Datasets:
Cheukting
/
mathqa

Dataset card Data Studio Files
xet
 Community
1
Problem
stringlengths
5
967
Rationale
stringlengths
1
2.74k
options
stringlengths
37
300
correct
stringclasses
5 values
annotated_formula
stringlengths
7
6.48k
linear_formula
stringlengths
8
925
category
stringclasses
6 values
12.5 * 5.2 * 8.8 = ?
"e 572 ? = 12.5 * 5.2 * 8.8 = 572"
a ) 352 , b ) 435 , c ) 224 , d ) 646 , e ) 572
e
multiply(12.5, 5.2)
multiply(n0,n1)|
general
if x + ( 1 / x ) = 5 , what is the value of t = x ^ 2 + ( 1 / x ) ^ 2 ?
"squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c"
a ) t = 21 , b ) t = 22 , c ) t = 23 , d ) 24 , e ) 27
c
subtract(power(5, 2), 2)
power(n1,n2)|subtract(#0,n2)|
general
a certain company had a total annual expenditure of 2.1 ∗ 10 ^ 7 on employee salaries last year . if the company employed 420 people , what was the average employee salary ?
"given : total annual expenditure of 2.1 ∗ 10 ^ 7 on employee salaries total employees = 420 observe that 420 * 5 = 2100 therefore try to bring the numerator in terms of 2100 average salary = ( 2100 * 10 ^ 4 ) / 420 = 5 * 10 ^ 4 = 50,000 option e"
a ) $ 20,000 , b ) $ 25,000 , c ) $ 35,000 , d ) $ 40,000 , e ) $ 50,000
e
multiply(divide(multiply(2.1, multiply(10, 7)), 420), const_100)
multiply(n1,n2)|multiply(n0,#0)|divide(#1,n3)|multiply(#2,const_100)|
general
on dividing 181 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?
"d = ( d - r ) / q = ( 181 - 1 ) / 9 = 180 / 9 = 20 e )"
a ) a ) 12 , b ) b ) 15 , c ) c ) 16 , d ) d ) 17 , e ) e ) 20
e
floor(divide(181, 9))
divide(n0,n1)|floor(#0)|
general
the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 9 . the true discount is
"solution t . d = [ b . g x 100 / r x t ] = rs . ( 9 x 100 / 12 x 1 ) = rs . 75 . answer a"
a ) rs . 75 , b ) rs . 36 , c ) rs . 54 , d ) rs . 50 , e ) none
a
divide(9, divide(12, const_100))
divide(n1,const_100)|divide(n2,#0)|
gain
a farmer spent $ 36 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have s...
"a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 36 = $ 14.4 on chicken feed , thus he spent the remaining 36 - 14.4 = $ 21.6 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 14.4 - - > x = $ 18 therefore if the farmer had paid full price...
a ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.60 , e ) $ 40.60
d
add(multiply(36, divide(40, const_100)), 36)
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
a garrison of 300 men had a provision for 31 days . after 28 days 280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ?
"400 - - - 31 400 - - - 3 120 - - - ? 300 * 3 = 20 * x = > x = 45 days answer : a"
a ) 45 days , b ) 55 days , c ) 35 days , d ) 25 days , e ) 40 days
a
divide(multiply(subtract(31, 28), 300), 280)
subtract(n1,n2)|multiply(n0,#0)|divide(#1,n3)|
other
what is the speed of the stream if a canoe rows upstream at 6 km / hr and downstream at 10 km / hr
sol . speed of stream = 1 / 2 ( 10 - 6 ) kmph = 2 kmph . answer d
a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph
d
divide(subtract(10, 6), const_2)
subtract(n1,n0)|divide(#0,const_2)
physics
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 25 % . what was its profit on the items sold in february ?
"assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.75 * 150 x = 112.5 x hence total profit = 12.5 % option a"
a ) 12.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 40 %
a
subtract(multiply(divide(subtract(const_100, 25), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)|
gain
a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ?
"explanation : sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 5 = rs . 100 . answer : d"
a ) 21 , b ) 28 , c ) 90 , d ) 100 , e ) 11
d
subtract(divide(8925, 85), 5)
divide(n1,n0)|subtract(#0,n2)|
physics
the length of a rectangle is halved , while its breadth is tripled . wa tis the % changein area ?
let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c
['a ) 34 %', 'b ) 45 %', 'c ) 50 %', 'd ) 60 %', 'e ) 67 %']
c
multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100)
divide(const_1,const_2)|multiply(#0,const_3)|subtract(#1,const_1)|divide(#2,const_1)|multiply(#3,const_100)
geometry
if k ^ 3 is divisible by 84 , what is the least possible value of integer k ?
"k ^ 3 = 84 * x = 2 ^ 2 * 3 * 7 * x the factors of k must at minimum include 2 * 3 * 7 = 42 . the answer is b ."
a ) 36 , b ) 42 , c ) 48 , d ) 56 , e ) 60
b
divide(divide(84, const_2), const_2)
divide(n1,const_2)|divide(#0,const_2)|
general
the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 5 hrs if the speed of the bike is twice speed of the tractor ?
"sol . average speed of a tractor = 25 km / h the speed of a bike in an hour = 25 × 2 = 50 km the speed of a car in an hour = 9 / 5 * 50 = 90 km so , the distance covered by car in 5 h is 90 × 5 = 450 km ans . ( a )"
a ) 450 km , b ) 500 km , c ) 360 km , d ) 550 km , e ) 600 km
a
multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 5)
divide(n1,n2)|divide(n3,n4)|add(n0,#0)|multiply(#1,const_2)|multiply(#2,#3)|multiply(n5,#4)|
general
a train 650 m long is running at a speed of 117 km / hr . in what time will it pass a bridge 325 m long ?
"speed = 117 * 5 / 18 = 65 / 2 m / sec total distance covered = 650 + 325 = 975 m required time = 975 * 2 / 65 = 30 sec answer : a"
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50
a
divide(650, multiply(subtract(117, 325), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 363 sq m , then what is the breadth of the rectangular plot ?
