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 ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | c | multiply(subtract(divide(add(divide(10, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(20, 14), 20))), const_1), const_100) | during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 14 candies . the store still makes a profit of 10 % on this sale . what is the mark up percentage on each candy ? | "hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the child bought it for 14 $ so , he got a discount of 6 / 20 * 100 = 30 % so , the latest s . p . = 70 % of 80 % of m = 0.7 * 0.8 m now , we are given that the shopkeeper still makes a profit of 12 % . so we have , 0.7 * 0.8 * m = 1.12 c . p so , we get , m = 2 c . p . i . e . marked price was kept 100 % above c . p . c" | a = 10 / 100
b = a + 1
c = 20 / 100
d = 1 - c
e = 20 - 14
f = e / 20
g = d * f
h = b / g
i = h - 1
j = i * 100
|
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30 | b | subtract(multiply(14, const_2), const_1) | a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 14 multiples of 14 , how many multiples of 7 are there in q ? | "halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 14 + 13 = 27 . the answer is b ." | a = 14 * 2
b = a - 1
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a ) 1500 , b ) 266 , c ) 299 , d ) 750 , e ) 261 | a | divide(divide(multiply(180, const_1000), divide(const_60, const_1)), const_2) | the length of a train and that of a platform are equal . if with a speed of 180 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 180 * 5 / 18 ] m / sec = 50 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 50 = > x = 50 * 60 / 2 = 1500 answer : a" | a = 180 * 1000
b = const_60 / 1
c = a / b
d = c / 2
|
a ) 114 , b ) 108 , c ) 192 , d ) 750 , e ) 777 | d | multiply(multiply(divide(240, 8), 5), 5) | running at the same constant rate , 8 identical machines can produce a total of 240 pens per minute . at this rate , how many pens could 5 such machines produce in 5 minutes ? | "let ' s take the approach that uses the answer choices to eliminate wasted time . 240 / 8 = 30 pens per minute per machine . 5 machines = 150 per minute . 5 minutes worth = 750 pens . looking at the answers it is clear . . . we can only choose ( d ) . the correct answer is d ." | a = 240 / 8
b = a * 5
c = b * 5
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(add(const_2, const_3), const_2) | the product of the squares of two positive integers is 81 . how many pairs of positive integers satisfy this condition ? | "ans : c - 2 pairs ( x ˆ 2 ) ( y ˆ 2 ) = 81 [ square root both sides ] xy = 9 9 = 1 x 9 , 9 x 1 , 3 x 3 cancel the repeats this leaves us with exactly 2 options . hence , c" | a = 2 + 3
b = a - 2
|
a ) 36 , b ) 42 , c ) 54 , d ) 60 , e ) 64 | a | subtract(subtract(multiply(4, add(subtract(multiply(4, 3), 4), subtract(multiply(4, 2), 2))), subtract(multiply(4, 2), 2)), add(subtract(multiply(4, 3), 4), subtract(multiply(4, 2), 2))) | before 2 years , dog a ’ s age was 4 times of dog b ’ s age and after 4 years , dog a ’ s age will be 3 times of dog b ’ s age . what is the difference of dog a ’ s age and dog b ’ s now ? | a - 2 = 4 ( b - 2 ) - - > a - 4 b = - 6 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 14 - - > a = 3 ( 18 ) = 50 a - b = 50 - 14 = 36 answer : a | a = 4 * 3
b = a - 4
c = 4 * 2
d = c - 2
e = b + d
f = 4 * e
g = 4 * 2
h = g - 2
i = f - h
j = 4 * 3
k = j - 4
l = 4 * 2
m = l - 2
n = k + m
o = i - n
|
a ) 250 , b ) 150 , c ) 500 , d ) 100 , e ) 200 | a | add(divide(500, const_2), const_1) | how many integers between 500 and 1000 are there such that their unit digit is even ? | 500 numbers between - 500 and 1000 out of which half would be even , half odd . number of even unit digit number = 250 . correct option is a | a = 500 / 2
b = a + 1
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a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 70 % | c | multiply(divide(subtract(add(add(100, 50), 100), multiply(2, 100)), 100), 100) | a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 50 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ? | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.5 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 50 % income the savings increased by 100 % . or s = . 5 i or s = 50 % of income c is the answer" | a = 100 + 50
b = a + 100
c = 2 * 100
d = b - c
e = d / 100
f = e * 100
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(multiply(24, const_4), add(const_4, const_1)) | the cost price of 24 articles is the same as the selling price of x articles . if the profit is 20 % , what is x ? | "let the cost price = y the cost price of 24 articles = 24 y the selling price of x articles = 1.20 y * x 1.20 y * x = 24 y x = 24 / 1.2 = 20 the answer is d ." | a = 24 * 4
b = 4 + 1
c = a / b
|
a ) 20 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 84.6 % | e | subtract(multiply(divide(240, subtract(240, 110)), const_100), const_100) | on increasing the number of lines in a page by 110 , they become 240 . what is the % of increase in the no . of lines in the page ? | "explanation : number of pages increased = 110 now , the number of pages of book = 240 number of pages of the books before increase = 240 – 110 = 130 % increase in the number of pages in the book = 110 / 130 x 100 % = 84.6 % e" | a = 240 - 110
b = 240 / a
c = b * 100
d = c - 100
|
a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 350 | e | divide(subtract(multiply(200, divide(16, const_100)), 18), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 18 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 18 gallons then 0.12 a + 0.16 ( 200 - a ) = 18 - - > a = 350 . answer : e ." | a = 16 / 100
b = 200 * a
c = b - 18
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) $ 130.00 , b ) $ 285.00 , c ) $ 200.00 , d ) $ 258.00 , e ) $ 210.00 | b | multiply(500, divide(add(40, 5), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 40 percent off the suggested retail price . if , during a special sale , an additional 5 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 500 before any discount ? | "original price : 500 $ max first discount = - 40 % thus : 500 â ˆ — ( 1 â ˆ ’ 40 / 100 ) = 300 second discount on the discounted price = - 20 % thus : 300 â ˆ — ( 1 â ˆ ’ 5 / 100 ) = 285 answer b ." | a = 40 + 5
