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a train 320 m in length crosses a telegraph post in 12 seconds . the speed of the train is ?
"s = 320 / 12 * 18 / 5 = 96 kmph answer : e"
a ) 77 kmph , b ) 95 kmph , c ) 94 kmph , d ) 98 kmph , e ) 96 kmph
e
multiply(const_3_6, divide(320, 12))
divide(n0,n1)|multiply(#0,const_3_6)|
physics
12 + 13 + 14 + . . . 51 + 52 + 53 + 54 = ?
"sum = 12 + 13 + 14 + . . . 51 + 52 + 53 + 54 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 54 positive integers = 55 * 54 / 2 sum of first 11 positive integers = 11 * 12 / 2 sum = 12 + 13 + 14 + . . . 51 + 52 + 53 + 54 = 55 * 54 / 2 - 11 * 12 / 2 = 1419 answer : a"
a ) 1419 , b ) 1362 , c ) 1363 , d ) 1364 , e ) 1365
a
divide(multiply(51, add(51, 12)), 13)
add(n0,n3)|multiply(n3,#0)|divide(#1,n1)|
general
2 x + y = 13 | y | < = 13 for how many ordered pairs ( x , y ) that are solutions of the system above are x and y both integers ?
hi - the questions says | y | < = 13 so the values of y will range from 0 to 13 . so the values of y will be 0,1 , 2,3 , 4,5 , 6,7 , 8,9 , 10,11 , 12,13 . so 14 ( e ) is the answer . remember sign does n ' t matter for y as y will always be positive .
a ) 7 , b ) 10 , c ) 12 , d ) 13 , e ) 14
e
add(divide(add(13, 13), const_2), const_1)
add(n1,n1)|divide(#0,const_2)|add(#1,const_1)
general
mary decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month . if mary ' s savings by the end of the year from these monthly savings were 7 times the amount she spent per month , what should be the fraction of her salary that she spent each month ?
saving = s , expenditure = e , income = i at the end of 12 months ; 12 s = 7 e = 7 ( i - s ) = > 19 s = 7 i = > savings = 7 / 19 i hence expenditure = 1 - ( 7 / 19 ) = 12 / 19 answer ( e )
a ) 1 / 4 , b ) 3 / 4 , c ) 10 / 19 , d ) 11 / 19 , e ) 12 / 19
e
divide(divide(const_12, 7), add(divide(const_12, 7), const_1))
divide(const_12,n0)|add(#0,const_1)|divide(#0,#1)
general
what is the total number of integers between 20 and 70 that are divisible by 2 ?
"22 , 24 , 26 , . . . , 56,58 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 70 - 22 ) / ( 2 ) + 1 = 48 / 2 + 1 = 24 + 1 = 25 answer is b"
a ) 19 , b ) 25 , c ) 24 , d ) 26 , e ) 20
b
add(divide(subtract(70, 20), 2), const_1)
subtract(n1,n0)|divide(#0,n2)|add(#1,const_1)|
general
9 men went to a hotel . 8 of them spent 3 each over their meals and the ninth spent 5 more than the average expenditure of all the 9 . the total money spent by all of them was
let the average expenditure of all the ninte be x then , 3 Γ— 8 + x + 5 = 9 x β‡’ x = 3.625 ∴ total money spent = 9 x = 9 Γ— 3.625 = 32.625 answer d
a ) 26 , b ) 40 , c ) 29.25 , d ) 32.625 , e ) none of the above
d
multiply(divide(add(multiply(8, 3), 5), 8), 9)
multiply(n1,n2)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)
general
when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 4 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ?
"say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 4 ) ( 4 + 4 x ) or , x = 1 answer a"
a ) 1 , b ) 2 / 5 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5
a
divide(4, subtract(multiply(subtract(6, 4), 4), 4))
subtract(n1,n0)|multiply(n4,#0)|subtract(#1,n0)|divide(n0,#2)|
general
find the value of 72519 x 9999 = m .
"72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 e"
a ) 233243413 , b ) 354545451 , c ) 343435451 , d ) 565656767 , e ) 725117481
e
multiply(subtract(9999, const_4), 72519)
subtract(n1,const_4)|multiply(#0,n0)|
general
john bought 9.25 m of cloth for $ 444 . find the cost price per metre .
"cloth bought by john = 9.25 m cost of 9.25 m = $ 444 cost per metre = 444 Γ· 9.25 the cost of the cloth per metre = $ 48 answers : a"
a ) 48 , b ) 47 , c ) 58 , d ) 56 , e ) 54
a
divide(444, 9.25)
divide(n1,n0)|
physics
an woman swims downstream 81 km and upstream 36 km taking 9 hours each time ; what is the speed of the current ?
81 - - - 9 ds = 9 ? - - - - 1 36 - - - - 9 us = 4 ? - - - - 1 s = ? s = ( 9 - 4 ) / 2 = 2.5 answer : c
a ) 0.5 , b ) 1.5 , c ) 2.5 , d ) 3.5 , e ) 4.5
c
divide(subtract(divide(81, 9), divide(36, 9)), const_2)
divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)
physics
an escalator moves towards the top level at the rate of 11 ft . sec and its length is 126 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length .
"time taken to cover the entire length = tot . dist / resultant speed = 126 / ( 11 + 3 ) = 9 sec answer : e"
a ) 14 sec , b ) 10 sec , c ) 12 sec , d ) 8 sec , e ) 9 sec
e
divide(126, add(11, 3))
add(n0,n2)|divide(n1,#0)|
gain
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 80,000 as total savings , how much has he saved in public provident fund ?
let savings in national savings certificate = x and savings in public provident fund = ( 180000 - x ) 1 / 3 x = 1 / 2 ( 180000 βˆ’ x ) β‡’ 2 x = 3 ( 180000 βˆ’ x ) β‡’ 2 x = 540000 βˆ’ 3 x β‡’ 5 x = 540000 β‡’ x = 540000 / 5 = 108000 savings in public provident fund = ( 180000 - 108000 ) = 72000 answer is d .
a ) 81000 , b ) 85000 , c ) 75000 , d ) 72000 , e ) 77000
d
multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(const_1, add(divide(const_3, const_2), 1)))
divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(const_1,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)
general
what will be the cost of building a fence around a square plot with area equal to 49 sq ft , if the price per foot of building the fence is rs . 58 ?
