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a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter q of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) q = 7 * 3 = 21 . c" | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
a ) s . 650 , b ) s . 690 , c ) s . 738 , d ) s . 700 , e ) s . 718 | c | subtract(855, divide(multiply(subtract(854, 855), 3), 4)) | a sum of money at simple interest amounts to rs . 855 in 3 years and to rs . 854 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 3 years = rs . ( 39 x 3 ) = rs . 117 . principal = rs . ( 855 - 117 ) = rs . 738 . answer : option c" | a = 854 - 855
b = a * 3
c = b / 4
d = 855 - c
|
a ) 90 , b ) 103 , c ) 130 , d ) 115 , e ) 120 | c | add(100, divide(subtract(multiply(multiply(multiply(add(const_4, const_3), multiply(const_3, const_2)), const_100), multiply(add(const_4, const_1), const_2)), multiply(multiply(multiply(const_100, const_100), divide(3, const_100)), 100)), multiply(multiply(const_100, const_100), divide(4, const_100)))) | a salesperson received a commission of 3 percent of the sale price for each of the first 100 machines that she sold and 4 percent of the sale price for each machine that she sold after the first 100 . if the sale price of each machine was $ 10,000 and the salesperson received a $ 42,000 commission , how many machines did she sell ? | first 100 machines = 3 % commission = 0.03 * 100 * 10000 = 30000 commission from sale of next machines = 42000 - 30000 = 12000 so 30 more machines . . total = 130 machines imo c . . . | a = 4 + 3
b = 3 * 2
c = a * b
d = c * 100
e = 4 + 1
f = e * 2
g = d * f
h = 100 * 100
i = 3 / 100
j = h * i
k = j * 100
l = g - k
m = 100 * 100
n = 4 / 100
o = m * n
p = l / o
q = 100 + p
|
a ) 1 / 2 , b ) 2 / 3 , c ) 2 / 1 , d ) 3 / 2 , e ) 4 / 3 | c | divide(subtract(divide(1, 2), divide(1, 6)), divide(1, 6)) | in a class of students , 1 / 2 of the number of girls is equal to 1 / 6 of the total number of students . what is the ratio of boys to girls in the class ? | "( 1 / 2 ) g = ( 1 / 6 ) ( b + g ) 6 g = 2 b + 2 g 4 g = 2 b b / g = 2 / 1 . the answer is c ." | a = 1 / 2
b = 1 / 6
c = a - b
d = 1 / 6
e = c / d
|
a ) 33.33 % , b ) 25 % , c ) 81.81 % , d ) 66.66 % , e ) none of these | c | multiply(subtract(divide(const_100, subtract(const_100, 45)), const_1), const_100) | if x is less than y by 45 % then y exceed x by : | "using formula ( x / ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 25 % ) = > 45 / ( 100 - 45 ) * 100 = 81.81 % answer : c" | a = 100 - 45
b = 100 / a
c = b - 1
d = c * 100
|
a ) 11 , b ) 17 , c ) 21 , d ) 29 , e ) 33 | d | add(multiply(divide(77, 2), 2), 2) | if the least common multiple of two prime numbers x and y is 77 , where x > y , then the value of 2 x + y is | "lcm ( x , y ) = 77 and both x an y are prime . the only values of x and y can be 11 and 7 ( x = 11 and y = 7 ) 2 x + y = 2 * 11 + 7 = 29 correct option : d" | a = 77 / 2
b = a * 2
c = b + 2
|
a ) 18 / 50 , b ) 16 / 50 , c ) 7 / 20 , d ) 19 / 50 , e ) none | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.35), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 0.35 in to a vulgar fraction ? | "answer 0.35 = 35 / 100 = 7 / 20 correct option : c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 0
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 98 , b ) 99 , c ) 100 , d ) 101 , e ) 97 | a | add(divide(subtract(1000, 10), 10), const_1) | how many multiples of 10 are there between 10 and 1000 , exclusive ? | "98 multiples of 5 between 1 and 36 exclusive . from 10 * 2 upto 10 * 99 , ( 2 , 3,4 , . . . , 99 ) . hence , 98 multiples ! correct option is a" | a = 1000 - 10
b = a / 10
c = b + 1
|
a ) 17 , b ) 20 , c ) 25 , d ) 30 , e ) 15 | a | divide(add(add(6, const_4), subtract(29, const_4)), const_2) | find the average of all the numbers between 6 and 29 which are divisible by 5 ? | average = ( 10 + 15 + 20 + 25 ) 4 = 70 / 4 = 17 answer is a | a = 6 + 4
b = 29 - 4
c = a + b
d = c / 2
|
a ) 320 , b ) 360 , c ) 400 , d ) 440 , e ) 480 | b | multiply(divide(multiply(30, const_2), subtract(36, 30)), 36) | train a leaves the station traveling at 30 miles per hour . two hours later train Π² leaves the same station traveling in the same direction at 36 miles per hour . how many miles from the station was train a overtaken by train b ? | "after two hours , train a is ahead by 60 miles . train b can catch up at a rate of 6 miles per hour . the time to catch up is 60 / 6 = 10 hours . in 10 hours , train a travels another 30 * 10 = 300 miles for a total of 360 miles . the answer is b ." | a = 30 * 2
b = 36 - 30
c = a / b
d = c * 36
|
a ) 1 hr , b ) 2 hr , c ) 6 hr , d ) 8 hr , e ) 7 hr | a | divide(600, divide(multiply(200, 30), const_10)) | 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 x ( 2 x + 1 ) = 3 2 x 2 + x - 3 = 0 ( 2 x + 3 ) ( x - 1 ) = 0 x = 1 hr . answer : a | a = 200 * 30
b = a / 10
c = 600 / b
|
a ) 30 % , b ) 40 % , c ) 85 % , d ) 19 % , e ) 29 % | a | divide(const_100, add(const_1, 9)) | solve the quickfire maths brain teaser β 9 % = ? | "a 30 % β 9 % = > β 9 / β 100 = > 3 / 10 = > 30 / 100 = > 30 %" | a = 1 + 9
b = 100 / a
|
a ) 1711 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(789, multiply(const_4, const_100)), add(multiply(subtract(789, multiply(const_4, const_100)), 2), const_1)) | balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 789 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did . how many balls made up the triangle ? | "as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 789 ) ( x - 8 ) ^ 2 = t + 789 - - - - - ( b ) now the hardest part of the question will be to solve these 2 equations and this looks like time consuming but the easy way will be plug and play . also , we ' ve to find a value of t ( from 5 optiosn given below ) which can make a square of a a number . one we know this , it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - 1711 + 789 = 2500 = 50 ^ 2 = ( x - 8 ) ^ 2 hence , x = 58 . cross check by putting in eq ( a ) = x ( x + 1 ) / 2 = t = > 58 * 59 / 2 = 1711 hence , answer is a ." | a = 4 * 100
b = 789 - a
c = 4 * 100
d = 789 - c
e = d * 2
f = e + 1
g = b * f
|
a ) 40 , b ) 26 , c ) 60 , d ) 97 , e ) 12 | c | add(30, divide(subtract(36, 26), 20)) | the average of 20 numbers is calculated as 30 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "20 * 30 + 36 β 26 = 160 / 10 = 60 answer : 60 answer is c" | a = 36 - 26
b = a / 20
