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a ) 450 min , b ) 528 min , c ) 610 min , d ) 714 min , e ) 359 min | b | add(4.5, 3.75) | the jogging track in a sports complex is 726 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 726 m apart to be 4.5 + 3.75 = 8.25 km apart , they take 1 hour to be 726 m apart , they take 100 / 825 * 726 / 1000 = 242 / 2750 * 60 = 528 min . answer is b" | a = 4 + 5
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a ) 6 : 8 , b ) 6 : 1 , c ) 6 : 5 , d ) 6 : 7 , e ) 6 : 3 | d | divide(multiply(4, 3), multiply(7, 2)) | the marks obtained by vijay and amith are in the ratio 4 : 7 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ? | "4 : 7 3 : 2 - - - - - - - 12 : 21 : 14 12 : 14 6 : 7 answer : d" | a = 4 * 3
b = 7 * 2
c = a / b
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a ) 1000 , b ) 450 , c ) 720 , d ) 180 , e ) 400 | a | multiply(multiply(subtract(10, const_1), 10), divide(10, const_2)) | there are , in a certain league , 10 teams , and each team face another team for a total of 10 times . how many games are played in the season ? | "by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 450 option a ." | a = 10 - 1
b = a * 10
c = 10 / 2
d = b * c
|
a ) 35 , b ) 37 , c ) 38 , d ) 41 , e ) 42 | c | add(multiply(const_2, 10), add(8, 10)) | in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 8 years older than b , the present age of b is | "explanation : let b ' s age = x years . then , as age = ( x + 8 ) years . ( x + 8 + 10 ) = 2 ( x — 10 ) hence x = 38 . present age of b = 38 years answer : option c" | a = 2 * 10
b = 8 + 10
c = a + b
|
a ) 38 , b ) 40 , c ) 42 , d ) 45 , e ) 50 | d | divide(multiply(60, const_2), add(divide(60, 36), const_1)) | an assembly line produces 36 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this whole time ? | the time to produce the first 60 cogs is 60 / 36 = 5 / 3 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / ( 8 / 3 ) hours = 45 cogs per hour . the answer is d . | a = 60 * 2
b = 60 / 36
c = b + 1
d = a / c
|
a ) 4 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | a | divide(subtract(add(multiply(7, 7), 5), 2), 13) | if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 2 is subtracted and divided by 13 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 2 ) / 13 = ( 54 - 2 ) / 13 = 4 answer : option a" | a = 7 * 7
b = a + 5
c = b - 2
d = c / 13
|
a ) 4 , b ) 5 , c ) 8 , d ) 10.72 , e ) 15 | d | sqrt(subtract(power(14, const_2), power(subtract(sqrt(subtract(power(14, const_2), power(5, const_2))), 4), const_2))) | a ladder 14 feet long is leaning against a wall that is perpendicular to level ground . the bottom of the ladder is 5 feet from the base of the wall . if the top of the ladder slips down 4 feet , how many feet will the bottom of the ladder slip ? | "14 ^ 2 - 5 ^ 2 = 171 it means that the height is equal to 13.07 ~ = 13 . since the top of the ladder slips down 4 feet , then the height of the wall = 13 - 4 = 9 the bottom = sqrt ( 14 ^ 2 - 9 ^ 2 ) = sqrt ( 196 - 81 ) = 10.72 ans is d" | a = 14 ** 2
b = 14 ** 2
c = 5 ** 2
d = b - c
e = math.sqrt(d)
f = e - 4
g = f ** 2
h = a - g
i = math.sqrt(h)
|
a ) rs . 9009 , b ) rs . 9008 , c ) rs . 9002 , d ) rs . 9202 , e ) rs . 9001 | a | multiply(multiply(multiply(add(multiply(multiply(multiply(2, const_3), const_100), const_100), multiply(multiply(multiply(const_3, const_3), const_100), multiply(add(5, 2), 2))), divide(add(multiply(15, 5), 2), 5)), divide(multiply(5, 5), multiply(2, multiply(2, 5)))), divide(const_1, const_100)) | find the simple interest on rs . 78000 at 15 ( 2 / 5 ) % per annum for 9 months . | "explanation : p = rs . 78000 , r = 77 / 5 % p . a and t = 9 / 12 years = ¾ years therefore , s . i = ( p * r * t ) / 100 = ( 78000 * 77 / 5 * ¾ * 1 / 100 ) = rs . 9009 answer : a" | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 5 + 2
g = f * 2
h = e * g
i = c + h
j = 15 * 5
k = j + 2
l = k / 5
m = i * l
n = 5 * 5
o = 2 * 5
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) 1 / 12 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 4 / 9 | e | subtract(add(divide(4, 9), divide(2, 3)), divide(6, 12)) | the instructions state that cheryl needs 4 / 9 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 6 / 12 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 4 / 9 + 2 / 3 left part 6 / 12 - - - > 2 / 3 so used part 4 / 9 + 2 / 3 - 2 / 3 = 4 / 9 ans e" | a = 4 / 9
b = 2 / 3
c = a + b
d = 6 / 12
e = c - d
|
a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | c | divide(32, 51) | how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 32 / 51 ? | "there is no certain way to solve 2 unknown with 1 equation . the best way is to look at the question and retrospect the most efficient way . in this question , a and b are only positive integers . so that is a big relief . now , we can start with putting a = 1,2 , . . and so on till the time we are confident about one of the options . so , we start with a = 1 , we get b as - ve . out a = 2 , we get b as 6 . yes ( now ( a , b ) = ( 2,6 ) . we can directly see that ( a , b ) = ( 6,2 ) will also satisfy . so we have 2 possible solutions ) a = 3 , we get b as 3 . yes ( now we have 3 possible solutions ) a = 4 , we get b as fraction . out a = 5 , we get b again as some fraction . out a = 6 already taken . we have a , b options left . c , d , e are out . a is 6 . to have 6 as the answer , we will need one more pair like 2,6 and one more solution where a = b . when a = b , we have only 1 solution = 2 . so , one more solution , where a = b is not possible . so , answer will be c ." | a = 32 / 51
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a ) 360 , b ) 227 , c ) 268 , d ) 198 , e ) 176 | a | multiply(divide(870, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 870 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should b pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 12 / 29 * 870 = 360 answer : a" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 870 / e
g = 16 * 9
h = f * g
|
a ) 3 cm , b ) 4 cm , c ) 6 cm , d ) 8 cm , e ) none | a | sqrt(divide(multiply(multiply(const_pi, multiply(6, divide(6, const_2))), const_2), multiply(const_pi, const_4))) | the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 6 cm each . the radius of the sphere is | "solution 4 î r 2 = 2 î 3 x 6 â ‡ ’ r 2 = ( 3 x 6 / 2 ) â ‡ ’ 9 â ‡ ’ r = 3 cm . answer a" | a = 6 / 2
b = 6 * a
c = math.pi * b
d = c * 2
e = math.pi * 4
f = d / e
g = math.sqrt(f)
|