"let the side of the square plot be a ft . a 2 = 49 = > a = 7 length of the fence = perimeter of the plot = 4 a = 28 ft . cost of building the fence = 28 * 58 = rs . 1624 . answer : b"
a ) 3944 , b ) 1624 , c ) 2999 , d ) 2667 , e ) 2121
b
multiply(square_perimeter(sqrt(49)), 58)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 11 years older than b , the present age of b is
"explanation : let b ' s age = x years . then , as age = ( x + 11 ) years . ( x + 11 + 10 ) = 2 ( x β€” 10 ) hence x = 41 . present age of b = 41 years answer : option d"
a ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42
d
add(multiply(const_2, 10), add(11, 10))
add(n0,n3)|multiply(n0,const_2)|add(#0,#1)|
general
the area of a parallelogram is 200 sq m and its altitude is twice the corresponding base . then the length of the base is ?
"2 x * x = 200 = > x = 10 answer : d"
a ) 8 , b ) 9 , c ) 7 , d ) 10 , e ) 5
d
sqrt(divide(200, const_2))
divide(n0,const_2)|sqrt(#0)|
geometry
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8210 / - . find out the amount that he borrowed .
"explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8210 = p + ( 54 p / 100 ) 8210 = ( 1...
a ) s . 5331 , b ) s . 5269 , c ) s . 5228 , d ) s . 5218 , e ) s . 52192
a
divide(8210, add(const_1, divide(multiply(6, 9), const_100)))
multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)|
gain
in a 200 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is :
"b runs 56 m in 7 sec . = > b runs 200 m in 7 / 56 * 200 = 25 seconds since a beats b by 7 seconds , a runs 200 m in ( 25 - 7 ) = 18 seconds hence , a ' s time over the course = 18 seconds answer : d"
a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds
d
subtract(multiply(divide(7, 56), 200), 7)
divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)|
physics
find the l . c . m of 2 / 15 , 3 / 10 , 6 / 5
"explanation : lcm of fractions = lcm of numerators / hcf of denominators ( lcm of 2 , 3 , 6 ) / ( hcf of 15 , 10 , 5 ) = 6 / 5 answer : option a"
a ) 6 / 5 , b ) 2 / 3 , c ) 7 / 5 , d ) 8 / 5 , e ) 9 / 5
a
multiply(10, const_3)
multiply(n3,const_3)|
physics
a horse chases a pony 3 hours after the pony runs . horse takes 4 hours to reach the pony . if the average speed of the horse is 35 kmph , what s the average speed of the pony ?
"pony take 7 hours and horse take 4 hours . . . then distance chased by them is 35 * 4 . so speed of pony is ( 35 * 4 ) / 7 = 20 kmph . answer : a"
a ) 20 kmph , b ) 22 kmph , c ) 24 kmph , d ) 26 kmph , e ) 28 kmph
a
divide(multiply(4, 35), add(4, 3))
add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|
physics
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 150 marks , the no of questions he attempts correctly is :
"explanation : let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 60 – x ) = 150 = > 5 x = 210 = > x = 42 answer : d"
a ) 35 , b ) 38 , c ) 40 , d ) 42 , e ) 44
d
divide(add(150, 60), add(4, 1))
add(n2,n3)|add(n0,n1)|divide(#0,#1)|
physics
the area of a square field is 900 km 2 . how long will it take for a horse to run around at the speed of 12 km / h ?
explanation area of field = 900 km 2 . then , each side of field = √ 900 = 30 km distance covered by the horse = perimeter of square field = 30 Γ— 4 = 120 km ∴ time taken by horse = distances / peed = 120 / 12 = 10 h answer b
['a ) 12 h', 'b ) 10 h', 'c ) 8 h', 'd ) 6 h', 'e ) none of these']
b
divide(multiply(const_4, sqrt(900)), 12)
sqrt(n0)|multiply(#0,const_4)|divide(#1,n2)
geometry
if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 60 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that...
"case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 60 ) = ( r + 5 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 2 ) x = 120 ans b"
a ) 100 , b ) 120 , c ) 140 , d ) 150 , e ) 160
b
add(multiply(subtract(60, 5), 2), multiply(2, 10))
multiply(n3,n4)|subtract(n2,n1)|multiply(n3,#1)|add(#2,#0)|
physics
98 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ?
"let b be the number of boys in the school . 98 = xb / 100 b = 0.5 x 9800 = 0.5 x ^ 2 x ^ 2 = 19600 x = 140 the answer is b ."
a ) 110 , b ) 140 , c ) 220 , d ) 250 , e ) 500
b
sqrt(divide(multiply(98, const_100), divide(50, const_100)))
divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)|
gain
two numbers are respectively 17 % and 30 % more than a third number . the percentage that is first of the second is ?
"i ii iii 117 130 100 130 - - - - - - - - - - 117 100 - - - - - - - - - - - ? = > 90 % answer : e"
a ) 75 % , b ) 80 % , c ) 85 % , d ) 70 % , e ) 90 %
e
subtract(const_100, multiply(divide(add(17, const_100), add(30, const_100)), const_100))
add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|
gain
16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 12 girls will take to construct ?
"explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 12 girls = 8 Γ— 6 / 4 + 12 = 12 + 12 = 24 girls 6 days to complete the work answer : option c"
a ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days
c
divide(const_1, add(multiply(12, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16))))
divide(const_1,n2)|divide(#0,n1)|divide(#0,n0)|multiply(n4,#1)|multiply(n3,#2)|add(#3,#4)|divide(const_1,#5)|
physics
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg will be ?
"since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . cost of 1 kg tea of ...
a ) 122.50 per kg , b ) 266.50 per kg , c ) 175.50 per kg , d ) 172.50 per kg , e ) 176.50 per kg
c
divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2)
add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)|
other
one - third of 1206 is what percent of 400 ?
