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a ) a ) 3 , b ) b ) 1 , c ) 2 , d ) d ) 21 , e ) e ) 28 | c | add(4, 4) | two positive integers differ by 4 , and sum of their reciprocals is 4 . then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 4 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 4 or ( n ^ 2 + 4 n ) * 4 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 2 . hence , c" | a = 4 + 4
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a ) 71 , b ) 72 , c ) 73 , d ) 74 , e ) 75 | e | subtract(110, divide(subtract(110, 5), const_3)) | if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 110 kilograms . what is the student ' s present weight in kilograms ? | let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 110 3 x = 105 x = 35 kg then the student ' s weight is 75 kg . the answer is e . | a = 110 - 5
b = a / 3
c = 110 - b
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a ) 315 , b ) 345 , c ) 425 , d ) 335 , e ) none of them | c | multiply(multiply(const_100.0, divide(12, 1428)), 3) | what annual installment will discharge a debt of rs . 1428 due in 3 years at 12 % simple interest ? | "let each installment be rs . x then , ( x + ( ( x * 12 * 1 ) / 100 ) ) + ( x + ( ( x * 12 * 2 ) / 100 ) ) + x = 1428 = ( ( 28 x / 25 ) + ( 31 x / 25 ) + x ) = 1428 ï ƒ › ( 28 x + 31 x + 25 x ) = ( 1428 * 25 ) x = ( 1428 * 25 ) / 84 = rs . 425 . therefore , each installment = rs . 425 . answer is c ." | a = 12 / 1428
b = 100 * 0
c = b * 3
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a ) 10 , b ) 12 , c ) 8 , d ) 12 , e ) 14 | c | divide(16, const_2) | in a group of ducks and cows , the total number of legs are 16 more than twice the no . of heads . find the total no . of buffaloes . | let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 16 = > 2 x = 16 = > x = 8 c | a = 16 / 2
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a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | d | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 4) | x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 4 . if x invested rs . 40,000 . the amount invested by y is | "solution suppose y invested rs . y then , 40000 / y = 2 / 4 â € ¹ = â € º y = ( 40000 ã — 4 / 2 ) . â € ¹ = â € º y = 80000 . answer d" | a = 1 + 4
b = a * 1000
c = b * 2
d = c / 4
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a ) 80 , b ) 55 , c ) 56 , d ) 16 , e ) 14 | b | multiply(divide(110, add(add(2, 3), 5)), 5) | the amounts of time that three secretaries worked on a special project are in the ratio of 2 to 3 to 5 . if they worked a combined total of 110 hours , how many hours did the secretary who worked the longest spend on the project ? | 10 x = 110 = > x = 11 therefore the secretary who worked the longest spent 11 x 5 = 55 hours on the project option ( b ) | a = 2 + 3
b = a + 5
c = 110 / b
d = c * 5
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a ) 98.5 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | b | divide(multiply(add(const_1, 3), 155), 7) | 3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 155 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 155 / 7 = 88.5 kgs answer : b" | a = 1 + 3
b = a * 155
c = b / 7
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a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | c | subtract(add(negate(const_2.0), multiply(subtract(18, const_1), const_2.0)), 12) | set j consists of 18 consecutive even numbers . if the smallest term in the set is - 12 , what is the range of the positive integers in set j ? | "since there are only 18 integers , another approach is the just list all 18 . we get : - 12 , - 10 , - 8 , - 6 , - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 range of positive integers = 22 - 2 = 20 answer : c" | a = negate + (
b = 18 - 1
c = b * 2
d = a - c
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a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | d | inverse(subtract(divide(const_1, 3), divide(const_1, 4))) | bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne ’ s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many s hours does it currently take anne to clean the house on her own ? | "lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 / a + 1 / b portion of the work which equals 1 / 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 / a + 2 / b which is equal to 1 / 3 ( as the work is completed in s = 3 hours ) solving these 2 equations we can find b as 12 so , d" | a = 1 / 3
b = 1 / 4
c = a - b
d = 1/(c)
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a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 640,000 | e | divide(const_3600, const_10) | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 56 years later ? | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 56 years , the investment will get doubled 56 / 14 = 4 times . so the amount invested will get doubled thrice . so , 40000 * 2 ^ 4 = 640000 hence , the answer is e . | a = 3600 / 10
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a ) $ 2.00 , b ) $ 2.25 , c ) $ 2.50 , d ) $ 2.75 , e ) $ 3.00 | e | divide(multiply(multiply(2, 5), 1.20), const_4) | having received his weekly allowance , a student spent 2 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.20 at the candy store . what is this student ’ s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 3 / 5 ) x = 120 cents ( 2 / 5 ) x = 120 x = $ 3.00 the answer is e ." | a = 2 * 5
b = a * 1
c = b / 4
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a ) 6 . , b ) 16 . , c ) 18 , d ) 21 , e ) 26 | d | add(add(15, 5), const_1) | ashley paid 5 dollars for 1 notebook and 1 pencil . if both prices were integers , how many pencils did ashley buy if she paid 93 dollars for the pencils and for 15 notebooks ? | 5 = 1 p + 1 n 93 = xp + 15 n , where x is the number of pencils possible prices for 1 p & 1 n ( integers ) : n = 1 p = 4 then 15 n = 15 , 4 x = 93 - 15 = 78 , x = 19.5 ( not an integer ) n = 2 p = 3 then 15 n = 30 , 3 x = 93 - 30 = 63 , x = 21 ( the answer we were looking for , no need to continue calculations ) n = 3 p = 2 n = 4 p = 1 answer : d | a = 15 + 5
b = a + 1
