options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 10.5 , b ) 11 , c ) 11.6 , d ) 12 , e ) 12.5 | c | subtract(16.2, multiply(2, 2.3)) | the arithmetic mean and standard deviation of a certain normal distribution are 16.2 and 2.3 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 16.2 two standard deviations is 2.3 + 2.3 = 4.6 there could be two calues for this . mean + two standard deviations = 20.8 mean - two standard deviations = 11.6 answer choice has 11.6 and so c is the answer ." | a = 2 * 2
b = 16 - 2
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a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 2 , e ) 17 : 5 | e | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 85,000 and rs . 25,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 25000 = 17 : 5 . answer : e" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
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a ) 22 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | a | divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 10), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | a shopkeeper has 280 kg of apples . he sells 40 % of these at 10 % profit and remaining 60 % at 30 % profit . find his % profit on total . | if the total quantity was 100 then 40 x 10 % + 60 x 30 % = 22 this profit will remain same for any total quantity unless the % of products remains the same . hence ' a ' is the answer | a = 280 * 40
b = a / 100
c = 100 + 10
d = c / 100
e = b * d
f = 280 * 60
g = f / 100
h = 100 + 30
i = h / 100
j = g * i
k = e + j
l = k - 280
m = l * 100
n = m / 280
|
a ) rs . 600 , b ) rs . 900 , c ) 1200 rupees , d ) rs . 1500 , e ) none | c | add(multiply(multiply(divide(600, 10), 5), const_3), multiply(divide(600, 10), 5)) | the simple interest on a sum of money will be rs . 600 after 10 years . if the principal is trebled after 5 years , what will be the total interest at the end of the tenth year ? | solution 22.22 let the sum be rs . x . now , s . i . = rs . 600 , t = 10 years . rate = % = ( 100 x 600 / x x 10 ) % = ( 6000 / x ) % s . i . for first 5 years = rs . ( x x 5 x 6000 / x x 100 ) = rs . 300 . s . i . for first 5 years = rs . ( 3 x x 5 x 6000 / x x 100 ) = rs . 900 . ∴ total interest = rs . 1200 . answer c | a = 600 / 10
b = a * 5
c = b * 3
d = 600 / 10
e = d * 5
f = c + e
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value w of the ratio a / b ? | "w - > a / b = 20 * 21 * … … * 39 / 21 * 22 * … . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a = 2020 / 2020
b = 2020 / 2020
c = 2020 / 2020
d = b + c
e = a / d
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(add(add(multiply(const_2, const_2), multiply(3, const_4)), 5), subtract(8, add(3, const_2))) | out of 3 consecutive odd numbers 8 times the first number is equal to addition of thrice the third number and adding 5 to twice the second . what is the first number ? | description : = > 8 x = 2 ( x + 2 ) + 5 + 3 ( x + 4 ) = > 3 x = 21 , x = 7 = > x + 4 = 11 x = 11 - 4 = 7 answer c | a = 2 * 2
b = 3 * 4
c = a + b
d = c + 5
e = 3 + 2
f = 8 - e
g = d / f
|
a ) 535 m , b ) 178 m , c ) 186 m , d ) 168 m , e ) 500 m | e | multiply(divide(multiply(60, const_1000), const_3600), 30) | a train running at the speed of 60 km / hr crosses a pole in 30 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 30 = 500 m answer : e" | a = 60 * 1000
b = a / 3600
c = b * 30
|
a ) 15 , b ) 17 , c ) 19 , d ) 21 , e ) 23 | d | add(add(11, divide(subtract(11, 5), 3)), 8) | a , b , c , d and e are 5 consecutive points on a straight line . if bc = 3 cd , de = 8 , ab = 5 and ac = 11 , what is the length of ae ? | "ac = 11 and ab = 5 , so bc = 6 . bc = 3 cd so cd = 2 . the length of ae is ab + bc + cd + de = 5 + 6 + 2 + 8 = 21 the answer is d ." | a = 11 - 5
b = a / 3
c = 11 + b
d = c + 8
|
a ) 1572 , b ) 1782 , c ) 1992 , d ) 2122 , e ) 2342 | a | add(lcm(lcm(9, 35), lcm(25, 21)), 3) | what is the smallest number which when increased by 3 is divisible by 9 , 35 , 25 and 21 ? | "when increased by 3 , the number must include at least 3 ^ 2 * 5 ^ 2 * 7 = 1575 the answer is a ." | a = math.lcm(9, 35)
b = math.lcm(25, 21)
c = math.lcm(a, b)
d = c + 3
|
a ) - 2 , b ) 2 , c ) 0 , d ) - 3 , e ) 3 | c | subtract(const_10, const_10) | if | r + s | = | r - s | , then r * s must be equal to : | solution 1 : distance perspective | r - s | = | r + s | = = > the distance of r and s is equal to the distance of r and - s . < = = = = = = = ( - s ) = = = = = = = 0 = = = = = = = ( s ) = = = = = = > only 0 is the value that has a distance equal to s and - s . solution 2 : | r - s | = | r + s | ( square both ) r ^ 2 - 2 rs + s ^ 2 = r ^ 2 + 2 rs + s ^ 2 4 rs = 0 rs = 0 answer : c | a = 10 - 10
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a ) 16 / 3 , b ) 7 / 3 , c ) 5 / 2 , d ) 4 / 3 , e ) 3 | a | multiply(divide(8, add(4, 8)), 8) | 8 cups of milk are to be poured into a 4 - cup bottle and a 8 - cup bottle . if each bottle is to be filled to the same fraction of its capacity , how many cups of milk should be poured into the 8 - cup bottle ? | let x be the # of cups going into the 8 cup bottle . so . . . . x / 8 = ( ( 8 - x ) / 4 ) 64 - 8 x = 4 x 64 = 12 x x = 16 / 3 . answer : a | a = 4 + 8
b = 8 / a
c = b * 8
|
a ) 45 , b ) 27 , c ) 28 , d ) 26 , e ) 91 | a | divide(6, multiply(multiply(subtract(const_1, divide(1, 3)), subtract(const_1, divide(2, 5))), subtract(const_1, divide(2, 3)))) | there are some people in party , 1 / 3 rd left the party . then 2 / 5 th of the remaining left the party , then 2 / 3 rd of the remaining left the party . at last 6 were remaining . how many people were in total ? | sol : 45 if x persons were there in total , then x × ( 1 – 1 / 3 ) × ( 1 – 2 / 5 ) × ( 1 – 2 / 3 ) = 6 x × 2 / 3 × 3 / 5 × 1 / 3 = 6 x = 6 × 5 × 3 / 2 = 45 answer : a | a = 1 / 3
b = 1 - a
c = 2 / 5
d = 1 - c
e = b * d
f = 2 / 3
g = 1 - f
h = e * g
i = 6 / h
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a ) 3 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 3 , e ) 1 : 1 | e | divide(subtract(72, 67), subtract(67, 62)) | in what ratio must tea at rs . 62 per kg be mixed with tea at rs . 72 per kg so that the mixture must be worth rs . 67 per kg ? | "required ratio = 500 : 500 = 3 : 1 answer e" | a = 72 - 67
b = 67 - 62
c = a / b
|
