A Peek Beyond the Point Class 7 Extra Questions Maths Chapter 3

During revision, students quickly go through Class 7 Maths Extra Questions Chapter 3 A Peek Beyond the Point Class 7 Extra Questions with Answers for clarity.

Class 7 A Peek Beyond the Point Extra Questions

Class 7 Maths Chapter 3 A Peek Beyond the Point Extra Questions

Class 7 Maths Chapter 3 Extra Questions – A Peek Beyond the Point Extra Questions Class 7

Question 1.
The lengths of the body parts of an ant are as follows:
Head = (1\(\frac{4}{10}\)) units,
Thorax = (2\(\frac{6}{10}\)) units
Abdomen = (3\(\frac{7}{10}\)) units.

Find its total length of the ant.
Solution:
Total length of ant = 1 \(\frac{4}{10}\) + 2\(\frac{6}{10}\) + 3\(\frac{7}{10}\)
= (1 + 2 + 3) +(\(\frac{4}{10}+\frac{6}{10}+\frac{7}{10}\)) = 6 + \(\frac{17}{10}\)
= 6 + \(\frac{10}{10}+\frac{7}{10}\)
= 6 + 1 + \(\frac{7}{10}\)
= 7 + \(\frac{7}{10}\)

So, the total length is 7 units and 7 one-tenths,
i.e., 7\(\frac{7}{10}\) units

Question 2.
(a) Find the value of 7\(\frac{9}{100}\) – 3\(\frac{5}{100}\)?
Solution:

(b) What will be the sum of 6\(\frac{4}{10} \frac{8}{100}\) and 14\(\frac{5}{10} \frac{3}{100}\)?
Answer:

Question 3.
Arrange the following numbers in descending order:
(a) 10.98, 10.089, 10.809, 10.908, 10.981
(b) 22.31, 22.13, 22.331, 22.313, 22.133
Solution:
(a) We will write the given decimal numbers as:
10.980, 10.089, 10.809, 10.908, 10.981
Hence, the given decimals in descending order are:
10.981, 10.98, 10.908, 10.809, 10.089

(b) We will write the given decimal numbers as:
22.310, 22.130, 22.331, 22.313, 22.133
Hence, the given decimals in descending order are:
22.331, 22.313, 22.31, 22.133, 22.13

Question 4.
Ravi delivers 2.5 kg, 3.2 kg, and 4.1 kg of vegetables to a store in the first three days. In 7 days, he delivers 20 kg of vegetables. What is the total quantity of vegetables delivered in the last four days?
Solution:
The total vegetables delivered in the first 3 days:
2.5 kg + 3.2 kg + 4.1 kg = 9.8 kg
Total vegetables delivered in 7 days = 20 kg (Given)
Vegetables delivered in the last 4 days = 20 kg – 9.8 kg = 10.2 kg
Ravi delivered 10.2 kg of vegetables in the last four days. 5.

Question 5.
In the number line given below, what decimal numbers do the boxes labelled ‘u’, ‘v’, ‘w’, and ‘x’ represent?
Solution:
There are 10 divisions between 6.1 and 6.6, so each division is a tenth part of 0.5 or \(\frac{5}{10}\) , i.e., \(\frac{5}{100}\) = 0.05 units. Therefore, the first division after 6.1, denoted by V, represents the number 6.15, while the fourth division, denoted by V, represents the number 6.3, the seventh division, denoted by ‘w’, represents the number 6.45, and the first division after 6.6, denoted by ‘x’ represents the number 6.65.

