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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(subtract(13, 5), const_1) | if 5 < x < 8 < y < 13 , then what is the greatest possible positive integer difference of x and y ? | "let x = 5.1 and y = 12.1 greatest possible difference = 12.1 - 5.1 = 7 answer e" | a = 13 - 5
b = a - 1
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a ) 20 , b ) 8 , c ) 12 , d ) 14 , e ) 16 | a | divide(500, subtract(26, const_1)) | in a garden , 26 trees are planted at equal distances along a yard 500 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "26 trees have 25 gaps between them . length of each gap = 500 / 25 = 20 i . e . , distance between two consecutive trees = 20 answer is a ." | a = 26 - 1
b = 500 / a
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a ) 5 % , b ) 10 % , c ) 7 % , d ) 8 % , e ) 9 % | b | multiply(divide(2200, add(multiply(5000, 2), multiply(3000, 4))), const_100) | a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 2200 in all from both of them as interest . the rate of interest per annum is : | "explanation : let the rate of interest per annum be r % simple interest for rs . 5000 for 2 years at rate r % per annum + simple interest for rs . 3000 for 4 years at rate r % per annum = rs . 2200 β 5000 Γ r Γ 2 / 100 + 3000 Γ r Γ 4 / 100 = 2200 β 100 r + 120 r = 2200 β 220 r = 2200 β r = 10 i . e , rate = 10 % . answer : option b" | a = 5000 * 2
b = 3000 * 4
c = a + b
d = 2200 / c
e = d * 100
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a ) 100 , b ) 120 , c ) 190 , d ) 160 , e ) 154 | e | add(multiply(add(11, 4), 10), 4) | 12 + 6 = 792 10 + 2 = 110 1 + 9 = 9 2 + 7 = 16 11 + 4 = ? ? solve it ? | x + y = x [ y + ( x - 1 ) ] = x ^ 2 + xy - x 12 + 6 = 12 [ 6 + ( 12 - 1 ) ] = 792 10 + 2 = 10 [ 2 + ( 10 - 1 ) ] = 110 1 + 9 = 1 [ 9 + ( 1 - 1 ) ] = 9 2 + 7 = 2 [ 7 + ( 2 - 1 ) ] = 16 11 + 4 = 11 [ 4 + ( 11 - 1 ) ] = 154 answer : e | a = 11 + 4
b = a * 10
c = b + 4
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a ) 1 / 2 , b ) 3 / 8 , c ) 5 / 16 , d ) 1 / 4 , e ) 7 / 16 | e | divide(add(const_1, divide(3, 4)), add(3, 1)) | at a certain high school , the senior class is 3 times the size of the junior class . if 1 / 3 of the seniors and 3 / 4 of the juniors study japanese , what fraction of the students in both classes study japanese ? | start by deciding on a number of students to represent the number of students in the senior class . for this example i will choose 300 students . that would make the number of students in the junior class 100 . then we can find out how many students are taking japanese in each grade and add them together . ( 1 / 3 ) * 300 = 100 and ( 3 / 4 ) * 100 = 75 . 100 + 75 = 175 . there are a total of 400 students in the junior class and senior class combined ( 100 + 300 = 300 ) , and there are 175 total students in japanese , so 175 students in japanese / 400 total students equals 7 / 16 of the students in both classes that study japanese . answer : e | a = 3 / 4
b = 1 + a
c = 3 + 1
d = b / c
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a ) 9 feet , b ) 12 feet , c ) 13 feet , d ) 15 feet , e ) 18 feet | a | multiply(divide(12, 4), 3) | a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 4 feet . how many feet does the squirrel travels if the post is 12 feet tall and 3 feet in circumference ? | "total circuit = 12 / 4 = 3 total feet squirrel travels = 3 * 3 = 9 feet answer : a" | a = 12 / 4
b = a * 3
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a ) 269.72 , b ) 268.22 , c ) 248.21 , d ) 248.22 , e ) 268.21 | d | divide(2, divide(22, 2)) | evaluate 2 + 22 + 222 + 2.22 | "2 + 22 + 222 + 2.22 = 248.22 option d" | a = 22 / 2
b = 2 / a
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a ) 857 , b ) 859 , c ) 869 , d ) 4320 , e ) none of these | a | subtract(lcm(lcm(lcm(24, 32), 36), 54), 7) | the least number which when increased by 7 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 7 = 864 - 7 = 857 . answer a" | a = math.lcm(24, 32)
b = math.lcm(a, 36)
c = math.lcm(b, 54)
d = c - 7
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a ) 288 , b ) 72 , c ) 200 , d ) 112 , e ) 178 | b | divide(square_area(12), const_2) | what is the area of a square field whose diagonal of length 12 m ? | "d 2 / 2 = ( 12 * 12 ) / 2 = 72 answer : b" | a = square_area / (
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a ) 33 % , b ) 40 % , c ) 55 % , d ) 57.14 % , e ) 66.14 % | d | multiply(divide(multiply(divide(3,500, 2), add(1, divide(1, 7))), 3,500), const_100) | at a contest with 3,500 participants , 1 / 2 of the people are aged 18 to 22 . next year , the number of people aged 18 to 22 will increase by 1 / 7 . after this change , what percentage of the total 3,500 people will the 18 - to 22 - year - olds represent ? | "i just wanted to mention a couple of things here : * this is a pure ratio question ; the number 3,500 is completely irrelevant , and you can ignore it if you like . when we increase something by 1 / 7 , we are multiplying it by 1 + 1 / 7 = 8 / 7 , so the answer here must be ( 1 / 2 ) * ( 8 / 7 ) = 4 / 7 = 57.14 % . answer : d" | a = 3 / 500
b = 1 / 7
c = 1 + b
d = a * c
e = d / 3
f = e * 100
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | min(min(divide(15, add(2, divide(1, 2))), floor(divide(16, add(divide(3, 4), 2)))), divide(8, add(divide(1, 3), 1))) | a recipe requires 2 1 / 2 ( mixed number ) cups of flour 2 3 / 4 ( mixed number ) cups of sugar and 1 1 / 3 ( mixed number ) cups of milk to make one cake . victor has 15 cups if flour , 16 cups of sugar and 8 cups of milk . what is the greatest number of cakes jerry can make using this recipe ? | "less work up front : go through each item and see what the greatest number of cakes you can make with each . the lowest of these will be the right answer . flour : 15 cups , we need 2.5 cups each . just keep going up the line to see how many cakes we can make : that means i can make 2 cakes with 5 cups , so 6 cakes overall with 15 cups . i ' ve already got the answer narrowed to either a or b . sugar : 16 cups , we need 2.75 cups each . same principle . i can make 2 cups with 5.5 cups , so to make 6 cakes i ' d need 16.5 cups . i do n ' t have that much sugar , so we ' re limited to 5 cakes . no need to even do milk because we ' re already at 5 . sugar will be the limiting factor . answer is a" | a = 1 / 2
