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a ) 12 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(220, multiply(add(60, 6), const_0_2778)) | a train 220 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 220 * 3 / 55 = 12 sec . answer : option a" | a = 60 + 6
b = a * const_0_2778
c = 220 / b
|
a ) 35 . , b ) 45 . , c ) 40 . , d ) 30 . , e ) 25 . | a | add(multiply(divide(subtract(const_100, 20), const_100), 50), multiply(divide(20, const_100), 50)) | in the hillside summer camp there are 50 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 20 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 50 students , 90 % of 50 = 45 boys and remaining 5 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to or 20 % . . if we add 35 to existing 45 then the count will be 80 and the girls number will be 20 as it . now boys are 80 % and girls are 20 % . ( out of 100 students = 80 boys + 20 girls ) . imo option a is correct ." | a = 100 - 20
b = a / 100
c = b * 50
d = 20 / 100
e = d * 50
f = c + e
|
a ) 8.58 , b ) 8.76 , c ) 9.04 , d ) 9.12 , e ) 9.25 | c | divide(add(8, 11), add(divide(8, 11), divide(11, 8))) | a cyclist rides a bicycle 8 km at an average speed of 11 km / hr and again travels 11 km at an average speed of 8 km / hr . what is the average speed for the entire trip ? | "distance = 19 km time = 8 / 11 + 11 / 8 = ( 64 + 121 ) / 88 = 185 / 88 hours average speed = ( 19 * 88 ) / 185 = 9.04 km / h the answer is c ." | a = 8 + 11
b = 8 / 11
c = 11 / 8
d = b + c
e = a / d
|
a ) 40 % , b ) 8 % , c ) 10.4 % , d ) 16.66 % , e ) 9 % | c | subtract(add(multiply(multiply(const_4, 8), divide(20, const_100)), multiply(const_4, 8)), const_100) | the balance of a trader weighs 8 % less than it should . still the trader marks up his goods to get an overall profit of 20 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 92 gms because it weighs 8 % less . say , the cost price is $ 92 ( $ 1 / gm ) . since he gets a profit of 20 % , the selling price must be 92 + ( 20 / 100 ) * 92 = $ 110.4 since the cost price is actually supposed to be $ 100 ( for 100 gms ) and the selling price is $ 110.4 , the mark up is simply 10.4 % . ans : c" | a = 4 * 8
b = 20 / 100
c = a * b
d = 4 * 8
e = c + d
f = e - 100
|
a ) 5 , b ) 6 , c ) 9 , d ) 11 , e ) 12 | b | add(divide(subtract(multiply(floor(divide(76, 10)), 10), multiply(add(floor(divide(13, 10)), const_1), 10)), 10), const_1) | how many numbers from 13 to 76 are exactly divisible by 10 ? | "option ' b ' 13 / 10 = 1 and 76 / 10 = 7 = = > 7 - 1 = 6 numbers" | a = 76 / 10
b = math.floor(a)
c = b * 10
d = 13 / 10
e = math.floor(d)
f = e + 1
g = f * 10
h = c - g
i = h / 10
j = i + 1
|
a ) 46 , b ) 42 , c ) 56 , d ) 70 , e ) 58 | d | divide(subtract(divide(30, divide(15, const_100)), 60), 2) | among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 15 % of listco ' s sales staff . listco employs 60 level - 1 employees , 30 of whom are college graduates . how many sales staff members without college degrees are level - 2 employees ? | i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 60 and that c and l 1 = 0.15 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.15 x = 30 , therefore the grand total is 200 . now we know that l 2 is 200 - 60 = 140 . we also learn that c and no c are equally represented thus 100 each . therefore no c and no l 2 will be 100 - 30 = 70 . thus d is the correct answer choice | a = 15 / 100
b = 30 / a
c = b - 60
d = c / 2
|
a ) 9000 , b ) 24000 , c ) 26682 , d ) 29973 , e ) 12312 | a | multiply(divide(const_100, 90), 8100) | 90 % of the population of a village is 8100 . the total population of the village is ? | "x * ( 90 / 100 ) = 8100 x = 90 * 100 x = 9000 answer : a" | a = 100 / 90
b = a * 8100
|
a ) 25 hr , b ) 35 hr , c ) 40 hr , d ) 20 hr , e ) 28 hr | e | multiply(add(add(multiply(const_2, const_2), const_2), const_1), 4) | a tank is filled by 3 pipes a , b , c in 4 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 / 4 7 / x = 1 / 4 x = 28 hours answer is e" | a = 2 * 2
b = a + 2
c = b + 1
d = c * 4
|
a ) 40 , b ) 50 , c ) 60 , d ) 80 , e ) 100 | d | multiply(divide(240, multiply(add(const_2, const_1), const_2)), const_2) | a rectangular garden is to be twice as long as it is wide . if 240 yards of fencing , including the gate , will completely enclose the garden , what will be the length of the garden , in yards ? | "alternate approach backsolving ( using answer options to reach the correct answer ) can work wonders here if one is fast in calculations . given perimeter is 240 so , 2 ( l + b ) = 240 or , l + b = 120 now use the answer options ( given length ; breath will be half the length ) ( a ) 40 l = 40 ; b = 20 l + b = 60 ( b ) 50 l = 50 ; b = 25 l + b = 75 ( c ) 60 l = 60 ; b = 30 l + b = 90 ( d ) 80 l = 80 ; b = 40 l + b = 120 ( e ) 100 l = 100 ; b = 50 l + b = 150 thus you see no , need of any calculations , u can reach the correct option only by checking options ; correct answer will be ( d )" | a = 2 + 1
b = a * 2
c = 240 / b
d = c * 2
|
a ) 4.58 % , b ) 5.96 % , c ) 6.52 % , d ) 4.98 % , e ) none of these | b | multiply(divide(divide(subtract(514, 415), 415), 4), const_100) | at what rate percent on simple interest will rs . 415 amount to rs . 514 in 4 years ? | "explanation : difference in amount = 514 - 415 = 99 99 = ( 415 x 4 x r ) / 100 r = 5.96 % answer is b" | a = 514 - 415
b = a / 415
c = b / 4
d = c * 100
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | b | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 90 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "7 ^ 2 + 5 ^ 2 + 4 ^ 2 = 49 + 25 + 16 = 90 7 + 5 + 4 = 16 hence answer is b" | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
|
a ) 60 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 55 % | a | multiply(divide(75, 125), const_100) | 75 is what percent of 125 ? | 75 / 125 × 100 = 50 % answer : a | a = 75 / 125
b = a * 100
|
a ) 482 , b ) 1205 , c ) 992 , d ) 837 , e ) 923 | b | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 50000), multiply(power(add(divide(20, const_100), const_1), const_2), 50000)) | how much more would rs . 50000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | "50000 ( 11 / 10 ) 4 - 50000 ( 6 / 5 ) 2 = 1205 answer : b" | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 4
e = d * 50000
f = 20 / 100
g = f + 1
h = g ** 2
i = h * 50000
