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 ) 3 , b ) 7 , c ) 189 , d ) 227 , e ) 234 | c | multiply(multiply(multiply(subtract(add(add(subtract(4, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(4, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(4, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | a β sophie germain β prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 4 is | "in that case , the sophie prime numbers greater than 5 are 7 , 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be c ." | a = 4 - 1
b = 2 + 1
c = a + b
d = 2 + 1
e = c + d
f = e - 10
g = 4 - 1
h = 2 + 1
i = g + h
j = 2 * i
k = j + 1
l = k - 10
m = f * l
n = 2 + 1
o = m * n
p = 4 - 1
q = 2 + 1
r = p + q
s = 2 + 1
t = r + s
u = 2 + 1
v = t + u
w = 2 * v
x = w + 1
y = 2 * 10
z = x - y
A = o * z
|
a ) 1 : 2 , b ) 9 : 2 , c ) 1 : 3 , d ) 3 : 1 , e ) 1 : 1 | b | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 8 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ£ β 12 : 30000 Γ£ β 4 = 45 Γ£ β 12 : 30 Γ£ β 4 = 3 Γ£ β 12 : 2 Γ£ β 4 = 9 : 2 answer : option b" | a = 45000 * 12
b = 4 + 3
c = 30000 * b
d = a / c
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a ) 22 , b ) 50 , c ) 88 , d ) 52 , e ) 60 | e | multiply(12, 5) | what number has a 5 : 1 ratio to the number 12 ? | "5 : 1 = x : 12 x = 60 answer : e" | a = 12 * 5
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a ) 23077 , b ) 24000 , c ) 24005 , d ) 24009 , e ) 24002 | a | subtract(50000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 50000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 50000 share of b = 6 / 13 ( 50000 ) = rs . 23076.92 ~ 23077 answer : a" | a = const_60 * 100
b = 50000 - a
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a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 | e | divide(multiply(130, 250), multiply(500, 65)) | if a * b * c = 130 , b * c * d = 65 , c * d * e = 500 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a Γ’ Λ β b Γ’ Λ β c / b Γ’ Λ β c Γ’ Λ β d = 130 / 65 = > a / d = 2 d Γ’ Λ β e Γ’ Λ β f / c Γ’ Λ β d Γ’ Λ β e = 250 / 500 = > f / c = 1 / 2 a / d * f / c = 2 * 1 / 2 = 1 answer : e" | a = 130 * 250
b = 500 * 65
c = a / b
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a ) 7 , b ) 9 , c ) 14 , d ) 20 , e ) 35 | c | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | how many internal diagonals does a heptagon ( seven sided polygon ) have ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 7 no . of diagonals = 7 ( 7 - 3 ) / 2 = 14 ans c" | a = 2 * 4
b = a - 3
c = 2 * 4
d = c / 2
e = b * d
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a ) 13 , b ) 14 , c ) 9 , d ) 10 , e ) 12 | e | multiply(multiply(7, 7), divide(5, 7)) | in the coordinate plane , points ( x , 5 ) and ( 7 , y ) are on line k . if line k passes through the origin and has slope 5 / 7 , then x + y = | "line k passes through the origin and has slope 5 / 7 means that its equation is y = 5 / 7 * x . thus : ( x , 5 ) = ( 7,5 ) and ( 7 , y ) = ( 7,5 ) - - > x + y = 7 + 5 = 12 . answer : e" | a = 7 * 7
b = 5 / 7
c = a * b
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a ) 5 , 500,000 , b ) 2 , 200,000 , c ) 55,000 , d ) 28,000 , e ) 280 | b | subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 2)), 2)), volume_cube(100)) | a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 2.5 grams per cubic centimeter and mineral y weighs 2 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 2 gms in 600,000 cm ^ 3 , y = 2 * 600,000 = 1 , 200,000 gms total gms in 1 m ^ 3 = 1 , 000,000 + 1 , 200,000 = 2 , 200,000 answer : b" | a = volume_cube / (
b = a * 10
c = b * 2
d = c + 2
e = volume_cube / (
f = e * 10
g = 2 * 2
h = f * g
i = d - h
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a ) 10 , b ) 12 , c ) 3 , d ) 11 , e ) 9 | c | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 56)) | if the wheel is 56 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 56 * x = 1056 = > x = 3 answer : c" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 56
h = 1056 / g
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a ) 1440 , b ) 1200 , c ) 2832 , d ) 1299 , e ) 1236 | a | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 12) | a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 12 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | "explanation : l * 30 = 1200 Γ¨ l = 40 40 + 30 + 50 = 120 120 * 12 = 1440 answer : option a" | a = 1200 / 30
b = 30 + a
c = 30 ** 2
d = 1200 / 30
e = d ** 2
f = c + e
g = math.sqrt(f)
h = b + g
i = h * 12
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a ) 1 kmph , b ) 2 kmph , c ) 1.75 kmph , d ) 2.5 kmph , e ) 3.5 kmph | c | divide(subtract(multiply(divide(1, 12), const_60), multiply(divide(1, 40), const_60)), const_2) | a boat moves upstream at the rate of 1 km in 40 minutes and down stream 1 km in 12 minutes . then the speed of the current is : | "rate upstream = ( 1 / 40 * 60 ) = 1.5 kmph rate dowm stream = 1 / 12 * 60 = 5 kmph rate of the current = Β½ ( 5 - 1.5 ) = 1.75 kmph answer : c" | a = 1 / 12
b = a * const_60
c = 1 / 40
d = c * const_60
e = b - d
f = e / 2
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a ) 8 , b ) 7 , c ) 5 , d ) 2.0 , e ) 1 | d | divide(multiply(multiply(multiply(10, 10), 3), 32), multiply(multiply(30, 8), 20)) | if 10 bulls can plough 20 identical fields in 3 days working 10 hours a day , then in how many days can 30 bulls plough 32 same identical fields working 8 hours a day ? | explanation : m 1 * d 1 * w 2 = m 2 * d 2 * w 1 10 * 3 * 10 * 32 = 30 * d * 8 * 20 d = 2 days answer : d | a = 10 * 10
b = a * 3
c = b * 32
d = 30 * 8
e = d * 20
f = c / e
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a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | add(divide(subtract(multiply(floor(divide(7, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | how many numbers from 2 to 7 are exactly divisible by 2 ? | "2 / 2 = 1 and 7 / 2 = 3 3 - 1 = 2 2 + 1 = 3 numbers . b )" | a = 7 / 2
b = math.floor(a)
