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 ) 35 , b ) 42 , c ) 100 , d ) 105 , e ) 140 | b | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 31), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 31 % iodine solution ? | "solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 31 / 100 40 x + 700 = 31 x + 1085 9 x = 385 x = 42 lts correct option : b" | a = 35 * 20
b = a / 100
c = b * 31
d = 35 * 20
e = d / 100
f = e * 20
g = c - f
h = 40 / 100
i = 20 * h
j = 35 * 20
k = j / 100
l = i - k
m = g / l
n = m + 35
|
a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(180, const_2), const_2) | if k ^ 3 is divisible by 180 , what is the least possible value of integer k ? | "180 = 2 ^ 2 * 3 ^ 2 * 5 therefore k must include at least 2 * 3 * 5 = 30 . the answer is b ." | a = 180 / 2
b = a / 2
|
a ) 45 minutes , b ) 55 minutes , c ) 35 minutes , d ) 25 minutes , e ) 50 minutes | e | multiply(add(const_3, 6), 5) | a clock shows the time as 8 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 6 p . m . ? | "there are 10 hours in between 8 a . m . to 6 p . m . 10 * 5 = 50 minutes . answer : e" | a = 3 + 6
b = a * 5
|
a ) s . 1000 , b ) s . 875 , c ) s . 1007 , d ) s . 1006 , e ) s . 1002 | b | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 12))) | a watch was sold at a loss of 12 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "explanation : 88 % 104 % - - - - - - - - 16 % - - - - 140 100 % - - - - ? = > rs . 875 answer : b" | a = 140 * 100
b = 100 + 4
c = 100 - 12
d = b - c
e = a / d
|
a ) 7 , b ) 6 , c ) 5 , d ) 9 , e ) 1 | c | multiply(6, add(const_1, divide(20, const_100))) | a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 20 % ? | "explanation : c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 120 % of re . 1 = rs . 6 / 5 for rs . 6 / 5 , toffees sold = 6 . for re . 1 . toffees sold = 6 * 5 / 6 = 5 answer : c" | a = 20 / 100
b = 1 + a
c = 6 * b
|
a ) 12 , b ) 15 , c ) 20 , d ) 22 , e ) 23 | a | divide(multiply(subtract(26, 10), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 10 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 10 = 26 = > x = 4 so deepak age is = 3 ( 4 ) = 12 answer : option a" | a = 26 - 10
b = a * 3
c = b / 4
|
a ) 9,6 , b ) 7,3 , c ) 9,3 , d ) 6,6 , e ) none of these | b | divide(subtract(10, 4), const_2) | a man can row downstream at the rate of 10 km / hr and upstream at 4 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 10 + 4 ) = 7 km / hr rate of current = 1 / 2 ( 10 - 4 ) = 3 km / hr answer : option b" | a = 10 - 4
b = a / 2
|
a ) 40 days , b ) 24 / 5 days , c ) 39 / 9 days , d ) 30 / 9 days , e ) 60 / 9 days | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 12))) | worker a takes 8 hours to do a job . worker b takes 12 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "a ' s one hour work = 1 / 8 . b ' s one hour work = 1 / 12 . ( a + b ) ' s one hour work = 1 / 8 + 1 / 12 = 5 / 24 . both a & b can finish the work in 24 / 5 days b" | a = 1 / 8
b = 1 / 12
c = a + b
d = 1 / c
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | divide(add(4, 6), subtract(3, const_1)) | denominator of a number is 4 less than the numerator . if 6 is added to the numerator , it becomes 3 times the denominator . find the denominator . | let the numerator be x and denominator be y . then = > x = y + 4 and 6 + x = 3 * y = > 6 + y + 4 = 3 * y = > 2 y = 10 = > y = 5 option c | a = 4 + 6
b = 3 - 1
c = a / b
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | a | divide(3, add(2, const_1)) | if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 3 - m ) then m = ? | "( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 3 - m ) = 4 ^ ( ( 3 - m ) / 2 ) therefore , m = ( 3 - m ) / 2 2 m = 3 - m m = 1 answer a" | a = 2 + 1
b = 3 / a
|
['a ) 140', 'b ) 388', 'c ) 238', 'd ) 119', 'e ) 112'] | a | multiply(subtract(34, multiply(const_10, const_2)), const_10) | a rectangular paper when folded into two congruent parts had a perimeter of 34 cm foer each part folded along one set of sides and the same is 38 cm . when folded along the other set of sides . what is the area of the paper ? | when folded along the breadth we have 2 ( l / 2 + b ) = 34 or l + 2 b = 34 . . . . . . . . . . . ( 1 ) when folded along the length , we have 2 ( l + b / 2 ) = 38 or 2 l + b = 38 . . . . . ( 2 ) from 1 & 2 we get l = 14 and b = 10 area of the paper = 14 * 10 = 140 sq cmv answer : a | a = 10 * 2
b = 34 - a
c = b * 10
|
a ) 130 , b ) 120 , c ) 140 , d ) 100 , e ) 300 | b | add(divide(multiply(add(divide(multiply(5.76, subtract(const_100, 12)), 12), 5.76), subtract(const_100, 40)), 40), add(divide(multiply(5.76, subtract(const_100, 12)), 12), 5.76)) | given a certain number ( x ) , find x given that 5.76 is 12 % of 40 % of x ? | using elimination method to pick the correct option it will be discovered that option b i . e 120 is the correct option because 40 % of 120 is 48 and 12 % of 48 is 5.76 . answer : b | a = 100 - 12
b = 5 * 76
c = b / 12
d = c + 5
e = 100 - 40
f = d * e
g = f / 40
h = 100 - 12
i = 5 * 76
j = i / 12
k = j + 5
l = g + k
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | d | divide(multiply(12.60, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 12.60 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 12.60 when c . p = ( 100 * 12.60 ) / 21 = 60 answer : d" | a = 12 * 60
b = 25 + 100
c = 30 + 100
d = 100 - 20
e = c * d
f = e / 100
g = b - f
h = a / g
|
a ) 230 , b ) 200 , c ) 250 , d ) 300 , e ) 500 | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 270) | a 270 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "answer : option a explanation : relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 270 ) / 9 = 500 / 9 = > x = 230 . answer a" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 270
|
a ) 2 : 5 , b ) 1 : 2 , c ) 4 : 5 , d ) 3 : 7 , e ) 5 : 6 | c | divide(add(const_100, 20), add(const_100, 50)) | two numbers are in respectively 20 % and 50 % more than a third number . the ratio of the two numbers is ? | "let the 3 rd number be x then , first number = 120 % of x = 120 x / 100 = 6 x / 5 second number = 150 % of x = 150 x / 100 = 3 x / 2 ratio of first two numbers = 6 x / 5 : 3 x / 2 = 12 x : 15 x = 4 : 5 answer is c" | a = 100 + 20
