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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | divide(log(45), log(add(const_4, const_1))) | for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 45 true ? | "answer b ( a + b ) ( a - b ) = 45 3 cases for ( a + b ) , ( a - b ) 45 , 1 15 , 3 9 , 5 answer b" | a = math.log(45)
b = 4 + 1
c = math.log(b)
d = a / c
|
a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 5 | d | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | if the number 892 , 142,24 x is divisible by 11 , what must be the value of x ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 142,24 x sum of digits at odd places = 8 + 2 + 4 + 2 + x = 16 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 4 = 16 ( ii ) ( i ) - ( ii ) = 16 + x - 16 = x - 0 hence x should be = 0 to make this a multiple of 11 ( 0 ) option d" | a = 3 * 10
b = a + 1
c = b * 2
d = c * 4
e = d * 11
|
a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio w of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } = ( 1 / 2 ) ^ 2 = 1 / 4 therefore w , w _ r = { w _ o } / 4 = 8 / 4 = 2 litres . b" | a = 3 + 2
b = 3 / a
c = 20 - 10
d = b * c
e = d + 10
f = e / 20
g = 20 - 10
h = f * g
i = h + 10
j = 3 + 2
k = 2 / j
l = 20 - 10
m = k * l
n = m / 20
o = 20 - 10
p = n * o
q = i / p
|
a ) 68.99 % , b ) 66.55 % , c ) 91.23 % , d ) 77.77 % , e ) 87.37 % | e | multiply(const_100, divide(divide(multiply(add(50, const_100), 60), const_100), add(const_100, 3))) | of the families in city x in 1992 , 60 percent owned a personal computer . the number of families in city x owning a computer in 1993 was 50 percent greater than it was in 1992 , and the total number of families in city x was 3 percent greater in 1993 than it was in 1992 . what percent of the families in city x owned a personal computer in 1993 ? | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 60 number of families owning computer in 1998 = 60 * 150 / 100 = 90 number of families in 1998 = 103 the percentage = 90 / 103 * 100 = 87.37 % . answer : e" | a = 50 + 100
b = a * 60
c = b / 100
d = 100 + 3
e = c / d
f = 100 * e
|
a ) 2.65 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | a | add(divide(sqrt(1.21), sqrt(0.81)), divide(sqrt(1), sqrt(0.49))) | find the value of ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.00 ) / ( √ 0.49 ) is | ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.00 ) / ( √ 0.49 ) 11 / 9 + 10 / 7 = > 2.65 answer is a | a = math.sqrt(1)
b = math.sqrt(0)
c = a / b
d = math.sqrt(1)
e = math.sqrt(0)
f = d / e
g = c + f
|
a ) 333 , b ) 383 , c ) 402 , d ) 433 , e ) 483 | c | divide(add(1000, add(add(multiply(multiply(1000, 3), divide(10, const_100)), multiply(multiply(multiply(1000, 3), divide(10, const_100)), divide(10, const_100))), const_1)), divide(add(add(multiply(multiply(1000, 3), divide(10, const_100)), multiply(multiply(multiply(1000, 3), divide(10, const_100)), divide(10, const_100))), const_1), const_100)) | louie takes out a 3 - month loan of $ 1000 . the lender charges him 10 % interest per month compounded monthly . the terms of the loan state that louie must repay the loan in 3 equal monthly payments . to the nearest dollar , how much does louis have to pay each month ? | after 1 st month : ( 1000 ) ( 1.1 ) - x = 1100 - x after 2 nd month : ( 1100 - x ) ( 1.1 ) - x = 1210 - 2.21 x after 3 rd month : ( 1210 - 2.21 x ) ( 1.1 ) - x = 1331 - 3.31 x now , the amount after the last payment in 3 rd month must bring the total to 0 . hence : 1331 - 3.31 x = 0 x = 1331 / 3.31 = 402.11 the answer is c | a = 1000 * 3
b = 10 / 100
c = a * b
d = 1000 * 3
e = 10 / 100
f = d * e
g = 10 / 100
h = f * g
i = c + h
j = i + 1
k = 1000 + j
l = 1000 * 3
m = 10 / 100
n = l * m
o = 1000 * 3
p = 10 / 100
q = o * p
r = 10 / 100
s = q * r
t = n + s
u = t + 1
v = u / 100
w = k / v
|
a ) 299 m , b ) 777 m , c ) 200 m , d ) 100 m , e ) 128 m | d | subtract(divide(600, const_2), 200) | if the perimeter of a rectangular garden is 600 m , its length when its breadth is 200 m is ? | "2 ( l + 200 ) = 600 = > l = 100 m answer : d" | a = 600 / 2
b = a - 200
|
a ) s 400 , b ) s 280 , c ) s 300 , d ) s 350 , e ) s 310 | c | divide(subtract(multiply(3000, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3000 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | explanation : average rate = ( 144 / 3000 ) * 100 = 4.8 ratio = 2 : 18 so , first part = ( 2 / 20 ) * 3000 = rs 300 . answer : c | a = 5 / 100
b = 3000 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 55 , b ) 53 , c ) 54 , d ) 52 , e ) 50 | c | multiply(divide(add(multiply(1, const_60), 15), 25), 18) | a machine , working at a constant rate , manufactures 18 dies in 25 minutes . how many dies does it make in 1 hr 15 min ? | "change 1 hr 15 min to 75 min . for this , we need to set up a simple proportion of dies per time 18 / 25 = s / 75 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of 25 in the two denominators . 18 / 1 = s / 3 s = 3 * 18 s = 54 the machine would be 54 dies in 1 hr 15 min . answer : c" | a = 1 * const_60
b = a + 15
c = b / 25
d = c * 18
|
['a ) 176', 'b ) 192', 'c ) 184', 'd ) 162', 'e ) 172'] | a | divide(multiply(multiply(5.44, const_100), multiply(3.74, const_100)), multiply(add(add(add(const_10, const_10), const_10), const_4), add(add(add(const_10, const_10), const_10), const_4))) | a room 5.44 m long and 3.74 m broad is to be paved with square tiles . the least number of square tiles required to cover the floor is : | area of the room = 544 * 374 sq . cm size of largest square tile = hcf of 544 cm & 374 cm area of 1 tile = 34 * 34 sq . cm therefore , number of tiles = ( 544 * 374 / 34 * 34 ) = 176 answer : a | a = 5 * 44
b = 3 * 74
c = a * b
d = 10 + 10
e = d + 10
f = e + 4
g = 10 + 10
h = g + 10
i = h + 4
j = f * i
k = c / j
|
a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(divide(50, const_10), const_3) | find the maximum value of n such that 50 ! is perfectly divisible by 2520 ^ n | 2520 = 2 ^ 3 * 3 ^ 2 * 5 * 7 here , 7 is the highest prime , so find the no . of 7 ' s in 50 ! only no . of 7 ' s in 50 ! = [ 50 / 7 ] + [ 50 / 7 ^ 2 ] = 7 + 1 = 8 answer : b | a = 50 / 10
b = a + 3
|
['a ) 16 : 9', 'b ) 9 : 16', 'c ) 9 : 12', 'd ) 16 : 12', 'e ) 4 : 12'] | a | divide(multiply(4, 4), multiply(3, 3)) | in two triangle , the ratio of the areas is 4 : 3 and that of their heights is 3 : 4 the ratio of their bases is | explanation : given a 1 / a 2 = 4 / 3 , h 1 / h 2 = 3 / 4 also , we know that a 1 / a 2 = ( 1 / 2 × b 1 × h 1 ) / ( 1 / 2 × b 2 × h 2 ) on substituting in above equation , we get b 1 / b 2 = 16 / 9 therefore b 1 : b 2 = 16 : 9 answer : option a | a = 4 * 4
