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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(const_4, add(floor(divide(50, add(const_4, const_3))), const_1)) | company z has 50 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | "say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 50 . now , plug options for x . only a and e give an integer value for y . but only for e x > y as needed . answer : e ." | a = 4 + 3
b = 50 / a
c = math.floor(b)
d = c + 1
e = 4 + d
|
a ) 37.5 % , b ) 48 % , c ) 50.5 % , d ) 28.125 % , e ) 56 % | d | multiply(subtract(divide(subtract(const_100, 18), 64), const_1), const_100) | the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 18 % ? | "explanation : let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 82 gain % = 18 / 64 * 100 = 28.125 % . answer : option d" | a = 100 - 18
b = a / 64
c = b - 1
d = c * 100
|
a ) 35 years , b ) 45 years , c ) 51 years , d ) 54 years , e ) none of these | d | add(39, const_1) | the average age of 39 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "explanation : age of the teacher = ( 40 * 15 - 39 * 14 ) years = 54 years . answer : d" | a = 39 + 1
|
a ) 616 , b ) 611 , c ) 661 , d ) 116 , e ) 666 | a | divide(multiply(multiply(multiply(const_12, const_2), 2), subtract(multiply(const_12, const_4), const_4)), multiply(const_12, const_2)) | how many times are the hands of a clock at right angle in 2 weeks ? | "in 1 day , they are at right angles 44 times . in 14 days , they are at right angles 616 times . answer : option a" | a = 12 * 2
b = a * 2
c = 12 * 4
d = c - 4
e = b * d
f = 12 * 2
g = e / f
|
a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 9 , d ) 1 / 24 , e ) 1 / 12 | e | multiply(divide(3, add(add(3, 4), 2)), divide(2, subtract(add(add(3, 4), 2), const_1))) | the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee ’ s report , what is the probability that the two members selected will both be maths teachers ? | probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( e ) | a = 3 + 4
b = a + 2
c = 3 / b
d = 3 + 4
e = d + 2
f = e - 1
g = 2 / f
h = c * g
|
a ) 38 , b ) 40 , c ) 26 , d ) 25 , e ) 11 | b | divide(add(125, 75), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 75 questions and secures 125 marks , the number of questions he attempts correctly , is : | "explanation : let the number of correct answers be x . then numbers of incorrect answers will be 75 – x we get 4 x – ( 75 – x ) × 1 = 125 on solving the equation we get x = 40 answer : b" | a = 125 + 75
b = 4 + 1
c = a / b
|
a ) 53.33 , b ) 46.33 , c ) 15.36 , d ) 15.45 , e ) 15.48 | a | add(add(const_3, multiply(add(const_1, const_4), const_10)), divide(add(multiply(3, const_10), const_3), const_100)) | the value of x . 320 ã · 2 ã · 3 = x | 320 ã · 2 ã · 3 = 320 ã — 1 / 2 ã — 1 / 3 = 160 / 3 = 53.33 correct answer : a | a = 1 + 4
b = a * 10
c = 3 + b
d = 3 * 10
e = d + 3
f = e / 100
g = c + f
|
['a ) 100 < v < 300', 'b ) 300 < v < 500', 'c ) 500 < v < 700', 'd ) 700 < v < 900', 'e ) 900 < v < 1100'] | d | multiply(multiply(const_pi, power(3.5, const_2)), multiply(3.5, divide(650, const_100))) | v is the volume of a cylinder ; the radius of the cylinder is 3.5 . the height of the cylinder is 650 % more than the radius . which of the following is true ? | as we see the answers are in form of range we can use approximation volume of cylinder is π r ^ 2 h given π = 22 / 7 and r = 3.5 so r ^ 2 ~ 12 and h = 6.5 * 3.5 ~ 23 so 22 / 7 * 12 * 23 ~ 868 so answer should be d . 700 < v < 900 answer : d | a = 3 ** 5
b = math.pi * a
c = 650 / 100
d = 3 * 5
e = b * d
|
a ) s . 263 , b ) s . 283 , c ) s . 293 , d ) s . 300 , e ) s . 383 | a | divide(multiply(580, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 580 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 580 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 580 ⇒ y [ 12 / 10 + 1 ] = 580 ⇒ 22 y / 10 = 580 ⇒ 22 y = 5800 ⇒ y = 5800 / 22 = rs . 263 a )" | a = 1 + 4
b = a * 2
c = 580 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 21 sec , b ) 32 sec , c ) 82 sec , d ) 20 sec , e ) 89 sec | a | divide(add(120, 300), multiply(add(42, 30), const_0_2778)) | two trains of length 120 m and 300 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 300 = 420 m . the time required = d / s = 420 / 20 = 21 sec . answer : a | a = 120 + 300
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 18 sec , b ) 70 sec , c ) 21 sec , d ) 20 sec , e ) 16 sec | e | divide(add(100, 220), multiply(add(42, 30), const_0_2778)) | two trains of length 100 m and 220 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 220 = 320 m . the time required = d / s = 320 / 20 = 16 sec . answer : e" | a = 100 + 220
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 9 | e | sqrt(power(power(18, divide(const_1, const_2)), const_3)) | what positive number , when squared , is equal to the cube of the positive square root of 18 ? | "let the positive number be x x ^ 2 = ( ( 18 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 81 = > x = 9 answer e" | a = 1 / 2
b = 18 ** a
c = b ** 3
d = math.sqrt(c)
|
a ) 40 sec , b ) 11 sec , c ) 88 sec , d ) 33.33 sec , e ) 10 sec | d | divide(360, multiply(subtract(54, 140), const_0_2778)) | a train 360 m long is running at a speed of 54 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 54 * 5 / 18 = 15 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 1 / 15 = 33.33 sec answer : d" | a = 54 - 140
b = a * const_0_2778
c = 360 / b
|