"let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 363 3 b 2 = 363 b 2 = 121 = 11 ( b > 0 ) b = 11 m . answer : a"
a ) 11 , b ) 17 , c ) 18 , d ) 101 , e ) 1322
a
sqrt(divide(363, const_3))
divide(n0,const_3)|sqrt(#0)|
geometry
in a class of 40 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ?
"answer let a be the set of students who have enrolled for english and b be the set of students who have enrolled for german . then , ( a u b ) is the set of students who have enrolled at least one of the two subjects . as the students of the class have enrolled for at least one of the two subjects , a u b = 40 we know...
a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32
c
subtract(subtract(add(40, 12), 22), 12)
add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)|
other
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 minutes , then the area of the park ( in sq . m ) is :
"explanation : question seems to be typical , but trust me it is too easy to solve , before solving this , lets analyse how we can solve this . we are having speed and time so we can calculate the distance or perimeter in this question . then by applying the formula of perimeter of rectangle we can get value of length ...
a ) 152600 m 2 , b ) 153500 m 2 , c ) 153600 m 2 , d ) 153800 m 2 , e ) none of these
c
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2))
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
physics
each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 98 % water by weight . what is the new weight of the cucumbers , in pounds ?
"out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after some water evaporates the cucumbers become 98 % water and 2 % of non - water , so now 1 pound of non - water composes 2 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.02 = 50 pounds . answer : b ."
a ) 2 , b ) 50 , c ) 92 , d ) 96 , e ) 98
b
multiply(divide(subtract(100, 99), subtract(100, 98)), 100)
subtract(n0,n1)|subtract(n0,n2)|divide(#0,#1)|multiply(#2,n0)|
gain
how much 60 % of 50 is greater than 45 % of 30 ?
"( 60 / 100 ) * 50 – ( 45 / 100 ) * 30 30 - 13.5 = 16.5 answer : c"
a ) 18 , b ) 91 , c ) 16.5 , d ) 17 , e ) 12
c
subtract(divide(multiply(60, 50), const_100), divide(multiply(45, 30), const_100))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)|
gain
rectangular tile each of size 50 cm by 40 cm must be laid horizontally on a rectangular floor of size 120 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f...
"area of tile = 50 * 40 = 2000 area of floor = 120 * 150 = 18000 no of tiles = 18000 / 2000 = 9 so , the no of tile = 9 answer : a"
a ) 9 , b ) 15 , c ) 12 , d ) 11 , e ) 10
a
divide(multiply(120, 150), multiply(50, 40))
multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)|
geometry
it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 60 pages ?
"if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b ...
a ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 30
b
divide(multiply(60, add(multiply(4, const_2), 4)), 40)
multiply(n0,const_2)|add(n0,#0)|multiply(n4,#1)|divide(#2,n1)|
physics
two trains of length 200 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ?
"relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 200 + 280 = 480 m . the time required = d / s = 480 / 20 = 24 sec . answer : e"
a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 24
e
divide(add(200, 280), multiply(add(42, 30), const_0_2778))
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
a woodworker normally makes a certain number of parts in 24 days . but he was able to increase his productivity by 5 parts per day , and so he not only finished the job in only 22 days but also he made 80 extra parts . how many parts does the woodworker normally makes per day and how many pieces does he make in 24 days...
let x be the number of parts the woodworker normally makes daily . in 24 days he makes 24 ⋅ x pieces . his new daily production rate is x + 5 pieces and in 22 days he made 22 ⋅ ( x + 5 ) parts . this is 80 more than 24 ⋅ x . therefore the equation is : 24 ⋅ x + 80 = 22 ( x + 5 ) 30 = 2 x x = 15 normally he makes 15 par...
a ) 360 , b ) 350 , c ) 340 , d ) 370 , e ) 380
a
multiply(divide(subtract(multiply(22, 5), 80), subtract(24, 22)), 24)
multiply(n1,n2)|subtract(n0,n2)|subtract(#0,n3)|divide(#2,#1)|multiply(n0,#3)
general
julian owes his classmate jenny 20 dollars . if he borrows 8 dollars how much will he owe her ?
add how he has already borrowed and how much he is going to borrow - 20 + - 8 correct answer c ) - 28
a ) 28 , b ) 12 , c ) - 28 , d ) 12 , e ) 0
c
add(20, 8)
add(n0,n1)
general
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 20 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 10 % . if ron imported a $ 20,000 imported car and...
"let t be the tier price , p be total price = 20000 per the given conditions : 0.20 t + 0.10 ( p - t ) = 2500 0.20 t + 0.10 * 20000 - 0.10 t = 2500 0.10 t + 2000 = 2500 0.10 t = 2500 - 2000 = 500 t = 500 / 0.10 = 5000 answer c"
a ) $ 6000 , b ) $ 5500 , c ) $ 5000 , d ) $ 5700 , e ) $ 5800
c
divide(subtract(2500, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(10, const_100))), subtract(divide(20, const_100), divide(10, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)|
general
the sum of ages of 5 children born at the intervals of 8 years each is 90 years . what is the age of the youngest child ?
"let the ages of children be x , ( x + 8 ) , ( x + 16 ) , ( x + 24 ) and ( x + 32 ) years . then , x + ( x + 8 ) + ( x + 16 ) + ( x + 24 ) + ( x + 32 ) = 90 5 x = 10 x = 2 . age of the youngest child = x = 2 years . answer : a"
a ) 2 , b ) 9 , c ) 6 , d ) 4 , e ) 0
a
subtract(subtract(divide(90, 5), 8), 8)
divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|
general
if a and b are the roots of the equation x 2 - 9 x + 20 = 0 , find the value of a 2 + b 2 + ab ?
"a 2 + b 2 + ab = a 2 + b 2 + 2 ab - ab i . e . , ( a + b ) 2 - ab from x 2 - 9 x + 20 = 0 , we have a + b = 9 and ab = 20 . hence the value of required expression ( 9 ) 2 - 20 = 61 . option c"
a ) 57 , b ) 59 , c ) 61 , d ) 63 , e ) 65
c
add(power(divide(subtract(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2), power(divide(add(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2))
multiply(n1,const_4)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|add(n1,#3)|subtract(n1,#3)|divide(#5,n0)|divide(#4,n0)|power(#6,n0)|power(#7,n0)|add(#8,#9)|
general
for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 96 θ 33 ) θ 17 ) - ( 94 θ ( 33 θ 17 ) ) ?
