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 ) 26 , b ) 32 , c ) 33 , d ) 34 , e ) 47 | e | subtract(add(floor(divide(subtract(99, 29), 3)), divide(subtract(99, 29), 2)), floor(divide(subtract(99, 29), multiply(2, 3)))) | if w is the set of all the integers between 29 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 29 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 28 = [ 99 / 2 ] - [ 28 / 2 ] = 49 - 14 = 35 multiples of 3 from 29 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 28 = [ 99 / 3 ] - [ 28 / 3 ] = 33 - 9 = 24 multiples of 2 and 3 bothi . e . 6 from 29 to 99 = multiples of 6 from 1 to 99 - multiples of 6 from 1 to 28 = [ 99 / 6 ] - [ 28 / 6 ] = 16 - 4 = 12 these 8 numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3 i . e . total numbers in w = 35 + 24 - 12 = 47 answer option e" | a = 99 - 29
b = a / 3
c = math.floor(b)
d = 99 - 29
e = d / 2
f = c + e
g = 99 - 29
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
|
a ) 86.25 , b ) 84 , c ) 85 , d ) 88.8 , e ) 80 | d | divide(add(multiply(86, 3), multiply(90, 7)), add(7, 3)) | for a certain exam , a score of 86 was 7 standard deviations below mean and a score of 90 was 3 standard deviations above mean . what was the mean score for the exam ? | "mean - 7 sd = 86 mean + 3 sd = 90 by solving above equations we get , sd ( absolute value ) = 0.4 mean = 88.8 ans . d" | a = 86 * 3
b = 90 * 7
c = a + b
d = 7 + 3
e = c / d
|
a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38 | c | divide(multiply(72, 24), 36) | in a division , a student took 72 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 72 = 24 . x = 24 * 72 . so correct answer would be , ( 24 * 72 ) / 36 = 48 . answer : c" | a = 72 * 24
b = a / 36
|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | add(divide(factorial(5), multiply(factorial(subtract(5, const_2)), factorial(const_2))), 6) | a scientist used a unique two - color code to identify each of the test subjects involved in a certain study . if the scientist found that choosing from among 5 colors produced enough color codes to identify all but 6 of the test subjects , how many test subjects were in the study ? ( assume that the order of the colors in the codes does not matter . ) | 5 c 2 = 10 the number of test subjects is 10 + 6 = 16 the answer is c . | a = math.factorial(5)
b = 5 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
g = f + 6
|
a ) 100 % , b ) 16 2 / 3 % , c ) 200 % , d ) 253 % , e ) 183 1 / 3 % | d | multiply(divide(subtract(const_100, divide(85, 3)), divide(85, 3)), const_100) | enrollment in city college in 1980 was 85 / 3 percent of enrollment in 1990 . what was the percent increase in the college ’ s enrollment from 1980 to 1990 ? | assume 100 enrollments present in 1990 85 / 3 % of 1990 = enrollments on 1980 enrollments on 1980 = 28.33 % change = ( enrollment in 1990 - enrollment in 1980 ) * 100 / ( enrollment in 1980 ) = ( 100 - 28.33 ) * 100 / 28.333 = 7166.67 / 28.333 = 253 % ans - d | a = 85 / 3
b = 100 - a
c = 85 / 3
d = b / c
e = d * 100
|
a ) 776 , b ) 1944 , c ) 2199 , d ) 2157 , e ) 2125 | b | divide(multiply(divide(288, divide(subtract(54, subtract(const_100, 54)), const_100)), 54), const_100) | there were two candidates in an election . winner candidate received 54 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 54 % l = 46 % 54 % - 46 % = 8 % 8 % - - - - - - - - 288 54 % - - - - - - - - ? = > 1944 answer : b" | a = 100 - 54
b = 54 - a
c = b / 100
d = 288 / c
e = d * 54
f = e / 100
|
a ) 40 , b ) 300 / 11 , c ) 243 / 7 , d ) 279 / 11 , e ) 100 / 207 | e | multiply(divide(subtract(multiply(104, 20), multiply(115, 18)), multiply(115, 18)), const_100) | a person bought 115 glass bowls at a rate of rs . 18 per bowl . he sold 104 of them at rs . 20 and the remaining broke . what is the percentage gain for a ? | "cp = 115 * 18 = 2070 and sp = 104 * 20 = 2080 gain % = 100 * ( 2080 - 2070 ) / 2070 = 100 / 207 answer : e" | a = 104 * 20
b = 115 * 18
c = a - b
d = 115 * 18
e = c / d
f = e * 100
|
a ) 5 , b ) 8 , c ) 11 , d ) 12 , e ) 14 | a | divide(10, subtract(3, const_1)) | lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 10 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 10 3 c = c + 10 c = 5 the answer is a ." | a = 3 - 1
b = 10 / a
|
a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(901, 449), multiply(45, const_2)), const_1) | how many even multiples of 45 are there between 449 and 901 ? | 450 = 10 * 45 900 = 20 * 45 the even multiples are 45 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b . | a = 901 - 449
b = 45 * 2
c = a / b
d = c + 1
|
a ) 286 , b ) 342 , c ) 422 , d ) 540 , e ) 618 | b | multiply(19, subtract(19, const_1)) | 19 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "2 * 19 c 2 = 2 * 171 = 342 the answer is b ." | a = 19 - 1
b = 19 * a
|
a ) 260 , b ) 150 , c ) 817 , d ) 716 , e ) 616 | a | divide(585, add(add(divide(3, 4), divide(3, 6)), const_1)) | rs . 585 is divided amongst a , b , c so that 4 times a ' s share , 6 times b ' s share and 3 times c ' s share are all equal . find c ' s share ? | a + b + c = 585 4 a = 6 b = 3 c = x a : b : c = 1 / 4 : 1 / 6 : 1 / 3 = 3 : 2 : 4 4 / 9 * 585 = rs . 260 answer : a | a = 3 / 4
b = 3 / 6
c = a + b
d = c + 1
e = 585 / d
|
a ) 37 , b ) 38 , c ) 39 , d ) 40 , e ) 41 | b | add(add(multiply(const_2, 10), 10), 8) | in 10 years , a will be twice as old as b was 10 years ago . if a is now 8 years older than b the present age of b is | "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 8 = > 2 b - b - 8 = 30 b = 38 so the present age of b is 38 years answer : b" | a = 2 * 10
b = a + 10
c = b + 8
|
a ) 1 / 36 , b ) 1 / 9 , c ) 1 / 7 , d ) 1 / 24 , e ) 2 / 36 | a | divide(const_1, 12) | a die is rolled twice . what is the probability of getting a sum equal to 12 ? | "total number of outcomes possible when a die is rolled = 6 ( ∵ any one face out of the 6 faces ) hence , total number of outcomes possible when a die is rolled twice , n ( s ) = 6 × 6 = 36 e = getting a sum of 12 when the two dice fall = { ( 6 , 6 ) } hence , n ( e ) = 1 p ( e ) = n ( e ) / n ( s ) = 1 / 36 = 1 / 36 answer : option a" | a = 1 / 12
