Reduce 261 432 to the lowest terms.

To reduce 261 432 to the lowest terms. A fraction is said to be in the lowest term if its numerator and the denominator have no common factor other than 1 . So, we need to find the highest common factor of 261 and 432 . 261 = 3 2 × 29 432 = 2 4 × 3 3 So, the HCF of 261 , 432 = 3 2 = 9 Now, 261 ÷ 9 432 ÷ 9 = 29 48 Therefore, 29 48 is the lowest term of the given fraction.

Reduce 289 391 to the lowest term.

To reduce 289 391 to the lowest term. A fraction is said to be in the lowest term if its numerator and the denominator have no common factor other than 1 . So, we need to find the highest common factor of 289 and 391 . 289 = 17 2 391 = 17 × 23 So, the HCF of 289 , 391 = 17 Now, 289 ÷ 17 391 ÷ 17 = 17 23 Therefore, 17 23 is the lowest term of the given fraction.

Reduce 296 481 to the lowest term.

To reduce 296 481 to the lowest term. A fraction is said to be in the lowest term if its numerator and the denominator have no common factor other than 1 . So, we need to find the highest common factor of 296 and 481 . 296 = 2 3 × 37 481 = 13 × 37 So, the HCF of 296 , 481 = 37 Now, 296 ÷ 37 481 ÷ 37 = 8 13 Therefore, 8 13 is the lowest term of the given fraction.

Reduce 336 672 to the lowest term.

To reduce 336 672 to the lowest term. A fraction is said to be in the lowest term if its numerator and the denominator have no common factor other than 1 . So, we need to find the highest common factor of 336 and 672 . 336 = 2 4 × 3 × 7 672 = 2 5 × 3 × 7 So, the HCF of 336 , 672 = 2 4 × 3 × 7 = 336 Now, 336 ÷ 336 672 ÷ 336 = 1 2 Therefore, 1 2 is the lowest term of the given fraction.

E M B I B E

New Global Mathematics - 6>Chapter 4 - Factors and Multiples>EXERCISE 4.5

Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Factors and Multiples, Exercise 5: EXERCISE 4.5

Author:Dr Pooranmal & Ashfaq A Khan

Dr Pooranmal Mathematics Solutions for Exercise - Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Factors and Multiples, Exercise 5: EXERCISE 4.5

Attempt the free practice questions on Chapter 4: Factors and Multiples, Exercise 5: EXERCISE 4.5 with hints and solutions to strengthen your understanding. New Global Mathematics - 6 solutions are prepared by Experienced Embibe Experts.

Questions from Dr Pooranmal and Ashfaq A Khan Solutions for Chapter: Factors and Multiples, Exercise 5: EXERCISE 4.5 with Hints & Solutions

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6th CBSE

IMPORTANT

New Global Mathematics - 6>Chapter 4 - Factors and Multiples>EXERCISE 4.5>Q 8

Three different tanks contain $450$ litres, $430$ litres and $540$ litres of water respectively. Find the maximum capacity (in litres) of the container which can measure the water of all the tank in exact number of times.