options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 6 , b ) 5 , c ) 2 , d ) 3 , e ) 4 | d | subtract(5, add(1, 1)) | if the integer n has exactly 5 positive divisors , including 1 and n , how many positive divisors does √ n have ? | take the example of 16 . . . it has 3 positive divisors ( 1 , 2,4 , 8,16 ) now , take the example of 4 . . . it has only 3 divisors . . so d is the ans | a = 1 + 1
b = 5 - a
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a ) 249 , b ) 498 , c ) 239 , d ) 823 , e ) 1,002 | c | divide(subtract(480, 2), 2) | a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 480 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 480 y = 2 x + 2 478 = 2 x x = 239 answer : c . 239" | a = 480 - 2
b = a / 2
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a ) a ) 985 , b ) b ) 1050 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1075 | a | add(multiply(7, 70), multiply(9, 55)) | bruce purchased 7 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | cost of 7 kg grapes = 70 × 7 = 490 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 490 + 495 = 985 a | a = 7 * 70
b = 9 * 55
c = a + b
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a ) 108 , b ) 119 , c ) 128 , d ) 135 , e ) 168 | e | multiply(divide(subtract(26, divide(52, 26)), divide(52, 26)), add(divide(subtract(26, divide(52, 26)), divide(52, 26)), divide(52, 26))) | if the sum of two positive integers is 26 and the difference of their squares is 52 , what is the product of the two integers ? | let the 2 positive numbers x and y x + y = 26 - - 1 x ^ 2 - y ^ 2 = 52 = > ( x + y ) ( x - y ) = 52 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 14 y = 12 product = 14 * 12 = 168 answer e | a = 52 / 26
b = 26 - a
c = 52 / 26
d = b / c
e = 52 / 26
f = 26 - e
g = 52 / 26
h = f / g
i = 52 / 26
j = h + i
k = d * j
|
a ) $ 300000 , b ) $ 320000 , c ) $ 335000 , d ) $ 336500 , e ) $ 337500 | e | multiply(9000, divide(900000, add(9000, add(7000, 8000)))) | fonzie , aunt bee and lapis paid $ 7000 , $ 8000 and $ 9000 respectively to buy a map to capt . plaidstache ' s treasure . when they found it , the treasure turned out to be worth a cool $ 900000 . how much of it should lapis get ? | a = 7000 b = 8000 c = 9000 a share 7 parts , b share 8 parts & c share 9 parts total 24 parts - - - - - > 900000 - - - - > 1 part - - - - - - - > 37500 c share = 9 parts - - - - - > 337500 e | a = 7000 + 8000
b = 9000 + a
c = 900000 / b
d = 9000 * c
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a ) [ 108 ] , b ) [ 54 ] , c ) [ 36 ] , d ) [ 27 ] , e ) [ 18 ] | a | divide(multiply(divide(4, 2), multiply(3, 9)), 3) | for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 9 ] * [ 4 ] equivalent to ? | "[ 9 ] * [ 4 ] = 27 * 2 = 54 = ( 1 / 2 ) ( 108 ) = [ 108 ] the answer is a ." | a = 4 / 2
b = 3 * 9
c = a * b
d = c / 3
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a ) 3.5 , b ) 4 , c ) 4.5 , d ) 5 , e ) 5.5 | d | multiply(subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 15)), 4)), 15) | two pipes p and q can fill a cistern in 10 and 15 minutes respectively . both are opened together , but at the end of 4 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | "let x be the total time it takes for the cistern to fill . 4 / 10 + x / 15 = 1 x / 15 = 3 / 5 x = 9 after the first pipe is turned off , it takes 5 more minutes to fill the cistern . the answer is d ." | a = 1 / 10
b = 1 / 15
c = a + b
d = c * 4
e = 1 - d
f = e * 15
|
a ) 23 , b ) 88 , c ) 90 , d ) 72 , e ) 21 | d | multiply(divide(20, const_1000), const_3600) | express 20 mps in kmph ? | 20 * 18 / 5 = 72 kmph answer : d | a = 20 / 1000
b = a * 3600
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a ) 35 % , b ) 33 % , c ) 36 % , d ) 38 % , e ) 40 % | b | subtract(40, divide(40, 5)) | you hold some gold in a vault as an investment . over the past year the price of gold increases by 40 % . in order to keep your gold in the vault , you must pay 5 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 40 % ) * ( 100 % - 5 % ) = 140 * 0.95 = 133 % an increase of 33 % your gold holdings have increased in value by 33 % . the answer is b" | a = 40 / 5
b = 40 - a
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a ) 30 , b ) 15 , c ) 20 , d ) 25 , e ) 50 | a | multiply(divide(const_1, add(inverse(40), add(inverse(60), inverse(40)))), const_2) | a large tanker can be filled by two pipes a and b in 60 and 40 minutes respectively . how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half ? | suppose the tank is filled in x minutes . then , x / 2 ( 1 / 24 + 1 / 40 ) = 1 x / 2 * 1 / 15 = 1 = > x = 30 min answer a | a = 1/(40)
b = 1/(60)
c = 1/(40)
d = b + c
e = a + d
f = 1 / e
g = f * 2
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | divide(subtract(add(27, add(27, 1)), multiply(const_3.0, const_3.0)), const_2) | the captain of a cricket team of 11 members is 27 years old and the wicket keeper is 1 year older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 27 + 28 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is d ." | a = 27 + 1
b = 27 + a
c = 3 * 0
d = b - c
e = d / 2
|
a ) 30 , b ) 28 , c ) 80 , d ) 26 , e ) 25 | c | add(multiply(const_4, 6.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 6.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 6.9421 + 1 ) = 10 ( 7.9421 ) = 79.421 x = antilog ( 79.421 ) therefore , number of digits in x = 80 . answer : c" | a = 4 * 6
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 90 , b ) 131 , c ) 150 , d ) 320 , e ) 450 | b | add(90, multiply(divide(subtract(787, 90), add(90, 80)), subtract(90, 80))) | two heavily loaded sixteen - wheeler transport trucks are 787 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , maintaining an average speed of 80 kilometers per hour . how many kilometers farther than driver b , will driver a have driven when they meet and pass each other on the highway ? | i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 787 - 90 = 697 k ratio of speeds 9 : 8 - > 697 / 17 = 41 truck a = 90 + 41 * 9 = 459 truck b = 41 * 8 = 328 delta = 131 km answer b | a = 787 - 90
b = 90 + 80
c = a / b
d = 90 - 80
e = c * d
f = 90 + e
|
