Large Numbers Around UsClass 7 Mathematics NCERT Solutions

Solution: To find the number of times the +1000 button should be pressed, we divide the target number by 1000. (b) $10,000 \div 1000 = 10$ times. (c) Fifty three thousand = $53,000$. $53,000 \div 1000 = 53$ times. (d) $90,000 \div 1000 = 90$ times. (e) One Lakh = $1,00,000$. $1,00,000 \div 1000 = 100$ times. (f) If the button is pressed 153 times, the number shown is $153 \times 1000 = 1,53,000$. (g) To make one lakh (1,00,000), we need $1,00,000 \div 1000 = 100$ thousands.

Solution: To find the number of times the +10 button should be pressed, we divide the target number by 10. (a) Five hundred = 500. $500 \div 10 = 50$ times. (b) $780 \div 10 = 78$ times. (c) $1000 \div 10 = 100$ times. (d) $3700 \div 10 = 370$ times. (e) $10,000 \div 10 = 1000$ times. (f) One lakh = $1,00,000$. $1,00,000 \div 10 = 10,000$ times. (g) If the button is pressed 435 times, the number shown is $435 \times 10 = 4,350$.

Solution: To find the number of times the +100 button should be pressed, we divide the target number by 100. (a) Four hundred = 400. $400 \div 100 = 4$ times. (b) $3,700 \div 100 = 37$ times. (c) $10,000 \div 100 = 100$ times. (d) Fifty three thousand = $53,000$. $53,000 \div 100 = 530$ times. (e) $90,000 \div 100 = 900$ times. (f) $97,600 \div 100 = 976$ times. (g) $1,00,000 \div 100 = 1000$ times. (h) If the button is pressed 582 times, the number shown is $582 \times 100 = 58,200$. (i) To make ten thousand (10,000), we need $10,000 \div 100 = 100$ hundreds. (j) To make one lakh (1,00,000), we need $1,00,000 \div 100 = 1000$ hundreds. (k) The statement is: "There are some numbers which Tedious Tens and Thoughtful Thousands can't show but I can."

Given: Two ways to get 5072 are shown: (a) $(50 \times 100)+(7 \times 10)+(2 \times 1)=5072$ (b) $(3 \times 1000)+(20 \times 100)+(72 \times 1)=5072$

To Find: A different way to get 5072.

Solution: We can break down the number 5072 in many ways. One different way is to use the standard place value expansion, which also corresponds to the minimum number of button presses for a calculator like 'Systematic Sippy'.

5072 can be expressed as:

This can be written as an expression: $(5 \times 1000) + (0 \times 100) + (7 \times 10) + (2 \times 1) = 5000 + 0 + 70 + 2 = 5072$

Another creative way could be: $(4 \times 1000) + (10 \times 100) + (7 \times 10) + (2 \times 1) = 4000 + 1000 + 70 + 2 = 5072$

Final Answer: A different way to get 5072 is $(4 \times 1000) + (10 \times 100) + (7 \times 10) + (2 \times 1) = 5072$.

To Find: Whether a person can taste all 1,00,000 varieties in a 100-year lifetime.

Solution: First, we calculate the total number of days in a 100-year lifetime, ignoring leap years. Number of days in 1 year = 365 Number of days in 100 years = $100 \times 365 = 36,500$ days.

Next, we calculate the total number of rice varieties that can be tasted in this period. Varieties tasted per day = 3 Total varieties tasted in 36,500 days = $36,500 \times 3 = 1,09,500$.

Now, we compare the number of varieties tasted with the total number of varieties available. Total varieties available = 1,00,000 Total varieties tasted = 1,09,500

Since $1,09,500 > 1,00,000$, the person will be able to taste all the lakh varieties.

Final Answer: Yes, they will be able to taste all the lakh varieties in a 100-year lifetime, as they can taste up to 1,09,500 varieties in that time.

Let: Let us choose $y = 274$ years (as hinted in the text).

Solution: We calculate the total number of days for $y = 274$ years. Total days = $365 \times y$ Total days = $365 \times 274$

Calculation: $365 \times 274 = 1,00,010$

Now, we compare this result with one lakh (1,00,000). The number of days is 1,00,010, which is very close to 1,00,000. The difference is $1,00,010 - 1,00,000 = 10$.

Final Answer: For a choice of $y = 274$ years, the number of days is 1,00,010, which is extremely close to one lakh, with a difference of only 10 days.

To Find: How much less than one lakh is 75,000.

Solution: We need to find the difference between one lakh and 75,000. Difference = $1,00,000 - 75,000$ Difference = 25,000

Final Answer: 75,000 is 25,000 less than one lakh.

To Find: How much more than one lakh is 1,06,000.

Solution: We need to find the difference between 1,06,000 and one lakh. Difference = $1,06,000 - 1,00,000$ Difference = 6,000

Final Answer: 1,06,000 is 6,000 more than one lakh.

To Find: The increase in population from 2011 to 2024.

Solution: We subtract the 2011 population from the 2024 population. Increase = Population in 2024 - Population in 2011 Increase = $1,06,000 - 75,000$ Increase = 31,000

Final Answer: The population of Chintamani increased by 31,000 from 2011 to 2024.