b = 100 + 50
c = a / b
|
a ) 22 , b ) 60 , c ) 28 , d ) 76 , e ) 21 | b | divide(multiply(multiply(6, 5), 4), const_2) | a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 4 rows or 5 rows or 6 rows , then no tree will be left . find the least number of trees required | explanation : the least number of trees that are required = lcm ( 4 , 5 , 6 ) = 60 . answer : b | a = 6 * 5
b = a * 4
c = b / 2
|
a ) 143 , b ) 144 , c ) 145 , d ) 146 , e ) 147 | b | divide(subtract(multiply(15, multiply(24, const_2)), multiply(12, 24)), 3) | a batsman ' s avg in 12 innings is 24.00 . if his avg is to be double of the no of innings ( 15 innings ) , what should he score in the remaining 3 innings ( avg ) ? | let total runs done by a batsman in 12 innings be x therefore , x / 12 = 24 that means x = 12 * 24 = 288 runs for 15 innings , avg was doubled i . e . . 48.00 therefore , ( x + y ) / 15 = 48 where y is the no of runs done in last 3 innings y = 432 therefore average is 432 / 3 = 144 answer : b | a = 24 * 2
b = 15 * a
c = 12 * 24
d = b - c
e = d / 3
|
a ) 3 , b ) 5 , c ) 4 , d ) 6.4 , e ) 7.2 | e | divide(80, multiply(40, const_0_2778)) | in what time will a railway train 80 m long moving at the rate of 40 kmph pass a telegraph post on its way ? | "t = 80 / 40 * 18 / 5 = 7.2 sec answer : e" | a = 40 * const_0_2778
b = 80 / a
|
a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people z who purchased only book a ? | "this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b and this is 1 / 2 of those that purchased a . so , 1500 purchased a . less the 500 that purchased both , z = 1000 purchased a only . ( this is much simpler to solve using the venn diagram ) . correct answer is d . 1000" | a = 500 / 2
b = 500 + a
c = b * 2
d = c - 500
|
a ) 40 , b ) 44 , c ) 100 , d ) 88 , e ) 48 | c | divide(subtract(power(20, const_2), 200), const_2) | if the sum of two numbers is 20 and the sum of their squares is 200 , then the product of the numbers is | "according to the given conditions x + y = 20 and x ^ 2 + y ^ 2 = 200 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 20 ^ 2 = 200 + 2 xy so xy = 200 / 2 = 100 answer : c" | a = 20 ** 2
b = a - 200
c = b / 2
|
a ) 16 : 15 , b ) 4 : 5 , c ) 7 : 12 , d ) 4 : 9 , e ) 6 : 5 | c | divide(multiply(2, multiply(divide(2, subtract(multiply(divide(add(const_100, 50), const_100), 2), divide(7, 3))), divide(7, 3))), add(divide(2, subtract(multiply(divide(add(const_100, 50), const_100), 2), divide(7, 3))), 2)) | ratio of two numbers x and y is 3 : 7 . if x is increased by 50 % and y is increased by 2 then the new ratio becomes 1 : 2 . what is the ratio 2 y : ( 2 x + 6 ) | "let : x = 3 n y = 7 n 4.5 n / ( 7 n + 5 ) = 1 / 2 9 n = 7 n + 2 n = 1 so , x = 3 ; y = 7 2 y / ( 2 x + 6 ) = 14 / 12 = 7 : 6 answer = c" | a = 100 + 50
b = a / 100
c = b * 2
d = 7 / 3
e = c - d
f = 2 / e
g = 7 / 3
h = f * g
i = 2 * h
j = 100 + 50
k = j / 100
l = k * 2
m = 7 / 3
n = l - m
o = 2 / n
p = o + 2
q = i / p
|
a ) a ) 1040 , b ) b ) 1050 , c ) c ) 1000 , d ) d ) 1065 , e ) e ) 1075 | c | add(multiply(8, 70), multiply(8, 55)) | bruce purchased 8 kg of grapes at the rate of 70 per kg and 8 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 × 8 = 560 . cost of 8 kg of mangoes = 55 × 8 = 440 . total cost he has to pay = 560 + 440 = 1000 c" | a = 8 * 70
b = 8 * 55
c = a + b
|
a ) 22 , b ) 18 , c ) 4.5 , d ) 99 , e ) 38 | c | divide(subtract(divide(36, divide(4, 3)), multiply(subtract(4, const_1), 3)), 3) | the sum of the ages of 4 children born at the intervals of 3 years each is 36 years . what is the age of the youngest child ? | "let x = the youngest child . each of the other four children will then be x + 3 , x + 6 , x + 9 we know that the sum of their ages is 36 so , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) = 36 therefore the youngest child is 4.5 years old answer : c" | a = 4 / 3
b = 36 / a
c = 4 - 1
d = c * 3
e = b - d
f = e / 3
|
a ) 1 , b ) 3 , c ) 5 , d ) 4 , e ) 2 | c | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 20))) | a pump can fill a tank with water in 4 hours . because of a leak , it took 20 hours to fill the tank . the leak can drain all the water of the tank in ? | p - - > pump , l - - - > leak pipe - - - - - - - - - - p - - - - - - - - - l - - - - - - - - - - - - ( p - l ) - - time - - - - - - - - - 4 - - - - - - - - - x - - - - - - - - - - - - - 20 * rate - - - - - - - - - 5 - - - - - - - - - x - - - - - - - - - - - - - 1 work - - - - - - - 20 - - - - - - - - - 20 - - - - - - - - - - - 20 time required to empty the tank if leakage present = 20 / ( ( 5 ) - 1 ) = ( 20 ) / ( 4 ) = 5 ans - c | a = 1 / 4
b = 1 / 20
c = a - b
d = 1 / c
|
a ) 9 , b ) 10 , c ) 11 , d ) 5 , e ) 13 | d | divide(15, divide(15, 5)) | suppose 5 monkeys take 5 minutes to eat 5 bananas . how many monkeys would it take to eat 15 bananas in 15 minutes ? | "one monkey takes 5 min to eat 1 banana , so in 15 mins 1 monkey will eat 3 bananas , so for 15 bananas in 15 min we need 15 / 3 = 5 monkeys answer : d" | a = 15 / 5
b = 15 / a
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.0002 , e ) 0.5 | d | power(divide(1, 4), 6) | what is the decimal equivalent of ( 1 / 4 ) ^ 6 ? | "( 1 / 4 ) ^ 6 = 1 / 4096 = 0.0002 answer : d" | a = 1 / 4
b = a ** 6
|
a ) 3 / 7 , b ) 3 / 10 , c ) 3 / 18 , d ) 3 / 16 , e ) 3 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 2), 2), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | 3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 2 men and 2 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 2 m + 2 w = 6 w answer is 6 / 14 = 3 / 7 answer : a" | a = 8 - 2
b = 6 - 3
c = a / b
d = c * 2
e = d + 2
f = 8 - 2
g = 6 - 3
h = f / g
i = 6 * h
j = i + 2
k = e / j
|
a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | what is the remainder when the number 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a = 5 / 5
|
