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 ) 2 , b ) 48 / 15 , c ) 7 / 9 , d ) 10 , e ) 24 / 7 | b | divide(const_1, add(add(divide(const_1, 16), divide(const_1, add(const_4, const_2))), divide(const_1, multiply(const_2, add(const_4, const_2))))) | a , b , c can complete a piece of work in 16 , 6,12 days . working together , they complete the same work in how many days ? | a + b + c 1 day work = 1 / 16 + 1 / 6 + 1 / 12 = 15 / 48 a , b , c together will complete the job in 48 / 15 days answer is b | a = 1 / 16
b = 4 + 2
c = 1 / b
d = a + c
e = 4 + 2
f = 2 * e
g = 1 / f
h = d + g
i = 1 / h
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 2) | the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 100,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 100 ( 100,000 / 1,000 = 100 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 100 , where x represents the number of times the population doubles ... | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 2
|
a ) 5 % , b ) 10 % , c ) 20 % , d ) 25 % , e ) 50 % | d | multiply(divide(subtract(divide(const_100, divide(subtract(const_100, 10), const_100)), divide(const_100, divide(add(const_100, 20), const_100))), divide(const_100, divide(subtract(const_100, 10), const_100))), const_100) | certain stocks in january were 10 % less than they were in february and 20 % greater than they were in march . what was the percentage decrease in the stocks from february to march ? | let j , f , m be the values of the stock in jan , feb and march . thus , per the question , j = 0.9 f = 1.2 m - - - - > m = 0.75 f thus the % decrease from f to m = ( f - m ) / ( f ) * 100 = ( f - 0.75 f ) / f * 100 = 0.25 * 100 = 25 % , d is the correct answer . | a = 100 - 10
b = a / 100
c = 100 / b
d = 100 + 20
e = d / 100
f = 100 / e
g = c - f
h = 100 - 10
i = h / 100
j = 100 / i
k = g / j
l = k * 100
|
a ) 23 , b ) 22 , c ) 21 , d ) 20 , e ) 19.2 | e | divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 15)) | to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 15 hours a day ? | "man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 15 = 300 therefore number of days = 5760 / 300 = 19.2 days ans = e" | a = 30 * 12
b = a * 8
c = b * 2
d = 20 * 15
e = c / d
|
a ) 70 minutes , b ) 72 minutes , c ) 50 minutes , d ) 66 minutes , e ) 67 minutes | c | divide(4000, 80) | a scuba diver descends at a rate of 80 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 4000 feet below the sea level . . how long will he take to reach the ship ? | "time taken to reach = 4000 / 80 = 50 minutes answer : c" | a = 4000 / 80
|
a ) 38 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 . | c | subtract(multiply(sqrt(divide(784, 4)), 4), sqrt(divide(784, 4))) | the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 784 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 784 x ^ 2 = 196 x = 14 length = 4 * 14 = 56 difference = 56 - 14 = 42 c is the answer" | a = 784 / 4
b = math.sqrt(a)
c = b * 4
d = 784 / 4
e = math.sqrt(d)
f = c - e
|
a ) 8.5 , b ) 10.5 , c ) 12.5 , d ) 14.5 , e ) 16.5 | c | multiply(add(1.5, const_1), 5) | two water pumps , working simultaneously at their respective constant rates , took exactly 5 hours to fill a certain swimming pool . if the constant rate of one pump was 1.5 times the constant rate of the other , how many hours would it have taken the slower pump to fill the pool if it had worked alone at its constant ... | let x be the rate of the slower pump . then 1.5 x is the rate of the faster pump . both pumps together can fill 1 / 5 of the pool each hour . 2.5 x = 1 / 5 x = 1 / 12.5 = 2 / 25 the slower pump could fill the pool in 25 / 2 = 12.5 hours . the answer is c . | a = 1 + 5
b = a * 5
|
a ) 1 / 8 , b ) 1 / 12 , c ) 1 / 14 , d ) 1 / 18 , e ) 1 / 28 | c | divide(const_4, divide(factorial(8), multiply(factorial(2), factorial(subtract(8, 2))))) | a certain box has 8 cards and each card has one of the integers from 1 to 8 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on t... | "the average of the numbers is 4.5 the total number of ways to choose 2 cards from 8 cards is 8 c 2 = 28 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } the probability is 2 / 28 = 1 / 14 the answer is c ." | a = math.factorial(8)
b = math.factorial(2)
c = 8 - 2
d = math.factorial(c)
e = b * d
f = a / e
g = 4 / f
|
a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 1 / 2 | c | divide(4, 3) | a positive number x is multiplied by 4 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | sq rt ( 4 x / 3 ) = x = > 4 x / 3 = x ^ 2 = > x = 4 / 3 ans - c | a = 4 / 3
|
a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | d | divide(subtract(multiply(100, divide(60, const_100)), multiply(100, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 100 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 100 total = 100 + 100 = 200 answer ( d )" | a = 60 / 100
b = 100 * a
c = 55 / 100
d = 100 * c
e = b - d
f = 55 / 100
g = 50 / 100
h = f - g
i = e / h
|
a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 250 m | c | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 270) | a 270 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | "relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 270 / 9 = 500 / 9 x = 230 ans is 230 m answer : c" | a = 120 + 80
b = a * const_0_2778
c = 9 * b
d = c - 270
|
a ) 320 , b ) 375.2 , c ) 400 , d ) 408 , e ) 440 | d | multiply(multiply(subtract(const_1, divide(20, const_100)), 51), 10) | car z travels 51 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per ho... | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 51 miles / gallon and at 60 miles / hour , that distance decreases by 20 %... | a = 20 / 100
b = 1 - a
c = b * 51
d = c * 10
|
a ) 4 % , b ) 3 6 / 7 % , c ) 2 6 / 7 % , d ) 3.47 % , e ) 6 % | d | multiply(divide(divide(subtract(2000, 1800), 1800), 5), const_100) | at what rate percent on simple interest will rs . 1800 amount to rs . 2000 in 5 years ? | "explanation : 250 = ( 1800 x 5 xr ) / 100 r = 3.47 % answer : option d" | a = 2000 - 1800
b = a / 1800
c = b / 5
d = c * 100
|
