options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 3.9 , b ) 4.6 , c ) 5.2 , d ) 6 , e ) 8.1 | a | divide(subtract(600, 20), 150) | a straight line in the xy - plane has y - intercept of 20 . on this line the x - coordinate of the point is 150 and y - coordinate is 600 then what is the slope of the line ? | "eq of line = y = mx + c c = 20 x = 150 y = 150 m + 20 , substitute y by 600 as given in question . 600 = 150 m + 20 , m = 3.9 correct option is a" | a = 600 - 20
b = a / 150
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a ) 400 bc , b ) 400 b / c , c ) 800 bc , d ) 800 b / c , e ) 800 / bc | c | multiply(2, 400) | a case contains c cartons . each carton contains b boxes , and each box contains 400 paper clips . how many paper clips are contained in 2 cases ? | "2 cases * c cartons / case * b boxes / carton * 400 clips / box = 800 bc paper clips the answer is c ." | a = 2 * 400
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a ) 238 , b ) 200 , c ) 287 , d ) 117 , e ) 198 | b | subtract(divide(divide(5300, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5300 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 5300 [ 4 l - 40 ] = 5300 / 26.50 = 200 4 l = 240 l = 240 / 4 = 60 meters . answer : b" | a = 5300 / 26
b = a / 2
c = 2 * 20
d = b - c
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a ) 12500 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17000 | a | multiply(divide(subtract(15000, divide(multiply(15000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 15000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 15000 - 10 / 100 ( 15000 ) = 15000 - 1500 = $ 13500 = > 1.08 x = 13500 = > x = 12500 a" | a = 15000 * 10
b = a / 100
c = 15000 - b
d = 100 + 8
e = c / d
f = e * 100
|
a ) 56 seconds , b ) 3515 seconds , c ) 41.15 seconds , d ) 30 seconds , e ) 36 seconds | c | divide(add(250, 150), divide(35, const_3_6)) | calculate the time it will take for a train that is 250 meter long to pass a bridge of 150 meter length , if the speed of the train is 35 km / hour ? | "speed = 35 km / hr = 35 * ( 5 / 18 ) m / sec = 9.72 m / sec total distance = 250 + 150 = 400 meter time = distance / speed = 400 * ( 1 / 9.72 ) = 41.15 seconds answer : c" | a = 250 + 150
b = 35 / const_3_6
c = a / b
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a ) 200 , b ) 300 , c ) 400 , d ) 500 , e ) 600 | c | divide(480, add(const_1, divide(20, const_100))) | a number increased by 20 % gives 480 . the number is ? | formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 480 ( 120 Γ 4 = 480 ) 100 % - - - - - - - > 400 ( 100 Γ 4 = 400 ) option ' c ' | a = 20 / 100
b = 1 + a
c = 480 / b
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a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | d | divide(divide(1620, const_2), const_2) | if k ^ 3 is divisible by 1620 , what is the least possible value of integer k ? | "1620 = 2 ^ 2 * 3 ^ 4 * 5 therefore k must include at least 2 * 3 ^ 2 * 5 = 90 . the answer is d ." | a = 1620 / 2
b = a / 2
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a ) a ) 4 , b ) b ) 1 , c ) c ) 6 , d ) d ) 3 , e ) e ) 5 | c | subtract(23, reminder(1052, 23)) | what least number should be added to 1052 , so that the sum is completely divisible by 23 | "explanation : ( 1052 / 23 ) gives remainder 17 17 + 6 = 23 , so we need to add 6 answer : option c" | a = 23 - reminder
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a ) 17608 , b ) 17606 , c ) 17604 , d ) 17600 , e ) 18560 | e | divide(multiply(add(const_100, 16), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 16 % if no discount was offered ? | price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 16 % if no discount was offered = 16000 * 116 / 100 = rs . 18560 . answer : e | a = 100 + 16
b = 12500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
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a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 6 / 17 | d | divide(subtract(7, multiply(const_2, const_3)), 7) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 7 | "total number of cases = 6 * 6 = 36 favourable cases = [ ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,4 ) , ( 5,6 ) , ( 5,3 ) , ( 5,5 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) ] = 15 so probability = 15 / 36 = 5 / 12 answer is d" | a = 2 * 3
b = 7 - a
c = b / 7
|
a ) 15.6 , b ) 16.0 , c ) 17.5 , d ) 18.6 , e ) 19.1 | d | add(3.00, multiply(subtract(divide(8, divide(1, 5)), 1), 0.40)) | a certain taxi company charges $ 3.00 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 $ 3.00 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 $ 3.00 for the first 0.2 miles plus ( 8 - 0.2 ) / 0.2 * 0.4 = $ 3.0 + $ 15.6 = $ 18.6 . answer : d ." | a = 1 / 5
b = 8 / a
c = b - 1
d = c * 0
e = 3 + 0
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a ) 80 , 8284 , b ) 65 , 6871 , c ) 45 , 4742 , d ) 100 , 10249 , e ) 87 , 8889 | a | divide(subtract(246, add(const_2, const_4)), 3) | the sum of 3 consecutive even numbers is 246 . what are the numbers ? | first x make the first x second x + 2 even numbers , sowe add 2 to get the next third x + 4 add 2 more ( 4 total ) to get the third f + s + t = 246 summeansaddfirst ( f ) plussecond ( s ) plusthird ( t ) ( x ) + ( x + 2 ) + ( x + 4 ) = 246 replace each f , s , and t withwhatwe labeled them x + x + 2 + x + 4 = 246 here the parenthesis are not needed 3 x + 6 = 246 combine like terms x + x + x and 2 + 4 β 6 β 6 subtract 6 fromboth sides 3 x = 240 the variable ismultiplied by 3 3 3 divide both sides by 3 x = 80 our solution for x first 80 replace x in the origional listwith 80 . second ( 80 ) + 2 = 82 the numbers are 80 , 82 , and 84 . third ( 80 ) + 4 = 84 correct answer a | a = 2 + 4
b = 246 - a
c = b / 3
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a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | a | divide(36, multiply(divide(48, multiply(4, 2)), 6)) | if 4 men can colour 48 m long cloth in 2 days , then 6 men can colour 36 m long cloth in | "the length of cloth painted by one man in one day = 48 / 4 Γ 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 Γ 6 = 1 day . a" | a = 4 * 2
