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a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 7 hours to fill the tank . the leak can drain all the water of the tank in ?
"work done by the tank in 1 hour = ( 1 / 2 - 2 1 / 7 ) = 1 / 30 leak will empty the tank in 30 hrs . answer : e"
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 24 hr , e ) 30 hr
e
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 7), 1), 7))))
divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)|
physics
a ranch has both horses and ponies . exactly 5 / 7 of the ponies have horseshoes , and exactly 2 / 3 of the ponies with horseshoes are from iceland . if there are 3 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ?
"5 / 7 * p are ponies with horseshoes , so p is a multiple of 7 . 2 / 3 * 5 / 7 * p = 10 / 21 * p are icelandic ponies with horseshoes , so p is a multiple of 21 . the minimum value of p is 21 . then h = p + 3 = 24 . the minimum number of horses and ponies is 45 . the answer is d ."
a ) 33 , b ) 37 , c ) 41 , d ) 45 , e ) 49
d
add(add(lcm(7, multiply(divide(7, 2), 3)), 3), lcm(7, multiply(divide(7, 2), 3)))
divide(n1,n2)|multiply(n3,#0)|lcm(n1,#1)|add(n4,#2)|add(#3,#2)|
general
there are 13 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl or 2 boys are selected ?
"n ( s ) = sample space = 23 c 3 = 1771 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 13 or 1 girl from 10 = 13 c 2 + 10 c 1 = 88 ans - a"
a ) 88 , b ) 90 , c ) 92 , d ) 96 , e ) 98
a
multiply(choose(13, 2), choose(10, 1))
choose(n0,n3)|choose(n1,n2)|multiply(#0,#1)|
probability
a student gets 70 % in one subject , 80 % in the other . to get an overall of 75 % how much should get in third subject .
"a student gets 70 % in one subject , 80 % in the other . average of 2 subjects is 75 % , if they have equal weightage . to get an overall of 75 % , he should get 75 % in third subject to maintain same % age . . if they have equal weightage . answer : a"
a ) 75 % , b ) 25 % , c ) 45 % , d ) 55 % , e ) 65 %
a
subtract(multiply(75, const_3), add(70, 80))
add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)|
gain
carina has 85 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ?
"lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 85 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 85 - > y = 75 / 15 . = 5 a"
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
a
divide(subtract(85, multiply(5, 2)), add(10, 5))
add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0)|
general
how much water must be added to 56 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 Γ— 2 ⁄ 3 ) = 40 ⁄ 3 ∴ ratio of water and milk = 8 ⁄ 3 : 32 ⁄ 3 = 8 : 32 = 1 : 4 ∴ quantity of water to be added to 56 litres of milk = ( 1 ⁄ 4 Γ— 56 ) litres = 14 litres . answer c"
a ) 10 litres , b ) 12 litres , c ) 14 litres , d ) 18 litres , e ) none of these
c
divide(20, add(1, divide(1, 2)))
divide(n1,n3)|add(n1,#0)|divide(n4,#1)|
general
if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2000 less than the principal . what was the principal ?
"p - 2000 = ( p * 5 * 4 ) / 100 p = 2500 answer : b"
a ) 1500 , b ) 2500 , c ) 2507 , d ) 3200 , e ) 11500
b
divide(2000, subtract(const_1, divide(multiply(4, 5), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
gain
if two integers x , y ( x > y ) are selected from - 10 to 10 ( inclusive ) , how many possible cases are there ?
"if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 21 c 2 = 21 * 20 / 2 = 210 . therefore , the answer is d"
a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 240
d
add(add(add(add(add(add(add(10, 10), add(10, const_2)), add(10, const_1)), 10), 10), const_2), const_1)
add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)|
probability
jolene entered an 18 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 12 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ?
"if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 926 . answer : e ."
a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24
e
add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))))
divide(n1,const_100)|divide(n3,const_100)|divide(n5,const_100)|power(const_100,n1)|multiply(#0,#3)|add(#4,#3)|multiply(#5,#1)|add(#5,#6)|multiply(#7,#2)|add(#8,#6)|
gain
the sum of the numbers is 110 . if the first number be twice the second and third number be one - third of the first , then the second number is :
"let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 110 11 x / 3 = 110 x = 30 answer : a"
a ) 30 , b ) 54 , c ) 72 , d ) 84 , e ) 27
a
divide(multiply(110, const_3), add(const_10, const_1))
add(const_1,const_10)|multiply(n0,const_3)|divide(#1,#0)|
general
total dinning bill of 3 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared .
"dinner bill of 3 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) / 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 / 3 = $ 50.96 answer : b"
a ) 21.84 , b ) 50.96 , c ) 53.84 , d ) 24.84 , e ) 50.26
b
divide(multiply(139.00, add(divide(const_1, 10), const_1)), 3)
divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)|
general
the ratio of the radius of two circles is 5 : 3 , and then the ratio of their areas is ?
"r 1 : r 2 = 5 : 3 Ο€ r 12 : Ο€ r 22 r 12 : r 22 = 25 : 9 answer : c"
a ) 1 : 8 , b ) 1 : 6 , c ) 25 : 9 , d ) 1 : 3 , e ) 1 : 2
c
divide(circle_area(5), circle_area(3))
circle_area(n0)|circle_area(n1)|divide(#0,#1)|
geometry
the population of locusts in a certain swarm doubles every two hours . if 4 hours ago there were 1,000 locusts in the swarm , in approximately how many hours will the swarm population exceed 512,000 locusts ?
"- 4 hours : 1,000 - 2 hours : 2,000 now : 4,000 + 2 hours : 8,000 + 4 hours : 16,000 + 6 hours : 32,000 + 8 hours : 64,000 + 10 hours : 128,000 + 12 hours : 256,000 + 14 hours : 512,000 answer : d"
a ) 18 , b ) 15 , c ) 13 , d ) 14 , e ) 12
d
subtract(multiply(log(divide(power(4, 4), const_2)), const_2), 4)
power(n0,n0)|divide(#0,const_2)|log(#1)|multiply(#2,const_2)|subtract(#3,n0)|
general
if the price of gasoline increases by 20 % and a driver intends to spend only 14 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ?
let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.2 * p ) y = 1.14 ( p * x ) y = ( 1.14 / 1.2 ) x = 0.95 x which is a reduction of 5 % . the answer is d .
a ) 1 % , b ) 2 % , c ) 4 % , d ) 5 % , e ) 6 %
d
multiply(divide(subtract(add(20, const_100), add(14, const_100)), add(20, const_100)), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)
general
a cube is divided into 343 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ?