b = 3 * 3
c = a / b
|
a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 48 | e | divide(multiply(30, 8), subtract(8, 3)) | a number exceeds by 30 from its 3 / 8 part . then the number is ? | "x – 3 / 8 x = 30 x = 48 answer : e" | a = 30 * 8
b = 8 - 3
c = a / b
|
a ) 13 , b ) 18 , c ) 21 , d ) 24 , e ) 38 | a | multiply(12, 480) | the h . c . f . of two numbers is 12 and their l . c . m . is 520 . if one of the numbers is 480 , then the other is : | "other number = ( 12 x 520 ) / 480 = 13 . answer : a" | a = 12 * 480
|
['a ) 1584', 'b ) 1854', 'c ) 1458', 'd ) 1485', 'e ) none of them'] | a | multiply(circumface(12), 21) | the radius of a cylinder is 12 m , height 21 m . the lateral surface area of the cylinder is : | lateral surface area = 2 π rh = 2 × 22 / 7 × 12 × 21 = 44 × 36 = 1584 m ( power 2 ) answer is a . | a = circumface * (
|
a ) 600 , b ) 900 , c ) 1200 , d ) 1500 , e ) 1800 | c | divide(multiply(240, const_100), subtract(const_100, 80)) | in an election between two candidates , the first candidate got 80 % of the votes and the second candidate got 240 votes . what was the total number of votes ? | "let v be the total number of votes . 0.2 v = 240 v = 1200 the answer is c ." | a = 240 * 100
b = 100 - 80
c = a / b
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(29, 19) | how many of the positive factors of 19 are not factors of 29 ? | "factors of 19 - 1 , 19 factors of 29 - 1 , 29 comparing both , we have three factors of 19 which are not factors of 29 - 19 , the answer is b" | a = 29 / 19
|
a ) 10 % , b ) 50 % , c ) 25 % , d ) 17.6 % , e ) none of these | d | subtract(multiply(divide(300, add(225, 30)), const_100), const_100) | a retailer buys a radio for rs 225 . his overhead expenses are rs 30 . he sellis the radio for rs 300 . the profit percent of the retailer is | "explanation : cost price = ( 225 + 30 ) = 255 sell price = 300 gain = ( 45 / 255 ) * 100 = 17.6 % . answer : d" | a = 225 + 30
b = 300 / a
c = b * 100
d = c - 100
|
a ) 27 , b ) 33 , c ) 54 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 6) | the number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "explanations 1 ) this tells us that the number of boxes is evenly divisible by both 6 and 27 ; in other words , it ’ s a common multiple of 6 and 27 . the question says : what ’ s the smallest value it could have ? in other words , what ’ s the lcm of 6 and 27 ? ( this question is one example of a real - world set - up where the question is actually asking for the lcm . ) step ( a ) : 6 = 2 * 3 27 = 3 * 3 * 3 step ( b ) : 6 = 2 * 3 27 = 3 * 3 * 3 gcf = 3 step ( c ) : 6 = 3 * 2 27 = 3 * 9 step ( d ) lcm = 3 * 2 * 9 = 54 thus , 54 is the lcm of 6 and 27 . answer : c ." | a = 2 * 2
b = a * 2
c = b * 3
d = c * 6
|
a ) 290 , b ) 320 , c ) 350 , d ) 380 , e ) 410 | c | divide(210, divide(60, const_100)) | if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 210 responses are needed , what is the minimum number of questionnaires that should be mailed ? | let x be the minimum number of questionnaires to be mailed . 0.6 x = 210 x = 350 the answer is c . | a = 60 / 100
b = 210 / a
|
a ) ( a ) 7 / 16 , b ) ( b ) 7 / 15 , c ) ( c ) 10 / 21 , d ) ( d ) 7 / 31 , e ) ( e ) 1 / 2 | d | divide(multiply(subtract(const_10, multiply(divide(4, 5), const_10)), multiply(divide(7, 6), multiply(divide(4, 5), const_10))), add(multiply(multiply(divide(4, 5), const_10), multiply(divide(4, 5), const_10)), multiply(subtract(const_10, multiply(divide(4, 5), const_10)), multiply(divide(7, 6), multiply(divide(4, 5), const_10))))) | a lemonade stand sold only small and large cups of lemonade on tuesday . 4 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | "a simpler way i guess would be to think that in total 5 cups were sold . out of which 4 are small and 1 is large . now let the small ones cost $ 6 . so the large ones would cost $ 7 . so , 4 * 6 = 24 and 1 * 7 = 7 . total revenue was 24 + 7 = 31 and large cup sales as found above is 7 therefore answer is 7 / 31 d" | a = 4 / 5
b = a * 10
c = 10 - b
d = 7 / 6
e = 4 / 5
f = e * 10
g = d * f
h = c * g
i = 4 / 5
j = i * 10
k = 4 / 5
l = k * 10
m = j * l
n = 4 / 5
o = n * 10
p = 10 - o
q = 7 / 6
r = 4 / 5
s = r * 10
t = q * s
u = p * t
v = m + u
w = h / v
|
a ) 30000 , b ) 28000 , c ) 24000 , d ) 25000 , e ) 25500 | c | multiply(divide(const_100, 96), 23040) | 96 % of the population of a city is 23040 . find the total population of the city ? | "population consider as x x * ( 96 / 100 ) = 23040 x = 240 * 100 = = > 24000 answer c" | a = 100 / 96
b = a * 23040
|
a ) 80 kg , b ) 83 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(58, multiply(10, add(2, divide(const_1, const_2)))) | the average weight of 10 men is increased by 2 ½ kg when one of the men who weighs 58 kg is replaced by a new man . what is the weight of the new man ? | since the average has increased by 2.5 kg , the weight of the man who stepped in must be equal to 58 + 10 x 2.5 58 + 25 = 83 kg ans : ' b ' | a = 1 / 2
b = 2 + a
c = 10 * b
d = 58 + c
|
a ) 10 , b ) 11 , c ) 7 , d ) 12 , e ) 9 | d | power(multiply(4, power(5, 2)), 6) | if 6 log ( 4 * 5 ^ 2 ) = x , find x | 6 ( log 2 ^ 2 * 5 ^ 2 ) = x 6 log ( 5 * 2 ) ^ 2 = x 6 * 2 log ( 5 * 2 ) = x 12 log 10 = x log 10 base 10 = 1 so 12 * 1 = x x = 12 answer : d | a = 5 ** 2
b = 4 * a
c = b ** 6
|
a ) 29997 , b ) 28088 , c ) 27098 , d ) 37500 , e ) 2799 | d | subtract(40000, 2500) | the price of a t . v . set worth rs . 40000 is to be paid in 20 installments of rs . 2500 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 2500 balance payment = ( 40000 - 2500 ) = rs . 37500 . answer : d" | a = 40000 - 2500
|
a ) $ 250 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550 | a | multiply(divide(multiply(2.4, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.50) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 2.4 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 400,000 / 4800 = 500 boxes total cost = 500 × $ 0.5 = $ 250 the answer is a ." | a = 1000 * 1000
b = 2 * 4
c = 20 * 20
d = c * 12
e = b / d
f = e * 0
|
