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 ) 40 sec , b ) 50 sec , c ) 44 sec , d ) 49 sec , e ) 60 sec | c | divide(410, multiply(subtract(45, 140), const_0_2778)) | a train 410 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 410 + 140 = 550 m required time = 550 * 2 / 25 = 44 sec answer : c" | a = 45 - 140
b = a * const_0_2778
c = 410 / b
|
a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | c | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 4)) | the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 4 of that total . what was the price of the basic computer ? | "let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 4 of that new total price : p = 1 / 4 * $ 3,000 = $ 750 . plug this value in the first equation : c + 750 = $ 2,500 - - > c = $ 1,750 answer : c ." | a = const_0_25 * 1000
b = a * 10
c = const_0_25 * 1000
d = c * 10
e = 500 + d
f = e / 4
g = b - f
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a ) 0.004 , b ) 0.08333 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(3, 0.03), divide(3, 0.37))) | 3 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 3 / 36 = . 08 something answer ( b )" | a = 3 / 0
b = 3 / 0
c = a + b
d = 1/(c)
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a ) 16 , b ) 14 , c ) 18 , d ) 6 , e ) 9 | e | divide(power(9, 2), 9) | n ^ ( n / 2 ) = 6 is true when n = 6 in the same way what is the value of n if n ^ ( n / 2 ) = 9 ? | n ^ ( n / 2 ) = 9 apply log n / 2 logn = log 9 nlogn = 2 log 9 = log 9 ^ 2 = log 81 logn = log 81 now apply antilog n = 81 / n now n = 9 . answer : e | a = 9 ** 2
b = a / 9
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a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | d | divide(add(58, 26), const_2) | y and z are in a relay race . y runs the first leg of the course in 58 seconds . z runs the second leg of the course in 26 seconds . what was the average time they took to run a leg of the course ? | ( 58 seconds + 26 seconds ) / 2 = 42 seconds correct option is : d | a = 58 + 26
b = a / 2
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a ) 3 / 8 , b ) 1 / 2 , c ) 9 / 16 , d ) 5 / 7 , e ) 3 / 4 | c | divide(add(const_1, divide(subtract(multiply(divide(5, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 5 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the median of set r ? | the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ) , the median is given to be 3 / 4 * b : ( a + b ) / 2 = ( 3 / 4 ) * b a = b / 2 for set q , consisting of numbers ( b , b + 1 , . . . , c ) , the median is given to be 5 / 8 * c : ( b + c ) / 2 = ( 5 / 8 ) * c b = ( 1 / 4 ) * c for set r , consisting of numbers ( a , a + 1 , . . . c ) , the median needs to be found : a = b / 2 = ( 1 / 4 * c ) / 2 = ( 1 / 8 ) * c median = ( a + c ) / 2 = ( 1 / 8 * c + c ) / 2 = ( 9 / 8 ) * c / 2 = ( 9 / 16 ) * c ( answer c ) | a = 5 / 8
b = a * 2
c = b - 1
d = 3 / 4
e = d * 2
f = e - 1
g = 1 / f
h = c / g
i = 1 + h
j = i / 2
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a ) 151 , b ) 149 , c ) 152 , d ) 148 , e ) 180.67 | e | divide(add(multiply(30, 180), subtract(155, 135)), 30) | the mean of 30 values was 180 . it was detected on rechecking that one value 155 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 180 Γ 30 β 135 + 155 / 30 = 5400 β 135 + 155 / 30 = 5420 / 30 = 180.67 answer e" | a = 30 * 180
b = 155 - 135
c = a + b
d = c / 30
|
['a ) 0', 'b ) 2', 'c ) 6', 'd ) 8', 'e ) 12'] | a | min(min(15, 17), 20) | three interviewers , x , y , and z are interviewing 40 applicants . only with three interviewers ' admission can an applicant be admitted . if interviewer x admitted 15 applicants , y admitted 17 applicants , and z admitted 20 applicants , at least how many applicants get the admission ? | if x admitted 15 are overlapping with y admission of 17 but z does not overlap with anybody . then no student will get nod from all the 3 . hence 0 student will get admission . answer : a | a = min(15)
b = min(a)
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a ) rs 14000 , b ) rs 15000 , c ) rs 16000 , d ) rs 18000 , e ) none of these | d | divide(multiply(6480, const_100), multiply(12, 3)) | a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 6480 interest . the principal amount borrowed by him was . | "explanation : s . i . = p Γ’ Λ β r Γ’ Λ β t / 100 = > p = s . i . Γ’ Λ β 100 / r Γ’ Λ β t = > p = 6480 Γ’ Λ β 100 / 12 Γ’ Λ β 3 = rs 18000 option d" | a = 6480 * 100
b = 12 * 3
c = a / b
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a ) 12 hours , b ) 111 hours , c ) 15 hours , d ) 14 hours , e ) 11 hours | a | inverse(subtract(inverse(3), inverse(add(3, const_1)))) | how long will take a leak at the bottom of a tank to empty it if it will take 3 hours to fill it without the leak , but due to the leak it will take one additional hour to be filled ? | part filled without leak in 1 hour = 1 / 3 part filled with leak in 1 hour = 1 / 4 work done by leak in 1 hour = 1 / 3 Γ’ Λ β 1 / 4 = 12 hours answer : a | a = 1/(3)
b = 3 + 1
c = 1/(b)
d = a - c
e = 1/(d)
|
a ) 120 , b ) 240 , c ) 360 , d ) 720 , e ) 1440 | c | multiply(factorial(const_3), divide(factorial(5), const_2)) | the security gate at a storage facility requires a 5 - digit lock code . if the lock code must consist only of digits from 1 through 7 , inclusive , with no repeated digits , and the first and last digits of the code must be even , how many lock codes are possible ? | x - x - x - x - x - - > there are 3 even digits from 1 to 7 , inclusive thus options for the first and the last x ' s are : 3 - x - x - x - 2 . other x ' s can take following values : 3 - 5 - 4 - 3 - 2 - - > 3 * 5 * 4 * 3 * 2 = 360 . answer : c . | a = math.factorial(3)
b = math.factorial(5)
c = b / 2
d = a * c
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a ) 65 , b ) 18 , c ) 20 , d ) 25 , e ) 33 | a | multiply(const_4.0, power(const_3, 5)) | if 5400 mn = k ^ 4 , where m , n , and k are positive integers , what is the least possible value of 5 m + n ? | "k ^ 4 = 5400 β mn k ^ 4 = 3 ^ 3.2 ^ 3.5 ^ 2 . mn in order to make rhs a perfect power of 4 , we require it to be multiplied by 3 , 2 & 5 ^ 2 mn = 3 . 2.5 ^ 2 mn = 150 = 10 * 15 ( least possible ) answer = 5 * 10 + 15 = 65 answer : a" | a = 3 ** 5
