options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | floor(sqrt(divide(107800, multiply(multiply(multiply(11, 7), 5), 2)))) | in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the competition and finished first , second , third , or fourth each time . the product of all the points he received was 107800 . how many times did he participate in the competition ? | "107800 = 2 * 2 * 2 * 5 * 5 * 7 * 7 * 11 john participated 8 times . the answer is e ." | a = 11 * 7
b = a * 5
c = b * 2
d = 107800 / c
e = math.sqrt(d)
f = math.floor(e)
|
a ) 29 / 30 , b ) 4 / 49 , c ) 2 / 7 , d ) 45 / 49 , e ) 13 / 14 | a | divide(subtract(choose(25, const_2), choose(subtract(25, multiply(25, divide(4, 5))), const_2)), choose(25, const_2)) | according to a recent student poll , 4 / 5 out of 25 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "20 students are interested , 5 are not interested prob = 1 - 5 c 2 / 25 c 2 = 1 - ( 5 * 4 / ( 25 * 24 ) ) = 1 - 1 / 30 = 29 / 30 answer : a" | a = math.comb(25, 2)
b = 4 / 5
c = 25 * b
d = 25 - c
e = math.comb(d, 2)
f = a - e
g = math.comb(25, 2)
h = f / g
|
a ) a . 72 , b ) b . 50 , c ) c . 48 , d ) d . 46 , e ) e . 23 | e | multiply(divide(11, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 11 seconds . at 12 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 11 seconds . thus x = 1 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 intervals so the total time is ( 12 + 11 ) x = 23 x = 23 seconds . answer e" | a = 1 + 4
b = 6 + a
c = 11 / b
d = 1 + 4
e = 6 + d
f = 12 + e
g = c * f
|
a ) 2436 , b ) 2801 , c ) - 2801 , d ) - 660 , e ) none of them | d | multiply(subtract(const_1, const_2), subtract(multiply(24, 29), 36)) | - 24 x 29 + 36 = ? | "given exp . = - 24 x ( 30 - 1 ) + 36 = - ( 24 x 30 ) + 24 + 36 = - 720 + 60 = - 660 answer is d" | a = 1 - 2
b = 24 * 29
c = b - 36
d = a * c
|
a ) 6 , b ) 9 , c ) 12 , d ) 24 , e ) 48 | b | divide(multiply(divide(70, const_100), 20), divide(70, const_100)) | an alloy weighing 20 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 70 percent gold ? | "in 24 ounces , gold is 20 * ( 70 / 100 ) = 14 ounces . now we add x ounces of pure gold to make it 90 % gold . so 14 + x = ( 24 + x ) * 70 / 100 = > x = 9 . answer is b ." | a = 70 / 100
b = a * 20
c = 70 / 100
d = b / c
|
a ) 6 , b ) 7 , c ) 5 , d ) 4 , e ) 1 | a | inverse(add(divide(const_1, 10), divide(const_1, multiply(9, const_2)))) | a can do a piece of work in 10 days . he works at it for 4 days and then b finishes it in 9 days . in how many days can a and b together finish the work ? | "4 / 10 + 9 / x = 1 = > x = 15 1 / 10 + 1 / 15 = 1 / 6 = > 6 days answer : a" | a = 1 / 10
b = 9 * 2
c = 1 / b
d = a + c
e = 1/(d)
|
a ) 289 , b ) 1289 , c ) 1389 , d ) 1281 , e ) 128 | b | add(add(multiply(const_100, const_4), const_100), add(multiply(12, const_4), add(const_4, const_3))) | if a wholesaler distributes items to several outlets ( a , b , c and d ) in the ratio of 1 / 2 : 1 / 11 : 1 / 12 : 1 / 13 then find the total number of items the wholesaler distributes ? | "here , a : b : c : d = 1 / 2 : 1 / 11 : 1 / 12 : 1 / 13 1 ) l . c . m of 2 : 11 : 12 : 13 is 1716 2 ) find the number of books each friend received - - - - - - - - - ( to find no . of books each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 2 ) x 1716 = 858 b = ( 1 / 11 ) x 1716 = 156 c = ( 1 / 12 ) x 1716 = 143 d = ( 1 / 13 ) x 1716 = 132 3 ) total number of toys = ( 858 x + 156 x + 143 x + 132 x ) = 1289 x minimum number of pens ( x ) = 1 therefore , total number of items = 1289 items . correct option : b" | a = 100 * 4
b = a + 100
c = 12 * 4
d = 4 + 3
e = c + d
f = b + e
|
a ) 17 , b ) 19 , c ) 16 , d ) 18 , e ) 21 | d | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 60), 5) | the sum of ages of 5 children born 3 years different each is 60 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 60 5 x = 30 x = 6 . x + 12 = 6 + 12 = 18 d" | a = 3 + 4
b = 3 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 60
g = f / 5
|
a ) 10 , b ) 8 , c ) 7 , d ) 3 , e ) 11 | d | add(add(3, 3), 3) | given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p - 3 ? | "p is a positive even integer with a positive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit of p - 3 is 6 - 3 = 3 . answer : d ." | a = 3 + 3
b = a + 3
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a ) 229 , b ) 144 , c ) 278 , d ) 126 , e ) 112 | b | multiply(divide(240, 6), const_3_6) | a 240 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 240 / 6 * 18 / 5 = 144 kmph answer : b" | a = 240 / 6
b = a * const_3_6
|
a ) 1 km , b ) 500 mts , c ) 660 mts , d ) 2 km , e ) 250 mts | c | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 2), divide(const_1, 6)), divide(const_1, 6))), const_3), const_1000) | a person travels equal distances with speeds of 2 km / hr , 6 km / hr , 6 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 2 ) + ( x / 6 ) + ( x / 6 ) = 11 / 60 x = 0.22 total distance = 3 * 0.22 = 0.66 km = 660 meters correct option is c" | a = 11 / const_60
b = 1 / 2
c = 1 / 6
d = b + c
e = 1 / 6
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 1.75 , b ) 2.25 , c ) 2.5 , d ) 2.75 , e ) 3.25 | b | divide(divide(multiply(add(4, 2), 3), 2), 4) | natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 3 km / hour x / 3 = 3 x = 9 km the average speed while climbing = distance / time = 9 / 4 = 2.25 km / h the answer is b . | a = 4 + 2
b = a * 3
c = b / 2
d = c / 4
|
a ) p = 26.7 , b ) p = 30.0 , c ) p = 40.0 , d ) p = 53.3 , e ) p = 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his average rate p in miles per hour for the entire trip ? | "veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is not 200 % of the first half . ) the overall time is 4 hours + 12 hours or 16 hours . since the definition of average rate = total distance traveled / total time of travel , mike ' s average rate = 640 / 16 or 40 miles per hour . answer choice c is correct ." | a = 640 / 2
b = a / 80
c = b * 2
d = 640 / 2
e = d / 80
f = c + e
g = 640 / 2
h = g / 80
i = f + h
j = 640 / i
|
a ) 33 , b ) 62 , c ) 49 , d ) 53 , e ) 86 | b | add(add(multiply(5, const_2), 1), add(multiply(5, multiply(const_2, 1)), 1)) | a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 1 students left over . what is the sum of the two smallest possible values of n ? | "4 x + 1 = 5 y + 1 . . . . . . . . . . . ie : 4 x - 5 y = 2 x , y must be > 1 and y is even ie ( 2 , 4,6 , . . etc ) if y = 2 thus x = fraction ( not possible ) if y = 4 thus x = 5 n = 21 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 41 21 + 41 = 62 . . . . . b" | a = 5 * 2
