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 |
|---|---|---|---|---|---|
a ) 1.6 km , b ) 2 km , c ) 5.1 km , d ) 4 km , e ) none of these | c | multiply(divide(14, const_60), add(18, 4)) | the speed of a boat in still water is 18 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 14 minutes is | "explanation : speed downstreams = ( 18 + 4 ) kmph = 22 kmph . distance travelled = ( 22 x 14 / 60 ) km = 5.1 km option c" | a = 14 / const_60
b = 18 + 4
c = a * b
|
a ) rs . 150.50 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 158 , e ) none | c | divide(169, power(add(divide(4, const_100), const_1), 2)) | the present worth of rs . 169 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 169 / ( 1 + 4 / 100 ) Β² ] = rs . ( 169 x 25 / 26 x 25 / 26 ) = rs . 156.25 answer c" | a = 4 / 100
b = a + 1
c = b ** 2
d = 169 / c
|
a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 40 | c | divide(factorial(subtract(add(const_4, 17), const_1)), multiply(factorial(17), factorial(subtract(const_4, const_1)))) | how many positive integers less than 350 are there such that they are multiples of 17 or multiples of 18 ? | "350 / 17 = 20 ( plus remainder ) so there are 20 multiples of 17 350 / 18 = 19 ( plus remainder ) so there are 19 multiples of 18 we need to subtract 1 because 17 * 18 is a multiple of both so it was counted twice . the total is 20 + 19 - 1 = 38 the answer is c ." | a = 4 + 17
b = a - 1
c = math.factorial(b)
d = math.factorial(17)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 87.8 % , b ) 87.9 % , c ) 37.8 % , d ) 87.7 % , e ) 22.8 % | a | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(8, const_100)))), add(600, 400)), const_100) | a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 8 % of bananas were rotten . find the percentage of fruits in good condition . | "otal number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ 600 = 9000 / 100 = 90 number of rotten bananas = 8 % of 400 = 8 / 100 Γ 400 = 3200 / 100 = 32 therefore , total number of rotten fruits = 90 + 32 = 122 therefore number of fruits in good condition = 1000 - 122 = 878 therefore percentage of fruits in good condition = ( 878 / 1000 Γ 100 ) % = ( 87800 / 1000 ) % = 87.8 % answer : a" | a = 600 + 400
b = 15 / 100
c = 600 * b
d = 8 / 100
e = 400 * d
f = c + e
g = a - f
h = 600 + 400
i = g / h
j = i * 100
|
a ) 5.925 % , b ) 5.253 % , c ) 7 % , d ) 6.52 % , e ) 5.200 % | a | sqrt(divide(632, divide(1800, const_100))) | reema took a loan of rs 1800 with simple interest for as many years as the rate of interest . if she paid rs . 632 as interest at the end of the loan period , what was the rate of interest . | "explanation : let rate = r % then time = r years . = > 1800 β r β r / 100 = 632 = > r 2 = 35.11 = > r = 5.925 % option a" | a = 1800 / 100
b = 632 / a
c = math.sqrt(b)
|
['a ) 334', 'b ) 377', 'c ) 495', 'd ) 766', 'e ) 261'] | c | divide(factorial(12), multiply(factorial(const_4), factorial(subtract(12, const_4)))) | 12 points lie on a circle . how many cyclic quadrilaterals can be drawn by using these points ? | explanation : for any set of 4 points we get a cyclic quadrilateral . number of ways of choosing 4 points out of 12 points is \ inline { \ color { black } 12 c _ 4 { } } = 495 . therefore , we can draw 495 quadrilaterals answer : c ) 495 | a = math.factorial(12)
b = math.factorial(4)
c = 12 - 4
d = math.factorial(c)
e = b * d
f = a / e
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | subtract(15, divide(180, 15)) | there are 15 members in a family . it takes 180 days for a women to complete the work , however man completes in 120 days . they take 17 days to complete the work if men work on alternate days and women work every third day . if all of them started together on the 1 st day , then how many men are there in the family | "let the number of men be m and so the number of women = 15 - m out of 17 days , men and women work together on 1,4 , 7,10 , 13,16 th days ( total of 6 days ) from the start . men work on alternate days , that is 1 , 3,5 , 7,9 , 11,13 , 15,17 th days ( total of 9 days ) . let the total work be 360 units ( lcm of 180 and 120 ) . 1 man does 360 / 120 = 3 units per day 1 woman does 360 / 180 = 2 units per day based on above facts , total work by men + women in these 17 days = 360 units m * 9 * 3 + ( 15 - m ) * 6 * 2 = 360 , 27 m + 180 - 12 m = 360 , solving m = 12 . num of men = 12 answer : d" | a = 180 / 15
b = 15 - a
|
a ) 200 , b ) 250 , c ) 300 , d ) 350 , e ) 400 | c | multiply(divide(multiply(divide(multiply(divide(480, 4), 8), 3), 15), 16), 4) | there is a train and car . the ratio between the speed of a train & a car is 16 : 15 respectively . also , a bus covered a distance of 480 km in 8 hours . the speed of the bus is 3 / 4 th of the speed of the train . how many kilometers will the car cover in 4 hours ? | "the speed of the bus is 480 / 8 = 60 km / hr the speed of the train is ( 60 * 4 ) / 3 = 80 km / hr the speed of the car is 80 / 16 * 15 = 75 km / hr the distance covered by the car in 4 hours is 75 Γ 4 = 300 km the answer is c ." | a = 480 / 4
b = a * 8
c = b / 3
d = c * 15
e = d / 16
f = e * 4
|
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) 16.667 % | e | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 75) | farm tax is levied on the 75 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 75 ) = 16.667 % the correct answer is ( e ) ." | a = 480 / 3840
b = a * 100
c = b * 100
d = c / 75
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | subtract(subtract(14, 2), multiply(2, 2)) | what is x if 2 x - y = 14 and y = 2 ? | substitute y by 2 in 2 x - y = 14 2 x - 2 = 14 2 x = 16 if we substitute x by 8 , we have 2 ( 8 ) = 16 . hence x = 8 correct answer b | a = 14 - 2
b = 2 * 2
c = a - b
|
['a ) 36 %', 'b ) 37 %', 'c ) 35 %', 'd ) 38 %', 'e ) 39 %'] | a | subtract(const_100, multiply(const_100, divide(circle_area(4), circle_area(5)))) | the radius of the circle is reduced from 5 cm to 4 cm then the % change of area . | for 5 cm - > pi * r ^ 2 - - > 3.14 * 5 ^ 2 - > 78.539 for 4 cm - > pi * r ^ 2 - - > 3.14 * 4 ^ 2 - > 50.265 % change - > ( 1 - 50.265 / 78.539 ) * 100 = 36 ie 36 % answer : a | a = circle_area / (
b = 100 * a
c = 100 - b
|
a ) 2.5 sec , b ) 2.8 sec , c ) 8.5 sec , d ) 3.3 sec , e ) 4.5 sec | d | divide(200, multiply(216, const_0_2778)) | in what time will a train 200 m long cross an electric pole , it its speed be 216 km / hr ? | "speed = 216 * 5 / 18 = 60 m / sec time taken = 200 / 60 = 3.3 sec . answer : d" | a = 216 * const_0_2778