|
a ) 1 kmph , b ) 3 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | e | divide(subtract(14, 4), const_2) | a man can row his boat with the stream at 14 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 14 us = 4 s = ? s = ( 14 - 4 ) / 2 = 5 kmph answer : e" | a = 14 - 4
b = a / 2
|
a ) t = 200 , b ) t = 600 , c ) t = 800 , d ) t = 1600 , e ) t = 50 | d | multiply(power(const_2, 4), const_100) | cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost is what percent t of the original cost ? | "original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600" | a = 2 ** 4
b = a * 100
|
a ) 100 , b ) 300 , c ) 400 , d ) 3000 , e ) 4000 | c | multiply(divide(12, subtract(99, 96)), const_100) | in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many q eggs does the plant process per day ? | straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there q = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c . | a = 99 - 96
b = 12 / a
c = b * 100
|
a ) 250 , b ) 3200 / 3 , c ) 3150 / 3 , d ) 4002 / 3 , e ) 550 | b | multiply(400, const_2) | on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 ⁄ 2 of the total distance and again after she traveled 1 ⁄ 4 of the distance remaining between her first stop and her destination . she then drove the remaining 400 miles and arrived safely at her destination . what was the total distance , in miles , from traci ’ s starting point to seal beach ? | "let d = total distance traci traveled 1 / 2 = d / 2 i . e . remaining distance = d / 2 she traveled 1 / 4 th of d / 2 = d / 8 thus : d = ( d / 2 ) + ( d / 8 ) + 400 d = 3200 / 3 answer : b" | a = 400 * 2
|
a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 5,000 | a | divide(multiply(1.2, power(10, 11)), multiply(240, power(10, add(const_4, const_2)))) | a certain country had a total annual expenditure of $ 1.2 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ? | "total expenditure / population = per capita expenditure hence , ( 1,2 x 10 ^ 11 ) / 240 000 000 = ( 12 x 10 ^ 10 ) / ( 2,4 x 10 ^ 8 ) = 5 x 10 ^ ( 10 - 8 ) = 5 x 10 ^ 2 = 500 . answer is a ." | a = 10 ** 11
b = 1 * 2
c = 4 + 2
d = 10 ** c
e = 240 * d
f = b / e
|
a ) 39,30 , b ) 41,32 , c ) 42,33 , d ) 43,34 , e ) 35,36 | c | add(divide(subtract(multiply(9, const_100), multiply(56, 9)), subtract(multiply(56, const_2), const_100)), 9) | two students appeared at an examination . one of them secured 9 marks more than the other and his marks was 56 % of the sum of their marks . the marks obtained by them are : | let their marks be ( x + 9 ) and x . then x + 9 = 56 / 100 ( x + 9 + x ) 3 x = 99 x = 33 their marks 42 and 33 answer c | a = 9 * 100
b = 56 * 9
c = a - b
d = 56 * 2
e = d - 100
f = c / e
g = f + 9
|
a ) 2 , b ) 10 , c ) 12 , d ) 8 , e ) 15 | d | divide(multiply(multiply(const_2.0, const_4.0), 3), multiply(4, 8)) | aaron will jog from home at 4 miles per hour and then walk back home by the same route at 8 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ? | "xyt / ( x + y ) x = 4 , y = 8 t = 3 4 * 8 * 3 / 4 + 8 = 96 / 12 = 8 answer : d" | a = 2 * 0
b = a * 3
c = 4 * 8
d = b / c
|
a ) 4 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | d | add(add(add(7, 2), divide(5, 5)), divide(3, 3)) | if x is the product of the positive integers from 1 to 8 , inclusive , and if i , k , m , and p are positive integers such that x = 2 ^ i ∗ 3 ^ k ∗ 5 ^ m ∗ 7 ^ p , then i + k + m + p = | "here the idea is to go expressing each factor from one to eight , decomposed into its primes : 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 1 * 2 * 3 * ( 2 * 2 ) * 5 + ( 2 * 3 ) * 7 * ( 2 * 2 * 2 ) = 2 ^ 7 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1 then i = 7 k = 2 m = 1 p = 1 i + k + m + p = 11 correct answer d" | a = 7 + 2
b = 5 / 5
c = a + b
d = 3 / 3
e = c + d
|
['a ) 32', 'b ) 64', 'c ) 25', 'd ) 16', 'e ) none'] | b | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 40), subtract(const_100, 40))), multiply(const_100, const_100)), const_100) | each side of a rectangular field diminished by 40 % . by how much per cent is the area of the field diminished ? | solution : let the original length of the rectangle be 20 unit and breadth be 10 unit . then original area = length * breadth = 20 * 10 = 200 square unit . 40 % decrease in each side , then length = ( 20 - 40 % of 20 ) = 12 unit . breadth = ( 10 - 40 % of 10 ) = 6 unit . now , area = 12 * 6 = 72 square unit . decrease in area = 200 - 72 = 128 square unit . % decrease in area = ( 128 / 200 ) * 100 = 64 % mind calculation method : let the original area be 100 square unit . 100 = = = 40 % ↓ ( decrease in length ) = = = > 60 = = = 40 % ↓ = = = ( decrease in breadth ) = = = > 36 . diminished in area = 100 - 36 = 64 % answer : option b | a = 100 * 100
b = 100 - 40
c = 100 - 40
d = b * c
e = a - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) a ) 16 , b ) b ) 15 , c ) c ) 19 , d ) d ) 22 , e ) e ) 24 | b | add(14, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 14 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 14 + 36 – 26 = 150 = > 150 / 10 = 15 b )" | a = 36 - 26
b = a / 10
c = 14 + b
|
a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | d | multiply(divide(add(divide(multiply(20, 20), const_100), divide(multiply(80, 15), const_100)), add(20, 80)), const_100) | there are two groups of students in the sixth grade . there are 20 students in group a , and 80 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 15 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their homework ? | "number of students in group a = 20 students who forget homework in group a = 20 % of 20 = 4 students number of students in group b = 80 students who forget homework in group b = 15 % of 80 = 12 students total number of students = 20 + 80 = 100 students who forgot homework = 4 + 12 = 16 students percentage of students who forget homework = 16 / 100 * 100 = 16 % answer : d" | a = 20 * 20
b = a / 100
c = 80 * 15
d = c / 100
e = b + d
