Datasets:

math_qa

like 67

on dividing 136 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?

"d = ( d - r ) / q = ( 136 - 1 ) / 9 = 135 / 9 = 15 b )"

a ) a ) 12 , b ) b ) 15 , c ) c ) 16 , d ) d ) 17 , e ) e ) 18

b

floor(divide(136, 9))

divide(n0,n1)|floor(#0)|

general

a 72 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ?

"amount of salt = 7.2 assume x gallons of water are added . 7.2 / 72 + x = 8 / 100 720 = 8 x + 576 8 x = 144 x = 18 correct option : c"

a ) 8 , b ) 12 , c ) 18 , d ) 14 , e ) 16

c

divide(multiply(72, subtract(10, 8)), 8)

subtract(n1,n2)|multiply(n0,#0)|divide(#1,n2)|

general

the radius of a cylindrical vessel is 9 cm and height is 3 cm . find the whole surface of the cylinder ?

"r = 9 h = 3 2 π r ( h + r ) = 2 * 22 / 7 * 9 ( 12 ) = 678.8 answer : e"

a ) 308 sq cm , b ) 220 sq cm , c ) 440 sq cm , d ) 132 sq cm , e ) 678.8 sq cm

e

surface_cylinder(9, 3)

surface_cylinder(n0,n1)|

geometry

the simple interest on a sum of money will be rs . 600 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ?

"p - - - 10 - - - - 600 p - - - 5 - - - - - 300 3 p - - - 5 - - - - - 900 - - - - - - = > 1200 answer : c"

a ) 8981 , b ) 3799 , c ) 1200 , d ) 2693 , e ) 1732

c

add(multiply(multiply(divide(600, 10), 5), const_3), multiply(divide(600, 10), 5))

divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)|

general

a rectangular floor that measures 24 meters by 64 meters is to be covered with carpet squares that each measure 8 meters by 8 meters . if the carpet squares cost $ 24 apiece , what is the total cost for the number of carpet squares needed to cover the floor ?

the width of the rectangular floor ( 24 m ) is a multiple of one side of the square ( 8 m ) , and the length of the floor ( 64 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 24 / 8 ) * ( 64 / 8 ) = 24 . the total cost is 24 * 24 = $ 576 . the answer is , therefore , b .

['a ) $ 200', 'b ) $ 576', 'c ) $ 480', 'd ) $ 960', 'e ) $ 1,920']

b

multiply(24, 24)

multiply(n0,n0)

geometry

what least number must be subtracted from 427751 so that remaining no . is divisible by 10

"explanation : on dividing 427751 by 10 we get the remainder 1 , so 1 should be subtracted option b"

a ) 3 , b ) 1 , c ) 6 , d ) 8 , e ) 9

b

subtract(427751, multiply(floor(divide(427751, 10)), 10))

divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|

general

jim is able to sell a hand - carved statue for $ 660 which was a 20 % profit over his cost . how much did the statue originally cost him ?

"cost price = 660 / 120 * 100 = 550 answer : c"

a ) $ 496.30 , b ) $ 512.40 , c ) $ 550 , d ) $ 574.90 , e ) $ 588.20

c

divide(660, add(divide(20, const_100), const_1))

divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|

gain

the minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time . how much does the clock gain or lose in 12 hours ?

after 60 minute rotation the minute hand will make an angle of 30 degree with hour hand . after that let the time required to make an overtake is t minute in this time angle covered by minute hand = ( 360 / 60 ) t = 6 t similarly angle covered by hour hand = ( 30 / 60 ) t = t / 2 so for overtake 6 t = 30 + t / 2 or t = 60 / 11 min = 5 min and 5 / 11 min = 5 min and 300 / 11 sec = 5 min and 27.273 sec so clock loses 27.273 seconds answer : a

a ) 27.273 seconds , b ) 29.273 seconds , c ) 37.273 seconds , d ) 47.273 seconds , e ) 17.273 seconds

a

multiply(subtract(divide(multiply(multiply(const_2, const_3), const_60), divide(subtract(multiply(multiply(const_2, const_3), const_60), divide(multiply(multiply(const_2, const_3), const_60), multiply(const_3, const_4))), const_60)), 65), const_60)

multiply(const_2,const_3)|multiply(const_3,const_4)|multiply(#0,const_60)|divide(#2,#1)|subtract(#2,#3)|divide(#4,const_60)|divide(#2,#5)|subtract(#6,n0)|multiply(#7,const_60)

physics

students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 184 cms . what was the ratio f of men to women ?

( a ) f = 184 x 5 + 170 x 2 = 1260 . a

a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 4 : 1 , e ) 3 : 1

a

divide(subtract(180, 170), subtract(184, 180))

subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)|

general

average monthly income of a family of 3 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 650 . the income of the deceased was ?

"answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 3 - 650 x 2 rs . = 905 rs . correct option : b"

a ) rs . 692.80 , b ) rs . 905 , c ) rs . 990 , d ) rs . 1385 , e ) none

b

subtract(multiply(735, 3), multiply(650, subtract(3, const_1)))

multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)|

general

a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 52000 , then the share of b is ?

"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 52000 share of b = 6 / 13 ( 52000 ) = rs . 24000 answer : b"

a ) 2888 , b ) 24000 , c ) 27777 , d ) 29997 , e ) 27712

b

subtract(52000, multiply(const_60, const_100))

multiply(const_100,const_60)|subtract(n3,#0)|

gain

a tank can supply water to a village for 20 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 15 days only . for how many days will the supply last if the leak drains out 20 liters per day ?

"losing 10 liters per day results in a loss of 150 liters in 15 days . so , those 150 liters were for 5 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 20 = 600 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last 600 / 50 = 12 days . the answer is d ."

a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14

d

divide(multiply(divide(multiply(10, 15), subtract(20, 15)), 20), add(20, divide(multiply(10, 15), subtract(20, 15))))

multiply(n1,n2)|subtract(n0,n2)|divide(#0,#1)|add(n3,#2)|multiply(n0,#2)|divide(#4,#3)|

physics

31 of the scientists that attended a certain workshop were wolf prize laureates , and 14 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 50 of the scientists attended that workshop , how many of them were nobel prize laureates ?

"lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 13 . . so total who won n = 11 + 14 = 25 . . c"

a ) a ) 11 , b ) b ) 18 , c ) c ) 25 , d ) d ) 29 , e ) d ) 36

c

add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 14)

subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)|

physics

a committee is reviewing a total of 20 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction x of the selected films are in color ?

