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 ) 12 liters , b ) 10 liters , c ) 4 liters , d ) 2 liters , e ) 3 liters | d | divide(subtract(multiply(divide(12, const_100), 70), multiply(divide(10, const_100), 70)), subtract(const_1, divide(12, const_100))) | a mixture of 70 liters of wine and water contains 10 % water . how much water must be added to make water 12 Β½ % of the total mixture ? | "70 * ( 10 / 100 ) = 7 wine water 87 1 / 2 % 12 1 / 2 % 87 1 / 2 % - - - - - - - 63 12 1 / 2 % - - - - - - - ? = > 9 - 7 = 2 answer : d" | a = 12 / 100
b = a * 70
c = 10 / 100
d = c * 70
e = b - d
f = 12 / 100
g = 1 - f
h = e / g
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a ) 17.5 minutes . , b ) 35 minutes . , c ) 52.5 minutes . , d ) 70 minutes . , e ) 105 minutes . | a | divide(35, const_2) | danny and steve are running towards each other , each one from his own house . danny can reach steve ' s house in 35 minutes of running , which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time , how much time longer will it take steve to reach the halfway point between their houses than danny to reach the halfway point between their houses ? | "danny ' s time is 35 minutes . the time to reach halfway is 17.5 minutes . steve ' s time is 70 minutes . the time to reach halfway is 35 minutes . the time difference is 35 - 17.5 = 17.5 minutes the answer is a ." | a = 35 / 2
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a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 28 % . | e | multiply(divide(add(multiply(divide(20, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 20 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "20 % of 2 litres = 0.4 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.8 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 28 % answer : e" | a = 20 / 100
b = a * 2
c = 40 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
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a ) 199 sec , b ) 156.7 sec , c ) 726 sec , d ) 127 sec , e ) 128 sec | b | subtract(divide(multiply(const_1, const_1000), divide(60, 10)), 10) | in a kilometer race , a beats b by 60 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 60 = 166.7 sec . time taken by a = 166.7 - 10 = 156.7 sec . answer : b" | a = 1 * 1000
b = 60 / 10
c = a / b
d = c - 10
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a ) 18 , b ) 24 , c ) 17 , d ) 12 , e ) 13 | b | divide(120, multiply(add(15, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 120 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 120 m = 120 / 5 = 24 seconds . answer : b" | a = 15 + 3
b = a * const_0_2778
c = 120 / b
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a ) 99.33 km , b ) 133.33 km , c ) 125.33 km , d ) 156.66 km , e ) none | b | divide(divide(20, const_60), divide(subtract(100, 80), multiply(100, 80))) | when a train moves at an average speed of 100 kmph it reaches its destination on time . when its average speed becomes 80 kmph , then it reaches its destination 20 minutes late . find the length of journey rounded to two decimal places . | sol . difference between timings = 20 min = 1 / 3 hr . let the length of journey be x km . then , x / 80 - x / 100 = 1 / 3 Γ’ β‘ β 5 x - 4 x = 133.33 Γ’ β‘ β x = 133.33 km . answer b | a = 20 / const_60
b = 100 - 80
c = 100 * 80
d = b / c
e = a / d
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a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19 | e | subtract(20, divide(20, 20)) | a certain bacteria colony doubles in size every day for 20 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat β s limit ? | "simultaneous grow = same amount of time in 20 days we should expect to have p ( 2 ) ^ 20 so if we have two things doing the job of getting us there then ; 2 ^ x + 2 ^ x = 20 2 ( 2 ) ^ x = 2 2 ^ ( x + 1 ) = 2 ^ 20 x + 1 = 20 x = 19 ; answer : e" | a = 20 / 20
b = 20 - a
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a ) 600 , b ) 720 , c ) 1080 , d ) 1200 , e ) 1440 | d | divide(120, divide(multiply(const_2, subtract(126, 120)), 120)) | julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 120 and $ 126 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same , what was the amount of julie ' s initial savings ? | "$ 120 for 2 years = $ 60 per year . extra $ 6 yearned with the compound interest is the percent yearned on percent . so , $ 6 is yearned on $ 60 , which means that the interest = 10 % . this on the other hand means that half of the savings = 60 * 10 = $ 600 . twice of that = $ 1,200 . answer : d ." | a = 126 - 120
b = 2 * a
c = b / 120
d = 120 / c
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a ) 1234 , b ) 1540 , c ) 1650 , d ) 1632 , e ) 1635 | e | multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder ? | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e" | a = 1365 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) $ 190 , b ) $ 210 , c ) $ 230 , d ) $ 250 , e ) $ 270 | c | divide(add(600, 90), const_3) | joan and karl each bought a sofa and the sum of their purchases was $ 600 . if twice of what joan paid was $ 90 more than what karl paid , what did joan pay for her sofa ? | j + k = 600 so k = 600 - j 2 j = k + 90 2 j = ( 600 - j ) + 90 3 j = 690 j = 230 the answer is c . | a = 600 + 90
b = a / 3
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a ) 36 , b ) 40 , c ) 45 , d ) 51 , e ) 56 | a | subtract(choose(8, 5), choose(subtract(8, 2), 2)) | a meeting has to be conducted with 5 managers . find the number of ways in which the managers may be selected from among 8 managers , if there are 2 managers who refuse to attend the meeting together . | "the total number of ways to choose 5 managers is 8 c 5 = 56 we need to subtract the number of groups which include the two managers , which is 6 c 3 = 20 . 56 - 20 = 36 the answer is a ." | a = math.comb(8, 5)
b = 8 - 2
c = math.comb(b, 2)
d = a - c
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a ) 72.15 , b ) 27.24 , c ) 24.25 , d ) 28.46 , e ) 20.9 | c | divide(divide(multiply(1080, 18), const_100), 8) | a reduction of 18 % in the price of oil enables a house wife to obtain 8 kgs more for rs . 1080 , what is the reduced price for kg ? | "1080 * ( 18 / 100 ) = 194 - - - - 8 ? - - - - 1 = > rs . 24.25 answer : c" | a = 1080 * 18
b = a / 100
c = b / 8
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a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 152 minw , e ) none of these | d | multiply(add(const_1, const_4), 38) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 38 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 38 = > 4 / x = 1 / 38 x = 38 β 4 = 152 mins option d" | a = 1 + 4