a ) 0.247 , b ) 2.47 , c ) 24.7 , d ) 0.0247 , e ) 0.00247 | b | divide(multiply(0.02, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 0.02 percent of 12,356 ? | since , percent = 1 / 100 , what = something ( s ) , and is : = . we can write the question as s = 0.02 ( 1 / 100 ) 12,356 . the answer is 2.47 . hence , the correct answer is b . | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 0 * 2
p = o / 100
|
a ) 6 , b ) 8 , c ) 4 , d ) 3 , e ) 2 | b | add(subtract(add(21, 27), subtract(45, 5)), subtract(27, 21)) | of 45 applicants for a job , 21 had at least 4 years ' experience , 27 had degrees , and 5 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "45 - 5 = 40 40 - 21 - 27 = - 8 then 8 are in the intersection between 4 years experience and degree . answer : b" | a = 21 + 27
b = 45 - 5
c = a - b
d = 27 - 21
e = c + d
|
a ) 176 , b ) 186 , c ) 184 , d ) 174 , e ) 164 | b | multiply(divide(12,21, 48,87), const_100) | 12,21 , 48,87 , __ | "21 = 12 * 2 - 3 48 = 21 * 2 + 6 87 = 48 * 2 - 9 so next number is 87 * 2 + 12 = 186 answer : b" | a = 12 / 21
b = a * 100
|
a ) 7 kmph , b ) 5 kmph , c ) 2 kmph , d ) 8 kmph , e ) 1 kmph | a | divide(subtract(divide(90, 3), divide(5, 3)), const_2) | a man rows his boat 90 km downstream and 5 ` km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 51 / ( 3 ) = 17 kmph the speed of the stream = ( 30 - 17 ) / 2 = 7 kmph answer : a" | a = 90 / 3
b = 5 / 3
c = a - b
d = c / 2
|
a ) 29 % , b ) 31 % , c ) 33 % , d ) 35 % , e ) 37 % | a | multiply(const_100, divide(add(divide(45, const_100), multiply(4, divide(25, const_100))), add(const_1, 4))) | because he β s taxed by his home planet , mork pays a tax rate of 45 % on his income , while mindy pays a rate of only 25 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ? | "let x be mork ' s income , then mindy ' s income is 4 x . the total tax paid is 0.45 x + 1.0 x = 1.45 x 1.45 x / 5 x = 0.29 the answer is a ." | a = 45 / 100
b = 25 / 100
c = 4 * b
d = a + c
e = 1 + 4
f = d / e
g = 100 * f
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | d | sqrt(divide(704, divide(1100, const_100))) | reema took a loan of rs 1100 with simple interest for as many years as the rate of interest . if she paid rs . 704 as interest at the end of the loan period , what was the rate of interest . | "explanation : let rate = r % then time = r years . = > 1100 β r β r / 100 = 704 = > r 2 = 64 = > r = 8 % option d" | a = 1100 / 100
b = 704 / a
c = math.sqrt(b)
|
a ) 30 , b ) 52 , c ) 66 , d ) 68 , e ) 84 | a | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | how many positive integers less than 300 can be formed using the numbers 1 , 2 , 3 a for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 number of possibilities = 3 * 3 * 3 = 27 . then , just add up the number of 1 digits numbers = 3 , so total is 27 + 3 = 30 . answer : a" | a = 4 + 1
b = a - 1
c = math.factorial(b)
d = math.factorial(1)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 65 % , b ) 70 % , c ) 75 % , d ) 80 % , e ) 85 % | c | multiply(const_100, divide(subtract(subtract(const_100, 52), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | in a company , 52 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | "the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 48 % of all employees are women , so non - union women are 48 % - 18 % = 30 % 40 % of all employees are non - union . the percent of non - union employees who are women is 30 % ... | a = 100 - 52
b = 70 / 100
c = 60 * b
d = 60 - c
e = a - d
f = 100 - 60
g = e / f
h = 100 * g
|
a ) 15 , b ) 17 , c ) 15 , d ) 16 , e ) 18 | e | divide(add(200, 200), multiply(80, const_0_2778)) | how long does a lorry 200 m long traveling at 80 kmph takes to cross a bridge of 200 m in length ? | d = 200 + 200 = 400 m s = 80 * 5 / 18 = 200 / 9 t = 400 * 9 / 200 = 18 sec answer : e | a = 200 + 200
b = 80 * const_0_2778
c = a / b
|
a ) 1925 , b ) 600 , c ) 925 , d ) 1325 , e ) 900 | c | divide(add(160, 25), divide(20, const_100)) | pradeep has to obtain 20 % of the total marks to pass . he got 160 marks and failed by 25 marks . the maximum marks are | "explanation : let their maximum marks be x . then , 20 % of x = 160 + 25 = > 20 / 100 x = 185 x = ( 18500 / 20 ) x = 925 . answer : c" | a = 160 + 25
b = 20 / 100
c = a / b
|
a ) 9.9 , b ) 8.0 , c ) 12.5 , d ) 12.2 , e ) 12.1 | a | divide(multiply(14, 1480), add(1480, 460)) | 1480 men have provisions for 14 days . if 460 more men join them , for how many days will the provisions last now ? | "1480 * 13 = 1940 * x x = 9.9 answer : a" | a = 14 * 1480
b = 1480 + 460
c = a / b
|
a ) rs . 2.04 , b ) rs . 2.09 , c ) rs . 2.06 , d ) rs . 2.22 , e ) rs . 2.08 | a | subtract(multiply(5000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(5000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | what is the difference between the c . i . on rs . 5000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 5000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 5000 * 26 / 25 * 51 / 50 = rs . 5304 c . i . when interest is compounded half - yearly = [ 5000 * ( 1 + 2 / 100 ) 2 ] = ( 5000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 5306.04 difference = ( 5306.04 - 5304 ) = rs . 2.04... | a = 2 / 100
b = 1 + a
c = 2 / 100
d = 1 + c
e = b * d
f = 2 / 100
g = 1 + f
h = e * g
i = 5000 * h
j = 2 / 100
k = 1 + j
l = 4 / 100
m = 1 + l
n = k * m
o = 5000 * n
p = i - o
|
a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) none | c | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(di... | the sum of three consecutive even numbers multiples of 3 is 72 . what is the largest number ? | "solution let the number be x and x + 2 . then , ( x + 2 ) 2 - x 2 = 84 β 4 x + 4 = 84 β 4 x = 80 β x = 20 . β΄ required sum = x + ( x + 2 ) = 2 x + 2 = 42 . answer c" | a = 3 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 3 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 3 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 3 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 144 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 15 km | a | divide(multiply(multiply(subtract(10, 2), add(10, 2)), 30), add(subtract(10, 2), add(10, 2))) | a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 30 hour to row to a place and come back , how far is the place ? | "speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 30 2 x + 3 x = 720 x = 144 km answer is a" | a = 10 - 2
