options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 5363 , b ) 8160 , c ) 2368 , d ) 1987 , e ) 2732 | b | multiply(divide(120, 10), 680) | in order to obtain an income of rs . 680 from 10 % stock at rs . 120 , one must make an investment of | "to obtain rs . 10 , investment = rs . 120 . to obtain rs . 680 , investment = = rs . 8160 . answer : b" | a = 120 / 10
b = a * 680
|
a ) 14 , b ) 12 , c ) 13 , d ) 15 , e ) 16 | d | divide(add(sqrt(add(multiply(multiply(105, const_2), const_4), const_1)), const_1), const_2) | in a party every person shakes hands with every other person . if there are 105 hands shakes , find the number of person in the party | "let n be the number of persons in the party . number of hands shake = 105 ; total number of hands shake is given by nc 2 . now , according to the question , nc 2 = 105 ; or , n ! / [ 2 ! * ( n - 2 ) ! ] = 105 ; or , n * ( n - 1 ) / 2 = 105 ; or , n 2 - n = 210 ; or , n 2 - n - 210 = 0 ; or , n = 15 , - 14 ; but , we can not take negative value of n . so , n = 15 i . e . number of persons in the party = 15 . answer d" | a = 105 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
a ) 0.2 % , b ) 0.8 % , c ) 1.2 % , d ) 1.6 % , e ) 2 % | c | multiply(subtract(multiply(add(divide(10, const_100), const_1), subtract(const_1, divide(8, const_100))), const_1), const_100) | the price of stock decreased by 8 % last year and rose by 10 % this year . what is the net percentage change in the price of the stock ? | ( 100 % - 8 % ) * ( 100 % + 10 % ) = 0.92 * 1.10 = 1.012 = 101.2 % . the net percentage change in the price of the stock is ( + ) 1.2 % the answer is c | a = 10 / 100
b = a + 1
c = 8 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 5 days , b ) 3 days , c ) 2 days , d ) 9 days , e ) 10 days | a | divide(const_1, multiply(multiply(4, 7), divide(subtract(divide(3, multiply(3, 7)), divide(4, multiply(5, 7))), 4))) | two men and three women working 7 hours a day finish a work in 5 days . four men and four women working 3 hours a day complete the work in 7 days . the number of days in which only 7 men working 4 hours a day will finish the work is ? | answer : option a 2 m + 3 w - - - - - 35 h 4 m + 4 w - - - - - - - 21 h 7 m - - - - - - - ? d 70 m + 105 w = 84 m + 84 m 21 w = 14 m = > 2 m = 3 w 4 * 35 = 7 * x = > x = 20 hours 20 / 4 = 5 days | a = 4 * 7
b = 3 * 7
c = 3 / b
d = 5 * 7
e = 4 / d
f = c - e
g = f / 4
h = a * g
i = 1 / h
|
a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 220 metres , e ) 250 metres | d | multiply(19.8, multiply(40, const_0_2778)) | a train is running at a speed of 40 km / hr and it crosses a post in 19.8 seconds . what is the length of the train ? | "speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 19.8 s distance covered , d = vt = ( 400 / 36 ) Γ£ β 19.8 = 220 m distance covered is equal to the length of the train = 220 m correct answer is 220 metres d" | a = 40 * const_0_2778
b = 19 * 8
|
a ) 79 , b ) 86 , c ) 95 , d ) 72 , e ) 80 | b | subtract(subtract(multiply(20, 15), multiply(5, 14)), multiply(9, 16)) | the average age of 20 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ? | "age of the 15 th student = 20 * 15 - ( 14 * 5 + 16 * 9 ) = 300 - 214 = 86 years answer is b" | a = 20 * 15
b = 5 * 14
c = a - b
d = 9 * 16
e = c - d
|
a ) 144 , b ) 36 , c ) 72 , d ) 56 , e ) 112 | c | multiply(add(8, 1), add(1, 7)) | in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 8 , 1 ) , ( 1 , 1 ) , ( 1 , - 7 and ( - 8 , - 7 ) ? | "length of side 1 = 8 + 1 = 9 length of side 2 = 7 + 1 = 8 area of rectangle = 9 * 8 = 72 c is the answer" | a = 8 + 1
b = 1 + 7
c = a * b
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | d | subtract(subtract(multiply(6, divide(add(multiply(15, 3), multiply(15, 1)), subtract(multiply(6, 1), 3))), 15), add(divide(add(multiply(15, 3), multiply(15, 1)), subtract(multiply(6, 1), 3)), 15)) | on a certain farm the ratio of horses to cows is 6 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 3 : 1 . after the transaction , how many more horses than cows would the farm own ? | "originally , there were 6 k horses and k cows . ( 6 k - 15 ) = 3 ( k + 15 ) 6 k - 3 k = 45 + 15 3 k = 60 k = 20 the difference between horses and cows is ( 6 k - 15 ) - ( k + 15 ) = 5 k - 30 = 70 the answer is d ." | a = 15 * 3
b = 15 * 1
c = a + b
d = 6 * 1
e = d - 3
f = c / e
g = 6 * f
h = g - 15
i = 15 * 3
j = 15 * 1
k = i + j
l = 6 * 1
m = l - 3
n = k / m
o = n + 15
p = h - o
|
a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) none of these | c | divide(multiply(195, 250), multiply(1000, 65)) | if a * b * c = 195 , b * c * d = 65 , c * d * e = 1000 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | explanation : a Γ’ Λ β b Γ’ Λ β c / b Γ’ Λ β c Γ’ Λ β d = 195 / 65 = > a / d = 3 d Γ’ Λ β e Γ’ Λ β f / c Γ’ Λ β d Γ’ Λ β e = 250 / 1000 = > f / c = 1 / 4 a / d * f / c = 3 * 1 / 4 = 3 / 4 answer : c | a = 195 * 250
b = 1000 * 65
c = a / b
|
a ) 7 , b ) 8 , c ) 9 , d ) 1 , e ) 2 | d | divide(divide(divide(lcm(2, 5478), 5478), const_4), const_4) | what is the least value of x , so that 2 x 5478 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 8 = 26 + x least value of x may be ' 1 ' , so that the total 26 + 1 = 27 is divisible by 9 . answer : option d" | a = math.lcm(2, 5478)
b = a / 5478
c = b / 4
d = c / 4
|
a ) 5.29 min , b ) 4.32 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(594, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 594 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 594 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 594 m apart , they take ( 100 / 825 * 594 / 1000 ) hrs = ( 594 / 8250 * 60 ) min = 4.32 min . answer : b" | a = 594 / 1000
b = 4 + 5
c = a / b
d = c * const_60
|
a ) 30 days , b ) 45 days , c ) 80 days , d ) 72 1 / 2 days , e ) 72 days | e | multiply(inverse(add(divide(const_1, 45), divide(const_1, 30))), 4) | a can do a job in 45 days and b can do it in 30 days . a and b working together will finish 4 times the amount of work in - - - - - - - days ? | 1 / 45 + 1 / 30 = 5 / 90 = 1 / 18 18 / 1 = 18 * 4 = 72 days answer : e | a = 1 / 45
b = 1 / 30
c = a + b
d = 1/(c)
e = d * 4
|
a ) 4 , b ) 3 , c ) 8 , d ) 7 , e ) 2 | b | multiply(subtract(divide(power(37, const_2), 357), floor(divide(power(37, const_2), 357))), 357) | on dividing a number by 357 , we get 37 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 37 = ( 17 * 21 * y ) + ( 17 * 2 ) + 3 = 17 * ( 21 y + 2 ) + 3 . required number = 3 . answer is b" | a = 37 ** 2
