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a ) 160400 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 128111 | a | divide(multiply(1, 801), const_4) | what is the sum of all even numbers from 1 to 801 ? | "explanation : 800 / 2 = 400 400 * 401 = 160400 answer : a" | a = 1 * 801
b = a / 4
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a ) 300 , b ) 400 , c ) 562.5 , d ) 600 , e ) none of these | c | divide(225, divide(2, 5)) | there are 225 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 225 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 225 = 2 / 5 t , or 2 / 5 ( t - m ) = 225 , from here we get that t - m = 562.5 , that would be total number of female employees and the answer ( c ) | a = 2 / 5
b = 225 / a
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a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6 | d | divide(const_4, const_10) | if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 8 } . what is the probability that the median of the set will increase but the range still remains the same ? | "we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the same . the median of the set is ( 3 + 4 ) / 2 = 3.5 , thus the number selected must be 4 or greater . for the range to remain the same , the number must be between 2 and 8 inclusive . to satisfy both conditions , the number selected must be 4 , 5 , 6 , 7 , or 8 . the probability is 5 / 10 = 0.5 the answer is d ." | a = 4 / 10
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a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | b | divide(multiply(divide(multiply(divide(40, const_100), 300), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 300 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 remaning amount is 40 % i . e ( 2 / 5 ) x - - - - - - because school has already received 60 % and the remaining people are 60 % i . e ( 3 / 5 ) . t - - - - - because 40 % of the people are already solicited so , average contribution required from the remaining targeted people is ( 2 / 5 ) x = ( amount required ) . ( 3 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 divide eqn 1 by eqn 2 amount required = $ 177.78 b" | a = 40 / 100
b = a * 300
c = 60 / 100
d = b / c
e = 40 / 100
f = d * e
g = 60 / 100
h = f / g
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | floor(const_2) | how many prime numbers are between 15 / 7 and 131 / 6 ? | "15 / 7 = 3 - 131 / 6 = 22 - prime numbers between 3 and 22 are 5 , 7 , 11 , 13 , 17 , and 19 - sign signifies that the number is marginally less . answer d" | a = math.floor(2)
|
a ) 29 Β½ days , b ) 22 Β½ days , c ) 92 Β½ days , d ) 28 Β½ days , e ) 92 Β½ days | b | add(divide(const_1, 18), divide(const_1, 30)) | a can do a job in 18 days and b can do it in 30 days . a and b working together will finish twice the amount of work in days ? | "1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 45 / 4 = 11 ΒΌ * 2 = 22 Β½ days answer : b" | a = 1 / 18
b = 1 / 30
c = a + b
|
a ) 7.67 % , b ) 6 % , c ) 0.54 % , d ) 9.54 % , e ) 8.54 % | b | subtract(add(add(multiply(divide(4, add(add(4, 3), 2)), add(const_100, 10)), multiply(divide(3, add(add(4, 3), 2)), add(const_100, 8))), multiply(divide(2, add(add(4, 3), 2)), subtract(const_100, 5))), const_100) | the cost of manufacturing a popular model car is made up of 3 items : cost of raw material , labour and overheads - in a year the cost of 3 items were in the ration of 4 : 3 : 2 . next year the cost of the raw material rose by 10 % , labour cost increased by 8 % but overhead reduced by 5 % . then % increase int the price of the car ? | before increase total cost = 4 + 3 + 2 = 9 , after increasing the cost = 9.54 so increase of . 54 over 9 , equal to 6 % answer : b | a = 4 + 3
b = a + 2
c = 4 / b
d = 100 + 10
e = c * d
f = 4 + 3
g = f + 2
h = 3 / g
i = 100 + 8
j = h * i
k = e + j
l = 4 + 3
m = l + 2
n = 2 / m
o = 100 - 5
p = n * o
q = k + p
r = q - 100
|
a ) 40 , b ) 60 , c ) 135 , d ) 80 , e ) 100 | c | multiply(90, sqrt(divide(9, 4))) | two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 90 kmph , find the rate at which the train b runs . | "if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in β a β and β b β hours respectively ( i . e . a takes β a hrs β to travel from the meeting point to his destination and b takes β b hrs β to travel from the meeting point to his destination ) , then the ratio of their speeds is given by : sa / sb = β ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = β ( 4 / 9 ) = 2 / 3 this gives us that the ratio of the speed of a : speed of b as 2 : 3 . since speed of a is 80 kmph , speed of b must be 80 * ( 3 / 2 ) = 135 kmph c" | a = 9 / 4
b = math.sqrt(a)
c = 90 * b
|
a ) 73.9 , b ) 72.9 , c ) 73.0 , d ) 73.5 , e ) 92.2 | b | subtract(subtract(90, multiply(const_3, const_3)), multiply(divide(subtract(90, multiply(const_3, const_3)), 90), multiply(const_3, const_3))) | a vessel of capacity 90 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] n * t for the given problem , t = 90 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 90 - 9 ) / 90 ] 2 ( 90 ) = 72.9 litres . answer : option b" | a = 3 * 3
b = 90 - a
c = 3 * 3
d = 90 - c
e = d / 90
f = 3 * 3
g = e * f
h = b - g
|
a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 10 , d ) 1 / 12 , e ) 1 / 15 | b | divide(divide(multiply(divide(divide(377, 13), 29), 1), 4), 2) | 377 Γ· 13 Γ· 29 Γ 1 / 4 Γ· 2 = ? | explanation : 377 Γ· 13 Γ· 29 Γ 1 / 4 Γ· 2 = ? = > ? = ( 337 / ( 13 Γ 29 Γ 4 Γ 2 ) ) = 1 / 8 answer : option b | a = 377 / 13
b = a / 29
c = b * 1
d = c / 4
e = d / 2
|
a ) 2 ^ 15 , b ) 2 ^ 14 , c ) 2 ^ 13 , d ) 2 ^ 12 , e ) 2 ^ 11 | e | multiply(power(2, 8), 8) | if x = 2 ^ 8 and x ^ x = 2 ^ k , what is k ? | "solution : we know that x = 2 ^ 8 which implies x ^ x = ( 2 ^ 8 ) ^ ( 2 ^ 8 ) = 2 ^ ( 8 * 2 ^ 8 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 2 ^ ( 2 ^ 3 * 2 ^ 8 ) = 2 ^ ( 2 ^ ( 3 + 8 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 2 ^ ( 2 ^ 11 ) given that x ^ x = 2 ^ k so 2 ^ ( 2 ^ 11 ) = 2 ^ k since the base is same the exponent will also be same therefore k = 2 ^ 11 answer : e" | a = 2 ** 8
b = a * 8
|
a ) 22.8 kg , b ) 25.6 kg , c ) 28 kg , d ) 26.5 kg , e ) none of these | a | divide(multiply(6, 42.75), 11.25) | if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 6 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 6 : : 42.75 : x = > 11.25 x x = 6 x 42.75 = > x = ( 6 x 42.75 ) / 11.25 = > x = 22.8 answer : a" | a = 6 * 42
b = a / 11
|
