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 ) 20 . , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many apples did each person get before 45 people joined the feast ? | if 1430 is divisible by anyone of the answer choices . a . 1430 / 20 = 143 / 2 b 1430 / 21 = 1430 / 21 c 1430 / 22 = 65 if 1430 apple was divided among 65 people , each would have received 22 . after addition of 45 people the answer should be 13 . 1430 / 110 = 13 . c is the answer . | a = 10 + 10
b = a + 2
c = 1430 / b
d = c + 45
e = 1430 / d
f = e + 9
|
a ) $ 0 , b ) $ 3 , c ) $ 6 , d ) $ 12 , e ) $ 15 | c | subtract(multiply(divide(42, subtract(const_1, divide(30, const_100))), subtract(const_1, divide(20, const_100))), 42) | a merchant purchased a jacket for $ 42 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 30 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant ’ s gross profit on this ... | actual cost = $ 42 sp = actual cost + mark up = actual cost + 30 % sp = 42 * 100 / 70 on sale sp = 80 / 100 ( 42 * 100 / 70 ) = 48 gross profit = $ 6 answer is c | a = 30 / 100
b = 1 - a
c = 42 / b
d = 20 / 100
e = 1 - d
f = c * e
g = f - 42
|
a ) 67 % . , b ) 70 % . , c ) 60 % . , d ) 68 % . , e ) 80 % . | e | multiply(divide(1, 5), const_100) | if two positive numbers are in the ratio 1 / 9 : 1 / 5 , then by what percent is the second number more than the first ? | "given ratio = 1 / 9 : 1 / 5 = 5 : 9 let first number be 5 x and the second number be 9 x . the second number is more than first number by 4 x . required percentage = 4 x / 5 x * 100 = 80 % . answer : e" | a = 1 / 5
b = a * 100
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a ) 8 % , b ) 15 % , c ) 35 % , d ) 42 % , e ) 56 % | d | multiply(divide(3, 25), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 25 million in sales of and then $ 9 million in royalties on the next $ 130 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 25 million in sales to the next $ 130 million in sales ? | "( 9 / 130 ) / ( 3 / 25 ) = 15 / 26 = 57.7 % it means that 9 / 130 represents only 57.7 % . therefore a decrease of 42 % . answer d" | a = 3 / 25
b = a * 100
|
a ) 4 , b ) 6 , c ) 8 , d ) 30 , e ) 12 | d | multiply(multiply(5, 2), 3) | running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 5 w widgets . | "i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget ... | a = 5 * 2
b = a * 3
|
a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 300 | b | divide(multiply(616, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 616 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 616 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 616 ⇒ y [ 12 / 10 + 1 ] = 616 ⇒ 22 y / 10 = 616 ⇒ 22 y = 6160 ⇒ y = 6160 / 22 = 560 / 2 = rs . 280 b" | a = 1 + 4
b = a * 2
c = 616 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 25 , b ) 23 , c ) 22 , d ) 15 , e ) 18 | d | divide(rectangle_area(5, 24), 8) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 8 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 8 w = 120 = > w = 15 answer d" | a = rectangle_area / (
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a ) 191 , b ) 355 , c ) 737 , d ) 840 , e ) 1,560 | d | divide(multiply(546, const_100), subtract(const_100, 35)) | a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 546 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 546 = > x = ( 546 * 100 ) / 65 = 840 answer d" | a = 546 * 100
b = 100 - 35
c = a / b
|
a ) 3 , b ) 7 , c ) 9 , d ) 21 , e ) 25 | c | add(4, subtract(add(14, 9), multiply(const_2, 9))) | jacob is now 14 years younger than michael . if 9 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | "jacob = x years , michael = x + 14 years 9 years from now , 2 ( x + 9 ) = x + 23 2 x + 18 = x + 23 x = 5 x + 4 = 9 years answer c" | a = 14 + 9
b = 2 * 9
c = a - b
d = 4 + c
|
a ) 11 / 12 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | divide(add(subtract(60, add(5, 9)), 9), 60) | a bag consists of 60 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | bag consists of 60 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 60 - 5 - 9 = 46 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 60 + 46 / 60 = 5... | a = 5 + 9
b = 60 - a
c = b + 9
d = c / 60
|
a ) 2 days , b ) 3 days , c ) 2.5 days , d ) 4 days , e ) 1 day | e | divide(100, multiply(5, divide(360, multiply(6, 3)))) | if 6 women can color 360 m long cloth in 3 days , then 5 women can color 100 m long cloth in ? | "the length of cloth painted by one woman in one day = 360 / 6 × 3 = 20 m no . of days required to paint 100 m cloth by 5 women = 100 / 5 × 20 = 1 day answer : e" | a = 6 * 3
b = 360 / a
c = 5 * b
d = 100 / c
|
a ) 2000 , b ) 2888 , c ) 2667 , d ) 2999 , e ) 2122 | a | add(add(990, multiply(990, divide(10, const_100))), subtract(990, multiply(990, divide(10, const_100)))) | ram sold two bicycles , each for rs . 990 . if he made 10 % profit on the first and 10 % loss on the second , what is the total cost of both bicycles ? | ( 10 * 10 ) / 100 = 1 % loss 100 - - - 99 ? - - - 1980 = > rs . 2000 answer : a | a = 10 / 100
b = 990 * a
c = 990 + b
d = 10 / 100
e = 990 * d
f = 990 - e
g = c + f
|
a ) 5 : 8 , b ) 7 : 9 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 1 | b | divide(3, 2) | the simple form of the ratio 7 / 6 : 3 / 2 is ? | "7 / 6 : 3 / 2 = 7 : 9 answer : b" | a = 3 / 2
|
a ) 2000 , b ) 4500 , c ) 5000 , d ) 8000 , e ) 9000 | d | divide(72000, 9) | a company produces 72000 bottles of water everyday . if a case can hold 9 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 9 . number of cases required to hold 72000 bottles = 72000 / 9 = 8000 cases . so the answer is d = 8000" | a = 72000 / 9
|