b = 10 + 2
c = a * b
d = c * 30
e = 10 - 2
f = 10 + 2
g = e + f
h = d / g
|
a ) q = 26.7 , b ) q = 30.0 , c ) q = 40.0 , d ) q = 53.3 , e ) q = 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his a... | "veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is ... | a = 640 / 2
b = a / 80
c = b * 2
d = 640 / 2
e = d / 80
f = c + e
g = 640 / 2
h = g / 80
i = f + h
j = 640 / i
|
a ) 156 , b ) 220 , c ) 130 , d ) 240 , e ) none | b | add(100, divide(add(multiply(100, 5), 100), 5)) | 6 friends went to a hotel and decided to pay the bill amount equally . but 5 of them could pay rs . 100 each as a result 6 th has to pay rs . 100 extra than his share . find the amount paid by him . | "explanation : average amount paid by 5 persons = rs . 100 increase in average due to rs . 120 paid extra by the 6 th men = rs . 100 / 5 = rs . 20 therefore , average expenditure of 6 friends = rs . 100 + rs . 20 = rs . 120 therefore , amount paid by the 6 th men = rs . 120 + rs . 100 = rs . 220 correct option : b" | a = 100 * 5
b = a + 100
c = b / 5
d = 100 + c
|
a ) 12 , b ) 24 , c ) 84 , d ) 48 , e ) 42 | c | multiply(sqrt(divide(multiply(84, 21), 4)), 4) | the h . c . f and l . c . m of two numbers are 84 and 21 respectively . if the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is : | "let the numbers be x and 4 x . then , x * 4 x = 84 * 21 x 2 = ( 84 * 21 ) / 4 = x = 21 . hence , larger number = 4 x = 84 . answer : c" | a = 84 * 21
b = a / 4
c = math.sqrt(b)
d = c * 4
|
a ) 15060000 , b ) 0.001506 , c ) 0.01506 , d ) 1.506 e - 07 , e ) none of these | d | multiply(divide(15.06, 0.00000001), const_100) | 15.06 * 0.00000001 = ? | "explanation : clearly after decimal 10 digits should be there . option d" | a = 15 / 6
b = a * 100
|
a ) 232 , b ) 242 , c ) 252 , d ) 262 , e ) 234 | e | divide(28, divide(450, 28)) | evaluate 28 % of 450 + 45 % of 240 | "explanation : = ( 28 / 100 ) * 450 + ( 45 / 100 ) * 240 = 126 + 108 = 234 answer : option e" | a = 450 / 28
b = 28 / a
|
a ) 27 , b ) 16 , c ) 29.7 , d ) 28 , e ) 18 | d | divide(subtract(7, 12), 4) | a straight line in the xy - plane has y - intercept of 12 and a slope of 4 / 7 . given the x - coordinate of the point is 49 , find it ' s y - coordinate . | "eq of line = y = mx + c m = 4 / 7 c = 12 x = 49 substitute the givens : y = ( 4 / 7 * 49 ) + 12 , y = 16 + 12 = 28 correct option is d" | a = 7 - 12
b = a / 4
|
a ) 120 , b ) 100 , c ) 75 , d ) 90 , e ) none of these | a | divide(36, multiply(divide(50, const_100), divide(3, 5))) | if 50 % of 3 / 5 of a number is 36 , then the number is ? | let the number be x . then 50 % of 3 / 5 of x = 36 50 / 100 * 3 / 5 * x = 36 x = ( 36 * 10 / 3 ) = 120 required number = 120 . correct option : a | a = 50 / 100
b = 3 / 5
c = a * b
d = 36 / c
|
a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9 | c | divide(const_5, 6) | a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability t that both selected dogs are not littermates ? | "we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one ... | a = 5 / 6
|
a ) rs 222 , b ) rs 216 , c ) rs 220 , d ) rs 210 , e ) rs 217 | a | add(add(180, divide(500, 100)), multiply(divide(20, const_100), add(180, divide(500, 100)))) | the manufacturing cost of a shoe is rs . 180 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | explanation : total cost of a watch = 180 + ( 500 / 100 ) = 185 . gain = 20 % = > sp = 1.2 cp = 1.2 x 185 = 222 answer : a | a = 500 / 100
b = 180 + a
c = 20 / 100
d = 500 / 100
e = 180 + d
f = c * e
g = b + f
|
a ) 72 % , b ) 70 % , c ) 52 % , d ) 64 % , e ) 28 % | d | subtract(80, multiply(divide(80, const_100), 20)) | a shirt goes on sale for 80 % of its original price . one week later , the sale price is marked down 20 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 80 then it is marked down by 20 % = 80 - 16 = 64 . hence it is 64 % od the original price . hence answer is d ." | a = 80 / 100
b = a * 20
c = 80 - b
|
a ) 62 % , b ) 57 % , c ) 52 % , d ) 22 % , e ) 48 % | e | subtract(const_100, add(multiply(30, divide(70, const_100)), multiply(divide(30, const_100), 70))) | in a factory , there are 30 % technicians and 70 % non - technicians . if the 30 % of the technicians and 70 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ? | "total = 100 t = 30 nt = 70 30 * ( 70 / 100 ) = 21 70 * ( 30 / 100 ) = 21 21 + 21 = 42 = > 100 - 42 = 58 % answer : e" | a = 70 / 100
b = 30 * a
c = 30 / 100
d = c * 70
e = b + d
f = 100 - e
|
a ) 24 , b ) 23 , c ) 24 , d ) 20.57 , e ) 22 | d | divide(multiply(multiply(multiply(30, 12), 8), 2), multiply(20, 14)) | to asphalt 1 km road , 30 men spent 12 days working 8 hours per day . how many days , 20 men will spend to asphalt a road of 2 km working 14 hours a day ? | man - hours required to asphalt 1 km road = 30 * 12 * 8 = 2880 man - hours required to asphalt 2 km road = 2880 * 2 = 5760 man - hours available per day = 20 * 14 = 280 therefore number of days = 5760 / 280 = 20.57 days ans = d | a = 30 * 12
b = a * 8
c = b * 2
d = 20 * 14
e = c / d
|
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 29 | a | multiply(3, 5) | if a * b = 2 a β 3 b + ab , then 3 * 5 + 5 * 3 is equal to : | "solution 3 Γ 5 + 5 Γ 3 = ( 2 Γ 3 - 3 Γ 5 + 3 Γ 5 ) + 2 ( 2 Γ 5 - 3 Γ 3 + 5 Γ 3 ) = ( 6 + 10 - 9 + 15 ) = 22 . answer a" | a = 3 * 5
|
a ) 36.7 Β° c , b ) 38.6 Β° c , c ) 39.8 Β° c , d ) 41.9 Β° c , e ) 51.9 Β° c | d | add(multiply(6, 6), add(subtract(multiply(40.3, 4), multiply(39.1, 4)), divide(const_0_33, const_3))) | average temperature of first 4 days of a week is 6 Β° c and that of the last 4 days is 40.3 Β° c . if the average temperature of the week be 39.1 Β° c , the temperature on 4 th day is ? | let temperature on 4 th day be x Β° c therefore , 4 x 38.6 + 4 x 40.3 - x = 7 x 39.1 = > x = 41.9 therefore , temperature on 4 th day = 41.9 Β° c . answer : d | a = 6 * 6