b = a * 38
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a ) 5 / 6 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) 1 / 24 | b | divide(24, add(24, multiply(24, multiply(divide(1, 6), 24)))) | a satellite is composed of 24 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 6 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ? | "let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 6 . the number of non - upgraded sensors on the whole satellite is 24 ( x / 6 ) = 4 x . the fraction of sensors which have been upgraded is x / ( x + 4 x ) = x / 5 x = 1 / 5 the answer is b ." | a = 1 / 6
b = a * 24
c = 24 * b
d = 24 + c
e = 24 / d
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a ) 90 , b ) 2,700 , c ) 4,500 , d ) 2,550 , e ) 324,000 | d | divide(multiply(add(30, 55), const_60), const_2) | a copy machine , working at a constant rate , makes 30 copies per minute . a second copy machine , working at a constant rate , makes 55 copies per minute . working together at their respective rates , how many copies do the two machines make in half an hour ? | "together the two machines make 30 + 55 = 85 copies per minute . so , in half an hour they will make 85 * 30 = 2,550 copies . answer : d ." | a = 30 + 55
b = a * const_60
c = b / 2
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a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 320 | b | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 5 , and the digit 2 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 5 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 2 = 160 numbers with two 2 - s , 2 x 2 x 1 * 5 * 1 * 2 = 10 . thus there are 160 - 10 = 150 such numbers . answer : b ." | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 * 4
e = c * d
f = 4 + 1
g = 4 + 1
h = f * g
i = h * 4
j = e - i
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a ) 400 , b ) 420 , c ) 430 , d ) 495 , e ) 510 | d | divide(subtract(multiply(15, 450), multiply(15, 120)), subtract(120, 110)) | the average salary of the employees in a office is rs . 120 / month . the avg salary of officers is rs . 450 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office . | let no . of non - officers be x 15 * 450 + x * 110 = ( x + 15 ) 120 x = 495 d | a = 15 * 450
b = 15 * 120
c = a - b
d = 120 - 110
e = c / d
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a ) 10 , b ) 30 , c ) 105 , d ) 60 , e ) 90 | c | divide(multiply(15, subtract(15, const_1)), const_2) | there are 15 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | 15 players are there . two players play one game with one another . so 15 c 2 = 15 x 14 / 2 = 105 so option c is correct | a = 15 - 1
b = 15 * a
c = b / 2
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a ) 90 , b ) 130 , c ) 135 , d ) 320 , e ) 450 | c | add(90, multiply(divide(subtract(855, 90), add(90, 80)), subtract(90, 80))) | two heavily loaded sixteen - wheeler transport trucks are 855 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 = 855 - 90 = 765 k ratio of speeds 9 : 8 - > 765 / 17 = 45 truck a = 90 + 45 * 9 = 495 truck b = 45 * 8 = 360 delta = 135 km answer c" | a = 855 - 90
b = 90 + 80
c = a / b
d = 90 - 80
e = c * d
f = 90 + e
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a ) 273232 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273207 | e | multiply(divide(5357, 51), const_100) | 5357 x 51 = ? | "5357 x 51 = 5357 x ( 50 + 1 ) = 5357 x 50 + 5357 x 1 = 267850 + 5357 = 273207 e )" | a = 5357 / 51
b = a * 100
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a ) s 200 , b ) s 750 , c ) s 300 , d ) s 450 , e ) s 550 | b | divide(300, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4))) | a person spends 1 / 3 rd of the money with him on clothes , 1 / 5 th of the remaining on food and 1 / 4 th of the remaining on travel . now , he is left with rs 300 . how much did he have with him in the beginning ? | "suppose the amount in the beginning was rs β x β money spent on clothes = rs 1 x / 3 balance = rs 2 x / 3 money spent on food = 1 / 5 of 2 x / 3 = rs 2 x / 15 balance = 2 x / 3 - 2 x / 15 = rs 8 x / 15 money spent on travel = 1 / 4 of 8 x / 15 = rs 2 x / 15 = 8 x / 15 - 2 x / 15 = 6 x / 15 = rs 2 x / 5 therefore 2 x / 5 = 300 = 750 answer : b" | a = 1 / 3
b = 1 - a
c = 1 / 3
d = 1 - c
e = d / 5
f = b - e
g = 1 / 3
h = 1 - g
i = 1 / 3
j = 1 - i
k = j / 5
l = h - k
m = l / 4
n = f - m
o = 300 / n
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | d | subtract(add(16, 6), const_1) | a student is ranked 16 th from right and 6 th from left . how many students are there in totality ? | "from right 16 , from left 6 total = 16 + 6 - 1 = 21 answer : d" | a = 16 + 6
b = a - 1
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a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum w of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : w = 2550 + ( 100 * 50 ) = 7550 . b" | a = 200 + 102
b = a / 2
c = 200 - 102
d = c / 2
e = d + 1
f = b * e
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a ) 143 , b ) 150 , c ) 123.75 , d ) 140.55 , e ) none | c | multiply(multiply(125, add(const_1, divide(10, const_100))), divide(subtract(const_100, 10), const_100)) | if a number x is 10 % less than another number y and y is 10 % more than 125 , then x is equal to | solution y = 125 + 10 % of 125 = 125 + 12.50 = 137.50 . x = 137.50 - 10 % of 137.50 = 137.50 - 13.75 = 123.75 . answer c | a = 10 / 100
b = 1 + a
c = 125 * b
d = 100 - 10
e = d / 100
f = c * e
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a ) 983.578 , b ) 1043.65 , c ) 1033.65 , d ) 656.112 , e ) 456.512 | b | multiply(575.104, power(15.98, 9.001)) | 575.104 x 15.98 Γ£ Β· 9.001 + 21.25 = ? | "explanation : ? = 575.104 x 15.98 Γ£ Β· 9.001 + 21.25 Γ’ β° Λ ( 575.104 x 16 / 9 ) + 21.25 Γ’ β° Λ 1022.40 + 21.25 Γ’ β° Λ 1043.65 answer : option b" | a = 15 ** 98
b = 575 * 104
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a ) 35 kmh , b ) 45 kmh , c ) 60 kmh , d ) 75 kmh , e ) 90 kmh | c | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(90, 30)), const_4), power(30, const_2))), 30), const_2) | if a car had traveled 30 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 90 km , at what speed did it travel ? | "time = distance / speed difference in time = 1 / 2 hrs 90 / x - 90 / ( x + 30 ) = 1 / 2 substitute the value of x from the options . - - > x = 60 - - > 90 / 60 - 90 / 90 = 3 / 2 - 1 = 1 / 2 answer : c" | a = 90 * 30
