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a ) 31 , b ) 40 , c ) 22 , d ) 27 , e ) 25 | a | add(divide(subtract(62, 2), 2), const_1) | find the number of terms in an arithmetic progression with the first term 2 and the last term being 62 , given that common difference is 2 . | the n th term = a + ( n - 1 ) * d 62 = 2 + ( n - 1 ) * 2 , 62 β 2 = ( n - 1 ) * 2 , n = 60 / 2 + 1 = 31 . answer : a | a = 62 - 2
b = a / 2
c = b + 1
|
a ) 50 seconds , b ) 29 seconds , c ) 40 seconds , d ) 11 seconds , e ) 12 seconds | a | divide(add(360, 140), divide(multiply(36, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 36 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 36 km / hr = 36 * ( 5 / 18 ) m / sec = 10 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 10 ) = 50 seconds answer : a" | a = 360 + 140
b = 36 * 1000
c = b / 3600
d = a / c
|
a ) 13 , b ) 8 , c ) 11 , d ) 12 , e ) 14 | b | divide(add(15, subtract(power(15, const_2), multiply(56, const_4))), const_2) | 15 times a positive integer is more than its square by 56 , then the positive integer is | explanation : let the number be x . then , 15 x = x 2 + 56 = > x 2 - 15 x + 56 = 0 = > ( x - 7 ) ( x - 8 ) = 0 = > x = 7 or 8 answer : option b | a = 15 ** 2
b = 56 * 4
c = a - b
d = 15 + c
e = d / 2
|
a ) 24 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | b | divide(multiply(subtract(add(multiply(divide(multiply(280, 35), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | a shopkeeper has 280 kg of apples . he sells 35 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total . | "if the total quantity was 100 then 35 x 20 % + 60 x 30 % = 25 this profit will remain same for any total quantity unless the % of products remains the same . hence ' b ' is the answer" | a = 280 * 35
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = 280 * 60
g = f / 100
h = 100 + 30
i = h / 100
j = g * i
k = e + j
l = k - 280
m = l * 100
n = m / 280
|
a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 15 hrs , e ) 6 hrs | a | multiply(divide(1, 5), 10) | a train running at 1 / 5 of its own speed reached a place in 10 hours . how much time could be saved if the train would have run at its own speed ? | "new speed = 1 / 5 of usual speed new time = 5 / 1 of usual time 5 / 1 of usual time = 10 hrs usual time = 10 * 1 / 5 = 2 hrs time saved = 10 - 2 = 8 hrs answer is a" | a = 1 / 5
b = a * 10
|
a ) 50 , b ) 60 , c ) 80 , d ) 320 , e ) none | a | divide(40, subtract(const_1, divide(20, const_100))) | a number exceeds 20 % of itself by 40 . the number is | "explanation : let the answer be ' a ' then a β 20 / 100 a = 40 β 5 a β a = 200 β a = 50 β a = 50 correct option : a" | a = 20 / 100
b = 1 - a
c = 40 / b
|
a ) 3 / 4 , b ) 1 , c ) 4 / 3 , d ) 2 , e ) 7 / 2 | c | divide(4, 3) | a line that passes through ( β 1 , β 4 ) and ( 3 , k ) has a slope = k . what is the value of k ? | "this is a considerably more difficult one , which will involve some algebra . let β s say that the β first β point is ( β 1 , β 4 ) and the β second , β ( 3 , k ) . the rise = k + 4 , which involves a variable . the run = 3 β ( β 1 ) = 4 . there the slope is ( k + 4 ) / 4 , and we can set this equal to k and solve for k . { { k + 4 } / 4 } = k k + 4 = 4 k 4 = 3 k k = 4 / 3 answer = c" | a = 4 / 3
|
a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 40 | e | subtract(94, subtract(add(41, 22), 9)) | in a class of 94 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 94 54 + not = 94 not = 94 - 54 = 40 so answer is e" | a = 41 + 22
b = a - 9
c = 94 - b
|
a ) a . 720 , b ) b . 256 , c ) c . 1 , d ) d . 12 , e ) e . 36 | a | multiply(multiply(multiply(multiply(multiply(6, subtract(6, const_1)), const_4), const_3), const_2), const_1) | the faces of storage box are to be painted by 6 different colors . in how many ways can be this be done ? | if i have to paint 6 sides with 6 different colour . . . first face can have 6 c 1 options , 2 nd would have 5 c 1 , and subsequent ones would have 4 c 1 , 3 c 1 , 2 c 1 and 1 options respectively . total options = 6 c 1 x 5 c 1 x 4 c 1 x 3 c 1 x 2 c 1 x 1 = 720 distinct ways correct answer - a | a = 6 - 1
b = 6 * a
c = b * 4
d = c * 3
e = d * 2
f = e * 1
|
a ) 68 % , b ) 74 % , c ) 76 % , d ) 62 % , e ) 89 % | e | multiply(divide(subtract(1000, add(add(add(20, 120), add(200, 250)), 300)), 1000), const_100) | john had a stock of 1000 books in his bookshop . he sold 20 on monday , 120 on tuesday , 200 on wednesday , 250 on thursday and 300 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 20 + 120 + 200 + 250 + 300 = 890 let m be the books not sold m = 1000 - n = 1200 - 890 = 798 percentage books not sold / total number of books = 890 / 1000 = 0.89 = 89 % correct answer e" | a = 20 + 120
b = 200 + 250
c = a + b
d = c + 300
e = 1000 - d
f = e / 1000
g = f * 100
|
a ) 112.5 , b ) 122.5 , c ) 132.5 , d ) 114.5 , e ) 45 | e | subtract(divide(multiply(multiply(1000, 11.5), 3), const_100), divide(multiply(multiply(1000, 10), 3), const_100)) | if a lends rs . 1000 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 1000 * 1.5 * 3 ) / 100 = > 45 answer : e" | a = 1000 * 11
b = a * 3
c = b / 100
d = 1000 * 10
e = d * 3
f = e / 100
g = c - f
|
a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | b | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | in n is a positive integer less than 200 , and 14 n / 60 is an integer , then n has how many different positive prime factors r ? | i like to put the numbers in prime factors so it is easier and faster to visualize . 14 * n / 60 if we write the factors of 14 - - > 2 , 7 , and the factors of 60 - - > 2 , 2 , 3 , 5 , we have ( 2 * 7 * n ) / ( 2 ^ 2 * 3 * 5 ) simplifying 7 * n / ( 2 * 3 * 5 ) the only way the equation above has an integer value is if n has at least the factors 2 , 3 and 5 , so we can simplify again and we have the number 7 . the number could be 2 * 3 * 5 , or 2 * 3 * 5 * 2 , or 2 * 3 * 5 * . . . . . however to be less than 200 we can not add any prime number . 2 * 3 * 5 = 120 if we added the next prime factor 7 , we would have r = 2 * 3 * 5 * 7 = 840 thus , answer b | a = 1 + 4
b = 60 / 2
c = b / 2
d = c / 3
e = a / d
f = e + 2
|