c = b * 2
d = 2 / 2
e = math.floor(d)
f = e + 1
g = f * 2
h = c - g
i = h / 2
j = i + 1
|
a ) 20 , b ) 23 , c ) 169 , d ) 172 , e ) 198 | b | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | the product of two numbers is 120 . the sum of their squares is 289 . the sum of the two numbers is : | "explanation : let the number be x and y . we know that , ( x + y ) 2 = ( x 2 + y 2 ) + 2 xy = 289 + 2 Γ 120 = 289 + 240 = 529 β x + y = β 529 = 23 correct option : b" | a = 2 * 120
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 120
f = d + e
g = math.sqrt(f)
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a ) 448 , b ) 488 , c ) 542 , d ) 548 , e ) 1088 | e | multiply(54, const_10) | the least number , which when divided by 8 , 15 , 20 and 54 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 8 , 15 , 20 , 54 ) + 8 = 1080 + 8 = 1088 . answer : e" | a = 54 * 10
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a ) 2 / 19 , b ) 3 / 31 , c ) 4 / 37 , d ) 5 / 41 , e ) 6 / 53 | b | multiply(divide(10, add(add(10, 15), 6)), divide(subtract(10, const_1), subtract(add(add(10, 15), 6), const_1))) | there are 10 slate rocks , 15 pumice rocks , and 6 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "10 / 31 * 9 / 30 = 3 / 31 the answer is b ." | a = 10 + 15
b = a + 6
c = 10 / b
d = 10 - 1
e = 10 + 15
f = e + 6
g = f - 1
h = d / g
i = c * h
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a ) 2.3 , b ) 2.6 , c ) 3.6 , d ) 3.7 , e ) 4.6 | d | divide(subtract(multiply(6, 2.8), add(multiply(2, 2.4), multiply(2, 2.3))), 2) | the average of 6 no . ' s is 2.80 . the average of 2 of them is 2.4 , while the average of the other 2 is 2.3 . what is the average of the remaining 2 no ' s ? | sum of the remaining two numbers = ( 2.80 * 6 ) - [ ( 2.4 * 2 ) + ( 2.3 * 2 ) ] = 16.80 - ( 4.8 + 4.6 ) = 16.80 - 9.40 = 7.40 required average = ( 7.4 / 2 ) = 3.7 answer : d | a = 6 * 2
b = 2 * 2
c = 2 * 2
d = b + c
e = a - d
f = e / 2
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a ) - 12 , b ) - 18 , c ) - 24 , d ) - 36 , e ) - 48 | d | multiply(negate(add(const_2, const_3)), subtract(12, add(const_2, const_3))) | if x and y are integers and | x - y | = 12 , what is the minimum possible value of xy ? | "we are given | x - y | = 12 , minimum possible value of xy would be maximum numeric value with a - ive sign . . so one of x or y will be negative and other negative . . | x - y | = 12 in this case means that the numeric sum of x and y is 12 . . various combinations could be - 1 and 11 , - 2 and 10 , - 6 and 6 , - 11 and 1 and so on . . when the sum of two numbers is given , the max product is when both x and y are same numeric value . . . so xy will have max numeric value when numeric value of x and y = 12 / 2 = 6 . . so numeric value of xy = 36 but one of x and y is - ive . . so answer is - 36 answer : d" | a = 2 + 3
b = negate * (
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['a ) 900', 'b ) 1,600', 'c ) 2,500', 'd ) 3,600', 'e ) 4,900'] | a | add(multiply(multiply(multiply(4, power(const_3, const_2)), 10), const_2), multiply(10, 18)) | what is the smallest positive perfect square that is divisible by 4 , 10 , and 18 ? | the number needs to be divisible by 2 ^ 2 , 2 * 5 , and 2 * 3 ^ 2 . the smallest such perfect square is 2 ^ 2 * 3 ^ 2 * 5 ^ 2 = 900 the answer is a . | a = 3 ** 2
b = 4 * a
c = b * 10
d = c * 2
e = 10 * 18
f = d + e
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a ) 63 , b ) 72 , c ) 81 , d ) 90 , e ) 99 | c | divide(multiply(divide(subtract(const_100, 19), const_100), 9), divide(subtract(const_100, 91), const_100)) | if grapes are 91 % water and raisins are 19 % water , then how many kilograms did a quantity of raisins , which currently weighs 9 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transformation . ) | "let x be the original weight . the weight of the grape pulp was 0.09 x . since the grape pulp is 81 % of the raisins , 0.09 x = 0.81 ( 9 kg ) . then x = 9 * 9 = 81 kg . the answer is c ." | a = 100 - 19
b = a / 100
c = b * 9
d = 100 - 91
e = d / 100
f = c / e
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a ) 20 coins , b ) 50 coins , c ) 100 coins , d ) 120 coins , e ) none of these | d | divide(210, add(add(inverse(const_4), inverse(const_2)), const_1)) | a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 210 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 Γ x + . 50 Γ x + . 25 x = 210 β 1.75 x = 210 β x = 120 coins answer d" | a = 1/(4)
b = 1/(2)
c = a + b
d = c + 1
e = 210 / d
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a ) 116 cm , b ) 144 cm , c ) 168 cm , d ) 173 cm , e ) 189 cm | b | divide(multiply(const_4, divide(power(12, const_3), power(4, const_2))), const_3) | a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 4 cm . what is the length of the wire ? | "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . Ο ( 4 ) ^ 2 * h = ( 4 / 3 ) Ο ( 12 ) ^ 3 = > h = 144 cm answer : b" | a = 12 ** 3
b = 4 ** 2
c = a / b
d = 4 * c
e = d / 3
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a ) 150 m , b ) 188 m , c ) 267 m , d ) 268 m , e ) 287 m | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(150, divide(20, divide(15, const_3)))) | a train crosses a platform of 150 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be β x β x + 150 / 15 = x + 250 / 20 4 x + 600 = 3 x + 750 x = 150 m answer : a" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 150 * e
g = c - f
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a ) 76 , b ) 5776 , c ) 304 , d ) 1296 , e ) none | b | power(multiply(4, 19), const_2) | find β ? / 19 = 4 ? | "answer let β n / 19 = 4 then β n = 19 x 4 = 76 β΄ n = 76 x 76 = 5776 . correct option : b" | a = 4 * 19
b = a ** 2