j = e - i
|
a ) $ 120 , b ) $ 135 , c ) $ 150 , d ) $ 174 , e ) $ 192 | d | add(multiply(divide(subtract(375, 40), 5), 2), 40) | a certain psychologist charges $ 40 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 375 , what is the total charge to a patient who receives only 2 hours of therapy ? | let the charge for first hour = x + 40 then charge for each additional hour = x x + 40 + 4 x = 375 = > 5 x = 335 = > x = 67 total charge for patient for 3 hours of therapy = x + 40 + x = 2 x + 40 = 174 $ answer d | a = 375 - 40
b = a / 5
c = b * 2
d = c + 40
|
a ) 1 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | a | divide(divide(multiply(multiply(4864, 9), 2), 12), divide(multiply(multiply(4864, 9), 2), 12)) | if the product 4864 * 9 p 2 is divisible by 12 , the value of p ? | explanation : clearly 4864 is divisible by 4 so 9 p 2 must be divisible by 3 . so ( 9 + p + 2 ) must be divisible by 3 . so p = 1 . answer : a ) 1 | a = 4864 * 9
b = a * 2
c = b / 12
d = 4864 * 9
e = d * 2
f = e / 12
g = c / f
|
a ) 4730 lt , b ) 4356 lt , c ) 7665 lt , d ) 2314 lt , e ) 6545 lt | a | subtract(8100, divide(subtract(9100, multiply(add(const_1, divide(10, const_100)), 8100)), subtract(add(const_1, divide(15, const_100)), add(const_1, divide(10, const_100))))) | the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked the second cow in the second year ? | "let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 − x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % or 10 % so 8100 + 15 / 100 ⋅ x + 10 / 100 ⋅ ( 8100 − x ) = 9100 therefore 8100 + 320 x + 110 ( 8100 − x ) = 9100 120 x = 190 x = 3800 therefore , the cows produced 3800 and 4300 litres of milk the first year , and 4370 and 4730 litres of milk the second year , respectively . correct answer a" | a = 10 / 100
b = 1 + a
c = b * 8100
d = 9100 - c
e = 15 / 100
f = 1 + e
g = 10 / 100
h = 1 + g
i = f - h
j = d / i
k = 8100 - j
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(divide(subtract(112, 20), 4), const_1) | how many multiples of 4 are there between 20 and 112 , inclusive ? | "the multiples of 4 are from 4 * 5 up to 4 * 28 . 28 - 5 + 1 = 24 . the answer is d ." | a = 112 - 20
b = a / 4
c = b + 1
|
a ) 515 . , b ) 572 . , c ) 618 . , d ) 644 . , e ) 666 . | b | add(226, 423) | in the faculty of reverse - engineering , 226 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 90 % of the total ? | "answer is b : 572 solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 90 % of total is 515 , so 100 % is approx . 572 ." | a = 226 + 423
|
a ) 1000 , b ) 1220 , c ) 1250 , d ) 1060 , e ) 1110 | a | divide(750, subtract(const_1, divide(multiply(5, 5), const_100))) | a fellow borrowed a certain sum of money at 5 % per annum at simple interest and in 5 years the interest amounted to rs . 750 less than the sum lent . what was the sum lent ? | p - 750 = ( p * 5 * 5 ) / 100 p = 1000 answer : a | a = 5 * 5
b = a / 100
c = 1 - b
d = 750 / c
|
a ) 18 , b ) 29 , c ) 37 , d ) 44 , e ) 85 | e | multiply(add(2, const_1), subtract(25, 2)) | the product x of two prime numbers is between 15 and 90 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25 , then x = | "option bc can be ruled out as they themselves are prime numbers 18 = 2 * 9 = 3 * 6 > > ignore 44 = 2 * 22 = 4 * 11 > > ignore 85 = 5 * 17 > > answer answer = e" | a = 2 + 1
b = 25 - 2
c = a * b
|
a ) 34880 , b ) 37780 , c ) 36000 , d ) 18000 , e ) 28810 | c | multiply(multiply(const_3, const_60), const_60) | if an object travels at ten feet per second , how many feet does it travel in one hour ? | "explanation : if an object travels at 5 feet per second it covers 5 x 60 feet in one minute , and 5 x 60 x 60 feet in one hour . answer = 36000 answer : c ) 36000" | a = 3 * const_60
b = a * const_60
|
a ) 1.5 , b ) 0.5 , c ) 0.32 , d ) 0.25 , e ) 6.32 | c | subtract(6.32, floor(6.32)) | for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.32 – 6.32 * ? | since z * is defined as the greatest positive even integer less than or equal to z , then 6.32 * = 6 ( the greatest positive even integer less than or equal to 6.32 is 6 ) . hence , 6.32 – 6.32 * = 6.32 - 6 = 0.32 answer : c . | a = math.floor(6, 32)
b = 6 - 32
|
a ) 3 , b ) 7 , c ) 5 , d ) 6 , e ) 4 | d | divide(add(add(2, 4), 6), 4) | the area of a triangle will be when a = 2 m , b = 4 m , c = 6 m , a , b , c being lengths of respective sides ? | "s = ( 2 + 4 + 6 ) / 2 = 6 answer : d" | a = 2 + 4
b = a + 6
c = b / 4
|
a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 20 % lemonade syrup ? | let the total solution is 150 l with 80 l water 70 l syrup . to make 20 % syrup solution , the result solution must have 120 l syrup and 30 l syrup . therefore we are taking 40 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 40 l syrup in 85.7 l solution we started by multiplying 10 now to get to the result we need to divide by 40 = > amount of solution to be replaced with water = ( 85.7 / 40 ) = 2.14 . correct option : c | a = 8 + 7
b = 7 / a
c = 2 + 3
d = 2 / c
e = b - d
f = 8 + 7
g = 7 / f
h = e / g
i = 8 + 7
j = h * i
|
a ) 12 , b ) 27 , c ) 21 , d ) 18 , e ) 11 | d | multiply(18, const_1) | the total age of a and b is 18 years more than the total age of b and c . c is how many years younger than a ? | ( a + b ) - ( b + c ) = 18 a - c = 18 . c is younger than a by 18 years . answer : d | a = 18 * 1
|
a ) 32 , b ) 36 , c ) 34 , d ) 39 , e ) 42 | a | subtract(divide(16, divide(2, 4)), 16) | a certain lab experiments with white and brown mice only . in one experiment , 2 / 4 of the mice are white . if there are 16 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 4 m number of brown mice = 2 / 4 m = 16 = > m = 32 answer a" | a = 2 / 4
b = 16 / a
c = b - 16
|
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | divide(divide(add(add(multiply(4, 7), const_10), const_10), 4), const_2) | if x is divided by 7 , the remainder is 5 . what is the remainder if 4 x is divided by 7 ? | x = 7 q + 5 4 x = 7 * 4 q + 20 4 x = 7 * 4 q + 7 * 2 + 6 4 x = 7 ( 4 q + 2 ) + 6 4 x = 7 k + 6 ( k = 4 q + 2 ) answer d | a = 4 * 7
b = a + 10
c = b + 10
d = c / 4
e = d / 2
|
a ) 1.6 km , b ) 2 km , c ) 6.3 km , d ) 4 km , e ) none of these | c | multiply(divide(13, const_60), add(21, 8)) | the speed of a boat in still water is 21 km / hr and the rate of current is 8 km / hr . the distance travelled downstream in 13 minutes is | "explanation : speed downstreams = ( 21 + 8 ) kmph = 18 kmph . distance travelled = ( 29 x 13 / 60 ) km = 6.3 km option c" | a = 13 / const_60
