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a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | c | divide(divide(choose(42, const_1), 42), power(const_3, const_2)) | each factor of 210 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 42 ? | "210 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 1 / 8 . answer : c ." | a = math.comb(42, 1)
b = a / 42
c = 3 ** 2
d = b / c
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a ) 1 , b ) 3 , c ) 2 , d ) 4 , e ) 5 | c | divide(const_1, add(divide(const_1, 3), divide(const_1, 6))) | a can do a work in 3 days . b can do in 6 days . if both a & b are working together in how many days they can finish the work ? | "1 day work of a = 1 / 3 1 day work of b = 1 / 6 1 day work of a & b = 1 / 3 + 1 / 6 = 1 / 2 a & b finish the work in 2 days answer is c" | a = 1 / 3
b = 1 / 6
c = a + b
d = 1 / c
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a ) 5000 , b ) 4860 , c ) 5400 , d ) 4800 , e ) 5120 | b | multiply(6000, power(divide(subtract(const_100, 10), const_100), 2)) | the population of a village is 6000 . it decreases annually at the rate of 10 % p . a . what will be its population after 2 years ? | formula : ( after = 100 denominator ago = 100 numerator ) 6000 ã — 90 / 100 ã — 90 / 100 = 4860 b | a = 100 - 10
b = a / 100
c = b ** 2
d = 6000 * c
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a ) − 0.3 , b ) 0 , c ) 0.05 , d ) 1.08 , e ) 2.46 | c | subtract(multiply(divide(divide(subtract(power(3, 2), power(2, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(2, 0.3)), const_1000), const_1000)) | what is the value of 3 x ^ 2 − 2 x + 0.3 for x = 0.5 ? | "3 x ^ 2 - 2 x + 0.3 for x = 0.6 = 3 ( 0.5 * 0.5 ) - 4 * 0.5 * ( 0.5 ) + 0.3 = - 0.5 * 0.5 + 0.3 = - 0.25 + 0.3 = 0.05 correct option : c" | a = 3 ** 2
b = 2 ** 0
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 2 ** 0
i = g - h
j = i / 1000
k = j / 1000
l = f - k
|
a ) 15 min , b ) 10 min , c ) 12 min , d ) 20 min , e ) 18 min | a | subtract(const_60, multiply(divide(48, 64), const_60)) | excluding the stoppages , the speed of a bus is 64 km / hr and including the stoppages the speed of the bus is 48 km / hr . for how many minutes does the bus stop per hour ? | "speed of the bus without stoppage = 64 km / hr speed of the bus with stoppage = 48 km / hr difference in speed = 16 km / hr so , the time taken in the stoppages = time taken to cover 16 km = ( 16 / 64 ) hr = 1 / 4 hr = 15 min answer : a" | a = 48 / 64
b = a * const_60
c = const_60 - b
|
a ) $ 4000 , b ) $ 5000 , c ) $ 9000 , d ) $ 11000 , e ) $ 13000 | d | subtract(12000, const_1000) | an analysis of the monthly incentives received by 5 salesmen : the mean and median of the incentives is $ 7000 . the only mode among the observations is $ 12000 . incentives paid to each salesman were in full thousands . what is the difference z between the highest and the lowest incentive received by the 5 salesmen in the month ? | break down the question in to steps : step # 1 : incentives received by 5 salesmen - - > abcde step # 2 : mean and median of the incentives is $ 7000 : via number properties - - > total incentives = 7000 * 5 = 35000 ab 7000 de step # 3 : only mode among the observations is $ 12000 : mode is the value that appears most often in a set of data . therefore 12000 must occur more then once and since 12000 is larger than the mean ( 7000 ) the most it can occur is twice in our list . if we place the numbers in ascending order we haveab 70001200012000 step # 4 : what is the difference z between the highest and the lowest incentive : - the total paid has to equal mean * # of numbers = 7000 * 5 = 35 000 . - the three values we have so far ( 12000 12000 and 7000 ) equal 31000 . - therefore the first two numbers ( ab ) must equal 4000 - there is only one mode which is 12000 , therefore 4000 must be equal to a set of two numbers that are not the and a < b if a = 1000 b = 3000 - - > difference between the highest and the lowest incentive = 12000 - 1000 = 11000 no other options for a or b to fit the criteria final list : 1000300070001200012000 answer d | a = 12000 - 1000
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a ) 30 , b ) 15 , c ) 20 , d ) 36 , e ) 46 | b | subtract(negate(60), multiply(subtract(1200, 300), divide(subtract(1200, 300), subtract(6000, 1200)))) | 6000 , 1200 , 300 , 60 . . . | "6000 / 5 = 1200 1200 / 4 = 300 300 / 5 = 60 60 / 4 = 15 answer : b" | a = negate - (
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a ) 1100 , b ) 3000 , c ) 2100 , d ) 1000 , e ) 2000 | e | sqrt(divide(100, add(power(20, 10), add(power(2, 10), power(10, 10))))) | the ratio of 2 numbers is 10 : 20 and their h . c . f . is 100 . their l . c . m . is ? | "let the numbers be 10 x and 20 x their h . c . f . = 100 so the numbers are 10 * 100 , 20 * 100 = 1000 , 2000 l . c . m = 2000 answer is e" | a = 20 ** 10
b = 2 ** 10
c = 10 ** 10
d = b + c
e = a + d
f = 100 / e
g = math.sqrt(f)
|
a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 97 , d ) 1 , e ) 2 | c | divide(1, 97) | if the numbers 1 to 97 are written on 97 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | "there are 25 primes , 71 composite numbers from 1 to 97 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 97 . answer : c" | a = 1 / 97
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a ) 0.35 % , b ) 3.5 % , c ) 35 % , d ) 350 % , e ) 0.0035 % | d | multiply(3.5, const_100) | 3.5 can be expressed in terms of percentage as ? | "answer : option d 3.5 = 35 / 10 = ( 35 / 10 * 100 ) = 350 %" | a = 3 * 5
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a ) 34 , b ) 36 , c ) 23 , d ) 40 , e ) 42 | c | divide(subtract(94, subtract(power(multiply(4, const_2), const_2), power(multiply(2, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 2 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 4 inches wide on each side , the total area would have been ( m + 94 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 4 ) ( b + 4 ) = m - - - - ( 2 ) and ( l + 8 ) ( b + 8 ) = m + 94 - - - - - - - - - > ( l + 8 ) ( b + 8 ) - 94 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 4 ) ( b + 4 ) = ( l + 8 ) ( b + 8 ) - 94 lb + 4 l + 4 b + 16 = lb + 8 l + 8 b + 64 - 94 simplify 4 l + 4 b = 46 - - - - - - > 2 ( l + b ) = 23 ( check eq ( 1 ) ) answer is c" | a = 4 * 2