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a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | "ratio ; white : black : gray = 3 : 5 : ( 3 + 5 ) = 3 : 5 : 8 if 2 gallons of the mixture is needed < = = > need 2 gallons of gray paint . then , 3 : 5 : 8 = x : y : 2 3 / 8 = x / 2 ; x = 3 / 4 5 / 8 = y / 2 ; y = 5 / 4 you need 3 / 4 gallon of white . you have to buy 1 gallons . you need 5 / 4 gallons of black . you have to buy 1 gallon and another 1 / 2 gallon of black . the least amount of paint = 1 + 1 + 1 / 2 = 2.5 . answer : b" | a = const_0_25 * 2
b = 2 * 5
c = a + b
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['a ) 4', 'b ) 8', 'c ) 12', 'd ) 15', 'e ) 20'] | c | subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2)) | if the radius of a circle that centers at the origin is 5 , how many j points on the circle have integer coordinates ? | i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : j = c | a = 5 * 5
b = 5 + 4
c = a - b
d = 2 ** 2
e = c - d
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a ) 1105 , b ) 1245 , c ) 1275 , d ) 1215 , e ) 1150 | e | add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(50, multiply(50, divide(20, const_100))), multiply(50, divide(20, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(50, divide(20, const_100))), multiply(25, multiply(50, divide(20, const_100))))) | an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 50 packages of gaskets . if 20 percent of the gaskets went to company x , 15 percent to company y , and the rest to company z , what was the total amount , in dollars , that the parts supplier received in payment for the gaskets ? | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 / 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 20 % of the gaskets to x = 10 gaskets = 25 * 10 = 250 2 . he sold 30 % of the gaskets to y = 15 gaskets = 25 * 10 + 20 * 5 = 350 3 . he sold remaining 50 % to z = 25 gaskets = 25 * 10 = 250 + 20 * 15 = 550 thus , total money earned 250 + 350 + 550 = 1150 answer is e" | a = 4 / 5
b = 25 * a
c = 20 / 100
d = 50 * c
e = 50 - d
f = 20 / 100
g = 50 * f
h = e - g
i = h - 10
j = b * i
k = 25 * 10
l = j + k
m = 20 / 100
n = 50 * m
o = 25 * n
p = 20 / 100
q = 50 * p
r = 25 * q
s = o + r
t = l + s
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a ) 11 days , b ) 9 days , c ) 8 days , d ) 12 days , e ) 10 days | e | divide(const_1, add(inverse(15), inverse(30))) | ravi can do a piece of work in 15 days while prakash can do it in 30 days . in how many days will they finish it together ? | "1 / 15 + 1 / 30 = 3 / 30 10 / 1 = 10 days answer : e" | a = 1/(15)
b = 1/(30)
c = a + b
d = 1 / c
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a ) 10 , b ) 100 , c ) 1000 , d ) 100000 , e ) none of these | d | multiply(1000, 10) | ( 1000 ) 7 Γ· ( 10 ) 16 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 16 = ( 10 ) 21 / ( 10 ) 16 = 10 ( 5 ) = 100000 option d" | a = 1000 * 10
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a ) 5 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | e | subtract(multiply(multiply(1425, 1427), 1429), subtract(multiply(multiply(1425, 1427), 1429), const_3)) | what is remainder of the division ( 1425 * 1427 * 1429 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : e" | a = 1425 * 1427
b = a * 1429
c = 1425 * 1427
d = c * 1429
e = d - 3
f = b - e
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a ) 7 km , b ) 6 km , c ) 6 7 / 8 km , d ) 9 km , e ) 5 km | c | add(multiply(add(5, divide(const_1, const_2)), subtract(add(5, divide(const_3, 4)), add(4, divide(const_1, const_2)))), const_2) | two men a and b start from place x walking at 4 Β½ kmph and 5 ΒΎ kmph respectively . how many km apart they are at the end of 5 Β½ hours if they are walking in the same direction ? | "rs = 5 ΒΎ - 4 Β½ = 1 ΒΌ t = 3 Β½ h . d = 5 / 4 * 11 / 2 = 55 / 8 = 6 7 / 8 km answer : c" | a = 1 / 2
b = 5 + a
c = 3 / 4
d = 5 + c
e = 1 / 2
f = 4 + e
g = d - f
h = b * g
i = h + 2
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a ) 1521 , b ) 1492 , c ) 1667 , d ) 1254 , e ) 1112 | c | multiply(divide(multiply(10, const_1000), const_60), 10) | a man walking at a rate of 10 km / hr crosses a bridge in 10 minutes . the length of the bridge is ? | "speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 10 minutes = 50 / 18 * 10 * 60 = 1667 m answer is c" | a = 10 * 1000
b = a / const_60
c = b * 10
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a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | e | multiply(divide(add(40, divide(40, const_2)), 6), const_2) | the distance from steve ' s house to work is 40 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 40 km = > speed = 20 kmph e" | a = 40 / 2
b = 40 + a
c = b / 6
d = c * 2
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a ) 75 , b ) 65 , c ) 87 , d ) 95 , e ) 80 | c | add(multiply(8, 2.5), 67) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 67 kg . what is the weight of the new person ? | "total increase in weight = 8 Γ 2.5 = 20 if x is the weight of the new person , total increase in weight = x β 67 = > 20 = x - 67 = > x = 20 + 67 = 87 answer is c ." | a = 8 * 2
b = a + 67
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a ) 80 , b ) 8 , c ) 15 , d ) 21 , e ) 35 | a | add(subtract(multiply(3, 8), 8), power(8, 2)) | if a # b = ab β b + b ^ 2 , then 3 # 8 = | "solution - simply substitute 3 and 8 in equation in the place of a and b respectively . 3 # 8 = 3 * 8 - 8 + 8 ^ 2 = 24 - 8 + 64 = 80 . ans a" | a = 3 * 8
b = a - 8
c = 8 ** 2
d = b + c
|
a ) 3 , b ) z = 4 , c ) z = 5 , d ) z = 6 , e ) z = 8 | e | divide(240, 30) | a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations z are possible ? | "the combinations could be { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pairs , and we are left only with . { ( 8,30 , ( 10,24 ) , ( 12,20 ) , ( 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) } hence 8 . e" | a = 240 / 30
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a ) 10 , b ) 40 , c ) 200 , d ) 100 , e ) 540 | d | add(divide(divide(10, divide(divide(divide(divide(divide(10, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(10, divide(divide(divide(divide(divide(10, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 10 odd numbers | "explanation : n 2 = 102 = 100 answer : option d" | a = 10 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 10 / e
