options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21 | b | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3)) | a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction w of the sum of the 21 numbers in the list ? | "this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction w of n to the total would be 4 a / 24 a or 1 / 6 answer b" | a = 1 * 1
b = 20 / 4
c = b + 21
d = c / 4
e = d * 2
f = e - 4
g = f - 3
h = a / g
|
a ) $ 18,714 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 12,300 | a | divide(subtract(20000, divide(add(multiply(4, subtract(16000, 20000)), multiply(5, subtract(25000, 20000))), subtract(16, add(5, 4)))), const_1000) | the average salary of 16 people in the shipping department at a certain firm is $ 20000 . the salary of 5 of the employees is $ 25000 each and the salary of 4 of the employees is $ 16000 each . what is the average salary of the remaining employees ? | total salary . . . 16 * 20 k = 320 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 7 emp sal = 320 k - 125 k - 64 k = 131 k average = 131 k / 7 = 18714 ans : a | a = 16000 - 20000
b = 4 * a
c = 25000 - 20000
d = 5 * c
e = b + d
f = 5 + 4
g = 16 - f
h = e / g
i = 20000 - h
j = i / 1000
|
a ) 19 , b ) 19.7 , c ) 21.3 , d ) 31.5 , e ) 34.7 | e | subtract(40, add(multiply(divide(30, const_100), divide(const_100, 6)), divide(30, const_100))) | jerry went to a shop and bought things worth rs . 40 , out of which 30 % went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 40 given that out of this rs . 40 , 30 % is given as tax = > total tax incurred = 30 % = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % β΄ ( 40 β 30 / 100 β x ) 6 / 100 = 30 / 100 β 6 ( 40 β 0.3 β x ) = 30 β ( 40 β 0.3 β x ) = 5 β x = 40 β 0.3 β 5 = 34.7 e )" | a = 30 / 100
b = 100 / 6
c = a * b
d = 30 / 100
e = c + d
f = 40 - e
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | c | divide(add(142, 110), 14) | a no . when divided by 142 gives a remainder 110 , what remainder will be obtainedby dividingthe same no . 14 ? | 142 + 110 = 252 / 14 = 18 ( remainder ) c | a = 142 + 110
b = a / 14
|
a ) 11 , b ) 10 , c ) 9 , d ) 12 , e ) 15 | c | add(8, divide(subtract(110, 20), add(25, 20))) | two stations a and b are 110 km apart on a straight line . one train starts from a at 6 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 6 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 110 45 x = 135 x = 3 . so , they meet at 9 a . m . answer : c" | a = 110 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) 100 min , b ) 150 min , c ) 200 min , d ) 250 min , e ) 300 min | c | multiply(add(const_1, const_4), 40) | one pipe can fill a tank four times as fast as another pipe . if together the two pipes can fill the tank in 40 minutes , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x minutes then , faster pipe will fill it in x / 4 minutes 1 / x + 4 / x = 1 / 40 5 / x = 1 / 40 x = 200 min answer is c" | a = 1 + 4
b = a * 40
|
a ) 20 inches , b ) 77 inches , c ) 16 inches , d ) 97 inches , e ) 66 inches | c | divide(add(multiply(6, const_12), 8), 5) | a scale 6 ft . 8 inches long is divided into 5 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 5 parts = 80 / 5 = 16 inches answer : c" | a = 6 * 12
b = a + 8
c = b / 5
|
['a ) 15 cm', 'b ) 20 cm', 'c ) 25 cm', 'd ) 30 cm', 'e ) none of these'] | d | divide(multiply(150, const_2), 10) | the area of rhombus is 150 cm square . the length of one of the its diagonals is 10 cm . the length of the other diagonal is : | explanation : we know the product of diagonals is 1 / 2 * ( product of diagonals ) let one diagonal be d 1 and d 2 so as per question 1 / 2 β d 1 β d 2 = 150 1 / 2 β 10 β d 2 = 150 d 2 = 150 / 5 = 30 option d | a = 150 * 2
b = a / 10
|
a ) 3 / 2 , b ) 4 / 3 , c ) 5 / 4 , d ) 6 / 5 , e ) 8 / 7 | b | inverse(add(add(inverse(4), inverse(3)), inverse(6))) | machine a can finish a job in 4 hours , machine Π² can finish the job in 3 hours , and machine Ρ can finish the job in 6 hours . how many hours will it take for a , b , and Ρ together to finish the job ? | "the combined rate is 1 / 4 + 1 / 3 + 1 / 6 = 9 / 12 of the job per hour . the time to complete the job is 12 / 9 = 4 / 3 hours . the answer is b ." | a = 1/(4)
b = 1/(3)
c = a + b
d = 1/(6)
e = c + d
f = 1/(e)
|
a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 58.66 , d ) rs . 55.33 , e ) none of the above | c | divide(add(multiply(10, 55), multiply(5, 66)), add(10, 5)) | if 10 litres of an oil of rs . 55 per litres be mixed with 5 litres of another oil of rs . 66 per litre then what is the rate of mixed oil per litre ? | 55 * 10 = 550 66 * 5 = 330 880 / 15 = 58.66 answer : c | a = 10 * 55
b = 5 * 66
c = a + b
d = 10 + 5
e = c / d
|
a ) rs . 8400 , b ) rs . 11,900 , c ) rs . 13,600 , d ) rs . 14,700 , e ) none of these | d | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , a receives : | explanation : let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . a ' s share = rs . ( 35000 x 21 / 50 ) = rs . 14,700 . answer is d | a = 10 * 5000
b = a - 5000
c = 4000 + 5000
d = b - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 10 * 5000
i = g / h
j = 3 + 4
k = j * 5000
l = i * k
m = l / 1000
n = math.floor(m)
o = n - 1
|
a ) 2160 , b ) 2350 , c ) 2000 , d ) 2300 , e ) 1800 | e | subtract(divide(multiply(add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 700), const_100), subtract(const_100, 10)), add(add(add(add(add(5000, 1500), 4500), 2500), 2000), 700)) | after spending rs . 5000 on rent , rs . 1500 on milk , rs . 4500 on groceries , rs . 2500 on childrens education rs . 2000 on petrol and rs . 700 on miscellaneous expenses , mr . kishore saved 10 % of his monthly salary . how much did he save in rs . ? | "explanation : total exp = 5000 + 1500 + 4500 + 2500 + 2000 + 700 = 16200 exp in % = 100 - 10 = 90 % , 16200 = 90 % saving = 10 % = 16200 x 10 / 90 = rs . 1800 answer : e" | a = 5000 + 1500
b = a + 4500
c = b + 2500
d = c + 2000
e = d + 700
f = e * 100
g = 100 - 10
h = f / g
i = 5000 + 1500
j = i + 4500