a ) 19 , b ) 18 , c ) 16 , d ) 17 , e ) 14 | c | power(divide(96, divide(subtract(96, 24), sqrt(9))), const_2) | the speed of a railway engine is 96 km per hour when no compartment is attached , and the reduction in speed is directly proportional to the square root of the number of compartments attached . if the speed of the train carried by this engine is 24 km per hour when 9 compartments are attached , the maximum number of compartments that can be carried by the engine is : | "the reduction in speed is directly proportional to the square root of the number of compartments attached doesreductionmean amount subtracted ? or percentage decrease ? there are at least two interpretations , and the wording does not provide a clear interpretation between them . evidently what the question intends is the subtraction interpretation . what is subtracted from the speed is directly proportional to the square root of the number of compartments attached . in other words , if s = speed , and n = number of compartments , then s = 42 - k * sqrt ( n ) wherekis a constant of the proportionality . in general , if a is directly proportional to b , we can write a = k * b and solve for k . if n = 9 , then s = 24 24 = 96 - k * sqrt ( 9 ) = 96 - 3 k k = 24 now , we need to know : what value of n makes s go to zero ? 0 = 96 - 24 * sqrt ( n ) 24 * sqrt ( n ) = 96 sqrt ( n ) = 4 n = 4 ^ 2 = 16 thus , 16 is the maximum number of cars the engine can pull and still move . c" | a = 96 - 24
b = math.sqrt(9)
c = a / b
d = 96 / c
e = d ** 2
|
a ) 800 , b ) 1400 , c ) 1000 , d ) 1100 , e ) 1200 | b | divide(multiply(subtract(870, 800), 4), divide(20, const_100)) | average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 870 . what was original rent of friend whose rent is increased ? | "0.2 x = 4 ( 870 - 800 ) 0.2 x = 280 x = 1400 answer b" | a = 870 - 800
b = a * 4
c = 20 / 100
d = b / c
|
a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 5 / 3 , e ) 40 / 3 | e | divide(log(16), log(power(2, 0.3))) | if n = 2 ^ 0.3 and n ^ b = 16 , b must equal | "30 / 100 = 3 / 10 n = 2 ^ 3 / 10 n ^ b = 2 ^ 4 ( 2 ^ 3 / 10 ) ^ b = 2 ^ 4 b = 40 / 3 answer : e" | a = math.log(16)
b = 2 ** 0
c = math.log(b)
d = a / c
|
a ) 25 , b ) 35 , c ) 23 , d ) 22 , e ) 33 | e | divide(subtract(multiply(95, 3), multiply(3, 40)), subtract(95, 90)) | the average mark of the students of a class in a particular exam is 90 . if 3 students whose average mark in that exam is 40 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 90 x . total marks of ( x - 3 ) students = 80 ( x - 3 ) 90 x - ( 3 * 40 ) = 95 ( x - 3 ) 165 = 5 x = > x = 33 answer : e" | a = 95 * 3
b = 3 * 40
c = a - b
d = 95 - 90
e = c / d
|
a ) 16.5 % , b ) 14 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(6.5, 3)), const_100) | a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 6.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.065 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 65 f - - > s = 6 f - - > f / s = 1 / 6 - - > f / ( s + f ) = 1 / ( 1 + 6 ) = 1 / 7 = 0.14 . answer : b . | a = 3 - 5
b = 6 - 5
c = a / b
d = c * 100
|
a ) 8 , b ) 21 , c ) 24 , d ) 27 , e ) 30 | b | multiply(divide(const_1, add(add(add(divide(const_1, 7), multiply(3, divide(const_1, 7))), add(divide(const_1, 7), multiply(3, divide(const_1, 7)))), multiply(3, add(divide(const_1, 7), multiply(3, divide(const_1, 7)))))), const_60) | machine a and machine b process the same work at different rates . machine c processes work as fast as machines a and b combined . machine d processes work 3 times as fast as machine c ; machine d ’ s work rate is also exactly 4 times machine b ’ s rate . assume all 4 machines work at fixed unchanging rates . if machine a works alone on a job , it takes 7 hours . if all 4 machines work together on the same job simultaneously , how many minutes will it take all of them to complete it ? | c = a + b d = 3 c = 3 ( a + b ) = 4 b then b = 3 a and c = 4 a the combined rate of the four machines is a + 3 a + 4 a + 12 a = 20 a machine a can complete the work in 420 minutes , so its rate is 1 / 420 of the work per minute . the combined rate is 20 / 420 = 1 / 21 so the work will be completed in 21 minutes . the answer is b . | a = 1 / 7
b = 1 / 7
c = 3 * b
d = a + c
e = 1 / 7
f = 1 / 7
g = 3 * f
h = e + g
i = d + h
j = 1 / 7
k = 1 / 7
l = 3 * k
m = j + l
n = 3 * m
o = i + n
p = 1 / o
q = p * const_60
|
a ) 10 , b ) 12 , c ) 18 , d ) 20 , e ) 15 | e | divide(divide(6000, 2), 200) | in a company the manager wants to give some gifts to all of the workers . in each block there are about 200 workers are there . the total amount for giving the gifts for all the workers is $ 6000 . the worth of the gift is 2 $ . how many blocks are there in the company ? | "each employee will get a gift worth of = $ 2 total employees = 6000 / 2 = 3000 total blocks = 3000 / 200 = 15 correct option is e" | a = 6000 / 2
b = a / 200
|
a ) 500 , b ) 289 , c ) 350 , d ) 882 , e ) 281 | a | subtract(multiply(divide(300, 18), 48), 300) | a 300 meter long train crosses a platform in 48 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 48 = 50 / 3 3 ( x + 300 ) = 2400 è x = 500 m . answer : a" | a = 300 / 18
b = a * 48
c = b - 300
|
a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | c | multiply(divide(const_100, add(const_100, 12)), 45) | from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 12 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 : - 12 x = 45 - - > 28 / 25 * x = 45 - - > x = 45 * 25 / 28 = 1125 / 28 = ~ 40 answer : c ." | a = 100 + 12
b = 100 / a
c = b * 45
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a ) 100 , b ) 150 , c ) 140 , d ) 200 , e ) 250 | d | divide(divide(multiply(add(100, 200), add(divide(subtract(200, 100), 100), const_1)), const_2), add(divide(subtract(200, 100), 100), const_1)) | what is the average ( arithmetic mean ) of the numbers 100 , 150 , 200 , 200 , 250 , and 300 ? | "{ 100 , 150 , 200 , 200 , 250 , 300 } = { 200 - 100,200 - 50 , 200 , 200,200 + 50,200 + 100 } - - > the average = 200 . answer : d ." | a = 100 + 200
b = 200 - 100
c = b / 100
d = c + 1
e = a * d
f = e / 2
g = 200 - 100
h = g / 100
i = h + 1
j = f / i
|
a ) 1000 , b ) 1376 , c ) 1456 , d ) 1900 , e ) 1566 | b | add(multiply(14, 54), multiply(10, 62)) | andrew purchased 14 kg of grapes at the rate of 54 per kg and 10 kg of mangoes at the rate of 62 per kg . how much amount did he pay to the shopkeeper ? | "cost of 14 kg grapes = 54 × 14 = 756 . cost of 10 kg of mangoes = 62 x 10 = 620 . total cost he has to pay = 756 + 620 = 1376 b" | a = 14 * 54
b = 10 * 62
c = a + b
|
a ) 100 m , b ) 75 m , c ) 120 m , d ) 50 m , e ) 70 m | b | multiply(multiply(subtract(50, 32), const_0_2778), 15) | two trains move in the same direction at speeds 50 kmph and 32 kmph respectively . a man in the slower train observes that 15 seconds elapse before the faster train completely passes by him . what is the length of faster train ? | since both trains move in same direction so : average speed = 50 - 32 = 18 kmph = 5 mps speeed = length of train / time length of train = 5 * 15 = 75 m answer : b | a = 50 - 32