b = a / 100
c = 500 * b
|
a ) 33 , b ) 32 , c ) 31 , d ) 30 , e ) 29 | a | divide(add(subtract(multiply(floor(divide(200, 3)), 3), add(3, multiply(3, floor(divide(100, 3))))), 3), 3) | what is the total number of d integers between 100 and 200 that are divisible by 3 ? | yes there is a different way of arriving at that answer . . . . u can also use airthmetic progression to get the answer since the first term to be divisble by 3 is 102 . . take that as a . . the starting no and since 198 is the last digit to be divisible by 3 take that as n . . . since the difference is 3 take that as d no u have to find what term is 198 take that as nth term the formula for that is n = a + ( n - 1 ) * d 198 = 102 + ( n - 1 ) * 3 from this u get n = 33 | a = 200 / 3
b = math.floor(a)
c = b * 3
d = 100 / 3
e = math.floor(d)
f = 3 * e
g = 3 + f
h = c - g
i = h + 3
j = i / 3
|
a ) 6 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) none of these | b | inverse(add(divide(5, multiply(10, 7)), divide(10, multiply(10, 14)))) | 10 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | explanation : 1 woman ' s 1 day ' s work = 1 / 70 1 child ' s 1 day ' s work = 1 / 140 5 women and 10 children 1 day work = ( 570 + 10140 ) = 17 so 5 women and 10 children will finish the work in 7 days answer : b | a = 10 * 7
b = 5 / a
c = 10 * 14
d = 10 / c
e = b + d
f = 1/(e)
|
a ) a ) 4 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 7 | e | add(subtract(38, add(30, 1)), 1) | the average weight of a group of boys is 30 kg . after a boy of weight 38 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 38 kg joins the group , total weight of boys = 30 x + 38 so 30 x + 38 = 31 ( x + 1 ) = > x = 7 . answer : e" | a = 30 + 1
b = 38 - a
c = b + 1
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | b | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 20), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 20), const_100))), const_100) | the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 20 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "let the output of a factory is 100 first , increased by 10 % = 110 due to holiday rush , it increase by 20 % = 110 + 22 = 132 132 - ( x / 100 * 132 ) = 100 24.24 answer : b" | a = 100 + 10
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = e - 1
g = 100 + 10
h = g / 100
i = 100 + 20
j = i / 100
k = h * j
l = f / k
m = l * 100
|
a ) 6 kg , b ) 5 kg , c ) 12 kg , d ) 24 kg , e ) 20 kg | a | subtract(10, multiply(2, 2)) | maria sold 10 kg of apples in her first hour at the market , but only 2 kg of apples in the second hour . on average , how many kg of apples did she sell in two hours at the market ? | ( 10 kg + 2 kg ) / 2 = 6 kg correct answer is : a | a = 2 * 2
b = 10 - a
|
a ) 11 days , b ) 12 days , c ) 14 days , d ) 10 days , e ) 15 days | b | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 30), const_2))), divide(const_1, 30)) | a can finish a piece of work in 15 days . b can do it in 30 days . they work together for 6 days and then a goes away . in how many days will b finish the work ? | "6 / 15 + ( 6 + x ) / 30 = 1 = > x = 12 days answer : b" | a = 1 / 4
b = a * 2
c = 1 / 30
d = c * 2
e = b + d
f = 1 - e
g = 1 / 30
h = f / g
|
a ) 98765 , b ) 98907 , c ) 99944 , d ) 99954 , e ) 99968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(5, const_10)), const_3), 88) | find the largest 5 digit number which is exactly divisible by 88 ? | "largest 5 digit number is 99999 after doing 99999 ÷ 88 we get remainder 31 hence largest 5 digit number exactly divisible by 88 = 99999 - 31 = 99968 e" | a = 100 * 100
b = 100 * 10
c = a + b
d = 100 * 3
e = c + d
f = 5 * 10
g = e + f
h = g + 3
i = h * 88
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | b | subtract(multiply(15, 8.00), multiply(6.00, 15)) | a club wants to mix 15 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ? | "let number of pounds of 5 $ candy to be used be w 6 = ( 15 * 8 + 5 * w ) / ( 15 + w ) = > 90 + 6 w = 120 + 5 w = > w = 30 answer b" | a = 15 * 8
b = 6 * 0
c = a - b
|
a ) 820 , b ) 340 , c ) 420 , d ) 209 , e ) none of these | b | add(multiply(5, 8), 300) | 300 + 5 × 8 = ? | 300 + 5 × 8 = ? or , ? = 300 + 40 = 340 answer b | a = 5 * 8
b = a + 300
|
a ) 18 kmph , b ) 17 kmph , c ) 24 kmph , d ) 19 kmph , e ) 12 kmph | c | divide(multiply(40, 3), add(divide(40, 40), divide(multiply(2, 40), 20))) | a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 3 x / ( x / 8 ) = 24 kmph answer : c" | a = 40 * 3
b = 40 / 40
c = 2 * 40
d = c / 20
e = b + d
f = a / e
|
a ) 81 , b ) 65 , c ) 75 , d ) 89 , e ) 90 | a | subtract(multiply(70, const_4), subtract(multiply(68, const_4), add(3, subtract(multiply(70, const_4), multiply(70, 3))))) | the avg weight of a , b & c is 70 kg . if d joins the group , the avg weight of the group becomes 70 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 68 kg . what is the weight of a ? | "a + b + c = 3 * 70 = 210 a + b + c + d = 4 * 70 = 280 - - - - ( i ) so , d = 70 & e = 70 + 3 = 73 b + c + d + e = 68 * 4 = 272 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 280 – 272 = 8 a = e + 8 = 73 + 8 = 81 answer : a" | a = 70 * 4
b = 68 * 4
c = 70 * 4
d = 70 * 3
e = c - d
f = 3 + e
g = b - f
h = a - g
|
a ) 50 % , b ) 60 % , c ) 40 % , d ) 30 % , e ) 45 % | a | multiply(divide(subtract(30, 20), 20), const_100) | john paid a sum of money for purchasing 30 pens , which he recovered in full when he sold 20 of them . what was his percentage of profit or loss per pen ? | "a 50 % if the sum he paid whilst purchasing 30 pens = a , then the cost price of each pen = a / 30 . since the amount he got whilst selling 20 pens is also = a then the selling price of each pen = a / 20 . since selling price > cost price , he made a profit . profit per pen = selling price - cost price = a / 20 - a / 30 = a / 60 . profit percentage per pen = profit per pen / cost per pen x 100 = ( a / 60 ) / ( a / 30 ) x 100 = 50 %" | a = 30 - 20
b = a / 20
c = b * 100
|
a ) 25.5 , b ) 2.5 , c ) 40 , d ) . 25 , e ) none of these | c | divide(1, 0.025) | 1 / 0.025 is equal to | "explanation : 1 / 0.025 = ( 1 * 1000 ) / 25 = 1000 / 25 = 40 option c" | a = 1 / 0
