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 ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | e | divide(const_1, add(multiply(4, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 4 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 4 girls = 8 Γ£ β 6 / 4 + 12 = 12 + 4 = 16 girls 9 days to complete the work answer : option e" | a = 1 / 6
b = a / 24
c = 4 * b
d = 1 / 6
e = d / 16
f = 8 * e
g = c + f
h = 1 / g
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 16 | e | divide(subtract(630, multiply(75, 2.00)), 30) | 30 pens and 75 pencils were purchased for 630 . if the average price of a pencil was 2.00 , find the average price of a pen . | "since average price of a pencil = 2 β΄ price of 75 pencils = 150 β΄ price of 30 pens = ( 630 β 150 ) = 480 β΄ average price of a pen = 480 β 60 = 16 answer e" | a = 75 * 2
b = 630 - a
c = b / 30
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a ) 5.55 % , b ) 5.65 % , c ) 5.75 % , d ) 5.85 % , e ) 5.95 % | e | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 1,600 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 8400 = 5 / 84 which is about 5.95 % the answer is e ." | a = 100 * 100
b = 5 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 215 , b ) 212 , c ) 278 , d ) 279 , e ) 222 | b | multiply(circumface(divide(34, const_2)), 2) | find the cost of fencing around a circular field of diameter 34 m at the rate of rs . 2 a meter ? | "2 * 22 / 7 * 17 = 106 106 * 2 = rs . 212 answer : b" | a = 34 / 2
b = circumface * (
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a ) 33 , b ) 37 , c ) 40 , d ) 38 , e ) 27 | c | multiply(divide(add(multiply(2, divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))), multiply(2, divide(40, const_60))), divide(40, const_60)), divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))) | a man covered a certain distance at some speed . if he had moved 3 kmph faster , he would have taken 40 minutes less . if he had moved 2 kmph slower , he would have taken 40 minutes more . what is the the distance in km ? | "let the distance be x km , the speed in which he moved = v kmph time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes β xv β xv + 3 = 4060 β x [ 1 v β 1 v + 3 ] = 23 β x [ v + 3 β vv ( v + 3 ) ] = 23 β 2 v ( v + 3 ) = 9 x . . . . . . . . . . . . . . . . ( equation 1 ) time taken when moving 2 kmph slower - time taken when moving at normal speed = 40 minutes β xv β 2 β xv = 4060 β x [ 1 v β 2 β 1 v ] = 23 β x [ v β v + 2 v ( v β 2 ) ] = 23 β x [ 2 v ( v β 2 ) ] = 23 β x [ 1 v ( v β 2 ) ] = 13 β v ( v β 2 ) = 3 x . . . . . . . . . . . . . . . . ( equation 2 ) equation 1 equation 2 β 2 ( v + 3 ) ( v β 2 ) = 3 β 2 v + 6 = 3 v β 6 β v = 12 substituting this value of v in equation 1 β 2 Γ 12 Γ 15 = 9 x = > x = 2 Γ 12 Γ 159 = 2 Γ 4 Γ 153 = 2 Γ 4 Γ 5 = 40 hence distance = 40 km answer : c" | a = 40 / const_60
b = 2 * a
c = 40 / const_60
d = 3 * c
e = b + d
f = 3 - 2
g = e / f
h = 2 * g
i = 40 / const_60
j = 2 * i
k = h + j
l = 40 / const_60
m = k / l
n = 40 / const_60
o = 2 * n
p = 40 / const_60
q = 3 * p
r = o + q
s = 3 - 2
t = r / s
u = m * t
|
a ) 200 , b ) 240 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(280, 200), const_2) | if x + y = 280 , x - y = 200 , for integers of x and y , y = ? | "x + y = 280 x - y = 200 2 x = 80 x = 40 y = 240 answer is b" | a = 280 + 200
b = a / 2
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a ) 2 / 9 , b ) 2 / 5 , c ) 7 / 9 , d ) 4 / 5 , e ) 8 / 9 | c | subtract(const_1, divide(const_2, 10)) | there are 10 students named alphabetically from a to j . what is the probability that a and d do not sit together if all 10 sit around a circular table ? | number of students = 10 number of ways 10 students can sit around a circular table = ( 10 - 1 ) ! = 9 ! number of ways a and d sit together ( consider a and d as one entity ) = ( 9 - 1 ) ! = 8 ! * 2 number of ways a and d do not sit together = 9 ! - ( 8 ! * 2 ) probability = ( 9 ! - ( 8 ! * 2 ) ) / 9 ! = 1 - 2 / 9 = 7 / 9 answer : c | a = 2 / 10
b = 1 - a
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a ) 2 / 3 , b ) 3 / 7 , c ) 8 / 15 , d ) 3 / 8 , e ) 4 / 7 | c | divide(multiply(divide(2, 3), 8), 10) | a pipe can empty 2 / 3 rd of a cistern in 10 mins . in 8 mins , what part of the cistern will be empty ? | "2 / 3 - - - - 10 ? - - - - - 8 = = > 8 / 15 c" | a = 2 / 3
b = a * 8
c = b / 10
|
a ) $ 18.33 , b ) $ 22.33 , c ) $ 28.33 , d ) $ 26.23 , e ) $ 16.23 | a | multiply(divide(multiply(const_2, 10), add(110, 10)), 110) | if $ 10 be allowed as true discount on a bill of $ 110 due at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is : | s . i . on $ ( 110 - 10 ) for a certain time = $ 10 . s . i . on $ 100 for double the time = $ 20 . t . d . on $ 120 = $ ( 120 - 100 ) = $ 20 . t . d . on $ 110 = $ ( 20 / 120 * 100 ) = $ 18.33 answer : a | a = 2 * 10
b = 110 + 10
c = a / b
d = c * 110
|
a ) 12 , b ) 13 , c ) 15 , d ) 18 , e ) 20 | d | divide(subtract(40, 4), const_2) | if you multiply two integers together and then add 4 , the result is 40 . which of the following could not be the sum of the two numbers ? | let the two integers equal x and y , and then create the following equation and simplify : xy + 4 = 40 xy = 36 so x and y are a pair of integers that equal 36 . try adding all possible combinations of two integers that multiply out to 36 : 1 Γ 36 = 36 1 + 36 = 37 2 Γ 18 = 36 2 + 18 = 20 3 Γ 12 = 36 3 + 12 = 15 4 Γ 9 = 36 4 + 9 = 13 6 Γ 6 = 36 6 + 6 = 12 this list of sums includes 12 , 13 , 15 , and 20 , but not 18 . thus , no pair of integers both satisfies the original equation and adds up to 18 correct answer d ) 18 | a = 40 - 4
b = a / 2
|
a ) 40 , b ) 87 , c ) 48 , d ) 21 , e ) 14 | c | divide(460, multiply(subtract(45, 140), const_0_2778)) | a train 460 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 460 + 140 = 600 m required time = 600 * 2 / 25 = 48 sec answer : c" | a = 45 - 140
b = a * const_0_2778
c = 460 / b
|
a ) 65 kg , b ) 90 kg , c ) 85 kg , d ) data inadequate , e ) none of these | a | add(multiply(8, 2.5), 45) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 45 kg . what might be the weight of the new person ? | "a 65 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 64 + 20 ) kg = 65 kg ." | a = 8 * 2
b = a + 45
|
