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 ) 20 / 8 sec , b ) 20 / 4 sec , c ) 20 / 7 sec , d ) 22 / 7 sec , e ) 60 / 7 sec | c | divide(100, multiply(add(54, 72), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 72 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 54 + 72 ) * 5 / 18 = 7 * 5 = 35 mps . the time required = d / s = 100 / 35 = 20 / 7 sec . answer : c" | a = 54 + 72
b = a * const_0_2778
c = 100 / b
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a ) 90 , b ) 75 , c ) 70 , d ) 80 , e ) 85 | d | divide(const_1, divide(subtract(divide(4, 5), divide(3, 4)), 4)) | an oil cylinder was 3 / 4 th full . when 4 bottles of oil is poured into it , it is 4 / 5 th full . how many bottles of oil can the full cylinder hold ? | ( 4 / 5 - 3 / 4 ) cylinder is filled by 4 bottles 0.8 - 0.75 = 0.05 cylinder is filled by 4 bottles 1 cylinder is filled by 4 / 0.05 = 80 bottles answer : d | a = 4 / 5
b = 3 / 4
c = a - b
d = c / 4
e = 1 / d
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a ) 2014 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2013) | if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2013 ) = ? | "f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 ⇒ ⇒ f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 ⇒ ⇒ f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 ⇒ ⇒ f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2013 ) = 2014 ans : a" | a = 1 + 2013
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a ) $ 752 , b ) $ 755 , c ) $ 765 , d ) $ 788 , e ) $ 775 | d | divide(add(multiply(add(750, 50), 50), multiply(6, 750)), 26) | last year manfred received 26 paychecks . each of his first 6 paychecks was $ 750 ; each of his remaining paychecks was $ 50 more than each of his first 6 paychecks . to the nearest dollar , what was the average ( arithmetic mean ) amount of his pay checks for the year ? | "= ( 750 * 6 + 800 * 20 ) / 26 = 788 answer is d . posted from my mobile device" | a = 750 + 50
b = a * 50
c = 6 * 750
d = b + c
e = d / 26
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a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 9 | e | subtract(subtract(subtract(110, 25), const_4), const_2) | a certain no . when divided by 110 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 110 + 25 = 135 / 15 = 9 ( remainder ) e" | a = 110 - 25
b = a - 4
c = b - 2
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a ) 80 , b ) 110 , c ) 160 , d ) 600 , e ) 400 | d | divide(30, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 30 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 30 grades are d ' s . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a ’ s + # b ’ s + # c ’ s + # d ’ s = total number of students in the class 1 / 5 ( t ) + ¼ ( t ) + ½ ( t ) + 30 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 600 = 20 t 19 t + 600 = 20 t 600 = t there are a total of 600 students in the math class . answer is d ." | a = 1 / 5
b = 1 / 4
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 30 / f
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a ) 2.9 , b ) 2.8 , c ) 2.4 , d ) 2.8 , e ) 2.2 | c | multiply(divide(multiply(add(3, 1.2), subtract(3, 1.2)), add(add(3, 1.2), subtract(3, 1.2))), const_2) | a man can row 3 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 3 s = 1.2 ds = 3.6 us = 1.8 x / 3.6 + x / 1.8 = 1 x = 1.2 d = 1.2 * 2 = 2.4 answer : c" | a = 3 + 1
b = 3 - 1
c = a * b
d = 3 + 1
e = 3 - 1
f = d + e
g = c / f
h = g * 2
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a ) $ 33.75 , b ) $ 47.25 , c ) $ 51.75 , d ) $ 54.00 , e ) $ 67.50 | e | add(multiply(7.5, 4.50), multiply(multiply(subtract(10.5, 7.5), 2.5), 4.50)) | lloyd normally works 7.5 hours per day and earns $ 4.50 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 2.5 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ? | "daily working hour * regular rate + overtime * increased rate 7.5 * 4.5 + 3 * 4.5 * 2.5 = 67.5 answer e" | a = 7 * 5
b = 10 - 5
c = b * 2
d = c * 4
e = a + d
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a ) 40 % , b ) 80 % , c ) 96 % , d ) 112 % , e ) 124 % | c | multiply(subtract(power(add(const_1, divide(40, const_100)), const_2), const_1), const_100) | if a large pizza has a radius that is 40 % larger that that of a medium pizza , what is the percent increase in area between a medium and a large pizza ? | let the radius of medium pizza be r . then the radius of large pizza is 1.4 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.4 * r ) ^ 2 = 1.96 * pi * r ^ 2 , an increase of 96 % . the answer is c . | a = 40 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) $ 30.60 , b ) $ 72.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60 | b | add(60.50, divide(multiply(60.50, 20), const_100)) | if tim had lunch at $ 60.50 and he gave 20 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 60.50 = ( 20 / 100 ) * 60.50 = $ 12.1 total spent 60.50 + 12.1 = $ 72.6 correct answer b" | a = 60 * 50
b = a / 100
c = 60 + 50
|
a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8600 , d ) rs . 9000 , e ) none | c | divide(add(add(add(add(8000, 5000), 14000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 14000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | answer average salary = 8000 + 5000 + 14000 + 7000 + 9000 / 5 = rs . 8600 correct option : c | a = 8000 + 5000
b = a + 14000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
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a ) 16 , b ) 72 , c ) 112 , d ) 128 , e ) 144 | e | multiply(2, divide(162, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 162 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is in 162 grams of water . 2 x + 16 x = 162 18 x = 162 x = 9 since x is 9 , we know that there are 16 x 9 = 144 grams of oxygen in 162 grams of water . answer e ." | a = 2 + 16
b = 162 / a
c = 2 * b
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a ) 13 , b ) 16 , c ) 25 , d ) 14 , e ) 15 | e | add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(30), inverse(10))), multiply(add(inverse(10), add(inverse(20), inverse(30))), 2))), inverse(30)), 4) | a can do a piece of work in 20 days and b can do it in 30 days and c can do it 10 days . they started the work together and a leaves after 2 days and c leaves after 4 days from the beginning . how long will work lost ? | 2 / 20 + x / 30 + 4 / 10 = 1 x = 300 / 20 = 15 answer : e | a = 4 - 2