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a ) 370 , b ) 380 , c ) 390 , d ) 400 , e ) 410 | d | divide(18, subtract(divide(18, 3), 4)) | a train covers a distance of 18 km in 3 min . if it takes 4 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 3 * 60 ) km / hr = ( 360 * 5 / 18 ) m / sec = 100 m / sec . length of the train = 100 * 4 = 400 m . answer : option d" | a = 18 / 3
b = a - 4
c = 18 / b
|
a ) 20 % , b ) 40 % , c ) 56.25 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 25)) | a merchant sells an item at a 20 % discount , but still makes a gross profit of 25 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | "original sp = x cost = c current selling price = . 8 x ( 20 % discount ) . 8 x = 1.5 c ( 25 % profit ) x = 1.25 / . 8 * c x = 12.5 / 8 c original selling price is 1.5625 c which is 56.25 % profit answer c" | a = 100 - 20
b = a - 25
c = 100 - b
|
a ) 49 , b ) 55 , c ) 69 , d ) 70 , e ) 82 | d | subtract(subtract(4, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | evaluate : | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = | "according to order of operations , inner brackets first . hence | 4 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 4 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 4 + 72 | - | 5 - 11 | = | 76 | - | - 6 | = 76 - 6 = 70 correct answer d ) 70" | a = 3 - 12
b = 8 * a
c = 4 - b
d = 5 - 11
e = c - negate
|
a ) 50 sec , b ) 26 sec , c ) 80 sec , d ) 82 sec , e ) 81 sec | a | divide(add(200, 300), multiply(subtract(72, 36), const_0_2778)) | how much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m , moving at 36 kmph in the same direction ? | the distance to be covered = sum of their lengths = 200 + 300 = 500 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 500 / 10 = 50 sec . answer : a | a = 200 + 300
b = 72 - 36
c = b * const_0_2778
d = a / c
|
a ) 330 , b ) 340 , c ) 350 , d ) 360 , e ) 370 | c | divide(36, subtract(divide(36, 12), 7)) | a train covers a distance of 36 km in 12 min . if it takes 7 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 36 / 12 * 60 ) km / hr = ( 180 * 5 / 18 ) m / sec = 50 m / sec . length of the train = 50 * 7 = 350 m . answer : option c" | a = 36 / 12
b = a - 7
c = 36 / b
|
a ) 11 , b ) 30 , c ) 28 , d ) 24 , e ) 82 | b | multiply(divide(subtract(3146, 2420), 2420), const_100) | a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 3146 in 3 years . find the rate percent ? | "explanation : 2420 - - - 726 100 - - - ? = > 30 % answer : option b" | a = 3146 - 2420
b = a / 2420
c = b * 100
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a ) 25 / 3 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32 | d | divide(divide(800, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2) | the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 800 meters per second , what was the speed , in meters per second , when the stopwatch was started ? | let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 800 x = ( 2 ^ 5 * 25 ) / 2 ^ 9 = 25 / 16 the answer is d . | a = const_60 / 2
b = a + const_60
c = b / 10
d = c - 1
e = 2 ** d
f = 800 / e
g = f / 2
|
a ) 11028 , b ) 14172 , c ) 14284 , d ) 14015 , e ) 14397 | a | add(divide(add(multiply(add(add(80, 2), 224), divide(subtract(224, add(80, 2)), const_2)), add(add(80, 2), 224)), const_2), add(add(2, 2), 8)) | if i equals the sum of the even integers from 2 to 224 , inclusive , and k equals the sum of the even integers from 8 to 80 , inclusive , what is the value of i - k ? | use following formulae for such problems : sum of evenly spaced integers = ( # of integers ) * ( mean of integers ) # of integers = [ ( last - first ) / 2 ] + 1 mean of integers = ( last + first ) / 2 in above problem : # of integers = [ ( 224 - 2 ) / 2 ] + 1 = 112 and [ ( 80 - 8 ) / 2 ] + 1 = 37 mean of integers = ( 224 + 2 ) / 2 = 113 and ( 80 + 8 ) / 2 = 44 sum of integers = ( 112 * 113 ) = 12656 and ( 37 * 44 ) = 1628 thus their difference ( i - k ) = 12656 - 1628 = 11028 answer : a | a = 80 + 2
b = a + 224
c = 80 + 2
d = 224 - c
e = d / 2
f = b * e
g = 80 + 2
h = g + 224
i = f + h
j = i / 2
k = 2 + 2
l = k + 8
m = j + l
|
a ) 660 , b ) 738 , c ) 837 , d ) 840 , e ) 83 | a | divide(subtract(multiply(1500, divide(12, const_100)), 15), divide(25, const_100)) | if 25 % of x is 15 less than 12 % of 1500 , then x is ? | "25 % of x = x / 4 ; 12 % of 1500 = 12 / 100 * 1500 = 180 given that , x / 4 = 180 - 15 = > x / 4 = 165 = > x = 660 . answer : a" | a = 12 / 100
b = 1500 * a
c = b - 15
d = 25 / 100
e = c / d
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a ) 3 , b ) 33 , c ) 43 , d ) 23 , e ) 13 | e | add(11, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | evaluate : 11 + sqrt ( - 4 + 6 × 4 ÷ 3 ) = . . . . | according to order of operations , inner brackets first where 6 × 4 ÷ 3 is first calculated since it has a multiplication and a division . 6 × 4 ÷ 3 = 24 ÷ 3 = 8 hence 11 + sqrt ( - 4 + 6 × 4 ÷ 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer is e ) 13 | a = 6 * 4
b = a / 3
c = b - 4
d = math.sqrt(c)
e = 11 + d
|
a ) 60 , b ) 100 , c ) 160 , d ) 240 , e ) 300 | b | multiply(multiply(subtract(6, const_4), const_10), const_10) | a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 800,000 ? | "the first digit and last digit are the same so the only possibility is 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 1 * 10 * 10 = 100 the answer is b ." | a = 6 - 4
b = a * 10
c = b * 10
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a ) $ 414.78 , b ) $ 424.78 , c ) $ 534.78 , d ) $ 434.78 , e ) $ 435.78 | d | divide(multiply(25, const_100), multiply(subtract(add(const_1, divide(5, const_100)), subtract(const_1, divide(18, const_100))), const_100)) | a trader sold an item at a loss of 18 % . if he had increased the price by $ 25 he would have made a gain of 5 % . what is the cost price of the item ? | let c . p . be $ x then 105 % of x - 82 % of x = 100 23 % of x = 100 23 / 100 * x = 100 23 x = 10000 x = 434.78 answer is d | a = 25 * 100
b = 5 / 100
c = 1 + b
d = 18 / 100
e = 1 - d
f = c - e
g = f * 100
h = a / g
|
a ) 2 days , b ) 3 days , c ) 5 days , d ) 1 day , e ) 4 days | a | divide(300, multiply(5, divide(180, multiply(3, 2)))) | if 3 women can color 180 m long cloth in 2 days , then 5 women can color 300 m long cloth in ? | "the length of cloth painted by one woman in one day = 180 / 3 × 2 = 30 m no . of days required to paint 300 m cloth by 5 women = 300 / 5 × 30 = 2 days answer : a" | a = 3 * 2
