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 ) 640 , b ) 540 , c ) 691 , d ) 385 , e ) 789 | b | multiply(add(12,15, const_1), const_2) | what is the least common multiple of 12,15 , 18,27 | "factors of 12 = 2 * 2 * 3 factors of 27 = 3 * 3 * 3 factors of 15 = 3 * 5 factors of 18 = 2 * 3 * 3 thus the l . c . m is 2 * 3 * 2 * 3 * * 1 * 5 * 3 = 540 the answer is b" | a = 12 + 15
b = a * 2
|
a ) 15 / 16 , b ) 9 / 16 , c ) 1 / 2 , d ) 1 / 4 , e ) 1 / 8 | b | power(divide(const_3, 4), const_2) | given a spinner with 4 sections of equal size labeled e , b , c , and d , what is the probability of not getting an e after spinning the spinner two times ? | the probability of not getting an e after spinning the spinner two times is 3 / 4 * 3 / 4 = 9 / 16 ( so getting any of the remaining 3 letters out of 4 ) . answer : b . | a = 3 / 4
b = a ** 2
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a ) 2.4 . , b ) 4.8 . , c ) 4.2 . , d ) 5.5 . , e ) 6.4 | b | multiply(multiply(divide(divide(2, divide(4, 2)), 2), 3), 4) | two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 3 carpenters build in 4 hours ? | "w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 Γ r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 3 carpenters in 4 hrs = 3 Γ rate of each carpenter x time = 3 Γ 2 / 5 Γ 4 = 4.8 desks b is the correct answer ." | a = 4 / 2
b = 2 / a
c = b / 2
d = c * 3
e = d * 4
|
a ) 9 : 2 , b ) 4 : 5 , c ) 9 : 3 , d ) 17 : 9 , e ) 17 : 4 | a | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 45,000 and rs . 10,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 45000 : 10000 = 9 : 2 answer : a" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 189 cm 2 , b ) 150 cm 2 , c ) 240 cm 2 , d ) 177 cm 2 , e ) 187 cm 2 | c | multiply(multiply(divide(const_1, const_2), add(6, 10)), 30) | find the area of the quadrilateral of one of its diagonals is 30 cm and its off sets 10 cm and 6 cm ? | "1 / 2 * 30 ( 10 + 6 ) = 240 cm 2 answer : c" | a = 1 / 2
b = 6 + 10
c = a * b
d = c * 30
|
a ) 3 / 40,000 , b ) 1 / 3,600 , c ) 9 / 2,000 , d ) 1 / 60 , e ) 1 / 16,000 | e | divide(1, const_3) | a certain junior class has 1,200 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs , each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected at will be a sibling pair ? | "total number of ways of choosing one student from each group is = 800 * 1200 number of cases in which a sibling pair will be got is = 60 thus the probability that the 2 students selected will be a sibling pair is = 60 / ( 800 * 1200 ) = 1 / 16,000 e" | a = 1 / 3
|
a ) 42 , b ) 77 , c ) 48 , d ) 99 , e ) 11 | a | subtract(divide(multiply(1.1, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.65, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.65 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.65 = 1.75 km = 1750 m . required time = 1750 * 3 / 125 = 42 sec . answer : a | a = 1 * 1
b = 60 * 1000
c = b / 3600
d = a / c
e = 0 * 65
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 300 , b ) 100 , c ) 150 , d ) 170 , e ) 270 | b | divide(multiply(20, 10), const_2) | the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 10 terms of the arithmetic progression ? | n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 10 . . . we get 10 / 2 [ 2 a + 14 d ] = 10 [ a + 7 d ] = 10 * 10 = 100 . . . answer is b . . . | a = 20 * 10
b = a / 2
|
a ) 119 , b ) 185 , c ) 190 , d ) 116 , e ) 200 | d | divide(divide(multiply(add(multiply(3000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3))) | a salt manufacturing company produced a total of 3000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ? | "total production of salt by the company in that year = 3000 + 3100 + 3200 + . . . . + 4100 = 42600 . average monthly production of salt for that year = 42600 / 365 Γ’ β° Λ 116 answer : d" | a = 3000 * 2
b = 12 - 1
c = b * 100
d = a + c
e = d * 12
f = e / 2
g = 3 * 100
h = 2 * 3
i = h * 10
j = g + i
k = 2 + 3
l = j + k
m = f / l
|
a ) 5 , b ) 8 , c ) 4 , d ) 1 , e ) 9 | d | add(subtract(15, 10), 10) | set x consists of the integers from 10 to 15 , inclusive , while set y consists of the integers from 5 to 10 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "x = { 10 , 11 , 12,13 , 14,15 } y = { 5,6 , 7 , 8 , 9 , 10 } common elements = { 10 } = 1 element answer : option d ." | a = 15 - 10
b = a + 10
|
a ) 29 , b ) 8 , c ) 2 , d ) 4 , e ) 6 | c | subtract(subtract(8, const_1), subtract(subtract(9, const_1), subtract(4, const_1))) | in a class of boys and girls vikas ' s rank is 9 th and tanvi ' s rank is 17 th . vikas ' s rank among the boys in that class is 4 th from the top and 18 th from the bottom and tanvi ' s rank among the girls is 8 th from top and 21 st from bottom . in the order of rank , how many girls are there between tanvi and vikas ? | explanation : vikas ' s rank in the class is 9 . so there are 8 people before him . his rank among boys is 4 . so 3 boys are before him . so there are 8 β 3 = 5 girls before him . tanvi ' s rank among the girls is 8 . so there are 7 girls before her . so number of girls between vikas and tanvi is 7 β 5 = 2 answer : c | a = 8 - 1
b = 9 - 1
c = 4 - 1
d = b - c
e = a - d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), const_12), multiply(multiply(multiply(const_2, const_100), const_100), const_12)), multiply(multiply(10, const_100), const_10)) | arjun started a business with rs . 5,000 and is joined afterwards by anoop with rs . 10 , 000 . after how many months did anoop join if the profits at the end of the year are divided equally ? | "suppose anoop joined after 3 months . then , 5000 * 12 = 10000 * ( 12 β x ) = > x = 6 . answer : d" | a = 4 * 100
b = a * 100
c = b * 12
d = 2 * 100
e = d * 100
f = e * 12
g = c - f
h = 10 * 100
i = h * 10
j = g / i
|
a ) a ) 1110 , b ) b ) 1045 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1075 | a | add(multiply(8, 70), multiply(910, 55)) | tom purchased 8 kg of apples at the rate of 70 per kg and 910 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 Γ 8 = 560 . cost of 10 kg of mangoes = 55 Γ 10 = 550 . total cost he has to pay = 560 + 550 = 1110 . a )" | a = 8 * 70