b = 48 / a
c = b * 6
d = 36 / c
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['a ) 75.00 %', 'b ) 56.80 %', 'c ) 110 %', 'd ) 150 %', 'e ) 180 %'] | b | multiply(divide(subtract(subtract(square_area(add(add(const_1, divide(100, 100)), divide(multiply(add(const_1, divide(100, 100)), 40), 100))), const_1), const_4), add(const_1, square_area(add(const_1, divide(100, 100))))), 100) | the length of each side of square a is increased by 100 percent to make square b . if the length of the side of square b is increased by 40 percent to make square c , by what percent is the area of square c greater than the sum of the areas of squares a and b ? | let length of each side of square a be 10 area of a = 10 ^ 2 = 100 since , length of each side of square a is increased by 100 percent to make square b length of each side of square b = 2 * 10 = 20 area of b = 20 ^ 2 = 400 since , length of the side of square b is increased by 40 percent to make square c length of each side of square c = 1.4 * 20 = 28 area of c = 28 ^ 2 = 784 difference in areas of c and cummulative areas of a and b = 784 - ( 400 + 100 ) = 284 percent is the area of square c greater than the sum of the areas of squares a and b = ( 284 / 500 ) * 100 % = 56.80 % answer b | a = 100 / 100
b = 1 + a
c = 100 / 100
d = 1 + c
e = d * 40
f = e / 100
g = b + f
h = square_area - (
i = h - 1
j = i / 4
k = 100 / 100
l = 1 + k
m = 1 + square_area
n = j * m
|
a ) 152 , b ) 120 , c ) 130 , d ) 160 , e ) 210 | a | divide(228, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100)))) | a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 228 for it , the cost price of the cricket bat for a is : | "a 152 125 % of 120 % of a = 228 125 / 100 * 120 / 100 * a = 228 a = 228 * 2 / 3 = 152 ." | a = 20 / 100
b = 1 + a
c = 25 / 100
d = 1 + c
e = b * d
f = 228 / e
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a ) 1.5 % , b ) 1.9 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(300, 0.8), multiply(700, 1.8)), const_1000) | by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 300 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 300 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 300 ) / 100 + ( 1.8 * 700 ) / 100 = 15 gms % of liquid x in resultant mixture = ( 15 / 1000 ) * 100 = 1.5 % answer : 1.5 % answer : a" | a = 300 * 0
b = 700 * 1
c = a + b
d = c / 1000
|
a ) 2 hrs , b ) 6 hrs , c ) 5 hrs , d ) 4 hrs , e ) 1 hr | b | divide(const_1, add(divide(const_1, 10), divide(const_1, 15))) | two pipes a and b can fill a tank in 10 hour and 15 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 10 part filled by b in 1 hour = 1 / 15 part filled by a + b in 1 hour = 1 / 10 + 1 / 15 = 1 / 6 both the pipes will together will fill the tank in 6 hours . answer is b" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / c
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a ) 4 / 3 , b ) 1 / 3 , c ) 1 / 27 , d ) - 1 / 12 , e ) - 4 / 3 | c | divide(power(divide(power(negate(1), 2), 2), 2), negate(3)) | if x # y is defined to equal x ^ 2 / y for all x and y , then ( - 1 # 3 ) # 3 = | "( - 1 ) ^ 2 / 3 = 1 / 3 ( 1 / 3 ) ^ 2 / 3 = 1 / 27 so c is my answer" | a = negate ** (
b = a / 2
c = b ** 2
d = c / 2
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a ) 4,888 , b ) 4,898 , c ) 4,889 , d ) 4,869 , e ) 4,896 | c | subtract(852,755, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 362,855 and 852,755 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 362,855 and 852,755 is 852,755 - 362,855 = 489,900 - one number per each hundred gives 133,900 / 100 = 4,889 numbers . answer : c ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 852 - 755
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a ) 250 m , b ) 300 m , c ) 120 m , d ) 200 m , e ) 166 m | b | divide(12, subtract(divide(12, 10), 15)) | a train covers a distance of 12 km in 10 min . if it takes 15 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 15 = 300 m . answer : b" | a = 12 / 10
b = a - 15
c = 12 / b
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['a ) β 3 : 1', 'b ) β 7 : 1', 'c ) β 2 : 1', 'd ) 2 : 1', 'e ) 3 : 1'] | c | sqrt(2) | two right circular cylinders of equal volumes have their heights in the ratio 1 : 2 . find the ratio of their radii . | explanation : let their heights be h and 2 h and radii be r and r respectively then . Ο r 2 h = Ο r 2 ( 2 h ) = > r 2 / r 2 = 2 h / h = 2 / 1 = > r / r = β 2 / 1 = > r : r = β 2 : 1 answer is c | a = math.sqrt(2)
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a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 5,000 | c | divide(multiply(4.8, power(10, 11)), multiply(240, power(10, add(const_4, const_2)))) | a certain country had a total annual expenditure of $ 4.8 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ? | "total expenditure / population = per capita expenditure hence , ( 4,8 x 10 ^ 11 ) / 240 000 000 = ( 4,8 x 10 ^ 11 ) / ( 2,4 x 10 ^ 8 ) = 2 x 10 ^ ( 11 - 8 ) = 2 x 10 ^ 3 = 2000 . answer is c ." | a = 10 ** 11
b = 4 * 8
c = 4 + 2
d = 10 ** c
e = 240 * d
f = b / e
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a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 3 , d ) 4 / 9 , e ) 5 / 9 | c | divide(2, add(2, 4)) | a waitress ' s income consists of her salary and tips . during one week , her tips were 2 / 4 of her salary . what fraction of her income for the week came from tips ? | "her tips were 2 / 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 2 / 4 ) ( $ 4 ) = $ 2 so , her total income = $ 4 + $ 2 = $ 6 what fraction of her income for the week came from tips $ 2 / $ 6 = 1 / 3 = c" | a = 2 + 4
b = 2 / a
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a ) 122821 , b ) 281228 , c ) 90300 , d ) 122850 , e ) 128111 | c | divide(multiply(1, 601), const_4) | what is the sum of all even numbers from 1 to 601 ? | "explanation : 600 / 2 = 300 300 * 301 = 90300 answer : c" | a = 1 * 601
b = a / 4