b = 4 * 0
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a ) 678 , b ) 944 , c ) 456 , d ) 323 , e ) 899 | b | multiply(59, 16) | the h . c . f . of two numbers is 59 and the other two factors of their l . c . m . are 13 and 16 . the larger of the two numbers is : | : explanation : clearly , the numbers are ( 59 x 13 ) and ( 59 x 16 ) . { \ color { blue } \ therefore } larger number = ( 59 x 16 ) = 944 . answer : b ) 944 | a = 59 * 16
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a ) 0.0006 , b ) 0.06 , c ) 0.6 , d ) 0.006 , e ) 6.0 | d | multiply(divide(0.012, 0.5), const_100) | 0.012 x 0.5 = ? | "12 x 5 = 60 . sum of decimal places = 4 0.012 x 0.5 = 0.006 answer : option d" | a = 0 / 12
b = a * 100
|
a ) 1,500 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,540 | a | divide(540, 540) | the rate of interest on a sum of money is 5 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | "explanation : i 1 = ( p x 3 x 5 ) / 100 = 3 p / 20 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 3 p / 20 + 4 p / 25 + p / 20 = 540 9 p / 25 = 540 p = 1500 answer : option a" | a = 540 / 540
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a ) 2,858 , b ) 18,667 , c ) 21,429 , d ) 35,000 , e ) 56,000 | e | divide(multiply(divide(divide(multiply(add(add(const_3, const_10), 15), power(const_100, const_2)), subtract(15, 10)), multiply(const_4, const_2)), const_3), const_1000) | a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 10 per item . if the manufacturer β s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ? | let the items manufactured or sold bex 130000 + 10 x = 15 x - 150000 5 x = 280000 x = 56000 ans : e | a = 3 + 10
b = a + 15
c = 100 ** 2
d = b * c
e = 15 - 10
f = d / e
g = 4 * 2
h = f / g
i = h * 3
j = i / 1000
|
a ) 8.25 , b ) 4.25 , c ) 9.25 , d ) 3.25 , e ) 2.25 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1700) | the difference between simple and compound interest on rs . 1700 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1700 * 10 * 1 ) / 100 = rs . 170 c . i . = [ 1700 * ( 1 + 5 / 100 ) 2 - 1700 ] = rs . 174.25 difference = ( 174.25 - 174 ) = rs . 4.25 . answer : b" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = 10 / 100
f = e + 1
g = d - f
h = g * 1700
|
a ) 1 : 3 , b ) 2 : 8 , c ) 1 : 6 , d ) 2 : 3 , e ) 2 : 5 | a | divide(divide(subtract(divide(const_100.0, const_2), const_10), const_2), add(divide(24, const_2), const_10)) | two whole numbers whose sum is 24 is the ratio | a ) 1 : 3 | a = 100 / 0
b = a - 10
c = b / 2
d = 24 / 2
e = d + 10
f = c / e
|
a ) 7450 , b ) 14900 , c ) 5625 , d ) 7500 , e ) none of these | c | divide(1800, divide(subtract(add(const_100, 16), subtract(const_100, 16)), const_100)) | vijay sells a cupboard at 16 % below cost price . had he got rs . 1800 more , he would have made a profit of 16 % . what is the cost price of the cupboard ? | explanation : cost price = 1800 / ( 0.16 + 0.16 ) = 1800 / 0.32 = rs . 5625 answer c | a = 100 + 16
b = 100 - 16
c = a - b
d = c / 100
e = 1800 / d
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a ) 28 : 37 , b ) 18 : 37 , c ) 16 : 53 , d ) 20 : 63 , e ) 15 : 48 | d | divide(multiply(4, 5), multiply(7, 9)) | if a : b = 4 : 7 and b : c = 5 : 9 , then a : c = ? | a / b = 4 / 7 and b / c = 5 / 9 a / b * b / c = 4 / 7 * 5 / 9 = 20 / 63 so a / c = 20 / 63 20 : 63 answer d | a = 4 * 5
b = 7 * 9
c = a / b
|
a ) 19 / 44 , b ) 5 / 44 , c ) 3 / 41 , d ) 13 / 44 , e ) 7 / 44 | e | divide(choose(5, 3), choose(add(7, 5), 3)) | there are 7 red balls and 5 blue balls in a jar . if 3 balls are selected from the jar , what is the probability that 3 balls selected are red balls ? | "probability = favorable outcomes / total outcomes probability = 7 c 3 / 12 c 3 or ( 7 / 12 ) * ( 6 / 11 ) * ( 5 / 10 ) = 7 / 55 answer option e" | a = math.comb(5, 3)
b = 7 + 5
c = math.comb(b, 3)
d = a / c
|
a ) 22.04 % , b ) 16.12 % , c ) 12.5 % , d ) 18.25 % , e ) 10 % | c | multiply(3, const_4) | on the first of the year , james invested x dollars at proudstar bank in an account that yields 3 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ? | per quarter , interest = 3 % so for a year , interest = 12 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 12 % answer = 12.5 % = answer : c | a = 3 * 4
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a ) rs . 12000 , b ) rs . 19500 , c ) rs . 18000 , d ) rs . 16500 , e ) rs . 17500 | d | multiply(800, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of rs . 800 per sq . meter . | "explanation : area = 5.5 Γ 3.75 sq . metre . cost for 11 sq . metre . = rs . 800 hence , total cost = 5.5 Γ 3.75 Γ 800 = 5.5 Γ 3000 = rs . 16500 answer : option d" | a = 5 * 5
b = 800 * a
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a ) 42 . , b ) 46 . , c ) 50 . , d ) 54 . , e ) 58 . | b | add(add(multiply(7, const_2), 2), add(multiply(7, multiply(const_2, 2)), 2)) | a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 7 with 2 students left over . what is the sum of the two smallest possible values of n ? | "n = 4 k + 1 = 7 j + 2 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 7 j + 2 . 1 , 5 , 9 = 7 ( 1 ) + 2 the next such number is 9 + 4 * 7 = 37 . 9 + 37 = 46 the answer is b ." | a = 7 * 2
b = a + 2
c = 2 * 2
d = 7 * c
e = d + 2
f = b + e
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['a ) 420', 'b ) 430', 'c ) 280', 'd ) 360', 'e ) 480'] | e | multiply(sqrt(divide(multiply(3200, 9), 8)), 8) | the breath of a rectangular landscape is 8 times its length . there is a playground in it whose area is 3200 square mtr & which is 1 / 9 rd of the total landscape . what is the breath of the landscape ? | sol . 8 x * x = 9 * 3200 x = 60 length = 8 * 60 = 480 e | a = 3200 * 9
b = a / 8
c = math.sqrt(b)
d = c * 8