a ) 8 , b ) 12 , c ) 9 , d ) 6 , e ) 5 | b | divide(subtract(78, 54), subtract(54, 52)) | rahul played weel in this season . his current batting averagge is 52 . if he score 78 runs in today match . his batting average will become 54 . how many matches had he played in this season . | "52 x + 78 = 54 ( x + 1 ) = > 2 x = 24 = > x = 12 answer : b" | a = 78 - 54
b = 54 - 52
c = a / b
|
a ) 1 : 2 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 3 | a | divide(add(multiply(12, divide(1, add(1, 2))), multiply(9, divide(1, add(1, 2)))), subtract(add(12, 9), add(multiply(12, divide(1, add(1, 2))), multiply(9, divide(1, add(1, 2)))))) | a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 9 ounces in the ratio 1 : 2 or , x + 2 x = 9 gives x ( alcohol ) = 3 and remaining water = 6 mixing both we have alcohol = 4 + 3 and water = 8 + 6 ratio thus alcohol / water = 7 / 14 = 1 / 2 answer a" | a = 1 + 2
b = 1 / a
c = 12 * b
d = 1 + 2
e = 1 / d
f = 9 * e
g = c + f
h = 12 + 9
i = 1 + 2
j = 1 / i
k = 12 * j
l = 1 + 2
m = 1 / l
n = 9 * m
o = k + n
p = h - o
q = g / p
|
a ) 1 / 3 , b ) 1 / 2 , c ) 2 / 3 , d ) 3 / 5 , e ) 4 / 7 | b | divide(35, add(50, 20)) | some of the 50 % solution of acid was removed and this was replaced with an equal amount of 20 % solution of acid . as a result , a 35 % solution of acid was obtained . what fraction of the original solution was replaced ? | let x be the fraction of the original solution that was replaced . 0.5 * ( 1 - x ) + 0.2 ( x ) = 0.35 0.3 x = 0.15 x = 1 / 2 the answer is b . | a = 50 + 20
b = 35 / a
|
a ) 93 , b ) 94 , c ) 92 , d ) 96 , e ) 95 | e | subtract(add(floor(divide(202, const_3)), floor(divide(202, add(const_1, const_4)))), multiply(floor(divide(202, multiply(const_3, add(const_1, const_4)))), const_2)) | there are 202 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off position at the beggining , how many lights are switched on by the end of this operation ? | "editing my solution : number of switches = 202 number of switches turned on by a : 3 , 6 , . . . 201 = 67 number of switches turned on by b : 5 , 10 , . . . . 200 = 40 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b as they are turned off . number of switches that are turned on = ( 67 - 6 ) + ( 40 - 6 ) = 95 answer : e" | a = 202 / 3
b = math.floor(a)
c = 1 + 4
d = 202 / c
e = math.floor(d)
f = b + e
g = 1 + 4
h = 3 * g
i = 202 / h
j = math.floor(i)
k = j * 2
l = f - k
|
a ) 4 , b ) 8 , c ) 16 , d ) 1911 , e ) none of these | a | gcd(1048, 828) | the maximum number of students among them 1048 pens and 828 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "explanation : required number of students = h . c . f of 1048 and 828 = 4 . answer : a" | a = math.gcd(1048, 828)
|
a ) 124 % , b ) b ) 120 % , c ) c ) 96 % , d ) d ) 80 % , e ) e ) 84 % | e | multiply(divide(add(const_100, 40), multiply(divide(const_100, subtract(const_100, 40)), const_100)), const_100) | marts income is 40 percent more than tims income and tims income is 40 percent less than juans income . what percentage of juans income is marts income | "m = ( 140 / 100 ) t t = ( 60 / 100 ) j = > m = ( 84 / 100 ) j answer e ." | a = 100 + 40
b = 100 - 40
c = 100 / b
d = c * 100
e = a / d
f = e * 100
|
a ) 330 , b ) 340 , c ) 350 , d ) 360 , e ) 370 | a | add(multiply(multiply(subtract(9, 6), const_10), const_10), multiply(subtract(9, 6), const_10)) | if all the 6 are replaced by 9 , then the algebraic sum of all the numbers from 1 to 100 ( both inclusive ) varies by | 1 11 21 . . . . . . . 91 2 12 22 . . . . . . . . 92 3 13 23 . . . . . . . . 93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . there are total 20 occurrence of 6 . once place at 6 th row in every column and tens place in 6 th column and diff of 6 and 9 is 3 so 3 * 1 * 10 + 3 * 10 * 10 = 330 answer : a | a = 9 - 6
b = a * 10
c = b * 10
d = 9 - 6
e = d * 10
f = c + e
|
a ) 1 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | a | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 12)) | a boy is travelling from his home to school at 3 km / hr and reached 7 min late . next day he traveled at 12 km / hr and reached 8 min early . distance between home and school ? | "let the distance be x t 1 = x / 3 hr t 2 = x / 12 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 3 - x / 12 = 1 / 4 x / 4 = 1 / 4 x = 1 km answer is a" | a = 7 / const_60
b = 8 / const_60
c = a + b
d = 1 / 12
e = c / d
|
a ) 104 , b ) 60 , c ) 62 , d ) 50 , e ) 25 | c | add(subtract(20, multiply(power(2, 2), 2)), multiply(2, power(5, 2))) | if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 20 , what is the value of f ( 5 ) ? | "f ( x ) = 2 x ^ 2 + y f ( 2 ) = 20 = > 2 * ( 2 ) ^ 2 + y = 20 = > 8 + y = 20 = > y = 12 f ( 5 ) = 2 * ( 5 ) ^ 2 + 12 = 62 answer c" | a = 2 ** 2
b = a * 2
c = 20 - b
d = 5 ** 2
e = 2 * d
f = c + e
|
a ) 21 , b ) 47 , c ) 45 , d ) 43 , e ) 41 | a | add(multiply(multiply(divide(25, const_100), 20), multiply(divide(25, const_100), 20)), divide(subtract(100, 20), 20)) | in a 100 member association consisting of men and women , exactly 20 % of men and exactly 25 % women are homeowners . what is the least number of members who are homeowners ? | "solution simple out of 100 20 % are male i . e 20 and 25 % are female i . e 25 , so total homeowner is 45 . now min number homeowner is 20 and max is 45 so question ask us to find least and 21 has least value among all option . so ans is 21 . answer : a" | a = 25 / 100
b = a * 20
c = 25 / 100
d = c * 20
e = b * d
f = 100 - 20
g = f / 20
h = e + g
|
a ) 3500 , b ) 4500 , c ) 5500 , d ) 5250 , e ) 5000 | e | divide(450, multiply(divide(subtract(subtract(const_100, 10), 80), const_100), divide(90, const_100))) | of the land owned by a farmer , 90 percent was cleared for planting . of the cleared land , 10 percent was planted with grapes and 80 percent of the cleared land was planted with potato . if the remaining 450 acres of cleared land was planted with tomato , how many acres did the farmer own ? | 10 % od 90 % = 9 % , 80 % 0 f 90 % = 72 % so the remaining 90 - 9 - 72 = 9 % = 450 acres or 10 % of 90 % = 9 % - - > 450 / 9 * 100 = 5000 acres answer ( e ) | a = 100 - 10
b = a - 80
c = b / 100
d = 90 / 100
e = c * d
f = 450 / e
|