b = 2 + a
c = 15 / b
d = 3 / 4
e = d + 2
f = 16 / e
g = math.floor(f)
h = min(c)
i = 1 / 3
j = i + 1
k = 8 / j
l = min(h)
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a ) 19.7 , b ) 20 , c ) 21.3 , d ) 21.5 , e ) 22 | a | subtract(subtract(25, divide(30, const_100)), divide(30, 6)) | daniel went to a shop and bought things worth rs . 25 , 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 . 25 given that out of this rs . 25 , 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 % β΄ ( 25 β 30 / 100 β x ) 6 / 100 = 30 / 100 β 6 ( 25 β 0.3 β x ) = 30 β ( 25 β 0.3 β x ) = 5 β x = 25 β 0.3 β 5 = 19.7 a" | a = 30 / 100
b = 25 - a
c = 30 / 6
d = b - c
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a ) and 27 , b ) and 24 , c ) and 23 , d ) and 29 , e ) of these | c | subtract(40, divide(add(40, 11), const_3)) | the sum of the present age of henry and jill is 40 . what is their present ages if 11 years ago henry was twice the age of jill ? | "let the age of jill 11 years ago be x , age of henry be 2 x x + 11 + 2 x + 11 = 40 x = 6 present ages will be 17 and 23 answer : c" | a = 40 + 11
b = a / 3
c = 40 - b
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a ) 2.87 , b ) 2.965 , c ) 3.876 , d ) 3.9 , e ) 4.526 | c | multiply(0.3010, divide(0.3010, 0.4771)) | if log 2 = 0.3010 and log 3 = 0.4771 , the value of log 5 512 is : | "log 5 512 = log 512 / log 5 = log 2 ^ 9 / log ( 10 / 2 ) = 9 log 2 / log 10 - log 2 = ( 9 * 0.3010 ) / 1 - 0.3010 = 2.709 / 0.699 = 2709 / 699 = 3.876 answer c" | a = 0 / 3010
b = 0 * 3010
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a ) 23 , b ) 22 , c ) 27 , d ) 36 , e ) 48 | c | multiply(18, divide(const_3, const_2)) | a is half good a work man as b and together they finish a job in 18 days . in how many days working alone b finish the job ? | "c 27 wc = 1 : 2 2 x + x = 1 / 18 = > x = 1 / 54 2 x = 1 / 27 = > 27 days" | a = 3 / 2
b = 18 * a
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['a ) 8 cm', 'b ) 10 cm', 'c ) 12 cm', 'd ) 14 cm', 'e ) none'] | b | divide(divide(96, const_3), const_pi) | the radius and height of a right circular cone are in the ratio 3 : 4 . if its volume is 96 β cm Β³ , what is its slant height ? | sol . let the radius and the height of the cone be 3 x and 4 x respectively . then , 1 / 3 * β * ( 3 x ) Β² * 4 x = 96 β β 36 x Β³ = ( 96 * 3 ) β x Β³ = [ 96 * 3 / 36 ] = 8 β x = 2 & ther 4 ; radius = 6 cm , height = 8 cm . slant height = β 6 Β² + 8 Β² cm = β 100 cm = 10 cm . answer b | a = 96 / 3
b = a / math.pi
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a ) 299 , b ) 466 , c ) 578 , d ) 600 , e ) 677 | d | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 84 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 84 5 x / 100 + 90 β 6 x / 100 = 84 x / 100 = 6 = > x = 600 answer : d" | a = 5 + 6
b = a * 100
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a ) 24 , b ) 28 , c ) 20 , d ) 32 , e ) 40 | a | subtract(60, multiply(sqrt(36), divide(subtract(60, 48), sqrt(4)))) | an engine moves at the speed of 60 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 4 coaches are attached speed decreases to 48 kmph . what will be the speed of train when 36 coaches are attached . | "1 . no . of coaches = 4 sqr root = 2 speed decreases by 12 12 = k * 2 k = 6 no . of coaches = 36 swr root = 6 decrease = 6 * 6 = 36 new speed = 60 - 36 = 24 a" | a = math.sqrt(36)
b = 60 - 48
c = math.sqrt(4)
d = b / c
e = a * d
f = 60 - e
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a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 6 , e ) $ 10 | e | add(divide(subtract(15, 5), const_2), 5) | you and your friend spent a total of $ 15 for lunch . your friend spent $ 5 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 5 ( l ) + ( l + 5 ) = 15 l + l + 5 - 5 = 15 - 5 2 l = 10 l = 5 my friends lunch l + 5 = 5 + 5 = $ 10 , the answer is e" | a = 15 - 5
b = a / 2
c = b + 5
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a ) 225 , b ) 90 , c ) 45 , d ) 37.5 , e ) 27 | c | divide(multiply(divide(add(330, 66), 6), const_3600), add(multiply(add(const_4, const_1), const_1000), subtract(multiply(const_3, const_100), multiply(const_2, const_10)))) | the rear β most end of a 66 foot truck exits a 330 foot tunnel exactly 6 seconds after the front β most end of the truck entered the tunnel . if the truck traveled the entire tunnel at a uniform speed , what is the speed of the truck in miles per hour ( 1 mile = 5,280 feet ) ? | total length will be 330 + 66 = 396 time = 6 secs speed = 396 / 6 = 66 feet / sec conversion to miles per hour = 66 * 3600 / 5280 = 45 . answer c | a = 330 + 66
b = a / 6
c = b * 3600
d = 4 + 1
e = d * 1000
f = 3 * 100
g = 2 * 10
h = f - g
i = e + h
j = c / i
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a ) rs . 15000 , b ) rs . 15550 , c ) rs . 15600 , d ) rs . 16500 , e ) none of these | d | multiply(800, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 800 per sq . metre . | "solution area of the floor = ( 5.5 Γ 3.75 ) m 2 = 20.625 m 2 β΄ cost of paving = rs . ( 800 Γ 20.625 ) = 16500 . answer d" | a = 5 * 5
b = 800 * a
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a ) 4 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | a | multiply(factorial(2), factorial(2)) | 2 men and 2 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | the list should be wmwmw . hence , from women 2 ! and men 2 ! , we get ( 2 ! ) ( 2 ! ) = 4 . therefore , the correct answer is a . | a = math.factorial(2)
b = math.factorial(2)
c = a * b