b = 1/(30)
c = 1/(10)
d = b + c
e = a * d
f = 1/(10)
g = 1/(20)
h = 1/(30)
i = g + h
j = f + i
k = j * 2
l = e + k
m = 1 - l
n = 1/(30)
o = m / n
p = o + 4
|
a ) 2845 , b ) 3250 , c ) 3740 , d ) 4150 , e ) 5140 | c | add(14,28, lcm(18,32, 48,52)) | find the least number which when divided by 18,32 , 48,52 leaves remainder 14,28 , 44,48 respectively . | "the difference of 18 - 14 = 4 , 32 - 28 = 4 , 48 - 44 = 4 , 52 - 48 = 4 lcm of 18,32 , 48,52 = 3744 required number = 3744 - 4 = 3740 answer is c" | a = math.lcm(18, 32)
b = 14 + 28
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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 ’ s earnings were 40 percent of his family ’ s total income . in march wilson earned 25 percent less than in february . if the rest of his family ’ s income was the same in both months , then , in march , wilson ’ s earnings were approximately what percent of his family ’ 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 = 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
|
a ) a ) 1.12 , b ) b ) 1.2 , c ) c ) 0.95 , d ) d ) 0.85 , e ) e ) 0.9 | d | divide(multiply(0.75, 8), 7) | if 0.75 : x : : 7 : 8 , then x is equal to : | "( x * 7 ) = ( 0.75 * 8 ) x = 6 / 7 x = 0.85 answer = d" | a = 0 * 75
b = a / 7
|
a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36 | b | divide(const_1, subtract(divide(add(const_1, const_2), 36), divide(const_2, 60))) | tom drives from town r to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between r and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per hour ? | let ' s assume that it takes 4 hours to go from point r to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b . | a = 1 + 2
b = a / 36
c = 2 / 60
d = b - c
e = 1 / d
|
a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(multiply(divide(1, 3), subtract(1, divide(1, 3))), add(multiply(divide(1, 3), subtract(1, divide(1, 3))), subtract(1, divide(1, 3)))), const_100) | of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 3 , what percent of the total number of the remaining employees would then be clerical ? | "let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 3 is 1200 * 1 / 3 so it reduced 400 people , so there is 800 clerical people left but since 400 people left , it also reduced from the total of 3600 so there are 3200 people total since 800 clerical left / 3200 people total you get ( a ) 25 %" | a = 1 / 3
b = 1 / 3
c = 1 - b
d = a * c
e = 1 / 3
f = 1 / 3
g = 1 - f
h = e * g
i = 1 / 3
j = 1 - i
k = h + j
l = d / k
m = l * 100
|
a ) 12.6 days , b ) 14.4 days , c ) 15.2 days , d ) 16.8 days , e ) 18.2 days | d | divide(multiply(6, const_3), subtract(divide(add(divide(multiply(6, const_3), 8), add(divide(multiply(6, const_3), 6), divide(multiply(6, const_3), 7))), const_2), divide(multiply(6, const_3), 6))) | a and b can do a piece of work in 6 days , b and c in 7 days , c and a in 8 days . how long will c take to do it ? | "2 c = 1 / 7 + 1 / 8 – 1 / 6 = 20 / 168 = 10 / 84 c = 5 / 84 = > 84 / 5 = 16.8 days the answer is d ." | a = 6 * 3
b = 6 * 3
c = b / 8
d = 6 * 3
e = d / 6
f = 6 * 3
g = f / 7
h = e + g
i = c + h
j = i / 2
k = 6 * 3
l = k / 6
m = j - l
n = a / m
|
a ) 56 , b ) 48 , c ) 47 , d ) 26 , e ) 24 | e | multiply(divide(add(1.1, 0.9), add(210, 90)), const_3600) | two trains are moving in opposite directions with speed of 210 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | explanation : relative speed = 210 + 90 = 300 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 300 hr = 1 / 150 hr = 3600 / 150 seconds = 24 seconds answer : option e | a = 1 + 1
b = 210 + 90
c = a / b
d = c * 3600
|
a ) 18 , b ) 36 , c ) 120 , d ) 90 , e ) 108 | c | multiply(divide(216, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1)) | pat , kate and mark charged a total of 216 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate . | "let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 216 - total hours charged for , x = 24 . mat charged 6 x - x or 5 x for more hours than kate , or for 120 hours . c is correct" | a = 1 + 2
b = 2 * 3
c = a + b
d = 216 / c
e = 2 * 3
f = e - 1
g = d * f
|
a ) 896 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 5), const_100))) | find the principle on a certain sum of money at 5 % per annum for 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 5 ) / 100 ] p = 896 answer : a" | a = 5 * 5
b = a / 100
c = 1 + b
d = 1120 / c
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a ) 45 % , b ) 55 % , c ) 65 % , d ) 75 % , e ) 85 % | e | multiply(add(divide(subtract(add(60, 40), multiply(10, const_2)), const_100), divide(divide(10, const_2), const_100)), const_100) | 10 percent of the women in a college class are science majors , and the non - science majors make up 60 % of the class . what percentage of the men are science majors if 40 % of the class are men ? | science majors make up 0.4 of the class . 60 % of the class are women and 0.1 * 0.6 = 0.06 of the class are female science majors . then 0.34 of the class are male science majors . 0.4 x = 0.34 x = 0.85 = 85 % the answer is e . | a = 60 + 40
b = 10 * 2
c = a - b
d = c / 100
e = 10 / 2
f = e / 100
g = d + f
h = g * 100
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a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | e | multiply(divide(add(multiply(40, divide(5, const_100)), 6.5), add(40, add(6.5, 3.5))), const_100) | a 40 - liter solution of alcohol and water is 5 percent alcohol . if 6.5 liters of alcohol and 3.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 6.5 ) / 50 = 8.5 / 50 = 17 % the answer is e ." | a = 5 / 100
b = 40 * a
c = b + 6
d = 6 + 5
e = 40 + d
f = c / e
g = f * 100
|
a ) 4 , b ) 9 , c ) 10 , d ) 4 , e ) 12 | b | sqrt(divide(243, const_2)) | the area of a parallelogram is 243 sq m and its altitude is three times the corresponding base . what is the length of the base ? | "3 x * x = 243 = > x = 9 answer : b" | a = 243 / 2
b = math.sqrt(a)