b = 21 + 8
c = a * b
|
a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24 | d | add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2)))) | jolene entered an 16 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 12 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ? | "if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 920.24 answer : d ." | a = 3 / 100
b = 2 / 100
c = 100 ** 2
d = b * c
e = 100 ** 2
f = d + e
g = a * f
h = 2 / 100
i = 100 ** 2
j = h * i
k = 100 ** 2
l = j + k
m = g + l
n = 4 / 100
o = m * n
p = 3 / 100
q = 2 / 100
r = 100 ** 2
s = q * r
t = 100 ** 2
u = s + t
v = p * u
w = o + v
|
a ) $ 50 , b ) $ 500 , c ) $ 150 , d ) $ 250 , e ) $ 600 | c | divide(600, 1) | divide $ 600 among a , b in the ratio 1 : 3 . how many $ that a get ? | "sum of ratio terms = 1 + 3 = 4 a = 600 * 1 / 4 = $ 150 answer is c" | a = 600 / 1
|
a ) 45.12 % , b ) 50 % , c ) 57 % , d ) 71.42 % , e ) 65 % | d | multiply(divide(20000, add(add(3000, 5000), 20000)), const_100) | 3 candidates in an election and received 3000 , 5000 and 20000 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 3000 + 5000 + 20000 ) = 28000 so , required percentage = 11628 / 20400 * 100 = 71.42 % d" | a = 3000 + 5000
b = a + 20000
c = 20000 / b
d = c * 100
|
a ) 5 , b ) 25 , c ) 125 , d ) 625 , e ) 3125 | e | multiply(power(5, 5), power(5, 3)) | what number times ( 1 ⁄ 5 ) ^ 2 will give the value of 5 ^ 3 ? | "x * ( 1 / 5 ) ^ 2 = 5 ^ 3 x = 5 ^ 2 * 5 ^ 3 = 5 ^ 5 = 3125 the answer is e ." | a = 5 ** 5
b = 5 ** 3
c = a * b
|
a ) 30 min , b ) 35 min , c ) 45 min , d ) 50 min , e ) 55 min | b | multiply(add(const_3, const_4), 5) | a clock shows the time as 9 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 4 p . m . ? | there are 7 hours in between 9 a . m . to 4 p . m . 7 * 5 = 35 minutes . answer : b | a = 3 + 4
b = a * 5
|
a ) 75 kg , b ) 82 kg , c ) 86 kg , d ) data inadequate , e ) none of these | b | add(multiply(4, 3), 70) | the average weight of 4 person ' s increases by 3 kg when a new person comes in place of one of them weighing 70 kg . what might be the weight of the new person ? | "total weight increased = ( 4 x 3 ) kg = 12 kg . weight of new person = ( 70 + 12 ) kg = 82 kg . answer : option b" | a = 4 * 3
b = a + 70
|
a ) 20 , b ) 35 , c ) 30 , d ) 40 , e ) 42 | c | subtract(multiply(add(7, const_1), const_4), const_2) | the no . of girls in a class are 7 times the no . of boys , which value cannever be the of total students ? | let the boys are x , then girls are 7 x , total = x + 7 x = 8 x so it should be multiple of 8 , 30 is not a multiple of 8 . c | a = 7 + 1
b = a * 4
c = b - 2
|
a ) 34 , b ) 40 , c ) 85 , d ) 88 , e ) none | c | add(multiply(divide(650, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 650 sq . feet , how many feet of fencing will be required ? | "explanation we have : l = 20 ft and lb = 650 sq . ft . so , b = 32.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 65 ) ft = 85 ft . answer c" | a = 650 / 20
b = a * 2
c = b + 20
|
a ) 20 % , b ) 30 % , c ) 50 % , d ) 70 % , e ) 80 % | c | subtract(100, 50) | john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 50 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it ’ s 100 – 50 = 50 . the “ original ” is our starting point ; in this case , it ’ s 100 . ( 50 / 100 ) * 100 = ( 0.5 ) * 100 = 50 % . c" | a = 100 - 50
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a ) 288 , b ) 3200 , c ) 800 , d ) 267 , e ) 121 | b | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 10) | 10 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 10 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 10 x = 320 x = 3200 answer : b" | a = 650 * 20
b = a / 100
c = b + 190
d = c * 100
e = d / 10
|
a ) $ 500 , b ) $ 467.5 , c ) $ 935 , d ) $ 525.5 , e ) $ 465.5 | b | divide(multiply(125, divide(multiply(add(const_100.0, 70), divide(add(const_100.0, 25), const_2)), 125)), subtract(divide(add(125, 25), const_2), 25)) | a clock store sold a certain clock to a collector for 25 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 55 percent of what the collector had paid . the shop then sold the clock again at a profit of 70 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 125 , for how much did the shop sell the clock the second time ? | "now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 25 % profit . this means the clocks ' selling price was c ( 1.25 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 55 % the price at which the collector bought it . thus , you get = 1.25 * 0.55 * c = 0.6875 c furthermore , the store sells the clock for the second time for 70 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.7 = 1.7 * 0.6875 c finally given that c - 0.6875 c = 125 - - - - > c = 400 $ thus , the cost of the clock the second time around = 1.7 * 0.6875 c = 1.7 * 0.6875 * 400 = 467.5 $ . hence b is the correct answer ." | a = 100 + 0
b = 100 + 0
c = b / 2
d = a * c
e = d / 125
f = 125 * e
g = 125 + 25
h = g / 2
i = h - 25
j = f / i
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | e | multiply(divide(63, subtract(9, 2)), 2) | the ratio of spinsters to cats is 2 to 9 . if there are 63 more cats than spinsters , how many spinsters are there ? | "let 2 x be the number of spinsters . then 9 x is the number of cats . 9 x - 2 x = 63 x = 9 and the number of spinsters is 2 ( 9 ) = 18 . the answer is e ." | a = 9 - 2
b = 63 / a
c = b * 2
|
a ) 16.5 % , b ) 9 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(8.5, 3)), const_100) | a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 8.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.085 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 85 f - - > s = 10 f - - > f / s = 1 / 10 - - > f / ( s + f ) = 1 / ( 1 + 10 ) = 1 / 11 = 0.09 . answer : b ." | a = 3 - 5
b = 8 - 5
c = a / b
d = c * 100
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | b | divide(add(negate(10), sqrt(subtract(power(negate(10), 2), multiply(const_4, negate(divide(multiply(600, 10), 2)))))), 2) | the time it took car p to travel 600 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | "let speed of car r be = x then speed of car p = x + 10 a / q , ( 600 / x ) - ( 600 / ( x + 10 ) ) = 2 solving for x = 50 miles \ hr answer : b" | a = negate + (
