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 ) 81 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(1, multiply(const_4, 1)) | xy = 1 then what is ( 3 ^ ( x + y ) ^ 2 ) / ( 3 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 3 ^ 4 = 81 answer a" | a = 4 * 1
b = 1 ** a
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a ) 7.9 s , b ) 2.5 s , c ) 7.5 s , d ) 7.6 s , e ) 7.4 s | c | multiply(divide(divide(125, const_1000), 60), const_3600) | how much time does a train 125 metres long running at 60 km / hr take to pass a pole ? | "explanation : 60 km / hr = 60 * 5 / 18 = 16.67 m / s speed = distance / time ; v = d / t 16.67 = 125 / t t = 7.5 s answer : c" | a = 125 / 1000
b = a / 60
c = b * 3600
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a ) 130 , b ) 140 , c ) 156 , d ) 184 , e ) 196 | d | add(lcm(lcm(5, 6), lcm(9, 12)), 4) | what is the least number which when divided by 5 , 6 , 9 and 12 leaves remainder 4 in each care ? | explanation : lcm of 5 , 6 , 9 and 12 is 180 required number = 180 + 4 = 184 answer : option d | a = math.lcm(5, 6)
b = math.lcm(9, 12)
c = math.lcm(a, b)
d = c + 4
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a ) 15 , b ) 20 , c ) 18 , d ) 25 , e ) 10 | b | divide(25, add(const_1, divide(25, const_100))) | sakshi can do a piece of work in 25 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work : | "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 25 : x ⇒ x = ( 25 x 4 / 5 ) ⇒ x = 20 days . hence , tanya takes 16 days is complete the work . answer b" | a = 25 / 100
b = 1 + a
c = 25 / b
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a ) 64.54 , b ) 62.6 , c ) 62.1 , d ) 62.7 , e ) 62.2 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 8)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 8 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 92 / 100 ) = 59.375 mp = 64.54 answer : a" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 8
f = d / e
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a ) 60 , b ) 50 , c ) 42 , d ) 84 , e ) 61 | e | add(add(20, 20), add(12, 12)) | in a bag , there are two 6 - sided dice ( numbered 1 to 6 ) , 3 12 - sided dice ( numbered 1 to 12 ) , and two 20 - sided dice ( numbered 1 to 20 ) . if 4 of these dice are selected at random from the bag , and then the 4 are rolled and we find the sum of numbers showing on the 4 dice , how many different possible totals are there for this sum ? | as ' complex - looking ' as this question might appear , it ' s actually rather simple . pay careful attention to what the specific questions asks for - the number of different possible sums from 4 dice . since we ' re dealing with some ' special ' dice ( some 12 - sided and 20 - sided dice ) , we have to adjust out math accordingly , but the possibilities are rather limited : 1 ) the minimum number on any given die is 1 2 ) the maximum possible sum would only occur if we took the 4 biggest possible dice and rolled the highest possible number on each . with 4 dice , we could end up with any sum between : 4 ( if we rolled 1 s on all 4 dice ) to 64 ( if we rolled two 20 s on the 20 - sided dice and two 12 s on the 12 - sided dice ) . thus , there are only 61 possible sums . e | a = 20 + 20
b = 12 + 12
c = a + b
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a ) 25630 yards , b ) 35200 yards , c ) 88000 yards , d ) 42560 yards , e ) 41520 yards | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 50), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 50 miles into yards ? | "1 mile = 1760 yards 50 miles = 50 * 1760 = 88000 yards answer is c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 50
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 120 m , b ) 180 m , c ) 240 m , d ) 200 m , e ) 280 m | d | multiply(divide(multiply(const_1000, 12), multiply(const_60, 5)), 5) | a bus covers a distance of 12 km in 5 min . if it takes 5 sec to pass a telegraph post , then the length of the bus is ? | speed = ( 12 / 5 * 60 ) km / hr = ( 144 * 5 / 18 ) m / sec = 40 m / sec . length of the bus = 40 * 5 = 200 m . answer : d | a = 1000 * 12
b = const_60 * 5
c = a / b
d = c * 5
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a ) 3.5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | d | divide(subtract(69, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if the resultant is trebled , it becomes 69 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 69 2 x = 14 = > x = 7 answer : d" | a = 3 * 9
b = 69 - a
c = 3 * 2
d = b / c
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a ) 5 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | c | divide(subtract(const_1, multiply(10, divide(const_1, 15))), divide(const_1, 18)) | x can finish a work in 18 days . y can finish the same work in 15 days . yworked for 10 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 18 work done by y in 1 day = 1 / 15 work done by y in 10 days = 10 / 15 = 2 / 3 remaining work = 1 – 2 / 3 = 1 / 3 number of days in which x can finish the remaining work = ( 1 / 3 ) / ( 1 / 18 ) = 6 c" | a = 1 / 15
b = 10 * a
c = 1 - b
d = 1 / 18
e = c / d
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a ) 8 , b ) 14 , c ) 10 , d ) 20 , e ) 40 | d | divide(power(20, 2), 20) | n ^ ( n / 2 ) = 8 is true when n = 8 in the same way what is the value of n if n ^ ( n / 2 ) = 20 ? | n ^ ( n / 2 ) = 20 apply log n / 2 logn = log 20 nlogn = 2 log 20 = log 20 ^ 2 = log 400 logn = log 400 now apply antilog n = 400 / n now n = 20 . answer : d | a = 20 ** 2
b = a / 20
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a ) 600 m , b ) 200 m , c ) 300 m , d ) 400 m , e ) 100 m | d | divide(multiply(12, multiply(1.6, const_1000)), 48) | amar takes as much time in running 12 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.6 km ? | "d 400 m distance covered by amar = 12 / 4.8 ( 1.6 km ) = 2.5 ( 1600 ) = 400 m answer is d" | a = 1 * 6
b = 12 * a
c = b / 48
