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 |
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a ) 320 , b ) 250 , c ) 560 , d ) 400 , e ) 450 | e | divide(multiply(150, 3), 1) | a train covers a distance at a speed of 150 kmph in 3 hours . to cover the same distance in 1 hours , it must travel at a speed of ? | distance = 150 * 3 = 450 km required speed = 450 / 1 = 450 km / hr answer is e | a = 150 * 3
b = a / 1
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a ) 0.5 , b ) 0.3 , c ) 0.9 , d ) 0.1 , e ) 0.7 | e | divide(0.63, 9) | find the quotient : 0.63 / 9 | "63 / 9 = 7 . dividend contains 2 places decimal . 0.63 / 9 = 0.7 answer is e ." | a = 0 / 63
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['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 272 degree', 'e ) e ) 192 degree'] | e | multiply(add(divide(divide(2, add(2, 3)), const_3), divide(multiply(divide(const_3, add(2, 3)), const_2), const_3)), const_360) | the ratio of boys to girls in a class is 2 : 3 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one third of the boys and two - third of the girls in the class ? | let the common ratio be x . . so b = 2 x and g = 3 x and total = 5 x 1 / 3 of m = 2 x / 3 and 2 / 3 of f = 2 x . . total preferring that carrer = 2 x / 3 + 2 x = 8 x / 3 . . now 5 x = 360 , so x = 360 / 5 = 72 . . so x * 8 / 3 = 72 * 8 / 3 = 192 ans 192 e | a = 2 + 3
b = 2 / a
c = b / 3
d = 2 + 3
e = 3 / d
f = e * 2
g = f / 3
h = c + g
i = h * 360
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a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | e | subtract(power(2, const_4), const_4) | x , y , and z are different prime numbers . the product x ^ 2 * y * z is divisible by how many different positive numbers ? | the exponents of x ^ 2 * y * z are 2 , 1 , and 1 . the number of factors is ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 12 the answer is e . | a = 2 ** 4
b = a - 4
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a ) a ) 6.24 , b ) b ) 13.1 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | b | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 25), divide(10, 55))), const_4) | due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 25 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 25 = 10 * 12 / 5 = 24 time difference = 13.1 ans : b" | a = 55 + 10
b = a - 1
c = 10 / 25
d = 10 / 55
e = c - d
f = b * e
g = max(f)
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a ) 229 , b ) 108 , c ) 278 , d ) 72 , e ) 112 | d | multiply(divide(120, 6), const_3_6) | a 120 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | "s = 120 / 6 * 18 / 5 = 72 kmph answer : d" | a = 120 / 6
b = a * const_3_6
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a ) $ 68.25 , b ) zero , c ) $ 6825.00 , d ) $ 6.82 , e ) $ 682.50 | e | add(multiply(const_0_25, const_1000), add(multiply(multiply(multiply(100000, divide(divide(12, 12), const_100)), divide(divide(12, 12), const_100)), add(multiply(multiply(100000, divide(divide(12, 12), const_100)), divide(divide(12, 12), const_100)), const_1)), multiply(const_3, const_100))) | peter takes a loan of $ 100000 with 12 % annual interest : the interest is paid once , at the end of the year . martha takes a loan of $ 100000 with 12 % annual interest , compounding monthly at the end of each month . at the end of one full year , compared to peter ' s loan interest , approximately how much more does martha have to repay ? | peters interest = $ 100000 * 0.12 = $ 12000 or $ 1000 each month . martha ’ s interest , 12 % / 12 = 1 % each month : for the 1 st month = $ 100000 * 0.01 = $ 1000 ; for the 2 nd month = $ 1000 + 1 % of 1000 = $ 1010 , so we would have interest earned on interest ( very small amount ) ; for the 3 rd month = $ 1010 + 1 % of 1010 = ~ $ 1020 ; for the 4 th month = $ 1020 + 1 % of 1020 = ~ $ 1030 ; . . . for the 12 th month = $ 1100 + 1 % of 1100 = ~ $ 1110 . the difference between peters interest and martha ’ s interest = ~ ( 10 + 20 + . . . + 110 ) = $ 660 . answer : e . | a = const_0_25 * 1000
b = 12 / 12
c = b / 100
d = 100000 * c
e = 12 / 12
f = e / 100
g = d * f
h = 12 / 12
i = h / 100
j = 100000 * i
k = 12 / 12
l = k / 100
m = j * l
n = m + 1
o = g * n
p = 3 * 100
q = o + p
r = a + q
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a ) 32.5 , b ) 35 , c ) 48 , d ) 56.8 , e ) 67.5 | d | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(15, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(15, const_100))))) | the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 15 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 63.75 . 4 ) now 63.75 has to be added byx % in order to equal the original price . 63.75 + ( x % ) 63.75 = 100 . solving this eq for x , we get x = 56.8 ans is d ." | a = 100 - 25
b = 100 - 25
c = 15 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 25
h = 100 - 25
i = 15 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
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a ) 3 / 40 , b ) 28 / 243 , c ) 3 / 25 , d ) 33 / 100 , e ) 64 / 125 | c | divide(1, 8) | 3 numbers are randomly selected , with replacement , from the set of integers { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . if the first number selected is w , the second number selected is x , and the third number is y , what is the probability that w < x < y ? | solution : total possible ways = 10 * 10 * 10 = 1000 case 1 : w = 0 . if x = 1 , then y can have 8 ways . if x = 2 , then y can have 7 ways and so on . so , no . of ways = 8 + 7 + . . + 1 case 2 : w = 1 . if x = 2 , then y can have 7 ways . if x = 3 , then y can have 6 ways and so on . so , no . of ways = 7 + 6 + . . + 1 there will be 8 cases in this way and they will follow a pattern which looks like this : no . of ways for case 1 : 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 no . of ways for case 2 : 1 + 2 + 3 + 4 + 5 + 6 + 7 no . of ways for case 3 : 1 + 2 + 3 + 4 + 5 + 6 no . of ways for case 4 : 1 + 2 + 3 + 4 + 5 no . of ways for case 5 : 1 + 2 + 3 + 4 no . of ways for case 6 : 1 + 2 + 3 no . of ways for case 7 : 1 + 2 no . of ways for case 8 : 1 ____________________________________________________________ total number of ways : 8 ( 1 ) + 7 ( 2 ) + 6 ( 3 ) + 5 ( 4 ) + 4 ( 5 ) + 3 ( 6 ) + 2 ( 7 ) + 1 ( 8 ) = 2 ( 4 ( 5 ) + 3 ( 6 ) + 2 ( 7 ) + 1 ( 8 ) ) = 2 ( 60 ) = 120 therefore , ans is 120 / 1000 = 3 / 25 . option c . | a = 1 / 8