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a ) rs 30 , b ) rs 40 , c ) rs 70 , d ) rs 210 , e ) rs 310 | b | subtract(multiply(multiply(4, 3.5), divide(730, add(add(multiply(3, 3.5), multiply(4, 3.5)), multiply(4, 3)))), multiply(multiply(4, 3), divide(730, add(add(multiply(3, 3.5), multiply(4, 3.5)), multiply(4, 3))))) | rs . 730 were divided among punith , michael , suresh in such a way that if punith gets rs . 3 , then michael gets rs . 4 and if michael gets rs . 3.50 then suresh gets rs . 3 . the share of michael exceeds that of suresh by | explanation : let a = punith , b = michael , c = suresh . a : b = 3 : 4 and b : c = 7 / 2 : 3 = ( 8 / 7 ) * ( 7 / 2 ) * ( 8 / 7 ) * 3 = 4 : ( 24 / 7 ) a : b : c = 3 : 4 : 24 / 7 = 21 : 28 : 24 . bs share = rs . [ 730 * ( 28 / 73 ) ] = rs . 280 . cs share = rs . [ 730 * ( 24 / 73 ) ] = rs . 240 . difference of their shares = 40 answer : b | a = 4 * 3
b = 3 * 3
c = 4 * 3
d = b + c
e = 4 * 3
f = d + e
g = 730 / f
h = a * g
i = 4 * 3
j = 3 * 3
k = 4 * 3
l = j + k
m = 4 * 3
n = l + m
o = 730 / n
p = i * o
q = h - p
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a ) 3 , b ) - 3 , c ) 1 / 3 , d ) - 1 / 3 , e ) none of these | b | log(divide(log(subtract(1, multiply(add(const_4, const_1), const_1000))), log(add(const_4, const_1)))) | the value of log 5 ( 1 / 125 ) is : | "log 5 ( 1 / 125 ) = log 5 ( 1 / 5 ) ^ 3 = 3 log 5 ( 1 / 5 ) = 3 ( log 5 1 - log 5 5 ) = 3 ( 0 - 1 ) = - 3 answer : b" | a = 4 + 1
b = a * 1000
c = 1 - b
d = math.log(c)
e = 4 + 1
f = math.log(e)
g = d / f
h = math.log(g)
|
a ) 150 , b ) 148 , c ) 130 , d ) 160 , e ) 210 | b | divide(222, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100)))) | a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 222 for it , the cost price of the cricket bat for a is : | "125 % of 120 % of a = 222 125 / 100 * 120 / 100 * a = 222 a = 222 * 2 / 3 = 148 . answer b" | a = 20 / 100
b = 1 + a
c = 25 / 100
d = 1 + c
e = b * d
f = 222 / e
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 37.5 | a | divide(subtract(add(40, 90), 75), subtract(divide(add(40, 90), 40), divide(75, 90))) | a car traveled 75 % of the way from town a to town b at an average speed of 90 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 40 miles per hour . what is s ? | "total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip → 75 / 90 = 0.833 2 nd part of the trip → 25 / s = t total trip → ( 75 + 25 ) / 40 = 0.833 + t » 100 / 40 = 0.833 + t » 2.5 = 0.833 + t » t = 1.667 back to 2 nd part of the trip formula : 25 / s = 1.667 » s = 15 ans a" | a = 40 + 90
b = a - 75
c = 40 + 90
d = c / 40
e = 75 / 90
f = d - e
g = b / f
|
a ) 72 , b ) 85 , c ) 94 , d ) 105 , e ) 108 | d | subtract(add(subtract(190, 59), 23), 49) | there are 190 items that are members of set u . of these items , 49 are members of set b , 59 are not members of either of set a or set b , and 23 are members of both sets a and b . how many of the members of set u are members of set a ? | "you had the answer almost right . the x = 82 refers to only set a . however what ' s being asked is how many members are part of set a . this will include : 1 . only set a 2 . set a and set b so the answer is set a = 82 + set ab = 82 + 23 = 105 d" | a = 190 - 59
b = a + 23
c = b - 49
|
a ) 23 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds | a | subtract(multiply(divide(7, 56), 240), 7) | in a 240 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is : | "b runs 56 m in 7 sec . = > b runs 240 m in 7 / 56 * 240 = 30 seconds since a beats b by 7 seconds , a runs 240 m in ( 30 - 7 ) = 23 seconds hence , a ' s time over the course = 23 seconds answer : a" | a = 7 / 56
b = a * 240
c = b - 7
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a ) 31 years , b ) 32 years , c ) 33 years , d ) 34 years , e ) 35 years | d | divide(subtract(40, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 40 years older than his son . in six years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 40 . = > 2 ( x + 6 ) = ( x + 40 + 6 ) = > 2 x + 12 = x + 46 = > x = 34 years answer : option d" | a = 2 * 2
b = a - 2
c = 40 - b
d = 2 - 1
e = c / d
|
a ) 22 , b ) 28 , c ) 98 , d ) 38.5 , e ) 13 | d | divide(198, add(const_2, const_pi)) | the perimeter of a semi circle is 198 cm then the radius is ? | "36 / 7 r = 198 = > r = 38.5 answer : d" | a = 2 + math.pi
b = 198 / a
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a ) 0.35 , b ) 0.5 , c ) 6.25 , d ) 0.25 , e ) 6.0 | a | subtract(6.35, floor(6.35)) | for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.35 – 6.35 * ? | "since z * is defined as the greatest positive even integer less than or equal to z , then 6.35 * = 6 ( the greatest positive even integer less than or equal to 6.35 is 6 ) . hence , 6.35 – 6.35 * = 6.35 - 6 = 0.35 answer : a ." | a = math.floor(6, 35)
b = 6 - 35
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a ) 42 , b ) 53 , c ) 50 , d ) 54 , e ) 56 | b | divide(add(360, 120), add(divide(360, 60), divide(120, 40))) | joe drives 360 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ? | t 1 = 360 / 60 = 6 hours t 2 = 120 / 40 = 3 hours t = t 1 + t 2 = 9 hours avg speed = total distance / t = 480 / 9 = 53 mph = b | a = 360 + 120
b = 360 / 60
c = 120 / 40
d = b + c
e = a / d
|
a ) 161 , b ) 152 , c ) 154 , d ) 158 , e ) 144 | a | add(multiply(17, 9), 8) | 12 . what is the dividend . divisor 17 , the quotient is 9 and the remainder is 8 . | "d = d * q + r d = 17 * 9 + 8 d = 153 + 8 d = 161 answer a" | a = 17 * 9
b = a + 8
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a ) 3 : 7 , b ) 4 : 9 , c ) 11 : 7 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(55, 8), multiply(70, 4)) | car a runs at the speed of 55 km / hr and reaches its destination in 8 hours . car b runs at the speed of 70 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 55 × 8 = 440 km car b travels 70 × 4 = 280 km the ratio is 440 : 280 = 44 : 28 = 11 : 7 the answer is c ." | a = 55 * 8
b = 70 * 4
c = a / b
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a ) 700 , b ) 770 , c ) 707 , d ) 710 , e ) 720 | a | multiply(divide(multiply(70, const_1000), const_3600), 36) | a train running at the speed of 70 km / hr crosses a pole in 36 seconds . what is the length of the train ? | "speed = ( 70 * 5 / 18 ) m / sec = ( 175 / 9 ) m / sec length of the train = ( speed x time ) = ( 175 / 9 * 36 ) m = 700 m . answer : a" | a = 70 * 1000
b = a / 3600
c = b * 36
|
