options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | b | add(divide(subtract(18, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 18 ? | "the sum of three consecutive integers can be written as n + ( n + 1 ) + ( n + 2 ) = 3 n + 3 if the sum is 24 , we need to solve the equation 3 n + 3 = 18 ; = > 3 n = 15 ; = > n = 5 the greatest of the three numbers is therefore 5 + 2 = 7 answer : b" | a = 18 - 3
b = a / 3
c = b + 2
|
a ) 1 kmph , b ) 3 kmph , c ) 6 kmph , d ) 7 kmph , e ) 5 kmph | c | divide(subtract(16, 4), const_2) | a man can row his boat with the stream at 16 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 16 us = 4 s = ? s = ( 16 - 4 ) / 2 = 6 kmph answer : c" | a = 16 - 4
b = a / 2
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a ) $ 960 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 1,800 | e | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 6.0), const_1000) | a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 6.0 . if the company ’ s latest order required a truck with at least 288,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "total volume is 288000 given lbh = 8 * 10 * 12 . the number of hats inside it = 288000 / 10 * 8 * 12 = 300 . price of each hat is 6 $ then total value is 300 * 6.0 = 1800 . imo option e is correct answer . ." | a = 3 * 100
b = 8 * 10
c = a + b
d = c + 4
e = d * 1000
f = 8 * 10
g = f * 12
h = e / g
i = h * 6
j = i / 1000
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | c | inverse(add(divide(subtract(35, 25), subtract(45, 35)), const_1)) | a certain quantity of 45 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | "let ' s say that the total original mixture a is 100 ml the original mixture a thus has 45 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 45 ml and 25 ml is 20 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 20 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 10 % . therefore , 10 out of 20 is 1 / 2 and c is the answer ." | a = 35 - 25
b = 45 - 35
c = a / b
d = c + 1
e = 1/(d)
|
a ) 7 , b ) 33 , c ) 12 , d ) 24 , e ) 25 | d | add(16, 8) | there are 16 bees in the hive , then 8 more fly . how many bees are there in all ? | 16 + 8 = 24 . answer is d . | a = 16 + 8
|
a ) 33.33 , b ) 27.33 , c ) 28.38 , d ) 29.37 , e ) 28.31 | a | multiply(divide(add(3, const_2), add(subtract(multiply(3, divide(add(3, const_2), subtract(3, divide(3, 2)))), add(3, const_2)), subtract(20, multiply(3, divide(add(3, const_2), subtract(3, divide(3, 2))))))), const_100) | let raj be 3 years older than ravi and hema be two years younger than ravi . raj is 3 times as old as rahul who is hema ’ s brother . the ratio of the ages of hema and her brother is 3 : 2 . find by how much percentage raj ’ s age is more than hema ’ s when raj will be 20 years old . | answer : 33.33 | a = 3 + 2
b = 3 + 2
c = 3 / 2
d = 3 - c
e = b / d
f = 3 * e
g = 3 + 2
h = f - g
i = 3 + 2
j = 3 / 2
k = 3 - j
l = i / k
m = 3 * l
n = 20 - m
o = h + n
p = a / o
q = p * 100
|
a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 4500 , e ) $ 5200 | d | subtract(6000, divide(6000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | the salaries of a and b together amount to $ 6000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 6000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 6000 - x ) x = $ 4500 answer is d" | a = 100 - 85
b = 100 - 95
c = a / b
d = c + 1
e = 6000 / d
f = 6000 - e
|
a ) 27 , b ) 36 , c ) 29 , d ) 10 , e ) 70 | e | subtract(multiply(divide(subtract(90, 50), subtract(const_12, 9)), const_12), 90) | gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 50 and the turban . find the price of the turban . | "let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 50 . thus , 3434 ( 90 + x ) = 50 + x or 270 + 3 x = 200 + 4 x or x = 70 answer : e" | a = 90 - 50
b = 12 - 9
c = a / b
d = c * 12
e = d - 90
|
['a ) 18', 'b ) 77', 'c ) 625', 'd ) 276', 'e ) 191'] | a | multiply(const_2, sqrt(add(add(20, 20), add(add(20, 20), const_1)))) | the diagonal of a rectangle is cm and its area is 20 sq . cm . the perimeter of the rectangle must be : | explanation : ( or ) { \ color { black } l ^ { 2 } + b ^ { 2 } = 41 } also , { \ color { black } ( l + b ) ^ { 2 } = l ^ { 2 } + b ^ { 2 } + 2 lb } = 41 + 40 = 81 ( l + b ) = 9 . perimeter = 2 ( l + b ) = 18 cm . answer : a ) 18 | a = 20 + 20
b = 20 + 20
c = b + 1
d = a + c
e = math.sqrt(d)
f = 2 * e
|
a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 106 | e | divide(subtract(multiply(214, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 214 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | say there are a gallons of fuel a in the tank , then there would be 214 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 214 - a gallons of fuel b is 0.16 ( 214 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 214 - a ) = 30 - - > a = 106 . answer : e . | a = 16 / 100
b = 214 * a
c = b - 30
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) 638 , b ) 330 , c ) 550 , d ) 430 , e ) 880 | a | multiply(divide(880, const_100), subtract(const_100, 27.5)) | if 27.5 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "students enrolled in biology are 27.5 % and therefore not enrolled are 72.5 % . so of 880 is 880 * . 225 = 638 answer is a 638" | a = 880 / 100
b = 100 - 27
c = a * b
|
a ) 3 , b ) 5 , c ) 10 , d ) 15 , e ) 21 | d | multiply(divide(3, subtract(3, 2)), 5) | the ratio of local and international calls made by amy this week is 5 to 2 . if the ratio changes to 5 to 3 after amy makes 3 more international calls , how many local calls did amy make this week ? | the ratio of local and international calls made by amy this week is 5 to 2 let the current ratio be 5 x / 2 x the ratio changes to 5 to 3 after amy makes three more international calls 5 x / ( 2 x + 3 ) = 5 / 3 solving we get x = 3 initial ratio 15 : 6 ( 5 : 2 ) final ratio 15 : ( 6 + 3 ) = 15 : 9 = 5 : 3 so total number of local calls = 15 answer : d | a = 3 - 2
b = 3 / a
c = b * 5
|