a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | b | subtract(subtract(subtract(210, 90), 60), divide(subtract(subtract(210, 90), 60), const_3)) | 210 college students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | 120 = 60 ( mac ) + x ( window ) + 20 ( both ) = > x = 40 answer : b | a = 210 - 90
b = a - 60
c = 210 - 90
d = c - 60
e = d / 3
f = b - e
|
a ) 18 , b ) 24 , c ) 60 , d ) 36 , e ) 48 | c | multiply(divide(divide(600, 1000), 36), const_3600) | if a truck is traveling at a constant rate of 36 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 36 km / hr = > 36,000 m / hr in one minute = > 36000 / 60 = 600 meters in one sec = > 600 / 60 = 10 meters time = total distance need to be covered / avg . speed = > 600 / 10 = 60 and hence the answer : c" | a = 600 / 1000
b = a / 36
c = b * 3600
|
a ) 11 , b ) 12 , c ) 5 , d ) 14 , e ) 18 | c | add(3, subtract(3, subtract(7, subtract(12, 6)))) | solve this 6 + 7 = 12 8 + 9 = 16 5 + 6 = 10 7 + 8 = 14 then , 3 + 3 = ? ? | 5 answer : c | a = 12 - 6
b = 7 - a
c = 3 - b
d = 3 + c
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(multiply(add(4, 3), const_2), 8) | | x + 3 | – | 4 - x | = | 8 + x | how many solutions zwill this equation have ? | "z = | x | = x when x > = 0 ( x is either positive or 0 ) | x | = - x when x < 0 ( note here that you can put the equal to sign here as well x < = 0 because if x = 0 , | 0 | = 0 = - 0 ( all are the same ) so the ' = ' sign can be put with x > 0 or with x < 0 . we usually put it with ' x > 0 ' for consistency . a" | a = 4 + 3
b = a * 2
c = b / 8
|
a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84 | e | divide(1.26, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 1.26 more than jane , what was the original price of the dish ? | the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 1.26 - - > 15 p - 13.5 p = 126 - - > p = 84 . answer : e . | a = 15 / 100
b = 10 / 100
c = 1 - b
d = 15 / 100
e = c * d
f = a - e
g = 1 / 26
|
a ) 5 / 24 , b ) 6 / 24 , c ) 7 / 24 , d ) 8 / 24 , e ) 9 / 24 | a | subtract(divide(3, 8), divide(3, 18)) | if 3 / p = 8 & 3 / q = 18 then p - q = ? | "p = 3 / 8 , q = 3 / 18 = > q = 1 / 6 therefore p - q = ( 3 / 8 ) - ( 1 / 6 ) = 5 / 24 answer : a" | a = 3 / 8
b = 3 / 18
c = a - b
|
a ) 42 , b ) 70 , c ) 147 , d ) 165 , e ) 315 | c | multiply(multiply(7, 3), 7) | a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 7 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "ans : 147 7 c 1 * 7 c 2 answer c )" | a = 7 * 3
b = a * 7
|
a ) 16000 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17000 | c | multiply(divide(subtract(18000, divide(multiply(18000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 18000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "c 15000 let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 18000 - 10 / 100 ( 18000 ) = 18000 - 1800 = $ 16200 = > 1.08 x = 16200 = > x = 15000" | a = 18000 * 10
b = a / 100
c = 18000 - b
d = 100 + 8
e = c / d
f = e * 100
|
a ) 13 , b ) 10 , c ) 16 , d ) 50 , e ) 23 | b | inverse(add(divide(const_1, 30), multiply(divide(const_1, 30), const_2))) | a is twice as fast as b . if b alone can do a piece of work in 30 days , in what time can a and b together complete the work ? | "option b explanation : a can do the work in 30 / 2 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 30 = ( 2 + 1 ) / 30 = 1 / 10 so a and b together can do the work in 10 days ." | a = 1 / 30
b = 1 / 30
c = b * 2
d = a + c
e = 1/(d)
|
a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability c that out of the two , one is white and one is blue ? | "the required probability c = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ." | a = 3 * 5
b = 3 + 5
c = 5 + 2
d = b * c
e = d / 2
f = a / e
|
a ) $ 200 , b ) $ 300 , c ) $ 400 , d ) $ 500 , e ) $ 1000 | e | multiply(multiply(5, 4), divide(50, subtract(5, 4))) | the total cost of a vacation was divided among 4 people . if the total cost of the vacation had been divided equally among 5 people , the cost per person would have been $ 50 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 4 * p c = 5 * p - 250 substituting the value of p from the first equation onto the second we get p = 250 . plugging in the value of p in the first equation , we get c = 1000 . which leads us to answer choice e" | a = 5 * 4
b = 5 - 4
c = 50 / b
d = a * c
|
a ) $ 110.66 , b ) $ 132.66 , c ) $ 116.66 , d ) $ 126.66 , e ) $ 118.66 | c | divide(subtract(multiply(add(2, 3), 177), multiply(multiply(2, 200), add(const_1, divide(18, const_100)))), multiply(3, add(const_1, divide(18, const_100)))) | by mixing two brands of coffee and selling the mixture at the rate of $ 177 per kg . a shopkeeper makes a profit of 18 % . if to every 2 kg of one brand costing $ 200 per kg , 3 kg of the other brand is added , then how much per kg does the other brand cost ? | c $ 116.66 let the cost of the brand be $ x per kg . c . p . of 5 kg = ( 2 * 200 + 3 * x ) = $ ( 400 + 3 x ) s . p of 5 kg = $ ( 5 * 177 ) = $ 885 [ 885 - ( 400 + 3 x ) ] / ( 400 + 3 x ) * 100 = 18 24250 - 150 x = 3600 + 27 x 177 x = 20650 = > x = 116 2 / 3 so , cost of the other brand = $ 116.66 . | a = 2 + 3
b = a * 177
c = 2 * 200
d = 18 / 100
e = 1 + d
f = c * e
g = b - f
h = 18 / 100
i = 1 + h
j = 3 * i
k = g / j
|
a ) 1280 , b ) 1145 , c ) 1040 , d ) 950 , e ) 260 | a | divide(add(subtract(1.75, divide(65, 100)), multiply(divide(divide(10, 100), 100), 250)), divide(divide(10, 100), 100)) | a courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof . what could be the weight in grams of a package for which the charge is $ 1.75 ? | "the charge is 65 cents for the first 250 grams . this leaves a charge of $ 1.75 - $ 0.65 = $ 1.10 the charge for the next 1000 grams is $ 1.00 which leaves a charge of $ 0.10 the weight is somewhere between 1250 and 1350 . the answer is a ." | a = 65 / 100
