Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase r in sales ? | last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase r = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c . | a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100) | general | C |
find the value of x from the below equation : x ^ 2 β 7 x + 10 = 0 | "here we need to find out a and b such that a + b = - 7 and ab = + 10 a = - 5 and b = - 2 satisfies the above condition . hence x ^ 2 β 7 x + 10 = ( x β 5 ) ( x β 2 ) x 2 β 7 x + 10 = ( x β 5 ) ( x β 2 ) x ^ 2 β 7 x + 10 = 0 β ( x β 5 ) ( x β 2 ) = 0 x 2 β 7 x + 10 = 0 β ( x β 5 ) ( x β 2 ) = 0 step 3 : equate each factor to 0 and solve the equations ( x β 5 ) ( x β 2 ) = 0 β ( x β 5 ) = 0 or ( x β 2 ) = 0 β x = 5 or 2 a" | a ) 5 , b ) 7 , c ) 6 , d ) 11 , e ) 12 | a | divide(subtract(7, sqrt(subtract(power(7, 2), multiply(10, 10)))), 2) | multiply(n2,n2)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|subtract(n1,#3)|divide(#4,n0)| | general | A |
each machine of type a has 2 steel parts and 3 chrome parts . each machine of type b has 3 steel parts and 5 chrome parts . if a certain group of type a and type b machines has a total of 40 steel parts and 32 chrome parts , how many machines are in the group | look at the below representation of the problem : steel chrome total a 2 3 40 > > no . of type a machines = 40 / 5 = 8 b 3 5 32 > > no . of type b machines = 32 / 8 = 4 so the answer is 12 i . e a . hope its clear . | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | add(divide(32, add(5, 3)), divide(40, add(2, 3))) | add(n1,n3)|add(n0,n1)|divide(n5,#0)|divide(n4,#1)|add(#2,#3) | general | A |
what is the least common multiple of 15 , 18 , and 34 ? | "let us first write the numbers in the form of prime factors : 15 = 3 * 5 18 = 2 * 3 ^ 2 34 = 2 * 17 ^ 1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 1530 option e" | a ) 60 , b ) 120 , c ) 240 , d ) 360 , e ) 1530 | e | lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 18) | add(const_10,const_2)|multiply(const_10,const_3)|subtract(#1,const_3)|lcm(#0,#2)|lcm(n1,#3)| | general | E |
if 8 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ? | "parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 8 c 2 for vertical 8 c 2 total parallelogram is 10 c 2 * 8 c 2 = 28 * 28 = 784 answer : a" | a ) 784 , b ) 763 , c ) 120 , d ) 160 , e ) 1260 | a | multiply(divide(multiply(8, subtract(8, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2)) | subtract(n0,const_1)|subtract(n1,const_1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,const_2)|divide(#3,const_2)|multiply(#4,#5)| | physics | A |
to deliver an order on time , a company has to make 25 parts a day . after making 25 parts per day for 3 days , the company started to produce 5 more parts per day , and by the last day of work 100 more parts than planned were produced . how many parts the company made ? | "let x be the number of days the company worked . then 25 x is the number of parts they planned to make . at the new production rate they made : 3 β
25 + ( x β 3 ) β
30 = 75 + 30 ( x β 3 ) therefore : 25 x = 75 + 30 ( x β 3 ) β 100 25 x = 75 + 30 x β 90 β 100 190 β 75 = 30 x β 25 115 = 5 x x = 23 so the company worked 23 days and they made 23 β
25 + 100 = 675 pieces . correct answer is a ) 675" | a ) 675 , b ) 778 , c ) 654 , d ) 412 , e ) 589 | a | add(multiply(divide(subtract(add(multiply(add(25, 5), 3), 100), multiply(25, 3)), 5), 25), 100) | add(n0,n3)|multiply(n0,n2)|multiply(n2,#0)|add(n4,#2)|subtract(#3,#1)|divide(#4,n3)|multiply(n0,#5)|add(n4,#6)| | physics | A |
if a book is sold at 7 % profit instead of 7 % loss , it would have brought rs 14 more . find out the cost price of the book | "let c . p . of the book be rs . β x β given , 1.07 x - 0.93 x = 14 = > 0.14 x = 14 = 14 / 0.14 = rs 100 answer : b" | a ) 75 , b ) 100 , c ) 60 , d ) 70 , e ) 80 | b | divide(multiply(const_100, divide(7, const_2)), 7) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0)| | gain | B |
6000 , 1200 , 300 , 60 . . . | "6000 / 5 = 1200 1200 / 4 = 300 300 / 5 = 60 60 / 4 = 15 answer : b" | a ) 30 , b ) 15 , c ) 20 , d ) 36 , e ) 46 | b | subtract(negate(60), multiply(subtract(1200, 300), divide(subtract(1200, 300), subtract(6000, 1200)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | B |
2 cow β s and 8 goats are brought for rs . 1400 . if the average price of a goat be rs . 60 . what is the average price of a cow . | explanation : average price of a goat = rs . 60 total price of 8 goats = 8 * 60 = rs . 480 but total price of 2 cows and 8 goats = rs . 1400 total price of 2 cows is = 1400 - 480 = 980 average price of a cow = 980 / 2 = rs . 460 answer : option d | a ) rs . 480 , b ) rs . 960 , c ) rs . 960 , d ) rs . 460 , e ) rs . 560 | d | divide(subtract(1400, multiply(8, 60)), 2) | multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0) | general | D |
an article is bought for rs . 675 and sold for rs . 1200 , find the gain percent ? | "675 - - - - 525 100 - - - - ? = > = 78 % answer : c" | a ) 87 % , b ) 88 % , c ) 78 % , d ) 75 % , e ) 73 % | c | subtract(const_100, divide(multiply(1200, const_100), 675)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain | C |
the length of a rectangle is twice the breadth . if the area is 128 cm 2 , determine the length and the breadth . | we are given length l = 2 b and l Γ b = 128 β΄ 2 b Γ b = 128 2 b ^ 2 = 128 b ^ 2 = 64 b = Β± 8 but breadth must be positive , therefore b = 8 cm , and l = 2 b = 16 cm . answer is b . | ['a ) b = 2 , l = 9', 'b ) b = 8 , l = 16', 'c ) b = 9 , l = 9', 'd ) b = 4 , l = 14', 'e ) b = 3 , l = 9'] | b | multiply(sqrt(divide(128, const_2)), const_2) | divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2) | geometry | B |
if 15 students in a class average 80 % on an english exam and 10 students average 90 % on the same exam , what is the average in percent for all 25 students ? | the average percentage = ( 15 x 80 % + 10 x 90 % ) / 25 = ( 15 x 80 % + 10 x 80 % + 10 x 10 % ) / 25 . . . . . . ( as 90 % = 10 % + 80 % ) = ( 25 x 80 % + 10 x 10 % ) / 25 = 80 % + 10 x 10 % / 25 = 80 % + 4 % = 84 % answer : c | a ) 83 % , b ) 83 1 β 2 % , c ) 84 % , d ) 85 % , e ) 86 2 β 3 % | c | divide(add(multiply(15, 80), multiply(10, 90)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4) | general | C |
