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 ) 16 , 35,36 , b ) 12 , 28,36 , c ) 16 , 28,27 , d ) 16 , 28,36 , e ) none of these | d | multiply(divide(add(add(add(56, 8), 8), 8), add(add(4, 7), 9)), 3) | ratio of ages of 3 persons is 4 : 7 : 9 , 8 years ago , the sum of their ages was 56 . find their present ages | explanation : let the present ages are 4 x , 7 x , 9 x . = > ( 4 x - 8 ) + ( 7 x - 8 ) + ( 9 x - 8 ) = 56 = > 20 x = 80 = > x = 4 so their present ages are : 16 , 28,36 answer : option d | a = 56 + 8
b = a + 8
c = b + 8
d = 4 + 7
e = d + 9
f = c / e
g = f * 3
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a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000 | a | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 12,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,000 and in another 9 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a ." | a = 2 + 3
b = a * 1000
c = b * 12
d = c / 2
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(divide(12, 3), divide(3, divide(12, 3))) | if 12 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "this question is asking how many power of 3 exists in 12 ! easiest way is to identify factor of three 12 - > 4 * 3 = > 1 power of 3 9 - > 3 * 3 = > 2 power of 3 6 - > 2 * 3 = > 1 power of 3 3 - > 1 * 3 = > 1 power of 3 total equal 5 answer : c" | a = 12 / 3
b = 12 / 3
c = 3 / b
d = a + c
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a ) 2 . , b ) 6 . , c ) 5 . , d ) 7 . , e ) 9 . | b | divide(subtract(13, const_1), const_2) | arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 13 from the product of their ages a year from today . what is their age today ? | ad = ( a + 1 ) ( d + 1 ) - 13 0 = a + d - 12 a + d = 12 a = d ( as they are twin brothers ) a = d = 6 b is the answer | a = 13 - 1
b = a / 2
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a ) 84 , b ) 72 , c ) 120 , d ) 144 , e ) 180 | a | divide(350, add(divide(100, const_60), divide(150, const_60))) | a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 350 bolts ? | old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 350 bolts 350 * 6 / 25 = 84 mins ans a | a = 100 / const_60
b = 150 / const_60
c = a + b
d = 350 / c
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a ) 26 , b ) 32 , c ) 18 , d ) 20 , e ) 30 | e | multiply(subtract(32, 26), 5) | what is the minimum possible range in scores of the 3 test - takers ? 3 people each took 5 tests . if the ranges of their scores in the 5 practice tests were 18 , 26 and 32 . | i simply looked at the 3 different possible scores for each individual test : 18 , 32,26 we have to find the minimum range : 32 - 18 = 14 32 - 26 = 6 26 - 18 = 8 the find the minimum range , you have to make the set of the 5 scores as small as possible . which means that 4 of the 5 scores of each individual person is zero . 6 * 5 = 30 answer : e | a = 32 - 26
b = a * 5
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a ) 7580 , b ) 7960 , c ) 8290 , d ) 8640 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(8, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 8 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 8 - 1 / 12 ) = 1 / 24 . work done by the inlet in 1 min . = ( 1 / 24 × 1 / 60 ) = 1 / 1440 . volume of 1 / 1440 part = 6 litres . therefore , volume of whole = ( 1440 × 6 ) ‹ = › 8640 litres . answer d" | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 8 * const_60
e = c / d
f = e - 1
g = b / f
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a ) 365 , b ) 52 , c ) 1024 , d ) 8760 , e ) no one knows | d | multiply(add(divide(const_3600, const_10), divide(const_10, const_2)), multiply(multiply(const_4, const_3), const_2)) | how many hours in one year ? | the day has 24 hours , and the year contains 365 days . so the hours exist in one year is the multiplication of 24 by 365 . so the answer = 365 * 24 = 8760 hours so the correct answer is d | a = 3600 / 10
b = 10 / 2
c = a + b
d = 4 * 3
e = d * 2
f = c * e
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a ) 0 , b ) 1 , c ) 37 , d ) 118 , e ) 513 | b | subtract(power(3, subtract(const_1, const_1)), power(subtract(const_1, const_1), 3)) | if k is a non - negative integer and 15 ^ k is a divisor of 759,325 then 3 ^ k - k ^ 3 = | "at first we should understand whether 759325 divisible by 15 . 15 is equal to 3 * 5 so 759325 should be divisible on these integers 759325 is divisible by 5 because of last digit 5 but not divisible by 3 because sum of its digits does n ' t divisible by 3 : 7 + 5 + 9 + 3 + 2 + 5 = 31 and 31 does n ' t divisible by 3 so 759325 can be divisible by 15 ^ k only if k = 0 and 15 ^ k = 1 so 3 ^ k - k ^ 3 = 1 - 0 = 1 answer is b" | a = 1 - 1
b = 3 ** a
c = 1 - 1
d = c ** 3
e = b - d
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a ) 1 / 3 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 5 | d | divide(const_1, power(const_2, 3)) | when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 3 times , what is the probability that it will land once on heads and twice tails ? | "must be once on heads and twice on tails 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 answer : d" | a = 2 ** 3
b = 1 / a
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a ) 141 , b ) 180 , c ) 130 , d ) 1260 , e ) 1420 | d | add(floor(divide(20, const_3)), const_1) | what is the smallest integer that is multiple of 7 , 9 and 20 | "correct answer : d it is the lcm of 7 , 9 and 20 which is 1260" | a = 20 / 3
b = math.floor(a)
c = b + 1
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a ) 1 , b ) 8 , c ) 9 , d ) 24 , e ) 3 | d | divide(multiply(subtract(const_100, 4), 36), add(const_100, 44)) | if a man lost 4 % by selling oranges at the rate of 36 a rupee at how many a rupee must he sell them to gain 44 % ? | "96 % - - - - 36 144 % - - - - ? 96 / 144 * 36 = 24 answer : d" | a = 100 - 4
b = a * 36
c = 100 + 44
d = b / c
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a ) 13 , b ) 14 , c ) 15 , d ) 41 , e ) 16 | e | subtract(add(4, 13), const_1) | the fisherman sale , all of the prices of the fishes sold were different . if the price of a radio sold at the fisherman sale was both the 4 th highest price and the 13 th lowest price among the prices of the fishes sold , how many fishes were sold at the fisherman sale ? | 3 + 12 + 1 = 16 answer : e | a = 4 + 13