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a ) 3 , b ) 4.5 , c ) 4 , d ) d ) 12 , e ) e ) 5 | d | subtract(40, 30) | 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 six 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 6 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 6 + t ) = 40 * t t = 18 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 18 / 60 = 12 hrs time taken by a = 6 + t = 6 + 18 = 24 hrs time taken by k = 12 hrs so k starts 24 - 12 = 12 hrs after a . ( answer d )" | a = 40 - 30
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a ) 99 , b ) 18 , c ) 16 , d ) 10 , e ) 231 | e | divide(multiply(70, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 70 m long and 16 1 / 2 m board ? | "70 * 33 / 2 = 5 / 2 * 2 * x = > x = 231 answer : e" | a = 1 / 2
b = 16 + a
c = 70 * b
d = 1 / 2
e = 2 + d
f = e * 2
g = c / f
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a ) 3.3 % , b ) 7 % , c ) 9 % , d ) 2 % , e ) 4 % | a | divide(multiply(const_100, 160), multiply(1200, 4)) | what is the rate percent when the simple interest on rs . 1200 amount to rs . 160 in 4 years ? | "160 = ( 1200 * 4 * r ) / 100 r = 3.3 % answer : a" | a = 100 * 160
b = 1200 * 4
c = a / b
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a ) 5 : 0 , b ) 5 : 3 , c ) 5 : 1 , d ) 5 : 2 , e ) 5 : 7 | d | divide(2, 5) | the simple form of the ratio 2 / 3 : 2 / 5 is ? | "2 / 3 : 2 / 5 = 10 : 6 = 5 : 3 answer : d" | a = 2 / 5
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a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | on a map , 1 inch represents 28 miles . how many inches would be necessary to represent a distance of 383.6 miles ? | "28 miles represented by 1 inch 1 mile represented by ( 1 / 28 ) inch 383.6 miles represented by ( 1 / 28 ) * 383.6 approximately = 14 we should use estimation here as the answer options are not close . ( 28 * 10 = 280 and 28 * 20 = 560 , hence we select the only option between 10 and 20 ) answer c" | a = 383 / 6
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a ) 1 : 2 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | a | divide(subtract(15.8, 15.5), 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 15.5 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 * 15.5 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.5 ) < = > k = 0.3 / 0.6 = 1 / 2 . required ratio = 1 / 2 : 1 = 1 : 2 . answer : a" | a = 15 - 8
b = 16 - 4
c = a / b
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a ) 8 , b ) 16 , c ) 21 , d ) 22 , e ) 27 | a | divide(336, divide(subtract(462, 336), 3)) | a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 3 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 3 ) x solve for r , r = 11 11 - 3 = 8 mpg a | a = 462 - 336
b = a / 3
c = 336 / b
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a ) 100 , b ) 208 , c ) 625 , d ) 832 , e ) 833 | a | add(divide(subtract(50, 5000), 5), const_1) | how many multiples of 5 are less than 5000 , and also multiples of 50 ? | the lcm of 5 and 50 is 50 . divide 5000 / 50 = 100 . xxx . so a is your answer . | a = 50 - 5000
b = a / 5
c = b + 1
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a ) 3.622 , b ) 3.911 , c ) 4.938 , d ) 2.986 , e ) 2.999 | a | add(add(3.15, divide(014, const_1000)), divide(458, const_1000)) | solution for 3.15 + . 014 + . 458 | "3.15 + . 014 + . 458 = 0 0 = 0 - 3.15 - 0.014 - 0.458 0 = - 3.622 answer : a" | a = 14 / 1000
b = 3 + 15
c = 458 / 1000
d = b + c
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a ) 2997 , b ) 1900 , c ) 2887 , d ) 2898 , e ) 27912 | b | divide(multiply(subtract(7400, divide(multiply(5, 7400), const_100)), multiply(10000, 3)), add(add(multiply(6500, 6), multiply(8400, 5)), multiply(10000, 3))) | a , b and c are entered into a partnership . a invested rs . 6500 for 6 months , b invested rs . 8400 for 5 months and c invested for rs . 10000 for 3 months . a is a working partner and gets 5 % of the total profit for the same . find the share of c in a total profit of rs . 7400 . | "65 * 6 : 84 * 5 : 100 * 3 26 : 28 : 20 c share = 74000 * 95 / 100 = 7030 * 20 / 74 = > 1900 . answer : b" | a = 5 * 7400
b = a / 100
c = 7400 - b
d = 10000 * 3
e = c * d
f = 6500 * 6
g = 8400 * 5
h = f + g
i = 10000 * 3
j = h + i
k = e / j
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a ) 43 , b ) 53 , c ) 63 , d ) 65 , e ) 78 | c | subtract(negate(9), multiply(subtract(7,12, 18,25), divide(subtract(7,12, 18,25), subtract(3, 7,12)))) | 3 , 7,12 , 18,25 . . . . . . . . . . . . . . 9 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 33 + 9 = 42 42 + 10 = 52 52 + 11 = 63 answer : c" | a = negate - (
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a ) 33 % , b ) 10 % , c ) 5 % , d ) 15 % , e ) 25 % | a | multiply(divide(subtract(60, 30), add(60, 30)), const_100) | if 60 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ? | "60 % of ( x - y ) = 30 % of ( x + y ) ( 60 / 100 ) ( x - y ) = ( 30 / 100 ) ( x + y ) 6 ( x - y ) = 3 ( x + y ) 3 x = 9 y x = 3 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / 3 y ) x 100 ) = 33 % answer is a ." | a = 60 - 30
b = 60 + 30
c = a / b
d = c * 100
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a ) 5 , b ) 4.5 , c ) 6 , d ) 7 , e ) 8 | a | inverse(add(inverse(subtract(10, const_1)), inverse(add(10, 2)))) | working alone at their respective constant rates , a can complete a task in β a β days and b in β b β days . they take turns in doing the task with each working 2 days at a time . if a starts they finish the task in exactly 10 days . if b starts , they take a day more . how long does it take to complete the task if they both work together ? | work done by a & b in a day = x & y respectively . when a starts : no . of days when a works = 6 no . of days when b works = 4 β 6 x + 4 y = 1 when b starts : no . of days when b works = 5 no . of days when a works = 5 β 5 x + 5 y = 1 solving the above two equations for xy x = 1 / 10 y = 1 / 10 β total work done by ab in a day = 1 / 10 + 1 / 10 = 2 / 10 = 1 / 5 β no . of days to complete the work when both work together = 5 answer : a | a = 10 - 1
b = 1/(a)
c = 10 + 2
d = 1/(c)
e = b + d
f = 1/(e)