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | add(divide(14, 14), const_3) | find the unit ’ s digit in the product ( 76 ^ 41 ) * ( 41 ^ 14 ) * ( 14 ^ 87 ) * ( 87 ^ 76 ) | units digit of 87 ^ 76 = 1 because 87 ^ 3 n + 1 has units place as 1 units digit of 14 ^ 87 = 4 because 14 ^ ( some odd no . ) has units place as 4 units digit of 41 ^ 14 = 1 because 41 ^ ( some even no . ) has units place as 1 units digit of 76 ^ 41 = 6 because 76 ^ ( any positive no . ) has units place as 6 so , units digit of the product is = ` ` 4 ` ` ( units digit of 1 * 4 * 1 * 6 ) answer : d | a = 14 / 14
b = a + 3
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(subtract(divide(multiply(100, const_2), 5), multiply(6, const_4)), const_2) | a man ate 100 bananas in 5 days , each day eating 6 more than the previous day . how many bananas did he eat on the first day ? | let the number of bannanas in first day be x for consecutive five days it will be like x , x + 6 , x + 12 , x + 18 , x + 24 the sum of those are 100 , 5 x + 60 = 100 so , x = 8 answer : c | a = 100 * 2
b = a / 5
c = 6 * 4
d = b - c
e = d / 2
|
a ) 2 , b ) 5 , c ) 1 , d ) 6 , e ) 7 | a | subtract(add(add(multiply(1894, const_10), const_1), const_1), multiply(1894, const_10)) | what is the least value of x . so that 1894 x is divisible by 3 ? | the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 1 + 8 + 9 + 4 + x = 22 + x least value of x may be 2 therefore 22 + 2 = 24 is divisible by 3 . a | a = 1894 * 10
b = a + 1
c = b + 1
d = 1894 * 10
e = c - d
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a ) 130 minutes , b ) 140 minutes , c ) 150 minutes , d ) 160 minutes , e ) 170 minutes | c | divide(multiply(25.5, const_100), 17) | if takes 25.5 minutes to download a file using modem a . modem b work at 17 % of the speed of modem a . how long does it take to download a file using a modem b ? | let speed of modem a be k so speed of modem b is 17 / 100 * k time taken by modem a = 25.5 minutes let time taken by modem b = x k * 25.5 = 17 / 100 * k * x 0.17 * x = 25.5 x = 25.5 / 0.17 = 150 minutes answer : c | a = 25 * 5
b = a / 17
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a ) 9 , b ) 17 , c ) 13 , d ) 10 , e ) 12 | a | multiply(multiply(divide(7, 5), divide(9, 7)), 5) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 9 . if p invested the money for 5 months , find for how much time did q invest the money ? | "7 * 5 : 5 * x = 7 : 9 x = 9 answer : a" | a = 7 / 5
b = 9 / 7
c = a * b
d = c * 5
|
a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salaries of all the employees at company e after july 1 last year was what percent w of thatbeforejuly 1 last year ? | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before w = [ 0.9 n ( 1.1 x ) ] / xn = 0.99 or 99 % = b" | a = 10 * 10
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a ) 25.6 , b ) 50.2 , c ) 30.3 , d ) 45 , e ) 56 | c | multiply(multiply(100, divide(10, const_100)), 3) | what annual installment will discharge a debt of $ 100 due in 3 years at 10 % simple interest ? | let each installment be x [ x + ( x * 10 * 1 / 100 ) ] + [ x + ( x * 10 * 2 / 100 ) ] + x = 100 11 x / 10 + 12 x / 10 + x = 100 x = 30.3 answer is c | a = 10 / 100
b = 100 * a
c = b * 3
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a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | e | divide(125, add(20, 5)) | a boat can travel with aspeed of 20 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 125 km downstream . | speed downstream = ( 20 + 5 ) km / hr = 25 km / hr . time taken to travel 125 km downstream = 125 / 25 hrs = 5 hrs . answer : e | a = 20 + 5
b = 125 / a
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a ) 46 , b ) 47 , c ) 58 , d ) 56 , e ) 43 | e | divide(397.75, 9.25) | john bought 9.25 m of cloth for $ 397.75 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 397.75 cost per metre = 397.75 ÷ 9.25 the cost of the cloth per metre = $ 43 answers : e" | a = 397 / 75
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a ) 6178.85 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these | a | divide(multiply(const_100, 4016.25), multiply(13, 5)) | a sum fetched total simple interest of 4016.25 at the rate of 13 p . c . p . a . in 5 years . what is the sum ? | "let the sums be p . now , 65 % of p = 4016.25 or , p = 6178.85 answer a" | a = 100 * 4016
b = 13 * 5
c = a / b
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a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 70 | b | multiply(3, const_4) | what could be the range of a set consisting of odd multiples of 3 ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 3 * ( 2 n + 1 ) where n is any positive integer . range = 3 * ( 2 m + 1 ) - 3 * ( 2 n + 1 ) = 3 * 2 * ( m - n ) = 6 * ( m - n ) . m , n - any positive integers the answer must be divisible by 6 , which is only 24 . the correct answer is b" | a = 3 * 4
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a ) 12 mins , b ) 10 mins , c ) 19 mins , d ) 15 mins , e ) 20 mins | c | subtract(add(subtract(55, 37), divide(37, const_2)), subtract(55, 37)) | i walk a certain distance and ride back taking a total time of 37 minutes . i could walk both ways in 55 minutes . how long would it take me to ride both ways ? | let the distance be x km . then , ( time taken to walk x km ) + ( time taken to ride x km ) = 37 min . ( time taken to walk 2 x km ) + ( time taken to ride 2 x km ) = 74 min . but , the time taken to walk 2 x km = 55 min . time taken to ride 2 x km = ( 74 - 55 ) min = 19 min . ans : c | a = 55 - 37
b = 37 / 2
c = a + b
d = 55 - 37
e = c - d
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a ) $ 14,755 , b ) $ 11,495 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | b | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 35 % next year and that the price of type b products will increase by 15 % next year . this year , the total amount paid for type a products was $ 4000 and the total price paid for type b products was $ 5300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | cost of type a products next year = 1.35 * 4000 = 5400 cost of type b products next year = 1.15 * 5300 = 6095 total 5400 + 6095 = 11495 answer : b | a = 3 / 4
b = a * 1000
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a ) 8 , b ) 9 , c ) 7 , d ) 11 , e ) 2 | d | sqrt(divide(242, const_2)) | the area of a parallelogram is 242 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 242 = > x = 11 answer : d" | a = 242 / 2