a ) 75 , b ) 27.6 , c ) 66.6 , d ) 16.5 , e ) 11 | c | divide(add(add(add(add(70, 60), 78), 60), 65), add(const_2, const_3)) | david obtained 70 , 60 , 78 , 60 and 65 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | "explanation : average = ( 70 + 60 + 78 + 60 + 65 ) / 5 = 333 / 5 = 66.6 . answer : c" | a = 70 + 60
b = a + 78
c = b + 60
d = c + 65
e = 2 + 3
f = d / e
|
a ) 36 mph , b ) 40 mph , c ) 44 mph , d ) 42 mph , e ) 58 mph | d | divide(add(60, 65), add(divide(60, 30), divide(65, 65))) | a car drives 60 miles on local roads at 30 mph , and 65 miles on the highway at 65 mph , what is the average speed of the entire trip ? | "so the answer is plainly d . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 60 miles , and v = 30 mph so we can obtain time : 30 * t = 60 or t = 60 / 30 = 2 hours the needed time to cover 60 miles in the same way we should divide 65 by 65 to find the needed time to cover 65 miles , so t = 1 hours so the total time for covering total distance would be 2 + 1 = 3 hours and total distance is 60 + 65 = 125 miles final stage : average speed is total distance divide to total time : 125 / 3 = 41.6 miles per hour that rounds up to 42 . . . ." | a = 60 + 65
b = 60 / 30
c = 65 / 65
d = b + c
e = a / d
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(7, 4) | set a consists of the integers from 4 to 15 , inclusive , while set b consists of the integers from 7 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 4 , 5,6 , 7,8 , 9,10 , 11,12 } b = { 6 , 7,8 , 9,10 , 11,12 . . . } thus we see that there are 9 distinct integers that are common to both . d is the correct answer ." | a = 7 + 4
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a ) 25 % , b ) 30 % , c ) 50 % , d ) 65 % , e ) 75 % | e | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | the length of a rectangle is halved while its width is also halved . what is the % change in area ? | "the original area is l * w the new area is 0.5 l * 0.5 w = 0.25 * l * w = l * w - 0.75 * l * w the area decreased by 75 % . the answer is e ." | a = 1 / 2
b = 3 * a
c = b - 1
d = c / 1
e = d * 100
|
['a ) 124 m 2', 'b ) 120 m 2', 'c ) 108 m 2', 'd ) 58 m 2', 'e ) none of these'] | a | subtract(rectangle_area(add(multiply(2, const_2), 15), add(12, multiply(2, const_2))), rectangle_area(15, 12)) | the floor of a rectangular room is 15 m long and 12 m wide . the room is surrounded by a verandah of width 2 m on all its sides . the area of the verandah is : | area of the outer rectangle = 19 Γ 16 = 304 m 2 area of the inner rectangle = 15 Γ 12 = 180 m 2 required area = ( 304 β 180 ) = 124 m 2 answer a | a = 2 * 2
b = a + 15
c = 2 * 2
d = 12 + c
e = rectangle_area - (
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a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | d | multiply(6600, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 6600 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 6600 x ( 2 / 11 ) ) = rs . 1200 . answer is d ." | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 6600 * f
|
a ) 1 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | b | multiply(divide(divide(multiply(40, 40), const_100), 64), const_12) | a reduction of 40 % in the price of apples would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "b 40 * ( 40 / 100 ) = 16 - - - 64 ? - - - 12 = > rs . 3" | a = 40 * 40
b = a / 100
c = b / 64
d = c * 12
|
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | c | add(add(11, divide(subtract(11, 5), 3)), 7) | a , b , c , d and e are 5 consecutive points on a straight line . if bc = 3 cd , de = 7 , ab = 5 and ac = 11 , what is the length of ae ? | ac = 11 and ab = 5 , so bc = 6 . bc = 3 cd so cd = 2 . the length of ae is ab + bc + cd + de = 5 + 6 + 2 + 7 = 20 the answer is c . | a = 11 - 5
b = a / 3
c = 11 + b
d = c + 7
|
a ) 6.28 % , b ) 7.14 % , c ) 7.86 % , d ) 8.25 % , e ) 8.63 % | b | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 3,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 7000 = 1 / 14 which is about 7.14 % the answer is b ." | a = 100 * 100
b = 5 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 57 : 11 , b ) - 57 : 11 , c ) 11 : 10 , d ) - 57 : 25 , e ) - 1 : 10 | d | divide(add(multiply(5, 3), multiply(7, 6)), subtract(3, multiply(4, 7))) | if a : b : : 3 : 7 , then what is ( 5 a + 6 b ) : ( a - 4 b ) ? | "a / b = 3 / 7 dividing numerator & denominator of ' ( 5 a + 6 b ) / ( a - 2 b ) ' by b , [ 5 ( a / b ) + 6 ] / [ ( a / b ) - 4 ] = [ 5 * ( 3 / 7 ) + 6 ] / [ ( 3 / 7 ) - 4 ] = - 57 / 25 answer : d" | a = 5 * 3
b = 7 * 6
c = a + b
d = 4 * 7
e = 3 - d
f = c / e
|
a ) 18 , b ) 100 , c ) 90 , d ) 45 , e ) 120 | c | divide(multiply(const_3, 165), divide(add(multiply(const_4, const_2), 3), const_2)) | of 3 numbers , the third is 4 times the second and the second is two times the first . if their average is 165 , the second number of the 3 numbers is : | explanation : let first number be x . so , 2 nd no . = 2 x & 3 rd no . = 8 x . so , x + 2 x + 8 x = 165 Γ 3 = 495 11 x = 495 / 11 x = 495 / 11 hence , first number is x = 45 so , second number is = 2 x 45 = 90 answer : c | a = 3 * 165
b = 4 * 2
c = b + 3
d = c / 2
e = a / d
|
a ) 223.28 , b ) 293.28 , c ) 293.78 , d ) 291.28 , e ) 193.28 | b | divide(multiply(16, sqrt(subtract(power(multiply(const_2, 20), const_2), power(16, const_2)))), const_2) | the side of a rhombus is 20 m and length of one of its diagonals is 16 m . the area of the rhombus is ? | "area of the rhombus = 1 / 2 * p * β 4 ( a ) 2 - ( p ) 2 a = 20 ; p = 16 a = 1 / 2 * 16 * β 4 ( 20 ) 2 - ( 16 ) 2 = 1 / 2 * 16 * β 1600 - 256 = 1 / 2 * 16 * β 1344 a = 293.28 answer : b" | a = 2 * 20
b = a ** 2
c = 16 ** 2
d = b - c
e = math.sqrt(d)
f = 16 * e
g = f / 2
|
a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 210 | a | multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(50, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(50, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 50 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ? | "20 % of the workerswithoutretirement plan are women , this means 80 % of the workerswithoutretirement plan are men right ? makes 100 % = no retirement plan . then 40 % of those wo have a retirement plan are men , which means 60 % of those who have a retirement plan are women . these percentages are always refering to the subgroupsretirement plan = yesandno retirement plan . make the total workers equal x and work from there and it should be fine > x / 3 have no retirement plan at all . i struggled with the question a lot , it was hard for me although usually i have no problems with sets questions like this . i believe it must be a 700 + question . yes , you are right . i read the question incorrectly ! i glazed overnotfor some reason . if 20 % of the workers who do not have a retirement plan are women , 80 % who do not have retirement plan are men . also 40 % who do have retirement plan are men . since 1 / 3 do not have retirement plan and 2 / 3 do have , it becomes a simple weighted average question . cavg = ( 80 * 1 + 50 * 2 ) / 3 = 60 % are men total = 120 * 100 / 60 = 200 no . of women = 200 - 120 = 80 a" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 50 / 100