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a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | b | multiply(divide(20, 300), multiply(divide(20, const_100), 200)) | 60 percent of movie theatres in town x have 3 screens or less . 20 % of those theatres sell an average of more than $ 200 worth of popcorn per showing . 56 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell an average of more than $ 300 worth of popcorn per day ? | "lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 200 = 20 % of 60 = 12 number of movie theaters selling popcorn at $ 300 or less = 56 = > number of movie theaters selling popcorn at more than $ 300 = 100 - 56 = 44 of these 44 theaters , 12 are those with 3 screens or less therefore 18 ( 44 - 12 ) must be those with four screens or more b is the answer" | a = 20 / 300
b = 20 / 100
c = b * 200
d = a * c
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a ) 400 , b ) 625 , c ) 1,750 , d ) 2,500 , e ) 10,000 | c | divide(70, divide(2, 50)) | in a certain pond , 70 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "this is a rather straight forward ratio problem . 1 . 70 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 70 / x thus , x = 1750 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . c" | a = 2 / 50
b = 70 / a
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a ) 75 , b ) 100 , c ) 125 , d ) 175 , e ) 165 | e | divide(subtract(multiply(divide(990, const_3), const_4), 990), const_2) | there are 990 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 990 - x x / 2 + 990 - x / 4 = 1 / 3 * ( 990 ) = 330 x = 330 x / 2 = 165 is supposed to be the answer m is missing something correct option e" | a = 990 / 3
b = a * 4
c = b - 990
d = c / 2
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a ) 10 , b ) 50 , c ) 40 , d ) 20 , e ) 60 | d | add(multiply(5, 3), 5) | 5 years ago amith was 3 times as old as his daughter and 10 years hence , he will be two times as old as his daughter . then amith β s current age is , | d 20 amith β s age = a and daughter β s age = b then , a β 5 = 3 ( b β 5 ) = > a β 3 b + 10 = 0 β¦ β¦ ( 1 ) and , a + 10 = 2 ( a + 10 ) = > a β 2 b β 10 = 0 β¦ β¦ ( 2 ) solving the 2 eqns , we will get a = 50 , b = 20 . therefore , amith β s age is 50 and his daughter β s age is 20 | a = 5 * 3
b = a + 5
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a ) $ 1220 , b ) $ 1280 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 7), 30), multiply(subtract(100, 30), 10)) | at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 7 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ? | "i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 7 ) = $ 280 total revenue : 1000 + 280 = $ 1280 the answer is b ." | a = 10 + 7
b = a * 30
c = 100 - 30
d = c * 10
e = b + d
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a ) 10 % , b ) 50 % , c ) 21.4 % , d ) 52 % , e ) none of these | c | subtract(multiply(divide(300, add(232, 15)), const_100), const_100) | a retailer buys a radio for rs 232 . his overhead expenses are rs 15 . he sellis the radio for rs 300 . the profit percent of the retailer is | explanation : cost price = ( 232 + 15 ) = 247 sell price = 300 gain = ( 53 / 247 ) * 100 = 21.4 % . answer : c | a = 232 + 15
b = 300 / a
c = b * 100
d = c - 100
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a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 9 / 7 | c | divide(multiply(subtract(divide(2, 3), divide(1, 4)), const_12), add(divide(const_12, 3), multiply(subtract(divide(2, 3), divide(1, 4)), const_12))) | a batch of cookies was divided among 3 tins : 2 / 3 of all the cookies were placed in either the blue tin or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction q of the cookies that were placed in the other tins were placed in the green tin ? | blue tin or red tin : 2 / 3 ( n ) red tin : ( 1 / 3 ) n blue tin : ( 1 / 4 ) n what the last statment meant , is it wants this fraction : ( # of cookies in green tin ) / ( # of cookies in red and green tin ) # of cookies in green tin = 2 n / 3 - n / 4 = 8 n - 3 n / 12 = 5 n / 12 # of cookies in red and green tin = n / 3 + 5 n / 12 = 9 n / 12 fraction q = 5 n / 12 * 12 / 9 n = 5 / 9 ( c ) | a = 2 / 3
b = 1 / 4
c = a - b
d = c * 12
e = 12 / 3
f = 2 / 3
g = 1 / 4
h = f - g
i = h * 12
j = e + i
k = d / j
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a ) 3 , b ) 6 , c ) 2 , d ) 5 , e ) 8 | b | divide(multiply(multiply(3, 6), 3), multiply(3, 6)) | aaron will jog from home at 3 miles per hour and then walk back home by the same route at 6 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ? | "xyt / ( x + y ) x = 3 , y = 6 , t = 3 3 * 6 * 3 / 3 + 6 = 54 / 9 = 6 answer : b" | a = 3 * 6
b = a * 3
c = 3 * 6
d = b / c
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a ) rs . 40 , b ) rs . 30 , c ) rs . 25 , d ) rs . 20 , e ) rs . 45 | a | multiply(divide(subtract(multiply(8, 710), multiply(multiply(const_3, const_4), 460)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 710 and a uniform after a year service . he served only for 8 months and got rs . 460 and a uniform . find the price of the uniform ? | "8 / 12 = 2 / 3 * 710 = 473.33 460.00 - - - - - - - - - - - - - - - 13.33 1 / 3 uniform 13.33 1 - - - - - - - - - - - - - - - ? = > rs . 40 answer : a" | a = 8 * 710
b = 3 * 4
c = b * 460
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
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a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | c | divide(subtract(add(add(26, 3), 26), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 26 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 26 + 29 ) = 9 ( x - 1 ) = > 11 x - 9 x = 46 = > 2 x = 46 = > x = 23 . so , average age of the team is 23 years . c" | a = 26 + 3
b = a + 26
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
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a ) 450 , b ) 300 , c ) 500 , d ) 600 , e ) 175 | d | divide(add(125, 73), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 73 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 125 + 73 33 x / 100 = 198 x = 600 answer is d" | a = 125 + 73
b = 33 / 100
c = a / b