b = a * const_0_2778
c = b * 15
|
a ) 1 / 4 , b ) 1 / 5 , c ) 3 / 10 , d ) 1 / 11 , e ) none of above | c | divide(circle_area(divide(30, const_2)), const_2) | what will be the fraction of 30 % | "explanation : it will 30 * 1 / 100 = 3 / 10 option c" | a = 30 / 2
b = circle_area / (
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a ) 63 , b ) 64 , c ) 65 , d ) 76 , e ) 67 | c | multiply(39, 3) | jacob is 39 years old . he is 3 times as old as his brother . how old will jacob be when he is twice as old ? | "j = 39 ; j = 3 b ; b = 39 / 3 = 13 ; twice as old so b = 13 ( now ) + ( 13 ) = 26 ; jacob is 39 + 26 = 65 answer : c" | a = 39 * 3
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a ) 21.4 sec , b ) 77 sec , c ) 25 sec , d ) 18 sec , e ) 17 sec | a | divide(add(150, 100), multiply(42, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 42 kmph ? | "d = 100 + 150 = 250 s = 42 * 5 / 18 = 11.7 mps t = 250 / 11.7 = 21.4 sec answer : a" | a = 150 + 100
b = 42 * const_0_2778
c = a / b
|
a ) 10 sec , b ) 30 sec , c ) 40 sec , d ) 20 s , e ) 50 sec | d | divide(add(280, 120), multiply(add(42, 30), const_0_2778)) | two cars of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | d relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 s . | a = 280 + 120
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 10 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | a | subtract(add(6, 5), const_1) | a student is ranked 6 th from right and 5 th from left . how many students are there in totality ? | "from right 6 , from left 5 total = 6 + 5 - 1 = 10 answer : a" | a = 6 + 5
b = a - 1
|
a ) 120 , b ) 360 , c ) 240 , d ) 182 , e ) 1000 | c | multiply(60, multiply(divide(12, 6), divide(20, 10))) | if 6 men can reap 60 acres of land in 10 days , how many acres of land can 12 men reap in 20 days ? | "6 men 60 acres 10 days 12 men ? 20 days 60 * 12 / 6 * 20 / 10 60 * 2 * 2 60 * 4 = 240 answer : c" | a = 12 / 6
b = 20 / 10
c = a * b
d = 60 * c
|
a ) 98 , b ) 165 , c ) 180 , d ) 253 , e ) none of these | b | multiply(15, divide(165, 15)) | solve for x and check : 15 x = 165 | solution : dividing each side by 15 , we obtain ( 15 x / 15 ) = ( 165 / 15 ) therefore : x = 11 check : 15 x = 165 ( 15 * 11 ) = 165 165 = 165 answer : b | a = 165 / 15
b = 15 * a
|
a ) 5 , b ) 8 , c ) 4 , d ) 7 , e ) 6 | e | inverse(add(divide(const_1, 10), divide(const_1, 15))) | if ram and gohul can do a job in 10 days and 15 days independently . how many days would they take to complete the same job working simultaneously ? | "if total work is x . ram rate of working = x / 10 per day . gohul rate of working = x / 15 per day . rate of work = ( x / 10 ) + ( x / 15 ) = 30 x / 5 x = 6 days answer is option e" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1/(c)
|
a ) 64.19 , b ) 64.12 , c ) 40.5 , d ) 64.1 , e ) 64.11 | c | subtract(subtract(multiply(1000, power(add(divide(8, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(8, const_100)), 4)) | what will be the difference between simple and compound interest at 8 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 8 * 4 ) / 100 = rs . 320 c . i . = [ 1000 * ( 1 + 8 / 100 ) 4 - 1000 ] = rs . 360.5 difference = ( 360.5 - 320 ) = rs . 40.5 answer : c" | a = 8 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 8 / 100
g = 1000 * f
h = g * 4
i = e - h
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 67.5 | e | multiply(divide(divide(add(divide(120, const_2), 120), 40), const_4), divide(120, const_2)) | a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 40 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | "a - b = 120 miles b - c = 60 miles avg speed = 40 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 40 = 180 / 4 t t = 67.5 b - c = 67.5 mph answer e" | a = 120 / 2
b = a + 120
c = b / 40
d = c / 4
e = 120 / 2
f = d * e
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(divide(add(add(add(add(add(add(add(add(add(add(50, 50), 97), 97), 97), 120), 125), 155), 199), 199), 239), add(const_10, const_1)), divide(add(add(add(add(add(add(add(add(add(add(50, 50), 97), 97), 97), 120), 125), 155), 199), 199), 239), add(const_10, const_1))) | company x sells a selection of products at various price points . listed below are unit sales made for one particular day . how many unit sales on that day were greater than the mean sale price but less than the median sale price ? $ 50 , $ 50 , $ 97 , $ 97 , $ 97 , $ 120 , $ 125 , $ 155 , $ 199 , $ 199 , $ 239 | taking the prices of products in ascending order ( already arranged ) $ 50 , $ 50 , $ 97 , $ 97 , $ 97 , $ 120 , $ 125 , $ 155 , $ 199 , $ 199 , $ 239 we see that median value is the 6 th value as there in total 11 values given arithmetic mean = total / number of entries = 1428 / 11 = 129.8181 we are asked to find how many unit sales on that day were greater than the mean sale price but less than the median sale price as we can clearly see that there is one value between $ 120 and $ 129.81 , answer is 1 unit correct answer - b | a = 50 + 50
b = a + 97
c = b + 97
d = c + 97
e = d + 120
f = e + 125
g = f + 155
h = g + 199
i = h + 199
j = i + 239
k = 10 + 1
l = j / k
m = 50 + 50
n = m + 97
o = n + 97
p = o + 97
q = p + 120
r = q + 125
s = r + 155
t = s + 199
u = t + 199
v = u + 239
w = 10 + 1
x = v / w
y = l / x
|
a ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | c | divide(subtract(multiply(15, subtract(40, 4)), multiply(15, 32)), 4) | the average age of an adult class is 40 years . 15 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 15 x 32 = ( y + 15 ) x 36 â ‡ ’ 40 y + 480 = 36 y + 540 â ‡ ’ 4 y = 60 â ˆ ´ y = 15 c" | a = 40 - 4
b = 15 * a
c = 15 * 32
d = b - c
e = d / 4
|
a ) 100 / 6 , b ) 289 / 4 , c ) 128 / 7 , d ) 456 / 6 , e ) 1000 / 301 | e | divide(const_1, 0.301) | if log 10 2 = 0.3010 , then log 2 10 is equal to : | explanation : log 2 10 = 1 / log 102 = 1 / 0.3010 = 10000 / 3010 = 1000 / 301 answer e | a = 1 / 0
|
a ) 64 and 15,625 , b ) 32 and 3,125 , c ) 64 and 15,620 , d ) 64 and 15,635 , e ) 64 and 16,625 | b | add(power(divide(log(multiply(const_100, const_1000)), log(const_10)), subtract(divide(log(multiply(const_100, const_1000)), log(const_10)), const_1)), 0) | find two integers , neither of which ends in a zero , and whose product is exactly 00000 | 1 , 00,000 = 10 ^ 5 = 10 x 10 x 10 x 10 x 10 = ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) = ( 2 ^ 5 ) x ( 5 ^ 5 ) = 32 x 3125 so the numbers are 32 and 3,125 answer : b | a = 100 * 1000
b = math.log(a)
c = math.log(10)
d = b / c
e = 100 * 1000
f = math.log(e)