b = a ** 2
c = 2 * 2
d = c ** 2
e = b - d
f = 94 - e
g = f / 2
|
a ) a ) 100 , b ) b ) 66.7 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160 | b | divide(multiply(10, const_100), 15) | an inspector rejects 15 % of the meters as defective . how many will he examine to reject 10 ? | "then , 10 % of x = 15 ( 15 / 100 ) x = 10 x = ( 10 * 100 * ) / 15 = 66.7 answer is b" | a = 10 * 100
b = a / 15
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a ) 16 , b ) 17 , c ) 19 , d ) 22 , e ) 26 | b | divide(add(subtract(multiply(16, 10), 25), 35), 10) | the average of 10 numbers is calculated as 16 . it is discovered later on that while calculating the average , the number 35 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | the total sum of the numbers should be increased by 10 . then the average will increase by 10 / 10 = 1 . the correct average is 17 . the answer is b . | a = 16 * 10
b = a - 25
c = b + 35
d = c / 10
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a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 3,120 , e ) $ 1,400 | d | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 20), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | a gambler bought $ 4,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 16 : x + x - 2 = 16 - - > x = 9 : 9 $ 20 chips were lost and 9 - 2 = 7 $ 100 chips were lost . total worth of chips lost is 9 * 20 + 7 * 100 = $ 880 , so the gambler kept $ 4,000 - $ 880 = $ 3,120 . answer : d" | a = 3 * 1000
b = 16 + 2
c = b / 2
d = c * 20
e = 16 + 2
f = e / 2
g = f - 2
h = g * 100
i = d + h
j = a - i
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a ) 100 , b ) 200 , c ) 250 , d ) 300 , e ) 400 | d | divide(subtract(multiply(600, divide(99, const_100)), multiply(600, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | there are 600 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ? | "there are 594 managers and 6 others . the 6 others would compose 2 % of the total number of people if there were 300 people in the room . thus 300 managers must leave . the answer is d ." | a = 99 / 100
b = 600 * a
c = 98 / 100
d = 600 * c
e = b - d
f = 98 / 100
g = 1 - f
h = e / g
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 3 / 4 , e ) 2 / 3 | a | divide(const_4, power(const_2, 3)) | when 3 fair coins are tossed together , what is the probability of getting 2 tails ? | we can get 2 tails in 3 c 2 ways . we can get 3 tails in exactly one way . also , there are a total out 8 possible outcomes . thus , the required probability = ( 3 + 1 ) / 8 = 1 / 2 answer : a | a = 2 ** 3
b = 4 / a
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a ) 3 : 2 , b ) 4 : 3 , c ) 5 : 4 , d ) 6 : 5 , e ) 5 : 3 | a | divide(multiply(divide(5, 3.3), 8), 8) | the ratio of two cow a , b milk is 5 : 3.3 kg milk of cow a get distributed and 1 kg milk of cow b get distributed . now , 8 kg milk of cow b left . now what is the ratio of cow a and b left milk ? | left milk of cow b = 3 x - 1 = 8,3 x = 9 , x = 3 . left milk of cow a = 15 - 3 = 12 12 / 8 = 3 : 2 . answer a | a = 5 / 3
b = a * 8
c = b / 8
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a ) 25 hr , b ) 35 hr , c ) 56 hr , d ) 20 hr , e ) 50 hr | c | multiply(add(add(multiply(const_2, const_2), const_2), const_1), 8) | a tank is filled by 3 pipes a , b , c in 8 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ? | "suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / 2 and x / 4 hours respectively to fill the tank . 1 / x + 2 / x + 4 / x = 1 / 8 7 / x = 1 / 8 x = 56 hours answer is c" | a = 2 * 2
b = a + 2
c = b + 1
d = c * 8
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a ) 425 miles , b ) 625 miles , c ) 325 miles , d ) 375 miles , e ) 625 miles | d | multiply(75, 5) | a car travels at a speed of 75 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 75 + 75 + 75 + 75 + 75 = 5 * 65 = 375 miles correct answer d" | a = 75 * 5
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a ) 2 , b ) 3 , c ) 4 , d ) 10 , e ) 12 | d | multiply(divide(subtract(150, divide(200, const_2)), 10), const_2) | angelina walked 150 meters from her home to the grocery at a constant speed . she then walked 200 meters to the gym at double the speed . she spent 10 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | let the speed be x . . . so time taken from home to grocery = 150 / x . . the speed to gym = 2 x . . so time taken = 200 / 2 x = 100 / x . . its given 150 / x - 100 / x = 10 50 / x = 10 x = 5 m / secs . . so grocery to gym = 2 * 5 = 10 m / s . . . answer : d | a = 200 / 2
b = 150 - a
c = b / 10
d = c * 2
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a ) 25 % , b ) 50 % , c ) 60 % , d ) 72 % , e ) 100 % | d | multiply(divide(divide(3, 5), divide(5, 6)), const_100) | a cylinder of height h is 5 / 6 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 5 / 6 new cylinder 3 / 5 so 3 / 5 / 5 / 6 = 3 / 5 * 6 / 5 = 0.72 or 72 % . answer d" | a = 3 / 5
b = 5 / 6
c = a / b
d = c * 100
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a ) 285 , b ) 227 , c ) 271 , d ) 187 , e ) 199 | a | 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 . the average number of visitors per day in a month of 30 days beginning with a sunday is ? | "since the month begins with a sunday , so there will be five sundays in the month , required average = ( 510 * 5 + 240 * 25 ) / 30 = 8550 / 30 = 285 . answer : a" | 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
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a ) 75 , b ) 65 , c ) 25 , d ) 288 , e ) 212 | a | multiply(divide(multiply(30, const_1000), const_3600), 9) | a train running at the speed of 30 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 30 * ( 5 / 18 ) m / sec = 25 / 3 m / sec length of train ( distance ) = speed * time ( 25 / 3 ) * 9 = 75 meter answer : a" | a = 30 * 1000
b = a / 3600
c = b * 9
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a ) 9 days , b ) 11 days , c ) 13 days , d ) 15 days , e ) 17 days | a | inverse(add(divide(6, multiply(8, 15)), divide(11, multiply(12, 15)))) | if 8 men or 12 women can do a piece of work in 15 days , in how many days can the same work be done by 6 men and 11 women ? | "8 men = 12 women ( i . e 2 men = 3 women ) 12 women 1 day work = 1 / 15 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 15 = 1 / 180 so , 20 women work = 20 / 180 = 1 / 9 ans : 9 days answer : a" | a = 8 * 15