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a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024 | c | power(5, 4) | a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many q ways can the test be completed if every question is unanswered ? | 5 choices for each of the 4 questions , thus total q of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c . | a = 5 ** 4
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a ) a ) 3 , b ) b ) 98 , c ) c ) 34 , d ) d ) 35 , e ) e ) 62 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 3 : 2 . the first number is : | "( 3 x - 9 ) : ( 5 x - 9 ) = 3 : 2 x = 1 = > 3 x = 3 answer : a" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
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a ) 3 / 25 , b ) 11 / 36 , c ) 7 / 12 , d ) 2 / 3 , e ) 5 / 6 | e | divide(divide(multiply(60, 5), 30), 12) | at a speed of 60 miles per hour , a certain car uses 1 gallon of gasoline every 30 miles . if the car starts with a full 12 gallon tank of gasoline and travels for 5 hours at 60 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | "gas used = ( 5 hours ) * ( 60 miles / hour ) * ( 1 gallon / 30 miles ) = 10 gallons portion used = ( 10 ) / 12 = 5 / 6 ans e" | a = 60 * 5
b = a / 30
c = b / 12
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a ) 226 , b ) 228 , c ) 230 , d ) 224 , e ) 232 | d | multiply(divide(38, 2.54), divide(30, 2)) | on a map , 2 inches represent 30 miles . how many miles approximately is the distance if you measured 38 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 2 inch = 2.54 * 2 cm 5.08 cm = 30 miles 38 cms = 30 / 5.08 * 38 = 224 miles answer : d" | a = 38 / 2
b = 30 / 2
c = a * b
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a ) 2 / 7 , b ) 1 / 2 , c ) 3 / 4 , d ) 1 , e ) 5 / 4 | a | subtract(divide(subtract(61, const_1), add(59, const_1)), divide(add(49, const_1), subtract(71, const_1))) | a is an integer greater than 49 but less than 61 , b is an integer greater than 59 but less than 71 , what is the range of a / b ? | "min value of a / b will be when b is highest and a is lowest - - - > a = 50 and b = 70 so , a / b = 5 / 7 max value of a / b will be when b is lowest and a is highest - - - > a = 60 and b = 60 so , a / b = 1 range is 1 - ( 5 / 7 ) = 2 / 7 . answer should be a ." | a = 61 - 1
b = 59 + 1
c = a / b
d = 49 + 1
e = 71 - 1
f = d / e
g = c - f
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a ) 10 , 5 , b ) 8 , 7 , c ) 9 , 6 , d ) 11 , 4 , e ) 12 , 3 | a | subtract(15, divide(subtract(15, 5), const_2)) | find the value of x , y by solving the below equations x + y = 15 x - y = 5 | x + y = 15 - - - ( i ) x - y = 5 - - - - - ( ii ) by adding ( i ) and ( ii ) - - - - - - - - - - - - 2 x = 20 = = > x = 20 / 2 = 10 by replacing the value of x in ( i ) we get 10 + y = 15 = = > y = 15 - 10 = 5 . so , x = 10 , y = 5 answer a ) 10 , 5 | a = 15 - 5
b = a / 2
c = 15 - b
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a ) 23 % , b ) 36 % , c ) 30 % , d ) 34 % , e ) 42 % | b | subtract(const_100, add(add(add(subtract(const_100, 90), subtract(const_100, 86)), subtract(const_100, 80)), subtract(const_100, 80))) | in a urban village of india named ` ` owlna ' ' , 90 % people have refrigerator , 86 % people have television , 80 % people got computers and 80 % got air - conditionor . how many people ( minimum ) got all these luxury . | "b 10 % 100 - [ ( 100 - 90 ) + ( 100 - 86 ) + ( 100 - 80 ) + ( 100 - 80 ) ] = 100 - ( 10 + 14 + 20 + 20 ) = 100 - 64" | a = 100 - 90
b = 100 - 86
c = a + b
d = 100 - 80
e = c + d
f = 100 - 80
g = e + f
h = 100 - g
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a ) 22 % , b ) 20 % , c ) 25 % , d ) 23 % , e ) 21 % | c | divide(multiply(27, const_100), add(27, const_100)) | the annual interest rate earned by an investment increased by 8 percent from last year to this year . if the annual interest rate earned by the investment this year was 27 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.08 i ( last year ) = 1.08 i ( last year ) 27 = 1.08 x i ( last year ) i ( last year ) = 27 / 1.08 = 2700 / 108 = 25 % answer : c" | a = 27 * 100
b = 27 + 100
c = a / b
|
a ) 0 and 2 , b ) 2 and 3 , c ) 3 and 4 , d ) 4 and 5 , e ) 5 and 8 | e | add(multiply(floor(power(140, inverse(const_3))), const_10), add(floor(power(140, inverse(const_3))), const_1)) | if a and b are positive numbers , and a ^ 3 + b ^ 3 = 140 , then the greatest possible value of a is between : | if a = 5.1 and b is a bit less than 2 , then a ^ 3 + b ^ 3 can equal 140 . if a > 8 , then a ^ 3 + b ^ 3 > 140 . the answer is e . | a = 1/(3)
b = 140 ** a
c = math.floor(b)
d = c * 10
e = 1/(3)
f = 140 ** e
g = math.floor(f)
h = g + 1
i = d + h
|
a ) 40 sec , b ) 50 sec , c ) 44 sec , d ) 24 sec , e ) 60 sec | d | divide(160, multiply(subtract(45, 140), const_0_2778)) | a train 160 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 160 + 140 = 300 m required time = 300 * 2 / 25 = 24 sec answer : d" | a = 45 - 140
b = a * const_0_2778
c = 160 / b
|
a ) 39 % , b ) 40 % , c ) 41 % , d ) 42 % , e ) 43 % | b | multiply(divide(add(1, 3), add(add(4, 2), add(1, 3))), const_100) | the proportion of water to alcohol in solution a is 4 : 1 and the proportion of water to alcohol in solution b is 2 : 3 . if an equal amount of each solution is mixed together , what is the concentration of alcohol in the new solution ? | let v be the total volume of the new solution . then a volume of v / 2 was added from each solution a and b . the amount of alcohol added to the new solution was : ( 1 / 5 ) ( v / 2 ) + ( 3 / 5 ) ( v / 2 ) = v / 10 + 3 v / 10 = 4 v / 10 = 2 v / 5 . the concentration of alcohol is 2 / 5 = 40 % the answer is b . | a = 1 + 3
b = 4 + 2
c = 1 + 3
d = b + c
e = a / d
f = e * 100
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a ) 6 % , b ) 2 % , c ) 4 % , d ) 5 % , e ) 3 % | d | multiply(divide(divide(subtract(900, 750), 750), 4), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 900 in 4 years ? | "150 = ( 750 * 4 * r ) / 100 r = 5 % answer : d" | a = 900 - 750
b = a / 750
c = b / 4
d = c * 100
|
