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mathqa_shuffle_question_test

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what is the angle between the 2 hands of the clock at 8 : 24 pm ?
"without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 38 % , thus he must pay slightly more than $ 62 . answer : e ."
a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 65.00
e
multiply(subtract(10, divide(multiply(30, 8), const_100)), 10)
multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,n0)|
gain
a train 120 m long is running with a speed of 62 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
each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 180 ã · 40 = 9.5 ã — 1 gallon = 9.5 gallons correct answer is b ) 9.5 gallons
a ) 3.5 gallons , b ) 9.5 gallons , c ) 8.7 gallons , d ) 4.5 gallons , e ) 9.2 gallons
b
divide(190, 20)
divide(n1,n0)
physics
little john had $ 8.50 . he spent $ 1.25 on sweets and gave to his two friends $ 1.20 and $ 2.20 . how much money was left ?
"amount = [ 35000 * ( 1 + 12 / 100 ) 3 ] = 35000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 49172.48 c . i . = ( 49172.48 - 35000 ) = rs : 14172.48 answer : b"
a ) s : 10123.19 , b ) s : 14172.48 , c ) s : 10123.20 , d ) s : 10123.28 , e ) s : 10123.12
b
subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100))
divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)|
gain
on a certain transatlantic crossing , 20 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 50 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ?
"48 / 3 = 16 a"
a ) a ) 3 , b ) b ) 5 , c ) c ) 9 , d ) d ) 7 , e ) e ) 11
a
sqrt(48)
sqrt(n0)|
general
a is twice as good a work man as b and together they finish the work in 10 days . in how many days a alone can finish the work ?
"66 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 5.45 % answer : c"
a ) 3.45 % , b ) 4.45 % , c ) 5.45 % , d ) 6.45 % , e ) 7.45 %
c
multiply(divide(subtract(subtract(66, multiply(66, divide(10, const_100))), 56.16), subtract(66, multiply(66, divide(10, const_100)))), const_100)
divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)|
gain
find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 30 as remainder
"sol . let c . p . = rs . 100 . then , marked price = rs . 160 , s . p . = rs . 99 . ∴ discount % = [ 11 / 160 * 100 ] % = 6.8 % answer c"
a ) 10 % , b ) 10.5 % , c ) 6.8 % , d ) 12.5 % , e ) none
c
multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 60)))
add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)|
gain
3 different containers contain 50 litres , 100 litres and 150 litres of mixture of milk and water respectively . what is the biggest measure can measure all the different quantities exactly ?
"explanation : 243 + 26 = 269 / 14 = 3 ( remainder ) answer : e"
a ) 7 , b ) 6 , c ) 9 , d ) 2 , e ) 3
e
subtract(26, multiply(14, const_2))
multiply(n2,const_2)|subtract(n1,#0)|
general
6 persons in an organization including x and y were to be divided in two groups of 3 members each . the total number of groups containing both x and y is what fraction of the total number of groups which can be formed ?
"actual price = 200 + 10 = $ 210 saving = 10 / 210 * 100 = 100 / 21 = 5 % approximately answer is d"
a ) 8 % , b ) 10 % , c ) 2 % , d ) 5 % , e ) 6 %
d
add(floor(multiply(divide(10, add(10, 200)), const_100)), const_1)
add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|floor(#2)|add(#3,const_1)|
general
what is the remainder when 49 ^ 74 - 5 ^ 74 is divided by 24 ?
"average age of 36 students in a group is 14 sum of the ages of 36 students = 36 × 14 when teacher ' s age is included to it , the average increases by one = > average = 15 sum of the ages of 36 students and the teacher = 37 × 15 hence teachers age = 37 × 15 - 36 × 14 = 37 × 15 - 14 ( 37 - 1 ) = 37 × 15 - 37 × 14 + 14 = 37 ( 15 - 14 ) + 14 = 37 + 14 = 51 answer is e ."
a ) 50 , b ) 49 , c ) 53 , d ) 54 , e ) 51
e
add(36, const_1)
add(n0,const_1)|
general
at 12 : 30 , the hour hand and the minute hand of a clock form an angle of
"8 x = 128 = > x = 16 therefore the secretary who worked the longest spent 16 x 5 = 90 hours on the project option ( a )"
a ) 90 , b ) 70 , c ) 56 , d ) 16 , e ) 14
a
multiply(divide(128, add(add(1, 2), 5)), 5)
add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|
physics
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 12 pumps work to empty the tank in 1 day ?
a can finish 1 work in 20 days b can finish 1 / 1.5 work in 20 days - since a is 1.5 faster than b this means b can finish 1 work in 20 * 1.5 days = 30 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 20 * 30 / ( 20 + 30 ) = 20 * 30 / 50 = 12 days so answer is c
a ) 23 , b ) 22 , c ) 12 , d ) 24 , e ) 25
c
divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 1.5)))
divide(const_1,n1)|divide(#0,n0)|add(#0,#1)|divide(const_1,#2)
physics
how many three - digit numbers are divisible by 6 in all ?
"explanation : probability that a speaks truth is 65 / 100 = 0.65 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) × p ( b ) = 0.65 × 0.6 = 0.39 answer : a"
a ) 0.39 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.411
a
multiply(divide(65, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5)))
multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)|
gain
the mean of 50 observations was 40 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is
"c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 6 : 24 pm . the correct answer is e ."
a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24
e
subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 35)))
add(n16,const_60)|add(n3,const_60)|multiply(n23,const_10)|subtract(const_60,n16)|subtract(const_60,n1)|divide(#0,const_60)|divide(#1,const_60)|divide(#3,const_60)|multiply(#6,#7)|multiply(#5,#8)|multiply(#9,#4)|subtract(#2,#10)|
physics
what is the angle between the hands of a clock when time is 10 : 30 ?
if a = 1 , then putting values in equation = - [ ( 1 ) ^ 2 + ( 1 ) ^ 3 + ( 1 ^ 4 ) + ( 1 ^ 5 ) ] = - [ 1 + 1 + 1 + 1 ] = - 4 answer = b = - 4
a ) - 14 , b ) - 4 , c ) 0 , d ) 4 , e ) 14
b
negate(add(add(add(power(1, 2), power(1, 3)), power(1, 4)), power(1, 5)))
power(n0,n1)|power(n0,n2)|power(n0,n3)|power(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|negate(#6)
general
find the length of the wire required to go 14 times round a square field containing 5625 m 2 .
"sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 70 2 / 3 sp 1 = 70 sp 1 = 105 100 - - - 105 = > 5 % answer : e"
a ) 20 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 5 %
e
subtract(divide(subtract(const_100, 30), divide(2, 3)), const_100)
divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|
gain
andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.82 . what is the median gasoline price ?
we can determine quickly that total number should range between 49 / 8 < = n < = 49 / 1.5 , so ans should be between 6 and 33 . now solving the expression 8 a + 1.5 b = 49 decreasing 49 in multiple of 8 and checking divisibility of that number by 1.5 . this way we get 2 red giants , 22 white dwarfs we get 49 , but 2 + 22 = 24 and 24 is not an option . next we get 5 red giants and 6 white dwarfs to get 49 , 5 * 8 + 6 * 1.5 = 49 hence total number is 5 + 6 = 11 ans b
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
b
add(divide(subtract(49, multiply(1.5, add(const_2, const_4))), 8), add(const_2, const_4))
add(const_2,const_4)|multiply(n1,#0)|subtract(n2,#1)|divide(#2,n0)|add(#0,#3)
general
the population of a town increased from 1 , 75,000 to 2 , 62,500 in a decade . the average percent increase of population per year is :
a = b + 300 . a = 4 ( b - 600 ) . 4 ( b - 600 ) = b + 300 . 3 b = 2700 . b = 900 . the answer is a .
a ) 900 , b ) 1000 , c ) 1100 , d ) 1200 , e ) 1300
a
divide(add(multiply(600, 4), 300), subtract(4, const_1))
multiply(n1,n2)|subtract(n2,const_1)|add(n0,#0)|divide(#2,#1)
general
how many odd numbers between 10 and 1,000 are the squares of integers ?
"explanation : 2 c + 3 t = 1500 - - - ( 1 ) 3 c + 2 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 300 = > t - c = 300 answer : e"
a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 300
e
subtract(divide(subtract(multiply(3, 1500), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1500), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3))
multiply(n2,const_3)|multiply(n3,const_2)|multiply(n1,const_3)|multiply(n0,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n0,#6)|subtract(n3,#7)|divide(#8,n1)|subtract(#6,#9)|
general
find the greatest number which leaves the same remainder when it divides 21 , 57 and 105 .
"ci = 4851 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 4851 = p [ 21 / 20 ] ^ 2 4851 [ 20 / 21 ] ^ 2 4400 answer : c"
a ) s . 4000 , b ) s . 5000 , c ) s . 4400 , d ) s . 4800 , e ) s . 5800
c
divide(4851, power(add(divide(5, const_100), const_1), 2))
divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)|
gain
a pupil ' s marks were wrongly entered as 73 instead of 40 . due to the average marks for the class got increased by half . the number of pupils in the class is ?
"manager ' s monthly salary = rs . ( 1700 * 11 - 1600 * 10 ) = rs . 2700 answer : c"
a ) rs . 3601 , b ) rs . 3618 , c ) rs . 2700 , d ) rs . 3619 , e ) rs . 3610
c
subtract(multiply(add(1600, 100), add(10, const_1)), multiply(1600, 10))
add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
general
6 students wrote science exam . their average marks are 70 . 5 students got 65 , 75 , 55 , 72 and 69 marks respectively . therefore what is the marks of the sixth student ?
"( 1200 * 8 * 2 ) / 100 = > 192 answer : c"
a ) 190 , b ) 188 , c ) 192 , d ) 145 , e ) 188
c
subtract(divide(multiply(multiply(1200, 18), 2), const_100), divide(multiply(multiply(1200, 10), 2), const_100))
multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|
gain
solving a linear equation with several occurrences of the variable , solve for w . simplify answer as much as possible . ( 7 w + 6 ) / 6 + ( 9 w + 8 ) / 2 = 22
i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 6 = integer , therefore we have 48 / 96 numbers divisible by 6 o e o / 6 = not integer we can not forget multiples of 6 from 1 to 96 we have 16 numbers that are multiple of 8 therefore , 48 / 96 + 16 / 96 = 64 / 96 = 2 / 3 answer : a
a ) 2 / 3 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4
a
divide(add(divide(96, 2), divide(96, 6)), 96)
divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)
general
a man can row 11 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?
"the numbers should be of the form 5 c + 3 . the minimum is 3 when c = 0 . the maximum is 48 when c = 9 . there are 10 such numbers . the answer is e ."
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
e
divide(const_100.0, const_10)
divide(const_100.0,const_10)|
general
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 15 % profit ?
explanation total marks of 5 students = ( 65 + 75 + 55 + 72 + 69 ) = 336 required marks = [ ( 70 x 6 ) – 336 ] = ( 420 – 336 ) = 84 answer a
a ) 84 , b ) 68 , c ) 85 , d ) 75 , e ) 42
a
subtract(multiply(70, 6), add(add(add(add(65, 75), 55), 72), 69))
add(n3,n4)|multiply(n0,n1)|add(n5,#0)|add(n6,#2)|add(n7,#3)|subtract(#1,#4)
general
jar a has 6 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ?
"a relative speed = ( 54 + 90 ) * 5 / 18 = 8 * 5 = 40 mps . the time required = d / s = ( 100 + 100 + 200 ) / 35 = 400 / 40 = 10 sec ."
a ) 10 sec , b ) 11 sec , c ) 12 sec , d ) 60 / 7 sec , e ) 90 / 7 sec
a
divide(100, multiply(add(54, 90), const_0_2778))
add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
a student chose a number , multiplied it by 2 , then subtracted 180 from the result and got 104 . what was the number he chose ?