b = 6 / a
c = 12 * 15
d = 11 / c
e = b + d
f = 1/(e)
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a ) 443 m , b ) 354 m , c ) 450 m , d ) 350 m , e ) 300 m | e | subtract(multiply(25, multiply(72, const_0_2778)), 200) | a train 200 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ? | "e 300 e = 72 * 5 / 18 = 25 = 500 â € “ 200 = 300" | a = 72 * const_0_2778
b = 25 * a
c = b - 200
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a ) 3.7 % , b ) 5.7 % , c ) 6.7 % , d ) 8.7 % , e ) 9.7 % | a | divide(multiply(const_100, 455), multiply(9100, 3)) | what is the rate percent when the simple interest on rs . 9100 amount to rs . 455 in 3 years ? | "455 = ( 4100 * 3 * r ) / 100 r = 3.7 % answer : a" | a = 100 * 455
b = 9100 * 3
c = a / b
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a ) 227 , b ) 299 , c ) 1050 , d ) 750 , e ) 211 | c | divide(divide(multiply(126, const_1000), divide(const_60, const_1)), const_2) | the length of a train and that of a platform are equal . if with a speed of 126 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 126 * 5 / 18 ] m / sec = 35 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 35 è x = 35 * 60 / 2 = 1050 answer : c" | a = 126 * 1000
b = const_60 / 1
c = a / b
d = c / 2
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a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.45 , d ) $ 38.10 , e ) $ 38.80 | c | add(multiply(divide(7, const_100), 35), 35) | laura took out a charge account at the general store and agreed to pay 7 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 7 % interest = ( 7 / 100 ) * 35 = 2.45 $ total amount that laura owes a year later = 35 + 2.45 = 37.45 $ answer c" | a = 7 / 100
b = a * 35
c = b + 35
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a ) 22 % , b ) 25 % , c ) 17.37 % , d ) 99 % , e ) 12 % | c | subtract(multiply(divide(const_100, 852), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 852 grams per kg , what is his percent ? | "852 - - - 148 100 - - - ? = > 17.37 % answer : c" | a = 100 / 852
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
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a ) 300 , b ) 318 , c ) 322 , d ) 324 , e ) 390 | e | add(multiply(6, const_4), multiply(divide(60, 10), const_60)) | a man walks at a rate of 10 mph . after every ten miles , he rests for 6 minutes . how much time does he take to walk 60 miles ? | "to cover 60 miles the man needs ( time ) = ( distance ) / ( rate ) = 60 / 10 = 6 hours = 360 minutes . he will also rest 5 times ( after 10 , 20 , 30,40 and 50 miles ) , so total resting time = 5 * 6 = 30 minutes . total time = 360 + 30 = 390 minutes . answer : e ." | a = 6 * 4
b = 60 / 10
c = b * const_60
d = a + c
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a ) 4500 , b ) 7500 , c ) 5000 , d ) 6000 , e ) none of these | d | divide(15, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 15 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2100 = 10100 × 5 × 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 − 100100 ( 1 + 5100 ) 2 − 100 ∴ = 100 × 21 × 2120 × 20 − 100 = 414 = 100 × 21 × 2120 × 20 − 100 = 414 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 15 , the sum = 400 × 15 = 6000 answer d" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 5 / 100
e = 2 * d
f = 1 + e
g = c - f
h = 15 / g
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a ) none , b ) three , c ) four , d ) five , e ) seven | b | subtract(subtract(40, const_4), const_4) | r is the set of positive odd integers less than 40 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "squares < 40 { 1,4 , 9,16 , 25,36 } s = { 1 , 9,25 } r = { 1,3 , . . . . . 40 } hence b ." | a = 40 - 4
b = a - 4
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['a ) 9 : 8', 'b ) 15 : 8', 'c ) 27 : 8', 'd ) 8 : 9', 'e ) 8 : 15'] | c | divide(subtract(multiply(const_3, const_10), const_3), multiply(const_4, const_2)) | a square and an equilateral triangle have the same perimeter . what is the ratio of the area of the circle circumscribing the square to the area of the circle inscribed in the triangle ? | let x be side of square perimeter of square = 4 x = perimeter of triangle = 3 * side of triangle so side of eq . triangle = ( 4 / 3 ) * x diameter of circle circumscribing the square = sqrt ( 2 ) * x area of circle circumscribing the square = pi * ( sqrt ( 2 ) * x ) ^ 2 / 4 = ( pi / 2 ) * x ^ 2 - - - - ( 1 ) to find radius of the circle inscribed in the triangle area of triangle = r * s = sqrt ( 3 ) / 4 * ( 4 x / 3 ) ^ 2 now s = ( 4 / 3 ) * x + ( 4 / 3 ) * x + ( 4 / 3 ) * x / 2 = 2 x so sqrt ( 3 ) / 4 * ( 4 x / 3 ) ^ 2 = r * 2 x gives r = { 2 / 3 * ( 3 ^ 1 / 2 ) } * x area of the circle inscribed in the triangle = pi * [ { 2 / 3 * ( 3 ^ 1 / 2 ) } * x ] ^ 2 = pi * ( 4 / 27 ) * x ^ 2 - - - - - - - ( 2 ) so reqd ratio = eqn ( 1 ) / eqn ( 2 ) = [ ( pi / 2 ) * x ^ 2 ] / [ pi * ( 4 / 27 ) * x ^ 2 ] = 27 / 8 so reqd ratio = 27 : 8 answer : c | a = 3 * 10
b = a - 3
c = 4 * 2
d = b / c
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a ) 75 , b ) 70 , c ) 35 , d ) 170 , e ) 50 | b | divide(add(10, 25), divide(50, const_100)) | 10 is subtracted from 50 % of a number , the result is 25 . find the number ? | "( 50 / 100 ) * x â € “ 10 = 25 5 x = 350 x = 70 answer : b" | a = 10 + 25
b = 50 / 100
c = a / b
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a ) 1 / 10 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | b | divide(choose(3, 3), choose(add(const_2.0, 3), const_2)) | a bag holds 3 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "given : 3 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 3 red and 3 green marbles after one red has been picked , we need to pick 1 red from a total of 2 red and 3 green marbles . p ( both red ) = ( 3 / 6 ) * ( 2 / 5 ) = 1 / 5 option b" | a = math.comb(3, 3)
b = 2 + 0
c = math.comb(b, 2)
d = a / c
