options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) $ 400 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | a | divide(multiply(10, subtract(60, 40)), divide(50, const_100)) | on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 50 percent . what was the total rent the motel actually charged for that night ? | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 50 % . 5 x = 200 = > x = 400 answer a" | a = 60 - 40
b = 10 * a
c = 50 / 100
d = b / c
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a ) rs . 16241.875 , b ) rs . 15041.85 , c ) rs . 15401.875 , d ) rs . 15041.875 , e ) none of these | d | divide(divide(multiply(6016.75, const_100), 8), 5) | a sum fetched a total simple interest of rs . 6016.75 at the rate of 8 p . c . p . a . in 5 years . what is the sum ? | "explanation : principal = rs 100 x 6016.75 / 8 x 5 = rs . 601675 / 40 = rs . 15041.875 answer : option d" | a = 6016 * 75
b = a / 8
c = b / 5
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a ) 1 : 2 , b ) 9 : 5 , c ) 1 : 5 , d ) 3 : 1 , e ) 1 : 1 | b | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 2 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ£ β 12 : 30000 Γ£ β 10 = 45 Γ£ β 12 : 30 Γ£ β 10 = 3 Γ£ β 12 : 2 Γ£ β 10 = 9 : 5 answer : option b" | a = 45000 * 12
b = 4 + 3
c = 30000 * b
d = a / c
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a ) 40 , b ) 99 , c ) 88 , d ) 77 , e ) 42 | e | divide(360, multiply(subtract(45, 165), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 165 m long ? | ": speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 165 = 525 m required time = 525 * 2 / 25 = 42 sec answer : e" | a = 45 - 165
b = a * const_0_2778
c = 360 / b
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a ) 30 minutes , b ) 17 minutes , c ) 15 minutes , d ) 10 minutes , e ) 14 minutes | e | subtract(add(divide(const_1, 8), divide(const_1, 5)), divide(const_1, 12)) | two pipes can fill a tank in 8 minutes and 5 minutes . an outlet pipe can empty the tank in 12 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 8 + 1 / 5 - 1 / 12 = ( 15 + 24 - 10 ) / 120 = 29 / 120 so , the tank becomes full in 14 minutes . answer : e" | a = 1 / 8
b = 1 / 5
c = a + b
d = 1 / 12
e = c - d
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a ) rs . 6725 , b ) rs . 7239.13 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 6728 | b | divide(8325, add(const_1, divide(15, const_100))) | the owner of a furniture shop charges his customer 15 % more than the cost price . if a customer paid rs . 8325 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8325 ( 100 / 115 ) = rs . 7239.13 . answer : b" | a = 15 / 100
b = 1 + a
c = 8325 / b
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a ) 200 km , b ) 180 km , c ) 160 km , d ) 220 km , e ) none of these | b | divide(19, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2)))) | a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c midway between a and b . if the velocity of the stream is 4 km / h and the speed of the boat in still water is 14 km / h , what is the distance between a and b ? | "speed of boat for downstream = 14 + 4 = 18 km / hr speed of boat for upstream = 14 β 4 = 10 km / hr distance = x x / 18 + ( x / 2 ) / 10 = 19 x = 180 km answer : b" | a = 14 + 4
b = 1 / a
c = 14 - 4
d = c * 2
e = 1 / d
f = b + e
g = 19 / f
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a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | d | add(divide(80, divide(const_10, const_2)), const_3) | if 73 ! has 16 zeroes at the end , how many zeroes will 80 ! have at the end ? | prime factoring : you need a 2 and a 5 to make a 10 ( a ` ` zero ' ' ) , and there are tons of 2 ' s so let ' s skip these and focus on the ( rarer ) 5 s : 80 ! = 1 * 2 * 3 * 4 * 5 * 6 . . . * 78 * 79 * 80 since there are 80 consecutive numbers , there are 16 multiples of 5 in there , but if we ' re prime factoring , we need to remember that some multiples of 5 actually contain more than just one 5 . which ? 25 comes to mind - - it ' s got two of them ! so all the multiples of 25 actually contain two 5 ' s ( ie : 50 and 75 ) so , to recap , we have 16 of them , plus 3 more ( the additional 5 ' s in 25 , 50 , and 75 ) , so that makes 19 , and since we have more than enough 2 ' s , we know our number will have exactly 19 zeros at the end . answer : d | a = 10 / 2
b = 80 / a
c = b + 3
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a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | a | add(1990, multiply(10, multiply(const_2, const_3))) | in 1990 the budgets for projects q and v were $ 700,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | "let the no of years it takes is x . 700 + 30 x = 780 - 10 x - - > 40 x = 80 and x = 2 . thus , it happens in 1992 . a ." | a = 2 * 3
b = 10 * a
c = 1990 + b
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a ) 15 , b ) 16 , c ) 30 , d ) 31 , e ) none of these | d | divide(multiply(60, const_60), multiply(add(const_10, const_2), const_10)) | 6 bells commence tolling together and toll at the intervals of 24 , 68 , 1012 seconds resp . in 60 minutes how many times they will toll together . | explanation : lcm of 2 - 4 - 6 - 8 - 10 - 12 is 120 seconds , that is 2 minutes . now 60 / 2 = 30 adding one bell at the starting it will 30 + 1 = 31 answer : option d | a = 60 * const_60
b = 10 + 2
c = b * 10
d = a / c
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a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 2 , d ) 3 / 4 , e ) 7 / 8 | d | subtract(const_1, divide(13, 52)) | in a full deck of 52 cards , there are 13 spades . a game is played whereby a card from is chosen at random from the deck . if the card is a spade , a winner is declared . if the card is not a spade , it is returned to the deck , the deck is reshuffled and another card is drawn . this process is repeated until a spade is drawn . what is the probability that one will pick the first spade on the second draw or later ? | favorable case = the spade is picked in the third draw or later unfavorable case = the spade is picked in either first draw or in second draw probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case : probability of spade picked in first draw = 13 / 52 = 1 / 4 i . e . , favorable probability = 1 - ( 1 / 4 ) = 3 / 4 answer option : d | a = 13 / 52
b = 1 - a