"total amount of water evaporated each day during a 50 - day period = . 008 * 50 = . 010 * 100 / 2 = 1.0 / 2 = . 5 percent of the original amount of water evaporated during this period = ( . 5 / 10 ) * 100 % = 5 % answer d"
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 5 % , e ) 40 %
d
multiply(divide(multiply(50, 0.010), 10), const_100)
multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|
gain
coconuts were purchased at 150 per 100 and sold at 2 per coconut . if 2000 coconuts were sold , what was the total profit made ?
"prime factors ’ number , as i assume , for a number x = a ^ n * b ^ m * c ^ o * d ^ p . . . is = n + m + o + p . . . so , 28 = 2 ^ 2 * 7 ^ 1 prime factors ’ number will be 2 + 1 = 3 . hence , answer is b ."
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
b
add(add(add(const_1, add(const_1, const_1)), const_1), const_1)
add(const_1,const_1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|
other
if taxi fares were $ 2.00 for the first 1 / 5 mile and $ 0.40 for each 1 / 5 mile there after , then the taxi fare for a 4 - mile ride was
t 7 + t 23 = t 8 + t 15 + t 13 = > a + 6 d + a + 22 d = a + 7 d + a + 14 d + a + 12 d = > a + 5 d = 0 = > t 6 = 0 i . e . 6 th term is zero . answer : a
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
a
subtract(add(13, add(8, 15)), add(7, 23))
add(n2,n3)|add(n0,n1)|add(n4,#0)|subtract(#2,#1)
general
how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 1 m x 2 m x 20 cm ?
"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 30 more than the number who did not . that means , total number of people who registered and paid for lunch = 30 + 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 ( 30 + x ) = 2160 + 72 x now , total amount received was 75360 . thus , from the above data , 50 x + 2160 + 72 x = 75360 122 x = 75360 - 2160 122 x = 73200 x = 600 . hence the correct ans is c"
a ) 700 , b ) 800 , c ) 600 , d ) 1,500 , e ) 1,800
c
multiply(const_1, const_1)
multiply(const_1,const_1)|
general
a man invests some money partly in 12 % stock at 105 and partly in 8 % stock at 88 . to obtain equal dividends from both , he must invest the money in the ratio :
"sol : breadth of the rectangular plot is = 5 ^ 2 - 4 ^ 2 = 3 m therefore , perimeter of the rectangular plot = 2 ( 4 + 3 ) = 14 m c ) 14 m"
a ) 20 m , b ) 15 m , c ) 14 m , d ) 10 m , e ) 25 m
c
divide(add(add(sqrt(subtract(power(5, const_2), power(4, const_2))), 4), add(sqrt(subtract(power(5, const_2), power(4, const_2))), 4)), 4)
power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|add(n1,#3)|add(#4,#4)|divide(#5,n1)|
geometry
the sum of all consecutive odd integers from − 19 to 29 , inclusive , is
"on dividing 709 by 9 , we get remainder = 7 therefore , required number to be subtracted = 7 answer : c"
a ) a ) 2 , b ) b ) 3 , c ) c ) 7 , d ) d ) 5 , e ) e ) 6
c
subtract(709, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9))
add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)|
general
how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?
prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 33 sum , c = 193 c / 3 = 64.3 answer c
a ) 155 , b ) 129 , c ) 64.3 , d ) 47 , e ) 43
c
add(divide(3, const_10), power(const_2, add(const_2, const_4)))
add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)
general
a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ?
speed downstream = ( 22 + 5 ) = 27 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time × speed = 2 / 5 × 27 = 10.8 km answer is c .
a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0
c
multiply(add(22, 5), divide(24, const_60))
add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)
physics
a man can row 11 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?
"volume of water displaced = ( 3 x 3 x 0.01 ) m 3 = 0.09 m 3 . mass of man = volume of water displaced x density of water = ( 0.09 x 1000 ) kg = 90 kg . answer : d"
a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 90 kg , e ) 110 kg
d
multiply(multiply(multiply(3, 3), divide(1, const_100)), const_1000)
divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|
physics
a train with 120 wagons crosses john who is going in the same direction , in 36 seconds . it travels for half an hour from the time it starts ove ( who is also riding on his horse ) coming from the opposite direction in 24 seconds . in how much time after the train has crossed the mike do the john meets to mike ? rtaking the john ( he is riding on the horse ) before it starts overtaking the mike
"80 % - - - 12 120 % - - - ? 80 / 120 * 12 = 8 answer : a"
a ) 8 , b ) 76 , c ) 17 , d ) 7 , e ) 77
a
multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 12)
add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)|
gain
what percent is 7 gm of 1 kg ?
"old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 35 = 10 * 12 / 7 = 17.14 time difference = 6.24 ans : a"
a ) 6.24 , b ) 8 , c ) 10 , d ) 15 , e ) 24
a
max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 35), divide(10, 55))), const_4)
add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)|
physics
find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 4 ) is
"speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 51 - - - - - - - - 7 4 - - - - - - - - - ? è 28 answer : c"
a ) 16 min , b ) 26 min , c ) 28 min , d ) 20 min , e ) 12 min
c
multiply(divide(7, divide(5, 4)), 5)
divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)|
physics
there are 3 prizes to be distributed among 10 students . if no students gets more than one prize , then this can be done in ?