i = g * h
j = e + i
k = 120 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 50 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) 22 sec , b ) 27 sec , c ) 25 sec , d ) 15 sec , e ) 11 sec | d | divide(add(80, 70), multiply(36, const_0_2778)) | how many seconds will a train 70 meters long take to cross a bridge 80 meters long if the speed of the train is 36 kmph ? | "explanation : d = 70 + 80 = 150 s = 36 * 5 / 18 = 10 mps t = 150 / 10 = 15 sec answer : option d" | a = 80 + 70
b = 36 * const_0_2778
c = a / b
|
a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 3 | e | subtract(divide(subtract(90, 30), subtract(35, 30)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 30 units . if today ' s production of 90 units raises the average to 35 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 30 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 30 n + 90 = 35 * ( n + 1 ) - - > n = 3 . answer : e ." | a = 90 - 30
b = 35 - 30
c = a / b
d = c - 1
|
a ) 4 , b ) 8 , c ) 11 , d ) 12 , e ) 15 | b | divide(divide(80, 5), const_2) | how many unique positive odd integers less than 80 are equal to the product of a positive multiple of 5 and an odd number ? | "the question basically asks how many positive odd integers less than 80 are odd multiples of 5 so we have 5,15 , 25,35 , 45,55 , 65,75 = 8 b" | a = 80 / 5
b = a / 2
|
a ) 228 , b ) 744 , c ) 490 , d ) 199 , e ) 231 | c | divide(multiply(divide(280, divide(subtract(70, subtract(const_100, 70)), const_100)), 70), const_100) | there were two candidates in an election . winner candidate received 70 % of votes and won the election by 280 votes . find the number of votes casted to the winning candidate ? | "w = 70 % l = 30 % 70 % - 30 % = 40 % 40 % - - - - - - - - 280 70 % - - - - - - - - ? = > 280 / 40 * 70 = 490 answer : c" | a = 100 - 70
b = 70 - a
c = b / 100
d = 280 / c
e = d * 70
f = e / 100
|
a ) 9.5 , b ) 8.5 , c ) 7.2 , d ) 7 , e ) 8 | c | add(divide(multiply(16, 36), const_100), divide(multiply(13, 45), const_100)) | add 16 % of 36 and 13 % of 45 . | "16 % of 36 + 13 % of 45 36 * 16 / 100 + 45 * 13 / 100 5.8 + 1.4 = 7.2 answer c" | a = 16 * 36
b = a / 100
c = 13 * 45
d = c / 100
e = b + d
|
a ) 11 days , b ) 13 days , c ) 20 3 / 17 days , d ) 21 days , e ) none of these | b | inverse(add(divide(const_1, 23), divide(const_1, multiply(add(divide(30, const_100), const_1), 23)))) | a is 30 % more efficient than b . how much time will they , working together , take to complete a job which a alone could have done in 23 days ? | "explanation : ratio of times taken by a and b = 100 : 130 = 10 : 13 suppose b takes x days to do the work . x = ( 23 * 13 ) / 10 = 299 / 10 a ' s 1 day work = 1 / 23 ; b ' s 1 day work = 10 / 299 ( a + b ) ' s 1 day work = ( 1 / 23 + 10 / 299 ) = 1 / 13 a and b together can complete the job in 13 days . answer is b" | a = 1 / 23
b = 30 / 100
c = b + 1
d = c * 23
e = 1 / d
f = a + e
g = 1/(f)
|
a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | b | multiply(subtract(multiply(12000, divide(subtract(multiply(const_3, const_4), 3), multiply(const_3, const_4))), multiply(9000, divide(subtract(multiply(const_3, const_4), 4), multiply(const_3, const_4)))), divide(3850, add(add(18000, multiply(12000, divide(subtract(multiply(const_3, const_4), 3), multiply(const_3, const_4)))), multiply(9000, divide(subtract(multiply(const_3, const_4), 4), multiply(const_3, const_4)))))) | suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3850 . what is the difference between rohan β s and sudhir β s share in the profit ? | suresh : rohan : sudhir ratio of their investments = 18000 Γ 12 : 12000 Γ 9 : 9000 Γ 8 = 6 : 3 : 2 the difference between rohan β s and sudhir β s share = 1 share : . i . e . = rs . 3850 Γ 1 / 11 = rs . 350 . b | a = 3 * 4
b = a - 3
c = 3 * 4
d = b / c
e = 12000 * d
f = 3 * 4
g = f - 4
h = 3 * 4
i = g / h
j = 9000 * i
k = e - j
l = 3 * 4
m = l - 3
n = 3 * 4
o = m / n
p = 12000 * o
q = 18000 + p
r = 3 * 4
s = r - 4
t = 3 * 4
u = s / t
v = 9000 * u
w = q + v
x = 3850 / w
y = k * x
|
a ) 229 , b ) 240 , c ) 288 , d ) 277 , e ) 320 | e | multiply(20, multiply(72, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 72 km / hr . what is the length of the platform ? | "speed = 72 * 5 / 18 = 20 m / sec . length of the train = 20 * 20 = 400 m . let the length of the platform be x m . then , ( x + 400 ) / 36 = 20 = > x = 320 m . answer : e" | a = 72 * const_0_2778
b = 20 * a
|
a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 57 % | b | add(multiply(60, divide(75, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 75 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ? | "registered voters = 100 d = 60 r = 40 75 % of d ( 60 ) = 45 20 % of r ( 40 ) = 8 total voter % of registered voters = 53 / 100 53 % answer : b" | a = 75 / 100
b = 60 * a
c = 100 - 60
d = 20 / 100
e = c * d
f = b + e
|
a ) 89 , b ) 20 , c ) 36 , d ) 41 , e ) 34 | d | divide(add(200, 210), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 200 m ahead of the engine of a 210 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 200 + 210 = 410 m . time taken = 410 / 10 = 41 sec . answer : option d" | a = 200 + 210
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 5 / 24 , b ) 6 / 24 , c ) 7 / 24 , d ) 1 / 6 , e ) 9 / 24 | d | subtract(divide(6, 8), divide(4, 18)) | if 4 / p = 8 & 6 / q = 18 then p - q = ? | "p = 4 / 8 , q = 6 / 18 = > q = 1 / 3 therefore p - q = ( 1 / 2 ) - ( 1 / 3 ) = 1 / 6 answer : d" | a = 6 / 8
b = 4 / 18
c = a - b
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a ) 40 , b ) 25 , c ) 30 , d ) 50 , e ) 35 | b | divide(add(30, 20), const_2) | a box contain the number of balls which is as much greater than 20 is less than 30 . the number of balls is ? | "answer let the number be x . then , x - 20 = 30 - x Γ’ β‘ β 2 x = 30 + 20 = 50 Γ’ β‘ β 2 x = 50 Γ’ Λ Β΄ x = 25 correct option : b" | a = 30 + 20
b = a / 2