b = 200 / a
|
a ) 29 , b ) 30 , c ) 31 , d ) 32 , e ) 33 | e | divide(97, subtract(const_100, 97)) | if t = 1 / ( 1 * 2 ) + 1 / ( 2 * 3 ) + 1 / ( 3 * 4 ) + 1 / ( 4 * 5 ) + . . . . . . . . . 1 / [ n ( n + 1 ) ] , for some positive integer n , what is the smallest value of n such that t > 097 ? | we can break the t as : ( n + 1 - n ) / n ( n + 1 ) - - - - for n = 1 to n 1 / n - 1 / n + 1 plug value , 1,2 . . n we will get a pattern 1 - 1 / 2 - - - for 1 1 / 2 - 1 / 3 - - for 2 1 / 3 - 1 / 4 - - for 4 now work with options . . for 31 , t = 1 - 1 / 32 = 30 / 31 = 0.968 for 32 , t = 1 - 1 / 33 = 0,9696 for 33 , t = 1 - 1 / 34 = 33 / 34 - 0.97 so e is correct answer | a = 100 - 97
b = 97 / a
|
a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 15 , d ) 1 / 28 , e ) 2 / 3 | d | inverse(divide(factorial(8), multiply(factorial(2), factorial(6)))) | jack and jill work at a hospital with 6 other workers . for an internal review , 2 of the 8 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "1 / 8 c 2 = 1 / 28 . answer : d ." | a = math.factorial(8)
b = math.factorial(2)
c = math.factorial(6)
d = b * c
e = a / d
f = 1/(e)
|
a ) 18 , b ) 82 , c ) 26 , d ) 27 , e ) 29 | c | multiply(subtract(83, 70), const_2) | a pupil ' s marks were wrongly entered as 83 instead of 70 . due to the average marks for the class got increased by half . the number of pupils in the class is ? | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 83 - 70 ) = > x / 2 = 13 = > x = 26 . answer : c" | a = 83 - 70
b = a * 2
|
a ) 36.9 , b ) 38.5 , c ) 18.5 , d ) 32.5 , e ) 28.5 | b | multiply(divide(add(add(floor(divide(18, 7)), const_1), floor(divide(57, 7))), const_2), 7) | calculate the average of all the numbers between 18 and 57 which are divisible by 7 . | "explanation : numbers divisible by 7 are 21,28 , 35,42 , 49,56 , average = ( 21 + 28 + 35 + 42 + 49 + 56 , ) / 6 = 231 / 6 = 38.5 answer : b" | a = 18 / 7
b = math.floor(a)
c = b + 1
d = 57 / 7
e = math.floor(d)
f = c + e
g = f / 2
h = g * 7
|
a ) a ) 30 , b ) b ) 31 , c ) c ) 40 , d ) d ) 33 , e ) e ) 34 | c | divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 are neither multiples of 2 or 5 . | "to answer this q we require to know 1 ) multiples of 2 till 100 = 100 / 2 = 50 2 ) multiples of 5 till 100 = 100 / 5 = 20 add the two 50 + 20 = 70 ; subtract common terms that are multiple of both 2 and 5 . . lcm of 2 and 5 = 10 multiples of 10 till 100 = 100 / 10 = 10 so total multiples of 2 and 5 = 70 - 10 = 6 ans = 100 - 60 = 40 c" | a = 4 + 2
b = a - 1
c = math.factorial(b)
d = math.factorial(2)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 22 , b ) 99 , c ) 25 , d ) 66 , e ) 887 | c | divide(multiply(50, 3), add(divide(50, 50), divide(multiply(2, 50), 20))) | a trained covered x km at 50 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 50 + 2 x / 20 hours = 6 x / 50 = 3 x / 25 hours average speed = 3 x / ( 3 x / 25 ) = 25 kmph answer : c" | a = 50 * 3
b = 50 / 50
c = 2 * 50
d = c / 20
e = b + d
f = a / e
|
a ) 3 . , b ) 2 . , c ) 1 / 2 . , d ) 1 / 4 , e ) there is n ' t enough data to answer the question . | d | add(3, divide(multiply(3, 1), 3)) | two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores is 1 / 3 , what is the value of y / x ? | "answer is d : 1 / 4 x - y = ( x + y ) / 3 solving for y / x = 1 / 4" | a = 3 * 1
b = a / 3
c = 3 + b
|
a ) 135 , b ) 137 , c ) 143 , d ) 310 , e ) 380 | c | add(divide(368, gcd(gcd(60, 144), 368)), add(divide(60, gcd(gcd(60, 144), 368)), divide(144, gcd(gcd(60, 144), 368)))) | a drink vendor has 60 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | "the number of liters in each can = hcf of 60 , 144 and 368 = 4 liters . number of cans of maaza = 60 / 4 = 15 number of cans of pepsi = 144 / 4 = 36 number of cans of sprite = 368 / 4 = 92 the total number of cans required = 15 + 36 + 92 = 143 cans . answer : c" | a = math.gcd(60, 144)
b = math.gcd(a, 368)
c = 368 / b
d = math.gcd(60, 144)
e = math.gcd(d, 368)
f = 60 / e
g = math.gcd(60, 144)
h = math.gcd(g, 368)
i = 144 / h
j = f + i
k = c + j
|
a ) 50 , b ) 200 , c ) 396 , d ) 398 , e ) 400 | c | multiply(inverse(10), multiply(multiply(const_100, 10), add(const_4, const_4))) | when 1 / 10 % of 4,000 is subtracted from 1 / 10 of 4,000 , the difference is | "1 / 10 % of 4000 = 4 1 / 10 of 4000 = 400 400 - 4 = 396 ans : c" | a = 1/(10)
b = 100 * 10
c = 4 + 4
d = b * c
e = a * d
|
a ) 4000 , b ) 3050 , c ) 4400 , d ) 4500 , e ) none of these | c | divide(99, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100))) | if 15 % of 30 % of 50 % of a number is 99 , then what is the number ? | let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 99 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 99 = > a = 10 * 20 * 10 * 2 = 4400 . answer : c | a = 50 / 100
b = 30 / 100
c = a * b
d = 15 / 100
e = c * d
f = 99 / e
|
a ) 600 , b ) 480 , c ) 1000 , d ) 1200 , e ) 1500 | b | multiply(divide(4, subtract(divide(60, 24), const_1)), 60) | working together , printer a and printer b would finish the task in 24 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 4 pages a minute more than printer a ? | answer : b . | a = 60 / 24
b = a - 1
c = 4 / b
d = c * 60
|
a ) 600 , b ) 490 , c ) 720 , d ) 1400 , e ) 1679 | d | multiply(divide(add(40, multiply(2, 85)), subtract(100, 85)), 100) | if a farmer wants to plough a farm field on time , he must plough 100 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ? | "let x be the number of days in the initial plan . therefore , the whole field is 100 Γ’ βΉ β¦ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 100 x = 85 ( x + 2 ) + 40 15 x = 210 x = 14 so the farmer planned to have the work done in 6 days , and the area of the farm field is 100 ( 14 ) = 1400 hectares correct answer d" | a = 2 * 85
b = 40 + a
c = 100 - 85
d = b / c
e = d * 100
|
a ) 72 , b ) 88 , c ) 95 , d ) 120 , e ) 135 | b | multiply(divide(330, multiply(5, 3)), 4) | in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 330 baseball fans in the town , each of whom is a fan of exactly one of those 3 teams , how many ny mets fans are there in this town ? | the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 / 15 of the population . ( 4 / 15 ) * 330 = 88 the answer is b . | a = 5 * 3
b = 330 / a
c = b * 4
|