a ) 80 , b ) 110 , c ) 112 , d ) 120 , e ) 160 | a | multiply(divide(const_100, 10), 8) | a 8 % stock yielding 10 % is quoted at : | "solution to earn rs . 10 , money invested = rs . 100 . to earn rs . 8 , money invested = rs . ( 100 / 10 x 8 ) = rs . 80 . â ˆ ´ market value of rs . 100 stock = rs . 80 answer a" | a = 100 / 10
b = a * 8
|
a ) 400 , b ) 420 , c ) 480 , d ) 500 , e ) 600 | e | divide(300, divide(50, const_100)) | if it is assumed that 50 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 50 % = 300 / 0.5 = 600 option e" | a = 50 / 100
b = 300 / a
|
a ) 9 years , b ) 11 years , c ) 15 years , d ) 21 years , e ) 25 years | c | subtract(multiply(15, 15), add(multiply(7, 14), multiply(7, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 7 students is 14 years and that of the other 7 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 7 + 16 x 7 ) ] = ( 225 - 210 ) = 15 years . answer c" | a = 15 * 15
b = 7 * 14
c = 7 * 16
d = b + c
e = a - d
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a ) 10 . , b ) 25 . , c ) 30 . , d ) 38 , e ) 50 . | e | divide(1000, add(multiply(subtract(18, 12), 3), subtract(18, 16))) | car a travels at 3 times the average speed of car b . car a started to travel at 12 : 00 o ' clock , car b started to travel at 16 : 00 o ' clock . what is the speed of car b ( in km / h ) if the total distance that both cars traveled until 18 : 00 was 1000 km ? | total distance - 1000 km time ratio a : b - 3 : 1 speed ratio a : b - 3 : 1 so distance ( speed * time ) ratio will be - 3 * 3 : 1 * 1 - 9 : 1 so total distance traveled by b - 100 km time taken by b ( 18 : 00 - 16 : 00 = 2 ) - 2 hours speed of b - 50 kmph answer : e | a = 18 - 12
b = a * 3
c = 18 - 16
d = b + c
e = 1000 / d
|
a ) a ) 40 , b ) b ) 14 , c ) c ) 50 , d ) d ) 55 , e ) e ) 60 | b | subtract(choose(6, 4), choose(4, 4)) | there are 6 books on a shelf , of which 2 are paperbacks and 4 are hardbacks . how many possible selections of 4 books from this shelf include at least one paperback ? | "approach 1 at - least 1 paper back = total - no paper back 6 c 4 - 4 c 4 = 14 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 4 c 3 + 2 c 2 * 4 c 2 = 14 answer is b" | a = math.comb(6, 4)
b = math.comb(4, 4)
c = a - b
|
a ) 25 , b ) 76 , c ) 29 , d ) 12 , e ) 20 | e | divide(subtract(subtract(subtract(931, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 931 . find the cost of each ice - cream cup ? | "explanation : let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 931 96 + 225 + 490 + 6 x = 931 6 x = 120 = > x = 20 . answer : e" | a = 16 * 6
b = 931 - a
c = 5 * 45
d = b - c
e = 7 * 70
f = d - e
g = f / 6
|
a ) 80 kg , b ) 93 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(68, multiply(10, add(2, divide(const_1, 2)))) | the average weight of 10 men is increased by 2 ½ kg when one of the men who weighs 68 kg is replaced by a new man . what is the weight of the new man ? | "since the average has increased by 1.5 kg , the weight of the man who stepped in must be equal to 68 + 10 x 2.5 68 + 25 = 93 kg ans : ' b '" | a = 1 / 2
b = 2 + a
c = 10 * b
d = 68 + c
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a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 154 | e | multiply(divide(176, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 176 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 176 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 176 = 16 / 30 x x = 330 330 - 176 = 154 answer e" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 176 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 24 | e | multiply(4, 6) | elena ’ s bread recipe calls for 3 ounces of butter for each 4 cups of flour used . she needs to make 6 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ? | "solving through algebra route : 3 b + 4 f = x amount if we multiply this equation with 6 we get : 18 b + 24 f = 6 x therefore , we got 18 ounces of butter and 6 x amount of quantity when we use 24 ounces of floor . ans : e" | a = 4 * 6
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['a ) 56', 'b ) 80', 'c ) 100', 'd ) 120', 'e ) 144'] | e | divide(factorial(6), 5) | in how many w ways can 5 people from a group of 6 people be seated around a circular table | w = 6 c 5 * ( 5 - 1 ) ! ( select 5 out of 6 and arrange them in circular manner ) = 6 * 4 ! = 6 * 24 = 144 answer - e | a = math.factorial(6)
b = a / 5
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a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 262.8 , d ) rs . 241.7 , e ) none | d | multiply(multiply(const_0_25, const_100), 10) | suganya and suriya are partners in a business . suganya invests rs . 35,000 for 8 months and suriya invests rs . 42,000 for 10 months . out of a profit of rs . 30,570 . suganya ' s share is | solution ratio of their shares = ( 35000 ã — 8 ) : ( 42000 ã — 10 ) = 128 : 125 suganya ' s share = rs . ( 30570 ã — 2 / 253 ) = rs . 241.7 . answer d | a = const_0_25 * 100
b = a * 10
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a ) 480 , b ) 2,880 , c ) 4,800 , d ) 28,800 , e ) 32,400 | e | multiply(9, const_3600) | a space shuttle orbits the earth at about 9 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 9 * 3600 = 32,400 answer : e" | a = 9 * 3600
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a ) 17.5 % , b ) 17.8 % , c ) 18.1 % , d ) 18.4 % , e ) 18.7 % | e | multiply(divide(add(multiply(7, divide(22, const_100)), multiply(divide(11, const_100), multiply(7, divide(22, const_100)))), add(7, const_2.0)), const_100) | 7 liters of a 22 percent solution of alcohol in water are mixed with 3 liters of an 11 percent alcohol in water solution . what is the percentage of alcohol in the new solution ? | "the total amount of alcohol is 0.22 ( 7 ) + 0.11 ( 3 ) = 1.87 liters . the percentage is 1.87 / 10 = 18.7 / 100 which is 18.7 % the answer is e ." | a = 22 / 100
b = 7 * a
c = 11 / 100
d = 22 / 100
e = 7 * d
f = c * e
g = b + f
h = 7 + 2
i = g / h
j = i * 100
|
a ) 46.2 % , b ) 54.54 % , c ) 55.88 % , d ) 70 % , e ) none of these | c | multiply(divide(subtract(136, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 136), const_100) | a cricketer scored 136 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 136 - ( 12 x 4 + 2 x 6 ) = 136 - ( 60 ) = 76 now , we need to calculate 72 is what percent of 136 . = > 76 / 136 * 100 = 55.88 % answer : c" | a = 12 * 4
b = 2 * 3
c = 2 * b
d = a + c