b = negate ** (
c = b - 2
d = 600 * 10
e = d / 2
f = 4 * negate
g = math.sqrt(c)
h = a / g
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a ) 276 , b ) 300 , c ) 500 , d ) 700 , e ) 900 | e | multiply(60, 15) | the h . c . f . of two numbers is 60 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is : | "the numbers are ( 60 x 11 ) and ( 60 x 15 ) . larger number = ( 60 x 15 ) = 900 . answer : e" | a = 60 * 15
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a ) 32 , b ) 42 , c ) 60 , d ) 25 , e ) 52 | c | subtract(multiply(multiply(add(add(const_2, const_3), const_1), add(const_3, const_1)), add(const_2, const_1)), multiply(add(const_3, const_1), add(const_2, const_1))) | the number of even factors of 21600 is | make a prime factorization of the number : 21,600 = 25 ∗ 33 ∗ 52 according to the above the number of factors is ( 5 + 1 ) ( 3 + 1 ) ( 2 + 1 ) = 72 now , get rid of powers of 2 as they give even factors - - > you ' ll have 33 ∗ 52 which has ( 3 + 1 ) ( 2 + 1 ) = 12 factors . all the remaining factors will be odd , therefore 21,600 has 72 - 12 = 60 even factors . answer : c . | a = 2 + 3
b = a + 1
c = 3 + 1
d = b * c
e = 2 + 1
f = d * e
g = 3 + 1
h = 2 + 1
i = g * h
j = f - i
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a ) 3 / 25 , b ) 5 / 6 , c ) 7 / 12 , d ) 2 / 3 , e ) 25 / 36 | b | divide(divide(multiply(50, 5), 30), 10) | at a speed of 50 miles per hour , a certain car uses 1 gallon of gasoline every 30 miles . if the car starts with a full 10 gallon tank of gasoline and travels for 5 hours at 50 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | "gas used = ( 5 hours ) * ( 50 miles / hour ) * ( 1 gallon / 30 miles ) = 8 + 1 / 3 gallons portion used = ( 8 + 1 / 3 ) / 10 = 5 / 6 ans b" | a = 50 * 5
b = a / 30
c = b / 10
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a ) 2 , b ) 3 , c ) 6 , d ) 10 , e ) 11 | d | subtract(2590, add(multiply(multiply(multiply(9, 11), 13), const_2), 6)) | what least number should be subtracted from 2590 so that the remainder when divided by 9 , 11 , and 13 will leave in each case the same remainder 6 ? | "the lcm of 9 , 11 , and 13 is 1,287 . the next multiple is 2 * 1,287 = 2,574 . 2,574 + { remainder } = 2,574 + 6 = 2,580 , which is 10 less than 2,590 . answer : d ." | a = 9 * 11
b = a * 13
c = b * 2
d = c + 6
e = 2590 - d
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a ) 35 , b ) 45 , c ) 55 , d ) 65 , e ) 75 | c | add(40, 15) | the second angle of a triangle is double the first ( 110 ) . the third angle is 40 less than the first ( 15 ) . find the first angle . | first x with nothing given about the firstwemake that x second 2 x the second is double the first , third x − 40 the third is 40 less than the first f + s + t = 180 all three angles add to 180 ( x ) + ( 2 x ) + ( x − 40 ) = 180 replace f , s , and t with the labeled values . x + 2 x + x − 40 = 180 here the parenthesis are not needed . 4 x − 40 = 180 combine like terms , x + 2 x + x + 40 + 40 add 40 to both sides 4 x = 220 the variable ismultiplied by 4 4 4 divide both sides by 4 x = 55 our solution for x first 55 replace xwith 55 in the original list of angles second 2 ( 55 ) = 110 our angles are 55 , 110 , and 15 third ( 55 ) − 40 = 15 correct answer c | a = 40 + 15
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a ) 2532.6 , b ) 2552.2 , c ) 2524.1 , d ) 5674.5 , e ) 2512.9 | d | subtract(multiply(36,000, power(add(const_1, divide(9, const_100)), 20)), 36,000) | find the compound interest on $ 36,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 36000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 36000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 41674.5 ci . = $ ( 41674.5 - 36000 ) = $ 5674.5 answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 36 * 0
e = d - 36
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a ) 22 % , b ) 24 % , c ) 25 % , d ) 27 % , e ) 28 % | a | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(25, const_100), subtract(800, 500))), 800), const_100) | for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 25 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman ’ s commission was approximately what percent of the total amount of the sale ? | "total sales = 800 comission = ( 20 / 100 ) * 500 + ( 25 / 100 ) * 300 = 100 + 75 = 175 % comission = ( 175 / 800 ) * 100 = 21.8 ~ 22 % answer is a" | a = 20 / 100
b = a * 500
c = 25 / 100
d = 800 - 500
e = c * d
f = b + e
g = f / 800
h = g * 100
|
a ) 21.2 kmph , b ) 12.1 kmph , c ) 23.4 kmph , d ) 20.4 kmph , e ) none of these | a | divide(add(divide(add(6, 20), const_2), 30), const_2) | mike took a bus from home to market , that travels at 30 kmph . while walking back at 6 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 20 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 30 , by walking = x / 6 , by cycling = x / 20 hours : . average speed = total distance / total time = 6 x / x / 15 + x / 6 + x / 20 = 6 * 60 / 4 + 10 + 3 = 21.2 answer : a | a = 6 + 20
b = a / 2
c = b + 30
d = c / 2
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a ) a ) 6 , b ) b ) 8 , c ) c ) 10 , d ) d ) 12 , e ) e ) 18 | e | divide(divide(6, subtract(divide(30, const_60), divide(10, const_60))), const_3) | circular gears l and r start to rotate at the same time at the same rate . gear l makes 10 complete revolutions per minute and gear r makes 30 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 6 more revolutions than gear l ? | "gear l - - 10 rotations per 60 seconds - - 1 rotation per 6 seconds . gear r - - 40 rotations per 60 seconds - - 3 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 2 rotations hence every 6 seconds the difference between the number of rotations of r and l gear is 2 units . required net difference should be 6 rotations = > 3 ( 6 seconds later ) = = > 18 seconds . answer : e" | a = 30 / const_60
b = 10 / const_60
c = a - b
d = 6 / c
e = d / 3
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a ) 100 , b ) 75 , c ) 25 , d ) 50 , e ) 125 | d | subtract(add(add(50, 25), 75), 100) | in the youth summer village there are 100 people , 50 of them are not working , 25 of them have families and 75 of them like to sing in the shower . what is the largest possible number of people in the village , which are working , that do n ' t have families and that are singing in the shower ? | "total = 100 not working = 50 having family = 25 like to sing in shower = 75 working = 100 - 50 = 50 not having family = 100 - 25 = 75 like to sing in shower = 75 largest possible number is the lowest possible among the above thus 50 d" | a = 50 + 25
