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in a tree , 3 / 7 of the birds are robins while the rest are bluejays . if 1 / 3 of the robins are female and 3 / 5 of the bluejays are female , what fraction of the birds in the tree are male ?
"the fraction of birds that are male robins is ( 2 / 3 ) ( 3 / 7 ) = 2 / 7 . the fraction of birds that are male bluejays is ( 2 / 5 ) ( 4 / 7 ) = 8 / 35 . the total fraction of male birds is 2 / 7 + 8 / 35 = 18 / 35 . the answer is c ."
a ) 11 / 35 , b ) 14 / 35 , c ) 18 / 35 , d ) 31 / 70 , e ) 37 / 70
c
add(multiply(divide(3, 7), divide(const_2.0, 5)), multiply(divide(3, 7), divide(1, 3)))
divide(n0,n1)|divide(n0,n5)|divide(n5,n1)|divide(n2,n3)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5)|
general
the length of a rectangular landscape is 8 times its breadth . there is a playground in it whose area is 1200 square mtr & which is 1 / 6 rd of the total landscape . what is the length of the landscape ?
"sol . x * 8 x = 6 * 1200 x = 30 length = 8 * 30 = 240 e"
a ) 180 , b ) 270 , c ) 340 , d ) 140 , e ) 240
e
multiply(multiply(const_4.0, 6), const_10)
multiply(const_4.0,n3)|multiply(#0,const_10)|
geometry
ratio and proportion 215 : 474 : : 537 : ?
2 + 1 + 5 + 4 + 7 + 4 = 23 5 + 3 + 7 + x = 23 = > x = 8 combination that match 8 is 26 since 2 + 6 = 8 answer : a
a ) 26 , b ) 27 , c ) 25 , d ) 22 , e ) 23
a
add(add(const_4, const_2), add(const_10, const_10))
add(const_2,const_4)|add(const_10,const_10)|add(#0,#1)
other
sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 288 per week . how much does she earn in dollars per hour ?
"let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 288 per week 36 x = 288 x = 8 correct option : d"
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7
d
divide(288, add(multiply(8, const_3), multiply(6, const_2)))
multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)|
physics
a statue is being carved by a sculptor . the original piece of marble weighed 200 kg . in the first week 20 percent is cut away . in the second week 25 percent of the remainder is cut away . in the third week the statue is completed when 35 percent of the remainder is cut away . what is the weight of the final statue ?
"a 97.5 kg 250 ã — 0.8 ã — 0.75 ã — 0.65 = 97.5 kg ."
a ) 97.5 kg , b ) 103 kg , c ) 108 kg , d ) 125 kg , e ) 117 kg
a
multiply(subtract(const_1, divide(35, const_100)), multiply(subtract(const_1, divide(25, const_100)), multiply(200, subtract(const_1, divide(20, const_100)))))
divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(n0,#3)|multiply(#6,#4)|multiply(#7,#5)|
gain
an offshore company sells two products , l and m . last year , seventy percent of the units sold were product l , and the price of product l was 25 percent greater than the price of product m . approximately what percent of the total revenue the company received last year was from the sale of product l ?
percentage of total units that is product l = 70 % price of product l = 25 percent greater than price of product m . now , we can consider this as a weighted average scenario . if the price of both the products l and m was equal , then contribution of product l to total sales would be = 70 % but since , price of product l is greater than m , then contribution of product l to total sales will be greater than 70 % the only option is 74 % answer e
a ) 15 % , b ) 25 % , c ) 35 % , d ) 65 % , e ) 74 %
e
multiply(divide(multiply(add(const_3, const_4), const_10), add(multiply(subtract(const_100, multiply(add(const_3, const_4), const_10)), subtract(const_1, divide(25, const_100))), multiply(add(const_3, const_4), const_10))), const_100)
add(const_3,const_4)|divide(n0,const_100)|multiply(#0,const_10)|subtract(const_1,#1)|subtract(const_100,#2)|multiply(#4,#3)|add(#5,#2)|divide(#2,#6)|multiply(#7,const_100)
general
if x , y , and z are positive integers , and 4 x = 5 y = 8 z , then the least possible value of x + y + z is
"take lcm of 4,5 and 8 = 40 now 4 x = 40 = > x = 10 5 y = 40 = > y = 8 8 z = 40 = > z = 5 10 + 8 + 5 = 23 . option b ."
a ) 15 , b ) 23 , c ) 37 , d ) 42 , e ) 60
b
add(add(divide(divide(multiply(multiply(4, 5), 8), const_2), 4), divide(divide(multiply(multiply(4, 5), 8), const_2), 5)), divide(divide(multiply(multiply(4, 5), 8), const_2), 8))
multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|divide(#2,n0)|divide(#2,n1)|divide(#2,n2)|add(#3,#4)|add(#6,#5)|
general
find the principal which yields a simple interest of rs . 20 and compound interest of rs . 28 in two years , at the same percent rate per annum ?
"explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 28 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 28 – 20 ) / 20 ] * 100 = 40 % p . a . let the principal be rs . x time = t = 2 years % rate = 40 % p . a . si = ( prt / 100 ) 20 = ( x * 40 * 2 ) / 100 x = rs . 25 answer : a"
a ) s . 25 , b ) s . 48 , c ) s . 42 , d ) s . 20 , e ) s . 60
a
divide(multiply(subtract(28, 20), const_100), const_2)
subtract(n1,n0)|multiply(#0,const_100)|divide(#1,const_2)|
gain
joshua and jose work at an auto repair center with 2 other workers . for a survey on health care insurance , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that joshua and jose will both be chosen ?
"two methods 1 ) probability of chosing josh first = 1 / 4 probability of chosing jose second = 1 / 3 total = 1 / 12 probability of chosing jose first = 1 / 4 probability of chosing josh second = 1 / 3 total = 1 / 12 final = 1 / 12 + 1 / 12 = 1 / 6 d"
a ) 1 / 15 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3
d
divide(divide(factorial(6), multiply(factorial(2), factorial(2))), factorial(6))
factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|divide(#4,#0)|
physics
in a certain accounting class of 100 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 60 % , and the students on the make - up date had an average score of 90 % , what was the average score for the entire class ?
"70 % of the class scored 60 % and 30 % of the class scored 90 % . the difference between 60 % and 90 % is 30 % . the average will be 60 % + 0.3 ( 30 % ) = 69 % . the answer is c ."
a ) 65 % , b ) 67 % , c ) 69 % , d ) 72 % , e ) 75 %
c
divide(add(multiply(70, 60), multiply(90, subtract(100, 70))), 100)
multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|add(#0,#2)|divide(#3,n0)|
general
of 90 applicants for a job , 42 had at least 4 years ' experience , 54 had degrees , and 8 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ?
"90 - 8 = 82 82 - 42 - 54 = - 14 then 14 are in the intersection between 4 years experience and degree . answer : c"
a ) 10 , b ) 8 , c ) 14 , d ) 16 , e ) 12
c
add(subtract(add(42, 54), subtract(90, 8)), subtract(54, 42))
add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)|
general
a shopkeeper sold an article at $ 1110 and gained a 20 % profit . what was the cost price ?