e = 136 - d
f = e / 136
g = f * 100
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a ) s . 9600 , b ) s . 7500 , c ) s . 5640 , d ) s . 6528 , e ) s . 6760 | d | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1700, multiply(const_4, const_100))) | by investing in 1623 % stock at 64 , one earns rs . 1700 . the investment made is | "explanation : market value = rs . 64 face value is not given and hence take it as rs . 100 16 2 / 3 % of the face value = 50 / 3 ie , to earn 50 / 3 , investment = rs . 64 hence , to earn rs . 1700 , investment needed = 64 × 3 × 1700 / 50 = 6528 answer : option d" | a = 4 * 2
b = 4 * 2
c = a * b
d = c * 3
e = 100 / 2
f = d / e
g = 4 * 100
h = 1700 - g
i = f * h
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a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | c | divide(divide(divide(945, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 945 what is the least possible value of k ? | "945 = 3 * 3 * 3 * 5 * 7 thus k must include numbers at least up to the number 9 so that there are at least three appearances of 3 ( that is : 3 , 6 , and 9 ) . the answer is c ." | a = 945 / 3
b = a / 3
c = 1 + 4
d = b / c
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a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 2 | e | multiply(3, 1) | if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 2 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 2 = e" | a = 3 * 1
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a ) 97 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 160)) | there are 6 people in the elevator . their average weight is 160 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 160 ) = 1057 - 960 = 97 . answer a" | a = 151 * 7
b = 6 * 160
c = a - b
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a ) 5 / 11 , b ) 6 / 11 , c ) 4 / 11 , d ) 7 / 11 , e ) 8 / 11 | b | divide(multiply(3, 2), add(multiply(3, 2), add(2, 3))) | what least fraction should be added to 1 / ( 2 × 3 ) + 1 / ( 3 × 4 ) + 1 / ( 4 × 5 ) + . . . . . . . + 1 / ( 21 × 22 ) so that the result is unity ? | explanation : given expression = 1 / ( 2 × 3 ) + 1 / ( 3 × 4 ) + 1 / ( 4 × 5 ) + . . . . . . + 1 / ( 21 × 22 ) = ( 1 / 2 - 1 / 3 ) + ( 1 / 3 - 1 / 4 ) + ( 1 / 4 - 1 / 5 ) + . . . . . . . . . . = ( 1 / 21 - 1 / 22 ) = 1 / 2 - 1 / 22 = 10 / 22 = 5 / 11 thus , least fraction to be added = 1 - 5 / 11 = 6 / 11 answer : option b | a = 3 * 2
b = 3 * 2
c = 2 + 3
d = b + c
e = a / d
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a ) 18 , b ) 19 , c ) 20 , d ) 23 , e ) 25 | d | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | how many positive integers less than 160 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 160 = 13 * 12 for 12 : 12 . . . 156 = 12 * 13 but there is one integer 13 * 12 . so n = ( 12 - 1 ) + ( 13 - 1 ) = 23 d" | a = 4 + 13
b = a - 1
c = math.factorial(b)
d = math.factorial(13)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 4224 , b ) 2224 , c ) 5374 , d ) 2434 , e ) 4254 | a | multiply(24, 22) | 24 * 22 = ? / 8 | "a 4224 ? / 8 = 24 * 22 ? = 24 * 22 * 8 ? = 4224" | a = 24 * 22
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a ) 44648 , b ) 27844 , c ) 28642 , d ) 16864 , e ) 32458 | c | multiply(power(multiply(const_4, const_2), subtract(5, const_1)), 7) | how many 5 - digit numbers that do not contain the digits 4 or 7 are there ? | we can have 7 digits ( 1 , 2,3 , 5,6 , 8,9 ) for the first place ( ten thousand ' s place ) . and similarly 8 digits for thousand ' s , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 5,6 , 8,9 ) so in total 7 * 8 * 8 * 8 * 8 = 28672 hence c | a = 4 * 2
b = 5 - 1
c = a ** b
d = c * 7
|
a ) 632 , b ) 642 , c ) 652 , d ) 675 , e ) 572 | e | divide(75, divide(450, 75)) | evaluate 75 % of 450 + 45 % of 750 | "explanation : = ( 75 / 100 ) * 450 + ( 45 / 100 ) * 750 = 675 option e" | a = 450 / 75
b = 75 / a
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a ) 19 , b ) 18 , c ) 17 , d ) 16 , e ) 15 | e | add(add(9, const_1), add(2, const_3)) | how many positive integers i less than 20 are either a multiple of 2 , an odd multiple of 9 , or the sum of a positive multiple of 2 and a positive multiple of 9 ? | we ' re asked to deal with the positive integers less than 20 . there are only 19 numbers in that group ( 1 to 19 , inclusive ) . we ' re asked to find all of the numbers that fit one ( or more ) of the given descriptions . looking at the answer choices , we have every value form 15 to 19 , inclusive , so most ( if not all ) of the numbers from 1 to 19 fit one ( or more ) of the descriptions . how long would it take you to find the ones that do n ' t fit . . . . . ? the first several should be pretty easy to find - the prompt wants us to focus on multiples of 2 and 9 ( and sums of those multiples ) . so what odd numbers are less than 9 ? 1 , 3 , 5 , 7 none of these values fit the given descriptions . that ' s 4 out of 19 that we know for sure do not fit . 19 - 4 = 15 = e | a = 9 + 1
b = 2 + 3
c = a + b
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a ) 228 , b ) 240 , c ) 887 , d ) 195 , e ) 1811 | d | multiply(20, multiply(54, const_0_2778)) | a train passes a station platform in 33 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 33 = 15 = > x = 195 m . answer : d" | a = 54 * const_0_2778
b = 20 * a
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a ) 45 kg , b ) 38 kg , c ) 66 kg , d ) 36 kg , e ) of these | d | multiply(divide(45, subtract(add(4, 6), 5)), 4) | the weights of 3 bags are in the ratio 4 : 5 : 6 . if the sum of the weights of the heaviest and the lightest bag is 45 kg more than the weight of the third bag , what is the weight of the lightest bag ? | let the weights of the three boys be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 45 = > 5 k = 45 = > k = 9 therefore the weight of the lightest boy = 4 k = 4 ( 9 ) = 36 kg . answer : d | a = 4 + 6
b = a - 5
c = 45 / b
d = c * 4
|
a ) 2399 , b ) 6204 , c ) 2999 , d ) 5808 , e ) 2888 | b | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.70) | the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.70 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.70 ) = rs . 6204 . answer : b ) 6204" | a = 13 / 86
b = math.sqrt(a)
c = b * 100
d = circumface * (
