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a ) 24 , b ) 23.5 , c ) 22 , d ) 21.2 , e ) 25.32 | e | multiply(divide(subtract(40, 2), add(2, 4)), 4) | one hour after yolanda started walking from x to y , a distance of 40 miles , bob started walking along the same road from y to x . if yolanda Γ’ s walking rate was 2 miles per hour and bob Γ’ s was 4 miles per hour , how many miles had bob walked when they met ? | "let t be the number of hours that bob had walked when he met yolanda . then , when they met , bob had walked 4 t miles and yolanda had walked 2 ( t + 1 ) miles . these distances must sum to 40 miles , so 4 t + 2 ( t + 1 ) = 40 , which may be solved for t as follows 4 t + 2 ( t + 1 ) = 40 4 t + 2 t + 2 = 40 6 t = 38 t = 6.33 ( hours ) therefore , bob had walked 4 t = 4 ( 7.6 ) = 25.32 miles when they met . the best answer is e ." | a = 40 - 2
b = 2 + 4
c = a / b
d = c * 4
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a ) 53 , b ) 47 , c ) 48 , d ) 59 , e ) 45 | d | subtract(negate(37), multiply(subtract(29, 31), divide(subtract(29, 31), subtract(23, 29)))) | 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , ( . . . ) | "explanation : all are prime numbers in their order , starting from 23 hence , next number is 59 answer : d" | a = negate - (
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a ) 5 % , b ) 7 % , c ) 9 % , d ) 2 % , e ) 4 % | a | divide(multiply(const_100, 160), multiply(800, 4)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 160 in 4 years ? | "160 = ( 180 * 4 * r ) / 100 r = 5 % answer : a" | a = 100 * 160
b = 800 * 4
c = a / b
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a ) 3.92 days , b ) 1.92 days , c ) 2.12 days , d ) 2.02 days , e ) 2.92 days | e | multiply(5, subtract(const_1, add(inverse(5), inverse(add(const_4, divide(const_2, const_3)))))) | a and b can do a piece of work in 4 2 / 3 days and 5 days respectively . they work together for 1 days and then a leaves . in how many days after that b will complete the work alone . | 3 / 14 * 1 + ( 1 + x ) / 5 = 1 x = 2.92 days answer : e | a = 1/(5)
b = 2 / 3
c = 4 + b
d = 1/(c)
e = a + d
f = 1 - e
g = 5 * f
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['a ) 30 pi', 'b ) 40 pi', 'c ) 50 pi', 'd ) 60 pi', 'e ) 70 pi'] | c | divide(multiply(power(divide(add(divide(subtract(add(subtract(power(sqrt(add(power(2, const_2), power(6, const_2))), const_2), const_4), power(sqrt(add(power(2, const_2), power(6, const_2))), const_2)), power(2, const_2)), power(2, const_2)), const_2), const_2), const_2), const_pi), const_2) | a semicircle is drawn with ab as its diameter . from c , a point on ab , a line perpendicular to ab is drawn , meeting the circumference of the semicircle at d . given that ac = 2 cm and cd = 6 cm , the area of the semicircle in square cm will be ? | let o be center of circle and radius be x . c can be anywhere on ab . let us suppose that c is somewhere between a and o . ac = 2 cm . . . . so co = ( x - 2 ) do = x , cd = 6 cm cdo form a right angled triangle with right angle at c . using pythagoras theorem , x ^ 2 = 6 ^ 2 + ( x - 2 ) ^ 2 4 x = 40 x = 10 cm area = [ pi ( x ) ^ 2 ] / 2 area = 50 pi answer : c | a = 2 ** 2
b = 6 ** 2
c = a + b
d = math.sqrt(c)
e = d ** 2
f = e - 4
g = 2 ** 2
h = 6 ** 2
i = g + h
j = math.sqrt(i)
k = j ** 2
l = f + k
m = 2 ** 2
n = l - m
o = 2 ** 2
p = n / o
q = p + 2
r = q / 2
s = r ** 2
t = s * math.pi
u = t / 2
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a ) 4 : 9 , b ) 4 : 3 , c ) 1 : 3 , d ) 4 : 8 , e ) 4 : 5 | c | divide(sqrt(8), sqrt(24)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 24 hours and 8 hours respectively . the ratio of their speeds is | let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 8 : 24 = 1 : 3 . answer : c | a = math.sqrt(8)
b = math.sqrt(24)
c = a / b
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a ) 50 m , b ) 60 m , c ) 65 m , d ) 75 m , e ) 80 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 20)), const_4) | length of a rectangular plot is 20 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 20 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 20 ) + x ] = 200 2 x + 20 = 100 2 x = 80 x = 40 . hence , length = x + 20 = 60 m b" | a = 5300 / 26
b = 2 * 20
c = a + b
d = c / 4
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a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 250 metres | a | multiply(17.1, multiply(40, const_0_2778)) | a train is running at a speed of 40 km / hr and it crosses a post in 17.1 seconds . what is the length of the train ? | "speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 17.1 s distance covered , d = vt = ( 400 / 36 ) Γ£ β 17.1 = 190 m distance covered is equal to the length of the train = 190 m correct answer is 190 metres a" | a = 40 * const_0_2778
b = 17 * 1
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a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | subtract(subtract(160, 50), 50) | of the 160 people at a party , 70 were women , and 50 women tried the appetizer . if 50 people did not try the appetizer , what is the total number of men who tried the appetizer ? | "total people at party = 160 women = 70 so men 160 - 70 = 90 no . of pple who tried appetizer = 160 - 50 ( given info ) = 110 no of women who tried appetizer = 50 so remaining ppl ( men ) who tried the appetizer = 110 - 50 = 60 correct option c" | a = 160 - 50
b = a - 50
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a ) 389 m , b ) 350 m , c ) 285 m , d ) 299 m , e ) 219 m | c | subtract(multiply(speed(300, 20), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 20 sec . what is the length of the platform ? | "speed = 300 / 20 = 15 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 15 = > x = 285 m . answer : c" | a = speed * (
b = a - 39
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a ) 78 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | c | divide(multiply(multiply(49, 42), 14), volume_cube(divide(14, const_2))) | a box measuring 49 inches long by 42 inches wide by 14 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 7 is the biggest number that could divide all three , 49 , 42 and 14 . thus side of cube must be 7 , and total number of cubes = 49 / 7 * 42 / 7 * 14 / 7 = 84 ans c ." | a = 49 * 42
b = a * 14
c = 14 / 2
d = b / volume_cube
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a ) 30 % , b ) 33 % , c ) 36 % , d ) 39 % , e ) 42 % | c | add(20, multiply(divide(20, const_100), 80)) | if 20 liters of chemical x are added to 80 liters of a mixture that is 20 % chemical x and 80 % chemical y , then what percentage of the resulting mixture is chemical x ? | "the amount of chemical x in the solution is 20 + 0.2 ( 80 ) = 36 liters . 36 liters / 100 liters = 36 % the answer is c ." | a = 20 / 100