b = a + 75
c = b - 100
|
a ) 1 , b ) 2 , c ) 4 , d ) 9 , e ) 3 | e | divide(multiply(3245613, 11), 3245613) | what is the smallest no . which must be added to 3245613 so as to obtain a sum which is divisible by 11 ? | "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 3245613 , difference = ( 3 + 4 + 6 + 3 ) - ( 2 + 5 + 1 ) = 16 - 8 = 8 . the units digit is at odd place . so we add 3 to the number = > 3245613 + 3 = 3245616 now , ( 3 + 4 + 6 + 6 ) - ( 2 + 5 + 1 ) = 19 - 8 = 11 = > 11 is a multiple of 11 and hence 3245616 is also divisible by 11 e" | a = 3245613 * 11
b = a / 3245613
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a ) 4 , b ) 7 , c ) 5 , d ) 9 , e ) 8 | b | multiply(const_3600, divide(divide(130, const_1000), add(60, 6))) | a train 130 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 130 * 3 / 55 ] sec = 7 sec answer : b" | a = 130 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
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a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent j of the distribution is less than m + d ? | "d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , j = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d . d" | a = 100 - 68
b = a / 2
c = 100 - b
|
a ) 24 , b ) 52 , c ) 96 , d ) 144 , e ) 648 | b | add(add(add(factorial(4), divide(factorial(4), factorial(const_2))), divide(factorial(4), factorial(const_2))), divide(factorial(4), factorial(3))) | in a 4 person race , medals are awarded to the fastest 3 runners . the first - place runner receives a gold medal , the second - place runner receives a silver medal , and the third - place runner receives a bronze medal . in the event of a tie , the tied runners receive the same color medal . ( for example , if there is a two - way tie for first - place , the top two runners receive gold medals , the next - fastest runner receives a silver medal , and no bronze medal is awarded ) . assuming that exactly 3 medals are awarded , and that the 3 medal winners stand together with their medals to form a victory circle , how many z different victory circles are possible ? | possible scenarios are : 1 . gold / silver / bronze / no medal ( no ties ) - 4 ! = 24 ; 2 . gold / gold / silver / no medal - 4 ! / 2 ! = 12 ; 3 . gold / silver / silver / no medal - 4 ! / 2 ! = 12 ; 4 . gold / gold / gold / no medal - 4 ! / 3 ! = 4 . total z : 24 + 12 + 12 + 4 = 52 answer : b . | a = math.factorial(4)
b = math.factorial(4)
c = math.factorial(2)
d = b / c
e = a + d
f = math.factorial(4)
g = math.factorial(2)
h = f / g
i = e + h
j = math.factorial(4)
k = math.factorial(3)
l = j / k
m = i + l
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a ) $ 442 , b ) $ 470 , c ) $ 472 , d ) $ 484 , e ) $ 491 | a | multiply(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(16, const_100)), 470)) | an air - conditioning unit costs $ 470 . on december there was a discount for christmas of 16 % . 6 months later , the holiday season was over so the company raised the price of the air - conditioning by 12 % . how much will an air - conditioning unit cost in november ? | if its previous november ( before discount ) then price is $ 470 . but if its november of next year then 16 % discount on $ 470 = 470 ( 1 - 16 / 100 ) = $ 394.8 again a corrected raised price of 12 % over $ 394.8 = 394.8 ( 1 + 12 / 100 ) = 442.176 ~ $ 4442 ans a | a = 12 / 100
b = a + 1
c = 16 / 100
d = 1 - c
e = d * 470
f = b * e
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a ) 50 , b ) 45 . , c ) 40 . , d ) 30 . , e ) 25 . | a | add(multiply(divide(subtract(const_100, 5), const_100), 50), multiply(divide(5, const_100), 50)) | in the hillside summer camp there are 50 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 50 students , 90 % of 50 = 45 boys and remaining 5 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 5 % . . if we add 50 to existing 45 then the count will be 95 and the girls number will be 5 as it . now boys are 95 % and girls are 5 % . ( out of 100 students = 95 boys + 5 girls ) . answer : option a is correct ." | a = 100 - 5
b = a / 100
c = b * 50
d = 5 / 100
e = d * 50
f = c + e
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a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | e | add(5, const_1) | the average of first five multiples of 5 is | "solution average = 5 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 75 / 5 . = 15 answer e" | a = 5 + 1
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a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability e , that a book selected at random is either a hard cover book or a chemistry book ? | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , e = 9 provide the desired outcome , making the answer e ." | a = 2 + 4
b = 2 + a
c = b + 12
d = 2 / c
e = 2 + 4
f = 2 + e
g = 2 + 4
h = 2 + g
i = h + 12
j = f / i
k = d + j
l = 2 + 4
m = 2 + l
n = m + 12
o = 2 / n
p = 2 + 4
q = 2 + p
r = 2 + 4
s = 2 + r
t = s + 12
u = q / t
v = o * u
w = k - v
|
a ) 102 , b ) 104 , c ) 110 , d ) 90 , e ) 84 | b | divide(multiply(subtract(17, const_1), subtract(subtract(17, const_1), const_3)), const_2) | how many diagonals does a polygon with 17 sides have , if one of its vertices does not connect to any diagonal ? | "if i calculate it using the formulae , # diagonals = n ( n - 3 ) / 2 each vertex sends of n - 3 diagonals n = 17 - 1 then 16 * ( 16 - 3 ) / 2 = 104 correct option : b" | a = 17 - 1
b = 17 - 1
c = b - 3
d = a * c
e = d / 2
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a ) 26 , b ) 27 , c ) 46 , d ) 57 , e ) none | b | add(divide(72, 3), 3) | the sum of 3 consecutive multiples of 3 is 72 . what is the larget number ? | sol . let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 72 ⇔ 9 x = 63 ⇔ x = 7 . ∴ largest number = 3 x + 6 = 27 . answer b | a = 72 / 3
b = a + 3
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a ) 0 , b ) 1 / 2 , c ) 1 , d ) 3 / 2 , e ) 2 | c | multiply(75, 15) | find the value of sin 75 ° + sin 15 ° = ? | "sin a + sin b = 2 sin [ ( a + b ) / 2 ] cos [ ( a - b ) / 2 ] = 2 sin [ ( 75 + 15 ) / 2 ] cos [ ( 75 - 15 ) / 2 ] = 2 sin 90 cos 60 = 2 * 1 * ( 1 / 2 ) = 1 answer : c" | a = 75 * 15