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a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | divide(multiply(add(const_4, 2), 40), 20) | machine c takes 2 more hours than machine b to make 20 widgets . if working together , the machines can make 25 widgets in 3 hours , how long will it take machine c to make 40 widgets ? | i approached this one by plugging in numbers . . . started with c . if 40 are made in 8 hours , then 20 are made in 4 hours . so time of c is 4 , and time of b is 2 . rate together : 20 / 4 + 20 / 2 = 5 + 10 = 15 . so in 1 hour , together make 15 widgets . in 3 hours = 45 . way too much . we can eliminate right away c , b , and a - because b and c reduces the time - the total # of widgets made will be even higher . now between d and e - > try only one . . if it does n ' t work , then the other one is the answer . i picked e : 12 h to make 40 widgets , and 6 hours to make 20 . this is the time of c . time of b = 4 hours . 20 / 6 + 20 / 4 = 10 / 3 + 20 / 4 find lcm of 3 and 4 = 12 . multiply first by 4 , and second by 3 : 40 + 60 / 12 = 100 / 12 divide by 4 : 25 / 3 so this is the rate given . e is the correct answer | a = 4 + 2
b = a * 40
c = b / 20
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a ) $ 496.30 , b ) $ 536.00 , c ) $ 555.40 , d ) $ 574.90 , e ) $ 588.20 | b | divide(670, add(divide(25, const_100), const_1)) | jim is able to sell a hand - carved statue for $ 670 which was a 25 % profit over his cost . how much did the statue originally cost him ? | "670 = 1.25 * x x = 670 / 1.25 = 536 $ 536 , which is ( b ) ." | a = 25 / 100
b = a + 1
c = 670 / b
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a ) $ 260 , b ) $ 290 , c ) $ 320 , d ) $ 350 , e ) $ 380 | c | divide(224, divide(subtract(const_100, 30), const_100)) | sandy had $ 224 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 = 224 x = 320 the answer is c ." | a = 100 - 30
b = a / 100
c = 224 / b
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a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 4 | e | multiply(3, 1) | if ( 18 ^ a ) * 9 ^ ( 3 a β 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a β 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a β 1 ) = ( 2 ^ 4 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 4 = e" | a = 3 * 1
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a ) 24 % , b ) 16 % , c ) 11 % , d ) 7 % , e ) 5 % | c | divide(multiply(subtract(const_100, 72), multiply(add(divide(const_2, const_0_25), const_2), const_4)), const_100) | each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , forty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 72 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on are incandescent ? | let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.4 i + 0.8 f = 0.72 ( i + f ) 0.08 f = 0.32 i f = 4 i this means that for every 1 incandescent bulb , there are 4 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.4 i / ( 0.4 i + 0.8 f ) = 0.4 i / ( 0.4 i + 0.8 * 4 i ) = 0.4 i / 3.6 i = 1 / 9 which is about 11 % . the answer is c . | a = 100 - 72
b = 2 / const_0_25
c = b + 2
d = c * 4
e = a * d
f = e / 100
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a ) 1 : 2 , b ) 1 : 5 , c ) 4 : 7 , d ) 1 : 1 , e ) 1 : 8 | c | divide(divide(224, divide(224, 32)), divide(224, subtract(divide(224, 32), const_3))) | a motorcyclist goes from bombay to pune , a distance of 224 kms at an average of 32 kmph speed . another man starts from bombay by car 2 Β½ hours after the first , and reaches pune Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | "t = 224 / 32 = 7 h t = 7 - 3 = 4 time ratio = 7 : 4 speed ratio = 4 : 7 answer : c" | a = 224 / 32
b = 224 / a
c = 224 / 32
d = c - 3
e = 224 / d
f = b / e
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a ) 16 , b ) 42 , c ) 28 , d ) 54 , e ) 45 | a | inverse(divide(const_3, multiply(12, const_4))) | if a is thrice as fast as b and together can do a work in 12 days . in how many days a alone can do the work ? | "a β s one day β s work = 1 / x b β s one day β s work = 1 / 3 x a + b β s one day β s work = 1 / x + 1 / 3 x = 1 / 12 = 3 + 1 / 3 x = 4 / 3 x = 1 / 12 x = 12 * 4 / 3 = 16 answer : a" | a = 12 * 4
b = 3 / a
c = 1/(b)
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a ) 75 , b ) 80 , c ) 90 , d ) 95 , e ) 100 | c | divide(multiply(60, divide(multiply(45, 60), multiply(subtract(45, 60), 60))), divide(60, const_10)) | a car travels from point a to point b . the average speed of the car is 60 km / hr and it travels the first half of the trip at a speed of 45 km / hr . what is the speed of the car in the second half of the trip ? | "let d be the distance and let v be the speed in the second half . the total time = t 1 + t 2 d / 60 = d / 90 + ( d / 2 ) / v d / 180 = d / 2 v and so v = 90 km / hr the answer is c ." | a = 45 * 60
b = 45 - 60
c = b * 60
d = a / c
e = 60 * d
f = 60 / 10
g = e / f
|
a ) 5 : 2 , b ) 4 : 3 , c ) 3 : 2 , d ) 2 : 1 , e ) 4 : 1 | d | add(subtract(add(const_4.0, 3), add(5, 1)), const_1) | in what ratio p : q should the mixture p of milk and water in the ratio of 5 : 3 be mixed with another mixture q of milk and water in the ratio 1 : 3 so that the resultant mixture contains equal quantities of milk and water ? | "( 5 / 8 ) * p + ( 1 / 4 ) * q = ( 3 / 8 ) * p + ( 3 / 4 ) * q 2 p = 4 q p / q = 2 / 1 the answer is d ." | a = 4 + 0
b = 5 + 1
c = a - b
d = c + 1
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a ) 33 , b ) 77 , c ) 48 , d ) 84 , e ) 21 | d | divide(add(multiply(80, add(const_1, divide(10, const_100))), 80), const_2) | a person travels from p to q a speed of 80 km / hr and returns by increasing his speed by 10 % . what is his average speed for both the trips ? | "speed on return trip = 110 % of 80 = 88 km / hr . average speed of trip = 80 + 88 / 2 = 168 / 2 = 84 km / hr answer : d" | a = 10 / 100
b = 1 + a
c = 80 * b
d = c + 80
e = d / 2
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a ) 92500 , b ) 75230 , c ) 65990 , d ) 45280 , e ) 87500 | e | divide(910, divide(multiply(divide(4, 5), 1.3), const_100)) | a tempo is insured to an extent of 4 / 5 of its original value . if the premium on it at the rate of 1.3 % amounts to $ 910 , the original value of the tempo is ? | let the original value of the tempo is $ x 1.3 % of 4 / 5 of x = 910 ( 13 / 10 ) * ( 1 / 100 ) * ( 4 / 5 ) x = 910 x = $ 87500 answer is e | a = 4 / 5
b = a * 1
c = b / 100
d = 910 / c
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | b | multiply(60, divide(18, 45)) | a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 60 meters , what is the height of the building ( in meters ) ? | the height : length ratio will be equal in both cases . 18 / 45 = x / 60 x = 24 the answer is b . | a = 18 / 45
