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 ) 15 hrs , b ) 20 hrs , c ) 25 hrs , d ) 30 hrs , e ) 32 hrs | d | divide(const_1, add(divide(const_1, 50), divide(const_1, 75))) | two pipes a and b can fill a tank in 50 hour and 75 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | part filled by a in 1 hour = 1 / 50 part filled by b in 1 hour = 1 / 75 part filled by a + b in 1 hour = 1 / 50 + 1 / 75 = 1 / 30 both the pipes will together will fill the tank in 30 hours . answer is d | a = 1 / 50
b = 1 / 75
c = a + b
d = 1 / c
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(50, const_10) | how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 7 ? | "the numbers should be of the form 7 c + 3 . the minimum is 3 when c = 0 . the maximum is 45 when c = 6 . there are 7 such numbers . the answer is c ." | a = 50 / 10
|
a ) 35 , b ) 64 , c ) 66.67 , d ) 60 , e ) 84 | c | multiply(divide(add(multiply(divide(75, const_100), 160), multiply(subtract(const_1, divide(40, const_100)), 200)), add(160, 200)), const_100) | addison high school ’ s senior class has 160 boys and 200 girls . if 75 % of the boys attended the college and 40 % of the girls did not attend the college , what percentage of the total class attended college ? | number of boys attending the college : 0.75 * 160 = 120 number of girls not attending the college : 0.40 * 200 = 80 number of girls attending the college : 200 - 80 = 120 total attending = 120 + 120 = 240 total people = 160 + 200 = 360 % of people attending = 240 / 360 = 66.67 % . c is the correct answer | a = 75 / 100
b = a * 160
c = 40 / 100
d = 1 - c
e = d * 200
f = b + e
g = 160 + 200
h = f / g
i = h * 100
|
a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | a | subtract(multiply(multiply(const_100, const_10), const_2), divide(add(multiply(multiply(10, 8), const_2), 92), add(divide(10, const_100), divide(8, const_100)))) | if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 92 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | 0.1 x = 0.08 ( 2000 - x ) + 92 0.18 x = 252 x = 1400 then the amount invested at 8 % is $ 2000 - $ 1400 = $ 600 the answer is a . | a = 100 * 10
b = a * 2
c = 10 * 8
d = c * 2
e = d + 92
f = 10 / 100
g = 8 / 100
h = f + g
i = e / h
j = b - i
|
a ) rs . 2000 , b ) rs . 4000 , c ) rs . 3000 , d ) rs . 5000 , e ) rs . 6000 | c | divide(3432, add(const_1, divide(add(subtract(30, 20), 10), const_100))) | there is a 30 % increase in the price of an article in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . if the final price of the article is rs . 3432 , then what was the price of the article initially ? | "let the price of the article , four years age be rs . 100 in the 1 st year , price of the article = 100 + 30 = rs . 130 . in the 2 nd year , price = 130 - 20 % of 130 = 130 - 26 = rs . 104 . in the 3 rd year , price = 104 + 10 % of 104 = 104 + 10.4 = rs . 114.40 . but present price of the article is rs . 3432 for 114.4 - - - > 100 ; 2288 - - - > ? required price = ( 3432 * 100 ) / 114.4 = 30 * 100 = rs . 3000 . answer : c" | a = 30 - 20
b = a + 10
c = b / 100
d = 1 + c
e = 3432 / d
|
a ) 40 % , b ) 8 % , c ) 25 % , d ) 16.66 % , e ) 26 % | e | subtract(add(multiply(multiply(const_4, 10), divide(40, const_100)), multiply(const_4, 10)), const_100) | the balance of a trader weighs 10 % less than it should . still the trader marks up his goods to get an overall profit of 40 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 90 gms because it weighs 10 % less . say , the cost price is $ 90 ( $ 1 / gm ) . since he gets a profit of 40 % , the selling price must be 90 + ( 40 / 100 ) * 90 = $ 126 since the cost price is actually supposed to be $ 100 ( for 100 gms ) and the selling price is $ 126 , the mark up is simply 26 % . ans : e" | a = 4 * 10
b = 40 / 100
c = a * b
d = 4 * 10
e = c + d
f = e - 100
|
a ) 1187.5 , b ) 1234 , c ) 1289 , d ) 1425 , e ) none of these | d | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | jaclyn buys $ 60 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 = 60000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 1425 answer : d" | a = 100 * 100
b = a * 9
c = b / 100
d = c / 4
|
a ) a ) 87 , b ) b ) 99 , c ) c ) 68 , d ) d ) 82 , e ) e ) 73 | e | divide(add(add(add(add(76, 65), 82), 67), 75), divide(const_10, const_2)) | danny obtained 76 , 65 , 82 , 67 and 75 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | average = ( 76 + 65 + 82 + 67 + 75 ) / 5 = 365 / 5 = 73 . answer : e | a = 76 + 65
b = a + 82
c = b + 67
d = c + 75
e = 10 / 2
f = d / e
|
a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 138 | e | add(add(const_1, 69), 69) | in the land of oz only one or two - letter words are used . the local language has 69 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 69 letters is 69 ; the number of 2 letter words ( xx ) that can be made from 69 letters is 69 * 69 , since each x can take 69 values . total : 69 + 69 * 69 . similarly : the number of 1 letter words ( x ) that can be made from 68 letters is 68 ; the number of 2 letter words ( xx ) that can be made from 68 letters is 68 * 68 , since each x can take 68 values . total : 68 + 68 * 68 . the difference is ( 69 + 69 * 69 ) - ( 68 + 68 * 68 ) = 138 . answer : e ." | a = 1 + 69
b = a + 69
|
a ) 7700 , b ) 8000 , c ) 8300 , d ) 8600 , e ) 8900 | b | divide(2400, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | a candidate got 35 % of the votes and lost to the rival by 2400 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % = 2400 10 % = 800 100 % of the votes = 8000 answer : b" | a = 35 / 100
b = 1 - a
c = 35 / 100
d = b - c
e = 2400 / d
|
a ) 8 / 17 , b ) 2 / 5 , c ) 3 / 5 , d ) 11 / 15 , e ) 7 / 15 | a | add(multiply(divide(8, add(9, 8)), divide(subtract(8, const_1), subtract(add(9, 8), const_1))), multiply(divide(9, add(9, 8)), divide(subtract(9, const_1), subtract(add(9, 8), const_1)))) | a bag contains 9 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "drawing two balls of same color from nine green balls can be done in 9 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 9 c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 8 / 17 answer : a" | a = 9 + 8
b = 8 / a