f = 20 + 80
g = e / f
h = g * 100
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | divide(subtract(multiply(6, 8), multiply(4, 5)), 2) | the average of 6 quantities is 8 . the average of 4 of them is 5 . what is the avg . of the remaining 2 numbers ? | "sum of 8 quantities = total no of quantities x average of them = 6 x 8 = 48 sum of 4 quantities = 4 x 5 = 20 sum of 4 quantities + sum of 2 quantities = sum of 6 quantities 20 + sum of 2 quantities = 48 sum of 2 quantities = 28 average of 2 quantities = sum of 2 quantities / 2 = 28 / 2 = 14 answer : d" | a = 6 * 8
b = 4 * 5
c = a - b
d = c / 2
|
a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | e | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(4048, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 4048 . what is the difference between rohan ’ s and sudhir ’ s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8 = 6 : 3 : 2 the difference between rohan ’ s and sudhir ’ s share = 1 share : . i . e . = rs . 4048 × 1 / 11 = rs . 368 . e" | a = 3 * 4
b = a - 3
c = 3 * 4
d = b / c
e = 12000 * d
f = 3 * 4
g = f - 4
h = 3 * 4
i = g / h
j = 9000 * i
k = e - j
l = 3 * 4
m = l - 3
n = 3 * 4
o = m / n
p = 12000 * o
q = 18000 + p
r = 3 * 4
s = r - 4
t = 3 * 4
u = s / t
v = 9000 * u
w = q + v
x = 4048 / w
y = k * x
|
a ) 15 , b ) 21 , c ) 17 , d ) 57 , e ) 77 | e | divide(subtract(multiply(500, 90), 10350), subtract(500, 50)) | a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 90 . find the number of notes of rs . 50 denomination . | total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 90 - x now , 50 * x + 500 * ( 90 - x ) = 10350 50 x + 45000 - 500 x = 10350 - 450 x = - 34650 x = 77 . no . of 50 rupees note = 77 . answer : option e | a = 500 * 90
b = a - 10350
c = 500 - 50
d = b / c
|
a ) 435,000 , b ) 440,000 , c ) 450,000 , d ) 480,000 , e ) 432,000 | e | multiply(150, divide(subtract(2007, 2003), const_2)) | during 2003 , a company produced an average of 4,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ? | "company produced 12 * 4000 = 48,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 48,000 . the gives the average of ( x + 48,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 48,000 + 150 % ( 48,000 ) = 120,000 . so : ( x + 48,000 ) / 4 = 120,000 x + 48,000 = 480,000 x = 432,000 the answer is e ." | a = 2007 - 2003
b = a / 2
c = 150 * b
|
a ) 10 , b ) 12 , c ) 16 , d ) 20 , e ) 22 | b | divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 32)), 4) | the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . thereforedecreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 ⇒ 40 y + 384 = 36 y + 432 ⇒ 4 y = 48 ∴ y = 12 b" | a = 40 - 4
b = 12 * a
c = 12 * 32
d = b - c
e = d / 4
|
a ) 2.5 , b ) 5 , c ) 7.5 , d ) 10 , e ) 12 | d | multiply(5, const_2) | the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 5 ) , how many units is p from the y - axis ? | "the x - axis is 5 units from the point p . thus the y - axis is 10 units from the point p . the answer is d ." | a = 5 * 2
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a ) 34 % , b ) 68 % , c ) 51 % , d ) 75 % , e ) 83 % | d | subtract(const_100, multiply(divide(add(20, const_100), add(60, const_100)), const_100)) | two numbers are respectively 20 % and 60 % more than a third number . the percentage that is first of the second is ? | "i ii iii 120 160 100 160 - - - - - - - - - - 120 100 - - - - - - - - - - - ? = > 75 % answer : d" | a = 20 + 100
b = 60 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 47 , b ) 53 , c ) 56 , d ) 61 , e ) 64 | b | subtract(90, subtract(397, multiply(90, 4))) | we bought a total of 90 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 397 . how many math books did we buy ? | "let m be the number of math books and let h be the number of history books . m + h = 90 . h = 90 - m . 4 m + 5 h = 397 . 4 m + 5 ( 90 - m ) = 397 . - m + 450 = 397 . m = 450 - 397 = 53 . the answer is b ." | a = 90 * 4
b = 397 - a
c = 90 - b
|
a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 232 , e ) s . 280 | d | divide(580, add(divide(150, const_100), const_1)) | two employees a and b are paid a total of rs . 580 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 580 but x = 150 % of y = 150 y / 100 = 15 y / 10 ∴ 15 y / 10 + y = 580 ⇒ y [ 15 / 10 + 1 ] = 580 ⇒ 25 y / 10 = 580 ⇒ 25 y = 5800 ⇒ y = 5800 / 25 = rs . 232 d )" | a = 150 / 100
b = a + 1
c = 580 / b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(11, 7), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 7 pencils and 11 charts have been brought to the meeting , how many people are present ? | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 7 total charts are 11 so a + 2 b = 11 add both : 3 a + 3 b = 18 so a + b = 6 total number of people = 6 a" | a = 11 + 7
b = 2 + 1
c = a / b
|
a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number r ? | "( 1 + 1 / 8 ) x - ( 1 - 1 / 4 ) x = 30 ( 9 / 8 ) x - ( 3 / 4 ) x = 30 x = 80 = r answer : b" | a = 12 / 5
b = 1 + a
c = 25 / 100
d = 1 - c
e = b - d
f = 30 / e
|
a ) 180.27 cm , b ) 180.29 cm , c ) 978.29 cm , d ) 178.89 cm , e ) 176.29 cm | b | floor(divide(add(subtract(multiply(35, 182), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 182 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? | "calculated average height of 35 boys = 182 cm . wrong total height of 35 boys = 182 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 182 cm - ( 166 cm - 106 cm ) / 35 = 182 cm - 60 / 35 cm = 182 cm - 1.71 cm = 180.29 cm . answer : b" | a = 35 * 182
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
|
a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 2250 , e ) $ 1200 | d | subtract(3000, divide(3000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | the salaries of a and b together amount to $ 3000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 3000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 3000 - x ) x = $ 2250 answer is d" | a = 100 - 85
b = 100 - 95
c = a / b
d = c + 1
e = 3000 / d
f = 3000 - e
|
