options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 70 % | e | multiply(divide(252, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 252 ° of the circle , what percentage of megatek employees are in manufacturing ? | "answer : e 252 ° divided by 360 ° equals 0.7 , therefore the sector is equal to 70 % of the total" | a = 3600 / 10
b = 252 / a
c = b * 100
|
a ) 3 , b ) 7 , c ) 8 , d ) 6 , e ) 0 | d | subtract(180, multiply(inverse(multiply(divide(100, subtract(100, 13)), divide(subtract(100, 10), 100))), 180)) | in a 100 m race , a beats b by 10 m and c by 13 m . in a race of 180 m , b will beat c by : | a : b = 100 : 90 . a : c = 100 : 87 . \ inline \ fn _ jvn { \ color { black } \ therefore \ frac { b } { c } = \ left ( \ frac { b } { a } \ times \ frac { a } { c } \ right ) = \ left ( \ frac { 90 } { 100 } \ times \ frac { 100 } { 87 } \ right ) = \ frac { 30 } { 29 } } when b runs 30 m , c runs 29 m . when b runs 180 m , c runs \ inline \ fn _ jvn { \ color { black } \ left ( \ frac { 29 } { 30 } \ times 180 \ right ) m } = 174 m \ inline \ fn _ jvn { \ color { blue } \ therefore } b beats c by ( 180 - 174 ) m = 6 m . answer : d ) 6 m | a = 100 - 13
b = 100 / a
c = 100 - 10
d = c / 100
e = b * d
f = 1/(e)
g = f * 180
h = 180 - g
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a ) 114 , b ) 116 , c ) 118 , d ) 119 , e ) 120 | d | divide(multiply(subtract(multiply(divide(add(6, 2), const_2.0), 4), 5), add(multiply(divide(add(6, 5), 5), subtract(4, 5)), 3)), 5) | the points a ( 0 , 0 ) , b ( 0 , 4 a - 5 ) and c ( 2 a + 3 , 2 a + 6 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ? | "1 / 2 bh = 1 / 2 ( 2 a + 3 ) ( 2 a + 6 ) now 4 a - 5 = 2 a + 6 2 a = 11 therefore , a ( 0,0 ) ; b ( 0,17 ) ; c ( 14,17 ) 1 / 2 * 14 * 17 = 119 answer : d" | a = 6 + 2
b = a / 2
c = b * 4
d = c - 5
e = 6 + 5
f = e / 5
g = 4 - 5
h = f * g
i = h + 3
j = d * i
k = j / 5
|
a ) 280 , b ) 300 , c ) 240 , d ) 350 , e ) 400 | c | add(multiply(divide(15, const_100), multiply(multiply(3, 8), 40)), multiply(multiply(3, 8), 4)) | jill works as a waitress at the local diner where she earns an hourly wage of $ 4.00 per hour and a standard tip rate of 15 % of the cost of the orders she serves . if she worked 3 8 - hour shifts this week and averaged $ 40 in orders per hour , how much did jill earn this week ? | jill earns 4 dollars / hour and the hourly tip is ( 3 / 20 ) * 40 . jill thus earns 4 * 8 + 8 * 2 * 3 per day ( or 4 ( 8 ) + 6 ( 8 ) = 10 ( 8 ) = 80 ) . jill has worked for 3 days - > 80 * 3 = 240 . this matches answer choice c . | a = 15 / 100
b = 3 * 8
c = b * 40
d = a * c
e = 3 * 8
f = e * 4
g = d + f
|
a ) 0.49 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.51 | e | multiply(divide(85, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 85 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time | "explanation : probability that a speaks truth is 85 / 100 = 0.85 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) × p ( b ) = 0.85 × 0.6 = 0.51 answer : e" | a = 4 * 5
b = a * 5
c = 85 / b
d = 4 * 5
e = d * 5
f = 60 / e
g = c * f
|
a ) 455 , b ) 525 , c ) 1400 , d ) 180 , e ) 560 | a | multiply(multiply(subtract(14, const_1), 5), divide(14, const_2)) | there are , in a certain league , 14 teams , and each team face another team for a total of 5 times . how many games are played in the season ? | "by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 455 option a ." | a = 14 - 1
b = a * 5
c = 14 / 2
d = b * c
|
a ) 21 , b ) 20 , c ) 25 , d ) 30 , e ) 45 | c | subtract(divide(multiply(add(divide(88, const_2), 5), 5), 7), 10) | 10 is added to a certain number , the sum is multiplied by 7 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 88 . what is the number ? | "let number is x . when 10 added to it , = ( x + 10 ) 7 multiplied to sum , = 7 * ( x + 10 ) now , = [ { 7 * ( x + 10 ) } / 5 ] and , = [ { 7 * ( x + 10 ) } / 5 ] - 5 according to question , [ { 7 * ( x + 10 ) } / 5 ] - 5 = half of 88 [ ( 7 x + 70 ) / 5 ) = 44 + 5 7 x + 70 = 49 * 5 x + 10 = 7 * 5 x + 10 = 35 x = 35 - 10 x = 25 so , required number is : 25 . answer : c" | a = 88 / 2
b = a + 5
c = b * 5
d = c / 7
e = d - 10
|
a ) 33.3 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 62.5 % | e | divide(25, divide(subtract(const_100, 60), const_100)) | on a certain transatlantic crossing , 25 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 60 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ? | "0.25 p = rt + c 0.6 ( rt ) = no c = > 0.40 ( rt ) had c 0.25 p = 0.40 ( rt ) rt / p = 62.5 % answer - e" | a = 100 - 60
b = a / 100
c = 25 / b
|
a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.2 , e ) 2 | c | divide(10, subtract(50, 40)) | two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 40 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 10 miles ahead of the black car , how many hours will it take the black car to overtake the red car ? | "option c 10 + 40 t = 50 t t = 1" | a = 50 - 40
b = 10 / a
|
a ) 74 , b ) 75 , c ) 103 , d ) 78 , e ) 45 | c | subtract(subtract(multiply(25, 19), multiply(12, 17)), multiply(12, 14)) | the average of 25 results is 19 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ? | "solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 19 or , x = 103 . answer : option c" | a = 25 * 19
b = 12 * 17
c = a - b
d = 12 * 14
e = c - d
|
a ) 46 % , b ) 52 % , c ) 58 % , d ) 64 % , e ) 70 % | e | multiply(divide(subtract(70, 21), 70), const_100) | in town p , 70 percent of the population are employed , and 21 percent of the population are employed males . what percent of the employed people in town p are females ? | the percent of the population who are employed females is 70 - 21 = 49 % the percent of employed people who are female is 49 % / 70 % = 70 % . the answer is e . | a = 70 - 21
b = a / 70
c = b * 100
|
a ) 20 , b ) 60 , c ) 108 , d ) 144 , e ) 180 | a | divide(multiply(10, 3), subtract(2, divide(3, 6))) | mysoon collects glass ornaments . 10 more than 1 / 6 of the ornaments in her collection are handmade , and 1 / 2 of the handmade ornaments are antiques . if 1 / 3 of the ornaments in her collection are handmade antiques , how many ornaments are in her collection ? | the number of ornaments = a ten more than 1 / 6 of the ornaments in her collection are handmade = > handmade = 10 + a / 6 1 / 2 of the handmade ornaments are antiques = > handmade ornaments = 1 / 2 * ( 10 + a / 6 ) = 5 + a / 12 1 / 3 of the ornaments in her collection are handmade antiques = > handmade ornaments = a / 3 = > 5 + a / 12 = a / 3 = > a = 200 ans : a | a = 10 * 3