b = 910 * 55
c = a + b
|
a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio q of revenue from car sales in 1996 to revenue from truck sales in 1996 ? | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a = 100 + 7
b = 100 + 1
c = a - b
d = 100 + 1
e = 100 - 11
f = d - e
g = c / f
|
a ) 2.4 , b ) 3.2 , c ) 4.0 , d ) 4.8 , e ) 5.6 | b | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 50 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 4.8 hours to travel 240 km , so c leaves 3.2 hours after a . the answer is b ." | a = 40 - 30
|
a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, const_2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint r , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , lets further reduce 0.75 : 1.25 gives 1 + 1.5 = 2.5 gallons of mixture . this looks ok , but lets reduce further just to be sure 0.375 : 0.625 gives 0.5 + 1 = 1.5 gallons of mixture , thats less than 2 gallons of mixture , so not acceptable . so correct ans is 2.5 gallons . b | a = const_0_25 * 2
b = 2 * 5
c = a + b
|
a ) 3 , b ) 5 , c ) 7 , d ) 8 , e ) 9 | c | subtract(652543, multiply(floor(divide(652543, 12)), 12)) | what least number must be subtracted from 652543 so that remaining no . is divisible by 12 | "explanation : on dividing 652543 by 12 we get the remainder 7 , so 7 should be subtracted option c" | a = 652543 / 12
b = math.floor(a)
c = b * 12
d = 652543 - c
|
a ) 65 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | c | multiply(112, divide(multiply(25, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 25 men in 3 days ? | "20 men is 6 days can build 112 metres 25 men in 3 days can build = 112 * ( 25 / 20 ) x ( 3 / 6 ) = 70 meters answer : c ." | a = 25 * 3
b = 20 * 6
c = a / b
d = 112 * c
|
a ) 144 , b ) 772 , c ) 288 , d ) 266 , e ) 2848 | a | divide(multiply(24, 12), const_2) | if the sides of a triangle are 26 cm , 24 cm and 12 cm , what is its area ? | "the triangle with sides 26 cm , 24 cm and 12 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 24 * 12 = 144 cm 2 answer : a" | a = 24 * 12
b = a / 2
|
a ) 22 % , b ) 23 % , c ) 24.4 % , d ) 25 % , e ) 34 % | c | divide(22, divide(subtract(const_100, 10), const_100)) | in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 10 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 10 % of the customers who sample the candyare notcaught , then 90 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.90 = 0.22 ; { % of customers who sample candy } = 0.25 . answer : c ." | a = 100 - 10
b = a / 100
c = 22 / b
|
a ) 2 , b ) 2 1 / 4 , c ) 11 6 / 2 , d ) 4 1 / 2 , e ) 5 | c | multiply(divide(const_60, add(divide(const_1, 7), divide(const_1, 4))), add(divide(const_1, 7), divide(const_1, 4))) | carl can wash all the windows of his house in 7 hours . his wife maggie can wash all the windows in 4 hours . how many hours will it take for both of them working together to wash all the windows ? | work hrs = ab / ( a + b ) = 28 / 11 = 11 6 / 2 answer is c | a = 1 / 7
b = 1 / 4
c = a + b
d = const_60 / c
e = 1 / 7
f = 1 / 4
g = e + f
h = d * g
|
a ) $ 2400 , b ) $ 3200 , c ) $ 6000 , d ) $ 3600 , e ) $ 9600 | d | multiply(multiply(const_100, add(const_3, const_2)), divide(108, 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 $ 108 dollar , which she typically uses for gifts and charitable causes . what is jill β s net monthly salary ? | "let x be the monthly salary 15 % of 1 / 5 * x = 108 x = 3600 answer d" | a = 3 + 2
b = 100 * a
c = 30 + 20
d = c + 35
e = 100 - d
f = 108 / e
g = b * f
|
a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 , e ) 3 | e | divide(1, divide(add(divide(2, 3), multiply(divide(2, 3), divide(1, 3))), const_2)) | if a certain toy store ' s revenue in november was 2 / 3 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "let dec rev = 100 then nov rev is 2 / 3 ( 100 ) = > 67 therefore jan rev = 1 / 3 ( nov rev ) = 1 / 3 ( 67 ) = > 23 hence dec rev = x * ( nov rev + jan rev ) / 2 100 = x * ( 67 + 23 ) / 2 x = 100 / 45 = > 2.22 = 3 ans ) e" | a = 2 / 3
b = 2 / 3
c = 1 / 3
d = b * c
e = a + d
f = e / 2
g = 1 / f
|
a ) s . 800 , b ) s . 810 , c ) s . 825 , d ) s . 900 , e ) s . 920 | c | multiply(2475, divide(1, 3)) | a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2475 ? | "2475 * 1 / 3 = 825 option c" | a = 1 / 3
b = 2475 * a
|
a ) 4.0 m , b ) 4.1 m , c ) 4.2 m , d ) 4.3 m , e ) none of these | c | divide(multiply(multiply(5, 4.5), 2.1), subtract(multiply(13.5, 2.5), multiply(5, 4.5))) | a rectangular tank measuring 5 m Γ 4.5 m Γ 2.1 m is dug in the centre of the field measuring 13.5 m Γ 2.5 . the earth dug out is spread evenly over the remaining portion of a field . how much is the level of the field raised ? | area of the field = 13.5 Γ 2.5 = 33.75 m 2 area covered by the rectangular tank = 5 Γ 4.5 = 22.50 m 2 area of the field on which the earth dug out is to be spread = 33.75 β 22.50 = 11.25 m 2 let the required height be h . then , 11.25 Γ h = 5 Γ 4.5 Γ 2.1 or h = 4.2 m answer c | a = 5 * 4
b = a * 2
c = 13 * 5
d = 5 * 4
e = c - d
f = b / e
|
a ) 40 , b ) 50 , c ) 62 , d ) 30 , e ) 56 | d | add(subtract(80, multiply(11, 5)), 5) | a batsman makes a score of 80 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 80 = 11 x x = 80 - 50 = 30 answer is d" | a = 11 * 5
b = 80 - a
c = b + 5
|
a ) 100 , b ) 110 , c ) 120 , d ) 210 , e ) 200 | c | divide(multiply(add(50, 10), const_100), 50) | paul has to secure 50 % marks to clear his exam of class 7 th . he got 50 marks and failed by 10 marks . what is the maximum marks ? | c 120 to pass the exam ravish needs 50 + 10 = 60 marks . = > ( 60 / 50 ) * 100 = 120 | a = 50 + 10
b = a * 100
c = b / 50