Solution: Here are two different expressions for each number:

Solution: (a) Largest and Smallest 3-digit number with 30 clicks Let the number of presses for +100, +10, and +1 be $a, b, c$ respectively. We are given $a+b+c=30$. The number formed is $N = 100a + 10b + c$.

To find the largest 3-digit number: We need to maximize $N = 100a + 10b + c$ subject to $a+b+c=30$ and $N < 1000$. To maximize $N$, we should maximize $a$ first. Let's try $a=9$. Then $b+c = 30-9=21$. The number is $N = 900 + 10b + c$. We need $10b+c < 100$. Substituting $c = 21-b$, we get $10b + (21-b) < 100 \Rightarrow 9b < 79 \Rightarrow b \le 8$. To maximize $N$, we choose the largest possible $b$, which is $b=8$. Then $c = 21-8=13$. This gives clicks $a=9, b=8, c=13$ (total 30 clicks). The number is $N = 100(9) + 10(8) + 13 = 900 + 80 + 13 = 993$.

To find the smallest 3-digit number: We need to minimize $N = 100a + 10b + c$ subject to $a+b+c=30$ and $N \ge 100$. To minimize $N$, we should minimize $a$ first. Let's try $a=0$. Then $b+c=30$. The number is $N = 10b+c$. We need $10b+c \ge 100$. Substituting $c=30-b$, we get $10b+(30-b) \ge 100 \Rightarrow 9b \ge 70 \Rightarrow b \ge 7.77$. The smallest integer value for $b$ is $b=8$. Then $c=30-8=22$. This gives clicks $a=0, b=8, c=22$ (total 30 clicks). The number is $N = 10(8) + 22 = 80 + 22 = 102$. If we try $a=1$, then $b+c=29$. The smallest number is when $b=0, c=29$, giving $N = 100(1) + 10(0) + 29 = 129$. Since $102 < 129$, the minimum is 102.

Final Answer for (a): The largest 3-digit number is 993. The smallest 3-digit number is 102.

(b) Can you make 997 with a different number of clicks? The standard way is using place value: $9 \times 100 + 9 \times 10 + 7 \times 1$. This requires $9+9+7 = 25$ clicks. Let's find another combination. Let the clicks be $a, b, c$ for +100, +10, +1 respectively. We need $100a + 10b + c = 997$. From this equation, we can see that $c$ must be a number ending in 7. Let's try $c=17$. Then $100a + 10b + 17 = 997 \Rightarrow 100a + 10b = 980 \Rightarrow 10a + b = 98$. For this equation, the only integer solution is $a=9, b=8$. So, we have a new combination of clicks: $a=9, b=8, c=17$. The total number of clicks is $9+8+17 = 34$.

Final Answer for (b): Yes, 997 can be made with a different number of clicks. For example, it can be made with 34 clicks ($9 \times 100, 8 \times 10, 17 \times 1$).

To Find: The approximate height of the building.

Solution: First, we calculate the height of a single floor. Height of one floor = $4 \times$ Somu's height = $4 \times 1 \text{ m} = 4 \text{ m}$.

Next, we calculate the total height of the 11-floor building. Total height of the building = Number of floors $\times$ Height of one floor Total height = $11 \times 4 \text{ m} = 44 \text{ m}$.

Final Answer: The approximate height of the building is 44 metres.

To Find: Which is taller and by how much.

Solution: First, we compare the heights. $180 \text{ m} > 44 \text{ m}$. So, the Statue of Unity is taller than the building.

Next, we find the difference in their heights. Difference = Height of Statue of Unity - Height of building Difference = $180 \text{ m} - 44 \text{ m} = 136 \text{ m}$.

Final Answer: The Statue of Unity is taller than the building by 136 metres.

To Find: How much taller the waterfall is than the building.

Solution: We find the difference in their heights. Difference = Height of waterfall - Height of building Difference = $450 \text{ m} - 44 \text{ m} = 406 \text{ m}$.

Final Answer: The Kunchikal waterfall is 406 metres taller than Somu's building.

To Find: The number of floors required to match the waterfall's height.

Solution: We divide the total height of the waterfall by the height of a single floor. Number of floors = Total height $\div$ Height per floor Number of floors = $450 \div 4 = 112.5$.

Since we cannot have half a floor, the building would need approximately 113 floors to be at least as high as the waterfall.

Final Answer: Somu's building should have about 113 floors to be as high as the waterfall.

Solution: (a) $3,00,600$ is written as: Three lakh six hundred. (b) $5,04,085$ is written as: Five lakh four thousand eighty-five. (c) $27,30,000$ is written as: Twenty-seven lakh thirty thousand. (d) $70,53,138$ is written as: Seventy lakh fifty-three thousand one hundred thirty-eight.

Solution: (a) One lakh twenty three thousand four hundred and fifty six: $1,23,456$ (b) Four lakh seven thousand seven hundred and four: $4,07,704$ (c) Fifty lakhs five thousand and fifty: $50,05,050$ (d) Ten lakhs two hundred and thirty five: $10,00,235$