a ) - 5 , b ) - 4 , c ) - 3 , d ) - 2 , e ) - 1 | e | subtract(1, const_2) | if m is not equal to zero , and m + 1 / m = 1 , then what is the value of m ^ 4 + ( 1 / m ) ^ 4 ? | m + 1 / m = 1 we square both sides so we have m ^ 2 + 1 / m ^ 2 + 2 = 1 or m ^ 2 + 1 / m ^ 2 = - 1 squaring again we have m ^ 4 + 1 / m ^ 4 + 2 = 1 or m ^ 4 + 1 / m ^ 4 = - 1 answer = - 1 ( e ) | a = 1 - 2
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a ) s . 2890 , b ) s . 2330 , c ) s . 1190 , d ) s . 1620 , e ) s . 3179 | e | multiply(divide(multiply(divide(539, 7), 17), 7), 17) | the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 539 , krishan has ? | "ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 539 : n & there n = 289 x 539 / 49 = rs . 3179 answer : e" | a = 539 / 7
b = a * 17
c = b / 7
d = c * 17
|
a ) 33 , b ) 40 , c ) 99 , d ) 28 , e ) 72 | b | subtract(const_100, multiply(divide(subtract(const_100, 20), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 20 per cent . the two reductions together are equal to a single reduction of | "explanation : let the original price of the cycle be 100 . after the first reduction the price will be 75 . this new price is then reduced by 20 % = 0.8 x 75 = 60 60 represents a reduction of 40 percent on the original . answer : b ) 40 %" | a = 100 - 20
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
|
a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 360 , e ) s . 368 | d | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(3960, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3960 . what is the difference between rohan β s and sudhir β s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 Γ 12 : 12000 Γ 9 : 9000 Γ 8 = 6 : 3 : 2 the difference between rohan β s and sudhir β s share = 1 share : . i . e . = rs . 3960 Γ 1 / 11 = rs . 360 . d" | a = 3 * 4
b = a - 3
c = 3 * 4
d = b / c
e = 12000 * d
f = 3 * 4
g = f - 4
h = 3 * 4
i = g / h
j = 9000 * i
k = e - j
l = 3 * 4
m = l - 3
n = 3 * 4
o = m / n
p = 12000 * o
q = 18000 + p
r = 3 * 4
s = r - 4
t = 3 * 4
u = s / t
v = 9000 * u
w = q + v
x = 3960 / w
y = k * x
|
a ) 72 , b ) 85 , c ) 64 , d ) 51 , e ) 45 | a | add(56, divide(56, const_2)) | in a ratio which is equal to 7 : 9 , if the antecedent is 56 , then the consequent is ? | "we have 7 / 9 = 56 / x 7 x = 56 * 9 x = 72 consequent = 72 answer is a" | a = 56 / 2
b = 56 + a
|
a ) 220 , b ) 210 , c ) 240 , d ) 250 , e ) 260 | c | multiply(divide(60, divide(20, const_100)), divide(80, const_100)) | if 20 % of a certain number is 60 , then what is 80 % of that number ? | "explanation : 20 % = 20 * 3 = 60 80 % = 80 * 3 = 240 answer : option c" | a = 20 / 100
b = 60 / a
c = 80 / 100
d = b * c
|
a ) 2 / 24 , b ) 6 / 18 , c ) 2 / 22 , d ) 5 / 12 , e ) 9 / 10 | e | divide(5, add(multiply(5, divide(5, 2)), multiply(6, divide(5, 3)))) | if 2 men or 3 women can reap a field in 5 days how long will 5 men and 6 women take to reap it ? | "explanation : 2 men reap 2 / 5 field in 1 day 1 man reap 1 / ( 2 x 5 ) 3 women reap 1 / 43 field in 1 day 1 woman reap 1 / ( 5 x 3 ) 5 men and 6 women reap ( 5 / ( 2 x 5 ) + 6 / ( 3 x 5 ) ) = 9 / 10 in 1 day 5 men and 6 women will reap the field in 9 / 10 days answer : option e" | a = 5 / 2
b = 5 * a
c = 5 / 3
d = 6 * c
e = b + d
f = 5 / e
|
a ) $ 9.84 , b ) $ 10.68 , c ) $ 11.26 , d ) $ 12.72 , e ) $ 13.54 | d | subtract(subtract(30, 1.28), multiply(8, const_2)) | a customer went to a shop and paid a total of $ 30 , out of which $ 1.28 was for sales tax on taxable purchases . if the tax rate was 8 % , then what was the cost of the tax free items ? | the total cost was $ 30 . the tax was $ 1.28 let the original price of the taxable items = x given that tax rate = 8 % 0.08 x = 1.28 x = $ 16 the cost of the tax free items was $ 30 - $ 16 - $ 1.28 = $ 12.72 the answer is d . | a = 30 - 1
b = 8 * 2
c = a - b
|
a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 16 , d ) 2 ^ 12 , e ) 2 ^ 37 | d | divide(multiply(2, add(2, 2)), 2) | 2 + 2 + 2 Β² + 2 Β³ . . . + 2 ^ 11 | "2 + 2 = 2 ^ 2 2 ^ 2 + 2 ^ 2 = ( 2 ^ 2 ) * ( 1 + 1 ) = 2 ^ 3 2 ^ 3 + 2 ^ 3 = ( 2 ^ 3 ) * ( 1 + 1 ) = 2 ^ 4 so you can notice the pattern . . . in the end you will have 2 ^ 11 + 2 ^ 11 , which will give you 2 ^ 12 answer d" | a = 2 + 2
b = 2 * a
c = b / 2
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | subtract(divide(subtract(add(166, 1), 5), 2), subtract(divide(subtract(add(166, 1), 5), 2), 1)) | if a number p is prime , and 2 p + 5 = 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 / 7 = 014285 . . . ( a repeating pattern one digit long ) a | a = 166 + 1
b = a - 5
c = b / 2
d = 166 + 1
e = d - 5
f = e / 2
g = f - 1
h = c - g
|
a ) 8 , b ) 16 , c ) 42 , d ) 48 , e ) 54 | a | divide(24, subtract(divide(const_1, divide(25, const_100)), const_1)) | walking at 25 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 25 % of speed meand 1 / 4 s takes 4 t . it takes 24 minutes extra to cover the distance . then 4 t = t + 24 3 t = 24 t = 8 . option a is correct" | a = 25 / 100
b = 1 / a
c = b - 1
d = 24 / c
|
a ) 303 , b ) 403 , c ) 203 , d ) 453 , e ) 203 | b | divide(1277, 2) | the lowest number which should be added to 1277 so that the sum is exactly divisible by 3 , 2 , 5 , 4 and 7 is : | "l . c . m . of 5 , 6 , 4 and 3 = 420 . on dividing 1277 by 420 , the remainder is 17 . number to be added = ( 420 - 17 ) = 403 . answer : option ' b '" | a = 1277 / 2
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a ) 2 times , b ) 2.5 times , c ) 3 times , d ) 3.8 times , e ) 4 times | a | divide(add(multiply(const_4, 8), add(8, 8)), add(8, add(8, 8))) | father is aged 3 times more than his son ronit . after 8 years , he would be two and half times if ronit ' s age . after further 8 years , how many times would he be of ronit ' s age ? | ronit ' s present age be x years . then , father ' s present age = ( x + 3 x ) years = 4 x years therefore ( 4 x + 8 ) = 5 / 2 ( x + 8 ) 8 x + 16 = 5 x + 40 3 x = 24 , x = 8 , ratio = ( 4 x + 16 ) / ( x + 16 ) = 48 / 24 = 2 , correct answer ( a ) | a = 4 * 8
b = 8 + 8
c = a + b
d = 8 + 8
e = 8 + d
f = c / e