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a ) 200 , b ) 210 , c ) 205 , d ) 225 , e ) 240 | e | multiply(divide(multiply(const_60.0, const_3), 10), 40) | if 10 typists can type 30 letters in 30 minutes , then how many letters will 40 typists working at the same rate complete in 1 hour ? | "no . of letters typing by 10 typists in 30 minutes = 30 no . of letters typing by 10 typists in 60 minutes = 30 * 2 = 60 no . of letters typing by 40 typists in 60 minutes = 60 / 10 * 40 = 240 answer : e" | a = const_60 * 0
b = a / 10
c = b * 40
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a ) 40 , b ) 50 , c ) 53 , d ) 55 , e ) 60 | a | divide(80, divide(add(multiply(subtract(10, 9), const_60), subtract(20, 10)), const_60)) | a car going at 20 miles per hour set out on an 80 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 20 m per hour = 1.5 hours distance travelled by car a = 20 * 1.5 = 30 miles since car b catches up car a at 10 : 30 , time = 80 mins = 4 / 3 hour speed of car b = 30 / ( 4 / 3 ) = 40 miles per hour answer a" | a = 10 - 9
b = a * const_60
c = 20 - 10
d = b + c
e = d / const_60
f = 80 / e
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a ) 45 , b ) 50 , c ) 60 , d ) 75 , e ) 90 | c | multiply(divide(90, const_3), 2) | the ratio of number of boys and girls in a school is 1 : 2 . if there are 90 students in the school , find the number of girls in the school ? | let the number of boys and girls be x and 2 x total students = 100 number of girls in the school = 2 * 90 / 3 = 60 answer is c | a = 90 / 3
b = a * 2
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a ) 386 , b ) 378 , c ) 396 , d ) 395 , e ) 394 | b | multiply(divide(1638, add(add(2, 3), 4)), 4) | an amount of rs . 1638 was divided among a , b and c , in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of c ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 / 12 : 4 / 12 : 3 / 12 = 6 : 4 : 3 . share of c = 3 / 13 * 1638 = rs . 378 answer : b" | a = 2 + 3
b = a + 4
c = 1638 / b
d = c * 4
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a ) $ 160 , b ) $ 96 , c ) $ 90 , d ) $ 80 , e ) $ 100 | e | divide(divide(720, divide(add(20, const_100), const_100)), add(3, 3)) | a group of 3 investors and 3 clients recently frequented the chinese luyang restaurant . the total bill for the meal , including 20 % gratuity , came to $ 720 . on average , how much did the meal of each individual cost before gratuity ? | 3 investors and 3 clients - so total 6 people the bill $ 720 includes 20 % gratuity . . . so the actual cost of dinner was $ 600 now , the cost per person will be $ 600 / 6 which is $ 100 option e | a = 20 + 100
b = a / 100
c = 720 / b
d = 3 + 3
e = c / d
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a ) 2 , b ) 8 , c ) 4 , d ) 6 , e ) 5 | c | subtract(10, divide(30, 5)) | a man can row a distance of 5 km in 60 min with the help of the tide . the direction of the tide reverses with the same speed . now he travels a further 30 km in 10 hours . how much time he would have saved if the direction of tide has not changed ? | "explanation : he covered 5 km in 1 hour , so he might cover 30 km in 6 hours . but he took 10 hours . he would have saved 10 Γ’ β¬ β 6 = 4 hours . answer : c" | a = 30 / 5
b = 10 - a
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a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 48 / 7 minutes | e | divide(subtract(10, divide(10, divide(add(4, 10), subtract(10, 4)))), const_1) | a man cycling along the road noticed that every 10 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ? | "let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 10 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 10 , d = 10 b - 10 c ; every 4 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = 4 , d = 4 b + 4 c ; d = 10 b - 10 c = 4 b + 4 c , - - > b = 7 c / 3 , - - > d = 12 b - 12 b * 3 / 7 = 48 / 7 b . interval = \ frac { d } { b } = \ frac { 48 / 7 b } { b } = 48 / 7 answer : e ( 48 / 7 minutes ) ." | a = 4 + 10
b = 10 - 4
c = a / b
d = 10 / c
e = 10 - d
f = e / 1
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a ) 10 : 6 , b ) 5 : 1 , c ) 15 : 3 , d ) 25 : 3 , e ) 30 : 3 | b | divide(2, 5) | the simple form of the ratio 4 / 2 : 2 / 5 is ? | "4 / 2 : 2 / 5 = 20 : 4 = 5 : 1 answer : b" | a = 2 / 5
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a ) 33 , b ) 21 , c ) 38 , d ) 40 , e ) 92 | d | add(37, divide(multiply(3, 35), 35)) | the average score of 35 students in a class is 37 . if every student is given 3 grace marks , the new average of the class is : | explanation : average score = 37 grace mark 3 is given to 35 student then its average will be 3 . hence new average = 37 + 3 = 40 answer : d | a = 3 * 35
b = a / 35
c = 37 + b
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a ) 1 / 7 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 11 | e | divide(3, add(multiply(4, 2), 3)) | chris mixed 3 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 2 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ? | 1 lbs of raisin = $ 1 3 lbs of raisin = $ 3 1 lbs of nuts = $ 2 4 lbs of nuts = $ 8 total value of mixture = 8 + 3 = 11 fraction of the value of raisin = 3 / 11 ans : e | a = 4 * 2
b = a + 3
c = 3 / b
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(volume_rectangular_prism(82, 83, 84), volume_cube(89)) | 82 * 83 * 84 * . . 89 what is the digit in units place ? | "in the product 82 * 83 * 84 * 85 * 86 . . . . . . * 89 last ( unit ) digit will be zero as 82 * 85 will give zero as last ( unit ) digit and further multiplication with any number will also give zero at unit place . answer : a" | a = volume_rectangular_prism / (
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a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | d | divide(choose(20, 3), choose(add(20, 20), 3)) | a bag contains 20 red jellybeans and 20 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 3 / 20 answer : option d" | a = math.comb(20, 3)
b = 20 + 20
c = math.comb(b, 3)
d = a / c
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a ) 800 , b ) 900 , c ) 100 , d ) 700 , e ) 500 | c | add(multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(3, 10)), multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(1, 8))) | a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on conveyence . if he has rs . 1400 left with him , find his expenditure on food and conveyence . | "let the salary be x , therefore house rent = 2 x / 5 , food = 3 x / 10 , conveyence = x / 8 . it is given that left over salary is 1400 / - so , x - ( 2 x / 5 + 3 x / 10 + x / 10 ) = 1400 by solving x = 8000 expenditure on food = 3200 and conveyence = 1000 answer : c" | a = 2 / 5
b = 3 / 10
c = a + b
d = 1 / 8
e = c + d
f = 1 - e
g = 1400 / f
h = 3 / 10
i = g * h
j = 2 / 5
k = 3 / 10