"answer let one - third of 1206 is n % of 400 . ∡ 1206 / 3 = ( n x 400 ) / 100 ∴ n = ( 402 x 100 ) / 400 = 100.5 correct option : d"
a ) 3 , b ) 30 , c ) 300 , d ) 100.5 , e ) 201.2
d
multiply(const_100, divide(divide(1206, const_3), 400))
divide(n0,const_3)|divide(#0,n1)|multiply(#1,const_100)|
gain
if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 60 minutes ?
"since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 60 minutes the population increases by 60 x 4 = 240 people . answer . d ."
a ) 80 , b ) 100 , c ) 150 , d ) 240 , e ) 300
d
multiply(divide(const_60, 15), 60)
divide(const_60,n0)|multiply(n1,#0)|
physics
1 / 3 rd of work is completed by kiran in 6 days . in how many days he can finish the remaining work ?
1 / 3 rd work finished in 6 days 2 / 3 rd work = 2 / 3 * 3 * 6 = 12 days answer is c
a ) 6 days , b ) 10 days , c ) 12 days , d ) 15 days , e ) 20 days
c
divide(multiply(multiply(6, 3), const_2), const_3)
multiply(n1,n2)|multiply(#0,const_2)|divide(#1,const_3)
physics
find the middle one when the sum of 3 consecutive even numbers is 57 ?
3 consecutive numbers can be a - 1 , a , a + 1 so sum of numbers = 3 a = 57 . hence a = 19 . c
a ) 10 , b ) 12 , c ) 19 , d ) 15 , e ) 16
c
add(divide(subtract(57, add(3, 3)), 3), const_2)
add(n0,n0)|subtract(n1,#0)|divide(#1,n0)|add(#2,const_2)
physics
a number is increased by 20 % and then decreased by 20 % , the final value of the number is ?
"here , x = 20 and y = - 20 therefore , the net % change in value = ( x + y + xy / 100 ) % = [ 20 - 20 + ( 20 x - 20 ) / 100 ] % or - 4 % since the sign is negative , there is a decrease in value by 4 % . answer : d"
a ) does not change , b ) decreases by 2 % , c ) increases by 4 % , d ) decreases by 4 % , e ) none of these
d
divide(multiply(add(20, const_100), subtract(const_100, 20)), const_100)
add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|
general
circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 10 more revolutions than gear p ?
"gear q makes 30 more revolutions each 60 seconds . the time to make 10 = 30 / 3 more revolutions is 60 / 3 = 20 seconds . the answer is d ."
a ) 10 , b ) 12 , c ) 15 , d ) 20 , e ) 25
d
divide(10, subtract(divide(40, const_60), divide(10, const_60)))
divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|
physics
find 18 / 6 * 4
"answer = 18 / 6 * 4 = 18 / 24 = 0.75 option e is correct"
a ) 1.25 , b ) 4.53 , c ) 2.69 , d ) 0.89 , e ) 0.75
e
divide(18, 6)
divide(n0,n1)|
general
if 2 x = 3 y = 10 , then 9 xy = ?
"2 x = 10 ; x = 5 3 y = 10 ; y = 10 / 3 multiply : 9 xy = 9 * 5 * 10 / 3 = 150 answer : c ."
a ) 1,200 , b ) 200 , c ) 150 , d ) 40 , e ) 20
c
divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2))
multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)|
general
the average of first 16 even numbers is ?
"sum of 16 even numbers = 16 * 17 = 272 average = 272 / 16 = 17 answer : e"
a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 17
e
add(16, const_1)
add(n0,const_1)|
general
if 5 a + 7 b = g , where a and b are positive integers , what is the largest possible value of g for which exactly one pair of integers ( a , b ) makes the equation true ?
5 * a 1 + 7 * b 1 = g 5 * a 2 + 7 * b 2 = g 5 * ( a 1 - a 2 ) = 7 * ( b 2 - b 1 ) since we are dealing with integers we can assume that a 1 - a 2 = 7 * q and b 2 - b 1 = 5 * q where q is integer , so whenever we get a pair for ( a ; b ) we can find another one by simply adding 7 to a and subtracting 5 from b or vice ve...
a ) 35 , b ) 48 , c ) 69 , d ) 70 , e ) 74
d
add(multiply(5, 7), multiply(7, 5))
multiply(n0,n1)|add(#0,#0)
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 $ 48 ?
every two days , the net total is $ 6 . after 8 days , the worker will have $ 24 . on day 9 , the worker will receive $ 24 for a net total of $ 48 . the answer is c .
a ) 6 , b ) 7 , c ) 9 , d ) 12 , e ) 16
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
the length of the bridge , which a train 200 metres long and travelling at 60 km / hr can cross in 30 seconds , is ?
"speed = [ 60 x 5 / 18 ] m / sec = [ 50 / 3 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 130 + x ) / 30 = 50 / 3 = > ( 130 + x ) / 30 = 370 = > x = 370 m . answer : d"
a ) 720 , b ) 470 , c ) 270 , d ) 370 m . , e ) 570
d
subtract(multiply(divide(multiply(60, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 200)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
the average monthly income of p and q is rs . 2050 . the average monthly income of q and r is rs . 5250 and the average monthly income of p and r is rs . 6200 . the monthly income ofq is :
"explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 2050 x 2 ) = 4100 . . . . ( i ) q + r = ( 5250 x 2 ) = 10500 . . . . ( ii ) p + r = ( 6200 x 2 ) = 12400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 27000 or p + q + r = 13500 . . . ...
a ) 1100 , b ) 2100 , c ) 1200 , d ) 3100 , e ) 1300
a
subtract(add(2050, 6200), 5250)
add(n0,n2)|subtract(#0,n1)|
general
riya and priya set on a journey . riya moves eastward at a speed of 24 kmph and priya moves westward at a speed of 35 kmph . how far will be priya from riya after 45 minutes
"total eastward distance = 24 kmph * 3 / 4 hr = 18 km total westward distance = 35 kmph * 3 / 4 hr = 26 km total distn betn them = 18 + 26 = 44 km ans 44 km answer : e"
a ) 25 kms , b ) 10 kms , c ) 50 kms , d ) 30 kms , e ) 44 kms
e
multiply(speed(add(24, 35), const_60), 45)
add(n0,n1)|speed(#0,const_60)|multiply(n2,#1)|
physics
0.009 / x = 0.01 . find the value of x
"x = 0.009 / 0.01 = 0.9 answer : c"
a ) 0.0009 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90
c
divide(0.009, 0.01)
divide(n0,n1)|
general
if the mean of numbers 28 , x , 70 , 88 and 104 is 67 , what is the mean of 50 , 62 , 97 , 124 and x ?