b = 2 * a
c = b * 4
d = 30 ** 2
e = c + d
f = math.sqrt(e)
g = f - 30
h = g / 2
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['a ) 12', 'b ) 45', 'c ) 36', 'd ) 48', 'e ) 60'] | b | add(const_4, rectangle_area(const_10, const_4)) | the length and width of a rectangle are integer values . what is the area of the smallest such rectangle that can be inscribed in a circle whose radius is also an integer | if we consider all pythagorean triplets with integral values , we have the following results : - ( 3 , 4,5 ) , ( 6 , 8,10 ) . . . . . . . . the diameter of the circle is 5 in the 1 st case and 10 in the 2 nd . but the question says that radius of the circle is also an integer hence the 1 st triplet will be ruled out . we are left with the 2 nd triplet where radius of the circle = 5 , hence the sides of the rectangle are 6 and 8 making the area = 45 . b | a = 4 + rectangle_area
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a ) 30 % gain , b ) 30 % loss , c ) 50 % gain , d ) 50 % loss , e ) 60 % loss | c | subtract(8, 12) | if the selling price of 8 articles is same as the cost price of 12 articles . find the gain or loss percentage ? | "explanation : let the c . p of each article be re 1 . then , s . p of 8 articles = c . p of 12 articles = rs . 12 / - now , c . p of 8 articles = rs . 8 / - , s . p of 8 articles = rs 12 / - gain = rs ( 12 - 10 ) = rs 4 / - . gain % = ( 4 / 8 * 100 ) % = 50 % gain answer : option c" | a = 8 - 12
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a ) 62 % , b ) 52 % , c ) 57 % , d ) 56 % , e ) 58 % | e | multiply(add(multiply(divide(70, const_100), 7), multiply(divide(30, const_100), 3)), const_10) | janet invited her boy and girl friends to her party at the ratio of 3 : 7 . if 30 % of the boys and 70 % of the girls will be coming for the party from texas , how many janet invited friends will be coming from texas . | let total no of janet invited friends be 100 , boys 30 and girls 70 , so total number of janet ' s invited guests coming from texas for both boys and girls respectively 30 * 30 / 100 = 9 and 70 * 70 / 100 = 49 so total 58 of janet ' s invited guest out of the 100 invited guests will be coming from texas . . so required % is 58 % answer : e | a = 70 / 100
b = a * 7
c = 30 / 100
d = c * 3
e = b + d
f = e * 10
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a ) 6500 , b ) 3500 , c ) 2500 , d ) 4500 , e ) 2000 | d | multiply(multiply(power(divide(multiply(multiply(2, 4), 500), 500), const_0_33), 500), power(divide(multiply(multiply(2, 4), 500), 500), const_0_33)) | david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 500 and then he left the lab . he returned in time to take a reading at 4 p . m . , by which point the population had grown to 13,500 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ? | "let the rate be x , then population of the bacteria after each hour can be given as 500,500 x , 500 ( x ^ 2 ) , 500 ( x ^ 3 ) now population at 4 pm = 13,500 thus we have 500 ( x ^ 3 ) = 13,500 = 27 thus x = 3 therefore population at 3 pm = 500 ( 9 ) = 4,500 answer : d" | a = 2 * 4
b = a * 500
c = b / 500
d = c ** const_0_33
e = d * 500
f = 2 * 4
g = f * 500
h = g / 500
i = h ** const_0_33
j = e * i
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a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 4 | e | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 3), divide(const_1, 4)))) | a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , while a and c together can do it in 3 hours . how long will b alone take to do it ? | "a ' s 1 hour ' s work = 1 / 4 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 3 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 4 + 1 / 3 ) = 7 / 12 . b ' s 1 hour ' s work = ( 7 / 12 - 1 / 3 ) = 1 / 4 . therefore a alone will take 4 hours to do the work . e" | a = 1 / 3
b = 1 / 3
c = 1 / 4
d = b - c
e = a - d
f = 1 / e
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a ) 5 , b ) 8 , c ) 26 , d ) 7 , e ) 12 | c | subtract(negate(17), multiply(subtract(5, 10), divide(subtract(5, 10), subtract(2, 5)))) | 2 , 5 , 10 , 17 , . . | "explanation : numbers are ( 1 * 1 ) + 1 = 2 ( 2 * 2 ) + 1 = 5 ( 3 * 3 ) + 1 = 10 ( 4 * 4 ) + 1 = 17 ( 5 * 5 ) + 1 = 26 answer : c" | a = negate - (
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a ) 200 , b ) 600 , c ) 800 , d ) 1600 , e ) 50 | d | multiply(power(const_2, 4), const_100) | cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost is what percent of the original cost ? | "original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600" | a = 2 ** 4
b = a * 100
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a ) a 17 , b ) b ) 28 , c ) c ) 35 , d ) d ) 45 , e ) e ) 80 | d | add(add(35, 5), 5) | 3 people each took 5 tests . if the ranges of their scores in the 5 practice tests were 17 , 28 and 35 , 45 what is the minimum possible range in scores of the 3 test - takers ? | it is finding the minimum range between all their scores , if all test taker scores are between 0 and maximum range we will have : a - 0 and 17 b - 0 and 28 c - 0 and 35 d - 0 and 45 therefore the minimum range is 45 , it cant be any lower however you play with the numbers . d | a = 35 + 5
b = a + 5
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a ) $ 2000 , b ) $ 5000 , c ) $ 8000 , d ) $ 1500 , e ) $ 1000 | e | divide(multiply(divide(multiply(add(1000, 1000), const_2.0), const_4.0), 3), 3) | a sum of money is distributed among a , b , c , d in the proportion of 5 : 3 : 2 : 3 . if a gets $ 1000 more than b , what is the c ' s share ? | "let the shares of a , b , c , d are 5 x , 3 x , 2 x , 3 x 5 x - 3 x = 1000 x = 500 c ' s share = 2 x = $ 1000 answer is e" | a = 1000 + 1000
b = a * 2
c = b / 4
d = c * 3
e = d / 3
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a ) 22 , b ) 50 , c ) 99 , d ) 35 , e ) 12 | d | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 15 seconds . the speed of the train is : | speed of the train relative to man = ( 125 / 15 ) m / sec = ( 25 / 3 ) m / sec . [ ( 25 / 3 ) * ( 18 / 5 ) ] km / hr = 30 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 30 = = > x = 35 km / hr . answer : d | a = 5 * const_0_2778
b = a * 5
c = 125 - b
d = c / 5
e = d / const_0_2778
