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 ) 15 % , b ) 16 1 / 6 % , c ) 16 2 / 3 % , d ) 20 % , e ) 30 % | c | subtract(const_100, multiply(divide(5, const_4), const_100)) | on an order of 5 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 6 dozens free . equivalent discount = 1 / 6 * 100 = 16 2 / 3 % . answer : c" | a = 5 / 4
b = a * 100
c = 100 - b
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a ) 10 , b ) 12 , c ) 18 , d ) 19 , e ) 20 | c | add(17, 1) | seller selling an apple for rs . 17 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 17 loss = cp 18 loss = cp β sp = cp β 17 β cp 18 = cp β 17 β 17 cp 18 = 17 β cp 18 = 1 β cp = 18 c" | a = 17 + 1
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a ) 25 , b ) 26 , c ) 30 , d ) 24 , e ) 28 | e | divide(multiply(16, 7), const_4) | there are two positive numbers in the ratio 7 : 11 . if the larger number exceeds the smaller by 16 , then find the smaller number ? | let the two positive numbers be 7 x and 11 x respectively . 11 x - 7 x = 16 4 x = 16 = > x = 4 = > smaller number = 7 x = 28 . answer : e | a = 16 * 7
b = a / 4
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a ) 10 , b ) 12.5 , c ) 25 , d ) 12 , e ) 15 | e | multiply(divide(120, multiply(400, 2)), const_100) | what is rate of interest if principal . amount be 400 , simple interest 120 and time 2 year . | "s . i = ( p * r * t ) / 100 120 = 800 r / 100 r = 120 / 8 = 15 % answer e" | a = 400 * 2
b = 120 / a
c = b * 100
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a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(15, const_2), choose(15, const_3)), choose(15, const_4)), choose(15, 15)), power(const_2, 15)) | a fair coin is tossed 15 times . what is the probability of getting more heads than tails in 15 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ." | a = math.comb(15, 2)
b = math.comb(15, 3)
c = a + b
d = math.comb(15, 4)
e = c + d
f = math.comb(15, 15)
g = e + f
h = 2 ** 15
i = g / h
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a ) 8 , b ) 9 , c ) 11 , d ) 7 , e ) 10 | e | subtract(subtract(multiply(6, 3), 6), const_2) | a bowler can take max . 3 wickets in a over . if he bowls 6 overs in an innings , how many maximum wickets can he take ? | 10 because after 10 wickets , the innings is complete . answer : e | a = 6 * 3
b = a - 6
c = b - 2
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a ) $ 3500 , b ) $ 5000 , c ) $ 4240 , d ) $ 7000 , e ) $ 10000 | c | multiply(divide(42400, const_100), subtract(42, 32)) | if the personal income tax rate is lowered from 42 % to 32 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 42400 ? | "saving = ( 42 - 32 ) % of 42400 = 4240 . answer : c" | a = 42400 / 100
b = 42 - 32
c = a * b
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a ) 256 , b ) 4 , c ) β 2 , d ) 16 , e ) none | a | power(add(power(divide(5568, 87), divide(1, 3)), power(multiply(72, 2), divide(1, 2))), 2) | ( 5568 / 87 ) 1 / 3 + ( 72 x 2 ) 1 / 2 = ( ? ) 1 / 2 ? | answer ? ) 1 / 2 = ( 5568 / 87 ) 1 / 3 + ( 72 x 2 ) 1 / 2 = ( 64 ) 1 / 3 + ( 144 ) 1 / 2 β΄ ? = ( 4 + 12 ) 2 = 256 correct option : a | a = 5568 / 87
b = 1 / 3
c = a ** b
d = 72 * 2
e = 1 / 2
f = d ** e
g = c + f
h = g ** 2
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 33 % , e ) 40 % | b | multiply(subtract(divide(add(const_100, 50), add(const_100, 20)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 20 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 50 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | say price at the beginning of year = 100 end of 1 st quarter = 100 + 20 = 120 end of 2 nd quarter = 100 + 50 = 150 percentage increase between 1 st & 2 nd quarter = 150 β 120 / 120 β 100 = 25 answer = b | a = 100 + 50
b = 100 + 20
c = a / b
d = c - 1
e = d * 100
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a ) s . 50 , b ) s . 200 , c ) s . 100 , d ) s . 80 , e ) s . 60 | b | multiply(800, divide(25, const_100)) | find the 25 % of rs . 800 . | "explanation : 25 % of 800 = > 25 / 100 * 800 = rs . 200 answer : b" | a = 25 / 100
b = 800 * a
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a ) 6 , b ) 7 , c ) 9 , d ) 11 , e ) 4 | e | divide(48, 88) | what is the 50 th digit to the right of the decimal point in the decimal form of 48 / 88 ? | "we have to use some long division . this long division allows us to get 48 / 88 in decimal form , which is 0.545454 β¦ where β 54 β is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of the decimal point is a 4 . in other words , each odd - positioned digit is a 5 , and each even - positioned digit is a 4 . then about the 50 digit to the right of the decimal point and we see that 50 is even , we know that the 50 th digit is a 4 . answer e ." | a = 48 / 88
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a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 12 | c | divide(divide(multiply(multiply(8, 12), 5), 12), 8) | a crate measures 5 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 5 for the surface 8 * 12 face . volume = 125 pi answer c" | a = 8 * 12
b = a * 5
c = b / 12
d = c / 8
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a ) 12 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | a | multiply(divide(4, 50), const_100) | a pharmaceutical company received $ 4 million in royalties on the first $ 50 million in sales of and then $ 12 million in royalties on the next $ 170 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 50 million in sales to the next $ 170 million in sales ? | "( 12 / 170 ) / ( 4 / 50 ) = 15 / 17 = 88 % it means that 12 / 170 represents only 88 % . therefore a decrease of 12 % . answer a" | a = 4 / 50
b = a * 100
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a ) 187 , b ) 180 , c ) 190 , d ) 195 , e ) 197 | a | multiply(divide(add(multiply(const_100, 5), 78), add(multiply(multiply(const_4, const_4), const_2), const_2)), divide(add(multiply(const_100, 3), 74), add(multiply(multiply(const_4, const_4), const_2), const_2))) | what is the minimum number of square tiles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm ? | length = width = = > square = 5 m 78 cm and 3 m 74 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm hcf of 578 and 374 = 34 square is 34 = 578 * 374 / 34 * 34 = = 17 * 11 = 187 answer a | a = 100 * 5