b = a / 357
c = 37 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
|
a ) $ 15,360 , b ) $ 17,360 , c ) $ 25,560 , d ) $ 21,360 , e ) $ 27,360 | c | add(multiply(const_100, 3), const_60) | a certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360 . what amount of total annual earnings would result in a 55 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) | "for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12000 - 0.55 * 12000 = 12000 - 6600 x = 5400 earning becomes 9360 + 3 x = 9360 + 3 * 5400 = 25560 ans : c" | a = 100 * 3
b = a + const_60
|
a ) 28.8 , b ) 72.9 , c ) 38.3 , d ) 78.3 , e ) 79.3 | a | subtract(subtract(45, multiply(const_3, const_3)), multiply(divide(subtract(45, multiply(const_3, const_3)), 45), multiply(const_3, const_3))) | a vessel of capacity 45 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 45 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 45 - 9 ) / 45 ] ^ 2 ( 45 ) = 28.8 litres . answer : option a" | a = 3 * 3
b = 45 - a
c = 3 * 3
d = 45 - c
e = d / 45
f = 3 * 3
g = e * f
h = b - g
|
a ) 16 , b ) 8 , c ) 3 , d ) 2 , e ) 1 | c | divide(divide(12, const_2), const_2) | if circles x and y have the same area and circle x has a circumference of 12 Ο , half of the radius of circle y is : | "x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 12 * pi x = 6 thus y = 6 y / 2 = 3 ans : c" | a = 12 / 2
b = a / 2
|
a ) 3.5 kmph . , b ) 2.5 kmph . , c ) 1.2 kmph . , d ) 1.54 kmph . , e ) 1.9 kmph . | d | multiply(const_3_6, divide(18, 42)) | convert the 18 / 42 m / s into kilometers per hour ? | "18 / 42 m / s = 18 / 42 * 18 / 5 = 1 ( 27 / 50 ) = 1.54 kmph . answer : d" | a = 18 / 42
b = const_3_6 * a
|
a ) 12 , b ) 14 , c ) 7 , d ) 11 , e ) 10 | d | divide(multiply(subtract(const_100, 1), 16), add(const_100, 44)) | if a man lost 1 % by selling oranges at the rate of 16 a rupee at how many a rupee must he sell them to gain 44 % ? | "99 % - - - - 16 144 % - - - - ? 99 / 144 * 16 = 11 answer : d" | a = 100 - 1
b = a * 16
c = 100 + 44
d = b / c
|
a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 35 % , e ) 50 % | d | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 34))), multiply(8, divide(const_1, 34))), multiply(8, divide(const_1, 34))), const_100) | a certain car uses one gallon of gasoline every 34 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | 4 miles on the highway = 4 / 34 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 34 + 4 / 20 = 27 / 85 gallons . 8 miles on the highway = 4 / 17 gallons . the % change = ( 27 / 85 - 4 / 17 ) / ( 4 / 17 ) = 0.35 . answer : d . | a = 1 / 20
b = a * 4
c = 1 / 34
d = 4 * c
e = b + d
f = 1 / 34
g = 8 * f
h = e - g
i = 1 / 34
j = 8 * i
k = h / j
l = k * 100
|
a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.28 % | e | subtract(subtract(9, 8), divide(multiply(9, 8), const_100)) | in measuring the sides of a rectangle , one side is taken 9 % in excess , and the other 8 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 12 / 11 ) x ( 23 / 25 ) y = ( 358 / 357 ) ( xy ) error in measurement = ( 358 / 357 ) xy - xy = ( 1 / 357 ) xy error percentage = [ ( 1 / 357 ) xy ( 1 / xy ) 100 ] % = ( 7 / 25 ) % = 0.28 % . answer is e ." | a = 9 - 8
b = 9 * 8
c = b / 100
d = a - c
|
a ) 2160 , b ) 3120 , c ) 17988 , d ) 6240 , e ) 5240 | c | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(50, 6)), 3)), multiply(multiply(add(const_1, divide(divide(50, 6), const_100)), add(const_1, divide(divide(50, 6), const_100))), add(const_1, divide(divide(50, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(50, 6)), 3))) | there is 50 % increase in an amount in 6 years at s . i . what will be the c . i . of rs . 12,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 50 and t = 6 years . r = ( 100 * 50 ) / ( 100 * 6 ) = 10 % p . a . now , p = rs . 12000 , t = 3 years and r = 8.33 % p . a . c . i . = [ 12000 * { ( 1 + 8.33 / 100 ) 3 - 1 } ] = rs . 17988 answer : c" | a = 100 * 100
b = 50 / 6
c = 100 * b
d = c * 3
e = a + d
f = 50 / 6
g = f / 100
h = 1 + g
i = 50 / 6
j = i / 100
k = 1 + j
l = h * k
m = 50 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 50 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
|
a ) 5 % , b ) 11 % , c ) 16 % , d ) 21 % , e ) 19 % | c | multiply(divide(subtract(const_100, 86), 86), const_100) | if the cost price is 86 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 86 profit = $ 14 profit % = 14 / 86 * 100 = 25 / 6 = 16 % approximately answer is c" | a = 100 - 86
b = a / 86
c = b * 100
|
a ) 1250 , b ) 2489 , c ) 6398 , d ) 15840 , e ) 2487 | d | multiply(square_perimeter(square_edge_by_area(69696)), 15) | find the length of the wire required to go 15 times round a square field containing 69696 m ^ 2 . | "explanation : a 2 = 69696 = > a = 264 4 a = 1056 1056 * 15 = 15840 answer : d" | a = square_perimeter * (
|
a ) 14 , b ) 10 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 8), multiply(3, 4)), 2) | the average of 5 quantities is 8 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | "answer : a ( 5 x 8 - 3 x 4 ) / 2 = 14" | a = 5 * 8
b = 3 * 4
c = a - b
d = c / 2
|
['a ) 5 cm', 'b ) 13 β 2 cm', 'c ) 12 cm', 'd ) 6 cm', 'e ) 8 cm'] | b | sqrt(multiply(add(power(divide(48, const_4), const_2), power(divide(20, const_4), const_2)), const_2)) | the perimeter of one square is 48 cm and that of another is 20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | explanation : 4 a = 48 4 a = 20 a = 12 a = 5 a ^ 2 = 144 a ^ 2 = 25 combined area = a ^ 2 = 169 = > a = 13 d = 13 β 2 cm answer : b | a = 48 / 4
b = a ** 2
c = 20 / 4
d = c ** 2
e = b + d
f = e * 2
g = math.sqrt(f)
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 14), 2)), 1), 3), add(multiply(12, 1), multiply(14, 2)))) | in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 14 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | the class borrowed a total of 40 * 2 = 80 books . the 28 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 14 * 2 = 40 . to maximize the number of books borrowed by 1 student , let ' s assume that 11 students borrowed 3 books and 1 student borrowed the rest . 80 - 40 - 3 * 11 = 7 the maximum number of books borrowed by any student is 7 . the answer is e . | a = 40 * 2