"( ( 96 θ 33 ) θ 17 ) the remainder of 96 divided by 33 is 30 ; the remainder of 30 divided by 17 is 13 ; ( 97 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 94 divided by 16 is 1 . 13 - 2 = 11 . answer : d ."
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
d
subtract(reminder(reminder(96, 33), 17), reminder(94, reminder(33, 17)))
reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3)|
general
a man bought 40 shares of rs . 60 at 6 discount , the rate of dividend being 12 1 / 2 % the rate of interest obtained is
"explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 6 = rs . 54 number of shares = 40 dividend = 12 1 / 2 % = 25 / 2 % dividend per share = 60 × 25 / 2 × 100 = rs . 7.5 total dividend = ( 40 × 7.5 ) ie , he got a dividend of ( 40 × 7.5 ) for an investment of rs . ( 40 × 54 ) interest o...
a ) 13.89 % , b ) 15.5 % , c ) 14 % , d ) 14.25 % , e ) 14.95 %
a
divide(multiply(multiply(40, 60), divide(12, const_100)), multiply(40, subtract(60, 6)))
divide(n3,const_100)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#1)|multiply(n0,#2)|divide(#3,#4)|
gain
the unit digit in the product ( 224 * 607 * 214 * 863 ) is :
"explanation : unit digit in the given product = unit digit in ( 4 * 7 * 4 * 3 ) = 6 answer : c"
a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10
c
subtract(multiply(multiply(multiply(224, 607), 214), 863), subtract(multiply(multiply(multiply(224, 607), 214), 863), add(const_4, const_4)))
add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)|
general
a number is increased by 70 % and then decreased by 70 % . find the net increase or decrease per cent .
let the number be 100 . increase in the number = 70 % = 70 % of 100 = ( 70 / 100 ã — 100 ) = 70 therefore , increased number = 100 + 70 = 170 this number is decreased by 70 % therefore , decrease in number = 70 % of 170 = ( 70 / 100 ã — 170 ) = 11900 / 100 = 119 therefore , new number = 170 - 119 = 51 thus , net decrea...
a ) 49 % , b ) 18 % , c ) 17 % , d ) 13 % , e ) 16 %
a
divide(multiply(add(70, const_100), subtract(const_100, 70)), const_100)
add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)
general
a train which has 400 m long , is running 50 kmph . in what time will it cross a person moving at 14 kmph in same direction ?
"time taken to cross a moving person = length of train / relative speed time taken = 400 / ( ( 50 - 14 ) ( 5 / 18 ) = 400 / 36 * ( 5 / 18 ) = 400 / 10 = 40 sec answer : a"
a ) 40 sec , b ) 35 sec , c ) 33 sec , d ) 42 sec , e ) 41 sec
a
divide(400, subtract(divide(50, const_3_6), divide(divide(14, const_2), const_3_6)))
divide(n1,const_3_6)|divide(n2,const_2)|divide(#1,const_3_6)|subtract(#0,#2)|divide(n0,#3)|
physics
how many even 4 - digit integers greater than 8000 are there such that all their digits are different ?
case 1 : numbers starting with 8 followed by even number the hundreds digit can take 4 numbers ( 0,2 , 4,6 ) so 4 ways the unit digit can take remaining 3 even numbers left after using 2 , to be even so 3 ways . the tens digit can take remaining 7 numbers left after using 3 numbers so 7 ways total = 4 * 3 * 7 = 84 case...
a ) 240 , b ) 369 , c ) 469 , d ) 520 , e ) 599
c
subtract(add(multiply(divide(factorial(multiply(4, const_2)), factorial(subtract(multiply(4, const_2), const_2))), 4), multiply(add(4, const_1), divide(factorial(multiply(4, const_2)), factorial(subtract(multiply(4, const_2), const_2))))), multiply(add(4, const_1), add(const_3, const_4)))
add(n0,const_1)|add(const_3,const_4)|multiply(n0,const_2)|factorial(#2)|multiply(#0,#1)|subtract(#2,const_2)|factorial(#5)|divide(#3,#6)|multiply(n0,#7)|multiply(#0,#7)|add(#8,#9)|subtract(#10,#4)
general
a certain farmer pays $ 60 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 1210 feet ? ( 43,560 square feet = 1 acre )
basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 600 ( d )
a ) $ 5,330 , b ) $ 3,360 , c ) $ 1,350 , d ) $ 600 , e ) $ 150
d
multiply(60, divide(multiply(360, 1210), divide(multiply(360, 1210), const_10)))
multiply(n1,n2)|divide(#0,const_10)|divide(#0,#1)|multiply(n0,#2)|
geometry
today joelle opened an interest - bearing savings account and deposited $ 6,000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ?
interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d
a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315
d
add(add(multiply(multiply(multiply(const_3, const_2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, const_2), const_100), const_10), 5), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, const_2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, ...
multiply(const_2,const_3)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)
gain
what is x if x + 2 y = 100 and y = 25 ?
"substitute y by 25 in x + 2 y = 100 x + 2 ( 25 ) = 100 x + 50 = 100 if we substitute x by 50 in x + 50 = 100 , we have 50 + 50 = 100 . hence x = 50 correct answer b"
a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 4
b
subtract(100, multiply(2, 25))
multiply(n0,n2)|subtract(n1,#0)|
general
a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 3 days . in how many days b alone can do the work ?
