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 ) 23 years , b ) 22 years , c ) 24 years , d ) 20 years , e ) 19 years | c | divide(subtract(26, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 26 years older than his son . in two years , his age will be twice the age of his son . what is the present age of his son ? | "let present age of the son = x years then , present age the man = ( x + 26 ) years given that , in 2 years , man ' s age will be twice the age of his son Γ’ β‘ β ( x + 26 ) + 2 = 2 ( x + 2 ) Γ’ β‘ β x = 24 answer : c" | a = 2 * 2
b = a - 2
c = 26 - b
d = 2 - 1
e = c / d
|
a ) 15.5 days , b ) 19.5 days , c ) 17.5 days , d ) 16.5 days , e ) 30 days | e | inverse(subtract(5, divide(5, 6))) | a and b can do a piece of work in 6 days . with the help of c they finish the work in 5 days . c alone can do that piece of work in ? | "c = 1 / 5 β 1 / 6 = 1 / 30 = > 30 days answer : e" | a = 5 / 6
b = 5 - a
c = 1/(b)
|
a ) 520 , b ) 620 , c ) 820 , d ) 740 , e ) 720 | a | divide(divide(multiply(120, 5), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 120 kmph in 5 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 120 x 5 ) = 600 km . speed = distance / time speed = 600 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 600 x 3 / 5 ) km / hr = 360 km / hr answer a ) 360 km / hr" | a = 120 * 5
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
|
a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 8 | b | divide(add(divide(12, 3), divide(18, 3)), const_2) | a man swims downstream 18 km and upstream 12 km taking 3 hours each time , what is the speed of the man in still water ? | "18 - - - 3 ds = 6 ? - - - - 1 12 - - - - 3 us = 4 ? - - - - 1 m = ? m = ( 6 + 4 ) / 2 = 5 answer : b" | a = 12 / 3
b = 18 / 3
c = a + b
d = c / 2
|
a ) 600 m , b ) 325 m , c ) 300 m , d ) 400 m , e ) 100 m | b | divide(multiply(18, multiply(1.3, const_1000)), 48) | amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.3 km ? | "b 325 m distance covered by amar = 18 / 4.8 ( 1.3 km ) = 3 / 8 ( 1300 ) = 325 m answer is b" | a = 1 * 3
b = 18 * a
c = b / 48
|
a ) 1629 , b ) 1648 , c ) 2677 , d ) 2986 , e ) 2679 | b | subtract(2678, divide(multiply(multiply(3, 5), 2678), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2678 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2678 - x ) * 3 * 5 ) / 100 24 x / 100 = 40170 / 100 - 15 x / 100 39 x = 40170 = > x = 1030 second sum = 2678 Γ’ β¬ β 1030 = 1648 answer : b" | a = 3 * 5
b = a * 2678
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2678 - f
|
a ) 250 , b ) 280 , c ) 230 , d ) 290 , e ) 250 | b | divide(subtract(148, power(24, const_2)), const_2) | the sum of two numbers is 24 and their product is 148 . find the sum of the squares of that numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 24 ab = 148 so , 24 ^ 2 = a ^ 2 + b ^ 2 + 2 * 148 576 = a ^ 2 + b ^ 2 + 296 a ^ 2 + b ^ 2 = 280 ans b | a = 24 ** 2
b = 148 - a
c = b / 2
|
a ) 24 , b ) 34 , c ) 36 , d ) 54 , e ) 64 | c | divide(multiply(54, 8), 12) | two numbers n and 12 have lcm = 54 and gcf = 8 . find n . | the product of two integers is equal to the product of their lcm and gcf . hence . 12 Γ n = 54 Γ 8 n = 54 Γ 8 / 12 = 36 correct answer c | a = 54 * 8
b = a / 12
|
a ) 16 , b ) 3 β 2 , c ) 8 , d ) 2 β 2 , e ) ( β 2 ) / 3 | b | sqrt(divide(multiply(6, 6), const_2)) | the two lines y = x and x = - 6 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 6 , - 6 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 6 Γ 6 ) = 18 ; so z = 18 . the next step requires us to find the length of a side of a cube that has a face area equal to 18 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 18 . since the area of a square is equal to s Β² , where s is the length of one of its side , we can write and solve the equation s Β² = 18 . clearly s = β 18 = 3 β 2 , oranswer choice ( b ) ." | a = 6 * 6
b = a / 2
c = math.sqrt(b)
|
a ) 14 years , b ) 17 years , c ) 19 years , d ) 21 years , e ) none | c | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 14 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . β΄ required average = 266 / 14 = 19 years . answer c" | a = 35 * 16
b = a + 21
c = b + 35
d = 35 * 16
e = c - d
|
['a ) 100', 'b ) 169', 'c ) 324', 'd ) 196', 'e ) 225'] | c | multiply(multiply(6, 3), multiply(6, 3)) | in the coordinate plane , one of the vertices of a square is the point ( - 6 , - 4 ) . if the diagonals of that square intersect at point ( 3 , 2 ) , what is the area of that square ? | one point ( - 6 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 6 - 3 = - 9 is the midpoint of the square - - > whole side 18 , 18 * 18 = 324 c | a = 6 * 3
b = 6 * 3
c = a * b
|
a ) 100000 , b ) 12000 , c ) 15000 , d ) 13000 , e ) 65000 | a | divide(12000, multiply(divide(divide(const_100, const_2), const_100), multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100))))) | mr . jones gave 40 % of the money he had to his wife . he also gave 20 % of the remaining amount to his 3 sons . half of the amount now left was spent on miscellaneous items and the remaining amount of rs . 12000 was deposited in the bank . how much money did mr . jones have initially ? | "let the initial amount with mr . jones be rs . x then , ( 1 / 2 ) [ 100 - ( 3 * 20 ) ] % of x = 12000 ο³ ( 1 / 2 ) * ( 40 / 100 ) * ( 60 / 100 ) * x = 12000 ο³ x = ( ( 12000 * 25 ) / 3 ) = 100000 answer a" | a = 100 / 2
b = a / 100
c = 40 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = d * f
h = b * g
i = 12000 / h
|
a ) six , b ) seven , c ) eight , d ) nine , e ) ten | b | subtract(multiply(3, 4), add(4, 1)) | if n = 3 * 4 * p where p is a prime number greater than 3 , how many different positive non - prime divisors does n have , excluding 1 and n ? | n = 3 β 2 ^ 2 β p number of divisors = 2 * 3 * 2 = 12 the 12 divisors includes 1 , n , 3 , 2 and p number of different non prime divisors excluding 1 and n = 12 - 5 = 7 answer : b | a = 3 * 4