a ) 3 / 25 , b ) 3 / 125 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 5 | a | divide(subtract(divide(multiply(800, 1), 5), 64), 800) | in a group of 800 people , 1 / 5 play at least one instrument , 64 play two or more . what is the probability that one student play exactly one instrument ? | "p ( playing 2 or more instruments ) = 64 / 800 = 2 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 1 / 5 - 2 / 25 = 3 / 25 . answer a ." | a = 800 * 1
b = a / 5
c = b - 64
d = c / 800
|
a ) 65 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 30 deg | b | subtract(multiply(30, multiply(const_3, const_2)), 8) | what is the angle between the hands of a clock when time is 8 : 30 ? | "angle between two hands = 30 h - 11 / 2 m = 30 * 8 - 30 * 11 / 2 = 240 - 165 = 75 deg answer : b" | a = 3 * 2
b = 30 * a
c = b - 8
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a ) 7900 , b ) 8,000 , c ) 10,000 , d ) 8,300 , e ) 9,300 | c | multiply(multiply(1280, 3), 3) | if the difference between compound interest ( interest compounded yearly ) and simple interest on a certain sum at the rate 20 % p . a . after 3 years is rs . 1280 then what is the principal ? | "let p is the principal , so p ( [ 120 / 100 ] ^ 3 - 1 ) - p * 20 / 100 * 3 = 1280 p [ 728 / 1000 - 3 / 5 ] = 1280 p = 1280 * 5 * 1000 / 728 * 5 - 3000 = 10000 answer : c" | a = 1280 * 3
b = a * 3
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a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | c | divide(subtract(add(5, 4), 5), 2) | in the xy - coordinate system , if ( m , n ) and ( m + 4 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 4 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 1 - - - > 5 = 2 k + 1 - - - > k = 2 . the answer is , therefore , ( c ) ." | a = 5 + 4
b = a - 5
c = b / 2
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a ) 21000 , b ) 19800 , c ) 16500 , d ) 15800 , e ) 8700 | c | multiply(divide(subtract(25200, 5400), add(5, const_1)), 5) | during the first week of performances of a certain play , 5400 tickets were sold , all at reduced price . during the remaining weeks of performances , 5 times as many tickets were sold at full price as were sold at reduced price . if the total number of tickets r sold was 25200 , how many of them were sold at full price ? | given : the question tells us about the number of tickets sold in the first week at reduced price = 5400 . it also tells us that during the remaining weeks , the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price . the total tickets sold was 25200 , and we are asked to find the number of tickets that where sold at full price . approach : the question asks us to find the number of ticket sold at full price . we know that the number of tickets sold at full price was 5 times the no . of tickets sold at reduced price in remaining weeks . hence , assuming tickets sold at reduced price in remaining weeks to be x would give us the no . of tickets sold at full price to be 5 x . tickets sold in first week + tickets sold in remaining weeks = total no . of tickets soldi . e . tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold we know the no . of tickets sold during first week and the total no . of tickets sold . we will use this relation to get to our answer . working out : tickets sold in first week + tickets sold at full price in remaining weeks + tickets sold at reduced price in remaining weeks = total no . of tickets sold substituting values in the equation , we get 5400 + 5 x + x = 25200 x = 3300 tickets sold at full price = 5 x = 16500 answer : option c | a = 25200 - 5400
b = 5 + 1
c = a / b
d = c * 5
|
a ) 79754 , b ) 87689 , c ) 59875 , d ) 11400 , e ) 87648 | d | subtract(19000, multiply(divide(4, 10), 19000)) | income and expenditure of a person are in the ratio 10 : 4 . if the income of the person is rs . 19000 , then find his savings ? | "let the income and the expenditure of the person be rs . 10 x and rs . 4 x respectively . income , 10 x = 19000 = > x = 1900 savings = income - expenditure = 10 x - 4 x = 6 x = 6 ( 1900 ) so , savings = rs . 11400 . answer : d" | a = 4 / 10
b = a * 19000
c = 19000 - b
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a ) 621 , b ) 276 , c ) 236 , d ) 800 , e ) 211 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 2000) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 2000 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 2000 = > 5 parts out of 15 parts is rs . 2000 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 2000 ) = rs . 800 answer : d" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 2000
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a ) 7 / 13 , b ) 15 / 26 , c ) 9 / 12 , d ) 8 / 13 , e ) 29 / 52 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | if a card is drawn from a well shuffled deck of cards , what is the probability of drawing a black card or a face card ? | "p ( b á ´ œ f ) = p ( b ) + p ( f ) - p ( b â ˆ © f ) , where b denotes black cards and f denotes face cards . p ( b á ´ œ f ) = 26 / 52 + 12 / 52 - 6 / 52 = 8 / 13 answer : d" | a = 3 / const_52
b = const_52 / 4
c = b / const_52
d = a + c
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a ) 23 m 2 , b ) 20 m 2 , c ) 52 m 2 , d ) 24 m 2 , e ) 55 m 2 | a | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 2), multiply(add(divide(25, const_100), 1), 4))), multiply(2, 4)) | a cistern 4 m long and 2 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 2 x 1.25 + 4 x 1.25 ) + 4 x 2 ] m 2 = 23 m 2 . answer : a" | a = 25 / 100
b = a + 1
c = b * 2
d = 25 / 100
e = d + 1
f = e * 4
g = c + f
h = 2 * g
i = 2 * 4
j = h + i
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 62), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 62 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 64.8 bears per 36 hours a week or 1.8 bears / hr ( [ 40 bears * 1.62 ] / [ 40 hrs * . 90 ] ) . [ ( 1.8 - 1 ) / 1 ] * 100 % = 80 % answer : b" | a = 100 - 62
b = a - 10
c = 10 / b
d = c * 100
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a ) 25 , b ) 20 , c ) 45 , d ) 30 , e ) 35 | c | divide(add(add(add(multiply(4, const_3), add(4, multiply(4, const_2))), multiply(4, const_4)), multiply(add(const_4, const_1), 4)), 4) | find the average of all numbers between 5 and 37 which are divisible by 4 | "explanation : average = ( 4 + 8 + 12 + 16 + 20 + 24 + 28 + 32 + 36 ) / 4 ) = 180 / 4 = 45 option c" | a = 4 * 3