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a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 75 | e | subtract(multiply(multiply(divide(600, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(600, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), subtract(const_1, divide(25, const_100)))) | of 600 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 600 = 1.25 b + b - 0.25 b - - > b = 300 , a = 1.25 b = 375 and { both } = 0.25 b = 75 . the number of students who read only book a is { a } - { both } = 375 - 75 = 300 ; the number of students who read only book b is { b } - { both } = 300 - 75 = 225 ; the difference is 300 - 225 = 75 . answer : e . | a = 25 / 100
b = 20 / 100
c = a / b
d = c + 1
e = 25 / 100
f = d - e
g = 600 / f
h = 25 / 100
i = 20 / 100
j = h / i
k = g * j
l = 20 / 100
m = 1 - l
n = k * m
o = 25 / 100
p = 20 / 100
q = o / p
r = q + 1
s = 25 / 100
t = r - s
u = 600 / t
v = 25 / 100
w = 1 - v
x = u * w
y = n - x
|
a ) 600 , b ) 700 , c ) 840 , d ) 900 , e ) none of these | c | multiply(4340, 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 4340 , r will receive | explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 4340 * ( 6 / 31 ) = 140 * 6 = 840 . answer : option c | a = 4 + 1
b = 10 + a
c = b + 10
d = c + 6
e = 6 / d
f = 4340 * e
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a ) 1 / 3 , b ) 1 / 6 , c ) 5 / 12 , d ) 5 / 24 , e ) 7 / 24 | b | divide(choose(4, 3), choose(add(6, 4), 3)) | there are 6 red balls and 4 blue balls in a jar . if 3 balls are selected from the jar , what is the probability that all 3 balls selected are red balls ? | "the number of ways of choosing 3 balls from the jar is 10 c 3 = 120 . the number of ways of choosing 3 red balls is 6 c 3 = 20 . p ( 3 red balls ) = 20 / 120 = 1 / 6 . the answer is b ." | a = math.comb(4, 3)
b = 6 + 4
c = math.comb(b, 3)
d = a / c
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a ) 30 , b ) 28 , c ) 27 , d ) 25 , e ) 26 | d | subtract(floor(divide(subtract(100, 7), 4)), floor(divide(subtract(1, 7), 4))) | for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 100 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 100 4 n < 93 n < 23 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 23 ) combining the two - 1.5 < n < 23.25 n = 1 to 23 ( 23 integers ) and n = - 1 and 0 so 25 integers . d is the answer" | a = 100 - 7
b = a / 4
c = math.floor(b)
d = 1 - 7
e = d / 4
f = math.floor(e)
g = c - f
|
a ) 20 , b ) 10 , c ) 30 , d ) 40 , e ) 5 | c | subtract(multiply(25, const_2), multiply(10, const_2)) | if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 25 , what is the value of r - p ? | "arithmetic mean expression for p and q : ( p + q ) / 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) / 2 = 20 ; q + r = 50 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 30 hence , the correct answer is c" | a = 25 * 2
b = 10 * 2
c = a - b
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a ) 100 , b ) 120 , c ) 144 , d ) 180 , e ) 160 | c | multiply(divide(const_60, 25), 1) | if the population of a certain country increases at the rate of one person every 25 seconds , by how many persons does the population increase in 1 hour ? | "answer = 2.4 * 60 = 144 answer is c" | a = const_60 / 25
b = a * 1
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a ) 456 , b ) 744 , c ) 912 , d ) 1200 , e ) 775 | e | divide(multiply(divide(300, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 300 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 300 62 % - - - - - - - - ? = > 775 answer : e" | a = 100 - 62
b = 62 - a
c = b / 100
d = 300 / c
e = d * 62
f = e / 100
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a ) rs . 18000 , b ) rs . 20000 , c ) rs . 25600 , d ) rs . 32000 , e ) none of these | c | multiply(multiply(divide(multiply(multiply(8, const_100), multiply(divide(1.92, const_12), const_2)), multiply(const_100, 10)), const_100), const_1000) | 10 percent of ram ' s monthly salary is equal to 8 percent of shyam ' s monthly salary . shyam ' s monthly salary is twice abhinav ' s monthly salary . if abhinav ' s annual salary is rs . 1.92 lakhs , find ram ' s monthly salary ? | let the monthly salaries of ram and shyam be rs . r and rs . s respectively . 10 / 100 r = 8 / 100 s r = 4 / 5 s monthly salary of abhinav = ( 1.92 lakhs ) / 12 = rs . 0.16 lakhs s = 2 ( 0.16 lakhs ) = 0.32 lakhs r = 4 / 5 ( 0.32 lakhs ) = rs . 25600 answer : c | a = 8 * 100
b = 1 / 92
c = b * 2
d = a * c
e = 100 * 10
f = d / e
g = f * 100
h = g * 1000
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a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | b | multiply(subtract(divide(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(20, 10), 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 10 candies . the store still makes a profit of 12 % 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 . b" | a = 12 / 100
b = a + 1
c = 20 / 100
d = 1 - c
e = 20 - 10
f = e / 20
g = d * f
h = b / g
i = h - 1
j = i * 100
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a ) 45 % , b ) 23 % , c ) 20 % , d ) 60 % , e ) 50 % | e | subtract(50, 25) | if the selling price of 50 articles is equal to the cost price of 25 articles , then the loss or gain percent is : | "c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 25 . loss % = 25 / 50 * 100 = 50 % answer e" | a = 50 - 25
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a ) 64 , b ) 48 , c ) 54 , d ) 46 , e ) can not be determined | e | subtract(negate(multiply(divide(16, add(3, 1)), const_2)), subtract(6, multiply(divide(16, add(3, 1)), const_3))) | when the bus started from the first bus stop , the number of male passengers to the number of female passengers was 1 : 3 . at the first stop , 16 passengers got down and 6 more female passengers got in . the ratio of the male to female passengers now became 1 : 2 . what was the total number of passengers in the bus when it started from the first stop ? | let , initially the number of males and females in the bus be β x β and β 3 x β respectively . at the first stop , suppose β m β males and β f β females left the bus . therefore at the first stop the number of males is β x - m β and the number of females is β ( 3 x - f ) + 6 β therefore ( x - m ) / [ ( 3 x - f ) + 6 ] = 1 / 2 = > 2 x - 2 m = 3 x - f + 6 also m + f = 16 x = 26 - f as we have 3 variable and 2 equations , a definite value can not be determined . answer : e | a = 3 + 1
b = 16 / a
c = b * 2
d = negate - (
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a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | c | divide(divide(subtract(const_100, 15), const_100), divide(50, const_100)) | when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 15 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 85 % ( 100 - 15 ) . therefore 0.5 x amount of flower - nectar = 0.85 x amount of honey = 0.85 x 1 kg therefore amount of flower - nectar needed = ( 0.85 / 0.51 ) kg = 1.7 kgs answer : c" | a = 100 - 15