"p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 150 15000 answer : a"
a ) s . 15000 , b ) s . 15100 , c ) s . 15800 , d ) s . 16000 , e ) s . 16200
a
divide(150, multiply(divide(10, const_100), divide(10, const_100)))
divide(n0,const_100)|multiply(#0,#0)|divide(n2,#1)|
gain
the speed of a boat in still water is 15 km / hr and the rate of the current is 3 km / hr . the distance travelled downstream in 12 minutes is
"solution speed in still water = 1 / 2 ( 8 + 2 ) km / hr = 5 kmph . answer b"
a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10
b
divide(add(8, 2), const_2)
add(n0,n1)|divide(#0,const_2)|
gain
a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c which is at midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ?
c . p . for one coconut = 150 ⁄ 100 = 3 ⁄ 2 s . p . for one coconut = 2 profit on one coconut = 2 - 3 ⁄ 2 = 1 ⁄ 2 ∴ profit on 2000 coconut = 1 ⁄ 2 × 2000 = 1000 answer b
a ) 500 , b ) 1000 , c ) 1500 , d ) 2000 , e ) none of these
b
multiply(2000, subtract(2, divide(150, 100)))
divide(n0,n1)|subtract(n2,#0)|multiply(n3,#1)
gain
s is a set of 85 consecutive multiples of 5 . if the smallest number in s is 90 , then the greatest number in s is
"c 120 120 ( 5 x 4 x 3 x 2 x 1 ) ."
a ) 1 , b ) 60 , c ) 120 , d ) 130 , e ) 180
c
circle_area(divide(5, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
other
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 9 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?
"rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * e = 1 = > x = 1 / e when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * 3 / 4 = 1 = > w + x = 4 / 3 = > 1 / 2 + 1 / e = 4 / 3 = > 1 / e = ( 8 - 3 ) / 6 = 5 / 6 = > e = 6 / 5 = 1 [ 1 / 5 ] answer b"
a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2
b
add(subtract(4, 2), divide(const_1, add(2, 3)))
add(n0,n1)|subtract(n2,n0)|divide(const_1,#0)|add(#2,#1)|
physics
a can give b 120 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ?
"explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 435 + 140 = 575 meter time = distance / speed = 575 ∗ 2 / 25 = 46 seconds option c"
a ) 20 seconds , b ) 30 seconds , c ) 46 seconds , d ) 50 seconds , e ) none of these
c
divide(add(435, 140), divide(multiply(45, const_1000), const_3600))
add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|
physics
a certain quantity of 50 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ?
"h . c . f = ( product of the numbers ) / ( their l . c . m ) = 22500 / 450 = 50 . answer : a"
a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these
a
divide(22500, 450)
divide(n1,n0)|
physics
the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the sum and the number if the ratio between the digits of the number is 1 : 2 ?
let base = x cm height = 2 x cm area = x ã — 2 x = 2 x ^ 2 area = x ã — 2 x = 2 x ^ 2 area is given as 72 cm ^ 2 2 x ^ 2 = 72 x ^ 2 = 36 x = 6 cm answer : c
['a ) 1 cm', 'b ) 3 cm', 'c ) 6 cm', 'd ) 4 cm', 'e ) 2 cm']
c
sqrt(divide(72, const_2))
divide(n0,const_2)|sqrt(#0)
geometry
alex and brian start a business with rs . 7000 each , and after 8 months , brian withdraws half of his capital . how should they share the profits at the end of the 18 months ?
pencil + notebook = 80 notebook + eraser = 115 pencil + eraser = 75 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 270 cents the cost to buy 3 of each would be ( 3 / 2 ) ( 270 ) = 405 the answer is e .
a ) 325 , b ) 345 , c ) 365 , d ) 385 , e ) 405
e
multiply(divide(add(add(multiply(1.15, const_100), 80), 75), const_2), 3)
multiply(n1,const_100)|add(n0,#0)|add(n2,#1)|divide(#2,const_2)|multiply(n3,#3)
gain
a train 250 m long passes a man , running at 10 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is :
"f 200 / x leaves a reminder 3 then ( 200 - 3 ) i . e . 197 is divisible by x so ( 200 + 197 ) / x leaves a reminder rem ( 200 / x ) + rem ( 197 / x ) = > 3 + 0 = 3 answer : b"
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8
b
subtract(const_100.0, subtract(297, 200))
subtract(n2,const_100.0)|subtract(n0,#0)|
general
the cyclist going at a constant rate of 18 miles per hour is passed by a motor - cyclist traveling in the same direction along the same path at 48 miles per hour . the motor - cyclist stops to wait for the cyclist 15 minutes after passing cyclist , while the cyclist continues to travel at constant rate , how many minutes must the motor - cyclist wait until the cyclist catches up ?
we can determine quickly that total number should range between 1020 / 160 < = n < = 1020 / 90 , so ans should be between 6 and 12 . now solving the expression 160 a + 90 b = 1020 decreasing 1020 by multiples of 160 and checking divisibility of that number by 9 , we get fast song plays for 3 minutes and slow somg plays for 6 minutes , 3 * 160 + 6 * 90 = 1020 hence total number of minutes stream of music plays is 3 + 6 = 9 minutes ans d
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
d
add(floor(multiply(divide(1020, add(160, 90)), const_2)), const_1)
add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)|floor(#2)|add(#3,const_1)
physics
there are 6 people in the elevator . their average weight is 150 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person .
"explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 190 n ( n - 1 ) / 2 = 190 â ˆ ´ n = 20 answer : option e"
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 20
e
divide(add(sqrt(add(multiply(multiply(190, 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
a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ?
"sol . required ratio = 6 * 1 * 1 / 6 * 7 * 7 = 1 / 49 = 1 : 49 . answer b"
a ) 1 : 25 , b ) 1 : 49 , c ) 1 : 52 , d ) 1 : 522 , e ) none
b
divide(const_4, const_100)
divide(const_4,const_100)|
geometry
the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 10 students is 16 years . the age of the 15 th student is
"present age is 4 x and 3 x , = > 4 x + 6 = 30 = > x = 6 so deepak age is = 3 ( 6 ) = 18 answer : a"
a ) 18 , b ) 15 , c ) 77 , d ) 266 , e ) 182
a
divide(multiply(subtract(30, 6), 3), 4)
subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|
other
a total of 30 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 25 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ]
"12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 480 = 149 answer : d"
a ) 270 , b ) 199 , c ) 676 , d ) 149 , e ) 122
d
multiply(divide(480, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9))
multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)|
general
the length of a rectangular floor is more than its breadth by 200 % . if rs . 324 is required to paint the floor at the rate of rs . 3 per sq m , then what would be the length of the floor ?
"remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : d"
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
d
subtract(multiply(multiply(1525, 1527), 1529), subtract(multiply(multiply(1525, 1527), 1529), const_3))
multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)|
general
r is the set of positive odd integers less than 100 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ?
solution : this is a percent decrease problem . we will use the formula : percent change = ( new – old ) / old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 20 million royalties / sales = 5 / 20 = 1 / 4 next 108 million royalties / sales = 9 / 108 = 1 / 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties / sales = ( 5 / 20 ) x 12 = 3 next 108 million royalties / sales = 9 / 108 = ( 1 / 12 ) x 12 = 1 we can plug 15 and 5 into our percent change formula : ( new – old ) / old x 100 [ ( 1 – 3 ) / 3 ] x 100 - 200 / 3 x 100 at this point we can stop and consider the answer choices . since we know that 200 / 3 is just a bit less than ½ , we know that - 200 / 3 x 100 is about a 67 % decrease . answer e .
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 67 %
e
multiply(divide(subtract(multiply(divide(5, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(5, 20), const_100)), const_100)
divide(n0,n1)|divide(n2,n3)|multiply(#0,const_100)|multiply(#1,const_100)|subtract(#2,#3)|divide(#4,#2)|multiply(#5,const_100)
general
three pipes a , b and c can fill a tank from empty to full in 30 minutes , 20 minutes and 10 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution r in the liquid in the tank after 3 minutes ?
"10 overs - run rate = 5.2 runs scored in first 10 overs = 52 remaining overs 40 total runs to be scored = 282 52 runs already scored 282 - 52 = 230 230 runs to be scored in 40 overs let required runrate be x 40 * x = 230 x = 230 / 40 x = 5.75 the required runrate is 5.75 answer : d"
a ) 6.25 , b ) 6.5 , c ) 6.75 , d ) 5.75 , e ) 8
d
divide(subtract(282, multiply(10, 5.2)), 40)
multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|
gain
circular gears l and r start to rotate at the same time at the same rate . gear l makes 10 complete revolutions per minute and gear r makes 40 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 9 more revolutions than gear l ?
"{ total } = { french } + { russian } - { both } + { neither } { total } = 30 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 20 . answer : a ."
a ) 20 , b ) 96 , c ) 108 , d ) 120 , e ) 150
a
divide(subtract(32, 30), subtract(divide(20, const_100), divide(10, const_100)))
divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)|
other
two numbers a and b are such that the sum of 5 % of a and 2 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b .
let number of cars sold in 1 st quarter = x number of cars sold in 2 nd quarter = 32 % greater than the number sold during the first quarter = ( 1 + 32 / 100 ) x = 1.32 x 1.32 x = 3 , 976,000 = > x = 3 , 012,121 so , answer will be d
a ) 714,240 , b ) 2 , 261,760 , c ) 2 , 400,000 , d ) 3 , 012,121 , e ) 3 , 915,790
d
multiply(multiply(divide(divide(divide(add(multiply(3, multiply(const_1000, const_1000)), 976000), add(divide(32, const_100), const_1)), const_1000), const_100), 3), 3)
divide(n3,const_100)|multiply(const_1000,const_1000)|add(#0,const_1)|multiply(n1,#1)|add(n2,#3)|divide(#4,#2)|divide(#5,const_1000)|divide(#6,const_100)|multiply(n1,#7)|multiply(n1,#8)
gain
10 men and 15 women together can complete a work in 4 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ?
"0.2 a = 0.08 b - > a / b = 0.08 / 0.20 = 8 / 20 = 2 / 5 : . a : b = 2 : 5 answer : c"
a ) 2 : 3 , b ) 3 : 4 , c ) 2 : 5 , d ) 20 : 3 , e ) 30 : 7
c
divide(multiply(0.08, const_100), multiply(0.2, const_100))
multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)|
other
for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers t are there between 6 * + 2 and 6 * + 6 , inclusive ?
explanation : let distance = x km and usual rate = y kmph . then , x / y - x / ( y + 3 ) = 40 / 60 - - > 2 y ( y + 3 ) = 9 x - - - - - ( i ) also , x / ( y - 2 ) - x / y = 40 / 60 - - > y ( y - 2 ) = 3 x - - - - - - - - ( ii ) on dividing ( i ) by ( ii ) , we get : x = 40 km . answer : c
a ) 27 , b ) 87 , c ) 40 , d ) 18 , e ) 17
c
multiply(multiply(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), divide(40, const_60)), add(const_1, divide(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), 3)))
divide(n1,const_60)|multiply(n0,n2)|subtract(n0,n2)|multiply(n2,#1)|divide(#3,#2)|divide(#4,n0)|multiply(#4,#0)|add(#5,const_1)|multiply(#7,#6)
physics
two employees x and y are paid a total of rs . 440 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?
w + b + c + 14 + 15 = 12 * 5 = 60 = > w + b + c = 60 - 29 = 31 w + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
d
divide(add(subtract(subtract(multiply(add(const_1, const_4), 12), 15), 14), 29), const_4)
add(const_1,const_4)|multiply(n2,#0)|subtract(#1,n1)|subtract(#2,n0)|add(n3,#3)|divide(#4,const_4)
general
if the radius of a cylinder is doubled and so is the height , what is the new volume of the cylinder divided by the old one ?
this is maximum - minimum . hence , 40 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 32 and 40 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 11 . so , 32 - 11 = 21 . the correct answer is a .
a ) 21 , b ) 29 , c ) 23 , d ) 25 , e ) 26
a
subtract(40, add(add(8, const_2), 8))
add(n0,const_2)|add(n0,#0)|subtract(n1,#1)|
general
if x is equal to the sum of the integers from 40 to 50 , inclusive , and y is the number of even integers from 40 to 50 , inclusive , what is the value of x + y ?