a ) 12 , b ) 29 , c ) 20 , d ) 15 , e ) 99 | c | divide(subtract(135, multiply(const_3, 5)), multiply(const_3, const_2)) | a number is doubled and 5 is added . if the resultant is trebled , it becomes 135 . what is that number ? | "explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 135 6 x + 15 = 135 6 x = 120 x = 20 answer : c" | a = 3 * 5
b = 135 - a
c = 3 * 2
d = b / c
|
a ) 77500 , b ) 1 , 860000 , c ) 2 , 480000 , d ) 3 , 720000 , e ) 39 , 680000 | e | multiply(multiply(floor(divide(add(divide(multiply(power(const_2, divide(12, divide(120, const_60))), multiply(310000, const_2)), const_1000), multiply(multiply(const_3, const_3), const_1000)), const_1000)), const_100), const_100) | on average , activated yeast cells divide once every 120 minutes ; that is , each individual cell divides into two individual cells over that time span . given that a certain yeast colony consisted of 310000 cells two hours ago , approximately how many cells will be populating the colony 12 hours from now ? | 310000 × 2 × 2 × 2 × 2 x 2 × 2 × 2 = 310000 × 128 = e | a = 120 / const_60
b = 12 / a
c = 2 ** b
d = 310000 * 2
e = c * d
f = e / 1000
g = 3 * 3
h = g * 1000
i = f + h
j = i / 1000
k = math.floor(j)
l = k * 100
m = l * 100
|
a ) 5100 , b ) 5120 , c ) 7200 , d ) 5400 , e ) 5500 | c | subtract(subtract(8000, multiply(8000, divide(10, const_100))), multiply(subtract(8000, multiply(8000, divide(10, const_100))), divide(10, const_100))) | the population of a town is 8000 . it decreases annually at the rate of 10 % p . a . what will be its population after 1 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8000 ã — 90 / 100 = 7200 answer : c" | a = 10 / 100
b = 8000 * a
c = 8000 - b
d = 10 / 100
e = 8000 * d
f = 8000 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(add(add(10, 30), 50), add(add(multiply(8, const_3), 20), 40)) | the average ( arithmetic mean ) of 10 , 30 , and 50 is 8 more than the average of 20 , 40 , and x . what is x ? | "the average of 10 , 30 , and 50 is 30 . the average of 20 , 40 and x is 22 . then 20 + 40 + x = 66 . x = 6 . the answer is c ." | a = 10 + 30
b = a + 50
c = 8 * 3
d = c + 20
e = d + 40
f = b - e
|
a ) 4 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 7 / 8 , e ) 8 / 7 | a | divide(multiply(add(add(const_100, const_60), const_1), 7), const_100) | what is the value of ( p + q ) / ( p - q ) if p / q is 7 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 7 + 1 ) / ( 7 - 1 ) = 8 / 6 = 4 / 3 answer : a" | a = 100 + const_60
b = a + 1
c = b * 7
d = c / 100
|
a ) 9 , b ) 36 , c ) 5 , d ) 6 , e ) 1 | c | divide(sqrt(625), 5) | what is the square root of 625 , divided by 5 ? | square root is a number times itself square root of 625 = 25 , 25 / 5 = 5 ( c ) 5 | a = math.sqrt(625)
b = a / 5
|
a ) 277 , b ) 175 , c ) 288 , d ) 266 , e ) 121 | b | multiply(350, divide(10, const_100)) | find the simple interest on rs . 350 for 10 months at 5 paisa per month ? | "i = ( 350 * 10 * 5 ) / 100 = 175 answer : b" | a = 10 / 100
b = 350 * a
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a ) 30 days , b ) 35 days , c ) 40 days , d ) 45 days , e ) 50 days | a | divide(subtract(10, add(divide(10, 60), divide(10, 60))), divide(10, 60)) | a and b can do a piece of work in 60 days and 60 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ? | "explanation : ( a + b ) ’ s 10 days work = 10 [ 1 / 60 + 1 / 60 ] = 10 [ 1 + 1 / 60 ] = 1 / 3 a complete remaining work in 1 / 3 * 60 = 20 total work = 10 + 20 = 30 days answer : option a" | a = 10 / 60
b = 10 / 60
c = a + b
d = 10 - c
e = 10 / 60
f = d / e
|
a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 16 , d ) 2 ^ 35 , e ) 2 ^ 37 | b | divide(multiply(2, add(2, 2)), 2) | 2 + 2 + 2 ² + 2 ³ . . . + 2 ^ 9 | "2 + 2 = 2 ^ 2 2 ^ 2 + 2 ^ 2 = ( 2 ^ 2 ) * ( 1 + 1 ) = 2 ^ 3 2 ^ 3 + 2 ^ 3 = ( 2 ^ 3 ) * ( 1 + 1 ) = 2 ^ 4 so you can notice the pattern . . . in the end you will have 2 ^ 9 + 2 ^ 9 , which will give you 2 ^ 10 answer b" | a = 2 + 2
b = 2 * a
c = b / 2
|
a ) 6 : 49 , b ) 1 : 3 , c ) 3 : 21 , d ) 1 : 7 , e ) 3 : 49 | a | divide(4, const_60) | what is the ratio of 4 / 7 to the product 2 * ( 7 / 3 ) ? | "4 / 7 / 14 / 3 = 6 / 49 . . . imo option a ." | a = 4 / const_60
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a ) 780 , b ) 891 , c ) 812 , d ) 847 , e ) 890 | a | multiply(add(39, const_1), divide(39, const_2)) | calculate the sum of first 39 natural numbers . | "solution we know that ( 1 + 2 + 3 + . . . . . + 39 ) = n ( n + 1 ) / 2 therefore ( 1 + 2 + 3 + . . . . + 39 ) = ( 39 × 40 / 2 ) = 780 . answer a" | a = 39 + 1
b = 39 / 2
c = a * b
|
a ) 0 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 40 % | a | divide(add(14, add(10, const_3)), const_2) | operation # is defined as adding a randomly selected two digit multiple of 14 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated 10 times , what is the probability that it will yield at least two integers ? | "any multiple of 14 is even . any two - digit prime number is odd . ( even + odd ) / 2 is not an integer . thus # does not yield an integer at all . therefore p = 0 . answer : a ." | a = 10 + 3
b = 14 + a
c = b / 2
|
a ) 120 cm 2 , b ) 130 cm 2 , c ) 312 cm 2 , d ) 315 cm 2 , e ) none of these | a | divide(multiply(24, 10), const_2) | if the sides of a triangle are 26 cm , 24 cm and 10 cm , what is its area ? | "the triangle with sides 26 cm , 24 cm and 10 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 24 * 10 = 120 cm 2 answer : a" | a = 24 * 10
b = a / 2
|
a ) 59 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | b | subtract(subtract(const_100, multiply(subtract(6, 1), const_10)), const_1) | n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 4,5 and 6 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | "you have 6 digits : 1,2 , 3 , 4 , 5 , 6 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small... | a = 6 - 1