"mean = ( sum of all no . in series ) / ( no . in series ) m 1 = 67 = ( x + 28 + 70 + 88 + 104 ) / ( 5 ) 67 x 5 = 290 + x x = 45 therefore mean of 2 nd series m 2 = ( x + 50 + 62 + 97 + 124 ) / 5 m 2 = 75.6 answer = b"
a ) 76.5 , b ) 75.6 , c ) 73.5 , d ) 79 , e ) 77
b
divide(add(add(add(add(subtract(multiply(67, add(const_4, const_1)), add(add(add(28, 70), 88), 104)), 50), 62), 97), 124), add(const_4, const_1))
add(const_1,const_4)|add(n0,n1)|add(n2,#1)|multiply(n4,#0)|add(n3,#2)|subtract(#3,#4)|add(n5,#5)|add(n6,#6)|add(n7,#7)|add(n8,#8)|divide(#9,#0)|
general
elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.80 apiece . if elena purchased a total of 10 of these pens for $ 42.00 , how many brand x pens did she purchase ?
"4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 5 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : b ."
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
b
subtract(multiply(4.00, 10), 42.00)
multiply(n0,n2)|subtract(#0,n3)|
general
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is :
"let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 9 = 4 : 3 . answer b"
a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these
b
divide(sqrt(16), sqrt(9))
sqrt(n1)|sqrt(n0)|divide(#0,#1)|
physics
if the speed of a man is 45 km per hour , then what is the distance traveled by him in 30 seconds ?
"the distance traveled in 30 sec = 45 * ( 5 / 18 ) * 30 = 375 m answer : c"
a ) 275 m , b ) 360 m , c ) 375 m , d ) 420 m , e ) 480 m
c
multiply(multiply(45, const_0_2778), 30)
multiply(n0,const_0_2778)|multiply(n1,#0)|
physics
a participated in cycling contest and he drove the lap at the rate of 6 kmph , 12 kmph , 18 kmph , 24 kmph . . what is his average speed ?
= 4 x / ( x / 6 + x / 12 + x / 18 + x / 24 ) = 11.52 answer : d
a ) 15 kmph , b ) 18 kmph , c ) 14.25 kmph , d ) 11.52 kmph , e ) 16 kmph
d
divide(subtract(multiply(const_100, const_3), const_12), add(24, const_1))
add(n3,const_1)|multiply(const_100,const_3)|subtract(#1,const_12)|divide(#2,#0)
physics
a watch was sold at a loss of 20 % . if it was sold for rs . 520 more , there would have been a gain of 6 % . what is the cost price ?
"80 % 106 % - - - - - - - - 26 % - - - - 520 100 % - - - - ? = > rs . 2000 answer : b"
a ) rs . 1500 , b ) rs . 2000 , c ) rs . 3000 , d ) rs . 4200 , e ) rs . 2500
b
divide(multiply(520, const_100), subtract(add(const_100, 6), subtract(const_100, 20)))
add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)|
gain
total dinning bill for 5 people was $ 211.00 . if they add 15 % tip and divided the bill evenly , approximate . what was each persons find share
"211 * 15 = 3165 / 100 = 31.65 211 + 31.65 = 242.65 242.65 / 5 = 48.53 answer : b"
a ) $ 30.14 , b ) 48.53 , c ) 34.66 , d ) 32.29 , e ) 33.16
b
divide(add(211.00, divide(multiply(15, 211.00), const_100)), 5)
multiply(n1,n2)|divide(#0,const_100)|add(n1,#1)|divide(#2,n0)|
general
kathleen can paint a room in 2 hours , and anthony can paint an identical room in 5 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ?
"( 1 / 2 + 1 / 5 ) t = 2 t = 20 / 7 answer : a"
a ) 20 / 7 , b ) 4 / 3 , c ) 15 / 8 , d ) 9 / 4 , e ) 15 / 4
a
multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 5))), 2)
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,n0)|
physics
in a school of 320 students , 85 students are in the band , 200 students are on sports teams , and 60 students participate in both activities . how many students are involved in either band or sports ?
25 + 60 + 140 = 225 there are 225 students involved in either band or sports . b )
a ) 220 , b ) 225 , c ) 230 , d ) 240 , e ) 245
b
add(60, add(subtract(85, 60), subtract(200, 60)))
subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)
other
if the cost price of 20 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ?
"explanation : let cost price of 1 article be re . 1 . therefore , cost price of 20 articles = rs . 20 . selling price of 16 articles = rs . 20 therefore , selling price of 20 articles is : - = > 20 / 16 Γ— 20 = > 25 . therefore , profit = selling price - cost price . = > 25 βˆ’ 20 = 5 . hence , the percentage of profit =...
a ) 20 % loss , b ) 25 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these
b
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 20)), divide(multiply(const_100, 16), 20)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
the positive numbers w , x , y , and z are such that x is 20 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ?
y = 100 ( start with yy , it makes getting x easier ) . x = y * 1.2 = 120 z = y / 1.2 = 83.33 w = x * 0.8 = 96 then for the answer : ( w - z ) / z = ( 96 - 83.33 ) / 83.33 = 15.2 % answer : a
a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 %
a
multiply(const_100, subtract(multiply(multiply(divide(add(20, const_100), const_100), divide(add(20, const_100), const_100)), divide(subtract(const_100, 20), const_100)), const_1))
add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#3,#3)|multiply(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)
general
on the coordinate plane , points p and u are defined by the coordinates ( - 1,0 ) and ( 3,3 ) , respectively , and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( 25 / 4 ) Ο€ , what are the coordinates of the center of the circle ?
although it took me 3 mins to solve this question using all those equations , later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is...