"let x be the cost price . 1.2 x = 1110 x = 1110 / 1.2 = 925 the answer is d ."
a ) $ 850 , b ) $ 875 , c ) $ 900 , d ) $ 925 , e ) $ 950
d
multiply(const_100.0, divide(const_100, add(1110, 20)))
add(n1,const_100)|divide(const_100,#0)|multiply(n0,#1)|
gain
two trains running in opposite directions cross a pole placed on the platform in 47 seconds and 31 seconds respectively . if they cross each other in 33 seconds , what is the ratio of their speeds ?
explanation : let the speed of the trains be x and y respectively length of train 1 = 47 x length of train 2 = 31 y relative speed = x + y time taken to cross each other = 33 s = > = 33 = > ( 47 x + 31 y ) = 33 ( x + y ) = > 14 x = 2 y = > x / y = 2 / 14 = 1 / 7 = 1 : 7 answer : a
a ) 1 : 7 , b ) 8 : 7 , c ) 6 : 7 , d ) 6 : 5 , e ) 4 : 7
a
divide(subtract(33, 31), subtract(47, 33))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)
physics
if 25 % of a class averages 80 % on a test , 50 % of the class averages 65 % on the test , and the remainder of the class averages 90 % on the test , what is the overall class average ?
"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 25 % - 50 % = 25 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.25 x 80 + 0.50 x 65 + 0.25 x 90 = 75 the class average is 75 % final answer d ) 75 %"
a ) 60 % , b ) 65 % , c ) 70 % , d ) 75 % , e ) 80 %
d
divide(add(add(multiply(25, 80), multiply(50, 65)), multiply(subtract(const_100, add(25, 50)), 90)), const_100)
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)|
general
a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord s , in feet ?
"because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . s = pi * r = 30 . round pi = 3 , then r = 10 . chord is about 10 feet long . d"
a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5
d
divide(30, const_3)
divide(n0,const_3)|
general
if xy > 0 , 1 / x + 1 / y = 6 , and 1 / xy = 12 , then ( x + y ) / 6 = ?
"( 1 / x + 1 / y ) = 6 canbe solved as { ( x + y ) / xy } = 12 . substituting for 1 / xy = 12 , we get x + y = 6 / 12 = = > ( x + y ) / 6 = 6 / ( 12 * 6 ) = 1 / 12 . a"
a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 5 , d ) 5 , e ) 6
a
divide(divide(6, 12), 12)
divide(n3,n5)|divide(#0,n5)|
general
how many 4 digit numbers have no repeat digits , do not contain zero , and have a sum of digits d equal to 28 ?
first , look for all 4 digits without repeat that add up to 28 . to avoid repetition , start with the highest numbers first . start from the largest number possible 9874 . then the next largest number possible is 9865 . after this , you ' ll realize no other solution . clearly the solution needs to start with a 9 ( cuz otherwise 8765 is the largest possible , but only equals 26 ) . with a 9 , you also need an 8 ( cuz otherwise 9765 is the largest possible , but only equals 27 ) . with 98 __ only 74 and 65 work . so you have two solutions . each can be rearranged in 4 ! = 24 ways . so d = 24 + 24 = 48 . d
a ) 14 , b ) 24 , c ) 28 , d ) 48 , e ) 96
d
divide(multiply(factorial(const_4), 4), const_2)
factorial(const_4)|multiply(n0,#0)|divide(#1,const_2)
general
a basket contains 8 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?
"the total number of ways to choose 2 apples is 8 c 2 = 28 the number of ways that include the spoiled apple is 7 c 1 = 7 p ( the spoiled apple is included ) = 7 / 28 = 1 / 4 the answer is c ."
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6
c
divide(choose(subtract(8, 1), 1), choose(8, 2))
choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)|
probability
a train is 460 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length
"explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 460 + 140 = 600 meter time = distance / speed = 600 ∗ 2 / 25 = 48 seconds option d"
a ) 20 seconds , b ) 30 seconds , c ) 40 seconds , d ) 48 seconds , e ) none of these
d
divide(add(460, 140), divide(multiply(45, const_1000), const_3600))
add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|
physics
of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout , the number of males is 45 more than twice the number of females . if the ratio of female speckled trout to male rainbow trout is 4 : 3 and the ratio of male rainbow trout to all trout is 3 : 20 , how many female rainbow trout are there ?
it will be easier if you draw a table for this one . let t be the total trout . total speckled trout ( st ) = 645 if female speckled trout = fst then male speckled trout = 45 + 2 fst therefore 45 + 2 fst + fst = 645 we solve and get fst = 200 and mst = 445 ratio of fst to mrt = 4 : 3 we have fst = 200 solve ratio to get mrt = 150 now ratio of mrt to t = 3 : 20 we have mrt = 150 solve ratio to get t = 1000 now unless youve drawn the table it might get a little complex . from above we have mst and mrt mst + mrt = 445 + 150 = 595 so t - 595 = total female trouts 1000 - 595 = 405 female trouts we have from above that fst = 200 so ( whew ! ) frt = 405 - 200 = 205 answer is d
a ) 192 , b ) 195 , c ) 200 , d ) 205 , e ) 208
d
subtract(subtract(divide(multiply(divide(multiply(divide(subtract(645, 45), const_3), 3), 4), 20), 3), 645), divide(multiply(divide(subtract(645, 45), const_3), 3), 4))
subtract(n0,n1)|divide(#0,const_3)|multiply(n3,#1)|divide(#2,n2)|multiply(n5,#3)|divide(#4,n3)|subtract(#5,n0)|subtract(#6,#3)
general
find compound interest on rs . 8000 at 15 % per annum for 2 years 4 months , compounded annually
"time = 2 yrs 4 mos = 2 ( 4 / 12 ) yrs = 2 ( 1 / 3 ) yrs . amount = rs . [ 8000 x ( 1 + ( 15 / 100 ) ) ^ 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = rs . 11109 . : . c . i . = rs . ( 11109 - 8000 ) = rs . 3109 . answer a"
a ) 3109 , b ) 3115 , c ) 3250 , d ) 3500 , e ) 4000
a
subtract(add(add(8000, divide(multiply(8000, 15), const_100)), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 8000)
multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)|
gain
it costs $ 3 for the first 1 / 3 hour to use the laundry machine at the laundromat . after the first ¼ hour it costs $ 12 per hour . if a certain customer uses the laundry machine for 2 hours and 35 minutes , how much will it cost him ?
"2 hrs 35 min = 155 min first 20 min - - - - - - > $ 3 time left is 135 min . . . now , 60 min costs $ 12 1 min costs $ 12 / 60 155 min costs $ 12 / 60 * 155 = > $ 31 so , total cost will be $ 31 + $ 3 = > $ 34 hence answer will be c"
a ) $ 30 . , b ) $ 31 , c ) $ 32 . , d ) $ 33 . , e ) $ 34 .
c
add(add(multiply(12, 2), 3), multiply(divide(12, const_60), divide(35, 3)))
divide(n3,const_60)|divide(n5,n2)|multiply(n3,n4)|add(n0,#2)|multiply(#0,#1)|add(#3,#4)|
physics
which number should replace both the asterisks in ( * / 21 ) x ( * / 84 ) = 1 ?