b = 4 + 1
c = a - b
|
a ) 1 / 15 , b ) 1 / 5 , c ) 9 / 20 , d ) 1 / 3 , e ) 2 / 5 | c | subtract(divide(1, 2), divide(1, multiply(2, const_10))) | of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 1 / 2 times the fraction of first - year students who have declared a major , what fraction of all the students in the dormitory are second - year students who have not declared a major ? | "tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 1 / 2 ( 1 / 5 ( x / 2 ) ) = 1 / 20 ( x ) - - - > non major = x / 2 - 1 / 20 ( x ) = 9 / 20 ( x ) hence 9 / 20 c" | a = 1 / 2
b = 2 * 10
c = 1 / b
d = a - c
|
a ) 86 , b ) 38 , c ) 62 , d ) 122 , e ) 672 | a | multiply(2, divide(divide(1419, add(const_10, const_1)), const_3)) | a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * β¦ * n until the product is divisible by 1419 , what is the value of n ? | factorise 1419 . . 3 * 11 * 43 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 43 = 86 . . ans : a | a = 10 + 1
b = 1419 / a
c = b / 3
d = 2 * c
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a ) 76 , b ) 67 , c ) 176 , d ) 671 , e ) 0 | a | subtract(multiply(17, 4), 8) | what is the dividend ? the divisor is 17 , the quotient is 4 and the remainder is 8 . | "divided = divisor * quotient + remainder ? = 17 * 4 + 8 68 + 8 76 ( e . g . answer : a )" | a = 17 * 4
b = a - 8
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a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | e | divide(add(divide(subtract(4, 8), 2), const_1), divide(add(2, 6), 2)) | line m lies in the xy - plane . the y - intercept of line m is - 1 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 8 ) . what is the slope of line m ? | ans : e solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 2 ) as we can see that the y axis of intercept point is ( 0 , - 1 ) means line m is parallel to x axis slope m = 0 ans : e | a = 4 - 8
b = a / 2
c = b + 1
d = 2 + 6
e = d / 2
f = c / e
|
a ) 12.6 , b ) 12.5 , c ) 12.9 , d ) 12.2 , e ) 12.1 | c | add(10, const_1) | the average of first 10 prime numbers is ? | "sum of 10 prime no . = 129 average = 129 / 10 = 12.9 . answer : c" | a = 10 + 1
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a ) 104 , b ) 60 , c ) 92 , d ) 50 , e ) 25 | c | add(subtract(50, multiply(power(2, 2), 2)), multiply(2, power(5, 2))) | if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 50 , what is the value of f ( 5 ) ? | "f ( x ) = 2 x ^ 2 + y f ( 2 ) = 50 = > 2 * ( 2 ) ^ 2 + y = 50 = > 8 + y = 50 = > y = 42 f ( 5 ) = 2 * ( 5 ) ^ 2 + 42 = 92 answer c" | a = 2 ** 2
b = a * 2
c = 50 - b
d = 5 ** 2
e = 2 * d
f = c + e
|
a ) 4 , b ) 33.8 , c ) 10 , d ) 30 , e ) 37.3 | e | add(add(add(multiply(divide(3000, 305), const_3), const_3), const_1), const_4) | a man has an investment of $ 3000 which yields a fixed 305 for every $ 500 invested . if the man takes out $ 12.6 for every $ 500 how long will it take for him to double his investment assuming no compounding of interest . | annual increase is ( 500 + 30.5 - 12.6 ) * 6 = 3107.4 hence every year there is an increase of 107.4 for his saving to double he needs additional $ 4000 therefore $ 4000 / 107.4 = 37.3 correct option is e ) 37.3 | a = 3000 / 305
b = a * 3
c = b + 3
d = c + 1
e = d + 4
|
a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | d | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 47 hours . if harry and annie were paid the same amount last week , how many hours did harry work last week ? | "annie earned 40 x + 7 ( 2 x ) = 54 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 54 x ( 1.5 x ) ( h ) = 69 x h = 46 hours the answer is d ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
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a ) 11 : 5 , b ) 11 : 8 , c ) 2 : 1 , d ) 15 : 7 , e ) 16 : 9 | c | divide(multiply(80, 5), multiply(100, 2)) | car a runs at the speed of 80 km / hr & reaches its destination in 5 hr . car b runs at the speed of 100 km / h & reaches its destination in 2 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 80 Γ 5 = 400 km distance travelled by car b = 100 Γ 2 = 200 km ratio = 400 / 200 = 2 : 1 answer : c" | a = 80 * 5
b = 100 * 2
c = a / b
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a ) 0 , b ) 2 , c ) 2.5 , d ) 4.67 , e ) 10 | b | divide(subtract(32, subtract(multiply(4, divide(10, 2)), 10)), 10) | the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 β 12 x + 9 . if the sum of the perimeters of both squares is 32 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 5 ) ^ 2 & a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 & l 2 = 2 x - 3 p 1 = 4 ( x + 5 ) & p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 32 4 ( x + 5 ) + 4 ( 2 x - 3 ) = 32 . . . . . . . . . . . . . . > x = 2 answer : b" | a = 10 / 2
b = 4 * a
c = b - 10
d = 32 - c
e = d / 10
|
a ) $ 270 , b ) $ 290 , c ) $ 310 , d ) $ 330 , e ) $ 350 | c | divide(217, divide(subtract(const_100, 30), const_100)) | sandy had $ 217 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ? | "let the money sandy took for shopping be x . 0.7 x = 217 x = 310 the answer is c ." | a = 100 - 30
b = a / 100
c = 217 / b
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['a ) 0.6', 'b ) 0.06', 'c ) 0.9', 'd ) 0.2', 'e ) 0.61'] | b | power(divide(divide(216, const_1000), const_1000), divide(const_1, const_3)) | the cube root of . 000216 is : | answer : b ) . 06 | a = 216 / 1000
b = a / 1000
c = 1 / 3
d = b ** c
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a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | e | divide(multiply(subtract(multiply(12, 12), multiply(4, 12)), const_100), multiply(12, 12)) | at a special sale , 12 tickets can be purchased for the price of 4 tickets . if 12 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 12 tickets ? | "let the price of a ticket be rs . 100 , so 4 tickets cost 400 & 12 tickets cost 1200 12 tickets purchased at price of 4 tickets ie . , for 400 , so amount saved s rs . 800 , % of 5 tickets = ( 800 / 1200 ) * 100 = 66.6 % answer : e" | a = 12 * 12