b = a * 80
c = 20 + b
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a ) 25 days , b ) 30 days , c ) 60 days , d ) 65 days , e ) 36 days | c | multiply(const_3, 20) | working together , jose and jane can complete an assigned task in 20 days . however , if jose worked alone and completed half the work and then jane takes over and completes the second half , the task will be completed in 45 days . how long will jose take to complete the task if he worked alone ? assume that jane is more efficient than jose . | "explanatory answer step 1 : assign variables and frame equations let jose take ' x ' days to complete the task if he worked alone . let jane take ' y ' days to complete the task if she worked alone . statement 1 of the question : they will complete the task in 20 days , if they worked together . in 1 day , jose will complete 1 / x of the task . in 1 day , jane will complete 1 / yof the task . together , in 1 day they will complete 1 / 20 of the task . therefore , 1 / x + 1 / y = 1 / 20 . . . . ( 1 ) statement 2 of the question : if jose worked alone and completed half the work and then jane takes over and completes the second half , the task will be completed in 45 days . jose will complete half the task in x / 2 days . jane will complete half the task in y / 2 days . β΄ x / 2 + y / 2 = 45 or , x + y = 90 or x = 90 - y . . . . ( 2 ) step 2 : solve the two equations and find x and y substitute the value of x as ( 90 - y ) in the first equation 1 / 90 β y + 1 / y = 1 / 20 or y 2 - 90 + 1800 = 0 . the quadratic equation factorizes as ( y - 60 ) ( y - 30 ) = 0 so , y = 60 or y = 30 . if y = 60 , then x = 90 - y = 90 - 60 = 30 and if y = 30 , then x = 90 - y = 90 - 30 = 60 . the question clearly states that jane is more efficient than jose . therefore , jane will take lesser time than jose . hence , jose will take 60 days to complete the task if he worked alone and jane will take 30 days to complete the same task . choice c" | a = 3 * 20
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a ) 420 , b ) 770 , c ) 840 , d ) 165 , e ) 315 | c | multiply(multiply(10, 4), 7) | a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 3 of 10 candidates eligible to fill 3 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 4 candidates are there to fill the 4 positions ? | "1 of 7 will be chosen for the math 3 of 10 will be chosen for the computer none of the 4 chosen people can be in more than one departments . we can choose any of the 7 candidates for the math dep . , which gives as 7 selections . we can choose 3 of the 10 candidates for the computer dep . , which gives us 3 selections and 7 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 3 ! * 7 ! = 120 we are multiplying our individual selections : 7 * 120 = 840 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans c" | a = 10 * 4
b = a * 7
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a ) 20 mph , b ) 25 mph , c ) 30 mph , d ) 35 mph , e ) 40 mph | b | divide(subtract(divide(900, 2), divide(900, add(2, divide(15, const_60)))), const_2) | a boeing 757 flies a direct route from dallas , tx , to phoenix , az and then returns to dallas , tx . the flight is 900 miles one way . it took 2 hrs and 15 minutes for the flight to phoenix and 2 hrs for the flight back to dallas . what was the speed of the wind in mph ? | explanation : change the times in the problem to minutes first . speed of airplane from dallas to phoenix : ( 900 / 135 ) x 60 = 400 mph speed of airplane from phoenix to dallas : ( 900 / 120 ) x 60 = 450 mph wind speed : 1 / 2 ( 450 - 400 ) = 50 / 2 = 25 mph answer : option b | a = 900 / 2
b = 15 / const_60
c = 2 + b
d = 900 / c
e = a - d
f = e / 2
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['a ) 10 %', 'b ) 80 %', 'c ) 100 %', 'd ) 120 %', 'e ) 125 %'] | a | multiply(multiply(power(divide(4, 10), const_2), divide(5, 8)), const_100) | tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 5 meters and a circumference of 4 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | for a , r = 4 / 2 pi . its capacity = ( 2 pi ) ^ 2 * 5 = 20 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 20 pi / 200 pi = 0.1 a | a = 4 / 10
b = a ** 2
c = 5 / 8
d = b * c
e = d * 100
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a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 90 | e | divide(180, const_2) | he total marks obtained by a student in physics , chemistry and mathematics is 180 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 180 + p c + m = 180 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 180 / 2 = 90 . answer : e | a = 180 / 2
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a ) 28 , b ) 18 , c ) 27 , d ) 25 , e ) 21 | d | divide(add(581, 594), add(27, 20)) | rahim bought 27 books for rs . 581 from one shop and 20 books for rs . 594 from another . what is the average price he paid per book ? | "average price per book = ( 581 + 594 ) / ( 27 + 20 ) = 1175 / 47 = rs . 25 answer : d" | a = 581 + 594
b = 27 + 20
c = a / b
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a ) 2999 , b ) 8000 , c ) 6000 , d ) 2889 , e ) 6612 | b | multiply(multiply(const_1, const_12), divide(24000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 24000 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 24000 = 8000 answer : b" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 24000 / h
j = a * i
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | multiply(divide(2, 6), multiply(const_4, 3)) | the ratio of flour to water to sugar in a recipe is 10 : 6 : 3 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 2 cups of water , how much sugar is required ? | "the ratio of flour to water is 10 : 3 . the ratio of flour to sugar is 5 : 3 = 10 : 6 . the new ratio of flour to water to sugar is 10 : 3 : 6 if we need 2 cups of water , then we need 4 cups of sugar . the answer is b ." | a = 2 / 6
b = 4 * 3
c = a * b
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a ) 1 / pi , b ) sqrt ( 8 / pi ) , c ) 1 , d ) 2 / sqrt ( pi ) , e ) pi / 2 | c | sqrt(divide(divide(square_area(4), 2), const_pi)) | an artist wishes to paint a circular region on a square poster that is 4 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | "area of the poster is 4 x 4 = 16 1 / 2 the area = 8 pi * r ^ 2 = 8 r ^ 2 = 8 / pi r = sqrt ( 8 / pi ) answer : c" | a = square_area / (
b = a / 2
c = math.sqrt(b)