"explanation : a ’ s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 3 x = 4.5 days answer : option d"
a ) 5 days , b ) 7 days , c ) 12 days , d ) 4.5 days , e ) 10 days
d
inverse(multiply(inverse(3), subtract(const_1, multiply(5, inverse(15)))))
inverse(n2)|inverse(n0)|multiply(n1,#1)|subtract(const_1,#2)|multiply(#0,#3)|inverse(#4)|
physics
a tiger is 50 of its own leaps behind a deer . the tiger takes 5 leaps per minutes to deer ’ s 4 . if the tiger and the deer cover 8 m and 5 m per leap respectively , what distance will the tiger have to run before it catches the deer ?
tiger takes 5 leaps / m and deer takes 4 leaps / m thus , tiger runs at the speed of 5 x 8 = 40 m / min and deer runs at the speed of 4 x 5 = 20 m / min tiger is 50 leaps behind , which is 50 x 8 = 400 m relative speed of tiger and deer is = 20 m / min thus , 400 / 20 = 20 m thus , the tiger will catch deer = 20 x 40 =...
a ) 600 m , b ) 700 m , c ) 800 m , d ) 1000 m , e ) 1200 m
c
multiply(divide(multiply(50, 8), subtract(multiply(8, 5), multiply(4, 5))), multiply(8, 5))
multiply(n0,n3)|multiply(n1,n3)|multiply(n1,n2)|subtract(#1,#2)|divide(#0,#3)|multiply(#4,#1)
physics
a man can row upstream at 25 kmph and downstream at 39 kmph , and then find the speed of the man in still water ?
"us = 25 ds = 39 m = ( 39 + 25 ) / 2 = 32 answer : c"
a ) 22 , b ) 37 , c ) 32 , d ) 27 , e ) 18
c
divide(add(25, 39), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
if the sum of two numbers is 60 and the h . c . f and l . c . m of these numbers are 3 and 120 respectively , then the sum of the reciprocal of the numbers is equal to :
"let the numbers be a and b . then , a + b = 60 and ab = 3 * 120 = 360 . required sum = 1 / a + 1 / b = ( a + b ) / ab = 60 / 360 = 1 / 6 . answer : c"
a ) 2 / 9 , b ) 4 / 5 , c ) 1 / 6 , d ) 6 / 6 , e ) 3 / 9
c
divide(add(3, 120), multiply(3, 120))
add(n1,n2)|multiply(n1,n2)|divide(#0,#1)|
general
last year department store x had a sales total for december that was 2 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ?
"let avg for 11 mos . = 10 therefore , dec = 20 year total = 11 * 10 + 20 = 130 answer = 20 / 130 = 2 / 13 = c"
a ) 1 / 4 , b ) 4 / 15 , c ) 2 / 13 , d ) 4 / 11 , e ) 4 / 5
c
divide(2, add(subtract(const_12, const_1), 2))
subtract(const_12,const_1)|add(n0,#0)|divide(n0,#1)|
general
a worker earns $ 24 on the first day and spends $ 18 on the second day . the worker earns $ 24 on the third day and spends $ 18 on the fourth day . if this pattern continues , on which day will the worker first reach a net total of $ 72 ?
"every two days , the net total is $ 6 . after 16 days , the worker will have $ 48 . on day 17 , the worker will receive $ 24 for a net total of $ 72 . the answer is c ."
a ) 9 , b ) 12 , c ) 17 , d ) 21 , e ) 24
c
add(multiply(divide(24, subtract(24, 18)), const_2), const_1)
subtract(n0,n1)|divide(n0,#0)|multiply(#1,const_2)|add(#2,const_1)|
physics
in a class of students , 2 / 3 of the number of girls is equal to 1 / 5 of the total number of students . what is the ratio of boys to girls in the class ?
( 2 / 3 ) g = ( 1 / 5 ) ( b + g ) 10 g = 3 b + 3 g 7 g = 3 b b / g = 7 / 3 . the answer is e .
a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3
e
divide(subtract(divide(2, 3), divide(1, 5)), divide(1, 5))
divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#1)
general
jaya two digits age when appended with the two digits age of amitabh the 4 digits number formed is a perfect square . what is the the sum of the 4 digits no . ?
min no could be = 7 ( 1024 ) let jaya has two digits number = 10 and amitabh has two digits number = 24 after appending = 1024 which is perfect square of 32 and sum of four digits number 1024 = 7 answer : b
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
b
add(add(const_1, const_2), 4)
add(const_1,const_2)|add(n0,#0)
general
pure milk costs 3.60 per litre . a milkman adds water to 25 litres of pure milk and sells the mixture at 3 per litre . how many litres of water does he add ?
in mixture , quantity of pure milk / quantity of water = 3 − 0 / 3.6 − 3 = 3 / 0.6 = 5 / 1 since in every 5 litres of milk , he adds 1 litre of water . ∴ in every 25 litres of milk , he adds 5 litres of water . answer b
a ) 2 litres , b ) 5 litres , c ) 7 litres , d ) 11 litres , e ) none of these
b
divide(subtract(multiply(25, 3.6), multiply(25, 3)), 3)
multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n2)
general
machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 12000 envelopes .
for 1 hour - machine a rate - 2000 envelopes machine b + c rate - 2400 envelopes since a + c = 3000 envelopes a ' s rate is 2000 envelopes as above , c has a rate of 1000 envelopes per hour . which makes machine b ' s rate as 1400 envelopes per hour . thus , it will take 8 hours to manufacture 12000 envelopes . answer ...
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 60 / 7
e
divide(12000, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3))))
divide(n0,n2)|divide(n3,n4)|divide(n0,n1)|subtract(#1,#2)|subtract(#0,#3)|divide(n5,#4)
physics
what is 2 4 / 3 - 3 1 / 4 divided by 2 / 3 - 1 / 6 ?
"2 4 / 3 - 3 1 / 4 = 10 / 3 - 13 / 4 = ( 40 - 39 ) / 12 = 1 / 12 2 / 3 - 1 / 6 = ( 4 - 1 ) / 6 = 3 / 6 = 1 / 2 so 1 / 12 / 1 / 2 = 2 / 12 = 1 / 6 answer - c"
a ) 5 / 4 , b ) 2 / 3 , c ) 1 / 6 , d ) 5 / 2 , e ) 1 / 8
c
subtract(divide(add(multiply(const_10, 2), 2), 3), divide(add(const_10, 3), 4))
add(n3,const_10)|multiply(const_10,n0)|add(n0,#1)|divide(#0,n5)|divide(#2,n2)|subtract(#4,#3)|
general
the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 6 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ?
"a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 6 = 74 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 74 + 4 = 78 d"
a ) 45 , b ) 65 , c ) 75 , d ) 78 , e ) 90
d
subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(const_3.0, subtract(multiply(80, const_4), multiply(84, 6)))))
multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,const_3)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5)|
general
the length of the bridge , which a train 160 metres long and travelling at 45 km / hr can cross in 30 seconds , is ?
"speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 160 + x ) / 30 = 25 / 2 = > 2 ( 160 + x ) = 750 = > x = 215 m . answer : e"
a ) 255 , b ) 267 , c ) 245 , d ) 277 , e ) 215
e
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
32 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ?
"x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 32 = 14 answer : e"
a ) 66 , b ) 26 , c ) 42 , d ) 27 , e ) 14
e
divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 32), const_2)
add(const_3,const_4)|add(#0,const_2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_2)|
general
pipe a can fill a tank in 6 hours . due to a leak at the bottom , it takes 9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ?
"let the leak can empty the full tank in x hours 1 / 6 - 1 / x = 1 / 9 = > 1 / x = 1 / 6 - 1 / 9 = ( 3 - 2 ) / 18 = 1 / 18 = > x = 18 . answer : c"
a ) 33 , b ) 77 , c ) 18 , d ) 99 , e ) 66
c
divide(multiply(9, 6), subtract(9, 6))
multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|
physics
if x / y = 3 and ( 2 a - x ) / ( 5 b - y ) = 3 , then the value of a / b is ?
"x = 3 y 2 a - 3 y = 3 ( 5 b - y ) 2 a - 3 y = 15 b - 3 y 2 a = 15 b a / b = 15 / 2 answer : a"
a ) 15 / 2 , b ) - 2 , c ) 1 , d ) 2 , e ) 3
a
divide(multiply(3, 3), 2)
multiply(n0,n0)|divide(#0,n1)|
general
what could be the range of a set consisting of odd multiples of 7 ?
"a set of odd multiples of 7 would be something like 7 , 21 , 35 , 49 . . . . the range would always be an even number multiple of 7 . answer : e"
a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 70
e
multiply(7, const_4)
multiply(n0,const_4)|
general
a man invests some money partly in 4 % stock at 48 and partly in 10 % stock at 110 . to obtain equal dividends from both , he must invest the money in the ratio :
"solution for an income of rs . 1 in 9 % stock at 96 , investment = rs . ( 48 / 4 ) = rs . 12 . for an income of rs . 1 in 12 % stock at 120 , investment = rs . ( 110 / 10 ) = rs . 10 . ∴ ratio of investments = 12 : 10 = 6 : 5 answer d"
a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 5 , d ) 6 : 5 , e ) none
d
divide(multiply(48, const_2), multiply(110, const_3))
multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)|
other
what least number must be added to 1096 , so that the sum is completely divisible by 23 ?
"1096 / 23 = 47.6 - - that is 23 * 47 = 1081 now , 1096 - 1081 = 15 to get completely divided by 23 add 8 to 1096 ans - e"
a ) 2 , b ) 1 , c ) 3 , d ) 5 , e ) 8
e
subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1096)
multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)|
general
the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 5 : 4 . if the oil bill for february had been $ 45 more , the corresponding ratio would have been 3 : 2 . how much was the oil bill for january ?
5 : 4 = 15 : 12 and 3 : 2 = 18 : 12 . an increase in $ 45 increases the ratio by 3 : 12 . therefore an increase in $ 15 increases the ratio by 1 : 12 . therefore , january ' s bill was 12 ( $ 15 ) = $ 180 . the answer is e .
a ) $ 60 , b ) $ 90 , c ) $ 120 , d ) $ 150 , e ) $ 180
e
divide(45, subtract(divide(3, 2), divide(5, 4)))
divide(n3,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2)
general
trains a and b start simultaneously from stations 450 miles apart , and travel the same route toward each other on adjacent parallel tracks . if train a and train b travel at a constant rate of 50 miles per hour and 50 miles per hour , respectively , how many miles will train a have traveled when the trains pass each o...
"since we know the distance ( 450 ) and the combined rate ( 100 ) , we plug it into the formula : distance = rate * time 450 = 100 * time we can solve for the time they will meet cause we added the rate of train a and train b together . so the time will be 450 / 100 from dividing 100 on both sides to isolate time in th...
a ) 112 , b ) 133 , c ) 150 , d ) 167 , e ) 225
e
multiply(50, divide(450, add(50, 50)))
add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|
physics
a theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 80 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of dayti...
"deviation from the mean for the daytime staff = 80 - 64 = 16 . deviation from the mean for the evening staff = 64 - 60 = 4 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 4 : 16 = 1 : 4 . the answer is b ."
a ) 1 : 2 , b ) 1 : 4 , c ) 3 : 7 , d ) 7 : 12 , e ) 4 : 5
b
divide(subtract(64, 60), subtract(80, 64))
subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|
general
12 welders work at a constant rate they complete an order in 3 days . if after the first day , 9 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ?
1 . we need to find out the time taken by 3 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 12 * 3 = 36 days 4 . but on day 1 twelve workers had already worked finishing 1 / 3 of the job . so 3 workers have to finish only 2 / 3 of the job . 5 . tot...
a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 6
c
divide(divide(subtract(const_1, multiply(divide(const_1, multiply(12, 3)), 12)), subtract(12, 9)), divide(const_1, multiply(12, 3)))
multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)
physics
a rectangular cube has sides measuring 15 inches long by 12 inches wide by 2 inches high . if the surface area of the rectangle is the same as a cube , what do the sides / walls of the cube measure ? round to the nearest whole number .
first calculate the surface area of the rectangle by multiplying the length and width of the rectangle together , then multiply by 2 to get both sides of the rectangle . this calculates to 360 inches . find the surface area of the sides of the rectangular cube . multiply the height by the length of the rectangle . mult...
a ) 7 , b ) 5 , c ) 9 , d ) 6 , e ) 4
c
floor(divide(surface_rectangular_prism(const_3.0, 12, 2), multiply(12, 15)))
multiply(const_2,n0)|surface_rectangular_prism(n0,n1,n2)|divide(#1,#0)|floor(#2)|
geometry
what is the units digit of the solution to 177 ^ 28 − 133 ^ 23 ?