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a ) 17 / 36 , b ) 36 / 17 , c ) 17 / 6 , d ) 85 / 2 , e ) 51 / 4 | d | subtract(divide(add(multiply(const_10, 5), 2), 3), divide(add(const_10, 1), 4)) | what is 2 2 / 3 - 1 1 / 4 divided by 1 / 5 - 1 / 6 ? | "2 2 / 3 - 1 1 / 4 = 8 / 3 - 5 / 4 = ( 32 - 15 ) / 12 = 17 / 12 1 / 5 - 1 / 6 = ( 6 - 5 ) / 30 = 1 / 30 so 17 / 12 / 1 / 30 = 17 / 12 * 30 = 85 / 2 answer - d" | a = 10 * 5
b = a + 2
c = b / 3
d = 10 + 1
e = d / 4
f = c - e
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a ) 540 , b ) 288 , c ) 200 , d ) 340 , e ) 712 | a | multiply(divide(multiply(142, const_1000), const_3600), 12) | a train running at the speed of 142 km / hr crosses a pole in 12 seconds . find the length of the train . | "speed = 162 * ( 5 / 18 ) m / sec = 45 m / sec length of train ( distance ) = speed * time 45 * 12 = 540 meter answer : a" | a = 142 * 1000
b = a / 3600
c = b * 12
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a ) 10 , b ) 20 , c ) 17.5 , d ) 43 , e ) 50 | c | multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100) | in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 10 inches . find the % correction made by the candidate from his stated height to his actual height ? | "his height was = 5 feet 5 inch = 10 + 60 = 70 inch . required % correction = 70 * ( 1.25 - 1 ) = 17.5 c" | a = const_0_25 + const_0_25
b = 25 + a
c = b - 5
d = const_0_25 + const_0_25
e = 25 + d
f = e - 5
g = 100 + f
h = c / g
i = h * 100
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a ) 9 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | subtract(multiply(divide(multiply(16, divide(5, add(5, 3))), divide(3, add(5, 3))), divide(6, add(5, 3))), subtract(16, multiply(16, divide(5, add(5, 3))))) | at deluxe paint store , fuchsia paint is made by mixing 5 parts of red paint with 3 parts of blue paint . mauve paint is made by mixing 3 parts of red paint with 6 parts blue paint . how many liters of blue paint must be added to 16 liters of fuchsia to change it to mauve paint ? | "in 16 liters , red = 5 / 8 * 16 = 10 and blue = 6 so , 10 / ( 6 + x ) = 3 / 6 or , x = 14 ( answer c )" | a = 5 + 3
b = 5 / a
c = 16 * b
d = 5 + 3
e = 3 / d
f = c / e
g = 5 + 3
h = 6 / g
i = f * h
j = 5 + 3
k = 5 / j
l = 16 * k
m = 16 - l
n = i - m
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a ) 9 % , b ) 8 % , c ) 7 % , d ) 6 % , e ) 5 % | c | sqrt(98) | the difference between c . i . and s . i . on an amount of $ 20,000 for 2 years is $ 98 . what is the rate of interest per annum ? | "$ 98 is the interest on the first year of interest . let x be the interest rate . the interest after the first year is 20000 * x . the interest on the first year ' s interest is 20000 * x * x 20000 * x ^ 2 = 98 x = 0.07 the answer is c ." | a = math.sqrt(98)
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a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 3.1 , e ) 1.01 | e | divide(divide(37.3, 3), multiply(4, 3)) | in 1979 approximately 1 / 3 of the 37.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 3 the number that used kennedy airport and 4 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ? | "number of passengers using kennedy airport = 37 / 3 = ~ 12.43 passengers using miami airport = 12.43 / 3 = ~ 4.14 passengers using logan airport = 4.14 / 4 = ~ 1.01 so e" | a = 37 / 3
b = 4 * 3
c = a / b
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a ) 1865113 , b ) 1775123 , c ) 1764813 , d ) 1675123 , e ) none of them | c | multiply(4300431, power(add(const_4, const_1), const_4)) | ( 4300431 ) - ? = 2535618 | "let 4300431 - x = 2535618 then x = 4300431 - 2535618 = 1764813 answer is c" | a = 4 + 1
b = a ** 4
c = 4300431 * b
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a ) 8 , b ) 9 , c ) 10 , d ) 89 , e ) 81 | b | multiply(divide(const_1, multiply(add(const_100, 15), divide(const_1, subtract(const_100, 15)))), 18) | by selling 18 pencils for a rupee a man loses 15 % . how many for a rupee should he sell in order to gain 15 % ? | "85 % - - - 18 115 % - - - ? 85 / 115 * 18 = 13 answer : b" | a = 100 + 15
b = 100 - 15
c = 1 / b
d = a * c
e = 1 / d
f = e * 18
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a ) 1 , b ) 5 , c ) 7 , d ) 8 , e ) 2 | b | subtract(divide(add(multiply(100, 35), multiply(add(const_2, const_3), multiply(100, const_2))), const_100), 40) | a contractor undertakes to do a piece of work in 40 days . he engages 100 men at the begining and 100 more after 35 days and completes the work in stipulated time . if he had not engaged the additional men , how many days behind schedule would it be finished ? | explanation : \ inline \ fn _ jvn [ ( 100 \ times 35 ) + ( 200 \ times 5 ) ] men can finish the work in 1 day \ inline \ fn _ jvn \ therefore 4500 men can finish the work in 1 day . 100 men can finish it in \ inline \ fn _ jvn \ frac { 4500 } { 100 } = 45 days . this is 5 days behind schedule answer : b | a = 100 * 35
b = 2 + 3
c = 100 * 2
d = b * c
e = a + d
f = e / 100
g = f - 40
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(subtract(divide(88, 9), 4), const_1) | how many integers are between 4 and 88 / 9 , inclusive ? | "88 / 9 = 9 . xx we are not concerned about the exact value of 88 / 9 as we just need the integers . the different integers between 4 and 88 / 9 would be 4 , 5 , 6 , 7 , 8 , 9 total number of integers = 6 option d" | a = 88 / 9
b = a - 4
c = b + 1
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) greater than 10 | d | divide(100, const_10) | how many integers from 1 to 100 ( both inclusive ) have odd number of factors ? | all perfect squares have odd # of factors . squaring integers 1 - 10 produces 1 , 4 , 9 , 16 , . . . 100 which all have odd # of factors . answer is d | a = 100 / 10
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a ) 131 , b ) 135 , c ) 139 , d ) 147 , e ) 188 | e | add(multiply(divide(subtract(278, 8), const_3), const_2), 8) | if jake loses 8 pounds , he will weigh twice as much as his sister . together they now weigh 278 pounds . what is jake ’ s present weight , in pounds ? | "consider sister weight now as x jakes weight ( 2 times + 8 ) = 2 x + 8 ( 2 x + 8 ) + ( x ) = 278 x = 90 jake ' s weight is . . . . 188 correct answer is e" | a = 278 - 8
b = a / 3
c = b * 2
d = c + 8