b = a / 100
c = 50 / 100
d = b / c
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a ) 30 % , b ) 40 % , c ) 44.44 % , d ) 60 % , e ) 80 % | c | multiply(divide(subtract(90, add(multiply(2, 7), multiply(7, 2))), 90), const_100) | a batsman scored 90 runs which included 2 boundaries and 7 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 90 - ( 2 x 4 + 7 x 6 ) = 90 - ( 50 ) = 40 now , we need to calculate 40 is what percent of 90 . = > 40 / 90 * 100 = 44.44 % option c" | a = 2 * 7
b = 7 * 2
c = a + b
d = 90 - c
e = d / 90
f = e * 100
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a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability w that all 3 jellybeans removed from the bag are blue ? | method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability w = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b | a = math.comb(10, 3)
b = 10 + 10
c = math.comb(b, 3)
d = a / c
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a ) 12 % , b ) 12.5 % , c ) 6.67 % , d ) 6.25 % , e ) 8.33 % | e | multiply(divide(divide(subtract(400, 300), 5), subtract(300, multiply(divide(subtract(400, 300), 5), 3))), const_100) | peter invested a certain sum of money in a simple interest bond whose value grew to $ 300 at the end of 3 years and further to $ 400 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | answer initial amount invested = $ x amount at the end of year 3 = $ 300 amount at the end of year 8 ( another 5 years ) = $ 400 therefore , the interest earned for the 5 year period between the 3 rd year and 8 th year = $ 400 - $ 300 = $ 100 as the simple interest earned for a period of 5 years is $ 100 , interest earned per year = $ 20 . therefore , interest earned for 3 years = 3 * 20 = $ 60 . hence , initial amount invested x = amount after 3 years - interest for 3 years = 300 - 60 = $ 240 . rate of interest = ( interest per year / principal ) * 100 = 20 / 240 * 100 = 8.33 % choice is ( e ) | a = 400 - 300
b = a / 5
c = 400 - 300
d = c / 5
e = d * 3
f = 300 - e
g = b / f
h = g * 100
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a ) 30 , b ) 42 , c ) 60 , d ) data inadequate , e ) none of these | b | divide(multiply(13.5, 28), 9) | 28 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 28 Γ£ β 13.5 = 378 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 378 Γ’ Β β 9 = 42 answer b" | a = 13 * 5
b = a / 9
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a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | subtract(subtract(multiply(16, 4), 2), multiply(const_12, add(3, 2))) | the average ( arithmetic mean ) of 4 different positive integers is 16 . if the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer , what is the least possible value of the fourth integer ? | let the second integer be x and the fourth be a . then [ 3 x + x + ( x + 2 ) + a ] / 4 = 16 = > 5 x + 2 + a = 64 = > 5 x + a = 62 = > a = 62 - 5 x from the above equation we can see that a is minimum when x is maximum , provided both are positive the maximum value that x can take in the above equation while still keeping a positive is x = 12 this gives us a = 62 - 60 = 2 therefore the minimum value that the fourth integer can have is 2 . option d . | a = 16 * 4
b = a - 2
c = 3 + 2
d = 12 * c
e = b - d
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a ) 3 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | a | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 3)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 3))) | 36 welders work at a constant rate they complete an order in 3 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 3 = 108 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 3 of the job . so 24 workers have to finish only 2 / 3 of the job . 5 . total time taken by 24 workers can be got from formula used at ( 2 ) . i . e . , 24 * total time taken = 108 . total time taken by 6 workers to finish the complete job is 108 / 24 = 4.5 days . 6 . time taken by 24 workers to finish 2 / 3 of the job is 2 / 3 * 4.5 = 3 days . the answer is choice a" | a = 36 * 3
b = 1 / a
c = b * 36
d = 1 - c
e = 36 - 12
f = d / e
g = 36 * 3
h = 1 / g
i = f / h
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a ) 14 , b ) 47 , c ) 50 , d ) 180 , e ) 240 | c | add(add(divide(200, 40), divide(200, 40)), 40) | frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 200 square feet . how many feet offence does frank use ? | "area = length x breadth 200 = 40 x breadth so , breadth = 5 units fencing required is - breadth + breadth + length 5 + 5 + 40 = > 50 feet answer must be ( c ) 50" | a = 200 / 40
b = 200 / 40
c = a + b
d = c + 40
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a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 44 | b | divide(add(120, 60), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 120 marks , the no of questions he attempts correctly is : | "explanation : let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 4 x β ( 60 β x ) = 120 = > 5 x = 180 = > x = 36 answer : b" | a = 120 + 60
b = 4 + 1
c = a / b
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a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | d | multiply(add(add(20, 4), 5), divide(4, const_60)) | a certain car increased its average speed by 4 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 20 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | "in the third time interval the average speed of the car was 20 + 4 + 5 = 29 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 29 * 1 / 12 = 2.5 miles . answer : d ." | a = 20 + 4
b = a + 5
c = 4 / const_60
d = b * c
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a ) 15 , b ) 20 , c ) 25 , d ) 28 , e ) 30 | e | subtract(divide(multiply(multiply(30, subtract(10, 2)), 5), multiply(2, subtract(15, 5))), 30) | an engineer undertakes a project to build a road 10 km long in 15 days and employs 30 men for the purpose . after 5 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 10 days 2 km done hence remaining is 8 km also , work has to be completed in next 10 days ( 15 - 10 = 5 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 30 * 8 * 5 ) / ( 2 * 10 ) thus , x = 60 thus , more men needed to finish the task = 60 - 30 = 30 answer : e" | a = 10 - 2
b = 30 * a
c = b * 5
d = 15 - 5
e = 2 * d
f = c / e
g = f - 30
|