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a ) 1008 , b ) 1012 , c ) 1022 , d ) 1032 , e ) 1043 | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 4) | the smallest number which when diminished by 4 , is divisible by 12 , 16 , 18 , 21 and 28 is | "required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 4 = 1008 + 4 = 1012 answer : b" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 4
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a ) 15 , b ) 33 , c ) 58 , d ) 56 , e ) 91 | b | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 is subtracted from each , the new numbers are in the ratio 12 : 23 . the smaller number is : | "let the numbers be 3 x and 5 x . then , ( 3 x - 9 ) / ( 5 x - 9 ) = 12 / 23 9 x = 99 = > x = 11 the smallest number = 3 * 11 = 33 . answerb" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
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a ) 15.5 sec , b ) 8.4 sec , c ) 33.6 sec , d ) 31.11 sec , e ) 16.8 sec | d | divide(multiply(divide(10, divide(15, 20)), add(20, 15)), 15) | two cars , car 1 and car 2 move towards each other from q and y respectively with respective speeds of 20 m / s and 15 m / s . after meeting each other car 1 reaches y in 10 seconds . in how many seconds does car 2 reach q starting from y ? | q - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - - - - - - - - - - - - - - - - - - - y car a ( 20 mps ) - - - - - - - - - - - - - - - - - - - - - - - - - > p < - - - - - - - - - - - - - - - car b ( 15 mps ) let 2 cars meet each other at point p in t seconds . car 1 covers distance = 20 t . car 2 covers distance = 15 t . so , total distance qy = 35 t . from p , car 1 reaches onto y in 10 secs . so it covers 15 t further . so , 15 t / 20 = 10 so t = 40 / 3 sec and total distance = ( 35 * 40 ) / 3 hence car 2 will cover total distance in ( 35 * 40 ) / ( 3 * 15 ) = 31.11 sec approx . answer d | a = 15 / 20
b = 10 / a
c = 20 + 15
d = b * c
e = d / 15
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a ) 200 , b ) 251 , c ) 250 , d ) 262 , e ) 254 | b | add(multiply(add(multiply(8, const_3), 3), divide(add(multiply(8, const_3), 3), 3)), 8) | in a division sum , the remainder is 8 and the divisor is 3 times the quotient and is obtained by adding 3 to the thrice of the remainder . the dividend is : | "diver = ( 8 * 3 ) + 3 = 27 3 * quotient = 27 quotient = 9 dividend = ( divisor * quotient ) + remainder dividend = ( 27 * 9 ) + 8 = 251 b" | a = 8 * 3
b = a + 3
c = 8 * 3
d = c + 3
e = d / 3
f = b * e
g = f + 8
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a ) 8.78 , b ) 11.25 , c ) 8.75 , d ) 8.98 , e ) 8.28 | b | multiply(divide(const_1, const_2), multiply(5, 4.5)) | the area of a sector of a circle of radius 5 cm formed by an arc of length 4.5 cm is ? | "( 5 * 4.5 ) / 2 = 11.25 answer : b" | a = 1 / 2
b = 5 * 4
c = a * b
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a ) 358 , b ) 348 , c ) 368 , d ) 388 , e ) 378 | e | subtract(multiply(power(subtract(negate(5), 8), const_2), 2), multiply(negate(5), 8)) | if x = - 5 and y = 8 , what is the value of 2 ( x - y ) ^ 2 - xy ? | x = - 5 and y = 8 x - y = - 5 - 8 = - 13 x * y = - 5 * 8 = - 40 now we apply it in the equation 2 ( x - y ) ^ 2 - xy = 2 ( - 13 ) ^ 2 - ( - 40 ) = = > 2 * 169 + 40 = 338 + 40 = 378 answer : e | a = negate - (
b = a ** 8
c = b * 2
d = c - 2
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a ) 1 / 13 , b ) 1 / 15 , c ) 11 / 9 , d ) 1 / 10 , e ) 1 / 25 | a | divide(7, choose(14, 2)) | kim has 7 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 14 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 14 shoes = 14 / 14 = 1 probability of selecting the next shoe ( out of 14 available ) having the same color = 1 / 14 ( as after selecting the 1 st one , there is only 1 another shoe left with the same color ) . thus the total probability = 1 * 1 / 13 = 1 / 13 . a is the correct answer ." | a = math.comb(14, 2)
b = 7 / a
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a ) 1 km , b ) 3 km , c ) 6 km , d ) 5 km , e ) 6 km | c | divide(multiply(36, divide(multiply(10, const_1000), const_60)), const_1000) | find the distance covered by a man walking for 36 min at a speed of 10 km / hr ? | distance = 10 * 36 / 60 = 6 km answer is c | a = 10 * 1000
b = a / const_60
c = 36 * b
d = c / 1000
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a ) 8 days , b ) 5 days , c ) 6 days , d ) 7 days , e ) 9 days | d | divide(subtract(const_1, add(multiply(divide(const_1, 5), const_2), multiply(divide(const_1, 10), const_2))), divide(const_1, 10)) | a can finish a piece of work in 5 days . b can do it in 10 days . they work together for 1 day and then a goes away . in how many days will b finish the work ? | "1 / 5 + ( 1 + x ) / 10 = 1 = > x = 7 days answer : d" | a = 1 / 5
b = a * 2
c = 1 / 10
d = c * 2
e = b + d
f = 1 - e
g = 1 / 10
h = f / g
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a ) 1 / 2 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 5 / 2 | c | divide(multiply(subtract(5, 4), const_3), 4) | in an xy - coordinate plane , a line is defined by y = kx + 1 . if ( 4 , b ) , ( a , 5 ) , and ( a , b + 1 ) are 3 points on the line , where a and b are unknown , then k = ? | b = 4 k + 1 . . . ( 1 ) b + 1 = ak + 1 . . . ( 2 ) 5 = ak + 1 . . . ( 3 ) taking ( 2 ) and ( 3 ) 5 = b + 1 b = 4 taking ( 1 ) 4 = 4 k + 1 k = 3 / 4 answer : c | a = 5 - 4
b = a * 3
c = b / 4
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a ) $ 200 , b ) $ 400 , c ) $ 500 , d ) $ 800 , e ) $ 1,200 | c | subtract(divide(add(add(multiply(const_100, const_10), 500), 500), const_2), 500) | david has $ 1,500 at the beginning of his trip , after spending money , he still has exactly $ 500 less than he spent on the trip . how much money does john still have ? | "suppose total money spent = x not spend ( money he still has ) = x - 500 x + x - 500 = 1500 x = 1000 money not spend = 1000 - 500 = 500 answer : c" | a = 100 * 10
b = a + 500
c = b + 500
d = c / 2
e = d - 500
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a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | c | divide(divide(multiply(multiply(18, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5)) | a student was asked to find 4 / 5 of a number . but the student divided the number by 4 / 5 , thus the student got 18 more than the correct answer . find the number . | "let the number be x . ( 5 / 4 ) * x = ( 4 / 5 ) * x + 18 25 x = 16 x + 360 9 x = 360 x = 40 the answer is c ." | a = 4 / 5
b = 18 * a
c = 4 / 5
d = b * c
e = 4 / 5
f = 4 / 5
g = e * f
h = 1 - g
i = d / h
j = 4 / 5
k = i / j
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a ) 5 : 1 , b ) 10 : 5 , c ) 15 : 2 , d ) 20 : 2 , e ) 25 : 2 | e | divide(multiply(5, 5), multiply(1, 3)) | jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 5 to 3 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ? | "u / i = 5 / 3 i / b = 5 / 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 25 : 15 : 2 i . e . u : b = 25 : 2 answer : option e" | a = 5 * 5