a ) 33.6 , b ) 47 , c ) 50.4 , d ) 52.4 , e ) 47.9 | a | subtract(add(divide(multiply(2, 18), subtract(18, const_1)), 18), 2) | the ages of 2 persons differ by 18 years . if 12 years ago the elder one be 6 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 18 6 ( x - 12 ) = x + 18 - 12 x = 15.6 age of elder person = 15.6 + 18 = 33.6 answer is a" | a = 2 * 18
b = 18 - 1
c = a / b
d = c + 18
e = d - 2
|
a ) 75 , b ) 45 , c ) 87 , d ) 165 , e ) 11 | b | divide(add(add(add(add(36, 35), 42), 57), 55), add(const_2, const_3)) | david obtained 36 , 35 , 42 , 57 and 55 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | "explanation : average = ( 36 + 35 + 42 + 57 + 55 ) / 5 = 225 / 5 = 45 . answer : b" | a = 36 + 35
b = a + 42
c = b + 57
d = c + 55
e = 2 + 3
f = d / e
|
a ) 11020 , b ) 11030 , c ) 11040 , d ) 11060 , e ) 11080 | c | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 40,56) | find the smallest number of five digits exactly divisible by 32 , 40,56 and 64 . | "smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 32,40 , 56,64 i . e 2240 , on dividing 10000 by 2240 , we get 1200 as remainder . therefore , required number = 10000 + ( 2240 â € “ 1200 ) = 11040 . answer is c ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 40
|
a ) 6109 , b ) 6209 , c ) 6218 , d ) 6107 , e ) 6100 | c | add(add(multiply(16000, divide(15, const_100)), multiply(add(16000, multiply(16000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(16000, multiply(16000, divide(15, const_100))), multiply(add(16000, multiply(16000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const_3))) | find compound interest on $ 16000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = $ [ 16000 x ( 1 + ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = $ [ 16000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = $ 22218 . : . c . i . = rs . ( 22218 - 16000 ) = $ 6218 answer c ." | a = 15 / 100
b = 16000 * a
c = 15 / 100
d = 16000 * c
e = 16000 + d
f = 15 / 100
g = e * f
h = b + g
i = 15 / 100
j = 16000 * i
k = 16000 + j
l = 15 / 100
m = 16000 * l
n = 16000 + m
o = 15 / 100
p = n * o
q = k + p
r = 15 / 100
s = r / 3
t = q * s
u = h + t
|
a ) 99 , b ) 100 , c ) 87 , d ) 95 , e ) 90 | a | subtract(subtract(subtract(subtract(subtract(multiply(10, 60), multiply(subtract(60, 5), 5)), subtract(60, 5)), add(subtract(60, 5), const_1)), add(subtract(60, 5), const_2)), add(subtract(60, 5), const_3)) | the average score in an examination of 10 students of a class is 60 . if the scores of the top 5 students are not considered , the average score of the remaining students falls by 5 . the pass mark was 40 and the maximum mark was 100 . it is also known that none of the students failed . if each of the top 5 scorers had distinct integral scores , the maximum possible score of the topper is . . . . . . | detailed solution 10 students have scored 600 marks amongst them , and no one is allowed to score lesser than 40 or higher than 100 . the idea now is to maximize what the highest scorer gets . the 5 least scores have an average of 55 , which means that they have scored 55 x 5 = 275 marks amongst them . this leaves 325 marks to be shared amongst the top 5 students . lets call them a , b , c , d and e . now , in order to maximize what the top scorer “ e ” gets , all the others have to get the least possible scores ( and at the same time , they should also get distinct integers . ) the least possible score of the top 5 should be at least equal to the highest of the bottom 5 . now we want to make sure that the highest of the bottom 5 is the least possible . this can be done by making all scores equal to 55 . if some scores are less than 55 , some other scores have to be higher than 55 to compensate and make the average 55 . thus the highest score is the least only when the range is 0 . so now , we have the lowest value that the top 5 can score , which is 55 . the others have to get distinct integer scores , and as few marks as possible , so that “ e ” gets the maximum . so , 55 + 56 + 57 + 58 + e = 325 e = 99 marks . answer choice ( a ) | a = 10 * 60
b = 60 - 5
c = b * 5
d = a - c
e = 60 - 5
f = d - e
g = 60 - 5
h = g + 1
i = f - h
j = 60 - 5
k = j + 2
l = i - k
m = 60 - 5
n = m + 3
o = l - n
|
a ) 253547 , b ) 143681 , c ) 143547 , d ) 147614 , e ) 255547 | c | add(add(multiply(add(multiply(multiply(7, 2), const_10), 3), const_1000), multiply(5, const_100)), subtract(add(multiply(7, 5), multiply(7, 2)), 2)) | 5 + 3 + 2 = 151022 9 + 2 + 4 = 183652 8 + 6 + 3 = 482466 5 + 4 + 5 = 202541 then ; 7 + 2 + 5 = ? | c 143547 | a = 7 * 2
b = a * 10
c = b + 3
d = c * 1000
e = 5 * 100
f = d + e
g = 7 * 5
h = 7 * 2
i = g + h
j = i - 2
k = f + j
|
a ) 174.2 , b ) 212 , c ) 283 , d ) 296 , e ) 282.4 | c | multiply(divide(45, 2.54), divide(24, 1.5)) | on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 45 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 45 cm . : 45 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 45 / ( 3.81 ) = 283 answer will be c ." | a = 45 / 2
b = 24 / 1
c = a * b
|
a ) 3 days , b ) 4 days , c ) 5 days , d ) 6 days , e ) 7 days | c | divide(subtract(50, add(37, divide(50, const_100))), add(divide(50, const_100), const_2)) | a labourer was engaged for 25 days on the condition that for every day , he works , he will be paid rs . 2 and for every day , he is absent he will be fined 50 p . if he receives only rs . 37 • 50 , find the no . of days he was absent is _____ | if a labour worked for 25 days he should get 50 rs . but in this he is getting 13 rs less . if he do n ' t go for work for 1 day he will loose 2.5 rs then no of days ( absent ) = 13 / 2.5 = 5.2 . means he was absent for 5 days answer : c | a = 50 / 100
b = 37 + a
c = 50 - b
d = 50 / 100
e = d + 2
f = c / e
|
a ) 70 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 80 % | b | multiply(divide(subtract(1210, 930), 930), const_100) | a cycle is bought for rs . 930 and sold for rs . 1210 , find the gain percent ? | "explanation : 930 - - - - 280 100 - - - - ? = > 30 % answer : b" | a = 1210 - 930
b = a / 930
c = b * 100
|