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a ) 8 and 9 , b ) 8 and 24 , c ) 25 and 41 , d ) 12 and 29 , e ) 17 and 29 | c | add(multiply(subtract(add(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 4), multiply(4, multiply(24, 4))))), 7), const_10), subtract(subtract(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 4), multiply(4, multiply(24, 4))))), 7)) | if x / 4 + 24 / x = 5 , what are the values of 4 x - 7 ? | i got the same thing c is the answer 25 or 41 | a = 24 - 4
b = 24 - 4
c = b ** 4
d = 24 * 4
e = 4 * d
f = c - e
g = math.sqrt(f)
h = a + g
i = h - 7
j = i * 10
k = 24 - 4
l = 24 - 4
m = l ** 4
n = 24 * 4
o = 4 * n
p = m - o
q = math.sqrt(p)
r = k - q
s = r - 7
t = j + s
|
a ) 107 , b ) 126 , c ) 117 , d ) 136 , e ) 147 | b | divide(multiply(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)), gcd(add(multiply(6, const_100), 72), add(multiply(4, const_100), 32)))) | a room is 6 meters 72 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 = 6 meters and 72 centimeters = 672 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 672 and 432 = 48 thus , the number of square tiles required = ( 672 x 432 ) / ( 48 x 48 ) = 14 x 9 = 126 answer : b | a = 6 * 100
b = a + 72
c = 4 * 100
d = c + 32
e = b * d
f = 6 * 100
g = f + 72
h = 4 * 100
i = h + 32
j = math.gcd(g, i)
k = 6 * 100
l = k + 72
m = 4 * 100
n = m + 32
o = math.gcd(l, n)
p = j * o
q = e / p
|
a ) 3 : 8 , b ) 3 : 9 , c ) 3 : 4 , d ) 3 : 12 , e ) 4 : 12 | d | divide(multiply(multiply(4, const_3.0), 3), multiply(multiply(4, 4), 4)) | a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 4 , and the ratio of soap to salt is four times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that time , what is the ratio of water to salt in the solution ? | "water : soap = 3 : 4 soap : salt = 12 : 16 = > for 12 soap , salt = 16 = > for 4 soap , salt = ( 16 / 12 ) * 4 = 16 / 12 = 64 / 12 = 18 / 3 = 6 / 1 so , water : soap : salt = 3 : 4 : 6 = 3 : 4 : 6 after open container , water : soap : salt = 1.5 : 4 : 6 so , water : salt = 1.5 : 6 = 3 : 12 answer : d" | a = 4 * 3
b = a * 3
c = 4 * 4
d = c * 4
e = b / d
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | divide(30, divide(12, 16)) | in a group of people , if 30 people were made to stand in each column , 16 columns could be formed . if 12 people were made to stand in a column , how many columns could be formed ? | "16 * 30 = 12 * n n = 40 the answer is c ." | a = 12 / 16
b = 30 / a
|
a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, const_1)) | if x is a positive integer with fewer than 3 digits , what is the probability q that x * ( x + 1 ) is a multiple of either 4 or 5 ? | interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability q that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) . | a = 100 / 4
b = a * 2
c = 100 / 5
d = c * 2
e = b + d
f = e - 4
g = 100 - 1
h = f / g
|
a ) 1.42 % , b ) 1.9 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(200, 0.8), multiply(700, 1.8)), const_1000) | by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 200 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 200 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 200 ) / 100 + ( 1.8 * 700 ) / 100 = 14.2 gms % of liquid x in resultant mixture = ( 14.2 / 1000 ) * 100 = 1.42 % a" | a = 200 * 0
b = 700 * 1
c = a + b
d = c / 1000
|
a ) 2197 , b ) 2800 , c ) 1750 , d ) 2267 , e ) 1262 | b | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(3, divide(5, const_100))) | the s . i . on a certain sum of money for 3 years at 5 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ? | "explanation : c . i . = [ 4000 * ( 1 + 10 / 100 ) 2 - 4000 ] = ( 4000 * 11 / 10 * 11 / 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) / ( 3 * 5 ) = rs . 2800 answer : b" | a = 10 / 100
b = 1 + a
c = b ** 2
d = 4000 * c
e = d - 4000
f = e / 2
g = 5 / 100
h = 3 * g
i = f / h
|
a ) 50 sec , b ) 58 sec , c ) 24 sec , d ) 22 sec , e ) 21 sec | c | divide(500, multiply(add(45, 30), const_0_2778)) | two train each 500 m long , are running in opposite directions on parallel tracks . if their speeds are 45 km / hr and 30 km / hr respectively , the time taken by the slower train to pass the driver of the faster one is | "explanation : relative speed = 45 + 30 = 75 km / hr = 750 / 36 m / s = 125 / 6 m / s we are calculating the time taken by the slower train to pass the driver of the faster one . hence the distance = length of the smaller train = 500 m time = distance / speed = 500 / ( 125 / 6 ) = 24 sec answer : option c" | a = 45 + 30
b = a * const_0_2778
c = 500 / b
|
a ) 8 / 27 , b ) 1 1 / 3 , c ) 2 2 / 3 , d ) 3 / 3 , e ) none of these | c | add(subtract(multiply(power(add(divide(26, 27), 18), divide(const_1, const_3)), const_3), power(add(divide(26, 27), 18), divide(const_1, const_3))), const_2) | if a fraction is multiplied by itself and then divided by the reciprocal of the same fraction , the result is 18 26 / 27 . find the fraction . | answer if the required fraction be p according to the question ( p x p ) / ( 1 / p ) = 1826 / 27 ⇒ p 3 = 512 / 27 ∴ p = 8 / 3 = 2 2 / 3 correct option : c | a = 26 / 27
b = a + 18
c = 1 / 3
d = b ** c
e = d * 3
f = 26 / 27
g = f + 18
h = 1 / 3
i = g ** h
j = e - i
k = j + 2
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | b | divide(630, multiply(add(90, 72), const_0_2778)) | two trains of length 100 meters and 200 meters are 630 meters apart . they are moving towards each other on parallel tracks , at speeds of 90 km / h and 72 km / h . after how many seconds will the trains meet ? | the speeds are 90000 / 3600 = 25 m / s and 72000 / 3600 = 20 m / s the relative speed is 45 m / s . time = 630 / 45 = 14 seconds the answer is b . | a = 90 + 72
b = a * const_0_2778
c = 630 / b
|
a ) 20 , b ) 22 1 / 2 , c ) 22 1 / 0 , d ) 22 1 / 9 , e ) 22 1 / 1 | a | add(divide(const_1, 15), divide(const_1, 30)) | a can do a job in 15 days and b can do it in 30 days . a and b working together will finish twice the amount of work in days ? | "1 / 15 + 1 / 30 = 3 / 30 = 1 / 10 10 = 10 * 2 = 20 days answer : a" | a = 1 / 15
b = 1 / 30
c = a + b
|
a ) 2450 , b ) 2560 , c ) 2650 , d ) 3500 , e ) 5184 | e | divide(multiply(power(12, 3), power(6, 4)), 432) | find the value of y from ( 12 ) ^ 3 x 6 ^ 4 ÷ 432 = y ? | "given exp . = ( 12 ) 3 x 64 = ( 12 ) 3 x 64 = ( 12 ) 2 x 62 = ( 72 ) 2 = 5184 432 12 x 62 e" | a = 12 ** 3
b = 6 ** 4
c = a * b
d = c / 432
|