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a ) 5.2 hrs , b ) 2.9 hrs , c ) 1.9 hrs , d ) 7.5 hrs , e ) 1.4 hrs | d | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 5))) | a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 5 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 3 - 1 / 5 ) = 2 / 15 the cistern will be filled in 15 / 2 hrs i . e . , 7.5 hrs . answer : d" | a = 1 / 3
b = 1 / 5
c = a - b
d = 1 / c
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a ) 328897 , b ) 120000 , c ) 877888 , d ) 277768 , e ) 188871 | b | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 24,000 is made ? | "the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 24,000 is made = 96,000 + 24,000 = rs . 1 , 20,000 answer : b" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 20 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
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a ) 50 , b ) 60 , c ) 65 , d ) 70 , e ) 72 | b | divide(multiply(divide(multiply(10, 75), subtract(100, 75)), 100), add(20, divide(multiply(10, 75), subtract(100, 75)))) | a tank can supply water to a village for 100 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 75 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 10 liters per day results in a loss of 750 liters in 75 days . so , those 750 liters were for 25 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 100 = 3000 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last 3000 / 50 = 60 days . the answer is b ." | a = 10 * 75
b = 100 - 75
c = a / b
d = c * 100
e = 10 * 75
f = 100 - 75
g = e / f
h = 20 + g
i = d / h
|
a ) - 4 . , b ) - 2 . , c ) - 1.7 . , d ) 3 . , e ) 2.5 . | a | divide(21, const_10) | if x + y = 2 x + 2 z , x - 2 y = 2 z and x + y + z = 21 , what is the value of y / z ? | x + y = 2 x + 2 z y = x + 2 z - - - - - - - - - - 1 x - 2 y = 2 z x - 2 z = 2 y - - - - - - - - - 2 subtracting equation 1 from equation 2 - 4 z = y y / z = - 4 a is the answer | a = 21 / 10
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a ) 300 m , b ) 278 m , c ) 350 m , d ) 228 m , e ) 282 m | a | subtract(multiply(divide(300, 18), 36), 300) | a 300 meter long train crosses a platform in 36 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 36 = 50 / 3 3 ( x + 300 ) = 1800 è x = 300 m . answer : a" | a = 300 / 18
b = a * 36
c = b - 300
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a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | d | add(multiply(9, 5), 9) | the ratio of pens to pencils is 5 to 6 . there are 9 more pencils than pens . how many pencils are there ? | let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 9 x = 9 the number of pencils is 54 . the answer is d . | a = 9 * 5
b = a + 9
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a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | c | divide(subtract(divide(70, const_2), sqrt(subtract(multiply(divide(70, const_2), divide(70, const_2)), multiply(const_4, 300)))), const_2) | the area of a rectangular field is equal to 300 square meters . its perimeter is equal to 70 meters . find the width of this rectangle . | "l * w = 300 : area , l is the length and w is the width . 2 l + 2 w = 70 : perimeter l = 35 - w : solve for l ( 35 - w ) * w = 300 : substitute in the area equation w = 15 and l = 20 : solve for w and find l using l = 35 - w . correct answer c" | a = 70 / 2
b = 70 / 2
c = 70 / 2
d = b * c
e = 4 * 300
f = d - e
g = math.sqrt(f)
h = a - g
i = h / 2
|
a ) 119 , b ) 117 , c ) 116 , d ) 113 , e ) 115 | d | divide(subtract(multiply(10, 80), multiply(6, 58)), 4) | if the average of 10 digits is 80 and the average of 6 of them is 58 , calculate the average of the remaining 4 numbers ? | "explanation : total of the 10 digits - 10 * 80 = 800 total of the 6 digits - 6 * 58 = 348 total of the remaining 4 digits - 800 - 348 = 452 average of the remaining 4 numbers = 452 / 4 = 113 answer : d" | a = 10 * 80
b = 6 * 58
c = a - b
d = c / 4
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a ) 150 % , b ) 200 % , c ) 250 % , d ) 300 % , e ) 400 % | a | divide(1, 2) | 1 / 2 is what percent of 1 / 3 | "explanation : 1 / 2 / 1 / 3 * 100 = 1 / 2 * 3 / 1 * 100 = 150 % answer : option a" | a = 1 / 2
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a ) 11 , b ) 12 , c ) 13 , d ) 15 , e ) 14 | c | divide(multiply(multiply(39, 12), 5), multiply(30, 6)) | 39 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "first codition = total hours to complete the work = 39 persons * 12 days * 5 hr second codition = total hours to complete the work = 30 persons * x days * 6 hr since the work is still the same so 39 * 12 * 5 = 30 * x * 6 x = 13 answer : c" | a = 39 * 12
b = a * 5
c = 30 * 6
d = b / c
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a ) 25 , b ) 28 , c ) 22 , d ) 29 , e ) 30 | d | multiply(23, divide(2, 3)) | 2 : 3 : : 23 : ? | "d 29 3 is the next prime number after 2 . similarly , 29 is the next prime number after 23 ." | a = 2 / 3
b = 23 * a
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a ) a ) 23 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35 | c | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 220))), const_2), multiply(const_4, 220))) | the product of two numbers is 220 and the sum of their squares is 289 . the sum of the number is ? | "let the numbers be x and y . then , xy = 220 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 220 ) = 729 x + y = 27 . option c" | a = 2 * 220
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 220
f = d + e
g = math.sqrt(f)
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a ) 421 m , b ) 422 m , c ) 423 m , d ) 424 m , e ) 425 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 300), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 300 resolutions . | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 300 resolutions . = 300 * 2 * 22 / 7 * 22.4 = 42200 cm = 422 m answer : b" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 300
i = h / 100
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a ) 7 / 9 , b ) 14 / 9 , c ) 7 / 18 , d ) 11 / 13 , e ) it can not be determined | e | divide(multiply(7, 9), multiply(9, 7)) | the ratio between x and y is 7 / 9 ; x is increased by 4 and y is decreased by 4 , what is the ratio between the new values of x and y ? | "ratio = 7 k / 9 k = 7 / 9 , 14 / 18 , etc . x is increased by 4 and y is decreased by 4 - - > ( 7 k + 4 ) / ( 9 k - 4 ) new ratio can be 11 / 15 , 18 / 14 , etc . answer : e" | a = 7 * 9