l = j + k
m = 1 / 8
n = l + m
o = 1 - n
p = 1400 / o
q = 1 / 8
r = p * q
s = i + r
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a ) 78 , b ) 80 , c ) 87 , d ) 91 , e ) 94 | e | add(78, 16) | there are 17 students in a certain class . on the day the test was given , taqeesha was absent . the other 16 students took the test , and their average was 77 . the next day , taqeesha took the test , and with her grade included , the new average is 78 . what is taqeesha β s grade on the test ? | explanations 1 ) the average of the first 16 students is 77 . this means , the sum of these 16 scores is sum = ( average ) * ( number of scores ) = 77 * 16 = 1232 once taqeesha takes her test , the average of all 17 scores is 78 . this means , the sum of these 17 scores is : sum = ( average ) * ( number of scores ) = 78 * 17 = 1326 once we had the sum of the 16 scores , all we had to do was add taqeesha β s score to that total to get the sum of all 17 . therefore , the difference in these two sums is taqeesha β s score . 1326 β 1232 = 94 . answer : e . | a = 78 + 16
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a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 52 | e | multiply(divide(const_100, add(const_100, 15)), 60) | from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 15 percent . the enrollment on january 1 , 1993 , was 60 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "1.15 x = 60 - - > 23 / 20 * x = 60 - - > x = 60 * 20 / 23 = 1200 / 23 = ~ 52 . answer : e ." | a = 100 + 15
b = 100 / a
c = b * 60
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a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | c | divide(divide(multiply(800, 30), const_100), 5) | a reduction of 30 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 30 / 100 ) = 240 - - - - 5 ? - - - - 1 = > rs . 48 answer : c" | a = 800 * 30
b = a / 100
c = b / 5
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a ) 2652 , b ) 2578 , c ) 2970 , d ) 3354 , e ) 4234 | c | multiply(add(divide(subtract(109, subtract(4, const_1)), const_2), const_1), add(add(divide(subtract(109, subtract(4, const_1)), const_2), const_1), const_1)) | a , b , c and d are 4 consecutive numbers . if the sum of a and dis 109 , what is the product of b and c ? | c 2970 here d = a + 3 a + a + 3 = 109 2 a = 106 a = 53 so , numbers are 53 , 54 , 55 and 56 β΄ b Γ c = 54 Γ 55 = 2970 | a = 4 - 1
b = 109 - a
c = b / 2
d = c + 1
e = 4 - 1
f = 109 - e
g = f / 2
h = g + 1
i = h + 1
j = d * i
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a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent q ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . q = [ ( 2 - 1 ) / 1 ] * 100 % = 100 %" | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
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a ) 120 , b ) 140 , c ) 160 , d ) 180 , e ) 200 | c | divide(40, subtract(subtract(const_1, inverse(25)), divide(const_1, const_2))) | a student traveled 25 percent of the distance of the trip alone , continued another 40 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.25 x + 40 miles + 0.5 x = x 40 miles = 0.25 x x = 160 miles the answer is c ." | a = 1/(25)
b = 1 - a
c = 1 / 2
d = b - c
e = 40 / d
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a ) 1 / 15 , b ) 4 / 5 , c ) 4 / 15 , d ) 1 / 5 , e ) 2 / 5 | d | subtract(const_1, divide(4, 5)) | of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 4 times the fraction of first - year students who have declared a major , what fraction of all the students in the dormitory are second - year students who have not declared a major ? | tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 4 ( 1 / 5 ( x / 2 ) ) = 2 / 5 ( x ) - - - > non major = x / 2 - 2 / 5 ( x ) = 1 / 5 ( x ) hence 1 / 5 d | a = 4 / 5
b = 1 - a
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a ) $ 3500 , b ) $ 5000 , c ) $ 4830 , d ) $ 7000 , e ) $ 10000 | c | multiply(divide(34500, const_100), subtract(42, 28)) | if the personal income tax rate is lowered from 42 % to 28 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 34500 ? | "saving = ( 42 - 28 ) % of 34500 = 4830 . answer : c" | a = 34500 / 100
b = 42 - 28
c = a * b
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a ) 4 , b ) 5 , c ) 7 , d ) 9 , e ) 3 | e | divide(add(14, 4), add(5, 5)) | solve the equation for x : 5 x - 14 + 3 x = 4 + 9 - x | "e 3 8 x + x = 13 + 14 9 x = 27 = > x = 3" | a = 14 + 4
b = 5 + 5
c = a / b
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a ) 4 , b ) 3 , c ) 5 , d ) 2 , e ) 1 | c | add(reminder(2, 6), reminder(3, 6)) | the positive integers a and b leave remainders of 2 and 3 , respectively , when divided by 6 . a > b . what is the remainder when a β b is divided by 6 ? | let ' s test out some values of m and n that satisfy the given information . a leaves are remainder of 2 when divided by 6 so , a could equal 8 b leaves are remainder of 3 when divided by 6 so , b could equal 3 what is the remainder when a β b is divided by 6 ? so , a - b = 8 - 3 = 5 , and when we divide 5 by 6 , we get 0 with remainder 5 answer : c | a = reminder + (
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a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | a | add(multiply(35, 2), multiply(3, 55)) | john left home and drove at the rate of 35 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 35 * 2 + 3 * 55 = 235 miles . answer a" | a = 35 * 2
b = 3 * 55
c = a + b
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a ) 60 , b ) 90 , c ) 120 , d ) 150 , e ) 180 | c | divide(multiply(30, divide(40, const_100)), subtract(divide(80, const_100), divide(70, const_100))) | a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 70 percent of its games that season , what was the total number of games that the team played ? | "70 % is 30 % - points above 40 % and 10 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 1 : 3 . so the team played a total of 30 + 90 = 120 games . the answer is c ." | a = 40 / 100
b = 30 * a
c = 80 / 100
d = 70 / 100
e = c - d
f = b / e
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a ) 36 , b ) 42 , c ) 72 , d ) 400 , e ) 56 | a | multiply(divide(120, 10), 3) | mangoes , oranges and apples in a fruit shop are in the ratio of 10 : 2 : 3 . if there are 120 mangoes , the number of apples in the shop is : . | explanation : let mangoes = 10 x , oranges = 2 x & apples = 3 x . now , 10 x = 120 hence x = 12 . number of apples = 3 x = 36 . answer : a | a = 120 / 10
b = a * 3