b = 3 / 6
c = 2 - b
d = a / c
|
a ) 384 , b ) 360 , c ) 324 , d ) 390 , e ) 395 | a | multiply(160, subtract(const_2, const_1)) | a train speeds past a pole in 12 seconds and a platform 160 m long in 17 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 12 = > y = x / 12 x + 160 / 17 = x / 12 x = 384 m . answer : a" | a = 2 - 1
b = 160 * a
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a ) 240 , b ) 75 , c ) 110 , d ) 90 , e ) 200 | d | add(divide(subtract(add(divide(subtract(240, 40), const_2), 40), 40), const_2), 40) | a picnic attracts 240 persons . there are 40 more men than women , and 40 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 40 i . e . y + ( y + 40 ) = 240 i . e . y = 100 i . e . adult = 100 + 40 = 140 adults include only men and women i . e . men + women = 140 let women , w = x then men , m = x + 40 i . e . x + ( x + 40 ) = 2 x + 40 = 140 i . e . x = 54 i . e . men , m = 54 + 40 = 90 answer : option d" | a = 240 - 40
b = a / 2
c = b + 40
d = c - 40
e = d / 2
f = e + 40
|
a ) 2 : 1 , b ) 4 : 1 , c ) 3 : 1 , d ) 2 : 3 , e ) 4 : 3 | c | divide(add(multiply(4, divide(add(7, 3), add(4, 1))), 7), add(multiply(1, divide(add(7, 3), add(4, 1))), 3)) | two vessels contains equal number of mixtures milk and water in the ratio 4 : 1 and 7 : 3 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | the ratio of milk and water in the new vessel is = ( 4 / 5 + 7 / 10 ) : ( 1 / 5 + 3 / 10 ) = 3 / 2 : 1 / 2 = 3 : 1 answer is c | a = 7 + 3
b = 4 + 1
c = a / b
d = 4 * c
e = d + 7
f = 7 + 3
g = 4 + 1
h = f / g
i = 1 * h
j = i + 3
k = e / j
|
a ) 2 : 5 , b ) 3 : 5 , c ) 1 : 5 , d ) 4 : 5 , e ) 1 : 3 | c | divide(1, 5) | if x : y is 1 : 2 and y : z is 2 : 5 then x : z is equal to | "the two ratios given are having the same number 2 for y in both the ratios . hence - x : y = 1 : 2 y : z = 2 : 5 = > x : z = 1 : 5 answer c" | a = 1 / 5
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a ) 10 , b ) 15 , c ) - 16 , d ) 16 , e ) 5 | c | subtract(add(2, 16), subtract(2, 16)) | if | x - 16 | = 2 x , then x = ? | "| x - 16 | = 2 x . . . ( given ) x ^ 2 - 32 x + 256 = 4 x ^ 2 3 * x ^ 2 + 32 * x - 256 = 0 . . . . ( by solving above eq . we get ) , x = - 16 or 5.33 = = = > ans - c" | a = 2 + 16
b = 2 - 16
c = a - b
|
a ) 47 sec , b ) 40 sec , c ) 45 sec , d ) 33 sec , e ) 35 sec | d | subtract(multiply(divide(7, 35), 200), 7) | in a 200 meter race a beats b by 35 meters in 7 seconds . a ' s time over the cause is ? | b covers 35 m in 7 seconds b take time = ( 200 * 7 ) / 35 = 40 a takes time = 40 - 7 = 33 sec answer : d | a = 7 / 35
b = a * 200
c = b - 7
|
a ) 36 / 165 , b ) 25 / 166 , c ) 26 / 165 , d ) 32 / 165 , e ) 33 / 165 | b | divide(multiply(0.02, power(const_5, const_2)), add(add(multiply(0.02, power(const_5, const_2)), multiply(0.03, multiply(add(const_3, const_4), const_2))), multiply(power(const_4, const_2), 0.15))) | a life insurance company insured 25,000 young boys , 14,000 young girls and 16,000 young adults . the probability of death within 10 years of a young boy , young girl and a young adult are 0.02 , 0.03 and 0.15 respectively . one insured person dies . what is the probability that the dead person is a young boy ? | the number of young boys that will die = 0.02 x 25,000 = 500 the number of young girls that will die = 0.03 x 14,000 = 420 the number of young adults that will die = 0.15 x 16,000 = 2400 the required probability = the number of young boys who will die / the total number of people who will die . = 500 / ( 500 + 420 + 2400 ) = 25 / 166 answer : b | a = 5 ** 2
b = 0 * 2
c = 5 ** 2
d = 0 * 2
e = 3 + 4
f = e * 2
g = 0 * 3
h = d + g
i = 4 ** 2
j = i * 0
k = h + j
l = b / k
|
['a ) 9 % less', 'b ) 1 % less', 'c ) equal to each other', 'd ) 1 % more', 'e ) 9 % more'] | b | divide(triangle_area(add(floor(const_100), divide(multiply(floor(const_100), 10), const_100)), subtract(floor(const_100), divide(multiply(floor(const_100), 10), const_100))), triangle_area(subtract(floor(const_100), divide(multiply(floor(const_100), 10), const_100)), add(floor(const_100), divide(multiply(floor(const_100), 10), const_100)))) | triangle a ’ s base is 10 % greater than the base of triangle b , and a ’ s height is 10 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | base of a = 11 / 10 * base of b height of a = 9 / 10 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 11 / 10 * 9 / 10 * area of b = 99 / 100 * area of b area of a is 1 % less than the area of b . answer ( b ) | a = math.floor(100)
b = math.floor(100)
c = b * 10
d = c / 100
e = a + d
f = math.floor(100)
g = math.floor(100)
h = g * 10
i = h / 100
j = f - i
k = triangle_area / (
|
a ) 1 : 5 , b ) 3 : 2 , c ) 1 : 2 , d ) 3 : 5 , e ) 2 : 3 | b | divide(subtract(const_1, divide(const_2.0, 5)), divide(3, 5)) | peter and tom shared the driving on a certain trip . if peter and tom both drove for the same amount of time , but peter only drove 3 / 5 of the total distance , what was the ratio of peter ' s average speed to tom ' s average speed ? | "the answer is suppose to be b . 3 : 2 . it ' s from the gmatprep answer : option b" | a = 2 / 0
b = 1 - a
c = 3 / 5
d = b / c
|
a ) 41 / 25 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | a | divide(divide(42, 2.5), divide(42, 4.2)) | jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 2.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 42 / ( 5 / 2 ) = 84 / 5 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 / 5 ) / 10 = 84 / 50 = 41 / 25 answer a" | a = 42 / 2
b = 42 / 4
c = a / b
|
a ) $ 1 , b ) $ 2 , c ) $ 3 , d ) $ 4 , e ) $ 5 | b | subtract(divide(900, 45), divide(360, 20)) | in a store , the total price for 20 shirts is $ 360 and the total price for 45 sweaters is $ 900 . by how much does the average ( arithmetic mean ) price of a sweater exceed that of a shirt in this store ? | "the average price of a shirt is : $ 360 / 20 = $ 18 . the average price of a sweater is : $ 900 / 45 = $ 20 . the difference in price is : $ 20 - $ 18 = $ 2 . the answer is b ." | a = 900 / 45
b = 360 / 20
c = a - b
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a ) 17.5 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 38 % | d | subtract(multiply(add(const_100, 20), divide(add(const_100, 10), const_100)), const_100) | a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 10 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | "suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.10 = 132 increase in 1993 from 1991 = 132 - 100 = 32 answer : d" | a = 100 + 20
