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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | add(floor(divide(22, const_3)), const_1) | what is the smallest integer k for which 64 ^ k > 4 ^ 22 ? | "64 ^ k > 4 ^ 22 4 ^ ( 3 k ) > 4 ^ 22 3 k > 22 k = 8 the answer is e ." | a = 22 / 3
b = math.floor(a)
c = b + 1
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a ) ½ , b ) 27 / 40 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | b | subtract(0.8, divide(1, 8)) | the number 0.8 is how much greater than 1 / 8 ? | let x be the difference then . 8 - 1 / 8 = x 8 / 10 - 1 / 8 = x x = 27 / 40 ans b | a = 1 / 8
b = 0 - 8
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a ) 22 , b ) 099 , c ) 75 , d ) 26 , e ) 18 | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(50, const_100)), const_2)), const_100) | if the radius of a circle is decreased by 50 % , find the percentage decrease | "let original radius = r . new radius = = original area = and new area = decrease in area = = 75 % answer : c" | a = 100 / 100
b = 100 / 100
c = 50 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
|
['a ) 3,360', 'b ) 307,500', 'c ) 621,500', 'd ) 704,000', 'e ) 2 , 816,000'] | b | add(rectangle_area(300, 500), rectangle_area(250, 630)) | guadalupe owns 2 rectangular tracts of land . one is 300 m by 500 m and the other is 250 m by 630 m . the combined area of these 2 tracts is how many square meters ? | one is 300 m by 500 m and the other is 250 m by 630 m : ( 300 * 500 ) + ( 250 * 630 ) = 150000 + 157500 307500 answer : b | a = rectangle_area + (
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a ) 899600 , b ) 856200 , c ) 875200 , d ) 896200 , e ) none | a | subtract(multiply(const_10, 9), 9) | the difference between the place values of 9 and 4 in the number 927435 is | sol . = ( place value of 9 ) – ( place value of 4 ) = ( 900000 - 400 ) = 899600 answer a | a = 10 * 9
b = a - 9
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(divide(subtract(31, multiply(5, const_3)), 8), divide(multiply(5, const_3), 5)) | a certain hobby shop sold model cars for $ 5.00 each and model trains for $ 8.00 each . if a customer purchased both cars and trains from the store for a total of $ 31.00 , what total number of cars and trains did the customer purchase ? | we can take numbers , as price of car 5 , 8 for train and 31 total for ease . now we can determine quickly that total number should range between 31 / 8 < = n < = 31 / 5 , so ans should be between 3 and 7 . now solving the expression 5 a + 8 b = 31 decreasing 31 in multiple of 5 and checking divisibility of that number by 8 . this way we get 3 cars , 2 trains we get $ 31.00 total 3 * 5 + 2 * 8 = 31 hence total number is 3 + 2 = 5 ans c | a = 5 * 3
b = 31 - a
c = b / 8
d = 5 * 3
e = d / 5
f = c + e
|
a ) $ 35,000 , b ) $ 45,000 , c ) $ 55,000 , d ) $ 65,000 , e ) $ 75,000 | c | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 180,000 . the 75 associates have an average salary of $ 30,000 . 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 = ( 6 a * 1 + a * 5 ) / 6 = 11 a / 6 = 55000 answer is c . $ 55,000" | a = 3 * 3
b = a * 10
c = 100 * 10
d = b * c
e = 3 * 10
f = 100 * 10
g = e * f
h = d + g
i = 75 + 15
j = h / i
|
a ) 1 / 45 , b ) 1 / 35 , c ) 1 / 25 , d ) 1 / 15 , e ) 1 / 10 | a | multiply(2, multiply(divide(const_1, subtract(10, const_1)), divide(const_1, 10))) | john and david work at a hospital with 10 other workers . for an internal review , 2 of the 10 workers will be randomly chosen to be interviewed . what is the probability that john and david will both be chosen ? | "probability that john and david will both be chosen out of 8 workers = ( 2 / 10 ) * ( 1 / 9 ) = 1 / 45 answer a" | a = 10 - 1
b = 1 / a
c = 1 / 10
d = b * c
e = 2 * d
|
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) none | a | add(multiply(10, 2), const_4) | in a two - digit , if it is known that its unit ' s digit exceeds its 10 ' s digit by 2 and that the product of the given number and the sum of its digits is equal to 144 , then the number is ? | let the ten ' s digit be x . then , unit ' s digit = x + 2 . number = 10 x + ( x + 2 ) = 11 x + 2 . sum of digits = x + ( x + 2 ) = 2 x + 2 . ( 11 x + 2 ) ( 2 x + 2 ) = 144 22 x 2 + 26 x - 140 = 0 11 x 2 + 13 x - 70 = 0 ( x - 2 ) ( 11 x + 35 ) = 0 x = 2 . hence , required number = 11 x + 2 = 24 . option a | a = 10 * 2
b = a + 4
|
a ) 30 , b ) 28 , c ) 26 , d ) 22 , e ) 19 | d | subtract(add(add(25, 22), 5), 30) | mrs . evans gave a test to her freshmen economics class , which has 30 students enrolled and 25 of them answered question 1 correctly . if 22 answered question 2 correctly and 5 did not take the test then how many answered both questions correctly ? | total number of enrolled students = 30 number of students who did not take test = 5 hence , number of students who took test = 30 - 5 = 25 number of students who answered q 2 correctly = 22 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 22 . number of students who answered both q 1 & q 2 correctly = 22 . answer : d | a = 25 + 22
b = a + 5
c = b - 30
|
a ) 276 , b ) 295 , c ) 230 , d ) 225 , e ) 245 | a | add(lcm(lcm(8, 12), lcm(22, 24)), 12) | what is the smallest number which when diminished by 12 , is divisible 8 , 12 , 22 and 24 ? | "required number = ( lcm of 8 , 12 , 22 and 24 ) + 12 = 264 + 12 = 276 answer : option a" | a = math.lcm(8, 12)
b = math.lcm(22, 24)
c = math.lcm(a, b)
d = c + 12
|
a ) 9 , b ) 8 , c ) 7 , d ) 6 , e ) 5 | b | add(add(3, const_4), const_1) | k is a 3 digit number such that the the ratio of the number to the sum of its digits is least . what is the difference between the hundreds and tens digit ? | explanation : this is only possible only when the hundreds digit has the least value and tens and units has the highest value . so we could get the highest possible sum but possible lowest number . so by this logic this number could only be 199 . so the difference is 8 . note : remember a number given as abc i . e abc = ax 100 + bx 10 + c . so 123 = 1 x 100 + 2 x 10 + 3 . answer : b | a = 3 + 4