b = 4 * 2
c = 4 + b
d = a + c
e = 4 * 4
f = d + e
g = 4 + 1
h = g * 4
i = f + h
j = i / 4
|
a ) 3 , b ) 6 , c ) 7.2 , d ) 7.8 , e ) 9 | e | multiply(divide(20, const_100), 20) | uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 18 ounces of chocolate . how many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ? | "answer is e . x / x + 36 = 1 / 5 x = 9 18 - 9 = 9" | a = 20 / 100
b = a * 20
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a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 80 % | b | multiply(subtract(multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 50 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | "assume the broker invested $ 100 . after year 1 ( 50 % increase in wealth ) : $ 100 * 0.5 = $ 50 increase + $ 100 = $ 150 after year 2 ( 30 % decrease in wealth ) : $ 150 * 0.3 = $ 45 ; $ 150 - $ 45 decrease = $ 105 net increase = $ 105 / $ 100 - 1 = 5 % answer is b ." | a = 50 / 100
b = 1 + a
c = 30 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
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a ) 10 , b ) 30 , c ) 40 , d ) 25 , e ) none of these | d | add(multiply(sqrt(divide(subtract(225, 200), const_2)), const_100), sqrt(subtract(225, divide(subtract(225, 200), const_2)))) | the sum of the squares of three numbers is 225 , while the sum of their products taken two at a time is 200 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 225 xy + yz + zx = 200 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 225 + ( 2 * 200 ) ( x + y + z ) ^ 2 = 625 so x + y + z = 25 answer : d" | a = 225 - 200
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 225 - 200
f = e / 2
g = 225 - f
h = math.sqrt(g)
i = d + h
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a ) 3 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | e | subtract(multiply(add(6, const_1), const_3), divide(480, 40)) | a train travels from new york to chicago , a distance of approximately 480 miles , at an average rate of 40 miles per hour and arrives in chicago at 6 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) | "6 : 00 in evening in chicago = 7 : 00 in evening in new york . so , the train was in chicago 7 : 00 in the evening , new york time . the trip took t = d / r = 480 / 40 = 12 hours . therefore , the train depart from new york at 7 : 00 - 12 hours = 7 : 00 in the morning , new york time . answer : e ." | a = 6 + 1
b = a * 3
c = 480 / 40
d = b - c
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a ) 300 , b ) 301 , c ) 301.5 , d ) 302.5 , e ) 306 | c | divide(add(add(multiply(9, floor(divide(100, 9))), multiply(divide(floor(divide(subtract(500, 100), 9)), const_2), 9)), add(add(multiply(9, floor(divide(100, 9))), multiply(divide(floor(divide(subtract(500, 100), 9)), const_2), 9)), 9)), const_2) | if m is the set of all consecutive multiples of 9 between 100 and 500 , what is the median of m ? | since 100 / 9 = 11.11 , 500 / 9 = 55.55 , m can be written as { 9 * 12 , 9 * 13 , . . . . , 9 * 54 , 9 * 55 } . the number of elements of m is , therefore , 44 ( = 55 - 12 + 1 ) . since the number of elements is even number the median is the average of the 22 nd element ( = 9 * 33 ) and 23 rd element ( = 9 * 34 ) . so the median is ( 9 * 33 + 9 * 34 ) / 2 = 301.5 . the answer is c . | a = 100 / 9
b = math.floor(a)
c = 9 * b
d = 500 - 100
e = d / 9
f = math.floor(e)
g = f / 2
h = g * 9
i = c + h
j = 100 / 9
k = math.floor(j)
l = 9 * k
m = 500 - 100
n = m / 9
o = math.floor(n)
p = o / 2
q = p * 9
r = l + q
s = r + 9
t = i + s
u = t / 2
|
a ) 99 , b ) 289 , c ) 350 , d ) 882 , e ) 400 | e | subtract(multiply(divide(300, 18), 42), 300) | a 300 meter long train crosses a platform in 42 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 42 = 50 / 3 3 ( x + 300 ) = 2100 è x = 400 m . answer : e" | a = 300 / 18
b = a * 42
c = b - 300
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a ) 500 , b ) 550 , c ) 600 , d ) 700 , e ) 975 | e | divide(780, subtract(const_1, divide(20, const_100))) | shop offered 20 % offer for every shirt , smith bought a shirt at rs . 780 . and what was the shop ' s original selling price ? | "sp * ( 80 / 100 ) = 780 sp = 9.75 * 100 = > cp = 975 answer : e" | a = 20 / 100
b = 1 - a
c = 780 / b
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a ) 2 , b ) b ) 1 , c ) c ) 5 , d ) d ) 21 , e ) e ) 28 | a | add(1, const_1) | two positive integers differ by 4 , and sum of their reciprocals is 1 . then one of the numbers is | algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 1 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 1 or n ^ 2 + 4 n = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 2 . hence , a | a = 1 + 1
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a ) 10 , b ) 15 , c ) 20 , d ) 24 , e ) 27 | c | divide(subtract(sqrt(add(multiply(multiply(multiply(360, 4), 4), 3), power(multiply(3, 4), const_2))), multiply(3, 4)), multiply(const_2, 3)) | john has rs 360 for his expenses . if he exceeds his days by 4 days he must cut down daily expenses by rs 3 . the number of days of john ' s tour program is | let john under takes a tour of x days . then , expenses for each day = 360 x 360 x + 4 = 360 x − 3 x = 20 and − 24 hence , x = 20 days . c | a = 360 * 4
b = a * 4
c = b * 3
d = 3 * 4
e = d ** 2
f = c + e
g = math.sqrt(f)
h = 3 * 4
i = g - h
j = 2 * 3
k = i / j
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 1200 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 1200 x = 20 the numbers are 3 * 20 , 4 * 20 , 5 * 20 hence required h . c . f . = 20 answer is a" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
|
