Fractions And Decimals(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

FRACTIONS AND DECIMALS MCQs

Fractions, Decimals

Total Questions : 227 | Page 20 of 23 pages

Question 191.

Express 3460 grams in kilograms.
Now express $17 \frac{1}{2}$ kg in grams. [2 MARKS]

Discuss Question

Answer: Option A. ->
:

In kilograms: 1 Mark
In grams: 1 Mark
1 Kilogram = 1000 grams
Or 1 gram= $\frac{1}{1000}$ kilograms
Let x kilograms = 3460 g
$= (\frac{3460}{1000})$ kilograms
= 3.460 kilograms
3460 grams = 3.460 kilograms
Now $17 \frac{1}{2} kg = \frac{17 \times 2 + 1}{2}$
= $\frac{34 + 1}{2}$
= $\frac{35}{2} kg$
= $\frac{35}{2} \times 1000 gms$
= $(35 \times 500) gm$
= $17500 gms$

Question 192.

Kartik ate $\frac{3}{5}$ part of an apple and Bharat ate the remaining part of the apple. What fraction of the apple did Bharat eat? Who had the larger share and by how much? [2 MARKS]

Discuss Question

Answer: Option A. ->
:
Each answer: 1 Mark
Given that:
Fraction of the apple Kartik ate

= \frac{3}{5}

Fraction of the apple uneaten

= 1 - \frac{3}{5}

= \frac{5 - 3}{5}

= \frac{2}{5}

This part was eaten by his brother Bharath

∴

Bharath ate

\frac{2}{5}

th part of the apple.
Comparing two fractions we have

\frac{3}{5} & \frac{2}{5}

Clearly

\frac{3}{5} > \frac{2}{5}

\Rightarrow

Karthik had the larger share which was larger by

\frac{3}{5} - \frac{2}{5} = \frac{1}{5}

Question 193.

How much less is 18km compared to 29000 metres ? [3 MARKS]

Discuss Question

Answer: Option A. ->
:

Conversion Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
We know that
1 km=1000 m
Converting kilometers into meters
we have
18 km = 18 $\times$ 1000=18000 metres

Difference = 29000-18000

=11000 metres

29000 metres= 29 km

Difference = 29-18

=11 km

Question 194.

Given that length of a side of an equilateral triangle is 3.4m. Find its perimeter in cm? [3 MARKS]

Discuss Question

Answer: Option A. ->
:

Conversion Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
In an equilateral triangle, all the sides are equal.

Let the side of an equilateral triangle be= a cm where a= 3.4 m ( given)

Now Perimeter = a+ a+a=(3a) m

= (3)*(3.4)

=10.2m
Hence perimeter of the equilateral triangle = 10.2m.
Now, 1 m= 100cm
$∴$ 10.2m=10.2 $\times$ 100 = 1020 cm.
Hence, the perimeter of the triangle is 1020 cm.

Question 195.

Mohan finished colouring a picture in $\frac{7}{12}$ hour. Keerthi finished colouring the same picture in $\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer? [3 MARKS]

Discuss Question

Answer: Option A. ->
:
Steps: 2 Mark
Answer: 1 Mark
Given that
Mohan finishes the work in

\frac{7}{12}

of an hour.
Also given that
Keerthi finishes the work in

\frac{3}{4}

of an hour.
Now,

\frac{7}{12}

of an hour

\Rightarrow \frac{7}{12} \times 60 m i n u t e s

\Rightarrow 35 m i n s

Again

\frac{3}{4}

of an hour

\Rightarrow \frac{3}{4} \times 60

\Rightarrow 45 m i n u t e s

Clearly, Keerthi worked more by 45 - 35 = 10 more minutes.
Now, when expressed as a fraction

\frac{10}{60} = \frac{1}{6}

of an hour.
Alternative Way
Method II
Now,

\frac{7}{12} a n d \frac{3}{4}

Making them Like fractions we have

\frac{7 \times 1}{12 \times 1} & \frac{3 \times 3}{4 \times 3}

i.e.