a ) s . 247 , b ) s . 248 , c ) s . 264 , d ) s . 329 , e ) s . 412 | c | add(divide(187, subtract(const_1, divide(15, const_100))), multiply(divide(187, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | a shopkeeper loses 15 % , if an article is sold for rs . 187 . what should be the selling price of the article to gain 20 % ? | "given that sp = rs . 187 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 187 ) / 85 = 20 * 6 = rs . 220 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 220 * 120 ) / 100 = rs . 264 answer : c" | a = 15 / 100
b = 1 - a
c = 187 / b
d = 15 / 100
e = 1 - d
f = 187 / e
g = 20 / 100
h = f * g
i = c + h
|
a ) - 11 , b ) - 5 , c ) 0 , d ) 5 , e ) 11 | c | multiply(negate(multiply(divide(65, 2), 2)), 11) | if 9 a - b = 10 b + 65 = - 12 b - 2 a , what is the value of 11 a + 11 b ? | "( i ) 9 a - 11 b = 65 ( ii ) 2 a + 22 b = - 65 adding ( i ) and ( ii ) : 11 a + 11 b = 0 the answer is c ." | a = 65 / 2
b = a * 2
c = negate * (
|
a ) 40 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 1 and 76 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 + 70 + 75 ) / 15 = 600 / 15 = 40 answer : option a" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
a ) 40 , b ) 60.5 , c ) 52 , d ) 55 , e ) 36 | b | add(subtract(134, multiply(3.5, 22)), 3.5) | a cricketer makes a score of 134 runs in the 22 nd inning and thus increases his average by 3.5 . find his average after 22 nd inning . | explanation : let the average after 22 nd innings = x then average after 21 th innings = ( x - 3.5 ) therefore 21 ( x - 3.5 ) + 134 = 22 x therefore x = 60.5 answer : b | a = 3 * 5
b = 134 - a
c = b + 3
|
a ) 22 % , b ) 24 percent , c ) 25 % , d ) 28 % , e ) 27 % | b | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(30, const_100), subtract(800, 500))), 800), const_100) | for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 30 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman β s commission was approximately what percent of the total amount of the sale ? | "total sales = 800 commission = ( 20 / 100 ) * 500 + ( 30 / 100 ) * 300 = 100 + 90 = 190 % commission = ( 190 / 800 ) * 100 = 23.7 ~ 24 % answer is b" | a = 20 / 100
b = a * 500
c = 30 / 100
d = 800 - 500
e = c * d
f = b + e
g = f / 800
h = g * 100
|
a ) 21 , b ) 20 , c ) 22 , d ) 19 , e ) 24 | b | divide(subtract(subtract(subtract(subtract(135, const_1), const_2), const_3), const_4), 6) | in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 6 rows in all and a total of 135 bricks in the wall , how many bricks does the bottom row contain ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 + x - 5 = 135 6 x - 15 = 135 6 x = 120 x = 20 answer : b" | a = 135 - 1
b = a - 2
c = b - 3
d = c - 4
e = d / 6
|
a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 59 , e ) 52.9 | d | divide(add(multiply(25, 50), multiply(40, 65)), add(25, 40)) | the average marks of a class of 25 students is 50 and that of another class of 40 students is 65 . find the average marks of all the students ? | "sum of the marks for the class of 25 students = 25 * 50 = 1250 sum of the marks for the class of 40 students = 40 * 65 = 2600 sum of the marks for the class of 65 students = 1250 + 2600 = 3850 average marks of all the students = 4850 / 65 = 59 . answer : d" | a = 25 * 50
b = 40 * 65
c = a + b
d = 25 + 40
e = c / d
|
a ) 18 , b ) 750 , c ) 23 , d ) 120 , e ) none of these | c | divide(circle_area(divide(34, multiply(2, const_pi))), 4) | how many plants will be there in a circular bed whose outer edge measure 34 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 34 cm area of circular bed = ( 34 ) 2 Γ’ Β β 4 Γ― β¬ space for each plant = 4 cm 2 Γ’ Λ Β΄ required number of plants = ( 34 ) 2 Γ’ Β β 4 Γ― β¬ Γ£ Β· 4 = 22.98 = 23 ( approx ) answer c" | a = 2 * math.pi
b = 34 / a
c = circle_area / (
|
a ) 550 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(10, const_100))) | a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 10 % ? | "cost price = rs . 500 profit = 10 % of 500 = rs . 50 selling price = cost price + profit = 500 + 50 = 550 answer : a" | a = 10 / 100
b = 500 * a
c = 500 + b
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a ) 5 miles , b ) 8 miles , c ) 6 miles , d ) 13 miles , e ) 12 miles | d | multiply(divide(subtract(30, divide(multiply(add(5, 1), 24), const_60)), add(5, add(5, 1))), 5) | stacy and helon are 30 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour faster than helon ' s constant rate of 5 miles / hour . if helon starts her journey 24 minutes after stacy , how far from the original destination has helon walked when the two meet ? | original distance between s and h = 30 miles . speed of s = 5 + 1 = 6 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 24 / 60 = t + 2 / 5 hours . now , the total distances traveled by s and h = 20 miles - - - > 6 * ( t + 2 / 5 ) + 5 * t = 30 - - - > t = 138 / 55 hours . thus h has traveled for 138 / 55 hours giving you a total distance for h = 5 * 138 / 55 = 13 miles . d is thus the correct answer . p . s . : based on the wording of the question , you should calculatehow far from theoriginal destination has heather walkedwhen the two meet . ' original destination ' for h does not make any sense . original destination for h was situated at a distance of 20 miles . | a = 5 + 1
b = a * 24
c = b / const_60
d = 30 - c
e = 5 + 1
f = 5 + e
g = d / f
h = g * 5
|
a ) 14.05 , b ) 14.02 , c ) 277 , d ) 288 , e ) 222 | a | multiply(0.30103, divide(0.30103, 0.4771)) | if log 2 = 0.30103 and log 3 = 0.4771 , find the number of digits in ( 648 ) 5 | "log ( 648 ) 5 = 5 log ( 648 ) = 5 log ( 81 Γ 8 ) = 5 [ log ( 81 ) + log ( 8 ) ] = 5 [ log ( 34 ) + log ( 23 ) ] = 5 [ 4 log ( 3 ) + 3 log ( 2 ) ] = 5 [ 4 Γ 0.4771 + 3 Γ 0.30103 ] = 5 ( 1.9084 + 0.90309 ) = 5 Γ 2.81149 β 14.05 answer : a" | a = 0 / 30103
b = 0 * 30103
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['a ) 15 : 12', 'b ) 15 : 14', 'c ) 15 : 16', 'd ) 15 : 22', 'e ) none of these'] | c | sqrt(divide(225, 256)) | if the ratio of the areas of two squares is 225 : 256 , then the ratio of their perimeters is : | explanation : a 2 / b 2 = 225 / 256 = 15 / 16 < = > 4 a / 4 b = 4 β 15 / 4 β 16 = 15 / 16 = 15 : 16 option c | a = 225 / 256
b = math.sqrt(a)
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a ) 49 , b ) 30 , c ) 29 , d ) 31 , e ) 32 | b | subtract(12702, multiply(floor(divide(12702, 99)), 99)) | what least number must be subtracted from 12702 to get number exactly 99 ? | "explanation : divide the given number by 99 and find the remainder . if you subtract the remainder from the given number then it is exactly divisible by 99 . 99 ) 12702 ( 128 99 280 198 822 792 30 required number is 30 . answer is b" | a = 12702 / 99