b = a + 78
c = 4 * 4
d = c * 2
e = d + 2
f = b / e
g = 100 * 3
h = g + 74
i = 4 * 4
j = i * 2
k = j + 2
l = h / k
m = f * l
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a ) 18 square meters , b ) 20 square meters , c ) 24 square meters , d ) 28 square meters , e ) 30 square meters | b | divide(subtract(subtract(204, 140), 24), const_2) | three rugs have a combined area of 204 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ? | "total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 204 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 20 . answer : b ." | a = 204 - 140
b = a - 24
c = b / 2
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a ) 276 , b ) 290 , c ) 304 , d ) 318 , e ) 332 | c | multiply(divide(multiply(divide(multiply(divide(320, 5), 5), 2), 19), 24), 4) | there is a train and car . the ratio between the speed of a train & a car is 24 : 19 respectively . also , a bus covered a distance of 320 km in 5 hours . the speed of the bus is 2 / 3 rd of the speed of the train . how many kilometers will the car cover in 4 hours ? | "the speed of the bus is 320 / 5 = 64 km / hr the speed of the train is ( 64 * 3 ) / 2 = 96 km / hr the speed of the car is 96 / 24 * 19 = 76 km / hr the distance covered by the car in 4 hours is 76 Γ 4 = 304 km the answer is c ." | a = 320 / 5
b = a * 5
c = b / 2
d = c * 19
e = d / 24
f = e * 4
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a ) 45 , b ) 47 , c ) 59 , d ) 51 , e ) 53 | c | subtract(100, 41) | in a group of 100 cars , 37 cars do not have air conditioning . if at least 41 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ? | "lets assume ac = 63 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 41 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 41 . . we have to assign atleast 4 tocars with ac and racing stripes . hence ac = 63 - 4 = 59 . the answer is" | a = 100 - 41
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a ) 2 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph | a | divide(subtract(12, 8), const_2) | a man goes downstream at 12 kmph , and upstream 8 kmph . the speed of the stream is | "speed of the stream = 1 / 2 ( 12 - 8 ) kmph = 2 kmph . correct option a" | a = 12 - 8
b = a / 2
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a ) 25 days , b ) 100 days , c ) 120 days , d ) 110 days , e ) 90 days | a | inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 18))), 10)))) | micheal and adam can do together a piece of work in 20 days . after they have worked together for 18 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately . | "rate of both = 1 / 20 together they do = 1 / 20 * 18 = 9 / 10 left work = 1 - 9 / 10 = 1 / 10 adam completes 1 / 10 work in 10 day so he took 10 * 10 = 100 days to complete the left work alone . thus the rate of adam is 1 / 100 rate of micheal = 1 / 20 - 1 / 100 = 1 / 25 thus micheal takes 25 days to complete the whole work . ans . a ." | a = 1/(20)
b = 1/(20)
c = b * 18
d = 1 - c
e = 1/(d)
f = e * 10
g = 1/(f)
h = a - g
i = 1/(h)
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a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | a | divide(80, multiply(7, 1)) | how many 7 in between 1 to 80 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78,79 18 7 ' s between 1 to 80 answer : a" | a = 7 * 1
b = 80 / a
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a ) 42 , b ) 43 , c ) 44 , d ) 82 , e ) 46 | d | add(subtract(115, multiply(12, 3)), 3) | a batsman in his 12 th innings makes a score of 115 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been β not out β ? | "let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 4 ) + 115 β΄ x = 82 answer d" | a = 12 * 3
b = 115 - a
c = b + 3
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a ) 60 % , b ) 40 % , c ) 80 % , d ) 125 % , e ) none | a | multiply(divide(3, 5), const_100) | the ratio 3 : 5 expressed as a percent equals | "solution 3 : 5 = 3 / 5 = ( 3 / 5 x 100 ) % . = 60 % . answer a" | a = 3 / 5
b = a * 100
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a ) 300 , b ) 375 , c ) 400 , d ) 460 , e ) 500 | c | divide(28, subtract(142.07, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 28 . if n / j = 142.07 , what is value of j ? | when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 142.07 here 142 is the quotient . given that remainder = 28 so , 142.07 = 142 + 28 / j so , j = 400 answer - c | a = 4 * 10
b = a + 2
c = 100 + b
d = 142 - 7
e = 28 / d
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a ) 16000 , b ) 16500 , c ) 17000 , d ) 17500 , e ) 18000 | b | add(add(divide(multiply(multiply(1200, 50), 10), const_100), divide(multiply(multiply(3000, 50), add(divide(const_1, const_2), 3)), const_100)), divide(multiply(multiply(3000, 50), add(divide(const_1, const_2), 3)), const_100)) | find the annual dividend received by nishita from 1200 preferred shares and 3000 common shares both of par value rs . 50 each if the dividend paid on preferred shares is 10 % and semi - annual dividend of 3 Β½ % is declared on common shares . | total number of preferred shares = 1200 face value = rs . 50 dividend paid on preferred shares is 10 % dividend per share = 50 Γ 10 / 100 = rs . 5 total dividend = 1200 Γ 5 = 6000 total number of common shares = 3000 face value = rs . 50 semi - annual dividend of 3 Β½ % is declared on common shares . semi - annual dividend per share = 50 Γ 7 / 2 Γ 100 = rs . 74 total semi - annual dividend = 7 / 4 Γ 3000 = rs . 5250 annual dividend = rs . 5250 Γ 2 = rs . 10500 total dividend on all all shares ( preferred and common ) = 6000 + 10500 = rs . 16500 answer is b . | a = 1200 * 50
b = a * 10
c = b / 100
d = 3000 * 50
e = 1 / 2
f = e + 3
g = d * f