b = a + 1
c = 2 * 1
d = 5 * c
e = d + 1
f = b + e
|
a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 10 % , e ) 10.9 % | e | divide(multiply(12, const_100), add(12, const_100)) | the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 12 percent , what was the annual interest rate last year ? | "12 = 1.1 * x x = 10.90 % answer e )" | a = 12 * 100
b = 12 + 100
c = a / b
|
a ) 10 , b ) 32 , c ) 15 , d ) 18 , e ) 19 | b | multiply(multiply(add(6, divide(subtract(sqrt(100), 6), 2)), divide(subtract(sqrt(100), 6), 2)), 2) | if a - b = 6 and a ^ 2 + b ^ 2 = 100 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 100 - 36 = 64 = > ab = 32 answer : b" | a = math.sqrt(100)
b = a - 6
c = b / 2
d = 6 + c
e = math.sqrt(100)
f = e - 6
g = f / 2
h = d * g
i = h * 2
|
a ) s . 13 , b ) s . 12 , c ) s . 15 , d ) s . 22 , e ) s . 14 | e | divide(subtract(multiply(20, 7), 56), subtract(7, const_1)) | the average amount with a group of 7 numbers is rs . 20 . if the newly joined member has rs . 56 with him , what was the average amount with the group before his joining the group ? | total members in the group = 7 average amount = rs . 20 total amount with them = 7 * 20 = rs . 140 one number has rs . 56 . so , the amount with remaining 6 people = 140 - 56 = rs . 84 the average amount with them = 84 / 6 = rs . 14 . answer : e | a = 20 * 7
b = a - 56
c = 7 - 1
d = b / c
|
a ) 8.55 , b ) 6.55 , c ) 8.75 , d ) 7.75 , e ) 4.25 | e | divide(multiply(2.125, const_100), 50) | find the missing figures : ? % of 50 = 2.125 | ( i ) let x % of 50 = 2.125 . then , ( x / 100 ) * 50 = 2.125 x = ( 2.125 * 2 ) = 4.25 answer is e . | a = 2 * 125
b = a / 50
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a ) 167.5 , b ) 150 , c ) 200 , d ) 112.5 , e ) 212.5 | c | divide(subtract(divide(multiply(multiply(5000, 8), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 8 % p . a for 2 years . find his gain in the transaction per year . | explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 4 × 2 / 100 = 400 he also lends it at 8 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 8 × 2 / 100 = 800 his overall gain in 2 years = rs . 800 - rs . 400 = rs . 400 his overall gain in 1 year = 400 / 2 = rs . 200 answer : option c | a = 5000 * 8
b = a * 2
c = b / 100
d = 5000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 27 , b ) 26 , c ) 42 , d ) 64 , e ) 147 | b | multiply(divide(48, 114), 63) | the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 114 ounces , what is the volume , in cubic inches , of 63 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 63 ounces of a substance has a volume of ( 48 / 114 ) * 63 = 26 cubic inches answer b" | a = 48 / 114
b = a * 63
|
['a ) 5 / 12', 'b ) 2 / 5', 'c ) 0.5451', 'd ) 3 / 4', 'e ) 5 / 6'] | c | divide(circle_area(10), square_area(24)) | a circular mat with radius 10 inches is placed on a square tabletop , each of whose sides is 24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? | c . it is a circle inscribed in a square . square side = 24 - - - > square ( table ) area = 24 ^ 2 circle diameter = 20 - - - > circle area = pir ^ 2 = 100 pi ( where pi = ~ 3.14 ) covered fraction = 100 * 3.14 / 24 * 24 = ~ 314 / 24 * 24 = 0.5451 c | a = circle_area / (
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a ) $ 3,750 , b ) $ 5,600 , c ) $ 13,500 , d ) $ 15,000 , e ) $ 22,500 | c | multiply(1500, power(const_3, divide(28, divide(112, 8)))) | money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 1500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 1500 triples to $ 4500 , in 28 years , it will again triple to $ 4500 * 3 = $ 13,500 answer c" | a = 112 / 8
b = 28 / a
c = 3 ** b
d = 1500 * c
|
a ) 7 , b ) 9 , c ) 16 , d ) 18 , e ) 21 | e | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 75), 5) | the sum of ages of 5 children born 3 years different each is 75 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 75 5 x = 45 x = 9 . x + 12 = 9 + 12 = 21 e" | a = 3 + 4
b = 3 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 75
g = f / 5
|
a ) 88 , b ) 90 , c ) 96 , d ) 98 , e ) 102 | c | multiply(subtract(multiply(const_2, multiply(2, 2)), 2), multiply(const_4, const_4)) | in how many ways can a teacher write an answer key for a mini - quiz that contains 3 true - false questions followed by 2 multiples - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | 2 multiples - choice questions can be answered in = 4 x 4 = 16 ways 3 true - false questions can be answered in = 2 x 2 x 2 = 8 ways but out of the 8 ways , 2 ways [ ( true - true - true ) ( false - false - false ) ] will contain same answers thus 3 true - false questions can be answered in = 2 x 2 x 2 = 6 ways total ways to answer the quiz = 16 x 6 = 96 answer c | a = 2 * 2
b = 2 * a
c = b - 2
d = 4 * 4
e = c * d
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a ) 5.5 % , b ) 6 % , c ) 6 1 / 3 % , d ) 6 2 / 3 % , e ) 7 % | e | multiply(divide(add(multiply(50, divide(5, const_100)), 1.5), add(50, add(1.5, 8.5))), const_100) | a 50 - liter solution of alcohol and water is 5 percent alcohol . if 1.5 liters of alcohol and 8.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "50 l * . 05 = 2.5 l of alc , 50 l - 2.5 l = 47.5 l of water 2.5 + 1.5 = 4 l of alcohol in new solution 47.5 l + 8.5 l = 56 l of water 4 l / 56 l = 1 / 14 this is ~ 7 % answer : e" | a = 5 / 100
b = 50 * a
c = b + 1
d = 1 + 5
e = 50 + d
f = c / e
g = f * 100
|
a ) 73 , b ) 74 , c ) 75 , d ) 76 , e ) 77 | e | add(add(divide(72, const_2), const_12), divide(add(add(divide(72, const_2), const_12), const_10), const_2)) | there are 3 departments having students 72 , 5824 . in an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only ( no mixing of departments ) . find the minimum number of rooms required ? | we need to take gcd which is 2 thus all the rooms will have 2 students of the same dept 1 ) 72 / 2 = 36 2 ) 58 / 2 = 29 3 ) 24 / 2 = 12 total no . of min rooms reqd = 36 + 12 + 29 = 77 answer : e | a = 72 / 2
b = a + 12
c = 72 / 2
d = c + 12
e = d + 10
f = e / 2
g = b + f
|
a ) 121.5 , b ) 122 , c ) 122.5 , d ) 113 , e ) 112.5 | e | divide(1, divide(add(multiply(const_3600, divide(1, 120)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 120 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "e 120 * t = 1 km = > t = 1 / 120 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 120 + 2 / 3600 ) = 1 = > v = 112.5 km / h" | a = 1 / 120
