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

Question 211.

Question 6
Lipika reads a book for $1 \frac{3}{4}$ hours in a day. If she read the entire book in 6 days how many hours did she take to read the book? In how many days will she be able to complete the book if she would have read for $\frac{1}{2}$ hours each day? [3 Marks]

Discuss Question

Answer: Option A. ->
:
Steps: 1 Mark
Calculation: 2 Marks
Time taken by Lipika to read a book

= 1 \frac{3}{4} h o u r s

She reads the entire book in 6 days.
Now, total number of hours she took to read the entire book

= 1 \frac{3}{4} \times 6 = \frac{7}{4} \times 6 = \frac{21}{2} = 10 \frac{1}{2} h o u r s

Now if she reads for

\frac{1}{2}

hours daily, time required by her to complete the book

= 10 \frac{1}{2} \div \frac{1}{2} = \frac{21}{2} \div \frac{1}{2} = 21

Hence, if she reads for half an hour daily she will finish reading the book in 21 days.

Question 212.

The length of the shaded rectangle is $(\frac{1}{10})$ of the bigger rectangle and its breadth is 0.001 times the bigger rectangle. If the area of the bigger rectangle is $10000 m^{2}$ . Find the area of the shaded rectangle. [4 MARKS]

Discuss Question

Answer: Option A. ->
:

Steps: [3 Marks]
Answer: [1 Mark]
Let the length and breadth of the bigger rectangle be $a$ meters and $b$ meters respectively
As per the question.
The length of the shaded rectangle is $(\frac{1}{10})$ of the bigger rectangle.

The length of the shaded rectangle is $0.1 \times a$ meters.
The breadth of the shaded rectangle is 0.001 times the bigger rectangle.

The breadth of the shaded rectangle is $0.001 \times b$ meters.

Area of the given bigger rectangle $= a \times b$ metres $= 10000 m^{2}$

Area of the shaded rectangle $= (0.1 a) \times (0.001 b)$

$= (0.0001) \times (a b)$

$= (0.0001) \times (10000)$

= 1 m^{2}

So, the area of the shaded rectangle is

1 m^{2}

.

Question 213.

A tap fills one half of a bottle in 1 minute. Three fourth of the remaining in next 1 minute. If the remaining volume is 7 $m^{3}$ , then find the volume of the bottle.
[4 MARKS]

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Answer: Option A. ->
:

Steps: [2 Marks]
Application: [1 Mark]
Answer: [1 Mark]
Let the total volume of the bottle be V $m^{3}$
Volume filled in the first minute is $\frac{V}{2}$
Volume of the bottle remaining= $V - \frac{V}{2} = \frac{V}{2}$
Volume filled in the second minute is $(\frac{3}{4}) \times (\frac{V}{2}) m^{3}$
Volume filled in the third minute is $V - (\frac{V}{2}) - (\frac{3}{4}) \times (\frac{V}{2})$
$= (\frac{V}{8}) m^{3}$
Then the remaining volume $= 7 m^{3}$
$(\frac{V}{8}) = 7$
$V = 56 m^{3}$
The volume of the bottle is $56 m^{3}$

Question 214.

Ram purchased 1.5 kgs of tomato for 35 rupees. If 5.6 kg of tomato is required to prepare 5 kg tomato pickle. What is the cost of 1 kg tomato pickle, given that only tomatoes are used for the preparation of the pickle? [4 MARKS]

Discuss Question

Answer: Option A. ->
:

Steps: 3 Marks
Result: 1 Mark
Given that
Ram purchased 1.5 kgs of tomato for 35 rupees.
Also, 5.6 kg of tomato is required to prepare 5 kg of tomato pickle.
The first step is to find per kg cost of tomato.
Now, cost of 1 kg of tomato = $\frac{35}{1.5}$ rupees = $\frac{35 \times 2}{1.5 \times 2}$ = $\frac{70}{3}$ rupees
Given that 5.6 kg of tomato is required to prepare 5 kg tomato pickle
Then tomatoes required for making of 1 kg of pickle = $\frac{5.6}{5}$ kgs = 1.12 kgs
Cost of 1kg of tomato pickle $= (1.12) \times \frac{70}{3}$ = 26.133 rupees

Question 215.