the average mark obtained by 22 candidates in an examination is 48 . the average of the first 10 is 55 while the last 11 is 40 . the marks obtained by the 11 th candidate is ? | it is clear that 22 x 48 = 10 x 55 + k + 11 x 40 β β k = 66 answer : e | a ) 22 , b ) 0 , c ) 49 , d ) 58 , e ) 66 | e | subtract(multiply(22, 48), add(multiply(10, 55), multiply(11, 40))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3) | general | E |
a certain box has 8 cards and each card has one of the integers from 1 to 8 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on the 8 cards ? | "the average of the numbers is 4.5 the total number of ways to choose 2 cards from 8 cards is 8 c 2 = 28 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } the probability is 2 / 28 = 1 / 14 the answer is c ." | a ) 1 / 8 , b ) 1 / 12 , c ) 1 / 14 , d ) 1 / 18 , e ) 1 / 28 | c | divide(const_4, divide(factorial(8), multiply(factorial(2), factorial(subtract(8, 2))))) | factorial(n0)|factorial(n3)|subtract(n0,n3)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)|divide(const_4,#5)| | general | C |
a , b and c enter into a partnership and theirs shares are in the ratio 1 / 2 : 1 / 3 : 1 / 4 . after two months , a withdraws half of his capital and after 10 months , a profit of rs . 378 is divided among them . what is b β s share ? | ratio of initial investments = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 . let their initial investments be 6 x , 2 x and 3 x respectively . a : b : c = ( 6 x * 2 + 3 x * 10 ) : ( 4 x * 12 ) : ( 3 x * 12 ) = 42 : 48 : 36 = 7 : 8 : 6 . therefore , b β s share = rs . 378 * 8 / 21 = rs . 144 . answer : b | a ) rs . 129 , b ) rs . 144 , c ) rs . 156 , d ) rs . 168 , e ) rs . 178 | b | divide(multiply(378, multiply(2, 4)), multiply(add(3, 4), 3)) | add(n3,n5)|multiply(n1,n5)|multiply(n7,#1)|multiply(n3,#0)|divide(#2,#3) | general | B |
j is 25 % less than p and 20 % less than t . t is e % less than p . what is the value of e ? | "usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( e / 100 ) ] * p i . e . 100 - e = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . e = 6.25 answer : option d" | a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | multiply(n0,n0)|divide(#0,const_100)| | gain | D |
shekar scored 76 , 65 , 82 , 67 and 55 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 67 + 55 ) / 5 = 375 / 5 = 69 hence average = 69 answer : b" | a ) 65 , b ) 69 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 55), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | B |
the sum of three consecutive even numbers is 93 . find the middle number of the three ? | "middle value = 93 / 3 = 31 ans d" | a ) 14 , b ) 35 , c ) 39 , d ) 31 , e ) 12 | d | add(add(power(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | D |
in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 70 % of calculus students are enrolled in the beginner course . if 4 / 5 of students are in the beginner courses , and one student is selected at random , what is the probability that a beginner trigonometry student is selected ? | "let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.7 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.7 x = 1.3 x the percentage in beginner trigonometry is 1.3 x / 2.5 x = 52 % the answer is d ." | a ) 40 % , b ) 44 % , c ) 48 % , d ) 52 % , e ) 56 % | d | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 70), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | divide(n2,n3)|multiply(n0,const_2)|add(n0,#1)|add(#2,#1)|multiply(#3,#0)|subtract(#4,n1)|divide(#5,#3)|multiply(#6,#1)| | gain | D |
on a game show , there are 3 tables . each table has 3 boxes ( one zonk ! , one cash prize , and one grand prize ) . the contestant can choose one box from each table for a total of 3 boxes . if any of the boxes are a zonk ! , the contestant loses everything . what is the probability of getting no zonk ! from any of the 3 boxes chosen ? | no zonk ! : 1 st box = no zonk ! = 2 / 3 2 nd box = no zonk ! = 2 / 3 3 rd box = no zonk ! = 2 / 3 ( 2 / 3 ) ( 2 / 3 ) ( 2 / 3 ) = 8 / 27 8 / 27 is the probability of no zonk ! , so . . . answer : c | a ) 1 / 5 , b ) 1 / 3 , c ) 8 / 27 , d ) 1 / 27 , e ) 7 / 8 | c | divide(power(const_2, 3), power(3, const_3)) | power(const_2,n0)|power(n0,const_3)|divide(#0,#1) | general | C |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 5 meters and a circumference of 4 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | for a , r = 4 / 2 pi . its capacity = ( 2 pi ) ^ 2 * 5 = 20 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 20 pi / 200 pi = 0.1 a | ['a ) 10 %', 'b ) 80 %', 'c ) 100 %', 'd ) 120 %', 'e ) 125 %'] | a | multiply(multiply(power(divide(4, 10), const_2), divide(5, 8)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100) | physics | A |
95 white and black tiles will be used to form a 10 x 10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | answer = c please refer diagram below 95 - 10 = 85 | a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | c | subtract(95, 10) | subtract(n0,n1) | general | C |
a rectangular courtyard 3.78 m lang and 5.25 m broad is to be paved exactly with square tiles , all of the same size . the minimum number of such tiles is : | l = 378 cm and b = 525 cm maximum length of a square tile = hcf of ( 378,525 ) = 21 cm number of tiles = ( 378 Γ 525 ) / ( 21 Γ 21 ) = ( 18 Γ 25 ) = 450 answer is d . | ['a ) 425', 'b ) 430', 'c ) 440', 'd ) 450', 'e ) 460'] | d | divide(multiply(multiply(3.78, const_100), multiply(5.25, const_100)), power(multiply(add(const_3, const_4), const_3), const_2)) | add(const_3,const_4)|multiply(n0,const_100)|multiply(n1,const_100)|multiply(#1,#2)|multiply(#0,const_3)|power(#4,const_2)|divide(#3,#5) | geometry | D |
a retailer sold 2 houses at each $ 15000 . on 1 st house he got 20 % profit and on 2 nd house he loss 5 % then find his net profit or loss ? | "a = 20 % b = - 5 % profit % = 20 - 5 + ( - 100 / 100 ) = 14 % answer is c" | a ) 10 % , b ) 9 % , c ) 14 % , d ) 15 % , e ) 12 % | c | multiply(subtract(multiply(subtract(1, divide(5, const_100)), 2), divide(5, const_100)), 5) | divide(n5,const_100)|subtract(n2,#0)|multiply(#1,n0)|subtract(#2,#0)|multiply(n5,#3)| | gain | C |
find the circumference and area of radius 4 cm . | "area of circle = Ο r Β² = 22 / 7 Γ 4 Γ 4 cm Β² = 50 cm Β² answer : c" | a ) 24 cm Β² , b ) 44 cm Β² , c ) 50 cm Β² , d ) 84 cm Β² , e ) 94 cm Β² | c | circle_area(4) | circle_area(n0)| | geometry | C |
