MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 4

The above diagram shows a number line.
The above number line represents the solution for

0%
$-3\leq x$ and $x> 4$
0%
$-3< x< 4$
0%
$-3\leq x< 4$
0%
$-3< x$ and $x\leq 4$

Explanation

From the given diagram, the number line is representing the value

$\ge -3$ and

$< 4$ .

Hence, C will be correct answer.

Solve the system of inequalities:

$\displaystyle \frac{x+7}{x-8}> 2, \frac{2x+1}{7x-1} > 5$

0%
$(-\dfrac{16}{7}, \dfrac{5}{7})$
0%
$(-\dfrac{7}{8}, \dfrac{2}{7})$
0%
(2,5)
0%
No solution

Explanation

$\frac{x+7}{x-8}> 2, \frac{2x+1}{7x-1} > 5$
Solving equation (1)

$x + 7 > 2x - 16$ and

$x - 8 > 0$ for

$x > 0$

$x < 23$ and

$x > 8$

$\therefore, x \in (8,23)$
Solving equation 2

$\frac{2x+1}{7x-1} > 5$

$2x +1 > 35x - 5$ and

$7x -1 > 0$ for

$x > 0$

$x < \frac{6}{33}$ and

$x > \frac{1}{7}$

$\therefore, x \in (1/7,6/33)$
From both the solutions, there is no solution for x to satisfy both the equations.

The solution set for:

$\dfrac{\left | x \right |-1}{\left | x \right |-2}, x\neq \pm 2$

0%
$(-2, 2)$
0%
$(-1, 2)\cup (3, \infty )$
0%
$(-\infty , 2)\cup (2, \infty )$
0%
$x\:\epsilon\:(-\infty,\infty)-\big\{-2,2\big\}$

Explanation

let

$f(x)=\frac{|x|-1}{|x|-2}$
Therefore the domain for the function implies

$|x|-2\neq2$

$x\neq\pm2$
Therefore

$x\:\epsilon\:(-\infty,\infty)-\big\{-2,2\big\}$

The solution set for:

$\displaystyle \left | 3x-2 \right |\leq \frac{1}{2}$ is

0%
$\displaystyle \left [\frac{1}{2},\frac{5}{6} \right ]$
0%
$\displaystyle \left [\frac{2}{3},\frac{1}{2} \right ]$
0%
$\displaystyle \left [\frac{5}{6},\frac{1}{2} \right ]$
0%
$\displaystyle \left [\frac{1}{2},\frac{2}{3} \right ]$

Explanation

$| 3x-2|\leq \frac{1}{2}$

$3x - 2 \leq \frac{1}{2}$ or

$3x -2 \geq \frac{-1}{2}$

$x \leq \frac{5}{6}$ or

$x \geq \frac{1}{2}$

$\therefore \frac{1}{2} \leq x \leq \frac{5}{6}$

State True or False.

Check whether the following half planes contain the corresponding point:

7x≥2y+x,7x≥2y+x, point (1, 0)

0%
True
0%
False

Explanation

To check whether the half plane contains

$(1,0)$ , substitute,

$(1,0)$ in the given inequality

$7x \ge 2y + x$

$7 \ge 0 + 1$

$7 \ge 1$ which is true.

Hence, given half plane contains

$(1,0)$

State True or False.

Check whether the following half planes contain origin (0,0) or not:

$2x+5y\geq x+12$

0%
True
0%
False

Explanation

To check whether the half plane contains the origin, substitute,

$(0,0)$ in the given inequality

$2x +5y \ge x + 12$

$0 + 0 \ge 0 + 12$

$0 \ge 12$ which is not true.

Hence, given half plane does not contain origin.