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | on a two - dimensional coordinate plane , the line s = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | "apparently it ' s s = x ^ 2 - x ^ 3 instead of s = x ^ 2 - s ^ 3 . in this case : the x - intercept is the value ( s ) of x for s = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c ." | a = max(3)
b = a - 1
|
a ) 22 , b ) 27 , c ) 13 , d ) 19 , e ) 17 | c | sqrt(add(power(subtract(11, 6), const_2), power(12, const_2))) | two vertical ladders length of 6 m and 11 m are kept vertically at a distance of 12 m . find the top distance of both ladders ? | so distance between the top points = ad = 122 + 52 β β β β β β β β = 13122 + 52 = 13 so in paper b he did well . answer : c | a = 11 - 6
b = a ** 2
c = 12 ** 2
d = b + c
e = math.sqrt(d)
|
['a ) 27', 'b ) 36', 'c ) 42', 'd ) 64', 'e ) 147'] | b | multiply(divide(48, 112), 84) | the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 84 ounces of this substance ? | 112 ounces of a substance has a volume of 48 cubic inches 84 ounces of a substance has a volume of ( 48 / 112 ) * 84 = 36 cubic inches answer b | a = 48 / 112
b = a * 84
|
a ) 40 , b ) 50 , c ) 62 , d ) 45 , e ) 56 | d | add(subtract(95, multiply(11, 5)), 5) | a batsman makes a score of 95 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 95 = 11 x x = 95 - 50 = 45 answer is d" | a = 11 * 5
b = 95 - a
c = b + 5
|
a ) 24 , b ) 36 , c ) 72 , d ) 1 , e ) 216 | e | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | of the three - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "assume : a , b , c is the digit and a , b , c not = 0 thus a , b , c could be 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 9 count the possible way to get three digit number abc . a = b , and c must be different from a , b thus , there are 3 possible ways of digit arrangement : aac , aca , caa case i : aac = > ( digit 1 st ) x ( digit 2 nd ) x ( digit 3 rd ) = > 9 x 1 x 8 { pick any number from group = 9 possible ways } x { pick number the same as the first pick = 1 way } x { pick any number from the rest = 8 possible ways } = 9 x 1 x 8 = 72 possible ways case ii : aca = > same as case i you have 72 possible ways case iii : caa = > same as case i you have 72 possible ways total of this set of number = 72 + 72 + 72 = 216 e ) is the answer" | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
|
a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 17.2 hrs , e ) 31.5 hrs | e | divide(const_1, subtract(divide(const_1, 7), divide(const_1, 9))) | a cistern can be filled by a tap in 7 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 7 - 1 / 9 = 2 / 63 therefore the cistern will be filled in 63 / 2 hours or 31.5 hours . answer : e" | a = 1 / 7
b = 1 / 9
c = a - b
d = 1 / c
|
a ) 40.09 , b ) 4.06 , c ) 40.06 , d ) 400.9 , e ) 400.6 | d | divide(16.036, 0.04) | 16.036 divided by 0.04 gives : | = 16.036 / 0.04 = 1603.6 / 4 = 400.9 answer is d . | a = 16 / 36
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a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25 | e | add(multiply(divide(3, 5), divide(3, 5)), add(multiply(divide(3, 5), divide(const_2, 5)), multiply(divide(3, 5), divide(const_2, 5)))) | a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also 3 / 5 . what is the probability r that you can still row the canoe ? | "simply look at the question from the other side . what is the probability that you can β t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability of something happening is 1 β the probability that it doesn β t happen , you can use the following equation to reach the right answer r : 1 β 4 / 25 = 21 / 25 . answer choice e ." | a = 3 / 5
b = 3 / 5
c = a * b
d = 3 / 5
e = 2 / 5
f = d * e
g = 3 / 5
h = 2 / 5
i = g * h
j = f + i
k = c + j
|
a ) 10 , b ) 12 , c ) 16 , d ) 24 , e ) 32 | c | divide(multiply(8, const_4), 2) | a is the average ( arithmetic mean ) of the first 7 positive multiples of 8 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 β b ^ 2 is zero , what is the value of n ? | "if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 32 , which also equals b . b is the second positive multiple of n , thus n = 32 / 2 = 16 . the answer is c ." | a = 8 * 4
b = a / 2
|
a ) 27 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | multiply(log(divide(multiply(multiply(add(const_4, const_1), const_1000), const_100), const_1000)), 3) | the population of a bacteria culture doubles every 3 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 3 minutes so answer is 9 * 3 = 27 minutes answer ( a ) | a = 4 + 1
b = a * 1000
c = b * 100
d = c / 1000
e = math.log(d)
f = e * 3
|
a ) 5.04 % , b ) 4.04 % , c ) 3.04 % , d ) 6.04 % , e ) 7.04 % | b | subtract(multiply(multiply(add(const_100, 36), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | shopkeeper rise price by 36 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ? | "let d initial price be 100 36 % rise now price = 136 / 100 * 100 = 136 10 % discount then price = 136 * 90 / 100 = 122.4 15 % discount then price = 122.4 * 85 / 100 = 104.04 so gain = 104.04 - 100 = 4.04 gain % = gain * 100 / cp = = > 4.04 * 100 / 100 = 4.04 % answer : b" | a = 100 + 36
b = 100 - 10
c = b / 100
d = a * c
e = 100 - 15
f = e / 100
g = d * f
h = g - 100
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | e | subtract(add(negate(22), multiply(subtract(18, const_1), 22)), 22) | set j consists of 18 consecutive even numbers . if the smallest term in the set is - 22 , what is the range of the positive integers in set j ? | "since there are only 18 integers , another approach is the just list all 18 . we get : - 22 , - 20 , - 18 , - 16 , - 14 , - 12 , - 10 , - 8 , - 6 , - 4 , - 2 , 0 , 2 , 4 , 6 , 8 , 10 , 12 range of positive integers = 12 - 2 = 10 answer : e" | a = negate + (
b = 18 - 1
c = b * 22
d = a - c
|