"let ' s reduce the clutter and make it easy to solve so , 7 ^ 28 - 3 ^ 23 will have the same units digit as the big numbers above both 7 and 3 have a cyclicity of 4 , i . e . their powers repeat the units digit after every 4 th power so , 7 ^ 28 has the same units digit as 7 ^ 4 , which is 1 similarly , 3 ^ 23 has the...
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9
c
add(add(const_4, const_3), const_2)
add(const_3,const_4)|add(#0,const_2)|
general
a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 3 % . what is the cost price ?
"90 % 103 % - - - - - - - - 13 % - - - - 140 100 % - - - - ? = > rs . 1077 answer : c"
a ) 1000 , b ) 2876 , c ) 1077 , d ) 2778 , e ) 2711
c
divide(multiply(140, const_100), subtract(add(const_100, 3), subtract(const_100, 10)))
add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)|
gain
if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 6 children have two children of the same sex and one of the opposite sex ?
"no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 6 = 6 c 2 total possible outcomes - 2 ^ 6 ( each child can be either a girl or a boy ) probability = 2 c 1 * 6 c 2 / 2 ^ 6 = 2 * 6 / 2 * 2 * 2 * 2 * 2 * 2 = 12 / 64 = 3 / 16 ans = d"
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 3 / 16 , e ) 3 / 5
d
divide(const_2, 6)
divide(const_2,n0)|
general
what is 25 % of 25 % equal to
"solution 25 % of 25 % = 25 / 100 x 25 / 100 = 1 / 16 = 0.0625 answer d"
a ) 0.625 , b ) 0.0625 , c ) 6.25 , d ) 0.00625 , e ) none
d
divide(multiply(25, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
when positive integer x is divided by 12 , the quotient is y and the remainder is 3 . when 2 x is divided by 11 , the quotient is 3 y and the remainder is 6 . what is the value of 21 y – x ?
"( 1 ) x = 12 y + 3 ( 2 ) 2 x = 33 y + 6 ( 2 ) - ( 1 ) : x = 21 y + 3 21 y - x = - 3 the answer is e ."
a ) 3 , b ) 2 , c ) - 1 , d ) - 2 , e ) - 3
e
subtract(multiply(21, divide(subtract(multiply(2, 3), 6), subtract(multiply(11, 3), multiply(2, 12)))), add(multiply(12, divide(subtract(multiply(2, 3), 6), subtract(multiply(11, 3), multiply(2, 12)))), 3))
multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n2)|subtract(#0,n5)|subtract(#1,#2)|divide(#3,#4)|multiply(n6,#5)|multiply(n0,#5)|add(n1,#7)|subtract(#6,#8)|
general
the slant height of a cone is 18 cm and radius of the base is 8 cm , find the curved surface of the cone ?
"π * 18 * 8 = 452 answer : b"
a ) 26 , b ) 452 , c ) 450 , d ) 440 , e ) 28
b
multiply(multiply(const_pi, 8), 18)
multiply(n1,const_pi)|multiply(n0,#0)|
geometry
a car has a 20 - gallon fuel to travel 500 miles . at this rate , how many gallons of fuel are needed for a 1000 mile car ?
number of gallons of fuel per mile = 500 / 20 = 25 gallons per mile number of gallons of fuel for a 1000 mile car = 1000 / 25 = 40 answer : d
a ) 10 , b ) 30 , c ) 20 , d ) 40 , e ) 50
d
add(add(add(const_3, const_2), divide(500, 20)), const_10)
add(const_2,const_3)|divide(n1,n0)|add(#0,#1)|add(#2,const_10)
gain
nicky and cristina are running a 300 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ?
"the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . c"
a ) 40 , b ) 50 , c ) 30 , d ) 60 , e ) 45
c
add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3)))
multiply(n1,n3)|subtract(n2,n3)|divide(#0,n3)|divide(#0,#1)|add(#2,#3)|
physics
nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 13 cm ?
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 13 ) = 247 cm 2 answer : d"
a ) 216 cm 2 , b ) 238 cm 2 , c ) 240 cm 2 , d ) 247 cm 2 , e ) 250 cm 2
d
divide(multiply(13, add(20, 18)), const_2)
add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|
physics
after decreasing 24 % in the price of an article costs rs . 760 . find the actual cost of an article ?
"cp * ( 76 / 100 ) = 760 cp = 10 * 100 = > cp = 1000 answer : c"
a ) 1218 , b ) 2777 , c ) 1000 , d ) 2688 , e ) 1991
c
divide(760, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
the salaries of a and b together amount to $ 7000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ?
"let a ' s salary is x b ' s salary = 7000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 7000 - x ) x = $ 5250 answer is b"
a ) $ 1000 , b ) $ 5250 , c ) $ 2500 , d ) $ 4500 , e ) $ 1200
b
subtract(7000, divide(7000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1)))
subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4)|
gain
what is the least number should be added to 1056 , so the sum of the number is completely divisible by 28 ?
"( 1056 / 28 ) gives remainder 20 20 + 8 = 28 , so we need to add 8 e"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 8
e
subtract(multiply(add(floor(divide(1056, 28)), const_1), 28), 1056)
divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)|
general
shopkeeper rise price by 34 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ?
let d initial price be 100 34 % rise now price = 134 / 100 * 100 = 134 10 % discount then price = 134 * 90 / 100 = 120.6 15 % discount then price = 120.6 * 85 / 100 = 102.51 so gain = 102.51 - 100 = 2.51 gain % = gain * 100 / cp = = > 2.51 * 100 / 100 = 2.51 % answer : a
a ) 2.51 % , b ) 4.51 % , c ) 3.51 % , d ) 6.51 % , e ) 7.51 %
a
subtract(multiply(multiply(add(const_100, 34), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100)
add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)
gain
find the smallest number which should be multiplied with 1152 to make it a perfect square .
1152 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 required smallest number = 2 2 is the smallest number which should be multiplied with 1152 to make it a perfect square . answer : a
['a ) 2', 'b ) 1', 'c ) 3', 'd ) 5', 'e ) 6']
a
divide(divide(divide(divide(divide(1152, const_3), const_3), const_4), const_4), const_4)
divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)|divide(#2,const_4)|divide(#3,const_4)
geometry
the sector of a circle has radius of 21 cm and central angle 45 o . find its perimeter ?