Question 6.
Case Based Question
A digital wallet is an electronic alternative to a traditional wallet, enabling you to make payments for a wide range of products and services. Whether shopping or dining out, many people now prefer using digital wallets over other payment methods.
Kritika and Ajay are friends, and they use digital wallets. Kritika had ₹ 256.45 in her wallet, Ajay wallet’s had ₹ 1121.77. On Sunday, they went a mall for shopping. –
Based on the above information, answer the following questions:
(a) Kritika bought a toothpaste for ₹ 71.75, soap for ₹ 41.25, and shoe polish for ₹ 52. Find the money left in her wallet.
(b) Ajay bought 3 notebooks for ₹ 106.50 and 4 bottles of juice for ₹ 873.00. Find the money left in his wallet.
(c) After shopping, find the sum of the money left in their wallets.
Solution:
(a) Kritika’s total expense = ₹ 71.75 + ₹ 41.25 + ₹ 52 = ₹ 165.00
Money left = ₹ 256.45 – ₹ 165.00 = ₹ 91.45

(b) Ajay’s total expense = ₹ 106.50 + ₹ 873.00 = ₹ 979.50
Money left = ₹ 1121.77 – ₹ 979.50 = ₹ 142.27

(c) Sum of the money left = ₹ 91.45 + ₹ 142.27 = ₹233.72

Very Short Answer Type Questions

Question 1.
Evaluate \(\frac{5}{10}\)+\(\frac{6}{100}\).
Answer:

Question 2.
Evaluate 3 \(\frac{8}{10}\)+8 \(\frac{3}{100}\).
Answer:

Question 3.
Evaluate \(\frac{36}{100}\)+6 \(\frac{3}{100}\)-5 \(\frac{45}{100}\).
Answer:

Question 4.
Evaluate 8 \(\frac{8}{10}\)-4 \(\frac{4}{10}\).
Answer:

Question 5.
Extend the pattern.
(i) \(\frac{3}{10}\), \(\frac{6}{10}\), \(\frac{9}{10}\), ……. , ……… .
(ii) 2 \(\frac{2}{10}\), 1 \(\frac{9}{10}\), 1 \(\frac{6}{10}\), ……. , ……. ,
(iii) 5.9,6.2,6.5, …….
Answer:

Question 6.
Write each of the following as decimals.
(i) 23+\(\frac{2}{10}\)+\(\frac{6}{1000}\)
(ii) 700+20+5+\(\frac{9}{100}\)
Answer:

Question 7.
Compare the decimal numbers.
700+20+3+\(\frac{0}{10}\)+\(\frac{4}{100}\) 700+20+3
+\(\frac{0}{10}\)+\(\frac{0}{100}\)+\(\frac{4}{1000}\)
Answer:

Question 8.
Write 6 mm in cm using decimal and 60 m in km using decimal.
Answer:
We know that 10 mm=1 cm
1 mm = \(\frac{1}{10}\) cm → 6 mm = \(\frac{6}{10}\) cm=0.6 cm

We know that 1000 m = 1 km
1 m = \(\frac{1}{1000}\) km → 60 m =\(\frac{60}{1000}\) km = 0.060 km

Question 9.
How will you write 2 kg 9 g in kg using decimal?
Answer:

Question 10.
If Meera gave 726 paise to Ajay, then how much money in rupees did she give?
Answer:
We know that,
1 rupee = 100 paise
i.e. 1 paisa = \(\frac{1}{100}\) rupee
Since, Meera gave 726 paise to Ajay.
∴ 726 paise = \(\frac{726}{100}\) = 7.26 rupees
Hence, Meera gave 7.26 rupees to Ajay.

Short Answer Type Questions

Question 1.
Arrange the following in an increasing order?
(i) 7 \(\frac{3}{10}\) \(\frac{4}{100}\)
(ii) 4 \(\frac{37}{100}\)
(iii) 0.37
(iv) 34 \(\frac{7}{10}\)
Answer:

Now, on comparing the numerators of all these, as their denominator is same in each case, we get

Question 2.
Mukesh is measuring some of his body parts. The total length of his arm is 7 \(\frac{5}{10}\) units and that of his upper arm is 4 \(\frac{6}{10}\) units. Then, find the length of his lower arm.
Answer:
Given, the total length of Mukesh’s arm = 7 \(\frac{5}{10}\) units and the length of his upper arm = 4 \(\frac{6}{10}\) units.
Then, the length of his lower arm = Total length – Length of upper arm

Question 3.
Write the following decimals in the place value table.
(i) 0.29
(ii) 2.08
(iii) 19.60
(iv) 148.32
(v) 200.812
Answer:

Now, the place value table is given below.