a ) 10 days , b ) 12 days , c ) 6 days , d ) 7 days , e ) 8 days | a | multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 2))), 3) | a can finish a work in 24 days , b in 9 days and c in 2 days , b and c start the work but are forced to leave after 3 days . the remaining work was done by a in ? | "b + c 1 day work = 1 / 9 + 1 / 12 = 7 / 36 work done by b and c in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 24 work is done by a in 1 day 5 / 12 work is done by a in 24 * 5 / 12 = 10 days answer is a" | a = 1 / 9
b = 1 / 2
c = a + b
d = 1 / c
e = d * 3
|
a ) s . 42 , b ) s . 46 , c ) s . 40 , d ) s . 80 , e ) s . 34 | d | divide(divide(multiply(800, 50), const_100), 5) | a reduction of 50 % 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 * ( 50 / 100 ) = 400 - - - - 5 ? - - - - 1 = > rs . 80 answer : d" | a = 800 * 50
b = a / 100
c = b / 5
|
a ) 8 and 9 , b ) 20 and 32 , c ) 17 and 19 , d ) 12 and 29 , e ) 17 and 29 | b | add(multiply(subtract(add(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 3), multiply(4, multiply(24, 4))))), 4), const_10), subtract(subtract(subtract(24, 4), sqrt(subtract(power(subtract(24, 4), 3), multiply(4, multiply(24, 4))))), 4)) | if x / 4 + 24 / x = 5 , what are the values of 3 x - 4 ? | i got the same thing b is the answer 20 or 32 | a = 24 - 4
b = 24 - 4
c = b ** 3
d = 24 * 4
e = 4 * d
f = c - e
g = math.sqrt(f)
h = a + g
i = h - 4
j = i * 10
k = 24 - 4
l = 24 - 4
m = l ** 3
n = 24 * 4
o = 4 * n
p = m - o
q = math.sqrt(p)
r = k - q
s = r - 4
t = j + s
|
a ) 150 , b ) 200 , c ) 250 , d ) 275 , e ) 285 | b | add(multiply(divide(add(50, 25), subtract(divide(45, const_100), divide(30, const_100))), divide(30, const_100)), 50) | a candidate who gets 30 % of the marks fails by 50 marks . but another candidate who gets 45 % marks gets 25 marks more than necessary for passing . find the number of marks for passing ? | 30 % - - - - - - - - - - - - 50 45 % - - - - - - - - - - - - 25 - - - - - - - - - - - - - - - - - - - - - - 15 % - - - - - - - - - - - - - 75 30 % - - - - - - - - - - - - - - ? 150 + 50 = 200 marks answer : b | a = 50 + 25
b = 45 / 100
c = 30 / 100
d = b - c
e = a / d
f = 30 / 100
g = e * f
h = g + 50
|
a ) 70 , b ) 74 , c ) 78 , d ) 82 , e ) 86 | c | add(add(add(add(add(add(add(6, 6), add(6, const_2)), add(6, const_1)), 6), 6), const_2), const_1) | if two integers x , y ( x > y ) are selected from - 6 to 6 ( inclusive ) , how many cases are there ? | "there are 13 integers from - 6 to 6 inclusive . 13 c 2 = 78 . the answer is c ." | a = 6 + 6
b = 6 + 2
c = a + b
d = 6 + 1
e = c + d
f = e + 6
g = f + 6
h = g + 2
i = h + 1
|
a ) 8 , b ) 10 , c ) 18 , d ) 4 , e ) 12 | d | multiply(multiply(divide(7, 5), divide(10, 7)), 2) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 2 months , find for how much time did q invest the money ? | "7 * 5 : 2 * x = 7 : 10 x = 4 answer : 4 answer is d" | a = 7 / 5
b = 10 / 7
c = a * b
d = c * 2
|
a ) 12 , b ) 14 , c ) 10 , d ) 11 , e ) 13 | b | inverse(add(inverse(subtract(14, const_1)), inverse(add(14, 2)))) | working alone at their respective constant rates , a can complete a task in β a β days and b in β b β days . they take turns in doing the task with each working 2 days at a time . if a starts they finish the task in exactly 14 days . if b starts , they take a day more . how long does it take to complete the task if they both work together ? | "work done by ab in a day = xy respectively . when a starts : no . of days when a works = 8 no . of days when b works = 6 β 8 x + 6 y = 1 when b starts : no . of days when a works = 7 no . of days when a works = 7 β 7 x + 7 y = 1 solving the above two equations for xy x = 1 / 14 y = 1 / 14 β total work done by ab in a day = 1 / 14 + 1 / 14 = 2 / 28 = 1 / 14 β no . of days to complete the work when both work together = 14 answer : b" | a = 14 - 1
b = 1/(a)
c = 14 + 2
d = 1/(c)
e = b + d
f = 1/(e)
|
a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 120,000 | e | add(subtract(multiply(12, 12), 50), const_1) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 15500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | answer equals e in 48 years . i thought by 50 th year it would reach 120,000 . options should have been separated more widely for clarity . | a = 12 * 12
b = a - 50
c = b + 1
|
a ) 75 , b ) 65 , c ) 55 , d ) 95 , e ) 80 | c | add(multiply(8, 2.5), 35) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 35 kg . what is the weight of the new person ? | "total increase in weight = 8 Γ 2.5 = 20 if x is the weight of the new person , total increase in weight = x β 35 = > 20 = x - 35 = > x = 20 + 35 = 55 answer is c ." | a = 8 * 2
b = a + 35
|
a ) 120,50 , b ) 180,60 , c ) 180,90 , d ) 120,40 , e ) none of them | a | add(divide(subtract(170, 15), add(const_1, divide(const_1, const_2))), 15) | a third of arun Γ’ β¬ β’ s marks in mathematics exceed a half of his marks in english by 15 . if he got 170 marks in two subjects together how many marks did he got in english ? | let arun Γ’ β¬ β’ s marks in mathematics and english be x and y then ( 1 / 3 ) x - ( 1 / 2 ) y = 15 2 x - 3 y = 90 Γ’ β¬ Β¦ Γ’ β¬ Β¦ > ( 1 ) x + y = 170 Γ’ β¬ Β¦ Γ’ β¬ Β¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 120 and y = 50 answer is a . | a = 170 - 15
b = 1 / 2
c = 1 + b
d = a / c
e = d + 15
|
a ) 0.125 % , b ) 0.2 % , c ) 0.16 % , d ) 1.25 % , e ) 2.0 % | c | multiply(4, divide(4, const_100)) | in the manufacture of a certain product , 4 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | "percent of defective produced = 4 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 4 / 100 ) * 100 % = ( 16 / 10000 ) * 100 % = ( 16 / 100 ) % = . 16 % answer c" | a = 4 / 100
b = 4 * a
|
a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 12 , d ) 1 / 3 ! , e ) 1 / 24 | d | divide(add(subtract(4.4, divide(multiply(const_4, const_10), const_100)), const_2), multiply(add(subtract(4.4, divide(multiply(const_4, const_10), const_100)), const_2), add(subtract(4.4, divide(multiply(const_4, const_10), const_100)), const_2))) | if you roll a fair - sided die twice , what is the probability of getting a ( 4.4 ) double ? | total no . in two dice ( 1,1 ) , ( 2,2 ) . . . . . ( 6 , ) total number of events = 6 * 6 total events of getting same pair ( 4 , 4 ) = 6 probability = 6 / 6 * 6 = 1 / 6 = 1 / 3 ! d | a = 4 * 10
b = a / 100
c = 4 - 4
d = c + 2
e = 4 * 10
f = e / 100
g = 4 - 4
h = g + 2
i = 4 * 10
j = i / 100
k = 4 - 4
l = k + 2