a ) 0.4 , b ) 0.44 , c ) 0.48 , d ) 0.52 , e ) 0.56 | d | subtract(const_1, add(multiply(divide(40, const_100), divide(60, const_100)), multiply(divide(40, const_100), subtract(const_1, divide(40, const_100))))) | in a graduating class , 40 percent of the students are male . in this class , 60 percent of the male students and 40 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "let x be the total number of students . the number students who are younger than 25 is 0.4 * 0.4 x + 0.6 * 0.6 x = 0.52 x the answer is d ." | a = 40 / 100
b = 60 / 100
c = a * b
d = 40 / 100
e = 40 / 100
f = 1 - e
g = d * f
h = c + g
i = 1 - h
|
a ) $ 1220 , b ) $ 1240 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 8), 30), multiply(subtract(100, 30), 10)) | at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 8 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ? | i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 8 ) = $ 320 total revenue : 1000 + 320 = $ 1320 the answer is b . | a = 10 + 8
b = a * 30
c = 100 - 30
d = c * 10
e = b + d
|
a ) 5 , b ) 30 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(multiply(2, multiply(1, const_10)), multiply(divide(3, 2), multiply(1, const_10))), 1) | audrey 4 hours to complete a certain job . ferris can do the same job in 3 hours . audrey and ferris decided to collaborate on the job , working at their respective rates . while audrey worked continuously , ferris took 1 breaks of equal length . if the two completed the job together in 2 hours , how many minutes long was each of ferris ’ breaks ? | "audery and ferris collective work rate : 1 / 4 + 1 / 3 = 7 / 12 collective work time = 12 / 7 = 1.7 hrs job was actually done in = 2 ( includes breaks ) breaks = actual time taken - collective work time = 2 - 1.7 = . 3 hrs = 1 / 2 so ferrais took 1 breaks = . 3 / 1 = . 3 hrs = 30 m so answer is b ) 30 mins" | a = 1 * 10
b = 2 * a
c = 3 / 2
d = 1 * 10
e = c * d
f = b - e
g = f / 1
|
a ) rs . 22500 , b ) rs . 25500 , c ) rs . 26000 , d ) rs . 28000 , e ) none of these | b | add(add(add(11000, 5000), 1000), multiply(divide(add(add(11000, 5000), 1000), const_100), 50)) | sahil purchased a machine at rs 11000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it . | "explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 11000 + 5000 + 1000 = 17000 150 % of 17000 = 150 / 100 * 17000 = 25500 option b" | a = 11000 + 5000
b = a + 1000
c = 11000 + 5000
d = c + 1000
e = d / 100
f = e * 50
g = b + f
|
a ) 2 hours . , b ) 3 hours . , c ) 4 hours . , d ) 5 hours . , e ) 6 hours . | b | divide(60, add(16, 4)) | a boat can travel with a speed of 16 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 60 km downstream . | "speed downstream = ( 16 + 4 ) km / hr = 20 km / hr . time taken to travel 60 km downstream = 60 / 20 = 3 hours . answer : b" | a = 16 + 4
b = 60 / a
|
a ) 22 , b ) 99 , c ) 27 , d ) 28 , e ) 20 | e | multiply(4, 5) | walking 5 / 4 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 5 / 4 = 4 : 5 time ratio = 5 : 4 1 - - - - - - - - 5 4 - - - - - - - - - ? 20 m . answer : e" | a = 4 * 5
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 46.5 meters . how many times did the ball bounce ? | "going down = 16 m going up = 8 - - > total = 24 going down = 8 - - > total = 32 going up = 4 - - > total = 36 going down = 4 - - > total = 40 going up = 2 - - > total = 42 going down = 2 - - > total = 44 going up = 1 - - > total = 45 going down = 1 - - > total = 46 going up = 0.5 - - > total = 46.5 ( caught ) no of going downs = 5 . . answer : a" | a = 16 / 2
b = a / 2
|
a ) 900 , b ) 500 , c ) 5750 , d ) 5000 , e ) none | c | subtract(multiply(divide(6000, const_100), 5000), multiply(divide(const_1, const_3), multiply(divide(6000, const_100), 5000))) | 6000 - 5000 ÷ 20.00 = ? | "answer given expression = 6000 - 5000 ÷ 20.00 = 6000 - 250 = 5750 correct option : c" | a = 6000 / 100
b = a * 5000
c = 1 / 3
d = 6000 / 100
e = d * 5000
f = c * e
g = b - f
|
a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 50 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 50))), 30)) | mahesh can do a piece of work in 50 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 50 days = 20 * 1 / 50 = 2 / 5 remaining work = 1 - 2 / 5 = 3 / 5 3 / 5 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 5 / 3 = 50 days answer is e" | a = 1 / 50
b = 20 * a
c = 1 - b
d = c / 30
e = 1 / d
|
['a ) 5 : 78', 'b ) 5 : 11', 'c ) 5 : 12', 'd ) 5 : 19', 'e ) 5 : 16'] | e | divide(subtract(sqrt(4096), 24), multiply(sqrt(4096), const_2)) | the area of a square is 4096 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4096 = 212 a = ( 212 ) 1 / 2 = 26 = 64 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 40 : 128 = 5 : 16 answer : e | a = math.sqrt(4096)
b = a - 24
c = math.sqrt(4096)
d = c * 2
e = b / d
|
a ) 6 , b ) 3 , c ) 8 , d ) 9 , e ) 10 | b | add(add(1, const_2), const_2) | if x < y < z and y - x > 1 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we have : 1 ) x < y < z 2 ) y - x > 1 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 1 , the least value for y - x must be 3 , the least value for x must be 2 , and , thus , the least possible value for y must be 3 ( y - 2 = 1 , y = 3 ) 2 < 3 < z , since z is odd , the least possible value for z is 5 z - x = 5 - 2 = 3 answer b" | a = 1 + 2
b = a + 2
|
a ) 165 , b ) 170 , c ) 175 , d ) 150 , e ) 185 | d | multiply(divide(const_100, 8), 12) | a 12 % stock yielding 8 % is quoted at ? | "assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 12 , money invested = 100 × 12 / 8 = rs . 150 ie , market value of the stock = rs . 150 answer is d ." | a = 100 / 8
b = a * 12
|
a ) 12 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | a | add(multiply(divide(multiply(divide(13.00, multiply(subtract(0.75, 0.10), const_2)), const_2), const_10), const_2), multiply(divide(13.00, multiply(subtract(0.75, 0.10), const_2)), const_2)) | the toll for crossing a certain bridge is $ 0.75 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 13.00 and then pay only $ 0.10 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day , what is the least value of x for which this driver can save money by using the sticker ? | "option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than 0.6 x 14 + 13 ( 21.4 ) . . . it ' s pretty close so let ' s see if one more day will make it worth it . . . if we raise the number of days to 15 , the sticker option looks like a better deal . . . 1.5 x 15 ( 22.5 ) vs 0.6 x 15 + 13 ( 22 ) . answer : a" | a = 0 - 75