"it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose x = 60 / 62 = 30 / 31 of the selected films . answer : e ."

a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31

e

divide(6, add(divide(20, const_100), 6))

divide(n0,const_100)|add(n1,#0)|divide(n1,#1)|

general

if the average ( arithmetic mean ) of x + 1 , x + 4 , and x + 7 is 0 , then x =

"( x + 1 + x + 4 + x + 7 ) / 3 = 0 = > 3 x + 12 = 0 = > x = - 4 answer a"

a ) – 4 , b ) – 3 , c ) – 2 , d ) – 1 , e ) 0

a

subtract(multiply(4, const_2), multiply(1, const_2))

multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|

general

for how many integers pair ( x , y ) satisfies the result ( 1 / x ) + ( ( 1 / y ) = 1 / 16

it seems that the accepted answer is wrong , according to wolframalpha . the answer should be 31 , which is n ' t even listed as a selectable answer . e

a ) a ) 12 , b ) b ) 6 , c ) c ) 10 , d ) d ) 16 , e ) e ) 19

e

add(add(add(16, 1), 1), 1)

add(n0,n3)|add(n0,#0)|add(n0,#1)

general

of the female students at barkely university , 3 / 5 are on the honor roll . of the male students , 2 / 5 are on the honor roll . if 5 / 9 of the students are female , what fraction of all the students are on the honor roll ?

"let the total students be 45 given 5 / 9 of the students are females = 25 then males = 4 / 9 = 20 3 / 5 of the females are on honor roll = 15 males on the honor roll = 2 / 5 = 8 total students on honor roll = 15 + 8 = 23 fraction = 23 / 45 c"

a ) 5 / 9 , b ) 3 / 15 , c ) 23 / 45 , d ) 43 / 60 , e ) 53 / 90

c

divide(add(multiply(multiply(3, 2), 2), 2), multiply(2, multiply(5, 3)))

multiply(n0,n2)|multiply(n0,n3)|multiply(n2,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)|

general

abcd is a square . e and f are the midpoints of sides cd and bc , respectively . what is the ratio of the shaded region area to the unshaded region ?

let ' s join the midpoints as shown in the figure then we ' ll get four small squares . . . aodx be 1 and similarly all squares . . and we can split the unit 1 into half and half as per the mid points . . . then shaded will be 1 + 1 / 2 + 1 / 2 + 1 / 2 = 5 / 2 and unshaded is 1 / 2 + 1 / 2 + 1 / 2 = 3 / 2 . . shaded : unshaded is 5 : 3 . . ans option d .

['a ) 1 : 1', 'b ) 2 : 1', 'c ) 3 : 1', 'd ) 5 : 3', 'e ) 8 : 3']

d

divide(add(add(add(const_1, divide(const_1, const_2)), divide(const_1, const_2)), divide(const_1, const_2)), add(add(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)))

divide(const_1,const_2)|add(#0,const_1)|add(#0,#0)|add(#1,#0)|add(#2,#0)|add(#3,#0)|divide(#5,#4)

geometry

a certain clock marks every hour by striking a number of times equal to the hour , and the time require for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds . at 7 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ?

"at 6 ' o clock , there would be 6 strikes . first strike , then a short interval , the second strike , then a short interval and so on till the 6 th strike . so there would be in all 5 intervals between 6 strikes . similarly , between 7 strikes , there would be 6 intervals . according to the question , the time spent in the strike and the interval is same . at 6 ' o clock , the 6 strikes and the 5 intervals together take 22 sec so each strike and each interval takes 2 secs . at 7 ' o clock , the 7 strikes and 6 intervals will take 2 * ( 7 + 6 ) = 26 secs a"

a ) 26 , b ) 29 , c ) 19 , d ) 30 , e ) 40

a

multiply(add(subtract(7, const_1), 7), divide(22, add(subtract(6, const_1), 6)))

subtract(n3,const_1)|subtract(n0,const_1)|add(n3,#0)|add(n0,#1)|divide(n2,#3)|multiply(#2,#4)|

general

a dishonest dealer professes to sell goods at the cost price but uses a weight of 600 grams per kg , what is his percent ?

"600 - - - 400 100 - - - ? = > 66.7 % answer : b"

a ) 28 % , b ) 66.7 % , c ) 55 % , d ) 28 % , e ) 55 %

b

subtract(multiply(divide(const_100, 600), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)

add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|

gain

15 men can complete a piece of work in 25 days . in how many days can 20 men complete that piece of work ?

"15 * 25 = 20 * x = > x = 18.75 days answer : b"

a ) 55 days , b ) 18.75 days , c ) 20 days , d ) 88 days , e ) 44 days

b

divide(multiply(25, 15), 20)

multiply(n0,n1)|divide(#0,n2)|

physics

if n is a prime number greater than 5 , what is the remainder when n ^ 2 is divided by 23 ?

"there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than 5 , square it and see what would be the remainder upon division of it by 23 . n = 5 - - > n ^ 2 = 49 - - > remainder upon division 49 by 23 is 3 . answer : d ."

a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5

d

subtract(power(add(5, 2), 2), multiply(23, const_4))

add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)|

general

the sum of all solutions for x in the equation x ^ 2 – 8 x + 21 = | x – 2 | + 5 is equal to :

"x ^ 2 - 8 x + 16 = | x - 2 | rhs can be - ve or + ve x ^ 2 - 9 x + 18 = 0 x ^ 2 - 7 x + 14 = 0 x = 7,6 , 3,2 we test all 3 values in original equation , all ok . thus , sum = 7 + 6 + 3 + 2 = 18 ans ( e )"

a ) – 7 , b ) 7 , c ) 10 , d ) 12 , e ) 18

e

multiply(5, 2)

multiply(n3,n4)|

general

80 is what percent of 160 ?

80 / 160 × 100 = 50 % answer : d

a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 55 %

d

multiply(divide(80, 160), const_100)

divide(n0,n1)|multiply(#0,const_100)|

gain

the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 40 cents more than the commodity y ?