m = h * l
n = d / m
|
a ) 63 , b ) 88 , c ) 66 , d ) 44 , e ) 35 | a | divide(770, multiply(subtract(46.5, 2.5), const_0_2778)) | how many seconds will a 770 m long train take to cross a man walking with a speed of 2.5 km / hr in the direction of the moving train if the speed of the train is 46.5 km / hr ? | "speed of train relative to man = 46.5 - 2.5 = 44 km / hr . = 44 * 5 / 18 = 110 / 9 m / sec . time taken to pass the man = 770 * 9 / 110 = 63 sec . answer : a" | a = 46 - 5
b = a * const_0_2778
c = 770 / b
|
a ) 36 , b ) 72 , c ) 122 , d ) 144 , e ) 180 | c | multiply(subtract(divide(multiply(const_2, 30), subtract(70, multiply(const_2, 30))), divide(30, subtract(70, 30))), const_60) | buses a and b start from a common bus stop x . bus a begins to travel in a straight line away from bus b at a constant rate of 30 miles per hour . one hour later , bus b begins to travel in a straight line in the exact opposite direction at a constant rate of 70 miles per hour . if both buses travel indefinitely , what is the positive difference , in minutes , between the amount of time it takes bus b to cover the exact distance that bus a has covered and the amount of time it takes bus b to cover twice the distance that bus a has covered ? | "1 st part : - in 1 hr , bus a covers 30 miles . relative speed of bus abus b is ( 80 - 30 ) = 50 mph . so time required for bus b to cover the exact distance as a is 50 * t = 30 t = 3 / 5 = 36 min 2 nd part 80 * t = 2 d - b has to cover twice the distance 30 * ( t + 1 ) = d - a traveled 1 hr more and has to travel only only d so d / 30 - 2 d / 80 = 1 d = 120 t = 3 hrs = 180 min question asks for + ve difference between part 1 and part 2 in minutes = 180 - 36 = 122 min c" | a = 2 * 30
b = 2 * 30
c = 70 - b
d = a / c
e = 70 - 30
f = 30 / e
g = d - f
h = g * const_60
|
a ) 24 , b ) 36 , c ) 48 , d ) 72 , e ) 132 | c | divide(84, divide(subtract(11, 4), 4)) | at a certain zoo , the ratio of sea lions to penguins is 4 to 11 . if there are 84 more penguins than sea lions at the zoo , how many sea lions are there ? | the ratio as mentioned = 4 / 11 also sea lion + 84 = penguins or , 4 x + 84 = 11 x x = 84 / 7 = 12 number of sea lions = 12 * 4 = 48 answer : c | a = 11 - 4
b = a / 4
c = 84 / b
|
a ) 10 sec , b ) 16 sec , c ) 19 sec , d ) 7.2 sec , e ) 13 sec | d | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 36), inverse(multiply(const_0_2778, add(60, 40)))) | two trains of equal length , running with the speeds of 60 and 40 kmph , take 36 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 36 d = 36 * 100 / 18 = 200 rs = 60 + 40 = 100 * 5 / 18 t = 200 * 18 / 500 = 7.2 sec answer : d | a = 60 - 40
b = const_0_2778 * a
c = b * 36
d = 60 + 40
e = const_0_2778 * d
f = 1/(e)
g = c * f
|
a ) - 4.5 . , b ) - 2 . , c ) - 1.7 . , d ) 3 . , e ) 6 . | e | add(negate(divide(divide(21, 2), negate(divide(21, const_10)))), const_1) | if x + y = 2 x + z , x - 2 y = 4 z and x + y + z = 21 , what is the value of y / z ? | x + y = 2 x + 2 z y = x + 2 z - - - - - - - - - - 1 x - 2 y = 4 z x - 4 z = 2 y - - - - - - - - - 2 subtracting equation 1 from equation 2 - 6 z = y y / z = - 6 e is the answer | a = 21 / 2
b = 21 / 10
c = a / negate
d = negate + (
|
a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | e | divide(multiply(30, 36), subtract(36, 30)) | the bankers discount of a certain sum of money is rs . 36 and the true discount on the same sum for the same time is rs . 30 . the sum due is : | "sum = ( b . d * t . d ) / ( b . d - t . d ) ( 36 * 30 ) / 36 - 30 ; 180 answer : e" | a = 30 * 36
b = 36 - 30
c = a / b
|
a ) 190 , b ) 132 , c ) 210 , d ) 220 , e ) 225 | b | multiply(12, subtract(12, const_1)) | 12 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "though 2 * ( 12 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 11 , so he ' ll play 22 matches . similarly , second guy has to play with the rest of 10 as his 2 games with the first guy are already played . so he plays 20 matches . this continues like this and the total matches are 22 + 20 + 18 . . . + 2 22 + 20 + . . . + 2 = 2 ( 11 + 10 + . . . + 1 ) = 2 ( ( 11 * 12 ) / 2 ) = 11 * 12 = 132 . answer : b" | a = 12 - 1
b = 12 * a
|
a ) 8 miles , b ) 12 miles , c ) 10 miles , d ) 11 miles , e ) 14 miles | b | multiply(3, add(3, 1)) | one hour after adrienne started walking the 60 miles from x to y , harold started walking from x to y as well . adrienne walks 3 miles per hour and harold walks 1 mile per hour faster than adrienne . how far from x will harold be when he catches up to adrienne ? | first , determine how far adrienne has walked in the one hour . she has walked three miles which means she is three miles ahead of harold when he sets off . harold walks at four miles / hour which means that every hour , harold will get one mile closer to adrienne . if he gets one mile closer every hour , it will take him three hours to catch up to her which means he travels 3 hours * 4 miles / hour = 12 miles and she travels 4 hours * 3 miles / hour = 12 miles . he will be 12 miles from x when he catches up to her . a slightly different way to solve . . . we do n ' t know how long they will walk before they catch up to one another but we do know that a walks for one hour more than j . j = t and a = t + 1 . we are looking for the distance at which they meet up which means the distance will be the same . d = r * t so , r * t ( harold ) = r * t ( adrienne ) r * ( t ) = r * ( t + 1 ) 4 t = 3 t + 3 t = 3 d = r * t d = 4 * 3 d = 12 b ) 12 miles | a = 3 + 1
b = 3 * a
|
a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38 | b | divide(multiply(48, 24), 36) | in a division , a student took 48 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 48 = 24 . x = 24 * 48 . so correct answer would be , ( 24 * 48 ) / 36 = 32 . answer : b" | a = 48 * 24
b = a / 36
|
a ) 32 kmph , b ) 34 kmph , c ) 30 kmph , d ) 45 kmph , e ) 40 kmph | e | divide(add(20, 60), const_2) | a man can row upstream at 20 kmph and downstream at 60 kmph , and then find the speed of the man in still water ? | "us = 20 ds = 60 m = ( 20 + 60 ) / 2 = 40 answer : e" | a = 20 + 60
b = a / 2
|