b = 12 * 1
c = b + 14
d = c + 2
e = 40 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 14 * 2
j = h + i
k = g + j
l = a - k
|
a ) 240 , b ) 75 , c ) 85 , d ) 130 , e ) 200 | c | add(divide(subtract(add(divide(subtract(240, 30), const_2), 30), 30), const_2), 30) | a picnic attracts 240 persons . there are 30 more men than women , and 30 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 30 i . e . y + ( y + 30 ) = 210 i . e . y = 115 i . e . adult = 115 + 30 = 145 adults include only men and women i . e . men + women = 145 let women , w = x then men , m = x + 30 i . e . x + ( x + 30 ) = 2 x + 30 = 145 i . e . x = 55 i . e . men , m = 55 + 30 = 85 answer : option c" | a = 240 - 30
b = a / 2
c = b + 30
d = c - 30
e = d / 2
f = e + 30
|
a ) 40 , b ) 400 , c ) 800 , d ) 1600 , e ) none | a | divide(80, 20) | find 80 / ? = ? / 20 | "answer let 80 / y = y / 20 then , y 2 = 80 x 20 = 1600 β΄ y = β 1600 = 40 . option : a" | a = 80 / 20
|
a ) r = 5 , b ) r = 9 , c ) r = 10 , d ) r = 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | what is the greatest positive integer r such that 3 ^ r is a factor of 9 ^ 10 ? | "what is the greatest positive integer r such that 3 ^ r is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a = 9 - 10
b = a * 10
|
a ) 15 , b ) 70 , c ) 20 , d ) 60 , e ) none | d | divide(add(add(40, const_4), subtract(80, const_4)), const_2) | find the average of all the numbers between 40 and 80 which are divisible by 3 . | "sol . average = ( 42 + 45 + 48 + 51 + 54 + 57 + 60 + 63 + 66 + 69 + 72 + 75 + 78 / 13 ) = 780 / 5 = 60 . answer d" | a = 40 + 4
b = 80 - 4
c = a + b
d = c / 2
|
a ) 1 / 9 , b ) 4 / 45 , c ) 2 / 7 , d ) 6 / 3 , e ) 5 / 2 | b | add(divide(const_1, 18), divide(const_1, 30)) | a can do a job in 18 days and b can do it in 30 days . in how many days will they finish it together ? | explanation : 1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 so , both together will finish the work in 4 / 45 days . answer : b | a = 1 / 18
b = 1 / 30
c = a + b
|
['a ) 1.33', 'b ) 1.25', 'c ) 8.0', 'd ) 12.5', 'e ) 80.0'] | a | subtract(divide(power(const_100, const_3), multiply(300, const_1000)), const_2) | the mass of 1 cubic meter of a substance is 300 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | density is mass divided by volume . so density of the given substance will be mass / volume = 800 kg / 1 m ^ 3 = 300 kg / m ^ 3 or 1 g / 1.33 cm ^ 3 = 0.3 g / cm ^ 3 . next , ask yourself if 300,000 g is equivalent to 1 , 000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters ? - - > 1 g - 1 , 000,000 / 300,000 = 10 / 3 = 1.33 cubic centimeters . answer is a | a = 100 ** 3
b = 300 * 1000
c = a / b
d = c - 2
|
a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | add(multiply(multiply(divide(15, const_100), 80), multiply(divide(15, const_100), 80)), divide(subtract(300, 80), 80)) | in a 300 member association consisting of men and women , exactly 80 % of men and exactly 15 % women are homeowners . what is the least number of members who are homeowners ? | "solution simple out of 300 80 % are male i . e 240 and 15 % are female i . e 45 , so total homeowner is 295 . now min number homeowner is 45 and max is 240 so question ask us to find least and 46 has least value among all option . so ans is 46 . ans : a" | a = 15 / 100
b = a * 80
c = 15 / 100
d = c * 80
e = b * d
f = 300 - 80
g = f / 80
h = e + g
|
['a ) between 9 and 10', 'b ) between 10 and 11', 'c ) between 11 and 12', 'd ) between 12 and 13', 'e ) between 13 and 14'] | a | sqrt(add(power(divide(add(5, sqrt(add(power(5, const_2), multiply(36, const_4)))), const_2), const_2), power(subtract(divide(add(5, sqrt(add(power(5, const_2), multiply(36, const_4)))), const_2), 5), const_2))) | mark has a rectangular driveway with a line painted diagonally across . the length of the driveway is 5 feet longer than its width . if the total area of the driveway is 36 square feet , what is the length of the painted line ( in feet ) ? | to solve this , we first need to set up an equation for the area of the room . if x is the width , then we have x ( x + 5 ) = 36 . by putting the equation in standard form , we get x ^ 2 + 5 x - 36 = 0 . by using the quadratic formula , we get roots of 4 and - 9 . we know that x is the width , and x + 5 is the length , so by using the roots , we get 4 as the width ( x ) , and 9 as the length ( x + 5 ) . once we have this , we can use the pythagorean theorem to solve for the diagonal . plugging in the length and width , we will get d ^ 2 = 4 ^ 2 + 9 ^ 2 = 16 + 81 = 97 the square root of 97 is less than 10 , since 10 ^ 2 = 100 . going down a number , we can compute 9 ^ 2 = 81 < 97 . therefore , the length of the diagonal must be between 9 and 10 . the correct answer is a . | a = 5 ** 2
b = 36 * 4
c = a + b
d = math.sqrt(c)
e = 5 + d
f = e / 2
g = f ** 2
h = 5 ** 2
i = 36 * 4
j = h + i
k = math.sqrt(j)
l = 5 + k
m = l / 2
n = m - 5
o = n ** 2
p = g + o
q = math.sqrt(p)
|
a ) 6 : 26 , b ) 6 : 56 , c ) 7 : 26 , d ) 7 : 56 , e ) 8 : 26 | c | divide(add(4, divide(multiply(add(subtract(4, 3), divide(subtract(26, 20), const_60)), 30), subtract(41, 30))), 26) | tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 41 pages per hour . if tom starts reading a novel at 3 : 20 , and jan begins reading an identical copy of the same book at 4 : 26 , at what time will they be reading the same page ? | since tom reads an average of 1 page every 2 minutes , tom will read 33 pages in the first 66 minutes . jan can catch tom at a rate of 11 pages per hour , so it will take 3 hours to catch tom . the answer is c . | a = 4 - 3
b = 26 - 20
c = b / const_60
d = a + c
e = d * 30
f = 41 - 30
g = e / f
h = 4 + g
i = h / 26
|
a ) 90 , b ) 115 , c ) 120 , d ) 130 , e ) 220 | b | subtract(divide(subtract(multiply(12, 355), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(355, divide(subtract(multiply(12, 355), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)))) | a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 355 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ? | "orchestra seats - a balcony seats - b a + b = 355 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 355 = 3320 - 2840 = 480 i . e . a = 120 and b = 355 - 120 = 235 more seats in balcony than orchestra = b - a = 235 - 120 = 115 answer : option b" | a = 12 * 355
b = 3 * 1000
