Tag: types of ratios

Questions Related to types of ratios

If a line with direction ratio $2:2:1$ intersects the line $\dfrac {x-7}{3}=\dfrac {y-5}{2}=\dfrac {z-3}{1}$ and $\dfrac {x-1}{2}=\dfrac {y+1}{4}=\dfrac {z+1}{3}$ at $A$ and $B$ then $AB=$

  1. $\sqrt {2}\ units$

  2. $2\ units$

  3. $\sqrt {3}\ units$

  4. $3\ units$


Correct Option: D
Explanation:

$\begin{array}{l} \frac { { x-7 } }{ 3 } =\frac { { y-5 } }{ 2 } =\frac { { z-3 } }{ 1 } \, \, \, \, \, \, \, \, Let\, \, A=\left( { 3{ r _{ 1 } }+7,\, \, 2{ r _{ 1 } }+5,\, \, { r _{ 1 } }+3 } \right)  \ \frac { { x-1 } }{ 2 } =\frac { { y+1 } }{ 4 } =\frac { { z+1 } }{ 3 } \, \, \, \, \, and,\, Let\, B=\left( { 2{ r _{ 2 } }+1,\, \, 4{ r _{ 2 } }-1,\, \, 3{ r _{ 2 } }-1 } \right)  \ Direction\, ratios\, of\, AB=\left( { 2{ r _{ 2 } }-3{ r _{ 1 } }-6,\, \, \, 4{ r _{ 2 } }-2{ r _{ 1 } }-6,\, \, \, 3{ r _{ 2 } }-{ r _{ 1 } }-4 } \right)  \ \frac { { 2{ r _{ 2 } }-3{ r _{ 1 } }-6 } }{ 2 } =\frac { { 4{ r _{ 2 } }-2{ r _{ 1 } }-6 } }{ 2 } =\frac { { 3{ r _{ 2 } }-{ r _{ 1 } }-4 } }{ 1 }  \ { r _{ 1 } }+2{ r _{ 2 } }=0\, \, \, \, \therefore { r _{ 1 } }=-2 \ and, \ 4{ r _{ 1 } }-2{ r _{ 1 } }-6=6{ r _{ 2 } }-2{ r _{ 1 } }-8 \ 2=2{ r _{ 2 } } \ \therefore { r _{ 2 } }=1 \ A\equiv \left( { 1,\, 1,\, 1 } \right) \, \, \, \, \, B=\left( { 3,\, 3,\, 2 } \right)  \ AB=\sqrt { 4+4+1 }  \ =3 \ Hence,\, Option\, D\, is\, the\, correct\, answer. \end{array}$

The L.C.M. of two numbers is 48. The numbers are in the ratio 2:3. Then sum of the number is:

  1. 40

  2. 50

  3. 60

  4. 35


Correct Option: A
Explanation:

Let the numbers be $2x$ and $3x $

Then their L.C.M. = $6x$ So $6x = 48$ or $x = 8$ 
The numbers are $16$ and $24$  Hence required sum $= (16 + 24) = 40$

The sides of a triangle are $10$ cm,$10$ cm and $12$ cm. If each of the two equal sides is increased in the ratio $5\colon4$ but the third side remains unchanged, in what ratio has its perimeter been increased?

  1. $\;5\colon32$

  2. $\;5\colon37$

  3. $\;37\colon32$

  4. $\;32\colon37$


Correct Option: C
Explanation:

Given, sides of the triangle $=10cm, 10cm, 12cm$


$\therefore $ perimeter of the given triangle $=10+10+12=32  cm$

As the two equal sides each $10 cm$ increased in the ratio $5:4$

So, the corresponding sides of the new triangle$=10\times \displaystyle \frac{5}{4}=12.5   cm$ 

Hence the 3 sides of the new triangles are $12.5 cm,12.5 cm,12 cm$.

Then the perimeter of the new triangle $=12.5+12.5+12=37   cm$

$\therefore$The ratio in which the perimeter of the new triangle is increased

$=\displaystyle \frac{perimeter \ of \   the \ new \  triangle}{perimeter \  of given \  triangle} =\frac{37}{32}$

Hence the ratio $=37:32$

Present ages of Amit and his father are in the ratio 2:5 respectively. Four years hence the ratio of their ages becomes 5:11 respectively. What was the father's age five years ago?