c = 8 - 1
d = 9 + 8
e = d - 1
f = c / e
g = b * f
h = 9 + 8
i = 9 / h
j = 9 - 1
k = 9 + 8
l = k - 1
m = j / l
n = i * m
o = g + n
|
a ) 27.35 hours , b ) 26.35 hours , c ) 25.35 hours , d ) 24.35 hours , e ) 23.35 hours | b | add(multiply(add(add(8, 0.5), subtract(8, 0.5)), 105), multiply(subtract(add(divide(105, add(8, 0.5)), divide(105, subtract(8, 0.5))), add(add(8, 0.5), subtract(8, 0.5))), const_60)) | speed of a boat in standing water is 8 kmph and speed of the stream is 0.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 7.5 kmph speed downstream = 8.5 kmph total time taken = 105 / 7.5 + 105 / 8.5 = 26.35 hours answer is b" | a = 8 + 0
b = 8 - 0
c = a + b
d = c * 105
e = 8 + 0
f = 105 / e
g = 8 - 0
h = 105 / g
i = f + h
j = 8 + 0
k = 8 - 0
l = j + k
m = i - l
n = m * const_60
o = d + n
|
a ) rs . 3168 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | a | multiply(8640, divide(add(divide(subtract(90000, add(6000, 3000)), const_3), 6000), 90000)) | a , b and c started a business with a total investment of rs . 90000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find a ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 90000 = > x + ( x - 3000 ) + ( x + 3000 ) = 90000 = > 3 x = 90000 = > x = 30000 hence , a ' s investment = rs . 33000 b ' s investment = rs . 27000 c ' s investment = rs . 30000 ratio of the capitals of a , b and c = 33000 : 27000 : 30000 = 11 : 9 : 10 a ' s share = rs . [ ( 11 / 30 ) ã — 8640 ] = rs . 3168 answer : option a" | a = 6000 + 3000
b = 90000 - a
c = b / 3
d = c + 6000
e = d / 90000
f = 8640 * e
|
a ) $ 0.15 , b ) $ 0.26 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | b | divide(multiply(add(const_1, const_1), 0.4), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.40 per glass on the first day , what was the price per glass on the second day ? | on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.4 = 3 * x - - > x = $ 0.26 . answer : b . | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
|
['a ) 8000', 'b ) 9261', 'c ) 9999', 'd ) none of these', 'e ) can not be determined'] | b | power(add(multiply(const_10, const_2), const_1), const_3) | the largest four - digit number which is a perfect cube , is | solution clearly , 9261 is a perfect cube satisfying the given property . answer b | a = 10 * 2
b = a + 1
c = b ** 3
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | divide(multiply(6, 3), 6) | a number when divided by 6 leaves a remainder 3 . when the square of the same number is divided by 6 , the remainder is | "sol . let x = 6 q + 3 . then x 2 = ( 6 q + 3 ) 2 = 36 q 2 + 36 q + 9 = 6 ( 6 q 2 + 6 q + 1 ) + 3 . so , when x 2 is divided by 6 , remainder = 3 . answer d" | a = 6 * 3
b = a / 6
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 30 | e | subtract(add(40, 30), divide(40, divide(40, const_100))) | the contents of a certain box consist of 40 apples and 30 oranges . how many oranges must be added to the box so that exactly 40 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.4 40 = ( 30 + 40 + x ) * 0.4 x = 30 answer : e" | a = 40 + 30
b = 40 / 100
c = 40 / b
d = a - c
|
a ) 25 and 40 , b ) 20 and 32 , c ) 30 and 48 , d ) 15 and 24 , e ) 16 and 25 | b | add(multiply(multiply(5, 4), const_100), multiply(8, 4)) | two numbers are in the ratio 5 : 8 and thier h . c . f is 4 . find the numbers . | explanation : let two numbers 5 x , 8 x their h . c . f = 4 therefore h . c . f is 4 i . e , x in this case . therefore the numbers are 5 x 4 = 20 8 x 4 = 32 answer : option b | a = 5 * 4
b = a * 100
c = 8 * 4
d = b + c
|
a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 74 % | e | multiply(subtract(const_1, multiply(divide(subtract(const_100, 60), const_100), divide(13, 20))), const_100) | a ' s speed is 20 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 60 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 40 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 20 / 13 of b , therefore , b ' s speed is 13 / 20 of a distance covered by b = speed x time = ( 13 / 20 ) x ( 0.4 ) = 0.26 ( which means b would have covered 0.26 of the race length during the time in which a has covered 0.4 of the race length . therefore to win , b needs a headstart of ( 1 - 0.26 = ) 0.74 of the race length . so the answer is e" | a = 100 - 60
b = a / 100
c = 13 / 20
d = b * c
e = 1 - d
f = e * 100
|
a ) 91 , b ) 87 , c ) 85 , d ) 81 , e ) none of them | d | subtract(divide(multiply(multiply(multiply(subtract(const_1, divide(4, 7)), 33), 8), 117), multiply(multiply(divide(4, 7), subtract(46, 33)), 9)), 117) | a contract is to be completed in 46 days sad 117 men were said to work 8 hours a day . after 33 days , ( 4 / 7 ) of the work is completed . how many additional men may be employed so that the work may be completed in time , each man now working 9 hours a day ? | remaining work = ( 1 - ( 4 / 7 ) = ( 3 / 7 ) remaining period = ( 46 - 33 ) days = 13 days let the total men working at it be x . less work , less men ( direct proportion ) less days , more men ( indirect proportion ) more hours per day , less men ( indirect proportion ) work ( 4 / 7 ) : ( 3 / 7 ) days 13 : 33 } : : 117 : x hrs / day 9 : 8 therefore ( 4 / 7 ) x 13 x 9 x x = ( 3 / 7 ) x 33 x 8 x 117 or x = ( 3 x 33 x 8 x 117 ) / ( 4 x 13 x 9 ) = 198 additional men to be employed = ( 198 - 117 ) = 81 . answer is d . | a = 4 / 7
b = 1 - a
c = b * 33
d = c * 8
e = d * 117
f = 4 / 7
g = 46 - 33
h = f * g
i = h * 9
j = e / i
k = j - 117
|
a ) 20068 , b ) 20050 , c ) 20000 , d ) 25000 , e ) 28000 | c | multiply(1000000, divide(const_1, const_100)) | a person borrows 1000000 for 7 years at 3 % p . a . simple interest . he immediately lends it to another person at 5 % p . a . for 7 years . find his gain in the transaction per year . | "gain in 7 years = [ ( 1000000 ã — 5 ã — 7 / 100 ) â ˆ ’ ( 1000000 ã — 7 ã — 3 / 100 ) ] = ( 350000 â € “ 210000 ) = 140000 â ˆ ´ gain in 1 year = ( 140000 / 7 ) = 20000 answer c" | a = 1 / 100
b = 1000000 * a
|
a ) 13 , b ) 14 , c ) 30 , d ) 31 , e ) 40 | a | multiply(31, divide(12, 21)) | 12 : 21 : : 31 : ? | "ans 13 reverse of 31 answer : a" | a = 12 / 21
b = 31 * a
|
