options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 11 , b ) 17 , c ) 13 , d ) 23 , e ) none of these | c | subtract(add(8, 9), 4) | when 242 is divided by a certain divisor the remainder obtained is 8 . when 698 is divided by the same divisor the remainder obtained is 9 . when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 4 . what is the value of the divisor ? | "explanatory answer when 242 is divided by a certain divisor the remainder obtained is 8 . let the divisor be d . when 242 is divided by d , let the quotient be ' x ' . the remainder is 8 . therefore , 242 = xd + 8 when 698 is divided by the same divisor the remainder obtained is 9 . let y be the quotient when 698 is divided by d . then , 698 = yd + 9 . when the sum of the two numbers , 242 and 698 , is divided by the divisor , the remainder obtained is 4 . 242 + 698 = 940 = xd + yd + 8 + 9 940 = xd + yd + 17 as xd and yd are divisible by d , the remainder when 940 is divided by d should have been 17 . however , because we know that the remainder is 4 , it would be possible only when 17 / d leaves a remainder of 4 . if the remainder obtained is 4 when 17 is divided by ' d ' , then ' d ' has to be 13 . choice c" | a = 8 + 9
b = a - 4
|
['a ) 135', 'b ) 155', 'c ) 125', 'd ) 145', 'e ) 115'] | c | power(sqrt(divide(100, const_4)), const_3) | the lateral surface area of cube is 100 sq . units . find the volume of cube ? | lateral surface = 4 a ( power ) 2 = 100 sq . units a ( power ) 2 = 25 a = 5 . cube volume = a ( power ) 3 = > 125 m ( power ) 3 answer is c . | a = 100 / 4
b = math.sqrt(a)
c = b ** 3
|
a ) 23 days , b ) 46 days , c ) 22 days , d ) 29 days , e ) 50 days | e | divide(multiply(35, 10), 7) | 10 men can complete a piece of work in 35 days . in how many days can 7 men complete that piece of work ? | e 50 days 10 * 35 = 7 * x = > x = 50 days | a = 35 * 10
b = a / 7
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a ) 720 , b ) 120 , c ) 300 , d ) 30 , e ) 333 | c | subtract(subtract(subtract(divide(divide(divide(factorial(12), factorial(subtract(12, 3))), factorial(3)), const_2), 12), 12), const_10) | mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 12 candidates . she plans to have one working interview with 3 of the 12 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ? | "300 . answer c" | a = math.factorial(12)
b = 12 - 3
c = math.factorial(b)
d = a / c
e = math.factorial(3)
f = d / e
g = f / 2
h = g - 12
i = h - 12
j = i - 10
|
a ) 60 , b ) 74 , c ) 84 , d ) 54 , e ) 104 | c | divide(factorial(subtract(10, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(10, const_1), subtract(const_4, const_1))))) | the number of positive integer solutions for the equation x + y + z + t = 10 is | "the number of positive integer solutions for the equatio fx 1 + x 2 + ⋯ + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 10 - 1 c 4 - 1 = 9 c 3 = 84 answer : c" | a = 10 - 1
b = math.factorial(a)
c = 4 - 1
d = math.factorial(c)
e = 10 - 1
f = 4 - 1
g = e - f
h = math.factorial(g)
i = d * h
j = b / i
|
a ) 32 , b ) 36 , c ) 39 , d ) 38 , e ) 35 | e | divide(10.5, subtract(5.6, 5.3)) | two boys starts from the same place walking at the rate of 5.3 kmph and 5.6 kmph respectively in the same direction . what time will they take to be 10.5 km apart ? | "explanation : relative speed = 5.6 - 5.3 = 0.3 kmph ( because they walk in the same direction ) distance = 10.5 km time = distance / speed = 10.5 / 0.3 = 35 hr answer : e" | a = 5 - 6
b = 10 / 5
|
a ) 22 kmph , b ) 77 kmph , c ) 54 kmph , d ) 72 kmph , e ) 88 kmph | d | multiply(const_3_6, divide(320, 16)) | a train 320 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 320 / 16 * 18 / 5 = 72 kmph answer : d" | a = 320 / 16
b = const_3_6 * a
|
a ) 64 , b ) 32 , c ) 16 , d ) 4 , e ) 1 | a | power(2, multiply(const_2, const_3)) | if x is a positive number and 1 / 2 the square root of x is the cube root of x , then x = | 1 / 2 the square root of x is cube root of x . if x = 64 . . then 1 / 2 the square root of x = 4 and cube of x is 64 . option a . | a = 2 * 3
b = 2 ** a
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10 | e | add(const_1, divide(32, 4)) | if k is the greatest positive integer such that 4 ^ k is a divisor of 32 ! then k = | "32 / 4 = 8 32 / 16 = 2 8 + 2 = 10 = k answer : e" | a = 32 / 4
b = 1 + a
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | c | subtract(divide(multiply(multiply(30, subtract(10, 2)), 100), multiply(2, subtract(300, 100))), 30) | an engineer undertakes a project to build a road 10 km long in 300 days and employs 30 men for the purpose . after 100 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | 30 workers working already let x be the total men required to finish the task in next 200 days 2 km done hence remaining is 8 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 30 * 8 * 100 ) / ( 2 * 200 ) thus , x = 60 thus , more men needed to finish the task = 60 - 30 = 30 answer : c | a = 10 - 2
b = 30 * a
c = b * 100
d = 300 - 100
e = 2 * d
f = c / e
g = f - 30
|
a ) 50 , b ) 57 , c ) 89 , d ) 56 , e ) 60 | e | add(50, const_1) | the average of first four prime numbers greater than 50 is ? | "53 + 59 + 61 + 67 = 240 / 4 = 60 answer : e" | a = 50 + 1
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a ) 2 % , b ) 5 % , c ) 28 % , d ) 30 % , e ) 45 % | d | multiply(divide(subtract(subtract(const_100, 80), multiply(divide(multiply(const_10, const_3), const_100), subtract(const_100, 40))), 40), const_100) | thirty percent of the women in a college class are science majors , and the non - science majors make up 80 % of the class . what percentage of the women are science majors if 40 % of the class are men ? | "3 / k + 2 / m = 6 / t assuming total # is 100 : [ science - women ] will have - 0.3 * 60 = 18 [ non - science - women ] will have - 42 [ science - men ] will have = 20 - 18 = 2 s 0 18 / 60 * 100 = 30 % answer - d" | a = 100 - 80
b = 10 * 3
c = b / 100
d = 100 - 40
e = c * d
f = a - e
g = f / 40
h = g * 100
|
a ) 24 , b ) 34 , c ) 37.8 , d ) 42 , e ) 84 | b | divide(0.51, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 0.51 more than jane , what was the original price of the dish ? | "the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 0.51 - - > 15 p - 13.5 p = 51 - - > p = 34 . answer : b ." | a = 15 / 100
b = 10 / 100
c = 1 - b
d = 15 / 100
e = c * d
f = a - e