['a ) 7', 'b ) 21', 'c ) 19', 'd ) 15', 'e ) 17'] | e | divide(add(const_1, sqrt(add(const_1, multiply(const_4, 272)))), const_2) | if a number is subtracted from its square we get 272 . find the number ? | x ^ 2 â € “ x = 272 = > x = 17 answer e | a = 4 * 272
b = 1 + a
c = math.sqrt(b)
d = 1 + c
e = d / 2
|
a ) 87 kmph , b ) 65 kmph , c ) 54 kmph , d ) 61 kmph , e ) 18 kmph | d | subtract(multiply(20, multiply(220, const_0_2778)), 120) | a train 120 m long crosses a platform 220 m long in 20 sec ; find the speed of the train ? | "d = 120 + 220 = 340 t = 20 s = 340 / 20 * 18 / 5 = 61 kmph answer : d" | a = 220 * const_0_2778
b = 20 * a
c = b - 120
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(reminder(reminder(88, 33), 17), reminder(90, reminder(33, 17))) | for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 88 θ 33 ) θ 17 ) - ( 90 θ ( 33 θ 17 ) ) ? | "( ( 88 θ 33 ) θ 17 ) the remainder of 88 divided by 33 is 22 ; the remainder of 22 divided by 17 is 5 ; ( 98 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 98 divided by 16 is 2 . 5 - 2 = 3 . answer : d ." | a = reminder - (
|
a ) 111.7 , b ) 118.7 , c ) 113.7 , d ) 218.7 , e ) 418.7 | b | add(add(multiply(10, 6), multiply(10, 4)), multiply(6, const_3)) | if solomon distributes his toys in the ratio of 1 / 4 : 1 / 6 : 1 / 10 : 1 / 9 between his 4 friends a , b , c and d , then find the total number of toys solomon should have ? | here , a : b : c : d = 1 / 4 : 1 / 6 : 1 / 10 : 1 / 9 1 ) l . c . m of 4 : 6 : 10 : 9 is 180 2 ) find the number of pens each friend received - - - - - - - - - ( to find no . of pens each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 4 ) x 180 = 45 b = ( 1 / 6 ) x 180 = 30 c = ( 1 / 10 ) x 180 = 18 d = ( 1 / 7 ) x 180 = 25.7 3 ) total number of toys = ( 45 x + 30 x + 18 x + 25.7 x ) = 118.7 x minimum number of pens ( x ) = 1 therefore , total number of pens = 118.7 pens . correct option : b | a = 10 * 6
b = 10 * 4
c = a + b
d = 6 * 3
e = c + d
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a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 2888 | d | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.40) | the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.40 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.40 ) = rs . 5808 . answer : d ) 5808" | a = 13 / 86
b = math.sqrt(a)
c = b * 100
d = circumface * (
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a ) 34 % , b ) 44 % , c ) 48 % , d ) 32 % , e ) 46 % | c | subtract(const_100, divide(multiply(1000, const_100), 675)) | an article is bought for rs . 675 and sold for rs . 1000 , find the gain percent ? | "675 - - - - 325 100 - - - - ? = > = 48 % answer : c" | a = 1000 * 100
b = a / 675
c = 100 - b
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a ) rs . 72 , b ) rs . 92 , c ) rs . 162.50 , d ) rs . 116.50 , e ) none of these | c | multiply(divide(const_100, 8), 13) | a 13 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 13 , investment = rs . ( 100 / 8 x 13 ) = rs . 162.50 ∴ market value of rs . 100 stock = rs . 162.50 answer c" | a = 100 / 8
b = a * 13
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a ) 15 , b ) 26 , c ) 22 , d ) 27 , e ) 11 | a | divide(multiply(6, sqrt(add(9, 16))), const_2) | from a container , 6 liters milk was drawn out and was replaced by water . again 6 liters of mixture was drawn out and was replaced by the water . thus the quantity of milk and water in the container after these two operations is 9 : 16 . the quantity of mixture is : | explanation : let quantity of mixture be x liters . suppose a container contains x units of liquid from which y units are taken out and replaced by water . after operations , the quantity of pure liquid = units , where n = no of operations . so , quantity of milk = given that , milk : water = 9 : 16 milk : ( milk + water ) = 9 : ( 9 + 16 ) milk : mixture = 9 : 25 answer : a ) 15 | a = 9 + 16
b = math.sqrt(a)
c = 6 * b
d = c / 2
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a ) 10 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | a | subtract(multiply(add(5, const_1), const_3), divide(480, 60)) | a train travels from new york to chicago , a distance of approximately 480 miles , at an average rate of 60 miles per hour and arrives in chicago at 5 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) | 5 : 00 in evening in chicago = 6 : 00 in evening in new york . so , the train was in chicago 6 : 00 in the evening , new york time . the trip took t = d / r = 480 / 60 = 8 hours . therefore , the train depart from new york at 6 : 00 - 8 hours = 10 : 00 in the morning , new york time . answer : a . | a = 5 + 1
b = a * 3
c = 480 / 60
d = b - c
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a ) 4 , b ) 55 , c ) 15 , d ) 48 , e ) 63 | b | subtract(add(power(7, 2), multiply(7, 2)), add(power(2, 2), multiply(2, 2))) | the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds than it was moving after 2 seconds ? | "given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 2 ) = 2 ^ 2 + 2 * 2 = 8 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 8 = 55 km / hr option b" | a = 7 ** 2
b = 7 * 2
c = a + b
d = 2 ** 2
e = 2 * 2
f = d + e
g = c - f
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a ) 1960 , b ) 1965 , c ) 1970 , d ) 1975 , e ) 1990 | e | add(1950, add(1.6, add(4, 3))) | in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.6 times as old as robert ? | "in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 1950 ro = 10 , ri = 40 1990 ro = 50 , ri = 80 = 1.6 times thus ans : e" | a = 4 + 3
b = 1 + 6
c = 1950 + b
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a ) 107 , b ) 153 , c ) 178 , d ) 198 , e ) 165 | b | divide(multiply(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)), gcd(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)))) | a room is 8 meters 16 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 8 meters and 16 centimeters = 816 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 816 and 432 = 48 thus , the number of square tiles required = ( 816 x 432 ) / ( 48 x 48 ) = 17 x 9 = 153 answer : b" | a = 8 * 100