a ) 7 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(add(const_4, const_4), const_1) | if ( n + 2 ) ! / n ! = 72 , n = ? | ( n + 2 ) ! / n ! = 72 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 72 from here , we might just test the answer choices . since ( 9 ) ( 8 ) = 72 , we can see that n = 7 a | a = 4 + 4
b = a - 1
|
a ) 176 , b ) 170 , c ) 160 , d ) 179 , e ) 174 | a | divide(multiply(add(multiply(5, const_100), 55), add(multiply(3, const_100), 74)), multiply(subtract(55, add(multiply(const_2, const_4), const_2)), subtract(55, add(multiply(const_2, const_4), const_2)))) | a room 5 m 55 cm long and 3 m 74 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor | "let x and y be the sides of the rectangle . then , correct area = xy . area of the room = ( 544 x 374 ) cm 2 . size of largest square tile = h . c . f . of 544 cm and 374 cm = 34 cm . area of 1 tile = ( 34 x 34 ) cm 2 . number of tiles required = ( 544 * 374 ) / ( 34 * 34 ) = 176 answer a 176" | a = 5 * 100
b = a + 55
c = 3 * 100
d = c + 74
e = b * d
f = 2 * 4
g = f + 2
h = 55 - g
i = 2 * 4
j = i + 2
k = 55 - j
l = h * k
m = e / l
|
a ) 20 years . , b ) 30 years . , c ) 40 years . , d ) 50 years . , e ) 60 years . | b | subtract(add(divide(multiply(16, 6), subtract(6, const_1)), 6), 16) | the ages of two persons differ by 16 years . if 6 years agao , the elder one be 3 times as old as the younger one , find their present ages . | "sol . let the age of the younger person be x years . then , age of the elder perosn = ( x + 16 ) years . ∴ 3 ( x - 6 ) = ( x + 16 - 6 ) ⇔ 3 x - 18 = x + 10 ⇔ 2 x = 28 ⇔ x = 14 hence , their present ages are 14 years and 30 years . answer b" | a = 16 * 6
b = 6 - 1
c = a / b
d = c + 6
e = d - 16
|
a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | d | multiply(divide(subtract(const_1, multiply(add(inverse(4), inverse(15)), 3)), inverse(4)), const_60) | working at a constant rate , p can finish a job in 4 hours . q , also working at a constant rate , can finish the same job in 15 hours . if they work together for 3 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | "each hour they complete 1 / 4 + 1 / 15 = 19 / 60 of the job . in 3 hours , they complete 3 ( 19 / 60 ) = 19 / 20 of the job . the time for p to finish is ( 1 / 20 ) / ( 1 / 4 ) = ( 1 / 5 ) hour = 12 minutes the answer is d ." | a = 1/(4)
b = 1/(15)
c = a + b
d = c * 3
e = 1 - d
f = 1/(4)
g = e / f
h = g * const_60
|
a ) 20 , b ) 31 , c ) 42 , d ) 43 , e ) 64 | d | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 256 are multiple of 4 but not multiples of 6 ? | from here , it might be easier to go up in bounds of 60 , so we know that 61 - 120 gives 10 more numbers . 121 - 180 and 181 - 240 as well . this brings us up to 240 with 40 numbers . a cursory glance at the answer choices should confirm that it must be 42 , as all the other choices are very far away . the numbers 244 and 248 will come and complete the list that ’ s ( naughty or nice ) under 250 . answer choice d is correct here . | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) $ 10079.44 , b ) e = $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much e does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a = 3 / 96
b = a / 2
c = 1 + b
d = c ** 4
e = 10000 * d
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 35 % | a | subtract(const_100, multiply(divide(add(19, const_100), add(40, const_100)), const_100)) | two numbers are respectively 19 % and 40 % more than a third number . the percentage that is first of the second is ? | i ii iii 119 140 100 140 - - - - - - - - - - 119 100 - - - - - - - - - - - ? = > 15 % answer : a | a = 19 + 100
b = 40 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 35 % , b ) 20 % , c ) 5 % , d ) 25 % , e ) 75 % | e | multiply(subtract(const_1, divide(5, 20)), const_100) | the c . p of 5 books is equal to the s . p of 20 books . find his gain % or loss % ? | "explanation : 5 cp = 20 sp 20 - - - 15 cp loss 100 - - - ? = > 75 % loss answer : e" | a = 5 / 20
b = 1 - a
c = b * 100
|
a ) 10 , b ) 8 , c ) 6 , d ) 4 , e ) 2 | d | subtract(reminder(reminder(90, 33), 17), reminder(99, reminder(33, 17))) | for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 90 θ 33 ) θ 17 ) - ( 99 θ ( 33 θ 17 ) ) ? | "( ( 90 θ 33 ) θ 17 ) the remainder of 90 divided by 33 is 24 ; the remainder of 24 divided by 17 is 7 ; ( 99 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; 7 - 3 = 4 . answer : d ." | a = reminder - (
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a ) 2.9 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1 kmph , e ) 5.7 kmph | d | multiply(const_3_6, divide(10, 36)) | convert the 10 / 36 m / s into kilometers per hour ? | "10 / 36 m / s = 10 / 36 * 18 / 5 = 1 / 1 = 1 kmph . answer : d" | a = 10 / 36
b = const_3_6 * a
|
a ) 0 , b ) 3 , c ) 2 , d ) 5 , e ) none of the above | a | divide(subtract(power(9, 1), power(9, 1)), 9) | what is the remainder when 9 ^ 1 + 9 ^ 2 + 9 ^ 3 + . . . + 9 ^ 9 is divided by 3 ? | "notice that in the brackets we have the sum of 8 odd multiples of 3 , which yields remainder of 0 upon division by 3 . answer : a" | a = 9 ** 1
b = 9 ** 1
c = a - b
d = c / 9
|
a ) 5 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 15 % | d | multiply(divide(subtract(add(25, const_100), add(10, const_100)), add(25, const_100)), const_100) | if the price of gasoline increases by 25 % and a driver intends to spend only 10 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ? | "let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.25 * p ) y = 1.1 ( p * x ) y = ( 1.1 / 1.25 ) * x = 0.88 x which is a reduction of 12 % . the answer is d ." | a = 25 + 100
b = 10 + 100
c = a - b
d = 25 + 100
e = c / d
f = e * 100
|
['a ) 2399', 'b ) 1600', 'c ) 2679', 'd ) 1697', 'e ) 1711'] | b | subtract(power(divide(add(81, 1), const_2), const_2), 81) | a square was given . inside the square there are white tiles and black tiles . black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1 cm x 1 cm . if there are 81 black tiles in the square . then find the no of white tiles in it . | in a square , number of squares on the diagonal is equal to the tiles on a single row . if there are even number of square on a side , then total squares on the diagonal is 2 n - 1 , otherwise 2 n . as the total tiles on the diagonal are given as 81 , then number of tiles on a side = 2 n - 1 = 81 so n = 41 . so number of white tiles = 412 − 81 = 1681 − 81412 − 81 = 1681 − 81 = 1600 answer : b | a = 81 + 1