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a ) 15 , b ) 25 , c ) 35 , d ) 13 , e ) 55 | d | subtract(add(add(20, 40), 60), add(add(multiply(9, const_3), 10), 70)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 9 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 9 = 31 sum of second list = 31 * 3 = 93 therefore the number = 93 - 80 = 13 answer : d" | a = 20 + 40
b = a + 60
c = 9 * 3
d = c + 10
e = d + 70
f = b - e
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a ) 9 . , b ) 12 . , c ) 15 . , d ) 16 , e ) 18 . | c | add(const_4.0, multiply(6, 2)) | if 4 xz + yw = 3 and xw + yz = 6 , what is the value of the expression ( 2 x + y ) ( 2 z + w ) ? | "expand this to get : ( 2 x + y ) ( 2 z + w ) = 4 xz + 2 xw + 2 yz + yw = 4 xz + yw + 2 xw + 2 yz [ i rearranged the terms ] = 3 + 2 xw + 2 yz [ replaced 4 xz + yw with 3 ] = 3 + 2 ( xw + yz ) factored out a 2 = 3 + 2 ( 6 ) = 3 + 12 = 15 answer : c" | a = 6 * 2
b = 4 + 0
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a ) 14 , b ) 66 , c ) 88 , d ) 99 , e ) 12 | a | add(add(power(add(add(divide(subtract(subtract(42, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(42, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(42, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(42, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 42 . find the middle number of the three ? | "three consecutive even numbers ( 2 p - 2 ) , 2 p , ( 2 p + 2 ) . ( 2 p - 2 ) + 2 p + ( 2 p + 2 ) = 42 6 p = 42 = > p = 7 . the middle number is : 2 p = 14 . answer : a" | a = 42 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 42 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 42 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 42 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 64 % , b ) 68 % , c ) 72 % , d ) 76 % , e ) 80 % | a | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "say may income = 100 l ' s income = 60 and rest of the family = 40 in june l ' s income = 60 * 120 / 100 = 72 so 72 / 72 + 40 = 64 % answer : a" | a = 100 + 20
b = 100 + 20
c = b + 100
d = a / c
e = d * 100
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(105, add(15, 20)) | two cars start at the same time from opposite ends of a highway that is 105 miles long . one car is riding at 15 mph and the second car is riding at 20 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 20 + 15 = 35 mph total distance to be covered = 105 miles . time taken would be : 105 miles / 35 mph = 3 hours c is the answer ." | a = 15 + 20
b = 105 / a
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a ) 2 year , b ) 8 year , c ) 10 year , d ) none of these , e ) 9 year | a | divide(subtract(divide(40, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | the sum of the ages of 5 children born at the intervals of 3 year each is 40 year . what is the age of the youngest child ? | "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 40 Γ’ β‘ β 5 x = 10 Γ’ β‘ β x = 2 . Γ’ Λ Β΄ age of the youngest child = x = 2 years . answer a" | a = 5 / 3
b = 40 / a
c = 5 - 1
d = c * 3
e = b - d
f = e / 3
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | c | subtract(multiply(15, 15), add(multiply(6, 14), multiply(9, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 6 students is 14 years and that of the other 9 students is 16 years , the age of the 15 th student is | "explanation : age of the 15 th student = [ 15 * 15 - ( 14 * 6 + 16 * 8 ) ] = 13 years . answer : c" | a = 15 * 15
b = 6 * 14
c = 9 * 16
d = b + c
e = a - d
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a ) 550 , b ) 500 , c ) 250 , d ) 300 , e ) 400 | b | divide(multiply(6800, 10), 136) | find the annual income derived by investing $ 6800 in 10 % stock at 136 . | "by investing $ 136 , income obtained = $ 10 . by investing $ 6800 , income obtained = $ [ ( 10 / 136 ) * 6800 ] = $ 500 . answer b ." | a = 6800 * 10
b = a / 136
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a ) 470 , b ) 468 , c ) 465 , d ) 463 , e ) 528 | e | add(divide(divide(32, divide(divide(divide(divide(divide(32, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(32, divide(divide(divide(divide(divide(32, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 32 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 32 ( 32 + 1 ) / 2 = 32 ( 33 ) / 2 = 528 answer : option e" | a = 32 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 32 / e
g = f / 2
h = 32 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 32 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
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a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 380 | e | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | how many 4 digit numbers are there , if it is known that the first digit is odd , the second is even , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 2 can be used only once ? | "4 options for the first digit : 1 , 3 , 5 , 7 , 9 ; 5 options for the second digit : 0 , 2 , 4 , 6 , 8 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 5 * 5 * 4 * 4 = 400 numbers with two 2 - s , 2 x 2 x 5 * 1 * 1 * 4 = 20 . thus there are 400 - 20 = 380 such numbers . answer : e ." | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 * 4
e = c * d
f = 4 + 1
g = 4 + 1
h = f * g
i = h * 4
j = e - i
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a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | e | subtract(subtract(60, 16), 18) | there are 60 people and 3 clubs m , s , and z in a company . 16 people joined m , 18 people joined s and 11 people joined z . if the members of m did not join any other club , at most , how many people of the company did not join any club ? | answer e . key is ' atmost ' here . so 16 people joined m . 18 people joined s and 11 people joined z . to get maximum people who did not join any club , 11 people who joined z are among 18 people who joined s . so 18 + 16 = 34 60 - 34 = 26 | a = 60 - 16
b = a - 18
|
a ) 125 , b ) 126 , c ) 130 , d ) 148 , e ) 152 | e | add(add(add(multiply(3, 14), multiply(3, 15)), multiply(3, 18)), 11) | 3 * 14 + 3 * 15 + 3 * 18 + 11 = ? | "3 * 14 + 3 * 15 + 3 * 18 + 11 = 42 + 45 + 54 + 11 = 152 the answer is e ." | a = 3 * 14
b = 3 * 15
c = a + b
d = 3 * 18
e = c + d
f = e + 11
|
a ) 3 , b ) 35 , c ) 50 , d ) 99 , e ) 101 | a | floor(add(const_1, multiply(divide(log(2), log(const_10)), 10))) | how many digits 2 ^ 10 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 therefore 3 digits would be my best guess a" | a = math.log(2)
b = math.log(10)
c = a / b
d = c * 10
e = 1 + d
f = math.floor(e)
|
a ) 545 , b ) 525 , c ) 535 , d ) 625 , e ) 105 | a | subtract(550, divide(104, 20.8)) | 550 β ( 104 Γ· 20.8 ) = ? | "explanation : = 550 β ( 104 / 208 x 10 ) = 550 β 5 = 545 option a" | a = 104 / 20