b = 100 + 10
c = b / 100
d = a * c
e = d - 100
|
a ) 80 , b ) 85 , c ) 75 , d ) 95 , e ) 100 | d | divide(add(add(multiply(85, 8), multiply(60, 4)), 30), add(8, 4)) | the average expenditure of a labourer for 8 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income is | "income of 8 months = ( 8 × 85 ) – debt = 680 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 950 average monthly income = 950 ÷ 10 = 95 answer d" | a = 85 * 8
b = 60 * 4
c = a + b
d = c + 30
e = 8 + 4
f = d / e
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a ) 122.6 $ , b ) 128.9 $ , c ) 243.7 $ , d ) 298.85 $ , e ) 314.45 $ | e | add(multiply(multiply(add(65, divide(multiply(65, 140), const_100)), 2), 0.6), multiply(multiply(65, 3), 0.6)) | in a fuel station the service costs $ 2.05 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 65 liters and a truck ' s tank is 140 % bigger and they are all empty - ? | "service cost of 3 van and 2 truck = 2.05 * ( 3 + 2 ) = 10.5 fuel in 3 van = 3 * 65 = 195 litre fuel in 2 trucks = 2 * 65 ( 1 + 140 / 100 ) = 312 total fuel ( van + truck ) = 507 litre total fuel cost = 507 * 0.6 = 304.2 total cost = fuel + service = 304.2 + 10.25 = 314.45 answer is e" | a = 65 * 140
b = a / 100
c = 65 + b
d = c * 2
e = d * 0
f = 65 * 3
g = f * 0
h = e + g
|
a ) 66 , b ) 84 , c ) 80 , d ) 72 , e ) 68 | b | divide(66, multiply(power(divide(1, const_2), const_2), const_pi)) | 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be : | "let the length of the wire be h . radius = 1 / 2 mm = 1 / 20 cm . then , ( 22 / 7 ) x ( 1 / 20 ) x ( 1 / 20 ) x h = 66 . h = ( 66 x 20 x 20 x 7 ) / 22 = 8400 cm = 84 m . answer b" | a = 1 / 2
b = a ** 2
c = b * math.pi
d = 66 / c
|
a ) 12 % , b ) 16 % , c ) 20 % , d ) 24 % , e ) 28 % | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 58)), divide(multiply(const_100, 50), 58))) | if the cost price of 58 articles is equal to the selling price of 50 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 50 y = 58 x y = 1.16 x the answer is b ." | a = 100 * 50
b = a / 58
c = 100 - b
d = 100 * 50
e = d / 58
f = c / e
g = 100 * f
|
a ) 180 km , b ) 200 km , c ) 160 km , d ) 220 km , e ) 240 km | c | multiply(20, 8) | a person is traveling at 20 km / hr and reached his destiny in 8 hr then find the distance ? | "t = 8 hrs d = t * s = 20 * 8 = 160 km answer is c" | a = 20 * 8
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a ) 3,108 , b ) 3,100 , c ) 3,108 , d ) 3,124 , e ) 3,250 | e | multiply(divide(300, 22.95), 250) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 300 ? | "to maximize number of hot dogs with 300 $ total number of hot dogs bought in 250 - pack = 22.95 * 13 = 298.35 $ amount remaining = 300 - 298.35 = 1.65 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 300 $ = 250 * 13 = 3250 answer e" | a = 300 / 22
b = a * 250
|
a ) 87 kmph , b ) 28 kmph , c ) 72 kmph , d ) 18 kmph , e ) 29 kmph | c | divide(divide(400, const_1000), divide(20, const_3600)) | a train 400 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 400 / 20 s = 20 m / sec speed = 20 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 72 kmph answer : c" | a = 400 / 1000
b = 20 / 3600
c = a / b
|
a ) 10 sec , b ) 30 sec , c ) 40 sec , d ) 20 s , e ) 50 sec | d | divide(add(120, 280), divide(multiply(add(42, 30), const_1000), const_3600)) | two diesel trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | d relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 sec . | a = 120 + 280
b = 42 + 30
c = b * 1000
d = c / 3600
e = a / d
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a ) 72 , b ) 74 , c ) 76 , d ) 78 , e ) 80 | e | multiply(divide(180, 5), divide(180, add(divide(180, 5), divide(180, 4)))) | two trains start simultaneously from opposite ends of a 180 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 180 - km trip in 5 hours . train y , travelling at a constant rate , completes the 180 - km trip in 4 hours . how many kilometers had train x traveled when it met train y ? | if the two trains cover a total distance d , then train x travels ( 4 / 9 ) * d while train y travels ( 5 / 9 ) * d . if the trains travel 180 km to the meeting point , then train x travels ( 4 / 9 ) * 180 = 80 km . the answer is e . | a = 180 / 5
b = 180 / 5
c = 180 / 4
d = b + c
e = 180 / d
f = a * e
|
a ) $ 10 , b ) $ 20 , c ) $ 30 , d ) $ 40 , e ) $ 50 | c | multiply(add(10, 5), 2) | maria bought 10 notebooks and 5 pens costing 2 dollars each . how much did maria pay ? | 2 × ( 10 + 5 ) = 2 × 10 + 2 × 5 = 20 + 10 = 30 dollars correct answer is c ) $ 30 | a = 10 + 5
b = a * 2
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a ) 67 sec , b ) 89 sec , c ) 36 sec , d ) 87 sec , e ) 40 sec | e | divide(add(280, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 280 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 280 + 120 = 400 m . time taken = 400 / 10 = 40 sec . answer : e" | a = 280 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) s . 3944 , b ) s . 3948 , c ) s . 3942 , d ) s . 3965 , e ) s . 3740 | e | multiply(square_perimeter(sqrt(289)), 55) | what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 55 ? | "let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 55 = rs . 3740 . answer : e" | a = math.sqrt(289)
b = square_perimeter * (
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a ) 16 years , b ) 88 years , c ) 39 years , d ) 66 years , e ) 77 years | c | add(multiply(const_2, 10), add(9, 10)) | in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 9 years older than b , the present age of b is | "let b ' s age = x years . then , as age = ( x + 9 ) years . ( x + 9 + 10 ) = 2 ( x — 10 ) hence x = 39 . present age of b = 39 years answer : c" | a = 2 * 10
b = 9 + 10
c = a + b
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a ) 3 times , b ) 2 times , c ) 4 times , d ) 5 times , e ) once | b | divide(multiply(multiply(const_3, const_4), const_2), multiply(const_3, const_4)) | how many times hour hand covers full circle in a day ? | 2 times answer : b | a = 3 * 4
b = a * 2
c = 3 * 4
d = b / c
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a ) 7 , b ) 11 , c ) 12 , d ) 14 , e ) 15 | a | subtract(power(2, 2), 2) | if x ^ 2 + ( 1 / x ^ 2 ) = 3 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 3 ^ 2 - 2 = 7 . thus , the answer is a ." | a = 2 ** 2