b = a + 1
|
a ) 1.05 , b ) 1.04 , c ) - 0.8 , d ) - 1.0 , e ) - 0.1 | c | subtract(2.423, 3.421) | ♠ n denotes the number obtained when n is rounded to the nearest tenth . for example ♠ 4.31 = 4.3 ♠ 2.423 - ♠ 3.421 = | ♠ 2.423 - ♠ 3.421 2.4 - 3.2 = - 0.8 answer : c | a = 2 - 423
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a ) 15 / 2 , b ) 20 / 1 , c ) 5 / 6 , d ) 3 / 10 , e ) 1 / 15 | b | add(divide(8, 1), divide(5, 1)) | if 1 tic equals 5 tacs and 2 tacs equal 8 tocs , what is the ratio of one tic to one toc ? | "tic = 5 * tac and 2 * tac = 8 * toc ; 2 * tic = 10 * tac and 10 * tac = 40 * toc - - > 2 * tic = 40 * toc - - > tic / toc = 40 / 2 = 20 / 1 . answer : b ." | a = 8 / 1
b = 5 / 1
c = a + b
|
a ) 31 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | e | multiply(divide(subtract(65, 5), add(5, 7)), 7) | one hour after yolanda started walking from x to y , a distance of 65 miles , bob started walking along the same road from y to x . if yolanda â s walking rate was 5 miles per hour and bob â s was 7 miles per hour , how many miles had bob walked when they met ? | "let t be the number of hours that bob had walked when he met yolanda . then , when they met , bob had walked 4 t miles and yolanda had walked 5 ( t + 1 ) miles . these distances must sum to 65 miles , so 7 t + 5 ( t + 1 ) = 65 , which may be solved for t as follows 7 t + 5 ( t + 1 ) = 65 7 t + 5 t + 5 = 65 12 t = 60 t = 5 ( hours ) therefore , bob had walked 7 t = 7 ( 5 ) = 35 miles when they met . the best answer is e ." | a = 65 - 5
b = 5 + 7
c = a / b
d = c * 7
|
a ) 40 , b ) 30 , c ) 60 , d ) 80 , e ) 50 | a | divide(subtract(multiply(divide(30, const_100), 100), 10), divide(50, const_100)) | if 30 % of 100 is greater than 50 % of a number by 10 , then the number is ? | "let the number be x 30 % of 100 - 50 % of x = 10 50 / 100 x = ( 30 / 100 * 100 ) - 10 x / 2 = 20 x = 40 answer is a" | a = 30 / 100
b = a * 100
c = b - 10
d = 50 / 100
e = c / d
|
a ) 125 , b ) 321 , c ) 561 , d ) 852 , e ) 420 | e | subtract(multiply(divide(add(42, 42), subtract(42, 35)), 35), add(subtract(subtract(42, divide(add(42, 42), subtract(42, 35))), divide(add(42, 42), subtract(42, 35))), const_2)) | there were 35 students in a hostel . due to the admission of 7 new students , he expenses of the mess were increased by rs . 42 per day while the average expenditure per head diminished by rs 1 . what was the original expenditure of the mess ? | "let the original average expenditure be rs . x . then , 42 ( x - 1 ) - 35 x = 42 ; 7 x = 84 ; x = 12 . original expenditure = rs . ( 35 x 12 ) = rs . 420 . . ans : e" | a = 42 + 42
b = 42 - 35
c = a / b
d = c * 35
e = 42 + 42
f = 42 - 35
g = e / f
h = 42 - g
i = 42 + 42
j = 42 - 35
k = i / j
l = h - k
m = l + 2
n = d - m
|
a ) 3 / 25 , b ) 5 / 12 , c ) 7 / 12 , d ) 2 / 3 , e ) 25 / 36 | b | divide(divide(multiply(40, 5), 40), 12) | at a speed of 40 miles per hour , a certain car uses 1 gallon of gasoline every 40 miles . if the car starts with a full 12 gallon tank of gasoline and travels for 5 hours at 40 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | "gas used = ( 5 hours ) * ( 40 miles / hour ) * ( 1 gallon / 40 miles ) = 5 gallons portion used = ( 5 ) / 12 = 5 / 12 ans b" | a = 40 * 5
b = a / 40
c = b / 12
|
a ) 84.71 kmph . , b ) 85.74 kmph . , c ) 83.71 kmph . , d ) 85.71 kmph . , e ) 84.71 kmph . | d | subtract(subtract(const_100, const_10), const_4) | niharika covers a distance in 30 min , if she drives at a speed of 60 km / h on an average . find the speed at which she must drive at to reduce the time of the journey by 30 % . | given speed = 60 kmph . it means niharika covered 60 km in 60 min ( 1 hour ) . so , in 30 min he will cover 30 km . 30 % time reduced = 30 - 30 % of 30 = 21 min . thus , niharika needs to cover 30 km in 21 min ( 0.35 hour ) . speed * time = distance . speed * 0.35 = 30 speed = 85.71 kmph . answer : option d | a = 100 - 10
b = a - 4
|
a ) 116 seconds , b ) 150 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | multiply(divide(230, add(230, 234)), 230) | rahul ' s housing colony have one tap , working alone at its maximum constant rate , can fill a certain tank in 230 seconds . a second tap , working alone at its maximum constant rate , can fill the same tank in 234 seconds . if both taps work together at their respective maximum constant rates , the time it will take to fill the tank is closest to | tap 1 : 230 secs tap 2 : 234 secs considering the average of these 2 taps : 232 secs . so 1 tap can fill the tank in 232 secs , so 2 taps can fill the tank in 232 / 2 = 116 secs . closest answer is a . | a = 230 + 234
b = 230 / a
c = b * 230
|
a ) 3 / 40,000 , b ) 1 / 10,000 , c ) 9 / 2,000 , d ) 1 / 60 , e ) 1 / 15 | b | divide(1, const_3) | a certain junior class has 1,000 students and a certain senior class has 600 students . among these students , there are 60 siblings pairs , each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected at will be a sibling pair ? | "total number of ways of choosing one student from each group is = 600 * 1000 number of cases in which a sibling pair will be got is = 60 thus the probability that the 2 students selected will be a sibling pair is = 60 / ( 600 * 1000 ) = 1 / 10,000 b" | a = 1 / 3
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a ) 4 , b ) 3 , c ) 2 , d ) none of the above , e ) 5 | d | divide(subtract(power(8, 1), power(8, 1)), 8) | what is the remainder when 8 ^ 1 + 8 ^ 2 + 8 ^ 3 + . . . + 8 ^ 9 is divided by 2 ? | "notice that in the brackets we have the sum of 9 even multiples of 2 , which yields remainder of 0 upon division by 2 . answer : d" | a = 8 ** 1
b = 8 ** 1
c = a - b
d = c / 8
|
a ) 30 days , b ) 25 days , c ) 20 days , d ) 10 days , e ) 15 days | e | divide(subtract(const_1, multiply(10, divide(const_1, 20))), divide(const_1, 30)) | a can do a work in 20 days , b in 30 days . a alone start the work after 10 days he left work . in how many days b can finish the work ? | "10 days work of a = 10 / 20 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 b can finish 1 / 2 work = 30 * 1 / 2 = 15 days answer is e" | a = 1 / 20
