Question

10 years ago the age of mother is thrice of her daughter. If the some of their present ages is 60, find out their present ages.​

Answer 1

Answer:

$\huge \: \bf \: Given$

• 10 years ago the age of mother is thrice of her daughter
• If the some of their present ages is 60,

$\huge \bf \: To \: Find$

Present age

$\huge \bf \: Solution$

Let ,

Daughter age = x

Mother age = 60 - x

10 years ago

Daughter age = x - 10

Mother age = 60 - 10 - x = 50 - x

ATQ

$\sf \implies \: 50 - x = 3(x - 10)$

$\sf \implies \: 50 - x = 3x - 30$

$\sf \implies \: 50 + 30 = 3x + x$

$\sf \implies \: 80 = 4x$

$\sf \implies \: x \: = \cancel \dfrac{80}{4}$

$\sf \implies \: x \: = 20$

Hence :-

Daughter age = 20 years

Mother's age = x - 20 = 60 - 20 = 40 years

Answer 2

Question and answer -

Let's understand the concept 1st -

★ This question says that 10 years ago the age of mother is thrice (3) of her daughter. If the sum of their present ages is 60, we have to find out their (mother and daughter) present age's.

Given that -

❥ 10 years ago the age of mother is thrice of her daughter

❥ The sum of their present ages is 60.

To find -

❥ Present age of mother and daughter.

Solution -

❥ Present age of mother = 40

❥ Present age of daughter = 20

Assumption -

Now,

❥ Age of mother = a

❥ Age of daughter = 60 - a

10 years ago

❥ Mother age

= 60 - 10 - a

= 50 - a

❥ Daughter age = a - 10

Full solution -

~ From all the above data let's work according to the question now,

◕ 50 - a = 3(a-10)

◕ 50 - a = 3a - 30

◕ 50 + 30 = 3a + a

◕ 80 = 4a

◕ 80/4 = a

◕ 20 = a

◕ a = 20

Means,

• Daughter age = 20 years

• Mother's age = 40 years

= a - 20

= 60 - 20

= 40 years

Knowledge booster -

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cuboid \: = \: \sqrt 3l}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cuboid \: = \: 12 \times Sides}}}$