b = 60 * a
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a ) 75 kgs , b ) 64 kgs , c ) 72 kgs , d ) 65 kgs , e ) 70 kgs | a | add(divide(multiply(30, subtract(const_100, 25)), const_100), 70) | fresh grapes contain 70 % water by weight and raisins obtained by drying fresh grapes contain 25 % water by weight . how many kgs of fresh grapes are needed to get 30 kgs of raisins ? | "the weight of non - water in 30 kg of dried grapes ( which is 100 - 25 = 75 % of whole weight ) will be the same as the weight of non - water in x kg of fresh grapes ( which is 100 - 70 = 30 % of whole weight ) , so 30 * 0.75 = x * 0.3 - - > x = 75 . answer : a ." | a = 100 - 25
b = 30 * a
c = b / 100
d = c + 70
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a ) 3 , b ) 5 , c ) 8 , d ) 13 , e ) 15 | b | subtract(multiply(38, 2), add(multiply(subtract(subtract(38, add(add(multiply(12, 1), 10), 2)), 1), 3), add(multiply(12, 1), multiply(10, 2)))) | in a class of 38 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 10 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "total # of students = 38 avg # of books per student = 2 total # of books = 38 * 2 = 76 # of student borrowed at least 3 books = 38 - 2 - 12 - 10 = 14 # of books borrowed by above 14 students = 76 - ( 12 * 1 ) + ( 10 * 2 ) = 44 considering that 13 out of above 14 students borrowed only 3 books , # of books borrowed = 13 * 3 = 39 so maximum # of books borrowed by any single student = 44 - 39 = 5 option b" | a = 38 * 2
b = 12 * 1
c = b + 10
d = c + 2
e = 38 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 10 * 2
j = h + i
k = g + j
l = a - k
|
a ) 20 , b ) 15 , c ) 12 , d ) 18 , e ) 24 | c | multiply(subtract(divide(add(multiply(add(100, 25), 500), 8500), 25), 500), add(100, 25)) | in a hostel there were 100 students . to accommodate 25 more students the average is decreased by some rupees . but total expenditure increased by rs . 500 . if the total expenditure of the hostel now 8500 , find decrease of average budget ? | "let average is x 100 x + 500 = 8500 x = 80 let decrease = y 125 ( 80 β y ) = 8500 y = 12 answer : c" | a = 100 + 25
b = a * 500
c = b + 8500
d = c / 25
e = d - 500
f = 100 + 25
g = e * f
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(multiply(multiply(8, 7), 18), multiply(12, 12)) | 8 men , working 7 hours a day can complete a work in 18 days . how many hours a day must 12 men work to complete the same work in 12 days ? | "the number of hours required to complete the work is 8 * 7 * 18 = 1008 12 Γ 12 Γ ( x ) = 1008 x = 7 the answer is b ." | a = 8 * 7
b = a * 18
c = 12 * 12
d = b / c
|
a ) 6048 , b ) 3648 , c ) 7248 , d ) 1448 , e ) 2168 | a | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | of the four - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "of the four - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for p - 9 ways ; choosing the digit for q - 8 ways ; choosing the digit for r - 7 ways ; # of permutations of 3 digits in ppqr - 4 ! / 2 ! total : 9 * 8 * 7 * 4 ! / 2 ! = 6048 . answer : a ." | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
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a ) 420 , b ) 260 , c ) 200 , d ) 300 , e ) 400 | d | multiply(power(60, 0.75), multiply(power(5, 1.25), power(12, 0.25))) | ( 5 ) 1.25 Γ ( 12 ) 0.25 Γ ( 60 ) 0.75 = ? | explanation : ( 5 ) 1.25 Γ ( 12 ) 0.25 Γ ( 60 ) 0.75 = ( 5 ) 1.25 Γ ( 12 ) 0.25 Γ ( 12 Γ 5 ) 0.75 = ( 5 ) 1.25 Γ ( 12 ) 0.25 Γ ( 12 ) 0.75 Γ ( 5 ) 0.75 = ( 5 ) ( 1.25 + 0.75 ) Γ ( 12 ) ( 0.25 + 0.75 ) = ( 5 ) 2 Γ ( 12 ) 1 = 25 Γ 12 = 300 answer : option d | a = 60 ** 0
b = 5 ** 1
c = 12 ** 0
d = b * c
e = a * d
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a ) 2.25 , b ) 3.125 . , c ) 4.5 , d ) 5.225 , e ) 6.25 | b | divide(divide(4200, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 4200 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 4200 / ( 28 * 24 * 2 ) = 3.125 answer b" | a = 4200 / 28
b = 28 - 4
c = b * 2
d = a / c
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a ) - 4 , b ) - 2 , c ) 2 and 6 , d ) 4 , e ) can not be determined . | c | divide(power(12, const_2), 4) | a number x is multiplied with itself and then added to the product of 4 and x . if the result of these two operations is 12 , what is the value of x ? | "a number x is multiplied with itself - - > x ^ 2 added to the product of 4 and x - - > x ^ 2 + 4 x if the result of these two operations is - 4 - - > x ^ 2 + 4 x = 4 i . e x ^ 2 + 4 x - 12 = 0 is the quadratic equation which needs to be solved . ( x - 2 ) ( x + 6 ) = 0 hence x = 2 , x = - 6 imo c" | a = 12 ** 2
b = a / 4
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a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | c | divide(subtract(multiply(add(14, 20), divide(70, const_100)), 14), divide(70, const_100)) | a bowl of fruit contains 14 apples and 20 oranges . how many oranges must be removed so that 70 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 20 let number of oranges that must be removed so that 70 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 14 + ( 20 - x ) = 34 - x 14 / ( 34 - x ) = 7 / 10 = > 20 = 34 - x = > x = 14 answer c" | a = 14 + 20
b = 70 / 100
c = a * b
d = c - 14
e = 70 / 100
f = d / e
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a ) 7 h , b ) 10 h , c ) 8 h , d ) 6 h , e ) none of these | a | divide(multiply(const_4, sqrt(1225)), 20) | the area of a square field is 1225 km 2 . how long will it take for a horse to run around at the speed of 20 km / h ? | explanation area of field = 1225 km 2 . then , each side of field = β 1225 = 35 km distance covered by the horse = perimeter of square field = 35 Γ 4 = 140 km β΄ time taken by horse = distances / peed = 140 / 20 = 7 h answer a | a = math.sqrt(1225)
b = 4 * a
c = b / 20
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a ) 40 , b ) 60 , c ) 48 , d ) 70 , e ) 80 | c | add(add(add(12, 12), 12), 12) | xavier starts from p towards q at a speed of 50 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 52 km , then how much time does he take to cover the distance ? | first 12 min = 50 * 12 / 60 = 10 km 2 nd 12 min = 60 * 12 / 60 = 12 km 3 rd 12 min = 70 * 12 / 60 = 14 km 4 th 12 min = 80 * 12 / 60 = 16 km total time 12.4 = 48 min c | a = 12 + 12