h = g / 100
i = c + h
j = 3000 * 50
k = 1 / 2
l = k + 3
m = j * l
n = m / 100
o = i + n
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a ) 56 m , b ) 66 m , c ) 76 m , d ) 86 m , e ) 96 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 32)), const_4) | length of a rectangular plot is 32 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 32 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 32 ) + x ] = 200 2 x + 32 = 100 2 x = 68 x = 34 . hence , length = x + 32 = 66 m b" | a = 5300 / 26
b = 2 * 32
c = a + b
d = c / 4
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a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | b | divide(120, subtract(divide(120, 40), divide(30, const_60))) | the distance from city a to city b is 120 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | the time it takes bob to drive to city b is 120 / 40 = 3 hours . alice needs to take less than 2.5 hours for the trip . alice needs to exceed a constant speed of 120 / 2.5 = 48 miles per hour . the answer is b . | a = 120 / 40
b = 30 / const_60
c = a - b
d = 120 / c
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a ) 108.45 , b ) 110.45 , c ) 106.45 , d ) 109.45 , e ) none of them | a | multiply(12.05, divide(5.4, 0.6)) | solve : 12.05 * 5.4 + 0.6 | = 12.05 * ( 5.4 / 0.6 ) = ( 12.05 * 9 ) = 108.45 answer is a . | a = 5 / 4
b = 12 * 5
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a ) $ 30,000 , b ) $ 30,400 , c ) $ 31,300 , d ) $ 32,500 , e ) $ 35,100 | a | multiply(divide(225, divide(9, multiply(const_3, const_4))), const_100) | an investment yields an interest payment of $ 225 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | "let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 225 = > p = ( 225 * 4 * 10 ^ 2 ) / 3 = 75 * 4 * 10 ^ 2 = 300 * 10 ^ 2 = 30000 answer a" | a = 3 * 4
b = 9 / a
c = 225 / b
d = c * 100
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a ) 5 , b ) 7 , c ) 8 , d ) q = 9 , e ) 10 | d | add(6, 3) | list i : { y , 2 , 4 , 7 , 10 , 11 } list ii : { 3 , 3 , 4 , 6 , 7 , 10 } if the median q of list i is equal to the sum of the median of list ii and the mode of list ii , then y equals | mode of list ii = 3 median q of list ii = 4 + 6 / 2 = 5 sum of mode + mean = 3 + 5 = 8 now to make 8 as the median we need to find a value of y such that if the no . of terms in list 1 are odd then y = 8 else if even then 7 + y / 2 = 8 here its even so 7 + y / 2 = 8 from this y = 9 ( d ) | a = 6 + 3
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a ) 562 , b ) 351 , c ) 452 , d ) 416 , e ) 512 | b | add(multiply(divide(8, 5), 135), 135) | in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 135 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 135 x = 27 total number of students = 13 x = 13 * 27 = 351 answer is b" | a = 8 / 5
b = a * 135
c = b + 135
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a ) 333 , b ) 500 , c ) 887 , d ) 720 , e ) 132 | d | divide(18, divide(2.5, const_100)) | an agent , gets a commission of 2.5 % on the sales of cloth . if on a certain day , he gets rs . 18 as commission , the cloth sold through him on that day is worth | "explanation : let the total sale be rs . x . then , 2.5 % . of x = 18 < = > ( 25 / 10 * 1 / 100 * x ) = 18 < = > x = 720 . answer : d" | a = 2 / 5
b = 18 / a
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a ) - 29 , b ) - 19 , c ) 19 , d ) 29 , e ) 39 | c | subtract(58, subtract(45, add(subtract(28, 37), 15))) | if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 58 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 58 = > 45 - [ 28 - { 37 - 15 + * } ] = 58 45 - [ 28 - 37 + 15 - * ] = 58 = > 45 [ 43 - 37 - * ] = 58 45 - [ 6 - * ] = 58 = > 45 - 6 + * = 58 39 + * = 58 = > * = 58 - 39 = 19 answer : c" | a = 28 - 37
b = a + 15
c = 45 - b
d = 58 - c
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a ) 17 , b ) 19 , c ) 17 , d ) 12 , e ) 91 | d | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | if the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 14 * x = 1056 = > x = 12 answer : d" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 14
h = 1056 / g
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a ) 512 , b ) 612 , c ) 712 , d ) 513 , e ) 613 | a | subtract(multiply(multiply(multiply(13, 7), const_3), const_2), multiply(const_10, const_4)) | a man walked a certain distance south and then the same distance plus 7 km due west . he is now 13 km from his starting point . what are the distances south and west that he walked ? | if a man walked distance ( in km ) ' x ' towards south & then ' x + 7 ' towards west . distance from starting point = hypotenuse of a right triangle with sides x & ( x + 7 ) so 13 ^ 2 = x ^ 2 + ( x + 7 ) ^ 2 x ^ 2 + 7 x - 60 = 0 or ( x + 12 ) ( x - 5 ) = 0 , so x = 5 distance towards west = x + 7 = 5 + 7 = 12 km . answer : a | a = 13 * 7
b = a * 3
c = b * 2
d = 10 * 4
e = c - d
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a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many r applicants majored in political science and had a gpa higher than 3.0 ? | "total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) r = 20 - 15 = 5 answer : a" | a = 10 + 15
b = 40 - a
c = 20 - b
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a ) 210 , b ) 150 , c ) 280 , d ) 300 , e ) 420 | b | divide(multiply(multiply(25, 42), 60), multiply(multiply(7, 12), 5)) | a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 12 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 42 * 60 dimensions of the soap = 5 * 12 * 7 placing the 7 inch side along 42 inch side we get 6 soaps in a line and in a similar way 5 along 25 and 6 along 60 we get = 5 x 6 x 5 = 150 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( b )" | a = 25 * 42
b = a * 60
c = 7 * 12
d = c * 5
e = b / d
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a ) 872 , b ) 738 , c ) 900 , d ) 840 , e ) 83 | c | divide(subtract(multiply(1600, divide(15, const_100)), 15), divide(25, const_100)) | if 25 % of x is 15 less than 15 % of 1600 , then x is ? | "25 % of x = x / 4 ; 15 % of 1600 = 15 / 100 * 1600 = 240 given that , x / 4 = 240 - 15 = > x / 4 = 225 = > x = 900 . answer : c" | a = 15 / 100