b = a * 10
c = 100 - b
d = c - 1
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | subtract(subtract(150, 50), 40) | of the 150 people at a party , 70 were women , and 40 women tried the appetizer . if 50 people did not try the appetizer , what is the total number of men who tried the appetizer ? | "total people at party = 150 women = 70 so men 150 - 70 = 80 no . of pple who tried appetizer = 150 - 50 ( given info ) = 100 no of women who tried appetizer = 40 so remaining ppl ( men ) who tried the appetizer = 100 - 40 = 60 correct option c" | a = 150 - 50
b = a - 40
|
a ) 520 , b ) 420 , c ) 320 , d ) 550 , e ) 450 | b | multiply(sqrt(divide(multiply(4200, 7), 6)), 6) | the breath of a rectangular landscape is 6 times its length . there is a playground in it whose area is 4200 square mtr & which is 1 / 7 th of the total landscape . what is the breath of the landscape ? | "sol . 6 x * x = 7 * 4200 x = 70 length = 6 * 70 = 420 b" | a = 4200 * 7
b = a / 6
c = math.sqrt(b)
d = c * 6
|
a ) 28 days . , b ) 30 days . , c ) 37.5 days . , d ) 40 days . , e ) 36 days . | c | multiply(25, divide(75, 25)) | it was calculated that 75 men could complete a piece of work in 25 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "one day work = 1 / 25 one man ’ s one day work = 1 / ( 25 * 75 ) now : no . of workers = 50 one day work = 50 * 1 / ( 25 * 75 ) the total no . of days required to complete the work = ( 75 * 25 ) / 50 = 37.5 answer : c" | a = 75 / 25
b = 25 * a
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a ) 100 , b ) 140 , c ) 180 , d ) 160 , e ) 120 | b | add(const_10, add(multiply(multiply(7, 4), const_4), const_12)) | a farmer divides his herd of x cows among his 4 sons so that one son gets one half of the herd , the second gets one - fourth , the third gets one - fifth and the fourth gets 7 cows . then x is equal to | no . of cows : : x 1 st son : x / 2 2 nd son : x / 4 3 rd son : x / 5 4 th son : 7 ( x ) + ( x / 4 ) + ( x / 5 ) + 7 = x = > x - ( 19 x / 20 ) = 7 = > ( 20 x - 19 x ) / 20 = 7 = > x = 140 answer : b | a = 7 * 4
b = a * 4
c = b + 12
d = 10 + c
|
a ) 2 , b ) 3 , c ) 1 , d ) - 2 , e ) 4 | d | multiply(5, divide(3, 5)) | solve for x and check : x + 5 = 3 | "solution : x + 5 - 5 = 3 - 5 x = - 2 check : x + 5 = 3 - 2 + 5 = 3 3 = 3 answer : d" | a = 3 / 5
b = 5 * a
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a ) 11.33 kmph , b ) 12.33 kmph , c ) 13.33 kmph , d ) 14.33 kmph , e ) 23.33 kmph | c | divide(multiply(40, 3), add(divide(40, 40), divide(multiply(4, 40), 20))) | a trained covered x km at 40 kmph and another 4 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 40 + 4 x / 20 hours = 9 x / 40 hours average speed = 3 x / ( 9 x / 40 ) = 13.33 kmph answer : c" | a = 40 * 3
b = 40 / 40
c = 4 * 40
d = c / 20
e = b + d
f = a / e
|
a ) 6 , b ) 10 , c ) 15 , d ) 40 , e ) 21 | e | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(56, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add... | a rectangular block 8 cm by 24 cm by 56 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 8 * 24 * 56 = 10752 cm ^ 3 side of the largest cube = h . c . f of 8 , 24,56 = 8 cm volume of the cube = 8 * 8 * 8 = 512 cm ^ 3 number of cubes = 10752 / 512 = 21 answer is e" | a = rectangle_area / (
b = a - 10
c = 3 * 2
d = 1000 * c
e = b + d
f = 3 * 1000
g = 56 * 10
h = f + g
i = e / h
|
a ) $ 83 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 104 | b | subtract(multiply(90, 15), add(multiply(87, 7), multiply(93, 7))) | the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 93 per day . what was his wage on the 8 th day ? | "average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 93 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 93 * 7 ) = 1350 - ( 609 + 651 ) = 1350 - 1260 = 90 answer b" | a = 90 * 15
b = 87 * 7
c = 93 * 7
d = b + c
e = a - d
|
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 5 : 3 | e | multiply(divide(2, 3), multiply(divide(2, 3), divide(5, 3))) | find the compound ratio of ( 2 : 3 ) , ( 5 : 11 ) and ( 11 : 2 ) is | "required ratio = 2 / 3 * 5 / 11 * 11 / 2 = 2 / 1 = 5 : 3 answer is e" | a = 2 / 3
b = 2 / 3
c = 5 / 3
d = b * c
e = a * d
|
a ) 60 , b ) 120 , c ) 240 , d ) 360 , e ) 1530 | e | lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 18) | what is the least common multiple of 15 , 18 , and 34 ? | "let us first write the numbers in the form of prime factors : 15 = 3 * 5 18 = 2 * 3 ^ 2 34 = 2 * 17 ^ 1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 1530 option e" | a = 10 + 2
b = 3 * 10
c = b - 3
d = math.lcm(a, c)
e = math.lcm(d, 18)
|
a ) 48 , b ) 55 , c ) 58 , d ) 62 , e ) 70 | b | subtract(choose(8, 4), choose(subtract(8, 2), 2)) | a meeting has to be conducted with 4 managers . find the number of ways in which the managers may be selected from among 8 managers , if there are 2 managers who refuse to attend the meeting together . | "the total number of ways to choose 4 managers is 8 c 4 = 70 we need to subtract the number of groups which include the two managers , which is 6 c 2 = 15 . 70 - 15 = 55 the answer is b ." | a = math.comb(8, 4)
b = 8 - 2
c = math.comb(b, 2)
d = a - c
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a ) 40 , b ) 50 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | e | power(const_10, subtract(9, 3)) | on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 9 is how many times as great as the frequency corresponding to a r... | "if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 3 to 4 it will be 10 if richter scale reading goes from 4 to 5 it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading goes from 6 to 7 it will be 10 and if richter scale read... | a = 9 - 3