k = j + 2500
l = k + 2000
m = l + 700
n = h - m
|
['a ) 75', 'b ) 79', 'c ) 72', 'd ) 70', 'e ) 80'] | a | subtract(const_100, multiply(power(divide(50, const_100), const_2), const_100)) | if the radius of a circle decreased by 50 % its area is decreased by : | original area = Ο r ( power ) 2 new area = Ο ( r / 2 ) ( power ) 2 = ( Ο r ( power ) 2 ) / 4 reduction in area = [ ( Ο r ( power ) 2 - ( Ο r ( power ) 2 ) / 4 ] 2 = ( 3 Ο r ( power ) 2 ) / 4 reduction percent = ( ( 3 Ο r ( power ) 2 ) / 4 Γ 1 / ( Ο r ( power ) 2 Γ 100 ) % = 75 % answer is a . | a = 50 / 100
b = a ** 2
c = b * 100
d = 100 - c
|
a ) 11 / 14 , b ) 13 / 18 , c ) 4 / 7 , d ) 3 / 7 , e ) 3 / 14 | a | divide(subtract(divide(2, add(1, 2)), subtract(divide(2, 9), multiply(divide(2, 9), divide(3, 4)))), divide(subtract(9, 2), 9)) | when 2 / 9 of the votes on a certain resolution have been counted , 3 / 4 of those counted are in favor of the resolution . what fraction e of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution ? | if we use variable for total votes there will be too many fractions to manipulate with , so pick some smart # : let set total # of votes is 18 . 2 / 9 of the votes on a certain resolution have been counted - - > 4 counted and 18 - 4 = 14 votes left to be counted ; 3 / 4 of those counted are in favor of the resolution - - > 3 in favor and 1 against ; ratio of those who voted against to those who voted for to be 2 to 1 there should be total of 18 * 2 / 3 = 12 people who voted against , so in the remaining 14 votes there should be 12 - 1 = 11 people who voted against . thus e = 11 / 14 of the remaining votes must be against . answer : a . | a = 1 + 2
b = 2 / a
c = 2 / 9
d = 2 / 9
e = 3 / 4
f = d * e
g = c - f
h = b - g
i = 9 - 2
j = i / 9
k = h / j
|
a ) 64 days . , b ) 24 days . , c ) 52 days . , d ) 35 days . , e ) 28 days . | a | multiply(add(const_3, const_1), divide(16, const_3)) | a is thrice as good a work man as b and together they finish a piece of work in 16 days . the number of days taken by b alone to finish the work is : | "solution ( a β s 1 day β s work ) : ( b β s 1 day β s work ) = 3 : 1 . ( a + b ) ' s 1 day β s work = 1 / 16 divide 1 / 14 in the ratio 3 : 1 . β΄ b β s 1 day β s work = ( 1 / 16 x 1 / 4 ) = 1 / 64 hence , b alone can finish the work in 64 days . answer a" | a = 3 + 1
b = 16 / 3
c = a * b
|
a ) 25 % , b ) 50 % , c ) 75 % , d ) 125 % , e ) 150 % | d | multiply(subtract(power(add(const_1, divide(50, const_100)), const_2), const_1), const_100) | if a large pizza has a radius that is 50 % 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 1.5 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.5 * r ) ^ 2 = 2.25 * pi * r ^ 2 , an increase of 125 % . the answer is d ." | a = 50 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 5 hours , b ) 4 hours , c ) 3 hours , d ) 2 hours , e ) 1 hour | b | divide(multiply(10, 6), 15) | if it takes 10 kids 6 hours to wear out their teacher , how long would it take 15 kids ? | 10 * 6 = 15 * x x = 4 answer : b | a = 10 * 6
b = a / 15
|
a ) 0.1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | d | divide(multiply(1,000, 1,000), multiply(500, 1,000)) | the mass of 1 cubic meter of a substance is 500 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "500 kg - 1 cubic meter ; 500,000 g - 1 cubic meter ; 500,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 500,000 = 10 / 5 = 2 cubic centimeters . answer : d ." | a = 1 * 0
b = 500 * 1
c = a / b
|
a ) 53 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 13) | john found that the average of 15 numbers is 40 . if 13 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 53 option a" | a = 40 + 13
|
a ) 2 , b ) 5.5 , c ) 6.5 , d ) 3.2 , e ) 9.9 | e | divide(subtract(1000, 10), 100) | a straight line in the xy - plane has y - intercept of 10 . on this line the x - coordinate of the point is 100 and y - coordinate is 1000 then what is the slope of the line ? | eq of line = y = mx + c c = 10 x = 100 y = 100 m + 10 , substitute y by 1000 as given in question . 1000 = 100 m + 10 , m = 9.9 . correct option is e | a = 1000 - 10
b = a / 100
|
a ) 1280 , b ) 1250 , c ) 1320 , d ) 1340 , e ) 1350 | b | multiply(divide(multiply(5, const_1000), const_60), 15) | a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . what is the length of the bridge ( in metres ) ? | speed = 5 km / hr time = 15 minutes = 1 / 4 hour length of the bridge = distance travelled by the man = speed Γ time = 5 Γ 1 / 4 km = 5 Γ ( 1 / 4 ) Γ 1000 metre = 1250 metre answer is b | a = 5 * 1000
b = a / const_60
c = b * 15
|
a ) 22.4 , b ) 22 , c ) 20 , d ) 19.2 , e ) none of these | a | add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 40), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20))) | a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 40 % , what should be the sale price of the mixed tea ? | c . p . of mixture = 80 Γ 15 + 20 Γ 20 / 80 + 20 = 16 β΄ s . p . = ( 100 + 40 ) / 100 Γ 16 = 22.4 answer a | a = 80 * 15
b = 20 * 20
c = a + b
d = 80 + 20
e = c / d
f = e * 40
g = f / 100
h = 80 * 15
i = 20 * 20
j = h + i
k = 80 + 20
l = j / k
m = g + l
|
a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 | d | divide(const_2, 3) | a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability z that the first ball picked up is odd numbered ? | "answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability z is 2 / 3 . d" | a = 2 / 3
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a ) 10 , b ) 388 , c ) 37 , d ) 29 , e ) 2 | e | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 30), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 40), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 30 % and then increased by 40 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 30 % = 70 then increased 40 % on rs 70 = 70 + 28 = 98 net change in price = 100 - 98 = 2 answer : e" | a = 100 * 30
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 40