"perimeter of the sector = length of the arc + 2 ( radius ) = ( 45 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 58.5 cm answer : option e"
a ) 91.5 , b ) 92 , c ) 93 , d ) 94 , e ) 58.5
e
multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21)
add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)|
physics
excluding stoppages , the speed of a train is 55 kmph and including stoppages it is 30 kmph . of how many minutes does the train stop per hour ?
"explanation : t = 25 / 55 * 60 = 27 answer : option b"
a ) 82 , b ) 27 , c ) 12 , d ) 82 , e ) 18
b
subtract(const_60, multiply(const_60, divide(30, 55)))
divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)|
physics
find the average of all the numbers between 16 and 34 which are divisible by 5 ?
"average = ( 20 + 25 + 30 ) 3 = 85 / 3 = 28 answer is c"
a ) 10 , b ) 20 , c ) 28 , d ) 30 , e ) 15
c
divide(add(add(16, const_4), subtract(34, const_4)), const_2)
add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|
general
what is the square root of 261,121 ?
"1 ) checking the last digits in the answer options a , c , e are out since last digit of the square is known to be 1 . 2 ) b = 489 ^ 2 and d = 511 ^ 2 b = ( 500 - 11 ) ^ 2 and d = ( 500 + 11 ) ^ 2 since we need the answer less than 250000 = > b is out . thus we have d as the ans ."
a ) 476 , b ) 489 , c ) 497 , d ) 511 , e ) 524
d
circle_area(divide(261,121, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
other
a man can row 10 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?
m = 10 s = 1.2 ds = 11.2 us = 8.8 x / 11.2 + x / 8.8 = 1 x = 4.93 d = 4.93 * 2 = 9.86 answer : c
a ) 6.24 km , b ) 6 km , c ) 9.86 km , d ) 5.66 km , e ) 10 km
c
multiply(divide(multiply(add(10, 1.2), subtract(10, 1.2)), add(add(10, 1.2), subtract(10, 1.2))), const_2)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)
physics
what is the tens digit of 36 ^ 10 ?
"36 ^ 10 = 6 ^ 20 if you type powers of six they end in 6 36 16 96 76 56 the pattern is 3 - - > 1 - - > 9 - - > 7 - - > 5 so for 36 we start with 3 - - > 9 - - > 5 - - > 1 - - > 7 and repeat . 36 ^ 10 will come at 7 . answer d ."
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
d
floor(divide(reminder(power(36, reminder(10, add(const_4, const_1))), const_100), const_10))
add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)|
general
the ratio between x and y is 8 / 5 ; x and y are decreased by 5 , what is the ratio between the new values of x and y ?
"ratio = 8 k / 5 k = 8 / 5 , 16 / 10 , etc . x and y are decreased by 5 - - > ( 8 k - 5 ) / ( 5 k - 5 ) new ratio can be 3 / 0 , 11 / 5 , etc . answer : e"
a ) 8 / 5 , b ) 5 / 8 , c ) 4 , d ) 5 , e ) it can not be determined
e
divide(multiply(8, 5), multiply(5, 8))
multiply(n0,n1)|divide(#0,#0)|
general
# p is defined as 2 p + 20 for any number p . what is p , if # ( # ( # p ) ) = - 4 ?
# p = 2 p + 20 - - - > # ( # p ) = 2 ( 2 p + 20 ) + 20 = 4 p + 60 and thus # ( 4 p + 60 ) = 2 ( 4 p + 60 ) + 20 = 8 p + 140 = - 4 - - - > 8 p = - 144 - - - > p = - 18 , d is the correct answer .
a ) – 108 , b ) – 44 , c ) 10 , d ) - 18 , e ) 18
d
divide(add(add(add(multiply(multiply(20, const_2), 2), multiply(20, const_2)), 20), 4), multiply(4, const_2))
multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,#0)|add(#2,#0)|add(n1,#3)|add(n2,#4)|divide(#5,#1)
general
the average student age of a certain class which has 30 students is 10 . if one student aged 11 go to another class and the age of the class teacher is included the average changes to 11 . what is the age of class teacher ?
"sum of ages of class before replacement = 30 x 10 = 300 sum of ages of class without teacher = 300 - 11 = 289 sum of ages of class after replacement = 30 x 11 = 330 age of teacher = 330 - 289 = 41 answer : b"
a ) 32 , b ) 41 , c ) 45 , d ) 48 , e ) 50
b
subtract(multiply(30, 11), subtract(multiply(30, 10), 11))
multiply(n0,n3)|multiply(n0,n1)|subtract(#1,n2)|subtract(#0,#2)|
general
find the number of divisors of 1728 . ?
"1728 = 2 ^ 6 * 3 ^ 3 hence the number of factors = ( 6 + 1 ) x ( 3 + 1 ) = 7 x 4 = 28 . we know that if a number represented in standard form ( a ^ m * b ^ n ) , then the number of factors is given by ( m + 1 ) ( n + 1 ) . answer is 28 answer : c"
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30
c
add(const_2, const_2)
add(const_2,const_2)|
general
the diagonals of two squares are in the ratio of 2 : 5 then find the ratio of its areas ?
"let the diagonals of the squares be 2 x and 5 x ratio of their areas = 1 / 2 ( 2 x ) ^ 2 : 1 / 2 ( 5 x ) ^ 2 = 4 x ^ 2 : 25 x ^ 2 = 4 : 25 answer is b"
a ) 2 : 12 , b ) 4 : 25 , c ) 5 : 31 , d ) 2 : 19 , e ) 3 : 17
b
divide(const_4, power(5, 2))
power(n1,n0)|divide(const_4,#0)|
geometry
ifaequals the sum of the even integers from 2 to 60 , inclusive , andbequals the sum of the odd integers from 1 to 59 , inclusive , what is the value of a - b ?