"each side of the cube has 7 x 7 = 49 cubelets . only the interior cubelets are painted one colour . on each side , 5 x 5 = 25 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 25 = 150 the answer is b ."
a ) 125 , b ) 150 , c ) 175 , d ) 200 , e ) 225
b
divide(subtract(343, multiply(multiply(const_4, const_2), const_3)), const_2)
multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)|
geometry
the average of 7 consecutive numbers is 20 . the largest of these numbers is :
"explanation : let the numbers be x , x + 1 , x + 2 , x + 3 , x + 4 , x + 5 and x + 6 , then ( x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) + ( x + 6 ) ) / 7 = 20 . or 7 x + 21 = 140 or 7 x = 119 or x = 17 . latest number = x + 6 = 23 . answer : c"
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25
c
add(divide(subtract(multiply(20, 7), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), const_1))), 7), add(add(add(add(const_1, const_2), const_1), const_1), const_1))
add(const_1,const_2)|multiply(n0,n1)|add(#0,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,const_1)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#8,#9)|subtract(#1,#10)|divide(#11,n0)|add(#4,#12)|
general
a number x is multiplied with itself and then added to the product of 4 and x . if the result of these two operations is 5 , what is the value of x ?
"a number x is multiplied with itself - - > x ^ 2 added to the product of 4 and x - - > x ^ 2 + 4 x if the result of these two operations is - 4 - - > x ^ 2 + 4 x = 5 i . e x ^ 2 + 4 x - 5 = 0 is the quadratic equation which needs to be solved . ( x - 1 ) ( x + 5 ) = 0 hence x = 1 . c = - 5 imo d"
a ) - 4 , b ) - 2 , c ) 2 , d ) 1 and - 5 , e ) can not be determined .
d
divide(power(5, const_2), 4)
power(n1,const_2)|divide(#0,n0)|
general
12 spheres of the same size are made from melting a solid cylinder of 8 cm diameter and 1 cm height . what is the diameter of each sphere ?
"volume of cylinder = pi * r ^ 2 * h volume of a sphere = 4 * pi * r ^ 3 / 3 12 * 4 * pi * r ^ 3 / 3 = pi * r ^ 2 * h r ^ 3 = r ^ 2 * h / 16 = 1 cm ^ 3 r = 1 cm d = 2 cm the answer is a ."
a ) 2 cm , b ) 4 cm , c ) 8 cm , d ) 10 cm , e ) 12 cm
a
multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(8, const_2), 1), const_pi), const_3), const_4), 12), const_4), const_4), const_2)
divide(n1,const_2)|volume_cylinder(#0,n2)|divide(#1,const_pi)|multiply(#2,const_3)|divide(#3,const_4)|divide(#4,n0)|divide(#5,const_4)|divide(#6,const_4)|multiply(#7,const_2)|
physics
9 neighbors are to split the cost of topsoil . the cost of the 5 truck loads of dirt was $ 690.47 . what is the least amount of money ( in whole number of dollars ) that they must add to bill if they wants to split this money evenly among his 9 neighbors ?
if the bill was $ 690.47 dollars , how much money should be removed with 1 cent as the smallest unit ? this is equivalent to finding the first number that is divisible by 9 that occurs after 690.47 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 6 + 9 + 0 + 4 + 7 = 26 . if you add 1 , the number is divisible by 9 ( 26 + 1 ) . 27 is divisible by 9 . hence , we need to add 1 cent from this number for it to be divisible by 9 . correct option : a
a ) 690.48 , b ) 690.49 , c ) 690.5 , d ) 690.51 , e ) 690.52
a
min(690.47, 690.47)
min(n2,n2)
general
evaluate : 11 + sqrt ( - 4 + 6 Γ— 4 Γ· 3 )
"according to order of operations , inner brackets first where 6 Γ— 4 Γ· 3 is first calculated since it has a multiplication and a division . 6 Γ— 4 Γ· 3 = 24 Γ· 3 = 8 hence 11 + sqrt ( - 4 + 6 Γ— 4 Γ· 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer e ) 13"
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
e
add(11, sqrt(subtract(divide(multiply(6, 4), 3), 4)))
multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3)|
general
16 people can write 52 book in 12 days working 8 hour a day . then in how many day 206 can be written by 64 people ?
"work per day epr hour per person = 52 / ( 12 * 8 * 16 ) / / eq - 1 people = 64 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 206 / ( p * 8 * 64 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 309 / 26 answer : a"
a ) 309 / 26 , b ) 309 / 28 , c ) 309 / 22 , d ) 319 / 26 , e ) 339 / 26
a
divide(multiply(52, 12), divide(64, const_2))
divide(n5,const_2)|multiply(n1,n2)|divide(#1,#0)|
physics
total dinning bill for 7 people was $ 211.00 . if they add 15 % tip and divided the bill evenly , approximate . what was each persons find share
"211 * 15 = 3165 / 100 = 31.65 211 + 31.65 = 242.65 242.65 / 7 = 34.66 answer : c"
a ) $ 30.14 , b ) 45.14 , c ) 34.66 , d ) 32.29 , e ) 33.16
c
divide(add(211.00, divide(multiply(15, 211.00), const_100)), 7)
multiply(n1,n2)|divide(#0,const_100)|add(n1,#1)|divide(#2,n0)|
general
what is the area of square field whose side of length 20 m ?
"20 * 20 = 400 sq m answer : e"
a ) 200 sq m , b ) 186 sq m , c ) 586 sq m , d ) 287 sq m , e ) 400 sq m
e
square_area(20)
square_area(n0)|
geometry
a cyclist bikes x distance at 8 miles per hour and returns over the same path at 812 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ?
"distance = d 1 = x miles speed = s 1 = 8 miles per hour time = t 1 = distance / speed = x / 8 2 . going from b to a distance = d 2 = x miles speed = s 2 = 12 miles per hour time = t 2 = distance / speed = x / 12 3 . average speed = total distance / total time total distance = x + x = 2 x total time = x / 12 + x / 8 = x ( 1 / 12 + 1 / 8 ) = = 5 x / 24 speed = 2 x / ( 5 x / 24 ) = 48 / 5 = 9.6 answer : d"
a ) 8.1 , b ) 8.3 , c ) 8.6 , d ) 9.6 , e ) 9.0
d
divide(add(8, 812), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
in a survey of 345 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ?
"- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 291 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 345 we have to find not part time and not uninsured . in other words not part time and insured = x / 345 = ( 291 - 104 + 13 ) / 345 = 40 / 69 answer is e ."
a ) 7 / 12 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 40 / 69
e
divide(subtract(345, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 345)
multiply(n1,n3)|multiply(const_1,const_100)|divide(#0,#1)|subtract(n2,#2)|add(n1,#3)|subtract(n0,#4)|divide(#5,n0)|
physics
two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team w . if team a ' s average speed was 5 mph greater than team w ' s , what was team w ' s average mph ?
"this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team w ' s speed was s . then team w ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team w . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team w is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct ."
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25
d
divide(divide(300, 5), 3)
divide(n0,n2)|divide(#0,n1)|
physics
if a ( a - 5 ) = 6 and b ( b - 5 ) = 6 where a β‰  b , then a + b =
"i . e . if a = - 1 then b = 6 or if a = 6 then b = - 1 but in each case a + b = - 1 + 6 = 5 answer : option e"
a ) βˆ’ 48 , b ) βˆ’ 2 , c ) 2 , d ) 9 , e ) 5
e
subtract(subtract(subtract(subtract(add(add(5, 6), subtract(5, 6)), const_1), const_1), const_1), const_1)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)|
general
how many positive integers less than 100 have a remainder of 6 when divided by 13 ?
