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 ) 25 , b ) 35 , c ) 75 , d ) 80 , e ) 79 | e | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(const_3.0, subtract(multiply(80, const_4), multiply(84, 7))))) | the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 7 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ? | "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 7 = 75 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 â € “ 316 = 4 a = e + 4 = 75 + 4 = 79 e" | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 7
e = c - d
f = 3 + 0
g = b - f
h = a - g
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | add(4, 4) | if there are 4 peanuts in a box and mary puts 4 more peanuts inside , how many peanuts are in the box ? | "4 + 4 = 8 correct answer is a ) 8" | a = 4 + 4
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a ) 16 , b ) 18 , c ) 20 , d ) 33.3 , e ) 30 | d | multiply(3, divide(50, add(3, 6))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 50 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 50 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 3 = 9 kms / hr time taken = 50 / 9 = 5.55 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 5.55 = 33.3 kms . . . answer - d" | a = 3 + 6
b = 50 / a
c = 3 * b
|
a ) 55.65 , b ) 57.5 , c ) 62.5 , d ) 45.25 , e ) 50 | b | divide(add(subtract(multiply(60, 30), 90), 15), 30) | the average marks of 30 students in a class is 60 . but a student mark is wrongly noted as 90 instead of 15 then find the correct average marks ? | correct avg marks = 60 + ( 15 - 90 ) / 30 avg = 60 - 2.5 = 57.5 answer is b | a = 60 * 30
b = a - 90
c = b + 15
d = c / 30
|
a ) 2 hours , b ) 2.5 hours , c ) 3 hours , d ) 3.5 hours , e ) 4 hours | a | subtract(10, subtract(9, subtract(2, 1))) | there are two inlets and one outlet to a cistern . one of the inlets takes 2 hours to fill up the cistern and the other inlet takes twice as much time to fill up the same cistern . both of the inlets are turned on at 9 : 00 am with the cistern completely empty , and at 10 : 00 am , the outlet is turned on and it takes 1 more hour to fill the cistern completely . how much time does the outlet working alone takes to empty the cistern when the cistern is full ? | the combined inflow rate of the two inlets is 1 / 2 + 1 / 4 = 3 / 4 cistern / hour . thus , working together , it takes 4 / 3 hours ( time is reciprocal of rate ) to fill the cistern . from 9 : 00 am to 10 : 00 am , so in 1 hours , the inlet pipes will fill ( time ) * ( rate ) = 1 * 3 / 4 = 3 / 4 th of the cistern . then the outlet is turned on and the remaining 1 / 4 th of the cistern is filled in 1 hour . letting x to be the rate of the outlet , we would have : 3 / 4 - x = 1 / 4 - - > x = 1 / 2 cistern / hour , which means that it takes 2 hours the outlet working alone to empty the cistern . answer : a | a = 2 - 1
b = 9 - a
c = 10 - b
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a ) 0.33365 , b ) 0.3368 , c ) 0.3396 , d ) 0.3369 , e ) 0.3469 | c | multiply(0.3364, divide(0.3364, 0.3332)) | if log 0.318 = 0.3364 and log 0.317 = 0.33320 then log 0.319 = ? | log 0.318 / log 0.317 = log ( 0.318 - 0.317 ) = log 0.001 = 0.3364 / 0.33320 = 1.0096 now , log 0.318 * log 0.001 = log ( 0.318 + 0.001 ) = log 0.319 = 0.3364 * 1.0096 = 0.3396 so option c answer : c | a = 0 / 3364
b = 0 * 3364
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 2) | the population of a bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 50,000 bacteria | "the question basically asks how many minutes it takes for a population to increase by factor 50 ( 50,000 / 1,000 = 50 ) . now you know that every two minutes the population doubles , i . e . is multiplied by 2 . so the equation becomes : 2 ^ x > = 50 , where x represents the number of times the population doubles . a lot of people remember that 2 ^ 10 = 1,024 . hence , 2 ^ 6 = 64 , i . e . the population has to double 6 times . since it takes the population 2 minutes to double once it takes 6 * 2 minutes = 12 minutes to double nine times . thus , solution b = 12 is correct ." | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 2
|
a ) 200 sec , b ) 210 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | b | divide(add(1200, 900), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 900 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2100 m t = d / s t = 2100 / 10 t = 210 sec answer : b" | a = 1200 + 900
b = 1200 / 120
c = a / b
|
a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | b | subtract(80, subtract(368, multiply(80, 4))) | we bought a total of 80 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 368 . how many math books did we buy ? | m + h = 80 h = 80 - m 4 m + 5 h = 368 4 m + 5 * ( 80 - m ) = 368 m = 32 the answer is b . | a = 80 * 4
b = 368 - a
c = 80 - b
|
a ) 25 , b ) 63 , c ) 40 , d ) 45 , e ) 60 | b | divide(420, multiply(const_0_2778, subtract(30, 6))) | how many seconds will a 420 metre long train take to cross a man running with a speed of 6 km / hr in the direction of the moving train if the speed of the train is 30 km / hr ? | "explanation : speed of train relatively to man = ( 30 - 6 ) km / hr = 24 km / hr = ( 24 x 5 / 18 ) m / sec = 6.66 m / sec time taken to pass the man = ( 420 / 6.66 ) sec = 63 sec . answer : b" | a = 30 - 6
b = const_0_2778 * a
c = 420 / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | sqrt(divide(25, 25)) | if x + y = 25 and x 2 y 3 + y 2 x 3 = 25 , what is the value of xy ? | xy = 1 as x + y = 25 x 2 y 3 + y 2 x 3 = 25 x 2 y 2 ( y + x ) = 25 substituting x + y x 2 y 2 = 1 xy = 1 answer : a | a = 25 / 25
b = math.sqrt(a)
|
a ) rs . 2008 , b ) rs . 2022 , c ) rs . 2000 , d ) rs . 2029 , e ) rs . 2021 | c | divide(2288, add(const_1, divide(add(subtract(30, 20), 10), const_100))) | there is a 30 % increase in the price of an article in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . if the final price of the article is rs . 2288 , then what was the price of the article initially ? | let the price of the article , four years age be rs . 100 in the 1 st year , price of the article = 100 + 30 = rs . 130 . in the 2 nd year , price = 130 - 20 % of 130 = 130 - 26 = rs . 104 . in the 3 rd year , price = 104 + 10 % of 104 = 104 + 10.4 = rs . 114.40 . but present price of the article is rs . 2288 for 114.4 - - - > 100 ; 2288 - - - > ? required price = ( 2288 * 100 ) / 114.4 = 20 * 100 = rs . 2000 . answer : c | a = 30 - 20