g = math.log(10)
h = f / g
i = h - 1
j = d ** i
k = j + 0
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | e | add(6, 4) | set a consists of the integers from 4 to 15 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | a = { 4,5 , 6,7 , 8,9 , 10,11 , 12,13 , 14,15 } b = { 6 , 7,8 , 9,10 , 11,12 . . . 20 } thus we see that there are 10 distinct integers that are common to both . e is the correct answer . | a = 6 + 4
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a ) 1 / 64 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 3 | a | multiply(divide(8, power(8, const_2)), divide(8, power(8, const_2))) | what is the probability that the sum of two dice will yield a 9 , and then when both are thrown again , their sum will again yield a 9 ? assume that each die has 8 sides with faces numbered 1 to 8 . | "solution - rolling dices is an independent event . the combinations to get 9 are ( 1,8 ) , ( 8,1 ) , ( 2,7 ) , ( 7,2 ) , ( 3,6 ) , ( 6,3 ) , ( 4,5 ) , ( 5,4 ) , and total combinations of both dices is 64 . the probability of getting 9 in first attempt is 8 / 64 = 1 / 8 . probability of getting 9 again in second attempt = ( 1 / 8 ) * ( 1 / 8 ) = 1 / 64 . ans a" | a = 8 ** 2
b = 8 / a
c = 8 ** 2
d = 8 / c
e = b * d
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a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 % | e | subtract(100, 20) | john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 20 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it ’ s 100 – 20 = 80 . the “ original ” is our starting point ; in this case , it ’ s 100 . ( 80 / 100 ) * 100 = ( 0.8 ) * 100 = 80 % . e" | a = 100 - 20
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a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | d | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent q of the list price is the lowest possible sale price ? | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which is q = 40 % of 80 hence , d is the answer" | a = 80 / 2
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(add(1, multiply(17, 4)), multiply(17, 4)) | how many two - digit numbers yield a remainder of 1 when divided by both 4 and 17 ? | easier to start with numbers that are of the form 17 p + 1 - - - > 18 , 35,52 , 69,86 . out of these , there is only one ( 69 ) is also of the form 4 q + 1 . thus 1 is the answer . b is the correct answer . | a = 17 * 4
b = 1 + a
c = 17 * 4
d = b - c
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a ) 33 , b ) 11 , c ) 96 , d ) 36 , e ) 91 | c | divide(multiply(120, 20), const_100) | the cost of an article is decreased by 20 % . if the original cost is $ 120 , find the decrease cost . | "original cost = $ 120 decrease in it = 20 % of $ 120 = 20 / 100 ã — 120 = 2400 / 100 = $ 24 therefore , decrease cost = $ 120 - $ 24 = $ 96 answer : c" | a = 120 * 20
b = a / 100
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a ) 1 , b ) undentified , c ) 3 , d ) 7 , e ) 5 | b | divide(log(multiply(3, 3)), log(const_10)) | 3 log 3 ( - 5 ) = ? | "since - 5 is not in the domain of function log 3 ( x ) , 3 log 3 ( - 5 ) is undefined correct answer b" | a = 3 * 3
b = math.log(a)
c = math.log(10)
d = b / c
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['a ) 13', 'b ) 12', 'c ) 4', 'd ) 16', 'e ) 18'] | a | divide(volume_cylinder(6, 12), volume_cylinder(const_2, 8)) | a cylinder with 6 meter radius and 12 meter height is filled to capacity with water . if the content of the cylinder is used to fill several smaller cylinders of 4 meter diameter and 8 meter height , how many smaller cylinders will be filled to capacity ? | calculate the volume of the larger cylinder and divide it by the volume of the smaller cylinder . volume of cylinder = π r 2 h larger cylinder volume = 1357.17 smaller cylinder volume = 100.53 therefore the number of cylinders b that can be filled to capacity = 1357.17 / 100.53 = 13.5 answer is a only 13 smaller cylinders can be filled to capacity . | a = volume_cylinder / (
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a ) 20940 , b ) 21009 , c ) 23000 , d ) 23450 , e ) 30000 | a | add(multiply(multiply(add(divide(3, const_100), divide(divide(subtract(19500, 12000), 4), 12000)), 12000), 4), 12000) | sonika deposited rs . 12000 which amounted to rs . 19500 after 4 years at simple interest . had the interest been 3 % more . she would get how much ? | "( 12000 * 4 * 3 ) / 100 = 1440 19500 - - - - - - - - 20940 answer : a" | a = 3 / 100
b = 19500 - 12000
c = b / 4
d = c / 12000
e = a + d
f = e * 12000
g = f * 4
h = g + 12000
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 17 | a | floor(divide(divide(subtract(add(multiply(16, const_100), 6), add(add(multiply(8, const_100), 8), 88)), const_10), add(const_10, const_1))) | 8 k 8 + k 88 - - - - - - - - 16 y 6 if k and y represent non - zero digits within the integers above , what is y ? | 8 k 8 k 88 - - - - - - - - 16 y 6 trial and error or just plug - in method might be the shortest way to solve this problem . though you can narrow down the possible values of k to just two : 7 and 8 - - > 8 * * + 7 * * = 16 * * or 8 * * + 8 * * = 16 * * ( k can not be less than 7 or 9 , as the result wo n ' t be 16 * * ) . after that it ' s easy to get that k = 7 and y = 6 . answer : a . | a = 16 * 100
b = a + 6
c = 8 * 100
d = c + 8
e = d + 88
f = b - e
g = f / 10
h = 10 + 1
i = g / h
j = math.floor(i)
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a ) 40 , b ) 50 , c ) 60 , d ) 200 , e ) 300 | e | multiply(divide(900, multiply(add(const_2, const_1), const_2)), const_2) | a rectangular garden is to be twice as long as it is wide . if 900 yards of fencing , including the gate , will completely enclose the garden , what will be the length of the garden , in yards ? | "alternate approach backsolving ( using answer options to reach the correct answer ) can work wonders here if one is fast in calculations . given perimeter is 900 so , 2 ( l + b ) = 900 or , l + b = 450 now use the answer options ( given length ; breath will be half the length ) ( a ) 40 l = 40 ; b = 20 l + b = 60 ( b ) 50 l = 50 ; b = 25 l + b = 75 ( c ) 60 l = 60 ; b = 30 l + b = 90 ( d ) 200 l = 200 ; b = 100 l + b = 300 ( e ) 300 l = 300 ; b = 150 l + b = 450 thus you see no , need of any calculations , u can reach the correct option only by checking options ; correct answer will be ( e )" | a = 2 + 1
b = a * 2
c = 900 / b
d = c * 2
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a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | floor(divide(log(divide(2, 2.134)), log(10))) | if x is an integer and 2.134 × 10 ^ x is less than 2 , 100,000 , what is the greatest possible value for x ? | "if x = 6 2.134 × 10 ^ 6 = 2 , 134,000 > 2 , 100,000 so , x = 5 answer : c" | a = 2 / 2
