n0)|divide(#0,n2)
|
general
|
B
|
the tax on a commodity is diminished by 10 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ?
|
"100 * 100 = 10000 80 * 120 = 9600 10000 - - - - - - - 400 100 - - - - - - - ? = 4 % answer : b"
|
a ) 18 % , b ) 4 % , c ) 32 % , d ) 12 % , e ) 52 %
|
b
|
subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 10)), const_100))
|
add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)|
|
general
|
B
|
a house wife saved $ 2.75 in buying an item on sale . if she spent $ 20 for the item , approximately how much percent she saved in the transaction ?
|
"actual price = 20 + 2.75 = $ 22.75 saving = 2.75 / 22.75 * 100 = 12 % approximately answer is e"
|
a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 12 %
|
e
|
multiply(divide(2.75, add(20, 2.75)), const_100)
|
add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|
|
general
|
E
|
if 7 persons can do 7 times of a particular work in 7 days , then , 9 persons can do 9 times of that work in ?
|
that is , 1 person can do one time of the work in 7 days . therefore , 9 persons can do 9 times work in the same 7 days itself . option ' e '
|
a ) 3 days , b ) 8 days , c ) 9 days , d ) 11 days , e ) 7 days
|
e
|
divide(divide(multiply(7, 7), divide(7, 9)), 9)
|
divide(n0,n3)|multiply(n0,n0)|divide(#1,#0)|divide(#2,n3)
|
physics
|
E
|
find the smallest number of five digits exactly divisible by 22 , 33,66 and 44 .
|
"smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 22,33 , 66,44 i . e 132 , on dividing 10000 by 132 , we get 32 as remainder . therefore , required number = 10000 + ( 132 â € “ 32 ) = 10100 . answer is e ."
|
a ) 10101 , b ) 11000 , c ) 10110 , d ) 10111 , e ) 10100
|
e
|
add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 33,66)
|
multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)|
|
general
|
E
|
each child has 3 crayons and 12 apples . if there are 6 children , how many crayons are there in total ?
|
3 * 6 = 18 . answer is c .
|
a ) 22 , b ) 65 , c ) 18 , d ) 36 , e ) 10
|
c
|
multiply(6, 3)
|
multiply(n0,n2)|
|
general
|
C
|
the difference between a number and its two - fifth is 510 . what is 15 % of that number ?
|
"let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 15 % of 850 = 127.5 . answer : e"
|
a ) 152.01 , b ) 85 , c ) 76 , d ) 168.2 , e ) 127.5
|
e
|
divide(multiply(510, add(const_4, const_1)), add(const_1, const_2))
|
add(const_1,const_4)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|
|
general
|
E
|
if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number a of integers that can be negative ?
|
minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning 5 out of 6 also possible as you will see a = 5 or 1 will give you same equation . . ans d
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
|
d
|
add(divide(const_10, const_2), 0)
|
divide(const_10,const_2)|add(n0,#0)
|
general
|
D
|
find large number from below question the difference of two numbers is 1000 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder
|
"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1000 = 6 x + 15 5 x = 985 x = 197 large number = 197 + 1365 = 1197 b"
|
a ) 1345 , b ) 1197 , c ) 1540 , d ) 1600 , e ) 1635
|
b
|
multiply(divide(subtract(1000, 15), subtract(6, const_1)), 6)
|
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|
|
general
|
B
|
find the probability that a number selected from numbers 1 , 2 , 3 , . . . , 50 is a prime number , when each of the given numbers is equally likely to be selected ?
|
let x be the event of selecting a prime number . x = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 } n ( x ) = 15 , n ( s ) = 50 hence , the required probability is 15 / 50 . answer : b
|
a ) 10 / 50 , b ) 15 / 50 , c ) 8 / 50 , d ) 3 / 50 , e ) 7 / 50
|
b
|
divide(multiply(const_3, const_5), 50)
|
multiply(const_3,const_5)|divide(#0,n3)
|
probability
|
B
|
12 : 6 seconds : : ? : 8 minutes
|
"12 * 8 = 6 * x x = 16 answer : c"
|
a ) 10 , b ) 15 , c ) 16 , d ) 25 , e ) 30
|
c
|
multiply(8, divide(12, 6))
|
divide(n0,n1)|multiply(n2,#0)|
|
physics
|
C
|
the largest natural number which exactly divides the product of any 4 consecutive natural numbers is
|
sol . required number = 1 × 2 × 3 × 4 = 24 . answer c
|
a ) 6 , b ) 12 , c ) 24 , d ) 120 , e ) none
|
c
|
factorial(4)
|
factorial(n0)
|
general
|
C
|
find the lowest common multiple of 24 , 36 and 42 .
|
"lcm = 2 * 2 * 2 * 3 * 3 * 7 = 504 . answer is b"
|
a ) 360 , b ) 504 , c ) 510 , d ) 320 , e ) 280
|
b
|
lcm(lcm(24, 36), 42)
|
lcm(n0,n1)|lcm(n2,#0)|
|
general
|
B
|
machine – a produces 40 % of the total output and machine - b produces 60 % of the total output . an average of 9 units out of a 1000 goods manufactured by machine - a and one unit of 50 units produced by machine - b prove to be defective . what is the probability that a unit chosen at random from the total daily output of the factory is defective ?