b = 40 * 3
c = 39 * 1
d = b - c
e = const_0_33 / 3
f = d + e
g = a + f
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a ) 1000 , b ) 1030 , c ) 1100 , d ) none of these , e ) 1101 | a | multiply(multiply(4600, divide(10, 4600)), 4) | what annual installment will discharge a debt of rs . 4600 due in 4 years at 10 % simple interest ? | "let the 1 st installment = 100 / - it is 3 yrs before due iind is 2 yr before due iii 3 rd is 1 yr before due ivth is 0 yrs before due on 1 st installment interest will be paid for 3 yrs 2 nd installment interest will be paid for 2 yrs 3 rd installment interest will be paid for 1 yr 4 th installment interest will be p... | a = 10 / 4600
b = 4600 * a
c = b * 4
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a ) $ 21,700 , b ) $ 19,500 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | a | multiply(const_2, const_10) | a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 3,000 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price... | price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 1500 net price of the car in 02 = 18700 +... | a = 2 * 10
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a ) 1370 , b ) 1530 , c ) 1790 , d ) 1950 , e ) 2110 | b | divide(multiply(multiply(18, subtract(18, const_1)), 10), const_2) | there are 18 teams in the hockey league , and each team faces all the other teams 10 times each . how many games are played in the season ? | "the number of ways to choose two teams is 18 c 2 = 18 * 17 / 2 = 153 the total number of games in the season is 10 * 153 = 1530 . the answer is b ." | a = 18 - 1
b = 18 * a
c = b * 10
d = c / 2
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a ) 5 , b ) 4 , c ) 6 , d ) 7 , e ) 3 | a | add(divide(subtract(50, 1), 9), const_1) | how many multiples of 9 are there between 1 and 50 , exclusive ? | "5 multiples of 9 between 1 and 50 exclusive . from 9 * 1 upto 9 * 5 , ( 1 , 2,3 , 4,5 ) . hence , 5 multiples ! correct option is a" | a = 50 - 1
b = a / 9
c = b + 1
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a ) rs . 500 , b ) rs . 840 , c ) rs . 650 , d ) rs . 720 , e ) none | b | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, const_100)), add(100, multiply(100, divide(20, const_100))))), divide(7, const_100)) | a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 7 % dividend at the end of the year , then how much does he get ? | solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 7 / 100 x 12000 ) = rs . 840 . answer b | a = 10 * 1000
b = 4 * 1000
c = a + b
d = 4 * 100
e = c + d
f = 20 / 100
g = 100 * f
h = 100 + g
i = e / h
j = 100 * i
k = 7 / 100
l = j * k
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a ) 11 , b ) 77 , c ) 16 , d ) 10 , e ) 98 | a | add(9, divide(130, add(25, 30))) | the distance between delhi and mathura is 130 kms . a starts from delhi with a speed of 25 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 30 kmph at 9 p . m . from delhi . when will they meet ? | "d = 130 β 25 = 105 rs = 30 + 25 = 55 t = 105 / 55 = 2 hours 8 a . m . + 2 = 11 a . m . . answer : a" | a = 25 + 30
b = 130 / a
c = 9 + b
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a ) 0.5 , b ) 0.05 , c ) 0.005 , d ) 0.0005 , e ) 5 | c | divide(divide(1, 2), const_100) | how is 1 / 2 % expressed as a decimal fraction ? | as 1 / 2 = 0.5 and its percent value will be 0.5 / 100 = 0.005 answer : c | a = 1 / 2
b = a / 100
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a ) 35 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | a | divide(multiply(250, 28), subtract(250, 50)) | a hostel had provisions for 250 men for 28 days . if 50 men left the hostel , how long will the food last at the same rate ? | a hostel had provisions for 250 men for 28 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 28 250 Γ 28 = 200 x 5 Γ 28 = 4 ... | a = 250 * 28
b = 250 - 50
c = a / b
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a ) 2 / 5 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 4 , e ) 3 / 2 | c | multiply(divide(subtract(const_100, 20), add(const_100, 25)), divide(5, 4)) | the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 25 percent of a , and m equals b decreased by 20 percent of b , what is the value of m / x ? | "x equals a increased by 25 percent of a = > x = a + 25 % a = 1.25 a m equals b decreased by 20 percent of b = > m = b - 20 % b = 0.8 b m / x = 0.8 b / 1.25 a putting the value of b / a = 5 / 4 we get m / x = 4 / 5 answer : c" | a = 100 - 20
b = 100 + 25
c = a / b
d = 5 / 4
e = c * d
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a ) s . 575 , b ) s . 595 , c ) s . 590 , d ) s . 570 , e ) s . 585 | d | subtract(multiply(subtract(540, 480), 6), subtract(540, 480)) | if rs . 480 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | "80 = ( 480 * 4 * r ) / 100 r = 3.125 % i = ( 480 * 6 * 3.125 ) / 100 = 90 480 + 90 = 570 answer : d" | a = 540 - 480
b = a * 6
c = 540 - 480
d = b - c
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a ) 52 , b ) 54 , c ) 55 , d ) 65 , e ) 48 | b | divide(multiply(180, 120), multiply(25, 16)) | rectangular tile each of size 25 cm by 16 cm must be laid horizontally on a rectangular floor of size 180 cm by 120 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 25 * 16 = 400 area of floor = 180 * 120 = 21600 no of tiles = 21600 / 400 = 54 so , the no of tile = 54 answer : b" | a = 180 * 120
b = 25 * 16
c = a / b
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a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 12 % | e | multiply(divide(2.75, add(20, 2.75)), const_100) | a house wife saved $ 2.75 in buying an item on sale . if she spent $ 20 for the item , approximately how much percent she saved in the transaction ? | "actual price = 20 + 2.75 = $ 22.75 saving = 2.75 / 22.75 * 100 = 12 % approximately answer is e" | a = 20 + 2
b = 2 / 75
c = b * 100
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a ) 60 , b ) 75 , c ) 85 , d ) 95 , e ) 105 | a | multiply(divide(40, subtract(3, 1)), 3) | the ratio of buses to cars on river road is 1 to 3 . if there are 40 fewer buses than cars on river road , how many cars are on river road ? | "b / c = 1 / 3 c - b = 40 . . . . . . . . . > b = c - 40 ( c - 40 ) / c = 1 / 3 testing answers . clearly eliminate bcde put c = 60 . . . . . . . . . > ( 60 - 40 ) / 60 = 20 / 60 = 1 / 3 answer : a" | a = 3 - 1
