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 ) 80 % , b ) 100 % , c ) 116.7 % , d ) 120 % , e ) 140 % | d | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(10, subtract(10, const_1)), subtract(const_100, 10)), const_100), 10), 25)), const_100), multiply(subtract(const_1, divide(add(10, divide(multiply(add(10, const_4), subtract(const_100, 10)), const_100)), 20)), const_100)), const_100) | during a special promotion , a certain filling station is offering a 10 percent discount on gas purchased after the first 10 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella β s total per - gallon discount is what percent of kim β s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 10 + 0.9 * 10 = 19 gallons , so the overall discount she got was 1 / 20 = 5 % . isabella purchased 25 gallons of gas . she paid for 10 + 0.9 * 15 = 23.5 gallons , so the overall discount she got was 1.5 / 25 = 6 % . 6 is 6 / 5 * 100 = 120 % of 5 . answer : d ." | a = 10 - 1
b = 10 + a
c = 100 - 10
d = b * c
e = d / 100
f = e + 10
g = f / 25
h = 1 - g
i = h * 100
j = 10 + 4
k = 100 - 10
l = j * k
m = l / 100
n = 10 + m
o = n / 20
p = 1 - o
q = p * 100
r = i / q
s = r * 100
|
a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 2 | e | divide(subtract(6, 4), subtract(10, 6)) | cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 4 % sugar by weight . to make a delicious and healthy mixture that is 6 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 6 - 4 ) / ( 10 - 6 ) = 2 / 4 = 1 / 2 so they should be mixed in the ratio 1 : 2 answer - e" | a = 6 - 4
b = 10 - 6
c = a / b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | subtract(floor(divide(subtract(580, 190), 45)), const_2) | how many numbers between 190 and 580 are divisible by 45 and 6 ? | every such number must be divisible by l . c . m of 4 , 5,6 i . e , 60 . such numbers are 240,300 , 360,420 , 480,540 . clearly , there are 6 such numbers answer : a | a = 580 - 190
b = a / 45
c = math.floor(b)
d = c - 2
|
a ) 22 , b ) 65 , c ) 78.57 , d ) 33 , e ) 25 | c | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 28), 50)), divide(multiply(const_100, 28), 50))) | if the cost price of 50 articles is equal to the selling price of 28 articles , then the gain or loss percent is ? | "given that , cost price of 50 article is equal to selling price of 28 articles . let cost price of one article = rs . 1 selling price of 28 articles = rs . 50 but cost price of 28 articles = rs . 28 therefore , the trader made profit . \ percentage of profit = 22 / 28 * 100 = 78.57 % answer : c" | a = 100 * 28
b = a / 50
c = 100 - b
d = 100 * 28
e = d / 50
f = c / e
g = 100 * f
|
a ) 4.5 % , b ) 8.3 % , c ) 10.2 % , d ) 12.5 % , e ) 15.3 % | a | multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100) | a block of wood has dimensions 10 cm x 10 cm x 100 cm . the block is painted red and then cut evenly at the 50 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? | "the area of each half is 100 + 4 ( 500 ) + 100 = 2200 the area that is not painted is 100 . the fraction that is not painted is 100 / 2200 = 1 / 22 = 4.5 % the answer is a ." | a = 4 * 100
b = a * 4
c = b + 100
d = c + 100
e = 100 / d
f = e * 100
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) none of them | d | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 75 sq . cm . find the length of the rectangle . | "let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 5 ) - 2 x * x = 75 < = > 5 x - 25 = 75 < = > x = 20 . therefore , length of the rectangle = 20 cm . answer is d" | a = 100 / 3
|
a ) 120 , b ) 150 , c ) 130 , d ) 180 , e ) 220 | c | subtract(add(175, 325), subtract(420, 50)) | out of 420 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 420 - 50 = 370 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 370 = 130 . answer is c" | a = 175 + 325
b = 420 - 50
c = a - b
|
a ) 36 , b ) 37 , c ) 38 , d ) 39 , e ) 40 | c | floor(divide(multiply(24, 8), 5)) | in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 5 : 24 . in the company , what is the maximum number of non - managers in a department that has 8 managers : | "m / nm > 5 x / 24 x m = 8 8 / nm > 5 x / 24 x 8 / nm - 5 x / 24 x > 0 ( 192 x - 5 x * nm ) / ( 24 x * nm ) > 0 192 x - 5 x * nm > 0 5 x * nm < 192 x nm < 192 x / 5 x nm < 38.4 maximum number of non managers = 38 answer : c" | a = 24 * 8
b = a / 5
c = math.floor(b)
|
a ) 31.56 , b ) 25.65 , c ) 21.43 , d ) 28.36 , e ) 27.96 | c | divide(add(multiply(30, 15), multiply(15, 20)), add(15, 20)) | the average runs scored by a batsman in 15 matches is 30 . in the next 20 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | total score of the batsman in 15 matches = 450 . total score of the batsman in the next 20 matches = 300 . total score of the batsman in the 35 matches = 750 . average score of the batsman = 750 / 35 = 21.43 . answer : c | a = 30 * 15
b = 15 * 20
c = a + b
d = 15 + 20
e = c / d
|
a ) 900 km , b ) 540 km , c ) 720 km , d ) 820 km , e ) 920 km | a | add(multiply(multiply(2, 180), 2), 180) | a goes to place x and b goes to place y . after a travelled 180 km more than b they together meet at a point between xy . they together had their lunch for half an hour . after that a took 2 hrs to reach x and b took 4.5 hrs to reach y . find distance between x and y ? | let speed of a be - - - - - - - sa let speed of b be - - - - - - - - sb let the distance covered by b while going to position xy be - - - - - - - - - - - x so distance covered by a in same time while going to xy is - - - - - - - - - - 180 + x ( it ' s given in the question ) so total distance b / w x and y is = x + ( x + 180 ) while going to position xy : - time taken by a = time taken by b ( x + 180 ) / sa = x / sb sa / sb = ( x + 180 ) / x - - - - - - - - - - - - - - - - - - - - - 1 after lunch : - a . c . t ( according to question ) sa = x / 2 sb = ( x + 180 ) / 4.5 dividing sa / sb = ( x * 4.5 ) / ( ( x + 180 ) * 2 ) - - - - - - - - - - - - - - - - - - - - 2 equating equations 1 and 2 9 / 4 = ( x + 180 ) ^ 2 / x ^ 2 now taking square root on both sides we get 3 / 2 = ( x + 180 ) / x x = 360 therefore distance between x and y = x + ( x + 180 ) = 900 km answer : a | a = 2 * 180
b = a * 2
c = b + 180
|
a ) 15 seconds , b ) 24 seconds , c ) 28 seconds , d ) 30 seconds , e ) 35 seconds | b | divide(add(140, 260), multiply(60, const_0_2778)) | a train 140 m long is running at 60 kmph . in how much time will it pass a platform 260 m long ? | "distance travelled = 140 + 260 m = 400 m speed = 60 * 5 / 8 = 50 / 3 m time = 400 * 3 / 50 = 24 second answer : b" | a = 140 + 260