b = math.floor(a)
c = b * 99
d = 12702 - c
|
a ) 48 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 108 | a | add(40, multiply(divide(20, const_100), 40)) | if x is 20 percent greater than 40 , then x = | "x is 20 % greater than 40 means x is 1.2 times 40 ( in other words 40 + 20 / 100 * 40 = 1.2 * 40 ) therefore , x = 1.2 * 40 = 48 answer : a" | a = 20 / 100
b = a * 40
c = 40 + b
|
a ) 770 , b ) 780 , c ) 790 , d ) 800 , e ) 810 | a | multiply(divide(add(14, 56), const_2), divide(add(subtract(56, 14), 2), 2)) | in a theater , the first row has 14 seats and each row has 2 more seats than previous row . if the last row has 56 seats , what is the total number of seats in the theater ? | "the number of seats in the theater is 14 + ( 14 + 2 ) + . . . + ( 14 + 42 ) = 22 ( 14 ) + 2 ( 1 + 2 + . . . + 21 ) = 22 ( 14 ) + 2 ( 21 ) ( 22 ) / 2 = 22 ( 14 + 21 ) = 22 ( 35 ) = 770 the answer is a ." | a = 14 + 56
b = a / 2
c = 56 - 14
d = c + 2
e = d / 2
f = b * e
|
a ) 36 , b ) 96 , c ) 100 , d ) 76 , e ) 72 | b | multiply(multiply(divide(60, subtract(const_1, divide(2, 3))), divide(2, 3)), divide(4, 5)) | a certain automobile company β s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 2 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 60 vehicles that are not speedsters , how many speedster convertibles are there ? | "total vehicle = 2 / 3 of speedster + 1 / 3 of others . speedster convertibles = 2 / 3 total vehicle * 4 / 5 given : 1 / 3 constitutes 60 vehicles . hence 2 / 3 constitutes 120 speedster convertibls = 120 * 4 / 5 = 96 b" | a = 2 / 3
b = 1 - a
c = 60 / b
d = 2 / 3
e = c * d
f = 4 / 5
g = e * f
|
a ) 20 lb , b ) 18 lb , c ) 12 lb , d ) 15 lb , e ) 5 lb | c | divide(12, const_1) | a bag of potatoes weighs 12 lbs divided by half of its weight . how much does the bag of potatoes weight ? | "sol . 12 Γ· 1 = 12 . answer : c" | a = 12 / 1
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a ) 620 , b ) 610 , c ) 630 , d ) 625 , e ) 635 | a | add(multiply(divide(1320, add(60, multiply(75, divide(80, 120)))), 25), multiply(multiply(divide(1320, add(60, multiply(75, divide(80, 120)))), divide(80, 120)), 40)) | the price of 80 apples is equal to that of 120 oranges . the price of 60 apples and 75 oranges together is rs . 1320 . the total price of 25 apples and 40 oranges is | "let the price of one apple = a and price of one orange = b the price of 80 apples is equal to that of 120 oranges 80 a = 120 b = > 2 a = 3 b β b = 2 a / 3 - - - - - ( equation 1 ) price of 60 apples and 75 oranges together is rs . 1320 = > 60 a + 75 b = 1320 = > 4 a + 5 b = 88 β 4 a + 5 ( 2 a ) / 3 = 88 ( β΅ substituted the value of b from equation 1 ) = > 12 a + 10 a = 88 Γ 3 = > 6 a + 5 a = 44 Γ 3 = > 11 a = 44 Γ 3 = > a = 4 Γ 3 = 12 b = 2 a / 3 = ( 2 Γ 12 ) / 3 = 8 total price of 25 apples and 40 oranges = 25 a + 40 b = ( 25 Γ 12 ) + ( 40 Γ 8 ) = 300 + 320 = 620 answer is a ." | a = 80 / 120
b = 75 * a
c = 60 + b
d = 1320 / c
e = d * 25
f = 80 / 120
g = 75 * f
h = 60 + g
i = 1320 / h
j = 80 / 120
k = i * j
l = k * 40
m = e + l
|
a ) 4 : 3 , b ) 3 : 4 , c ) 5 : 6 , d ) 7 : 9 , e ) none | d | divide(7, add(2, 7)) | two vessels a and b contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively . find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ration 8 : 5 ? | let the c . p . of spirit be re . 1 litre . spirit in 1 litre mix . of a = 5 / 7 litre , c . p . of 1 litre mix . in a = re . 5 / 7 spirit in 1 litre mix . of b = 7 / 13 litre , c . p . of 1 litre mix . in b = re . 7 / 13 spirit in 1 litre mix . of c = 8 / 13 litre , mean price = re . 8 / 13 . by the rule of alligation , we have : required ratio = 1 / 13 : 9 / 91 = 7 : 9 . answer d | a = 2 + 7
b = 7 / a
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a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 18 inches , e ) 66 inches | d | divide(add(multiply(7, const_12), 6), 5) | a scale 7 ft . 6 inches long is divided into 5 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 7 * 12 ) + 6 = 90 inches length of each of the 5 parts = 90 / 5 = 18 inches answer : d" | a = 7 * 12
b = a + 6
c = b / 5
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['a ) 18', 'b ) 6', 'c ) 27', 'd ) 48', 'e ) 36'] | e | multiply(multiply(multiply(const_2, const_2), const_3), const_3) | a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of three cubic feet . if a large block has twice the width , twice the depth , and three times the length of a normal block , what is the volume of cheese in a large block in cubic feet ? | volume of cube = lbh = 3 new cube l , b , h are increases of 3 l , 2 b , 2 h new volume of cube = 3 l * 2 b * 2 h = 12 lbh = 12 * 3 = 36 answer : e | a = 2 * 2
b = a * 3
c = b * 3
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a ) $ 2.25 , b ) $ 2.75 , c ) $ 3.00 , d ) $ 3.50 , e ) $ 3.75 | a | divide(multiply(multiply(3, 5), 0.60), const_4) | having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.60 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 60 cents ( 4 / 15 ) x = 60 x = $ 2.25 the answer is a ." | a = 3 * 5
b = a * 0
c = b / 4
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a ) 91 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | a | divide(multiply(14, subtract(14, const_1)), const_2) | there are 14 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "10 players are there . two players play one game with one another . so 14 c 2 = 14 * 13 / 2 = 91 so option a is correct" | a = 14 - 1
b = 14 * a
c = b / 2
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a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | divide(multiply(30, divide(40, const_100)), subtract(divide(80, const_100), divide(60, const_100))) | a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 60 percent of its games that season , what was the total number of games that the team played ? | "60 % is 20 % - points above 40 % and 20 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 1 : 1 . so the team played a total of 30 + 30 = 60 games . the answer is b ." | a = 40 / 100
b = 30 * a
c = 80 / 100
d = 60 / 100
e = c - d
f = b / e
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a ) 2 : 5 , b ) 1 : 3 , c ) 2 : 7 , d ) 3 : 4 , e ) 1 : 5 | c | divide(subtract(4, 2), subtract(11, 4)) | cereal a is 11 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "2 % is 2 % - points below 4 % and 11 % is 7 % - points above 4 % . the ratio of a : b should be 2 : 7 . the answer is c ." | a = 4 - 2