"let ( y / 21 ) x ( y / 84 ) = 1 y ^ 2 = 21 x 84 = 21 x 21 x 4 y = ( 21 x 2 ) = 42 the answer is b ."
a ) 21 , b ) 42 , c ) 63 , d ) 72 , e ) 168
b
sqrt(multiply(84, 21))
multiply(n0,n1)|sqrt(#0)|
general
a train 140 m in length crosses a telegraph post in 16 seconds . the speed of the train is ?
"s = 140 / 16 * 18 / 5 = 31.5 kmph answer : b"
a ) 30.5 kmph , b ) 31.5 kmph , c ) 32.5 kmph , d ) 33.5 kmph , e ) 21.5 kmph
b
multiply(const_3_6, divide(140, 16))
divide(n0,n1)|multiply(#0,const_3_6)|
physics
how much water must be added to 52 litres of milk at 1 1 ⁄ 2 litres for 20 so as to have a mixture worth 10 2 ⁄ 3 a litre ?
c . p . of 1 litre of milk = ( 20 × 2 ⁄ 3 ) = 40 ⁄ 3 ∴ ratio of water and milk = 8 ⁄ 3 : 32 ⁄ 3 = 8 : 32 = 1 : 4 ∴ quantity of water to be added to 52 litres of milk = ( 1 ⁄ 4 × 52 ) litres = 13 litres . answer b
a ) 10 litres , b ) 13 litres , c ) 15 litres , d ) 18 litres , e ) none of these
b
divide(20, add(1, divide(1, 2)))
divide(n1,n3)|add(n1,#0)|divide(n4,#1)
general
a train 500 m long takes 10 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ?
"let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 500 / [ ( x + 5 ) * 5 / 18 ] = 10 10 ( x + 5 ) = 1800 x = 175 kmph answer is a"
a ) 175 kmph , b ) 150 kmph , c ) 162 kmph , d ) 145 kmph , e ) 100 kmph
a
subtract(divide(divide(500, 10), const_0_2778), 5)
divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2)|
physics
a card is drawn from a pack of 52 cards . the probability of getting a face card ?
"clearly in the 52 cards out of which there are 16 face cards . probability of getting a face card = 16 / 52 = 4 / 13 correct option is d"
a ) 2 / 15 , b ) 2 / 13 , c ) 1 / 15 , d ) 4 / 13 , e ) 5 / 7
d
divide(subtract(52, multiply(const_4, const_4)), 52)
multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)|
probability
in what time will a railway train 90 m long moving at the rate of 44 kmph pass a telegraph post on its way ?
"t = 90 / 44 * 18 / 5 = 7.4 sec answer : e"
a ) 3 , b ) 5 , c ) 4.3 , d ) 6.5 , e ) 7.4
e
divide(90, multiply(44, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 72 per kg ?
"( 96 - 72 ) / ( 72 - 68 ) = 24 / 4 = 6 / 1 answer : a"
a ) 6 : 1 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8
a
divide(divide(subtract(96, 72), subtract(96, 68)), subtract(const_1, divide(subtract(96, 72), subtract(96, 68))))
subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|
other
john was 43 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ?
"assume betty ' s age on marriage = x years . john ' s age on marriage = 43 john ' s age after 5 years = 48 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 48 ) = 42 therefore betty ' s current age = 42 option c"
a ) 24 , b ) 26 , c ) 42 , d ) 30 , e ) 32
c
multiply(divide(7, 8), add(43, const_4))
add(n0,const_4)|divide(n1,n2)|multiply(#0,#1)|
general
what is 120 % of 13 / 24 of 720 ?
"120 % * 13 / 24 * 360 = 1.2 * 13 * 30 = 468 the answer is c ."
a ) 372 , b ) 434 , c ) 468 , d ) 512 , e ) 564
c
divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 700,000 ?
"the first digit and last digit are the same so the only possibility is 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 1 * 10 * 10 = 100 the answer is b ."
a ) 60 , b ) 100 , c ) 180 , d ) 240 , e ) 300
b
multiply(multiply(subtract(6, const_4), const_10), const_10)
subtract(n2,const_4)|multiply(#0,const_10)|multiply(#1,const_10)|
general
the area of a side of a box is 120 sq cm . the area of the other side of the box is 72 sq cm . if the area of the upper surface of the box is 60 sq cm then find the volume of the box .
volume of the box = √ 120 × 72 × 60 = 720 cm 3 answer c
['a ) 259200 cm 3', 'b ) 86400 cm 3', 'c ) 720 cm 3', 'd ) can not be determined', 'e ) none of these']
c
volume_rectangular_prism(sqrt(divide(multiply(60, 72), 120)), divide(60, sqrt(divide(multiply(60, 72), 120))), multiply(divide(120, 60), sqrt(divide(multiply(60, 72), 120))))
divide(n0,n2)|multiply(n1,n2)|divide(#1,n0)|sqrt(#2)|divide(n2,#3)|multiply(#0,#3)|volume_rectangular_prism(#4,#5,#3)
geometry
mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 16 miles , how many miles was mike ' s ride ?
"the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 16 ) = $ 11.50 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 11.5 0.25 * x = 9 x = 36 miles the answer is b ."
a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48
b
divide(subtract(add(add(2.50, 5.00), multiply(0.25, 16)), 2.50), 0.25)
add(n0,n3)|multiply(n1,n5)|add(#0,#1)|subtract(#2,n0)|divide(#3,n1)|
general
if a man walks at a rate of 5 kmph , he misses a train by 7 minutes . however , if he walks at the rate of 6 kmph , he reaches the station 5 minutes before the arrival of the train . find the distance covered by him to reach the station .
let the required distance x km difference in the times taken at two speeds = 12 min = 1 / 5 hr ( x / 5 ) - ( x / 6 ) = 1 / 5 x = 6 the required distance is 6 km answer is d
a ) 8 , b ) 4 , c ) 15 , d ) 6 , e ) 12
d
multiply(add(add(multiply(divide(7, const_60), 5), multiply(divide(5, const_60), 6)), divide(7, const_60)), 5)
divide(n1,const_60)|divide(n0,const_60)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|add(#4,#0)|multiply(n0,#5)
physics
the milk level in a rectangular box measuring 62 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons )
"6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 62 * 25 * 1 / 2 = 775 feet ^ 3 of milk must be removed , which equals to 775 * 7.5 = 5812.5 gallons . answer : d ."
a ) 100 , b ) 250 , c ) 750 , d ) 5812.5 , e ) 5835.5
d
multiply(multiply(multiply(62, 25), divide(1, const_2)), 7.5)
divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)|
general
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 320 resolutions ?