a ) 2 , b ) 5 , c ) 10 , d ) 16 , e ) 60 | e | subtract(add(50, 60), 50) | two family reunions are happening at the same hotel , the dates reunion and the hall reunion . all 50 guests at the hotel attend at least one of the reunions . if 50 people attend the dates reunion and 60 people attend the hall reunion , how many people attend both reunions ? | no of people in dates reunion = 50 no of people in hall reunion = 60 attending both = x all guests attend at least one . therefore , 50 = 50 + 60 - ( both ) both = 60 answer e | a = 50 + 60
b = a - 50
|
a ) 1 km , b ) 1500 mts , c ) 1800 mts , d ) 2 km , e ) 1250 mts | c | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 6), divide(const_1, 12)), divide(const_1, 18))), const_3), const_1000) | a person travels equal distances with speeds of 6 km / hr , 12 km / hr , 18 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 6 ) + ( x / 12 ) + ( x / 18 ) = 11 / 60 x = 0.6 total distance = 3 * 0.6 = 1.8 km = 1800 meters correct option is c" | a = 11 / const_60
b = 1 / 6
c = 1 / 12
d = b + c
e = 1 / 18
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 3 minutes , b ) 4 minutes , c ) 5 minutes , d ) 6 minutes , e ) 10 minutes | e | divide(multiply(5, 10), add(speed(10, 10), speed(multiply(2, 10), 5))) | working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 2 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 5 x boxes ? | "rate = work / time given rate of machine a = x / 10 min machine b produces 2 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 4 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / 10 min + rate of machine b = 4 x / 10 = 5 x / 10 now combine time = combine work needs to be done / combine rate = 5 x / 5 x * 10 = 10 min ans : e" | a = 5 * 10
b = speed + (
c = 2 * 10
d = a / b
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a ) 6 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(11, 3) | if ( a β b ) is 11 more than ( c + d ) and ( a + b ) is 3 less than ( c β d ) , then ( a β c ) is : | "( a β b ) β ( c + d ) = 11 and ( c β d ) β ( a + b ) = 3 = > ( a β c ) β ( b + d ) = 11 and ( c β a ) β ( b + d ) = 3 = > ( b + d ) = ( a β c ) β 11 and ( b + d ) = ( c β a ) β 3 = > ( a β c ) β 11 = ( c β a ) β 3 = > 2 ( a β c ) = 8 = > ( a β c ) = 4 answer : d" | a = 11 / 3
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a ) 10.7 , b ) 10.9 , c ) 10.24 , d ) 12.24 , e ) 18.8 | d | divide(add(180, 160), multiply(add(60, 40), const_0_2778)) | two trains 180 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 180 + 160 = 340 m . required time = 340 * 9 / 250 = 12.24 sec . answer : d" | a = 180 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
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a ) 178 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(267, subtract(20, 8)), 8) | a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 267 metres long in 20 seconds . the length of the train ( in metres ) is : | explanation : let the length of train be l m . acc . to question ( 267 + l ) / 20 = l / 8 2136 + 8 l = 20 l l = 2136 / 12 = 178 m answer a | a = 20 - 8
b = 267 / a
c = b * 8
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a ) 1.5 , b ) 1.7 , c ) 2.3 , d ) 2.5 , e ) 3.0 | c | divide(divide(multiply(15, 50), 50), const_2) | mixture a is 15 percent alcohol , and mixture b is 50 percent alcohol . if the two are poured together to create a 4 - gallon mixture that contains 30 percent alcohol , approximately how many gallons of mixture a are in the mixture ? | "ratio of a to b in the final mixture = 50 - 30 : 30 - 15 = 4 : 3 gallons of a used = 4 / 7 * 4 = 2.28 answer : c" | a = 15 * 50
b = a / 50
c = b / 2
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a ) 600 , b ) 320 , c ) 500 , d ) 520 , e ) 720 | b | multiply(160, 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 160 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 - 160 ) = 2 + 2 ( extra ) = > 2 x - 320 = x = > x = 320 . answer : b" | a = 160 * 2
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | multiply(divide(3, add(0.35, 0.25)), const_2) | a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 3.00 , what total number of pens and pencils did the customer purchase ? | answer : algebraically , the question looks like this : 3.0 = 0.35 x + 0.25 y there are six possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 4 pens : $ 1.40 5 pens : $ 1.75 balance 1.25 5 pencils cost $ 1.25 that ' s 5 pens and 5 pencils choice ( c ) is correct . | a = 0 + 35
b = 3 / a
c = b * 2
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a ) 4800 , b ) 4860 , c ) 5000 , d ) 5500 , e ) 5550 | b | add(100, const_1) | what is the sum of natural numbers between 20 and 100 | "a = first number l = last number sn = n / 2 [ a + l ] between 20 and 100 numbers = 81 = > 100 - 20 = 80 + 1 = 81 sn = 81 / 2 Γ 120 = 81 Γ 60 = 4860 b )" | a = 100 + 1
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a ) 6 , b ) 9 , c ) 12 , d ) 15 , e ) 18 | b | multiply(6, multiply(6, divide(4, multiply(4, 4)))) | 4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 6 weavers in 6 days ? | 1 weaver can weave 1 mat in 4 days . 6 weavers can weave 6 mats in 4 days . 6 weavers can weave 9 mats in 6 days . the answer is b . | a = 4 * 4
b = 4 / a
c = 6 * b
d = 6 * c
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a ) 160 , b ) 180 , c ) 200 , d ) 240 , e ) none | d | divide(add(6, 6), subtract(divide(const_1, const_4), divide(const_1, add(const_1, const_4)))) | a number whose fifth part increased by 6 is equal to its fourth part diminished by 6 is ? | "answer let the number be n . then , ( n / 5 ) + 6 = ( n / 4 ) - 6 Γ’ β‘ β ( n / 4 ) - ( n / 5 ) = 12 Γ’ β‘ β ( 5 n - 4 n ) / 20 = 12 Γ’ Λ Β΄ n = 240 option : d" | a = 6 + 6
b = 1 / 4
c = 1 + 4
d = 1 / c
e = b - d
f = a / e
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a ) 3 , b ) 9 , c ) 10 , d ) 12 , e ) 14 | d | subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(32, 4), 4), 4), 4), const_1), 3), const_1), 4), 3), 4) | if 3 a β 2 b β 4 c = 32 and β 3 a - β ( 2 b + 2 c ) = 4 , what is the value of a + b + c ? | "when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a β 2 b β 2 c = 32 can be written as β 3 a ^ 2 - β ( 2 b + 2 c ) ^ 2 = 32 { β 3 a - β ( 2 b + 2 c ) } { β 3 a + β ( 2 b + 2 c ) } = 32 . . or 4 * β 3 a + β ( 2 b + 2 c ) = 32 . . or β 3 a + β ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) β 3 a - β ( 2 b + 2 c ) = 4 . . 2 ) β 3 a + β ( 2 b + 2 c ) = 8 . . a ) add the two eq . . β 3 a + β ( 2 b + 2 c ) + β 3 a - β ( 2 b + 2 c ) = 12 . . 2 β 3 a = 12 . . or β 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . β 3 a + β ( 2 b + 2 c ) - β 3 a + β ( 2 b + 2 c ) = 4 . . 2 β ( 2 b + 2 c ) = 4 . . β ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . d" | a = 32 - 4