b = 180 / a
c = 5 * b
d = 300 / c
|
a ) $ 6 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | a | multiply(6, const_1) | if $ 5,000 is invested in an account that earns 6 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "ans a solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.06 / 2 ) ^ 2 = 5406 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 8 / 100 ) = 5400 difference = 5406 - 5400 = 6 $ a" | a = 6 * 1
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a ) 150 sec , b ) 130 sec , c ) 190 sec , d ) 200 sec , e ) 100 sec | c | divide(add(1200, 700), divide(1200, 120)) | a train with 1200 m crosses a tree in 120 sec , how much time will i take to pass a platform 700 m long ? | s = 1200 / 120 s - 10 m / sec total length = 1900 m t = d / s t = 1900 / 10 t = 190 sec answer c | a = 1200 + 700
b = 1200 / 120
c = a / b
|
['a ) 2 * pi', 'b ) pi', 'c ) 2', 'd ) 4 * pi', 'e ) 4'] | e | sqrt(subtract(power(divide(multiply(subtract(const_1, const_pi), const_2), const_pi), const_2), multiply(const_4, negate(divide(10, const_pi))))) | the area of a circle is added to its diameter , and the circumference is then subtracted from this total , the result is 10 . what is the radius of the circle ? | the equation is ; diameter + area - circumference = d + a - c = d + pi * r ^ 2 - 2 * pi * r = 2 r + pi * r ( r - 2 ) . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : e | a = 1 - math.pi
b = a * 2
c = b / math.pi
d = c ** 2
e = 10 / math.pi
f = 4 * negate
g = d - f
h = math.sqrt(g)
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a ) 15 , b ) 77 , c ) 30 , d ) 88 , e ) 34 | a | divide(add(5, 25), const_2) | a man can row upstream at 5 kmph and downstream at 25 kmph , and then find the speed of the man in still water ? | "us = 5 ds = 25 m = ( 5 + 25 ) / 2 = 15 answer : a" | a = 5 + 25
b = a / 2
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a ) 2398 , b ) 2399 , c ) 2400 , d ) 2401 , e ) 2402 | d | multiply(multiply(49, const_2), divide(49, const_2)) | sum of 49 odd numbers is ? | sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 49 odd numbers = 49 ^ 2 = 2401 answer : d | a = 49 * 2
b = 49 / 2
c = a * b
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a ) 1200 km , b ) 1500 km , c ) 2000 km , d ) 2500 km , e ) 3600 km | d | multiply(30, 40) | a walks at 30 kmph and 30 hours after his start , b cycles after him at 40 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 40 = 30 x = 3600 km answer is d" | a = 30 * 40
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a ) 16.66 % , b ) 33.33 % , c ) 44.44 % , d ) 55.55 % , e ) 19.92 % | a | multiply(subtract(const_1, divide(const_100, add(const_100, 20))), const_100) | the product of x and y is a constant . if the value of x is increased by 20 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.2 x ( 120 ) i . e . 120 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 120 = 83.33 i . e . y decreases from 100 to 83.33 i . e . decrease of 16.66 % answer : option a" | a = 100 + 20
b = 100 / a
c = 1 - b
d = c * 100
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a ) 10 days , b ) 30 days , c ) 20 days , d ) 80 days , e ) 40 days | c | inverse(add(divide(9, multiply(12, 27)), divide(12, multiply(20, 27)))) | if 12 men or 20 women can do a piece of work in 27 days , then in how many days can 9 men and 12 women together do the work ? | "c 20 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 27 days . so , 27 women can do it in ( 20 * 27 ) / 27 = 20 days ." | a = 12 * 27
b = 9 / a
c = 20 * 27
d = 12 / c
e = b + d
f = 1/(e)
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a ) a ) 7 , b ) b ) 6 , c ) c ) 9 , d ) d ) 11 , e ) e ) none of the above | a | divide(490, 70) | how many trucks are there if each truck carrying 70 packages and total of 490 packages ? | sol . total packages 490 each truck carries 70 packages = 490 / 70 = 7 answer : a | a = 490 / 70
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a ) 212 , b ) 219 , c ) 217 , d ) 213 , e ) 216 | e | multiply(42120, 195) | calculate 42120 ÷ ? = 195 | "answer let 42120 / x = 195 then x = 42120 / 195 = 216 option : e" | a = 42120 * 195
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a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | e | subtract(multiply(subtract(14, 1), add(6, 1)), multiply(14, 6)) | the average of 6 observations is 14 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 84 + x ) / 7 = 13 = > x = 7 . hence , the seventh observation is 7 . answer : e" | a = 14 - 1
b = 6 + 1
c = a * b
d = 14 * 6
e = c - d
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a ) 2520 , b ) 3520 , c ) 4520 , d ) 1520 , e ) 1750 | c | multiply(add(divide(20, 90), 50), 90) | find the value of ( 50 + 20 / 90 ) × 90 | "( 4500 + 20 ) / 90 * 90 = 4520 answer : c" | a = 20 / 90
b = a + 50
c = b * 90
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a ) 0 and 3 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 7 , e ) 7 and 9 | a | power(20, divide(const_1, 3)) | if a and b are positive real numbers , and a ^ 3 + b ^ 3 = 20 , then the greatest possible value of a is between : | "if a > 3 , then a ^ 3 + b ^ 3 > 20 . the answer is a ." | a = 1 / 3
b = 20 ** a
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a ) 30 , b ) 42.5 , c ) 45 , d ) 49 , e ) 47.9 | d | subtract(add(divide(multiply(2, 32), subtract(32, const_1)), 32), 2) | the ages of 2 persons differ by 32 years . if 9 years ago the elder one be 5 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 32 5 ( x - 9 ) = x + 32 - 9 x = 17 age of elder person = 17 + 32 = 49 answer is d" | a = 2 * 32
b = 32 - 1
c = a / b
d = c + 32
e = d - 2
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a ) 20 cm , b ) 23 cm , c ) 25 cm , d ) 10 cm , e ) 28 cm | d | multiply(50, const_4) | what is the greatest possible length which can be used to measure exactly the lengths 18 m , 40 m and 6 m 50 cm ? | "required length = hcf of 1800 cm , 4000 cm , 650 cm = 50 cm answer is d" | a = 50 * 4