b = a / 2
c = b ** 2
d = c - 81
|
a ) 30.25 , b ) 32.25 , c ) 38.25 , d ) 37.25 , e ) 41.25 | b | multiply(const_100, divide(subtract(power(add(const_100, 15), const_3), power(const_100, const_3)), power(const_100, const_3))) | if each edge of cube increased by 15 % , the percentage increase in | "100 × ( 115 ) / 100 × ( 115 ) / 100 = 132.25 = > 32.25 % answer is b ." | a = 100 + 15
b = a ** 3
c = 100 ** 3
d = b - c
e = 100 ** 3
f = d / e
g = 100 * f
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a ) 11 , b ) 66 , c ) 18 , d ) 16 , e ) 42 | e | subtract(multiply(16, 16), add(multiply(5, 14), multiply(9, 16))) | the average age of 16 students of a class is 16 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 12 th student is | "explanation : age of the 16 th student = [ 16 * 16 - ( 14 * 5 + 16 * 9 ) ] = ( 256 - 214 ) = 42 years . answer : e" | a = 16 * 16
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
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a ) 973 , b ) 6973 , c ) 5994 , d ) 6084 , e ) none of these | e | subtract(multiply(const_10, 8), 8) | the difference between the place value and the face value of 8 in the numeral 856973 is | "( place value of 8 ) - ( face value of 8 ) = ( 800000 - 8 ) = 799992 answer : option e" | a = 10 * 8
b = a - 8
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a ) 400 km , b ) 500 km , c ) 360 km , d ) 540 km , e ) 600 km | d | multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 6) | the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 6 hrs if the speed of the bike is twice speed of the tractor ? | "sol . average speed of a tractor = 25 km / h the speed of a bus in an hour = 25 × 2 = 50 km the speed of a car in an hour = 9 / 5 * 50 = 90 km so , the distance covered by car in 6 h is 90 × 6 = 540 km ans . ( d )" | a = 4 / 5
b = 1 + a
c = 575 / 23
d = 2 * c
e = b * d
f = e * 6
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a ) 22 , b ) 77 , c ) 36 , d ) 25 , e ) 13 | a | divide(multiply(18, 33), 27) | 33 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 33 : : 18 : x [ please pay attention , we have written 27 : 33 rather than 33 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { 33 \ times 18 } { 27 } } x = 22 so 22 days will be required to get work done by 27 men . answer : a" | a = 18 * 33
b = a / 27
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a ) 7.69 % , b ) 16.66 % , c ) 17.8 % , d ) 19 % , e ) 21 % | a | multiply(divide(subtract(70, 65), 65), const_100) | john makes $ 65 a week from his job . he earns a raise andnow makes $ 70 a week . what is the % increase ? | "increase = ( 5 / 65 ) * 100 = ( 1 / 13 ) * 100 = 7.69 % . a" | a = 70 - 65
b = a / 65
c = b * 100
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a ) 11 , b ) 12 , c ) 77 , d ) 33 , e ) 88 | b | add(add(6, 2), 4) | in kaya ' s teacher ' s desk there are 6 pink highlighters , 2 yellow highlighters , and 4 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 6 + 2 + 4 = 12 . answer is b ." | a = 6 + 2
b = a + 4
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a ) 0.82 , b ) 0.84 , c ) 1.5 , d ) 1.2 , e ) 1.4 | b | multiply(divide(20, multiply(multiply(const_4, const_5), const_5)), divide(80, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 20 % of the times , b speaks the truth 80 % of the times . what is the probability that at least one will tell the truth | "probability of a speaks truth p ( a ) = 2 / 10 ; false = 8 / 10 probability of b speaks truth p ( b ) = 8 / 10 ; false = 2 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 8 / 10 ) * ( 2 / 10 ) ] = 1 - 16 / 100 = 1 - 0.16 = 0.84 answer : b" | a = 4 * 5
b = a * 5
c = 20 / b
d = 4 * 5
e = d * 5
f = 80 / e
g = c * f
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a ) 13 , b ) 9 , c ) 15 , d ) 8 , e ) 10 | e | divide(multiply(15, 6), subtract(15, 6)) | a man can do a piece of work in 15 days , but with the help of his son , he can do it in 6 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 6 ) - ( 1 / 15 ) = 3 / 30 = 1 / 10 the son alone can do the work in 10 days answer is e" | a = 15 * 6
b = 15 - 6
c = a / b
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(16, 45) | how many of the positive factors of 45 , 16 and how many common factors are there in numbers ? | "factors of 45 - 1 , 3 , 9 , 15 and 45 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 1 answer ( a )" | a = 16 / 45
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a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 8820 multiplied by y is the square of an integer , then y must be | "8820 = 2 ^ 2 * 3 ^ 2 * 5 * 7 ^ 2 to be perfect square , we need to multiply by at least 5 . the answer is b ." | a = 2 + 3
b = a * 2
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a ) 240 , b ) 200 , c ) 270 , d ) 300 , e ) 310 | c | multiply(divide(multiply(60, const_3), 20), 30) | if 20 typists can type 60 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "no . of letters typing by 20 typists in 20 minutes = 60 no . of letters typing by 20 typists in 60 minutes = 60 * 3 = 180 no . of letters typing by 30 typists in 60 minutes = 180 / 20 * 30 = 270 answer : c" | a = 60 * 3
b = a / 20
c = b * 30
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a ) 19 , b ) 29 , c ) 37 , d ) 49 , e ) 59 | c | add(subtract(85, multiply(17, 3)), 3) | a batsman makes a score of 85 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 85 = 17 x therefore x = 37 answer : c" | a = 17 * 3
b = 85 - a
c = b + 3
