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 ) 46 , b ) 49 , c ) 40 , d ) 21 , e ) 19 | c | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . how old id the mother at present ? | "let the mother ' s present age be x years . then , the person ' s present age = 2 / 5 x years . ( 2 / 5 x + 8 ) = 1 / 2 ( x + 8 ) 2 ( 2 x + 40 ) = 5 ( x + 8 ) = > x = 40 answer : c" | a = 2 + 3
b = 8 - a
c = 1 / 2
d = 2 + 3
e = 2 / d
f = c - e
g = b / f
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a ) 0.45 , b ) 0.3505 , c ) 0.3509 , d ) 0.351 , e ) 0.3527 | a | multiply(divide(4.509, 10.02), const_100) | 4.509 / 10.02 = | "4.509 / 10.02 4509 / 1002 = 4.5 move the comma two places to the left as we have 2 decimal places from the divisor : 0.45 . answer : a" | a = 4 / 509
b = a * 100
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a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | divide(subtract(add(26, add(26, 5)), multiply(const_3.0, const_3.0)), const_2) | the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 26 + 31 ) = 9 ( x - 1 ) 11 x - 9 x = 48 2 x = 48 x = 24 . the average age of the team is 24 years . the answer is d ." | a = 26 + 5
b = 26 + a
c = 3 * 0
d = b - c
e = d / 2
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a ) 6,12 , b ) 10,6 , c ) 10,8 , d ) 8,4 , e ) 6,8 | c | divide(divide(multiply(5, 80), 4), const_10) | the ratio of type of pen and type of notebook is 5 : 4 . and in 80 ways one can start writing . tell me the no . of pen and no . of notebook . | let the ratio constant be x . no . of pen = 5 x , no . of notebook = 4 x . total no . of ways of start writing = 5 x * 4 x = 20 ( square of x ) = 80 x = â ˆ š 4 = 2 pen = 10 , notebook = 8 answer c | a = 5 * 80
b = a / 4
c = b / 10
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a ) 20 , b ) 35 , c ) 55 , d ) 65 , e ) 80 | b | subtract(multiply(50, 4), multiply(55, const_3)) | joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 50 . he was allowed to drop his lowest score . after doing so , his average test score improved to 55 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 50 . so what we can assume is that we have 4 test scores , each 50 . he dropped his lowest score and the avg went to 55 . this means that the lowest score was not 50 and other three scores had given the lowest score 5 each to make it up to 50 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 50 . so the lowest score = 50 - 15 = 35 answer ( b )" | a = 50 * 4
b = 55 * 3
c = a - b
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(divide(divide(2420, 11), 11), const_10) | jamshid and irwin decided to make up their own version of basketball . in addition to being able to score 2 - point baskets , they created ways to score 5 - , 11 - , and 13 - point baskets . if at halftime the product of the point values of irwin ’ s baskets was 2420 , how many 11 - point baskets e did irwin score ? | first thing i noticed was that all these numbers are prime . . . prime factorization gives the following step 1 : 2 * 1210 step 2 : 2 * 11 * 110 step 4 : 2 * 11 * 11 * 10 step 5 : 2 * 11 * 11 * 2 * 5 so he scored e = 2 11 - point baskets in my opinion . answer c | a = 2420 / 11
b = a / 11
c = b / 10
|
a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 250 metres | c | multiply(18, multiply(40, const_0_2778)) | a train is running at a speed of 40 km / hr and it crosses a post in 18 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 = 18 s distance covered , d = vt = ( 400 / 36 ) × 18 = 200 m distance covered is equal to the length of the train = 200 m correct answer is 200 metres | a = 40 * const_0_2778
b = 18 * a
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a ) 48 , b ) 68 , c ) 54 , d ) 60 , e ) none of these | c | divide(multiply(70, 300), 25) | 70 % of 300 + 25 % of 400 - ? = 256 | "explanation : solution : let 70 % of 300 + 25 % of 400 - x = 256 . then , x = ( 70 / 100 * 300 ) + ( 25 / 100 * 400 ) - 256 = 210 + 100 - 256 = 54 . answer : c" | a = 70 * 300
b = a / 25
|
a ) 25 kms , b ) 60 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | b | multiply(speed(add(30, 50), const_60), 45) | riya and priya set on a journey . riya moves eastward at a speed of 30 kmph and priya moves westward at a speed of 50 kmph . how far will be priya from riya after 45 minutes | "total eastward distance = 30 kmph * 3 / 4 hr = 22.50 km total westward distance = 50 kmph * 3 / 4 hr = 37.5 km total distn betn them = 22.5 + 37.5 = 60 km ans 60 km answer : b" | a = 30 + 50
b = speed * (
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a ) 155 , b ) 129 , c ) 61 , d ) 25.3 , e ) 43 | d | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | let c be defined as the sum of all prime numbers between 0 and 23 . what is c / 3 | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 sum , c = 77 c / 3 = 25.3 answer d" | a = 3 / 10
b = 2 + 4
c = 2 ** b
d = a + c
|
a ) 50 hours , b ) 80 hours , c ) 100 hours , d ) 150 hours , e ) 110 hours | e | inverse(subtract(divide(const_1, 10), divide(const_1, const_10))) | a cistern which could be filled in 10 hours takes 2 hour more to be filled owing to a leak in its bottom . if the cistern is full , in what time will the leak empty it ? | "let the leak empty the full cistern in x hours 10 x / x - 10 = 10 + 1 x = 110 hours answer is e" | a = 1 / 10
b = 1 / 10
c = a - b
d = 1/(c)
|
a ) 72 , b ) 71 , c ) 70 , d ) 30 , e ) 45 | c | divide(7, subtract(86.1, floor(86.1))) | when positive integer x is divided by positive integer y , the remainder is 7 . if x / y = 86.1 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 7 - - > x = qy + 7 ; x / y = 86.1 - - > x = 86 y + 0.1 y ( so q above equals to 86 ) ; 0.1 y = 7 - - > y = 70 . answer : c ." | a = math.floor(86, 1)