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a ) 240 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | d | subtract(multiply(divide(multiply(180, const_1000), const_3600), 30), multiply(divide(multiply(180, const_1000), const_3600), 12)) | a train traveling at 180 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length of the train plus the length of the platform . the extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover . i . e . , length of the platform . compute length of platform length of the platform = speed of train * extra time taken to cross the platform . length of platform = 180 kmph * 12 seconds convert 180 kmph into m / sec 1 kmph = 5 / 18 m / s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . ∴ 180 kmph = 5 / 18 ∗ 180 = 50 m / sec therefore , length of the platform = 50 m / s * 12 sec = 600 meters . choice d" | a = 180 * 1000
b = a / 3600
c = b * 30
d = 180 * 1000
e = d / 3600
f = e * 12
g = c - f
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a ) 20 cs , b ) cs / 2 , c ) 84 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(7, const_12) | a certain school implemented a reading program for its students , with the goal of getting each student to read 7 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | ans : c solution : simple multiplication s students , c classes , 7 books / month = 84 books a year total number of books = 84 cs | a = 7 * 12
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a ) 1 day , b ) 4 days , c ) 19 days , d ) 41 days , e ) 52 days | d | subtract(multiply(12, const_4), add(const_3, const_4)) | a man , a woman and a boy can complete a job in 3,4 and 12 days respectively . how many boys must assist 1 man and 1 woman to complete the job in ¼ of a day ? | ( 1 man + 1 woman ) ' s 1 days work = 1 / 3 + 1 / 4 = 7 / 12 work done by 1 man and 1 women n 1 / 4 day = ( ( 7 / 12 ) * ( 1 / 4 ) ) = 7 / 48 remaining work = 1 - 7 / 48 = 41 / 48 work done by 1 boy in ¼ day = ( ( 1 / 12 ) * ( 1 / 4 ) ) = 1 / 48 therefore , number of boys required = ( ( 41 / 48 ) * 48 ) = 41 days answer d 41 days | a = 12 * 4
b = 3 + 4
c = a - b
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a ) 177 , b ) 168 , c ) 100 , d ) 120 , e ) 122 | c | divide(12, subtract(divide(12, 12), 6)) | a train covers a distance of 12 km in 12 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 12 * 60 ) km / hr = ( 60 * 5 / 18 ) m / sec = 16.6 m / sec . length of the train = 16.6 * 6 = 100 m . answer : c" | a = 12 / 12
b = a - 6
c = 12 / b
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a ) 7.0 , b ) 5.0 , c ) 6.0 , d ) 4.0 , e ) 8.0 | d | add(subtract(25, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 25 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 25 kg joins the group , total weight of boys = 20 x + 25 so 20 x + 25 = 21 ( x + 1 ) = > x = 4 . answer : d" | a = 20 + 1
b = 25 - a
c = b + 1
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a ) 80 % , b ) 100 % , c ) 120 % , d ) 60 % , e ) 90 % | b | multiply(subtract(divide(divide(divide(add(const_100, 35), const_100), subtract(const_1, divide(subtract(20, 15), 20))), divide(subtract(const_100, 10), const_100)), const_1), const_100) | a dealer offers a cash discount of 10 % and still makes a profit of 35 % when he further allows 20 articles to be sold at the cost price of 15 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | "given cash discount - 10 % profit - 35 % items sold - 20 price sold at = list price of 15 assume list price = $ 20 total invoice = $ 300 - 10 % cash discount = $ 270 let cost price of 20 items be x so total cost = 20 * x given the shopkeeper had a profit of 20 % 20 * x * 135 / 100 = 270 or x = $ 10 which means his products were listed at $ 20 which is a 100 % markup over $ 10 answer b" | a = 100 + 35
b = a / 100
c = 20 - 15
d = c / 20
e = 1 - d
f = b / e
g = 100 - 10
h = g / 100
i = f / h
j = i - 1
k = j * 100
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a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 18 % | e | multiply(subtract(divide(22, const_100), divide(subtract(9, multiply(divide(22, const_100), multiply(20, 2))), subtract(multiply(15, 3), multiply(20, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 20 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 22 percent , what is the discount rate on pony jeans ? | "x discount on pony jeans , ( 0.22 - x ) discount on fox jeans . set the equation : 3 * 15 ( 0.22 - x ) + 2 * 20 x = 9 - - > x = 0.18 = 18 % answer : e ." | a = 22 / 100
b = 22 / 100
c = 20 * 2
d = b * c
e = 9 - d
f = 15 * 3
g = 20 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | a | divide(log(divide(multiply(power(5, 7), power(9, 4)), 3)), log(15)) | if ( 5 ^ 7 ) ( 9 ^ 4 ) = 3 ( 15 ^ x ) , what is the value of x ? | "( 5 ^ 7 ) ( 9 ^ 4 ) = 3 ( 15 ^ x ) = > 5 ^ 7 * 3 ^ 8 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 7 * 3 ^ 8 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 7 answer a" | a = 5 ** 7
b = 9 ** 4
c = a * b
d = c / 3
e = math.log(d)
f = math.log(15)
g = e / f
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a ) 4 , b ) 7 , c ) 12 , d ) 15 , e ) 20 | c | divide(multiply(8, const_60), subtract(multiply(1, const_60), 20)) | henrikh lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bicycle to work . if it takes him exactly 8 minutes more to walk to work than to ride his bicycle , then x equals | the best way would be . . . per km , he takes 40 secs extra if he walks so he will take 8 mins or 8 * 60 secs in 8 * 60 / 40 = 12 km answer : c | a = 8 * const_60
b = 1 * const_60
c = b - 20
d = a / c
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | a | divide(subtract(60, subtract(8, 2)), const_3) | george is 8 years more than christopher and ford is 2 years younger than christopher . the sum of their ages is 60 . find the ages of christopher . | christopher age = x george age , y = x + 8 - - - - - - - - - - > ( 1 ) ford age , z = x - 2 - - - - - - - - - - - - > ( 2 ) sum of their ages , x + y + z = 60 - - - - > ( 3 ) substitute z and y values in equation ( 3 ) therefore , x + ( x + 8 ) + ( x - 2 ) = 60 = > 3 x + 8 - 2 = 60 = > 3 x = 60 - 6 = > 3 x = 54 = > x = 54 / 3 x = 18 answer : a | a = 8 - 2
b = 60 - a
c = b / 3