b = a - 1
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a ) 102 , b ) 112 , c ) 122 , d ) 132 , e ) 142 | a | subtract(600, add(add(multiply(divide(45, const_100), 600), multiply(divide(23, const_100), 600)), multiply(divide(15, const_100), 600))) | in a school of 600 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "45 + 23 + 15 = 83 % 100 – 83 = 17 % 600 * 17 / 100 = 102 the answer is a ." | a = 45 / 100
b = a * 600
c = 23 / 100
d = c * 600
e = b + d
f = 15 / 100
g = f * 600
h = e + g
i = 600 - h
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a ) 80 , b ) 120 , c ) 160 , d ) 270 , e ) 110 | c | multiply(80, const_2) | the average of the marks of 10 students in a class is 80 . if the marks of each student are doubled , find the new average ? | "sum of the marks for the 10 students = 10 * 80 = 800 . the marks of each student are doubled , the sum also will be doubled . the new sum = 800 * 2 = 1600 . so , the new average = 1600 / 10 = 160 . answer : c" | a = 80 * 2
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a ) 150 , b ) 750 , c ) 1,250 , d ) 1,500 , e ) 2,500 | e | multiply(50, 5) | in a forest 250 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 250 tagged deers comprise 10 % of total # of deers - - > total # of deers = 250 * 10 = 2,500 . answer : e" | a = 50 * 5
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a ) 13.78 , b ) 13.67 , c ) 13.75 , d ) 13.98 , e ) 13.28 | c | multiply(divide(const_1, const_2), multiply(5, 5.5)) | the area of a sector of a circle of radius 5 cm formed by an arc of length 5.5 cm is ? | "( 5 * 5.5 ) / 2 = 13.75 answer : c" | a = 1 / 2
b = 5 * 5
c = a * b
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a ) 70 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | a | multiply(divide(10, subtract(subtract(const_100, 60), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 60 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 ? | "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 70 % a" | a = 100 - 60
b = a - 10
c = 10 / b
d = c * 100
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | c | subtract(100, add(add(add(subtract(100, 83), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 83 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | 100 - ( 100 - 83 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 17 - 25 - 15 - 30 = 100 - 87 = 13 answer : c | a = 100 - 83
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
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a ) 600 , b ) 525 , c ) 360 , d ) 370 , e ) 380 | b | subtract(multiply(speed(450, 18), 39), 450) | a 450 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 450 / 18 = 25 m / sec . let the length of the platform be x meters . then , ( x + 450 ) / 39 = 25 = > x = 975 m . l = 975 - 450 = 525 answer : option b" | a = speed * (
b = a - 39
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a ) 3 . , b ) 7 . , c ) 10 . , d ) 12 , e ) 15 . | d | multiply(4, 4) | the distance between west - town to east - town is 15 kilometers . two birds start flying simultaneously towards one another , the first leaving from west - town at a speed of 4 kilometers per minute and the second bird , leaving from east - town , at a speed of 1 kilometers per minute . what will be the distance , in kilometers , between the meeting point and west - town ? | "by end of 1 st min : distance covered by 1 st bird is 4 km and 2 nd is 1 km . total distance = 5 km . by end of 2 nd min : distance covered by 1 st bird is 8 km and 2 nd is 2 km . total distance = 10 km . by end of 3 rd min : distance covered by 1 st bird is 12 km and 2 nd is 3 km . total distance = 15 km . so , meeting point is 12 kms from west - town ( from where 1 st bird started ) hence , answer will be d ." | a = 4 * 4
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a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 59 / 94 | e | divide(add(divide(94, 2), divide(94, 8)), 94) | if an integer n is to be chosen at random from the integers 1 to 94 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? | "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 96 , there are 47 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 47 + 12 ) / 94 = 59 / 94 = 5 / 8 ans is e" | a = 94 / 2
b = 94 / 8
c = a + b
d = c / 94
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a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | b | subtract(10, multiply(2, 1)) | what is x if x + 2 y = 10 and y = 1 ? | x = 10 - 2 y x = 10 - 2 . x = 8 answer : b | a = 2 * 1
b = 10 - a
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a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | b | divide(50, const_2) | what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 50 ? | "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 50 / 2 = 25 answer is b" | a = 50 / 2
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 20 | d | subtract(add(negate(10), multiply(subtract(15, const_1), const_2.0)), 10) | set j consists of 15 consecutive even numbers . if the smallest term in the set is - 10 , what is the range of the positive integers in set j ? | "since there are only 15 integers , another approach is the just list all 15 . we get : - 10 , - 8 , - 6 , - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 range of positive integers = 18 - 2 = 16 answer : d" | a = negate + (
b = 15 - 1
c = b * 2
d = a - c
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a ) 21 / 40 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30 | a | divide(add(floor(divide(40, 2)), divide(40, 21)), 40) | a number is selected at random from the first 40 natural numbers . what is the probability that the number is a multiple of either 2 or 21 ? | "number of multiples of 2 from 1 through 40 = 40 / 2 = 20 number of multiples of 21 from 1 through 40 = 40 / 21 = 1 number of multiples of 2 and 21 both from 1 through 40 = number of multiples of 21 * 2 ( = 42 ) = 0 total favourable cases = 20 + 1 - 0 = 21 probability = 21 / 40 answer : option a" | a = 40 / 2
b = math.floor(a)
c = 40 / 21
d = b + c
e = d / 40