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a ) 10 s , b ) 6 s , c ) 14 s , d ) 8 s , e ) 12 s | c | divide(add(200, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | two trains a and b are 200 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 ∗ 1000 / 60 ∗ 60 = 15 m / s speed of b = 36 ∗ 1000 / 60 ∗ 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross each other ? the length that needs to be crossed = 200 + 150 = 350 m . time taken = 350 / 25 = 14 s . correct answer c ." | a = 200 + 150
b = 54 * 1000
c = b / 3600
d = 36 * 1000
e = d / 3600
f = c + e
g = a / f
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a ) 80 % , b ) 74 % , c ) 58 % , d ) 56 % , e ) 50 % | c | divide(add(add(multiply(20, 80), multiply(50, 60)), multiply(subtract(const_100, add(20, 50)), 40)), const_100) | if 20 % of a class averages 80 % on a test , 50 % of the class averages 60 % on the test , and the remainder of the class averages 40 % on the test , what is the overall class average ? | 2 students scored 80 % 5 students scored 60 % 3 students scored 40 % ( 2 ) ( 80 ) + 5 ( 60 ) + 3 ( 40 ) = 160 + 300 + 120 = 580 580 / 10 students = 58 - - > 58 % average answer : c | a = 20 * 80
b = 50 * 60
c = a + b
d = 20 + 50
e = 100 - d
f = e * 40
g = c + f
h = g / 100
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a ) 113 , b ) 114 , c ) 115 , d ) 116 , e ) 117 | c | divide(subtract(multiply(floor(12.4), 6), 26), subtract(12.4, floor(12.4))) | a man whose bowling average is 12.4 , takes 6 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 6 + x ) 12 solve equation x = 115 answer : c" | a = math.floor(12, 4)
b = a * 6
c = b - 26
d = math.floor(12, 4)
e = 12 - 4
f = c / e
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a ) 60 , b ) 80 , c ) 100 , d ) 130 , e ) 150 | a | subtract(200, subtract(add(90, 70), 20)) | of the 200 stamps in a collection , 90 are foreign and 70 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ? | 20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 50 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 50 ) = 60 the answer is a . | a = 90 + 70
b = a - 20
c = 200 - b
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a ) 28 , b ) 24 , c ) 20 , d ) 22 , e ) 18 | c | subtract(add(add(multiply(20, 5), 1), 20), multiply(20, 5)) | the average age of 20 students in a class is 5 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ? | "total age of 50 students = 20 * 5 = 100 total age of 51 persons = 20 * 6 = 120 age of teacher = 120 - 100 = 20 years answer is c" | a = 20 * 5
b = a + 1
c = b + 20
d = 20 * 5
e = c - d
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a ) 15 , b ) 19 , c ) 23 , d ) 27 , e ) 31 | b | add(divide(80, add(const_4, const_1)), const_2) | how many zeroes are there at the end of the number n , if n = 80 ! + 160 ! ? | "the number of zeroes at the end of 80 ! will be less than the number of zeroes at the end of 160 ! hence it is sufficient to calculate the number of zeroes at the end of 80 ! the number of zeroes = [ 80 / 5 ] + [ 80 / 25 ] + [ 80 / 125 ] = 16 + 3 + 0 = 19 the answer is b ." | a = 4 + 1
b = 80 / a
c = b + 2
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a ) 288 , b ) 267 , c ) 200 , d ) 4096 , e ) 2771 | d | divide(4913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | the principal that amounts to rs . 4913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 4913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 4096 . answer : d" | a = 1 / 4
b = 6 + a
c = b / 100
d = 1 + c
e = d ** 3
f = 4913 / e
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a ) 11 , b ) 18 , c ) 45 , d ) 17 , e ) 12 | c | multiply(divide(75, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by c are ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 15 / 25 * 75 = 45 answer : c" | a = 2 / 3
b = 5 / 2
c = a + b
d = c + 2
e = 75 / d
f = e * 5
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a ) 7 , b ) 6 , c ) 9 , d ) 12 , e ) 7 | b | divide(9, const_2) | if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 9 , what is the least possible value of n ? | we need at least 2 factors of three , which means n must be at least 6 , so b is the correct answer . | a = 9 / 2
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a ) 17 , b ) 15 , c ) 12 , d ) - 8 / 5 , e ) none | c | subtract(sqrt(289), divide(sqrt(625), sqrt(sqrt(625)))) | √ 289 - √ 625 ÷ √ 25 is equal to ? | answer √ 289 - √ 625 ÷ √ 25 = √ 17 x 17 - √ 25 x 25 ÷ √ 5 x 5 = 17 - 25 ÷ 5 = 17 - 5 = 12 correct option : c | a = math.sqrt(289)
b = math.sqrt(625)
c = math.sqrt(625)
d = math.sqrt(c)
e = b / d
f = a - e
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a ) 8 , b ) 9 , c ) 10 , d ) 21 , e ) 28 | e | divide(subtract(multiply(90, 8), multiply(83, 8)), subtract(92, 90)) | the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 83 , and the average of the female students ’ grades was 92 , how many female students took the test ? | total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 83 thus m = 664 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 664 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 28 ans : e | a = 90 * 8
b = 83 * 8
c = a - b
d = 92 - 90
e = c / d
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a ) 810.91 , b ) 840.91 , c ) 880.91 , d ) 887.91 , e ) 980.91 | c | multiply(9500, subtract(power(divide(add(divide(6, const_2), const_100), const_100), multiply(1, const_2)), const_1)) | what is the compound interest on rs . 9500 at 6 % p . a . compounded half - yearly for 1 1 / 2 years . | "compound interest : a = p ( 1 + r / n ) nt a = 10 , 380.91 c . i . > > 10 , 380.91 - 9500 > > rs . 880.91 answer : c" | a = 6 / 2
b = a + 100
c = b / 100
d = 1 * 2
e = c ** d
f = e - 1
g = 9500 * f
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a ) 2777 , b ) 1000 , c ) 2789 , d ) 2776 , e ) 2881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | a and b ’ s salaries together amount to rs . 4,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is b ’ s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 1000 4 b = 1000 = > b = 1000 answer b" | a = 2 + 3
b = 4 * a
c = 2 * b
d = c * 100
e = 85 / 100
f = 1 - e
g = d * f
h = 95 / 100
i = 1 - h
j = 85 / 100
k = 1 - j
l = i + k
m = g / l