a ) rs . 320 , b ) rs . 450 , c ) rs . 550 , d ) rs . 640 , e ) rs . 680 | c | add(divide(450, subtract(const_1, divide(10, const_100))), multiply(divide(450, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | boy sells a book for rs . 450 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "find selling price to gain 10 % . now , we are asked to find selling price to gain 10 % profit . hint : selling price = ( 100 + gain % ) × c . p . 100 selling price = ( 100 + 10 ) × 500 100 selling price = ( 110 ) × 500 100 therefore , selling price = rs . 550 c" | a = 10 / 100
b = 1 - a
c = 450 / b
d = 10 / 100
e = 1 - d
f = 450 / e
g = 10 / 100
h = f * g
i = c + h
|
a ) 30 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 % | a | add(divide(multiply(25, subtract(const_100, 20)), const_100), subtract(20, 10)) | in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 25 % of the people who do not have the job of their choice have a university diploma . if 20 % of the people have the job of their choice , what percent of the people have a university diploma ? | "setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = 10 % job of choice total = 20 % not job of choice with diploma = . 25 x not job of choice w / o diploma = . 75 x total not job of choice = x total with diploma total without diploma total citizen = 100 if 20 % of people have their job of choice , then 80 % of people do not have their job of choice . 25 % of 80 % = 20 % . we can also see that 10 % of the people have their job of choice and a diploma ( 20 % - 10 % = 10 % ) . 10 % + 20 % = 30 % . therefore 30 % of the people in country z have a diploma . ans a" | a = 100 - 20
b = 25 * a
c = b / 100
d = 20 - 10
e = c + d
|
a ) 19 , b ) 29 , c ) 36 , d ) 49 , e ) 59 | c | add(subtract(84, multiply(17, 3)), 3) | a batsman makes a score of 84 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 84 = 17 x therefore x = 36 answer : c" | a = 17 * 3
b = 84 - a
c = b + 3
|
a ) 40 , b ) 48 , c ) 50 , d ) 52 , e ) 56 | a | subtract(multiply(120, divide(40, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | of the 120 passengers on flight 750 , 40 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "number of passengers on flight = 120 number of female passengers = . 4 * 120 = 48 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 48 - 8 = 40 answer a" | a = 40 / 100
b = 120 * a
c = 10 / 100
d = 120 * c
e = 10 / 100
f = 120 * e
g = f / 3
h = d - g
i = b - h
|
a ) 41 , b ) 910 , c ) 1001 , d ) 1911 , e ) none | a | subtract(451, 410) | the maximum numbers of students among them 451 pens and 410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is | "olution required number of students . = h . c . f of 451 and 410 . â € ¹ = â € º 41 . answer a" | a = 451 - 410
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a ) 17.5 , b ) 20 , c ) 24 , d ) 30 , e ) 32 | a | divide(add(add(4, const_4), subtract(32, const_4)), const_2) | find the average of all the numbers between 4 and 32 which are divisible by 5 . | "solution average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 ) = 105 / 6 = 17.5 . answer a" | a = 4 + 4
b = 32 - 4
c = a + b
d = c / 2
|
['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) none'] | a | sqrt(divide(divide(multiply(volume_cylinder(8, 2), const_3), 6), const_pi)) | a cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm . the radius of the cone will be : | sol . let the radius of the cone be r cm . then , 1 / 3 ∏ * r ² * 6 = ∏ * 8 * 8 * 2 ⇔ r ² = [ 8 * 8 * 2 * 3 / 6 ] = 64 ⇔ r = 8 cm . answer a | a = volume_cylinder * (
b = a / 3
c = b / 6
d = math.sqrt(c)
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | floor(divide(18, subtract(5, const_1))) | if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 18 ? | "@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 18 . then 4 n < 18 and n < 4.5 the largest possible integer is n = 4 . the answer is d ." | a = 5 - 1
b = 18 / a
c = math.floor(b)
|
a ) 120 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | a | multiply(60, const_2) | a runs twice as fast as b and gives b a start of 60 m . how long should the racecourse be so that a and b might reach in the same time ? | ratio of speeds of a and b is 2 : 1 b is 60 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 60 m is 60 / 1 = 60 m so the total time taken by a and b to reach = 2 * 60 = 120 m answer : a | a = 60 * 2
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a ) 10.7 sec , b ) 10.9 sec , c ) 10.1 sec , d ) 15.1 sec , e ) 12.7 sec | b | divide(add(100, 142), multiply(80, const_0_2778)) | how long does a train 100 m long running at the speed of 80 km / hr takes to cross a bridge 142 m length ? | "speed = 80 * 5 / 18 = 22.2 m / sec total distance covered = 100 + 142 = 242 m . required time = 242 / 22.2 ' = 10.9 sec . answer : b" | a = 100 + 142
b = 80 * const_0_2778
c = a / b
|
a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | b | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 5 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , b can do 1 / 6 of the project . thus a and b must complete 5 / 6 of the project , which takes 10 days . the total number of days is 10 + 5 = 15 . the answer is b ." | a = 1 / 30
b = a * 5
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 5
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a ) 23,500 , b ) 24,500 , c ) 25,000 , d ) 26,500 , e ) 27,500 | c | floor(divide(divide(subtract(580, multiply(1,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 1,000)) | angelo and isabella are both salespersons . in any given week , angelo makes $ 580 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "let the weekly sales of both = x 580 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 25000 answer : c" | a = 8 / 100
b = 1 * 0
c = 580 - b
d = 10 / 100
e = 8 / 100
f = d - e
g = c / f
h = g / 1
i = math.floor(h)
|
a ) 1.5 , b ) 2.14 , c ) 3 , d ) 4.5 , e ) 5 | b | divide(divide(subtract(add(50, 50), 50), subtract(50, 30)), const_2) | a dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce . a chef accidentally makes 15 cups of the sauce with 30 % melted chocolate and 70 % raspberry puree instead . how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50 % of each ? | "yes , we assume that the mix is homogeneous . otherwise , we will not be able to solve the question . you have 15 cups of sauce with 30 % chocolate . you also have unlimited amount of pure chocolate sauce . now you need to mix these two in such a way that you get total 15 cups of sauce with 50 % chocolate . using scale method : w 1 / w 2 = ( 100 - 50 ) / ( 50 - 30 ) = 5 / 2 w 1 - amount of 30 % chocolate sauce w 2 - amount of pure chocolate sauce so for every 5 / 2 cups of 30 % chocolate sauce , we need 2 cup of pure chocolate sauce . this will give us 7 cups of 50 % chocolate sauce . but we need 15 cups of 50 % chocolate sauce . so we need to mix 5 * 15 / 7 = 12.5 cups of 40 % chocolate sauce with 1 * 15 / 7 = 2.14 cups of pure chocolate sauce . answer ( b )" | a = 50 + 50