b = a - 2
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a ) 250 , b ) 150 , c ) 300 , d ) 400 , e ) 450 | c | divide(subtract(1000, 400), const_2) | how many integers between 400 and 1000 are there such that their unit digit is odd ? | there are 600 numbers from 401 to 1000 ( inclusive ) . half of the numbers are odd , so there are 300 odd numbers . the answer is c . | a = 1000 - 400
b = a / 2
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a ) 25 , b ) 31 , c ) 18 , d ) 19 , e ) 01 | b | divide(multiply(divide(multiply(15.5, 50), const_100), const_100), 25) | a company pays 15.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 15.5 * 50 ) / 100 = rs . 7.75 rs . 25 is income on an investment of rs . 100 rs . 7.75 is income on an investment of rs . ( 7.75 * 100 ) / 25 = rs . 31 answer : b" | a = 15 * 5
b = a / 100
c = b * 100
d = c / 25
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a ) 10.11 , b ) 11.11 , c ) 12.11 , d ) 6.83 , e ) 14.11 | d | add(6, const_1) | the average of first 6 prime numbers is ? | "sum of 6 prime no . = 41 average = 41 / 6 = 6.83 answer : d" | a = 6 + 1
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a ) 49 th , b ) 48 th , c ) 47 th , d ) 50 th , e ) none of these | a | add(subtract(53, 5), const_1) | jaya ranks 5 th in a class of 53 . what is her rank from the bottom in the class ? | there are 4 students above her . when takan from bottom there are 4 students below her rank . so her rank wud be 53 - 4 i . e . 49 th . answer : a | a = 53 - 5
b = a + 1
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a ) 7.2 kg . , b ) 10.8 kg . , c ) 14.04 kg . , d ) 18.0 kg , e ) none | c | divide(multiply(6, 23.4), 10) | if the weight of 10 meters long rod is 23.4 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 10 m long rod = 23.4 kg ∴ weight of 1 m long rod = 23.4 / 10 kg ∴ weight of 6 m long rod = 23.4 x 6 / 10 = 14.04 kg option : c" | a = 6 * 23
b = a / 10
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a ) 75000 , b ) 160000 , c ) 170000 , d ) 60000 , e ) 70000 | d | divide(30000, multiply(divide(2, 3), divide(3, 4))) | a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 30000 rs , what is the value of buzness ? | if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 30000 - > x = 60000 answer : d | a = 2 / 3
b = 3 / 4
c = a * b
d = 30000 / c
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a ) 5 / 8 , b ) 4 / 5 , c ) 3 / 8 , d ) 1 / 2 , e ) 3 / 4 | a | divide(const_3, multiply(const_2, const_4)) | the center of a circle lies on the origin of the coordinate plane . if a point ( x , y ) is randomly selected inside of the circle , what is the probability that y > x or x < 0 ? | "the line y = x divides the circle into two equal areas . all the points above the line y = x satisfy the condition that y > x . all the points to the left of the y - axis satisfy the condition that x < 0 . the union of these two areas is 5 / 8 of the circle . the answer is a ." | a = 2 * 4
b = 3 / a
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a ) 1.4 % , b ) 5.9 % , c ) 11.1 % , d ) 6.67 % , e ) 23.6 % | d | multiply(subtract(divide(16, 15), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 15 . if the price per share of stock k was $ 16 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 15 closing = 16 rise in price = 1 so , percent increase = 1 / 15 * 100 = 6.67 answer : d" | a = 16 / 15
b = a - 1
c = b * 100
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a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | b | subtract(70, add(add(subtract(50, 35), subtract(40, 35)), 35)) | of the 70 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | "neither car nor garage = total - garage - ( swim - common ) = 70 - 50 - ( 40 - 35 ) = 70 - 55 = 15 answer b" | a = 50 - 35
b = 40 - 35
c = a + b
d = c + 35
e = 70 - d
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a ) 32 % , b ) 34.4 % , c ) 35 % , d ) 35.6 % , e ) 36.4 % | d | multiply(divide(add(multiply(divide(30, const_100), 57000), multiply(divide(40, const_100), 72000)), add(72000, 57000)), const_100) | john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 57000 and ingrid makes $ 72000 , what is their combined tax rate ? | "( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 72000 / 129000 = 24 / 43 . 24 / 43 is a little above 24 / 48 = 1 / 2 . thus , our answer is just a little above 35 . answer : d" | a = 30 / 100
b = a * 57000
c = 40 / 100
d = c * 72000
e = b + d
f = 72000 + 57000
g = e / f
h = g * 100
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a ) 7 , b ) 8.5 , c ) 10 , d ) 12 , e ) 14 | c | divide(subtract(29, power(3, 2)), 2) | if a - b = 3 and a 2 + b 2 = 29 , then find the value of ab | explanation : 2 ab = ( a 2 + b 2 ) − ( a − b ) 2 = > 2 ab = 29 - 9 = 20 = > ab = 10 option c | a = 3 ** 2
b = 29 - a
c = b / 2
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a ) $ 255 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 663 | e | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 1.30) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 1.30 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "total no . of boxes = 3060000 / ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 1.30 = $ 663 answer e" | a = 1000 * 1000
b = 3 * 6
c = 20 * 20
d = c * 15
e = b / d
f = e * 1
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a ) 40 % , b ) 35 % , c ) 30 % , d ) 25 % , e ) 20 % | d | multiply(divide(subtract(divide(multiply(4, subtract(const_100, 50)), const_100), multiply(0.5, 4)), subtract(4, multiply(0.5, 4))), const_100) | a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 50 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 4 hours ? | "the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.5 ( 6 ) = 3 gallons at the start , the amount of water is 0.5 ( 6 ) = 3 gallons after 4 hours , the amount of water is 3 - 0.5 ( 4 ) = 1 gallon the concentration of water is 1 / ( 3 + 1 ) = 1 / 4 = 25 % the answer is d ." | a = 100 - 50
b = 4 * a
c = b / 100
d = 0 * 5
e = c - d
f = 0 * 5
g = 4 - f
h = e / g
i = h * 100
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a ) 2 : 6 , b ) 3 : 7 , c ) 3 : 8 , d ) 2 : 1 , e ) 8 : 3 | b | inverse(subtract(const_1, divide(const_3.0, 4))) | he ratio between the sale price and the cost price of an article is 4 : 7 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 7 x and s . p . = rs . 4 x . then , gain = rs . 3 x required ratio = 3 x : 7 x = 3 : 7 . answer : b" | a = 3 / 0