b = 4 * 12
c = a - b
d = c * 100
e = 12 * 12
f = d / e
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(6, subtract(12, 6)) | a person can swim in still water at 12 km / h . if the speed of water 6 km / h , how many hours will the man take to swim back against the current for 6 km ? | "m = 12 s = 6 us = 12 - 6 = 6 d = 36 t = 36 / 6 = 6 answer : d" | a = 12 - 6
b = 6 / a
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a ) 77 kmph , b ) 95 kmph , c ) 94 kmph , d ) 98 kmph , e ) 96 kmph | e | multiply(const_3_6, divide(320, 12)) | a train 320 m in length crosses a telegraph post in 12 seconds . the speed of the train is ? | "s = 320 / 12 * 18 / 5 = 96 kmph answer : e" | a = 320 / 12
b = const_3_6 * a
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a ) 1419 , b ) 1362 , c ) 1363 , d ) 1364 , e ) 1365 | a | divide(multiply(51, add(51, 12)), 13) | 12 + 13 + 14 + . . . 51 + 52 + 53 + 54 = ? | "sum = 12 + 13 + 14 + . . . 51 + 52 + 53 + 54 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 54 positive integers = 55 * 54 / 2 sum of first 11 positive integers = 11 * 12 / 2 sum = 12 + 13 + 14 + . . . 51 + 52 + 53 + 54 = 55 * 54 / 2 - 11 * 12 / 2 = 1419 answer : a" | a = 51 + 12
b = 51 * a
c = b / 13
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a ) 7 , b ) 10 , c ) 12 , d ) 13 , e ) 14 | e | add(divide(add(13, 13), const_2), const_1) | 2 x + y = 13 | y | < = 13 for how many ordered pairs ( x , y ) that are solutions of the system above are x and y both integers ? | hi - the questions says | y | < = 13 so the values of y will range from 0 to 13 . so the values of y will be 0,1 , 2,3 , 4,5 , 6,7 , 8,9 , 10,11 , 12,13 . so 14 ( e ) is the answer . remember sign does n ' t matter for y as y will always be positive . | a = 13 + 13
b = a / 2
c = b + 1
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a ) 1 / 4 , b ) 3 / 4 , c ) 10 / 19 , d ) 11 / 19 , e ) 12 / 19 | e | divide(divide(const_12, 7), add(divide(const_12, 7), const_1)) | mary decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month . if mary ' s savings by the end of the year from these monthly savings were 7 times the amount she spent per month , what should be the fraction of her salary that she spent each month ? | saving = s , expenditure = e , income = i at the end of 12 months ; 12 s = 7 e = 7 ( i - s ) = > 19 s = 7 i = > savings = 7 / 19 i hence expenditure = 1 - ( 7 / 19 ) = 12 / 19 answer ( e ) | a = 12 / 7
b = 12 / 7
c = b + 1
d = a / c
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a ) 19 , b ) 25 , c ) 24 , d ) 26 , e ) 20 | b | add(divide(subtract(70, 20), 2), const_1) | what is the total number of integers between 20 and 70 that are divisible by 2 ? | "22 , 24 , 26 , . . . , 56,58 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 70 - 22 ) / ( 2 ) + 1 = 48 / 2 + 1 = 24 + 1 = 25 answer is b" | a = 70 - 20
b = a / 2
c = b + 1
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a ) 26 , b ) 40 , c ) 29.25 , d ) 32.625 , e ) none of the above | d | multiply(divide(add(multiply(8, 3), 5), 8), 9) | 9 men went to a hotel . 8 of them spent 3 each over their meals and the ninth spent 5 more than the average expenditure of all the 9 . the total money spent by all of them was | let the average expenditure of all the ninte be x then , 3 Γ 8 + x + 5 = 9 x β x = 3.625 β΄ total money spent = 9 x = 9 Γ 3.625 = 32.625 answer d | a = 8 * 3
b = a + 5
c = b / 8
d = c * 9
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a ) 1 , b ) 2 / 5 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | a | divide(4, subtract(multiply(subtract(6, 4), 4), 4)) | when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 4 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 4 ) ( 4 + 4 x ) or , x = 1 answer a" | a = 6 - 4
b = a * 4
c = b - 4
d = 4 / c
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a ) 233243413 , b ) 354545451 , c ) 343435451 , d ) 565656767 , e ) 725117481 | e | multiply(subtract(9999, const_4), 72519) | find the value of 72519 x 9999 = m . | "72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 e" | a = 9999 - 4
b = a * 72519
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a ) 48 , b ) 47 , c ) 58 , d ) 56 , e ) 54 | a | divide(444, 9.25) | john bought 9.25 m of cloth for $ 444 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 444 cost per metre = 444 Γ· 9.25 the cost of the cloth per metre = $ 48 answers : a" | a = 444 / 9
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a ) 0.5 , b ) 1.5 , c ) 2.5 , d ) 3.5 , e ) 4.5 | c | divide(subtract(divide(81, 9), divide(36, 9)), const_2) | an woman swims downstream 81 km and upstream 36 km taking 9 hours each time ; what is the speed of the current ? | 81 - - - 9 ds = 9 ? - - - - 1 36 - - - - 9 us = 4 ? - - - - 1 s = ? s = ( 9 - 4 ) / 2 = 2.5 answer : c | a = 81 / 9
b = 36 / 9
c = a - b
d = c / 2
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a ) 14 sec , b ) 10 sec , c ) 12 sec , d ) 8 sec , e ) 9 sec | e | divide(126, add(11, 3)) | an escalator moves towards the top level at the rate of 11 ft . sec and its length is 126 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length . | "time taken to cover the entire length = tot . dist / resultant speed = 126 / ( 11 + 3 ) = 9 sec answer : e" | a = 11 + 3
b = 126 / a
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a ) 81000 , b ) 85000 , c ) 75000 , d ) 72000 , e ) 77000 | d | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(const_1, add(divide(const_3, const_2), 1))) | one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 80,000 as total savings , how much has he saved in public provident fund ? | let savings in national savings certificate = x and savings in public provident fund = ( 180000 - x ) 1 / 3 x = 1 / 2 ( 180000 β x ) β 2 x = 3 ( 180000 β x ) β 2 x = 540000 β 3 x β 5 x = 540000 β x = 540000 / 5 = 108000 savings in public provident fund = ( 180000 - 108000 ) = 72000 answer is d . | a = 100 * 10