Look at the figure above. This is a staircase consisting of 5 steps. The height of each in-step(green area) is 6 cms.
(i) What is the total length of the pink line if the breadth of each step is 32 cms.
(ii) Write the ratio of the height of the instep to its total length of the pink line in the form of a fraction.
[4 MARKS]

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Answer: Option A. ->
:
(i) Formula: 1 Mark
Result: 1 Mark
(ii) Formula: 1 Mark
Result: 1 Mark
(i) Given: The height of the staircase = 6 cms
The breadth of the staircase = 32 cms

Look At The Figure Above. This Is A Staircase Consisting Of ...

So, length of the pink line will be = (breadth + height) of five steps.

\Rightarrow

5 (6 cms) + 5 (32 cms)

\Rightarrow 5 [6 + 32] c m s

\Rightarrow 5 [38] c m s

\Rightarrow 190 c m s

(ii)

\frac{T h e t o t a l l e n g t h o f t h e i n - s t e p}{T h e t o t a l l e n g t h o f t h e p i n k l i n e}

\Rightarrow \frac{5 (6) c m s}{190 c m s}

\Rightarrow \frac{30}{190} \Rightarrow \frac{3}{19}

Question 216.

A train got derailed and ${\frac{11}{14}}^{t h}$ of the length of the train was found off the track. If the length of the train was 322 m, find out what length of the train was found off the track? What length of the train was still on track? [4 MARKS]

Discuss Question

Answer: Option A. ->
:
Steps: 2 Marks
Answers: 2 Marks

A Train Got Derailed And 1114th Of The Length Of The Train W...

Given the length of the train is 322 m
Let x metres be the length that was found off the track. Then using unitary method we have

Or,

x \times \frac{14}{14} = 322 \times \frac{11}{14}

Or,

14 x = 322 \times 11

∴ x = \frac{322 \times 11}{14}

= \frac{322}{14} \times 11

= 23 \times 11

= 253 m e t r e s

The total length of the train = 322 m
Total length of the train off the track = 253 m
So, the length of the train which was still on track = 322 - 253 = 69 m.

Question 217.

Shivaji galloped with his horse for 21 km. He crossed the river which was $\frac{1}{3}$ the length he covered with the horse. He walked the remaining distance which was 4 times the breadth of the river.
Calculate the total distance covered by him. [4 MARKS]

Discuss Question

Answer: Option A. ->
:
Concept: 1 Mark
Steps: 2 Marks
Result: 1 Mark

Shivaji Galloped With His Horse For 21 Km. He Crossed The Ri...

The breadth of the river

= \frac{1}{3} \times 21

= 7 k m

Length covered by walking

= 4 (\frac{1}{3} o f 21) k m

\Rightarrow 4 (7 k m)

\Rightarrow 28 k m

Hence, the total distance covered by Shivaji = 21km + 7km + 28km

\Rightarrow

28km + 28km
= 56km
The total distance covered by Shivaji is 56km.

Question 218.

$(0.89345 \times 1000) + 8.9345 - 893.45 =$

$89.345$
$8.9345$
$893.45$
$8934.50$

Discuss Question

Answer: Option B. ->

8.9345

:
B

Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
$0.89345 \times 1000 = 893.45$

$∴ (0.89345 \times 1000) + 8.9345 - 893.45$
$= 893.45 + 8.9345 - 893.45$
$= 0 + 8.9345$
$= 8.9345$

Question 219.

$(12.4 + 9.6) \times (0.1 + 0.001)$ _____ $(1.24 \times 10.96) \times 0.2$

$>$
$<$
$=$
$can't say$

Discuss Question

Answer: Option B. ->

<

:
B

$(12.4 + 9.6) \times (0.1 + 0.001)$

$= 22.0 \times 0.101$

$= 2.222$

And, $(1.24 \times 10.96) \times 0.2$

$= 13.59 \times 0.2$

$= 2.718$

$2.718$ is greater than $2.222$
Therefore, $(12.4 + 9.6) \times (0.1 + 0.001)$ < $(1.24 \times 10.96) \times 0.2$

Question 220.

$(60 \times 0.06 \times 6) - (0.0006 \times 600 \times 0.6)$ is equal to:

$21.384$
$213.84$
$2.1834$
$20.484$

Discuss Question

Answer: Option A. ->

21.384

:
A

(60 \times 0.06 \times 6) - (0.0006 \times 600 \times 0.6)

= (21.6) - (0.216)

= 21.384

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