"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 320 resolutions . = 320 * 2 * 22 / 7 * 22.4 = 45056 cm = 450.6 m answer : d"
a ) 724 m , b ) 704 m , c ) 287 m , d ) 450.6 m , e ) 927 m
d
divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 320), 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
what is 10 - 8 + 6 - 4 + . . . + ( - 12 ) ?
"the expression considers all even numbers between 10 and - 12 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , - 10 , and - 12 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) - ( - 12 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 = 42 - 30 = 12 hence the correct answer choice is c ."
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16
c
subtract(6, 8)
subtract(n2,n1)|
general
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 36 kilometers , maxwell ' s walking speed is 2 km / h , and brad ' s running speed is 4 km / h . what is the distance traveled by maxwell when they meet in the middle ?
"consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 2 xt = 36 - 4 xt ( i . e distance = speed x time ) = > 6 t = 36 hence t = 6 oa = 2 x 6 = 12 km answer : c"
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 9
c
multiply(2, divide(36, add(2, 4)))
add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|
physics
if 20 % of ( x - y ) = 15 % of ( x + y ) , then what percent of x is y ?
"20 % of ( x - y ) = 15 % of ( x + y ) 20 / 100 ( x - y ) = 15 / 100 ( x + y ) 5 x = 35 y required percentage = y / x * 100 = 5 y / 35 y * 100 = 14.28 % answer is d"
a ) 50.5 % , b ) 44.4 % , c ) 22.2 % , d ) 14.3 % , e ) 25 %
d
multiply(divide(subtract(20, 15), add(20, 15)), const_100)
add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|
general
if a tap could fill entire tank in 18 hrs due to leakage , then in how much time tank can be emptied by leakage if tap can fill entire tank in 12 hrs without leakage ?
if a tap could fill entire tank in 18 hrs due to leakage , tap can fill entire tank in 12 hrs without leakage . tank emptied in 0 ne hr by leakage = 1 / 12 - 1 / 18 = ( 6 - 4 ) / 72 = 2 / 72 = 1 / 36 . full tank can be emptied in 36 hrs . answer : b
a ) 35 hrs , b ) 36 hrs , c ) 37 hrs , d ) 38 hrs , e ) 39 hrs
b
inverse(subtract(divide(const_1, 12), divide(const_1, 18)))
divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)
general
what is the percentage increase in the area of a rectangle , if each of its sides is increased by 20 % ?
let original length = l original breadth = b then original area = lb length is increased by 20 % ⇒ new length = l × 120100 = 1.2 l breadth is increased by 20 % ⇒ new breadth = b × 120100 = 1.2 b new area = 1.2 l × 1.2 b = 1.44 lb increase in area = new area - original area = 1.44 lb − lb = 0.44 lb percentage increase in area = increase in areaoriginal area × 100 = 0.44 lblb × 100 = 44 % answer is d .
a ) 42 % , b ) 43 % , c ) 45 % , d ) 44 % , e ) 46 %
d
divide(subtract(multiply(add(const_100, 20), add(const_100, 20)), multiply(const_100, const_100)), const_100)
add(n0,const_100)|multiply(const_100,const_100)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,const_100)
geometry
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 252 ° of the circle , what percentage of megatek employees are in manufacturing ?
"answer : e 252 ° divided by 360 ° equals 0.7 , therefore the sector is equal to 70 % of the total"
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 70 %
e
multiply(divide(252, divide(const_3600, const_10)), const_100)
divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)|
physics
in a 100 m race , a beats b by 10 m and c by 13 m . in a race of 180 m , b will beat c by :
a : b = 100 : 90 . a : c = 100 : 87 . \ inline \ fn _ jvn { \ color { black } \ therefore \ frac { b } { c } = \ left ( \ frac { b } { a } \ times \ frac { a } { c } \ right ) = \ left ( \ frac { 90 } { 100 } \ times \ frac { 100 } { 87 } \ right ) = \ frac { 30 } { 29 } } when b runs 30 m , c runs 29 m . when b runs 180 m , c runs \ inline \ fn _ jvn { \ color { black } \ left ( \ frac { 29 } { 30 } \ times 180 \ right ) m } = 174 m \ inline \ fn _ jvn { \ color { blue } \ therefore } b beats c by ( 180 - 174 ) m = 6 m . answer : d ) 6 m
a ) 3 , b ) 7 , c ) 8 , d ) 6 , e ) 0
d
subtract(180, multiply(inverse(multiply(divide(100, subtract(100, 13)), divide(subtract(100, 10), 100))), 180))
subtract(n0,n2)|subtract(n0,n1)|divide(n0,#0)|divide(#1,n0)|multiply(#2,#3)|inverse(#4)|multiply(n3,#5)|subtract(n3,#6)
physics
the points a ( 0 , 0 ) , b ( 0 , 4 a - 5 ) and c ( 2 a + 3 , 2 a + 6 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ?
"1 / 2 bh = 1 / 2 ( 2 a + 3 ) ( 2 a + 6 ) now 4 a - 5 = 2 a + 6 2 a = 11 therefore , a ( 0,0 ) ; b ( 0,17 ) ; c ( 14,17 ) 1 / 2 * 14 * 17 = 119 answer : d"
a ) 114 , b ) 116 , c ) 118 , d ) 119 , e ) 120
d
divide(multiply(subtract(multiply(divide(add(6, 2), const_2.0), 4), 5), add(multiply(divide(add(6, 5), 5), subtract(4, 5)), 3)), 5)
add(n4,n8)|subtract(n3,n5)|divide(#0,n4)|multiply(#2,#1)|multiply(n3,#2)|add(#3,n6)|subtract(#4,n4)|multiply(#5,#6)|divide(#7,n4)|
geometry
in a rectangular coordinate system , what is the area of a triangle whose vertices have the coordinates ( 3 , 0 ) , ( 6 , 3 ) , and ( 6 , - 3 ) ?
the triangle is symmetric about the x - axis . the part above the x - axis forms a triangle with a base of 3 and a height of 3 . the area of this part is ( 1 / 2 ) ( 3 ) ( 3 ) . we can double this to find the area of the whole triangle . the total area is ( 2 ) ( 1 / 2 ) ( 3 ) ( 3 ) = 9 . the answer is a .
['a ) 9', 'b ) 10', 'c ) 11', 'd ) 12', 'e ) 13']
a
triangle_area(6, 3)
triangle_area(n0,n2)
geometry
jill works as a waitress at the local diner where she earns an hourly wage of $ 4.00 per hour and a standard tip rate of 15 % of the cost of the orders she serves . if she worked 3 8 - hour shifts this week and averaged $ 40 in orders per hour , how much did jill earn this week ?
jill earns 4 dollars / hour and the hourly tip is ( 3 / 20 ) * 40 . jill thus earns 4 * 8 + 8 * 2 * 3 per day ( or 4 ( 8 ) + 6 ( 8 ) = 10 ( 8 ) = 80 ) . jill has worked for 3 days - > 80 * 3 = 240 . this matches answer choice c .
a ) 280 , b ) 300 , c ) 240 , d ) 350 , e ) 400
c
add(multiply(divide(15, const_100), multiply(multiply(3, 8), 40)), multiply(multiply(3, 8), 4))
divide(n1,const_100)|multiply(n2,n3)|multiply(n4,#1)|multiply(n0,#1)|multiply(#0,#2)|add(#4,#3)
general
if a speaks the truth 85 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time
"explanation : probability that a speaks truth is 85 / 100 = 0.85 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 ) = 0.85 × 0.6 = 0.51 answer : e"
a ) 0.49 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.51
e
multiply(divide(85, 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
there are , in a certain league , 14 teams , and each team face another team for a total of 5 times . how many games are played in the season ?