b = 86 - 1
c = 7 / b
|
a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 800 , e ) $ 1000 | d | multiply(10000, divide(8, const_100)) | find the simple interest on $ 10000 at 8 % per annum for 12 months ? | "p = $ 10000 r = 8 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 8 * 1 / 100 = $ 800 answer is d" | a = 8 / 100
b = 10000 * a
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a ) 24 , b ) 12 , c ) 9 , d ) 6 , e ) 4 | a | subtract(multiply(6, const_2), multiply(2, const_2)) | a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 6 : 5 : 2 , respectively , by weight . if 78 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ? | "we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 6 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : 6 x + 5 x + 2 x = 78 13 x = 78 x = 6 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( 2 ) ( 6 ) = 12 pounds of apples = ( 6 ) ( 6 ) = 36 thus we know that there are 36 - 12 = 24 more pounds of apples than grapes . answer is a ." | a = 6 * 2
b = 2 * 2
c = a - b
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a ) 16 . , b ) 8 . , c ) 5 . , d ) 2 . , e ) - 2 . | c | subtract(add(11, 3), 9) | if ( a + b ) = 11 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 11 = > a = 11 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 11 - b + 3 - c = > 14 - ( b + c ) = > 14 - 9 = 5 . option c . . ." | a = 11 + 3
b = a - 9
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a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | d | subtract(1, add(divide(1, 3), multiply(divide(3, 4), subtract(const_1, divide(1, 3))))) | from a vessel on the first day , 1 / 3 rd of the liquid evaporates . on the second day 3 / 4 th of the remaining liquid evaporates . what fraction of the volume is present at the end of the 2 day | let x be the volume . . . after 1 st day volume remaining = ( x - x / 3 ) = 2 x / 3 after 2 st day volume remaining = ( 2 x / 3 ) - ( ( 2 x / 3 ) * ( 3 / 4 ) ) = ( 2 x / 3 ) ( 1 - 3 / 4 ) = ( 2 x / 3 ) * ( 1 / 4 ) = x / 6 answer : d | a = 1 / 3
b = 3 / 4
c = 1 / 3
d = 1 - c
e = b * d
f = a + e
g = 1 - f
|
a ) $ 16800 , b ) $ 16500 , c ) $ 16000 , d ) $ 16300 , e ) $ 16200 | c | multiply(8, 2000) | the ratio of the incomes of uma and bala is 8 : 7 and the ratio of their expenditure is 7 : 6 . if at the end of the year , each saves $ 2000 then the income of uma is ? | let the income of uma and bala be $ 8 x and $ 7 x let their expenditures be $ 7 y and $ 6 y 8 x - 7 y = 2000 - - - - - - - 1 ) 7 x - 6 y = 2000 - - - - - - - 2 ) from 1 ) and 2 ) x = 1000 uma ' s income = 8 x = 8 * 2000 = $ 16000 answer is c | a = 8 * 2000
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a ) 25 , b ) 26 , c ) 18 , d ) 19 , e ) 10 | a | subtract(multiply(3, 35), subtract(multiply(6, 30), multiply(4, 25))) | the average of 6 numbers is 30 . if the average of first 4 is 25 and that of last 3 is 35 , the fourth number is : | let the six numbers be , a , b , c , d , e , f . a + b + c + d + e + f = 30 × 6 = 180 - - - - ( 1 ) a + b + c + d = 25 × 4 = 100 - - - - ( 2 ) d + e + f = 35 × 3 = 105 - - - - ( 3 ) add 2 nd and 3 rd equations and subtract 1 st equation from this . d = 25 answer : a | a = 3 * 35
b = 6 * 30
c = 4 * 25
d = b - c
e = a - d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(45, 15) | how many of the positive factors of 15 , 45 and how many common factors are there in numbers ? | factors of 15 - 1 , 3 , 5 , and 15 factors of 45 - 1 , 3 , 9 , 15 and 45 comparing both , we have three common factors of 45,16 - 3 answer c | a = 45 / 15
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a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 3200 , 28800 , e ) 12000 , 20001 | d | multiply(subtract(9, const_2), divide(32000, add(1, subtract(9, const_2)))) | divide rs . 32000 in the ratio 1 : 9 ? | "1 / 10 * 32000 = 3200 9 / 10 * 32000 = 28800 answer : d" | a = 9 - 2
b = 9 - 2
c = 1 + b
d = 32000 / c
e = a * d
|
a ) 30 kmh , b ) 40 kmh , c ) 45 kmh , d ) 50 kmh , e ) 55 kmh | d | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(75, 25)), const_4), power(25, const_2))), 25), const_2) | if a car had traveled 25 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 75 km , at what speed did it travel ? | time = distance / speed difference in time = 1 / 2 hrs 75 / x - 75 / ( x + 25 ) = 1 / 2 substitute the value of x from the options . - - > x = 50 - - > 75 / 50 - 75 / 75 = 3 / 2 - 1 = 1 / 2 answer : d | a = 75 * 25
b = 2 * a
c = b * 4
d = 25 ** 2
e = c + d
f = math.sqrt(e)
g = f - 25
h = g / 2
|
a ) 515 . , b ) 545 . , c ) 661 . , d ) 644 . , e ) 666 . | c | add(240, 423) | in the faculty of reverse - engineering , 240 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | "answer is c : 661 solution : total number of students studying both are 423 + 240 - 134 = 529 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 529 , so 100 % is approx . 661 ." | a = 240 + 423
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a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 7.7 | e | divide(770, divide(multiply(multiply(10, 770), divide(add(const_100, 10), const_100)), subtract(multiply(770, divide(add(const_100, 10), const_100)), 770))) | machine a and machine b are each used to manufacture 770 sprockets . it takes machine a 10 hours longer to produce 770 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 770 qt = 700 for a : ( t + 10 ) ( q ) = 770 qt + 10 q = 770 700 + 10 q = 770 q = 7 so a can produce 7 / hour . then b can produce = 7 ( 1.1 ) = 7.7 / hour . e" | a = 10 * 770
