Class 6 Mathematics â Mixed Practice Questions : āĻļā§ā§°ā§āĻŖā§ ā§Ŧ āĻāĻŖāĻŋāĻ¤ â āĻŽāĻŋāĻļā§ā§° āĻ āĻ¨ā§āĻļā§āĻ˛āĻ¨ā§ āĻĒā§ā§°āĻļā§āĻ¨
Q1. A school has 420 students. If 3/7 of the students are boys, how many are girls ? āĻĒā§ā§°āĻļā§āĻ¨ ā§§: āĻāĻāĻ¨ āĻŦāĻŋāĻĻā§āĻ¯āĻžāĻ˛āĻ¯āĻŧāĻ¤ 420 āĻāĻ¨ āĻļāĻŋāĻā§āĻˇāĻžā§°ā§āĻĨā§ āĻāĻā§āĨ¤ āĻ¯āĻĻāĻŋ āĻļāĻŋāĻā§āĻˇāĻžā§°ā§āĻĨā§ā§° 3/7 āĻ āĻāĻļ āĻ˛'ā§°āĻž āĻšāĻ¯āĻŧ, āĻ¤ā§āĻ¨ā§āĻ¤ā§ āĻā§ā§ąāĻžāĻ˛ā§ā§° āĻ¸āĻāĻā§āĻ¯āĻž āĻāĻŋāĻŽāĻžāĻ¨ ?
Solution / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
Number of boys / āĻ˛'ā§°āĻžā§° āĻ¸āĻāĻā§āĻ¯āĻž
3/7 × 420 = 1260/7 = 180
Girls (āĻā§ā§ąāĻžāĻ˛ā§ā§° āĻ¸āĻāĻā§āĻ¯āĻž) = Total students (āĻŽā§āĻ āĻļāĻŋāĻā§āĻˇāĻžā§°ā§āĻĨā§) − Boys (āĻ˛'ā§°āĻž)
= 420 − 180 = 240
Ans / āĻāĻ¤ā§āĻ¤ā§°: 240 girls (240 āĻāĻ¨ āĻā§ā§ąāĻžāĻ˛ā§)
Q2. The length of a rectangle is 12 cm and its width is 8 cm. Find its perimeter and area. āĻĒā§ā§°āĻļā§āĻ¨ ā§¨: āĻāĻāĻ¨ āĻāĻ¯āĻŧāĻ¤āĻā§āĻˇā§āĻ¤ā§ā§°ā§° āĻĻā§ā§°ā§āĻā§āĻ¯ 12 cm āĻā§°ā§ āĻĒā§ā§°āĻ¸ā§āĻĨ 8 cmāĨ¤ āĻāĻ¯āĻŧāĻžā§° āĻĒā§°āĻŋāĻ¸ā§āĻŽāĻž āĻā§°ā§ āĻā§āĻˇā§āĻ¤ā§ā§°āĻĢāĻ˛ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻāĨ¤
Solution / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
Perimeter / āĻĒā§°āĻŋāĻ¸ā§āĻŽāĻž
Perimeter = 2(l + w)
= 2(12 + 8)
= 2 × 20
= 40 cm
Area / āĻā§āĻˇā§āĻ¤ā§ā§°āĻĢāĻ˛
Area = l × w
= 12 × 8
= 96 cm²
Ans / āĻāĻ¤ā§āĻ¤ā§°:
- Perimeter (āĻĒā§°āĻŋāĻ¸ā§āĻŽāĻž) = 40 cm
- Area (āĻā§āĻˇā§āĻ¤ā§ā§°āĻĢāĻ˛) = 96 cm²
Q3. Simplify: āĻĒā§ā§°āĻļā§āĻ¨ ā§Š: āĻ¸ā§°āĻ˛ āĻā§°āĻžāĨ¤
(5x2 − 3x + 7) − (2x2 + x − 4)
Solution / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
= 5x2 − 3x + 7 − 2x2 − x + 4
= (5x2 − 2x2) + (− 3x − x) + (7 + 4)
= 3x2 − 4x + 11
Ans / āĻāĻ¤ā§āĻ¤ā§°: 3x2 − 4x + 11
Q4. Find the value of x. āĻĒā§ā§°āĻļā§āĻ¨ ā§Ē: x-ā§° āĻŽāĻžāĻ¨ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻāĨ¤
3x − 7 = 2x + 9
Solution / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
3x − 7 = 2x + 9
Subtract 2x from both sides / āĻĻā§āĻ¯āĻŧā§āĻĢāĻžāĻ˛ā§° āĻĒā§°āĻž 2x āĻŦāĻŋāĻ¯āĻŧā§āĻ āĻā§°ā§āĻ
3x − 2x = 9 + 7
x = 16
Ans / āĻāĻ¤ā§āĻ¤ā§°: x = 16
Q5. A pen costs âš120 and a notebook costs âš280. If Jane bought 3 pens and 4 notebooks, how much did she spend ? āĻĒā§ā§°āĻļā§āĻ¨ ā§Ģ: āĻāĻāĻž āĻāĻ˛āĻŽā§° āĻŽā§āĻ˛ā§āĻ¯ âš120 āĻā§°ā§ āĻāĻāĻž āĻāĻžāĻ¤āĻžā§° āĻŽā§āĻ˛ā§āĻ¯ âš280āĨ¤ āĻ¯āĻĻāĻŋ Jane-āĻ 3āĻāĻž āĻāĻ˛āĻŽ āĻā§°ā§ 4āĻāĻ¨ āĻāĻžāĻ¤āĻž āĻāĻŋāĻ¨ā§, āĻ¤ā§āĻ¨ā§āĻ¤ā§ āĻŽā§āĻ āĻāĻŋāĻŽāĻžāĻ¨ āĻāĻāĻž āĻā§°āĻ āĻā§°āĻŋāĻ˛ā§ ?