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a ) 18 , b ) 16 , c ) 19 , d ) 15 , e ) 34 | b | divide(32, subtract(7.5, 5.5)) | two boys starts from the same place walking at the rate of 5.5 kmph and 7.5 kmph respectively in the same direction . what time will they take to be 32 km apart ? | "explanation : relative speed = 7.5 - 5.5 = 2 kmph ( because they walk in the same direction ) distance = 32 km time = distance / speed = 32 / 2 = 16 hr answer : b" | a = 7 - 5
b = 32 / a
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a ) 5 / 6 , b ) 1 / 4 , c ) 2 / 3 , d ) 3 / 4 , e ) 11 / 12 | a | subtract(const_1, multiply(subtract(const_1, divide(const_2, const_3.0)), subtract(const_1, divide(1, 2)))) | two thirds of the roads from a to b are at least 5 miles long , and 1 / 2 of the roads from b to c are at least 5 miles long . if you randomly pick a road from a to b and then randomly pick a road from b to c , what is the probability that at least one of the roads you pick is at least 5 miles long ? | "so please : find the probability of the event thatnoneof the roads you pick will be at least 5 miles long and subtract from 1 to get the probability thatat least oneof the roads you pick will be at least 5 miles long : p = 1 - 1 / 3 * 1 / 2 = 5 / 6 . answer : a ." | a = 2 / 3
b = 1 - a
c = 1 / 2
d = 1 - c
e = b * d
f = 1 - e
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a ) 152.01 , b ) 85 , c ) 76 , d ) 168.2 , e ) 127.5 | e | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | the difference between a number and its two - fifth is 510 . what is 15 % of that number ? | "let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 15 % of 850 = 127.5 . answer : e" | a = 4 + 1
b = 510 * a
c = 1 + 2
d = b / c
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a ) 20 , b ) 21 , c ) 42 , d ) 23 , e ) 24 | c | add(7, divide(multiply(7, subtract(18000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 18000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 18000 x 7 ) + 6000 ( x - 7 ) x = 42 answer c" | a = 18000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
a ) 23 , b ) 87 , c ) 84 , d ) 26 , e ) 19 | c | divide(427, add(divide(7, 7), add(divide(7, 3), divide(7, 4)))) | a sum of rs . 427 is to be divided among a , b and c in such a way that 3 times a β s share , 4 times b β s share and 7 times c β s share are all equal . the share of c is ? | given total sum = rs . 427 and given that 3 times a β s share , 4 times b β s share and 7 times c β s share are all equal . = > 3 a = 4 b = 7 c but given = > a + b + c = 427 now express a and b in terms of c i . e = > ( 7 c / 3 ) + ( 7 c / 4 ) + c = 427 = > c = 84 answer : c | a = 7 / 7
b = 7 / 3
c = 7 / 4
d = b + c
e = a + d
f = 427 / e
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a ) β 5 , b ) 1 , c ) 13 , d ) 17 , e ) 551 | d | add(add(4, 11), 2) | if x and t are integers and 2 x β t = 11 , then 4 x + t can not be | 2 x - t = 11 . . . . t = 2 x - 11 4 x + t = 4 x + 2 x - 11 = 6 x - 11 6 x - 11 = - 5 . . . x = 1 6 x - 11 = 1 . . . x = 2 6 x - 11 = 13 . . . x = 4 6 x - 11 = 17 . . x is not integer 6 x - 11 = 551 . . x is not integer i think the choice e is 55 not 551 . otherwise both de can not be solution = d | a = 4 + 11
b = a + 2
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a ) 10 min , b ) 12 min , c ) 18 min , d ) 15 min , e ) 20 min | d | multiply(const_60, divide(subtract(40, 30), 40)) | excluding stoppages , the speed of a bus is 40 kmph and including stoppages , it is 30 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 10 km less . time taken to cover 10 km = ( 10 / 40 x 60 ) min = 15 min answer : d" | a = 40 - 30
b = a / 40
c = const_60 * b
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a ) 15 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 72 % | a | multiply(divide(54, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 54 Β° of the circle , what percentage of megatek employees are in manufacturing ? | answer : a 54 Β° divided by 360 Β° equals 0.15 , therefore the sector is equal to 15 % of the total | a = 3600 / 10
b = 54 / a
c = b * 100
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a ) 30 , b ) 35 , c ) 40 , d ) 25 , e ) 60 | a | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | nicky and cristina are running a 100 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . a" | a = 3 * 12
b = a / 3
c = 3 * 12
d = 5 - 3
e = c / d
f = b + e
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a ) 5 , b ) 6 , c ) 3 , d ) 8 , e ) 9 | c | divide(divide(20, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 20 - meter platform and caught it after it had traveled 55 meters . how many times did the ball bounce ? | "going down = 20 m going up = 10 - - > total = 30 going down = 10 - - > total = 40 going up = 5 - - > total = 45 going down = 5 - - > total = 50 going up = 2.5 - - > total = 52.5 going down = 2.5 - - > total = 55 ( caught ) no of bouncing = 3 answer : c" | a = 20 / 2
b = a / 2
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a ) 58210607666 , b ) 58310607896 , c ) 58295817036 , d ) 58591650836 , e ) 58310692816 | c | multiply(subtract(99999, const_4), 582964) | find the value of 582964 x 99999 = m ? | "582964 x 99999 = 582964 x ( 10000 - 1 ) = 582964 x 100000 - 582964 x 1 = 58296400000 - 582964 = 58295817036 c" | a = 99999 - 4
b = a * 582964
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a ) 36 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 60 liters | b | divide(18, subtract(divide(3, 4), divide(30, const_100))) | a big container is 30 % full with water . if 18 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 18 liters of water is added , the container becomes 75 % full . hence these 18 liters account for 45 % of the container , which means that the capacity of it is 18 / 0.45 = 40 liters . or : if the capacity of the container is x liters then : 0.3 x + 18 = 0.75 x - - > x = 40 liters . answer : b ." | a = 3 / 4
b = 30 / 100
c = a - b
d = 18 / c
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a ) 12132 , b ) 14678 , c ) 12708 , d ) 14663 , e ) 15688 | d | multiply(18.6, power(add(const_4, const_1), const_4)) | ( 18.6 x 17.2 Γ· 0.6 ) x 27.5 = ? | "explanation : ( 319.92 Γ· 0.6 ) x 27.5 = 533.2 x 27.5 = 14663 answer : option d" | a = 4 + 1
b = a ** 4
c = 18 * 6
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a ) 287 , b ) 350 , c ) 828 , d ) 277 , e ) 550 | e | subtract(multiply(speed(300, 18), 51), 300) | a 300 m long train crosses a platform in 51 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 5 a = 50 / 3 3 x + 900 = 2550 = > x = 550 m . answer : e" | a = speed * (
b = a - 51