a ) 11.00 sec , b ) 12.50 sec , c ) 15.00 sec , d ) 15.23 sec , e ) 5.23 sec | c | divide(add(100, 150), multiply(60, const_0_2778)) | chandigarh express of 100 m runs at a speed of 60 km / hr . what will be the time taken to cross a platform of 150 meters long ? | explanation : given : length of train = 100 m , speed of train = 60 km / hr , length of platform = 150 m 1 ) always remember first step is the conversion of units . convert 60 km / hr into m / s by multiplying it with ( 5 / 18 ) speed of the train = 60 x 5 / 18 = 16.66 m / s 2 ) distance covered by the train in passing the platform = ( length of train + length of platform ) = ( 100 + 150 ) = 250 m therefore , the time taken = distance / speed = 250 / 16.66 = 15 sec answer is c | a = 100 + 150
b = 60 * const_0_2778
c = a / b
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a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16 | a | multiply(const_4, power(27, divide(const_1, const_3))) | a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many p cubes are painted on only 2 faces ? | "1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all on the edge of the cube , in themiddleof the edge . there are 4 in front , 4 in back and 4 more on thestripthat runs around the left / top / right / bottom of the cube . p = 4 + 4 + 4 = 12 . answer a" | a = 1 / 3
b = 27 ** a
c = 4 * b
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a ) 50 , b ) 60 , c ) 10 , d ) 20 , e ) 30 | c | divide(divide(70, subtract(subtract(subtract(const_1, divide(50, const_100)), divide(1, 20)), divide(1, 10))), 20) | in a school 50 % of the students are younger than 10 , 1 / 20 are 10 years old and 1 / 10 are older than 10 but younger than 12 , the remaining 70 students are 12 years or older . how many students are 10 years old ? | let us write the fraction for each group of students group a : younger than 10 : 50 % = 50 / 100 = 1 / 2 group b : 10 years old : 1 / 20 group c : older that 10 but younger than 12 : 1 / 10 group d : 12 years or older : 70 students the fraction for group a , b and c together is given by 1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator = 13 / 20 , add numerators the fraction for group d is given by 20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students if x is the total number of students , then 7 / 20 of x = 70 or ( 7 / 20 ) * x = 70 solve for x x = 70 * ( 20 / 7 ) = 200 students who are 10 years old have a fraction of 1 / 20 of the total x and their number is equal to ( 1 / 20 ) of 200 = 10 students correct answer c | a = 50 / 100
b = 1 - a
c = 1 / 20
d = b - c
e = 1 / 10
f = d - e
g = 70 / f
h = g / 20
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a ) 180 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(270, subtract(20, 8)), 8) | a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 270 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 270 + l ) / 20 = l / 8 2160 + 8 l = 20 l l = 2160 / 12 = 180 m answer a" | a = 20 - 8
b = 270 / a
c = b * 8
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a ) 80 , b ) 120 , c ) 160 , d ) 240 , e ) 560 | e | multiply(multiply(2, 7), multiply(multiply(2, 4), 5)) | if the operation β¬ is defined for all x and y by the equation x β¬ y = 2 * x * y , then 7 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 2 * 4 * 5 = 40 7 β¬ 40 = 2 * 3 * 40 = 560 hence , answer is e" | a = 2 * 7
b = 2 * 4
c = b * 5
d = a * c
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a ) 1600 , b ) 2490 , c ) 3780 , d ) 4235 , e ) 6179 | c | multiply(divide(add(40, multiply(2, 85)), subtract(90, 85)), 90) | if a farmer wants to plough a farm field on time , he must plough 90 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ? | "let x be the number of days in the initial plan . therefore , the whole field is 90 Γ’ βΉ β¦ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 90 x = 85 ( x + 2 ) + 40 5 x = 210 x = 42 so the farmer planned to have the work done in 6 days , and the area of the farm field is 90 ( 42 ) = 3780 hectares correct answer c" | a = 2 * 85
b = 40 + a
c = 90 - 85
d = b / c
e = d * 90
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a ) 160 , b ) 220 , c ) 240 , d ) 360 , e ) 420 | d | add(divide(multiply(divide(55, const_100), 30), subtract(divide(60, const_100), divide(55, const_100))), 30) | the workforce of company x is 60 % female . the company hired 30 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 30 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 30 ) 0.05 x = 16.5 x = 330 - this is quantity of employees before adding 30 men so after adding it will be 360 answer is d" | a = 55 / 100
b = a * 30
c = 60 / 100
d = 55 / 100
e = c - d
f = b / e
g = f + 30
|
a ) $ 8,617 , b ) $ 8,710 , c ) $ 8,810 , d ) $ 8,835 , e ) $ 8,915 | b | multiply(multiply(const_2, const_3), const_100) | a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 4,500 meters of fabric ? | for first 2000 metres he does not get any discount amount = 2 * 2000 = $ 4000 for next 1500 metres he gets 5 % discount amount = 2 * 1500 - ( 1 / 20 ) * 3000 = $ 2850 for the nest 1000 metres he gets 7 % discount amount = 2 * 1000 - 7 * 20 = $ 1860 total amount = $ 4000 + $ 2850 + $ 1860 = $ 8710 answer : b | a = 2 * 3
b = a * 100
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a ) 16.67 % , b ) 23 % , c ) 18.3 % , d ) 19.75 % , e ) 17.4 % | e | multiply(divide(subtract(add(20, 3), add(multiply(divide(subtract(const_100, 20), const_100), 20), 3)), add(20, 3)), const_100) | 20 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 20 litres of mix . = 20 % of 20 litres = ( 20 * 20 / 100 ) = 4 litres water in it = 20 - 3 = 17 litres new quantity of mix . = 20 + 3 = 23 litres quantity of alcohol in it = 4 litres percentage of alcohol in new mix . = 4 * 100 / 23 = 17.4 % answer is e" | a = 20 + 3
b = 100 - 20
c = b / 100
d = c * 20
e = d + 3
f = a - e
g = 20 + 3
h = f / g
i = h * 100
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a ) 4 m , b ) 4.5 m , c ) 5 m , d ) 6 m , e ) none | a | divide(sqrt(divide(16128, multiply(multiply(6, 7), 6))), const_2) | the heights of a wall is 6 times its width and the length of the wall is 7 times its height . if volume of the wall be 16128 cu . m , its width is | solution let the width of the wall be x metres . then , height = ( 6 x ) m and length = ( 42 x ) m . 42 x Γ x Γ 6 x = 16128 x 3 βΉ = βΊ ( 16128 / 42 Γ 6 ) = 64 βΉ = βΊ x = 4 . answer a | a = 6 * 7
b = a * 6
c = 16128 / b
d = math.sqrt(c)
e = d / 2
|