a ) 17 % , b ) 25.6 % , c ) none % , d ) 33.3 % , e ) 50 % | d | multiply(subtract(const_1, divide(const_1, divide(add(const_100, 50), const_100))), const_100) | the price of food in the area where the sims family lives is scheduled to increase by 50 % next year . since the sims family can not afford an increase in their food bill , how much will they have to reduce consumption to keep their cost the same ? | solution : let the current food expense be represented by rs . 100 . the cost of food rises 50 % . so , to buy same amount of food , they need to increase their expense , = ( 100 + 50 % of 100 ) = rs . 150 . but , they want to keep food expense the same , so they have to cut rs . by 50 to keep it to rs . = 100 . the % decrease in consumption is , ( 50 / 150 ) * 100 = 33.3 % . mental calculation method ; 100 - - - - - 50 % ↑ - - - → 150 - - - - - - x % ↓ - - - → 100 . here , x = ( 50 / 125 ) * 100 = 33.3 % . answer : option d | a = 100 + 50
b = a / 100
c = 1 / b
d = 1 - c
e = d * 100
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a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 710 , e ) $ 1000 | b | multiply(10000, divide(5, const_100)) | find the simple interest on $ 10000 at 5 % per annum for 12 months ? | "p = $ 10000 r = 5 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 5 * 1 / 100 = $ 500 answer is b" | a = 5 / 100
b = 10000 * a
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a ) 20 % , b ) 25 % , c ) 18 % , d ) 50 % , e ) none of these | d | multiply(divide(divide(8, const_100), divide(16, const_100)), const_100) | if the given two numbers are respectively 8 % and 16 % of a third number , then what percentage is the first of the second ? | "here , l = 8 and m = 16 therefore , first number = l / m x 100 % of second number = 8 / 16 x 100 % of second number = 50 % of second number answer : d" | a = 8 / 100
b = 16 / 100
c = a / b
d = c * 100
|
a ) 5 / 5 , b ) 4 / 5 , c ) 3 / 5 , d ) 2 / 5 , e ) 1 / 5 | d | divide(const_4, add(const_4, const_6)) | if you keep rolling a pair of dice together till a sum of 5 or 7 is obtained , then what is the probability that a sum of 5 comes before a sum of 7 ? | this can be solved in a generic and complex way but let us not go into all that . there can be four ways through which the pair of dice results in a sum of 5 . there can be six ways through which the pair of dice can result in a sum of 7 . now , we want the probability of the pair of dice resulting in a sum of 5 before a sum of 7 . thus probability = 4 / ( 4 + 6 ) = 4 / 10 or 2 / 5 . correct answer is d ) 2 / 5 | a = 4 + 6
b = 4 / a
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a ) 18 , b ) 82 , c ) 20 , d ) 27 , e ) 29 | c | multiply(subtract(73, 63), const_2) | a pupil ' s marks were wrongly entered as 73 instead of 63 . 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 = ( 73 - 63 ) = > x / 2 = 10 = > x = 20 . answer : c" | a = 73 - 63
b = a * 2
|
a ) 196 , b ) 125 , c ) 140 , d ) 128 , e ) 134 | a | multiply(multiply(3, 4), multiply(3, 4)) | in the coordinate plane , one of the vertices of a square is the point ( - 3 , - 4 ) . if the diagonals of that square intersect at point ( 4 , 2 ) , what is the area of that square ? | "one point ( - 3 - 4 ) , intersection ( 4,2 ) so the distance from the first point - 3 - 4 = - 7 is the midpoint of the square - - > whole side 14 , 14 * 14 = 196 a" | a = 3 * 4
b = 3 * 4
c = a * b
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a ) 50 , b ) 60 , c ) 490 , d ) 500 , e ) 980 | b | divide(588, 9.8) | a sports equipment store sold ping pong rackets for a total of $ 588 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 588 x = 60 pairs were sold . answer : b" | a = 588 / 9
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | power(add(multiply(1, 2), 2), 2) | if a is a positive integer , and if the units digit of a ^ 2 is 1 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 1 , then the units digit of a is either 1 or 9 . if the units digit of ( a + 1 ) ^ 2 is 4 , then the units digit of a + 1 is either 2 or 8 . to satisfy both conditions , the units digit of a must be 1 . then a + 2 has the units digit of 3 , thus the units digit of ( a + 2 ) ^ 2 will be 9 . the answer is e ." | a = 1 * 2
b = a + 2
c = b ** 2
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a ) $ 5 , b ) $ 8 , c ) $ 9 , d ) $ 3 , e ) $ 7 | d | subtract(subtract(multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 1200), 1200), divide(multiply(1200, 10), const_100)) | the difference between simple interest and compound interest on $ 1200 for one year at 10 % per annum reckoned half - yearly is : | "s . i = $ [ ( 1200 * 10 * 1 ) / 100 ] = $ 120 . c . i = $ [ 1200 * ( 1 + ( 5 / 100 ) ^ 2 ) - 1200 ] = $ 123 . difference = $ ( 123 - 120 ) = $ 3 . answer ( d )" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 1200
f = e - 1200
g = 1200 * 10
h = g / 100
i = f - h
|
a ) 45 kmph , b ) 50 kmph , c ) 55 kmph , d ) 61 kmph , e ) 70 kmph | d | subtract(divide(110, multiply(6, const_0_2778)), 5) | a train 110 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 / 18 m / sec . therefore 110 / ( ( x + 5 ) × 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 1980 < = > x = 61 speed of the train is 61 kmph . answer : option d | a = 6 * const_0_2778
b = 110 / a
c = b - 5
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a ) $ 4.50 , b ) $ 5.10 , c ) $ 5.30 , d ) $ 5.50 , e ) $ 5.60 | e | add(3.20, multiply(divide(subtract(5.80, 3.20), add(5, subtract(sqrt(5), 2))), 5)) | the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 5.80 . the price of corn is increasing at a constant rate of 5 x 5 x cents per day while the price of wheat is decreasing at a constant rate of 2 √ ∗ x − x 2 ∗ x − x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ? | "let yy be the # of days when these two bushels will have the same price . first let ' s simplify the formula given for the rate of decrease of the price of wheat : 2 √ ∗ x − x = 1.41 x − x = 0.41 x 2 ∗ x − x = 1.41 x − x = 0.41 x , this means that the price of wheat decreases by 0.41 x 0.41 x cents per day , in yy days it ' ll decrease by 0.41 xy 0.41 xy cents ; as price of corn increases 5 x 5 x cents per day , in yy days it ' ll will increase by 5 xy 5 xy cents ; set the equation : 320 + 5 xy = 580 − 0.41 xy 320 + 5 xy = 580 − 0.41 xy , solve for xyxy - - > xy = 48 xy = 48 ; the cost of a bushel of corn in yy days ( the # of days when these two bushels will have the same price ) will be 320 + 5 xy = 320 + 5 ∗ 48 = 560320 + 5 xy = 320 + 5 ∗ 48 = 560 or $ 5.6 . answer : e ." | a = 5 - 80
b = math.sqrt(5)
c = b - 2
d = 5 + c
e = a / d