a ) 15 , b ) 24 , c ) 27 , d ) 36 , e ) 40 | a | subtract(const_60, multiply(divide(const_2, add(const_2, const_4)), const_60)) | bob finishes the first half of an exam in two - sixth the time it takes him to finish the second half . if the whole exam takes him an hour , how many minutes does he spend on the first half of the exam ? | many times , it is easiest to think of problems like this conceptually ( as opposed to formulaically ) . conceptually , if the first half of the exam takes 2 / 6 ' s the time of the second half ( 6 / 6 ' s ) , we can see that the hour the entire exam took to finish can be broken down into eight ' s . ( another way to look at this problem is via ratio ' s - first half is 2 : 8 and the second half is 6 : 8 ) . with each sixth of an hour being 15 / 2 minutes ( 60 / 8 ) , the first half of the exam would have taken 15 minutes . correct answer is a . | a = 2 + 4
b = 2 / a
c = b * const_60
d = const_60 - c
|
a ) 26 , b ) 24 , c ) 40 , d ) 48 , e ) 28 | e | subtract(negate(244), multiply(subtract(10, 82), divide(subtract(10, 82), subtract(4, 10)))) | 4 , 10 , ( ? ) , 82 , 244 , 730 | "4 × 3 - 2 = 10 10 × 3 - 2 = 28 28 × 3 - 2 = 82 82 × 3 - 2 = 244 244 × 3 - 2 = 730 answer is e ." | a = negate - (
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a ) 1.4 metres , b ) 2.8 metres , c ) 28 metres , d ) 14 metres , e ) none of these | e | multiply(sqrt(divide(multiply(divide(multiply(246.4, const_1000), multiply(4, const_100)), add(const_3, const_4)), add(divide(const_60, const_3), const_2))), const_2) | the capacity of a cylindrical tank is 246.4 litres . if the height is 4 metres , what is the diameter of the base ? | capacity ( volume ) of a cylindrical tank = π r 2 h ( here r = radius and h = height of the tank ) now , from the question , 246.4 × 0.001 = 22 ⁄ 7 × r 2 × 4 [ ∵ 1 litre = 1000 cm 3 = 0.001 m 3 ] or , 0.2464 × 7 / 22 × 4 = r 2 or , r = 0.14 m or , diameter = 2 r = 0.28 m answer e | a = 246 * 4
b = 4 * 100
c = a / b
d = 3 + 4
e = c * d
f = const_60 / 3
g = f + 2
h = e / g
i = math.sqrt(h)
j = i * 2
|
a ) 1 : 1 , b ) 3 : 1 , c ) 4 : 5 , d ) 4 : 3 , e ) 4 : 1 | e | divide(80, 20) | calculate the ratio between x and y if 80 % of x equal to 20 % of y ? | explanation : 80 x = 20 y x : y = 80 : 20 = 4 : 1 answer : e | a = 80 / 20
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a ) 76 hrs , b ) 99 hrs , c ) 55 hrs , d ) 90 hrs , e ) 11 hrs | d | inverse(subtract(divide(const_1, 9), divide(const_1, const_10))) | a cistern which could be filled in 9 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? | "1 / 9 - 1 / x = 1 / 10 = > 90 hrs answer : d" | a = 1 / 9
b = 1 / 10
c = a - b
d = 1/(c)
|
a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 21 | d | add(10, const_4) | how many odd numbers between 10 and 1600 are the squares of integers ? | "the numbers are the squares of 5 , 7 , 9 , . . . , 39 which includes 18 numbers . the answer is d ." | a = 10 + 4
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a ) 12 , b ) 14 , c ) 16 , d ) 15 , e ) 11 | a | divide(multiply(multiply(6, 12), 3), multiply(6, 12)) | aaron will jog from home at 6 miles per hour and then walk back home by the same route at 12 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ? | "xyt / ( x + y ) x = 6 , y = 12 t = 3 6 * 12 * 3 / 6 + 12 = 12 answer : a" | a = 6 * 12
b = a * 3
c = 6 * 12
d = b / c
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(divide(6, const_2), const_1) | if n is the greatest positive integer for which 2 ^ n is a factor of 6 ! , then n = ? | "6 ! = 720 e . 2 ^ 10 = 512 ( 720 / 512 ) - this is not a factor of 6 ! d . 2 ^ 8 = 256 ( 720 / 256 ) - this is not a factor of 6 ! c . 2 ^ 6 = 64 ( 720 / 64 ) - this is not a factor of 6 ! b . 2 ^ 4 = 16 ( 720 / 16 ) - this is a factor of 6 ! b is answer" | a = 6 / 2
b = a - 1
|
a ) $ 1 , b ) $ 1.85 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15 | b | min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.85), 7), 1.85) | andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.85 . what is the median gasoline price ? | "ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.85 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.85 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b" | a = 1 + 75
b = a + 1
c = b + 2
d = c + 1
e = d + 2
f = e + 1
g = f / 7
h = min(g)
|
a ) 28.3 , b ) 1.45 , c ) 2.78 , d ) 6.84 , e ) 7.23 | a | subtract(divide(power(negate(9), 2), 3), 2) | find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 9 x + 3 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 9 / 1 = - 9 ab = 3 / 1 = 3 hence a / b + b / a = [ ( - 9 ) 2 - 2 ( 3 ) ] / 3 = 56 / 4 = 28.3 . a )" | a = negate ** (
b = a / 2
c = b - 3
|
a ) 40 m , b ) 44 m , c ) 47 m , d ) 49 m , e ) 50 m | d | multiply(divide(56, multiply(20, 6)), multiply(35, 3)) | if 20 men can build a water fountain 56 metres long in 6 days , what length of a similar water fountain can be built by 35 men in 3 days ? | explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 6 : 3 : : 56 : x therefore ( 20 x 6 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 120 = 49 hence , the required length is 49 m . answer : d | a = 20 * 6
b = 56 / a
c = 35 * 3
d = b * c
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a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | e | divide(divide(choose(46, const_1), 46), power(const_3, const_2)) | each factor of 230 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 46 ? | "so the # of factors 230 has 8 ( see below ) ; so out of 8 factors only two are multiples of 46 : 46 and 230 , itself ; so , the probability is 2 / 8 = 1 / 4 . answer : e ." | a = math.comb(46, 1)
b = a / 46
c = 3 ** 2
d = b / c
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a ) $ 115 , b ) $ 135 , c ) $ 150 , d ) $ 165 , e ) $ 192 | a | add(multiply(divide(subtract(250, 25), 5), 2), 25) | a certain psychologist charges $ 25 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 250 , what is the total charge to a patient who receives only 2 hours of therapy ? | "let the charge for first hour = x + 25 then charge for each additional hour = x x + 25 + 4 x = 250 = > 5 x = 225 = > x = 45 total charge for patient for 3 hours of therapy = x + 25 + x = 2 x + 25 = 115 $ answer a" | a = 250 - 25