|
let total production be 10000 units . . . . a produces 4000 units and 36 units are defective b produces 6000 units and 120 units are defective so , out of total 10,000 units 156 units are defective . . . so the required probability = 156 / 10000 = > 0.0156 answer will be ( c )
|
a ) a . 0.156 , b ) b . 0.01506 , c ) c . 0.0156 , d ) d . 0.0001566 , e ) e . 0.00001566
|
c
|
divide(add(divide(divide(multiply(1000, 60), 40), 50), 9), add(divide(multiply(1000, 60), 40), 1000))
|
multiply(n1,n3)|divide(#0,n0)|add(n3,#1)|divide(#1,n4)|add(n2,#3)|divide(#4,#2)
|
general
|
C
|
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 110000 , find the cost of a camel ?
|
"cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 110000 p = rs . [ ( 24 * 4 * 4 * 110000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 42240000 / 9600 ) = > p = rs . 4400 answer : d"
|
a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 4400 , e ) s . 6880
|
d
|
divide(multiply(multiply(multiply(24, 4), 4), 110000), multiply(multiply(multiply(10, 16), 6), 10))
|
multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)|
|
general
|
D
|
in a hostel , the number of students decreased by 5 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ?
|
cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 5 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.95 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.95 * 1.2 ) = > f ( new ) = 0.877 f therefore the new cost of food must be 87.7 % of the old cost , or the cost of food must decrease by 12.28 % ( optionb )
|
a ) 19 % , b ) 12.3 % , c ) 25 % , d ) 40 % , e ) 9.4 %
|
b
|
multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 5), add(const_100, 20)))), const_100)
|
add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)
|
general
|
B
|
a train speeds past a pole in 50 seconds and a platform 500 m long in 100 seconds . its length is :
|
"let the length of the train be x meters and its speed be y m / sec . they , x / y = 50 = > y = x / 50 x + 500 / 100 = x / 50 x = 500 m . answer : option e"
|
a ) 550 m . , b ) 300 m . , c ) 600 m . , d ) 400 m . , e ) 500 m .
|
e
|
multiply(500, subtract(const_2, const_1))
|
subtract(const_2,const_1)|multiply(n1,#0)|
|
physics
|
E
|
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 231 games will be played during the tournament . what is the number of participants ?
|
"let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 231 n * ( n - 1 ) = 462 = 22 * 21 ( trial and error ) the answer is d ."
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24
|
d
|
divide(add(sqrt(add(multiply(multiply(231, const_2), const_4), const_1)), const_1), const_2)
|
multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|
|
general
|
D
|
in a 500 member association consisting of men and women , exactly 20 % of men and exactly 50 % women are homeowners . what is the least number of members who are homeowners ?
|
"solution simple out of 500 20 % are male i . e 100 and 50 % are female i . e 250 , so total homeowner is 350 . now min number homeowner is 100 and max is 250 so question ask us to find least and 101 has least value among all option . so ans is 101 . ans : c"
|
a ) 149 , b ) 147 , c ) 101 , d ) 143 , e ) 141
|
c
|
add(multiply(multiply(divide(50, const_100), 20), multiply(divide(50, const_100), 20)), divide(subtract(500, 20), 20))
|
divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)|
|
general
|
C
|
there were totally 100 men . 83 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ?
|
100 - ( 100 - 83 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 17 - 25 - 15 - 30 = 100 - 87 = 13 answer : c
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
|
c
|
subtract(100, add(add(add(subtract(100, 83), subtract(100, 75)), subtract(100, 85)), subtract(100, 70)))
|
subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)
|
general
|
C
|
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 51200 ?
|
51200 * 9 / 8 * 9 / 8 = 64800 . answer : e
|
a ) 81000 , b ) 81028 , c ) 27772 , d ) 29889 , e ) 64800
|
e
|
add(add(51200, multiply(divide(1, 8), 51200)), multiply(divide(1, 8), add(51200, multiply(divide(1, 8), 51200))))
|
divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)|
|
general
|
E
|
4 - fifths of 3 - eighths of a number is 24 . what is 250 per cent of that number ?
|
let the number = x . 4 ⁄ 5 × 3 ⁄ 8 x = 24 or x = 24 × 2 × 5 / 3 = 80 ∴ 250 per cent of the number = 250 ⁄ 100 × 80 = 200 answer d
|
a ) 100 , b ) 160 , c ) 120 , d ) 200 , e ) none of these
|
d
|
multiply(divide(24, multiply(divide(4, add(const_1, const_4)), divide(3, add(const_4, const_4)))), divide(250, const_100))
|
add(const_1,const_4)|add(const_4,const_4)|divide(n3,const_100)|divide(n0,#0)|divide(n1,#1)|multiply(#3,#4)|divide(n2,#5)|multiply(#6,#2)
|
general
|
D
|
at the end of the day , february 14 th , a florist had 150 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 20 of which were white and 15 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ?