b = a + 10
c = b / 100
d = 1 + c
e = 2288 / d
|
a ) 7 / 25 , b ) 4 / 25 , c ) 8 / 25 , d ) 2 / 13 , e ) 3 / 17 | b | multiply(divide(2, 5), divide(2, 5)) | if p ( a ) = 2 / 5 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events . | "p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 2 / 5 . 2 / 5 p ( a n b ) = 4 / 25 . b" | a = 2 / 5
b = 2 / 5
c = a * b
|
a ) - 16 , b ) - 18 , c ) 0 , d ) 14 , e ) 16 | b | add(sqrt(64), sqrt(64)) | if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 64 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | "( x + 1 ) ^ 2 < = 64 x < = 7 x > = - 9 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 9 × - 6 = 54 minimum possible value of xy is - 9 × 8 = - 72 - 72 + 54 = - 18 answer : b" | a = math.sqrt(64)
b = math.sqrt(64)
c = a + b
|
a ) rs . 12000 , b ) rs . 15000 , c ) rs . 13000 , d ) rs . 10000 , e ) rs . 14000 | b | multiply(multiply(150, 10), 10) | if the difference between compound interest ( interest compounded yearly ) and simple interest on a sum for 2 years at 10 % p . a . is rs . 150 then sum is | compund interest = p [ 1 + r / 100 ] ^ t - p ci = p [ 21 / 100 ] simple interest = ptr / 100 si = p [ 20 / 100 ] difference p [ 21 / 100 ] - p [ 20 / 100 ] = 150 p = 15000 answer : b | a = 150 * 10
b = a * 10
|
a ) 7580 , b ) 7960 , c ) 8290 , d ) 6048 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(7, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 7 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 7 - 1 / 12 ) = 5 / 84 work done by the inlet in 1 min . = ( 5 / 84 × 1 / 60 ) = 0.000992 volume of 0.000992 part = 6 litres . therefore , volume of whole = ( ( 1 / 0.000992 ) × 6 ) ‹ = › 6048 litres . answer d" | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 7 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 340 , b ) 346 , c ) 352 , d ) 365 , e ) 370 | d | multiply(multiply(multiply(500, subtract(1, divide(const_1, 10))), subtract(1, divide(const_1, 10))), subtract(1, divide(const_1, 10))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 500 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | number of newborns that can die in first month = 1 / 10 * 500 = 50 survived = 450 number of newborns that can die in second month = 1 / 10 * 450 = 45 survived = 405 number of newborns that can die in third month = 1 / 10 * 405 = 40 survived = 365 answer : d | a = 1 / 10
b = 1 - a
c = 500 * b
d = 1 / 10
e = 1 - d
f = c * e
g = 1 / 10
h = 1 - g
i = f * h
|
a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25 | c | divide(5, choose(10, 2)) | kim has 5 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 10 shoes , what is the probability that she will select 2 shoes of the same color ? | "total pairs = 10 c 2 = 45 ; same color pairs = 5 c 1 * 1 c 1 = 5 ; prob = 1 / 9 or 2 / 10 * 1 / 9 * 5 = 1 / 9 ans c" | a = math.comb(10, 2)
b = 5 / a
|
a ) 35.24 , b ) 36.14 , c ) 36.24 , d ) 36.44 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 45 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 45 - 23 ) = 1822 . â ˆ ´ correct mean = 1822 / 50 = 36.44 . answer d" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 7 1 / 2 days , b ) 6 1 / 2 days , c ) 5 1 / 2 days , d ) 4 1 / 2 days , e ) 3 1 / 2 days | a | divide(multiply(5, 3), subtract(5, 3)) | a man can do a piece of work in 5 days , but with the help of his son he can do it in 3 days . in what time can the son do it alone ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as , son ' s one day work = ( 1 / 3 − 1 / 5 ) = ( 5 − 3 / 15 ) = 2 / 15 so son will do whole work in 15 / 2 days which is = 7 1 / 2 days option a" | a = 5 * 3
b = 5 - 3
c = a / b
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | multiply(divide(subtract(subtract(43, const_1), multiply(subtract(15, const_1), const_2)), subtract(multiply(subtract(15, const_1), const_4), multiply(subtract(15, const_1), const_2))), subtract(15, const_1)) | a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 15 heads and 43 legs . how many cats does the ship host ? | "sa ' s + co + ca + cats = 15 . sa ' s + 1 + 1 + cats = 15 or sa ' s + cats = 13 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 43 sa ' s * 2 + cats * 4 = 40 or sa ' s + cats * 2 = 20 or 13 - cats + cat * 2 = 20 then cats = 7 d" | a = 43 - 1
b = 15 - 1
c = b * 2
d = a - c
e = 15 - 1
f = e * 4
g = 15 - 1
h = g * 2
i = f - h
j = d / i
k = 15 - 1
l = j * k
|
a ) 3377 , b ) 2678 , c ) 5460 , d ) 7500 , e ) 1671 | d | divide(multiply(divide(multiply(5000, add(const_100, 20)), const_100), add(const_100, 25)), const_100) | find the amount on rs . 5000 in 2 years , the rate of interest being 20 % per first year and 25 % for the second year ? | "5000 * 120 / 100 * 125 / 100 = > 7500 answer : d" | a = 100 + 20
b = 5000 * a
c = b / 100
d = 100 + 25
e = c * d
f = e / 100
|
a ) 26 sec , b ) 76 sec , c ) 48 sec , d ) 27 sec , e ) 22 sec | c | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.9, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.9 = 2 km = 2000 m . required time = 2000 * 3 / 125 = 48 sec . answer : c" | a = 1 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 0 * 9
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 3377 , b ) 2899 , c ) 3750 , d ) 2778 , e ) 1991 | c | divide(multiply(300, const_100), 2) | a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 300 when the rate to profit rises from 5 to 7 percent . find the capital of c ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 300 c ' s capital = 22500 * 1 / 6 = 3750 answer : c" | a = 300 * 100
b = a / 2
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a ) − 2 π , b ) - 1 , c ) − π , d ) 2 π , e ) 4 π | c | subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(add(const_10, const_4), const_2)), const_1000), const_10), const_3), subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(add(const_10, const_4), const_2)), const_1000), const_10), const_3), negate(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(add(const_10, const_4), const_2)), const_1000), const_10), const_3)))) | define a * by the equation a * = π - x . then ( ( − π ) * ) * = | for a * f ( f ( − π ) ) = f ( π − ( − π ) ) = f ( π + π ) = f ( 2 π ) = π − 2 π = − π = c | a = 10 + 4
b = a * 100
c = 10 + 4
d = c + 2
e = b + d
f = e / 1000
g = f / 10
h = g + 3
i = 10 + 4
j = i * 100
k = 10 + 4