"the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 40 = 60 cents on commodity y , which will take 6 years . the answer is a ."

a ) 2007 , b ) 2008 , c ) 2009 , d ) 2010 , e ) 2011

a

add(2001, divide(add(divide(40, const_100), subtract(4.40, 4.20)), subtract(divide(30, const_100), subtract(4.40, 4.20))))

divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)|

general

a 300 metre long train crosses a platform in 39 seconds while it crosses a post in 18 seconds . what is the length of the platform ?

explanation : length of the train = distance covered in crossing the post = speed × time = speed × 18 speed of the train = 300 / 18 m / s = 50 / 3 m / s time taken to cross the platform = 39 s ( 300 + x ) / ( 50 / 3 ) = 39 s where x is the length of the platform 300 + x = ( 39 × 50 ) / 3 = 650 meter x = 650 - 300 = 350 meter answer : option b

a ) 150 m , b ) 350 m , c ) 420 m , d ) 600 m , e ) 700 m

b

divide(subtract(multiply(divide(300, divide(18, const_3)), 39), multiply(const_3, 300)), const_3)

divide(n2,const_3)|multiply(n0,const_3)|divide(n0,#0)|multiply(n1,#2)|subtract(#3,#1)|divide(#4,const_3)

physics

what is the remainder when 1034 * 1047 * 1050 * 1053 is divided by 33 ?

"take the remainder from each of 1034 / 33 , 1047 / 33 and so on . . 1034 / 33 gives remainder = 11 1047 / 33 gives remainder = 24 1050 / 33 gives remainder = 27 1053 / 33 gives remainder = 30 the net remainder is the product of above individual remainders . i . e = 11 * 24 * 27 * 30 break them into pairs 11 * 24 / 33 gives remainder 0 and 27 * 30 / 33 gives remainder 18 so 0 * 18 / 33 gives remainder 0 . a"

a ) 0 , b ) 27 , c ) 30 , d ) 21 , e ) 18

a

reminder(multiply(1047, 1034), 1050)

multiply(n0,n1)|reminder(#0,n2)|

general

39 persons can repair a road in 24 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ?

"let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 39 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 39 x 5 x 24 x = ( 39 x 5 x 24 ) / ( 30 x 6 ) x = 26 . answer : e"

a ) 10 , b ) 13 , c ) 14 , d ) 15 , e ) 26

e

divide(multiply(multiply(39, 24), 5), multiply(30, 6))

multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)|

physics

two boats are heading towards each other at constant speeds of 4 miles / hr and 20 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ?

"the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 4 + 20 = 25 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 24 * 1 / 60 = 6 / 15 miles apart . answer : b ."

a ) 1 / 12 , b ) 6 / 15 , c ) 1 / 6 , d ) 1 / 3 , e ) 1 / 5

b

divide(add(20, 4), const_60)

add(n0,n1)|divide(#0,const_60)|

physics

1 / 0.04 is equal to

"explanation : 1 / 0.04 = ( 1 * 100 ) / 4 = 100 / 4 = 25 option c"

a ) 25.5 , b ) 2.5 , c ) 25 , d ) . 25 , e ) none of these

c

divide(1, 0.04)

divide(n0,n1)|

general

a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 30 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ?

"wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 30 emlpoyee paid = . 7 * 240 = 168 $ answer e"

a ) $ 198 , b ) $ 216 , c ) $ 220 , d ) $ 230 , e ) $ 168

e

subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 30))

divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4)|

gain

if an integer n is to be chosen at random from the integers 1 to 92 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?

"n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 96 , there are 46 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 46 + 12 ) / 92 = 58 / 96 = 29 / 46 ans is d"

a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 29 / 46 , e ) 3 / 4

d

divide(add(divide(92, 2), divide(92, 8)), 92)

divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|

general

on dividing a number by 56 , we get 27 as remainder . on dividing the same number by 8 , what will be the remainder ?

"number = 56 x + 27 ( ∵ since the number gives 27 as remainder on dividing by 56 ) = ( 7 × 8 × x ) + ( 3 × 8 ) + 3 hence , if the number is divided by 8 , we will get 3 as remainder . answer : b"

a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6

b

multiply(subtract(divide(power(27, const_2), 56), floor(divide(power(27, const_2), 56))), 56)

power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)|

general

what sum of money will produce rs . 90 as simple interest in 4 years at 3 1 / 2 percent ?

"90 = ( p * 4 * 7 / 2 ) / 100 p = 642.9 answer : d"

a ) 263 , b ) 500 , c ) 367 , d ) 642.9 , e ) 339

d

divide(90, divide(multiply(4, add(3, divide(1, 2))), const_100))

divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)|

gain

a and b enter into a partnership . a contributed rs . 5000 for 8 months and b rs . 6000 for 5 months . find a ' s share in a total profit of rs . 8400 .

explanation : ratio of capitals of a and b = ratio of their investments = 5000 × 8 : 6000 × 5 = 40000 : 30000 = 4 : 3 . : a ' s share = rs . [ ( 4 / 7 ) × 8400 ] = rs . 4800 answer : option a

a ) rs . 4800 , b ) rs . 2400 , c ) rs . 3600 , d ) rs . 5600 , e ) none of these

a

multiply(8400, divide(multiply(5000, 8), add(multiply(5000, 8), multiply(6000, 5))))

multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#0,#2)|multiply(n4,#3)

gain

two vessels a and b contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively . find the ratio in which these mixtures be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8 : 5 ?

"explanation : let cost price ( cp ) of 1 litre spirit be rs . 1 quantity of spirit in 1 litre mixture from vessel a = 5 / 7 cost price ( cp ) of 1 litre mixture from vessel a = rs . = 5 / 7 quantity of spirit in 1 litre mixture from vessel b = 7 / 13 cost price ( cp ) of 1 litre mixture from vessel b = rs . = 7 / 13 quantity of spirit to be obtained in 1 litre mixture from vessel c = 8 / 13 cost price ( cp ) of 1 litre mixture from vessel c ( mean price ) = rs . = 8 / 13 = > mixture from vessel a : mixture from vessel b = 1 / 13 : 9 / 91 = 7 : 9 answer is d"

a ) 3 : 4 , b ) 4 : 3 , c ) 9 : 7 , d ) 7 : 9 , e ) 8 : 1

d

divide(7, add(2, 7))

add(n1,n2)|divide(n2,#0)|

other

company a imported 10,000 widgets made of either brass or aluminum . the widgets are painted blue , red or green . if 20 percent of the widgets are made of brass and of those 20 percent are painted green and 40 percent are painted red how many brass widgets painted blue were imported ?

