Tag: maths

Questions Related to maths

The fourth proportional of 5, 6 and 7 correct to two places of decimal is 8.40

  1. True

  2. False


Correct Option: A
Explanation:

In $ a:b::c:d;   d $ is the fourth proportional.





For, $ 5:6 :: 7 :d $, product of extremes $ = $ product of means





$ 5 \times d = 6 \times 7 $


$ d = \frac {42}{5} = 8.40 $

The third proportional to $(x^2\, -\, y^2)$ and $(x - y)$ is

  1. $(x+y)$

  2. $\displaystyle \frac {x + y}{x - y}$

  3. $\displaystyle \frac {x - y}{x + y}$

  4. $(x^2\, -\, y^2)$


Correct Option: C
Explanation:

Let third proportional is x
$\left ( x^{2}-y^{2} \right ):\left ( x-y \right )=\left ( x-y \right ):x$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left ( x^{2}-y^{2} \right )}$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left (x+y  \right )\left (x-y  \right )}$
$\Rightarrow x=\frac{\left ( x-y \right )}{\left (x+y  \right )}$
 

Find the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$.

  1. $\displaystyle \frac{x+y}{x-y}$

  2. $x-y$

  3. $\displaystyle \frac{x-y}{x+y}$

  4. 1


Correct Option: A
Explanation:

Let the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$ be A.

Then,
$\displaystyle \left ( x^{2}-y^{2} \right ):\left ( x+y \right )::\left ( x+y \right ):A$
$\displaystyle \Rightarrow (x^{2}-y^{2})\times A=(x+y)^{2}$

$\Rightarrow A=\cfrac{(x+y)^{2}}{x^{2}-y^{2}}$
$\Rightarrow A=\cfrac{(x+y)^{2}}{(x+y)(x-y)}=\cfrac{x+y}{x-y}$

By mistake instead of dividing Rs. $117$ among $A,B$ and $C$ into the ratio $\displaystyle \frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ , it was divided in the ratio of $2:3:4$ . Who gains the most and by how much?

  1. $A, Rs. 28$

  2. $B, Rs. 3$

  3. $C, Rs. 20$

  4. $C, Rs. 25$


Correct Option: D
Explanation:

To convert $\displaystyle \frac{1}{2}:\frac{1}{3}:\frac{1}{4}$ into normal ratio, multiply each fraction by the LCM$(2,3,4) = 12$

$\displaystyle= 12\times \frac{1}{2}:12\times\frac{1}{3}:12\times\frac{1}{4}$
= $\displaystyle 6:4:3 $
Thus, A would have got $\displaystyle \frac{6}{6+4+3} \times 117 = 54$

B would have got $\displaystyle \frac{4}{6+4+3} \times 117 = 36$

And C would have got $\displaystyle \frac{3}{6+4+3} \times 117 = 27$

Instead Rs. $117$ have been divided in the ratio of $2:3:4$

A gets $\displaystyle \frac {2}{2+3+4} \times 117 = 26$

B gets $\displaystyle \frac {3}{2+3+4} \times 117 = 39$

C gets $\displaystyle \frac {4}{2+3+4} \times 117 = 52$

Thus, by changing the ratio,
A gained $26 - 54$ = -$28$
B gained $39 - 36$ = $3$
And C gained $52 - 27 = 25$
Thus, C's gain of $25$ is the most.

Divide Rs 7053 into three parts so that the amount after 2, 3 and 4 years respectively may be equal, the rates of interest being 4% per annum. 

  1. Rs 2500, Rs 3500, Rs 1053

  2. Rs 2436, Rs 2349, Rs 2268

  3. Rs 2568, Rs 3200, Rs1285

  4. Rs 2360, Rs 2289, Rs 2404


Correct Option: B
Explanation:

Take the parts as x, y & z.

Then x+y+z=7053.
Obtain the the respective amounts for x,y & z, which are equal, for the 
given years.
Now solve for x, y & z.

If 7676 is divided into four parts proportional to $7, 5, 3, 4$, then the smallest part is:

  1. $12$

  2. $15$

  3. $16$

  4. $19$


Correct Option: A
Explanation:

Let the parts be $7x,5x,3x$ and $4x.$ 

$\therefore 7x+5x+3x+4x=76$ 
$\Rightarrow 19x=76$
$\Rightarrow x =4$ 
Hence, the smallest part $=3x=3 \times 4=12$

The third proportional to 8 and 10 correct to two places of decimal is 12.50

  1. True

  2. False


Correct Option: A
Explanation:

If a : b :: b : c, then we say that a, b, c are in continued

proportion, and c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ or $ c = \frac{{b}^{2}}{a} $





So, for $ 8, 10 $, the third proportional is $ c = \frac{{b}^{2}}{a} = \frac {{10}^{2}}{8} = 12.50 $

If 150 is the third proportional to 6 and x; find the value of x.

  1. 30

  2. 60

  3. 63

  4. 34


Correct Option: A
Explanation:

If a : b :: b : c, then we say that a, b, c are in continued proportion, and

c is the third proportional of a and b.





Here, $ {b}^{2} = ac $ 





Given  the third proportional of $ 6, x $ is 150 

$ => {x}^{2} = 6 \times 150 = 6 \times 6 \times 25 $

$ => x = \sqrt { 6 \times 6 \times 25 } $

$ x = 6 \times 5 = 30 $


A leak in the bottom of a tank can empty the full tank in $6$ hours. An inlet pipe fills water at the rate of $4$ litres per minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in $8$ hours, then the capacity of the tank is:

  1. $5260$ L

  2. $5760$ L

  3. $5846$ L

  4. $6970$ L


Correct Option: B
Explanation:

Work done by the inlet in 1 hour $= \dfrac{1}{6}.\dfrac{1}{4}=\dfrac{1}{24}$

Work done by inlet in 1 min $=\dfrac{1}{24}\times \dfrac{1}{60}$
                                           
                                              $=\dfrac{1}{1440}$

Volume of$ \dfrac{1}{1440}$ part$ = 4$ litres
Volume of whole = $(1440 \times 4)$litres=$5760$  litres

Find the fourth proportion to $2,\,3,\,6$.

  1. $8$

  2. $9$

  3. $4$

  4. $6$


Correct Option: B
Explanation:

$2:3 : :6:x$
$\Longrightarrow \dfrac{2}{3}=\dfrac{6}{x}$

$\Longrightarrow x=\dfrac{18}{2}$
$\Longrightarrow\;x=9$