Find the sum : \$\displaystyle \frac{-9}{10}+\frac{22}{15}\$

Given, \$-\dfrac{9}{10}+\dfrac{22}{15}\$ \$=-\dfrac{9\times 3}{10\times 3}+\dfrac{22\times 2}{15\times 2}\$ \$=-\dfrac{27}{30}+\dfrac{44}{30}\$ \$=\dfrac{-27+44}{30}\$ \$=\dfrac{17}{30}\$

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>>Class 8
>>Maths
>>Playing with Numbers
>> $Numbers in General Form$

Numbers in General Form

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Generalised Form of a Number

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Applying the General Form of Numbers

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Q: Some students planned a picnic. The budget for food was \$Rs.\ 480\$. But eight of these failed to go and thus the cost of food for each member increased by \$Rs.\ 10\$. How many students attended the picnic?

Let \$x be the number students in the picnic. Given that, Total budget \$=Rs\,\,480\$ Thus, the cost of food per person\$=\dfrac{480}{x}\$ Eight of student failed to join. Thus, number of students\$=x-8\$ Also, Given that the new cost has been increased by Rs. \$10\$ So, \$ \dfrac{480}{x-8}=\dfrac{480}{x}+10 \$ \$ \Rightarrow \dfrac{480}{x-8}=\dfrac{480+10x}{x} \$ \$ \Rightarrow 480x=\left( x-8 \right)\left( 480+10x \right) \$ \$ \Rightarrow 480x=480x+10{{x}^{2}}-3840-80x \$ \$ \Rightarrow 10{{x}^{2}}-80x-3840=0 \$ \$ \Rightarrow {{x}^{2}}-8x-384=0 \$ \$ \Rightarrow {{x}^{2}}-24x+16x-384=0 \$ \$ \Rightarrow x\left( x-24 \right)+16\left( x-24 \right)=0 \$ \$ \Rightarrow \left( x-24 \right)\left( x+16 \right)=0 \$ If, \$ x+16=0 \$ \$ x=-16 \$ If, \$ x-24=0 \$ \$ \Rightarrow x=24 \$ Number of student cannot be negative. Therefore, Total number of students is \$24\$Then, going student of picnic \$=x-8=24-8=16 Hence, this is the answer.

Q: Show that the square of any positive integer cannot be of the form \$(5q+2)\$ or \$(5q+3)\$ for any integer \$q\$.

Let a be any positive integerBy Euclid's division lemma\$a = bm + r\$\$a = 5m + r\$ (when \$b = 5\$)So, \$r\$ can be \$0,1,2,3,4\$ Case 1:\$\therefore a = 5m\$ (when \$r = 0\$)\$a^2 = 25m^2\$\$a^2 = 5 (5m^2)=5q\$where \$q = 5m^2\$Case 2:when \$r=1\$\$a = 5m+1\$\$a^2 = (5m + 1)^2 = 25m^2 + 10m + 1\$\$a^2 = 5(5m^2 + 2m)+1\$\$= 5q+1\$ where \$q = 5m^2 + 2m\$Case 3:\$a = 5m+2\$\$a^2 = 25m^2 + 20m + 4\$\$a^2 = 5(5m^2 + 4m)+4\$\$5q + 4\$where \$q = 5m^2 + 4m\$Case 4:\$a = 5m + 3\$\$a^2 = 25m^2 + 30m + 9\$\$=25m^2 + 30m + 5+4\$\$= 5(5m^2 + 6m +1)+4\$\$= 5q+4\$where \$q = 5m^2+ 6m+1\$Case 5:\$a = 5m+4\$\$a^2 = 25m^2 + 40m + 16 = 25m^2 + 40m + 15 + 1\$\$= 5(5m^2 + 8m + 3)+1\$\$= 5q+1\$ where \$q = 5m^2 + 8m + 3\$From these cases, we see that square of any positive no can't be of the form \$5q +2, \, 5q+3\$.