b = 1 - a
c = 1/(b)
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a ) 48 km , b ) 56 km , c ) 64 km , d ) 96 km , e ) 106 km | d | divide(multiply(add(4, const_2), multiply(add(divide(add(const_60, 4), 4), 4), divide(add(const_60, 4), 4))), add(divide(add(const_60, 4), 4), 4)) | a man covers a certain distance on bike . had he moved 4 kmph faster , he would have taken 72 minutes less . if he moves slow by 4 kmph , he will take hours more to cover the distance . what is the distance travelled by him ? | explanation : let the distance be ' d ' and the speed be ' x ' kmph . d / x + 4 = d / x - 72 / 60 d / x - d / ( x - y ) = 6 / 5 = d ( x + 4 - x ) / x ( x - 4 ) = 6 / 5 ∴ d / x ( x + 4 ) = 3 / 10 . . . . . . ( i ) d / x - 4 - d / x = 2 ∴ dx - dx + 4 d / x ( x - 4 ) = 2 ∴ d / x ( x - 4 ) = 1 / 2 . . . . . ( ii ) from equation ( i ) and ( ii ) : 3 / 10 x ( x + 4 ) = 1 / 2 x ( x - 4 ) ∴ 3 ( x + 4 ) = 5 ( x - 4 ) = 3 x + 12 = 5 x - 20 2 x = 32 ∴ x = 16 kmn - 1 ∴ d = x ( x - 4 ) / 2 = 16 x 12 / 2 = 16 x 6 = 96 km answer : option d | a = 4 + 2
b = const_60 + 4
c = b / 4
d = c + 4
e = const_60 + 4
f = e / 4
g = d * f
h = a * g
i = const_60 + 4
j = i / 4
k = j + 4
l = h / k
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a ) a ) 0.96 , b ) b ) 1.2 , c ) c ) 1.25 , d ) d ) 1.3 , e ) e ) 0.8 | a | divide(multiply(0.6, 8), 5) | if 0.6 : x : : 5 : 8 , then x is equal to : | "( x * 5 ) = ( 0.6 * 8 ) x = 4.8 / 5 x = 0.96 answer = a" | a = 0 * 6
b = a / 5
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a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 15 min | d | multiply(const_60, divide(subtract(50, 36), 50)) | excluding stoppages , the speed of a bus is 50 km / hr and including stoppages , it is 36 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 14 km less . time taken to cover 14 km = 14 / 50 * 60 = 16 min . answer : d" | a = 50 - 36
b = a / 50
c = const_60 * b
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a ) 20 , b ) 18 , c ) 11 , d ) 5 , e ) 50 | d | add(multiply(2, subtract(1, multiply(3, divide(subtract(7, 19), subtract(multiply(7, 3), const_1))))), divide(subtract(7, 19), subtract(multiply(7, 3), const_1))) | 7 x + y = 19 , and x + 3 y = 1 . find the value of 2 x + y | add these two equations 8 x + 4 y = 20 divide by 2 ( to get 2 x + y ) answer will be d . 10 | a = 7 - 19
b = 7 * 3
c = b - 1
d = a / c
e = 3 * d
f = 1 - e
g = 2 * f
h = 7 - 19
i = 7 * 3
j = i - 1
k = h / j
l = g + k
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a ) 19 , b ) 22 , c ) 52 , d ) 36 , e ) 38 | c | divide(multiply(12, 16), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 12 and 16 ? | "prime factorization of the given numbers 12 = 2 ^ 2 * 3 48 = 2 ^ 4 * 3 greatest common factor = 4 lowest common multiple = 48 sum = 4 + 48 = 52 answer c" | a = 12 * 16
b = a / 4
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a ) a ) 2 , b ) b ) 3 , c ) c ) 0 , d ) d ) 5 , e ) e ) 6 | c | subtract(909, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | the least number which must be subtracted from 909 to make it exactly divisible by 9 is : | "on dividing 909 by 9 , we get remainder = 0 therefore , required number to be subtracted = 0 answer : c" | a = 4 + 1
b = a * 10
c = 4 + 2
d = b + c
e = d * 9
f = 909 - e
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a ) $ 2000 , b ) $ 1200 , c ) $ 2500 , d ) $ 1800 , e ) $ 1600 | b | multiply(divide(3, add(add(2, 3), 4)), 3600) | a person want to give his money of $ 3600 to his 3 children a , b , c in the ratio 2 : 3 : 4 . what is the b ' s share ? | "b ' s share = 3600 * 3 / 9 = $ 1200 answer is b" | a = 2 + 3
b = a + 4
c = 3 / b
d = c * 3600
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a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | b | add(divide(subtract(add(40, 2), 24), 1.5), 24) | each week , harry is paid x dollars per hour for the first 24 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "42 x = 24 x + 1.5 x ( h - 24 ) = = > 42 = 24 + 1.5 ( h - 24 ) = = > h - 24 = 18 / 1.5 = 12 = > h = 36 answer is b" | a = 40 + 2
b = a - 24
c = b / 1
d = c + 24
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a ) 1100 , b ) 1200 , c ) 1300 , d ) 1400 , e ) 1700 | a | multiply(add(5, 6), const_100) | rs . 1600 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 85 . how much was lent at 5.00001 % ? | ( x * 5 * 1 ) / 100 + [ ( 1600 - x ) * 6 * 1 ] / 100 = 85 5 x / 100 + ( 9600 – 6 x ) / 100 = 85 = > x = 1100 answer : a | a = 5 + 6
b = a * 100
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a ) 12 , b ) 15 , c ) 17 , d ) 18 , e ) 21 | d | add(16, const_2) | a box contains 10 tablets of medicine a and 16 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted . | "the worst case scenario will be if we remove all 16 tablets of medicine b first . the next 2 tablets we remove have to be of medicine a , so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 16 + 2 = 18 tablets . answer : d ." | a = 16 + 2
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a ) 2820 , b ) 3440 , c ) 4860 , d ) 5560 , e ) 6720 | c | add(divide(subtract(subtract(171, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 171 ? | "100 + 102 + . . . + 170 = 100 * 36 + ( 2 + 4 + . . . + 70 ) = 100 * 36 + 2 * ( 1 + 2 + . . . + 35 ) = 100 * 36 + 2 ( 35 ) ( 36 ) / 2 = 100 * 36 + 35 * 36 = 135 ( 36 ) = 4860 the answer is c ." | a = 171 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
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a ) 90 , b ) 95 , c ) 100 , d ) 105 , e ) 110 | a | divide(divide(3000, const_1000), divide(120, const_3600)) | a train 3000 m long can cross an electric pole in 120 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 3000 / 120 s = 25 m / sec speed = 25 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 90 kmph answer : a" | a = 3000 / 1000
b = 120 / 3600
c = a / b
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a ) 1,108 , b ) 4,100 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 500) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 500 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 500 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 500 x 8 = 4000 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 4100 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 4100 ( answer b )" | a = 200 / 22
b = a * 500