b = 1 - 75
c = 10 / 100
d = c / 100
e = d * 250
f = b + e
g = 10 / 100
h = g / 100
i = f / h
|
a ) $ 67.50 , b ) $ 54 , c ) $ 28.50 , d ) $ 12.50 , e ) $ 9.60 | a | multiply(6.75, 10) | johnny makes $ 6.75 per hour at his work . if he works 10 hours , how much money will he earn ? | 6.75 * 10 = 67.50 . answer is a . | a = 6 * 75
|
a ) 4 , b ) 8 , c ) 12 , d ) 18 , e ) 24 | e | divide(multiply(12, divide(add(6, 3), 3)), 3) | machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 3 hours . how many hours does it take machines x and y , working simultaneously , to produce 12 k bottles ? | "x rate = k / 6 y rate = k / 3 k / 6 + k / 3 = 12 k / t solving t = 24 answer e" | a = 6 + 3
b = a / 3
c = 12 * b
d = c / 3
|
a ) 33 , b ) 82 , c ) 99 , d ) 89 , e ) 27 | d | subtract(negate(67,78), multiply(subtract(45, 56), divide(subtract(45, 56), subtract(34, 45)))) | 34 , 45 , 56 , 67,78 , _ ? | answer : d | a = negate - (
|
a ) 19 , b ) 12 , c ) 11 , d ) 24 , e ) 114 | d | multiply(add(add(const_1, const_2), const_3), 4) | a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 4 days . b can do the work alone in ? | "suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 4 6 / x = 1 / 4 = > x = 24 so , b takes 12 hours to finish the work . answer : d" | a = 1 + 2
b = a + 3
c = b * 4
|
a ) 750 , b ) 277 , c ) 878 , d ) 450 , e ) 272 | a | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(3, 150))) | a man purchased 3 blankets @ rs . 100 each , 3 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 3 * 100 + 3 * 150 = 750 1500 – 750 = 750 answer : a" | a = 10 * 150
b = 3 * 100
c = 3 * 150
d = b + c
e = a - d
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | a | multiply(divide(subtract(18522, 17640), 17640), const_100) | an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 18522 / - in 3 years at the same rate of interest . find the rate percentage ? | "the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 18522 / - - 17640 / - = rs . 882 / - rate of interest = ( 882 / 17640 ) × ( 100 / 1 ) = > 8820 / 1764 = 5 % principal = amount / ( 1 + r / 100 ) n = 17640 / ( 1 + 5 / 100 ) 2 = 17640 / ( 21 / 20 × 21 / 20 ) = 17640 / ( 1.05 × 1.05 ) = 17640 / 1.1025 = 16000 thus the principal is rs . 16000 / - and the rate of interest is 5 % a" | a = 18522 - 17640
b = a / 17640
c = b * 100
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | e | divide(1000, const_10) | how many integers from 101 to 1000 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 9 = 9 second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 9 * 10 = 90 answer is e ." | a = 1000 / 10
|
a ) 3 , b ) 4 , c ) 12 , d ) 6 , e ) 35 | d | subtract(37, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 37 ? | "n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 37 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 6 more to make it 37 m + 36 answer - d" | a = 37 - reminder
|
a ) 40 , b ) 80 , c ) 20 , d ) 100 , e ) 60 | a | divide(12, subtract(divide(12, 10), 2)) | a train covers a distance of 12 km in 10 min . if it takes 2 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 2 = 40 m . answer : option a" | a = 12 / 10
b = a - 2
c = 12 / b
|
a ) 4 : 7 , b ) 7 : 3 , c ) 4 : 9 , d ) 4 : 4 , e ) 4 : 2 | b | divide(sqrt(49), sqrt(9)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 49 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 49 : â ˆ š 9 = 7 : 3 . answer : b" | a = math.sqrt(49)
b = math.sqrt(9)
c = a / b
|
a ) 65.9 , b ) 62.3 , c ) 62.7 , d ) 62.2 , e ) 62.9 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 10)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 90 / 100 ) = 59.375 mp = 65.9 answer : a" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 10
f = d / e
|
a ) 1 : 1 , b ) 2 : 1 , c ) 3 : 1 , d ) 5 : 1 , e ) 6 : 1 | a | power(1, 3) | two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 1 : 2 and 6 : 4 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | "vessel a = 300 gallons - - > milk = 100 , water = 200 ; vessel b = 500 gallons - - > milk = 300 , water = 200 ; vessel a + b = 800 gallons - - > milk = 400 , water 400 the ratio = 400 / 400 - - > 1 : 1 answer : a" | a = 1 ** 3
|
a ) 32.5 , b ) 58.7 , c ) 48 , d ) 65 , e ) 67.5 | b | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(30, const_100)))), subtract(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(30, const_100))))) | the price of a jacket is reduced by 10 % . during a special sale the price of the jacket is reduced another 30 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 10 % , therefore bringing down the price to $ 90 . 3 ) again it is further discounted by 30 % , therefore bringing down the price to $ 63 . 4 ) now 67.5 has to be added byx % in order to equal the original price . 63 + ( x % ) 63 = 100 . solving this eq for x , we get x = 58.7 ans is b" | a = 100 - 10
b = 100 - 10
c = 30 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 10
h = 100 - 10
i = 30 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 2 : 8 , b ) 2 : 1 , c ) 2 : 2 , d ) 2 : 4 , e ) 2 : 5 | e | divide(subtract(7, 5), 5) | the ratio between the sale price and the cost price of an article is 7 : 5 . what is the ratio between the profit and the cost price of that article ? | "explanation : let c . p . = rs . 5 x and s . p . = rs . 7 x . then , gain = rs . 2 x required ratio = 2 x : 5 x = 2 : 5 answer : e" | a = 7 - 5
b = a / 5
|