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a ) 10 , b ) 12 , c ) 14 , d ) 54 , e ) 18 | d | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 6) | the population of a bacteria culture doubles every 6 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | "this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 6 minutes so answer is 9 * 6 = 54 minutes answer ( d )" | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 6
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a ) 68 , b ) 78 , c ) 88 , d ) 98 , e ) 56 | d | multiply(add(subtract(7, const_1), 8), 7) | can you solve it ? 2 + 3 = 8 , 3 + 7 = 27 , 4 + 5 = 32 , 5 + 8 = 60 , 6 + 7 = 72 , 7 + 8 = ? ? | 2 + 3 = 2 * [ 3 + ( 2 - 1 ) ] = 8 3 + 7 = 3 * [ 7 + ( 3 - 1 ) ] = 27 4 + 5 = 4 * [ 5 + ( 4 - 1 ) ] = 32 5 + 8 = 5 * [ 8 + ( 5 - 1 ) ] = 60 6 + 7 = 6 * [ 7 + ( 6 - 1 ) ] = 72 therefore 7 + 8 = 7 * [ 8 + ( 7 - 1 ) ] = 98 x + y = x [ y + ( x - 1 ) ] = x ^ 2 + xy - x correct answer is d ) 98 | a = 7 - 1
b = a + 8
c = b * 7
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | divide(14, 7) | if the remainder is 10 when positive integer n is divided by 14 , what is the remainder when n is divided by 7 ? | "assume x is quotient here , n = 14 x + 7 - - - - - - - - - - ( 1 ) and n = 7 x + ? we can also write equation ( 1 ) as : n = ( 14 x + 7 ) + 3 . ie 7 ( 2 x + 1 ) + 3 ie the first term is perfectly divisible by 7 . so , the remainder left is 3 . so , answer ( d ) is right choice ." | a = 14 / 7
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a ) 10 , b ) 2 , c ) 4 , d ) 16 , e ) 18 | b | subtract(power(2, const_2), const_2) | if x ^ 2 + 1 / x ^ 2 = 2 , what is the value of x ^ 4 + 1 / x ^ 4 ? | important : i notice that if we square x ² , we get x ⁴ , and if we square 1 / x ² , we get 1 / x ⁴ , so let ' s see what happens if we take the equation x ² + 1 / x ² = 2 andsquareboth sides : ( x ² + 1 / x ² ) ² = 4 so , ( x ² + 1 / x ² ) ( x ² + 1 / x ² ) = 4 expand to get : x ⁴ + 1 + 1 + 1 / x ⁴ = 4 simplify : x ⁴ + 1 / x ⁴ = 2 answer : b | a = 2 ** 2
b = a - 2
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a ) 443 , b ) 450 , c ) 460 , d ) 468 , e ) 475 | c | divide(add(multiply(12, 410), multiply(add(410, 65), subtract(52, 12))), 52) | a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 12 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 12 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? | "total number of installments = 52 payment per installment for the first 12 installments = 410 payment per installment for the remaining 32 installments = 410 + 65 = 475 average = ( 12 * 410 + 40 * 475 ) / 52 = 460 answer c" | a = 12 * 410
b = 410 + 65
c = 52 - 12
d = b * c
e = a + d
f = e / 52
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a ) 6.54 % , b ) 4.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 5.54 % | b | divide(const_100, 22) | at what rate percent of simple interest will a sum of money double itself in 22 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 22 ) = 100 / 22 = 4.54 % answer : b" | a = 100 / 22
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['a ) 2', 'b ) 3', 'c ) 4', 'd ) 16', 'e ) 41'] | a | divide(power(const_1, const_2), power(sqrt(divide(const_1, add(const_1, const_1))), const_2)) | consider a square of diagonal length d . let another square be formed with d as its side . find the ratio of the area of the bigger square to that of the smaller square . | d ^ 2 / ( 1 / 2 ) * d ^ 2 = 2 answer : a | a = 1 ** 2
b = 1 + 1
c = 1 / b
d = math.sqrt(c)
e = d ** 2
f = a / e
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a ) 3572021 , b ) 3570221 , c ) 3572012 , d ) 3752012 , e ) none of them | a | divide(1635773, divide(1936248, const_100)) | find the missing value : ? - 1936248 = 1635773 | "let x - 1936248 = 1635773 . then , x = 1635773 + 1936248 = 3572021 answer is a ." | a = 1936248 / 100
b = 1635773 / a
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a ) 2.1 % , b ) 8.1 % , c ) 5.1 % , d ) 6.1 % , e ) 1.1 % | e | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(14, const_100))), const_100)) | the tax on a commodity is diminished by 14 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 86 * 115 = 9890 - - - - - - - - - - - 10000 - - - - - - - - - - - 110 100 - - - - - - - - - - - ? = > 1.1 % decrease answer : e" | a = 15 / 100
b = 1 + a
c = 14 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 7 , b ) 8 , c ) 10 , d ) none of these , e ) can not be determined | a | divide(subtract(divide(58, divide(4, 5)), multiply(subtract(4, const_1), 5)), 5) | the sum of the ages of 4 children born at the intervals of 5 years each is 58 years . what is the age of the youngest child ? | "explanation : let x = the youngest child . each of the other four children will then be x + 5 , x + 10 , x + 15 we know that the sum of their ages is 58 . so , x + ( x + 5 ) + ( x + 10 ) + ( x + 15 ) = 58 x = 7 the youngest child is 4 years old . answer : a" | a = 4 / 5
b = 58 / a
c = 4 - 1
d = c * 5
e = b - d
f = e / 5
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | c | divide(add(sqrt(add(multiply(multiply(136, const_2), const_4), const_1)), const_1), const_2) | in a party every person shakes hands with every other person . if there were a total of 136 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 136 n ( n - 1 ) / 2 = 136 â ˆ ´ n = 17 answer : c" | a = 136 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
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a ) 12 , b ) 18 , c ) 16.6 , d ) 17 , e ) 23 | b | subtract(subtract(32, const_2), add(add(12, const_4), const_1)) | set a consists of all the prime numbers between 12 and 32 . what is the range of set a ? | "the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 31 lowest number = 13 range = highest - lowest = 31 - 13 = 18 option b" | a = 32 - 2
b = 12 + 4
c = b + 1
d = a - c
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a ) 5 , b ) 10 , c ) 20 , d ) 40 , e ) 49 | e | divide(subtract(74.50, 50), subtract(1.50, 1.00)) | caleb spends $ 74.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | "solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 74.50 . by solving the equations y = 49 . ans e ." | a = 74 - 50
b = 1 - 50
c = a / b
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a ) 8 , b ) 9 , c ) 10 , d ) 12 , e ) 16 | c | subtract(42, subtract(multiply(48, const_4), multiply(40, const_4))) | the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 40 degrees . if the temperature on monday was 42 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 40 = 160 m = 42 tu + w + th = 192 - 42 = 150 f = 160 – 150 = 10 answer : c" | a = 48 * 4