b = a - 4
c = b - 4
d = c - 4
e = d - 1
f = e - 3
g = f - 1
h = g + 4
i = h - 3
j = i - 4
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a ) 1000 , b ) 1100 , c ) 1200 , d ) 900 , e ) 10000 | d | multiply(multiply(3, const_1000), 3) | find the product of the localvalue and absolutevalue of 3 in 20368 ? | "local value of 3 = 3 x 100 = 300 place value of 3 = 3 there fore = 3 x 300 = 900 d" | a = 3 * 1000
b = a * 3
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a ) 22 sec , b ) 32 sec , c ) 25 sec , d ) 24 sec , e ) 31 sec | b | divide(add(200, 120), multiply(36, const_0_2778)) | how many seconds will a train 120 meters long take to cross a bridge 200 meters long if the speed of the train is 36 kmph ? | "explanation : d = 120 + 200 = 320 s = 36 * 5 / 18 = 10 mps t = 250 / 10 = 32 sec answer : option b" | a = 200 + 120
b = 36 * const_0_2778
c = a / b
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a ) 6 cm , b ) 7 cm , c ) 9 cm , d ) 11 cm , e ) none | c | divide(subtract(32, multiply(2, 2)), const_4) | the difference of the areas of two squares drawn on two line segments of different lengths is 32 sq . cm . find the length of the greater line segment if one is longer than the other by 2 cm . | "solution let the lengths of the line segments be x cm and ( x + 2 ) cm . then ( x + 2 ) 2 - x 2 = 32 < = > x 2 + 4 x + 4 - x 2 = 32 < = > 4 x = 28 therefore x = 7 . length of longer line segment = ( 7 + 2 ) cm = 9 cm . answer c" | a = 2 * 2
b = 32 - a
c = b / 4
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a ) 1 : 4 , b ) 1 : 25 , c ) 1 : 52 , d ) 1 : 522 , e ) none | a | divide(const_4, const_100) | a cube of edge 6 cm is cut into cubes each of edge 3 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | "sol . required ratio = 6 * 3 * 3 / 6 * 6 * 6 = 1 / 4 = 1 : 4 . answer a" | a = 4 / 100
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a ) 4 , b ) 7 , c ) 9 , d ) 5 , e ) 2 | a | subtract(7, 3) | robert ate 7 chocolates , nickel ate 3 chocolates . how many more chocolates did robert ate than nickel ? | 7 - 3 = 4 . answer is a | a = 7 - 3
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a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | d | divide(1, divide(add(multiply(const_3600, divide(1, 36)), 20), const_3600)) | a car traveling at a certain constant speed takes 20 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 36 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "36 * t = 1 km = > t = 1 / 36 km / h v * ( t + 20 / 3600 ) = 1 v ( 1 / 36 + 20 / 3600 ) = 1 v ( 120 / 3600 ) = 1 v = 30 km / h the answer is d ." | a = 1 / 36
b = 3600 * a
c = b + 20
d = c / 3600
e = 1 / d
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a ) 50 % , b ) 40 % , c ) 33 1 β 3 % , d ) 25 % , e ) 12 1 β 2 % | c | multiply(subtract(const_1, divide(const_100, add(const_100, 50))), const_100) | the product of x and y is a constant . if the value of x is increased by 50 % , by what percentage must the value of y be decreased ? | "product of x and y = xy if the value of x is increased by 50 % , value of y needs to be = xy / ( 1.5 x ) = 2 / 3 y decrease in value of y = y - 2 / 3 y = 1 / 3 y % decrease in value of y = ( 1 / 3 y ) / y * 100 % = 33 1 β 3 % answer c" | a = 100 + 50
b = 100 / a
c = 1 - b
d = c * 100
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a ) 10 cm , b ) 12 cm , c ) 14 cm , d ) 16 cm , e ) 18 cm | a | divide(150, const_10) | the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 150 sq . cm , what is the width of the rectangle ? | "2 l + 2 w = 5 w l = 3 w / 2 w * l = 150 3 w ^ 2 / 2 = 150 w ^ 2 = 100 w = 10 the answer is a ." | a = 150 / 10
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a ) 30 , b ) 36 , c ) 40 , d ) 45 , e ) 48 | b | multiply(divide(subtract(const_1, multiply(add(inverse(3), inverse(15)), const_2)), inverse(3)), const_60) | working at a constant rate , p can finish a job in 3 hours . q , also working at a constant rate , can finish the same job in 15 hours . if they work together for 2 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | each hour they complete 1 / 3 + 1 / 15 = 2 / 5 of the job . in 2 hours , they complete 2 ( 2 / 5 ) = 4 / 5 of the job . the time for p to finish is ( 1 / 5 ) / ( 1 / 3 ) = ( 3 / 5 ) hour = 36 minutes the answer is b . | a = 1/(3)
b = 1/(15)
c = a + b
d = c * 2
e = 1 - d
f = 1/(3)
g = e / f
h = g * const_60
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a ) 251 , b ) 261 , c ) 262 , d ) 269 , e ) 278 | c | add(multiply(subtract(10, const_1), subtract(30, const_2)), 10) | what is the sum of the digits of integer k , if k = ( 10 ^ 30 - 54 ) | "there are 31 digits in 10 ^ 30 when we subtract 54 from it , there will be 30 digits left . 10 ^ 30 can be written as 9999999 . . . . ( 30 times ) + 1 so , 10 ^ 30 - 54 = 9999999 . . . . ( 30 times ) + 1 - 54 = 9999999 . . . . ( 30 times ) - 53 consider the last 2 digits , 99 - 53 = 46 the last 2 digits will be 46 . and our number would be 99999 . . . . . . 99946 with 2 less 9 s after subtraction . number of 9 s left are 28 and the last two digits are 46 the sum of the digits will be ( 28 * 9 ) + 4 + 6 = 262 answer : - c" | a = 10 - 1
b = 30 - 2
c = a * b
d = c + 10
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a ) 88900 , b ) 87700 , c ) 88970 , d ) 88000 , e ) 89000 | c | divide(multiply(add(multiply(6, const_100), 20), add(multiply(5, const_100), 74)), power(divide(add(multiply(6, const_100), 20), power(const_2, const_4)), const_2)) | a room of 6 m 20 cm long and 5 m 74 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 = 620 * 574 sq cm size of largest square tile = h . c . f of 620 cm and 574 cm = 2 cm area of 1 tile = 2 * 2 sq cm no . of tiles required = ( 620 * 574 ) / ( 2 * 2 ) = 88970 answer : c ) 88970" | a = 6 * 100
b = a + 20
c = 5 * 100
d = c + 74
e = b * d
f = 6 * 100
g = f + 20
h = 2 ** 4
i = g / h
j = i ** 2
k = e / j