The solution set for:

$\left |\dfrac{2(3-x)}{5} \right |< \dfrac{3}{5}$

0%
$(\dfrac{1}{4}, \dfrac{3}{4})$
0%
$(\dfrac{3}{2}, \dfrac{9}{2})$
0%
$(\dfrac{1}{2}, \dfrac{3}{2})$
0%
$(\dfrac{3}{2}, \dfrac{5}{2})$

Explanation

$|\frac{2(3-x)}{5}|<\frac{3}{5}$

$\frac{-3}{5}<\frac{2(3-x)}{5}<\frac{3}{5}$

$-3<6-2x<3$
By adding

$(-6)$ we get

$-9<-2x<-3$
Multiplying by -ve sign or -1 (inequality sign gets reversed)

$9>2x>3$

$\frac{9}{2}>x>\frac{3}{2}$ or

$\frac{3}{2}<x<\frac{9}{2}$
Hence

$x\:\epsilon\:(\frac{3}{2},\frac{9}{2})$
The answer is B

Solve:

$\left | 2x-5 \right |< 1$

0%
$x\in (4, 5)$
0%
$x\in (3, 5)$
0%
$x\in (2, 3)$
0%
$x\in (1, 2.5)$

Explanation

Let $2x -5 = y$

$=> 2x - 5 < 1$

$=> x < \dfrac {6}{2} = 3$ -- (1)

And when $y < 0$ , we have $\left | y \right | = - y$

So, $-y-1 < 0$

$y > -1$

$=> 2x -5 > -1$

$=> x > 2$ -- (2)
From $(1), (2) ; x \in (2, 3)$

$\cos x-y^{2}-\sqrt{y-x^{2}-1} \geq 0,$ then

0%
$y \geq 1$
0%
$x \in R$
0%
$y=1$
0%
$x=0$

Explanation

Given,

$\cos x-y^{2}-\sqrt{y-x^{2}-1} \geq 0$ ....(1)

Now,

$\sqrt{y-x^{2}-1}$ is defined when

$y-x^{2}-1 \geq 0$ or

$y \geq x^{2}+1 .$

So, minimum value of

$y$ is

$1$

From (1)

$\cos x-y^{2} \geq \sqrt{y-x^{2}-1}$

where

$\cos x-y^{2} \leq 0$ [as when

$\cos x$ is maximum

$(=1)$ and

$y^{2}$ is minimum

$(=1), \text { so } \cos x-y^{2}$ is maximum.

$\quad \cos x-y^{2}=0$

Also,

$\sqrt{y-x^{2}-1} \geq 0$

Hence,

$\quad \cos x-y^{2}=\sqrt{y-x^{2}-1}=0$

$\Rightarrow \quad y=1 \text { and } \cos x=1, y=x^{2}+1$

$\Rightarrow \quad x=0, y=1$

The range of the values of

$x$ which satisfies the inequation

$\left(\dfrac{3x+2}{2x+3}\right )\geq 4$ is

0%
$x\in \left [ -2, -\dfrac{3}{2} \right )$
0%
$x\in \left ( -2, -\dfrac{3}{2} \right )$
0%
$x\in \left [ -\dfrac{2}{3}, -\dfrac{3}{2} \right ]$
0%
$x\in \left [ -2, \dfrac{3}{2} \right ]$

Explanation

$\dfrac {3x+2}{2x+3} \ge 4$

$=> \dfrac {3x+2}{2x+3} - 4 \ge 0$

$=> \dfrac {3x+2 -8x-12 }{2x + 3} \ge 0$

$\dfrac {-5x-10}{2x + 3 } \ge 0$

$\dfrac {5x + 10}{2x + 3 } \le 0$

Now, $5x+ 10 \ge 0$ and $2x + 3 < 0$

=> $x \ge -2$ and $x < -\dfrac {3}{2}$

Thus, $x\in \left [ -2, -\dfrac{3}{2} \right )$

Form a new inequality from the given inequality by performing the operation stated in brackets.
(a)

$4 >2\left [ \times 5 \right ]$
(b)

$-4x\geq 32\left [ \div (-4) \right ]$

0%

$(a)\quad 20 > 10$
$(b)\quad x\le -8\quad$
0%

$(a)\quad 20 < 10$
$(b)\quad x > 8\quad$
0%
$(a)\quad 20\le 10$
$(b)\quad x\ge -8\quad$
0%
$(a)\quad 20\ge 10$
$(b)\quad x\ge -8\quad$

Explanation

$\quad 4>2\left[ \times 5 \right] \\ \Longrightarrow 20>10\quad (by\quad a)\quad \& \\ -4x>32\left[ \div -4 \right] \\ \Longrightarrow x<-8\quad (by\quad b).\quad$

ans- Option A.