if a rectangular billboard has an area of 104 square feet and a perimeter of 42 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 104 and a perimeter of 42 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 104 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 104 , so the correct answer is probably a or c . let ' s test answer c : 8 if . . . the shorter side = 8 . . . the area = 104 . . . . 104 / 8 = 13 = the longer side perimeter = 8 + 8 + 13 + 13 = 42 d" | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(42, const_2), sqrt(subtract(power(divide(42, const_2), const_2), multiply(const_4, 104)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry | D |
two numbers are in respectively 20 % and 50 % more than a third number . the ratio of the two numbers is ? | "let the 3 rd number be x then , first number = 120 % of x = 120 x / 100 = 6 x / 5 second number = 150 % of x = 150 x / 100 = 3 x / 2 ratio of first two numbers = 6 x / 5 : 3 x / 2 = 12 x : 15 x = 4 : 5 answer is c" | a ) 2 : 5 , b ) 1 : 2 , c ) 4 : 5 , d ) 3 : 7 , e ) 5 : 6 | c | divide(add(const_100, 20), add(const_100, 50)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | other | C |
it takes 3 workers a total of 9 hours to build a giant machine with each worker working at the same rate . if 3 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 2 : 00 pm , at what time will the machine be complete ? | "3 workers build 1 / 9 of a machine in one hour . 1 worker builds 1 / 27 of a machine in one hour . in the first 3 hours , 3 workers build 3 * ( 1 / 27 ) * 3 = 9 / 27 of a machine from 2 : 00 to 3 : 00 , 4 workers build another 4 / 27 . the total is 13 / 27 . from 3 : 00 to 4 : 00 , 5 workers build another 5 / 27 . the total is 18 / 27 . from 4 : 00 to 5 : 00 , 6 workers build another 6 / 27 . the total is 24 / 27 . to build another 3 / 27 , 7 workers need ( 3 / 7 ) * 60 minutes which is about 26 minutes . the machine is complete at about 5 : 26 . the answer is b ." | a ) 5 : 15 , b ) 5 : 26 , c ) 5 : 37 , d ) 5 : 48 , e ) 5 : 59 | b | subtract(divide(multiply(3, 9), 9), const_0_33) | multiply(n0,n1)|divide(#0,n1)|subtract(#1,const_0_33)| | general | B |
the cost of an article is decreased by 50 % . if the original cost is $ 200 , find the decrease cost . | original cost = $ 200 decrease in it = 50 % of $ 200 = 50 / 100 Γ£ β 200 = 10000 / 100 = $ 100 therefore , decrease cost = $ 200 - $ 100 = $ 100 answer : d | a ) 330 , b ) 200 , c ) 150 , d ) 100 , e ) 91 | d | divide(multiply(200, 50), const_100) | multiply(n0,n1)|divide(#0,const_100) | gain | D |
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 18 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 / 18 = 7 / 18 of the job . in 2 hours , they complete 2 ( 7 / 18 ) = 7 / 9 of the job . the time for p to finish is ( 2 / 9 ) / ( 1 / 3 ) = ( 2 / 3 ) hour = 40 minutes the answer is c ." | a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48 | c | multiply(divide(subtract(const_1, multiply(add(inverse(3), inverse(18)), 2)), inverse(3)), const_60) | inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|divide(#4,#0)|multiply(#5,const_60)| | physics | C |
a room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 6 meters and 24 centimeters = 624 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 624 and 432 = 48 thus , the number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 answer : b" | a ) 107 , b ) 117 , c ) 127 , d ) 137 , e ) 147 | b | divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)))) | multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)| | general | B |
the average student age of a certain class which has 30 students is 10 . if one student aged 11 go to another class and the age of the class teacher is included the average changes to 11 . what is the age of class teacher ? | "sum of ages of class before replacement = 30 x 10 = 300 sum of ages of class without teacher = 300 - 11 = 289 sum of ages of class after replacement = 30 x 11 = 330 age of teacher = 330 - 289 = 41 answer : b" | a ) 32 , b ) 41 , c ) 45 , d ) 48 , e ) 50 | b | subtract(multiply(30, 11), subtract(multiply(30, 10), 11)) | multiply(n0,n3)|multiply(n0,n1)|subtract(#1,n2)|subtract(#0,#2)| | general | B |
a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 555 in all . for how many days was he absent ? | "30 * 25 = 750 455 - - - - - - - - - - - 195 25 + 7.50 = 32.5 195 / 32.5 = 6 d" | a ) 8 , b ) 10 , c ) 15 , d ) 6 , e ) 19 | d | subtract(30, divide(add(multiply(7.50, 30), 555), add(7.50, 25))) | add(n1,n2)|multiply(n0,n2)|add(n3,#1)|divide(#2,#0)|subtract(n0,#3)| | physics | D |
in the rectangular coordinate system , points ( 2 , 0 ) and ( β 2 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ? | "the answer is a it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 2" | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above | a | sqrt(power(2, const_2)) | power(n0,const_2)|sqrt(#0)| | geometry | A |
4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 10 weavers in 10 days ? | "1 weaver can weave 1 mat in 4 days . 10 weavers can weave 10 mats in 4 days . 10 weavers can weave 25 mats in 10 days . the answer is d ." | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | d | multiply(10, multiply(10, divide(4, multiply(4, 4)))) | multiply(n0,n0)|divide(n0,#0)|multiply(n3,#1)|multiply(n3,#2)| | gain | D |
in a group of cows and chickens , the number of legs was 10 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 10 more than twice the number of heads . therefore , 2 Γ ( x + y ) + 10 = 4 x + 2 y . or , 2 x + 2 y + 10 = 4 x + 2 y . or , 2 x + 10 = 4 x [ subtracting 2 y from both sides ] . or , 10 = 4 x β 2 x [ subtracting 2 x from both sides ] . or , 10 = 2 x . or , x = 5 [ dividing by 2 on both sides ] . therefore , the number of cows = 5 . correct answer : a ) 5" | a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | a | divide(10, subtract(const_4, const_2)) | subtract(const_4,const_2)|divide(n0,#0)| | general | A |
ashok borrowed some money at the rate of 6 % p . a . for the first two years , at the rate of 9 % p . a . for the next 3 years and at the rate of 14 % p . a . for the period beyond 5 years . if he pays a total interest of 11400 / - at the end of 9 years , how much money did he borrow ? | we have , si = p Γ r Γ t / 100 β΄ 11400 = p Γ 6 Γ 2 / 100 + p Γ 9 Γ 3 / 100 + p Γ 14 Γ 4 / 100 or , 12 p + 27 p + 56 p = 11400 Γ 100 or , 95 p = 11400 Γ 100 β΄ p = 12000 answer d | a ) 16060 , b ) 14000 , c ) 18000 , d ) 12000 , e ) none of these | d | divide(11400, add(divide(multiply(14, subtract(subtract(9, const_3), const_2)), const_100), add(divide(multiply(6, const_2), const_100), divide(multiply(9, 3), const_100)))) | multiply(n0,const_2)|multiply(n1,n2)|subtract(n1,const_3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,const_2)|add(#3,#4)|multiply(n3,#5)|divide(#7,const_100)|add(#6,#8)|divide(n5,#9) | general | D |