Question 4.
Which one is greater 1 m 40 cm + 60 cm or 2.6 m ?
Answer:
We have, 1 m 40 cm + 60 cm = 1 m + 40 cm + 60 cm
= 1 m+100 cm
We know that 1 cm = \(\frac{1}{100}\) m
∴ 1 m 40 cm + 60 cm = 1 m + \(\frac{100}{100}\) m
=1 m + 1 m = 2.0 m
On comparing 2.0 m and 2.6 m, we get 2.6>2
Hence, 2.6 is greater than 2.

Question 5.
Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80 then find the total amount given to Radhika by her parents.
Answer:

i.e ₹ 26.30.

Question 6.
Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Answer:
Given, cost of book bought by Raju = ₹ 35.65
and money gave to shopkeeper = ₹ 50
∴ Money get back from shopkeeper = ₹(50-35.65)
= ₹ 14.35

Question 7.
Express in kilometres, using decimals.
(i) 15 km 245 m
(ii) 19 km 48 m
Answer:
We know that, 1000 m = 1 km
(i) We have,
15 km 245 m = 15 km + \(\frac{245}{1000}\) km
= 15 km + 0.245 km
= 15.245 km

(ii) We have,
19 km 48 m =19 km + 48 m
= 19 km + \(\frac{48}{1000}\) km
= 19 km + 0.048 km
= 19.048 km

Long Answer Type Questions

Question 1.
The total length of the body parts of a honeybee are given as,
Head: 3 \(\frac{2}{10}\) units
Thorax : 4 \(\frac{8}{10}\) units
Abdomen : 4 \(\frac{6}{10}\) units
Find the total length of honey bee’s body.
Answer:

Question 2.
Arrange the following in decreasing order.
(i) 201 \(\frac{8}{10}\)
(ii) 34 \(\frac{2}{10}\) \(\frac{3}{100}\)
(iii) 45 \(\frac{5}{10}\)
(iv) 36 \(\frac{36}{100}\)
(v) 20.12
Answer:

Now, arranging all the numerators in decreasing order, as their denominators are same. We get,

Question 3.
Write the numbers given in the following place value table in decimal form.

Answer:

(iii) Do same as above
Required decimal = 30.025

(iv) Do same as above
Required decimal = 211.902

(v) Do same as above
Required decimal = 12.241

Question 4.
Arrange 12.142,12.124,12.104,12.401 and 12.214 in ascending order.
Answer:

Again, hundredths part of 12.142 = \({4}{100}\)
∴ Hundredths part of 12.124 = \({2}{100}\)
and hundredths part of 12.104 = \({0}{100}\)
∴ \({4}{100}\) > \({2}{100}\)>\({0}{100}\)
∴ 12.142 > 12.124 > 12.104
Hence, the ascending order of given numbers are
12.104 < 12.124 < 12.142 < 12.214 < 12.401

Question 5.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Answer:
Here, whole part of all numbers are same and tenths part of 12.142, 12.124 and 12.104 are same.
Now, tenths part of 12.401= \(\frac{4}{10}\)
and tenths part of 12.214= \(\frac{2}{10}\)
∵ \(\frac{4}{10} >[latex]\frac{2}{10}\)
∴ 12.401 >12.214
Given, the distance travelled by bus
= 15 km 268 m
= 15 km + 268 m
= 15 km + 268 × \(\frac{1}{1000}\) km [∵ 1 m = \(\frac{1}{1000}\) km]
= 15 km + \(\frac{268}{1000}\) km
= (15 + 0.268) km
= 15.268 km
distance travelled by car
= 7 km 7 m= 7 km + 7 m
= 7 km + 7 × \(\frac{1}{1000}\) km [∵ 1 m = \(\frac{1}{1000}\) km]
= 7 km + \(\frac{7}{1000}\) km
= 7 km + 0.007 km
= (7 + 0.007) km
= 7.007 km
and distance travelled by foot
= 500 m
= 500 × \(\frac{1}{1000}\) km [∵ 1 m= \(\frac{1}{1000}\) km]
= \(\frac{500}{1000}\) km = 0.500 km
∴ Total distance travelled by Sunita

Hence, total distance from her residence to her school is 22.775 km.

