Simplify the following: 5 7 10 - 3 2 5 + 4 1 15

Given, 5 7 10 - 3 2 5 + 4 1 15 Converting the mixed fractions to improper fractions, we have: 5 7 10 = 57 10 , 3 2 5 = 17 5 and 4 1 15 = 61 15 Taking the LCM of the denominators, we get: LCM 5 , 10 , 15 = 30 Making equivalent fractions with denominator 30 , we have: 57 10 → 57 × 3 10 × 3 = 171 30 17 5 → 17 × 6 5 × 6 = 102 30 61 15 → 61 × 2 15 × 2 = 122 30 Therefore, simplifying like fractions, we get: 171 30 - 102 30 + 122 30 = 171 - 102 + 122 30 = 191 30 Hence, the required answer is 6 11 30 .

Vibha bought a notebook for ₹ 13 2 5 , and eraser for ₹ 1 3 4 and a pen for ₹ 11 2 3 . What is the total amount she paid?

Given, Cost of a notebook = ₹ 13 2 5 = ₹ 67 5 Cost of an eraser = ₹ 1 3 4 = ₹ 7 4 Cost of a pen = ₹ 11 2 3 = ₹ 35 3 Taking the LCM of denominators, we have: LCM 5 , 4 , 3 = 60 Making equivalent fractions with denominator 60 , we get: 67 5 → 67 × 12 5 × 12 = 804 60 7 4 → 7 × 15 4 × 15 = 105 60 35 3 → 35 × 20 3 × 20 = 700 60 Therefore, we get: The total amount she paid = 804 60 + 105 60 + 700 60 = 804 + 105 + 700 60 = ₹ 1609 60 Hence, the required answer is ₹ 26 49 60 .

A labourer earned ₹ 152 2 3 and out of this money he spent ₹ 82 1 3 . How much money is left with him?

Given, The money he earned = ₹ 152 2 3 = ₹ 458 3 The money he spent = ₹ 82 1 3 = ₹ 247 3 Here, obtained improper fractions are like fractions. Therefore, we get: The money left with him = ₹ 458 3 - ₹ 247 3 = ₹ 458 - 247 3 = ₹ 211 3 Hence, the required answer is ₹ 70 1 3 .

What should be added to 7 3 11 to get 9 2 7 ?

Let the number to be added be x . Therefore, we get: x + 7 3 11 = 9 2 7 x = 9 2 7 - 7 3 11 . . . ( i ) Converting the mixed fractions in to improper fractions, we get: 7 3 11 = 80 11 and 9 2 7 = 65 7 Taking the LCM of the denominators, we have: LCM 11 , 7 = 77 Making equivalent fractions with denominator 77 , we get: 80 11 → 80 × 7 11 × 7 = 560 77 65 7 → 65 × 11 7 × 11 = 715 77 Substituting these equivalent fractions in equation ( i ) , we get: x = 715 77 - 560 77 = 715 - 560 77 = 155 77 Hence, the required number to be added is 2 1 77 .

What should be added to 19 2 3 to get 23 1 2 ?

Let the number to be added be x . Therefore, we get: x + 19 2 3 = 23 1 2 x = 23 1 2 - 19 2 3 . . . ( i ) Converting the mixed fractions in to improper fractions, we get: 19 2 3 = 59 3 and 23 1 2 = 47 2 Taking the LCM of the denominators, we have: LCM 3 , 2 = 6 Making equivalent fractions with denominator 6 , we get: 59 3 → 59 × 2 3 × 2 = 118 6 47 2 → 47 × 3 2 × 3 = 141 6 Substituting these equivalent fractions in equation ( i ) , we get: x = 141 6 - 118 6 = 141 - 118 6 = 23 6 Hence, the required number to be added is 3 5 6 .

Subtract the sum of 4 2 3 and 1 2 9 from the sum of 2 5 7 and 3 2 3 .