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a ) 12100 , b ) 15240 , c ) 12456 , d ) 11452 , e ) 14520 | e | multiply(multiply(divide(add(10, const_100), const_100), 12000), divide(add(10, const_100), const_100)) | if the annual increase in the population of a town is 10 % and the present number of people is 12000 , what will the population be in 2 years ? | "the required population is = 12000 ( 1 + 10 / 100 ) ^ 2 = 12000 * 11 / 10 * 11 / 10 = 14520 answer is e" | a = 10 + 100
b = a / 100
c = b * 12000
d = 10 + 100
e = d / 100
f = c * e
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a ) 24,602 , b ) 26,200 , c ) 24,600 , d ) 24,628 , e ) 24,6012 | b | divide(65.50, divide(const_4, 4)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 65.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 65.50 32 x - 31 x = 6550 * 4 x = 26,200 . answer : b" | a = 4 / 4
b = 65 / 50
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a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | d | subtract(divide(19, subtract(3, 2)), 2) | a monkey start climbing up a tree 19 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | "if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 16 hrs it wil be 16 ft . bt since the height of the tree is 19 ft , so if the monkey hops up the tree in the next hr i . e 17 th hr then it reaches at the top of the tree . hence it takes 17 hrs for monkey to reach at the top answer : d" | a = 3 - 2
b = 19 / a
c = b - 2
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a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 72 | e | divide(multiply(power(12, 2), power(6, 3)), 432) | find the value of y from ( 12 ) ^ 2 x 6 ^ 3 ÷ 432 = y ? | "72 e" | a = 12 ** 2
b = 6 ** 3
c = a * b
d = c / 432
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a ) 1 / 3 , b ) 1 / 30 , c ) 1 / 10 , d ) 1 / 78 , e ) 30 | d | divide(const_1, multiply(divide(6.5, 5), 60)) | before leaving home for the town of madison , pete checks a map which shows that madison is 5 inches from his current location , gardensquare . pete arrives in madison 6.5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ? | "pete covered 6.5 * 60 = 390 miles which correspond to 5 inches on the map - - > scale in inches per mile is 5 / 390 = 1 / 78 . answer : d ." | a = 6 / 5
b = a * 60
c = 1 / b
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a ) 64 , b ) 72 , c ) 80 , d ) 88 , e ) 96 | d | add(add(multiply(12, 4), multiply(const_2, multiply(4, add(1, divide(25, const_100))))), multiply(const_2, multiply(12, add(1, divide(25, const_100))))) | a cistern 12 meters long and 4 meters wide contains water up to a depth of 1 meter 25 cm . what is the total area of the wet surface ? | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 12 x 1.25 ) + 12 x 4 ] = 88 the answer is d ." | a = 12 * 4
b = 25 / 100
c = 1 + b
d = 4 * c
e = 2 * d
f = a + e
g = 25 / 100
h = 1 + g
i = 12 * h
j = 2 * i
k = f + j
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a ) 1 / 12 , b ) 1 / 9 , c ) 1 / 8 , d ) 1 1 / 9 , e ) 2 1 / 9 | c | subtract(add(divide(2, 9), divide(1, 8)), divide(4, 18)) | the instructions state that cheryl needs 2 / 9 square yards of one type of material and 1 / 8 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 4 / 18 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | total bought = 2 / 9 + 1 / 8 left part 4 / 18 - - - > 2 / 9 so used part 2 / 9 + 1 / 8 - 2 / 9 = 1 / 8 answer : c | a = 2 / 9
b = 1 / 8
c = a + b
d = 4 / 18
e = c - d
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a ) 62 , b ) 66 , c ) 18 , d ) 75 , e ) 12 | d | subtract(multiply(17, 17), add(multiply(5, 14), multiply(9, 16))) | the average age of 17 students of a class is 17 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 17 th student is | "explanation : age of the 17 th student = [ 17 * 17 - ( 14 * 5 + 16 * 9 ) ] = ( 289 - 214 ) = 75 years . answer : d" | a = 17 * 17
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
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['a ) 1', 'b ) 3 / 2', 'c ) 2', 'd ) 3', 'e ) 6'] | d | divide(volume_rectangular_prism(sqrt(2), sqrt(2), 3), const_2) | right triangle abc is the base of the prism in the figure above . if ab = ac = √ 2 and the height of the prism is 3 , what is the volume of the prism ? | volume of prism = area of base * height = 1 / 2 * ( square root of 2 ) * ( square root of 2 ) * 3 = 3 answer : d | a = math.sqrt(2)
b = math.sqrt(2)
c = volume_rectangular_prism / (
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a ) 1 , b ) 40 , c ) 20 , d ) 26 , e ) 30 | c | multiply(divide(20, 20), 20) | in a dairy farm , 20 cows eat 20 bags of husk in 20 days . in how many days one cow will eat one bag of husk ? | "explanation : one bag of husk = 20 cows per day ⇒ 20 × 1 × 20 = 1 × 20 × x for one cow = 20 days answer : c" | a = 20 / 20
b = a * 20
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a ) 5 , b ) 19 , c ) 29 , d ) 30 , e ) 33 | a | subtract(subtract(subtract(multiply(68, 4), 90), subtract(90, const_1)), subtract(90, const_2)) | the average ( arithmetic mean ) of 4 different integers is 68 . if the largest integer is 90 , what is the least possible value of the smallest integer ? | "total of integers = 68 * 4 = 272 lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer . but all integers are distinct . so if the largest integer is 90 , then the middle 2 will be 88 and 89 lowest of least possible integer = 272 - ( 90 + 89 + 88 ) = 272 - 267 = 5 answer : a" | a = 68 * 4
b = a - 90
c = 90 - 1
d = b - c
e = 90 - 2
f = d - e
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a ) 482 , b ) 725 , c ) 992 , d ) 1446 , e ) 923 | d | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 60000), multiply(power(add(divide(20, const_100), const_1), const_2), 60000)) | how much more would rs . 60000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | "60000 ( 11 / 10 ) 4 - 60000 ( 6 / 5 ) 2 = 1446 answer : d" | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 4
e = d * 60000
f = 20 / 100
g = f + 1
h = g ** 2
i = h * 60000
j = e - i
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a ) 16 % , b ) 17 % , c ) 40 % , d ) 58 % , e ) 13 % | c | multiply(divide(subtract(80000, add(45000, 12000)), add(45000, 12000)), const_100) | ramu bought an old car for rs . 45000 . he spent rs . 12000 on repairs and sold it for rs . 80000 . what is his profit percent ? | "total cp = rs . 45000 + rs . 12000 = rs . 57000 and sp = rs . 80000 profit ( % ) = ( 80000 - 57000 ) / 57000 * 100 = 40 % answer : c" | a = 45000 + 12000