\frac{7}{12} < \frac{9}{12}

Clearly, Keerthi who takes

{\frac{3}{4}}^{t h}

of an hour takes longer than Mohan who takes

{\frac{7}{12}}^{t h}

of an hour.
So, subtracting we have

\frac{9}{12} - \frac{7}{12} = \frac{2}{12} = \frac{1}{6}

of an hour.

Question 196.

A circle is divided into 8 equal parts. Given that the area of the shaded portion is $\frac{10}{4}$ $m^{2}$ . Find the area of the unshaded part of the circle.

[4 MARKS]

Discuss Question

Answer: Option A. ->
:

Steps: 2 Marks
Application: 1 Mark
Result: 1 Mark
Given that
Area of the shaded portion = $\frac{10}{4}$ $m^{2}$

The circle is divided into 8 equal parts

Area of the circle = (8) $\times$ $\frac{10}{4}$

= 20 $m^{2}$
Area of unshaded part = Total area - shaded area
   = 20 - $\frac{10}{4}$ $m^{2}$
   = $\frac{80 - 10}{4} m^{2}$
   = $\frac{70}{4}$ = 17.5 $m^{2}$
Hence, area of the unshaded portion is 17.5 $m^{2}$

Question 197.

In cricket, run rate is defined as the number of runs scored divided by the number of overs bowled. Team 'A' scores 260 runs in 50 overs batting first. The other team batting second scores at half the run rate of the team batting first and plays entire 50 overs. How many runs did the team batting second score? [3 MARKS]

Discuss Question

Answer: Option A. ->
:

Steps: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Given that
Team 'A' scored 260 runs in 50 overs.
So, run rate of the team batting first =(260/50)

= 5.2 runs per over
The team batting second scored at half the run rate of the team batting first

Run rate of team the batting second
= (5.2/2)

= 2.6 runs per over

Runs scored by the team batting second
$= (50) \times (2.6)$

= 130
Hence, the team batting second scored 130 runs.

Question 198.

Express $\frac{(0.8 \times 0.7) - (0.65 \times 0.6)}{(0.65 \times 0.6) - (0.62 \times 0.6)}$ in the form of a fraction.
[4 MARKS]

Discuss Question

Answer: Option A. ->
:

Bodmas operation: 1 Mark
Simplifying the numerator and the denominator: 1 Mark
Answer: 2 Marks
Multiply and divide the equation by 100
= $\frac{(8 \times 7) - (6.5 \times 6)}{(6.5 \times 6) - (6.2 \times 6)}$
= $\frac{(56 - 39)}{(39 - 37.2)}$
= $\frac{17}{1.8}$
= $\frac{170}{18}$
= $\frac{85}{9}$

Question 199.

Add the following decimals and then express it as a fraction.
7.205, 8.012, 3.1, 4.9, 27.3, 3.0009, 8.0102. [4 MARKS]

Discuss Question

Answer: Option A. ->
:
Making them into like decimals: 1 Mark
Mathematical operation: 1 Mark
Fraction expression: 2 Marks
Now converting the decimals given into like decimals and adding it.

0 - 7.205 0 -

0 - 8. 012 0 -

0 - 3. 1 0 - 0 - 0 -

0 - 4. 9 0 - 0 - 0 -

2 7. 3 0 - 0 - 0 -

0 - 3. 0009

0 - 8. 0102

We get

6 1. 4281

(II) Now, 61. 4281 can be rounded up
to 61.428
to 61. 43
to 61. 4
Now

61.4 = \frac{61.4 \div 2}{100 \div 2} = \frac{307}{50} = 6 \frac{7}{50}

Question 200.

Anubha's father weighs 91 kg. Anubha's mother is 5 times Anubha's dog's weight. If Anubha's dog weighs $\frac{1}{7}$ of Anubha's father's weight, find Anubha's mother's weight. [4 MARKS]