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a ) 49.66 % , b ) 52.66 % , c ) 28.22 % , d ) 33.33 % , e ) 25.92 % | e | multiply(const_100, divide(divide(multiply(add(40, const_100), 20), const_100), add(const_100, 8))) | of the families in city x in 1998 , 20 percent owned a personal computer . the number of families in city x owning a computer in 2002 was 40 percent greater than it was in 1998 , and the total number of families in city x was 8 percent greater in 2002 than it was in 1998 . what percent of the families in city x owned a personal computer in 2002 ? | "say a 100 families existed in 1998 then the number of families owning a computer in 1998 - 20 number of families owning computer in 2002 = 20 * 140 / 100 = 28 number of families in 2002 = 108 the percentage = 28 / 108 * 100 = 25.92 % . answer : e" | a = 40 + 100
b = a * 20
c = b / 100
d = 100 + 8
e = c / d
f = 100 * e
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a ) 30 , b ) 20 , c ) 40 , d ) 60 , e ) 70 | b | multiply(2, 4) | three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 240 . their h . c . f . is : | "explanation : let the number be 2 x , 3 x and 4 x lcm of 2 x , 3 x and 4 x = 12 x 12 x = 240 = > x = 240 / 12 = 20 hcf of 2 x , 3 x and 4 x = x = 20 answer : option b" | a = 2 * 4
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a ) 44 % , b ) 120 % , c ) 144 % , d ) 40 % , e ) 21 % | e | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(10, const_100)), const_2)), const_100) | if the radius of a circle is increased by 10 % then the area is increased by : | "initially a = pi * r 2 now r 2 = 110 / 100 r then area = pi * ( 110 r / 100 ) 2 area = 121 r / 100 that is area increases from 100 to 121 = increase in area = 21 % answer : e" | a = 100 / 100
b = 100 / 100
c = 10 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
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a ) 79698 , b ) 80578 , c ) 80698 , d ) 81268 , e ) none of them | b | multiply(217, power(217, 183)) | 217 x 217 + 183 x 183 = ? | "= ( 217 ) ^ 2 + ( 183 ) ^ 2 = ( 200 + 17 ) ^ 2 + ( 200 - 17 ) ^ 2 = 2 [ ( 200 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 40000 + 289 ] = 2 x 40289 = 80578 answer is b" | a = 217 ** 183
b = 217 * a
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a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 11 , e ) 1 / 25 | d | divide(6, choose(12, 2)) | kim has 6 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 12 shoes , what is the probability that she will select 2 shoes of the same color ? | total pairs = 12 c 2 = 66 ; same color pairs = 6 c 1 * 1 c 1 = 6 ; prob = 1 / 11 ans d | a = math.comb(12, 2)
b = 6 / a
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a ) 290 sec , b ) 190 sec , c ) 199 sec , d ) 192 sec , e ) 240 sec | e | subtract(divide(multiply(const_1, const_1000), divide(40, 10)), 10) | in a kilometer race , a beats b by 40 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 40 = 250 sec . time taken by a = 250 - 10 = 240 sec . answer : e" | a = 1 * 1000
b = 40 / 10
c = a / b
d = c - 10
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a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | d | divide(divide(multiply(100, multiply(15, const_2)), const_1000), divide(15, const_10)) | a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 100 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | the distance between the nest and the ditch is 100 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 100 = 3000 . d = st 3000 / 1.5 = t , i think we can take 3000 meters as 3 km , then only we get t = 2 . ( 1000 meters = 1 km ) d ) | a = 15 * 2
b = 100 * a
c = b / 1000
d = 15 / 10
e = c / d
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a ) 998 , b ) 297 , c ) 999 , d ) 902 , e ) 900 | e | add(108, multiply(subtract(100, const_1), 8)) | s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is | "last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is first term = 102 ; total terms = 100 ; difference = 3 108 + ( 99 ) 8 = 900 ans e" | a = 100 - 1
b = a * 8
c = 108 + b
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a ) 179 , b ) 208 , c ) 210 , d ) 225 , e ) 229 | d | divide(add(150, subtract(multiply(62, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 62 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 62 * 46 = 2852 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2852 – 2552 = 300 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 300 2 x = 450 x = 225 runs answer d" | a = 62 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
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a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 120 | d | subtract(100, subtract(power(2, divide(100, 50)), const_1)) | what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 50 ? | put in simple words , we need to find the highest power of 2 in 100 ^ 50 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 50 = ( 2 ^ 2 * 5 ^ 2 ) ^ 50 = 2 ^ 100 * 5 ^ 100 answer : d | a = 100 / 50
b = 2 ** a
c = b - 1
d = 100 - c
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a ) 28 % , b ) 30 % , c ) 44 % , d ) 54 % , e ) 64 % | a | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 10), subtract(const_100, 20))), const_100) | a towel , when bleached , lost 10 % of its length and 20 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( − 10 − 20 + ( 10 × 20 ) / 100 ) % = − 28 % i . e . , area is decreased by 28 % answer : a" | a = 100 * 100
b = 100 - 10
c = 100 - 20
d = b * c
e = a - d
f = e / 100
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(log(multiply(power(5, 5), power(5, 7))), log(power(5, 4))) | if 5 ^ 5 × 5 ^ 7 = ( 5 ^ 4 ) ^ x , then what is the value of x ? | 5 ^ 5 × 5 ^ 7 = ( 5 ^ 4 ) ^ x 5 ^ ( 5 + 7 ) = 5 ^ ( 12 ) with multiplying numbers with exponents it ' s the same as adding the exponents 5 ^ 12 = ( 5 ^ 4 ) ^ x since base is same , we can equate powers : when raising an exponent to an exponent , it ' s the same as multiplying them , therefore : 12 = 4 x x = 3 ans . b ) 3 | a = 5 ** 5
b = 5 ** 7
c = a * b
d = math.log(c)
e = 5 ** 4
f = math.log(e)
g = d / f
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a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | subtract(16, const_4) | in a group of cows and hens , the number of legs are 16 more than twice the number of heads . the number of cows is | "explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 16 4 x + 2 y = 2 x + 2 y + 16 2 x = 16 x = 8 . answer : b" | a = 16 - 4