Q: Show that any positive integer is of the form 3q or, 3q+1 or,3q+2 for some integer q.

Let a be any positive integer and b=3. Applying division lemma with a and b=3, we havea=3q+r ,where,0 is less than or equal to r and r is less than 3 and q is some integer.a=3q+0 or,3q+1 or, 3q+2a=3q or,a=3q+1 or, a=3q+2 for some integer q.

Q: Give five examples where the number of things counted would be more than \$6\$-digit number.

\$1\$. The population of Chandigarh.\$2\$. Numbers of stars in the sky.\$3\$. Numbers of cars in India.\$4\$. Numbers of trees in India.\$5\$. Numbers of mobiles in India.

Q: Find \$1\$% of \$1\$ hour.

\$1\$% of \$1\$ hours \$=1\$% of \$60\$ minutes \$=1\$% of \$(60\times 60)\$ seconds\$=\cfrac{1}{100}\times 60\times 60=6\times 6=36\$ seconds.

Q: The sum of digits of a 2-digit number is \$ 11 \$. If the number obtained by reversing the digits is \$ 9 \$ less than the original number, find the number.

Given:Sum of two digit number = \$11 \$Let unit’s digit be \$ ‘x’ \$and tens digit be \$ ‘y’, \$then \$ x + y = 11 … (i) \$and number = \$ x + 10y \$By reversing the digits,Unit digit be \$ ‘y’ \$and tens digit be \$ ‘x’ \$and number \$ = y + 10x + 9 \$Now by equating both numbers,\$ y + 10x + 9 = x + 10y \$\$ 10x + y – 10y – x = -9 \$ \$ 9x – 9y = -9 \$\$ x – y = -1 … (ii) \$Adding (i) and (ii), we get \$ 2x = 10 \$\$ x = 10/2 \$ \$ = 5 \$\$ \therefore y = 1 + 5 = 6 \$By substituting the vales of x and y, we getNumber = \$ x + 10y \$\$ = 5 + 10 × 6 \$\$ = 5 + 60 \$\$ = 65 \$\$ \therefore \$ The number is \$ 65 \$.

Q: The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.

Let the units digit of the two-digit number be \$y\$ and the tens digit be \$x\$.Hence, the number will be \$10x+y\$.Given that the sum of digits is \$12\$.Hence, \$x+y=12\quad .......(i)\$It is also given that the new number formed by reversing the digits is greater than the original number by \$54\$.On reversing the digits, \$x\$ becomes the units digit and \$y\$ becomes the tens digit.Hence, the new number will be \$10y+x\$.\$\therefore \ 10y+x=10x+y+54\$\$\Rightarrow 9y-9x=54\$\$\Rightarrow y-x=6\$\$\Rightarrow -x+y=6\quad .........(ii)\$Adding \$(i)\$ and \$(ii)\$, we get:\$(x+y)+(-x+y)=12+6\$\$\Rightarrow 2y=18\$\$\therefore \ y=\dfrac{18}{2}=9\$Substituting \$y=9\$ in \$(i)\$, we get:\$x+9=12\$\$\therefore \ x=12-9=3\$The units digit is \$3\$ and the tens digit is \$9\$.Hence, the number is \$39\$.Checking the solution:Original number \$=39\$Hence, number obtained by reversing the digits \$=93\$\$93=39+54\$Hence, the number obtained by reversing the digits is greater than the original number by \$54\$. Hence, the answer is correct.

Q: Read and expand the numbers wherever there are blanks. Number Number NameExpansion 20000 twenty thousand\$2 \times 10000\$ 26000 twenty six thousand\$2 \times 10000 + 6 \times 1000\$ 38400thirty eight thousand four hundred \$3\times 10000 + 8 \times 1000 + 4 \times 100\$ 65740sixty five thousand seven hundred forty \$6\times 10000+5 \times 1000 + 7\times 100 + 4\times 10\$ 89324 eighty nine thousand three hundred twenty four \$8\times 10000 + 9 \times 1000 + 3 \times 100+ 2 \times 10 + 4 \times 1\$ 50000 ----- ---- 41000 ---- ---- 47300 ---- ---- 57630 ---- ----29485 ---- ---- 29085 ---- ---- 20085 ---- ----Write five more 5-digit numbers, read them and expand them.