b = 11 - 4
c = a / b
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a ) 30 , b ) 35 , c ) 37 , d ) 41 , e ) 43 | b | add(multiply(3, divide(subtract(10, divide(10, const_2)), subtract(3, divide(4, const_2)))), multiply(4, divide(subtract(10, divide(10, const_2)), subtract(3, divide(4, const_2))))) | 10 years ago a was half of b in age . if the ratio of their present ages is 3 : 4 , what will be the total of their present ages | explanation : let a ' s age 10 years ago = x years . then , b ' s age 10 years ago = 2 x years . ( x + 10 ) / ( 2 x + lo ) = 3 / 4 = > x = 5 . so , the total of their present ages = ( x + 10 + 2 x + 10 ) = ( 3 x + 20 ) = 35 years . answer : option b | a = 10 / 2
b = 10 - a
c = 4 / 2
d = 3 - c
e = b / d
f = 3 * e
g = 10 / 2
h = 10 - g
i = 4 / 2
j = 3 - i
k = h / j
l = 4 * k
m = f + l
|
['a ) 5', 'b ) 6', 'c ) 9', 'd ) 13', 'e ) 28'] | c | divide(126, divide(add(negate(4), sqrt(add(power(4, const_2), multiply(4, multiply(126, 2))))), const_2)) | a rectangular tiled patio is composed of 126 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 126 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 126 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 126 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 126 ( 2 r - 4 ) * r = 126 2 r ^ 2 - 4 r - 126 = 0 r can not be negative so r = 9 and c = 14 so , answer will be c | a = negate + (
b = 4 ** 2
c = 126 * 2
d = 4 * c
e = b + d
f = math.sqrt(e)
g = a / f
h = 126 / g
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | subtract(39, subtract(add(26, 20), 17)) | in a class of 39 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 39 - 29 = 10 play neither answer is c" | a = 26 + 20
b = a - 17
c = 39 - b
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a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 2 | e | divide(divide(18, const_3), const_3) | if a * b denotes the greatest common divisor of a and b , then ( ( 12 * 16 ) * ( 18 * 12 ) ) = ? | "the greatest common divisor of 12 and 16 is 4 . hence 12 * 16 = 4 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 12 is 6 . hence 18 * 12 = 6 . hence ( ( 12 * 16 ) * ( 18 * 12 ) ) = 4 * 6 . the greatest common divisor of 4 and 6 is 2 . answer ; e ." | a = 18 / 3
b = a / 3
|
a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | a | divide(multiply(24, 15), 18) | if 24 men take 15 days to to complete a job , in how many days can 18 men finish that work ? | ans . 20 days | a = 24 * 15
b = a / 18
|
a ) 1200 ft , b ) 800 ft , c ) 900 ft , d ) 1000 ft , e ) 1500 ft | a | multiply(40, add(divide(multiply(40, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | the circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long . what is the distance travelled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | "total distance - x x / 40 - x / 48 = 5 x = 1200 ft answer a" | a = 10 / 2
b = 40 * a
c = b / 3
d = 10 / 2
e = c + d
f = 40 * e
|
a ) 110 kmph , b ) 120 kmph , c ) 108 kmph , d ) 100 kmph , e ) 98 kmph | c | multiply(divide(600, subtract(30, 10)), const_3_6) | a train requires 10 seconds to pass a pole while it requires 30 seconds to cross a stationary train which is 600 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 30 sec the train crosses one more stationary train in 20 sec the train travels a distance of 600 mtrs speed = 600 / 20 = 30 m / s = 30 ( 3600 / 1000 ) = 30 * 18 / 5 = 108 kmph answer : c" | a = 30 - 10
b = 600 / a
c = b * const_3_6
|
a ) 76 , b ) 80 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 36)), multiply(10, 36)) | the average of runs of a cricket player of 10 innings was 36 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 36 so , total = average x no . of innings = 36 x 10 = 360 . now increase in avg = 4 runs . so , new avg = 36 + 4 = 40 runs total runs = new avg x new no . of innings = 40 x 11 = 440 runs made in the 11 th inning = 440 - 360 = 80 answer : b" | a = 10 + 1
b = 4 + 36
c = a * b
d = 10 * 36
e = c - d
|
a ) 78 % , b ) 66 % , c ) 62 % , d ) 64 % , e ) 60 % | b | multiply(divide(subtract(800, add(add(add(62, 62), add(60, 48)), 40)), 800), const_100) | john had a stock of 800 books in his bookshop . he sold 62 on monday , 62 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 62 + 62 + 60 + 48 + 40 = 272 let m be the books not sold m = 800 - n = 1400 - 272 = 528 percentage books not sold / total number of books = 528 / 800 = 0.66 = 66 % correct answer b" | a = 62 + 62
b = 60 + 48
c = a + b
d = c + 40
e = 800 - d
f = e / 800
g = f * 100
|
a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 15,000 , e ) e ) 11,100 | d | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 180,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 180 k x = 15,000 answer : d" | a = 4 * 2
b = a * 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(const_3, const_3) | a palindrome is a number that reads the same front - to - back as it does back - to - front ( e . g . 202 , 575 , 1991 , etc . ) p is the smallest integer greater than 100 that is both a prime and a palindrome . what is the sum of the digits of p ? | "given that p is smallest integer greater than 200 - assume there is a 3 - digit that satisfies the above conditions . let the number be xyx ; question asks us the values of 2 x + y we can straight away cross out options a ) and d ) - sum of digits 3 or 6 implies it is divisible by 3 - - - > we know that p is a prime number coming to option b ) 2 x + y = 4 - - > only x = 2 and y = 0 satisfy this equation ( x > 2 will never give sum of digits = 4 ) ; but 202 is divisible by 2 ; we know that p is a prime number similarly option c ) 2 x + y = 5 - - > only x = 2 and y = 1 satisfy this equation ( x > 2 will never give sum of digits = 5 ) ; but 212 is divisible by 2 ; we know that p is a prime number therefore answer option should be e - - - > can be verified by taking 2 x + y = 7 - - - > x = 3 and y = 1 ; gives 313 c" | a = 3 + 3
|
a ) a ) 6 , b ) b ) 8 , c ) c ) 10 , d ) d ) 12 , e ) e ) 15 | a | divide(divide(6, subtract(divide(40, const_60), divide(20, const_60))), const_3) | circular gears l and r start to rotate at the same time at the same rate . gear l makes 20 complete revolutions per minute and gear r makes 40 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 6 more revolutions than gear l ? | gear l - - 20 rotations per 60 seconds - - 2 rotation per 6 seconds . gear r - - 40 rotations per 60 seconds - - 4 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 2 rotations hence every 6 seconds the difference between the number of rotations of r and l gear is 2 units . required net difference should be 6 rotations = > 3 ( 6 seconds later ) = = > 18 seconds . answer : a | a = 40 / const_60