b = a * 100
c = 100 * 10
d = c * 100
e = 2 * 100
f = e * 100
g = d - f
h = b + g
i = 3 / 2
j = i + 1
k = 1 / j
l = h * k
|
a ) 3944 , b ) 1624 , c ) 2999 , d ) 2667 , e ) 2121 | b | multiply(square_perimeter(sqrt(49)), 58) | what will be the cost of building a fence around a square plot with area equal to 49 sq ft , if the price per foot of building the fence is rs . 58 ? | "let the side of the square plot be a ft . a 2 = 49 = > a = 7 length of the fence = perimeter of the plot = 4 a = 28 ft . cost of building the fence = 28 * 58 = rs . 1624 . answer : b" | a = math.sqrt(49)
b = square_perimeter * (
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a ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42 | d | add(multiply(const_2, 10), add(11, 10)) | in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 11 years older than b , the present age of b is | "explanation : let b ' s age = x years . then , as age = ( x + 11 ) years . ( x + 11 + 10 ) = 2 ( x β 10 ) hence x = 41 . present age of b = 41 years answer : option d" | a = 2 * 10
b = 11 + 10
c = a + b
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a ) 8 , b ) 9 , c ) 7 , d ) 10 , e ) 5 | d | sqrt(divide(200, const_2)) | the area of a parallelogram is 200 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 200 = > x = 10 answer : d" | a = 200 / 2
b = math.sqrt(a)
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a ) s . 5331 , b ) s . 5269 , c ) s . 5228 , d ) s . 5218 , e ) s . 52192 | a | divide(8210, add(const_1, divide(multiply(6, 9), const_100))) | mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8210 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8210 = p + ( 54 p / 100 ) 8210 = ( 100 p + 54 p ) / 100 8210 = 154 p / 100 p = ( 8210 * 100 ) / 154 = rs . 5331.169 answer : a" | a = 6 * 9
b = a / 100
c = 1 + b
d = 8210 / c
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a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds | d | subtract(multiply(divide(7, 56), 200), 7) | in a 200 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is : | "b runs 56 m in 7 sec . = > b runs 200 m in 7 / 56 * 200 = 25 seconds since a beats b by 7 seconds , a runs 200 m in ( 25 - 7 ) = 18 seconds hence , a ' s time over the course = 18 seconds answer : d" | a = 7 / 56
b = a * 200
c = b - 7
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a ) 6 / 5 , b ) 2 / 3 , c ) 7 / 5 , d ) 8 / 5 , e ) 9 / 5 | a | multiply(10, const_3) | find the l . c . m of 2 / 15 , 3 / 10 , 6 / 5 | "explanation : lcm of fractions = lcm of numerators / hcf of denominators ( lcm of 2 , 3 , 6 ) / ( hcf of 15 , 10 , 5 ) = 6 / 5 answer : option a" | a = 10 * 3
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a ) 20 kmph , b ) 22 kmph , c ) 24 kmph , d ) 26 kmph , e ) 28 kmph | a | divide(multiply(4, 35), add(4, 3)) | a horse chases a pony 3 hours after the pony runs . horse takes 4 hours to reach the pony . if the average speed of the horse is 35 kmph , what s the average speed of the pony ? | "pony take 7 hours and horse take 4 hours . . . then distance chased by them is 35 * 4 . so speed of pony is ( 35 * 4 ) / 7 = 20 kmph . answer : a" | a = 4 * 35
b = 4 + 3
c = a / b
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a ) 35 , b ) 38 , c ) 40 , d ) 42 , e ) 44 | d | divide(add(150, 60), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 150 marks , the no of questions he attempts correctly is : | "explanation : let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 4 x β ( 60 β x ) = 150 = > 5 x = 210 = > x = 42 answer : d" | a = 150 + 60
b = 4 + 1
c = a / b
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['a ) 12 h', 'b ) 10 h', 'c ) 8 h', 'd ) 6 h', 'e ) none of these'] | b | divide(multiply(const_4, sqrt(900)), 12) | the area of a square field is 900 km 2 . how long will it take for a horse to run around at the speed of 12 km / h ? | explanation area of field = 900 km 2 . then , each side of field = β 900 = 30 km distance covered by the horse = perimeter of square field = 30 Γ 4 = 120 km β΄ time taken by horse = distances / peed = 120 / 12 = 10 h answer b | a = math.sqrt(900)
b = 4 * a
c = b / 12
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a ) 100 , b ) 120 , c ) 140 , d ) 150 , e ) 160 | b | add(multiply(subtract(60, 5), 2), multiply(2, 10)) | if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 60 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day ? | "case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 60 ) = ( r + 5 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 2 ) x = 120 ans b" | a = 60 - 5
b = a * 2
c = 2 * 10
d = b + c
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a ) 110 , b ) 140 , c ) 220 , d ) 250 , e ) 500 | b | sqrt(divide(multiply(98, const_100), divide(50, const_100))) | 98 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ? | "let b be the number of boys in the school . 98 = xb / 100 b = 0.5 x 9800 = 0.5 x ^ 2 x ^ 2 = 19600 x = 140 the answer is b ." | a = 98 * 100
b = 50 / 100
c = a / b
d = math.sqrt(c)
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a ) 75 % , b ) 80 % , c ) 85 % , d ) 70 % , e ) 90 % | e | subtract(const_100, multiply(divide(add(17, const_100), add(30, const_100)), const_100)) | two numbers are respectively 17 % and 30 % more than a third number . the percentage that is first of the second is ? | "i ii iii 117 130 100 130 - - - - - - - - - - 117 100 - - - - - - - - - - - ? = > 90 % answer : e" | a = 17 + 100
b = 30 + 100
c = a / b
d = c * 100
e = 100 - d