c = 30 + b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(2, reminder(4, 5)) | when n is divided by 12 , the remainder is 2 . what is the remainder when 4 n is divided by 5 ? | "let n = 2 ( leaves a remainder of 2 when divided by 15 ) 4 n = 4 ( 2 ) = 8 , which leaves a remainder of 3 when divided by 5 . answer c" | a = 2 - reminder
|
a ) 198 , b ) 297 , c ) 320 , d ) 402 , e ) 405 | c | add(68, multiply(subtract(64, const_1), 4)) | s is a set of 64 consecutive multiples of 4 . if the smallest number in s is 68 , then the greatest number in s is | last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 64 consecutive multiples of 4 . if the smallest number in s is 68 , then the greatest number in s is first term = 68 ; total terms = 64 ; difference = 4 68 + ( 63 ) 4 = 320 ans c | a = 64 - 1
b = a * 4
c = 68 + b
|
a ) 3 / 8 , b ) 13 / 36 , c ) 17 / 36 , d ) 19 / 36 , e ) 23 / 36 | b | divide(add(add(add(6, 1), 3), 3), multiply(6, 6)) | the β s - number β of a number x is defined as the ones digit of 2 ^ x . antony rolls a die with 6 sides labeled with the integers from 1 to 6 , each of which has an equal probability of landing face - up . he then takes 3 ^ s , where s is the s - number of the result of his die roll , and plots 3 ^ s on a number line as the point a . finally , he repeats this entire process , this time plotting the result as the point b . what is the probability that the distance between a and b is greater than the value of b ? | if you calculate 3 ^ s for 1 st roll , all 6 results will be 9 , 81 , 6561 , 729 , 9 , 81 . this result is the same for 2 nd roll . 9 , 81 , 6561 , 729 , 9 , 81 . about distance : if the first result is 9 and the second is also 9 , the distance is 9 - 9 = 0 which is smaller than 9 . if the first result is 9 and the second is 81 , the distance is 81 - 9 = 72 which is also smaller than b which has the value of 81 . if the first result is 81 and the second is 9 , the distance will be greater than b . distance 81 - 9 = 72 > 9 . on the first roll , the probability of getting result 9 is 2 / 6 . in this case no other alternative values for second roll which would make the distance greater than b . so probability is 0 . so next estimations are : probability of getting 81 on the first roll ( 2 / 6 ) * probability of getting 9 on the second roll ( 2 / 6 ) = 1 / 9 probability of getting 729 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 on the second roll ( 4 / 6 ) = 1 / 9 probability of getting 6561 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 , 729 on the first roll ( 5 / 6 ) = 5 / 36 all together : 1 / 9 + 1 / 9 + 5 / 36 = 13 / 36 = b | a = 6 + 1
b = a + 3
c = b + 3
d = 6 * 6
e = c / d
|
a ) 25 % , b ) 60 % , c ) 80 % , d ) 55 % , e ) 74 % | c | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | two numbers are respectively 20 % and 50 % more than a third number . what percentage is the first of the second ? | "required answer = ( 100 + x ) / ( 100 + y ) * 100 = 120 * 100 / 150 = 80 % answer is c" | a = 20 + 100
b = 50 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 1.5 hrs , b ) 2.5 hrs , c ) 3.5 hrs , d ) 1.2 hrs , e ) 2.9 hrs | b | add(divide(reminder(subtract(multiply(10, const_60), multiply(multiply(10, divide(3, 4)), const_60)), const_60), const_100), floor(divide(subtract(multiply(10, const_60), multiply(multiply(10, divide(3, 4)), const_60)), const_60))) | calculate how much time could be saved if the train would run at its usual speed , given that it ran at 3 / 4 of its own speed and got to its destination in 10 hours ? | "new speed = 3 / 4 of usual speed new time = 3 / 4 of usual time 3 / 4 of usual time = 10 hrs usual time = 10 * 3 / 4 = 7.5 hrs time saved = 10 - 7.5 = 2.5 hrs answer is b" | a = 10 * const_60
b = 3 / 4
c = 10 * b
d = c * const_60
e = a - d
f = reminder / (
g = f + 100
|
a ) 12 , b ) 23 , c ) 34 , d ) 48 , e ) 52 | d | multiply(divide(subtract(1200, divide(480, const_2)), 40), const_2) | angelina walked 1200 meters from her home to the grocery at a constant speed . she then walked 480 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | "let the speed be x . . . so time taken from home to grocery = 1200 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 1200 / x - 240 / x = 40 . . 960 / x = 40 . . x = 24 m / secs . . so grocery to gym = 2 * 24 = 48 m / s . . . d" | a = 480 / 2
b = 1200 - a
c = b / 40
d = c * 2
|
a ) 4 / 9 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 7 , e ) 6 / 9 | d | divide(divide(5, 2), add(divide(5, 2), const_1)) | a waiter ' s salary consists of his salary and tips . during one week his tips were 5 / 2 of his salary . what fraction of his income came from tips ? | income = salary ( s ) + tips = s + s * 5 / 2 = s * 7 / 2 tips = s * 7 / 2 fraction of his income came from tips = ( s * 5 / 2 ) / ( s * 7 / 2 ) = 5 / 7 answer : d | a = 5 / 2
b = 5 / 2
c = b + 1
d = a / c
|
a ) 12 , b ) 5 , c ) 6 , d ) 16 , e ) 18 | a | add(5, 7) | if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 7 is divided by 41 ? | "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . when k = 0 , 1,2 etc x + 7 = 82 k + 5 + 7 = 82 k + 12 or x + 7 = 12,94 , 176 etc . k = 0 , 1,2 when divided by 41 , we will get the remainder 12 . a" | a = 5 + 7
|
a ) 1 , b ) 3 , c ) 4 , d ) 7 , e ) 10 | c | multiply(divide(divide(multiply(50, 55), const_100), 80), const_12) | a reduction of 50 % in the price of apples would enable a man to obtain 80 more for rs . 55 , what is reduced price per dozen ? | c 55 * ( 50 / 100 ) = 28 - - - 64 ? - - - 12 = > rs . 4 | a = 50 * 55
b = a / 100
c = b / 80
d = c * 12
|
a ) 2928 , b ) 3638 , c ) 4098 , d ) 4418 , e ) 5378 | c | subtract(subtract(subtract(multiply(258, const_10), const_100), 3), 3) | what is the next number : 3 , 18 , 258 , __ | "4 ^ 0 + 2 = 3 4 ^ 2 + 2 = 18 4 ^ 4 + 2 = 258 4 ^ 6 + 2 = 4098 the answer is c ." | a = 258 * 10
b = a - 100
c = b - 3
d = c - 3
|
a ) $ 0.00 , b ) $ 1.00 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.25 | e | subtract(multiply(85, divide(add(const_100, 25), const_100)), divide(85, divide(subtract(const_100, 15), const_100))) | cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 85 after it went on sale for 15 % off , what is the difference between the original price and the final price ? | "0.85 * { original price } = $ 85 - - > { original price } = $ 100 . { final price } = $ 85 * 1.25 = $ 106.25 . the difference = $ 106.25 - $ 100 = $ 6.25 . answer : e ." | a = 100 + 25
b = a / 100
c = 85 * b
d = 100 - 15
e = d / 100
f = 85 / e
g = c - f