f = e * 5
g = 3 + 20
|
a ) 17 , b ) 15 , c ) 12 , d ) 8 , e ) 3 | e | add(const_2, gcd(60, 17)) | find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number | "f the number is x , then , x + 17 = 60 / x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 , - 20 , so x = 3 ( as 3 is positive ) answer : e" | a = math.gcd(60, 17)
b = 2 + a
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | sqrt(divide(588, const_3)) | the length of a rectangular garden is three times its width . if the area of the rectangular garden is 588 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 588 x ^ 2 = 196 x = 14 the answer is c ." | a = 588 / 3
b = math.sqrt(a)
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a ) 40 sec , b ) 29 sec , c ) 25 sec , d ) 27 sec , e ) 34 sec | c | divide(360, multiply(subtract(72, 140), const_0_2778)) | a train 360 m long is running at a speed of 72 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 1 / 20 = 25 sec answer : c" | a = 72 - 140
b = a * const_0_2778
c = 360 / b
|
a ) 63 , b ) 72 , c ) 144 , d ) 216 , e ) 400 | d | multiply(divide(multiply(48, const_3), 20), 30) | if 20 typists can type 48 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 48 letters , so 30 typists can type = 48 * 30 / 20 48 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 48 * 30 * 60 / 20 * 20 = 216 d is the answer" | a = 48 * 3
b = a / 20
c = b * 30
|
a ) 48 , b ) 50 , c ) 52 , d ) 54 , e ) 56 | a | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 15), const_3600)) | a car traveling at a certain constant speed takes 15 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 60 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "60 * t = 1 km = > t = 1 / 60 km / h v * ( t + 15 / 3600 ) = 1 v ( 1 / 60 + 15 / 3600 ) = 1 v ( 75 / 3600 ) = 1 v = 48 km / h the answer is a ." | a = 1 / 60
b = 3600 * a
c = b + 15
d = c / 3600
e = 1 / d
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a ) 17 : 6 , b ) 24 : 7 , c ) 25 : 6 , d ) 21 : 3 , e ) 17 : 4 | b | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 96,000 and rs . 28,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 96000 : 28000 = 24 : 7 answer : b" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 190 , b ) 200 , c ) 180 , d ) 881 , e ) 271 | a | subtract(multiply(18, multiply(56, const_0_2778)), 90) | a train 90 m long running at 56 kmph crosses a platform in 18 sec . what is the length of the platform ? | "d = 56 * 5 / 18 = 25 = 280 – 90 = 190 answer : a" | a = 56 * const_0_2778
b = 18 * a
c = b - 90
|
a ) 1 / 13 , b ) 2 / 13 , c ) 1 / 26 , d ) 1 / 52 , e ) none of these | c | divide(subtract(52, multiply(const_4, const_4)), 52) | a card is drawn from a pack of 52 cards . the probability of ge ƫ ng a queen of club or a king of heart is | "explanation : total number of cases = 52 favourable cases = 2 probability = 2 / 56 = 1 / 26 answer : c" | a = 4 * 4
b = 52 - a
c = b / 52
|
a ) 26 days , b ) 27 days , c ) 23 days , d ) 35 days , e ) 24 days | d | add(multiply(divide(const_1, const_2), 14), multiply(14, const_2)) | this topic is locked . if you want to discuss this question please re - post it in the respective forum . matt and peter can do together a piece of work in 20 days . after they have worked together for 12 days matt stops and peter completes the remaining work in 14 days . in how many days peter complete the work separately . | together they complete the job in 20 days means they complete 12 / 20 of the job after 12 days . peter completes the remaining ( 8 / 20 ) of the job in 14 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 14 < = > x = 14 / ( 8 / 20 ) = 35 thus the answer is d . | a = 1 / 2
b = a * 14
c = 14 * 2
d = b + c
|
a ) 23 , b ) 27 , c ) 31 , d ) 35 , e ) 39 | b | subtract(80, subtract(373, multiply(80, 4))) | we bought a total of 80 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 373 . how many math books did we buy ? | "m + h = 80 h = 80 - m 4 m + 5 h = 373 4 m + 5 * ( 80 - m ) = 373 m = 27 the answer is b ." | a = 80 * 4
b = 373 - a
c = 80 - b
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number s of integers that can be negative ? | "minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning s = 5 out of 6 also possible as you will see 5 or 1 will give you same equation . . ans d" | a = 10 / 2
b = a + 0
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a ) 25 th , b ) 22 nd , c ) 23 rd , d ) 24 th , e ) none of these | a | subtract(multiply(2, 14), 1) | a monkey ascends a greased pole 14 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 12 metres , he ascends in 24 minutes . â ˆ ´ he reaches the top in 25 th minute . answer a" | a = 2 * 14
b = a - 1
|
a ) 7 , b ) 8 , c ) 8.5 , d ) 9 , e ) 9.5 | b | divide(multiply(divide(divide(25, 4), const_100), 120), divide(subtract(const_100, divide(25, 4)), const_100)) | due to a reduction of 25 / 4 % in the price of sugar , a man is able to buy 1 kg more for rs . 120 . find the original rate of sugar . | rate reduced by 6.25 % , so amount rs 120 reduced by = ( 6.25 / 100 ) * 120 = 7.5 so in rs 7.5 extra sugar obtained is 1 kg i . e . new rate is rs . 7.5 per kg thus original rate = [ 7.5 / ( 100 – 6.25 ) ] / 100 = 7.5 / 0.9375 = rs . 8 answer : option b | a = 25 / 4
b = a / 100
c = b * 120
d = 25 / 4
e = 100 - d
f = e / 100
g = c / f
|
a ) 100 , b ) 90 , c ) 80 , d ) 70 , e ) 60 | a | subtract(multiply(82, 10), multiply(9, 80)) | on a test average ( arithmetic mean ) test score for 9 students is 80 . what must be 10 th student ' s score that average score for 10 students to be 82 ? | "( 9 * 80 + x ) / 10 = 82 x = ( 10 * 82 ) - ( 9 * 80 ) x = 820 - 720 total score required 820 - 720 = 100 correct answer is a" | a = 82 * 10
b = 9 * 80
c = a - b
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a ) 7.13 , b ) 6.19 , c ) 4.12 , d ) 3.19 , e ) 4.21 | b | multiply(divide(multiply(add(4, 1.8), subtract(4, 1.8)), add(add(4, 1.8), subtract(4, 1.8))), const_2) | a man can row 4 kmph in still water . when the river is running at 1.8 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 4 s = 1.8 ds = 5.8 us = 2.2 x / 5.8 + x / 2.2 = 1 x = 1.59 d = 1.59 * 2 = 3.19 answer : b" | a = 4 + 1
b = 4 - 1
c = a * b
d = 4 + 1