b = 1600 * a
c = b - 15
d = 25 / 100
e = c / d
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a ) 1 / 200 , b ) 1 / 100 , c ) 1 / 50 , d ) 1 / 40 , e ) 23 / 1800 | e | divide(23, multiply(30, 60)) | 30 ladies and 60 gentlemen are present at a party . there are 23 couples among them . if a lady and a gentleman is selected at random , what is the probability that they will be a couple ? | in how many ways we can select a woman and a man from 30 women and 60 men ? in 30 * 60 = 1800 ways . we have a total of 23 couples so , the probability of selecting a couple is 23 / 1800 = 23 / 1800 . ans - e | a = 30 * 60
b = 23 / a
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a ) 1 / 4 , b ) 2 / 3 , c ) 3 / 5 , d ) 2 / 5 , e ) 2 / 7 | d | subtract(const_1, add(divide(10, multiply(multiply(const_5, const_5), const_4)), add(divide(30, multiply(multiply(const_5, const_5), const_4)), divide(20, multiply(multiply(const_5, const_5), const_4))))) | in a class , 30 % of the students speaks truth , 20 % speaks lie and 10 % speaks both . if a student is selected at random , what is the probability that he has speak truth or lie ? | d ) 2 / 5 | a = 5 * 5
b = a * 4
c = 10 / b
d = 5 * 5
e = d * 4
f = 30 / e
g = 5 * 5
h = g * 4
i = 20 / h
j = f + i
k = c + j
l = 1 - k
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a ) 100 , b ) 250 , c ) 750 , d ) 4687.5 , e ) 5635.5 | d | multiply(multiply(multiply(50, 25), divide(1, const_2)), 7.5) | the milk level in a rectangular box measuring 50 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 50 * 25 * 1 / 2 = 625 feet ^ 3 of milk must be removed , which equals to 625 * 7.5 = 4687.5 gallons . answer : d ." | a = 50 * 25
b = 1 / 2
c = a * b
d = c * 7
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) none of them | b | subtract(subtract(25, const_4), const_1) | in the new budget , the price of kerosene oil rose by 25 % . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? | reduction in consumption = [ ( ( r / ( 100 + r ) ) * 100 ] % = [ ( 25 / 125 ) * 100 ] % = 20 % . answer is b . | a = 25 - 4
b = a - 1
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a ) 15.27 , b ) 14.15 , c ) 17.27 , d ) 18.27 , e ) 19.27 | b | multiply(divide(add(multiply(const_2, 10), const_2), add(const_3, const_4)), multiply(multiply(5.2, 5.2), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | the length of minute hand of a clock is 5.2 cm . what is the area covered by this in 10 minutes | "area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.2 ^ 2 ) * ( 60 / 360 ) = 14.15 cm ^ 2 answer : b" | a = 2 * 10
b = a + 2
c = 3 + 4
d = b / c
e = 5 * 2
f = 1 * const_60
g = 100 * const_3_6
h = f / g
i = e * h
j = d * i
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a ) 715 , b ) 825 , c ) 286 , d ) 582 , e ) 465 | c | divide(multiply(2310, 26), 210) | the lcm of two numbers is 2310 and hcf is 26 . if one of the numbers is 210 . then what is the other number ? | "first number * second number = lcm * hcf other number = 2310 * 26 / 210 = 11 * 26 = 286 answer : c" | a = 2310 * 26
b = a / 210
|
a ) 3337 , b ) 2782 , c ) 2788 , d ) 4266 , e ) 2321 | d | multiply(add(multiply(multiply(add(const_3, const_4), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), multiply(subtract(multiply(add(const_2, const_3), const_2), const_2), const_100)), subtract(power(add(const_1, divide(13.5, const_100)), const_2), const_1)) | find the compound interest accrued on an amount of rs . 14,800 at 13.5 % p . a at the end of two years . ( round off your answer to nearest integer ) | ci = 14800 { [ 1 + 13.5 / 100 ] 2 - 1 } = 14800 { [ 1 + 27 / 200 ] 2 - 1 = 14800 { 2 + 27 / 200 } { 27 / 200 } = ( 74 ) [ 2 + 27 / 200 ] ( 27 ) = 1998 [ 2 + 27 / 200 ] = 3996 + 269.73 = rs . 4266 answer : d | a = 3 + 4
b = a * 2
c = 2 + 3
d = c * 2
e = 100 * d
f = b * e
g = 2 + 3
h = g * 2
i = h - 2
j = i * 100
k = f + j
l = 13 / 5
m = 1 + l
n = m ** 2
o = n - 1
p = k * o
|
a ) 4 , b ) 6 , c ) 12 , d ) 16 , e ) 24 | c | multiply(4, const_3) | if all of the telephone extensions in a certain company must be even numbers , and if each of the extensions uses all 4 of the digits 1 , 2 , 3 , and 8 , what is the greatest number of 4 - digit extensions that the company can have ? | since the phone number must be even , the unit ' s digit can be either 2 or 8 . when the unit ' s digit is 2 - - > number of possibilities is 3 ! = 6 when the unit ' s digit is 8 - - > number of possibilities is 3 ! = 6 largest number of extensions = 6 + 6 = 12 answer : c | a = 4 * 3
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a ) 4350 , b ) 4725 , c ) 4328 , d ) 4329 , e ) 4829 | b | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 7))), rectangle_area(75, 55)), 7) | a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 7 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 7 = rs . 4725 answer : b" | a = 2 * 5
b = 75 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 14 , b ) 3 , c ) 9 , d ) 7 , e ) 5 | b | add(subtract(add(24, 15), subtract(40, 4)), subtract(15, 24)) | of 40 applicants for a job , 24 had at least 4 years ' experience , 15 had degrees , and 4 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 40 - 4 = 36 aub = a + b - aib aib = 15 + 24 - 36 = 3 answer b" | a = 24 + 15
b = 40 - 4
c = a - b
d = 15 - 24
e = c + d
|
a ) 2103 , b ) 2106 , c ) 1053 , d ) 1252 , e ) 1535 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 3), add(const_2, const_4)) | what is the sum of all the multiples of 3 between 30 and 80 ? | "you first have to know all the multiples of 3 between 30 and 80 . they are 3 , 6,9 , 12,15 , 18,21 , 24,27 , 30,33 , 36,39 , 42,45 , 48,51 , 54,57 , 60,63 , 66,69 , 72,75 , and 78 . if you add all these numbers together , you get 1053 . final answer : c" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 3
t = 2 + 4
u = s + t