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | max(multiply(subtract(add(55, 12), const_1), subtract(divide(12, 35), divide(12, 55))), const_4) | due to construction , the speed limit along an 12 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "12 / 35 - 12 / 55 = 12 / 5 * ( 11 - 7 ) / 77 = 12 / 5 * 4 / 77 * 60 min = 12 * 12 * 4 / 77 = 576 / 77 ~ 7.4 answer - d" | a = 55 + 12
b = a - 1
c = 12 / 35
d = 12 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 52.6 , b ) 52.9 , c ) 67.9 , d ) 52.1 , e ) 52.2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(54, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 54 Γ’ Β° is ? | "54 / 360 * 22 / 7 * 12 * 12 = 67.9 m 2 answer : c" | a = 12 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 54 / g
i = f * h
|
a ) 12 , b ) 14 , c ) 45 , d ) 6 , e ) 7 | d | inverse(add(divide(divide(const_100, subtract(const_100, 18)), 10), divide(multiply(divide(divide(const_100, subtract(const_100, 18)), 10), 30), const_100))) | by selling 10 pens for a rupee a woman loses 18 % . how many for a rupee should he sell in order to gain 30 % ? | "d 82 % - - - 10 130 % - - - ? 82 / 130 * 10 = 6" | a = 100 - 18
b = 100 / a
c = b / 10
d = 100 - 18
e = 100 / d
f = e / 10
g = f * 30
h = g / 100
i = c + h
j = 1/(i)
|
a ) 9800 , b ) 9600 , c ) 9400 , d ) 9500 , e ) 9200 | b | multiply(floor(divide(multiply(const_100, const_100), lcm(lcm(lcm(15, 25), 40), 75))), lcm(lcm(lcm(15, 25), 40), 75)) | what is the greatest number of 4 digits which is divisible by 15 , 25 , 40 and 75 ? | greatest number of four digits = 9999 lcm of 15 , 25 , 40 and 75 = 600 9999 Γ· 600 = 16 , remainder = 399 hence , greatest number of four digits which is divisible by 15 , 25 , 40 and 75 = 9999 - 399 = 9600 answer : b | a = 100 * 100
b = math.lcm(15, 25)
c = math.lcm(b, 40)
d = math.lcm(c, 75)
e = a / d
f = math.floor(e)
g = math.lcm(15, 25)
h = math.lcm(g, 40)
i = math.lcm(h, 75)
j = f * i
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | c | add(3, 4) | a one - foot stick is marked in 1 / 3 and 1 / 4 portion . how many total markings will there be , including the end points ? | "lcm of 12 = 12 1 / 3 marking are ( table of 4 ) 0 . . . . . . 4 . . . . . . 8 . . . . . 12 ( total = 4 ) 1 / 4 marking are ( table of 3 ) 0 . . . . . . . 3 . . . . . . 6 . . . . . . 9 . . . . . . . . 12 ( total = 5 ) overlapping markings are 0 . . . . . . . . 12 ( total = 2 ) total markings = 4 + 5 - 2 = 7 answer = c" | a = 3 + 4
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a ) 140 , b ) 321 , c ) 390 , d ) 500 , e ) 130 | c | multiply(multiply(13, 3), 10) | a company has a hierarchical system where for every 10 workers , there is one team lead , and for every 3 teams leads , there is one supervisor . if the company has 13 supervisors , how many workers does it have ? | some students may initially answer 130 to this question , especially if answering similar problems . however , that would be incorrect , as this is not a straight - forward ratio of 1 : 3 : 10 . the ratio of supervisors to team leads is 1 : 3 , and the ratio of team leads to workers is 1 : 10 . therefore , the ratio of supervisors to workers is 1 : 30 . therefore , 13 * 30 = 390 answer : c ) | a = 13 * 3
b = a * 10
|
a ) 4.92 , b ) 6.92 , c ) 7.92 , d ) 4.92 , e ) 2.92 | b | divide(120, multiply(add(const_60.0, 12), const_0_2778)) | a train 120 m long is running with a speed of 66 km / hr . in what time will it pass a man who is roller skating at 12 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 66 + 12 = 78 km / hr . = 78 * 5 / 18 = 65 / 3 m / sec . time taken to pass the man = 150 * 3 / 65 = 6.92 sec . answer : b" | a = const_60 + 0
b = a * const_0_2778
c = 120 / b
|
a ) 26 / 27 , b ) 13 / 17 , c ) 13 / 19 , d ) 12 / 19 , e ) 11 / 19 | a | divide(add(multiply(40, const_100), multiply(40, subtract(const_100, 20))), add(multiply(20, const_100), multiply(add(20, 20), 40))) | paul ' s income is 20 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 60 % of his income to sam and 40 % of his income to quentin , quentin ' s new income would be what fraction of sam ' s new income ? | "make r = 10 p = 0.8 r = 8 q = 0.8 p = 6.4 s = 0.6 p = 4.8 for that we get s = 10.8 and q 10.4 so 10.4 / 10.8 = 2.6 / 2.7 ans : a" | a = 40 * 100
b = 100 - 20
c = 40 * b
d = a + c
e = 20 * 100
f = 20 + 20
g = f * 40
h = e + g
i = d / h
|
a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 2 | e | divide(divide(18, const_3), const_3) | if a * b denotes the greatest common divisor of a and b , then ( ( 20 * 16 ) * ( 18 * 24 ) ) = ? | "the greatest common divisor of 20 and 16 is 4 . hence 20 * 16 = 4 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 24 is 6 . hence 18 * 24 = 6 . hence ( ( 20 * 16 ) * ( 18 * 24 ) ) = 4 * 6 . the greatest common divisor of 4 and 6 is 2 . answer ; e ." | a = 18 / 3
b = a / 3
|
a ) 1 , b ) 8 , c ) 12 , d ) 9 , e ) 7 | b | multiply(2, const_4) | how many quarters are equal to 2 dollars ? | b . 8 quarters | a = 2 * 4
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a ) 1600 , b ) 1700 , c ) 1550 , d ) 1650 , e ) 1750 | e | multiply(7000, divide(const_1, const_100)) | a person borrows 7000 for 3 years at 6 % p . a . simple interest . he immediately lends it to another person at 31 % p . a . for 3 years . find his gain in the transaction per year . | "gain in 3 years = [ ( 7000 Γ£ β 31 Γ£ β 3 / 100 ) Γ’ Λ β ( 7000 Γ£ β 6 Γ£ β 3 / 100 ) ] = ( 6510 Γ’ β¬ β 1260 ) = 5250 . Γ’ Λ Β΄ gain in 1 year = ( 5250 Γ’ Β β 3 ) = 1750 answer e" | a = 1 / 100
b = 7000 * a
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a ) 75 , b ) 276 , c ) 87 , d ) 85 , e ) 11 | d | divide(add(add(add(add(86, 85), 82), 87), 85), add(const_2, const_3)) | david obtained 86 , 85 , 82 , 87 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | "explanation : average = ( 86 + 85 + 82 + 87 + 85 ) / 5 = 425 / 5 = 85 . answer : d" | a = 86 + 85
b = a + 82
c = b + 87
d = c + 85
e = 2 + 3
f = d / e
|
a ) 2 , b ) 4 , c ) 3 , d ) 5 , e ) 1 | c | divide(96, 56) | how many of the positive factors of 56 , 96 and how many common factors are there in numbers ? | "factors of 56 - 1 , 2 , 4 , 7 , 8 , 14 , 28 , 56 factors of 96 - 1 , 2 , 3 , 4 , 6 , 8 , 12 , 16 , 24 , 32 , 48 , 96 comparing both , we have three common factors of 56 and 96 - 1 , 2,4 answer ( c )" | a = 96 / 56