b = math.log(a)
c = math.log(10)
d = b / c
e = math.floor(d)
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a ) 5 , b ) 5 1 / 2 , c ) 4 4 / 5 , d ) 6 , e ) 9 1 / 2 | c | divide(multiply(const_4.0, 3), subtract(8, 3)) | a man can do a piece of work in 8 days , but with the help of his son , he can finish it in 3 days . in what time can the son do it alone ? | "son ' s 1 day work = 1 / 3 - 1 / 8 = 5 / 24 son alone can do the work in 24 / 5 days = 4 4 / 5 days answer is c" | a = 4 * 0
b = 8 - 3
c = a / b
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a ) 10 and 12 , b ) 9 and 11 , c ) 20 and 22 , d ) 8 and 10 , e ) 19 and 21 | c | add(add(multiply(10, 2), 2), multiply(multiply(10, 10), multiply(10, 2))) | caleb and kyle built completed the construction of a shed in 10 and half days . if they were to work separately , how long will it take each for each of them to build the shed , if it will take caleb 2 day earlier than kyle ? | work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here , we ' re told that ( working together ) the two workers would complete a job in 12 days . this means that ( individually ) each of them would take more than 10 days to do the job . answers e , a and c are illogical , since the individual times must both be greater than 10 days . so we can test the values for answers b and d . using the values for answers b and d . . . answer b : ( 20 ) ( 22 ) / ( 20 + 22 ) = 440 / 42 = 10.5 this is a match final answer : c | a = 10 * 2
b = a + 2
c = 10 * 10
d = 10 * 2
e = c * d
f = b + e
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a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | b | divide(divide(multiply(multiply(36, 12), 8), 72), 8) | in a manufacturing plant , it takes 36 machines 8 hours of continuous work to fill 8 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ? | "the choices give away the answer . . 36 machines take 4 hours to fill 8 standard orders . . in next eq we aredoubling the machines from 36 to 72 , but thework is not doubling ( only 1 1 / 2 times ) , = 8 * 36 / 72 * 12 / 8 = 6 ans b" | a = 36 * 12
b = a * 8
c = b / 72
d = c / 8
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a ) 6 , b ) 12 , c ) 9 , d ) 5 , e ) 2 | e | add(const_3, const_4) | what is the smallest positive integer x such that 864 x is the cube of a positive integer | "given 864 x is a perfect cube so we will take 1728 = 12 * 12 * 12 864 x = 1728 x = 1728 / 864 = 2 correct option is e" | a = 3 + 4
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a ) 70 , b ) 120 , c ) 100 , d ) 90 , e ) 110 | b | multiply(multiply(10, 4), 3) | at the beginning of the year , the ratio of boys to girls in high school x was 3 to 4 . during the year , 10 boys and twice as many girls transferred to another high school , while no new students joined high school x . if , at the end of the year , the ratio of boys to girls was 4 to 5 , how many boys were there in high school x at the beginning of the year ? | let the total number of boys and girls at the beginning of the year be 3 x and 4 x respectively . now 10 boys and 20 girls are transferred to another school . thus no . of boys and girls students left in the school x are 3 x - 10 and 4 x - 20 respectively . the ratio of these boys and girls students = 4 / 5 thus we have ( 3 x - 10 ) / ( 4 x - 20 ) = 4 / 5 15 x - 50 = 16 x - 80 x = 30 thus total no . of boys at the beginning of the year = 4 ( 30 ) = 120 answer is option b | a = 10 * 4
b = a * 3
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a ) 972 , b ) 990 , c ) 1098 , d ) 1305 , e ) 1405 | a | subtract(subtract(multiply(360, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | the total number of digits used in numbering the pages of a book having 360 pages is | "total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 261 ) = ( 9 + 180 + 783 ) = 972 . answer : a" | a = 360 * 3
b = 100 - 1
c = a - b
d = 10 - 1
e = c - d
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a ) 260 , b ) 258 , c ) 252 , d ) 250 , e ) 244 | c | subtract(negate(234), multiply(subtract(224, 228), divide(subtract(224, 228), subtract(222, 224)))) | 222 , 224 , 228 , 234 , 242 , ( . . . . ) | "explanation : the pattern is 2 , 4 , 6 , 8 , 10 , etc . hence 10 = 252 answer : c" | a = negate - (
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a ) 8 , b ) 12 , c ) 18 , d ) 24 , e ) 48 | b | divide(multiply(add(add(6, 4), 2), divide(2000, 2)), const_1000) | a 2000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ? | "in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 1,000 by that net inflow to get the minutes ) - 12 min . answer b ." | a = 6 + 4
b = a + 2
c = 2000 / 2
d = b * c
e = d / 1000
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a ) 900 , b ) 980 , c ) 1600 , d ) 1240 , e ) 1400 | c | subtract(divide(4800, 2), divide(4800, 6)) | share rs . 4800 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ? | amount received by sanjay . 4 / 12 x 4800 = 1600 = ( related ratio / sum of ratio ) x total amount so , the amount received by sanjay is 1600 . c | a = 4800 / 2
b = 4800 / 6
c = a - b
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a ) 5.7 , b ) 6.0 , c ) 4.0 , d ) 9.7 , e ) 18.0 | c | divide(add(divide(divide(25, const_3), const_3), divide(multiply(divide(25, const_3), const_2), const_3)), const_2) | the total circumference of two circles is 25 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 π r + 2 π r = 25 where 2 r = r thus : 2 π r + 4 π r = 25 6 π r = 25 r = approx 1.33 π r + 2 r π = 25 3 π r = 25 r = approx 2.65 r + r = approx 3.98 = 4.0 answer : c" | a = 25 / 3
b = a / 3
c = 25 / 3
d = c * 2
e = d / 3
f = b + e
g = f / 2
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a ) 757 , b ) 758 , c ) 718 , d ) 1200 , e ) 738 | d | divide(divide(multiply(144, 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 144 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 144 * 5 / 18 ] m / sec = 40 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 = 40 è x = 40 * 60 / 2 = 1200 answer : d" | a = 144 * 1000
b = const_60 / 1
c = a / b
d = c / 2
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a ) 347.4 , b ) 987.4 , c ) 877.4 , d ) 637.4 , e ) 667.4 | c | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(20.50, 6), 2))), multiply(3, divide(multiply(20.50, 6), 2))), multiply(5, 20.50)) | the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 20.50 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | "c $ 877.40 let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 20.50 = 2 r a = 12 / 5 r and r = 61.5 a = 147.6 required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5 = 590.4 + 184.5 + 102.5 = $ 877.40" | a = 24 / 10
b = 20 * 50