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 770 * f
h = g - 770
i = d / h
j = 770 / i
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(5, const_3), const_2), const_4), 5), add(multiply(multiply(multiply(5, const_3), const_2), const_4), 3)), 1) | for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 5 + 3 and * 5 + 5 , inclusive ? | generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 5 would be divisible by all numbers from 1 to 5 . = > * 5 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 5 is divisible by 3 ) in fact adding anyprimenumber between 1 to 5 to * 5 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 5 + 3 and * 5 + 11 then the answer would be 1 . for * 5 + 3 and * 5 + 13 , it is 2 and so on . . . a | a = 5 * 3
b = a * 2
c = b * 4
d = c + 5
e = 5 * 3
f = e * 2
g = f * 4
h = g + 3
i = d - h
j = i - 1
|
a ) 40 , b ) 30 , c ) 25 , d ) data inadequate , e ) none of these . | a | divide(add(60, 20), const_2) | the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a" | a = 60 + 20
b = a / 2
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a ) 95 % , b ) 93 % , c ) 90 % , d ) 80 % , e ) none of these | b | subtract(const_100, multiply(divide(add(40, const_100), add(30, const_100)), const_100)) | two numbers are respectively 40 % and 30 % more than a third number . the second number expressed in terms of percentage of the first is ? | "here , x = 40 and y = 30 ; therefore second number = [ [ ( 100 + y ) / ( 100 + x ) ] x 100 ] % of first number = [ [ ( 100 + 30 ) / ( 100 + 40 ) ] x 100 ] % of first number = 92.8 % of the first answer : b" | a = 40 + 100
b = 30 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 13 , 3 , b ) 18 , 4 , c ) 15 , 3 , d ) 14 , 4 , e ) none of these | b | divide(divide(add(22, 14), const_2), const_2) | a man can row downstream at 22 kmph and upstream at 14 kmph . find the speed of the man in still water and the speed of stream respectively ? | "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 14 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 4 . answer : option b" | a = 22 + 14
b = a / 2
c = b / 2
|
a ) 426.2 , b ) 426.4 , c ) 106.6 , d ) 422.4 , e ) 522.5 | b | multiply(8.5, 6.4) | 8.5 × 6.4 + 4.5 × 11.6 = ? ÷ 4 | "explanation : ( 54.4 + 52.2 ) × 4 = 426.4 answer : option b" | a = 8 * 5
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['a ) 1 / 3', 'b ) 2 / 3', 'c ) 3 / 4', 'd ) 2 / 5', 'e ) 5 / 6'] | d | divide(2, divide(50, const_10)) | a cubical tank is filled with water to a level of 2 feet . if the water in the tank occupies 50 cubic feet , to what fraction of its capacity is the tank filled with water ? | the volume of water in the tank is h * l * b = 50 cubic feet . since h = 2 , then l * b = 25 and l = b = 5 . since the tank is cubical , the capacity of the tank is 5 * 5 * 5 = 125 . the ratio of the water in the tank to the capacity is 50 / 125 = 2 / 5 the answer is d . | a = 50 / 10
b = 2 / a
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a ) 88 , b ) 45 , c ) 36 , d ) 80 , e ) 12 | b | divide(add(240, 210), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 210 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 210 = 450 m . time taken = 450 / 10 = 45 sec . answer : b" | a = 240 + 210
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 10 hours , b ) 84 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours | b | inverse(subtract(divide(const_1, 12), divide(const_1, add(12, 2)))) | a cistern is normally filled in 12 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 12 - 1 / x = 1 / 14 , solving 1 / x = 1 / 84 so in 84 hours full cistern will be empty . answer : b" | a = 1 / 12
b = 12 + 2
c = 1 / b
d = a - c
e = 1/(d)
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a ) 1 , b ) 3 , c ) 0 , d ) 5 , e ) 7 | d | divide(divide(lcm(100, const_1), const_10), const_2) | what is the units digit of the product of the first 100 odd numbers ? | 1 * 3 * 5 * 7 . . . . . . . . . . . . . . . . . . . will end up in 5 in the units place answer : d | a = math.lcm(100, 1)
b = a / 10
c = b / 2
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a ) 201 , b ) 193 , c ) 200 , d ) 199 , e ) 195 | d | divide(subtract(2000, 20), 10) | how many positive integers between 20 and 2000 ( both are inclusive ) are there such that they are multiples of 10 ? | "multiples of 10 = 20 , 30,40 - - - - - , 1990,2000 number of multiples of 10 = > 2000 - 20 / 10 + 1 = 199 answer is d" | a = 2000 - 20
b = a / 10
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a ) rs . 5280 , b ) rs . 5500 , c ) rs . 6000 , d ) rs . 6480 , e ) rs . 7680 | d | multiply(multiply(54, 24), divide(50, 10)) | what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 54 m & 24 m respectively ? | area of the parallelogram = length of the base * perpendicular height = 54 * 24 = 1296 m . total cost of levelling = rs . 6480 d | a = 54 * 24
b = 50 / 10
c = a * b
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 5 | a | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 12))) | a and b together can complete a piece of work in 4 days . if a alone can complete the same work in 12 days , in how many days , in how many days can b alone complete that work ? | "( a + b ) ' s 1 day ' s work = 1 / 4 a ' s 1 day ' s work = 1 / 12 b ' s 1 days ' s work = ( 1 / 4 ) - ( 1 / 12 ) = 1 / 6 hence , b alone can complete the work in 6 days . answer is a" | a = 1 / 4
b = 1 / 12
c = a - b
d = 1 / c
|
['a ) 5', 'b ) 4', 'c ) 3', 'd ) 2', 'e ) 1'] | b | power(divide(4, const_2), const_2) | the area of a square is added to one of its sides , and the perimeter is then subtracted from this total , the result is 4 . what is the length of one side ? | the equation is ; side + area - perimeter = s + a - p = s + s ^ 2 - 4 s = s ( 1 + s - 4 ) . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : b | a = 4 / 2