b = a + 12
c = b + 12
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a ) 92.5 % , b ) 89.0 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(5, const_100)))), add(600, 400)), const_100) | a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 5 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ 600 = 9000 / 100 = 90 number of rotten bananas = 5 % of 400 = 20 therefore , total number of rotten fruits = 90 + 20 = 110 therefore number of fruits in good condition = 1000 - 110 = 890 therefore percentage of fruits in good condition = ( 890 / 1000 Γ 100 ) % = ( 89000 / 1000 ) % = 89.0 % answer : b" | a = 600 + 400
b = 15 / 100
c = 600 * b
d = 5 / 100
e = 400 * d
f = c + e
g = a - f
h = 600 + 400
i = g / h
j = i * 100
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['a ) 24,000', 'b ) 16 , 000,000', 'c ) 2 , 400,000', 'd ) 24 , 000,000', 'e ) 240 , 000,000'] | b | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(5, 100)), multiply(8, 100)) | if a rectangular room measures 8 meters by 5 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | b . 160 , 000,000 8 * 100 * 5 * 100 * 4 * 100 = 160 , 000,000 | a = 4 * 100
b = 1 / 10
c = a * b
d = 5 * 100
e = c * d
f = 8 * 100
g = e * f
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a ) 75 , b ) 105 , c ) 117 , d ) 150 , e ) 162 | c | add(multiply(divide(subtract(20, multiply(30, divide(12, const_60))), 8), const_60), 12) | on a 20 mile course , pat bicycled at an average rate of 30 miles per hour for the first 12 minutes and without a break , ran the rest of the distance at an average rate of 8 miles per hour . how many minutes did pat take to cover the entire course ? | at an average rate of 30 miles per hour in 12 minute ( 1 / 5 hours ) pat covers ( distance ) = ( time ) * ( rate ) = 1 / 5 * 30 = 6 miles , thus she should cover the remaining distance of 20 - 6 = 14 miles at an average rate of 8 miles per hour . to cover 14 miles at an average rate of 8 miles per hour pat needs ( time ) = ( distance ) / ( rate ) = 14 / 8 = 1.75 hours = 105 minutes . therefore pat needs total of 12 + 105 = 117 minutes to cover the entire course . answer : c . | a = 12 / const_60
b = 30 * a
c = 20 - b
d = c / 8
e = d * const_60
f = e + 12
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a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 47 | e | add(add(negate(25), const_1), add(add(negate(25), const_1), const_1)) | the sum of all the integers k such that β 25 < k < 24 is | "- 24 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 24 - 23 sum of which is - 47 . hence option d . e" | a = negate + (
b = a + 1
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a ) 13505 , b ) 12004 , c ) 15003 , d ) 14500 , e ) 16400 | a | add(multiply(multiply(add(divide(3, const_100), divide(divide(subtract(12200, 14500), 3), 14500)), 14500), 3), 14500) | sonika deposited rs . 14500 which amounted to rs . 12200 after 3 years at simple interest . had the interest been 3 % more . she would get how much ? | "( 14500 * 3 * 3 ) / 100 = 1305 12200 - - - - - - - - 13505 answer : a" | a = 3 / 100
b = 12200 - 14500
c = b / 3
d = c / 14500
e = a + d
f = e * 14500
g = f * 3
h = g + 14500
|
a ) 321 , b ) 223 , c ) 245 , d ) 265 , e ) 162 | b | subtract(subtract(298, divide(multiply(298, 12), const_100)), divide(multiply(subtract(298, divide(multiply(298, 12), const_100)), 15), const_100)) | the sale price sarees listed for rs . 298 after successive discount is 12 % and 15 % is ? | "explanation : 298 * ( 88 / 100 ) * ( 85 / 100 ) = 223 answer : b" | a = 298 * 12
b = a / 100
c = 298 - b
d = 298 * 12
e = d / 100
f = 298 - e
g = f * 15
h = g / 100
i = c - h
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a ) 97.1 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 170), 7) | 3 friends a , b , c went for week end party to mcdonald β s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 170 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 170 / 7 = 97.1 kgs answer : a" | a = 1 + 3
b = a * 170
c = b / 7
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['a ) ( β 3 ) / 2', 'b ) β 3', 'c ) β 6', 'd ) 3', 'e ) 2 β 3'] | b | divide(multiply(0.75, const_4), sqrt(const_3)) | if the area of the equilateral triangle above is 0.75 , what is the area of the adjacent square ? | area of equilateral triangle : a ^ 2 * ( square root of 3 ) / 4 = 0.75 hence , area of square = a ^ 2 = 0.75 * 4 / ( square root of 3 ) = 3 / ( square root of 3 ) = square root of 3 answer : b | a = 0 * 75
b = math.sqrt(3)
c = a / b
|
a ) 60 , b ) 50 , c ) 40 , d ) 70 , e ) 45 | a | add(divide(const_100, const_2), divide(multiply(20, divide(const_100, const_2)), const_100)) | a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 20 percent of the drive from the center , what percent of the round - trip has the technician completed ? | "round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 20 % of 50 % i . e 10 % . so he completed 50 + 10 = 60 % of total trip a" | a = 100 / 2
b = 100 / 2
c = 20 * b
d = c / 100
e = a + d
|
['a ) 3 hours', 'b ) 4 hours 15 minutes', 'c ) 6 hours', 'd ) 2 hours 5 minutes', 'e ) 8 hours'] | a | divide(15, 5) | it has been raining at the rate of 5 centimeters per hour . if the rain filled a cylindrical drum with a depth of 15 centimeters , and area 300 square centimeters , how long did it take to fill the drum completely ? | answer is : a , 3 hours the volume of the drum is irrelevant and only height matters since rain fell all over the city . thus , it takes only 15 / 5 = 3 hours of rain to fill the drum | a = 15 / 5
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a ) 3387 , b ) 1000 , c ) 2866 , d ) 2787 , e ) 1400 | e | multiply(multiply(28, 10), 5) | in digging a pond 28 m * 10 m * 5 m the volumes of the soil extracted will be ? | "28 * 10 * 5 = 1400 answer : e" | a = 28 * 10
b = a * 5
