# Aptitude Decimal Fraction Test Paper 1

1) Find the missing term of the given expression:
18.834 + 818.34 -? = 618.43

1. 217.644
2. 218.744
3. 217.744
4. 217.844

Explanation:

18.834 + 818.34 - ? = 618.43

Or, 18.834 + 818.340 - 618.43 =?

837.174 - 618.430 = 218.744

2) What is the value of (25.732)2 - (15.732)2?

1. 414.64
2. 414.256
3. 414.128
4. 414.52

Explanation:

We know that, a2 - b2 = (a+b) (a - b)

Or, (25.732)2 - (15.732)2 = (25.732 + 15.732) (25.732 - 15.732)

Or, the value = 41.464 * 10 = 414.64

3) Approximate value of 0.4*0.4 + 0.04*0.04 + 0.04 is

1. 0.20
2. 0.38
3. 0.16
4. 0.46

Explanation:

0.4 * 0.4 = 0.16

0.04*0.04 = 0.0016

Hence, 0.16+0.0016+0.04 = 0.2016, or the approximate value = 0.20

4) The value of 0.3̅4̅2̅1 is

1. 3421/10000
2. 3421/9.999
3. 3421/9999
4. None of these

Explanation:

Solution 1:

We have 0.3̅4̅2̅1 = 0.342134213421........ (i)

Multiply both sides by 10, we get

10 * 0.3̅4̅2̅1 = 3.421................... (ii)

But on comparing both equations, the repeating number should be the same after the decimal point.

So, multiply equation 2 by 1000, we get
10000* 0.3̅4̅2̅1 = 3421.342134213421.... ....... (iii)

Now subtract equation i from iii, we get
(10000 - 1) * 0.3̅4̅2̅1 = 3421.342134213421... - 0.342134213421.....
Or, 9999 * 0.3̅4̅2̅1= 3421
Or, 0.3̅4̅2̅1 = 3421/9999

Solution 2:

Quick method:

Note: the period of 0.7̅ is 7, so the numerator of the fraction is 7, and there is one digit in the period, the denominator will have one nine.

Therefore, the vulgar fraction = 7/9

Similarly, the period of 0.3̅4̅2̅1 is 3421, so the numerator of the fraction is 3421, and there is four digit in the period, the denominator will have four nine.

Hence, the vulgar fraction will be 3421/9999.

5) In the given expression (1.05)2 *x = 44.1, find the value of x.

1. 40.00
2. 40.01
3. 42.00
4. 42.01

Explanation:

Given, (1.05)2 * x =44.1
Or, x = 44.1/ (1.05)2 = 44.1/ (1.05 * 1.05)
Hence, x = 40.00

Aptitude Decimal Fraction Test Paper 2
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Decimal Fraction Concepts

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