a ) s . 6750 , b ) s . 7750 , c ) s . 8750 , d ) s . 9372 , e ) s . 9750 | a | multiply(multiply(multiply(multiply(const_2, const_3), const_4), multiply(multiply(const_2, add(const_2, const_3)), const_100)), divide(multiply(multiply(const_3, multiply(const_2, const_3)), multiply(multiply(const_2, add(const_2, const_3)), const_100)), add(multiply(multiply(multiply(const_2, const_3), const_4), multiply(multiply(const_2, add(const_2, const_3)), const_100)), multiply(add(multiply(const_4, multiply(const_2, add(const_2, const_3))), const_2), multiply(multiply(const_2, add(const_2, const_3)), const_100))))) | suresh and ramesh started a business investing rs . 24,000 and rs . 40,000 respectively . out of total profit of rs . 18,000 , what is suresh ' s share ? | explanation : ratio of suresh and ramesh ' s share = 24,000 : 40,000 = 3 : 5 suresh ' s share = rs . ( 18000 x 3 / 8 ) = rs . 6750 answer : a | a = 2 * 3
b = a * 4
c = 2 + 3
d = 2 * c
e = d * 100
f = b * e
g = 2 * 3
h = 3 * g
i = 2 + 3
j = 2 * i
k = j * 100
l = h * k
m = 2 * 3
n = m * 4
o = 2 + 3
p = 2 * o
q = p * 100
r = n * q
s = 2 + 3
t = 2 * s
u = 4 * t
v = u + 2
w = 2 + 3
x = 2 * w
y = x * 100
z = v * y
A = r + z
B = l / A
C = f * B
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a ) 9 , b ) 36 , c ) 122 , d ) 6 , e ) 1 | e | divide(sqrt(36), 6) | what is the square root of 36 , divided by 6 ? | "square root is a number times itself square root of 36 = 6 , 6 / 6 = 1 ( e ) 1" | a = math.sqrt(36)
b = a / 6
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a ) 9 , b ) 8 , c ) 6 , d ) 7 , e ) 2 | d | divide(add(divide(24, 4), divide(32, 4)), const_2) | a man swims downstream 32 km and upstream 24 km taking 4 hours each time , what is the speed of the man in still water ? | "32 - - - 4 ds = 8 ? - - - - 1 24 - - - - 4 us = 6 ? - - - - 1 m = ? m = ( 8 + 6 ) / 2 = 7 answer : d" | a = 24 / 4
b = 32 / 4
c = a + b
d = c / 2
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a ) 9 % , b ) 10 % , c ) 11 % , d ) 18 % , e ) 90 % | d | divide(multiply(15, const_100), subtract(const_100, 15)) | during a sale , the price of a pair of shoes is marked down 15 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | assume the price = 100 price during sale = 85 price after sale = 100 percent increase = 15 / 85 * 100 = 18 % approx . correct option : d | a = 15 * 100
b = 100 - 15
c = a / b
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(6, const_3), 9) | a and b can do a piece of work in 18 and 9 days respectively . a does the work for a few days and left , then b continued for 6 days . how many days did a and b work together ? | a and b work together for 2 days . a 1 day work 1 / 18 b 1 day work 1 / 9 let a work x days . then equation form is ( x / 18 ) + ( ( x + 6 ) / 9 ) = 1 from this equation x value is 2 . answer : b | a = 6 * 3
b = a / 9
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a ) 20 days , b ) 48 days , c ) 98 days , d ) 31 days , e ) 22 days | a | inverse(subtract(inverse(10), inverse(20))) | a and b can finish a work in 10 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 10 β 1 / 20 = 1 / 20 = > 20 days answer : a" | a = 1/(10)
b = 1/(20)
c = a - b
d = 1/(c)
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a ) 20 , b ) 60 , c ) 600 , d ) 1,000 , e ) 1,200 | d | divide(divide(multiply(subtract(multiply(12, const_2), 12), const_1000), 12), const_1000) | a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 12 minutes , a combustion reaction creates 12,000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | "solution : 12,000 carbon dioxide molecules are created over a period of 12 minutes . therefore 12,000 / 12 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 Γ 1,000 = 1,000 carbon atoms and 2 Γ 1,000 = 2,000 oxygen atoms . the difference is 2,000 β 1,000 = 1,000 . the correct answer is d ." | a = 12 * 2
b = a - 12
c = b * 1000
d = c / 12
e = d / 1000
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a ) 36 , b ) 46 , c ) 56 , d ) 66 , e ) 96 | c | divide(factorial(8), factorial(subtract(8, 2))) | 8 ! / ( 8 - 2 ) ! = ? | 8 ! / ( 8 - 2 ) ! = 8 ! / 6 ! = 8 * 7 = 56 . hence , the correct answer is c . | a = math.factorial(8)
b = 8 - 2
c = math.factorial(b)
d = a / c
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a ) $ 135875 , b ) $ 52700 , c ) $ 127900 , d ) $ 105500 , e ) $ 105000 | d | divide(527500, add(const_4, const_1)) | sally , mark , sarah , steve and jim are inheriting the money from their grandmother . she left them $ 527500 to split equally between all of them . if the money is split equally , how much money will each person receive ? | there are 5 people inheriting the money . we have to take the total of the inheritance and divide it by 5 for each person receiving money . $ 527500 / 5 = $ 105500 . therefore the correct answer is d . | a = 4 + 1
b = 527500 / a
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(const_100, divide(150, subtract(165, 150))) | if the cost price of 165 chocolates is equal to the selling price of 150 chocolates , the gain percent is | "let c . p . of each chocolate be re . 1 . then , c . p . of 150 chocolates = rs . 150 ; s . p . of 150 chocolates = rs . 165 . gain % = 15 / 150 * 100 = 10 % answer : c" | a = 165 - 150
b = 150 / a
c = 100 / b
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a ) 1 / 3 , b ) 1 / 6 , c ) 1 / 15 , d ) 3 / 8 , e ) 2 / 3 | b | inverse(divide(factorial(4), multiply(factorial(2), factorial(2)))) | jack and jill work at a hospital with 2 other workers . for an internal review , 2 of the 4 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "1 / 4 c 2 = 1 / 6 . answer : b ." | a = math.factorial(4)
b = math.factorial(2)
c = math.factorial(2)
d = b * c
e = a / d
f = 1/(e)
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a ) 5000 , b ) 5600 , c ) 5400 , d ) 4500 , e ) none of these | b | divide(126, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100))) | if 15 % of 30 % of 50 % of a number is 126 , then what is the number ? | "let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 126 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 126 = > a = 10 * 20 * 10 * 2 = 5600 . answer : b" | a = 50 / 100
b = 30 / 100
c = a * b
d = 15 / 100