b = 40 * 4
c = a - b
d = 42 - c
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a ) 45 , b ) 60 , c ) 75 , d ) 96 , e ) 100 | d | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | if eight 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 8 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 16 / 3 = 10 / 3 * x x = 8 / 5 hour so 96 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 ) 60 , b ) 120 , c ) 160 , d ) 240 , e ) 840 | a | multiply(divide(15, const_2), 8) | if the sum of the 4 th term and the 12 th term of an arithmetic progression is 8 , what is the sum of the first 15 terms of the progression ? | "let a = first term , d = common difference . 4 th term = a + 3 d 12 th term = a + 11 d a + 3 d + a + 11 d = 8 2 a + 14 d = 8 a + 7 d = 4 . sum of the first 15 terms = 15 / 2 ( 2 a + 14 d ) = 15 ( a + 7 d ) = 15 ( 4 ) = 60 . ans ( a )" | a = 15 / 2
b = a * 8
|
a ) 101.02 kmph , b ) 121.62 kmph , c ) 100.62 kmph , d ) 111.00 kmph , e ) 121.00 kmph | b | multiply(divide(const_1, divide(50, const_100)), 60) | niki covers a distance in 45 min , if she drives at a speed of 60 km / h on an average . find the speed at which she must drive at to reduce the time of the journey by 50 % . | given speed = 60 kmph . it means niki covered 60 km in 60 min ( 1 hour ) . so , in 45 min he will cover 45 km . 50 % time reduced = 45 - 50 % of 45 = 22.5 min . thus , niki needs to cover 45 km in 22.5 min ( 0.37 hour ) . speed * time = distance . speed * 0.37 = 45 speed = 121.62 kmph . answer : option b | a = 50 / 100
b = 1 / a
c = b * 60
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a ) 220 km , b ) 240 km , c ) 230 km , d ) 232 km , e ) 234 km | b | multiply(const_2, divide(multiply(multiply(10, 15), 20), add(10, 15))) | a man complete a journey in 20 hours . he travels first half of the journey at the rate of 10 km / hr and second half at the rate of 15 km / hr . find the total journey in km . | "0.5 x / 10 + 0.5 x / 15 = 20 - - > x / 10 + x / 15 = 40 - - > 5 x = 30 x 40 - - > x = ( 30 x 40 ) / 5 = 240 km . answer : b ." | a = 10 * 15
b = a * 20
c = 10 + 15
d = b / c
e = 2 * d
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a ) 5 / 6 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) 1 / 24 | d | divide(24, add(24, multiply(24, multiply(divide(1, 4), 24)))) | a satellite is composed of 24 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 4 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ? | "let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 4 . the number of non - upgraded sensors on the whole satellite is 24 ( x / 4 ) = 6 x . the fraction of sensors which have been upgraded is x / ( x + 6 x ) = x / 7 x = 1 / 7 the answer is d ." | a = 1 / 4
b = a * 24
c = 24 * b
d = 24 + c
e = 24 / d
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a ) 2 . , b ) 4 . , c ) b = 5 . , d ) b = 6 . , e ) b = 8 . | c | divide(multiply(3, 20), 12) | 20 beavers , working together in a constant pace , can build a dam in 3 hours . how many b hours will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it could have taken double b = 6 hrs . . so closest to that option is 5 . c" | a = 3 * 20
b = a / 12
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(multiply(add(2, const_3.0), const_2), 8) | | x + 3 | – | 2 - x | = | 8 + x | how many solutions will this equation have ? | "you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , so it becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 2 c" | a = 2 + 3
b = a * 2
c = b / 8
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a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | d | multiply(divide(const_100, add(const_100, 10)), 45) | from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 10 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "1.10 x = 45 - - > 11 / 10 * x = 45 - - > x = 45 * 10 / 11 = 450 / 11 = ~ 41 . answer : d ." | a = 100 + 10
b = 100 / a
c = b * 45
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a ) 50 % , b ) 100 % , c ) 150 % , d ) 200 % , e ) 250 % | b | multiply(const_100, divide(multiply(surface_cube(1), surface_cube(2)), surface_cube(2))) | if a 2 cm cube is cut into 1 cm cubes , then what is the percentage increase in the surface area of the resulting cubes ? | "the area a of the large cube is 2 * 2 * 6 = 24 square cm . the area of the 8 small cubes is 8 * 6 = 48 = 2 a , an increase of 100 % . the answer is b ." | a = surface_cube * (
b = a / surface_cube
c = 100 * b
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a ) $ 160 , b ) $ 70 , c ) $ 90 , d ) $ 80 , e ) $ 55 | b | divide(divide(756, divide(add(20, const_100), const_100)), add(4, 5)) | a group of 4 investment bankers and 5 clients recently frequented the bonbon ribs restaurant . the total bill for the meal , including 20 % gratuity , came to $ 756 . on average , how much did the meal of each individual cost before gratuity ? | 4 ibs and 5 clients - so total 9 people the bill $ 756 includes 20 % gratuity . . . so the actual cost of dinner was $ 630 now , the cost per person will be $ 630 / 9 which is $ 70 option b | a = 20 + 100
b = a / 100
c = 756 / b
d = 4 + 5
e = c / d
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a ) $ 150 , b ) $ 345 , c ) $ 365 , d ) $ 715 , e ) $ 730 | b | subtract(divide(subtract(multiply(500, 7), add(add(add(406, 413), add(436, 420)), 395)), const_2), 370) | a salesman ' s income consists of a commission and a base salary of $ 370 per week . over the past 5 weeks , his weekly income totals have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 over the 7 - week period ? | "total weekly income over 5 weeks = $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2070 = $ 1430 from this , we subtract base salary for 2 weeks i . e $ 370 * 2 = $ 740 therefore , commission = $ 1430 - $ 740 = $ 690 for 2 weeks avg weekly commission = $ 345 answer b" | a = 500 * 7
b = 406 + 413
c = 436 + 420
d = b + c
e = d + 395
f = a - e
g = f / 2
h = g - 370
|
a ) 100 m , b ) 136 m , c ) 180 m , d ) 200 m , e ) 250 m | b | multiply(divide(60, const_3_6), 8) | a car is running at a speed of 60 kmph . what distance will it cover in 8 sec ? | "speed = 60 kmph = 60 * 5 / 18 = 17 m / s distance covered in 8 sec = 17 * 8 = 136 m answer is b" | a = 60 / const_3_6
b = a * 8