b = 9 * 7
c = a / b
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a ) 8600 litres , b ) 12800 litres , c ) 12100 litres , d ) 11200 litres , e ) 13200 litres | b | multiply(divide(multiply(multiply(8, const_60), add(10, 6)), 6), 10) | an outlet pipe empties a tank which is full in 10 hours . if the inlet pipe is kept open , which lets liquid in at the rate of 8 litres / min then outlet pipe would take 6 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 10 hours = ( x - 480 ) * ( 10 + 6 ) hours 10 x = ( x - 480 ) * 16 - - > x = 1280 - - > capacity = 10 x = 12800 liters . answer : b" | a = 8 * const_60
b = 10 + 6
c = a * b
d = c / 6
e = d * 10
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a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many y miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance traveled by tom when karen is 4 miles ahead of him y : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) = 24 miles . d is the correct answer ." | a = 45 / 60
b = a * 4
c = b + 4
d = 60 / 60
e = 45 / 60
f = d - e
g = c / f
h = g - 4
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a ) 380 , b ) 350 , c ) 331 , d ) 310 , e ) none of these | c | multiply(3108, power(add(const_4, const_1), const_4)) | ( 3108 + 6160 ) / 28 | "explanation : as per bodmas rule , first we will solve the equation in bracket then we will go for division = ( 9268 ) / 28 = 331 option c" | a = 4 + 1
b = a ** 4
c = 3108 * b
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a ) $ 20960 , b ) $ 21920 , c ) $ 22680 , d ) $ 28500 , e ) $ 29160 | c | add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 3), 8), multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 4)) | for a certain art exhibit , a museum sold admission tickets to a group of 30 people every 5 minutes from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive . the price of a regular admission ticket was $ 8 and the price of a student ticket was $ 4 . if on one day 3 times as many regular admission tickets were sold as student tickets , what was the total revenue from ticket sales that day ? | "from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive there are 9 * 12 = 108 five - minute intervals , thus total of 108 * 30 tickets were sold . say x student and 3 x regular tickets were sold , then x + 3 x = 108 * 30 - - > x = 27 * 30 and 3 x = 3 * ( 27 * 30 ) = 27 * 90 . therefore , the total revenue from ticket sales that day was 27 * 30 * 4 + 27 * 90 * 8 = $ 22,680 . answer : c ." | a = 12 + 5
b = a - 9
c = b + 1
d = c * 12
e = d * 30
f = 3 + 1
g = e / f
h = g * 3
i = h * 8
j = 12 + 5
k = j - 9
l = k + 1
m = l * 12
n = m * 30
o = 3 + 1
p = n / o
q = p * 4
r = i + q
|
a ) $ 1.50 , b ) $ 2.50 , c ) $ 4.10 , d ) $ 4.70 , e ) $ 8.20 | a | divide(subtract(9.85, 6.85), const_2) | a train ride from two p to town q costs $ 6.85 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.85 . what is the cost of a bus ride from town p to town q ? | "let x be the cost of a bus ride . x + ( x + 685 ) = 985 2 x = 300 x = $ 1.50 the answer is a ." | a = 9 - 85
b = a / 2
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a ) 54 , b ) 185 , c ) 250 , d ) 270 , e ) 322 | b | divide(multiply(222, 10), add(2, 10)) | compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 222 grams of compound x ? | "total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 222 gms of x have 222 * 10 / 12 = 185 gms of b and 222 - 185 = 37 gms of a . cross check : - a / b = 37 / 185 = 2 / 10 ( as given ) ans b" | a = 222 * 10
b = 2 + 10
c = a / b
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a ) 6 sec , b ) 9 sec , c ) 12 sec , d ) 15 sec , e ) 18 sec | a | divide(100, multiply(subtract(68, 8), const_0_2778)) | a train 100 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 100 m at 50 / 3 m / sec = 100 * 3 / 50 sec = 6 sec answer : a" | a = 68 - 8
b = a * const_0_2778
c = 100 / b
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a ) 25 , b ) 28 , c ) 62 , d ) 83 , e ) 89 | a | add(divide(subtract(64, 1600), 8), const_1) | how many multiples of 8 are less than 1600 , and also multiples of 64 ? | the lcm of 8 and 64 is 64 . divide 1600 / 64 = 25 . xxx . so a is your answer . | a = 64 - 1600
b = a / 8
c = b + 1
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a ) 100 , b ) 200 , c ) 300 , d ) 600 , e ) 900 | a | multiply(subtract(36, 50), 50) | what is the greatest positive integer n such that 3 ^ n is a factor of 36 ^ 50 ? | "36 = 3 ^ 2 * 2 ^ 2 . 36 ^ 50 = 3 ^ 100 * 2 ^ 100 the answer is a ." | a = 36 - 50
b = a * 50
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a ) 6 , b ) 7 , c ) 8 , d ) 4 , e ) 3 | c | divide(32, 4) | find k if 32 / k = 4 . | "since 32 / k = 4 and 32 / 8 = 4 , then k = 8 correct answer c" | a = 32 / 4
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a ) 21 , b ) 35 , c ) 39 , d ) 42 , e ) 65 | d | divide(multiply(multiply(7, 14), 3), 7) | exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 14 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ? | "the total number of the people in the room must be a multiple of both 7 and 14 ( in order 3 / 7 and 5 / 14 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 14 , which is 98 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 98 people in the room . therefore there are 3 / 7 * 98 = 42 people in the room under the age of 21 . answer : d ." | a = 7 * 14
b = a * 3
c = b / 7
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | divide(36, const_2) | in a group of ducks and cows , the total number of legs are 36 more than twice the number of heads . find the total number of cows . | "let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 36 more than twice the number of heads = > 2 ( d + 2 c ) = 36 + 2 ( c + d ) = > d + 2 c = 18 + c + d = > 2 c = 18 + c = > c = 18 i . e . , total number of cows = 18 answer is d ." | a = 36 / 2
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a ) 32.7 , b ) 32.4 , c ) 22.4 , d ) 33.9 , e ) 32.1 | d | add(divide(circumface(6.6), const_2), multiply(6.6, const_2)) | the radius of a semi circle is 6.6 cm then its perimeter is ? | "36 / 7 r = 6.6 = 33.9 answer : d" | a = circumface / (
b = a + 2