['a ) ( 1.5 , 1 )', 'b ) ( 2 , - 5 )', 'c ) ( 0,0 )', 'd ) ( 1 , 1.5 )', 'e ) ( 2,2 )']
d
add(multiply(const_3, const_3), sqrt(divide(25, 4)))
divide(n2,n3)|multiply(const_3,const_3)|sqrt(#0)|add(#1,#2)
geometry
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 1.25 km respectively . the time taken by the slower train to cross the faster train in seconds is ?
"relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 1.25 = 2 km = 2250 m . required time = 2250 * 3 / 125 = 54 sec . answer : d"
a ) 12 , b ) 77 , c ) 48 , d ) 54 , e ) 11
d
subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.25, const_1000), divide(multiply(90, const_1000), const_3600)))
multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)|
physics
paul and jack are 2 mechanics who work 8 hours per day changing oil in cars at the oil spot . if paul can change oil in 2 cars per hour , and jack can change oil in 3 cars per hour . what is the least number of cars per work day the 2 men can finish ?
the proportion equation can be used . let x = number cars and c = least number of cars per day the 2 men can finish ( c = x / ( 8 hours ) , p = cars per hour paul can change , and j = cars per hour jack can change . note : 8 hours = a work day p + j = c ( 2 cars / hour ) + ( 3 cars / hour ) = x / ( 8 hours ) proportion...
a ) 50 , b ) 40 , c ) 35 , d ) 10 , e ) 90
b
add(multiply(3, 8), multiply(const_2, 8))
multiply(n1,n3)|multiply(n1,const_2)|add(#0,#1)
physics
the edge of three cubes of metal is 3 dm , 4 dm and 5 dm . they are melted and formed into a single cube . find the edge of the new cube ?
33 + 43 + 53 = a 3 = > a = 6 answer : d
['a ) 2', 'b ) 8', 'c ) 7', 'd ) 6', 'e ) 9']
d
multiply(const_2, const_3)
multiply(const_2,const_3)
geometry
if a farmer sells 5 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 10 more chickens , he will run out of feed 3 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ?
"say farmer has n chicken and he is good for d days . : - we have 3 equations given in question : - ( n - 5 ) * d + 4 = ( n + 10 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 10 d - 3 n = 30 4 n - 5 d = 20 = > n = 20 ans c it is !"
a ) 12 , b ) 24 , c ) 20 , d ) 55 , e ) 60
c
multiply(add(const_2, 4), multiply(3, 4))
add(const_2,n1)|multiply(n1,n3)|multiply(#0,#1)|
general
if a solid sphere of radius 10 cms is moulded into 8 spherical solid balls of equal radius , then surface area of each ball ( in cm 2 ) is ?
4 / 3 Ο€ * 10 * 10 * 10 = 8 * 4 / 3 Ο€ r 3 r = 5 4 Ο€ * 5 * 5 = 100 Ο€ answer : a
['a ) 100', 'b ) 272', 'c ) 287', 'd ) 189', 'e ) 289']
a
multiply(const_4, power(power(divide(multiply(multiply(10, 10), 10), 8), const_0_33), const_2))
multiply(n0,n0)|multiply(n0,#0)|divide(#1,n1)|power(#2,const_0_33)|power(#3,const_2)|multiply(#4,const_4)
geometry
at the end of the month , a certain ocean desalination plant ’ s reservoir contained 26 million gallons of water . this amount is eighth the normal level . if this amount represents 65 % of the reservoir ’ s total capacity , how many million gallons short of total capacity is the normal level ?
"the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 26 mg = 65 % of total . . total = 26 / . 65 = 40 mg . . normal level = 1 / 8 of 40 = 5 mg . . shortage of normal level = 40 - 5 = 35 mg . . c"
a ) 15 , b ) 25 , c ) 35 , d ) 45 , e ) 55
c
divide(divide(26, const_2), divide(65, const_100))
divide(n0,const_2)|divide(n1,const_100)|divide(#0,#1)|
gain
if t = 5 / 9 * ( k - 32 ) , and if t = 20 , then what is the value of k ?
"k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 20 ) / 5 + 32 = 68 the answer is a ."
a ) 68 , b ) 70 , c ) 72 , d ) 74 , e ) 76
a
add(divide(multiply(20, 9), 5), 32)
multiply(n1,n3)|divide(#0,n0)|add(n2,#1)|
general
someone on a skateboard is traveling 10 miles per hour . how many feet does she travel in 20 seconds ? ( 1 mile = 5280 feet )
per second = > 10 * 5280 ft / 60 * 60 = 14.67 ft 20 seconds = > 14.67 * 20 = 293.4 ft answer : b
a ) 292 ft , b ) 293.4 ft , c ) 295 ft , d ) 296 ft , e ) 297 ft
b
multiply(20, divide(multiply(10, 5280), const_3600))
multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)
physics
a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 8 feet and a heig...
"the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 64 * 10 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 640 / 10,000 = 64 / 1000 = 0.064 ft the answer is c ."
a ) 0.008 , b ) 0.08 , c ) 0.064 , d ) 0.64 , e ) 6.4
c
divide(volume_cylinder(8, 10), circle_area(100))
circle_area(n0)|volume_cylinder(n2,n3)|divide(#1,#0)|
geometry
at the end of the first quarter , the share price of a certain mutual fund was 30 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 75 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of...
"another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 75 = 30 + b + 30 b / 100 b = 34.6 answer ( d )"
a ) 20 % , b ) 25 % , c ) 30 % , d ) 34.6 % , e ) 40 %
d
multiply(subtract(divide(add(const_100, 75), add(const_100, 30)), const_1), const_100)
add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100)|
gain
solve : 18800 + 470 + 20
"18800 + 470 + 20 = ( 18800 / 470 ) + 20 = 40 / 20 = 2 . answer is a ."
a ) 2 , b ) 5 , c ) 4 , d ) 0 , e ) 1
a
multiply(18800, divide(470, 20))
divide(n1,n2)|multiply(n0,#0)|
general
how many even integers n , such that 20 < = n < = 240 are of the form 3 k + 4 , where k is any natural number ?