Question 6.
Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Answer:
Given, the total length of cloth
= 20 m 5 cm= 20 m + 5 cm
= 20 m + 5 × \(\frac{1}{100}\) m [∵ 1 cm = \(\frac{1}{100}\) m]
= 20 m + \(\frac{5}{100}\) m
= (20 + 0.05) m
= 20.05 m
and length of cloth cut by Tina
= 4 m 50 cm
= 4 m + 50 cm
= 4 m + 50 × \(\frac{1}{100}\) m [∵ 1 cm= \(\frac{1}{100}\) m]
= 4 m + \(\frac{50}{100}\) m
= (4 + 0.50) m= 4.50 m
∴ Length of cloth left with Tina
= (20.05-4.50) m
= 15.55 m

Question 7.
Veenu purchased 2 kg 300 g tomatoes, 350 g dhania, 6 kg 400 g onion, 800 g palak and 4 kg 700 g potatoes. Find the total weight of her purchases in kilograms.
Answer:
Given, weight of tomatoes = 2 kg 300 gm
= 2.300 kg,
weight of dhania = 350 g
= 0.350 kg,
weight of onion = 6 kg 400 g
= 6.400 kg,
weight of palak = 800 g
= 0.800 kg
weight of potatoes = 4 kg 700 g
= 4.700 kg
∴ Total weight of all vegetables
= (2.300 + 0.350 + 6.400 + 0.800 + 4.700) kg
= 14.550 kg
∴ Total weight of his purchases in kilograms
= Weight of potatoes + Weight of dhania + Weight of onion + Weight of palak + Weight of tomatoes
= 1.200 kg + 0.250 kg + 5.300 kg + 0.500 kg + 2.600 kg
= (1200 + 0.250 + 5.300 + 0.500 + 2.600) kg
= 9.850 kg

Question 8.
Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.
Answer:

Question 9.
Seema has ₹ 2000 , she bought readymade garments for ₹ 987.50, medicines for ₹ 210.25, groceries for ₹ 53025 . She donated ₹ 200 for charity.
(i) How much money is left with her?
(ii) Mention the value you depict from this.
Answer:
(i) Given, the total money that Seema has = ₹ 2000,
cost of readymade garments = ₹ 987.50,
cost of medicines = ₹ 210.25
and cost of groceries = ₹ 530.25
∴ Total cost = ₹(987.50 + 210.25 + 530.25)
= ₹ 1728.00
Since, money donated for charity = ₹ 200
So, total money spent = ₹ 1728.00 + ₹ 200
= ₹ 1928
Therefore, money left with Seema
= ₹ 2000-₹ 1928
= ₹ 72
So, ₹ 72 are left with her.
(ii) Humanity, helpfulness.

Question 10.
Mark bought a note book for ₹ 13.85 , a pen for ₹ 9.75 and a marker for ₹ 8.35 from book shop. He gave a ₹ 50 note to the shopkeeper. What amount did he get back?
Answer:
Given, the cost of a notebook = ₹ 13.85,
the cost of a pen = ₹9.75
and the cost of a marker = ₹8.35
Then, the total cost = ₹(13.85 + 9.75 + 8.35)
= ₹ 31.95
Since, Mark gave ₹50 note to shopkeeper.
Then, the shopkeeper returned him the amount of
₹(50-31.95)= ₹ 18.05.

Skill Based Questions

Direction In questions 1 and 2 , replace ‘?’ with appropriate fraction.

Question 1.

Answer:
500

Question 2.

Answer:
0.00001

Question 3.
A student compared – \(\frac{1}{4}\) and -0.3 . He changed – \(\frac{1}{4}\) to the decimal -0.25 and wrote, ‘Since 0.3 is greater than 0.25 , -0.3 is greater than -0.25’. What was the student’s error?
Answer:
Do yourself

Question 4.
State whether the answer is greater than 1 or less than 1. Put a ‘✓’ mark in appropriate box.

Answer:
Do yourself

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