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
|
a ) 30.6 % , b ) 30.5 % , c ) 30.7 % , d ) 30.8 % , e ) 30.9 % | c | subtract(const_100, divide(multiply(1000, const_100), 765)) | an article is bought for rs . 765 and sold for rs . 1000 , find the gain percent ? | "765 - - - - 235 100 - - - - ? = > 30.7 % answer : c" | a = 1000 * 100
b = a / 765
c = 100 - b
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a ) 45 % , b ) 40 % , c ) 35 % , d ) 77.5 % , e ) 30 % | d | subtract(const_100, multiply(divide(subtract(const_100, 70), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 70 per cent . the two reductions together are equal to a single reduction of | "price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 70 percent is reduced which means the new price is merely 30 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 3 / 10 ) p = 9 / 40 p is the new price after both deduction which is 22.5 percent of the original value p . this implies this entire series of deduction is worth having discounted 77.5 % of p . so answer is d = 77.5 %" | a = 100 - 70
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
|
a ) 2080 , b ) 2778 , c ) 2299 , d ) 2778 , e ) 2771 | a | multiply(divide(add(const_100, 35), const_100), divide(add(1920, 1280), const_2)) | the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 35 % profit ? | let c . p . be rs . x . then , ( 1920 - x ) / x * 100 = ( x - 1280 ) / x * 100 1920 - x = x - 1280 2 x = 3200 = > x = 1600 required s . p . = 130 % of rs . 1600 = 130 / 100 * 1600 = rs . 2080 . answer : a | a = 100 + 35
b = a / 100
c = 1920 + 1280
d = c / 2
e = b * d
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a ) 648 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none | b | multiply(multiply(divide(270, 6), 4), 10) | running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the number of required bottles be x machine 6 : 10 time 1 : 4 = : : 270 : x therefore 6 x 1 x x = 10 x 4 x 270 = > x = ( 10 x 4 x 270 ) / ( 6 ) = > x = 1800 answer : option b" | a = 270 / 6
b = a * 4
c = b * 10
|
a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140 | c | divide(add(40, 30), divide(subtract(const_100, 30), const_100)) | thirty percent of the members of a swim club have passed the lifesaving test . among the members who havenotpassed the test , 40 have taken the preparatory course and 30 have not taken the course . how many members are there in the swim club ? | "30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 30 + 40 = 70 members have not passed the test , thus 0.7 * total = 70 - - > total = 100 . answer : c ." | a = 40 + 30
b = 100 - 30
c = b / 100
d = a / c
|
a ) 50 , b ) 45 , c ) 44 , d ) 60 , e ) 65 | c | add(lcm(lcm(9, 6), lcm(12, 18)), 8) | what is the smallest number which when diminished by 8 , is divisible by 9 , 6 , 12 and 18 ? | "explanation : required number = lcm of ( 9 , 6 , 12 and 18 ) + 8 = 36 + 8 = 44 answer : option c" | a = math.lcm(9, 6)
b = math.lcm(12, 18)
c = math.lcm(a, b)
d = c + 8
|
a ) 5 : 6 , b ) 25 : 27 , c ) 15 : 16 , d ) 20 : 21 , e ) it can not be determined from the information given | e | divide(add(1, 15), add(2, 15)) | the ratio of two quantities is 1 to 2 . if each of the quantities is increased by 15 , what is the ratio of these 2 new quantities ? | "the ratio can not be straight way added any quantity . . . 1 : 2 means 1 x : 2 x . . . so when you add a quantity , it becomes 1 x + 15 : 2 x + 15 . . so value of x is must . . ans e" | a = 1 + 15
b = 2 + 15
c = a / b
|
a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 7 , d ) 1 / 5 , e ) 1 / 4 | c | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets rain cloud , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction of the total distance did he cover while the sun was shining on him ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is 2 / 14 = 1 / 7 . answer : c | a = math.lcm(2, 3)
b = a / 2
c = b - 2
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['a ) 120', 'b ) 150', 'c ) 200', 'd ) 270', 'e ) 420'] | e | add(multiply(divide(divide(factorial(10), factorial(subtract(10, const_2))), factorial(const_2)), divide(factorial(6), factorial(subtract(6, const_1)))), multiply(divide(factorial(10), factorial(subtract(const_10, const_1))), divide(divide(factorial(6), const_2), factorial(subtract(6, const_2))))) | two tracks are parallel . the first track has 6 checkpoints and the second one has 10 checkpoints . in how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle ? | to make a triangle , you need 2 checkpoints from one track and 1 from the other . you can not have all 3 from the same track since then the points will be in a line ( assuming straight line of track ) you select 2 checkpoints from the first track and one from the second or two from the second track and one from the first . 6 c 2 * 10 c 1 + 10 c 2 * 6 c 1 = 150 + 270 = 420 answer ( e ) | a = math.factorial(10)
b = 10 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 1
i = math.factorial(h)
j = g / i
k = f * j
l = math.factorial(10)
m = 10 - 1
n = math.factorial(m)
o = l / n
p = math.factorial(6)
q = p / 2
r = 6 - 2
s = math.factorial(r)
t = q / s
u = o * t
v = k + u
|
a ) a . 43 , b ) b . 45 , c ) c . 55 , d ) d . 68 , e ) e . 83 | e | subtract(divide(multiply(multiply(55, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 55) | an engineer undertakes a project to build a road 15 km long in 300 days and employs 55 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "55 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 55 * 12.5 * 100 ) / ( 2.5 * 200 ) thus , x = 137.5 that is approximately 138 thus , more men needed to finish the task = 138 - 55 = 83 hence answer is e" | a = 15 - 2
b = 55 * a
c = b * 100
d = 300 - 100
e = 2 * 5
f = c / e
g = f - 55
|
a ) $ 500 , b ) $ 550 , c ) $ 600 , d ) $ 650 , e ) $ 700 | b | subtract(multiply(2000, add(const_1, divide(12, const_100))), multiply(1000, power(add(const_1, divide(30, const_100)), const_2))) | bob invested $ 2000 in fund a and $ 1000 in fund b . over the next two years , the money in fund a earned a total interest of 12 percent for the two years combined and the money in fund b earned 30 percent annual interest compounded annually . two years after bob made these investments . bob ' s investment in fund a was worth how much more than his investment in fund b ? | bob earned 2000 * ( 1 + 0.12 ) in total by fund a and earned 1000 * ( 1 + 0.3 ) ^ 2 in total by fund b . so 2000 * ( 1.12 ) - 1000 * ( 1.69 ) = 550 . the answer is , therefore , ( b ) . | a = 12 / 100