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a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | b | multiply(divide(16.02, 0.0001), const_100) | 16.02 Γ£ β 0.0001 = ? | "16.02 Γ£ β 0.0001 = 0.001602 the answer is b ." | a = 16 / 2
b = a * 100
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a ) 2150 , b ) 2600 , c ) 3120 , d ) 1500 , e ) 1895 | b | divide(add(1000, 100), subtract(const_1, divide(multiply(30, 2), const_100))) | at an election 2 candidates are participated and a candidate got 30 % of votes and defeated by 1000 . and 100 votes are invalid . find the total polled votes ? | "winner votes = 100 - 30 = 70 polled votes = [ ( 100 * 1000 ) / 2 * 70 - 100 ] + 100 = 2600 answer is b" | a = 1000 + 100
b = 30 * 2
c = b / 100
d = 1 - c
e = a / d
|
a ) 25.8 , b ) 26.8 , c ) 38.4 , d ) 34.4 , e ) 29.8 | c | divide(multiply(multiply(const_2, 32), 48), add(32, 48)) | find avrg speed if a man travels at speed of 32 km / hr up and 48 km / hr dawn at an altitude of 230 m . | "avg speed = 2 * x * y / ( x + y ) = 2 * 32 * 48 ( 32 + 48 ) = 38.4 answer : c" | a = 2 * 32
b = a * 48
c = 32 + 48
d = b / c
|
a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) 2 | c | add(0.5, negate(1)) | what is the least integer greater than β 1 + 0.5 ? | "this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first , let ' s take care of the arithmetic : ( - 1 ) + ( 0.5 ) = - 0.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from - 1 to - 0.5 ) . next , the question asks for the least integer that is greater than - 0.5 again , we can use a number line . numbers become greater as you move to the right . the first integer to the right of - 0.5 is 0 . final answer : c" | a = 0 + 5
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | subtract(multiply(15, 15), add(multiply(7, 14), multiply(7, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 7 students is 14 years and that of the other 7 students is 16 years , the age of the 15 th student is | "explanation : age of the 15 th student = [ 15 * 15 - ( 14 * 7 + 16 * 7 ) ] = 15 years . answer : e" | a = 15 * 15
b = 7 * 14
c = 7 * 16
d = b + c
e = a - d
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a ) 4.76 % , b ) 5.88 % , c ) 6.33 % , d ) 7.75 % , e ) 8.33 % | b | multiply(divide(divide(subtract(500, 400), 5), subtract(400, multiply(divide(subtract(500, 400), 5), 3))), const_100) | joe invested a certain sum of money in a simple interest bond whose value grew to $ 400 at the end of 3 years and to $ 500 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 400 - $ 60 = 340 . the interest rate is $ 20 / $ 340 = 1 / 17 which is about 5.88 % the answer is b . | a = 500 - 400
b = a / 5
c = 500 - 400
d = c / 5
e = d * 3
f = 400 - e
g = b / f
h = g * 100
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a ) 8 % , b ) 9 % , c ) 50 % , d ) 12.5 % , e ) 14.8 % | d | multiply(divide(subtract(multiply(divide(2, 15), 120), 24), multiply(divide(2, 15), 120)), const_100) | a doctor prescribed 24 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage per 15 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 15 ) = 2 * 8 = 16 c . c dosage prescribed by doctor for 120 pound patient = 24 c . c % prescribed dosage greater than the typical dosage = ( 24 - 16 / 16 ) * 100 % = ( 8 / 16 ) * 100 % = 50 % answer d" | a = 2 / 15
b = a * 120
c = b - 24
d = 2 / 15
e = d * 120
f = c / e
g = f * 100
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a ) 15 % , b ) 17 % , c ) 24 % , d ) 30 % , e ) 33 % | e | multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100) | in february wilson β s earnings were 40 percent of his family β s total income . in march wilson earned 25 percent less than in february . if the rest of his family β s income was the same in both months , then , in march , wilson β s earnings were approximately what percent z of his family β s total income ? | "lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb = 60 x thus wilson ' s earning as % of total family income in march z = 30 x / ( 30 + 60 ) x = 30 x / 90 x = 33.33 % thus answer is e" | a = 40 / 100
b = 25 / 100
c = 40 / 100
d = b * c
e = a - d
f = 25 / 100
g = 40 / 100
h = f * g
i = 1 - h
j = e / i
k = j * 100
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a ) 9 , b ) 16 , c ) 25 , d ) 36 , e ) 49 | d | add(power(const_2, const_2), const_2) | how many different positive integers are factors of 324 ? | "18 Γ 18 = 6 ^ 2 Γ 3 ^ 2 so total factors = ( 5 + 1 ) ( 5 + 1 ) = 36 answer : d" | a = 2 ** 2
b = a + 2
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a ) $ 11 , b ) $ 5 , c ) $ 45 , d ) $ 400 , e ) $ 5.5 | e | divide(subtract(250, multiply(subtract(const_1, divide(10, const_100)), 250)), subtract(5, divide(const_1, const_2))) | a reduction in the price of petrol by 10 % enables a motorist to buy 5 gallons more for $ 250 . find the original price of petrol ? | "price decreased by 10 % , so 9 / 10 times , which means that original gallons bought increased 10 / 9 times . since this increase equals to 5 gallons then 45 gallons were bought originally ( 45 * 10 / 9 = 50 - - > increase 5 gallons ) . hence original price was 250 / 45 = $ 5.5 answer : e ." | a = 10 / 100
b = 1 - a
c = b * 250
d = 250 - c
e = 1 / 2
f = 5 - e
g = d / f
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a ) a ) 125 , b ) b ) 175 , c ) c ) 225 , d ) d ) 375 , e ) e ) 544 | e | divide(multiply(divide(258, const_100), 1265), 6) | ( 258 % of 1265 ) Γ· 6 = ? | "explanation : ? = ( 258 x 1265 / 100 ) Γ· 6 = 326370 / 600 = 544 answer : option e" | a = 258 / 100
b = a * 1265
c = b / 6
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a ) 40 , b ) 45 , c ) 35 , d ) 30 , e ) 25 | c | add(subtract(35, 10), 10) | if ( 1 / 5 ) ^ n * ( 1 / 4 ) ^ 18 = 1 / ( 2 * ( 10 ) ^ 35 ) , then n = | ( 1 / 5 ) ^ m ( 1 / 4 ) ^ 18 = 1 / { 2 ( 10 ) ^ 35 } 2 ( 10 ) ^ 35 / 4 ^ 18 = 5 ^ m [ 2 ( 2 ^ 35 ) ( 5 ^ 35 ) ] / [ 2 ^ 36 ] = 5 ^ m 5 ^ 35 = 5 ^ m n = 35 c | a = 35 - 10
b = a + 10
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | b | divide(power(10, const_2), power(const_2, const_2)) | let exp ( m , n ) = m to the power n . if exp ( 10 , m ) = n exp ( 22 ) where m and n are integers then n = ___ ? | 10 ^ m = n * ( 2 ^ 2 ) ( 2 * 5 ) ^ m = n * ( 2 ^ 2 ) ( 2 ^ m ) * ( 5 ^ m ) = n * ( 2 ^ 2 ) now equate both side we get m = 2 and ( 5 ^ m ) = n ( 5 ^ 2 ) = n = > n = 25 answer : b | a = 10 ** 2
b = 2 ** 2
c = a / b