|
"r + w + p = 150 s + l = 150 1 / 3 * 150 = 50 short - stemmed white = 20 short - stemmed pink = 15 = > short - stemmed red = 15 15 / p = 15 / r = > r = p so total long stemmed = 100 and long stemmed red + long stemmed pink = 100 so long stemmed red / long stemmed = ? total white = 20 ( as no long stemmed white ) = > r + r + 20 = 150 = > r = 130 and r = 65 long stemmed r = 65 - 15 = 50 so long stemmed red / r = 50 / 100 = 50 % answer - a"
|
a ) 50 % , b ) 25 % , c ) 250 % , d ) 75 % , e ) 80 %
|
a
|
multiply(divide(divide(multiply(subtract(150, divide(150, const_3)), subtract(subtract(divide(150, const_3), 20), 15)), add(15, subtract(subtract(divide(150, const_3), 20), 15))), subtract(150, divide(150, const_3))), const_100)
|
divide(n1,const_3)|subtract(n1,#0)|subtract(#0,n2)|subtract(#2,n3)|add(n3,#3)|multiply(#1,#3)|divide(#5,#4)|divide(#6,#1)|multiply(#7,const_100)|
|
general
|
A
|
in a group of 27 junior high school students , 7 take french , 10 take spanish , and 4 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ?
|
"a 7 add 7 + 10 + 4 to get 21 . then subtract 21 from the total students ⇒ 27 – 21 = 6 . answer is e"
|
a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 6
|
e
|
subtract(27, add(7, 10))
|
add(n1,n2)|subtract(n0,#0)|
|
other
|
E
|
the ratio of boarders to day students at a school was originally 5 to 12 . however , after a number of new boarders join the initial 220 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ?
|
"let x be the number of new boarders . the ratio changed from 5 : 12 up to 1 : 2 = 6 : 12 . 220 / ( 220 + x ) = 5 / 6 x = 44 the answer is a ."
|
a ) 44 , b ) 64 , c ) 70 , d ) 80 , e ) 84
|
a
|
subtract(divide(multiply(divide(220, 5), 12), 2), 220)
|
divide(n2,n0)|multiply(n1,#0)|divide(#1,n4)|subtract(#2,n2)|
|
other
|
A
|
by selling 8 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 10 % ?
|
"90 % - - - 8 110 % - - - ? 90 / 110 * 8 = 6.54 answer : a"
|
a ) 6.54 , b ) 9.45 , c ) 7.34 , d ) 6.67 , e ) 5.09
|
a
|
multiply(divide(const_1, multiply(add(const_100, 10), divide(const_1, subtract(const_100, 10)))), 8)
|
add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)|
|
gain
|
A
|
300 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees
|
26 trees have 25 gaps between them , required distance ( 300 / 25 ) = 12 b
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
|
b
|
divide(300, add(subtract(26, 2), const_1))
|
subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)
|
physics
|
B
|
in an institute , 60 % of the students are boys and the rest are girls . further 15 % of the boys and 7.5 % of the girls are getting a fee waiver . if the number of those getting a fee waiver is 90 , find the total number of students getting 50 % concessions if it is given that 50 % of those not getting a fee waiver are eligible to get half fee concession ?
|
solution : let us assume there are 100 students in the institute . then , number of boys = 60 and , number of girls = 40 further , 15 % of boys get fee waiver = 9 boys 7.5 % of girls get fee waiver = 3 girls total = 12 students who gets fee waiver but , here given 90 students are getting fee waiver . so we compare 12 = 90 so , 1 = 90 / 12 = 7.5 now number of students who are not getting fee waiver = 51 boys and 37 girls . 50 % concession = 25.5 boys and 18.5 girls ( i . e . total 44 ) . hence , required students = 44 * 7.5 = 330 . answer : option d
|
a ) 360 , b ) 280 , c ) 320 , d ) 330 , e ) 350
|
d
|
multiply(divide(90, add(multiply(divide(60, const_100), divide(15, const_100)), multiply(subtract(const_1, divide(60, const_100)), divide(7.5, const_100)))), add(divide(multiply(subtract(const_1, divide(15, const_100)), divide(60, const_100)), const_2), divide(multiply(subtract(const_1, divide(7.5, const_100)), subtract(const_1, divide(60, const_100))), const_2)))
|
divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|subtract(const_1,#1)|subtract(const_1,#2)|subtract(const_1,#0)|multiply(#0,#4)|multiply(#5,#6)|multiply(#2,#6)|add(#3,#9)|divide(#7,const_2)|divide(#8,const_2)|add(#11,#12)|divide(n3,#10)|multiply(#13,#14)
|
gain
|
D
|
9 log 9 ( 4 ) = ?
|
"exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 4 ) = 4 correct answer d"
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
|
d
|
divide(log(multiply(9, 9)), log(const_10))
|
log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)|
|
other
|
D
|
the ratio of the number of females to males at a party was 1 : 2 but when 6 females and 6 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 - 6 ) = 2 x - 6 x = 12 there were 3 x = 36 people at the party originally . the answer is d ."
|
a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40
|
d
|
add(divide(subtract(multiply(multiply(6, 2), 3), multiply(6, 2)), 2), subtract(multiply(multiply(6, 2), 3), multiply(6, 2)))
|
multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,n1)|add(#3,#2)|
|
other
|
D
|
if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 27 , find the value of ab .
|
"2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 27 - 9 = 18 ab = 9 . answer is b ."
|
a ) 5 , b ) 9 , c ) 4 , d ) 10 , e ) 3
|
b
|
divide(subtract(27, power(3, 2)), 2)
|
power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|
|
general
|
B
|
the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 56 years later ?
|
the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 56 years , the investment will get doubled 56 / 14 = 4 times . so the amount invested will get doubled thrice . so , 40000 * 2 ^ 4 = 640000 hence , the answer is e .