g = f / 2
h = 10 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 10 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 140 sec , b ) 132 sec , c ) 192 sec , d ) 252 sec , e ) none | a | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(44, 40))) | two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 44 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "solution relative speed = ( 44 - 40 ) kmph = 4 kmph = ( 4 x 5 / 18 ) m / sec = ( 20 / 18 ) m / sec time taken = ( 350 x 18 / 20 ) sec = 140 sec . answer a" | a = 200 + 150
b = a / 1000
c = 44 - 40
d = b / c
e = 3600 * d
|
a ) 7 : 26 , b ) 7 : 56 , c ) 8 : 26 , d ) 8 : 56 , e ) 9 : 26 | d | divide(add(4, divide(multiply(add(subtract(4, 4), divide(subtract(56, 00), const_60)), 30), subtract(37, 30))), 56) | tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 37 pages per hour . if tom starts reading a novel at 4 : 00 , and jan begins reading an identical copy of the same book at 4 : 56 , at what time will they be reading the same page ? | "since tom reads an average of 1 page every 2 minutes , tom will read 28 pages in the first 56 minutes . jan can catch tom at a rate of 7 pages per hour , so it will take 4 hours to catch tom . the answer is d ." | a = 4 - 4
b = 56 - 0
c = b / const_60
d = a + c
e = d * 30
f = 37 - 30
g = e / f
h = 4 + g
i = h / 56
|
a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 27 | e | divide(multiply(subtract(42, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 42 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 42 = > x = 9 so deepak age is = 3 ( 9 ) = 27 option e" | a = 42 - 6
b = a * 3
c = b / 4
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a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 9 , d ) 4 / 15 , e ) 9 / 20 | d | divide(multiply(8, divide(2, 3)), 20) | an outlet pipe can empty 2 / 3 of a cistern in 20 minutes . in 8 minutes , what part of the cistern will be emptied ? | "8 / 20 * 2 / 3 = 4 / 15 the answer is d ." | a = 2 / 3
b = 8 * a
c = b / 20
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | the number of diagonals of a polygon of n sides is given by the formula f = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | "f = n ( n - 3 ) f = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b" | a = 3 + 2
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a ) 287 m , b ) 431.25 m , c ) 267 m , d ) 287.25 m , e ) 656 m | b | subtract(multiply(speed(300, 16), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 16 sec . what is the length of the platform ? | "speed = 300 / 16 = 75 / 4 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 75 / 4 = > x = 731.25 m answer : b ( 431.25 )" | a = speed * (
b = a - 39
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a ) 12 m , b ) 5 m , c ) 13 m , d ) 11 m , e ) 12 m | c | sqrt(add(power(12, const_2), power(subtract(11, 6), const_2))) | two poles of height 6 meters and 11 meters stand on a plane ground . if the distance between their feet is 12 meters then find the difference in the distance between their tops : | distance between their tops = sqrt ( 12 ^ 2 + ( 11 - 6 ) ^ 2 ) = sqrt 169 = 13 m answer : c | a = 12 ** 2
b = 11 - 6
c = b ** 2
d = a + c
e = math.sqrt(d)
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind loe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of loe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . loe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena can cover 7.5 miles in 30 mins teena will cover 15 miles in 60 mins so answer 30 + 60 = 90 mins . d" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 1 / 33 , b ) 1 / 22 , c ) 1 / 12 , d ) 1 / 44 , e ) 1 / 55 | e | divide(choose(4, 3), choose(add(add(4, 5), 3), 3)) | a bag contains 4 red , 5 blue and 3 green balls . if 3 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 4 c 3 / 12 c 3 = 1 / 55 e" | a = math.comb(4, 3)
b = 4 + 5
c = b + 3
d = math.comb(c, 3)
e = a / d
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | c | divide(divide(divide(507, const_3), const_3), const_4) | the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 507 sq meters , then what is the width ( in meters ) of the rectangular plot ? | "area = l * w = 3 w ^ 2 = 507 w ^ 2 = 169 w = 13 the answer is c ." | a = 507 / 3
b = a / 3
c = b / 4
|
a ) 32 hours , b ) 40 hours , c ) 35 hours , d ) 30 hours , e ) 36 hours | d | subtract(multiply(divide(140, add(add(3, 5), 6)), 6), multiply(divide(140, add(add(3, 5), 6)), 3)) | the amount of time that three people worked on a special project was in the ratio of 3 to 5 to 6 . if the project took 140 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 3 * 140 / 14 = 30 hours b = 5 * 140 / 14 = 50 hours c = 6 * 140 / 14 = 60 hours c is the hardest worker and a worked for the least number of hours . so the difference is 60 - 30 = 30 hours . answer : d" | a = 3 + 5
b = a + 6
c = 140 / b
d = c * 6
e = 3 + 5
f = e + 6
g = 140 / f
h = g * 3
i = d - h
|
a ) 75 % , b ) 25 % , c ) 85 % , d ) 55 % , e ) 65 % | c | subtract(multiply(75, const_3), add(60, 80)) | a student gets 60 % in one subject , 80 % in the other . to get an overall of 75 % how much should get in third subject . | let the 3 rd subject % = x 60 + 80 + x = 3 * 75 140 + x = 225 x = 225 - 140 = 85 answer : c | a = 75 * 3
b = 60 + 80
c = a - b
|
a ) 15 , b ) 34 , c ) 50 , d ) 67 , e ) 100 | a | divide(multiply(6, 200), divide(200, const_10)) | according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 20 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 20 oz can = 300 oz / 20 oz = 15 answer a , 15 cans" | a = 6 * 200
b = 200 / 10
c = a / b
|
a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84 | a | divide(0.36, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.36 more than jane , what was the original price of the dish ? | "the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.36 - - > 15 p - 13.5 p = 36 - - > p = 24 . answer : a ." | a = 15 / 100