a ) 10 % , b ) 10.5 % , c ) 12 % , d ) none of these , e ) can not be determined | d | divide(multiply(const_100, subtract(subtract(696.30, divide(660, 2)), divide(660, 2))), divide(660, 2)) | on a sum of money , the simple interest for 2 years is rs . 660 , while the compound interest is rs . 696.30 , the rate of interest being the same in both the cases . the rate of interest is : | "solution difference in c . i . and s . i . for 2 years = rs . ( 696.30 - 660 ) = rs . 36.30 . s . i . for one year = rs . 330 . ∴ s . i . on rs . 330 for 1 year = rs . 36.30 . ∴ rate ( 100 x 36.30 / 330 x 1 ) % = 11 % . answer d" | a = 660 / 2
b = 696 - 30
c = 660 / 2
d = b - c
e = 100 * d
f = 660 / 2
g = e / f
|
a ) 3 : 8 , b ) 3 : 6 , c ) 3 : 7 , d ) 4 : 1 , e ) 3 : 3 | d | divide(4, divide(4, 4)) | what is the ratio between perimeters of two squares one having 4 times the diagonal then the other ? | "d = 4 d d = d a √ 2 = 4 d a √ 2 = d a = 4 d / √ 2 a = d / √ 2 = > 4 : 1 answer : d" | a = 4 / 4
b = 4 / a
|
a ) 1000 km , b ) 700 km , c ) 800 km , d ) 1400 km , e ) 1200 km | e | multiply(150, 8) | what is the distance covered by a train if it travels with a speed of 150 kmh for 8 hours ? | "distance = time x speed distance = 150 x 8 = 1200 answer : e" | a = 150 * 8
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a ) 11 . , b ) 12 . , c ) 16 . , d ) 14 . , e ) 14.5 | c | add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(12, const_100), add(const_100, 50))) | following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 12 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "price of candy before price increase = 10 / 1.25 = 8 price of soda before price increase = 12 / 1.5 = 8 total price = 8 + 8 = 16 c is the answer" | a = 10 * 100
b = 100 + 25
c = a / b
d = 12 * 100
e = 100 + 50
f = d / e
g = c + f
|
a ) 44 , b ) 48 , c ) 50 , d ) 52 , e ) 58 | e | subtract(multiply(120, divide(55, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | of the 120 passengers on flight 750 , 55 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | number of passengers on flight = 120 number of female passengers = . 5 * 120 = 66 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 66 - 8 = 58 answer e | a = 55 / 100
b = 120 * a
c = 10 / 100
d = 120 * c
e = 10 / 100
f = 120 * e
g = f / 3
h = d - g
i = b - h
|
a ) 10 , b ) 72 , c ) 79 , d ) 16 , e ) 18 | c | subtract(power(2, 2), 2) | if x ^ 2 + 1 / x ^ 2 = 9 , what is the value of x ^ 4 + 1 / x ^ 4 ? | "important : i notice that if we square x ² , we get x ⁴ , and if we square 1 / x ² , we get 1 / x ⁴ , so let ' s see what happens if we take the equation x ² + 1 / x ² = 9 andsquareboth sides : ( x ² + 1 / x ² ) ² = 81 so , ( x ² + 1 / x ² ) ( x ² + 1 / x ² ) = 81 expand to get : x ⁴ + 1 + 1 + 1 / x ⁴ = 81 simplify : x ⁴ + 1 / x ⁴ = 79 answer : c" | a = 2 ** 2
b = a - 2
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | c | subtract(subtract(subtract(65, 25), const_4), const_2) | a certain no . when divided by 65 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 65 + 25 = 90 / 15 = 6 ( remainder ) c" | a = 65 - 25
b = a - 4
c = b - 2
|
a ) 70000 , b ) 60000 , c ) 80000 , d ) 92000 , e ) 50000 | d | add(add(multiply(divide(24000, 36000), 36000), multiply(divide(32000, 36000), 36000)), 36000) | a , b and c started a partnership business by investing rs . 24000 , rs . 32000 , rs . 36000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | "a : b : c = 24000 : 32000 : 36000 = 6 : 8 : 9 let total profit = p then p ã — 9 / 23 = 36000 p = ( 36000 ã — 23 ) / 9 = 92000 answer is d ." | a = 24000 / 36000
b = a * 36000
c = 32000 / 36000
d = c * 36000
e = b + d
f = e + 36000
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | e | divide(60, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 60 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 500 gm ) of tea and 1 / 2 kg ( 500 gm ) of coffee ( equal quantities ) = 50 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 60 = > x = 60 e" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = c + 1
e = b * d
f = 60 / e
|
a ) 250 , b ) 300 , c ) 350 , d ) 700 , e ) 500 | d | add(multiply(const_2, 300), 150) | brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 300 meters . they next meet after sally has run 150 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ? | "nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 300 meters , thus second time she runs 2 * 300 = 600 meters . since second time ( when they run full circumference ) brenda runs 600 meters and sally runs 150 meters , thus the circumference is 600 + 150 = 750 meters . answer : d ." | a = 2 * 300
b = a + 150
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a ) 6 km , b ) 3 km , c ) 7 km , d ) 5 km , e ) 2 km | c | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 8), divide(const_1, 9)), divide(const_1, 10))), const_3), const_1000) | a person travels equal distances with speeds of 8 km / hr , 9 km / hr and 10 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 8 + x / 9 + x / 10 = 47 / 60 x / 3 = 47 / 60 = > x = 2.33 . total distance = 3 * 2.33 = 6.99 km . answer : c" | a = 47 / const_60
b = 1 / 8
c = 1 / 9
d = b + c
e = 1 / 10
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 300 , b ) 320 , c ) 370 , d ) 400 , e ) 450 | b | divide(16, subtract(127.05, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 16 . if n / j = 127.05 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 127.05 here 127 is the quotient . given that remainder = 16 so , 127.05 = 127 + 16 / j so , j = 320 answer : b" | a = 4 * 10
b = a + 2
c = 100 + b
d = 127 - 5
e = 16 / d
|
a ) 22 : 17 , b ) 9 : 11 , c ) 5 : 4 , d ) 4 : 5 , e ) 8 : 11 | a | divide(add(multiply(4, divide(28, add(4, 3))), 5), add(multiply(3, divide(28, add(4, 3))), 5)) | the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 28 years . the ratio of their ages after 5 years will be | "let mini ’ s age = 4 x and minakshi ’ s age = 3 x then 4 x + 3 x = 28 x = 4 mini ’ s age = 16 years and minakshi ’ s age = 12 years ratio of their ages after 8 years = ( 16 + 5 ) : ( 12 + 5 ) = 22 : 17 answer : a" | a = 4 + 3
b = 28 / a
c = 4 * b
d = c + 5
e = 4 + 3
f = 28 / e
g = 3 * f
h = g + 5
i = d / h
|