e = c * d
f = 126 / e
|
a ) 2.5 , b ) 0.025 , c ) 0.05 , d ) 0.03 , e ) 3 | d | divide(volume_cylinder(5, 12), circle_area(100)) | a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 5 feet and a height of 12 feet , how far ( in feet ) did the oil level drop in the stationary tank ? | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 25 * 12 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 300 / 10,000 = 3 / 100 = 0.03 ft the answer is d ." | a = volume_cylinder / (
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a ) 121 , b ) 125 , c ) 132 , d ) 134 , e ) 138 | d | divide(100, subtract(5, divide(const_1, 1))) | company x provides bottled water to its employees in 5 liter bottles , $ 2 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 100 fixed monthly rate and $ 1 for 10 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 10 l bottles will be x / 2 ( to equate the vol . ) since the total cost will be equal , 2 x = 100 + x / 2 so x = 133.33 or 134 . answer is ( d ) . | a = 1 / 1
b = 5 - a
c = 100 / b
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a ) 17 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 35)), const_100), const_100) | when the price of an article was reduced by 35 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 35 % , the new price is 0.65 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.65 p ) ( 1.8 n ) = 1.17 np , which is 1.17 times the old revenue , so is 17 % greater . answer : a" | a = 80 + 100
b = 100 - 35
c = a * b
d = c / 100
e = d - 100
|
a ) 225 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) none | a | multiply(multiply(add(36, 45), const_0_2778), 10) | two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 10 seconds . the length of the faster train is | "solution relative speed = ( 36 + 45 ) km / hr = ( 81 x 5 / 18 ) m / sec = ( 45 / 2 ) m / sec length of the train = ( 45 / 2 x 10 ) m = 225 m . answer a" | a = 36 + 45
b = a * const_0_2778
c = b * 10
|
a ) r = 350 , b ) r = 353 , c ) r = 354 , d ) r = 356 , e ) 357 | b | add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits r are in s ( 99 ) ? | "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = r = 353 . b" | a = const_60 + 4
b = a + 4
c = const_60 + 4
d = c + 4
e = d * 4
f = b + e
g = 3 * 1
h = 3 * 2
i = h * 2
j = g + i
k = f + j
|
a ) 187 km , b ) 480 km , c ) 720 km , d ) 297 km , e ) 671 km | c | divide(1, 2) | with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 6 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 6 2 x / 15 - x / 8 = 6 = > x = 720 km . answer : c" | a = 1 / 2
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a ) 8 , b ) 4 , c ) 16 , d ) 32 , e ) 64 | a | multiply(power(2, 2), 2) | if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 2 , what is the value of 2 ^ a * 2 ^ b ? | "2 ^ ab = 2 ^ 2 therefore ab = 2 either a = 1 or 2 or b = 2 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 3 = 8 a" | a = 2 ** 2
b = a * 2
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a ) 5 , b ) 17 , c ) 13 , d ) 23 , e ) none of these | a | subtract(add(4, 8), 7) | when 242 is divided by a certain divisor the remainder obtained is 4 . when 698 is divided by the same divisor the remainder obtained is 8 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 7 . what is the value of the divisor ? | "let that divisor be x since remainder is 4 or 8 it means divisor is greater than 8 . now 242 - 4 = 238 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 8 = 690 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 238 + 690 ) + 4 + 8 = x ( k + l ) + 12 when we divide this number by x then remainder will be equal to remainder of ( 12 divided by x ) = 7 hence x = 12 - 7 = 5 hence a" | a = 4 + 8
b = a - 7
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a ) 440 , b ) 520 , c ) 620 , d ) 720 , e ) 820 | a | add(200, multiply(divide(200, 5), 6)) | in a college , the ratio of the number of boys to girls is 6 : 5 . if there are 200 girls , the total number of students in the college is | "explanation : let the boy are 6 x and girls are 5 x = > 5 x = 200 = > x = 40 total students = 6 x + 5 x = 11 x = 11 ( 40 ) = 440 option a" | a = 200 / 5
b = a * 6
c = 200 + b
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a ) a ) 90 , b ) b ) 100 , c ) c ) 110 , d ) d ) 70 , e ) 80 | e | divide(1280, multiply(power(const_2, const_4), const_2)) | what is the greater of the two numbers whose product is 1280 , given that the sum of the two numbers exceeds their difference by 32 ? | "let the greater and the smaller number be g and s respectively . g * s = 1288 g + s exceeds g - s by 32 i . e . , g + s - ( g - s ) = 32 i . e . , 2 s = 32 = > s = 16 . g = 1280 / s = 80 . answer : e" | a = 2 ** 4
b = a * 2
c = 1280 / b
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a ) rs . 3200 , b ) rs . 3600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | a | multiply(divide(8000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | p , q and r have rs . 8000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 8000 - r ) = > 3 r = 16000 - 2 r = > 5 r = 16000 = > r = 3200 . answer : a" | a = 2 / 3
b = 1 + a
c = 8000 / b
d = 2 / 3
e = c * d
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a ) 122.9 m , b ) 127.5 m . , c ) 122.2 m , d ) 222.9 m , e ) 212.5 m | e | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 1000 m ? | when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 1000 m , the number of meters by which a beats c = ( 1000 * 212.5 ) / 1000 = 212.5 m . answer : e | a = 1000 - 100
b = 800 - 100
c = a * b
d = c / 800
e = 1000 - d
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a ) 31 , b ) 33 , c ) 35 , d ) 36 , e ) 23 | b | divide(subtract(300, 1), 9) | how many positive integers between 1 and 300 are there such that they are multiples of 9 ? | "multiples of 9 = 9 , 18,27 , - - - - - 297 number of multiples of 15 = > 9 * 33 = 297 answer is b" | a = 300 - 1
b = a / 9