b = 20 / const_60
c = a - b
d = 6 / c
e = d / 3
|
a ) 12 , b ) 3 , c ) 6 , d ) 9 , e ) 10 | b | add(multiply(4, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 4), divide(const_1, 12)), const_2)), 4), const_60)) | two pipes a and b can fill a tank in 4 and 12 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 14 part filled by b in 1 min . = 1 / 12 part filled by ( a + b ) in 1 min . = 1 / 4 + 1 / 12 = 1 / 3 . both the pipes can fill the tank in 3 minutes . answer : b" | a = 4 * 100
b = 1 / 4
c = 1 / 12
d = b + c
e = d * 2
f = 1 - e
g = f * 4
h = g * const_60
i = a + h
|
a ) 130 , b ) 132 , c ) 134 , d ) 136 , e ) 138 | c | divide(1206, subtract(43, 34)) | a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1206 . find the number to be multiplied . | "let the required number be x . then , 43 x β 34 x = 1206 or 9 x = 1206 or x = 134 . required number = 134 . answer : c" | a = 43 - 34
b = 1206 / a
|
a ) 25 , b ) 66 , c ) 18 , d ) 37 , e ) 01 | d | divide(multiply(divide(multiply(18.5, 50), const_100), const_100), 25) | a company pays 18.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 18.5 * 50 ) / 100 = rs . 9.25 rs . 25 is income on an investment of rs . 100 rs . 9.25 is income on an investment of rs . ( 9.25 * 100 ) / 25 = rs . 37 answer : d" | a = 18 * 5
b = a / 100
c = b * 100
d = c / 25
|
a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 12.2 , e ) 12.1 | e | divide(multiply(20, 1000), add(1000, 650)) | 1000 men have provisions for 20 days . if 650 more men join them , for how many days will the provisions last now ? | "1000 * 20 = 1650 * x x = 12.1 answer : e" | a = 20 * 1000
b = 1000 + 650
c = a / b
|
a ) 8000 , b ) 8500 , c ) 9000 , d ) 9500 , e ) 100000 | e | divide(multiply(90000, const_100), subtract(const_100, 10)) | david ' s bank ' s saving amount is decreased 10 % due to loan payment and current balance is rs . 90000 . find the actual balance before deduction ? | 10 % decreased 90 % balance = 90000 100 % = 90000 / 90 * 100 = 100000 answer : e | a = 90000 * 100
b = 100 - 10
c = a / b
|
a ) s . 800 , b ) s . 2400 , c ) s . 4000 , d ) s . 5500 , e ) s . 4200 | d | multiply(subtract(multiply(divide(2200, 2), 3), 2200), 5) | the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 2200 , then what is the income of p 1 ? | "let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x β 3 y = 2200 β¦ ( i ) and 4 x β 2 y = 2200 β¦ β¦ . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 2200 - > x = 1100 p 1 β s income = rs 5 * 1100 = rs . 5500 answer : d" | a = 2200 / 2
b = a * 3
c = b - 2200
d = c * 5
|
a ) 81 , b ) 81.5 , c ) 82 , d ) 84.5 , e ) none of these | b | divide(add(multiply(83, subtract(79, 76)), add(multiply(85, subtract(81, 76)), multiply(76, subtract(83, 79)))), add(subtract(81, 76), add(subtract(83, 79), subtract(79, 76)))) | 3 math classes : x , y , and z , take an algebra test . the average score in class x is 83 . the average score in class y is 76 . the average score in class z is 85 . the average score of all students in classes x and y together is 79 . the average score of all students in classes y and z together is 81 . what is the average score for all the 3 classes , taken together ? | explanation : let the number of students in classes x , y and z be a , b and c respectively . then total of x = 83 a total of y = 76 b total of z = 85 c and , ( 83 a + 76 b ) / ( a + b ) = 79 . i . e 4 a = 3 b . also , ( 76 b + 85 c ) / ( b + c ) = 81 . i . e 4 c = 5 b . hence , b = ( 4 / 3 ) a , c = ( 5 / 4 ) b = ( 5 / 4 ) x ( 4 / 3 ) a = ( 5 / 3 ) a . average of x , y and z = ( 83 a + 76 b + 85 c ) / ( a + b + c ) . = > 83 a + 76 β ( 4 / 3 ) a + 85 β ( 5 / 3 ) a / a + ( 4 / 3 ) a + ( 5 / 3 ) a . = > 978 / 12 = > 81.5 answer : b | a = 79 - 76
b = 83 * a
c = 81 - 76
d = 85 * c
e = 83 - 79
f = 76 * e
g = d + f
h = b + g
i = 81 - 76
j = 83 - 79
k = 79 - 76
l = j + k
m = i + l
n = h / m
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 14), const_2))), divide(const_1, 14)) | a can finish a piece of work in 4 days . b can do it in 14 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 14 = 1 = > x = 5 days answer : b" | a = 1 / 4
b = a * 2
c = 1 / 14
d = c * 2
e = b + d
f = 1 - e
g = 1 / 14
h = f / g
|
a ) 4.37 % , b ) 7.3 % , c ) 7.6 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 34,000 to 2 , 32,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 232500 - 134000 ) = 98500 . increase % = ( 98500 / 134000 x 100 ) % = 73 % . required average = ( 73 / 10 ) % = 7.3 % . answer : option b" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | subtract(divide(30, const_2), 1) | if r is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of r ? | answer is c . numbers in format of 3 ^ k in the series from 1 to 30 inclusive r are : 3 * 1 , 3 * 2 , 3 * 3 , 3 * 4 , 3 * 5 , 3 * 3 * 2 , 3 * 7 , 3 * 8 , 3 * 3 * 3 , 3 * 10 . total number of 3 = 13 . so k = 14 . | a = 30 / 2
b = a - 1
|
a ) 33 , b ) 30 , c ) 36 , d ) 28 , e ) 26 | a | divide(subtract(460, multiply(8, 41)), const_4) | rs . 460 was divided among 41 boys and girls such that each boy rs . 12 and each girl got rs . 8 . what is the number of boys ? | explanation : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - assume that the number of boys = b and number of girls is g number of boys + number of girls = 41 = > b + g = 41 - - - - - - - - - - - - ( equation 1 ) given that each boy got rs . 12 and each girl got rs . 8 and total amount = rs . 460 = > 12 b + 8 g = 460 - - - - - - - - ( equation 2 ) now we need solve equation 1 and equation 2 to get b and g ( equation 1 ) Γ 8 = > 8 b + 8 g = 8 Γ 41 = 328 β β β β β β β β ( equation 3 ) ( equation 2 ) - ( equation 3 ) = 4 b = 460 - 328 = 132 = > b = 1324 = 33 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - given that amount received by a boy = rs . 12 and amount received by a girl = rs . 8 total amount = 460 given that number of boys + number of girls = 41 hence mean amount = 460 / 41 by the rule of alligation , we have amount received by a boy amount received by a girl 12 8 mean amount 460 / 41 460 / 41 - 8 = 132 / 41 12 - 460 / 41 = 32 / 41 number of boys : number of girls = 132 / 41 : 32 / 41 = 132 : 32 = 66 : 16 = 33 : 8 given that number of boys + number of girls = 41 hence number of boys = 41 Γ 3341 = 33 answer : option a | a = 8 * 41