l = k + 2
m = j + l
n = m / 1000
o = n / 10
p = o + 3
q = 10 + 4
r = q * 100
s = 10 + 4
t = s + 2
u = r + t
v = u / 1000
w = v / 10
x = w + 3
y = p - negate
z = h - y
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | divide(subtract(35, 17), subtract(17, 15)) | the average age of a group of n people is 15 years old . one more person aged 35 joins the group and the new average is 17 years old . what is the value of n ? | 15 n + 35 = 17 ( n + 1 ) 2 n = 18 n = 9 the answer is b . | a = 35 - 17
b = 17 - 15
c = a / b
|
['a ) 1.6 units', 'b ) 6 units', 'c ) 3 units', 'd ) 5 units', 'e ) 12 units'] | a | divide(triangle_area_three_edges(4, 11, 12), divide(triangle_perimeter(4, 11, 12), const_2)) | what is the measure of the radius of the circle inscribed in a triangle whose sides measure 4 , 11 and 12 units ? | sides are 4 , 11 and 12 . . . thus it is right angle triangle since 12 ^ 2 = 4 ^ 2 + 11 ^ 2 therefore , area = 1 / 2 * 11 * 4 = 22 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 12 + 11 + 4 / 2 = 13,5 thus , 22 = 13,5 * r and hence r = in - radius = 1.6 option a | a = triangle_area_three_edges / (
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a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | e | subtract(10, multiply(2, 4)) | what is x if x + 2 y = 10 and y = 4 ? | "x = 10 - 2 y x = 10 - 8 . x = 2 answer : e" | a = 2 * 4
b = 10 - a
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a ) 47 , b ) 54 , c ) 56 , d ) 61 , e ) 64 | b | subtract(90, subtract(396, multiply(90, 4))) | we bought a total of 90 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 396 . how many math books did we buy ? | "m + h = 90 h = 90 - m 4 m + 5 h = 396 4 m + 5 * ( 90 - m ) = 396 m = 54 the answer is b ." | a = 90 * 4
b = 396 - a
c = 90 - b
|
a ) 900 , b ) 1,600 , c ) 2,500 , d ) 3,600 , e ) 4,900 | a | add(multiply(multiply(multiply(12, power(const_3, const_2)), 15), const_2), multiply(15, 18)) | what is the smallest positive perfect square that is divisible by 12 , 15 , and 18 ? | "take the lcm of 12 , 15,18 that will come 180 . the smallest positive perfect square given in the option that can be divided with 180 is 900 . so the ans is ( a )" | a = 3 ** 2
b = 12 * a
c = b * 15
d = c * 2
e = 15 * 18
f = d + e
|
a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7 | e | divide(const_2, add(const_3, const_4)) | find the probability that a leap year selected at random will have 53 mondays | there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 mondays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 mondays we should have either { mon , tuesday } or { sun , mon } case . the probability of that is 2 / 7 . answer : e . | a = 3 + 4
b = 2 / a
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a ) 240 , b ) 288 , c ) 277 , d ) 360 , e ) 422 | d | multiply(20, multiply(81, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 81 km / hr . what is the length of the platform ? | "speed = 81 * 5 / 18 = 22.5 m / sec . length of the train = 22.5 * 20 = 450 m . let the length of the platform be x m . then , ( x + 450 ) / 36 = 22.5 = > x = 360 m . answer : d" | a = 81 * const_0_2778
b = 20 * a
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a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 % | b | multiply(divide(divide(subtract(40, 32), subtract(32, 30)), add(divide(subtract(40, 32), subtract(32, 30)), const_1)), const_100) | solution x is 30 % chemical a and 70 % chemical b by volume . solution y is 40 % chemical a and 60 % chemical b by volume . if a mixture of x and y is 32 % chemical a , what percent of the mixture is solution x ? | the volume of the mixture be x + y . 0.3 x + 0.4 y = 0.32 ( x + y ) x = 4 y x / ( x + y ) = 4 / 5 = 80 % . the answer is b . | a = 40 - 32
b = 32 - 30
c = a / b
d = 40 - 32
e = 32 - 30
f = d / e
g = f + 1
h = c / g
i = h * 100
|
a ) 18.75 % , b ) 23 % , c ) 15 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(add(const_100, 50), add(const_100, 30)), add(const_100, 30)), const_100) | the wages earned by robin is 30 % more than that earned by erica . the wages earned by charles is 50 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? | "let wage of erica = 10 wage of robin = 1.3 * 10 = 13 wage of charles = 1.5 * 10 = 15 percentage by which wage earned by charles is more than that earned by robin = ( 15 - 13 ) / 13 * 100 % = 2 / 13 * 100 % = 15 % answer c" | a = 100 + 50
b = 100 + 30
c = a - b
d = 100 + 30
e = c / d
f = e * 100
|
a ) 30 , b ) 45 , c ) 50 , d ) 31 , e ) 21 | d | divide(subtract(multiply(8, 5.4), multiply(0.6, 10)), 1.2) | 8 x 5.4 - 0.6 x 10 / 1.2 = ? | "given expression = ( 43.2 - 6 ) / 1.2 = 37.2 / 1.2 = 31 answer is d ." | a = 8 * 5
b = 0 * 6
c = a - b
d = c / 1
|
a ) 96 , b ) 108 , c ) 110 , d ) 90 , e ) 93 | a | subtract(subtract(subtract(182, 22), subtract(65, const_1)), const_4) | veena ranks 65 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | "total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 65 + 1 = 96 answer : a" | a = 182 - 22
b = 65 - 1
c = a - b
d = c - 4
|
a ) 45,000 , b ) 360,000 , c ) 364,500 , d ) 363,600 , e ) 396,900 | d | multiply(9, divide(subtract(440000, 36000), add(9, const_1))) | there were 36000 hardback copies of a certain novel sold before the paperback version was issued . from the time the first paperback copy was sold until the last copy of the novel was sold 9 times as many paperback copies as hardback copies were sold . if a total of 440000 copies of the novel were sold in all , how many paperback copies were sold ? | say x was the # of hardback copies sold from the time the first paperback copy was sold . then the total # of paperback copies sold was 9 x ; hence the total # of copies sold was ( hardback ) + ( paperback ) = ( 36 + x ) + ( 9 x ) = 440 - - > x = 40.4 . so , the total # of paperback copies sold was 9 x = 9 * 40.4 = 363.6 or 363,600 . answer : d . | a = 440000 - 36000
b = 9 + 1
c = a / b
d = 9 * c
|
a ) $ 18,400 , b ) $ 19,500 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | c | multiply(const_2, const_10) | a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 1,500 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price in 2002 ? | "important point to notice - 2.5 % decrease from the original price 2.5 % of 22,000 = 550 total reduction in 6 years = 550 * 6 = 3300 final price = 22,000 + 1500 - 3300 = 20,200 c is the answer" | a = 2 * 10