b = 1 + a
c = 2000 * b
d = 30 / 100
e = 1 + d
f = e ** 2
g = 1000 * f
h = c - g
|
a ) 2 : 9 , b ) 1 : 5 , c ) 3 : 6 , d ) 2 : 0 , e ) 2 : 1 | b | divide(subtract(6, 5), 5) | the ratio between the sale price and the cost price of an article is 6 : 5 . what is the ratio between the profit and the cost price of that article ? | let c . p . = rs . 5 x and s . p . = rs . 6 x . then , gain = rs . x required ratio = x : 5 x = 1 : 5 answer : b | a = 6 - 5
b = a / 5
|
a ) 31 , b ) 35 , c ) 50 , d ) 99 , e ) 310 | e | floor(add(const_1, multiply(divide(log(2), log(const_10)), 1000))) | how many digits 2 ^ 1000 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 = > 2 ^ 1000 = ( 1.024 ) ^ 100 * 10 ^ 300 therefore 310 digits would be my best guess e" | a = math.log(2)
b = math.log(10)
c = a / b
d = c * 1000
e = 1 + d
f = math.floor(e)
|
a ) 22 , b ) 87 , c ) 29 , d ) 26 , e ) 191 | b | divide(choose(const_2, const_1), factorial(3)) | a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose their own dress ? | answer : b | a = math.comb(2, 1)
b = math.factorial(3)
c = a / b
|
a ) 120 , b ) 150 , c ) 240 , d ) 220 , e ) 200 | e | multiply(divide(const_60, 60), 100) | if the population of a certain country increases at the rate of two person every 60 seconds , by how many persons does the population increase in 100 minutes ? | "answer = 2 * 100 = 200 answer = e" | a = const_60 / 60
b = a * 100
|
a ) 100 , b ) 105 , c ) 109 , d ) 120 , e ) 90 | a | divide(divide(multiply(add(multiply(2500, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3))) | a salt manufacturing company produced a total of 2500 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ? | "total production of salt by the company in that year = 2500 + 2600 + 2700 + . . . . + 3600 = 36600 . average monthly production of salt for that year = 36600 / 365 â ‰ ˆ 100 answer : a" | a = 2500 * 2
b = 12 - 1
c = b * 100
d = a + c
e = d * 12
f = e / 2
g = 3 * 100
h = 2 * 3
i = h * 10
j = g + i
k = 2 + 3
l = j + k
m = f / l
|
a ) 12.9 , b ) 12.0 , c ) 12.5 , d ) 12.2 , e ) 13.1 | e | divide(multiply(18, 1200), add(1200, 450)) | 1200 men have provisions for 18 days . if 450 more men join them , for how many days will the provisions last now ? | "1200 * 18 = 1650 * x x = 13.1 answer : e" | a = 18 * 1200
b = 1200 + 450
c = a / b
|
a ) 18.75 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | when a train travels at a speed of 90 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 90 km / h 50 km / h - - - - > x + 15 distance is equal so 90 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 90 x = 50 x + 750 so the would be in km and x = 18.75 answer : a" | a = 15 / const_60
b = a * 50
c = b / 50
d = const_60 * c
|
a ) 180 , b ) 200 , c ) 220 , d ) 240 , e ) 260 | b | divide(add(80, 40), divide(60, const_100)) | a student needs 60 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 40 marks , find the maximum marks set for the test . | "60 % = 120 marks 1 % = 2 marks 100 % = 200 marks the answer is b ." | a = 80 + 40
b = 60 / 100
c = a / b
|
a ) 122821 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 40200 | e | divide(multiply(1, 401), const_4) | what is the sum of all even numbers from 1 to 401 ? | "explanation : 400 / 2 = 200 200 * 201 = 40200 answer : e" | a = 1 * 401
b = a / 4
|
a ) 42 , b ) 40 , c ) 38 , d ) 36 , e ) 48 | b | subtract(divide(power(21, const_2), const_2), divide(power(19, const_2), const_2)) | the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 21 - inch flat - screen television than a square 19 - inch flat - screen television ? | "if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 21 - inch flat - screen television the diagonal ( hypotenuse ) = 21 so , we can apply the pythagorean theorem to get x ² + x ² = 21 ² simplify : 2 x ² = 21 ² divide both sides by 2 to get : x ² = 21 ² / 2 since the area of the square = x ² , we can see that the area of this square is 21 ² / 2 square 19 - inch flat - screen television the diagonal ( hypotenuse ) = 19 so , we can apply the pythagorean theorem to get x ² + x ² = 19 ² simplify : 2 x ² = 19 ² divide both sides by 2 to get : x ² = 19 ² / 2 since the area of the square = x ² , we can see that the area of this square is 19 ² / 2 difference in areas = 21 ² / 2 - 19 ² / 2 = ( 21 + 19 ) ( 21 - 19 ) / 2 = 40 b" | a = 21 ** 2
b = a / 2
c = 19 ** 2
d = c / 2
e = b - d
|
a ) 37 kg , b ) 35 kg , c ) 85 kg , d ) 45 kg , e ) 25 kg | b | multiply(divide(7, add(7, 13)), 100) | the proportion of copper and zinc in the brass is 13 : 7 . how much zinc will there be in 100 kg of brass ? | 7 / 20 * 100 = 35 answer : b | a = 7 + 13
b = 7 / a
c = b * 100
|
a ) 33.56 , b ) 23.68 , c ) 11.42 , d ) 9.56 , e ) 23.55 | c | divide(multiply(subtract(const_100, 8), 18), add(const_100, 45)) | if a man lost 8 % by selling oranges at the rate of 18 a rupee at how many a rupee must he sell them to gain 45 % ? | "92 % - - - - 18 145 % - - - - ? 92 / 145 * 18 = 11.42 answer : c" | a = 100 - 8
b = a * 18
c = 100 + 45
d = b / c
|
a ) 23 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(add(negate(10), sqrt(subtract(power(negate(10), 2), multiply(const_4, negate(divide(multiply(150, 10), 2)))))), 2) | the time it took car p to travel 150 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | "let speed of car r be = x then speed of car p = x + 10 a / q , ( 150 / x ) - ( 150 / ( x + 10 ) ) = 2 solving for x = 23 miles \ hr . a" | a = negate + (
b = negate ** (
c = b - 2
d = 150 * 10
e = d / 2
f = 4 * negate
g = math.sqrt(c)
h = a / g
|
a ) 0.98 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 2), const_100) | subtracting 2 % of a from a is equivalent to multiplying a by how much ? | answer let a - 2 % of a = ab . ⇒ ( 98 x a ) / 100 = ab ∴ b = 0.98 correct option : a | a = 100 - 2
b = a / 100
|
a ) 42 m sqaure , b ) 49 m sqaure , c ) 25 m sqaure , d ) 28 m sqaure , e ) none of these | c | add(multiply(const_2, add(multiply(add(divide(10, const_100), 1), 2), multiply(add(divide(10, const_100), 1), 5))), multiply(2, 5)) | a cistern 5 m long and 2 m wide contains water up to a breadth of 1 m 10 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 2 * 1.1 + 5 * 1.1 ) ] + 5 * 2 = 25 m square option c" | a = 10 / 100
