options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 120 metres , b ) 180 metres , c ) 100 metres , d ) 228 metres , e ) 150 metres | c | multiply(divide(multiply(60, const_1000), const_3600), 6) | a train running at the speed of 60 km / hr crosses a pole in 6 seconds . what is the length of the train ? | "speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 6 m = 100 m . option c" | a = 60 * 1000
b = a / 3600
c = b * 6
|
a ) 88 , b ) 56 , c ) 89 , d ) 86 , e ) 96 | a | gcd(subtract(3815, 31), subtract(4521, 33)) | find the greatest number which on dividing 3815 and 4521 , leaves a reminder of 31 and 33 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 3815 - 31 ) and ( 4521 - 33 ) = hcf ( 3784 , 4488 ) = 88 option a" | a = 3815 - 31
b = 4521 - 33
c = math.gcd(a, b)
|
a ) 55 days , b ) 77 days , c ) 22.4 days , d ) 88 days , e ) 44 days | c | divide(multiply(28, 16), 20) | 16 men can complete a piece of work in 28 days . in how many days can 20 men complete that piece of work ? | "16 * 28 = 20 * x = > x = 22.4 days answer : c" | a = 28 * 16
b = a / 20
|
a ) 4 : 9 , b ) 4 : 3 , c ) 5 : 4 , d ) 4 : 1 , e ) 4 : 2 | c | divide(sqrt(25), sqrt(16)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 25 : √ 26 = 5 : 4 answer : c" | a = math.sqrt(25)
b = math.sqrt(16)
c = a / b
|
['a ) a ) 2 : 3', 'b ) b ) 3 : 4', 'c ) c ) 4 : 3', 'd ) d ) 3 : 2', 'e ) e ) 2 : 1'] | b | divide(multiply(subtract(4, const_1), const_2), subtract(power(subtract(4, const_1), const_2), const_1)) | two sides of a triangle have lengths x and w and meet at a right angle . if the perimeter of the triangle is 4 x , what is the ratio of x to w ? | ahhhh . . . . my bad . thank you , you ' re correct . 8 x = 6 w - - > x / w = 3 / 4 . actually b is the right answer . | a = 4 - 1
b = a * 2
c = 4 - 1
d = c ** 2
e = d - 1
f = b / e
|
a ) 25.5 , b ) 30.4 , c ) 31.8 , d ) 32.5 , e ) 33.8 | a | multiply(divide(subtract(69.0, 55.0), 69.0), const_100) | the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 55.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 55 = 14 $ ( 14 / 55 ) * 100 = 25.5 % answer : a" | a = 69 - 0
b = a / 69
c = b * 100
|
a ) 25630 yards , b ) 35200 yards , c ) 52800 yards , d ) 42560 yards , e ) 41520 yards | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 30), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 30 miles into yards ? | "1 mile = 1760 yards 30 miles = 30 * 1760 = 52800 yards answer is c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 30
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 1977 , b ) 1893 , c ) 1979 , d ) 3000 , e ) 1278 | d | subtract(divide(subtract(multiply(2000, 65), multiply(2000, 15)), 20), 2000) | a garrison of 2000 men has provisions for 65 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ? | "2000 - - - - 65 2000 - - - - 50 x - - - - - 20 x * 20 = 2000 * 50 x = 5000 2000 - - - - - - - 3000 answer : d" | a = 2000 * 65
b = 2000 * 15
c = a - b
d = c / 20
e = d - 2000
|
a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 2 sec | b | divide(70, multiply(36, const_0_2778)) | in what time will a railway train 70 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 70 / 36 * 18 / 5 = 7 sec answer : b" | a = 36 * const_0_2778
b = 70 / a
|
a ) $ 198 , b ) $ 216 , c ) $ 192 , d ) $ 230 , e ) $ 240 | c | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 20)) | a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 20 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ? | "wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 20 emlpoyee paid = . 8 * 240 = 192 $ answer c" | a = 200 / 100
b = a * 20
c = 200 + b
d = 200 / 100
e = d * 20
f = 200 + e
g = f / 100
h = g * 20
i = c - h
|
a ) 897 , b ) 98,910 , c ) 64,749 , d ) 49,700 , e ) 56,720 | b | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 5), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 5 ? | find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 5 seeing the options after dividing and finding the reminder of 2 my answer was b | a = 1000 - 3
b = 3 * 3
c = b * 10
d = 3 * 3
e = c + d
f = e + 3
g = a - f
h = g / 5
i = h + 1
j = i / 2
k = 1000 - 3
l = 3 * 3
m = l * 10
n = 3 * 3
o = m + n
p = o + 3
q = k + p
r = j * q
|
a ) 20 kmph , b ) 28 kmph , c ) 25 kmph , d ) 22 kmph , e ) 24 kmph | e | multiply(divide(20, const_60), 72) | the speed of a train is 72 kmph . what is the distance covered by it in 20 minutes ? | "72 * 20 / 60 = 24 kmph answer : e" | a = 20 / const_60
b = a * 72
|
['a ) 9', 'b ) 10', 'c ) 11', 'd ) 12', 'e ) 13'] | b | subtract(subtract(divide(divide(multiply(27, 8), const_10), const_2), const_0_25), const_0_25) | a math teacher has 27 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 27 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 27 - 17 = 10 answer ( b ) | a = 27 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
|
a ) 47 , b ) 76 , c ) 28 , d ) 66 , e ) 99 | e | multiply(divide(45, subtract(6, const_1)), subtract(12, const_1)) | at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 45 seconds . how long does it tick at 12 ′ o clock | "explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 45 / 5 = 9 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 9 = 99 secs . answer : e" | a = 6 - 1
b = 45 / a
c = 12 - 1
d = b * c
|
a ) 40 , b ) 45 , c ) 27 , d ) 28 , e ) 22 | b | divide(add(70, 20), const_2) | the total marks obtained by a student in mathematics and physics is 70 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 70 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 70 + 20 ) / 2 = 45 . answer : b" | a = 70 + 20
b = a / 2
|
a ) $ 36 , b ) $ 42 , c ) $ 60 , d ) $ 75 , e ) $ 105 | e | multiply(divide(multiply(divide(36, 6), 7), divide(36, 6)), 15) | the ratio of the money with ravi and giri is 6 : 7 and that with giri and kiran is 6 : 15 . if ravi has $ 36 , how much money does kiran have ? | rita : sita : kavita 6 : 7 6 : 15 36 : 42 : 105 the ratio of money with ravi , giri and kiran is 36 : 42 : 105 we see that 36 = $ 36 then 105 = $ 105 answer is e | a = 36 / 6
