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Prepared by:
SABJA RAJAN
MATHEMATICS
MTTC,PATHANAPURAM
PROBLEM1
The sides of a rectangle are in the ratio
4:5.
i.By what fraction should the shorter
side be increased to make it a
square?
ii.By what fraction should the longer
side be decreased to make it a
square?
SOLUTION
π‘Ήπ’‚π’•π’Šπ’ 𝒐𝒇 π’”π’Šπ’…π’†π’” = πŸ’ ∢ πŸ“
∴ 𝑺𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = πŸ’π’™
π‘³π’‚π’“π’ˆπ’†π’“ π’”π’Šπ’…π’† = πŸ“π’™
i. 𝑳𝒆𝒕 π’Œ 𝒑𝒂𝒓𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’†
𝒃𝒆 π’Šπ’π’„π’“π’†π’‚π’”π’†π’….
∴ π’Œ 𝒑𝒂𝒓𝒕 𝒐𝒇 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = π’Œ Γ— πŸ’π’™
= πŸ’π’Œπ’™
∴ π‘΅π’†π’˜ 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = πŸ’π’™ + πŸ’π’Œπ’™
∴ πŸ’π’™ + πŸ’π’Œπ’™ ∢ πŸ“π’™ = 𝟏 ∢ 𝟏
πŸ’π’™ + πŸ’π’Œπ’™
πŸ“π’™
=
𝟏
𝟏
πŸ’π’™ + πŸ’π’Œπ’™ = πŸ“π’™
πŸ’π’Œπ’™ = πŸ“π’™ βˆ’ πŸ’π’™
πŸ’π’Œπ’™ = 𝒙
π’Œ =
𝒙
πŸ’π’™
∴ π’Œ =
𝟏
πŸ’
, 𝒕𝒉𝒆 π’“π’†π’’π’–π’Šπ’“π’†π’… π’‡π’“π’‚π’„π’•π’Šπ’π’.
ii. 𝐿𝑒𝑑 π‘˜ π‘π‘Žπ‘Ÿπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒
𝑏𝑒 π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘.
∴ π‘˜ π‘π‘Žπ‘Ÿπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒
= π‘˜ Γ— 5π‘₯
= 5π‘˜π‘₯
∴ 𝑁𝑒𝑀 π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒 = 5π‘₯ βˆ’ 5π‘˜π‘₯
4π‘₯ ∢ 5π‘₯ βˆ’ 5π‘˜π‘₯ = 1 ∢ 1
πŸ’π’™
πŸ“π’™ βˆ’ πŸ“π’Œπ’™
=
𝟏
𝟏
∴ πŸ’π’™ = πŸ“π’™ βˆ’ πŸ“π’Œπ’™
πŸ“π’Œπ’™ = πŸ“π’™ βˆ’ πŸ’π’™
πŸ“π’Œπ’™ = 𝒙
π’Œ =
𝒙
πŸ“π’™
∴ π’Œ =
𝟏
πŸ“
, π’Šπ’” 𝒕𝒉𝒆 π’“π’†π’’π’–π’Šπ’“π’†π’… π’‡π’“π’‚π’„π’•π’Šπ’π’.
PROBLEM 2
Two quantities are in the ratio 3:5.
i. If the smaller alone is made four times
the original, what would the ratio be?
ii.If the smaller is doubled and the larger
is halved, what would the ratio be?
SOLUTION
π‘Ήπ’‚π’•π’Šπ’ 𝒐𝒇 π’’π’–π’‚π’π’•π’Šπ’•π’Šπ’†π’” = πŸ‘ ∢ πŸ“
∴ π‘Ίπ’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ‘π’™
π‘³π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ“π’™
i. π‘΅π’†π’˜ π’”π’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ’ Γ— πŸ‘π’™
= πŸπŸπ’™
π‘³π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ“π’™
∴ π‘΅π’†π’˜ π’“π’‚π’•π’Šπ’ = πŸπŸπ’™ ∢ πŸ“π’™
= 𝟏𝟐 ∢ πŸ“
ii. π‘΅π’†π’˜ π’”π’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = 𝟐 Γ— πŸ‘π’™
π‘΅π’†π’˜ π’π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š =
πŸ“π’™
𝟐
π‘΅π’†π’˜ π’“π’‚π’•π’Šπ’ = πŸ”π’™ ∢
πŸ“π’™
𝟐
= πŸ” ∢
πŸ“
𝟐
= πŸ” Γ— 𝟐 ∢
πŸ“
𝟐
Γ— 𝟐
= 𝟏𝟐 ∢ πŸ“
Problems & Solutions involving Ratio

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Problems & Solutions involving Ratio

  • 2. PROBLEM1 The sides of a rectangle are in the ratio 4:5. i.By what fraction should the shorter side be increased to make it a square? ii.By what fraction should the longer side be decreased to make it a square?