b = 1 * 3
c = a / b
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a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 95 % | e | multiply(divide(multiply(10, subtract(const_1, divide(5, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 5 percent , but profits were 10 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "the profit 0 f 2009 interms of 2008 = 0.95 * 10 / 10 * 100 = 95 % e" | a = 5 / 100
b = 1 - a
c = 10 * b
d = c / 10
e = d * 100
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a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 12 % | c | multiply(divide(3.50, add(32.50, 3.50)), const_100) | a house wife saved $ 3.50 in buying an item on sale . if she spent $ 32.50 for the item , approximately how much percent she saved in the transaction ? | "actual price = 32.50 + 3.50 = $ 36 saving = 3.50 / 36 * 100 = 10 % approximately answer is c" | a = 32 + 50
b = 3 / 50
c = b * 100
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a ) 1000 , b ) 2250 , c ) 60 , d ) 45 , e ) 1250 | e | subtract(divide(subtract(multiply(1000, 60), multiply(1000, 15)), 20), 1000) | a garrison of 1000 men has provisions for 60 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ? | "1000 - - - - 60 1000 - - - - 45 x - - - - - 20 x * 20 = 1000 * 45 x = 2250 1000 - - - - - - - 1250 answer : e" | a = 1000 * 60
b = 1000 * 15
c = a - b
d = c / 20
e = d - 1000
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a ) 0.1 % , b ) 0.5 % , c ) 1 % , d ) 5 % , e ) 10 % | c | subtract(const_100, multiply(divide(99, const_4), const_100)) | on an order of 99 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 100 dozens free . equivalent discount = 1 / 100 * 100 = 1 % . answer c ) 1 %" | a = 99 / 4
b = a * 100
c = 100 - b
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a ) 12 , b ) 13 , c ) 10 , d ) 18 , e ) 13 | b | multiply(multiply(add(3, divide(subtract(sqrt(35), 3), 2)), divide(subtract(sqrt(35), 3), 2)), 2) | if a - b = 3 and a ^ 2 + b ^ 2 = 35 , find the value of ab . | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 35 - 9 = 26 ab = 13 . answer is b ." | a = math.sqrt(35)
b = a - 3
c = b / 2
d = 3 + c
e = math.sqrt(35)
f = e - 3
g = f / 2
h = d * g
i = h * 2
|
a ) 600 , b ) 640 , c ) 500 , d ) 520 , e ) 720 | b | multiply(320, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 320 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 320 ) = 2 + 2 ( extra ) = > 2 x - 640 = x = > x = 640 . answer : b" | a = 320 * 2
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a ) 17 , b ) 97 , c ) 18 , d ) 55 , e ) 72 | d | subtract(divide(multiply(divide(multiply(45, 8), 30), 50), 6), 45) | 45 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 45 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 100 100 β 45 = 55 answer : d" | a = 45 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 45
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a ) 6 miles , b ) 8,4 miles , c ) 19 miles , d ) 19,6 miles , e ) 24 miles | e | multiply(divide(38, add(3, 2)), 2) | one hour before john started walking from p to q , a distance of 38 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | "ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 38 - 3 = 35 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 35 miles so it will take them 35 / 5 = 7 hours to meet . in 7 hrs , ann would have covered 7 hrs * 3 miles per hour = 21 miles . adding this to the 3 miles she covered before john , ann covered a total of 3 + 21 = 24 miles . answer ( e )" | a = 3 + 2
b = 38 / a
c = b * 2
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a ) 5865863355 , b ) 5665863355 , c ) 4865863355 , d ) 4665863355 , e ) none of these | a | multiply(586645, 9999) | simplify 586645 * 9999 | explanation : although it is a simple question , but the trick is to save time in solving this . rather than multiplying it we can do as follows : 586645 * ( 10000 - 1 ) = 5866450000 - 586645 = 5865863355 option a | a = 586645 * 9999
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a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1575 , d ) rs . 2268 , e ) none of these | b | add(divide(189, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 189) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 189 . the amount of the bill is : | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 189 . β΄ x 16 x 9 / 12 x 1 / 100 = 189 or x = 1575 . β΄ p . w . = rs . 1575 . β΄ sum due = p . w . + t . d . = rs . ( 1575 + 189 ) = rs . 1764 . answer b" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 189 / d
f = e + 189
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a ) 25 % , b ) 23 % , c ) 16.67 % , d ) 33.33 % , e ) none of these | b | multiply(subtract(const_1, divide(const_100, add(const_100, 30))), const_100) | if the price of petrol increases by 30 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 30 ) i . e 130 . now , his consumption should be reduced to : - = ( 130 β 100 ) / 130 β 100 . hence , the consumption should be reduced to 23 % . answer : b" | a = 100 + 30
b = 100 / a
c = 1 - b
d = c * 100
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a ) 36 mins , b ) 35 mins , c ) 40 mins , d ) 32 mins , e ) 30 mins | e | divide(subtract(const_1, multiply(30, divide(divide(const_1, const_60), const_2))), add(divide(const_1, const_60), divide(divide(const_1, const_60), const_2))) | pipe a that can fill a tank in two hour and pipe b that can fill the tank in an hour are opened simultaneously when the tank is empty . pipe b is shut 30 minutes before the tank overflows . when will the tank overflow ? | the last 30 minutes only pipe a was open . since it needs 2 hour to fill the tank , then in 30 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 3 = 1 / 4 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : e | a = 1 / const_60
b = a / 2
c = 30 * b
d = 1 - c
e = 1 / const_60
f = 1 / const_60
g = f / 2
h = e + g
i = d / h
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a ) 2 , b ) 5 , c ) 10 , d ) 16 , e ) 28 | e | subtract(150, add(52, 70)) | two family reunions are happening at the taj hotel , the oates reunion and the hall reunion . all 150 guests at the hotel attend at least one of the reunions . if 70 people attend the oates reunion and 52 people attend the hall reunion , how many people attend both reunions ? | no of people in oates reunion = 70 no of people in hall reunion = 52 attending both = x all guests attend at least one . therefore , 150 = 70 + 52 - ( both ) both = 28 answer e | a = 52 + 70