a ) 313 , b ) 314 , c ) 315 , d ) 316 , e ) 317 | c | subtract(multiply(multiply(const_12, const_10), const_3), subtract(multiply(7, 30), multiply(divide(30, add(const_4, const_1)), divide(multiply(multiply(const_12, const_10), const_3), const_12)))) | there is an antique clock exhibition . so many clock are piled up with distinct numbers tagged with them . the person managing it hangs different clocks numbers periodically on the main wall . at 9 : 20 am , he hung the clock number 200 ; at 11 : 00 am , he hung the clock number 30 ; at 4 : 00 pm , he hung the clock number 240 . which numbered clock will he hang on the wall when the time is 7 : 30 pm ? | solution : he will hang clock number 315 at 7 : 30 pm . the person has been hanging the clock number which corresponds with the angle between the hour hand and the minute hand of the clock at that time . answer c | a = 12 * 10
b = a * 3
c = 7 * 30
d = 4 + 1
e = 30 / d
f = 12 * 10
g = f * 3
h = g / 12
i = e * h
j = c - i
k = b - j
|
a ) 151 , b ) 149 , c ) 152 , d ) 148 , e ) none of the above | a | divide(add(multiply(30, 150), subtract(165, 135)), 30) | the mean of 30 values was 150 . it was detected on rechecking that one value 165 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 150 × 30 − 135 + 165 / 30 = 4500 − 135 + 165 / 30 = 4530 / 30 = 151 answer a" | a = 30 * 150
b = 165 - 135
c = a + b
d = c / 30
|
a ) 5.5 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(550, divide(multiply(multiply(10, 550), divide(add(const_100, 10), const_100)), subtract(multiply(550, divide(add(const_100, 10), const_100)), 550))) | machine a and machine b are each used to manufacture 550 sprockets . it takes machine a 10 hours longer to produce 660 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 550 qt = 500 for a : ( t + 10 ) ( q ) = 550 qt + 10 q = 550 500 + 10 q = 550 q = 5 so a can produce 5 / hour . then b can produce = 5 ( 1.1 ) = 5.5 / hour . a" | a = 10 * 550
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 550 * f
h = g - 550
i = d / h
j = 550 / i
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a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 5 / 8 | e | add(subtract(1, divide(2, 3)), subtract(divide(2, 3), divide(1, 4))) | a batch of cookies was divided amomg 4 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin | "this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 4 b = 1 / 4 we can solve for g which is 5 / 12 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 5 / 12 + 1 / 4 ) = 5 / 12 x = 5 / 8 answer : e" | a = 2 / 3
b = 1 - a
c = 2 / 3
d = 1 / 4
e = c - d
f = b + e
|
a ) 150 , b ) 121 , c ) 144 , d ) 180 , e ) 1200 | c | multiply(multiply(subtract(8, const_1), 12), divide(8, const_2)) | there are , in a certain league , 8 teams , and each team face another team for a total of 12 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 : 144 option c ." | a = 8 - 1
b = a * 12
c = 8 / 2
d = b * c
|
a ) 2025 , b ) 2125 , c ) 2225 , d ) 2325 , e ) 2425 | b | divide(add(add(add(multiply(1700, 3), multiply(1550, 4)), multiply(1800, 5)), 5200), const_12) | a man spends rs . 1700 per month on an average for the first 3 months , rs 1550 for next 4 months and rs . 1800 per month for the last 5 months and saves rs . 5200 a year . what is his average monthly income ? | explanation : total expenditure for the first 3 months = 3 ã — 1700 = 5100 total expenditure for 4 months = 4 ã — 1550 = 6200 total expenditure for 5 months = 5 ã — 1800 = 9000 total expenditure and saving ( which is income for one year ) = 5100 + 6200 + 9000 + 5200 = rs . 25500 so , average monthly income = 25500 / 12 = rs . 2125 answer b | a = 1700 * 3
b = 1550 * 4
c = a + b
d = 1800 * 5
e = c + d
f = e + 5200
g = f / 12
|
a ) 725 , b ) 720 , c ) 600 , d ) 525 , e ) none of them | d | multiply(25, power(10, 25)) | 25 x 10 + 25 x 11 = ? | "= 25 x ( 10 + 11 ) ( by distributive law ) = 25 x 21 = 525 answer is d" | a = 10 ** 25
b = 25 * a
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a ) 740 , b ) 750 , c ) 690 , d ) 780 , e ) 708 | e | subtract(subtract(multiply(25, 45), multiply(12, 17)), multiply(12, 14)) | the average of 25 results is 45 . 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 * 45 or , x = 708 . answer : option e" | a = 25 * 45
b = 12 * 17
c = a - b
d = 12 * 14
e = c - d
|
a ) 297 , b ) 272 , c ) 342 , d ) 762 , e ) 269 | b | subtract(subtract(400, divide(multiply(400, 20), const_100)), divide(multiply(subtract(400, divide(multiply(400, 20), const_100)), 15), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 20 % and 15 % is ? | "400 * ( 80 / 100 ) * ( 85 / 100 ) = 272 answer : b" | a = 400 * 20
b = a / 100
c = 400 - b
d = 400 * 20
e = d / 100
f = 400 - e
g = f * 15
h = g / 100
i = c - h
|
a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability q that out of the two , one is white and one is blue ? | the required probability q = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer . | a = 3 * 5
b = 3 + 5
c = 5 + 2
d = b * c
e = d / 2
f = a / e
|
a ) 3 / 5 , b ) 3 / 2 , c ) 5 / 3 , d ) 5 / 7 , e ) 7 / 3 | e | divide(subtract(3.60, 3.25), subtract(3.25, 3.10)) | in what proportion must rice at rs 3.10 per kg be mixed with rice at rs 3.60 per kg , so that the mixture be worth rs 3.25 a kg ? | "c . p of 1 kg of cheaper rice = rs 3.10 c . p of 1 kg of expensive rice = rs 3.60 the mixture be worth for 1 kg = rs 3.25 by the alligation rule : quantity of cheaper rice / quantity of expensive rice = ( 3.6 - 3.25 ) / ( 3.25 - 3.10 ) = ( 0.35 ) / ( 0.15 ) = 7 / 3 e" | a = 3 - 60
b = 3 - 25
c = a / b
|
a ) 50 % , b ) 52 % , c ) 56 % , d ) 70 % , e ) 74 % | a | multiply(const_100, divide(divide(multiply(add(30, const_100), 40), const_100), add(const_100, 4))) | of the families in city x in 1994 , 40 percent owned a personal computer . the number of families in city x owning a computer in 1998 was 30 percent greater than it was in 1994 , and the total number of families in city x was 4 percent greater in 1998 than it was in 1994 . what percent of the families in city x owned a personal computer in 1998 ? | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 40 number of families owning computer in 1998 = 40 * 130 / 100 = 52 number of families in 1998 = 104 the percentage = 52 / 104 * 100 = 50 % . answer : a" | a = 30 + 100