a ) $ 28.50 , b ) $ 30.50 , c ) $ 33.33 , d ) $ 40.50 , e ) $ 50.00 | c | divide(subtract(multiply(30, 5), add(30, 20)), 3) | a retailer sells 5 shirts . he sells first 2 shirts for $ 30 and $ 20 . if the retailer wishes to sell the 5 shirts for an overall average price of over $ 30 , what must be the minimum average price of the remaining 3 shirts ? | first 2 shirts are sold for $ 30 and $ 20 = $ 50 . to get average price of $ 30 , total sale should be 5 * $ 30 = $ 150 so remaining 3 shirts to be sold for $ 150 - $ 50 = $ 100 answer should be 100 / 3 = $ 33.33 that is c | a = 30 * 5
b = 30 + 20
c = a - b
d = c / 3
|
a ) 12 , b ) 15 , c ) 20 , d ) 22 , e ) 24 | a | divide(multiply(subtract(26, 10), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 10 years rahul age will be 26 years . what is deepak present age . | explanation : present age is 4 x and 3 x , = > 4 x + 10 = 26 = > x = 16 so deepak age is = 3 ( 4 ) = 12 option a | a = 26 - 10
b = a * 3
c = b / 4
|
a ) 2640 , 1000 , b ) 3660 , 4400 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 4660 , 3000 | b | multiply(divide(7.5, 12.5), divide(1760, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1760 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1760 5 * y / 3 β y = 1760 2 * y / 3 = 1760 y = [ ( 1760 * 3 ) / 2 ] = 2640 . one number = 2640 , second number = 5 * y / 3 = 4400 . answer b ." | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1760 / c
e = a * d
|
a ) 33 β 2 , b ) 15 , c ) 12 , d ) 22 , e ) none of these | b | multiply(subtract(6, const_1), divide(33, subtract(12, const_1))) | if a clock strikes 12 in 33 seconds , it will strike 6 in how many seconds ? | in order to strike 12 , there are 11 intervals of equal time = 33 β 11 = 3 seconds each therefore , to strike 6 it has 5 equal intervals , it requires 5 Γ 3 = 15 sec . answer b | a = 6 - 1
b = 12 - 1
c = 33 / b
d = a * c
|
a ) 4 : 1 , b ) 11 : 8 , c ) 13 : 7 , d ) 15 : 7 , e ) 16 : 9 | a | divide(multiply(50, 8), multiply(25, 4)) | car a runs at the speed of 50 km / hr & reaches its destination in 8 hr . car b runs at the speed of 25 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 50 Γ 8 = 400 km distance travelled by car b = 25 Γ 4 = 100 km ratio = 400 / 100 = 4 : 1 answer : a" | a = 50 * 8
b = 25 * 4
c = a / b
|
a ) 36 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | a | multiply(divide(const_100, add(const_100, 25)), 45) | from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 25 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "1.25 x = 45 - - > 5 / 4 * x = 45 - - > x = 45 * 4 / 5 = 180 / 5 = 36 . answer : a ." | a = 100 + 25
b = 100 / a
c = b * 45
|
a ) 20 , b ) 25 , c ) 30 , d ) 24 , e ) 19 | d | divide(const_1, subtract(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 20)), const_2), divide(const_1, 15))) | if 3 people a , b , c can do a work simultaneously in 6 days . a and b can do work in 10 days . b and c can do work in 15 days . c and a can do the same work in 20 days . in how many days that a can complete the same work alone ? | a + b + c = 1 / 6 ( one day work ) a + b = 1 / 10 ; b + c = 1 / 15 ; c + a + 1 / 20 b = 1 / 15 - c ; c = 1 / 20 - a ; - - - - - - - - - ( 2 ) substitute equation 2 in a + b = 1 / 10 a = 1 / 10 - b ; a = 1 / 10 - 1 / 15 + c ; a = 1 / 10 - 1 / 15 + 1 / 20 - a ; 2 a = 1 / 10 - 1 / 15 + 1 / 20 ; a = 1 / 24 ( one day work ) that is the no of days that a can work alone is 24 days the option d is answer | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / 20
e = c + d
f = e / 2
g = 1 / 15
h = f - g
i = 1 / h
|
a ) 30 , b ) 33 , c ) 55 , d ) 77 , e ) 88 | a | subtract(divide(170, add(divide(10, const_100), divide(3, 4))), 170) | a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 170 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 170 = > . 1 x + 3 / 4 x = 170 = > x = 200 200 - 170 = pure gala = 30 . answer a" | a = 10 / 100
b = 3 / 4
c = a + b
d = 170 / c
e = d - 170
|
a ) 44 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 16 years | a | divide(subtract(46, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 46 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 46 ) + 2 = 2 ( x + 2 ) x + 48 = 2 x + 4 x = 44 . answer : a" | a = 2 * 2
b = a - 2
c = 46 - b
d = 2 - 1
e = c / d
|
a ) 37 mins , b ) 35 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins | a | subtract(add(divide(multiply(add(divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), divide(multiply(const_1, const_60), const_2)), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), const_3) | pipe a that can fill a tank in an hour and pipe b that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe b is shut 10 minutes before the tank overflows . when will the tank overflow ? | the last 10 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 10 minutes it fills 1 / 5 th of the tank , thus 5 / 6 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 5 / 6 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 5 / 6 ) / 3 = 5 / 18 hours = 27 minutes . total time = 10 + 27 = 37 minutes . answer : a . | a = 1 * const_60
b = 1 * const_60
c = b / 2
d = a * c
e = 1 * const_60
f = e / 2
g = d / f
h = 1 * const_60
i = 1 * const_60
j = i / 2
k = h * j
l = 1 * const_60
m = k / l
n = g + m
o = n * 10
p = 1 * const_60
q = 1 * const_60
r = q / 2
s = p * r
t = 1 * const_60
u = s / t
v = o / u
w = v + 10
x = w - 3
|
a ) 15 , b ) 66 , c ) 25 , d ) 88 , e ) 17 | a | subtract(power(negate(3), 2), multiply(negate(3), 2)) | what is the maximum value of vx - yz . if the value of v , x , y , z have to be chosen from the set a where a ( - 3 , - 2 , - 1 , 01 , 23 ) | explanation : to maximize the value of vx - yz , we make yz negative and vx as maximum as possible using given value . vx β yz = ( β 3 ) 2 β ( β 3 Γ 2 ) vx β yz = ( β 3 ) 2 β ( β 3 Γ 2 ) = 15 answer : a | a = negate ** (
b = a - 2
|
a ) 34 , b ) 43.2 , c ) 40 , d ) 38 , e ) 29 | b | divide(add(add(subtract(54, const_10), const_100), add(subtract(54, const_10), const_100)), add(divide(add(subtract(54, const_10), const_100), 54), divide(add(subtract(54, const_10), const_100), 36))) | x and y are two towns . ganesh covers the distance from x to y at an average speed of 54 km / hr . however , he covers the distance from y to x at an average speed of 36 km / hr . his average speed during the whole journey in km / hr . is : | "solution : average speed = 2 xy / x + y = 2 * 54 * 36 / 54 + 36 = 43.2 answer : b" | a = 54 - 10
b = a + 100
c = 54 - 10