a ) 48 kmph , b ) 252 kmph , c ) 293 kmph , d ) 263 kmph , e ) 265 kmph | c | divide(440, multiply(divide(1, 2), 3)) | a jeep takes 3 hours to cover a distance of 440 km . how much should the speed in kmph be maintained to cover the same direction in 1 / 2 th of the previous time ? | "time = 3 distance = 340 1 / 2 of 3 hours = 6 * 1 / 2 = 1.5 hours required speed = 440 / 1.5 = 293 kmph c )" | a = 1 / 2
b = a * 3
c = 440 / b
|
a ) 2442 , b ) 2552 , c ) 2662 , d ) 2772 , e ) 2882 | a | multiply(add(add(multiply(add(add(5, 4), 2), const_100), multiply(add(add(const_4.0, 4), 2), const_10)), add(add(5, 4), 2)), 4) | what is the sum of all possible 3 - digit numbers that can be constructed using the digits 2 , 4 , and 5 if each digit can be used only once in each number ? | "there are 6 possible arrangements of the three numbers . then each number will be in the hundreds , tens , and ones place two times each . the sum is 2 ( 222 ) + 2 ( 444 ) + 2 ( 555 ) = 2442 the answer is a ." | a = 5 + 4
b = a + 2
c = b * 100
d = 4 + 0
e = d + 2
f = e * 10
g = c + f
h = 5 + 4
i = h + 2
j = g + i
k = j * 4
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a ) 30 , b ) 35 , c ) 25 , d ) 28 , e ) 32 | b | add(subtract(20, subtract(20, add(const_3, const_2))), multiply(subtract(20, add(const_3, const_2)), const_2)) | the difference between the ages of two persons is 20 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the younger person is ? | let their ages be x years and ( x + 20 ) years then , ( x + 20 ) - 15 = 2 ( x - 15 ) x + 5 = 2 x - 30 x = 35 answer is b | a = 3 + 2
b = 20 - a
c = 20 - b
d = 3 + 2
e = 20 - d
f = e * 2
g = c + f
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a ) 105 , b ) 168 , c ) 342 , d ) 660 , e ) 1050 | e | divide(multiply(70, 600), 40) | if 70 percent of 600 is 40 percent of x , then x = ? | "given : 0.7 * 600 = 0.4 x - - > 420 = 4 / 10 * x - - > x = 1,050 . answer : e ." | a = 70 * 600
b = a / 40
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['a ) 21', 'b ) 22', 'c ) 23', 'd ) 24', 'e ) 25'] | a | add(divide(divide(544, divide(divide(divide(divide(divide(544, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(544, divide(divide(divide(divide(divide(544, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of divisors of 544 which are perfect squares | perfect squares divisiors are 1 , 2 ^ 2 & 2 ^ 4 only sum of divisors of 544 which are perfect squares = 1 + 2 ^ 2 + 2 ^ 4 = 21 answer : a | a = 544 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 544 / e
g = f / 2
h = 544 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 544 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
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a ) 2331.75 , b ) 2209.75 , c ) 2108.75 , d ) 2107.75 , e ) 2100.75 | a | add(add(multiply(6000, divide(15, const_100)), multiply(add(6000, multiply(6000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(6000, multiply(6000, divide(15, const_100))), multiply(add(6000, multiply(6000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const_3))) | find compound interest on $ 6000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = $ [ 6000 x ( 1 + ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = $ [ 6000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = $ 8331.75 . : . c . i . = rs . ( 8331.75 - 6000 ) = $ 2331.75 answer a ." | a = 15 / 100
b = 6000 * a
c = 15 / 100
d = 6000 * c
e = 6000 + d
f = 15 / 100
g = e * f
h = b + g
i = 15 / 100
j = 6000 * i
k = 6000 + j
l = 15 / 100
m = 6000 * l
n = 6000 + m
o = 15 / 100
p = n * o
q = k + p
r = 15 / 100
s = r / 3
t = q * s
u = h + t
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a ) 24 % , b ) 30 % , c ) 44 % , d ) 58 % , e ) 64 % | d | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), subtract(const_100, 40))), const_100) | a towel , when bleached , lost 30 % of its length and 40 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( − 30 − 40 + ( 30 × 40 ) / 100 ) % = − 58 % i . e . , area is decreased by 58 % answer : d" | a = 100 * 100
b = 100 - 30
c = 100 - 40
d = b * c
e = a - d
f = e / 100
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a ) rs . 16000 , b ) rs . 15000 , c ) rs . 15250 , d ) rs . 15200 , e ) rs . 15300 | d | multiply(divide(280, 70), 3800) | in order to obtain an income of rs . 3800 from 70 % stock at rs . 280 , one must make an investment of | "explanation : market value = rs . 280 required income = rs . 3800 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 70 ( ie , 70 % of the face value 100 ) , investment = rs . 280 to obtain rs . 3800 , investment = 280 / 70 ã — 3800 = rs . 15200 answer : option d" | a = 280 / 70
b = a * 3800
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a ) 65 kmph , b ) 70 kmph , c ) 72 kmph , d ) 75 kmph , e ) 78 kmph | c | divide(divide(300, const_1000), divide(15, const_3600)) | a train 300 m long can cross an electric pole in 15 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 300 / 15 s = 20 m / sec speed = 20 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 72 kmph answer : c" | a = 300 / 1000
b = 15 / 3600
c = a / b
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a ) 2 / 3 , b ) 5 / 18 , c ) 4 / 9 , d ) 3 / 8 , e ) 4 / 7 | b | divide(multiply(divide(2, 3), 5), 12) | a pipe can empty 2 / 3 rd of a cistern in 12 mins . in 5 mins , what part of the cistern will be empty ? | 2 / 3 - - - - 12 ? - - - - - 5 = = > 5 / 18 b | a = 2 / 3
b = a * 5
c = b / 12
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a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40 | d | add(add(choose(8, const_1), choose(8, const_1)), choose(const_4, const_1)) | jane and thomas are among the 8 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 8 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 3 people from 8 is 8 c 3 = 56 . the number of committees without jane or thomas is 6 c 3 = 20 . there are 56 - 20 = 36 possible committees which include jane and / or thomas . the answer is d ." | a = math.comb(8, 1)
b = math.comb(8, 1)
c = a + b
d = math.comb(4, 1)
e = c + d
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a ) $ 225 , b ) $ 150 , c ) $ 200 , d ) $ 250 , e ) $ 450 | a | add(subtract(multiply(add(6, const_1), 75), multiply(add(6, const_1), 50)), 50) | if john makes a contribution to a charity fund at school , the average contribution size will increase by 50 % reaching $ 75 per person . if there were 6 other contributions made before john ' s , what is the size of his donation ? | "cavg = average contribution before john cavg * 1.5 = 75 , therefore the average cont is $ 50 before john . if he needs to increase the average contribution by $ 25 , he must put in $ 25 for each of the 6 people . so $ 150 . but , he also has to put in the average for himself ( the seventh person ) , so add $ 75 . so $ 225 is your answer . answer a" | a = 6 + 1