b = a * 2
c = 13 / 0
d = c * 2
e = d / 10
f = e * 2
g = 0 - 75
h = g * 2
i = 13 / 0
j = i * 2
k = f + j
|
a ) 1 / 9 , b ) 1 / 8 , c ) 1 / 7 , d ) 7 / 8 , e ) 8 / 7 | b | divide(subtract(divide(const_1, subtract(9, const_1)), divide(const_1, 9)), divide(const_1, 9)) | 9 people are planning to share equally the cost of a rental van . if one person withdraws from the arrangement and the others share equally the cost of the rental van , then the share of each of the remaining people will increase by . . . ? | let p = total cost of the rental van . the original share per person is p / 9 . the new share per person is p / 8 . p / 8 = p / 9 * 9 / 8 = ( 1 + 1 / 8 ) * original share the answer is b . | a = 9 - 1
b = 1 / a
c = 1 / 9
d = b - c
e = 1 / 9
f = d / e
|
a ) 125 % , b ) 150 % , c ) 100 % , d ) 90 % , e ) 85 % | a | multiply(subtract(power(add(const_1, divide(50, const_100)), const_2), const_1), const_100) | what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 50 % ? | "the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 50 % increase a surface = 6 * ( ( 1.5 a ) ^ 2 ) = 6 * 2.25 * a ^ 2 the increase in the surface area = ( 6 * 2.25 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 2.25 - 1 ) ) / ( 6 * a ^ 2 ) = 2.25 - 1 = 1.25 = 125 % answer : a" | a = 50 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 55 % , b ) 60 % , c ) 70 % , d ) 72 % , e ) 95 % | e | multiply(subtract(const_1, divide(5, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))), subtract(const_1, divide(5, multiply(multiply(const_2, const_5), multiply(const_2, const_5))))) | there is a 5 % chance that tigers will not win at all during the whole season . there is a 5 % chance that federer will not play at all in the whole season . what is the greatest possible probability that the tigers will win and federer will play during the season ? | "there is a 5 % chance that tigers will not win at all during the whole season we can infer that there is 95 % chance tigers will win . similarly there is a 5 % chance that federer will not play at all in the whole season we can also infer that there is 95 % chance that federer will play . answer e" | a = 2 * 5
b = 2 * 5
c = a * b
d = 5 / c
e = 1 - d
f = 2 * 5
g = 2 * 5
h = f * g
i = 5 / h
j = 1 - i
k = e * j
|
a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 300 | e | subtract(divide(negate(subtract(multiply(25, 8), multiply(40, 8))), divide(40, const_100)), subtract(multiply(25, 8), multiply(40, 8))) | at a florist shop on a certain day , all corsages sold for either $ 25 or $ 40 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 40 percent . what was the store ' s actual revenue from corsages that day ? | "i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 25 = x , no . of corsages @ $ 40 = y and revenue = r so , 25 x + 40 y = r . . . . . . . . . ( 1 ) now , given the situation , 25 ( x + 8 ) + 40 ( y - 8 ) = r - . 4 r = > 25 x + 200 + 40 y - 320 = . 6 r = > 25 x + 40 y = . 8 r + 120 . . . . . . . . . . . . ( 2 ) so , r = . 6 r + 120 = > r = 300 the answer is e ." | a = 25 * 8
b = 40 * 8
c = a - b
d = negate / (
e = 40 / 100
f = d - e
|
a ) 900 , b ) 1200 , c ) 1500 , d ) 1800 , e ) 2100 | b | divide(multiply(480, const_100), subtract(const_100, 60)) | in an election between two candidates , the first candidate got 60 % of the votes and the second candidate got 480 votes . what was the total number of votes ? | "let v be the total number of votes . 0.4 v = 480 v = 1200 the answer is b ." | a = 480 * 100
b = 100 - 60
c = a / b
|
a ) 67 sec , b ) 89 sec , c ) 38 sec , d ) 87 sec , e ) 45 sec | c | divide(add(260, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 260 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 260 + 120 = 380 m . time taken = 380 / 10 = 38 sec . answer : c" | a = 260 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 40 mins , b ) 35 mins , c ) 30 mins , d ) 32 mins , e ) 36 mins | c | add(multiply(divide(divide(const_3, const_4), const_3), const_60), 15) | pipe a that can fill a tank in an hour and pipe b that can fill the tank in 15 are opened simultaneously when the tank is empty . pipe b is shut 15 minutes before the tank overflows . when will the tank overflow ? | the last 15 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 4 = 5 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 5 = 3 / 20 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : c | a = 3 / 4
b = a / 3
c = b * const_60
d = c + 15
|
a ) 15 , b ) 18 , c ) 20 , d ) 22 , e ) 17 | b | multiply(divide(add(31, 5), const_4), const_2) | john has exactly 31 pens , and each pen is either bule , black , or red . if he has 5 more black pens than red pens , and twice as many blue pens as black pens , how many blue pens does john has ? | x = the number of blue pens y = the number of black pens z = the number of red pens from the first sentence we have equation # 1 : x + y + z = 31 . . . he has 5 more black pens than red pens . . . equation # 2 : y = 5 + z . . . twice as many blue pens as black pens . . . equation # 3 : x = 2 y solve equation # 1 for z : z = y - 5 now , we can replace both x and z with y in equation # 1 2 y + y + ( y - 5 ) = 31 4 y - 5 = 31 4 y = 36 y = 9 there are 9 black pens . this is 5 more than the number of red pens , so z = 4 . hence the answer is . x = 18 , and just for check 18 + 9 + 4 = 31 . answer = 18 , ( b ) | a = 31 + 5
b = a / 4
c = b * 2
|
a ) 4.2 , b ) 6.5 , c ) 6 , d ) 7 , e ) 8 | d | divide(divide(multiply(add(5, 7), add(divide(subtract(7, 6), 5), const_1)), const_2), add(divide(subtract(7, 5), 5), const_1)) | what is the average ( arithmetic mean ) of the numbers 5 , 6 , 7 , 7 , 8 , and 9 ? | "{ 5 , 6 , 7 , 7 , 8 , 9 } = { 7 - 2,7 - 1 , 7 , 7,7 + 1,7 + 2 } - - > the average = 7 . answer : d ." | a = 5 + 7
b = 7 - 6
c = b / 5
d = c + 1
e = a * d
f = e / 2
g = 7 - 5
h = g / 5
i = h + 1
j = f / i
|
a ) rs . 16003 , b ) rs . 16029 , c ) rs . 1288 , d ) rs . 16108 , e ) rs . 12000 | e | subtract(multiply(add(1500, 500), add(20, const_1)), multiply(1500, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 500 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 2000 * 21 - 1500 * 20 ) = rs . 12000 . answer : e" | a = 1500 + 500
b = 20 + 1
c = a * b