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a ) 6 2 / 3 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(4, subtract(divide(multiply(const_2, 4), 5), 1)) | it takes joey the postman 1 hours to run a 4 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 4 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 4 * x / 4 + x = 5 8 x = 5 x + 20 = > 3 x = 20 = > x = 6 2 / 3 a" | a = 2 * 4
b = a / 5
c = b - 1
d = 4 / c
|
a ) 39.55 $ , b ) 40.63 $ , c ) 41.63 $ , d ) 42.15 $ , e ) 43.15 $ | b | divide(50, add(divide(add(7, 15), const_100), const_1)) | a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 50 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ? | "let x be the amount the most their food can cost . x + 0.07 x + 0 . 15 x = 50 solving for x , x = 40.98 $ , this is the maximum amount that can be used for food . hence 40.63 $ is the maximum value less than 40.98 $ hence option b" | a = 7 + 15
b = a / 100
c = b + 1
d = 50 / c
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | e | add(divide(add(37, 7), const_2), 7) | the sum of present age of abe and the age before 7 years is 37 . find the present age of abe . what will be his age after 7 years ? | "present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 37 2 x - 7 = 37 2 x = 37 + 7 x = 44 / 2 x = 22 z = x + 7 = 22 + 7 = 29 answer : e" | a = 37 + 7
b = a / 2
c = b + 7
|
a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 160 | c | divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price w of the machine ? | "since the wholesale price was $ 90 and the profit was 20 % of the wholesale price ( [ . 2 ] [ 90 ] = $ 18 ) , the retail price would have to be above $ 108 , but not that much greater than that . let ' s start by testing answer c : $ 120 . . . . if . . . . . retail price w = $ 120 10 % discount off = $ 120 - ( . 1 ) ( 120 ) = 120 - 12 = 108 20 % profit on wholesale price = 90 + ( . 2 ) ( 90 ) = 90 + 18 = 108 these two numbers match , so this must be the answer ! final answer : [ reveal ] spoiler : c" | a = 90 * 20
b = a / 100
c = 90 + b
d = c * 100
e = 3 * 3
f = e * 10
g = d / f
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a ) a ) 6.7 , b ) b ) 2.3 , c ) c ) 9.6 , d ) d ) 12.5 , e ) e ) 7.9 | b | divide(subtract(12, 10), subtract(const_1, divide(12, 100))) | how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 12 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 12 / 100 250 + 25 x = 300 + 3 x 22 x = 50 x = 2.3 answer is b" | a = 12 - 10
b = 12 / 100
c = 1 - b
d = a / c
|
a ) 12 , b ) 10 , c ) 11 , d ) 14 , e ) 13 | a | add(divide(subtract(222, 111), 9), const_1) | what is the total number of integers between 111 and 222 that are divisible by 9 ? | "117 , 126 , 135 , . . . , 207,216 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 216 - 117 ) / ( 9 ) + 1 = 99 / 9 + 1 = 11 + 1 = 12 answer is a" | a = 222 - 111
b = a / 9
c = b + 1
|
a ) 725 , b ) 625 , c ) 225 , d ) 525 , e ) 425 | d | add(350, multiply(350, divide(50, const_100))) | 350 is increased by 50 % . find the final number . | explanation final number = initial number + 50 % ( original number ) = 350 + 50 % ( 350 ) = 350 + 175 = 525 . answer d | a = 50 / 100
b = 350 * a
c = 350 + b
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a ) 77 kmph , b ) 55 kmph , c ) 54 kmph , d ) 58 kmph , e ) 76 kmph | c | multiply(const_3_6, divide(240, 16)) | a train 240 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 240 / 16 * 18 / 5 = 54 kmph answer : c" | a = 240 / 16
b = const_3_6 * a
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a ) s . 300 , b ) s . 600 , c ) s . 420 , d ) s . 400 , e ) s . 480 | a | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 2400) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 2400 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 2400 Γ ( 1 / 8 ) = 300 answer : option a" | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 2400
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a ) 19 % , b ) 21 % , c ) 23 % , d ) 25 % , e ) 27 % | b | multiply(const_100, divide(add(divide(45, const_100), multiply(4, divide(15, const_100))), add(const_1, 4))) | because he β s taxed by his home planet , mork pays a tax rate of 45 % on his income , while mindy pays a rate of only 15 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ? | "let x be mork ' s income , then mindy ' s income is 4 x . the total tax paid is 0.45 x + 0.6 x = 1.05 x 1.05 x / 5 x = 0.21 the answer is b ." | a = 45 / 100
b = 15 / 100
c = 4 * b
d = a + c
e = 1 + 4
f = d / e
g = 100 * f
|
['a ) 11 / 20', 'b ) 3 / 5', 'c ) 13 / 20', 'd ) 18 / 25', 'e ) 18 / 26'] | c | divide(triangle_perimeter(6, divide(multiply(multiply(4, const_2), 6), sqrt(add(power(multiply(4, const_2), const_2), power(6, const_2)))), divide(multiply(6, 6), multiply(4, const_2))), triangle_perimeter(multiply(4, const_2), 6, sqrt(add(power(multiply(4, const_2), const_2), power(6, const_2))))) | pr is tangent to a circle at point p . q is another point on circle such that pq is diameter of circle and rq cuts circle at m . if radius of circle is 4 units and pr = 6 units . find ratio of triangle pmr to pqr . | let pm = y qm = x mr = 10 - x in triangle pqm , pm ^ 2 + qm ^ 2 = pq ^ 2 = > x ^ 2 + y ^ 2 = 64 . . . . . ( 1 ) in triangle pmr , pm ^ 2 + mr ^ 2 = pr ^ 2 y ^ 2 + ( 10 - x ) ^ 2 = 36 . . . . . . ( 2 ) solving for x and y x = 6.4 y = 4.8 taking permineter of triangle pqr = 24 perimeter of triangle pmr = 14.4 take ratio 14.4 / 24 = 0.6 answer : c | a = 4 * 2
b = a * 6
c = 4 * 2
d = c ** 2
e = 6 ** 2
f = d + e
g = math.sqrt(f)
h = b / g
i = 6 * 6
j = 4 * 2
k = i / j
l = triangle_perimeter / (
|
a ) 6 , b ) 12 , c ) 24 , d ) 6 only , e ) 6 and 12 both | e | add(multiply(6, const_100), multiply(2, 6)) | if n is a natural number , then ( 6 n ^ 2 + 6 n ) is always divisible by : | explanation : ( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . e | a = 6 * 100
b = 2 * 6
c = a + b
|