what is the square root of 4,624 ? | "a square root of a number is a number times itself equals the original number . ex ) the square root of 81 is 9 because 9 times itself ( 9 ) equals the original number ( 81 ) . 68 * 68 = 4,624 4624 divided by x = 68 x = 68 ( c ) 68" | a ) a ) 18 , b ) b ) 9 , c ) c ) 68 , d ) d ) 62 , e ) e ) 81 | c | circle_area(divide(4,624, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other | C |
a train 550 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 = [ 550 * 3 / 55 ] sec = 30 sec answer : a | a ) 30 , b ) 60 , c ) 25 , d ) 40 , e ) 10 | a | multiply(const_3600, divide(divide(550, const_1000), add(60, 6))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600) | physics | A |
an escalator is descending at constant speed . a walks down and takes 50 steps to reach the bottom . b runs down and takes 90 steps in the same time as a takes 10 steps . how many steps are visible when the escalator is not operating ? | lets suppose that a walks down 1 step / min and escalator moves n steps / min it is given that a takes 50 steps to reach the bottom in the same time escalator would have covered 50 n steps so total steps on escalator is 50 + 50 n . again it is given that b takes 90 steps to reach the bottom and time taken by him for this is equal to time taken by a to cover 10 steps i . e 10 minutes . so in this 10 min escalator would have covered 10 n steps . so total steps on escalatro is 90 + 10 n again equating 50 + 50 n = 90 + 10 n we get n = 1 hence total no . of steps on escalator is 100 . answer : c | a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | c | add(50, multiply(50, divide(subtract(90, 50), subtract(90, 50)))) | subtract(n1,n0)|divide(#0,#0)|multiply(n0,#1)|add(n0,#2) | physics | C |
a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 400 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 40 % of them donated $ 16000 ( 400 * 40 ) $ 16000 is 60 % of total amount . so total amount = 16000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 16000 * ( 100 / 60 ) * ( 40 / 100 ) = 32000 / 3 this amount has to be divided by 60 ( remaining people are 60 ) so per head amount is 32000 / 3 / 60 = 32000 / 180 = 1600 / 9 = $ 177.78 ; answer : b" | a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | b | divide(multiply(divide(multiply(divide(40, const_100), 400), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general | B |
what is the cp of rs 100 stock at 4 discount , with 1 / 5 % brokerage ? | explanation : use the formula , cp = 100 β discount + brokerage % cp = 100 - 4 + 1 / 5 96.2 thus the cp is rs 96.2 . answer : c | a ) 96.9 , b ) 96.3 , c ) 96.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 4), divide(1, 5)) | divide(n2,n3)|subtract(n0,n1)|add(#0,#1) | gain | C |
in a renowned city , the average birth rate is 10 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 8 persons are added ( 10 - 2 ) . every second 4 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 4 = 345600 option c" | a ) 32,300 , b ) 172,800 , c ) 345,600 , d ) 338,200 , e ) 259,200 | c | multiply(multiply(subtract(10, 2), const_3600), const_12) | subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)| | general | C |
if x is equal to the sum of the integers from 20 to 40 , inclusive , and y is the number of even integers from 20 to 40 , inclusive , what is the value of x + y ? | "x = 20 + 21 + . . . + 40 = 21 ( 30 ) = 630 y = 11 x + y = 641 the answer is e ." | a ) 601 , b ) 611 , c ) 621 , d ) 631 , e ) 641 | e | add(multiply(divide(add(20, 40), const_2), add(subtract(40, 20), const_1)), add(divide(subtract(40, 20), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)| | general | E |
a β s speed is 30 / 17 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a β s speed and b β s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 30 meters , b covers 17 meters in that time . so if the race is 30 meters long , when a reaches the finish line , b would be 13 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 13 meters so that he doesn β t need to cover that . in that case , the time required by a ( to cover 30 meters ) would be the same as the time required by b ( to cover 17 meters ) to reach the finish line . so b should get a head start of 13 / 30 th of the race . answer ( b )" | a ) 1 / 17 , b ) 13 / 30 , c ) 17 / 30 , d ) 11 / 30 , e ) 11 / 13 | b | divide(subtract(30, 17), 30) | subtract(n0,n1)|divide(#0,n0)| | general | B |
in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 50 per square m ? | "length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 50 * 950 = rs . 47500 e" | a ) s . 43000 , b ) s . 50000 , c ) s . 57000 , d ) s . 60000 , e ) s . 47500 | e | add(add(multiply(multiply(multiply(20, 15), const_2), 50), multiply(multiply(multiply(20, 5), const_2), 50)), multiply(multiply(multiply(15, 5), const_2), 50)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#8)| | general | E |
walking with 4 / 5 of my usual speed , i arrive at the bus stop 6 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 6 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 24 the answer is e ." | a ) 20 min , b ) 22 min , c ) 15 min , d ) 18 min , e ) 24 min | e | divide(6, subtract(divide(5, 4), const_1)) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)| | physics | E |
j is 25 % less than p and 20 % less than t . t is q % less than p . what is the value of q ? | "usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( q / 100 ) ] * p i . e . 100 - q = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . q = 6.25 answer : option d" | a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | multiply(n0,n0)|divide(#0,const_100)| | gain | D |
lloyd normally works 7.5 hours per day and earns $ 4.50 per hour . for each hour he works in excess of 7.5 hours on a given day , he is paid 2.5 times his regular rate . if lloyd works 10.5 hours on a given day , how much does he earn for that day ? | "daily working hour * regular rate + overtime * increased rate 7.5 * 4.5 + 3 * 4.5 * 2.5 = 67.5 answer e" | a ) $ 33.75 , b ) $ 47.25 , c ) $ 51.75 , d ) $ 54.00 , e ) $ 67.50 | e | add(multiply(7.5, 4.50), multiply(multiply(subtract(10.5, 7.5), 2.5), 4.50)) | multiply(n0,n1)|subtract(n4,n0)|multiply(n3,#1)|multiply(n1,#2)|add(#0,#3)| | physics | E |