b = 60 * const_0_2778
c = a / b
|
a ) 9000 cubic meters , b ) 10500 cubic meters , c ) 11750 cubic meters , d ) 12000 cubic meters , e ) 12500 cubic meters | d | multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1)))) | with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity r of the pool ? | "d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) / ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or , 120 x = 80 ( x + 50 ) . or x = 100 . hence , the capacity of pool = 120 x = 12000 cubic meters ." | a = 48 / const_60
b = 4 + 1
c = a * b
d = const_60 * c
e = 50 * d
|
a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 15 , d ) 4 / 5 , e ) 7 / 6 | c | divide(multiply(2, 1), multiply(3, 5)) | if the ratio of a to b is 2 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 2 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 2 : 15 answer c" | a = 2 * 1
b = 3 * 5
c = a / b
|
a ) $ 180,000 , b ) $ 185,000 , c ) $ 180,000 , d ) $ 200,000 , e ) $ 192,500 | e | multiply(add(divide(1, 25), 1), divide(multiply(1000, divide(add(add(multiply(add(multiply(divide(const_3, const_2), multiply(5, 25)), multiply(40, 5)), subtract(25, add(const_3, 5))), multiply(multiply(40, 5), 5)), multiply(multiply(5, 25), const_3)), 25)), 1000)) | a factory that employs 1000 assembly line workers pays each of these workers $ 5 per hour for the first 40 hours worked during a week and 1 Β½ times that rate for hours worked in excess of 40 . what was the total payroll for the assembly - line workers for a week in which 30 percent of them worked 25 hours , 50 percent worked 40 hours , and the rest worked 50 hours ? | "30 % of 1000 = 300 worked for 20 hours payment @ 5 / hr total payment = 300 * 25 * 5 = 37500 50 % of 1000 = 500 worked for 40 hours payment @ 5 / hr total payment = 500 * 40 * 5 = 100000 remaining 200 worked for 50 hours payment for first 40 hours @ 5 / hr payment = 200 * 40 * 5 = 40000 payment for next 10 hr @ 7.5 / hr payment = 200 * 10 * 7.5 = 15000 total payment = 37500 + 100000 + 40000 + 15000 = 192500 hence , answer is e" | a = 1 / 25
b = a + 1
c = 3 / 2
d = 5 * 25
e = c * d
f = 40 * 5
g = e + f
h = 3 + 5
i = 25 - h
j = g * i
k = 40 * 5
l = k * 5
m = j + l
n = 5 * 25
o = n * 3
p = m + o
q = p / 25
r = 1000 * q
s = r / 1000
t = b * s
|
a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 250 m | a | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 290) | a 290 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | "relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 290 / 9 = 500 / 9 x = 210 ans is 210 m answer : a" | a = 120 + 80
b = a * const_0_2778
c = 9 * b
d = c - 290
|
a ) 20.23 , b ) 20.13 , c ) 30.93 , d ) 25 , e ) 10.93 | d | multiply(divide(100, add(add(divide(2, 4), divide(5, 2)), const_1)), 5) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 4 and b : c = 2 : 5 . if the total runs scored by all of them are 100 , the runs scored by b are ? | "a : b = 2 : 4 b : c = 2 : 5 a : b : c = 4 : 8 : 20 8 / 32 * 100 = 25 answer : d" | a = 2 / 4
b = 5 / 2
c = a + b
d = c + 1
e = 100 / d
f = e * 5
|
a ) 510 , b ) 540 , c ) 480 , d ) 520 , e ) 589 | c | divide(820, multiply(subtract(78, 1), const_0_2778)) | a train 820 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 820 + x ) / 60 = 65 / 3 x = 480 m . answer : option c" | a = 78 - 1
b = a * const_0_2778
c = 820 / b
|
a ) $ 1.00 , b ) $ 1.25 , c ) $ 5.00 , d ) $ 8.00 , e ) $ 20.00 | d | subtract(20.00, multiply(subtract(5, 3.00), 3.00)) | if 5 people contributed a total of $ 20.00 toward a gift and each of them contributed at least $ 3.00 , then the maximum possible amount any one person could have contributed is | "d for me 4 people with 3 $ each - > maximum = 8" | a = 5 - 3
b = a * 3
c = 20 - 0
|
a ) 1600 , b ) 1500 , c ) 1200 , d ) 1388 , e ) 1211 | a | multiply(multiply(multiply(5, const_4.0), 10), 8) | a man bought an article and sold it at a gain of 5 % . if he had bought it at 5 % less and sold it for re 8 less , he would have made a profit of 10 % . the c . p . of the article was | "explanation : let original cost price is x its selling price = ( 105 / 100 ) * x = 21 x / 20 new cost price = ( 95 / 100 ) * x = 19 x / 20 new selling price = ( 110 / 100 ) * ( 19 x / 20 ) = 209 x / 200 [ ( 21 x / 20 ) - ( 209 x / 200 ) ] = 8 = > x = 1600 answer : a ) rs 1600" | a = 5 * 4
b = a * 10
c = b * 8
|
a ) 3 , b ) 7 , c ) 15 , d ) 5 , e ) 25 | d | add(4, subtract(add(12, 11), multiply(const_2, 11))) | jacob is now 12 years younger than michael . if 11 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | jacob = x years , mike = x + 12 years 11 years from now , 2 ( x + 11 ) = x + 23 2 x + 22 = x + 23 x = 1 x + 4 = 5 years answer d | a = 12 + 11
b = 2 * 11
c = a - b
d = 4 + c
|
a ) 6 : 8 , b ) 3 : 1 , c ) 6 : 5 , d ) 3 : 2 , e ) 3 : 5 | e | divide(multiply(2, 3), multiply(5, 2)) | the marks obtained by vijay and amith are in the ratio 2 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ? | "2 : 5 3 : 2 - - - - - - - 6 : 15 : 10 6 : 10 3 : 5 answer : e" | a = 2 * 3
b = 5 * 2
c = a / b
|
a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | d | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent e of the list price is the lowest possible sale price ? | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which e is 40 % of 80 hence , d is the answer" | a = 80 / 2
|
['a ) 22 metres', 'b ) 18 metres', 'c ) 16 metres', 'd ) 24 metres', 'e ) none of these'] | c | divide(multiply(6, 144), subtract(add(144, 54), 144)) | the area of a rectangular field is 144 m 2 . if the length had been 6 metres more , the area would have been 54 m 2 more . the original length of the field is | let the length and breadth of the original rectangular field be x m and y m respectively . area of the original field = x Γ y = 144 m 2 β΄ x = 144 β y . . . . . ( i ) if the length had been 6 m more , then area will be ( x + 6 ) y = 144 + 54 β ( x + 6 ) y = 198 β¦ ( ii ) putting the value of x from eq ( i ) in eq ( ii ) , we get ( 144 β y + 6 ) y = 198 β 144 + 6 y = 198 β 6 y = 54 β y = 9 m putting the value of y in eq ( i ) we get x = 16 m answer c | a = 6 * 144
b = 144 + 54
c = b - 144
d = a / c
|
