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['a ) 5 / 3', 'b ) 9 / 5', 'c ) 9 / 25', 'd ) 3 / 5', 'e ) 25 / 9'] | c | divide(power(3, const_2), power(5, const_2)) | rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 3 / 5 , what is the ratio of rectangle a β s area to rectangle b β s area ? | the area of rectangle a is ab . c = 5 a / 3 and d = 5 b / 3 . the area of rectangle b is cd = 25 ab / 9 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 25 ab / 9 ) = 9 / 25 . the answer is c . | a = 3 ** 2
b = 5 ** 2
c = a / b
|
a ) 22 , b ) 14 , c ) 6 , d ) 8 , e ) 16 | b | add(divide(24, const_4), divide(24, const_3)) | j is faster than p . j and p each walk 24 km . sum of the speeds of j and p is 7 kmph . sum of time taken by them is 14 hours . then j speed is equal to | given j > p j + p = 7 , only options are ( 6 , 1 ) , ( 5 , 2 ) , ( 4 , 3 ) from the given options , if j = 4 the p = 3 . times taken by them = 244 + 243 = 14 answer : b | a = 24 / 4
b = 24 / 3
c = a + b
|
a ) 34545 , b ) 65657 , c ) 65567 , d ) 45677 , e ) 56782 | c | divide(power(2, 8), power(2, 31)) | 2 ^ 8 Γ 39 + 31 = ? | "256 * 39 + 31 65567 c" | a = 2 ** 8
b = 2 ** 31
c = a / b
|
a ) 5832.75 , b ) 5839.75 , c ) 5837.75 , d ) 5222.75 , e ) 5835.75 | a | divide(multiply(multiply(5, 35), multiply(2, const_1000)), multiply(const_1, const_60)) | a river 5 m deep and 35 m wide is flowing at the rate of 2 kmph , calculate the amount of water that runs into the sea per minute ? | "rate of water flow - 2 kmph - - 2000 / 60 - - 33.33 m / min depth of river - - 5 m width of river - - 35 m vol of water per min - - 33.33 * 5 * 35 - - - 5832.75 answer a" | a = 5 * 35
b = 2 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 11.7 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 30 % | a | divide(const_100, multiply(multiply(divide(10, const_100), divide(15, const_100)), const_100)) | on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 15 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "answer is a . this question is in the og and thus well explained by ets . those who exceed : x so x = 10 % + 0,15 x id est x = 11,7 %" | a = 10 / 100
b = 15 / 100
c = a * b
d = c * 100
e = 100 / d
|
a ) 288 , b ) 195 , c ) 881 , d ) 1277 , e ) 121 | b | multiply(23, multiply(54, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 23 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 23 = 345 m . let the length of the platform be x m . then , ( x + 345 ) / 36 = 15 = > x = 195 m . answer : b" | a = 54 * const_0_2778
b = 23 * a
|
['a ) 98 m', 'b ) 102 m', 'c ) 95 m', 'd ) 105 m', 'e ) 96 m'] | a | multiply(sqrt(4802), sqrt(2)) | the area of a square field is 4802 m ( power ) 2 the length of its diagonal is : | let the diagonal be d metres , then 1 / 2 d ( power ) 2 = 4802 d 2 = 9604 d = β 9604 d = 98 m answer is a . | a = math.sqrt(4802)
b = math.sqrt(2)
c = a * b
|
a ) 0 , b ) 1 , c ) 27 , d ) 0 , e ) can not be determined | d | divide(multiply(multiply(multiply(const_3, const_2), multiply(const_3, const_2)), multiply(const_3, const_2)), const_4) | for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.567 ] = 1 and [ 1.999 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | k = [ lcm of 80 and 45 ] * ( any integer ) however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer for value of k ( 60 ) * any integer unit value will be always zero . d | a = 3 * 2
b = 3 * 2
c = a * b
d = 3 * 2
e = c * d
f = e / 4
|
a ) 340 , b ) 335 , c ) 370 , d ) 360 , e ) 350 | e | add(100, 100) | in the faculty of reverse - engineering , 100 second year students study numeric methods , 100 second year students study automatic control of airborne vehicles and 25 second year students study them both . how many students are there in the faculty if the second year students are approximately 50 % of the total ? | "total number of students studying both are 100 + 100 - 25 = 175 ( subtracting the 25 since they were included in the both the other numbers already ) . so 50 % of total is 175 , so 100 % is 350 answer is e" | a = 100 + 100
|
a ) 7 / 30 , b ) 11 / 40 , c ) 17 / 50 , d ) 19 / 60 , e ) 33 / 80 | d | subtract(add(divide(3, 4), subtract(const_1, divide(3, 5))), divide(5, 6)) | the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 3 / 5 . if the probability of getting at least one contract is 5 / 6 , what is the probability that it will get both the contracts ? | let , a β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 3 / 5 ) = 2 / 5 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ) . = > 5 / 6 = ( 3 / 4 ) + ( 2 / 5 ) - p ( ab ) . = > p ( ab ) = 19 / 60 . hence , the required probability is 19 / 60 . the answer is d . | a = 3 / 4
b = 3 / 5
c = 1 - b
d = a + c
e = 5 / 6
f = d - e
|
a ) 52 kmph . , b ) 62 kmph . , c ) 72 kmph . , d ) 80 kmph . , e ) none | a | subtract(multiply(divide(280, 9), const_3_6), 60) | a man sitting in a train which is travelling at 60 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ? | "solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 60 ) kmph = 52 kmph . answer a" | a = 280 / 9
b = a * const_3_6
c = b - 60
|
a ) 200 , b ) 250 , c ) 100 , d ) 300 , e ) 220 | a | subtract(900, multiply(300, 2)) | a straight line in the xy - plane has slope 2 . on this line the x - coordinate of the point is 300 and y - coordinate is 900 then what is the y intercept of the plane ? | "eq of line = y = mx + c m = 2 x = 300 y = 300 * 2 + c , substitute y by 900 as given in question . 900 = 600 + c , c = 200 correct option is a" | a = 300 * 2
b = 900 - a
|
a ) 1260 , b ) 2100 , c ) 3600 , d ) 3200 , e ) 5200 | d | multiply(divide(add(multiply(multiply(3, 3), const_1000), const_100), 7), 2) | a marketing survey of anytown found that the ratio of trucks to sedans to motorcycles was 3 : 7 : 2 , respectively . given that there are 11,200 sedans in anytown , how many motorcycles are there ? | "let the total number of trucks = 3 x total number of sedans = 7 x total number of motorcycles = 2 x total number of sedans = 11200 = > 7 x = 11200 = > x = 1600 total number of motorcycles = 2 x = 2 * 1600 = 3200 answer d" | a = 3 * 3
b = a * 1000
c = b + 100
d = c / 7
e = d * 2
|