|
a ) 500 % , b ) 600 % , c ) 800 % , d ) 1100 % , e ) 1200 % | d | subtract(add(multiply(multiply(const_100, divide(300, const_100)), divide(300, const_100)), multiply(const_100, divide(300, const_100))), const_100) | the first half of the 20 th century , the population of a particular country increased by 200 percent . in the second half of the century , the population increased by 300 percent . what was the percent increase for the 20 th century as a whole ? | say initially population was 100 . what is 200 % of 100 ? it is 200 / 100 * 100 = 200 . an increase of 200 % means the new population became 100 + 200 = 300 what is 300 % of 300 ? it is 300 / 100 * 300 = 900 an increase of 300 % means the new population now is 300 + 900 = 1200 so from 100 , the population increased to 1200 i . e . an increase of 1100 . 1100 is what percent of 100 ? 1100 = x / 100 * 100 i . e . it is 1100 % d | a = 300 / 100
b = 100 * a
c = 300 / 100
d = b * c
e = 300 / 100
f = 100 * e
g = d + f
h = g - 100
|
a ) 2 , b ) 5 , c ) 9 , d ) 15 , e ) 30 | b | subtract(divide(subtract(120, 70), subtract(70, 60)), divide(subtract(120, const_100), const_100)) | at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 60 . if the restaurant serves 120 customers today , raising the average to 70 customers per day , what is the value of x ? | "withoutusing the formula , we can see that today the restaurant served 50 customers above the average . the total amount above the average must equal total amount below the average . this additional 50 customers must offset the β deficit β below the average of 70 created on the x days the restaurant served only 60 customers per day . 50 / 10 = 5 days . choice ( a ) . withthe formula , we can set up the following : 70 = ( 60 x + 120 ) / ( x + 1 ) 70 x + 70 = 60 x + 120 10 x = 50 x = 5 answer choice ( b )" | a = 120 - 70
b = 70 - 60
c = a / b
d = 120 - 100
e = d / 100
f = c - e
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | c | subtract(const_100, subtract(add(60, 40), 20)) | 100 students appeared in 2 tests . 60 students passed 1 st test . 40 students passed in the 2 nd test . 20 students passed in both 1 and 2 tests . what is the probability of the students who failed in both tests ? | 20 student passed both two test 40 student passed only 1 st test 20 student passed only 2 nd test so 100 - ( 20 + 40 + 20 ) = 20 student failed in both sub so ans is 20 % answer : c | a = 60 + 40
b = a - 20
c = 100 - b
|
a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 6), 6) | how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 6 | 560 , 562 , 564 , 566 , and 568 , so total 5 . hence option a . | a = 3 + 6
b = a + 6
|
a ) 20 % , b ) 24 % , c ) 28 % , d ) 32 % , e ) 36 % | e | multiply(subtract(const_1, divide(const_100, add(add(const_100, 30), divide(multiply(add(const_100, 30), 20), const_100)))), const_100) | there has been successive increases of 30 % and then 20 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | "let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.2 * 1.3 * p = x * p y = x / ( 1.2 * 1.3 ) which is about 0.64 x which is a decrease of about 36 % . the answer is e ." | a = 100 + 30
b = 100 + 30
c = b * 20
d = c / 100
e = a + d
f = 100 / e
g = 1 - f
h = g * 100
|
a ) 550 m . , b ) 300 m . , c ) 600 m . , d ) 400 m . , e ) 500 m . | e | multiply(500, subtract(const_2, const_1)) | a train speeds past a pole in 50 seconds and a platform 500 m long in 100 seconds . its length is : | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 50 = > y = x / 50 x + 500 / 100 = x / 50 x = 500 m . answer : option e" | a = 2 - 1
b = 500 * a
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(24, add(2, const_1)) | if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 24 - m ) then m = ? | "2 m = 24 - m 3 m = 24 m = 8 the answer is b ." | a = 2 + 1
b = 24 / a
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 24 | e | multiply(divide(16, 2), const_3) | a seller of used cars has 16 cars to sell and each of his clients selected 2 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ? | ifno caris selected more than once then the number of clients = 16 / 2 = 8 but since every car is being selected three times so no . of clients must be thrice as well = 8 * 3 = 24 answer : option e | a = 16 / 2
b = a * 3
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a ) 36 , b ) 76 , c ) 98 , d ) 27 , e ) 24 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number is ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
|
a ) 30 / 7 , b ) 30 / 9 , c ) 30 / 2 , d ) 30 / 3 , e ) 30 / 6 | a | divide(multiply(6, 5), 7) | the smallest fraction , which each of 6 / 7 , 5 / 14 , 10 / 21 will divide exactly is ? | "required fraction = l . c . m of 6 / 7 , 5 / 14 , 10 / 21 = ( l . c . m of 6 , 5 , 10 ) / ( h . c . f of 7 , 14 , 21 ) = 30 / 7 answer : a" | a = 6 * 5
b = a / 7
|
a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 40 % lemonade syrup ? | "let the total solution is 150 l with 80 l water & 70 l syrup . to make 40 % syrup solution , the result solution must have 90 l syrup and 60 l syrup . therefore we are taking 10 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 10 l syrup in 21.4 l solution we started by multiplying 10 now to get to the result we need to divide by 10 = > amount of solution to be replaced with water = ( 21.4 / 10 ) = 2.14 . correct option : c" | a = 8 + 7
b = 7 / a
c = 2 + 3
d = 2 / c
e = b - d
f = 8 + 7
g = 7 / f
h = e / g
i = 8 + 7
j = h * i
|
a ) 19.6 m . , b ) 20.0 m . , c ) 19.3 m . , d ) 18.5 m . , e ) 18.9 m . | a | multiply(sqrt(divide(divide(640, 5), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 640 is required to paint the floor at the rate of rs . 5 per sq m , then what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 3 b = 3 b area of the floor = 640 / 5 = 128 sq m l b = 128 i . e . , l * l / 3 = 128 l 2 = 384 = > l = 19.6 m . answer : a" | a = 640 / 5