b = a + 16
c = 4 * 100
d = c + 32
e = b * d
f = 8 * 100
g = f + 16
h = 4 * 100
i = h + 32
j = math.gcd(g, i)
k = 8 * 100
l = k + 16
m = 4 * 100
n = m + 32
o = math.gcd(l, n)
p = j * o
q = e / p
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a ) 34 , b ) 35 , c ) 36 , d ) 37 , e ) 38 | c | subtract(factorial(subtract(8, 5)), add(multiply(5, const_3), const_3)) | there are 8 executives , including the ceo and cfo , that are asked to form a small team of 5 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | "the total number of ways to form a team of 5 is 8 c 5 = 56 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 6 c 3 = 20 . the number of ways to form an acceptable team is 56 - 20 = 36 . the answer is c ." | a = 8 - 5
b = math.factorial(a)
c = 5 * 3
d = c + 3
e = b - d
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a ) 334 , b ) 500 , c ) 376 , d ) 288 , e ) 271 | b | divide(460, divide(92, const_100)) | victor gets 92 % marks in examinations . if these are 460 marks , find the maximum marks . | "let the maximum marks be m then 92 % of m = 460 ⇒ 92 / 100 × m = 460 ⇒ m = ( 460 × 100 ) / 92 ⇒ m = 46000 / 92 ⇒ m = 500 therefore , maximum marks in the examinations are 500 . answer : b" | a = 92 / 100
b = 460 / a
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a ) 70000 , b ) 73000 , c ) 75000 , d ) 77000 , e ) 79000 | c | divide(82500, add(divide(10, const_100), const_1)) | max planned to buy bike , sales man advised him to buy within this month , if he purchase next month means state tax will be increasing 10 % than its cost and it would be 82500 . what is the bike ' s selling price in this month | sp * ( 110 / 100 ) = 82500 sp = 750 * 100 = > sp = 75000 answer : c | a = 10 / 100
b = a + 1
c = 82500 / b
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a ) 3 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | c | multiply(divide(5, add(divide(3, 2), const_1)), 3) | a boy goes to his school from his house at a speed of 3 km / hr and returns at a speed of 2 km / hr . if he takes 5 hours in going and coming . the distance between his house and school is ? | "average speed = ( 2 * 3 * 2 ) / ( 3 + 2 ) = 12 / 5 km / hr . distance traveled = 12 / 5 * 5 = 12 km . distance between house and school = 12 / 2 = 6 km . answer : c" | a = 3 / 2
b = a + 1
c = 5 / b
d = c * 3
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a ) 500 m , b ) 1000 m , c ) 1500 m , d ) 2000 m , e ) 2500 m | a | multiply(500, const_1) | in a 500 m race , ravi beats ramesh by 5 seconds or 100 m . they decide to run another race and this time ravi gives ramesh a head tart of 200 m . if ravi ’ s speed is twice his previous speed and ramesh ’ s speed is one and half times his previous speed , how far from the starting point should the winning post be so that they finish at the same time ? | ravi beats ramesh by 5 sec or 100 mtr . . . . means ramesh cover 100 mtr in 5 sec . . . . by this we can get ramesh speed in 20 m / sec 20 . . . . . . . . . . . . . in 1 sec 500 . . . . . . . . in 500 / 20 = 25 sec ( ramesh ' s time ) so ravi take 20 sec to cover 500 mtr . . . . . . by this we can get that ravis speed is 25 m / sec now assume x be the distance from staring line where we need to put the winning post . according to question ravi ’ s speed will be 25 * 2 = 50 m / sec & ramesh ’ s speed will be 20 * 1.5 = 30 m / sec now assume x be the distance from staring line where we need to put the winning post . they will take same time to reach the wining post . as ravi gives ramesh a 200 m ahead start so he has to accede x meter and ramesh has to accede ( x - 200 ) meter hence we may write , x / 50 = ( x - 200 ) / 30 x = 500 meter answer : a | a = 500 * 1
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a ) $ 14,755 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 25 % next year and that the price of type b products will increase by 10 % next year . this year , the total amount paid for type a products was $ 4500 and the total price paid for type b products was $ 8300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.25 * 4500 = 5625 cost of type b products next year = 1.1 * 8300 = 9130 total 5625 + 9130 = 14755 answer : a" | a = 3 / 4
b = a * 1000
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a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 2 / 3 | d | divide(sqrt(25), sqrt(49)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 25 : 49 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 25 / 49 - - > height / height = 5 / 7 . answer : d ." | a = math.sqrt(25)
b = math.sqrt(49)
c = a / b
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a ) 35 , b ) 45 , c ) 25 , d ) 72 , e ) 55 | d | multiply(280, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 280 , then how much amount should c pay as his share of rent ? | a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 280 × 9 / 35 = 8 × 9 = 72 answer is d | a = 15 * 3
b = 10 * 7
c = 12 * 5
d = b + c
e = 15 * 3
f = d + e
g = a / f
h = 280 * g
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a ) 20 % , b ) 42 % , c ) 60 % , d ) 80 % , e ) 84 % | c | multiply(subtract(const_1, divide(30, subtract(const_100, 25))), const_100) | exactly 30 % of the reporters for a certain wire service cover local politics in country x . if 25 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | let ' s assume there are 100 reporters - - > 30 reporters cover local politics . now , as 25 % of the reporters who cover all politics do not cover local politics then the rest 75 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 75 % of them equal to 30 ( # of reporters who cover local politics ) : 0.75 x = 30 - - > x = 40 , hence 40 reporters cover politics and the rest 100 - 40 = 60 reporters do not cover politics at all . answer : c . | a = 100 - 25
b = 30 / a
c = 1 - b
d = c * 100