b = a * 75
c = 6 + 1
d = c * 50
e = b - d
f = e + 50
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a ) 4 , b ) 5 , c ) 1 , d ) 2 , e ) 7 | d | divide(divide(divide(lcm(30, 55), 55), const_4), const_4) | what is the least value of x . so that 30 x 55 is divisible by 3 ? | "the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 3 + 0 + x + 5 + 5 = 13 + x least value of x may be 2 therefore 13 + 2 = 15 is divisible by 3 . d" | a = math.lcm(30, 55)
b = a / 55
c = b / 4
d = c / 4
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a ) 100 , b ) 120 , c ) 140 , d ) 150 , e ) 160 | c | add(multiply(subtract(70, 4), 2), multiply(2, 10)) | if a motorist had driven 1 hour longer on a certain day and at an average rate of 4 miles per hour faster , he would have covered 70 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day ? | "case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 70 ) = ( r + 4 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 2 ) x = 140 ans c" | a = 70 - 4
b = a * 2
c = 2 * 10
d = b + c
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a ) 2 , b ) 2 / 3 , c ) 1 , d ) 0 , e ) - 2 | d | subtract(multiply(3, 2), multiply(2, 2)) | 2 ^ ( 2 x / 3 ) = 2 ^ ( x / 2 ) what is the value of x ? | 2 x / 3 = x / 2 4 x = 3 x 4 x - 3 x = 0 ; x = 0 ans : d | a = 3 * 2
b = 2 * 2
c = a - b
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a ) 17 , b ) 18 , c ) 54 , d ) 56 , e ) 864 | d | divide(multiply(multiply(35, 20), 10), volume_cube(divide(10, const_2))) | a box measuring 35 inches long by 20 inches wide by 10 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 5 is the biggest number that could divide all three , 35 , 20 and 10 . thus side of cube must be 5 , and total number of cubes = 35 / 5 * 20 / 5 * 10 / 5 = 56 ans d it is ." | a = 35 * 20
b = a * 10
c = 10 / 2
d = b / volume_cube
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a ) 2 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | multiply(const_4, 1) | how many prime numbers between 1 and 100 are factors of 1771 ? | "factor of 1771 = 7 * 11 * 23 - - - 3 prime numbers e" | a = 4 * 1
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 3 | d | add(add(const_4, 4), subtract(4, const_4)) | what is the sum of all possible solutions to | x - 4 | ^ 2 + | x - 4 | = 20 ? | "denote | x - 4 | as y : y ^ 2 + y = 20 - - > y = - 5 or y = 4 . discard the first solution since y = | x - 4 | , so it ' s an absolute value and thus can not be negative . y = | x - 4 | = 4 - - > x = 8 or x = - 1 . the sum = 7 . answer : d ." | a = 4 + 4
b = 4 - 4
c = a + b
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a ) a ) 99 , b ) b ) 84 , c ) c ) 51 , d ) d ) 65 , e ) e ) 57 | c | multiply(add(14, const_3), const_3) | if p represents the product of the first 14 positive integers , then p is not a multiple of : | a ) 99 = 9 * 11 b ) 84 = 4 * 7 * 3 c ) 51 = 17 * 3 d ) 65 = 5 * 13 e ) 57 = 19 * 3 since 17 is not there in first 15 positive numbers it is the only possibility c | a = 14 + 3
b = a * 3
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a ) 90 kg , b ) 100 kg , c ) 95 kg , d ) 85 kg , e ) 92 kg | b | add(multiply(10, 5), 50) | the average weight of 10 girls increases by 5 kg when a new girl comes in place of one of them weighing 50 kg . what might be the weight of the new girl ? | "total weight increased = 10 x 5 kg = 50 kg . weight of new person = 50 + 50 kg = 100 kg answer : b" | a = 10 * 5
b = a + 50
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a ) a ) 11 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | a | floor(divide(144, 13)) | on dividing 144 by a number , the quotient is 13 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 144 - 1 ) / 13 = 143 / 13 = 11 a )" | a = 144 / 13
b = math.floor(a)
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a ) 26 , b ) b ) 10 , c ) c ) 11 , d ) d ) 15 , e ) e ) 23.5 | a | add(20, 5) | a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 1100 . how many workers were part of the latter production run ? | the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1100 . . thus for the current period = 1100 - 500 = 600 ( 5 n + 45 ) k = 600 k = 600 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 600 / 5 n + 45 500 ( 5 n + 45 ) = 600 ( 5 n + 20 ) 2500 n + 22500 = 3000 n + 12000 - 500 n = - 10500 n = 21 thus n + 5 = 26 hence a | a = 20 + 5
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a ) 298 cm 2 , b ) 384 cm 2 , c ) 279 cm 2 , d ) 297 cm 2 , e ) 278 cm 2 | b | multiply(24, 16) | find the area of a parallelogram with base 24 cm and height 16 cm ? | "area of a parallelogram = base * height = 24 * 16 = 384 cm 2 answer : b" | a = 24 * 16
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a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | c | divide(4893, power(add(const_1, divide(add(6, divide(1, 4)), const_100)), 3)) | the principal that amounts to rs . 4893 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 4893 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 4085 . answer : c | a = 1 / 4
b = 6 + a
c = b / 100
d = 1 + c
e = d ** 3
f = 4893 / e
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a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | d | subtract(divide(8, const_2), multiply(46, 46)) | what is the remainder when 46 * 49 is divided by 8 ? | "we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 46 * 49 ) / 8 } } = remainder of ( 46 / 8 ) * remainder of ( 49 / 8 ) = 6 * 1 = 6 answer : d" | a = 8 / 2
b = 46 * 46
c = a - b
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a ) 24 , b ) 26 , c ) 30 , d ) 25 , e ) 40 | d | floor(25) | an old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 25 times the difference between the two numbers . how many coins did the old man have ? | "let the number of coins one son got be x and the number of coins another got be y . total = x + y . x ^ 2 - y ^ 2 = 25 ( x - y ) - - > x + y = 25 . answer : d ." | a = math.floor(25)