b = a * 40
c = b / 100
d = 100 + 4
e = c / d
f = 100 * e
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a ) 15 , b ) 30 , c ) 35 , d ) 42 , e ) 45 | d | subtract(multiply(divide(35, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(35, const_100), 180))) | one - third less than 35 % of 180 is equal to : | lots of ways to tackle this . 35 % of 180 = 63 1 / 3 of 63 = 21 so , 1 / 3 less than 63 is equal to 63 - 21 = 42 answer : d | a = 35 / 100
b = a * 180
c = 1 / 3
d = 35 / 100
e = d * 180
f = c * e
g = b - f
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a ) $ 214 million , b ) $ 1120 million , c ) $ 1144 million , d ) $ 1240 million , e ) $ 1080 million | e | subtract(multiply(420, divide(const_12, const_2)), multiply(1.44, const_1000)) | country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 420 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 420 mill / 2 month i . e . $ 210 mill / month difference / month = 210 - 120 = 90 difference / year = $ 90 mill * 12 = $ 1080 mill answer : e" | a = 12 / 2
b = 420 * a
c = 1 * 44
d = b - c
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a ) 2000 , b ) 4500 , c ) 5000 , d ) 8000 , e ) 9000 | c | divide(65000, 13) | a company produces 65000 bottles of water everyday . if a case can hold 13 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 13 number of cases required to hold 65000 bottles = 65000 / 13 = 5000 cases . so the answer is c = 5000" | a = 65000 / 13
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a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | a | divide(circle_area(divide(5, const_2)), const_2) | what will be the fraction of 5 % | "explanation : 5 * 1 / 100 = 1 / 20 . option a" | a = 5 / 2
b = circle_area / (
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a ) 0 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 40 % | a | divide(add(4, add(10, const_3)), const_2) | operation # is defined as adding a randomly selected two digit multiple of 4 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated 10 times , what is the probability that it will yield at least two integers ? | "any multiple of 4 is even . any two - digit prime number is odd . ( even + odd ) / 2 is not an integer . thus # does not yield an integer at all . therefore p = 0 . answer : a ." | a = 10 + 3
b = 4 + a
c = b / 2
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['a ) 36 cm 2', 'b ) 35 cm 2', 'c ) 30 cm 2', 'd ) 32 cm 2', 'e ) 31 cm 2'] | c | multiply(divide(12, const_2), 5) | calculate the area of a triangle , if the sides are 13 cm , 12 cm and 5 cm , what is its area ? | the triangle with sides 13 cm , 12 cm and 5 cm is right angled , where the hypotenuse is 13 cm . area of the triangle = 1 / 2 * 12 * 5 = 30 cm 2 answer : c | a = 12 / 2
b = a * 5
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a ) 1356 , b ) 1557 , c ) 1688 , d ) 1700 , e ) 2500 | e | subtract(negate(50), multiply(subtract(3,5, 7,9), divide(subtract(3,5, 7,9), subtract(1, 3,5)))) | 1 , 3,5 , 7,9 , . . 50 find term of sequnce for this . | "this is an arithmetic progression , and we can write down a = 1 a = 1 , d = 2 d = 2 , n = 50 n = 50 . we now use the formula , so that sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 98 ) = 25 × ( 2 + 98 ) = 2500 = 2500 . e" | a = negate - (
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | add(multiply(sqrt(177), const_10), const_3) | if a ^ 2 + b ^ 2 = 177 and ab = 54 then find the value of a + b / a - b ? | "( a + b ) ^ 2 = a ^ 2 + b ^ 2 + 2 ab = 117 + 2 * 24 = 225 a + b = 15 ( a - b ) ^ 2 = a ^ 2 + b ^ 2 - 2 ab = 117 - 2 * 54 a - b = 3 a + b / a - b = 15 / 3 = 5 answer is c ." | a = math.sqrt(177)
b = a * 10
c = b + 3
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a ) 150 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 210), add(5, 2)) | a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 210 pounds of the mixture ? | "almonds : walnuts = 5 : 2 total mixture has 7 parts in a 210 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 210 = 150 pounds answer ( a )" | a = 5 * 210
b = 5 + 2
c = a / b
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a ) 8981 , b ) 3799 , c ) 1200 , d ) 2400 , e ) 1732 | d | add(multiply(multiply(divide(1200, 10), 5), const_3), multiply(divide(1200, 10), 5)) | the simple interest on a sum of money will be rs . 1200 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 1200 p - - - 5 - - - - - 600 3 p - - - 5 - - - - - 1800 - - - - - - = > 2400 answer : d" | a = 1200 / 10
b = a * 5
c = b * 3
d = 1200 / 10
e = d * 5
f = c + e
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a ) 9 / 29 , b ) 8 / 23 , c ) 3 / 4 , d ) 17 / 29 , e ) 3 / 8 | e | divide(multiply(const_3.0, const_3.0), add(multiply(multiply(4, 4), const_2), multiply(4, 4))) | pipe p can drain the liquid from a tank in 4 / 4 the time that it takes pipe q to drain it and in 2 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | "suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 4 / 4 ) rq ] = 4 / 4 = 1 also , rp = rr / ( 2 / 3 ) = > 1 = rr / ( 2 / 3 ) = > rr = 2 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 1 + 2 / 3 = 8 / 3 = 1 / ( 3 / 8 ) thus running simultaneously , pipe q will drain 3 / 8 of the liquid . thus answer = e ." | a = 3 * 0
b = 4 * 4
c = b * 2
d = 4 * 4
e = c + d
f = a / e
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a ) 160 , b ) 180 , c ) 200 , d ) 220 , e ) none | a | divide(add(4, 4), subtract(divide(const_1, 4), divide(const_1, add(const_1, 4)))) | a number whose fifth part increased by 4 is equal to its fourth part diminished by 4 is ? | "answer let the number be n . then , ( n / 5 ) + 4 = ( n / 4 ) - 4 â ‡ ’ ( n / 4 ) - ( n / 5 ) = 8 â ‡ ’ ( 5 n - 4 n ) / 20 = 8 â ˆ ´ n = 160 option : a" | a = 4 + 4
b = 1 / 4
c = 1 + 4
d = 1 / c
e = b - d
f = a / e
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a ) $ 100 , b ) $ 200 , c ) $ 250 , d ) $ 300 , e ) $ 500 | a | subtract(300, subtract(600, 400)) | a , b , c together can earn $ 600 per day . while a and c together earn $ 400 and b and c together can earn $ 300 . the daily earnings of c ? | b ' s daily earnings = 600 - 400 = $ 200 a ' s daily earnings = 600 - 300 = $ 300 c ' s daily earnings = 600 - 200 - 300 = $ 100 answer is a | a = 600 - 400