b = 460 - a
c = b / 4
|
a ) 12 , b ) 27 , c ) 29 , d ) 55 , e ) 95 | e | add(divide(18000, 200), 5) | a shopkeeper sells 200 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 200 = rs . 90 loss per metre = rs . 5 cp per metre = 90 + 5 = rs . 95 . answer : e | a = 18000 / 200
b = a + 5
|
a ) 29 , b ) 27 , c ) 28 , d ) 24 , e ) 21 | a | add(add(multiply(2, 11), 5), 2) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 . | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 2 = 2 ^ 22 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer is a ." | a = 2 * 11
b = a + 5
c = b + 2
|
a ) 3.0 , b ) 5.0 , c ) 5.5 , d ) 6.5 , e ) 6.75 | e | multiply(45, divide(15, 100)) | a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . if 45 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | "construct an equation : 15 / 100 = x / 45 - > x = 6.75 answer : e ." | a = 15 / 100
b = 45 * a
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a ) 22 , b ) 12 , c ) 7.5 , d ) 99 , e ) 21 | c | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 5))) | two trains of equal lengths take 5 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 5 = 24 m / sec . speed of the second train = 120 / 15 = 8 m / sec . relative speed = 24 + 8 = 32 m / sec . required time = ( 120 + 120 ) / 32 = 7.5 sec . answer : c" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 6 , b ) 6.6 , c ) 8.8 , d ) 100 , e ) 110 | c | divide(880, divide(multiply(multiply(10, 880), divide(add(const_100, 10), const_100)), subtract(multiply(880, divide(add(const_100, 10), const_100)), 880))) | machine a and machine b are each used to manufacture 880 sprockets . it takes machine a 10 hours longer to produce 880 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 880 qt = 800 for a : ( t + 10 ) ( q ) = 880 qt + 10 q = 880 800 + 10 q = 880 q = 8 so a can produce 8 / hour . then b can produce = 8 ( 1.1 ) = 8.8 / hour . c" | a = 10 * 880
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 880 * f
h = g - 880
i = d / h
j = 880 / i
|
a ) 160 , b ) 150 , c ) 100 , d ) 66 , e ) 50 | d | divide(subtract(multiply(204, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 204 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 204 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 204 - a gallons of fuel b is 0.16 ( 204 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 204 - a ) = 30 - - > a = 66 . answer : d ." | a = 16 / 100
b = 204 * a
c = b - 30
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) 4676 , b ) 4678 , c ) 9950 , d ) 9504 , e ) 9936 | c | multiply(floor(divide(power(const_10, 4), 50)), 50) | what is the largest 4 digit number exactly divisible by 50 ? | "largest 4 digit number = 9999 9999 Γ· 50 = 199 , remainder = 49 hence largest 4 digit number exactly divisible by 50 = 9999 - 49 = 9950 answer : c" | a = 10 ** 4
b = a / 50
c = math.floor(b)
d = c * 50
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind poe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of poe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . poe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena can cover 7.5 miles in 30 mins teena will cover 15 miles in 60 mins so answer 30 + 60 = 90 mins = d" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | d | divide(multiply(subtract(1512, 1400), const_100), 1400) | the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1512 respectively . what is the rate of interest ? | "1512 - 1400 = 112 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 112 ) / ( 1400 ) = 8 % the answer is d ." | a = 1512 - 1400
b = a * 100
c = b / 1400
|
a ) 86 , b ) 90 , c ) 92 , d ) 94 , e ) 96 | a | add(multiply(add(multiply(6, const_3), 2), divide(add(multiply(6, const_3), 2), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 2 = 20 5 * quotient = 20 quotient = 4 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 4 ) + 6 = 86 . a" | a = 6 * 3
b = a + 2
c = 6 * 3
d = c + 2
e = d / 5
f = b * e
g = f + 6
|
a ) 5 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | a | divide(multiply(44, 432), 31) | a number when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 31 | "explanation : p Γ· 44 = 432 = > p = 432 * 44 = 19008 p / 31 = 19008 / 31 = 613 , remainder = 5 option a" | a = 44 * 432
b = a / 31
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a ) $ 190 , b ) $ 180 , c ) $ 200 , d ) $ 240 , e ) $ 250 | e | add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(160, 120), 120), 120), 3)), 120)) | if $ 120 invested at a certain rate of simple interest amounts to $ 160 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ? | "120 amounts to 160 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 160 120 + 120 * ( r / 100 ) * ( 3 ) = 160 = > r = 100 / 9 150 in 6 years = principal + interest = 150 + 150 * ( r / 100 ) * ( 6 ) 250 answer is e ." | a = 150 * 6
b = 160 - 120
c = b * 120
d = c / 120
e = d / 3
f = a * e
g = f / 120
h = 150 + g
|
a ) 4.5 % , b ) 5.5 % , c ) 6.5 % , d ) 8.75 % , e ) none | a | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(1, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(1, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 33,400 to 1 , 93,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 193500 - 133400 ) = 60100 . increase % = ( 60100 / 133400 x 100 ) % = 45 % . required average = ( 45 / 10 ) % = 4.5 % . answer : option a" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 1 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 1 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) a ) 1200000 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 500000 , e ) e ) 652000 | a | multiply(divide(60000, subtract(const_100, add(add(multiply(20, 3), 30), 5))), const_100) | a person distributed 20 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 60000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 30 % orphan house = 5 % total = 60 + 30 + 5 = 95 % remaining = 100 - 95 = 5 % 5 % = 60000 100 % = 60000 * 100 / 5 = $ 1200000 answer is a" | a = 20 * 3
b = a + 30
c = b + 5
d = 100 - c
e = 60000 / d
f = e * 100
|