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a ) 20 , b ) 40 , c ) 80 , d ) 160 , e ) 320 | c | multiply(multiply(divide(50, 16), divide(50, 16)), const_1000) | positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 16 . what is the value of x ? | "y = 0.5 * 0.5 * x = x / 4 y % * x = 16 ( y / 100 ) * x = 16 ( x / 400 ) * x = 16 x ^ 2 = 16 * 400 x = 80 the answer is c ." | a = 50 / 16
b = 50 / 16
c = a * b
d = c * 1000
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a ) 16 , b ) 17 , c ) 15 , d ) 18 , e ) 12 | c | subtract(const_60, multiply(const_60, divide(36, 48))) | excluding stoppages , the speed of a train is 48 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 12 / 48 * 60 = 15 answer : c" | a = 36 / 48
b = const_60 * a
c = const_60 - b
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a ) 315 : 26 , b ) 315 : 25 , c ) 315 : 30 , d ) 415 : 35 , e ) 425 : 35 | a | divide(const_3, 2) | difference between the length & breadth of a rectangle is 26 m . if its perimeter is 208 m , then the ratio of area and perimeter is ? ? we have : ( l - b ) = 26 and 2 ( l + b ) = 208 or ( l + b ) = 104 ? | "solving the two equations , we get : l = 63 and b = 40 . area = ( l x b ) = ( 63 x 40 ) m 2 = 2520 m ^ 2 the required ratio is 2520 : 208 = 315 : 26 a" | a = 3 / 2
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a ) 14 , b ) 12 , c ) 11 , d ) 15 , e ) 16 | d | divide(divide(multiply(105, const_2), const_3), const_4) | everyone shakes hands with everyone else in a room . total number of handshakes is 105 . number of persons = ? | "in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 105 or n ( n - 1 ) = 210 or n = 15 answer is ( d )" | a = 105 * 2
b = a / 3
c = b / 4
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a ) 29997 , b ) 10800 , c ) 27098 , d ) 19000 , e ) 2799 | b | subtract(12000, 1200) | the price of a t . v . set worth rs . 12000 is to be paid in 20 installments of rs . 1200 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | money paid in cash = rs . 1200 balance payment = ( 12000 - 1200 ) = rs . 10800 . answer : b | a = 12000 - 1200
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a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 21 | e | divide(add(140, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 140 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 140 + 280 = 420 m . the time required = d / s = 420 / 20 = 21 sec . answer : e" | a = 140 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
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a ) 18 % , b ) 19 % , c ) 32 % , d ) 12 % , e ) 52 % | b | subtract(const_100, divide(multiply(add(const_100, 12), subtract(const_100, 28)), const_100)) | the tax on a commodity is diminished by 28 % but its consumption is increased by 12 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 72 * 112 = 8064 10000 - - - - - - - 1936 100 - - - - - - - ? = 19.36 % answer : b" | a = 100 + 12
b = 100 - 28
c = a * b
d = c / 100
e = 100 - d
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a ) 7 / 4 , b ) 4 / 3 , c ) 14 / 5 , d ) 10 / 3 , e ) 17 / 5 | c | max(divide(subtract(const_1, multiply(add(divide(const_1, add(const_4, const_2)), divide(const_1, 5)), const_2)), divide(const_1, 5)), const_3) | two mechanics were working on your car . one can complete the given job in 5 hours , but the new guy takes eight hours . they worked together for the first two hours , but then the first guy left to help another mechanic on a different job . how long will it take the new guy to finish your car ? | "rate ( 1 ) = 1 / 5 rate ( 2 ) = 1 / 8 combined = 13 / 40 work done in 2 days = 13 / 20 work left = 7 / 20 rate * time = work left 1 / 8 * time = 7 / 20 time = 14 / 5 c" | a = 4 + 2
b = 1 / a
c = 1 / 5
d = b + c
e = d * 2
f = 1 - e
g = 1 / 5
h = f / g
i = max(h)
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a ) 45 , b ) 42 , c ) 52 , d ) 43 , e ) 51 | d | add(divide(subtract(196, 20), 4), const_1) | how many multiples of 4 are there between 20 and 196 ? | "it should be mentioned whether 20 and 196 are inclusive . if 20 and 196 are inclusive , then the answer is ( 196 - 20 ) / 4 + 1 = 45 . if 20 and 196 are not inclusive , then the answer is ( 192 - 24 ) / 4 + 1 = 43 . since oa is d , then we have not inclusive case ." | a = 196 - 20
b = a / 4
c = b + 1
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a ) 15 , b ) 6 , c ) 40 , d ) 10 , e ) 8 | e | divide(multiply(4, 12), subtract(12, 6)) | a group of men decided to do a work in 6 days , but 4 of them became absent . if the rest of the group did the work in 12 days , find the original number of men ? | "original number of men = 4 * 12 / ( 12 - 6 ) = 8 answer is e" | a = 4 * 12
b = 12 - 6
c = a / b
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a ) 85 , b ) 74 , c ) 67 , d ) 53 , e ) 207 | e | subtract(500, add(multiply(4, 61.5), 47)) | after 4 games , team b had an average of 61.5 points per game . if it got only 47 points in game 5 , how many more points does it need to score to get its total above 500 ? | "( 4 * 61.5 ) + 47 + x > 500 246 + 47 + x > 500 293 + x > 500 = > x > 207 option e" | a = 4 * 61
b = a + 47
c = 500 - b
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a ) 5.29 min , b ) 5.28 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(726, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 726 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 726 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 726 m apart , they take ( 100 / 825 * 726 / 1000 ) hrs = ( 242 / 2750 * 60 ) min = 5.28 min . answer : b" | a = 726 / 1000
b = 4 + 5
c = a / b
d = c * const_60
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a ) 3 : 1 , b ) 4 : 3 , c ) 3 : 8 , d ) 3 : 25 , e ) 3 : 4 | b | divide(subtract(35, 32), subtract(32, 28)) | two trains running in opposite directions cross a man standing on the platform in 35 seconds and 28 seconds respectively and they cross each other in 32 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 35 x meters , and length of the second train = 28 y meters . ( 35 x + 28 y ) / ( x + y ) = 32 = = > 35 x + 28 y = 32 x + 32 y = = > 3 x = 4 y = = > x / y = 4 / 3 answer : option b" | a = 35 - 32
b = 32 - 28
c = a / b