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a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 10 | a | subtract(add(3.4, 12.7), 5.6) | if [ x ] is the greatest integer less than or equal to x , what is the value of [ - 5.6 ] + [ 3.4 ] + [ 12.7 ] ? | "you are asked what the closest lesser integer value to [ x ] is . [ - 5.6 ] = - 6.0 [ 3.4 ] = 3.0 [ 12.7 ] = 12.0 therefore , answer is : - 6.0 + 3.0 + 12.0 = 9.0 option d ." | a = 3 + 4
b = a - 5
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a ) 150 , b ) 750 , c ) 1,250 , d ) 1,500 , e ) 2,500 | d | multiply(50, 5) | in a forest 150 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "given 150 deers were caught and they are tagged in the first slot . in the second slot , 50 were caught and 5 we tagged then 10 % of them are tagged . given this 10 % approximates the percentage of tagged deer in the forest . . here number is 150 . . then 100 % is 1500 . . is the number of deers population in the forest . . answer : d is correct answer . ." | a = 50 * 5
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a ) 33 , b ) 88 , c ) 70 , d ) 123 , e ) 18 | e | add(add(power(add(add(divide(subtract(subtract(84, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(84, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(84, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(84, const_10), const_2), const_4), const_2), const_2))) | the sum of four consecutive even numbers is 84 . what would be the smallest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 84 = > x = 11 smallest number is : 2 ( x - 2 ) = 18 . answer : e" | a = 84 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 84 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 84 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 84 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 30 , b ) 19 , c ) 10 , d ) 9 , e ) 7 | b | subtract(divide(subtract(90, 50), subtract(52, 50)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 90 units raises the average to 52 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 50 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 50 n + 90 = 52 * ( n + 1 ) - - > n = 19 . or as 40 extra units increased the average for n + 1 days by 2 units per day then 40 / ( n + 1 ) = 2 - - > n = 19 . answer : b ." | a = 90 - 50
b = 52 - 50
c = a / b
d = c - 1
|
a ) 14 sec , b ) 10 sec , c ) 16 sec , d ) 8 sec , e ) 9 sec | c | divide(160, add(8, 2)) | an escalator moves towards the top level at the rate of 8 ft . sec and its length is 160 feet . if a person walks on the moving escalator at the rate of 2 feet per second towards the top level , how much time does he take to cover the entire length . | "time taken to cover the entire length = tot . dist / resultant speed = 160 / ( 8 + 2 ) = 16 sec answer : c" | a = 8 + 2
b = 160 / a
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a ) 10 , b ) 100 , c ) 1000000 , d ) 10000 , e ) none of these | c | multiply(1000, 10) | ( 1000 ) 7 Γ· ( 10 ) 15 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 15 = ( 10 ) 21 / ( 10 ) 15 = 10 ( 6 ) = 1000000 option c" | a = 1000 * 10
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a ) 20000 , b ) 6000 , c ) 8000 , d ) 10000 , e ) none of these | a | divide(multiply(3000, const_100), 5) | a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 5 % on conveyance . if his savings at the end of the month is rs . 3000 , then his salary per month in rupees is : | "total expenditure = 40 + 25 + 15 + 5 = 85 % saving = ( 100 - 85 ) = 15 % 15 / 100 Γ salary = 3000 , salary = 20000 rs . answer : a" | a = 3000 * 100
b = a / 5
|
a ) 22 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | divide(multiply(8, 48), 3) | the l . c . m of two numbers is 48 . the numbers are in the ratio 8 : 3 . the sum of numbers is : | "let the numbers be 8 x and 3 x . then , their l . c . m = 24 x . so , 24 x = 48 or x = 2 . the numbers are 16 and 6 . hence , required sum = ( 16 + 6 ) = 22 . answer : a" | a = 8 * 48
b = a / 3
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(subtract(13, 5), add(5, 2)) | when positive integer e is divided by 13 , the remainder is 2 . when n is divided by 8 , the remainder is 5 . how many such values are less than 180 ? | the equation that can be formed e is 13 x + 2 = 8 y + 5 . . 13 x - 3 = 8 y . . . as we can see x can take only odd values as the rhs will always be even . . also x can take values till 13 as 13 * 14 > 180 . . now we have to substitue x as 1 , 35 , 79 , 1113 . . . once we find 7 fitting in , any other value need not be checked as every 4 th value will give us answer so next value will be 15 . . ans 1 . . b | a = 13 - 5
b = 5 + 2
c = a - b
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a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.32 , e ) $ 0.45 | d | divide(multiply(add(const_1, const_1), 0.48), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.48 per glass on the first day , what was the price per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.48 = 3 * x - - > x = $ 0.32 . answer : d ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
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a ) rs . 415 , b ) rs . 325 , c ) rs . 352 , d ) rs . 315 , e ) rs . 425 | d | add(divide(406, add(const_1, divide(16, const_100))), multiply(divide(10, const_100), divide(406, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 406 per week . if instead , her weekly salary were to decreased by 10 percent , how much would she earn per week ? | "soln : - ( 406 / 116 ) 90 = 315 in this case long division does not take much time . ( 406 / 116 ) = 3.5 3.5 * 90 = rs . 315 answer : d" | a = 16 / 100
b = 1 + a
c = 406 / b
d = 10 / 100
e = 16 / 100
f = 1 + e
g = 406 / f
h = d * g
i = c + h
|
a ) 1 / 36 , b ) 1 / 30 , c ) 5 / 36 , d ) 1 / 24 , e ) 1 / 28 | a | divide(multiply(divide(const_1, 8), divide(const_1, 6)), add(multiply(divide(const_1, 8), divide(const_1, 6)), multiply(divide(5, 6), divide(add(6, const_1), 8)))) | amit , sumit and pramit go to a seaside town to spend a vacation there and on the first day everybody decides to visit different tourist locations . after breakfast , each of them boards a different tourist vehicle from the nearest bus - depot . after 3 hours , sumit who had gone to a famous beach , calls on the mobile of pramit and claims that he has observed a shark in the waters . pramit learns from the local guide that at that time of the year , only 8 sea - creatures ( including a shark ) are observable and the probability of observing any creature is equal . however , amit and pramit later recall during their discussion that sumit has a reputation for not telling the truth 5 out of 6 times . what is the probability that sumit actually observed a shark in the waters ? | explanation : the probability that sumit actually sees a shark , given that he claimed to have seen one . = > p ( he actually sees the shark and reports truth ) / p ( he claims of seeing a shark ) . = > p ( sees the shark ) p ( reports truth ) / ( p ( he sees a shark ) p ( reports truth ) + p ( does n ' t see ) p ( reports false ) . = > ( 1 / 8 Γ 1 / 6 ) / ( 1 / 8 Γ 1.6 + 7 / 8 Γ 5 / 6 ) = > 1 / 36 . answer : a | a = 1 / 8