"the first number is 22 = 16 + 6 ( 1 ) . we can continue adding 6 to make a list : 22 , 28 , 34 , . . . the last number is 238 = 16 + 6 ( 37 ) there are 37 numbers in the list . the answer is d ."
a ) 31 , b ) 33 , c ) 35 , d ) 37 , e ) 39
d
add(divide(subtract(subtract(240, const_2), add(20, const_2)), add(const_2, 4)), const_1)
add(n0,const_2)|add(const_2,n3)|subtract(n1,const_2)|subtract(#2,#0)|divide(#3,#1)|add(#4,const_1)|
general
a person crosses a 600 m long street in 2 minutes . what is his speed in km per hour ?
distance = 600 meter time = 2 minutes = 2 x 60 seconds = 120 seconds speed = distance / time = 600 / 120 = 5 m / s = 5 Γ£ β€” 18 / 5 km / hr = 18 km / hr answer : c
a ) 12 , b ) 24 , c ) 18 , d ) 42 , e ) 64
c
divide(divide(600, const_1000), divide(multiply(2, const_60), const_3600))
divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)
physics
a train ride from two p to town q costs $ 2.35 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.85 . what is the cost of a bus ride from town p to town q ?
let x be the cost of a bus ride . x + ( x + 235 ) = 985 2 x = 750 x = $ 3.75 the answer is c .
a ) $ 1.25 , b ) $ 2.50 , c ) $ 3.75 , d ) $ 4.70 , e ) $ 8.20
c
divide(subtract(9.85, 2.35), const_2)
subtract(n1,n0)|divide(#0,const_2)
general
in a can , there is a mixture of milk and water in the ratio 5 : 3 . if the can is filled with an additional 8 liters of milk , the can would be full and the ratio of milk and water would become 2 : 1 . find the capacity of the can ?
"let c be the capacity of the can . ( 5 / 8 ) * ( c - 8 ) + 8 = ( 2 / 3 ) * c 15 c - 120 + 192 = 16 c c = 72 the answer is e ."
a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72
e
add(add(multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 5))), 5), 8), multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 5))), 3))
multiply(n2,n4)|multiply(n1,n3)|multiply(n0,n4)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)|multiply(n1,#4)|add(n2,#5)|add(#7,#6)|
general
in what time will a railway train 30 m long moving at the rate of 36 kmph pass a telegraph post on its way ?
"t = 30 / 36 * 18 / 5 = 3 sec answer : c"
a ) 5 sec , b ) 4 sec , c ) 3 sec , d ) 6 sec , e ) 1 sec
c
divide(30, multiply(36, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
a 50 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ?
1 : 2 / 5 = 5 : 2 2 / 7 * 50 = 14.28 answer : b
a ) 24.28 , b ) 14.28 , c ) 16.28 , d ) 18.28 , e ) 12.28
b
subtract(50, divide(50, add(divide(2, 5), const_1)))
divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)
physics
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 46 meters . how many times did the ball bounce ?
division of total distance traveled will be 16 + 16 + 8 + 4 + 2 ans : a
a ) 4 , b ) 6 , c ) 7 , d ) 8 , e ) 9
a
divide(divide(16, const_2), const_2)
divide(n0,const_2)|divide(#0,const_2)
general
( 786 Γ— 74 ) Γ· ? = 1938.8
explanation : 58164 / x = 1938.8 = > x = 58164 / 1938.8 = 30 answer : option c
a ) a ) 4.8 , b ) b ) 48 , c ) c ) 30 , d ) d ) 68 , e ) e ) 48
c
divide(multiply(786, 74), 1938.8)
multiply(n0,n1)|divide(#0,n2)
general
in a 100 m race between rose and diane , rose lost to diane by 0.75 m . if diane was given a 7 m head start , how many meters more will it take before rose overtakes diane ?
distance rose ran before completion of race - - - 100 m - 0.75 m = 99.25 m distance gained on diane over 99.25 m - - - 7 m - 0.75 m = 6.25 then rose gains 99.25 / 6.25 m = 1 m on kelly every 15.88 meters . therefore 15.88 divided by 0.75 of 1 meter . answer : b . 11.91 m
a ) 12.91 m , b ) 11.91 m , c ) 12.5 m , d ) 10.91 m , e ) 13.91 m
b
multiply(divide(subtract(100, 0.75), subtract(7, 0.75)), 0.75)
subtract(n0,n1)|subtract(n2,n1)|divide(#0,#1)|multiply(n1,#2)
physics
if ramola ranks 14 th in a class of 26 , what is her rank from the last ?
when considered from last , 12 ( i . e 26 - 14 ) students are ranked before her . so she is 13 th rank from last . answer : b
a ) 11 th , b ) 13 th , c ) 12 th , d ) 14 th , e ) 15 th
b
add(subtract(26, 14), const_1)
subtract(n1,n0)|add(#0,const_1)
other
the ratio of three numbers is 1 : 2 : 5 and the sum of their squares is 4320 . the sum of the numbers is ?
"let the numbers be x , 2 x , 5 x then , x ^ 2 + 4 x ^ 2 + 25 x ^ 2 = 4320 30 x ^ 2 = 4320 x ^ 2 = 144 x = 12 answer is b"
a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 14 , e ) e ) 9
b
sqrt(divide(4320, add(power(5, 2), add(power(1, 2), power(2, 2)))))
power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)|
other
the average salary per head of the entire staff of an office including the officers and clerks is rs . 90 . the average salary of officers is rs . 600 and that of the clerks is rs . 84 . if the number of officers is 2 , find the number of officers in the office ?
6 * 22 = 132 7 * 19 = 133 - - - - - - - - - - - - - - 1 . answer : a
a ) 1 , b ) 8 , c ) 7 , d ) 6 , e ) 2
a
divide(90, add(84, add(const_3, const_3)))
add(const_3,const_3)|add(n2,#0)|divide(n0,#1)
general
the parameter of a square is equal to the perimeter of a rectangle of length 20 cm and breadth 16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ?
"let the side of the square be a cm . parameter of the rectangle = 2 ( 20 + 16 ) = 72 cm parameter of the square = 72 cm i . e . 4 a = 72 a = 18 diameter of the semicircle = 18 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 18 ) = 1 / 2 ( 22 / 7 ) ( 18 ) = 396 / 14 = 28.28 cm to two decimal places answer : a"
a ) 28.28 cm , b ) 25.45 cm , c ) 26.45 cm , d ) 20.18 cm , e ) 32.14 cm
a
divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(20, 16)), const_2)), const_2)
rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)|
geometry
100 Γ— 10 – 100 + 2000 Γ· 100 = ?
