GSEB Class 6 Maths Solutions Chapter 1 Knowing Our Numbers Exercise 1.2

Question 1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Answer:
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day = 1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the fourth day = 2751
Therefore, the total number of tickets sold on all four days = \(1094 + 1812 + 2050 + 2751 = 7707\).
In simple words: We simply add up the number of tickets sold each day to find the total for the whole event.

Exam Tip: For problems asking for a "total number," always perform addition. Make sure to align the numbers correctly when adding to prevent calculation errors.

Question 2. Shekhar is a famous cricket player. He has so far scored 6,980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Answer:
Total runs Shekhar has scored so far = 6,980 runs
Target runs to be achieved = 10,000 runs
Therefore, the number of additional runs needed = \(10,000 - 6,980 = 3,020\) runs.
In simple words: To find out how many more runs Shekhar needs, we subtract his current score from his goal.

Exam Tip: When a question asks "how many more," it indicates a subtraction problem. Carefully set up the subtraction with the larger number at the top.

Question 3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Answer:
Number of votes secured by the successful candidate = 5,77,500 votes
Number of votes secured by his nearest rival = 3,48,700 votes
The margin of votes by which the successful candidate won = \(5,77,500 - 3,48,700 = 2,28,800\) votes.
In simple words: To find the difference in votes, we subtract the rival's votes from the winner's votes.

Exam Tip: The term "margin" in such questions always implies finding the difference between two quantities, which requires subtraction.

Question 4. Kirjj bookstore sold books worth Rs. 2,85,891 in the first week of June and books worth Rs. 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Answer:
Sale of books in the first week = Rs. 2,85,891
Sale of books in the second week = Rs. 4,00,768
Total sale for the two weeks together = \( \text{Rs. } 2,85,891 + \text{Rs. } 4,00,768 = \text{Rs. } 6,86,659 \).
Clearly, the sale was greater in the second week.
Difference in sale = \( \text{Rs. } 4,00,768 - \text{Rs. } 2,85,891 = \text{Rs. } 1,14,877 \).
In simple words: First, add up sales from both weeks to get the total. Then, compare the two weekly sales and subtract the smaller from the larger to find the difference.

Exam Tip: Break down multi-part questions into smaller, manageable steps. First, calculate the total, then compare and find the difference. Remember to include units (Rs.) in your answer.

Question 5. Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Answer:
The given digits are 6, 2, 7, 4, and 3.
To form the greatest number, arrange the digits in descending order: 76,432
To form the smallest number, arrange the digits in ascending order: 23,467
The difference between the greatest and the least number = Greatest number - Smallest number
\( = 76,432 - 23,467 = 52,965 \).
In simple words: Arrange the given digits from biggest to smallest to make the largest number, and from smallest to biggest to make the smallest number. Then, subtract the smaller number from the larger one.

Exam Tip: Remember that to form the greatest number, digits are arranged in descending order, and for the smallest number, they are arranged in ascending order. Pay close attention to digit placement, especially if a zero is involved (though not in this specific question).

Question 6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Answer:
Number of days in January 2006 = 31 days
Number of screws manufactured in one day = 2,825 screws
Therefore, the number of screws manufactured in 31 days = \(31 \times 2,825 = 87,575\) screws.
Thus, the machine manufactured 87,575 screws in the month of January 2006.
In simple words: Multiply the number of screws made per day by the total number of days in January to get the total production for the month.

Exam Tip: Always recall the number of days in each month to correctly solve such problems. January always has 31 days.

Question 7. A merchant had Rs. 78,592 with her. She placed an order for purchasing 40 radio sets at Rs. 1200 each. How much money will remain with her after the purchase?
Answer:
Total money the merchant had = Rs. 78,592
Number of radio sets to be purchased = 40
Cost of one radio set = Rs. 1200
Total cost of 40 radio sets = \(40 \times \text{Rs. } 1200 = \text{Rs. } 48,000\)
Money left with the merchant after the purchase = Total money - Cost of radio sets
\( = \text{Rs. } 78,592 - \text{Rs. } 48,000 = \text{Rs. } 30,592 \).
In simple words: First, find the total cost of all the radio sets by multiplying the number of sets by the price of each. Then, subtract this total cost from the initial amount of money the merchant had to find the remaining balance.

Exam Tip: In problems involving purchases, always calculate the total expenditure first before subtracting it from the initial amount to find the remaining balance.

Question 8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? Hint: Do you need to do both the multiplications?
Answer:
The multiplication done by the student = \(7236 \times 65\)
The required (correct) multiplication = \(7236 \times 56\)
The difference between the student's answer and the correct answer = \( (7236 \times 65) - (7236 \times 56) \)
We can use the distributive property to simplify this: \( 7236 \times (65 - 56) \)
\( = 7236 \times 9 \)
\( = 65124 \)
Thus, the student's answer is 65124 more than the correct answer.
In simple words: Instead of doing two long multiplications, you can simply find the difference between the two multipliers (65 and 56) and then multiply that difference by the original number (7236). This will directly tell you how much greater the incorrect answer was.