b = a - 50
c = 50 - 30
d = b / c
e = d / 2
|
a ) 2006 , b ) 2007 , c ) 2008 , d ) 2009 , e ) 2010 | c | add(2001, divide(add(divide(35, const_100), subtract(7.30, 5.20)), subtract(divide(45, const_100), subtract(7.30, 5.20)))) | the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 35 cents less than the price of commodity y ? | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 35 cents less than commodity y . $ 1.75 / 25 cents = 7 years the answer is 2001 + 7 years = 2008 . the answer is c ." | a = 35 / 100
b = 7 - 30
c = a + b
d = 45 / 100
e = 7 - 30
f = d - e
g = c / f
h = 2001 + g
|
a ) 30 , b ) 28 , c ) 27 , d ) 26 , e ) 25 | a | add(multiply(const_4, 1.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 1.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 1.9421 + 1 ) = 10 ( 2.9421 ) = 29.421 x = antilog ( 29.421 ) therefore , number of digits in x = 30 . answer : a" | a = 4 * 1
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 25000 , b ) 26000 , c ) 27000 , d ) 28000 , e ) 29000 | b | divide(multiply(multiply(4, 65), multiply(6, const_1000)), multiply(const_1, const_60)) | a river 4 m deep and 65 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is ? | "rate of water flow - 6 kmph - - 6000 / 60 - - 100 m / min depth of river - - 4 m width of river - - 65 m vol of water per min - - 100 * 4 * 65 - - - 26000 answer b" | a = 4 * 65
b = 6 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 20 m , b ) 22 m , c ) 34 m , d ) 26 m , e ) 28 m | c | add(add(multiply(subtract(17, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 17 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is | "explanation : each row contains 17 plants . there are 15 gapes between the two corner trees ( 16 x 2 ) metres and 1 metre on each side is left . therefore length = ( 32 + 2 ) m = 34 m . answer : c" | a = 17 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
|
a ) 46,800 , b ) 66,800 , c ) 56,800 , d ) 26,800 , e ) 76,800 | a | divide(add(multiply(add(const_4, const_1), const_100), 70), sqrt(const_100)) | a person spent rs . 6,040 from his salary on food and 8,000 on house rent . after that he was left with 70 % of his monthly salary . what is his monthly salary ? | "total money spent on food and house rent = 6,040 + 8,000 = 14,040 which is 100 - 70 = 30 % of his monthly salary ∴ his salary = 14040 x 100 / 30 = 46800 answer : a" | a = 4 + 1
b = a * 100
c = b + 70
d = math.sqrt(100)
e = c / d
|
a ) 50 , b ) 55 , c ) 110 , d ) 75 , e ) 80 | c | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(40, 3), 3), 3), 3), add(add(add(add(add(add(40, divide(40, 3)), divide(40, 3)), divide(divide(40, 3), 3)), divide(divide(40, 3), 3)), divide(divide(divide(40, 3), 3), 3)), divide(divide(divide(40, 3), 3), 3))), 3), divide(divide(divide(40, 3), 3), 3)), divide(divide(divide(40, 3), 3), 3)), divide(divide(divide(40, 3), 3), 3)), divide(divide(divide(40, 3), 3), 3)) | a basketball is dropped from a height of 40 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 3 full bounces . | "initial distance = 40 feet first bounce = 20 feet up + 20 feet down = 40 feet second bouche = 10 feet up + 10 feet down = 20 feet third bounce = 5 feet up and 5 feet down = 10 feet total distance covered = 40 + 40 + 20 + 10 = 110 answer is c" | a = 40 / 3
b = a / 3
c = b / 3
d = c / 3
e = 40 / 3
f = 40 + e
g = 40 / 3
h = f + g
i = 40 / 3
j = i / 3
k = h + j
l = 40 / 3
m = l / 3
n = k + m
o = 40 / 3
p = o / 3
q = p / 3
r = n + q
s = 40 / 3
t = s / 3
u = t / 3
v = r + u
w = d + v
x = w * 3
y = 40 / 3
z = y / 3
A = z / 3
B = x - A
C = 40 / 3
D = C / 3
E = D / 3
F = B - E
G = 40 / 3
H = G / 3
I = H / 3
J = F - I
K = 40 / 3
L = K / 3
M = L / 3
N = J - M
|
a ) 900 , b ) 2440 , c ) 750 , d ) 244 , e ) 960 | b | divide(multiply(add(80, divide(370, 860)), 1550), const_100) | 80 370 860 1550 ? 3530 | "10 ^ 2 - 20 = 80 20 ^ 2 - 30 = 370 30 ^ 2 - 40 = 860 40 ^ 2 - 50 = 1550 50 ^ 2 - 60 = 2440 60 ^ 2 - 70 = 3530 . answer : b" | a = 370 / 860
b = 80 + a
c = b * 1550
d = c / 100
|
a ) 9005 , b ) 8925 , c ) 2345 , d ) 6474 , e ) 8723 | a | divide(divide(multiply(4052.25, const_100), 9), 5) | a sum fetched a total simple interest of 4052.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | principal = ( 100 x 4052.25 ) / ( 9 x 5 ) = 405225 / 45 = 9005 . answer a | a = 4052 * 25
b = a / 9
c = b / 5
|
a ) 176 , b ) 192 , c ) 202 , d ) 218 , e ) 284 | b | divide(divide(multiply(60, 4), 2.5), divide(50, const_100)) | a rotameter is a device that measures flow of liquid and gases . when measuring liquid phase flows , 2.5 inches represent 60 liters per minute of liquid . with gas measurements the rotameter moves 50 % of the movement he moves with the liquid phase . how many liters of gas passed through the rotameter if it measured 4 inches ? | in case of liquid - 2.5 inches represents 60 lit / min . in case of gas - 50 % of 2.5 inches represents 60 lit / min 1.25 inches represents 60 lit / min 4 inches will represent 60 * 4 / 1.25 = 192 b is the answer | a = 60 * 4
b = a / 2
c = 50 / 100
d = b / c
|
a ) 1.44 % , b ) 1.74 % , c ) 1.84 % , d ) 1.47 % , e ) 1.00 % | e | divide(multiply(10, 10), const_100) | if a trader sold two cars each at rs . 325475 and gains 10 % on the first and loses 10 % on the second , then his profit or loss percent on the whole is ? | "sp of each car is rs . 325475 , he gains 10 % on first car and losses 10 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 10 ) ( 10 ) / 100 % = 1.00 % answer : e" | a = 10 * 10
b = a / 100
|
a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 28 | c | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | how many positive integers less than 130 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 130 = 13 * 10 for 12 : 12 . . . 130 = 12 * 10 but there is one integer 13 * 12 . so n = ( 10 ) + ( 10 ) = 20 c" | a = 4 + 13
b = a - 1
c = math.factorial(b)
d = math.factorial(13)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 3 : 2 , b ) 5 : 4 , c ) 7 : 6 , d ) 9 : 8 , e ) 11 : 10 | b | divide(multiply(5, 3), multiply(6, 2)) | the marks obtained by polly and sandy are in the ratio 5 : 6 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 5 : 6 sandy : willy = 3 : 2 = 6 : 4 polly : sandy : willy = 5 : 6 : 4 polly : willy = 5 : 4 the answer is b ." | a = 5 * 3