|
a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 640,000
|
e
|
divide(const_3600, const_10)
|
divide(const_3600,const_10)
|
general
|
E
|
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6 hours , then the distance covered at 40 kmph is
|
"let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6 or , x / 40 + ( 250 - x ) / 60 = 6 or , ( 3 x + 500 - 2 x ) / / 120 = 6 or 500 + x = 720 x = 220 answer : a"
|
a ) 220 km , b ) 120 km , c ) 100 km , d ) 140 km , e ) none of these
|
a
|
multiply(divide(subtract(multiply(60, 6), 250), negate(subtract(40, 60))), 40)
|
multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)|
|
physics
|
A
|
on a certain test , bill scored 20 more points than john , but half as many points as sue . if the sum of the scores of the 3 students was 160 points , how many points did bill receive ?
|
consider john ' s points as x so bill ' s points are x + 20 so sue ' s points are 2 x + 40 total of all three students is 4 x + 60 , but this total is given as 160 x = 25 so bill ' s points = 25 + 20 = 45 correct answer option b
|
a ) 30 , b ) 45 , c ) 50 , d ) 80 , e ) 100
|
b
|
divide(add(160, 20), add(add(const_1, const_1), const_2))
|
add(n0,n2)|add(const_1,const_1)|add(#1,const_2)|divide(#0,#2)
|
general
|
B
|
what percent of 16 is 16 percent of 1 ?
|
"16 % of 1 = ( 16 / 100 ) * 1 = 16 / 100 to determine what percentage of 16 this is : [ 16 ] [ / 100 * 16 ] * 100 = 1 % ans : b"
|
a ) 0.001 , b ) 1 , c ) 0.1 , d ) 100 , e ) 101 .
|
b
|
p_after_gain(16, 16)
|
p_after_gain(n1,n0)|
|
gain
|
B
|
let a be the event that a randomly selected two digit number is divisible by 10 and let b be the event that a randomly selected two digit number is divisible by 5 . what is p ( a and b ) ?
|
p ( a and b ) = 1 / 10 * 1 / 5 = 1 / 50 the answer is a .
|
a ) 1 / 50 , b ) 1 / 30 , c ) 1 / 15 , d ) 1 / 10 , e ) 1 / 5
|
a
|
multiply(divide(subtract(10, const_1), multiply(subtract(10, const_1), 10)), divide(multiply(subtract(10, const_1), const_2), multiply(subtract(10, const_1), 10)))
|
subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)
|
physics
|
A
|
anna left for city a from city b at 5.20 a . m . she traveled at the speed of 80 km / hr for 2 hrs 15 min . after that the speed was reduced to 60 km / hr . if the distance between two cities is 350 kms , at what time did anna reach city a ?
|
distance covered in 2 hrs 15 min i . e . , 2 1 / 4 hrs = 80 * 9 / 4 = 180 hrs . time taken to cover remaining distance = ( 350 - 180 ) / 60 = 17 / 6 hrs = 2 5 / 6 = 2 hrs 50 min total time taken = ( 2 hrs 15 min + 2 hrs 50 min ) = 5 hrs 5 min . so , anna reached city a at 10.25 a . m . answer : e
|
a ) 10.21 , b ) 10.27 , c ) 10.25 , d ) 10.23 , e ) 60.25
|
e
|
add(60, divide(multiply(subtract(divide(divide(subtract(350, divide(multiply(80, multiply(const_3, const_3)), const_4)), 60), 2), const_1), const_60), const_100))
|
multiply(const_3,const_3)|multiply(n1,#0)|divide(#1,const_4)|subtract(n5,#2)|divide(#3,n4)|divide(#4,n2)|subtract(#5,const_1)|multiply(#6,const_60)|divide(#7,const_100)|add(n4,#8)
|
physics
|
E
|
a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and captain ) to a nearby port . altogether , these passengers have 15 heads and 42 legs . how many cats does the ship host ?
|
"sa ' s + co + ca + cats = 15 . sa ' s + 1 + 1 + cats = 15 or sa ' s + cats = 13 . sa ' s ( 2 ) + 2 + 2 + cats * 4 = 42 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 13 - cats + cat * 2 = 19 then cats = 6 b"
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
|
b
|
multiply(divide(subtract(subtract(42, const_1), multiply(subtract(15, const_1), const_2)), subtract(multiply(subtract(15, const_1), const_4), multiply(subtract(15, const_1), const_2))), subtract(15, const_1))
|
subtract(n1,const_1)|subtract(n0,const_1)|multiply(#1,const_2)|multiply(#1,const_4)|subtract(#0,#2)|subtract(#3,#2)|divide(#4,#5)|multiply(#6,#1)|
|
general
|
B
|
find the 25 % of rs . 300 .
|
"explanation : 25 % of 300 = > 25 / 100 * 300 = rs . 75 answer : e"
|
a ) s . 50 , b ) s . 70 , c ) s . 100 , d ) s . 80 , e ) s . 75
|
e
|
multiply(300, divide(25, const_100))
|
divide(n0,const_100)|multiply(n1,#0)|
|
gain
|
E
|
what is the decimal equivalent of ( 1 / 6 ) ^ 2 ?