b = 150 - a
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a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | e | sqrt(1024) | you buy a piece of land with an area of Γ’ Λ Ε‘ 1024 , how long is one side of the land plot ? | "try filling the numbers into the answer y x y = find the closest to 1024 . answer e" | a = math.sqrt(1024)
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a ) s . 800 , b ) s . 400 , c ) s . 600 , d ) s . 180 , e ) s . 900 | d | divide(720, const_3) | divide rs . 720 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 720 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 720 = > 180 answer : d" | a = 720 / 3
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a ) 115 , b ) 230 , c ) 460 , d ) 575 , e ) 690 | b | multiply(divide(690, multiply(3, const_10)), multiply(2, add(const_2, const_3))) | the least common multiple of positive integer e and 3 - digit integer n is 690 . if n is not divisible by 3 and e is not divisible by 2 , what is the value of n ? | the lcm of n and e is 690 = 2 * 3 * 5 * 23 . e is not divisible by 2 , thus 2 goes to n n is not divisible by 3 , thus 3 goes to e . from above : n must be divisible by 2 and not divisible by 3 : n = 2 * . . . in order n to be a 3 - digit number it must take all other primes too : n = 2 * 5 * 23 = 230 . answer : b . | a = 3 * 10
b = 690 / a
c = 2 + 3
d = 2 * c
e = b * d
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a ) 8 , b ) 20 , c ) 10 , d ) 11 , e ) 12 | b | multiply(subtract(15, 5), const_2) | half a number plus 5 is 15 . what is the number ? | "let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 5 = 15 ( 1 / 2 ) x = 15 - 5 ( 1 / 2 ) x = 10 x = 20 correct answer is b" | a = 15 - 5
b = a * 2
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a ) 48 kmph , b ) 50 kmph , c ) 52 kmph , d ) 56 kmph , e ) 70 kmph | e | divide(630, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 630 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 630 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 630 / 9 = 70 kmph e" | a = 6 * 3
b = a / 2
c = 630 / b
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a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 2384525 / 2 | e | multiply(1100000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1100000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 2384525 / 2 e" | a = 15 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1100000 * h
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a ) 1250 m , b ) 1110 m , c ) 950 m , d ) 750 m , e ) 1300 m | d | multiply(divide(multiply(15, const_1000), const_60), 15) | a man walking at a rate of 15 km / hr crosses a bridge in 15 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 15 minutes = 15 / 18 * 15 * 60 = 750 m answer is d" | a = 15 * 1000
b = a / const_60
c = b * 15
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a ) 38636 , b ) 38640 , c ) 38647 , d ) 38591 , e ) 38675 | d | divide(multiply(add(multiply(4, const_100), 47), add(multiply(7, const_100), 77)), multiply(subtract(47, add(multiply(const_2, const_4), const_2)), subtract(47, add(multiply(const_2, const_4), const_2)))) | a room 4 m 47 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = ( 447 x 777 ) cm 2 . size of largest square tile = h . c . f . of 447 cm and 777 cm = 3 cm . area of 1 tile = ( 3 x 3 ) cm 2 . number of tiles required = ( 447 Γ 777 ) / ( 3 Γ 3 ) = 38591 answer : option d" | a = 4 * 100
b = a + 47
c = 7 * 100
d = c + 77
e = b * d
f = 2 * 4
g = f + 2
h = 47 - g
i = 2 * 4
j = i + 2
k = 47 - j
l = h * k
m = e / l
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a ) 31.5 % , b ) 35.9 % , c ) 37.5 % , d ) 39.5 % , e ) 30.5 % | b | multiply(subtract(multiply(divide(const_100, 64), divide(subtract(const_100, 13), const_100)), const_1), const_100) | the cost price of an book is 64 % of the marked price . calculate the gain percent after allowing a discount of 13 % ? | "let marked price = $ 100 . then , c . p . = $ 64 , s . p . = $ 87 gain % = 23 / 64 * 100 = 35.9 % . b" | a = 100 / 64
b = 100 - 13
c = b / 100
d = a * c
e = d - 1
f = e * 100
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a ) r = 16 , b ) r = 8 β 2 , c ) r = 8 , d ) r = 2 β 2 , e ) ( β 2 ) / 3 | d | sqrt(divide(multiply(4, 4), const_2)) | the two lines y = x and x = - 4 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length r of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 4 , - 4 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 4 Γ 4 ) = 8 ; so z = 8 . the next step requires us to find the length of a side of a cube that has a face area equal to 8 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 8 . since the area of a square is equal to s Β² , where s is the length of one of its side , we can write and solve the equation s Β² = 8 . clearly s = β 8 = 2 β 2 , oranswer choice ( d ) ." | a = 4 * 4
b = a / 2
c = math.sqrt(b)
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a ) 7500 , b ) 2028 , c ) 2775 , d ) 5496 , e ) 8000 | e | divide(4000, subtract(subtract(const_1, divide(25, const_100)), divide(25, const_100))) | a candidate got 25 % of the votes polled and he lost to his rival by 4000 votes . how many votes were cast ? | "25 % - - - - - - - - - - - l 75 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 50 % - - - - - - - - - - 4000 100 % - - - - - - - - - ? = > 8000 answer : e" | a = 25 / 100
b = 1 - a
c = 25 / 100
d = b - c
e = 4000 / d
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a ) 27 , b ) 68 , c ) 107 , d ) 108 , e ) 28 | a | subtract(negate(36), multiply(subtract(24, 12), divide(subtract(24, 12), subtract(8, 24)))) | 8 , 24 , 12 , 36 , 18 , 54 , ( . . . . ) | "8 Γ 3 = 24 24 Γ· 2 = 12 12 Γ 3 = 36 36 Γ· 2 = 18 18 Γ 3 = 54 54 Γ· 2 = 27 answer is a" | a = negate - (
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a ) 50 % , b ) 125 % , c ) 25 % , d ) none of above , e ) 30 % | c | multiply(divide(5, 20), const_100) | the ratio 5 : 20 expressed as percent equals to | "explanation : actually it means 5 is what percent of 20 , which can be calculated as , ( 5 / 20 ) * 100 = 5 * 5 = 25 option c" | a = 5 / 20
b = a * 100
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a ) 50 / 9 , b ) 50 / 7 , c ) 50 / 6 , d ) 50 / 8 , e ) 50 / 3 | c | multiply(divide(subtract(divide(7, 6), const_1), 2), const_100) | a sum of money becomes 7 / 6 of itself in 2 years at a certain rate of simple interest . the rate per annum is ? | "let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 2 years . rate = ( 100 * x ) / ( x * 6 * 2 ) = 50 / 6 % . answer : c" | a = 7 / 6
b = a - 1
c = b / 2
d = c * 100