b = 6 * 2
c = a / b
|
a ) 5 , b ) 14 , c ) 10 , d ) 20 , e ) 30 | b | multiply(subtract(9, 7), 7) | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ? | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ? 9 ^ 7 = ( 3 ^ 2 ) ^ 7 = 3 ^ 14 b . 14 | a = 9 - 7
b = a * 7
|
a ) 45 % , b ) 40 % , c ) 35 % , d ) 30 % , e ) 25 % | a | subtract(add(add(70, 55), 20), const_100) | if 70 percent of a class answered the first question on a certain test correctly , 55 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "70 % answered the first question correctly and 20 % answered neither correctly . then 10 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 55 % - 10 % = 45 % . the answer is a ." | a = 70 + 55
b = a + 20
c = b - 100
|
a ) 11 , b ) 21 , c ) 31 , d ) 61 , e ) 51 | b | add(divide(66, add(4, 2)), divide(50, add(3, 2))) | each machine of type a has 3 steel parts and 2 chrome parts . each machine of type b has 2 steel parts and 4 chrome parts . if a certain group of type a and type b machines has a total of 50 steel parts and 66 chrome parts , how many machines are in the group | "look at the below representation of the problem : steel chrome total a 3 2 50 > > no . of type a machines = 50 / 5 = 10 b 2 4 66 > > no . of type b machines = 66 / 6 = 11 so the answer is 21 i . e b . hope its clear ." | a = 4 + 2
b = 66 / a
c = 3 + 2
d = 50 / c
e = b + d
|
a ) 42 , b ) 43 , c ) 44 , d ) 64 , e ) 46 | d | add(subtract(75, multiply(12, 1)), 1) | a batsman in his 12 th innings makes a score of 75 and thereby increases his average by 1 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 1 ) + 75 ∴ x = 64 answer d" | a = 12 * 1
b = 75 - a
c = b + 1
|
a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 5 : 4 , e ) 3 : 1 | d | divide(subtract(180, 170), subtract(188, 180)) | students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 188 cms . what was the ratio of men to women ? | "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 188 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 188 m ) / ( w + m ) = 180 170 w + 188 m = 180 w + 180 m 8 m = 10 w 4 m = 5 w m / w = 5 / 4 the ratio of men to women is 5 t ổ 4 . d" | a = 180 - 170
b = 188 - 180
c = a / b
|
a ) 350 , b ) 475 , c ) 300 , d ) 425 , e ) 400 | e | subtract(divide(add(100, 200), const_2), divide(add(150, 950), const_2)) | the average ( arithmetic mean ) of the integers from 100 to 200 , inclusive , is how much lesser than the average of the integers from 150 to 950 , inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 150 to 950 , inclusive = ( 150 + 950 ) / 2 = 550 average of numbers between 100 to 200 , inclusive = ( 100 + 200 ) / 2 = 150 difference = 550 - 150 = 400 answer is e" | a = 100 + 200
b = a / 2
c = 150 + 950
d = c / 2
e = b - d
|
a ) 20 , 35,45 , b ) 28 , 49,63 , c ) 16 , 28,36 , d ) 16 , 28,46 , e ) none of these | b | add(multiply(4, divide(add(multiply(3, 8), 116), add(add(4, 7), 9))), 8) | the present ages of 3 persons are in the proportion of 4 : 7 : 9 . 8 years ago , the sum of their ages was 116 . find their present ages . | let the present ages of three persons be 4 k , 7 k and 9 k respectively . ( 4 k - 8 ) + ( 7 k - 8 ) + ( 9 k - 8 ) = 116 20 k = 140 k = 7 therefore , then present ages are 28 , 49,63 . answer : b | a = 3 * 8
b = a + 116
c = 4 + 7
d = c + 9
e = b / d
f = 4 * e
g = f + 8
|
a ) 70 , b ) 80 , c ) 74 , d ) 75 , e ) 78 | b | divide(1, divide(add(multiply(const_3600, divide(1, 90)), 5), const_3600)) | a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 90 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 90 hours = 3,600 / 90 seconds = 40 seconds ; time to cover 1 kilometer at regular speed is 40 + 5 = 45 seconds = 45 / 3,600 hours = 1 / 80 hours ; so , we get that to cover 1 kilometer 1 / 80 hours is needed - - > regular speed 80 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : b ." | a = 1 / 90
b = 3600 * a
c = b + 5
d = c / 3600
e = 1 / d
|
a ) 6.66 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 12.5 % | b | multiply(subtract(const_1, divide(add(const_1, divide(15, const_100)), add(const_1, divide(25, const_100)))), const_100) | if the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline , by how much percent should he reduce the quantity of gasoline that he buys ? | "let price of original price of gasoline = 10 $ and let original consumption of gasoline = 10 amount spend on gasoline originally = 10 * 10 = 100 $ the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline increased price of gasoline = 12.5 $ now ron intends to spend on gasoline = 1.15 * 100 = 115 $ consumption of gasoline needs to be = 115 / 12.5 = 1150 / 125 = 46 / 5 = 9.2 reduction in consumption of gasoline = 10 - 9.2 = . 8 % reduction in consumption of gasoline = . 8 / 10 * 100 % = 8 % answer b" | a = 15 / 100
b = 1 + a
c = 25 / 100
d = 1 + c
e = b / d
f = 1 - e
g = f * 100
|
a ) 3.19 , b ) 3.28 , c ) 3.35 , d ) 3.43 , e ) 3.56 | d | multiply(divide(subtract(7, 1), add(add(7, 1), subtract(7, 1))), add(7, 1)) | a rower can row 7 km / h in still water . when the river is running at 1 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ? | "let x be the distance to big rock . time = x / 6 + x / 8 = 1 x = 48 / 14 = 3.43 km the answer is d ." | a = 7 - 1
b = 7 + 1
c = 7 - 1
d = b + c
e = a / d
f = 7 + 1
g = e * f
|
a ) 33 , b ) 88 , c ) 40 , d ) 99 , e ) 12 | e | inverse(subtract(divide(const_1, 4), divide(const_1, add(4, const_2)))) | a cistern is normally filled in 4 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 4 - 1 / x = 1 / 6 x = 12 answer : e" | a = 1 / 4
b = 4 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) 1.012526333 , b ) 0.012625333 , c ) 0.125263333 , d ) 0.126253333 , e ) 0.016833333 | e | divide(25.25, 1500) | 25.25 / 1500 is equal to : | "25.25 / 2000 = 2525 / 200000 = 0.016833333 answer : e" | a = 25 / 25
|
a ) 131 , b ) 135 , c ) 156 , d ) 177 , e ) 188 | c | add(multiply(divide(subtract(224, 20), const_3), const_2), 20) | if jake loses 20 pounds , he will weigh twice as much as his sister . together they now weigh 224 pounds . what is jake ’ s present weight , in pounds ? | "lets say j is the weight of jack and s is the wt of his sister . if he loses 20 pounds , he s twice as heavy as his sister . j - 20 = 2 * s also , together they weight 224 pounds j + s = 224 solvong the 2 equation , we get j = 156 pounds ! c" | a = 224 - 20
b = a / 3
c = b * 2
d = c + 20
|