a ) 276 , b ) 299 , c ) 391 , d ) 345 , e ) 355 | c | multiply(23, 17) | the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 17 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 13 ) and ( 23 x 17 ) . larger number = ( 23 x 17 ) = 391 . answer : option c" | a = 23 * 17
|
['a ) a . 3 / 16', 'b ) 1 / 4', 'c ) c . 1 / 2', 'd ) 25 / 32', 'e ) 7 / 8'] | d | divide(subtract(64, multiply(2, 7)), 64) | a rectangular rug with side lengths of 2 feet and 7 feet is placed on a square floor that has an area of 64 square feet . if the surface of the rug does not extend beyond the area of the floor , what fraction of the area of the floor is not covered by the rug ? | area of the rectangular rug = 2 * 7 = 14 fraction not covered by the rug = ( total area - rug area ) / total area = ( 64 - 14 ) / 64 = 25 / 32 = d | a = 2 * 7
b = 64 - a
c = b / 64
|
a ) 4423 , b ) 4403 , c ) 4413 , d ) 2403 , e ) 4375 | b | multiply(add(divide(113, 78), 55), 78) | find the value of ( 55 + 113 / 78 ) Γ 78 | "= ( 55 + 113 / 78 ) Γ 78 = ( 4290 + 113 ) / 78 Γ 78 = 4403 / 78 Γ 78 = 4403 answer is b ." | a = 113 / 78
b = a + 55
c = b * 78
|
a ) 3 , b ) 4 , c ) 10 , d ) 12 , e ) 36 | c | divide(divide(25, divide(const_1, const_2)), 5) | if two typists can type two pages in two minutes , how many typists will it take to type 25 pages in 5 minutes ? | in 2 minutes 2 typists type 2 pages which means that in 5 minutes they will type 5 pages but to type 25 pages ( 5 times ) we need 5 times more typists i . e . 2 x 5 = 10 typists . c | a = 1 / 2
b = 25 / a
c = b / 5
|
a ) 14 years , b ) 12 years , c ) 56 years , d ) 66 years , e ) 11 years | e | multiply(11, const_1) | the total age of a and b is 11 years more than the total age of b and c . c is how many year younger than | "given that a + b = 11 + b + c = > a Γ’ β¬ β c = 11 + b Γ’ β¬ β b = 11 = > c is younger than a by 11 years answer : e" | a = 11 * 1
|
a ) 12 , b ) 6 , c ) 14 , d ) 16 , e ) 10 | a | divide(power(12, 2), 12) | n ^ ( n / 2 ) = 2 is true when n = 2 in the same way what is the value of n if n ^ ( n / 2 ) = 12 ? | n ^ ( n / 2 ) = 12 apply log n / 2 logn = log 12 nlogn = 2 log 12 = log 12 ^ 2 = log 144 logn = log 144 now apply antilog n = 144 / n now n = 12 answer : a | a = 12 ** 2
b = a / 12
|
a ) rs . 670 , b ) rs . 690 , c ) rs . 600 , d ) rs . 625 , e ) rs . 654 | c | subtract(800, divide(multiply(subtract(850, 800), 3), 4)) | a sum of money at simple interest amounts to rs . 800 in 3 years and to rs . 850 in 4 years . the sum is : | "explanation : s . i . for 1 year = rs . ( 850 - 800 ) = rs . 50 . s . i . for 3 years = rs . ( 50 x 3 ) = rs . 150 . principal = rs . ( 800 - 150 ) = rs . 600 . answer : option c" | a = 850 - 800
b = a * 3
c = b / 4
d = 800 - c
|
a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | e | add(150, multiply(3, const_10)) | according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 19 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounces of water , that would mean we would need 150 packets of smoothie mix ( along with 12 ( 150 ) ounces of water for a total of 150 packets . however , we know , after a more careful reading of the stimulus , that the 3 ounces are not included in the 12 ounces . as such , the answer has to be less than 150 packets , since 150 would be too much powder considering you already have 150 ( 12 ) ounces of water and need less packets than water to make a smoothie . as such , the only answer less than 150 is 120 , a . does this make sense ? or am i way off base ? e" | a = 3 * 10
b = 150 + a
|
['a ) 2', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 10'] | e | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 4410 multiplied by y is the square of an integer , then y must be | 4410 = 2 * 3 ^ 2 * 5 * 7 ^ 2 to be perfect square , we need to multiply by at least 2 * 5 = 10 . the answer is e . | a = 2 + 3
b = a * 2
|
a ) 35 , b ) 42 , c ) 44 , d ) 46 , e ) none | a | subtract(add(20, 17), const_2) | if p and q are positive integers each greater than 1 , and 17 ( p + 1 ) = 20 ( q + 1 ) , what is the least possible value of p + q ? | "17 ( p + 1 ) = 29 ( q + 1 ) - - > ( p + 1 ) / ( q + 1 ) = 20 / 17 - - > the least positive value of p + 1 is 20 , so the least value of p is 19 and the least positive value of q + 1 is 17 , so the least value of q is 16 - - > the least value of p + q is 19 + 16 = 35 . answer : a" | a = 20 + 17
b = a - 2
|
a ) 1 / 3 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | b | divide(subtract(80, 50), subtract(80, 20)) | a portion of the 80 % solution of chemicals was replaced with an equal amount of 20 % solution of chemicals . as a result , 50 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.8 ( 1 - x ) + 0.2 * x = 0.5 - - > x = 1 / 2 . answer : b ." | a = 80 - 50
b = 80 - 20
c = a / b
|
a ) 16 , b ) 18 , c ) 36 , d ) 19 , e ) 32 | a | gcd(1008, 928) | the maximum number of students among them 1008 pens and 928 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 1008 number of pencils = 928 required number of students = h . c . f . of 1008 and 928 = 16 answer is a" | a = math.gcd(1008, 928)
|
a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | c | add(10, 12) | lilly has 10 fish and rosy has 12 fish . in total , how many fish do they have in all ? | 10 + 12 = 22 the answer is c . | a = 10 + 12
|
a ) 72 , b ) 120 , c ) 240 , d ) 360 , e ) 720 | a | lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 18) | what is the least common multiple of 6 , 18 , and 24 ? | "let us first write the numbers in the form of prime factors : 6 = 2 * 3 18 = 2 * 3 ^ 2 24 = 2 ^ 3 * 3 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 72 option a" | a = 10 + 2
b = 3 * 10
c = b - 3
d = math.lcm(a, c)
e = math.lcm(d, 18)
|
a ) 150 , b ) 250 , c ) 300 , d ) 370 , e ) 280 | c | multiply(divide(420, const_3), const_2.0) | the ratio of number of boys and girls in a school is 2 : 5 . if there are 420 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 2 x and 5 x total students = 420 number of girls in the school = 5 * 420 / 7 = 300 answer is c" | a = 420 / 3
b = a * 2
|
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