b = a ** 2
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a ) 240 , b ) 270 , c ) 276 , d ) 271 , e ) 272 | a | multiply(divide(870, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 870 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should a pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 8 / 29 * 870 = 240 answer : a" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 870 / e
g = 16 * 9
h = f * g
|
a ) 5 , b ) 10 , c ) 37 , d ) 40 , e ) 45 | c | divide(subtract(68.5, 50), subtract(1.5, 1)) | caleb spends $ 68.50 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 = 68.50 . by solving the equations y = 37 . ans c . | a = 68 - 5
b = 1 - 5
c = a / b
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a ) 144 , b ) 115 , c ) 113 , d ) 88 , e ) 31 | b | subtract(119, subtract(add(144, 119), 234)) | in a graduating class of 234 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 234 = 29 greatest - smallest is 144 - 29 = 115 so answer must be b . 115" | a = 144 + 119
b = a - 234
c = 119 - b
|
a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec | c | divide(add(150, 100), multiply(add(60, 70), const_0_2778)) | two trains are moving at 60 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ? | "70 + 60 = 130 * 5 / 18 = 325 / 9 mps d = 150 + 100 = 250 m t = 250 * 9 / 325 = 7 sec answer : c" | a = 150 + 100
b = 60 + 70
c = b * const_0_2778
d = a / c
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a ) 24 % , b ) 34 % , c ) 22 % , d ) 31 % , e ) 8.5 % | d | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(11, const_100), const_4), divide(11, const_100))), const_100) | one fourth of a solution that was 11 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.11 + 25 * x = 100 * 0.16 - - > x = 0.31 . answer : d ." | a = 16 / 100
b = a * 4
c = 11 / 100
d = c * 4
e = 11 / 100
f = d - e
g = b - f
h = g * 100
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a ) a ) 1682209 , b ) b ) 1951601 , c ) c ) 1951602 , d ) d ) 1951603 , e ) e ) 1951604 | a | multiply(divide(1297, 1297), const_100) | 1297 x 1297 = ? | "1297 x 1297 = ( 1297 ) 2 = ( 1300 - 3 ) 2 = ( 1300 ) 2 + ( 3 ) 2 - ( 2 x 1300 x 3 ) = 1690000 + 9 - 7800 = 1690009 - 7800 = 1682209 . answer : a" | a = 1297 / 1297
b = a * 100
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a ) 0.45 , b ) 0.55 , c ) 0.65 , d ) 0.32 , e ) 0.35 | d | multiply(divide(80, multiply(multiply(const_5, const_5), const_4)), divide(40, multiply(multiply(const_5, const_5), const_4))) | 80 percent of the members of a study group are women , and 40 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ? | say there are 100 people in that group , then there would be 0.8 * 0.40 * 100 = 32 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable / total = 32 / 100 . answer : d | a = 5 * 5
b = a * 4
c = 80 / b
d = 5 * 5
e = d * 4
f = 40 / e
g = c * f
|
a ) 8 , b ) 64 , c ) 16 , d ) 32 , e ) 4 | b | multiply(power(2, 2), 2) | if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 5 , what is the value of 2 ^ a * 2 ^ b ? | "2 ^ ab = 2 ^ 5 therefore ab = 5 either a = 1 or 5 or b = 5 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 6 = 64 b" | a = 2 ** 2
b = a * 2
|
a ) $ 20000 , b ) $ 15000 , c ) $ 32000 , d ) $ 10000 , e ) $ 9000 | c | multiply(multiply(8000, const_2), const_2) | if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 8000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , when pat is ready for college ? | "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 8,000 * 2 ) * 2 = $ 32,000 ( after 9 years investment will become $ 8,000 * 2 = $ 16,000 and in another 9 years it ' ll become $ 16,000 * 2 = $ 32,000 ) . answer : c ." | a = 8000 * 2
b = a * 2
|
a ) 10 , b ) 15 , c ) 20 , d ) 30 , e ) 60 | d | subtract(divide(add(60, 0), const_2), divide(add(120, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 120 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 60 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 120 is 120 / 2 = 60 the average of the even numbers from 0 to 60 is 60 / 2 = 30 the answer is d ." | a = 60 + 0
b = a / 2
c = 120 + 0
d = c / 2
e = b - d
|
a ) 45 ( 4 / 11 ) % , b ) 50 % , c ) 45 ( 5 / 11 ) % , d ) 44 ( 5 / 11 ) % , e ) none of these | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | a batsman scored 120 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : total runs scored = 120 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 120 - 60 = 60 required % = ( 60 / 120 ) × 100 = 50 % answer : option b" | a = 3 * 8
b = 8 * 3
c = a + b
d = 120 - c
e = d / 120
f = e * 100
|
a ) 37 , b ) 26 , c ) 34 , d ) 18 , e ) 11 | c | divide(add(add(24, 26), multiply(6, 3)), const_2) | the average age of 6 men increases by 3 years when two women are included in place of two men of ages 24 and 26 years . find the average age of the women ? | explanation : 24 + 26 + 6 * 3 = 68 / 2 = 34 answer : c | a = 24 + 26
b = 6 * 3
c = a + b
d = c / 2
|
a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | a | subtract(subtract(9, 4), subtract(14, 12)) | | 9 - 4 | - | 12 - 14 | = ? | "| 9 - 4 | - | 12 - 14 | = | 5 | - | - 2 | = 5 - 2 = 3 correct answer a" | a = 9 - 4
b = 14 - 12
c = a - b
|
