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a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 102 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 102 million in sales ?
"( 9 / 102 ) / ( 3 / 20 ) = 30 / 54 = 58,5 % it means that 9 / 108 represents only 58,5 % . therefore a decrease of 45 % . answer c"
a ) 8 % , b ) 15 % , c ) 40 % , d ) 52 % , e ) 56 %
c
multiply(divide(3, 20), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 55 , then how long is the base of isosceles triangle ?
"the base of the isosceles triangle is 55 - 20 - 20 = 15 units the answer is c ."
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25
c
subtract(subtract(55, divide(60, const_3)), divide(60, const_3))
divide(n0,const_3)|subtract(n1,#0)|subtract(#1,#0)|
geometry
one fourth of one third of two fifth of a number is 10 . what will be 40 % of that number
"explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 10 then x = 10 * 30 = 300 40 % of 300 = 120 answer : option c"
a ) a ) 140 , b ) b ) 150 , c ) c ) 120 , d ) d ) 200 , e ) e ) 220
c
divide(multiply(divide(10, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100)
add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)|
gain
in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 52 , what is the greatest possible value of b Γ— a ?
"you should notice that 52 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 22 ( next number wo n ' t have 110 as the first 3 digit...
a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 22
e
subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 52), const_1), 52), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 52), const_1), 52), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000),...
power(n0,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,n0)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)|
general
find the average of first 8 multiples of 5 ?
"average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 ) / 8 = 22.5 answer is d"
a ) 10 , b ) 15 , c ) 12.5 , d ) 22.5 , e ) 21
d
divide(add(add(add(8, const_1), add(add(const_4.0, const_1), const_2)), add(subtract(5, 8), subtract(5, const_2))), 8)
add(n0,const_1)|subtract(n1,const_4.0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)|
general
a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 16 heads and 45 legs . how many cats does the ship host ?
"a ' s + co + ca + cats = 16 . sa ' s + 1 + 1 + cats = 16 or sa ' s + cats = 14 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 45 sa ' s * 2 + cats * 4 = 42 or sa ' s + cats * 2 = 21 or 14 - cats + cat * 2 = 21 then cats = 7 e"
a ) 3 , b ) 6 , c ) 9 , d ) 8 , e ) 7
e
multiply(divide(subtract(subtract(45, const_1), multiply(subtract(16, const_1), const_2)), subtract(multiply(subtract(16, const_1), const_4), multiply(subtract(16, const_1), const_2))), subtract(16, const_1))
subtract(n1,const_1)|subtract(n0,const_1)|multiply(#1,const_2)|multiply(#1,const_4)|subtract(#0,#2)|subtract(#3,#2)|divide(#4,#5)|multiply(#6,#1)|
general
how many 1 x 1 x 1 cubes can you fit in a 15 x 16 x 13 box ?
the answer is d ) 3,120 . a 1 x 1 x 1 cube has an area of 1 . the 15 x 16 x 13 box has an area of 3,120 . if you divide 3,120 by 1 , you get 3,120 . therefore the answer is d .
['a ) 3,200', 'b ) 3,150', 'c ) 3,448', 'd ) 3,120', 'e ) 2,980']
d
volume_rectangular_prism(15, 16, 13)
volume_rectangular_prism(n3,n4,n5)
geometry
the lcm of two numbers is 2310 and hcf is 83 . if one of the numbers is 210 . then what is the other number ?
"first number * second number = lcm * hcf other number = 2310 * 83 / 210 = 11 * 83 = 913 answer : a"
a ) 913 , b ) 825 , c ) 330 , d ) 582 , e ) 465
a
divide(multiply(2310, 83), 210)
multiply(n0,n1)|divide(#0,n2)|
physics
nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days , they completed 1 / 3 part of work . than how many more employees should nhai hire to finish it in time working 10 hours a day ?
explanation : the given problem can be written in a tabular form like below : - men days hour work planned 100 50 8 1 actual 100 25 8 1 / 3 remaining ? 25 10 2 / 3 we can apply chain rule now . total men required to complete the remaining work = 100 Γ— ( 25 / 25 ) Γ— ( 8 / 10 ) Γ— ( 2 / 3 ) / ( 1 / 3 ) = 160 . so addition...
a ) 55 , b ) 60 , c ) 70 , d ) 50 , e ) 65
b
subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100)
multiply(n0,n4)|multiply(n4,n7)|multiply(n3,#0)|multiply(n1,#2)|divide(#3,#1)|subtract(#4,n0)
physics
a man is walking at a speed of 6 km per hour . after every km he takes rest for 6 minutes . how much time will he take to cover a distance of 12 km .
"explanation : for 12 km he takes 11 times rest . 11 x 6 = 66 minutes rest = 1 hr 6 min for 12 km , 2 hr is required + rest = 2 + 1 hr 6 m = 3 hr . 6 min answer : option d"
a ) 3 hr , b ) 5 hr 42 min , c ) 4 hrs . 10 min , d ) 3 hr . 6 min , e ) none of these
d
add(multiply(divide(6, 6), speed(const_60, const_1)), multiply(const_4, 6))
divide(n1,n0)|multiply(n1,const_4)|speed(const_60,const_1)|multiply(#0,#2)|add(#3,#1)|
physics
what least number must be subtracted from 9671 so that the remaining number is divisible by 2 ?
"on dividing 9671 by 2 , we get remainder = 1 . required number be subtracted = 1 answer : b"
a ) 3 , b ) 1 , c ) 16 , d ) 11 , e ) 15
b
subtract(9671, multiply(floor(divide(9671, 2)), 2))
divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|
general
find the value of ( 200 + 200 / 90 ) Γ— 90
"( 18000 + 200 ) / 90 * 90 = 18200 answer : b"
a ) 18500 , b ) 18200 , c ) 17500 , d ) 16500 , e ) 13500
b
multiply(add(divide(200, 90), 200), 90)
divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|
general
mr . loyd wants to fence his square shaped land of 130 sqft each side . if a pole is laid every 10 ft how many poles do he need ?