Solution / āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨:
Cost of 3 pens / 3āĻāĻž āĻāĻ˛āĻŽā§° āĻŽā§āĻ˛ā§āĻ¯
= 3 × âš120
= âš360
Cost of 4 notebooks / 4āĻāĻ¨ āĻāĻžāĻ¤āĻžā§° āĻŽā§āĻ˛ā§āĻ¯
= 4 × âš280
= âš1120
Total cost / āĻŽā§āĻ āĻā§°āĻ
= âš360 + âš1120
= âš1480
Ans / āĻāĻ¤ā§āĻ¤ā§°: âš1480
Tricks / āĻā§āĻļāĻ˛
- Fraction problems: Find the given fraction first, then subtract if required. (āĻāĻā§āĻ¨āĻžāĻāĻļā§° āĻĒā§ā§°āĻļā§āĻ¨: āĻĒā§ā§°āĻĨāĻŽā§ āĻĻāĻŋāĻ¯āĻŧāĻž āĻāĻā§āĻ¨āĻžāĻāĻļā§° āĻŽāĻžāĻ¨ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻāĻ, āĻ¤āĻžā§° āĻĒāĻŋāĻāĻ¤ āĻĒā§ā§°āĻ¯āĻŧā§āĻāĻ¨ āĻš'āĻ˛ā§ āĻŦāĻŋāĻ¯āĻŧā§āĻ āĻā§°āĻāĨ¤)
- Rectangle: Perimeter = 2(l + w), Area = l × w. (āĻāĻ¯āĻŧāĻ¤āĻā§āĻˇā§āĻ¤ā§ā§°: āĻĒā§°āĻŋāĻ¸ā§āĻŽāĻž = 2(āĻĻā§ā§°ā§āĻā§āĻ¯ + āĻĒā§ā§°āĻ¸ā§āĻĨ), āĻā§āĻˇā§āĻ¤ā§ā§°āĻĢāĻ˛ = āĻĻā§ā§°ā§āĻā§āĻ¯ × āĻĒā§ā§°āĻ¸ā§āĻĨāĨ¤)
- Algebra: Remove brackets carefully and combine like terms. (āĻŦā§āĻāĻāĻŖāĻŋāĻ¤: āĻŦā§ā§°ā§āĻā§āĻ āĻ¸āĻžā§ąāĻ§āĻžāĻ¨ā§ āĻāĻāĻ¤ā§°āĻžāĻ āĻāĻā§ āĻ§ā§°āĻŖā§° āĻĒāĻĻ āĻāĻā§āĻ˛āĻ āĻā§°āĻāĨ¤)
- Linear equations: Bring variables to one side and numbers to the other. (āĻ¸āĻŽā§āĻā§°āĻŖ: āĻāĻ˛āĻāĻŦā§ā§° āĻāĻĢāĻžāĻ˛ā§ āĻā§°ā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻŦā§ā§° āĻāĻ¨āĻĢāĻžāĻ˛ā§ āĻāĻ¨āĻāĨ¤)
- Word problems: Calculate each item's cost separately, then add. (āĻļāĻŦā§āĻĻ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž: āĻĒā§ā§°āĻ¤āĻŋāĻā§ āĻŦāĻ¸ā§āĻ¤ā§ā§° āĻŽā§āĻ˛ā§āĻ¯ āĻĒā§āĻĨāĻā§ āĻāĻ˛āĻŋāĻ¯āĻŧāĻžāĻ āĻļā§āĻˇāĻ¤ āĻ¯ā§āĻ āĻā§°āĻāĨ¤)