d = c + 100
e = b + d
f = 54 - 10
g = f + 100
h = g / 54
i = 54 - 10
j = i + 100
k = j / 36
l = h + k
m = e / l
|
a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 450 | b | add(divide(multiply(50, const_100), 20), 50) | p runs 20 % faster than q so p gives q a 50 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ? | let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.2 v = ( d - 50 ) / v d = 1.2 d - 60 0.2 d = 60 d = 300 meters . the answer is b . | a = 50 * 100
b = a / 20
c = b + 50
|
a ) 9 , b ) 17 , c ) 11 , d ) 10 , e ) 13 | d | divide(70, divide(70, 10)) | suppose 10 monkeys take 10 minutes to eat 10 bananas . how many monkeys would it take to eat 70 bananas in 70 minutes ? | "one monkey takes 10 min to eat 1 banana , so in 70 mins 1 monkey will eat 7 bananas , so for 70 bananas in 70 min we need 70 / 7 = 10 monkeys answer : d" | a = 70 / 10
b = 70 / a
|
a ) 300 , b ) 318 , c ) 320 , d ) 324 , e ) 330 | c | add(multiply(5, const_4), multiply(divide(50, 10), const_60)) | a man walks at a rate of 10 mph . after every ten miles , he rests for 5 minutes . how much time does he take to walk 50 miles ? | to cover 50 miles the man needs ( time ) = ( distance ) / ( rate ) = 50 / 10 = 5 hours = 300 minutes . he will also rest 4 times ( after 10 , 20 , 30 and 40 miles ) , so total resting time = 4 * 5 = 20 minutes . total time = 300 + 20 = 320 minutes . answer : c . | a = 5 * 4
b = 50 / 10
c = b * const_60
d = a + c
|
a ) 1000 , b ) 1100 , c ) 1200 , d ) 1250 , e ) 1300 | a | multiply(divide(254, 2.54), divide(45, 4.5)) | on a map , 4.5 inches represent 45 miles . how many miles approximately is the distance if you measured 254 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 4.5 inch = 2.54 * 4.5 cm 11.43 cm = 45 miles 254 cms = 45 / 11.43 * 254 = 1000 miles answer : a" | a = 254 / 2
b = 45 / 4
c = a * b
|
a ) 100 , b ) 120 , c ) 150 , d ) 160 , e ) 180 | d | subtract(subtract(400, 1), subtract(add(add(divide(subtract(subtract(400, const_2), const_2), const_2), 1), add(divide(subtract(subtract(400, divide(400, const_100)), divide(400, const_100)), divide(400, const_100)), 1)), add(divide(subtract(subtract(400, const_10), const_10), const_10), 1))) | what is the total number of positive integers that are less than 400 and that have no positive factor in common with 400 other than 1 ? | "since 400 = 2 ^ 4 * 5 ^ 2 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 200 odd numbers from 1 to 400 . we then need to eliminate the 40 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 395 . there are a total of 200 - 40 = 160 such numbers between 1 and 400 . the answer is d ." | a = 400 - 1
b = 400 - 2
c = b - 2
d = c / 2
e = d + 1
f = 400 / 100
g = 400 - f
h = 400 / 100
i = g - h
j = 400 / 100
k = i / j
l = k + 1
m = e + l
n = 400 - 10
o = n - 10
p = o / 10
q = p + 1
r = m - q
s = a - r
|
a ) 46 , b ) 47 , c ) 45 , d ) 56 , e ) 54 | c | divide(416.25, 9.25) | john bought 9.25 m of cloth for $ 416.25 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 416.25 cost per metre = 416.25 Γ· 9.25 the cost of the cloth per metre = $ 45 answers : c" | a = 416 / 25
|
a ) 6 , b ) 1 , c ) 0 , d ) 8 , e ) 7 | b | subtract(add(12, power(add(const_1, const_4), 2)), multiply(12, 3)) | if p is a prime number greater than 3 , find the remainder when p ^ 2 + 12 is divided by 12 . | "every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 13 = 36 n ^ 2 + 12 n + 1 + 12 = 36 n ^ 2 + 12 n + 12 + 1 if p = 6 n - 1 , then p ^ 2 + 13 = 36 n ^ 2 - 12 n + 1 + 12 = 36 n ^ 2 - 12 n + 12 + 1 when divided by 12 , it must leave a remainder of 1 . the answer is b ." | a = 1 + 4
b = a ** 2
c = 12 + b
d = 12 * 3
e = c - d
|
a ) 432 , b ) 444 , c ) 2881 , d ) 287 , e ) 221 | a | add(multiply(divide(60, subtract(21, 15)), 15), multiply(divide(60, subtract(21, 15)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 15 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 15 + 21 = 36 t = 12 d = 36 * 12 = 432 answer : a" | a = 21 - 15
b = 60 / a
c = b * 15
d = 21 - 15
e = 60 / d
f = e * 21
g = c + f
|
a ) 7.14 , b ) 6.14 , c ) 5.14 , d ) 3.14 , e ) 2.14 | a | divide(multiply(const_100, 260), multiply(910, 4)) | what is the rate percent when the simple interest on rs . 910 amount to rs . 260 in 4 years ? | "260 = ( 910 * 4 * r ) / 100 r = 7.14 % answer : a" | a = 100 * 260
b = 910 * 4
c = a / b
|
a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 3.1 , e ) 1.03 | e | divide(divide(37.3, 3), multiply(3, 4)) | in 1979 approximately 1 / 3 of the 37.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 4 the number that used kennedy airport and 3 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ? | "number of passengers using kennedy airport = 37 / 3 = ~ 12.43 passengers using miami airport = 12.43 / 4 = ~ 3.1 passengers using logan airport = 3.1 / 3 = ~ 1.03 so e" | a = 37 / 3
b = 3 * 4
c = a / b
|
a ) 54.54 % , b ) 55.55 % , c ) 56.56 % , d ) 57.57 % , e ) 58.58 % | a | divide(792, add(5, 6)) | if $ 792 are divided between worker a and worker b in the ratio 5 : 6 , what is the share that worker b will get ? | "worker b will get 6 / 11 = 54.54 % the answer is a ." | a = 5 + 6
b = 792 / a
|
a ) 674 , b ) 574 , c ) 174 , d ) 274 , e ) 374 | c | subtract(1182, multiply(multiply(12, 3), 2)) | evaluate : 1182 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 1182 - 12 * 3 * 2 = 1182 - 8 = 174 correct answer c" | a = 12 * 3
b = a * 2
c = 1182 - b
|
a ) β 48 , b ) β 4 , c ) 4 , d ) 9 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 5), subtract(4, 5)), const_1), const_1), const_1), const_1) | if a ( a - 4 ) = 5 and b ( b - 4 ) = 5 , where a β b , then a + b = | "i . e . if a = - 1 then b = 5 or if a = 5 then b = - 1 but in each case a + b = - 1 + 5 = 4 answer : option c" | a = 4 + 5
b = 4 - 5
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | a | subtract(subtract(const_1, divide(2, 8)), divide(2, 4)) | in a software company 4 / 10 of people know c + + , 2 / 4 of them know java , 2 / 8 of them know neither , find the total possibility to know c + + ? | assume 100 persons in a company person knows c + + = 4 / 10 = 100 * 4 / 100 = 20 persons person knows java = 2 / 4 = 100 * 2 / 4 = 50 persons neither = 2 / 8 = 100 * 2 / 8 = 25 persons according to our assumption 100 person = 20 + 50 + 25 = 95 person covered ( 5 remaining ) so , total possibility = 20 person ( c + + ) + 5 ( remaining ) = 25 persons which is 25 / 100 = 1 / 4 answer : a | a = 2 / 8