g = 0 / 51
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | floor(divide(122, 6)) | on dividing 122 by a number , the quotient is 6 and the remainder is 2 . find the divisor ? | "d = ( d - r ) / q = ( 122 - 2 ) / 6 = 122 / 6 = 20 d )" | a = 122 / 6
b = math.floor(a)
|
a ) 426 , b ) 526 , c ) 626 , d ) 726 , e ) 826 | a | subtract(434, multiply(multiply(12, 3), 2)) | evaluate : 434 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 434 - 12 * 3 * 2 = 434 - 8 = 426 correct answer a" | a = 12 * 3
b = a * 2
c = 434 - b
|
a ) 1,108 , b ) 2,100 , c ) 2,628 , d ) 2,124 , e ) 2,256 | c | multiply(divide(250, 22.95), 250) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 250 ? | "to maximize number of hot dogs with 250 $ total number of hot dogs bought in 250 - pack = 22.95 * 10 = 229.5 $ amount remaining = 250 - 229.5 = 20.5 $ total number of hot dogs bought in 20 - pack = 3.05 * 6 = 18.3 $ amount remaining = 20.5 - 18.3 = 2.2 $ total number of hot dogs bought in 8 - pack = 1.55 * 1 = 1.55 $ amount remaining = 2.2 - 1.55 = 0.65 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 250 $ = 250 * 10 + 20 * 6 + 8 * 1 = 2628 answer c" | a = 250 / 22
b = a * 250
|
a ) 2 / 3 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 1 / 5 | a | divide(subtract(divide(40, const_100), divide(60, const_100)), subtract(divide(30, const_100), divide(60, const_100))) | some of 60 % - intensity red paint is replaced with 30 % solution of red paint such that the new paint intensity is 40 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 60 ( 1 - x ) + 30 x = 40 x = 2 / 3 answer : a" | a = 40 / 100
b = 60 / 100
c = a - b
d = 30 / 100
e = 60 / 100
f = d - e
g = c / f
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 4)) | a bucket full of nuts was discovered by the crow living in the basement . the crow eats a fifth of the total number of nuts in 4 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? | "in one hour , the crow eats 1 / 20 of the nuts . ( 1 / 4 ) / ( 1 / 20 ) = 5 hours the answer is b ." | a = 1 / 4
b = 2 + 3
c = 1 / b
d = c / 4
e = a / d
|
a ) 60 , b ) 60.75 , c ) 58.11 , d ) 15.18 , e ) 16.21 | d | subtract(negate(6.75), multiply(subtract(3, 4.5), divide(subtract(3, 4.5), subtract(2, 3)))) | 2 , 3 , 4.5 , 6.75 , 10.12 , ( . . . ) | "2 ( 2 ã — 3 ) ã · 2 = 3 ( 3 ã — 3 ) ã · 2 = 4.5 ( 4.5 ã — 3 ) ã · 2 = 6.75 ( 6.75 ã — 3 ) ã · 2 = 10.12 ( 10.12 ã — 3 ) ã · 2 = 15.78 answer is d" | a = negate - (
|
a ) 45 , b ) 15 , c ) 25 , d ) 5 , e ) 30 | a | add(add(20, 10), 15) | there are 20 brown dogs , 10 white dogs , 15 black dogs . totally how many dogs are there in all ? | 20 + 10 + 15 = 45 . answer is a | a = 20 + 10
b = a + 15
|
a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 70.20 % , e ) none of these | d | multiply(divide(subtract(2000, add(divide(multiply(900, 32), const_100), divide(multiply(subtract(2000, 900), 28), const_100))), 2000), const_100) | in an examination , there were 2000 candidates , out of which 900 candidates were girls and rest were boys . if 28 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | girls = 900 , boys = 1100 passed = ( 28 % of 1100 ) + ( 32 % of 900 ) = 308 + 288 = 596 failed = 2000 - 596 = 1404 failed % = [ ( 1404 / 2000 ) x 100 ] % = 70.2 % . answer : d | a = 900 * 32
b = a / 100
c = 2000 - 900
d = c * 28
e = d / 100
f = b + e
g = 2000 - f
h = g / 2000
i = h * 100
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | multiply(594,323, const_1) | if 3 ^ x * 4 ^ y = 1 , 594,323 and x – y = 13 , then x = ? | "since 1 , 594,323 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 13 and y = 0 , then x = 13 . the answer is d ." | a = 594 * 323
|
a ) 0.8 , b ) 0.92 , c ) 1.02 , d ) 1.2 , e ) 1.3 | b | multiply(divide(80, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5))) | if a speaks the truth 80 % of the times , b speaks the truth 60 % of the times . what is the probability that at least one will tell the truth | "probability of a speaks truth p ( a ) = 8 / 10 ; false = 2 / 10 probability of b speaks truth p ( b ) = 6 / 10 ; false = 4 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 2 / 10 ) * ( 4 / 10 ) ] = 1 - 8 / 10 = 1 - 0.08 = 0.92 answer : b" | a = 4 * 5
b = a * 5
c = 80 / b
d = 4 * 5
e = d * 5
f = 60 / e
g = c * f
|
a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | c | multiply(divide(divide(subtract(560, 460), 5), subtract(460, multiply(divide(subtract(560, 460), 5), 3))), const_100) | joe invested a certain sum of money in a simple interest bond whose value grew to $ 460 at the end of 3 years and to $ 560 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 460 - $ 60 = 400 . the interest rate is $ 20 / $ 400 = 5 % the answer is c ." | a = 560 - 460
b = a / 5
c = 560 - 460
d = c / 5
e = d * 3
f = 460 - e
g = b / f
h = g * 100
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent t ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . t = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % . c" | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
|
['a ) 11', 'b ) 12', 'c ) 13', 'd ) 15', 'e ) 18'] | b | divide(36, add(1, 2)) | 36 people { a 1 , a 2 , . . . , a 36 } meet and shake hands in a circular fashion . in other words , there are totally 36 handshakes involving the pairs , { a 1 , a 2 } , { a 2 , a 3 } , . . . , { a 35 , a 36 } , { a 36 , a 1 } . then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is | if we take person a 1 in the set then it will cover a 36 and a 2 for shaking hand . similarly if we take person a 4 in the set then it will cover a 5 and a 3 for shaking hand and so on , so minimum 12 persons are required . answer : b | a = 1 + 2
b = 36 / a
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a ) 4055 , b ) 4500 , c ) 5000 , d ) 9000 , e ) 4000 | e | divide(power(const_10, divide(8, const_2)), const_2) | a palindrome is a number that reads the same forward and backward , such as 787 . how many even , 8 - digit numbers are palindromes ? | "first recognize you only need to consider the first four digits ( because the second four are just the first four flipped ) there are 9000 possibilities for the first four digits of a 8 digit number , 1000 - 9999 inclusive . everything starting with a 2,4 , 6,8 will be odd , which is 4 / 9 ths of the combinations . 4 / 9 * 9000 = 4000 answer : e" | a = 8 / 2
b = 10 ** a
c = b / 2
|