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a ) 6 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | e | subtract(add(const_4, 2), divide(divide(add(16, 4), const_2), 2)) | on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 2 cups of tea . last week mo drank a total of 16 cups of tea and hot chocolate together . if during that week mo drank 4 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 16 t - c = 4 - > t = 10 . c = 6 . mo drinks 5 cups of tea a day then number of days that are not rainy = 10 / 2 = 5 so number of rainy days = 7 - 5 = 2 e is the answer ." | a = 4 + 2
b = 16 + 4
c = b / 2
d = c / 2
e = a - d
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a ) 4 % increase , b ) 10 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | a | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 30)), const_100), const_100) | if the price of a tv is first decreased by 20 % and then increased by 30 % , then the net change in the price will be : | explanation : solution : let the original price be rs . 100 . new final price = 130 % of ( 80 % of 100 ) = rs . 130 / 100 * 80 / 100 * 100 = rs . 104 . . ' . increase = 4 % answer : a | a = 100 - 20
b = 100 + 30
c = a * b
d = c / 100
e = d - 100
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a ) 1 / 7 , b ) 2 / 5 , c ) 1 / 2 , d ) 7 / 12 , e ) 2 / 3 | b | divide(const_2, 5) | equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 5 full and other jar 1 / 4 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 5 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 5 + 1 / 5 = 2 / 5 full . answer : b ." | a = 2 / 5
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a ) – 4 , b ) – 3 , c ) – 2 , d ) – 1 , e ) 0 | a | subtract(multiply(4, const_2), multiply(3, const_2)) | if the average ( arithmetic mean ) of x + 3 , x + 4 , and x + 5 is 0 , then x = | "( x + 3 + x + 4 + x + 5 ) / 3 = 0 = > 3 x + 12 = 0 = > x = - 4 answer a" | a = 4 * 2
b = 3 * 2
c = a - b
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | multiply(subtract(const_1, divide(6, 15)), 10) | dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 6 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 15 of the job per hour . in 6 hours , dan completes 6 ( 1 / 15 ) = 2 / 5 of the job . annie can complete 1 / 10 of the job per hour . to complete the job , annie will take 3 / 5 / 1 / 10 = 6 hours . the answer is c ." | a = 6 / 15
b = 1 - a
c = b * 10
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a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.6 kg , e ) none of these | d | divide(divide(subtract(const_100, 20), const_100), divide(50, const_100)) | when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 20 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 80 % ( 100 - 20 ) . therefore 0.5 x amount of flower - nectar = 0.80 x amount of honey = 0.80 x 1 kg therefore amount of flower - nectar needed = ( 0.80 / 0.51 ) kg = 1.6 kgs answer : d" | a = 100 - 20
b = a / 100
c = 50 / 100
d = b / c
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a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these | a | divide(subtract(const_6, 2), const_6) | in a throw of dice what is the probability of ge æ « ing number greater than 2 | "explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : a" | a = 6 - 2
b = a / 6
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a ) 900 cm cube , b ) 910 cm cube , c ) 920 cm cube , d ) 930 cm cube , e ) 960 cm cube | e | multiply(multiply(8, 10), 12) | find the area of a cuboid of length 8 cm , breadth 10 cm . and height 12 cm . | "area of a cuboid = lxbxh = 8 cm x 10 cm x 12 cm = 960 cm cube answer : e" | a = 8 * 10
b = a * 12
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a ) 38 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh | a | divide(const_3, add(add(divide(const_1, 60), divide(const_1, 24)), divide(const_1, 48))) | if a car went the first third of the distance at 60 kmh , the second third at 24 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 240 ( this number is convenient because it is divisible by 60 , 24 and 48 ) so : 240 = 60 * t 1 = 4 hrs 240 = 24 * t 2 = 10 hrs 240 = 48 * t 3 = 5 hrs t = t 1 + t 2 + t 3 = 19 hrs d = rt ( 240 * 3 ) = r * 19 r = 38 answer : a" | a = 1 / 60
b = 1 / 24
c = a + b
d = 1 / 48
e = c + d
f = 3 / e
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a ) 13 , b ) 15 , c ) 25 , d ) 17 , e ) 19 | b | divide(divide(multiply(900, 10), const_100), 6) | a reduction of 10 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 900 , what is the reduced price for kg ? | "900 * ( 10 / 100 ) = 90 - - - - 6 ? - - - - 1 = > rs . 15 answer : b" | a = 900 * 10
b = a / 100
c = b / 6
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a ) 27 : 125 , b ) 1 : 62 , c ) 1 : 34 , d ) 3 : 64 , e ) 5 : 64 | a | divide(power(const_2.0, 5), power(5, 5)) | the triplicate ratio of 3 : 5 is ? | "3 ^ 3 : 5 ^ 3 = 27 : 125 answer : a" | a = 2 ** 0
b = 5 ** 5
c = a / b
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a ) 3377 , b ) 2899 , c ) 2500 , d ) 2778 , e ) 10000 | e | divide(multiply(200, 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 . 200 when the rate to profit rises from 5 to 7 percent . find the capital of a ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 200 a capital = 15000 * 2 / 3 = 10000 answer : e" | a = 200 * 100
b = a / 2
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a ) 16 , b ) 11 , c ) 31 , d ) 29 , e ) 19 | a | power(divide(7, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | find out the square of a number which when doubled exceeds its quarter by 7 ? | "a 16 explanation : let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 4 ) + 7 = > 8 p = p + 28 = > p = 4 p ^ 2 = 4 ^ 2 = 16 ." | a = 4 + 1
b = 1 / a
c = 2 - b
d = 7 / c
e = d ** 2
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a ) 50 , b ) 100 , c ) 490 , d ) 500 , e ) 980 | a | divide(490, 9.8) | a sports equipment store sold ping pong rackets for a total of $ 490 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 490 x = 50 pairs were sold . answer : a" | a = 490 / 9