b = 1 - a
c = 2 / 4
d = b - c
|
a ) 87 , b ) 98 , c ) 30 , d ) 28 , e ) 23 | e | divide(add(add(20, 10), multiply(8, 2)), const_2) | the average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 10 years . find the average age of the women ? | "20 + 10 + 8 * 2 = 46 / 2 = 23 answer : e" | a = 20 + 10
b = 8 * 2
c = a + b
d = c / 2
|
a ) Β½ , b ) 3 / 10 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | b | subtract(0.8, divide(1, 2)) | the number 0.8 is how much greater than 1 / 2 ? | "let x be the difference then . 8 - 1 / 2 = x 8 / 10 - 1 / 2 = x x = 3 / 10 ans b" | a = 1 / 2
b = 0 - 8
|
a ) 1 : 1 , b ) 3 : 2 , c ) 3 : 8 , d ) 3 : 25 , e ) 3 : 4 | a | divide(subtract(20, 19), subtract(19, 18)) | two trains running in opposite directions cross a man standing on the platform in 20 seconds and 18 seconds respectively and they cross each other in 19 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 20 x meters , and length of the second train = 18 y meters . ( 20 x + 15 y ) / ( x + y ) = 19 = = > 20 x + 18 y = 19 x + 19 y = = > x = y = = > x / y = 1 / 1 answer : option a" | a = 20 - 19
b = 19 - 18
c = a / b
|
a ) $ 0.00 , b ) $ 1.00 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.80 | d | subtract(multiply(68, divide(add(const_100, 25), const_100)), divide(68, divide(subtract(const_100, 15), const_100))) | cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 68 after it went on sale for 15 % off , what is the difference between the original price and the final price ? | 0.85 * { original price } = $ 68 - - > { original price } = $ 80 . { final price } = $ 68 * 1.25 = $ 85 . the difference = $ 85 - $ 85 = $ 5 . answer : d . | a = 100 + 25
b = a / 100
c = 68 * b
d = 100 - 15
e = d / 100
f = 68 / e
g = c - f
|
a ) $ 4 , b ) $ 0.4 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | subtract(const_1, divide(60, const_100)) | the cost per pound of milk powder and coffee were the same in june . in july , the price of coffee shot up by 200 % and that of milk powder dropped by 60 % . if in july , a mixture containing equal quantities of milk powder and coffee costs $ 5.10 for 3 lbs , how much did a pound of milk powder cost in july ? | "lets assume price of coffee in june = 100 x price of tea in june = 100 x price of coffee in july = 300 x ( because of 200 % increase in price ) price of tea in july = 40 x ( because of 60 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of tea in july will be = 450 x + 60 x = 510 x as per question 510 x = 5.10 $ x = 0.01 s so the price of tea in july = 40 x = 40 x 0.01 = 0.4 $ / pound answer b" | a = 60 / 100
b = 1 - a
|
['a ) 7', 'b ) 11', 'c ) 13', 'd ) 16', 'e ) 38'] | d | divide(multiply(55, divide(60, const_100)), const_2) | each of the 55 points is placed either inside or on the surface of a perfect sphere . if 60 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | maximum number of points on the surface is 60 % * 55 = 33 now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 33 points , so at best we can form 16 pairs ( 32 ) . so , answer is ( d ) | a = 60 / 100
b = 55 * a
c = b / 2
|
a ) 40 , b ) 50 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | e | power(const_10, subtract(2, 8)) | on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 2 is how many times as great as the frequency corresponding to a reading of 8 ? | "if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 2 to 3 it will be 10 if richter scale reading goes from 3 to 4 it will be 10 if richter scale reading goes from 4 to 5 it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading goes from 6 to 7 it will be 10 and if richter scale reading goes from 7 to 8 it will be 10 so it will from 2 to 8 i . e 3,4 , 5,6 , 7,8 = 10 * 10 * 10 * 10 * 10 * 10 = 10 ^ 6 answer is e" | a = 2 - 8
b = 10 ** a
|
a ) 187 , b ) 279 , c ) 120 , d ) 278 , e ) 379 | c | multiply(divide(multiply(40, const_1), subtract(60, 40)), 60) | a train leaves mumabai at 9 am at a speed of 40 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 60 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 40 * 1 = 40 km so , the distance between first train and second train is 40 km at 10.00 am time taken by the trains to meet = distance / relative speed = 40 / ( 60 - 40 ) = 2 hours so , the two trains meet at 12 a . m . the two trains meet 2 * 60 = 120 km away from mumbai . answer : c" | a = 40 * 1
b = 60 - 40
c = a / b
d = c * 60
|
a ) 1.5 , b ) 0.5 , c ) 6.25 , d ) 0.25 , e ) 0.3 | e | subtract(6.30, floor(6.30)) | for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.30 β 6.30 * ? | "since z * is defined as the greatest positive even integer less than or equal to z , then 6.30 * = 6 ( the greatest positive even integer less than or equal to 6.30 is 6 ) . hence , 6.30 β 6.30 * = 6.30 - 6 = 0.30 answer : e ." | a = math.floor(6, 30)
b = 6 - 30
|
a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | d | add(floor(divide(406, 2.5)), const_1) | the guests at a football banquet consumed a total of 406 pounds of food . if no individual guest consumed more than 2.5 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 162 * 2.5 ( max possible amount of food a guest could consume ) = 405 pounds , so there must be more than 162 guests , next integer is 163 . answer : d ." | a = 406 / 2
b = math.floor(a)
c = b + 1
|
a ) 70 , b ) 30 , c ) 25 , d ) 100 , e ) 110 | b | subtract(add(200, 230), 600) | in a group of 600 readers who read science fiction or literacy works or both , 200 read science fiction and 230 read literacy works . how many read both science fiction and literacy works ? | "consider total number of reader n ( s u l ) = 600 people who read science fiction n ( s ) = 400 people who read literacy works n ( l ) = 230 both science fiction and literacy n ( s Γ’ Λ Β© l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s Γ’ Λ Β© l ) 600 = 400 + 230 - n ( s Γ’ Λ Β© l ) n ( s Γ’ Λ Β© l ) = 630 - 600 n ( s Γ’ Λ Β© l ) = 30 so people read both science fiction and literacy works are 30 answer : b" | a = 200 + 230
b = a - 600
|
a ) 13 : 2 , b ) 11 : 2 , c ) 13 : 3 , d ) 13 : 4 , e ) 13 : 5 | a | divide(add(power(2, 4), 4), add(add(3, 5), 4)) | if x : y = 3 : 5 , find the value of ( 2 x + 4 y ) : ( 3 x β y ) | "explanation : given : x / y = 3 / 5 ( 2 x + 4 y ) / ( 3 x β y ) = ( 2 * 3 + 4 * 5 ) : ( 3 * 3 β 5 ) = 26 : 4 = 13 : 2 answer : a" | a = 2 ** 4
b = a + 4
c = 3 + 5
d = c + 4
e = b / d