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a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 43 | e | multiply(divide(const_100, add(const_100, 5)), 45) | from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 5 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 5 x = 45 - - > 21 / 20 * x = 45 - - > x = 45 * 20 / 21 = 300 / 7 = ~ 43 . answer : e ." | a = 100 + 5
b = 100 / a
c = b * 45
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a ) 50 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(42, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 42 is : | "solution solution let the number be x . x - 16 % of x = 42 x - 16 / 100 x = 42 x - 4 / 25 x = 42 21 / 25 x = 42 x = ( 42 x 25 / 21 ) = 50 answer a" | a = 42 * 100
b = 100 - 16
c = a / b
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a ) 31 , b ) 35 , c ) 21 , d ) 40 , e ) 45 | c | divide(subtract(62.50, 10), 2.5) | in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 10 and the other one , that is cheaper , costs $ 2.5 . the factory holds a $ 62.50 worth stock of 22 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 2.50 modules must be 21 so that the leftover 1 modules are of $ 10 which will give a total value $ 62.50 . 21 * 2.50 + 1 * 10 = 52.50 + 10 = 62.50 answer : c" | a = 62 - 50
b = a / 2
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a ) 25 minutes , b ) 30 minutes , c ) 10 minutes , d ) 15 minutes , e ) 7 minutes | b | add(add(const_3, const_12), multiply(divide(multiply(30, divide(add(const_3, const_12), const_60)), subtract(60, 30)), const_60)) | mary goes into labor at her local grocery store and is rushed to a hospital in an ambulance traveling 60 mph . her husband don drives after the ambulance at an average speed of 30 mph . mary reaches the hospital fifteen minutes later . how long does it take don to get there from the store ? | distance covered by the ambulance in 15 minutes = 15 miles 60 mph is one mile per minute , therefore 30 mph will be 1 / 2 a mile per minute . so , he reaches the hospital half an hour after leaving the store . correct answer : b | a = 3 + 12
b = 3 + 12
c = b / const_60
d = 30 * c
e = 60 - 30
f = d / e
g = f * const_60
h = a + g
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a ) 3 : 7 , b ) 3 : 5 , c ) 3 : 8 , d ) 7 : 1 , e ) 3 : 5 | d | divide(7, divide(7, 7)) | what is the ratio between perimeters of two squares one having 7 times the diagonal then the other ? | "d = 7 d d = d a √ 2 = 7 d a √ 2 = d a = 7 d / √ 2 a = d / √ 2 = > 7 : 1 answer : d" | a = 7 / 7
b = 7 / a
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a ) 0.25 , b ) 0.5 , c ) 0.75 , d ) 0.1 , e ) 0.3 | c | divide(subtract(65, power(divide(16, const_4), const_2)), 65) | square a has an area of 65 square centimeters . square b has a perimeter of 16 centimeters . if square b is placed within square a and a random point is chosen within square a , what is the probability the point is not within square b ? | "i guess it ' s mean that square b is placed within square aentirely . since , the perimeter of b is 16 , then its side is 16 / 4 = 4 and the area is 4 ^ 2 = 16 ; empty space between the squares is 65 - 16 = 48 square centimeters , so if a random point is in this area then it wo n ' t be within square b : p = favorable / total = 48 / 64 = 0.75 . answer : c" | a = 16 / 4
b = a ** 2
c = 65 - b
d = c / 65
|
a ) 330000 , b ) 340000 , c ) 380800 , d ) 356000 , e ) 357000 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(80, const_100)) | in an election , candidate a got 80 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 80 % therefore , the number of valid votes polled in favour of candidate a = 80 % of 476000 = 80 / 100 × 476000 = 38080000 / 100 = 380800 c )" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 80 / 100
e = c * d
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a ) 34 , b ) 40 , c ) 68 , d ) 76 , e ) 92 | d | add(multiply(divide(560, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 560 sq . feet , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 560 breadth = 28 feet area to be fenced = 2 b + l = 2 ( 28 ) + 20 = 76 feet answer : d" | a = 560 / 20
b = a * 2
c = b + 20
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a ) 40 rs , b ) 45 rs , c ) 50 rs , d ) 55 rs , e ) 60 rs | e | add(multiply(divide(add(const_3, const_2), const_2), multiply(const_4, const_4)), 20) | if a man earns rs 20 on first day and spends rs 15 on second day , and earns rs 20 on third day and spends rs 15 on fourth day and goes on on which day he will be having rs 60 . | earns rs 20 on first day and spends rs 15 on second day means , end of day 2 , he have 5 rs . therefore , in 16 days he have 40 rs 2 * 8 = 16 days 5 rs * 8 = 40 rs and on the 17 th day again he earn rs . 20 , 40 + 20 = 60 rs . answer : e | a = 3 + 2
b = a / 2
c = 4 * 4
d = b * c
e = d + 20
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a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 17 | d | divide(multiply(multiply(4, 8), 8), multiply(4, 4)) | 4 mat - weaves can weaves 4 mats in 4 days . at the same rate , how many mats would be woven by 8 mat - weaves in 8 days ? | solution let the required number of mats be x . more weaves , more mats ( direct proportion ) more days , more mats ( direct proportion ) ∴ 4 × 4 × x = 8 × 8 × 4 ⇔ x = ( 8 x 8 x 4 ) / ( 4 x 4 ) = 16 . answer d | a = 4 * 8
b = a * 8
c = 4 * 4
d = b / c
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a ) 31 , b ) 36 , c ) 41 , d ) 51 , e ) none | d | add(36, const_1) | the average age of 36 students in a group is 14 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "sol . age of the teacher = ( 37 × 15 – 36 × 14 ) years = 51 years . answer d" | a = 36 + 1
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a ) 20 , b ) 25 , c ) 35 , d ) 50 , e ) 65 | b | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 18), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 18 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 18 seconds , the relative difference in distance traveled is 18 * 25 / 9 = 50 meters this distance is twice the length of each train . the length of each train is 50 / 2 = 25 meters the answer is b ." | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 18
e = d / 2