b = 550 - a
|
a ) 13 , b ) 17 , c ) 19 , d ) 23 , e ) 29 | a | floor(divide(4, divide(6, const_2))) | what is the greatest prime factor of 4 ^ 6 - 1 ? | "4 ^ 6 - 1 = ( 4 ^ 3 - 1 ) ( 4 ^ 3 + 1 ) = 63 * 65 = 7 * 9 * 5 * 13 the answer is a ." | a = 6 / 2
b = 4 / a
c = math.floor(b)
|
a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | b | inverse(add(divide(const_12, 4), const_1)) | a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 4 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay , what fraction of her take - home pay did she save each month ? | "let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 4 ( 1 - x ) p 12 xp = 4 p - 4 xp 16 xp = 4 p x = 1 / 4 the answer is b ." | a = 12 / 4
b = a + 1
c = 1/(b)
|
a ) 277 , b ) 620 , c ) 640 , d ) 678 , e ) 121 | b | multiply(620, divide(10, const_100)) | find the simple interest on rs . 620 for 10 months at 10 paisa per month ? | "i = ( 620 * 10 * 10 ) / 100 = 620 answer : b" | a = 10 / 100
b = 620 * a
|
a ) 1 hour , b ) 1.1 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours | b | divide(const_1, subtract(const_1, divide(const_1, multiply(4, const_2)))) | one pump drains one - half of a pond in 4 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? | "first pump drains 1 / 2 of the tank in 4 hours so 8 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 / 8 + 1 / a ) part in 1 hour son in 1 / 2 hour they drain 1 / 2 * ( 1 / 8 + 1 / a ) part of the tank given that in 1 / 2 hour they drain 1 / 2 of the tank hence we can say 1 / 2 * ( 1 / 8 + 1 / a ) = 1 / 2 solving u get a = 8 / 7 = 1.1 hence answer is b" | a = 4 * 2
b = 1 / a
c = 1 - b
d = 1 / c
|
a ) 0.1 , b ) 0.4 , c ) 0.82 , d ) 0.8 , e ) none | b | divide(multiply(0.60, 4), 6) | if 0.60 : x : : 6 : 4 , then x is equal to | "sol . ( x Γ 6 ) = ( 0.60 Γ 4 ) β x = 2.4 / 6 = 0.40 . answer b" | a = 0 * 60
b = a / 6
|
a ) $ 420 , b ) $ 440 , c ) $ 450 , d ) $ 460 , e ) $ 480 | d | divide(subtract(multiply(400, 10), multiply(360, 6)), 4) | in a small snack shop , the average ( arithmetic mean ) revenue was $ 400 per day over a 10 - day period . during this period , if the average daily revenue was $ 360 for the first 6 days , what was the average daily revenue for the last 4 days ? | the average revenue was $ 400 per day over a 10 - day period - - > total revenue over 10 - day period is 10 * $ 400 = $ 4000 ; the average daily revenue was $ 360 for the first 6 days - - > total revenue for the first 6 days is 6 * $ 360 = 2160 ; total revenue for the last 4 days is $ 4000 - 2160 = $ 1840 ; average daily revenue for the last 4 days is $ 1840 / 4 = $ 460 . answer : d . | a = 400 * 10
b = 360 * 6
c = a - b
d = c / 4
|
a ) 101 Β° f , b ) 108 Β° f , c ) 116 Β° f , d ) 122 Β° f , e ) 127 Β° f | d | add(multiply(divide(subtract(212, 32), 100), 50), 32) | water boils at 212 Β° f or 100 Β° c and ice melts at 32 Β° f or 0 Β° c . if the temperature of a pot of water is 50 Β° c , what is the temperature of the pot of water in Β° f ? | let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) f = 9 c / 5 + 32 f = 9 ( 50 ) / 5 + 32 = 122 Β° f the answer is d . | a = 212 - 32
b = a / 100
c = b * 50
d = c + 32
|
a ) 333 , b ) 383 , c ) 402 , d ) 433 , e ) 483 | c | divide(multiply(multiply(multiply(1000, add(const_1, divide(10, const_100))), add(const_1, divide(10, const_100))), add(const_1, divide(10, const_100))), add(const_1, multiply(add(const_1, add(const_1, divide(10, const_100))), add(const_1, divide(10, const_100))))) | louie takes out a 3 - month loan of $ 1000 . the lender charges him 10 % interest per month compunded 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 louie have to pay each month ? | let the monthly payment be xx . after the 1 st month there will be 1000 β 1.1 β x dollars left to repay ; after the 2 nd month there will be ( 1000 β 1.1 β x ) β 1.1 β x = 1210 β 2.1 x dollars left to repay ; after the 3 rd month there should be 0 dollars left to repay : ( 1210 β 2.1 x ) β 1.1 β x = 0 - - > 1331 = 3.31 x - - > x β 402 x β 402 answer : c . | a = 10 / 100
b = 1 + a
c = 1000 * b
d = 10 / 100
e = 1 + d
f = c * e
g = 10 / 100
h = 1 + g
i = f * h
j = 10 / 100
k = 1 + j
l = 1 + k
m = 10 / 100
n = 1 + m
o = l * n
p = 1 + o
q = i / p
|
a ) 75 , b ) 65 , c ) 60 , d ) 130 , e ) 30 | b | divide(26, subtract(150.40, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 26 . if n / j = 150.40 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 150.40 here 150 the quotient . given that remainder = 26 so , 150.40 = 150 + 26 / j so , j = 65 answer : b" | a = 4 * 10
b = a + 2
c = 100 + b
d = 150 - 40
e = 26 / d
|
a ) 2 / 5 , b ) 11 / 5 , c ) 1 / 9 , d ) 1 / 17 , e ) 1 / 5 | d | divide(9, choose(18, 2)) | kim has 9 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 18 shoes , what is the probability that she will select 2 shoes of the same color ? | "can be tackled in this way as well : probability of selecting any 1 out of 10 shoes = 18 / 18 = 1 probability of selecting the next shoe ( out of 18 available ) having the same color = 1 / 18 ( as after selecting the 1 st one , there is only 1 another shoe left with the same color ) . thus the total probability = 1 * 1 / 17 = 1 / 17 . d is the correct answer ." | a = math.comb(18, 2)
b = 9 / a
|
a ) $ 6 , b ) $ 5 , c ) $ 5.5 , d ) $ 6.5 , e ) $ 7 | a | divide(add(multiply(4, 5), 10), 5) | rahim bought 4 pillows with the average cost of $ 5 . if rahim also buys the fifth pillow with the price of $ 10 , what is the average ( arithmetic mean ) price of those 5 pillows ? | 4 pillowss - > total cost = 4 * avg cost = 4 * 5 = 20 fifth pillow cost = 10 total cost for 5 pillows = 20 + 10 = 30 am = 30 / 5 = 6 hence a | a = 4 * 5
b = a + 10
c = b / 5
|
a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total e of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) 6 : 4 : 3 : 13 - - - - > this means , the total number e must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) ." | a = 3000 + 3
|
a ) 2436 , b ) 2801 , c ) - 1496 , d ) - 2071 , e ) none of them | c | multiply(subtract(const_1, const_2), subtract(multiply(84, 19), 100)) | - 84 x 19 + 100 = ? | "given exp . = - 84 x ( 20 - 1 ) + 100 = - ( 84 x 20 ) + 84 + 100 = - 1680 + 184 = - 1496 answer is c" | a = 1 - 2
b = 84 * 19
c = b - 100
d = a * c
|
a ) 30 % , b ) 40 % , c ) 50 percent , d ) 60 % , e ) 70 % | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | the length of a rectangle is halved , while its breadth is tripled . whatis the % change in area ? | "let original length = x and original breadth new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c" | a = 1 / 2