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a ) 67 , b ) 45 , c ) 60 , d ) 90 , e ) 120 | c | multiply(15, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 15 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "i believe there is a better way to do it . basically one of the options should satisfy the given criteria . 60 did 1 / 5 * 60 = 12 2 / 3 * 60 = 40 so total distance 52 + remaining 15 = 67" | a = 1 / 5
b = 2 / 3
c = a + b
d = 1 - c
e = 1/(d)
f = 15 * e
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | add(divide(18, 3), divide(3, divide(18, 3))) | if 18 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "18 - 3 * 3 * 2 15 - 5 * 3 12 - 4 * 3 9 - 3 * 3 6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 8 is allowed . imo e ." | a = 18 / 3
b = 18 / 3
c = 3 / b
d = a + c
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a ) 220 , b ) 210 , c ) 252 , d ) 236 , e ) 260 | b | sqrt(divide(10, add(power(7, 3), add(power(2, 3), power(3, 3))))) | the ratio of 2 numbers is 3 : 7 and their h . c . f . is 10 . their l . c . m . is ? | "let the numbers be 3 x and 7 x their h . c . f . = 10 so the numbers are 3 * 10 , 7 * 10 = 30,70 l . c . m . = 210 answer is b" | a = 7 ** 3
b = 2 ** 3
c = 3 ** 3
d = b + c
e = a + d
f = 10 / e
g = math.sqrt(f)
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a ) 15 sec , b ) 18.3 sec , c ) 19.6 sec , d ) 20.4 sec , e ) 22.5 sec | c | divide(add(500, 700), multiply(add(100, 120), const_0_2778)) | two trains are moving in opposite directions @ 100 kmph and 120 kmph . their length are 500 m and 700 m . the time taken by the slower train to cross faster train in second is ? | relative speed = 100 + 120 = 220 * 5 / 18 = 550 / 9 m / s distance covered = 500 + 700 = 1200 m required time = 1200 * 9 / 550 = 19.6 sec answer is c | a = 500 + 700
b = 100 + 120
c = b * const_0_2778
d = a / c
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a ) 70 % , b ) 90 % , c ) 60 % , d ) 50 % , e ) 65 % | c | multiply(divide(1, 5), const_100) | if two positive numbers are in the ratio 1 / 8 : 1 / 5 , then by what percent is the second number more than the first ? | "given ratio = 1 / 8 : 1 / 5 = 5 : 8 let first number be 5 x and the second number be 8 x . the second number is more than first number by 3 x . required percentage = 3 x / 5 x * 100 = 60 % . answer : c" | a = 1 / 5
b = a * 100
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a ) 100 % , b ) 150 % , c ) 155 % , d ) 300 % , e ) 355 % | e | add(multiply(25, const_0_33), add(const_100, const_0_33)) | if the price of a certain bond on may 1 st was 1 / 4 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 = 1 / 4 * 12 = 3 ; the price on july 1 st = 3 * 1.25 = 3.75 . the average price of the bond on may 1 st and july 1 st = ( 3 + 3.75 ) / 2 = 3.375 . the price of the bond on june 1 st ( 12 ) is 32 / 9 times ( 355.55 % ) the average price of the bond on may 1 st and july 1 st . answer : e ." | a = 25 * const_0_33
b = 100 + const_0_33
c = a + b
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a ) 12 , b ) 30 , c ) 40 , d ) 15 , e ) 60 | d | divide(250, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 250 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 250 * 3 / 50 = 15 sec . answer : option d" | a = 63 - 3
b = a * const_0_2778
c = 250 / b
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a ) 20 , b ) 50 , c ) 30 , d ) 40 , e ) 10 | b | multiply(add(const_4, const_1), const_10) | the last two - digits in the multiplication 122 × 123 × 125 × 127 × 129 will be | solution the answer will be 50 since , ( 125 × 122 ) will give 50 as the last two digits . answer b | a = 4 + 1
b = a * 10
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a ) 25 kms , b ) 10 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | a | multiply(speed(add(20, 30), const_60), 30) | riya and priya set on a journey . riya moves eastward at a speed of 20 kmph and priya moves westward at a speed of 30 kmph . how far will be priya from riya after 30 minutes | "total eastward distance = 20 kmph * 1 / 2 hr = 10 km total westward distance = 30 kmph * 1 / 2 hr = 15 km total distn betn them = 10 + 15 = 25 km ans 25 km answer : a" | a = 20 + 30
b = speed * (
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a ) 187 m , b ) 350 m , c ) 267 m , d ) 287 m , e ) 870 m | e | subtract(multiply(speed(300, 10), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 10 sec . what is the length of the platform ? | "speed = 300 / 10 = 30 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 30 = > x = 870 m . answer : e" | a = speed * (
b = a - 39
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a ) 22 , b ) 20 , c ) 21 , d ) 18 , e ) 17 | c | subtract(multiply(10, 2), 2) | how many digits are in the number 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 2 ? | "whenever you are asked to find the number of digits , try to bring the number in multiples of 10 . this way , we can wasily calculate the umber of 0 ' s through the powers of 10 50 ^ 8 × 8 ^ 3 × 11 ^ 2 x 10 ^ 2 = ( 5 ^ 2 * 2 ) ^ 8 * 2 ^ 9 * 11 ^ 2 x 10 ^ 2 = 5 ^ 16 * 2 ^ 17 * 11 ^ 2 x 10 ^ 2 = 2 * 11 ^ 2 * 10 ^ 16 x 10 ^ 2 = 242 * 10 ^ 18 hence we would have 16 trailing 0 ' s and the three digits from 242 total digits = 3 + 18 = 21 correct option : c" | a = 10 * 2
b = a - 2
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a ) 1 : 1 , b ) 2 : 3 , c ) 5 : 2 , d ) 4 : 3 , e ) 7 : 9 | a | divide(add(multiply(4, 5), multiply(3, divide(add(5, 3), const_2))), add(multiply(4, 3), multiply(3, divide(add(5, 3), const_2)))) | two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2 , respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio 4 : 3 . what is the ratio of the composition of the two basic elements in alloy x ? | "mixture a has a total of 5 + 3 = 8 parts . if in the final mixture this represents 4 parts , then the total number of parts in mixture b should be ( 8 / 4 ) * 3 = 6 . so , we should take of mixture b a quantity with 2 and 4 parts , respectively . this will give us in the final mixture ( 5 + 2 ) : ( 3 + 4 ) , which means 7 : 7 , or 1 : 1 . answer a ." | a = 4 * 5
b = 5 + 3
c = b / 2
d = 3 * c
e = a + d
f = 4 * 3
g = 5 + 3
h = g / 2
i = 3 * h
j = f + i
k = e / j