"answer a . we are told that 10 % of all imported widgets are made of brass and of those , 20 % are green and 40 % are red . since we know that there are only three colors , the remaining 40 % must be blue . 40 % blue of 20 % brass widgets leads to 8 % blue brass widgets out of the total 10,550 widgets . 10,000 / 100 * 8 = 800 . answer c ."

a ) 420 , b ) 840 , c ) 800 , d ) 1680 , e ) 2100

c

multiply(multiply(multiply(multiply(divide(20, const_100), divide(40, const_100)), divide(add(20, const_2), 20)), const_100), const_100)

add(n1,const_2)|divide(n1,const_100)|divide(n3,const_100)|divide(#0,n1)|multiply(#1,#2)|multiply(#3,#4)|multiply(#5,const_100)|multiply(#6,const_100)|

gain

a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 6 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ?

"let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 6 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 3 m / s a"

a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 10

a

divide(add(multiply(6, 3), 6), 10)

multiply(n1,n3)|add(n2,#0)|divide(#1,n0)|

physics

a rectangular lawn of dimensions 80 m * 40 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ?

"area = ( l + b â € “ d ) d ( 80 + 40 â € “ 10 ) 10 = > 1100 m 2 1100 * 3 = rs . 3300 answer : c"

a ) 2288 , b ) 2779 , c ) 3300 , d ) 3900 , e ) 2781

c

multiply(multiply(subtract(add(80, 40), 10), 10), 3)

add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)|

geometry

there is enough provisions for 1200 girls in a hostel for 50 days . if there were 600 girls less , how long will the provision last ?

"we have , m 1 d 1 = m 2 d 2 1200 * 50 = 600 * d 2 d 2 = 1200 * 50 / 600 = 100 days . answer : e"

a ) 75 , b ) 80 , c ) 90 , d ) 95 , e ) 100

e

add(divide(multiply(1200, 600), const_10), 50)

multiply(n0,n2)|divide(#0,const_10)|add(n1,#1)|

physics

a 50 kg metal bar made of alloy of tin and silver lost 5 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ?

"you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 5 kg when 50 kg alloy is immersed . this is a loss of ( 5 / 50 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e"

a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3

e

divide(divide(subtract(5, multiply(divide(0.375, 5), 50)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(50, divide(subtract(5, multiply(divide(0.375, 5), 50)), subtract(divide(1.375, 10), divide(0.375, 5)))))

divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)|

other

a , b and c have rs . 300 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ?

"a + b + c = 300 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 300 - - - - - - - - - - - - - - - - c = 250 answer : a"

a ) a ) 250 , b ) b ) 249 , c ) c ) 248 , d ) d ) 247 , e ) e ) 246

a

subtract(add(200, 350), 300)

add(n1,n2)|subtract(#0,n0)|

general

united telephone charges a base rate of $ 7.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ?

"lets take number of minutesx . given that , 7 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 100 minutes ans b"

a ) 200 minutes , b ) 100 minutes , c ) 120 minutes , d ) 140 minutes , e ) 160 minutes

b

divide(subtract(12.00, 7.00), subtract(0.25, 0.20))

subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)|

general

there are 50 students in a class . if 12 % are absent on a particular day , find the number of students present in the class .

"number of students absent on a particular day = 12 % of 50 i . e . , 12 / 100 × 50 = 6 therefore , the number of students present = 50 - 6 = 44 students . answer : e"

a ) 43 , b ) 36 , c ) 28 , d ) 129 , e ) 44

e

multiply(subtract(const_1, divide(12, const_100.0)), 50)

divide(n1,const_100.0)|subtract(const_1,#0)|multiply(#1,n0)|

gain

how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 50 percent butterfat to obtain milk that is 20 percent butterfat ?

"equate the fat : 0.1 x + 0.5 * 8 = 0.2 ( x + 8 ) - - > x = 24 . answer : d ."

a ) 6 , b ) 12 , c ) 14 , d ) 24 , e ) 28

d

divide(multiply(subtract(50, 20), 8), 10)

subtract(n2,n3)|multiply(n1,#0)|divide(#1,n0)|

general

can c and can в are both right circular cylinders . the radius of can c is twice the radius of can b , while the height of can c is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can c with the same brand of gasoline ?

let x be the radius of b and 2 h be the height of b . therefore , radius of c = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e

a ) $ 1 , b ) $ 2 , c ) $ 4 , d ) $ 8 , e ) $ 16

e

multiply(multiply(const_2, 4), const_2)

multiply(n0,const_2)|multiply(#0,const_2)

geometry

a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 3 years ago ?

"let ' s say age of richard is ` ` r ' ' age of david is ` ` d ' ' age of scott is ` ` s ' ' now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6 = ( s + 8 ) + 6 = s + 14 therefore , 2 s + 8 = s + 14 i . e . s = 6 i . e . r = 20 i . e . d = 14 now , how old was david 3 years ago ? i . e . d - 4 = 14 - 3 = 11 years answer : a"

a ) 11 , b ) 10 , c ) 12 , d ) 14 , e ) 16

a

subtract(add(6, 8), 3)

add(n1,n2)|subtract(#0,n4)|

general

6 years ago , the ratio of the ages of vimal and saroj was 6 : 5 . 4 years hence , the ratio of their ages will be 11 : 10 . what is saroj ' s age at present ?

given that , six years ago , the ratio of the ages of vimal and saroj = 6 : 5 hence we can assume that age of vimal six years ago = 6 x age of saroj six years ago = 5 x after 4 years , the ratio of their ages = 11 : 10 ⇒ ( 6 x + 10 ) : ( 5 x + 10 ) = 11 : 10 ⇒ 10 ( 6 x + 10 ) = 11 ( 5 x + 10 ) ⇒ 5 x = 10 ⇒ x = 105 = 2 saroj ' s present age = ( 5 x + 6 ) = 5 × 2 + 6 = 16 answer is c .

a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 15

c

add(multiply(5, divide(subtract(multiply(11, 10), multiply(const_10, 10)), subtract(multiply(10, 6), multiply(11, 5)))), 6)

multiply(n4,n5)|multiply(n5,const_10)|multiply(n0,n5)|multiply(n2,n4)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n2,#6)|add(n0,#7)

other

a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 70 km downstream .