|
"( 1 / 6 ) ² = ( 1 / 6 ) ( 1 / 6 ) = 1 / 36 approach # 1 : use long division to divide 36 into 1 to get 1 / 36 = 0.0278 d"
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.278 , e ) 0.5
|
d
|
power(divide(1, const_4.0), 2)
|
divide(n0,const_4.0)|power(#0,n2)|
|
general
|
D
|
a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 18 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ?
|
"c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 18 - 2 cmb = 100 b - cb - mb - 2 cmb = 28 hence the probability = 28 / 100 = 7 / 25 . d"
|
a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 2 , d ) 7 / 25 , e ) 9 / 10
|
d
|
divide(subtract(100, subtract(add(50, 40), 18)), 100)
|
add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|divide(#2,n0)|
|
other
|
D
|
tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 45000 after completion of one year . what will be jose ' s share of profit ?
|
"sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 45000 = 25000 c"
|
a ) 20000 , b ) 10000 , c ) 25000 , d ) 34000 , e ) 30000
|
c
|
multiply(45000, subtract(const_1, divide(multiply(30000, multiply(2, multiply(2, const_3))), add(multiply(45000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(30000, multiply(2, multiply(2, const_3)))))))
|
multiply(n1,const_3)|multiply(#0,n1)|multiply(n0,#1)|subtract(#1,n1)|multiply(n2,#3)|add(#4,#2)|divide(#2,#5)|subtract(const_1,#6)|multiply(n3,#7)|
|
gain
|
C
|
john purchased 1325 large bottles at $ 1.89 per bottle and 750 small bottles at $ 1.38 per bottle . what was the approximate average price paid per bottle ?
|
"( 1325 * 1.89 + 750 * 1.38 ) / ( 1325 + 750 ) = ~ 1.71 option ( b )"
|
a ) $ 1.63 , b ) $ 1.71 , c ) $ 1.68 , d ) $ 1.72 , e ) $ 1.76
|
b
|
divide(add(multiply(1325, 1.89), multiply(750, 1.38)), add(1325, 750))
|
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|
|
general
|
B
|
in a certain corporation , there are 300 male employees and 100 female employees . it is known that 20 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 400 employees is chosen at random , what is the probability this employee has an advanced degree or is female ?
|
"in this corporation , there are 400 total employees . there are 100 women . of the 300 men , 20 % have advanced degrees — - 10 % of 300 must be 30 , so 20 % of 300 must be 60 . add the women and the men with advanced degrees : 100 + 60 = 160 . this is the or region , full set of individuals that satisfy the condition “ has an advanced degree or is female . ” of the 400 employees , what ’ s the probability of picking one of the 160 in this particular group ? p = 160 / 400 = 16 / 40 = 4 / 10 = 2 / 5 answer = d"
|
a ) 1 / 20 , b ) 1 / 10 , c ) 1 / 5 , d ) 2 / 5 , e ) 3 / 4
|
d
|
add(divide(multiply(subtract(const_1, divide(40, multiply(20, 20))), 100), 400), divide(add(multiply(divide(20, multiply(20, 20)), 300), multiply(divide(40, multiply(20, 20)), 100)), 400))
|
multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9)|
|
other
|
D
|
calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 15 % and its breadth is decreased by 10 % ?
|
"let l and b be 100 each 100 * 100 = 10000 l increase by 15 % = 115 b decrease by 10 % = 90 115 * 90 = 10350 3.5 percent increase answer : e"
|
a ) 3.5 percent increase , b ) 9.5 percent increase , c ) 1.5 percent increase , d ) 3.9 percent increase , e ) 3.5 percent increase
|
e
|
multiply(divide(subtract(multiply(add(const_100, 15), subtract(const_100, 10)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100)
|
add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)|
|
geometry
|
E
|
a lemonade stand sold only small and large cups of lemonade on tuesday . 1 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ?
|
let the total # of cups be 10 . # of small cups sold 1 / 5 * 10 = 2 ; # of large cups sold 10 - 2 = 8 ; let the price of small cup be $ 6 , then the price of larges cup would be 7 / 6 * 6 = $ 7 ; revenue from small cups : 2 * $ 6 = $ 12 ; revenue from large cups cups : 8 * $ 7 = $ 56 ; fraction of total revenue from large cups : 56 / ( 56 + 12 ) = 14 / 17 . answer : d .
|
a ) 7 / 16 , b ) 7 / 15 , c ) 10 / 21 , d ) 14 / 17 , e ) 1 / 2
|
d
|
divide(multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), const_10))), add(multiply(multiply(divide(1, 5), const_10), multiply(divide(1, 5), const_10)), multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), const_10)))))
|
divide(n2,n3)|divide(n0,n1)|multiply(#1,const_10)|multiply(#0,#2)|multiply(#2,#2)|subtract(const_10,#2)|multiply(#3,#5)|add(#4,#6)|divide(#6,#7)
|
general
|
D
|
how many consecutive zeros are there at the end of 100 ! ( 100 factorial ) ?
|
number of zeros will be 24 reason 100 / 5 = 20 20 / 5 = 4 so the total number of 5 will be 20 + 4 = 24 answer : d
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
|
d
|
add(divide(100, add(const_4, const_1)), divide(100, multiply(add(const_4, const_1), add(const_4, const_1))))
|
add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3)
|
other
|
D
|
the maximum number of students among them 640 pens and 520 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ?