b = 10 / 100
c = 1 - b
d = 15 / 100
e = c * d
f = a - e
g = 0 / 36
|
a ) rs 690 , b ) rs 790 , c ) rs 890 , d ) rs 990 , e ) none of these | c | multiply(divide(8000, add(add(8000, 4000), 8000)), 4005) | a , b and c invested rs . 8000 , rs . 4000 and rs . 8000 respectively in a business . a left after six months . if after eight months , there was a gain of rs . 4005 , then what will be the share of b ? | "explanation : a : b : c = ( 8000 * 6 ) : ( 4000 * 8 ) : ( 8000 * 8 ) = 48 : 32 : 64 = 3 : 2 : 4 so b share = ( 2 / 9 ) * 4005 = rs 890 option c" | a = 8000 + 4000
b = a + 8000
c = 8000 / b
d = c * 4005
|
a ) 32 , b ) 34 , c ) 36 , d ) 38 , e ) 40 | e | subtract(48, divide(48, 6)) | two pumps are connected to an empty tank . pump x fills the tank with water at a constant rate , while pump y drains water out of the tank at a constant rate . the two pumps finish filling the tank in 6 times the duration it would take pump x alone to fill the tank . if pump y alone can empty a whole tank in 48 minutes , then how many minutes does it take pump x alone to fill the tank ? | let v be the volume of the tank . let r be the rate per minute that pump x fills the tank . let t be the time it takes pump x to fill the tank . the rate at which pump y empties the tank is v / 48 per minute . ( r - v / 48 ) * 6 t = v = rt . ( r - v / 48 ) * 6 = r 5 r = v / 8 . r = v / 40 . it takes pump x 40 minutes to fill the tank . the answer is e . | a = 48 / 6
b = 48 - a
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 80 | e | divide(multiply(16.80, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 16.80 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 16.80 when c . p = ( 100 * 16.80 ) / 21 = 50 answer : e" | a = 16 * 80
b = 25 + 100
c = 30 + 100
d = 100 - 20
e = c * d
f = e / 100
g = b - f
h = a / g
|
a ) 27 , b ) 16 , c ) 38 , d ) 61 , e ) 83 | a | add(multiply(const_2, 10), add(divide(subtract(sqrt(subtract(multiply(add(multiply(5, add(10, const_1)), 10), add(multiply(5, add(10, const_1)), 10)), multiply(multiply(subtract(multiply(5, 5), add(multiply(const_60, const_4), const_3)), multiply(add(10, const_1), const_2)), const_4))), add(multiply(5, add(10, const_1)), 10)), multiply(multiply(add(10, const_1), const_2), const_2)), 5)) | given a two - digit number , the unit ' s digit exceeds its 10 ' s digit by 5 and the product of the given number and the sum of its digits is equal to 90 , which of the options is the number ? | using the elimination method the option that fits this description is 27 7 - 2 = 5 ( unit ' s digit that exceeds its ten ' s digit by 3 ) 27 * 9 = 243 ( the product of the given number and the sum of its digits is equal to 175 ) answer : a | a = 2 * 10
b = 10 + 1
c = 5 * b
d = c + 10
e = 10 + 1
f = 5 * e
g = f + 10
h = d * g
i = 5 * 5
j = const_60 * 4
k = j + 3
l = i - k
m = 10 + 1
n = m * 2
o = l * n
p = o * 4
q = h - p
r = math.sqrt(q)
s = 10 + 1
t = 5 * s
u = t + 10
v = r - u
w = 10 + 1
x = w * 2
y = x * 2
z = v / y
A = z + 5
B = a + A
|
a ) 14.5 , b ) 13.5 , c ) 12.5 , d ) 17.5 , e ) 11.5 | a | add(multiply(divide(60, 40), divide(240, 80)), 10) | if Γ’ β¬ Ε * Γ’ β¬ Β is called Γ’ β¬ Ε + Γ’ β¬ Β , Γ’ β¬ Ε / Γ’ β¬ Β is called Γ’ β¬ Ε * Γ’ β¬ Β , Γ’ β¬ Ε - Γ’ β¬ Β is called Γ’ β¬ Ε / Γ’ β¬ Β , Γ’ β¬ Ε + Γ’ β¬ Β is called Γ’ β¬ Ε - Γ’ β¬ Β . 240 - 80 / 60 - 40 * 10 = ? | explanation : given : 240 - 80 / 60 - 40 * 10 = ? substituting the coded symbols for mathematical operations , we get , 240 / 80 * 60 / 40 + 10 = ? 3 * 1.5 + 10 = 14.5 answer : a | a = 60 / 40
b = 240 / 80
c = a * b
d = c + 10
|
a ) 427 , b ) 859 , c ) 869 , d ) 861 , e ) none of these | d | subtract(lcm(lcm(lcm(24, 32), 36), 54), 3) | the least number which when increased by 3 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 3 = 864 - 3 = 861 . answer d" | a = math.lcm(24, 32)
b = math.lcm(a, 36)
c = math.lcm(b, 54)
d = c - 3
|
a ) 65 sec , b ) 30 sec , c ) 48 sec , d ) 33 sec , e ) 12 sec | b | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.15, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.15 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.15 = 1.25 km = 1250 m . required time = 1250 * 3 / 125 = 30 sec . answer : b" | a = 1 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 0 * 15
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 1341.69 , b ) 1342.69 , c ) 1343.69 , d ) 1344.69 , e ) 1345.69 | a | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(4, const_3)))), const_100) | what is the compound interest on rs : 80,000 for 4 months at the rate of 5 % per annum | "it is monthly compound rate = 5 / 12 % per month 80000 * ( 1 + 5 / 1200 ) ^ 4 - 80000 = 1341.69 answer : a" | a = 3 * 100
b = a * 100
c = 4 * 3
d = 4 / c
e = 5 * d
f = b * e
g = f / 100
|
a ) 1 / 2 , b ) 1 / 2 , c ) 8 / 15 , d ) 9 / 20 , e ) 9 / 24 | d | divide(3, const_10) | tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ? | "explanation : here , s = { 1 , 2 , 3 , 4 , . . . . , 19 , 20 } . let e = event of getting a multiple of 3 or 5 = { 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 20 } . p ( e ) = n ( e ) / n ( s ) = 9 / 20 answer : d" | a = 3 / 10
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(770, divide(add(14, 8), const_2)) | the cross - section of a water channel is a trapezium in shape . if the channel is 14 meters wide at the top and 8 meters wide at the bottom and the area of cross - section is 770 square meters , what is the depth of the channel ( in meters ) ? | "1 / 2 * d * ( 14 + 8 ) = 770 d = 70 the answer is c ." | a = 14 + 8
b = a / 2
c = 770 / b
|
a ) 2.5 , b ) 6.5 , c ) 5 , d ) 6.0 , e ) 7 | b | divide(triangle_area_three_edges(5, 12, 13), divide(triangle_perimeter(5, 12, 13), const_2)) | what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 5 , 12 and 13 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 13 / 2 = 6.5 ans . b" | a = triangle_area_three_edges / (