b = 300 - a
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a ) 6 days , b ) 5 days , c ) 10 days , d ) 9 days , e ) 8 days | c | divide(divide(const_1, add(divide(const_1, multiply(3, 5)), divide(const_1, multiply(3, 5)))), divide(3, 4)) | a can do a piece of work in 3 days of 5 hours each and b alone can do it in 5 days of 3 hours each . how long will they take it to do working together 3 / 4 hours a day ? | a ' s work per hour = 1 / 15 b ' s work per hour = 1 / 15 a & b ' s work per hour together = ( 1 / 15 ) + ( 1 / 15 ) = 2 / 15 so a & b together complete the work in 15 / 2 hours . . . if they work 2 1 / 2 = 5 / 2 hours a day , it will take ( 15 / 2 ) / ( 3 / 4 ) days = 10 days . . . answer : c | a = 3 * 5
b = 1 / a
c = 3 * 5
d = 1 / c
e = b + d
f = 1 / e
g = 3 / 4
h = f / g
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a ) 50 % , b ) 114.28 % , c ) 120 % , d ) 133 1 / 3 % , e ) 150 % | b | add(multiply(50, const_0_33), add(const_100, const_0_33)) | if the price of a certain bond on may 1 st was 3 / 4 the price of the bond on june 1 st and the price of the bond on july 1 st was 50 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1 st ? | the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 3 / 4 * 12 = 9 ; the price on july 1 st = 8 * 1.5 = 12 . the average price of the bond on may 1 st and july 1 st = ( 9 + 12 ) / 2 = 10.5 . the price of the bond on june 1 st ( 12 ) is 24 / 21 times ( 114.28 % ) the average price of the bond on may 1 st and july 1 st . answer : b . | a = 50 * const_0_33
b = 100 + const_0_33
c = a + b
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a ) 288 , b ) 350 , c ) 889 , d ) 367 , e ) 234 | d | subtract(multiply(speed(300, 18), 40), 300) | a 300 m long train crosses a platform in 40 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 40 = 50 / 3 3 x + 900 = 2000 = > x = 367 m . answer : d" | a = speed * (
b = a - 40
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a ) 40 , b ) 41 , c ) 52 , d ) 62 , e ) 73 | c | subtract(multiply(add(25, 1), add(26, 1)), multiply(25, 26)) | the average age of a class of 25 students is 26 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "total age of all students = 25 ã — 26 total age of all students + age of the teacher = 26 ã — 27 age of the teacher = 27 ã — 26 â ˆ ’ 25 ã — 26 = 26 ( 27 â ˆ ’ 25 ) = 26 ã — 2 = 52 answer is c ." | a = 25 + 1
b = 26 + 1
c = a * b
d = 25 * 26
e = c - d
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a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) 340 | b | divide(subtract(1356, 15), subtract(6, const_1)) | the difference between two numbers is 1356 . when the larger number is divided by the smaller one , the quotient is 6 and the remainder is 15 . the smaller number is | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 smaller number = 270 answer : b" | a = 1356 - 15
b = 6 - 1
c = a / b
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 3 / 5 , e ) 4 / 5 | a | divide(subtract(96, 94), subtract(94, 90)) | on a test , the boys in the class averaged 90 points and the girls in the class averaged 96 points . if the overall class average was 94 points , what is the ratio of boys to girls in the class ? | ( 96 g + 90 b ) / ( g + b ) = 94 96 g + 90 b = 94 ( g + b ) 2 g = 4 b b / g = 1 / 2 the answer is a . | a = 96 - 94
b = 94 - 90
c = a / b
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a ) 1 , b ) 3 , c ) 6 , d ) 9 , e ) 12 | b | add(subtract(9, divide(3, divide(1, 3))), 3) | 9 - 3 ÷ 1 / 3 + 3 = ? | 9 - 3 1 / 3 + 3 = 9 - 3 ÷ 1 / 3 + 3 = 9 - ( 3 x 3 ) + 3 = 9 - 9 + 3 = 3 correct answer : b | a = 1 / 3
b = 3 / a
c = 9 - b
d = c + 3
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a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | c | multiply(5, 2) | if n divided by 7 has a remainder of 2 , what is the remainder when 5 times n is divided by 7 ? | "as per question = > n = 7 p + 2 for some integer p hence 5 n = > 35 q + 10 = > remainder = > 10 for some integer q alternatively = > n = 2 > 5 n = > 10 = > 10 divided by 7 will leave a remainder 3 hence c" | a = 5 * 2
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a ) 8 , b ) 6 , c ) 5 , d ) 3 , e ) 2 | a | multiply(multiply(8, 2), divide(2, 2)) | in the coordinate plane , points ( x , 2 ) and ( 8 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x + y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 2 ) = ( 4 , 2 ) and ( 8 , y ) = ( 8,4 ) - - > x + y = 4 + 4 = 8 . answer : a" | a = 8 * 2
b = 2 / 2
c = a * b
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a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | d | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(355, multiply(floor(divide(355, const_4)), const_4))), multiply(const_2, const_10)) | find the ones digit of 73 ^ 355 | the units digit of 73 ^ 355 will be the same as the units digit of 3 ^ 355 . 3 ^ 1 = 3 - - > the units digit is 3 ; 3 ^ 2 = 9 - - > the units digit is 9 ; 3 ^ 3 = 27 - - > the units digit is 7 ; 3 ^ 4 = 81 - - > the units digit is 1 ; 3 ^ 5 = 243 - - > the units digit is 3 again ; . . . so , as you can see the units digit repeats in blocks of 4 : { 3 , 9 , 7 , 1 } , { 3 , 9 , 7 , 1 } , . . . now , since 355 = 352 + 3 = ( multiple of 4 ) + 2 , then the units digit of 3 ^ 355 will be the third number in the pattern thus 7 . answer : d . | a = 3 + 4
b = a * 10
c = 73 - b
d = 355 / 4
e = math.floor(d)
f = e * 4
g = 355 - f
h = c ** g
i = 2 * 10
j = h - i
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a ) 6.15 , b ) 5.15 , c ) 3.15 , d ) 1.15 , e ) 7.15 | b | multiply(divide(divide(140, const_1000), 98), const_3600) | how long will it take a train travelling at 98 kmph to pass an electric pole if the train is 140 m long | "sol . speed = [ 98 x 5 / 18 ] m / sec = 27.2 m / sec . time taken = ( 140 / 27.2 ) sec = 5.15 sec . answer b" | a = 140 / 1000
b = a / 98
c = b * 3600