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 6 minutes , b ) 9 minutes , c ) 45 minutes , d ) 3 minutes , e ) 4 minutes | b | divide(45, add(const_2, const_3)) | two pipes a and b can fill a cistern in 37 Β½ min and 45 minutes respectively . both the pipes are opened . the cistern will be filled in just half an hour , if the pipe b is turned off after | if pipe b is turned off after x mins , then ( 2 * 30 ) / 75 + x / 45 = 1 x / 45 = 1 - 60 / 75 = 1 / 5 x = 45 / 5 = 9 mins answer : b | a = 2 + 3
b = 45 / a
|
a ) 1264 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(158, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 158 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ? | "a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 158 liters was transferred from container c to container b , then both the containers would have equal quantities of milk . 5 k - 158 = 3 k + 158 = > 2 k = 316 = > k = 158 the initial quantity of milk in a = 8 k = 8 * 158 = 1264 liters ." | a = 1/(2)
b = 62 / 5
c = 1 - b
d = a - c
e = 158 / d
|
a ) 0.12 , b ) 0.15 , c ) 0.17 , d ) 0.19 , e ) 0.25 | d | divide(add(multiply(6, const_3), 1), const_100) | two spies agreed to meet at a gas station between noon and 1 pm , but they have both forgotten the arranged time . each arrives at a random time between noon and 1 pm and stays for 6 minutes unless the other is there before the 6 minutes are up . assuming all random times are equally likely , what is the probability that they will meet within the hour ( noon to 1 pm ) ? | the probability that they do not meet is represented by the total of the areas of the two outer triangles in the figure below , which is 0.81 . so the probability of a meeting is 1 - 0.81 = 0 . 19 . correct answer d | a = 6 * 3
b = a + 1
c = b / 100
|
a ) 600 , b ) 480 , c ) 750 , d ) 650 , e ) 560 | d | subtract(subtract(multiply(45, 12), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 12 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 12 * 45 - 15 * 22 + 20 * 22 = 540 - 330 + 440 = 650 answer is d" | a = 45 * 12
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
|
a ) 20 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | add(10, add(const_3, const_4)) | how many digits are in ( 8 Γ 10 ^ 8 ) ( 10 Γ 10 ^ 10 ) ? | "the question simplfies to ( 8 Γ 10 ^ 8 ) ( 10 ^ 11 ) = > 8 * 10 ^ 19 = > will contain 19 zeros + 1 digit 8 = > 20 ans a" | a = 3 + 4
b = 10 + a
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a ) 100 , b ) 250 , c ) 750 , d ) 1200 , e ) 4687.5 | e | multiply(volume_rectangular_prism(50, 25, divide(6, add(const_10, const_2))), 7.5) | the water level in a rectangular swimming pool measuring 50 feet by 25 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 50 * 25 * 1 / 2 = 625 feet ^ 3 of water must be removed , which equals to 625 * 7.5 = 4687.5 gallons . answer : e ." | a = 10 + 2
b = 6 / a
c = volume_rectangular_prism * (
|
a ) 65 , b ) 68 , c ) 61 , d ) 55 , e ) 74 | b | divide(multiply(34, 40), 20) | if 34 men do a work in 40 days , in how many days will 20 men do it ? | "34 * 40 = 20 * x x = 68 days answer : b" | a = 34 * 40
b = a / 20
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a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 10 | d | divide(multiply(divide(multiply(20, add(add(1, 4), 10)), 10), 4), add(add(1, 4), 10)) | susan finds that she spends her free time during summer vacation between swimming , reading , and hanging out with friends in a ratio of 1 : 4 : 10 . how many hours did she spend reading if she hung out with her friends for 20 hours ? | the ratio is 1 hour swimming : 4 hours reading : 10 hours hanging out with friends . divide 20 by 10 to find 1 ` ` part ' ' of the ratio . 20 / 10 = 2 multiply this by 4 to find the hours spent reading . 2 * 4 = 8 the answer is d . | a = 1 + 4
b = a + 10
c = 20 * b
d = c / 10
e = d * 4
f = 1 + 4
g = f + 10
h = e / g
|
a ) 23 % , b ) 35 % , c ) 33 % , d ) 40 % , e ) 45 % | c | multiply(const_100, divide(multiply(divide(42, const_100), 11), add(3, 11))) | 3 ltr of water is added with 11 ltr of a solution containing 42 % of alcohol in the water . the % of alcohol in the new mixture is ? | "we have a 11 litre solution containing 42 % of alcohol in the water . = > quantity of alcohol in the solution = 11 Γ 42 / 100 now 3 litre of water is added to the solution . = > total quantity of the new solution = 11 + 3 = 14 percentage of alcohol in the new solution = 11 Γ 42 / 100 14 Γ 100 = 11 Γ 4210014 Γ 100 = 11 Γ 3 / 100 = 33 % c" | a = 42 / 100
b = a * 11
c = 3 + 11
d = b / c
e = 100 * d
|
a ) 320 $ , b ) 255 $ , c ) 420 $ , d ) 450 $ , e ) 480 $ | b | multiply(multiply(0.65, 30), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 30 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 30 ) = 21 + 234 = > 255 hence answer will be ( b )" | a = 0 * 65
b = a * 12
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a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | e | divide(1.44, subtract(96.12, floor(96.12))) | when positive integer x is divided by positive integer y , the remainder is 1.44 . if x / y = 96.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 1.44 - - > x = qy + 1.44 ; x / y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 1.44 - - > y = 12 . answer : e ." | a = math.floor(96, 12)
b = 96 - 12
c = 1 / 44
|
a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | d | inverse(subtract(divide(const_1, 3), divide(const_1, 4))) | bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne β s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many e hours does it currently take anne to clean the house on her own ? | lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 / a + 1 / b portion of the work which equals 1 / 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 / a + 2 / b which is equal to 1 / 3 ( as the work is completed in e = 3 hours ) solving these 2 equations we can find b as 12 so , d | a = 1 / 3
b = 1 / 4
c = a - b
d = 1/(c)
|