$\left |\dfrac{8-n}{3} \right |\geq 12$

0%
$28 >n >-28$
0%
$-28\leq n\leq 28$
0%
$n\leq -28$
0%
$n\geq 44$

Explanation

$\left |\frac{8-n}{3} \right |\geq 12$

$|8-n| \geq 36$

$8 - n \geq 36$ or

$8 - n \leq -36$

$n \leq 28$ or

$n \geq 44$

Which region is described by the shade in the graph given above?

0%
$2x+3y=3$
0%
$2x+3y< 3$
0%
$2x+3y> 3$
0%
$-2x+3y< 3$

Explanation

The line represents

$2x+3y=3$
Writing it in intercept form, we get

$\dfrac{x}{3/2}+\dfrac{y}{1}=1$
Hence,

$x$ intercept is

$1.5$ units and

$y$ intercept is

$1$ unit.
Consider any arbitrary point in the shaded region such that it does not lie on the line

$2x+3y=3$ .
Let us consider

$(4,4)$ :

$(4,4)$ clearly does not lie on the line but it does lie in the shaded region.

$2(4)+3(4)=20$ and

$20>3$
Therefore in the shaded region

$2x+3y>3$
Since the line on the graph is not dotted, therefore the shaded region includes

$2x+3y=3$ or

$2x+3y>3$
Hence,

$2x+3y\geq3$

Identify the region described by the shaded part in the graph above.

0%
$y=4x-6$
0%
$y\neq 4x-6$
0%
$y< 4x-6$
0%
$y> 4x-6$

Explanation

The line represents the linear equation

$y=4x-6$
Any other region on the given figure represents

$y\neq4x-6$ ..

$(i)$
Now writing the given equation in intercept form, we get

$4x-y=6$

$\Rightarrow \dfrac{x}{\frac{6}{4}}-\dfrac{y}{6}=1$
Therefore the line cuts the

$x$ -axis at

$x=\dfrac{6}{4}$ and

$y$ -axis at

$-6$ .
Now consider an arbitrary point in the shaded region

$(x,y)$ which does not lie on the line to check the inequality.
Consider point

$(10,10)$

$4x-6=40-6=34$ and

$y=-6$ .

Hence,

$y<4x-6$
Similarly, we can check the inequality for the shaded region by substituting points in that region, which do not lie on the line, and by comparing

$y$ and

$4x-6$
Hence, the correct answers are options B and C.

For any positive real number 'a', the following results hold.
(i)

$\left | x \right |\leq a\Rightarrow -a\leq x\leq a$
(ii)

$\left | x \right |\geq a\Rightarrow x\geq a$ or

$x\leq a$
Using the above indentities, evaluate the following:

$\left |\dfrac{4m-8}{2}\ \right |\leq 6$

0%
$1\leq m\leq 5$
0%
$-1\leq m\leq 5$
0%
$1\leq m\leq -5$
0%
$-1\leq m\leq -5$

Explanation

$|\frac{4m-8}{2}|\leq6$

$-6\leq\frac{4m-8}{2}\leq6$

$-6\leq2m-4\leq6$
Adding 4 we get

$-2\leq2m\leq10$

$-1\leq m\leq5$
Hence answer is B

Given:

$A:B = 3:4$ and

$B:C = 6:7$ , so

$A: C is 9: 14$
State true or false.

0%
True
0%
False

Explanation

In the given ratios "B" is the

common term, and the values of B in both ratios are not equal. To make them

equal, find the L.C.M. of values corresponding to B i.e., $4$ and $6$ .

L.C.M. of $4$ and $6 = 12$

Therefore, an equivalent ratio of $A:B = 3:4 = 3 \times 3:4 \times 3 = 9:12$

Similarly, an equivalent ratio of $B:C = 6:7 = 6 \times 2:7 \times 2 = 12:14$

Therefore, $A:C = 9:14$

$A : B = 3 : 4$ and

$B : C = 6 : 7$ , then

$A : B : C$ is

$9 : 12 : 14$

State true or false.