"by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 455 option a ."
a ) 455 , b ) 525 , c ) 1400 , d ) 180 , e ) 560
a
multiply(multiply(subtract(14, const_1), 5), divide(14, const_2))
divide(n0,const_2)|subtract(n0,const_1)|multiply(#1,n1)|multiply(#0,#2)|
general
10 is added to a certain number , the sum is multiplied by 7 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 88 . what is the number ?
"let number is x . when 10 added to it , = ( x + 10 ) 7 multiplied to sum , = 7 * ( x + 10 ) now , = [ { 7 * ( x + 10 ) } / 5 ] and , = [ { 7 * ( x + 10 ) } / 5 ] - 5 according to question , [ { 7 * ( x + 10 ) } / 5 ] - 5 = half of 88 [ ( 7 x + 70 ) / 5 ) = 44 + 5 7 x + 70 = 49 * 5 x + 10 = 7 * 5 x + 10 = 35 x = 35 - 10 x = 25 so , required number is : 25 . answer : c"
a ) 21 , b ) 20 , c ) 25 , d ) 30 , e ) 45
c
subtract(divide(multiply(add(divide(88, const_2), 5), 5), 7), 10)
divide(n4,const_2)|add(n2,#0)|multiply(n2,#1)|divide(#2,n1)|subtract(#3,n0)|
general
on a certain transatlantic crossing , 25 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 60 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 ?
"0.25 p = rt + c 0.6 ( rt ) = no c = > 0.40 ( rt ) had c 0.25 p = 0.40 ( rt ) rt / p = 62.5 % answer - e"
a ) 33.3 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 62.5 %
e
divide(25, divide(subtract(const_100, 60), const_100))
subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|
gain
two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 40 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 10 miles ahead of the black car , how many hours will it take the black car to overtake the red car ?
"option c 10 + 40 t = 50 t t = 1"
a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.2 , e ) 2
c
divide(10, subtract(50, 40))
subtract(n1,n0)|divide(n2,#0)|
physics
the average of 25 results is 19 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ?
"solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 19 or , x = 103 . answer : option c"
a ) 74 , b ) 75 , c ) 103 , d ) 78 , e ) 45
c
subtract(subtract(multiply(25, 19), multiply(12, 17)), multiply(12, 14))
multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)|
general
in town p , 70 percent of the population are employed , and 21 percent of the population are employed males . what percent of the employed people in town p are females ?
the percent of the population who are employed females is 70 - 21 = 49 % the percent of employed people who are female is 49 % / 70 % = 70 % . the answer is e .
a ) 46 % , b ) 52 % , c ) 58 % , d ) 64 % , e ) 70 %
e
multiply(divide(subtract(70, 21), 70), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)
gain
mysoon collects glass ornaments . 10 more than 1 / 6 of the ornaments in her collection are handmade , and 1 / 2 of the handmade ornaments are antiques . if 1 / 3 of the ornaments in her collection are handmade antiques , how many ornaments are in her collection ?
the number of ornaments = a ten more than 1 / 6 of the ornaments in her collection are handmade = > handmade = 10 + a / 6 1 / 2 of the handmade ornaments are antiques = > handmade ornaments = 1 / 2 * ( 10 + a / 6 ) = 5 + a / 12 1 / 3 of the ornaments in her collection are handmade antiques = > handmade ornaments = a / 3 = > 5 + a / 12 = a / 3 = > a = 200 ans : a
a ) 20 , b ) 60 , c ) 108 , d ) 144 , e ) 180
a
divide(multiply(10, 3), subtract(2, divide(3, 6)))
divide(n6,n2)|multiply(n0,n6)|subtract(n4,#0)|divide(#1,#2)
general
a train speeds past a pole in 12 seconds and a platform 160 m long in 17 seconds . its length is ?
"let the length of the train be x meters and its speed be y m / sec . they , x / y = 12 = > y = x / 12 x + 160 / 17 = x / 12 x = 384 m . answer : a"
a ) 384 , b ) 360 , c ) 324 , d ) 390 , e ) 395
a
multiply(160, subtract(const_2, const_1))
subtract(const_2,const_1)|multiply(n1,#0)|
physics
a picnic attracts 240 persons . there are 40 more men than women , and 40 more adults than children . how many men are at this picnic ?
"adult + children = 240 let , children = y then , adult = y + 40 i . e . y + ( y + 40 ) = 240 i . e . y = 100 i . e . adult = 100 + 40 = 140 adults include only men and women i . e . men + women = 140 let women , w = x then men , m = x + 40 i . e . x + ( x + 40 ) = 2 x + 40 = 140 i . e . x = 54 i . e . men , m = 54 + 40 = 90 answer : option d"
a ) 240 , b ) 75 , c ) 110 , d ) 90 , e ) 200
d
add(divide(subtract(add(divide(subtract(240, 40), const_2), 40), 40), const_2), 40)
subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|subtract(#2,n1)|divide(#3,const_2)|add(n1,#4)|
general
two vessels contains equal number of mixtures milk and water in the ratio 4 : 1 and 7 : 3 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ?
the ratio of milk and water in the new vessel is = ( 4 / 5 + 7 / 10 ) : ( 1 / 5 + 3 / 10 ) = 3 / 2 : 1 / 2 = 3 : 1 answer is c
a ) 2 : 1 , b ) 4 : 1 , c ) 3 : 1 , d ) 2 : 3 , e ) 4 : 3
c
divide(add(multiply(4, divide(add(7, 3), add(4, 1))), 7), add(multiply(1, divide(add(7, 3), add(4, 1))), 3))
add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)
other
if x : y is 1 : 2 and y : z is 2 : 5 then x : z is equal to
"the two ratios given are having the same number 2 for y in both the ratios . hence - x : y = 1 : 2 y : z = 2 : 5 = > x : z = 1 : 5 answer c"
a ) 2 : 5 , b ) 3 : 5 , c ) 1 : 5 , d ) 4 : 5 , e ) 1 : 3
c
divide(1, 5)
divide(n0,n3)|
general
if | x - 16 | = 2 x , then x = ?