b = a * 7
c = 36 / 6
d = b / c
e = d * 15
|
a ) 3 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | add(add(multiply(3, 2), multiply(7, 2)), subtract(multiply(2, 2), multiply(7, 3))) | ( 3 x + 2 ) ( 2 x - 7 ) = ax ^ 2 + kx + n . what is the value of a - n + k ? | expanding we have 6 x ^ 2 - 21 x + 4 x - 14 6 x ^ 2 - 17 x - 14 taking coefficients , a = 6 , k = - 17 , n = - 14 therefore a - n + k = 6 - ( - 14 ) - 17 = 20 - 17 = 3 the answer is a . | a = 3 * 2
b = 7 * 2
c = a + b
d = 2 * 2
e = 7 * 3
f = d - e
g = c + f
|
a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(subtract(300, 200), const_1), add(subtract(add(add(add(floor(divide(add(subtract(300, 200), const_1), 2)), const_1), add(floor(divide(subtract(300, 200), 3)), const_1)), add(divide(subtract(300, 200), 5), const_1)), add(add(floor(divide(divide(subtract(300, 200), const_2), const_3)), add(floor(divide(divide(subtract(300, 200), 5), 3)), const_1)), add(divide(subtract(300, 200), const_10), const_1))), add(floor(divide(divide(const_100, const_3), const_10)), const_1))) | how many positive integers h between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e . | a = 300 - 200
b = a + 1
c = 300 - 200
d = c + 1
e = d / 2
f = math.floor(e)
g = f + 1
h = 300 - 200
i = h / 3
j = math.floor(i)
k = j + 1
l = g + k
m = 300 - 200
n = m / 5
o = n + 1
p = l + o
q = 300 - 200
r = q / 2
s = r / 3
t = math.floor(s)
u = 300 - 200
v = u / 5
w = v / 3
x = math.floor(w)
y = x + 1
z = t + y
A = 300 - 200
B = A / 10
C = B + 1
D = z + C
E = p - D
F = 100 / 3
G = F / 10
H = math.floor(G)
I = H + 1
J = E + I
K = b - J
|
a ) 120 cm 2 , b ) 1470 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2 | b | divide(multiply(84, 35), const_2) | if the sides of a triangle are 91 cm , 84 cm and 35 cm , what is its area ? | "the triangle with sides 91 cm , 84 cm and 35 cm is right angled , where the hypotenuse is 91 cm . area of the triangle = 1 / 2 * 84 * 35 = 1470 cm 2 answer : b" | a = 84 * 35
b = a / 2
|
a ) 2 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | d | subtract(divide(28, 2), const_1) | for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 28 ) ? | "f ( 28 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 * 28 the greatest prime factor in this list is 13 . the answer is d ." | a = 28 / 2
b = a - 1
|
a ) 643 , b ) 677 , c ) 760 , d ) 767 , e ) 832 | c | divide(add(212, 16), divide(30, const_100)) | mike needs 30 % to pass . if he scored 212 marks and falls short by 16 marks , what was the maximum marks he could have got ? | if mike had scored 16 marks more , he could have scored 30 % therefore , mike required 212 + 16 = 228 marks let the maximum marks be m . then 30 % of m = 228 ( 30 / 100 ) × m = 228 m = ( 228 × 100 ) / 30 m = 22800 / 30 m = 760 answer : c | a = 212 + 16
b = 30 / 100
c = a / b
|
a ) 4 : 3 , b ) 5 : 4 , c ) 6 : 5 , d ) 7 : 6 , e ) 8 : 7 | d | divide(3, const_4) | a student completes a 3 part exam as follows . the student completes sections a in 18 minutes and takes an equal time to complete the two remaining sections . what is the ratio of time spent on section c to section a if the total duration is 1 hour ? | "the time spent on section b and section c is 21 minutes each . the ratio of c to a is 21 : 18 = 7 : 6 the answer is d ." | a = 3 / 4
|
a ) 104 , b ) 55 , c ) 148 , d ) 150 , e ) 159 | c | divide(742, 7) | . a car covers a distance of 742 km in 7 hours . find its speed ? | "742 / 5 = 148 kmph answer : c" | a = 742 / 7
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a ) 12250 , b ) 13400 , c ) 12500 , d ) 14340 , e ) none of these | a | divide(multiply(multiply(10, 8), 24.5), divide(divide(multiply(multiply(20, 10), 8), const_100), const_100)) | how many bricks each measuring 20 cm x 10 cm x 8 cm , will be needed to build a wall 10 m x 8 m x 24.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 1000 x 800 x 24.5 ) / ( 20 x 10 x 8 ) = 12250 answer : a" | a = 10 * 8
b = a * 24
c = 20 * 10
d = c * 8
e = d / 100
f = e / 100
g = b / f
|
a ) 19,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(2,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 2,000)) | angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 2,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo ’ s and isabella ’ s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal base salary plus commission . so angelo ’ s earnings are 550 + ( 0.08 ) ( x – 2,000 ) , and isabella ’ s are 0.10 x . set up the equation and solve : 550 + ( 0.08 ) ( x – 2,000 ) = 0.10 x distribute the 0.08 : 550 + 0.08 x – 160 = 0.10 x combine terms and subtract 0.08 x from both sides : 390 = 0.02 x divide both sides by 0.02 : 19,500 = x your answer is a ." | a = 8 / 100
b = 2 * 0
c = 550 - b
d = 10 / 100
e = 8 / 100
f = d - e
g = c / f
h = g / 2
i = math.floor(h)
|
a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50 | a | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(120, const_100))), const_100) | if 45 % of z is 120 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 120 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 120 / 100 ) * ( 3 / 4 ) x i . e . z = ( 120 * 3 ) x / ( 45 * 4 ) i . e . z = ( 2 ) x = ( 200 / 100 ) x i . e . z is 200 % of x answer : option a" | a = 75 / 100
b = 45 / 100
c = 120 / 100
d = b / c
e = a / d
f = e * 100
|
a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000 | d | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 13,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | answer equals d in 48 years . i thought by 50 th year it would reach 100,500 . options should have been separated more widely for clarity . | a = 2 + 3
b = a * 1000
c = b * 12
d = c / 2
|