|
"number of pens = 640 number of pencils = 520 required number of students = h . c . f . of 640 and 520 = 40 answer is a"
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60
|
a
|
gcd(640, 520)
|
gcd(n0,n1)|
|
general
|
A
|
16 men can complete a piece of work in 28 days . in how many days can 20 men complete that piece of work ?
|
"16 * 28 = 20 * x = > x = 22.4 days answer : c"
|
a ) 55 days , b ) 77 days , c ) 22.4 days , d ) 88 days , e ) 44 days
|
c
|
divide(multiply(28, 16), 20)
|
multiply(n0,n1)|divide(#0,n2)|
|
physics
|
C
|
jar y has 10 % more marbles than jar x . what percent of marbles from jar y need to be moved into x so that both jars have equal marbles
|
no ; the choices will be approximated only when gmat question explicitly mentions that ; here the answer is very close 4.54 , not 5 . jar y has 10 % more marbles than x y = 1.1 x if k marbles are taken out of y . the remaining marbles in jar y = 1.1 x - k if k marbles are added to x . the new marbles ' count in jar x = x + k and we know they are equal ; x + k = 1.1 x - k 2 k = 0.1 x k = 0.05 x the marbles taken out of y = 0.05 x question is : 0.05 x is what percent of marbles that was originally there in y i . e . 1.1 x 0.05 x 1.1 x ∗ 1000.05 x 1.1 x ∗ 100 0.05 x 1.1 x ∗ 1000.05 x 1.1 x ∗ 100 5011 = 4.54 ¯ ≈ 4.545011 = 4.54 ¯ ≈ 4.54 answer : a
|
a ) 4.54 % , b ) 9.09 % , c ) 10 % , d ) 11 % , e ) 12.25 %
|
a
|
multiply(divide(subtract(add(const_100, 10), divide(add(const_100, add(const_100, 10)), const_2)), add(const_100, 10)), const_100)
|
add(n0,const_100)|add(#0,const_100)|divide(#1,const_2)|subtract(#0,#2)|divide(#3,#0)|multiply(#4,const_100)
|
general
|
A
|
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 400 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ?
|
"i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 400 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 400 ) / 100 + ( 1.8 * 700 ) / 100 = 15.8 gms % of liquid x in resultant mixture = ( 15.8 / 1000 ) * 100 = 1.58 % a"
|
a ) 1.58 % , b ) 1.98 % , c ) 10 % , d ) 15 % , e ) 19 %
|
a
|
divide(add(multiply(400, 0.8), multiply(700, 1.8)), const_1000)
|
multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)|
|
gain
|
A
|
x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 4 will be equal to :
|
"explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 4 ^ 2 = 9 / 16 answer : e"
|
a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 16
|
e
|
divide(multiply(1, power(3, const_2)), power(4, const_2))
|
power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)|
|
general
|
E
|
when positive integer w is divided by 13 , the remainder is 2 . when n is divided by 8 , the remainder is 5 . how many such values are less than 180 ?
|
the equation that can be formed w is 13 x + 2 = 8 y + 5 . . 13 x - 3 = 8 y . . . as we can see x can take only odd values as the rhs will always be even . . also x can take values till 13 as 13 * 14 > 180 . . now we have to substitue x as 1 , 35 , 79 , 1113 . . . once we find 7 fitting in , any other value need not be checked as every 4 th value will give us answer so next value will be 15 . . ans 1 . . b
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
|
b
|
subtract(reminder(multiply(13, add(5, 2)), 8), 2)
|
add(n1,n3)|multiply(n0,#0)|reminder(#1,n2)|subtract(#2,n1)
|
general
|
B
|
a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 17 seconds . what is the length of the platform in meters ?
|
"speed of the train in metres / sec = 72000 / 3600 = 20 distance travelled by train to cross the platform = 30 * 20 = 600 = length of train + length of platform distance travelled by train to cross the man = 17 * 20 = 340 = length of train length of platform = 600 - 340 = 260 answer : b"
|
a ) 240 meters , b ) 260 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined
|
b
|
subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 17))
|
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)|
|
physics
|
B
|
the sum of two numbers is 12 and the product of the numbers is 35 . find sum of the squares of that numbers .
|
let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 12 ab = 35 so , 12 ^ 2 = a ^ 2 + b ^ 2 + 2 * 35 144 = a ^ 2 + b ^ 2 + 70 a ^ 2 + b ^ 2 = 74 ans e
|
a ) 84 , b ) 87 , c ) 96 , d ) 79 , e ) 74
|
e
|
divide(subtract(35, power(12, const_2)), const_2)
|
power(n0,const_2)|subtract(n1,#0)|divide(#1,const_2)|
|
general
|
E
|
anil had to do a multiplication . instead of taking 35 as one of the multipliers , he took 53 . as a result , the product went up by 540 . what is the new product ?
|
explanation : let the number that anil wanted to multiply be ' y ' . he was expected to find the value of 35 y . instead , he found the value of 53 y . the difference between the value that he got ( 53 y ) and what he was expected to get ( 35 y ) is 540 . = > 53 y - 35 y = 540 = > ( 53 - 35 ) * y = 540 = > y = 30 therefore , the correct product = 53 * 30 = 1590 answer : a
|
a ) 1590 , b ) 1600 , c ) 1950 , d ) 1630 , e ) 1760
|
a
|
add(subtract(add(multiply(35, 53), multiply(const_3, const_10)), multiply(const_4, const_100)), const_100)
|
multiply(n0,n1)|multiply(const_10,const_3)|multiply(const_100,const_4)|add(#0,#1)|subtract(#3,#2)|add(#4,const_100)
|
general
|
A
|
nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 1.5 screws and 4.5 nails ?