b = 10 * a
c = 1 - b
d = 1 / 30
e = c / d
|
a ) 1 - 9 , b ) 1 - 7 , c ) 2 - 8 , d ) 1 - 3 , e ) 3 - 10 | d | divide(27, 24) | there are 24 students in chad â € ™ s class . he brought 27 cookies to pass out for his birthday . how many cookies will each student get ? will there be any cookies left over ? | 27 / 24 = 1 r 3 chad will give each student 1 cookie and there will be 3 cookies left over correct answer d | a = 27 / 24
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a ) 12 , b ) 27 , c ) 29 , d ) 45 , e ) 21 | d | add(divide(18000, 450), 5) | a shopkeeper sells 450 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 450 = rs . 40 loss per metre = rs . 5 cp per metre = 40 + 5 = rs . 45 . answer : d | a = 18000 / 450
b = a + 5
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a ) 37 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 51 % . | a | multiply(divide(add(multiply(divide(20, const_100), 2), multiply(divide(55, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 20 % of alcohol and another vessel of capacity 6 litre had 55 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "20 % of 2 litres = 0.4 litres 55 % of 6 litres = 3.3 litres therefore , total quantity of alcohol is 3.7 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 37 % a" | a = 20 / 100
b = a * 2
c = 55 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
|
a ) 38.9 kmph , b ) 38.8 kmph , c ) 38.1 kmph , d ) 37.1 kmph , e ) 37.8 kmph | c | divide(add(35, 28), const_2) | a man goes from a to b at a speed of 35 kmph and comes back to a at a speed of 28 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 35 ) + ( d / 28 ] = ( 2 d ) / [ 11 d / 140 ) = > 38.1 kmph . answer : c" | a = 35 + 28
b = a / 2
|
a ) 23.56 , b ) 33.0 , c ) 23.55 , d ) 23.53 , e ) 23.57 | b | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(22, 20)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 22 cm and breadth 20 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . | "let the side of the square be a cm . parameter of the rectangle = 2 ( 22 + 20 ) = 84 cm parameter of the square = 84 cm i . e . 4 a = 84 a = 21 diameter of the semicircle = 21 cm circimference of the semicircle = 1 / 2 ( â ˆ ) 21 = 1 / 2 ( 22 / 7 ) ( 21 ) = 66 / 2 = 33 cm answer : b" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) 20 , b ) 12 , c ) 15 , d ) 18 , e ) 16 | a | subtract(divide(12, divide(2, 5)), 12) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 5 of the mice are brown . if there are 12 white mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of brown mice = 2 / 5 m number of white mice = 3 / 5 m = 12 = > m = 20 answer a" | a = 2 / 5
b = 12 / a
c = b - 12
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['a ) w = 15 and l = 20', 'b ) w = 25 and l = 30', 'c ) w = 35 and l = 40', 'd ) w = 45 and l = 50', 'e ) w = 55 and l = 60'] | a | divide(subtract(divide(70, const_2), sqrt(subtract(power(divide(70, const_2), const_2), multiply(const_4, 300)))), const_2) | the area of a rectangular field is equal to 300 square meters . its perimeter is equal to 70 meters . find the length and width of this rectangle . | l * w = 300 : area , l is the length and w is the width . 2 l + 2 w = 70 : perimeter l = 35 - w : solve for l ( 35 - w ) * w = 300 : substitute in the area equation w = 15 and l = 20 : solve for w and find l using l = 35 - w . correct answer a | a = 70 / 2
b = 70 / 2
c = b ** 2
d = 4 * 300
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
|
a ) 1 / 4 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | a | divide(subtract(0.75, 0.5), subtract(0.6, 0.5)) | last month , john rejected 0.5 % of the products that he inspected and jane rejected 0.6 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.6 ( x ) + 0.5 ( 1 - x ) = 0.75 0.1 x = 0.75 - 0.5 x = 0.25 / 0.1 x = 1 / 4 therefore the answer is a : 1 / 4 ." | a = 0 - 75
b = 0 - 6
c = a / b
|
a ) 732 , b ) 990 , c ) 1098 , d ) 930 , e ) 1405 | d | subtract(subtract(multiply(346, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | the total number of digits used in numbering the pages of a book having 346 pages is | "total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 247 ) = ( 9 + 180 + 741 ) = 930 . answer : d" | a = 346 * 3
b = 100 - 1
c = a - b
d = 10 - 1
e = c - d
|
a ) 143 , b ) 145 , c ) 156 , d ) 134 , e ) 100 | a | multiply(110, add(const_1, divide(30, const_100))) | a light has a rating of 110 watts , it is replaced with a new light that has 30 % higher wattage . how many watts does the new light have ? | "final number = initial number + 30 % ( original number ) = 110 + 30 % ( 110 ) = 110 + 33 = 143 . answer a" | a = 30 / 100
b = 1 + a
c = 110 * b
|
a ) 100 , b ) 70 , c ) 95 , d ) 90 , e ) 85 | d | subtract(multiply(10, multiply(100, const_0_2778)), 150) | a train 150 m long crosses a platform 100 m long in 10 sec ; find the speed of the train ? | "d = 150 + 100 = 250 t = 10 s = 250 / 10 * 18 / 5 = 90 kmph . answer : d" | a = 100 * const_0_2778
b = 10 * a
c = b - 150
|
a ) 21.64 seconds , b ) 22.64 seconds , c ) 23.64 seconds , d ) 26.64 seconds , e ) 35.64 seconds | d | divide(add(240, 130), divide(50, const_3_6)) | calculate the time it will take for a train that is 240 meter long to pass a bridge of 130 meter length , if the speed of the train is 50 km / hour ? | "speed = 35 km / hr = 50 * ( 5 / 18 ) m / sec = 13.89 m / sec total distance = 240 + 130 = 370 meter time = distance / speed = 370 * ( 1 / 13.89 ) = 26.64 seconds answer : d" | a = 240 + 130
b = 50 / const_3_6
c = a / b
|
a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | b | multiply(divide(200, add(4, 6)), 6) | there are 200 doctors and nurses at a hospital . if the ratio of doctors to nurses is 4 to 6 , how many nurses are at the hospital ? | "the number of nurses at the hospital is ( 6 / 10 ) * 200 = 120 . the answer is b ." | a = 4 + 6
b = 200 / a
c = b * 6
|