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | a | subtract(const_60, multiply(divide(200, 300), const_60)) | without stoppages , a train travels certain distance with an average speed of 300 km / h , and with stoppages , it covers the same distance with an average speed of 200 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 100 km less . time taken to cover 100 km = 100 â „ 300 h = 1 â „ 3 h = 1 â „ 3 ã — 60 min = 20 min answer a" | a = 200 / 300
b = a * const_60
c = const_60 - b
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a ) 60 , b ) 80 , c ) 50 , d ) 100 , e ) 40 | c | divide(20, divide(subtract(const_100, add(25, 35)), const_100)) | a fruit drink is made of grapefruit , lemon , and orange juice , where 25 percent of the drink is grapefruit juice and 35 percent is lemon juice . if the drink is made with 20 ounces of orange juice , how many ounces is the drink total ? | let the total number of ounces in the drink be x % of orange = 25 % % of watermelon = 35 % % of grape = 100 % - 60 % = 40 % but this number is given as 20 ounces so 40 % of x = 20 and x = ( 20 ) ( 100 / 40 ) = ( 20 ) ( 5 / 2 ) x = 50 therefore there a total of 50 ounces in the drink . correct answer - c | a = 25 + 35
b = 100 - a
c = b / 100
d = 20 / c
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a ) 300 , b ) 320 , c ) 340 , d ) 360 , e ) 380 | d | multiply(divide(subtract(1240, 40), const_10), 3) | ` ` how old are you , alchemerion ? ' ' asked one of the wizards appearances the wizards answer with a riddle , ` ` i am still very young as wizards go . i am only 3 times my son ' s age . my father is 40 year more than twice of my age . together the 3 of us are a 1240 year old ' ' . how old is alchemerion | let alchemerion be a let son be s let father be f a = 3 * s f = 40 + ( 2 * a ) therefore 1240 = ( s + f + a ) 1240 = ( a / 3 ) + ( 40 + ( 2 * a ) ) + a a = 360 answer : d | a = 1240 - 40
b = a / 10
c = b * 3
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a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 128 % | e | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(16, const_100)), divide(10, const_100)), const_100) | in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 16 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,128 r = x / 100 * 0.1 r answer e" | a = 20 / 100
b = 1 - a
c = 16 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 50 km , b ) 56 km , c ) 70 km , d ) 80 km , e ) 90 km | a | multiply(10, divide(20, subtract(14, 10))) | if a person walks at 14 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance travelled by him is | let the actual distance travelled be x km . then x / 10 = ( x + 20 ) / 14 - - > 14 x = 10 x + 200 - - > 4 x = 200 - - > x = 50 km . answer : a . | a = 14 - 10
b = 20 / a
c = 10 * b
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a ) 1,108 , b ) 2,500 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 300) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 300 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 300 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 300 x 8 = 2400 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 2500 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 2500 ( answer b )" | a = 200 / 22
b = a * 300
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a ) s . 50 , b ) s . 70 , c ) s . 100 , d ) s . 80 , e ) s . 75 | e | multiply(300, divide(25, const_100)) | find the 25 % of rs . 300 . | "explanation : 25 % of 300 = > 25 / 100 * 300 = rs . 75 answer : e" | a = 25 / 100
b = 300 * a
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a ) 1 , b ) 3 , c ) 6 , d ) 7 , e ) 10 | c | multiply(divide(divide(multiply(50, 50), const_100), 50), const_12) | a reduction of 50 % in the price of apples would enable a man to obtain 50 more for rs . 50 , what is reduced price per dozen ? | "c 50 * ( 50 / 100 ) = 25 - - - 50 ? - - - 12 = > rs . 6" | a = 50 * 50
b = a / 100
c = b / 50
d = c * 12
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a ) 6 , b ) 12 , c ) 18 , d ) 24 , e ) 48 | a | divide(multiply(add(add(6, 4), 2), divide(1000, 2)), const_1000) | a 1000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ? | "in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 500 by that net inflow to get the minutes ) - 6 min . answer a ." | a = 6 + 4
b = a + 2
c = 1000 / 2
d = b * c
e = d / 1000
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a ) 24 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | c | multiply(multiply(4, 2), divide(6, 3)) | working simultaneously and independently at an identical constant rate , 4 machines of a certain type can produce a total of x units of product p in 6 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 2 x units of product p in 3 days ? | "4 machines do x units in 6 days we have x / 6 = > rate of the 4 machines we know that we need to have 2 x units in 3 days therefore , we need to get to 2 x / 3 rate of the machines . rate of one machine is x / 6 * 1 / 4 = x / 24 . now , we need to know how many machines need to work simultaneously , to get 2 x done in 3 days . 2 x / 3 work needs to be done by machines that work at x / 24 rate . let ' s assign a constant y for the number of machines : ( x / 24 ) * y = 2 x / 3 y = 2 x / 3 * 24 / x cancel 3 with 24 , and x with x and get - > 16 . answer choice c" | a = 4 * 2
b = 6 / 3
c = a * b
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a ) 20 lb , b ) 30 lb , c ) 10 lb , d ) 15 lb , e ) 36 lb | e | divide(36, const_1) | a bag of potatoes weighs 36 lbs divided by half of its weight . how much does the bag of potatoes weight ? | sol . 36 ÷ 3 = 12 . answer : e | a = 36 / 1
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a ) s . 50 , b ) s . 70 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | subtract(add(divide(600, 5), divide(910, 7)), divide(1800, 9)) | p , q and r together earn rs . 1800 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ? | explanation : amount earned by p , q and r in 1 day = 1800 / 9 = 200 - - - ( 1 ) amount earned by p and r in 1 day = 600 / 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 / 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 130 - 200 = 50 = > amount earned by r in 1 day = 50 answer : option a | a = 600 / 5
b = 910 / 7
c = a + b
d = 1800 / 9
e = c - d