e = 4 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 25 % , b ) 24 % , c ) 23 % , d ) 22 % , e ) 21 % | a | divide(multiply(26.25, const_100), add(26.25, const_100)) | the annual interest rate earned by an investment increased by 5 percent from last year to this year . if the annual interest rate earned by the investment this year was 26.25 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.05 i ( last year ) = 1.05 i ( last year ) 26.25 = 1.05 x i ( last year ) i ( last year ) = 26.25 / 1.05 = 2625 / 105 = 25 % answer : a" | a = 26 * 25
b = 26 + 25
c = a / b
|
['a ) $ 5,330', 'b ) $ 3,360', 'c ) $ 1,350', 'd ) $ 300', 'e ) $ 150'] | d | multiply(30, divide(multiply(360, 1210), divide(multiply(360, 1210), const_10))) | a certain farmer pays $ 30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 1210 feet ? ( 43,560 square feet = 1 acre ) | basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 300 ( d ) | a = 360 * 1210
b = 360 * 1210
c = b / 10
d = a / c
e = 30 * d
|
a ) 48 , b ) 47 , c ) 46 , d ) 45 , e ) 44 | b | add(divide(subtract(343, 21), 7), const_1) | how many multiples of 7 are there between 21 and 343 , inclusive ? | "7 * 3 = 21 7 * 49 = 343 total multiples = ( 49 - 3 ) + 1 = 47 include 21 and 343 = 47 answer is b" | a = 343 - 21
b = a / 7
c = b + 1
|
a ) 690 , b ) 674 , c ) 672 , d ) 960 , e ) none | c | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1600) | in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1600 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1600 female employees , 60 % of 1600 . = ( 60 * 1600 ) / 100 = 960 . then male employees , = 640 50 % of male are computer literate , = 320 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1600 ) / 100 = 992 computer literate . thus , female computer literate = 992 - 320 = 672 . answer : option c" | a = 62 / 100
b = 60 / 100
c = 1 - b
d = 50 / 100
e = c * d
f = a - e
g = f * 1600
|
a ) 500 , b ) 550 , c ) 600 , d ) 700 , e ) 750 | d | divide(560, subtract(const_1, divide(20, const_100))) | shop offered 20 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | "sp * ( 80 / 100 ) = 560 sp = 70 * 100 = > cp = 700 answer : d" | a = 20 / 100
b = 1 - a
c = 560 / b
|
a ) $ 1200 , b ) $ 1850 , c ) $ 2010 , d ) $ 3200 , e ) $ 2500 | e | divide(3600, power(add(divide(20, const_100), const_1), 2)) | the present worth of $ 3600 due in 2 years at 20 % per annum compound interest is ? | present worth = 3600 / ( 1 + 20 / 100 ) ^ 2 = 3600 * 5 / 6 * 5 / 6 = $ 2500 answer is e | a = 20 / 100
b = a + 1
c = b ** 2
d = 3600 / c
|
a ) 11 , b ) 14 , c ) 18 , d ) 10 , e ) 19 | d | divide(1156, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 19)) | if the wheel is 19 cm then the number of revolutions to cover a distance of 1156 cm is ? | 2 * 22 / 7 * 19 * x = 1156 = > x = 9.7 answer : d | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 19
h = 1156 / g
|
a ) a ) 1055 , b ) b ) 1075 , c ) c ) 1065 , d ) d ) 1070 , e ) e ) 1100 | e | add(multiply(8, 70), multiply(9, 60)) | harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 60 per kg . how much amount did he pay to the shopkeeper ? | cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 60 × 9 = 540 . total cost he has to pay = 560 + 540 = 1100 . e ) | a = 8 * 70
b = 9 * 60
c = a + b
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a ) . 5 hours , b ) 1 hour , c ) 1.5 hours , d ) 2 hours , e ) 2.5 hours | d | divide(200, multiply(divide(200, 5), divide(5, const_2))) | the commuter rail between scottsdale and sherbourne is 200 km of track , and one train makes a round trip in 5 hours . if harsha boards the train at the forest grove station , which is located one fifth of the track ' s length out from scottsdale , how long will it take her to get to sherbourne ? | distance between scottsdale and sherbourne = 200 km time for a one - way trip between scottsdale and sherbourne : 5 / 2 = 2.5 hours distance between forest grove and scottsdale : 200 / 5 = 40 km distance between forest grove and sherbourne : 40 x 4 = 160 km time between scottsdale and forest grove : 2.5 / 5 = . 5 hours time between forest grove and sherborne : . 5 x 4 = 2 hours . | a = 200 / 5
b = 5 / 2
c = a * b
d = 200 / c
|
a ) 73 , b ) 83 , c ) 12 , d ) 13 , e ) 28 | d | subtract(const_60, multiply(const_60, divide(35, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 35 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 10 / 45 * 60 = 13 answer : option d" | a = 35 / 45
b = const_60 * a
c = const_60 - b
|
a ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 150 % | d | add(multiply(25, const_0_33), add(const_100, const_0_33)) | if the price of a certain bond on may 1 st was 2 / 3 the price of the bond on june 1 st and the price of the bond on july 1 st was 25 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1 st ? | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 2 / 3 * 12 = 8 ; the price on july 1 st = 8 * 1.25 = 10 . the average price of the bond on may 1 st and july 1 st = ( 8 + 10 ) / 2 = 9 . the price of the bond on june 1 st ( 12 ) is 4 / 3 times ( 134 % ) the average price of the bond on may 1 st and july 1 st . answer : d ." | a = 25 * const_0_33
b = 100 + const_0_33
c = a + b
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a ) 28.8 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | a | divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 10)) | to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 10 hours a day ? | "man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 10 = 200 therefore number of days = 5760 / 200 = 28.8 days ans = a" | a = 30 * 12
b = a * 8
c = b * 2
d = 20 * 10
e = c / d
|
a ) 12.2 sec , b ) 24.9 sec , c ) 86.4 sec , d ) 60.2 sec , e ) none | c | multiply(multiply(900, inverse(multiply(add(45, 30), const_0_2778))), const_2) | two good train each 900 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 900 + 900 ) m = 1000 m . required time = ( 1800 x 6 / 125 ) sec = 86.4 sec . answer c" | a = 45 + 30
b = a * const_0_2778
c = 1/(b)
d = 900 * c
e = d * 2
|
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number e of tickets sold by the male members of the association | given that , total average e sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c . | a = 1 + 2