c = 3 * 100
d = b + c
e = 2 * 10
f = d + e
g = a - f
h = 12 - 8
i = g / h
j = 12 * 355
k = 3 * 1000
l = 3 * 100
m = k + l
n = 2 * 10
o = m + n
p = j - o
q = 12 - 8
r = p / q
s = 355 - r
t = i - s
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | subtract(add(6, 8), 6) | a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 6 years ago ? | "let ' s say age of richard isr age of david isd age of scott iss now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6 = ( s + 8 ) + 6 = s + 14 therefore , 2 s + 8 = s + 14 i . e . s = 6 i . e . r = 20 i . e . d = 14 now , how old was david 6 years ago ? i . e . d - 6 = 14 - 6 = 8 years answer : option a" | a = 6 + 8
b = a - 6
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a ) 1 / 5 , b ) 29 / 35 , c ) 5 / 6 , d ) 35 / 29 , e ) 35 / 12 | d | inverse(add(inverse(divide(5, 2)), inverse(divide(7, 3)))) | a type t machine can complete a job in 5 hours and a type b machine can complete the job in 7 hours . how many hours will it take 2 type t machines and 3 type b machines working together and independently to complete the job ? | now d should be the answer . t need 5 hours to complete and b needs 7 hours to compete so 2 t + 3 b will complete 2 / 5 + 3 / 7 or 29 / 35 portion of the job in 1 hour so the whole job will take 35 / 29 hours . . . . = d | a = 5 / 2
b = 1/(a)
c = 7 / 3
d = 1/(c)
e = b + d
f = 1/(e)
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a ) 15 , b ) 21 , c ) 18 , d ) 20 , e ) 24 | b | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | on a certain date , pat invested $ 10,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 40,000 , in how many years , the total value of the investment plus interest will increase to $ 120,000 ? | "if i were to choose during the test , would go for 18 or 20 . probably 18 cuz it wont take too long to get the value doubled . . . . i found a method : rule of 72 . given an x % return , it takes 10,000 to quadralope 12 years . so according to the rule : 72 / x is the no of years 10 , 000.00 took to double 20 , 000.00 . again , 20 , 000.00 took to double 40 , 000.00 same ( 72 / x ) no of years . 72 / x + 72 / x = 12 x = 12 % ( though rate here is not very much required ) . again , 40 , 000.00 takes the same ( 72 / x ) no of years to double 120 , 000.00 . 72 / x = 6 years . so altogather : 10,000 - 20,000 = 6 years 20,000 - 40,000 = 6 years 40,000 - 80,000 = 6 years 80,000 - 120,000 = 3 years total 21 years . answer b" | a = 3 * 10
b = 4 + 1
c = a / b
d = math.log(c)
e = 2 * 10
f = 4 + 1
g = e / f
h = 1 / 12
i = g ** h
j = math.log(i)
k = d / j
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a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 28 | e | divide(subtract(power(16, const_2), 200), const_2) | if the sum of two numbers is 16 and the sum of their squares is 200 , then the product of the numbers is | "according to the given conditions x + y = 16 and x ^ 2 + y ^ 2 = 200 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 16 ^ 2 = 200 + 2 xy so xy = 56 / 2 = 28 answer : e" | a = 16 ** 2
b = a - 200
c = b / 2
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a ) 33 , b ) 878 , c ) 30 , d ) 88 , e ) 12 | e | inverse(subtract(6, divide(6, 12))) | a and b can do a piece of work in 12 days . with the help of c they finish the work in 6 days . c alone can do that piece of work in ? | "c = 1 / 6 Γ’ β¬ β 1 / 12 = 1 / 12 = > 12 days answer : e" | a = 6 / 12
b = 6 - a
c = 1/(b)
|
a ) 2788 , b ) 5000 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(4000, subtract(7, 3))) | an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 4000 , what will be the difference between q and r ' s share ? | "4 - - - 4000 5 - - - ? = > 5000 answer : b" | a = 12 - 7
b = 7 - 3
c = 4000 / b
d = a * c
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a ) 648 , b ) 1800 , c ) 2600 , d ) 10800 , e ) 10900 | c | multiply(multiply(divide(390, 6), 4), 10) | running at the same constant rate , 6 identical machines can produce a total of 390 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 390 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 390 x = ( 10 x 4 x 390 ) / ( 6 ) x = 2600 . answer : c" | a = 390 / 6
b = a * 4
c = b * 10
|
a ) 16 , b ) 10.34 , c ) 12.46 , d ) 11.2 , e ) 14 | b | multiply(const_100, divide(subtract(32, multiply(14.5, 2)), multiply(14.5, 2))) | a certain pair of used shoes can be repaired for $ 14.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 32.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? | 1 ) cost of repairing = 14.5 ( for one year ) , therefore for 2 years it would be $ 29 . 2 ) cost of new pair which will last for 2 years is $ 32 . percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoeinitial value would be the cost of repairing the old shoe . i . e ( 32 - 29 ) / ( 29 ) * 100 = 10.34 % . ans is b | a = 14 * 5
b = 32 - a
c = 14 * 5
d = b / c
e = 100 * d
|
['a ) it will have no effect', 'b ) 1 % increase', 'c ) 1 % decrease', 'd ) 2 % increase', 'e ) 2 % decrease'] | c | subtract(add(divide(multiply(10, 10), const_100), 10), 10) | if length of rectangle is increased by 10 % , and breadth is decreased by 10 % . what will it effect on area of rectangle ? | let initial length = l , and breadth = b area = l * b new length = l * 110 % = [ l * 110 / 100 ] = 11 l / 10 , new breadth = 9 b / 10 new area = 11 l / 10 * 9 b / 10 = 99 / 100 lb decrease in area = [ lb - 99 / 100 lb ] = lb / 100 = % decrease = [ lb / 100 * 1 / lb * 100 ] = 1 % answer c | a = 10 * 10
b = a / 100
c = b + 10
d = c - 10
|
a ) 1 / 4 , b ) 3 / 4 , c ) 1 / 5 , d ) 2 / 2 , e ) 3 / 1 | c | divide(12, const_60) | what is the ratio of 12 minute to 1 hour ? | c 1 / 5 | a = 12 / const_60
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(8, 8) | 8 * 8 * 8 * 8 = 2 ^ ? | "solution : 2 ^ 3 * 2 ^ 3 * 2 ^ 3 * 2 ^ 3 = 2 ^ ( 3 + 3 + 3 + 3 ) = 2 ^ 12 answer : 12 option : e" | a = 8 * 8
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a ) 5 , b ) 4 , c ) 0 , d ) 1 , e ) 2 | c | divide(multiply(53403977, 11), 53403977) | what is the smallest no . which must be added to 53403977 so as to obtain a sum which is divisible by 11 ? | "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 53403977 , difference = ( 5 + 4 + 3 + 7 ) - ( 3 + 0 + 9 + 7 ) = 19 - 19 = 0 . = > 53403977 is also divisible by 11 c" | a = 53403977 * 11
b = a / 53403977
|