b = 10 ** a
|
a ) 68 , b ) 78 , c ) 88 , d ) 138 , e ) 108 | d | subtract(add(72, 30), divide(72, divide(30, const_100))) | the contents of a certain box consist of 72 apples and 30 oranges . how many oranges must be added to the box so that exactly 30 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.3 72 = ( 30 + 72 + x ) * 0.3 x = 138 answer : d" | a = 72 + 30
b = 30 / 100
c = 72 / b
d = a - c
|
a ) 4436 toys , b ) 1500 toys , c ) 6113 toys , d ) 2354 toys , e ) 1375 toys | b | divide(6000, 4) | a factory produces 6000 toys per week . if the workers at this factory work 4 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 4 days ) by 4 . 6000 / 4 = 1500 toys correct answer b" | a = 6000 / 4
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a ) 7 / 4 , b ) 32 / 9 , c ) 15 / 4 , d ) 10 / 3 , e ) 17 / 5 | b | max(divide(subtract(const_1, multiply(add(divide(const_1, add(const_4, const_2)), divide(const_1, 9)), const_2)), divide(const_1, 9)), const_3) | two mechanics were working on your car . one can complete the given job in six hours , but the new guy takes 9 hours . they worked together for the first two hours , but then the first guy left to help another mechanic on a different job . how long will it take the new guy to finish your car ? | "rate ( 1 ) = 1 / 6 rate ( 2 ) = 1 / 9 combined = 5 / 18 work done in 2 days = 5 / 9 work left = 4 / 9 rate * time = work left 1 / 8 * time = 4 / 9 time = 32 / 9 b" | a = 4 + 2
b = 1 / a
c = 1 / 9
d = b + c
e = d * 2
f = 1 - e
g = 1 / 9
h = f / g
i = max(h)
|
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | c | add(power(divide(1, const_2), const_3), multiply(multiply(3, power(divide(1, const_2), const_2)), divide(1, const_2))) | a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box , what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ? | the sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls ( odd + odd + odd = odd ) or two even numbered balls and one odd numbered ball ( even + even + odd = odd ) ; p ( ooo ) = ( 1 / 2 ) ^ 3 ; p ( eeo ) = 3 * ( 1 / 2 ) ^ 2 * 1 / 2 = 3 / 8 ( you should ... | a = 1 / 2
b = a ** 3
c = 1 / 2
d = c ** 2
e = 3 * d
f = 1 / 2
g = e * f
h = b + g
|
a ) 33 : 1 , b ) 33 : 7 , c ) 33 : 8 , d ) 33 : 5 , e ) 33 : 2 | b | divide(add(divide(multiply(62.5, 2), const_100), divide(multiply(87.5, 8), const_100)), add(subtract(2, divide(multiply(62.5, 2), const_100)), subtract(8, divide(multiply(87.5, 8), const_100)))) | two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 2 litres from vessel p is mixed with 8 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 2 = 5 / 4 litres quantity of alcohol in vessel q = 87.5 / 100 * 4 = 7 / 1 litres quantity of alcohol in the mixture formed = 5 / 4 + 7 / 1 = 33 / 4 = 8.25 litres as 10 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 8.25 : 1.75 = 33 : 7 . ... | a = 62 * 5
b = a / 100
c = 87 * 5
d = c / 100
e = b + d
f = 62 * 5
g = f / 100
h = 2 - g
i = 87 * 5
j = i / 100
k = 8 - j
l = h + k
m = e / l
|
a ) 20 , b ) 23 , c ) 25 , d ) 21 , e ) 22 | c | subtract(multiply(add(39, 3), add(15, 3)), multiply(39, 15)) | the average age of a class of 39 students is 15 years . if the age of the teacher be included , then the average increases by 3 months . find the age of the teacher . | "total age of 39 persons = ( 39 x 15 ) years = 585 years . average age of 40 persons = 15 yrs 3 months = 61 / 4 years . total age of 40 persons = ( 61 / 4 ) x 40 ) years = 610 years . : . age of the teacher = ( 610 - 585 ) years = 25 years . answer is c ." | a = 39 + 3
b = 15 + 3
c = a * b
d = 39 * 15
e = c - d
|
a ) 2.91 , b ) 3.48 , c ) 2.98 , d ) 3.78 , e ) 4.21 | d | multiply(divide(multiply(add(5, 2.3), subtract(5, 2.3)), add(add(5, 2.3), subtract(5, 2.3))), const_2) | a man can row 5 kmph in still water . when the river is running at 2.3 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 5 s = 2.3 ds = 6.3 us = 2.7 x / 6.3 + x / 2.7 = 1 x = 1.89 d = 1.89 * 2 = 3.78 answer : d" | a = 5 + 2
b = 5 - 2
c = a * b
d = 5 + 2
e = 5 - 2
f = d + e
g = c / f
h = g * 2
|
a ) 9 , b ) 13.04 , c ) 10 , d ) 11 , e ) 12 | b | multiply(divide(const_1, multiply(divide(add(15, const_100), const_100), 15)), const_100) | a part - time employee whose hourly wage was increased by 15 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | "let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.15 x now we need to find number of hours worked so that x * y = 1.15 x * z i . e z = 1 / 1.15 y % decrease = ( y - 1 / 1.15 y ) / y * 100 = 13.04 % thus my answer is b ." | a = 15 + 100
b = a / 100
c = b * 15
d = 1 / c
e = d * 100
|
a ) 300 , b ) 600 , c ) 900 , d ) 800 , e ) 500 | c | multiply(4650, divide(6, add(add(add(10, add(4, const_1)), 10), 6))) | if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4650 , how much r will receive ? | "let p ' s capital be p , q ' s capital be q , and r ' s capital be r then 4 p = 6 q = 10 r 2 p = 3 q = 5 r ⋯ ( a ) from ( a ) , q = 2 p / 3 ⋯ ( 1 ) r = 2 p / 5 ⋯ ( 2 ) p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 4650 × 6 / 31 = 150 × 6 = 900 answer is c ." | a = 4 + 1
b = 10 + a
c = b + 10
d = c + 6
e = 6 / d
f = 4650 * e
|
a ) 240 mts , b ) 270 mts , c ) 260 mts , d ) 250 mts , e ) none of these | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 260) | a 260 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | explanation : relative speed = 120 + 80 = 200 km / hr . = 200 x 5 / 18 = 500 / 9 m / sec . let the length of the other train be l mts . then , ( l + 260 ) / 9 = 500 / 9 = > l = 240 mts . answer is a | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 260