"if each side is 120 feet . . then total perimeter is 130 * 4 = 520 poles every 10 feet hence no of poles = 520 / 10 = 52 answer : e"
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 52
e
divide(divide(multiply(130, const_4), const_10), const_2)
multiply(n0,const_4)|divide(#0,const_10)|divide(#1,const_2)|
geometry
a beekeeper has 600000 bees living in 4 beehives . among these beehives , the sizes of the beehives are proportional to the numbers of bees living in them . no beehive has less than 80 % the number of bees of any other beehive . what is the maximum possible number of bees that the greatest beehive could have ?
x be the # of bees in the hive with the max # of bees . let other hives have same # of bees . so , x + 3 * ( 0.8 ) * x = 600,000 = > 3.4 x = 600,000 = > x = 176470 . b is the answer .
a ) 20,000 , b ) 176470 , c ) 10000 , d ) 13000 , e ) 56600
b
add(subtract(80, const_10), add(add(divide(600000, const_100), add(add(multiply(const_1000, const_10), divide(600000, 4)), multiply(const_1000, const_10))), multiply(const_100, const_4)))
divide(n0,n1)|divide(n0,const_100)|multiply(const_10,const_1000)|multiply(const_100,const_4)|subtract(n2,const_10)|add(#0,#2)|add(#5,#2)|add(#6,#1)|add(#7,#3)|add(#8,#4)
general
if 36 men can do a piece of work in 25 hours , in how many hours will 10 men do it ?
explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 10 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 10 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 10 } = 90 hence , ...
a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 90
e
divide(multiply(36, 25), 10)
multiply(n0,n1)|divide(#0,n2)
physics
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 10 percent on food , and 40 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what perce...
"let amount spent by jill = 100 clothing = 50 , food = 10 , others = 40 tax on clothing = 2 tax on others = 3.2 percentage = 5.2 / 100 = 5.2 % answer : d"
a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 %
d
multiply(divide(add(multiply(50, divide(4, const_100)), multiply(40, divide(8, const_100))), const_100), const_100)
divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)|
general
if the average ( arithmetic mean ) of 4 x - 5 , 8 x - 6 , and 9 x + 5 is 279 , what is the value of x ?
"am of 4 x - 5 , 8 x - 6 , and 9 x + 5 = 4 x - 5 + 8 x - 6 + 9 x + 5 / 3 = 31 x / 3 given that 31 x / 3 = 279 x = 837 / 31 = 27 x = 27 answer is c"
a ) 18 , b ) 25 , c ) 27 , d ) 29 , e ) 31
c
add(subtract(5, 4), const_1)
subtract(n1,n0)|add(#0,const_1)|
general
a retailer bought a hat at wholesale and marked it up 70 % to its initial price of $ 34 . by how many more dollars does he need to increase the price to achieve a 100 % markup ?
let x be the wholesale price . then 1.7 x = 34 and x = 34 / 1.7 = 20 . to achieve a 100 % markup , the price needs to be $ 40 . the retailer needs to increase the price by $ 6 more . the answer is c .
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
c
subtract(multiply(divide(34, divide(add(const_100, 70), const_100)), const_2), 34)
add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(#2,const_2)|subtract(#3,n1)
general
an exam consists of 8 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 90 % is a passing grade , what is the probability that brian passes ?
"if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure ...
a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 219 / 256 , e ) 15 / 16
e
add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), mult...
divide(const_1,const_2)|multiply(n0,const_3)|power(#0,n0)|multiply(n0,#2)|multiply(#1,#2)|add(#3,#2)|add(#5,#4)|subtract(const_1,#6)|divide(#7,const_10)|add(#8,#7)|
general
a fruit seller had some oranges . he sells 30 % oranges and still has 420 oranges . how many oranges he had originally ?
"explanation : he sells 30 % of oranges and still there are 420 oranges remaining = > 70 % of oranges = 420 β‡’ ( 70 Γ— total oranges ) / 100 = 420 β‡’ total oranges / 100 = 6 β‡’ total oranges = 6 Γ— 100 = 600 answer : option b"
a ) 420 , b ) 600 , c ) 220 , d ) 400 , e ) none of these
b
add(420, multiply(420, divide(30, const_100)))
divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|
gain
70 is subtracted from 30 % of a number , the result is 20 . find the number ?
( 30 / 100 ) * x – 70 = 20 3 x = 900 x = 300 answer : d
a ) 150 , b ) 997 , c ) 266 , d ) 300 , e ) 271
d
divide(add(70, 20), divide(30, const_100))
add(n0,n2)|divide(n1,const_100)|divide(#0,#1)
gain
sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 0.5 % more . she would get how much ?
"( 8000 * 3 * 0.5 ) / 100 = 120 9200 - - - - - - - - 9320 answer : a"
a ) 9320 , b ) 96288 , c ) 26667 , d ) 1662 , e ) 2882
a
add(multiply(multiply(add(divide(0.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000)
divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)|
gain
the average of 11 numbers is 60 . out of 11 numbers the average of first 6 no . is 88 , and last 6 numbers is 65 then find 6 th number ?
"6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 88 + 6 * 65 - 11 * 60 = 258 answer is a"
a ) a ) 258 , b ) b ) 65 , c ) c ) 58 , d ) d ) 62 , e ) e ) 48
a
subtract(add(multiply(6, 88), multiply(6, 65)), multiply(11, 60))
multiply(n3,n4)|multiply(n3,n6)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)|
general
what is the area of a square field whose sides have a length of 13 meters ?
"13 * 13 = 169 sq m the answer is b ."
a ) 225 sq m , b ) 169 sq m , c ) 128 sq m , d ) 267 sq m , e ) 231 sq m
b
divide(square_area(13), const_2)
square_area(n0)|divide(#0,const_2)|
geometry
what is the cp of rs 100 stock at 6 discount , with 1 / 5 % brokerage ?
"explanation : use the formula , cp = 100 Γ’ € β€œ discount + brokerage % cp = 100 - 6 + 1 / 5 94.2 thus the cp is rs 94.2 . answer : c"
a ) 96.9 , b ) 96.3 , c ) 94.2 , d ) 96.7 , e ) 96.21
c
add(subtract(100, 6), divide(1, 5))
divide(n2,n3)|subtract(n0,n1)|add(#0,#1)|
gain
the edges of cuboid are 4 cm ; 5 cm and 6 cm . find its surface area ?
2 ( 4 * 5 + 5 * 6 + 4 * 6 ) = 148 answer : b
['a ) 228', 'b ) 148', 'c ) 992', 'd ) 772', 'e ) 222']
b
add(multiply(multiply(const_2, const_4), 5), add(multiply(multiply(const_2, 6), const_4), multiply(6, multiply(const_2, 5))))
multiply(n2,const_2)|multiply(n1,const_2)|multiply(const_2,const_4)|multiply(#0,const_4)|multiply(n2,#1)|multiply(n1,#2)|add(#3,#4)|add(#6,#5)
geometry
x does a work in 15 days . y does the same work in 30 days . in how many days they together will do the same work ?