b = math.sqrt(a)
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a ) 9.09 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | multiply(divide(const_1, multiply(divide(add(10, const_100), const_100), const_10)), const_100) | a part - time employee whose hourly wage was increased by 10 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.1 x now we need to find number of hours worked so that x * y = 1.1 x * z i . e z = x * y / 1.1 x = 1 / 1.1 y % decrease = ( y - 1 / 1.1 y ) / y * 100 = 1 - 1 / 1.1 * 100 = 9.09 % thus my answer is a . | a = 10 + 100
b = a / 100
c = b * 10
d = 1 / c
e = d * 100
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a ) $ 35,000 , b ) $ 45,000 , c ) $ 50,000 , d ) $ 65,000 , e ) $ 75,000 | c | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 150,000 . the 75 associates have an average salary of $ 30,000 . what is the average salary for the company ? | "another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 5 a * 1 + a * 5 ) / 6 = 10 a / 6 = 50000 answer is c . $ 50,000" | a = 3 * 3
b = a * 10
c = 100 * 10
d = b * c
e = 3 * 10
f = 100 * 10
g = e * f
h = d + g
i = 75 + 15
j = h / i
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['a ) a . 40', 'b ) b . 100', 'c ) c . 400', 'd ) d . 1,800', 'e ) e . 10,000'] | d | divide(const_180, const_1000) | the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 2304 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 2304 pi r = 12 so area = 4 . pi . r ^ 2 = 576 . pi = 576 x 3.14 = 1800 ( approx ) d | a = const_180 / 1000
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a ) 0 , b ) 10 , c ) 20 , d ) 30 , e ) 35 | e | multiply(add(const_3, const_4), add(const_2, const_3)) | in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 35 , what is the greatest possible value of b x a ? | "i also was confused when i was looking forabove number : d as far as i understood , 35 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 35 x 1 ( a = 3 , b = 5 respectively we have = 15 ) . but we do n ' t have this number in answer choices , move on . ( avoid even multiples of 2,4 , 6,8 etc ( we will have 0 in units thus making our result 0 ) 35 x 3 = 105 ( a = 0 b = 5 respectively . hey ! that ' s 0 now . 35 x 5 = 175 ( a = 7 , b = 5 respectively . increasing trend , we have 35 now ) . this is the greatest possible value of b x a . imo e ." | a = 3 + 4
b = 2 + 3
c = a * b
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a ) 15 , b ) 12 , c ) 18 , d ) 10 , e ) 14 | b | add(multiply(2, 3), 6) | if 2 x - y = 4 then 6 x - 3 y = ? | 6 x - 3 y = 3 * ( 2 x - y ) = 3 * 4 = 12 answer : b | a = 2 * 3
b = a + 6
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a ) 18.44 , b ) 18.48 , c ) 18.42 , d ) 18.49 , e ) 18.41 | a | divide(3486, 189) | find 3486 / 189 | first we make the denominator 200 . i . e . , we are introducing an error of 11 which is 11189 = 11711189 = 117 th of the denominator . now 3486200 = 17.433486200 = 17.43 but this answer is less than actual answer . so we have to increase it by 1 / 17 th part of it . 117 × 17.43 ≃ 1117 × 17.43 ≃ 1 so final answer = 17.43 + 1 = 18.43 actual answer = 18.44 answer : a | a = 3486 / 189
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a ) $ 8500 , b ) $ 9500 , c ) $ 7000 , d ) $ 7500 , e ) $ 6500 | d | divide(1000, multiply(subtract(divide(600, const_1000), divide(400, const_1000)), subtract(const_1, divide(const_1, const_3)))) | mary and mike enter into a partnership by investing $ 600 and $ 400 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if mary received $ 1000 more than mike did , what was the profit made by their business in that year ? | "explanatory answer let the profit made during the year be $ 3 x therefore , $ x would have been shared equally and the remaining $ 2 x would have been shared in the ratio 6 : 4 . i . e . , 60 % of 2 x would go to mary and 40 % of 2 x would go to mike . hence , mary would get ( 60 - 40 ) % of 2 x more than mike or 20 % of 2 x = $ 1000 i . e . , ( 20 / 100 ) * 2 x = 1000 or 2 x = 5000 . hence , the profit made by the company during the year $ 3 x = $ 7500 . answer : d" | a = 600 / 1000
b = 400 / 1000
c = a - b
d = 1 / 3
e = 1 - d
f = c * e
g = 1000 / f
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a ) 10 % , b ) 11 % , c ) 12.5 % , d ) 13 % , e ) 14 % | c | multiply(subtract(const_1, divide(multiply(20, const_2), add(add(20, 27), const_1))), const_100) | a bookstore has a shelf that contains biographies which normally sell for $ 20 each and mysteries that normally sell for $ 12 each . during a sale , the biographies and mysteries are discounted at different rates so that a customer saves a total of $ 19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries . if the sum of the discount rates for the two types of books is 27 percent , what is the discount rate on mysteries ? | "let b be the discount on biographies and m be the discount on mysteries so . , b + m = 0.27 - - - - - ( 1 ) and ( 20 * 5 + 12 * 3 ) - ( 20 * 5 * ( 1 - b ) + 12 * 3 * ( 1 - m ) ) = 19 - - > 100 ( 1 - ( 1 - b ) ) + 36 ( 1 - ( 1 - m ) = 19 100 b + 36 m = 19 - - - - - - ( 2 ) solving 12 . , we get m = 0.125 = 12.5 % c" | a = 20 * 2
b = 20 + 27
c = b + 1
d = a / c
e = 1 - d
f = e * 100
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a ) 45 % , b ) 48 % , c ) 50 % , d ) 58 % , e ) 65 % | c | multiply(divide(200, divide(add(120, 200), divide(80, const_100))), const_100) | after a storm deposits 120 billion gallons of water into the city reservoir , the reservoir is 80 % full . if the original contents of the reservoir totaled 200 billion gallons , the reservoir was approximately what percentage full before the storm ? | "when the storm deposited 120 billion gallons , volume of water in the reservoir = 200 + 120 = 320 billion gallons if this is only 80 % of the capacity of the reservoir , the total capacity of the reservoir = 320 / 0.8 = 400 billion gallons therefore percentage of reservoir that was full before the storm = ( 200 / 400 ) * 100 = 50 % option c" | a = 120 + 200
b = 80 / 100
c = a / b
d = 200 / c
e = d * 100
