Fractions and Decimals Test

Result

Fractions and Decimals Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Find the sum of $$1.2$$ and $$2.3$$.

A
$$35$$

B
$$3.5$$

C
$$0.35$$

D
$$1.23$$

Solution

The sum of $$1.2$$ and $$2.3$$ will be $$1.2+2.3=3.5$$
Question 2
1 / -0

$$5.6 \times 4.5 + 3.4= ?$$

A
123.42

B
2613.76

C
243.12

D
28.6

Solution

$$5.6\times4.5=25.2$$
so, $$5.6\times4.5 +3.4=25.2+3.4=28.6$$
Question 3
1 / -0

Solve: $$2 \dfrac { 1 } { 2 } \mathrm { } \text { of } 10 \mathrm { cm }$$

A
30 cm

B
25 cm

C
20 cm

D
50 cm

Solution

Given $$2 \dfrac { 1 } { 2 } \mathrm { } \text { of } 10 \mathrm { cm }$$

$$=2\dfrac{1}{2}\times 10$$

$$=\dfrac{2\cdot2+1}{2}\times 10$$

$$=\dfrac{5}{2}\times 10$$

$$=5\times 5=25$$
Question 4
1 / -0

Simplify the expression $$2\dfrac{1}{4}\times \dfrac{5}{12}+\dfrac{1}{2}$$

A
$$\dfrac{23}{16}$$

B
$$5\dfrac{5}{2}$$

C
$$4\dfrac{3}{3}$$

D
$$3\dfrac{1}{5}$$

Solution

$$2\cfrac { 1 }{ 4 } \times \cfrac { 5 }{ 12 } +\cfrac { 1 }{ 2 } $$

$$=\cfrac { 9 }{ 4 } \times \cfrac { 5 }{ 12 } +\cfrac { 1 }{ 2 } $$

$$ =\cfrac { 15 }{ 16 } +\cfrac { 1 }{ 2 } $$

$$ =\cfrac { 15+8 }{ 16 } $$

$$ =\cfrac { 23 }{ 16 } $$
Question 5
1 / -0

A proper fraction with denominator $$10$$ is

A
$$\dfrac{8}{7}$$

B
$$\dfrac{4}{7}$$

C
$$\dfrac{6}{10}$$

D
$$\dfrac{11}{7}$$

Solution

In a proper fraction numerator of the fraction is less than the denominator of the fraction. Also, the denominator must be $$10$$ according to the question.

Hence, option $$C$$ is correct.
Question 6
1 / -0

Which one of the following is an improper fraction

A
$$\dfrac{11}{4}$$

B
$$\dfrac{3}{8}$$

C
$$\dfrac{3}{4}$$

D
$$\dfrac{9}{11}$$

Solution

We have,
let, $$2\dfrac{3}{4}$$
$$=\dfrac{(4\times 2)+3}{4}$$
$$=\dfrac{8+3}{4}$$
$$=\dfrac{11}{4}$$
Hence, this is answer.
Question 7
1 / -0

The improper fraction is :

A
$$\dfrac { 12 }{ 15 } $$

B
$$\dfrac { 13 }{ 17 } $$

C
$$\dfrac { 16 }{ 21 } $$

D
$$\dfrac { 25 }{ 11 } $$

Solution

A fraction in which the numerator is greater than the denominator is an improper fraction.
$$(a)\dfrac{12}{15}:$$ Here $$12<15\Rightarrow$$Numerator $$<$$ Denominator.Hence it is not an improper fraction.

$$(b)\dfrac{13}{17}:$$ Here $$13<17\Rightarrow$$Numerator $$<$$ Denominator.Hence it is not an improper fraction.

$$(c)\dfrac{16}{21}:$$ Here $$16<21\Rightarrow$$Numerator $$<$$ Denominator.Hence it is not an improper fraction.

$$(d)\dfrac{25}{11}:$$ Here $$25>11\Rightarrow$$Numerator $$>$$ Denominator.Hence it is an improper fraction.
Question 8
1 / -0

Which of the following is a proper fraction?

A
$$5/3$$

B
$$5$$

C
$$2$$

D
none of these

Solution

If the numerator is less than the denominator then the fraction is called as proper fraction.
Hence none of these are proper fractions.
Question 9
1 / -0

The product of $$\frac{11}{13}$$ and 4 is:

A
$$3\frac {5}{13}$$

B
$$5\frac{3}{13}$$

C
$$13\frac{3}{13}$$

D
$$13\frac{5}{13}$$

Solution
Question 10
1 / -0

$$3.\overline{25}$$ is equal to

A
$$ 2+\dfrac{25}{99} $$

B
$$ 3+\dfrac{24}{99} $$

C
$$ 3+\dfrac{25}{99} $$

D
$$ 3+\dfrac{25}{98} $$

Solution

$$3.\bar { 25 } =3+25*0.\bar { 01 } =3+25*\dfrac { 1 }{ 99 } =3+\dfrac { 25 }{ 99 } $$

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Fractions and Decimals Test - 27

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Fractions and Decimals Test - 27

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SHARING IS CARING

Question 1

$$35$$

$$3.5$$

$$0.35$$

$$1.23$$

Solution

Question 2

123.42

2613.76

243.12

28.6

Solution

Question 3

30 cm

25 cm

20 cm

50 cm

Solution

Question 4

$$\dfrac{23}{16}$$

$$5\dfrac{5}{2}$$

$$4\dfrac{3}{3}$$

$$3\dfrac{1}{5}$$

Solution

Question 5

$$\dfrac{8}{7}$$

$$\dfrac{4}{7}$$

$$\dfrac{6}{10}$$

$$\dfrac{11}{7}$$

Solution

Question 6

$$\dfrac{11}{4}$$

$$\dfrac{3}{8}$$

$$\dfrac{3}{4}$$

$$\dfrac{9}{11}$$

Solution

Question 7

$$\dfrac { 12 }{ 15 } $$

$$\dfrac { 13 }{ 17 } $$

$$\dfrac { 16 }{ 21 } $$

$$\dfrac { 25 }{ 11 } $$

Solution

Question 8

$$5/3$$

$$5$$

$$2$$

none of these

Solution

Question 9

$$3\frac {5}{13}$$

$$5\frac{3}{13}$$

$$13\frac{3}{13}$$

$$13\frac{5}{13}$$

Solution

Question 10

$$ 2+\dfrac{25}{99} $$

$$ 3+\dfrac{24}{99} $$

$$ 3+\dfrac{25}{99} $$

$$ 3+\dfrac{25}{98} $$

Solution

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