|
"let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 1.5 s and 4.5 n = 1.5 s + 4.5 * 1.5 s = 8.25 s . . ratio = 15 s / 8.25 s = 15 / 8.25 = 20 / 11 e"
|
a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 20 / 11
|
e
|
divide(add(multiply(3, const_4), 3), const_10)
|
multiply(n1,const_4)|add(n1,#0)|divide(#1,const_10)|
|
general
|
E
|
the largest four digit number which is a multiple of 8 , 10,12 and 15 is :
|
"lcm = 120 9999 / 120 = 83 83 * 120 = 9960 answer : a"
|
a ) 9960 , b ) 9950 , c ) 9940 , d ) 9930 , e ) 9920
|
a
|
square_area(const_pi)
|
square_area(const_pi)|
|
general
|
A
|
a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 7,000 meters of fabric ?
|
"for first 2000 meters he does not get any discount . the price is 2 * 2000 = $ 4000 for next 1500 meters , he gets a 5 % discount . the price is 1.9 * 1500 = $ 2850 for the next 1500 meters , he gets a 7 % discount . the price is 1.86 * 3500 = $ 6510 the total price is $ 4000 + $ 2850 + $ 6510 = $ 13,360 the answer is a ."
|
a ) $ 13,360 , b ) $ 14,450 , c ) $ 15,720 , d ) $ 16,780 , e ) $ 17,240
|
a
|
multiply(multiply(2, const_3), const_100)
|
multiply(n4,const_3)|multiply(#0,const_100)|
|
gain
|
A
|
a rotameter is a device that measures flow of liquid and gases . when measuring liquid phase flows , 2.5 inches represent 60 liters per minute of liquid . with gas measurements the rotameter moves 50 % of the movement he moves with the liquid phase . how many liters of gas passed through the rotameter if it measured 4 inches ?
|
in case of liquid - 2.5 inches represents 60 lit / min . in case of gas - 50 % of 2.5 inches represents 60 lit / min 1.25 inches represents 60 lit / min 4 inches will represent 60 * 4 / 1.25 = 192 b is the answer
|
a ) 176 , b ) 192 , c ) 202 , d ) 218 , e ) 284
|
b
|
divide(divide(multiply(60, 4), 2.5), divide(50, const_100))
|
divide(n2,const_100)|multiply(n1,n3)|divide(#1,n0)|divide(#2,#0)
|
physics
|
B
|
at veridux corporation , there are 210 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ?
|
"well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 210 â € “ 90 = 120 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 210 â € “ 40 = 170 must be associates . now , in the â € œ associate â € row , 135 + e = 170 , which means e = 35 â € ” the other 35 associates must be female . now , to find b , which is what the question is asking , we need only look at the sum in the â € œ female â € column : b + 35 = 90 , which means b = 55 . there are fifteen female managers in this company . thus , the answer = ( e ) ."
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 55
|
e
|
multiply(40, const_1)
|
multiply(n2,const_1)|
|
general
|
E
|
sandy walked 20 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 20 meters . what distance is she from the starting point and in which direction ?
|
"the net distance is 20 + 20 = 40 meters to the east . the answer is d ."
|
a ) 20 m east , b ) 40 m north , c ) 30 m west , d ) 40 m east , e ) 20 m west
|
d
|
add(20, 20)
|
add(n0,n3)|
|
physics
|
D
|
andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.86 . what is the median gasoline price ?
|
"ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.86 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.86 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b"
|
a ) $ 1 , b ) $ 1.86 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15
|
b
|
min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.86), 7), 1.86)
|
add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|add(n6,#3)|add(n7,#4)|divide(#5,n0)|min(n7,#6)|
|
general
|
B
|
if 6 a = 7 b = 20 , then 84 ab =
|
"6 a * 7 b = 20 * 20 = 400 42 ab = 400 i . e . 84 ab = 800 answer : option d"
|
a ) 50 , b ) 100 , c ) 625 , d ) 800 , e ) 1250
|
d
|
multiply(84, multiply(divide(20, 6), divide(20, 7)))
|
divide(n2,n0)|divide(n2,n1)|multiply(#0,#1)|multiply(n3,#2)|
|
general
|
D
|
what is the smallest positive integer that leaves a remainder of 1 when divided by 2 , remainder of 2 when divided by 3 , a remainder of 3 when divided by 4 , and a remainder of 9 when divided by 10 ?
|
remainder of 1 when divided by 2 : ( 2 - 1 = 1 ) remainder of 2 when divided by 3 : ( 3 - 2 = 1 ) remainder of 3 when divided by 4 : ( 4 - 3 = 1 ) remainder of 9 when divided by 10 : ( 2 - 1 = 1 ) so n + 1 should be divisible by 23 , 410 ( where n is desired number ) l . c . m . of 23 , 410 = 60 ( = n + 1 ) n = 59 answer : b
|
a ) 58 , b ) 59 , c ) 60 , d ) 61 , e ) 62
|
b
|
subtract(lcm(lcm(lcm(2, 3), 4), 10), const_1)
|
lcm(n1,n3)|lcm(n5,#0)|lcm(n7,#1)|subtract(#2,const_1)
|
general
|
B
|
a train running at the speed of 60 km / hr crosses a pole in 42 seconds . what is the length of the train ?