"x ' s 1 day ' s work = 1 / 15 y ' s 1 day ' s work = 1 / 30 ( x + y ) ' s 1 day ' s work = ( 1 / 15 + 1 / 30 ) = 1 / 10 both together will finish the work in 10 days . correct option is a"
a ) 10 , b ) 12 , c ) 20 , d ) 30 , e ) 15
a
inverse(add(divide(const_1, 15), divide(const_1, 30)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|
physics
two trains of equal lengths take 10 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ?
"speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 12 + 6 = 18 m / sec . required time = ( 120 + 120 ) / 18 = 13 sec . answer : b"
a ) 16 sec , b ) 13 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec
b
divide(multiply(120, const_2), add(speed(120, 20), speed(120, 10)))
multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)|
physics
set a contains all the even numbers between 4 and 50 inclusive . set b contains all the even numbers between 104 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ?
"set a contains 4 , 6 . . . 50 set b contains 104 , 106 . . . 150 number of terms in each set = 24 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 24 = 2400 answer a"
a ) 2400 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325
a
multiply(subtract(104, 4), add(divide(subtract(50, 4), const_2), const_1))
subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)|
general
for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 8.2 – 8.2 * ?
"since y * is defined as the greatest positive even integer less than or equal to y , then 8.2 * = 4 ( the greatest positive even integer less than or equal to 5.2 is 4 ) . hence , 8.2 – 8.2 * = 8.2 - 6 = 2.2 answer : d ."
a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0
d
subtract(8.2, subtract(floor(8.2), const_1))
floor(n0)|subtract(#0,const_1)|subtract(n0,#1)|
general
( 1000 ) 7 Γ· ( 10 ) 20 = ?
"explanation : = ( 103 ) 7 / ( 10 ) 20 = ( 10 ) 21 / ( 10 ) 20 = 10 ( 1 ) = 10 option a"
a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these
a
multiply(1000, 10)
multiply(n0,n2)|
general
a number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34 . such a least possible number v is
i solved this question by plugging in numbers from the answer choices . a . ) 74 starting with answer choice a , i immediately eliminated it because 74 is not even divisible by 5 . b . ) 75 i divide 75 / 5 and get 15 as an answer . i divide 75 / 34 and get a remainder of 7 . 15 - 7 = 8 so i know the correct answer isb
a ) 74 , b ) v = 75 , c ) v = 175 , d ) v = 680 , e ) 690
b
add(multiply(34, const_2), divide(subtract(multiply(34, const_2), multiply(8, 5)), subtract(5, const_1)))
multiply(n2,const_2)|multiply(n0,n1)|subtract(n0,const_1)|subtract(#0,#1)|divide(#3,#2)|add(#4,#0)
general
in the hillside summer camp there are 50 children . 80 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 10 % of the total number of children in the camp . how many more boys must she bring to make that happen ?
"given there are 50 students , 80 % of 50 = 40 boys and remaining 10 girls . now here 80 % are boys and 20 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 20 or 20 % . . if we add 50 to existing 40 then the count will be 90 and the girls number will be 10 as ...
a ) 55 , b ) 50 , c ) 40 . , d ) 30 . , e ) 25 .
b
add(multiply(divide(subtract(const_100, 10), const_100), 50), multiply(divide(10, const_100), 50))
divide(n2,const_100)|subtract(const_100,n2)|divide(#1,const_100)|multiply(n0,#0)|multiply(n0,#2)|add(#4,#3)|
general
a man invests rs . 8,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ?
explanation : interest on rs . 8000 at 5 % per annum = ( 8000 Γ— 5 Γ— 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ— 8 Γ— 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) Γ— 6 Γ— 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) Γ— 6 Γ— 1...
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these
d
multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6)))
multiply(n2,const_1000)|subtract(n3,n1)|subtract(n2,n3)|divide(#1,#2)|multiply(#3,#0)
gain
how many 3 digit odd numbers do not use any digit more than once
"the way i solved it is a b c ( hundreds , tens , units ) c can be 1 3 5 7 9 ( any of the 5 digits ) a can be anything except ( d or 0 ) so 8 possibilitie b can be anything execpt ( a d c ) so 7 possibilities total ways are 7 * 8 * 5 = 280 ans : e"
a ) 1720 , b ) 216 , c ) 224 , d ) 2460 , e ) 280
e
multiply(multiply(multiply(add(3, const_1), subtract(subtract(const_10, const_1), const_1)), subtract(subtract(const_10, const_1), const_1)), subtract(subtract(subtract(const_10, const_1), const_1), const_1))
add(n0,const_1)|subtract(const_10,const_1)|subtract(#1,const_1)|multiply(#0,#2)|subtract(#2,const_1)|multiply(#3,#2)|multiply(#5,#4)|
general
a sum of 25 was paid for a work which a can do in 32 days , b in 20 days , b and c in 12 days and d in 24 days . how much did c receive if all the 4 work together ?
a ' s one day ' s work = 1 ⁄ 32 b ' s one day ' s work = 1 ⁄ 20 ( b + c ) ' s one day ' s work = 1 ⁄ 12 ∴ c ' s one day ' s work = 1 ⁄ 12 - 1 ⁄ 20 = 1 ⁄ 30 d ' s one day ' s work = 1 ⁄ 24 ∴ ( a + b + c + d ) ' s one day ' s work 1 / 32 + 1 / 20 + 1 / 30 + 1 / 24 = 75 + 120 + 80 + 100 / 2400 = 375 ⁄ 2400 = 15 ⁄ 96 = 5 ⁄...
a ) 14 ⁄ 3 , b ) 16 ⁄ 3 , c ) 15 ⁄ 3 , d ) 17 ⁄ 3 , e ) none of these
b
multiply(divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20)))), divide(25, add(add(add(divide(lcm(lcm(32, 20), 24), 32), divide(lcm(lcm(32, 20), 24), 20)), divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20))))), divide(lcm(lcm(32, 2...
divide(const_1,n3)|divide(const_1,n2)|lcm(n1,n2)|lcm(n4,#2)|subtract(#0,#1)|divide(const_1,#4)|divide(#3,n1)|divide(#3,n2)|divide(#3,n4)|add(#6,#7)|divide(#3,#5)|add(#9,#10)|add(#11,#8)|divide(n0,#12)|multiply(#10,#13)
physics
a room 13 m 47 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor .
"explanation : area of the room = ( 1347 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1347 cm and 777 cm = 3 cm . area of 1 tile = ( 3 x 3 ) cm 2 . number of tiles required = ( 1347 Γ— 777 ) / ( 3 Γ— 3 ) = 116291 answer : option d"
a ) 116636 , b ) 116406 , c ) 116647 , d ) 116291 , e ) 116675
d
divide(multiply(add(multiply(13, const_100), 47), add(multiply(7, const_100), 77)), multiply(subtract(47, add(multiply(const_2, const_4), const_2)), subtract(47, add(multiply(const_2, const_4), const_2))))
multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)|
physics
if a number p is prime , and 2 p + 3 = q , where q is also prime , then the decimal expansion of 1 / q will produce a decimal with q - 1 digits . if this method produces a decimal with 166 digits , what is the units digit of the product of p and q ?
"1 / 5 = 0.2 . . ( a repeating pattern one digit long ) a"
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
a
subtract(divide(subtract(add(166, 1), 3), 2), subtract(divide(subtract(add(166, 1), 3), 2), 1))
add(n2,n4)|subtract(#0,n1)|divide(#1,n0)|subtract(#2,n2)|subtract(#2,#3)|
general
what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 60 % ?
"the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 60 % increase a surface = 6 * ( ( 1.6 a ) ^ 2 ) = 6 * 2.56 * a ^ 2 the increase in the surface area = ( 6 * 2.56 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 2.56 - 1 ) ) / ( 6 * a ^ 2 ) = 2.56 - 1 = 1.56 = 156 % answer : c"
a ) 150 % , b ) 160 % , c ) 156 % , d ) 158 % , e ) 162 %
c
multiply(subtract(power(add(const_1, divide(60, const_100)), const_2), const_1), const_100)
divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|
geometry
the workforce of company x is 60 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ?
let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 20 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 20 ) 0.05 x = 11 x = 220 - this is quantity of employees before adding 2020 men so after adding it will be 240 answer...
a ) 160 , b ) 220 , c ) 240 , d ) 360 , e ) 420
c
add(divide(multiply(divide(55, const_100), 20), subtract(divide(60, const_100), divide(55, const_100))), 20)
divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)
gain
if | 20 x - 10 | = 90 , then find the product of the values of x ?
"| 20 x - 10 | = 90 20 x - 10 = 90 or 20 x - 10 = - 90 20 x = 100 or 20 x = - 80 x = 5 or x = - 4 product = - 4 * 5 = - 20 answer is e"
a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 20
e
subtract(subtract(subtract(90, 10), add(90, 10)), 10)
add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)|
general
two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores is 1 / 3 , what is the value of x / y ?
answer is d : 4 x - y = ( x + y ) / 2 solving for x / y = 4
a ) 3 . , b ) 2 . , c ) 1 / 2 . , d ) 4 , e ) there is n ' t enough data to answer the question .
d
add(3, divide(multiply(3, 1), 3))
multiply(n0,n1)|divide(#0,n1)|add(n1,#1)
general
let a be the event that a randomly selected two digit number is divisible by 2 and let b be the event that a randomly selected two digit number is divisible by 10 . what is p ( a and b ) ?
"p ( a and b ) = 1 / 2 * 1 / 10 = 1 / 20 the answer is a ."
a ) 1 / 20 , b ) 1 / 15 , c ) 1 / 12 , d ) 1 / 8 , e ) 1 / 2
a
multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2)))
subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)|
physics
in a flight of 600 km , an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by 200 km / hr and the time of flight increased by 30 minutes . the duration of the flight is :
"let the duration of the flight be x hours then 600 / x - 600 / x ( 1 / 2 ) = 200 = 600 / x - 1200 / 2 x + 1 = 200 = x ( 2 x + 1 ) = 3 = 2 xsquare + x - 3 = 0 = ( 2 x + 3 ) ( x - 1 ) = 0 = x = 1 hr ( neglecting the - ve value of x ) answer : option a"
a ) 1 hour , b ) 2 hour , c ) 3 hour , d ) 4 hour , e ) 5 hour
a
divide(600, divide(multiply(200, 30), const_10))
multiply(n1,n2)|divide(#0,const_10)|divide(n0,#1)|
general
a person travels equal distances with speeds of 50 km / hr , 100 km / hr and 120 km / hr and takes a total time of 23 minutes . the total distance is ?
"let the total distance be 3 x km x / 50 + x / 100 + x / 120 = 23 / 60 23 x / 600 = 23 / 60 23 x = 230 x = 10 km total distance = 3 x = 30 km answer is b"
a ) 15 km , b ) 30 km , c ) 45 km , d ) 50 km , e ) 60 km
b
multiply(multiply(divide(divide(23, const_60), add(add(divide(const_1, 50), divide(const_1, 100)), divide(const_1, 120))), 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
two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 25 hr . if all pipes are opened simultaneously , then the cistern will be filled in
"solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 βˆ’ 1 / 25 = 1 / 7 hence , tank will be filled in 6.98 hour option ( b )"
a ) 7.5 hr , b ) 6.98 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these
b
inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 25)))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)|
physics
if 5 + 3 + 2 = 151022 , 9 + 2 + 4 = 183652 , then 7 + 2 + 5 = ?
if the given number is a + b + c then a . b | a . c | a . b + a . c - b β‡’ β‡’ 5 + 3 + 2 = 5.3 | 5.2 | 5.3 + 5.2 - 3 = 151022 β‡’ β‡’ 9 + 2 + 4 = 9.2 | 9.4 | 9.2 + 9.4 - 2 = 183652 7 + 2 + 5 = 7.2 | 7.5 | 7.2 + 7.5 - 2 = 143547 answer : b
a ) 223888 , b ) 143547 , c ) 2607778 , d ) 126997 , e ) 127811
b
add(multiply(add(multiply(multiply(7, 2), const_100), multiply(7, 5)), const_100), subtract(add(multiply(7, 2), multiply(7, 5)), 2))
multiply(n2,n8)|multiply(n0,n8)|add(#0,#1)|multiply(#0,const_100)|add(#3,#1)|subtract(#2,n2)|multiply(#4,const_100)|add(#6,#5)
general
the inner circumference of a circle race track 18 m wide is 880 m . find the radius of the outer circle .
let inner radius be r metres then 2 Ο€ r = 640 2 Γ— 22 / 7 r = 880 44 / 7 r = 880 r = 880 Γ— 7 / 44 = 140 m answer is d .
