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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 sale ? | 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
|
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 in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . so , he will take 30 days to doing 5 w widgets . answer : d" | 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 / (
|
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 = 55 / 60 = 11 / 12 hence , answer will be a . | 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
|
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
|
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 as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 1 and 6 in the hundreds places and reserve them for the tens places now what are the options ? try and make a pair with ( 2 * * and 3 * * ) . make the 2 * * number as large as possible and make the 3 * * number as small as possible . 265 and 314 ( difference is 49 ) or try and make a pair with ( 4 * * and 5 * * ) . make the 4 * * number as large as possible and make the 5 * * number as small as possible . we get 463 and 512 ( difference is 49 ) b" | 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
|
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
|
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(multiply(const_3, const_1000), multiply(56, const_10))), const_10)) | 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
|
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 reading of 3 ? | "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 reading goes from 7 to 8 it will be 10 and if richter scale reading goes from 8 to 9 it will be 10 so it will from 3 to 9 i . e 4,5 , 6,7 , 8,9 = 10 * 10 * 10 * 10 * 10 * 10 = 10 ^ 6 answer is e" | 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
|
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 multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways : eeo , eoe , or oee ) ; so finally p = 1 / 8 + 3 / 8 = 1 / 2 . answer : c . | 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 . answer : b" | 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 it can not be a prime number , . . . the same way for all numbers between 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 , inclusive . which means that there are no primes r in this range . answer : a ." | 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 to 11 . the number of possible lines is 33 + 11 + 1 = 45 the answer is c ." | 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 amount the university must spend on boxes ? | "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 times ) = 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 . the answer is c . | 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 − 5 = 29.7 e )" | 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 together to complete the work = 2 / ( 8 / 15 ) = 30 / 8 = 15 / 4 hence answer = e" | 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 the second purchase compared to the first purchase ? | 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 . hence , the correct answer is d . | 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
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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 mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 25 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 45 % . therefore , 45 out of 55 is 9 / 11 and e is the answer ." | 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 other , to the nearest mile ? | 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 equation above . time will be 5 hours so now you can plug that in for train a ’ s distance . distance = rate * time distance = 40 * 5 distance = 200 according to answer choice e . | 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 / 12 x = 1 / 2 . so d | 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 or - 8 thus has 2 for every combination hence 4 x 2 = 8 answer d | a = 2 ** 3
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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 compensation , how many total hours did he work that week ? | "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 = 2 + 1 + 3 + 12 + 5 = 23 answer : d | a = 6 + 15
b = 6 + a
c = b - 3
d = c - 2
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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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