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a ) 120 kmph , b ) 150 kmph , c ) 50 kmph , d ) 80 kmph , e ) 75 kmph | a | divide(multiply(subtract(multiply(12.5, const_2), const_1), multiply(12.5, const_2)), add(const_4, const_1)) | a train can travel 50 % faster than a car . both start from point a at the same time and reach point b 75 kms away from a at the same time . on the way , however , the train lost about 12.5 minutes while stopping at the stations . the speed of the car is : | "let speed of the car be x kmph . then , speed of the train = 150 / 100 x = 3 / 2 x kmph . 75 / x - 75 / ( 3 / 2 ) x = 125 / 10 x 60 75 / x - 50 / x = 5 / 24 x = 25 x 24 / 5 = 120 kmph . answer : a" | a = 12 * 5
b = a - 1
c = 12 * 5
d = b * c
e = 4 + 1
f = d / e
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a ) a ) 3 , b ) b ) 4 , c ) c ) 11 , d ) d ) 9 , e ) e ) 7 | c | sqrt(22) | from below option 22 is divisible by which one ? | "22 / 11 = 2 c" | a = math.sqrt(22)
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a ) 732 % , b ) 560 % , c ) 641 % , d ) 370 % , e ) 300 % | e | multiply(divide(subtract(2, divide(1, 2)), divide(1, 2)), const_100) | by approximately what percent is x greater than 1 / 2 if ( 1 / 2 ) ( x ) = 1 ? | "what percent is x greater than 1 / 2 if ( 1 / 2 ) ( x ) = 1 ? = > x = 2 % change = [ ( 2 - 1 / 2 ) / ( 1 / 2 ) ] * 100 = ( 4 - 1 ) * 100 = 300 % approx ans , e" | a = 1 / 2
b = 2 - a
c = 1 / 2
d = b / c
e = d * 100
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a ) 84 , b ) 89 , c ) 90 , d ) 92 , e ) 91 | e | subtract(subtract(100, const_3), 6) | set x consists of all two - digit primes and set y consists of all positive multiples of 6 less than 100 . if the two sets are combined into one , what will be the range of the new set ? | "set x = { 11 , 13 , 17 , . . . . . . . . . . . . . , 83 , 89 , 97 } set y = { 6 , 12 , 18 , . . . . . . . . . . . . . . . , 84 , 90 , 96 } combining two sets , say set z set z = { 6 , 11 , 12 , 13 , 17 , 18 , . . . . . . . . . . . . . . . . . . . , 83 , 84 , 89 , 90 , 96,97 } range = max value - min value range ( z ) = 97 - 6 = 91 oa e is the answer ." | a = 100 - 3
b = a - 6
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a ) 33 , b ) 28 , c ) 22.5 , d ) 24 , e ) 23 | c | divide(add(23, 26), const_2) | if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 23 , 30 , 22 , 26 , 19 , 33 and 35 ? | "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 23 , 30 , 22 , 26 , 19 , 33 and 35 19 , 22 , 23 , 26 , 30 , 33 , 35 , 42 . . . numbers less than median are 19 , 22 , 23 , 26 . . the median of these numbers = center of 22 and 23 = 22.5 c" | a = 23 + 26
b = a / 2
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a ) 40 , b ) 60 , c ) 70 , d ) 80 , e ) 20 | a | divide(divide(multiply(800, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "a 800 * ( 25 / 100 ) = 200 - - - - 5 ? - - - - 1 = > rs . 40" | a = 800 * 25
b = a / 100
c = b / 5
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a ) 20 % , b ) 30 % , c ) 25 % , d ) 37.5 % , e ) 38 % | c | multiply(divide(subtract(20, 15), 20), const_100) | 37 . if the cost price of 15 tables be equal to the selling price of 20 tables , the loss per cent is ? | let c . p . of each table = re . 1 c . p . of 20 tables = rs . 20 s . p . of 20 table = c . p . of 15 tables = rs . 15 loss = ( 5 / 20 ) x 100 % = 25 % answer : c | a = 20 - 15
b = a / 20
c = b * 100
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a ) 17 , b ) 16 , c ) 15 , d ) 10 , e ) 13 | d | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 46 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "i think brute force with some common sense should be used to solve this problem . write down all perfect squares less than 46 : 1 , 4 , 9 , 16 , 25 , 36 . now , 46 should be the sum of 3 of those 8 numbers . also to simplify a little bit trial and error , we can notice that as 46 is an odd numbers then either all three numbers must be odd ( odd + odd + odd = odd ) or two must be even and one odd ( even + even + odd = odd ) . we can find that 46 equals to 1 + 9 + 36 = 1 ^ 2 + 3 ^ 2 + 6 ^ 2 = 46 - - > 1 + 3 + 6 = 10 . answer : d ." | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
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['a ) 80', 'b ) 96', 'c ) 108', 'd ) 126', 'e ) 132'] | d | divide(factorial(9), multiply(factorial(divide(const_10, const_2)), factorial(4))) | in a certain circle there are 9 points . what is the number of the triangles connecting 4 points of the 9 points ? | imo : d here we have to select 4 points out of 9 points . order is not important so the answer will be 9 c 4 = 126 answer d | a = math.factorial(9)
b = 10 / 2
c = math.factorial(b)
d = math.factorial(4)
e = c * d
f = a / e
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a ) 3377 , b ) 2678 , c ) 5565 , d ) 1976 , e ) 1671 | c | divide(multiply(divide(multiply(5000, add(const_100, 5)), const_100), add(const_100, 6)), const_100) | find the amount on rs . 5000 in 2 years , the rate of interest being 5 % per first year and 6 % for the second year ? | "5000 * 105 / 100 * 106 / 100 = > 5565 answer : c" | a = 100 + 5
b = 5000 * a
c = b / 100
d = 100 + 6
e = c * d
f = e / 100
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 22 | b | add(divide(add(50, const_1), add(const_1, const_4)), const_2) | how many trailing zeroes does 49 ! + 50 ! have ? | "49 ! + 50 ! = 51 * 49 ! no of trailing 0 ' s = no of 5 * 2 no of 2 ' s = 49 / 2 + 49 / 4 + 49 / 8 + 49 / 16 + 49 / 32 = d no of 5 ' s = 49 / 5 + 49 / 25 = 10 = p now since p < d no of trailing 0 ' s = 10 answer : b" | a = 50 + 1
b = 1 + 4
c = a / b
d = c + 2
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a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | d | subtract(multiply(30, 8.00), multiply(6.00, 30)) | a club wants to mix 30 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ? | "let number of pounds of 5 $ candy to be used be w 6 = ( 30 * 8 + 5 * w ) / ( 30 + w ) = > 180 + 6 w = 240 + 5 w = > w = 60 answer d" | a = 30 * 8
b = 6 * 0
c = a - b
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a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years | d | divide(subtract(divide(70, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | the sum of the ages of 5 children born at the intervals of 3 years each is 70 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 70 5 x = 40 = > x = 8 . age of youngest child = x = 8 years . answer : d" | a = 5 / 3