"solution given exp . = 100 Γ— 10 - 100 + 20 = 1000 - 100 + 20 = 1020 - 100 = 920 . answer c"
a ) 29 , b ) 780 , c ) 920 , d ) 979 , e ) none of these
c
multiply(100, 10)
multiply(n0,n1)|
general
what percent is 1 gm of 1 kg ?
"1 kg = 1000 gm 1 / 1000 Γ— 100 = 100 / 1000 = 1 / 10 = 0.1 % a )"
a ) 0.1 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 3 %
a
multiply(divide(1, 1), const_100)
divide(n0,n1)|multiply(#0,const_100)|
gain
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8700 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ?
"sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8700 = 34800 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 2800 salary of may is rs . 6500 salary of jan...
a ) s . 3700 , b ) s . 4570 , c ) s . 4500 , d ) s . 4550 , e ) s . 2500
a
subtract(multiply(8000, const_4), subtract(multiply(8700, const_4), 6500))
multiply(n0,const_4)|multiply(n1,const_4)|subtract(#1,n2)|subtract(#0,#2)|
general
one pump drains one - half of a pond in 5 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ?
"the tricky part here , i believed is one half hour = 1 / 2 . then everything would be easy . we have the 1 st pump working rate / hour = 1 / 2 : 5 = 1 / 10 working rate of 2 pumps : 1 / 2 : 1 / 2 = 1 . working rate of 2 nd pump : 1 - 1 / 10 = 9 / 10 - - > time taken for the 2 nd pump to finish : 1 : 9 / 10 = 10 / 9 = ...
a ) 1 hour , b ) 1.2 hour , c ) 3 hours , d ) 5 hours , e ) 1.1 hours
e
divide(const_1, subtract(const_1, divide(const_1, multiply(5, const_2))))
multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)|
physics
if a jogger jogs 16 km / hr instead of 12 km / hr , he would have jogged 10 km more . the actual distance he jogged is
explanation : the actual distance jogged be d . = > d / 12 = ( d + 10 ) / 16 = > 16 d = 12 d + 120 = > 4 d = 120 = > d = 30 answer : c
a ) 50 km , b ) 40 km , c ) 30 km , d ) 20 km , e ) none of these
c
multiply(12, divide(10, subtract(16, 12)))
subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)
general
when a certain number x is divided by 84 , the remainder is 25 . what is the remainder when x is divided by 14 ?
"let possible value of x is 109 least possible value of x / 14 is 109 / 14 = > 7 quotient with remainder 11 thus answer is ( e ) 11"
a ) 10 , b ) 8 , c ) 9 , d ) 5 , e ) 11
e
reminder(25, 14)
reminder(n1,n2)|
general
how many multiples of 100 are there between 100 and 1000 ( both are inclusive ) ?
"the answer is ( 1000 - 100 ) / 100 + 1 = 10 answer is e"
a ) 2 , b ) 8 , c ) 4 , d ) 5 , e ) 10
e
add(divide(subtract(1000, 100), 100), const_1)
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
other
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 50 % ?
"cost price = rs . 500 profit = 50 % of 500 = rs . 250 selling price = cost price + profit = 500 + 250 = 750 answer : a"
a ) 750 , b ) 882 , c ) 772 , d ) 652 , e ) 271
a
add(500, multiply(500, divide(50, const_100)))
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , what was the number of valid votes that the other candidate got ?
"explanation : total number of votes = 7500 given that 20 % of percentage votes were invalid = > valid votes = 80 % total valid votes = 7500 Γ— ( 80 / 100 ) 1 st candidate got 55 % of the total valid votes . hence the 2 nd candidate should have got 45 % of the total valid votes = > valid votes that 2 nd candidate got = ...
a ) 2800 , b ) 2700 , c ) 2100 , d ) 2500 , e ) none of these
b
multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500)
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|
gain
a and b can do a piece of work in 30 days , while b and c can do the same work in 24 days and c and a in 20 days . they all work together for 10 days when b and c leave . how many days more will a take to finish the work ?
explanation : 2 ( a + b + c ) ' s 1 day ' s work = ( 1 / 30 + 1 / 24 + 1 / 20 ) = 15 / 120 = 1 / 8 work done by a , b , c in 10 days = 10 / 6 = 5 / 8 . a ' s 1 day ' s work = ( 1 / 16 - 1 / 24 ) = 1 / 48 now , 1 / 48 work is done by a in 1 day so , 3 / 8 work will be done by a in ( 48 x 3 / 8 ) = 18 days . answer is a
a ) 18 days , b ) 24 days , c ) 30 days , d ) 36 days , e ) 35 days
a
divide(subtract(const_1, multiply(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), 10)), subtract(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), divide(const_1, 24)))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|multiply(n3,#5)|subtract(#5,#1)|subtract(const_1,#6)|divide(#8,#7)
physics
the h . c . f and l . c . m of two numbers are 84 and 21 respectively . if the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is
explanation : let the numbers be x and 4 x . then , \ inline \ fn _ jvn x \ times 4 x = 84 \ times 21 \ : \ : \ leftrightarrow \ : \ : x ^ { 2 } = \ frac { 84 \ times 21 } { 4 } \ : \ : \ leftrightarrow \ : \ : x = 21 hence larger number = 4 x = 84 answer : c ) 84
a ) 32 , b ) 37 , c ) 84 , d ) 29 , e ) 21
c
multiply(sqrt(divide(multiply(84, 21), 4)), 4)
multiply(n0,n1)|divide(#0,n3)|sqrt(#1)|multiply(n3,#2)
other
if it takes a tub 2 minutes to drain 5 / 7 of its content , how much more time will it take for the tub to be empty ?