a ) $ 200 , b ) 100 , c ) $ 105 , d ) $ 250 , e ) $ 300 | b | divide(multiply(divide(multiply(divide(40, const_100), 600), divide(80, const_100)), divide(40, const_100)), divide(80, const_100)) | a school has received 80 % of the amount it needs for a new building by receiving a donation of $ 600 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 40 % of them donated $ 24000 ( 600 * 40 ) $ 24000 is 80 % of total amount . so total amount = 24000 * 100 / 80 remaining amount is 20 % of total amount . 20 % of total amount = 24000 * ( 100 / 80 ) * ( 20 / 100 ) = 6000 this amount has to be divided by 60 ( remaining people are 60 ) so per head amount is 6000 / 60 = $ 100 answer : b" | a = 40 / 100
b = a * 600
c = 80 / 100
d = b / c
e = 40 / 100
f = d * e
g = 80 / 100
h = f / g
|
a ) 400 , b ) 540 , c ) 100 , d ) 830 , e ) 420 | a | multiply(power(5, 2), power(4, 2)) | if 4 ^ k = 5 , then 4 ^ ( 2 k + 2 ) = | "4 ^ k = 5 4 ^ 2 k = 5 ^ 2 4 ^ 2 k = 25 4 ^ ( 2 k + 2 ) = 4 ^ 2 k * 4 ^ 2 = 25 * 16 = 400 answer : a" | a = 5 ** 2
b = 4 ** 2
c = a * b
|
a ) 288 , b ) 267 , c ) 261 , d ) 211 , e ) 306 | e | multiply(add(16, const_1), add(add(16, const_1), const_1)) | there are 16 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | the total number of stations = 18 from 18 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 18 p ₂ ways . 18 p ₂ = 18 * 17 = 306 . answer : e | a = 16 + 1
b = 16 + 1
c = b + 1
d = a * c
|
a ) $ 115,000 , b ) $ 160,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(15, 10), 10) | an auction house charges a commission of 15 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | say the price of the house was $ x , then 0.15 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 215,000 ( 15 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c | a = 15 * 10
b = a + 10
|
a ) 168 km , b ) 864 km , c ) 200 km , d ) 240 km , e ) 460 km | c | add(multiply(divide(13, const_3), 20), multiply(divide(13, const_3), 25)) | a man travelled for 13 hours . he covered the first half of the distance at 20 kmph and remaining half of the distance at 25 kmph . find the distance travelled by the man ? | let the distance travelled be x km . total time = ( x / 2 ) / 20 + ( x / 2 ) / 25 = 13 = > x / 40 + x / 50 = 13 = > ( 5 x + 4 x ) / 200 = 13 = > x = 200 km answer : c | a = 13 / 3
b = a * 20
c = 13 / 3
d = c * 25
e = b + d
|
a ) 600 , b ) 1200 , c ) 1500 , d ) 1600 , e ) 3000 | e | multiply(divide(5, 4), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 4 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | sugar : flour = 5 : 4 = 25 : 20 ; flour : soda = 10 : 1 = 20 : 2 ; thus we have that sugar : flour : soda = 25 x : 20 x : 2 x . also given that 20 x / ( 2 x + 60 ) = 8 / 1 - - > x = 120 - - > sugar = 25 x = 3,000 answer : e . | a = 5 / 4
b = 8 * 60
c = 10 - 8
d = b / c
e = 10 * d
f = a * e
|
a ) 100,200 , b ) 400,100 , c ) 300,150 , d ) 300,200 , e ) 250,250 | d | divide(multiply(2, 3), add(2, 3)) | a can do a work in 2 days . b can do the same work in 3 days . both a & b together will finish the work and they got $ 500 from that work . find their shares ? | "ratio of their works a : b = 2 : 3 ratio of their wages a : b = 3 : 2 a ' s share = ( 3 / 5 ) 500 = 300 b ' s share = ( 2 / 5 ) 500 = 200 correct option is d" | a = 2 * 3
b = 2 + 3
c = a / b
|
a ) 18 , b ) 20 , c ) 77 , d ) 66 , e ) 41 | b | multiply(divide(subtract(1890, 1500), 1890), const_100) | the cost price of a radio is rs . 1890 and it was sold for rs . 1500 , find the loss % ? | "1890 - - - - 390 100 - - - - ? = > 20 % answer : b" | a = 1890 - 1500
b = a / 1890
c = b * 100
|
a ) 25 % , b ) 20 % , c ) 16.67 % , d ) 33.33 % , e ) none of these | e | multiply(subtract(const_1, divide(const_100, add(const_100, 44))), const_100) | if the price of petrol increases by 44 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 44 ) i . e 144 . now , his consumption should be reduced to : - = ( 144 − 100 ) / 144 ∗ 100 . hence , the consumption should be reduced to 30.6 % . answer : e" | a = 100 + 44
b = 100 / a
c = 1 - b
d = c * 100
|
a ) rs . 7000 , b ) rs . 9000 , c ) rs . 14000 , d ) rs . 17000 , e ) rs . 27000 | c | divide(840, divide(multiply(subtract(15, 12), const_2), const_100)) | a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 840 . find the sum ? | let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 840 = > 30 x / 100 - 24 x / 100 = 840 = > 6 x / 100 = 840 = > x = 14000 . answer : c | a = 15 - 12
b = a * 2
c = b / 100
d = 840 / c
|
a ) 44 , b ) 55 , c ) 66 , d ) 77 , e ) 88 | d | divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 35 % of 4 / 13 of 715 ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 715 , or 1 / 9 of 715 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate , the errors will partially cancel each other out . our estimate will be relatively accurate . the number 715 is about 720 , so ( 1 / 9 ) * 715 will be a bit less than 80 . keeping track not only of your current estimate , but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently . the closest answer is 77 , so this is the answer to choose . the answer is d ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 35 * n
p = o / 100
|
a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 16 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(16, 8)), 16)), subtract(16, 9)), divide(const_1, multiply(16, 8))) | 16 welders work at a constant rate they complete an order in 8 days . if after the first day , 9 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 7 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 16 * 8 = 128 days 4 . but on day 1 sixteen workers had already worked finishing 1 / 8 of the job . so 7 workers have to finish only 7 / 8 of the job . 5 . total time taken by 7 workers can be got from formula used at ( 2 ) . i . e . , 7 * total time taken = 128 . total time taken by 7 workers to finish the complete job is 128 / 7 = 18.286 days . 6 . time taken by 7 workers to finish 7 / 8 of the job is 7 / 8 * 18.286 = 16 days . the answer is choice e" | a = 16 * 8