b = 80000 - a
c = 45000 + 12000
d = b / c
e = d * 100
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 6 | e | floor(subtract(divide(300, 40), divide(50, 10))) | subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 15 pages . | "subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 15 / 2.5 ) = 6 hrs answer : e" | a = 300 / 40
b = 50 / 10
c = a - b
d = math.floor(c)
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a ) 8 , b ) 5 , c ) 9 , d ) 6 , e ) 13 | e | divide(subtract(198, multiply(3, 40)), multiply(3, const_2)) | a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 198 , how many hours overtime did he work ? | "at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 198 - $ 120 = $ 78 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 78 / $ 6 = 13 ans is e" | a = 3 * 40
b = 198 - a
c = 3 * 2
d = b / c
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a ) 4 : 30 , b ) 5 : 00 , c ) 5 : 30 , d ) 6 : 00 , e ) 6 : 30 | d | divide(add(70, multiply(70, divide(const_1, const_2))), subtract(84, 70)) | a train sets off at 9 : 00 am at the speed of 70 km / h . another train starts at 10 : 30 am in the same direction at the rate of 84 km / h . at what time will the second train catch the first train ? | "in one hour and thirty minutes the first train travels 105 km . the second train catches the first train at a rate of 84 km / h - 70 km / h = 14 km / h . the second train will catch the first train in 105 / 14 = 7.5 hours , so at 6 : 00 pm . the answer is d ." | a = 1 / 2
b = 70 * a
c = 70 + b
d = 84 - 70
e = c / d
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a ) 1 : 7 , b ) 1 : 5 , c ) 1 : 4 , d ) 1 : 2 , e ) 2 : 5 | c | divide(const_1, const_4) | a grocery store bought some apples at a rate of 5 for a dollar . they were separated into two stacks , one of which was sold at a rate of 3 for a dollar and the other at a rate of 6 for a dollar . what was the ratio of the number of apples in the two stacks if the store broke even after having sold all of its apples ? | to avoid fractions , assume there were 45 apples . so the store bought them for 45 / 5 = 9 dollars . to break even , the selling price should be $ 9 too . 45 mangoes can be split into 1 : 4 , 2 : 3 or 1 : 2 . so let ' s try these . 45 split in the ratio 1 : 4 gives 9 and 36 . 9 apples split into 3 apples each will give $ 3 . 36 apples split into 6 apples each will give $ 6 . they add up to $ 9 so we have hit the right answer . answer ( c ) | a = 1 / 4
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a ) 1 minutes , b ) 11 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(11, 11), 11) | if 11 lions can kill 11 deers in 11 minutes how long will it take 100 lions to kill 100 deers ? | "we can try the logic of time and work , our work is to kill the deers so 11 ( lions ) * 11 ( min ) / 11 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 11 answer : b" | a = 11 / 11
b = a * 11
|
a ) 19 : 2 , b ) 18 : 3 , c ) 29 : 11 , d ) 29 : 5 , e ) 29 : 4 | c | divide(add(divide(multiply(62.5, 6), const_100), divide(multiply(87.5, 4), const_100)), add(subtract(6, divide(multiply(62.5, 6), const_100)), subtract(4, divide(multiply(87.5, 4), const_100)))) | two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 6 litres from vessel p is mixed with 4 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 6 = 15 / 4 litres quantity of alcohol in vessel q = 87.5 / 100 * 4 = 7 / 2 litres quantity of alcohol in the mixture formed = 15 / 4 + 7 / 2 = 29 / 4 = 7.25 litres as 10 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 7.25 : 2.75 = 29 : 11 . answer : c" | a = 62 * 5
b = a / 100
c = 87 * 5
d = c / 100
e = b + d
f = 62 * 5
g = f / 100
h = 6 - g
i = 87 * 5
j = i / 100
k = 4 - j
l = h + k
m = e / l
|
a ) 123.56 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356 | a | divide(multiply(1, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 1 percent of 12,356 ? | since , percent = 1 / 100 , what = something ( s ) , and is : = . we can write the question as s = 1 ( 1 / 100 ) 12,356 . the answer is 123.56 . hence , the correct answer is a . | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 1 * n
p = o / 100
|
a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | b | divide(power(2, 2), multiply(choose(4, 2), choose(2, 1))) | if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 } , what is the probability that xy will be even ? | probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 1 / 3 . answer : b . | a = 2 ** 2
b = math.comb(4, 2)
c = math.comb(2, 1)
d = b * c
e = a / d
|
a ) 48.3 % , b ) 52.8 % , c ) 55.5 % , d ) 58.3 % , e ) 61.4 % | d | multiply(divide(70, multiply(const_100, const_2)), const_100) | sue ' s trail mix is 30 % nuts and 70 % dried fruit . jane ' s trail mix is 60 % nuts and 40 % chocolate chips . if the combined mixture of sue and jane ' s trails mix contains 35 % nuts , what percent of the combined mixture is dried fruit ? | "35 % is 5 % - points above 30 % and 25 % - points below 60 % . so the ratio of sue ' s mix to jane ' s mix is 5 : 1 . 5 / 6 * 70 % = 58.3 % the answer is d ." | a = 100 * 2
b = 70 / a
c = b * 100
|
a ) 3 / 16,000 , b ) 1 / 8,100 , c ) 7 / 48,000 , d ) 1 / 90 , e ) 2 / 45 | c | multiply(divide(70, 800), divide(const_1, 600)) | a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 70 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ? | "solution provided by stanford 2012 is correct : 70 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 70 / 800 * 1 / 600 = 7 / 48,000 . answer : c ." | a = 70 / 800
b = 1 / 600
c = a * b
|
a ) 9 , b ) 5 , c ) 7 , d ) 6 , e ) 10 | e | multiply(divide(24, const_60), add(20, 5)) | the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 24 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 24 / 60 ) ) km = 10 km . answer : e" | a = 24 / const_60
b = 20 + 5
c = a * b
|
a ) 1 / 12 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 2 1 / 9 | c | subtract(add(divide(4, 9), divide(2, 3)), divide(8, 18)) | the instructions state that cheryl needs 4 / 9 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 8 / 18 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 4 / 9 + 2 / 3 left part 8 / 18 - - - > 4 / 9 so used part 4 / 9 + 2 / 3 - 4 / 9 = 2 / 3 answer : c" | a = 4 / 9
