std 8 chp 2 mathas practice work 2.4
Answers
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1.
Amina thinks of a number and subtracts \frac { 5 }{ 2 } from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the required number be x.
Condition I: x – \frac { 5 }{ 2 }
Condition II: 8 × (x – \frac { 5 }{ 2 })
Condition III: 8 × (x – \frac { 5 }{ 2 }) = 3x
⇒ 8x – \frac { 5 }{ 2 } × 8 = 3x (Solving the bracket)
⇒ 8x – 20 = 3x
⇒ 8x – 3x = 20 (Transposing 3x to LHS and 20 to RHS)
⇒ 5x = 20
⇒ x = 20 ÷ 5 = 4 (Transposing 5 to RHS)
Thus, x = 4 is the required number.
2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other number. What are the numbers?
Solution:
Let the positive number be x.
Other number = 5x
Condition I: x + 21 and 5x + 21
Condition II: 5x + 21 = 2 (x + 21)
⇒ 5x + 21 = 2x + 42 (Solving the bracket)
⇒ 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)
⇒ 3x = 21
⇒ x = 21 ÷ 3 = 7 (Transposing 3 to RHS)
Thus, the required numbers are 7 and 7 × 5 = 35
3.
Sum of the digits of a two digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution:
Let unit place digit be x.
Ten’s place digit = 9 – x
Original number = x + 10(9 – x)
Condition I: 10x + (9 – x) (Interchanging the digits)
Condition II: New number = original number + 27
⇒ 10x + (9 – x) = x + 10(9 – x) + 27
⇒ 10x + 9 – x = x + 90 – 10x + 27 (solving the brackets)
⇒ 9x + 9 = -9x + 117 (Transposing 9x to LHS and 9 to RHS)
⇒ 9x + 9x = 117 – 9
⇒ 18x = 108
⇒ x = 108 ÷ 18 (Transposing 18 to RHS)
⇒ x = 6
Unit place digit = 6
Ten’s place digit = 9 – 6 = 3
Thus, the required number = 6 + 3 × 10 = 6 + 30 = 36
Answer:
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