Revision: Comparing Quantities Maths Secondary School (English Medium) (5 to 8) Class 7 CBSE

A ratio is the relationship between two quantities of the same kind with the same unit, obtained by dividing the first by the second.

Example:
The ratio between 15 kg and 20 kg 
15 kg : 20 kg = `15/20` = `3/4` = 3:4.

Equivalent Ratios: If two ratios have the same value when simplified, then they are called Equivalent Ratios. Thus, 3: 2 is equivalent to 6: 4 or 12: 8.

Four non-zero quantities, a, b, c, and d, are said to be in proportion (or are proportional) if:

a : b = c : d.
The above equation is expressed as a : b :: c : d 

This is read as “a is to b as c is to d.”

The unitary method is a process used to find the value of a single unit from the value of multiple units and then find the value of multiple units from the value of a single unit.

• Cost Price (C.P.): The amount for which an article is bought is called its Cost Price (C.P.).
• Selling Price (S.P.): The price at which a product is sold is known as its selling price (S.P.).
• Profit or Gain: When the S.P. is more than the C.P., then there is a profit or gain.
• Loss: When the S.P. is less than the C.P., then there is a loss.
• Discount: A Discount is the reduction given on the marked price of an article by the seller, usually to attract customers.
• Marked Price (M.P.): Marked Price (also called Tag Price) is the price printed or written on an article by the shopkeeper, which is usually higher than the cost price

Percentage = `"value"/"Total value"` × 100

% (Percentage) = `"Part"/"Whole"` × 100

Percent is the numerator of a fraction with a denominator of 100. 

• Profit = Selling Price - Cost Price, which means CP < SP.
• Loss = Cost Price - Selling Price, which means CP > SP.
• Selling Price = Marked Price – Discount
• Discount% = `"Discount"/"Marked Price"` × 100%

Profit = Selling Price - Cost Price, which means CP < SP.

Profit Percent = `"Profit"/"Cost Price" xx 100`.

Loss = Cost Price - Selling Price, which means CP > SP.

Loss Percent = `"Loss"/"Cost Price" xx 100`.

`x/a = y/b` = k (say)

x = ak, y = bk

L.H.S. = `(x^4 + a^4)/(x^3 + a^3) + (y^4 + b^4)/(y^3 + b^3)`

= `(a^4k^4 + a^4)/(a^3k^3 + a^3) + (b^4k^4 + b^4)/(b^3k^3 + b^3)`

= `(a^4(k^4 + 1))/(a^3(k^3 + 1)) + (b^4(k^4 + 1))/(b^3(k^3 + 1)`

= `(a(k^4 + 1))/(k^3 + 1) + (b(k^4 + 1))/(k^3 + 1)`

= `(a(k^4 + 1) + b(k^4 + 1))/(k^3 + 1)`

= `((k^4 + 1)(a + b))/(k^3 + 1)`

R.H.S. = `((x + y)^4 + (a + b)^4)/((x+ y)^3 + (a + b)^3`

= `((ak + bk)^4 + (a + b)^4)/((ak + bk)^3 + (a + b)^3`

= `(k^4(a + b)^4 + (a - b)^4)/(k^3(a + b)^3(a + b)^3`

= `((a + b)^4(k^4 + 1))/((a + b)^3(k^3 + 1)`

= `((a + b)(k^4 + 1))/(k^3 + 1)`

= `((k^4 + 1)(a + b))/(k^3 + 1)`

∴ L.H.S. = R.H.S.

Hence proved

`x/a = y/b = z/c` = k(say)
x = ak, y = bk, z = ck

L.H.S. = `(3x^3  5y^3 + 4z^3)/(3a^3  5b^3 + 4c^3)`

= `(3a^3k^3 - 5b^3k^3 + 4c^3k^3)/(3a^3 - 5b^3 + 4ac^3)`

= `(k^3(3a^3 - 5b^3 + 4c^3))/(3a^3 - 5b^3 + 4c^3`
= k³
R.H.S. = `((3x - 5y + 4z)/(3a - 5b + 4c))^3`

= `((3ak - 5bk + 4ck)/(3a - 5b + ac))^3`

= `((k(3a - 5b + 4c))/(3a - 5b + 4c))^3`
= (k)³
= k³
∴ L.H.S. = R.H.S.

a, b, c, d are in continued proportion

∴ `a/b = b/c = c/d` = k(say)

∴ c = dk, b = dk², a = bk = dk². k = dk³

L.H.S.

= `((a - b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2`

= `((dk^3 - dk^2)/(dk) + (dk^3 - dk)/(dk^2))^2 - ((d - dk^2)/(dk) + (d - dk)/(dk^2))^2`

= `((dk^2(k - 1))/(dk) + (dk(k^2 - 1))/(dk^2))^2 - ((d(1 - k^2))/(dk) + (d( 1 - k^2))/(dk^2))^2`

= `((k(k - 1) + (k^2 - 1))/k)^2 - ((1 - k^2)/k + (1 - k)/k^2)^2`

= `((k^2(k - 1) + (k^2 - 1))/k)^2 - ((k (1- k^2) + 1 - k)/k^2)^2`

= `((k^3 - 1)^2)/k^2 - (-k^3 + 1)^2/k^4`

= `(k^3 - 1)^2/k^2 - (1 - k^3)^2/k^4`

= `((k^3 - 1)/k^2)^2 ((1 - 1)/k^2)`

= `((k^3 - 1)^2(k^2 - 1))/k^4`

= `((k^3 - 1)^2(k^2 - 1))/k^4`

R.H.S.