a ) 180 , b ) 200 , c ) 250 , d ) 270 , e ) 300 | d | divide(90, subtract(divide(const_2, const_3), subtract(const_1, divide(const_2, const_3)))) | sophia finished 23 of a book . she calculated that she finished 90 more pages than she has yet to read . how long is her book ? | "let x be the total number of pages in the book , then she finished 23 β
x pages . then she has x β 23 β
x = 13 β
x pages left . 23 β
x β 13 β
x = 90 13 β
x = 90 x = 270 . so answer is d ." | a = 2 / 3
b = 2 / 3
c = 1 - b
d = a - c
e = 90 / d
|
a ) 871 , b ) 881 , c ) 891 , d ) 904 , e ) 987 | b | multiply(multiply(9, 9), 11) | on dividing a n by 9 , remainder is 8 . the quotient obtained when divided by 11 , leaves remainder 9 . now the quotient when divided by 13 , leaves remainder 8 . find the remainder when when the n is divided by 1287 | take from the last step division suppose no . is n which is divided by 13 and remainder 8 = 13 n + 8 now 13 n + 8 will be the no . which is divided by 11 and remainder is 9 = [ 11 * ( 13 n + 8 ) ] + 9 now this no . is used for the very first step which is divider is 9 and remainder is 8 = { 9 * [ 11 * ( 13 n + 8 ) ] + 9 } + 8 solve and ans is 881 answer : b | a = 9 * 9
b = a * 11
|
a ) 17.5 , b ) 130 , c ) 175 , d ) 195 , e ) 220 | c | multiply(divide(multiply(7, 50), subtract(11, 7)), const_2) | a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 50 extra yellow balls and making ratio of white balls to yellow balls 7 / 11 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 50 ) = 7 : 11 - - > 11 x = 7 x + 350 - - > x = 87.5 . the total # of balls originally x + x = 87.5 + 87.5 = 175 . answer : c ." | a = 7 * 50
b = 11 - 7
c = a / b
d = c * 2
|
a ) 0.1 , b ) 0.5 , c ) 1 , d ) 1.2 , e ) 2 | b | divide(20, subtract(50, 10)) | two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 10 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 20 miles ahead of the black car , how many hours will it take the black car to overtake the red car ? | "option b 20 + 10 t = 50 t t = 0.5" | a = 50 - 10
b = 20 / a
|
['a ) 18', 'b ) 27', 'c ) 36', 'd ) 48', 'e ) 64'] | c | multiply(power(3, const_2), 4) | an equilateral triangle and three squares are combined as shown above , forming a shape of area 48 + 4 β 3 . what is the perimeter of the shape formed by the triangle and squares ? | triangle area = root ( 3 ) s ^ 2 / 4 area of 3 squares together = 3 s ^ 2 root ( 3 ) s ^ 2 / 4 + 3 s ^ 2 = 48 + 4 root ( 3 ) root ( 3 ) / 4 s ^ 2 = 4 root ( 3 ) s ^ 2 = 16 = > s = 4 there are 3 sides of each square = 3 ( 4 ) ( 3 ) = 36 option c | a = 3 ** 2
b = a * 4
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a ) rs . 49.17 , b ) rs . 58 , c ) rs . 54.17 , d ) rs . 55.33 , e ) none of the above | b | divide(add(multiply(10, 54), multiply(5, 66)), add(10, 5)) | if 10 litres of an oil of rs . 54 per litres be mixed with 5 litres of another oil of rs . 66 per litre then what is the rate of mixed oil per litre ? | "54 * 10 = 540 66 * 5 = 330 870 / 15 = 58 answer : b" | a = 10 * 54
b = 5 * 66
c = a + b
d = 10 + 5
e = c / d
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a ) 20 hr , b ) 52 hr , c ) 70 hr , d ) 66 hr , e ) 48 hr | c | multiply(add(add(multiply(const_2, const_2), const_2), const_1), 10) | a tank is filled by 3 pipes a , b , c in 10 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? | "suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / 2 and x / 4 hours respectively to fill the tank . 1 / x + 2 / x + 4 / x = 1 / 10 7 / x = 1 / 10 x = 70 hours answer is c" | a = 2 * 2
b = a + 2
c = b + 1
d = c * 10
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a ) 4 / 5 , b ) 7 / 11 , c ) 7 / 13 , d ) 7 / 15 , e ) 8 / 11 | a | divide(const_4, subtract(9, const_4)) | the difference between a positive proper fraction and its reciprocal is 9 / 20 . the fraction is | let the required fraction be x . then 1 - x = 9 x 20 1 - x 2 = 9 x 20 20 - 20 x 2 = 9 x 20 x 2 + 9 x - 20 = 0 20 x 2 + 25 x - 16 x - 20 = 0 5 x ( 4 x + 5 ) - 4 ( 4 x + 5 ) = 0 ( 4 x + 5 ) ( 5 x - 4 ) = 0 x = 4 / 5 a ) | a = 9 - 4
b = 4 / a
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a ) $ 50 , b ) $ 40 , c ) $ 60 , d ) $ 100 , e ) $ 900 | e | multiply(divide(2, add(3, 2)), 2250) | 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 $ 2250 . what is the share of rahul ? | "rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 2250 * 2 / 5 = $ 900 answer is e" | a = 3 + 2
b = 2 / a
c = b * 2250
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a ) 1 / 6 , b ) 3 / 10 , c ) 1 / 2 , d ) 5 / 6 , e ) 8 / 9 | a | divide(subtract(subtract(3, 1), 3), multiply(3, 3)) | a can complete the job in 3 hours and b can complete the same job in 3 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete | 1 / 6 = a | a = 3 - 1
b = a - 3
c = 3 * 3
d = b / c
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a ) 36 , b ) 66 , c ) 132 , d ) 264 , e ) 220 | e | divide(multiply(20, 396), 36) | hcf and lcm two numbers are 20 and 396 respectively . if one of the numbers is 36 , then the other number is ? | 20 * 396 = 36 * x x = 220 answer : e | a = 20 * 396
b = a / 36
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a ) 160 , b ) 180 , c ) 200 , d ) 220 , e ) none of these | c | divide(add(5, 5), subtract(divide(const_1, 5), divide(const_1, add(const_1, 5)))) | a number whose fifth part increased by 5 is equal to its fourth part diminished by 5 , is | "x / 5 + 5 = x / 4 - 5 = > x / 5 - x / 4 = 10 x / 20 = 10 = > x = 200 answer : c ." | a = 5 + 5
b = 1 / 5
c = 1 + 5
d = 1 / c
e = b - d
f = a / e
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a ) 187 m , b ) 250 m , c ) 876 m , d ) 150 m , e ) 267 m | b | multiply(divide(multiply(60, const_1000), const_3600), 15) | a train running at the speed of 60 km / hr crosses a pole in 15 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 15 ) m = 250 m . answer : b" | a = 60 * 1000
b = a / 3600
c = b * 15
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a ) 14 , b ) 18 , c ) 22 , d ) 26 , e ) 30 | b | add(3, add(4, add(6, add(5, multiply(5, const_2))))) | what is the least number which should be added to 2982 so that the sum is exactly divisible by 5 , 6 , 4 , and 3 ? | "l . c . m . of 5 , 6 , 4 and 3 = 60 . when dividing 2982 by 60 , the remainder is 42 . the number to be added = 60 - 42 = 18 . the answer is b ." | a = 5 * 2