"if 5 / 7 of tub ' s content is drained 2 / 7 th of tub still needs to be drained . if it takes 2 minutes to drain 5 / 7 th of tub it takes 2 * ( 7 / 5 ) minutes to drain the entire tub and 2 * ( 7 / 5 ) * ( 2 / 7 ) min to drain 2 / 7 th of the tub which is 4 / 5 minutes or 48 minutes so answer is a"
a ) 48 seconds , b ) 1 minute , 12 seconds , c ) 1 minute , 50 seconds , d ) 2 minutes , 14 seconds , e ) 4 minutes , 12 seconds
a
add(subtract(const_1, divide(5, 7)), divide(5, 7))
divide(n1,n2)|subtract(const_1,#0)|add(#0,#1)|
general
the true discount on a bill due 9 months hence at 16 % per annum is rs . 210 . the amount of the bill is
"explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 210 . x Γ£ β€” 16 Γ£ β€” ( 9 / 12 ) Γ£ β€” ( 1 / 100 ) = 210 or x = 1750 p . w . = rs . 1750 sum due = p . w . + t . d . = rs . ( 1750 210 ) = rs . 1960 . answer : b"
a ) 1200 , b ) 1960 , c ) 1600 , d ) 1354 , e ) none of these
b
add(divide(210, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 210)
multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)|
gain
how many numbers between 100 and 630 are divisible by 2 , 3 , and 7 together ?
"explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 600 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 630 : 630 / 42 = 13 hence , 42 * 15 = 630 is the last number divisible by 42 within 630 h...
a ) 112 , b ) 77 , c ) 267 , d ) 13 , e ) 99
d
subtract(divide(630, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7)))
multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)|
general
a and b together can complete a work in 12 days . a alone can complete it in 20 days . if b does the work only for half a day daily , then in how many days a and b together will complete the work ?
b ' s 1 day ' s work = ( 1 / 12 - 1 / 20 ) = 2 / 60 = 1 / 30 now , ( a + b ) ' s 1 day ' s work = ( 1 / 20 + 1 / 60 ) = 4 / 60 = 1 / 15 so , a and b together will complete the work in 15 days . correct option : c
a ) 10 days , b ) 11 days , c ) 15 days , d ) 20 days , e ) none of these
c
divide(inverse(subtract(inverse(12), inverse(20))), const_2)
inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)|divide(#3,const_2)
physics
marketing executives for a certain chewing gum company projected a 20 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ?
"last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 120 . actual / projected * 100 = 75 / 120 * 100 = 62.5 % . answer : c"
a ) 53 % , b ) 58 % , c ) 62.5 % , d ) 64 % , e ) 75 %
c
multiply(divide(subtract(const_100, 25), add(20, const_100)), const_100)
add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)|multiply(#2,const_100)|
general
what is the tens digit of 7 ^ 1033 ?
"7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 1033 = 4 * 258 + 1 the ten ' s digit will be 0 . answer c"
a ) 0 , b ) 1 , c ) 0 , d ) 3 , e ) 4
c
floor(divide(reminder(power(7, reminder(1033, 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
country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 360 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ?
"convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 360 mill / 2 month i . e . $ 180 mill / month difference / month = 180 - 120 = 60 difference / year = $ 60 mill * 12 = $ 720 mill answer : e"
a ) $ 24 million , b ) $ 120 million , c ) $ 644 million , d ) $ 540 million , e ) $ 720 million
e
subtract(multiply(360, divide(const_12, const_2)), multiply(1.44, const_1000))
divide(const_12,const_2)|multiply(n0,const_1000)|multiply(n2,#0)|subtract(#2,#1)|
general
a man is walking at the rate of 6 km / hr crosses a bridge in 15 minutes . the length of the bridge is
"explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . speed = 6 Γ’ Λ† β€” 5 / 18 = 5 / 3 m / sec time = 15 Γ’ Λ† β€” 60 seconds = 900 seconds distance = time Γ’ Λ† β€” spe...
a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 1500 meters , e ) none of these
d
multiply(multiply(6, divide(15, const_60)), const_1000)
divide(n1,const_60)|multiply(n0,#0)|multiply(#1,const_1000)|
gain
2 minute is taken to travel the diagonal of square field at 3 km / hour . calculate the area of field ?
speed of man = [ 3 * 5 / 18 ] m / sec . = 5 / 6 m / sec . , distance travel in 2 minute = { 5 / 6 * 2 * 60 } m . = 100 m . diagonal of square field = 100 m area of field = 1 / 2 * diagonal ^ 2 = [ 1 / 2 * 100 * 100 ] sq . m = 5000 sq . m = 5000 / 100 = 50 are ( 1 are = 100 sq . m ) answer c
['a ) 25 are', 'b ) 30 are', 'c ) 50 are', 'd ) 60 are', 'e ) 45 are']
c
divide(divide(power(multiply(multiply(2, multiply(multiply(2, 3), const_10)), divide(multiply(3, const_1000), multiply(multiply(const_180, const_10), const_2))), const_2), 2), const_100)
multiply(n1,const_1000)|multiply(const_10,const_180)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_10)|divide(#0,#3)|multiply(n0,#4)|multiply(#5,#6)|power(#7,const_2)|divide(#8,n0)|divide(#9,const_100)
geometry
a 400 meter long train crosses a platform in 45 seconds while it crosses a signal pole in 30 seconds . what is the length of the platform ?
"speed = [ 400 / 30 ] m / sec = 40 / 3 m / sec . let the length of the platform be x meters . then , x + 400 / 45 = 40 / 3 3 ( x + 400 ) = 1800 Γ¨ x = 200 m . answer : a"
a ) 200 , b ) 400 , c ) 350 , d ) 1800 , e ) 45
a
subtract(multiply(divide(400, 30), 45), 400)
divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|
physics
a and b complete a work in 8 days . a alone can do it in 14 days . if both together can do the work in how many days ?
"1 / 8 + 1 / 14 = 11 / 56 56 / 11 = 5 days answer : a"
a ) 5 days , b ) 8 days , c ) 6 days , d ) 3.15 days , e ) 2.75 days
a
inverse(add(inverse(8), inverse(14)))
inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)|
physics
if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 27 , what is the value of r - p ?
"arithmetic mean expression for p and q : ( p + q ) / 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) / 2 = 20 ; q + r = 54 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 34 hence , the correct answer is c"
a ) 20 , b ) 10 , c ) 34 , d ) 40 , e ) 5
c
subtract(multiply(27, const_2), multiply(10, const_2))
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
general