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a ) 300 , b ) 277 , c ) 266 , d ) 200 , e ) 121 | d | divide(120, divide(120, const_100)) | 40 % of a number is added to 120 , the result is the same number . find the number ? | ": ( 40 / 100 ) * x + 120 = x 2 x = 400 x = 200 answer : d" | a = 120 / 100
b = 120 / a
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a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | a | multiply(divide(16.02, 0.01), const_100) | 16.02 ã — 0.01 = ? | "16.02 ã — 0.01 = 0.1602 the answer is a ." | a = 16 / 2
b = a * 100
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a ) s . 800 , b ) s . 810 , c ) s . 805 , d ) s . 900 , e ) s . 920 | c | multiply(2415, 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 . 2415 ? | 2415 * 1 / 3 = 805 option c | a = 1 / 3
b = 2415 * a
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a ) 39 and 31 , b ) 40 and 33 , c ) 42 and 33 , d ) 37 and 31 , e ) 35 and 32 | c | multiply(divide(multiply(9, const_100), subtract(56, subtract(const_100, 56))), 56) | 2 students appeared at an examination . 1 of them secured 9 marks more than the other one and hismarks was 56 % of the sum of their marks . find the marks obtained by them are ? | let their marks be ( x + 9 ) and x . then , x + 9 = 56 / 100 ( x + 9 + x ) = > 25 ( x + 9 ) = > 14 ( 2 x + 9 ) = > 3 x = 99 = > x = 33 . so , their marks are 42 and 33 c | a = 9 * 100
b = 100 - 56
c = 56 - b
d = a / c
e = d * 56
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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 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
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a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 8 / 45 , e ) 9 / 45 | b | divide(2, 5) | what is the hcf of 2 / 3 , 4 / 9 and 6 / 5 | "explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 9 , 5 ) = 2 / 45 answer : option b" | a = 2 / 5
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a ) 6 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(multiply(multiply(3, 8), 2), 8) | 3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 8 pumps work to empty the tank in 1 day ? | "3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 8 pumps then 48 / 8 = 6 hrs . answer : a" | a = 3 * 8
b = a * 2
c = b / 8
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a ) 2 , b ) 4 , c ) 5 , d ) 1 11 / 13 , e ) 9 | d | inverse(add(inverse(8), add(inverse(4), inverse(6)))) | a , b and c can do a piece of work in 4 days , 6 days and 8 days respectively . how long will they taken , if all the three work together ? | "1 / 4 + 1 / 6 + 1 / 8 = 13 / 24 = > 24 / 13 = > 1 11 / 13 days answer : d" | a = 1/(8)
b = 1/(4)
c = 1/(6)
d = b + c
e = a + d
f = 1/(e)
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a ) 300 , b ) 250 , c ) 280 , d ) 320 , e ) 400 | e | divide(add(150, 30), divide(45, const_100)) | a student has to obtain 45 % of the total marks to pass . he got 150 marks and failed by 30 marks . the maximum marks are ? | "let the maximum marks be x then , 45 % of x = 150 + 30 45 x / 100 = 180 45 x = 180 * 100 = 18000 x = 400 answer is e" | a = 150 + 30
b = 45 / 100
c = a / b
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | b | add(reminder(multiply(reminder(46, const_4), 93), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 46 ) is : | "step 1 : ( 35 ) ^ ( 87 ) can be broken up into ( 5 ) ^ 87 x ( 7 ) ^ 87 . each power of 7 ends in a units digit of either a 7 , 9,3 or 1 . each power of 5 ends in a 5 . when you multiply 5 by any odd number you will end up with a units digit of 5 . step 2 : ( 93 ) ^ ( 46 ) can be broken up into ( 3 ) ^ 46 x ( 31 ) ^ 46 . each power of 31 ends in a units digit of 1 . each power of 3 ends in a units digit of either 3 , 9,7 and 1 . since there is a pattern here where every 4 th power of 3 ends in a units digit of 3 , the 46 th power of 3 would end in a units digit of 9 . when you multiply 1 by 9 you end up with a units digit of 9 . therefore , the units digit we are looking for is 5 + 9 = 14 . units digit will be 4 . answer b ." | a = reminder * (
b = reminder + (
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a ) 16.5 % , b ) 20 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(5.5, 3)), const_100) | a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 5.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | "in my view easiest method would be , eliminating the percentage sign let 3.5 be 35 5.5 be 55 and 3 be 30 then first eq : s + f = e 130 s + 155 f = 135 e 5 s = 20 f therefore , s = 4 f and then e = 5 f f / e = f / 5 f = 20 % therefore , answer is b" | a = 3 - 5
b = 5 - 5
c = a / b
d = c * 100
|
a ) 104 , b ) 60 , c ) 52 , d ) 50 , e ) 222 | e | add(subtract(30, multiply(power(2, 2), 2)), multiply(2, power(10, 2))) | if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 30 , what is the value of f ( 10 ) ? | f ( x ) = 2 x ^ 2 + y f ( 2 ) = 30 = > 2 * ( 2 ) ^ 2 + y = 30 = > 8 + y = 30 = > y = 22 f ( 10 ) = 2 * ( 10 ) ^ 2 + 22 = 222 answer e | a = 2 ** 2
b = a * 2
c = 30 - b
d = 10 ** 2
e = 2 * d
f = c + e
|
a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 1 / 32 | e | power(divide(1, 2), 5) | if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 2 flips but not on the last 3 flips ? | "p ( hhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is e ." | a = 1 / 2
b = a ** 5
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a ) 1.3 % , b ) 1.2 % , c ) 1.1 % , d ) 1.0 % , e ) 0.9 % | a | multiply(divide(add(const_1, divide(multiply(1.5, 800), const_100)), const_1000), const_100) | by weight , liquid x makes up 0.5 percent of solution p and 1.5 percent of solution q . if 200 grams of solution p are mixed with 800 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | the number of grams of liquid x is 0.5 ( 200 ) / 100 + 1.5 ( 800 ) / 100 = 1 + 12 = 13 grams . 13 / 1000 = 1.3 % the answer is a . | a = 1 * 5
b = a / 100
c = 1 + b
d = c / 1000
e = d * 100
|
a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total q of the book ? | first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) = q 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) . | a = 3000 + 3