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a ) 287 , b ) 278 , c ) 698 , d ) 279 , e ) 275 | c | subtract(815, divide(multiply(subtract(854, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 854 in 4 years . the sum is ? | "s . i . for 1 year = ( 854 - 815 ) = rs . 39 s . i . for 3 years = 39 * 3 = rs . 117 principal = ( 815 - 117 ) = rs . 698 . answer : c" | a = 854 - 815
b = a * 3
c = b / 4
d = 815 - c
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a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | subtract(7, 5) | a football team lost 5 yards and then gained 7 . what is the team ' s progress ? | for lost , use negative . for gain , use positive . progress = - 5 + 7 = 2 yards a | a = 7 - 5
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a ) 26 , b ) 28 , c ) 29 , d ) 30 , e ) 27 | e | divide(add(102, const_1), const_2) | the sum of four consecutive numbers is 102 . the greatest among these three number is : | "let the numbers be x , x + 1 , x + 2 and x + 3 then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 102 4 x = 96 x = 24 greatest number , ( x + 3 ) = 27 answer : e" | a = 102 + 1
b = a / 2
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a ) 72 , b ) 120 , c ) 240 , d ) 360 , e ) 720 | a | lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 18) | what is the least common multiple of 12 , 18 , and 24 ? | "let us first write the numbers in the form of prime factors : 12 = 2 ^ 2 * 3 ^ 1 18 = 2 ^ 1 * 3 ^ 2 24 = 2 * 17 ^ 1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 72 option a" | a = 10 + 2
b = 3 * 10
c = b - 3
d = math.lcm(a, c)
e = math.lcm(d, 18)
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a ) 40 , b ) 50 , c ) 62 , d ) 45 , e ) 56 | a | add(subtract(90, multiply(11, 5)), 5) | a batsman makes a score of 90 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 90 = 11 x x = 90 - 50 = 40 answer is a" | a = 11 * 5
b = 90 - a
c = b + 5
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a ) 28 years , b ) 19 years , c ) 29 years , d ) 10 years , e ) 12 years | a | divide(multiply(subtract(72, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 72 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 72 5 x = 70 = > x = 14 hence , b ' s age = 2 x = 28 years . answer : a" | a = 72 - 2
b = a * 2
c = 4 + 1
d = b / c
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a ) 2372 , b ) 4634 , c ) 4542 , d ) 2343 , e ) 3972 | b | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3))) | there is 60 % increase in an amount in 6 years at si . what will be the ci of rs . 14,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 14000 . t = 3 years and r = 10 % p . a . c . i . = rs . 14000 x 1 + 10 3 - 1 100 = rs . 14000 x 331 1000 = 4634 . b" | a = 100 * 100
b = 60 / 6
c = 100 * b
d = c * 3
e = a + d
f = 60 / 6
g = f / 100
h = 1 + g
i = 60 / 6
j = i / 100
k = 1 + j
l = h * k
m = 60 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 60 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
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a ) 3 % , b ) 5 % , c ) 10 % , d ) 15 % , e ) 18 % | d | multiply(subtract(const_1, subtract(const_2, add(divide(15, const_100), const_1))), const_100) | dick and jane each saved $ 1,000 in 1989 . in 1990 dick saved 15 percent more than in 1989 , and together he and jane saved a total of $ 2,000 . approximately what percent less did jane save in 1990 than in 1989 ? | 1990 dick saved = $ 1150 jane saved = $ 850 ( jane saved $ 150 less than she did the prior year ) jane saved approximately $ 150 / 1000 $ ( 15 % ) less in 1990 answer : d | a = 15 / 100
b = a + 1
c = 2 - b
d = 1 - c
e = d * 100
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a ) 40 days , b ) 40 / 9 days , c ) 39 / 9 days , d ) 6 days , e ) 60 / 9 days | d | divide(const_1, add(divide(const_1, 10), divide(const_1, 15))) | worker a takes 10 hours to do a job . worker b takes 15 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "a ' s one hour work = 1 / 10 . b ' s one hour work = 1 / 15 . ( a + b ) ' s one hour work = 1 / 10 + 1 / 15 = 5 / 30 = 1 / 6 . both a & b can finish the work in 6 days d" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / c
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a ) 25 , b ) 30 , c ) 31 , d ) 45 , e ) 60 | c | divide(620, multiply(const_0_2778, subtract(80, 8))) | how many seconds will a 620 metre long train take to cross a man running with a speed of 8 km / hr in the direction of the moving train if the speed of the train is 80 km / hr ? | "explanation : speed of train relatively to man = ( 80 - 8 ) km / hr = 72 km / hr = ( 72 x 5 / 18 ) m / sec = 20 m / sec time taken to pass the man = ( 620 / 20 ) sec = 31 sec . answer : c" | a = 80 - 8
b = const_0_2778 * a
c = 620 / b
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a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6000 | b | divide(1040, divide(subtract(60, subtract(const_100, 60)), const_100)) | in an election a candidate who gets 60 % of the votes is elected by a majority of 1040 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1040 20 x / 100 = 1040 x = 1040 * 100 / 20 = 5200 answer is b" | a = 100 - 60
b = 60 - a
c = b / 100
d = 1040 / c
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a ) 260 km , b ) 240 km , c ) 280 km , d ) 340 km , e ) 350 km | b | multiply(40, 6) | a person is traveling at 40 km / hr and reached his destiny in 6 hr then find the distance ? | "t = 6 hrs d = t * s = 40 * 6 = 240 km answer is b" | a = 40 * 6
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a ) $ 0.60 , b ) $ 0.70 , c ) $ 0.86 , d ) $ 0.90 , e ) $ 1.00 | c | add(multiply(multiply(multiply(0.07, const_2), const_2), const_2), 0.30) | a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.07 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the fifth day it is overdue ? | "1 st day fine - 0.07 2 nd day fine - 0.07 * 2 = 0.14 ( as doubling gives lower value ) 3 rd day fine - 0.14 * 2 = 0.28 ( as doubling gives lower value ) 4 rd day fine - 0.28 * 2 = 0.56 ( as doubling gives lower value ) 5 th day fine - 0.56 + 0.3 = 0.86 ( as doubling gives higher value we add 0.3 this time ) answer : c ." | a = 0 * 7
b = a * 2
c = b * 2
d = c + 0
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a ) 342 , b ) 125 , c ) 289 , d ) 564 , e ) 36 | a | subtract(negate(124), multiply(subtract(26, 63), divide(subtract(26, 63), subtract(7, 26)))) | 7 , 26 , 63 , 124 , ( . . . . ) | "explanation : numbers are 2 ^ 3 - 1 = 7 3 ^ 3 - 1 = 26 4 ^ 3 - 1 = 63 5 ^ 3 - 1 = 124 6 ^ 3 - 1 = 342 answer : a" | a = negate - (