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a ) 63.83 , b ) 111.67 , c ) 111.64 , d ) 111.11 , e ) 101.12 | a | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 120)), subtract(multiply(const_100, const_10), 60))) | a can give b 60 meters start and c 120 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 940 m and c runs 880 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 880 ) / 940 = 936.17 m . b can give c = 1000 - 936.17 = 63.83 m . answer : a" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 120
e = b * d
f = 100 * 10
g = f - 60
h = e / g
i = a - h
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['a ) 27', 'b ) 43', 'c ) 11', 'd ) 32', 'e ) 45'] | d | subtract(59, multiply(multiply(const_3, const_3), const_3)) | you have 59 cubic blocks . what is the minimum number that needs to be taken away in order to construct a solid cube with none left over ? | d 32 the next cube number below 64 ( 4 × 4 × 4 ) is 27 ( 3 × 3 × 3 ) . in order to construct a solid cube , therefore , with none left over , 59 – 27 = 32 blocks need to be taken away . | a = 3 * 3
b = a * 3
c = 59 - b
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a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 12 | a | subtract(divide(multiply(const_3, multiply(8, 3)), add(8, 1)), 3) | from an island , it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for 3 hours at 8 kilometers per hour , while ferry q takes a route that is three times longer . if ferry p is slower than ferry q by 1 kilometer per hour , how many hours longer is the journey of ferry q compared with the journey of ferry p ? | "the distance traveled by ferry p is 24 km . then the distance traveled by ferry q is 72 km . ferry q travels at a speed of 9 kph . the time of the journey for ferry q is 72 / 9 = 8 hours , which is 5 hours more than ferry p . the answer is a ." | a = 8 * 3
b = 3 * a
c = 8 + 1
d = b / c
e = d - 3
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a ) 62 , b ) 63 , c ) 64 , d ) 66.66 , e ) 65 | d | multiply(add(divide(1, 3), divide(1, 3)), const_100) | n a certain flower shop , which stocks 4 types of flowers , there are 1 / 3 as many violets as carnations , and 1 / 4 as many tulips as violets . if there are equal numbers of roses and tulips , what percent of the flowers in the shop are carnations ? | given : - violets - c / 3 carnations - c tulip - c / 12 rose - c / 12 total flowers in terms of c = c / 3 + c + c / 12 + c / 12 = 18 c / 12 percentage of carnations = c / 3 c / 2 * 100 = 66.66 % answer d | a = 1 / 3
b = 1 / 3
c = a + b
d = c * 100
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a ) 33.25 % , b ) 40.25 % , c ) 50.25 % , d ) 55.25 % , e ) 56.25 % | e | multiply(divide(multiply(divide(500, 2), add(1, divide(1, 8))), 500), const_100) | at a contest with 500 participants , 1 / 2 of the people are aged 28 to 32 . next year , the number of people aged 28 to 32 will increase by 1 / 8 . after this change , what percentage of the total 500 people will the 28 - to 32 - year - olds represent ? | "i just wanted to mention a couple of things here : * this is a pure ratio question ; the number 500 is completely irrelevant , and you can ignore it if you like . when we increase something by 1 / 8 , we are multiplying it by 1 + 1 / 8 = 9 / 8 , so the answer here must be ( 1 / 2 ) * ( 9 / 8 ) = 9 / 16 = 56.25 % . answer is e" | a = 500 / 2
b = 1 / 8
c = 1 + b
d = a * c
e = d / 500
f = e * 100
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a ) 23.5 % , b ) 24.5 % , c ) 23.1 % , d ) 25.5 % , e ) 26.5 % | c | multiply(divide(subtract(add(multiply(10, 12), 30), multiply(10, 12)), 130), const_100) | on a purchase of $ 130 , a store offered a payment plan consisting of a $ 30 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | 12 * 10 + 30 = 150 . ( 30 / 130 ) * 100 = 23.1 answer : c | a = 10 * 12
b = a + 30
c = 10 * 12
d = b - c
e = d / 130
f = e * 100
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a ) 38 , b ) 27 , c ) 92 , d ) 17 , e ) 80 | c | subtract(multiply(77, 6), multiply(74, 5)) | ashok secured average of 77 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 77 therefore total marks in 6 subjects = 77 * 6 = 462 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 462 – 370 = 92 answer : c" | a = 77 * 6
b = 74 * 5
c = a - b
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a ) 156 , b ) 162 , c ) 316 , d ) 324 , e ) 325 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 7.30))) | roses can be purchased individually for $ 7.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | "buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 7.3 = ~ 4 roses for that amount . total = 312 + 4 = 316 . answer : c ." | a = 680 / 50
b = math.floor(a)
c = 12 * 2
d = b * c
e = 680 / 50
f = math.floor(e)
g = f * 50
h = 680 - g
i = h / 7
j = math.floor(i)
k = d + j
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['a ) x ^ 2 + y ^ 2 + 6 x – 4 y = 23', 'b ) x ^ 2 + y ^ 2 - 6 x + 4 y = 23', 'c ) x ^ 2 + y ^ 2 + 6 x + 4 y = 23', 'd ) x ^ 2 + y ^ 2 - 6 x – 4 y = - 12', 'e ) x ^ 2 + y ^ 2 - 6 x – 4 y = 12'] | e | subtract(subtract(power(5, const_2), power(3, const_2)), power(2, const_2)) | what is the equation of a circle of radius 5 units centered at ( 3 , 2 ) ? | the equation of a circle with center at ( a , b ) and radius r is ( x - a ) ^ 2 + ( y - b ) ^ 2 = r ^ 2 = > answer = ( x - 3 ) ^ 2 + ( y - 2 ) ^ 2 = 5 ^ 2 solving , , we get the equation in answer choice e . | a = 5 ** 2
b = 3 ** 2
c = a - b
d = 2 ** 2
e = c - d
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a ) 5 % , b ) 10 % , c ) 15 % , d ) 14 % , e ) 4 % | e | divide(multiply(circle_area(20), const_100), circle_area(const_100)) | if the diameter of circle r is 20 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ? | "let diameter of circle r , dr = 20 and diameter of circle s , ds = 100 radius of circle r , rr = 10 radius of circle s , rs = 50 area of circle r / area of circle s = ( pi * rr ^ 2 ) / ( pi * rs ^ 2 ) = ( 10 / 50 ) ^ 2 = ( 2 / 10 ) ^ 2 = 4 % answer : e" | a = circle_area * (
b = a / 100
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a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40 | d | divide(9, subtract(divide(5, 4), const_1)) | walking with 4 / 5 of my usual speed , i arrive at the bus stop 9 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 9 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 36 the answer is d ." | a = 5 / 4
b = a - 1