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a ) 3 , b ) 2 , c ) 1 , d ) 1 / 2 , e ) 1 / 3 | c | divide(2, 2) | if a , b , c and d are positive integers less than 4 , and 4 ^ a + 3 ^ b + 2 ^ c + 1 ^ d = 78 then what is the value of b / c ? | since a , b and c are positive integers less than 4 we have the following possibilities : 4 ^ a = 416 , or 64 3 ^ b = 39 , or 27 2 ^ c = 24 , or 8 trial and error gives us quite quick the solution of 4 ^ a = 64 3 ^ b = 9 2 ^ c = 4 64 + 9 + 4 = 77 i . e . c = 2 and b = 2 - - - - > b / c = 1 / 1 = 1 the correct answer is c | a = 2 / 2
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a ) 28 , b ) 27 , c ) 11 , d ) 15 , e ) 19 | d | divide(multiply(subtract(112, 32), 6), 32) | on a school â € ™ s annual day sweets were to be equally distributed amongst 112 children . but on that particular day , 32 children were absent . thus the remaining children got 6 extra sweets . how many sweets was each child originally supposed to get ? | explanation : let ' k ' be the total number of sweets . given total number of students = 112 if sweets are distributed among 112 children , let number of sweets each student gets = ' l ' = > k / 112 = l . . . . ( 1 ) but on that day students absent = 32 = > remaining = 112 - 32 = 80 then , each student gets ' 6 ' sweets extra . = > k / 80 = l + 6 . . . . ( 2 ) from ( 1 ) k = 112 l substitute in ( 2 ) , we get 112 l = 80 l + 480 32 l = 480 l = 15 therefore , 15 sweets were each student originally supposed to get . answer : d | a = 112 - 32
b = a * 6
c = b / 32
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a ) 4500 , b ) 7500 , c ) 5000 , d ) 6400 , e ) none of these | d | divide(16, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 16 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2 / 100 = 10 and ci = 100 ( 1 + 5 / 100 ) 2 − 100 ∴ = 100 × 21 × 21 / 20 × 20 − 100 = 41 / 4 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 16 , the sum = 400 × 16 = 6400 answer d" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 5 / 100
e = 2 * d
f = 1 + e
g = c - f
h = 16 / g
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a ) a ) 1 / 2 , b ) b ) 5 / 8 , c ) c ) 6 , d ) d ) 6 / 7 , e ) e ) 7 | a | divide(subtract(divide(5, const_4), const_1), divide(50, const_100)) | there are two numbers . if 50 % of the first number is added to the second number , then the second number increases to its 5 - fourth . find the ratio of the first number to the second number ? | let the two numbers be x and y . 50 / 100 * x + y = 5 / 4 y = > 1 / 2 x = 1 / 4 y = > x / y = 1 / 2 a ) | a = 5 / 4
b = a - 1
c = 50 / 100
d = b / c
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a ) 7000 , b ) 10500 , c ) 14000 , d ) 12000 , e ) 3500 | b | multiply(subtract(2000, 500), add(const_4, const_3)) | if you wish to live to be 100 years old ( hypothetically ) , you must consume 500 calories less than your average daily allowance for your age . if you are in your 60 ' s , and your average daily allowance is 2000 calories per day , how many calories are you allowed in a week ? | to determine the amount of calories allowed per day , you need to subtract 500 from 2000 , for a total of 1500 to determine the amount of calories allowed per week , you need to multiply the daily amount allowed ( 1500 ) by the number of days in a week ( 7 ) for a total of 10500 the correct answer is b | a = 2000 - 500
b = 4 + 3
c = a * b
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 5 | e | divide(250, add(subtract(51, 2), const_1)) | 250 metres long yard , 51 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "51 trees have 50 gaps between them , required distance ( 250 / 50 ) = 15 e" | a = 51 - 2
b = a + 1
c = 250 / b
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a ) 15 , b ) 10 , c ) 28 , d ) 24 , e ) 82 | a | multiply(divide(subtract(2783, 2420), 2420), const_100) | a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 2783 in 3 years . find the rate percent ? | "explanation : 2420 - - - 363 100 - - - ? = > 15 % answer : option a" | a = 2783 - 2420
b = a / 2420
c = b * 100
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a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | b | divide(const_2, 16) | if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 16 ? | "because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 and , thus , a probability of 1 / 8 that it will be divisible by 16 1 * 1 / 8 = 1 / 8 answer : b" | a = 2 / 16
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(add(divide(50, add(const_4, const_1)), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros are there in 50 ! | "to find the number of trailing zeroes in 50 ! factorial we must find the number of integral values in : 50 / 5 + 50 / 5 ^ 2 + 50 / 5 ^ 3 + . . . . . . . . . = 10 + 2 + 0 + 0 . . . . = 12 answer : c" | a = 4 + 1
b = 50 / a
c = 4 + 1
d = 50 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 50 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
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a ) 8 / 3 , b ) 2 / 1 , c ) 8 / 5 , d ) 5 / 8 , e ) 5 / 3 | b | divide(divide(450, 3), divide(300, 4)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 450 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 600 km time taken by eddy = 3 hours average speed of eddy = 450 / 3 = 150 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 150 / 75 = 2 / 1 answer b" | a = 450 / 3
b = 300 / 4
c = a / b
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a ) 35 mins , b ) 30 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins | b | multiply(add(divide(subtract(const_1, divide(15, multiply(const_60, const_1))), add(const_1, const_2)), divide(15, multiply(const_60, const_1))), const_60) | pipe x that can fill a tank in an hour and pipe y that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe y is shut 15 minutes before the tank overflows . when will the tank overflow ? | the last 15 minutes only pipe x was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 3 = 1 / 4 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : b | a = const_60 * 1
b = 15 / a
c = 1 - b
d = 1 + 2
e = c / d
f = const_60 * 1
g = 15 / f
h = e + g
i = h * const_60