|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | divide(multiply(20, const_4), add(const_4, const_1)) | the cost price of 20 articles is the same as the selling price of x articles . if the profit is 25 % then determine the value of x . | "explanation : let the cost price 1 article = re 1 cost price of x articles = x s . p of x articles = 20 gain = 20 - x = > 25 = ( 20 β x / x β 100 ) = > 2000 β 100 x = 25 x = > x = 16 option c" | a = 20 * 4
b = 4 + 1
c = a / b
|
a ) 20 , b ) 31 , c ) 45 , d ) 53 , e ) 64 | c | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 260 are multiple of 4 but not multiples of 6 ? | "260 / 4 = 65 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 65 - 20 = 45 answer : c" | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 36.7 , b ) 52.9 , c ) 52.8 , d ) 52.1 , e ) 52.2 | a | multiply(multiply(power(10, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 10 metro and whose angle at the center is 42 Γ’ Β° is ? | "42 / 360 * 22 / 7 * 10 * 10 = 36.7 m 2 answer : a" | a = 10 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 42 / g
i = f * h
|
a ) 0.2 % , b ) 2 % , c ) 5 % , d ) 150 % , e ) 500 % | d | multiply(divide(120, 80), const_100) | 120 is what percent of 80 ? | "120 = x * 80 / 100 x = 150 % ans ; d" | a = 120 / 80
b = a * 100
|
a ) 32 , b ) 36 , c ) 40 , d ) 46 , e ) 256 | e | subtract(multiply(add(subtract(divide(140, const_10), const_2), const_10), add(divide(140, const_10), subtract(divide(140, const_10), const_2))), 140) | a rectangular room has the rectangular shaped rug shown as above figure such that the rug β s area is 140 square feet and its length is 4 feet longer than its width . if the uniform width between the rug and room is 4 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | "rug ' s area = 140 which is ( x ) x ( 4 + x ) = 140 so x = 10 rug maintains a uniform distance of 4 feet so room has dimension 10 + 8 and 14 + 8 i . e . 18 and 22 area of room 18 x 22 = 396 area covered is 140 so uncovered area is 396 - 140 = 256 ( answer e )" | a = 140 / 10
b = a - 2
c = b + 10
d = 140 / 10
e = 140 / 10
f = e - 2
g = d + f
h = c * g
i = h - 140
|
a ) 10 , b ) 6 , c ) 7 , d ) 4 , e ) 14 | a | divide(subtract(add(25, 15), 20), const_2) | at a meeting , 25 attendants used a pencil and 15 attendants used a pen . if exactly 20 attendants used only one of the two types of writing tools , how many attendants wrote with both types of writing tools ? | say x attendants wrote with both writing tools . ( 25 - x ) + ( 15 - x ) = 20 - - > x = 10 . answer : a . | a = 25 + 15
b = a - 20
c = b / 2
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(divide(subtract(18, 15), divide(60, multiply(const_100, const_1))), subtract(15, divide(subtract(18, 15), divide(60, multiply(const_100, const_1))))) | working together , tim and tom can type 15 pages in one hour . if they would be able to type 18 pages in one hour if tom increases his typing speed by 60 % , what is the ratio of tom ' s normal typing speed to that of tim ? | "lets say tim types x pages an hour and tom types y pages an hour . we know that x + y = 15 tom increase speed by 60 % means he will type 1.6 y pages an hour . so we get x + 1.6 y = 18 we need to know the ratio of tom ' s speed to tim ' s speed . this is going to be proportional to the number of pages each can type in an hour , hence ( y / x ) . subtracting both : 0.6 y = 3 so y = 5 . . . so x = 10 ( y / x ) = 5 / 10 = 1 / 2 answer is ( a )" | a = 18 - 15
b = 100 * 1
c = 60 / b
d = a / c
e = 18 - 15
f = 100 * 1
g = 60 / f
h = e / g
i = 15 - h
j = d / i
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 32 | e | add(10, add(const_3, const_4)) | how many digits are in ( 12 Γ 10 ^ 14 ) ( 10 Γ 10 ^ 15 ) ? | "the question simplfies to ( 12 Γ 10 ^ 14 ) ( 10 ^ 16 ) = > 12 * 10 ^ 30 = > will contain 30 zeros + 2 digit 12 = > 32 ans e" | a = 3 + 4
b = 10 + a
|
a ) 23 , b ) 27 , c ) 24 , d ) 27 , e ) 11 | c | divide(multiply(30, 20), 25) | 20 men can complete a piece of work in 30 days . in how many days can 25 men complete that piece of work ? | "20 * 30 = 25 * x = > x = 24 days answer : c" | a = 30 * 20
b = a / 25
|
['a ) 248', 'b ) 312.5', 'c ) 346.5', 'd ) 392.5', 'e ) 424'] | c | divide(circle_area(21), const_4) | a lamp is put on one corner of a square plot of side 50 m . it ' s light reaches 21 m . find the area of that plot that is lit by that lamp ? | area covered by lamp = pi * r ^ 2 / 4 ( here we divide by 4 because lamp is put in the corner of the plot and only 1 / 4 part lit is of the plot ) where r = 21 m = length of part lit so area = ( 22 / 7 ) * 21 * 21 / 4 = 346.5 sq m answer : c | a = circle_area / (
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a ) 25 , b ) 24 , c ) 30 , d ) 27 , e ) 28 | a | divide(multiply(12.5, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 12.5 % of 4 / 12 of 600 ? | "12.5 % = 12.5 / 100 = 1 / 8 of 4 / 12 = 1 / 8 * 4 / 12 = 1 / 24 of 600 = 1 / 24 * 600 = 25 ans - a" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 12 * 5
p = o / 100
|
a ) 1 , b ) 2 , c ) - 6 , d ) 4 , e ) 5 | c | divide(1, 6) | find the slope of the line perpendicular to the line y = ( 1 / 6 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 6 . if m is the slope of the line perpendicular to the given line , then m Γ ( 1 / 6 ) = - 1 solve for m m = - 6 correct answer c ) - 6" | a = 1 / 6
|
a ) a ) 42 , b ) b ) 61 , c ) c ) 63 , d ) d ) 65 , e ) e ) 67 | a | divide(add(36, 49), const_2) | find the mean proportional between 36 & 49 ? | "formula = β a Γ b a = 36 and b = 49 β 36 Γ 49 = 6 Γ 7 = 42 a" | a = 36 + 49
b = a / 2
|
a ) 20.23 , b ) 20.13 , c ) 75 , d ) 20.93 , e ) 10.93 | c | multiply(divide(95, add(add(divide(1, 3), divide(5, 1)), const_1)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 1 : 3 and b : c = 1 : 5 . if the total runs scored by all of them are 95 , the runs scored by c are ? | a : b = 1 : 3 b : c = 1 : 5 a : b : c = 1 : 3 : 15 15 / 19 * 95 = 75 answer : c | a = 1 / 3
b = 5 / 1
c = a + b
d = c + 1
e = 95 / d
f = e * 5
|