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a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 42 | e | multiply(const_3, divide(60, const_10)) | jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 2 percent solution will be required ? | "use weighted average : 2 % and 12 % solutions mix to give 5 % solution . w 1 / w 2 = ( a 2 - avg ) / ( avg - a 1 ) = ( 12 - 5 ) / ( 5 - 2 ) = 7 / 3 you need 7 parts of 2 % solution and 3 parts of 12 % solution to get 10 parts of 5 % solution . if total 5 % solution is actually 60 litres , you need 7 * 6 = 42 litres of 2 % solution and 3 * 6 = 18 litres of 12 % solution . answer ( e )" | a = 60 / 10
b = 3 * a
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a ) 150 s , b ) 450 s , c ) 750 s , d ) 825 s , e ) 925 s | c | divide(multiply(const_1000, add(const_1, const_4)), divide(divide(40, const_2), const_3)) | in a kilo meter race , if a gives b a 40 m start , a wins by 19 s . but if a gives b a 30 s start , b wins by 40 m . find the time taken by b to run 5,000 m ? | when b is given a 40 m start then b runs 960 m in t secs and a runs 1000 m in t - 19 secs . when b is given a 30 second start then b runs 1000 m in t secs and a runs 960 m in t - 30 secs . assume that the speed of a ( va ) and b ( vb ) is constant in both situations . then vb = 960 ÷ t = 1000 ÷ t , and va = 1000 ÷ ( t - 19 ) = 960 ÷ ( t - 30 ) so 960 t = 1000 t , and 1000 t - 30000 = 960 t - 18240 . then 40 t = - 40 t + 11760 , t = - t + 294 . substituting for t in 960 t = 1000 t gives : - 960 t + 282240 = 1000 t 1960 t = 282240 t = 144 b ' s speed is 960 / t = 960 / 144 = 6.67 metres per second . assuming the same constant speed to run 5000 metres then the time taken is 5000 / 6.67 = 750 seconds answer : c | a = 1 + 4
b = 1000 * a
c = 40 / 2
d = c / 3
e = b / d
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a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.1111 | d | multiply(divide(0.9999, 0.1111), const_100) | 0.9999 + 0.1111 = ? | "0.9999 + 0.1111 = 0.9999 + 0.111 + 0.0001 = ( 0.9999 + 0.0001 ) + 0.111 = 1 + 0.111 = 1.111 d" | a = 0 / 9999
b = a * 100
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a ) 20 % , b ) 42 % , c ) 44 % , d ) 70 % , e ) 84 % | d | multiply(subtract(const_1, divide(18, subtract(const_100, 40))), const_100) | exactly 18 % of the reporters for a certain wire service cover local politics in country x . if 40 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "let ' s assume there are 100 reporters - - > 18 reporters cover local politics . now , as 40 % of the reporters who cover all politics do not cover local politics then the rest 60 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 60 % of them equal to 18 ( # of reporters who cover local politics ) : 0.6 x = 18 - - > x = 30 , hence 30 reporters cover politics and the rest 100 - 30 = 70 reporters do not cover politics at all . answer : d ." | a = 100 - 40
b = 18 / a
c = 1 - b
d = c * 100
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a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 5 | e | subtract(200, subtract(395, 200)) | when 200 is divided by positive integer x , the remainder is 5 . what is the remainder when 395 is divided by x ? | "if 200 / x leaves a reminder 5 then ( 200 - 5 ) i . e . 195 is divisible by x so ( 200 + 195 ) / x leaves a reminder rem ( 200 / x ) + rem ( 195 / x ) = > 5 + 0 = 5 answer : e" | a = 395 - 200
b = 200 - a
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a ) 2401 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(7, multiply(const_4, 1)) | xy = 1 then what is ( 7 ^ ( x + y ) ^ 2 ) / ( 7 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 7 ^ 4 = 2401 answer a" | a = 4 * 1
b = 7 ** a
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a ) - 3 , b ) - 4 , c ) - 6 , d ) - 8 , e ) - 2 | e | add(divide(80, const_10), divide(80, divide(80, const_10))) | if a ( a + 2 ) = 80 and b ( b + 2 ) = 80 , where a ≠ b , then a + b = | i . e . if a = 8 then b = - 10 or if a = - 10 then b = 8 but in each case a + b = - 10 + 8 = - 2 answer : e | a = 80 / 10
b = 80 / 10
c = 80 / b
d = a + c
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a ) 50 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 72 kmph | e | subtract(multiply(divide(280, 9), const_3_6), 40) | a man sitting in a train which is traveling at 40 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | "relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 40 ) kmph = 72 kmph . answer : e" | a = 280 / 9
b = a * const_3_6
c = b - 40
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a ) 62 , b ) 52 , c ) 37 , d ) 12 , e ) 22 | c | subtract(45, multiply(multiply(12, 3), 2)) | evaluate : 45 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 45 - 12 * 3 * 2 = 45 - 8 = 37 correct answer c" | a = 12 * 3
b = a * 2
c = 45 - b
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a ) rs . 209.20 , b ) rs . 309.20 , c ) rs . 409.20 , d ) rs . 509.20 , e ) none of these | c | multiply(divide(55, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 55 paise per sq m is | explanation : area to be plastered = [ 2 ( l + b ) ã — h ] + ( l ã — b ) = [ 2 ( 25 + 12 ) ã — 6 ] + ( 25 ã — 12 ) = 744 sq m cost of plastering = 744 ã — ( 55 / 100 ) = rs . 409.20 answer : c | a = 55 / 100
b = 25 * 12
c = 25 * 6
d = 2 * c
e = 12 * 6
f = e * 2
g = d + f
h = b + g
i = a * h
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(2, const_1) | let p be a prime number greater than 2 and let n = 14 p . how many even numbers divide n ? | answer : a . there is exactly one . the prime factorization of 14 is 14 = 2 * 7 , so n = 2 * 7 * p = 7 * 2 * p = 7 * 2 p , so 2 p is the only even number that divides n . | a = 2 - 1
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a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 788 , e ) none of the above | d | subtract(multiply(5000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 5000) | compound interest of rs . 5000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 5000 [ 1 + ( 5 / 100 ) ] ^ 3 - 5000 = 5000 [ ( 21 / 20 ) ^ 3 - 1 ] = 5000 ( 9261 - 8000 ) / 8000 = 5 * 1261 / 8 = 788 answer : d" | a = 10 / 2
b = a / 100
c = 1 + b
d = 1 / 2
e = 1 + d
f = e * 2
g = c ** f
h = 5000 * g
i = h - 5000
|