a ) none , b ) one , c ) two , d ) three , e ) four | c | subtract(15, multiply(3, const_4)) | a = 5 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "a = 5 ^ 15 - 625 ^ 3 = > 5 ^ 15 - ( 5 ^ 4 ) ^ 3 = > 5 ^ 15 - 5 ^ 12 = 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 * 124 124 = 31 * 4 a / x is integer for condition of 2 < p < x only 5 and 31 satisfies this hence answer is c" | a = 3 * 4
b = 15 - a
|
a ) 60 , b ) 50 , c ) 40 , d ) 70 , e ) 65 | c | divide(360, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 360 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 360 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 360 / 9 = 40 kmph answer c ." | a = 6 * 3
b = a / 2
c = 360 / b
|
a ) 1 / 33 , b ) 2 / 33 , c ) 1 / 3 , d ) 16 / 33 , e ) 47 / 84 | e | multiply(multiply(multiply(divide(multiply(7, const_2), multiply(7, const_2)), divide(multiply(4, 4), subtract(multiply(7, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | if 4 people are selected from a group of 7 married couples , what is the probability that none of them would be married to each other ? | "if we are to select 4 people from 7 couples without any restriction , how many ways can we make the selection ? 14 ! / 4 ! 10 ! = 1001 if we are to select 4 people from 7 couples with restriction that no married couple can both make it to the group , only a representative ? 7 ! / 4 ! 3 ! = 35 but we know that to select a person from each couple , take 2 possibilities 35 * 2 * 2 * 2 * 2 = 560 probability = desired / all possibilities = 560 / 1001 = 47 / 84 answer : e" | a = 7 * 2
b = 7 * 2
c = a / b
d = 4 * 4
e = 7 * 2
f = e - 1
g = d / f
h = c * g
i = 4 * 4
j = i - 2
k = 4 * 4
l = j / k
m = h * l
n = 4 * 4
o = n - 2
p = o - 2
q = 4 * 4
r = q - 1
s = p / r
t = m * s
|
a ) 27.5 % , b ) 25.6 % , c ) 31.5 % , d ) 35.9 % , e ) 29.5 % | a | divide(multiply(subtract(add(multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 25), const_100)), multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 30), const_100))), 100), const_100), 100) | a shopkeeper has 100 kg of apples . he sells 50 % of these at 25 % profit and remaining 50 % at 30 % profit . find his % profit on total . | "total number of apples = 100 let the cost price be x selling price at 25 % profit = 1.25 x selling price at 30 % profit = 1.3 x profit % = ( ( sp - cp ) / cp ) * 100 profit % = ( ( 1 / 2 ) * 100 * 1.25 x + ( 1 / 2 ) * 100 * 1.3 x - 100 x ) / 100 x * 100 = ( 255 - 200 ) / 2 = 27.5 % answer is a" | a = 100 * 50
b = a / 100
c = 100 + 25
d = c / 100
e = b * d
f = 100 * 50
g = f / 100
h = 100 + 30
i = h / 100
j = g * i
k = e + j
l = k - 100
m = l * 100
n = m / 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(multiply(multiply(multiply(const_4, 4), 2), 1), multiply(multiply(4, 2), 1)) | a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β high β and β low β cards to earn 6 points if you must draw exactly 4 β low β cards ? | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 3 times . as a result , the formula is : 5 ! / 4 ! which represents ( 5 * 4 * 3 * 2 * 1 ) / ( 4 * 3 * 2 * 1 ) which simplifies to just 5 , giving you answer e ." | a = 4 * 4
b = a * 2
c = b * 1
d = 4 * 2
e = d * 1
f = c / e
|
a ) 3.09 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(970, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 970 gm . for a kg . his gain is β¦ % . | "his percentage gain is 100 * 30 / 970 as he is gaining 30 units for his purchase of 970 units . so 3.09 % . answer : a" | a = 4 + 1
b = a * 2
c = b * 100
d = 970 / c
e = 1/(d)
f = e - 1
g = f * 100
|
a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | c | divide(multiply(const_12, log(2)), log(2)) | if 2 ^ ( 2 w ) = 8 ^ ( w β 6 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w β 6 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w β 6 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 18 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 18 w = 18 the answer is c ." | a = math.log(2)
b = 12 * a
c = math.log(2)
d = b / c
|
a ) 12.45 mph , b ) 11.25 mph , c ) 10.95 mph , d ) 10.91 mph , e ) 10.56 mph | d | divide(20, add(divide(10, subtract(20, 10)), divide(10, 12))) | tom traveled the entire 20 miles trip . if he did the first 10 miles of at a constant rate 12 miles per hour and the remaining trip of at a constant rate 10 miles per hour , what is the his average speed , in miles per hour ? | avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 10 + 10 ) / ( ( 10 / 12 ) + ( 10 / 10 ) = 120 / 11 = 10.91 mph d | a = 20 - 10
b = 10 / a
c = 10 / 12
d = b + c
e = 20 / d
|
a ) a ) 2 , b ) b ) 6 , c ) c ) 8 , d ) d ) 15 , e ) e ) 20 | d | multiply(3, 5) | a , b are positive integers . the remainder of a to be divided by 8 is 3 and the remainder of b to be divided by 6 is 5 . which is possible to be the remainder of a * b to be divided by 48 | two ways to do it . . . a = 8 x + 3 . . b = 6 y + 5 . . 1 ) convenient way . . take x and y as 0 , and you will get a * b as 3 * 5 = 15 answer : d | a = 3 * 5
|
a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 7.2 , e ) 9 | e | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.25))), 1.25) | one pipe can fill a pool 1.25 times faster than a second pipe . when both pipes are opened , they fill the pool in five hours . how long would it take to fill the pool if only the faster pipe is used ? | say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.25 r = 5 r / 4 . since when both pipes are opened , they fill the pool in five hours , then their combined rate is 1 / 5 pool / hour . thus we have that r + 5 r / 4 = 1 / 5 - - > r = 4 / 45 pool / hour , faster pipe fills at 1.25 r which is 5 / 4 * 4 / 45 = 1 / 9 - - > time is reciprocal of rate thus it ' s 9 / 1 = 9 hours . answer : e . | a = 3 + 2
b = 1/(a)
c = 1 + 1
d = b / c
e = 1/(d)
f = e / 1
|
a ) 10 , b ) 15 , c ) 20 , d ) 30 , e ) 35 | c | sqrt(multiply(4, const_100)) | the revenue from sales items in 1996 increases by x percent , compared in 1995 and the revenue in 1997 decreases by x percent , compared in 1996 . if the revenue from sales items in 1997 decreases by 4 percent , compared in 1995 , what is the value of x ? | 1995 - let the value be z 1996 - x % increase - z ( 1 + x / 100 ) 1997 - x % decrease - z ( 1 + x / 100 ) ( 1 - x / 100 ) from 95 to 97 decrease is 4 % hence , [ z - { z ( 1 + x / 100 ) ( 1 - x / 100 ) } ] / z = 4 / 100 solving x = 20 % c is the answer | a = 4 * 100
b = math.sqrt(a)
|