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a ) 16 , b ) 12 , c ) 15 , d ) 20 , e ) 10 | b | multiply(12, const_1) | the total age of a and b is 12 years more than the total age of b and c . c is how many years younger than a . ? | "( a + b ) - ( b - c ) = 12 a - c = 12 answer is b" | a = 12 * 1
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a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | a | divide(subtract(const_100, add(35, 20)), const_3) | a polling company surveyed a certain country , and it found that 35 % of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that 20 % had a favorable impression only of party w . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country ’ s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "s = 100 not ( w and b ) = 35 only w = 20 ( w and b ) / b = 1 / 2 let ( w and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans" | a = 35 + 20
b = 100 - a
c = b / 3
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a ) 0.5 , b ) 1.0 , c ) 1.5 , d ) 2.0 , e ) 2.5 | d | inverse(add(divide(const_1, 6), divide(2, 6))) | pipe a fills a tank in 6 minutes . pipe b can fill the same tank 2 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | "a ' s rate is 1 / 6 and b ' s rate is 1 / 3 . the combined rate is 1 / 6 + 1 / 3 = 1 / 2 the pipes will fill the tank in 2 minutes . the answer is d ." | a = 1 / 6
b = 2 / 6
c = a + b
d = 1/(c)
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a ) s . 4100 , b ) s . 4270 , c ) s . 4500 , d ) s . 4550 , e ) s . 2500 | a | subtract(multiply(8000, const_4), subtract(multiply(8600, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8600 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8600 = 34400 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 2400 salary of may is rs . 6500 salary of january = rs . 4100 answer : a" | a = 8000 * 4
b = 8600 * 4
c = b - 6500
d = a - c
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 0 | e | floor(divide(reminder(power(7, reminder(545, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 7 ^ 545 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 546 = 4 * 136 + 1 the ten ' s digit will be 0 . answer e" | a = 4 + 1
b = 7 ** reminder
c = reminder / (
d = math.floor(c, 100)
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a ) 16 , b ) 34 , c ) 16.6 , d ) 17 , e ) 23 | b | subtract(subtract(42, const_2), add(add(6, const_4), const_1)) | set a consists of all the prime numbers between 6 and 42 . 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 = 41 lowest number = 7 range = highest - lowest = 41 - 7 = 34 option b" | a = 42 - 2
b = 6 + 4
c = b + 1
d = a - c
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | multiply(divide(2000, add(200, 230)), const_2) | a certain elevator has a safe weight limit of 2000 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 200 pounds and the average weight of the others being 230 pounds ? | lets assume there are 2 x people . half of them have average weight of 200 and other half has 230 . maximum weight is = 2000 so 200 * x + 230 * x = 2000 = > 430 x = 2000 = > x is approximately equal to 5 . so total people is 2 * 5 = 10 we are not taking 10 as answer because say 10 th person has minimum of 200 weight then 200 * 5 + 230 * 5 = 2150 ( which is more than 2000 ) answer c . | a = 200 + 230
b = 2000 / a
c = b * 2
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a ) s . 380 , b ) s . 600 , c ) s . 420 , d ) s . 400 , e ) s . 480 | c | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3360) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3360 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 3360 × ( 1 / 8 ) = 420 answer : option c | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 3360
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a ) 100 , b ) 110 , c ) 135 , d ) 140 , e ) 160 | c | add(90, multiply(90, divide(50, const_100))) | 90 is increased by 50 % . find the final number . | "final number = initial number + 50 % ( original number ) = 90 + 50 % ( 90 ) = 90 + 45 = 135 . answer c" | a = 50 / 100
b = 90 * a
c = 90 + b
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a ) s . 17 , b ) s . 18 , c ) s . 16 , d ) s . 11 , e ) s . 10 | c | divide(add(1500, 340), add(55, 60)) | rahim bought 55 books for rs . 1500 from one shop and 60 books for rs . 340 from another . what is the average price he paid per book ? | "average price per book = ( 1500 + 340 ) / ( 55 + 60 ) = 1840 / 115 = rs . 16 answer : c" | a = 1500 + 340
b = 55 + 60
c = a / b
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | a | divide(add(multiply(10, 16), subtract(46, 26)), 10) | the average of 10 numbers was calculated as 16 . it is discovered later on that while calculating the average , one number , namely 46 , was incorrectly read as 26 . what is the correct average ? | "10 * 16 - 26 + 46 = 180 180 / 10 = 18 the answer is a ." | a = 10 * 16
b = 46 - 26
c = a + b
d = c / 10
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a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 2 miles , e ) 6 1 / 3 miles | e | divide(divide(multiply(subtract(4.90, 1.25), const_100), const_3), const_4) | a taxi company charges $ 1.25 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ? | "if we start out with $ 4.90 and have to spend $ 1.25 for the first quarter - mile , we will have $ 3.65 left to spend on quarter - mile intervals . since $ 3.65 / $ 0.15 = 24.33 we can buy 24.33 more quarter - miles , and will travel 25.33 quarter miles in all : 25.33 × 1 / 4 = 6 1 / 3 miles . the correct answer is choice ( e ) ." | a = 4 - 90
b = a * 100
c = b / 3
d = c / 4
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a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 16 days , e ) 12 days | a | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 20), divide(const_1, 30))), 10) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "answer a can complete a project in 20 days . so , a will complete 1 / 20 th of the project in a day . b can complete a project in 30 days . so , b will complete 1 / 30 th of the project in a day . let the total number of days taken to complete the project be x days . b worked for all x days . however , a worked for ( x - 10 ) days because a quits 10 days before the project is completed . in a day , a completes 1 / 20 th of the project . therefore , a would have completed x − 10 / 20 th of the project in ( x - 10 ) days . in a day , b completes 130130 th of the project . therefore , b would have completed x / 30 h of the project in x days . ∴ x − 10 / 20 + x / 30 = 1 or x = 18 . choice a" | a = 1 / 30