a ) 11.34 , b ) 12.34 , c ) 13.34 , d ) 14.34 , e ) 15.34 | c | subtract(add(5.6, 10.3), 2.56256) | ( 9 ^ 5.6 x 9 ^ 10.3 ) Γ· 9 ^ 2.56256 = 9 ^ ? | 5.6 + 10.3 = 15.9 i . e . . 15.9 - 2.56 = 13.34 so ur answer is 9 ^ 13.34 answer : c | a = 5 + 6
b = a - 2
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a ) 382550 , b ) 384560 , c ) 385690 , d ) 389650 , e ) 381150 | e | multiply(divide(3850, 99), const_100) | 3850 * 99 | "explanation : 3850 * ( 100 - 1 ) = 385000 - 3850 = 381150 option e" | a = 3850 / 99
b = a * 100
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a ) 22 , b ) 25 , c ) 27 , d ) 29 , e ) 11 | b | multiply(divide(subtract(multiply(add(add(add(20, 5), const_10), add(divide(5, const_2), const_3)), 18), multiply(20, 18)), multiply(20, 18)), const_100) | a milk man has 20 liters of milk . if he mixes 5 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is : | "explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 20 liters of milk . therefore , profit percentage = \ inline \ frac { 5 } { 20 } \ times 100 = 25 % % [ profit % = \ inline \ frac { profit } { cost \ : price } \ times 100 ] answer : b ) 25 %" | a = 20 + 5
b = a + 10
c = 5 / 2
d = c + 3
e = b + d
f = e * 18
g = 20 * 18
h = f - g
i = 20 * 18
j = h / i
k = j * 100
|
a ) 8 , b ) 55 , c ) 87 , d ) 90 , e ) 2 | b | multiply(divide(10, 45), const_100) | 10 + 45 | b | a = 10 / 45
b = a * 100
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a ) 1 / 7 , b ) 2 / 7 , c ) 1 / 2 , d ) 7 / 12 , e ) 1 / 4 | e | divide(const_2, 8) | equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 8 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 8 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 8 + 1 / 8 = 1 / 4 full . answer : e ." | a = 2 / 8
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a ) 46 , b ) 40 , c ) 44 , d ) 48 , e ) 49 | a | add(40, divide(subtract(1000, multiply(20, 40)), divide(multiply(20, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 20 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 $ 1000 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 20 = 800 excess = 1000 - 800 = 200 for extra hours = . 75 ( 20 ) = 15 + 20 = 35 number of extra hrs = 200 / 35 = 5.71 = 6 approx . total hrs = 40 + 6 = 46 answer a 46" | a = 20 * 40
b = 1000 - a
c = 100 + 75
d = 20 * c
e = d / 100
f = b / e
g = 40 + f
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a ) 20 hours , b ) 24 hours , c ) 26 hours , d ) 18 hours , e ) none of these | b | inverse(subtract(inverse(6), inverse(add(2, 6)))) | a cistern normally takes 6 hours to be filled by a tap but because of a leak , 2 hours more . in how many hours will the leak empty a full cistern ? | β΅ cistern fill in 6 hours . β΄ in 1 hour , filled part = 1 β 6 th now , due to leakage , filled part in 1 hour = 1 β 8 th part of the cistern emptied , due to leakage in 1 hour = 1 β 6 - 1 β 8 = 1 β 24 th β΄ the leakage will empty the full cistern in 24 hrs . answer b | a = 1/(6)
b = 2 + 6
c = 1/(b)
d = a - c
e = 1/(d)
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | floor(divide(20, divide(10, const_2))) | what is the greatest integer m for which the number 20 ! / 10 ^ m is an integer ? | "10 ^ m = 2 ^ m * 5 ^ m . let ' s figure out how many 5 ' s are in the prime factorization of 20 ! the multiples of 5 are : 5 , 10 , 15 , 20 . thus 5 ^ 4 will divide 20 ! but 5 ^ 5 will not . clearly 2 ^ 4 will divide 20 ! so m = 4 is the largest possible integer . the answer is b ." | a = 10 / 2
b = 20 / a
c = math.floor(b)
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a ) 5 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 7 / 8 , e ) 8 / 7 | a | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of ( p + q ) / ( p - q ) if p / q is 4 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 4 + 1 ) / ( 4 - 1 ) = 5 / 3 = 5 / 3 answer : a" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
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a ) 25 , b ) 30 , c ) 10 , d ) 15 , e ) 20 | c | divide(50, 5) | doughnuts , cookies and muffins in a pastry shop are in the ratio of 5 : 3 : 1 . if there are 50 doughnuts , the number of muffins in the shop is : . | explanation : let doughnuts = 5 x , cookies = 3 x & muffins = 1 x . now , 5 x = 50 hence x = 10 . number of muffins = 1 x which is 10 . answer : c | a = 50 / 5
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a ) 3 , b ) 1 , c ) 16 , d ) 4 , e ) 15 | d | subtract(9679, multiply(floor(divide(9679, 15)), 15)) | what least number must be subtracted from 9679 so that the remaining number is divisible by 15 ? | "on dividing 9679 by 15 , we get remainder = 4 . required number be subtracted = 4 answer : d" | a = 9679 / 15
b = math.floor(a)
c = b * 15
d = 9679 - c
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a ) 33 , 13 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | a | add(divide(add(multiply(5, 8), subtract(20, 8)), subtract(5, const_1)), 20) | the ages of two person differ by 20 years . if 8 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | let their ages be x and ( x + 20 ) years . then , 5 ( x - 8 ) = ( x + 20 - 8 ) = > 4 x = 52 = > x = 13 their present ages are 33 years and 13 year . answer : a | a = 5 * 8
b = 20 - 8
c = a + b
d = 5 - 1
e = c / d
f = e + 20
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a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is = . 25 / 1 * 100 = 25 % d is the answer ." | a = 9 / 12
b = a * 100
c = 100 - b
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a ) 4500 , b ) 7500 , c ) 5000 , d ) 6800 , e ) none of these | d | divide(17, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 17 . find the sum : | sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the difference is 17 , the sum = 400 Γ 17 = 6800 answer d | a = 5 / 100
b = 1 + a
c = b ** 2
d = 5 / 100
e = 2 * d
f = 1 + e
g = c - f
h = 17 / g
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a ) 188 m , b ) 876 m , c ) 251 m , d ) 150 m , e ) 400 m | e | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 20 seconds and a platform 100 m long in 25 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 20 = > y = x / 20 x + 100 / 25 = x / 20 x = 400 m . answer : e" | a = 2 - 1
b = 100 * a
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a ) s . 738 , b ) s . 750 , c ) s . 650 , d ) s . 730 , e ) s . 735 | b | subtract(825, divide(multiply(subtract(850, 825), 3), 4)) | a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 850 in 4 years . the sum is ? | "s . i . for 1 year = ( 850 - 825 ) = rs . 25 s . i . for 3 years = 25 * 3 = rs . 75 principal = ( 825 - 75 ) = rs . 750 . answer : b" | a = 850 - 825