b = a / 5
c = b * 2
d = c + 25
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(subtract(23, multiply(8, const_2)), subtract(subtract(12, 8), const_1)) | there are 8 stores in town that had a total of 23 visitors on a particular day . however , only 12 people went shopping that day ; some people visited more than one store . if 8 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visited ? | "8 people visited 2 stores each for 16 visits . to maximize the number of stores that one person visited , let ' s assume that 3 people visited 1 store each . the number of remaining visits is 23 - 16 - 3 = 4 , which is the maximum that one person could have visited . the answer is c ." | a = 8 * 2
b = 23 - a
c = 12 - 8
d = c - 1
e = b - d
|
a ) s . 6100 , b ) s . 5100 , c ) s . 5800 , d ) s . 6000 , e ) s . 6200 | a | divide(61, multiply(divide(10, const_100), divide(10, const_100))) | if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 61 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 61 6100 answer : a" | a = 10 / 100
b = 10 / 100
c = a * b
d = 61 / c
|
a ) 659 , b ) 698 , c ) 780 , d ) 910 , e ) none | a | subtract(815, divide(multiply(subtract(854, 815), 4), 5)) | a sum of money at simple interest amounts to rs . 815 in 4 years and to rs . 854 in 5 years . the sum is : | "sol . s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 4 years = rs . ( 39 * 4 ) = rs . 156 . â ˆ ´ principal = rs . ( 815 - 156 ) = rs . 659 answer a" | a = 854 - 815
b = a * 4
c = b / 5
d = 815 - c
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | divide(divide(multiply(120, 6), add(4, 1)), 12) | according to the directions on a can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 4 cans of water to make orange juice . how many 12 ounce cans of concentrate are required to prepare 120 6 - ounce servings of orange juice ? | 1 x 12 ounce can of concentrate + 4 x 12 ounce cans of water = 60 ounces of the mixture . 60 ounces of mixture gives ( 60 / 6 ) = 10 servings . thus 1 x 12 ounce can of concentrate is used to produce 10 servings of the mixture . to make 120 servings of the mixture , we need 120 / 10 = 12 cans of the concentrate . the answer is b . | a = 120 * 6
b = 4 + 1
c = a / b
d = c / 12
|
a ) $ 1.50 , b ) $ 3.00 , c ) $ 4.00 , d ) $ 5.50 , e ) $ 7.00 | a | subtract(add(divide(const_1, 5), subtract(add(divide(multiply(7.35, 5), const_10), divide(multiply(7.35, 5), const_10)), 5)), const_1) | at a certain restaurant , the price of a sandwich is $ 4.00 more than the price of a cup of coffee . if the price of a sandwich and a cup of coffee is $ 7.35 , including a sales tax of 5 % , what is the price of a cup of coffee , excluding the sales tax ? | let the price of sandwich = s price of coffee = c s = c + 4 = > s - c = 4 - - 1 the price of a sandwich and a cup of coffee is $ 7.35 , including a sales tax of 5 % , 1.05 ( s + c ) = 7.35 = > s + c = 7 - - 2 from equations 1 and 2 , we get s = 5.5 $ c = 1.5 $ answer a | a = 1 / 5
b = 7 * 35
c = b / 10
d = 7 * 35
e = d / 10
f = c + e
g = f - 5
h = a + g
i = h - 1
|
['a ) 70 %', 'b ) 69 %', 'c ) 80 %', 'd ) 82 %', 'e ) 55 %'] | b | multiply(subtract(power(add(const_1, divide(30, const_100)), const_2), const_1), const_100) | what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 30 % ? | the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 30 % increase a surface = 6 * ( ( 1.3 a ) ^ 2 ) = 6 * 1.69 * a ^ 2 the increase in the surface area = ( 6 * 1.69 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 1.69 - 1 ) ) / ( 6 * a ^ 2 ) = 1.69 - 1 = 0.69 = 69 % answer : b | a = 30 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number a of integers that can be negative ? | minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning 5 out of 6 also possible as you will see a = 5 or 1 will give you same equation . . ans d | a = 10 / 2
b = a + 0
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a ) 12.5 % , b ) 20.83 % , c ) 25 % , d ) 50 % , e ) none of these | b | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | ritesh and co . generated revenue of rs . 1,500 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 1,500 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1250 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) × 1500 . = > 12,000 . hence , the total revenue by end of 2007 is rs . 12,000 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 12000 ) × 100 . = > 20.83 % . answer : b" | a = 1 + 4
b = 2 * a
c = b * 100
d = 2 * c
e = 1 + 4
f = e * 100
g = d + f
h = 1 + 4
i = 2 * h
j = i * 100
k = 2 + 4
l = k * 100
m = j + l
n = 12 / 5
o = m / n
p = g / o
q = p * 100
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a ) 5 , b ) 4 , c ) 1 , d ) - 4 , e ) - 5 | c | divide(add(4, 5), add(2, 5)) | if x is a number such that x ^ 2 + 4 x - 5 = 0 and x ^ 2 - 5 x + 4 = 0 , then x = | "x ^ 2 + 4 x - 5 = ( x + 5 ) ( x - 1 ) = 0 then x = - 5 or x = 1 . x ^ 2 - 5 x + 4 = ( x - 4 ) ( x - 1 ) = 0 then x = 4 or x = 1 . thus x = 1 . the answer is c ." | a = 4 + 5
b = 2 + 5
c = a / b
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | divide(180, multiply(3, 4)) | if the ratio of two number is 3 : 4 and lcm of the number is 180 then what is the number . | "product of two no = lcm * hcf 3 x * 4 x = 180 * x x = 15 answer : a" | a = 3 * 4
b = 180 / a
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a ) 71.6 , b ) 70.6 , c ) 80.6 , d ) 60.6 , e ) 50.6 | b | divide(add(add(add(56, 60), add(72, 85)), 80), add(const_2, const_3)) | calculate the average marks of a student who obtained 56 , 60 , 72 , 85 and 80 marks ( out of 100 ) in geography , history and government , art , computer science and modern literature ? | explanation : average = ( 56 + 60 + 72 + 85 + 80 ) / 5 = 70.6 . answer : b ) 70.6 | a = 56 + 60
b = 72 + 85
c = a + b
d = c + 80
e = 2 + 3
f = d / e
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a ) 100 , b ) 102 , c ) 103 , d ) 150 , e ) 46 | c | add(multiply(subtract(21, const_1), 5), 3) | find the 21 th term of an arithmetic progression whose first term is 3 and the common difference is 5 . | "n th term of a . p = a + ( n - 1 ) * d = 3 + ( 21 - 1 ) * 5 , = 3 + 100 = 103 . answer : c" | a = 21 - 1