"( 6 * 65 ) + 47 + x > 500 390 + 47 + x > 500 437 + x > 500 = > x > 63 option d"
a ) 85 , b ) 74 , c ) 67 , d ) 63 , e ) 28
d
subtract(500, add(multiply(6, 65), 47))
multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)|
general
a rainstorm increased the amount of water stored in state j reservoirs from 130 billion gallons to 160 billion gallons . if the storm increased the amount of water in the reservoirs to 80 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ?
"solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 16 more transactions than cal . jade = 66 + 16 = 82 jade handled = 82 transactions . answer : c"
a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84
c
add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 16)
add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)|
general
a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 50 square feet of the wall , and no tiles overlap , what is the area of the entire wall ?
"c . p . = $ 100 s . p . = $ 125 gain = $ 25 gain % = 25 / 100 * 100 = 25 % answer is c"
a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 %
c
subtract(divide(125, divide(100, const_100)), const_100)
divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)|
gain
if a and b are positive integers , and a = 20 b - 15 , the greatest common divisor of a and b can not be
"this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 780 , or 1 / 9 of 780 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate , the errors will partially cancel each other out . our estimate will be relatively accurate . the number 780 is between 720 and 810 , so ( 1 / 9 ) * 780 will be between 80 and 90 . keeping track not only of your current estimate , but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently . the closest answer is 84 , so this is the answer to choose . the answer is c ."
a ) 62 , b ) 73 , c ) 84 , d ) 95 , e ) 106
c
divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
diana is painting statues . she has 1 / 2 of a gallon of paint remaining . each statue requires 1 / 16 gallon of paint . how many statues can she paint ?
"let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 20 + v / 12 - v / 10 = v / 30 per minute the tank will be filled in 30 minutes . the answer is d ."
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32
d
subtract(add(divide(const_1, 20), divide(const_1, 12)), divide(const_1, 10))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|
physics
boy sells a book for rs . 630 he gets a loss of 10 % , to gain 10 % , what should be the sp ?
"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 20 = 31428.5 cm = 314.3 m answer : c"
a ) 708 m , b ) 704 m , c ) 314.3 m , d ) 714 m , e ) 744 m
c
divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 20), 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
determine the value of 3 * 27 / 31 + 81 / 93
since - 4 is the least integer in list a , then 7 is the largest integer in that list . thus the range of the positive integers in the list is 7 - 1 = 6 . answer : b .
a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12
b
subtract(subtract(12, add(4, const_1)), const_1)
add(n1,const_1)|subtract(n0,#0)|subtract(#1,const_1)
general
a train 150 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ?
"total paint initially = 360 gallons paint used in the first week = ( 1 / 2 ) * 360 = 180 gallons . remaning paint = 180 gallons paint used in the second week = ( 1 / 5 ) * 180 = 36 gallons total paint used = 216 gallons . option d"
a ) 18 , b ) 144 , c ) 175 , d ) 216 , e ) 250
d
add(multiply(divide(360, 2), 1), divide(subtract(360, multiply(divide(360, 2), 1)), 5))
divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)|
physics
a satellite is composed of 30 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 6 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ?
"1 feet = 12 inches 1 mile = 5280 feet 100 mile = 5280 * 12 * 100 = 6336000 ans : a"
a ) 6336000 , b ) 6542000 , c ) 5462300 , d ) 6213000 , e ) 6120330
a
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 100), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
each child has 4 crayons and 14 apples . if there are 9 children , how many crayons are there in total ?
"sum of decimal places = 7 . since the last digit to the extreme right will be zero ( since 5 x 4 = 20 ) so there will be 6 significant digits to the right of the decimal point . answer is e ."
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
e
subtract(subtract(const_100, 95.75), const_1)
subtract(const_100,n0)|subtract(#0,const_1)|
general
it costs $ 2 for the first 15 minutes to use the bumper cars at a fair ground . after the first 15 minutes it costs $ 6 per hour . if a certain customer uses the bumper cars for 3 hours and 25 minutes , how much will it cost him ?
"i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 4 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 4 = not integer we can not forget multiples of 8 from 1 to 96 we have 24 numbers that are multiple of 4 therefore , 48 / 96 + 24 / 96 = 72 / 96 = 3 / 4 answer : e"
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4
e
divide(add(divide(96, 2), divide(96, 4)), 96)
divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|
general
a train 120 m long running at 60 kmph crosses a platform in 35 sec . what is the length of the platform ?
"increase in house value = $ 24,000 - $ 20,000 = $ 4000 so , tax increase = 12 % of $ 4000 = $ 480 answer : e"
a ) $ 32 , b ) $ 50 , c ) $ 320 , d ) $ 400 , e ) $ 480
e
divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 12), const_100)
add(const_3,const_4)|add(const_1,const_4)|multiply(#0,const_4)|multiply(#1,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|subtract(#4,#5)|multiply(n0,#6)|divide(#7,const_100)|
general
in a class of 60 students , 20 did not opt for math . 15 did not opt for science and 5 did not opt for either . how many students of the class opted for both math and science ?
y = 288 * a + 44 = ( 24 * 12 ) * a + 24 + 20 the answer is a .
a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26
a
reminder(44, 24)
reminder(n1,n2)
general
machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 60 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 6 minutes ?
"1 of 5 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more than one departments . we can choose any of the 5 candidates for the math dep . , which gives as 5 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections and 8 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 2 ! * 8 ! = ( 9 * 10 ) / 2 = 90 / 2 = 45 . we are multiplying our individual selections : 5 * 45 = 225 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans e"
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 225
e
multiply(multiply(10, 3), 5)
multiply(n3,n5)|multiply(n1,#0)|
other
the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 25 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ?