a ) 22 , b ) 56 , c ) 12 , d ) 36 , e ) 10 | b | multiply(7, 8) | each child has 8 crayons and 15 apples . if there are 7 children , how many crayons are there in total ? | 8 * 7 = 56 . answer is b . | a = 7 * 8
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a ) 79 , b ) 89 , c ) 95 , d ) 81.81 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 60)), add(6, 5))) | a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 60 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 60 * 5 = 300 student ' s average = ( 600 + 300 ) / 11 = 81.81 answer d" | a = 6 * 100
b = 5 * 60
c = a + b
d = 6 + 5
e = c / d
f = math.floor(e)
|
a ) 53 km , b ) 55 km , c ) 52 km , d ) 60 km , e ) 50 km | e | multiply(20, 2.5) | a person is traveling at 20 km / hr and reached his destiny in 2.5 hr then find the distance ? | "t = 2.5 hrs = 5 / 2 hrs d = t * s = 20 * 5 / 2 = 50 km answer is e" | a = 20 * 2
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a ) 22 cm , b ) 47.14 cm , c ) 84.92 cm , d ) 94.94 cm , e ) 23.57 cm | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(16, 12)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 12 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 16 + 12 ) = 56 cm parameter of the square = 56 cm i . e . 4 a = 56 a = 14 diameter of the semicircle = 14 cm circimference of the semicircle = 1 / 2 ( β ) ( 14 ) = 1 / 2 ( 22 / 7 ) ( 14 ) = 308 / 14 = 22 cm answer : a" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) 2 : 8 , b ) 1 : 2 , c ) 2 : 5 , d ) 6 : 9 , e ) 6 : 2 | b | divide(divide(192, divide(192, 32)), divide(192, subtract(divide(192, 32), const_3))) | a motorcyclist goes from bombay to pune , a distance of 192 kms at an average of 32 kmph speed . another man starts from bombay by car 2 Β½ hours after the first , and reaches pune Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | b 1 : 2 t = 192 / 32 = 6 h t = 6 - 3 = 3 time ratio = 6 : 3 = 2 : 1 speed ratio = 1 : 2 | a = 192 / 32
b = 192 / a
c = 192 / 32
d = c - 3
e = 192 / d
f = b / e
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a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 86 % | e | multiply(divide(subtract(divide(add(30, 8), 150), divide(8, 50)), divide(8, 50)), const_100) | a corporation paid $ 8 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 million in profits ? | "difference in ratios = ( 30 / 150 ) - ( 8 / 50 ) = ( 6 / 50 ) % change = ( change ( 6 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 86 % answer - e" | a = 30 + 8
b = a / 150
c = 8 / 50
d = b - c
e = 8 / 50
f = d / e
g = f * 100
|
a ) 6 , b ) 4 , c ) 24.5 , d ) 20.5 , e ) 12.5 | d | divide(triangle_area_three_edges(9, 40, 41), divide(triangle_perimeter(9, 40, 41), const_2)) | what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9 , 40 and 41 ? | "explanatory answer 9 , 40 and 41 is pythagorean triplet . so , the triangle is right angled . key property about right triangles in a right angled triangle , the radius of the circle that circumscribes the triangle is half the hypotenuse . in the given triangle , the hypotenuse = 41 . therefore , the radius of the circle that circumscribes the triangle = 41 / 2 = 20.5 units . choice d" | a = triangle_area_three_edges / (
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a ) one , b ) two , c ) three , d ) four , e ) five | b | add(subtract(75, floor(subtract(const_100, 25.4))), const_1) | 70 , 75 , 80 , 85 , 90 , 105 , 105 , 130 , 130 , 130 the list shown consists of the times , in seconds , that it took each of 10 schoolchildren to run a distance of 400 on of meters . if the standard devastation of the 10 running times is 25.4 seconds , rounded to the nearest tenth of a second , how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times ? | the most time consuming part in this question is to define the mean . under exam pressure and time pressure it is very easy to make mistake . it is easier to group numbers : 130 * 3 = 390 ; 105 * 2 = 210 ; 75 + 85 = 160 ; 70 + 80 = 150 ; 90 ; next stage combine results , again using more convenient ways to calculate : 390 + 210 = 600 ; 160 + 150 = 310 ; 90 . 600 + 310 + 90 = 1000 . since there are 10 numbers the mean is 100 . questions asks to find the quantity of numbers one sd below the mean , which is 100 - 25,4 = 74,6 . there are only two numbers below 74,6 . the answer is b | a = 100 - 25
b = math.floor(a)
c = 75 - b
d = c + 1
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a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | d | divide(power(divide(400, const_2), const_2), subtract(410, 400)) | if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 400 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 410 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 20,000 . now , put the value of x = 4000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 4000 + 5 % of 4000 = 4200 . 2 nd year , amount = 4200 + 5 % of 4200 = 4410 . we can see after 2 years interest = 4410 - 4000 = 410 . hence , it satisfies the question . hence d is the correct answer" | a = 400 / 2
b = a ** 2
c = 410 - 400
d = b / c
|
a ) 24 % , b ) 25.5 % , c ) 33.33 % , d ) 36.3 % , e ) 40 % | c | multiply(divide(225, subtract(900, 225)), const_100) | a cricket bat is sold for $ 900 , making a profit of $ 225 . the profit percentage would be | "225 / ( 900 - 225 ) = 33.33 % . answer : c" | a = 900 - 225
b = 225 / a
c = b * 100
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a ) 5 : 4 , b ) 1 : 4 , c ) 3 : 1 , d ) 3 : 5 , e ) 3 : 4 | e | multiply(divide(12, const_100), 16) | a part of certain sum of money is invested at 16 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 16 = 3 : 4 answer : e" | a = 12 / 100
b = a * 16
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a ) 14 / 13 , b ) 13 / 17 , c ) 16 / 17 , d ) 13 / 18 , e ) none of these | c | divide(divide(divide(4128, const_2), const_3), divide(divide(4386, const_2), const_3)) | reduce 4128 / 4386 to its lowest terms | explanation : hcf of 4128 and 4386 4128 ) 4386 ( 1 4128 - - - - - 258 ) 4128 ( 16 4128 - - - - 0 hence hcf of 4128 and 4386 is 258 . 4128 / 258 = 16 4386 / 258 = 17 4128 / 4386 = 16 / 17 answer : option c | a = 4128 / 2
b = a / 3
c = 4386 / 2
d = c / 3
e = b / d
