Simplification- Aptitude Questions and Answers - in Hindi

if 5668X25Y is divisible by 48 then it should be divisible by co-prime of 48.
There, 3*16 =>3*2*8 =>3*2*2*4=48 where 2, 3, 4, 8 and 16 are co-prime numbers of 48.

5668X25Y divisible by 8 , that means last 3 digit, 25Y should be divisible by 8.
So, Y should be 6

5668X256 to be divisible by 3, that means sum of digits 5+6+6+8+X+2+5+6 should be divisible by 3
So, 38+X is divisible by 3, that means X will be = 1 or 4 or 7

5668X256 divisible by 16,that means last four digits X256 should be divisible by 16
So, suppose X=1, then last four digits 1256 is not divisible by 16.
suppose X=4, then last four digits 4256 is divisible by 16.
suppose X=7, then last four digits 7256 is not divisible by 16.

Hence the correct number is 56684256,which is divisible by 48.
Therefore, least value of X+Y=4+6=10

Let the digits of number x and y, with x being the tens digit that means the number would be 10x+y.
as per question product of digits x*y = 12 ---------------eq1

When 36 is added to this number the digits inter change their places, so:
10x + y + 36 = 10y + x
9x + 36 = 9y
x + 4 = y ---------------- eq2

Now subsitute eq2 into eq1:
x*y = 12
x*(x + 4) = 12
x² + 4x - 12 = 0
(x + 6)(x - 2) = 0
x = -6 or 2
ignore the -ve value then x=2 and y=6. Hence number is 26 (there 26+36=62, which is reversed of 26).

as given, if time taken by Shailendra to do any work is X, then Harish would be do that in X/3 ,and Vipin would be do in X/12

Shailendra takes 156 minutes So X=156 minutes
thus Vipin would take X/12 => 156/12 => 13 minutes

Let the digits of number x and y, where x is the tens place, y is the ones place that means Number is: 10x + y

As given the sum of the digits is 8.
x + y = 8 -------------eq1
If the digits are reversed, the new number is 18 greater than the original number.
10y + x = 18 + 10x + y
y - x = 2
y = 2 + x --------------eq2
Substitute value of eq2 in eq1
x + 2 + x = 8
2x = 6
x = 3
So, y = 2 + x = 2 + 3 = 5

Hence number is: 10x + y = 10*3 + 5 = 35

Let 54 is divided into two parts (54 - x) and x.
As given, 10 (54 - x) + 22x = 780
12x = 240
x = 20 and (54 - x) = 34
thus product of the two numbers 20*34=>680

let number is x
then according to question
x/3+x/2+x/4 = x+12
(6x+4x+3x) / 12 = x+12
13x = 12(x+12)
13x = 12x+144
13x-12x = 144

hence number is x=144.

((144)² ÷ 48 × 18 ÷ 36)
=> (144*144*18) / (48*36)
=> (144*144*18) / (48*36)
=>216
=>√(216*216)
=>√46656

Let equal part be x and three parts are A ,B and C there
A *1/2= B *1/3=C *1/6= x
A = 2x, B = 3x, C =6x
A : B : C = 2 : 3 : 6
Thus,first part = 2x ,second part = 3x and third part = 6x
as sum total of all parts = 935
therefore 2x+3x+6x = 935
11x = 935
x = 935/11 = 85

hence third part = 6x =6*85=510

(1 + (27/169))^1/2=1+(x/13)
=>(1 + (27/169))^1/2=1+(x/13)
=>(196/169)^1/2=1+(x/13)
=>14/13=1+(x/13)
=>(14/13)-1=x/13
=>1/13=x/13

Hence x=1

(2x + 3y) : (3x + 5y) = 18 : 29
58x + 87y = 54x + 90y
58x - 54x = 90y - 87y
4x = 3y
Thus, x : y = 3 : 4

Let the digits of number x and y, then
x+y = 18 --------------eq1
x²+y² = 164 ------------eq2
Substitute value of x from eq1 to in eq2
x = (18-y)
(18-y)² + y² = 164
324 - 36y + y² + y² =164
324 - 36y + 2y² = 164
2y² -36y + 160 =0
2(y² -18y + 80)=0
(y-10)(y-8)=0
y = 8 or 10

from eq1, if y= 8, then x+y = x+8 = 18
x = 10
if y= 10, then x+y = x+10 = 18
x = 8

So the numbers are 8 and 10.and smallest number is 8

(7 + 7 + 7 / 7) / (5 + 5 + 5 / 5)
= (14 + 1) / (10 + 1)
=15 / 11

Let x be the number of 5 paise coins, then (26 - x) will be the number of 2 paise coins.
So, 5*x + 2*(26 - x)= 100
5x + 52 - 2x = 100
3x = 48
x = 16
Number of 5 paise coins = 16
Number of 2 paise coins = 26 - 16 = 10

A - B = (x + 1)/(x - 1) - (x - 1)/(x + 1)
=(x + 1)² - (x - 1)² / (x² - 1²)
=(x² + 1 + 2x - x² - 1 + 2x) / (x² - 1)
=4x / (x² - 1)

total number of workers = 125*9 =1125

As given,
A+C=94 -----eq1
B+C=76 -----eq2
A+B+C=150-----eq3

from eq2 and eq3 ,
A+76 =150
A=74
now from eq1,
74+C=94
C=20