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a ) 60 , b ) 45 , c ) 20 , d ) 15 , e ) none | b | multiply(divide(180, multiply(3, 4)), 3) | a ratio between two numbers is 3 : 4 and their l . c . m . is 180 . the first number is | "sol . let the required numbers be 3 x and 4 x . then , their l . c . m . is 12 x . β΄ 12 x = 180 β x = 15 . hence , the first number is 45 . answer b" | a = 3 * 4
b = 180 / a
c = b * 3
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a ) 110 , b ) 120 , c ) 140 , d ) 130 , e ) 112 | c | subtract(add(multiply(3, 40), 50), 30) | mike , jim and bob are all professional fisherman . mike can catch 30 fish in one hour , jim can catch twice as much and bob can catch 50 % more than jim . if the 3 started to fish together and after 40 minutes mike and bob left , how many fish did the 3 fishermen catch in one hour ? | all of them catch fishes in relation to number 30 . . . . 2 / 3 * 30 + 2 * 30 + 2 * 1.5 * 30 * 2 / 3 = 140 answer is c | a = 3 * 40
b = a + 50
c = b - 30
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a ) 41 / 50 , b ) 1 / 221 , c ) 1 / 216 , d ) 1 / 36 , e ) 1 / 42 | d | divide(const_1, power(subtract(divide(22, const_3), const_1), const_2)) | in a certain game of dice , the player β s score is determined as a sum of two throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if john plays this game against 22 other players , what is the probability of the minimum score that will guarantee john some monetary payoff ? | "to guarantee that john will get some monetary payoff he must score the maximum score of 6 + 6 = 12 , because if he gets even one less than that so 11 , someone can get 12 and john will get nothing . p ( 12 ) = 1 / 6 ^ 2 = 1 / 36 . answer : d ." | a = 22 / 3
b = a - 1
c = b ** 2
d = 1 / c
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a ) 10 days . , b ) 20 days . , c ) 25 days . , d ) 12.5 days . , e ) 15 days . | e | multiply(10, divide(75, 25)) | it was calculated that 75 men could complete a piece of work in 10 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "one day work = 1 / 10 one man β s one day work = 1 / ( 10 * 75 ) now : no . of workers = 50 one day work = 50 * 1 / ( 10 * 75 ) the total no . of days required to complete the work = ( 75 * 10 ) / 50 = 15 answer : e" | a = 75 / 25
b = 10 * a
|
a ) 0 , b ) 1 / 14 , c ) 2 / 15 , d ) 1 / 10 , e ) 2 / 10 | b | multiply(divide(1, 7), divide(subtract(const_1, divide(2, 3)), divide(2, 3))) | two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 7 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | suppose the water in each jug is l liters cx x ( 1 / 7 ) = l cx = 7 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 7 l = 1 / 14 answer will be b | a = 1 / 7
b = 2 / 3
c = 1 - b
d = 2 / 3
e = c / d
f = a * e
|
a ) 1 : 4 , b ) 1 : 5 , c ) 2 : 3 , d ) 1 : 2 , e ) 1 : 2 | e | divide(subtract(divide(const_1, 3), divide(const_1, 4)), subtract(divide(const_1, 4), divide(const_1, multiply(const_2, const_4)))) | a grocery store bought some mangoes at a rate of 4 for a dollar . they were separated into two stacks , one of which was sold at a rate of 3 for a dollar and the other at a rate of 6 for a dollar . what was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes ? | "the cost price of a mango = 1 / 4 dollars . the selling price of a mango from the first stack = 1 / 3 dollars - - > the profit from one mango = 1 / 3 - 1 / 4 = 1 / 12 dollars . the selling price of a mango from the second stack = 1 / 6 dollars - - > the loss from one mango = 1 / 3 - 1 / 6 = 1 / 6 dollars . the profit from one mango from the first stack is 4 times the loss from one mango from the second stack . the ratio is 1 / 12 * 6 / 1 = 1 / 2 = 1 : 2 e" | a = 1 / 3
b = 1 / 4
c = a - b
d = 1 / 4
e = 2 * 4
f = 1 / e
g = d - f
h = c / g
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a ) 7.2 kg . , b ) 10.8 kg . , c ) 12.4 kg . , d ) 6.18 kg , e ) none | d | divide(multiply(6, 13.4), 13) | if the weight of 13 meters long rod is 13.4 kg . what is the weight of 6 meters long rod ? | answer β΅ weight of 13 m long rod = 13.4 kg β΄ weight of 1 m long rod = 13.4 / 13 kg β΄ weight of 6 m long rod = 13.4 x 6 / 13 = 6.18 kg option : d | a = 6 * 13
b = a / 13
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a ) 284 % , b ) 276 % , c ) 265.25 % , d ) 284.5 % , e ) 275 % | e | multiply(subtract(power(multiply(divide(50, 50), const_2), const_2), const_1), const_100) | if each side of a rectangle is increased by 50 % with the length being double the width , find the percentage change in its area ? | "area = 2 a x a new length = 250 a / 100 = 5 a / 2 new width = 150 a / 100 = 3 a / 2 new area = ( 5 a x 3 a ) / ( 2 x 2 ) = ( 15 a Β² / 4 ) increased area = = ( 15 a Β² / 4 ) - a Β² increase % = [ ( 11 a Β² / 4 ) x ( 1 / a Β² ) x 100 ] % = 275 % answer : e" | a = 50 / 50
b = a * 2
c = b ** 2
d = c - 1
e = d * 100
|
a ) 22488 , b ) 16500 , c ) 27889 , d ) 23778 , e ) 29982 | b | 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 paying the floor by slabs at the rate of rs . 800 per sq . metre | "explanation : area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = rs . 800 hence total cost = 5.5 Γ 3.75 Γ 800 = 5.5 Γ 3000 = rs . 16500 answer : b ) rs . 16500" | a = 5 * 5
b = 800 * a
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a ) 13 % , b ) 18 % , c ) 36 % , d ) 64 % , e ) none of these | a | multiply(subtract(const_1, sqrt(divide(subtract(const_100, 25), const_100))), const_100) | if the area of a circle decreases by 25 % , then the radius of a circle decreases by | "if area of a circle decreased by x % then the radius of a circle decreases by ( 100 β 10 β 100 β x ) % = ( 100 β 10 β 100 β 25 ) % = ( 100 β 10 β 75 ) % = 100 - 87 = 13 % answer a" | a = 100 - 25
b = a / 100
c = math.sqrt(b)
d = 1 - c
e = d * 100
|
a ) 5.45 % , b ) 6.23 % , c ) 1.75 % , d ) 8.12 % , e ) 10 % | c | multiply(divide(subtract(5800, add(4700, 1000)), add(4700, 1000)), const_100) | alfred buys an old scooter for $ 4700 and spends $ 1000 on its repairs . if he sells the scooter for $ 5800 , his gain percent is ? | "c . p . = 4700 + 1000 = $ 5700 s . p . = $ 5800 gain = 5800 - 5700 = $ 100 gain % = 100 / 5700 * 100 = 1.75 % answer is c" | a = 4700 + 1000