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a ) 1 / 10 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | a | divide(choose(2, 2), choose(add(2, 3), const_2)) | a bag holds 2 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | probability of selecting first red marble = 2 / 5 probability of selecting second red marble without replacement = 1 / 4 final probability = 2 / 5 * 1 / 4 = 1 / 10 correct answer - a | a = math.comb(2, 2)
b = 2 + 3
c = math.comb(b, 2)
d = a / c
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | subtract(floor(divide(subtract(40, 7), 4)), floor(divide(subtract(1, 7), 4))) | for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 40 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 40 4 n < 33 n < 8 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 8 ) combining the two - 1.5 < n < 8.25 n = 1 to 8 ( 48 integers ) and n = - 1 and 0 so 10 integers . a" | a = 40 - 7
b = a / 4
c = math.floor(b)
d = 1 - 7
e = d / 4
f = math.floor(e)
g = c - f
|
a ) 15 , b ) 6 , c ) 13 , d ) 9 , e ) none of these | a | divide(multiply(15, 6), subtract(15, 9)) | a contractor undertook to do a piece of work in 9 days . he employed certain number of laboures but 6 of them were absent from the very first day and the rest could finish the work in only 15 days . find the number of men originally employed . | let the number of men originally employed be x . 9 x = 15 ( x β 6 ) or x = 15 answer a | a = 15 * 6
b = 15 - 9
c = a / b
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a ) 145 , b ) 146 , c ) 147 , d ) 148 , e ) 149 | c | subtract(add(multiply(5, 49), multiply(7, 52)), multiply(11, 42)) | the average of 11 results is 42 , if the average of first 5 results is 49 and that of the last 7 is 52 . find the fifth result ? | 1 to 11 = 11 * 42 = 462 1 to 5 = 5 * 49 = 245 5 to 11 = 7 * 52 = 364 5 th = 245 + 364 β 462 = 147 answer : c | a = 5 * 49
b = 7 * 52
c = a + b
d = 11 * 42
e = c - d
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a ) 400 m , b ) 450 m , c ) 560 m , d ) 600 m , e ) 680 m | d | multiply(45, multiply(48, const_0_2778)) | a train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph , in 12 seconds . it also passes a railway platform in 45 seconds . the length of the platform is | length of the ist train = 2 x m and the length of the other train be x m ( 1 / 2 ) the relative speed of the train ( 48 + 42 ) = 90 kmph or 25 m / s time taken to cross one another = 12 s hence , 12 = 2 x + x / 25 solving , x = 100 m and 200 m now to determine the length of the platform 45 = 100 + length of the platform / speed of the ist train 45 = 100 + l / 240 / 18 ( 48 kmph = 240 / 118 m / s ) solving , we get l = 600 m answer : d | a = 48 * const_0_2778
b = 45 * a
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a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | e | divide(power(2, 2), multiply(choose(5, 2), choose(2, 1))) | if x is to be chosen at random from the set { 1 , 2 , 3 } and y is to be chosen at random from the set { 5 , 6 , 7 } , 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 = 5 / 6 . answer : e ." | a = 2 ** 2
b = math.comb(5, 2)
c = math.comb(2, 1)
d = b * c
e = a / d
|
a ) 45 , b ) 60 , c ) 75 , d ) 84 , e ) 100 | d | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | if seven machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 7 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 14 / 3 = 10 / 3 * x x = 7 / 5 hour so 84 minutes answer : d" | a = 3 / 5
b = 3 / 4
c = b / 5
d = c / 30
e = d * 2
f = a / e
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a ) 0 % , b ) 20 % increase , c ) 36 % decrease , d ) 1 % decrease , e ) insufficient data | c | subtract(const_100, divide(multiply(add(const_100, 60), subtract(const_100, 60)), const_100)) | what is the % change in the area of a rectangle when its length increases by 60 % and its width decreases by 60 % ? | "( 16 / 10 ) * ( 4 / 10 ) = 64 / 100 of original area 64 / 100 is a 36 % decrease from 100 / 100 - > c" | a = 100 + 60
b = 100 - 60
c = a * b
d = c / 100
e = 100 - d
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a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 7 | c | divide(add(subtract(const_1, divide(4, 5)), subtract(const_1, divide(3, 5))), add(divide(3, 5), divide(4, 5))) | initially two cups of same volume are present with milk filled upto 3 / 5 th and 4 / 5 th of their volumes . water is then filled . then two mixtures are mixed . find the ratio of water to milk in the mixture | water content can b in 1 st and second is 2 / 5 and 1 / 5 total : 3 / 5 . the total milk content is 7 / 5 . the ratio is ( ( 3 / 5 ) / ( 7 / 5 ) ) = 3 / 7 . answer : c | a = 4 / 5
b = 1 - a
c = 3 / 5
d = 1 - c
e = b + d
f = 3 / 5
g = 4 / 5
h = f + g
i = e / h
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a ) 784596 , b ) 845796 , c ) 965604 , d ) 784596 , e ) 864520 | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 600), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 600 miles into meters ? | "1 mile = 1609.34 meters 600 mile = 600 * 1609.34 = 965604 meters answer is c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 600
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
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a ) 3.75 liters . , b ) 2.5 liters . , c ) 8.5 liters . , d ) 2.6 liters . , e ) 2.1 liters . | a | divide(multiply(divide(subtract(const_100, 90), const_100), 30), divide(subtract(const_100, 20), const_100)) | heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 30 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . Γ’ β‘ β Γ’ β‘ β 10 % ( 30 ) = 80 % ( x ) Γ’ β‘ β Γ’ β‘ β x = 3.75 liters . answer : a" | a = 100 - 90
b = a / 100
c = b * 30
d = 100 - 20
e = d / 100
f = c / e
|
a ) 7 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | a | add(add(3, const_2), const_2) | if x < y < z and y - x > 3 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we have : 1 ) x < y < z 2 ) y - x > 3 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 3 , the least value for y - x must be 5 , the least value for x must be 2 , and , thus , the least possible value for y must be 7 ( y - 2 = 5 , y = 7 ) 2 < 7 < z , since z is odd , the least possible value for z is 9 z - x = 9 - 2 = 7 answer a" | a = 3 + 2
b = a + 2
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a ) rs 1950 , b ) rs 2000 , c ) rs 745 , d ) rs 765 , e ) none | a | divide(42900, 22) | if an amount of rs 42900 is distributed equally amongst 22 persons , how much amount would each person get ? | required amount = 42900 / 22 = rs 1950 answer a | a = 42900 / 22