a ) 41 , b ) 28 , c ) 30 , d ) 34 , e ) 36 | a | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(6, const_2)), add(add(6, const_2), const_2)), add(add(add(6, const_2), const_2), const_4)), add(add(add(add(6, const_2), const_2), const_4), const_2)), add(add(add(add(add(6, const_2), const_2), const_4), const_2), const_4)) | find a sum for 1 st 6 prime number ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 + 13 ) = 41 note : 1 is not a prime number option a" | a = 2 + 3
b = 2 + 3
c = a + b
d = 2 + 3
e = d + 2
f = c + e
g = 6 + 2
h = f + g
i = 6 + 2
j = i + 2
k = h + j
l = 6 + 2
m = l + 2
n = m + 4
o = k + n
p = 6 + 2
q = p + 2
r = q + 4
s = r + 2
t = o + s
u = 6 + 2
v = u + 2
w = v + 4
x = w + 2
y = x + 4
z = t + y
|
a ) 1 / 7 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 19 | e | divide(3, add(multiply(4, 4), 3)) | chris mixed 3 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 4 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ? | "1 lbs of raisin = $ 1 3 lbs of raisin = $ 3 1 lbs of nuts = $ 4 4 lbs of nuts = $ 16 total value of mixture = 16 + 3 = 19 fraction of the value of raisin = 3 / 19 ans : e" | a = 4 * 4
b = a + 3
c = 3 / b
|
a ) 16.2 , b ) 4 , c ) 5 , d ) 9 , e ) 11 | d | divide(multiply(divide(45, const_100), 900), 45) | 45 x ? = 45 % of 900 | "answer let 45 x a = ( 25 x 900 ) / 100 β΄ a = ( 45 x 9 ) / 45 = 9 correct option : d" | a = 45 / 100
b = a * 900
c = b / 45
|
a ) 2 , b ) 9 , c ) 8 , d ) 6 , e ) 3 | b | add(3, const_1) | the average of first five multiples of 3 is ? | "average = 3 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 45 / 5 = 9 . answer : b" | a = 3 + 1
|
a ) s . 30 , b ) s . 70 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | subtract(add(divide(600, 5), divide(910, 7)), divide(1980, 9)) | p , q and r together earn rs . 1980 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ? | "explanation : amount earned by p , q and r in 1 day = 1980 / 9 = 220 - - - ( 1 ) amount earned by p and r in 1 day = 600 / 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 / 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 130 - 220 = 30 = > amount earned by r in 1 day = 30 answer : option a" | a = 600 / 5
b = 910 / 7
c = a + b
d = 1980 / 9
e = c - d
|
a ) 18 % , b ) 12 % , c ) 22 % , d ) 35 % , e ) 21 % | d | multiply(divide(subtract(1390, 890), 1390), const_100) | the cost price of a radio is rs . 1390 and it was sold for rs . 890 , find the loss % ? | "1390 - - - - 500 100 - - - - ? = > 35 % answer : d" | a = 1390 - 890
b = a / 1390
c = b * 100
|
a ) 48 , b ) 93 , c ) 24 , d ) 23 , e ) 12 | c | add(60, 90) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 60 km / hr and 90 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 60 + 90 = 150 km / hr . 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 125 = 24 sec . answer : c" | a = 60 + 90
|
['a ) 30 cm', 'b ) 60 cm', 'c ) 72 cm', 'd ) 132 cm .', 'e ) 142 cm .'] | b | multiply(5, divide(circumface(42), multiply(const_2, add(6, 5)))) | a circular wire of radius 42 cm is cut and bent into the form of a rectangle whose sides are in the ratio of 6 : 5 . the smaller side of the rectangle is : | length of wire = circumference of circle of radius 42 cm = ( 2 * 22 / 7 * 42 ) = 264 cm . therefore , perimeter of rectangle = 264 cm . let , length = 6 x cm & breadth = 5 x cm . therefore , 2 ( 6 x + 5 x ) = 264 or x = 12 . therefore , smaller side = 60 cm answer : b | a = circumface / (
b = 6 + 5
c = 2 * b
d = 5 * a
|
a ) 48 , b ) 60 , c ) 84 , d ) 96 , e ) 69 | b | add(multiply(subtract(5, 2), 18), 6) | if 6 years are subtracted from the present age of arun and the remainder is divided by 18 , then the present age of his grandson gokul is obtained . if gokul is 2 years younger to madan whose age is 5 years , then what is the age of arun ? | arun age x . ( x - 6 ) / 18 = y y is gokul age which is 3 x = 60 answer : b | a = 5 - 2
b = a * 18
c = b + 6
|
a ) 68225 , b ) 75625 , c ) 76569 , d ) 45854 , e ) 45858 | b | multiply(121, power(add(const_4, const_1), const_4)) | solve for the given equation 121 x 54 = ? | explanation : this question can be solved by 2 approaches : = > direct multiplication = > numerical operations direct multiplication is quite time consuming and involves a lot of calculations . an easier approach is based on numerical operations . = > 121 x ( 10 / 2 ) 4 = 121 x 10000 / 16 = 75625 answer : b | a = 4 + 1
b = a ** 4
c = 121 * b
|
a ) 72 , b ) 54 , c ) 64 , d ) 51 , e ) 45 | b | add(42, divide(42, const_2)) | in a ratio which is equal to 7 : 9 , if the antecedent is 42 , then the consequent is ? | "we have 7 / 9 = 42 / x 7 x = 42 * 9 x = 54 consequent = 54 answer is b" | a = 42 / 2
b = 42 + a
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | d | subtract(10, const_3) | john has 10 pairs of matched socks . if he loses 11 individual socks , what is the greatest number of pairs of matched socks he can have left ? | "because we have to maximize the pair of matched socks , we will remove 5 pairs ( 10 socks ) out of 10 pairs 1 sock from the 6 th pair . thus the no of matching socks pair remaining = 10 - 6 = 4 answer d" | a = 10 - 3
|
a ) 18 , b ) 21 , c ) 25 , d ) 30 , e ) 35 | c | subtract(choose(7, 4), choose(subtract(7, 2), 2)) | a meeting has to be conducted with 4 managers . find the number of ways in which the managers may be selected from among 7 managers , if there are 2 managers who refuse to attend the meeting together . | the total number of ways to choose 4 managers is 7 c 4 = 35 we need to subtract the number of groups which include the two managers , which is 5 c 2 = 10 . 35 - 10 = 25 the answer is c . | a = math.comb(7, 4)
b = 7 - 2
c = math.comb(b, 2)
d = a - c
|
a ) 112 cm , b ) 120 cm , c ) 150 cm , d ) 210 cm , e ) 254 cm | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 100), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 100 inches into centimeter ? | "1 inch = 2.54 cm 100 inches = 100 * 2.54 = 254 cm answer is e" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 100
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 33 , b ) 38 , c ) 32 , d ) 28 , e ) 19 | b | subtract(60, 23) | nitin ranks 23 th in a class of 60 students . what is rank from the last ? | explanation : number students behind the nitin in rank = ( 60 - 23 ) = 37 nitin is 38 nd from the last answer : b ) 38 | a = 60 - 23
|
a ) 144 , b ) 288 , c ) 24 , d ) 256 , e ) none | c | multiply(factorial(4), factorial(1)) | in how many ways 4 boys and 1 girls can be seated in a row so that they are alternate . | solution : let the arrangement be , b b b b g 4 boys can be seated in 4 ! ways . girl can be seated in 1 ! ways . required number of ways , = 4 ! * 1 ! = 24 . answer : option c | a = math.factorial(4)