a ) 0.4 , b ) 0.04 , c ) 0.004 , d ) 0.0004 , e ) 4 | b | divide(4, const_100) | how is 4 % expressed as a decimal fraction ? | "4 / 100 = 0.04 answer : b" | a = 4 / 100
|
a ) 27 , b ) 36 , c ) 45 , d ) 72 , e ) 80 | e | subtract(divide(128, subtract(const_1, divide(5, 13))), 128) | in a certain school , the ratio of boys to girls is 5 to 13 . if there are 128 more girls than boys , how many boys are there ? | "the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 128 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 128 . so 1 part corresponds to 128 / 8 = 16 and so 5 parts correspond to 5 * 16 = 80 . e" | a = 5 / 13
b = 1 - a
c = 128 / b
d = c - 128
|
a ) 5624 , b ) 5625 , c ) 5626 , d ) 5627 , e ) none | b | power(75, const_2) | i remember during the school days , the teacher asked the class ` ` can you tell me the sum of the first 50 odd numbers ? ` ` . i ran quickly to the teacher and told her ` ` the answer is 2500 ' ' . the teacher replied ` ` lucky guess ' ' . she then asked me ` ` can you tell me the sum of first 75 odd numbers ? ` ` . i wait for approx 10 seconds and replied with the correct answer . how can i answer so quickly and whats the correct answer ? | solution : 5625 n ^ 1 75 * 75 = 5625 ( sum of first 75 odd numbers ) . 50 * 50 = 2500 ( sum of first 50 odd numbers ) . answer b | a = 75 ** 2
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a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | d | divide(add(11, 7), const_2) | in one hour , a boat goes 11 km along the stream and 7 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 7 ) km / hr = 9 kmph . answer d" | a = 11 + 7
b = a / 2
|
a ) 16.6 , b ) 12.9 , c ) 31.3 , d ) 19.2 , e ) 15.7 | a | add(10, const_1) | the average of first five prime numbers greater than 10 is ? | "11 + 13 + 17 + 19 + 23 = 83 / 5 = 16.6 answer : a" | a = 10 + 1
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a ) 11 . , b ) 12 . , c ) 13 . , d ) 14 . , e ) 16 | e | add(divide(multiply(15, const_100), add(const_100, 25)), divide(multiply(6, const_100), add(const_100, 50))) | following an increase in prices , the price of a candy box was 15 pounds and the price of a can of soda was 6 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "price of candy before price increase = 15 / 1.25 = 12 price of soda before price increase = 6 / 1.5 = 4 total price = 12 + 4 = 16 e is the answer" | a = 15 * 100
b = 100 + 25
c = a / b
d = 6 * 100
e = 100 + 50
f = d / e
g = c + f
|
a ) 5 : 00 , b ) 5 : 30 , c ) 6 : 00 , d ) 6 : 20 , e ) 7 : 00 | d | divide(add(4, const_2), 20) | city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 4 : 20 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? | "train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 4 : 20 pm + 2 hrs = 6 : 20 pm answer : d . 6 : 20" | a = 4 + 2
b = a / 20
|
a ) 73.41 , b ) 72.28 , c ) 72.43 , d ) 73.43 , e ) can not be determined | b | divide(subtract(multiply(35, 72), subtract(56, 46)), 35) | a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 46 instead of 56 then find the correct average marks up to two decimal places . | "total marks = 35 x 72 = 2520 corrected total marks = 2520 - 46 + 56 = 2530 correct average = 2530 / 35 = 72.28 answer : b" | a = 35 * 72
b = 56 - 46
c = a - b
d = c / 35
|
a ) 12 , b ) 13 , c ) 34 , d ) 37 , e ) 59 | d | add(21, 11) | a number when divided by a divisor leaves a remainder of 21 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 21 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 34 as d is also integer and 34 is a prime number , the d should be 34 to satisfy the above equation . hence answer is ' d '" | a = 21 + 11
|
a ) 4063 , b ) 5325 , c ) 5385 , d ) 6000 , e ) 6154 | c | divide(multiply(5000, divide(7, const_100)), divide(6.5, const_100)) | last year a certain bond price with a face value of 5000 yielded 7 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ? | "interest = 0.07 * 5000 = 0.065 * selling price - - > selling price = 0.07 * 5000 / 0.065 - - > selling price = ~ 5,385 answer : c ." | a = 7 / 100
b = 5000 * a
c = 6 / 5
d = b / c
|
['a ) 2 √ 2', 'b ) 2 √ 5', 'c ) 5', 'd ) 4 √ 2', 'e ) 8'] | e | divide(32, const_4) | 32 meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ? | area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 32 = > x / 360 * 2 * pi * r + 2 r = 32 = > 2 a / r + 2 r = 32 = > a = r 16 - r ^ 2 = r 16 - r ^ 2 we will now max using derivations max value of a will found at a = 0 i . e 16 - 2 r = 0 r = 8 e | a = 32 / 4
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a ) 287 , b ) 350 , c ) 600 , d ) 277 , e ) 122 | c | subtract(multiply(speed(300, 18), 54), 300) | a 300 m long train crosses a platform in 54 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 54 = 50 / 3 3 x + 900 = 2700 = > x = 600 m . answer : c" | a = speed * (
b = a - 54
|
['a ) 24', 'b ) 27', 'c ) 25', 'd ) 26', 'e ) 29'] | a | multiply(sqrt(36), const_4) | compute the perimeter of a square with area 36 . | if the length of the square is s , then s 2 = 36 so s = 6 . the perimeter is therefore 4 s = 24 . correct answer a | a = math.sqrt(36)
b = a * 4
|
a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 96 m | e | add(multiply(4, divide(divide(72, 4), subtract(4, const_1))), 72) | a can run 4 times as fast as b and gives b a start of 72 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 72 m he gains 72 * ( 4 / 3 ) i . e 96 m answer : e" | a = 72 / 4
b = 4 - 1
c = a / b
d = 4 * c
e = d + 72
|
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | b | subtract(30, multiply(multiply(12, 3), 2)) | evaluate : 30 - 12 ÷ 3 × 2 = ? | "according to order of operations , 12 ÷ 3 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 3 × 2 = 4 × 2 = 8 hence 30 - 12 ÷ 3 × 2 = 30 - 8 = 22 correct answer b ) 22" | a = 12 * 3
b = a * 2
c = 30 - b
|