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a ) 4 , b ) 10 , c ) 15 , d ) 16 , e ) 26 | e | divide(10, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 10)) | six bells commence tolling together and toll at intervals of 2 , 4 , 6 , 8 10 and 12 seconds respectively . in 50 minutes , how many times do they toll together ? | "l . c . m of 2,4 , 6,8 , 10,12 is 120 . i . e after each 2 min they will toll together . so in 50 min they will toll 25 times . as they have initially tolled once , the answer will be 25 + 1 = 26 . answer : e" | a = 2 + 3
b = a * 2
c = b * 3
d = 2 * c
e = d * 2
f = e / 10
g = 10 / f
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a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | subtract(9671, multiply(floor(divide(9671, 5)), 5)) | what least number must be subtracted from 9671 so that the remaining number is divisible by 5 ? | "on dividing 9671 by 5 , we get remainder = 1 . required number be subtracted = 1 answer : e" | a = 9671 / 5
b = math.floor(a)
c = b * 5
d = 9671 - c
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a ) 90 , b ) 45 , c ) 60 , d ) 90 , e ) 26 | c | divide(add(add(add(50, const_1), add(add(50, const_1), const_2)), add(subtract(70, 50), subtract(70, const_2))), 50) | find the average of all prime numbers between 50 and 70 | "prime numbers between 50 and 70 are 53 , 59 , 61 , 67 required average = ( 53 + 59 + 61 + 67 ) / 4 = 240 / 4 = 60 answer is c" | a = 50 + 1
b = 50 + 1
c = b + 2
d = a + c
e = 70 - 50
f = 70 - 2
g = e + f
h = d + g
i = h / 50
|
a ) 3 years , b ) 4 years , c ) 6 years , d ) 7 years , e ) 8 years | e | subtract(subtract(divide(70, 5), 3), 3) | the sum of 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 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 the youngest child = x = 8 years . e )" | a = 70 / 5
b = a - 3
c = b - 3
|
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | e | subtract(floor(divide(subtract(120, 7), 4)), floor(divide(subtract(1, 7), 4))) | for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 120 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 120 4 n < 113 n < 28 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 48 ) combining the two - 1.5 < n < 48.25 n = 1 to 28 ( 48 integers ) and n = - 1 and 0 so 30 integers . e is the answer" | a = 120 - 7
b = a / 4
c = math.floor(b)
d = 1 - 7
e = d / 4
f = math.floor(e)
g = c - f
|
a ) 47 , b ) 45 , c ) 25 , d ) 37 , e ) 44 | b | subtract(multiply(add(add(6, 6), const_1), 9), add(multiply(6, 5), multiply(6, 7))) | the average of thirteen numbers is 9 . the average of first 6 numbers is 5 and the average of last 6 numbers is 7 . what is the middle number ? | the total of thirteen numbers = 13 x 9 = 117 the total of first 6 and last 6 numbers is = 6 x 5 + 6 x 7 = 72 so , the middle number is ( 117 - 72 ) = 45 b | a = 6 + 6
b = a + 1
c = b * 9
d = 6 * 5
e = 6 * 7
f = d + e
g = c - f
|
a ) 90 , b ) 130 , c ) 150 , d ) 320 , e ) 145 | e | add(90, multiply(divide(subtract(1025, 90), add(90, 80)), subtract(90, 80))) | two heavily loaded sixteen - wheeler transport trucks are 1025 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , maintaining an average speed of 80 kilometers per hour . how many kilometers farther than driver b , will driver a have driven when they meet and pass each other on the highway ? | "i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 1025 - 90 = 935 k ratio of speeds 9 : 8 - > 935 / 17 = 55 truck a = 90 + 55 * 9 = 585 truck b = 55 * 8 = 440 delta = 145 km answer e" | a = 1025 - 90
b = 90 + 80
c = a / b
d = 90 - 80
e = c * d
f = 90 + e
|
a ) 1565 , b ) 1665 , c ) 1300 , d ) 1448 , e ) 1465 | b | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 3)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 3), add(const_100, const_4)), 3), const_1), const_2)) | find the sum of all 2 digit numbers divisible by 3 . | "all 2 digit numbers divisible by 3 are 12,15 , 18,21 , β¦ β¦ β¦ . . 99 this is an a . p with a = 12 , and d = 3 , let it contain n terms then , 12 + ( n β 1 ) * 3 = 99 , or n = ( 99 - 12 ) / 3 + 1 = 30 required sum = 30 / 2 * ( 12 + 99 ) = 15 * 111 = 1665 answer : b" | a = 100 + 4
b = 100 * 10
c = b - 3
d = a + c
e = 100 * 10
f = e - 3
g = 100 + 4
h = f - g
i = h / 3
j = i + 1
k = j / 2
l = d * k
|
a ) - 6 , b ) 8 , c ) - 2 , d ) 4 , e ) 10 | a | subtract(negate(0), multiply(subtract(3, 6), divide(subtract(3, 6), subtract(2, 3)))) | 2 , 3 , 6 , 0 , 10 , - 3 , 14 , ( . . . ) | "there are two series 2 , 6 , 10 , 14 , . . . ( adding 4 ) 3 , 0 , - 3 , . . . ( subtracting 3 ) hence , next term is - 3 - 3 = - 6 answer is a" | a = negate - (
|
a ) 9,6 , b ) 6,3 , c ) 9,3 , d ) 10,2 , e ) none of these | d | divide(subtract(12, 8), const_2) | a man can row downstream at the rate of 12 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 12 + 8 ) = 10 km / hr rate of current = 1 / 2 ( 12 - 8 ) = 2 km / hr answer : option d" | a = 12 - 8
b = a / 2
|
a ) 1000 , b ) 2000 , c ) 1563 , d ) 2153 , e ) 1245 | c | add(1000, divide(multiply(1000, 25), const_100)) | the present population of a town is 1000 . population increase rate is 25 % p . a . find the population of town after 2 years ? | "p = 1000 r = 25 % required population of town = p ( 1 + r / 100 ) ^ t = 1000 ( 1 + 25 / 100 ) ^ 2 = 1000 ( 5 / 4 ) ^ 2 = 1563 ( approximately ) answer is c" | a = 1000 * 25
b = a / 100
c = 1000 + b
|
a ) s . 1014 , b ) s . 1332 , c ) s . 999 , d ) s . 1085 , e ) s . 1020 | b | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 2.0), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 2.0 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 β 6 = 666 * 2.0 = 1332 answer : b" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 2
f = e * 3
|
a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | d | divide(605, add(divide(120, const_100), const_1)) | two employees m and n are paid a total of $ 605 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 605 2.2 n = 605 n = 275 the answer is d ." | a = 120 / 100
b = a + 1
c = 605 / b
|