d = 1500 * 20
e = c - d
|
a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | b | divide(50, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(50, 2))))), 2)) | a rectangular tiled patio is composed of 50 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 50 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | "r * c = 50 and ( 6 + 4 ) ( 7 - 2 ) = 50 - - > r = 6 and c = 7 . answer : b ." | a = negate + (
b = 4 ** 2
c = 50 * 2
d = 4 * c
e = b + d
f = math.sqrt(e)
g = a / f
h = 50 / g
|
a ) 12 % , b ) 14 % , c ) 16 % , d ) 20 % , e ) 8 % | e | subtract(const_100, divide(multiply(add(const_100, 15), subtract(const_100, 20)), const_100)) | the tax on a commodity is diminished by 20 % but its consumption is increased by 15 % . find the decrease percent in the revenue derived from it ? | "explanation : 100 * 100 = 10000 80 * 115 = 9200 10000 - - - - - - - 800 100 - - - - - - - ? = 8 % e )" | a = 100 + 15
b = 100 - 20
c = a * b
d = c / 100
e = 100 - d
|
a ) 20 , b ) 30 , c ) 70 , d ) 60 , e ) 90 | c | multiply(4, divide(lcm(add(5, 3), add(4, 3)), add(4, 3))) | the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 15 of her stamps , the ration of the number of kaye had to the number of alberto had was 4 : 3 . as a result of the gift , kaye had how many more stamps than alberto ? | "c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 15 a 2 = 3 x + 15 k 2 / a 2 = 7 / 5 ( 5 x - 15 ) / ( 3 x + 15 ) = 4 / 3 3 * ( 5 x - 15 ) = 4 * ( 3 x + 15 ) 15 x - 45 = 12 x + 60 3 x = 105 x = 35 k 2 = 5 * 35 - 15 = 160 a 2 = 3 * 35 + 15 = 90 k 2 - a 2 = 70 answer : c" | a = 5 + 3
b = 4 + 3
c = math.lcm(a, b)
d = 4 + 3
e = c / d
f = 4 * e
|
a ) 240 cm 2 , b ) 384 cm 2 , c ) 279 cm 2 , d ) 200 cm 2 , e ) 297 cm 2 | a | multiply(24, 10) | find the area of a parallelogram with base 24 cm and height 10 cm ? | "area of a parallelogram = base * height = 24 * 10 = 240 cm 2 answer : a" | a = 24 * 10
|
a ) 25 miles , b ) 20.4 miles , c ) 30 miles , d ) 32 miles , e ) 33 miles | b | sqrt(add(power(add(5, 15), const_2), power(4, const_2))) | beginning in town a , biker bob rides his bike 8 miles west , 5 miles north , 4 miles east , and then 15 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 15 + 5 = 20 m other being the base ie distance traveled west - distance traveled eat = 8 - 4 = 4 m now this third side or the distance between town a and town b = 20 ^ 2 + 4 ^ 2 = sq root 416 = 20.4 miles answer : b" | a = 5 + 15
b = a ** 2
c = 4 ** 2
d = b + c
e = math.sqrt(d)
|
a ) 44 ft , b ) 88 ft , c ) 74 ft , d ) 11 ft , e ) 66 ft | c | add(multiply(divide(680, 40), const_2), 40) | a rectangular field is to be fenced on three sides leaving a side of 40 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 40 x breadth = 680 breadth = 17 feet area to be fenced = 2 b + l = 2 ( 17 ) + 40 = 74 feet answer : c ) 74 ft" | a = 680 / 40
b = a * 2
c = b + 40
|
a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | e | add(divide(subtract(140, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 140 ? | "let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 140 5 x - 20 = 140 5 x = 160 x = 32 the answer is e ." | a = 2 + 2
b = 2 + a
c = 2 + 2
d = c + 2
e = b + d
f = 2 + 2
g = f + 2
h = g + 2
i = e + h
j = 140 - i
k = j / 5
l = 2 + 2
m = l + 2
n = m + 2
o = k + n
|
a ) 6 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | a | multiply(subtract(divide(60, const_2), divide(multiply(60, 45), const_100)), const_2) | a chemical supply company has 60 liters of a 45 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 45 % hno 3 solution means hno 3 = 27 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 27 + x = 50 % of ( 60 + x ) or x = 6 . hence , a" | a = 60 / 2
b = 60 * 45
c = b / 100
d = a - c
e = d * 2
|
a ) 10 % , b ) 15 % , c ) 11.2 % , d ) 7 % , e ) 13 % | c | subtract(multiply(multiply(subtract(const_1, divide(20, const_100)), add(const_1, divide(40, const_100))), const_100), const_100) | a furniture store owner decided to drop the price of her recliners by 20 % to spur business . by the end of the week she had sold 40 % more recliners . what is the percentage increase of the gross ? | "say a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 % , she will earn $ 80 per recliner and she sells 40 % more ie . , 140 recliners hence her sales tields 140 * 80 = $ 11200 increase in sales = 11200 - 10000 = $ 1120 so % increase = 1120 * 100 / 10000 = 11.20 % c is the answer" | a = 20 / 100
b = 1 - a
c = 40 / 100
d = 1 + c
e = b * d
f = e * 100
g = f - 100
|
a ) $ 2,250 , b ) $ 2,500 , c ) $ 2,100 , d ) $ 2,150 , e ) $ 2,700 | c | floor(divide(subtract(subtract(multiply(500, 8), multiply(5, 100)), multiply(subtract(500, 100), 3.5)), const_1000)) | company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 8.00 each , what was company c ’ s gross profit ? | cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 8 = $ 4000 profit : 4000 - 1900 = $ 2100 option c is correct | a = 500 * 8
b = 5 * 100
c = a - b
d = 500 - 100
e = d * 3
f = c - e
g = f / 1000
h = math.floor(g)
|
a ) 30000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) 70000 | a | divide(subtract(divide(multiply(50000, 20), const_100), 7000), divide(10, const_100)) | rs 50000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 7000 find money given by 10 % ? | "let first parrt is x and second part is y then x + y = 50000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 7000 = ( x * 10 * 1 ) / 100 + ( y * 20 * 1 ) / 100 70000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 70000 = 50000 + y so y = 20000 then x = 50000 - 20000 = 30000 first part = 30000 answer : a" | a = 50000 * 20
b = a / 100
c = b - 7000
d = 10 / 100
e = c / d
|
a ) 22 , b ) 88 , c ) 77 , d ) 40 , e ) 99 | d | subtract(multiply(const_1, const_60), multiply(divide(20, 60), const_60)) | excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 20 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 20 km with stoppages . stoppage time = time take to travel ( 60 - 20 ) km i . e 40 km at 60 km / hr . stoppage time = 40 / 60 hrs = 40 min . answer : d" | a = 1 * const_60
b = 20 / 60
c = b * const_60
d = a - c
|
a ) 11 , b ) 50 , c ) 60 , d ) 27 , e ) 18 | c | add(25, 35) | laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 25 kmph and prasanna moves southward at a speed of 35 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 25 kmph x 1 hr = 25 km total southward prasanna ' s distance = 35 kmph x 1 hr = 35 km total distance between prasanna and laxmi is = 25 + 35 = 60 km . answer : c" | a = 25 + 35