a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 5 : 2 : 2 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 4 : 3 , respectively . how many more acres of land were planted with tobacco under the new system ? | "originally ( 2 / 9 ) * 1350 = 300 acres were planted with tobacco . in the new system ( 3 / 9 ) * 1350 = 450 acres were planted with tobacco . the answer is b ." | a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | b | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(3, add(add(3, 2), 2))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(2, add(add(5, 2), 2)))) | add(n1,n5)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n1,const_10)|add(#1,#2)|add(n2,#0)|add(#4,#3)|divide(n1,#5)|divide(n2,#5)|multiply(#6,#7)|multiply(#6,#8)|subtract(#9,#10)| | other | B |
if the cost price of 50 articles is equal to the selling price of 46 articles , then the gain or loss percent is ? | given that , cost price of 50 article is equal to selling price of 46 articles . let cost price of one article = rs . 1 selling price of 46 articles = rs . 50 but cost price of 46 articles = rs . 46 therefore , the trader made profit . \ percentage of profit = 4 / 46 * 100 = 8.69 % answer : e | a ) 22 , b ) 65 , c ) 78 , d ) 33 , e ) 8.69 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 46), 50)), divide(multiply(const_100, 46), 50))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100) | gain | E |
a person incurs 10 % loss by selling a watch for $ 100 . at what price should the watch be sold to earn 10 % profit ? | "let the new selling price be $ x ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) ( 100 - 10 ) / 100 = ( 100 + 10 ) / x x = 110 * 100 / 90 = 122 approximately answer is b" | a ) $ 100 , b ) $ 122 , c ) $ 150 , d ) $ 210 , e ) $ 170 | b | divide(100, subtract(const_1, divide(10, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | B |
a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio r of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } Β = ( 1 / 2 ) ^ 2 Β = 1 / 4 therefore r , w _ r Β = { w _ o } / 4 = 8 / 4 Β = 2 litres . b" | a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | add(n1,n2)|subtract(n0,n3)|divide(n1,#0)|divide(n2,#0)|multiply(#2,#1)|multiply(#3,#1)|add(n3,#4)|divide(#5,n0)|divide(#6,n0)|multiply(#7,#1)|multiply(#8,#1)|add(n3,#10)|divide(#11,#9)| | general | B |
a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 15 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 15 ) x solve for r , r = 55 55 - 15 = 40 mpg e" | a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 40 | e | divide(336, divide(subtract(462, 336), 15)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics | E |
the length of the bridge , which a train 130 metres long and travelling at 45 km / hr can cross in 30 seconds , is : | "given , length of the train = 130 meters speed of the train = 45 km / hr = 45 * 518 m / sec = 252 m / sec time = 30 secs let the length of the bridge be x meters . the distance travelled by the train while crossing the bridge will be sum of the length of train and length of the bridge = ( x + 130 ) meters so , 252 = x + 13030 = 30 * 25 = 2 x + 260 = 750 - 260 = 2 x = 490 = 2 x = x = 245 meters . answer c" | a ) 200 meters , b ) 225 meters , c ) 245 meters , d ) 250 meters , e ) 300 meters | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics | C |
in the quadratic equation ax 2 - x - 12 = 0 , if the sum of two roots is 1 , what is the product of the two roots ? | explanation : the sum of the roots of the quadratic equation ax 2 + bx + c = 0 are ( - b / a ) and the product of the roots are ( c / a ) . thus , in the equation ax 2 - 11 x + 40 = 0 , where a = a , b = - 1 and c = - 12 . we get , sum of the roots = 1 / a = 1 a = 1 product of the roots = - 12 / 1 = - 12 answer : a | a ) - 12 , b ) - 6 , c ) 12 , d ) 6 , e ) 24 | a | multiply(divide(12, const_3), negate(divide(12, const_4))) | divide(n1,const_3)|divide(n1,const_4)|negate(#1)|multiply(#0,#2) | general | A |
in a hostel , the number of students decreased by 10 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ? | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 8 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.9 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.9 * 1.2 ) = > f ( new ) = 0.9259 f therefore the new cost of food must be 92.59 % of the old cost , or the cost of food must decrease by 7.4 % ( option e )" | a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 7.4 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 10), add(const_100, 20)))), const_100) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general | E |
the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 50 % profit ? | "c $ 960 let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 150 % of $ 640 = 150 / 100 * 640 = $ 960 ." | a ) $ 480 , b ) $ 450 , c ) $ 960 , d ) $ 870 , e ) $ 660 | c | add(divide(multiply(divide(add(832, 448), const_2), 50), const_100), divide(add(832, 448), const_2)) | add(n0,n1)|divide(#0,const_2)|multiply(n2,#1)|divide(#2,const_100)|add(#3,#1)| | gain | C |
a 40 - liter solution of alcohol and water is 5 percent alcohol . if 3.5 liters of alcohol and 6.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 3.5 ) / 50 = 5.5 / 50 = 11 % the answer is b ." | a ) 8 % , b ) 11 % , c ) 14 % , d ) 17 % , e ) 20 % | b | multiply(divide(add(multiply(40, divide(5, const_100)), 3.5), add(40, add(3.5, 6.5))), const_100) | add(n2,n3)|divide(n1,const_100)|add(n0,#0)|multiply(n0,#1)|add(n2,#3)|divide(#4,#2)|multiply(#5,const_100)| | general | B |
a batsman scored 140 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "number of runs made by running = 140 - ( 3 x 4 + 8 x 6 ) = 140 - ( 60 ) = 80 now , we need to calculate 80 is what percent of 140 . = > 80 / 140 x 100 = 57 % answer : d" | a ) 50 % , b ) 40 % , c ) 60 % , d ) 57 % , e ) 80 % | d | multiply(divide(subtract(140, add(multiply(3, 8), multiply(8, 3))), 140), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general | D |