a ) 2 , b ) 9 , c ) 15 , d ) 21 , e ) 30 | c | divide(subtract(0.60, multiply(0.06, 5)), subtract(0.08, 0.06)) | a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 5 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company charges the same amount under plan a and under plan b ? | "let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 5 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 5 ) = 0.08 t - - > 60 + 6 ( t - 5 ) = 8 t - - > t = 15 . answer : c ." | a = 0 * 6
b = 0 - 60
c = 0 - 8
d = b / c
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a ) 52 , b ) 27 , c ) 99 , d ) 61 , e ) 11 | a | divide(subtract(500, multiply(60, 30)), negate(subtract(30, 5))) | an employer pays rs . 30 for each day a worker works , and forfeits rs . 5 for each day he is idle . at the end of 60 days , a worker gets rs . 500 . for how many days did the worker remain idle ? | explanation : suppose the worker remained idle for m days . then , he worked for ( 60 - m ) days . 30 ( 60 - m ) – 5 m = 500 1800 – 25 m = 500 25 m = 1300 m = 52 so , the worker remained idle for 52 days . answer : a | a = 60 * 30
b = 500 - a
c = 30 - 5
d = b / negate
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a ) 3 days , b ) 5 days , c ) 4 days , d ) 6 days , e ) 7 days | e | divide(subtract(const_1, add(multiply(divide(const_1, 5), const_2), multiply(divide(const_1, 15), const_2))), divide(const_1, 15)) | a can finish a piece of work in 5 days . b can do it in 15 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 5 + ( 2 + x ) / 15 = 1 = > x = 7 days answer : e" | a = 1 / 5
b = a * 2
c = 1 / 15
d = c * 2
e = b + d
f = 1 - e
g = 1 / 15
h = f / g
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a ) 96.7 m , b ) 120 m , c ) 130 m , d ) 150 m , e ) none of these | a | subtract(180, multiply(divide(5, const_60), const_1000)) | a policeman noticed a criminal from a distance of 180 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 5 minutes ? | "explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 5 / 60 ) km = 1 / 12 km = 83.3 m . . ' . distance between the criminal and policeman = ( 180 - 83.3 ) m = 96.7 m . answer : a" | a = 5 / const_60
b = a * 1000
c = 180 - b
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a ) 85432864 , b ) 82673462 , c ) 86985434 , d ) 68954365 , e ) 65486954 | b | multiply(divide(8269, 9998), const_100) | 8269 ã — 9998 = ? | "8269 ã — ( 10000 - 2 ) 82690000 - 16538 = 82673462 answer b" | a = 8269 / 9998
b = a * 100
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a ) 45 , 63 , b ) 50 , 70 , c ) 55 , 77 , d ) 60 , 84 , e ) 65 , 91 | a | subtract(multiply(multiply(subtract(multiply(17, 12), multiply(11, 12)), 5), divide(7, 5)), multiply(subtract(multiply(17, 12), multiply(11, 12)), 5)) | two numbers are in the ratio of 5 : 7 . if 12 is subtracted from each number , they are in the ratio of 11 : 17 . what are the two numbers ? | "( 5 x - 12 ) : ( 7 x - 12 ) = 11 y : 17 y 5 x - 12 = 11 y 7 x - 12 = 17 y 84 - 60 = 85 y - 77 y 24 = 8 y y = 3 5 x - 12 = 33 x = 9 the two numbers are 5 x and 7 x which are 45 and 63 . the answer is a ." | a = 17 * 12
b = 11 * 12
c = a - b
d = c * 5
e = 7 / 5
f = d * e
g = 17 * 12
h = 11 * 12
i = g - h
j = i * 5
k = f - j
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a ) 227 , b ) 513 , c ) 342 , d ) 787 , e ) 191 | b | subtract(subtract(600, divide(multiply(600, 10), const_100)), divide(multiply(subtract(600, divide(multiply(600, 10), const_100)), 5), const_100)) | the sale price sarees listed for rs . 600 after successive discount is 10 % and 5 % is ? | "600 * ( 90 / 100 ) * ( 95 / 100 ) = 513 answer : b" | a = 600 * 10
b = a / 100
c = 600 - b
d = 600 * 10
e = d / 100
f = 600 - e
g = f * 5
h = g / 100
i = c - h
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a ) 1888 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2184 | e | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 6720), divide(multiply(15, 6720), const_100)), const_2) | in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 6720 votes were polled , then how many valid votes did b get ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 6720 32.5 k = 32.5 k / 100 k * 6720 = 2184 the number of valid votes polled in favour of b is 2184 . answer : e" | a = 20 / 100
b = 1 - a
c = b * 6720
d = 15 * 6720
e = d / 100
f = c - e
g = f / 2
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a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 11 | e | divide(multiply(5, const_1), add(multiply(5, const_1), multiply(5, 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 5 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 5 * x his total tip for the 7 months = 6 * ( average tip for the months other than august ) + 5 x = 11 x august tips as a fraction of total tips = 5 x / 11 x = 5 / 11 . so e" | a = 5 * 1
b = 5 * 1
c = 5 * 1
d = b + c
e = a / d
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a ) $ 149.55 , b ) $ 134.56 , c ) $ 175.43 , d ) $ 158.65 , e ) $ 255.45 | d | divide(multiply(180, divide(multiply(add(180, 55), divide(add(180, 45), const_2)), 180)), subtract(divide(add(180, 45), const_2), 45)) | a clock store sold a certain clock to a collector for 45 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 25 percent of what the collector had paid . the shop then sold the clock again at a profit of 55 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 $ 180 , 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 45 % profit . this means the clocks ' selling price was c ( 1.45 ) 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 25 % the price at which the collector bought it . thus , you get = 1.45 * 0.25 * c = 0.3625 c furthermore , the store sells the clock for the second time for 55 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.55 = 1.55 * 0.3625 c finally given that c - 0.3625 c = 180 - - - - > c = 282.35 $ thus , the cost of the clock the second time around = 1.55 * 0.3625 c = 1.55 * 0.3625 * 282.35 = 158.65 $ . hence d is the correct answer ." | a = 180 + 55