"this is a solution from beatthegmat : even numbers : ( 60 - 2 ) / 2 + 1 = 30 even integers . ( 60 + 2 ) / 2 = 31 is the average of the even set . sum = avg * ( # of elements ) = 31 * 30 = 930 = a odd numbers : ( 59 - 1 ) / 2 + 1 = 30 odd integers . ( 59 + 1 ) / 2 = 30 is the average of the odd set . sum = avg * ( # of...
a ) 1 , b ) 10 , c ) 30 , d ) 20 , e ) 21
c
subtract(multiply(divide(60, 2), add(divide(60, 2), 1)), multiply(divide(add(59, 1), 2), add(divide(subtract(59, 1), 2), 1)))
add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)|
general
a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 310 . how much does c have ?
"a + b + c = 500 a + c = 200 b + c = 310 - - - - - - - - - - - - - - a + b + 2 c = 510 a + b + c = 500 - - - - - - - - - - - - - - - - c = 10 answer : e"
a ) 50 , b ) 78 , c ) 267 , d ) 29 , e ) 10
e
subtract(add(200, 310), 500)
add(n1,n2)|subtract(#0,n0)|
general
a car takes 6 hours to cover a distance of 288 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ?
"time = 6 distence = 288 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 288 / 9 = 32 kmph c )"
a ) 50 kmph , b ) 60 kmph , c ) 32 kmph , d ) 70 kmph , e ) none
c
divide(288, divide(multiply(6, 3), 2))
multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)|
physics
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if jill paid a 4 percent tax on the clothing , no tax on the food , and an 8 percent tax on all other items , then the total tax that she paid was what perce...
"let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2 tax on others = 2.4 percentage = 4.4 / 100 4.4 % answer : c"
a ) 2.8 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 %
c
multiply(divide(add(multiply(50, divide(4, const_100)), multiply(30, divide(8, const_100))), const_100), const_100)
divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)|
general
of the 200 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 30 percent of the employees prefer city y and 70 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ...
"140 prefer x ( group 1 ) ; 60 prefer y ( group 2 ) . city y needs 150 people : letall 60 who prefer y ( entire group 2 ) be relocated there , the rest 90 will be those who prefer x from group 1 ; city x needs 50 people : 140 - 90 = 50 from group 1 will be relocated to x , which they prefer . so , the highest possible ...
a ) 65 , b ) 110 , c ) 115 , d ) 130 , e ) 135
b
add(multiply(30, const_2), 70)
multiply(n3,const_2)|add(n4,#0)|
gain
united telephone charges a base rate of $ 8.00 for service , plus an additional charge of $ 0.25 per minute . atlantic call charges a base rate of $ 12.00 for service , plus an additional charge of $ 0.20 per minute . for what number of minutes would the bills for each telephone company be the same ?
"lets take number of minutesx . given that , 8 + 0.25 x = 12 + 0.2 x - > 0.05 x = 2 - > x = 80 minutes ans d"
a ) 25 minutes , b ) 10 minutes , c ) 20 minutes , d ) 80 minutes , e ) 60 minutes
d
divide(subtract(12.00, 8.00), subtract(0.25, 0.20))
subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)|
general
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 600 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ?
"i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 600 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 600 ) / 100 + ( 1.8 * 700 ) / 100 = 17.4 gms % of liquid x in resultant...
a ) 1.74 % , b ) 1.94 % , c ) 10 % , d ) 15 % , e ) 19 %
a
divide(add(multiply(600, 0.8), multiply(700, 1.8)), const_1000)
multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)|
gain
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 . 12200 after a year ?
"6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12200 = 3660 . answer : a"
a ) 3660 , b ) 2881 , c ) 2887 , d ) 9977 , e ) 2212
a
multiply(divide(6300, add(add(6300, 4200), 10500)), 12200)
add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|
gain
a right triangle with sides 3 cm , 4 cm and 5 cm is rotated about the side of 3 cm to form a cone . the volume of the cone so formed is
solution clearly , we have r = 3 cm and h = 4 cm . ∴ volume = 1 / 3 π r 2 h = ( 1 / 3 x π x 32 x 4 ) cm 3 = 12 π cm 3 . answer a
['a ) 12 π cm 3', 'b ) 15 π cm 3', 'c ) 16 π cm 3', 'd ) 20 π cm 3', 'e ) none']
a
volume_cone(3, 4)
volume_cone(n0,n1)
geometry
on a certain transatlantic crossing , 20 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 80 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ?
"let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.2 x had round trip tickets and took their cars . 0.2 x = 0.2 t x = t the answer is e ."
a ) 20 % , b ) 40 % , c ) 60 % , d ) 80 % , e ) 100 %
e
divide(20, subtract(const_1, divide(80, const_100)))
divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|
gain
the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 686 , krishan has ?
"ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 686 : n & there n = 289 x 686 / 49 = rs . 4046 answer : b"
a ) s . 2890 , b ) s . 4046 , c ) s . 1190 , d ) s . 1620 , e ) s . 2680
b
multiply(divide(multiply(divide(686, 7), 17), 7), 17)
divide(n4,n0)|multiply(n1,#0)|divide(#1,n0)|multiply(n1,#2)|
other
a no . when divided by 220 gives a remainder 43 , what remainder will be obtainedby dividingthe same no . 17 ?
"221 + 43 = 264 / 17 = 9 ( remainder ) c"
a ) 5 , b ) 6 , c ) 9 , d ) 11 , e ) 15
c
divide(add(220, 43), 17)
add(n0,n1)|divide(#0,n2)|
general
two stations a and b are 110 km apart on a straight line . one train starts from a at 8 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 8 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 11 a . m . answer : a"
a ) 11 , b ) 10 , c ) 8 , d ) 12 , e ) 15
a
add(8, divide(subtract(110, 20), add(25, 20)))
add(n2,n4)|subtract(n0,n2)|divide(#1,#0)|add(n3,#2)|
physics
it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ?
"let his speed for one half of the journey be 2 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 2 * x / 2 + x = 5 4 x = 5 x + 10 = > x = 10 a"
a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15
a
divide(2, subtract(divide(multiply(const_2, 2), 5), 1))
multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)|
physics
a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days w will the work get over ?
"we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 there...
a ) 60 , b ) w = 70 , c ) w = 75 , d ) w = 80 , e ) w = 100
c
divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2))
multiply(n1,n2)|subtract(const_1,const_0_25)|subtract(n1,n3)|divide(#0,const_0_25)|multiply(#3,#1)|divide(#4,#2)|
physics