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | e | divide(14400, multiply(20, multiply(const_2, divide(21600, multiply(40, 30))))) | if daily wages of a man is double to that of a woman , how many men should work for 20 days to earn rs . 14400 ? given that wages for 40 women for 30 days are rs . 21600 . | "explanation : wages of 1 woman for 1 day = 21600 / 40 Γ£ β 30 wages of 1 man for 1 day = 21600 Γ£ β 2 / 40 Γ£ β 30 wages of 1 man for 20 days = 21600 Γ£ β 2 Γ£ β 20 / 40 Γ£ β 30 number of men = 14400 / ( 21600 Γ£ β 2 Γ£ β 20 / 40 Γ£ β 30 ) = 144 / ( 216 Γ£ β 40 / 40 Γ£ β 30 ) = 20 answer : option e" | a = 40 * 30
b = 21600 / a
c = 2 * b
d = 20 * c
e = 14400 / d
|
a ) 0.3995 , b ) 0.4995 , c ) 0.5995 , d ) 0.6995 , e ) 0.7995 | b | divide(subtract(const_1, divide(1, 1000)), const_2) | persons a and b . person a picks a random no . from 1 to 1000 . then person b picks a random no . from 1 to 1000 . what is the probability of b getting no . greater then what a has picked . | probability of a choosing 1 and b greater = ( 1 / 1000 ) * ( 999 / 1000 ) ; probability of a choosing 2 and b greater = ( 1 / 1000 ) * ( 998 / 1000 ) ; probability of a choosing 3 and b greater = ( 1 / 1000 ) * ( 997 / 1000 ) ; . . . ; probability of a choosing 998 and b greater = ( 1 / 1000 ) * ( 2 / 1000 ) ; probability of a choosing 999 and b greater = ( 1 / 1000 ) * ( 1 / 1000 ) ; total = ( 1 / 1000 ) * ( 1 / 1000 ) * ( 999 + 998 + 997 + . . . + 2 + 1 ) = ( 1 / 1000 ) * ( 1 / 1000 ) * ( 999 * 1000 / 2 ) = 0.4995 answer : b | a = 1 / 1000
b = 1 - a
c = b / 2
|
a ) 66.5 , b ) 70.5 , c ) 72.5 , d ) 75.5 , e ) 82.5 | e | multiply(divide(multiply(110, 2), add(50, 110)), const_60) | cole drove from home to work at an average speed of 50 kmh . he then returned home at an average speed of 110 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 50 time from work to home = x / 110 total time = 2 hrs ( x / 50 ) + ( x / 110 ) = 2 solving for x , we get x = 275 / 4 time from home to work in minutes = ( 275 / 4 ) * 60 / 50 = 82.5 minutes ans = e" | a = 110 * 2
b = 50 + 110
c = a / b
d = c * const_60
|
a ) 59 , b ) 65 , c ) 70 , d ) 77 , e ) 98 | a | add(add(add(divide(lcm(lcm(lcm(5, 1), 1), 3), 5), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 3)) | john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | "rakesh : rahul : ranjan : rohit = 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 5 x 60 ) = 12 rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : total number of pencils are ( 12 x + 20 x + 15 x + 12 x ) = 59 x . for minimum number of pencils x = 1 . the person should have at least 59 pencils . a )" | a = math.lcm(5, 1)
b = math.lcm(a, 1)
c = math.lcm(b, 3)
d = c / 5
e = math.lcm(5, 1)
f = math.lcm(e, 1)
g = math.lcm(f, 3)
h = g / 1
i = d + h
j = math.lcm(5, 1)
k = math.lcm(j, 1)
l = math.lcm(k, 3)
m = l / 1
n = i + m
o = math.lcm(5, 1)
p = math.lcm(o, 1)
q = math.lcm(p, 3)
r = q / 3
s = n + r
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3 | b | subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5))) | an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how many g meters long ? | "let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m , leaving 1 m of empty space . thus , i believe the answer is b ) 1" | a = 4 * 2
b = 4 * 1
c = a + b
d = 15 - c
|
a ) - 3200 , b ) 3120 , c ) - 3120 , d ) 3200 , e ) 3208 | c | add(multiply(negate(88), 39), 312) | - 88 Γ 39 + 312 = ? | - 88 Γ 39 + 312 = - 88 Γ ( 40 - 1 ) + 312 = - 88 Γ 40 + 88 + 312 = - 3520 + 88 + 312 = - 3120 answer is c | a = negate * (
b = a + 39
|
a ) 1160 , b ) 1251 , c ) 1102 , d ) 1352 , e ) 1450 | a | multiply(square_perimeter(sqrt(25)), 58) | what will be the cost of building a fence around a square plot with area equal to 25 sq ft , if the price per foot of building the fence is rs . 58 ? | "let the side of the square plot be a ft . a 2 = 25 = > a = 5 length of the fence = perimeter of the plot = 4 a = 20 ft . cost of building the fence = 20 * 58 = rs . 1160 . answer : a" | a = math.sqrt(25)
b = square_perimeter * (
|
a ) 15 sec . , b ) 16 sec . , c ) 18 sec . , d ) 20 sec . , e ) none | b | multiply(divide(280, multiply(63, const_1000)), const_3600) | a train 280 m long , running with a speed of 63 km / hr will pass a tree in | "solution speed = ( 63 x 5 / 18 ) m / sec = 35 / 2 msec time taken = ( 280 x 2 / 35 ) m / sec = 16 sec . answer b" | a = 63 * 1000
b = 280 / a
c = b * 3600
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 100 % , e ) 200 % | e | multiply(divide(10, subtract(15, 10)), const_100) | a shop owner sells 15 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | "here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 15 m cloth ) β ( c . p . of 15 m cloth ) selling price of 5 m cloth = selling price of 15 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 5 m cloth = rs . 500 and s . p . of 5 m cloth = rs . rs . 1500 profit % = 1000 / 500 Γ 100 = 200 % profit of 200 % was made by the merchant . e" | a = 15 - 10
b = 10 / a
c = b * 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | e | add(divide(power(2, 2), 2), const_1) | if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 7 k , what is one possible value for k ? | "first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 12 - 4 / 2 = 10 . so the notation f ( 2 k ) means that we are going to plug a 2 k in for x everywhere in the formula for f ( x ) . that would be : f ( 2 k ) = 12 - ( 2 k ) ^ 2 / 2 = 12 - 2 k ^ 2 . remember that we have to square both the 2 and the k , to get 4 k 2 . now , this expression , the output , we will set equal to 2 k . 12 - 2 k ^ 2 = 2 k - - > k = - 3 or k = 7 . all the answers are positive , so we choose k = 2 . answer = e" | a = 2 ** 2
b = a / 2
c = b + 1
|
a ) 456 , b ) 744 , c ) 899 , d ) 1200 , e ) 1400 | c | divide(multiply(divide(348, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 348 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 348 62 % - - - - - - - - ? = > 899 answer : c" | a = 100 - 62
b = 62 - a
c = b / 100
d = 348 / c
e = d * 62
f = e / 100
|
a ) 5 , b ) t = 9 , c ) t = 10 , d ) t = 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | what is the greatest positive integer t such that 3 ^ t is a factor of 9 ^ 10 ? | "what is the greatest positive integer t such that 3 ^ t is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a = 9 - 10
b = a * 10
|