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a ) 10 days , b ) 12 days , c ) 15 days , d ) 8 days , e ) 16 days | b | add(multiply(subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 10)), 2)), 15), 2) | a and b can complete a work in 15 days and 10 day . they started doing the work together but after 2 days b had to leave and a alone completed the remaining work . the whole work was completed in ? | "a + b 1 day work = 1 / 15 + 1 / 10 = 1 / 6 work done by a and b in 2 days = 1 / 6 * 2 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 now 1 / 15 work is done by a in 1 day 2 / 3 work will be done by a in 15 * 2 / 3 = 10 days total time taken = 10 + 2 = 12 days answer is b" | a = 1 / 15
b = 1 / 10
c = a + b
d = c * 2
e = 1 - d
f = e * 15
g = f + 2
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a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,500 | d | multiply(divide(6000, 2), divide(const_1, 2)) | the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 6000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ? | "x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 6000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 6000 ) * t = = = > 3 d = ts + 6000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 6000 t - 2 d = - 2 st + 6000 t 2 d = 2 st - 6000 t d = st - 3000 t x 13 : d = s * t st - 3000 t = s * t s - 3000 = s - 1500 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 6000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 6000 2 = 3000 d = s * t d = 3000 * ( 1 / 2 ) d = 1500 answer : d" | a = 6000 / 2
b = 1 / 2
c = a * b
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a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 120 / 7 , e ) 11 / 4 | d | divide(12, divide(const_1, add(divide(const_1, 30), divide(const_1, 40)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 30 , and 40 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 30 + 1 / 40 = 7 / 120 therefore they take 120 / 7 ratio = 120 / 7 = d" | a = 1 / 30
b = 1 / 40
c = a + b
d = 1 / c
e = 12 / d
|
a ) 32 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 . | a | subtract(multiply(sqrt(divide(768, 3)), 3), sqrt(divide(768, 3))) | the roof of an apartment building is rectangular and its length is 3 times longer than its width . if the area of the roof is 768 feet squared , what is the difference between the length and the width of the roof ? | "answer is a : 42 let w be the width , so length is 3 w . therefore : w * 4 w = 768 , solving for , w = 16 , so 3 w - w = 2 w = 2 * 16 = 32" | a = 768 / 3
b = math.sqrt(a)
c = b * 3
d = 768 / 3
e = math.sqrt(d)
f = c - e
|
a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15 | a | multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33)) | the sum of two numbers is 528 and their h . c . f . is 33 . the number of pairs of numbers satisfying the above conditions is : | "explanation : let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 . = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . â ˆ ´ required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 x 7 , 33 x 9 ) . the number of such pairs is 4 . answer is a" | a = 528 + 33
b = 1 + 1
c = a / b
d = 528 + 33
e = 1 + 1
f = d / e
g = f - 33
h = c * g
|
a ) 21 / 4 , b ) 35 , c ) 28 , d ) 26 , e ) 21 | a | divide(add(multiply(divide(5, 2), 3), 3), 2) | a certain fraction is equivalent to 2 / 5 . if the numerator of the fraction is increased by 3 and the denominator is doubled , the new fraction is equivalent to 1 / 3 . what is the sum of the numerator and denominator of the original fraction ? | x / y = 2 / 5 - > 1 ( x + 3 ) / 2 y = 1 / 3 - > 2 divide 1 by 2 : = > 2 x / ( x + 3 ) = 6 / 5 = > 10 x = 6 x + 18 = > x = 3 / 2 = > y = 5 / 2 * 3 / 2 = 15 / 4 so x + y = 21 / 4 the answer is a | a = 5 / 2
b = a * 3
c = b + 3
d = c / 2
|
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | c | subtract(30, log(64)) | it takes 30 days to fill a laboratory dish with bacteria . if the size of the bacteria doubles each day , how long did it take for the bacteria to fill 1 / 64 of the dish ? | the bacteria doubles each day , so after 29 days , the dish was half full . after 28 days , the dish was one quarter full . after 27 days , the dish was one eighth full . after 26 days , the dish was one sixteenth full . after 25 days , the dish was 1 / 32 full . after 24 days , the dish was 1 / 64 full . the answer is c . | a = math.log(64)
b = 30 - a
|
a ) 1.6 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | a | divide(95, multiply(214, const_0_2778)) | in what time will a train 95 m long cross an electric pole , it its speed be 214 km / hr ? | "speed = 214 * 5 / 18 = 59 m / sec time taken = 95 / 59 = 1.6 sec . answer : a" | a = 214 * const_0_2778
b = 95 / a
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | e | add(add(add(add(add(2, 1), 2), 3), 4), 2) | q - 2 : how many two digit numbers of distinct digits can be formed by using digits 1 , 2 , 3 , 4 , 5 , 6 and 7 such that the numbers are divisible by 3 ? | concept : a number will be divisible by 3 if sum of all the digits of the number is divisible by 3 here the possible sum of the two distinct digits can be 3 , 6 , 9 and 12 only in order to satisfy the given condition if sum of digits = 3 , no . = 12 , 21 - - - 2 cases if sum of digits = 6 , no . = 15 , 24 , 42 , 51 - - - 4 cases if sum of digits = 9 , no . = 27 , 36 , 45 , 54 , 63 , 72 - - - 6 cases if sum of digits = 12 , no . = 57 , 75 - - - 2 cases total cases = 2 + 4 + 6 + 2 = 14 cases answer : option e | a = 2 + 1
b = a + 2
c = b + 3
d = c + 4
e = d + 2
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a ) 15 , b ) 20 , c ) 40 , d ) 47 , e ) 49 | e | multiply(35, divide(multiply(divide(72, const_60), 35), subtract(65, 35))) | car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 65 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? | "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 65 t t = 1.4 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 1.4 = 49 miles answer : - e" | a = 72 / const_60
b = a * 35
c = 65 - 35
d = b / c
e = 35 * d
|
a ) 28 , b ) 32 , c ) 35 , d ) 53 , e ) 54 | c | add(multiply(divide(42, multiply(2, 3)), 3), multiply(divide(42, multiply(2, 3)), 2)) | the l . c . m . of two numbers is 42 . the numbers are in the ratio 2 : 3 . then sum of the number is : | let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 42 or x = 7 . the numbers are 14 and 21 . hence , required sum = ( 14 + 21 ) = 35 . answer : option c | a = 2 * 3
b = 42 / a
c = b * 3
d = 2 * 3
e = 42 / d
f = e * 2
g = c + f
|