b = 2 / 3
c = a + b
d = 8 / 18
e = c - d
|
a ) 5 % , b ) 4 % , c ) - 4 % , d ) - 12 % , e ) 2 % | b | multiply(subtract(multiply(add(const_1, divide(30, const_100)), subtract(const_1, divide(20, const_100))), const_1), const_100) | a sells a bicycle to b at a profit of 30 % and b sells it to c at a loss of 20 % . find the resultant profit or loss . | the resultant profit or loss = 30 - 20 - ( 30 * 20 ) / 100 = 4 % profit = 4 % answer is b | a = 30 / 100
b = 1 + a
c = 20 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 5.33 km , b ) 6.33 km , c ) 5.63 km , d ) 7.33 km , e ) 5.93 km | a | divide(multiply(subtract(6, 2), const_3), 6) | a man can row at 6 kmph in still water . if the velocity of current is 2 kmph and it takes him 2 hour to row to a place and come back , how far is the place ? | "speed in still water = 6 kmph speed of the current = 2 kmph speed downstream = ( 6 + 2 ) = 8 kmph speed upstream = ( 6 - 2 ) = 4 kmph let the required distance be x km total time taken = 2 hour x / 8 + x / 4 = 2 x + 2 x / 8 = 2 3 x / 8 = 2 3 x = 16 x = 5.33 km answer : a" | a = 6 - 2
b = a * 3
c = b / 6
|
a ) 38 / 67 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 41 / 91 | a | divide(subtract(335, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 335) | in a survey of 335 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ? | "- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 281 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 335 we have to find not part time and not uninsured . in other words not part time and insured = x / 335 = ( 281 - 104 + 13 ) / 335 = 38 / 67 answer is a ." | a = 12 * 5
b = 1 * 100
c = a / b
d = 54 - c
e = d + 104
f = 335 - e
g = f / 335
|
a ) 9 % , b ) 10 % , c ) 10 5 / 8 % , d ) 11 % , e ) 12 % | a | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(8.5, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | a $ 500 investment and a $ 1,500 investment have a combined yearly return of 8.5 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1 , 500 investment have ? | "the ratio of investments is 500 : 1,500 = 1 : 3 . the deviation from the average return from $ 500 investment and $ 1,500 investment must be in the ration 3 : 1 . $ 500 investment return has the deviation from the mean of 8.5 - 7 = 1.5 % , thus $ 1,500 investment return must have the deviation from the mean equal to 0.5 % , which means that $ 1 , 500 investment has 8.5 + 0.5 = 9 % yearly return . answer : a ." | a = 3 * 500
b = 500 + a
c = 8 / 5
d = b * c
e = 7 / 100
f = 500 * e
g = d - f
h = 3 * 500
i = g / h
j = i * 100
|
a ) 10 , b ) 8 , c ) 20 , d ) 25 , e ) 30 | b | multiply(4, divide(8, 4)) | 8 : 4 seconds : : ? : 4 minutes | "8 * 4 = 4 * x x = 8 answer : b" | a = 8 / 4
b = 4 * a
|
a ) 4500 m 3 , b ) 4580 m 3 , c ) 10500 m 3 , d ) 4900 m 3 , e ) 4700 m 3 | c | divide(multiply(multiply(2, 45), multiply(7, const_1000)), multiply(const_1, const_60)) | a river 2 m deep and 45 m wide is flowing at the rate of 7 kmph the amount of water that runs into the sea per minute is ? | "explanation : ( 7000 * 2 * 45 ) / 60 = 10500 m 3 answer : option c" | a = 2 * 45
b = 7 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 83 % , b ) 80 % , c ) 20 % , d ) 17 % , e ) 12 % | c | subtract(25, const_1) | at a local appliance manufacturing facility , the workers received a 25 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | "let ' s say he works usually 10 hours and earns 100 per hour . 10 * 100 = 1000 10 * 125 = 1250 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 125 = 8 so he can reduce his work by 2 hours . which is 20 % . answer c" | a = 25 - 1
|
a ) - 4 , b ) - 1 / 4 , c ) 0 , d ) 1 / 4 , e ) 4 | a | divide(const_1, 4) | if 625 ^ ( - x ) + 5 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 15 , what is the value of x ? | "we ' re told that 625 ^ ( - x ) + 25 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 15 . we ' re asked for the value of x . since each of the calculated terms must be positive ( regardless of what the exponent is ) , we can use thebasesto our advantage . . . . . with answer a , we ' d have 625 ^ 4 , which is much bigger than 15 ( and we ' d be adding to that big number ) . eliminate a . with answer e , we ' d have 625 ^ ( - 4 ) , which would create a tiny fraction ( and we ' d add some other fractions to it , so the total would be much too small ) . eliminate e . with answer d , we ' d have 625 ^ ( - 1 / 4 ) , which will also be a fraction ( just not as tiny as the one in answer e ) , but the total would still be too small . eliminate d . with answer c , anything to the ' 0 power ' is 1 , so we ' d have 1 + 1 + 1 = 3 . this is not 15 . eliminate c . a" | a = 1 / 4
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a ) s . 4529 , b ) s . 4586 , c ) s . 4597 , d ) s . 4530 , e ) s . 5436 | e | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 6) | the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 6 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | "area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 6 = rs . 5436 answer : e" | a = 2 * 12
b = 15 + 25
c = a * b
d = 6 * 3
e = c - d
f = 4 * 3
g = 3 * f
h = e - g
i = h * 6
|
a ) 15 % , b ) 18 % , c ) 20 % , d ) 21 % , e ) 22 % | c | multiply(divide(add(multiply(40, divide(10, const_100)), multiply(80, divide(25, const_100))), add(40, 80)), const_100) | a car dealership has 40 cars on the lot , 10 % of which are silver . if the dealership receives a new shipment of 80 cars , 25 % of which are not silver , what percentage of total number of cars are silver ? | the number of silver cars is 0.1 * 40 + 0.25 * 80 = 24 the percentage of cars which are silver is 24 / 120 = 20 % the answer is c . | a = 10 / 100
b = 40 * a
c = 25 / 100
d = 80 * c
e = b + d
f = 40 + 80
g = e / f
h = g * 100
|
a ) 3.69 , b ) 3.66 , c ) 3.6 , d ) 150 , e ) 3.61 | d | multiply(divide(subtract(333, multiply(subtract(33, 30), divide(subtract(366, 333), subtract(36, 33)))), 30), 15) | apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 333 and for 36 kgs of apples is 366 then the cost of first 15 kgs of apples is | "ans : by framing equations we get 30 l + 3 q = 333 30 l + 6 q = 366 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first 60 l + 6 q = 666 30 l + 6 q = 366 30 l = 300 = > l = 10 then we get l = 10 cost of 15 kgs of apples = 15 x 10 = 150 answer : d" | a = 33 - 30