a ) 29 , b ) 72 , c ) 39 , d ) 37 , e ) 75 | b | divide(add(add(add(add(61, 65), 82), 67), 85), divide(const_10, const_2)) | dacid obtained 61 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 61 + 65 + 82 + 67 + 85 ) / 5 = 72 answer : b" | a = 61 + 65
b = a + 82
c = b + 67
d = c + 85
e = 10 / 2
f = d / e
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a ) 7500 , b ) 7525 , c ) 7550 , d ) 8000 , e ) none of these | b | add(7500, divide(1250, 50)) | 7500 + ( 1250 / 50 ) | explanation : as per bodmas rule first we will solve the terms in the bracket then other . = 7500 + ( 25 ) = 7525 option b | a = 1250 / 50
b = 7500 + a
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a ) 50 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | a | divide(50, 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 50 / 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 ) = 50 = > x = 50 thus proved . . . option a ." | a = 20 / 100
b = 1 - a
c = 20 / 100
d = c + 1
e = b * d
f = 50 / e
|
a ) 48 , b ) 52 , c ) 56 , d ) 60 , e ) 64 | d | multiply(divide(320, add(add(multiply(2, 1), 3), 7)), 3) | in a certain town , the ratio of ny yankees fans to ny mets fans is 2 : 1 , and the ratio of ny mets fans to boston red sox fans is 3 : 7 . if there are 320 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ? | "the ratio of yankees : mets : red sox = 6 : 3 : 7 the mets fans are 3 / 16 of the population . ( 3 / 16 ) * 320 = 60 the answer is d ." | a = 2 * 1
b = a + 3
c = b + 7
d = 320 / c
e = d * 3
|
a ) - 8 , b ) - 6 , c ) 9 , d ) 8 , e ) - 7 | b | divide(negate(add(11, 1)), 2) | solve below question 2 x + 1 = - 11 | "2 x + 1 = - 11 x = - 6 b" | a = 11 + 1
b = negate / (
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 4 , e ) 8 | a | divide(7, add(7, 7)) | if 7 / w + 7 / x = 7 / y and wx = y , then the average ( arithmetic mean ) of w and x is | given : 7 / w + 7 / x = 7 / ywx = y find : ( w + x ) / 2 = ? 7 ( 1 / w + 1 / x ) = 7 ( 1 / y ) - divide both sides by 7 ( 1 / w + 1 / x ) = 1 / y ( x + w ) / wx = 1 / wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) / 2 = 1 / 2 ans : a | a = 7 + 7
b = 7 / a
|
a ) 5.02 , b ) 4.92 , c ) 4.82 , d ) 4.72 , e ) 4.61 | b | multiply(divide(multiply(add(10, 1.2), subtract(10, 1.2)), add(add(10, 1.2), subtract(10, 1.2))), const_2) | a man can row 10 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 10 s = 1.2 ds = 10 + 1.2 = 11.2 us = 10 - 1.2 = 8.8 x / 11.2 + x / 8.8 = 1 x = 4.92 . answer : b" | a = 10 + 1
b = 10 - 1
c = a * b
d = 10 + 1
e = 10 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 225 , b ) 280 , c ) 300 , d ) 385 , e ) none | d | divide(subtract(700, multiply(divide(700, const_2), divide(const_1, add(const_1, const_4)))), const_2) | a large field of 700 hectares is divided into two parts . the difference of the areas of the two parts is one β fifth of the average o fthe two areas . what is the area of the smaller part in hectares ? | "solution let the areas of the two parts be x and ( 700 - x ) hectares = [ x - ( 700 - x ) ] = 1 / 5 x [ x + ( 700 - x ) / 2 ] = 2 x - 700 = 70 x = 385 . answer d" | a = 700 / 2
b = 1 + 4
c = 1 / b
d = a * c
e = 700 - d
f = e / 2
|
a ) 7724 , b ) 7804 , c ) 4144 , d ) 7844 , e ) none | c | divide(multiply(add(multiply(4, const_100), 44), multiply(13, const_100)), power(44, const_2)) | what is the least number of square tiles required to pave the floor of a room 13 m 44 cm long and 4 m 44 cm broad ? | "solution length of largest tile = h . c . f . of 1344 cm & 444 cm = 12 cm . area of each tile = ( 12 x 12 ) cm 2 β΄ required number of tiles = [ 1344 x 444 / 12 x 12 ] = 4144 . answer c" | a = 4 * 100
b = a + 44
c = 13 * 100
d = b * c
e = 44 ** 2
f = d / e
|
a ) 80 days , b ) 100 / 3 days , c ) 120 days , d ) 110 days , e ) 90 days | b | inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 16))), 10)))) | micheal and adam can do together a piece of work in 20 days . after they have worked together for 16 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately . | "rate of both = 1 / 20 together they do = 1 / 20 * 16 = 4 / 5 left work = 1 - 4 / 5 = 1 / 5 adam completes 1 / 5 work in 10 day so he took 10 * 5 = 50 days to complete the left work alone . thus the rate of adam is 1 / 50 rate of micheal = 1 / 20 - 1 / 50 = 1 / ( 100 / 3 ) thus micheal takes 100 / 3 days to complete the whole work . ans . b ." | a = 1/(20)
b = 1/(20)
c = b * 16
d = 1 - c
e = 1/(d)
f = e * 10
g = 1/(f)
h = a - g
i = 1/(h)
|
a ) 405 , b ) 450 , c ) 400 , d ) 900 , e ) 2500 | c | divide(power(const_10, divide(6, const_2)), const_2) | a palindrome is a number that reads the same forward and backward , such as 171 . how many even , 6 - digit numbers are palindromes ? | "first recognize you only need to consider the first three digits ( because the second three are just the first three flipped ) there are 900 possibilities for the first three digits of a 6 digit number , 100 - 999 inclusive . everything starting with a 2,4 , 6,8 will be even , which is 4 / 9 ths of the combinations . 4 / 9 * 900 = 400 answer : c" | a = 6 / 2
b = 10 ** a
c = b / 2
|
a ) 30 , b ) 80 , c ) 100 , d ) 130 , e ) 150 | b | subtract(200, subtract(add(90, 50), 20)) | of the 200 stamps in a collection , 90 are foreign and 50 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ? | "20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 30 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 30 ) = 80 the answer is b ." | a = 90 + 50
b = a - 20
c = 200 - b
|
a ) 1 / 15 , b ) 2 / 15 , c ) 4 / 15 , d ) 2 / 5 , e ) none of these | c | subtract(const_1, add(divide(3, 5), multiply(subtract(const_1, divide(3, 5)), divide(1, 3)))) | after doing 3 / 5 of the biology homework on monday night , sanjay did 1 / 3 of the remaining homework on tuesday night . what fraction of the original homework would sanjay have to do on wednesday night to complete the biology assignment ? | sol . ( c ) remaining homework on monday night = 1 - ( 3 / 5 ) = 2 / 5 work done on tuesday night = 1 / 3 of 2 / 5 = 2 / 15 remaining homework to complete the biology assignment = 2 / 5 - 2 / 15 = ( 6 - 2 ) / 15 = 4 / 15 answer c | a = 3 / 5