a ) 31 , b ) 33 , c ) 35 , d ) 37 , e ) 39 | a | divide(add(add(25, 31), add(35, 33)), 4) | the weight of 4 dogs is determined to be 25 pounds , 31 pounds , 35 pounds and 33 pounds respectively . the weight of a fifth dog is determined to be y pounds . if the average ( arithmetic mean ) weight of the first 4 dogs is the same as that of all 5 dogs what is the value of y ? | total weight of the 4 dogs = ( 25 + 31 + 35 + 33 ) = 124 avg = 124 / 4 = 31 total weight of 5 dogs = 124 + y or 4 ( 31 ) + y average of 5 dogs as per question = 31 equation : 4 ( 31 ) + y = 5 ( 31 ) , or y = 31 . choose a | a = 25 + 31
b = 35 + 33
c = a + b
d = c / 4
|
a ) 16.67 , b ) 40 , c ) 50 , d ) 60.33 , e ) 70 | b | divide(subtract(multiply(divide(40, const_100), 40), multiply(divide(20, const_100), 40)), subtract(divide(40, const_100), divide(20, const_100))) | how many ounces of a 60 % salt solution must be added to 40 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ? | "let x = ounces of 60 % salt solution to be added . 2 * 40 + . 6 x = . 4 ( 40 + x ) x = 40 amswer b" | a = 40 / 100
b = a * 40
c = 20 / 100
d = c * 40
e = b - d
f = 40 / 100
g = 20 / 100
h = f - g
i = e / h
|
a ) 3 : 7 , b ) 4 : 9 , c ) 13 : 7 , d ) 5 : 3 , e ) 6 : 11 | d | divide(multiply(50, 8), multiply(60, 4)) | car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 60 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 50 Γ 8 = 400 km car b travels 60 Γ 4 = 240 km the ratio is 400 : 240 = 40 : 24 = 5 : 3 the answer is d ." | a = 50 * 8
b = 60 * 4
c = a / b
|
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))) | 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 = 100 * 46
b = a / 50
c = 100 - b
d = 100 * 46
e = d / 50
f = c / e
g = 100 * f
|
a ) 2 miles , b ) 4 miles , c ) 6.5 miles , d ) 8 miles , e ) 10 miles | c | multiply(divide(const_1, add(divide(const_1, 5), divide(const_1, 21))), const_1_6) | johnny travels a total of one hour to and from school . on the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 21 miles per hour . how far is it to the school ? | "answer : c ) 6.5 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 5 * 21 / ( 5 + 21 ) = 6.5 m / hr the distance between schoolhome should be half of that . ie . 6.5 miles answer c" | a = 1 / 5
b = 1 / 21
c = a + b
d = 1 / c
e = d * const_1_6
|
a ) 2 , b ) 48 , c ) 0 , d ) 10 , e ) 3 | b | multiply(3, power(subtract(40, multiply(divide(add(40, 20), add(3, 2)), 3)), const_2)) | if 3 x + y = 40 , 2 x - y = 20 , for integers of x and y , 3 y ^ 2 = ? | 3 x + y = 40 2 x - y = 20 5 x = 60 x = 12 y = 4 3 y ^ 2 = 3 * 16 = 48 answer is b | a = 40 + 20
b = 3 + 2
c = a / b
d = c * 3
e = 40 - d
f = e ** 2
g = 3 * f
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | add(const_2, const_3) | how many distinct 3 digit number less than 10 | "000,001 , 002,003 , 004,005 , 006,007 , 008,009 . 10 distinct 3 digit numbers less than 10 . answer : c" | a = 2 + 3
|
a ) 10 , b ) 20 , c ) 37 , d ) 26 , e ) 22 | b | divide(subtract(multiply(15.5, 20), multiply(15, 20)), subtract(16, 15.5)) | the average age of a group of persons going for picnic is 16 years . 20 new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years . the number of persons initially going for picnic is ? | let the initial number of persons be x . then , 16 x + 20 * 15 = 15.5 ( x + 20 ) < = > 0.5 x = 10 < = > x = 20 . answer : b | a = 15 * 5
b = 15 * 20
c = a - b
d = 16 - 15
e = c / d
|
a ) 20000 , b ) 25000 , c ) 24000 , d ) 23000 , e ) 22000 | a | divide(2000, divide(10, const_100)) | in an election between two candidates , the winner has a margin of 10 % of the votes polled . if 2000 people change their mind and vote for the loser , the loser would have won by a margin of 10 % of the votes polled . find the total number of votes polled in the election ? | "winner - looser 55 % - 45 % if 2000 people change their mind and vote for the loser : winner - looser 45 % - 55 % thus 2,000 people compose 10 % of all voters , which means that the total number of votes is 20,000 . answer : a ." | a = 10 / 100
b = 2000 / a
|
a ) 19,104 , b ) 19,903 , c ) 19 , 356.732 , d ) 19,502 , e ) 19,880 | b | add(100, const_1) | if 99 < x < 199 and 10 < y < 100 , then the product xy can not be equal to : | correct answer : ( b ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 99 * 10 . the largest value must be less than 199 * 100 . this means that 990 < xy < 19,900 . ( b ) is outside of this range , so it is not a possible product of xy . | a = 100 + 1
|
a ) 14700 , b ) 14500 , c ) 14900 , d ) 16380 , e ) 14000 | d | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 39,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 β 3 x + 14000 = 50000 β 3 x = 50000 β 14000 = 36000 β x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit Γ 21 / 50 = 39000 Γ 21 / 50 = 16380 answer is d" | a = 5000 * 10
b = 4000 + 5000
c = b + 5000
d = a - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 3 * 12
i = h * 1000
j = 5000 * 10
k = i / j
l = g * k
|
a ) 80 , b ) 120 , c ) 160 , d ) 540 , e ) 660 | d | multiply(multiply(3, 3), multiply(multiply(3, 4), 5)) | if the operation β¬ is defined for all x and y by the equation x β¬ y = 3 * x * y , then 3 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 3 * 4 * 5 = 60 3 β¬ 60 = 3 * 3 * 60 = 540 hence , answer is d" | a = 3 * 3
b = 3 * 4
c = b * 5
d = a * c
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(8, subtract(8, 4)) | a person can swim in still water at 8 km / h . if the speed of water 4 km / h , how many hours will the man take to swim back against the current for 8 km ? | "m = 8 s = 4 us = 8 - 4 = 4 d = 4 t = 8 / 4 = 2 answer : a" | a = 8 - 4
b = 8 / a
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | a | divide(subtract(multiply(multiply(30, 4), divide(3, 4)), 40), subtract(multiply(30, 4), 20)) | thomas and matt has some money with them in the ratio of 3 : 4 . if their father gives rs . 30 to each of them then the ratio becomes 4 : 5 . what would be the ratio if thomas spend rs . 40 and matt spend rs . 20 from what they have now ? | let money of thomas and matt be x and y respectively . x / y = 3 / 4 = > x = 3 / 4 y ( x + 30 ) / ( y + 30 ) = 4 / 5 = > 5 x + 150 = 4 y + 120 = > 5 x = 4 y - 30 but x = 3 / 4 y 5 * 3 / 4 y = 4 y - 30 15 y = 16 y - 120 = > y = 120 = > x = 3 * 120 / 4 = > x = 90 ratio of their money , if thomas spend rs . 40 and matt spend rs . 20 = > ( 90 - 40 ) / ( 120 - 20 ) = 50 / 100 . = > 1 : 2 answer : a | a = 30 * 4
b = 3 / 4
c = a * b
d = c - 40
e = 30 * 4
f = e - 20
g = d / f
|
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