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a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 12 | d | divide(multiply(5, 12), add(5, const_1)) | a little girl went to the farm to buy fruits . she bought a bag of apples and grabbed a bag of oranges which was 5 times as heavy as the apples . she brought home a total of 12 pounds of fruits . how many pounds was the oranges ? | pounds of apples = x pounds of oranges = 5 x x + 5 x = 12 6 x = 12 x = 12 / 6 = 2 oranges = 5 x = 5 * 2 = 10 answer : d | a = 5 * 12
b = 5 + 1
c = a / b
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a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 75 % | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | a batsman scored 120 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % answer is b" | a = 3 * 8
b = 8 * 3
c = a + b
d = 120 - c
e = d / 120
f = e * 100
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a ) 7 , b ) 4 , c ) 5 , d ) 6 , e ) 3 | a | subtract(subtract(15, 7), const_1) | if 7 < x < 9 < y < 15 , then what is the greatest possible positive integer difference of x and y ? | "let x = 7.1 and y = 14.1 greatest possible difference = 14.1 - 7.1 = 7 answer a" | a = 15 - 7
b = a - 1
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a ) – 18 , b ) – 10 , c ) – 6 , d ) 10 , e ) 14 | b | subtract(negate(2), multiply(const_2, subtract(2, negate(2)))) | on the number line , the number p is twice as many units from - 2 as - 2 is from 2 . if p is less than – 2 , what is the value of p ? | distance between - 2 and 2 = 4 since number p is twice as many units from - 2 as - 2 is from 2 , therefore p can be - 10 since , p is less than - 2 , the value of p = - 10 answer b | a = negate - (
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a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | b | subtract(100, subtract(197, 100)) | when 100 is divided by positive integer x , the remainder is 3 . what is the remainder when 197 is divided by x ? | if 100 / x leaves a reminder 3 then ( 100 - 3 ) i . e . 97 is divisible by x so ( 100 + 97 ) / x leaves a reminder rem ( 100 / x ) + rem ( 97 / x ) = > 3 + 0 = 3 answer : b | a = 197 - 100
b = 100 - a
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a ) 11 / 3 , b ) 4 ^ 11 - 1 , c ) 242 , d ) 3 ^ 10 , e ) 3 ^ 11 - 1 | b | subtract(power(4, add(9, const_2)), const_1) | if 4 ^ k + 4 ^ k = ( 4 ^ 9 ) ^ ( 4 ^ 9 ) - 4 ^ k , then k = ? | 4 ^ k + 4 ^ k = ( 4 ^ 9 ) ^ 4 ^ 9 - 4 ^ k 4 * ( 4 ^ k ) = 4 ^ ( 16 * 4 ^ 9 ) = 4 ^ ( 4 ^ 2 * 4 ^ 9 ) = 4 ( 4 ^ 11 ) 4 ^ k + 1 = 4 ^ ( 4 ^ 11 ) so k + 1 = 4 ^ 11 so k = 4 ^ 11 - 1 answer is b | a = 9 + 2
b = 4 ** a
c = b - 1
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a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | d | multiply(sqrt(divide(212, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 212 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose 12 at first , but then changed to 48 . i ' m not a native speaker , so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) , so is its largest factor . please correct if i ' m wrong here . i know that if n = 12 , n ^ 2 = 144 = 2 * 72 ( satisfy the condition ) . when n = 12 , the largest factor of n is n itself , which is 12 . check : 12 is the largest positive number that must divide 12 - - > true however if n = 48 , n ^ 2 = 48 * 48 = 32 * 72 ( satisfy the condition too ) . when n = 48 , the largest factor of n is n itself , which is 48 . check : 48 is the largest positive number that must divide 48 - - > true so , i also notice that the keyword ismust , notcould . the question is , why is 48 notmust divide 48 , but instead onlycould divide 48 ? i ' m not clear right here . why is 12 must divide 12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ 2 is divisible by 72 . the least value of n for which n ^ 2 is divisible by 72 is 12 , thus nmustbe divisible by 12 ( n is in any case divisible by 12 ) . for all other values of n , for which n ^ 2 is divisible by 72 , n will still be divisible by 12 . this means that n is always divisible by 12 if n ^ 2 is divisible by 72 . now , ask yourself : if n = 36 , is n divisible by 48 ? no . so , n is not always divisible by 48 . d" | a = 212 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 16.5 % , b ) 20 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(5.5, 3)), const_100) | a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 5.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.055 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 55 f - - > s = 4 f - - > f / s = 1 / 4 - - > f / ( s + f ) = 1 / ( 1 + 4 ) = 1 / 5 = 0.2 . answer : b ." | a = 3 - 5
b = 5 - 5
c = a / b
d = c * 100
|
a ) 16 km , b ) 10 km , c ) 12 km , d ) 30 km , e ) 25 km | d | multiply(3, 5) | a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 5 km on foot . his total journey is | "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 5 = x 5 x + 30 = 6 x x = 30 km answer : option d" | a = 3 * 5
|
a ) 12 - 14 , b ) 13 - 13 , c ) 10 - 16 , d ) 9 - 17 , e ) 8 - 18 | a | add(add(power(add(add(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(26, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive even integers is 26 . what are the two numbers ? | "let 2 n be the first even integer and let 2 n + 2 be the second integer 2 n + 2 n + 2 = 26 4 n + 2 = 26 4 n + 2 - 2 = 26 - 2 4 n = 24 n = 6 so the first even integer is 2 n = 2 × 6 = 12 and the second is 12 + 2 = 14 correct answer is a ) 12 - 14" | a = 26 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 26 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 26 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 26 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 10 , b ) 12 , c ) 15 , d ) 25 , e ) 18 | d | add(add(divide(3, 4), multiply(divide(3, 4), 25)), multiply(const_0_25, 25)) | total 75 cows 25 cow gives each 2 liter milk 25 cow gives each 3 / 4 liter milk 25 cow gives each 1 / 4 liter milk this is split into 3 son per each 25 cows & 25 liter milk how ? | "25 cow 2 liter each = 50 liter 25 cow 3 / 4 liter each = 3 / 4 = 0.75 * 25 = 18.75 25 cow 1 / 4 liter each = 1 / 4 = 0.25 * 25 = 6.25 add 50 + 18.75 + 6.25 = 75 milk split into 3 son each 25 liter then 75 / 3 = 25 answer : d" | a = 3 / 4
b = 3 / 4
c = b * 25
d = a + c
e = const_0_25 * 25
f = d + e
|