c = 9 / b
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a ) 60 m , b ) 65 m , c ) 63.5 m , d ) 64 m , e ) 62.5 m | e | multiply(multiply(75, const_0_2778), 3) | what will be the length of the mini - train , if it crosses a pole in 3 sec while travelling at a speed of 75 kmph ? | explanation : d = 75 * 5 / 18 * 3 = 62.5 m answer : option e | a = 75 * const_0_2778
b = a * 3
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a ) 52 , b ) 58 , c ) 84 , d ) 122 , e ) 180 | e | divide(multiply(multiply(8, 12), 4), const_4) | if 4 x = 8 y = 12 z , then what is a possible sum of positive integers x , y , and z ? | "4 x = 8 y = 12 z x = 2 y = 3 z 3 ( 2 * 3 ) = 2 ( 3 * 1 ) = 3 ( 1 * 2 ) addition = 18 + 6 + 6 = 30 answer would be multiple of 30 which is 180 answer : e" | a = 8 * 12
b = a * 4
c = b / 4
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a ) 41 , b ) 46 , c ) 42 , d ) 46 , e ) 47 | a | multiply(multiply(6, 6), add(6, const_4)) | find a two digit number , given that the sum of the digits is 12 and the difference of the digits is 6 . ? | "using elimination method find which of the options fit the description of the number . . . from the option only 41 meets this description sum of digits - - - 4 + 1 = 5 difference of digits - - - 4 - 1 = 3 answer a ." | a = 6 * 6
b = 6 + 4
c = a * b
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 18 | e | divide(3, subtract(divide(multiply(const_2, 3), 6), 1)) | it takes joey the postman 1 hours to run a 3 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 6 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 3 miles an hour let the other half be x miles an hour now , avg speed = 6 mile an hour 2 * 3 * x / 3 + x = 6 6 x = 6 x + 18 = > x = 18 e" | a = 2 * 3
b = a / 6
c = b - 1
d = 3 / c
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a ) 1936372 , b ) 2029272 , c ) 1896172 , d ) 1923472 , e ) none of them | b | multiply(divide(2056, 987), const_100) | 2056 x 987 = ? | "= 2056 x 987 = 2056 x ( 1000 - 13 ) = 2056 x 1000 - 2056 x 13 = 2056000 - 26728 = 2029272 answer is b" | a = 2056 / 987
b = a * 100
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a ) 1 hour , b ) 30 minutes , c ) 3 hours , d ) 2 hours 30 min , e ) 2 hours | c | divide(multiply(multiply(30, divide(6, 7)), inverse(subtract(const_1, divide(6, 7)))), const_60) | a train is moving at 6 / 7 of its usual speed . the train is 30 minutes too late . what is the usual time for the train to complete the journey ? | "new time = d / ( 6 v / 7 ) = 7 / 6 * usual time 30 minutes represents 1 / 6 of the usual time . the usual time is 3 hours . the answer is c ." | a = 6 / 7
b = 30 * a
c = 6 / 7
d = 1 - c
e = 1/(d)
f = b * e
g = f / const_60
|
a ) a ) 36 , b ) b ) 20 , c ) c ) 40 , d ) d ) 42 , e ) e ) 44 | b | subtract(530, multiply(85, 6)) | we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 530 . how many green hats did we buy ? | let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 530 . 6 ( 85 - g ) + 7 g = 530 . 510 - 6 g + 7 g = 530 . g = 530 - 510 = 20 . the answer is b . | a = 85 * 6
b = 530 - a
|
a ) 16.16 % , b ) 15.15 % , c ) 14.95 % , d ) 13.33 % , e ) 12.52 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(5, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | a volunteer organization is recruiting new members . in the fall they manage to increase their number by 5 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 5 % ) * ( 100 % - 19 % ) = 1.05 * . 81 = 0.8505 . 1 - 0.8505 = 14.95 % lost = - 14.95 % the answer is c the organization has lost 14.95 % of its total volunteers from fall to spring ." | a = 5 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 6 , b ) 6.5 , c ) 7 , d ) 7.5 , e ) 8 | b | divide(add(multiply(90, const_4), multiply(divide(multiply(const_4, 90), add(const_1, const_4)), const_4)), add(90, divide(multiply(const_4, 90), add(const_1, const_4)))) | two cars are driving toward each other . the first car is traveling at a speed of 90 km / h , which is 50 % faster than the second car ' s speed . if the distance between the cars is 975 km , how many hours will it take until the two cars meet ? | "the speed of the first car is 90 km / h . the speed of the second car is 90 / 1.5 = 60 km / h . the two cars complete a total of 150 km each hour . the time it takes the cars to meet is 975 / 150 = 6.5 hours . the answer is b ." | a = 90 * 4
b = 4 * 90
c = 1 + 4
d = b / c
e = d * 4
f = a + e
g = 4 * 90
h = 1 + 4
i = g / h
j = 90 + i
k = f / j
|
a ) 550 , b ) 2000 , c ) 250 , d ) 3000 , e ) 400 | b | divide(multiply(6800, 40), 136) | find the annual income derived by investing $ 6800 in 40 % stock at 136 . | "by investing $ 136 , income obtained = $ 40 . by investing $ 6800 , income obtained = $ [ ( 40 / 136 ) * 6800 ] = $ 2000 . answer b ." | a = 6800 * 40
b = a / 136
|
a ) 3 % , b ) 16 2 / 3 % , c ) 25 % , d ) 33 1 / 3 % , e ) 60 % | d | subtract(const_100, divide(multiply(subtract(14.0, 12.0), const_100), subtract(18.8, 15.6))) | in 1982 and 1983 , company b ’ s operating expenses were $ 12.0 million and $ 14.0 million , respectively , and its revenues were $ 15.6 million and $ 18.8 million , respectively . what was the percent increase in company b ’ s profit ( revenues minus operating expenses ) from 1982 to 1983 ? | "profit in 1982 = 15.6 - 12.0 = 3.6 profit in 1983 = 18.8 - 14.0 = 4.8 increase in profit = 4.8 - 3.6 = 1.2 percentage increase 12 * 100 / 36 = 33.33 ; answer : d" | a = 14 - 0
b = a * 100
c = 18 - 8
d = b / c
e = 100 - d
|
a ) 693 , b ) 656 , c ) 691 , d ) 9890 , e ) 10000 | a | subtract(multiply(7, const_100), 7) | what is the difference between local value & face value of 7 in the numeral 65793 ? | ( local value of 7 ) - ( face value of 7 ) = ( 700 - 7 ) = 693 a | a = 7 * 100
b = a - 7
|
a ) 28 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | d | divide(subtract(add(add(2.5, 5), multiply(0.25, 14)), 2.5), 0.25) | mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 14 miles , how many miles was mike ' s ride ? | the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 14 ) = $ 11 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 11 0.25 * x = 8.5 x = 34 miles the answer is d . | a = 2 + 5
b = 0 * 25
c = a + b
d = c - 2
e = d / 0
|