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a ) 1 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | a | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(356, multiply(floor(divide(356, const_4)), const_4))), multiply(const_2, const_10)) | find the ones digit of 73 ^ 356 | "the units digit of 73 ^ 356 will be the same as the units digit of 3 ^ 356 . 3 ^ 1 = 3 - - > the units digit is 3 ; 3 ^ 2 = 9 - - > the units digit is 9 ; 3 ^ 3 = 27 - - > the units digit is 7 ; 3 ^ 4 = 81 - - > the units digit is 1 ; 3 ^ 5 = 243 - - > the units digit is 3 again ; . . . so , as you can see the units digit repeats in blocks of 4 : { 3 , 9 , 7 , 1 } , { 3 , 9 , 7 , 1 } , . . . now , since 356 = 356 + 0 = ( multiple of 4 ) + 0 , then the units digit of 3 ^ 356 will be thus 1 . answer : a ." | a = 3 + 4
b = a * 10
c = 73 - b
d = 356 / 4
e = math.floor(d)
f = e * 4
g = 356 - f
h = c ** g
i = 2 * 10
j = h - i
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a ) 1 / 6 , b ) 4 / 120 , c ) 29 / 30 , d ) 1 / 9 , e ) none of these | c | divide(subtract(choose(10, 3), choose(subtract(10, 6), 3)), choose(10, 3)) | a classroom has 3 electric lamps . from a collocation of 10 electric bulbs of which 6 are good , 3 are selected at random and put in the lamps . find the probability that the room is lighted . | explanation : out of 10 electric bulbs , 3 are selected in 10 c 3 ways . so , number of possible outcomes = t = 10 c 3 = 120 . the room is lighted i . e at least one fo the 3 bulbs chosen at random is a good bulb . wherever ' at least one ' is present , it is easier to do the problem using complement of the event ( i . e room is dark and all 3 bulbs are bad ) . p ( room is dark ) = f / t = p ( none of the bulbs is good ) = > 4 c 3 / 120 . [ since 4 bulbs are bad ] . p ( room is lighted ) = 1 - ( 4 c 3 / 120 ) . [ room is lighted if at least one bulb is good ] . or p ( at least one ) = 1 - ( 4 / 120 ) . = 29 / 30 . answer : c | a = math.comb(10, 3)
b = 10 - 6
c = math.comb(b, 3)
d = a - c
e = math.comb(10, 3)
f = d / e
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a ) 108 % , b ) 105 % , c ) 111 % , d ) 112 % , e ) 114 % | c | multiply(divide(const_100, 90), const_100) | if a population of women in a town is 90 % of men . what is the population of men as a % of population of women ? | we ' re told that the number of women in a town is equal to 90 % of the number of men in that town . if . . . . men = 10 women = 9 we ' re asked for the number of men , as a percentage of the number of women . m / w = 10 / 9 = 1 1 / 9 = 1.111111 = 111 % answer : c | a = 100 / 90
b = a * 100
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a ) 7.18 , b ) 7.12 , c ) 7.3 , d ) 7.45 , e ) 7.51 | d | divide(add(135, 165), multiply(add(80, 65), const_0_2778)) | two trains 135 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 135 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.45 answer : d" | a = 135 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
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a ) 480 , b ) 239 , c ) 270 , d ) 282 , e ) 112 | a | multiply(divide(1344, add(add(add(2, 3), 4), 5)), 5) | 4 numbers are in the ratio 2 : 3 : 4 : 5 add up to give a sum of 1344 . find the biggest number . | answer : a | a = 2 + 3
b = a + 4
c = b + 5
d = 1344 / c
e = d * 5
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a ) rs 200 , b ) rs 300 , c ) rs 400 , d ) rs 500 , e ) rs 600 | c | multiply(divide(const_4, add(const_4, add(const_1, const_2))), 700) | rs . 700 is divided among a , b , c so that a receives half as much as b and b half as much as c . then c ' s share is : | let c = x . then b = x / 2 and a = x / 4 a : b : c = 1 : 2 : 4 . c ' s share rs . [ ( 4 / 7 ) x 700 ) = 400 answer : c | a = 1 + 2
b = 4 + a
c = 4 / b
d = c * 700
|
a ) 25 days , b ) 88 days , c ) 21 days , d ) 11 days , e ) 24 days | e | multiply(divide(multiply(8, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 8 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 8 x = 1 / 24 = > 24 days answer : e" | a = 2 + 1
b = 8 * a
c = b / 2
d = c * 2
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a ) 19 , b ) 17 , c ) 32 , d ) 9 , e ) 8 | c | add(const_12, multiply(5, const_4)) | find the greatest number that will divide 150 , 230 and 175 leaving 50 , 5 and 25 as remainders respectively | explanation : answer will be hcf of ( 150 - 50 , 230 - 5 , 175 - 25 ) hcf of ( 100 , 225 , 150 ) = 32 option c | a = 5 * 4
b = 12 + a
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a ) 16.8 , b ) 17.8 , c ) 15.9 , d ) 13.2 , e ) 12.4 | a | divide(add(110, 170), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ? | "d = 110 + 170 = 280 m s = 60 * 5 / 18 = 50 / 3 t = 280 * 3 / 50 = 16.8 sec answer a" | a = 110 + 170
b = 60 * const_0_2778
c = a / b
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a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum x of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum x = average * total count = 200 * 101 = 20,200 answer is b" | a = 301 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
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a ) 24 miles , b ) 25 miles , c ) 26 miles , d ) 27 miles , e ) 28 miles | d | multiply(divide(3, const_4), add(27, 9)) | in track last week , the boys ran 27 laps . the girls ran 9 more laps . each lap is a 3 - fourth of a mile . how many miles did the girls run ? | the girls ran 27 + 9 = 36 laps . 36 x 3 / 4 = 108 / 4 , which reduces to 27 . the girls ran 27 miles correct answer d | a = 3 / 4
b = 27 + 9
c = a * b
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a ) $ 30.60 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 60.24 | e | add(50.20, divide(multiply(50.20, 20), const_100)) | if tim had lunch at $ 50.20 and he gave 20 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.20 = ( 20 / 100 ) * 50.20 = $ 10.04 total spent 50.20 + 10.04 = $ 60.24 correct answer e" | a = 50 * 20
b = a / 100
c = 50 + 20
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a ) 80 , b ) 120 , c ) 160 , d ) 640 , e ) 760 | d | multiply(multiply(2, 8), multiply(multiply(2, 4), 5)) | if the operation € is defined for all x and y by the equation x € y = 2 * x * y , then 8 € ( 4 € 5 ) = | "working inside out , ( 4 € 5 ) = 2 * 4 * 5 = 40 8 € 40 = 2 * 8 * 40 = 640 hence , answer is d" | a = 2 * 8