a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability y that the next couple the maitre ' d seats will not order dessert ? | "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . y = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 %" | a = 2 * 5
b = 2 * 5
c = a * b
d = 60 / c
e = 2 * 5
f = 2 * 5
g = e * f
h = 20 / g
i = 1 - h
j = d / i
k = 1 - j
l = 2 * 5
m = 2 * 5
n = l * m
o = k * n
|
a ) 14 , 18 , b ) 14 , 6 , c ) 14 , 14 , d ) 6 , 6 , e ) 14 , 10 | b | divide(add(divide(96, 12), divide(200, 10)), const_2) | if a boat is rowed downstream for 200 km in 10 hours and upstream for 96 km in 12 hours , what is the speed of the boat and the river ? | "explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 20 ; us = 8 x = ( 20 + 8 ) / 2 = 28 / 2 = 14 km / hr y = ( 20 - 8 ) / 2 = 12 / 2 = 6 km / hr answer : b" | a = 96 / 12
b = 200 / 10
c = a + b
d = c / 2
|
a ) 8 sec , b ) 1 sec , c ) 9 sec , d ) 6 sec , e ) 2 sec | b | divide(10, multiply(36, const_0_2778)) | in what time will a car 10 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | t = 10 / 36 * 18 / 5 = 1 sec answer : b | a = 36 * const_0_2778
b = 10 / a
|
a ) $ 21 , b ) $ 27 , c ) $ 31 , d ) $ 61 , e ) it can not be determined | d | subtract(add(130, subtract(164.50, 130)), multiply(subtract(164.50, 130), 3)) | at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 130 . at the same place it would cost $ 164.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | "let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 130 4 b + 10 s + c = 164.5 subtracting the first equation from the second gives us b + 3 s = 34.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 61 . in any case , there is no necessity to know each item ' s price , just the sum . answer : d" | a = 164 - 50
b = 130 + a
c = 164 - 50
d = c * 3
e = b - d
|
a ) 16 , b ) 77 , c ) 25 , d ) 17 , e ) 171 | d | floor(sqrt(306)) | if the sum of a number and its square is 306 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 306 x 2 + x - 306 = 0 ( x + 18 ) ( x – 17 ) = 0 x = 17 answer : d" | a = math.sqrt(306)
b = math.floor(a)
|
a ) 29997 , b ) 28088 , c ) 27098 , d ) 13000 , e ) 2799 | d | subtract(15000, 2000) | the price of a t . v . set worth rs . 15000 is to be paid in 20 installments of rs . 2000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 2000 balance payment = ( 15000 - 2000 ) = rs . 13000 . answer : d" | a = 15000 - 2000
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a ) 208 , b ) 277 , c ) 350 , d ) 881 , e ) 271 | a | subtract(multiply(20, multiply(70, const_0_2778)), 180) | a train 180 m long running at 70 kmph crosses a platform in 20 sec . what is the length of the platform ? | "d = 70 * 5 / 18 = 20 = 388 – 180 = 208 answer : a" | a = 70 * const_0_2778
b = 20 * a
c = b - 180
|
a ) 4000 , b ) 6000 , c ) 18000 , d ) 10000 , e ) none of these | c | divide(multiply(2700, const_100), 5) | a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 5 % on conveyance . if his savings at the end of the month is rs . 2700 , then his salary per month in rupees is : | "total expenditure = 40 + 25 + 15 + 5 = 85 % saving = ( 100 - 85 ) = 15 % 15 / 100 × salary = 2700 , salary = 18000 rs . answer : c ." | a = 2700 * 100
b = a / 5
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a ) 77 sec , b ) 66 sec , c ) 48 sec , d ) 55 sec , e ) 36 sec | e | divide(multiply(250, const_2), multiply(add(30, 20), const_0_2778)) | two trucks each 250 m long are running in opposite directions on parallel paths . their speeds are 30 km / hr and 20 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | relative speed = 20 + 30 = 50 km / hr . 50 * 5 / 18 = 125 / 9 m / sec . distance covered = 250 + 250 = 500 m . required time = 500 * 9 / 125 = 36 sec . answer : e | a = 250 * 2
b = 30 + 20
c = b * const_0_2778
d = a / c
|
a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 1 / 3 , e ) none of these | d | divide(multiply(130, 150), multiply(1000, 65)) | if a * b * c = 130 , b * c * d = 65 , c * d * e = 1000 and d * e * f = 150 the ( a * f ) / ( c * d ) = ? | "explanation : a â ˆ — b â ˆ — c / b â ˆ — c â ˆ — d = 130 / 65 = > a / d = 2 d â ˆ — e â ˆ — f / c â ˆ — d â ˆ — e = 150 / 1000 = > f / c = 1 / 4 a / d * f / c = 2 * 15 / 10 = 1 / 3 answer : d" | a = 130 * 150
b = 1000 * 65
c = a / b
|
a ) 6 % , b ) 8 % , c ) 9 % , d ) 11 % , e ) 12 % | a | subtract(multiply(divide(subtract(37, 25), subtract(const_100, 25)), const_100), const_10) | two numbers are less than third number by 25 % and 37 % respectively . how much percent is the second number less than by the first | let the third number is x . then first number = ( 100 - 25 ) % of x = 75 % of x = 75 x / 100 second number is ( 63 x / 100 ) difference = 75 x / 100 - 63 x / 100 = 3 x / 25 so required percentage is , difference is what percent of first number ( 3 x / 25 * 100 / 75 x * 100 ) % = 6 % answer : a | a = 37 - 25
b = 100 - 25
c = a / b
d = c * 100
e = d - 10
|
a ) 5 , b ) 5 / 4 , c ) 4 / 5 , d ) 1 / 4 , e ) 1 / 5 | d | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 6 * / 4 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . based on this definition . . . . if . . . . k = 2 k * = ( 1 / 1 ) ( 1 / 2 ) = 1 / 2 if . . . . k = 3 k * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) = 1 / 6 we ' re asked to find the value of 5 * / 4 * now that we know how the symbol ' works ' , solving this problem should n ' t be too difficult . you can actually choose to do the math in a couple of different ways . . . . 5 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) do n ' t calculate this just yet though . . . . since we ' re dividing by 4 * , many of those fractions will ' cancel out . ' 4 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) we ' re looking for the value of : ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) / ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) since the first four fraction in the numerator and denominator cancel out , we ' re left with just one fraction : 1 / 4 d" | a = 1 / 3
b = a / 3
c = 1 + 4
d = b / c
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | floor(divide(10, subtract(5, const_1))) | if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 10 ? | "the product of n and 5 = 5 n . . the difference between an integer n and the product of n and 5 = 5 n - n = 4 n . . the outcome of the binary operation of n is less than 10 ? so 4 n < 10 . . . max value that n can take as an integer is 2 answer : b" | a = 5 - 1