"1 / 2 * d ( 14 + 4 ) = 380 d = 42 answer : d"
a ) 39 , b ) 28 , c ) 27 , d ) 42 , e ) 71
d
divide(divide(divide(380, divide(add(14, 4), const_2)), 4), const_2)
add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)|
physics
sum of 36 odd numbers is ?
"cp * ( 76 / 100 ) = 1140 cp = 15 * 100 = > cp = 1500 answer : a"
a ) 1500 , b ) 6789 , c ) 1200 , d ) 6151 , e ) 1421
a
divide(1140, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? a . 15 b . 18
"60 students total 20 did not opt for math 15 did not opt for science 5 did not opt for either total of 40 students in math and 10 did not opt for sci but did for math 40 - 10 = 30 30 students of the class opted for both math and science answer : d . 30"
a ) 23 , b ) 25 , c ) 27 , d ) 30 , e ) 48
d
subtract(subtract(60, 20), subtract(subtract(60, 15), 5))
subtract(n0,n1)|subtract(n0,n2)|subtract(#1,n3)|subtract(#0,#2)|
other
the original price of a suit is $ 100 . the price increased 20 % , and after this increase , the store published a 20 % off coupon for a one - day sale . given that the consumers who used the coupon on sale day were getting 20 % off the increased price , how much did these consumers pay for the suit ?
"time = 6 distance = 540 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 540 / 9 = 60 kmph c )"
a ) 48 kmph , b ) 52 kmph , c ) 6 o kmph , d ) 63 kmph , e ) 65 kmph
c
divide(540, multiply(divide(3, 2), 6))
divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)|
physics
if 4 spiders make 4 webs in 4 days , then 1 spider will make 1 web in how many days ?
"length = 5 m 44 cm = 544 cm breadth = 3 m 74 cm = 374 cm area = 544 * 374 hcf = 34 area of square = 34 * 34 cm 2 no of tiles req = 544 * 374 / 34 * 34 = 16 * 11 = 176 answer a"
a ) 176 , b ) 124 , c ) 224 , d ) 186 , e ) 190
a
divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), multiply(subtract(44, add(multiply(const_2, const_4), const_2)), subtract(44, add(multiply(const_2, const_4), const_2))))
multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)|
physics
find the area of the quadrilateral of one of its diagonals is 50 cm and its off sets 15 cm and 5 cm ?
"let the numbers be 2 x and 3 x their h . c . f . = 10 so the numbers are 2 * 10 , 3 * 10 = 20,30 l . c . m . = 60 answer is b"
a ) 30 , b ) 60 , c ) 20 , d ) 10 , e ) 40
b
sqrt(divide(10, add(power(3, 2), add(power(2, 2), power(2, 2)))))
power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)|
other
marty ' s pizza shop guarantees that their pizzas all have at least 75 % of the surface area covered with toppings , with a crust of uniform width surrounding them . if you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust ?
"oa is ' c ' . oe : take the remainder from each of 1250 / 18 , 1090 / 18 and so on . . 1250 / 18 gives remainder = 8 1090 / 18 gives remainder = 10 1045 / 18 gives remainder = 1 1055 / 18 gives remainder = 11 the net remainder is the product of above individual remainders . i . e = 8 * 10 * 1 * 11 break them into pairs 8 * 10 / 18 gives remainder 8 and 1 * 11 / 18 gives remainder 11 so 8 * 11 / 18 gives remainder 16 . answer : c"
a ) 34 , b ) 19 , c ) 16 , d ) 14 , e ) 10
c
reminder(multiply(1090, 1250), 1045)
multiply(n0,n1)|reminder(#0,n2)|
general
a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 325,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ?
"explanation : required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 8 = 540 + 8 = 548 . answer : option d"
a ) 504 , b ) 536 , c ) 544 , d ) 548 , e ) none of these
d
multiply(54, const_10)
multiply(n3,const_10)|
general
one fourth of a solution that was 10 % sugar by weight was replaced with by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ?
"( i ) 7 a - 13 b = 50 ( ii ) 2 a + 22 b = - 50 adding ( i ) and ( ii ) : 9 a + 9 b = 0 the answer is c ."
a ) - 9 , b ) - 6 , c ) 0 , d ) 6 , e ) 9
c
divide(const_0_33, const_1000)
divide(const_0_33,const_1000)|
general
the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 90 and rs . 80 respectively . the sum is :
"one monkey takes 20 min to eat 1 banana , so in 80 mins 1 monkey will eat 4 bananas , so for 80 bananas in 80 min we need 80 / 4 = 20 monkeys answer : d"
a ) 9 , b ) 10 , c ) 11 , d ) 20 , e ) 13
d
divide(const_3.0, divide(80, 10))
divide(n3,n1)|divide(n3,#0)|
physics
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 19404 / - in 3 years at the same rate of interest . find the rate percentage ?
"if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 11 hence answer should be a"
a ) 11 , b ) - 1 , c ) 3 , d ) 5 , e ) 13
a
multiply(6, const_2)
multiply(n1,const_2)|
general
two cars start from the opposite places of a main road , 140 km apart . first car runs for 25 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run only 35 km along the main road . what would be the distance between two cars at this point ?
"3 x + y = 40 2 x - y = 20 5 x = 60 x = 12 y = 4 4 y ^ 2 = 4 * 16 = 64 answer is e"
a ) 2 , b ) 4 , c ) 0 , d ) 10 , e ) 64
e
multiply(3, power(subtract(40, multiply(divide(add(40, 20), add(3, 2)), 3)), 2))
add(n1,n3)|add(n0,n2)|divide(#0,#1)|multiply(n0,#2)|subtract(n1,#3)|power(#4,n2)|multiply(n0,#5)|
general
in what time will a train 100 metres long cross an electic pole , if its speed be 144 km / hr ?