b = math.factorial(1)
c = a * b
|
a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | a | divide(1, add(10, 2)) | { - 10 , - 6 , - 5 , - 4 , - 2.5 , - 1 , 0 , 2.5 , 4 , 6 , 7 , 10 } a number is to be selected at random from the set above . what is the probability that the number will be a solution to the equation ( x 3 ) ( x + 14 ) ( 2 x + 5 ) = 0 ? | x = - 4 prob = 1 / 12 answer - a | a = 10 + 2
b = 1 / a
|
a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) 40.1 | e | divide(subtract(add(multiply(40.2, 10), add(13, 19)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 19 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 19 = 401 hence , new average = 401 β 10 = 40.1 answer e" | a = 40 * 2
b = 13 + 19
c = a + b
d = c - 31
e = d / 10
|
a ) 327.9 , b ) 325.9 , c ) 347.7 , d ) 357.9 , e ) 327.9 | a | divide(multiply(16, sqrt(subtract(power(multiply(const_2, 22), const_2), power(16, const_2)))), const_2) | the side of a rhombus is 22 m and length of one of its diagonals is 16 m . the area of the rhombus is ? | "area of the rhombus = 1 / 2 * p * β 4 ( a ) 2 - ( p ) 2 a = 22 ; p = 16 a = 1 / 2 * 16 * β 4 ( 22 ) 2 - ( 16 ) 2 = 1 / 2 * 16 * β 1936 - 256 = 1 / 2 * 16 * β 1680 a = 327.9 answer : a" | a = 2 * 22
b = a ** 2
c = 16 ** 2
d = b - c
e = math.sqrt(d)
f = 16 * e
g = f / 2
|
a ) 20,088 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | sonika bought a v . c . r . at the list price of 18,600 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,600 rate of sales tax = 8 % β΄ sales tax = 8 % of 18,600 = 8 β 100 Γ 18600 = 1488 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,600 + 1488 = 20,088 . answer a" | a = 1 + 4
b = a * 2
c = b + 8
d = c * 100
e = 1 + 4
f = e * 2
g = d * f
h = 1 + 4
i = h * 100
j = g + i
k = 100 + 8
l = j * k
m = l / 100
n = 1 + 4
o = n * 100
p = o * 2
q = m / p
r = math.floor(q)
|
a ) 2.5 , b ) 6.5 , c ) 12.5 , d ) 14 , e ) 16 | c | divide(triangle_area_three_edges(7, 24, 25), divide(triangle_perimeter(7, 24, 25), const_2)) | what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 7 , 24 and 25 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 25 / 2 = 12.5 ans . c" | a = triangle_area_three_edges / (
|
a ) a . 4 , b ) b . 8 , c ) c . 10 , d ) d . 12 , e ) e . 16 | c | subtract(divide(multiply(20, 36), multiply(divide(const_3, const_4), 36)), 20) | if 20 machine can finish a job in 36 days , then how many more machines would be needed to finish the job in one - third less time ? | "you might think of this in a management context - we can use the principle of ' person - hours ' to solve any problem where we have identical workers . so , using simpler numbers , suppose you know that 6 identical employees , working simultaneously , would finish a job in 5 hours . then that job requires 6 * 5 = 30 total hours of person - work . if instead you wanted the job done in 3 hours , you ' d assign 30 / 3 = 10 employees to do the job , because you want to get a total of 30 hours of work from the employees . we can solve this problem identically . if 20 machines ( identical ones , i assume ) work simultaneously for 36 days , they will do a total of 20 * 36 machine - days of work . so the job requires 20 * 36 days of machine work in total . we instead want the job done in 1 / 3 less time , so in 24 days . so we ' ll need 20 * 36 / 24 = 30 machines , or 10 additional machines . c" | a = 20 * 36
b = 3 / 4
c = b * 36
d = a / c
e = d - 20
|
a ) rs . 1435 , b ) rs . 12,628 , c ) rs . 1685 , d ) rs . 18,942 , e ) none | a | multiply(multiply(const_0_25, const_100), 10) | suganya and suriya are partners in a business . suganya invests rs . 38,000 for 8 months and suriya invests rs . 40,000 for 10 months . out of a profit of rs . 31,570 . suganya ' s share is | "solution ratio of their shares = ( 35000 Γ£ β 8 ) : ( 42000 Γ£ β 10 ) = 2 : 44 . suganya ' s share = rs . ( 31570 Γ£ β 2 / 44 ) = rs . 1435 . answer a" | a = const_0_25 * 100
b = a * 10
|
a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | a | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(3795, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3795 . what is the difference between rohan β s and sudhir β s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 Γ 12 : 12000 Γ 9 : 9000 Γ 8 = 6 : 3 : 2 the difference between rohan β s and sudhir β s share = 1 share : . i . e . = rs . 3795 Γ 1 / 11 = rs . 345 . a" | a = 3 * 4
b = a - 3
c = 3 * 4
d = b / c
e = 12000 * d
f = 3 * 4
g = f - 4
h = 3 * 4
i = g / h
j = 9000 * i
k = e - j
l = 3 * 4
m = l - 3
n = 3 * 4
o = m / n
p = 12000 * o
q = 18000 + p
r = 3 * 4
s = r - 4
t = 3 * 4
u = s / t
v = 9000 * u
w = q + v
x = 3795 / w
y = k * x
|
a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 3 , d ) 4 / 9 , e ) 2 / 3 | d | multiply(subtract(1, divide(1, 3)), subtract(1, divide(subtract(1, divide(const_2.0, 3)), 1))) | for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 1 / 3 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ? | "probability of not burning out during the first 6 months 1 - 1 / 3 = 2 / 3 probability of not burning out during the next 6 months 2 / 3 / 2 = 1 / 3 , hence probability of burning out 1 - 1 / 3 = 2 / 3 . probability of burning out during the period from 6 months to 1 year = probability of not burning out in first 6 months * probability of burning out in next 6 months = 2 / 3 * 2 / 3 = 4 / 9 answer : d ." | a = 1 / 3
b = 1 - a
c = 2 / 0
d = 1 - c
e = d / 1
f = 1 - e
g = b * f
|
a ) 4902 , b ) 4922 , c ) 5922 , d ) 5924 , e ) 5928 | c | subtract(multiply(add(5, const_1), 5900), add(add(add(add(5921, 5468), 5568), 6088), 6433)) | a grocer has a sale of rs . 5921 , rs . 5468 , rs . 5568 , rs . 6088 and rs . 6433 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5900 ? | "total sale for 5 months = rs . ( 5921 + 5468 + 5568 + 6088 + 6433 ) = rs . 29478 . required sale = rs . [ ( 5900 x 6 ) - 29478 ] = rs . ( 35400 - 29478 ) = rs . 5922 . answer : c" | a = 5 + 1
b = a * 5900
c = 5921 + 5468
d = c + 5568
e = d + 6088
f = e + 6433
g = b - f
|
a ) 2.145 , b ) 6.125 , c ) 8.725 , d ) 11.885 , e ) 13.485 | b | subtract(add(multiply(3, divide(multiply(2, 2), 3)), 5), multiply(2, power(divide(multiply(2, 2), 3), 2))) | if x is real , find the maximum value of the expression - 2 x ^ 2 + 3 x + 5 . | "this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 3 x + 5 = ( - 2 x + 5 ) ( x + 1 ) the roots are 5 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 0.75 the maximum value is - 2 ( 0.75 ) ^ 2 + 3 ( 0.75 ) + 5 = 6.125 the answer is b ." | a = 2 * 2