a ) $ 642986 , b ) $ 642987 , c ) $ 642988 , d ) $ 642989 , e ) $ 642990 | b | add(642986, divide(9, 9)) | peter has $ 642986 in his savings account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among his 9 children ? | to find the least amount the man should add to his saving account to split the money evenly among his 9 children , he needs to make the total divisible by 9 simply add the individual digits of the total = 6 + 4 + 2 + 9 + 8 + 6 = 35 if you add 1 , the number is divisible by 9 ( 35 + 1 ) correct option : b | a = 9 / 9
b = 642986 + a
|
a ) 11.12 , b ) 10.11 , c ) 72 , d ) 10.66 , e ) 9.2 | d | multiply(divide(const_4, 3), power(3, 3)) | the measurement of a rectangular box with lid is 25 cmx 4 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of Ο cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 4 , r = 2 ; volume of the largest sphere = 4 / 3 Ο r 3 = 4 / 3 * Ο * 2 * 2 * 2 = 10.66 Ο cm 3 answer : d" | a = 4 / 3
b = 3 ** 3
c = a * b
|
a ) 425 miles , b ) 625 miles , c ) 300 miles , d ) 225 miles , e ) 625 miles | c | multiply(60, 5) | a car travels at a speed of 60 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 60 + 60 + 60 + 60 + 60 = 5 * 65 = 300 miles correct answer c" | a = 60 * 5
|
a ) 12 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | a | divide(multiply(72, divide(multiply(10, const_1000), const_60)), const_1000) | find the distance covered by a man walking for 72 min at a speed of 10 km / hr ? | "distance = 10 * 72 / 60 = 12 km answer is a" | a = 10 * 1000
b = a / const_60
c = 72 * b
d = c / 1000
|
a ) 276 , b ) 299 , c ) 312 , d ) 322 , e ) none | d | multiply(23, 14) | the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is | "solution clearly , the numbers are ( 23 x 13 ) and ( 23 x 14 ) . larger number = ( 23 x 14 ) = 322 . answer d" | a = 23 * 14
|
a ) 0.5 % , b ) 0.2 % , c ) 1.5 % , d ) 2 % , e ) 3 % | b | multiply(divide(2, 1), const_100) | what percent is 2 gm of 1 kg ? | "1 kg = 1000 gm 2 / 1000 Γ 100 = 200 / 1000 = 1 / 5 = 0.2 % b )" | a = 2 / 1
b = a * 100
|
a ) 71.11 , b ) 71.12 , c ) 72.4 , d ) 71.17 , e ) 71.13 | c | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 70)))) | a car travels first 160 km at 75 km / hr and the next 160 km at 70 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 70 km / hr time taken to travel next 160 km = distancespeed = 160 / 70 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 160 / 75 + 160 / 70 average speed = total distance traveled / total time taken = 320 / ( 160 / 75 + 160 / 70 ) = 72.4 km / hr answer : c" | a = 160 / 75
b = 160 / 70
c = a + b
d = 1 / c
e = 320 * d
|
a ) 62 m , b ) 54 m , c ) 50 m , d ) 55 m , e ) 56 m | c | multiply(9, subtract(subtract(multiply(divide(multiply(4, const_1000), const_3600), 10), multiply(divide(multiply(2, const_1000), const_3600), 9)), divide(multiply(2, const_1000), const_3600))) | a train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . what is the length of the train ? | explanation : let x is the length of the train in meter and v is its speed in kmph x / 9 = ( v - 2 ) ( 10 / 36 ) - - - ( 1 ) x / 10 = ( v - 4 ) ( 10 / 36 ) - - - ( 2 ) dividing equation 1 with equation 2 10 / 9 = ( v - 2 ) / ( v - 4 ) = > 10 v - 40 = 9 v - 18 = > v = 22 substituting in equation 1 , x / 9 = 200 / 36 = > x = 9 Γ 200 / 36 = 50 m answer : option c | a = 4 * 1000
b = a / 3600
c = b * 10
d = 2 * 1000
e = d / 3600
f = e * 9
g = c - f
h = 2 * 1000
i = h / 3600
j = g - i
k = 9 * j
|
a ) 7.6 , b ) 8.7 , c ) 9.1 , d ) 10.2 , e ) 11.6 | d | divide(multiply(15, 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 15 percent of 68 ? | "( 15 / 100 ) * 68 = 10.2 the answer is d ." | 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 = 15 * n
p = o / 100
|
a ) a ) 10.5 , b ) b ) 12 , c ) c ) 15 , d ) d ) 18 , e ) e ) 20 | a | divide(subtract(25, power(2, 2)), 2) | if a - b = 2 and a 2 + b 2 = 25 , find the value of ab . | "explanation : 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 25 - 4 = 21 ab = 10.5 answer : a" | a = 2 ** 2
b = 25 - a
c = b / 2
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | c | sqrt(divide(98, const_2)) | the area of a parallelogram is 98 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 98 = > x = 7 answer : c" | a = 98 / 2
b = math.sqrt(a)
|
a ) 40 , b ) 72 , c ) 84 , d ) 90 , e ) 108 | a | multiply(subtract(divide(multiply(const_2, const_2), subtract(8, multiply(const_2, 2))), divide(const_2, subtract(8, 2))), const_60) | tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 2 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour . if both tom and linda travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes tom to cover half of the distance that linda has covered and the amount of time it takes tom to cover twice the distance that linda has covered ? | "a is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 2 t ) = 8 t = = > t = 1 / 3 = = > 20 minutes to travel double distance , 2 ( 2 + 2 t ) = 8 t = = > 1 = = > 60 minutes difference , 40 minutes a" | a = 2 * 2
b = 2 * 2
c = 8 - b
d = a / c
e = 8 - 2
f = 2 / e
g = d - f
h = g * const_60
|
a ) 240 , b ) 388 , c ) 379 , d ) 277 , e ) 320 | e | multiply(divide(840, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8)))) | a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 840 find the share of a . | "explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 / 21 * 840 = 320 answer : e" | a = 3000 * 8
b = 3000 - 1000
c = 12 - 8
d = b * c
e = a + d
f = 4000 * 8
g = 4000 + 1000
h = 12 - 8
i = g * h
j = f + i
k = e + j
l = 840 / k
m = 3000 * 8
n = 3000 - 1000
o = 12 - 8
p = n * o
q = m + p
r = l * q
|
a ) 1 / 6 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 / 3 , e ) 2 / 3 | e | subtract(1, multiply(divide(const_2, add(const_2, const_3)), divide(add(const_2, const_3), add(add(const_2, const_3), const_1)))) | set # 1 = { a , b , o , d , e } set # 2 = { k , l , m , n , u , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? | at least questions are best solved by taking the opposite scenario and subtracting it from 1 . probability of choosing no vowel from set 1 is 2 / 5 and set 2 is 5 / 6 . multiply these to get 1 / 3 . therefore , probability of picking at least one vowel = 1 - 1 / 3 = 2 / 3 . answer = e | a = 2 + 3
b = 2 / a
c = 2 + 3
d = 2 + 3
e = d + 1
f = c / e
g = b * f
h = 1 - g
|
a ) a ) 36 , b ) b ) 12 , c ) c ) 18 , d ) d ) 64 , e ) e ) 10 | b | add(multiply(3, 3), 3) | the value of x + ( xx ) when x = 3 is : | x + ( xx ) put the value of x = 2 in the above expression we get , 3 + ( 33 ) = 3 + ( 3 Γ 3 ) = 3 + ( 9 ) = 3 + 9 = 12 b | a = 3 * 3
b = a + 3
|
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