  1. 40 years

  2. 45 years

  3. 30 years

  4. 35 years


Correct Option: D
Explanation:

 Let the present ages of Amit and his father be 2x and 5x years respectively 
Four years hence 
Amit's Age = ( 2x+ 4) years
Father's age = (5x + 4) years
Given $\displaystyle \frac{2x+4}{5x+4}=\frac{5}{11}\Rightarrow (2x+4)=5(5x+4)$


$\displaystyle \Rightarrow 22x+44=25x+20\Rightarrow 3x=24\Rightarrow x=8$

$\displaystyle \therefore $ Father's present age = ( 5 x 8) years = 40 years Father's age 5 years ago = ( 40 - 5)= 35 years

The difference of the squares is of two numbers is 80% of the sum of their squares The ratio of the larger number to the smaller number is

  1. 5 : 2

  2. 2 : 5

  3. 3 : 1

  4. 1 : 3


Correct Option: C
Explanation:

Let the two numbers be x and y Then $\displaystyle x^{2}-y^{2}=80\%$ of $\displaystyle (x^{2}+y^{2})$
$\displaystyle \Rightarrow x^{2}-y^{2}=\frac{4}{5}(x^{2}+y^{2})\Rightarrow x^{2}-\frac{4}{5}x^{2}=\frac{4}{5}y^{2}+y^{2}$
$\displaystyle \Rightarrow \frac{1}{5}x^{2}=\frac{9}{5}y^{2}\Rightarrow \frac{x^{2}}{y^{2}}=\frac{9}{1}\Rightarrow \frac{x}{y}=\frac{3}{1}\Rightarrow x:y=3:1$

If a, b, c, d are positive real number such that $\frac {a}{3}=\frac {a+b}{4}=\frac {a+b+c}{5}=\frac {a+b+c+d}{6}$, then $\frac {a}{b+2c+3d}$ is

  1. $\frac {1}{2}$

  2. 1

  3. 2

  4. not determinable


Correct Option: A
Explanation:

$a =3k$
$b=k$
$c= 5k-4k =k$
$d =6k-5k =k$
$\frac {a}{b+2c+3d}=\frac {3k}{k+2k+3k}=\frac {1}{2}$

The incomes of A, B and C are in the ratio 7 : 9 : 12 and their spendings are in the ratio 8 : 9 : 15. If A saves $\displaystyle \left ( 1/4 \right )^{th}$ of his income then?

  1. $56 : 99 : 69$

  2. $69 : 56 : 99$

  3. $99 : 56 : 69$

  4. $99 : 69 : 56$


Correct Option: A
Explanation:

Solution:

Let income of $A=7x$
Income of $B=9x$
Income of $C=12x$
and  Spendings of $A=8y$
Spendings of $B=9y$
Spendings of $C=15y$
Income=Savings + Expenditures
Savings of $A=\cfrac14\times 7x=\cfrac{7x}4$
or, $7x=\cfrac{7x}{4}+8y$
or, $21x=32y$
or, $x=\cfrac{32}{21}y$
Now,
Income of $A=7x=7\times \cfrac{32}{21}y=\cfrac{32}{3}y$
Income of $B=9x=9\times \cfrac{32}{21}y=\cfrac{96}{7}y$
Income of $C=7x=12\times \cfrac{32}{21}y=\cfrac{128}{7}y$
Now, 
Savings of $A=7x-8y=\cfrac{32}3y-8y=\cfrac83y$
Savings of $B=9x-9y=\cfrac{96}7y-9y=\cfrac{33}7y$
Savings of $C=12x-15y=\cfrac{128}7y-15y=\cfrac{23}7y$
So, Savings of $A,B$ and $C$ in ratio
$\cfrac83:\cfrac{33}{7}:\cfrac{23}{7}::56:99:69$
Hence, A is the correct option.

The least whole number which when subtracted from both the terms of the ratio  $6:7$  gives a ratio less than $16:21.$

  1. $2$

  2. $3$

  3. $4$

  4. $6$


Correct Option: B
Explanation:

Let the whole number is X.
Now, according to question,
(6-X) / (7-X) < 16/21
21 *(6-X) < 16 *(7-X)
126 - 21X < 112 - 16X
126 - 112 < -16X + 21X
14 < 5X
5X > 14
X > 2.8
So, Least such whole number would be 3.

The length of the ribbon was originally $30cm$. It was reduced in the ratio $5:3$. What is its length now?

  1. $15$

  2. $18$

  3. $20$

  4. $25$


Correct Option: B
Explanation:

Length of ribbon originally $=30cm$
Let the original length be $5x$ and reduced length be $3x$.
But $5x=30cm$
$\Longrightarrow x=\dfrac{30}{5}cm=6cm$
Therefore, reduced length $=3\times6cm=18cm$

State whether true or false:
The following operation will increase the value of the original fraction:
Multiply a positive proper fraction by $\cfrac{3}{8}$.

  1. True

  2. False


Correct Option: B
Explanation:

Decrease: Multiplying a  proper fraction by a value less than 1 (0 < x < 1) decreases the number.

So here $3/8 = 0.375 < 1$ So, the value of original fraction decreases.