a ) 62.5 , b ) 65 , c ) 61 , d ) 64 , e ) 60 | a | divide(multiply(25, add(const_4, const_1)), const_2) | to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 25 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 25 x ) / ( 2 x / 5 ) = ( 25 x ) x 5 / 2 x = 125 / 2 = 62.5 answer is a ." | a = 4 + 1
b = 25 * a
c = b / 2
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | c | divide(subtract(multiply(divide(20, const_60), 5), multiply(divide(4, const_60), 5)), divide(4, const_60)) | the walker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at 20 miles per hour . the cyclist stops to wait for the hiker 5 minutes after passing her , while the walker continues to walk at her constant rate , how many minutes must the cyclist wait until the walker catches up ? | after passing the hiker the cyclist travels for 5 minutes at a rate of 20 miles / hour . in those 5 mins the cyclist travels a distance of 5 / 3 miles . in those 5 mins the hiker travels a distance of 1 / 3 miles . so the hiker still has to cover 4 / 3 miles to meet the waiting cyclist . the hiker will need 1 / 3 hours or 20 mins to cover the remaining 4 / 3 miles . so the answer is c . | a = 20 / const_60
b = a * 5
c = 4 / const_60
d = c * 5
e = b - d
f = 4 / const_60
g = e / f
|
a ) 98 , b ) 84 , c ) 90 , d ) 120 , e ) 135 | a | divide(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), subtract(divide(multiply(add(multiply(multiply(const_4, const_2), const_10), multiply(const_100, const_4)), const_3), 94), add(multiply(const_4, const_2), const_3))) | a train traveled the first d miles of its journey it an average speed of 60 miles per hour , the next d miles of its journey at an average speed of y miles per hour , and the final d miles of its journey at an average speed of 160 miles per hour . if the train β s average speed over the total distance was 94 miles per hour , what is the value of y ? | "average speed = total distance traveled / total time taken 3 d / d / 60 + d / y + d / 160 = 94 solving for d and y , 15 y = 11 y + 480 5 y = 480 y = 98 answer a" | a = 4 * 2
b = a * 10
c = 100 * 4
d = b + c
e = 4 * 2
f = e * 10
g = 100 * 4
h = f + g
i = h * 3
j = i / 94
k = 4 * 2
l = k + 3
m = j - l
n = d / m
|
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | d | subtract(subtract(1000, 1), subtract(add(add(divide(subtract(subtract(1000, const_2), const_2), const_2), 1), add(divide(subtract(subtract(1000, divide(1000, const_100)), divide(1000, const_100)), divide(1000, const_100)), 1)), add(divide(subtract(subtract(1000, const_10), const_10), const_10), 1))) | what is the total number of positive integers that are less than 1000 and that have no positive factor in common with 1000 other than 1 ? | "since 1000 = 2 ^ 3 * 5 ^ 3 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 500 odd numbers from 1 to 1000 . we then need to eliminate the 100 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 995 . there are a total of 500 - 100 = 400 such numbers between 1 and 1000 . the answer is d ." | a = 1000 - 1
b = 1000 - 2
c = b - 2
d = c / 2
e = d + 1
f = 1000 / 100
g = 1000 - f
h = 1000 / 100
i = g - h
j = 1000 / 100
k = i / j
l = k + 1
m = e + l
n = 1000 - 10
o = n - 10
p = o / 10
q = p + 1
r = m - q
s = a - r
|
a ) rs . 420000 , b ) rs . 180000 , c ) rs . 436800 , d ) rs . 504000 , e ) none of these | c | multiply(multiply(multiply(13000, add(const_1, divide(12, const_100))), divide(5, 2)), 12) | the monthly incomes of a and b are in the ratio 5 : 2 . b ' s monthly income is 12 % more than c ' s monthly income . if c ' s monthly income is rs . 13000 , then find the annual income of a ? | "b ' s monthly income = 13000 * 112 / 100 = rs . 14560 b ' s monthly income = 2 parts - - - - > rs . 14560 a ' s monthly income = 5 parts = 5 / 2 * 14560 = rs . 36400 a ' s annual income = rs . 36400 * 12 = rs . 436800 answer : c" | a = 12 / 100
b = 1 + a
c = 13000 * b
d = 5 / 2
e = c * d
f = e * 12
|
a ) 125 , b ) 126 , c ) 130 , d ) 143 , e ) 151 | d | add(add(add(multiply(3, 13), multiply(3, 14)), multiply(3, 17)), 11) | 3 * 13 + 3 * 14 + 3 * 17 + 11 = ? | "3 * 13 + 3 * 14 + 3 * 17 + 11 = 39 + 42 + 51 + 11 = 143 the answer is d ." | a = 3 * 13
b = 3 * 14
c = a + b
d = 3 * 17
e = c + d
f = e + 11
|
a ) 5 / 6 , b ) 3 / 4 , c ) 7 / 8 , d ) 1 / 8 , e ) 1 / 2 | c | add(add(divide(1, 4), divide(1, 2)), divide(1, 8)) | in mrs . susna ' s class , 1 / 4 of her students are getting an a , 1 / 2 are getting a b , 1 / 8 are getting a c , 1 / 12 are getting a d , and 1 / 24 are getting a f . what fraction of mrs . susna ' s class getting a passing grade of c or higher . | if mrs . susna ' s class has 24 students , the number of students receiving as , bs , or cs , are 6 , 12 , and 3 respectively . that means 21 / 24 of her students are receiving passing grades or 7 / 8 . ( answer : c ) | a = 1 / 4
b = 1 / 2
c = a + b
d = 1 / 8
e = c + d
|
['a ) 45 m 3', 'b ) 40 m 3', 'c ) 60 m 3', 'd ) 600 m 3', 'e ) 300 m 3'] | b | divide(multiply(2, multiply(20, 10)), 10) | the area of the house of a hall is 20 m 2 that of a longer wall 10 m 2 and of the shorter wall 8 m 2 , find the edge of the new cube ? | lb = 20 ; lh = 10 ; fh = 8 ( lbh ) 2 = 20 * 10 * 8 = > lbh = 40 m 3 answer : b | a = 20 * 10
b = 2 * a
c = b / 10
|
a ) 0 , b ) 4 , c ) 6 , d ) 7 , e ) 10 | e | add(10, 2) | there is a 45 cm line marked at each centimeter and an insect is placed at every centimeter . 9 frogs are trained to jump a constant distance . the first one jumps 2 cm in every leap , the second one jumps 3 cm and so on until the 9 th one jumps 10 cm in every leap and they eat any insect that is available at that spot . if all of them start from start line and finish the entire 45 cm , how many insects were still left after the race was over ? | "only the prime numbers greater than 10 and less than 45 were left . that is 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , and 43 . the total is 10 . the answer is e ." | a = 10 + 2
|
a ) 2500 , b ) 2800 , c ) 3500 , d ) 3600 , e ) 2050 | b | add(4032, divide(multiply(4032, 20), const_100)) | the present population of a town is 4032 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 4032 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 4032 / ( 1 + 20 / 100 ) ^ 2 = 4032 / ( 6 / 5 ) ^ 2 = 2800 ( approximately ) answer is b" | a = 4032 * 20