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) can not be determined | e | multiply(4000, const_1) | how many books each of volume 200 meter cube can be packed into a crate of volume 4000 meter cube ? | "gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 4000 m 3 can be 400 * 10 or 40 * 100 or just 4000 * 1 we do n ' t know , so we cant calculate answer : e" | a = 4000 * 1
|
a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31 | e | divide(6, add(divide(20, const_100), 6)) | a committee is reviewing a total of 20 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction q of the selected films are in color ? | it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose q = 60 / 62 = 30 / 31 of the selected films . answer : e . | a = 20 / 100
b = a + 6
c = 6 / b
|
a ) 80 kmph , b ) 60 kmph , c ) 72 kmph , d ) 90 kmph , e ) none of these | c | divide(add(125, 270), add(2.5, 3)) | a train travels 125 km in 2.5 hours and 270 km in 3 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 125 + 270 = 395 km thus , total speed = 5.5 hrs or , average speed = 5 = 395 / 5.5 or , 72 kmph . answer : c" | a = 125 + 270
b = 2 + 5
c = a / b
|
a ) z / 2 , b ) 5 z / 12 , c ) z / 3 , d ) 3 z / 5 , e ) z / 9 | b | divide(subtract(divide(multiply(4, const_100), const_2), const_2), add(divide(multiply(4, const_100), const_2), const_2)) | if 4 x = 6 y = z , what is x + y , in terms of z ? | "4 x = 6 y = z x = z / 4 and y = z / 6 x + y = z / 4 + z / 6 = 5 z / 12 answer is b" | a = 4 * 100
b = a / 2
c = b - 2
d = 4 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) [ 22 ] , b ) [ 44 ] , c ) [ 75 ] , d ) [ 88 ] , e ) [ 90 ] | c | multiply(add(divide(7, 2), 1), add(multiply(2, 8), 1)) | when x is even , [ x ] = x / 2 + 1 , when x is odd [ x ] = 2 x + 1 then [ 7 ] * [ 8 ] = ? | "[ 7 ] * [ 8 ] = ( 7 * 2 + 1 ) ( 8 / 2 + 1 ) = [ 75 ] . ans - c" | a = 7 / 2
b = a + 1
c = 2 * 8
d = c + 1
e = b * d
|
a ) 40 % , b ) 50 % , c ) 55 % , d ) 80 % , e ) 90 % | b | multiply(subtract(divide(divide(divide(add(const_100, 20), const_100), subtract(const_1, divide(subtract(45, 40), 45))), divide(subtract(const_100, 10), const_100)), const_1), const_100) | a dealer offers a cash discount of 10 % and still makes a profit of 20 % when he further allows 45 articles to be sold at the cost price of 40 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | "given cash discount - 10 % profit - 20 % items sold - 45 price sold at = list price of 40 assume list price = $ 10 total invoice = $ 400 - 10 % cash discount = $ 360 let cost price of 45 items be x so total cost = 45 * x given the shopkeeper had a profit of 20 % 45 * x * 120 / 100 = 360 or x = $ 20 / 3 = $ 6 + ( 2 / 3 ) which means his products were listed at $ 10 which is a 50 % markup over $ 6 + ( 2 / 3 ) answer b" | a = 100 + 20
b = a / 100
c = 45 - 40
d = c / 45
e = 1 - d
f = b / e
g = 100 - 10
h = g / 100
i = f / h
j = i - 1
k = j * 100
|
a ) 171 , b ) 855 , c ) 870 , d ) 788 , e ) 684 | b | divide(multiply(57, 3), divide(20, const_100)) | at the end of year x , automobile installment credit accounted for 20 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 57 billion of credit , or 1 / 3 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at that time ? | "system of equations a = ( 20 / 100 ) c ( 1 / 3 ) a = 57 - - > a = 171 substitution 171 = ( 20100 ) c c = ( 100 / 20 ) 171 - calculate 171 / 20 * 100 the correct answer b . the answer is 855 ." | a = 57 * 3
b = 20 / 100
c = a / b
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a ) 22 , b ) 37 , c ) 30 , d ) 27 , e ) 40 | e | divide(add(32, 48), const_2) | a man can row upstream at 32 kmph and downstream at 48 kmph , and then find the speed of the man in still water ? | "us = 32 ds = 48 m = ( 48 + 32 ) / 2 = 40 answer : e" | a = 32 + 48
b = a / 2
|
a ) 71 % , b ) 51 % , c ) 91 % , d ) 81 % , e ) 61 % | c | subtract(power(10, const_2), power(3, const_2)) | the radius of the two circular fields is in the ratio 3 : 10 the area of the first field is what percent less than the area of the second ? | "r = 3 Γ― β¬ r 2 = 9 r = 10 Γ― β¬ r 2 = 100 100 Γ― β¬ Γ’ β¬ β 9 Γ― β¬ = 91 100 - - - - ? = > 91 % . answer : c" | a = 10 ** 2
b = 3 ** 2
c = a - b
|
a ) 6 , b ) 9 , c ) 12 , d ) 15 , e ) 18 | b | divide(multiply(24, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 30), multiply(divide(add(24, 60), 4), 3)), multiply(divide(8, 2), 3))) | 3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 24 days . if 60 more cows and 30 more buffaloes are brought in , how many days will the fodder last ? | "2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 24 days . 60 cows + 30 buffaloes = 30 oxen + 20 oxen = 50 oxen the new total is equal to 80 oxen instead of 30 oxen . 30 * 24 / 80 oxen = 9 days the answer is b ." | a = 24 / 4
b = a * 3
c = 15 + b
d = 8 / 2
e = d * 3
f = c + e
g = 24 * f
h = 15 + 30
i = 24 + 60
j = i / 4
k = j * 3
l = h + k
m = 8 / 2
n = m * 3
o = l + n
p = g / o
|
a ) 1.6 , b ) 1.8 , c ) 3.5 , d ) 1.5 , e ) 2.5 | e | divide(subtract(divide(64, 8), divide(24, 8)), const_2) | an woman swims downstream 64 km and upstream 24 km taking 8 hours each time ; what is the speed of the current ? | "64 - - - 8 ds = 8 ? - - - - 1 24 - - - - 8 us = 3 ? - - - - 1 s = ? s = ( 8 - 3 ) / 2 = 2.5 answer : e" | a = 64 / 8
b = 24 / 8
c = a - b
d = c / 2
|
a ) 61 , b ) 63 , c ) 65 , d ) 67 , e ) 69 | b | divide(multiply(30, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 30 % of 210 ? | "x / 210 = 30 / 100 100 x = 210 x 30 x = ( 210 x 30 ) / 100 x = 6300 / 100 = 63 answer = b = 63" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 30 * n
p = o / 100
|
a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 27 | e | divide(50, subtract(1, divide(const_1, 1))) | company x provides bottled water to its employees in 1 liter bottles , $ 2 each . at least how many 1 l bottles must company x buy monthly , so that new contract with $ 50 fixed monthly rate and $ 1 for 10 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 1 liter bottles be x , so the no . of 10 l bottles will be x / 10 ( to equate the vol . ) since the total cost will be equal , 2 x = 50 + x / 10 so x = 26.31 or 27 . answer is ( e ) . | a = 1 / 1
b = 1 - a
c = 50 / b
|
a ) 7 : 3 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 5 : 6 | e | divide(subtract(15.8, 15.3), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.3 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.3 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.3 ) < = > k = 0.5 / 0.6 = 5 / 6 . required ratio = 5 / 6 : 1 = 5 : 6 . answer : e" | a = 15 - 8
b = 16 - 4
c = a / b
|