b = 366 - 333
c = 36 - 33
d = b / c
e = a * d
f = 333 - e
g = f / 30
h = g * 15
|
a ) 4 : 6 , b ) 1 : 2 , c ) 1 : 3 , d ) 1 : 4 , e ) 2 : 3 | a | divide(sqrt(16), sqrt(36)) | the sub - duplicate ratio of 16 : 36 is | "root ( 16 ) : root ( 36 ) = 4 : 6 answer : a" | a = math.sqrt(16)
b = math.sqrt(36)
c = a / b
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 50 , d ) d ) 60 , e ) e ) 100 | e | multiply(factorial(2), factorial(4)) | there are 2 red chips and 4 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | "using anagram method : 6 _ 5 _ 4 _ 3 _ 2 _ 1 r _ r _ b _ b _ b _ b so . . 6 ! / number of repeated letters ( 2 ! ) ( 4 ! ) = 10 ans : e" | a = math.factorial(2)
b = math.factorial(4)
c = a * b
|
a ) 139.78 , b ) 139.13 , c ) 139.22 , d ) 111.0 , e ) 111.12 | a | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 70))) | a can give b 70 meters start and c 200 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "a runs 1000 m while b runs 930 m and c runs 800 m . the number of meters that c runs when b runs 1000 m , = ( 1000 * 800 ) / 930 = 860.21 m . b can give c = 1000 - 860.21 = 139.78 m . answer : a" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 200
e = b * d
f = 100 * 10
g = f - 70
h = e / g
i = a - h
|
a ) 36 mph , b ) 40 mph , c ) 44 mph , d ) 52 mph , e ) 58 mph | c | divide(add(40, 180), add(divide(40, 20), divide(180, 60))) | a car drives 40 miles on local roads at 20 mph , and 180 miles on the highway at 60 mph , what is the average speed of the entire trip ? | "explanations 1 ) in phase # 1 of the trip , the car traveled 40 mi at 20 mph . that time of this phase was : time = distance / rate = ( 40 mi ) / ( 20 mph ) = 2 hr in phase # 2 of the trip , the car traveled 180 mi at 60 mph . that time of this phase was : time = distance / rate = ( 180 mi ) / ( 60 mph ) = 3 hr the total distance of the trip = 40 mi + 180 mi = 220 mi the total time of the trip = 2 hr + 3 hr = 5 hr the average speed of trip is given by speed = distance / time = ( 220 mi ) / ( 5 hr ) = 44 mph answer : c ." | a = 40 + 180
b = 40 / 20
c = 180 / 60
d = b + c
e = a / d
|
a ) rs . 600 , b ) rs . 800 , c ) rs . 500 , d ) rs . 900 , e ) rs . 970 | b | divide(16800, add(18, 3)) | a man sold 18 toys for rs . 16800 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 16800 . gain = sp – cp 3 x = 16800 – 18 x 21 x = 16800 x = rs . 800 . answer : b" | a = 18 + 3
b = 16800 / a
|
a ) 6.5 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | a | multiply(multiply(10, 4), divide(1, 4)) | in the coordinate plane , points ( x , 1 ) and ( 10 , y ) are on line k . if line k passes through the origin and has slope 1 / 4 , then x + y = | "line k passes through the origin and has slope 1 / 4 means that its equation is y = 1 / 4 * x . thus : ( x , 1 ) = ( 4 , 1 ) and ( 10 , y ) = ( 10 , 2.5 ) - - > x + y = 4 + 2.5 = 6.5 answer : a ." | a = 10 * 4
b = 1 / 4
c = a * b
|
a ) 25 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | a | multiply(divide(subtract(multiply(divide(2, 15), 120), 20), multiply(divide(2, 15), 120)), const_100) | a doctor prescribed 20 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage per 15 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 15 ) = 2 * 8 = 16 c . c dosage prescribed by doctor for 120 pound patient = 20 c . c % prescribed dosage greater than the typical dosage = ( 20 - 16 / 16 ) * 100 % = ( 2 / 16 ) * 100 % = 25 % answer a" | a = 2 / 15
b = a * 120
c = b - 20
d = 2 / 15
e = d * 120
f = c / e
g = f * 100
|
a ) 3 : 4 , b ) 3 : 5 , c ) 7 : 8 , d ) 4 : 3 , e ) none of these | c | divide(add(4, add(4, divide(multiply(6, 4), subtract(7, 6)))), add(4, divide(multiply(6, 4), subtract(7, 6)))) | the ratio between the present age of p and q is 6 : 7 . if q is 4 years old then p , what will be the ratio of the ages of p and q after 4 year ? | "solution let p ' s age and q ' s age be 6 x years and 7 x years respectively . then , 7 x - 6 x = 4 ⇔ x = 4 . ∴ required ratio = ( 6 x + 4 ) : ( 7 x + 4 ) = 28 : 32 = 7 : 8 . answer c" | a = 6 * 4
b = 7 - 6
c = a / b
d = 4 + c
e = 4 + d
f = 6 * 4
g = 7 - 6
h = f / g
i = 4 + h
j = e / i
|
a ) $ 72 , b ) $ 85 , c ) $ 99 , d ) $ 101 , e ) $ 202 | c | divide(add(add(28, 72), 98), const_2) | 3 people have $ 28 , $ 72 , and $ 98 , respectively . if they pool their money then redistribute it among them , what is the maximum value for the median amount of money ? | solution - total money distributed is $ 198 . in order to maximize median , one person has to accept $ 0 and remaining two people share $ 99 each . hence median is $ 99 . ans c | a = 28 + 72
b = a + 98
c = b / 2
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a ) 52 / 45 , b ) 23 / 47 , c ) 60 / 41 , d ) 51 / 42 , e ) 41 / 60 | e | subtract(divide(subtract(11, const_1), add(11, const_1)), divide(add(2, const_1), subtract(21, const_1))) | if a is an integer greater than 2 but less than 11 and b is an integer greater than 11 but less than 21 , what is the range of a / b ? | "the way to approach this problem is 2 < a < 11 and 11 < b < 21 minimum possible value of a is 3 and maximum is 10 minimum possible value of b is 12 and maximum is 20 range = max a / min b - min a / max b ( highest - lowest ) 10 / 12 - 3 / 20 = 41 / 60 hence e" | a = 11 - 1
b = 11 + 1
c = a / b
d = 2 + 1
e = 21 - 1
f = d / e
g = c - f
|
a ) 131.58 , b ) 62.0 , c ) 62.6 , d ) 62.1 , e ) 62.2 | a | divide(multiply(add(100, divide(multiply(100, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 100 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 100 sp = 100 * ( 125 / 100 ) = 125 mp * ( 95 / 100 ) = 125 mp = 131.58 answer : a" | a = 100 * 25
b = a / 100
c = 100 + b
d = c * 100
e = 100 - 5
f = d / e
|
a ) 9.8 sec , b ) 12.1 sec , c ) 12.42 sec , d ) 14.3 sec , e ) 24.3 sec | b | divide(add(110, 132), multiply(72, const_0_2778)) | how long does a train 110 m long running at a speed of 72 kmph take to cross a bridge 132 m in length ? | "72 km / hr = 72 * 5 / 18 m / sec = 20 m / sec time = d / s = 110 + 132 / 20 = 242 / 20 = 12.1 sec answer : b" | a = 110 + 132