a ) 10.9 , b ) 10.7 , c ) 10.3 , d ) 10.8 , e ) 9.72 | e | divide(add(110, 160), multiply(add(60, 40), const_0_2778)) | two trains 110 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 110 + 160 = 270 m . required time = 270 * 9 / 250 = 243 / 25 = 9.72 sec . ' answer : e" | a = 110 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | b | divide(add(divide(multiply(divide(multiply(add(5, 3), 5), const_2), 3), 5), 4), add(divide(multiply(add(5, 3), 5), const_2), 4)) | thomas ' s age and matt ' s age are in the ratio 3 : 5 . two years ago the ratio of their ages was 5 : 9 . find the ratio of their ages 4 years hence ? | let ages of thomas and matt be x and y respectively . x / y = 3 / 5 = > x = 3 / 5 y ( x - 2 ) / ( y - 2 ) = 5 / 9 = > 9 x - 18 = 5 y - 10 = > 9 x = 5 y + 8 but x = 3 / 5 y 9 * 3 / 5 y = 5 y + 8 27 y = 25 y + 40 = > y = 20 years = > x = 3 * 20 / 5 = > 12 years ratio of their ages four years hence , ( 12 + 4 ) / ( 20 + 4 ) = 16 / 24 . = > 2 : 3 answer : b | a = 5 + 3
b = a * 5
c = b / 2
d = c * 3
e = d / 5
f = e + 4
g = 5 + 3
h = g * 5
i = h / 2
j = i + 4
k = f / j
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a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | b | divide(add(16, multiply(8, 4)), subtract(4, const_1)) | dan ' s age after 16 years will be 4 times his age 8 years ago . what is the present age of dan ? | let dan ' s present age be x . x + 16 = 4 ( x - 8 ) 3 x = 48 x = 16 the answer is b . | a = 8 * 4
b = 16 + a
c = 4 - 1
d = b / c
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a ) 428 , b ) 429 , c ) 430 , d ) 431 , e ) 432 | e | multiply(subtract(12, multiply(6, const_2)), multiply(18, 14)) | how much space , in cubic units , is left vacant when maximum number of 6 x 6 x 6 cubes are fitted in a rectangular box measuring 12 x 18 x 14 ? | "no of cubes that can be accommodated in box = ( 12 * 18 * 14 ) / ( 6 * 6 * 6 ) 12 * 16 in numerator can be perfectly divided by 6 * 6 in denominator . side with length 14 ca n ' t be perfectly divided by 6 and hence is the limiting factor . closet multiple of 6 less that 14 is 12 . so vacant area in cube = = 18 * 12 * ( 14 - 12 ) = 18 * 12 * 2 = 432 ans - e" | a = 6 * 2
b = 12 - a
c = 18 * 14
d = b * c
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a ) 2 days , b ) 3 days , c ) 4 days , d ) 5 1 / 3 days , e ) 6 days | d | inverse(add(inverse(16), multiply(const_2, inverse(16)))) | a work as fast as b . if b can complete a work in 16 days independently , the number of days in which a and b can together finish the work in ? | "ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 8 b ' s 1 day work = 1 / 16 a + b 1 day work = 1 / 8 + 1 / 16 = 3 / 16 = > 16 / 3 = 5 1 / 3 a and b can finish the work in 5 1 / 3 days answer is d" | a = 1/(16)
b = 1/(16)
c = 2 * b
d = a + c
e = 1/(d)
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a ) 100 , b ) 102 , c ) 104 , d ) 106 , e ) 108 | d | add(add(power(add(add(divide(subtract(subtract(11, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(11, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(11, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(11, const_10), const_2), const_4), const_2), const_2))) | the sum of 11 consecutive integers is 1111 . what is the greatest integer in the set ? | "let x be the first integer in the set , then x + 10 is the largest integer . the sum is : x + ( x + 1 ) + ( x + 2 ) + . . . + ( x + 10 ) = 11 x + 10 * 11 / 2 = 11 ( x + 5 ) then x + 5 = 101 x = 96 the largest integer in the set is 96 + 10 = 106 the answer is d ." | a = 11 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 11 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 11 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 11 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | divide(add(factorial(6), 6), add(factorial(6), 6)) | for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers q are there between 6 * + 2 and 6 * + 6 , inclusive ? | given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so it can not be a prime number , . . . the same way for all numbers between 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 , inclusive . which means that there are no primes q in this range . answer : a . | a = math.factorial(6)
b = a + 6
c = math.factorial(6)
d = c + 6
e = b / d
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a ) 4520 , b ) 4350 , c ) 4210 , d ) 4100 , e ) 4970 | e | subtract(5020, divide(1004, 20.08)) | 5020 − ( 1004 ÷ 20.08 ) = ? | "explanation : = 5020 − ( 1004 / 2008 × 100 ) = 5020 − 50 = 4970 option e" | a = 1004 / 20
b = 5020 - a
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a ) 20 min , b ) 25 min , c ) 26 min , d ) 36 min , e ) 30 min | d | add(divide(60, 2), divide(300, 50)) | it takes 60 identical printing machines 2 hours to print 300 cards . how much more time would it take 50 of these machines to print the same total ? | 50 % of 60 = 30 % or 0.30 2 hr x 60 min = 120 min 120 min x 0.30 = 36 min it will take 36 min longer to print them with only 50 printers . answer is d | a = 60 / 2
b = 300 / 50
c = a + b
|
a ) s . 83.33 , b ) s . 110 , c ) s . 112 , d ) s . 120 , e ) s . 150 | e | multiply(divide(const_100, 12), 18) | a 18 % stock yielding 12 % is quoted at : | "income of rs 12 on investment of rs 100 income of rs 18 on investment of ? = ( 18 * 100 ) / 12 = 150 answer : e" | a = 100 / 12
b = a * 18
|
a ) 1223441288 , b ) 2142579288 , c ) 2142339288 , d ) 2142339277 , e ) 1680579288 | e | multiply(23341379, power(add(const_4, const_1), const_4)) | ( 23341379 × 72 ) = ? | "23341379 × 72 = 23341379 ( 70 + 2 ) = ( 23341379 × 70 ) + ( 23341379 × 2 ) = 1633896530 + 46682758 = 1680579288 answer is e ." | a = 4 + 1
b = a ** 4
c = 23341379 * b
|