"speed of boat in still water = 13 km / hr speed of the stream = 5 km / hr speed downstream = ( 13 + 5 ) = 18 km / hr time taken to travel 70 km downstream = 70 ⠁ „ 18 = 3.8 hours answer is b"

a ) 4.25 hr , b ) 3.8 hr , c ) 8.25 hr , d ) 2.25 hr , e ) 2.50 hr

b

divide(70, add(13, 5))

add(n0,n1)|divide(n2,#0)|

physics

if 60 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ?

"60 % of ( x - y ) = 20 % of ( x + y ) 60 / 100 ( x - y ) = 20 / 100 ( x + y ) 4 x = 8 y required percentage = y / x * 100 = 4 y / 8 y * 100 = 50 % answer is a"

a ) 50 % , b ) 44.4 % , c ) 22.2 % , d ) 33.3 % , e ) 25 %

a

multiply(divide(subtract(60, 20), add(60, 20)), const_100)

add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|

general

a and b undertake to do a piece of work for rs . 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . ! find the share of each .

c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a ’ s share = rs . ( 600 * 4 / 8 ) = rs . 300 , b ' s share = rs . ( 600 * 3 / 8 ) = rs . 225 . c ' s share = rs . [ 600 - ( 300 + 225 » ) = rs . 75 . answer is c

a ) 25 , b ) 50 , c ) 75 , d ) 80 , e ) none of them

c

multiply(divide(const_1, add(add(const_4, const_3), const_1)), 600)

add(const_3,const_4)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2)

physics

how many even integers n , such that 20 < = n < = 220 are of the form 3 k + 4 , where k is any natural number ?

"the first number is 22 = 16 + 6 ( 1 ) . we can continue adding 6 to make a list : 22 , 28 , 34 , . . . the last number is 220 = 16 + 6 ( 34 ) there are 34 numbers in the list . the answer is c ."

a ) 28 , b ) 31 , c ) 34 , d ) 37 , e ) 40

c

add(divide(subtract(subtract(220, const_2), add(20, const_2)), add(const_2, 4)), const_1)

add(n0,const_2)|add(const_2,n3)|subtract(n1,const_2)|subtract(#2,#0)|divide(#3,#1)|add(#4,const_1)|

general

the edges of three metal cubes are 1 cm , 3 cm , and 4 cm respectively . a new cube is made by melting these three cubes together . what is the edge of the new cube ( in centimeters ) ?

the total volume is 1 ^ 3 + 3 ^ 3 + 4 ^ 3 = 92 the edge of the new cube is the cube root of 92 which is about 4.5 cm . the answer is c .

a ) 3.9 , b ) 4.2 , c ) 4.5 , d ) 4.8 , e ) 5.1

c

power(add(power(4, const_3), add(1, power(3, const_3))), const_0_33)

power(n1,const_3)|power(n2,const_3)|add(n0,#0)|add(#2,#1)|power(#3,const_0_33)

physics

an inspector rejects 10 % of the meters as defective . how many will he examine to reject 12 ?

"then , 10 % of x = 12 ( 10 / 100 ) x = 12 x = ( 12 * 100 * ) / 10 = 120 answer is b"

a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160

b

divide(multiply(12, const_100), 10)

multiply(n1,const_100)|divide(#0,n0)|

gain

a survey was sent to 60 customers , 7 of whom responded . then the survey was redesigned and sent to another 72 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ?

"rate of first survey = 7 / 60 rate of second survey = 9 / 72 % response rate increase ( 9 / 72 - 7 / 60 ) / ( 7 / 60 ) = 7 % answer is b = 7 %"

a ) 2 % , b ) 7 % , c ) 14 % , d ) 28 % , e ) 63 %

b

floor(multiply(subtract(divide(9, 72), divide(7, 60)), const_100))

divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)|

gain

if the sides of a square are multiplied by 5 , the area of the original square is how many times as large as the area of the resultant square ?

"let x be the original length of one side . then the original area is x ^ 2 . the new square has sides of length 5 x , so the area is 25 x ^ 2 . the area of the original square is 1 / 25 = 4 % times the area of the new square . the answer is a ."

a ) 4 % , b ) 25 % , c ) 50 % , d ) 200 % , e ) 400 %

a

divide(multiply(5, 5), const_100)

multiply(n0,n0)|divide(#0,const_100)|

geometry

car a runs at the speed of 50 km / hr & reaches its destination in 6 hr . car b runs at the speed of 100 km / h & reaches its destination in 1 h . what is the respective ratio of distances covered by car a & car b ?

"sol . distance travelled by car a = 50 × 6 = 300 km distance travelled by car b = 100 × 1 = 100 km ratio = 300 / 100 = 3 : 1 answer : b"

a ) 11 : 5 , b ) 3 : 1 , c ) 13 : 7 , d ) 15 : 7 , e ) 16 : 9

b

divide(multiply(50, 6), multiply(100, 1))

multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)|

physics

a cubical block of metal weighs 8 pounds . how much will another cube of the same metal weigh if its sides are twice as long ?

"for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 6 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 8 pounds = 64 pounds so answer is e and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 8 * 8 = 64 so answer is e"

a ) 48 , b ) 32 , c ) 24 , d ) 18 , e ) 64

e

multiply(8, multiply(const_2, const_4))

multiply(const_2,const_4)|multiply(n0,#0)|

geometry

of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 20 grades are d ' s . what is the number of students in the course ?

"we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 20 grades are d ' s . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a ’ s + # b ’ s + # c ’ s + # d ’ s = total number of students in the class 1 / 5 ( t ) + ¼ ( t ) + ½ ( t ) + 20 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 400 = 20 t 19 t + 400 = 20 t 400 = t there are a total of 400 students in the math class . answer is e ."

a ) 80 , b ) 110 , c ) 160 , d ) 200 , e ) 400

e

divide(20, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2))))

divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)|

general

how many pieces of 0.85 meteres can be cut from a rod 42.5 meteres long

explanation : we need so simple divide 42.5 / 0.85 , = ( 4250 / 85 ) = 50 option c

a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70

c

divide(42.5, 0.85)

divide(n1,n0)

physics

ratio between rahul and deepak is 5 : 2 , after 6 years rahul age will be 26 years . what is deepak present age .

"explanation : present age is 5 x and 2 x , = > 5 x + 6 = 26 = > x = 4 so deepak age is = 2 ( 4 ) = 8 option c"

a ) 14 , b ) 15 , c ) 8 , d ) 22 , e ) 24

c

divide(multiply(subtract(26, 6), 2), 5)

subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|

other

in what time will a train 400 m long cross an electric pole , it its speed be 144 km / hr ?