b = 5800 - a
c = 4700 + 1000
d = b / c
e = d * 100
|
a ) 25 , b ) 77 , c ) 20 , d ) 28 , e ) 10 | a | divide(add(add(11, const_4), subtract(36, const_4)), const_2) | find the average of all the numbers between 11 and 36 which are divisible by 5 . | "explanation : average = ( 15 + 20 + 25 + 30 + 35 ) / 5 = 125 / 5 = 25 answer : a" | a = 11 + 4
b = 36 - 4
c = a + b
d = c / 2
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a ) 120 , b ) 240 , c ) 360 , d ) 540 , e ) 720 | b | multiply(factorial(5), const_2) | how many e ways can jason sit with his 5 friends in a row of 6 seats with an aisle on either side of the row , if jason insists on sitting next to one of the aisles ? | jason can select his seat in 2 ways ( two aisles ) his 1 st of 4 friends have 5 seats to select = > his 2 nd of remaining 3 friends will have 4 seat to chose from . . . and so on total ways e = > 2 * 5 * 4 * 3 * 2 * 1 = 240 . b | a = math.factorial(5)
b = a * 2
|
a ) 49 , b ) 84 , c ) 86 , d ) 64 , e ) 56 | c | multiply(68, divide(56, 65)) | 56 : 65 : : 68 : ? | "ans 86 reverse of 68 answer : c" | a = 56 / 65
b = 68 * a
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a ) 10 , b ) 25 , c ) 35 , d ) 40 , e ) 55 | b | divide(multiply(multiply(const_1000, const_100), 0.037), divide(148000, const_1000)) | in the biology lab of ` ` jefferson ' ' high school there are 0.037 * 10 ^ 5 germs , equally divided among 148000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ? | "0.037 * 10 ^ 5 can be written as 3700 148000 * 10 ^ ( - 3 ) can be written as 148 required = 3700 / 148 = 25 answer : b" | a = 1000 * 100
b = a * 0
c = 148000 / 1000
d = b / c
|
a ) 8 , b ) 9 , c ) 10 , d ) 16 , e ) 13 | e | divide(add(add(add(add(1.5, const_4), add(1.5, const_4)), add(const_4, const_4)), 50), 5) | the sum of ages of 5 children born 1.5 years different each is 50 years . what is the age of the elder child ? | "let the ages of children be x , ( x + 1.5 ) , ( x + 3 ) , ( x + 4.5 ) and ( x + 6 ) years . then , x + ( x + 1.5 ) + ( x + 3 ) + ( x + 4.5 ) + ( x + 6 ) = 50 5 x = 35 x = 7 . x + 6 = 7 + 6 = 13 answer : e" | a = 1 + 5
b = 1 + 5
c = a + b
d = 4 + 4
e = c + d
f = e + 50
g = f / 5
|
a ) 190 , b ) 195 , c ) 200 , d ) 205 , e ) 55 | e | divide(divide(multiply(add(10, 100), add(divide(subtract(100, 10), 10), const_1)), const_2), add(divide(subtract(100, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 100 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 100 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 10 ( 10 ) = 100 so we know that there are 10 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 100 = 110 20 + 90 = 110 30 + 80 = 110 etc . since there are 10 total terms , this pattern will create 5 ' pairs ' of 110 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 5 ) ( 110 ) / ( 10 ) = 55 answer : e" | a = 10 + 100
b = 100 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 100 - 10
h = g / 10
i = h + 1
j = f / i
|
a ) 7.5 , b ) 20 , c ) 30 , d ) 32 , e ) 40 | b | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 5))) | a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 5 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 5 ) = 1 / 20 the cistern will be filled in 20 / 1 hrs i . e . , 20 hrs . answer : b" | a = 1 / 4
b = 1 / 5
c = a - b
d = 1 / c
|
a ) 123,750 , b ) 164,850 , c ) 164,749 , d ) 149,700 , e ) 156,720 | a | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 4), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 4 ? | "find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 4 seeing the options after dividing and finding the reminder of 2 my answer was a" | a = 1000 - 3
b = 3 * 3
c = b * 10
d = 3 * 3
e = c + d
f = e + 3
g = a - f
h = g / 4
i = h + 1
j = i / 2
k = 1000 - 3
l = 3 * 3
m = l * 10
n = 3 * 3
o = m + n
p = o + 3
q = k + p
r = j * q
|
a ) 4500 , b ) 5200 , c ) 6900 , d ) 7520 , e ) 6000 | e | divide(1200, divide(subtract(60, subtract(const_100, 60)), const_100)) | in an election a candidate who gets 60 % of the votes is elected by a majority of 1200 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1200 20 x / 100 = 1200 x = 1200 * 100 / 20 = 6000 answer is e" | a = 100 - 60
b = 60 - a
c = b / 100
d = 1200 / c
|
a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(5000, const_2), const_2) | if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 5000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , when pat is ready for college ? | "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,000 and in another 9 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a ." | a = 5000 * 2
b = a * 2
|
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 9 , e ) 1 / 12 | c | divide(6, multiply(6, 6)) | tomeka is playing a dice game . if she rolls the same number on her second roll as she rolls on her first , she wins . each roll is with two , fair , 6 - sided dice . if tomeka rolled a 7 on her first roll , what is the probability that she will win on her second roll ? | there are 6 ways to roll a seven : 1 and 6 , 6 and 1 , 2 and 5 , 5 and 2 , 3 and 4 , 4 and 3 . there are 6 * 6 = 36 ways to roll two six - sided dice . thus the probability of winning by rolling a seven on the second roll given a seven on the first roll is 6 / 36 = 1 / 6 c | a = 6 * 6
b = 6 / a
|
a ) $ 35,000 , b ) $ 40,000 , c ) $ 55,000 , d ) $ 65,000 , e ) $ 75,000 | b | divide(add(multiply(15, 90000), multiply(75, 30000)), add(15, 75)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 90000 . the 75 associates have an average salary of $ 30000 . what is the average salary for the company ? | another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 3 a * 1 + a * 5 ) / 6 = 8 a / 6 = 40000 answer is b . $ 40,000 | a = 15 * 90000
b = 75 * 30000
c = a + b
d = 15 + 75
e = c / d
|
a ) 29 , b ) 92 , c ) 30 , d ) 32 , e ) 45 | e | divide(add(37, 53), const_2) | a man can row upstream at 37 kmph and downstream at 53 kmph , and then find the speed of the man in still water ? | "us = 37 ds = 53 m = ( 53 + 37 ) / 2 = 45 answer : e" | a = 37 + 53
b = a / 2
|