a ) 1 , b ) 10 , c ) 11 , d ) 19 , e ) none of these | a | gcd(1802, 1203) | the maximum number of student amoung them 1802 pens and 1203 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | solution required number of student = h . c . f of 1802 and 1203 = 1 . answer a | a = math.gcd(1802, 1203)
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a ) 2 , b ) 4 , c ) 3 , d ) 5 , e ) 1 | c | divide(150, 25) | how many of the positive factors of 25 , 150 and how many common factors are there in numbers ? | "factors of 25 - 1 , 5 , 25 factors of 150 - 1 , 2 , 3 , 5 , 6 , 10 , 15 , 25 , 30 , 50 , 75 , 150 comparing both , we have three common factors of 25 and 150 - 1 , 5,25 answer ( c )" | a = 150 / 25
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a ) 66.55 % , b ) 66.66 % , c ) 66 % , d ) 55 % , e ) 66.33 % | b | divide(const_100, divide(21, subtract(35, 21))) | if the cost price of 35 chocolates is equal to the selling price of 21 chocolates , the gain percent is : | "explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 21 chocolates = rs . 21 ; s . p . of 21 chocolates = rs . 35 . . ' . gain % = 14 * 100 / 21 = 66.66 % answer : b" | a = 35 - 21
b = 21 / a
c = 100 / b
|
['a ) 766 cm 2', 'b ) 535 cm 2', 'c ) 285 cm 2', 'd ) 324 cm 2', 'e ) 235 cm 2'] | c | divide(multiply(15, add(20, 18)), const_2) | nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 15 cm ? | area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 15 ) = 285 cm 2 answer : c | a = 20 + 18
b = 15 * a
c = b / 2
|
a ) 12 , b ) 18 , c ) 22 , d ) 28 , e ) 36 | c | multiply(11, const_2) | if ( 10 ^ 4 * 3.456789 ) ^ 11 is written as a single term , how many digits would be to the right of the decimal place ? | "3.456789 ^ 11 has 6 * 11 = 66 decimal places . 10 ^ 44 moves the decimal place to the right 44 places . ( 10 ^ 4 * 3.456789 ) ^ 11 has 66 - 44 = 22 digits after the decimal point . the answer is c ." | a = 11 * 2
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a ) 288 , b ) 300 , c ) 889 , d ) 276 , e ) 234 | b | subtract(multiply(speed(300, 18), 36), 300) | a 300 m long train crosses a platform in 36 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 ) / 36 = 50 / 3 3 x + 900 = 1800 = > x = 300 m . answer : b" | a = speed * (
b = a - 36
|
a ) 70.2 , b ) 50.1 , c ) 75.2 , d ) 78.4 , e ) 52.5 | e | divide(add(add(multiply(85, 3), multiply(60, 4)), 30), add(3, 4)) | the average expenditure of a labourer for 3 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income is | "income of 3 months = ( 3 × 85 ) – debt = 255 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 525 average monthly income = 525 ÷ 10 = 52.5 answer e" | a = 85 * 3
b = 60 * 4
c = a + b
d = c + 30
e = 3 + 4
f = d / e
|
a ) a ) 10 , b ) b ) 25 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | d | subtract(multiply(16, 62.5), multiply(subtract(16, const_1), 63)) | the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 62.5 . when one student dropped the class , the average of the remaining scores increased to 63.0 . what is the quiz score of the student who dropped the class ? | total score of 16 students is 16 * 62.50 = 1000 total score of 15 students is 15 * 63 = 945 so , the score of the person who left is 55 ( 1000 - 945 ) answer will be ( d ) | a = 16 * 62
b = 16 - 1
c = b * 63
d = a - c
|
a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(const_10, add(const_3, const_4)))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5))) | how many positive integers w between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e . | a = 300 - 200
b = a + 1
c = 300 / 2
d = 200 / 2
e = c - d
f = e + 1
g = b - f
h = 300 / 3
i = 200 / 3
j = h - i
k = j + 1
l = 3 + 4
m = 10 + l
n = k - m
o = g - n
p = 300 / 5
q = 200 / 5
r = p - q
s = r + 1
t = 10 + 5
u = s - t
v = o - u
|
a ) 37.5 % , b ) 37.6 % , c ) 38.5 % , d ) 17.1 % , e ) 37.2 % | d | multiply(subtract(divide(subtract(const_100, 25), 64), const_1), const_100) | the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 25 % ? | "let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 75 gain % = 11 / 64 * 100 = 17.1 % . answer : d" | a = 100 - 25
b = a / 64
c = b - 1
d = c * 100
|
a ) 3 , b ) 2 , c ) 9 , d ) 8 , e ) 7 | b | divide(subtract(subtract(add(add(divide(40, const_4), divide(40, const_4)), multiply(divide(40, 8), 5)), 40), const_1), const_2) | one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , and 5 / 8 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are 40 workers in the palabras bookstore , how many of them have read both books ? | there are total 40 workers . one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , so 10 have read saramago . 5 / 8 of the workers have read the latest book by h . kureishi . so ( 5 / 8 ) * 40 = 25 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books , 20 - b have read saramago but not kureishi . so , ( 10 - b - 1 ) have read neither . total = n ( a ) + n ( b ) - both + neither 40 = 10 + 25 - b + ( 10 - b - 1 ) b = 2 answer ( b ) | a = 40 / 4
b = 40 / 4
c = a + b
d = 40 / 8
e = d * 5
f = c + e
g = f - 40
h = g - 1
i = h / 2
|
a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 60 | e | divide(subtract(75, 57), subtract(const_1, divide(70, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 70 percent of books that were loaned out are returned and there are 57 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "i did n ' t understand how did we get 100 ? total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 70 % = 30 % of books that were loaned out are not returned . now , there are 57 books , thus 76 - 68 = 7 books are not returned . { loaned out } * 0.30 = 18 - - > { loaned out } = 60 . answer : e ." | a = 75 - 57