a ) 1 / 18 , b ) 1 / 6 , c ) 2 / 3 , d ) 1 / 2 , e ) 5 / 6 | c | subtract(add(12, divide(12, const_3.0)), divide(const_2.0, 6)) | in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 12 or larger is drawn is 1 / 2 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 1 / 2 + p ( set 2 ) - 1 / 6 = 2 / 3 p ( set 2 ) = 2 / 3 answer ( c )" | a = 12 / 3
b = 12 + a
c = 2 / 0
d = b - c
|
a ) 32.25 % , b ) 40 % , c ) 37.50 % , d ) 36.75 % , e ) 32 % | a | multiply(subtract(power(add(divide(15, const_100), const_1), const_2), const_1), const_100) | if the sides of a rectangle are increased by 15 % , what is the percentage increase in the area ? | "say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . the area of new rectangle would be 115 * 115 = 13,225 , which is 32.25 % greater than the old area . answer : a ." | a = 15 / 100
b = a + 1
c = b ** 2
d = c - 1
e = d * 100
|
a ) 65 % , b ) 70 % , c ) 75 % , d ) 85 % , e ) 96 % | e | multiply(subtract(divide(multiply(multiply(const_100, add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100), const_1), const_100) | a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 40 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ? | "find the product of the two increases : ( 14 / 10 ) * ( 14 / 10 ) which is 1.96 and a 96 % increase . e" | a = 40 / 100
b = 1 + a
c = 100 * b
d = 40 / 100
e = 1 + d
f = c * e
g = f / 100
h = g - 1
i = h * 100
|
a ) 1 / 9 , b ) 29 / 180 , c ) 26 / 143 , d ) 1 / 5 , e ) 31 / 220 | e | add(divide(const_1, divide(multiply(50, 4), 10)), divide(const_1, divide(multiply(30, 5,5), 15))) | if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 5,5 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ? | "50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 / 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 5.5 * 2 = 11 hours ; in 1 hour 15 journeymen can finish 1 / 11 of the job . therefore , in 1 hour 10 apprentices and 15 journeymen can finish 1 / 20 + 1 / 11 = 31 / 220 of the job . answer : e" | a = 50 * 4
b = a / 10
c = 1 / b
d = 30 * 5
e = d / 15
f = 1 / e
g = c + f
|
a ) 15 , b ) 30 , c ) 20 , d ) 40 , e ) 10 | b | multiply(multiply(1, 6), multiply(5, 3)) | calculate the l . c . m of 1 / 5 , 6 / 7 , 5 / 6 , 3 / 5 is : | "required l . c . m = l . c . m . of 1 , 6 , 5 , 3 / h . c . f . of 5 , 7 , 6 , 5 = 30 / 1 = 30 answer is b" | a = 1 * 6
b = 5 * 3
c = a * b
|
a ) 2.5 , b ) 1.35 , c ) 4.25 , d ) 6.12 , e ) 7.13 | c | divide(add(add(add(1, const_1), add(add(1, const_1), const_2)), add(subtract(10, 1), subtract(10, const_2))), 1) | find the average of all prime numbers between 1 and 10 | "prime numbers between 1 and 10 are 2,3 , 5,7 required average = ( 2 + 3 + 5 + 7 ) / 4 = 17 / 4 = 4.25 answer is c" | a = 1 + 1
b = 1 + 1
c = b + 2
d = a + c
e = 10 - 1
f = 10 - 2
g = e + f
h = d + g
i = h / 1
|
a ) 40 years , b ) 41 years , c ) 42 years , d ) 43 years , e ) 44 years | b | subtract(multiply(add(20, 2), add(19, 2)), multiply(20, 19)) | the average age of a class of 20 students is 19 years . the average increased by 2 when the teacher ' s age also included . what is the age of the teacher ? | "if age of the teacher was 19 , average would not have changed . since average increased by 2 , age of the teacher = 19 + 22 × 1 = 41 answer : b" | a = 20 + 2
b = 19 + 2
c = a * b
d = 20 * 19
e = c - d
|
a ) 520 , b ) 720 , c ) 425 , d ) 625 , e ) 420 | e | subtract(multiply(divide(add(42, 42), subtract(42, 35)), 35), add(subtract(subtract(42, divide(add(42, 42), subtract(42, 35))), divide(add(42, 42), subtract(42, 35))), const_2)) | there were 35 students in a hostel . due to the admission of 7 new students , ; he expenses of the mess were increased by rs . 42 per day while the average expenditure per head diminished by rs 1 . wbat was the original expenditure of the mess ? | "let the original average expenditure be rs . x . then , 42 ( x - 1 ) - 35 x = 42 7 x = 84 x = 12 . original expenditure = rs . ( 35 x 12 ) = rs . 420 . answer is e ." | a = 42 + 42
b = 42 - 35
c = a / b
d = c * 35
e = 42 + 42
f = 42 - 35
g = e / f
h = 42 - g
i = 42 + 42
j = 42 - 35
k = i / j
l = h - k
m = l + 2
n = d - m
|
a ) 40 , b ) 50 , c ) 73 , d ) 70 , e ) 80 | c | subtract(subtract(7, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | evaluate : | 7 - 8 ( 3 - 12 ) | - | 5 - 11 | = ? | "according to order of operations , inner brackets first . hence | 7 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 7 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 7 + 72 | - | 5 - 11 | = | 79 | - | - 6 | = 79 - 6 = 73 correct answer c ) 73" | a = 3 - 12
b = 8 * a
c = 7 - b
d = 5 - 11
e = c - negate
|
a ) 1 minutes , b ) 14 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(14, 14), 14) | if 14 lions can kill 14 deers in 14 minutes how long will it take 100 lions to kill 100 deers ? | "we can try the logic of time and work , our work is to kill the deers so 14 ( lions ) * 14 ( min ) / 14 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 14 answer : b" | a = 14 / 14
b = a * 14
|
a ) 4 seconds , b ) 25.33 seconds , c ) 29 seconds , d ) 21 seconds , e ) 6.25 seconds | b | divide(subtract(100, 24), const_3) | in a 100 m race , sam beats john by 3 seconds . on the contrary , if sam allowed john to start 24 m ahead of sam , then sam and john reach the finishing point at the same time . how long does sam take to run the 100 m race ? | their difference is 3 second but this difference is 0 if john allows sam to start the race from 24 m ahead . that means jhon was 24 m away from finishing line when they started together . so he will cover 24 m in 3 seconds . so his speed = 24 / 3 = 8 metre / second . so time taken = 100 / 3 = 33.33 seconds . so sam took = 25.33 seconds . correct answer = b | a = 100 - 24
b = a / 3
|
a ) 0 , b ) - 100 , c ) - 50 , d ) 50 , e ) 100 | d | subtract(add(50, 25), subtract(50, 25)) | if | x - 25 | = 50 what is the sum of all the values of x . | there will be two cases x - 25 = 50 and x - 25 = - 50 solve for x = > x = 50 + 25 = > x = 75 or x = - 50 + 25 = > x = - 25 the sum of both values will be 75 + - 25 = 50 answer is d | a = 50 + 25
b = 50 - 25
c = a - b
|
a ) 1 , b ) 117 , c ) 116 , d ) equal to their h . c . f , e ) can not be calculated | b | multiply(117, const_1) | product of two co - prime numbers is 117 . their l . c . m should be : | "h . c . f of co - prime numbers is 1 . so , l . c . m = 117 / 1 = 117 . answer : b" | a = 117 * 1
|