a certain company β s profit in 1996 was 17 percent greater than its profit in 1995 , and its profit in 1997 was 25 percent greater than its profit in 1996 . the company β s profit in 1997 was what percent greater than its profit in 1995 ? | "profit in 1995 - 100 profit in 1996 - 117 % increae profit in 1997 in comparison to 1995 = 17 + 117 * 25 % = 46.25 correct option : e" | a ) 5 % , b ) 18 % , c ) 33 % , d ) 35 % , e ) 46.25 % | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(const_1, divide(17, const_100))), const_1), const_100) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain | E |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 42 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 42 + 30 = 72 km / hr . 72 * 5 / 18 = 20 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 / 20 = 50 sec . answer : c" | a ) 22 , b ) 27 , c ) 50 , d ) 99 , e ) 12 | c | add(42, 30) | add(n1,n2)| | physics | C |
a merchant has selected two items to be placed on sale , one of which currently sells for 30 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 70 ; we must increase $ 70 to $ 100 , so by $ 30 , which is approximately 40 % increase : ( 100 - 70 ) / 70 = 3 / 7 = ~ 0.42 . answer : c ." | a ) 22 % , b ) 30 % , c ) 42 6 / 7 % , d ) 70 % , e ) 130 % | c | multiply(divide(divide(30, const_100), subtract(const_1, divide(30, const_100))), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(#0,#1)|multiply(#2,const_100)| | general | C |
two cars are driving toward each other . the first car is traveling at a speed of 100 km / h , which is 25 % faster than the second car ' s speed . if the distance between the cars is 720 km , how many hours will it take until the two cars meet ? | the speed of the first car is 100 km / h . the speed of the second car is 100 / 1.25 = 80 km / h . the two cars complete a total of 180 km each hour . the time it takes the cars to meet is 720 / 180 = 4 hours . the answer is c . | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | c | divide(add(multiply(100, const_4), multiply(divide(multiply(const_4, 100), add(const_1, const_4)), const_4)), add(100, divide(multiply(const_4, 100), add(const_1, const_4)))) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|add(n0,#2)|multiply(#2,const_4)|add(#1,#4)|divide(#5,#3) | physics | C |
how long does a train 165 meters long running at the rate of 72 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 165 ) / 72 * 18 / 5 t = 41.25 answer : b" | a ) 28 , b ) 41.25 , c ) 55 , d ) 18 , e ) 12 | b | divide(add(165, 660), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | B |
a shopkeeper sells his goods at cost price but uses a weight of 800 gm instead of kilogram weight . what is his profit percentage ? | "100 ( 1000 - 800 ) / 800 = 25 % profit percentage = 25 % answer a" | a ) 25 % , b ) 18 % , c ) 20 % , d ) 40 % , e ) 45 % | a | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 800), 800), const_100) | add(const_1,const_4)|add(#0,#0)|multiply(#1,const_100)|subtract(#2,n0)|divide(#3,n0)|multiply(#4,const_100)| | gain | A |
if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 246 m , what is its area ? | "length = breadth + 23 . therefore , 4 Γ breadth + 2 Γ 23 = 246 m β breadth = 50 m length = 50 + 23 = 73 m area = 73 Γ 50 = 3650 m 2 answer is b ." | a ) 2510 , b ) 3650 , c ) 2530 , d ) 2515 , e ) 2520 | b | rectangle_area(add(divide(subtract(246, multiply(const_2, 23)), const_4), 23), divide(subtract(246, multiply(const_2, 23)), const_4)) | multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)| | geometry | B |
1 + 1 | e | a ) 9 , b ) 8 , c ) 3 , d ) 0 , e ) 2 | e | multiply(divide(1, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
20 is added to a certain number , the sum is multiplied by 2 , the product is divided by 2 and 2 is subtracted from the quotient . the remainder left is half of 88 . what is the number ? | "let number is x . when 20 added to it , = ( x + 20 ) 2 multiplied to sum , = 2 * ( x + 20 ) now , = [ { 2 * ( x + 20 ) } / 2 ] and , = [ { 2 * ( x + 20 ) } / 2 ] - 2 according to question , [ { 2 * ( x + 20 ) } / 2 ] - 2 = half of 88 [ ( 2 x + 40 ) / 2 ) = 44 + 2 2 x + 40 = 46 * 2 x + 20 = 46 x = 46 - 20 x = 26 so , required number is : 26 . answer : c" | a ) 21 , b ) 20 , c ) 26 , d ) 30 , e ) 45 | c | subtract(divide(multiply(add(divide(88, const_2), 2), 2), 2), 20) | divide(n4,const_2)|add(n2,#0)|multiply(n2,#1)|divide(#2,n1)|subtract(#3,n0)| | general | C |
a miniature roulette wheel is divided into 9 equal sectors , each bearing a distinct integer from 1 to 9 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun four times , approximately what is the probability that the product of the four winning sectors β integers will be even ? | "the only way to have an odd product is if all 4 integers are odd . p ( odd product ) = 5 / 9 * 5 / 9 * 5 / 9 * 5 / 9 = 625 / 6561 p ( even product ) = 1 - 625 / 6561 = 5936 / 6561 which is about 90 % the answer is d ." | a ) 50 % , b ) 67 % , c ) 80 % , d ) 90 % , e ) 95 % | d | multiply(subtract(1, power(divide(divide(9, const_2), 9), const_2)), const_100) | divide(n0,const_2)|divide(#0,n0)|power(#1,const_2)|subtract(n1,#2)|multiply(#3,const_100)| | general | D |
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 350 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 350 = > x = 17500 answer : d" | a ) 20029 , b ) 20000 , c ) 20289 , d ) 17500 , e ) 20026 | d | divide(350, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)| | gain | D |
` ` i am 8 times as old as you were when i was as old as you are ` ` , said a man to his son . find out their present ages if the sum of the their ages is 75 years . | present age of boy x and previous age of boy y so present age of dad : 8 y and previous age is x so the gap between duration must be same so 8 y - x = x - y - - > x = 9 y / 2 and x + 8 y = 75 ( given ) - - > 9 y / 2 + y = 75 so y = 6 and and x = 27 therefore fathers age is 8 y = 8 * 6 = 48 and child ' s age = x = 27 answer : e | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | e | subtract(75, divide(75, add(divide(add(8, const_1), multiply(8, const_2)), const_1))) | add(n0,const_1)|multiply(n0,const_2)|divide(#0,#1)|add(#2,const_1)|divide(n1,#3)|subtract(n1,#4) | general | E |
harkamal purchased 3 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 3 kg grapes = 70 Γ 3 = 210 . cost of 9 kg of mangoes = 55 Γ 9 = 490 . total cost he has to pay = 210 + 490 = 700 a" | a ) a ) 700 , b ) b ) 1065 , c ) c ) 1070 , d ) d ) 1075 , e ) e ) 1080 | a | add(multiply(3, 70), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | A |