"we have to include 6 also . as 13 * 0 + 6 = 6 if somebody says to divide 6 by 13 , we will be telling we have 0 quotient and remainder as 6 . answer is c"
a ) 2 , b ) 7 , c ) 6 , d ) 9 , e ) 10
c
divide(factorial(subtract(add(const_4, 6), const_1)), multiply(factorial(6), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
general
given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] – 1 = f ( 3 x – 6 )
"explanations we have the function f ( x ) = 3 x – 5 , and we want to some sophisticated algebra with it . let ’ s look at the two sides of the prompt equation separately . the left side says : 2 * [ f ( x ) ] – 1 β€” - this is saying : take f ( x ) , which is equal to its equation , and multiply that by 2 and then subtract 1 . 2 * [ f ( x ) ] – 1 = 2 * ( 3 x – 5 ) – 1 = 6 x – 10 – 1 = 6 x – 11 the right side says f ( 3 x – 6 ) β€” this means , take the algebraic expression ( 3 x – 6 ) and plug it into the function , as discussed above in the section β€œ how a mathematician things about a function . ” this algebraic expression , ( 3 x – 6 ) , must take the place of x on both sides of the function equation . f ( 3 x – 6 ) = 3 * [ 3 x – 6 ] – 5 = 9 x – 18 – 5 = 9 x – 23 now , set those two equal and solve for x : 9 x – 23 = 6 x – 11 9 x = 6 x – 11 + 23 9 x = 6 x + 12 9 x – 6 x = 12 3 x = 12 x = 4 answer = b"
a ) 0 , b ) 4 , c ) 6 , d ) 7 , e ) 13
b
divide(subtract(add(multiply(2, 5), 1), add(multiply(3, 3), 5)), subtract(multiply(2, 3), multiply(3, const_1)))
multiply(n1,n2)|multiply(n4,n0)|multiply(n0,n2)|multiply(n0,const_1)|add(n3,#0)|add(n1,#1)|subtract(#2,#3)|subtract(#4,#5)|divide(#7,#6)|
general
in n is a positive integer less than 200 , and 18 n / 60 is an integer , then n has how many different positive prime factors ?
"( a ) . 18 n / 60 must be an integer . = > 3 n / 10 must be an integer . hence n must be a multiple of 2 * 5 . = > n has 2 different prime integers ."
a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8
a
add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2)
add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)|
general
the ratio of numbers is 3 : 4 and their h . c . f is 8 . their l . c . m is :
let the numbers be 3 x and 4 x . then their h . c . f = x . so , x = 8 . so , the numbers are 24 and 32 . l . c . m of 24 and 32 = 96 . answer : b
a ) 12 , b ) 96 , c ) 24 , d ) 48 , e ) 98
b
lcm(multiply(3, 8), multiply(4, 8))
multiply(n0,n2)|multiply(n1,n2)|lcm(#0,#1)
other
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 59000 ?
"59000 * 9 / 8 * 9 / 8 = 74671.875 answer : c"
a ) 81000 , b ) 81007 , c ) 74671.875 , d ) 81066 , e ) 81022
c
add(add(59000, multiply(divide(1, 8), 59000)), multiply(divide(1, 8), add(59000, multiply(divide(1, 8), 59000))))
divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)|
general
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 30 % ?
"cost price = rs . 500 profit = 30 % of 500 = rs . 150 selling price = cost price + profit = 500 + 150 = 650 answer : a"
a ) 650 , b ) 882 , c ) 772 , d ) 652 , e ) 271
a
add(500, multiply(500, divide(30, const_100)))
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
if the sum and difference of two numbers are 20 and 8 respectively , then the difference of their square is :
"let the numbers be x and y . then , x + y = 20 and x - y = 8 x 2 - y 2 = ( x + y ) ( x - y ) = 20 * 8 = 160 . answer : c"
a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 18
c
subtract(power(divide(add(20, 8), const_2), const_2), power(subtract(20, divide(add(20, 8), const_2)), const_2))
add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4)|
general
raja complete a journey in 10 hours . he travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km .
consider x - - > ( 1 / 2 ) x / 21 + ( 1 / 2 ) x / 24 = 10 = = > x / 21 + x / 24 = 20 15 x = 168 * 20 = = > 224 km answer c
a ) 200 km , b ) 250 km , c ) 224 km , d ) 255 km , e ) 260 km
c
add(multiply(divide(10, const_2), 24), multiply(divide(10, const_2), 21))
divide(n0,const_2)|multiply(n2,#0)|multiply(n1,#0)|add(#1,#2)
physics
the percentage increase in the area of a rectangle , if each of its sides is increased by 20 % is :
"let original length = x metres and original breadth = y metres . original area = ( xy ) m 2 . new length = 120 x m = 6 x m . 100 5 new breadth = 120 y m = 6 y m . 100 5 new area = 6 x x 6 y m 2 = 36 xy m 2 . 5 5 25 the difference between the original area = xy and new - area 36 / 25 xy is = ( 36 / 25 ) xy - xy = xy ( 36 / 25 - 1 ) = xy ( 11 / 25 ) or ( 11 / 25 ) xy increase % = 11 xy x 1 x 100 % = 44 % . 25 xy a"
a ) 44 % , b ) 46 % , c ) 48 % , d ) 50 % , e ) 51 %
a
multiply(subtract(multiply(divide(add(const_100, 20), const_100), divide(add(const_100, 20), const_100)), const_1), const_100)
add(n0,const_100)|divide(#0,const_100)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|
geometry
a person travels equal distances with speeds of 2 km / hr , 4 km / hr , 6 km / hr . and takes a total time of 11 minutes . find the total distance ?
"let the each distance be x km total distance = 3 x then total time , ( x / 2 ) + ( x / 4 ) + ( x / 6 ) = 11 / 60 x = 0.2 total distance = 3 * 0.2 = 0.6 km = 600 meters correct option is c"
a ) 1 km , b ) 500 mts , c ) 600 mts , d ) 2 km , e ) 250 mts
c
multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 2), divide(const_1, 4)), divide(const_1, 6))), const_3), const_1000)
divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)|
physics
the ratio of the radius of two circles is 1 : 9 , and then the ratio of their areas is ?
"r 1 : r 2 = 1 : 9 Ο€ r 12 : Ο€ r 22 r 12 : r 22 = 1 : 81 answer : e"
a ) 1 : 8 , b ) 1 : 6 , c ) 1 : 9 , d ) 1 : 3 , e ) 1 : 81
e
divide(circle_area(1), circle_area(9))
circle_area(n0)|circle_area(n1)|divide(#0,#1)|
geometry
the average salary of all the workers in a workshop is rs . 9000 . the average salary of 6 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ?