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a ) 32 % , b ) 34.4 % , c ) 35 % , d ) 35.6 % , e ) 36.4 % | d | multiply(divide(add(multiply(divide(30, const_100), 56000), multiply(divide(40, const_100), 74000)), add(74000, 56000)), const_100) | john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 56000 and ingrid makes $ 74000 , what is their combined tax rate ? | "( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 74000 / 128000 = 74 / 128 = 37 / 65 is a little above 1 / 2 . thus , our answer is just a little above 35 . answer : d" | a = 30 / 100
b = a * 56000
c = 40 / 100
d = c * 74000
e = b + d
f = 74000 + 56000
g = e / f
h = g * 100
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a ) 1 , b ) 10 , c ) 100 , d ) 1000 , e ) 10000 | b | power(10, subtract(65, 64)) | 10 ^ ( 65 ) ã · 10 ^ ( 64 ) = ? | "10 ^ ( 65 ) ã · 10 ^ ( 64 ) = 10 ^ ( 65 - 64 ) = 10 ^ 1 = 10 answer : b" | a = 65 - 64
b = 10 ** a
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a ) 69 , b ) 70 , c ) 71 , d ) 72 , e ) 75 | d | multiply(36, const_2) | the average of the marks of 11 students in a class is 36 . if the marks of each student are doubled , find the new average ? | sum of the marks for the 11 students = 11 * 36 = 396 . the marks of each student are doubled , the sum also will be doubled . the new sum = 396 * 2 = 792 . so , the new average = 792 / 11 = 72 . answer : d | a = 36 * 2
|
a ) 2 : 5 , b ) 2 : 3 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 9 | b | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is : | "let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 2 / 3 required ratio = 2 / 3 : 1 = 2 : 3 . answer : b" | a = 15 - 8
b = 16 - 4
c = a / b
|
a ) 30 % , b ) 53 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(140, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(140, const_100)), const_100))) | in a restaurant , the profit is 140 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 140 , s . p . = rs . 240 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 240 . profit = rs . ( 240 - 112 ) = rs . 128 . required percentage = ( 128 / 240 * 100 ) % = 53 % appox answer : b" | a = 140 / 100
b = 100 * a
c = b + 100
d = 12 / 100
e = 1 + d
f = 100 * e
g = c - f
h = 140 / 100
i = 100 * h
j = i + 100
k = g / j
l = 100 * k
|
['a ) 55.83 cm 3', 'b ) 56.83 cm 3', 'c ) 113.67 cm 3', 'd ) 58.83 cm 3', 'e ) none of these'] | b | divide(subtract(volume_sphere(add(4, 0.5)), volume_sphere(4)), const_2) | a hemispherical bowl is made of 0.5 cm thick steel . inner radius of the bowl is 4 cm . the volume of steel used in making the bowl is | inner r = 4 outer r = 4.5 formula = 2 / 3 * 22 / 7 ( r ^ 2 - r ^ 2 ) ( 2 / 3 ) * ( 22 / 7 ) * ( 4.5 * 4.5 * 4.5 - 4 * 4 * 4 ) = 56.83 answer : b | a = 4 + 0
b = volume_sphere - (
c = b / volume_sphere
|
a ) a ) 220 , b ) b ) 490 , c ) c ) 380 , d ) d ) 450 , e ) e ) 504 | e | multiply(18, 28) | a large box contains 18 small boxes and each small box contains 28 chocolate bars . how many chocolate bars are in the large box ? | "the only thing you have to do is a simple multiply we have : 18 * 28 = 504 correct answer is : e ) 504" | a = 18 * 28
|
a ) 65 , b ) 79 , c ) 68 , d ) 87 , e ) 90 | a | add(multiply(2, power(5, 2)), multiply(3, 5)) | if one root of the equation 2 x ^ 2 + 3 x – k = 0 is 5 , what is the value of k ? | "we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 5 ^ 2 + 3 * 5 - k = 0 k = 50 + 15 = 65 the answer is a" | a = 5 ** 2
b = 2 * a
c = 3 * 5
d = b + c
|
a ) rs 2125 , b ) rs 2040 , c ) rs 2100 , d ) rs 2200 , e ) rs 2,300 | a | divide(multiply(1377, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(10, subtract(const_100, 20)), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(10, subtract(const_100, 20)), const_100)), 10), const_100))) | from the salary of rahul , 20 % is deducted as house rent , 10 % of the rest he spends on children ’ s education and 10 % of the balance , he spends on clothes . after this expenditure he is left with rs 1377 . his salary is ? | suppose that his salary = rs 100 house rent = rs 20 , balance = rs 80 expenditure on education = rs ( 10 x 80 ) / 100 = rs 8 balance = rs 72 expenditure on clothes rs ( 10 x 72 ) / 100 = rs 7.2 balance now = rs 64.8 if balance is rs 64.8 , salary = rs 100 if balance is rs 1,377 , salary = rs ( 100 x 1377 ) / 64.8 rs 2,125 answer : a | a = 1377 * 100
b = 100 - 20
c = 100 - 20
d = 10 * c
e = d / 100
f = b - e
g = 100 - 20
h = 100 - 20
i = 10 * h
j = i / 100
k = g - j
l = k * 10
m = l / 100
n = f - m
o = a / n
|
a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | the closest approximation of q = ( 69.28 × 0.004 ) / 0.03 is | "q = ( 69.28 × 0.004 ) / 0.03 1 . 0.004 = 4 × 10 ^ ( - 3 ) 2 . 0.03 = 3 × 10 ^ ( - 2 ) 3 . ( a × b ) / c = a × ( b / c ) 4 . 0.004 / 0.03 = 4 × 10 ^ ( - 3 ) / ( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 × 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ) therefore , ( 69.28 × 0.004 ) / 0.03 = 69.28 × ( 0.004 / 0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33 / 10 = 6.928 * 1.33 now , 7 × 2 = 14 7 × 1 = 7 or better : 6.9 × 1 = 6.9 6.9 × 2 = 13.8 hence , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c" | a = 0 / 4
b = a * 69
|
a ) 120 , b ) 160 , c ) 132 , d ) 130 , e ) none of these | a | multiply(divide(multiply(800, 25), const_100), divide(subtract(const_100, 40), const_100)) | in an office , totally there are 800 employees and 25 % of the total employees are males . 40 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ? | "number of male employees = 800 * 25 / 100 = 200 required number of male employees who are less than 50 years old = 200 * ( 100 - 40 ) % = 200 * 60 / 100 = 120 . answer : a" | a = 800 * 25
b = a / 100
c = 100 - 40
d = c / 100
e = b * d
|
a ) 3 , b ) 18 , c ) 6 , d ) 16 , e ) 5 | b | add(5, 13) | if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 13 is divided by 41 ? | "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 13 = 82 k + 5 + 13 = 82 k + 18 or x + 13 = 18,100 , 182 etc . when divided by 41 , we will get the remainder 18 . b" | a = 5 + 13
|
a ) 221 , b ) 309 , c ) 400 , d ) 288 , e ) 171 | b | divide(341, power(add(const_1, divide(5, const_100)), 2)) | find the sum lend at c . i . at 5 p . c per annum will amount to rs . 341 in 2 years ? | "explanation : 341 = p ( 21 / 20 ) 2 p = 309.29 answer : b" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 341 / c