b = 70 / 100
c = 1 - b
d = a / c
|
a ) 18 sec , b ) 15 sec , c ) 4 sec , d ) 16 sec , e ) 12 sec | a | multiply(const_3600, divide(divide(330, const_1000), add(60, 6))) | a train 330 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 330 * 3 / 55 ] sec = 18 sec answer : a" | a = 330 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 45 % , b ) 23 % , c ) 20 % , d ) 60 % , e ) 56 % | c | subtract(50, 40) | if the selling price of 50 articles is equal to the cost price of 40 articles , then the loss or gain percent is : | "c c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 40 . loss % = 10 / 50 * 100 = 20 %" | a = 50 - 40
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a ) s . 147 , b ) s . 148 , c ) s . 144 , d ) s . 189 , e ) s . 192 | e | add(divide(136, subtract(const_1, divide(15, const_100))), multiply(divide(136, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | a shopkeeper loses 15 % , if an article is sold for rs . 136 . what should be the selling price of the article to gain 20 % ? | "given that sp = rs . 136 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 136 ) / 85 = rs . 160 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 160 * 120 ) / 100 = rs . 192 answer : e" | a = 15 / 100
b = 1 - a
c = 136 / b
d = 15 / 100
e = 1 - d
f = 136 / e
g = 20 / 100
h = f * g
i = c + h
|
a ) 95 , b ) 99 , c ) 26 , d ) 73 , e ) none of the above | d | add(multiply(const_10, add(subtract(15, 5), const_1)), 5) | what two - digit number is less than the sum of the square of its digits by 15 and exceeds their doubled product by 5 ? | "let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 15 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 15 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . . thus the two numbers can be 15 or 73 thus the answer is d" | a = 15 - 5
b = a + 1
c = 10 * b
d = c + 5
|
a ) 174.2 , b ) 212 , c ) 288.1 , d ) 296 , e ) 282.4 | d | multiply(divide(47, 2.54), divide(24, 1.5)) | on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 47 centimeters assuming that 1 - inch is 2.54 centimeters ? | 1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 47 cm . : 47 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 47 / ( 4.81 ) = 296 answer will be d . | a = 47 / 2
b = 24 / 1
c = a * b
|
a ) 87 kmph , b ) 97 kmph , c ) 72 kmph , d ) 144 kmph , e ) 120 kmph | d | divide(divide(800, const_1000), divide(20, const_3600)) | a train 800 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 800 / 20 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : d" | a = 800 / 1000
b = 20 / 3600
c = a / b
|
a ) 40 , b ) 45 , c ) 49 , d ) 53 , e ) 58 | c | multiply(14, multiply(14, divide(4, multiply(4, 4)))) | 4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 14 weavers in 14 days ? | "1 weaver can weave 1 mat in 4 days . 14 weavers can weave 14 mats in 4 days . 14 weavers can weave 49 mats in 14 days . the answer is c ." | a = 4 * 4
b = 4 / a
c = 14 * b
d = 14 * c
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a ) 10 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | d | divide(divide(7, 8), divide(1, 16)) | diana is painting statues . she has 7 / 8 of a gallon of paint remaining . each statue requires 1 / 16 gallon of paint . how many statues can she paint ? | number of statues = all the paint ÷ amount used per statue = 7 / 8 ÷ 1 / 16 = 7 / 8 * 16 / 1 = 112 / 8 = 14 . answer is d . | a = 7 / 8
b = 1 / 16
c = a / b
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a ) $ 200 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550 | a | multiply(divide(multiply(2.4, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.40) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.40 for every box , and if the university needs 2.4 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 400,000 / 4800 = 500 boxes total cost = 500 × $ 0.4 = $ 200 the answer is a ." | a = 1000 * 1000
b = 2 * 4
c = 20 * 20
d = c * 12
e = b / d
f = e * 0
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a ) 13426 , b ) 14236 , c ) 13436 , d ) 14263 , e ) 15263 | a | divide(79523, add(const_3, const_3)) | if pintu is coded as 79523 in a certain code language , how would you code jumbo in the same language ? | 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . jumbo is 13426 . . . answer : a | a = 3 + 3
b = 79523 / a
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a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) none of these | a | multiply(subtract(const_1, divide(9, 15)), 10) | suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 10 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | "the part of job that suresh completes in 9 hours = 9 ⁄ 15 = 3 ⁄ 5 remaining job = 1 - 3 ⁄ 5 = 2 ⁄ 5 remaining job can be done by ashutosh in 2 ⁄ 5 × 10 = 4 hours answer a" | a = 9 / 15
b = 1 - a
c = b * 10
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a ) 15 / 8 , b ) 3 / 8 , c ) 5 / 8 , d ) 7 / 15 , e ) 8 / 15 | a | divide(multiply(5, 3), add(3, 5)) | in an electric circuit , two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is 3 ohms and y is 5 ohms ? | "1 / r = 1 / x + 1 / y 1 / r = 1 / 3 + 1 / 5 = 8 / 15 r = 15 / 8 the answer is a ." | a = 5 * 3
b = 3 + 5
c = a / b
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a ) 5 , b ) 6 , c ) 10 , d ) 3 , e ) 8 | c | divide(52.416, 0.12510168) | when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 52.416 is divided by 0.12510168 ? | "for the 1 st no . there are 2 digits after decimal for the 2 nd no . there are 9 digits after decimal total no . of decimals = 11 req . no . of digits = ( n - 1 ) = ( 11 - 1 ) = 10 answer : c" | a = 52 / 416
