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T3 Rev WS6

T3 Rev WS6. Algebraic Word Problems. Q1). Joe and Ahmad have 80 marbles altogether. Ahmad has 4 times as many marbles as Joe. How many marbles does each boy have?. Q1). Joe and Ahmad have 80 marbles altogether. Ahmad has 4 times as many marbles as Joe. How many marbles does each boy have?.

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T3 Rev WS6

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  1. T3 Rev WS6 Algebraic Word Problems

  2. Q1) Joe and Ahmad have 80 marbles altogether. Ahmad has 4 times as many marbles as Joe. How many marbles does each boy have?

  3. Q1) Joe and Ahmad have 80 marbles altogether. Ahmad has 4 times as many marbles as Joe. How many marbles does each boy have? Let the no. of marbles Joe has = x  The no. of marbles Ahmad has = 4x 4x + x = 80 x = 80 / 5 x = 16  Joe has 16 marbles and Ahmad has (4 x 16) = 64 marbles.

  4. Q2) Amy is 4 years older than Peter. Alison is 2 years younger than Peter. If the sum of their ages is 47, what are their ages?

  5. Amy is 4 years older than Peter. Alison is 2 years younger than Peter. If the sum of their ages is 47, what are their ages? Let the age of Amy = x years old The age of Peter = (x – 4) years old & the age of Alison = (x – 4 – 2) = (x – 6) years old x + (x – 4) + (x – 6) = 47 3x – 10 = 47 x = (47 + 10) / 3 x = 19 The age of Amy, Peter and Alison are 19, 15 and 13 years old respectively. Q2)

  6. Q3) Peter has 25 sweets and Lilian has 55. How many sweets must Peter give Lilian so that Lilian will have 4 times as many sweets as Peter?

  7. Q3) Let the no. of sweets Peter gives to Lilian = x 55 + x = 4(25 – x) 55 + x = 100 – 4x 4x + x = 100 – 55 x = 45 / 5 x = 9 Peter must give Lilian 9 sweets Peter has 25 sweets and Lilian has 55. How many sweets must Peter give Lilian so that Lilian will have 4 times as many sweets as Peter?

  8. Q4) The numerator of a fraction is 5 less than the denominator. If 1 is added to both the numerator and denominator, the fraction would become 2/3. Find the fraction.

  9. Q4) The numerator of a fraction is 5 less than the denominator. If 1 is added to both the numerator and denominator, the fraction would become 2/3. Find the fraction. • Let the numerator of the fraction = x • the denominator = x + 5 The fraction is 9 / 14

  10. Q5) Tim, Dom and Harry share $256. Dom’s share is four times as much as Tim’s and Tim’s share is one-third of Harry’s. How much is each of their share?

  11. Q5) Tim, Dom and Harry share $256. Dom’s share is four times as much as Tim’s and Tim’s share is one-third of Harry’s. How much is each of their share? • Let the amt of money Harry has = $ x • Amt of money Tim has = $ (x/3) & amt of money Dom has = $ (4x/3) Tim, Dom and Harry has $32, $128 and $96 respectively.

  12. Q6) A man normally takes 5 hours to travel at a certain speed from city A to city B. One day, he increases his speed by 4 km/h and finds that the journey from A to B takes half an hour less than the normal time. Find his normal speed.

  13. Q6) A man normally takes 5 hours to travel at a certain speed from city A to city B. One day, he increases his speed by 4 km/h and finds that the journey from A to B takes half an hour less than the normal time. Find his normal speed. 5 hours at x km/h City A 4.5 hours at (x + 4) km/h City B Let his normal speed = x km/h Distance between City A and City B = speed x time = 5x km Using speed = (x + 4) km/h & time = 4.5 h, Distance between City A and City B = speed x time = 4.5(x + 4)km 5x = 4.5(x + 4) … (DIY)… x = 36 His normal speed is 36 km/h

  14. Q7) The sum of the ages of Farmi and Mumud is 38. Seven years ago, Farmi was three times as old as Mumud. Find their present ages.

  15. Q7) The sum of the ages of Farmi and Mumud is 38. Seven years ago, Farmi was three times as old as Mumud. Find their present ages. • Let the present age of Farmi = x years old • The present age of Mumud = (38 – x) yr old Seven years ago, we’ll have, x – 7 = 3(38 – x – 7) … (DIY) … x = 25 The present age of Farmi & Mumud are 25 and 13 years old respectively.

  16. Q8) Aramugam has enough money to buy 24 apples. If the price if each apple is reduced by 5 cent, he will be able to buy an extra 6 apples with the same sum of money. Find the original cost of each apple.

  17. Q8) Aramugam has enough money to buy 24 apples. If the price if each apple is reduced by 5 cent, he will be able to buy an extra 6 apples with the same sum of money. Find the original cost of each apple. Let the original cost of each apple = x cents Since he can buy 24 apples,  Total amt he has = 24x cents Given that if price of each apple = (x – 5) cents, he can buy (24 + 6) = 30 apples, 30(x – 5) = 24x … (DIY) … x = 25 The original cost of each apple = 25 cents

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