"speed = 144 * 5 / 18 = 40 m / sec time taken = 400 / 40 = 10.0 sec . answer : e"

a ) 2.5 sec , b ) 12.9 sec , c ) 11.3 sec , d ) 21.9 sec , e ) 10.0 sec

e

divide(400, multiply(144, const_0_2778))

multiply(n1,const_0_2778)|divide(n0,#0)|

physics

set a consists of all the prime numbers between 62 and 85 . what is the range of set a ?

the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 83 lowest number = 67 range = highest - lowest = 83 - 67 = 16 option d

a ) 12 , b ) 13 , c ) 16.6 , d ) 16 , e ) 23

d

subtract(subtract(85, const_2), add(add(62, const_4), const_1))

add(n0,const_4)|subtract(n1,const_2)|add(#0,const_1)|subtract(#1,#2)

general

a fruit seller had some apples . he sells 60 % apples and still has 300 apples . originally , he had :

"a fruit seller had some apples . he sells 60 % apples and still has 300 apples . originally , he had : a . 656 apples b . 700 apples c . 750 apples d . 780 apples 800 apples answer : option d explanation : suppose originally he had x apples . then , ( 100 - 60 ) % of x = 300 . 40 / 100 x x = 300 100 x = ( 300 x 1000 ) / 40 = 750 . answer : c"

a ) 656 apples , b ) 700 apples , c ) 750 apples , d ) 780 apples , e ) 800 apples

c

original_price_before_loss(60, 300)

original_price_before_loss(n0,n1)|

gain

s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits w are in s ( 99 ) ?

"focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 . b"

a ) w = 350 , b ) w = 353 , c ) w = 354 , d ) 356 , e ) 357

b

add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2)))

add(n6,const_60)|multiply(n0,n4)|multiply(n2,n4)|add(n6,#0)|multiply(n2,#2)|add(#1,#4)|multiply(n6,#3)|add(#3,#6)|add(#7,#5)|

general

a candidate got 35 % of the votes and lost to the rival by 1650 votes . how many votes were cast ?

"35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % = 1650 10 % = 550 100 % of the votes = 5500 answer : b"

a ) 5000 , b ) 5500 , c ) 6000 , d ) 6500 , e ) 7000

b

divide(1650, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100)))

divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|

gain

a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 8 miles per gallon , and from town b to town c , the car averaged 18 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ?

"ans is b given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 8 ) + ( x / 18 ) ) = 15.5 ( formula avg speed = total distance / total time )"

a ) 13 , b ) 15.5 , c ) 14 , d ) 14.5 , e ) 15

b

divide(add(multiply(18, const_10), divide(multiply(18, const_10), const_2)), add(divide(multiply(18, const_10), 8), divide(divide(multiply(18, const_10), const_2), 18)))

multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)|

general

the total age of a and b is 10 years more than the total age of b and c . c is how many years younger than a . ?

"( a + b ) - ( b - c ) = 10 a - c = 10 answer is c"

a ) 16 , b ) 12 , c ) 10 , d ) 20 , e ) 10

c

multiply(10, const_1)

multiply(n0,const_1)|

general

if the area of a circle is 25 pi square feet , find its circumference .

"the area is given by pi * r * r . hence pi * r * r = 25 pi r * r = 25 ; hence r = 25 feet the circumference is given by 2 * pi * r = 2 * pi * 5 = 10 pi feet correct answer a"

a ) 10 pi feet , b ) 28 pi feet , c ) 38 pi feet , d ) 48 pi feet , e ) 58 pi feet

a

circumface(sqrt(25))

sqrt(n0)|circumface(#0)|

geometry

excluding stoppages , the speed of a train is 67 kmph and including stoppages it is 51 kmph . of how many minutes does the train stop per hour ?

"explanation : t = 16 / 67 * 60 = 14 answer : option c"

a ) 82 , b ) 17 , c ) 14 , d ) 82 , e ) 18

c

subtract(const_60, multiply(const_60, divide(51, 67)))

divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|

physics

the sides of a square region , measured to the nearest centimeter , are 9 centimeters long . the least possible value of the actual area of the square region is

"though there might be some technicalities concerning the termnearest ( as 7.5 is equidistant from both 7 and 8 ) the answer still should be : 7.5 ^ 2 = 56.25 . answer : c"

a ) 60.25 sq cm , b ) 66.25 sq cm , c ) 56.25 sq cm , d ) 72.25 sq cm , e ) 58.25 sq cm

c

power(subtract(subtract(9, const_0_25), const_0_25), const_2)

subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2)|

geometry

the average runs of a cricket player of 8 innings was 24 . how many runs must he make in his next innings so as to increase his average of runs by 8 ?

"explanation : average after 9 innings = 32 required number of runs = ( 32 * 9 ) – ( 24 * 8 ) = 288 – 192 = 96 answer d"

a ) 12 , b ) 42 , c ) 64 , d ) 96 , e ) 84

d

subtract(multiply(add(8, const_1), add(24, 8)), multiply(24, 8))

add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|

general

if jack walked 4 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ?

distance walked in 1 hour and 15 mins = 4 miles speed per hour = distance / time = 4 / ( 5 / 4 ) = 3.2 miles per hour answer a

a ) 3.2 , b ) 4.5 , c ) 6 , d ) 6.25 , e ) 15

a

divide(multiply(4, const_60), add(15, const_60))

add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0)

physics

yesterday it took robert 6 hours to drive from city a to city b . today it took robert 4.5 hours to drive back from city в to city a along the same route . if he had saved 30 minutes in both trips , the speed for the round trip would be 90 miles per hour . what is the distance between city a and city b ?