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | divide(add(factorial(6), 6), add(factorial(6), 6)) | for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers r are there between 6 * + 2 and 6 * + 6 , inclusive ? | "given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so... | a = math.factorial(6)
b = a + 6
c = math.factorial(6)
d = c + 6
e = b / d
|
a ) 1628.4 , b ) 1534 , c ) 1492 , d ) 158.25 , e ) none of these | d | multiply(divide(add(multiply(15, 5), multiply(subtract(25, 15), 4)), subtract(25, subtract(25, 15))), 25) | 25 people went to a hotel for combine dinner party 15 of them spent rs . 5 each on their dinner and rest spent 4 more than the average expenditure of all the 25 . what was the total money spent by them . | "solution : let average expenditure of 25 people be x . then , 25 x = 15 * 5 + 10 * ( x + 4 ) ; or , 25 x = 15 * 5 + 10 x + 20 ; or , x = 6.33 ; so , total money spent = 6.33 * 25 = rs . 158.25 . answer : option d" | a = 15 * 5
b = 25 - 15
c = b * 4
d = a + c
e = 25 - 15
f = 25 - e
g = d / f
h = g * 25
|
a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | a | subtract(multiply(multiply(const_4, const_4), const_2), const_3) | find the number of divisors of 1080 excluding the divisors which are perfect squares . | 1080 = 2 ^ 3 * 3 ^ 3 * 5 ^ 1 total no . of divisors = ( 3 + 1 ) * ( 3 + 1 ) * ( 1 + 1 ) = 32 only 4 divisors 1 , 2 ^ 2 = 4 , 3 ^ 2 = 9 & 2 ^ 2 * 3 ^ 2 = 36 are perfect squares so , number of divisors excluding perfect squares divisors = 32 - 4 = 28 answer : a | a = 4 * 4
b = a * 2
c = b - 3
|
a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | c | add(subtract(divide(60, 5), const_2), add(divide(60, 3), divide(60, 5))) | line q has the equation 5 y – 3 x = 60 . if line s is perpendicular to q , has an integer for its y - intercept , and intersects q in the second quadrant , then how many possible line s ’ s exist ? ( note : intersections on one of the axes do not count . ) | "5 y - 3 x = 60 and so y = 3 x / 5 + 12 when x = 0 , then y = 12 . when y = 0 , then x = - 20 the slope is 3 / 5 , so the slope of line s is - 5 / 3 . through the point ( - 20 , 0 ) , 0 = - 5 ( - 20 ) / 3 + c the y - intercept is c = - 100 / 3 < - 33 . thus the perpendicular line s can have y - intercepts from - 33 up ... | a = 60 / 5
b = a - 2
c = 60 / 3
d = 60 / 5
e = c + d
f = b + e
|
a ) $ 255 , b ) $ 357 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550 | b | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 0.70) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.70 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum... | "total no . of boxes = 3060000 / ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 0.7 = $ 357 answer b" | a = 1000 * 1000
b = 3 * 6
c = 20 * 20
d = c * 15
e = b / d
f = e * 0
|
a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number y of races we need to conduct to get the top 3 fastest horses ? | "y = 7 is the correct answer . good solution buneul . b" | a = 25 / 5
b = a + 2
|
a ) 70 minutes , b ) 2 hours , c ) 2 / 3 hours , d ) 5 / 6 hours , e ) 85 minutes | d | divide(10, subtract(inverse(divide(5, 6)), const_1)) | walking at 5 / 6 of its usual speed , a train is 10 minutes too late . what is usual time to cover the journey ? | "et speed be s and time to cover the journey be t s * t = d ( distance of journey ) - - - 1 now as per the question stem we have ( 5 / 6 ) s * ( t + 10 ) = d - - - - 2 equating value of d from ( 1 ) in ( 2 ) we get t = 50 mins or 5 / 6 hrs . answer : d" | a = 5 / 6
b = 1/(a)
c = b - 1
d = 10 / c
|
a ) rs 7.41 , b ) rs 9.81 , c ) rs 10.41 , d ) rs 11.81 , e ) none of these | a | divide(multiply(6, add(const_100, 5)), subtract(const_100, 15)) | a fruit seller sells mangoes at the rate of rs . 6 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 85 : 6 = 105 : x x = ( 6 × 105 / 85 ) = rs 7.41 option a" | a = 100 + 5
b = 6 * a
c = 100 - 15
d = b / c
|
a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 3500 , e ) $ 3750 | e | subtract(5000, divide(5000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | the salaries of a and b together amount to $ 5000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 5000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 5000 - x ) x = $ 3750 answer is e" | a = 100 - 85
b = 100 - 95
c = a / b
d = c + 1
e = 5000 / d
f = 5000 - e
|
a ) 1 / 4 , b ) 1 / 6 , c ) 1 / 8 , d ) 1 / 12 , e ) 1 / 16 | c | add(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), 4)) | when tossed , a certain coin has an equal probability of landing on either side . if the coin is tossed 4 times , what is the probability that it will land on the same side each time ? | on the first toss , the coin will land on one side or the other . on the second toss , the probability of landing on the same side is 1 / 2 . on the third toss , the probability of landing on the same side is 1 / 2 . on the fourth toss , the probability of landing on the same side is 1 / 2 . p ( same side all four time... | a = 1 / 2
b = a ** 4
c = 1 / 2
d = c ** 4
e = b + d
|
a ) 3 % , b ) 5 % , c ) 8 % , d ) 9 % , e ) 12 % | d | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(9, const_60), 42), add(multiply(10, const_60), 40)))) | bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 40 seconds and his sister ’ s time is currently 9 minutes 42 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | "bob ' s time = 640 secs . his sis ' time = 582 secs . percent increase needed = ( 640 - 582 / 640 ) * 100 = 58 / 640 * 100 = 9 % . ans ( d ) ." | a = 10 * 10
b = 9 * const_60
c = b + 42
d = 10 * const_60
e = d + 40
f = c / e
g = 1 - f
h = a * g
|
a ) 784596 , b ) 845796 , c ) 804670 , d ) 482802 , e ) 864520 | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 300), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 300 miles into meters ? | "1 mile = 1609.34 meters 300 mile = 300 * 1609.34 = 482802 meters answer is d" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 300