a ) 22 , b ) 28 , c ) 120 , d ) 287 , e ) 278 | c | divide(multiply(subtract(multiply(12.5, const_2), const_1), multiply(12.5, const_2)), add(const_4, const_1)) | a train can travel 50 % faster than a car . both start from point a at the same time and reach point b 75 kms away from a at the same time . on the way , however , the train lost about 12.5 minutes while stopping at the stations . what is the speed of the car ? | let speed of the car = x kmphthen speed of the train = ( 100 + 50 ) 100 x = 150100 x = 32 x kmphtime taken by the car to travel from a to b = 75 x hourstime taken by the train to travel from a to b = 75 ( 32 x ) + 12.560 hourssince both start from a at the same time and reach point b at the same time 75 x = 75 ( 32 x ) + 12.56025 x = 12.560 x = 25 Γ 6012.5 = 2 Γ 60 = 120 answer : c | a = 12 * 5
b = a - 1
c = 12 * 5
d = b * c
e = 4 + 1
f = d / e
|
a ) 120 , b ) 772 , c ) 288 , d ) 160 , e ) 2848 | d | divide(multiply(32, 10), const_2) | if the sides of a triangle are 39 cm , 32 cm and 10 cm , what is its area ? | "the triangle with sides 39 cm , 32 cm and 10 cm is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 32 * 10 = 160 cm 2 answer : d" | a = 32 * 10
b = a / 2
|
a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3 | c | power(3, add(200, const_1)) | the remainder when q = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 . | q = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 - 1 ) / 2 divided by 13 3 ^ 201 - 1 / 26 wkt , 3 ^ 3 = 27 = 26 + 1 { ( 26 + 1 ) ^ 67 - 1 } / 26 { 1 - 1 } / 26 = > 0 . c | a = 200 + 1
b = 3 ** a
|
a ) 3500 , b ) 5600 , c ) 2200 , d ) 2400 , e ) none | e | multiply(divide(5600, 5), 2) | p and q invested in a shop . the profits were divided in the ratio of 5 : 2 respectively . if p invested rs . 5600 , the amount invested by q is : | suppose q invested rs . y . then , 5600 / y = 5 / 2 or y = [ 5600 x 2 / 5 ] = 2240 answer e | a = 5600 / 5
b = a * 2
|
a ) 8.5 gallons , b ) 5 gallons , c ) 7 gallons , d ) 5.5 gallons , e ) 6 gallons | e | divide(120, 20) | a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 120 kilometers ? | "each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 120 kilometers ? 120 / 20 = 4.5 * 1 gallon = 6 gallons correct answer e" | a = 120 / 20
|
a ) 6 , b ) 9 , c ) 24 , d ) 30 , e ) 35 | c | divide(multiply(multiply(6, 9), 12), multiply(multiply(divide(6, const_2), divide(6, const_2)), divide(6, const_2))) | a cuboidal block 6 cm x 9 cm x 12 cm is cut up into an exact number of equal cubes . the least possible number of equal cubes will be | explanation : volume of block = ( 6 x 9 x 12 ) cm 3 = 648 cm 3 side of largest cube = h . c . f of 6 , 912 = 3 cm volume of the cube = ( 3 x 3 x 3 ) = 27 cm 3 number of cubes = ( 648 / 27 ) = 24 answer : c | a = 6 * 9
b = a * 12
c = 6 / 2
d = 6 / 2
e = c * d
f = 6 / 2
g = e * f
h = b / g
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(multiply(25, 2), add(multiply(subtract(subtract(25, add(add(multiply(12, 1), 4), 2)), 1), 3), add(multiply(12, 1), multiply(4, 2)))) | in a class of 25 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 4 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the total number of books the students borrowed is 25 * 2 = 50 . the students who borrowed zero , one , or two books borrowed 12 * 1 + 4 * 2 = 20 books . the 7 students who borrowed at least three books borrowed 50 - 20 = 30 books . if 6 of these students borrowed exactly 3 books , then the maximum that one student could have borrowed is 30 - 18 = 12 books . the answer is e ." | a = 25 * 2
b = 12 * 1
c = b + 4
d = c + 2
e = 25 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 4 * 2
j = h + i
k = g + j
l = a - k
|
a ) 91.5 , b ) 92 , c ) 108 , d ) 94 , e ) 95 | c | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 180 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 108 cm answer : option c" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 27 miles , b ) 25 miles , c ) 24 miles , d ) 28.3 miles , e ) 23 miles | d | sqrt(add(power(add(6, 18), const_2), power(15, const_2))) | beginning in town a , biker bob rides his bike 30 miles west , 6 miles north , 15 miles east , and then 18 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 18 + 6 = 24 m other being the base ie distance traveled west - distance traveled eat = 30 - 15 = 15 m now this third side or the distance between town a and town b = 24 ^ 2 + 15 ^ 2 = sq root 801 = 28.3 m answer : d" | a = 6 + 18
b = a ** 2
c = 15 ** 2
d = b + c
e = math.sqrt(d)
|
a ) 16 , b ) 28 , c ) 32 , d ) 44 , e ) 48 | b | divide(subtract(15, multiply(divide(20, const_100), 68)), subtract(divide(25, const_100), divide(20, const_100))) | a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 68 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | "lets say x ounces of p is mixed with q . = > 68 - x ounces of q is present in the mixture ( as the total = 68 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 68 - x ) = 15 = > x = 40 = > 68 - 40 = 28 ounces of q is present in the mixture . answer is b ." | a = 20 / 100
b = a * 68
c = 15 - b
d = 25 / 100
e = 20 / 100
f = d - e
g = c / f
|
a ) 10 sec , b ) 15 sec , c ) 18 sec , d ) 20 sec , e ) 22 sec | b | divide(add(200, 300), multiply(add(70, 50), const_0_2778)) | two trains 200 m and 300 m long run at the speed of 70 kmph and 50 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 70 + 50 = 120 kmph * 5 / 18 = 100 / 3 m / s distance covered in crossing each other = 200 + 300 = 500 m required time = 500 * 3 / 100 = 15 sec answer is b" | a = 200 + 300
b = 70 + 50
c = b * const_0_2778
d = a / c
|
a ) 297 cm 2 , b ) 384 cm 2 , c ) 120 cm 2 , d ) 267 cm 2 , e ) 186 cm 2 | c | multiply(12, 10) | find the area of a parallelogram with base 12 cm and height 10 cm ? | "area of a parallelogram = base * height = 12 * 10 = 120 cm 2 answer : c" | a = 12 * 10
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | floor(divide(multiply(const_1, const_3600), lcm(lcm(7, 8), lcm(lcm(3, 4), lcm(5, 6))))) | 6 lights begin to flash together at 12 : 00 noon and flash respectively at intervals of 3 , 4 , 5 , 6 , 7 , and 8 seconds . not including the flash at 12 : 00 noon , how many more times will all 6 lights flash together before 1 : 00 pm ( i . e . one hour later ) ? | the least common multiple is 2 * 2 * 2 * 3 * 5 * 7 = 840 . 3600 seconds / 840 = 4 + remainder . the answer is a . | a = 1 * 3600
b = math.lcm(7, 8)
c = math.lcm(3, 4)
d = math.lcm(5, 6)
e = math.lcm(c, d)
f = math.lcm(b, e)
g = a / f
h = math.floor(g)
|