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a ) 8 hrs , b ) 9 hrs , c ) 2 hrs , d ) 2 hrs , e ) 9 hrs | c | multiply(4, divide(30, const_60)) | in a covering a certain distance , the speeds of a and b are in the ratio of 3 : 4 . a takes 30 minutes more than b to reach the destination . the time taken by a to reach the destination is ? | ratio of speeds = 3 : 4 ratio of times taken = 4 : 3 suppose a takes 4 x hrs and b takes 3 x hrs to reach the destination . then , 4 x - 3 x = 30 / 60 = > x = 1 / 2 time taken by a = 4 x hrs = 4 * 1 / 2 = 2 hrs . answer : c | a = 30 / const_60
b = 4 * a
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 19 | b | divide(24, const_2) | in a group of ducks and cows , the total number of legs are 24 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 24 = > 2 x = 24 = > x = 12 b" | a = 24 / 2
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a ) rs . 12500 , b ) rs . 1000 , c ) rs . 8000 , d ) rs . 6000 , e ) rs . 15000 | a | multiply(2500, add(const_4, const_1)) | rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 2500 . then his monthly salary is | "sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 2500 â ‡ ” 20 / 100 x = 2500 â ‡ ” x = 2500 ã — 5 = 12500 . answer a" | a = 4 + 1
b = 2500 * a
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a ) 220 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 200 km | e | multiply(const_2, divide(multiply(multiply(20, 20), 10), add(20, 20))) | a man complete a journey in 10 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 20 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 20 = 10 - - > x / 20 + x / 20 = 20 - - > 2 x = 20 x 20 - - > x = ( 20 x 20 ) / 2 = 200 km . answer : e ." | a = 20 * 20
b = a * 10
c = 20 + 20
d = b / c
e = 2 * d
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a ) 10 / 15 , b ) 11 / 15 , c ) 12 / 15 , d ) 13 / 15 , e ) 14 / 15 | d | divide(add(const_12, const_1), multiply(3, 5)) | the probability of a student possessing a ball point pen in exam is 3 / 5 & possessing an ink pen is 2 / 3 . find his probability of possessing at least one of them | the probability of a student possessing a ball point pen in exam is 3 / 5 . the probability of a student not possessing a ball point pen in exam is 2 / 5 & possessing an ink pen is 2 / 3 . & not possessing an ink pen is 1 / 3 . his probability of possessing none of them = 2 / 5 * 1 / 3 = 2 / 15 his probability of possessing one of them = 1 - 2 / 15 = 13 / 15 answer : d | a = 12 + 1
b = 3 * 5
c = a / b
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a ) 40 , b ) 42 , c ) 44 , d ) 48 , e ) 49 | e | add(divide(divide(200, add(const_4, const_1)), add(const_4, const_1)), divide(200, add(const_4, const_1))) | if 200 ! / 10 ^ n is an integer , what is the largest possible value of n ? | "we have to basically count the number of 5 s withing 200 200 / 5 = 40 200 / 5 ^ 2 = 8 200 / 5 ^ 3 = 1 we know for sure that there will be at least 49 even numbers ( multiples of 2 ) within 200 so there are 49 10 s in 200 ! answer : e" | a = 4 + 1
b = 200 / a
c = 4 + 1
d = b / c
e = 4 + 1
f = 200 / e
g = d + f
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a ) a ) 11 , b ) b ) 15 , c ) c ) 20 , d ) d ) 21 , e ) e ) 22 | b | divide(subtract(50, 1), 3) | if you write down all the numbers from 1 to 50 , then how many times do you write 3 ? | "explanation : explanation : clearly , from 1 to 50 , there are ten numbers with 3 as the unit ' s digit - 3 , 13 , 23 , 33 , 43 , and ten numbers with 3 as the ten ' s digit - 30 , 31 , 32 , 33 , 34 , 35 , 36 , 37 , 38 , 39 . so , required number = 5 + 10 = 15 . answer : b" | a = 50 - 1
b = a / 3
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a ) 14 , b ) 26 , c ) 16 , d ) 36 , e ) 46 | c | multiply(const_2, subtract(multiply(64, const_4), 248)) | evaluate 248 + 64 − − √ − − − − − − − − − √ | explanation : = 248 + 64 − − √ − − − − − − − − − √ = 248 + 8 − − − − − − √ = 256 − − − √ = 16 answer : c | a = 64 * 4
b = a - 248
c = 2 * b
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a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 300 | c | divide(multiply(638, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 638 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 638 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 638 ⇒ y [ 12 / 10 + 1 ] = 638 ⇒ 22 y / 10 = 638 ⇒ 22 y = 6380 ⇒ y = 6380 / 22 = 580 / 2 = rs . 290 c" | a = 1 + 4
b = a * 2
c = 638 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
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a ) 14 , b ) 12 , c ) 10 , d ) 16 , e ) 18 | d | add(add(4, 3), add(add(4, 4), 1)) | for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 920 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 1000 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 920 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 920 so , 3 y < 408 = = > y < 136 so , y can take any value which is less than 136 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . d" | a = 4 + 3
b = 4 + 4
c = b + 1
d = a + c
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['a ) 1694 m 2', 'b ) 1700 m 2', 'c ) 1598 m 2', 'd ) 1500 m 2', 'e ) none of thes'] | a | subtract(circle_area(add(35, 7)), circle_area(35)) | a circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside . find the area of path . | radius of a circular grass lawn ( without path ) = 35 m ∴ area = π r 2 = π ( 35 ) 2 radius of a circular grass lawn ( with path ) = 35 + 7 = 42 m ∴ area = π r 2 = π ( 42 ) 2 ∴ area of path = π ( 42 ) 2 – π ( 35 ) 2 = π ( 422 – 352 ) = π ( 42 + 35 ) ( 42 – 35 ) answer a | a = 35 + 7
b = circle_area - (
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a ) 9660 , b ) 6560 , c ) 7560 , d ) 9560 , e ) 8560 | d | add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 6000), 3), 6000)), 6000), 3), 6000) | sonika deposited rs . 6000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ? | "( 6000 * 3 * 2 ) / 100 = 360 9200 - - - - - - - - 9560 answer : d" | a = 2 / 100
b = 9200 - 6000
c = b / 3
d = c / 6000
e = a + d
f = e * 6000
g = f * 3
h = g + 6000
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a ) a ) 19.42 , b ) b ) 20.42 , c ) c ) 21.42 , d ) d ) 22.42 , e ) e ) 23.42 | c | subtract(const_60, multiply(const_60, divide(27, 42))) | excluding stoppages , the speed of a train is 42 kmph and including stoppages it is 27 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 15 / 42 * 60 = 21.42 answer : option c" | a = 27 / 42