b = a / 100
c = 4032 + b
|
a ) 480 , b ) 240 , c ) 289 , d ) 270 , e ) 927 | b | multiply(24, 20) | a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 2 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 2 = 240 answer : b" | a = 24 * 20
|
a ) 10000 , b ) 11100 , c ) 15000 , d ) 12100 , e ) 14520 | d | add(10000, divide(multiply(10000, 10), const_100)) | the present population of a town is 10000 . population increase rate is 10 % p . a . find the population of town after 2 years ? | "p = 10000 r = 10 % required population of town = p ( 1 + r / 100 ) ^ t = 10000 ( 1 + 10 / 100 ) ^ 2 = 10000 ( 11 / 10 ) ^ 2 = 12100 answer is d" | a = 10000 * 10
b = a / 100
c = 10000 + b
|
a ) 87 , b ) 90 , c ) 92 , d ) 94 , e ) 97 | d | subtract(multiply(86, const_2), 82) | john has taken 4 ( 4 ) tests that have an average of 82 . in order to bring his course grade up to a β b β , he will need to have a final average of 86 . what will he need to average on his final two tests to achieve this grade ? | traditional method : total scored till now 82 * 4 = 328 total score to avg 86 in 6 tests = 86 * 6 = 516 total to be scored on 2 tests = 516 - 328 = 188 avg on 2 tests = 188 / 2 = 94 answer : d | a = 86 * 2
b = a - 82
|
a ) 47 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(38, multiply(2, 10)), 38), 2) | a car started running at a speed of 38 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | a 47 km the total distance covered by the car in the first 10 hours = 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 = sum of 10 terms in ap whose first term is 30 and last term is 56 = 10 / 2 [ 38 + 56 ] = 470 km . | a = 2 * 10
b = 38 + a
c = b + 38
d = c / 2
|
a ) 53 : 115 , b ) 55 : 115 , c ) 115 : 55 , d ) 5 : 115 , e ) 115 : 53 | a | divide(add(add(multiply(divide(1, add(1, 5)), 5), multiply(divide(3, add(3, 5)), 4)), multiply(divide(5, add(5, 7)), 5)), subtract(add(add(5, 4), 5), add(add(multiply(divide(1, add(1, 5)), 5), multiply(divide(3, add(3, 5)), 4)), multiply(divide(5, add(5, 7)), 5)))) | 3 containers a , b and c are having mixtures of milk and water in the ratio of 1 : 5 and 3 : 5 and 5 : 7 respectively . if the capacities of the containers are in the ratio of all the 3 containers are in the ratio 5 : 4 : 5 , find the ratio of milk to water , if the mixtures of all the 3 containers are mixed together . | assume that there are 500400 and 500 liters respectively in the 3 containers . then , we have , 83.33 , 150 and 208.33 liters of milk in each of the three containers . thus , the total milk is 441.66 liters . hence , the amount of water in the mixture is 1400 - 441.66 = 958.33 liters . hence , the ratio of milk to water is 441.66 : 958.33 = > 53 : 115 ( using division by . 3333 ) the calculation thought process should be ( 441 * 2 + 2 ) : ( 958 * 3 + 1 ) = 1325 : 2875 dividing by 25 = > 53 : 115 . answer a 53 : 115 | a = 1 + 5
b = 1 / a
c = b * 5
d = 3 + 5
e = 3 / d
f = e * 4
g = c + f
h = 5 + 7
i = 5 / h
j = i * 5
k = g + j
l = 5 + 4
m = l + 5
n = 1 + 5
o = 1 / n
p = o * 5
q = 3 + 5
r = 3 / q
s = r * 4
t = p + s
u = 5 + 7
v = 5 / u
w = v * 5
x = t + w
y = m - x
z = k / y
|
a ) 15 % , b ) 20 % , c ) 28 % , d ) 35 % , e ) 40 % | c | multiply(divide(subtract(const_1, divide(subtract(const_100, 22), const_100)), divide(subtract(const_100, 22), const_100)), const_100) | the length of a rectangle is reduced by 22 % . by what % would the width have to be increased to maintainthe original area ? | "sol . required change = ( 22 * 100 ) / ( 100 - 22 ) = 28 % c" | a = 100 - 22
b = a / 100
c = 1 - b
d = 100 - 22
e = d / 100
f = c / e
g = f * 100
|
a ) 25 / 6 , b ) 42 / 25 , c ) 6 / 5 , d ) 5 / 6 , e ) 25 / 36 | b | multiply(divide(divide(1, const_3.0), divide(1, 5)), divide(7, 5)) | if 5 x = 7 y and xy β 0 , what is the ratio of 1 / 5 * x to 1 / 6 * y ? | "5 x = 7 y = > x / y = 7 / 5 1 / 5 * x to 1 / 6 * y = x / y * 6 / 5 = ( 7 / 5 ) * ( 6 / 5 ) = 42 / 25 ans : b" | a = 1 / 3
b = 1 / 5
c = a / b
d = 7 / 5
e = c * d
|
a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 54.17 , d ) rs . 55.66 , e ) none of the above | d | divide(add(multiply(10, 50), multiply(5, 67)), add(10, 5)) | if 10 litres of an oil of rs . 50 per litres be mixed with 5 litres of another oil of rs . 67 per litre then what is the rate of mixed oil per litre ? | "50 * 10 = 500 67 * 5 = 335 835 / 15 = 55.66 answer : d" | a = 10 * 50
b = 5 * 67
c = a + b
d = 10 + 5
e = c / d
|
a ) 1035 , b ) 1040 , c ) 1042 , d ) 1045 , e ) 1050 | a | divide(multiply(45, add(45, const_1)), const_2) | the sum of first 45 natural numbers is : | "let sn = ( 1 + 2 + 3 + . . . + 45 ) . this is an a . p . in which a = 1 , d = 1 , n = 45 . sn = n [ 2 a + ( n - 1 ) d ] = 45 x [ 2 x 1 + ( 45 - 1 ) x 1 ] = 45 x 46 = ( 45 x 23 ) 2 2 2 = 45 x ( 20 + 3 ) = 45 x 20 + 45 x 3 = 900 + 135 = 1035 . shorcut method : sn = n ( n + 1 ) = 45 ( 45 + 1 ) = 1035 . 2 2 a" | a = 45 + 1
b = 45 * a
c = b / 2
|
a ) 42 % , b ) 48 % , c ) 54 % , d ) 60 % , e ) 66 % | c | divide(add(30, 60), multiply(multiply(const_5, const_5), const_4)) | there is a 30 % chance sandy will visit china this year , while there is a 60 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | p ( china and not malaysia ) = 0.3 * 0.4 = 0.12 p ( malaysia and not china ) = 0.6 * 0.7 = 0.42 total probability = 0.12 + 0.42 = 0.54 = 54 % the answer is c . | a = 30 + 60
b = 5 * 5
c = b * 4
d = a / c
|
a ) 25 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | a | multiply(divide(10.5, 14), const_100) | mike earns $ 14 per hour and phil earns $ 10.5 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "let it be x % less , then = 14 ( 1 - x / 100 ) = 10.5 1 - x / 100 = 10.5 / 14 x = 350 / 14 x = 25 % answer : a" | a = 10 / 5
b = a * 100
|
a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | e | subtract(2500, divide(add(2500, 500), 6)) | the total price of a basic computer and printer are $ 2500 . 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 / 6 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 / 6 of that new total price : p = 1 / 6 * $ 3,000 = $ 500 . plug this value in the first equation : c + 500 = $ 2,500 - - > c = $ 2000 answer : e . | a = 2500 + 500