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(reminder(gcd(subtract(272738, 2), subtract(232342, 13)), const_10), const_2) | the numbers 272738 and 232342 , when divided by n , a 2 digit number leave a remainder 13 and 17 respectively . find the sum of digits of n ? | "we have to find the h . c . f of both 1.272738 - 13 2.232325 - 17 h . c . f is 25 ans is 7 answer : d" | a = 272738 - 2
b = 232342 - 13
c = math.gcd(a, b)
d = reminder + (
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a ) 27 seconds , b ) 29 seconds , c ) 40 seconds , d ) 11 seconds , e ) 20 seconds | e | divide(add(360, 140), divide(multiply(90, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 90 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 90 km / hr = 90 * ( 5 / 18 ) m / sec = 25 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 25 ) = 20 seconds answer : e" | a = 360 + 140
b = 90 * 1000
c = b / 3600
d = a / c
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a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 22 / 10 , e ) 4 | d | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fourth of his earnings to charity . if the worker ' s charitable contributions totaled one - eighth of his earnings for the entire year , what is the value of n ? | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 4 = 1 / 8 ( 11 x + nx ) ; 11 / 10 + n / 4 = 1 / 8 ( 11 + n ) = > 44 + 10 n / 40 = 11 + n / 8 352 + 80 n = 440 + 40 n = > 40 n = 88 n = 88 / 40 = 22 / 10 answer : d ." | a = 12 - 1
b = 12 - 3
c = 1/(b)
d = 1/(10)
e = c - d
f = a * e
g = 1 + 4
h = 1/(g)
i = 12 - 3
j = 1/(i)
k = h - j
l = f / k
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a ) $ 374 , b ) $ 382 , c ) $ 385 , d ) $ 360 , e ) $ 399 | d | add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(20, const_100), divide(348, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 20 percent , how much would she earn per week ? | "( 348 / 116 ) 120 = 360 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 120 = 360 ( 300 + 60 ) answer d" | a = 16 / 100
b = 1 + a
c = 348 / b
d = 20 / 100
e = 16 / 100
f = 1 + e
g = 348 / f
h = d * g
i = c + h
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a ) 4 / 15 , b ) 11 / 15 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 6 | b | divide(multiply(11, 1), multiply(3, 5)) | if the ratio of a to b is 11 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 11 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 11 : 15 answer b" | a = 11 * 1
b = 3 * 5
c = a / b
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a ) 12 , b ) 17 , c ) 18 , d ) 4 , e ) 26 | d | subtract(5, reminder(2496, 5)) | what should be the least number to be added to the 2496 number to make it divisible by 5 ? | "answer : 4 option : d" | a = 5 - reminder
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a ) 125 , b ) 70 , c ) 40 , d ) 25 , e ) none of these | c | divide(multiply(35, 8), subtract(multiply(8, 4), multiply(5, 5))) | 1 ⁄ 5 of a number is equal to 5 ⁄ 8 of the second number . if 35 is added to the first number then it becomes 4 times of second number . what is the value of the second number ? | let x be the first number and y be the second number . 1 ⁄ 5 x = 5 ⁄ 8 y \ x ⁄ y = 25 ⁄ 8 . . . . . . ( i ) x + 35 = 4 y or , 25 ⁄ 8 y + 35 = 4 y \ y = 40 answee c | a = 35 * 8
b = 8 * 4
c = 5 * 5
d = b - c
e = a / d
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a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | e | subtract(75, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 75 . | "first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 , 10 + 1 , 15 + 1 , 20 + 1 , 25 + 1 , 30 + 1 , 35 + 1 etc and n = 7 b + 1 e . g . 7 + 1 , 14 + 1 , 21 + 1 , 28 + 1 , 35 + 1 etc so when it is divided by the lcm , 35 , it will give 1 as remainder ( as is apparent above ) next , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3 , we can say the number is of the form n = 5 a + 1 and n = 7 b + 3 now , the only thing you should try to understand here is that when n is divided by 5 and if i say the remainder is 1 , it is the same as saying the remainder is - 4 . e . g . when 6 is divided by 5 , remainder is 1 because it is 1 more than a multiple of 5 . i can also say it is 4 less than the next multiple of 5 , ca n ' t i ? 6 is one more than 5 , but 4 less than 10 . therefore , we can say n = 5 x - 4 and n = 7 y - 4 ( a remainder of 3 when divided by 7 is the same as getting a remainder of - 4 ) now this question is exactly like the question above . so when you divide n by 75 , remainder will be - 4 i . e . n will be 4 less than a multiple of 75 . so you must add 35 to n to make it a multiple of 75 e" | a = 75 - reminder
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a ) 2.5 sec , b ) 3.25 sec , c ) 5 sec , d ) 12.5 sec , e ) 6 sec | b | divide(130, multiply(144, const_0_2778)) | in what time will a train 130 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 130 / 40 = 3.25 sec . answer : b" | a = 144 * const_0_2778
b = 130 / a
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a ) 40 minutes , b ) 45 minutes , c ) 38 minutes , d ) 48 minutes , e ) 30 minutes | d | divide(multiply(2, const_1000), divide(add(divide(multiply(2, const_1000), 40), divide(multiply(2, const_1000), 60)), 2)) | donald duck can can swim his pool downstream ( with the pool current helping time ) in exact 40 seconds and upstream ( against the pool current ) in a pool in exact 60 seconds . the length of pool is 2 kilometers . how long donald duck can cover distance of one side at a still pool ( with no current ) . | d 48 minutes . donald duck ' s speed = x km / seconds pool current speed = y km / seconds 2 / ( x + y ) = 40 2 / ( x - y ) = 60 solving the simultaneous equations gives x = 1 / 24 therefore , to cover 2 km will take 2 / x = 48 seconds | a = 2 * 1000
b = 2 * 1000
c = b / 40
d = 2 * 1000
e = d / 60
f = c + e
g = f / 2
h = a / g
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a ) does not change , b ) decreases by 1 % , c ) increases by 1 % , d ) increases by 0.1 % , e ) none of these | b | divide(multiply(add(11, const_100), subtract(const_100, 11)), const_100) | a number is increased by 11 % and then reduced by 10 % . after these operations , the number : | "let the original number be 100 . then , the new number = 100 × 1.1 × 0.9 = 99 i . e . the number decreases by 1 % . answer b" | a = 11 + 100