b = a + 1
c = b * 2
d = 10 / 100
e = d + 1
f = e * 5
g = c + f
h = 2 * g
i = 2 * 5
j = h + i
|
a ) 24 , b ) 27 , c ) 30 , d ) 33 , e ) 36 | b | multiply(divide(subtract(42, 6), 4), 3) | right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 42 years . what is molly ' s age right now ? | "now , sandy is 42 - 6 = 36 molly ' s age is ( 3 / 4 ) * 36 = 27 the answer is b ." | a = 42 - 6
b = a / 4
c = b * 3
|
a ) 40 , b ) 35 , c ) 20 , d ) 25 , e ) 30 | c | divide(240, const_12) | how many boxes do you need if you have to pack 240 pairs ornamental bangles into boxes that each hold 2 dozens of bangles ? | "c 20 240 pairs of bangles = 480 bangles = 40 dozens . 40 ÷ 2 = 20 boxes ." | a = 240 / 12
|
a ) 355600 , b ) 355800 , c ) 404600 , d ) 356800 , e ) 357000 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(85, const_100)) | in an election , candidate a got 85 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate ? | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 85 % therefore , the number of valid votes polled in favour of candidate a = 85 % of 476000 = 85 / 100 × 476000 = 40460000 / 100 = 404600 c )" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 85 / 100
e = c * d
|
a ) 7.5 % , b ) 5 % , c ) 12 % , d ) 8 % , e ) 10 % | a | multiply(subtract(divide(98.9, 92), const_1), const_100) | jerry bought a bottle of perfume for a gift for his wife . the perfume cost $ 92 before tax . if the total price including tax was $ 98.90 , find the tax rate | total price including tax is $ 98.90 perfume cost before tax is = 92 ie 92 * 7.5 % + 98.90 answer is 7.5 % | a = 98 / 9
b = a - 1
c = b * 100
|
a ) 175.5 , b ) 182.5 , c ) 170.0 , d ) 180.0 , e ) 190.0 | a | add(add(add(add(153, const_10), const_10), 2), add(const_0_25, const_0_25)) | tea worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , what is the price of the third variety per kg ? | "tea worth rs . 126 ratio 1 : 1 average price = ( 126 + 135 ) / 2 = 130.5 mean price = ( x - 153 ) : 22.50 = > x - 153 = 22.50 x = 175.5 answer a" | a = 153 + 10
b = a + 10
c = b + 2
d = const_0_25 + const_0_25
e = c + d
|
a ) 33 , b ) 36 , c ) 39 , d ) 42 , e ) 45 | e | add(add(add(add(divide(3.6, divide(8, const_60)), 3), 3), 3), 3) | the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 3.6 miles in the seventh 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | "( 3.6 miles / 8 minutes ) * 60 minutes / hour = 27 mph let x be the original speed . x - 6 ( 3 ) = 27 x = 45 mph the answer is e ." | a = 8 / const_60
b = 3 / 6
c = b + 3
d = c + 3
e = d + 3
f = e + 3
|
a ) 11 , b ) 12 , c ) 10 , d ) 22 , e ) 44 | a | add(divide(factorial(add(3, 2)), multiply(factorial(3), factorial(2))), const_1) | a 7 - digit number comprises of only 2 ' s and 3 ' s . how many of these are multiples of 12 ? | detailed solution number should be a multiple of 3 and 4 . so , the sum of the digits should be a multiple of 3 . we can either have all seven digits as 3 , or have three 2 ' s and four 3 ' s , or six 2 ' s and a 3 . ( the number of 2 ' s should be a multiple of 3 ) . for the number to be a multiple of 4 , the last 2 digits should be 32 . now , let us combine these two . all seven 3 ' s - no possibility . three 2 ' s and four 3 ' s - the first 5 digits should have two 2 ' s and three 3 ' s in some order . no of possibilities = 5 ! 3 ! 2 ! = 10 six 2 ' s and one 3 - the first 5 digits should all be 2 ' s . so , there is only one number 2222232 . so , there are a total of 10 + 1 = 11 solutions . correct answer : a . | a = 3 + 2
b = math.factorial(a)
c = math.factorial(3)
d = math.factorial(2)
e = c * d
f = b / e
g = f + 1
|
a ) rs . 4,900 , b ) rs . 5,000 , c ) rs . 5,100 , d ) rs . 5266 , e ) none of these | d | divide(8110, add(const_1, divide(multiply(6, 9), const_100))) | mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8110 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8110 = p + ( 54 p / 100 ) 8110 = ( 100 p + 54 p ) / 100 8110 = 154 p / 100 p = ( 8110 * 100 ) / 154 = rs . 5266.233 answer d" | a = 6 * 9
b = a / 100
c = 1 + b
d = 8110 / c
|
a ) 5 % , b ) 10 % , c ) 25 % , d ) 40 % , e ) 55 % | e | add(divide(const_100, const_2), divide(multiply(10, divide(const_100, const_2)), const_100)) | a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 10 percent of the drive from the center , what percent of the round - trip has the technician completed ? | "assuming that the technician makes a round - trip of 40 miles ( each way is 20 miles ) , then the technician would have completed 20 miles + 2 miles ( 10 % of the remaining 20 miles ) . therefore , the total is 22 miles . 22 miles / 40 miles is 55 % of the entire trip . answer : e" | a = 100 / 2
b = 100 / 2
c = 10 * b
d = c / 100
e = a + d
|
a ) 210 , b ) 280 , c ) 360 , d ) 720 , e ) none | d | divide(multiply(60, 54), subtract(60, 54)) | the banker â € ™ s discount of a certain sum of money is rs . 60 and the true discount on the same sum for the same time is rs . 54 . the sum due is | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 60 * 54 / 60 - 54 ] = rs . [ 72 * 60 / 6 ] = rs . 720 answer d" | a = 60 * 54
b = 60 - 54
c = a / b
|
a ) 15 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(25, 10), 10), subtract(25, subtract(25, 10))) | a certain number of workers can do a work in 25 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | number of workers = 10 * ( 25 - 10 ) / 10 = 15 answer is a | a = 25 - 10
b = a * 10
c = 25 - 10
d = 25 - c
e = b / d
|
a ) 20 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 60 liters | a | divide(9, subtract(divide(3, 4), divide(30, const_100))) | a big container is 30 % full with water . if 9 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 9 liters of water is added , the container becomes 75 % full . hence these 9 liters account for 45 % of the container , which means that the capacity of it is 9 / 0.45 = 20 liters . or : if the capacity of the container is x liters then : 0.3 x + 9 = 0.75 x - - > x = 20 liters . answer : a ." | a = 3 / 4
b = 30 / 100
c = a - b
d = 9 / c
|