b = 180 + 45
c = b / 2
d = a * c
e = d / 180
f = 180 * e
g = 180 + 45
h = g / 2
i = h - 45
j = f / i
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a ) 30 , b ) 20 , c ) 35 , d ) 32 , e ) 40 | d | subtract(subtract(divide(5400, subtract(divide(subtract(5400, 400), 100), 10)), 100), const_3) | in a hostel there were 100 students . to accommodate some more students the average budget is decreased by rupees 10 . but total expenditure increased by rs . 400 . if the total expenditure of the hostel now 5400 , find the number of student joined ? | let average is x 100 x + 400 = 5400 x = 50 let the number of student joined is y ( 100 + y ) * ( 50 - 10 ) = 5400 y = 35 answer : d | a = 5400 - 400
b = a / 100
c = b - 10
d = 5400 / c
e = d - 100
f = e - 3
|
a ) 33 , b ) 66 , c ) 18 , d ) 100 , e ) 281 | b | subtract(divide(multiply(divide(multiply(54, 8), 30), 50), 6), 54) | 54 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 54 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 120 120 – 54 = 66 answer : b" | a = 54 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 54
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a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 16 , d ) 1 / 24 , e ) 1 / 32 | b | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting sum 10 with the help of three dice ? | "( 1 , 3,6 ) 3 ! = 6 ( 2 , 2,6 ) = 3 ! / 2 = 3 ( 1 , 4,5 ) 3 ! = 6 ( 2 , 4,4 ) = 3 ( 2 , 3,5 ) 3 ! = 6 ( 3 , 3,4 ) = 3 prob . of getting sum 10 = ( 6 + 6 + 6 + 3 + 3 + 3 ) / 216 = 1 / 8 ans answer : b" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) 1 / 20 , b ) 1 / 25 , c ) 1 / 30 , d ) 1 / 35 , e ) 1 / 40 | d | divide(1, 18) | a certain fraction has the same ratio to 1 / 18 , as 2 / 5 does to 7 / 9 . what is this certain fraction ? | "x / ( 1 / 18 ) = ( 2 / 5 ) / ( 7 / 9 ) x = 2 * 9 * 1 / 18 * 5 * 7 = 1 / 35 the answer is d ." | a = 1 / 18
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a ) 36 , b ) 39 , c ) 23 , d ) 45 , e ) none of the above | c | add(subtract(87, multiply(17, 4)), 4) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his average by 4 . find his average after 17 th inning . | "let the average after 17 th inning = x . then , average after 16 th inning = ( x – 4 ) . ∴ 16 ( x – 4 ) + 87 = 17 x or x = ( 87 – 64 ) = 23 . answer c" | a = 17 * 4
b = 87 - a
c = b + 4
|
['a ) 3 : 1', 'b ) 8 : 1', 'c ) 9 : 1', 'd ) 12 : 1', 'e ) 9 : 1'] | c | divide(surface_cube(3), surface_cube(1)) | . if the sides of two cubes are in the ratio 3 : 1 the ratio of their total surface area is ? | a 1 : a 2 = 3 : 1 6 a 12 : 6 a 22 = 9 : 1 answer : c | a = surface_cube / (
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a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 1,496 | e | divide(divide(multiply(multiply(16, add(16, const_1)), add(multiply(16, 2), const_1)), 6), divide(multiply(multiply(16, add(16, const_1)), add(multiply(16, 2), const_1)), 6)) | the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n ^ 3 / 3 + c * n ^ 2 + n / 6 , where c is a constant . what is the sum of the first 16 positive perfect squares ? | "first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 16 and c = 1 / 2 into the equation and simplify to get the answer 1496 option e ." | a = 16 + 1
b = 16 * a
c = 16 * 2
d = c + 1
e = b * d
f = e / 6
g = 16 + 1
h = 16 * g
i = 16 * 2
j = i + 1
k = h * j
l = k / 6
m = f / l
|
a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | b | subtract(divide(add(add(3, 8), 10), const_3), divide(add(add(4, 9), 11), 3)) | set a : 3 , r , 8 , 10 set b : 4 , g , 9 , 11 the terms of each set above are given in ascending order . if the median of set a is equal to the median of set b , what is the value of g – r ? | so we have even no . of elements in the set so median is the average of middle two numbers ( r + 8 ) / 2 = ( g + 9 ) / 2 g - r = - 1 answer b | a = 3 + 8
b = a + 10
c = b / 3
d = 4 + 9
e = d + 11
f = e / 3
g = c - f
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a ) 20 , b ) 31 , c ) 35 , d ) 40 , e ) 45 | e | divide(200, const_10) | how many integers from 40 to 200 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 40 to 200 , inclusive is ( 189 - 42 ) / 21 + 1 = 8 ; 53 - 8 = 45 . answer : e ." | a = 200 / 10
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a ) 92 ° f , b ) 97 ° f , c ) 104 ° f , d ) 113 ° f , e ) 118 ° f | d | add(multiply(divide(subtract(212, 32), 100), 45), 32) | water boils at 212 ° f or 100 ° c and ice melts at 32 ° f or 0 ° c . if the temperature of a pot of water is 45 ° c , what is the temperature of the pot of water in ° f ? | "let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) f = 9 c / 5 + 32 f = 9 ( 45 ) / 5 + 32 = 113 ° f the answer is d ." | a = 212 - 32
b = a / 100
c = b * 45
d = c + 32
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a ) rs . 316 , b ) rs . 251 , c ) rs . 154 , d ) rs . 294 , e ) rs . 200 | a | subtract(415, divide(multiply(subtract(514, 415), 2), 4)) | a sum of money at simple interest amounts to rs . 415 in 2 years and to rs . 514 in 4 years . the sum is ? | "explanation : s . i . for 2 years = ( 514 - 415 ) = rs . 99 s . i . for 1 year = 99 / 2 principal = ( 415 - 99 ) = rs . 316 . answer is a" | a = 514 - 415
b = a * 2
c = b / 4
d = 415 - c
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a ) 1 kmph , b ) 7 kmph , c ) 98 kmph , d ) 11 kmph , e ) 4 kmph | d | divide(subtract(26, 4), const_2) | a man can row his boat with the stream at 26 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 26 us = 4 s = ? s = ( 26 - 4 ) / 2 = 11 kmph answer : d" | a = 26 - 4
b = a / 2