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) one , b ) two , c ) three , d ) seven , e ) ten | b | add(1, 2) | if d = 1 / ( 2 ^ 3 * 5 ^ 9 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | "another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 9 ) = multiply and divide by 2 ^ 4 = 2 ^ 6 / ( 2 ^ 3 * 2 ^ 6 * 5 ^ 9 ) = 2 ^ 6 / 10 ^ 9 = > non zero digits are 64 = > ans b" | a = 1 + 2
|
a ) 70 , b ) 86.5 , c ) 80 , d ) 88.5 , e ) 75 | a | add(multiply(5, 4), 50) | the average weight of 5 person ' s increases by 4 kg when a new person comes in place of one of them weighing 50 kg . what is the weight of the new person ? | "total increase in weight = 5 × 4 = 20 if x is the weight of the new person , total increase in weight = x − 50 = > 20 = x - 50 = > x = 20 + 50 = 70 answer : a" | a = 5 * 4
b = a + 50
|
a ) 141 , b ) 180 , c ) 130 , d ) 122 , e ) 420 | e | add(floor(divide(20, const_3)), const_1) | what is the smallest integer that is multiple of 7 , 12 and 20 | "correct answer : e it is the lcm of 7 , 12 and 20 which is 420" | a = 20 / 3
b = math.floor(a)
c = b + 1
|
a ) 9 years , b ) 11 years , c ) 17 years , d ) 21 years , e ) 25 years | c | subtract(multiply(15, 15), add(multiply(8, 14), multiply(6, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 8 students is 14 years and that of the other 6 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 8 + 16 x 6 ) ] = ( 225 - 208 ) = 17 years . answer c" | a = 15 * 15
b = 8 * 14
c = 6 * 16
d = b + c
e = a - d
|
a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 45 % | e | multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100) | the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 174,000 . what was the percent increase in the cost of the home ? | "increase = 174000 - 120000 = 54000 % increase = 54000 * 100 / 120000 = 45 % answer : option e" | a = 4 * 4
b = 12 * a
c = b * 1000
d = 12 * 1000
e = d * 10
f = c - e
g = 12 * 1000
h = g * 10
i = f / h
j = i * 100
|
a ) 834 , b ) 750 , c ) 633 , d ) 654 , e ) 812 | a | add(200, 400) | in the faculty of reverse - engineering , 200 second year students study numeric methods , 400 second year students study automatic control of airborne vehicles and 100 second year students study them both . how many students are there in the faculty if the second year students are approximately 60 % of the total ? | "total number of students studying both are 400 + 200 - 100 = 500 ( subtracting the 100 since they were included in the both the other numbers already ) . so 60 % of total is 500 , so 100 % is approx . 834 . answer is a" | a = 200 + 400
|
a ) 19 , b ) 19.7 , c ) 21.3 , d ) 21.5 , e ) 29.7 | e | subtract(35, add(multiply(divide(30, const_100), divide(const_100, 6)), divide(30, const_100))) | jerry went to a shop and bought things worth rs . 35 , out of which 30 % went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 35 given that out of this rs . 35 , 30 % is given as tax = > total tax incurred = 30 % = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % ∴ ( 35 − 30 / 100 − x ) 6 / 100 = 30 / 100 ⇒ 6 ( 35 − 0.3 − x ) = 30 ⇒ ( 35 − 0.3 − x ) = 5 ⇒ x = 35 − 0.3 ... | a = 30 / 100
b = 100 / 6
c = a * b
d = 30 / 100
e = c + d
f = 35 - e
|
a ) 82.1 sec , b ) 12.1 sec , c ) 24.2 sec , d ) 13.1 sec , e ) 12.15 sec | c | divide(add(110, 132), multiply(36, const_0_2778)) | how long does a train 110 m long running at the speed of 36 km / hr takes to cross a bridge 132 m length ? | "speed = 36 * 5 / 18 = 10 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 10 = 24.2 sec . answer : c" | a = 110 + 132
b = 36 * const_0_2778
c = a / b
|
a ) 8 / 15 , b ) 4 / 3 , c ) 15 / 8 , d ) 9 / 4 , e ) 15 / 4 | e | multiply(divide(const_1, add(divide(const_1, 3), divide(const_1, 5))), 3) | kathleen can paint a room in 3 hours , and anthony can paint an identical room in 5 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "kat - - - 3 hrs - - - 1 room ; so rate is work / time = 1 / 3 anthony - - - 5 hrs - - - 1 room ; rate = 1 / 5 rate of kat and anthony together = 1 / 3 + 1 / 5 = 8 / 15 but we are told that they both paint 2 rooms which are identical . so ; work is 2 rooms , rate is 8 / 15 , total time required for both of them togethe... | a = 1 / 3
b = 1 / 5
c = a + b
d = 1 / c
e = d * 3
|
a ) 12.85 , b ) 12.62 , c ) 12.5 , d ) 12.24 , e ) 12.1 | a | divide(multiply(15, 1200), add(1200, 200)) | 1200 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ? | "1200 * 15 = 1400 * x x = 12.85 . answer : a" | a = 15 * 1200
b = 1200 + 200
c = a / b
|
a ) 55 , b ) 65 , c ) 100 , d ) 109 , e ) 129 | e | subtract(multiply(50, 4), multiply(35, 2)) | the average ( arithmetic mean ) of 4 positive integers is 50 . if the average of 2 of these integers is 35 , what is the greatest possible value that one of the other 2 integers can have ? | "a + b + c + d = 200 a + b = 70 c + d = 130 greatest possible = 129 ( just less than 1 ) answer = e" | a = 50 * 4
b = 35 * 2
c = a - b
|
a ) rs . 8503.49 , b ) rs . 9720 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 21,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 21000 * ( 1 + 12 / 100 ) 3 ] = 21000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 29503.49 c . i . = ( 29503.49 - 21000 ) = rs . 8503.49 answer : a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 1.55 , b ) 1.85 , c ) 1.65 , d ) 1.35 , e ) 1.15 | d | divide(add(divide(45, 10), divide(72, 10)), const_2) | a man swims downstream 72 km and upstream 45 km taking 10 hours each time ; what is the speed of the current ? | "72 - - - 10 ds = 7.2 ? - - - - 1 45 - - - - 10 us = 4.5 ? - - - - 1 s = ? s = ( 7.2 - 4.5 ) / 2 = 1.35 answer : d" | a = 45 / 10