['a ) 120 m', 'b ) 145 m', 'c ) 130 m', 'd ) 140 m', 'e ) 122 m']
d
divide(divide(880, const_pi), const_2)
divide(n1,const_pi)|divide(#0,const_2)
physics
4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 14 mat - weavers in 14 days ?
"let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 14 : : 4 : x days 4 : 14 4 * 4 * x = 14 * 14 * 4 x = ( 14 * 14 * 4 ) / ( 4 x 4 ) x = 49 . answer is e ."
a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 49
e
multiply(multiply(4, divide(14, 4)), divide(14, 4))
divide(n3,n0)|multiply(n0,#0)|multiply(#0,#1)|
gain
find √ ? / 15 = 4 ?
"answer let √ n / 15 = 4 then √ n = 15 x 4 = 60 ∴ n = 60 x 60 = 3600 . correct option : d"
a ) 76 , b ) 5776 , c ) 304 , d ) 3600 , e ) none
d
power(multiply(4, 15), const_2)
multiply(n0,n1)|power(#0,const_2)|
general
if a ( a - 1 ) = 20 and b ( b - 1 ) = 20 , where a β‰  b , then a + b =
"i . e . if a = 5 then b = - 4 or if a = - 4 then b = 5 but in each case a + b = - 5 + 4 = - 1 answer : option b"
a ) βˆ’ 48 , b ) βˆ’ 1 , c ) 2 , d ) 46 , e ) 67
b
subtract(subtract(subtract(subtract(add(add(1, 20), subtract(1, 20)), 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
10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 3240 , how much was it in the beginning ?
"x * ( 90 / 100 ) * ( 80 / 100 ) = 3240 x = 4500 answer : d"
a ) a ) 3800 , b ) b ) 4200 , c ) c ) 4400 , d ) d ) 4500 , e ) e ) 4600
d
floor(divide(3240, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))))
subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|
general
the length of the bridge , which a train 130 meters long and travelling at 45 km / hr can cross in 30 seconds , is ?
speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 130 + x ) / 30 = 25 / 2 = = > 2 ( 130 + x ) = 750 = = > x = 245 m . answer : c
a ) 288 , b ) 266 , c ) 245 , d ) 277 , e ) 211
c
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
a sum of rs . 2743 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ?
"( x * 8 * 3 ) / 100 = ( ( 2743 - x ) * 3 * 5 ) / 100 24 x / 100 = 41145 / 100 - 15 x / 100 39 x = 41145 = > x = 1055 second sum = 2743 – 1055 = 1688 answer : d"
a ) 1642 , b ) 1640 , c ) 1632 , d ) 1688 , e ) 6386
d
subtract(2743, divide(multiply(multiply(3, 5), 2743), add(multiply(3, 5), multiply(8, 3))))
multiply(n2,n4)|multiply(n1,n2)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|subtract(n0,#4)|
gain
tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many t minutes it takes john to reach the station ?
since the question states β€œ after 15 minutes ” , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john ’ s speed i...
a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min
c
multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10)
divide(n0,const_60)|subtract(n0,n1)|multiply(#0,#1)|multiply(#2,const_60)|divide(#3,n1)|divide(#4,const_2)|subtract(#5,const_0_25)|multiply(n1,#6)|
physics
when positive integer n is divided by positive integer j , the remainder is 18 . if n / j = 134.08 , what is value of j ?
"when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 18 so , 134.08 = 134 + 18 / j so , j = 225 ans e"
a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 225
e
divide(18, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2))))
multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)|
general
a can finish a work in 18 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ?
b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 18 * 1 / 3 ) = 6 days . correct option : c
a ) 5 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these
c
divide(multiply(multiply(divide(const_1, 15), 10), 18), const_2)
divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)
physics
the difference of two numbers is 1375 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ?
"let the smaller number be x . then larger number = ( x + 1375 ) . x + 1375 = 6 x + 15 5 x = 1360 x = 272 smaller number = 270 . answer b"
a ) 240 , b ) 272 , c ) 295 , d ) 360 , e ) 252
b
divide(add(1375, 15), subtract(6, const_1))
add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|
general
the population of a town is 7600 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ?
"formula : ( after = 100 denominator ago = 100 numerator ) 7600 Γ£ β€” 100 / 90 Γ£ β€” 100 / 90 = 9382 a )"
a ) 9400 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000
a
subtract(subtract(7600, multiply(7600, divide(10, const_100))), multiply(subtract(7600, multiply(7600, divide(10, const_100))), divide(10, const_100)))
divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)|
gain
the average of first 10 prime numbers which are odd is ?
explanation : sum of first 10 prime no . which are odd = 158 average = 158 / 10 = 15.8 answer : d
a ) 22.2 , b ) 22.1 , c ) 15.3 , d ) 15.8 , e ) 14.6
d
divide(add(add(add(add(add(add(add(add(add(const_3, add(add(const_1, const_3), const_1)), add(add(const_1, const_3), const_3)), add(10, const_1)), add(10, const_3)), add(10, add(add(const_1, const_3), const_3))), add(add(10, add(add(const_1, const_3), add(add(const_1, const_3), const_1))), add(const_1, const_3))), add(...
add(const_1,const_3)|add(n0,const_1)|add(n0,const_3)|add(#0,const_1)|add(#0,const_3)|add(#3,const_3)|add(n0,#4)|add(#0,#3)|add(#5,#4)|add(n0,#7)|add(#7,#4)|add(#8,#1)|add(#9,#0)|add(n0,#9)|add(#10,#1)|add(#11,#2)|add(n0,#14)|add(#15,#6)|add(#17,#12)|add(#18,#13)|add(#19,#16)|add(#20,#9)|divide(#21,n0)
general
on increasing the price of t . v . sets by 60 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ?
explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 Γ— 100 ) = rs . 10000 new sale value = rs . ( 160 Γ— 80 ) = rs . 12800 increase % = 2800 / 10000 Γ— 100 = 28 % answer : d
a ) 4 , b ) 5 , c ) 56 , d ) 28 , e ) 7
d
subtract(add(const_100, 60), add(divide(multiply(add(const_100, 60), 20), const_100), const_100))
add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#2,const_100)|subtract(#0,#3)
gain
a meal cost $ 35.50 adn there was no tax . if the tip was more than 2 pc but less than 12 pc of the price , then the total amount paid should be :
"2 % ( 35.5 ) = 0.71 12 % ( 35.5 ) = 4.26 total amount could have been 35.5 + 0.71 and 35.5 + 4.26 = > could have been between 36.21 and 39.76 = > approximately between 36 and 40 answer is e ."
a ) 40 - 43 , b ) 39 - 42 , c ) 39 - 42 , d ) 38 - 41 , e ) 36 - 40
e
add(divide(multiply(35.50, 2), const_100), 35.50)
multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|
general
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1200 employees , then the no . of female employees who are computer literate ?
solution : total employees , = 1200 female employees , 60 % of 1200 . = ( 60 * 1200 ) / 100 = 720 . then male employees , = 480 50 % of male are computer literate , = 240 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1200 ) / 100 = 744 computer literate . thus , female computer liter...
a ) 504 , b ) 674 , c ) 672 , d ) 960 , e ) none
a
multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1200)
divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)
gain
a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 40 pints of perfect pink paint ?
"3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 40 pints will require 9 / 17 * 40 = 21 pints . . c"
a ) 9 , b ) 16 , c ) 21 , d ) 25 1 / 3 , e ) 28 1 / 2
c
multiply(40, divide(multiply(3, 2), multiply(4, 3)))
multiply(n0,n2)|multiply(n0,n1)|divide(#0,#1)|multiply(n4,#2)|
general
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 72 kmph and 90 kmph . in how much time will the trains cross each other ?
"relative speed = ( 72 + 90 ) * 5 / 18 = 9 * 5 = 45 mps . the time required = d / s = ( 100 + 100 + 200 ) / 45 = 400 / 45 = 80 / 9 sec . answer : b"
a ) 80 / 8 , b ) 80 / 9 , c ) 80 / 5 , d ) 80 / 3 , e ) 80 / 1
b
divide(100, multiply(add(72, 90), const_0_2778))
add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
a fruit seller sells mangoes at the rate of rs . 13 per kg and thereby loses 25 % . at what price per kg , he should have sold them to make a profit of 15 % ?
"solution 75 : 13 = 115 : x x = ( 13 Γ£ β€” 115 / 75 ) = rs . 19.93 hence , s . p per kg = rs . 19.93 answer b"
a ) rs . 11.81 , b ) rs . 19.93 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none
b
divide(multiply(13, add(const_100, 15)), subtract(const_100, 25))
add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)|
gain
mahesh can do a piece of work in 40 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ?
"work done by mahesh in 60 days = 20 * 1 / 40 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 1 / 2 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 2 = 60 days answer is e"
a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 60
e
divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 40))), 30))
divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)|
physics
a store ’ s selling price of $ 2240 for a certain printer would yield a profit of 40 percent of the store ’ s cost for the printer . what selling price would yield a profit of 50 percent of the printer ’ s cost ?
"1.4 x = 2240 x = 2240 / 1.4 so , 1.5 x = 2240 * 1.5 / 1.4 = 2400 answer : - e"
a ) $ 2800 , b ) $ 2464 , c ) $ 2650 , d ) $ 2732 , e ) $ 2400
e
multiply(add(const_1, divide(40, const_100)), original_price_before_gain(40, 2240))
divide(n1,const_100)|original_price_before_gain(n1,n0)|add(#0,const_1)|multiply(#2,#1)|
gain
a contractor undertakes to built a walls in 50 days . he employs 50 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?
"50 men complete 0.4 work in 25 days . applying the work rule , m 1 Γ— d 1 Γ— w 2 = m 2 Γ— d 2 Γ— w 1 we have , 50 Γ— 25 Γ— 0.6 = m 2 Γ— 25 Γ— 0.4 or m 2 = 50 Γ— 25 Γ— 0.6 / 25 Γ— 0.4 = 75 men answera"
a ) 25 , b ) 30 , c ) 35 , d ) 20 , e ) none of these
a
divide(multiply(50, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10))
divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0)|
physics
convert 500 miles into meters ?
"1 mile = 1609.34 meters 500 mile = 500 * 1609.34 = 804670 meters answer is c"
a ) 784596 , b ) 845796 , c ) 804670 , d ) 784596 , e ) 864520
c
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 500), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 12 , and k < m < r < s < t . if t is 20 , what is the greatest possible value of the median of the 5 integers ?
"we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 12 which means that in effect , all numbers are equal to 12 . if the largest number is 20 , it is 8 more than 12 . we need r to be max...
a ) 16 , b ) 18 , c ) 19 , d ) 20 , e ) 22
d
subtract(divide(subtract(multiply(12, 5), 20), const_2), const_2)
multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|subtract(#2,const_2)|
general
a work as fast as b . if b can complete a work in 36 days independently , the number of days in which a and b can together finish the work in ?
"ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 18 b ' s 1 day work = 1 / 36 a + b 1 day work = 1 / 18 + 1 / 36 = 3 / 36 = 1 / 12 a and b can finish the work in 12 days answer is a"
a ) 12 days , b ) 3 days , c ) 4 days , d ) 5 days , e ) 6 days
a
inverse(add(inverse(36), multiply(const_2, inverse(36))))
inverse(n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|
physics
a car covers a distance of 200 km in 2 hours 40 minutes , whereas a jeep covers the same distance in 2 hours . what is the ratio of t
a 3 : 4
a ) 3 : 4 , b ) 4 : 3 , c ) 4 : 5 , d ) 5 : 4 , e ) 8 : 7
a
divide(multiply(const_60, 2), add(multiply(const_60, 2), 40))
multiply(n1,const_60)|add(n2,#0)|divide(#0,#1)
physics
the average of 13 numbers is 60 . average of the first 7 of them is 54 and that of the last 7 is 61 . find the 8 th number ?