b = a / 3
c = 3 * b
d = c + 5
e = 2 * 2
f = e / 3
g = f ** 2
h = 2 * g
i = d - h
|
a ) $ 2225 , b ) $ 2782 , c ) $ 1625 , d ) $ 2629 , e ) $ 2625 | e | subtract(divide(875, divide(25, const_100)), 875) | nelly made a part payment of $ 875 toward buying a brand new refrigerator , which represents 25 % of the total cost of the refrigerator , how much remains to be paid ? | explanation : let ' s start with what the total price of the refrigerator would be . if 25 % is equal to $ 875 then 100 % equals $ x . we just have to multiply $ 875 by 4 to get total amount = $ 3500 . out of this amount we then need to deduct the amount already paid which was $ 875 so we have $ 3500 - $ 875 = $ 2625 answer : option e | a = 25 / 100
b = 875 / a
c = b - 875
|
a ) 16 , b ) 17.78 , c ) 17 , d ) 18.5 , e ) 18.23 | b | multiply(divide(16, 90), const_100) | 16 is what % of 90 ? | we assume that 90 is 100 % assume ' x ' is value we looking for here , 90 = 100 % and x % = 16 therefore , 100 / x = 90 / 16 100 / x = 5.625 x = 17.78 b | a = 16 / 90
b = a * 100
|
a ) 700 m , b ) 500 m , c ) 373 m , d ) 356 m , e ) 640 m | c | divide(multiply(12, multiply(1.4, const_1000)), 45) | amar takes as much time in running 12 meters as a car takes in covering 45 meters . what will be the distance covered by amar during the time the car covers 1.4 km ? | "distance covered by amar = 12 / 45 ( 1.4 km ) = 4 / 15 ( 1400 ) = 373 m answer : c" | a = 1 * 4
b = 12 * a
c = b / 45
|
a ) 18 , b ) 28 , c ) 38 , d ) 68 , e ) 58 | a | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(2, 2)), multiply(const_4, const_2))), const_10) | how many 2 digit number contain number 3 ? | "total 2 digit no . = 9 * 10 * = 90 not containing 3 = 8 * 9 = 72 total 2 digit number contain 3 = 90 - 72 = 18 answer : a" | a = 1000 - 10
b = 2 * 2
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 5 , b ) 8 , c ) 12 , d ) 16 , e ) 20 | d | divide(4, subtract(divide(28, multiply(add(13, const_1), 28)), divide(13, multiply(add(13, const_1), 28)))) | marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 28 laps around the track per hour and nick completes 13 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the track than nick ? | "maria ' s rate - 28 laps per hour - - > 28 / 60 laps / min nick ' s rate - 13 laps per hour - - > 13 / 60 laps / min lets set equations : 28 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 13 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 28 / 60 - 13 / 60 ) * t = 4 - 0 ( since nick was chasing maria ) t = 16 min needed maria to run 4 laps answer : d" | a = 13 + 1
b = a * 28
c = 28 / b
d = 13 + 1
e = d * 28
f = 13 / e
g = c - f
h = 4 / g
|
a ) 3630 , b ) 9232 , c ) 4260 , d ) 2387 , e ) 2813 | c | multiply(divide(6300, add(add(6300, 4200), 10500)), 14200) | a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 14200 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 14200 = 4260 answer : c" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 14200
|
a ) 1 / 2 , b ) 1 / 4 , c ) 2 / 3 , d ) 7 / 8 , e ) 3 / 8 | d | divide(2, power(const_2, 2)) | three coins are tossed . find the probability of at most 2 tails ? | "n ( s ) = 2 ^ 3 = 8 let e is the event of getting at most 2 tails n ( e ) = hhh , tth , htt , tht , hht , hth , thh = 7 p ( e ) = n ( e ) / n ( s ) = 7 / 8 ans - d" | a = 2 ** 2
b = 2 / a
|
a ) 52 , b ) 68 , c ) 78 , d ) 92 , e ) 65 | d | subtract(subtract(multiply(25, 8), multiply(12, 17)), multiply(12, 14)) | the average of 25 results is 8 . the average of first 12 of them is 14 and that of last 12 is 17 . find the 13 result ? | "13 th result = sum of 25 results - sum of 24 results 18 * 25 - 14 * 12 + 17 * 12 = 78 answer is d" | a = 25 * 8
b = 12 * 17
c = a - b
d = 12 * 14
e = c - d
|
a ) 5.02 % , b ) 50.2 % , c ) 209 % , d ) 502 % , e ) none of these | d | multiply(5.02, const_100) | 5.02 can be expressed in terms of percentage as | "explanation : while calculation in terms of percentage we need to multiply by 100 , so 5.02 * 100 = 502 answer : option d" | a = 5 * 2
|
a ) rs . 45,000 , b ) rs . 40,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | b | divide(multiply(multiply(add(const_1, const_4), const_1000), 1), 2) | x and y invested in a business . they earned some profit which they divided in the ratio of 1 : 2 . if x invested rs . 20,000 . the amount invested by y is | "solution suppose y invested rs . y then , 20000 / y = 1 / 2 Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = ( 20000 Γ£ β 2 / 1 ) . Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = 40000 . answer b" | a = 1 + 4
b = a * 1000
c = b * 1
d = c / 2
|
a ) 13 : 00 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 12 : 00 | e | add(divide(add(700, 50), subtract(500, divide(700, const_2))), 07) | sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 05 : 00 and starts climbing at a constant pace . ken starts climbing only at 07 : 00 , when sari is already 700 meters ahead of him . nevertheless , ken climbs at a constant pace of 500 meters per hour , and reaches the summit before sari . if sari is 50 meters behind ken when he reaches the summit , at what time did ken reach the summit ? | "both sari and ken climb in the same direction . speed of sari = 700 / 2 = 350 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 700 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / relative speed = ( 700 + 50 ) / ( 500 - 350 ) = 5 hrs starting from 7 : 00 , in 5 hrs , the time will be 12 : 00 answer ( e )" | a = 700 + 50
b = 700 / 2
c = 500 - b
d = a / c
e = d + 7
|
a ) 36.6 , b ) 36.1 , c ) 36.5 , d ) 36.2 , e ) 36.9 | c | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer : c" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 5 : 30 , b ) 5 : 45 , c ) 6 : 00 , d ) 6 : 15 , e ) 6 : 30 | d | subtract(divide(multiply(3, 12), 12), const_0_33) | it takes 3 workers a total of 12 hours to build a giant machine with each worker working at the same rate . if 4 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 3 : 00 pm , at what time will the machine be complete ? | "3 workers build 1 / 12 of a machine in one hour . 1 worker builds 1 / 36 of a machine in one hour . in the first 4 hours , 4 workers build 4 * ( 1 / 36 ) * 4 = 16 / 36 of a machine from 3 : 00 to 4 : 00 , 5 workers build another 5 / 36 . the total is 21 / 36 . from 4 : 00 to 5 : 00 , 6 workers build another 6 / 36 . the total is 27 / 36 . from 5 : 00 to 6 : 00 , 7 workers build another 7 / 36 . the total is 34 / 36 . to build another 2 / 36 , 8 workers need ( 2 / 8 ) * 60 minutes which is 15 minutes . the machine is complete at 6 : 15 . the answer is d ." | a = 3 * 12