b = a * 5
c = b + 3
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a ) 9 / 4 , b ) 12 / 5 , c ) 16 / 5 , d ) 15 / 4 , e ) 21 / 4 | e | divide(7, 6) | in δ pqs above , if pq = 6 and ps = 7 , then | "there are two ways to calculate area of pqs . area remains same , so both are equal . 6 * 7 / 2 = pr * 8 / 2 pr = 21 / 4 e" | a = 7 / 6
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a ) 20 days . , b ) 30 days . , c ) 25 days . , d ) 28 days . , e ) 45 days . | e | multiply(30, divide(75, 25)) | it was calculated that 75 men could complete a piece of work in 30 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "one day work = 1 / 30 one man ’ s one day work = 1 / ( 30 * 75 ) now : no . of workers = 50 one day work = 50 * 1 / ( 30 * 75 ) the total no . of days required to complete the work = ( 75 * 30 ) / 50 = 45 answer : e" | a = 75 / 25
b = 30 * a
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a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31 | e | divide(6, add(divide(20, const_100), 6)) | a committee is reviewing a total of 20 x black - and - white films and 6 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 w of the selected films are in color ? | "it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose w 60 / 62 = 30 / 31 of the selected films . answer : e ." | a = 20 / 100
b = a + 6
c = 6 / b
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a ) 367 , b ) 378 , c ) 365 , d ) 26 , e ) 28 | a | multiply(multiply(const_pi, 9), 13) | the slant height of a cone is 13 cm and radius of the base is 9 cm , find the curved surface of the cone ? | "π * 13 * 9 = 367 answer : a" | a = math.pi * 9
b = a * 13
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a ) 200 , b ) 400 , c ) 300 , d ) 450 , e ) 500 | b | divide(480, add(const_1, divide(20, const_100))) | a number increased by 20 % gives 480 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 480 ( 120 × 4 = 480 ) 100 % - - - - - - - > 400 ( 100 × 4 = 400 ) b )" | a = 20 / 100
b = 1 + a
c = 480 / b
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a ) s 200 , b ) s 1000 , c ) s 300 , d ) s 450 , e ) s 550 | b | divide(400, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4))) | a person spends 1 / 3 rd of the money with him on clothes , 1 / 5 th of the remaining on food and 1 / 4 th of the remaining on travel . now , he is left with rs 400 . how much did he have with him in the beginning ? | "suppose the amount in the beginning was rs ’ x ’ money spent on clothes = rs 1 x / 3 balance = rs 2 x / 3 money spent on food = 1 / 5 of 2 x / 3 = rs 2 x / 15 balance = 2 x / 3 - 2 x / 15 = rs 8 x / 15 money spent on travel = 1 / 4 of 8 x / 15 = rs 2 x / 15 = 8 x / 15 - 2 x / 15 = 6 x / 15 = rs 2 x / 5 therefore 2 x / 5 = 400 = 1000 answer : b" | a = 1 / 3
b = 1 - a
c = 1 / 3
d = 1 - c
e = d / 5
f = b - e
g = 1 / 3
h = 1 - g
i = 1 / 3
j = 1 - i
k = j / 5
l = h - k
m = l / 4
n = f - m
o = 400 / n
|
a ) 4.8 sec , b ) 5.9 sec , c ) 5.0 sec , d ) 5.4 sec , e ) 5.8 sec | e | divide(200, multiply(124, const_0_2778)) | in what time will a train 200 metres long cross an electric pole , if its speed be 124 km / hr ? | "solution speed = ( 124 x 5 / 18 ) m / sec = 34.44 m / sec time taken = ( 200 / 34.44 ) sec = 5.8 sec . answer e" | a = 124 * const_0_2778
b = 200 / a
|
a ) 338 , b ) 278 , c ) 322 , d ) 368 , e ) 121 | d | multiply(23, 16) | the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 15 and 16 . the larger of the two numbers is : | clearly , the numbers are ( 23 * 15 ) and ( 23 * 16 ) . larger number = ( 23 * 16 ) = 368 . answer : d | a = 23 * 16
|
a ) 6 , b ) 3 , c ) 4.2 , d ) 9 , e ) 4 | c | add(divide(multiply(add(45, const_1), 60), const_1000), 1.5) | a train of 45 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 45 * 60 + 1500 = 4200 m t = 4200 / 60 * 18 / 5 = 252 sec = 4.2 mins answer : c" | a = 45 + 1
b = a * 60
c = b / 1000
d = c + 1
|
a ) 343 , b ) 377 , c ) 750 , d ) 367 , e ) 232 | c | divide(add(212, 13), divide(30, const_100)) | mike needs 30 % to pass . if he scored 212 marks and falls short by 13 marks , what was the maximum marks he could have got ? | "if mike had scored 13 marks more , he could have scored 30 % therefore , mike required 212 + 13 = 225 marks let the maximum marks be m . then 30 % of m = 225 ( 30 / 100 ) × m = 225 m = ( 225 × 100 ) / 30 m = 22500 / 30 m = 750 answer : c" | a = 212 + 13
b = 30 / 100
c = a / b
|
a ) $ 13,360 , b ) $ 14,450 , c ) $ 15,720 , d ) $ 16,780 , e ) $ 17,240 | a | multiply(multiply(2, const_3), const_100) | a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 7,000 meters of fabric ? | "for first 2000 meters he does not get any discount . the price is 2 * 2000 = $ 4000 for next 1500 meters , he gets a 5 % discount . the price is 1.9 * 1500 = $ 2850 for the next 1500 meters , he gets a 7 % discount . the price is 1.86 * 3500 = $ 6510 the total price is $ 4000 + $ 2850 + $ 6510 = $ 13,360 the answer is a ." | a = 2 * 3
b = a * 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | subtract(100, subtract(196, 100)) | when 100 is divided by positive integer x , the remainder is 4 . what is the remainder when 196 is divided by x ? | "f 100 / x leaves a reminder 4 then ( 100 - 4 ) i . e . 96 is divisible by x so ( 100 + 96 ) / x leaves a reminder rem ( 100 / x ) + rem ( 96 / x ) = > 4 + 0 = 4 answer : c" | a = 196 - 100
b = 100 - a
|
a ) 72 % , b ) 45 % , c ) 32 % , d ) 12 % , e ) 22 % | b | subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 50), const_100))) | a fair price shopkeeper takes 10 % profit on his goods . he lost 50 % goods during theft . his loss percent is : | "explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 50 . s . p . of each item = rs . 1.10 total sale = 1.10 * 50 = rs . 55 hence , loss % = 45 / 100 * 100 = 45 % answer : b" | a = 100 + 10
b = 100 + 10
c = b * 50
d = c / 100
e = a - d
f = 100 - e
|