a ) 82 % , b ) 75 % , c ) 65 % , d ) 50 % , e ) 45 % | a | multiply(subtract(multiply(divide(add(const_100, 30), const_100), divide(add(40, const_100), const_100)), const_1), const_100) | the price of a tv was increased by 30 percent . the new price was then increased by 40 percent . a single increase of what percent is equivalent to these two successive increases ? | "consider base price - $ 100 30 % increase = 1.30 * 100 = $ 130 another 40 % increase on new price = 1.4 * 130 = $ 182 so final price of radio - $ 182 therefore a 82 % increase correct option - a" | a = 100 + 30
b = a / 100
c = 40 + 100
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 9 , b ) 11 , c ) 24 , d ) 48 , e ) 54 | d | multiply(multiply(const_2, 4), multiply(const_2, 3)) | terry is having lunch at a salad bar . there are two types of lettuce to choose from , as well as 3 types of tomatoes , and 4 types of olives . he must also choose whether or not to have one of the two types of soup on the side . if terry has decided to have the salad and soup combo and he picks one type of lettuce , one type of tomato , and one type of olive for his salad , how many total options does he have for his lunch combo ? | terry can pick one lunch salad bar . . . . we can choose lettuce 2 c 1 , tomatoes - 3 c 1 and olives - 4 c 1 . . . two types of soups , if he picks one then 2 c 1 ways . terry decided to have the combo : soup + salad . . . hmmm . . . using the above information , we get ( 2 c 1 * 2 c 1 * 3 c 1 * 4 c 1 ) = 48 . . ans option d is correct answer . . | a = 2 * 4
b = 2 * 3
c = a * b
|
a ) 72 , b ) 43 , c ) 91 , d ) 40 , e ) 30 | e | subtract(add(power(31, 31), 31), multiply(32, floor(divide(add(power(31, 31), 31), 32)))) | what will be the reminder when ( 31 ^ 31 + 31 ) is divided by 32 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 31 ^ 31 + 1 ) will be divisible by ( 31 + 1 ) ; ( 31 ^ 31 + 1 ) + 30 when divided by will give 30 as remainder . correct option : e" | a = 31 ** 31
b = a + 31
c = 31 ** 31
d = c + 31
e = d / 32
f = math.floor(e)
g = 32 * f
h = b - g
|
a ) 400 , b ) 5550 , c ) 6922 , d ) 3786 , e ) 5553 | d | add(power(divide(add(111, 1), 2), const_2), multiply(divide(50, 2), add(divide(50, 2), const_1))) | if m equals the sum of the odd integers from 1 to 111 , inclusive , and t equals the sum of the even integers from 2 to 50 , inclusive , what is the value of m + t ? | use following formulae for such problems : sum of evenly spaced integers = ( # of integers ) * ( mean of integers ) # of integers = [ ( last - first ) / 2 ] + 1 mean of integers = ( last + first ) / 2 in above problem : # of integers = [ ( 111 - 1 ) / 2 ] + 1 = 56 and [ ( 50 - 2 ) / 2 ] + 1 = 25 mean of integers = ( 111 + 1 ) / 2 = 56 and ( 50 + 2 ) / 2 = 26 sum of integers = ( 56 * 56 ) = 3136 and ( 25 * 26 ) = 650 thus their sum ( m + t ) = 3786 answer : d | a = 111 + 1
b = a / 2
c = b ** 2
d = 50 / 2
e = 50 / 2
f = e + 1
g = d * f
h = c + g
|
a ) 120 , b ) 110 , c ) 105 , d ) 140 , e ) 135 | d | subtract(choose(10, 6), choose(subtract(10, 2), 2)) | a meeting has to be conducted with 6 managers . find the number of ways in which the managers be selected from among 10 managers , if 2 managers will not attend the meeting together ? | "we can either choose all 6 people from 8 manager who have no problems or choose 5 from the 8 and 1 from the 2 managers who have a problem sitting together so 8 c 6 + ( 8 c 5 * 2 c 1 ) this is 28 + 112 = 140 answer : d" | a = math.comb(10, 6)
b = 10 - 2
c = math.comb(b, 2)
d = a - c
|
a ) 4,5 , b ) 3,4 , c ) 2,6 , d ) 5,6 , e ) 2,3 | b | add(2, 2) | the smallest value of n , for which 2 n + 3 is not a prime number , is | "( 2 ã — 1 + 3 ) = 5 . ( 2 ã — 2 + 3 ) = 7 . ( 2 ã — 3 + 3 ) = 9 . ( 2 ã — 4 + 3 ) = 15 . which is not prime , n = 3,4 . answer : b" | a = 2 + 2
|
a ) 40 years , b ) 34 years , c ) 42 years , d ) 43 years , e ) 44 years | b | subtract(multiply(add(12, 2), add(19, 2)), multiply(12, 19)) | the average age of a class of 12 students is 19 years . the average increased by 2 when the teacher ' s age also included . what is the age of the teacher ? | "if age of the teacher was 12 , average would not have changed . since average increased by 2 , age of the teacher = 12 + 22 × 1 = 34 answer : b" | a = 12 + 2
b = 19 + 2
c = a * b
d = 12 * 19
e = c - d
|
a ) 1 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | a | subtract(multiply(divide(3, 2), add(14, 6)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(14, 6)))), const_1)) | a certain basketball team that has played 2 / 3 of its games has a record of 14 wins and 6 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ? | "14 wins , 6 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 6 games so it can lose another 1 at most . answer ( a )" | a = 3 / 2
b = 14 + 6
c = a * b
d = 3 / 4
e = 3 / 2
f = 14 + 6
g = e * f
h = d * g
i = math.floor(h)
j = i + 1
k = c - j
|
a ) 250 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | a | subtract(multiply(multiply(45, const_0_2778), 30), 125) | a train , 125 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 125 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 125 + x ) / 30 m / s = > 45 ã — ( 10 / 36 ) = ( 125 + x ) / 30 = > 45 ã — 10 ã — 30 / 36 = 125 + x = > 45 ã — 10 ã — 10 / 12 = 125 + x = > 15 ã — 10 ã — 10 / 4 = 125 + x = > 15 ã — 25 = 125 + x = 375 = > x = 375 - 125 = 250 answer : option a" | a = 45 * const_0_2778
b = a * 30
c = b - 125
|
a ) 3066 , b ) 3586 , c ) 3086 , d ) 3968 , e ) 3286 | c | add(multiply(36, 85), 26) | in a division sum , the quotient is 36 , the divisor 85 and the remainder 26 , find the dividend ? | "explanation : 36 * 85 + 26 = 3086 answer : c" | a = 36 * 85
b = a + 26
|
a ) 46 , b ) q . 80 , c ) q . 90 , d ) q . 100 , e ) 200 | d | add(subtract(divide(add(add(subtract(const_100, 23), 23), 20), const_2), 20), divide(add(add(subtract(const_100, 23), 23), 20), const_2)) | a company conducted a survey about its two brands , a and b . x percent of respondents liked product a , ( x – 20 ) percent liked product b , 23 percent liked both products , and 23 percent liked neither product . what is the minimum number q of people surveyed by the company ? | "100 = x + x - 20 + 23 - 23 x = 60 , so , product a = 60 % , product b = 40 % , both = 23 % , neither = 23 % 23 % of the total no . of people should be an integer . so , a , bc are out . 60 % of d and 40 % of d are both integers . so , d satisfies all conditions . so , answer is d ." | a = 100 - 23
b = a + 23
c = b + 20
d = c / 2
e = d - 20
f = 100 - 23
g = f + 23
h = g + 20
i = h / 2
j = e + i