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a ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | c | divide(const_1, add(multiply(12, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 12 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 12 girls = 8 Γ 6 / 4 + 12 = 12 + 12 = 24 girls 6 days to complete the work answer : option c" | a = 1 / 6
b = a / 24
c = 12 * b
d = 1 / 6
e = d / 16
f = 8 * e
g = c + f
h = 1 / g
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a ) 122.50 per kg , b ) 266.50 per kg , c ) 175.50 per kg , d ) 172.50 per kg , e ) 176.50 per kg | c | divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2) | tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg will be ? | "since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . cost of 1 kg tea of 1 st kind cost of 1 kg tea of 2 nd kind x - 153 / 22.50 = 1 = > x - 153 = 22.50 = > x = 175.50 . hence , price of the third variety = rs . 175.50 per kg . answer : c" | a = 1 + 1
b = a + 2
c = 153 * b
d = 126 + 126
e = c - d
f = e / 2
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a ) 3 , b ) 30 , c ) 300 , d ) 100.5 , e ) 201.2 | d | multiply(const_100, divide(divide(1206, const_3), 400)) | one - third of 1206 is what percent of 400 ? | "answer let one - third of 1206 is n % of 400 . β΅ 1206 / 3 = ( n x 400 ) / 100 β΄ n = ( 402 x 100 ) / 400 = 100.5 correct option : d" | a = 1206 / 3
b = a / 400
c = 100 * b
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a ) 80 , b ) 100 , c ) 150 , d ) 240 , e ) 300 | d | multiply(divide(const_60, 15), 60) | if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 60 minutes ? | "since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 60 minutes the population increases by 60 x 4 = 240 people . answer . d ." | a = const_60 / 15
b = a * 60
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a ) 6 days , b ) 10 days , c ) 12 days , d ) 15 days , e ) 20 days | c | divide(multiply(multiply(6, 3), const_2), const_3) | 1 / 3 rd of work is completed by kiran in 6 days . in how many days he can finish the remaining work ? | 1 / 3 rd work finished in 6 days 2 / 3 rd work = 2 / 3 * 3 * 6 = 12 days answer is c | a = 6 * 3
b = a * 2
c = b / 3
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a ) 10 , b ) 12 , c ) 19 , d ) 15 , e ) 16 | c | add(divide(subtract(57, add(3, 3)), 3), const_2) | find the middle one when the sum of 3 consecutive even numbers is 57 ? | 3 consecutive numbers can be a - 1 , a , a + 1 so sum of numbers = 3 a = 57 . hence a = 19 . c | a = 3 + 3
b = 57 - a
c = b / 3
d = c + 2
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a ) does not change , b ) decreases by 2 % , c ) increases by 4 % , d ) decreases by 4 % , e ) none of these | d | divide(multiply(add(20, const_100), subtract(const_100, 20)), const_100) | a number is increased by 20 % and then decreased by 20 % , the final value of the number is ? | "here , x = 20 and y = - 20 therefore , the net % change in value = ( x + y + xy / 100 ) % = [ 20 - 20 + ( 20 x - 20 ) / 100 ] % or - 4 % since the sign is negative , there is a decrease in value by 4 % . answer : d" | a = 20 + 100
b = 100 - 20
c = a * b
d = c / 100
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a ) 10 , b ) 12 , c ) 15 , d ) 20 , e ) 25 | d | divide(10, subtract(divide(40, const_60), divide(10, const_60))) | circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 10 more revolutions than gear p ? | "gear q makes 30 more revolutions each 60 seconds . the time to make 10 = 30 / 3 more revolutions is 60 / 3 = 20 seconds . the answer is d ." | a = 40 / const_60
b = 10 / const_60
c = a - b
d = 10 / c
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a ) 1.25 , b ) 4.53 , c ) 2.69 , d ) 0.89 , e ) 0.75 | e | divide(18, 6) | find 18 / 6 * 4 | "answer = 18 / 6 * 4 = 18 / 24 = 0.75 option e is correct" | a = 18 / 6
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a ) 1,200 , b ) 200 , c ) 150 , d ) 40 , e ) 20 | c | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | if 2 x = 3 y = 10 , then 9 xy = ? | "2 x = 10 ; x = 5 3 y = 10 ; y = 10 / 3 multiply : 9 xy = 9 * 5 * 10 / 3 = 150 answer : c ." | a = 2 * 100
b = a / 2
c = b - 2
d = 2 * 100
e = d / 2
f = e + 2
g = c / f
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a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 17 | e | add(16, const_1) | the average of first 16 even numbers is ? | "sum of 16 even numbers = 16 * 17 = 272 average = 272 / 16 = 17 answer : e" | a = 16 + 1
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a ) 35 , b ) 48 , c ) 69 , d ) 70 , e ) 74 | d | add(multiply(5, 7), multiply(7, 5)) | if 5 a + 7 b = g , where a and b are positive integers , what is the largest possible value of g for which exactly one pair of integers ( a , b ) makes the equation true ? | 5 * a 1 + 7 * b 1 = g 5 * a 2 + 7 * b 2 = g 5 * ( a 1 - a 2 ) = 7 * ( b 2 - b 1 ) since we are dealing with integers we can assume that a 1 - a 2 = 7 * q and b 2 - b 1 = 5 * q where q is integer , so whenever we get a pair for ( a ; b ) we can find another one by simply adding 7 to a and subtracting 5 from b or vice versa , subtracting 7 from a and adding 5 to b . lets check how it works for our numbers , starting from the largest : e ) 74 = 5 * 12 + 7 * 2 ( a 1 = 12 , b 1 = 2 ) , subtract 7 from a and add 5 to b respectively , so a 2 = 5 and b 2 = 7 , second pair - bad d ) 70 = 5 * 7 + 7 * 5 ( a 1 = 7 , b 1 = 5 ) , if we add 7 to a we will have to subtract 5 from b but b ca n ' t be 0 , so - no pair , if we subtract 7 from a , we ' ll get a = 0 which also is n ' t allowed - no pair , thus this is the only pair for ( a ; b ) that works , good ! , thus d is the answer | a = 5 * 7
b = 7 * 5
c = a + b
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a ) 6 , b ) 7 , c ) 9 , d ) 12 , e ) 16 | c | add(multiply(divide(24, subtract(24, 18)), const_2), const_1) | a worker earns $ 24 on the first day and spends $ 18 on the second day . the worker earns $ 24 on the third day and spends $ 18 on the fourth day . if this pattern continues , on which day will the worker first reach a net total of $ 48 ? | every two days , the net total is $ 6 . after 8 days , the worker will have $ 24 . on day 9 , the worker will receive $ 24 for a net total of $ 48 . the answer is c . | a = 24 - 18