a ) 19.81 % , b ) 20 % , c ) 37.5 % , d ) 25 % , e ) 42.86 % | e | multiply(subtract(add(const_100, 11), add(9.09, subtract(const_100, 20))), const_2) | a man cheats while buying as well as while selling . while buying he takes 11 % more than what he pays for and while selling he gives 20 % less than what he claims to . find the profit percent , if he sells at 9.09 % below the cost price of the claimed weight . | there is a one step calculation method too . it requires more thought but is faster . the man takes 11 % more than what he pays for . so if he claims to take 100 pounds , he pays $ 100 but he actually takes 111 pounds for which he will take from the customer $ 111 . hence , in effect , there is a 11 % mark up . while selling , he sells 30 % less . this means , he claims to sell 100 pounds and gets $ 100 but actually sells only 70 pounds and should have got only $ 70 for it . so this is again a mark up of $ 30 on $ 70 . but he also sells at 9.09 % less i . e . gives a discount of 1 / 11 . ( 1 + m 1 % ) ( 1 + m 2 % ) ( 1 - d % ) = ( 1 + p % ) 11 / 10 * 10 / 7 * 10 / 11 = ( 1 + p % ) profit % = 42.86 % e | a = 100 + 11
b = 100 - 20
c = 9 + 9
d = a - c
e = d * 2
|
a ) 68 , b ) 84 , c ) 90 , d ) 110 , e ) 135 | d | divide(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), subtract(divide(multiply(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), const_3), 90), add(multiply(const_4, const_2), const_3))) | a train traveled the first d miles of its journey it an average speed of 60 miles per hour , the next d miles of its journey at an average speed of y miles per hour , and the final d miles of its journey at an average speed of 160 miles per hour . if the train ’ s average speed over the total distance was 90 miles per hour , what is the value of y ? | "average speed = total distance traveled / total time taken 3 d / d / 60 + d / y + d / 160 = 90 solving for d and y , 15 y = 11 y + 480 4 y = 440 y = 110 answer d" | a = 4 * 2
b = a * 10
c = 100 * 4
d = b + c
e = 4 * 2
f = e * 10
g = 100 * 4
h = f + g
i = h * 3
j = i / 90
k = 4 * 2
l = k + 3
m = j - l
n = d / m
|
a ) 8 , b ) 10 , c ) 5 , d ) 14 , e ) 7 | c | divide(subtract(75, multiply(const_3, 15)), multiply(const_3, const_2)) | a number is doubled and 15 is added . if resultant is trebled , it becomes 75 . what is that number | "explanation : = > 3 ( 2 x + 15 ) = 75 = > 2 x + 15 = 25 = > x = 5 option c" | a = 3 * 15
b = 75 - a
c = 3 * 2
d = b / c
|
a ) 1300 , b ) 1200 , c ) 1375 , d ) 1400 , e ) 8400 | b | divide(2800, const_3) | divide rs . 2800 among a , b and c so that a receives 3 / 4 as much as b and c together and b receives 1 / 4 as a and c together . a ' s share is ? | "a + b + c = 2800 a = 3 / 4 ( b + c ) ; b = 1 / 4 ( a + c ) a / ( b + c ) = 3 / 4 a = 1 / 7 * 8400 = > 1200 answer : b" | a = 2800 / 3
|
a ) 3675 , b ) 3575 , c ) 3475 , d ) 3375 , e ) 3275 | a | multiply(subtract(divide(multiply(69, add(69, const_1)), const_2), multiply(divide(subtract(69, const_1), const_2), add(divide(subtract(69, const_1), const_2), const_1))), const_3) | if 1 + 2 + 3 + . . . + n = n ( n + 1 ) , then 3 ( 1 + 3 + 5 + . . . . + 69 ) = ? | explanation : to solve this use the formula of ap , sn = ( n / 2 ) ( a + l ) . . . . . . . . . . . . . . . . ( 1 ) to find n , use = > tn = a + ( n - 1 ) d = > 69 = 1 + ( n - 1 ) 2 = > n = 35 use value of n in ( 1 ) then , sn = ( 35 / 2 ) ( 1 + 69 ) = 1225 ans : - 3 ( sn ) = 3675 answer : a | a = 69 + 1
b = 69 * a
c = b / 2
d = 69 - 1
e = d / 2
f = 69 - 1
g = f / 2
h = g + 1
i = e * h
j = c - i
k = j * 3
|
a ) a ) 407 , b ) b ) 518 , c ) c ) 519 , d ) d ) 521 , e ) e ) 525 | a | add(subtract(414, 8), const_1) | if the average ( arithmetic mean ) of 8 consecutive odd integers is 414 , then the least of these integers is | a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 414 . there are 8 numbers in the set , and in a set with an even number of terms the median is just the average of the two most median terms ( here the 4 th and 5 th numbers in the set ) . this means that numbers 4 and 5 in this set are 413 and 415 . because we know that number 4 is 413 , we know that the smallest number is 3 odd numbers below this , which means that it is 3 * 2 = 6 below this ( every odd number is every other number ) . therefore 413 - 6 = 407 , answer choice a | a = 414 - 8
b = a + 1
|
a ) 1 / 250 , b ) 1 / 84 , c ) 1 / 11 , d ) 1 / 9 , e ) 1 / 3 | d | multiply(divide(84, 250), divide(subtract(84, const_1), add(250, const_1))) | a shipment of 250 smartphones contains 84 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ? a . b . c . d . e . | probability of choosing one defective phone from a lot of 250 which ontains 84 defective phones is = ( 84 / 250 ) probability of choosing one defective phone from a lot of 249 ( we already picked one ) which ontains 83 ( we already picked one ) defective phones is = ( 83 / 249 ) combined probability of series of events = product of the probabilities = ( 84 / 250 ) * ( 83 / 249 ) 84 / 250 is close to ( 1 / 3 ) and ( 83 / 249 ) = ( 1 / 3 ) so answer is ( 1 / 3 ) * ( 1 / 3 ) = ( 1 / 9 ) answer : d | a = 84 / 250
b = 84 - 1
c = 250 + 1
d = b / c
e = a * d
|
a ) 11 , b ) 52 , c ) 83 , d ) 94 , e ) 99 | e | subtract(add(add(add(add(add(const_1, power(const_2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_4, const_1), const_2)), power(7, const_2)), add(add(add(const_4, const_3), 7), add(add(add(add(const_2, const_3), add(const_4, const_1)), add(const_4, const_3)), add(add(const_2, const_3), 7)))) | what is the positive difference between the sum of the squares of the first 7 positive integers and the sum of the prime numbers between the first square and fourth square ? | "forget conventional ways of solving math questions . in ps , ivy approach is the easiest and quickest way to find the answer . the sum of the squares of the first 4 positive integers = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 7 ^ 2 = 140 the sum of the prime numbers between the first square ( = 1 ) and fourth square ( = 16 ) = 2 + 3 + 5 + 7 + 11 + 13 = 41 . so the difference between 41 and 140 is 99 . so the answer is ( e ) ." | a = 2 ** 2
b = 1 + a
c = 3 ** 2
d = b + c
e = 4 ** 2
f = d + e
g = 4 + 1
h = g ** 2
i = f + h
j = 7 ** 2
k = i + j
l = 4 + 3
m = l + 7
n = 2 + 3
o = 4 + 1
p = n + o
q = 4 + 3
r = p + q
s = 2 + 3
t = s + 7
u = r + t
v = m + u
w = k - v
|
a ) 79698 , b ) 80578 , c ) 500578 , d ) 81268 , e ) none of them | c | multiply(517, power(517, 483)) | 517 x 517 + 483 x 483 = ? | "= ( 517 ) ^ 2 + ( 483 ) ^ 2 = ( 500 + 17 ) ^ 2 + ( 500 - 17 ) ^ 2 = 2 [ ( 500 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 250000 + 289 ] = 2 x 250289 = 500578 answer is c" | a = 517 ** 483
b = 517 * a
|