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a ) 596 , b ) 587 , c ) 533 , d ) 568 , e ) 599 | e | subtract(multiply(const_10, 6), 6) | the difference between the place values of 6 and 1 in the number 527631 is | sol . = ( place value of 6 ) – ( place value of 1 ) = ( 600 - 1 ) = 599 answer e | a = 10 * 6
b = a - 6
|
a ) 1 / 16 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 / 8 | e | 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 twice on heads and once on tails ? | "must be once on twice and once on tails 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 answer : e" | a = 2 ** 3
b = 1 / a
|
a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | c | multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100) | the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 186,000 . what was the percent increase in the cost of the home ? | "increase = 186000 - 120000 = 66000 % increase = 66000 * 100 / 120000 = 55 % answer : option c" | a = 4 * 4
b = 12 * a
c = b * 1000
d = 12 * 1000
e = d * 10
f = c - e
g = 12 * 1000
h = g * 10
i = f / h
j = i * 100
|
a ) 25 , b ) 20 , c ) 18 , d ) 16 , e ) 14 | a | divide(multiply(divide(divide(5, 2), divide(1, 2)), divide(5, 2)), divide(multiply(divide(5, 2), divide(1, 2)), divide(5, 2))) | [ ( 5 / 2 ÷ 1 / 2 × 5 / 2 ) ] / [ ( 5 / 2 × 1 / 2 ÷ 5 / 2 ) ] = ? | explanation : [ ( 5 / 2 ÷ 1 / 2 × 5 / 2 ) ] / [ ( 5 / 2 × 1 / 2 ÷ 5 / 2 ) ] = [ 5 / 2 × 2 / 1 × 5 / 2 ] / [ 5 / 2 × 1 / 2 × 2 / 5 ] = 25 / 2 / 1 / 2 = 25 / 2 x 2 / 1 = 25 answer : option a | a = 5 / 2
b = 1 / 2
c = a / b
d = 5 / 2
e = c * d
f = 5 / 2
g = 1 / 2
h = f * g
i = 5 / 2
j = h / i
k = e / j
|
a ) $ 12537.50 , b ) $ 13537.50 , c ) $ 14437.50 , d ) $ 16437.50 , e ) $ 18537.50 | c | multiply(700, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 700 per sq . metre . | "c area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 700 hence , total cost = 5.5 × 3.75 × 700 = $ 14437.50" | a = 5 * 5
b = 700 * a
|
a ) 11 , b ) 10 , c ) 8 , d ) 15 , e ) 12 | d | multiply(12, 480) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the numbers is 480 , then the other is : | "other number = ( 12 x 600 ) / 480 = 15 . answer : d" | a = 12 * 480
|
a ) 9263 , b ) 9244 , c ) 9636 , d ) 9623 , e ) 9637 | e | multiply(add(divide(32, 113), 85), 113) | find the value of ( 85 + 32 / 113 ) × 113 | "= ( 85 + 32 / 113 ) × 113 = ( 9605 + 32 ) / 113 × 113 = 9637 / 113 × 113 = 9637 answer is e ." | a = 32 / 113
b = a + 85
c = b * 113
|
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 24 | e | multiply(subtract(subtract(subtract(19, const_4), const_4), 1), const_3) | you collect baseball cards . suppose you start out with 19 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the end ? | "solution start with 19 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 19 + 1 which is 10 , so you ' re left with 19 - 10 = 9 . peter takes 1 baseball card from you : 9 - 1 = 8 baseball cards . paul triples the number of baseball cards you have : 8 ã — 3 = 24 baseball cards . so you have 24 at the end . correct answer : e" | a = 19 - 4
b = a - 4
c = b - 1
d = c * 3
|
a ) 23 / 30 , b ) 11 / 15 , c ) 47 / 60 , d ) 1 / 6 , e ) 2 / 15 | c | divide(add(12, 5), multiply(12, const_2)) | in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 12 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 5 ) = 1 / 15 therefore only video cassette recorder = 1 / 12 - 1 / 15 = 1 / 60 total = 1 / 5 + 1 / 60 + neither cable tv nor videocassette recorders 1 = 13 / 60 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = 1 - 13 / 60 = 47 / 60 hence c ." | a = 12 + 5
b = 12 * 2
c = a / b
|
a ) 23 , b ) 49 , c ) 22 , d ) 20 , e ) 21 | b | add(7, divide(multiply(7, subtract(20000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 20000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 20000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 98000 = x = 49 . answer : b" | a = 20000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
a ) t = 16 , b ) t = 8 √ 2 , c ) t = 8 , d ) t = 2 √ 2 , e ) ( √ 2 ) / 3 | d | sqrt(divide(multiply(4, 4), const_2)) | the two lines y = x and x = - 4 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length t of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 4 , - 4 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 4 × 4 ) = 8 ; so z = 8 . the next step requires us to find the length of a side of a cube that has a face area equal to 8 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 8 . since the area of a square is equal to s ² , where s is the length of one of its side , we can write and solve the equation s ² = 8 . clearly s = √ 8 = 2 √ 2 , oranswer choice ( d ) ." | a = 4 * 4
b = a / 2
c = math.sqrt(b)
|
a ) - 1 , b ) - 2 , c ) 1 , d ) 2 , e ) 3 | a | divide(subtract(2, sqrt(subtract(power(2, 3), multiply(5, 5)))), 3) | find the value of x from the below equation ? : 3 x ^ 2 + 5 x + 2 = 0 | "a = 3 , b = 5 , c = 2 x 1,2 = ( - 5 ± √ ( 52 - 4 × 3 × 2 ) ) / ( 2 × 3 ) = ( - 5 ± √ ( 25 - 24 ) ) / 6 = ( - 5 ± 1 ) / 6 x 1 = ( - 5 + 1 ) / 6 = - 4 / 6 = - 2 / 3 x 2 = ( - 5 - 1 ) / 6 = - 6 / 6 = - 1 a" | a = 2 ** 3
b = 5 * 5
c = a - b
d = math.sqrt(c)
e = 2 - d
f = e / 3
|
a ) 299 , b ) 278 , c ) 322 , d ) 231 , e ) 121 | a | multiply(23, 13) | the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 12 and 13 . the larger of the two numbers is : | "clearly , the numbers are ( 23 * 12 ) and ( 23 * 13 ) . larger number = ( 23 * 13 ) = 299 . answer : a" | a = 23 * 13
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['a ) a ) 2 : 3', 'b ) b ) 3 : 4', 'c ) c ) 4 : 3', 'd ) d ) 3 : 2', 'e ) e ) 2 : 1'] | b | divide(const_3, 4) | two sides of a triangle have lengths x and t and meet at a right angle . if the perimeter of the triangle is 4 x , what is the ratio of x to t ? | ahhhh . . . . my bad . thank you , you ' re correct . 8 x = 6 t - - > x / t = 3 / 4 . actually b is the right answer . | a = 3 / 4
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