b = 10 / a
c = math.floor(b)
|
a ) 287 , b ) 288 , c ) 900 , d ) 277 , e ) 121 | c | multiply(multiply(subtract(divide(400, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2) | how many seconds will a 400 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "let length of tunnel is x meter distance = 400 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 400 + x = ( 65 / 3 ) * 60 400 + x = 20 * 65 = 1300 x = 1300 - 400 = 900 meters answer : c" | a = 63 - 3
b = a * const_0_2778
c = 400 / b
d = c - 1
e = d * 10
f = e * 2
|
a ) 36 , b ) 48 , c ) 17 , d ) 6 , e ) 5 | b | add(power(divide(subtract(8, sqrt(subtract(power(8, 2), multiply(const_4, 8)))), 2), 2), power(divide(add(8, sqrt(subtract(power(8, 2), multiply(const_4, 8)))), 2), 2)) | if a and b are the roots of the equation x 2 - 8 x + 8 = 0 , then the value of a 2 + b 2 is : | "sol . ( b ) the sum of roots = a + b = 8 product of roots = ab = 8 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 64 - 16 = 48 answer b" | a = 8 ** 2
b = 4 * 8
c = a - b
d = math.sqrt(c)
e = 8 - d
f = e / 2
g = f ** 2
h = 8 ** 2
i = 4 * 8
j = h - i
k = math.sqrt(j)
l = 8 + k
m = l / 2
n = m ** 2
o = g + n
|
a ) 730848 , b ) 705265 , c ) 730846 , d ) 730596 , e ) 731848 | a | subtract(45678, multiply(multiply(8, 4), 8)) | evaluate : 45678 * 8 * 4 * 8 = ? | "according to order of operations , 8 ? 4 ? 8 ( division and multiplication ) is done first from left to right 8 / 4 = 2 * 8 = 16 hence 45678 * 8 * 4 * 8 = 45678 * 16 = 730848 correct answer a" | a = 8 * 4
b = a * 8
c = 45678 - b
|
a ) 60 , b ) 70 , c ) 80 , d ) 90 , e ) 100 | c | subtract(subtract(200, 1), subtract(add(add(divide(subtract(subtract(200, const_2), const_2), const_2), 1), add(divide(subtract(subtract(200, divide(200, const_100)), divide(200, const_100)), divide(200, const_100)), 1)), add(divide(subtract(subtract(200, const_10), const_10), const_10), 1))) | what is the total number of positive integers that are less than 200 and that have no positive factor in common with 200 other than 1 ? | "since 200 = 2 ^ 3 * 5 ^ 2 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 100 odd numbers from 1 to 200 . we then need to eliminate the 20 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 195 . there are a total of 100 - 20 = 80 such numbers between 1 and 200 . the answer is c ." | a = 200 - 1
b = 200 - 2
c = b - 2
d = c / 2
e = d + 1
f = 200 / 100
g = 200 - f
h = 200 / 100
i = g - h
j = 200 / 100
k = i / j
l = k + 1
m = e + l
n = 200 - 10
o = n - 10
p = o / 10
q = p + 1
r = m - q
s = a - r
|
a ) $ 23.45 , b ) $ 65.33 , c ) $ 40 , d ) $ 38.27 , e ) $ 39.33 | e | add(add(add(25, 8), 6), divide(add(25, 8), const_100)) | it costs $ 6 for the first 1 / 4 hour to use the copier at kinkos . after the first ¼ hour it costs $ 8 per hour . if a certain customer uses the copier for 4 hours and 25 minutes , how much will it cost him ? | 4 hrs 25 min = 265 min first 15 min - - - - - - > $ 6 time left is 250 min . . . now , 60 min costs $ 8 1 min costs $ 8 / 60 250 min costs $ 8 / 60 * 250 = > $ 33.33 so , total cost will be $ 33.33 + $ 6 = > $ 39.33 the answer will be ( e ) $ 39.33 | a = 25 + 8
b = a + 6
c = 25 + 8
d = c / 100
e = b + d
|
a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 2818075 / 2 | e | multiply(1300000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1300000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city at the end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 2818075 / 2 e" | a = 15 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1300000 * h
|
a ) 30000 , b ) 25375 , c ) 28680 , d ) 32000 , e ) 23125 | e | add(subtract(50000, multiply(multiply(multiply(multiply(10, 8), 11), 11), 3)), multiply(multiply(multiply(10, 8), 11), 3)) | a man invested rs . 50000 at the rate of 5 % p . a . for the first 3 years , 8 % p . a . for the next 5 years and 10 % p . a . for the period beyond 8 years . at the end of 11 years , he want to distribute total money among his 4 sons , how much money will each son get ? | let the interest be i . then , ( 50000 × 3 × 5 / 100 ) + ( 50000 × 5 × 8 / 100 ) + ( 50000 × 3 × 10 / 100 ) = i ⇒ 7500 + 20000 + 15000 = i ⇒ i = 42500 total money = 42500 + 50000 = 92500 each son will get = 92500 / 4 = 23125 answer : e | a = 10 * 8
b = a * 11
c = b * 11
d = c * 3
e = 50000 - d
f = 10 * 8
g = f * 11
h = g * 3
i = e + h
|
a ) 270 , b ) 279 , c ) 226 , d ) 231 , e ) 232 | e | multiply(divide(841, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 841 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should a pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 8 / 29 * 841 = 232 answer : e" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 841 / e
g = 16 * 9
h = f * g
|
a ) 1 / 144 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3 | b | multiply(divide(6, power(6, const_2)), divide(6, power(6, const_2))) | what is the probability that the sum of two dice will yield a 7 , and then when both are thrown again , their sum will again yield a 7 ? assume that each die has 6 sides with faces numbered 1 to 6 . | "in order to get the sum of the two dices to be 7 , there are six combinations that satisfy this requirement . 1 + 6 = 7 2 + 5 = 7 3 + 4 = 7 4 + 3 = 7 5 + 2 = 7 6 + 1 = 7 each combination has 1 / 36 of probability of occurrence , or in total 6 / 36 for the six combinations , which is 1 / 6 . in order to get the probability of this occurring twice in a row we multiply 1 / 6 * 1 / 6 and we get 1 / 36 . the answer is b ." | a = 6 ** 2
b = 6 / a
c = 6 ** 2
d = 6 / c
e = b * d
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 3))), const_10) | one fast typist type some matter in 2 hrs and another slow typist type the same matter in 3 hrs . if both do combine in how much time they will take to finish . | fast ' s 1 hr work = 1 / 2 slow ' s 1 hr work = 1 / 3 1 / 2 + 1 / 3 = 5 / 6 they ll finish in 6 / 5 hrs = 1 1 / 5 = = > 1 hr 12 mins answer : c | a = 1 / 2
b = 1 / 3
c = a + b
d = 1 / c
e = d * 10
|
a ) 25 , b ) 27 , c ) 40 , d ) 56 , e ) 80 | b | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | how many internal diagonals does a nonagon ( nine sided polygon ) have ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 9 no . of diagonals = 9 ( 9 - 3 ) / 2 = 27 ans b" | a = 2 * 4
b = a - 3
c = 2 * 4
d = c / 2
e = b * d
|