b = 72 / 10
c = a + b
d = c / 2
|
a ) 12 cm , b ) 14 cm , c ) 16 cm , d ) 18 cm , e ) 20 cm | c | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 65 sq . cm . find the length of the rectangle . | "explanation : let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 5 ) - 2 x * x = 65 = > 5 x - 25 = 65 = > x = 16 . length of the rectangle = 16 cm . answer : option c" | a = 100 / 3
|
a ) 5568 , b ) 6369 , c ) 5460 , d ) 5635 , e ) 6552 | e | divide(multiply(divide(multiply(6000, add(const_100, 4)), const_100), add(const_100, 5)), const_100) | find the amount on rs . 6000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ? | "6000 * 104 / 100 * 105 / 100 = > 6552 answer : e" | a = 100 + 4
b = 6000 * a
c = b / 100
d = 100 + 5
e = c * d
f = e / 100
|
a ) 520 , b ) 620 , c ) 820 , d ) 768 , e ) 720 | d | divide(divide(multiply(320, 4), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 320 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 240 x 5 ) = 1280 km . speed = distance / time speed = 1280 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 1280 x 3 / 5 ) km / hr = 768 km / hr answer d ) 768 km / hr" | a = 320 * 4
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
|
a ) 16 % , b ) 18 % , c ) 20 % , d ) 22 % , e ) 24 % | a | multiply(subtract(const_1, multiply(add(divide(20, const_100), const_1), divide(70, const_100))), const_100) | a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 20 % so that the customer could not buy the required amount of the product . the customer managed to buy only 70 % of the required amount . what is the difference in the amount of money that the customer paid for th... | let x be the amount of money paid for the first purchase . the second time , the customer paid 0.7 ( 1.2 x ) = 0.84 x . the difference is 16 % . the answer is a . | a = 20 / 100
b = a + 1
c = 70 / 100
d = b * c
e = 1 - d
f = e * 100
|
a ) 15 , b ) 20 , c ) 30 , d ) 25 , e ) 18 | c | divide(multiply(5, 12), subtract(12, 10)) | a group of men decided to do a work in 10 days , but 5 of them became absent . if the rest of the group did the work in 12 days , find the original number of men ? | original number of men = 5 * 12 / ( 12 - 10 ) = 30 answer is c | a = 5 * 12
b = 12 - 10
c = a / b
|
a ) 12 : 00 pm , b ) 1 : 00 pm , c ) 2 : 00 pm , d ) 3 : 00 pm , e ) 4 : 00 pm | b | subtract(add(multiply(inverse(add(divide(const_1, 10), divide(const_1, 12))), subtract(const_1, multiply(add(add(divide(const_1, 8), divide(const_1, 10)), divide(const_1, 12)), const_2))), 11), 12) | george can do a piece of work in 8 hours . paul can do the same work in 10 hours , hari can do the same work in 12 hours . george , paul and hari start the same work at 9 am , while george stops at 11 am , the remaining two complete the work . what time will the work complete ? | ( ( 1 / 8 ) + ( 1 / 10 ) + ( 1 / 12 ) ) 2 + ( ( 1 / 10 ) + ( 1 / 12 ) ) ( x - 2 ) = 1 x = 4 hrs work started at 9 am . . takes 4 hrs to complete 9 am + 4 hrs = 1 : 00 pm answer : b | a = 1 / 10
b = 1 / 12
c = a + b
d = 1/(c)
e = 1 / 8
f = 1 / 10
g = e + f
h = 1 / 12
i = g + h
j = i * 2
k = 1 - j
l = d * k
m = l + 11
n = m - 12
|
a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | if x and y are both odd prime numbers and x < y , how many distinct positive integer factors does 2 xy have ? | the solution should be the same for all numbers that fulfill the condition above . as x and y are odd prime numbers , they must be > 2 . also , they must be distinct as x < y . i choose x = 3 and y = 5 : how many factors does 2 x 3 x 5 = 30 have : 1 , 2 , 3 , 5 , 6 , 10 , 15 , 30 = 8 distinct positive integer factors .... | a = 1 + 1
b = 2 * a
c = 1 + 1
d = b * c
|
a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 5 minutes | c | multiply(3, const_1) | if 3 cats can kill 3 rats in 3 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 3 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 3 minutes , so 100 cats can kill 100 rats in 3 minutes answer c" | a = 3 * 1
|
a ) 1 / 4 , b ) 2 / 3 , c ) 1 , d ) 3 / 2 , e ) 2 | a | divide(subtract(6, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 5)) | hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of two hammers and three wrenches is one - third that of 6 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ? | "x be the weight of a hammer and y be the weight of a wrench . ( 2 x + 3 y ) = 1 / 3 * ( 7 x + 5 y ) 3 ( 2 x + 3 y ) = ( 7 x + 5 y ) 6 x + 9 y = 7 x + 5 y 4 y = x y = x / 4 ans - a" | a = 3 * 2
b = 6 - a
c = 3 * 2
d = c - 5
e = b / d
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 9 / 11 | e | inverse(add(divide(subtract(35, 25), subtract(80, 35)), const_1)) | a certain quantity of 80 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | "let ' s say that the total original mixture a is 100 ml the original mixture a thus has 50 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 80 ml and 25 ml is 55 ml per 100 ml of ... | a = 35 - 25
b = 80 - 35
c = a / b
d = c + 1
e = 1/(d)
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | the number of diagonals of a polygon of n sides is given by the formula c = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | c = n ( n - 3 ) c = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b | a = 3 + 2
|
a ) 272258 , b ) 272358 , c ) 283974 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 53), const_100) | 5358 x 53 = ? | "5358 x 51 = 5358 x ( 50 + 3 ) = 5358 x 50 + 5358 x 3 = 267900 + 16074 = 283974 . c )" | a = 5358 / 53
b = a * 100
|
a ) 100 , b ) 110 , c ) 120 , d ) 30 , e ) 160 | c | add(80, multiply(80, divide(50, const_100))) | 80 is increased by 50 % . find the final number . | "final number = initial number + 50 % ( original number ) = 80 + 50 % ( 80 ) = 80 + 40 = 120 . answer c" | a = 50 / 100