a ) 240 , b ) 248 , c ) 280 , d ) 288 , e ) 290 | d | divide(multiply(subtract(multiply(subtract(248, 12), const_2), 280), const_3), const_2) | a train starts full of passengers . at the first station , it drops one - third of the passengers and takes 280 more . at the second station , it drops one - half of the new total and takes 12 more . on arriving at the third station , it is found to have 248 passengers . find the number of passengers in the beginning . | let x be no . of passengers at beginning at , 1 st station is drops 1 / 3 rd so remaining 2 / 3 rd are in train only plus 280 i . e . , 2 x / 3 + 280 at 2 nd stop 1 / 2 of new total and 12 more i . e . , ( 2 x / 3 + 280 ) / 2 + 12 = 248 on solving above equation we get x as 288 answer : d | a = 248 - 12
b = a * 2
c = b - 280
d = c * 3
e = d / 2
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a ) 20.8 , b ) 24.6 , c ) 31.8 , d ) 32.5 , e ) 33.8 | b | multiply(divide(subtract(69.0, 52.0), 69.0), const_100) | the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 52.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 52 = 17 $ ( 17 / 69 ) * 100 = 24.6 % answer : b" | a = 69 - 0
b = a / 69
c = b * 100
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a ) rs 13123.20 , b ) rs 14103.48 , c ) rs 12172.48 , d ) rs 14172.48 , e ) none of these | d | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on rs . 35000 after 3 years at the rate of 12 % per annum | "explanation : ( 35000 × ( 1 + 12 / 100 ) 3 ) = > 35000 × 28 / 25 × 28 / 25 × 28 / 25 = > 49172.48 so compound interest will be 49172.48 - 35000 = rs 14172.48 option d" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
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a ) 3 : 4 , b ) 3 : 2 , c ) 3 : 5 , d ) 3 : 7 , e ) 2 : 3 | b | divide(subtract(3.08, 2.9), subtract(3.2, 3.08)) | in what ratio must wheat at rs . 3.20 pe rkg be mixed with wheat at rs . 2.90 per kg so that the mixture be worth rs . 3.08 per kg | if 2 ingredients are mixed , then the required ratio is given by , the rule of allegation c . p of a unit quantity of cheaper ( c ) c . p . of a unit quantity of dearer ( d ) mean price ( m ) d - m m - c required ratio is cheaper quantity : dearer quantity = ( d - m ) : ( m - c ) answer with explanation : given i . c . p of a unit quantity of i kind ( p ) = 3.20 ii . c . p of a unit quantity of ii kind ( p ) = 2.90 iii . mean price ( p ) = 3.08 iv . d – m = 3.08 – 2.90 = 0.18 v . m – c = 3.20 – 3.08 = 0.12 required ratio = 0.18 : 0.12 = 3 : 2 answer : b | a = 3 - 8
b = 3 - 2
c = a / b
|
a ) 12.5 days , b ) 16 days , c ) 18 days , d ) 11 days , e ) 38 days | a | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 10)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 10))), add(divide(const_1, 24), divide(const_1, 30)))), 10) | a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 10 days before the completion of the work . in how many days was the work completed ? | "one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 40 = 1 / 10 work done by a and b together in the last 10 days = 10 * ( 1 / 24 + 1 / 30 ) = 3 / 4 remaining work = 1 / 4 the number of days required for this initial work = 2.5 days . the total number of days required = 10 + 2.5 = 12.5 days . answer : a" | a = 1 / 24
b = 1 / 30
c = a + b
d = c * 10
e = 1 - d
f = 2 + 3
g = 2 * 10
h = f * g
i = 1 / h
j = 1 / 24
k = 1 / 30
l = j + k
m = i + l
n = e / m
o = n + 10
|
a ) 14 , b ) 20 , c ) 15 , d ) 32 , e ) 64 | c | sqrt(add(213, multiply(6, 2))) | if a 2 + b 2 + c 2 = 213 and ab + bc + ca = 6 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 213 and ab + bc + ca = 6 , ( a + b + c ) ^ 2 = 213 + 2 ( 6 ) = 225 = 15 ^ 2 therefore : a + b + c = 15 answer : c" | a = 6 * 2
b = 213 + a
c = math.sqrt(b)
|
a ) 8 , b ) 10 , c ) 12 , d ) 6 , e ) 19 | d | divide(subtract(48, power(6, 2)), 2) | if a - b = 6 and a 2 + b 2 = 48 , find the value of ab . | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 48 - 36 = 12 ab = 6 . answer : d" | a = 6 ** 2
b = 48 - a
c = b / 2
|
a ) 50 % , b ) 33.33 % , c ) 25 % , d ) 16.66 % , e ) 12.5 % | d | multiply(subtract(divide(2, 5), multiply(divide(2, 5), divide(1, 2))), const_100) | a certain article of clothing was discounted during a special sale to 2 / 5 of its original retail price . when the clothing did n ' t sell , it was discounted even further to 1 / 2 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale to the second sale ? | "say the original retail price of the item was $ 200 . the price after the first sale = 3 / 5 * $ 200 = $ 120 . the price after the second sale = 1 / 2 * $ 200 = $ 100 . the percent change from the first sale to the second = ( 120 - 100 ) / 120 = 1 / 3 = 16.66 % . answer : d ." | a = 2 / 5
b = 2 / 5
c = 1 / 2
d = b * c
e = a - d
f = e * 100
|
a ) 22 / 21 , b ) 20 / 21 , c ) 23 / 21 , d ) 25 / 21 , e ) 28 / 21 | b | divide(6, add(divide(30, const_100), 6)) | a committee is reviewing a total of 30 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ? | "say x = y = 10 . in this case we would have : 30 x = 300 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 10 % = 1 % of the black - and - white films , so 3 black - and - white films and all 60 color films , thus total of 63 films were selected . color films thus compose 60 / 63 = 20 / 21 of the selected films . answer : b" | a = 30 / 100
b = a + 6
c = 6 / b
|
a ) $ 13300 , b ) $ 13000 , c ) $ 12900 , d ) $ 10000 , e ) $ 9000 | b | divide(subtract(12000, multiply(divide(10, const_100), 3000)), subtract(const_1, divide(10, const_100))) | after paying a 10 percent tax on all income over $ 3000 , a person had a net income of $ 12000 . what was the income before taxes ? | let x be the income over 3000 then ( x - x / 10 ) + 3000 = 12000 = > x = 10000 therefore income before taxes = 10000 + 3000 = 13000 option ( b ) | a = 10 / 100
b = a * 3000
c = 12000 - b
d = 10 / 100
e = 1 - d
f = c / e
|
a ) 3200.0 , b ) 11520.0 , c ) 8820.0 , d ) 7354.0 , e ) 16537.11 | c | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 8000) | sam invested rs . 8000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 8000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 8000 * ( 1 + 5 / 100 ) 2 ] = ( 8000 * 21 / 20 * 21 / 20 ) = rs . 8820.00 answer : c" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 8000
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(subtract(11, 5), const_2) | half a number plus 5 is 11 . what is the number ? | "solution let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 5 = 11 ( 1 / 2 ) x + 5 - 5 = 11 - 5 ( 1 / 2 ) x = 6 2 × ( 1 / 2 ) x = 6 × 2 x = 12 answer c" | a = 11 - 5
b = a * 2
|