b = 1 / 6
c = a * b
d = 1 / 8
e = 1 / 6
f = d * e
g = 5 / 6
h = 6 + 1
i = h / 8
j = g * i
k = f + j
l = c / k
|
a ) s . 8400 , b ) s . 9000 , c ) s . 14000 , d ) s . 17000 , e ) s . 27000 | a | divide(840, divide(multiply(subtract(15, 10), const_2), const_100)) | a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 10 % p . a . for the same time period . therefore the interest received is more by rs . 840 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 10 * 2 ) / 100 = 840 = > 30 x / 100 - 20 x / 100 = 840 = > 10 x / 100 = 840 = > x = 8400 . answer : a" | a = 15 - 10
b = a * 2
c = b / 100
d = 840 / c
|
a ) 5.3 % , b ) 8.2 % , c ) 4.3 % , d ) 8.3 % , e ) 8.0 % | a | multiply(subtract(inverse(divide(950, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 950 gm . for a kg . his gain is β¦ % . | he sells 950 grams of pulses and gains 50 grams . if he sells 100 grams of pulses then he will gain ( 50 / 950 ) * 100 = 5.26 4 . a software engineer has the capability of thinking 100 lines of code in five minutes and can type 100 lines of code in 10 minutes . he takes a break for five minutes after every ten minutes . how many lines of codes will he complete typing after an hour ? answer : a | a = 4 + 1
b = a * 2
c = b * 100
d = 950 / c
e = 1/(d)
f = e - 1
g = f * 100
|
a ) 115 kg . , b ) 118 kg . , c ) 113 kg . , d ) 110 kg . , e ) 112 kg . | c | divide(add(multiply(7, 121), add(110, 60)), add(7, const_2)) | there are 7 players in a bowling team with an average weight of 121 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | the new average will be = ( 121 * 7 + 110 + 60 ) / 9 = 113 kgs c is the answer | a = 7 * 121
b = 110 + 60
c = a + b
d = 7 + 2
e = c / d
|
a ) 130 , b ) 120 , c ) 110 , d ) 180 , e ) 150 | c | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(70, 3)) | a student scored an average of 70 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 70 * 3 = 210 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 210 = 110 answer : c" | a = 90 * 2
b = 70 * 2
c = a + b
d = 70 * 3
e = c - d
|
a ) 3 / 10 , b ) 23 / 60 , c ) 7 / 12 , d ) 41 / 60 , e ) 5 / 6 | d | subtract(const_1, add(divide(add(3, 3), multiply(add(3, 3), const_10)), divide(add(const_10, const_2), multiply(add(3, 3), const_10)))) | a box contains 3 pairs of blue gloves and two pairs of green gloves . each pair consists of a left - hand glove and a right - hand glove . each of the gloves is separate from its mate and thoroughly mixed together with the others in the box . if 3 gloves are randomly selected from the box , what is the probability that a matched set ( i . e . , a left - and right - hand glove of the same color ) will be among the 3 gloves selected ? | you can use the ' calculating the reverse ' method used by bunuel above or if you would like to calculate the probability of getting a matched set in the usual way , you can think of it in this way : bleft ( 3 ) , bright ( 3 ) , gleft ( 2 ) , gright ( 2 ) bleft , bright , g get a bleft and bright in ( 3 / 10 ) * ( 3 / 9 ) ways . then get any green in 4 / 8 ways . probability of getting b pair and a g = ( 3 / 10 ) * ( 3 / 9 ) * ( 4 / 8 ) * 3 ! ( you multiply by 3 ! here because you could pick in some other order e . g . bright , bleft , g or bleft , g , bright etc ) bleft , bright , b get a bleft and bright in ( 3 / 10 ) * ( 3 / 9 ) ways . then get any blue in 4 / 8 ways . probability of getting b pair and another b = ( 3 / 10 ) * ( 3 / 9 ) * ( 4 / 8 ) * 3 ! / 2 ! ( you multiply by 3 ! here to account for the order e . g . bright , bleft , bleft or bleft , bright , bright etc but two gloves will be identical so you divide by 2 ! ) gleft , gright , b get a gleft and gright in ( 2 / 10 ) * ( 2 / 9 ) ways . then get any blue in 6 / 8 ways . probability of getting g pair and a b = ( 2 / 10 ) * ( 2 / 9 ) * ( 6 / 8 ) * 3 ! gleft , gright , g get a gleft and gright in ( 2 / 10 ) * ( 2 / 9 ) ways . then get any other g in 2 / 8 ways . probability of getting g pair and a g = ( 2 / 10 ) * ( 2 / 9 ) * ( 2 / 8 ) * 3 ! / 2 ! adding them all up , you get 41 / 60 . answer : d | a = 3 + 3
b = 3 + 3
c = b * 10
d = a / c
e = 10 + 2
f = 3 + 3
g = f * 10
h = e / g
i = d + h
j = 1 - i
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | c | multiply(divide(subtract(multiply(divide(add(const_100, const_10), const_100), divide(add(const_100, 30), const_100)), const_1), multiply(divide(add(const_100, const_10), const_100), divide(add(const_100, 30), const_100))), const_100) | the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 30 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 30 % then output equals 143 to restore new holidy demand output to original 100 . final - initial / final * 100 = 43 / 143 * 100 = 30 % approxiamately . option c is correct . | a = 100 + 10
b = a / 100
c = 100 + 30
d = c / 100
e = b * d
f = e - 1
g = 100 + 10
h = g / 100
i = 100 + 30
j = i / 100
k = h * j
l = f / k
m = l * 100
|
a ) 99 , b ) 289 , c ) 350 , d ) 882 , e ) 600 | e | subtract(multiply(divide(300, 18), 54), 300) | a 300 meter long train crosses a platform in 54 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 / 54 = 50 / 3 3 ( x + 300 ) = 2700 Γ¨ x = 350 m . answer : e | a = 300 / 18
b = a * 54
c = b - 300
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | 150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration e of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration e would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a = 150 * 30
b = a / 100
c = 400 * 12
d = c / 100
e = b + d
f = 150 + 400
g = e / f
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(floor(divide(19, const_3)), const_1) | what is the smallest integer k for which 64 ^ k > 4 ^ 19 ? | 64 ^ k > 4 ^ 19 4 ^ ( 3 k ) > 4 ^ 19 3 k > 19 k = 7 the answer is d . | a = 19 / 3
b = math.floor(a)
c = b + 1
|