b = 24 / a
c = b * 2
d = c + 1
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a ) 1100 , b ) 2100 , c ) 1200 , d ) 3100 , e ) 1300 | a | subtract(add(2050, 6200), 5250) | the average monthly income of p and q is rs . 2050 . the average monthly income of q and r is rs . 5250 and the average monthly income of p and r is rs . 6200 . the monthly income ofq is : | "explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 2050 x 2 ) = 4100 . . . . ( i ) q + r = ( 5250 x 2 ) = 10500 . . . . ( ii ) p + r = ( 6200 x 2 ) = 12400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 27000 or p + q + r = 13500 . . . . ( iv ) subtracting ( iii ) from ( iv ) , we get p = 1100 . p ' s monthly income = rs . 1100 . answer : a" | a = 2050 + 6200
b = a - 5250
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a ) 25 kms , b ) 10 kms , c ) 50 kms , d ) 30 kms , e ) 44 kms | e | multiply(speed(add(24, 35), const_60), 45) | riya and priya set on a journey . riya moves eastward at a speed of 24 kmph and priya moves westward at a speed of 35 kmph . how far will be priya from riya after 45 minutes | "total eastward distance = 24 kmph * 3 / 4 hr = 18 km total westward distance = 35 kmph * 3 / 4 hr = 26 km total distn betn them = 18 + 26 = 44 km ans 44 km answer : e" | a = 24 + 35
b = speed * (
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a ) 0.0009 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90 | c | divide(0.009, 0.01) | 0.009 / x = 0.01 . find the value of x | "x = 0.009 / 0.01 = 0.9 answer : c" | a = 0 / 9
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a ) 76.5 , b ) 75.6 , c ) 73.5 , d ) 79 , e ) 77 | b | divide(add(add(add(add(subtract(multiply(67, add(const_4, const_1)), add(add(add(28, 70), 88), 104)), 50), 62), 97), 124), add(const_4, const_1)) | if the mean of numbers 28 , x , 70 , 88 and 104 is 67 , what is the mean of 50 , 62 , 97 , 124 and x ? | "mean = ( sum of all no . in series ) / ( no . in series ) m 1 = 67 = ( x + 28 + 70 + 88 + 104 ) / ( 5 ) 67 x 5 = 290 + x x = 45 therefore mean of 2 nd series m 2 = ( x + 50 + 62 + 97 + 124 ) / 5 m 2 = 75.6 answer = b" | a = 4 + 1
b = 67 * a
c = 28 + 70
d = c + 88
e = d + 104
f = b - e
g = f + 50
h = g + 62
i = h + 97
j = i + 124
k = 4 + 1
l = j / k
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(multiply(4.00, 10), 42.00) | elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.80 apiece . if elena purchased a total of 10 of these pens for $ 42.00 , how many brand x pens did she purchase ? | "4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 5 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : b ." | a = 4 * 0
b = a - 42
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a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | b | divide(sqrt(16), 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 16 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 16 : β 9 = 4 : 3 . answer b" | a = math.sqrt(16)
b = math.sqrt(9)
c = a / b
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a ) 275 m , b ) 360 m , c ) 375 m , d ) 420 m , e ) 480 m | c | multiply(multiply(45, const_0_2778), 30) | if the speed of a man is 45 km per hour , then what is the distance traveled by him in 30 seconds ? | "the distance traveled in 30 sec = 45 * ( 5 / 18 ) * 30 = 375 m answer : c" | a = 45 * const_0_2778
b = a * 30
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a ) 15 kmph , b ) 18 kmph , c ) 14.25 kmph , d ) 11.52 kmph , e ) 16 kmph | d | divide(subtract(multiply(const_100, const_3), const_12), add(24, const_1)) | a participated in cycling contest and he drove the lap at the rate of 6 kmph , 12 kmph , 18 kmph , 24 kmph . . what is his average speed ? | = 4 x / ( x / 6 + x / 12 + x / 18 + x / 24 ) = 11.52 answer : d | a = 100 * 3
b = a - 12
c = 24 + 1
d = b / c
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a ) rs . 1500 , b ) rs . 2000 , c ) rs . 3000 , d ) rs . 4200 , e ) rs . 2500 | b | divide(multiply(520, const_100), subtract(add(const_100, 6), subtract(const_100, 20))) | a watch was sold at a loss of 20 % . if it was sold for rs . 520 more , there would have been a gain of 6 % . what is the cost price ? | "80 % 106 % - - - - - - - - 26 % - - - - 520 100 % - - - - ? = > rs . 2000 answer : b" | a = 520 * 100
b = 100 + 6
c = 100 - 20
d = b - c
e = a / d
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a ) $ 30.14 , b ) 48.53 , c ) 34.66 , d ) 32.29 , e ) 33.16 | b | divide(add(211.00, divide(multiply(15, 211.00), const_100)), 5) | total dinning bill for 5 people was $ 211.00 . if they add 15 % tip and divided the bill evenly , approximate . what was each persons find share | "211 * 15 = 3165 / 100 = 31.65 211 + 31.65 = 242.65 242.65 / 5 = 48.53 answer : b" | a = 15 * 211
b = a / 100
c = 211 + 0
d = c / 5
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a ) 20 / 7 , b ) 4 / 3 , c ) 15 / 8 , d ) 9 / 4 , e ) 15 / 4 | a | multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 5))), 2) | kathleen can paint a room in 2 hours , and anthony can paint an identical room in 5 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "( 1 / 2 + 1 / 5 ) t = 2 t = 20 / 7 answer : a" | a = 1 / 2
b = 1 / 5
c = a + b
d = 1 / c
e = d * 2
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a ) 220 , b ) 225 , c ) 230 , d ) 240 , e ) 245 | b | add(60, add(subtract(85, 60), subtract(200, 60))) | in a school of 320 students , 85 students are in the band , 200 students are on sports teams , and 60 students participate in both activities . how many students are involved in either band or sports ? | 25 + 60 + 140 = 225 there are 225 students involved in either band or sports . b ) | a = 85 - 60
b = 200 - 60