b = 66 * a
c = 70 * 2
d = b - c
|
a ) 4 , b ) 6 , c ) 8 , d ) 2 , e ) 10 | d | multiply(divide(40, const_100), 40) | uncle bruce is baking chocolate chip cookies . he has 72 ounces of dough ( with no chocolate ) and 50 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 40 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 60 % * total = 72 = > ( 6 / 10 ) total = 72 = > total = 720 / 6 = > total = 120 oz , from there , you must find 10 % of the total 40 oz of the mixture . 40 % * total = > ( 4 / 10 ) ( 120 ) = 48 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 50 - 48 oz = 2 oz chocolate left over . answer : d" | a = 40 / 100
b = a * 40
|
a ) 20198 , b ) 20098 , c ) 20000 , d ) 20088 , e ) 20085 | b | add(subtract(power(107, const_2), 107), subtract(power(93, const_2), 93)) | 107 ã — 107 + 93 ã — 93 = ? | "explanation : ( a + b ) 2 + ( a â ˆ ’ b ) 2 = 2 ( a 2 + b 2 ) ( reference : basic algebraic formulas ) 1072 + 932 = ( 100 + 7 ) 2 + ( 100 â ˆ ’ 7 ) 2 = 2 ( 1002 + 72 ) = 2 ( 10000 + 49 ) = 20098 . answer : option b" | a = 107 ** 2
b = a - 107
c = 93 ** 2
d = c - 93
e = b + d
|
a ) 50 , b ) 60 , c ) 80 , d ) 70 , e ) 90 | d | divide(800, const_10) | how many integers from 100 to 800 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 7 = 7 8 is not possible here because it would result in a number greater than 8 ( i . e 808 , 818 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 7 * 10 = 70 answer is d ." | a = 800 / 10
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a ) 9 , b ) 13 , c ) 17 , d ) 20 , e ) 33.33 | e | multiply(divide(const_1, multiply(divide(add(50, const_100), const_100), 50)), const_100) | a part - time employee whose hourly wage was increased by 50 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | "let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.5 x now we need to find number of hours worked so that x * y = 1.5 x * z i . e z = 1 / 1.5 y = 2 / 3 y % decrease = ( y - 2 / 3 y ) / y * 100 = 100 / 3 = 33.33 % . thus my answer is e ." | a = 50 + 100
b = a / 100
c = b * 50
d = 1 / c
e = d * 100
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed w ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this w we get x = 8 . b ." | a = 96 / 16
b = 96 / 16
c = 50 / 100
d = 1 + c
e = b / d
f = e * 4
g = f * 4
h = 96 / 16
i = h ** 2
j = g + i
k = math.sqrt(j)
l = a + k
m = l / 2
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a ) 649011 , b ) 540041 , c ) 654321 , d ) 653001 , e ) 643001 | d | multiply(divide(653, 1001), const_100) | 653 ã — 1001 = ? | "653 ã — ( 1000 + 1 ) 653000 + 1 653001 answer d" | a = 653 / 1001
b = a * 100
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed x ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this x we get x = 8 . b ." | a = 96 / 16
b = 96 / 16
c = 50 / 100
d = 1 + c
e = b / d
f = e * 4
g = f * 4
h = 96 / 16
i = h ** 2
j = g + i
k = math.sqrt(j)
l = a + k
m = l / 2
|
a ) 14 % , b ) 4 % , c ) 15 % , d ) 5 % , e ) 25 % | b | divide(multiply(divide(multiply(50, 40), const_100), 20), const_100) | at a certain college , 50 percent of the total number of students are freshmen . if 40 percent of the fresh - men are enrolled in the school of liberal arts and , of these , 20 percent are psychology majors , what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? | "let ' s say there is a total of 100 students at this college . 50 percent of the total number of students are freshmen . # of freshmen = 50 % of 100 = 50 40 percent of the fresh - men are enrolled in the school of liberal arts . . . number of liberal arts freshmen = 40 % of 50 = 20 . . . and , of these , 20 percent are psychology majors . . . number of liberal arts freshmen who are psychology majors = 20 % of 20 = 4 what percent of the students at the college are freshmen psychology majors enrolled in the school of liberal arts ? 4 / 100 = 4 % answer : b" | a = 50 * 40
b = a / 100
c = b * 20
d = c / 100
|
a ) rs . 31 , b ) rs . 32.10 , c ) rs . 40.40 , d ) rs . 64.10 , e ) rs . 74.10 | d | subtract(subtract(multiply(1000, power(add(divide(10, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(10, const_100)), 4)) | what will be the difference between simple and compound interest at 10 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 10 * 4 ) / 100 = rs . 400 c . i . = [ 1000 * ( 1 + 10 / 100 ) 4 - 1000 ] = rs . 464.10 difference = ( 464.10 - 400 ) = rs . 64.10 answer : d" | a = 10 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 10 / 100
g = 1000 * f
h = g * 4
i = e - h
|
a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | a | multiply(divide(20, const_100), multiply(divide(20, const_100), 300)) | 60 percent of movie theatres in town x have 3 screens or less . 20 % of those theatres sell an average of more than $ 300 worth of popcorn per showing . 50 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell an average of more than $ 100 worth of popcorn per day ? | lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 300 = 20 % of 60 = 12 number of movie theaters selling popcorn at $ 300 or less = 56 = > number of movie theaters selling popcorn at more than $ 300 = 100 - 56 = 44 of these 44 theaters , 12 are those with 3 screens or less therefore 12 ( 44 - 12 ) must be those with four screens or more a is the answer | a = 20 / 100
b = 20 / 100
c = b * 300
d = a * c
|
a ) 28 , b ) 50 , c ) 77 , d ) 22 , e ) 12 | c | divide(divide(subtract(200, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 200 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 200 / 10 ) m / sec = ( 20 ) m / sec . [ ( 20 ) * ( 18 / 5 ) ] km / hr = 72 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 72 = = > x = 77 km / hr . answer : c" | a = 5 * const_0_2778
b = a * 5
c = 200 - b
d = c / 5
e = d / const_0_2778
|
a ) 45.9 % , b ) 43.9 % , c ) 42.9 % , d ) 41.9 % , e ) 44.9 % | e | divide(multiply(70, const_100), 156) | if there are 156 laborers in a crew , and on a certain day , 70 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | "70 / 156 * 100 = 44.87 44.9 % correct answer e" | a = 70 * 100
b = a / 156
|