Given, To subtract the sum of 4 2 3 and 1 2 9 from the sum of 2 5 7 and 3 2 3 . Converting the mixed fractions in to improper fractions, we get: 4 2 3 = 14 3 , 1 2 9 = 11 9 , 2 5 7 = 19 7 and 3 2 3 = 11 3 Taking LCM of the denominators, we have: LCM 3 , 9 , 7 = 63 Making equivalent fractions with denominator 63 , we get: 14 3 → 14 × 21 3 × 21 = 294 63 11 9 → 11 × 7 9 × 7 = 77 63 19 7 = 19 × 9 7 × 9 = 171 63 11 3 = 11 × 21 3 × 21 = 231 63 Sum of first two terms = 294 63 + 77 63 = 294 + 77 63 = 371 63 Sum of last two terms = 171 63 + 231 63 = 171 + 231 63 = 402 63 Their difference = 402 63 - 371 63 = 402 - 371 63 = 31 63 Hence, the required answer is 31 63 .

An iron pipe of length 5 1 3 m long was cut in to two pieces. One piece is 3 1 7 m long. What is the length of the other piece?

Given, Total length of the iron pipe = 5 1 3 m = 16 3 m Length of one piece = 3 1 7 m = 22 7 m Finding LCM of denominators, we have: LCM 3 , 7 = 21 Making equivalent fractions with denominator 21 , we get: 16 3 → 16 × 7 3 × 7 = 112 21 22 7 → 22 × 3 7 × 3 = 66 21 Therefore, we get: The length of the other piece = 112 21 - 66 21 = 112 - 66 21 = 46 21 m Hence, the required answer is 2 4 21 m .

Anil takes 17 1 3 minutes to run around the school ground. Sunil takes 12 1 5 minutes to do the same. Who takes lesser time and by what fraction?

Given, Time taken by Anil to run = 17 1 3 minutes = 52 3 minutes Time taken by Sunil to run = 12 1 5 minutes = 61 5 minutes Taking LCM of the denominators, we have: LCM 3 , 5 = 15 Making equivalent fractions by keeping the denominator as 15 , we get: 52 3 → 52 × 5 3 × 5 = 260 15 61 5 → 61 × 3 5 × 3 = 183 15 Comparing the obtained like fractions, we have: 260 15 > 183 15 And, their difference = 260 15 - 183 15 = 260 - 183 15 = 77 15 minutes. Hence, we get: Sunil took lesser time by 5 2 15 minutes.

E M B I B E

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5

Jagat Jitendra Matto Solutions for Chapter: Fractions, Exercise 7: EXERCISE 5.5

Author:Jagat Jitendra Matto

Jagat Jitendra Matto Mathematics Solutions for Exercise - Jagat Jitendra Matto Solutions for Chapter: Fractions, Exercise 7: EXERCISE 5.5

Attempt the free practice questions on Chapter 5: Fractions, Exercise 7: EXERCISE 5.5 with hints and solutions to strengthen your understanding. Explore Mathematics 6 solutions are prepared by Experienced Embibe Experts.

Questions from Jagat Jitendra Matto Solutions for Chapter: Fractions, Exercise 7: EXERCISE 5.5 with Hints & Solutions

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 3

Simplify the following:

$5 \frac{7}{10} - 3 \frac{2}{5} + 4 \frac{1}{15}$

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 4

Vibha bought a notebook for $₹ 13 \frac{2}{5}$ , and eraser for $₹ 1 \frac{3}{4}$ and a pen for $₹ 11 \frac{2}{3}$ . What is the total amount she paid?

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 5

A labourer earned $₹ 152 \frac{2}{3}$ and out of this money he spent $₹ 82 \frac{1}{3}$ . How much money is left with him?

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 6

What should be added to $7 \frac{3}{11}$ to get $9 \frac{2}{7}$ ?

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 7

What should be added to $19 \frac{2}{3}$ to get $23 \frac{1}{2}$ ?

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 8

Subtract the sum of $4 \frac{2}{3}$ and $1 \frac{2}{9}$ from the sum of $2 \frac{5}{7}$ and $3 \frac{2}{3}$ .

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 9

An iron pipe of length $5 \frac{1}{3} m$ long was cut in to two pieces. One piece is $3 \frac{1}{7} m$ long. What is the length of the other piece?

MEDIUM

6th CBSE

IMPORTANT

Explore Mathematics 6>Chapter 5 - Fractions>EXERCISE 5.5>Q 10

Anil takes $17 \frac{1}{3}$ minutes to run around the school ground. Sunil takes $12 \frac{1}{5}$ minutes to do the same. Who takes lesser time and by what fraction?

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