a ) 124 , b ) 241 , c ) 394 , d ) 421 , e ) 842 | b | subtract(add(add(multiply(multiply(4, const_1), const_100), multiply(multiply(4, 2), const_10)), divide(multiply(4, const_1), 2)), add(add(multiply(divide(multiply(4, const_1), 2), const_100), multiply(multiply(4, const_1), const_10)), const_1)) | a is the hundreds digit of the 3 digit integer x , b is the tens digit of x , and c is the units digit of x . 2 a = b = 4 c , and a > 0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . | ratio of a : b : c = 2 : 4 : 1 two possible greatest single digit values for b are 8 and 4 if b is 8 , then x = 482 if b is 4 , then x = 241 difference = 482 - 241 = 241 b is the answer | a = 4 * 1
b = a * 100
c = 4 * 2
d = c * 10
e = b + d
f = 4 * 1
g = f / 2
h = e + g
i = 4 * 1
j = i / 2
k = j * 100
l = 4 * 1
m = l * 10
n = k + m
o = n + 1
p = h - o
|
a ) 1 , b ) 2 , c ) - 2 , d ) 4 , e ) 5 | c | divide(1, 2) | find the slope of the line perpendicular to the line y = ( 1 / 2 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 2 . if m is the slope of the line perpendicular to the given line , then m Γ ( 1 / 2 ) = - 1 solve for m m = - 2 correct answer c ) - 2" | a = 1 / 2
|
a ) 500 , b ) 277 , c ) 222 , d ) 448 , e ) 111 | d | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 25)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 25 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 125 = rs . 448 answer : d" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 25
g = e / f
|
a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 2 | b | divide(power(1, 2), power(const_3.0, 2)) | the triplicate ratio of 1 : 2 is ? | "13 : 23 = 1 : 8 answer : b" | a = 1 ** 2
b = 3 ** 0
c = a / b
|
a ) 2588 , b ) 2474 , c ) 2254 , d ) 2519 , e ) 8725 | d | subtract(multiply(multiply(lcm(lcm(9, 10), 7), 2), 2), const_1) | a number with an interesting property : when i divide it by 2 , the remainder is 1 . when i divide it by 3 , the remainder is 2 . when i divide it by 4 , the remainder is 3 . when i divide it by 5 , the remainder is 4 . when i divide it by 6 , the remainder is 5 . when i divide it by 7 , the remainder is 6 . when i divide it by 8 , the remainder is 7 . when i divide it by 9 , the remainder is 8 . when i divide it by 10 , the remainder is 9 . it ' s not a small number , but it ' s not really big , either . when i looked for a smaller number with this property i could n ' t find one . can you find it ? | d 2519 the number has to end in 9 . looked brute force for small numbers . 59 and 119 were promising , but no cigar . then looked for agreement among 39 + multiples of 40 , 69 + multiples of 70 and 89 + multiples of 90 smallest one was 2519 . | a = math.lcm(9, 10)
b = math.lcm(a, 7)
c = b * 2
d = c * 2
e = d - 1
|
a ) 100 , b ) 120 , c ) 140 , d ) 130 , e ) 115 | c | divide(multiply(multiply(21, 5), divide(const_100, 5)), subtract(multiply(divide(const_100, 5), 4), 65)) | 65 % of a number is 21 less than 4 / 5 th of that number . what is the number ? | let the number be x then , 4 x / 5 - ( 65 % of x ) = 21 4 x / 5 - 65 x / 100 = 21 x = 140 answer is c | a = 21 * 5
b = 100 / 5
c = a * b
d = 100 / 5
e = d * 4
f = e - 65
g = c / f
|
a ) 12,295 , b ) 13,000 , c ) 14,300 , d ) 14,410 , e ) 28,820 | d | divide(multiply(45, 175), const_4) | what is the sum of the integers 45 through 175 inclusive ? | "sum of n consecutive positive integers = n ( n + 1 ) / 2 . . in one case n = 44 and other 175 . . . subtract the sum to get the answer sum of first 175 + ive numbers = 175 * 176 / 2 = 15400 . . sum of first 45 + i ' ve numbers = 45 * 44 / 2 = 990 . . answer = 15400 - 990 = 14410 answer : d" | a = 45 * 175
b = a / 4
|
a ) 120 , b ) 126 , c ) 132 , d ) 138 , e ) 144 | b | divide(multiply(2, const_360), add(2, 3)) | the ratio of males to females in a class is 2 : 3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one - half of the males and one - fourth of the females in the class ? | "1 / 2 * 2 / 5 + 1 / 4 * 3 / 5 = 4 / 20 + 3 / 20 = 7 / 20 the number of degrees is 7 / 20 * 360 = 126 degrees the answer is b ." | a = 2 * 360
b = 2 + 3
c = a / b
|
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | c | add(divide(subtract(subtract(const_100, 60), multiply(divide(40, const_100), subtract(const_100, 60))), const_2), multiply(divide(40, const_100), subtract(const_100, 60))) | a vendor sells 60 percent of the apples he had and throws away 40 percent of the remainder . the next day , the vendor sells 50 percent of the remaining apples and throws away the rest . in total , what percent of his apples does the vendor throw away ? | "let x be the original number of apples . on day one , the vendor throws away ( 0.4 ) ( 0.4 ) x = 0.16 x . the remaining apples are ( 0.6 ) ( 0.4 ) x = 0.24 x . on day two , the vendor throws away ( 0.5 ) ( 0.24 ) x = 0.12 x . the vendor throws away a total of 0.16 x + 0.12 x = 0.28 x . the vendor throws away 28 percent of the apples . the answer is c ." | a = 100 - 60
b = 40 / 100
c = 100 - 60
d = b * c
e = a - d
f = e / 2
g = 40 / 100
h = 100 - 60
i = g * h
j = f + i
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | d | multiply(const_100, add(subtract(subtract(subtract(const_1, divide(80, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(80, const_100)))), multiply(subtract(subtract(const_1, divide(80, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(80, const_100)))), divide(80, const_100))), multiply(divide(50, const_100), subtract(const_1, divide(80, const_100))))) | a vendor sells 80 percent of the pears he had and throws away 50 percent of the remainder . the next day , the vendor sells 80 percent of the remaining pears and throws away the rest . in total , what percent of his pears does the vendor throw away ? | "let x be the original number of pears . on day one , the vendor throws away ( 0.5 ) ( 0.2 ) x = 0.1 x . the remaining pears are ( 0.5 ) ( 0.2 ) x = 0.1 x . on day two , the vendor throws away ( 0.2 ) ( 0.1 ) x = 0.02 x . the vendor throws away a total of 0.1 x + 0.02 x = 0.12 x . the vendor throws away 12 percent of the pears . the answer is d ." | a = 80 / 100
b = 1 - a
c = 50 / 100
d = 80 / 100
e = 1 - d
f = c * e
g = b - f
h = 80 / 100
i = 1 - h
j = 50 / 100
k = 80 / 100
l = 1 - k
m = j * l
n = i - m
o = 80 / 100
p = n * o
q = g - p
r = 50 / 100
s = 80 / 100
t = 1 - s
u = r * t
v = q + u
w = 100 * v
|
a ) 117 kmph , b ) 178 kmph , c ) 126 kmph , d ) 118 kmph , e ) 119 kmph | c | divide(divide(700, const_1000), divide(20, const_3600)) | a train 700 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 700 / 20 s = 35 m / sec speed = 35 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 126 kmph answer : c" | a = 700 / 1000
b = 20 / 3600
c = a / b
|
