Solve all the questions with proper method. - Class - 6th

                CLASS - 6th                             By Prem Sir 

1. Find the value of:

• (i) $| -9 | = 9$
• (ii) $7 + | -3 | = 7 + 3 = 10$
• (iii) $8 - | -7 | = 8 - 7 = 1$
• (iv) $- | -3 | = -3$

2. Write 5 integers less than -20:

• Examples: -21, -22, -23, -24, -25

3. A car traveled 60 km north and then 90 km south. How far from Patna was the car finally?

• Net distance = $90 - 60 = 30$ km south of Patna.

4. Subtract the sum of -250 and 138 from the sum of 136 and -272:

• Sum of -250 and 138: $-250 + 138 = -112$
• Sum of 136 and -272: $136 + (-272) = -136$
• Required subtraction: $-136 - (-112) = -136 + 112 = -24$

5. Simplify:

• (i) $\frac{7}{2} - \left[ \frac{1}{4} + \left\{ \frac{1}{4} - \frac{1}{2} \left( \frac{3}{2} - \frac{1}{6} \right) \right\} \right]$
  
  

(i) Solve

$$\frac{7}{2}-[\frac{1}{4}+\{\frac{1}{4}-\frac{1}{2}(\frac{3}{2}-\frac{1}{6})\}]$$

Step 1: Solve the innermost parentheses $(\frac{3}{2}-\frac{1}{6})$

Find a common denominator (LCM of 2 and 6 is 6):

$$\frac{3}{2}=\frac{9}{6},\frac{1}{6}=\frac{1}{6}$$$$\frac{3}{2}-\frac{1}{6}=\frac{9}{6}-\frac{1}{6}=\frac{8}{6}=\frac{4}{3}$$

Step 2: Multiply by $\frac{1}{2}$

$$\frac{1}{2}\times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}$$

Step 3: Compute $\frac{1}{4}-\frac{2}{3}$

Find a common denominator (LCM of 4 and 3 is 12):

$$\frac{1}{4}=\frac{3}{12},\frac{2}{3}=\frac{8}{12}$$$$\frac{1}{4}-\frac{2}{3}=\frac{3}{12}-\frac{8}{12}=\frac{-5}{12}$$

Step 4: Compute $\frac{1}{4}+(-\frac{5}{12})$

Find a common denominator (LCM of 4 and 12 is 12):

$$\frac{1}{4}=\frac{3}{12}$$$$\frac{3}{12}-\frac{5}{12}=\frac{-2}{12}=\frac{-1}{6}$$

Step 5: Compute the final subtraction

$$\frac{7}{2}-(-\frac{1}{6})$$

This is equivalent to:

$$\frac{7}{2}+\frac{1}{6}$$

Find a common denominator (LCM of 2 and 6 is 6):

$$\frac{7}{2}=\frac{21}{6}$$$$\frac{21}{6}+\frac{1}{6}=\frac{22}{6}=\frac{11}{3}$$

Final Answer:

$$\frac{11}{3}$$

• 
  
• (ii) $5x - [4y - \{ 7x - (3z - 2y) + 4z \}]$
  (Expand and simplify.)

(ii) Solve

$$5x-[4y-\{7x-(3z-2y)+4z\}]$$

Step 1: Solve the inner parentheses $(3z-2y)$

$$=3z-2y$$

Step 2: Solve $7x-(3z-2y)+4z$

$$=7x-3z+2y+4z$$$$=7x+2y+(4z-3z)$$$$=7x+2y+z$$

Step 3: Solve $4y-\{7x+2y+z\}$

Distribute the negative sign:

$$=4y-7x-2y-z$$$$=(4y-2y)-7x-z$$$$=2y-7x-z$$

Step 4: Compute $5x-[2y-7x-z]$

Distribute the negative sign:

$$5x-2y+7x+z$$$$(5x+7x)-2y+z$$$$=12x-2y+z$$

Final Answer:

$$\mathbf{12}\mathbf{x}-\mathbf{2}\mathbf{y}+\mathbf{z}\mathbf{<br>}$$

6. How much is $a + 2b - 3c$ greater than $2a - 3b + c$?

• Difference: $$(a + 2b - 3c) - (2a - 3b + c) = a + 2b - 3c - 2a + 3b - c = -a + 5b - 4c$$

7. By how much does $3x^2 - 5x + 6$ exceed $x^3 - x^2 + 4x - 1$?

• Difference: $$(3x^2 - 5x + 6) - (x^3 - x^2 + 4x - 1) = -x^3 + 3x^2 - (-x^2) - 5x - 4x + 6 - (-1) = -x^3 + 4x^2 - 9x + 7$$

8. Write True/False:

• (i) False (The smallest integer is not zero; it is negative and infinite.)
• (ii) True (-10 is greater than -16.)
• (iii) False (The absolute value is always non-negative, but not always greater than the integer. For example, |5| = 5.)
• (iv) True ($| -40 | + 40 = 40 + 40 = 0$)
• (v) False (-10 is to the left of -6 on the number line.)

Let me know if you need detailed solutions! 😊