"you may set up common equation like this : job / a + job / b + job / c = job / x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 30 / 40 + 30 / 30 + 30 / 24 = 30 / x if you solve this equation , you get the same answer b ( 10 )"
a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 18 minutes , e ) 20 minutes
b
divide(30, add(divide(30, 24), add(divide(30, 40), divide(30, 30))))
divide(n0,n1)|divide(n0,n2)|divide(n0,n3)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|
physics
the average age of a family of 6 members is 26 years . if the age of the youngest member is 10 years , what was the average age of the family at the birth of the youngest member ?
"original perimeter = x hence original side = x / 4 new side = 7 x / 4 new perimeter = 4 * 7 x / 4 = 7 x correct option : c"
a ) 3 x , b ) 4 x , c ) 7 x , d ) 12 x , e ) 27 x
c
square_perimeter(multiply(7, const_4))
multiply(n0,const_4)|square_perimeter(#0)|
geometry
for 2 consecutive yrs , my incomes are in the ratio of 4 : 7 and expenses in the ratio of 3 : 5 . if my income in the 2 nd yr is rs . 42000 & my expenses in the first yr in rs . 21000 , my total savings for the two - year is
"the worst case scenario will be if we remove all 17 tablets of medicine b first . the next 2 tablets we remove have to be of medicine a , so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 17 + 2 = 19 tablets . answer : d ."
a ) 12 , b ) 15 , c ) 17 , d ) 19 , e ) 21
d
add(17, const_2)
add(n1,const_2)|
general
the present worth of rs . 1014 due in 2 years at 4 % per annum compound interest is
"6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 60 * 25 * 1 / 2 = 750 feet ^ 3 of milk must be removed , which equals to 750 * 7.5 = 5625 gallons . answer : d ."
a ) 100 , b ) 250 , c ) 750 , d ) 5625 , e ) 5635
d
multiply(multiply(multiply(60, 25), divide(1, const_2)), 7.5)
divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)|
general
in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what percent z of all the people in smithtown are left - handed women ?
"the total number of ways to choose 2 children from 8 is 8 c 2 = 28 the number of ways to choose 1 boy and 1 girl is 4 * 4 = 16 p ( 1 boy and 1 girl ) = 16 / 28 = 4 / 7 the answer is d ."
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 9
d
divide(multiply(choose(4, const_2), choose(4, const_2)), choose(add(4, 4), 2))
add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)|
probability
how long does a truck of 200 m long traveling at 60 kmph takes to cross a bridge of 180 m in length ?
since jill owns 5 of the pen , the subset from which the 2 pens hould be chosen are the 2 pens not owned by jill fom the universe of 7 . the first pen can be one of the 2 from the 7 with probability 2 / 7 . the second pen can be one of the 1 from the 6 remaining with probability 1 / 6 , the total probability will be 2 / 7 × 1 / 6 . on cancellation , this comes to 2 / 42 . thus , the answer is b - 2 / 42 .
a ) 5 / 42 , b ) 2 / 42 , c ) 7 / 42 , d ) 2 / 7 , e ) 5 / 7
b
multiply(divide(subtract(7, 5), 7), divide(subtract(subtract(7, 5), const_1), subtract(7, const_1)))
subtract(n1,n2)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)
probability
what will be the difference between simple and compound interest at 14 % per annum on a sum of rs . 1000 after 4 years ?
"let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit w = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ."
a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 %
c
subtract(const_100, subtract(subtract(const_100, 20), 20))
subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)|
gain
chris age after 20 years will be 5 times his age 5 years back . what is the present age of chris ?
"equation is correct , so math must be a problem . 0.13 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 5,200 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 10,800 - - > x = 54,000 . answer : a ."
a ) $ 54,000 , b ) $ 56,000 , c ) $ 64,000 , d ) $ 66,667 , e ) $ 80,000
a
add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(13, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100)))
divide(n0,const_100)|divide(n2,const_100)|multiply(const_2,const_4)|multiply(const_10,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(#4,#6)|divide(#7,#1)|add(#8,#5)|
general
kathleen can paint a room in 2 hours , and anthony can paint an identical room in 7 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ?
"explanation : p ÷ 4 = 40 = > p = 40 * 4 = 160 p / 3 = 160 / 3 = 53 , remainder = 1 answer : option a"
a ) a ) 1 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7
a
divide(multiply(4, 40), 3)
multiply(n0,n1)|divide(#0,n3)|
general
in a group of ducks and cows , the total number of legs are 8 more than twice the no . of heads . find the total no . of buffaloes .
"6 a 2 = 294 = 6 * 49 a = 7 = > a 3 = 343 cc answer : d"
a ) 8 cc , b ) 9 cc , c ) 2 cc , d ) 343 cc , e ) 6 cc
d
volume_cube(sqrt(divide(294, add(const_2, const_4))))
add(const_2,const_4)|divide(n0,#0)|sqrt(#1)|volume_cube(#2)|
geometry
bookman purchased 55 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 460 , how many paperback copies were sold ?
"set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 30 - 5 = 25 aub = a + b - aib aib = 21 + 16 - 25 = 12 answer a"
a ) 12 , b ) 10 , c ) 9 , d ) 7 , e ) 5
a
add(subtract(add(16, 21), subtract(30, 5)), subtract(21, 16))
add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)|
general
what is the sum of the multiples of 4 between 38 and 127 inclusive ?
"correct avg marks = 100 + ( 10 - 50 ) / 20 avg = 100 - 2 = 98 answer is c"
a ) 78 , b ) 82 , c ) 98 , d ) 91 , e ) 85
c
divide(add(subtract(multiply(100, 20), 50), 10), 20)
multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|
general
find the average of first 3 multiples of 5 ?
"10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 20 = 240 $ 520 - 240 = 280 ans e"
a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 280
e
subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(20, const_12))
add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)|
general