in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 3 , - 1 ) , ( 3 , 8 ) , ( 12 , 2 ) , ( 12 , - 7 ) ? | "by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) Γ ( base 1 + base 2 ) Γ ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 Γ ( 9 + 9 ) Γ 9 = 81 . the answer is c ." | a ) 69 , b ) 75 , c ) 81 , d ) 87 , e ) 93 | c | divide(multiply(add(add(1, 8), add(2, 8)), subtract(12, 3)), 2) | add(n1,n3)|add(n3,n5)|subtract(n4,n0)|add(#0,#1)|multiply(#3,#2)|divide(#4,n5)| | geometry | C |
the average of 20 numbers is calculated as 30 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "20 * 30 + 36 β 26 = 160 / 10 = 60 answer : 60 answer is c" | a ) 40 , b ) 26 , c ) 60 , d ) 97 , e ) 12 | c | add(30, divide(subtract(36, 26), 20)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general | C |
in 10 years , a will be twice as old as b was 10 years ago . if a is now 8 years older than b the present age of b is | "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 8 = > 2 b - b - 8 = 30 b = 38 so the present age of b is 38 years answer : b" | a ) 37 , b ) 38 , c ) 39 , d ) 40 , e ) 41 | b | add(add(multiply(const_2, 10), 10), 8) | multiply(n0,const_2)|add(n0,#0)|add(n2,#1)| | general | B |
a salesman Γ’ β¬ β’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 700 more than that by the previous schema , his sales were worth ? | [ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 700 x = 8000 answer a | a ) s . 8,000 , b ) s . 9,000 , c ) s . 20,000 , d ) s . 10,000 , e ) s . 50,000 | a | subtract(multiply(5, const_4), const_12) | multiply(n0,const_4)|subtract(#0,const_12) | general | A |
what is the maximum number r of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ? | 27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can have are 4 * 3 * 2 = 24 l * w * h is 8 * 12 * 9 , then max . cube we can have are 2 * 4 * 3 = 24 l * w * h is 8 * 9 * 12 , then max . cube we can have are 2 * 3 * 4 = 24 in all cases we get r = 24 cubes . ans . c | ['a ) 36', 'b ) 32', 'c ) 24', 'd ) 21', 'e ) 15'] | c | multiply(multiply(divide(9, const_3), divide(12, cube_edge_by_volume(27))), floor(divide(8, cube_edge_by_volume(27)))) | cube_edge_by_volume(n0)|divide(n2,const_3)|divide(n1,#0)|divide(n3,#0)|floor(#2)|multiply(#1,#3)|multiply(#4,#5) | geometry | C |
the cost price of a radio is rs . 4800 and it was sold for rs . 4400 , find the loss % ? | "4800 - - - - 400 100 - - - - ? = > 8 % answer : a" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | a | multiply(divide(subtract(4800, 4400), 4800), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain | A |
an athlete runs 200 metres race in 20 seconds . what is his speed ? | "speed = distance / time = 200 / 20 = 10 m / s = 10 * 18 / 5 = 36 km / hr answer : d" | a ) 23 , b ) 38 , c ) 37 , d ) 36 , e ) 28 | d | multiply(divide(200, 20), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | D |
what will be the reminder when ( 67 ^ 67 + 67 ) is divided by 68 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 67 ^ 67 + 1 ) will be divisible by ( 67 + 1 ) ; ( 67 ^ 67 + 1 ) + 66 when divided by 68 will give 66 as remainder . correct option : c" | a ) 1 , b ) 63 , c ) 66 , d ) 67 , e ) none of these | c | subtract(add(power(67, 67), 67), multiply(68, floor(divide(add(power(67, 67), 67), 68)))) | power(n0,n0)|add(n0,#0)|divide(#1,n3)|floor(#2)|multiply(n3,#3)|subtract(#1,#4)| | general | C |
if two integers x , y ( x > y ) are selected from - 10 to 4 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 15 c 2 = 15 * 14 / 2 = 105 . therefore , the answer is e ." | a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 105 | e | add(add(add(add(add(add(add(4, 10), add(4, const_2)), add(4, const_1)), 4), 10), const_2), const_1) | add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)| | probability | E |
a sum fetched a total simple interest of 4025.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4025.25 ) / ( 9 x 5 ) = 402525 / 45 = 8945 . answer d" | a ) 5768 , b ) 8925 , c ) 2345 , d ) 8945 , e ) 8723 | d | divide(divide(multiply(4025.25, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain | D |
a certain experimental mathematics program was tried out in 2 classes in each of 22 elementary schools and involved 31 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and greatest possible values of n , respectively , are | "one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 22 * 2 = 44 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one class ( 44 - 31 = 13 ) . thus we get a number of the available extra - classes for teachers , and all that we need is just to count how many teachers could take 2 more classes , which is 13 / 2 = 6.5 . so the greatest possible value of the # of teachers who had 3 classes is 6 . only answer d has this option ." | a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 6 , e ) 2 and 8 | d | divide(subtract(multiply(22, 2), 31), 2) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,n0)| | general | D |
an apple orchard grows red and green apples . on the first day of harvest , 496 apples are collected . there are 3 times as many red apples as there are green apples . how many green apples were harvested ? | let x be the number of green apples . there were 496 total apples ; there are three times as many red apples as green apples . therefore , x + 3 x = 496 4 x = 496 x = 124 answer : d | a ) 99 , b ) 112 , c ) 118 , d ) 124 , e ) 130 | d | divide(496, add(3, const_1)) | add(n1,const_1)|divide(n0,#0) | physics | D |
a vendor sells 50 percent of the apples he had and throws away 20 percent of the remainder . the next day , the vendor sells 50 percent of the remaining apples and throws away the rest . in total , what percent of his apples does the vendor throw away ? | "let x be the original number of apples . on day one , the vendor throws away ( 0.2 ) ( 0.5 ) x = 0.1 x . the remaining apples are ( 0.8 ) ( 0.5 ) x = 0.4 x . on day two , the vendor throws away ( 0.5 ) ( 0.4 ) x = 0.2 x . the vendor throws away a total of 0.1 x + 0.2 x = 0.3 x . the vendor throws away 30 percent of the apples . the answer is a ." | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | a | add(divide(subtract(subtract(const_100, 50), multiply(divide(20, const_100), subtract(const_100, 50))), const_2), multiply(divide(20, const_100), subtract(const_100, 50))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|divide(#3,const_2)|add(#4,#2)| | general | A |