['a ) 0', 'b ) 1', 'c ) 3', 'd ) 4', 'e ) 2'] | d | multiply(subtract(divide(power(3, const_2), 5), floor(divide(power(3, const_2), 5))), 5) | on dividing a number by 5 , we get 3 as remainder . what will be the remainder when the square of this number is divided by 5 ? | let the number be x and on dividing x by 5 , we get k as quotient and 3 as remainder . x = 5 k + 3 x ^ 2 = ( 5 k + 3 ) ^ 2 = ( 25 k ^ 2 + 30 k + 9 ) = 5 ( 5 k ^ 2 + 6 k + 1 ) + 4 on dividing x ^ 2 by 5 , we get 4 as remainder . answer is d | a = 3 ** 2
b = a / 5
c = 3 ** 2
d = c / 5
e = math.floor(d)
f = b - e
g = f * 5
|
a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 45 % , e ) 50 % | d | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 38))), multiply(8, divide(const_1, 38))), multiply(8, divide(const_1, 38))), const_100) | a certain car uses one gallon of gasoline every 38 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | "4 miles on the highway = 4 / 38 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 38 + 4 / 20 = 29 / 95 gallons . 8 miles on the highway = 8 / 38 gallons . the % change = ( 29 / 95 - 8 / 38 ) / ( 8 / 38 ) = 0.45 . answer : d ." | a = 1 / 20
b = a * 4
c = 1 / 38
d = 4 * c
e = b + d
f = 1 / 38
g = 8 * f
h = e - g
i = 1 / 38
j = 8 * i
k = h / j
l = k * 100
|
a ) $ 50 , b ) $ 500 , c ) $ 150 , d ) $ 250 , e ) $ 100 | e | divide(300, 1) | divide $ 300 among a , b in the ratio 1 : 2 . how many $ that a get ? | "sum of ratio terms = 1 + 2 = 3 a = 300 * 1 / 3 = $ 100 answer is e" | a = 300 / 1
|
a ) 32 kmph , b ) 34 kmph , c ) 30 kmph , d ) 50 kmph , e ) 65 kmph | d | divide(add(45, 55), const_2) | a man can row upstream at 45 kmph and downstream at 55 kmph , and then find the speed of the man in still water ? | "us = 45 ds = 55 m = ( 45 + 55 ) / 2 = 50 answer : d" | a = 45 + 55
b = a / 2
|
['a ) 3 cm', 'b ) 4 cm', 'c ) 6 cm', 'd ) 8 cm', 'e ) none'] | b | sqrt(divide(multiply(multiply(const_pi, multiply(8, divide(8, const_2))), const_2), multiply(const_pi, const_4))) | the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 8 cm each . the radius of the sphere is | solution 4 î r 2 = 2 î 4 x 8 â ‡ ’ r 2 = ( 4 x 8 / 2 ) â ‡ ’ 16 â ‡ ’ r = 4 cm . answer b | a = 8 / 2
b = 8 * a
c = math.pi * b
d = c * 2
e = math.pi * 4
f = d / e
g = math.sqrt(f)
|
a ) 202 , b ) 217 , c ) 232 , d ) 247 , e ) 262 | d | add(add(add(add(add(add(add(30, const_1), add(add(30, const_1), const_1)), add(add(add(30, const_1), const_1), const_2)), add(add(add(add(30, const_1), const_1), const_2), const_1)), add(add(add(add(add(30, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(30, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(30, const_1), const_1), const_2), const_1), const_1), const_1), const_1)) | the sum of the non - prime numbers between 30 and 40 , non - inclusive , is | "sum of consecutive integers from 31 to 39 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 31 + 39 ) / 2 * 9 = 35 * 9 = 315 sum of non - prime numbers b / w 30 and 40 , non inclusive = = = > 315 - 68 ( i . e . , 31 + 37 , being the prime # s in the range ) = 247 answer : d" | a = 30 + 1
b = 30 + 1
c = b + 1
d = a + c
e = 30 + 1
f = e + 1
g = f + 2
h = d + g
i = 30 + 1
j = i + 1
k = j + 2
l = k + 1
m = h + l
n = 30 + 1
o = n + 1
p = o + 2
q = p + 1
r = q + 1
s = m + r
t = 30 + 1
u = t + 1
v = u + 2
w = v + 1
x = w + 1
y = x + 1
z = s + y
A = 30 + 1
B = A + 1
C = B + 2
D = C + 1
E = D + 1
F = E + 1
G = F + 1
H = z + G
|
a ) 17 , b ) 88 , c ) 14 , d ) 65 , e ) 89 | b | subtract(divide(power(negate(9), 2), 4), 2) | find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 9 x + 4 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 9 / 1 = - 9 ab = 4 / 1 = 4 hence a / b + b / a = [ ( - 9 ) 2 - 2 ( 4 ) ] / 4 = 73 / 4 = 18.25 . b )" | a = negate ** (
b = a / 2
c = b - 4
|
a ) 75 . , b ) 85 . , c ) 90 . , d ) 94 . , e ) 100 . | b | divide(subtract(multiply(6, 80), multiply(const_2, 70)), const_4) | 6 wires are by average 80 cm long each . if the average length of one third of the wires is 70 cm , what is the average of the other wires ? | edit : given ( x 1 + x 2 . . . + x 6 ) / 6 = 80 ( x 1 + x 2 . . . + x 6 ) = 480 - - > eq 1 . now given avg length of one third wires is 70 . that means out 6 / 3 = 2 wires . let the avg length of two wires be ( x 1 + x 2 ) / 2 = 70 . ( x 1 + x 2 ) = 140 . - - > eq 2 . now we are asked to find the average of the remaining i . e . ( x 3 + x 4 + x 5 + x 6 ) substitute eq 2 in eq 1 then we get 140 + x 3 + x 4 + x 5 + x 6 = 480 = > x 3 + x 4 + x 5 + x 6 = 340 now divide 340 by 4 we get 85 . = > ( x 3 + x 4 + x 5 + x 6 ) / 4 = 85 = avg length of remaining wires . imo correct option is b . | a = 6 * 80
b = 2 * 70
c = a - b
d = c / 4
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(add(subtract(15, 4), 4), 15) | when 15 is divided by integer u , the remainder is 4 . for how many values of u is this be true ? | when 15 is divided by u , the remainder is 4 i . e . 4 mangoes left over after grouping , so u must be greater than 4 . it also means that 11 is completely divisible by u . factors of 11 are 1 and 11 . out of these , u could be 11 . answer ( a ) | a = 15 - 4
b = a + 4
c = b / 15
|
a ) 4 : 5 , b ) 16 : 15 , c ) 9 : 13 , d ) 5 : 15 , e ) 9 : 17 | b | divide(multiply(4, 3), 5) | a dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare . compare their speeds ? | "let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare be y then 3 x = 4 y x = 4 / 3 y 4 x = 16 / 3 y ratio of speeds of dog and hare = ratio of distances covered by them in the same time = 4 x : 5 y = 16 / 3 y : 5 = 16 : 15 answer is b" | a = 4 * 3
b = a / 5
|
a ) 4 , b ) 9 , c ) 39 , d ) 48 , e ) 63 | c | subtract(add(power(7, 2), multiply(7, 2)), add(power(4, 2), multiply(4, 2))) | the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds than it was moving after 4 seconds ? | "given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 4 ) = 4 ^ 2 + 2 * 4 = 24 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 24 = 39 km / hr option c" | a = 7 ** 2
b = 7 * 2
c = a + b
d = 4 ** 2
e = 4 * 2
f = d + e
g = c - f
|
a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | if p # q denotes the least common multiple of p and q , then w = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , w = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c ." | a = 18 * 24
b = 12 * 16
c = a - b
d = c / 2
e = d + 24
|