|
"speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 42 m = 150 m . option b"
|
a ) 120 metres , b ) 700 metres , c ) 324 metres , d ) 828 metres , e ) 600 metres
|
b
|
multiply(divide(multiply(60, const_1000), const_3600), 42)
|
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
|
physics
|
B
|
what is the value of x ^ 2 yz − xyz ^ 2 , if x = 2 , y = 1 , and z = 2 ?
|
"4 * 1 * 2 - ( 2 * 1 * 4 ) = 8 + 8 = 0 ans : c"
|
a ) 20 , b ) 24 , c ) 0 , d ) 32 , e ) 48
|
c
|
multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1))
|
negate(n0)|multiply(n3,#0)|subtract(#0,n3)|multiply(n3,#1)|multiply(#3,#2)|
|
general
|
C
|
given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] – 16 = f ( x – 6 ) ?
|
"2 ( 3 x - 5 ) - 16 = 3 ( x - 6 ) - 5 3 x = 3 x = 1 the answer is a ."
|
a ) 1 , b ) 4 , c ) 6 , d ) 7 , e ) 13
|
a
|
divide(subtract(add(multiply(2, 5), 16), add(multiply(3, 6), 5)), subtract(multiply(2, 3), multiply(3, const_1)))
|
multiply(n1,n2)|multiply(n4,n0)|multiply(n0,n2)|multiply(n0,const_1)|add(n3,#0)|add(n1,#1)|subtract(#2,#3)|subtract(#4,#5)|divide(#7,#6)|
|
general
|
A
|
a 340 - liter solution of kola is made from 75 % water , 5 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 12 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ?
|
"denominator : 340 + 12 + 3.2 + 6.8 = 362 numerator : 340 ( 1 - . 75 - . 05 ) + 3.2 340 ( 0.2 ) + 3.2 68 + 3.2 71.2 ratio : 71.2 / 362 = 0.196 answer : c"
|
a ) 6 % . , b ) 7.5 % . , c ) 19.6 % . , d ) 10.5 % . , e ) 11 % .
|
c
|
multiply(divide(add(subtract(subtract(340, multiply(340, divide(75, const_100))), multiply(340, divide(5, const_100))), 3.2), add(add(add(340, 3.2), 12), 6.8)), const_100)
|
add(n0,n3)|divide(n1,const_100)|divide(n2,const_100)|add(n4,#0)|multiply(n0,#1)|multiply(n0,#2)|add(n5,#3)|subtract(n0,#4)|subtract(#7,#5)|add(n3,#8)|divide(#9,#6)|multiply(#10,const_100)|
|
gain
|
C
|
in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is 28160 , how many purple chips were selected ?
|
"28160 = 1 * 5 * 8 ^ 3 * 11 the factors of 8 must come from the purple point value , so there are 3 purple chips . the answer is c ."
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
|
c
|
subtract(5, const_2)
|
subtract(n1,const_2)|
|
general
|
C
|
operation # is defined as adding a randomly selected two digit multiple of 8 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated 10 times , what is the probability that it will yield at least two integers ?
|
any multiple of 8 is even . any two - digit prime number is odd . ( even + odd ) / 2 is not an integer . thus # does not yield an integer at all . therefore p = 0 . answer : a .
|
a ) 0 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 40 %
|
a
|
divide(add(8, add(10, const_3)), const_2)
|
add(n1,const_3)|add(n0,#0)|divide(#1,const_2)
|
general
|
A
|
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ?
|
"the time it takes printer x is 12 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 12 / ( 20 / 3 ) = 3 * 12 / 20 = 9 / 5 the answer is c ."
|
a ) 5 / 3 , b ) 7 / 4 , c ) 9 / 5 , d ) 11 / 6 , e ) 13 / 7
|
c
|
divide(12, divide(const_1, add(divide(const_1, 10), divide(const_1, 20))))
|
divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)|
|
general
|
C
|
the average of 45 results is 8 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ?
|
"23 th result = sum of 45 results - sum of 44 results 8 * 45 - 15 * 22 + 20 * 22 = 360 - 330 + 440 = 470 answer is b"
|
a ) 600 , b ) 470 , c ) 750 , d ) 650 , e ) 560
|
b
|
subtract(subtract(multiply(45, 8), multiply(22, 20)), multiply(22, 15))
|
multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)|
|
general
|
B
|
the remainder when e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 .
|
e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 - 1 ) / 2 divided by 13 3 ^ 201 - 1 / 26 wkt , 3 ^ 3 = 27 = 26 + 1 { ( 26 + 1 ) ^ 67 - 1 } / 26 { 1 - 1 } / 26 = > 0 . c
|
a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3
|
c
|
multiply(divide(add(200, const_1), const_3), add(add(reminder(1, 13), reminder(3, 13)), reminder(power(3, const_2), 13)))
|
add(n7,const_1)|power(n1,const_2)|reminder(n0,n8)|reminder(n1,n8)|add(#2,#3)|divide(#0,const_3)|reminder(#1,n8)|add(#4,#6)|multiply(#7,#5)
|
general
|
C
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.