"let the total number of workers be x . then , 9000 x = ( 12000 * 6 ) + 6000 ( x - 6 ) = > 3000 x = 36000 = x = 12 . answer : e"
a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 12
e
add(6, divide(multiply(6, subtract(12000, 9000)), subtract(9000, 6000)))
subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)|
general
the first flight out of phoenix airport had a late departure . if the next 3 flights departed on - time , how many subsequent flights need to depart from phoenix on - time , for the airport ' s on - time departure rate to be higher than 60 % ? i will see what is the quickest way to solve it then i will provide the explanation
the following approach might be the easiest one and less error prone . we need on - time departure rate to be higher than 6 / 10 , so it should be at least 7 / 11 , which means that 7 out of 11 flights must depart on time . since for now 3 out of 4 flights departed on time then 7 - 3 = 4 subsequent flights need to depart on - time . answer : d
a ) 6 , b ) 7 , c ) 9 , d ) 4 , e ) 11
d
subtract(add(multiply(divide(60, const_100), const_10), const_1), 3)
divide(n1,const_100)|multiply(#0,const_10)|add(#1,const_1)|subtract(#2,n0)
gain
6 ) a marketing firm determined that , of 300 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ?
"220 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 220 = > 4 x = 160 = > x = 40 answer - d"
a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 40 , e ) e ) 45
d
divide(subtract(subtract(300, 80), 60), const_4)
subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4)|
other
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 8 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?
"tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 15 / 24 ) = 8 / 15 hours peter worked for 23 / 15 hours so he painted 23 / 15 * 1 / 3 = 23 / 45 of the room . the answer is d ."
a ) 7 / 15 , b ) 11 / 25 , c ) 17 / 35 , d ) 23 / 45 , e ) 27 / 55
d
divide(const_4, add(multiply(const_4, 8), const_1))
multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)|
physics
a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 50,000 and at the same time n is a perfect square and a perfect cube ?
if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 , 5 ^ 6 = 15,625 , 6 ^ = 46,656 . the answer is d .
['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7']
d
divide(add(const_4, const_2), const_1)
add(const_2,const_4)|divide(#0,const_1)
geometry
at joel ’ s bookstore , the current inventory is 40 % historical fiction . of the historical fiction books , 40 % are new releases , while 40 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ?
"let there be 100 books in all historic fiction books = 40 % of total = 40 other books = 60 new historic fiction = 40 % of 40 = 16 other new books = 40 % of 60 = 24 total new books = 24 + 16 = 40 fraction = 16 / 40 = 2 / 5 ans : e"
a ) 4 / 25 , b ) 8 / 23 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 5
e
divide(divide(multiply(40, 40), const_100), add(divide(multiply(40, 40), const_100), divide(multiply(40, subtract(const_100, 40)), const_100)))
multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)|
gain
the price of lunch for 15 people was $ 209.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ?
"take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 209 is for 115 then we have to calculate for 100 for 115 209 for 100 x so by cross multiplication we get 115 x = 100 * 209 = > x = 100 * 209 / 110 by simplifying we get x as 190 which is the price of lunch before gratuity so the gratuity is 19 so as the question ask the average price person excluding gratuity is 190 / 15 = 12.66 so our answer is b )"
a ) $ 11.73 , b ) $ 12.66 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87
b
multiply(multiply(divide(209.00, add(const_100, 15)), const_100), divide(const_1, 15))
add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3)|
general
the average of 50 numbers id 44 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is :
"explanation : total of 50 numbers = ( 50 Γ— 44 ) = 2200 total of 48 numbers = ( 2200 - ( 45 + 55 ) ] = 2100 required average = 2100 / 48 = 43.75 answer : e"
a ) 22.35 , b ) 33.25 , c ) 22.25 , d ) 11.35 , e ) 43.75
e
divide(subtract(multiply(50, 44), add(45, 55)), subtract(50, const_2))
add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2)|
general
mike earns $ 14 per hour and phil earns $ 12 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ?
"what % less of 14 is 12 let it be x % less , then = 14 ( 1 - x / 100 ) = 12 1 - x / 100 = 12 / 14 x = 100 / 7 x = 14 % ans c"
a ) 25 % , b ) 32.5 % , c ) 14 % , d ) 37.5 % , e ) 40 %
c
multiply(divide(12, 14), const_100)
divide(n1,n0)|multiply(#0,const_100)|
general
what is the remainder when 45 * 49 is divided by 8 ?
"we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 45 * 49 ) / 8 } } = remainder of ( 45 / 8 ) * remainder of ( 49 / 8 ) = 5 * 1 = 7 answer : d"
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 7
d
subtract(divide(8, const_2), multiply(45, 45))
divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)|
general
6 persons in an organization including a and b were to be divided in two groups of 3 members each . the total number of groups containing both a and b is what fraction of the total number of groups which can be formed ?
the fraction is nothing but the probability . . number to choose 3 out of 6 = 6 c 3 number to choose a and b and 2 from remaining 4 = 4 c 2 . . prob of a and b choosen = 4 c 2 / 6 c 3 = 3 / 10 answer : a
a ) 3 / 10 , b ) 1 / 70 , c ) 3 / 14 , d ) 1 / 10 , e ) 11 / 14
a
divide(power(6, const_2), multiply(multiply(6, subtract(6, const_1)), subtract(subtract(6, const_1), const_1)))
power(n0,const_2)|subtract(n0,const_1)|multiply(n0,#1)|subtract(#1,const_1)|multiply(#2,#3)|divide(#0,#4)
general
solution x is 80 % chemical a and 20 % chemical b by volume . solution y is 50 % chemical a and 50 % chemical b by volume . if a mixture of x and y is 68 % chemical a , what percent of the mixture is solution x ?
"the volume of the mixture be x + y . 0.8 x + 0.5 y = 0.68 ( x + y ) 0.12 x = 0.18 y x = 3 y / 2 x / ( x + y ) = ( 3 y / 2 ) / ( 5 y / 2 ) = 3 / 5 = 40 % . the answer is e ."
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 %
e
multiply(divide(divide(subtract(50, 68), subtract(68, 80)), add(divide(subtract(50, 68), subtract(68, 80)), const_1)), const_100)
subtract(n2,n4)|subtract(n4,n0)|divide(#0,#1)|add(#2,const_1)|divide(#2,#3)|multiply(#4,const_100)|
gain
what number has a 5 : 1 ratio to the number 11 ?
"5 : 1 = x : 10 x = 55 answer : c"
a ) 22 , b ) 50 , c ) 55 , d ) 52 , e ) 12
c
multiply(11, 5)
multiply(n0,n2)|
other
in a class , if 50 % of the boys were girls , then there would be 50 % more girls than boys . what percentage of the overall class is girls ?
detailed solution for questions of this type , it is best to go from the final step . in the final state , the number of girls should be 1.5 * the number of boys . when 50 % of the boys are taken as girls , let the number of boys = x number of girls = 1.5 x total number of students = 2.5 x original number of boys = 2 x ( 50 % of boys = x ) original number of girls = 0.5 x girls form 20 % of the overall class . correct answer d .
a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 20 % , e ) 10 %
d
multiply(divide(subtract(multiply(add(const_1, divide(50, const_100)), divide(50, const_100)), divide(50, const_100)), add(subtract(multiply(add(const_1, divide(50, const_100)), divide(50, const_100)), divide(50, const_100)), const_1)), const_100)
divide(n0,const_100)|add(#0,const_1)|multiply(#1,#0)|subtract(#2,#0)|add(#3,const_1)|divide(#3,#4)|multiply(#5,const_100)
gain
a man can row downstream at the rate of 30 kmph and upstream at 10 kmph . find the man ’ s rate in still water and rate of current ?
"rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 30 + 10 ) = 20 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 30 - 10 ) = 1 / 2 ( 20 ) = 10 kmph answer is a ."
a ) 10 , b ) 8.0 , c ) 9.5 , d ) 9.0 , e ) 8.25
a
divide(subtract(30, 10), const_2)
subtract(n0,n1)|divide(#0,const_2)|
gain
the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 192,000 . what was the percent increase in the cost of the home ?
increase = 192000 - 120000 = 72000 % increase = 72000 * 100 / 120000 = 60 % answer : option a
a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 %
a
multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100)
multiply(const_4,const_4)|multiply(const_1000,const_12)|multiply(#0,const_12)|multiply(#1,const_10)|multiply(#2,const_1000)|subtract(#4,#3)|divide(#5,#3)|multiply(#6,const_100)
general
a dishonest dealer professes to sell goods at the cost price but uses a weight of 900 grams per kg , what is his percent ?
"900 - - - 100 100 - - - ? = > 11.11 % answer : a"
a ) 11.11 % , b ) 25 % , c ) 22 % , d ) 29 % , e ) 45 %
a
subtract(multiply(divide(const_100, 900), 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
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 5 ounces of corn syrup , how many ounces of water does it contain ?
"f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 5 * 180 / 12 = 75 ounces of water e"
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 75
e
multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 5)
divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)|
other
what is the value of x ^ 2 yz βˆ’ xyz ^ 2 , if x = - 1 , y = 1 , and z = 3 ?
"1 * 1 * 3 - ( - 1 * 1 * 9 ) = 3 + 9 = 12 ans : d"
a ) 20 , b ) 24 , c ) 30 , d ) 12 , e ) 48
d
multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1))
negate(n0)|multiply(n3,#0)|subtract(#0,n3)|multiply(n3,#1)|multiply(#3,#2)|
general
a batsman in his 19 th inning makes a score of 100 and their by increasing his average by 2 . what is his average after the 19 th inning ?
"18 x + 100 = 19 ( x + 2 ) x = 62 + 2 = 64 answer : d"
a ) 58 , b ) 60 , c ) 62 , d ) 64 , e ) 66
d
add(subtract(100, multiply(2, 19)), 2)
multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|
general
b takes 12 more days than a to finish a task . b and a start this task and a leaves the task 12 days before the task is finished . b completes 60 % of the overall task . how long would b have taken to finish the task if he had worked independently ?
detailed solution let us say a and b split their share of the task and started doing their respective shares simultaneously . let ’ s say a takes a days to finish the task . therefore , b takes a + 12 days to finish the entire task . a has to finish 40 % of the task , since b is doing the rest . so a will only take 2 a / 5 number of days . b only has to finish 60 % of the task , so b will take ( 3 ( a + 12 ) / 5 ) number of days . but as we know , b starts working along with a and finishes 12 days after a stops working . so , ( 3 ( a + 12 ) / 5 ) = ( ( 2 a ) / 5 + 12 ) 3 a + 36 = 2 a + 60 a = 24 ; b = 36 days . answer choice b
a ) 48 days , b ) 36 days , c ) 28 days , d ) 32 days , e ) 22 days
b
add(divide(multiply(12, const_10), add(const_4, const_1)), 12)
add(const_1,const_4)|multiply(n0,const_10)|divide(#1,#0)|add(n0,#2)
physics
a train 100 m long is running with a speed of 68 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going ?
"solution speed of the train relative to man = ( 68 - 8 ) = 60 kmph = 60 x 5 / 18 = 50 / 3 m / sec . time taken by it to cover 100 m at ( 50 / 3 ) m / sec = ( 100 x 3 / 50 ) sec = 6 sec . answer a"
a ) 6 sec . , b ) 7 sec . , c ) 9 sec . , d ) 11 sec . , e ) none
a
divide(100, multiply(add(68, 8), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 4 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ?
"let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 4 2 x / 15 - x / 8 = 4 = > x = 480 km . answer : b"
a ) 347 , b ) 480 , c ) 288 , d ) 278 , e ) 131
b
divide(1, 2)
divide(n3,n4)|
physics
an error 19 % in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ?
"percentage error in calculated area = ( 19 + 19 + ( 19 Γ£ β€” 19 ) / 100 ) % = 41.61 % answer : b"
a ) 6.64 % , b ) 41.61 % , c ) 45.64 % , d ) 46.64 % , e ) 10.64 %
b
divide(multiply(subtract(square_area(add(const_100, 19)), square_area(const_100)), const_100), square_area(const_100))
add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|
gain
in the first 20 overs of a cricket game , the run rate was only 4.2 . what should be the run rate in the remaining 30 overs to reach the target of 324 runs ?
"explanation : required run rate = ( 324 - 4.2 20 ) = 240 / 30 = 8 . answer : c"
a ) 7.25 , b ) 8.5 , c ) 8 , d ) 9.15 , e ) 10
c
divide(subtract(324, multiply(20, 4.2)), 30)
multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|
gain
x and y are integers . a + b < 11 , and a > 6 . what is the smallest possible value of a - b ?
focus on the transition points and plug in the values . a > 6 so a could be 7 , 8 , 9 , 10 , 11 etc look at a = 7 7 + b < 11 b < 4 b could be 3 , 2 , 1 , 0 , - 1 etc when b is 3 , we get a - b = 4 this must be the smallest value because as a increases , b reduces so a - b increases . take another example . a = 11 so b < 0 . a - b will give a value greater than 11 . answer ( c ) .
a ) 1 , b ) 2 , c ) 4 , d ) - 2 , e ) - 4
c
subtract(divide(add(subtract(add(11, 6), const_4), const_1), const_2), const_3)
add(n0,n1)|subtract(#0,const_4)|add(#1,const_1)|divide(#2,const_2)|subtract(#3,const_3)
general
a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 65 percent .
