Test: Fractions- 2 - Class 5 MCQ

When a fraction is multiplied by its reciprocal, the result is always 1.

For example, if the fraction is a/b, its reciprocal is b/a. Multiplying these two gives:

This property holds true for any non-zero fraction. Hence, the product of a fraction and its reciprocal is always 1.

Reciprocals of 9 and 2/9 are 1/9 and 9/2 respectively.
Their sum =

1 / 8 < 1 / 4. So, it should lie between 0 and 1.

To find which fraction is greater than 7/8, let's compare each option with 7/8 by converting them into decimals.

• 7/8 = 0.875
• 2/3 ≈ 0.6667
• 5/8 = 0.625
• 9/11 ≈ 0.8182
• 11/12 ≈ 0.9167

Now, compare these decimal values:

• 2/3 (0.6667) is less than 7/8 (0.875).
• 5/8 (0.625) is less than 7/8 (0.875).
• 9/11 (0.8182) is less than 7/8 (0.875).
• 11/12 (0.9167) is greater than 7/8 (0.875).

Thus, 11/12 is the fraction greater than 7/8.

On cancelling the factors common to numerator and denominator on both the sides,
          


Hence the missing number is 5.

Let the original fraction be x/y, where x is the numerator and y is the denominator. We know two things:

1. The sum of the numerator and the denominator is 67:
   x + y = 67
2. When 31 is added to the denominator, the fraction becomes 3/11:
   x / (y + 31) = 3 / 11

We can now solve the system of equations:

From the second equation:
x / (y + 31) = 3 / 11
Cross-multiply to get:
11x = 3(y + 31)
Expanding the right side:
11x = 3y + 93

Now, we have two equations:

1. x + y = 67
2. 11x = 3y + 93

From the first equation, solve for x:
x = 67 - y

Substitute this value of x into the second equation:
11(67 - y) = 3y + 93
Expanding the left side:
737 - 11y = 3y + 93

Now, move all terms involving y to one side:
737 - 93 = 3y + 11y
Simplifying:
644 = 14y
Solving for y:
y = 644 / 14 = 46

Now, substitute y = 46 into the equation x + y = 67:
x + 46 = 67
x = 67 - 46 = 21

So, the original fraction is 21/46.

Let the marks scored by Prateek be x.

According to the problem, Pavan scored 4/5​ of what Prateek scored. So, Pavan's marks are:

Now, solve for x:

So, Prateek scored 95 marks.

The total marks scored by both Pavan and Prateek is:

76 + 95 = 171

Thus, their total marks are 171 marks.

Amount Manish had = Rs.45
Fraction of the amount spent = 35
∴ Amount left = 

The equation is:
_____ + ⁴/₅ = ⁴/₅
To satisfy this equation, we need to add a number that keeps the total sum equal to 4/5​. The only number that can be added to 4/5 without changing its value is 0.

So, the blank should be filled with 0 to make the equation true.
0 + ⁴/₅ = ⁴/₅