= `(a - d)^2(1 / c^2 - 1/b^2)`

= `(dk^3 - d)^2(1 / (d^2k^2) - (1)/(d^2k^4))`

= `d^2(k^3 - 1)^2((k^2 - 1)/(d^2k^4))`

= `((k^3 - 1)^2(k^2 - 1))/k^4`
∴ L.H.S. = R.H.S.

Given: x, y and z are in continued proportion.

∴ `x/y = y/z`

⇒ y² = xz

To prove: `x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2) = 1/x^3 + 1/y^3 + 1/z^3`

Proof: Solving L.H.S.:

`x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2)`

⇒ `(x^3 + y^3 + z^3)/(x^2.y^2.z^2)`

⇒ `(x^3 + y^3 + z^3)/(x^2.xz.z^2)`

⇒ `(x^3 + y^3 + z^3)/(x^3.z^3)`

⇒ `x^3/(x^3.z^3) + y^3/(x^3.z^3) + z^3/(x^3.z^3)`

⇒ `1/z^3 + y^3/(xz)^3 + 1/x^3`

⇒ `1/z^3 + y^3/(y^2)^3 + 1/x^3`

⇒ `1/z^3 + y^3/y^6 + 1/x^3`

⇒ `1/z^3 + 1/y^3 + 1/x^3`

Since L.H.S. = R.H.S.

Hence proved.

`a/c = c/d = e/f` = k(say)

∴ a = bk, c = dk, e =fk

L.H.S.

= `(a^3 + c^3)^2/(b^3 + d^3)^2`

= `(b^3k^3 + d^3k^3)^2/(b^3 + d^3)^2`

= `[k^3(b^3 + d^3)]^2/(b^3 + a^3)^2`

= `(k^6(b^3 + d^3)^2)/(b^3 + d^3)^2`

= k⁶

R.H.S. = `e^6/f^6`

= `f^6k^6/f^6`

= k⁶

∴ L.H.S. = R.H.S.

a, b, c, d are in continued proportion

∴ `a/b = b/c = c/d` = k(say) 

∴ c = dk, b = ck = dk. k = dk²,

a = bk = dk². k = dk³ 

L.H.S. = (a + d)(b + c) – (a + c)(b + d)

= (dk³ + d) (dk² + dk) – (dk³ + dk) (dk² + d)

= d(k³ + 1) dk(k + 1) –  dk (k² + 1) d(k² + 1)

= d²k(k + 1) (k³ + 1) – d²k (k² + 1) (k² + 1)

= d²k[k⁴ + k³ + k + 1 – k⁴ - 2k² - 1]

= d²k[k³ – 2k² + k]

= d²k²[k² – 2k + 1]

= d²k²(k – 1)²

R.H.S. = (b – c)²

= (dk² – dk)²

= d²k²(k – 1)²

∴ L.H.S. = R.H.S.

Hence proved.

a + c = mb and `1/b + 1/d = m/c`

a + c = mb   ...(1)

`1/b + 1/d = m/c`   ...(2)

Step 1: Simplify the second condition

`1/b + 1/d = m/c`

Take LCM of b and d:

`(d + b)/(bd) = m/c`

c(d + b) = mbd

cd + cb = mbd   ...(3)

Step 2: Use the first condition

a + c = mb

Multiply both sides by d:

d(a + c) = mbd

ad + cd = mbd   ...(4)

Step 3: Compare equations (3) and (4)

cd + cb = mbd

ad + cd = mbd

ad + cd = cd + cb

Subtract cdcdcd from both sides:

ad = cb

Step 4: Convert to ratio form

ad = bc

Divide both sides by bd:

`a/b = c/d`

Thus,

a : b = c : d

Hence, a, b, c and d are proportional.

a, b, c, d are in continued proportion

∴ `a/b = b/c = c/d` = k(say)

∴ c = dk, b = ck = dk², a = bk = dk³ 

L.H.S.

= `(a^3 + b^3 + c^3)/(b^3 + c^3 + d^3)`

= `((dk^3)^3 + (dk^2)^3 + (dk)^3)/((dk^2)^3 + (dk)^3 + d^3)`

= `(d^3k^9 + d^3k^6 + d^3k^3)/(d^3k^6 + d^3k^3 + d^3)`

= `(d^3k^3(k^6 + k^3 + 1))/(d^3(k^6 + k^3 + 1)`

= k³

R.H.S.

= `a/d`

= `(dk^3)/d`

= k³

∴ L.H.S. = R.H.S.

x, y, z are in continued proportion

Let `x/y = y/z = k`

Then y = kz

x = yk

= kz × k

= k²z

Now L.H.S.

= `(x + y)^2/(y + z)^2`

= `(k^2 z + kz)^2/(kz + z)^2`

= `{kz(k + 1)}^2/{z(k + 1)}^2`

= `(k^2z^2(k + 1)^2)/(z^2(k + 1)^2)`

= k²

R.H.S. = `x/z`

= `(k^2z)/z`

= k²

∴ L.H.S. = R.H.S.