b = 80 * a
c = 80 + b
|
a ) 12 , b ) 81 , c ) 77 , d ) 66 , e ) 73 | a | multiply(12, const_1) | the total age of a and b is 12 years more than the total age of b and c . c is how many years younger than a ? | "( a + b ) - ( b + c ) = 12 a - c = 12 answer : a" | a = 12 * 1
|
a ) 900 , b ) 1600 , c ) 750 , d ) 890 , e ) 1010 | a | multiply(divide(17, add(add(15, 13), 17)), 2400) | in a bag , there are 2400 ball , and their colors are red , green blue . . the ratio of the balls are 15 : 13 : 17 . then how many red color balls are available in the bag ? | red : green : blue = 15 + 13 + 17 = 45 ; ratio of the red balls = 15 / 40 simplify = 3 / 8 * 2400 = 900 . answer = a | a = 15 + 13
b = a + 17
c = 17 / b
d = c * 2400
|
a ) 50 , b ) 55 , c ) 60 , d ) can not be determined , e ) none of these | a | add(39, const_1) | the average age of 39 students in a group is 10 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "age of the teacher = ( 40 × 11 – 39 × 10 ) years = 50 years . answer a" | a = 39 + 1
|
a ) 12 kg , b ) 8.3 kg , c ) 10.7 kg , d ) 15.5 kg , e ) 7 kg | c | divide(multiply(24, 4), 9) | an alloy is to contain copper and zinc in the ratio 9 : 4 . the zinc required to be melted with 24 kg of copper is ? | "let the required quantity of copper be x kg 9 : 4 : : 24 : x 9 x = 4 * 24 x = 10 2 / 3 kg = 32 / 3 = 10.7 kg answer is c" | a = 24 * 4
b = a / 9
|
a ) 10 , b ) 11.3 , c ) 13 , d ) 14 , e ) 15 | b | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 9))) | two trains of equal lengths take 9 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 / 9 = 13.3 m / sec . speed of the second train = 120 / 15 = 8 m / sec . relative speed = 13.3 + 8 = 21.3 m / sec . required time = ( 120 + 120 ) / 21.3 = 11.3 sec . answer : option b" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 112 , b ) 133 , c ) 150 , d ) 167 , e ) 200 | e | multiply(40, divide(350, add(40, 30))) | trains a and b start simultaneously from stations 350 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 40 miles per hour and 30 miles per hour , respectively , how many miles will train a have traveled when the trains pass each o... | since we know the distance ( 350 ) and the combined rate ( 70 ) , we plug it into the formula : distance = rate * time 350 = 70 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 350 / 70 from dividing 70 on both sides to isolate time in the equ... | a = 40 + 30
b = 350 / a
c = 40 * b
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 7 , d ) 1 / 2 , e ) 4 / 7 | d | divide(multiply(6, const_1), add(multiply(6, const_1), multiply(6, const_1))) | harold works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 6 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | the time from beginning of march to the end of september is 7 months . if x is the average monthly tip for all months other than august then his august month tip will be 6 * x his total tip for the 7 months = 6 * ( average tip for the months other than august ) + 6 x = 12 x august tips as a fraction of total tips = 6 x... | a = 6 * 1
b = 6 * 1
c = 6 * 1
d = b + c
e = a / d
|
a ) 0 , b ) 2 , c ) 4 , d ) 8 , e ) more than 8 | d | power(2, const_3) | for how many values of n , is | | | n - 5 | - 10 | - 5 | = 2 ? ( those ls are mods ) | i think its 8 | | | n - 5 | - 10 | - 5 | = 2 let | n - 5 | = a which makes above | | a - 10 | - 5 | = 2 let | a - 10 | = b which makes | b - 5 | = 2 now for the above b can take 3 , 7 for every b = 3 a can have 13 , 7 and for b = 7 a can have 17 and 3 so ' a ' has four solutions 13 , 7 , 17 and 3 for a = 13 ; x has 18 ... | a = 2 ** 3
|
a ) 10 % , b ) 12 % , c ) 14 % , d ) 17 % , e ) 20 % | c | multiply(divide(subtract(divide(50, const_100), divide(42, const_100)), subtract(const_1, divide(42, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 42 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "let me try a simpler one . lets assume that candidate got 42 % votes and total votes is 100 . candidate won = 42 remaining = 58 to get 50 % , candidate requires 8 votes from 100 which is 8 % and 8 votes from 58 . 8 / 58 = . 137 = 13.7 % which is approx 14 % . hence the answer is c ." | a = 50 / 100
b = 42 / 100
c = a - b
d = 42 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) 56 , b ) 51 , c ) 59 , d ) 54 , e ) 52 | d | add(40, divide(subtract(976, multiply(15, 40)), divide(multiply(15, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 15 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compens... | "for 40 hrs = 40 * 15 = 600 excess = 976 - 600 = 376 for extra hours = . 75 ( 15 ) = 11.25 + 15 = 26.25 number of extra hrs = 376 / 26.25 = 14.3 = 14 approx . total hrs = 40 + 14 = 54 answer d" | a = 15 * 40
b = 976 - a
c = 100 + 75
d = 15 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | subtract(subtract(add(6, add(6, 15)), 3), const_2) | in a class , 6 students can speak tamil , 15 can speak gujarati and 6 can speak h . if two students can speak two languages and one student can speak all the 3 languages , then how many students are there in the class ? | t = 6 g = 15 h = 6 students can speak two languages ( suppose tamil & gujarati ) = 2 student can speak all the three languages = 1 student can speak only tamil language = 6 - 2 - 1 = 3 student can speak only gujarati language = 15 - 2 - 1 = 12 student can speak only h language = 6 - 1 = 5 so , total number of student =... | a = 6 + 15
b = 6 + a
c = b - 3
d = c - 2
|
a ) 12 , b ) 17 , c ) 2 , d ) 77 , e ) 26 | c | subtract(8, reminder(3198, 8)) | what should be the least number to be added to the 3198 number to make it divisible by 8 ? | "answer : 2 option : c" | a = 8 - reminder
|
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