a ) 6 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 10 | e | divide(multiply(19, 90), const_100) | a can complete a certain job in 19 days . b is 90 % more efficient than a . in how many days can b complete the same job ? | "let , total work unit = 190 units a can finish in 19 days = 190 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 90 / 100 ) * 10 = 19 unit work days in which b will complete the work alone = 190 / 19 = 10 days answer : option e" | a = 19 * 90
b = a / 100
|
a ) 40 m , b ) 42 m , c ) 47 m , d ) 49 m , e ) 50 m | b | multiply(divide(56, multiply(20, 7)), multiply(35, 3)) | if 20 men can build a water fountain 56 metres long in 7 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 7 : 3 : : 56 : x therefore ( 20 x 7 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 140 = 42 hence , the required length is 42 m . answer : b" | a = 20 * 7
b = 56 / a
c = 35 * 3
d = b * c
|
a ) 5 , b ) 10 , c ) 22 , d ) 25 , e ) 30 | c | add(divide(lcm(20, 55), 20), const_10) | jaime earned enough money by selling seashells at 20 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ? | "let ' s test answer c : 22 seashells . . . with 22 seashells , jamie would have 22 ( 20 ) = 440 cents . this would allow him to buy 8 books for 440 cents total , so this is the correct answer . c" | a = math.lcm(20, 55)
b = a / 20
c = b + 10
|
a ) 20 , b ) 41 , c ) 42 , d ) 53 , e ) 64 | b | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 248 are multiple of 4 but not multiples of 6 ? | "244 / 4 = 61 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 61 - 20 = 41 answer : b" | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 34 , b ) 41 , c ) 68 , d ) 88 , e ) none | b | add(multiply(divide(210, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 210 sq . feet , how many feet of fencing will be required ? | "explanation we have : l = 20 ft and lb = 210 sq . ft . so , b = 10.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 21 ) ft = 41 ft . answer b" | a = 210 / 20
b = a * 2
c = b + 20
|
a ) - 4 , b ) 4 , c ) - 1 / 4 , d ) 1 / 4 , e ) 2 | d | divide(const_1, 4) | in the coordinate plane a slope of the line k is 4 times the x - intercept of the line k . what is the y - intercept of the line k ? | as 4 y = mx + c , from 4 y = 0 m + m we get y = 1 / 4 . hence , the correct answer choice is d . | a = 1 / 4
|
a ) 896 , b ) 1008 , c ) 1120 , d ) 1024 , e ) none of these | b | multiply(divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(112, const_1000), const_3), const_3), const_3)) | find the smallest four - digit number which is a multiple of 112 . | "the smallest four digit number is 1000 . if 1000 is divided by 112 , the remainder is 104 . 112 - 104 = 8 , if 8 is added to 1000 , it will become the smallest four digit number and a multiple of 112 . answer : b" | a = 112 / 1000
b = a / 3
c = b / 3
d = c / 3
e = 112 / 1000
f = e / 3
g = f / 3
h = g / 3
i = d * h
|
a ) 0 , b ) 6993 , c ) 699930 , d ) 996330 , e ) none of them | c | divide(add(69758472, const_1), const_2) | the difference between the place values of two sevens in the numerical 69758472 is | "required difference = ( 700000 - 70 ) = 699930 . answer is c" | a = 69758472 + 1
b = a / 2
|
a ) 2 , b ) 0 , c ) - 1 , d ) can not be determined , e ) 1 | a | subtract(subtract(subtract(add(add(9, 8), 3), 3), 5), 8) | a set s = { x , - 8 , - 5 , - 3 , 3 , 6 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ? | "median of the set = ( - 3 + 3 ) / 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 18 - 16 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer a" | a = 9 + 8
b = a + 3
c = b - 3
d = c - 5
e = d - 8
|
a ) $ 8,000 , b ) $ 4,000 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 800 | b | subtract(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 30), multiply(const_2, const_4)), multiply(divide(30, const_100), const_2))) | the market value of a certain machine decreased by 30 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 10,000 , what was its market value two years later ? | "b . market value in 1982 = $ 10000 market value in 1983 = $ 10000 - ( $ 10000 x 30 / 100 ) = 10000 - 3000 = $ 7000 market value in 1984 = market value in 1983 - ( 30 % of $ 10000 ) = 7000 - 3000 = $ 4000" | a = 100 * 30
b = 2 * 4
c = a * b
d = 100 * 30
e = 2 * 4
f = d * e
g = 30 / 100
h = g * 2
i = f * h
j = c - i
|
a ) 1 / 8 , b ) 4 , c ) 8 , d ) 16 , e ) 17 | b | log(divide(log(subtract(16, multiply(add(const_4, const_1), const_1000))), log(add(const_4, const_1)))) | the value of log 2 16 is | "solution let log 216 = n . then , 2 n = 16 = 24 ‹ = › n = 4 . answer b" | a = 4 + 1
b = a * 1000
c = 16 - b
d = math.log(c)
e = 4 + 1
f = math.log(e)
g = d / f
h = math.log(g)
|
a ) 35 ⁄ 18 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 9 ⁄ 20 , e ) 5 ⁄ 16 | a | multiply(divide(7, 3), divide(5, 6)) | dividing by 3 ⁄ 7 and then multiplying by 5 ⁄ 6 is the same as dividing by what number ? | "say x / 3 / 7 * 5 / 6 = x * 7 / 3 * 5 / 6 = x * 35 / 18 a" | a = 7 / 3
b = 5 / 6
c = a * b
|
a ) 19 , b ) 17 , c ) 13 , d ) 10 , e ) 12 | c | multiply(multiply(divide(7, 5), divide(13, 7)), 5) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 13 . if p invested the money for 5 months , find for how much time did q invest the money ? | "7 * 5 : 5 * x = 7 : 13 x = 13 answer : c" | a = 7 / 5
b = 13 / 7
c = a * b
d = c * 5
|
a ) 7 min , b ) 6 min , c ) 9 min , d ) 10 min , e ) 11 min | b | multiply(const_60, divide(subtract(50, 45), 50)) | excluding stoppages , the speed of a bus is 50 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 5 km less . time taken to cover 5 km = ( 5 / 50 ) x 60 = 6 min answer : b" | a = 50 - 45
b = a / 50
c = const_60 * b
|
a ) 28 , b ) 25 , c ) 24 , d ) 23 , e ) 21 | c | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 16) | by selling 16 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 16 120 % - - - ? 80 / 120 * 16 = 24 answer : c" | a = 100 + 20
b = 100 - 20
c = 1 / b
d = a * c
e = 1 / d
f = e * 16
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | multiply(divide(42, subtract(9, 2)), 2) | the ratio of spinsters to cats is 2 to 9 . if there are 42 more cats than spinsters , how many spinsters are there ? | "let 2 x be the number of spinsters . then 9 x is the number of cats . 9 x - 2 x = 42 x = 6 and the number of spinsters is 2 ( 6 ) = 12 . the answer is d ." | a = 9 - 2
b = 42 / a
c = b * 2
|
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