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a ) 116 kmph , b ) 106 kmph , c ) 186 kmph , d ) 126 kmph , e ) 176 kmph | b | divide(add(325, 470), add(3.5, 4)) | a train travels 325 km in 3.5 hours and 470 km in 4 hours . find the average speed of train ? | explanation : as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 325 + 470 = 795 km thus , total speed = 7.5 hrs average speed = 795 / 7.5 = > 106 kmph . answer : b | a = 325 + 470
b = 3 + 5
c = a / b
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 8)) | a bucket full of nuts was discovered by the crow living in the basement . the crow eats a sixth of the total number of nuts in 8 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? | "in one hour , the crow eats 1 / 48 of the nuts . ( 1 / 4 ) / ( 1 / 48 ) = 12 hours the answer is b ." | a = 1 / 4
b = 2 + 3
c = 1 / b
d = c / 8
e = a / d
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a ) 914.2 hours , b ) 900 hours , c ) 915 hours , d ) 905 hours , e ) 960 hours | e | add(divide(7560, add(16, 2)), divide(7560, subtract(16, 2))) | speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7560 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7560 / 18 + 7560 / 14 = 420 + 540 = 960 hours answer : option e" | a = 16 + 2
b = 7560 / a
c = 16 - 2
d = 7560 / c
e = b + d
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a ) $ 4 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 20 | e | subtract(150, add(add(multiply(150, divide(1, 5)), multiply(150, divide(1, 6))), multiply(150, divide(1, 2)))) | jennifer had $ 150 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | "1 / 5 x $ 150 = $ 30 for sandwich 1 / 6 x $ 150 = $ 25 for museum 1 / 2 x $ 150 = $ 75 for book $ 30 + $ 25 + $ 75 = $ 130 spent $ 150 - $ 130 = $ 20 left over correct answer e" | a = 1 / 5
b = 150 * a
c = 1 / 6
d = 150 * c
e = b + d
f = 1 / 2
g = 150 * f
h = e + g
i = 150 - h
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a ) 15 min , b ) 10 min , c ) 12 min , d ) 30 min , e ) 18 min | d | subtract(const_60, multiply(divide(16, 32), const_60)) | excluding the stoppages , the speed of a bus is 32 km / hr and including the stoppages the speed of the bus is 16 km / hr . for how many minutes does the bus stop per hour ? | "speed of the bus without stoppage = 32 km / hr speed of the bus with stoppage = 16 km / hr difference in speed = 16 km / hr so , the time taken in the stoppages = time taken to cover 16 km = ( 16 / 32 ) hr = 1 / 2 hr = 30 min answer : d" | a = 16 / 32
b = a * const_60
c = const_60 - b
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a ) 500 , b ) 620 , c ) 450 , d ) 340 , e ) 280 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | what is the sum of all the multiples of 10 between 0 and 75 ? | "the multiples of 10 between 0 and 99 are 10 , 20 , 30 , 40 , 50 , 60 , 70 . if these are all added together , the result is 280 . final answer : e" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 10
t = 2 + 4
u = s + t
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a ) 60 , b ) 75 , c ) 90 , d ) 105 , e ) 120 | b | add(add(add(add(add(add(add(10, const_1), add(add(10, const_1), const_1)), add(add(add(10, const_1), const_1), const_2)), add(add(add(add(10, const_1), const_1), const_2), const_1)), add(add(add(add(add(10, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(10, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(10, const_1), const_1), const_2), const_1), const_1), const_1), const_1)) | the sum of the non - prime numbers between 10 and 20 , non - inclusive , is | "sum of consecutive integers from 11 to 19 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 11 + 19 ) / 2 * 9 = 15 * 9 = 135 sum of non - prime numbers b / w 10 and 20 , non inclusive = = = > 135 - 60 ( i . e . , 11 + 13 + 17 + 19 , being the prime # s in the range ) = 75 answer : b" | a = 10 + 1
b = 10 + 1
c = b + 1
d = a + c
e = 10 + 1
f = e + 1
g = f + 2
h = d + g
i = 10 + 1
j = i + 1
k = j + 2
l = k + 1
m = h + l
n = 10 + 1
o = n + 1
p = o + 2
q = p + 1
r = q + 1
s = m + r
t = 10 + 1
u = t + 1
v = u + 2
w = v + 1
x = w + 1
y = x + 1
z = s + y
A = 10 + 1
B = A + 1
C = B + 2
D = C + 1
E = D + 1
F = E + 1
G = F + 1
H = z + G
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a ) 126 , b ) 156 , c ) 340 , d ) 321 , e ) 260 | c | multiply(add(divide(subtract(divide(divide(2210, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(2210, 6.5), const_2), 10), const_2), 10)), const_2) | the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 2210 . the perimeter of the plot is ? | "sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 2210 x = 80 required perimeter = 2 ( 80 + 90 ) = 340 c" | a = 2210 / 6
b = a / 2
c = b - 10
d = c / 2
e = 2210 / 6
f = e / 2
g = f - 10
h = g / 2
i = h + 10
j = d + i
k = j * 2
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a ) 81 : 121 , b ) 81 : 126 , c ) 81 : 189 , d ) 81 : 176 , e ) 81 : 117 | a | power(divide(729, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 729 : 1331 . what is the ratio of their total surface areas ? | "ratio of the sides = Β³ β 729 : Β³ β 1331 = 9 : 11 ratio of surface areas = 92 : 112 = 81 : 121 answer : a" | a = 729 / 1331
b = 1 / 3
c = a ** b
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a ) 4 % , b ) 3.57 % , c ) 2 6 / 7 % , d ) 5 % , e ) 6 % | b | multiply(divide(divide(subtract(2000, 1750), 1750), 4), const_100) | at what rate percent on simple interest will rs . 1750 amount to rs . 2000 in 4 years ? | "explanation : 250 = ( 1750 x 4 xr ) / 100 r = 3.57 % answer : option b" | a = 2000 - 1750
b = a / 1750
c = b / 4
d = c * 100
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a ) 72 , b ) 84 , c ) 88 , d ) 96 , e ) 108 | d | multiply(add(add(8, subtract(8, 2)), 2), add(subtract(add(8, subtract(8, 2)), divide(8, const_2)), 2)) | roy is now 8 years older than julia and half of that amount older than kelly . if in 2 years , roy will be three times as old as julia , then in 2 years what would be roy β s age multiplied by kelly β s age ? | "r = j + 8 = k + 4 r + 2 = 3 ( j + 2 ) ( j + 8 ) + 2 = 3 j + 6 j = 2 r = 10 k = 6 in 2 years ( r + 2 ) ( k + 2 ) = 12 * 8 = 96 the answer is d ." | a = 8 - 2
b = 8 + a
c = b + 2
d = 8 - 2
e = 8 + d
f = 8 / 2
g = e - f
h = g + 2
i = c * h
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a ) 28 , b ) 27 , c ) 23 , d ) 22 , e ) 24 | c | add(subtract(95, multiply(19, 4)), 4) | having scored 95 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | "explanation : let the average score of the first 18 innings be n 18 n + 95 = 19 ( n + 4 ) = > n = 19 so , average score after 19 th innings = x + 4 = 23 . answer : c" | a = 19 * 4
b = 95 - a
c = b + 4