  • 3. SOLUTION π‘Ήπ’‚π’•π’Šπ’ 𝒐𝒇 π’”π’Šπ’…π’†π’” = πŸ’ ∢ πŸ“ ∴ 𝑺𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = πŸ’π’™ π‘³π’‚π’“π’ˆπ’†π’“ π’”π’Šπ’…π’† = πŸ“π’™ i. 𝑳𝒆𝒕 π’Œ 𝒑𝒂𝒓𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† 𝒃𝒆 π’Šπ’π’„π’“π’†π’‚π’”π’†π’…. ∴ π’Œ 𝒑𝒂𝒓𝒕 𝒐𝒇 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = π’Œ Γ— πŸ’π’™ = πŸ’π’Œπ’™ ∴ π‘΅π’†π’˜ 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 π’”π’Šπ’…π’† = πŸ’π’™ + πŸ’π’Œπ’™
  • 4. ∴ πŸ’π’™ + πŸ’π’Œπ’™ ∢ πŸ“π’™ = 𝟏 ∢ 𝟏 πŸ’π’™ + πŸ’π’Œπ’™ πŸ“π’™ = 𝟏 𝟏 πŸ’π’™ + πŸ’π’Œπ’™ = πŸ“π’™ πŸ’π’Œπ’™ = πŸ“π’™ βˆ’ πŸ’π’™ πŸ’π’Œπ’™ = 𝒙 π’Œ = 𝒙 πŸ’π’™ ∴ π’Œ = 𝟏 πŸ’ , 𝒕𝒉𝒆 π’“π’†π’’π’–π’Šπ’“π’†π’… π’‡π’“π’‚π’„π’•π’Šπ’π’.
  • 5. ii. 𝐿𝑒𝑑 π‘˜ π‘π‘Žπ‘Ÿπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒 𝑏𝑒 π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘. ∴ π‘˜ π‘π‘Žπ‘Ÿπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒 = π‘˜ Γ— 5π‘₯ = 5π‘˜π‘₯ ∴ 𝑁𝑒𝑀 π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑠𝑖𝑑𝑒 = 5π‘₯ βˆ’ 5π‘˜π‘₯ 4π‘₯ ∢ 5π‘₯ βˆ’ 5π‘˜π‘₯ = 1 ∢ 1
  • 6. πŸ’π’™ πŸ“π’™ βˆ’ πŸ“π’Œπ’™ = 𝟏 𝟏 ∴ πŸ’π’™ = πŸ“π’™ βˆ’ πŸ“π’Œπ’™ πŸ“π’Œπ’™ = πŸ“π’™ βˆ’ πŸ’π’™ πŸ“π’Œπ’™ = 𝒙 π’Œ = 𝒙 πŸ“π’™ ∴ π’Œ = 𝟏 πŸ“ , π’Šπ’” 𝒕𝒉𝒆 π’“π’†π’’π’–π’Šπ’“π’†π’… π’‡π’“π’‚π’„π’•π’Šπ’π’.
  • 7. PROBLEM 2 Two quantities are in the ratio 3:5. i. If the smaller alone is made four times the original, what would the ratio be? ii.If the smaller is doubled and the larger is halved, what would the ratio be?
  • 8. SOLUTION π‘Ήπ’‚π’•π’Šπ’ 𝒐𝒇 π’’π’–π’‚π’π’•π’Šπ’•π’Šπ’†π’” = πŸ‘ ∢ πŸ“ ∴ π‘Ίπ’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ‘π’™ π‘³π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ“π’™ i. π‘΅π’†π’˜ π’”π’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ’ Γ— πŸ‘π’™ = πŸπŸπ’™ π‘³π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ“π’™ ∴ π‘΅π’†π’˜ π’“π’‚π’•π’Šπ’ = πŸπŸπ’™ ∢ πŸ“π’™ = 𝟏𝟐 ∢ πŸ“
  • 9. ii. π‘΅π’†π’˜ π’”π’Žπ’‚π’π’π’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = 𝟐 Γ— πŸ‘π’™ π‘΅π’†π’˜ π’π’‚π’“π’ˆπ’†π’“ π’’π’–π’‚π’π’•π’Šπ’•π’š = πŸ“π’™ 𝟐 π‘΅π’†π’˜ π’“π’‚π’•π’Šπ’ = πŸ”π’™ ∢ πŸ“π’™ 𝟐 = πŸ” ∢ πŸ“ 𝟐 = πŸ” Γ— 𝟐 ∢ πŸ“ 𝟐 Γ— 𝟐 = 𝟏𝟐 ∢ πŸ“