a ) 157.78 , b ) 157.98 , c ) 400 , d ) 420 , e ) 430 | c | subtract(divide(multiply(multiply(3200, 14.5), 5), const_100), divide(multiply(multiply(3200, 12), 5), const_100)) | if a lends rs . 3200 to b at 12 % per annum and b lends the same sum to c at 14.5 % per annum then the gain of b in a period of 5 years is ? | "( 3200 * 2.5 * 5 ) / 100 = > 400 answer : c" | a = 3200 * 14
b = a * 5
c = b / 100
d = 3200 * 12
e = d * 5
f = e / 100
g = c - f
|
a ) $ 40,000 , b ) $ 56,000 , c ) $ 64,000 , d ) $ 66,667 , e ) $ 80,000 | b | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(12, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | country x taxes each of its citizens an amount equal to 12 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ? | "equation is correct , so math must be a problem . 0.12 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 4,800 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 11,200 - - > x = 56,000 . answer : b ." | a = 4 * 10
b = a * 1000
c = 4 * 2
d = c * 1000
e = 12 / 100
f = 4 * 10
g = f * 1000
h = e * g
i = d - h
j = 20 / 100
k = i / j
l = b + k
|
a ) 212 , b ) 190 , c ) 277 , d ) 181 , e ) 122 | a | subtract(divide(multiply(const_1, const_1000), divide(54, 12)), 12) | in a kilometer race , a beats b by 54 meters or 12 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 12 ) / 54 = 222 sec . time taken by a = 224 - 12 = 212 sec . answer : a" | a = 1 * 1000
b = 54 / 12
c = a / b
d = c - 12
|
a ) 45 , b ) 55 , c ) 69 , d ) 65 , e ) 77 | e | divide(add(add(add(60, const_1), add(add(60, const_1), const_2)), add(subtract(90, 60), subtract(90, const_2))), 60) | find the average of all prime numbers between 60 and 90 | "prime numbers between 60 and 90 are 67 , 71 , 73 , 79 , 83 , 89 required average = ( 67 + 71 + 73 + 79 + 83 + 89 ) / 6 = 462 / 6 = 77 answer is e" | a = 60 + 1
b = 60 + 1
c = b + 2
d = a + c
e = 90 - 60
f = 90 - 2
g = e + f
h = d + g
i = h / 60
|
a ) 1 / 3 , b ) 2 / 3 , c ) 11 / 120 , d ) 11 / 30 , e ) 76 / 120 | d | subtract(add(subtract(inverse(factorial(const_2)), inverse(factorial(const_3))), inverse(factorial(const_4))), inverse(factorial(5))) | 5 letters have to be put in 5 different envelopes numbered 1 through 5 such that each of the letters go into only 1 envelope . the letter is said to be put in correct position if for example letter 1 goes into envelope 1 . now what is the probability that all letters be put into wrong envelopes ? | this q is based on derangement , a permutation in which all elements are in the wrong position . number of derangements = n ! ( 12 ! − 13 ! + 14 ! + . . . + ( ( − 1 ) n ) n ! ) n ! ( 12 ! − 13 ! + 14 ! + . . . + ( ( − 1 ) n ) n ! ) . . of course it can be derived but thats not required . . since there are 5 letters and 5 envelopes : - derangements = 5 ! ( 12 ! − 13 ! + 14 ! − 15 ! ) derangements = 5 ! ( 12 ! − 13 ! + 14 ! − 15 ! ) = > 120 ( 1 / 2 − 1 / 6 + 1 / 24 − 1 / 120 ) = 60 − 20 + 5 − 1 = 44.120 ( 1 / 2 − 1 / 6 + 1 / 24 − 1 / 120 ) = 60 − 20 + 5 − 1 = 44 . total possible ways = 5 ! = 120 . therefore required prob , p = 44 / 120 = 11 / 30 . ans d | a = math.factorial(2)
b = 1/(a)
c = math.factorial(3)
d = 1/(c)
e = b - d
f = math.factorial(4)
g = 1/(f)
h = e + g
i = math.factorial(5)
j = 1/(i)
k = h - j
|
a ) s . 3988 , b ) s . 3900 , c ) s . 3228 , d ) s . 3922 , e ) s . 3600 | e | multiply(multiply(subtract(add(70, 60), 10), 10), 3) | a rectangular lawn of dimensions 70 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "area = ( l + b – d ) d ( 70 + 60 – 10 ) 10 = > 1200 m 2 1200 * 3 = rs . 3600 answer : e" | a = 70 + 60
b = a - 10
c = b * 10
d = c * 3
|
a ) 44 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | a | divide(220, subtract(divide(220, 40), divide(30, const_60))) | the distance from city a to city b is 220 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | "the time it takes bob to drive to city b is 220 / 40 = 5.5 hours . alice needs to take less than 5 hours for the trip . alice needs to exceed a constant speed of 220 / 5 = 44 miles per hour . the answer is a ." | a = 220 / 40
b = 30 / const_60
c = a - b
d = 220 / c
|
a ) 34 % , b ) 24 % , c ) 42 % , d ) 18 % , e ) 8.5 % | c | multiply(divide(subtract(multiply(const_100, divide(18, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 18 percent sugar by weight . the second solution was what percent sugar by weight ? | "instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 10 % + 1 part of x ( unknown ) % = 4 parts of 18 % = > x % = 72 % - 30 % = 42 % ans c it is ." | a = 18 / 100
b = 100 * a
c = 1 / 4
d = c * 100
e = 100 - d
f = 10 / 100
g = e * f
h = b - g
i = 1 / 4
j = i * 100
k = h / j
l = k * 100
|
a ) 18 , b ) 19 , c ) 17 , d ) 16 , e ) 15 | e | divide(subtract(multiply(5, 800), multiply(650, 5)), subtract(700, 650)) | the average salary / head of all the workers in a workshop is rs . 700 , if the average salary / head of 5 technician is rs . 800 and the average salary / head of the rest is rs . 650 , the total no . of workers in the work - shop is ? | let the total number of workers be y . so sum of salary for all workers = sum of salary of 5 technician + sum of salary for other y - 5 workers . 5 x 800 + 650 ( y - 5 ) = 700 y ⇒ 4000 + 650 y - 3250 = 700 y ⇒ 50 y = 750 ∴ y = 15 so total number of workers = 15 e | a = 5 * 800
b = 650 * 5
c = a - b
d = 700 - 650
e = c / d
|
a ) 85 , b ) 87 , c ) 89 , d ) 91 , e ) 93 | d | subtract(factorial(subtract(9, 5)), add(multiply(5, const_3), const_3)) | there are 9 executives , including the ceo and cfo , that are asked to form a small team of 5 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | "the total number of ways to form a team of 5 is 9 c 5 = 126 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 7 c 3 = 35 . the number of ways to form an acceptable team is 126 - 35 = 91 . the answer is d ." | a = 9 - 5
b = math.factorial(a)
c = 5 * 3
d = c + 3
e = b - d
|
a ) 75 , b ) 100 , c ) 120 , d ) 175 , e ) 225 | c | divide(subtract(multiply(divide(720, const_3), const_4), 720), const_2) | there are 720 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 720 . hence , we have m + f = 720 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f / 2 + m / 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 720 / 3 = 240 solving the three equations yields the solution f = 240 , m = 480 , and d = 240 . the number of female democratic participants equals half the female participants equals 240 / 2 = 120 . answer : c" | a = 720 / 3