a ) 5 : 00 , b ) 5 : 30 , c ) 6 : 00 , d ) 6 : 30 , e ) 7 : 00 | c | divide(subtract(divide(335, 35), 1), add(1, divide(40, 35))) | at 1 : 00 pm , a truck left city p and headed toward city q at a constant speed of 35 km / h . one hour later , a car left city q and headed toward city p along the same road at a constant speed of 40 km / h . if the distance between city p and city q is 335 km , at what time will the truck and the car meet each other ? | "at 2 : 00 pm , the truck and the car are 300 km apart . the truck and the car complete a distance of 75 km each hour . the time it takes to meet is 300 / 75 = 4 hours . they will meet at 6 : 00 pm . the answer is c ." | a = 335 / 35
b = a - 1
c = 40 / 35
d = 1 + c
e = b / d
|
a ) 18 / 79 , b ) 1 / 6 , c ) 1 / 25 , d ) 1 / 50 , e ) 1 / 900 | d | divide(divide(multiply(3, const_60), 10), multiply(power(const_10, subtract(4, const_2)), power(const_3, const_2))) | a license plate in the country kerrania consists of 4 digits followed by two letters . the letters a , b , and c are used only by government vehicles while the letters d through z are used by non - government vehicles . kerrania ' s intelligence agency has recently captured a message from the country gonzalia indicating that an electronic transmitter has been installed in a kerrania government vehicle with a license plate starting with 79 . if it takes the police 10 minutes to inspect each vehicle , what is the probability that the police will find the transmitter within 3 hours ? | if it takes 10 minutes to inspect one vehicle , the # of vehicles that can be inspected in 3 hours ( 180 minutes ) = 180 / 10 = 18 . hence , for calculating the probability that the police will find the transmitter within three hours , the favorable cases = 18 . now , we need to figure out the total # of cases . the total # of cases = total # of such cars possible . the details given about the car is that it starts with 79 , which leaves 2 more digits , both of which can be filled by all 10 numbers ( 0 - 9 ) . in addition , we have 3 letters , each of which can be filled by any from the set { a , b , c } . hence the total # of such cars possible = 10 * 10 * 3 * 3 = 900 so , the probability that the police will find the transmitter within three hours = 18 / 900 = 1 / 50 . option d | a = 3 * const_60
b = a / 10
c = 4 - 2
d = 10 ** c
e = 3 ** 2
f = d * e
g = b / f
|
a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 55 | d | divide(subtract(75, 61), subtract(const_1, divide(65, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 61 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "the total number of books is 75 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 61 books , thus the number of un - returned books is 75 - 61 = 14 books . 0.35 x = 14 x = 40 the answer is d ." | a = 75 - 61
b = 65 / 100
c = 1 - b
d = a / c
|
a ) 1.8 % , b ) 2.9 % , c ) 4.1 % , d ) 5.6 % , e ) 6.4 % | c | subtract(multiply(divide(1, add(4, 1)), const_100), multiply(divide(4, add(4, 1)), multiply(divide(1, add(4, 1)), const_100))) | one - sixth of the light switches produced by a certain factory are defective . 4 - fifths of the defective switches are rejected and 1 / 15 of the non defective switches are rejected by mistake . if all the switches not rejected are sold , what percent of the switches sold by the factory are defective ? | 1 / 6 of the switches are defective . the defective switches that are not rejected are 1 / 5 * 1 / 6 = 1 / 30 = 3 / 90 of all switches . the non defective switches that are sold are 5 / 6 * 14 / 15 = 70 / 90 of all switches . the percent of switches sold that are defective is 3 / 73 which is about 4.1 % . the answer is c . | a = 4 + 1
b = 1 / a
c = b * 100
d = 4 + 1
e = 4 / d
f = 4 + 1
g = 1 / f
h = g * 100
i = e * h
j = c - i
|
a ) 1410 , b ) 1420 , c ) 1430 , d ) 1408 , e ) 1540 | d | divide(multiply(subtract(const_100, 12), 1600), const_100) | a man buys a cycle for rs . 1600 and sells it at a loss of 12 % . what is the selling price of the cycle ? | s . p . = 88 % of rs . 1600 = 88 / 100 x 1600 = rs . 1408 answer : d | a = 100 - 12
b = a * 1600
c = b / 100
|
a ) 8 / 17 , b ) 1 / 2 , c ) 8 / 9 , d ) 9 / 8 , e ) 17 / 8 | a | divide(multiply(multiply(divide(2, 3), 3), 4), add(multiply(3, 3), multiply(multiply(divide(2, 3), 3), 4))) | at milk factory , each employee working the second shift produced 2 / 3 as many widgets as each employee working the first shift . if the first shift has 3 / 4 as many employees , what fraction of the total widgets did the second shift produce ? | at milk factory , let the first shift have 3 employee and each produce 3 widgets , so the total number of widgets produced by the first shift is 3 * 3 = 9 ; then the second shift would have 4 employees and each second shift employee would produce 3 * 2 / 3 = 2 widgets , so the total number of widgets produced by the second shift employees would be 4 * 2 = 8 ; the ratio of the second shift production to the total is 8 / ( 9 + 8 ) = 8 / 17 . answer : a . | a = 2 / 3
b = a * 3
c = b * 4
d = 3 * 3
e = 2 / 3
f = e * 3
g = f * 4
h = d + g
i = c / h
|
a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54 | b | divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2) | the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if steve and david have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ? | "if steve and david submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combined ` ` no discount cost ' ' . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) ." | a = 80 * 0
b = 2 * a
c = 100 - 25
d = c / 100
e = 160 * d
f = e * 0
g = b - f
h = g / 2
|
a ) $ 1200 , b ) $ 1600 , c ) $ 2000 , d ) $ 2500 , e ) $ 3000 | d | multiply(multiply(subtract(4, 3), 500), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is a ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 a ' s share = 5 x = 5 * $ 500 = $ 2500 the answer is d ." | a = 4 - 3
b = a * 500
c = b * 3
|
a ) 3 / 10 , b ) 2 / 5 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | b | divide(4, subtract(multiply(subtract(6, 4), 7), 4)) | when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 7 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 7 ) ( 4 + 4 x ) or , x = 2 / 5 answer b | a = 6 - 4
b = a * 7
c = b - 4
d = 4 / c
|
a ) 50 , b ) 88 , c ) 77 , d ) 66 , e ) 51 | a | subtract(add(200, 350), 500) | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 350 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 350 - - - - - - - - - - - - - - a + b + 2 c = 550 a + b + c = 500 - - - - - - - - - - - - - - - - c = 50 answer : a" | a = 200 + 350
b = a - 500
|
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