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a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 30 % | a | multiply(divide(40, 100), const_100) | by selling 100 pens , a trader gains the cost of 40 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 100 pens = rs . 100 profit = cost of 40 pens = rs . 40 profit % = 40 / 100 * 100 = 40 % answer : a" | a = 40 / 100
b = a * 100
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a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 330 , e ) none | d | subtract(subtract(multiply(divide(1870, const_10), const_2), const_12), const_12) | a sum of rs . 1870 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1870 = > 17 x / 12 = 1870 = > 1870 x 12 / 17 = rs . 1320 hence , b β s share = rs . ( 1320 / 4 ) = rs . 330 . answer d" | a = 1870 / 10
b = a * 2
c = b - 12
d = c - 12
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a ) 50 , b ) 80 , c ) 69 , d ) 125 , e ) 250 | c | multiply(divide(subtract(power(add(divide(multiply(30, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | a circular logo is enlarged to fit the lid of a jar . the new diameter is 30 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 5.2 , so radius is 2.6 new area = 6.76 Ο increase in area is 2.76 Ο % increase in area = 2.76 / 4 * 100 so , % increase is 69 % answer will be ( c )" | a = 30 * 2
b = a / 1000
c = b + 3
d = c ** 2
e = d - 4
f = e / 4
g = f * 100
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a ) 1978 , b ) 2707 , c ) 7728 , d ) 4200 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(25000, add(15000, 25000)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 15000 and b rs . 25000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "15 : 25 = > 3 : 5 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 3 / 8 = 3240 + 960 = 4200 answer : d" | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 15000 + 25000
e = 25000 / d
f = c * e
g = 9600 - f
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a ) 5 , b ) 6 , c ) 9 , d ) 11 , e ) 12 | b | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(19, 11)), const_1), 11)), 11), const_1) | how many numbers from 19 to 79 are exactly divisible by 11 ? | "19 / 11 = 1 and 79 / 11 = 7 = = > 7 - 1 = 6 numbers answer : b" | a = 79 / 11
b = math.floor(a)
c = b * 11
d = 19 / 11
e = math.floor(d)
f = e + 1
g = f * 11
h = c - g
i = h / 11
j = i + 1
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a ) 7.0 , b ) 7.5 , c ) 8.0 , d ) 8.5 , e ) 9.0 | b | subtract(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 30)) | two pipes can fill a tank in 10 minutes and 15 minutes . an outlet pipe can empty the tank in 30 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 10 + v / 15 - v / 30 = 2 v / 15 per minute the tank will be filled in 15 / 2 = 7.5 minutes . the answer is b ." | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / 30
e = c - d
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a ) 20 % , b ) 42 % , c ) 44 % , d ) 75 % , e ) 84 % | d | multiply(subtract(const_1, divide(20, subtract(const_100, 20))), const_100) | exactly 20 % of the reporters for a certain wire service cover local politics in country x . if 20 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "you are correct . people who cover local politics are a subset of people who cover politics . 20 % of reporters who cover politics do not cover local politics so 80 % do cover local politics . reporters covering local politics = 80 % of reporters covering politics = 20 % ofall reporters reporters covering politics / all reporters = 20 / 80 = 1 / 4 reporters not covering politics / all reporters = 3 / 4 = 75 % = d" | a = 100 - 20
b = 20 / a
c = 1 - b
d = c * 100
|
a ) $ 1000 , b ) $ 1050 , c ) $ 1100 , d ) $ 1150 , e ) $ 1200 | e | divide(840, subtract(const_1, divide(30, const_100))) | we had $ 840 left after spending 30 % of the money that we took for shopping . how much money did we start with ? | let x be the amount of money we started with . 0.7 x = 840 x = 1200 the answer is e . | a = 30 / 100
b = 1 - a
c = 840 / b
|
a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 80 | e | subtract(power(divide(add(10, 8), const_2), const_2), power(subtract(10, divide(add(10, 8), const_2)), const_2)) | if the sum and difference of two numbers are 10 and 8 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 10 and x - y = 8 x 2 - y 2 = ( x + y ) ( x - y ) = 10 * 8 = 80 . answer : e" | a = 10 + 8
b = a / 2
c = b ** 2
d = 10 + 8
e = d / 2
f = 10 - e
g = f ** 2
h = c - g
|
a ) 300 , b ) 380 , c ) 180 , d ) 415 , e ) 264 | a | multiply(multiply(multiply(1, const_10), const_10), const_3) | how many times will the digit 5 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 5 ) 3000 / 10 = 300 times . answer : a ." | a = 1 * 10
b = a * 10
c = b * 3
|
a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 67 | b | add(30, const_1) | the average of first six prime numbers which are between 30 and 70 is | "explanation : first six prime numbers which are between 30 and 70 = 31 , 37 , 41 , 43 , 47 , 53 average = ( 31 + 37 + 41 + 43 + 47 + 53 ) / 6 = 42 answer : b" | a = 30 + 1
|
a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | a | add(add(const_4, const_3), const_2) | what is the units digit of ( 3 ^ 11 ) ( 4 ^ 13 ) ? | "- > the ones place of ( ~ 3 ) ^ n repeats after 4 times like 3 ο 9 ο 7 ο 1 ο 3 . the ones place of ( ~ 4 ) ^ n repeats after 2 times like 4 ο 6 ο 4 . then , 3 ^ 13 = 3 ^ 4 * 3 + 1 ο 3 ^ 1 . , 4 ^ 13 = 4 ^ 2 * 6 + 1 = 4 ^ 1 = ~ 4 which is ( 3 ^ 13 ) ( 4 ^ 13 ) ο ( 3 ^ 1 ) ( ~ 4 ) = ( ~ 3 ) ( ~ 4 ) = ~ 2 . therefore , the answer is a ." | a = 4 + 3
b = a + 2
|
a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 , e ) 20 . | d | multiply(divide(add(30, divide(30, const_2)), 6), const_2) | the distance from steve ' s house to work is 30 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 30 km = > speed = 15 kmph answer : d" | a = 30 / 2
b = 30 + a
c = b / 6
d = c * 2
|