\$50000\$ --- fifty thousand --- \$5\times 10000\$\$41000\$ --- forty one thousand ---\$4\times 10000+1\times 1000\$\$47300\$ --- forty seven thousand three hundred ----\$4\times 10000+7\times 1000+3\times 100\$\$57630\$ --- fifty seven thousand six hundred thirty ----\$5\times 10000+7\times 1000+6\times 100+3\times 10\$\$29485\$ --- twenty nine thousand four hundred eighty five ----\$2\times 10000+9\times 1000+4\times 100+8\times 10+5\times 1\$\$29085\$ --- twenty nine thousand eighty five ---\$2\times 10000+9\times 1000+8\times 10+5\times 1\$\$20085\$ - twenty thousand eighty five ----\$2\times 10000+8\times 10+5\times 1\$

General form of Numbers

4 mins

Simple Puzzles with Numbers

6 mins

Quick Summary With Stories

General form of a two and three digit numbers

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Read and expand the numbers wherever there are blanks.

Number Number Name Expansion
20000 twenty thousand $2 \times 10000$
26000 twenty six thousand $2 \times 10000 + 6 \times 1000$
38400 thirty eight thousand four hundred $3 \times 10000 + 8 \times 1000 + 4 \times 100$
65740 sixty five thousand seven hundred forty $6 \times 10000 + 5 \times 1000 + 7 \times 100 + 4 \times 10$
89324 eighty nine thousand three hundred twenty four $8 \times 10000 + 9 \times 1000 + 3 \times 100 + 2 \times 10 + 4 \times 1$
50000 ----- ----
41000 ---- ----
47300 ---- ----
57630 ---- ----
29485 ---- ----
29085 ---- ----
20085 ---- ----
Write five more 5-digit numbers, read them and expand them.

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Number	Number Name	Expansion
20000	twenty thousand	$2 \times 10000$
26000	twenty six thousand	$2 \times 10000 + 6 \times 1000$
38400	thirty eight thousand four hundred	$3 \times 10000 + 8 \times 1000 + 4 \times 100$
65740	sixty five thousand seven hundred forty	$6 \times 10000 + 5 \times 1000 + 7 \times 100 + 4 \times 10$
89324	eighty nine thousand three hundred twenty four	$8 \times 10000 + 9 \times 1000 + 3 \times 100 + 2 \times 10 + 4 \times 1$
50000	-----	----
41000	----	----
47300	----	----
57630	----	----
29485	----	----
29085	----	----
20085	----	----

Revise with Concepts

Generalised Form of a Number

Applying the General Form of Numbers

Learn with Videos

Quick Summary With Stories

General form of a two and three digit numbers

Applications on General Form of Numbers

Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic?

Show that the square of any positive integer cannot be of the form (5q+2) or (5q+3) for any integer q.

Show that any positive integer is of the form 3q or, 3q+1 or,3q+2 for some integer q.

Find the sum : 10−9​+1522​

Find the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8,15 and 21.

Give five examples where the number of things counted would be more than 6-digit number.

Find 1% of 1 hour.

The sum of digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.

The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.

Some students planned a picnic. The budget for food was $R s . 480$ . But eight of these failed to go and thus the cost of food for each member increased by $R s . 10$ . How many students attended the picnic?

Show that the square of any positive integer cannot be of the form $(5 q + 2)$ or $(5 q + 3)$ for any integer $q$ .

Find the sum : $\frac{- 9}{1 0} + \frac{2 2}{1 5}$

Give five examples where the number of things counted would be more than $6$ -digit number.

Find
$1$ % of $1$ hour.

The sum of digits of a 2-digit number is $11$ . If the number obtained by reversing the digits is $9$ less than the original number, find the number.