a ) 75 kg , b ) 65 kg , c ) 55 kg , d ) 85 kg , e ) 93 kg | e | add(65, multiply(8, 3.5)) | the average weight of 8 persons increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 3.5 ) kg = 28 kg . weight of new person = ( 65 + 28 ) kg = 93 kg . answer : e" | a = 8 * 3
b = 65 + a
|
a ) 29.32 sec , b ) 42.51 sec , c ) 39.25 sec , d ) 45.61 sec , e ) none of these | d | divide(52, 1.14) | an industrial loom weaves 1.14 meters of cloth every second . approximately , how much time will it take to weave 52 meters of cloth ? | explanation : given loom weaves 1.14 mts of cloth in one second then 52 mts of cloth can be weaved by loom in , 1.14 - - - - - 1 52.0 - - - - - - ? = > 52 / 1.14 = 45.61 sec answer is d | a = 52 / 1
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a ) 229 , b ) 787 , c ) 144 , d ) 148 , e ) 121 | d | multiply(add(const_1, const_4), 37) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 37 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 37 4 / x = 1 / 37 = > x = 148 min . answer : d" | a = 1 + 4
b = a * 37
|
a ) 9.5 , b ) 6.4 , c ) 2.3 , d ) 8.9 , e ) 12.5 | e | divide(100, multiply(divide(divide(20, 10), 10), 20)) | if 10 machines can produce 20 units in 10 hours , how long would it take 40 machines to produce 100 units ? | "10 machines would produce 100 units in 50 hours . increasing the amount of machines by 4 would mean dividing 50 hours by 4 . 50 / 4 = 12.5 answer : e" | a = 20 / 10
b = a / 10
c = b * 20
d = 100 / c
|
a ) 22 , b ) 56 , c ) 78 , d ) 200 , e ) 175 | d | divide(14, subtract(134.07, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 14 . if n / j = 134.07 , what is value of j ? | "1 ) we know that decimal part of decimal quotient = { remainder / divisor } so 0.07 , the decimal part of the decimal quotient , must equal the remainder , 14 , divided by the divisor j . 0.07 = 14 / j 0.07 * j = 14 j = 14 / 0.07 = 1400 / 7 = 200 so j = 200 , answer = d ." | a = 4 * 10
b = a + 2
c = 100 + b
d = 134 - 7
e = 14 / d
|
a ) 1 : 4 , b ) 1 : 3 , c ) 5 : 2 , d ) 3 : 2 , e ) 1 : 2 | e | divide(2, 4) | if m : n is 2 : 9 and n : p is 9 : 4 then m : p is equal to | the two ratios given are having the same number 9 for n in both the ratios . hence - m : n = 2 : 9 n : p = 9 : 4 = > m : p = 2 : 4 = > 1 : 2 answer e | a = 2 / 4
|
a ) 220 , b ) 1314 , c ) 1326 , d ) 7920 , e ) none | a | sqrt(add(power(sqrt(subtract(6, multiply(const_2, 1320))), const_2), multiply(const_4, 1320))) | the product of two numbers is 1320 and their h . c . f is 6 . the l . c . m of the numbers is | "solution l . c . m = product of numbers / h . c . f βΉ = βΊ 1320 / 6 βΉ = βΊ 220 . answer a" | a = 2 * 1320
b = 6 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 1320
f = d + e
g = math.sqrt(f)
|
a ) 15.6 , b ) 16.0 , c ) 18.4 , d ) 18.7 , e ) 19.1 | c | add(2.80, multiply(subtract(divide(8, divide(1, 5)), 1), 0.40)) | a certain taxi company charges $ 2.80 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? | "a certain taxi company charges $ 2.80 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? a . 15.60 b . 16.00 c . 17.50 d . 18.70 e . 19.10 1 / 5 miles = 0.2 miles . the cost of 8 miles long ride would be $ 2.80 for the first 0.2 miles plus ( 8 - 0.2 ) / 0.2 * 0.4 = $ 2.8 + $ 15.6 = $ 18.4 . answer : c ." | a = 1 / 5
b = 8 / a
c = b - 1
d = c * 0
e = 2 + 80
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | add(4, const_1) | the average of first five multiples of 4 is | "solution average = 4 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 60 / 5 . = 12 answer d" | a = 4 + 1
|
a ) 90 , b ) 82 , c ) 80 , d ) 45 , e ) 36 | e | add(divide(subtract(divide(200, const_2), const_1), 3), 3) | of the 3 - digit integers greater than 200 , how many have two digits that are equal to each other and the remaining digit different from the other two ? | let the three digit number be represented as x y z . there are 3 cases : case i . [ x = y ] z is not equal to xy : xxzoryyz x can be either 7 , 8 or 9 , so digit at x can be chosen in 3 ways after x is chosen , y can be chosen in 1 way after xy are chosen , z can be chosen in 9 ways thus , possible no of digits = ( 3 ways ) * ( 1 way ) * ( 9 ways ) = 27 . . . . ( 1 ) [ example numbers : 774,779 , 882,993 etc ] case ii . [ x = z ] y is not equal to xz : xyxorzyz x can be either 7 , 8 or 9 , so digit at x can be chosen in 3 ways after x is chosen , z can be chosen in 1 way after xz are chosen , z can be chosen in 9 ways thus , possible no of digits = ( 3 ways ) * ( 9 ways ) * ( 1 way ) = 27 . . . . ( 2 ) [ example numbers : 747,797 , 828,939 etc ] case iii . [ y = z ] x is not equal to yz : xyyorxzz x can be either 7 , 8 or 9 , so digit at x can be chosen in 3 ways after x is chosen , y can be chosen in 9 ways after y is chosen , z can have 1 way thus , possible no of digits = ( 3 ways ) * ( 1 way ) * ( 9 ways ) = 27 . . . . ( 3 ) [ example numbers : 744,799 , 822,933 etc ] therefore , total numbers of possible digits [ sum of case ( 1 ) , ( 2 ) ( 3 ) above ] = 27 + 27 + 27 - 1 = 80 one digit is subtracted from total number of possible digits to eliminate one possibility of xyz = 700 to satisfy the condition that digit > 200 . answer : ( e ) | a = 200 / 2
b = a - 1
c = b / 3
d = c + 3
|
['a ) 19 m', 'b ) 17 m', 'c ) 15 m', 'd ) 16 m', 'e ) 20 m'] | b | sqrt(subtract(power(divide(46, const_2), const_2), multiply(120, const_2))) | a rectangular mat has an area of 120 sq . metres and perimeter of 46 m . the length of its diagonal is : | rectangular area = l Γ b = 120 and perimeter = 2 ( l + b ) = 46 l + b = 23 ( l - b ) power 2 - 4 lb = ( 23 ) power 2 - 4 Γ 120 = ( 529 - 480 ) = 49 = l - b = 7 l + b = 23 , l - b = 7 , we get l = 15 , b = 8 diagonal = β 152 + 82 β 225 + 64 = > β 289 = 17 m answer is b . | a = 46 / 2
b = a ** 2
c = 120 * 2
d = b - c
e = math.sqrt(d)
|
a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many r acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn r = 4 * 94 = 376 correct option - c" | a = 1000 * 1
b = 10 * 3
c = b + 4
d = a + c
e = 5 + 2
f = e + 4
g = d / f
h = g * 4
|
a ) 18 % , b ) 72 % , c ) 25 % , d ) 12 % , e ) 52 % | c | subtract(const_100, divide(multiply(add(const_100, 25), subtract(const_100, 40)), const_100)) | the tax on a commodity is diminished by 40 % but its consumption is increased by 25 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 60 * 125 = 7500 10000 - - - - - - - 2500 100 - - - - - - - ? = 25 % answer : c" | a = 100 + 25
b = 100 - 40
c = a * b
d = c / 100
e = 100 - d
|