"| x - 16 | = 2 x . . . ( given ) x ^ 2 - 32 x + 256 = 4 x ^ 2 3 * x ^ 2 + 32 * x - 256 = 0 . . . . ( by solving above eq . we get ) , x = - 16 or 5.33 = = = > ans - c"
a ) 10 , b ) 15 , c ) - 16 , d ) 16 , e ) 5
c
subtract(add(2, 16), subtract(2, 16))
add(n0,n1)|subtract(n1,n0)|subtract(#0,#1)|
general
in a 200 meter race a beats b by 35 meters in 7 seconds . a ' s time over the cause is ?
b covers 35 m in 7 seconds b take time = ( 200 * 7 ) / 35 = 40 a takes time = 40 - 7 = 33 sec answer : d
a ) 47 sec , b ) 40 sec , c ) 45 sec , d ) 33 sec , e ) 35 sec
d
subtract(multiply(divide(7, 35), 200), 7)
divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)
physics
a life insurance company insured 25,000 young boys , 14,000 young girls and 16,000 young adults . the probability of death within 10 years of a young boy , young girl and a young adult are 0.02 , 0.03 and 0.15 respectively . one insured person dies . what is the probability that the dead person is a young boy ?
the number of young boys that will die = 0.02 x 25,000 = 500 the number of young girls that will die = 0.03 x 14,000 = 420 the number of young adults that will die = 0.15 x 16,000 = 2400 the required probability = the number of young boys who will die / the total number of people who will die . = 500 / ( 500 + 420 + 2400 ) = 25 / 166 answer : b
a ) 36 / 165 , b ) 25 / 166 , c ) 26 / 165 , d ) 32 / 165 , e ) 33 / 165
b
divide(multiply(0.02, power(const_5, const_2)), add(add(multiply(0.02, power(const_5, const_2)), multiply(0.03, multiply(add(const_3, const_4), const_2))), multiply(power(const_4, const_2), 0.15)))
add(const_3,const_4)|power(const_5,const_2)|power(const_4,const_2)|multiply(n4,#1)|multiply(#0,const_2)|multiply(n6,#2)|multiply(n5,#4)|add(#3,#6)|add(#7,#5)|divide(#3,#8)
probability
triangle a ’ s base is 10 % greater than the base of triangle b , and a ’ s height is 10 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ?
base of a = 11 / 10 * base of b height of a = 9 / 10 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 11 / 10 * 9 / 10 * area of b = 99 / 100 * area of b area of a is 1 % less than the area of b . answer ( b )
['a ) 9 % less', 'b ) 1 % less', 'c ) equal to each other', 'd ) 1 % more', 'e ) 9 % more']
b
divide(triangle_area(add(floor(const_100), divide(multiply(floor(const_100), 10), const_100)), subtract(floor(const_100), divide(multiply(floor(const_100), 10), const_100))), triangle_area(subtract(floor(const_100), divide(multiply(floor(const_100), 10), const_100)), add(floor(const_100), divide(multiply(floor(const_100), 10), const_100))))
floor(const_100)|multiply(n0,#0)|divide(#1,const_100)|add(#2,#0)|subtract(#0,#2)|triangle_area(#3,#4)|divide(#5,#5)
geometry
peter and tom shared the driving on a certain trip . if peter and tom both drove for the same amount of time , but peter only drove 3 / 5 of the total distance , what was the ratio of peter ' s average speed to tom ' s average speed ?
"the answer is suppose to be b . 3 : 2 . it ' s from the gmatprep answer : option b"
a ) 1 : 5 , b ) 3 : 2 , c ) 1 : 2 , d ) 3 : 5 , e ) 2 : 3
b
divide(subtract(const_1, divide(const_2.0, 5)), divide(3, 5))
divide(const_2.0,n1)|subtract(const_1,#0)|divide(#1,#0)|
general
jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 2.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill )
"average speed of jack = distance / time = 42 / ( 5 / 2 ) = 84 / 5 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 / 5 ) / 10 = 84 / 50 = 41 / 25 answer a"
a ) 41 / 25 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined
a
divide(divide(42, 2.5), divide(42, 4.2))
divide(n0,n1)|divide(n0,n2)|divide(#0,#1)|
physics
in a store , the total price for 20 shirts is $ 360 and the total price for 45 sweaters is $ 900 . by how much does the average ( arithmetic mean ) price of a sweater exceed that of a shirt in this store ?
"the average price of a shirt is : $ 360 / 20 = $ 18 . the average price of a sweater is : $ 900 / 45 = $ 20 . the difference in price is : $ 20 - $ 18 = $ 2 . the answer is b ."
a ) $ 1 , b ) $ 2 , c ) $ 3 , d ) $ 4 , e ) $ 5
b
subtract(divide(900, 45), divide(360, 20))
divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|
general
a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 10 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ?
"suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.10 = 132 increase in 1993 from 1991 = 132 - 100 = 32 answer : d"
a ) 17.5 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 38 %
d
subtract(multiply(add(const_100, 20), divide(add(const_100, 10), const_100)), const_100)
add(n1,const_100)|add(n4,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)|
gain
the average expenditure of a labourer for 8 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income is
"income of 8 months = ( 8 × 85 ) – debt = 680 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 950 average monthly income = 950 ÷ 10 = 95 answer d"
a ) 80 , b ) 85 , c ) 75 , d ) 95 , e ) 100
d
divide(add(add(multiply(85, 8), multiply(60, 4)), 30), add(8, 4))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)|
general
in a fuel station the service costs $ 2.05 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 65 liters and a truck ' s tank is 140 % bigger and they are all empty - ?
"service cost of 3 van and 2 truck = 2.05 * ( 3 + 2 ) = 10.5 fuel in 3 van = 3 * 65 = 195 litre fuel in 2 trucks = 2 * 65 ( 1 + 140 / 100 ) = 312 total fuel ( van + truck ) = 507 litre total fuel cost = 507 * 0.6 = 304.2 total cost = fuel + service = 304.2 + 10.25 = 314.45 answer is e"
a ) 122.6 $ , b ) 128.9 $ , c ) 243.7 $ , d ) 298.85 $ , e ) 314.45 $
e
add(multiply(multiply(add(65, divide(multiply(65, 140), const_100)), 2), 0.6), multiply(multiply(65, 3), 0.6))
multiply(n4,n5)|multiply(n2,n4)|divide(#0,const_100)|multiply(n1,#1)|add(n4,#2)|multiply(n3,#4)|multiply(n1,#5)|add(#6,#3)|
general
66 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be :
"let the length of the wire be h . radius = 1 / 2 mm = 1 / 20 cm . then , ( 22 / 7 ) x ( 1 / 20 ) x ( 1 / 20 ) x h = 66 . h = ( 66 x 20 x 20 x 7 ) / 22 = 8400 cm = 84 m . answer b"
a ) 66 , b ) 84 , c ) 80 , d ) 72 , e ) 68
b
divide(66, multiply(power(divide(1, const_2), const_2), const_pi))
divide(n1,const_2)|power(#0,const_2)|multiply(#1,const_pi)|divide(n0,#2)|
physics
if the cost price of 58 articles is equal to the selling price of 50 articles , then what is the percent profit ?
