CBSE Class 6 Mathematics Knowing our Numbers Assignment Set 04

Important Concepts

We often use different units to measure things in our daily lives, such as:

• Length - measured in metres
• Weight - measured in grams
• Capacity - measured in litres

To measure anything from the short length of a pencil to the long distance between two states, we use the following table of measurement of length:

Commonly Used Conversions

• 1 km = 1000 m
• 1 m = 100 cm
• 1 cm = 10 mm
• 1 m = 1000 mm
• 1 km = 1,00,000 cm

Abbreviations

• Kilometre is written as Km
• Metre is written as m
• Decimetre is written as dm
• Centimetre is written as cm
• Millimetre is written as mm

Weight Conversions

• 1 kilogram (Kg) = 1000 grams (gm)
• 1 gram = 1000 milligrams (mg)
• 1 kilogram = 1,000 × 1,000 milligrams = 10,00,000 milligrams (mg)

Capacity Conversions

• 1 kilolitre (kl) = 1000 litres
• 1 litre (l) = 1000 millilitres

Question. In a town there are 45679 men,39842 women and 24056 children.Find the total population of town?
Answer: To find the total population, we need to add the number of men, women, and children.
Number of men = 45,679
Number of women = 39,842
Number of children = 24,056
Let us perform the addition:
\( 45,679 + 39,842 + 24,056 \)
First, add the number of men and women:
\( 45,679 + 39,842 = 85,521 \)
Now, add the number of children to this sum:
\( 85,521 + 24,056 = 1,09,577 \)
So, the total population of the town is 1,09,577.
In simple words: We add all the people together. The total number of people in the town is 1,09,577.

Exam Tip: Always write down each group's number in a column and align the digits properly from right to left before adding them up.

Question. Raghav deposited Rs.14572, Rs.20450, Rs.2987 and 9852 in the bank in four different months. Find the total deposit in the bank.
Answer: To find the total deposit, we add all the amounts Raghav put in the bank.
Deposit in the first month = Rs. 14,572
Deposit in the second month = Rs. 20,450
Deposit in the third month = Rs. 2,987
Deposit in the fourth month = Rs. 9,852
Let us find the total sum:
\( \text{Total deposit} = \text{Rs. } 14,572 + \text{Rs. } 20,450 + \text{Rs. } 2,987 + \text{Rs. } 9,852 \)
Let us calculate step-by-step:
\( 14,572 + 20,450 = 35,022 \)
\( 35,022 + 2,987 = 38,009 \)
\( 38,009 + 9,852 = 47,861 \)
The total amount of money deposited is Rs. 47,861.
In simple words: Raghav put money into the bank four times. We add all four amounts to get a total of Rs. 47,861.

Exam Tip: Do not forget to write "Rs." before your final answer to show the unit of money clearly.

Question. A shopkeeper gives a discount of Rs.12.50 on an article which costs Rs.50.what is the selling price of the article?
Answer: The original cost of the article is Rs. 50.
The discount given by the shopkeeper is Rs. 12.50.
We subtract the discount from the cost to find the selling price.
\( \text{Selling Price} = \text{Cost} - \text{Discount} \)
\( \text{Selling Price} = \text{Rs. } 50.00 - \text{Rs. } 12.50 = \text{Rs. } 37.50 \)
So, the selling price of the article is Rs. 37.50.
In simple words: The shopkeeper reduces the price by Rs. 12.50. This means you only pay Rs. 37.50 to buy it.

Exam Tip: Align decimal points carefully when subtracting money so you get the correct paisa value.

Question. Roni went to the market with Rs.1000.She bought a saree for Rs.375,a pair of shoes for Rs.155 and a shawl for Rs.2.85.How much money is left with him?
Answer: Roni started with Rs. 1,000.
Let us find the total amount she spent on shopping.
Cost of saree = Rs. 375
Cost of shoes = Rs. 155
Cost of shawl = Rs. 2.85
\( \text{Total money spent} = 375 + 155 + 2.85 = \text{Rs. } 532.85 \)
Now, subtract this total from the starting money:
\( \text{Money left} = 1000 - 532.85 = \text{Rs. } 467.15 \)
So, the money left is Rs. 467.15.
In simple words: Roni spends Rs. 532.85 in total. We take this away from Rs. 1000 to find she has Rs. 467.15 left.

Exam Tip: Read the values in the question carefully. Even if a value like Rs. 2.85 seems small for a shawl, use the exact numbers given in the question to solve it.