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['a ) 10 feet', 'b ) 12 feet', 'c ) 13 feet', 'd ) 15 feet', 'e ) 18 feet'] | d | multiply(divide(25, 5), 3) | a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 5 feet . how many feet does the squirrel travels if the post is 25 feet tall and 3 feet in circumference ? | total circuit = 25 / 5 = 5 total feet squirrel travels = 5 * 3 = 15 feet answer : d | a = 25 / 5
b = a * 3
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a ) 3 kmph , b ) 0.5 kmph , c ) 13 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph | b | divide(add(divide(42, 6), divide(48, 6)), const_2) | a man swims downstream 48 km and upstream 42 km taking 6 hours each time ; what is the speed of the current ? | "explanation : 48 - - - 6 ds = 8 ? - - - - 1 42 - - - - 6 us = 7 ? - - - - 1 s = ? s = ( 8 - 7 ) / 2 = 0.5 answer : option b" | a = 42 / 6
b = 48 / 6
c = a + b
d = c / 2
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a ) a ) 48.55 , b ) b ) 49 , c ) c ) 51 , d ) d ) 61 , e ) e ) 62 | a | divide(add(multiply(16, 50.25), multiply(8, 45.15)), add(16, 8)) | the average weight of 16 girls in a class is 50.25 kg and that of the remaining 8 girls is 45.15 kg . find the average weights of all the girls in the class . | required average = ( 50.25 x 16 + 45.15 x 8 ) / ( 16 + 8 ) = ( 804 + 361.20 ) / 24 = 1165.20 / 24 = 48.55 kg answer is a | a = 16 * 50
b = 8 * 45
c = a + b
d = 16 + 8
e = c / d
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a ) 190 , b ) 204 , c ) 220 , d ) 230 , e ) none of these | b | multiply(782, divide(divide(const_12, 2), add(add(divide(const_12, 2), multiply(divide(const_12, 3), const_2)), multiply(divide(const_12, 4), const_3)))) | if rs . 782 is divided into 3 parts , proportional to 1 / 2 : 2 / 3 : 3 / 4 find the first part . | explanation : 1 / 2 : 2 / 3 : 3 / 4 = 6 : 8 : 9 first part = 782 β 6 / 23 = 204 option b | a = 12 / 2
b = 12 / 2
c = 12 / 3
d = c * 2
e = b + d
f = 12 / 4
g = f * 3
h = e + g
i = a / h
j = 782 * i
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a ) 462 , b ) 450 , c ) 480 , d ) 504 , e ) 600 | e | multiply(50, 12) | the hcf of two numbers is 50 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 50 hcf will be always a factor of lcm 50 is factor of lcm other two factors are 11 & 12 then the numbers are ( 50 * 11 ) and ( 50 x 12 ) = 550 and 600 answer : option e" | a = 50 * 12
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a ) 0 , b ) - 1 , c ) 1 , d ) 2 , e ) 3 | c | subtract(21, multiply(const_2, const_10)) | find the value for m ? 21 ( m + n ) + 21 = 21 ( - m + n ) + 21 | 21 m + 21 n - 21 = - 21 m + 21 n + 21 42 m = 42 = > m = 1 c | a = 2 * 10
b = 21 - a
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a ) 96 , b ) 94 , c ) 128 , d ) 74 , e ) 110 | c | divide(40, 320) | find 40 % of 320 | "we know that r % of m is equal to r / 100 Γ m . so , we have 40 % of 320 40 / 100 Γ 320 = 128 answer : c" | a = 40 / 320
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a ) 4 : 00 , b ) 4 : 30 , c ) 5 : 00 , d ) 5 : 30 , e ) 6 : 00 | a | divide(subtract(divide(230, 50), 1), add(1, divide(40, 50))) | at 1 : 00 pm , a truck left city p and headed toward city q at a constant speed of 50 km / h . one hour later , a car left city q and headed toward city p along the same road at a constant speed of 40 km / h . if the distance between city p and city q is 230 km , at what time will the truck and the car meet each other ? | "at 2 : 00 pm , the truck and the car are 180 km apart . the truck and the car complete a distance of 90 km each hour . the time it takes to meet is 180 / 90 = 2 hours . they will meet at 4 : 00 pm . the answer is a ." | a = 230 / 50
b = a - 1
c = 40 / 50
d = 1 + c
e = b / d
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a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 200 % | e | multiply(divide(subtract(150, 50), 50), const_100) | a certain telescope increases the visual range at a particular location from 50 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | original visual range = 50 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 50 ) / 50 * 100 % = 2 * 100 % = 200 % answer e | a = 150 - 50
b = a / 50
c = b * 100
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a ) 25 , b ) 90 , c ) 35 , d ) 20 , e ) none of these | b | divide(multiply(60, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | a contractor undertakes to built a walls in 50 days . he employs 60 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | "60 men complete 0.4 work in 25 days . applying the work rule , m 1 Γ d 1 Γ w 2 = m 2 Γ d 2 Γ w 1 we have , 60 Γ 25 Γ 0.6 = m 2 Γ 25 Γ 0.4 or m 2 = 60 Γ 25 Γ 0.6 / 25 Γ 0.4 = 90 men answerb" | a = 100 - 40
b = a / 100
c = 60 * b
d = 4 / 10
e = c / d
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a ) 12 min , b ) 18 min , c ) 16 min , d ) 20 min , e ) 26 min | a | subtract(multiply(const_1, const_60), multiply(divide(40, 50), const_60)) | excluding stoppages , the average speed of a bus is 50 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 50 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 50 - 40 ) km i . e 10 km at 50 km / hr . stoppage time = 10 / 50 hrs = 12 min answer : a" | a = 1 * const_60
b = 40 / 50
c = b * const_60
d = a - c
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a ) 40 , b ) 26 , c ) 20 , d ) 28 , e ) 22 | c | divide(add(20, 20), const_2) | the total marks obtained by a student in mathematics and physics is 20 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 20 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 20 + 20 ) / 2 = 20 . answer : c" | a = 20 + 20
b = a / 2
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a ) 900 , b ) 1000 , c ) 1110 , d ) 1200 , e ) 1400 | c | add(divide(subtract(multiply(floor(divide(10000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1) | how many numbers from 10 to 10000 are exactly divisible by 9 ? | "10 / 9 = 1 and 10000 / 9 = 1111 = = > 1111 - 1 = 1110 . therefore 1110 answer : c" | a = 10000 / 9