b = a * 3
c = b / 4
d = 825 - c
|
a ) 120 , b ) 100 , c ) 125 , d ) 150 , e ) none of these | b | divide(multiply(100, add(const_4, const_1)), add(const_1, const_2)) | the difference between a number and its three - fourth is 100 . what is 25 % of that number ? | "explanation : solution : let the number be x . then , x - 3 / 4 x = 100 = > x / 4 = 100 = > x = 400 . 25 % of x = 25 % of 400 = 25 / 100 * 400 = 100 . answer : b" | a = 4 + 1
b = 100 * a
c = 1 + 2
d = b / c
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a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | b | subtract(subtract(multiply(3, 16), add(subtract(11, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 11 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 11 a β d = 3 a + d = 11 2 d = 8 d = 4 answer b" | a = 3 * 16
b = 11 - 16
c = b + 3
d = a - c
e = d - 16
|
a ) 235 miles . , b ) 245 miles . , c ) 240 miles . , d ) 265 miles . , e ) 275 miles . | c | add(multiply(45, 2), multiply(3, 50)) | john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 50 mph to reach his destination . how many miles did john drive ? | the total distance d traveled by john is given by d = 45 * 2 + 3 * 50 = 240 miles . answer c | a = 45 * 2
b = 3 * 50
c = a + b
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a ) 3 , b ) 1 , c ) 5 , d ) 11 , e ) 15 | c | subtract(3830, multiply(floor(divide(3830, 15)), 15)) | what least number must be subtracted from 3830 so that the remaining number is divisible by 15 ? | "on dividing 3830 by 15 , we get remainder = 5 . required number be subtracted = 5 answer : c" | a = 3830 / 15
b = math.floor(a)
c = b * 15
d = 3830 - c
|
a ) 2 / 3 , b ) 5 / 8 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 8 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what isthe probability that theextracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initially in the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a = 1 / 2
b = 1 + 3
c = 1 / b
d = 1 + 3
e = 3 / d
f = c + e
g = a + f
h = 1 + g
i = 1 + 3
j = h / i
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a ) 25 , b ) 28 , c ) 30 , d ) 77 , e ) 36 | d | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(8, const_2)), add(add(8, const_2), const_2)), add(add(add(8, const_2), const_2), const_4)), add(add(add(add(8, const_2), const_2), const_4), const_2)), add(add(add(add(add(8, const_2), const_2), const_4), const_2), const_4)) | find a sum for 1 st 8 prime number ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 ) = 77 note : 1 is not a prime number option d" | a = 2 + 3
b = 2 + 3
c = a + b
d = 2 + 3
e = d + 2
f = c + e
g = 8 + 2
h = f + g
i = 8 + 2
j = i + 2
k = h + j
l = 8 + 2
m = l + 2
n = m + 4
o = k + n
p = 8 + 2
q = p + 2
r = q + 4
s = r + 2
t = o + s
u = 8 + 2
v = u + 2
w = v + 4
x = w + 2
y = x + 4
z = t + y
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a ) 100 , b ) 10 , c ) 1,000 , d ) 10,000 , e ) 100,000 | c | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | if 1,000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "given that 1,000 microns = 1 decimeter = 1 , 000,000 angstroms so , 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : c" | a = 100 * 100
b = 1 / a
c = 100 * 100
d = b * c
|
a ) 16 , b ) 20 , c ) 24 , d ) 28 , e ) 32 | b | multiply(50, divide(18, 45)) | a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 50 meters , what is the height of the building ( in meters ) ? | "the height : length ratio will be equal in both cases . 18 / 45 = x / 50 x = 20 the answer is b ." | a = 18 / 45
b = 50 * a
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a ) 3 , b ) 4 , c ) 2 , d ) 1 , e ) 7 | b | divide(add(subtract(7, const_4), subtract(7, const_2)), const_2) | if 7 ^ ( x - y ) = 343 and 7 ^ ( x + y ) = 16807 , what is the value of x ? | 7 ^ ( x - y ) = 343 = 7 ^ 3 = > x - y = 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 1 ) 7 ^ ( x + y ) = 16807 = 7 ^ 5 = > x + y = 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 2 ) ( equation 1 ) + ( equation 2 ) = > 2 x = 3 + 5 = 8 = > x = 8 / 2 = 4 answer is b | a = 7 - 4
b = 7 - 2
c = a + b
d = c / 2
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a ) 120 , b ) 150 , c ) 180 , d ) 200 , e ) 220 | e | sqrt(divide(multiply(242, const_100), divide(50, const_100))) | 242 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ? | "let b be the number of boys in the school . 242 = xb / 100 b = 0.5 x 24200 = 0.5 x ^ 2 x ^ 2 = 48400 x = 220 the answer is e ." | a = 242 * 100
b = 50 / 100
c = a / b
d = math.sqrt(c)
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a ) 65 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(75, 10), 10), subtract(75, subtract(75, 10))) | a certain number of workers can do a work in 75 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 75 - 10 ) / 10 = 65 answer is a" | a = 75 - 10
b = a * 10
c = 75 - 10
d = 75 - c
e = b / d
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a ) $ 8600 , b ) $ 2400 , c ) $ 1000 , d ) $ 3000 , e ) $ 3400 | e | add(multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(3, 10)), multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(1, 8))) | a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on conveyance . if he has $ 1400 left with him , find his expenditure on food and conveyance . | remaining part of salary left = 1 - ( 2 / 5 + 3 / 10 + 1 / 8 ) = 1 - ( 33 / 40 ) = 7 / 40 let the monthly salary be $ x then , 7 / 40 of x = $ 1400 x = ( 1400 * 40 / 7 ) = $ 8600 expenditure on food = $ ( 3 / 10 * 800 ) = $ 2400 expenditure on conveyance = $ ( 1 / 8 * 8000 ) = $ 1000 $ 2400 + $ 1000 = $ 3400 so the answer is option e ) $ 3400 | a = 2 / 5
b = 3 / 10
c = a + b
d = 1 / 8
e = c + d
f = 1 - e
g = 1400 / f
h = 3 / 10
i = g * h
j = 2 / 5
k = 3 / 10
l = j + k
m = 1 / 8
n = l + m
o = 1 - n
p = 1400 / o
q = 1 / 8
r = p * q
s = i + r
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a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 3 / 2 , e ) 3 / 5 | d | divide(subtract(divide(30, const_100), divide(15, const_100)), subtract(divide(25, const_100), divide(15, const_100))) | some of 15 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 15 ( 1 - x ) + 25 x = 30 x = 3 / 2 answer : d" | a = 30 / 100
b = 15 / 100
c = a - b
d = 25 / 100
e = 15 / 100
f = d - e
g = c / f
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a ) $ 120 , b ) $ 150 , c ) $ 548 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 2)), 1200) | find the compound interest on $ 1200 for 2 years at 20 % p . a . if ci is component yearly ? | a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 2 = $ 1728 ci = $ 548 answer is c | a = 20 / 100