b = const_60 * a
c = const_60 - b
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a ) 50 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | c | divide(80, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 80 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 800 gm ) of tea and 1 / 2 kg ( 800 gm ) of coffee ( equal quantities ) = 80 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 80 = > x = 80 thus proved . . . option c ." | a = 20 / 100
b = 1 - a
c = 20 / 100
d = c + 1
e = b * d
f = 80 / e
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a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 6), add(divide(25, const_100), const_1)), multiply(divide(const_1, 6), divide(20, const_100)))) | working alone , mary can pave a driveway in 6 hours and hillary can pave the same driveway in 6 hours . when they work together , mary thrives on teamwork so her rate increases by 25 % , but hillary becomes distracted and her rate decreases by 20 % . if they both work together , how many hours will it take to pave the driveway ? | "initial working rates : mary = 1 / 6 per hour hillary = 1 / 6 per hour rate when working together : mary = 1 / 6 + ( 1 / 4 * 1 / 6 ) = 1 / 5 per hour hillary = 1 / 6 - ( 1 / 5 * 1 / 6 ) = 2 / 15 per hour together they work 1 / 5 + 2 / 15 = 1 / 3 per hour so they will need 3 hours to complete the driveway . the correct answer is a ." | a = 1 / 6
b = 25 / 100
c = b + 1
d = a * c
e = 1 / 6
f = 20 / 100
g = e * f
h = d + g
i = 1/(h)
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a ) 9.5 , b ) 5.5 , c ) 8.5 , d ) 9.6 , e ) 9.7 | a | divide(9.5, subtract(5.5, 4.5)) | two boys starts from the same place walking at the rate of 4.5 kmph and 5.5 kmph respectively in the same direction . what time will they take to be 9.5 km apart ? | explanation : relative speed = 5.5 - 4.5 = 1 kmph ( because they walk in the same direction ) distance = 9.5 km time = distance / speed = 9.5 / 1 = 9.5 hr answer : a | a = 5 - 5
b = 9 / 5
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a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | e | divide(multiply(multiply(45, divide(4, 5)), inverse(subtract(const_1, divide(4, 5)))), const_60) | a train is moving at 4 / 5 of its usual speed . the train is 45 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | new time = d / ( 4 v / 5 ) = 5 / 4 * usual time 45 minutes represents 1 / 4 of the usual time . the usual time is 4 * 45 minutes = 3 hours . the answer is e . | a = 4 / 5
b = 45 * a
c = 4 / 5
d = 1 - c
e = 1/(d)
f = b * e
g = f / const_60
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a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | a | subtract(subtract(add(52, 12), 22), 12) | in a class of 52 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | total = english + german - both + neither - - > 52 = english + 22 - 12 + 0 - - > english = 42 - - > only english = english - both = 42 - 12 = 30 answer : a . | a = 52 + 12
b = a - 22
c = b - 12
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a ) 1 : 3 , b ) 2 : 5 , c ) 3 : 5 , d ) 5 : 8 , e ) 4 : 7 | d | divide(divide(5, 2), 4) | if the ratio of apples to bananas is 5 to 2 and the ratio of bananas to cucumbers is 1 to 4 , what is the ratio of apples to cucumbers ? | the ratio of bananas to cucumbers is 1 to 4 which equals 2 to 8 . the ratio of apples to bananas to cucumbers is 5 to 2 to 8 . the ratio of apples to cucumbers is 5 to 8 . the answer is d . | a = 5 / 2
b = a / 4
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a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 | d | divide(const_2, 3) | a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability l that the first ball picked up is odd numbered ? | answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability l is 2 / 3 . d | a = 2 / 3
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a ) 80 , b ) 90 , c ) 100 , d ) 120 , e ) 130 | c | divide(subtract(multiply(120, 35), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 and that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidates who passed the examination is ? | let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 ⇒ 24 y = 4200 - 1800 ∴ y = 2400 / 24 = 100 c | a = 120 * 35
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
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a ) 360 π , b ) 400 π , c ) 450 π , d ) 500 π , e ) 600 π | a | multiply(power(multiply(sqrt(90), const_2), const_2), const_pi) | in may , the groundskeeper at spring lake golf club built a circular green with an area of 90 π square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | "area = π r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 90 π ) = 360 π the answer is a ." | a = math.sqrt(90)
b = a * 2
c = b ** 2
d = c * math.pi
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a ) 35.4 , b ) 50.7 , c ) 45.7 , d ) 57 , e ) 67 | b | add(40, const_1) | the average of first six prime numbers which are between 40 and 80 is | "explanation : first six prime numbers which are between 40 and 80 = 41 , 43 , 47 , 53 , 59 , 61 average = ( 41 + 43 + 47 + 53 + 59 + 61 ) / 6 = 50.7 answer : b" | a = 40 + 1
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a ) 430 , b ) 880 , c ) 550 , d ) 590 , e ) 600 | b | add(divide(720, subtract(const_1, divide(10, const_100))), multiply(divide(720, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | boy sells a book for rs . 720 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "cost price = 720 / 90 x 100 = 800 to gain 10 % = 800 x 10 / 100 = 80 sp = cp + gain = 800 + 80 = 880 answer : b" | a = 10 / 100
b = 1 - a
c = 720 / b
d = 10 / 100
e = 1 - d
f = 720 / e
g = 10 / 100
h = f * g
i = c + h
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a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 20 | e | inverse(subtract(divide(const_1, 4), divide(const_1, 5))) | renu can do a piece of work in 5 days , but with the help of her friend suma , she can do it in 4 days . in what time suma can do it alone ? | "renu â € ™ s one day â € ™ s work = 1 / 5 suma â € ™ s one day â € ™ s work = 1 / 4 - 1 / 5 = 1 / 20 suma can do it alone in 20 days . answer : e" | a = 1 / 4
b = 1 / 5
c = a - b
d = 1/(c)
