Simplification- Aptitude Questions and Answers

Let the price of book is X and pencil is Y.

X+2*Y=165
X+10*Y=169

After resolving above equetion 8*Y=4
hence, price of pencil is = 0.5

As given, the ratio of the number of head and the number of feet =12 : 35 and the number of head is 48-
So 12*P=48
P=4

Hence, the number of feet = 35*4 =140
Now,Let the number of chickens is X and cows is Y.
So,
X+Y=48
2X+4Y=140
After resolving above equetion X=26

Let the weights of the pieces of diamond are x, 2x, 3x, 4x.
Total weight of original diamond = 1x + 2x + 3x + 4x = 10x
as is given Cost ∝ (weight)2
Total cost of original diamond = (10x)² = 100x² ..…(i)
Now, Cost of each piece = x ², 4x ², 9x ², 16x ²
Total cost pieces = x²(1 + 4 + 9 + 16) = 30x² …..(ii)

Total loss = 100x² - 30x² = 70x²
But total loss = 70000
70x² = 70000
x² = 1000

Total cost of original diamond = 100x²
= 100 * 1000 [ x² = 1000]
= 100000

Let the number are x and 5x
Now, (x )(5x ) = 320
5x² = 320
x² = 64
x = 8

Therefore Numbers are 8 and 40.
Hence, difference between their squares= (40)² - (8)² = 1600 - 64 = 1536

Let total property of man= x
His widow gets = x/3
His daughter gets = 3/5 *[x − (x/3)]
=3/5 *(2x/3)=2x/5

His son's property = x−[(x/3)+(2x/5)]
6400 = x−(11x/15)
6400 = (15x−11x)/15
6400*15 = 4x
x = 1600*15
x=24000

Hence, total property=24000

Let x be units digit and y be tens digit.then number is 10y + x
As given,
x + y = 11 ----eq1
10x + y = 10y + x + 27----eq2

from eq2
9x -9y = 27
x - y = 3 ----eq3

by adding eq1 and eq3
2x = 14
x = 7
Substitute x in eq1
7 + y =11
y = 4

Therefore, the number is 10x + y = 10*4 + 7 = 47

Let the number be x ,then
-1/5 * x = 9/25
x = 9*5 / 25*-1
x = 45/ -25
x = - 45/25
x = - 9/5
Hence the number is -9/5

5000g = 5kg

Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then
work done in one min by A+B=1/20, B+C=1/40, C+A=1/30
solving the above equations,
A+B+ B+C+ C+A= 1/20+ 1/40+ 1/30
2A + 2B + 2C = 6+3+4 /120
2A + 2B + 2C = 13/120
A + B + C = 13/240

work done in one min by C = (A+B+C) - (A+B)
=> 13/240 - 1/20
=> 1/240
So Charles alone can complete the work in 240 min.

As is given, 60*n = 50*(n+1)
=> n=5

∴Sum of the money = 60*n= 60*5=>300
So, the correct answer is 300

Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

Let the number be x y.
When we multiply the number by the sum of its digit 405 is achieved.

(10x+y)(x+y)=405....................1

If we multiply the number written in reverse order by its sum of digits we get 486.

(10y+x)(x+y)=486......................2

dividing 1 and 2
(10x+y)(x+y)/(10y+x)(x+y) = 405/486
10x+y / 10y+x = 5/6
60x+6y = 50y+5x
55x = 44y
5x = 4y
x = 4y/5
now put the value in eq 1
(10x+y)(x+y)=405....................1
(10*4y/5+ y)(4y/5 + y)=405
(9y)(9y/5)=405
81*y^2/5 = 405
y^2 = 25
y=5
so X=4y/5 = 4

The number is 45

Let the one digit numbers x,y,z
Sum of all possible two digit numbers
=>(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22.

Let the quotient be q .
n=3(5(8q+7)+4)+1
=3(40q+39)+1
=120q+117+1
=120q+118.
Now, if it is divided by 8.
We have 8(15q+14)+6, Remainder will be 6.
And if 15q+14 is divisible by 5. Remainder will be 4.
And if 3q+2 is divisible by 3. Remainder will be 2.
Hence, will be the remainder, 6,4,2.

Let the numbers be a,b where a is smaller than b.
The AM of two numbers is smaller by 24 than the larger of the two numbers.
AM=b-24

AM of two numbers is a+b/2
=>a+b/2 = b-24
a+b = 2b-48
a = b-48
b = a + 48...................1

The GM of the two numbers exceeds by 12 the smaller of the numbers
GM = a+12

GM of two numbers is (ab)^1/2
=>(ab)^1/2= a+12
=>a*b = a^2 + 24*a + 144

from eq...1 b=a+48
a*(a + 48) = a^2 + 24*a + 144
a^2 + 48*a = a^2 + 24*a + 144
24*a = 144
a=6

Therefore, b=a+48=54

Let the digit number be x y.
given that x^2+y^2=13

And, num(x, y)-9= num(y, x)
10x+y-9=10y+x
9x-9y=9
x-y=1 ............e1

(x-y)^2 =x^2+y^2-2xy
1 =13-2xy
2xy = 12
xy = 6

x^2+y^2=13
(x+y)^2-2xy=13
(x+y)^2=13+2*6
(x+y)^2=13+12
(x+y)^2=25
(x+y)=5 .............e2

from e2+e1
(x+y) + (x-y) = 5+1
x=3
from e2-e1
(x+y) - (x-y) = 5-1
y=2

Therefore the number is, xy = 32

Let n(x,y) be two digit number
n(x,y)=10x+y
(square of of digits in ten’s place)+(product of digits)=52
x^2+xy=52
x(x+y)=52....................1

(square of digit in unit place)+(product of digits)=117
y^2+xy= 117
y(x+y)=117.........................2

dividing 1 and 2
x(x+y)/y(x+y) = 52/117=4/9
x/y=4/9

it is clear that two digit number is n(x,y),10x+y = 49