a ) 40 , b ) 45 , c ) 48 , d ) 50 , e ) 55 | d | divide(multiply(60, divide(multiply(75, 60), multiply(subtract(75, 60), 60))), divide(60, const_10)) | a car travels from point a to point b . the average speed of the car is 60 km / hr and it travels the first half of the trip at a speed of 75 km / hr . what is the speed of the car in the second half of the trip ? | let d be the distance and let v be the speed in the second half . the total time = t 1 + t 2 d / 60 = d / 150 + ( d / 2 ) / v d / 100 = d / 2 v and so v = 50 km / hr the answer is d . | a = 75 * 60
b = 75 - 60
c = b * 60
d = a / c
e = 60 * d
f = 60 / 10
g = e / f
|
['a ) 4', 'b ) 8', 'c ) 16', 'd ) 32', 'e ) 64'] | d | add(add(sqrt(divide(128, const_2)), multiply(sqrt(divide(128, const_2)), const_2)), sqrt(divide(128, const_2))) | a rectangular table , kept against a wall has a three sides free and the wall along the fourth side . the side opposite the wall is twice the length of each of the other two free sides . if the area of the rectangular table is 128 square feet , what is the total length of the table free sides , in feet ? | two sides each = x the third = 2 x and the wall side is thus 2 x too x * 2 x = 2 x ^ 2 = 128 ie x ^ 2 = 64 ie x = 8 l = 16 w = 8 total lenght of table ' s free sides = 2 * 8 + 16 = 32 my answer is d | a = 128 / 2
b = math.sqrt(a)
c = 128 / 2
d = math.sqrt(c)
e = d * 2
f = b + e
g = 128 / 2
h = math.sqrt(g)
i = f + h
|
a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 80 kg , e ) 240 kg | e | multiply(multiply(multiply(8, 3), divide(1, const_100)), const_1000) | a boat having a length 8 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "volume of water displaced = ( 8 x 3 x 0.01 ) m 3 = 0.24 m 3 . mass of man = volume of water displaced x density of water = ( 0.24 x 1000 ) kg = 240 kg . answer : e" | a = 8 * 3
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 7.5 min , b ) 16 min , c ) 20 min , d ) 7 min , e ) 40 min | d | multiply(const_60, divide(subtract(86, 76), 86)) | excluding stoppages , the speed of a bus is 86 kmph and including stoppages , it is 76 kmph . for how many minutes does the bus stop per hour ? | "d 7 min due to stoppages , it covers 10 km less . time taken to cover 10 km = ( 10 / 86 x 60 ) min = 7 min" | a = 86 - 76
b = a / 86
c = const_60 * b
|
['a ) 22.78', 'b ) 23.54', 'c ) 24.5', 'd ) 25.55', 'e ) 23.51'] | a | floor(divide(circumface(divide(divide(multiply(const_2, add(14, 15)), const_4), const_2)), const_2)) | the perimeter of a square is equal to the perimeter of a rectangle of length 15 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | let the side of the square be a cm . perimeter of the rectangle = 2 ( 15 + 14 ) = 58 cm perimeter of the square = 58 cm i . e . 4 a = 58 a = 14.5 diameter of the semicircle = 14.5 cm circumference of the semicircle = 1 / 2 ( ∏ ) ( 14.5 ) = 1 / 2 ( 22 / 7 ) ( 14.5 ) = 22.78 cm to two decimal places answer : a | a = 14 + 15
b = 2 * a
c = b / 4
d = c / 2
e = circumface / (
f = math.floor(e, 2)
|
a ) 23 , b ) 77 , c ) 88 , d ) 48 , e ) 11 | d | lcm(multiply(3, 4), multiply(4, 4)) | the ratio of numbers is 3 : 4 and their h . c . f is 4 . their l . c . m is ? | "let the numbers be 3 x and 4 x . then their h . c . f = x . so , x = 4 . so , the numbers are 12 and 16 . l . c . m of 12 and 16 = 48 . answer : d" | a = 3 * 4
b = 4 * 4
c = math.lcm(a, b)
|
a ) 16 % , b ) 100 % , c ) 17 % , d ) 17.61 % , e ) 17.56 % | b | multiply(divide(subtract(120, 60), 60), const_100) | john makes $ 60 a week from his job . he earns a raise and now makes $ 120 a week . what is the % increase ? | "increase = ( 60 / 60 ) * 100 = ( 6 / 6 ) * 100 = 100 % . b" | a = 120 - 60
b = a / 60
c = b * 100
|
a ) 34 , b ) 32 , c ) 36 , d ) 38 , e ) 40 | b | divide(add(multiply(5, 20), multiply(5, 20)), subtract(add(5, 5), 5)) | a taxi leaves point a 5 hours after a bus left the same spot . the bus is traveling 20 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in three hours . | "let the speed of bus be v - 20 , speed of taxi be v the bus travelled a total of 8 hrs and taxi a total of 3 hrs . hence 8 * ( v - 20 ) = 3 v 8 v - 160 = 3 v 5 v = 160 v = 32 mph b" | a = 5 * 20
b = 5 * 20
c = a + b
d = 5 + 5
e = d - 5
f = c / e
|
a ) 4 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these | a | divide(multiply(multiply(divide(const_1, 15), 10), 12), const_2) | a can finish a work in 12 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 12 * 1 / 3 ) = 4 days . correct option : a" | a = 1 / 15
b = a * 10
c = b * 12
d = c / 2
|
a ) 56 , b ) 40 , c ) 47 , d ) 26 , e ) 25 | b | multiply(divide(add(1.10, 0.9), add(90, 90)), const_3600) | two trains are moving in opposite directions with speed of 90 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | "explanation : relative speed = 90 + 90 = 180 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 180 hr = 1 / 90 hr = 3600 / 90 seconds = 40 seconds answer : option b" | a = 1 + 10
b = 90 + 90
c = a / b
d = c * 3600
|
a ) 24 days , b ) 65 days , c ) 86 days , d ) 40 days , e ) 17 days | d | divide(10, subtract(const_1, divide(add(10, 10), 40))) | a can do a piece of work in 40 days ; b can do the same in 20 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 40 + 10 / 20 + 10 / x = 1 x = 40 days answer : d" | a = 10 + 10
b = a / 40
c = 1 - b
d = 10 / c
|
a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | b | subtract(5.2, const_4) | for any number s , s * is defined as the greatest positive even integer less than or equal to s . what is the value of 5.2 – 5.2 * ? | since s * is defined as the greatest positive even integer less than or equal to s , then 5.2 * = 4 ( the greatest positive even integer less than or equal to 5.2 is 4 ) . hence , 5.2 – 5.2 * = 5.2 - 4 = 1.2 answer : b . | a = 5 - 2
|
a ) 30 , b ) 31 , c ) 32 , d ) 35 , e ) 36 | c | multiply(divide(add(multiply(7, const_3), 59), add(add(5, 7), 8)), 8) | the present ages of a , b & c are in the ratio of 5 : 7 : 8 . 7 years ago , the sum oftheir ages was 59 . what is the present age of eldest one ? | let their present ages be 5 x , 7 x and 8 x . 7 years ago , sum of their ages was 59 . sum of their present ages = 59 + ( 3 x 7 ) = 80 sum of their present ages given is 5 x + 7 x + 8 x = 20 x 20 x = 80 = > x = 80 20 = 4 age of the eldest one = 8 x = 8 x 4 = 32 years . c | a = 7 * 3
b = a + 59
c = 5 + 7
d = c + 8
e = b / d
f = e * 8
|
a ) 277 , b ) 36 , c ) 64 , d ) 72 , e ) none of these | b | divide(multiply(129.6, const_100), 360) | ? % of 360 = 129.6 | "? % of 360 = 129.6 or , ? = 129.6 × 100 / 360 = 36 answer b" | a = 129 * 6
b = a / 360
|
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