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a ) 1800 , b ) 1825 , c ) 1850 , d ) 1855 , e ) 1875 | e | divide(subtract(800, multiply(12.50, 40)), 16) | suppose you work for a manufacturing plant that pays you $ 12.50 an hour and $ . 16 for each widget you make . how many widgets must you produce in a 40 hour week to earn $ 800 ( before payroll deductions ) ? | "hourly = $ 12.50 / hour totalhours = 40 piecework = $ 0.16 / widget totalpieces = ? ? ? ? total pay = ( hourly * totalhours ) + ( piecework * totalpieces ) so , put it all together : $ 800 = ( $ 12.50 * 40 ) + ( $ 0.16 * total pieces ) totalpieces = ( $ 800 - ( $ 12.50 * 40 ) ) / $ 0.16 = 1875 widgets e . 1875" | a = 12 * 50
b = 800 - a
c = b / 16
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a ) 125 , b ) 155 , c ) 130 , d ) 120 , e ) 100 | a | divide(volume_cylinder(divide(10, const_2), 5), const_pi) | the diameter of a cylindrical tin is 10 cm and height is 5 cm . find the volume of the cylinder ? | "r = 5 h = 5 Ο * 5 * 5 * 5 = 125 Ο cc answer : a" | a = 10 / 2
b = volume_cylinder / (
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a ) 10 , b ) 20 , c ) 12 , d ) 16 , e ) 18 | e | multiply(1, 8) | if 8 spiders make 4 webs in 9 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 8 : 1 ( webs ) 1 : 9 } : : x : 4 Γ’ β‘ β 8 Γ£ β 1 Γ£ β 9 = 1 Γ£ β 4 Γ£ β x Γ’ β‘ β x = 18 answer : option e" | a = 1 * 8
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a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 79 | e | subtract(power(2, 2), 2) | if x ^ 2 + ( 1 / x ^ 2 ) = 9 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 9 ^ 2 - 2 = 79 . thus , the answer is e ." | a = 2 ** 2
b = a - 2
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a ) a ) 125 , b ) b ) 175 , c ) c ) 271 , d ) d ) 375 , e ) e ) 524 | c | divide(multiply(divide(150, const_100), 1265), 7) | ( 150 % of 1265 ) Γ· 7 = ? | "explanation : ? = ( 150 x 1265 / 100 ) Γ· 7 = 189750 / 700 = 271 answer : option c" | a = 150 / 100
b = a * 1265
c = b / 7
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a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 1 , d ) 11 : 3 , e ) 15 : 4 | c | divide(add(multiply(7, divide(add(8, 1), add(7, 2))), 8), add(multiply(2, divide(add(8, 1), add(7, 2))), 1)) | two vessels contains equal number of mixtures milk and water in the ratio 7 : 2 and 8 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 7 / 9 + 8 / 9 ) : ( 2 / 9 + 1 / 9 ) = 15 / 9 : 3 / 9 = 5 : 1 answer is c" | a = 8 + 1
b = 7 + 2
c = a / b
d = 7 * c
e = d + 8
f = 8 + 1
g = 7 + 2
h = f / g
i = 2 * h
j = i + 1
k = e / j
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a ) 477 , b ) 396 , c ) 279 , d ) 276 , e ) 291 | b | add(multiply(divide(60, subtract(21, 12)), 12), multiply(divide(60, subtract(21, 12)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 12 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 12 + 21 = 33 t = 12 d = 33 * 12 = 396 . answer : b" | a = 21 - 12
b = 60 / a
c = b * 12
d = 21 - 12
e = 60 / d
f = e * 21
g = c + f
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a ) 4 : 7 , b ) 4 : 0 , c ) 4 : 3 , d ) 4 : 1 , e ) 4 : 2 | c | multiply(divide(12, const_100), 9) | a part of certain sum of money is invested at 9 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 9 = 4 : 3 answer : c" | a = 12 / 100
b = a * 9
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a ) 1 / 35 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | e | divide(58, multiply(6, 210)) | the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 58 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 6 x 210 = 1260 Γ’ Λ Β΄ 1 / m + 1 / n = ( m + n ) / mn = 58 / 1260 = 17 / 620 correct option : e" | a = 6 * 210
b = 58 / a
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a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(3, multiply(divide(0.375, 5), 30)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(30, divide(subtract(3, multiply(divide(0.375, 5), 30)), subtract(divide(1.375, 10), divide(0.375, 5))))) | a 30 kg metal bar made of alloy of tin and silver lost 3 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 3 kg when 30 kg alloy is immersed . this is a loss of ( 3 / 30 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e" | a = 0 / 375
b = a * 30
c = 3 - b
d = 1 / 375
e = 0 / 375
f = d - e
g = c / f
h = 0 / 375
i = h * 30
j = 3 - i
k = 1 / 375
l = 0 / 375
m = k - l
n = j / m
o = 30 - n
p = g / o
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a ) 33 , b ) 45 , c ) 66 , d ) 77 , e ) 98 | e | divide(volume_cylinder(divide(14, const_2), 2), const_pi) | the diameter of a cylindrical tin is 14 cm and height is 2 cm . find the volume of the cylinder ? | "r = 7 h = 5 Ο * 7 * 7 * 5 = 98 Ο cc answer : e" | a = 14 / 2
b = volume_cylinder / (
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a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | c | floor(multiply(const_100, divide(subtract(2552.36, 2400), 2400))) | a shopkeeper sold an article for rs 2552.36 . approximately what was his profit percent if the cost price of the article was rs 2400 | explanation : gain % = ( 152.36 * 100 / 2400 ) = 6.34 % = 6 % approx option c | a = 2552 - 36
b = a / 2400
c = 100 * b
d = math.floor(c)
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a ) 40 % , b ) 8 % , c ) 25 % , d ) 17 % , e ) 9 % | d | subtract(add(multiply(multiply(const_4, 10), divide(30, const_100)), multiply(const_4, 10)), const_100) | the balance of a trader weighs 10 % less than it should . still the trader marks up his goods to get an overall profit of 30 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 90 gms because it weighs 10 % less . say , the cost price is $ 90 ( $ 1 / gm ) . since he gets a profit of 30 % , the selling price must be 90 + ( 30 / 100 ) * 90 = $ 117 since the cost price is actually supposed to be $ 100 ( for 100 gms ) and the selling price is $ 117 , the mark up is simply 17 % . ans : d" | a = 4 * 10
b = 30 / 100
c = a * b
d = 4 * 10
e = c + d
f = e - 100
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a ) 1 / 2 , b ) 2 / 3 , c ) 1 , d ) 3 / 2 , e ) 1 / 7 | e | divide(subtract(3, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 4)) | hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of 3 hammers and 4 wrenches is one - third that of 10 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ? | "x be the weight of a hammer and y be the weight of a wrench . ( 3 x + 4 y ) = 1 / 3 * ( 10 x + 5 y ) 3 ( 3 x + 4 y ) = ( 10 x + 5 y ) 9 x + 12 y = 10 x + 5 y 7 y = x y = x / 7 ans - e" | a = 3 * 2