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) rs . 10 , b ) rs . 15 , c ) rs . 20 , d ) rs . 25 , e ) rs . 30 | d | divide(subtract(500, multiply(divide(subtract(const_100, 15), const_100), 500)), 3) | a reduction of 15 % in the price of wheat enables a house wife to obtain 3 kgs more for rs . 500 , what is the reduced price for kg ? | explanation : 500 * ( 15 / 100 ) = 75 - - - - 3 ? - - - - 1 = > rs . 25 answer : d | a = 100 - 15
b = a / 100
c = b * 500
d = 500 - c
e = d / 3
|
a ) 7 / 15 , b ) 3 / 10 , c ) 3 / 5 , d ) 2 / 5 , e ) 3 / 4 | a | add(divide(multiply(subtract(const_1, divide(40, multiply(20, 20))), 150), 450), divide(add(multiply(divide(20, multiply(20, 20)), 300), multiply(divide(40, multiply(20, 20)), 150)), 450)) | in a certain corporation , there are 300 male employees and 150 female employees . it is known that 20 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | "p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.2 * 300 / 450 = 60 / 450 = 2 / 15 the sum of the probabilities is 7 / 15 the answer is a ." | a = 20 * 20
b = 40 / a
c = 1 - b
d = c * 150
e = d / 450
f = 20 * 20
g = 20 / f
h = g * 300
i = 20 * 20
j = 40 / i
k = j * 150
l = h + k
m = l / 450
n = e + m
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a ) 5 : 2 , b ) 2 : 1 , c ) 11 : 7 , d ) 13 : 9 , e ) none of these | b | divide(add(multiply(7, sqrt(divide(756, multiply(7, 3)))), 6), add(multiply(3, sqrt(divide(756, multiply(7, 3)))), 6)) | the ration of the father β s age to his son β s age is 7 : 3 . the product of their ages is 756 . the ratio of their ages after 6 years will be : | solution let the present ages of the father and son be 7 x and 3 x years respectively . then , 7 x 3 x = 756 β 21 x 2 = 756 β x 2 = 36 β x = 6 . β΄ required ratio = ( 7 x + 6 ) : ( 3 x + 6 ) = 48 : 24 = 2 : 1 . answer b | a = 7 * 3
b = 756 / a
c = math.sqrt(b)
d = 7 * c
e = d + 6
f = 7 * 3
g = 756 / f
h = math.sqrt(g)
i = 3 * h
j = i + 6
k = e / j
|
a ) 25 , b ) 37.5 , c ) 55 , d ) 62.5 , e ) 75 | c | add(subtract(const_100, 68), subtract(90, 68)) | at a certain university , 68 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ? | total women = 68 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 22 % but question wants to know the percent of men that are tenured 22 % / 40 % = 55 % c | a = 100 - 68
b = 90 - 68
c = a + b
|
a ) 42 , b ) 44 , c ) 49 , d ) 41 , e ) 47 | b | multiply(multiply(subtract(const_12, const_1), const_2), const_2) | how many times are the hands of a clock at right angles in a day ? | in 12 hours , they are at right angles 22 times . = = > in 24 hours , they are at right angles 44 times . answer is b . | a = 12 - 1
b = a * 2
c = b * 2
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a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | b | divide(subtract(240, multiply(divide(subtract(12, 4), const_2), 10)), add(divide(subtract(12, 4), const_2), add(divide(subtract(12, 4), const_2), 4))) | a cyclist traveled for two days . on the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 240 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | solution : d = 280 mi t = 12 hrs Δ Γ’ y 1 time = t 1 d Γ’ y 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs d Γ y 1 rate = r 1 d Γ’ y 2 rate = r 2 r 1 - r 2 = 10 mph Γ . αΊΉ . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 Γ . αΊΉ . 280 = 8 r 2 + 4 ( 10 + r 2 ) Γ . αΊΉ . r 2 = 20 mph answer : b | a = 12 - 4
b = a / 2
c = b * 10
d = 240 - c
e = 12 - 4
f = e / 2
g = 12 - 4
h = g / 2
i = h + 4
j = f + i
k = d / j
|
a ) 27 days , b ) 54 days , c ) 56 days , d ) 68 days , e ) none of these | b | multiply(18, const_3) | a is twice as good as workman as b and together they finish a piece of work in 18 days . in how many days will b alone finish the work . | explanation : as per question , a do twice the work as done by b . so a : b = 2 : 1 also ( a + b ) one day work = 1 / 18 to get days in which b will finish the work , lets calculate work done by b in 1 day = = ( 118 β 13 ) = 154 [ please note we multiplied by 1 / 3 as per b share and total of ra Ι΅ o is 1 / 3 ] so b will finish the work in 54 days answer : b | a = 18 * 3
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a ) 0.8 , b ) 1.0 , c ) 1.2 , d ) 1.4 , e ) 1.6 | a | divide(divide(add(6, 2), const_2), divide(add(4, 6), const_2)) | in the xy - coordinate system , what is the slope of the line that goes through the origin and is equidistant from the two points p = ( 4 , 6 ) and q = ( 6 , 2 ) ? | first , get the middle coordinate between ( 46 ) and ( 62 ) . x = 4 + ( 6 - 4 ) / 2 = 5 y = 2 + ( 6 - 2 ) / 2 = 4 second , get the slope of ( 54 ) and ( 00 ) . m = 4 - 0 / 5 - 0 = 4 / 5 = 0.8 answer : a | a = 6 + 2
b = a / 2
c = 4 + 6
d = c / 2
e = b / d
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a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 80.00 | e | multiply(subtract(10, divide(multiply(15, 8), const_100)), 10) | an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 15 % . also , if a customer will β buy at least 10 β items they will deduct an additional 8 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 23 % , thus he must pay slightly more than $ 77 . answer : e ." | a = 15 * 8
b = a / 100
c = 10 - b
d = c * 10
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a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | d | multiply(multiply(power(divide(60, multiply(add(132, 7), 7)), 7), 132), 7) | the perimeter of a rectangle is 60 mete ( 132 ) 7 Γ ( 132 ) ? = ( 132 ) 11.5 . | "7 + x = 11.5 x = 11.5 - 7 x = 4.5 answer : d" | a = 132 + 7
b = a * 7
c = 60 / b
d = c ** 7
e = d * 132
f = e * 7
|
a ) 17 , b ) 18 , c ) 19 , d ) 21 , e ) 22 | a | divide(subtract(multiply(4, 21), add(add(4, 5), 8)), 4) | the youngest of 4 children has siblings who are 3 , 5 , and 8 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 21 , what is the age of the youngest sibling ? | "total age of the 4 sibling is 21 x 4 = 84 years . . we already have the total age of all the children is 4 y + 16 so , 4 y + 16 = 84 or , 4 y = 68 or , y = 17 so , age of the youngest child is 17 years . answer : a" | a = 4 * 21