b = 72 * const_0_2778
c = a / b
|
a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 4 sec , e ) 2 sec | b | multiply(const_3600, divide(divide(110, const_1000), add(60, 6))) | a train 110 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 110 * 3 / 55 ] sec = 6 sec answer : b" | a = 110 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 3 , b ) 5 , c ) 6 , d ) 67 , e ) 7 | a | divide(6, subtract(4, 2)) | a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 6 km ? | "m = 4 s = 2 us = 4 - 2 = 2 d = 6 t = 6 / 2 = 3 answer : a" | a = 4 - 2
b = 6 / a
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a ) 280000 , b ) 300000 , c ) 355000 , d ) 356000 , e ) 357000 | e | multiply(multiply(subtract(const_100, 15), 75), divide(divide(560000, const_100), const_100)) | in an election , candidate x got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate ? | total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate x = 75 % therefore , the number of valid votes polled in favour of candidate x = 75 % of 476000 = 75 / 100 × 476000 = 35700000 / 100 = 357000 e ) | a = 100 - 15
b = a * 75
c = 560000 / 100
d = c / 100
e = b * d
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | add(15, const_1) | the average of first 15 natural numbers is ? | "sum of 15 natural no . = 240 / 2 = 120 average = 120 / 15 = 8 answer : b" | a = 15 + 1
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a ) 77 , b ) 76 , c ) 28 , d ) 66 , e ) 11 | a | multiply(divide(35, subtract(6, const_1)), subtract(12, const_1)) | at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 35 seconds . how long does it tick at 12 ′ o clock | "explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 35 / 5 = 7 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 7 = 77 secs . answer : a" | a = 6 - 1
b = 35 / a
c = 12 - 1
d = b * c
|
a ) 122456 , b ) 220020 , c ) 220022 , d ) 235467 , e ) 220026 | b | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 20) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 20 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 20 = 220000 + 20 = 220020 b" | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 20
|
a ) 300 bc , b ) 300 b / c , c ) 600 bc , d ) 600 b / c , e ) 600 / bc | c | multiply(2, 300) | a case contains c cartons . each carton contains b boxes , and each box contains 300 paper clips . how many paper clips are contained in 2 cases ? | "2 cases * c cartons / case * b boxes / carton * 300 clips / box = 600 bc paper clips the answer is c ." | a = 2 * 300
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a ) 4 , b ) 5 , c ) 5.6 , d ) 9.2 , e ) 3.1 | c | divide(add(21, 4), add(2, 2)) | solve the equation for x : 2 x - 21 + 3 x = 4 + 9 - x | "c 5.6 5 x + x = 13 + 21 6 x = 34 = > x = 5.6" | a = 21 + 4
b = 2 + 2
c = a / b
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a ) 4,514 , b ) 4,477 , c ) 4,521 , d ) 4,428 , e ) 4,349 | b | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 15 nor 20 ? | "integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 20 5000 / 20 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 20 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 20 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4477 answer b ." | a = 4 + 15
b = a - 1
c = math.factorial(b)
d = math.factorial(15)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 13.6 % , b ) 28.6 % , c ) 35.6 % , d ) 42.3 % , e ) none | b | multiply(subtract(divide(subtract(const_100, 10), 70), const_1), const_100) | the cost price of an article is 70 % of the marked price . calculate the gain percent after allowing a discount of 10 % . | "sol . let marked price = rs . 100 . then , c . p . = rs . 70 . s . p = rs . 90 . â ˆ ´ gain % = [ 20 / 70 * 100 ] % = 28.6 % . answer b" | a = 100 - 10
b = a / 70
c = b - 1
d = c * 100
|
a ) 215 / 216 , b ) 216 , c ) 1 / 216 , d ) none , e ) 2 | a | subtract(const_1, multiply(multiply(divide(const_1, const_6), divide(const_1, const_6)), divide(add(4, const_1), const_6))) | lola rolls a die 3 times . what is the probability that she gets a 4 on the last one of the 3 rolls ? | the die has 6 sides and it was rolle 3 times , then 6 * 6 * 6 = 216 the number of ways to get 4 is 1 then 1 / 216 the probability would be 1 - 1 / 216 = 215 / 216 answer a | a = 1 / 6
b = 1 / 6
c = a * b
d = 4 + 1
e = d / 6
f = c * e
g = 1 - f
|
a ) 90 , b ) 72 , c ) 82 , d ) 54 , e ) 65 | c | subtract(add(48, 30), divide(48, divide(30, const_100))) | the contents of a certain box consist of 48 apples and 30 kiwis . how many kiwis must be added to the box so that exactly 30 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + kiwi + x ) * 0.3 48 = ( 30 + 48 + x ) * 0.3 x = 82 answer : c" | a = 48 + 30
b = 30 / 100
c = 48 / b
d = a - c
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(subtract(subtract(20, 5), 3), 7) | a group of people participate in some curriculum , 35 of them practice yoga , 20 study cooking , 15 study weaving , 7 of them study cooking only , 5 of them study both the cooking and yoga , 3 of them participate all curriculums . how many people study both cooking and weaving ? | "both cooking and weaving = 20 - ( 7 + 5 + 3 ) = 5 so , the correct answer is e ." | a = 20 - 5
b = a - 3
c = b - 7
|
a ) 10 days , b ) 12 days , c ) 6 days , d ) 20 days , e ) 8 days | d | multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 2))), 3) | a can finish a work in 48 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 / 48 work is done by a in 1 day 5 / 12 work is done by a in 48 * 5 / 12 = 20 days answer is d" | a = 1 / 9
b = 1 / 2
c = a + b
d = 1 / c
e = d * 3
|
a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | d | multiply(15, divide(16, subtract(11, 7))) | in a certain animal shelter , the ratio of the number of catsto the number of dogs is 15 to 7 . if 16 additional dogs were to be taken in by the shelter , the ratio of the number of cats to the number of dogs would be 15 to 11 . how many cats are in the shelter ? | "this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of cats to the number of dogs is 15 : 7 . we ' re then told that 16 more dogs are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of cats . algebraically , since the number of cats is a multiple of 15 and the number of dogs is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 16 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 16 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 16 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 16 ) ( 1 ) 11 x = 7 x + 16 4 x = 16 x = 4 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 4 ) = 60 final answer : d" | a = 11 - 7