b = 40 / a
c = b * 3
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a ) 28 , b ) 30 , c ) 36 , d ) 42 , e ) 45 | b | add(multiply(multiply(4, 6), const_100), multiply(5, 6)) | three numbers are in the ratio 4 : 5 : 6 and their average is 25 . the largest number is : | "explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 25 15 x = 75 x = 5 largest number = 6 x = 30 . answer b" | a = 4 * 6
b = a * 100
c = 5 * 6
d = b + c
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a ) 10 , b ) 8 , c ) 12 , d ) 24 , e ) 16 | d | divide(600, subtract(26, const_1)) | in a garden , 26 trees are planted at equal distances along a yard 600 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "26 trees have 25 gaps between them . length of each gap = 600 / 25 = 24 i . e . , distance between two consecutive trees = 24 answer is d ." | a = 26 - 1
b = 600 / a
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a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | floor(divide(const_2, multiply(add(divide(const_1, 6), divide(const_1, 7)), add(1, divide(36, const_60))))) | on a wedding catering service , an experienced chef can prepare a service for a wedding in 6 hours while an novice chef would finish the preparations in 7 hours . if the catering service employs the same number of novice and experienced chefs , then how many chefs would it take to prepare a wedding service in 1 hour an... | experienced chefs work = 1 wedding / 6 hours novice chefs work = 1 wedding / 7 hours since we do n ' t know the number of experienced or novice chefs but know that there is an equal number each , let the number of chefs for each group equalx 1 hr and 36 mins = 8 / 5 an hour x / 6 + x / 7 = 1 wedding / ( 8 / 5 ) x / 6 +... | a = 1 / 6
b = 1 / 7
c = a + b
d = 36 / const_60
e = 1 + d
f = c * e
g = 2 / f
h = math.floor(g)
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a ) 40 % , b ) 48 % , c ) 56 % , d ) 64 % , e ) 72 % | c | divide(add(20, 60), multiply(multiply(const_5, const_5), const_4)) | there is a 20 % chance sandy will visit china this year , while there is a 60 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | "p ( china and not malaysia ) = 0.2 * 0.4 = 0.08 p ( malaysia and not china ) = 0.6 * 0.8 = 0.48 total probability = 0.08 + 0.48 = 0.56 = 56 % the answer is c ." | a = 20 + 60
b = 5 * 5
c = b * 4
d = a / c
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a ) 10 , b ) 8 , c ) 7.5 , d ) 6 , e ) 3.5 | a | multiply(5, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quality of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x and 3 x 4 x / ( 3 x + 5 ) = 4 / 5 20 x = 4 ( 3 x + 5 ) x = 2.5 quantity of alcohol = 4 * 2.5 = 10 liters . answer is a" | a = 5 * 1
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a ) 220 meter , b ) 225 meter , c ) 230 meter , d ) 350 meter , e ) none of these | d | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 150) | a 150 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 150 / 9 = 500 / 9 = > x + 150 = 500 = > x = 350 so the length of the train is 350 meters opt... | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 150
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a ) 30.5 , b ) 26 , c ) 30 , d ) 18 , e ) 11 | a | divide(add(add(18, 22), multiply(7, 3)), const_2) | the average age of 7 men increases by 3 years when two women are included in place of two men of ages 18 and 22 years . find the average age of the women ? | "explanation : 18 + 22 + 7 * 3 = 61 / 2 = 30.5 answer : a" | a = 18 + 22
b = 7 * 3
c = a + b
d = c / 2
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a ) 600 , b ) 720 , c ) 1000 , d ) 1200 , e ) 1440 | c | divide(100, divide(multiply(const_2, subtract(105, 100)), 100)) | julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 100 and $ 105 from the simple interest account and the compound interest account respectively . if the interest rates for both account... | "$ 100 for 2 years = $ 50 per year . extra $ 5 yearned with the compound interest is the percent yearned on percent . so , $ 5 is yearned on $ 50 , which means that the interest = 10 % . this on the other hand means that half of the savings = 50 * 10 = $ 500 . twice of that = $ 1,000 . answer : c ." | a = 105 - 100
b = 2 * a
c = b / 100
d = 100 / c
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a ) 62.6 km , b ) 26.7 km , c ) 22.8 km , d ) 19.5 km , e ) none | b | multiply(4, 10) | a walks at 4 kmph and 4 hours after his start , b cycles after him at 10 kmph . how far from the start does b catch up with a ? | "sol . suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 4 hours . β΄ x / 4 - x / 10 = 4 or x = 26.7 km . answer b" | a = 4 * 10
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a ) 120 , b ) 600 , c ) 1000 , d ) 360 , e ) 240 | b | divide(subtract(multiply(divide(6, const_100), 2400), multiply(2400, divide(5, const_100))), subtract(divide(10, const_100), divide(6, const_100))) | barbata invests $ 2400 in the national bank at 5 % . how much additional money must she invest at 10 % so that the total annual income will be equal to 6 % of her entire investment ? | "let the additional invested amount for 10 % interest be x ; equation will be ; 2400 + 0.05 * 2400 + x + 0.10 x = 2400 + x + 0.06 ( 2400 + x ) 0.05 * 2400 + 0.10 x = 0.06 x + 0.06 * 2400 0.04 x = 2400 ( 0.06 - 0.05 ) x = 2400 * 0.01 / 0.04 = 600 ans : b" | a = 6 / 100
b = a * 2400
c = 5 / 100
d = 2400 * c
e = b - d
f = 10 / 100
g = 6 / 100
h = f - g
i = e / h
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a ) 100 , b ) 110 , c ) 150 , d ) 155 , e ) 160 | c | add(100, multiply(100, divide(50, const_100))) | 100 is increased by 50 % . find the final number . | "final number = initial number + 50 % ( original number ) = 100 + 50 % ( 100 ) = 100 + 50 = 150 . answer c" | a = 50 / 100
b = 100 * a
c = 100 + b