Question. The monthly fee for a student in a school is Rs.310.If there are 620 students in the school , find the total monthly collection of fees.
Answer: The monthly fee for one student is Rs. 310.
There are 620 students in the school.
To find the total fee collected, we multiply the fee by the number of students.
\( \text{Total monthly fee} = 310 \times 620 \)
Let us perform the multiplication:
\( 310 \times 620 = 1,92,200 \)
So, the total monthly fee collection is Rs. 1,92,200.
In simple words: We multiply the fee of one student by the total number of students. The school collects Rs. 1,92,200 every month.

Exam Tip: A quick way to multiply 310 by 620 is to multiply 31 by 62, and then add two zeros at the end of the answer.

Question. Gita has Rs.78,592 with her.She placed an order for purchasing 39 radio sets at Rs.1234 each.How much money will remain with her after the purchase?
Answer: The total money Gita has is Rs. 78,592.
She ordered 39 radio sets.
The price of one radio set is Rs. 1,234.
First, find the total cost of all 39 radio sets:
\( \text{Total cost of radio sets} = 39 \times 1,234 \)
\( 39 \times 1,234 = 48,126 \)
Now, subtract the total cost from Gita's money:
\( \text{Remaining money} = 78,592 - 48,126 = \text{Rs. } 30,466 \)
So, the money remaining with her is Rs. 30,466.
In simple words: Gita spends Rs. 48,126 on radios. Subtract this from her money to find she has Rs. 30,466 left.

Exam Tip: Make sure to do the multiplication step carefully before performing subtraction. Double-check both steps to avoid small errors.

Question. There are fifteen students in a row.If there are 737 students,calculate the number of rows in which they stand?
Answer: The number of students in each row is 15.
The total number of students is 737.
To find the number of rows, we divide 737 by 15.
\( 737 \div 15 \)
Let us find the quotient and remainder:
\( 737 = 15 \times 49 + 2 \)
This means we have 49 complete rows, and 2 students are left over.
These 2 leftover students will need one more row to stand in.
Total rows = 49 + 1 = 50 rows.
Therefore, the students stand in 50 rows.
In simple words: Dividing 737 by 15 gives 49 rows with 2 students left over. These 2 students will stand in a 50th row.

Exam Tip: In word problems like this, think about the leftover remainder. Even a few leftover students still need a row to stand in!

Question. Find the largest six digits number which is divisible by 120 exactly
Answer: The largest 6-digit number is 9,99,999.
We divide 9,99,999 by 120 to find the remainder.
\( 9,99,999 \div 120 \)
Let us do the division:
\( 9,99,999 = 120 \times 8333 + 39 \)
The remainder is 39.
To get a number exactly divisible by 120, we subtract the remainder from 9,99,999.
\( \text{Required number} = 9,99,999 - 39 = 9,99,960 \)
So, the largest six-digit number exactly divisible by 120 is 9,99,960.
In simple words: We find the remainder when dividing the biggest six-digit number by 120. Subtracting that remainder gives us the exact answer.

Exam Tip: Always subtract the remainder from the largest number to find the greatest multiple. Do not add anything, or it will become a 7-digit number!

Question. The distance between the park and house of a student is 1Km 575m.Everyday he walks both ways between the park and his house . Find the total distance covered by him in a week’s time?
Answer: Distance between the park and house = 1 km 575 m = 1575 m.
The student walks both ways every day.
Distance covered in one day = \( 2 \times 1575 \text{ m} = 3150 \text{ m} \)
A week has 7 days.
Total distance covered in a week = \( 7 \times 3150 \text{ m} \)
\( 7 \times 3150 = 22,050 \text{ m} \)
Now, convert the distance back into kilometres and metres:
\( 22,050 \text{ m} = 22 \text{ km } 50 \text{ m} \)
So, the total distance covered in a week is 22 km 50 m.
In simple words: The student walks to and from the park daily, covering 3150 m. Over 7 days, this adds up to 22 km 50 m.

Exam Tip: Remember that "both ways" means you must multiply the one-way distance by 2 before multiplying by 7 for the week.

Question. Find the sum of the greatest and the least numbers that can be written using the digits 9,8,0,6,4 each only once .
Answer: The given digits are 9, 8, 0, 6, and 4.
To make the greatest 5-digit number, arrange the digits from largest to smallest:
Greatest number = 98,640
To make the least 5-digit number, arrange the digits from smallest to largest. Since we cannot start with 0, we must place the smallest non-zero digit first:
Least number = 40,689
Now, let us find their sum:
\( \text{Sum} = 98,640 + 40,689 \)
\( 98,640 + 40,689 = 1,39,329 \)
So, the sum of the greatest and least numbers is 1,39,329.
In simple words: The biggest number is 98,640 and the smallest is 40,689. Adding them together gives 1,39,329.

Exam Tip: Never start a number with 0 when finding the least 5-digit number, as that would make it a 4-digit number instead of a 5-digit number.