|
a ) 8925 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these | b | divide(multiply(const_100, 4016.25), multiply(10, 5)) | a sum fetched total simple interest of 4016.25 at the rate of 10 p . c . p . a . in 5 years . what is the sum ? | let the sums be p . now , 50 % of p = 4016.25 or , p = 8032.5 answer b | a = 100 * 4016
b = 10 * 5
c = a / b
|
a ) 20 , b ) 36 , c ) 48 , d ) 50 , e ) 59 | a | divide(subtract(multiply(add(25, 5), 45), multiply(25, 50)), 5) | in a factory , an average of 50 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 45 sets / day . the average production per day for day last 5 days is ? | "production during these 5 days = total production in a month - production in first 25 days . = 30 x 45 - 25 x 50 = 100 ∴ average for last 5 days = 100 / 5 = 20 a" | a = 25 + 5
b = a * 45
c = 25 * 50
d = b - c
e = d / 5
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | a | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 42 ? | "21 # 42 = 21 * 24 * 27 * 30 * 33 * 36 * 39 * 42 = 3 ^ 11 * 7 ^ 2 * 2 ^ 7 * 5 ^ 1 * 11 ^ 1 * 13 ^ 1 so total is = 11 + 2 + 7 + 1 + 1 + 1 = 23 answer a" | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
|
a ) 0.68 , b ) 0.6 , c ) 0.8 , d ) 0.3 , e ) none of them | d | divide(subtract(power(1.2, 3), power(0.9, 3)), add(add(power(1.2, 2), 1.08), power(0.9, 2))) | ( 1.2 ) ( power 3 ) - ( 0.9 ) ( power 3 ) / ( 1.2 ) ( power 2 ) + 1.08 + ( 0.9 ) ( power 2 ) is : | "given expression = ( 1.2 ) ( power 3 ) - ( 0.9 ) ( power 3 ) / ( 1.2 ) ( power 2 ) + ( 1.2 x 0.9 ) + ( 0.9 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 1.2 - 0.9 ) = 0.3 answer is d" | a = 1 ** 2
b = 0 ** 9
c = a - b
d = 1 ** 2
e = d + 1
f = 0 ** 9
g = e + f
h = c / g
|
a ) 2812 , b ) 8231 , c ) 2734 , d ) 4222 , e ) 2749 | e | divide(10996, 4) | a volume of 10996 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 10996 l can be kept in ( 10996 / 4 ) hemispheres ans . 2749 answer : e" | a = 10996 / 4
|
a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | d | divide(200, subtract(const_1, divide(3, 4))) | linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 200 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 200 . so 1 / 4 of her savings is $ 200 . so her original savings are 4 times $ 200 = $ 800 correct answer d" | a = 3 / 4
b = 1 - a
c = 200 / b
|
a ) 14 % , b ) 15 % , c ) 12 % , d ) 13.75 % , e ) 14.56 % | d | multiply(divide(2200, add(multiply(4000, 2), multiply(2000, 4))), const_100) | a lent rs . 4000 to b for 2 years and rs . 2000 to c for 4 years on simple interest at the same rate of interest and received rs . 2200 in all from both of them as interest . the rate of interest per annum is : | explanation : let the rate of interest per annum be r % simple interest for rs . 4000 for 2 years at r % + simple interest for rs . 2000 for 4 years at r % = 2200 ( 4000 × r × 2 ) / 100 + ( 2000 × r × 4 ) / 100 = 2200 80 r + 80 r = 2200 160 r = 2200 16 r = 220 4 r = 55 r = 55 / 4 = 13.75 % answer : option d | a = 4000 * 2
b = 2000 * 4
c = a + b
d = 2200 / c
e = d * 100
|
a ) 3 , b ) 4 , c ) 8 , d ) 9 , e ) 6 | b | divide(36, 66) | what is the 23 th digit to the right of the decimal point in the decimal form of 36 / 66 ? | "we first have to use some long division . this long division allows us to get 36 / 66 in decimal form , which is 0.545454 … where “ 54 ” is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of the decimal point is a 4 . in other words , each odd - positioned digit is a 5 , and each even - positioned digit is a 4 . then about the 10 digit to the right of the decimal point and we see that 10 is even , we know that the 10 th digit is a 4 . answer b ." | a = 36 / 66
|
a ) 1 kmph , b ) 2 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph | b | divide(subtract(6, 2), const_2) | a man can row his boat with the stream at 6 km / h and against the stream in 2 km / h . the man ' s rate is ? | "ds = 6 us = 2 s = ? s = ( 6 - 2 ) / 2 = 2 kmph answer : b" | a = 6 - 2
b = a / 2
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | a | multiply(5, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "explanation : let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 5 ) = 4 / 5 20 x = 4 ( 3 x + 5 ) 8 x = 20 x = 2.5 quantity of alcohol = ( 4 x 2.5 ) litres = 10 litres . answer : a" | a = 5 * 1
|
a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | b | divide(18000, 600) | if the l . c . m of two numbers is 600 and their product is 18000 , find the h . c . f of the numbers . | "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18000 / 600 = 30 . answer : b" | a = 18000 / 600
|
a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48 | c | multiply(divide(subtract(const_1, multiply(add(inverse(3), inverse(18)), 2)), inverse(3)), const_60) | working at a constant rate , p can finish a job in 3 hours . q , also working at a constant rate , can finish the same job in 18 hours . if they work together for 2 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | "each hour they complete 1 / 3 + 1 / 18 = 7 / 18 of the job . in 2 hours , they complete 2 ( 7 / 18 ) = 7 / 9 of the job . the time for p to finish is ( 2 / 9 ) / ( 1 / 3 ) = ( 2 / 3 ) hour = 40 minutes the answer is c ." | a = 1/(3)
b = 1/(18)
c = a + b
d = c * 2
e = 1 - d
f = 1/(3)
g = e / f
h = g * const_60
|
a ) 96 , b ) 240 , c ) q = 256 , d ) q = 384 , e ) q = 480 | e | multiply(multiply(16, const_2), 15) | at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses q at the restaurant ? ( assume that all boxes are filled to capacity . ) | "most test takers would recognize the system of equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e" | a = 16 * 2
b = a * 15
|
a ) 70 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(7, 10), const_2), divide(10, 7)), const_100) | tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 7 meters , and the interior of tank b has a height of 7 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "the radius of tank a is 7 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 49 / ( 4 * pi ^ 2 ) = 245 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 7 * pi * 100 / ( 4 * pi ^ 2 ) = 350 / ( 2 * pi ) tank a / tank b = 245 / 350 = 7 / 10 = 70 % the answer is a ." | a = 7 / 10
b = a ** 2
c = 10 / 7
d = b * c
e = d * 100
|
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