a ) 4096 , b ) 4098 , c ) 5000 , d ) 5096 , e ) can not be determined | a | power(add(divide(subtract(127, 1), const_2), 1), const_2) | a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 127 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ? | "explanatory answer the shape of the area used for growing cabbage has remained a square in both the years . let the side of the square area used for growing cabbage this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbage last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbage grown has increased by 127 , the area would have increased by 127 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 127 ( x + y ) ( x - y ) = 127 . 127 is a prime number and hence it will have only two factors . i . e . , 127 and 1 . therefore , 127 can be expressed as product of 2 numbers in only way = 127 * 1 i . e . , ( x + y ) ( x - y ) = 127 * 1 so , ( x + y ) should be 127 and ( x - y ) should be 1 . solving the two equations we get x = 64 and y = 63 . therefore , number of cabbage produced this year = x 2 = 642 = 4096 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice b , c and d . as 4098,5000 and 5096 are not the square of any number . check choice a : if this year ' s produce is 4096 , last year ' s produce would have been 4096 - 127 = 3969 3969 is the square of 63 . so , 4096 is the answer . choice a" | a = 127 - 1
b = a / 2
c = b + 1
d = c ** 2
|
a ) 7 , b ) 16 , c ) 17 , d ) 20 , e ) 21 | a | subtract(20, 8) | a snail , climbing a 20 feet high wall , climbs up 8 feet on the first day but slides down 4 feet on the second . it climbs 8 feet on the third day and slides down again 4 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ? | "total transaction in two days = 8 - 4 = 4 feet in 7 days it will climb 20 feet thus reaching the top therefore , total no of days required = 7 a" | a = 20 - 8
|
a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many of q applicants majored in political science and had a gpa higher than 3.0 ? | total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) q = 20 - 15 = 5 answer : a | a = 10 + 15
b = 40 - a
c = 20 - b
|
a ) 2480 , b ) 3490 , c ) 6785 , d ) 8245 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1270) | the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1270 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | "you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 15 + 15 ) ^ 2 now we can expand each square ; they are all in the ( x + y ) ^ 2 = x ^ 2 + 2 xy + y ^ 2 pattern . = ( 15 ^ 2 + 2 * 15 + 1 ^ 2 ) + ( 15 ^ 2 + 4 * 15 + 2 ^ 2 ) + ( 15 ^ 2 + 6 * 15 + 3 ^ 2 ) + . . . + ( 15 ^ 2 + 30 * 15 + 15 ^ 2 ) now we have fifteen 15 ^ 2 terms , so adding these gives 15 * 15 ^ 2 = 15 ^ 3 = 3375 . we also have the entire sum 1 ^ 2 + 2 ^ 2 + . . . + 15 ^ 2 , which we know is equal to 1240 . finally adding the middle terms , we have : 2 * 15 + 4 * 15 + 6 * 15 + . . . + 30 * 15 = 15 ( 2 + 4 + 6 + . . . . + 30 ) = 15 * 2 * ( 1 + 2 + 3 + . . . + 15 ) = 15 * 2 * 8 * 15 = 3600 so the sum must be 3375 + 1240 + 3600 = 8215 alternatively , we can use a different factoring pattern . we want to find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 . well if we subtract 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 from this , the answer will be 1240 less than what we want to find . so if we can find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 - ( 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 ) then we can add 1240 to get the answer . now grouping the terms above to get differences of squares , we have = ( 30 ^ 2 - 15 ^ 2 ) + ( 29 ^ 2 - 14 ^ 2 ) + . . . + ( 16 ^ 2 - 1 ^ 2 ) and factoring each of these using x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) , we have = 45 * 15 + 43 * 15 + 41 * 15 + . . . + 17 * 15 = 15 ( 45 + 43 + 41 + . . . + 17 ) in brackets we have an equally spaced sum with fifteen terms , which we can evaluate using the familiar formula . so the above equals 15 * 15 * 62 / 2 = 6975 and adding back the 1270 , we get the answer of 8245 . ( ans d )" | a = 30 + 1
b = 30 * a
c = 2 * 30
d = c + 1
e = b * d
f = 3 + 3
g = e / f
h = g - 1270
|
a ) 100 % , b ) 160 % , c ) 180 % , d ) 200 % , e ) 220 % | c | divide(multiply(30, subtract(const_100, 10)), 15) | last week vartan spent 15 percent of his wages on recreation . this week , his wages are 10 percent less than last week ΚΌ s wages and he spent 30 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week ? | say vartan ' s wages last week were $ 100 , so he spent 0.15 * 100 = $ 15 on recreation ; this week ' s wages is 0.9 * 100 = $ 90 , so he spends 0.3 * 90 = $ 27 on recreation ; 27 / 15 = 1.8 , hence the amount he spends on recreation this week is 180 % of the amount he spent on recreation last week : 15 * 1.8 = 27 . answer : c . | a = 100 - 10
b = 30 * a
c = b / 15
|
a ) 29963 sq m , b ) 28937 sq m , c ) 43162 sq m , d ) 27688 sq m , e ) 17248 sq m | e | subtract(circle_area(divide(704, multiply(const_2, const_pi))), circle_area(divide(528, multiply(const_2, const_pi)))) | the circumferences of two circles are 528 meters and 704 meters . find the difference between the areas of the larger and the smaller circles ? | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 β s = 528 and 2 β l = 704 s = 528 / 2 β and l = 704 / 2 β difference between the areas = β l ^ 2 - β s ^ 2 = β { 264 ^ 2 / β ^ 2 - 352 ^ 2 / β ^ 2 } = 264 ^ 2 / β - 352 ^ 2 / β = ( 264 - 352 ) ( 264 + 352 ) / β = ( 88 ) ( 616 ) / ( 22 / 7 ) = 17248 sq m answer : e" | a = 2 * math.pi
b = 704 / a
c = circle_area - (
|
a ) 5 : 3 , b ) 4 : 3 , c ) 7 : 16 , d ) 12 : 17 , e ) 11 : 13 | a | divide(divide(25, const_100), divide(15, const_100)) | if 15 % of a is the same as 25 % of b , then a : b is : | "expl : 15 % of a i = 25 % of b = 15 a / 100 = 25 b / 100 = 5 / 3 = 5 : 3 answer : a" | a = 25 / 100
b = 15 / 100
c = a / b
|
a ) 23 , b ) 27 , c ) 12 , d ) 43 , e ) 37 | c | multiply(add(3, 3), const_2) | the ages of old and young total 48 . old is twice as old as young was when old was half as old as young will be when young is 3 times as old was when old was 3 times as old as young . how old is old ? | explanation : from the options itself , we can see that option c old = 30 young = 18 30 + 18 = 48 by reducing this years only by one before 6 years old = 24 ( half of young ) young = 12 ( twice of old ) answer : c | a = 3 + 3
b = a * 2
|
a ) 3000 , b ) 1230 , c ) 1750 , d ) 5600 , e ) 3400 | c | multiply(divide(add(const_100, 25), const_100), divide(add(1520, 1280), const_2)) | the percentage profit earned by selling an article for rs . 1520 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ? | "c . p . be rs . x . then , ( 1520 - x ) / x * 100 = ( x - 1280 ) / x * 100 1520 - x = x - 1280 2 x = 2800 = > x = 1400 required s . p . = 125 % of rs . 1400 = 125 / 100 * 1400 = rs . 1750 . answer c" | a = 100 + 25
