Tag: mathematical logic

Questions Related to mathematical logic

Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases $p\%$ then $y$ decreases by  

maths mixture types of ratios ratios in proportion mathematical logic

  1. $p\%$

  2. $\dfrac{p}{1+p}\%$

  3. $\dfrac{100}{p}\%$

  4. $\dfrac{100p}{100+p}\%$

Show answer
Correct Option: D
Explanation:

It is given that $x$ and $y$ are inversely proportional


$x\propto \cfrac{1}{y}$


$\Rightarrow$ $xy=$ constant

${x} _{1}{y} _{1}={x} _{2}{y} _{2}...(1)$

$x$ has $p$ percent increase

${x} _{2}={x} _{1}\left(1+\cfrac{p}{100}\right)$

${x} _{1}{y} _{1}={x} _{2}{y} _{2}$

${x} _{1}{y} _{1}={x} _{1}\left(1+\cfrac{p}{100}\right){y} _{2}$

${y} _{2}=\cfrac{100{y} _{1}}{100+p}$

${y} _{2}=\left(\cfrac{100+p}{100+p}\right){y} _{1}-\cfrac{p{y} _{1}}{100+p}$

${y} _{2}-{y} _{1}=\cfrac{p{y} _{1}}{100+p}$

$\cfrac{{y} _{2}-{y} _{1}}{{y} _{1}}=-\cfrac{p}{100+p}$

$\Rightarrow$ $\cfrac{{y} _{1}-{y} _{2}}{{y} _{1}}=\cfrac{p}{100+p}$

percentage decrease $=\cfrac{{y} _{1}-{y} _{2}}{{y} _{1}}\times 100=\cfrac{100p}{100+p}$%

A jar contained a mixture of two liquids A and B in the ratio 7 : 2. When 18 litres of mixture was taken out and 18 litres of liquid B was poured into the jar. This ratio became 2 : 3. The quantity of liquid A contained in the jar initially was: 

maths mixture types of ratios ratios in proportion mathematical logic

  1. $\frac { 450 } { 17 }$ litres

  2. $\frac { 490 } { 19 }$ litres

  3. $\frac { 490 } { 17 }$ litres

  4. $\frac { 450 } { 19 }$ litres

Show answer
Correct Option: A

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 $AB=$

maths mixture types of ratios ratios in proportion mathematical logic

  1. $\sqrt {2}\ units$

  2. $2\ units$

  3. $\sqrt {3}\ units$

  4. $3\ units$

Show answer
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:

maths mixture types of ratios ratios in proportion mathematical logic

  1. 40

  2. 50

  3. 60

  4. 35

Show answer
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?

maths mixture types of ratios ratios in proportion mathematical logic

  1. $\;5\colon32$

  2. $\;5\colon37$

  3. $\;37\colon32$

  4. $\;32\colon37$

Show answer
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?

maths mixture types of ratios ratios in proportion mathematical logic

  1. 40 years

  2. 45 years

  3. 30 years

  4. 35 years

Show answer
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

maths mixture types of ratios ratios in proportion mathematical logic

  1. 5 : 2

  2. 2 : 5

  3. 3 : 1

  4. 1 : 3

Show answer
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

maths mixture types of ratios ratios in proportion mathematical logic

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

  2. 1

  3. 2

  4. not determinable

Show answer
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?

maths mixture types of ratios ratios in proportion mathematical logic

  1. $56 : 99 : 69$

  2. $69 : 56 : 99$

  3. $99 : 56 : 69$

  4. $99 : 69 : 56$

Show answer
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 $16:21.$

maths mixture types of ratios ratios in proportion mathematical logic

  1. $2$

  2. $3$

  3. $4$

  4. $6$

Show answer
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.