b = 70 / a
c = 5 - 1
d = c * 3
e = b - d
f = e / 3
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a ) rs . 190 , b ) rs . 260 , c ) rs . 493 , d ) rs . 393 , e ) rs . 593 | c | multiply(4, divide(1070, add(add(4, 2), const_3))) | rs . 1070 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 1070 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1070 = rs . 493 answer : c" | a = 4 + 2
b = a + 3
c = 1070 / b
d = 4 * c
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a ) 50 , b ) 80 , c ) 70 , d ) 60 , e ) 71 | c | divide(add(add(add(add(66, 75), 52), 68), 89), add(const_4, const_1)) | nancy obtained 66 , 75 , 52 , 68 and 89 marks ( out of 100 ) in american literature , history , home economics , physical education and art . calculate her average marks ? | explanation : average = ( 66 + 75 + 52 + 68 + 89 ) / 5 = 70 . answer : c ) 70 | a = 66 + 75
b = a + 52
c = b + 68
d = c + 89
e = 4 + 1
f = d / e
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a ) 75142 , b ) 64851 , c ) 5149 , d ) 69993 , e ) none of them | d | subtract(multiply(7, const_1000), 7) | the difference between the local value and the face value of 7 in the numerical 32675149 is | "= ( local value of 7 ) - ( face value of 7 ) = ( 70000 - 7 ) = 69993 answer is d" | a = 7 * 1000
b = a - 7
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a ) 0.64202 , b ) 0.64204 , c ) 0.642022 , d ) 0.642075 , e ) none | d | divide(add(multiply(0.64206, 0.64207), multiply(0.64206, 0.64208)), 0.64206) | the average of numbers 0.64206 , 0.64207 , 0.64208 and 0.64209 is ? | "answer average = ( 0.64206 + 0.64207 + 0.64208 + 0.64209 ) / 4 = 2.5683 / 4 = 0.642075 correct option : d" | a = 0 * 64206
b = 0 * 64206
c = a + b
d = c / 0
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a ) 13000 , b ) 7000 , c ) 10000 , d ) 4375 , e ) none of these | d | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(80, 1), 80), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(80, 1), 80))), 75000)) | a textile manufacturing firm employees 80 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 80 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | "explanation : profit = 5 , 00,000 â ˆ ’ ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 ã — ( 79 / 80 ) â ˆ ’ 150000 ã — ( 79 / 80 ) â ˆ ’ 75000 . = > rs 2 , 70,625 . decrease in profit = > 2 , 75,000 â ˆ ’ 2 , 70,625 = > rs . 4,375 answer : d" | a = 5 * 1000
b = a * 100
c = 75000 * 2
d = c + 75000
e = b - d
f = 80 - 1
g = f / 80
h = 5 * 1000
i = h * 100
j = g * i
k = 75000 * 2
l = 80 - 1
m = l / 80
n = k * m
o = j - n
p = o - 75000
q = e - p
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a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | a | multiply(divide(subtract(multiply(12, divide(5, const_60)), multiply(4, divide(5, const_60))), 4), const_60) | a hiker walking at a constant rate of 4 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 12 kilometers per hour . the cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate . how many minutes must the cyclist wait until the hiker catches up ? | "in 5 minutes , the cyclist travels a distance of ( 5 / 60 ) * 12 = 1 km . the time it takes the hiker to complete this distance is 1 / 4 hours = 15 minutes the cyclist needs to wait 15 - 5 = 10 minutes the answer is a ." | a = 5 / const_60
b = 12 * a
c = 5 / const_60
d = 4 * c
e = b - d
f = e / 4
g = f * const_60
|
a ) 20 % , b ) 25 % , c ) 33.33 % , d ) 35 % , e ) none of these | d | multiply(divide(subtract(multiply(add(add(add(20, 5), const_10), add(divide(5, const_2), const_3)), 18), multiply(30, 18)), multiply(30, 18)), const_100) | a milk man has 30 liters of milk . if he mixes 5 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is : | explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 30 liters of milk . therefore , profit percentage = 35 % . answer : d | a = 20 + 5
b = a + 10
c = 5 / 2
d = c + 3
e = b + d
f = e * 18
g = 30 * 18
h = f - g
i = 30 * 18
j = h / i
k = j * 100
|
a ) 280 , b ) 400 , c ) 540 , d ) 900 , e ) 840 | d | divide(divide(divide(180, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(2, 3)) | of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 180 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 180 geese survived the first year : 1 / 5 * ( total ) = 180 - - > ( total ) = 900 . answer : d ." | a = 3 / 5
b = 1 - a
c = 180 / b
d = 3 / 4
e = c / d
f = 2 / 3
g = e / f
|
a ) 2 l , b ) 3 l , c ) 2.5 l , d ) 4 l , e ) 5 l | a | multiply(20, divide(const_10, const_100)) | 20 litres of mixture of acid and water contain 10 % water . how much water should be added so that percentage of water becomes 20 % in this mixture ? | total mixture = 20 ltr water contain = 10 % of 20 ltr = > 2 ltr water it means 18 ltr is acid in that mixture we have to make 20 % water so . . 20 % of 20 = 4 ltr water must be there in the new mixture we have already 2 ltr water so we have to add 2 ltr more answer : a | a = 10 / 100
b = 20 * a
|
a ) 12 , b ) 3 , c ) 16 , d ) 0 , e ) 9 | d | subtract(3, 8) | what will be the remainder when 8 - 3 + 3 ^ 5 + 2 ^ 10 is divided by 4 ? | "the multiplication ( e . g . 2 ^ 10 ) is done first and then the sum ( e . g . 15 + 20 ) and subtraction ( e . g . 8 - 3 ) , and after all this expression it should be divided by 4 and the answer is 0 , option d ." | a = 3 - 8
|
a ) 4 : 5 , b ) 8 : 3 , c ) 8 : 4 , d ) 4 : 8 , e ) 4 : 1 | b | divide(sqrt(64), sqrt(9)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 64 hours respectively . the ratio of their speeds is ? | let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 64 : √ 9 = 8 : 3 answer : b | a = math.sqrt(64)
b = math.sqrt(9)
c = a / b
|
a ) 119 , b ) 110 , c ) 112 , d ) 113 , e ) 115 | a | multiply(subtract(divide(divide(multiply(subtract(const_100, 15), add(const_100, 40)), const_100), const_100), const_1), const_100) | a trader bought a car at 15 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 85 s = 85 * ( 140 / 100 ) = 119 100 - 119 = 19 % answer : a" | a = 100 - 15
b = 100 + 40