b = 5 + a
c = 6 + b
d = 4 + c
e = 3 + d
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a ) 24 , b ) 36 , c ) 42 , d ) 72 , e ) 120 | c | subtract(add(subtract(multiply(7, 7), multiply(5, 3)), 12), const_4) | the ratio of toddlers to infants at a day care center is 7 to 3 . if 12 more infants join the day care to change the ratio to 7 to 5 , how many toddlers are there at this day care center ? | 5 x - 3 x = 12 so , 2 x = 6 hence no of toddlers is 6 * 7 = 42 answer will be ( c ) 42 | a = 7 * 7
b = 5 * 3
c = a - b
d = c + 12
e = d - 4
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(const_60, multiply(divide(360, 400), const_60)) | without stoppages , a train travels certain distance with an average speed of 400 km / h , and with stoppages , it covers the same distance with an average speed of 360 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 40 km less . time taken to cover 40 km = 40 Γ’ Β β 400 h = 1 Γ’ Β β 10 h = 1 Γ’ Β β 10 Γ£ β 60 min = 6 min answer d" | a = 360 / 400
b = a * const_60
c = const_60 - b
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a ) 10 mps , b ) 06 mps , c ) 09 mps , d ) 05 mps , e ) 11 mps | d | multiply(const_0_2778, 18) | express a speed of 18 kmph in meters per second ? | d 5 mps 18 * 5 / 18 = 5 mps | a = const_0_2778 * 18
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a ) 5 hours , b ) 6 hours , c ) 8 hours , d ) 16 hours , e ) 12 hours | c | divide(subtract(12, 10), 16) | two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 16 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 16 = 8 hours answer is c" | a = 12 - 10
b = a / 16
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a ) 8 , b ) 12 , c ) 18 , d ) 20 , e ) 24 | c | divide(subtract(add(multiply(10, divide(subtract(multiply(20, 30), multiply(30, 18)), subtract(20, 10))), multiply(20, subtract(30, divide(subtract(multiply(20, 30), multiply(30, 18)), subtract(20, 10))))), multiply(30, 15)), subtract(20, 15)) | each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 15 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes must be removed ? | "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 30 boxes , 6 are 10 - pound boxes and 24 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 15 , the ratio of 10 - pound boxes : 20 - pound boxes should be 1 : 1 . the number of 10 pound boxes remain the same so we still have 6 of them . to get a ratio of 1 : 1 , the number of 20 - pound boxes must be 6 . we need to remove 18 of the 20 - pound boxes . the answer is c ." | a = 20 * 30
b = 30 * 18
c = a - b
d = 20 - 10
e = c / d
f = 10 * e
g = 20 * 30
h = 30 * 18
i = g - h
j = 20 - 10
k = i / j
l = 30 - k
m = 20 * l
n = f + m
o = 30 * 15
p = n - o
q = 20 - 15
r = p / q
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a ) 17 sec , b ) 16 sec , c ) 18 sec , d ) 14 sec , e ) 12 sec | d | multiply(divide(280, multiply(72, const_1000)), const_3600) | a train 280 m long , running with a speed of 72 km / hr will pass a tree in ? | "speed = 72 * 5 / 18 = 20 m / sec time taken = 280 * 1 / 20 = 14 sec answer : d" | a = 72 * 1000
b = 280 / a
c = b * 3600
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a ) 3 , b ) 4 , c ) 8 , d ) 9 , e ) none of the above | c | divide(subtract(multiply(180, 4), multiply(60, 4)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 4 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 4 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | ( x + 4 ) Γ 60 + x Γ 120 + ( x β 4 ) Γ 180 / ( x + 4 ) + x + ( x β 4 ) = 100 β 360 x β 480 / 3 x = 100 β 60 x = 480 β x = 8 answer c | a = 180 * 4
b = 60 * 4
c = a - b
d = 60 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
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a ) 48 , b ) 22 , c ) 32 , d ) 36 , e ) 38 | a | divide(multiply(12, 36), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 12 and 36 ? | "prime factorization of 12 = 2 x 2 x 3 prime factorization of 36 = 2 x 2 x 3 x 3 gcf = 12 lcm = 36 sum = 48 . answer a ." | a = 12 * 36
b = a / 4
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a ) 126 , b ) 199 , c ) 198 , d ) 188 , e ) 122 | b | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 30), add(const_2, const_4)) | what is the sum of all the prime numbers greater than 30 but less than 50 ? | "required sum = ( 31 + 37 + 41 + 43 + 47 ) = 199 note : 1 is not a prime number answer b" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 30
t = 2 + 4
u = s + t
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a ) 3 , b ) 11 , c ) 8 , d ) 9 , e ) 10 | b | divide(add(16, 6), const_2) | in one hour , a boat goes 16 km along the stream and 6 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 16 + 6 ) kmph . = 11 kmph . answer b" | a = 16 + 6
b = a / 2
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a ) 865 , b ) 955 , c ) 1050 , d ) 765 , e ) 750 | d | divide(multiply(150, 155), const_4) | what is the sum of the integers from - 150 to 155 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 155 , a = - 150 , d = 1 hence , 155 = - 150 + ( n - 1 ) or n = 306 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 306 * ( - 150 + 155 ) / 2 sn = 306 * 5 / 2 = 765 answer : d" | a = 150 * 155
b = a / 4
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a ) 12.7 sec , b ) 13.9 sec , c ) 18.1 sec , d ) 17.1 sec , e ) 19.7 sec | b | divide(add(120, 150), multiply(70, const_0_2778)) | how long does a train 120 m long running at the speed of 70 km / hr takes to cross a bridge 150 m length ? | "speed = 70 * 5 / 18 = 19.4 m / sec total distance covered = 120 + 150 = 270 m . required time = 270 / 19.4 ' = 13.9 sec . answer : b" | a = 120 + 150
b = 70 * const_0_2778
c = a / b