a ) 3 : 1 , b ) 4 : 3 , c ) 2 : 3 , d ) 2 : 1 , e ) 2 : 4 | a | divide(divide(const_1, multiply(add(16, const_2), const_10)), divide(const_1, multiply(16, const_10))) | a work can be finished in 16 days by 30 women . the same work can be finished in 10 days by sixteen men . the ratio between the capacity of a man and a woman is | "work done by 30 women in 1 day = 1 / 16 work done by 1 woman in 1 day = 1 / ( 16 × 30 ) work done by 16 men in 1 day = 1 / 10 work done by 1 man in 1 day = 1 / ( 10 × 16 ) ratio of the capacity of a man and woman = 1 / ( 10 × 16 ) : 1 / ( 16 × 30 ) = 1 / 10 : 1 / 30 = 1 / 1 : 1 / 3 = 3 : 1 option a" | a = 16 + 2
b = a * 10
c = 1 / b
d = 16 * 10
e = 1 / d
f = c / e
|
a ) 80 , b ) 16 , c ) 24 , d ) 48 , e ) 98 | a | lcm(multiply(4, 4), multiply(5, 4)) | the ratio of numbers is 4 : 5 and their h . c . f is 4 . their l . c . m is : | "let the numbers be 4 x and 5 x . then their h . c . f = x . so , x = 4 . so , the numbers are 16 and 20 . l . c . m of 16 and 20 = 80 . answer : a" | a = 4 * 4
b = 5 * 4
c = math.lcm(a, b)
|
['a ) 56 m', 'b ) 60 m', 'c ) 80 m', 'd ) 82 m', 'e ) 84 m'] | a | divide(add(divide(5300, 26.5), multiply(const_2, 12)), const_4) | length of a rectangular plot is 12 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | let breadth = x metres . then , length = ( x + 12 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 12 ) + x ] = 200 2 x + 12 = 100 2 x = 88 x = 44 hence , length = x + 12 = 56 m a | a = 5300 / 26
b = 2 * 12
c = a + b
d = c / 4
|
a ) 585 km , b ) 767 km , c ) 276 km , d ) 178 km , e ) 176 km | a | multiply(add(20, 25), divide(65, subtract(25, 20))) | two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 65 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 65 km more than the other , 25 x â € “ 20 x = 65 5 x = 65 x = 13 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45 km / hr therefore , total distance = 45 * 13 = 585 km . answer : a" | a = 20 + 25
b = 25 - 20
c = 65 / b
d = a * c
|
a ) 205 , b ) 220 , c ) 235 , d ) 250 , e ) 265 | c | subtract(multiply(multiply(45, const_0_2778), 30), 140) | what is the length of a bridge ( in meters ) , which a train 140 meters long and travelling at 45 km / h can cross in 30 seconds ? | "speed = 45 km / h = 45000 m / 3600 s = 25 / 2 m / s in 30 seconds , the train can travel 25 / 2 * 30 = 375 meters 375 = length of train + length of bridge length of bridge = 375 - 140 = 235 meters the answer is c ." | a = 45 * const_0_2778
b = a * 30
c = b - 140
|
a ) 190 , b ) 88 , c ) 77 , d ) 62 , e ) 52 | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(140, divide(20, divide(15, const_3)))) | a train crosses a platform of 140 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | length of the train be â € ˜ x â € ™ x + 140 / 15 = x + 250 / 20 4 x + 560 = 3 x + 750 x = 190 m answer : a | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 140 * e
g = c - f
|
a ) 181 , b ) 171 , c ) 191 , d ) 201 , e ) 202 | b | divide(multiply(170, 171), const_4) | what is the sum of the integers from - 170 to 171 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 171 , a = - 170 , d = 1 hence , 171 = - 170 + ( n - 1 ) or n = 342 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 342 * ( - 170 + 171 ) / 2 sn = 342 * 1 / 2 sn = 171 answer : b" | a = 170 * 171
b = a / 4
|
a ) 21 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | a | divide(factorial(7), multiply(factorial(subtract(7, const_2)), factorial(const_2))) | if 7 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 7 * 6 / 2 = 21 answer : a" | a = math.factorial(7)
b = 7 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) s . 750 , b ) s . 900 , c ) s . 1000 , d ) s . 1200 , e ) s . 1500 | c | multiply(divide(divide(90, 3), 3), const_100) | a sum was put a simple interest at a certain rate for 3 years . had it been put at 3 % higher rate , it would have fetched rs . 90 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x ã — ( r + 3 ) ã — 3 / 100 â ˆ ’ x ã — r ã — 3 / 100 = 90 â ‡ ’ x ã — 3 ã — 3 / 100 = 90 â ‡ ’ x ã — 3 / 100 = 30 â ‡ ’ 3 x = 3000 â ‡ ’ x = 1000 answer : option c" | a = 90 / 3
b = a / 3
c = b * 100
|
a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | a | add(multiply(0.25, subtract(5, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 8.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ? | oa is definitely wrong . the answer should be a . | a = 5 - 1
b = 0 * 25
c = b + 0
|
a ) 6 : 7 , b ) 2 : 3 , c ) 2 : 1 , d ) 2 : 2 , e ) 2 : 8 | a | divide(subtract(6.30, 5.70), subtract(7.00, 6.30)) | find the ratio in which rice at rs . 7.00 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg | "by the rule of alligation : cost of 1 kg rice of 1 st kind cost of 1 kg rice of 2 nd kind required ratio = 60 : 70 = 6 : 7 answer : a" | a = 6 - 30
b = 7 - 0
c = a / b
|
a ) 18 , b ) 16 , c ) 14 , d ) 13 , e ) 15 | a | divide(multiply(36, 2), const_4) | if the lcm and hcf of 4 and another number is 36 and 2 respectively . find the other number ? | hcf x lcm = product of numbers 2 x 36 = 4 x the other number other number = ( 2 x 36 ) / 4 other number = 18 answer : a | a = 36 * 2
b = a / 4
|
a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 25 % , e ) 50 % | e | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 40))), multiply(8, divide(const_1, 40))), multiply(8, divide(const_1, 40))), const_100) | a certain car uses one gallon of gasoline every 40 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | "4 miles on the highway = 4 / 40 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 40 + 4 / 20 = 3 / 10 gallons . 8 miles on the highway = 8 / 40 gallons . the % change = ( 3 / 10 - 8 / 40 ) / ( 8 / 40 ) = 1 / 2 = 0.50 . answer : e ." | a = 1 / 20
b = a * 4
c = 1 / 40
d = 4 * c
e = b + d
f = 1 / 40
g = 8 * f
h = e - g
i = 1 / 40
j = 8 * i
k = h / j
l = k * 100
|
a ) s . 650 , b ) s . 690 , c ) s . 668 , d ) s . 700 , e ) s . 760 | c | subtract(815, divide(multiply(subtract(864, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 864 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 864 - 815 ) = rs . 49 . s . i . for 3 years = rs . ( 49 x 3 ) = rs . 147 . principal = rs . ( 815 - 147 ) = rs . 668 . answer : option c" | a = 864 - 815
b = a * 3
c = b / 4
d = 815 - c
|
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