a ) 4 % of a , b ) 8 % of a , c ) 12 % of a , d ) 9 % of a , e ) 1 % of a | a | divide(20, 20) | if 20 % of a = b , then b % of 20 is the same as : | "20 % of a = b b % of 20 = = = = 4 % of a . answer : a" | a = 20 / 20
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a ) 400 , b ) 450 , c ) 500 , d ) 550 , e ) 600 | b | subtract(subtract(multiply(add(add(3, 5), const_2), 150), multiply(5, 150)), multiply(3, 100)) | a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 150 . find the unknown rate of two towels ? | "10 * 150 = 1500 3 * 100 + 5 * 150 = 1050 1500 – 1050 = 450 b" | a = 3 + 5
b = a + 2
c = b * 150
d = 5 * 150
e = c - d
f = 3 * 100
g = e - f
|
a ) $ 125 , b ) $ 100 , c ) $ 200 , d ) $ 215 , e ) $ 225 | e | multiply(400, power(subtract(const_1, divide(25, const_100)), 2)) | a present value of a machine is $ 400 . its value depletiation rate is 25 % per annum then find the machine value after 2 years ? | "p = $ 400 r = 25 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 400 * 3 / 4 * 3 / 4 = $ 225 approximately answer is e" | a = 25 / 100
b = 1 - a
c = b ** 2
d = 400 * c
|
a ) 24,200 , b ) 28,888 , c ) 24,600 , d ) 24,628 , e ) 24,6012 | a | divide(60.50, divide(const_4, 4)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 60.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 60.50 32 x - 31 x = 6050 * 4 x = 24,200 . answer : a" | a = 4 / 4
b = 60 / 50
|
a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 15 | a | subtract(multiply(divide(60, subtract(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))), 60) | a merchant purchased a jacket for $ 60 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 20 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant ’ s gross profit on this sale ? | "actual cost = $ 60 sp = actual cost + mark up = actual cost + 20 % sp = 60 * 100 / 80 on sale sp = 80 / 100 ( 60 * 100 / 80 ) = 60 gross profit = $ 0 answer is a" | a = 20 / 100
b = 1 - a
c = 60 / b
d = 20 / 100
e = 1 - d
f = c * e
g = f - 60
|
a ) $ 2400 , b ) $ 2464 , c ) $ 2650 , d ) $ 2732 , e ) $ 2800 | a | multiply(2240, divide(add(const_100, 50), add(const_100, 40))) | a store ’ s selling price of $ 2240 for a certain computer would yield a profit of 40 percent of the store ’ s cost for the computer . what selling price would yield a profit of 50 percent of the computer ’ s cost ? | "2240 = x * 1.4 = x * ( 7 / 5 ) . therefore , x = 2240 * 5 / 7 . 2240 is easily divided by 7 to be 320 , which is then easily multiplied by 5 to give x = 1600 . to get 50 % profit , you need 1600 * 1.5 = 2400 . answer : a" | a = 100 + 50
b = 100 + 40
c = a / b
d = 2240 * c
|
['a ) 8 : 5', 'b ) 9 : 25', 'c ) 5 : 8', 'd ) 25 : 9', 'e ) can not be determined from the information provided'] | c | divide(circumface(divide(50, const_2)), circumface(divide(80, const_2))) | two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 80 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 40 / speed b since the times are same , 50 pi / speed a = 80 pi / speed b speeda / speed b = 50 pi / 80 pi = 5 / 8 correct option : c | a = 50 / 2
b = circumface / (
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a ) a ) 1 , b ) b ) 2 , c ) c ) 4 , d ) d ) 6 , e ) e ) 10 | b | multiply(subtract(subtract(3, const_2.0), 3), multiply(multiply(3, const_0_25), subtract(subtract(3, 3), 3))) | what is the area inscribed by the lines y = 3 , x = 2 , y = x + 3 on an xy - coordinate plane ? | "first , let ' s graph the lines y = 3 and x = 2 at this point , we need to find the points where the line y = x + 3 intersects the other two lines . for the vertical line , we know that x = 2 , so we ' ll plug x = 2 into the equation y = x + 3 to get y = 2 + 3 = 5 perfect , when x = 2 , y = 5 , so one point of intersection is ( 2,5 ) for the horizontal line , we know that y = 3 , so we ' ll plug y = 3 into the equation y = x + 3 to get 3 = x + 3 . solve to get : x = 0 so , when y = 3 , x = 0 , so one point of intersection is ( 0,3 ) now add these points to our graph and sketch the line y = x + 3 at this point , we can see that we have the following triangle . the base has length 2 and the height is 2 area = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 2 ) ( 2 ) = 2 answer : b" | a = 3 - 2
b = a - 3
c = 3 * const_0_25
d = 3 - 3
e = d - 3
f = c * e
g = b * f
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 50 | e | divide(subtract(sqrt(add(multiply(multiply(300, 10), const_4), power(10, const_2))), 10), const_2) | tough and tricky questions : distance / rate . on a reconnaissance mission , a state - of - the - art nuclear powered submarine traveled 300 miles to reposition itself in the proximity of an aircraft carrier . this journey would have taken 1 hour less if the submarine had traveled 10 miles per hour faster . what was the average speed , in miles per hour , for the actual journey ? | say , if speed is 60 , 300 / 50 = 6 hrs and 300 / 60 = 5 hrs ( a reduction of 1 hr - > correct answer ) answer ( e ) | a = 300 * 10
b = a * 4
c = 10 ** 2
d = b + c
e = math.sqrt(d)
f = e - 10
g = f / 2
|
a ) 2500 , b ) 3500 , c ) 4500 , d ) 2800 , e ) 4000 | e | multiply(multiply(const_1, const_12), divide(12000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 12000 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 12000 = 4000 answer : e" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 12000 / h
j = a * i
|
a ) 6 , b ) 10 , c ) 9 , d ) 11 , e ) 13 | b | subtract(divide(1300, 30), divide(100, 3)) | one computer can upload 100 megabytes worth of data in 3 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 30 seconds . how long would it take for the second computer , working on its own , to upload 100 megabytes of data ? | since the first computer can upload 100 megabytes worth of data in 3 seconds then in 3 * 10 = 30 seconds it can upload 10 * 100 = 1000 megabytes worth of data , hence the second computer in 30 seconds uploads 1300 - 1000 = 300 megabytes worth of data . the second computer can upload 100 megabytes of data in 10 seconds . answer : b . | a = 1300 / 30
b = 100 / 3
c = a - b
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a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 40 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 40), subtract(const_100, 10)), const_100), subtract(const_100, 10)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 40 % while buying and by 10 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 140 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 140 - 90 ) / 90 * 100 = 40 % answer : d ." | a = 100 + 40