b = 100 - 11
c = a * b
d = c / 100
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a ) 100 , b ) 720 , c ) 1100 , d ) 940 , e ) 860 | b | multiply(240, divide(5, divide(5, 3))) | an plane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 5 / 3 hours , it must travel at a speed of ? | we know distance = speed * time = 240 * 5 = 1200 km now to cover 1200 km in 5 / 3 hrs , speed = distance / time = 1200 / ( 5 / 3 ) = 720 kmph ans - b | a = 5 / 3
b = 5 / a
c = 240 * b
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a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 33 | e | subtract(multiply(multiply(4, 5), divide(132, add(add(multiply(3, const_3), multiply(const_3, 5)), multiply(4, 5)))), multiply(multiply(3, const_3), divide(132, add(add(multiply(3, const_3), multiply(const_3, 5)), multiply(4, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 4 . if the sum of their ages is 132 , what is the difference between the ages of patrick and monica ? | ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 4 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 20 x sum = 44 x = 132 x = 3 difference if ages of pa and mo = 20 x - 9 x = 11 x = 11 * 3 = 33 answer e | a = 4 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 4 * 5
f = d + e
g = 132 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 4 * 5
n = l + m
o = 132 / n
p = i * o
q = h - p
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a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 15 : 4 | d | divide(add(multiply(5, divide(add(6, 1), add(5, 2))), 6), add(multiply(2, divide(add(6, 1), add(5, 2))), 1)) | two vessels contains equal number of mixtures milk and water in the ratio 5 : 2 and 6 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 5 / 7 + 6 / 7 ) : ( 2 / 7 + 1 / 7 ) = 11 / 7 : 3 / 7 = 11 : 3 answer is d" | a = 6 + 1
b = 5 + 2
c = a / b
d = 5 * c
e = d + 6
f = 6 + 1
g = 5 + 2
h = f / g
i = 2 * h
j = i + 1
k = e / j
|
a ) 38 , b ) 27 , c ) 99 , d ) 110 , e ) 80 | d | subtract(multiply(80, 6), multiply(74, 5)) | ashok secured average of 80 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 80 therefore total marks in 6 subjects = 80 * 6 = 480 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 480 – 370 = 110 answer : d" | a = 80 * 6
b = 74 * 5
c = a - b
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a ) 120 , b ) 150 , c ) 100 , d ) 140 , e ) 220 | d | subtract(add(175, 325), subtract(410, 50)) | out of 410 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 410 - 50 = 360 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 360 = 140 . answer is d | a = 175 + 325
b = 410 - 50
c = a - b
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a ) 145 , b ) 253 , c ) 370 , d ) 352 , e ) 300 | b | add(add(multiply(subtract(divide(add(10, const_2), const_4), const_1), const_100), multiply(add(divide(add(10, const_2), const_4), subtract(divide(add(10, const_2), const_4), const_1)), const_10)), divide(add(10, const_2), const_4)) | a number consists of 3 digit whose sum is 10 . the middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed . the number is : | explanation : let the middle digit be x . then , 2 x = 10 or x = 5 . so , the number is either 253 or 352 . since the number increases on reversing the digits , so the hundred ' s digit is smaller than the unit ' s digit . hence , required number = 253 . answer is b | a = 10 + 2
b = a / 4
c = b - 1
d = c * 100
e = 10 + 2
f = e / 4
g = 10 + 2
h = g / 4
i = h - 1
j = f + i
k = j * 10
l = d + k
m = 10 + 2
n = m / 4
o = l + n
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a ) $ 5850 , b ) $ 5950 , c ) $ 6050 , d ) $ 6150 , e ) $ 6250 | c | multiply(5000, power(add(const_1, divide(10, const_100)), const_2)) | what amount does an investor receive if the investor invests $ 5000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 5000 = 1.21 * 5000 = 6050 the answer is c ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 5000 * c
|
a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | c | multiply(divide(multiply(divide(1, 4), subtract(1, divide(1, 4))), add(multiply(divide(1, 4), subtract(1, divide(1, 4))), subtract(1, divide(1, 4)))), const_100) | of the 3,600 employees of company x , 1 / 4 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "let ' s see , the way i did it was 1 / 4 are clerical out of 3600 so 900 are clerical 900 reduced by 1 / 4 is 900 * 1 / 4 so it reduced 225 people , so there is 675 clerical people left but since 225 people left , it also reduced from the total of 3600 so there are 3375 people total since 675 clerical left / 3375 people total you get ( c ) 20 %" | a = 1 / 4
b = 1 / 4
c = 1 - b
d = a * c
e = 1 / 4
f = 1 / 4
g = 1 - f
h = e * g
i = 1 / 4
j = 1 - i
k = h + j
l = d / k
m = l * 100
|
a ) 55 , b ) 66 , c ) 77 , d ) 88 , e ) 99 | b | divide(multiply(12, subtract(12, const_1)), const_2) | there are 12 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "12 c 2 = 66 the answer is b ." | a = 12 - 1
b = 12 * a
c = b / 2
|
a ) 14 , b ) 16 , c ) 17 , d ) 20 , e ) 26 | c | subtract(19, const_2) | tough and tricky questions : statistics . set x consists of prime numbers { 3 , 11 , 7 , a , 17 , 19 } . if integer y represents the product of all elements in set x and if 11 y is an even number , what is the range of set x ? | since 11 y = even therefore y has to beevensince 11 is a odd integer ( even * odd = even ) similarly , y is the product of all integers in set x but all integers in set x are odd except the unknown a and since x contains only prime numbers , a has to equal to 2 . . . ( 2 is the only even prime number and the product of all prime numbers in set x has to be even , even * odd = even ) since you know value of a you can calculate the range = largest integer in the set minus smallest integer in the set = 19 - 2 = 17 answer is c | a = 19 - 2
|
a ) 1 / 2 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 1 / 3 | e | divide(subtract(4, multiply(const_2, const_3)), 4) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 4 | "total number of cases = 3 * 3 = 9 favourable cases = [ ( 2,3 ) , ( 3,2 ) , ( 3,3 ) ] = 3 so probability = 3 / 9 = 1 / 3 answer is e" | a = 2 * 3
b = 4 - a
c = b / 4
|