b = a * 4
c = b - 720
d = c / 2
|
a ) 6.25 , b ) 6.28 , c ) 6.11 , d ) 5.53 , e ) 6.21 | d | divide(subtract(250, multiply(20, 4.2)), 30) | in the first 20 overs of a cricket game , the run rate was only 4.2 . what should be the rate in the remaining 30 overs to reach the target of 250 runs ? | "required run rate = [ 250 - ( 4.2 * 20 ) ] / 30 = 166 / 30 = 5.53 answer : d" | a = 20 * 4
b = 250 - a
c = b / 30
|
a ) 10 , b ) 20 , c ) 30 , d ) 60 , e ) 80 | e | divide(multiply(multiply(const_1000, const_100), 0.036), divide(45000, const_1000)) | in the biology lab of ` ` jefferson ' ' high school there are 0.036 * 10 ^ 5 germs , equally divided among 45000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ? | "0.036 * 10 ^ 5 can be written as 3600 45000 * 10 ^ ( - 3 ) can be written as 45 required = 3600 / 45 = 80 answer : e" | a = 1000 * 100
b = a * 0
c = 45000 / 1000
d = b / c
|
a ) 11 , b ) 14 , c ) 18 , d ) 19 , e ) 10 | d | divide(1200, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 10)) | if the wheel is 10 cm then the number of revolutions to cover a distance of 1200 cm is ? | "2 * 22 / 7 * 10 * x = 1200 = > x = 19 answer : d" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 10
h = 1200 / g
|
a ) 45 , b ) 70 , c ) 100 , d ) 105 , e ) 140 | a | subtract(subtract(441, divide(subtract(multiply(4, divide(441, 7)), multiply(5, divide(multiply(4, divide(441, 7)), add(5, 1)))), divide(1, 5))), multiply(5, divide(multiply(4, divide(441, 7)), add(5, 1)))) | in an apartment building that has 441 units , 4 out of every 7 units are currently rented , including 1 / 5 of the one - bedroom units . if , of the rented apartments , there is a 6 : 1 ratio of two - bedroom units to one - bedroom units , and the building only consists of two - bedroom and one - bedroom units , how many two - bedroom units are not rented ? | "rented = 4 / 7 * 441 = 252 . rented two - bedroom = 6 / 7 * 252 = 216 ; rented one - bedroom = 1 / 7 * 252 = 36 or 252 - 216 = 36 ) . rented one - bedroom units are 1 / 5 of all one - bedroom : 1 / 5 * { all one bedroom } = 36 - - > { all one - bedroom } = 180 . { all two - bedroom } = 441 - 180 = 261 . two - bedroom not rented = 261 - 216 = 45 . answer : a ." | a = 441 / 7
b = 4 * a
c = 441 / 7
d = 4 * c
e = 5 + 1
f = d / e
g = 5 * f
h = b - g
i = 1 / 5
j = h / i
k = 441 - j
l = 441 / 7
m = 4 * l
n = 5 + 1
o = m / n
p = 5 * o
q = k - p
|
a ) 24 , b ) 35 , c ) 28 , d ) 30 , e ) 32 | b | multiply(divide(7, 8), add(35, const_4)) | john was 35 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 35 john ' s age after 5 years = 40 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 40 ) = 35 therefore betty ' s current age = 35 option b" | a = 7 / 8
b = 35 + 4
c = a * b
|
a ) rs . 3840 , b ) rs . 5270 , c ) rs . 6275 , d ) rs . 6720 , e ) none of these | a | divide(4800, add(const_1, divide(25, const_100))) | the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 4800 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 4800 ( 100 / 125 ) = rs . 3840 . answer : a" | a = 25 / 100
b = 1 + a
c = 4800 / b
|
a ) 16 , b ) 17 , c ) 18 , d ) 15 , e ) 14 | d | sqrt(divide(675, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 675 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 675 3 b 2 = 675 b 2 = 225 b = 15 m . answer : option d" | a = 675 / 3
b = math.sqrt(a)
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(add(add(16, 19), 2), 30) | in a sports club with 30 members , 16 play badminton and 19 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 16 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 16 - x ) + ( 19 - x ) + x + 2 = 30 37 - 2 x + x = 30 37 - x = 30 x = 7 so 7 members play both badminton and tennis . a" | a = 16 + 19
b = a + 2
c = b - 30
|
a ) 10.9 sec , b ) 14.8 sec , c ) 6.12 sec , d ) 10.8 sec , e ) 20.8 sec | c | divide(add(50, 120), multiply(add(60, 40), const_0_2778)) | two trains 50 m and 120 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 50 + 120 = 170 m . required time = 170 * 9 / 250 = 6.12 = 6.12 sec . answer : c" | a = 50 + 120
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 205 , b ) 215 , c ) 225 , d ) 260 , e ) 270 | c | divide(subtract(500, multiply(divide(500, const_2), divide(const_1, add(const_1, const_4)))), const_2) | a large field of 500 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | "average of the two areas = 500 / 2 = 250 one - fifth of the average of the two areas = 250 / 5 = 50 ⇒ difference of the two areas = 50 let area of the smaller part = x hectares . then , area of the larger part = x + 50 hectares . = > x + x + 50 = 500 = > 2 x = 450 = > x = 225 answer : c" | a = 500 / 2
b = 1 + 4
c = 1 / b
d = a * c
e = 500 - d
f = e / 2
|
a ) 182 , b ) 260 , c ) 224 , d ) 254 , e ) 302 | b | divide(195, divide(subtract(const_100, 25), const_100)) | a small college reduced its faculty by approximately 25 percent to 195 professors . what was the original number of faculty members ? | "if x is the original number of faculty members , then after 25 % reduction in faculty members number is . 75 x but we are given . 75 x = 195 x = 260 so the original number of faculty members is 260 correct answer - b" | a = 100 - 25
b = a / 100
c = 195 / b
|
a ) s . 1350 , b ) s . 1327 , c ) s . 1328 , d ) s . 1397 , e ) s . 1918 | e | multiply(subtract(rectangle_area(add(75, multiply(3.5, 2)), add(55, multiply(3.5, 2))), rectangle_area(75, 55)), 2) | a rectangular grass field is 75 m * 55 m , it has a path of 3.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 3.5 * 2 ) 2 * 3.5 = > 959 959 * 2 = rs . 1918 answer : e" | a = 3 * 5
b = 75 + a
c = 3 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 28 sec , b ) 16 sec , c ) 39 sec , d ) 18 sec , e ) 17 sec | c | divide(add(240, 150), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 150 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 150 = 390 m . time taken = 390 / 10 = 39 sec . answer : c" | a = 240 + 150
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 2,000 , b ) 1,000 , c ) 400 , d ) 50 , e ) 1,500 | a | divide(20, 10) | a bag making machine can make 20 bags per hour . if the company adjusts the machine to produce twice the amount per hour to satisfy expected rise in demand . how many bags can the machine make in the next 5 days if it works 10 hours a day ? | 20 * 2 = 40 bags per hour . 5 days * 10 hours * 40 bags per hour = 2,000 bags . answer is option a . | a = 20 / 10