a ) 145 cm , b ) 135 cm , c ) 121 cm , d ) 141 cm , e ) 151 cm | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 270 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 270 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 99 + 42 = 141 cm answer : d" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 21 , b ) 27 , c ) 24 , d ) 22 , e ) 25 | d | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | how many times in a day the hands of a clock are straight ? | "in 12 hours , the hands coincide or are in oppesite direction 22 times . in 24 hours , they are at right angles 22 times . answer is d ." | a = 2 * 12
b = 2 * a
c = b - 4
d = c / 2
|
a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 9 / 16 , e ) 15 / 16 | c | subtract(const_1, power(divide(const_1, 2), 3)) | on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow β s neck . what is the probability that the rancher will be able to place a lasso around a cow β s neck at least once in 3 attempts ? | "p ( missing all 3 ) = ( 1 / 2 ) ^ 3 = 1 / 8 p ( success on at least one attempt ) = 1 - 1 / 8 = 7 / 8 the answer is c ." | a = 1 / 2
b = a ** 3
c = 1 - b
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | d | subtract(multiply(12, inverse(divide(8, 12))), 12) | twelve machines , each working at the same constant rate , together can complete a certain job in 12 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 8 days ? | another solution which is faster is since each machine works at a constant rate . the time needs to bought down from 12 to 8 . so the new time is 2 / 3 of the original time . thus to achieve this we need the rate to be 3 / 2 of original . so 3 / 2 * 12 = 18 so we need 18 - 12 = 6 more machines . answer : d | a = 8 / 12
b = 1/(a)
c = 12 * b
d = c - 12
|
a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) 2 | a | divide(log(divide(1, 49)), log(7)) | if 49 ( 7 ^ x ) = 1 then x = | "7 ^ x = 1 / 49 7 ^ x = 1 / 7 ^ 2 7 ^ x = 7 ^ - 2 x = - 2 a" | a = 1 / 49
b = math.log(a)
c = math.log(7)
d = b / c
|
a ) $ 420 , b ) $ 410 , c ) $ 400 , d ) $ 390 , e ) $ 380 | c | divide(add(add(add(add(200, 150), 750), 400), 500), 5) | a cab driver 5 days income was $ 200 , $ 150 , $ 750 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 200 + 150 + 750 + 400 + 500 ) / 5 = 400 answer is c" | a = 200 + 150
b = a + 750
c = b + 400
d = c + 500
e = d / 5
|
a ) 15 / 28 , b ) 17 / 28 , c ) 13 / 28 , d ) 19 / 28 , e ) 11 / 28 | b | divide(5, divide(multiply(14, 20), add(14, 20))) | x is able to do a piece of work in 14 days and y can do the same work in 20 days . if they can work together for 5 days , what is the fraction of work completed ? | explanation : amount of work x can do in 1 day = 1 / 14 amount of work y can do in 1 day = 1 / 20 amount of work x and y can do in 1 day = 1 / 14 + 1 / 20 = 17 / 140 amount of work x and y can together do in 5 days = 5 Γ ( 17 / 140 ) = 17 / 28 answer : option b | a = 14 * 20
b = 14 + 20
c = a / b
d = 5 / c
|
['a ) 450', 'b ) 600', 'c ) 700', 'd ) 2400', 'e ) 3200'] | c | multiply(5600, divide(1200, 9600)) | a football field is 9600 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 5600 square yards ? | answer c ) 9600 yards need 1200 lbs 1 yard will need 1200 / 9600 = 1 / 8 lbs 3600 yards will need 1 / 8 * 5600 yards = 700 lbs | a = 1200 / 9600
b = 5600 * a
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | multiply(divide(subtract(multiply(add(1, divide(20, const_100)), 15), 12), 12), const_100) | if a man buys 1 liter of milk for 12 rs . and mixes it with 20 % water and sells it for 15 rs then what is the % age of gain . . . | quantity after adding 20 % water = 1.2 liter sp of 1.2 lit . @ 15 rs . / liter = 15 * 1.2 = 18 rs cp = 12 rs / lit . % profit = 100 * ( 18 - 12 ) / 12 = 50 answer : c | a = 20 / 100
b = 1 + a
c = b * 15
d = c - 12
e = d / 12
f = e * 100
|
a ) 27 / 32 , b ) 1 / 2 , c ) 5 / 32 , d ) 5 / 27 , e ) . 2 / 3 | c | divide(50, 320) | a tightrope approximately 320 m long is suspended between two poles . during a performance , a break occurs in the line . assume that the line has an equal chance of breaking anywhere along its length . what is the probability that the break happened in the first 50 meters of the rope ? | simly 50 / 320 = 5 / 32 answer will be ( c ) | a = 50 / 320
|
a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | e | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 18) | on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 18 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 Γ 10 % = 90 + 90 Γ 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 Γ 99 = 66 jade handled 18 more transactions than cal . jade = 66 + 18 = 84 jade handled = 84 transactions . answer : e" | a = 90 / 100
b = 10 + 100
c = a * b
d = c * 2
e = d / 3
f = e + 18
|
a ) 2 , b ) 2.2 , c ) 2.4 , d ) 2.6 , e ) 2.8 | c | divide(240, add(divide(240, 4), divide(240, 6))) | while working alone at their constant rates , computer x can process 240 files in 4 hours , and computer y can process 240 files in 6 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ? | both computers together process files at a rate of 240 / 4 + 240 / 6 = 60 + 40 = 100 files per hour . the time required to process 240 files is 240 / 100 = 2.4 hours the answer is c . | a = 240 / 4
b = 240 / 6
c = a + b
d = 240 / c
|
a ) $ 2125.15 , b ) $ 2083.12 , c ) $ 2184.22 , d ) $ 2150.84 , e ) $ 2080.80 | e | multiply(const_100, const_100) | $ 2,000 is deposited in a savings account that pays 4 % annual interest compounded semiannually . to the nearest dollar , how much is in the account at the end of the year ? | "this is the case of semi - annual compoundingso , multiply time period by 2 and divide rate by 2 so , new time = 2 periods and new rate = 4 / 2 = 2 % now , ca = 2000 ( 1 + 2 / 100 ) ^ 2 = $ 2080.80 hence answer is e" | a = 100 * 100
|
a ) 35 , b ) 45 , c ) 25 , d ) 36 , e ) 55 | d | multiply(140, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 140 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 140 Γ 9 / 35 = 4 Γ 9 = 36 answer is d" | a = 15 * 3
b = 10 * 7
c = 12 * 5
d = b + c
e = 15 * 3
f = d + e
g = a / f
h = 140 * g
|
a ) 5 % , b ) 6 % , c ) 2 % , d ) 95 % , e ) 10 % | e | divide(multiply(const_100, 400), multiply(1000, 4)) | what is the rate percent when the simple interest on rs . 1000 amount to rs . 400 in 4 years ? | "interest for 4 yrs = 400 interest for 1 yr = 100 interest rate = 100 / 1000 x 100 = 10 % answer : e" | a = 100 * 400
b = 1000 * 4
c = a / b
|
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