b = a * 10
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 10
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a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | c | multiply(1200, multiply(800, divide(60, multiply(800, 600)))) | 60 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 60 = 80 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 80 = 120 litres . answer c | a = 800 * 600
b = 60 / a
c = 800 * b
d = 1200 * c
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a ) 2 , b ) 3 , c ) 1 , d ) 0 , e ) 4 | e | subtract(divide(5, const_2), multiply(8, 8)) | what is the remainder when 8 ^ 714 is divided by 5 ? | "8 ^ 0 = 1 8 ^ 1 = 8 8 ^ 2 = 64 8 ^ 3 = 512 8 ^ 4 = 4096 8 ^ 5 = . . . . . 8 and so on after ignoring the first last - digit of 8 ^ 0 = 1 , in a recurring last - digit pattern of 8,4 , 2,6 , 8,4 , 2,6 . . . . 715 / 4 = 178 ( with a remainder 2 ) so the last number is a 4 . so when 8 ^ 714 is divided by 5 the remainder is 4 , ans = e" | a = 5 / 2
b = 8 * 8
c = a - b
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a ) 1.01252677 , b ) 0.013466667 , c ) 0.125266677 , d ) 0.126256677 , e ) 0.12725776 | b | divide(20.20, 2000) | 20.20 / 2000 is equal to : | "20.20 / 2000 = 2525 / 200000 = 0.013466667 answer : b" | a = 20 / 20
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a ) 10 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | c | divide(subtract(sqrt(add(multiply(multiply(450, 5), const_4), power(5, const_2))), 5), const_2) | a bus trip of 450 miles would have taken 1 hour less if the average speed s for the trip had been greater by 5 miles per hour . what was the average speed s , in miles per hour , for the trip ? | "( s + 5 ) ( t - 1 ) = 450 s * t = 450 solving both the equations , we get : s = 45 or - 50 since speed should be positive , s = 45 hence , option c" | a = 450 * 5
b = a * 4
c = 5 ** 2
d = b + c
e = math.sqrt(d)
f = e - 5
g = f / 2
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3 | b | subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5))) | an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how many meters w long ? | let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m , leaving 1 m of empty space . thus , i believe the answer is b ) 1 | a = 4 * 2
b = 4 * 1
c = a + b
d = 15 - c
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a ) 198 mph , b ) 91 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph | b | divide(add(110, 72), const_2) | i flew my tiny seaplane to visit my mother . on the flight up , i flew at 110 mph . on the way home , i flew 72 mph . what was my average speed for the trip ? | "( 110 mph + 72 mph ) / 2 = 91 mph correct answer is : b" | a = 110 + 72
b = a / 2
|
a ) 22378 , b ) 279889 , c ) 2788 , d ) 9000 , e ) 2882 | d | divide(divide(multiply(multiply(35000, const_12), 3), multiply(subtract(const_12, 5), 2)), const_10) | a starts business with rs . 35000 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ’ s contribution in the capital ? | explanation : ratio in which profit is to be divided = 2 : 3 assume that b ' s contribution to the capital = b ⇒ 3500 × 12 : b × 7 = 2 : 3 ⇒ 3500 × 12 / 7 b = 2 / 3 ⇒ b = ( 3500 × 12 × 3 ) / ( 2 × 7 ) = 500 × 6 × 3 = 9000 answer : d | a = 35000 * 12
b = a * 3
c = 12 - 5
d = c * 2
e = b / d
f = e / 10
|
a ) - 29 , b ) - 19 , c ) 16 , d ) 29 , e ) 39 | c | subtract(55, subtract(45, add(subtract(28, 37), 15))) | if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 55 , then * is equal to : | 45 - [ 28 - { 37 - ( 15 - * ) } ] = 55 = > 45 - [ 28 - { 37 - 15 + * } ] = 55 45 - [ 28 - 37 + 15 - * ] = 55 = > 45 [ 43 - 37 - * ] = 55 45 - [ 6 - * ] = 55 = > 45 - 6 + * = 55 39 + * = 55 = > * = 55 - 39 = 16 answer : c | a = 28 - 37
b = a + 15
c = 45 - b
d = 55 - c
|
a ) 1 / 15 , b ) 3 / 4 , c ) 5 / 8 , d ) 9 / 6 , e ) 8 / 7 | b | multiply(divide(divide(3, 8), divide(1, 5)), divide(2, 5)) | find the fraction which has the same ratio to 2 / 5 that 3 / 8 has to 1 / 5 | "p : 2 / 5 = 3 / 8 : 1 / 5 as the product of the means is equal to the product of the extremes . p * 1 / 5 = 2 / 5 * 3 / 8 p * 1 / 5 = 6 / 40 p = 3 / 4 = > p = 3 / 4 answer : b" | a = 3 / 8
b = 1 / 5
c = a / b
d = 2 / 5
e = c * d
|
a ) a ) 47 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48 , e ) e ) 49.3 | e | divide(add(multiply(45, subtract(27, add(3, 5))), multiply(95, 5)), 27) | in a class of 27 students in an examination in maths 5 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ? | explanation : total marks obtained by a class of 25 students = 5 * 95 + 3 * 0 + 19 * 45 = 1330 average marks of whole class = 1330 / 27 = 49.3 answer : option e | a = 3 + 5
b = 27 - a
c = 45 * b
d = 95 * 5
e = c + d
f = e / 27
|
a ) 52 kmph , b ) 56 kmph , c ) 58 kmph , d ) 42 kmph , e ) 72 kmph | d | subtract(multiply(divide(410, 15), const_3_6), 56) | a man sitting in a train which is traveling at 56 kmph observes that a goods train , traveling in opposite direction , takes 15 seconds to pass him . if the goods train is 410 m long , find its speed | "explanation : relative speed = 410 / 15 m / sec = ( ( 410 / 15 ) × ( 18 / 5 ) ) kmph = 98 kmph . speed of goods train = ( 98 - 56 ) kmph = 42 kmph answer : option d" | a = 410 / 15
b = a * const_3_6
c = b - 56
|
a ) 10 % , b ) 10.5 % , c ) 11 % , d ) 12.5 % , e ) 9.8 | e | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 12))) | a trader marked the selling price of an article at 12 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . ∴ discount % = [ 11 / 112 * 100 ] % = 9.8 % answer e" | a = 2 + 3
b = a * 2
c = b + 1
d = 100 + 12
e = c / d
f = 100 * e
|
a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 24 hours , e ) none of these | d | add(divide(64, subtract(6, 2)), divide(64, add(6, 2))) | in a river flowing at 2 km / hr , a boat travels 64 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 64 km total time taken = time taken downstream + time taken upstream = ( 64 / 8 ) + ( 64 / 4 ) = = 24 hr . answer : option d" | a = 6 - 2
b = 64 / a
