MCQ Questions for Class 7 Maths Fractions And Decimals Quiz 7

State true or false

$3.68\times 5=18.4$

0%
True
0%
False

Explanation

Multiply

$3.68 \times 5$

$= \dfrac{{368}}{{100}} \times 5$

$= \dfrac{{1840}}{{100}}$

$18.4$

The sum of two numbers is

$31.021$ . If one of them is

$11.56$ , then the other number is _______ .

0%
$19.461$
0%
$17.461$
0%
$18.641$
0%
$19.561$

The product of

$3.92\times 0.1\times 0.0\times 6.3$ is _______

0%
$0.392$
0%
$0.1176$
0%
$0$
0%
$6.3$

State true or false:

$1.5\times 0.3 = 0.45$

0%
True
0%
False

Explanation

$1.5\times 0.3 = \dfrac{15}{10}\times \dfrac{3}{10}$

$= \dfrac{45}{100}$

$= 0.45$

Ans. True

If the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is?

0%
Always greater than the original fraction
0%
Always less than the original fraction
0%
Always equal to the original fraction
0%
None of the above

Explanation

If the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is always greater than the original fraction.

Let us understand and verify it with the help of couple of examples:

$(i)$ Let

$\dfrac{1}{2}$ be an original fraction.

$\dfrac{1}{2}=0.5$

If the numerator and denominator are increased by

$5$ , the fraction becomes:

$\dfrac{1+5}{2+5}=\dfrac{5}{7}=0.71$

$\therefore$

$\dfrac{1}{2}<\dfrac{5}{7}$

$(ii)$ Let

$\dfrac{7}{9}$ be an original fraction.

$\dfrac{7}{9}\approx 0.78$

If the numerator and denominator are increased by

$1$ , the fraction becomes:

$\dfrac{7+1}{9+1}=\dfrac{8}{10}=0.8$

$\therefore$

$\dfrac{7}{9}<\dfrac{8}{10}$

Hence, if the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is always greater than the original fraction.

Therefore, option A is correct.

$8m$

$57cm$ pole was put in a pond to measure its depth. If

$4m$

$66cm$ of pole remained outside water, then depth of pond is _____ .

0%
$3.91m$
0%
$3.00m$
0%
$3.45m$
0%
$3.23m$

Evaluate:

$3.7-1.9$

0%
$0.6$
0%
$7.3$
0%
$3.4$
0%
$1.8$

Explanation

$3.7-1.9$

$=1.8$

The simplification of

$33.6 - 2 { .05 } + 0.133$ is equal to

0%
$31.63$
0%
$29.643$
0%
$25. { 61 }3$
0%
None of these

A proper fraction with denominator

$3$ , which is less than 1 is

0%
$\frac{4}{3}$
0%
$\frac{2}{3}$
0%
$1 \frac{2}{3}$
0%
None

Explanation

Clearly from question denominator

$=3$

numerator

$=2$

Fraction

$=\dfrac{2}{3}$

A fraction whose numerator is greater than its denominator is fraction.

0%
an improper
0%
a proper
0%
a mixed
0%
None of these

Explanation

That will be an improper fraction

$\dfrac{6}{5}$

Multiply

$\dfrac{6}{13}$ by the reciprocal of

$\dfrac{-7}{16}$

0%
$\dfrac{-95}{91}$
0%
$\dfrac{-96}{91}$
0%
$\dfrac{96}{91}$
0%
None of these

Explanation

Reciprocal of

$\dfrac{-7}{16}$ is

$\dfrac{-16}{7}$

Now

$\dfrac{6}{13}\times$ Reciprocal of

$\dfrac{-7}{16}$

$=\dfrac{6}{13}\times\dfrac{-16}{7}$

$=\dfrac{6\times -16}{13\times 7}=\dfrac{-96}{91}$

Solve the following equation

$\dfrac25\times\dfrac{-3}7-\dfrac16\times\dfrac32+\dfrac1{14}\times\dfrac25$

0%
$7$
0%
$4$
0%
$\dfrac{-11}{28}$
0%
$\dfrac{4}{3}$

Explanation

Given,

$\dfrac{2}{5}\times \dfrac{-3}{7}-\dfrac{1}{6} \times \dfrac{3}{2}+\dfrac{1}{14} \times \dfrac{2}{5}$

$=-\dfrac{6}{35}-\dfrac{1}{4}+\dfrac{1}{35}$

$=-\dfrac{1}{7}-\dfrac{1}{4}$

$=\dfrac{-11}{28}$

Simplify:

$\displaystyle\frac{2.3\times2.3}{2.3\times2.3-2\times2.3\times2.3+2.3\times2.3+1}$

0%
$0$
0%
$3.24$
0%
$1.96$
0%
$5.29$

Explanation

We have,

$\dfrac { 2.3\times 2.3 }{ 2.3\times 2.3-2\times 2.3\times 2.3+2.3\times 2.3+1 }$

$\dfrac { 2.3\times 2.3 }{ 0+1 }$

$=2.3\times 2.3=5.29$

Hence, this is the answer.