b = 3 * a
c = b - 1
d = c / 1
e = d * 100
|
a ) $ 45.10 , b ) $ 34.31 , c ) $ 28.44 , d ) $ 26.66 , e ) $ 65.23 | d | divide(multiply(multiply(const_100, const_100), 15), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 15 , what was the price before the first discount ? | "let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 15 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 15 solve for x x = $ 26.66 correct answer d" | a = 100 * 100
b = a * 15
c = 100 - 25
d = c * 100
e = 100 - 25
f = e * 25
g = d - f
h = b / g
|
a ) 34 , b ) 42 , c ) 45.7 , d ) 57 , e ) 67 | a | add(20, const_1) | the average of first six prime numbers which are between 20 and 60 is | "explanation : first six prime numbers which are between 20 and 60 = 23 , 29 , 31 , 37 , 41 , 43 average = ( 23 + 29 + 31 + 37 + 41 + 43 ) / 6 = 34 answer : a" | a = 20 + 1
|
a ) 12 % , b ) 26 % , c ) 29 % , d ) 30 % , e ) 60 % | b | multiply(divide(add(multiply(divide(20, const_100), 300), multiply(divide(35, const_100), 200)), add(300, 200)), const_100) | for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 20 percent of the seeds in the first plot germinated and exactly 35 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "in the first plot 20 % of 300 seeds germinated , so 0.20 x 300 = 60 seeds germinated . in the second plot , 35 % of 200 seeds germinated , so 0.35 x 200 = 70 seeds germinated . since 60 + 70 = 130 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 130 / 500 ) x 100 % , or 26 % . answer : b ." | a = 20 / 100
b = a * 300
c = 35 / 100
d = c * 200
e = b + d
f = 300 + 200
g = e / f
h = g * 100
|
a ) 30 , b ) 45 , c ) 60 , d ) 150 , e ) 135 | d | divide(multiply(divide(multiply(multiply(3, 5), 30), 3), 5), 3) | in cliff β s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 3 / 5 are shiny and the rest are matte , while 1 / 5 of the sedimentary rocks are shiny . if there are 30 shiny igneous rocks , how many total rocks does cliff have ? | "we can start with the known quantity and then go on to find the others . shiny igneous ricks are 30 . these are ( 3 / 5 ) of total igneous rocks . ( 3 / 5 ) * total igneous rocks = 30 total igneous rocks = 30 * ( 5 / 3 ) = 50 total sedimentary rocks = 2 * total igneous rocks = 2 * 50 = 100 total number of rocks = 50 + 100 = 150 answer ( d )" | a = 3 * 5
b = a * 30
c = b / 3
d = c * 5
e = d / 3
|
a ) 38 , b ) 41 , c ) 28 , d ) 88 , e ) 22 | a | multiply(divide(22, 12), 20) | there are 20 poles with a constant distance between each pole . a car takes 22 second to reach the 12 th pole . how much will it take to reach the last pole . | "assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 22 seconds 1 pole ( 22 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 22 / 11 ) seconds = 38 seconds answer : a" | a = 22 / 12
b = a * 20
|
a ) 156,789 , b ) 188,889 , c ) 192,389 , d ) 241,689 , e ) 275,889 | b | add(add(add(subtract(const_10, 1), multiply(subtract(const_100, const_10), const_2)), multiply(subtract(const_100, const_10), const_3)), multiply(subtract(const_100, const_10), const_4)) | meena wrote all the numbers from 1 to 39,999 inclusive . how many digits did she write in total ? | 1 - 9 = > 1 * 9 digits 10 - 99 = > 2 * 90 = 180 ( numbers between 10 - 99 is 90 where each has 2 digits ) 100 - 999 = > 3 * 900 = 2700 1000 - 9999 = > 4 * 9000 = 36,000 10000 - 39999 = > 5 * 30,000 = 150,000 the answer is 188,889 the answer is b . | a = 10 - 1
b = 100 - 10
c = b * 2
d = a + c
e = 100 - 10
f = e * 3
g = d + f
h = 100 - 10
i = h * 4
j = g + i
|
a ) 6 : 4 , b ) 6 : 14 , c ) 2 : 3 , d ) 4 : 6 , e ) 3 : 7 | c | divide(divide(subtract(multiply(50, divide(70, const_100)), multiply(50, divide(50, const_100))), subtract(divide(70, const_100), divide(20, const_100))), subtract(50, divide(subtract(multiply(50, divide(70, const_100)), multiply(50, divide(50, const_100))), subtract(divide(70, const_100), divide(20, const_100))))) | solution a is 20 % salt and solution b is 70 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 70 - 50 ) / ( 50 - 20 ) = 2 / 3 so the volume of the two solutions will be equal . answer has to be 2 : 3 c" | a = 70 / 100
b = 50 * a
c = 50 / 100
d = 50 * c
e = b - d
f = 70 / 100
g = 20 / 100
h = f - g
i = e / h
j = 70 / 100
k = 50 * j
l = 50 / 100
m = 50 * l
n = k - m
o = 70 / 100
p = 20 / 100
q = o - p
r = n / q
s = 50 - r
t = i / s
|
a ) - 2 , b ) 10 , c ) 3 , d ) - 1 , e ) 0 | d | divide(subtract(negate(const_1), 3), subtract(subtract(power(negate(const_1), 2), multiply(8, negate(const_1))), 20)) | what is the smallest integer that satisfies the inequality ( x - 3 ) / ( x ^ 2 - 8 x - 20 ) > 0 > 0 ? | "explanatory answer let us factorize the denominator and rewrite the expression as x β 3 / ( x β 10 ) ( x + 2 ) > 0 equate each of the terms of the expression to zero to identify the values of x that are relevant to test whether the inequality holds good . the values that are relevant to us are x = 3 , x = 10 and x = - 2 . let us arrange these values in ascending order : - 2 , 3 and 10 . the quickest way to solve inequalities questions after arriving at these values is verifying whether the inequality holds good at the following intervals . interval 1 : x < - 2 . pick a value in that range and check whether the inequality holds good . let us take x = - 10 . when x = - 10 , x β 3 / ( x β 10 ) ( x + 2 ) < 0 ; the inequality does not hold good in this interval . interval 2 : - 2 < x < 3 . let us take x = 0 . when x = 0 , x β 3 / ( x β 10 ) ( x + 2 ) = 0 β 3 / ( 0 β 10 ) ( 0 + 2 ) > 0 ; the inequality holds good in this interval . the least integer value that x can take if x > - 2 is x = - 1 . so , the correct answer is - 1 . choice d . remember : we have to find out the least integer value . and we have arrived at - 1 . do , not waste time computing the entire range of values of x that satisfy the inequality . note : in any inequality question , when the question asks us to determine the intervals in which the inequality holds good , we have to eliminate values of x that will result in the denominator becoming zero . choice d" | a = negate - (
b = a / 3
|
a ) 20 m , b ) 25 m , c ) 22.5 m , d ) 9 m , e ) none of these | a | multiply(divide(100, 45), subtract(45, 36)) | in a 100 m race , a covers the distance in 36 seconds and b in 45 second . in this race a beats b by : | "solution distance covered by b in 9 sec . = ( 100 / 45 x 9 ) m = 20 m . β΄ a beats b by 20 metres . answer a" | a = 100 / 45
b = 45 - 36
c = a * b
|
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