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a ) 10 days , b ) 12 days , c ) 20 days , d ) 18 days , e ) 25 days | c | add(multiply(subtract(const_1, multiply(add(divide(const_1, 30), divide(const_1, 15)), 5)), 30), 5) | a and b can complete a work in 30 days and 15 day . they started doing the work together but after 5 days b had to leave and a alone completed the remaining work . the whole work was completed in ? | a + b 1 day work = 1 / 30 + 1 / 15 = 1 / 10 work done by a and b in 10 days = 1 / 10 * 5 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 now 1 / 30 work is done by a in 1 day 1 / 2 work will be done by a in 30 * 1 / 2 = 15 days total time taken = 15 + 5 = 20 days answer is c | a = 1 / 30
b = 1 / 15
c = a + b
d = c * 5
e = 1 - d
f = e * 30
g = f + 5
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a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, const_60), 4), 4), subtract(divide(60, const_60), divide(45, const_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 e 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 e : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) = 24 miles . d is the correct answer . | a = 45 / const_60
b = a * 4
c = b + 4
d = 60 / const_60
e = 45 / const_60
f = d - e
g = c / f
h = g - 4
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a ) 101 , b ) 45 , c ) 2044 , d ) 2048 , e ) 2064 | c | divide(subtract(28, power(73, const_2)), const_2) | the sum of two numbers is 73 and their difference is 28 . find the difference between their squares . | "explanation : let x and y be the two given numbers ( x + y ) = 73 ; ( x – y ) = 28 ( x 2 – y 2 ) = ( x + y ) ( x – y ) = 73 * 28 = 2044 answer : c" | a = 73 ** 2
b = 28 - a
c = b / 2
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a ) 10 min , b ) 15 min , c ) 30 min , d ) 25 min , e ) 42 min | c | multiply(divide(const_1, add(add(divide(const_1, 60), divide(const_1, 40)), divide(const_1, 40))), const_2) | a large tank can filled by a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tanker from empty state if b is used for half the time and a and b fill it together for the other half ? | part filled by a + b in 1 minute = 1 / 60 + 1 / 40 = 1 / 24 suppose the tank is filled in x minutes then , x / 2 ( 1 / 24 + 1 / 40 ) = 1 x / 2 * 1 / 15 = 1 x = 30 min answer is c | a = 1 / 60
b = 1 / 40
c = a + b
d = 1 / 40
e = c + d
f = 1 / e
g = f * 2
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['a ) 80', 'b ) 70', 'c ) 60', 'd ) 40', 'e ) 50'] | c | multiply(divide(300, 100), divide(40, const_2)) | the ratio , by volume , of bleach to detergent to water in a certain solution is 4 : 40 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 300 liters of water , how many liters of detergent will it contain ? | b : d : w = 4 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 4 / 40 ) = ( 1 / 30 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 300 dnew = wnew / 5 = 300 / 5 = 60 so , answer will be c | a = 300 / 100
b = 40 / 2
c = a * b
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a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 11 , d ) 11 : 3 , e ) 17 : 3 | e | divide(add(multiply(8, divide(add(9, 1), add(8, 2))), 9), add(multiply(2, divide(add(9, 1), add(8, 2))), 1)) | two vessels contains equal number of mixtures milk and water in the ratio 8 : 2 and 9 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 8 / 10 + 9 / 10 ) : ( 2 / 10 + 1 / 10 ) = 17 / 10 : 3 / 10 = 17 : 3 answer is e" | a = 9 + 1
b = 8 + 2
c = a / b
d = 8 * c
e = d + 9
f = 9 + 1
g = 8 + 2
h = f / g
i = 2 * h
j = i + 1
k = e / j
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers r are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a = 1 + 1
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a ) 3 / 10 , b ) 3 / 20 , c ) 2 / 5 , d ) 1 / 2 , e ) 1 / 3 | a | divide(3, const_10) | tickets numbered 1 to 20 are mixed up and then a ticked is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of 3 ? | s = { 1 , 23 , … . 20 } e = { 3 , 6 , 912 , 1518 } p ( e ) = 6 / 20 = 3 / 10 option a | a = 3 / 10
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['a ) 4 : 3', 'b ) 3 : 4', 'c ) 3 : 2', 'd ) 5 : 2', 'e ) 4 : 6'] | a | divide(multiply(const_4, multiply(power(const_2, const_2), const_pi)), multiply(const_3, multiply(power(const_2, const_2), const_pi))) | find the ratio between whole surfaces of a sphere and a hemisphere ? | 4 ï € r 2 : 3 ï € r 2 = > 4 : 3 answer a | a = 2 ** 2
b = a * math.pi
c = 4 * b
d = 2 ** 2
e = d * math.pi
f = 3 * e
g = c / f
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a ) 2 / 5 , b ) 2 / 3 , c ) 1 / 2 , d ) 1 / 3 , e ) 1 / 5 | c | divide(subtract(divide(15, const_100), divide(10, const_100)), subtract(divide(20, const_100), divide(10, const_100))) | some of 10 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 15 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 10 ( 1 - x ) + 20 x = 15 x = 1 / 2 answer : c" | a = 15 / 100
b = 10 / 100
c = a - b
d = 20 / 100
e = 10 / 100
f = d - e
g = c / f
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a ) 37 , b ) 39 , c ) 41 , d ) 43 , e ) 45 | d | divide(100, divide(add(divide(100, 25), divide(100, 150)), const_2)) | a person walks from one end to the other of a 100 - meter long moving walkway at a constant rate in 25 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 150 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | "let v be the speed of the person and let x be the speed of the walkway . 25 ( v + x ) = 100 then 150 ( v + x ) = 600 150 ( v - x ) = 100 when we add the two equations : 300 v = 700 v = 7 / 3 time = 100 / ( 7 / 3 ) = ( 300 / 7 ) seconds which is about 43 seconds the answer is d ." | a = 100 / 25
b = 100 / 150
c = a + b
d = c / 2
e = 100 / d
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a ) 55 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(65, 10), 10), subtract(65, subtract(65, 10))) | a certain number of workers can do a work in 65 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 65 - 10 ) / 10 = 55 answer is a" | a = 65 - 10