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a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 5 hrs , e ) 6 hrs | d | divide(50, add(5, 5)) | two cyclist start from the same places in opposite directions . one is going towards north at 5 kmph and the other is going towards south 5 kmph . what time will they take to be 50 km apart ? | "to be ( 5 + 5 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 10 * 50 = 5 hrs answer is d" | a = 5 + 5
b = 50 / a
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a ) 9 , b ) 10 , c ) 15 , d ) 12 , e ) 14 | a | divide(divide(multiply(add(2, 6), add(divide(subtract(6, 2), 2), const_1)), const_2), add(divide(subtract(6, 2), 2), const_1)) | what is the average ( arithmetic mean ) of the numbers 2 , 4 , 6 , 8 , 10 , 12 , 14 and 16 ? | "avg = sum of observations / number of observations avg = ( 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 ) / 8 = 9 answer is a" | a = 2 + 6
b = 6 - 2
c = b / 2
d = c + 1
e = a * d
f = e / 2
g = 6 - 2
h = g / 2
i = h + 1
j = f / i
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a ) 88 % , b ) 75 % , c ) 67 % , d ) 63 % , e ) 50 % | b | multiply(subtract(1, power(divide(divide(10, const_2), 10), const_2)), const_100) | a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun two times , approximately what is the probability that the product of the... | the only way to have an odd product is if both integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 = 1 / 4 p ( even product ) = 1 - 1 / 4 = 3 / 4 = 75 % the answer is b . | a = 10 / 2
b = a / 10
c = b ** 2
d = 1 - c
e = d * 100
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a ) 1 , b ) 3 , c ) 4 , d ) 7 , e ) 8 | a | subtract(2, 2) | when the number 2 y 31129 is exactly divisible by 11 , then what can be the smallest whole number in place of y ? | "the given number = 2 y 31129 sum of the odd places = 9 + 1 + 3 + 2 = 15 sum of the even places = 2 + 1 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 15 - ( 3 + y ) = divisible by 11 12 οΏ½ y = divisible by 11 . y must be 1 , to make given number divisible by 11 . a" | a = 2 - 2
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a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 2 : 5 , e ) 68 : 85 | e | divide(multiply(17, const_4), add(multiply(17, const_4), 17)) | a pet store holds cats and dogs . if the difference between the number of cats and the number of dogs is 17 . what could be the ratio of cats to dogs in the pet store ? | "say theratioof cats to dogs is a / b . then thenumberof cats would be ax and thenumberof dogs bx , for some positive integer x . we are told that ax - bx = 17 - - > x ( a - b ) = 17 . since 17 is a prime number it could be broken into the product of two positive multiples only in one way : x ( a - b ) = 1 * 17 . the a... | a = 17 * 4
b = 17 * 4
c = b + 17
d = a / c
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a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(15, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,15 = 15 / 100 x = 80 * 15 / 100 x = 12 12 / 10 = 1,2 = 120 % , answer c" | a = 20 / 100
b = 1 - a
c = 15 * b
d = c / 10
e = d * 100
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a ) 3600 , b ) 1800 , c ) 18 , d ) 18000 , e ) 1.8 | a | divide(multiply(36, 15), divide(multiply(3, 5), const_100)) | a hall 36 m long and 15 m broad is to be paved with stones , each measuring 3 dm by 5 dm . the number of stones required is : | "area of the hall = 3600 * 1500 area of each stone = ( 30 * 50 ) therefore , number of stones = ( 3600 * 1500 / 30 * 50 ) = 3600 answer : a" | a = 36 * 15
b = 3 * 5
c = b / 100
d = a / c
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a ) 333 , b ) 200 , c ) 230 , d ) 276 , e ) 1999 | c | subtract(divide(divide(6095, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 6095 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 6095 [ 4 l - 40 ] = 6095 / 26.50 = 230 4 l = 270 l = 270 / 4 = 67.5 meters . answer : c" | a = 6095 / 26
b = a / 2
c = 2 * 20
d = b - c
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a ) 3 , b ) 6 , c ) 48 , d ) 12 , e ) 14 | c | multiply(divide(28, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3) | jill has 28 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ? | let the number of each size of jar = wthen 1 / 4 w + 1 / 2 w + w = 28 1 3 / 4 w = 28 w = 16 the total number of jars = 3 w = 48 answer : c | a = 1 / 2
b = const_0_25 + a
c = 1 + b
d = 28 / c
e = d * 3
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a ) 330 , b ) 300 , c ) 210 , d ) 250 , e ) 350 | c | divide(multiply(30, 2310), 330) | the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 330 the other is | "the other number = l . c . m * h . c . f / given number = 2310 * 30 / 330 = 210 answer is c ." | a = 30 * 2310
b = a / 330
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a ) 110 , b ) 111 , c ) 115 , d ) 123 , e ) none | c | add(50, divide(add(multiply(60, 10), 50), 10)) | 11 friends went to a hotel and decided to pay the bill amount equally . but 10 of them could pay rs . 60 each as a result 11 th has to pay rs . 50 extra than his share . find the amount paid by him . | explanation : average amount paid by 10 persons = rs . 60 increase in average due to rs . 50 paid extra by the 11 th men = rs . 50 / 10 = rs . 5 therefore , average expenditure of 11 friends = rs . 60 + rs . 5 = rs . 65 therefore , amount paid by the 11 th men = rs . 65 + rs . 50 = rs . 115 correct option : c | a = 60 * 10
b = a + 50
c = b / 10
d = 50 + c
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a ) 24 , b ) 36 , c ) 48 , d ) 70 , e ) 132 | d | floor(multiply(divide(85, add(5, const_1)), 5)) | at a certain zoo , the ratio of sea horse to penguins is 5 to 11 . if there are 85 more penguins than sea horses at the zoo , how many sea horses are there ? | 5 / 11 = x / x + 85 5 * 85 = 6 x x = 70 d . 70 | a = 5 + 1
b = 85 / a
c = b * 5
d = math.floor(c)