c = a + b
d = 60 + c
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a ) 20 % loss , b ) 25 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 20)), divide(multiply(const_100, 16), 20))) | if the cost price of 20 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ? | "explanation : let cost price of 1 article be re . 1 . therefore , cost price of 20 articles = rs . 20 . selling price of 16 articles = rs . 20 therefore , selling price of 20 articles is : - = > 20 / 16 Γ 20 = > 25 . therefore , profit = selling price - cost price . = > 25 β 20 = 5 . hence , the percentage of profit = profit x 100 / c . p . = > 5 / 20 Γ 100 . = > 25 % profit . answer : b" | a = 100 * 16
b = a / 20
c = 100 - b
d = 100 * 16
e = d / 20
f = c / e
g = 100 * f
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a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | a | multiply(const_100, subtract(multiply(multiply(divide(add(20, const_100), const_100), divide(add(20, const_100), const_100)), divide(subtract(const_100, 20), const_100)), const_1)) | the positive numbers w , x , y , and z are such that x is 20 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | y = 100 ( start with yy , it makes getting x easier ) . x = y * 1.2 = 120 z = y / 1.2 = 83.33 w = x * 0.8 = 96 then for the answer : ( w - z ) / z = ( 96 - 83.33 ) / 83.33 = 15.2 % answer : a | a = 20 + 100
b = a / 100
c = 20 + 100
d = c / 100
e = b * d
f = 100 - 20
g = f / 100
h = e * g
i = h - 1
j = 100 * i
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['a ) ( 1.5 , 1 )', 'b ) ( 2 , - 5 )', 'c ) ( 0,0 )', 'd ) ( 1 , 1.5 )', 'e ) ( 2,2 )'] | d | add(multiply(const_3, const_3), sqrt(divide(25, 4))) | on the coordinate plane , points p and u are defined by the coordinates ( - 1,0 ) and ( 3,3 ) , respectively , and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( 25 / 4 ) Ο , what are the coordinates of the center of the circle ? | although it took me 3 mins to solve this question using all those equations , later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - 1 + 3 ) / 2 , ( 3 + 0 ) / 2 i . e . ( 1 , 1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that pu is the diameter of the circle . there can be a case when pu is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d | a = 3 * 3
b = 25 / 4
c = math.sqrt(b)
d = a + c
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a ) 12 , b ) 77 , c ) 48 , d ) 54 , e ) 11 | d | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.25, 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 1.25 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 + 1.25 = 2 km = 2250 m . required time = 2250 * 3 / 125 = 54 sec . answer : d" | a = 1 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 25
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
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a ) 50 , b ) 40 , c ) 35 , d ) 10 , e ) 90 | b | add(multiply(3, 8), multiply(const_2, 8)) | paul and jack are 2 mechanics who work 8 hours per day changing oil in cars at the oil spot . if paul can change oil in 2 cars per hour , and jack can change oil in 3 cars per hour . what is the least number of cars per work day the 2 men can finish ? | the proportion equation can be used . let x = number cars and c = least number of cars per day the 2 men can finish ( c = x / ( 8 hours ) , p = cars per hour paul can change , and j = cars per hour jack can change . note : 8 hours = a work day p + j = c ( 2 cars / hour ) + ( 3 cars / hour ) = x / ( 8 hours ) proportion : 2 / 1 + 3 / 1 / = x / 300 x = 8 ( 2 + 3 ) = 40 . at least 40 cars per work day . answer is b | a = 3 * 8
b = 2 * 8
c = a + b
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['a ) 2', 'b ) 8', 'c ) 7', 'd ) 6', 'e ) 9'] | d | multiply(const_2, const_3) | the edge of three cubes of metal is 3 dm , 4 dm and 5 dm . they are melted and formed into a single cube . find the edge of the new cube ? | 33 + 43 + 53 = a 3 = > a = 6 answer : d | a = 2 * 3
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a ) 12 , b ) 24 , c ) 20 , d ) 55 , e ) 60 | c | multiply(add(const_2, 4), multiply(3, 4)) | if a farmer sells 5 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 10 more chickens , he will run out of feed 3 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | "say farmer has n chicken and he is good for d days . : - we have 3 equations given in question : - ( n - 5 ) * d + 4 = ( n + 10 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 10 d - 3 n = 30 4 n - 5 d = 20 = > n = 20 ans c it is !" | a = 2 + 4
b = 3 * 4
c = a * b
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['a ) 100', 'b ) 272', 'c ) 287', 'd ) 189', 'e ) 289'] | a | multiply(const_4, power(power(divide(multiply(multiply(10, 10), 10), 8), const_0_33), const_2)) | if a solid sphere of radius 10 cms is moulded into 8 spherical solid balls of equal radius , then surface area of each ball ( in cm 2 ) is ? | 4 / 3 Ο * 10 * 10 * 10 = 8 * 4 / 3 Ο r 3 r = 5 4 Ο * 5 * 5 = 100 Ο answer : a | a = 10 * 10
b = a * 10
c = b / 8
d = c ** const_0_33
e = d ** 2
f = 4 * e
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a ) 15 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | c | divide(divide(26, const_2), divide(65, const_100)) | at the end of the month , a certain ocean desalination plant β s reservoir contained 26 million gallons of water . this amount is eighth the normal level . if this amount represents 65 % of the reservoir β s total capacity , how many million gallons short of total capacity is the normal level ? | "the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 26 mg = 65 % of total . . total = 26 / . 65 = 40 mg . . normal level = 1 / 8 of 40 = 5 mg . . shortage of normal level = 40 - 5 = 35 mg . . c" | a = 26 / 2
b = 65 / 100
c = a / b