a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 10 : 12 , e ) 1 : 2 | d | divide(divide(multiply(const_4, const_3.0), multiply(4, 4)), divide(multiply(4, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 5 : 4 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 15 : 10 . the ratio of economy cars to sport utility vehicles is 5 : 4 - - > e : s = 5 : 4 = 15 : 12 . thus , l : s = 10 : 12 . answer : d ." | a = 4 * 3
b = 4 * 4
c = a / b
d = 4 * 4
e = 2 * 4
f = d / e
g = c / f
|
a ) 40 % , b ) 35 % , c ) 30 % , d ) 20 % , e ) 28 % | d | divide(multiply(1,500, const_100), add(add(multiply(const_2, const_1000), const_100), 1,500)) | in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 1,500 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 6,000 higher that m , then what percentage of apartments in the building are two - bedroom apartments ? | "ratio of 2 bedroom apartment : 1 bedroom apartment = 1500 : 6000 - - - - - > 1 : 4 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 5 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 5 ) * 100 - - - > 20 % answer : d" | a = 1 * 500
b = 2 * 1000
c = b + 100
d = c + 1
e = a / d
|
a ) - 8 , b ) - 9 , c ) - 5 , d ) - 4 , e ) 9 | e | divide(negate(add(17, 1)), 2) | solve below question 2 x - 1 = 17 | 1 . add 1 to both sides : 2 x - 1 + 1 = 17 + 1 2 . simplify both sides : 2 x = 18 3 . divide both sides by 2 : 4 . simplify both sides : x = 9 e | a = 17 + 1
b = negate / (
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a ) 24 , b ) 34.8 , c ) 36 , d ) 42 , e ) 84 | c | divide(0.54, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.54 more than jane , what was the original price of the dish ? | "the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.54 - - > 15 p - 13.5 p = 54 - - > p = 36 . answer : c ." | a = 15 / 100
b = 10 / 100
c = 1 - b
d = 15 / 100
e = c * d
f = a - e
g = 0 / 54
|
a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.75 , e ) 1.0 | b | divide(multiply(divide(multiply(8, 5), const_100), 80), const_100) | 80 % of 5 / 8 = | "should be simple . 0.8 * 5 / 8 = 4 / 8 = 1 / 2 = 0.5 correct option : b" | a = 8 * 5
b = a / 100
c = b * 80
d = c / 100
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(72, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer t that must divide n is ? | "q : if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer t that must divide n is : a 6 , b 12 , c 24 , d 36 , e 48 n ^ 2 is divisible by 72 , but it must also be greater than 72 . if n is an integer , then n ^ 2 must be a perfect square . the factorization of 72 is ( 8 ) ( 9 ) , so if it is multiplied by 2 , it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144 , a perfect square . so n ^ 2 must be at least 144 or a multiple of 144 , which means that n must be 12 or a multiple of 12 . b" | a = 72 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 24 | e | multiply(divide(subtract(720, divide(480, const_2)), 40), const_2) | angelina walked 720 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 = 720 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 720 / x - 240 / x = 40 . . 480 / x = 40 . . x = 12 m / secs . . so grocery to gym = 2 * 12 = 24 m / s . . . e" | a = 480 / 2
b = 720 - a
c = b / 40
d = c * 2
|
a ) 11 , b ) 14 , c ) 15 , d ) 13 , e ) 18 | a | subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is : | "age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = ( 225 - 214 ) = 11 years . answer a" | a = 15 * 15
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
|
a ) 3 / 10 , b ) 4 / 11 , c ) 5 / 12 , d ) 6 / 13 , e ) 7 / 15 | c | divide(divide(factorial(subtract(9, const_2)), multiply(factorial(subtract(6, const_2)), factorial(subtract(subtract(9, const_2), subtract(6, const_2))))), divide(factorial(9), multiply(factorial(6), factorial(subtract(9, 6))))) | john and peter are among the 9 players a volleyball coach can choose from to field a 6 - player team . if all 6 players are chosen at random , what is the probability of choosing a team that includes john and peter ? | the total possible ways of selecting a 6 - member team is 9 c 6 = 84 the possible ways which include john and peter is 7 c 4 = 35 the probability of choosing both john and peter is 35 / 84 = 5 / 12 the answer is c . | a = 9 - 2
b = math.factorial(a)
c = 6 - 2
d = math.factorial(c)
e = 9 - 2
f = 6 - 2
g = e - f
h = math.factorial(g)
i = d * h
j = b / i
k = math.factorial(9)
l = math.factorial(6)
m = 9 - 6
n = math.factorial(m)
o = l * n
p = k / o
q = j / p
|
a ) a . 150 , b ) b . 300 , c ) c . 600 , d ) d . 1,200 , e ) e . 360 | e | multiply(0.6, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100))) | a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.6 per box per month for the record storage , what is the total amount that the company pays each month for record storage ? | "volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.6 = 360 answer : e" | a = 15 * 12
b = a * 10
c = b / 100
d = c / 100
e = d / 100
f = 1 / 8
g = 0 * 6
|
a ) 1829 , b ) 2356 , c ) 3458 , d ) 3843 , e ) 4820 | a | divide(factorial(62), multiply(factorial(subtract(62, const_2)), factorial(const_2))) | how many diagonals does a 62 - sided convex polygon have ? | "a 62 - sided convex polygon has 62 vertices . if we examine a single vertex , we can see that we can connect it with 59 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 62 vertices can be connected with 59 vertices to create a diagonal then the total number of diagonals would be ( 62 ) ( 59 ) = 3658 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3658 by 2 to get 1829 . the answer is a ." | a = math.factorial(62)
b = 62 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 % | d | multiply(const_100, add(divide(8, 20), divide(3, 10))) | if y > 0 , ( 8 y ) / 20 + ( 3 y ) / 10 is what percent of y ? | "can be reduced to 4 y / 10 + 3 y / 10 = 7 y / 10 = 70 % answer d" | a = 8 / 20
b = 3 / 10
c = a + b
d = 100 * c
|