what is the smallest number which when diminished by 8 , is divisible by 9 , 6 , 12 and 18 ? | "explanation : required number = lcm of ( 9 , 6 , 12 and 18 ) + 8 = 36 + 8 = 44 answer : option c" | a ) 50 , b ) 45 , c ) 44 , d ) 60 , e ) 65 | c | add(lcm(lcm(9, 6), lcm(12, 18)), 8) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general | C |
a and b together can do a piece of work in 3 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "b = 1 / 2 β 1 / 20 = 0.28 days answer : a" | a ) 0.28 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | a | inverse(subtract(inverse(3), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics | A |
find the middle one when the sum of 3 consecutive even numbers is 57 ? | 3 consecutive numbers can be a - 1 , a , a + 1 so sum of numbers = 3 a = 57 . hence a = 19 . c | a ) 10 , b ) 12 , c ) 19 , d ) 15 , e ) 16 | c | add(divide(subtract(57, add(3, 3)), 3), const_2) | add(n0,n0)|subtract(n1,#0)|divide(#1,n0)|add(#2,const_2) | physics | C |
a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . what is the length of the bridge ( in metres ) ? | speed = 5 km / hr time = 15 minutes = 1 / 4 hour length of the bridge = distance travelled by the man = speed Γ time = 5 Γ 1 / 4 km = 5 Γ ( 1 / 4 ) Γ 1000 metre = 1250 metre answer is b | a ) 1280 , b ) 1250 , c ) 1320 , d ) 1340 , e ) 1350 | b | multiply(divide(multiply(5, const_1000), const_60), 15) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1) | physics | B |
a car started running at a speed of 28 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | "a 37 km the total distance covered by the car in the first 10 hours = 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 = sum of 10 terms in ap whose first term is 28 and last term is 46 = 10 / 2 [ 28 + 46 ] = 370 km ." | a ) 37 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(28, multiply(2, 10)), 28), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)| | physics | A |
a and b undertake to do a piece of work for rs . 1200 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of c . | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a β s share = rs . ( 1200 * 4 / 8 ) = rs . 600 , b ' s share = rs . ( 1200 * 3 / 8 ) = rs . 450 c ' s share = rs . [ 1200 - ( 300 + 225 Β» ) = rs . 150 . answer is a" | a ) 150 , b ) 450 , c ) 750 , d ) 800 , e ) none of them | a | multiply(divide(const_1, add(add(const_4, 3), const_1)), 1200) | add(n3,const_4)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2)| | physics | A |
a man walking at the rate of 5 km / hr crosses a bridge in 15 minutes . the length of the bridge ( in metres ) is | "explanation : speed = ( 5 Γ 5 / 18 ) m / sec = 25 / 18 m / sec . distance covered in 15 minutes = ( 25 / 18 Γ 15 Γ 60 ) m = 1250 m . answer : d" | a ) 600 , b ) 750 , c ) 1000 , d ) 1250 , e ) none of these | d | multiply(divide(multiply(5, const_1000), const_60), 15) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | physics | D |
a car travels at a speed of 65 miles per hour . how far will it travel in 7 hours ? | "during each hour , the car travels 65 miles . for 7 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 = 7 Γ 65 = 455 miles correct answer is c ) 455 miles" | a ) 125 miles , b ) 225 miles , c ) 455 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 7) | multiply(n0,n1)| | physics | C |
if a ( a - 6 ) = 27 and b ( b - 6 ) = 27 , where a β b , then a + b = | "i . e . if a = - 3 then b = 9 or if a = 9 then b = - 3 but in each case a + b = - 3 + 9 = 6 answer : option c" | a ) β 48 , b ) β 2 , c ) 6 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(6, 27), subtract(6, 27)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general | C |
the ratio of the volumes of two cubes is 1728 : 2197 . what is the ratio of their total surface areas ? | "ratio of the sides = Γ’ Β³ Γ’ Λ Ε‘ 1728 : Γ’ Β³ Γ’ Λ Ε‘ 2197 = 12 : 13 ratio of surface areas = 144 : 169 answer : b" | a ) 12 : 18 , b ) 144 : 169 , c ) 44 : 22 , d ) 28 : 144 , e ) 96 : 12 | b | power(divide(1728, 2197), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry | B |
288 Γ£ Β· ? Γ£ β 15 + 270 = 405 | "explanation : 288 Γ£ Β· ? Γ£ β 15 = 405 - 270 = 135 ( 288 Γ£ β 15 ) / ? = 135 ? = ( 288 Γ£ β 15 ) / 135 = 32 answer : option c" | a ) 24 , b ) 25 , c ) 32 , d ) 39 , e ) 43 | c | add(add(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), divide(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), const_10)), const_1) | subtract(n3,n2)|add(n1,#0)|multiply(#1,const_2)|divide(#2,const_10)|divide(#3,const_10)|add(#3,#4)|add(#5,const_1)| | general | C |
a plant manager must assign 6 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ? | "whatever : my take selecting team of 2 out of 10 to assign to the shifts = 6 c 2 = 15 ways . now 2 out of 10 means total of 3 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 3 * 2 * 1 = 6 here alt and alt are the same : so 6 / 2 = 3 ways . total ways of selecting = ( selecting 2 out of 6 ) * arranging those teams in shifts = 15 * 3 = 45 ans : d" | a ) 24 , b ) 27 , c ) 33 , d ) 45 , e ) 54 | d | multiply(divide(factorial(divide(6, 2)), const_2), divide(factorial(6), multiply(factorial(subtract(6, 2)), factorial(2)))) | divide(n0,n1)|factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#0)|factorial(#3)|divide(#4,const_2)|multiply(#5,#2)|divide(#1,#7)|multiply(#6,#8)| | physics | D |
the sum of the ages of 5 children born at the intervals of 3 years each is 70 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 70 5 x = 40 = > x = 8 . age of youngest child = x = 8 years . answer : d" | a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years | d | divide(subtract(divide(70, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general | D |
a department store perceived a run on a lower priced garment and raised the price by 20 % to keep stock available until another shipment arrived . customers fumed and the department store manager allowed them to purchase the garments at 80 % of the new price to save face . what was the difference in the new retail price and the price the customers were allowed to pay instead ? | quantity x rate = price 1 x 1 = 1 0.8 x 1.20 = 0.960 decrease in price = ( 0.040 / 1 ) Γ£ β 100 = 4.0 % answer = option a | a ) 4.0 % , b ) 5.0 % , c ) 6.0 % , d ) 7.0 % , e ) 8.0 % | a | multiply(subtract(const_1, multiply(divide(80, const_100), divide(add(const_100, 20), const_100))), const_100) | add(n0,const_100)|divide(n1,const_100)|divide(#0,const_100)|multiply(#1,#2)|subtract(const_1,#3)|multiply(#4,const_100) | general | A |
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