b = 1 + a
c = b ** 2
d = 1200 * c
e = d - 1200
|
a ) 70 , b ) 20 , c ) 50 , d ) 60 , e ) 30 | d | multiply(divide(subtract(const_1, add(divide(const_1, 2), divide(const_1, 4))), divide(const_1, 4)), const_60) | it takes avery 2 hours to build a brick wall while tom can do it in 4 hour . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | avery takes 2 hours tom takes 4 hours efficiency of avery is 1 / 2 units / hr efficiency of tom is 1 / 4 units / hr combined efficiency of tom and avery is 1 / 2 + 1 / 4 = 3 / 4 units / hr since they worked for 1 hour they completed 3 / 4 units of work and 1 / 4 units of work is left which is to be completed by tom ( since avery left ) so time taken by tom to complete the remaining work will be 1 / 4 / 1 / 4 hours = > 1 * 60 = 60 minutes . . . answer will be ( d ) | a = 1 / 2
b = 1 / 4
c = a + b
d = 1 - c
e = 1 / 4
f = d / e
g = f * const_60
|
a ) 21 , b ) 20 , c ) 11 , d ) 10 , e ) 9 | d | divide(subtract(45, 25), const_2) | how many of the integers between 25 and 45 are even ? | number start between 25 to 45 is 20 numbers half of them is even . . which is 10 answer : d | a = 45 - 25
b = a / 2
|
a ) $ 200 , b ) $ 260 , c ) $ 320 , d ) $ 380 , e ) $ 440 | e | multiply(divide(28, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 28 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 28 0.07 = 28 p = 400 the original selling price was 400 * 1.1 = 440 . the correct answer is e ." | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 28 / h
j = 10 / 100
k = 1 + j
l = i * k
|
a ) 2 : 3 , b ) 7 : 40 , c ) 3 : 20 , d ) 20 : 3 , e ) none | b | divide(multiply(0.07, const_100), multiply(0.4, const_100)) | if 0.4 of a number is equal to 0.07 of another number , the ratio of the numbers i | "sol . 0.4 a = 0.08 b Γ’ β‘ β a / b = 0.07 / 0.40 = 7 / 40 . Γ’ Λ Β΄ a : b = 7 : 40 . answer b" | a = 0 * 7
b = 0 * 4
c = a / b
|
a ) 140 , b ) 165 , c ) 150 , d ) 142 , e ) 174 | a | divide(multiply(21, const_100), multiply(const_3, 5)) | sixty 5 percent of a number is 21 less than 4 fifth of that number . what is the number | let the number be x . then , 4 * x / 5 β ( 65 % of x ) = 21 4 x / 5 β 65 x / 100 = 21 5 x = 2100 x = 140 . answer a 140 | a = 21 * 100
b = 3 * 5
c = a / b
|
a ) 59.8 % , b ) 63.6 % , c ) 67.2 % , d ) 70.5 % , e ) 73.6 % | b | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(7)))) | a wooden cube whose edge length is 7 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ? | "there are a total of 7 * 7 * 7 = 343 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 5 * 5 * 5 = 125 cubes . the number of cubes with at least one side painted red is 343 - 125 = 218 cubes the probability that a cube has at least one side painted red is 218 / 343 which is about 63.6 % the answer is b ." | a = 1 * 4
b = volume_cube / (
c = 1 - b
d = 100 * c
|
a ) 2 : 1 , b ) 1 : 2 , c ) 4 : 9 , d ) 1 : 1 , e ) 3 : 2 | c | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(4, const_100)), subtract(divide(7, const_100), divide(multiply(divide(6, const_100), 4), const_3))) | two numbers a and b are such that the sum of 7 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "explanation : 7 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 7 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) β 7 a + 4 b = 2 / 3 ( 6 a + 8 b ) β 21 a + 12 b = 12 a + 16 b β 9 a = 4 b β ab = 4 / 9 β a : b = 4 : 9 answer : option c" | a = 8 / 100
b = a * 4
c = b / 3
d = 4 / 100
e = c - d
f = 7 / 100
g = 6 / 100
h = g * 4
i = h / 3
j = f - i
k = e / j
|
a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | a | add(add(50, 4), add(4, 1)) | the average weight of a class is x pounds . when a new student weighing 50 pounds joins the class , the average decreases by 1 pound . in a few months the student β s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students β weights changed . what is the value of x ? | "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 = 6 students ( including the new one ) . so , initially there were 5 student . total weight = 5 x + 80 = 6 ( x - 1 ) - - > x = 85 pounds . answer : a ." | a = 50 + 4
b = 4 + 1
c = a + b
|
a ) 72 kmph , b ) 75 kmph , c ) 30 kmph , d ) 90 kmph , e ) 82 kmph | d | divide(add(140, 40), const_2) | the speed of a car is 140 km in the first hour and 40 km in the second hour . what is the average speed of the car ? | "s = ( 140 + 40 ) / 2 = 90 kmph answer : d" | a = 140 + 40
b = a / 2
|
a ) 6 , b ) 3 , c ) 4 , d ) 4.5 , e ) 2 | d | add(divide(multiply(add(50, const_1), 60), const_1000), 1.5) | a train of 50 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 50 * 60 + 1500 = 3000 m t = 4500 / 60 * 18 / 5 = 270 sec = 4.5 mins answer : d" | a = 50 + 1
b = a * 60
c = b / 1000
d = c + 1
|
a ) 50 , b ) 150 , c ) 250 , d ) 350 , e ) 450 | e | multiply(subtract(10, multiply(const_4.0, const_2)), multiply(15, 18)) | how much space , in cubic units , is left vacant when maximum number of 5 x 5 x 5 cubes are fitted in a rectangular box measuring 10 x 15 x 18 ? | "no of cubes that can be accommodated in box = ( 10 * 15 * 18 ) / ( 5 * 5 * 5 ) 15 * 10 in numerator can be perfectly divided by 5 * 5 in denominator . side with length 18 ca n ' t be perfectly divided by 5 and hence is the limiting factor . closet multiple of 5 less that 18 is 15 . so vacant area in cube = = 15 * 10 * ( 18 - 15 ) = 15 * 10 * 3 = 450 ans - e" | a = 4 * 0
b = 10 - a
c = 15 * 18
d = b * c
|
a ) 39 , b ) 84 , c ) 80 , d ) 82 , e ) 86 | b | divide(multiply(49, 12), 7) | 12 buckets of water fill a tank when the capacity of each bucket is 49 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 7 litres ? | capacity of the tank = ( 12 Γ£ β 49 ) litre number of buckets required of capacity of each bucket is 17 litre = 12 Γ£ β 49 / 7 = 12 Γ£ β 7 = 84 answer is b | a = 49 * 12
b = a / 7
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | floor(sqrt(divide(8100, 101))) | if n is an integer and 101 n ^ 2 is less than or equal to 8100 , what is the greatest possible value of n ? | "101 * n ^ 2 < = 8100 n ^ 2 < = 8100 / 101 which will be less than 81 since 8100 / 100 = 81 which is the square of 9 next closest value of n where n ^ 2 < = 81 is 8 ans b" | a = 8100 / 101
b = math.sqrt(a)
c = math.floor(b)
|
a ) 338 m , b ) 778 m , c ) 375 m , d ) 276 m , e ) 971 m | c | subtract(divide(950, const_2), 100) | if the perimeter of a rectangular garden is 950 m , its length when its breadth is 100 m is ? | "2 ( l + 100 ) = 950 = > l = 375 m answer : c" | a = 950 / 2