"let x be the cost price of one article . let y be the selling price of one article . 50 y = 58 x y = 1.16 x the answer is b ."
a ) 12 % , b ) 16 % , c ) 20 % , d ) 24 % , e ) 28 %
b
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 58)), divide(multiply(const_100, 50), 58)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
a person is traveling at 20 km / hr and reached his destiny in 8 hr then find the distance ?
"t = 8 hrs d = t * s = 20 * 8 = 160 km answer is c"
a ) 180 km , b ) 200 km , c ) 160 km , d ) 220 km , e ) 240 km
c
multiply(20, 8)
multiply(n0,n1)|
physics
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 300 ?
"to maximize number of hot dogs with 300 $ total number of hot dogs bought in 250 - pack = 22.95 * 13 = 298.35 $ amount remaining = 300 - 298.35 = 1.65 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 300 $ = 250 * 13 = 3250 answer e"
a ) 3,108 , b ) 3,100 , c ) 3,108 , d ) 3,124 , e ) 3,250
e
multiply(divide(300, 22.95), 250)
divide(n6,n5)|multiply(n4,#0)|
general
a train 400 m long can cross an electric pole in 20 sec and then find the speed of the train ?
"length = speed * time speed = l / t s = 400 / 20 s = 20 m / sec speed = 20 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 72 kmph answer : c"
a ) 87 kmph , b ) 28 kmph , c ) 72 kmph , d ) 18 kmph , e ) 29 kmph
c
divide(divide(400, const_1000), divide(20, const_3600))
divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|
physics
two diesel trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?
d relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 sec .
a ) 10 sec , b ) 30 sec , c ) 40 sec , d ) 20 s , e ) 50 sec
d
divide(add(120, 280), divide(multiply(add(42, 30), const_1000), const_3600))
add(n0,n1)|add(n2,n3)|multiply(#1,const_1000)|divide(#2,const_3600)|divide(#0,#3)
physics
two trains start simultaneously from opposite ends of a 180 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 180 - km trip in 5 hours . train y , travelling at a constant rate , completes the 180 - km trip in 4 hours . how many kilometers had train x traveled when it met train y ?
if the two trains cover a total distance d , then train x travels ( 4 / 9 ) * d while train y travels ( 5 / 9 ) * d . if the trains travel 180 km to the meeting point , then train x travels ( 4 / 9 ) * 180 = 80 km . the answer is e .
a ) 72 , b ) 74 , c ) 76 , d ) 78 , e ) 80
e
multiply(divide(180, 5), divide(180, add(divide(180, 5), divide(180, 4))))
divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(n0,#2)|multiply(#0,#3)|
physics
maria bought 10 notebooks and 5 pens costing 2 dollars each . how much did maria pay ?
2 × ( 10 + 5 ) = 2 × 10 + 2 × 5 = 20 + 10 = 30 dollars correct answer is c ) $ 30
a ) $ 10 , b ) $ 20 , c ) $ 30 , d ) $ 40 , e ) $ 50
c
multiply(add(10, 5), 2)
add(n0,n1)|multiply(n2,#0)
general
a jogger running at 9 km / hr along side a railway track is 280 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 280 + 120 = 400 m . time taken = 400 / 10 = 40 sec . answer : e"
a ) 67 sec , b ) 89 sec , c ) 36 sec , d ) 87 sec , e ) 40 sec
e
divide(add(280, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))
add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|
general
what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 55 ?
"let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 55 = rs . 3740 . answer : e"
a ) s . 3944 , b ) s . 3948 , c ) s . 3942 , d ) s . 3965 , e ) s . 3740
e
multiply(square_perimeter(sqrt(289)), 55)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 9 years older than b , the present age of b is
"let b ' s age = x years . then , as age = ( x + 9 ) years . ( x + 9 + 10 ) = 2 ( x — 10 ) hence x = 39 . present age of b = 39 years answer : c"
a ) 16 years , b ) 88 years , c ) 39 years , d ) 66 years , e ) 77 years
c
add(multiply(const_2, 10), add(9, 10))
add(n0,n3)|multiply(n0,const_2)|add(#0,#1)|
general
how many times hour hand covers full circle in a day ?
2 times answer : b
a ) 3 times , b ) 2 times , c ) 4 times , d ) 5 times , e ) once
b
divide(multiply(multiply(const_3, const_4), const_2), multiply(const_3, const_4))
multiply(const_3,const_4)|multiply(#0,const_2)|divide(#1,#0)
physics
if x ^ 2 + ( 1 / x ^ 2 ) = 3 , x ^ 4 + ( 1 / x ^ 4 ) = ?
"- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 3 ^ 2 - 2 = 7 . thus , the answer is a ."
a ) 7 , b ) 11 , c ) 12 , d ) 14 , e ) 15
a
subtract(power(2, 2), 2)
power(n0,n0)|subtract(#0,n0)|
general
how many integers between 400 and 1000 are there such that their unit digit is odd ?
there are 600 numbers from 401 to 1000 ( inclusive ) . half of the numbers are odd , so there are 300 odd numbers . the answer is c .
a ) 250 , b ) 150 , c ) 300 , d ) 400 , e ) 450
c
divide(subtract(1000, 400), const_2)
subtract(n1,n0)|divide(#0,const_2)
general
how many 4 - digit numbers contain no . 2 ?
answer : b
a ) 22 , b ) 98 , c ) 28 , d ) 29 , e ) 22
b
rectangle_area(4, 2)
rectangle_area(n0,n1)|
general
a company pays 15.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ?
"explanation : dividend on 1 share = ( 15.5 * 50 ) / 100 = rs . 7.75 rs . 25 is income on an investment of rs . 100 rs . 7.75 is income on an investment of rs . ( 7.75 * 100 ) / 25 = rs . 31 answer : b"
a ) 25 , b ) 31 , c ) 18 , d ) 19 , e ) 01
b
divide(multiply(divide(multiply(15.5, 50), const_100), const_100), 25)
multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)|
gain
the average of first 6 prime numbers is ?
"sum of 6 prime no . = 41 average = 41 / 6 = 6.83 answer : d"
a ) 10.11 , b ) 11.11 , c ) 12.11 , d ) 6.83 , e ) 14.11
d
add(6, const_1)
add(n0,const_1)|
general
jaya ranks 5 th in a class of 53 . what is her rank from the bottom in the class ?
there are 4 students above her . when takan from bottom there are 4 students below her rank . so her rank wud be 53 - 4 i . e . 49 th . answer : a
a ) 49 th , b ) 48 th , c ) 47 th , d ) 50 th , e ) none of these
a
add(subtract(53, 5), const_1)
subtract(n1,n0)|add(#0,const_1)
other
if the weight of 10 meters long rod is 23.4 kg . what is the weight of 6 meters long rod ?
