Tag: maths

Questions Related to maths

A shopkeeper brought two $TV$ sets at $Rs. 10,000$ each. He sold one at a profit of $10\%$ and the another at a loss of $10\%$. Find his total profit or loss.

maths mixture types of ratios ratios in proportion mathematical logic

  1. $20 \%$

  2. $0 \%$

  3. $10 \%$

  4. $5 \%$

Show answer
Correct Option: B
Explanation:

Given that cost of two TV sets is $Rs.10,000$ each


One was sold at a profit of $10\%$ and

the other at a loss of $10\%$

Therefore, the total cost is $20,000+10000\times0.1-10000\times0.1=20,000$

Hence, the total profit or loss percentage is $0$

A year ago, the cost of Maruti and Figo are in the ratio of $3 : 4$. The ratio of present and past year costs of Maruti and Figo are $3 : 2$. Find the ratio of present costs

maths mixture types of ratios ratios in proportion mathematical logic

  1. $\dfrac98$

  2. $\dfrac38$

  3. $\dfrac58$

  4. $\dfrac97$

Show answer
Correct Option: A

One year ago, the ratio between Ram's and Shyam's salaries was $3 : 5$. The ratio of their individual salaries of last year and present year are $2 : 3$ and $4 : 5$ respectively. If their total salaries for the present year is $Rs. 8600$, find the present salary of Ram ?

maths mixture types of ratios ratios in proportion mathematical logic

  1. $3200$

  2. $3600$

  3. $4000$

  4. $4400$

Show answer
Correct Option: B

The ratio of the number of student studying in schools A, B and C is $6:8:7$ respectively if the number of student studying in each of the schools is increased by $20\%,15\%$ and $20\%$ respectively then what will be the new ratio of the number of student in school A, B and C ?

maths mixture types of ratios ratios in proportion mathematical logic

  1. $18:23:21$

  2. $16:28:30$

  3. $20:12:36$

  4. $22:21:12$

Show answer
Correct Option: A
Explanation:
The ratio of the number of students studying in school A, B, C.
$= 6 : 8 : 7$

Suppose number of student's in school
$A=6x$
$B=8x$
$C= 7x$

It will be Increased by
$A= 6x + 6x \times 20$ %

$= 6x+\dfrac {6x\times 20}{100}$

$= 6x+\dfrac {6x}{5}$

$= \dfrac {30x+6x}{5}$

$=\dfrac {36 x}{5}$

$B = 8x +\dfrac {8x \times 15}{100}$

$ =8x +\dfrac {6x}{5}$

$ = \dfrac {46x}{5}$

$C= 7x+\dfrac {7x\times 20}{100}$

$=\dfrac {35x +7x}{5}$

$=\dfrac {42 x}{5}$

New Ratio $= \dfrac {36 x}{5}:\dfrac {46 x}{5}: \dfrac {42 x}{5}$

$ =36:  46 : 42$

$ = 18  : 23  : 21$

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