c = 6 + 2
d = 64 / c
e = b + d
|
a ) 524.48 , b ) 556.02 , c ) 574.36 , d ) 181.04 , e ) 595.46 | d | multiply(128.5, power(add(const_4, const_1), const_4)) | ( 128.5 x 32 ) + ( 13.8 x 30 ) = ? x 25 | "explanation : ? = ( 128.5 x 32 ) + ( 13.8 x 30 ) / 25 = 4112 + 414 / 25 = 181.04 answer : option d" | a = 4 + 1
b = a ** 4
c = 128 * 5
|
a ) 3 , b ) 5 , c ) 6 , d ) 11 , e ) 16 | b | subtract(8, divide(24, 8)) | if ( m - 8 ) is a factor of m ^ 2 - pm - 24 , then p = | "( m - 8 ) ( m - a ) = m ^ 2 - pm - 24 a = - 3 p = 8 + a = 5 = b" | a = 24 / 8
b = 8 - a
|
a ) $ 9 , b ) $ 8 , c ) $ 4 , d ) $ 6 , e ) $ 5 | b | add(divide(subtract(15, 1), const_2), 1) | you and your friend spent a total of $ 15 for lunch . your friend spent $ 1 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 1 ( l ) + ( l + 1 ) = 15 l + l + 1 - 1 = 15 - 1 2 l = 14 l = 7 my friends lunch l + 1 = 7 + 1 = $ 8 , the answer is b" | a = 15 - 1
b = a / 2
c = b + 1
|
a ) 28 , b ) 37 , c ) 32 , d ) 35 , e ) 39 | b | add(divide(subtract(160, 8), 4), const_1) | how many multiples of 4 are there between 8 and 160 ? | "it should be mentioned whether 8 and 160 are inclusive . if 8 and 160 are inclusive , then the answer is ( 160 - 8 ) / 4 + 1 = 39 . if 8 and 160 are not inclusive , then the answer is ( 156 - 12 ) / 4 + 1 = 37 . since oa is b , then we have not inclusive case ." | a = 160 - 8
b = a / 4
c = b + 1
|
a ) 0.55 , b ) 0.96 , c ) 0.1 , d ) 0.45 , e ) 0.6 | d | divide(multiply(multiply(sqrt(7), sqrt(7)), 7), const_100) | √ 7 percent of 7 √ 7 = | "the other solutions are great , but here ' s one more approach . first , we scan the answer choices to see that they are quite spread apart . this means we can also use someapproximation all test - takers must memorize the following approximations : √ 2 ≈ 1.4 √ 3 ≈ 1.7 √ 5 ≈ 2.2 √ 6 ≈ 2.4 √ 7 ≈ 2.6 so , √ 7 percent of 7 √ 7 ≈ 2.6 percent of ( 7 ) ( 2.6 ) ≈ ( 2.6 / 100 ) ( 18.2 ) ≈ 47.32 / 100 ≈ 0.47 check the answer choices to see that d is the only answer that ' s close . answer : d" | a = math.sqrt(7)
b = math.sqrt(7)
c = a * b
d = c * 7
e = d / 100
|
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 7 , d ) 1 / 2 , e ) 7 / 13 | e | divide(multiply(7, const_1), add(multiply(7, const_1), multiply(7, const_1))) | harold works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 7 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "the time from beginning of march to the end of september is 7 months . if x is the average monthly tip for all months other than august then his august month tip will be 7 * x his total tip for the 7 months = 6 * ( average tip for the months other than august ) + 7 x = 13 x august tips as a fraction of total tips = 7 x / 13 x = 7 / 13 . so e" | a = 7 * 1
b = 7 * 1
c = 7 * 1
d = b + c
e = a / d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | d | divide(add(54, 6), add(4, 6)) | maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 54 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he meets up with brad ? | "total distance = 54 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 50 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken to meet brad after brad started = 50 / 10 = 5 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 5 = 20 + 4 = 24 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell / maxwell ' s speed = 24 / 4 = 6 hrs . . . answer d alternatively . . . total time taken by maxwell to meet brad = maxwell started 1 hour earlier to brad + total time taken to meet brad after brad started = 1 + 5 = 6 hrs . . . answer d" | a = 54 + 6
b = 4 + 6
c = a / b
|
a ) 13 , b ) 15 , c ) 14 , d ) 13.19 , e ) 16 | d | divide(subtract(sqrt(add(multiply(multiply(10, 5), const_4), power(10, const_2))), 10), const_2) | tom read a book containing 480 pages by reading the same number of pages each day . if he would have finished the book 5 days extra by reading 10 pages a day less , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p - 10 ) ( d + 5 ) = 480 = > pd + 5 p - 10 d - 50 = 480 as the 1 ) stated u can put 1 ) into 2 ) = > 480 + 5 p - 10 d - 50 = 480 = > 5 p - 10 d = 50 put the bold one into it = > 5 * 480 / d - 10 d = 50 the we get the final equation 2400 / d - 10 d = 50 = > d ^ 2 + 5 d - 240 = 0 ( d - 15 ) ( d + 10 ) = 0 so d = 13.19 days . ans : d" | a = 10 * 5
b = a * 4
c = 10 ** 2
d = b + c
e = math.sqrt(d)
f = e - 10
g = f / 2
|
a ) 4 , b ) 24 , c ) 40 , d ) 16 , e ) 20 | b | divide(600, multiply(const_0_25, const_100)) | what are the total number of divisors of 600 ( including 1 and 600 ) ? | write the prime factorization . 600 = ( 2 ^ 3 ) ( 3 ^ 1 ) ( 5 ^ 2 ) the number of divisors ( including 1 and the number itself ) is the product of one plus the exponents . ( 3 + 1 ) ( 1 + 1 ) ( 2 + 1 ) = 24 answer : b | a = const_0_25 * 100
b = 600 / a
|
a ) 38 , b ) 37 , c ) 34 , d ) 35 , e ) 33 | a | subtract(add(divide(93, 3), divide(divide(93, 3), 2)), const_10) | the overall ages of kumar , hathim and rahul is 93 years . 1 decade ago , the ratio of their ages was 2 : 3 : 4 . what is the current age of rahul ? | a 38 if the ages of kumar , hathim and rahul 1 decade ago be 2 a , 3 a and 4 a years correspondingly then , ( 2 a + 10 ) + ( 3 a + 10 ) + ( 4 a + 10 ) = 93 9 a = 93 - 30 9 a = 63 a = 7 rahul ’ s current age = ( 4 a + 10 ) = 38 years . | a = 93 / 3
b = 93 / 3
c = b / 2
d = a + c
e = d - 10
|
a ) 22 , b ) 60 , c ) 126 , d ) 76 , e ) 21 | c | divide(multiply(multiply(9, 7), 2), const_2) | a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 2 rows or 7 rows or 9 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 2 , 7 , 9 ) = 126 answer : c" | a = 9 * 7
b = a * 2
c = b / 2
|
a ) 1 / 6 , b ) 1 / 3 , c ) 7 / 12 , d ) 2 / 3 , e ) 5 / 6 | c | subtract(1, multiply(divide(const_1, const_2), divide(add(const_4, const_1), add(add(const_4, const_1), const_1)))) | set # 1 = { a , b } set # 2 = { k , l , m , n , o , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? | so not a vowel in set - 1 : 1 / 2 and not a vowel in ser - 2 : 5 / 6 now , 1 / 2 ∗ 5 / 6 = 5 / 12 this is for not a vowel . then for at least one vowel will be = 1 − 5 / 12 = 7 / 12 answer will be c . | a = 1 / 2
b = 4 + 1
c = 4 + 1
d = c + 1
e = b / d
f = a * e
g = 1 - f
|
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