Exam Tip: Look for opportunities to simplify calculations using properties like the distributive property. It often saves time and reduces the chance of errors, especially in "difference of products" problems.

Question 9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? Hint: Convert data in cm.
Answer:
Cloth required to stitch 1 shirt = 2 m 15 cm
Converting to cm: \(2 \times 100 \text{ cm} + 15 \text{ cm} = 200 \text{ cm} + 15 \text{ cm} = 215 \text{ cm}\).
Total cloth available = 40 m
Converting to cm: \(40 \times 100 \text{ cm} = 4000 \text{ cm}\).
To find the number of shirts and remaining cloth, we divide the total cloth by the cloth needed per shirt:
\( 4000 \text{ cm} \div 215 \text{ cm} \)
This division gives a quotient of 18 and a remainder of 130.
\( 215 ) 4000 ( 18 \)
\( \quad -215 \)
\( \quad 1850 \)
\( \quad -1720 \)
\( \quad \quad 130 \)
So, 18 shirts can be stitched, and 130 cm of cloth will remain.
Converting the remainder back to meters and centimeters: \(130 \text{ cm} = 1 \text{ m } 30 \text{ cm}\).
Thus, 18 shirts can be stitched, and 1 m 30 cm of cloth will be left over.
In simple words: First, change all lengths to centimeters. Then, divide the total cloth by the amount needed for one shirt to find how many shirts you can make. The leftover part of the division tells you how much cloth remains.

Exam Tip: Always convert all units to the smallest common unit (here, centimeters) before performing calculations involving mixed units. Clearly state the quotient as the number of items and the remainder as the leftover quantity, converting it back if necessary.

Question 10. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Answer:
Weight of one box of medicine = 4 kg 500 g
Converting to grams: \(4 \times 1000 \text{ g} + 500 \text{ g} = 4000 \text{ g} + 500 \text{ g} = 4500 \text{ g}\).
Maximum weight the van can carry = 800 kg
Converting to grams: \(800 \times 1000 \text{ g} = 800000 \text{ g}\).
To find the number of boxes, we divide the maximum carrying capacity by the weight of one box:
Number of boxes = \(800000 \text{ g} \div 4500 \text{ g}\)
This division gives a quotient of 177 and a remainder of 3500.
\( 4500 ) 800000 ( 177 \)
\( \quad -4500 \)
\( \quad 35000 \)
\( \quad -31500 \)
\( \quad \quad 35000 \)
\( \quad \quad -31500 \)
\( \quad \quad \quad 3500 \)
Since we can only load complete boxes, 177 boxes can be loaded in the van.
In simple words: First, change all weights to grams. Then, divide the van's total weight capacity by the weight of one box. The whole number part of the answer is how many boxes can be loaded.

Exam Tip: For loading problems, always convert all weight measurements to a single unit (usually the smallest, like grams here) before dividing. The number of boxes will always be the whole number (quotient) from the division; disregard the remainder as it represents incomplete items.

Question 11. The distance between the school and the house of a student is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Answer:
Distance between school and house = 1 km 875 m
Converting to meters: \(1 \times 1000 \text{ m} + 875 \text{ m} = 1000 \text{ m} + 875 \text{ m} = 1875 \text{ m}\).
Since she walks both ways (to school and back home), the distance covered in one day = \(2 \times 1875 \text{ m} = 3750 \text{ m}\).
Total distance covered in six days = \(6 \times 3750 \text{ m} = 22500 \text{ m}\).
Converting this back to kilometers and meters: \(22500 \text{ m} = 22 \text{ km } 500 \text{ m}\).
In simple words: Convert the distance from kilometers and meters to just meters. Double this distance to find how far she walks in one day (since she goes to school and back). Then, multiply this daily distance by six to get the total distance for six days. Finally, convert the answer back to kilometers and meters.

Exam Tip: Remember to account for "both ways" in such problems by doubling the one-way distance. Consistent unit conversion is crucial; convert to the smallest unit for calculations, and convert back for the final answer if required.

Question 12. A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Answer:
Total quantity of curd in the vessel = 4 liters 500 ml
Converting to milliliters: \(4 \times 1000 \text{ ml} + 500 \text{ ml} = 4000 \text{ ml} + 500 \text{ ml} = 4500 \text{ ml}\).
Capacity of one glass = 25 ml
Number of glasses that can be filled = Total quantity of curd \( \div \) Capacity of one glass
\( = 4500 \text{ ml} \div 25 \text{ ml} \)
\( = 180 \)
\( 25 ) 4500 ( 180 \)
\( \quad -25 \)
\( \quad 200 \)
\( \quad -200 \)
\( \quad \quad 00 \)
\( \quad \quad -00 \)
\( \quad \quad \quad 0 \)
Thus, 180 glasses can be filled.
In simple words: First, change the total amount of curd into milliliters. Then, divide this total amount by the size of one glass (in milliliters) to find out how many glasses you can fill completely.

Exam Tip: Ensure all volumes are in the same unit (milliliters is easiest here) before performing division. This is a direct division problem to find out how many equal parts can be made from a total quantity.