b = math.floor(a)
c = b * 9
d = 10 / 9
e = math.floor(d)
f = e + 1
g = f * 9
h = c - g
i = h / 9
j = i + 1
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a ) 21 years , b ) 71 years , c ) 51 years , d ) 91 years , e ) 11 years | c | add(36, const_1) | the average age of 36 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "explanation : age of the teacher = ( 37 * 15 - 36 * 14 ) years = 51 years . answer : c" | a = 36 + 1
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a ) 933 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 4), const_100))) | find the principle on a certain sum of money at 5 % per annum for 4 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 4 ) / 100 ] p = 933 answer : a" | a = 5 * 4
b = a / 100
c = 1 + b
d = 1120 / c
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a ) 1 / 3 , b ) 4 / 9 , c ) 5 / 6 , d ) 3 / 7 , e ) 1 / 4 | c | divide(subtract(divide(divide(factorial(10), factorial(subtract(10, const_4))), factorial(const_4)), divide(divide(factorial(subtract(10, 3)), factorial(subtract(subtract(10, 3), const_4))), factorial(const_4))), divide(divide(factorial(10), factorial(subtract(10, const_4))), factorial(const_4))) | out of 10 students in a class , 4 are from maharashtra , 3 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ? | "total possible ways of selecting 4 students out of 10 = 10 c 4 = ( 10 * 9 * 8 * 7 ) / ( 4 * 3 * 2 * 1 ) = 210 the number of ways of selecting a 4 students in which no student belongs to karnataka = 7 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 7 c 4 = 175 probability = 175 / 210 = 5 / 6 answer is c" | a = math.factorial(10)
b = 10 - 4
c = math.factorial(b)
d = a / c
e = math.factorial(4)
f = d / e
g = 10 - 3
h = math.factorial(g)
i = 10 - 3
j = i - 4
k = math.factorial(j)
l = h / k
m = math.factorial(4)
n = l / m
o = f - n
p = math.factorial(10)
q = 10 - 4
r = math.factorial(q)
s = p / r
t = math.factorial(4)
u = s / t
v = o / u
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a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900 | e | divide(81, multiply(divide(3, const_100), 3)) | a sum was put at simple interest at a certain rate for 3 years had it been put at 3 % higher rate , it would have fetched 81 more . find the sum . | "difference in s . i . = p Γ t / 100 ( r 1 β r 2 ) β 81 = p Γ 3 x 3 / 100 ( β΅ r 1 - r 2 = 2 ) β p = 81 Γ 100 / 3 x 3 = 900 answer e" | a = 3 / 100
b = a * 3
c = 81 / b
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(divide(subtract(multiply(21, 6), multiply(7, 16)), subtract(21, 16)), divide(const_2, const_10)) | the least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ratio less than 16 : 21 , is | "explanation : let x is subtracted . then , ( 6 β x ) / ( 7 β x ) < 16 / 21 21 ( 6 β x ) < 16 ( 7 β x ) = > 5 x > 14 = x > 2.8 least such number is 3 option a" | a = 21 * 6
b = 7 * 16
c = a - b
d = 21 - 16
e = c / d
f = 2 / 10
g = e + f
|
a ) $ . 90 , b ) $ 1.00 , c ) $ 1.20 , d ) $ 1.50 , e ) $ 1.60 | c | multiply(divide(20, 100), 6) | the total cost of 100 paper plates and 200 paper cups is $ 6.00 at the same rates what is the total cost of 20 of the plates and 40 of the cups ? | u dont need to go through all this what u have with u is 100 p + 200 c = $ 6.00 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 1.20 therefore oa is c | a = 20 / 100
b = a * 6
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 40 % | e | multiply(subtract(const_1, divide(multiply(const_1, 3), 5)), const_100) | if the price of sugar rises from rs . 3 per kg to rs . 5 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 3 = x Γ 5 = x = 60 kg . β΄ reduction in consumption = 40 % . answer e | a = 1 * 3
b = a / 5
c = 1 - b
d = c * 100
|
a ) 176 sec , b ) 190 sec , c ) 178 sec , d ) 150 sec , e ) 276 sec | d | divide(add(1200, 300), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 300 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1500 m t = d / s t = 1500 / 10 t = 150 sec answer : d" | a = 1200 + 300
b = 1200 / 120
c = a / b
|
a ) 30 , b ) 33.6 , c ) 39.8 , d ) 44.9 , e ) none | c | divide(add(add(add(30, const_1), add(add(const_4.0, const_1), const_2)), add(subtract(50, const_4.0), subtract(50, const_2))), 30) | find the average of all prime numbers between 30 and 50 . | "sol . there are five prime numbers between 30 and 50 . they are 31 , 37 , 41 , 43 and 47 β΄ required average = [ 31 + 37 + 41 + 43 + 47 / 5 ] = 199 / 5 = 39.8 answer c" | a = 30 + 1
b = 4 + 0
c = b + 2
d = a + c
e = 50 - 4
f = 50 - 2
g = e + f
h = d + g
i = h / 30
|
a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2 | d | multiply(multiply(divide(subtract(divide(50, 2), const_1), subtract(50, const_1)), divide(divide(50, 2), 50)), 2) | there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability w that the both positions will be taken by members of just one party ? | "w probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total candidates . since this result can happen for both parties , ( 12 / 49 ) + ( 12 / 49 ) = ( 24 / 49 ) answer : d" | a = 50 / 2
b = a - 1
c = 50 - 1
d = b / c
e = 50 / 2
f = e / 50
g = d * f
h = g * 2
|
a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 2,030 | e | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 30), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 30 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 30 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 30 chips lost and minimize # of $ 100 chips lost , which means that # of $ 30 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 30 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 16 : x + x - 2 = 16 - - > x = 9 : 9 $ 30 chips were lost and 9 - 2 = 7 $ 100 chips were lost . total worth of chips lost is 9 * 30 + 7 * 100 = $ 970 , so the gambler kept $ 3,000 - $ 970 = $ 2,030 . answer : e" | a = 3 * 1000
b = 16 + 2
c = b / 2
d = c * 30
e = 16 + 2
f = e / 2
g = f - 2
h = g * 100
i = d + h