1 hr = 20 percent . thus 15 min = 5 percent . now to charge 65 percent 195 min . answer : c
a ) 145 , b ) 150 , c ) 195 , d ) 160 , e ) 130
c
multiply(divide(65, 20), const_60)
divide(n2,n1)|multiply(#0,const_60)
physics
if x is the product of the positive integers from 1 to 8 , inclusive , and if i , k , m , and p are positive integers such that x = 2 ^ i * 3 ^ k * 5 ^ m * 7 ^ p , then i + k + m + p =
given that x = 8 ! = 2 ^ 7 βˆ— 3 ^ 2 βˆ— 5 βˆ— 7 x . hence , x = 2 ^ 7 βˆ— 3 ^ 2 βˆ— 5 ^ 1 βˆ— 7 ^ 1 = 2 ^ i βˆ— 3 ^ k βˆ— 5 ^ m βˆ— 7 p , since i , k , m , and p are positive integers , then we can equate the exponents , so we have that i = 7 , k = 2 , m = 1 , and p = 1 therefore , i + k + m + p = 7 + 2 + 1 + 1 = 11 answer : d .
a ) 4 , b ) 7 , c ) 8 , d ) 11 , e ) 12
d
add(add(add(7, 2), divide(5, 5)), divide(3, 3))
add(n2,n5)|divide(n4,n4)|divide(n3,n3)|add(#0,#1)|add(#3,#2)
general
find the base k of the number system , if ( 524 ) 8 = ( 664 ) k ?
5 * 8 ^ 2 + 2 * 8 ^ 1 + 4 * 8 ^ 0 = 6 * k ^ 2 + 6 * k ^ 1 + 4 * k ^ 0 320 + 16 + 4 = 6 ( k ^ 2 ) + 6 k + 4 340 = 6 ( k ^ 2 ) + 6 k + 4 6 ( k ^ 2 ) + 6 k + 4 - 340 = 0 6 ( k ^ 2 ) + 6 k - 336 = 0 dividing by 6 ; ( k ^ 2 ) + k - 56 = 0 solving this quadratic equation we get , k = 7 and k = - 8 thus k = 7 answer : d
['a ) 4', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 8']
d
divide(subtract(sqrt(subtract(const_1, multiply(const_4, divide(divide(subtract(const_4, add(const_4, add(multiply(const_2, 8), multiply(power(8, const_2), divide(const_10, const_2))))), const_3), const_2)))), const_1), const_2)
divide(const_10,const_2)|multiply(n1,const_2)|power(n1,const_2)|multiply(#0,#2)|add(#1,#3)|add(#4,const_4)|subtract(const_4,#5)|divide(#6,const_3)|divide(#7,const_2)|multiply(#8,const_4)|subtract(const_1,#9)|sqrt(#10)|subtract(#11,const_1)|divide(#12,const_2)
geometry
if n is the greatest number that will divide 1305 , 4665 and 6905 , leaving the same remainder in each case . what is the sum of the digits of n ?
n = h . c . f . of ( 4665 - 1305 ) , ( 6905 - 4665 ) and ( 6905 - 1305 ) = h . c . f . of 3360 , 2240 and 5600 = 1120 . sum of digits in n = ( 1 + 1 + 2 + 0 ) = 4 answer : e
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 4
e
add(add(add(floor(divide(divide(divide(gcd(subtract(6905, 1305), gcd(subtract(4665, 1305), subtract(6905, 4665))), const_10), const_10), const_10)), floor(divide(subtract(gcd(subtract(4665, 1305), subtract(6905, 4665)), multiply(floor(divide(divide(divide(gcd(subtract(6905, 1305), gcd(subtract(4665, 1305), subtract(6905, 4665))), const_10), const_10), const_10)), multiply(const_10, const_100))), const_100))), floor(divide(subtract(gcd(subtract(4665, 1305), subtract(6905, 4665)), multiply(floor(divide(divide(gcd(subtract(6905, 1305), gcd(subtract(4665, 1305), subtract(6905, 4665))), const_10), const_10)), const_100)), const_10))), subtract(gcd(subtract(4665, 1305), subtract(6905, 4665)), multiply(floor(divide(gcd(subtract(6905, 1305), gcd(subtract(4665, 1305), subtract(6905, 4665))), const_10)), const_10)))
multiply(const_10,const_100)|subtract(n1,n0)|subtract(n2,n1)|subtract(n2,n0)|gcd(#1,#2)|gcd(#4,#3)|divide(#5,const_10)|divide(#6,const_10)|floor(#6)|divide(#7,const_10)|floor(#7)|multiply(#8,const_10)|floor(#9)|multiply(#10,const_100)|subtract(#4,#11)|multiply(#12,#0)|subtract(#4,#13)|divide(#16,const_10)|subtract(#4,#15)|divide(#18,const_100)|floor(#17)|floor(#19)|add(#12,#21)|add(#22,#20)|add(#23,#14)
general
an area of a circularplace is 17 hectares . find the cost of fencing it at the rate of rs . 3 permeter approximately .
area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . Ο€ r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 Ο€ r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . c
['a ) rs . 4000', 'b ) rs . 4450', 'c ) rs . 4457', 'd ) rs . 4560', 'e ) rs . 5457']
c
multiply(circumface(multiply(sqrt(divide(17, const_pi)), const_100)), 3)
divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(n1,#3)
geometry
having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.90 at the candy store . what is john ’ s weekly allowance ?
"total allowance = x amount spent at the arcade = 3 / 5 x amount remaining = 2 / 5 x amount spent at the toy store = 2 / 5 * 1 / 3 x = 2 / 15 x amount remaining = 2 / 5 x - 2 / 15 x = 4 / 15 x now , 4 / 15 x = $ 0.9 therefore , x = $ 3.40 . answer c"
a ) $ 2.40 , b ) $ 3.00 , c ) $ 3.40 , d ) $ 3.60 , e ) $ 4.80
c
divide(0.90, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5))))))
divide(n0,n1)|divide(const_1,n0)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(const_1,#4)|divide(n2,#5)|
general
if a tire rotates at 400 revolutions per minute when the car is traveling 96 km / h , what is the circumference of the tire ?
"400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 96,000 m : c is the circumference c = 4 meters correct answer c"
a ) 7 meters , b ) 9 meters , c ) 4 meters , d ) 5 meters , e ) 3 meters
c
multiply(divide(96, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000)
multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)|
physics
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation f of the mean , what percent of the distribution is less than m + f ?
"16 % ________________________________________________ m + f 34 % ________________________________________________ m 34 % ________________________________________________ m - f 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below m - f and 16 % lie above m + f now below m + f = 16 + 34 + 34 = 84 % hence d"
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
d
subtract(const_100, divide(subtract(const_100, 68), const_2))
subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)|
general
the result when a number subtracted from 100 is the same as the number added to 40 . what is the number ?
"answer let the number be x . then , x - 40 = 100 - x Γ’ ‑ ’ 2 x = 100 + 40 = 140 Γ’ ‑ ’ 2 x = 140 Γ’ Λ† Β΄ x = 70 correct option : a"
a ) 70 , b ) 65 , c ) 80 , d ) 140 , e ) 60
a
add(divide(subtract(100, 40), const_2), 40)
subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|
general
if a / b = 6 / 5 , then ( 5 a + 4 b ) / ( 5 a - 4 b ) = ?