c = b / 2
d = a * c
e = 4 * d
f = 20 * 50
g = f / 2
h = 3 * g
i = e + h
j = 5 * 20
k = i + j
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a ) 23 , b ) 18 , c ) 21 , d ) 24 , e ) 25 | b | divide(const_1, add(divide(const_1, 30), divide(divide(const_1, 30), 1.5))) | a is 1.5 times as fast as b . a alone can do the work in 30 days . if a and b working together in how many days will the work be completed ? | "a can finish 1 work in 30 days b can finish 1 / 1.5 work in 30 days - since a is 1.5 faster than b this means b can finish 1 work in 30 * 1.5 days = 45 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 45 * 30 / ( 45 + 30 ) = 20 * 30 / 50 = 18 days so answer is b" | a = 1 / 30
b = 1 / 30
c = b / 1
d = a + c
e = 1 / d
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a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 95 , d ) 1 , e ) 2 | c | divide(1, 95) | if the numbers 1 to 95 are written on 95 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | "there are 25 primes , 69 composite numbers from 1 to 95 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 95 . answer : c" | a = 1 / 95
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a ) 8 hrs , b ) 20 hrs , c ) 12 hrs , d ) 15 hrs , e ) 6 hrs | b | multiply(divide(1, 6), 24) | a train running at 1 / 6 of its own speed reached a place in 24 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 6 * 24 = 4 hrs time saved = 24 - 4 = 20 hrs answer : b" | a = 1 / 6
b = a * 24
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a ) s : 1000 , b ) s : 1067 , c ) s : 1200 , d ) s : 1028 , e ) s : 1027 | c | divide(multiply(168, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 168 more , there would have been a gain of 4 % . what is the cost price ? | "90 % 104 % - - - - - - - - 14 % - - - - 168 100 % - - - - ? = > rs : 1200 answer : c" | a = 168 * 100
b = 100 + 4
c = 100 - 10
d = b - c
e = a / d
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(const_1, subtract(divide(const_1, 2), divide(const_1, 3))) | a pump can fill a tank with a water in 2 hours . because of a leak , it took 3 hours to fill the tank . the leak can drain all the water of the full tank in how many hours ? | the rate of the pump + leak = 1 / 3 1 / 2 - leak ' s rate = 1 / 3 leak ' s rate = 1 / 2 - 1 / 3 = 1 / 6 the leak will empty the tank in 6 hours . the answer is e . | a = 1 / 2
b = 1 / 3
c = a - b
d = 1 / c
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['a ) becomes 8 times', 'b ) becomes 9 times', 'c ) is double', 'd ) becomes 6 times', 'e ) none'] | a | power(const_2, const_3) | if each edge of a cube is doubled , then its volume : | sol . let original edge = a . then , volume = a ³ new edge = 2 a . so , new volume = ( 2 a ) ³ = 8 a ³ ∴ volume becomes 8 times answer a | a = 2 ** 3
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a ) s . 28028 , b ) s . 28000 , c ) s . 28003 , d ) s . 28029 , e ) s . 24029 | b | subtract(52000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 52000 , then the share of a is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 52000 share of b = 7 / 13 ( 52000 ) = rs . 28000 answer : b" | a = const_60 * 100
b = 52000 - a
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a ) 6.9 , b ) 7.2 , c ) 7.5 , d ) 7.8 , e ) 8.1 | c | divide(add(18, 12), subtract(7.2, add(divide(18, 9), divide(12, 10)))) | on a trip , a cyclist averaged 9 miles per hour for the first 18 miles and 10 miles per hour for the remaining 12 miles . if the cyclist returned immediately via the same route and took a total of 7.2 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 30 miles was 18 / 9 + 12 / 10 = 2 + 1.2 = 3.2 hours . the average speed for the return trip was 30 miles / 4 hours = 7.5 mph . the answer is c ." | a = 18 + 12
b = 18 / 9
c = 12 / 10
d = b + c
e = 7 - 2
f = a / e
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | c | multiply(divide(subtract(60, 42), 60), const_100) | in town p , 60 percent of the population are employed , and 42 percent of the population are employed males . what percent of the employed people in town p are females ? | "the percent of the population who are employed females is 60 - 42 = 18 % the percent of employed people who are female is 18 % / 60 % = 30 % . the answer is c ." | a = 60 - 42
b = a / 60
c = b * 100
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a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 2 / 5 | e | multiply(divide(divide(100, 5), 100), const_2.0) | an integer n between 1 and 100 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 5 ? | "n ( n + 1 ) to be divisible by 3 either n or n + 1 must be a multiples of 3 . in each following group of numbers : { 1 , 2 , 3 , 4 , 5 } , { 6 , 7 , 8 , 9 , 10 } , . . . , { 96 , 97 , 98 , 99,100 } there is exactly 1 numbers out of 5 satisfying the above condition . for example in { 1 , 2 , 3 , 4 , 5 } n can be : 4 or 5 . thus , the overall probability is 2 / 5 . answer : d ." | a = 100 / 5
b = a / 100
c = b * 2
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a ) 3 , b ) 4 , c ) 7 , d ) 32 , e ) 35 | c | subtract(38, 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 38 ? | "n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 38 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 7 more to make it 38 m + 38 answer - c" | a = 38 - reminder
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 25 | e | add(divide(18, const_2), subtract(34, 18)) | jane started baby - sitting when she was 18 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 34 years old , and she stopped baby - sitting 12 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | "check two extreme cases : jane = 18 , child = 9 , years ago = 34 - 18 = 16 - - > child ' s age now = 9 + 16 = 25 ; jane = 22 , child = 11 , years ago = 34 - 22 = 12 - - > child ' s age now = 11 + 12 = 23 . answer : e ." | a = 18 / 2
b = 34 - 18
c = a + b
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a ) 1 / 12 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 24 , e ) 5 / 9 | a | multiply(multiply(multiply(divide(add(5, const_1), add(9, const_1)), divide(subtract(add(5, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(5, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(5, const_1), const_1), const_1), const_1), subtract(subtract(subtract(add(9, const_1), const_1), const_1), const_1))) | each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 5 of the slips are the drawn , without replacement , what is the probability that all 5 have a odd number written on it ? | "key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 p ( + on 5 rd pick ) = 1 / 6 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 a" | a = 5 + 1
b = 9 + 1
c = a / b
d = 5 + 1
e = d - 1