a ) 12 , b ) 16 , c ) 20 , d ) 22 , e ) 25 | b | add(12, divide(12, const_2)) | in a ratio which is equal to 3 : 4 , if the antecedent is 12 , then the consequent is ? | "we have 3 / 4 = 12 / x 3 x = 48 x = 16 consequent = 16 answer is b" | a = 12 / 2
b = 12 + a
|
a ) 8.2 % , b ) 8 % , c ) 8.5 % , d ) 9 % , e ) 9.5 % | b | divide(multiply(const_100, 640), multiply(4000, 2)) | what is the rate percent when the simple interest on rs . 4000 amount to rs . 640 in 2 years ? | "interest for 1 year = 640 / 2 = 320 interest on rs 4000 p / a = 320 interest rate = 320 / 4000 * 100 = 8 % answer : b" | a = 100 * 640
b = 4000 * 2
c = a / b
|
a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 5 / 3 | e | divide(add(divide(subtract(multiply(7, 2), 8), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 8), subtract(multiply(2, 2), const_1))))), 3) | if 2 x + y = 7 and x + 2 y = 8 , then ( x + y ) / 3 = | "we have two equations : 2 x + y = 7 x + 2 y = 8 notice that something nice happens when we add them . we get : 3 x + 3 y = 15 divide both sides by 3 to get : x + y = 5 so , ( x + y ) / 3 = 5 / 3 answer : e" | a = 7 * 2
b = a - 8
c = 2 * 2
d = c - 1
e = b / d
f = 7 * 2
g = f - 8
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 7 - k
m = e + l
n = m / 3
|
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 13.88 % , e ) none | d | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 90) | farm tax is levied on the 90 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 90 ) = 13.88 % the correct answer is ( d ) ." | a = 480 / 3840
b = a * 100
c = b * 100
d = c / 90
|
a ) 2177 , b ) 2876 , c ) 4500 , d ) 2981 , e ) 2711 | c | subtract(multiply(8000, const_4), subtract(multiply(8500, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8500 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8500 = 34000 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 2000 salary of may is rs . 6500 salary of january = rs . 4500 . answer : c" | a = 8000 * 4
b = 8500 * 4
c = b - 6500
d = a - c
|
a ) 40 , b ) 460 , c ) 500 , d ) 540 , e ) 150 | d | subtract(add(divide(840, divide(8, const_100)), 840), add(divide(800, divide(8, const_100)), 800)) | don and his wife each receive an 8 percent annual raise . if don receives a raise rs . 800 and his wife receives a raise of rs . 840 , what is the difference between their annual income after their raises ? | don salary 8 % = 800 ; therefore , 100 % = 10000 his wife salary 8 % = 840 ; therefore , 100 % = 10500 after raise 8 % , don salary = 10000 + 800 = 10800 and his wife salary = 10500 + 840 = 11340 difference = 11340 - 10800 = 540 answer : d | a = 8 / 100
b = 840 / a
c = b + 840
d = 8 / 100
e = 800 / d
f = e + 800
g = c - f
|
a ) rs . 51.75 , b ) rs 51.50 , c ) rs 51.25 , d ) rs 51 , e ) none of these | c | subtract(add(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), 5), const_100)), divide(50, multiply(divide(5, const_100), 2))) | if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 50 , what will be the compound interest on same values | "explanation : s . i . = p β r β t / 100 p = 50 β 100 / 5 β 2 = 500 amount = 500 ( 1 + 5100 ) 2 500 ( 21 / 20 β 21 / 20 ) = 551.25 c . i . = 551.25 β 500 = 51.25 option c" | a = 5 / 100
b = a * 2
c = 50 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 50 / g
i = e + h
j = 5 / 100
k = j * 2
l = 50 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 50 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 50 / w
y = u - x
|
a ) 200 , b ) 250 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(300, 200), const_2) | if x + y = 300 , x - y = 200 , for integers of x and y , y = ? | "x + y = 300 x - y = 200 2 x = 100 x = 50 y = 250 answer is b" | a = 300 + 200
b = a / 2
|
a ) 22 , b ) 77 , c ) 18 , d ) 21.6 , e ) 88 | d | multiply(divide(subtract(66900, add(42000, 13000)), add(42000, 13000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 66900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 66900 profit ( % ) = ( 66900 - 55000 ) / 55000 * 100 = 21.6 % answer : d" | a = 42000 + 13000
b = 66900 - a
c = 42000 + 13000
d = b / c
e = d * 100
|
a ) 3 days , b ) 16 / 3 days , c ) 3 / 4 days , d ) 9 / 12 days , e ) 5 / 7 days | b | divide(multiply(8, 16), add(8, 16)) | a can do a work in 8 days . b can do the same work in 16 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 8 b rate = 1 / 16 ( a + b ) rate = ( 1 / 8 ) + ( 1 / 16 ) = 3 / 16 a & b finish the work in 16 / 3 days correct option is b" | a = 8 * 16
b = 8 + 16
c = a / b
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 3 , e ) 2 / 5 | a | divide(const_4, add(multiply(const_4, 6), const_1)) | tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 6 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ? | "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 4 / 6 ) = 1 / 2 hours peter worked for 3 / 2 hours so he painted 3 / 2 * 1 / 3 = 1 / 2 of the room . the answer is a ." | a = 4 * 6
b = a + 1
c = 4 / b
|
a ) 20 m , b ) 16 m , c ) 14 m , d ) 10 m , e ) 15 m | c | multiply(divide(subtract(25, 20), 25), 70) | if in a race of 70 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 70 * 5 ) / 25 = 14 m hence , a beats b by 14 m . answer c" | a = 25 - 20
b = a / 25
c = b * 70
|
a ) $ 15,360 , b ) $ 17,360 , c ) $ 18,000 , d ) $ 21,960 , e ) $ 27,360 | d | divide(multiply(divide(add(multiply(3, multiply(divide(subtract(65, multiply(const_3, const_10)), const_100), 12000)), 9360), add(multiply(3, multiply(divide(subtract(65, multiply(const_3, const_10)), const_100), 12000)), 9360)), const_3600), const_10) | a certain social security recipient will receive an annual benefit of $ 12000 provided he has annual earnings of $ 9360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9360 . what amount of total annual earnings would result in a 65 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) | for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960 ans : d | a = 3 * 10
b = 65 - a
c = b / 100
d = c * 12000
e = 3 * d
f = e + 9360
g = 3 * 10
h = 65 - g
i = h / 100
j = i * 12000
k = 3 * j
l = k + 9360
m = f / l
n = m * 3600
o = n / 10
|
a ) 10 days , b ) 15 days , c ) 16 days , d ) 17 days , e ) 20 days | d | subtract(multiply(inverse(multiply(4, divide(1, 3))), 40), 13) | jane and ashley take 13 1 / 3 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an 8 day leave from work . this led to jane ' s working for 4 extra days on her own to complete the project . how long did it take to finish the project ? | let us assume that the work is laying 40 bricks . jane = 3 bricks per day ashley = 1 brick per day together = 4 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 12 remaining bricks = 40 - 20 = 20 so together , they would take 20 / 4 = 5 total no . of days = 8 + 4 + 5 = 17 answer is d | a = 1 / 3