a ) 10 liters , b ) 15 liters , c ) 20 liters , d ) 18 liters , e ) 22 liters | b | subtract(divide(multiply(divide(5, const_100), 10), divide(2, const_100)), 10) | milk contains 5 % water . what content of pure milk should be added to 10 liters of milk to reduce this to 2 % ? | "quantity of water in 10 liters = 5 % of 10 liters = 0.5 liters let x liters of pure milk be added . then , 0.5 / ( 10 + x ) = 2 / 100 2 x = 30 x = 15 liters answer is b" | a = 5 / 100
b = a * 10
c = 2 / 100
d = b / c
e = d - 10
|
a ) 69 miles , b ) 85.5 miles , c ) 96 miles , d ) 100 miles , e ) 112 miles | b | divide(multiply(multiply(subtract(18, 4), add(18, 4)), 10), add(add(18, 4), subtract(18, 4))) | the current in a river is 4 mph . a boat can travel 18 mph in still water . how far up the river can the boat travel if the round trip is to take 10 hours ? | "upstream speed = 18 - 4 = 14 mph downstream speed = 18 + 4 = 22 mph d / 14 + d / 22 = 10 hours solving for d we get d = 85.5 answer : b" | a = 18 - 4
b = 18 + 4
c = a * b
d = c * 10
e = 18 + 4
f = 18 - 4
g = e + f
h = d / g
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a ) 40 % , b ) 50 % , c ) 57 % , d ) 60 % , e ) 80 % | c | multiply(divide(subtract(90, add(multiply(2, 5), multiply(5, 2))), 90), const_100) | a batsman scored 90 runs which included 2 boundaries and 5 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 90 - ( 2 x 4 + 5 x 6 ) = 90 - ( 38 = 52 now , we need to calculate 38 is what percent of 90 . = > 52 / 90 * 100 = 57 % option c" | a = 2 * 5
b = 5 * 2
c = a + b
d = 90 - c
e = d / 90
f = e * 100
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | c | divide(divide(divide(2835, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 2835 what is the least possible value of k ? | "2835 = 3 * 3 * 3 * 3 * 5 * 7 thus k must include numbers at least up to the number 9 so that there are at least four appearances of 3 ( that is : 3 , 6 , and 9 = 3 * 3 ) . the answer is c ." | a = 2835 / 3
b = a / 3
c = 1 + 4
d = b / c
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a ) 63 , b ) 95 , c ) 77 , d ) 87 , e ) 48 | d | subtract(subtract(120, 25), subtract(subtract(120, 10), 2)) | in a class of 120 students , 25 did not opt for math . 10 did not opt for science and 2 did not opt for either . how many students of the class opted for both math and science ? | "120 students total 25 did not opt for math 10 did not opt for science 2 did not opt for either total of 95 students in math and 8 did not opt for sci but did for math 95 - 8 = 87 87 students of the class opted for both math and science answer : d . 87" | a = 120 - 25
b = 120 - 10
c = b - 2
d = a - c
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 14 , e ) e ) 9 | e | sqrt(divide(1701, add(power(4, 2), add(power(1, 2), power(2, 2))))) | the ratio of three numbers is 1 : 2 : 4 and the sum of their squares is 1701 . the sum of the numbers is ? | "let the numbers be x , 2 x , 4 x then , x ^ 2 + 4 x ^ 2 + 16 x ^ 2 = 1701 21 x ^ 2 = 1701 x ^ 2 = 81 x = 9 answer is e" | a = 4 ** 2
b = 1 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 1701 / e
g = math.sqrt(f)
|
a ) $ 1.40 , b ) $ 2.40 , c ) $ 3.40 , d ) $ 4.40 , e ) $ 5.40 | c | multiply(subtract(const_1, multiply(divide(add(const_100, 20), const_100), divide(const_1, const_2))), 8.50) | at joes steakhouse the hourly wage for a chef is 20 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 8.50 per hour , how much less than a manager does a chef earn each hour ? | "manager wages per hour = $ 8.50 dishwasher wages per hour = half of manager ' s wages . = 1 / 2 ( $ 8.50 ) = = > $ 4.25 chef wages per hour = 20 % greater than dishwasher wages - - > 20 % of $ 4.25 = ( 20 * ( $ 4.25 ) ) / 100 - - > ( $ 85 ) / 100 - - > $ 0.85 therefore , chef wages per hour = $ 4.25 + $ 0.85 = = > $ 5.10 difference of wages between manager and chef = $ 8.50 - $ 5.10 = = > $ 3.40 answer : c" | a = 100 + 20
b = a / 100
c = 1 / 2
d = b * c
e = 1 - d
f = e * 8
|
a ) 10002 , b ) 20004 , c ) 20002 , d ) 20102 , e ) 20202 | c | add(subtract(power(101, const_2), 101), subtract(power(99, const_2), 99)) | 101 ã — 101 + 99 ã — 99 = ? | "explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1012 + 992 = ( 100 + 1 ) 2 + ( 100 â ˆ ’ 1 ) 2 = 2 ( 1002 + 12 ) = 2 ( 10000 + 1 ) = 20002 . answer : option c" | a = 101 ** 2
b = a - 101
c = 99 ** 2
d = c - 99
e = b + d
|
a ) 231 / 260 , b ) 117 / 130 , c ) 51 / 65 , d ) 23 / 32 , e ) 11 / 16 | a | divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(850, subtract(const_1, divide(40, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25))) | in a village of 2,600 people , 800 people are over 70 years old and 850 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a female or younger than 70 years old ? | "the number of people younger than 70 years old is 2600 - 800 = 1800 the number of females older than 70 years old is 0.6 * 850 = 510 the number of people who are either female or younger than 70 is 1800 + 510 = 2310 . p ( a person is younger than 70 or male ) = 2310 / 2600 = 231 / 260 the answer is a ." | a = 5 * 5
b = a / const_0_25
c = const_26 * b
d = 5 * 5
e = d / const_0_25
f = 40 / e
g = 1 - f
h = 850 * g
i = c - h
j = 5 * 5
k = j / const_0_25
l = const_26 * k
m = i / l
|
a ) 950 , b ) 1234 , c ) 1289 , d ) 1345 , e ) none of these | a | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | jaclyn buys $ 40 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 = 40000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 950 answer : a" | a = 100 * 100
b = a * 9
c = b / 100
d = c / 4
|
a ) 14.4 , b ) 180 / 7 , c ) 50.4 , d ) 60 , e ) 90 | d | divide(multiply(4, 9), divide(60, const_100)) | 60 percent of andrea ' s living room floor is covered by a carpet that is 4 feet by 9 feet . what is the area of her living room floor ? | "60 % of area of the floor = 4 * 9 square feet = 36 square feet i . e . 100 % area of floor = ( 36 / 60 ) * 100 = 60 square feet answer : option d" | a = 4 * 9
b = 60 / 100
c = a / b
|
