Reduce 161 207 to the lowest terms.

Given fraction to reduce to lowest terms: 161 207 Here, first we need to find out the HCF between the numerator and the denominator. The HCF of 161 and 207 can be found out by using division method. It can be carried out in the following steps: From the above steps, we observe that, HCF is 23 . Now, let us simplify the given fraction, by dividing its numerator and the denominator by the HCF. ⇒ 161 207 = 161 ÷ 23 207 ÷ 23 = 7 9 Hence, the reduced form of the fraction 161 207 to its lowest terms is 7 9 .

Reduce 517 799 to the lowest terms.

Given fraction to reduce with lowest terms: 517 799 To reduce to lowest terms, we should divide the numerator and denominator by their HCF. So, first we need to find out the HCF between the numerator and the denominator. The HCF of 517 and 799 can be found out by using division method. We can perform the division as follows. The division is carried out by considering the remainder of first step as the divisor of the second step and the divisor of the first step as the dividend for the second step, until a zero remainder is obtained. We know that the HCF of the numbers will be the last divisor for which the remainder is 0 in the division. From the division, we can observe that, the remainder becomes 0 for the divisor 47 . Hence, the HCF is 47 . Now, let us simplify the given fraction, by dividing its numerator and the denominator by the HCF. So, we have 517 799 = 517 ÷ 47 799 ÷ 47 = 11 17 Hence, the reduced form of the fraction 517 799 to its lowest terms is 11 17 .

Reduce 296 481 to the lowest terms.

Given fraction to reduce with the lowest terms: 296 481 To reduce to the lowest terms, we should divide the numerator and denominator by their HCF. So, first we need to find out the HCF of the numerator and the denominator. We can find the HCF of 296 and 481 by using the division method. We can perform the division as follows. The division is carried out by considering the remainder of first step as the divisor of the second step and the divisor of the first step as the dividend for the second step. This process is continued until a zero remainder is obtained. We know that the HCF of the numbers will be the last divisor for which the remainder is 0 in the division. From the division, we can observe that, the remainder becomes 0 for the divisor 37 . Hence, the HCF is 37 . Now, let us simplify the given fraction, by dividing its numerator and the denominator by the HCF. So, we have, 296 481 = 296 ÷ 37 481 ÷ 37 = 8 13 Hence, the reduced form of the fraction 296 481 to its lowest terms is 8 13 .

Reduce 289 391 to the lowest terms.

Given fraction to reduce to the lowest terms is 289 391 . Here, first we need to find out the HCF of the numerator and the denominator. The HCF of 289 and 391 can be found by using prime factorization method. 289 = 17 × 17 391 = 17 × 23 We observe that, the common factor between 289 & 391 is 17 . ∴ The HCF of 289 and 391 is 17 . Now, let us simplify the given fraction, by dividing its numerator and the denominator by the HCF. ⇒ 289 391 = 289 ÷ 17 391 ÷ 17 = 17 23 Hence, the reduced form of the fraction 289 391 to its lowest terms is 17 23 .

E M B I B E

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise

S K Khanuja Solutions for Chapter: Factors and Multiples, Exercise 5: Exercise

Author:S K Khanuja

S K Khanuja Mathematics Solutions for Exercise - S K Khanuja Solutions for Chapter: Factors and Multiples, Exercise 5: Exercise

Attempt the free practice questions on Chapter 3: Factors and Multiples, Exercise 5: Exercise with hints and solutions to strengthen your understanding. Tryout Maths A book on Mental Maths solutions are prepared by Experienced Embibe Experts.

Questions from S K Khanuja Solutions for Chapter: Factors and Multiples, Exercise 5: Exercise with Hints & Solutions

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 2

Find the HCF of the following numbers using the division method.

$1045, 1520$

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 3

Reduce $\frac{161}{207}$ to the lowest terms.

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 3

Reduce $\frac{517}{799}$ to the lowest terms.

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 3

Reduce $\frac{296}{481}$ to the lowest terms.

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 3

Reduce $\frac{289}{391}$ to the lowest terms.

HARD

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 4

Find the greatest number which divides $615$ and $963$ leaving the remainder $6$ in each case.

HARD

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 5

Find the greatest number that will divide $445, 572$ , and $699$ , leaving remainders $4, 5, 6$ respectively.

MEDIUM

6th CBSE

IMPORTANT

Tryout Maths A book on Mental Maths>Chapter 3 - Factors and Multiples>Exercise>Q 6

A rectangular courtyard is $18 m 72 cm$ long and $13 m 20 cm$ broad. It is to be paved with square tiles of the same size. Find the least number of such tiles.

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|