b = 1 / a
c = b * 16
d = 1 - c
e = 16 - 9
f = d / e
g = 16 * 8
h = 1 / g
i = f / h
|
a ) 25 % , b ) 50 % , c ) 75 % , d ) 95 % , e ) none of them | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(50, const_100)), const_2)), const_100) | if the radius of a circle is decreased by 50 % , find the percentage decrease in its area . | "let original radius = r . new radius = ( 50 / 100 ) r = ( r / 2 ) original area = 22 / 7 ( r ) 2 = and new area = ( ( r / 2 ) ) 2 = ( ( r ) 2 ) / 4 decrease in area = ( ( 3 ( 22 / 7 ) ( r ) 2 ) / 4 x ( 1 / ( 22 / 7 ) ( r ) 2 ) x 100 ) % = 75 % answer is c ." | a = 100 / 100
b = 100 / 100
c = 50 / 100
d = b - c
e = d ** 2
f = a - e
g = f * 100
|
a ) 14 , b ) 20 , c ) 40 , d ) 44 , e ) 48 | a | divide(rectangle_area(multiply(8, const_100), multiply(82, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | what is the least number of squares tiles required to pave the floor of a room 8 m 82 cm long and 2 m 52 cm broad ? | "length of largest tile = h . c . f . of 882 cm and 252 cm = 126 cm . area of each tile = ( 126 x 126 ) cm 2 . required number of tiles = 882 x 252 / ( 126 ^ 2 ) = 14 . answer : a" | a = 8 * 100
b = 82 * 100
c = rectangle_area / (
|
a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 32 , d ) 1 / 64 , e ) 1 / 128 | c | multiply(power(divide(const_1, const_2), 7), multiply(choose(8, 7), divide(const_1, const_2))) | a coin is tossed 8 times . what is the probability of getting exactly 7 heads ? | "the number of possible outcomes is 2 ^ 8 = 256 there are 8 ways to get exactly 7 heads . p ( exactly 7 heads ) = 8 / 256 = 1 / 32 the answer is c ." | a = 1 / 2
b = a ** 7
c = math.comb(8, 7)
d = 1 / 2
e = c * d
f = b * e
|
a ) 105 , b ) 45 , c ) 85 , d ) 95 , e ) 90.5 | e | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(30, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(30, const_100))))) | the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 30 % . 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 30 % , therefore bringing down the price to $ 52.5 . 4 ) now 52.5 has to be added byx % in order to equal the original price . 52.5 + ( x % ) 52.5 = 100 . solving this eq for x , we get x = 90.5 ans is e" | a = 100 - 25
b = 100 - 25
c = 30 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 25
h = 100 - 25
i = 30 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 20 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | c | divide(multiply(divide(multiply(multiply(4, const_3), const_100), 20), multiply(20, 4)), const_100) | tough and tricky questions : word problems . mike , sarah and david decided to club together to buy a present . each of them gave equal amount of money . luckily sarah negotiated a 20 % discount for the present so that each of them paid 4 dollars less . how much did they pay for a present ? | answer c . we know that sarah negotiated a discount of 20 % , so each of them paid $ 4 less . since there are three people , the 20 % of the original price amounts to $ 12 . 5 times 12 $ is 60 $ , so the original price , before sarah negotiated the discount , had been $ 60 . they paid $ 12 less than the base price , so they spent $ 48 . | a = 4 * 3
b = a * 100
c = b / 20
d = 20 * 4
e = c * d
f = e / 100
|
a ) 356,580 , b ) 358,680 , c ) 360,780 , d ) 362,880 , e ) 364,980 | d | factorial(6) | if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } is used exactly once , in how many ways can the digits be arranged ? | "9 ! = 362,880 the answer is d ." | a = math.factorial(6)
|
a ) 10 , b ) 12 , c ) 16 , d ) 18 , e ) 20 | c | subtract(17, 8) | 8 ^ 100 is divisible by 17 . find the remainder ? | "this is an extremely difficult problem to solve with out fermat ' s little theorem . by applying fermat ' s little theorem , we know that 816 when divided by 17 , the remainder is 1 . so divide 100 by 16 and find the remainder . remainder = 4 therefore , 100 = ( 16 × 6 ) + 4 now this problem can be written as 810017 = 816 × 6 + 417 = ( 816 ) 6 × 8417 now this problem simply boils down to ( 1 ) 6 × 8417 = 8417 84 = 82 × 82 , we need to find the remainder when 64 × 64 is divisible by 17 . or 13 × 13 = 169 . when 169 is divided by 17 , remainder is 16 . c" | a = 17 - 8
|
a ) 8000 , b ) 7540 , c ) 6500 , d ) 9100 , e ) 6000 | a | multiply(multiply(subtract(7, 6), const_100), 80) | in a competitive examination in a state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 80 more candidates got selected than a . what was the number of candidates appeared from each state ? | let the number of candidates appeared be x then , 7 % of x - 6 % of x = 80 1 % of x = 80 x = 80 * 100 = 8000 answer is a | a = 7 - 6
b = a * 100
c = b * 80
|
a ) 13 , b ) 25 , c ) 87 , d ) 49 , e ) 63 | a | add(divide(subtract(subtract(100, const_4), add(23, const_1)), 6), const_1) | how many natural numbers are there between 23 and 100 which are exactly divisible by 6 ? | explanation : required numbers are 24 , 30 , 36 , 42 , . . . , 96 this is an a . p . in which a = 24 , d = 6 and l = 96 let the number of terms in it be n . then tn = 96 a + ( n - 1 ) d = 96 24 + ( n - 1 ) x 6 = 96 ( n - 1 ) x 6 = 72 ( n - 1 ) = 12 n = 13 required number of numbers = 13 . a ) | a = 100 - 4
b = 23 + 1
c = a - b
d = c / 6
e = d + 1
|
a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 8 | c | divide(multiply(9, 4), add(multiply(9, 4), multiply(3, 4))) | at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 4 / 9 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 3 / 4 * 4 / 9 x = 1 / 3 x % loaded by day crew = x / ( x + 1 / 3 x ) = 3 / 4 answer c" | a = 9 * 4
b = 9 * 4