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a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | c | add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of two metres is left from all sides of the boundary of the garden . what is the length of the garden ? | "between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 2 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 Γ£ β 2 ) + 2 + 2 = 26 meter answer is c ." | a = 12 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
|
a ) 40 , b ) 45 , c ) 50 , d ) 90 , e ) 2500 | c | divide(power(const_10, divide(4, const_2)), const_2) | a palindrome is a number that reads the same forward and backward , such as 121 . how many odd , 4 - digit numbers are palindromes ? | first recognize you only need to consider the first two digits ( because the second two are just the first two flipped ) there are 90 possibilities for the first two digits of a 4 digit number , 10 - 99 inclusive . everything starting with a 1 , 3,5 , 7,9 will be odd , which is 5 / 9 ths of the combinations . 5 / 9 * 90 = 50 answer : c | a = 4 / 2
b = 10 ** a
c = b / 2
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a ) 27 , b ) 37 , c ) 17 , d ) 47 , e ) 07 | b | multiply(3, 12) | in an aquarium there is 14 purple fish , 8 orange fish , 12 pink fish , 3 golden fish . how many fish are there in an aquarium ? | 14 + 8 + 12 + 3 = 37 . answer is b | a = 3 * 12
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a ) 500 , b ) 1500 , c ) 2500 , d ) 2000 , e ) 1000 | d | multiply(100, multiply(2, const_10)) | last year , for every 100 million vehicles that traveled on a certain highway , 100 vehicles were involved in accidents . if 2 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 ) | "to solve we will set up a proportion . we know that β 100 million vehicles is to 100 accidents as 2 billion vehicles is to x accidents β . to express everything in terms of β millions β , we can use 2,000 million rather than 2 billion . creating a proportion we have : 100 / 100 = 2,000 / x cross multiplying gives us : 100 x = 2,000 * 100 x = 20 * 100 = 2000 answer : d" | a = 2 * 10
b = 100 * a
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a ) 5 hours , b ) 40 hours , c ) 8 hours , d ) 20 hours , e ) 30 hours | d | divide(subtract(12, 10), 40) | two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 40 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 40 = 20 hours answer is d" | a = 12 - 10
b = a / 40
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a ) 12 , b ) 8 , c ) 4 , d ) 10 , e ) 16 | d | multiply(divide(const_10, const_2), const_2) | the population of locusts in a certain swarm doubles every two hours . if 4 hours ago there were 1000 locusts in the swarm , in approximately how many hours will the swarm population exceed 128000 locusts ? | - 4 hours : 1,000 - 2 hours : 2,000 now : 4,000 + 2 hours : 8,000 + 4 hours : 16,000 + 6 hours : 32,000 + 8 hours : 64,000 + 10 hours : 128,000 answer : d | a = 10 / 2
b = a * 2
|
['a ) 4.00 %', 'b ) 4.04 %', 'c ) 4.16 %', 'd ) 4.30 %', 'e ) 5 %'] | b | divide(subtract(multiply(add(const_100, 2), add(const_100, 2)), multiply(const_100, const_100)), const_100) | an error 2 % in excess ismade while measuring the side ofa square . the % of error in the calculated area of the square is ? | 100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 102 ) 2 - ( 100 ) 2 ] = ( 102 + 100 ) x ( 102 - 100 ) = 404 cm 2 . percentage error = 404 x 100 % = 4.04 % 100 x 100 b | a = 100 + 2
b = 100 + 2
c = a * b
d = 100 * 100
e = c - d
f = e / 100
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a ) 4 , b ) 7 , c ) 8 , d ) 11 , e ) 13 | e | add(add(add(7, 2), divide(5, 5)), divide(3, 3)) | if x is the product of the positive integers from 1 to 9 , inclusive , and if i , k , m , and p are positive integers such that x = 2 i 3 k 5 m 7 px = 2 i 3 k 5 m 7 p , then i + k + m + p = | "x = 9 ! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = ( 3 ^ 2 ) * ( 2 ^ 3 ) * 7 * ( 2 * 3 ) * 5 * 2 ^ 2 * 3 * 2 * 1 = 2 ^ 7 * 3 ^ 4 * 5 * 7 = 2 ^ i * 3 ^ k * 5 ^ m * 7 ^ p i + k + m + p = 7 + 4 + 1 + 1 = 13 answer e" | a = 7 + 2
b = 5 / 5
c = a + b
d = 3 / 3
e = c + d
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a ) 72 , b ) 96 , c ) 100 , d ) 120 , e ) 150 | c | divide(subtract(32, 22), subtract(divide(20, const_100), divide(10, const_100))) | of the diplomats attending a summit conference , 22 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ? | "{ total } = { french } + { russian } - { both } + { neither } { total } = 22 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 100 . answer : c ." | a = 32 - 22
b = 20 / 100
c = 10 / 100
d = b - c
e = a / d
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | c | subtract(10, 5) | a football team lost 5 yards and then gained 10 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 10 = 5 yards c" | a = 10 - 5
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a ) 470 , b ) 210 , c ) 465 , d ) 463 , e ) 485 | b | add(divide(divide(20, divide(divide(divide(divide(divide(20, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(20, divide(divide(divide(divide(divide(20, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 20 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 20 ( 20 + 1 ) / 2 = 20 ( 21 ) / 2 = 210 answer : option b" | a = 20 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 20 / e
g = f / 2
h = 20 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 20 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 17 % , b ) 10 % , c ) 21 % , d ) 16 % , e ) none | a | add(15, multiply(subtract(15, 10), divide(2, 5))) | weights of two friends ram and shyam are in the ratio 2 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | "solution : given ratio of ram and shayam ' s weight = 2 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 2 / 5 or , x = 17 % . answer : option a" | a = 15 - 10
b = 2 / 5
c = a * b
d = 15 + c