a ) 8 / 3 , b ) 9 / 7 , c ) 7 / 4 , d ) 5 / 2 , e ) 6 / 7 | a | divide(subtract(10, 2), subtract(const_1, divide(10, 30))) | how many kg of pure salt must be added to 30 kg of 2 % solution of salt and water to increase it to a 10 % solution ? | "amount salt in 30 kg solution = 2 * 30 / 100 = 0.6 kg let x kg of pure salt be added then ( 0.6 + x ) / ( 30 + x ) = 10 / 100 60 + 100 x = 300 + 10 x 90 x = 240 x = 8 / 3 answer is a" | a = 10 - 2
b = 10 / 30
c = 1 - b
d = a / c
|
a ) 743 , b ) 154 , c ) 852 , d ) 741 , e ) 165 | e | divide(subtract(14698, 14), 89) | on dividing 14698 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 14698 - 14 ) / 89 = 165 answer ( e )" | a = 14698 - 14
b = a / 89
|
a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 18.0 sec , e ) 72.0 sec | d | divide(add(110, 190), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 190 m in length ? | "d = 110 + 190 = 300 m s = 60 * 5 / 18 = 50 / 3 t = 300 * 3 / 50 = 18.0 sec answer : d" | a = 110 + 190
b = 60 * const_0_2778
c = a / b
|
a ) 20 % , b ) 25 % , c ) 50 % , d ) 75 % , e ) 60 % | c | multiply(divide(subtract(subtract(add(const_1, divide(20, const_100)), multiply(add(const_1, divide(10, const_100)), divide(75, const_100))), subtract(const_1, divide(75, const_100))), subtract(const_1, divide(75, const_100))), const_100) | paulson spends 75 % of his income . his income is increased by 20 % and he increased his expenditure by 10 % . find the percentage increase in his savings ? | "let original income = $ 100 expenditure = $ 75 savings = $ 25 new income = $ 120 new expenditure = $ 110 / 100 * 75 = $ 165 / 2 new savings = 120 - 165 / 2 = $ 75 / 2 increase in savings = 75 / 2 - 25 = $ 25 / 2 increase % = 25 / 2 * 1 / 25 * 100 = 50 % answer is c" | a = 20 / 100
b = 1 + a
c = 10 / 100
d = 1 + c
e = 75 / 100
f = d * e
g = b - f
h = 75 / 100
i = 1 - h
j = g - i
k = 75 / 100
l = 1 - k
m = j / l
n = m * 100
|
a ) 27 , b ) 30 , c ) 33 , d ) 36 , e ) 39 | c | divide(multiply(divide(33, 4), 16), 4) | rice weighing 33 / 4 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces ) | "33 / 4 Γ· 4 = 33 / 16 pounds in each container 33 / 16 pounds * 16 ounces / pound = 33 ounces in each container the answer is c ." | a = 33 / 4
b = a * 16
c = b / 4
|
a ) 22 , b ) 15 , c ) 77 , d ) 266 , e ) 12 | e | divide(multiply(subtract(22, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 22 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 22 = > x = 4 so deepak age is = 3 ( 4 ) = 12 answer : e" | a = 22 - 6
b = a * 3
c = b / 4
|
a ) 80 , b ) 98 , c ) 105 , d ) 120 , e ) 210 | b | multiply(divide(112, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 112 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 112 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 112 = 16 / 30 x x = 210 210 - 112 = 98 answer b" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 112 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) 15.95 , b ) 17.28 , c ) 19.54 , d ) 21.73 , e ) 23.86 | e | divide(divide(70, 5280), multiply(2, divide(1, const_3600))) | if an object travels 70 feet in 2 seconds , what is the object β s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "70 feet / 2 seconds = 35 feet / second ( 35 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 23.86 miles / hour ( approximately ) the answer is e ." | a = 70 / 5280
b = 1 / 3600
c = 2 * b
d = a / c
|
a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | d | multiply(1200, multiply(800, divide(85, multiply(800, 600)))) | 85 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 85 = 113.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 113.33 = 170 litres . answer d" | a = 800 * 600
b = 85 / a
c = 800 * b
d = 1200 * c
|
a ) 232 , b ) 308 , c ) 252 , d ) 262 , e ) 282 | b | divide(28, divide(650, 28)) | evaluate 28 % of 650 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 650 + ( 45 / 100 ) * 280 = 182 + 126 = 308 answer : option b" | a = 650 / 28
b = 28 / a
|
a ) 111.12 , b ) 111.67 , c ) 428.57 , d ) 111.11 , e ) 101.12 | c | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 600)), subtract(multiply(const_100, const_10), 300))) | a can give b 300 meters start and c 600 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 700 m and c runs 400 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 400 ) / 700 = 571.43 m . b can give c = 1000 - 571.43 = 428.57 m . answer : c" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 600
e = b * d
f = 100 * 10
g = f - 300
h = e / g
i = a - h
|
a ) 1 , b ) 10 , c ) 100 , d ) 50 , e ) 20 | c | divide(multiply(divide(multiply(10, 2000), const_100), 50), const_100) | there are 2000 students in a school and among them 10 % of them attends chess class . 50 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "10 % of 2000 gives 200 . so 200 attends chess and 50 % of 200 gives 100 so 100 enrolled for swimming answer : c" | a = 10 * 2000
b = a / 100
c = b * 50
d = c / 100
|
a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15 | e | multiply(add(multiply(6, 3), 1), multiply(divide(1, 3), divide(1, 5))) | a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 6 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the purposes of renovating the hospital ' s original wing , all of the patients in the hospital ' s original beds must be transferred to beds in the new tower . if patients are neither admitted nor discharged during the transfer , what fraction of the beds in the new tower will be unoccupied once the transfer is complete ? | "i think e - 11 / 15 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 6 x so old = x , new = 5 x now 1 / 3 of x are occupied and 1 / 5 of 5 x are occupied which simplifies to ( 5 / 5 ) x we are shifting 1 / 3 of x to the new ward so there will now be : 1 / 3 of x plus 5 / 5 of x occupied in the new ward . add them up to get 4 / 3 of x there are 5 x beds in new tower so ratio is : ( 4 / 3 ) x / 5 x = 4 / 15 of x subtract that from 15 / 15 of x and you get the number of un - occupied beds to total capacity of new tower = 11 / 15 . e" | a = 6 * 3
b = a + 1
c = 1 / 3
d = 1 / 5
e = c * d
f = b * e
|