b = 4 + 5
c = b + 8
d = a - c
e = d / 4
|
a ) 15 , b ) 10 , c ) 12 , d ) 11 , e ) 13 | c | add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(15), inverse(15))), 2))), inverse(15)), 4) | a can do a piece of work in 15 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ? | "2 / 15 + 4 / 15 + x / 20 = 1 x = 12 answer : c" | a = 4 - 2
b = 1/(15)
c = 1/(20)
d = b + c
e = a * d
f = 1/(20)
g = 1/(15)
h = 1/(15)
i = g + h
j = f + i
k = j * 2
l = e + k
m = 1 - l
n = 1/(15)
o = m / n
p = o + 4
|
a ) 10 kmph , b ) 14 kmph , c ) 12 kmph , d ) 16 kmph , e ) 15 kmph | c | divide(36, add(const_1, const_2)) | the speed of a boat in still water is 36 kmph . what is the speed of the stream if the boat can cover 80 km downstream or 40 km upstream in the same time ? | "x = the speed of the stream ( 36 + x ) / ( 36 - x ) = 2 / 1 36 + x = 72 - 2 x 3 x = 36 x = 12 km / hour if the speed of the stream is 12 km / hour , then the ' downstream ' speed of the boat is 36 + 12 = 48 km / hour and the ' upstream ' speed of the boat is 36 - 12 = 24 km / hour . in that way , if the boat traveled for 2 hours , it would travel 2 x 48 = 96 km downstream and 2 x 24 = 48 km / hour upstream . answer : c" | a = 1 + 2
b = 36 / a
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a ) 1410 , b ) 1620 , c ) 1430 , d ) 1440 , e ) 1540 | b | divide(multiply(subtract(const_100, 10), 1800), const_100) | a man buys a cycle for rs . 1800 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1800 = 90 / 100 x 1800 = rs . 1620 answer : b" | a = 100 - 10
b = a * 1800
c = b / 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 9 , e ) 5 | e | multiply(5, 1) | if n divided by 11 has a remainder of 1 , what is the remainder when 5 times n is divided by 11 ? | "as per question = > n = 11 p + 1 for some integer p hence 5 n = > 55 q + 5 = > remainder = > 5 for some integer q hence e" | a = 5 * 1
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['a ) 1 / pi', 'b ) sqrt ( 2 / pi )', 'c ) 1', 'd ) 2 / sqrt ( pi )', 'e ) sqrt ( 5.78 / pi )'] | e | sqrt(divide(divide(square_area(3.4), 2), const_pi)) | an artist wishes to paint a circular region on a square poster that is 3.4 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | area of the poster is 3.4 x 3.4 = 11.56 1 / 2 the area = 5.78 pi * r ^ 2 = 5.78 r ^ 2 = 5.78 / pi r = sqrt ( 5.78 / pi ) answer ( e ) | a = square_area / (
b = a / 2
c = math.sqrt(b)
|
a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.12 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.567), multiply(multiply(7.57, 23.5), 0.0673)), const_10) | the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.5 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.5 * 7.57 ) = 25.4303 / 11.972 = 2.12 answer : d" | a = 34 * 31
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) 17 years , b ) 19 years , c ) 18 years , d ) 10 years , e ) 12 years | c | divide(multiply(subtract(47, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 = > x = 9 hence , b ' s age = 2 x = 18 years . answer : c" | a = 47 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 65842158943 , b ) 65839570421 , c ) 65821141683 , d ) 66821785904 , e ) 65821041783 | e | multiply(subtract(99999, const_4), 658217) | find the value of 658217 x 99999 = m ? | "658217 x 99999 = 658217 x ( 100000 - 1 ) = 658217 x 100000 - 658217 x 1 = 65821700000 - 658217 = 65821041783 e" | a = 99999 - 4
b = a * 658217
|
a ) 130 , b ) 132 , c ) 134 , d ) 136 , e ) 139 | e | divide(1251, 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 1251 . find the number to be multiplied . | "let the required number be x . then , 43 x β 34 x = 1251 or 9 x = 1251 or x = 139 . required number = 139 . answer : e" | a = 43 - 34
b = 1251 / a
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a ) 999955 , b ) 999900 , c ) 999845 , d ) 999790 , e ) 999735 | a | multiply(add(const_100, const_2), 55) | calculate the largest 6 digit number which is exactly divisible by 55 ? | "largest 4 digit number is 999999 after doing 999999 Γ· 55 we get remainder 44 hence largest 4 digit number exactly divisible by 88 = 999999 - 44 = 999955 a" | a = 100 + 2
b = a * 55
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a ) 15 , b ) 11 , c ) 18 , d ) 16 , e ) 20 | d | add(10, divide(12, const_2)) | 10 years ago , the age of peter was one - third the age of jacob at that time . the present age of jacob is 12 years more than the present age of peter . find the present age of peter ? | d 16 let the present ages of peter and jacob be ' a ' and ' b ' respectively . a - 10 = 1 / 3 ( b - 10 ) - - - ( 1 ) b = a + 12 substituting b = a + 12 in first equation , a - 10 = 1 / 3 ( a + 2 ) = > 3 a - 30 = a + 2 = > 2 a = 32 = > a = 16 . | a = 12 / 2
b = 10 + a
|
a ) $ 60 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 180 | d | divide(8, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2)))) | at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 8 on candy , how much did john spend at the supermarket ? | let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t . we are also given that he spent the remaining $ 8 on candy . since we know where all his money was allocated , we can sum these values together and set the sum to t . so we have : ( 1 / 2 ) t + ( 1 / 3 ) t + ( 1 / 10 ) t + 8 = t to get rid of the fractions we can multiply the entire equation by 30 , and we obtain : 15 t + 10 t + 3 t + 240 = 30 t 28 t + 240 = 30 t 240 = 2 t t = 120 john spent $ 90 at the supermarket . answer : d | a = 1 / 10
b = 1 / 3
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 8 / f
|
a ) 3 / 2 , b ) - 1 / 2 , c ) - 1 , d ) 1 / 2 , e ) 2 / 3 | d | divide(gcd(64, 25), const_2) | tough and tricky questions : exponents . if 5 ^ ( x + 1 ) * 4 ^ ( y - 1 ) = 25 ^ x * 64 ^ y , then x + y = | here is my solution . 5 ^ ( x + 1 ) * 4 ^ ( y - 1 ) = 25 ^ x * 64 ^ y here rhs 25 ^ x * 64 ^ y = 5 ^ ( 2 x ) * 4 ^ ( 3 y ) equating powers on both sides - - > x + 1 = 2 x , thus x = 1 and 2 y - 1 = 3 y giving y = - 1 / 2 so , x + y = 1 / 2 option : d | a = math.gcd(64, 25)