b = 100 - 10
c = a - b
d = c * 100
e = 100 - 10
f = d / e
|
a ) 10.9 , b ) 10.7 , c ) 10.3 , d ) 10.44 , e ) 10.8 | d | divide(add(130, 160), multiply(add(60, 40), const_0_2778)) | two trains 130 m and 160 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 = 130 + 160 = 290 m . required time = 290 * 9 / 250 = 261 / 25 = 10.44 sec . ' answer : d" | a = 130 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number w of employees who take the subway to work ? | p / 3 + f / 4 = p / 3 + ( 48 - p ) / 4 = 12 + p / 2 p / 3 + f / 3 = ( p + f ) / 3 = 48 / 3 = 16 p / 4 + f / 4 = 12 p / 3 + f / 3 > p / 3 + f / 4 > p / 4 + f / 4 - - > 16 > 12 + p / 12 > 12 greatest possible w : 12 + p / 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer | a = 48 * 3
b = 3 * 4
c = a / b
d = 3 * 4
e = 48 - d
f = 48 * 3
g = 3 * 4
h = f / g
i = e / h
j = c + i
|
a ) 30 , b ) 40 , c ) 76 , d ) none , e ) can not be determined | c | divide(57, 75) | how many pieces of 75 cm can be cut from a rope 57 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 57 meters long is = ( 57 meters ) / ( 75 cm ) = ( 57 meters ) / ( 0.75 meters ) = 76 answer : c" | a = 57 / 75
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a ) 2 ^ 3 , b ) 2 ^ 6 , c ) 3 ^ 3 , d ) 6 ^ 3 , e ) 12 ^ 2 | d | power(sqrt(multiply(const_4, power(divide(12, const_4), 2))), const_3) | if r is a positive integer and r ^ 2 is divisible by 12 , then the largest positive integer that must divide r ^ 3 is | since r is an integer so r can not have a 2 and sqrt 3 ( because squaring this will give us a 2 ^ 2 and 3 ( making the product as 12 , and making r ^ 2 as a multiple of 12 ) ) r ^ 2 is divisible by 12 ( 12 = 2 * 2 * 3 ) , so , r should have at least one 2 and one 3 so that r ^ 2 has a 2 ^ 2 and two 3 so , r will have a 2 and a 3 . or r will be a multiple of 6 so , largest possible integer than should divide r ^ 3 is 6 ^ 3 so , answer will be d | a = 12 / 4
b = a ** 2
c = 4 * b
d = math.sqrt(c)
e = d ** 3
|
a ) $ 374 , b ) $ 382 , c ) $ 385 , d ) $ 392 , e ) $ 399 | c | add(divide(406, add(const_1, divide(16, const_100))), multiply(divide(10, const_100), divide(406, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 406 per week . if instead , her weekly salary were to increase by 10 percent , how much would she earn per week ? | "soln : - ( 406 / 116 ) 110 = 385 in this case long division does not take much time . ( 406 / 116 ) = 3.5 35 * 11 = 385 ( 350 + 35 ) answer : c" | a = 16 / 100
b = 1 + a
c = 406 / b
d = 10 / 100
e = 16 / 100
f = 1 + e
g = 406 / f
h = d * g
i = c + h
|
a ) a ) 66 , b ) b ) 77 , c ) c ) 79 , d ) d ) 81 , e ) e ) 82 | a | multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 6) | what is the average of first 21 multiples of 6 ? | "required average = 6 ( 1 + 2 + . . . . + 21 ) / 21 ( 6 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 66 a" | a = 21 + 1
b = 21 * a
c = b / 2
d = c / 21
e = d * 6
|
a ) 6 , b ) 7.5 , c ) 5 , d ) 6 , e ) 7 | e | divide(multiply(14, 3), subtract(14, 8)) | a contractor undertook to do a piece of work in 8 days . he employed certain number of laboures but 3 of them were absent from the very first day and the rest could finish the work in only 14 days . find the number of men originally employed ? | "let the number of men originally employed be x . 8 x = 14 ( x â € “ 3 ) or x = 7 answer e" | a = 14 * 3
b = 14 - 8
c = a / b
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a ) 42 , b ) 84 , c ) 48 , d ) 24 , e ) 43 | a | add(divide(subtract(84, multiply(const_2, const_2)), multiply(const_2, const_2)), add(divide(subtract(84, multiply(const_2, const_2)), multiply(const_2, const_2)), const_2)) | what is the sum of two consecutive even numbers , the difference of whose squares is 84 ? | let the numbers be x and x + 2 . then , ( x + 2 ) 2 - x 2 = 84 4 x + 4 = 84 4 x = 80 x = 20 . the required sum = x + ( x + 2 ) = 2 x + 2 = 42 . answer a 42 | a = 2 * 2
b = 84 - a
c = 2 * 2
d = b / c
e = 2 * 2
f = 84 - e
g = 2 * 2
h = f / g
i = h + 2
j = d + i
|
a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 15 | e | divide(subtract(divide(90, 2), divide(40, 2)), const_2) | a man rows his boat 90 km downstream and 40 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 2 1 / 2 ) = 36 kmph speed upstream = d / t = 40 / ( 2 1 / 2 ) = 16 kmph the speed of the stream = ( 36 - 16 ) / 2 = 15 kmph answer : e" | a = 90 / 2
b = 40 / 2
c = a - b
d = c / 2
|
a ) 150 , b ) 125 , c ) 175 , d ) 200 , e ) 100 | e | subtract(add(add(200, 100), 125), 300) | in the youth summer village there are 300 people , 200 of them are not working , 100 of them have families and 125 of them like to sing in the shower . what is the largest possible number of people in the village , which are working , that do n ' t have families and that are singing in the shower ? | "total = 300 not working = 200 having family = 100 like to sing in shower = 125 working = 300 - 200 = 100 not having family = 300 - 100 = 200 like to sing in shower = 125 largest possible number is the lowest possible among the above thus 100 e" | a = 200 + 100
b = a + 125
c = b - 300
|
a ) 6 % , b ) 24 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | b | multiply(add(divide(const_1, 5), divide(const_1, 25)), const_100) | if x > 0 , x / 5 + x / 25 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 5 + 100 / 25 = 20 + 4 = 24 24 is 24 % of 100 = b" | a = 1 / 5
b = 1 / 25
c = a + b
d = c * 100
|
a ) $ 3400 , b ) $ 3200 , c ) $ 6000 , d ) $ 6400 , e ) $ 9600 | a | multiply(multiply(const_100, add(const_3, const_2)), divide(102, subtract(const_100, add(add(30, 20), 35)))) | each month , after jill pays for rent , utilities , food , and other necessary expenses , she has one fifth of her net monthly salary left as discretionary income . of this discretionary income , she puts 30 % into a vacation fund , 20 % into savings , and spends 35 % on eating out and socializing . this leaves her with $ 102 dollar , which she typically uses for gifts and charitable causes . what is jill ’ s net monthly salary ? | "let x be the monthly salary 15 % of 1 / 5 * x = 102 x = 3400 answer a" | a = 3 + 2
b = 100 * a
c = 30 + 20
d = c + 35
e = 100 - d