a ) s . 750 , b ) s . 1000 , c ) s . 1250 , d ) s . 1500 , e ) s . 1750 | b | multiply(divide(divide(120, 4), 3), const_100) | a sum was put a simple interest at a certain rate for 4 years . had it been put at 3 % higher rate , it would have fetched rs . 120 more . the sum is : | explanation : let the sub be rs . x and the initial rate be r % . then x ã — ( r + 3 ) ã — 4 / 100 â ˆ ’ x ã — r ã — 4 / 100 = 120 â ‡ ’ x ã — 3 ã — 4 / 100 = 120 â ‡ ’ x ã — 4 / 100 = 40 â ‡ ’ 4 x = 4000 â ‡ ’ x = 1000 answer : option b | a = 120 / 4
b = a / 3
c = b * 100
|
a ) 7 , b ) 14 , c ) 49 , d ) 21 , e ) none of these | b | sqrt(power(14, 2)) | √ ( 14 ) ^ 2 | "explanation √ ( 14 ) ^ 2 = ? or , ? = 14 answer b" | a = 14 ** 2
b = math.sqrt(a)
|
a ) 144 , b ) 119 , c ) 113 , d ) 117 , e ) 31 | d | subtract(119, subtract(add(144, 119), 236)) | in a graduating class of 236 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 236 = 27 greatest - smallest is 144 - 27 = 117 so answer must be d . 117" | a = 144 + 119
b = a - 236
c = 119 - b
|
a ) 1040 , b ) 1050 , c ) 1055 , d ) 1065 , e ) 1075 | c | add(multiply(8, 70), multiply(9, 55)) | bruce purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 560 + 490 = 1055 c" | a = 8 * 70
b = 9 * 55
c = a + b
|
a ) 12 , b ) 14 , c ) 15 , d ) 24 , e ) 36 | d | multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3)) | initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many r women are currently in the room ? | "the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d" | a = 14 - 2
b = 5 * a
c = b / 4
d = c - 3
e = 2 * d
|
a ) 0 , b ) 25 , c ) 50 , d ) 30 , e ) 40 | b | subtract(const_100, add(50, divide(multiply(50, 50), const_100))) | paul ' s salary was decreased by 50 % and subsequently increased by 50 % . how much percent does he lose ? | let the original salary = $ 100 new final salary = 150 % of ( 50 % of rs . 100 ) = $ ( ( 150 / 100 ) * ( 50 / 100 ) * 100 ) = $ 75 . decrease = 25 % answer option b ) 25 . | a = 50 * 50
b = a / 100
c = 50 + b
d = 100 - c
|
a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21 | b | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3)) | a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction u of the sum of the 21 numbers in the list ? | "this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction u of n to the total would be 4 a / 24 a or 1 / 6 answer b" | a = 1 * 1
b = 20 / 4
c = b + 21
d = c / 4
e = d * 2
f = e - 4
g = f - 3
h = a / g
|
a ) 7.25 , b ) 9.47 , c ) 11.58 , d ) 13.64 , e ) 15.92 | d | divide(divide(80, 5280), multiply(4, divide(1, const_3600))) | if an object travels 80 feet in 4 seconds , what is the object ’ s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "80 feet / 4 seconds = 20 feet / second ( 20 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 13.64 miles / hour ( approximately ) the answer is d ." | a = 80 / 5280
b = 1 / 3600
c = 4 * b
d = a / c
|
a ) 12 , b ) 15 , c ) 17 , d ) t = 18 , e ) 20 | d | multiply(multiply(3, const_2), 3) | two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer t between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | "since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 11,12 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11,13 ) , ( 7,17 ) , ( 5 , 19 ) 2 , 3 , 5 , 7 , 11 , 13 , 15,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 13 , 17 ) , ( 11 , 19 ) , ( 7 , 23 ) 2 , 3 , 5 , 7 , 11 , 13,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11 , 23 ) , ( 5 , 29 ) , ( 3 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 18,19 , 23 , 29 , 31 , 37 - four pairs ( 17 , 19 ) , ( 13 , 23 ) , ( 7 , 29 ) , ( 5 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 20,23 , 29 , 31 , 37 - definitely can not be more than 4 since there are only 4 primes more than 20 . so must be less than 4 pairs . ignore . answer ( d ) ." | a = 3 * 2
b = a * 3
|
a ) 3 , b ) 7 , c ) 11 , d ) 13 , e ) 79 | d | add(add(divide(add(multiply(42, 2), const_1), 6300), const_3), const_3) | what is the greatest prime factor of 42 ^ 2 + 75 ^ 2 + 6300 | "42 ^ 2 = ( 50 - 8 ) ( 50 - 8 ) = 1764 75 ^ 2 = ( 100 - 25 ) ( 100 - 25 ) = 5625 1764 + 5625 + 6300 = 13689 this number is divisible by 3 and 9 divide by 9 and get 1521 , the sum of this number is again a multiple of 9 , thus we can divide it one more time by 9 . we get 169 , and 169 is 13 ^ 2 . thus the answer is 13 . answer : d" | a = 42 * 2
b = a + 1
c = b / 6300
d = c + 3
e = d + 3
|
a ) 23 , b ) 27 , c ) 36 , d ) 27 , e ) 11 | c | divide(multiply(30, 12), 10) | 12 men can complete a piece of work in 30 days . in how many days can 10 men complete that piece of work ? | "12 * 30 = 10 * x = > x = 36 days answer : c" | a = 30 * 12
b = a / 10
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 32 | e | multiply(multiply(add(4, divide(subtract(sqrt(80), 4), 2)), divide(subtract(sqrt(80), 4), 2)), 2) | if a - b = 4 and a ^ 2 + b ^ 2 = 80 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 80 - 16 = 64 = > ab = 32 answer : e" | a = math.sqrt(80)
b = a - 4
c = b / 2
d = 4 + c
e = math.sqrt(80)
f = e - 4
g = f / 2
h = d * g
i = h * 2
|
a ) 56 minutes , b ) 52 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | b | divide(multiply(4.20, const_60), 5) | the timing of a college is from 11 p . m to 4.20 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | "explanation : total time a student spends in college = 5 hours 20 minutes = 280 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 280 â € “ 20 ) / 5 = 52 minutes answer b" | a = 4 * 20
b = a / 5
|
a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(sqrt(divide(subtract(power(10, const_2), power(const_2, const_2)), const_2)), const_2) | efrida and frazer who live 10 miles apart , meet at a restaurant that is directly north of efrida ' s home and directly east of frazer ' s home . if the restaurant is three miles closer to efrida ' s home , than to frazer ' s home , how many miles is the restaurant from frazer ' s home ? | "it ' s a simple geometry problem . look at the diagram below : answer : b ." | a = 10 ** 2
b = 2 ** 2
c = a - b
d = c / 2
e = math.sqrt(d)
f = e + 2
|
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