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a ) 1 , b ) 2 , c ) 4 , d ) 19 , e ) 20 | c | subtract(30, reminder(5, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 30 ? | "when positive integer n is divided by 5 , the remainder is 1 i . e . , n = 5 x + 1 values of n can be one of { 1 , 6 , 11 , 16 , 21 , 26 , 31 . . . . . . . . . . . . . 46 , 51 , 56,61 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 7 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can be one of { 5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 . . . . . . . . } combining both the sets we get n = { 19 , 52 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 30 or 30 x in case of n = 26 p = 4 c is the answer ." | a = 30 - reminder
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a ) 1 : 2 , b ) 5 : 9 , c ) 2 : 1 , d ) 11 : 24 , e ) none | c | multiply(multiply(divide(2, 3), divide(6, 11)), divide(11, 2)) | the compounded ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 2 ) is : | answer : option c 2 / 3 : 6 / 11 : 11 / 2 = 2 : 1 | a = 2 / 3
b = 6 / 11
c = a * b
d = 11 / 2
e = c * d
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a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number p and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neither } . { both } is minimized when { neither } is 0 . in this case { both } = 31 . the greatest possible number of students p that could have taken both geometry and biology , is 119 . thus , the answer is 119 - 31 = 88 . answer : d ." | a = 144 + 119
b = a - 232
c = 119 - b
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a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people v who purchased only book a ? | "this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b and this is 1 / 2 of those that purchased a . so , 1500 purchased a . less the 500 that purchased both , v = 1000 purchased a only . ( this is much simpler to solve using the venn diagram ) . correct answer is d . 1000" | a = 500 / 2
b = 500 + a
c = b * 2
d = c - 500
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a ) 50 , b ) 54 , c ) 56 , d ) 51 , e ) 63 | d | add(divide(const_100, const_2), 1) | in a recent election , geoff received 1 percent of the 6,000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 3000 more votes to win the election , what is the value of x ? | "word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 60 votes he needs 3571 more votes so : 60 + 3000 = 3060 now what ' s the problem wants ? ? a x % . . . . . . . . 3060 is what % of total votes 6000 . . . . . . . . translating : 3060 = x / 100 * 6000 - - - x = 51 % . . . . . . . . . . d" | a = 100 / 2
b = a + 1
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a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability e that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability e = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a = math.comb(4, 1)
b = a * 2
c = math.comb(6, 3)
d = b / c
e = 5 * 5
f = e * 4
g = d * f
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a ) 75 , b ) 100 , c ) 120 , d ) 30 , e ) 80 | e | multiply(divide(160, divide(40, const_100)), divide(20, const_100)) | if 40 % of a certain number is 160 , then what is 20 % of that number ? | "explanation : 40 % = 40 * 4 = 160 20 % = 20 * 4 = 80 answer : option e" | a = 40 / 100
b = 160 / a
c = 20 / 100
d = b * c
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a ) 300 , b ) 230 , c ) 500 , d ) 560 , e ) 590 | a | divide(multiply(multiply(subtract(3.15, 3), const_1000), const_100), 50) | workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.15 lacs . how many workers were they ? | "n * 50 = ( 315000 - 300000 ) = 15000 n = 15000 / 50 = 300 a" | a = 3 - 15
b = a * 1000
c = b * 100
d = c / 50
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a ) - 21 , b ) - 53 , c ) - 35 , d ) - 55 , e ) - 58 | d | power(5, 2) | 1 ^ 2 – 2 ^ 2 + 3 ^ 2 – 4 ^ 2 + 5 ^ 2 – 6 ^ 2 + 7 ^ 2 – 8 ^ 2 + 9 ^ 2 – 10 ^ 2 = ? | 1 - 4 + 9 - 16 + 25 - 36 + 49 - 64 + 81 - 100 = - 55 answer : d | a = 5 ** 2
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a ) 115 kg . , b ) 110 kg . , c ) 106 kg . , d ) 105 kg . , e ) 100 kg . | c | divide(add(multiply(7, 112), add(110, 60)), add(7, const_2)) | there are 7 players in a bowling team with an average weight of 112 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 112 * 7 + 110 + 60 ) / 9 = 106 kgs c is the answer" | a = 7 * 112
b = 110 + 60
c = a + b
d = 7 + 2
e = c / d
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a ) 2 kg , b ) 2.4 kg , c ) 3.12 kg , d ) 10 kg , e ) none of these | c | multiply(divide(divide(multiply(subtract(const_100, 90), 25), const_100), subtract(const_100, 20)), const_100) | fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 25 kg of fresh grapes ? | the weight of non - water in 25 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 25 â ˆ — 0.1 = x â ˆ — 0.8 - - > x = 3.12 answer : c . | a = 100 - 90
b = a * 25
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
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a ) 10.21 , b ) 10.27 , c ) 10.25 , d ) 10.23 , e ) 60.25 | e | add(60, divide(multiply(subtract(divide(divide(subtract(350, divide(multiply(80, multiply(const_3, const_3)), const_4)), 60), 2), const_1), const_60), const_100)) | anna left for city a from city b at 5.20 a . m . she traveled at the speed of 80 km / hr for 2 hrs 15 min . after that the speed was reduced to 60 km / hr . if the distance between two cities is 350 kms , at what time did anna reach city a ? | distance covered in 2 hrs 15 min i . e . , 2 1 / 4 hrs = 80 * 9 / 4 = 180 hrs . time taken to cover remaining distance = ( 350 - 180 ) / 60 = 17 / 6 hrs = 2 5 / 6 = 2 hrs 50 min total time taken = ( 2 hrs 15 min + 2 hrs 50 min ) = 5 hrs 5 min . so , anna reached city a at 10.25 a . m . answer : e | a = 3 * 3
b = 80 * a
c = b / 4
d = 350 - c
e = d / 60
f = e / 2
g = f - 1
h = g * const_60
i = h / 100
j = 60 + i
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | e | multiply(40, const_1) | at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 155 male associates , how many female managers are there ? | "250 employees : 90 male , 160 female 40 managers , 210 associates 155 male associates implies 55 female associates which means the remaining 35 females must be managers e . 35" | a = 40 * 1