b = 3 - a
c = 3 * 2
d = c - 4
e = b / d
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a ) 23 / 34 , b ) 57 / 60 , c ) 51 / 67 , d ) 19 / 71 , e ) 75 / 64 | b | subtract(divide(subtract(7, const_1), add(4, const_1)), divide(add(2, const_1), subtract(13, const_1))) | if a is an integer greater than 2 but less than 7 and b is an integer greater than 4 but less than 13 , what is the range of a / b ? | "the way to approach this problem is 2 < a < 7 and 4 < b < 13 minimum possible value of a is 3 and maximum is 6 minimum possible value of b is 5 and maximum is 12 range = max a / min b - min a / max b ( highest - lowest ) 6 / 5 - 3 / 12 = 57 / 60 hence b" | a = 7 - 1
b = 4 + 1
c = a / b
d = 2 + 1
e = 13 - 1
f = d / e
g = c - f
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a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | d | divide(multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_2), 3), multiply(multiply(const_2, const_2), const_2)) | all factors of a positive integer a are multiplied and the product obtained is a ^ 3 . if a is greater than 1 , how many factors does a have ? | answer = d = 6 number . . . . . . . . . . . . . . . . factors . . . . . . . . . . . . . . . . . . . . . . . . product 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 , 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 , 3 , 6 . . . . . . . . . . . . . . . . . . . . . . . . . 36 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 , 4 , 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 3 , 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 , 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 , 3 , 4 , 6 , 12 . . . . . . . . . . . . . . . . . 12 * 12 * 12 number which we are looking for = 12 answer = d | a = 2 * 2
b = a * 2
c = b * 2
d = c * 3
e = 2 * 2
f = e * 2
g = d / f
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a ) 85 , b ) 98 , c ) 89 , d ) 105 , e ) 129 | c | add(multiply(multiply(1, 1), const_10), 5) | when 5 + 1 = 56 , 6 + 1 = 67 , 7 + 1 = 78 , then 8 + 1 = ? | "5 + 1 = > 5 x 1 = 5 & 5 + 1 = 6 = > 5 & 6 = > 56 6 + 1 = > 6 x 1 = 6 & 6 + 1 = 7 = > 6 & 7 = > 67 7 + 1 = > 7 x 1 = 7 & 7 + 1 = 8 = > 7 & 8 = > 78 then 8 + 1 = > 8 x 1 = 8 & 8 + 1 = 9 = > 8 & 9 = > 89 answer : c" | a = 1 * 1
b = a * 10
c = b + 5
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a ) 79.2 kmph , b ) 89.2 kmph , c ) 79.6 kmph , d ) 99.2 kmph , e ) 69.2 kmph | a | multiply(divide(22, const_1000), const_3600) | express 22 mps in kmph ? | "22 * 18 / 5 = 79.2 kmph answer : a" | a = 22 / 1000
b = a * 3600
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a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 4 | e | subtract(divide(subtract(90, 50), subtract(58, 50)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 90 units raises the average to 58 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 50 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 50 n + 90 = 58 * ( n + 1 ) - - > n = 4 . or as 40 extra units increased the average for n + 1 days by 8 units per day then 40 / ( n + 1 ) = 8 - - > n = 4 . answer : e ." | a = 90 - 50
b = 58 - 50
c = a / b
d = c - 1
|
a ) 17 litres , b ) 21.3 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(32, 1399.45), multiply(32, 262.85)), subtract(3104.35, 1399.45)) | the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a Γ§ ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 32 litres of the mixed fruit juice . how many litres of the a Γ§ ai berry juice does he need to add ? | "262.85 ( 32 ) + 3 , 104.35 x = 1 , 399.45 ( 32 + x ) solve the equation . 262.85 ( 32 ) + 3 , 104.35 x = 1 , 399.45 ( 32 + x ) 8 , 411.2 + 3 , 104.35 x = 44 , 782.4 + 1 , 399.45 x 8 , 411.2 + 1 , 704.9 x = 44 , 782.4 1 , 704.9 x = 36 , 371.2 x β 21.3 answer is b ." | a = 32 * 1399
b = 32 * 262
c = a - b
d = 3104 - 35
e = c / d
|
a ) 20 , b ) 11 , c ) 21 , d ) 41 , e ) none of these | c | sqrt(add(241, multiply(100, const_2))) | the sum of the squares of 3 numbers is 241 , while the sum of their products taken two at a time is 100 . their sum is : | x ^ + y ^ 2 + z ^ 2 = 241 xy + yz + zx = 100 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 241 + ( 2 * 100 ) ( x + y + z ) ^ 2 = 441 so x + y + z = 21 answer : c | a = 100 * 2
b = 241 + a
c = math.sqrt(b)
|
a ) 22 sec , b ) 88 sec , c ) 48 sec , d ) 18 sec , e ) 72 sec | e | add(45, 30) | two goods trains each 720 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 750 + 750 = 1500 m . required time = 1500 * 6 / 125 = 72 sec . answer : e" | a = 45 + 30
|
a ) 4 / 5 , b ) 6 / 5 , c ) 7 / 5 , d ) 11 / 5 , e ) 9 / 5 | d | add(divide(divide(1, 1), divide(5, 6)), const_1) | if w / x = 1 / 6 and w / y = 1 / 5 , then ( x + y ) / y = | "w / x = 1 / 6 = > x = 6 w and w / y = 1 / 5 = > y = 5 w ( x + y ) / y = ( 6 w + 5 w ) / 5 w = 11 w / 5 w = 11 / 5 correct option : d" | a = 1 / 1
b = 5 / 6
c = a / b
d = c + 1
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | subtract(divide(multiply(divide(multiply(add(5, 7), divide(multiply(35, 4), add(4, 3))), 5), 7), add(5, 7)), divide(multiply(35, 3), add(4, 3))) | 35 liters of a mixture is created by mixing liquid p and liquid q in the ratio 4 : 3 . how many liters of liquid q must be added to make the ratio 5 : 7 ? | let x be the amount of liquid q to be added . ( 3 / 7 ) * 35 + x = ( 7 / 12 ) * ( 35 + x ) 1260 + 84 x = 1715 + 49 x 35 x = 455 x = 13 the answer is d . | a = 5 + 7
b = 35 * 4
c = 4 + 3
d = b / c
e = a * d
f = e / 5
g = f * 7
h = 5 + 7
i = g / h
j = 35 * 3
k = 4 + 3
l = j / k
m = i - l
|
a ) 12 , b ) 15 , c ) 32 , d ) 20 , e ) 25 | c | multiply(divide(multiply(add(multiply(3, 2), multiply(2, 2)), divide(20, subtract(multiply(3, 2), 1))), add(4, 1)), 4) | a jar contains a mixture of ab in the ratio 4 : 1 . when 20 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially . | "20 litres of mixture that is replaced will contain 16 litres of a and 4 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 16 ] / [ k - 4 + 20 ] = 2 / 3 = > 12 k - 48 = 2 k - 8 + 40 = > 10 k = 80 solve for k which is k = 8 so initial volume of liquid a = 4 k = 32 litres answer : c" | a = 3 * 2
b = 2 * 2
c = a + b
d = 3 * 2
e = d - 1
f = 20 / e
g = c * f
h = 4 + 1