b = 4 * a
c = 1/(b)
d = c * 40
e = d - 13
|
a ) 8 , b ) 9 , c ) 10 , d ) 21 , e ) 32 | e | divide(subtract(multiply(90, 8), multiply(82, 8)), subtract(92, 90)) | the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students β grades was 82 , and the average of the female students β grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 82 thus m = 656 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 656 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 32 ans : e" | a = 90 * 8
b = 82 * 8
c = a - b
d = 92 - 90
e = c / d
|
a ) 12 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | a | divide(24, const_2) | if the length of the longest chord of a certain circle is 24 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 24 = 2 * 12 so radius of the circle = 12 correct answer - a" | a = 24 / 2
|
a ) 5768 , b ) 5760 , c ) 5762 , d ) 5766 , e ) 5712 | b | divide(multiply(4, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 4 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 answer : b" | a = 8 * const_60
b = 4 * a
c = 8 * const_60
d = 6 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 30 , b ) 31 , c ) 45 , d ) 90 , e ) 20 | e | multiply(2, 10) | the class mean score on a test was 60 , and the standard deviation was 10 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ? | 1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 70 to 50 , where 70 is within sd above the mean and 50 within 1 sd below the mean 2 sd = 10 twice = 20 from the the mean , which is 80 to 40 , where 80 is within 2 sd above the mean and 20 is within 2 sd below the mean . answer = e | a = 2 * 10
|
a ) 34880 , b ) 3778 , c ) 12788 , d ) 18000 , e ) 28800 | e | multiply(multiply(const_3, const_60), const_60) | if an object travels at eight feet per second , how many feet does it travel in one hour ? | "explanation : if an object travels at 5 feet per second it covers 5 x 60 feet in one minute , and 5 x 60 x 60 feet in one hour . answer = 28800 answer : e ) 28800" | a = 3 * const_60
b = a * const_60
|
a ) 50 , b ) 40 , c ) 100 , d ) 30 , e ) 20 | c | multiply(divide(90, subtract(10, 1)), 10) | the ratio of buses to cars on river road is 1 to 10 . if there are 90 fewer buses than cars on river road , how many cars are on river road ? | "b / c = 1 / 10 c - b = 90 . . . . . . . . . > b = c - 90 ( c - 90 ) / c = 1 / 10 testing answers . clearly eliminate abde put c = 100 . . . . . . . . . > ( 100 - 90 ) / 100 = 10 / 100 = 1 / 10 answer : c" | a = 10 - 1
b = 90 / a
c = b * 10
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(divide(subtract(subtract(add(const_1, divide(75, const_100)), const_1), divide(57.5, const_100)), add(const_1, divide(75, const_100))), const_100) | a merchant marks goods up by 75 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 57.5 % . what % discount did the merchant offer ? | "let p be the original price of the goods and let x be the rate after the markup . ( 1.75 p ) * x = 1.575 p x = 1.575 / 1.75 = 0.9 which is a discount of 10 % . the answer is a ." | a = 75 / 100
b = 1 + a
c = b - 1
d = 57 / 5
e = c - d
f = 75 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 0 , b ) 24 / 625 , c ) 2 / 9 , d ) 1 / 3 , e ) 1 | b | multiply(factorial(5), power(divide(1, 5), 5)) | a certain roller coaster has 5 cars , and a passenger is equally likely to ride in any 1 of the 5 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 5 times , what is the probability that the passenger will ride in each of the 5 cars ? | "if he is to ride 5 times and since he can choose any of the 5 cars each time , total number of ways is = 5 * 5 * 5 * 5 * 5 = 3125 now the number of ways if he is to choose a different car each time is = 5 * 4 * 3 * 2 * 1 = 120 so the probability is = 120 / 3125 = 24 / 625 answer : b" | a = math.factorial(5)
b = 1 / 5
c = b ** 5
d = a * c
|
a ) 10.2 days , b ) 15.2 days , c ) 16.2 days , d ) 18 days , e ) 20 days | b | add(add(divide(subtract(subtract(const_1, multiply(const_4, divide(const_1, 10))), multiply(add(const_4, const_4), divide(const_1, 40))), add(divide(const_1, 10), divide(const_1, 40))), add(const_4, const_4)), const_4) | jane and ashley take 10 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an eight day leave from work . this led to jane ' s working for four extra days on her own to complete the project . how long did it take to finish the project ? | let us assume that the work is laying 40 bricks . jane = 4 bricks per day ashley = 1 brick per day together = 5 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 16 remaining bricks = 40 - 24 = 16 so together , they would take 16 / 5 = 3.2 total no . of days = 8 + 4 + 3.2 = 15.2 answer is b | a = 1 / 10
b = 4 * a
c = 1 - b
d = 4 + 4
e = 1 / 40
f = d * e
g = c - f
h = 1 / 10
i = 1 / 40
j = h + i
k = g / j
l = 4 + 4
m = k + l
n = m + 4
|
a ) 9 km , b ) 2 km , c ) 1 km , d ) 6 km , e ) 7 km | d | add(multiply(4, multiply(divide(const_2, const_3), add(1, divide(24, const_60)))), multiply(multiply(subtract(1, divide(const_2, const_3)), add(1, divide(24, const_60))), 5)) | if john covers a certain distance in 1 hr . 24 min . by covering two third of the distance at 4 kmph and the rest at 5 kmph , then find the total distance . | "explanation : let the total distance be y km . then , ( 2 / 3 ) y / 4 + ( 1 / 3 ) y / 5 = 7 / 5 y / 6 + y / 15 = 7 / 5 7 y = 42 y = 6 km answer : d" | a = 2 / 3
b = 24 / const_60
c = 1 + b
d = a * c
e = 4 * d
f = 2 / 3
g = 1 - f
h = 24 / const_60
i = 1 + h
j = g * i
k = j * 5
l = e + k
|
a ) 28 , b ) 224 , c ) 320 , d ) 512 , e ) 1,600 | a | gcd(10, const_4) | if m and n are positive integers and m ^ 2 + n ^ 2 = 10 , what is the value of m ^ 3 + n ^ 3 ? | "you need to integers which squared are equal 10 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 stop . the integers ca n ' t be greater than 3 or we will score above 10 . the second integer need to be picked up the same way . 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 the only pair that matches is 3 ^ 2 + 1 ^ 2 = 10 . so 3 ^ 3 + 1 ^ 3 = 28 . answer a . )" | a = math.gcd(10, 4)
|
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