b = a / 12
c = b - const_0_33
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | subtract(4, 2) | buffalo gives 4 liter milk , cow gives ( 1 / 2 ) liter milk and goat gives 1 / 4 liter milk . you have to get 20 liter milk by 20 animals . what is the number of cows ? | "assume number of respective animals are x , y , z . x + y + z = 20 - - - ( 1 ) as the total number of animal has to be 20 amt of milk will be 4 x + ( 1 / 2 ) y + ( 1 / 4 ) z = 20 - - - ( 2 ) solving equation 1 and 2 we get 15 x + y = 60 - - - - ( 3 ) since buffalo gives 4 litre and total milk is 20 , x < 5 but from eq 3 , x can not be more than 4 ; further if x = 1 or 2 ; y > 20 . . . not possible , since total animal is 20 thus , x = 3 , y = 15 , z = 2 15 cows answer : d" | a = 4 - 2
|
a ) 39 , b ) 50 , c ) 54 , d ) 57 , e ) 36 | e | add(subtract(80, 25), const_1) | claire has a total of 80 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male , and one - third of the hamsters are male . if there are 25 males altogether , how many gerbils does claire have ? | "g + h = 80 . . . 1 ; g / 4 + h / 3 = 25 . . . . 2 or 3 g + 4 h = 25 * 12 = 300 g = 80 - h or 3 ( 80 - h ) + 4 h = 300 h = 300 - 240 = 60 then g = 96 - 60 = 36 e" | a = 80 - 25
b = a + 1
|
a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | c | divide(55, divide(add(add(divide(55, 11), divide(55, 45)), add(divide(55, 11), divide(55, 55))), const_2)) | a , band c can do a piece of work in 11 days , 45 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | "( a + b ) ' s 1 day ' s work = 1 / 11 + 1 / 45 = 56 / 495 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 day ' s = 56 / 495 + 6 / 55 = 2 / 9 2 / 9 th work done in 2 days work done = 9 / 2 * 2 = 9 days answer : c" | a = 55 / 11
b = 55 / 45
c = a + b
d = 55 / 11
e = 55 / 55
f = d + e
g = c + f
h = g / 2
i = 55 / h
|
a ) 240 / 9 , b ) 309 / 28 , c ) 100 / 9 , d ) 200 / 9 , e ) 339 / 26 | c | divide(multiply(90, 20), divide(72, const_2)) | 15 people can write 90 book in 20 days working 7 hour a day . then in how many day 240 can be written by 72 people ? | "work per day epr hour per person = 90 / ( 20 * 7 * 15 ) / / eq - 1 people = 72 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 240 / ( p * 7 * 72 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 100 / 9 answer : c" | a = 90 * 20
b = 72 / 2
c = a / b
|
a ) 1 / 5 , b ) 7 / 9 , c ) 1 / 3 , d ) 2 / 3 , e ) 2 / 9 | e | divide(subtract(10, 7), 7) | a number , x is chosen at random from the set of positive integers less than 10 . what is the probability that ( 7 / x ) > x ? | "number x has to be chosen from numbers 1 - 9 ( 7 / x ) > x = > 7 > x ^ 2 = > x ^ 2 - 7 < 0 x can have 2 values only 1 , 2 therefore , probability = 2 / 9 answer e" | a = 10 - 7
b = a / 7
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(25, 17) | a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , with 17 balls per box . if 94 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | "there is no way to store 94 balls without leftovers : 94 β 0 β 25 = 94 - 94 β 2 β 25 = 44 , 94 β 3 β 25 = 19 are not divisible by 17 . 93 balls can be stored successfully : 93 β 1 β 25 = 68 is divisible by 17 . thus , 93 = 1 β 25 + 4 β 17 and we need 1 big box and 4 small boxes . answer : b" | a = 25 - 17
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(33, multiply(3, 10)) | what is x if x + 3 y = 33 and y = 10 ? | "substitute y by 10 in x + 3 y = 33 x + 3 ( 10 ) = 33 x + 30 = 33 if we substitute x by 3 in x + 30 = 33 , we have 3 + 30 = 33 . hence x = 3 correct answer c" | a = 3 * 10
b = 33 - a
|
a ) 20 % , b ) 22 % , c ) 25 % , d ) 30 % , e ) 35 % | a | multiply(divide(subtract(60, 50), 50), const_100) | a shopkeeper sold 50 articles at the cost price of 60 articles . then find the profit % or lost % | "here 50 articles selling price = 60 articles cost price so the difference = 60 - 50 = 10 % of profit = 10 * 100 / 50 = 20 % correct option is a" | a = 60 - 50
b = a / 50
c = b * 100
|
a ) 2500 , b ) 2772 , c ) 1991 , d ) 6725 , e ) 2099 | a | divide(3000, add(const_1, divide(20, const_100))) | the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 3000 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 3000 ( 100 / 120 ) = rs . 2500 . answer : a" | a = 20 / 100
b = 1 + a
c = 3000 / b
|
a ) 4 , b ) 5 , c ) 8 , d ) 12 , e ) none of these | c | multiply(subtract(const_1, divide(9, 15)), 20) | suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 20 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | the part of job that suresh completes in 9 hours = 9 Γ’ Β β 15 = 3 Γ’ Β β 5 remaining job = 1 - 3 Γ’ Β β 5 = 2 Γ’ Β β 5 remaining job can be done by ashutosh in 2 Γ’ Β β 5 Γ£ β 20 = 8 hours answer c | a = 9 / 15
b = 1 - a
c = b * 20
|
a ) 9680 , b ) 2277 , c ) 2999 , d ) 10700 , e ) 10740 | e | add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(10200, 9000), 3), 9000)), 9000), 3), 9000) | sonika deposited rs . 9000 which amounted to rs . 10200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ? | "( 9000 * 3 * 2 ) / 100 = 540 10200 - - - - - - - - 10740 answer : e" | a = 2 / 100
b = 10200 - 9000
c = b / 3
d = c / 9000
e = a + d
f = e * 9000
g = f * 3
h = g + 9000
|
a ) 97 , b ) 89 , c ) 90 , d ) 93 , e ) 95 | a | subtract(multiply(4, add(85, 3)), multiply(85, 3)) | jerry β s average ( arithmetic mean ) score on the first 3 of 4 tests is 85 . if jerry wants to raise his average by 3 points , what score must he earn on the fourth test ? | "total score on 3 tests = 85 * 3 = 255 jerry wants the average to be = 88 hence total score on 4 tests should be = 88 * 4 = 352 score required on the fourth test = 352 - 255 = 97 option a" | a = 85 + 3
b = 4 * a
c = 85 * 3
d = b - c
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(multiply(multiply(multiply(const_4, 2), const_2.0), 1), multiply(multiply(2, 2), 1)) | a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β high β and β low β cards to earn 4 points if you must draw exactly 2 β low β cards ? | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 3 times . as a result , the formula is : 3 ! / 2 ! which represents ( 3 * 2 * 1 ) / ( 2 * 1 ) which simplifies to just 4 , giving you answer c ." | a = 4 * 2
b = a * 2
c = b * 1
d = 2 * 2