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a ) 26 , b ) 36 , c ) 46 , d ) 56 , e ) 66 | b | multiply(divide(120, 10), 3) | pencils , pens and exercise books in a shop are in the ratio of 10 : 2 : 3 . if there are 120 pencils , the number of exercise books in the shop is : | "explanation : let pencils = 10 x , pens = 2 x & exercise books = 3 x . now , 10 x = 120 hence x = 12 . number of exercise books = 3 x = 36 . answer : b" | a = 120 / 10
b = a * 3
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a ) a ) 14 , b ) b ) 13 , c ) c ) 16 , d ) d ) 8 , e ) e ) 6 | b | divide(26, const_2) | in a group of ducks and cows , the total number of legs are 26 more than thrice the number of heads . find the total number of cows . | "explanation : let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 26 more than twice the number of heads = > 2 ( d + 2 c ) = 26 + 2 ( c + d ) = > d + 2 c = 13 + c + d = > 2 c = 13 + c = > c = 13 i . e . , total number of cows = 13 answer : b" | a = 26 / 2
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a ) 3 minutes , b ) 7.5 minutes , c ) 5 minutes , d ) 6 minutes , e ) 12 minutes | b | divide(multiply(3, 5), add(speed(5, 5), speed(multiply(2, 5), 10))) | working alone at its constant rate , machine a produces x boxes in 5 minutes and working alone at its constant rate , machine b produces 2 x boxes in 10 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 3 x boxes ? | "rate = work / time given rate of machine a = 2 x / 10 min machine b produces 2 x boxes in 10 min hence , machine b produces 2 x boxes in 10 min . rate of machine b = 2 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = 2 x / 10 min + rate of machine b = 2 x / 10 = 4 x / 10 now combine time = combine work needs to be done / combine rate = 3 x / 4 x * 10 = 7.5 min ans : b" | a = 3 * 5
b = speed + (
c = 2 * 5
d = a / b
|
a ) $ 180,000 , b ) $ 185,000 , c ) $ 190,000 , d ) $ 207,500 , e ) $ 205,000 | d | multiply(add(divide(1, 35), 1), divide(multiply(1000, divide(add(add(multiply(add(multiply(divide(const_3, const_2), multiply(5, 35)), multiply(40, 5)), subtract(35, add(const_3, 5))), multiply(multiply(40, 5), 5)), multiply(multiply(5, 35), const_3)), 35)), 1000)) | a factory that employs 1000 assembly line workers pays each of these workers $ 5 per hour for the first 40 hours worked during a week and 1 ½ times that rate for hours worked in excess of 40 . what was the total payroll for the assembly - line workers for a week in which 30 percent of them worked 35 hours , 50 percent worked 40 hours , and the rest worked 50 hours ? | "30 % of 1000 = 300 worked for 20 hours payment @ 5 / hr total payment = 300 * 35 * 5 = 52500 50 % of 1000 = 500 worked for 40 hours payment @ 5 / hr total payment = 500 * 40 * 5 = 100000 remaining 200 worked for 50 hours payment for first 40 hours @ 5 / hr payment = 200 * 40 * 5 = 40000 payment for next 10 hr @ 7.5 / hr payment = 200 * 10 * 7.5 = 15000 total payment = 52500 + 100000 + 40000 + 15000 = 207500 hence , answer is d" | a = 1 / 35
b = a + 1
c = 3 / 2
d = 5 * 35
e = c * d
f = 40 * 5
g = e + f
h = 3 + 5
i = 35 - h
j = g * i
k = 40 * 5
l = k * 5
m = j + l
n = 5 * 35
o = n * 3
p = m + o
q = p / 35
r = 1000 * q
s = r / 1000
t = b * s
|
a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | multiply(divide(subtract(divide(80, const_100), divide(50, const_100)), divide(60, const_100)), const_2) | solution a is 20 % sugar and solution b is 80 % sugar . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % sugar solution ? | forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 - 20 ) = 1 / 1 so the volume of the two solutions will be equal . answer has to be 4 : 4 = c | a = 80 / 100
b = 50 / 100
c = a - b
d = 60 / 100
e = c / d
f = e * 2
|
a ) 80 , b ) 160 , c ) 720 , d ) 1600 , e ) 1848 | e | multiply(divide(divide(factorial(12), factorial(subtract(12, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | 12 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | no . of ways of picking 2 biology books ( from 12 books ) = 12 c 2 = ( 12 * 11 ) / 2 = 66 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 66 * 28 = 1848 ( option e ) | a = math.factorial(12)
b = 12 - 2
c = math.factorial(b)
d = a / c
e = d / 2
f = math.factorial(8)
g = 8 - 2
h = math.factorial(g)
i = f / h
j = i / 2
k = e * j
|
a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | a | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 3)) | the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 3 of that total . what was the price of the basic computer ? | "let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 3 of that new total price : p = 1 / 3 * $ 3,000 = $ 1000 plug this value in the first equation : c + 1000 = $ 2,500 - - > c = $ 1,500 . answer : a ." | a = const_0_25 * 1000
b = a * 10
c = const_0_25 * 1000
d = c * 10
e = 500 + d
f = e / 3
g = b - f
|
a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 72 % | d | subtract(const_100, multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | a baseball card decreased in value 20 % in its first year and 20 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.8 = 80 after second year price = 80 * 0.8 = 64 final decrease = [ ( 100 - 64 ) / 100 ] * 100 = 36 % correct answer - d" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) none of these | b | multiply(divide(2, 50), const_100) | 2 is what percent of 50 | "explanation : 2 / 50 * 100 = 1 / 25 * 100 = 4 % option b" | a = 2 / 50
b = a * 100
|
a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | e | subtract(add(14, 26), divide(14, divide(70, const_100))) | the contents of a certain box consist of 14 apples and 26 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ? | "the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 26 oranges , so to get to 6 oranges , we should remove 26 - 6 = 20 oranges . answer e" | a = 14 + 26
b = 70 / 100
c = 14 / b
d = a - c
|
a ) 380 times , b ) 381 times , c ) 382 times , d ) 248 times , e ) 482 times | d | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 20) | light glows for every 20 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 20 ) = 248 times . answer : d" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 20
|
a ) 34 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | a | add(divide(subtract(add(40, 2), 18), 1.5), 18) | each week , harry is paid x dollars per hour for the first 18 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "42 x = 18 x + 1.5 x ( h - 18 ) = = > 42 = 18 + 1.5 ( h - 18 ) = = > h - 18 = 24 / 1.5 = 16 = = > h = 34 answer is a" | a = 40 + 2