"sum of all the 13 numbers = 13 * 60 = 780 sum of the first 7 of them = 7 * 54 = 378 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 378 - 780 = 25 . answer : a"
a ) 25 , b ) 83 , c ) 45 , d ) 53 , e ) 64
a
subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(54, 7)))
multiply(n5,n6)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3)|
general
a man is 22 years older than his son . in three years , his age will be thrice the age of his son . the present age of this son is
"let ' s son age is x , then father age is x + 22 . = > 3 ( x + 3 ) = ( x + 22 + 3 ) = > 3 x + 9 = x + 25 = > 2 x = 16 years = 8 years answer : d"
a ) 8 years , b ) 2 years , c ) 7 years , d ) 8 years , e ) 5 years
d
divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))
multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|
general
for each color copy , print shop x charges $ 1.20 and print shop y charges $ 1.70 . how much greater is the charge for 70 color copies at print shop y than at print shop x ?
the difference in the two prices is $ 1.70 - $ 1.20 = $ 0.50 for each color copy . each color copy will cost an extra $ 0.50 at print shop y . 70 * $ 0.50 = $ 35 the answer is e .
a ) $ 27 , b ) $ 29 , c ) $ 31 , d ) $ 33 , e ) $ 35
e
subtract(multiply(1.7, 70), multiply(1.2, 70))
multiply(n1,n2)|multiply(n0,n2)|subtract(#0,#1)
general
pipe a fills a tank of capacity 850 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute a...
"in one cycle they fill 40 + 30 - 20 = 50 liters 850 = 50 * n = > n = 17 here n = number of cycles . total time = 17 * 3 = 51 as in one cycle there are 3 minutes . thus 51 minutes answer : b"
a ) 42 minutes , b ) 51 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these
b
multiply(divide(850, subtract(add(40, 30), 20)), const_3)
add(n1,n2)|subtract(#0,n3)|divide(n0,#1)|multiply(#2,const_3)|
physics
a picnic attracts 240 persons . there are 80 more men than women , and 80 more adults than children . how many men are at this picnic ?
"adult + children = 240 let , children = y then , adult = y + 80 i . e . y + ( y + 80 ) = 240 i . e . y = 80 i . e . adult = 80 + 80 = 160 adults include only men and women i . e . men + women = 160 let women , w = x then men , m = x + 80 i . e . x + ( x + 20 ) = 2 x + 80 = 160 i . e . x = 40 i . e . men , m = 40 + 80 ...
a ) 240 , b ) 75 , c ) 110 , d ) 120 , e ) 200
d
add(divide(subtract(add(divide(subtract(240, 80), const_2), 80), 80), const_2), 80)
subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|subtract(#2,n1)|divide(#3,const_2)|add(n1,#4)|
general
tap a runs 3 liters of water per minute into a bucket that has a total volume of 36 liters . tap b fills a third of the bucket in 20 mins . working together how long will it take both taps to fill the bucket ?
pipe a has a rate of 3 liters per minute which equals a rate of 180 liters per hour . pipe b has a rate of 36 liters per hour . 180 + 36 ( a + b ) = together they have a rate of 2000 liters per hour . 36 ( the bucket ) / 216 = 1 / 6 = 10 minutes answer - a
a ) 10 minutes , b ) 1.5 hours , c ) 9 hours and 12 minutes , d ) 15 hours , e ) 1 hour and 48 minutes .
a
inverse(add(divide(const_1, divide(36, 3)), divide(const_1, multiply(20, 3))))
divide(n1,n0)|multiply(n0,n2)|divide(const_1,#0)|divide(const_1,#1)|add(#2,#3)|inverse(#4)
physics
what is the smallest integer that is multiple of 5 , 7,11
"it is the lcm of 5 , 7 and 11 which is 385 . the answer is c ."
a ) a ) 70 , b ) b ) 35 , c ) c ) 385 , d ) d ) 280 , e ) e ) 140
c
add(const_3, const_4)
add(const_3,const_4)|
general
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 22000 . what is the share of c ?
"explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 Γ— ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) Γ— 22000 ] = rs . 6000 answer : option c"
a ) rs . 2250 , b ) rs . 3375 , c ) rs . 6000 , d ) rs . 5625 , e ) none of these
c
multiply(22000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1)))
divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)|
gain
if the radius of a circle is decreased 40 % , what happens to the area ?
"area of square = pi * radius ^ 2 new radius = 0.6 * old radius so new area = ( 0.6 ) ^ 2 old area = > 4 / 25 of old area = > 36 % old area ans : e"
a ) 10 % decrease , b ) 20 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 64 % decrease
e
subtract(const_100, multiply(power(divide(40, const_100), const_2), const_100))
divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)|
geometry
what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 9 - sided polygon ) ?
"from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 6 = 1 / 3 . the answer is b ."
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 9
b
divide(9, divide(multiply(9, subtract(9, const_3)), const_2))
subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)|divide(n0,#2)|
probability
if 10 gallons of grape juice are added to 30 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ?
"official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 30 Γ— 0.10 = 3 gallons of grape juice . 30 Γ— 0.90 = 27 gallons of other components . if we add 10 gallons of grape juice , we will end up with 13 gallons of grape juice and 36 gallons of other components , and we will...
a ) 14 % , b ) 25 % , c ) 28 % , d ) 32.5 % , e ) 50 %
d
multiply(divide(add(multiply(divide(10, const_100), 30), 10), add(30, 10)), const_100)
add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|add(n0,#2)|divide(#3,#0)|multiply(#4,const_100)|
general
if ( 10 ^ 4 * 3.456789 ) ^ 12 is written as a single term , how many digits would be to the right of the decimal place ?
"3.456789 ^ 12 has 6 * 12 = 72 decimal places . 10 ^ 48 moves the decimal place to the right 48 places . ( 10 ^ 4 * 3.456789 ) ^ 12 has 72 - 48 = 24 digits after the decimal point . the answer is c ."
a ) 16 , b ) 20 , c ) 24 , d ) 28 , e ) 32
c
multiply(12, const_2)
multiply(n3,const_2)|
general
convert 150 inches into centimeter ?
"1 inch = 2.54 cm 150 inches = 110 * 2.54 = 381 cm answer is b"
a ) 312 cm , b ) 381 cm , c ) 350 cm , d ) 310 cm , e ) 354 cm
b
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 150), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
one day eesha started 30 min late from home and reached her office 50 min late while driving 25 % slower than her usual speed . how much time in min does eesha usually take to reach her office from home ?
we know that speed is inversely proportional to time while she drives 25 % slower means she drove at 3434 ( s ) we know that d = s x t when speed became 3434 ( s ) then time taken should be 4343 ( t ) i . e , she has taken 4343 ( t ) - t extra to cover the distance . extra time = t 3 t 3 = 20 min ( as 20 min late due t...
a ) 22 , b ) 60 , c ) 88 , d ) 26 , e ) 18
b
divide(multiply(subtract(const_1, divide(25, const_100)), subtract(50, 30)), divide(25, const_100))
divide(n2,const_100)|subtract(n1,n0)|subtract(const_1,#0)|multiply(#2,#1)|divide(#3,#0)
physics
the sum of three consecutive even numbers is 24 . find the middle number of the three ?