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a ) $ 2000 , b ) $ 1500 , c ) $ 2500 , d ) $ 1800 , e ) $ 1600 | b | multiply(divide(3, add(add(2, 3), 4)), 4500) | a person want to give his money of $ 4500 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ? | b ' s share = 4500 * 3 / 9 = $ 1500 answer is b | a = 2 + 3
b = a + 4
c = 3 / b
d = c * 4500
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a ) 39 , b ) 41 , c ) 43 , d ) 45 , e ) 47 | c | subtract(180, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 100 and 180 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 180 , inclusive , is 81 . 3 * 34 = 102 and 3 * 60 = 180 so the number of multiples of 3 is 27 . 5 * 20 = 100 and 5 * 36 = 180 so the number of multiples of 5 is 17 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 12 = 180 so the number of multiples of 15 is 6 . thus the total number is 81 - 27 - 17 + 6 = 43 . the answer is c ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 180 - e
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a ) 88 , b ) 82 , c ) 84 , d ) 44 , e ) 54 | a | divide(rectangle_area(multiply(2, const_100), multiply(42, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | what is the least number of squares tiles required to pave the floor of a room 2 m 42 cm long and 1 m 76 cm broad ? | "length of largest tile = h . c . f . of 242 cm and 176 cm = 22 cm . area of each tile = ( 22 x 22 ) cm 2 . required number of tiles = 242 x 176 / ( 22 ^ 2 ) = 88 . answer : a" | a = 2 * 100
b = 42 * 100
c = rectangle_area / (
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a ) $ 50 , b ) $ 42 , c ) $ 60 , d ) $ 100 , e ) $ 90 | b | multiply(divide(2, add(3, 2)), 105) | rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 105 . what is the share of rahul ? | "rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 105 * 2 / 5 = $ 42 answer is b" | a = 3 + 2
b = 2 / a
c = b * 105
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a ) 1 / 11 , b ) 2 / 9 , c ) 2 / 11 , d ) 2 / 3 , e ) 7 / 9 | c | divide(subtract(12, 9), 9) | a number , x is chosen at random from the set of positive integers less than 12 . what is the probability that ( 9 / x ) > x ? | "number x has to be chosen from numbers 1 - 11 ( 9 / x ) > x = > 9 > x ^ 2 = > x ^ 2 - 9 < 0 x can have 2 values only 1 , 2 therefore , probability = 2 / 11 answer c" | a = 12 - 9
b = a / 9
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a ) 3999 , b ) 7799 , c ) 2500 , d ) 7500 , e ) 2912 | d | divide(multiply(600, const_100), 2) | a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 600 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 600 b capital = 45000 * 1 / 6 = 7500 . answer : d" | a = 600 * 100
b = a / 2
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a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) none of these | a | subtract(427398, multiply(floor(divide(427398, 15)), 15)) | what least number must be subtracted from 427398 so that remaining number is divisible by 15 | "explanation : on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted answer : option a" | a = 427398 / 15
b = math.floor(a)
c = b * 15
d = 427398 - c
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a ) 4 ^ 12 , b ) 4 ^ 35 , c ) 17 ( 4 ^ 5 ) , d ) 8 ^ 12 , e ) 7 ( 4 ^ 5 ) | c | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of 4 ^ 5 + 4 ^ 7 ? | "4 ^ 5 + 4 ^ 7 = 4 ^ 5 ( 1 + 4 ^ 2 ) = 4 ^ 5 * 17 answer c" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
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a ) 119 , b ) 187 , c ) 181 , d ) 120 , e ) 891 | d | divide(multiply(divide(multiply(141, add(const_4, const_4)), 3), 32.08), const_100) | if 3 eighth of a number is 141 . what will be the approximately value of 32.08 % of this number ? | explanation : x * 3 / 8 = 141 = > x = 376 376 * 32.08 / 100 = 120 answer : d | a = 4 + 4
b = 141 * a
c = b / 3
d = c * 32
e = d / 100
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a ) fifty nine , b ) 57 , c ) 55 , d ) 58 , e ) 60 | a | divide(divide(add(floor(divide(subtract(multiply(4, 1000), 1), 17)), const_1), const_2), const_2) | the number of positive integers valued pairs ( x , y ) ssatisfying 4 x - 17 y = 1 and x < = 1000 ( x is less than or equal to 1000 ) is | 4 x - 17 y = 1 y = ( 4 x - 1 ) / 17 x has to be a positive integer such that 1 < = x < = 1000 ( 4 x - 1 ) is a factor of 17 so that y is also a positive integer so we need to know how many numbers in the set { 37 , 1115 , . . . . , 3999 } are divisible by 17 lets try to observe the pattern 1 . 3 mod 17 = 3 2 . 7 mod 17 = 7 3 . 11 mod 17 = 11 4 . 15 mod 17 = 15 5 . 19 mod 17 = 2 6 . 23 mod 17 = 6 7 . 27 mod 17 = 10 8 . 31 mod 17 = 14 9 . 35 mod 17 = 1 10 . 39 mod 17 = 5 11 . 43 mod 17 = 9 12 . 47 mod 17 = 13 13.51 mod 17 = 0 14 . 55 mod 17 = 4 15 . 59 mod 17 = 8 16 . 63 mod 17 = 12 17 . 67 mod 17 = 16 . . . and after this the remainders will repeat ( this is no co - incidence that the cyclicity is also 17 ) so the right choices are x = 1313 + 1713 + 17 + 17 , . . . . . the number of such numbers is 59 ( highest one being 999 ) a | a = 4 * 1000
b = a - 1
c = b / 17
d = math.floor(c)
e = d + 1
f = e / 2
g = f / 2
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a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 4 | e | divide(const_1, subtract(divide(const_1, 2), subtract(divide(const_1, 2), divide(const_1, 4)))) | a can do a piece of work in 4 hours ; b and c together can do it in 2 hours , while a and c together can do it in 2 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 / 2 ; ( a + c ) ' s 1 hour ' s work = 1 / 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 4 + 1 / 2 ) = 3 / 4 . b ' s 1 hour ' s work = ( 3 / 4 - 1 / 2 ) = 1 / 4 . therefore a alone will take 4 hours to do the work . e" | a = 1 / 2