b = a - 18
c = b / 1
d = c + 18
|
a ) 6 , b ) 8 , c ) 9 , d ) 5 , e ) 7 | a | add(2, const_1) | the average of first five multiples of 2 is ? | "average = 2 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 30 / 5 = 6 . answer : a" | a = 2 + 1
|
a ) 40 , b ) 44 , c ) 35 , d ) 49 , e ) 38 | b | subtract(157, add(add(32, 89), divide(add(32, subtract(111, add(add(89, 56), 32))), 2))) | in an intercollegiate competition that lasted for 3 days , 157 students took part on day 1 , 111 on day 2 and 98 on day 3 . if 89 took part on day 1 and day 2 and 56 took part on day 2 and day 3 and 32 took part on all three days , how many students took part only on day 1 ? | "day 1 & 2 = 89 ; only day 1 & 2 ( 89 - 32 ) = 57 , day 2 & 3 = 56 ; only day 2 & 3 ( 56 - 32 ) = 24 , only day 1 = 157 - ( 57 + 24 + 32 ) = 44 answer : b" | a = 32 + 89
b = 89 + 56
c = b + 32
d = 111 - c
e = 32 + d
f = e / 2
g = a + f
h = 157 - g
|
a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 9 , 2,23 | d | divide(multiply(4, const_3), const_3) | find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 21 ? | let the numbers be a , b and c . a : b : c = 3 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 21 = > 3 x + 4 x = 21 = > x = 3 a , b , c are 3 x , 2 x , 4 x a , b , c are 9 , 6 , 12 . answer : d | a = 4 * 3
b = a / 3
|
a ) 5 , b ) 2 , c ) 10 , d ) 1 , e ) 12 | c | divide(2000, multiply(20, 10)) | in a certain village , 20 litres of water are required per household per month . at this rate , if there are 10 households in the village , how long ( in months ) will 2000 litres of water last ? | i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 20 litres of water are required per household per month . at this rate , if there aren 10 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 20 * 10 = 200 ( m * n ) water available is 2000 ( p ) it will last 10 months ( p / m * n ) ans : c | a = 20 * 10
b = 2000 / a
|
a ) 1 , b ) - 1 , c ) 3 , d ) 12 , e ) 13 | d | multiply(6, const_2) | if x + | x | + y = 5 and x + | y | - y = 6 what is x + y = ? | if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 12 hence answer should be d | a = 6 * 2
|
a ) 87 , b ) 95 , c ) 102 , d ) 103 , e ) 115 | d | add(add(add(add(add(add(8, 1), add(add(8, 1), 1)), add(add(add(8, 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(add(8, 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(8, 1), 1), 1), 1), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1))) | a certain clock rings two notes at quarter past the hour , 4 notes at half past , and 6 notes at 3 - quarters past . on the hour , it rings 8 notes plus an additional number of notes equal to whatever hour it is . how many notes will the clock ring from 1 : 00 p . m . through 5 : 00 p . m . , including the rings at 1 : 00 and 5 : 00 ? | form 1 pm to 5 pm . excluding the actual hour chime we have 20 ( 1 pm ) + 20 ( 2 pm ) + 20 ( 3 pm ) + 20 ( 4 pm ) + 8 ( 5 pm ) = 88 now the hour chimes are 1 + 2 + 3 + 4 + 5 = 15 total = 88 + 15 = 103 answer d . | a = 8 + 1
b = 8 + 1
c = b + 1
d = a + c
e = 8 + 1
f = e + 1
g = f + 1
h = d + g
i = 8 + 1
j = i + 1
k = j + 1
l = k + 1
m = h + l
n = 8 + 1
o = n + 1
p = o + 1
q = p + 1
r = q + 1
s = m + r
t = 8 + 1
u = t + 1
v = u + 1
w = v + 1
x = 8 + 1
y = x + 1
z = y + 1
A = z + 1
B = w + A
C = 8 + 1
D = C + 1
E = D + 1
F = E + 1
G = B + F
H = 8 + 1
I = H + 1
J = I + 1
K = J + 1
L = G + K
M = s + L
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | subtract(const_100, multiply(divide(9, const_4), const_100)) | on an order of 9 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 10 dozens free . equivalent discount = 1 / 10 * 100 = 10 % . answer a ) 10 %" | a = 9 / 4
b = a * 100
c = 100 - b
|
a ) 13 , b ) 24 , c ) c is elder than a , d ) data inadequate , e ) none | a | multiply(13, const_1) | the total age of a and b is 13 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 13 â € ¹ = â € º a - c = 13 . answer a" | a = 13 * 1
|
a ) s 525 , b ) s 527 , c ) s 598 , d ) s 400 , e ) s 500 | d | multiply(5, divide(1200, add(add(5, 10), const_3))) | rs . 1200 is divided so that 5 times the first share , 10 times the 2 nd share and fifteen times third share amount to the same . what is the value of the second share ? | "a + b + c = 1200 5 a = 10 b = 15 c = x a : b : c = 1 / 5 : 1 / 10 : 1 / 15 = 3 : 2 : 1 2 / 6 * 1200 = rs 400 answer : d" | a = 5 + 10
b = a + 3
c = 1200 / b
d = 5 * c
|
a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 26.25 | e | multiply(const_3, divide(60, const_10)) | jackie has two solutions that are 4 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 4 percent solution will be required ? | "let a = amount of 4 % acid and b = amount of 12 % acid . now , the equation translates to , 0.04 a + . 12 b = . 05 ( a + b ) but a + b = 60 therefore . 04 a + . 12 b = . 05 ( 60 ) = > 4 a + 12 b = 300 but b = 60 - a therefore 4 a + 12 ( 60 - a ) = 300 = > 16 a = 420 hence a = 26.25 . answer : e" | a = 60 / 10
b = 3 * a
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | subtract(6, const_1) | if the product of 6 integers is negative , at most how many of the integers can be negative ? | "product of even number of - ve numbers is positive so it cant be 6 . answer : d" | a = 6 - 1
|
a ) 25.0 % , b ) 50.0 % , c ) 62.5 % , d ) 75.0 % , e ) 80.0 % | b | multiply(add(multiply(divide(subtract(power(multiply(const_2, const_5), const_2), 50), power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2))), multiply(divide(50, power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2)))), power(multiply(const_2, const_5), const_2)) | there is a 50 % chance jen will visit chile this year , while there is a 50 % chance that she will visit madagascar this year . what is the probability that jen will visit either chile or madagascar this year , but not both ? | "p ( chile and not madagascar ) = 0.5 * 0.5 = 0.25 p ( madagascar and not chile ) = 0.5 * 0.5 = 0.25 total probability = 0.25 + 0.25 = 0.5 = 50 % the answer is b ." | a = 2 * 5
b = a ** 2
c = b - 50
d = 2 * 5
e = d ** 2
f = c / e
g = 2 * 5
h = g ** 2
i = 50 / h
j = f * i
k = 2 * 5
l = k ** 2
m = 50 / l
n = 2 * 5
o = n ** 2
p = 50 / o
q = m * p
r = j + q
s = 2 * 5
t = s ** 2
u = r * t
|
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