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) 5 : 4 , b ) 3 : 0 , c ) 5 : 1 , d ) 3 : 2 , e ) 3 : 7 | c | divide(add(multiply(6, divide(add(6, 6), subtract(6, 3))), 6), subtract(multiply(3, divide(add(6, 6), subtract(6, 3))), 6)) | the ratio between the present ages of a and b is 6 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ? | "let the present ages of a and b be 6 x and 3 x years respectively . then , ( 6 x - 4 ) / ( 3 x + 4 ) = 1 / 1 3 x = 8 = > x = 2.67 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 20 : 4 = 5 : 1 . answer : c" | a = 6 + 6
b = 6 - 3
c = a / b
d = 6 * c
e = d + 6
f = 6 + 6
g = 6 - 3
h = f / g
i = 3 * h
j = i - 6
k = e / j
|
a ) 1000 , b ) 1500 , c ) 2778 , d ) 2788 , e ) 2991 | b | divide(multiply(225, const_100), subtract(add(const_100, 5), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 225 more , there would have been a gain of 5 % . what is the cost price ? | "90 % 105 % - - - - - - - - 15 % - - - - 225 100 % - - - - ? = > rs . 1500 answer : b" | a = 225 * 100
b = 100 + 5
c = 100 - 10
d = b - c
e = a / d
|
a ) 2.5 sec , b ) 2.8 sec , c ) 7.5 sec , d ) 2.3 sec , e ) 1.5 sec | e | divide(60, multiply(144, const_0_2778)) | in what time will a train 60 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 60 / 40 = 1.5 sec . answer : e" | a = 144 * const_0_2778
b = 60 / a
|
a ) 0 , b ) 67 , c ) 88 , d ) 12 , e ) 66 | a | subtract(multiply(40, divide(50, const_100)), multiply(divide(4, 5), 25)) | how much is 50 % of 40 is greater than 4 / 5 of 25 ? | "( 50 / 100 ) * 40 – ( 4 / 5 ) * 25 = 0 answer : a" | a = 50 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
|
a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 87.5 % | e | multiply(divide(subtract(150, 80), 80), const_100) | a certain telescope increases the visual range at a particular location from 80 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | "original visual range = 80 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 80 ) / 80 * 100 % = 7 / 8 * 100 % = 87.5 % answer e" | a = 150 - 80
b = a / 80
c = b * 100
|
['a ) 16', 'b ) 20', 'c ) 18', 'd ) 22', 'e ) 21'] | b | sqrt(add(multiply(16, 16), multiply(12, 12))) | 34 . the side surface of a cylinder is rolled with a rectangle . if the height of a cylinder is 16 feet and the perimeter of the circular base . is 12 feet , what is the diagonal of the rectangle ? | think of an aluminum can can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both , use pythagoreans theorem or properties of 3 - 4 - 5 triangles to solve for the hypothenuse , 20 . ( correct answer b ) | a = 16 * 16
b = 12 * 12
c = a + b
d = math.sqrt(c)
|
a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | c | multiply(multiply(7, 9), const_2) | if ( x + yi ) / i = ( 7 + 9 i ) , where x and y are real , what is the value of ( x + yi ) ( x - yi ) ? | ( x + yi ) / i = ( 7 + 9 i ) ( x + yi ) = i ( 7 + 9 i ) = - 9 + 7 i ( x + yi ) ( x - yi ) = ( - 9 + 7 i ) ( - 9 - 7 i ) = 81 + 49 = 130 correct answer is c 0130 | a = 7 * 9
b = a * 2
|
a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 21.00 sec , e ) 72.0 sec | d | divide(add(110, 240), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 240 m in length ? | "d = 110 + 240 = 350 m s = 60 * 5 / 18 = 50 / 3 t = 350 * 3 / 50 = 21.00 sec answer : d" | a = 110 + 240
b = 60 * const_0_2778
c = a / b
|
a ) 635 cm , b ) 620 cm , c ) 650 cm , d ) 610 cm , e ) 654 cm | a | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 250), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 250 inches into centimeter ? | "1 inch = 2.54 cm 250 inches = 250 * 2.54 = 635 cm answer is a" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 250
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 14 , b ) 15 , c ) 16 , d ) 18 , e ) 20 | c | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 12), 30)) | it takes 40 identical printing presses 12 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | "40 printing presses can do 1 / 12 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 12 = 1 / 16 of the job each hour . the answer is c ." | a = 1000 / 2
b = a * 1000
c = 1000 / 2
d = c * 1000
e = d / 40
f = e / 12
g = f * 30
h = b / g
|
a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 13 % | e | multiply(divide(subtract(add(const_1, divide(80, const_100)), add(const_1, divide(60, const_100))), add(const_1, divide(60, const_100))), const_100) | a certain company reported that the revenue on sales increased 60 % from 2000 to 2003 , and increased 80 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 160 and and in 2005 180 , so from 2003 to 2005 it increased by ( 180 - 160 ) / 160 = 20 / 160 = 13 % answer : e ." | a = 80 / 100
b = 1 + a
c = 60 / 100
d = 1 + c
e = b - d
f = 60 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 10 , b ) 388 , c ) 37 , d ) 29 , e ) 22 | a | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 25), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 20), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 25 % and then increased by 20 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . new final price = 120 % of ( 75 % of rs . 100 ) = rs . ( 120 / 100 * 75 / 100 * 100 ) = rs . 90 . decrease = 10 % answer : a ) 10" | a = 100 * 25
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 20
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 1500 , b ) 1600 , c ) 1700 , d ) 1800 , e ) 1900 | d | add(divide(multiply(300, const_100), 20), 300) | p runs 20 % faster than q so p gives q a 300 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ? | "let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.2 v = ( d - 300 ) / v d = 1.2 d - 360 0.2 d = 360 d = 1800 meters . the answer is d ." | a = 300 * 100
b = a / 20
c = b + 300
|
a ) 18 , b ) 28 , c ) 48 , d ) 38 , e ) 55 | e | divide(multiply(subtract(71, 11), 11), add(11, const_1)) | if a certain number is divided by 11 , the quotient , dividend , and divisor , added together , will amount to 71 . what is the number ? | "let x = the number sought . then x / 11 + x + 11 = 71 and x = 55 . e" | a = 71 - 11
b = a * 11
c = 11 + 1
d = b / c
|
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