i = g / h
j = i * 4
|
a ) 5.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 15 sec | c | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | two trains 140 m and 160 m long run at the speed of 60 kmph and 40 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 kmph * 5 / 18 = 250 / 9 m / s distance covered in crossing each other = 140 + 160 = 300 m required time = 300 * 9 / 250 = 54 / 5 = 10.8 sec answer is c" | a = 140 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | b | divide(add(multiply(60, 3), multiply(100, 2)), add(2, 3)) | for a certain exam , a score of 60 was 2 standard deviations below mean and a score of 100 was 3 standard deviations above mean . what was the mean score for the exam ? | "mean - 2 sd = 60 mean + 3 sd = 100 by solving above we get , sd ( absolute value ) = 8 mean = 76 ans . b" | a = 60 * 3
b = 100 * 2
c = a + b
d = 2 + 3
e = c / d
|
a ) and 27 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | d | subtract(48, divide(add(48, 9), const_3)) | the sum of the present age of henry and jill is 48 . what is their present ages if 9 years ago henry was twice the age of jill ? | "let the age of jill 9 years ago be x , age of henry be 2 x x + 9 + 2 x + 9 = 48 x = 10 present ages will be 19 and 29 answer : d" | a = 48 + 9
b = a / 3
c = 48 - b
|
a ) 30 , b ) 26 , c ) 16 , d ) 97 , e ) 12 | a | add(20, divide(subtract(36, 26), 15)) | the average of 15 numbers is calculated as 20 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "15 * 20 + 36 β 26 = 300 / 10 = 30 answer : a" | a = 36 - 26
b = a / 15
c = 20 + b
|
a ) 128 , b ) 289 , c ) 200 , d ) 112 , e ) 178 | a | divide(square_area(16), const_2) | what is the area of a square field whose diagonal of length 16 m ? | "d 2 / 2 = ( 16 * 16 ) / 2 = 128 answer : a" | a = square_area / (
|
a ) 8 , b ) 12 , c ) 15 , d ) 16 , e ) 19 | b | subtract(subtract(32, 12), subtract(40, 32)) | the avg age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . therefore decreasingthe avg by 4 year . find what was the original strength of class ? | let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 β 40 y + 384 = 36 y + 432 β 4 y = 48 β΄ y = 12 b | a = 32 - 12
b = 40 - 32
c = a - b
|
a ) 287 , b ) 600 , c ) 289 , d ) 276 , e ) 900 | e | multiply(multiply(108, const_0_2778), 30) | what distance will be covered by a bus moving at 108 kmph in 30 seconds ? | "108 kmph = 108 * 5 / 18 = 30 mps d = speed * time = 30 * 30 = 900 m . answer : e" | a = 108 * const_0_2778
b = a * 30
|
a ) 7450 , b ) 14900 , c ) 6250 , d ) 6000 , e ) none of these | c | divide(1500, divide(subtract(add(const_100, 12), subtract(const_100, 12)), const_100)) | vijay sells a cupboard at 12 % below cost price . had he got rs . 1500 more , he would have made a profit of 12 % . what is the cost price of the cupboard ? | "explanation : cost price = 1500 / ( 0.12 + 0.12 ) = 1500 / 0.24 = rs . 6250 answer c" | a = 100 + 12
b = 100 - 12
c = a - b
d = c / 100
e = 1500 / d
|
a ) 65 mtr . , b ) 84 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | b | multiply(112, divide(multiply(30, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 30 men in 3 days ? | 20 men is 6 days can build 112 metres 30 men in 3 days can build = 112 * ( 30 / 20 ) x ( 3 / 6 ) = 84 meters answer : b . | a = 30 * 3
b = 20 * 6
c = a / b
d = 112 * c
|
a ) 275 m , b ) 360 m , c ) 475 m , d ) 420 m , e ) 440 m | c | multiply(multiply(57, const_0_2778), 30) | if the speed of a man is 57 km per hour , then what is the distance traveled by him in 30 seconds ? | "the distance traveled in 30 sec = 57 * ( 5 / 18 ) * 30 = 475 m answer : c" | a = 57 * const_0_2778
b = a * 30
|
a ) 334 , b ) 500 , c ) 376 , d ) 450 , e ) 271 | d | divide(405, divide(90, const_100)) | victor gets 90 % marks in examinations . if these are 405 marks , find the maximum marks . | "let the maximum marks be m then 90 % of m = 405 β 90 / 100 Γ m = 405 β m = ( 405 Γ 100 ) / 90 β m = 40500 / 90 β m = 450 therefore , maximum marks in the examinations are 450 answer : d" | a = 90 / 100
b = 405 / a
|
a ) 90 , b ) 120 , c ) 45 , d ) 180 , e ) 200 | c | divide(multiply(multiply(3, const_4), multiply(6, 3)), power(factorial(2), 2)) | there are 3 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 3 c 2 number of ways of selecting 2 non fiction books = 6 c 2 3 c 2 * 6 c 2 = 3 * 15 = 45 answer : c" | a = 3 * 4
b = 6 * 3
c = a * b
d = math.factorial(2)
e = d ** 2
f = c / e
|
a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 200 | e | divide(16, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 16 . if n / j = 134.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 16 so , 134.08 = 134 + 16 / j so , j = 200 ans e" | a = 4 * 10
b = a + 2
c = 100 + b
d = 134 - 8
e = 16 / d
|
a ) 8.4 , b ) 840 , c ) 84.0 , d ) 0.84 , e ) 0.084 | b | divide(multiply(0.0048, 3.5), multiply(multiply(0.05, 0.1), 0.004)) | ( 0.0048 ) ( 3.5 ) / ( 0.05 ) ( 0.1 ) ( 0.004 ) = | ( 0.0048 ) ( 3.5 ) / ( 0.05 ) ( 0.1 ) ( 0.004 ) = 0.0048 * 350 / 5 ( 0.1 ) ( 0.004 ) = 0.048 * 70 / 1 ( 0.004 ) = 48 * 70 / 4 = 12 * 70 = 840 answer : b | a = 0 * 48
b = 0 * 5
c = b * 0
d = a / c
|
a ) 47 , b ) 37 , c ) 39 , d ) 43 , e ) 42 | b | add(subtract(85, multiply(3, 17)), 3) | a batsman in his 17 th innings makes a score of 85 , and thereby increases his average by 3 . what is his average after the 17 th innings ? he had never been β not out β . | average score before 17 th innings = 85 - 3 Γ 17 = 34 average score after 17 th innings = > 34 + 3 = 37 answer : b | a = 3 * 17
b = 85 - a
c = b + 3
|
a ) 190 , b ) 180 , c ) 150 , d ) 200 , e ) 300 | d | subtract(300, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(300, 15.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | a snooker tournament charges $ 40.00 for vip seats and $ 15.00 for general admission ( β regular β seats ) . on a certain night , a total of 300 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | "let no of sits in vip enclosure is x then x * 40 + 15 ( 300 - x ) = 7500 or 25 x = 7500 - 4500 , x = 3000 / 25 = 120 vip = 120 general 200 d" | a = 3 + 4
b = a * 1000
c = 2 + 3
d = c * 100
e = b + d
f = 300 * 15
g = e - f
h = 2 + 3
i = 2 + 3
j = h * i
k = g / j
l = 300 - k
|
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