b = a - 100
|
a ) 41 , b ) 42 , c ) 45 , d ) 47 , e ) 49 | d | add(subtract(multiply(13, const_2), const_1), subtract(multiply(11, const_2), const_1)) | a certain league has two divisions . the respective divisions had 11 and 13 teams qualify for the playoffs . each division held its own double - elimination tournament - - where a team is eliminated from the tournament upon losing two games - - in order to determine its champion . the 4 division champions then played in a single - elimination tournament - - where a team is eliminated upon losing one game - - in order to determine the overall league champion . assuming that there were no ties and no forfeits , what is the maximum number of games that could have been played in order to determine the overall league champion ? | overall , there can be 47 minuses . tus , it is d . | a = 13 * 2
b = a - 1
c = 11 * 2
d = c - 1
e = b + d
|
a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 6 | b | divide(multiply(5, 1), multiply(3, 5)) | if the ratio of a to b is 5 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 5 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 4 : 15 answer b" | a = 5 * 1
b = 3 * 5
c = a / b
|
a ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | e | divide(multiply(100, add(multiply(4, const_2), 4)), 40) | it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 100 pages ? | "if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 100 / 50 = 30 minutes to print 100 pages ( answer e )" | a = 4 * 2
b = a + 4
c = 100 * b
d = c / 40
|
a ) 12455 , b ) 14153 , c ) 14256 , d ) 13541 , e ) 14541 | d | subtract(add(multiply(78, 7), multiply(add(const_10, const_3), multiply(const_100, const_10))), add(const_4, const_1)) | what does the hexa number e 78 in radix 7 . | conversion of hexa decimal to decimal ( e 78 ) = ( 8 * 16 ^ 0 ) + ( 7 * 16 ^ 1 ) + ( 14 * 16 ^ 2 ) = 8 + 112 + 3584 = 3704 conversion of decimal to radix 7 3704 / 7 : reminder : 1 and dividend : 529 529 / 7 : reminder : 4 and dividend : 75 75 / 7 : reminder : 5 and dividend : 10 10 / 7 : reminder : 3 and dividend : 1 1 / 7 : reminder 1 then finally result is : 13541 answer : d | a = 78 * 7
b = 10 + 3
c = 100 * 10
d = b * c
e = a + d
f = 4 + 1
g = e - f
|
a ) rs . 720 , b ) rs . 920 , c ) rs . 537 , d ) rs . 540 , e ) rs . 637 | d | multiply(4, divide(1170, add(add(4, 2), const_3))) | rs . 1170 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | a + b + c = 1170 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1170 = rs . 540 answer : d | a = 4 + 2
b = a + 3
c = 1170 / b
d = 4 * c
|
a ) 29 , b ) 776 , c ) 2 , d ) 12 , e ) 99 | c | subtract(multiply(40, divide(55, const_100)), multiply(divide(4, 5), 25)) | how much is 55 % of 40 is greater than 4 / 5 of 25 ? | "( 55 / 100 ) * 40 β ( 4 / 5 ) * 25 22 - 20 = 2 answer : c" | a = 55 / 100
b = 40 * a
c = 4 / 5
d = c * 25
e = b - d
|
a ) 1209 , b ) 1615 , c ) 1245 , d ) 1300 , e ) 1635 | b | multiply(divide(subtract(1515, 15), subtract(16, const_1)), 16) | find large number from below question the difference of two numbers is 1515 . on dividing the larger number by the smaller , we get 16 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1515 ) . x + 1515 = 16 x + 15 15 x = 1500 x = 100 large number = 100 + 1515 = 1615 answer : b" | a = 1515 - 15
b = 16 - 1
c = a / b
d = c * 16
|
a ) 22678 , b ) 59000 , c ) 26788 , d ) 19000 , e ) 26711 | b | subtract(60000, 1000) | the price of a t . v . set worth rs . 60000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 1000 balance payment = ( 60000 - 1000 ) = rs . 59000 answer : b" | a = 60000 - 1000
|
a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | a | add(2, divide(subtract(const_1, divide(2, 15)), add(inverse(15), inverse(10)))) | p and q can complete a work in 15 days and 10 days respectively . they started the work together and then q left after 2 days . p alone completed the remaining work . the work was finished in - - - days . | "explanation : work done by p in 1 day = 1 / 15 work done by q in 1 day = 1 / 10 work done by p and q in 1 day = 1 / 15 + 1 / 10 = 1 / 6 work done by p and q in 2 days = 2 Γ ( 1 / 6 ) = 1 / 3 remaining work = 1 β 1 / 3 = 2 / 3 time taken by p to complete the remaining work 2 / 3 = ( 2 / 3 ) / ( 1 / 15 ) = 10 days total time taken = 2 + 10 = 12 days answer : option a" | a = 2 / 15
b = 1 - a
c = 1/(15)
d = 1/(10)
e = c + d
f = b / e
g = 2 + f
|
a ) 1350 , b ) 1971 , c ) 9676 , d ) 1679 , e ) 1250 | e | multiply(subtract(rectangle_area(add(65, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(65, 55)), 2) | a rectangular grass field is 65 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 65 + 55 + 2.5 * 2 ) 2 * 2.5 = > 625 625 * 2 = rs . 1250 answer : e" | a = 2 * 5
b = 65 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 2 , b ) 4 , c ) 6 , d ) 10 , e ) 20 | e | subtract(50, multiply(divide(50, const_100), 5)) | how many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 5 percent sugar solution ? | "3 % sugar solution means that currently there are 1.5 liters of sugar and this amount must be 5 % of a reduced final amount ( 50 - x ) 1.5 = 0.05 * ( 50 - x ) 0.05 x = 1 x = 20 answer e" | a = 50 / 100
b = a * 5
c = 50 - b
|
a ) $ 6.8 , b ) $ 12 . , c ) $ 13 . , d ) $ 14.5 . , e ) $ 15 . | a | add(add(multiply(3, 3), multiply(add(subtract(7, 4), divide(40, const_100)), 2)), divide(40, const_100)) | the mall charges 40 cents for the first hour of parking and $ 3 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 2 per hour . if mr . park parked his car in the mall for 7 hours and 30 minutes , how much is he going to pay ? | "0.40 + 2 * 2 + 1 * 2 + 0.40 = $ 6.8 answer : $ 6.8" | a = 3 * 3
b = 7 - 4
c = 40 / 100
d = b + c
e = d * 2
f = a + e
g = 40 / 100
h = f + g
|
a ) $ 5 , b ) $ 7 , c ) $ 6 , d ) $ 8 , e ) $ 9 | b | subtract(divide(350, const_10), add(subtract(20, 2), const_10)) | david spends $ 350 buying his favorite dolls . if he buys only small lemonhead dolls , which are $ 2 cheaper than the large lemonhead dolls , he could buy 20 more dolls than if he were to buy only large lemonhead dolls . how much does a large lemonhead doll cost ? | if large doll costs $ 5 , then he can buy 350 / 5 = 70 large dolls and 350 / 3 = 116 small dolls . difference is more than 20 . discard . if large doll costs $ 7 , then he can buy 350 / 7 = 50 large dolls and 350 / 5 = 70 small dolls . difference is 20 , which is we wanted . answer b . | a = 350 / 10
b = 20 - 2
c = b + 10
d = a - c
|
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