c = 3 * 4
d = b + c
e = a / d
|
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 40 % | c | multiply(divide(multiply(50, 0.0008), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.0008 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | total amount of water evaporated each day during a 50 - day period = . 0008 * 50 = . 0008 * 100 / 2 = . 08 / 2 = . 04 percent of the original amount of water evaporated during this period = ( . 04 / 10 ) * 100 % = 0.4 % answer c | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | e | power(add(sqrt(5), sqrt(5)), 2) | if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 5 ¤ √ 5 = | "x = √ 5 and y also = √ 5 applying the function ( √ 5 + √ 5 ) ^ 2 - ( √ 5 - √ 5 ) ^ 2 = ( 2 √ 5 ) ^ 2 - 0 = 4 x 5 = 20 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = √ 5 and y = √ 5 you get the answer 20 answer is e ." | a = math.sqrt(5)
b = math.sqrt(5)
c = a + b
d = c ** 2
|
a ) a ) 11 , b ) b ) 18 , c ) c ) 24 , d ) d ) 27 , e ) d ) 36 | d | add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 16) | 31 of the scientists that attended a certain workshop were wolf prize laureates , and 16 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 50 of the scientists attended that workshop , how many of them were nobel prize laureates ? | lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 13 . . so total who won n = 11 + 16 = 27 . . d | a = 50 - 31
b = a - 3
c = b / 2
d = 3 + c
e = d + 16
|
a ) 20 , b ) 40 , c ) 60 , d ) 90 , e ) 120 | c | sqrt(multiply(180, 20)) | which number should replace both the asterisks in ( * / 20 ) x ( * / 180 ) = 1 ? | "let ( y / 20 ) x ( y / 180 ) = 1 y ^ 2 = 20 x 180 = 20 x 20 x 9 y = ( 20 x 3 ) = 60 the answer is c ." | a = 180 * 20
b = math.sqrt(a)
|
a ) 80 sec , b ) 20 sec , c ) 40 sec , d ) 50 sec , e ) 60 sec | d | divide(add(150, 100), multiply(18, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 18 kmph ? | "d = 100 + 150 = 250 s = 18 * 5 / 18 = 5 mps t = 250 / 5 = 50 sec d ) 50 sec" | a = 150 + 100
b = 18 * const_0_2778
c = a / b
|
a ) 122821 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 128111 | d | divide(multiply(1, 701), const_4) | what is the sum of all even numbers from 1 to 701 ? | "explanation : 700 / 2 = 350 350 * 351 = 122850 answer : d" | a = 1 * 701
b = a / 4
|
a ) 140 , b ) 181 , c ) 160 , d ) 170 , e ) 180 | b | divide(1991, 11) | two numbers have a h . c . f of 11 and a product of two numbers is 1991 . find the l . c . m of the two numbers ? | "l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 1991 / 11 = 181 . answer : b" | a = 1991 / 11
|
a ) 72.85 , b ) 69.85 , c ) 62.85 , d ) 82.85 , e ) 60.85 | b | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 15)), const_100) | at what price must an book costing $ 47.50 be marked in order that after deducting 15 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 85 / 100 ) = 59.375 mp = 69.85 b" | a = 100 + 25
b = 47 * 50
c = b / 100
d = 100 - 15
e = c / d
f = e * 100
|
a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | d | multiply(divide(16.02, 0.001), const_100) | 16.02 × 0.001 = ? | "16.02 × 0.001 = ? or , ? = 0.01602 answer d" | a = 16 / 2
b = a * 100
|
a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 18 , 12,18 | e | divide(multiply(3, 3), 3) | find the numbers which are in the ratio 3 : 2 : 3 such that the sum of the first and the second added to the difference of the third and the second is 24 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 3 given , ( a + b ) + ( c - b ) = 24 = > a + c = 24 = > 3 x + 3 x = 24 = > x = 6 a , b , c are 3 x , 2 x , 3 x a , b , c are 18 , 12 , 18 . answer : e" | a = 3 * 3
b = a / 3
|
a ) 2 , b ) 4 , c ) 9 , d ) 12 , e ) 15 | c | divide(add(13, 5), const_2) | in one hour , a boat goes 13 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 13 + 5 ) kmph = 9 kmph . answer c" | a = 13 + 5
b = a / 2
|
a ) 290 , b ) 304 , c ) 285 , d ) 270 , e ) 275 | c | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 510 visitors on sundays and 240 on other days . what is the average number of visitors per day in a month of 30 days beginning with a sunday ? | "in a month of 30 days beginning with a sunday , there will be 4 complete weeks and another two days which will be sunday and monday . hence there will be 5 sundays and 25 other days in a month of 30 days beginning with a sunday average visitors on sundays = 510 total visitors of 5 sundays = 510 × 5 average visitors on other days = 240 total visitors of other 25 days = 240 × 25 total visitors = ( 510 × 5 ) + ( 240 × 25 ) total days = 30 average number of visitors per day = ( ( 510 × 5 ) + ( 240 × 25 ) ) / 30 = ( ( 51 × 5 ) + ( 24 × 25 ) ) / 3 = ( 17 × 5 ) + ( 8 × 25 ) = 85 + 200 = 285 answer is c ." | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 510
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | divide(add(sqrt(add(multiply(multiply(231, const_2), const_4), const_1)), const_1), const_2) | if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 231 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 231 n * ( n - 1 ) = 462 = 22 * 21 ( trial and error ) the answer is d ." | a = 231 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | floor(sqrt(divide(15400, multiply(multiply(multiply(11, 7), 5), 2)))) | in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the competition and finished first , second , third , or fourth each time . the product of all the points he received was 15400 . how many times did he participate in the competition ? | "15400 = 2 * 2 * 2 * 5 * 5 * 7 * 11 john participated 7 times . the answer is c ." | a = 11 * 7
b = a * 5
c = b * 2
d = 15400 / c
e = math.sqrt(d)