f = 9 + 1
g = f - 1
h = e / g
i = c * h
j = 5 + 1
k = j - 1
l = k - 1
m = 9 + 1
n = m - 1
o = n - 1
p = l / o
q = i * p
r = 5 + 1
s = r - 1
t = s - 1
u = t - 1
v = 9 + 1
w = v - 1
x = w - 1
y = x - 1
z = u / y
A = q * z
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a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5 % , e ) 6.0 % | d | multiply(divide(add(multiply(50, divide(4, const_100)), multiply(30, divide(10, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2 tax on others = 3 percentage = 5 / 100 = 5 % answer : d | a = 4 / 100
b = 50 * a
c = 10 / 100
d = 30 * c
e = b + d
f = e / 100
g = f * 100
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a ) 9 , b ) 8 , c ) 7 , d ) 10 , e ) 11 | a | multiply(divide(252, add(add(16, 28), 40)), const_3) | jeff has 252 ounces of peanut butter in 16 , 28 . and 40 ounce jars . he has an equal number of each sized jar . how many jars of peanut butter does jeff have ? | let p equal the number of each sized jar then 16 p + 28 p + 40 p = 252 84 p = 252 p = 3 therefore , the total number of jars of peanut butter jeff has = 3 p = 9 answer : a | a = 16 + 28
b = a + 40
c = 252 / b
d = c * 3
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a ) 189.2 mtrs , b ) 378.4 mtrs , c ) 478.4 mtrs , d ) 488.4 mtrs , e ) 578.4 mtrs | a | multiply(66, divide(multiply(86, 8), multiply(20, 12))) | if 20 men can build a wall 66 metres long in 12 days , what length of a similar can be built by 86 men in 8 days ? | "if 20 men can build a wall 66 metres long in 12 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) / ( 12 * 20 ) = 189.2 mtrs answer : a" | a = 86 * 8
b = 20 * 12
c = a / b
d = 66 * c
|
['a ) 3.17 feet', 'b ) 3.2 feet', 'c ) 3.3 feet', 'd ) 3.4 feet', 'e ) 3.5 feet'] | a | divide(add(add(multiply(add(2, 10), 2), const_10), 4), add(2, 10)) | carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 4 inches tall . how tall are the sculpture andbase together ? | we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 4 = 38 38 / 12 = 3.17 feet answer : a | a = 2 + 10
b = a * 2
c = b + 10
d = c + 4
e = 2 + 10
f = d / e
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a ) 2.3 , b ) 2.6 , c ) 3.6 , d ) 4.5 , e ) 5 | e | divide(subtract(multiply(6, 2.5), add(multiply(2, 1.1), multiply(2, 1.4))), 2) | the average of 6 no . ' s is 2.5 . the average of 2 of them is 1.1 , while the average of the other 2 is 1.4 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 2.5 * 6 ) - [ ( 1.1 * 2 ) + ( 1.4 * 2 ) ] = 15 - ( 2.2 + 2.8 ) = 15 - 5 = 10 required average = ( 10 / 2 ) = 5 answer : e" | a = 6 * 2
b = 2 * 1
c = 2 * 1
d = b + c
e = a - d
f = e / 2
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a ) 5,050 , b ) 7,500 , c ) 10,500 , d ) 11,700 , e ) 19,600 | d | multiply(divide(add(200, 101), const_2), add(divide(subtract(200, 101), const_2), const_1)) | the sum of the first 92 positive even integers is 2,550 . what is the sum of the odd integers from 101 to 200 , inclusive ? | "101 + 103 + . . . . . . . 199 if we remove 100 from each of these it will be sum of 1 st 100 odd numbers . so 101 + 103 + . . . . . . . 199 = 92 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) sum of 1 st 100 natural numbers = ( 100 * 101 ) / 2 = 5050 sum of 1 st 92 positive even integers = 2550 sum of 1 st 100 odd numbers = 5050 - 2550 = 2500 so 101 + 103 + . . . . . . . 199 = 92 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) = 9200 + 2500 = 11700 d is the answer ." | a = 200 + 101
b = a / 2
c = 200 - 101
d = c / 2
e = d + 1
f = b * e
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a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 1 , e ) 5 / 4 | c | subtract(divide(subtract(21, const_1), add(19, const_1)), divide(add(9, const_1), subtract(31, const_1))) | a is an integer greater than 9 but less than 21 , b is an integer greater than 19 but less than 31 , what is the range of a / b ? | min value of a / b will be when b is highest and a is lowest - - - > a = 10 and b = 30 so , a / b = 1 / 3 max value of a / b will be when b is lowest and a is highest - - - > a = 20 and b = 20 so , a / b = 1 range is 1 - ( 1 / 3 ) = 2 / 3 . answer should be c | a = 21 - 1
b = 19 + 1
c = a / b
d = 9 + 1
e = 31 - 1
f = d / e
g = c - f
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a ) 95 , b ) 90 , c ) 85 , d ) 78 , e ) 72 | e | divide(18, divide(subtract(100, 75), 100)) | if 18 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | "let x be the number of students in the school . 0.18 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.18 x / 0.25 = 0.72 x students went to the camping which is 72 % . the answer is e" | a = 100 - 75
b = a / 100
c = 18 / b
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a ) 1 / 2 , b ) 3 / 4 , c ) 1 , d ) 2 , e ) 3 | a | subtract(1, multiply(divide(factorial(2), factorial(1)), power(divide(1, 1), 2))) | a couple decides to have 2 children . if they succeed in having 2 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 1 girl and 1 boy ? | "sample space = 2 ^ 2 = 4 . favourable events = { bg } , { gb } , probability = 2 / 4 = 1 / 2 . ans ( a ) ." | a = math.factorial(2)
b = math.factorial(1)
c = a / b
d = 1 / 1
e = d ** 2
f = c * e
g = 1 - f
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a ) $ 28 , b ) $ 82 , c ) $ 110 , d ) $ 138 , e ) $ 195 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(5, const_100)), multiply(add(80, 5), 150)), 150) | a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 5 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | by the question , the equation would be 150 p - 85 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 195 . answer e for me . | a = 4 * 4
b = a * 1000
c = 5 * 100
d = b + c
e = 80 + 5
f = e * 150
g = d + f
h = g / 150
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a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 50 | e | divide(subtract(75, 60), subtract(const_1, divide(70, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 70 percent of books that were loaned out are returned and there are 60 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "i did n ' t understand how did we get 100 ? total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 70 % = 30 % of books that were loaned out are not returned . now , there are 60 books , thus 76 - 60 = 16 books are not returned . { loaned out } * 0.30 = 7 - - > { loaned out } = 50 . answer : e ." | a = 75 - 60
b = 70 / 100
c = 1 - b
d = a / c
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a ) 59 , b ) 58 , c ) 60 , d ) 61 , e ) 12 | c | multiply(divide(100, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 100 , the runs scored by c are ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 15 / 25 * 100 = 60 answer : c" | a = 2 / 3