b = 2 * 4
c = b * 5
d = a * c
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a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,500 , e ) s . 7,200 | e | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 24,000 earned at the end of 2 years from the startingof the business ? | "a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 24000 x 3 / 10 = rs . 7,200 . e" | a = 4 + 1
b = a * 4
c = 3 * 4
d = 2 * c
e = b * d
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a ) 87111 , b ) 81007 , c ) 81008 , d ) 81066 , e ) 81022 | a | add(add(64000, multiply(divide(1, 6), 64000)), multiply(divide(1, 6), add(64000, multiply(divide(1, 6), 64000)))) | every year an amount increases by 1 / 6 th of itself . how much will it be after two years if its present value is rs . 64000 ? | "64000 * 7 / 6 * 7 / 6 = 87111 answer : a" | a = 1 / 6
b = a * 64000
c = 64000 + b
d = 1 / 6
e = 1 / 6
f = e * 64000
g = 64000 + f
h = d * g
i = c + h
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a ) 400 , b ) 440 , c ) 550 , d ) 625 , e ) 700 | b | multiply(divide(600, 11), 8) | a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 305 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 305 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 605 a + b + c = 55 per hour so for eight hrs = 55 * 8 = 440 = answer = b | a = 600 / 11
b = a * 8
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(10, const_4) | how many odd numbers between 10 and 1200 are the squares of integers ? | "the numbers are the squares of 5 , 7 , 9 , . . . , 33 which includes 15 numbers . the answer is d ." | a = 10 + 4
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a ) 25 , b ) 30 , c ) 29 , d ) 12 , e ) 20 | b | divide(subtract(subtract(subtract(961, multiply(16, 5)), multiply(5, 45)), multiply(7, 70)), 5) | alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 5 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 961 . find the cost of each ice - cream cup ? | "explanation : let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 961 96 + 225 + 490 + 5 x = 961 5 x = 150 = > x = 30 . answer : b" | a = 16 * 5
b = 961 - a
c = 5 * 45
d = b - c
e = 7 * 70
f = d - e
g = f / 5
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a ) 12 : 8 , b ) 12 : 4 , c ) 13 : 8 , d ) 14 : 5 , e ) 13 : 5 | c | divide(multiply(subtract(15, const_1), 2), multiply(3, 2)) | 300 liters of a mixture contains milk and water in the ratio 3 : 2 if 15 liters of this mixture be replaced by 15 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 300 liters if mix = 300 * 3 / 5 = 180 liters quantity of milk in 315 liters of new mix = 180 + 15 = 195 liters quantity of water in it = 315 - 195 = 120 liters ratio of milk and water in new mix = 90 : 20 = 13 : 8 answer isc" | a = 15 - 1
b = a * 2
c = 3 * 2
d = b / c
|
a ) 342 km , b ) 352 km , c ) 642 km , d ) 742 km , e ) 492 km | e | divide(add(add(30, multiply(2, 12)), 30), 2) | a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 12 hours of the journey . | "the total distance covered by the car in the first 12 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 = sum of 12 terms in ap whose first term is 30 and last term is 52 = 12 / 2 [ 30 + 52 ] = 492 km . answer : e" | a = 2 * 12
b = 30 + a
c = b + 30
d = c / 2
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a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | e | multiply(subtract(divide(60, const_2), divide(multiply(60, 25), const_100)), const_2) | a chemical supply company has 60 liters of a 25 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 25 % hno 3 solution means hno 3 = 15 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 15 + x = 50 % of ( 60 + x ) or x = 30 . hence , e" | a = 60 / 2
b = 60 * 25
c = b / 100
d = a - c
e = d * 2
|
a ) 60 m , b ) 40 m , c ) 20 m , d ) 10 m , e ) 50 m | c | subtract(500, divide(multiply(subtract(500, 140), 4), 3)) | in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 140 m . then , a wins by | "solution to reach the winning post a will have to cover a distance of ( 500 - 140 ) m , i . e , 360 m . while a covers 3 m , b covers 4 m . while a covers 360 m , b covers ( 4 / 3 x 360 ) m = 480 m thus , when a reaches the winning post , b covers 480 m and therefore remains 20 m behind . ∴ a wins by 20 m . answer c" | a = 500 - 140
b = a * 4
c = b / 3
d = 500 - c
|
a ) 10 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | d | divide(multiply(3, 2), subtract(3, 2)) | pipe a can fill a tank in 2 hours . due to a leak at the bottom , it takes 3 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 2 - 1 / x = 1 / 3 = > 1 / x = 1 / 2 - 1 / 3 = ( 3 - 2 ) / 6 = 1 / 6 = > x = 6 . answer : d" | a = 3 * 2
b = 3 - 2
c = a / b
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a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | c | add(40, divide(subtract(752, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 752 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 752 - 640 = 112 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 112 / 28 = 28 / 7 = 4 total hrs = 40 + 4 = 44 answer c 44" | a = 16 * 40
b = 752 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
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a ) 7.2 kg . , b ) 10.8 kg . , c ) 12.4 kg . , d ) 6.1 kg , e ) none | d | divide(multiply(6, 12.2), 12) | if the weight of 12 meters long rod is 12.2 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 12 m long rod = 12.2 kg ∴ weight of 1 m long rod = 12.2 / 12 kg ∴ weight of 6 m long rod = 12.2 x 6 / 12 = 6.1 kg option : d" | a = 6 * 12
b = a / 12
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a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 4 : 3 | e | divide(multiply(45000, const_12), multiply(45000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 45000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 × 12 : 45000 × 9 = 45 × 12 : 45 × 9 = 3 × 12 : 2 × 9 = 4 : 3 answer is e ." | a = 45000 * 12
b = 4 + 3
c = 45000 * b
d = a / c