|
a ) 1 / 70 , b ) 1 / 35 , c ) 2 / 35 , d ) 1 / 12 , e ) 1 / 7 | b | divide(1, add(add(const_4, const_1), multiply(const_3, const_10))) | 0 - - - : - | - - - : - - | - : - - : - | - - : - - - | - : - - - - 1 on the number line above , the segment from 0 to 1 has been divided into fifths , as indicated by the large tick marks , and also into sevenths , as indicated by the small tick marks . what is the least possible distance z between any two of the tick marks ? | divisions are 0 1 / 7 2 / 7 3 / 7 4 / 7 5 / 7 6 / 7 1 1 / 5 2 / 5 3 / 5 4 / 5 5 / 5 expressing the same in terms of the lcm of 57 , i . e 35 0 7 / 35 14 / 35 21 / 35 28 / 35 35 / 35 5 / 35 10 / 35 15 / 35 20 / 35 25 / 35 30 / 35 by comparing the divisions with each other we can see 1 / 35 is the shortest possible distance between any two selected divisions . z = 1 / 35 answer is b . | a = 4 + 1
b = 3 * 10
c = a + b
d = 1 / c
|
a ) 1 : 2 , b ) 2 : 2 , c ) 2 : 5 , d ) 2 : 6 , e ) 2 : 7 | a | divide(subtract(divide(3, add(6, 2)), divide(6, 2)), subtract(divide(6, 2), divide(6, add(6, 2)))) | an alloy of copper and zinc contains copper and zinc in the ratio 3 : 5 . another alloy of copper and zinc contains copper and zinc in the ratio 6 : 2 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 3 x / 8 , and fraction of zinc in alloy _ 1 = 5 x / 8 . similarly , fraction of copper in alloy _ 2 = 6 y / 8 , and fraction of zinc in alloy _ 2 = 2 y / 8 . mixing them , we get copper = 3 x / 8 + 6 y / 8 ; zinc = 5 x / 8 + 2 y / 8 . so , 3 x + 6 y = 5 x + 2 y - > 2 x = 4 y - > x / y = 2 / 4 = 1 / 2 so , they must be mixed in the ratio 1 : 2 answer : a" | a = 6 + 2
b = 3 / a
c = 6 / 2
d = b - c
e = 6 / 2
f = 6 + 2
g = 6 / f
h = e - g
i = d / h
|
a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) 79 | d | divide(add(add(add(add(76, 65), 82), 67), 85), add(const_2, const_3)) | david obtained 76 , 65 , 82 , 67 and 85 marks ( out in 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "add all those marks and divide it by 5 becz they gave us 5 subjects for finding average marks we need to divide them add them 76 + 65 + 82 + 67 + 85 / 5 = 375 / 5 = 75 answer : d" | a = 76 + 65
b = a + 82
c = b + 67
d = c + 85
e = 2 + 3
f = d / e
|
a ) 3775 , b ) 3779 , c ) 3629 , d ) 3729 , e ) 3719 | a | divide(multiply(100, add(100, 51)), 52) | 51 + 52 + 53 + . . . . . . . . . + 100 ) is equal to | "51 + 52 + 53 + . . . . . . . . . . . + 100 = ( 1 + 2 + 3 + . . . . + 100 ) - ( ! + 2 + 3 + . . . . . . + 50 ) = = ( 5050 - 1275 ) = 3775 answer ; a" | a = 100 + 51
b = 100 * a
c = b / 52
|
a ) 6 , b ) 8 , c ) 10 , d ) 11 , e ) 13 | c | divide(log(add(divide(multiply(2952400, const_2), 100), const_1)), log(const_3)) | joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 2952400 , how many months will it take joan to pay back her mortgage ? | "joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 2952400 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 29524 so we have 1 , 3 , 9 , 27 . . . . . 2652400 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) / 1 - r using this and plugging in the information we get . . . 29524 = 1 - 3 ^ n / 1 - 3 ; 1 - 3 ^ n / - 2 cross multiplying we get 29524 x - 2 = 1 - 3 ^ n - 59048 = 1 - 3 ^ n - 59049 = - 3 ^ n 59049 = 3 ^ n ( negatives cancel out ) 59049 can also be re written as 3 ^ 10 therefore ; 3 ^ 10 = 3 ^ n thus n = 10 ( c )" | a = 2952400 * 2
b = a / 100
c = b + 1
d = math.log(c)
e = math.log(3)
f = d / e
|
a ) 3.0 , b ) 7.0 , c ) 5.0 , d ) 2.0 , e ) 1.0 | c | divide(const_100, 20) | a company decreased the price of its main product by 20 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | "for the total revenue to remain the same when the price is one fifth , the number of products sold must five times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 20 % = 5.0 answer : c" | a = 100 / 20
|
a ) 1152 , b ) 1987 , c ) 2012 , d ) 1856 , e ) 2546 | a | multiply(factorial(4), factorial(4)) | in how many ways 4 boys and 4 girls can be seated in a row so that boys and girls are alternate ? | "number of ways = 2 * 4 ! * 4 ! = 1152 answer is a" | a = math.factorial(4)
b = math.factorial(4)
c = a * b
|
a ) 3 , b ) 9 , c ) 12 , d ) 15 , e ) 17 | b | add(3, const_1) | the average of first five multiples of 3 is : | "solution average = 3 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 45 / 5 = 9 answer b" | a = 3 + 1
|
a ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | d | divide(subtract(multiply(17, subtract(40, 4)), multiply(17, 32)), 4) | the average age of an adult class is 40 years . 17 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 17 x 32 = ( y + 17 ) x 36 â ‡ ’ 40 y + 544 = 36 y + 612 â ‡ ’ 4 y = 68 â ˆ ´ y = 17 d" | a = 40 - 4
b = 17 * a
c = 17 * 32
d = b - c
e = d / 4
|
a ) 60 kmph , b ) 75 kmph , c ) 80 kmph , d ) 85 kmph , e ) 72 kmph | e | multiply(divide(300, subtract(27, 12)), const_3_6) | a train requires 12 seconds to pass a pole while it requires 27 seconds to cross a stationary train which is 300 mtrs long . find the speed of the train . | "in 12 s the train crosses the pole and in 27 sec the train crosses one more stationary train in 15 sec the train travels a distance of 300 mtrs speed = 300 / 15 = 20 m / s = 20 ( 3600 / 1000 ) = 20 * 18 / 5 = 72 kmph answer : e" | a = 27 - 12
b = 300 / a
c = b * const_3_6
|
a ) 560 , b ) 240 , c ) 329 , d ) 443 , e ) 544 | b | multiply(divide(360, const_3), const_2) | a salesman sold twice as much pears in the afternoon than in the morning . if he sold 360 kilograms of pears that day , how many kilograms did he sell in the morning and how many in the afternoon ? | "let x be the number of kilograms he sold in the morning . then in the afternoon he sold 2 xkilograms . so , the total is x + 2 x = 3 x this must be equal to 360 . 3 x = 360 x = 360 / 3 x = 120 therefore , the salesman sold 120 kg in the morning and 2 ⋅ 120 = 240 kg in the afternoon ." | a = 360 / 3
b = a * 2
|
a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) 62 sec | d | add(30, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 30 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 = 30 + 30 = 60 km / hr . 60 * 5 / 18 = 50 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 50 = 60 sec . answer : d" | a = 30 + 30
