Home
Class 7
MATHS
2 (3)/(5) + 1 (4)/(5) + 3 (2)/(5...

`2 (3)/(5) + 1 (4)/(5) + 3 (2)/(5`

A

`=7(4)/(7)`

B

`=7(4)/(5)`

C

`=7(3)/(5)`

D

`=5(4)/(5)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem \(2 \frac{3}{5} + 1 \frac{4}{5} + 3 \frac{2}{5}\), we will follow these steps: ### Step 1: Convert Mixed Numbers to Improper Fractions First, we convert each mixed number into an improper fraction. - For \(2 \frac{3}{5}\): \[ 2 \frac{3}{5} = \frac{2 \times 5 + 3}{5} = \frac{10 + 3}{5} = \frac{13}{5} \] - For \(1 \frac{4}{5}\): \[ 1 \frac{4}{5} = \frac{1 \times 5 + 4}{5} = \frac{5 + 4}{5} = \frac{9}{5} \] - For \(3 \frac{2}{5}\): \[ 3 \frac{2}{5} = \frac{3 \times 5 + 2}{5} = \frac{15 + 2}{5} = \frac{17}{5} \] ### Step 2: Add the Improper Fractions Now we add the three improper fractions together: \[ \frac{13}{5} + \frac{9}{5} + \frac{17}{5} \] Since all fractions have the same denominator, we can add the numerators directly: \[ \frac{13 + 9 + 17}{5} \] ### Step 3: Calculate the Sum of the Numerators Now we calculate the sum of the numerators: \[ 13 + 9 = 22 \] \[ 22 + 17 = 39 \] So, we have: \[ \frac{39}{5} \] ### Step 4: Convert to Mixed Number (if needed) To convert \(\frac{39}{5}\) back to a mixed number: - Divide 39 by 5: - \(39 \div 5 = 7\) remainder \(4\) Thus, we can express \(\frac{39}{5}\) as: \[ 7 \frac{4}{5} \] ### Final Answer The final answer is: \[ \frac{39}{5} \text{ or } 7 \frac{4}{5} \] ---
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • FRACTIONS

    ICSE|Exercise Try This|37 Videos
  • FRACTIONS

    ICSE|Exercise Warm-up Exercise|24 Videos
  • FRACTIONS

    ICSE|Exercise EXERCISE 3B|30 Videos
  • EXPONENTS AND POWERS

    ICSE|Exercise Unit practice paper -1|7 Videos
  • FUNDAMENTAL GEOMETRICAL CONCEPTS

    ICSE|Exercise Exercise |14 Videos

Similar Questions

Explore conceptually related problems

Simplify : ((3)/(5)+(1)/(3))/((2)/(5)+(2)/(5))

Evaluate : (ii) (2)/(3)+(-4)/(5)-(1)/(3)-(2)/(5)

Evaluate : (ii) (2)/(3)+(-4)/(5)+(1)/(3)+(2)/(5)

The sum of the infinite series, 1 ^(2) -(2^(2))/(5) + (3 ^(2))/(5 ^(2))+ (4^(2))/(5 ^(3))+ (5 ^(2))/(5 ^(4))-(6 ^(2))/(5 ^(5)) + ..... is:

Solve (a) 2/3 + 1/7 (b) 3/10 + 7/15 (c) 4/9 + 2/7 (d) 5/7 + 1/3 (e) 2/5 + 1/6 (f) 4/5 + 2/3 (g) (3/4) (1/3) (h) (5/6) (1/3) (1) 2/3 + 3/4 + 1/2 (1) 1/2 + 1/3 = 1/6 (k) (1) 1/3 + (3) 2/3 (1) (4) 2/3 +(3) 1/4 (m) ((16)/5))(7/5)

Show that 1 + 1/2.(3/5) + (1.3)/(2.4) (3/5)^2 + (1.3.5)/(2.4.6) (3/5)^3 + …… = sqrt([5/2])

Find the sum of 7 terms of the sequence (1/5+2/(5^2)+3/(5^3)),\ (1/(5^4)+2/(5^5)+3/(5^6)),\ (1/(5^7)+2/(5^8)+3/(5^9)),\ ,

The image of the line (x-1)/3=(y-3)/1=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line (1) (x+3)/3=(y-5)/1=(z-2)/(-5) (2) (x+3)/(-3)=(y-5)/(-1)=(z+2)/5 (3) (x-3)/3=(y+5)/1=(z-2)/(-5) (3) (x-3)/(-3)=(y+5)/(-1)=(z-2)/5

Sum up to n terms the series 1-(2)/(5)+(3)/(5^(2))-(4)/(5^(3)) + ...

The inverse of the matrix [[2,3],[-4,-5]] is: (a) [[-(5)/(2),4],[(9)/(2),1]] , (b) [[(5)/(2),2],[(4)/(2),2]] (c) [[-(5)/(2),-(3)/(2)],[2,1]] (d) [[1,3],[(2)/(3),4]]