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a ) 68 , b ) 70 , c ) 72 , d ) 74 , e ) 76 | a | add(divide(multiply(20, 9), 5), 32) | if t = 5 / 9 * ( k - 32 ) , and if t = 20 , then what is the value of k ? | "k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 20 ) / 5 + 32 = 68 the answer is a ." | a = 20 * 9
b = a / 5
c = b + 32
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a ) 292 ft , b ) 293.4 ft , c ) 295 ft , d ) 296 ft , e ) 297 ft | b | multiply(20, divide(multiply(10, 5280), const_3600)) | someone on a skateboard is traveling 10 miles per hour . how many feet does she travel in 20 seconds ? ( 1 mile = 5280 feet ) | per second = > 10 * 5280 ft / 60 * 60 = 14.67 ft 20 seconds = > 14.67 * 20 = 293.4 ft answer : b | a = 10 * 5280
b = a / 3600
c = 20 * b
|
a ) 0.008 , b ) 0.08 , c ) 0.064 , d ) 0.64 , e ) 6.4 | c | divide(volume_cylinder(8, 10), circle_area(100)) | a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 8 feet and a height of 10 feet , how far ( in feet ) did the oil level drop in the stationary tank ? | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 64 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 640 / 10,000 = 64 / 1000 = 0.064 ft the answer is c ." | a = volume_cylinder / (
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 34.6 % , e ) 40 % | d | multiply(subtract(divide(add(const_100, 75), add(const_100, 30)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 30 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 75 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | "another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 75 = 30 + b + 30 b / 100 b = 34.6 answer ( d )" | a = 100 + 75
b = 100 + 30
c = a / b
d = c - 1
e = d * 100
|
a ) 2 , b ) 5 , c ) 4 , d ) 0 , e ) 1 | a | multiply(18800, divide(470, 20)) | solve : 18800 + 470 + 20 | "18800 + 470 + 20 = ( 18800 / 470 ) + 20 = 40 / 20 = 2 . answer is a ." | a = 470 / 20
b = 18800 * a
|
a ) 31 , b ) 33 , c ) 35 , d ) 37 , e ) 39 | d | add(divide(subtract(subtract(240, const_2), add(20, const_2)), add(const_2, 4)), const_1) | how many even integers n , such that 20 < = n < = 240 are of the form 3 k + 4 , where k is any natural number ? | "the first number is 22 = 16 + 6 ( 1 ) . we can continue adding 6 to make a list : 22 , 28 , 34 , . . . the last number is 238 = 16 + 6 ( 37 ) there are 37 numbers in the list . the answer is d ." | a = 240 - 2
b = 20 + 2
c = a - b
d = 2 + 4
e = c / d
f = e + 1
|
a ) 12 , b ) 24 , c ) 18 , d ) 42 , e ) 64 | c | divide(divide(600, const_1000), divide(multiply(2, const_60), const_3600)) | a person crosses a 600 m long street in 2 minutes . what is his speed in km per hour ? | distance = 600 meter time = 2 minutes = 2 x 60 seconds = 120 seconds speed = distance / time = 600 / 120 = 5 m / s = 5 Γ£ β 18 / 5 km / hr = 18 km / hr answer : c | a = 600 / 1000
b = 2 * const_60
c = b / 3600
d = a / c
|
a ) $ 1.25 , b ) $ 2.50 , c ) $ 3.75 , d ) $ 4.70 , e ) $ 8.20 | c | divide(subtract(9.85, 2.35), const_2) | a train ride from two p to town q costs $ 2.35 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.85 . what is the cost of a bus ride from town p to town q ? | let x be the cost of a bus ride . x + ( x + 235 ) = 985 2 x = 750 x = $ 3.75 the answer is c . | a = 9 - 85
b = a / 2
|
a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72 | e | add(add(multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 5))), 5), 8), multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 5))), 3)) | in a can , there is a mixture of milk and water in the ratio 5 : 3 . if the can is filled with an additional 8 liters of milk , the can would be full and the ratio of milk and water would become 2 : 1 . find the capacity of the can ? | "let c be the capacity of the can . ( 5 / 8 ) * ( c - 8 ) + 8 = ( 2 / 3 ) * c 15 c - 120 + 192 = 16 c c = 72 the answer is e ." | a = 1 * 8
b = 3 * 2
c = 1 * 5
d = b - c
e = a / d
f = e * 5
g = f + 8
h = 1 * 8
i = 3 * 2
j = 1 * 5
k = i - j
l = h / k
m = l * 3
n = g + m
|
a ) 5 sec , b ) 4 sec , c ) 3 sec , d ) 6 sec , e ) 1 sec | c | divide(30, multiply(36, const_0_2778)) | in what time will a railway train 30 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 30 / 36 * 18 / 5 = 3 sec answer : c" | a = 36 * const_0_2778
b = 30 / a
|
a ) 24.28 , b ) 14.28 , c ) 16.28 , d ) 18.28 , e ) 12.28 | b | subtract(50, divide(50, add(divide(2, 5), const_1))) | a 50 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | 1 : 2 / 5 = 5 : 2 2 / 7 * 50 = 14.28 answer : b | a = 2 / 5
b = a + 1
c = 50 / b
d = 50 - c
|
a ) 4 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 46 meters . how many times did the ball bounce ? | division of total distance traveled will be 16 + 16 + 8 + 4 + 2 ans : a | a = 16 / 2
b = a / 2
|
a ) a ) 4.8 , b ) b ) 48 , c ) c ) 30 , d ) d ) 68 , e ) e ) 48 | c | divide(multiply(786, 74), 1938.8) | ( 786 Γ 74 ) Γ· ? = 1938.8 | explanation : 58164 / x = 1938.8 = > x = 58164 / 1938.8 = 30 answer : option c | a = 786 * 74
b = a / 1938
|
a ) 12.91 m , b ) 11.91 m , c ) 12.5 m , d ) 10.91 m , e ) 13.91 m | b | multiply(divide(subtract(100, 0.75), subtract(7, 0.75)), 0.75) | in a 100 m race between rose and diane , rose lost to diane by 0.75 m . if diane was given a 7 m head start , how many meters more will it take before rose overtakes diane ? | distance rose ran before completion of race - - - 100 m - 0.75 m = 99.25 m distance gained on diane over 99.25 m - - - 7 m - 0.75 m = 6.25 then rose gains 99.25 / 6.25 m = 1 m on kelly every 15.88 meters . therefore 15.88 divided by 0.75 of 1 meter . answer : b . 11.91 m | a = 100 - 0
b = 7 - 0
c = a / b
d = c * 0
|
a ) 11 th , b ) 13 th , c ) 12 th , d ) 14 th , e ) 15 th | b | add(subtract(26, 14), const_1) | if ramola ranks 14 th in a class of 26 , what is her rank from the last ? | when considered from last , 12 ( i . e 26 - 14 ) students are ranked before her . so she is 13 th rank from last . answer : b | a = 26 - 14
b = a + 1
|
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