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a ) 2 / 4 , b ) 6 / 78 , c ) 4 / 7 , d ) 3 / 5 , e ) 1 / 7 | e | multiply(divide(divide(3, 7), divide(6, 5)), divide(2, 5)) | find the fraction which has the same ratio to 2 / 5 that 3 / 7 has to 6 / 5 | "p : 2 / 5 = 3 / 7 : 6 / 5 as the product of the means is equal to the product of the extremes . p * 6 / 5 = 2 / 5 * 3 / 7 p * 6 / 5 = 6 / 35 p = 1 / 7 = > p = 1 / 7 answer : e" | a = 3 / 7
b = 6 / 5
c = a / b
d = 2 / 5
e = c * d
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a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 150 | b | divide(20, subtract(subtract(const_1, inverse(25)), divide(const_1, const_2))) | a student traveled 25 percent of the distance of the trip alone , continued another 20 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.25 x + 20 miles + 0.5 x = x 20 miles = 0.25 x x = 80 miles the answer is b ." | a = 1/(25)
b = 1 - a
c = 1 / 2
d = b - c
e = 20 / d
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a ) 75 , b ) 62.6 , c ) 87 , d ) 165 , e ) 11 | b | divide(add(add(add(add(72, 60), 35), 62), 84), add(const_2, const_3)) | david obtained 72 , 60 , 35 , 62 and 84 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | explanation : average = ( 72 + 60 + 35 + 62 + 84 ) / 5 = 313 / 5 = 62.6 . answer : b | a = 72 + 60
b = a + 35
c = b + 62
d = c + 84
e = 2 + 3
f = d / e
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a ) 16000 , b ) 25000 , c ) 10000 , d ) 18000 , e ) 17000 | c | multiply(divide(subtract(12000, divide(multiply(12000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 12000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 12000 - 10 / 100 ( 12000 ) = 12000 - 1200 = $ 10800 = > 1.08 x = 10800 = > x = 10000 c" | a = 12000 * 10
b = a / 100
c = 12000 - b
d = 100 + 8
e = c / d
f = e * 100
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a ) 6 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | multiply(divide(1.80, add(0.35, 0.25)), const_2) | a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 1.80 , what total number of pens and pencils did the customer purchase ? | "answer : algebraically , the question looks like this : 2.5 = 0.35 x + 0.25 y there are three possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 and 3 pencils for $ 0.25 each . that ' s 3 pens and 3 pencils for a total of 6 pens and pencils . choice ( a ) is correct ." | a = 0 + 35
b = 1 / 80
c = b * 2
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a ) 20 % , b ) 29 % , c ) 60 % , d ) 27 % , e ) 28 % | c | subtract(divide(subtract(const_100, 20), divide(1, 2)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 1 / 2 rd of that price , there would be a loss of 20 % ? | "sp 2 = 1 / 2 sp 1 cp = 100 sp 2 = 80 1 / 2 sp 1 = 80 sp 1 = 160 100 - - - 160 = > 60 % answer : c" | a = 100 - 20
b = 1 / 2
c = a / b
d = c - 100
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a ) 5 / 8 , b ) 7 / 8 , c ) 6 / 8 , d ) 2 / 9 , e ) 1 / 7 | a | divide(10, add(10, subtract(add(const_3, const_4), const_1))) | williams works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 10 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 10 x total tips for all months = 6 * x + 10 x = 16 x tips for august / total tips for all months = 10 x / 16 x = 5 / 8 answer : a" | a = 3 + 4
b = a - 1
c = 10 + b
d = 10 / c
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a ) . 1 , b ) 0.3 , c ) 0.6 , d ) 2 , e ) 1.6 | c | divide(21, multiply(divide(77, const_3600), multiply(subtract(add(77, 77), const_4), const_10))) | how many seconds will it take for a car that is traveling at a constant rate of 77 miles per hour to travel a distance of 21 yards ? ( 1 mile = 1,160 yards ) | "speed = 77 miles / hr = 37.64 yard / s distance = 21 yards time = distance / speed = 21 / 37.64 = 0.6 sec ans - c" | a = 77 / 3600
b = 77 + 77
c = b - 4
d = c * 10
e = a * d
f = 21 / e
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a ) 228 , b ) 320 , c ) 287 , d ) 480 , e ) 811 | b | divide(multiply(400, const_100), add(const_100, 25)) | by selling an article at rs . 400 , a profit of 25 % is made . find its cost price ? | "sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 25 ) ] = 400 * [ 100 / 125 ] = rs . 320 answer : b" | a = 400 * 100
b = 100 + 25
c = a / b
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a ) 800 , b ) 1,250 , c ) 80,000 , d ) 12,000 , e ) 80,000 | c | multiply(divide(multiply(divide(multiply(4, 10), 50), power(10, const_4)), const_1000), 4) | a certain galaxy is known to comprise approximately 4 x 10 ^ 12 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | "total no . of stars on galaxy = 4 * 10 ^ 12 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 4 * 10 ^ 12 / 50 * 10 ^ 6 = 40 * 10 ^ 4 / 5 = 80000 therefore answer is c" | a = 4 * 10
b = a / 50
c = 10 ** 4
d = b * c
e = d / 1000
f = e * 4
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a ) 48 , b ) 96 , c ) 24 , d ) none of these , e ) can not be determined | a | divide(multiply(divide(multiply(50, 192), const_100), 12.5), const_100) | 12.5 % of 192 = 50 % of ? | "answer let 12.5 % of 192 = 50 % of a , ⇒ ( 12.5 x 192 ) / 100 = ( 50 x a ) / 100 ∴ a = ( 12.5 x 192 ) / 50 = 48 correct option : a" | a = 50 * 192
b = a / 100
c = b * 12
d = c / 100
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a ) rs 45 , b ) rs 25 , c ) rs 40 , d ) rs 39 , e ) rs 36 | e | multiply(divide(const_100, 25), 9) | a 9 % stock yields 25 % . the market value of the stock is : | "explanation : for an income of rs . 25 , investment = rs . 100 . for an income of rs 9 , investment = rs . 100 / 25 x 9 = rs 36 market value of rs . 100 stock = rs 36 answer is e" | a = 100 / 25
b = a * 9
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a ) 1245 , b ) 1567 , c ) 1678 , d ) 1335 , e ) 1635 | e | multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 asquotient and the 15 as remainder ? | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e" | a = 1365 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) 5 , b ) 8 , c ) 6 , d ) 4 , e ) none of these | a | multiply(subtract(const_1, multiply(divide(const_1, 12), 3)), divide(multiply(12, 15), power(3, const_3))) | ‘ a ’ completes a work in 12 days . ‘ b ’ completes the same work in 15 days . ‘ a ’ started working alone and after 3 days b joined him . how many days will they now take together to complete the remaining work ? | work done by ‘ a ’ in 3 days = 1 ⁄ 12 × 3 = 1 ⁄ 4 ∴ remaining work = 1 - 1 ⁄ 4 = 3 ⁄ 4 work done by a and b together = 12 × 15 / 27 = 20 / 3 ∴ remaining work done by a and b together in = 3 ⁄ 4 × 20 ⁄ 3 = 5 days answer a | a = 1 / 12