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['a ) 4', 'b ) 7', 'c ) 8', 'd ) 13', 'e ) 26'] | c | divide(subtract(divide(42, const_2), sqrt(subtract(power(divide(42, const_2), const_2), multiply(const_4, 104)))), const_2) | if a rectangular billboard has an area of 104 square feet and a perimeter of 42 feet , what is the length of each of the shorter sides ? | assume the sides to be x and y . perimeter = 2 x + 2 y = 42 hence x + y = 21 - ( i ) area = xy = 104 - ( ii ) since these are the sides of a billboard , we are dealing with positive numbers only and the options tell us that they are integers . 104 can be written as a product of two numbers as 104 * 1 or 52 * 2 or 26 * 4 or 13 * 8 of these , only 13 * 8 satisfies the equation ( i ) hence the lengths are 13 and 8 length of shorter side = 8 answer : c | a = 42 / 2
b = 42 / 2
c = b ** 2
d = 4 * 104
e = c - d
f = math.sqrt(e)
g = a - f
h = g / 2
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a ) 3 / 5 , b ) 7 / 10 , c ) 11 / 15 , d ) 23 / 30 , e ) 31 / 45 | c | divide(subtract(multiply(const_26, divide(multiply(const_5, const_5), const_0_25)), multiply(1200, subtract(const_1, divide(40, divide(multiply(const_5, const_5), const_0_25))))), multiply(const_26, divide(multiply(const_5, const_5), const_0_25))) | in a village of 2,700 people , 900 people are over 70 years old and 1200 people are female . it is known that 40 percent of the females are younger than 70 years old . if no one in the village is 70 years old , what is the probability that a person chosen at random is either a male or younger than 70 years old ? | "the number of people younger than 70 years old is 2700 - 900 = 1800 the number of females older than 70 years old is 0.6 * 1200 = 720 the number of males older than 70 years old is 900 - 720 = 180 the number of people who are either male or younger than 70 is 1800 + 180 = 1980 . p ( a person is younger than 70 or male ) = 1980 / 2700 = 11 / 15 the answer is c ." | a = 5 * 5
b = a / const_0_25
c = const_26 * b
d = 5 * 5
e = d / const_0_25
f = 40 / e
g = 1 - f
h = 1200 * g
i = c - h
j = 5 * 5
k = j / const_0_25
l = const_26 * k
m = i / l
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a ) 3 : 2 , b ) 6 : 1 , c ) 6 : 5 , d ) 6 : 2 , e ) 6 : 3 | a | divide(multiply(4, 3), multiply(4, 2)) | the marks obtained by vijay and amith are in the ratio 4 : 4 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ? | "4 : 4 3 : 2 - - - - - - - 12 : 12 : 8 12 : 8 3 : 2 answer : a" | a = 4 * 3
b = 4 * 2
c = a / b
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a ) 20 , b ) 18 , c ) 11 , d ) 10 , e ) 5 | d | divide(add(19, 1), const_2) | 5 x + y = 19 , and x + 3 y = 1 . find the value of 3 x + 2 y | add these two equations 6 x + 4 y = 20 divide by 2 ( to get 3 x + 2 y ) answer will be d . 10 | a = 19 + 1
b = a / 2
|
a ) 25 % , b ) 18 % , c ) 17 % , d ) 13 % , e ) 16 % | a | divide(multiply(add(50, const_100), subtract(const_100, 50)), const_100) | a number is increased by 50 % and then decreased by 50 % . find the net increase or decrease per cent . | "let the number be 100 . increase in the number = 50 % = 50 % of 100 = ( 50 / 100 Γ£ β 100 ) = 50 therefore , increased number = 100 + 50 = 150 this number is decreased by 50 % therefore , decrease in number = 50 % of 150 = ( 50 / 100 Γ£ β 150 ) = 7500 / 100 = 75 therefore , new number = 150 - 75 = 75 thus , net decreases = 100 - 75 = 25 hence , net percentage decrease = ( 25 / 100 Γ£ β 100 ) % = ( 2500 / 100 ) % = 25 % answer : a" | a = 50 + 100
b = 100 - 50
c = a * b
d = c / 100
|
a ) 2 , b ) 7 , c ) 6 , d ) 0 , e ) 8 | d | subtract(multiply(multiply(multiply(115, 297), 196), 108), subtract(multiply(multiply(multiply(115, 297), 196), 108), add(const_4, const_4))) | the unit digit in the product ( 115 * 297 * 196 * 108 ) is : | "explanation : unit digit in the given product = unit digit in ( 5 * 7 * 6 * 8 ) = 0 answer : d" | a = 115 * 297
b = a * 196
c = b * 108
d = 115 * 297
e = d * 196
f = e * 108
g = 4 + 4
h = f - g
i = c - h
|
a ) 1.12 , b ) 1.2 , c ) 1.25 , d ) 1.3 , e ) none | b | divide(multiply(0.75, 8), 5) | if 0.75 : x : : 5 : 8 , then x is equal to : | "( x * 5 ) = ( 0.75 * 8 ) x = 6 / 5 x = 1.20 answer = b" | a = 0 * 75
b = a / 5
|
a ) 6.09 % , b ) 6.08 % , c ) 6.1 % , d ) 6.07 % , e ) 6.05 % | a | add(add(divide(6, const_2), divide(6, const_2)), divide(multiply(divide(6, const_2), divide(6, const_2)), const_100)) | the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half - yearly is : | "amount of rs . 100 for 1 year when compounded half - yearly = 100 x ( 1 + 3 / 100 ) ^ 2 = 106.09 therefore effective rate = ( 106.09 - 100 ) % = 6.09 % answer a ) 6.09 %" | a = 6 / 2
b = 6 / 2
c = a + b
d = 6 / 2
e = 6 / 2
f = d * e
g = f / 100
h = c + g
|
a ) s . 40 , b ) s . 46 , c ) s . 49 , d ) s . 41 , e ) s . 60 | e | divide(divide(multiply(1200, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 1200 , what is the reduced price for kg ? | "1200 * ( 25 / 100 ) = 300 - - - - 5 ? - - - - 1 = > rs . 60 answer : e" | a = 1200 * 25
b = a / 100
c = b / 5
|
a ) 2.5 % , b ) 10 % , c ) 5 % , d ) 15 % , e ) 43 % | e | multiply(divide(subtract(50, 20), add(50, 20)), const_100) | if 50 % of ( x - y ) = 20 % of ( x + y ) then what percent of x is y ? | "50 % of ( x - y ) = 20 % of ( x + y ) ( 50 / 100 ) ( x - y ) = ( 20 / 100 ) ( x + y ) 5 ( x - y ) = 2 ( x + y ) 3 x = 7 y x = 7 / 3 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / ( 7 / 3 ) y ) x 100 ) = 43 % answer is e ." | a = 50 - 20
b = 50 + 20
c = a / b
d = c * 100
|
a ) 11 , b ) 12 , c ) 15 , d ) 18 , e ) 16 | a | add(10, const_1) | ashwin rented a power tool from a rental shop . the rent for the tool was $ 25 for the first hour and $ 10 for each additional hour . if ashwin paid a total of $ 125 , excluding sales tax , to rent the tool , for how many hours did she rent it ? | 25 + 10 n = 125 n = 10 total time = n + 1 hrs = 10 + 1 hrs = 11 hrs answer : a | a = 10 + 1
|
a ) 237 , b ) 780 , c ) 197 , d ) 287 , e ) 720 | b | multiply(390, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 390 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 390 ) = 2 + 2 ( extra ) = > 2 x - 780 = x = > x = 780 . answer : b" | a = 390 * 2
|
a ) 65 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 112 mtr . , e ) 17 mtr . | d | multiply(112, divide(multiply(40, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 40 men in 3 days ? | "20 men is 6 days can build 112 metres 25 men in 3 days can build = 112 * ( 40 / 20 ) x ( 3 / 6 ) = 112 meters answer : d ." | a = 40 * 3
b = 20 * 6
c = a / b
d = 112 * c
|