b = 5 / 2
c = a + b
d = c + 2
e = 100 / d
f = e * 5
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a ) 10 , b ) 12 , c ) 42 , d ) 20 , e ) 22 | c | divide(subtract(multiply(12, subtract(50, 4)), multiply(12, 32)), 4) | the average age of an adult class is 50 years . 12 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 years . find what was the original average age of the class ? | "let original strength = y then , 50 y + 12 x 32 = ( y + 12 ) x 46 â ‡ ’ 50 y + 384 = 46 y + 552 â ‡ ’ 4 y = 168 â ˆ ´ y = 42 c" | a = 50 - 4
b = 12 * a
c = 12 * 32
d = b - c
e = d / 4
|
a ) $ 300 , b ) $ 400 , c ) $ 450 , d ) $ 625 , e ) $ 750 | e | add(multiply(3, 150), multiply(divide(const_2, 3), multiply(3, 150))) | anne and katherine are both saving money from their summer jobs to buy bicycles . if anne had $ 150 less , she would have exactly 1 / 3 as much as katherine . and if katherine had twice as much , she would have exactly 3 times as much as anne . how much money have they saved together ? ( | if anne had $ 150 less , katherine would have three times more than anne . make this statement into an equation and simplify : 3 ( a – 150 ) = k 3 a – 450 = k and if katherine had twice as much , she would have three times more than anne : 2 k = 3 a substitute 3 a – 450 for k into the last equation and solve for a 2 ( 3 a – 450 ) = 3 a 6 a – 900 = 3 a – 900 = – 3 a 300 = a now substitute 300 for a into the same equation and solve for k : 2 k = 3 ( 300 ) 2 k = 900 k = 450 thus , together anne and katherine have 300 + 450 = 750 correct answer e ) $ 750 | a = 3 * 150
b = 2 / 3
c = 3 * 150
d = b * c
e = a + d
|
a ) 145 , b ) 180 , c ) 181 , d ) 184 , e ) 150 | b | subtract(multiply(180, divide(15, divide(15, const_3))), multiply(120, divide(18, divide(15, const_3)))) | a train crosses a platform of 120 m in 15 sec , same train crosses another platform of length 180 m in 18 sec . then find the length of the train ? | "length of the train be ‘ x ’ x + 120 / 15 = x + 180 / 18 6 x + 720 = 5 x + 900 x = 180 m answer : option b" | a = 15 / 3
b = 15 / a
c = 180 * b
d = 15 / 3
e = 18 / d
f = 120 * e
g = c - f
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a ) 5 / 32 , b ) 5 / 27 , c ) 1 / 27 , d ) 1 / 32 , e ) 27 / 32 | a | divide(subtract(32, 27), 32) | a ’ s speed is 32 / 27 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 32 meters , b covers 27 meters in that time . so if the race is 32 meters long , when a reaches the finish line , b would be 5 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 5 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 32 meters ) would be the same as the time required by b ( to cover 27 meters ) to reach the finish line . so b should get a head start of 5 / 32 th of the race . answer ( a )" | a = 32 - 27
b = a / 32
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a ) 17725 cm , b ) 15625 cm , c ) 12786 cm , d ) 12617 cm , e ) 12187 cm | b | divide(volume_cube(1), volume_cube(divide(4, const_100))) | how many cubes of 4 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 4 * 4 * 4 = 15625 note : 1 m = 100 cm answer : b" | a = volume_cube / (
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a ) 509 , b ) 624 , c ) 756 , d ) 832 , e ) 947 | a | subtract(375,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 375,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 375,513 the number of integers is 3756 - 3247 = 509 the answer is a ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 375 - 600
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['a ) 5 / ( √ 6 )', 'b ) 5 · √ ( 2 / 3 )', 'c ) 5 · √ ( 3 / 2 )', 'd ) 5 · √ 3', 'e ) 5 · √ 6'] | b | multiply(5, const_1) | if the space diagonal of cube c is 5 inches long , what is the length , in inches , of the diagonal of the base of cube c ? | n cube c , let ' s say that each side has side length x so , the diagonal = √ ( x ² + x ² + x ² ) = √ ( 3 x ² ) here , we ' re told that the diagonal has length 5 , so we can write : 5 = √ ( 3 x ² ) square both sides to get : 25 = 3 x ² divide both sides by 3 to get : 25 / 3 = x ² square root both sides : √ ( 25 / 3 ) = x or . . . . ( √ 25 ) / ( √ 3 ) = x simplify : 5 / ( √ 3 ) = x answer : b | a = 5 * 1
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a ) 145 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 445 m | a | multiply(divide(multiply(58, const_1000), const_3600), 9) | a train running at the speed of 58 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 58 * 5 / 18 = 145 / 9 m / sec length of the train = speed * time = 145 / 9 * 9 = 145 m answer : a" | a = 58 * 1000
b = a / 3600
c = b * 9
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a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | if x and y are both odd prime numbers and x < y , how many distinct positive integer w factors does 2 xy have ? | since 2 xy prime w factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . | a = 1 + 1
b = 2 * a
c = 1 + 1
d = b * c
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a ) 5 hrs , b ) 7 hrs , c ) 8 hrs , d ) 9 hrs 50 mins , e ) 9 hrs | e | multiply(6, multiply(divide(4, 2), divide(6000, 8000))) | if it takes 4 identical printing presses exactly 6 hours to print 8000 newspapers , how long would it take 2 of these presses to print 6000 newspapers ? | 4 presses - 8,000 newspapers - 6 hours ; 2 presses - 4,000 newspapers - 6 hours ; ( 360 mins ) 2 presses - 6,000 newspapers - 360 / 4000 * 6000 = 540 mins = 9 hrs answer : e | a = 4 / 2
b = 6000 / 8000
c = a * b
d = 6 * c
|
a ) 16 , b ) 12 , c ) 2 , d ) 3 , e ) 14 | d | add(const_10, 2) | if ( t - 8 ) is a factor of t ^ 2 - kt - 43 , then k = | "t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 3 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ) ( t + m ) d" | a = 10 + 2
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a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | c | add(subtract(55, multiply(12, 1)), 1) | a batsman in his 12 th innings makes a score of 55 and thereby increases his average by 1 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 1 ) + 55 ∴ x = 44 answer c" | a = 12 * 1
b = 55 - a
c = b + 1
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a ) 1209 , b ) 1324 , c ) 1245 , d ) 1300 , e ) 2535 | e | multiply(divide(subtract(2415, 15), subtract(21, const_1)), 21) | find large number from below question the difference of two numbers is 2415 . on dividing the larger number by the smaller , we get 21 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 2415 ) . x + 2415 = 21 x + 15 20 x = 2400 x = 120 large number = 120 + 2415 = 2535 answer : e" | a = 2415 - 15
b = 21 - 1
c = a / b
d = c * 21
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