b = a / 2
|
['a ) 128', 'b ) 158', 'c ) 178', 'd ) 139', 'e ) 140'] | d | subtract(92555, multiply(floor(sqrt(92555)), floor(sqrt(92555)))) | what no should be subtracted from 92555 to make it a perfect square ? | root 92555 = 304.228 and 304 ^ 2 = 92416 92555 - 92416 = 139 should be subtracted from 92555 to make it a perfect square answer : d | a = math.sqrt(92555)
b = math.floor(a)
c = math.sqrt(92555)
d = math.floor(c)
e = b * d
f = 92555 - e
|
a ) rs . 72 , b ) rs . 36 , c ) rs . 54 , d ) rs . 50 , e ) none | d | divide(6, divide(12, const_100)) | the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 6 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 6 x 100 / 12 x 1 ) = rs . 50 . answer d" | a = 12 / 100
b = 6 / a
|
a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | a line has a slope of 3 / 4 and intersects the point q ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point q of intersection is ( 40,0 ) . a is the correct answer ." | a = negate - (
b = negate * (
c = 3 / 4
d = a / b
e = negate * (
|
a ) 79 , b ) 89 , c ) 95 , d ) 90.91 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 80)), add(6, 5))) | a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 80 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 80 * 5 = 400 student ' s average = ( 600 + 400 ) / 11 = 90.91 answer d | a = 6 * 100
b = 5 * 80
c = a + b
d = 6 + 5
e = c / d
f = math.floor(e)
|
a ) 0.004 , b ) 0.04 , c ) 4 , d ) 40 , e ) 400 | c | multiply(divide(8.008, 2.002), const_100) | 8.008 / 2.002 | "answer is 4 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 8008 / 2002 = 4 answer c" | a = 8 / 8
b = a * 100
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | d | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 86 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "7 ^ 2 + 6 ^ 2 + 1 ^ 2 = 49 + 36 + 1 = 86 7 + 6 + 1 = 14 hence answer is d" | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
|
a ) 40 , b ) 45 , c ) 50 , d ) 54 , e ) 60 | d | multiply(divide(divide(add(divide(120, const_2), 120), 50), const_4), divide(120, const_2)) | a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 50 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | "a - b = 120 miles b - c = 60 miles avg speed = 50 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 50 = 180 / 4 t t = 54 b - c = 54 mph answer d" | a = 120 / 2
b = a + 120
c = b / 50
d = c / 4
e = 120 / 2
f = d * e
|
a ) 83.33 , b ) 110 , c ) 112 , d ) 140 , e ) 160 | d | multiply(divide(const_100, 10), 14) | a 14 % stock yielding 10 % is quoted at : | "solution to earn rs . 10 , money invested = rs . 100 . to earn rs . 14 , money invested = rs . ( 100 / 10 x 14 ) = rs . 140 . Γ’ Λ Β΄ market value of rs . 100 stock = rs . 140 answer d" | a = 100 / 10
b = a * 14
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | floor(subtract(divide(300, 40), divide(50, 10))) | subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 30 pages . | "subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 30 / 2.5 ) = 12 hrs answer : c" | a = 300 / 40
b = 50 / 10
c = a - b
d = math.floor(c)
|
a ) s . 83.33 , b ) s . 110 , c ) s . 112 , d ) s . 125 , e ) s . 140 | d | multiply(divide(const_100, 20), 25) | a 25 % stock yielding 20 % is quoted at : | "income of rs 20 on investment of rs 100 income of rs 25 on investment of ? = ( 25 * 100 ) / 20 = 125 answer : d" | a = 100 / 20
b = a * 25
|
a ) 40 days , b ) 2 days , c ) 4 days , d ) 8 days , e ) 40 days | c | inverse(subtract(3, divide(3, 12))) | a and b can do a piece of work in 12 days . with the help of c they finish the work in 3 days . c alone can do that piece of work in ? | "c 30 days c = 1 / 3 β 1 / 12 = 1 / 4 = > 4 days" | a = 3 / 12
b = 3 - a
c = 1/(b)
|
a ) 71.5 , b ) 11.5 , c ) 81.5 , d ) 11.5 , e ) 22.5 | e | subtract(add(multiply(10.5, 6), multiply(11.4, 6)), multiply(9.9, 11)) | the average of 11 numbers is 9.9 . if the average of the first 6 numbers is 10.5 and that of the last 6 numbers is 11.4 , then the middle number is | explanation : middle numbers = [ ( 10.5 x 6 + 11.4 x 6 ) - 9.9 x 11 ] = 22.5 . answer : e | a = 10 * 5
b = 11 * 4
c = a + b
d = 9 * 9
e = c - d
|
a ) 164 , b ) 224 , c ) 280 , d ) 384 , e ) 476 | d | multiply(subtract(power(3, 3), 3), multiply(4, 4)) | in how many ways can an answer key for a quiz be written if the quiz contains 3 true - false questions followed by 3 multiple - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ? | "there are 2 ^ 3 = 8 possibilities for the true - false answers . however we need to remove two cases for ttt and fff . there are 4 * 4 * 4 = 64 possibilities for the multiple choice questions . the total number of possibilities is 6 * 64 = 384 . the answer is d ." | a = 3 ** 3
b = a - 3
c = 4 * 4
d = b * c
|
a ) 26 , b ) 39 , c ) 42 , d ) 144 , e ) 156 | d | multiply(multiply(multiply(power(2, 2), 3), divide(12, 2)), 2) | if 2 ^ 5 , 3 ^ 3 , and 12 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ? | "here 156 has three two ' s two three ' s and one 13 rest of them must be in w so w = 12 * 3 * 4 = 144 smash d" | a = 2 ** 2
b = a * 3
c = 12 / 2
d = b * c
e = d * 2
|
a ) 24 , b ) 26 , c ) 27 , d ) 400 , e ) 30 | d | divide(1000, add(add(divide(5, 4), divide(1, 4)), 1)) | the ratio of 3 numbers is 5 : 1 : 4 and their sum is 1000 . the last number of the 3 numbers is ? | 5 : 1 : 4 total parts = 10 10 parts - - > 1000 1 part - - - - > 100 the last number of the three numbers is = 4 * 100 = 400 answer : d | a = 5 / 4
b = 1 / 4
c = a + b
d = c + 1
e = 1000 / d
|
a ) 17 , b ) 20 , c ) 21 , d ) 24 , e ) 25 | b | subtract(21, const_1) | when average age of 21 members are 0 , how many members greater than 0 ? | "average of 21 numbers = 0 . sum of 21 numbers ( 0 x 21 ) = 0 . it is quite possible that 20 of these numbers may be positive and if their sum is a then 21 st number is ( - a ) answer is 20 ( b )" | a = 21 - 1
|
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