"answer dividing numerator as well as denominator by b , we get given exp . = ( 5 a + 4 b ) / ( 5 a - 4 b ) = ( 5 a / b + 4 ) / ( 5 a / b - 4 ) since a / b = 6 / 5 this implies that = [ ( 5 * 6 ) / 5 + 4 ] / [ ( 5 * 6 ) / 5 - 4 ) ] = ( 6 + 4 ) / ( 6 - 4 ) = 5 option : c"
a ) 4 , b ) 7 , c ) 5 , d ) 3 , e ) 6
c
divide(add(multiply(5, 6), multiply(4, 5)), subtract(multiply(5, 6), multiply(4, 5)))
multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)|subtract(#0,#1)|divide(#2,#3)|
general
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 1 w widgets .
"i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . answer : b"
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
b
multiply(multiply(1, 2), 3)
multiply(n0,n4)|multiply(n3,#0)|
general
if the number 761 * 829 is completely divisible by 9 , then the smallest whole number in place of * will b
sum of digits = ( 7 + 6 + 1 + x + 8 + 2 + 9 ) = ( 33 + x ) , which must be divisible by 9 . x = 3 . c )
a ) 4 , b ) 5 , c ) 3 , d ) 8 , e ) 9
c
subtract(add(add(add(add(add(add(add(const_4, const_3), add(const_3, const_3)), const_1), add(const_4, const_4)), const_2), 9), const_3), add(add(add(add(add(add(const_4, const_3), add(const_3, const_3)), const_1), add(const_4, const_4)), const_2), 9))
add(const_3,const_4)|add(const_3,const_3)|add(const_4,const_4)|add(#0,#1)|add(#3,const_1)|add(#4,#2)|add(#5,const_2)|add(n2,#6)|add(#7,const_3)|subtract(#8,#7)
general
teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 157 per kg , the price of the third variety per kg will be ?
explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 157 ) / \ x Γ’ Λ† ’ 157 22.50 = > x Γ’ Λ† ’ ( 157 / 22.50 ) = 1 . = > x Γ’ Λ† ’ 157 = 22.50 . = > x = 179.50 rs . answer : c
a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 179.50 , d ) rs . 258.50 , e ) none of these
c
divide(subtract(multiply(157, add(add(1, 1), 2)), add(126, 135)), 2)
add(n2,n2)|add(n0,n1)|add(n4,#0)|multiply(n5,#2)|subtract(#3,#1)|divide(#4,n4)
other
the sub - duplicate ratio of 9 : 25 is
root ( 9 ) : root ( 25 ) = 3 : 5 answer : c
a ) 4 : 3 , b ) 1 : 2 , c ) 3 : 5 , d ) 1 : 4 , e ) 2 : 3
c
divide(sqrt(9), sqrt(25))
sqrt(n0)|sqrt(n1)|divide(#0,#1)
other
sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 162 . if the ratio of his marks in english and maths is 3 : 5 , find his marks in science ?
"s : e = 1 : 3 e : m = 3 : 5 - - - - - - - - - - - - s : e : m = 3 : 9 : 15 3 / 27 * 162 = 18 answer : a"
a ) 18 , b ) 68 , c ) 67 , d ) 55 , e ) 56
a
add(add(add(multiply(3, const_3), 3), multiply(multiply(3, const_3), add(multiply(3, const_3), 3))), 3)
multiply(n1,const_3)|add(n1,#0)|multiply(#1,#0)|add(#1,#2)|add(n1,#3)|
general
a train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively . what is the speed of the train ?
let x is the length of the train and v is the speed time taken to move the post = 8 s = > x / v = 8 = > x = 8 v - - - ( 1 ) time taken to cross the platform 264 m long = 20 s ( x + 264 ) / v = 20 = > x + 264 = 20 v - - - ( 2 ) substituting equation 1 in equation 2 , we get 8 v + 264 = 20 v = > v = 264 / 12 = 22 m / s = 22 Γ— 36 / 10 km / hr = 79.2 km / hr answer : b
a ) 79.2 km / hr , b ) 69 km / hr , c ) 74 km / hr , d ) 61 km / hr , e ) none of these
b
multiply(speed(264, 8), const_2)
speed(n0,n1)|multiply(#0,const_2)
physics
of 120 applicants for a job , 56 had at least 4 years ' experience , 72 had degrees , and 12 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 ?
"120 - 12 = 108 108 - 56 - 72 = - 20 then 20 are in the intersection between 4 years experience and degree . answer : d"
a ) 10 , b ) 15 , c ) 22 , d ) 20 , e ) 24
d
add(subtract(add(56, 72), subtract(120, 12)), subtract(72, 56))
add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#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 . 39000 , then the share of b is ?
"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 39000 share of b = 6 / 13 ( 39000 ) = rs . 18000 answer : e"
a ) a ) 34500 , b ) b ) 24000 , c ) c ) 24098 , d ) d ) 19007 , e ) e ) 18000
e
subtract(39000, multiply(const_60, const_100))
multiply(const_100,const_60)|subtract(n3,#0)|
gain
divide rs . 32000 in the ratio 3 : 5 ?
"3 / 8 * 32000 = 12000 5 / 8 * 32000 = 20000 answer : a"
a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 12000 , 20001
a
multiply(subtract(5, const_2), divide(32000, add(3, subtract(5, const_2))))
subtract(n2,const_2)|add(n1,#0)|divide(n0,#1)|multiply(#2,#0)|
other
what is the hcf of 2 / 3 , 4 / 15 and 6 / 5
"explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 15 , 5 ) = 2 / 15 answer : option e"
a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 8 / 45 , e ) 2 / 15
e
divide(2, 5)
divide(n0,n5)|
general
n a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 600 m ?
when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 600 m , the number of meters by which a beats c = ( 600 * 212.5 ) / 1000 = 127.5 m . answer : ob
a ) 127.6 , b ) 127.5 , c ) 127.0 , d ) 127.2 , e ) 127.3
b
subtract(600, divide(multiply(divide(multiply(subtract(1000, 100), 600), 1000), subtract(800, 100)), 800))
subtract(n0,n1)|subtract(n2,n1)|multiply(n4,#0)|divide(#2,n0)|multiply(#3,#1)|divide(#4,n2)|subtract(n4,#5)
physics
3 , 14,39 , _ ?
"n ^ 3 + n ^ 2 + n soln : 1 ^ 3 + 1 ^ 2 + 1 = 3 2 ^ 3 + 2 ^ 2 + 2 = 8 + 4 + 2 = 14 3 ^ 3 + 3 ^ 2 + 3 = 27 + 9 + 3 = 39 4 ^ 3 + 4 ^ 2 + 4 = 64 + 16 + 4 = 84 answer : d"
a ) 81 , b ) 82 , c ) 83 , d ) 84 , e ) 85
d
multiply(divide(3, 14,39), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
if x : y = 3 : 5 , find the value of ( 2 x + 4 y ) : ( 4 x – y )
"explanation : given : a / b = 3 / 5 ( 2 * 3 + 4 * 5 ) : ( 4 * 3 – 5 ) = 26 : 7 answer : c"
a ) 29 : 7 , b ) 22 : 7 , c ) 26 : 7 , d ) 20 : 7 , e ) 26 : 5
c
divide(add(power(2, 4), 4), add(add(3, 5), 4))
add(n0,n1)|power(n2,n3)|add(n3,#1)|add(n3,#0)|divide(#2,#3)|
general
nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 14 cm ?
