any 5 rational numbers sums asked in cbse boards
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ANSWER .....
THESE ARE MAJOR SUMS THAT ARE ALWAYS ASKED IN CBSE EXAMS.........
The additive inverse of 2/5 is -2/5
Therefore, 4/7 - 2/5 = 4/7 + (-2/5)
⇒ 4/7 - 2/5 = 4 × 5/7 × 5 + (-2) × 7/5 × 7
= 20/35 + -14/35
= 20 + (-14)/35
= 6/35
Therefore, 4/7 - 2/5 = 6/35
2. Subtract -6/7 from -5/8.
Solution:
The additive inverse of -6/7 is 6/7
Therefore, -5/8 - (-6/7) = -5/8 + 6/7, [Since, -(-6/7) = 6/7)]
⇒ -5/8 - (-6/7) = -5 × 7/8 × 7 + 6 × 8/7 × 8
⇒ -5/8 - (-6/7) = -35/56 + 48/56
⇒ -5/8 - (-6/7) = -35 + 48/56
⇒ -5/8 - (-6/7) = 13/56
Therefore, -5/8 - (-6/7) = 13/56
3. Subtract -4/9 from 2/5
Solution:
The additive inverse of -4/9 is 4/9.
Therefore, 2/5 - (-4/9) = 2/5 + 4/9, [Since, -(-4/9) = 4/9)]
⇒ 2/5 - (-4/9) = 2 × 9/5 × 9 + 4 × 5/9 × 5
⇒ 2/5 - (-4/9) = 18/45 + 20/45
⇒ 2/5 - (-4/9) = 18 + 20/45
Therefore, 2/5 - (-4/9) = 38/45
4. The sum of two rational numbers is -3/5. If one of the number is -9/20, find the other.
Solution:
Sum other number = -3/5, One number = -9/20
Therefore, the other number = Sum of the two rational numbers - One of the given rational number.
= -3/5 - (-9/20)
= -3/5 + 9/20, [Since - (-9/20) = 9/20]
= (-3) × 4 + 9 × 1/20
= -12 + 9/20
= -3/20
Therefore, the required rational number is -3/20.
5. Which rational number should be added to -7/11 so as to get 4/7?
Solution:
Su of the given number and the required rational number = 4/7.
Given rational number = -7/11.
Therefore, the required number = Sum - Given number
= 4/7 + 7/11
= 4 × 11/7 ×11 + 7 × 7/11 × 7
= 44/77 + 49/77
= 44 + 49/77
= 93/77
Thus, the rational number 93/77 should be added to -7/11 so as to get 4/7.
6. What should be subtracted from -4/5 so as to get 6/15?
Solution:
Difference of the given rational number and the required rational number = 6/15.
Given rational number = -4/5.
Therefore the required rational number = -4/5 - 6/15
= -4/5 + -6/15
= (-4) × 3/5 × 3 + -6/15
= -12/15 + -6/15
= (-12) + (-6)/15
= -18/15
= -6/5
Thus, the rational number -6/5 subtracted from -4/5 so as to get 6/15.
I HOPE U GOT YOUR ANSWER.....
THESE ARE MAJOR SUMS THAT ARE ALWAYS ASKED IN CBSE EXAMS.........
The additive inverse of 2/5 is -2/5
Therefore, 4/7 - 2/5 = 4/7 + (-2/5)
⇒ 4/7 - 2/5 = 4 × 5/7 × 5 + (-2) × 7/5 × 7
= 20/35 + -14/35
= 20 + (-14)/35
= 6/35
Therefore, 4/7 - 2/5 = 6/35
2. Subtract -6/7 from -5/8.
Solution:
The additive inverse of -6/7 is 6/7
Therefore, -5/8 - (-6/7) = -5/8 + 6/7, [Since, -(-6/7) = 6/7)]
⇒ -5/8 - (-6/7) = -5 × 7/8 × 7 + 6 × 8/7 × 8
⇒ -5/8 - (-6/7) = -35/56 + 48/56
⇒ -5/8 - (-6/7) = -35 + 48/56
⇒ -5/8 - (-6/7) = 13/56
Therefore, -5/8 - (-6/7) = 13/56
3. Subtract -4/9 from 2/5
Solution:
The additive inverse of -4/9 is 4/9.
Therefore, 2/5 - (-4/9) = 2/5 + 4/9, [Since, -(-4/9) = 4/9)]
⇒ 2/5 - (-4/9) = 2 × 9/5 × 9 + 4 × 5/9 × 5
⇒ 2/5 - (-4/9) = 18/45 + 20/45
⇒ 2/5 - (-4/9) = 18 + 20/45
Therefore, 2/5 - (-4/9) = 38/45
4. The sum of two rational numbers is -3/5. If one of the number is -9/20, find the other.
Solution:
Sum other number = -3/5, One number = -9/20
Therefore, the other number = Sum of the two rational numbers - One of the given rational number.
= -3/5 - (-9/20)
= -3/5 + 9/20, [Since - (-9/20) = 9/20]
= (-3) × 4 + 9 × 1/20
= -12 + 9/20
= -3/20
Therefore, the required rational number is -3/20.
5. Which rational number should be added to -7/11 so as to get 4/7?
Solution:
Su of the given number and the required rational number = 4/7.
Given rational number = -7/11.
Therefore, the required number = Sum - Given number
= 4/7 + 7/11
= 4 × 11/7 ×11 + 7 × 7/11 × 7
= 44/77 + 49/77
= 44 + 49/77
= 93/77
Thus, the rational number 93/77 should be added to -7/11 so as to get 4/7.
6. What should be subtracted from -4/5 so as to get 6/15?
Solution:
Difference of the given rational number and the required rational number = 6/15.
Given rational number = -4/5.
Therefore the required rational number = -4/5 - 6/15
= -4/5 + -6/15
= (-4) × 3/5 × 3 + -6/15
= -12/15 + -6/15
= (-12) + (-6)/15
= -18/15
= -6/5
Thus, the rational number -6/5 subtracted from -4/5 so as to get 6/15.
I HOPE U GOT YOUR ANSWER.....
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