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a ) 9,00 , b ) 1,800 , c ) 2,700 , d ) 3,600 , e ) 4,800 | d | multiply(multiply(divide(45, 15), 8), 150) | running at the same constant rate , 15 identical machines can produce a total of 45 bags per minute . at this rate , how many bags could 150 such machines produce in 8 minutes ? | "let ' s take the approach that uses the answer choices to eliminate wasted time . 45 / 15 = 3 bags per minute per machine . 150 machines = 450 per minute . 8 minutes worth = 3600 bags . looking at the answers it is clear . . . we can only choose ( d ) . the correct answer is d ." | a = 45 / 15
b = a * 8
c = b * 150
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a ) 987 , b ) 271 , c ) 72 , d ) 27 , e ) 63 | e | divide(divide(700, const_1000), divide(40, const_3600)) | a train 700 m long can cross an electric pole in 40 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 700 / 40 s = 17.5 m / sec speed = 17.5 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 63 kmph answer : e" | a = 700 / 1000
b = 40 / 3600
c = a / b
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a ) 12 days , b ) 14 days , c ) 28 days , d ) 18 days , e ) 16 days | d | inverse(divide(inverse(12), add(const_2, const_1))) | a is twice as good a work man as b and together they finish the work in 12 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 12 = > x = 1 / 36 2 x = 1 / 18 a can do the work in 18 days . answer : d" | a = 1/(12)
b = 2 + 1
c = a / b
d = 1/(c)
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a ) 3 / 16,000 , b ) 1 / 4,800 , c ) 3 / 1,600 , d ) 1 / 90 , e ) 2 / 45 | b | multiply(divide(100, 800), divide(const_1, 600)) | a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 100 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ? | solution provided by stanford 2012 is correct : 100 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 100 / 800 * 1 / 600 = 1 / 4,800 . answer : b . | a = 100 / 800
b = 1 / 600
c = a * b
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a ) 85 , b ) 81 , c ) 53 , d ) 82 , e ) 84 | e | add(85, const_1) | total 85 matches are conducted in knockout match type . how many players will be participated in that tournament ? | 84 players answer : e | a = 85 + 1
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a ) 0.954 , b ) 0.93 , c ) 0.933 , d ) 0.96 , e ) 0.974 | a | multiply(add(const_3, const_4), 0.934) | if log 27 = 1.431 , then the value of log 9 is : 0.934 b . 0.945 c . 0.954 d . 0.958 | "if log 27 = 1.431 , then the value of log 9 is : a . 0.934 b . 0.945 c . 0.954 d . 0.958 answer : option c explanation : log 27 = 1.431 log ( 33 ) = 1.431 3 log 3 = 1.431 log 3 = 0.477 log 9 = log ( 32 ) = 2 log 3 = ( 2 x 0.477 ) = 0.954 . answer a" | a = 3 + 4
b = a * 0
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a ) rs . 10060.625 , b ) rs . 10040.62 , c ) rs . 12040.62 , d ) rs . 20040.625 , e ) rs . 10030.625 | b | divide(divide(multiply(4016.25, const_100), 8), 5) | a sum fetched a total simple interest of rs . 4016.25 at the rate of 8 % p . a . in 5 years . what is the sum ? | "principal = ( 100 * 4016.25 ) / ( 8 * 5 ) = rs . 10040.625 . answer : b" | a = 4016 * 25
b = a / 8
c = b / 5
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 45 | e | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 5) | the population of a bacteria culture doubles every 5 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | "this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 5 minutes so answer is 9 * 5 = 45 minutes answer ( e )" | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 5
|
a ) 334 km . , b ) 216 km . , c ) 314 km . , d ) 224 km . , e ) 560 km . | e | multiply(multiply(divide(multiply(25, 24), add(24, 21)), 21), const_2) | a woman complete a journey in 25 hours . she travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km . | 0.5 x / 21 + 0.5 x / 24 = 25 - - > x / 21 + x / 24 = 50 - - > x = 560 km . e | a = 25 * 24
b = 24 + 21
c = a / b
d = c * 21
e = d * 2
|
a ) 8 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | subtract(add(negate(const_2.0), multiply(subtract(10, const_1), const_2.0)), 4) | set j consists of 10 consecutive even numbers . if the smallest term in the set is - 4 , what is the range of the positive integers in set j ? | "since there are only 10 integers , another approach is the just list all 10 . we get : - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 range of positive integers = 14 - 2 = 12 answer : b" | a = negate + (
b = 10 - 1
c = b * 2
d = a - c
|
a ) 1.44 % , b ) 2.56 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | a | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 12)), 675958), multiply(divide(const_100, subtract(const_100, 12)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 12)), 675958), multiply(divide(const_100, subtract(const_100, 12)), 675958))), const_100) | a man two flats for $ 675958 each . on one he gains 12 % while on the other he loses 12 % . how much does he gain or lose in the whole transaction ? | "in such a case there is always a loss loss % = ( 12 / 10 ) ^ 2 = 36 / 25 = 1.44 % answer is a" | a = 100 + 12
b = 100 / a
c = b * 675958
d = 100 - 12
e = 100 / d
f = e * 675958
g = c + f
h = 675958 + 675958
i = g - h
j = 100 + 12
k = 100 / j
l = k * 675958
m = 100 - 12
n = 100 / m
o = n * 675958
p = l + o
q = i / p
r = q * 100
|
a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 91 | e | multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(10, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(10, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 10 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ? | 20 % of the workerswithoutretirement plan are women , this means 80 % of the workerswithoutretirement plan are men right ? makes 100 % = no retirement plan . then 40 % of those wo have a retirement plan are men , which means 60 % of those who have a retirement plan are women . these percentages are always refering to the subgroupsretirement plan = yesandno retirement plan . make the total workers equal x and work from there and it should be fine > x / 3 have no retirement plan at all . i struggled with the question a lot , it was hard for me although usually i have no problems with sets questions like this . i believe it must be a 700 + question . yes , you are right . i read the question incorrectly ! i glazed overnotfor some reason . if 20 % of the workers who do not have a retirement plan are women , 80 % who do not have retirement plan are men . also 40 % who do have retirement plan are men . since 1 / 3 do not have retirement plan and 2 / 3 do have , it becomes a simple weighted average question . cavg = ( 90 * 1 + 40 * 2 ) / 3 = 170 / 3 % are men total = 120 * 300 / 170 = 211 no . of women = 211 - 120 = 91 e | a = 1 / 3