"three consecutive even numbers ( 2 p - 2 ) , 2 p , ( 2 p + 2 ) . ( 2 p - 2 ) + 2 p + ( 2 p + 2 ) = 24 6 p = 24 = > p = 4 . the middle number is : 4 p = 16 . answer : b"
a ) 14 , b ) 16 , c ) 18 , d ) 24 , e ) 34
b
add(add(power(add(add(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), power(add...
subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)|
physics
two trains of equal length , running with the speeds of 50 and 40 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ?
"rs = 50 - 40 = 10 * 5 / 18 = 50 / 18 t = 50 d = 50 * 50 / 18 = 1250 / 9 rs = 50 + 40 = 90 * 5 / 18 = 25 t = 1250 / 9 * 1 / 25 = 5.55 sec answer : a"
a ) 5.55 sec , b ) 4.55 sec , c ) 5.95 sec , d ) 6.55 sec , e ) 9.55 sec
a
multiply(multiply(multiply(const_0_2778, subtract(50, 40)), 50), inverse(multiply(const_0_2778, add(50, 40))))
add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)|
physics
daniel went to a shop and bought things worth rs . 40 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ?
"total cost of the items he purchased = rs . 40 given that out of this rs . 40 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % ∴ ( 40 βˆ’ 30 / 100 βˆ’ x ) 6 / 100 = 30 / 100 β‡’ 6 ( 40 βˆ’ 0.3 βˆ’ x ) = 30 β‡’ ( 40 βˆ’ 0.3 βˆ’ x ) = 5 β‡’ x = 4...
a ) a ) 34.7 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.5 , e ) e ) 22
a
subtract(subtract(40, divide(30, const_100)), divide(30, 6))
divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)|
gain
if 6 men can colour 48 m long cloth in 2 days , then 2 men can colour 36 m long cloth in
"the length of cloth painted by one man in one day = 48 / 6 Γ— 2 = 4 m no . of days required to paint 36 m cloth by 6 men = 36 / 4 Γ— 2 = 4.5 day . d"
a ) 1 day , b ) 2.1 days , c ) 3 days , d ) 4.5 days , e ) 5 days
d
divide(36, multiply(divide(48, multiply(6, 2)), 2))
multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)|
physics
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts five hours after a . if b and k overtake a at the same instant , how many hours after a did k start ?
the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 5 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ...
a ) 3 , b ) 4.5 , c ) 4 , d ) d ) 10 , e ) e ) 5
d
subtract(40, 30)
subtract(n1,n0)
physics
the decimal 0.1 is how many times greater than the decimal ( 0.001 ) ^ 4 ?
"0.1 = 10 ^ - 1 ( 0.001 ) ^ 4 = ( 10 ^ - 3 ) ^ 4 = 10 ^ - 12 10 ^ 11 * 10 ^ - 12 = 10 ^ - 1 the answer is d ."
a ) 10 ^ 8 , b ) 10 ^ 9 , c ) 10 ^ 10 , d ) 10 ^ 11 , e ) 10 ^ 12
d
divide(0.1, power(0.001, 4))
power(n1,n2)|divide(n0,#0)|
general
in what time will a train 400 meters long cross an electric pole , if its speed is 144 km / hr
"explanation : first convert speed into m / sec speed = 144 * ( 5 / 18 ) = 40 m / sec time = distance / speed = 400 / 40 = 10 seconds answer : c"
a ) 5 seconds , b ) 4.5 seconds , c ) 10 seconds , d ) 2.5 seconds , e ) none of these
c
divide(400, multiply(144, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
p software has coding line 5 % more than n , n software has coding line 5 / 2 more than m . m software has 108 lines of coding . find p lines .
"m s / w has 108 line of code n s / w has = 108 + 108 * 5 / 2 = 378 line of code p s / w 5 % more n ' code 378 + 18.9 = 396.9 or 397 line of code answer : c"
a ) 106 , b ) 107 , c ) 397 , d ) 109 , e ) 110
c
add(108, divide(108, 2))
divide(n3,n2)|add(n3,#0)|
general
a certain list consists of 21 different numbers . if n is in the list and n is 2 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ?
"series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 2 * x hence , a 1 + a 2 + . . + a 20 + n = 22 x so , the fraction asked = 2 x / 22 x = 1 / 11 answer c"
a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 11 , d ) 4 / 21 , e ) 5 / 21
c
divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 2), 21), const_4.0), const_2), 2), const_3))
divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,const_4.0)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)|
general
the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 20,000 is made ?
"the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 20,000 is made = 96,000 + 20,000 = rs . 1 , 16,000 answer : d"
a ) 328897 , b ) 120000 , c ) 877888 , d ) 116000 , e ) 188871
d
add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100)))
add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)|
gain
mary works in a restaurant a maximum of 60 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ?
mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 40 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 40 ) = $ 400 . the total amount is $ 160 + $ 400 = $ 560 answer c 560 .
a ) 300 , b ) 420 , c ) 560 , d ) 320 , e ) 400
c
add(multiply(60, 8), multiply(subtract(60, 20), multiply(8, divide(25, const_100))))
divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4)
gain
how many seconds will a 600 m long train take to cross a man walking with a speed of 4 km / hr in the direction of the moving train if the speed of the train is 64 km / hr ?
"speed of train relative to man = 64 - 4 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 600 * 3 / 50 = 36 sec . answer : b"
a ) 30 , b ) 36 , c ) 50 , d ) 44 , e ) 60
b
divide(600, multiply(subtract(64, 4), const_0_2778))
subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 40 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ?
"same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 20...
a ) 4 : 5 , b ) 9 : 25 , c ) 5 : 4 , d ) 25 : 9 , e ) can not be determined from the information provided
c
divide(circumface(divide(50, const_2)), circumface(divide(40, const_2)))
divide(n1,const_2)|divide(n0,const_2)|circumface(#0)|circumface(#1)|divide(#2,#3)|
physics