a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 55 m 2 , e ) 57.5 m 2 | e | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 5), multiply(add(divide(25, const_100), 1), 6))), multiply(5, 6)) | a cistern 6 m long and 5 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 ( 5 x 1.25 + 6 x 1.25 ) + 6 x 5 ] m 2 = 57.5 m 2 . answer : option e" | a = 25 / 100
b = a + 1
c = b * 5
d = 25 / 100
e = d + 1
f = e * 6
g = c + f
h = 2 * g
i = 5 * 6
j = h + i
|
['a ) 6 p 5', 'b ) 1', 'c ) 5', 'd ) none of these', 'e ) can not be determined'] | d | divide(factorial(6), multiply(factorial(subtract(6, const_2)), factorial(const_2))) | there are 6 equally spaced points a , b , c , d , e and f marked on a circle with radius r . how many convex pentagons of distinctly different areas can be drawn using these points as vertices ? | solution : since , all the points are equally spaced ; hence the area of all the convex pentagons will be same . answer : option d | a = math.factorial(6)
b = 6 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 140 , b ) 142 , c ) 144 , d ) 146 , e ) 148 | c | divide(divide(1600, const_1000), divide(40, const_3600)) | a train 1600 m long can cross a pole in 40 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 1600 / 40 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : c" | a = 1600 / 1000
b = 40 / 3600
c = a / b
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | power(multiply(4, power(5, 2)), 4) | if 4 log ( 4 * 5 ^ 2 ) = x , find x | "4 ( log 2 ^ 2 * 5 ^ 2 ) = x 4 log ( 5 * 2 ) ^ 2 = x 4 * 2 log ( 5 * 2 ) = x 8 log 10 = x log 10 base 10 = 1 so 8 * 1 = x x = 8 answer : d" | a = 5 ** 2
b = 4 * a
c = b ** 4
|
a ) 12.125 , b ) 15.125 , c ) 18.125 , d ) 21.125 , e ) 24.125 | d | subtract(add(multiply(9, divide(multiply(2, 2), 9)), 11), multiply(2, power(divide(multiply(2, 2), 9), 2))) | if x is real , find the maximum value of the expression - 2 x ^ 2 + 9 x + 11 . | "this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 9 x + 11 = ( - 2 x + 11 ) ( x + 1 ) the roots are 11 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 2.25 the maximum value is - 2 ( 2.25 ) ^ 2 + 9 ( 2.25 ) + 11 = 21.125 the answer is d ." | a = 2 * 2
b = a / 9
c = 9 * b
d = c + 11
e = 2 * 2
f = e / 9
g = f ** 2
h = 2 * g
i = d - h
|
a ) 48 , b ) 96 , c ) 80 , d ) 44 , e ) 140 | a | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(24, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | find the l . c . m of 12 , 8 , 24 and 16 . | "explanation : 2 x 3 x 2 x 2 x 2 = 48 answer : option a" | a = 3 ** 3
b = 2 ** 3
c = 4 + 1
d = c ** 2
e = b * d
f = a * e
g = 24 / 2
h = g / 2
i = h / 3
j = 4 + 1
k = i / j
l = 4 + 1
m = k / l
n = f * m
|
a ) 3 , b ) 10 , c ) 13 , d ) 16 , e ) 17 | c | subtract(add(13.4, 12.7), 11.6) | if [ x ] is the greatest integer less than or equal to x , what is the value of [ - 11.6 ] + [ 13.4 ] + [ 12.7 ] ? | "you are asked what the closest lesser integer value to [ x ] is . [ - 11.6 ] = - 12.0 [ 13.4 ] = 13.0 [ 12.7 ] = 12.0 therefore , answer is : - 12.0 + 13.0 + 12.0 = 13.0 option a ." | a = 13 + 4
b = a - 11
|
a ) 88888888 , b ) 888888888 , c ) 898989898 , d ) 9999999998 , e ) none | b | multiply(12345679, power(add(const_4, const_1), const_4)) | ( 12345679 x 72 ) = ? | "explanation : 12345679 x 72 = 12345679 x ( 70 + 2 ) = 12345679 x 70 + 12345679 x 2 = 864197530 + 24691358 = 888888888 answer : option b" | a = 4 + 1
b = a ** 4
c = 12345679 * b
|
a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | d | divide(multiply(choose(5, 2), choose(3, 1)), choose(add(add(5, 4), 3), 3)) | a jar contains only red , yellow , and orange marbles . if there are 3 red , 5 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | 5 c 2 * 3 c 1 = 10 * 3 = 30 12 c 3 = 220 3 / 22 answer : d | a = math.comb(5, 2)
b = math.comb(3, 1)
c = a * b
d = 5 + 4
e = d + 3
f = math.comb(e, 3)
g = c / f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(const_1, 5) | if 10 ! - 2 * ( 5 ! ) ^ 2 is divisible by 10 ^ n , what is the greatest value of n ? | = 10 β 9 β 8 β 7 β 6 β 5 ! β 2 β 5 ! β 5 β 4 β 3 β 2 β 1 = 5 ! β 10 β 3 β 8 ( 125 ) = 5 β ( 2 β 2 ) β 3 β 2 β ( 5 β 2 ) β 24 β 5 β 5 β 5 = 5 β 2 β 5 β 2 β 5 β 2 β 5 β 2 β 24 β 5 β 3 = 10 ^ 4 β 2 ^ 4 β 5 β 3 = 10 ^ 4 β 5 β 2 β 12 β 3 = 10 ^ 5 β 36 answer : e | a = 1 * 5
|
a ) 867 , b ) 1179 , c ) 1196 , d ) 1098 , e ) 1197 | e | add(add(multiply(add(divide(27, 3), 2), const_100), multiply(divide(27, 3), const_10)), subtract(divide(27, 3), 2)) | in a 3 digit number , the 100 digit is 2 more than the tens digit and the units digit is 2 less than the tens digit . if the sum of the digits is 27 , find the number ? | e 1197 let the three digit numbers be 100 a + 10 b + c a = b + 2 c = b - 2 a + b + c = 3 b = 27 = > b = 9 so a = 11 and b = 7 hence the three digit number is : 1197 | a = 27 / 3
b = a + 2
c = b * 100
d = 27 / 3
e = d * 10
f = c + e
g = 27 / 3
h = g - 2
i = f + h
|
a ) 45 , b ) 41.2 , c ) 40 , d ) 50 , e ) 48 | b | divide(original_price_before_loss(10, 90), divide(original_price_before_gain(20, 60), 20)) | a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 10 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 90 * ( 100 % + 10 % ) = $ 99 number of articles to be sold for $ 108 to incur a 20 % loss : $ 99 / $ 2.40 = 41.2 thus , solution b is correct ." | a = original_price_before_loss / (
|
a ) 10 % , b ) 2 % , c ) 4 % , d ) 5 % , e ) 3 % | a | multiply(divide(divide(subtract(900, 750), 750), 2), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 900 in 2 years ? | "150 = ( 750 * 2 * r ) / 100 r = 10 % answer : a" | a = 900 - 750
b = a / 750
c = b / 2
d = c * 100
|
a ) 2 , b ) 3 , c ) 5 , d ) 4 , e ) 6 | d | subtract(multiply(add(multiply(const_4, const_10), const_2), 5), 686) | what least number must be added to 686 , so that the sum is completely divisible by 5 ? | "if we divide 686 by 5 remainder is 1 5 - 1 = 4 answer : d" | a = 4 * 10
b = a + 2
c = b * 5
d = c - 686
|
a ) 20 cm , b ) 25 cm , c ) 35 cm , d ) 50 cm , e ) none of these | e | divide(multiply(4, 45), multiply(40, 20)) | 45 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 4 m 3 , then the rise in the water level in the tank will be : | explanation : total volume of water displaced = ( 4 x 45 ) m 3 = 180 m 3 rise in water level = 180 / 40 Γ£ β 20 = 0.225 m = 22.5 cm answer : e | a = 4 * 45
b = 40 * 20
c = a / b
|
['a ) 9001', 'b ) 9621', 'c ) 9431', 'd ) 9801', 'e ) 9601'] | d | multiply(99, 99) | square of 99 = ? | square of ( 100 - 1 ) = 10000 + 1 - 200 9801 answer d | a = 99 * 99
|
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