a ) 500 bc , b ) 500 b / c , c ) 1000 bc , d ) 1000 b / c , e ) 1000 / bc | c | multiply(2, 500) | a case contains c cartons . each carton contains b boxes , and each box contains 500 paper clips . how many paper clips are contained in 2 cases ? | "2 cases * c cartons / case * b boxes / carton * 500 clips / box = 1000 bc paper clips the answer is c ." | a = 2 * 500
|
a ) 296 km , b ) 352 km , c ) 642 km , d ) 742 km , e ) 382 km | a | divide(add(add(30, multiply(2, 8)), 30), 2) | a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 8 hours of the journey . | "the total distance covered by the car in the first 8 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 = sum of 8 terms in ap whose first term is 30 and last term is 44 = 8 / 2 [ 30 + 44 ] = 296 km . answer : a" | a = 2 * 8
b = 30 + a
c = b + 30
d = c / 2
|
a ) 20000 , b ) 21500 , c ) 20500 , d ) 22000 , e ) 21000 | c | add(20000, divide(5000, const_10)) | a , b and c start a business with each investing rs 20000 . after 5 months a withdraws rs 5000 , b withdraws rs 4000 and c invests rs 6000 more . at the end of the year , a total profit of rs 69900 was recorded . find the share of a . | a : b : c = ( 20000 × 5 + 15000 × 7 ) : ( 20000 × 5 + 16000 × 7 ) : ( 20000 × 5 + 26000 × 7 ) = ( 20 × 5 + 15 × 7 ) : ( 20 × 5 + 16 × 7 ) : ( 20 × 5 + 26 × 7 ) = 205 : 212 : 282 a ' s share = 69900 × 205 / ( 205 + 212 + 282 ) = 69900 × 205 / 699 = 20500 answer is c . | a = 5000 / 10
b = 20000 + a
|
a ) 25 % loss , b ) 16 2 / 8 % loss , c ) 16 3 / 3 % loss , d ) 36 2 / 3 % loss , e ) 56 2 / 3 % loss | a | multiply(subtract(const_1, divide(15, 20)), const_100) | the c . p of 15 books is equal to the s . p of 20 books . find his gain % or loss % ? | "15 cp = 20 sp 20 - - - 5 cp loss 100 - - - ? = > 25 % loss answer : a" | a = 15 / 20
b = 1 - a
c = b * 100
|
a ) 2300 , b ) 2887 , c ) 2000 , d ) 1129 , e ) 1192 | a | divide(6647, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1)) | vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 6647 , how much amount did raghu invest ? | "let money invested by raghu = rs . x money invested by trishul = 9 / 10 x = 0.9 x money invested by vishal = 9 / 10 x * 110 / 100 = 0.99 x also , x + 0.9 x + 0.99 x = 6647 = x = 6647 / 2.89 = 2300 therefore , amount invested by raghu is rs . 2300 . answer : a" | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 - f
h = e + g
i = h + 1
j = 6647 / i
|
a ) 20,304 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | sonika bought a v . c . r . at the list price of 18,800 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,800 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,800 = 8 ⁄ 100 × 18800 = 1504 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,800 + 1504 = 20,304 . answer a" | a = 1 + 4
b = a * 2
c = b + 8
d = c * 100
e = 1 + 4
f = e * 2
g = d * f
h = 1 + 4
i = h * 100
j = g + i
k = 100 + 8
l = j * k
m = l / 100
n = 1 + 4
o = n * 100
p = o * 2
q = m / p
r = math.floor(q)
|
a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | a | divide(const_1, divide(20, const_2)) | a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 42 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 48 in what ratio should he mix the water and milk ? | "water = w ( liter ) milk = m ( liter ) = = > cost = price x quantity = 0.42 m = = > revenue = price x quantity = 0.48 ( m + w ) = = > profit = 0.48 ( m + w ) - 0.42 m = 0.2 * ( 0.42 m ) [ 20 % of cost ] = = > 0.48 m + 0.48 w - 0.42 m = 0.084 m = = > 0.024 m = 0.48 w = = > m / w = 0.48 / 0.024 = 20 - - or - - w / m = 1 / 20 a is correct ." | a = 20 / 2
b = 1 / a
|
a ) 2058 , b ) 2069 , c ) 2072 , d ) 2089 , e ) none of these | e | multiply(divide(4524, 2456), const_100) | 4524 - ? = 2456 | "explanation : none of these answer : option e" | a = 4524 / 2456
b = a * 100
|
a ) 225 , b ) 275 , c ) 300 , d ) 350 , e ) 400 | c | divide(add(600, 300), const_3) | in a weight - lifting competition , the total weight of joe ' s two lifts was 600 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | "this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 600 pounds . we sum the two variables to obtain : f + s = 600 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f – 300 = s we can now plug in ( 2 f – 300 ) for s into the first equation , so we have : f + 2 f – 300 = 600 3 f = 900 f = 300 answer is c" | a = 600 + 300
b = a / 3
|
a ) 0.11 % , b ) 0.7 % , c ) 0.44 % , d ) 0.6 % , e ) 0.8 % | c | subtract(subtract(8, 7), divide(multiply(8, 7), const_100)) | in measuring the sides of a rectangle , one side is taken 8 % in excess , and the other 7 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 27 / 25 ) x ( 40 / 43 ) y = ( 913 / 909 ) ( xy ) error in measurement = ( 913 / 909 ) xy - xy = ( 4 / 909 ) xy error percentage = [ ( 4 / 909 ) xy ( 1 / xy ) 100 ] % = ( 11 / 25 ) % = 0.44 % . answer is c ." | a = 8 - 7
b = 8 * 7
c = b / 100
d = a - c
|
a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 2 | a | divide(48, 11) | what is the 9 th digit to the right of the decimal point in the decimal equivalent of 48 / 11 ? | "48 / 11 = 4.3636 . . . . 36 is non - terminating repeating decimal . the 9 th digit to the right of decimal point will be 3 . answer a" | a = 48 / 11
|
a ) 25 , b ) 20 , c ) 15 , d ) 10 , e ) 5 | d | divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . the total of the ages of a , b and c is 27 . how old is b ? | "let age of c = x . then , age of b = 2 x age of a = 2 + 2 x total age of a , b and c = 27 ⇒ ( 2 + 2 x ) + 2 x + x = 27 ⇒ 5 x = 25 ⇒ x = 255 = 5 b ' s age = 2 x = 2 × 5 = 10 answer is d ." | a = 27 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 9 , b ) - 1 , c ) - 3 , d ) 3 , e ) - 9 | c | multiply(6, const_3) | find the value of x : x ² + 6 x + 9 . | that quadratic is factored as follows : x ² + 6 x + 9 = ( x + 3 ) ( x + 3 ) . lesson 17 . now , it is easy to see that that both factors will be 0 when x = - 3 . the solution is : x = - 3 c | a = 6 * 3
|
a ) 18 , b ) 120 , c ) 175 , d ) 216 , e ) 250 | b | add(multiply(divide(360, 6), 1), divide(subtract(360, multiply(divide(360, 6), 1)), 5)) | joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 6 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 6 ) * 360 = 60 gallons . remaning paint = 300 gallons paint used in the second week = ( 1 / 5 ) * 300 = 60 gallons total paint used = 120 gallons . option b" | a = 360 / 6