0%
True
0%
False

Explanation

Similarly, an equivalent ratio of $B:C = 6:7 = 6 \times 2:7 \times 2 = 12:14$

Therefore, $A: B:C = 9:12:14$

A boat can go across a lake and return in time

${ t }_{ 0 }$ at a speed

$u$ . On a rough day, there is uniform current at speed

$v$ to help the onward journey and impede the return journey. If the time taken to go across and return on the rough day be

$t$ , then

$\dfrac{t}{{ t }_{ 0 }}$ is equal to

0%
$1-\cfrac { { v }^{ 2 } }{ { u }^{ 2 } }$
0%
$\cfrac { 1 }{ 1-\left( \cfrac { { v }^{ 2 } }{ { u }^{ 2 } } \right) }$
0%
$1+\cfrac { { v }^{ 2 } }{ { u }^{ 2 } }$
0%
$\cfrac { 1 }{ 1+\left( \cfrac { { v }^{ 2 } }{ { u }^{ 2 } } \right) }$

Explanation

Let

$L$ be the length of lake so, for normal day time

${ t }_{ o }=\dfrac { 2L }{ u }$

and for rough day time

$\displaystyle t=\dfrac { L }{ u+v } +\dfrac { L }{ u-v } =\dfrac { 2Lu }{ { u }^{ 2 }-{ v }^{ 2 } }$

therefore

$\displaystyle \dfrac { t }{ { t }_{ o } } =\dfrac { { u }^{ 2 } }{ { u }^{ 2 }-{ v }^{ 2 } } =\dfrac { 1 }{ 1-{ \left( \dfrac { { v }^{ 2 } }{ { u }^{ 2 } }\right)}}$

A boat takes

$2$ hours to travel

$8\ km$ and back in still water lake. With water velocity of

$4\ { km }/{ h }$ , the time taken for going upstream of

$8\ km$ and coming back is

0%
$160\ minutes$
0%
$80\ minutes$
0%
$100\ minutes$
0%
$120\ minutes$

Explanation

Velocity of boat

$=\displaystyle\dfrac { 8+8 }{ 2 } =8\ { km }/{ h }$

Velocity of water

$=4\ { km }/{ h }$

$\implies t=\displaystyle\dfrac { 8 }{ 8-4 } +\displaystyle\dfrac { 8 }{ 8+4 }$

$=\displaystyle\dfrac { 8 }{ 3 } h$

$=160\ minutes$

In what ratio must two kinds of tea worth Rs 18 and Rs 28 per kg be mixed so that by selling the mixture at Rs 32 per kg, there may be a gain of 20%.

0%
2 : 13
0%
1 : 13
0%
3 : 14
0%
2 : 3

Explanation

Let the two kinds of tea be

$x$ and

$y$ .

Cost of the tea mixture will be

$18x + 28y$ .

Selling price after a profit will be 1.2(18x + 28y) = 21.6x + 33.6y

Equating both selling prices, 32x + 32y = 21.6x + 33.6y

Solving, we get

$\dfrac{x}{y}$ as

$\dfrac{1.6}{10.4}$ which is nothing but

$\dfrac{2}{13}$ .

So, the ratio is

$2 : 13.$

In an alloy, the ratio of copper to zinc is

$5\,\colon\,2$ . If

$1.250\:kg$ of zinc is mixed in

$17.500\:kg$ of alloy, then what will be the new ratio of copper to zinc?

0%
$\;2\,\colon\,1$
0%
$\;2\,\colon\,3$
0%
$\;3\,\colon\,2$
0%
$\;1\,\colon\,2$

Explanation

Amount of copper

$=\displaystyle\frac{5}{7}\times17.5\:kg=12.5\:kg$
Amount of zinc

$=17.5\:kg-12.5\:kg=5\:kg$ .

Now, the amount of zinc

$=(5+1.25)kg$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6.25\:kg$ .

$\therefore\,$ Required ratio

$=12.5\colon6.25=\displaystyle\frac{12.5}{6.25}=\displaystyle\frac{1250}{625}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle\frac{2}{1}=2 \colon1$

Wind is blowing west to east along two parallel tracks. Two trains moving with same speed in opposite directions have the relative velocity with respect to wind in the ratio

$1 : 2$ . The speed of each train is

0%
equal to that of wind
0%
double that of wind
0%
three times that of wind
0%
half that of wind

Explanation

Let

$v$ be velocity of wind and

$u$ be velocity of each train.
Relative velocity of one train with respect to wind

$= 2\times$ Relative velocity of other train with respect to wind

$u+v=2\left( u-v \right)$

$v+2v=2u-u=u$
i.e.,

$u=3v$ .