Mark against the correct answer in each of the following:

$( 1.007 - 0.7) = ?$

0%
$1$
0%
$0.37$
0%
$0.307$
0%
None of these

Explanation

$3.07$

First convert the given decimals into like decimals

$= (1.007 – 0.700)$

$= 0.307$

$5.2 - 3.6$ is equal to

0%
$0.16$
0%
$2.6$
0%
$0.26$
0%
$1.6$

Explanation

$5.2 - 3.6 = \dfrac{52}{10} - \dfrac{36}{10}$

$= \dfrac{16}{10}$

$=1.6$

$\dfrac{11}{7}$ can be expressed in the form

0%
$7 \dfrac{1}{4}$
0%
$4 \dfrac{1}{7}$
0%
$1 \dfrac{4}{7}$
0%
$11 \dfrac{1}{7}$

Explanation

Numbers of the type

$7 \dfrac{1}{4} , 4\dfrac{1}{7}, 1\dfrac{4}{7} , 11\dfrac{1}{7}$ …….. are called mixed fractions. A mixed fraction is a combination of whole and fractional parts.

To find a mixed fraction, divide: 11 by 7

$\dfrac{11}{7} = \dfrac{1 × 7 + 4}{7}$

Where, 1 is the quotient and 4 is the remainder

A mixed fraction is written as

$Quotient \dfrac{remainder}{divisor}$

$\dfrac{11}{7}=1 \dfrac{4}{7}$

$\therefore \textbf{Option C is correct.}$

$15.8-6.73$ is equal to

0%
$8.07$
0%
$9.07$
0%
$9.13$
0%
$9.25$

Explanation

To subtract two decimals, convert 15.8 into like decimals with 2 decimal places,

$15.8 =15.80$

Now, subtracting decimals we get,

$15.80 - 6.73 = 9.07$

$\therefore \textbf{Option B is correct.}$

$0.07 + 0.008$ is equal to

0%
$0.15$
0%
$0.015$
0%
$0.078$
0%
$0.78$

Explanation

To add two decimals, we need to convert 0.07 to 3 decimal places,

0.07 = 0.070

Now adding both the like decimals 0.070 and 0.008, we get

$\therefore \textbf{Option (C) is correct}$

The mixed fraction

$5 \dfrac{4}{7}$ can be expressed as

0%
$\dfrac{33}{7}$
0%
$\dfrac{39}{7}$
0%
$\dfrac{33}{4}$
0%
$\dfrac{39}{4}$

Explanation

An improper fraction is obtained by adding the whole part with a fractional part of a mixed fraction.

$5\dfrac{4}{7}$ can be expressed into improper fraction as ,

$= \dfrac{ 5×7+4}{7}$

$= \dfrac{35+4}{7}$

$= \dfrac{39}{7}$

$\therefore \textbf{Option B is correct.}$

The product of 7 and

$6\frac{3}{4}$ is:

0%
$42 \frac{1}{4}$
0%
$47\frac{1}{4}$
0%
$42\frac{3}{4}$
0%
$47\frac{3}{4}$

$42.28-3.19=39.09 \quad$

0%
True
0%
False

Explanation

$42.28 - 3.19 = 39.09$

$42.28$

$-3.19$
__________

$39.09$

$3.03+ 0.016=3.019$

0%
True
0%
False

Explanation

$\Rightarrow 3.030 + 0.016 = 3.046$

$\Rightarrow 3.046 \neq 3.019$

State whether the following statements are true

$\left(T\right)$ or false

$\left(F\right)$ :
A fraction in which the numerator is greater than is denominator is called an improper fraction

0%
True
0%
False

The product of

$0.03$ and

$0.9$ is:

0%
$2.7$
0%
$0.27$
0%
$0.027$
0%
$0.0027$

Explanation

We need to find the product of

$0.03$ and

$0.9$

$0.03\times 0.9=0.027$

Hence, the correct option is

$Op-C$

$24.58=2\times 10+4\times 1+5\times 10+8\times 100$

0%
True
0%
False

Explanation

False

$RHS=2\times 10+4\times 1+5\times 10+8\times 100=20+4+50+800=874\\ LHS\neq RHS$

$329.25=3\times 10^{2}+2\times 10^{1}+9\times 10^{0}+2\times 10^{-1}+5\times 10^{-2}$

0%
True
0%
False

Explanation

True

$RHS=3\times 10^{2}+2\times 10^{1}+9\times 10^{0}+2\times 10^{-1}+5\times 10^{-2}$

$=3\times 100+2\times 10+9\times 1+\dfrac{2}{10}+\dfrac{5}{10\times 10}$

$=3\times 100+2\times 10+9\times 1+0.2+0.05=329.25$
Therefore,

$LHS=RHS$

$0.31$ is greater than

0%
$0.32$
0%
$0.30$
0%
$0.34$
0%
$0.33$

$5.462$ is greater than

0%
$5.352$
0%
$5.468$
0%
$5.523$
0%
$5.469$

$1.461$ is less than

0%
$1.462$
0%
$1.469$
0%
$1.490$
0%
$1.450$

$0.046$ is less than

0%
$0.048$
0%
$0.045$
0%
$0.056$
0%
$0.050$

CBSE Questions for Class 7 Maths Fractions And Decimals Quiz 7 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Fractions And Decimals Quiz 7 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

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