e = d * 1
f = c / e
|
a ) 1.22 hrs , b ) 2.22 hrs , c ) 5.22 hrs , d ) 6.22 hrs , e ) 7.22 hrs | e | add(divide(reminder(subtract(multiply(13, const_60), multiply(multiply(13, divide(4, 9)), const_60)), const_60), const_100), floor(divide(subtract(multiply(13, const_60), multiply(multiply(13, divide(4, 9)), const_60)), const_60))) | calculate how much time could be saved if the train would run at its usual speed , given that it ran at 4 / 9 of its own speed and got to its destination in 13 hours ? | "new speed = 4 / 9 of usual speed new time = 4 / 9 of usual time 4 / 9 of usual time = 13 hrs usual time = 13 * 4 / 9 = 5.78 hrs time saved = 13 - 5.78 = 7.22 hrs answer is e" | a = 13 * const_60
b = 4 / 9
c = 13 * b
d = c * const_60
e = a - d
f = reminder / (
g = f + 100
|
a ) 5 : 6 , b ) 2 : 3 , c ) 2 : 5 , d ) 2 : 1 , e ) 2 : 4 | a | divide(subtract(15.8, 15.3), 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 15.3 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 * 15.3 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.3 ) = k = 0.5 / 0.6 = 5 / 6 required ratio = 5 / 6 : 1 = 5 : 6 . answer : a" | a = 15 - 8
b = 16 - 4
c = a / b
|
a ) 2 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | b | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 12)) | a boy is travelling from his home to school at 6 km / hr and reached 7 min late . next day he travelled at 12 km / hr and reached 8 min early . distance between home and school ? | "let the distance be x t 1 = x / 6 hr t 2 = x / 12 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 6 - x / 12 = 1 / 4 x / 12 = 1 / 4 x = 3 km answer is b" | a = 7 / const_60
b = 8 / const_60
c = a + b
d = 1 / 12
e = c / d
|
a ) 56 m , b ) 88 m , c ) 120 m , d ) 137 m , e ) none | a | divide(44, multiply(power(divide(1, const_2), const_2), const_pi)) | 44 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be : | "sol . let the length of the wire b h . radius = 1 / 2 mm = 1 / 20 cm . then , 22 / 7 * 1 / 20 * 1 / 20 * h = 44 β = [ 44 * 20 * 20 * 7 / 22 ] = 5600 cm = 56 m . answer a" | a = 1 / 2
b = a ** 2
c = b * math.pi
d = 44 / c
|
a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | divide(84, add(24, 4)) | a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 84 km downstream | "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 84 km time taken = distance / speed = 84 / 28 = 3 hours answer : option a" | a = 24 + 4
b = 84 / a
|
a ) 10 % , b ) ( 8 ) 20 % , c ) 50 % , d ) 60 % , e ) 65 % | c | subtract(add(add(75, 55), 20), const_100) | if 75 % of a boys answered the first question on a certain test correctly , 55 % answered the second question on the test correctly , and 20 % answered neither of the questions correctly , what % answered both correctly ? | { total } = { first } + { second } - { both } + { neither } 100 % = 75 % + 55 % - { both } + 20 % - - > { both } = 50 % . answer : c | a = 75 + 55
b = a + 20
c = b - 100
|
a ) 9 , b ) 8 , c ) 7 , d ) 6 , e ) 4 | e | subtract(subtract(divide(50, 5), 3), 3) | the sum of ages of 5 children born at the intervals of 3 years each is 50 years . find out the age of the youngest child ? | let the age of the youngest child = x then , the ages of 5 children can be written as x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 β 5 x + 30 = 50 β 5 x = 20 β x = 205 = 4 answer is e . | a = 50 / 5
b = a - 3
c = b - 3
|
a ) 40 kmph , b ) 45 kmph , c ) 48 kmph , d ) 50 kmph , e ) 55 kmph | c | divide(multiply(40, 6), add(divide(40, 40), divide(multiply(2, 40), 20))) | a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 6 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 6 x / ( x / 8 ) = 48 kmph answer : c" | a = 40 * 6
b = 40 / 40
c = 2 * 40
d = c / 20
e = b + d
f = a / e
|
a ) 300 , b ) 420 , c ) 410 , d ) 320 , e ) 400 | c | add(multiply(45, 8), multiply(subtract(45, 20), multiply(8, divide(25, const_100)))) | mary works in a restaurant a maximum of 45 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | "mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 25 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 25 ) = $ 250 . the total amount is $ 160 + $ 250 = $ 410 answer c 410 ." | a = 45 * 8
b = 45 - 20
c = 25 / 100
d = 8 * c
e = b * d
f = a + e
|
a ) 165 , b ) 170 , c ) 250 , d ) 180 , e ) 185 | c | multiply(divide(const_100, 8), 20) | a 20 % stock yielding 8 % is quoted at ? | "assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 20 , money invested = 100 Γ 20 / 8 = rs . 250 ie , market value of the stock = rs . 250 answer is c ." | a = 100 / 8
b = a * 20
|
a ) 11 , b ) 10 , c ) 8 , d ) 12 , e ) 15 | d | add(8, divide(subtract(110, 20), add(25, 20))) | two stations a and b are 110 km apart on a straight line . one train starts from a at 9 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 9 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 110 45 x = 135 x = 3 . so , they meet at 12 a . m . answer : d" | a = 110 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | c | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 25 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ? | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 25 minutes , sam rides 5 km so sam will be 3 km ahead . it will take annie 60 minutes to catch sam . the answer is c ." | a = 15 - 12
b = 40 / const_60
c = a * b
d = 12 / c
e = 15 - 12
f = 40 / const_60
g = e * f
h = d - g
i = 15 - 12
j = h / i
k = j * const_60
|
a ) 25 , b ) 50 , c ) 250 , d ) 300 , e ) none | a | multiply(3, multiply(divide(5, 5), divide(5, 3125))) | if 5 a = 3125 , then the value of 5 ( a - 3 ) is | "sol . 5 a = 3125 = 5 a = 55 a = 5 . 5 ( 1 - 3 ) = 5 ( 5 - 3 ) = 5 ( 2 ) = 25 . answer a" | a = 5 / 5
b = 5 / 3125
c = a * b
d = 3 * c
|
a ) 20 sec , b ) 17.23 sec , c ) 12.86 sec , d ) 9.5 sec , e ) 8.5 sec | c | divide(250, divide(70, const_3_6)) | a train of length 250 m runs at a speed of 70 km / hr . what will be the time taken to cross any stationary object standing at the railway station ? | explanation : length of train = 250 m , speed of train = 70 km / hr length of train is always considered as distance , and hence here distance = 250 m 1 ) first convert speed of km / hr into m / s speed of train = 70 x 5 = 19.44 m / s 18 2 ) we know that , speed = distance / time time taken to cross stationary object = 250 / 19.44 time taken to cross stationary object = 12.86 sec answer is c | a = 70 / const_3_6
b = 250 / a
|