b = a * 10
c = 65 - 10
d = 65 - c
e = b / d
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a ) 4 , b ) 4 / 3 , c ) 4 / 5 , d ) 4 / 9 , e ) 3 / 5 | e | divide(add(divide(divide(factorial(9), factorial(subtract(9, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 9)), factorial(subtract(add(6, 9), const_2))), factorial(const_2))) | a bag contains 6 black and 9 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 9 ) = 15 . number of white balls = 9 . p ( drawing a white ball ) = 9 / 15 = 3 / 5 hence answer is e" | a = math.factorial(9)
b = 9 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 6 + 9
o = math.factorial(n)
p = 6 + 9
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t
v = m / u
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a ) 100 min , b ) 110 min , c ) 140 min , d ) 120 min , e ) 150 min | d | multiply(divide(multiply(8, add(4, divide(45, const_60))), 19), const_60) | walking at the rate of 8 kmph a man cover certain distance in 4 hr 45 min . running at a speed of 19 kmph the man will cover the same distance in . | "distance = speed * time 8 * 19 / 4 = 38 km new speed = 19 kmph therefore time = d / s = 38 / 19 = 120 min answer : d ." | a = 45 / const_60
b = 4 + a
c = 8 * b
d = c / 19
e = d * const_60
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a ) 15 , b ) 16 , c ) 17 , d ) 20 , e ) 24 | e | multiply(divide(420, multiply(63, const_1000)), const_3600) | a train 420 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 420 * 2 / 35 = 24 sec answer : option e" | a = 63 * 1000
b = 420 / a
c = b * 3600
|
a ) 10 % , b ) 28 % , c ) 33 % , d ) 42 % , e ) 52 % | e | multiply(divide(subtract(add(divide(32, 2), multiply(subtract(100, 32), 2)), 100), 100), const_100) | every day daniel drives 100 miles back from work . on sunday , daniel drove all the way back from work at a constant speed of x miles per hour . on monday , daniel drove the first 32 miles back from work at ( 2 x ) miles per hour , and the rest of the way at ( x / 2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ? | "let ' s test x = 4 . . . . on sunday , daniel drove 100 miles at 4 miles / hour . d = ( r ) ( t ) 100 = ( 4 ) ( t ) 100 / 4 = 25 = t it takes 25 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles / hour and the rest of the way ( 68 miles ) at 4 / 2 = 2 miles / hour d = ( r ) ( t ) 32 = ( 8 ) ( t ) 32 / 8 = 4 = t it takes 4 hours for the first part d = ( r ) ( t ) 68 = ( 2 ) ( t ) 68 / 2 = 34 = t it takes 34 hours for the second part total time to drive home on monday = 4 + 34 = 38 hours we ' re asked by what percent 38 hours is greater than 25 hours . 38 / 25 = 1.52 , so it is 52 % greater . e" | a = 32 / 2
b = 100 - 32
c = b * 2
d = a + c
e = d - 100
f = e / 100
g = f * 100
|
a ) 5 : 3 , b ) 3 : 5 , c ) 2 : 3 , d ) 3 : 4 , e ) 1 : 2 | a | divide(subtract(50, 45), subtract(45, 42)) | in what ratio must tea of rs . 42 per kg be mixed with tea of rs . 50 per kg so that cost of mixture is rs . 45 per kg ? | 50 - 45 : 45 - 42 = 5 : 3 answer : a | a = 50 - 45
b = 45 - 42
c = a / b
|
a ) 69 . , b ) 71 , c ) 73 , d ) 75 , e ) 76 | c | add(subtract(add(32, 45), 13), 9) | in the biotechnology class of 2000 , there were x graduates . 32 of the graduates found a job , 45 continued on to their second degree and 13 did both . if only 9 people did n ' t do both , what is x equal to ? | graduates got job = 32 graduates continued to second degree = 45 graduates got job and continued to second degree = 13 graduates did n ' t get job and did n ' t continue to second degree = 9 job no job total second degree 13 32 45 no second degree 19 9 28 total 32 41 x therefore x = 73 answer c | a = 32 + 45
b = a - 13
c = b + 9
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | e | add(add(add(const_4, 3), add(3, const_2)), 3) | the number 61 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "i think brute force with some common sense should be used to solve this problem . write down all perfect squares less than 61 : 1 , 4 , 9 , 16 , 25 , 36 , 49 . now , 61 should be the sum of 3 of those 7 numbers . also to simplify a little bit trial and error , we can notice that as 61 is an odd numbers then either all three numbers must be odd ( odd + odd + odd = odd ) or two must be even and one odd ( even + even + odd = odd ) . we can find that 61 equals to 9 + 16 + 36 = 3 ^ 2 + 4 ^ 2 + 6 ^ 2 = 61 - - > 3 + 4 + 6 = 13 . answer : e ." | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 3
|
a ) 74 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | a | add(65, multiply(6, 1.5)) | the average weight of 6 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 6 × 1.5 = 9 kg so the weight of new person = 65 + 9 = 74 kg answer a" | a = 6 * 1
b = 65 + a
|
a ) 234 , b ) 116 , c ) 117 , d ) 118 , e ) 119 | a | add(add(multiply(divide(const_100, 45), 54), multiply(divide(50, 45), 54)), 54) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 54 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 54 39 - - - ? = > 234 answer : a" | a = 100 / 45
b = a * 54
c = 50 / 45
d = c * 54
e = b + d
f = e + 54
|
a ) 3.0 , b ) 3.36 , c ) 24.56 , d ) 25.0 , e ) 31.36 | c | divide(multiply(28, const_100), add(const_100, 14)) | from the sale of sleeping bags , a retailer made a gross profit of 14 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | "cost price * 1.14 = selling price - - > cost price * 1.14 = $ 28 - - > cost price = $ 24.56 . answer : c ." | a = 28 * 100
b = 100 + 14
c = a / b
|
a ) 20 m , b ) 25 m , c ) 22.5 m , d ) 9 m , e ) 30 m | e | multiply(divide(150, 45), subtract(45, 36)) | in 150 m race , a covers the distance in 36 seconds and b in 45 seconds . in this race a beats b by : | "distance covered by b in 9 sec . = 150 / 45 x 9 m = 30 m . a beats b by 30 metres . answer : option e" | a = 150 / 45
b = 45 - 36
c = a * b
|