b = 3 / 5
c = 1 - b
d = 1 / 3
e = c * d
f = a + e
g = 1 - f
|
a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2790 , 4650 , e ) 4660 , 3000 | d | multiply(divide(7.5, 12.5), divide(1860, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1860 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1860 5 * y / 3 β y = 1860 2 * y / 3 = 1860 y = [ ( 1860 * 3 ) / 2 ] = 2790 . one number = 2790 , second number = 5 * y / 3 = 4650 . answer d ." | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1860 / c
e = a * d
|
a ) 7 / 72 , b ) 1 / 6 , c ) 7 / 36 , d ) 5 / 28 , e ) 21 / 36 | d | divide(5, add(add(multiply(3, 5), const_10), const_3)) | a box contains 9 slips that are each labeled with one number : 1 , 2 , 3 , 5 , 8 , 13 , 21 , and 55 . two of the slips are drawn at random from the box without replacement . what is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box ? | probability = no : of desired outcomes / total no : of outcomes . you are picking two slips out of 8 slips . so total no : of outcomes = 8 c 2 = 28 desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box . how many such outcomes are there ? if you look at the numbers closely , you will see that the following pair of numbers will give you the desired outcome . ( 12 ) ( 23 ) ( 35 ) ( 58 ) ( 813 ) . there are 5 such pairs . if the two numbers which i pick is from any of these 5 pairs , then i get my desired outcome . so no : of desired outcomes = 5 probability = 5 / 28 answer : d | a = 3 * 5
b = a + 10
c = b + 3
d = 5 / c
|
a ) 24 , b ) 28 , c ) 31 , d ) 32 , e ) 34 | c | subtract(divide(multiply(add(const_100, 26), const_100), subtract(const_100, 4)), const_100) | a shopkeeper sold an article offering a discount of 4 % and earned a profit of 26 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 126 let marked price be rs . x . then , 96 / 100 x = 126 x = 12600 / 96 = rs . 131.25 now , s . p . = rs . 131.25 , c . p . = rs . 100 profit % = 31 % . answer : c" | a = 100 + 26
b = a * 100
c = 100 - 4
d = b / c
e = d - 100
|
a ) 126 , b ) 136 , c ) 146 , d ) 221 , e ) 266 | d | divide(multiply(17, 312), 24) | the reciprocal of the hcf and lcm of two are 1 / 17 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 17 and 1 / 312 so , hcf = 17 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 17 * 312 / 24 = 221 answer : d" | a = 17 * 312
b = a / 24
|
a ) $ 35 , b ) $ 25 , c ) $ 45 , d ) $ 50 , e ) $ 55 | b | divide(add(160, multiply(160, divide(25, const_100))), subtract(10, 2)) | a man bought 10 crates of mangoes for $ 160 total . if he lost 2 of the crates , at what price would he have to sell each of the remaining crates in order to earn a total profit of 25 percent of the total cost ? | "as given , after lost , the remaining 8 crates total cost = $ 160 so , 1 crate cost = 160 / 8 = 20 to get 25 % profit , 1 crate cost should be = 20 + 20 * 25 / 100 = $ 25 answer : b" | a = 25 / 100
b = 160 * a
c = 160 + b
d = 10 - 2
e = c / d
|
a ) 90 seconds , b ) 36 seconds , c ) 26 seconds , d ) 18 seconds , e ) 6.5 seconds | b | divide(90, multiply(add(6.5, 2.5), const_0_2778)) | the speed at which a man can row a boat in still water is 6.5 kmph . if he rows downstream , where the speed of current is 2.5 kmph , what time will he take to cover 90 metres ? | "speed of the boat downstream = 6.5 + 2.5 = 9 kmph = 9 * 5 / 18 = 2.5 m / s hence time taken to cover 90 m = 90 / 2.5 = 36 seconds . answer : b" | a = 6 + 5
b = a * const_0_2778
c = 90 / b
|
a ) - 1 , b ) 0 , c ) 5 , d ) 2 , e ) 3 | c | add(multiply(power(add(1, sqrt(2)), const_2), power(subtract(add(1, sqrt(2)), 2), const_2)), 4) | if x = 1 + β 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 4 ? | answer x = 1 + β 2 β΄ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 4 = x 2 ( x - 2 ) 2 + 4 = ( 1 + β 2 ) 2 ( 1 + β 2 - 2 ) 2 + 4 = ( β 2 + 1 ) 2 ( β 2 - 1 ) 2 + 4 = [ ( β 2 ) 2 - ( 1 ) 2 ] 2 + 4 = ( 2 - 1 ) 2 = 1 + 4 = 5 correct option : c | a = math.sqrt(2)
b = 1 + a
c = b ** 2
d = math.sqrt(2)
e = 1 + d
f = e - 2
g = f ** 2
h = c * g
i = h + 4
|
a ) 2878 , b ) 3375 , c ) 1000 , d ) 2997 , e ) 2701 | b | divide(power(power(1.5, const_2), const_3), power(1.5, const_3)) | a cube of side 1.5 meter length is cut into small cubes of side 10 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 150 / 10 = 15 along each edge 10 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 15 * 15 * 15 = 3375 answer : b" | a = 1 ** 5
b = a ** 3
c = 1 ** 5
d = b / c
|
a ) 25 , b ) 50 , c ) 90 , d ) 140 , e ) it can not be determined from the information given | b | subtract(multiply(70, const_2), multiply(45, const_2)) | if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 70 , what is the value of c β a ? | "the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 140 - - 2 subtracting 1 from 2 we have c - a = 50 ; answer : b" | a = 70 * 2
b = 45 * 2
c = a - b
|
a ) 80 deg , b ) 40 deg , c ) 35 deg , d ) 75 deg , e ) 60 deg | d | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between the minute hand and the hour hand of a clock when the time is 8 : 30 | "angle traced hr hand 17 / 2 hrs = 300 / 12 * 17 / 2 = 255 angle by min 30 min ( 360 / 60 * 30 ) = 180 req = 255 - 180 = 75 answer d" | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 30 min , b ) 20 min , c ) 25 min , d ) 40 min , e ) 45 min | e | multiply(divide(const_1, add(add(divide(const_1, 30), divide(const_1, 45)), divide(const_1, 45))), const_2) | a large tank can filled by a and b in 30 minutes and 45 minutes respectively . how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half ? | "part filled by a + b in 1 minute = 1 / 30 + 1 / 45 = 1 / 45 suppose the tank is filled in x minutes then , x / 2 ( 1 / 45 + 1 / 45 ) = 1 x / 2 * 2 / 45 = 1 x = 45 min answer is e" | a = 1 / 30
b = 1 / 45
c = a + b
d = 1 / 45
e = c + d
f = 1 / e
g = f * 2
|
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