2 d / 90 = 9.5 ( because time = 6 + 4.5 - 1 hrs ) = > d = 430 answer - e

a ) 90 , b ) 120 , c ) 150 , d ) 440 , e ) 430

e

divide(multiply(90, subtract(add(6, 4.5), const_1)), const_2)

add(n0,n1)|subtract(#0,const_1)|multiply(n3,#1)|divide(#2,const_2)

physics

a number consists of 3 digits whose sum is 10 . the middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed . the number is :

"let the middle digit be x . then , 2 x = 10 or x = 5 . so , the number is either 253 or 352 . since the number increases on reversing the digits , so the hundred ' s digits is smaller than the unit ' s digit . hence , required number = 253 . option b"

a ) 251 , b ) 253 , c ) 255 , d ) 257 , e ) 260

b

add(add(multiply(const_100, subtract(divide(10, const_2), 3)), multiply(divide(10, const_2), 10)), 3)

divide(n1,const_2)|multiply(n1,#0)|subtract(#0,n0)|multiply(#2,const_100)|add(#3,#1)|add(n0,#4)|

general

how many seconds will a train 110 meters long take to cross a railway platform 165 meters long if the speed of the train is 132 kmph ?

"d = 110 + 165 = 275 s = 132 * 5 / 18 = 110 / 3 mps t = 275 * 3 / 110 = 7.5 sec b ) 7.5 sec"

a ) 6 sec , b ) 7.5 sec , c ) 8 sec , d ) 3 sec , e ) 10 sec

b

divide(add(165, 110), multiply(132, const_0_2778))

add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)|

physics

2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations . what is the ratio of their speeds

"explanation : we know total distance is 200 km if both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [ 110 + 90 = 200 km ] so ratio of their speed = 110 : 90 = 11 : 9 option a"

a ) 11 : 9 , b ) 13 : 9 , c ) 17 : 9 , d ) 21 : 9 , e ) none of these

a

inverse(divide(subtract(200, 110), 110))

subtract(n2,n3)|divide(#0,n3)|inverse(#1)|

physics

50 percent of andrea ' s living room floor is covered by a carpet that is 4 feet by 9 feet . what is the area of her living room floor ?

"50 % of area of the floor = 4 * 9 square feet = 36 square feet i . e . 100 % area of floor = ( 36 / 50 ) * 100 = 64.8 square feet answer : option d"

a ) 14.4 , b ) 180 / 7 , c ) 50.4 , d ) 64.8 , e ) 90

d

divide(multiply(4, 9), divide(50, const_100))

divide(n0,const_100)|multiply(n1,n2)|divide(#1,#0)|

gain

a work can be finished in 13 days by ten women . the same work can be finished in fifteen days by ten men . the ratio between the capacity of a man and a woman is

work done by 10 women in 1 day = 1 / 13 work done by 1 woman in 1 day = 1 / ( 13 ã — 10 ) work done by 16 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 ã — 10 ) ratio of the capacity of a man and woman = 1 / ( 15 ã — 10 ) : 1 / ( 13 ã — 10 ) = 1 / 15 : 1 / 13 = 1 / 15 : 1 / 13 = 13 : 15 answer is b .

a ) 14 : 13 , b ) 13 : 15 , c ) 12 : 13 , d ) 13 : 12 , e ) 14 : 15

b

divide(divide(const_1, multiply(add(13, const_2), const_10)), divide(const_1, multiply(13, const_10)))

add(n0,const_2)|multiply(n0,const_10)|divide(const_1,#1)|multiply(#0,const_10)|divide(const_1,#3)|divide(#4,#2)

physics

if 213 × 16 = 3408 , then 1.6 × 213.0 is equal to :

solution 1.6 × 21.3 = ( 16 / 10 x 213 ) = ( 16 x 213 / 10 ) = 3408 / 10 = 340.8 . answer d

a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these

d

multiply(1.6, 213)

multiply(n0,n3)

general

a train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds . its length is

"let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 25 = x / 15 x = 150 m . answer : option d"

a ) 100 , b ) 120 , c ) 130 , d ) 150 , e ) 140

d

multiply(100, subtract(const_2, const_1))

subtract(const_2,const_1)|multiply(n1,#0)|

physics

the population of a bacteria colony doubles every day . if it was started 3 days ago with 3 bacteria and each bacteria lives for 12 days , how large is the colony today ?

"9 days ago - 2 8 days ago - 4 7 days ago - 8 6 days ago - 16 5 days ago - 32 4 days ago - 64 3 days ago - 128 2 days ago - 256 yesterday - 512 today - 1024 answer : b"

a ) 512 , b ) 1024 , c ) 768 , d ) 2048 , e ) 4096

b

subtract(power(3, add(3, const_1)), const_1)

add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|

physics

the h . c . f of two numbers is 15 and their l . c . m is 3300 . if one of the numbers is 225 , then the other is ?

"other number = ( 15 * 3300 ) / 225 = 220 . answer : c"

a ) 200 , b ) 210 , c ) 220 , d ) 230 , e ) 240

c

multiply(15, 225)

multiply(n0,n2)|

physics

a bag marked at $ 80 is sold for $ 68 . the rate of discount is ?

"rate of discount = 12 / 80 * 100 = 15 % answer is b"

a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 %

b

multiply(divide(subtract(80, 68), 80), const_100)

subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|

gain

{ 5 , 6 , 3 , 8 , 4 } if we want to get the median value of the set to be 10 which value should be inserted in the set of 5 values ?

median is the middle number of a sorted distribution . in case of even number of items , median will be the average of middle two values . the question asks fora new number that will cause the median of new set at 10 . mentionable that given set is already sorted and number of items of new set will be 5 + 1 = 6 . to get median 10 , the sum of 3 th 4 th term should be 20 i . e . 3 + 7 . so to keep the median 10 , the new item should be > = 7 answer is b

a ) 5 , b ) 7 , c ) 15 , d ) 12 , e ) 13

b

add(add(add(add(add(add(subtract(add(5, 6), 10), const_1), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10)), subtract(add(5, 6), 10))

add(n0,n1)|subtract(#0,n5)|add(#1,const_1)|add(#2,#1)|add(#3,#1)|add(#4,#1)|add(#5,#1)|add(#6,#1)

general

the area of a rhombus is equal to the area of a square whose side length is 8 cm . if one of the diagonals is 16 cm what is the length of other diagonal ?

area of the square = 8 x 8 = 64 let ' l ' be the unknown diagonal 0.5 x 16 x ' l ' = 64 which gives ' l ' = 8 cm answer : b

['a ) 4', 'b ) 8', 'c ) 12', 'd ) 16', 'e ) 20']

b

divide(multiply(square_area(8), const_2), 16)

square_area(n0)|multiply(#0,const_2)|divide(#1,n1)

geometry

the cost price of 7 articles is equal to the selling price of 5 articles . what is the profit percent ?