The ratio of the present ages of Sunita and Vinita is

$4\,\colon\,5$ . Six years hence, the ratio of their ages will be

$14\,\colon\,17$ . What will be the ratio of their age

$12$ years hence?

0%
$\;15\,\colon\,19$
0%
$\;13\,\colon\,15$
0%
$\;16\,\colon\,19$
0%
$\;17\,\colon\,19$

Explanation

Let the present ages of Sunita and Vinita be

$4x$ and

$5x$ respectively. Then,

$\;\;\;\;\;\;\displaystyle\frac{4x+6}{5x+6}=\displaystyle\frac{14}{17}\;\Rightarrow\;17(4x+6)=14(5x+6)$

$\;\;\;\;\;\;\;\Rightarrow\,68x+102=70x+84$

$\;\;\;\;\;\;\;\Rightarrow\,2x=18\;\;\Rightarrow\,x=9$

$\;\;\;\;\;\;\;\therefore\;Required\;ratio=\displaystyle\frac{4\times9+12}{5\times9+12}=\displaystyle\frac{48}{57}=\displaystyle\frac{16}{19}=16\,\colon\,19$ .

$A : B : C = 2 : 3 : 4,$ then what is

$\displaystyle \frac{A}{B} : \frac{B}{C} : \frac{C}{A}$ equal to?

0%
$4 : 5 : 15$
0%
$3 : 7 : 18$
0%
$6 : 2 : 21$
0%
$8 : 9 : 24$

Explanation

Let A = 2k, B = 3k, C = 4k.
Then,

$\displaystyle \frac{A}{B} = \dfrac{2k}{3k} = \frac{2}{3} , \dfrac{B}{C} = \frac{3k}{4k} = \dfrac{3}{4}, \frac{C}{A} = \dfrac{4k}{2k} = \dfrac{2}{1}$

$\therefore \displaystyle \frac{A}{B} : \frac{B}{C} : \dfrac{C}{A} = \dfrac{2}{3} : \dfrac{3}{4} : \dfrac{2}{1} (LCM = 12)$

Let A = 2k, B = 3k, C = 4k.

Then,

$\displaystyle \dfrac{A}{B} = \dfrac{2k}{3k} = \dfrac{2}{3} , \dfrac{B}{C} = \dfrac{3k}{4k} = \frac{3}{4}, \dfrac{C}{A} = \dfrac{4k}{2k} = \frac{2}{1}$

$\therefore \displaystyle \dfrac{A}{B} : \dfrac{B}{C} : \dfrac{C}{A} = \dfrac{2}{3} : \dfrac{3}{4} : \dfrac{2}{1} (LCM = 12)$

$=\displaystyle \frac{2}{3} \times 12 : \dfrac{3}{4} \times 12 : \dfrac{2}{1} \times 12 = 8 : 9 : 24.$

$=\displaystyle \dfrac{2}{3} \times 12 : \dfrac{3}{4} \times 12 : \dfrac{2}{1} \times 12 = 8 : 9 : 24.$

$a : b = 2 : 3$ and

$b : c = 4 : 5$ , find

$a^2 : b^2 : bc$ .

0%
$4 : 9 : 45$
0%
$16 : 36 : 45$
0%
$16 : 36 : 20$
0%
$4 : 36 : 20$

Explanation

$\displaystyle a : b = 2 : 3 \Rightarrow \frac{a}{b} = \frac{2}{3} \Rightarrow a = \frac{2b}{3} \Rightarrow a^2 = \frac{4b^2}{9}$

$\displaystyle b : c = 4 : 5 \Rightarrow \frac{b}{c} = \frac{4}{5} \Rightarrow b = \frac{4c}{5} \Rightarrow b^2 = \frac{4bc}{5} \Rightarrow bc = \frac{5b^2}{4}$

$\therefore \displaystyle a^2 : b^2 : bc = \frac{4b^2}{9} : b^2 : \frac{5b^2}{4} = \frac{4}{9}: 1: \frac{5}{4}$

$= \displaystyle \frac{4}{9} \times 36 : 1 \times 36 : \frac{5}{4} \times 36$

$(\because LCM = 36)$

$=16 : 36 : 45.$

Zinc and copper are in the ratio of

$5\,\colon\,3$ in

$200\:$ gm of an alloy. How much grams of copper must be added to make the ratio

$3\,\colon\,5$ ?