f = math.floor(e)
|
a ) rs . 5725 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6500 , e ) none of these | d | divide(8450, add(const_1, divide(30, const_100))) | the owner of a furniture shop charges his customer 30 % more than the cost price . if a customer paid rs . 8450 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8450 ( 100 / 130 ) = rs . 6500 . answer : d" | a = 30 / 100
b = 1 + a
c = 8450 / b
|
a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | c | subtract(add(const_4, 6), divide(divide(add(22, 14), const_2), 6)) | on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 6 cups of tea . last week mo drank a total of 22 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 22 t - c = 14 - > t = 18 . c = 2 . mo drinks 6 cups of tea a day then number of days that are not rainy = 18 / 6 = 3 so number of rainy days = 7 - 3 = 4 c is the answer ." | a = 4 + 6
b = 22 + 14
c = b / 2
d = c / 6
e = a - d
|
a ) 565350 , b ) 595650 , c ) 535950 , d ) 565350 , e ) 575350 | b | multiply(subtract(3, const_4), 85) | find the value of 85 p 3 . | "85 p 3 = 85 ! / ( 85 - 3 ) ! = 85 ! / 82 ! = 85 * 84 * 83 * 82 ! / 82 ! = 85 * 84 * 83 = 595650 answer : b" | a = 3 - 4
b = a * 85
|
a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 12,000 , e ) e ) 11,100 | d | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 144,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 142 k x = 12,000 answer : d | a = 4 * 2
b = a * 100
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | e | multiply(2, divide(171, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of hydrogen are there in 171 grams of water ? | "( 2 / 18 ) * 171 = 19 grams the answer is e ." | a = 2 + 16
b = 171 / a
c = 2 * b
|
a ) 2 : 5 , b ) 1 : 6 , c ) 2 : 4 , d ) 3 : 8 , e ) 2 : 40 | d | divide(divide(subtract(divide(const_100.0, const_2), const_10), const_2), add(divide(42, const_2), const_10)) | two whole numbers whose sum is 42 can not be in the ratio | d ) 3 : 8 | a = 100 / 0
b = a - 10
c = b / 2
d = 42 / 2
e = d + 10
f = c / e
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 24 | e | multiply(multiply(4, 2), 3) | running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 4 w widgets . | "i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . so , he will take 24 days to doing 4 w widgets . answer : e" | a = 4 * 2
b = a * 3
|
a ) 16.5 , b ) 17.5 , c ) 18.4 , d ) 15.4 , e ) 15.1 | a | add(32, const_1) | the average of first 32 natural numbers is ? | "sum of 32 natural no . = 1056 / 2 = 55 average = 528 / 32 = 16.5 answer : a" | a = 32 + 1
|
a ) $ 1100 , b ) $ 520 , c ) $ 2080 , d ) $ 1170 , e ) $ 630 | c | multiply(4680, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 4680 , what will be a ' s share of the earnings ? | "explanatory answer a , b and c will share the amount of $ 4680 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , the ratio of the work done by a : b : c when they work together will be equal to 1 / 6 : 1 / 8 : 1 / 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 / 6 : 24 / 8 : 24 / 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 4680 will be 4 : 3 : 2 . hence , a ' s share will be 4 * 4680 / 9 = 2080 . correct choice is ( c )" | a = 1/(8)
b = 1/(12)
c = 1/(6)
d = 1/(8)
e = c + d
f = b + e
g = a / f
h = 4680 * g
|
a ) 322 , b ) 480 , c ) 287 , d ) 192 , e ) 107 | b | divide(multiply(600, const_100), add(const_100, 25)) | by selling an article at rs . 600 , a profit of 25 % is made . find its cost price ? | "explanation : sp = 600 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 600 * [ 100 / ( 100 + 25 ) ] = 600 * [ 100 / 125 ] = rs . 480 answer : b" | a = 600 * 100
b = 100 + 25
c = a / b
|
a ) 600 , b ) 700 , c ) 1080 , d ) 900 , e ) 1000 | c | divide(multiply(9, 1200), add(1, 9)) | ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 1200 then loss of pyarelal ? | let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 1200 , loss of pyarelal = 1200 * 9 x / 10 x = 1080 answer : c | a = 9 * 1200
b = 1 + 9
c = a / b
|
a ) 76 , b ) 50 , c ) 27 , d ) 80 , e ) 25 | b | divide(divide(divide(800, divide(add(20, 12), const_2)), 12), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 20 m wide at the top and 12 m wide at the bottom and the area of cross - section is 800 sq m , the depth of cannel is ? | "1 / 2 * d ( 20 + 12 ) = 800 d = 50 answer : b" | a = 20 + 12
b = a / 2
c = 800 / b
d = c / 12
e = d / 2
|
a ) $ 76 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 104 | a | subtract(multiply(90, 15), add(multiply(87, 7), multiply(95, 7))) | the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 95 per day . what was his wage on the 8 th day ? | average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 95 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 95 * 7 ) = 1350 - ( 609 + 665 ) = 1350 - 1274 = 76 answer a | a = 90 * 15
b = 87 * 7
c = 95 * 7
d = b + c
e = a - d
|
a ) 3.33 % , b ) 6 % , c ) 2 % , d ) 95 % , e ) 1 % | a | divide(multiply(const_100, 250), multiply(1500, 5)) | what is the rate percent when the simple interest on rs . 1500 amount to rs . 250 in 5 years ? | "interest for 5 yrs = 250 interest for 1 yr = 50 interest rate = 50 / 1500 x 100 = 3.33 % answer : a" | a = 100 * 250
b = 1500 * 5
c = a / b
|
a ) 230 cm 2 , b ) 245 cm 2 , c ) 255 cm 2 , d ) 260 cm 2 , e ) 280 cm 2 | b | divide(multiply(14, add(20, 15)), const_2) | nd the area of trapezium whose parallel sides are 20 cm and 15 cm long , and the distance between them is 14 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 15 ) * ( 14 ) = 245 cm 2 answer : b" | a = 20 + 15
b = 14 * a
c = b / 2
|
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