b = a * 3
c = 1 - b
d = 12 * 15
e = 3 ** 3
f = d / e
g = c * f
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a ) 8 , b ) 9 , c ) 10 , d ) 16 , e ) 17 | d | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 50), 5) | the sum of ages of 5 children born 3 years different each is 50 years . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 5 x = 20 x = 4 . x + 12 = 4 + 12 = 16 d" | a = 3 + 4
b = 3 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 50
g = f / 5
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(multiply(180, 2), multiply(40, 2)), subtract(add(add(40, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 2 ) trees yield 40 nuts per year , x trees yield 120 nuts per year and ( x – 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) × 40 + x × 120 + ( x − 2 ) × 180 / ( x + 2 ) + x + ( x − 2 ) = 100 ⇒ 340 x − 280 / 3 x = 100 ⇒ 40 x = 280 ⇒ x = 7 answer e" | a = 180 * 2
b = 40 * 2
c = a - b
d = 40 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
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a ) 1 , b ) x = - 4 , c ) - 3 , d ) - 5 , e ) 5 | b | subtract(3, 1) | find value for x from given equation ? x + 1 = - 3 | "1 . subtract 1 from both sides : x + 1 - 1 = - 3 - 1 2 . simplify both sides : x = - 4 b" | a = 3 - 1
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a ) 50 , b ) 40 , c ) 25 , d ) 36 , e ) 45 | d | divide(multiply(150, 390), multiply(25, 65)) | rectangular tile each of size 25 cm by 65 cm must be laid horizontally on a rectangular floor of size 150 cm by 390 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 25 * 65 = 1625 area of floor = 150 * 390 = 58500 no of tiles = 58500 / 1625 = 36 so , the no of tile = 36 answer : d" | a = 150 * 390
b = 25 * 65
c = a / b
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a ) 13 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(multiply(divide(1, 6), subtract(1, divide(1, 4))), add(multiply(divide(1, 6), subtract(1, divide(1, 4))), subtract(1, divide(1, 6)))), const_100) | of the 3,600 employees of company x , 1 / 6 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 6 are clerical out of 3600 so 600 are clerical 600 reduced by 1 / 4 is 1200 * 1 / 4 so it reduced 150 people , so there is 450 clerical people left but since 150 people left , it also reduced from the total of 3600 so there are 3450 people total since 450 clerical left / 3450 people total you get ( a ) 13 % answer : a" | a = 1 / 6
b = 1 / 4
c = 1 - b
d = a * c
e = 1 / 6
f = 1 / 4
g = 1 - f
h = e * g
i = 1 / 6
j = 1 - i
k = h + j
l = d / k
m = l * 100
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a ) 3327 , b ) 3237 , c ) 3337 , d ) 2337 , e ) none of these | e | subtract(13200, 9886) | 9886 + x = 13200 , then x is ? | "answer x = 13200 - 9886 = 3314 option : e" | a = 13200 - 9886
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a ) 1 , b ) 16 , c ) 26 , d ) 52 , e ) 1014 | b | multiply(add(subtract(13, const_10), const_2), add(subtract(13, const_10), const_2)) | if x is a sum of all even integers on the interval 13 . . . 45 and y is their number , what is the gcd ( x , y ) ? | "x = 14 + 16 + . . . + 44 = ( largest + smallest ) / 2 * ( # of terms ) = ( 14 + 44 ) / 2 * 16 = 29 * 16 . gcd of 16 and 29 * 16 is 16 . answer : b ." | a = 13 - 10
b = a + 2
c = 13 - 10
d = c + 2
e = b * d
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a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) none of them | a | divide(add(6, 6), divide(33, const_60)) | in a stream running at 2 kmph , a motor boat goes 6 km upstream and back again to the starting point in 33 minutes . find the speed of the motorboat in still water . | let the speed of the motarboat in still water be x kmph . then , 6 / x + 2 + 6 / x - 2 = 33 / 60 11 x 2 - 240 x - 44 = 0 11 x 2 - 242 x + 2 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 . answer is a | a = 6 + 6
b = 33 / const_60
c = a / b
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a ) 60 , b ) 62 , c ) 64 , d ) 66 , e ) 68 | d | divide(1, subtract(divide(1, add(5, divide(1, 2))), divide(1, const_4))) | a cistern is filled by a tap in 5 1 / 2 hours . due to leak in the bottom of the cistern , it takes half an hour longer to fill the cistern . if the cistern is full how many hours will it take the leak to empty it ? | "filling rate - leak rate = net rate 1 / 5.5 - leak rate = 1 / 6 leak rate = 2 / 11 - 1 / 6 = 1 / 66 the answer is d ." | a = 1 / 2
b = 5 + a
c = 1 / b
d = 1 / 4
e = c - d
f = 1 / e
|
a ) rs 1.10 , b ) rs 11 , c ) rs 16.50 , d ) rs 27.50 , e ) none of these | d | subtract(multiply(15, add(10, 15)), add(multiply(15, 14.5), multiply(10, 13))) | ajay bought 15 kg of dal at the rate of rs 14.50 per kg and 10 kg at the rate of rs 13 per kg . he mixed the two and sold the mixture at the rate of rs 15 per kg . what was his total gain in this transaction ? | explanation : cost price of 25 kg = rs . ( 15 x 14.50 + 10 x 13 ) = rs . 347.50 . sell price of 25 kg = rs . ( 25 x 15 ) = rs . 375 . profit = rs . ( 375 — 347.50 ) = rs . 27.50 . answer : d | a = 10 + 15
b = 15 * a
c = 15 * 14
d = 10 * 13
e = c + d
f = b - e
|
a ) 24 hours , b ) 21 hr 30 min , c ) 20 hr 50 min , d ) 18 hr 15 min , e ) 22 hrs | c | add(multiply(add(add(10, 2), subtract(10, 2)), 100), multiply(subtract(add(divide(100, add(10, 2)), divide(100, subtract(10, 2))), add(add(10, 2), subtract(10, 2))), const_60)) | speed of a boat in standing water is 10 kmph and speed of the stream is 2 kmph . a man can rows to a place at a distance of 100 km and comes back to the starting point . the total time taken by him is ? | speed upstream = 8 kmph speed downstream = 12 kmph total time taken = 100 / 8 + 100 / 12 = 20 hours 50 min answer is c | a = 10 + 2
b = 10 - 2
c = a + b
d = c * 100
e = 10 + 2