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | b | multiply(6, divide(15, 13)) | john was thrice as old as tom 6 years ago . john will be 15 / 13 times as old as tom in 6 years . how old is tom today ? | j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 15 / 13 * ( t + 6 ) 13 j + 78 = 15 t + 90 13 ( 3 t - 12 ) + 78 = 15 t + 90 24 t = 168 t = 7 the answer is b . | a = 15 / 13
b = 6 * a
|
a ) 500 , b ) 768 , c ) 276 , d ) 280 , e ) 279 | a | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 50)), 1000), multiply(divide(const_100, add(const_100, 50)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 50)), 1000), multiply(divide(const_100, add(const_100, 50)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 50 % profit on the first article and 50 % loss on second article . find the net profit or loss percent ? | "profit on first article = 50 % of 1000 = 500 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : a" | a = 100 - 50
b = 100 / a
c = b * 1000
d = 100 + 50
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 50
l = 100 / k
m = l * 1000
n = 100 + 50
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 18 days , b ) 15 days , c ) 26.67 days , d ) 16 days , e ) 12 days | b | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 10), divide(const_1, 30))), 10) | a can complete a project in 10 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 10 ) * ( x - 10 ) + ( 1 / 30 ) * ( x ) = 1 ( x / 10 ) + ( x / 30 ) - 1 ) = 1 4 x / 30 = 2 x = 15 therefore , the answer is b : 15 ." | a = 1 / 30
b = a * 10
c = 1 - b
d = 1 / 10
e = 1 / 30
f = d + e
g = c / f
h = g + 10
|
a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 5 / 7 , e ) 3 / 4 | e | divide(add(const_1, divide(subtract(multiply(divide(8, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 8 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the median of set r ? | "the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ) , the median is given to be 3 / 4 * b : ( a + b ) / 2 = ( 3 / 4 ) * b a = b / 2 for set q , consisting of numbers ( b , b + 1 , . . . , c ) , the median is given to be 8 / 8 * c : ( b + c ) / 2 = ( 8 / 8 ) * c b = ( 1 ) * c for set r , consisting of numbers ( a , a + 1 , . . . c ) , the median needs to be found : a = b / 2 = ( 1 * c ) / 2 = ( 1 / 2 ) * c median = ( a + c ) / 2 = ( 1 / 2 * c + c ) / 2 = ( 3 / 2 ) * c / 2 = ( 3 / 4 ) * c ( answer e )" | a = 8 / 8
b = a * 2
c = b - 1
d = 3 / 4
e = d * 2
f = e - 1
g = 1 / f
h = c / g
i = 1 + h
j = i / 2
|
a ) 70 , b ) 72 , c ) 74 , d ) 65 , e ) 78 | d | divide(1, divide(add(multiply(const_3600, divide(1, 80)), 10), const_3600)) | a car traveling at a certain constant speed takes 10 seconds longer to travel 1 km than it would take to travel 1 km at 80 km / hour . at what speed , in km / hr , is the car traveling ? | time to cover 1 kilometer at 80 kilometers per hour is 1 / 80 hours = 3,600 / 80 seconds = 45 seconds ; time to cover 1 kilometer at regular speed is 45 + 10 = 55 seconds = 55 / 3,600 hours = 1 / 65 hours ; so , we get that to cover 1 kilometer 1 / 65 hours is needed - - > regular speed 65 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : d | a = 1 / 80
b = 3600 * a
c = b + 10
d = c / 3600
e = 1 / d
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | divide(subtract(sqrt(add(multiply(multiply(210, 5), const_4), power(5, const_2))), 5), const_2) | a bus trip of 210 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ? | "the time is the distance / speed . the time difference is 1 hour . 210 / v - 210 / ( v + 5 ) = 1 210 ( v + 5 ) - 210 v = ( v ) ( v + 5 ) 1050 = ( v ) ( v + 5 ) 30 * 35 = ( v ) ( v + 5 ) v = 30 mph the answer is b ." | a = 210 * 5
b = a * 4
c = 5 ** 2
d = b + c
e = math.sqrt(d)
f = e - 5
g = f / 2
|
a ) 17 , b ) 18 , c ) 20 , d ) 22 , e ) 33 | b | add(multiply(5, 3), 3) | if the 5 th date of a month is tuesday , what date will be 3 days after the 3 rd friday in the month ? | b 18 5 th date of a month is tuesday friday will be on = 5 + 3 = 8 th of a month 1 st friday is on 1 st of a month 2 nd friday is on 8 th of a month 3 rd friday will be on 15 th of a month 3 days after 15 th = 15 + 3 = 18 | a = 5 * 3
b = a + 3
|
a ) a ) 130 , b ) b ) 134 , c ) c ) 148 , d ) d ) 176 , e ) e ) 160 | d | add(multiply(19, 9), 5) | what is the dividend . divisor 19 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 19 * 9 + 5 d = 171 + 5 d = 176 answer d" | a = 19 * 9
b = a + 5
|
a ) 770 , b ) 712 , c ) 720 , d ) 740 , e ) 700 | e | divide(multiply(add(divide(multiply(150, 60), const_100), 120), const_100), 30) | 30 % of a number is more than 60 % of 150 by 120 . find the number ? | "( 30 / 100 ) * x – ( 60 / 100 ) * 650 = 190 2 / 7 x = 210 x = 700 answer : e" | a = 150 * 60
b = a / 100
c = b + 120
d = c * 100
e = d / 30
|
a ) 0 , b ) 4 , c ) 6 , d ) 7 , e ) 13 | c | divide(subtract(add(multiply(2, 5), 5), add(multiply(3, 3), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] + 5 = f ( 3 x – 6 ) | "answer = c = 6 f ( x ) = 3 x – 5 2 * [ f ( x ) ] + 5 = f ( 3 x – 6 ) 2 ( 3 x - 5 ) + 5 = 3 ( 3 x - 6 ) - 5 6 x - 5 = 9 x - 23 x = 6" | a = 2 * 5
b = a + 5
c = 3 * 3
d = c + 5
e = b - d
f = 2 * 3
g = 3 * 1
h = f - g
i = e / h
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | d | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 2), 3)), 3), divide(multiply(divide(multiply(4, multiply(multiply(4, 2), 3)), 3), 4), 2)), subtract(multiply(multiply(4, 2), 3), divide(multiply(3, multiply(multiply(4, 2), 3)), 4))), const_100) | one day a car rental agency rented 3 / 4 of its cars , including 2 / 3 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 3 ) ( 3 / 5 ) = 1 / 5 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 1 / 5 ) / ( 1 / 4 ) = 4 / 5 the answer is d ." | a = 4 * 2
b = a * 3
c = 4 * b
d = c / 3
e = 4 * 2
f = e * 3
g = 4 * f
h = g / 3
i = h * 4
j = i / 2
k = d - j
l = 4 * 2
m = l * 3
n = 4 * 2
o = n * 3
p = 3 * o
q = p / 4
r = m - q
s = k / r
t = s * 100
|
a ) 128 , b ) 177 , c ) 199 , d ) 178 , e ) 150 | e | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 25 = x / 15 x = 150 m . answer : e" | a = 2 - 1
b = 100 * a
|
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