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a ) 12 hours , b ) 24 hours , c ) 36 hours , d ) 48 hours , e ) none | b | add(divide(105, add(9, 1.5)), divide(105, subtract(9, 1.5))) | speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph . a man rows to place at a distance of 105 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 7.5 kmph ; speed downstream = 10.5 kmph . β΄ total time taken = [ 105 / 7.5 + 105 / 10.5 ] hours = 24 hours . answer b" | a = 9 + 1
b = 105 / a
c = 9 - 1
d = 105 / c
e = b + d
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(divide(100, const_2), multiply(7, 7)) | what is the remainder when 7 ^ 700 is divided by 100 ? | "( 7 ^ 700 ) mod 100 = [ 7 ^ ( 2 * 350 ) ] mod 100 = [ 49 ^ 350 ] mod 100 = [ ( 50 - 1 ) ^ 350 ] mod 100 = [ { 350 c 349 } * 50 * ( - 1 ) ^ 349 + 350 c 350 * ( - 1 ) ^ 350 ] mod 100 = [ - 350 * 50 + 1 ] mod 100 = 1 mod 100 so , evidently , the remainder will be 1 . answer : a" | a = 100 / 2
b = 7 * 7
c = a - b
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a ) 40 , b ) 300 / 11 , c ) 243 / 7 , d ) 279 / 11 , e ) 279 / 8 | b | multiply(divide(subtract(multiply(100, 14), multiply(110, 10)), multiply(110, 10)), const_100) | a person bought 110 glass bowls at a rate of rs . 10 per bowl . he sold 100 of them at rs . 14 and the remaining broke . what is the percentage gain for a ? | "cp = 110 * 10 = 1100 and sp = 100 * 14 = 1400 gain % = 100 * ( 1400 - 1100 ) / 1100 = 300 / 11 answer : b" | a = 100 * 14
b = 110 * 10
c = a - b
d = 110 * 10
e = c / d
f = e * 100
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a ) 76 , b ) 56 , c ) 88 , d ) 66 , e ) 75 | e | power(add(45, const_4), const_4) | the difference between a number and its two - fifth is 45 . what is the number ? | "explanation : let the number be x . x β ( 2 / 5 ) x = 45 ( 3 / 5 ) x = 45 x = 75 answer : e" | a = 45 + 4
b = a ** 4
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a ) 12 , b ) 16 , c ) 24 , d ) 48 , e ) 98 | d | lcm(multiply(3, 4), multiply(4, 4)) | the ratio of numbers is 3 : 4 and their h . c . f is 4 . their l . c . m is : | "let the numbers be 3 x and 4 x . then their h . c . f = x . so , x = 4 . so , the numbers are 12 and 16 . l . c . m of 12 and 16 = 48 . answer : d" | a = 3 * 4
b = 4 * 4
c = math.lcm(a, b)
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a ) 7 : 8 , b ) 9 : 2 , c ) 1 : 2 , d ) 5 : 4 , e ) 3 : 2 | d | divide(add(3, 2), add(const_3.0, 1)) | in the first m games of a team ' s season , the ratio of the team ' s wins to its losses was 3 : 1 . in the subsequent n games , the ratio of the team Β΄ s wins to losses was 2 : 3 . if m : n = 4 : 5 , what was the ratio of the team ' s wins to its losses for all m + n games ? | "m = 4 / 9 of total games n = 5 / 9 of total games wins = 3 / 4 * 4 / 9 + 2 / 5 * 5 / 9 = 3 / 9 + 2 / 9 = 5 / 9 losses = 1 - 5 / 9 = 4 / 9 the ratio of wins to losses is 5 : 4 . the answer is d ." | a = 3 + 2
b = 3 + 0
c = a / b
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a ) 9 , b ) 14 , c ) 25 , d ) 30 , e ) 44 | e | divide(divide(multiply(138, const_100), 82), const_2) | a rainstorm increased the amount of water stored in state j reservoirs from 124 billion gallons to 138 billion gallons . if the storm increased the amount of water in the reservoirs to 82 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | "since we need to find only an approximate value and the answer choices are quite widespread , then use : 80 % instead of 82 % ( notice that this approximation gives the bigger tank capacity ) ; 140 billion gallons instead of 138 billion gallons ( notice that this approximation also gives the bigger tank capacity ) ; 1... | a = 138 * 100
b = a / 82
c = b / 2
|
a ) β 6 , b ) β 2 , c ) 0 , d ) 2 , e ) 6 | a | subtract(add(power(negate(6), 2), 16), multiply(10, negate(6))) | for what value of x between β 6 and 6 , inclusive , is the value of x ^ 2 β 10 x + 16 the greatest ? | "we can see from the statement that two terms containing x , x ^ 2 will always be positive and - 10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive , and lower the value of x , greater is the equation . so - 6 will give the greatest value . . ans a" | a = negate ** (
b = a + 2
c = b - 16
|
a ) 22 , b ) 88 , c ) 12 , d ) 73 , e ) 29 | c | multiply(divide(subtract(12005, 9800), subtract(multiply(9800, 8), multiply(5, 12005))), const_100) | a sum of money amounts to rs . 9800 after 5 years and rs . 12005 after 8 years at the same rate of simple interest . the rate of interest per annum is : | "explanation : s . i . for 3 years = rs . ( 12005 - 9800 ) = rs . 2205 . s . i . for 5 years = rs . = rs . 3675 principle = rs . ( 9800 - 3675 ) = rs . 6125 hence , rate = = 12 % answer : c ) 12 %" | a = 12005 - 9800
b = 9800 * 8
c = 5 * 12005
d = b - c
e = a / d
f = e * 100
|
a ) 28 sec , b ) 23 sec , c ) 24 sec , d ) 25 sec , e ) 26 sec | e | divide(add(150, 110), multiply(36, const_0_2778)) | how many seconds will a train 110 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ? | "d = 110 + 150 = 260 s = 36 * 5 / 18 = 10 mps t = 260 / 10 = 26 sec e" | a = 150 + 110
b = 36 * const_0_2778
c = a / b
|
a ) 74 , b ) 75 , c ) 76 , d ) 77 , e ) 79 | e | divide(add(add(multiply(65, 4), multiply(90, 6)), multiply(77, 5)), add(add(4, 6), 5)) | a teacher gave the same test to 3 history classes : a , b , and c . the average ( arithmetic mean ) scores for the 3 classes were 65 , 90 , and 77 , respectively . the ratio of the numbers of students in each class who took the test was 4 to 6 to 5 , respectively . what was the average score for the 3 classes combined ... | ratio is 4 : 6 : 5 , numbers are 4 x , 6 x , 5 x total scores of each class is ( 65 * 4 x + 6 x * 90 + 77 * 5 x ) = 260 x + 540 x + 385 x = 1185 x total number of students = 15 x average = 1185 x / 15 x = 79 e is the answer | a = 65 * 4
b = 90 * 6
c = a + b
d = 77 * 5
e = c + d
f = 4 + 6
g = f + 5
h = e / g
|
a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 18 | e | subtract(multiply(divide(54, subtract(const_1, divide(40, const_100))), subtract(const_1, divide(20, const_100))), 54) | a merchant purchased a jacket for $ 54 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 40 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant β s gross profit on this ... | "actual cost = $ 54 sp = actual cost + mark up = actual cost + 40 % sp = 54 * 100 / 60 on sale sp = 80 / 100 ( 54 * 100 / 60 ) = 72 gross profit = $ 18 answer is e" | a = 40 / 100