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 14 ) = 266 cm 2 answer : c"
a ) 235 cm 2 , b ) 260 cm 2 , c ) 266 cm 2 , d ) 270 cm 2 , e ) 280 cm 2
c
divide(multiply(14, add(20, 18)), const_2)
add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|
physics
a β€œ 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 600,000 ?
"the first digit and last digit are the same so the 2 possibilities are 6 or 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 2 * 10 * 10 = 200 the answer is b ."
a ) 150 , b ) 200 , c ) 240 , d ) 300 , e ) 480
b
multiply(multiply(subtract(6, const_4), const_10), const_10)
subtract(n2,const_4)|multiply(#0,const_10)|multiply(#1,const_10)|
general
a reduction of 20 % in the price of salt enables a lady to obtain 4 kgs more for rs . 100 , find the original price per kg ?
"100 * ( 20 / 100 ) = 20 - - - 4 ? - - - 1 = > rs . 5 100 - - - 80 ? - - - 5 = > rs . 6.25 answer : c"
a ) 6.16 , b ) 6.21 , c ) 6.25 , d ) 2.4 , e ) 2.7
c
multiply(divide(divide(multiply(divide(20, const_100), 100), 4), multiply(divide(20, const_100), 100)), const_100)
divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)|
gain
kyle , david , and catherine each try independently to solve a problem . if their individual probabilities for success are 1 / 8 , 1 / 4 and 3 / 5 , respectively , what is the probability no one will solve the problem ?
"p ( kyle will not solve ) = 1 - 1 / 8 = 7 / 8 p ( david will not solve ) = 1 - 1 / 4 = 3 / 4 p ( catherine will not solve ) = 1 - 3 / 5 = 2 / 5 p = ( 7 / 8 ) * ( 3 / 4 ) * ( 2 / 5 ) = 42 / 160 = 21 / 80 answer : b"
a ) 3 / 80 , b ) 21 / 80 , c ) 5 / 16 , d ) 25 / 76 , e ) 9 / 80
b
divide(multiply(3, 3), add(multiply(const_60, 8), multiply(8, 8)))
multiply(n4,n4)|multiply(n1,const_60)|multiply(n1,n1)|add(#1,#2)|divide(#0,#3)|
general
if p and q are prime numbers , how many divisors does the product p ^ 6 * q ^ 7 have ?
"when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is b . 7 * 8 = 56"
a ) 40 , b ) 56 , c ) 60 , d ) 62 , e ) 70
b
multiply(add(6, const_1), add(7, const_1))
add(n0,const_1)|add(n1,const_1)|multiply(#0,#1)|
general
if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 4 children have two children of the same sex and one of the opposite sex ?
no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 4 = 4 c 2 total possible outcomes - 2 ^ 4 ( each child can be either a girl or a boy ) probability = 2 c 1 * 4 c 2 / 2 ^ 4 = 2 * 4 / 2 * 2 * 2 * 2 = 8 / 16 = 1 / 2 ans = d
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 2 , e ) 3 / 5
d
divide(const_2, 4)
divide(const_2,n0)
general
the measurement of a rectangular box with lid is 25 cmx 12 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of Ο€ cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere )
"d = 12 , r = 6 ; volume of the largest sphere = 4 / 3 Ο€ r 3 = 4 / 3 * Ο€ * 6 * 6 * 6 = 288 Ο€ cm 3 answer : a"
a ) 288 , b ) 48 , c ) 72 , d ) 864 , e ) 964
a
multiply(divide(const_4, 3), power(3, 3))
divide(const_4,n3)|power(n3,n3)|multiply(#0,#1)|
geometry
when a certain number x is divided by 72 , the remainder is 19 . what is the remainder when x is divided by 8 ?
"let possible value of x is 91 least possible value of x / 8 is 91 / 8 = > 11 quotient with remainder 3 thus answer is ( a ) 3"
a ) 3 , b ) 2 , c ) 4 , d ) 1 , e ) 5
a
reminder(19, 8)
reminder(n1,n2)|
general
in a tournament prize of 1 st , 2 nd , and 3 rd is given in same interval . if total amount of prize rs 4800 and prize of 1 st is rs 2000 , find the interval ? ?
let the interval is x . prize of 2 nd and 3 rd is 2000 - x and 2000 - 2 x 2000 + 2000 - x + 2000 - 2 x = 4800 x = 400 answer : e
a ) 600 , b ) 450 , c ) 350 , d ) 550 , e ) 400
e
divide(subtract(multiply(2000, 3), 4800), const_3)
multiply(n2,n5)|subtract(#0,n3)|divide(#1,const_3)
general
if n = 8 ^ 11 – 1 , what is the units digit of n ?
"8 ^ 11 - 8 = 8 ( 8 ^ 10 - 1 ) = = > 8 ( 2 ^ 30 - 1 ) last digit of 2 ^ 30 is 4 based on what explanation livestronger is saying . 2 ^ 30 - 1 yields 4 - 1 = 3 as the unit digit . now on multiply this with 1 , we get unit digit as 3 . answer : a"
a ) 3 , b ) 1 , c ) 0 , d ) 6 , e ) 4
a
divide(log(1), log(power(8, 11)))
log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)|
general
$ 600 will become $ 720 in 4 years find the rate of interest ?
"si = simple interest = a - p = 720 - 600 = $ 120 r = 100 si / pt = 100 * 120 / 600 * 4 = 5 % answer is c"
a ) 10 % , b ) 15 % , c ) 5 % , d ) 7 % , e ) 20 %
c
divide(subtract(720, 600), multiply(600, divide(4, const_100)))
divide(n2,const_100)|subtract(n1,n0)|multiply(n0,#0)|divide(#1,#2)|
gain
the area of a circle is increased by 300 % . by what percent has the diameter of the circle increased ?
the area of the circle is increased by 300 % , thus the area is increased 4 times . the area of a circle it proportional to the square of the diameter ( area = Ο€ d ^ 2 / 4 ) , therefore the diameter must increase 2 times ( diameter increase 2 times = area increase 4 times ) , which is increase by 100 % . answer : d .
['a ) 400 %', 'b ) 200 %', 'c ) 300 %', 'd ) 100 %', 'e ) 800 %']
d
multiply(const_100, divide(const_2, const_2))
divide(const_2,const_2)|multiply(#0,const_100)
geometry
18 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ?
"3 w = 2 m 18 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 18 * 21 * 8 = 14 * x * 6 x = 36 answer : c"
a ) 32 , b ) 87 , c ) 36 , d ) 99 , e ) 77
c
add(floor(divide(multiply(multiply(21, 8), multiply(18, 3)), multiply(multiply(21, 2), 6))), const_1)
multiply(n1,n2)|multiply(n0,n5)|multiply(n4,n6)|multiply(#0,#1)|multiply(n3,#2)|divide(#3,#4)|floor(#5)|add(#6,const_1)|
physics