0%
$\;\displaystyle\frac{400}{3}$
0%
$\;\displaystyle\frac{1}{200}$
0%
$\;72$
0%
$\;66$

Explanation

Quantity of Zinc in the alloy

$=\displaystyle\frac{5}{(5+3)}\times200\:g$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\displaystyle\frac{5}{8}\times200\:g=125\:g$
Quantity of Copper in the alloy

$=200\:g-125\:g$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=75\:g$
Let the quantity of Copper added

$=x\:g$ .

Then,

$\;\;\;\;\;\;\;\;\;\displaystyle\frac{125}{75+x}=\displaystyle\frac{3}{5}\;\;\;\;\Rightarrow\;\;625=225+3x$

$\;\;\;\;\;\;\;\;\;\;\Rightarrow\,3x=400\;\;\;\Rightarrow\;\;\;\;x=133\displaystyle\frac{1}{3}\:g$ .

$Rs.782$ be divided into

$3$ parts, proportional to

$\dfrac{1}{2}:\dfrac{2}{3}:\dfrac{3}4$ , then the first part is

0%
$Rs. 182$
0%
$Rs. 190$
0%
$Rs. 196$
0%
$Rs. 204$

Explanation

Given ratio

$=\dfrac{1}{2}:\dfrac{2}{3}:\dfrac{3}{4}=6:8:9$

$1$ st part

$=Rs.\begin{bmatrix}782\times \dfrac{6}{23}\end{bmatrix}=Rs. 204$

There are

$90$ multiple choice questions in a test. Suppose you get two marks for every correct answer and for every question you leave unattempted or answer wrongly one mark is deducted from your total score of correct answers. If you get

$60$ marks in the test, then how many questions did you answer correctly?

0%
$40$
0%
$60$
0%
$50$
0%
$48$

Explanation

Let the number of questions answered correctly by

$'x'$

$\therefore$ The number of questions answered correctly=

$90-x$

Total Marks obtained

$=60$

Marks obtained for correct answer

$=22$

Marks deducted for wrong answer=

$-1(90-2)=x-90$

$\therefore 2x+2-90=60$

$\Rightarrow$

$3x=150$

$\Rightarrow$

$x=\dfrac { 150 }{ 3 }$

$\Rightarrow$

$x=50$

Therefore the number of question answered correctly is 50

$a\,\colon\,b=5\,\colon\,7$ and

$c\,\colon\,d=2a\,\colon\,3b$ , then find

$ac\,\colon\,bd$ .

0%
$\;20\,\colon\,38$
0%
$\;50\,\colon\,147$
0%
$\;10\,\colon\,21$
0%
$\;50\,\colon\,151$

Explanation

$\displaystyle\frac{a}{b}=\displaystyle\frac{5}{7},\,\displaystyle\frac{c}{d}=\displaystyle\frac{2a}{3b}=\displaystyle\frac{2}{3}\times\displaystyle\frac{a}{b}=\displaystyle\frac{2}{3}\times\displaystyle\frac{5}{7}=\displaystyle\frac{10}{21}$

$\;\;\;\;\;\;\;\;\;\therefore\,\displaystyle\frac{a}{b}\times\displaystyle\frac{c}{d}=\displaystyle\frac{5}{7}\times\displaystyle\frac{10}{21}\;\Rightarrow\;\displaystyle\frac{ac}{bd}=\displaystyle\frac{50}{147}$

$\;\;\;\;\;\;\;\;\;\Rightarrow\,ac\,\colon\,bd=50\,\colon\,147$ .

The ratio of a father's age to his son's age is

$4:1.$ The product of their ages is

$196.$ What is the ratio of their ages after 5 years?

0%
$11:4$
0%
$5:3$
0%
$3:8$
0%
$6:7$

Explanation

Let the father's age and son's age be 4x years and x years respectively.
Given,

$\displaystyle 4x \times x = 196 \Rightarrow x^2 = \frac {196}{4} = 49 \Rightarrow x = 7$

$\displaystyle \therefore Required ratio = \frac {4x+5}{x+5} = \frac {4 \times 7 + 5}{7+5} = \frac {33}{12} = \frac {11}{4}$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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