b = a / 6
c = 2500 - b
|
a ) 10 , b ) 12.6 , c ) 22.5 , d ) 31.3 , e ) 40.8 | c | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(9, 4), subtract(9, const_2))), 4) | find the average of first 4 multiples of 9 ? | "average = ( 9 + 18 + 27 + 36 ) / 4 = 22.5 answer is c" | a = 4 + 1
b = 4 + 1
c = b + 2
d = a + c
e = 9 - 4
f = 9 - 2
g = e + f
h = d + g
i = h / 4
|
a ) 10 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 75 % | c | subtract(subtract(add(divide(multiply(divide(60, 30), 25), 10), 90), 60), 10) | company a ' s workforce consists of 10 percent managers and 90 percent software engineers . company b ' s workforce consists of 30 percent managers , 10 percent software engineers , and 60 percent support staff . the two companies merge , every employee stays with the resulting company , and no new employees are added . if the resulting company Γ s workforce consists of 25 percent managers , what percent of the workforce originated from company a ? | let say company a has x employes and b has y employees . now they merge and total no of employees = x + y employees . per the question company a ' s workforce consists of 10 percent managers and 90 percent software engineers . company b ' s workforce consists of 30 percent managers , 10 percent software engineers , and 60 percent support staff . we translate it into equation as follows : . 1 x + . 3 y = . 25 ( x + y ) = > x + 3 y = 2.5 ( x + y ) = > . 5 y = 1.5 x = > y = 3 x . now we know total employee = x + y . we need to find % age of x in total ( x + y ) ie x / ( x + y ) x 100 % = > x / ( 3 x + x ) [ substitute y = 3 x ] = > x / 4 x x 100 % = > . 25 x 100 % = > 25 % . answer : c | a = 60 / 30
b = a * 25
c = b / 10
d = c + 90
e = d - 60
f = e - 10
|
a ) a ) 30 , b ) b ) 31 , c ) c ) 32 , d ) d ) 33 , e ) e ) 67 | e | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 are neither multiples of 5 or 6 . | "to answer this q we require to know 1 ) multiples of 5 till 100 = 100 / 5 = 20 2 ) multiples of 6 till 100 = 100 / 6 = 16.66 = 16 add the two 20 + 16 = 36 ; subtract common terms that are multiple of both 5 and 6 . . lcm of 5 and 6 = 30 multiples of 30 till 100 = 100 / 30 = 3.3 = 3 so total multiples of 2 and 3 = 36 - 3 = 33 ans = 100 - 33 = 67 e" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 28 , b ) 27 , c ) 26 , d ) 25 , e ) 20 | e | add(add(2, 3), add(2, 3)) | a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 12 : 00 p . m . ? | "sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour . hence , they cross each other 20 times before 12 : 00 p . m . answer e" | a = 2 + 3
b = 2 + 3
c = a + b
|
a ) 8 , b ) 27 , c ) 81 , d ) 110 , e ) 125 | d | multiply(sqrt(divide(550, 3)), 3) | if n is a positive integer and n ^ 3 is a multiple of 550 , what is the least possible value of n ? | "cube of a number entails having 3 copies of the original prime factors . since 550 is a factor of n ^ 3 the least possible value of n is 5 ( 11 ) 2 = 110 the answer is d ." | a = 550 / 3
b = math.sqrt(a)
c = b * 3
|
a ) 40 , b ) 45 , c ) 48 , d ) 51 , e ) 36 | e | add(45, 30) | two goods trains each 375 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 375 + 375 = 750 m . required time = 750 * 6 / 125 = 36 sec . answer : option e" | a = 45 + 30
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a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | e | divide(power(divide(300, const_2), const_2), subtract(307.50, 300)) | if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 300 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 307.50 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 15,000 . now , put the value of x = 3000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 3000 + 5 % of 3000 = 3150 . 2 nd year , amount = 3150 + 5 % of 3150 = 3307.50 . we can see after 2 years interest = 3307.50 - 3000 = 307.50 . hence , it satisfies the question . hence e is the correct answer" | a = 300 / 2
b = a ** 2
c = 307 - 50
d = b / c
|
a ) 0.6 d , b ) 0.7 d , c ) 0.5 d , d ) 0.8 d , e ) 0.9 d | a | subtract(divide(subtract(const_100, 25), const_100), multiply(divide(subtract(const_100, 25), const_100), divide(20, const_100))) | a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 25 % . staff are allowed a further 20 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 25 - 20 + ( - 25 ) ( - 20 ) / 100 = - 40 sale price = d * ( 1 - 40 / 100 ) sale price = . 6 * d answer ( a )" | a = 100 - 25
b = a / 100
c = 100 - 25
d = c / 100
e = 20 / 100
f = d * e
g = b - f
|
a ) $ 130 , b ) $ 140 , c ) $ 150 , d ) $ 160 , e ) $ 170 | c | divide(108, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))) | what is the normal price of an article sold at $ 108 after two successive discounts of 10 % and 20 % ? | "0.8 * 0.9 * cost price = $ 108 cost price = $ 150 the answer is c ." | a = 100 - 10
b = a / 100
c = 100 - 20
d = c / 100
e = b * d
f = 108 / e
|
a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 15 | e | divide(8, choose(16, 2)) | kim has 8 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 16 shoes , what is the probability that she will select 2 shoes of the same color ? | "total pairs = 16 c 2 = 120 ; same color pairs = 8 c 1 * 1 c 1 = 8 ; prob = 1 / 15 ans e" | a = math.comb(16, 2)
b = 8 / a
|
a ) 2 , b ) 8 , c ) 4 , d ) 25 , e ) 26 | c | divide(16, multiply(const_10, const_2)) | how many factors does 16 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 2 * 2 = 4 answer c ." | a = 10 * 2
b = 16 / a
|
a ) 91 , b ) 98 , c ) 105 , d ) 112 , e ) 119 | a | divide(multiply(14, subtract(14, const_1)), const_2) | there are 14 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "14 c 2 = 91 the answer is a ." | a = 14 - 1
b = 14 * a
c = b / 2