a ) 11 : 9 , b ) 11 : 18 , c ) 21 : 19 , d ) 21 : 19 , e ) none | a | divide(divide(subtract(40, subtract(divide(40, const_2), const_2)), const_2), divide(subtract(divide(40, const_2), const_2), const_2)) | the sum of two numbers is 40 and their difference is 4 . the ratio of the numbers is | sol . let the numbers be x and y . then , x + y / x - y = 40 / 4 = 10 β ( x + y ) = 10 ( x - y ) β 9 x = 11 y β x / y = 11 / 9 . answer a | a = 40 / 2
b = a - 2
c = 40 - b
d = c / 2
e = 40 / 2
f = e - 2
g = f / 2
h = d / g
|
a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number q of races we need to conduct to get the top 3 fastest horses ? | q = 7 is the correct answer . good solution buneul . b | a = 25 / 5
b = a + 2
|
a ) 30 , b ) 48.5 , c ) 50.4 , d ) 62.6 , e ) 47.9 | e | subtract(add(divide(multiply(2, 21), subtract(21, const_1)), 21), 2) | the ages of 2 persons differ by 21 years . if 15 years ago the elder one be 8 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 21 8 ( x - 15 ) = x + 21 - 15 x = 18 age of elder person = 18 + 21 = 39 answer is e" | a = 2 * 21
b = 21 - 1
c = a / b
d = c + 21
e = d - 2
|
a ) 33.33 m , b ) 25 m , c ) 45 m , d ) 30 m , e ) none of these | a | multiply(divide(100, 45), subtract(45, 30)) | in a 100 m race , a covers the distance in 30 seconds and b in 45 second . in this race a beats b by : | solution distance covered by b in 15 sec . = ( 100 / 45 x 15 ) m = 33.33 m . β΄ a beats b by 33.33 metres . answer a | a = 100 / 45
b = 45 - 30
c = a * b
|
a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 14 | e | add(12, 2) | if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 12 ) , what is the value of x ? | i am guessing the question is : ( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 12 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 12 hence x = 14 . answer is e | a = 12 + 2
|
a ) rs . 240 , b ) rs . 340 , c ) rs . 260 , d ) rs . 280 , e ) rs . 440 | a | multiply(500, divide(inverse(15), add(add(inverse(multiply(const_2, const_10)), inverse(add(multiply(const_2, const_10), multiply(const_100, const_0_25)))), inverse(15)))) | a , b , c can do a work in 15 , 20,45 days respectively . they get rs 500 for their work . what is the share of a ? | lcm = 180 share of a = ( lcm / a x total amount ) / lcm / a + lcm / b + lcm / c = ( 180 / 15 ) / ( 180 / 15 + 180 / 20 + 180 / 45 ) = ( 12 / 25 ) * 500 = rs . 240 answer : a | a = 1/(15)
b = 2 * 10
c = 1/(b)
d = 2 * 10
e = 100 * const_0_25
f = d + e
g = 1/(f)
h = c + g
i = 1/(15)
j = h + i
k = a / j
l = 500 * k
|
a ) 10 mps , b ) 05 mps , c ) 09 mps , d ) 12 mps , e ) 70 mps | e | multiply(const_0_2778, 252) | express a speed of 252 kmph in meters per second ? | "e 70 mps 252 * 5 / 18 = 70 mps" | a = const_0_2778 * 252
|
a ) 67.5 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 40 minutes after it was poured was how many degrees fahrenheit ? | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 40 minutes after it was poured is : f = 120 ( 2 ^ - 40 / 10 ) + 60 f = 120 * 1 / 16 + 60 f = 15 / 2 + 60 f = 135 / 2 = 67.5 a" | a = 60 / 10
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
|
a ) 6 , b ) 2 , c ) 4 , d ) 8 , e ) 16 | a | multiply(const_2, sqrt(power(3, const_2))) | a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 3 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 3 ( pir ^ 2 ) pir ^ 2 = 6 pir r = 6 answer : a" | a = 3 ** 2
b = math.sqrt(a)
c = 2 * b
|
a ) 1 / 49 , b ) 2 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | b | subtract(const_1, sqrt(divide(25, 49))) | jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 25 / 49 , what is the probability that the next one she draws will be pink ? | "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 25 / 49 = ( 5 / 7 ) * ( 5 / 7 ) therefore probability that the next one she draws is pink = 2 / 7 option ( b )" | a = 25 / 49
b = math.sqrt(a)
c = 1 - b
|
['a ) 1 : 4', 'b ) 1 : 16', 'c ) 1 : 8', 'd ) 1 : 64', 'e ) 2 : 3'] | c | divide(1, multiply(4, const_2)) | the areas of the two spheres are in the ratio 1 : 4 . the ratio of their volume is ? | 4 Ο r 12 : 4 Ο r 22 = 1 : 4 r 1 : r 2 = 1 : 2 4 / 3 Ο r 13 : 4 / 3 Ο r 23 r 13 : r 23 = 1 : 8 answer : c | a = 4 * 2
b = 1 / a
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | c | divide(multiply(250, 40), subtract(250, 50)) | a hostel had provisions for 250 men for 40 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 40 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 40 250 Γ 40 = 200 x 5 Γ 40 = 4 x x = 5 Γ 10 = 50 answer c" | a = 250 * 40
b = 250 - 50
c = a / b
|
['a ) 18 * sqrt 3', 'b ) 18 * sqrt 2', 'c ) 18 * sqrt 4', 'd ) 19 * sqrt 3', 'e ) 17 * sqrt 3'] | a | multiply(multiply(const_2, const_3), divide(9, sqrt(const_3))) | the distance between parallel sides of the hexagon is 9 . determine the length of the hexagon | length of each side of regular hexagon will be 3 * sqrt 3 . total length ( perimeter ) of regular hexagon will be 18 * sqrt 3 answer : a | a = 2 * 3
b = math.sqrt(3)
c = 9 / b
d = a * c
|
a ) 70 m , b ) 16 m , c ) 60 m , d ) 80 m , e ) 88 m | c | divide(12, subtract(divide(12, 10), 3)) | a train covers a distance of 12 km in 10 min . if it takes 3 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 3 = 60 m . answer : c" | a = 12 / 10
b = a - 3
c = 12 / b
|
a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(2, multiply(2, 2)) | if g and f are both odd prime numbers andg < f , then how many different positive integer factors does 2 gfhave ? | g and f are both odd prime numbers - it means either g or f is not 2 and since prime numbers have only two factors - 1 and the number itself g and f each will have ( 1 + 1 ) = 2 factors hence 2 gf will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer | a = 2 * 2
b = 2 * a
|
a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 72 sec , e ) 189 sec | d | divide(600, subtract(multiply(60, const_0_2778), multiply(30, const_0_2778))) | a and b go around a circular track of length 600 m on a cycle at speeds of 30 kmph and 60 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 30 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 72 sec . answer : d" | a = 60 * const_0_2778
b = 30 * const_0_2778
c = a - b
d = 600 / c
|
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