b = 1 / 3
c = 10 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 120 / j
l = 1 / 3
m = 10 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 40 | e | divide(120, subtract(divide(120, 30), divide(60, const_60))) | the distance from city a to city b is 120 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 60 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ? | the time it takes cara to drive to city b is 120 / 30 = 4 hours . dan needs to take less than 3 hours for the trip . dan needs to exceed a constant speed of 120 / 3 = 40 miles per hour . the answer is e . | a = 120 / 30
b = 60 / const_60
c = a - b
d = 120 / c
|
a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | d | multiply(divide(150, add(7, 8)), 10) | jack and christina are standing 150 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 7 feet per second and christina walks in a straight line toward jack at a constant speed of 8 feet per second . lindy runs at a constant speed of 10 feet per second from christina to jack , back to christina , back to jack , and so forth . what is the total distance , in feet , that lindy has traveled when the three meet at one place ? | "the relative speed of jack and christina is 7 + 8 = 15 feet per second . the distance between them is 210 feet , hence they will meet in ( time ) = ( distance ) / ( relative speed ) = 150 / 15 = 10 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 10 * 10 = 100 feet . answer : d ." | a = 7 + 8
b = 150 / a
c = b * 10
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(add(const_4, const_3), const_1) | how many different positive integers exist between 10 ^ 4 and 10 ^ 5 , the sum of whose digits is equal to 2 ? | "10001 10010 10100 11000 20000 total no . = 5 c" | a = 4 + 3
b = a + 1
|
a ) 1 : 2 , b ) 2 : 3 , c ) 10 : 27 , d ) 5 : 4 , e ) 4 : 5 | b | divide(add(subtract(divide(192, 32), 2.5), divide(const_1, const_2)), divide(192, 32)) | a man by motorcycle goes from delhi to bharatpur , a distance of 192 km , at an average speed of 32 kmph . another man starts from delhi by car 2.5 h after the motorcyclist starts and reaches bharatpur half an hour late . what is the ratio of person on the motorcycle to the person going by car ? | a man by motorcycle goes from delhi to bharatpur the time take by the motorcyclist = 192 / 32 = 6 hours another man start from delhi be car after 2.5 hour the man starts = 6 - 2.5 = 3.5 and the man on can reaches half and hour late which is = 3.5 + . 5 = 4 thus , the ratio between motorcycle to car is 4 : 6 or , 2 : 3 . answer : b | a = 192 / 32
b = a - 2
c = 1 / 2
d = b + c
e = 192 / 32
f = d / e
|
a ) 3 / 7 , b ) 34 , c ) 1 , d ) 2 , e ) 3 | e | add(divide(18, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5)))) | if p / q = 4 / 5 , then the value of 18 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ? | "answer given exp . = 18 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 18 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 18 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 18 / 7 + 6 / 14 = 18 / 7 + 3 / 7 = 21 / 7 = 3 correct option : e" | a = 18 / 7
b = 4 / 5
c = 2 - b
d = 4 / 5
e = 2 + d
f = c / e
g = a + f
|
a ) 70 minutes , b ) 72 minutes , c ) 80 minutes , d ) 76 minutes , e ) 77 minutes | c | divide(2400, 30) | a scuba diver descends at a rate of 30 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 2400 feet below the sea level . . how long will he take to reach the ship ? | "time taken to reach = 2400 / 30 = 80 minutes answer : c" | a = 2400 / 30
|
a ) 28 , b ) 31 , c ) 34 , d ) 37 , e ) 40 | d | add(divide(150, add(const_4, const_1)), const_2) | how many zeroes are there at the end of the number n , if n = 150 ! + 300 ! ? | "the number of zeroes at the end of 150 ! will be less than the number of zeroes at the end of 300 ! hence it is sufficient to calculate the number of zeroes at the end of 150 ! the number of zeroes = [ 150 / 5 ] + [ 150 / 25 ] + [ 150 / 125 ] = 30 + 6 + 1 = 37 the answer is d ." | a = 4 + 1
b = 150 / a
c = b + 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | floor(const_2) | how many prime numbers are between 26 / 3 and 85 / 6 ? | "26 / 3 = 8 . xxx 85 / 6 = 14 . xxx so we need to find prime numbers between 8 ( exclusive ) - 13 ( inclusive ) there are 2 prime numbers 1113 hence answer will be ( b ) 2 b" | a = math.floor(2)
|
a ) a ) 70 , b ) b ) 76 , c ) c ) 78 , d ) d ) 80 , e ) e ) 87 | e | subtract(multiply(add(32, 5), add(10, const_1)), multiply(10, 32)) | average of 10 matches is 32 , how many runs one should should score to increase his average by 5 runs . | "explanation : average after 11 innings should be 37 so , required score = ( 11 * 37 ) - ( 10 * 32 ) = 407 - 320 = 87 answer : option e" | a = 32 + 5
b = 10 + 1
c = a * b
d = 10 * 32
e = c - d
|
a ) 8 , b ) 9 , c ) 6 , d ) 11 , e ) 12 | c | add(4, 2) | if there are 4 peanuts in a box and mary puts 2 more peanuts inside , how many peanuts are in the box ? | "2 + 4 = 6 correct answer is c ) 6" | a = 4 + 2
|
a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 868.6 hours , e ) 915 hours | d | add(divide(6840, add(16, 2)), divide(6840, subtract(16, 2))) | speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 6840 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 6840 / 18 + 6840 / 14 = 380 + 488.6 = 868.6 hours answer : option d" | a = 16 + 2
b = 6840 / a
c = 16 - 2
d = 6840 / c
e = b + d
|
a ) 40 , b ) 52 , c ) 68 , d ) 75 , e ) 98 | d | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(22, const_100), 120)), subtract(const_1, divide(52, const_100))) | in august , a cricket team that played 120 matches won 22 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 26.4 + x by solving we get x = 75 answer : d" | a = 52 / 100
b = a * 120
c = 22 / 100
d = c * 120
e = b - d
f = 52 / 100
g = 1 - f
h = e / g
|
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