f = 102 / e
g = b * f
|
a ) 2018 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2017) | if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2017 ) = ? | "f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 ⇒ ⇒ f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 ⇒ ⇒ f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 ⇒ ⇒ f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2017 ) = 2018 ans : a" | a = 1 + 2017
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a ) 448 , b ) 299 , c ) 421 , d ) 460 , e ) 365 | a | multiply(32, 14) | the h . c . f . of two numbers is 32 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is : | "clearly , the numbers are ( 32 x 13 ) and ( 32 x 14 ) . larger number = ( 32 x 14 ) = 448 . answer : option a" | a = 32 * 14
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a ) 1.8 , b ) 1.4 , c ) 1.2 , d ) 1.0 , e ) 0.8 | e | subtract(divide(multiply(divide(90, const_100), 4), divide(75, const_100)), 4) | a solution is 90 % glycerin . if there are 4 gallons of the solution , how much water , in gallons must be addded to make a 75 % glycerin solution | in 4 gallons of the solution there are 0.9 ∗ 4 = 3.80 gallons of glycerin . we want to add ww gallons of water to 4 gallons of solution so that these 3.6 gallons of glycerin to be 75 % of new solution : 0.9 ∗ 4 = 0.75 ( 4 + w ) - - > w = 0.8 answer : e . | a = 90 / 100
b = a * 4
c = 75 / 100
d = b / c
e = d - 4
|
a ) 44 , b ) 28 , c ) 50 , d ) 52 , e ) 56 | b | subtract(multiply(120, divide(30, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | of the 120 passengers on flight 750 , 30 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "number of passengers on flight = 120 number of female passengers = . 3 * 120 = 36 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 36 - 8 = 28 answer b" | a = 30 / 100
b = 120 * a
c = 10 / 100
d = 120 * c
e = 10 / 100
f = 120 * e
g = f / 3
h = d - g
i = b - h
|
a ) 12.9 , b ) 12.0 , c ) 9.7 , d ) 8.6 , e ) 12.1 | c | divide(multiply(12, 1240), add(1240, 300)) | 1240 men have provisions for 12 days . if 300 more men join them , for how many days will the provisions last now ? | "1240 * 12 = 1540 * x x = 9.7 answer : c" | a = 12 * 1240
b = 1240 + 300
c = a / b
|
a ) 37.5 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 66.6 % | a | divide(15, divide(subtract(const_100, 60), const_100)) | on a certain transatlantic crossing , 15 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 60 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ? | 0.15 p = rt + c 0.6 ( rt ) = no c = > 0.40 ( rt ) had c 0.15 p = 0.40 ( rt ) rt / p = 37.5 % answer - a | a = 100 - 60
b = a / 100
c = 15 / b
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | divide(150, multiply(const_0_2778, 36)) | in how many seconds will a train 150 meters long pass an oak tree , if the speed of the train is 36 km / hr ? | "speed = 36 * 5 / 18 = 10 m / s time = 150 / 10 = 15 seconds the answer is c ." | a = const_0_2778 * 36
b = 150 / a
|
a ) 35 , b ) 56 , c ) 76 , d ) 87 , e ) 12 | e | divide(multiply(48, 4), 16) | two numbers n and 16 have lcm = 48 and gcf = 4 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 * n = 48 * 4 n = 48 * 4 / 16 = 12 correct answer e" | a = 48 * 4
b = a / 16
|
a ) 81000 , b ) 81007 , c ) 81008 , d ) 81066 , e ) 82265.6 | e | add(add(65000, multiply(divide(1, 8), 65000)), multiply(divide(1, 8), add(65000, multiply(divide(1, 8), 65000)))) | every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 65000 ? | "65000 * 9 / 8 * 9 / 8 = 82265.6 answer : e" | a = 1 / 8
b = a * 65000
c = 65000 + b
d = 1 / 8
e = 1 / 8
f = e * 65000
g = 65000 + f
h = d * g
i = c + h
|
a ) 4 : 3 , b ) 3 : 5 , c ) 3 : 4 , d ) 50 : 3 , e ) none | c | divide(divide(550, const_60), divide(multiply(33, const_1000), multiply(45, const_60))) | a track covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minute . the ratio of their speeds is : | solution : speed of track = 550 per minute . speed of bus = 33 kms / 45 = 33000 / 45 = 733.33 m / minute ratio of their speeds = 550 / 733.33 = 3 : 4 . answer : option c | a = 550 / const_60
b = 33 * 1000
c = 45 * const_60
d = b / c
e = a / d
|
a ) 432 sq m , b ) 192 sq m , c ) 452 sq m , d ) 428 sq m , e ) 528 sq m | b | multiply(multiply(divide(64, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(64, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 64 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 64 l = 24 b = 8 lb = 24 * 8 = 192 answer : b" | a = 3 * 2
b = 1 * 2
c = a + b
d = 64 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 64 / h
j = e * i
|
a ) 1 / 12 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 / 3 , e ) 2 1 / 9 | d | subtract(add(divide(3, 8), divide(1, 3)), divide(15, 40)) | the instructions state that cheryl needs 3 / 8 square yards of one type of material and 1 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 15 / 40 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 3 / 8 + 1 / 3 left part 15 / 40 - - - > 3 / 8 so used part 3 / 8 + 1 / 3 - 3 / 8 = 1 / 3 ans d" | a = 3 / 8
b = 1 / 3
c = a + b
d = 15 / 40
e = c - d
|
a ) 10 , b ) 1 / 2 , c ) 2 , d ) 1 / 4 , e ) 20 | b | multiply(10, 1) | what number has a 1 : 20 ratio to the number 10 ? | "1 : 20 = x : 10 20 x = 10 ; x = 10 / 20 x = 1 / 2 answer : b" | a = 10 * 1
|
a ) 1344 , b ) 4200 , c ) 8400 , d ) 50400 , e ) 67200 | b | multiply(multiply(560, divide(10, 8)), 6) | running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 10 machines produce in 6 minutes ? | 8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 10 machine produce 560 * 6 * ( 10 / 8 ) in 6 minutes 560 * 6 * 10 / 8 = 4200 answer is b . | a = 10 / 8
b = 560 * a
c = b * 6
|
a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 891.4 hours , e ) 915 hours | d | add(divide(7020, add(16, 2)), divide(7020, subtract(16, 2))) | speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7020 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7020 / 18 + 7020 / 14 = 390 + 501.4 = 891.4 hours answer : option d" | a = 16 + 2
b = 7020 / a
c = 16 - 2
d = 7020 / c
e = b + d
|