a ) 2 , b ) 1.15 , c ) 2.05 , d ) 0.29 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.567), multiply(multiply(7.57, 23.25), 0.5)), const_10) | the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.5 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.567 ) / ( 0.5 * 23.25 * 7.57 ) = 25.4303 / 88.00125 = 0.29 answer : d" | a = 34 * 31
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) 78.4 , b ) 91.0 , c ) 88.0 , d ) 70.4 , e ) 71.2 | a | add(70, multiply(divide(12, const_100), 70)) | if x is 12 percent greater than 70 , then x = | "12 % of 70 = ( 70 * 0.11 ) = 8.4 12 % greater than 70 = 70 + 8.4 = 78.4 answer is clearly a ." | a = 12 / 100
b = a * 70
c = 70 + b
|
a ) 26 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) none | a | divide(subtract(28, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 28 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is | "solution let the son ' s present age be x years . then , man ' s present age = ( x + 28 ) years . then â € ¹ = â € º ( x + 28 ) + 2 = 2 ( x + 2 ) â € ¹ = â € º x + 30 = 2 x + 4 x = 26 . answer a" | a = 2 * 2
b = a - 2
c = 28 - b
d = 2 - 1
e = c / d
|
a ) 9 : 6 , b ) 9 : 8 , c ) 9 : 20 , d ) 9 : 9 , e ) 9 : 5 | c | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(10000, add(6, 6))) | a and b invests rs . 3000 and rs . 10000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 10 * 12 ) 54 : 120 = > 9 : 20 . answer : c" | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 10000 * e
g = d / f
|
a ) a ) 105 , b ) b ) 35 , c ) c ) 200 , d ) d ) 280 , e ) e ) 140 | a | add(const_3, const_4) | what is the smallest integer that is multiple of 3 , 5,7 | "it is the lcm of 3 , 5 and 7 which is 105 . the answer is a ." | a = 3 + 4
|
a ) 3 / 2 , b ) 3 / 4 , c ) 8 / 15 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 7))) | a call center has two teams . each member of team a was able to process 7 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | "let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 7 calls , so total calls for team a = 35 fraction for team b = 40 / ( 40 + 35 ) = 8 / 15 = answer = c" | a = 8 * 5
b = 8 * 5
c = 5 * 7
d = b + c
e = a / d
|
a ) 18 seconds , b ) 34 seconds , c ) 14 seconds , d ) 12 seconds , e ) 15 seconds | a | divide(60, multiply(add(10, 2), const_0_2778)) | the speed at which a man can row a boat in still water is 10 kmph . if he rows downstream , where the speed of current is 2 kmph , what time will he take to cover 60 metres ? | "speed of the boat downstream = 10 + 2 = 12 kmph = 12 * 5 / 18 = 10 / 3 m / s hence time taken to cover 60 m = 60 * 3 / 10 = 18 seconds . answer : a" | a = 10 + 2
b = a * const_0_2778
c = 60 / b
|
a ) 4 ^ 12 , b ) 5 ( 4 ^ 4 ) , c ) 17 ( 4 ^ 5 ) , d ) 8 ^ 12 , e ) 7 ( 4 ^ 5 ) | b | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of 4 ^ 5 + 4 ^ 4 ? | "4 ^ 5 + 4 ^ 4 = 4 ^ 4 ( 4 + 1 ) = 4 ^ 4 * 5 answer b" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
|
a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | b | multiply(const_100, subtract(multiply(multiply(divide(add(10, const_100), const_100), divide(add(10, const_100), const_100)), divide(subtract(const_100, 10), const_100)), const_1)) | the positive numbers w , x , y , and z are such that x is 10 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 16 - answer b" | a = 10 + 100
b = a / 100
c = 10 + 100
d = c / 100
e = b * d
f = 100 - 10
g = f / 100
h = e * g
i = h - 1
j = 100 * i
|
a ) 22 , b ) 45 , c ) 66 , d ) 88 , e ) 12 | b | divide(multiply(multiply(40, const_3), divide(30, const_60)), subtract(divide(multiply(multiply(40, const_3), divide(30, const_60)), 30), divide(40, const_60))) | a man misses a bus by 40 minutes if he travels at 30 kmph . if he travels at 40 kmph , then also he misses the bus by 10 minutes . what is the minimum speed required to catch the bus on time ? | let the distance to be travelled to catch the bus be x km x / 30 - x / 40 = 30 / 60 = > ( 4 x - 3 x ) / 120 = 1 / 2 = > x = 60 km by traavelling 30 kmph time taken = 60 / 30 = 2 hours by taking 2 hours , he is late by 40 min . so , he has to cover 60 km in at most speed = 60 / ( 4 / 3 ) = 45 kmph . answer : b | a = 40 * 3
b = 30 / const_60
c = a * b
d = 40 * 3
e = 30 / const_60
f = d * e
g = f / 30
h = 40 / const_60
i = g - h
j = c / i
|
a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 2 and 7 , e ) 2 and 9 | b | add(multiply(2, const_10), divide(add(44, 2), 7)) | 5 n + 2 > 12 and 7 n + 2 < 44 ; n must be between which numbers ? | "5 n + 2 > 12 5 n > 10 n > 2 7 n + 2 < 44 7 n < 42 n < 6 so n must be between 2 and 6 2 < n < 6 correct answer b" | a = 2 * 10
b = 44 + 2
c = b / 7
d = a + c
|
a ) 26 / 21 , b ) 28 / 21 , c ) 20 / 21 , d ) 40 / 21 , e ) 60 / 21 | c | divide(4, add(divide(20, const_100), 4)) | a committee is reviewing a total of 20 x black - and - white films and 4 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ? | "say x = y = 10 . in this case we would have : 20 x = 200 black - and - white films ; 4 y = 40 color films . y / x % = 10 / 10 % = 1 % of the black - and - white films , so 2 black - and - white films and all 40 color films , thus total of 42 films were selected . color films thus compose 40 / 42 = 20 / 21 of the selected films . answer : c" | a = 20 / 100
b = a + 4
c = 4 / b
|
a ) 3.6 sec , b ) 37 sec , c ) 36 sec , d ) 72 sec , e ) none of these | b | multiply(multiply(divide(divide(add(250, 120), const_1000), subtract(45, 9)), const_60), const_60) | a jogger running at 9 kmph along side a railway track is 250 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = ( 45 – 9 ) km / h = 36 km / h = ( 36 × 5 ⁄ 18 ) m / sec = 10 m / sec distance to be covered = ( 250 + 120 ) m = 370 m . ∴ time taken = ( 370 ⁄ 10 ) sec = 37 sec . answer b" | a = 250 + 120
b = a / 1000
c = 45 - 9
d = b / c
e = d * const_60
f = e * const_60
|
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