b = a / 100
c = 1520 + 1280
d = c / 2
e = b * d
|
a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares . | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 2.5), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 2.5 hectares in ares | "2.5 hectares in ares 1 hectare = 100 ares therefore , 2.5 hectares = 2.5 Γ 100 ares = 250 ares . answer - e" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 2
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6 | e | divide(const_4, const_10) | if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 9 } . what is the probability that the median of the set will increase but the range still remains the same ? | "we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the same . the median of the set is ( 3 + 4 ) / 2 = 3.5 , thus the number selected must be 4 or greater . for the range to remain the same , the number must be between 2 and 9 inclusive . to satisfy both conditions , the number selected must be 4 , 5 , 6 , 7 , 8 , or 9 . the probability is 6 / 10 = 0.6 the answer is e ." | a = 4 / 10
|
a ) 60 % , b ) 72 % , c ) 88 % , d ) 95 % , e ) 110 % | c | multiply(divide(divide(subtract(multiply(32, const_100), multiply(20, const_100)), subtract(const_100, 32)), 20), const_100) | keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 32 % of the total number of books in the library . if prior to the purchase , only 20 % of the books in keats library were biographies , by what percent has the number of biographies in the library increased ? | let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.32 ( 5 b + x ) = b + x 0.6 b = 0.68 x x = 0.88 b the answer is c . | a = 32 * 100
b = 20 * 100
c = a - b
d = 100 - 32
e = c / d
f = e / 20
g = f * 100
|
a ) 46 , b ) 49 , c ) 40 , d ) 38 , e ) 35 | c | divide(add(multiply(40, subtract(subtract(20, 2), 3)), multiply(2, 100)), 20) | in a class of 20 students in an examination in mathematics 2 students scored 100 marks each , 3 get zero each and the average of the rest was 40 . what is the average of the whole class ? | total marks obtained by a class of 20 students = 2 * 100 + 3 * 0 + 15 * 40 = 200 + 600 = 800 : average marks of whole class = 800 / 20 = 40 answer is c . | a = 20 - 2
b = a - 3
c = 40 * b
d = 2 * 100
e = c + d
f = e / 20
|
a ) 33 , b ) 32 , c ) 68 , d ) 36 , e ) 38 | b | divide(multiply(40, 20), const_100) | the cost of an article is decreased by 20 % . if the original cost is $ 40 , find the decrease cost . | "original cost = $ 40 decrease in it = 20 % of $ 40 = 20 / 100 Γ£ β 40 = 800 / 100 = $ 8 therefore , decrease cost = $ 40 - $ 8 = $ 32 answer : b" | a = 40 * 20
b = a / 100
|
a ) 90 min , b ) 48 min , c ) 54 min , d ) 120 min , e ) 130 min | b | inverse(subtract(add(inverse(24), inverse(48)), inverse(12))) | two pipes a and b can separately fill a cistern in 24 minutes and 48 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 12 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 12 - ( 1 / 24 + 1 / 48 ) = - 1 / 48 third pipe can empty in 48 minutes answer : b" | a = 1/(24)
b = 1/(48)
c = a + b
d = 1/(12)
e = c - d
f = 1/(e)
|
a ) 7 days , b ) 8.6 days , c ) 13 days , d ) 5.6 days , e ) 10 days | b | inverse(add(inverse(15), inverse(20))) | a can do a piece of work in 20 days , and b can do it in 15 days . how long will they take if both work together ? | a + b can do the work in 20 * 15 / 20 + 15 = 20 * 15 / 35 = 8.6 days answer is b | a = 1/(15)
b = 1/(20)
c = a + b
d = 1/(c)
|
a ) 120 cm 2 , b ) 156 cm 2 , c ) 178 cm 2 , d ) 189 cm 2 , e ) 176 cm 2 | b | divide(multiply(26, 12), const_2) | if the sides of a triangle are 28 cm , 26 cm and 12 cm , what is its area ? | "the triangle with sides 28 cm , 26 cm and 12 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 26 * 12 = 156 cm 2 answer : b" | a = 26 * 12
b = a / 2
|
a ) 22 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 10 kg , e ) none of these | d | subtract(multiply(add(29, const_1), 27.4), multiply(29, 28)) | the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.4 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.4 weight of the new student = ( 30 * 27.4 β 29 * 28 ) = 822 - 812 = 10 answer : d" | a = 29 + 1
b = a * 27
c = 29 * 28
d = b - c
|
a ) 33 , b ) 32 , c ) 60 , d ) data inadequate , e ) none of these | a | divide(multiply(13.5, 22), 9) | 22 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 22 Γ£ β 13.5 = 297 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 297 Γ’ Β β 9 = 33 answer a" | a = 13 * 5
b = a / 9
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | a | add(const_2, gcd(60, 17)) | find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number ? | let the number be x . then , x + 17 = 60 x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 option a | a = math.gcd(60, 17)
b = 2 + a
|
a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | e | subtract(17.5, multiply(2, 2.5)) | the arithmetic mean and standard deviation of a certain normal distribution are 17.5 and 2.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 17.5 two standard deviations is 2.5 + 2.5 = 5.0 there could be two calues for this . mean + two standard deviations = 22.5 mean - two standard deviations = 12.5 answer choice has 12.5 and so e is the answer ." | a = 2 * 2
b = 17 - 5
|
a ) 0.31 , b ) 0.55 , c ) 0.65 , d ) 0.75 , e ) 0.85 | a | divide(subtract(130, 90), 130) | in a group of 130 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ? | "number of people whose age is less than 20 is given by 130 - 90 = 40 probability p that a person selected at random from the group is less than 20 is given by 40 / 130 = 0.31 correct answer a" | a = 130 - 90
b = a / 130
|
a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase e in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase e = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a = 385 - 320
b = a / 320
c = b * 100
|
a ) 16,500 , b ) 16,600 , c ) 16,700 , d ) 16,800 , e ) 16,900 | b | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(7, const_1)), const_1)) | a city with a population of 126,160 is to be divided into 7 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 126,160 = p + 6 ( 1.1 ) p 7.6 p = 126,160 p = 16,600 the answer is b ." | a = 1000 * 100
b = 3600 - 100
c = b - 1000
d = a - c
e = d / 1000
f = 10 / 100
g = 1 + f
h = 7 - 1
i = g * h
j = i + 1
k = e / j
|
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