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | floor(divide(reminder(power(36, reminder(12, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 36 ^ 12 ? | "36 ^ 10 = 6 ^ 24 ( 6 ^ 2 ) = 6 * 6 = 36 ( 6 ^ 3 ) = 36 * 6 = . 16 ( 6 ^ 4 ) = . 16 * 6 = . . 96 ( 6 ^ 5 ) = . . 96 * 6 = . . 76 ( 6 ^ 6 ) = . . 76 * 6 = . . . 56 ( 6 ^ 7 ) = . . . . 56 * 6 = . . . . 36 if you see there is a pattern here in tens digits 3 , 1,9 , 7,5 , 3,1 and so on . . . continue the pattern up to 6 ^ 24 ( dont actually calculate full values ) and answer is e : 9" | a = 4 + 1
b = 36 ** reminder
c = reminder / (
d = math.floor(c, 100)
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a ) 1008 , b ) 1015 , c ) 488 , d ) 490 , e ) 590 | c | add(multiply(multiply(power(const_3, 8), power(8, const_4)), add(const_3, const_4)), 8) | the smallest number which when diminished by 8 , is divisible 8 , 16 , 24 , 32 and 40 is : | "required number = ( l . c . m . of 8 , 16 , 24 , 32 , 40 ) + 8 = 480 + 8 = 488 answer : option c" | a = 3 ** 8
b = 8 ** 4
c = a * b
d = 3 + 4
e = c * d
f = e + 8
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a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 7 , e ) e ) 8 | d | add(divide(subtract(multiply(floor(divide(15, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | how many numbers from 2 to 15 are exactly divisible by 2 ? | "2 / 2 = 1 and 15 / 2 = 7 7 - 1 = 6 6 + 1 = 7 numbers . answer : d" | a = 15 / 2
b = math.floor(a)
c = b * 2
d = 2 / 2
e = math.floor(d)
f = e + 1
g = f * 2
h = c - g
i = h / 2
j = i + 1
|
a ) 4691130840 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) none | a | multiply(469160, 9999) | calculate 469160 x 9999 = ? | "answer 469160 x 9999 = 469160 x ( 10000 - 1 ) = 4691600000 - 469160 = 4691130840 . option : a" | a = 469160 * 9999
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a ) 2 : 9 , b ) 2 : 3 , c ) 2 : 1 , d ) 2 : 2 , e ) 2 : 6 | b | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years , the ratio of the number of boys to the number of girls in the class is : | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) < = > k = 0.4 / 0.6 = 2 / 3 . required ratio = 2 / 3 : 1 = 2 : 3 . answer : b" | a = 15 - 8
b = 16 - 4
c = a / b
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a ) 2.35 % , b ) 5.95 % , c ) 4.35 % , d ) 5.33 % , e ) 6.33 % | a | multiply(divide(divide(subtract(950, 850), 850), 5), const_100) | at what rate percent on simple interest will rs . 850 amount to rs . 950 in 5 years ? | "100 = ( 850 * 5 * r ) / 100 r = 2.35 % answer : a" | a = 950 - 850
b = a / 850
c = b / 5
d = c * 100
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a ) 1 / 33 , b ) 2 / 85 , c ) 1 / 3 , d ) 56 / 85 , e ) 11 / 85 | d | multiply(multiply(multiply(divide(multiply(9, const_2), multiply(9, const_2)), divide(multiply(const_4, const_4), subtract(multiply(9, const_2), const_1))), divide(subtract(multiply(const_4, const_4), const_2), multiply(const_4, const_4))), divide(subtract(subtract(multiply(const_4, const_4), const_2), const_2), subtract(multiply(const_4, const_4), const_1))) | if 4 people are selected from a group of 9 married couples , what is the probability that none of them would be married to each other ? | if we are to select 4 people from 9 couples without any restriction , how many ways can we make the selection ? 18 ! / 4 ! 14 ! = 3060 if we are to select 4 people from 9 couples with restriction that no married couple can both make it to the group , only a representative ? 9 ! / 4 ! 5 ! = 126 but we know that to select a person from each couple , take 2 possibilities 126 * 2 * 2 * 2 * 2 = 2016 probability = desired / all possibilities = 2016 / 3060 = 56 / 85 answer : d | a = 9 * 2
b = 9 * 2
c = a / b
d = 4 * 4
e = 9 * 2
f = e - 1
g = d / f
h = c * g
i = 4 * 4
j = i - 2
k = 4 * 4
l = j / k
m = h * l
n = 4 * 4
o = n - 2
p = o - 2
q = 4 * 4
r = q - 1
s = p / r
t = m * s
|
a ) 3 , b ) 3.5 , c ) 3 , d ) 4.5 , e ) 5 | c | divide(subtract(198, 180), divide(subtract(222, 198), 4)) | joe went on a diet 4 months ago when he weighed 222 pounds . if he now weighs 198 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 180 pounds ? | 222 - 198 = 24 pounds lost in 4 months 24 / 4 = 6 , so joe is losing weight at a rate of 6 pounds per month . . . . in approximately how many months will he weigh 180 pounds ? a simple approach is to just list the weights . now : 198 lbs in 1 month : 192 lbs in 2 months : 186 lbs in 3 months : 180 lbs answer : c | a = 198 - 180
b = 222 - 198
c = b / 4
d = a / c
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a ) 6 , b ) 7 , c ) 8 , d ) 13 , e ) 10 | d | add(divide(40, 3), const_1) | how many integers are divisible by 3 between 10 ! and 10 ! + 40 inclusive ? | "d - 7 10 ! is divisible by 3 there are 12 numbers between 10 ! and 10 ! + 40 that are divisible by 3 . hence 13" | a = 40 / 3
b = a + 1
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a ) 500 , b ) 400 , c ) 300 , d ) 200 , e ) 100 | e | subtract(divide(800, const_2), 300) | if the perimeter of a rectangular stadium is 800 m , its length when its breadth is 300 m is ? | 2 ( l + 300 ) = 800 = > l = 100 m answer : e | a = 800 / 2
b = a - 300
|
a ) 7 : 8 , b ) 8 : 9 , c ) 4 : 6 , d ) 4 : 8 , e ) 5 : 8 | a | divide(add(multiply(6, 4), 4), add(add(multiply(6, 4), 4), 4)) | the ratio between the current age of a and b is 6 : 7 . if b is 4 years old than a . what will be the ratio of the ages of a and b after 4 years ? | a 7 : 8 if a β s age and b β s age will be 6 a years and 7 a years correspondingly then , 7 a β 6 a = 4 a = 4 needed ratio = ( 6 a + 4 ) : ( 7 a + 4 ) = 28 : 32 = 7 : 8 | a = 6 * 4
b = a + 4
c = 6 * 4
d = c + 4
e = d + 4
f = b / e
|
a ) 100 , b ) 250 , c ) 750 , d ) 5250 , e ) 5635 | d | multiply(multiply(multiply(56, 25), divide(1, const_2)), 7.5) | the milk level in a rectangular box measuring 56 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 56 * 25 * 1 / 2 = 700 feet ^ 3 of milk must be removed , which equals to 700 * 7.5 = 5250 gallons . answer : d ." | a = 56 * 25
b = 1 / 2
c = a * b
d = c * 7
|
a ) 122821 , b ) 281228 , c ) 281199 , d ) 62750 , e ) 128111 | d | divide(multiply(1, 501), const_4) | what is the sum of all even numbers from 1 to 501 ? | "explanation : 500 / 2 = 250 250 * 251 = 62750 answer : d" | a = 1 * 501