a ) 4 hrs 29 min , b ) 4 hrs 24 min , c ) 8 hrs 24 min , d ) 4 hrs 44 min , e ) 5 hrs 24 min | b | add(multiply(4, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 4), divide(const_1, 5)), const_2)), 4), const_60)) | two pipes a and b can fill a tank in 4 and 5 hours respectively . if they are turned up alternately for one hour each , the time taken to fill the tank is ? | 1 / 4 + 1 / 5 = 9 / 20 20 / 9 = 2 2 / 9 9 / 20 * 2 = 9 / 10 - - - - 4 hours wr = 1 - 9 / 10 = 1 / 10 1 h - - - - 1 / 4 ? - - - - - 1 / 10 2 / 5 * 60 = 24 = 4 hrs 24 min answer : b | a = 4 * 100
b = 1 / 4
c = 1 / 5
d = b + c
e = d * 2
f = 1 - e
g = f * 4
h = g * const_60
i = a + h
|
a ) β 48 , b ) β 2 , c ) 6 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(6, 27), subtract(6, 27)), const_1), const_1), const_1), const_1) | if a ( a - 6 ) = 27 and b ( b - 6 ) = 27 , where a β b , then a + b = | "i . e . if a = - 3 then b = 9 or if a = 9 then b = - 3 but in each case a + b = - 3 + 9 = 6 answer : option c" | a = 6 + 27
b = 6 - 27
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(power(subtract(add(5, 3), 1), 2), multiply(floor(divide(power(subtract(add(5, 3), 1), 2), 5)), 5)) | when n is divided by 5 the remainder is 3 . what is the remainder when ( n - 1 ) ^ 2 is divided by 5 ? | n = 5 x + 3 , for some integer x ( n - 1 ) ^ 2 = ( 5 x + 2 ) ^ 2 = 5 y + 4 , for some integer y when we divide this by 5 , the remainder is 4 . the answer is e . | a = 5 + 3
b = a - 1
c = b ** 2
d = 5 + 3
e = d - 1
f = e ** 2
g = f / 5
h = math.floor(g)
i = h * 5
j = c - i
|
a ) 350 , b ) 450 , c ) 500 , d ) 650 , e ) 700 | b | divide(add(37, divide(1, 2)), divide(1, 12)) | how many 1 / 12 s are there in 37 1 / 2 ? | "required number = ( 75 / 2 ) / ( 1 / 12 ) = ( 75 / 2 x 12 / 1 ) = 450 . answer : b" | a = 1 / 2
b = 37 + a
c = 1 / 12
d = b / c
|
a ) 2008 , b ) 2009 , c ) 2010 , d ) 2011 , e ) none of these | d | add(2001, divide(add(40, subtract(6.3, 4.2)), subtract(40, 15))) | . the price of commodity p increases by 40 paise every year , while the price of commodity q increases by 15 paise every year . if in 2001 , the price of commodity p was rs . 4.20 and that of q was rs . 6.30 , in which year commodity p will cost 40 paise more than the commodity q ? | explanation : let the commodity p costs 40 paise more than the commodity q after n years price of the commodity p in 2001 = rs . 4.20 since the price of the commodity p increases by rs 0.40 every year , price of the commodity p after n years from 2001 = rs . 4.20 + ( n Γ . 40 ) price of the commodity q in 2001 = rs . 6.30 since the price of the commodity q increases by rs 0.15 every year , price of the commodity q after n years from 2001 = rs . 6.30 + ( n Γ . 15 ) since the commodity p costs rs . 0.40 more that the commodity q after n years from 2001 , 4.20 + ( n Γ . 40 ) = 6.30 + ( n Γ . 15 ) + 0.40 = > ( 40 n - . 15 n ) = 6.30 - 4.20 + 0.40 = 2.5 = > . 25 n = 2.5 β n = 2.5 / . 25 = 250 / 25 = 101.15 Β― = > commodity p costs rs . 0.40 more that the commodity q after 10 years from 2001 . i . e . , in 2011 . answer : option d | a = 6 - 3
b = 40 + a
c = 40 - 15
d = b / c
e = 2001 + d
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | d | divide(multiply(subtract(7.00, 2.50), 100), 2.50) | a wholesaler wishes to sell 100 pounds of mixed nuts at $ 2.50 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 7.00 a pound . how many pounds of cashews must she use ? | "from the question stem we know that we need a mixture of 100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y , we get x + y = 100 . since the wholesaler wants to sell the mixture of 100 pounds @ $ 2.50 , we can write this as : $ 2.5 * ( x + y ) = $ 1.5 x + $ 4 y from the equation x + y = 100 , we can rewrite y as y = 100 - x and substitute this into our equation to get : $ 2.5 * ( x + 100 - x ) = $ 1.5 x + $ 4 ( 100 - x ) if you solve for x , you will get x = 60 , and therefore y = 40 . so the wholesaler must use 40 pounds of cashews . you can substitute into the original equation to see that : $ 250 = $ 1.5 ( 60 ) + $ 4 ( 40 ) answer is d ." | a = 7 - 0
b = a * 100
c = b / 2
|
a ) 3 , b ) 8 , c ) 12 , d ) 32 , e ) 35 | b | subtract(39, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 39 ? | "n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 39 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 8 more to make it 39 m + 39 answer - b" | a = 39 - reminder
|
a ) 40 , b ) 80 , c ) 100 , d ) 120 , e ) 200 | b | add(add(multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)), add(multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)), multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)))), divide(800, add(add(add(multiply(add(const_1, const_1), const_2), const_1), multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1))), add(multiply(add(const_1, const_1), const_2), const_1)))) | a rectangular region has a fence along three sides and a wall along the fourth side . the fenced side opposite the wall is twice the length of each of the other two fenced sides . if the area of the rectangular region is 800 square feet , what is the total length of the fence , in feet ? | "two sides each = x the third = 2 x and the wall length is thus 2 x too x * 2 x = 2 x ^ 2 = 800 ie x ^ 2 = 400 ie x = 20 l = 40 w = 20 total length of fence = 2 * 20 + 40 = 80 my answer is b" | a = 1 + 1
b = a * 2
c = b + 1
d = 2 * c
e = 1 + 1
f = e * 2
g = f + 1
h = 2 * g
i = 1 + 1
j = i * 2
k = j + 1
l = 2 * k
m = h + l
n = d + m
o = 1 + 1
p = o * 2
q = p + 1
r = 1 + 1
s = r * 2
t = s + 1
u = 2 * t
v = q + u
w = 1 + 1
x = w * 2
y = x + 1
z = v + y
A = 800 / z
B = n + A
|
a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 32 , d ) 1 / 64 , e ) 1 / 128 | d | power(divide(1, 2), 6) | if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 6 times , what is the probability that it will land heads up on the first 3 flips but not on the last 3 flips ? | "p ( hhhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 64 the answer is d ." | a = 1 / 2
b = a ** 6
|
a ) 20 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 19 % | a | subtract(subtract(add(50, const_100), divide(multiply(add(50, const_100), 20), const_100)), const_100) | a merchant marks his goods up by 50 % and then offers a discount of 20 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 150 after a 50 % markup . now a discount of 20 % on 150 . profit = 120 - 100 20 % answer a" | a = 50 + 100
b = 50 + 100
c = b * 20
d = c / 100
e = a - d
f = e - 100
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.