b = 1 / 2
c = 1 / 4
d = b - c
e = a - d
f = 1 / e
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a ) 700 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(40, const_100))) | a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 40 % ? | "cost price = rs . 500 profit = 40 % of 500 = rs . 200 selling price = cost price + profit = 500 + 200 = 700 answer : a" | a = 40 / 100
b = 500 * a
c = 500 + b
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a ) 25 , b ) 75 , c ) 125 , d ) 150 , e ) 175 | b | add(add(add(add(add(multiply(multiply(5, 2), 2), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) has two appearances of 3 ( in 12 ^ 2 ) and no appearances of 5 . thus n must include at least 3 * 5 ^ 2 = 75 the answer is b ." | a = 5 * 2
b = a * 2
c = 5 * 2
d = c * 2
e = b + d
f = 5 * 2
g = f * 2
h = e + g
i = h + 2
j = i + 4
k = j + 4
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a ) 2532 , b ) 2552 , c ) 6524 , d ) 8827 , e ) 9512 | d | subtract(multiply(56,000, power(add(const_1, divide(9, const_100)), 20)), 56,000) | find the compound interest on $ 56,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 56000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 56000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 64827 . ci . = $ ( 64827 - 56000 ) = $ 8827 . answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 56 * 0
e = d - 56
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a ) 22 % , b ) 32 % , c ) 34 % , d ) 64 % , e ) 36 % | d | subtract(power(4, const_2), power(5, const_2)) | the radius of the two circular fields is in the ratio 5 : 4 the area of the first field is what percent greater than the area of the second ? | "r = 5 ï € r 2 = 25 r = 4 ï € r 2 = 16 25 ï € â € “ 9 ï € 100 - - - - ? = > 36 % . answer : d" | a = 4 ** 2
b = 5 ** 2
c = a - b
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a ) $ 50 , b ) $ 40 , c ) $ 60 , d ) $ 100 , e ) $ 90 | d | multiply(divide(2, add(3, 2)), 250) | rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 250 . what is the share of rahul ? | rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 250 * 2 / 5 = $ 100 answer is d | a = 3 + 2
b = 2 / a
c = b * 250
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a ) 6 only , b ) 6 and 12 both , c ) 12 only , d ) 6 or 12 only , e ) none | b | add(multiply(6, const_100), multiply(2, 6)) | if n is a natural number , then ( 6 n 2 + 6 n ) is always divisible by ? | "( 6 n 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . b )" | a = 6 * 100
b = 2 * 6
c = a + b
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a ) 13 , b ) 12 , c ) 16 , d ) 17 , e ) 18 | b | add(divide(subtract(multiply(floor(divide(46, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1) | how many numbers from 10 to 46 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 . 12 numbers . 10 / 3 = 3 and 46 / 3 = 15 = = > 15 - 3 = 12 . therefore 12 digits b )" | a = 46 / 3
b = math.floor(a)
c = b * 3
d = 10 / 3
e = math.floor(d)
f = e + 1
g = f * 3
h = c - g
i = h / 3
j = i + 1
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a ) 732 , b ) 990 , c ) 1098 , d ) 1305 , e ) 1605 | b | subtract(subtract(multiply(366, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | the total number of digits used in numbering the pages of a book having 366 pages is | "explanation : total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 267 ) = ( 9 + 180 + 801 ) = 990 . answer : b" | a = 366 * 3
b = 100 - 1
c = a - b
d = 10 - 1
e = c - d
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a ) 45 , b ) 80 , c ) 85 , d ) 55 , e ) 40 | c | divide(add(252, const_1), const_2) | the sum of three consecutive numbers is 252 . the greatest among these three number is : | "let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 252 3 x = 252 x = 83 greatest number , ( x + 2 ) = 85 answer : c" | a = 252 + 1
b = a / 2
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a ) 54 , b ) 66 , c ) 80 , d ) 45 , e ) 96 | d | multiply(15, divide(divide(36, 2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 15 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "d 45 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 18 mph = 45 miles" | a = 36 / 2
b = a / 6
c = 15 * b
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a ) 305 , b ) 5 , c ) 365 , d ) 405 , e ) 495 | b | add(add(3, 7), 7) | how many 3 - digit even numbers are possible such that if one of the digits is 4 , the next / succeeding digit to it should be 7 ? | 470 , 472 , 474 , 476 , and 478 , so total 5 . hence option b . | a = 3 + 7
b = a + 7
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a ) 330 , b ) 110 , c ) 120 , d ) 140 , e ) 150 | a | divide(multiply(600, subtract(const_100, 45)), const_100) | in a public show 45 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | 75 % of 600 = 45 / 100 × 600 = 270 therefore , the number of vacant seats = 600 - 270 = 330 . answer : a | a = 100 - 45
b = 600 * a
c = b / 100
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a ) 4 , b ) 5 , c ) 6 , d ) 3 , e ) 2 | a | subtract(add(multiply(divide(1, 4), 10), multiply(divide(1, 5), 10)), divide(5, 10)) | if ( 1 / 5 ) ^ m * ( 1 / 4 ) ^ 2 = 1 / ( ( 10 ) ^ 4 ) , then m = | 1 / 5 ^ m * 1 / 4 ^ 2 = 1 / 2 ^ 4 * 5 ^ 4 2 ^ 4 * 5 ^ 4 / 2 ^ 4 = 5 ^ m or 5 ^ 4 = 5 ^ m therefore m = 4 a | a = 1 / 4
b = a * 10
c = 1 / 5
d = c * 10
e = b + d
f = 5 / 10
g = e - f
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