b = a / 4
|
a ) 450,000 , b ) 475,000 , c ) 500,000 , d ) 525,000 , e ) 550,000 | a | multiply(300, divide(subtract(2008, 2005), const_2)) | during 2005 , a company produced an average of 2,500 products per month . how many products will the company need to produce from 2006 through 2008 in order to increase its monthly average for the period from 2005 through 2008 by 300 % over its 2005 average ? | "company produced 12 * 2500 = 30,000 products in 2005 . if company produces x products from 2006 to 2008 , then total amount of product produced in 4 years ( 2005 through 2008 ) is x + 30,000 . the gives the average of ( x + 30,000 ) / 4 . this average needs to be 200 % higher than that in 2005 . in math terms , 30,000 + 300 % ( 30,000 ) = 120,000 . so : ( x + 30,000 ) / 4 = 120,000 x + 30,000 = 480,000 x = 450,000 the answer is a ." | a = 2008 - 2005
b = a / 2
c = 300 * b
|
a ) 1 , b ) 8 , c ) 12 , d ) 9 , e ) 7 | c | multiply(3, const_4) | how many quarters are equal to 3 dollars ? | "3 * 4 = 12 quarters answer : c" | a = 3 * 4
|
a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(60, const_2), const_2) | if k ^ 3 is divisible by 60 , what is the least possible value of integer k ? | "60 = 2 ^ 2 * 3 * 5 therefore k must include at least 2 * 3 * 5 = 30 . the answer is b ." | a = 60 / 2
b = a / 2
|
a ) 15 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | e | add(multiply(divide(12, 48), const_100), 20) | the purchase price of an article is $ 48 . in order to include 20 % of cost for overhead and to provide $ 12 of net profit , the markup should be | "cost price of article = 48 $ % of overhead cost = 20 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 20 = 45 % answer e" | a = 12 / 48
b = a * 100
c = b + 20
|
a ) 3 hr , b ) 3.10 hr , c ) 3.15 hr , d ) 3.20 hr , e ) 3.30 hr | d | multiply(divide(3, 4), inverse(add(divide(const_1, 10), divide(const_1, 8)))) | pipe a can fill a tank in 10 hrs and pipe b can fill it in 8 hrs . if both the pipes are opened in the empty tank . there is an outlet pipe in 3 / 4 th of the tank . in how many hours will it be fill 3 / 4 th of that tank ? | part filled a in 1 hr = ( 1 / 10 ) part filled b in 1 hr = ( 1 / 8 ) part filled by ( a + b ) together in 1 hr = ( 1 / 10 ) + ( 1 / 8 ) = 18 / 80 so , the tank will be full in 80 / 18 hrs . time taken to fill exact 3 / 4 th of the tank = ( 80 / 18 ) * ( 3 / 4 ) = 3.20 hrs answer : d | a = 3 / 4
b = 1 / 10
c = 1 / 8
d = b + c
e = 1/(d)
f = a * e
|
a ) 122443 , b ) 154546 , c ) 165454 , d ) 186545 , e ) 220030 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 30) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 30 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 30 = 220000 + 30 = 220030 e" | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 30
|
a ) 36 , b ) 77 , c ) 88 , d ) 99 , e ) 22 | a | divide(subtract(192, multiply(const_2, 60)), subtract(const_4, const_2)) | there are some rabbits and peacocks in a zoo . the total number of their heads is 60 and total number of their legs is 192 . find the number of total rabbits ? | let the number of rabbits and peacocks be ' r ' and ' p ' respectively . as each animal has only one head , so r + p = 60 - - - ( 1 ) each rabbit has 4 legs and each peacock has 2 legs . total number of legs of rabbits and peacocks , 4 r + 2 p = 192 - - - ( 2 ) multiplying equation ( 1 ) by 2 and subtracting it from equation ( 2 ) , we get = > 2 r = 72 = > r = 36 . answer : a | a = 2 * 60
b = 192 - a
c = 4 - 2
d = b / c
|
a ) 2 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | b | subtract(55, reminder(const_4.0, 11)) | when positive integer n is divided by 5 , the remainder is 2 . when n is divided by 11 , the remainder is 8 . what is the smallest positive integer k such that k + n is a multiple of 55 ? | "n = 5 p + 2 = 11 q + 8 n + 3 = 5 p + 5 = 11 q + 11 n + 3 is a multiple of 5 and 11 , so it is a multiple of 55 . the answer is b ." | a = 55 - reminder
|
a ) 24 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 48 km | e | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 10), add(subtract(10, 2), add(10, 2))) | a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 10 hour to row to a place and come back , how far is the place ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 10 2 x + 3 x = 240 x = 48 km answer is e" | a = 10 - 2
b = 10 + 2
c = a * b
d = c * 10
e = 10 - 2
f = 10 + 2
g = e + f
h = d / g
|
a ) 724 , b ) 804 , c ) 814 , d ) 844 , e ) none | c | divide(multiply(add(multiply(9, const_100), 2), multiply(15, const_100)), power(2, const_2)) | what is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ? | "solution length of largest tile = h . c . f . of 1517 cm & 902 cm = 41 cm . area of each tile = ( 41 x 41 ) cm 2 β΄ required number of tiles = [ 1517 x 902 / 41 x 41 ] = 814 . answer c" | a = 9 * 100
b = a + 2
c = 15 * 100
d = b * c
e = 2 ** 2
f = d / e
|
a ) 10.11 , b ) 9.625 , c ) 12.11 , d ) 13.11 , e ) 14.11 | b | add(8, const_1) | the average of first 8 prime numbers is ? | "sum of first 8 prime nos . = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77 average = 77 / 8 = 9.625 answer : b" | a = 8 + 1
|
a ) 1 / 10 , b ) 4 / 9 , c ) 1 / 2 , d ) 9 / 15 , e ) 2 / 3 | d | divide(multiply(choose(3, const_2), choose(3, const_2)), choose(add(3, 3), 2)) | from a group of 3 boys and 3 girls , 2 children are to be randomly selected . what is the probability that 1 boy and 1 girl will be selected ? | "the total number of ways to choose 2 children from 6 is 6 c 2 = 15 the number of ways to choose 1 boy and 1 girl is 3 * 3 = 9 p ( 1 boy and 1 girl ) = 9 / 15 the answer is d ." | a = math.comb(3, 2)
b = math.comb(3, 2)
c = a * b
d = 3 + 3
e = math.comb(d, 2)
f = c / e
|
a ) 6 , b ) 8 , c ) 9 , d ) 5 , e ) 21 | e | add(7, const_1) | the average of first five multiples of 7 is ? | "average = 7 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 105 / 5 = 21 . answer : e" | a = 7 + 1
|
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