b = a * 1
c = 360 / 6
d = c * 1
e = 360 - d
f = e / 5
g = b + f
|
a ) 1.8 m , b ) 6 m , c ) 7 m , d ) 9 m , e ) 12 m | b | divide(add(7, 5), subtract(7, 5)) | the e train leaves station a moving at a constant speed , and passes by stations b and c , in this order . it takes the e train 7 hours to reach station b , and 5 additional hours to reach station c . the distance between stations a and b is m kilometers longer than the distance between stations b and c . what is the distance between stations a and c in terms of m ? | the reason it is failing for you is that you chose incorrect numbers . if the question says e it took 7 hrs to reach from a to b and 5 hrs to reach from b to c at a constant speed . it shows that distance ab and bc should be in ratio of 7 / 5 . if you take such numbers you can solve problem . ab = 7 , bc = 5 therefore ab - bc = 2 but from question , ab - bc = m = > m = 2 now total distance = ab + bc = 12 substitute 12 to get answer in terms of m total distance = 12 = 6 m ans b | a = 7 + 5
b = 7 - 5
c = a / b
|
a ) 9800 , b ) 10000 , c ) 10200 , d ) 10400 , e ) 10600 | c | subtract(multiply(5000, 4), multiply(multiply(multiply(const_3, const_4), divide(multiply(5000, 4), const_100)), 4)) | dana borrows 5000 pounds annually for her college education . if dana gives her parents 4 % of that amount back each month , how much will she still owe her parents after 4 years of college ? | dana borrows 5000 and returns 4 % per month . so returns 200 per month . and 2400 per year . so dana owes 5000 - 2400 = 2600 per year . after 4 years she will owe 10400 . ans ( c ) | a = 5000 * 4
b = 3 * 4
c = 5000 * 4
d = c / 100
e = b * d
f = e * 4
g = a - f
|
a ) 86 , b ) 16 , c ) 76 , d ) 95 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(75, 3)) | a student scored an average of 75 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 75 * 3 = 225 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 225 = 95 answer : d" | a = 90 * 2
b = 70 * 2
c = a + b
d = 75 * 3
e = c - d
|
a ) 9 / 4 , b ) 3 / 2 , c ) 8 / 3 , d ) 2 / 3 , e ) 1 / 2 | c | divide(8, 3) | a positive number x is multiplied by 8 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "sq rt ( 8 x / 3 ) = x = > 8 x / 3 = x ^ 2 = > x = 8 / 3 ans - c" | a = 8 / 3
|
a ) 68 days , b ) 84 days , c ) 90 days , d ) 72 days , e ) 50 days | b | inverse(subtract(divide(const_1, 21), divide(const_1, 28))) | a and b working together could mow a field in 28 days and with the help of c they could have mowed it in 21 days . how long would c take by himself ? | required answer = 28 * 21 / 28 - 21 = 84 days answer is b | a = 1 / 21
b = 1 / 28
c = a - b
d = 1/(c)
|
a ) s . 268 , b ) s . 288 , c ) s . 298 , d ) s . 300 , e ) s . 388 | a | divide(multiply(590, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 590 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 590 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 590 ⇒ y [ 12 / 10 + 1 ] = 590 ⇒ 22 y / 10 = 590 ⇒ 22 y = 5900 ⇒ y = 5900 / 22 = 500 / 2 = rs . 268 a )" | a = 1 + 4
b = a * 2
c = 590 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 20000 , b ) 10000 , c ) 25000 , d ) 34000 , e ) 30000 | a | multiply(36000, subtract(const_1, divide(multiply(30000, multiply(2, multiply(2, const_3))), add(multiply(45000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(30000, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 36000 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 36000 = 20000 a" | a = 2 * 3
b = 2 * a
c = 30000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 45000 * f
h = 2 * 3
i = 2 * h
j = 30000 * i
k = g + j
l = c / k
m = 1 - l
n = 36000 * m
|
a ) 2 : 6 , b ) 2 : 1 , c ) 7 : 1 , d ) 2 : 8 , e ) 2 : 2 | b | divide(20, 10) | if 10 % of x = 20 % of y , then x : y is equal to ? | 10 % of x = 20 % of y 10 x / 100 = 20 y / 100 = > x / 10 = y / 5 x / y = 10 / 5 = 2 / 1 x : y = 2 : 1 . answer : b | a = 20 / 10
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a ) 40 % , b ) 44 % , c ) 48 % , d ) 52 % , e ) 56 % | d | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 70), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 70 % of calculus students are enrolled in the beginner course . if 4 / 5 of students are in the beginner courses , and one student is selected at random , what is the probability that a beginner trigonometry student is selected ? | "let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.7 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.7 x = 1.3 x the percentage in beginner trigonometry is 1.3 x / 2.5 x = 52 % the answer is d ." | a = 50 * 2
b = a + 50
c = 50 * 2
d = b + c
e = 4 / 5
f = d * e
g = f - 70
h = 50 * 2
i = h + 50
j = 50 * 2
k = i + j
l = g / k
m = 50 * 2
n = l * m
|
a ) 180 / 15 , b ) 180 / 10 , c ) 180 / 13 , d ) 180 / 15 , e ) 180 / 11 | c | divide(const_1, divide(add(add(inverse(12), inverse(30)), inverse(36)), const_2)) | a and b can do a work in 12 days , b and c in 30 days and c and a in 36 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 12 one day work of b and c = 1 / 30 one day work of c and a = 1 / 36 2 ( a + b + c ) = 1 / 12 + 1 / 30 + 1 / 36 2 ( a + b + c ) = 13 / 90 ( a + b + c ) = 13 / 180 number of days required = 180 / 13 days . answer : c" | a = 1/(12)
b = 1/(30)
c = a + b
d = 1/(36)
e = c + d
f = e / 2
g = 1 / f
|
a ) 33 , b ) 25 , c ) 27 , d ) 24 , e ) 23 | e | divide(add(24, 28), const_2) | if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 22 , 28 , 19 , 33 and 35 ? | "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 22 , 28 , 19 , 33 and 35 19 , 22 , 24 , 28 , 30 , 33 , 35 , 42 . . . numbers less than median are 19 , 22 , 24 , 28 . . the median of these numbers = center of 22 and 24 = 23 e" | a = 24 + 28
b = a / 2
|