"answer ∵ weight of 10 m long rod = 23.4 kg ∴ weight of 1 m long rod = 23.4 / 10 kg ∴ weight of 6 m long rod = 23.4 x 6 / 10 = 14.04 kg option : c"
a ) 7.2 kg . , b ) 10.8 kg . , c ) 14.04 kg . , d ) 18.0 kg , e ) none
c
divide(multiply(6, 23.4), 10)
multiply(n1,n2)|divide(#0,n0)|
physics
a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 30000 rs , what is the value of buzness ?
if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 30000 - > x = 60000 answer : d
a ) 75000 , b ) 160000 , c ) 170000 , d ) 60000 , e ) 70000
d
divide(30000, multiply(divide(2, 3), divide(3, 4)))
divide(n0,n1)|divide(n1,n3)|multiply(#0,#1)|divide(n4,#2)
general
the center of a circle lies on the origin of the coordinate plane . if a point ( x , y ) is randomly selected inside of the circle , what is the probability that y > x or x < 0 ?
"the line y = x divides the circle into two equal areas . all the points above the line y = x satisfy the condition that y > x . all the points to the left of the y - axis satisfy the condition that x < 0 . the union of these two areas is 5 / 8 of the circle . the answer is a ."
a ) 5 / 8 , b ) 4 / 5 , c ) 3 / 8 , d ) 1 / 2 , e ) 3 / 4
a
divide(const_3, multiply(const_2, const_4))
multiply(const_2,const_4)|divide(const_3,#0)|
physics
at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 15 . if the price per share of stock k was $ 16 at the closing of the day , what was the percent increase in the price per share of stock k for that day ?
"opening = 15 closing = 16 rise in price = 1 so , percent increase = 1 / 15 * 100 = 6.67 answer : d"
a ) 1.4 % , b ) 5.9 % , c ) 11.1 % , d ) 6.67 % , e ) 23.6 %
d
multiply(subtract(divide(16, 15), const_1), const_100)
divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)|
general
of the 70 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ?
"neither car nor garage = total - garage - ( swim - common ) = 70 - 50 - ( 40 - 35 ) = 70 - 55 = 15 answer b"
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30
b
subtract(70, add(add(subtract(50, 35), subtract(40, 35)), 35))
subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)|
other
john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 57000 and ingrid makes $ 72000 , what is their combined tax rate ?
"( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 72000 / 129000 = 24 / 43 . 24 / 43 is a little above 24 / 48 = 1 / 2 . thus , our answer is just a little above 35 . answer : d"
a ) 32 % , b ) 34.4 % , c ) 35 % , d ) 35.6 % , e ) 36.4 %
d
multiply(divide(add(multiply(divide(30, const_100), 57000), multiply(divide(40, const_100), 72000)), add(72000, 57000)), const_100)
add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|
gain
if a - b = 3 and a 2 + b 2 = 29 , then find the value of ab
explanation : 2 ab = ( a 2 + b 2 ) − ( a − b ) 2 = > 2 ab = 29 - 9 = 20 = > ab = 10 option c
a ) 7 , b ) 8.5 , c ) 10 , d ) 12 , e ) 14
c
divide(subtract(29, power(3, 2)), 2)
power(n0,n1)|subtract(n3,#0)|divide(#1,n1)
general
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 1.30 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ?
"total no . of boxes = 3060000 / ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 1.30 = $ 663 answer e"
a ) $ 255 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 663
e
multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 1.30)
multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)|
general
a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 50 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 4 hours ?
"the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.5 ( 6 ) = 3 gallons at the start , the amount of water is 0.5 ( 6 ) = 3 gallons after 4 hours , the amount of water is 3 - 0.5 ( 4 ) = 1 gallon the concentration of water is 1 / ( 3 + 1 ) = 1 / 4 = 25 % the answer is d ."
a ) 40 % , b ) 35 % , c ) 30 % , d ) 25 % , e ) 20 %
d
multiply(divide(subtract(divide(multiply(4, subtract(const_100, 50)), const_100), multiply(0.5, 4)), subtract(4, multiply(0.5, 4))), const_100)
multiply(n2,n3)|subtract(const_100,n0)|multiply(n3,#1)|subtract(n3,#0)|divide(#2,const_100)|subtract(#4,#0)|divide(#5,#3)|multiply(#6,const_100)|
gain
he ratio between the sale price and the cost price of an article is 4 : 7 . what is the ratio between the profit and the cost price of that article ?
"let c . p . = rs . 7 x and s . p . = rs . 4 x . then , gain = rs . 3 x required ratio = 3 x : 7 x = 3 : 7 . answer : b"
a ) 2 : 6 , b ) 3 : 7 , c ) 3 : 8 , d ) 2 : 1 , e ) 8 : 3
b
inverse(subtract(const_1, divide(const_3.0, 4)))
divide(const_3.0,n0)|subtract(const_1,#0)|inverse(#1)|
other
a man covers a certain distance on bike . had he moved 4 kmph faster , he would have taken 72 minutes less . if he moves slow by 4 kmph , he will take hours more to cover the distance . what is the distance travelled by him ?
explanation : let the distance be ' d ' and the speed be ' x ' kmph . d / x + 4 = d / x - 72 / 60 d / x - d / ( x - y ) = 6 / 5 = d ( x + 4 - x ) / x ( x - 4 ) = 6 / 5 ∴ d / x ( x + 4 ) = 3 / 10 . . . . . . ( i ) d / x - 4 - d / x = 2 ∴ dx - dx + 4 d / x ( x - 4 ) = 2 ∴ d / x ( x - 4 ) = 1 / 2 . . . . . ( ii ) from equation ( i ) and ( ii ) : 3 / 10 x ( x + 4 ) = 1 / 2 x ( x - 4 ) ∴ 3 ( x + 4 ) = 5 ( x - 4 ) = 3 x + 12 = 5 x - 20 2 x = 32 ∴ x = 16 kmn - 1 ∴ d = x ( x - 4 ) / 2 = 16 x 12 / 2 = 16 x 6 = 96 km answer : option d
a ) 48 km , b ) 56 km , c ) 64 km , d ) 96 km , e ) 106 km
d
divide(multiply(add(4, const_2), multiply(add(divide(add(const_60, 4), 4), 4), divide(add(const_60, 4), 4))), add(divide(add(const_60, 4), 4), 4))
add(n0,const_2)|add(n0,const_60)|divide(#1,n0)|add(n0,#2)|multiply(#3,#2)|multiply(#0,#4)|divide(#5,#3)
physics
if 0.6 : x : : 5 : 8 , then x is equal to :
"( x * 5 ) = ( 0.6 * 8 ) x = 4.8 / 5 x = 0.96 answer = a"
a ) a ) 0.96 , b ) b ) 1.2 , c ) c ) 1.25 , d ) d ) 1.3 , e ) e ) 0.8
a
divide(multiply(0.6, 8), 5)
multiply(n0,n2)|divide(#0,n1)|
general
excluding stoppages , the speed of a bus is 50 km / hr and including stoppages , it is 36 km / hr . for how many minutes does the bus stop per hour ?
"due to stoppages , it covers 14 km less . time taken to cover 14 km = 14 / 50 * 60 = 16 min . answer : d"
a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 15 min
d
multiply(const_60, divide(subtract(50, 36), 50))
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)|
physics