|
a ) 46 , b ) 80 , c ) f = 90 , d ) f = 100 , e ) 200 | d | add(subtract(divide(add(add(subtract(const_100, 23), 23), 20), const_2), 20), divide(add(add(subtract(const_100, 23), 23), 20), const_2)) | a company conducted a survey about its two brands , a and b . x percent of respondents liked product a , ( x – 20 ) percent liked product b , 23 percent liked both products , and 23 percent liked neither product . what is the minimum number f of people surveyed by the company ? | "100 = x + x - 20 + 23 - 23 x = 60 , so , product a = 60 % , product b = 40 % , both = 23 % , neither = 23 % 23 % of the total no . of people should be an integer . so , a , bc are out . 60 % of d and 40 % of d are both integers . so , d satisfies all conditions . so , answer is d ." | a = 100 - 23
b = a + 23
c = b + 20
d = c / 2
e = d - 20
f = 100 - 23
g = f + 23
h = g + 20
i = h / 2
j = e + i
|
a ) 50 , b ) 100 , c ) 150 , d ) 260 , e ) 250 | d | divide(subtract(multiply(130, divide(60, const_100)), multiply(130, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 130 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 130 total = 130 + 130 = 260 answer ( d )" | a = 60 / 100
b = 130 * a
c = 55 / 100
d = 130 * c
e = b - d
f = 55 / 100
g = 50 / 100
h = f - g
i = e / h
|
a ) 12 , b ) 14 , c ) 16 , d ) 12.3 , e ) 18 | d | divide(add(120, 120), add(divide(120, 16), divide(120, 10))) | two tains of equal lengths take 10 seconds and 16 seconds respectively to cross a telegraph post . if the length of each train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "sol . speed of the first train = [ 120 / 10 ] m / sec = 12 m / sec . speed of the second train = [ 120 / 16 ] m / sec = 7.5 m / sec . relative speed = ( 12 + 7.5 ) = m / sec = 19.5 m / sec . ∴ required time = ( 120 + 120 ) / 19.5 secc = 12.3 sec . answer d" | a = 120 + 120
b = 120 / 16
c = 120 / 10
d = b + c
e = a / d
|
['a ) 240', 'b ) 18', 'c ) 26', 'd ) 17', 'e ) 12'] | a | divide(rectangle_area(30, 16), rectangle_area(2, 1)) | how many paying stones , each measuring 2 m * 1 m are required to pave a rectangular court yard 30 m long and 16 m board ? | 30 * 16 = 2 * 1 * x = > x = 240 answer : a | a = rectangle_area / (
|
a ) 388.15 , b ) 2992 , c ) 4312.5 , d ) 2887.27 , e ) 5773.09 | e | subtract(circle_area(divide(380, multiply(const_2, const_pi))), circle_area(divide(268, multiply(const_2, const_pi)))) | the circumferences of two circles are 268 meters and 380 meters . find the difference between the areas of the larger and the smaller circles . | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 268 and 2 ∏ l = 380 s = 268 / 2 ∏ and l = 380 / 2 ∏ difference between the areas = ∏ l 2 - ∏ s 2 = ∏ { 1902 / ∏ 2 - 1342 / ∏ 2 } = 1902 / ∏ - 1342 / ∏ = ( 190 - 134 ) ( 190 + 134 ) / ∏ = ( 56 ) ( 324 ) / ( 22 / 7 ) = ( 18144 ) / ( 22 ) ( 7 ) = 5773.09 sq m . answer : e" | a = 2 * math.pi
b = 380 / a
c = circle_area - (
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind roe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of roe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . roe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena can cover 7.5 miles in 30 mins teena will cover 15 miles in 60 mins so answer 30 + 60 = 90 mins . d" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 5 | d | subtract(multiply(3, const_1000), 3) | the difference between the local value and the face value of 3 in the numeral 62975143 is | "explanation : ( local value of 3 ) - ( face value of 3 ) = ( 3 - 3 ) = 0 d )" | a = 3 * 1000
b = a - 3
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 36 ? | first , let ' s translate the expression 21 # 36 , using the definition given : 21 # 36 = 21 × 24 × 27 × 30 × 33 × 36 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 36 = 3 ^ 6 ( 7 × 8 × 9 × 10 × 11 × 12 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 36 = 3 ^ 6 ( 7 × 2 ^ 3 × 3 ^ 2 × ( 2 × 5 ) × 11 × ( 2 ^ 2 × 3 ) ) let ' s group the prime bases : 21 # 36 = 2 ^ 6 × 3 ^ 9 × 5 × 7 × 11 the sum of the exponents is 6 + 9 + 1 + 1 + 1 = 18 the answer is c . | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
|
a ) rs 741 , b ) rs 5991 , c ) rs 6001 , d ) rs 6991 , e ) none of these | a | subtract(multiply(add(5, const_1), 625), add(add(add(add(435, 927), 855), 230), 562)) | a grocer has a sale of rs . 435 , rs . 927 , rs . 855 , rs . 230 and rs . 562 for 5 consecutive days . how much sale must he have in the sixth day so that he gets an average sale of rs . 625 ? | "explanation : total sale for 5 days = rs . ( 435 + 927 + 855 + 230 + 562 ) = rs . 3009 . required sale = rs . [ ( 625 x 6 ) - 3009 ] = rs . ( 3750 - 3009 ) = rs . 741 . answer : a" | a = 5 + 1
b = a * 625
c = 435 + 927
d = c + 855
e = d + 230
f = e + 562
g = b - f
|
a ) 43 , b ) 56 , c ) 68 , d ) 360 , e ) 92 | d | multiply(add(const_10, const_2), divide(2400, rectangle_area(10, 8))) | the monthly rent of a shop of dimension 10 feet × 8 feet is rs . 2400 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 2400 / ( 10 * 8 ) = 30 & annual rent per square feet = 12 * 30 = 360 answer : d" | a = 10 + 2
b = 2400 / rectangle_area
c = a * b
|
a ) 4663.5 , b ) 4203.5 , c ) 4303.5 , d ) 4403.5 , e ) 4103.5 | a | add(add(multiply(12000, divide(15, const_100)), multiply(add(12000, multiply(12000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(12000, multiply(12000, divide(15, const_100))), multiply(add(12000, multiply(12000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const_3))) | find compound interest on $ 12000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = $ [ 12000 x ( 1 + ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = $ [ 12000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = $ 16663.50 . : . c . i . = rs . ( 16663.50 - 12000 ) = $ 4663.50 answer a ." | a = 15 / 100
b = 12000 * a
c = 15 / 100
d = 12000 * c
e = 12000 + d
f = 15 / 100
g = e * f
h = b + g
i = 15 / 100
j = 12000 * i
k = 12000 + j
l = 15 / 100
m = 12000 * l
n = 12000 + m
o = 15 / 100
p = n * o
q = k + p
r = 15 / 100
s = r / 3
t = q * s
u = h + t
|
a ) 400 , b ) 520 , c ) 651 , d ) 525 , e ) 421 | e | divide(320, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 320 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 320 cp = 4.21 * 100 = > cp = 421 answer : e" | a = 24 / 100
b = 1 - a
c = 320 / b
|
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