a ) 15 days , b ) 12 days , c ) 8 days , d ) 16 days , e ) 9 days | d | divide(const_1, divide(add(add(inverse(18), inverse(24)), inverse(36)), const_2)) | a and b can do a piece of work in 18 days ; band c can do it in 24 days a and c can do it in 36 days . in how many days will a , band c finish it together ? | sol . ( a + b ) ' s 1 day ' s work = ( 1 / 18 ) ( b + c ) ' s 1 day ' s work = ( 1 / 24 ) and ( a + c ) ' s 1 day ' s work = ( 1 / 36 ) adding , we get : 2 ( a + b + c ) ' s 1 day ' s work = ¬ ( 1 / 18 + 1 / 24 + 1 / 36 ) = 9 / 72 = 1 / 8 ( a + b + c ) ' s 1 day ' s work = 1 / 16 thus , a , band c together can finish the work in 16 days . ans : d | a = 1/(18)
b = 1/(24)
c = a + b
d = 1/(36)
e = c + d
f = e / 2
g = 1 / f
|
a ) 4 , b ) 5 , c ) 3 , d ) 2 , e ) 1 | b | add(add(const_4, 3), subtract(3, const_4)) | what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 30 ? | "denote | x - 3 | as y : y ^ 2 + y = 30 - - > y = - 6 or y = 5 . discard the first solution since y = | x - 3 | , so it ' s an absolute value and thus can not be negative . y = | x - 3 | = 5 - - > x = 8 or x = - 3 . the sum = 5 . answer : b ." | a = 4 + 3
b = 3 - 4
c = a + b
|
a ) 18 , b ) 20 , c ) 26 , d ) 28 , e ) 30 | a | divide(198, subtract(12, const_1)) | let the number which when multiplied by 12 is increased by 198 . | "solution let the number be x . then , 12 x - x = 198 ‹ = › 11 x = 198 x ‹ = › 18 . answer a" | a = 12 - 1
b = 198 / a
|
a ) 25 , b ) 50 , c ) 57 , d ) 60 , e ) 75 | b | subtract(subtract(subtract(200, 80), 70), divide(subtract(subtract(200, 80), 70), const_3)) | 200 college students were asked in a survey if they preferred windows or mac brand computers . 70 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 80 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | "200 = 70 ( mac ) + x ( window ) + 80 ( both ) = > x = 50 answer : b" | a = 200 - 80
b = a - 70
c = 200 - 80
d = c - 70
e = d / 3
f = b - e
|
a ) 8 ° , b ) 10 ° , c ) 18 ° , d ) 33 ° , e ) 52 ° | d | divide(multiply(subtract(const_100, add(add(add(add(14, 25), 15), 29), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 25 % home electronics ; 15 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "here all percentage when summed we need to get 100 % . as per data 14 + 25 + 15 + 29 + 8 = 91 % . so remaining 9 % is the balance for the astrophysics . since this is a circle all percentage must be equal to 360 degrees . 100 % - - - - 360 degrees then 9 % will be 33 degrees . . imo option d ." | a = 14 + 25
b = a + 15
c = b + 29
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
|
a ) 9 , b ) 9.6 , c ) 10 , d ) 10.8 , e ) 7 | d | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | two trains 140 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "explanation : relative speed = ( 60 + 40 ) km / hr = [ 100 x ( 5 / 18 ) ] m / sec = ( 250 / 9 ) m / sec . distance covered in crossing each other = ( 140 + 160 ) m = 300 m . required time = [ 300 x ( 9 / 250 ) ] sec = ( 54 / 5 ) sec = 10.8 sec . answer : d" | a = 140 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio t of revenue from car sales in 1996 to revenue from truck sales in 1996 ? | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a = 100 + 7
b = 100 + 1
c = a - b
d = 100 + 1
e = 100 - 11
f = d - e
g = c / f
|
a ) 50 , b ) 50.5 , c ) 56 , d ) 60 , e ) 63 | b | add(divide(const_100, const_2), 0.5) | in a recent election , geoff received 0.5 percent of the 6,000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 3,000 more votes to win the election , what is the value of x ? | word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 30 votes he needs 3571 more votes so : 30 + 3000 = 3030 now what ' s the problem wants ? ? a x % . . . . . . . . 3030 is what % of total votes 6000 . . . . . . . . translating : 3030 = x / 100 * 6000 - - - x = 50.5 % . . . . . . . . . . b | a = 100 / 2
b = a + 0
|
a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | c | multiply(divide(240, power(const_2, const_3)), const_2) | if a ^ 3 is divisible by 240 what is the least possible value of integer a ? | 240 can be written as ( 2 ^ 4 ) * 3 * 5 . for a ^ 3 to be divisible by 240 it should contain atleast 2 ^ 4 and 3 and 5 in its factors . we can leave out option a because 12 doesnt have 5 as one of its factor . now if we check for option b , 30 can be written as 2 * 3 * 5 , hence 30 ^ 3 will have 3 as the maximum power of 2 , so we can leave out this option too . option c is the right answer if we follow the same method as we followed for other two previous options . 60 = ( 2 ^ 2 ) * 3 * 5 ; 60 ^ 3 = ( 2 ^ 6 ) * ( 3 ^ 3 ) * ( 5 ^ 3 ) . so it shows that 60 ^ 3 is divisible by 240 and hence the answer . answer : c | a = 2 ** 3
b = 240 / a
c = b * 2
|
a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) 250 m | d | subtract(multiply(multiply(72, const_0_2778), 26), 250) | a goods train runs at a speed of 72 kmph and crosses a 250 m long platform in 26 seconds . what is the length of the goods train ? | "s = 250 + x / t 72 * 5 / 18 = 250 + x / 26 x = 270 answer : d" | a = 72 * const_0_2778
b = a * 26
c = b - 250
|
a ) 50 , b ) 25 , c ) 40 , d ) 45 , e ) 75 | a | subtract(subtract(multiply(60, add(const_2, const_3)), multiply(50, const_4)), 50) | the average of temperatures at noontime from monday to friday is 60 ; the lowest one is 50 , what is the possible maximum range of the temperatures ? | "average = 60 , sum of temperatures = 60 * 5 = 300 as the min temperature is 50 , max would be 300 - 4 * 50 = 100 - - > the range = 100 ( max ) - 50 ( min ) = 50 answer : a" | a = 2 + 3
b = 60 * a
c = 50 * 4
d = b - c
e = d - 50
|
a ) 600 cm 2 , b ) 500 cm 2 , c ) 516 cm 2 , d ) 416 cm 2 , e ) 100 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 26), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(26, 26)))), 26) | find the area of a rhombus one side of which measures 20 cm and one diagonal is 26 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 26 x 32 ) cm 2 = 416 cm 2 answer : option d" | a = 1 / 2
b = a * 26
c = 20 * 20
d = c * 4
e = 26 * 26
f = d - e
g = math.sqrt(f)
h = b * g
i = h + 26
|