b = 16 / a
c = 15 * b
|
a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | e | subtract(divide(add(add(add(2, 4), 7), 32), add(4, const_1)), 4) | for the positive integers x , x + 2 , x + 4 , x + 7 , and x + 32 , the mean is how much greater than the median ? | "mean = ( x + x + 2 + x + 4 + x + 7 + x + 32 ) / 5 = ( 5 x + 45 ) / 5 = x + 9 median = x + 4 thus mean - median = x + 9 - ( x + 4 ) = 5 answer = e" | a = 2 + 4
b = a + 7
c = b + 32
d = 4 + 1
e = c / d
f = e - 4
|
['a ) 18 : 21', 'b ) 81 : 13', 'c ) 8 : 15', 'd ) 16 : 21', 'e ) 1 : 13'] | d | power(divide(27, 64), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 27 : 64 . what is the ratio of their total surface areas ? | ratio of the sides = ³ √ 27 : ³ √ 64 = 3 : 4 ratio of surface areas = 32 : 42 = 16 : 21 answer : d | a = 27 / 64
b = 1 / 3
c = a ** b
|
a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | divide(multiply(subtract(1, divide(1, 3)), 18), subtract(1, subtract(1, divide(1, 3)))) | a driver would have reduced the time it took to drive from home to the store by 1 / 3 if the average speed had been increased by 18 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | "let r be the original speed and let t be the original time . since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 3 would give us : d = ( r ) ( t ) = ( 2 t / 3 ) ( x * r ) x = 3 / 2 since ( 2 t / 3 ) ( 3 r / 2 ) = ( r ) ( t ) = d 3 r / 2 = r + 18 r / 2 = 18 r = 36 the answer is a ." | a = 1 / 3
b = 1 - a
c = b * 18
d = 1 / 3
e = 1 - d
f = 1 - e
g = c / f
|
a ) 1 / 13 , b ) 1 / 20 , c ) 1 / 26 , d ) 1 / 50 , e ) 1 / 100 | c | divide(divide(factorial(subtract(multiply(const_3, const_4), const_1)), power(factorial(const_2), const_3)), divide(factorial(add(subtract(multiply(const_3, const_4), const_1), const_2)), multiply(factorial(const_3), power(factorial(const_2), const_3)))) | what is the probability q of randomly selecting an arrangement of the letters of the wordmediterraneanin which the first letter is e and the last letter is r ? | why do you calculate the total combinations and everything , it is much easier to calculate as follows : 3 / 13 * 1 * 1 * 1 * 1 . . . . . * 2 / 12 = 6 / 156 = 1 / 26 this is because 3 / 13 = probability that first letter will be e and 2 / 12 = probability that thefirstletter will be r . between everything else cuts down to 1 * 1 * 1 . . . e . g . probability q of the second letter is 11 / 11 = 1 , > we do not care what ' s in between . c | a = 3 * 4
b = a - 1
c = math.factorial(b)
d = math.factorial(2)
e = d ** 3
f = c / e
g = 3 * 4
h = g - 1
i = h + 2
j = math.factorial(i)
k = math.factorial(3)
l = math.factorial(2)
m = l ** 3
n = k * m
o = j / n
p = f / o
|
a ) 15 , b ) 66 , c ) 77 , d ) 30 , e ) 42 | d | multiply(divide(20, const_60), 90) | the speed of a train is 90 kmph . what is the distance covered by it in 20 minutes ? | "90 * 20 / 60 = 30 kmph answer : d" | a = 20 / const_60
b = a * 90
|
a ) 8.5 , b ) 10.0 , c ) 12.5 , d ) 11.0 , e ) 8.25 | d | divide(subtract(45, 23), const_2) | a man can row downstream at the rate of 45 kmph and upstream at 23 kmph . find the man ’ s rate in still water and rate of current ? | rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 45 + 23 ) = 34 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 45 - 23 ) = 1 / 2 ( 22 ) = 11 kmph answer is d . | a = 45 - 23
b = a / 2
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | b | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(9, 11)), const_1), 11)), 11), const_1) | how many numbers from 9 to 79 are exactly divisible by 11 ? | "9 / 11 = 0 and 79 / 11 = 7 = = > 7 - 0 = 7 numbers answer : b" | a = 79 / 11
b = math.floor(a)
c = b * 11
d = 9 / 11
e = math.floor(d)
f = e + 1
g = f * 11
h = c - g
i = h / 11
j = i + 1
|
a ) 79 kmph , b ) 40 kmph , c ) 34 kmph , d ) 23 kmph , e ) 14 kmph | b | divide(add(20, 60), const_2) | the speed of a car is 20 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 20 + 60 ) / 2 = 40 kmph answer : b" | a = 20 + 60
b = a / 2
|
a ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5 | e | subtract(23, reminder(1053, 23)) | what least number should be added to 1053 , so that the sum is completely divisible by 23 | "explanation : ( 1053 / 23 ) gives remainder 18 18 + 5 = 23 , so we need to add 5 answer : option e" | a = 23 - reminder
|
a ) 76 kg , b ) 80 kg , c ) 85 kg , d ) data inadequate , e ) none of these | b | add(multiply(6, 2.5), 65) | the average weight of 6 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 6 x 2.5 ) kg = 15 kg . weight of new person = ( 65 + 15 ) kg = 80 kg . answer : b" | a = 6 * 2
b = a + 65
|
a ) 45 cm , b ) 25 cm , c ) 24 cm , d ) 15 cm , e ) 10 cm | e | floor(divide(add(multiply(10, const_100), 40), add(multiply(14, const_100), 10))) | which greatest possible length can be used to measure exactly 10 meter 40 cm , 14 meter 10 cm and 7 meter 90 cm | "explanation : convert first all terms into cm . i . e . 1040 cm , 1410 cm , 790 cm . now whenever we need to calculate this type of question , we need to find the hcf . hcf of above terms is 10 . option e" | a = 10 * 100
b = a + 40
c = 14 * 100
d = c + 10
e = b / d
f = math.floor(e)
|
a ) s . 59 , b ) s . 58 , c ) s . 09 , d ) s . 50 , e ) s . 66 | e | add(divide(12000, 200), 6) | a shopkeeper sells 200 metres of cloth for rs . 12000 at a loss of rs . 6 per metre . find his cost price for one metre of cloth ? | "sp per metre = 12000 / 200 = rs . 60 loss per metre = rs . 6 cp per metre = 60 + 6 = rs . 66 answer : e" | a = 12000 / 200
b = a + 6
|
a ) 1 : 3 , b ) 9 : 4 , c ) 8 : 7 , d ) 11 : 9 , e ) 11 : 7 | c | divide(subtract(10.8, 10), subtract(10, 9.3)) | in what ratio must rice at $ 9.30 / kg be mixed with rice at $ 10.80 / kg so that the mixture is worth $ 10 / kg ? | 9.30 x + 10.80 y = 10 ( x + y ) . 80 y = . 70 x x / y = 8 / 7 c is the answer | a = 10 - 8
b = 10 - 9
c = a / b
|
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