a ) 27 , b ) 28 , c ) 29 , d ) 30 , e ) 25 | c | add(divide(subtract(250, 100), 5), const_1) | what is the total number of integers between 100 and 250 that are divisible by 5 ? | "105 , 110 , 115 , . . . , 240,245 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 245 - 105 ) / ( 5 ) + 1 = 140 / 5 + 1 = 28 + 1 = 29 answer is c" | a = 250 - 100
b = a / 5
c = b + 1
|
a ) 0 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 28 % | a | subtract(divide(subtract(const_100, 20), divide(4, 5)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 4 / 5 rd of that price , there would be a loss of 20 % ? | "sp 2 = 4 / 5 sp 1 cp = 100 sp 2 = 80 4 / 5 sp 1 = 80 sp 1 = 100 100 - - - 100 = > 0 % answer : a" | a = 100 - 20
b = 4 / 5
c = a / b
d = c - 100
|
a ) 15 % , b ) 14.25 % , c ) 12.5 % , d ) 10.5 % , e ) 11.5 % | c | divide(multiply(const_100, subtract(subtract(340, divide(320, 2)), divide(320, 2))), divide(320, 2)) | on a sum of money , the simple interest for 2 years is rs . 320 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is | explanation : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - simple interest for 2 years is rs . 320 = > simple interest for first year = 320 / 2 = 160 = > similarly , simple interest for second year is also 160 compound interest for first year = 160 compound interest for second year = 340 - 160 = 180 we can see that compound interest for second year is more than simple interest for second year by 180 - 160 = 20 i . e . , rs . 20 is the simple interest on rs . 160 for 1 year r = 100 × si / pt = ( 100 × 20 ) / ( 160 × 1 ) = 12.5 % - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r × si ) / ( 2 × 100 ) difference between the compound interest and simple interest = 340 - 320 = 20 ( r × si ) / ( 2 × 100 ) = 20 ( r × 320 ) / ( 2 × 100 ) = 20 r = 20 × 100 × 2320 = 12.5 % answer : option c | a = 320 / 2
b = 340 - a
c = 320 / 2
d = b - c
e = 100 * d
f = 320 / 2
g = e / f
|
a ) 120 , b ) 50 , c ) 60 , d ) 70 , e ) 90 | a | subtract(100, subtract(82, 62)) | if x , y , and z are positive real numbers such that x ( y + z ) = 62 , y ( z + x ) = 82 , and z ( x + y ) = 100 , then xyz is | "xy + xz = 62 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 ) yz + yx = 82 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ) xz + zy = 100 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 ) re - writing equation 3 as follows : xz + zy = 82 + 18 xz + zy = yz + yx + 18 xz = yx + 18 . . . . . . . . . . . . . . . ( 4 ) adding ( 1 ) ( 4 ) 2 xz = 80 xz = 40 xyz has to be multiple of 40 , only 120 fits in answer = a" | a = 82 - 62
b = 100 - a
|
a ) 43 % , b ) 42.85 % , c ) 45.25 % , d ) 46.23 % , e ) 41.66 % | b | multiply(const_100, divide(divide(multiply(add(50, const_100), 30), const_100), add(const_100, 5))) | of the families in city x in 1992 , 30 percent owned a personal computer . the number of families in city x owning a computer in 1999 was 50 percent greater than it was in 1992 , and the total number of families in city x was 5 percent greater in 1999 than it was in 1994 . what percent of the families in city x owned a personal computer in 1999 ? | "say a 100 families existed in 1992 then the number of families owning a computer in 1992 - 30 number of families owning computer in 1999 = 30 * 150 / 100 = 45 number of families in 1999 = 105 the percentage = 45 / 105 * 100 = 42.85 % . answer : b" | a = 50 + 100
b = a * 30
c = b / 100
d = 100 + 5
e = c / d
f = 100 * e
|
a ) 350 , b ) 375 , c ) 400 , d ) 425 , e ) 450 | a | divide(70, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | of the votes cast on a certain proposal , 70 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 70 0.2 x = 70 x = 350 the answer is a ." | a = 40 / 100
b = 1 - a
c = 40 / 100
d = b - c
e = 70 / d
|
a ) 12 days , b ) 6 days , c ) 5 days , d ) 4 days , e ) none of these | a | divide(multiply(4, 3), subtract(4, 3)) | a man can do a piece of work in 4 days , but with the help of his son he can do it in 3 days . in what time can the son do it alone ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 3 − 1 / 4 ) = ( 4 − 3 ) / 12 = 1 / 12 so son will do whole work in 12 days answer : a" | a = 4 * 3
b = 4 - 3
c = a / b
|
a ) 22 miles , b ) 20.62 miles , c ) 25 miles , d ) 26 miles , e ) 28 miles | b | sqrt(add(power(add(5, 15), const_2), power(5, const_2))) | beginning in town a , biker bob rides his bike 10 miles west , 5 miles north , 5 miles east , and then 15 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 15 + 5 = 20 m other being the base ie distance traveled west - distance traveled eat = 10 - 5 = 5 m now this third side or the distance between town a and town b = 20 ^ 2 + 5 ^ 2 = sq root 425 = 20.62 mile answer : b" | a = 5 + 15
b = a ** 2
c = 5 ** 2
d = b + c
e = math.sqrt(d)
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 100 | b | subtract(add(300, 150), 400) | a , b and c have rs . 400 between them , a and c together have rs . 300 and b and c rs . 150 . how much does c have ? | "a + b + c = 400 a + c = 300 b + c = 150 - - - - - - - - - - - - - - a + b + 2 c = 450 a + b + c = 400 - - - - - - - - - - - - - - - - c = 50 answer : b" | a = 300 + 150
b = a - 400
|
a ) 19.64 , b ) 14.64 , c ) 12.6 , d ) 15.64 , e ) 13.64 | c | divide(0.0002152, 0.000205) | 0.0002152 / 0.000205 x 12.05 = ? | "explanation : ? = 0.0002152 / 0.000205 x 12.05 = 12.6 answer : option c" | a = 0 / 2152
|
a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 4 : 1 , e ) 16 : 1 | e | multiply(8, const_2) | city x has a population 8 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | "x = 8 y , y = 2 * z x : y , y : z 8 : 1 , 2 : 1 16 : 2 , 2 : 1 so , x : z = 16 : 1 ( e )" | a = 8 * 2
|
a ) 3387 , b ) 1600 , c ) 2866 , d ) 2787 , e ) 1121 | b | multiply(multiply(20, 10), 8) | in digging a pond 20 m * 10 m * 8 m the volumes of the soil extracted will be ? | "20 * 10 * 8 = 1600 answer : b" | a = 20 * 10
b = a * 8
|
a ) 1502 , b ) 1900 , c ) 1250 , d ) 1750 , e ) 2000 | c | divide(divide(multiply(1500, const_100), 10), 12) | a man took loan from a bank at the rate of 12 % p . a . s . i . after 10 years he had to pay rs . 1500 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 1500 ) / ( 12 * 10 ) = rs . 1250 answer : c" | a = 1500 * 100
b = a / 10
c = b / 12
|
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