a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 16 | b | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 448 * 349 ) ^ 2 | "( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 8,9 ; 5 is missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 44 * 44 ) ^ 2 = 36 ^ 2 = 76 hence answer is b" | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) t = 1 / 6 | e | divide(const_1, divide(factorial(4), multiply(factorial(2), factorial(2)))) | if two of the 4 expressions x + y , x + 5 y , x - y , and 5 x - y are chosen at random , what is the probability t that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | only ( x + y ) ( x - y ) pair will give the form x ^ 2 - ( by ) ^ 2 the probability of selecting these two pairs are 1 / 4 * 1 / 3 = 1 / 12 , assuming x + y is picked first , then x - y , but x - y can be picked first followed by x + y . so the probability t = 1 / 12 * 2 = 1 / 6 ans e | a = math.factorial(4)
b = math.factorial(2)
c = math.factorial(2)
d = b * c
e = a / d
f = 1 / e
|
a ) - 0.0053 , b ) - 0.0003 , c ) 0.0007 , d ) 0.0047 , e ) 0.0033 | e | subtract(subtract(2.0767, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0767) | if n = 2.0767 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * – n ? | "n * = 2.08 n * - n = 2.08 - 2.0767 0.0033 answer : e" | a = 2 * 10
b = a + 3
c = b / 100
d = c / 100
e = 2 - 767
f = e - 2
|
a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12 | b | divide(12, const_2) | list f consists of 12 consecutive integers . if - 4 is the least integer in list f , what is the range of positive integers in list f ? | since - 4 is the least integer in list f , then 7 is the largest integer in that list . thus the range of the positive integers in the list is 7 - 1 = 6 . answer : b . | a = 12 / 2
|
a ) 16 , b ) 19 , c ) 15 , d ) 22 , e ) 30 | e | divide(subtract(32, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 32 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 32 ) + 2 = 2 ( x + 2 ) x + 34 = 2 x + 4 = > x = 30 . answer : e" | a = 2 * 2
b = a - 2
c = 32 - b
d = 2 - 1
e = c / d
|
a ) 3 kmph , b ) 2.5 kmph , c ) 3.5 kmph , d ) 4.3 kmph , e ) 5 kmph | b | divide(add(8, 13), const_2) | a man can row upstream at 8 kmph and downstream at 13 kmph . the speed of the stream is ? | "speed of stream = 1 / 2 ( 13 - 8 ) = 2.5 kmph answer is b" | a = 8 + 13
b = a / 2
|
a ) 14.7 , b ) 2.9 , c ) 38.3 , d ) 78.3 , e ) 79.3 | a | subtract(subtract(30, multiply(const_3, const_3)), multiply(divide(subtract(30, multiply(const_3, const_3)), 30), multiply(const_3, const_3))) | a vessel of capacity 30 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 30 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 30 - 9 ) / 30 ] ^ 2 ( 30 ) = 14.7 litres . answer : option a" | a = 3 * 3
b = 30 - a
c = 3 * 3
d = 30 - c
e = d / 30
f = 3 * 3
g = e * f
h = b - g
|
a ) 12 % , b ) 14 % , c ) 16 % , d ) 20 % , e ) 22 % | c | subtract(const_100, divide(multiply(add(const_100, 5), subtract(const_100, 20)), const_100)) | the tax on a commodity is diminished by 20 % but its consumption is increased by 5 % . find the decrease percent in the revenue derived from it ? | "explanation : 100 * 100 = 10000 80 * 105 = 8400 10000 - - - - - - - 1600 100 - - - - - - - ? = 16 % c )" | a = 100 + 5
b = 100 - 20
c = a * b
d = c / 100
e = 100 - d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(divide(15, 3), divide(3, divide(15, 3))) | if 15 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "15 - 5 * 3 12 - 4 * 3 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 6 is allowed . imo d ." | a = 15 / 3
b = 15 / 3
c = 3 / b
d = a + c
|
a ) 17 cm , b ) 10 cm , c ) 8 cm , d ) 6 cm , e ) 7 cm | c | divide(volume_cube(1), volume_cube(divide(50, const_100))) | how many cubes of 50 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 50 * 50 * 50 = 8 number of cubes = 100 â ˆ — 100 â ˆ — 10010 â ˆ — 10 â ˆ — 10 = 1000 note : 1 m = 100 cm answer : c" | a = volume_cube / (
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | divide(subtract(80, 20), const_2) | we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 20 , and the difference between the total number of strawberries and the total number of blueberries would increase by 80 . each blue box contains how many blueberries ? | let x be the number of blueberries in each blue box . then there are x + 20 strawberries in each red box . x + ( x + 20 ) = 80 x = 30 the answer is b . | a = 80 - 20
b = a / 2
|
a ) 5 / 7 , b ) 6 / 7 , c ) 9 / 5 , d ) 13 / 9 , e ) 13 / 6 | a | divide(subtract(90, 40), subtract(90, 20)) | a portion of the 90 % solution of chemicals was replaced with an equal amount of 20 % solution of chemicals . as a result , 40 % 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 : 0.90 ( 1 - x ) + 0.2 * x = 0.4 - - > x = 5 / 7 . answer : a . | a = 90 - 40
b = 90 - 20
c = a / b
|
a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what s percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c" | a = 3 + 1
b = 3 + 2
c = a / b
d = 2 * c
e = 3 + 1
f = 3 + 2
g = e / f
h = 3 * g
i = 3 - h
j = d - i
k = 3 + 1
l = j / k
m = l * 100
|
a ) 21 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 26 years | b | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 24 . = > 2 ( x + 2 ) = ( x + 24 + 2 ) = > 2 x + 4 = x + 26 = > x = 22 years option b" | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
|
a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these | c | divide(volume_cube(1), volume_cube(divide(10, const_100))) | how many cubes of 10 cm edge can be put in a cubical box of 1 m edge | "explanation : number of cubes = ( 100 x 100 x 100 ) / ( 10 x 10 x 10 ) = 1000 answer : c" | a = volume_cube / (
|
a ) 19 , b ) 12 , c ) 11 , d ) 10 , e ) 8 | e | add(divide(14, const_2), 1) | each of the integers from 1 to 14 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ? | "out of the 14 integers : 7 are odd and 7 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 7 odd numbers first 2 . 8 th number will be the first even number so we need to withdraw at least 8 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 8 . ( e )" | a = 14 / 2
b = a + 1
|
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