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | multiply(divide(subtract(divide(12, 30), divide(5, 15)), divide(5, 15)), const_100) | a technology company made a $ 5 million profit on its first $ 15 million in sales and a $ 12 million profit on its next $ 30 million in sales . by what percent did the ratio of profit to sales increase from the first $ 15 million in sales to the next $ 30 million in sales ? | solution : this is a percent increase problem . we will use the formula : percent change = ( new β old ) / old x 100 to calculate the final answer . we first set up the ratios of profits to sales . the first ratio will be for the first 15 million in sales , and the second ratio will be for the next 30 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 15 million profit / sales = 5 / 15 = 1 / 3 next 30 million profit / sales = 12 / 30 = 2 / 5 we can simplify each ratio by multiplying each by the lcm of the two denominators , which is 15 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 15 million royalties / sales = ( 1 / 3 ) x 15 = 5 next 30 million royalties / sales = ( 2 / 5 ) x 15 = 6 we can plug 5 and 6 into our percent change formula : ( new β old ) / old x 100 [ ( 6 β 5 ) / 5 ] x 100 1 / 5 x 100 , so a 20 % increase . answer b . | a = 12 / 30
b = 5 / 15
c = a - b
d = 5 / 15
e = c / d
f = e * 100
|
a ) 3.5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | b | divide(subtract(63, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if the resultant is trebled , it becomes 63 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 63 2 x = 12 = > x = 6 answer : b" | a = 3 * 9
b = 63 - a
c = 3 * 2
d = b / c
|
a ) 4 / 14 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | a | divide(choose(4, 4), choose(add(const_2.0, 3), const_2)) | a bag holds 4 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "given : 4 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 4 red and 3 green marbles after one red has been picked , we need to pick 1 red from a total of 3 red and 3 green marbles . p ( both red ) = ( 4 / 7 ) * ( 3 / 6 ) = 4 / 14 option a" | a = math.comb(4, 4)
b = 2 + 0
c = math.comb(b, 2)
d = a / c
|
a ) 60 m , b ) 70 m , c ) 100 m , d ) 120 m , e ) 150 m | e | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and speeds past a platform 100 meters long in 25 seconds . its length in meters is : | "let the length of the train be x metres and its speed be y metres / sec . then x / y = 15 y = x / 15 x + 100 / 25 now , ( x + 100 ) / 25 = x / 15 x = 150 m answer : e" | a = 2 - 1
b = 100 * a
|
a ) $ 500 , b ) $ 650 , c ) $ 420 , d ) $ 375 , e ) $ 625 | c | divide(add(add(add(add(300, 150), 750), 400), 500), 5) | a cab driver 5 days income was $ 300 , $ 150 , $ 750 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 300 + 150 + 750 + 400 + 500 ) / 5 = 420 answer is c" | a = 300 + 150
b = a + 750
c = b + 400
d = c + 500
e = d / 5
|
a ) 154 , b ) 124 , c ) 153 , d ) 163 , e ) 183 | d | divide(subtract(12401, 13), 76) | if 12401 is divided by any no . then quotient is 76 and remainder is 13 . what is divisor ? | divisor = ( dividend - remainder ) / quotient = ( 12401 - 13 ) / 76 = 12388 / 76 = 163 divisor = 163 answer d | a = 12401 - 13
b = a / 76
|
a ) 70 , b ) 50 , c ) 62 , d ) 49 , e ) 50 | b | add(add(12, 17), add(11, 10)) | in a games hour 4 different types of players came to the ground ? cricket 12 , hokey 17 , football 11 , softball 10 . how many players are present in the ground ? | "total number of players = 12 + 17 + 11 + 10 = 50 answer is b" | a = 12 + 17
b = 11 + 10
c = a + b
|
a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 2200 , e ) none of these | d | multiply(55, 40) | a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 55 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 β 55 = rs . 150 cp on 40 metres = 150 x 40 = rs . 6000 profit earned on 40 metres cloth = rs . 8200 β rs . 6000 = rs . 2200 . answer : option d | a = 55 * 40
|
a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10 | b | subtract(subtract(add(subtract(90, 11), 20), 44), 20) | in a neighborhood having 90 households , 11 did not have either a car or a bike . if 20 households had a both a car and a bike and 44 had a car , how many had bike only ? | "{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 20 + 11 - - > { bike } = 55 - - > # those who have bike only is { bike } - { both } = 55 - 20 = 35 . answer : b ." | a = 90 - 11
b = a + 20
c = b - 44
d = c - 20
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | add(add(const_4, const_3), const_2) | what is the units digit of 32 ! + 50 ! + 2 ! + 4 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 2 + 4 = 6 the units digit is 6 . the answer is d ." | a = 4 + 3
b = a + 2
|
a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 480,000 , e ) $ 540,000 | d | divide(const_3600, const_10) | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 60,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 60,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 42 years , the investment will get doubled 42 / 14 = 3 times . so the amount invested will get doubled thrice . so , 60000 * 2 ^ 3 = 480000 hence , the answer is d ." | a = 3600 / 10
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a ) 13 , b ) 12 , 45 , c ) 1 , 25 , d ) all of these , e ) 12 , 35 | e | multiply(multiply(3, 4), add(5, multiply(3, const_10))) | if a and b are multiples of 3 then which are all the multiples of 3 1 ) a / b 2 ) a ^ b 3 ) a + b 4 ) a - b 5 ) a * b | a * b a ^ b a + b a - b are multiples of 3 except a / b answer : e | a = 3 * 4
b = 3 * 10
c = 5 + b
d = a * c
|
a ) 65 , b ) 66 , c ) 67 , d ) 68 , e ) 69 | d | subtract(add(75, 83), subtract(const_100, 10)) | 75 persons major in physics , 83 major in chemistry , 10 not at major in these subjects u want to find number of students majoring in both subjects | consider the number of total students = n ( t ) = 100 number of persons major in physics = n ( p ) = 75 number of persons major in chemistry = n ( c ) = 83 according to the question ; 10 not at major in these subjects = n ( p ' β© c ' ) = 10 n ( p ' β© c ' ) = n ( p u c ) ' = 10 n ( p u c ) ' = n ( t ) - n ( p u c ) 10 = 100 - n ( p u c ) n ( p u c ) = 90 n ( p u c ) = n ( p ) + n ( c ) - n ( p β© c ) 90 = 75 + 83 - n ( p β© c ) n ( p β© c ) = 158 - 90 n ( p β© c ) = 68 number of students majoring in both subjects will be 68 answer : d | a = 75 + 83
b = 100 - 10
c = a - b
|
a ) 14 , b ) 47 , c ) 54 , d ) 56 , e ) 240 | d | add(add(divide(320, 40), divide(320, 40)), 40) | frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 320 square feet . how many feet offence does frank use ? | "area = length x breadth 320 = 40 x breadth so , breadth = 8 units fencing required is - breadth + breadth + length 8 + 8 + 40 = > 56 feet answer must be ( d ) 56" | a = 320 / 40
