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Aptitude Decimal Fraction Test Paper 5


21) Three of the following four options (a), (b), (c), and (d) are exactly equal. Which option is not equal to the other three options?

  1. 1.2* 10 /0.1*100
  2. 0.6* 0.5/0.01*0.5
  3. 2.4*1000/0.1*0.5
  4. 0.2*0.06/0.001*1000

Answer: B

Explanation:

Convert all the fractions in the decimal numbers
According to option a, 1.2 * 10/0.1*100 = 12000
According to option b, 0.6*0.5/0.01*0.5 = 15
According to option c, 2.4* 1000/0.1 * 0.5 = 12000
According to option d, 0.2*0.06/ 0.001*1000 = 12000

Clearly, we can say that option B has a different value from the other three options.


22) The value of (0.333...) * (0.444....) is

  1. 0.121212....
  2. 1.3333.....
  3. 0.777......
  4. 0.148148148.....

Answer: D

Explanation:

We can write (0.3333....) = 0.3̅
Period of 0.3̅ is 3, so the numerator of the fraction is 3, and as there is one digit in the period, the denominator will have one nine.
Therefore, the vulgar fraction = 3/9
Similarly, (0.444...) = 0.4̅
So, fraction = 4/9

Hence, (3/9) * (4/9) = 12/81 = 0.148148....


23) The value of the expression [(0.05)2 + (0.41)2 + (0.073)2]/ [(0.005)2 + (0.041)2 +(0.0073)2] is

  1. 100
  2. 10
  3. 1000
  4. None of these

Answer: A

Explanation:

The expression is [(0.05)2 + (0.41)2 + (0.073)2]/ [(0.005)2 + (0.041)2 + (0.0073)2]

Or, [(0.05)2 + (0.41)2 + (0.073)2]/ [(0.05/10)2 + (0.41/10)2 + (0.073/10)2]

Or, [(0.05)2 + (0.41)2 + (0.073)2]/ [(0.05)2+ (0.41)2 + (0.073)2] *(1/100)

We know that x/y/100 = x*100/ y

Similarly, [(0.05)2 + (0.41)2 + (0.073)2] * 100/ [(0.05)22 + (0.41)2 + (0.073)2]

Now, the numerator and denominator have the same value, so, they will cancel out each other.

Hence, the value of the expression will be 100.


24) The value of [(0.43)3 + (1.47)3 + (1.1)3 - 3*0.43*1.47*1.1]/ [(0.43) 2 + (1.47) 2+ (1.1) 2 - 0.43*1.43 - 0.43*1.1 - 1.47*1.1] is

  1. 1.90
  2. 2.87
  3. 3.47
  4. 3

Answer: D

Explanation:

We know that [a3 + b3 + c3 - 3abc]/ [a 2 + b 2 + c 2 -ab- bc-ca] = [a+b+c]
Now, compare the given expression with the formula, we get
a = 0.43, b = 1.47, and c = 1.1
So the required value = a + b + c = 0.43 + 1.47 + 1.1 = 3.



Aptitude Decimal Fraction Test Paper 1
Aptitude Decimal Fraction Test Paper 2
Aptitude Decimal Fraction Test Paper 3
Aptitude Decimal Fraction Test Paper 4
Decimal Fraction Concepts





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