a ) 0 ° , b ) 5 ° , c ) 10 ° , d ) 20 ° , e ) none | c | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between the minute hand and the hour hand of a clock when the time is 4.20 ° is | "sol . angle traced by hour hand in 13 / 3 hrs = ( 360 / 12 x 13 / 3 ) ° = 130 ° angle traced by min . hand in 20 min . = ( 360 / 60 x 20 ) ° = 120 ° ∴ required angle = ( 130 - 120 ) ° = 10 ° answer c" | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 5 , d ) 16 : 15 , e ) none | d | divide(multiply(96, const_2), multiply(120, const_3)) | a man invests some money partly in 9 % stock at 96 and partly in 12 % stock at 120 . to obtain equal dividends from both , he must invest the money in the ratio : | "solution for an income of rs . 1 in 9 % stock at 96 , investment = rs . ( 96 / 9 ) = rs . 32 / 3 . for an income of rs . 1 in 12 % stock at 120 , investment = rs . ( 120 / 12 ) = rs . 10 . ∴ ratio of investments = 32 / 3 : 10 = 32 : 30 = 16 : 15 answer d" | a = 96 * 2
b = 120 * 3
c = a / b
|
a ) 54 sec , b ) 48 sec , c ) 42 sec , d ) 70 sec , e ) 60 sec | c | multiply(subtract(8, const_1), divide(30, subtract(6, const_1))) | at 6 ' o clock clock ticks 6 times . the time between first and last ticks was 30 sec . how much time it takes at 8 ' o clock . | at 6 ' 0 clock , clock ticks 6 times . so , there must be 5 intervals between clock ticks . time between first and last ticks = 30 sec so , 1 interval = 30 / 5 = 6 sec so 6 ' o clock 5 * 6 = 30 sec 7 ' o clock 6 * 6 = 36 sec 8 ' o clock 7 * 6 = 42 sec so , 42 sec at 8 ' o clock . answer : c | a = 8 - 1
b = 6 - 1
c = 30 / b
d = a * c
|
a ) 72 , b ) 75 , c ) 81 , d ) 84 , e ) 88 | a | divide(factorial(9), factorial(subtract(9, const_2))) | 9 fencers participate in a fencing championship . assuming all competitors have an equal chance of winning , how many possibilities are there with respect to how a first - place and second - place medal can be awarded ? | 9 * 8 = 72 the answer is a . | a = math.factorial(9)
b = 9 - 2
c = math.factorial(b)
d = a / c
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | what is the units digit of 2222 ^ ( 333 ) * 3333 ^ ( 222 ) ? | "( 2222 ^ 333 ) * ( 3333 ^ 222 ) = 2222 ^ 111 * ( 2222 ^ 222 * 3333 ^ 222 ) here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 ( 2222 ^ 222 * 3333 ^ 222 ) . since 6 powered in any number more than 0 results in 6 as a units digit , as a result we have - 6 * 2222 ^ 111 2 has a cycle of 4 . 111 = 27 * 4 + 3 . 2 ^ 3 = 8 6 * 8 = 48 so the units digit is 8 , and the answer is e" | a = 4 + 3
b = a + 2
|
a ) 20 , b ) 36 , c ) 48 , d ) 50 , e ) 58 | e | divide(subtract(multiply(add(25, 5), 68), multiply(25, 70)), 5) | in a factory , an average of 70 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 68 sets / day . the average production per day for day last 5 days is ? | "production during these 5 days = total production in a month - production in first 25 days . = 30 x 68 - 25 x 70 = 290 ∴ average for last 5 days = 290 / 5 = 58 e" | a = 25 + 5
b = a * 68
c = 25 * 70
d = b - c
e = d / 5
|
a ) 3.42 , b ) 6.16 , c ) 8.32 , d ) 2.0 , e ) 1.75 | e | subtract(divide(add(add(add(22, 25), 30), 40), 4), divide(add(25, 30), const_2)) | alex has 4 pens worth of { 22 , 25 , 30 , 40 } , what is the total mean and median of the worth of pens ? | this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd , the middle term is the median . if the number of terms is even , the average of middle two terms is the median coming to this question , mean = ( 22 + 25 + 30 + 40 ) / 4 = 29.25 median = ( 25 + 30 ) / 2 = 27.5 total = 1.75 option e | a = 22 + 25
b = a + 30
c = b + 40
d = c / 4
e = 25 + 30
f = e / 2
g = d - f
|
a ) 320 , b ) 120 , c ) 400 , d ) 420 , e ) 514 | d | subtract(multiply(divide(add(42, 42), subtract(42, 35)), 35), add(subtract(subtract(42, divide(add(42, 42), subtract(42, 35))), divide(add(42, 42), subtract(42, 35))), const_2)) | there were 35 students in a hostel . due to the admission of 7 new students , ; he expenses of the mess were increased by rs . 42 per day while the average expenditure per head diminished by rs 1 . what was the original expenditure of the mess ? | "sol . let the original average expenditure be rs . x . then , 42 ( x - 1 ) - 35 x = 42 7 x = 84 x = 12 . original expenditure = rs . ( 35 x 12 ) = rs . 420 . . answer d" | a = 42 + 42
b = 42 - 35
c = a / b
d = c * 35
e = 42 + 42
f = 42 - 35
g = e / f
h = 42 - g
i = 42 + 42
j = 42 - 35
k = i / j
l = h - k
m = l + 2
n = d - m
|
a ) 1.25 , b ) 5.75 , c ) 3.25 , d ) 3.75 , e ) none | d | subtract(negate(0.75), multiply(subtract(0.5, 0.5), divide(subtract(0.5, 0.5), subtract(1, 0.5)))) | 1 , 0.5 , 0.5 , 0.75 , 1.5 , ____ | "1 , 0.5 , 0.5 , 0.75 , 1.5 , . . . . . 1 * 0.5 = 0.5 0.5 * 1 = 0.5 0.5 * 1.5 = 0.75 0.75 * 2 = 1.5 so , 1.5 * 2.5 = 3.75 answer : d" | a = negate - (
|
a ) 25 , b ) 66 , c ) 20 , d ) 19 , e ) 17 | c | multiply(divide(15, 30), const_100) | what percent of 15 kg is 30 gms ? | "explanation : required percentage = ( 30 / 15000 * 100 ) % = 1 / 5 % = 0.2 % answer : c ) . 20 %" | a = 15 / 30
b = a * 100
|
a ) $ 190 , b ) $ 180 , c ) $ 200 , d ) $ 240 , e ) $ 270 | e | add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(140, 100), const_100), 100), 3)), 100)) | if $ 100 invested at a certain rate of simple interest amounts to $ 140 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ? | 100 amounts to 140 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 140 100 + 100 * ( r / 100 ) * ( 3 ) = 140 = > r = 40 / 3 150 in 6 years = principal + interest = 150 + 150 * ( r / 100 ) * ( 6 ) = 270 answer is e . | a = 150 * 6
b = 140 - 100
c = b * 100
d = c / 100
e = d / 3
f = a * e
g = f / 100
h = 150 + g
|
a ) 800 , b ) 650 , c ) 500 , d ) 600 , e ) 250 | c | divide(1000, add(const_1, divide(100, const_100))) | the owner of a furniture shop charges his customer 100 % more than the cost price . if a customer paid rs . 1000 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 1000 ( 100 / 200 ) = rs . 500 . answer : c" | a = 100 / 100
b = 1 + a
c = 1000 / b
|
a ) 7.5 , b ) 8.5 , c ) 9 , d ) 9.5 , e ) 10 | b | subtract(divide(25, const_2), 4) | if a father said to his elder son , ` ` i was as old as you are at the present at the time of your birth ' ' . if the father ' s age is 25 years now , what was the son ' s age 4 years back ? | let son ' s present age be a years . then , ( 25 − a ) = a ⇒ 2 a = 25 ⇒ a = 25 / 2 = 12.5 son ' s age 4 years back = 12.5 - 4 = 8.5 answer : b | a = 25 / 2
b = a - 4
|
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