"5 * sp = 7 * cp sp = 1.4 * cp the profit percent is 40 % . the answer is d ."

a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %

d

divide(multiply(7, const_4), add(const_4, const_1))

add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|

gain

in a kilometer race , a beats b by 20 meters or 10 seconds . what time does a take to complete the race ?

"time taken by b run 1000 meters = ( 1000 * 10 ) / 20 = 500 sec . time taken by a = 500 - 10 = 490 sec . answer : e"

a ) 290 sec , b ) 190 sec , c ) 199 sec , d ) 192 sec , e ) 490 sec

e

subtract(divide(multiply(const_1, const_1000), divide(20, 10)), 10)

divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|

physics

a positive number when decreased by 4 is equal to 21 times the reciprocal of the number . the number is

"sol . let the numbers be x . then , x - 4 = 21 / x ⇔ x 2 - 4 x - 21 = 0 ⇔ ( x - 7 ) ( x + 3 ) = 0 ⇔ x = 7 . answer c"

a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 12

c

divide(add(sqrt(add(multiply(4, 21), power(4, const_2))), 4), const_2)

multiply(n0,n1)|power(n0,const_2)|add(#0,#1)|sqrt(#2)|add(n0,#3)|divide(#4,const_2)|

general

at what price must an book costing $ 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ?

"c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 95 / 100 ) = 59.375 mp = 62.5"

a ) 72.5 , b ) 55.5 , c ) 62.5 , d ) 82.5 , e ) 60.5

c

multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 5)), const_100)

add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)|

gain

a car consumes 210 liters of petrol weekly . for how many days will 1470 liters of petrol be sufficient ?

explanation : no explanation is available for this question ! answer : d

a ) 38 , b ) 20 , c ) 28 , d ) 49 , e ) 10

d

multiply(divide(1470, 210), add(const_4, const_3))

add(const_3,const_4)|divide(n1,n0)|multiply(#0,#1)

physics

one type of liquid contains 20 % of water and the second type of liquid contains 35 % of water . a glass is filled with 10 parts of 1 st liquid and 4 parts of 2 nd liquid . the percentage of water in the new mixture in the glass is ?

required percentage = ( 20 % of 10 + 35 % of 4 ) / 10 + 4 * 100 = 3.4 / 14 * 100 = 24.3 % answer is c

a ) 22.15 % , b ) 23.5 % , c ) 24.3 % , d ) 25 % , e ) 29.6 %

c

multiply(divide(const_100, add(10, 4)), add(divide(multiply(20, 10), const_100), divide(multiply(35, 4), const_100)))

add(n2,n4)|multiply(n0,n2)|multiply(n1,n4)|divide(#1,const_100)|divide(#2,const_100)|divide(const_100,#0)|add(#3,#4)|multiply(#6,#5)

gain

the sum of two consecutive integers is 61 . find the numbers .

"n + ( n + 1 ) = 61 2 n + 1 = 61 2 n = 60 n = 30 answer : b"

a ) 17 , 18 , b ) 30,31 , c ) 5 , 6 , d ) 1 , 2 , e ) 8 , 9

b

add(add(power(add(add(divide(subtract(subtract(61, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(61, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(61, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(61, const_10), const_2), const_4), const_2), const_2)))

subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)|

physics

cuboid volume and base area are 144 m ^ 3 and 18 m ^ 3 respectively . find the height of a cuboid ?

height of the cuboid = volume * base area = 144 / 18 = 8 metre answer is option d

['a ) 5 m', 'b ) 12 m', 'c ) 6 m', 'd ) 8 m', 'e ) 10 m']

d

divide(144, 18)

divide(n0,n2)

geometry

how many positive even integers less than 100 contain digits 7 or 9 ?

"two digit numbers : 7 at tens place : 70 , 72,74 , 76,78 9 at tens place : 90 , 92,94 , 96,98 if 7 and 9 is at units place , the number cant be even total : 5 + 5 = 10 answer e"

a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 10

e

multiply(7, const_2)

multiply(n1,const_2)|

general

in a group of 650 readers who read science fiction or literacy works or both , 250 read science fiction and 550 read literacy works . how many read both science fiction and literacy works ?

"consider total number of reader n ( s u l ) = 650 people who read science fiction n ( s ) = 250 people who read literacy works n ( l ) = 550 both science fiction and literacy n ( s â ˆ © l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s â ˆ © l ) 650 = 250 + 550 - n ( s â ˆ © l ) n ( s â ˆ © l ) = 800 - 650 n ( s â ˆ © l ) = 150 so people read both science fiction and literacy works are 150 answer : c"

a ) 650 , b ) 250 , c ) 150 , d ) 100 , e ) 400

c

subtract(add(250, 550), 650)

add(n1,n2)|subtract(#0,n0)|

other

if anangletis defined as 1 percent of 1 degree , then how many anglets are there in one half circle ?

"1 degree * 1 / 100 = 1 anglet so 1 degree = 100 anglets = > 180 degrees = 18,000 anglets answer - d"

a ) 0.36 , b ) 3.6 , c ) 360 , d ) 18,000 , e ) 36,000

d

multiply(divide(1, add(const_2, const_4)), multiply(divide(const_3600, const_10), const_100))

add(const_2,const_4)|divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)|multiply(#2,#3)|

geometry

the speed of a car is 80 km in the first hour and 40 km in the second hour . what is the average speed of the car ?

"s = ( 80 + 40 ) / 2 = 60 kmph answer : b"

a ) 72 kmph , b ) 60 kmph , c ) 30 kmph , d ) 80 kmph , e ) 82 kmph

b

divide(add(80, 40), const_2)

add(n0,n1)|divide(#0,const_2)|

physics

two trains 200 m and 250 m long run at the speed of 72 kmph and 18 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ?

"relative speed = 72 + 18 = 90 kmph * 5 / 18 = 25 m / s distance covered in crossing each other = 200 + 250 = 450 m required time = 450 * 1 / 25 = 18 sec answer is e"

a ) 5.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 18 sec

e

divide(add(200, 250), multiply(add(72, 18), const_0_2778))

add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|

physics

the “ length of integer x ” refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 × 2 × 3 × 5 . ) what is the greatest possible length of integer z if z < 3000 ?

"to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 11 = 2048 < 3000 , then the greatest possible length of integer z is 11 . the answer is c ."

a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15

c

log(power(2, const_10))

power(n3,const_10)|log(#0)|

general