|
a ) 7.16 , b ) 7.59 , c ) 7.12 , d ) 7.15 , e ) 7.11 | b | divide(add(141, 165), multiply(add(80, 65), const_0_2778)) | two trains 141 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 141 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.59 answer : b" | a = 141 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
|
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 72 % | e | multiply(divide(subtract(multiply(divide(6, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(6, 20), const_100)), const_100) | a pharmaceutical company received $ 6 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "solution : this is a percent decrease problem . we will use the formula : percent change = ( new – old ) / old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 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 20 million royalties / sales = 6 / 20 = 3 / 10 next 108 million royalties / sales = 9 / 108 = 1 / 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties / sales = ( 6 / 20 ) x 60 = 18 next 108 million royalties / sales = 9 / 108 = ( 1 / 12 ) x 60 = 5 we can plug 18 and 5 into our percent change formula : ( new – old ) / old x 100 [ ( 5 – 18 ) / 18 ] x 100 - 650 / 9 x 100 at this point we can stop and consider the answer choices . since we know that 650 / 9 is just a bit less than ½ , we know that - 650 / 9 x 100 is about a 72 % decrease . answer e ." | a = 6 / 20
b = a * 100
c = 9 / 108
d = c * 100
e = b - d
f = 6 / 20
g = f * 100
h = e / g
i = h * 100
|
a ) 81 min , b ) 108 min , c ) 144 min , d ) 192 min , e ) none | c | multiply(add(const_1, const_4), 86) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 86 minutes , then the slower pipe alone will be able to fill the tank in | "solution let the slower pipe alone fill the tank in x minutes . then , faster pipes will fill it in x / 3 minutes . therefore , 1 / x + 3 / x = 1 / 36 ‹ = › 4 / x = 1 / 36 ‹ = › x = 144 min . answer c" | a = 1 + 4
b = a * 86
|
a ) 80 , b ) 70 , c ) 56 , d ) 45 , e ) 14 | c | multiply(divide(90, add(add(2, 3), 5)), 5) | the amounts of time that three secretaries worked on a special project are in the ratio of 2 to 3 to 5 . if they worked a combined total of 90 hours , how many hours did the secretary who worked the longest spend on the project ? | "10 x = 90 = > x = 9 therefore the secretary who worked the longest spent 9 x 5 = 45 hours on the project option ( c )" | a = 2 + 3
b = a + 5
c = 90 / b
d = c * 5
|
a ) 3 : 3 , b ) 5 : 7 , c ) 5 : 4 , d ) 7 : 2 , e ) 8 : 5 | c | divide(multiply(45, const_0_2778), 10) | one train is travelling 45 kmph and other is at 10 meters a second . ratio of the speed of the two trains is ? | c 5 : 4 45 * 5 / 18 = 10 25 : 20 = > 5 : 4 | a = 45 * const_0_2778
b = a / 10
|
a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 2 / 5 | e | divide(const_2, add(divide(40, const_10), const_1)) | in the xy - plane , a triangle has vertices ( 00 ) , ( 40 ) and ( 410 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | the area of the right triangle is ( 1 / 2 ) * 4 * 10 = 20 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 20 = 2 / 5 the answer is e . | a = 40 / 10
b = a + 1
c = 2 / b
|
a ) 75 kmph , b ) 85 kmph , c ) 95 kmph , d ) 105 kmph , e ) 115 kmph | b | subtract(divide(divide(150, 6), const_0_2778), 5) | a train 150 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ? | let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 150 / [ ( x + 5 ) * 5 / 18 ] = 6 30 ( x + 5 ) = 2700 x = 85 kmph answer is b | a = 150 / 6
b = a / const_0_2778
c = b - 5
|
a ) 4 , b ) 6 , c ) 8 , d ) 7 , e ) 12 | d | divide(multiply(3, add(6, const_1)), 3) | working together , wayne and his son can shovel the entire driveway in 3 hours . if wayne can shovel 6 times as fast as his son can , how many hours would it take for his son to shovel the entire driveway on his own ? | w : the time for wyane to do the job s : the time for his son to do the job we have 1 / w + 1 / s = 1 / 6 and w = 6 s then we have 1 / ( 6 * s ) + 1 / s = 1 / 6 < = > 7 / ( 6 * s ) = 1 / 6 < = > s = 7 ans : d | a = 6 + 1
b = 3 * a
c = b / 3
|
a ) 347.8 , b ) 987.8 , c ) 877.8 , d ) 898.8 , e ) 667.4 | d | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(21, 6), 2))), multiply(3, divide(multiply(21, 6), 2))), multiply(5, 21)) | the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 21 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | "let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 21 = 2 r a = 12 / 5 r and r = 63 a = 151.2 required total cost = 4 * 151.2 + 3 * 63 + 5 * 21 = 604.8 + 189 + 105 = $ 898.80 d" | a = 24 / 10
b = 21 * 6
c = b / 2
d = a * c
e = 4 * d
f = 21 * 6
g = f / 2
h = 3 * g
i = e + h
j = 5 * 21
k = i + j
|
a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | c | divide(multiply(25.5, const_100), 85) | how many pieces of 85 cm length can be cut from a rod of 25.5 meters long ? | "number of pieces = 2550 / 85 = 30 the answer is c ." | a = 25 * 5
b = a / 85
|
a ) 540 , b ) 1040 , c ) 2040 , d ) 1030 , e ) 15,30 | b | add(const_1000, multiply(divide(20, const_2), 4)) | a father is 4 times as old as his son . in 20 years , he ' ll be twice as old . how old are they now ? | if f is age of father and s is age of son f = 4 s f + 20 = 2 * ( s + 20 ) 4 s + 20 = 2 s + 40 2 s = 20 s = 10 f = 40 answer : b | a = 20 / 2
b = a * 4
c = 1000 + b
|
a ) 24 , b ) 30 , c ) 48 , d ) 54 , e ) 72 | d | add(add(factorial(subtract(divide(divide(divide(210, add(const_4, const_1)), const_3), const_2), const_3)), factorial(subtract(divide(divide(divide(210, add(const_4, const_1)), const_3), const_2), const_3))), factorial(subtract(divide(divide(divide(210, add(const_4, const_1)), const_3), const_2), subtract(divide(divide(divide(210, add(const_4, const_1)), const_3), const_2), const_3)))) | how many positive integers h less than 10000 are such that the product of their digits is 210 ? | 210 is the answer when 2 , 3 , 5 and 7 are multiplied . 210 can also be arrive using 56 and 7 and 1 , 5 , 6 and 7 . so sum of arrangements of 2357 , 567 and 1567 . this translates to 4 ! + 3 ! + 4 ! , this equals to 24 + 6 + 24 = 54 , d is the answer . | a = 4 + 1
b = 210 / a
c = b / 3
d = c / 2
e = d - 3
f = math.factorial(e)
g = 4 + 1
h = 210 / g
i = h / 3
j = i / 2
k = j - 3
l = math.factorial(k)
m = f + l
n = 4 + 1
o = 210 / n
p = o / 3
q = p / 2
r = 4 + 1
s = 210 / r
t = s / 3
u = t / 2
v = u - 3
w = q - v
x = math.factorial(w)
y = m + x
|
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