f = 100 / e
g = 10 - 2
h = 100 / g
i = f + h
j = 10 + 2
k = 10 - 2
l = j + k
m = i - l
n = m * const_60
o = d + n
|
a ) 65 , b ) 77 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 95), add(const_1, const_4)) | shekar scored 76 , 65 , 82 , 67 and 95 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | explanation : average = ( 76 + 65 + 82 + 67 + 95 ) / 5 = 385 / 5 = 77 hence average = 77 answer : b | a = 76 + 65
b = a + 82
c = b + 67
d = c + 95
e = 1 + 4
f = d / e
|
a ) 20 m , b ) 16 m , c ) 11 m , d ) 22 m , e ) 15 m | d | multiply(divide(subtract(25, 20), 25), 110) | if in a race of 110 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 110 * 5 ) / 25 = 22 m hence , a beats b by 22 m . answer d" | a = 25 - 20
b = a / 25
c = b * 110
|
a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 244 | e | divide(195, divide(subtract(const_100, 20), const_100)) | a small company reduced its faculty by approximately 20 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 20 % reduction in employees number is . 80 x but we are given . 80 x = 195 x = 244 so the original number of employees is 244 correct answer - e" | a = 100 - 20
b = a / 100
c = 195 / b
|
a ) 4 : 6 , b ) 4 : 9 , c ) 5 : 3 , d ) 4 : 1 , e ) 4 : 2 | c | multiply(divide(15, const_100), 9) | a part of certain sum of money is invested at 9 % per annum and the rest at 15 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "15 : 9 = 5 : 3 answer : c" | a = 15 / 100
b = a * 9
|
a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | b | divide(43, add(multiply(7, divide(43, 3)), multiply(5, divide(43, 4)))) | if 3 men or 4 women can reap a field in 43 days how long will 7 men and 5 women take to reap it ? | "explanation : 3 men reap 1 / 43 field in 1 day 1 man reap 1 / ( 3 x 43 ) 4 women reap 1 / 43 field in 1 day 1 woman reap 1 / ( 43 x 4 ) 7 men and 5 women reap ( 7 / ( 3 x 43 ) + 5 / ( 4 x 43 ) ) = 1 / 12 in 1 day 7 men and 5 women will reap the field in 12 days answer : option b" | a = 43 / 3
b = 7 * a
c = 43 / 4
d = 5 * c
e = b + d
f = 43 / e
|
a ) 630 , b ) 633 , c ) 150 , d ) 140 , e ) 627 | e | subtract(add(multiply(divide(4590, 510), divide(105, 3)), multiply(divide(4590, 510), divide(105, 3))), const_3) | the smallest number when increased by ` ` 3 ` ` is exactly divisible by 510 , 4590 , and 105 is : | lcm = 630 630 - 3 = 627 answer : e | a = 4590 / 510
b = 105 / 3
c = a * b
d = 4590 / 510
e = 105 / 3
f = d * e
g = c + f
h = g - 3
|
['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(multiply(3, 2), power(divide(multiply(add(3, 2), const_2), const_4), const_2)) | if the perimeter of square region d and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of d | we know perimeter of a square ( pd ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration 2 k : 3 k = 5 k ; we can assume such a number ) therefore , pd = pr = 40 area of the square = 100 sq . units we know 2 ( length + breadth ) = 40 i . e . length + breadth = 20 ( or 5 k = 20 given that l : b ( or b : l ) = 2 : 3 ) therefore length = 8 , breath = 12 area of the rectangle = 8 * 12 = 96 sq . units question asked = area of the rectangle : area of the square = 96 : 100 = = > 24 : 25 = b | a = 3 * 2
b = 3 + 2
c = b * 2
d = c / 4
e = d ** 2
f = a / e
|
a ) 4 / 9 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 9 , e ) 6 / 9 | c | divide(divide(5, 3), add(divide(5, 3), const_1)) | a waiter ' s salary consists of his salary and tips . during one week his tips were 5 / 3 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 5 / 3 = s * 8 / 3 tips = s * 8 / 3 fraction of his income came from tips = ( s * 5 / 3 ) / ( s * 8 / 3 ) = 5 / 8 answer : c" | a = 5 / 3
b = 5 / 3
c = b + 1
d = a / c
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 3.33 , e ) 2.5 | d | multiply(subtract(5, const_2.0), 5) | if ( c - a ) / ( c - b ) = 3 , then ( 5 b - 5 a ) / ( c - a ) = | "let ' s say c = 7 , b = 5 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 3.33 answer d ( hopefully ) ) )" | a = 5 - 2
b = a * 5
|
a ) $ 36.35 , b ) $ 26.35 , c ) $ 16.35 , d ) $ 46.35 , e ) $ 56.35 | a | add(10, subtract(50, subtract(add(5, add(8, 15.65)), 5))) | rachel and sarah spent $ 8 for gasoline , $ 15.65 for their lunch , and $ 5 apiece for gifts for grandma . grandma gave each of them $ 10 . if the girls left home with a total of $ 50 , how much do they have for the return trip ? | $ 50 + $ 20 = $ 70 what they left with plus $ 10 to each $ 8 + $ 15.65 + $ 10 = $ 33.65 what they spent ( gifts are $ 5 + $ 5 ) $ 70 - $ 33.65 = $ 36.35 left correct answer a | a = 8 + 15
b = 5 + a
c = b - 5
d = 50 - c
e = 10 + d
|
a ) 69 . , b ) 73 . , c ) 81 , d ) 65 , e ) 108 | d | subtract(subtract(420, 320), subtract(250, divide(multiply(250, 86), const_100))) | in the fifth grade at parkway elementary school there are 420 students . 320 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "total students = 420 boys = 320 , girls = 100 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 100 - 35 = 65 . correct option : d" | a = 420 - 320
b = 250 * 86
c = b / 100
d = 250 - c
e = a - d
|
a ) 5000000 , b ) 4000000 , c ) 1000000 , d ) 6000000 , e ) none | c | multiply(divide(500000, 2), 4) | p and q invested in a shop . the profits were divided in the ratio of 2 : 4 respectively . if p invested rs . 500000 , the amount invested by q is : | "suppose q invested rs . y . then , 500000 / y = 2 / 4 or y = [ 500000 x 4 / 2 ] = 1000000 . answer c" | a = 500000 / 2
b = a * 4
|
a ) 1212120010 , b ) 1212120000 , c ) 1212120011 , d ) 1212121010 , e ) 1212121000 | b | multiply(divide(1939392, 625), const_100) | 1939392 * 625 | "explanation : trick : when multiplying with 5 n 5 n then put n zeros to the right of multiplicand and divide the number with 2 n 2 n so using this we can solve this question in much less time . 1939392 × 5 ( 4 ) = 1939392000016 = 1212120000 option b" | a = 1939392 / 625
b = a * 100
|
a ) 3 , b ) 5 , c ) 6 , d ) 11 , e ) 16 | b | subtract(8, divide(24, 8)) | if ( m - 8 ) is a factor of m ^ 2 - qm - 24 , then q = | "( m - 8 ) ( m - a ) = m ^ 2 - qm - 24 a = - 3 q = 8 + a = 5 = b" | a = 24 / 8
b = 8 - a
|
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