a ) 3 / 20,000 , b ) 1 / 3,600 , c ) 9 / 2,000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | a certain junior class has 1,000 students and a certain senior class has 400 students . among these students , there are 60 siblings pairs , each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected at will be a sibling pair ? | "total number of ways of choosing one student from each group is = 400 * 1000 number of cases in which a sibling pair will be got is = 60 thus the probability that the 2 students selected will be a sibling pair is = 60 / ( 400 * 1000 ) = 3 / 20,000 a" | a = 1 / 3
|
a ) 150 , b ) 88 , c ) 480 , d ) 288 , e ) 400 | e | multiply(divide(multiply(180, const_1000), const_3600), 8) | a train running at the speed of 180 km / hr crosses a pole in 8 seconds . find the length of the train . | "speed = 180 * ( 5 / 18 ) m / sec = 50 m / sec length of train ( distance ) = speed * time 50 * 8 = 400 meter answer : e" | a = 180 * 1000
b = a / 3600
c = b * 8
|
a ) 25 % , b ) 35 % , c ) 45 % , d ) 70 % , e ) 80 % | b | add(const_10, divide(add(25, 25), const_2)) | in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds b in the nature reserve are not hawks , paddyfield - warblers , or kingfishers ? | "1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if non hawks are 700 ( this is 70 % of the total , so total = 1000 ) , then paddy warbs are 2 / 5 x 700 = 1400 / 5 = 280 . kingfishers , therefore , are 280 / 4 = 70 . lets add them up . 300 hawks + 280 peddy warbs + 70 kingsifhers = 650 . so all others are 1000 - 650 = 350 or 35 % of total birds . the main job here to to identify the smart number to start the question with . this can be time consuming , but once identified , this question can be solved fairly quickly . 2 . another method : if x is total - - > non hawks = 0.7 x - - > warbs = 0.4 ( 0.7 x ) - - > kfs = 0.25 ( 0.4 ( 0.7 x ) ) . our job is to find out b : ( 0.3 x + 0.28 x + 0.07 x ) / x . or 0.65 x / x = 0.65 . we need to find 1 - 0.65 = 0.35 or 35 % . b" | a = 25 + 25
b = a / 2
c = 10 + b
|
a ) s . 11 , b ) s . 14 , c ) s . 15 , d ) s . 21 , e ) s . 29 | c | multiply(divide(divide(multiply(divide(20, const_100), 600), 10), multiply(divide(20, const_100), 600)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 600 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 600 - - - 80 ? - - - 2 = > rs . 15 answer : c" | a = 20 / 100
b = a * 600
c = b / 10
d = 20 / 100
e = d * 600
f = c / e
g = f * 100
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(3, 3)) | what is the remainder when 3 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 3 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? d | a = 5 / 2
b = 3 * 3
c = a - b
|
a ) 22 , b ) 18 , c ) 12 , d ) 88 , e ) 66 | c | subtract(42, multiply(10, 3)) | the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 42 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 42 + q = > q = 12 answer : c" | a = 10 * 3
b = 42 - a
|
['a ) 2870', 'b ) 2000', 'c ) 5650', 'd ) 6650', 'e ) 7650'] | a | divide(multiply(multiply(20, add(20, 1)), add(multiply(const_2, 20), 1)), multiply(const_2, const_3)) | what is the sum of the squares of the first 20 natural numbers ( 1 to 20 ) ? | n ( n + 1 ) ( 2 n + 1 ) / 6 20 ( 21 ) ( 21 ) / 6 = 2870 answer : a | a = 20 + 1
b = 20 * a
c = 2 * 20
d = c + 1
e = b * d
f = 2 * 3
g = e / f
|
a ) 27 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(7, 5)), multiply(7, 9)) | the average of 9 observations was 7 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 7 = 63 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 β 63 = 27 option a" | a = 10 * 5
b = 7 * 5
c = a + b
d = 7 * 9
e = c - d
|
a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | a | add(floor(divide(319, 2)), const_1) | the guests at a football banquet consumed a total of 319 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 159 * 2 ( max possible amount of food a guest could consume ) = 318 pounds , so there must be more than 159 guests , next integer is 160 . answer : a ." | a = 319 / 2
b = math.floor(a)
c = b + 1
|
a ) 12 , b ) 10 , c ) 8 , d ) 6 , e ) 5 | c | divide(100, multiply(30, divide(140, multiply(8, 42)))) | 8 persons can build a wall 140 m long in 42 days . in how many days can 30 persons complete a similar wall 100 m long ? | explanation : more persons , less days ( indirect proportion ) more length of the wall , more days ( direct proportion ) β 8 Γ 100 Γ 42 = 30 Γ 140 Γ x β x = ( 8 Γ 100 Γ 42 ) / ( 30 Γ 140 ) = ( 8 Γ 100 Γ 14 ) / ( 10 Γ 140 ) = ( 8 Γ 100 ) / ( 10 Γ 10 ) = 8 . answer : option c | a = 8 * 42
b = 140 / a
c = 30 * b
d = 100 / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(add(add(const_1, add(const_1, const_1)), const_1), const_1) | what is the prime factors β number of 48 ? | prime factors β number , as i assume , for a number x = a ^ n * b ^ m * c ^ o * d ^ p . . . is = n + m + o + p . . . so , 24 = 2 ^ 4 * 3 ^ 1 prime factors β number will be 4 + 1 = 5 . hence , answer is d . | a = 1 + 1
b = 1 + a
c = b + 1
d = c + 1
|
a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | c | subtract(20, const_4) | in a group of cows and hens , the number of legs are 20 more than twice the number of heads . the number of cows is | "explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 20 4 x + 2 y = 2 x + 2 y + 20 2 x = 20 x = 10 answer : c" | a = 20 - 4
|
a ) 98.5 , b ) 08.56 , c ) 98.56 , d ) 98.86 , e ) 98.46 | c | add(divide(multiply(832, divide(multiply(subtract(subtract(832, divide(800, 2)), divide(800, 2)), const_100), divide(800, 2))), const_100), subtract(subtract(832, divide(800, 2)), divide(800, 2))) | the compound interest on a sum of money for 2 years is rs . 832 and the simple interest on the same sum for the same period is rs . 800 . the difference between the compound interest and simple interest for 3 years | explanation : difference in c . i and s . i in 2 years = rs . 32 s . i for 1 year = rs . 400 s . i for rs . 400 for one year = rs . 32 rate = [ 100 * 32 ) / ( 400 * 1 ) % = 8 % difference between in c . i and s . i for 3 rd year = s . i on rs . 832 = rs . ( 832 * 8 * 1 ) / 100 = rs . 66.56 answer : c ) rs . 98.56 | a = 800 / 2
b = 832 - a
c = 800 / 2
d = b - c
e = d * 100
f = 800 / 2
g = e / f
h = 832 * g
i = h / 100
j = 800 / 2
k = 832 - j
l = 800 / 2
m = k - l
n = i + m
|
a ) 11 , b ) 21 , c ) 22 , d ) 23 , e ) 32 | b | divide(log(multiply(power(4, 11), power(5, 21))), log(2)) | if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n . what is the value of n ? | "if 5 ^ 21 x 4 ^ 11 = 2 x 10 ^ n what is the value of n 4 ^ 11 = ( 2 ^ 2 ) ^ 11 = 2 ^ 22 10 ^ n = ( 5 x 2 ) ^ n = 5 ^ n x 2 ^ n 5 ^ 21 x 2 ^ 22 = 2 x 2 ^ n x 5 ^ n = 2 ^ n + 1 x 5 ^ n = 2 ^ 22 x 5 ^ 21 n = 21 option b" | a = 4 ** 11
b = 5 ** 21
c = a * b
d = math.log(c)
e = math.log(2)
f = d / e
|
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