b = 320 / 40
c = a + b
d = c + 40
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | power(add(multiply(4, 2), 2), 2) | if a is a positive integer , and if the units digit of a ^ 2 is 4 and the units digit of ( a + 1 ) ^ 2 is 9 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 4 , then the units digit of a is either 2 or 8 . if the units digit of ( a + 1 ) ^ 2 is 9 , then the units digit of a + 1 is either 3 or 7 . to satisfy both conditions , the units digit of a must be 2 . then a + 2 has the units digit of 4 , thus the units digit of ( a + 2 ) ^ 2 will be 6 . the answer is d ." | a = 4 * 2
b = a + 2
c = b ** 2
|
a ) 0.22598 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.63435 | a | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 54,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 354000 x \ therefore required percentage = \ inline \ frac { 800 x } { 354000 } \ times 100 = 0.22598 a" | a = 60 / 100
b = 480 / a
c = 3 * 100
d = c * 1000
e = 4 * 10
f = e + 4
g = f * 1000
h = d + g
i = b / h
j = i * 100
|
a ) 78 % , b ) 79 % , c ) 80 % , d ) 81 % , e ) 82 % | a | divide(add(multiply(15, 70), multiply(10, 90)), 25) | if 15 students in a class average 70 % on an exam and 10 students average 90 % on the same exam , what is the average in percent for all 25 students ? | ( 15 * 70 + 10 * 90 ) / 25 = 78 % the answer is a . | a = 15 * 70
b = 10 * 90
c = a + b
d = c / 25
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | divide(subtract(multiply(90, 8), multiply(87, 8)), subtract(92, 90)) | the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students β grades was 87 , and the average of the female students β grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 87 thus m = 696 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 696 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 12 answer : e" | a = 90 * 8
b = 87 * 8
c = a - b
d = 92 - 90
e = c / d
|
a ) after 20 hours , b ) after 115 hours , c ) after 115 minutes , d ) after 20 minutes , e ) after 30 minutes | b | multiply(115, subtract(5, 4)) | ajith and rana walk around a circular course 115 km in circumference , starting together from the same point . if they walk at speed of 4 and 5 kmph respectively , in the same direction , when will they meet ? | rana is the faster person . he gains 1 km in 1 hour . so rana will gain one complete round over ajith in 115 hours . i . e . they will meet after 115 hours . answer : b | a = 5 - 4
b = 115 * a
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a ) 20 hrs , b ) 21 hrs , c ) 22 hrs , d ) 23 hrs , e ) 24 hrs | b | add(17, log(16)) | two pipes a and b fill at a certain rate . b is filled at 1020 , 4080 ( 10 in 1 hour , 20 in 2 hours , 40 in 3 hrs and so on ) . if 1 / 16 of b is filled in 17 hrs , what time it will take to get completely filled ? | 1 / 16 in 17 hrs so 1 / 8 in 18 hrs 1 / 4 in 19 hrs 1 / 2 in 20 hrs and complete tank in 21 hrs answer : b | a = math.log(16)
b = 17 + a
|
a ) 3 : 5 , b ) 4 : 5 , c ) 5 : 5 , d ) 4 : 3 , e ) 2 : 1 | b | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | two numbers are respectively 20 % and 50 % more than a third number . the ratio of the two numbers is : | "third no is x first no - 120 % of x = 6 x / 5 second no = 150 x / 100 = 3 x / 2 ration of first nos = 6 x / 5 : 3 x / 2 = 12 x : 15 x = 4 : 5 answer b" | a = 20 + 100
b = 50 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) $ 603 , b ) $ 624 , c ) $ 625 , d ) $ 626 , e ) $ 627 | b | add(divide(multiply(multiply(26, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(26, const_1000), subtract(const_1, divide(10, const_100))))) | a car is purchased on hire - purchase . the cash price is $ 26 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ? | "explanation : cash price = $ 26 000 deposit = 10 % Γ£ β $ 26 000 = $ 2600 loan amount = $ 26000 Γ’ Λ β $ 2600 number of payments = 60 = $ 23400 i = p * r * t / 100 i = 14040 total amount = 23400 + 14040 = $ 37440 regular payment = total amount / number of payments = 624 answer : b" | a = 26 * 1000
b = 10 / 100
c = 1 - b
d = a * c
e = d / 60
f = 12 / 100
g = f / 12
h = 26 * 1000
i = 10 / 100
j = 1 - i
k = h * j
l = g * k
m = e + l
|
a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | c | divide(15, 0.129) | a certain industrial loom weaves 0.129 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | "let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.129 : 15 : : 1 : x = > 0.129 * x = 15 = > x = 15 / 0.129 = > x = 116 answer : c" | a = 15 / 0
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a ) 24 , b ) 60 , c ) 28 , d ) 76 , e ) 21 | a | divide(multiply(multiply(8, 4), 6), const_2) | a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 6 rows or 4 rows or 8 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 6 , 4,8 ) = 24 answer : a" | a = 8 * 4
b = a * 6
c = b / 2
|
a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 8 / 15 , e ) 11 / 30 | d | divide(add(floor(divide(30, 2)), divide(30, 19)), 30) | a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 2 or 19 ? | "number of multiples of 2 from 1 through 30 = 30 / 2 = 15 number of multiples of 19 from 1 through 30 = 30 / 19 = 1 number of multiples of 2 and 19 both from 1 through 30 = number of multiples of 19 * 2 ( = 38 ) = 0 total favourable cases = 15 + 1 - 0 = 16 probability = 16 / 30 = 8 / 15 answer : option d" | a = 30 / 2
b = math.floor(a)
c = 30 / 19
d = b + c
e = d / 30
|
a ) 51 : 52 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | c | divide(add(const_100, 6), add(const_100, 8)) | the cash difference between the selling prices of an article at a profit of 6 % and 8 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 106 % of x ) / ( 108 % of x ) = 106 / 108 = 53 / 54 = 53 : 54 . answer : c" | a = 100 + 6
b = 100 + 8
c = a / b