a ) a ) 15 , b ) b ) 10 , c ) c ) 14 , d ) d ) 12 , e ) e ) 11 | e | divide(968, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | if the wheel is 14 cm then the number of revolutions to cover a distance of 968 cm is ? | "2 * 22 / 7 * 14 * x = 968 = > x = 11 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 14
h = 968 / g
|
a ) 16.92 sec , b ) 14.8 sec , c ) 10.68 sec , d ) 10.8 sec , e ) 20.8 sec | a | divide(add(210, 260), multiply(add(60, 40), const_0_2778)) | two trains 210 m and 260 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 = 210 + 260 = 470 m . required time = 470 * 9 / 250 = 16.92 = 16.92 sec . answer : a" | a = 210 + 260
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) s . 7000 , b ) s . 9000 , c ) s . 15000 , d ) s . 17000 , e ) s . 27000 | c | divide(900, divide(multiply(subtract(15, 12), const_2), const_100)) | a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 900 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 900 = > 30 x / 100 - 24 x / 100 = 900 = > 6 x / 100 = 900 = > x = 15000 . answer : c" | a = 15 - 12
b = a * 2
c = b / 100
d = 900 / c
|
a ) 45 % , b ) 500 / 11 , c ) 600 / 11 , d ) 55 % , e ) 60 % | e | multiply(divide(subtract(125, add(multiply(5, 5), multiply(5, 5))), 125), const_100) | a batsman scored 125 runs which included 5 boundaries and 5 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 125 − ( 5 × 4 + 5 × 6 ) . = > 125 - 50 = > 75 hence , the required percentage is : - = > 75 / 125 * 100 = > 60 % answer : e" | a = 5 * 5
b = 5 * 5
c = a + b
d = 125 - c
e = d / 125
f = e * 100
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 2 | e | subtract(multiply(divide(3, 2), add(15, 5)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(15, 5)))), const_1)) | a certain basketball team that has played 2 / 3 of its games has a record of 15 wins and 5 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ? | "15 wins , 3 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 5 games so it can lose another 2 at most . answer ( e )" | a = 3 / 2
b = 15 + 5
c = a * b
d = 3 / 4
e = 3 / 2
f = 15 + 5
g = e * f
h = d * g
i = math.floor(h)
j = i + 1
k = c - j
|
a ) s . 45 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 90 | e | divide(720, subtract(13, 5)) | on selling 13 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 13 balls ) - ( s . p . of 13 balls ) = ( c . p . of 5 balls ) c . p . of 8 balls = s . p . of 13 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 8 = rs . 90 . answer : option e" | a = 13 - 5
b = 720 / a
|
a ) $ 122 , b ) $ 152 , c ) $ 202 , d ) $ 107 , e ) $ 102 | e | add(add(multiply(add(8, const_12), 4), const_12), const_10) | 3 friends neha , sabi and mahi divide $ 1100 amongs them in such a way that if $ 5 , $ 8 and $ 4 are removed from the sums that neha , sabi and mahi received respectively , then the share of the sums that they got will be in the ratio of 2 : 8 : 6 . how much did mahi receive ? | explanatory answer let the sums of money received by a , b and c be x , y and z respectively . then x - 5 : y - 8 : z - 4 is 2 a : 8 a : 6 a when $ 5 , $ 8 and $ 4 are removed , we are removing a total of $ 17 from $ 1100 . therefore , 2 a + 8 a + 6 a = 1100 - 17 = 1093 i . e . , 16 a = 1093 or a = 68 . we know that z - 6 = 16 a = 6 * 16 = 96 therefore , z = 96 + 6 = $ 102 the correct choice is ( e ) | a = 8 + 12
b = a * 4
c = b + 12
d = c + 10
|
a ) 80 , b ) 79 , c ) 78 , d ) 81 , e ) 82 | b | gcd(2923, 3239) | the maximum number of girls among them 2923 bags and 3239 eyeliners can be distributed in such a way that each student gets the same number of bags and eyeliners . | max no of girls = hcf of 2923 and 3239 = 79 answer : b | a = math.gcd(2923, 3239)
|
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 w 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 w in this range . answer : a ." | a = math.factorial(6)
b = a + 6
c = math.factorial(6)
d = c + 6
e = b / d
|
['a ) 190 π', 'b ) 210 π', 'c ) 240 π', 'd ) 320 π', 'e ) 380 π'] | e | multiply(power(multiply(sqrt(95), const_2), const_2), const_pi) | in may , the groundskeeper at spring lake golf club built a circular green with an area of 95 π square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | area = π r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 95 π ) = 380 π the answer is e . | a = math.sqrt(95)
b = a * 2
c = b ** 2
d = c * math.pi
|
a ) 150 min , b ) 245 min , c ) 185 min , d ) 300 min , e ) 205 min | b | multiply(subtract(add(add(20, 20), const_10), const_1), 5) | i have a huge bed - sheet of 1 meter long . i want to cuts this role into pieces of 20 centimeter long cloth . how long will it take me to cut the bed sheet if each 20 centimeter cut took me 5 minutes ? | b it will take 49 cuts not 50 . so 49 * 5 = > 245 minutes | a = 20 + 20
b = a + 10
c = b - 1
d = c * 5
|
a ) 87 days , b ) 16 days , c ) 55 days , d ) 44 days , e ) 22 days | b | divide(subtract(multiply(30, 40), multiply(40, 14)), 40) | 30 men can do a work in 40 days . when should 14 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 14 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 14 * 40 ) = 1200 40 p = 640 = > p = 16 days answer : b" | a = 30 * 40
b = 40 * 14
c = a - b
d = c / 40
|
a ) 22 , b ) 56.25 , c ) 78 , d ) 33 , e ) 25 | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 32), 50)), divide(multiply(const_100, 32), 50))) | if the cost price of 50 articles is equal to the selling price of 32 articles , then the gain or loss percent is ? | "given that , cost price of 50 article is equal to selling price of 32 articles . let cost price of one article = rs . 1 selling price of 32 articles = rs . 50 but cost price of 32 articles = rs . 32 therefore , the trader made profit . \ percentage of profit = 18 / 32 * 100 = 56.25 % answer : b" | a = 100 * 32
b = a / 50
c = 100 - b
d = 100 * 32
e = d / 50
f = c / e
g = 100 * f
|
a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 7 | c | floor(divide(23, 7)) | on dividing 23 by a number , the quotient is 7 and the remainder is 2 . find the divisor . | "d = ( d - r ) / q = ( 23 - 2 ) / 7 = 21 / 7 = 3 c" | a = 23 / 7
b = math.floor(a)
|
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