b = 1 - a
c = 54 / b
d = 20 / 100
e = 1 - d
f = c * e
g = f - 54
|
a ) $ 27.50 , b ) $ 28.50 , c ) $ 29.50 , d ) $ 30.50 , e ) $ 31.50 | b | divide(add(multiply(divide(20, const_100), 19), 19), divide(subtract(const_100, 20), const_100)) | a distributor sells a product through an online store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 19 per item . what is the price that the buyer observers online if the distributor wants to maintain a 20 % profit on the co... | "let x be the price that buyers see online . the distributor wants to receive 1.2 ( original price ) which should be 80 % of x . 1.2 ( 19 ) = 0.8 x x = 1.2 ( 19 ) / 0.8 = 1.5 ( 19 ) = $ 28.50 the answer is b ." | a = 20 / 100
b = a * 19
c = b + 19
d = 100 - 20
e = d / 100
f = c / e
|
a ) 18 , b ) 8 , c ) 12 , d ) 6 , e ) 4 | a | inverse(multiply(divide(const_2, const_3), divide(const_1, 12))) | three pipes of same capacity can fill a tank in 12 hours . if there are only two pipes of same capacity , the tank can be filled in ? | "the part of the tank filled by three pipes in one hour = 1 / 12 = > the part of the tank filled by two pipes in 1 hour = 2 / 3 * 1 / 12 = 1 / 18 . the tank can be filled in 18 hours . answer : a" | a = 2 / 3
b = 1 / 12
c = a * b
d = 1/(c)
|
a ) 1 , b ) 16 , c ) 20 , d ) 71 , e ) 60 | b | subtract(multiply(const_100, add(const_10, multiply(const_3, const_2))), 800) | what is the least number to be subtracted from 800 to make it a perfect square ? | "the numbers less than 800 and are squares of certain number is 784 . the least number that should be subtracted from 800 to make it perfect square = 800 - 784 = 16 . answer : b" | a = 3 * 2
b = 10 + a
c = 100 * b
d = c - 800
|
a ) 150 , b ) 88 , c ) 77 , d ) 310 , e ) 52 | d | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(110, divide(20, divide(15, const_3)))) | a train crosses a platform of 110 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be Γ’ β¬ Λ x Γ’ β¬ β’ x + 110 / 15 = x + 250 / 20 4 x + 440 = 3 x + 750 x = 310 m answer : d" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 110 * e
g = c - f
|
a ) 30 , b ) 45 , c ) 60 , d ) 80 , e ) 100 | c | divide(const_1, add(divide(const_1, 80), divide(const_1, multiply(80, const_3)))) | bucket p has thrice the capacity as bucket q . it takes 80 turns for bucket p to fill the empty drum . how many turns it will take for both the buckets p and q , having each turn together to fill the empty drum ? | "explanation : let capacity of q = 1 litre . then , capacity of p = 3 litre . given that it takes 80 turns for bucket p to fill the empty drum . = > capacity of the drum = 80 Γ 3 = 240 litre . number of turns required if both p and q are used having each turn together 240 / 3 + 1 = 60 litre . answer : option c" | a = 1 / 80
b = 80 * 3
c = 1 / b
d = a + c
e = 1 / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 47 minutes . the total distance ( in km ) is : | "sol . let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 47 / 60 β 47 x / 60 = 47 / 60 β x = 1 . β΄ total distance = ( 3 * 1 ) km = 3 km . answer c" | a = 47 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 19 % , b ) 10 % , c ) 22 % , d ) 16 % , e ) none | c | add(15, multiply(subtract(15, 10), divide(7, 5))) | weights of two friends ram and shyam are in the ratio 7 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | "solution : given ratio of ram and shayam ' s weight = 7 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 7 / 5 or , x = 22 % . answer : option c" | a = 15 - 10
b = 7 / 5
c = a * b
d = 15 + c
|
a ) 16 % , b ) 66 % , c ) 77 % , d ) 88 % , e ) 52 % | c | multiply(divide(subtract(90, 20), 90), const_100) | in town x , 90 percent of the population are employed , and 20 percent of the population are employed males . what percent of the employed people in town x are females ? | "total employed people 90 % , out of which 20 are employed males , hence 70 % are employed females . ( employed females ) / ( total employed people ) = 70 / 90 = 7 / 9 = 77 % answer : c ." | a = 90 - 20
b = a / 90
c = b * 100
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | add(divide(add(multiply(3, 6), subtract(16, 6)), const_2), 16) | the ages of two persons differ by 16 years . 6 years ago , the elder one was 3 times as old as the younger one . what are their present ages of the elder person ? | let present age of the elder person = x and present age of the younger person = x β 16 ( x β 6 ) = 3 ( x β 16 β 6 ) x β 6 = 3 x β 66 2 x = 60 x = 60 / 2 = 30 answer : option c | a = 3 * 6
b = 16 - 6
c = a + b
d = c / 2
e = d + 16
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | d | divide(factorial(subtract(add(const_4, 6), const_1)), multiply(factorial(6), factorial(subtract(const_4, const_1)))) | how many positive integers less than 90 are multiples of 6 but not multiples of 9 ? | "the lcm of 6 and 9 is 18 . if x < 90 and x is divisible by 6 not by 9 - - > x is not divisible by 18 . from 1 - - > 90 , we have 5 numbers which is divisible by 18 : 18 , 36 , 54 , 72 , 90 . from 1 - - > 90 , we have ( 90 - 6 ) / 6 + 1 = 15 numbers divisible by 6 . therefore , our answer is 15 - 5 = 10 numbers . d" | a = 4 + 6
b = a - 1
c = math.factorial(b)
d = math.factorial(6)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 22 , b ) 12 , c ) 38 , d ) 15 , e ) 17 | d | subtract(multiply(multiply(subtract(multiply(32, 10), multiply(16, 10)), 6), divide(8, 6)), multiply(subtract(multiply(32, 10), multiply(16, 10)), 6)) | two numbers are in the ratio of 6 : 8 . if 10 is subtracted from each , the new numbers are in the ratio 16 : 32 . find the smaller number . | "explanation : let the number be 6 x and 8 x . then , ( 6 x β 10 ) / ( 8 x β 10 ) = 1 / 2 2 ( 6 x β 10 ) = ( 8 x β 10 ) 12 x β 20 = 8 x β 10 4 x = 10 x = 2.5 the smaller number = ( 6 x 2.5 ) = 15 answer d" | a = 32 * 10
b = 16 * 10
c = a - b
d = c * 6
e = 8 / 6
f = d * e
g = 32 * 10
h = 16 * 10
i = g - h
j = i * 6
k = f - j
|
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