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a ) 41 , b ) 44 , c ) 45 , d ) 47 , e ) 50 | c | multiply(multiply(30, 5), divide(3, 10)) | working simultaneously and independently at an identical constant rate , 30 machines of a certain type can produce a total of x units of product p in 3 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 5 x units of product p in 10 days ? | "the rate of 30 machines is rate = job / time = x / 3 units per day - - > the rate of 1 machine 1 / 30 * ( x / 3 ) = x / 90 units per day ; now , again as { time } * { combined rate } = { job done } then 10 * ( m * x / 90 ) = 5 x - - > m = 45 . answer : c ." | a = 30 * 5
b = 3 / 10
c = a * b
|
a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 435 | e | multiply(subtract(30, const_1), divide(30, const_2)) | 30 men shake hands with each other . maximum no of handshakes without cyclic handshakes . | "or , if there are n persons then no . of shakehands = nc 2 = 30 c 2 = 435 answer : e" | a = 30 - 1
b = 30 / 2
c = a * b
|
a ) 14 , b ) 18 , c ) 21 , d ) 22 , e ) 27 | b | divide(336, divide(subtract(448, 336), 6)) | a car traveled 448 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 448 / h = > 336 / c = 448 / ( c + 6 ) = > 336 c + 336 * 6 = 448 c = > c = 336 * 6 / 112 = 18 answer b ." | a = 448 - 336
b = a / 6
c = 336 / b
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a ) 3.33 % , b ) 5.93 % , c ) 4.33 % , d ) 5.33 % , e ) 6.33 % | d | multiply(divide(divide(subtract(950, 750), 750), 5), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 950 in 5 years ? | "200 = ( 750 * 5 * r ) / 100 r = 5.33 % answer : d" | a = 950 - 750
b = a / 750
c = b / 5
d = c * 100
|
a ) 3.0 , b ) 3.36 , c ) 24.34 , d ) 25.0 , e ) 31.36 | c | divide(multiply(28, const_100), add(const_100, 15)) | from the sale of sleeping bags , a retailer made a gross profit of 15 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | "cost price * 1.15 = selling price - - > cost price * 1.15 = $ 28 - - > cost price = $ 24.34 . answer : c ." | a = 28 * 100
b = 100 + 15
c = a / b
|
a ) 96 , b ) 94 , c ) 86 , d ) 74 , e ) 110 | a | divide(40, 240) | find 40 % of 240 | "we know that r % of m is equal to r / 100 Γ m . so , we have 40 % of 240 40 / 100 Γ 240 = 96 answer : a" | a = 40 / 240
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a ) 7 , b ) 8 , c ) 10 , d ) 24 , e ) 15 | d | divide(70, divide(add(3.5, 2.5), const_2)) | sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 3.5 per pound , and german chocolate cost $ 2.5 per pound . if the total the she spent on chocolate was $ 70 and both types of chocolate were purchased in whole number of pounds , how many total pounds of chocolate she purchased ? | if there were all the expensive ones , 3.5 . . . . there would be 70 / 3.5 or 20 of them but since 2.5 $ ones are also there , answer has to be > 20 . . . . if all were 2.5 $ ones , there will be 70 / 2.5 or 28 . . so only 24 is left ans d . . | a = 3 + 5
b = a / 2
c = 70 / b
|
a ) 750 , b ) 850 , c ) 900 , d ) 980 , e ) 950 | b | add(add(add(340, 170), 170), 170) | two men start from opposite banks of a river . they meet 340 meters away from one of the banks on forward journey . after that they meet at 170 meters from the other bank of the river on their backward journey . what will be the width of the river ( in meters ) ? | let p , q are two persons then their speeds be a , b - > m / hr x - - - - 340 - - - - - - - - - - - - - - - - - - - - - - | - - - ( d - 340 ) - - - - - - - - - y in forward journey time taken by p = 340 / a ; time taken by q = d - 340 / b ; in forward journey , time taken by p = time taken by q so 340 / a = d - 340 / b a / b = 340 / d - 340 - - - - - - - - - - - - ( 1 ) after that they continue their journeyand reach other banks now in backward journey , x - - - - - - - - - - - - - - - ( d - 170 ) - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - 170 - - - - - - y distance travelled by p from meet 1 to meet 2 = ( d - 340 ) + 170 distance travelled by p from meet 1 to meet 2 = 340 + ( d - 170 ) now time taken by p = time taken by q ( d - 340 ) + 170 / a = 340 + ( d - 170 ) / b ( d - 340 ) + 170 / 340 + ( d - 170 ) = a / b - - - - - - - - - ( 2 ) from 1 & 2 340 / ( d - 340 ) = ( d - 340 ) + 170 / 340 + ( d - 170 ) 340 / ( d - 340 ) = d - 170 / d + 170 340 d + 57800 = d ^ 2 - 340 d - 170 d + 57800 340 d = d ^ 2 - 510 d d ^ 2 - 850 d = 0 d ( d - 850 ) = 0 d = 0 ord = 850 so width of the river is 850 answer : b | a = 340 + 170
b = a + 170
c = b + 170
|
a ) 4 , b ) 7 , c ) 8 , d ) 2 , e ) 3 | d | divide(subtract(add(multiply(7, 7), 5), 34), 10) | if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 34 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 34 ) / 10 = ( 54 - 34 ) / 10 = 2 answer : option d" | a = 7 * 7
b = a + 5
c = b - 34
d = c / 10
|
a ) 96 kmph , b ) 64 kmph , c ) 52 kmph , d ) 86 kmph , e ) 76 kmph | b | multiply(divide(160, 9), const_3_6) | a 160 meter long train crosses a man standing on the platform in 9 sec . what is the speed of the train ? | "s = 160 / 9 * 18 / 5 = 64 kmph answer : b" | a = 160 / 9
b = a * const_3_6
|
a ) 625 , b ) 827 , c ) 657 , d ) 726 , e ) 634 | a | inverse(multiply(power(divide(4, const_100), 2), 1)) | the difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4 % per annum is re . 1 . the sum ( in rs . ) is ? | "let the sum be rs . x . then , [ x ( 1 + 4 / 100 ) 2 - x ] = ( 676 / 625 x - x ) = 51 / 625 x s . i . = ( x * 4 * 2 ) / 100 = 2 x / 25 51 x / 625 - 2 x / 25 = 1 or x = 625 . answer : a" | a = 4 / 100
b = a ** 2
c = b * 1
d = 1/(c)
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(42.5, 0.85) | how many pieces of 0.85 metres can be cut from a rod 42.5 metres long ? | "number of pieces = 42.5 / 0.85 = 42.50 / 0.85 = 4250 / 85 = 50 . answer : c" | a = 42 / 5
|
a ) 300 , b ) 400 , c ) 266 , d ) 99 , e ) 121 | b | divide(160, divide(160, const_100)) | 60 % of a number is added to 160 , the result is the same number . find the number ? | ": ( 60 / 100 ) * x + 160 = x 2 x = 800 x = 400 answer : b" | a = 160 / 100
b = 160 / a
|
a ) 125,150 , b ) 5124,4515 , c ) 4150,2490 , d ) 3250,4510 , e ) 1254,3210 | c | multiply(divide(1660, 7.5), divide(2, subtract(const_1, divide(1660, 7.5)))) | difference of 2 numbers is 1660 . if 7.5 % of one number is 12.5 % of the other number , find the 2 numbers ? | "let the numbers be x and y 7.5 % of x = 12.5 % of y x = 125 y / 75 = 5 y / 3 x - y = 1660 5 y / 3 - y = 1660 y = 2490 x = 5 y / 3 = 4150 answer is c" | a = 1660 / 7