j = a - i
|
a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 22 % | e | multiply(divide(multiply(divide(60, const_100), 445), multiply(const_100, power(const_4, const_2))), const_100) | an association of mathematics teachers has 1,260 members . only 445 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? | "total umber of members = 1260 number of members that cast votes = 445 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 445 = 267 percent of total membership that voted for winning candidate = ( 267 / 1260 ) * 100 = 21.19 % answer e" | a = 60 / 100
b = a * 445
c = 4 ** 2
d = 100 * c
e = b / d
f = e * 100
|
a ) 4 , b ) 6 , c ) 8 , d ) 25 , e ) 18 | d | add(const_3, const_4) | what is the smallest positive integer x , such that 5000 x is a perfect cube ? | "take out the factors of 5000 that will come 10 ^ 3 * 5 . for perfect cube you need every no . raise to the power 3 . for 5000 x to be a perfect cube , you need two 5 that means 25 . d is the answer ." | a = 3 + 4
|
a ) 7580 , b ) 7960 , c ) 8290 , d ) 4320 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(6, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 6 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 / 6 - 1 / 12 ) = 1 / 12 . work done by the inlet in 1 min . = ( 1 / 12 Γ 1 / 60 ) = 0.001389 volume of 0.001389 part = 6 litres . therefore , volume of whole = 1 / 0.001389 x 6 βΉ = βΊ 4320 litres . answer d" | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 6 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | divide(add(24, 6), add(4, 6)) | maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 24 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he meets up with brad ? | "total distance = 24 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 20 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken to meet brad after brad started = 20 / 10 = 2 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 2 = 8 + 4 = 12 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell / maxwell ' s speed = 12 / 4 = 3 hrs . . . answer a" | a = 24 + 6
b = 4 + 6
c = a / b
|
a ) 40 min , b ) 94 min , c ) 110 min , d ) 112 min , e ) 120 min | a | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 3 robots altogether , if the robots build only one robot at a time ? | "first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) and now we have 3 robots in 40 minutes answer - a" | a = 1/(1)
b = 1/(2)
c = a + b
d = 1/(2)
e = c + d
f = 1/(e)
g = 1/(1)
h = 1/(2)
i = g + h
j = 1/(i)
k = f + j
|
a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 75 | d | divide(20, divide(subtract(100, 75), 100)) | if 20 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | "let x be the number of students in the school . 0.2 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.2 x / 0.25 = 0.8 x students went to the camping which is 80 % . the answer is d" | a = 100 - 75
b = a / 100
c = 20 / b
|
a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | b | subtract(add(250, 230), 400) | in a group of 400 readers who read science fiction or literacy works or both , 250 read science fiction and 230 read literacy works . how many read both science fiction and literacy works ? | "consider total number of reader n ( s u l ) = 400 people who read science fiction n ( s ) = 250 people who read literacy works n ( l ) = 230 both science fiction and literacy n ( s β© l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s β© l ) 400 = 250 + 230 - n ( s β© l ) n ( s β© l ) = 480 - 400 n ( s β© l ) = 80 so people read both science fiction and literacy works are 80 answer : b" | a = 250 + 230
b = a - 400
|
a ) 2 : 5 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 2 | d | divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13))) | if 12 men and 16 boys can do a piece of work in 5 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy ? | 12 m + 16 b - - - - - 5 days 13 m + 24 b - - - - - - - 4 days 60 m + 80 b = 52 m + 96 b 8 m = 16 b = > 1 m = 2 b m : b = 2 : 1 answer : d | a = 4 * 24
b = 5 * 16
c = a - b
d = 5 * 12
e = 4 * 13
f = d - e
g = c / f
|
a ) 2 , b ) 4 , c ) 12 , d ) 40 , e ) 160 | a | subtract(divide(480, 40), 10) | the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 10 miles per gallon , will not get him there . by how many miles per gallon must the aircraft β s fuel efficiency be improved to make the flight to cleveland possible ? | "actual miles / gallon is = 480 / 40 = 12 miles / gallon . current engine miles / gallon is 10 miles / gallon . additional 2 miles / gallon is required to match the actual mileage . imo option a ." | a = 480 / 40
b = a - 10
|
a ) 60 , b ) 68 , c ) 65 , d ) 64 , e ) 61 | d | add(40, divide(subtract(976, multiply(12, 40)), divide(multiply(12, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 12 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 12 = 480 excess = 976 - 480 = 496 for extra hours = . 75 ( 12 ) = 9 + 12 = 21 number of extra hrs = 496 / 21 = 23.62 total hrs = 40 + 23.62 = 63.62 = 64 approx . answer d 64" | a = 12 * 40
b = 976 - a
c = 100 + 75
d = 12 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 30 , b ) 98 , c ) 27 , d ) 21 , e ) 45 | e | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their average is 36 . the largest number is : | "explanation : let the numbers be 3 x , 4 x and 5 x , then , ( 3 x + 4 x + 5 x ) / 3 = 36 = > 12 x = 36 * 3 = > x = 9 largest number 5 x = 5 * 9 = 45 answer : e" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(100, multiply(5, const_4)) | 5 cats can catch 5 mice in 5 minutes . how many cats does it take to catch 100 mice in 100 minutes ? | let x be no . of cat 5 cat catch 5 mice in 5 min 5 cat catch 1 mice in ( 5 / 5 = 1 min ) 1 cat catch 1 mice in 1 * 5 = 5 min 1 cat catch 100 mice in 5 * 100 min x cat catch 100 mice in ( 5 * 100 ) / x min a / q ( 5 * 100 ) / x = 100 = > 5 * 100 = x * 100 = > x = 5 answer : e | a = 5 * 4
b = 100 / a
|
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