b = 1660 / 7
c = 1 - b
d = 2 / c
e = a * d
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | e | add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5) | an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 2,395 , what was the total number of pairs of contact lenses that she sold ? | "( x + 5 ) * 150 + x * 85 = 2395 = > x = 7 total lens = 7 + ( 7 + 5 ) = 19 answer e" | a = 4 * 100
b = 5 * 10
c = a + b
d = c + 5
e = d + 1000
f = 150 * 5
g = e - f
h = 150 + 85
i = g / h
j = i * 2
k = j + 5
|
a ) 35 , b ) 50 , c ) 62 , d ) 75 , e ) 78 | a | multiply(divide(multiply(105, 1), add(75, 105)), const_60) | cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 1 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 1 hrs therefore , 1 = d / 75 + d / 105 lcm of 75 and 105 = 525 1 = d / 75 + d / 105 = > 1 = 7 d / 525 + 5 d / 525 = > d = 525 / 12 km therefore , t 1 = d / 75 = > t 1 = 525 / ( 12 x 75 ) = > t 1 = ( 7 x 60 ) / 12 - - in minutes = > t 1 = 35 minutes . a" | a = 105 * 1
b = 75 + 105
c = a / b
d = c * const_60
|
a ) 20 , b ) 21 , c ) 26 , d ) 28 , e ) 30 | b | sqrt(add(multiply(131, const_2), 179)) | sum of the squares of 3 no . ' s is 179 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 179 + 2 * 131 a + b + c = β 441 = 21 b" | a = 131 * 2
b = a + 179
c = math.sqrt(b)
|
a ) 16 , b ) 24 , c ) 20 , d ) 32 , e ) none of these | a | divide(subtract(multiply(divide(20, const_100), 30), 2), divide(25, const_100)) | if 20 % of 30 is greater than 25 % of a number by 2 , the number is : | explanation : = > 20 / 100 * 30 - 25 / 100 * x = 2 = > x / 4 = 4 so x = 16 answer : a | a = 20 / 100
b = a * 30
c = b - 2
d = 25 / 100
e = c / d
|
a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | c | floor(divide(subtract(subtract(5000, const_1), subtract(add(floor(divide(5000, 15)), floor(divide(5000, 23))), floor(divide(5000, lcm(15, 23))))), const_1000)) | how many positive integers less than 5000 are evenly divisible by neither 15 nor 23 ? | integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 23 5000 / 23 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 23 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 23 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4521 answer c . | a = 5000 - 1
b = 5000 / 15
c = math.floor(b)
d = 5000 / 23
e = math.floor(d)
f = c + e
g = math.lcm(15, 23)
h = 5000 / g
i = math.floor(h)
j = f - i
k = a - j
l = k / 1000
m = math.floor(l)
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(const_1, add(divide(const_1, 12), divide(divide(const_1, 12), const_3))) | p alone can complete a job in 12 days . the work done by q alone in one day is equal to one - half of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 12 q ' s rate is 1 / 24 the combined rate is 1 / 12 + 1 / 24 = 1 / 8 if they work together , the job will take 8 days . the answer is d ." | a = 1 / 12
b = 1 / 12
c = b / 3
d = a + c
e = 1 / d
|
a ) 15 , b ) 18 , c ) 21 , d ) 23 , e ) 25 | a | divide(subtract(460, multiply(22, divide(subtract(multiply(460, const_2), 560), subtract(multiply(22, const_2), 8)))), 16) | suzie β s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 460 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 560 . how much more do pairs of boots cost than pairs of shoes at suzie β s discount footwear ? | let x be pair of shoes and y be pair of boots . 22 x + 16 y = 460 . . . eq 1 8 x + 32 y = 560 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 920 8 x = 560 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 560 . . . then we get 32 y = 480 then y = 25 differenece between x and y is 15 answer : a | a = 460 * 2
b = a - 560
c = 22 * 2
d = c - 8
e = b / d
f = 22 * e
g = 460 - f
h = g / 16
|
a ) 16 . , b ) 8 . , c ) 7 , d ) 2 . , e ) - 2 | e | subtract(add(4, 3), 9) | if ( a + b ) = 4 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 4 = > a = 4 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 4 - b + 3 - c = > 7 - ( b + c ) = > 7 - 9 = - 2 . option e . . ." | a = 4 + 3
b = a - 9
|
a ) 3 , b ) 5 , c ) 296297 , d ) 888891 , e ) 2666673 | b | power(add(8, const_4), const_4) | the difference between a 8 digit number and the number formed by reversing its digit is not a multiple of | "another approach is to test a number . let ' s say the original number is 12000002 so , the reversed number is 20000021 the difference = 20000021 - 12000002 = 8000019 no check the answer choices 8000019 is a multiple of 3,296297 , 888891,2666673 5 is not a multiple of 8000019 answer ; b" | a = 8 + 4
b = a ** 4
|
a ) 11 years , b ) 10 years , c ) 18 years , d ) 189 years , e ) 28 years | b | divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 = > x = 5 hence , b ' s age = 2 x = 10 years . answer : b" | a = 27 - 2
b = a * 2
c = 4 + 1
d = b / c
|
a ) 6630 , b ) 6650 , c ) 6560 , d ) 6530 , e ) none of these | a | divide(multiply(add(15, divide(51, 216)), 1100), const_100) | 15 51 216 1100 ? 46452 | "the format of the series 15 * 3 + 6 = 51 51 * 4 + 12 = 216 216 * 5 + 20 = 1100 1100 * 6 + 30 = 6630 6630 * 7 + 42 = 46452 answer : a" | a = 51 / 216
b = 15 + a
c = b * 1100
d = c / 100
|
a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 3 | e | divide(divide(18, const_3), const_3) | if a * b denotes the greatest common divisor of a and b , then ( ( 12 * 15 ) * ( 18 * 24 ) ) = ? | "the greatest common divisor of 12 and 15 is 3 . hence 12 * 15 = 3 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 24 is 6 . hence 18 * 24 = 6 . hence ( ( 12 * 16 ) * ( 18 * 24 ) ) = 3 * 6 . the greatest common divisor of 3 and 6 is 3 . answer ; e ." | a = 18 / 3
b = a / 3
|
a ) 315 , b ) 345 , c ) 325 , d ) 375 , e ) none of them | d | multiply(multiply(const_100.0, divide(12, 1260)), 3) | what annual installment will discharge a debt of rs . 1260 due in 3 years at 12 % simple interest ? | "let each installment be rs . x then , ( x + ( ( x * 12 * 1 ) / 100 ) ) + ( x + ( ( x * 12 * 2 ) / 100 ) ) + x = 1260 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1260 Γ― Ζ βΊ ( 28 x + 31 x + 25 x ) = ( 1260 * 25 ) x = ( 1260 * 25 ) / 84 = rs . 375 . therefore , each installment = rs . 375 . answer is d ." | a = 12 / 1260
b = 100 * 0
c = b * 3
|
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