Formula:

\(b^2-4 a c>0\)

\(a=\left(a^2-b^2\right) \)

\(b=2(a c+b d) \)

\(c=\left(c^2-d^2\right)\)

\(b^2-4 a c \)

\(9{[2(a c+b d)]^2-\left[4\left(a^2-b^2\right)\left(c^2-d^2\right)\right]} \)

\(\Rightarrow 4 a^2 c^2+4 b^2 d^2+4 a c b d-4 a^2 c^2+4 a^2 d^2-4 b^2 d^2+4 b^2 c^2\)

\(\Rightarrow 4 a^2 d^2+4 b^2 c^2+4 a c b d \)

\(\Rightarrow 4\left(a^2 d^2+b^2 c^2+a c b d\right)\)

\(b^2-4 a c>0\)

So, Two different real roots

\(\therefore\) The quadratic equation

\(x^2\left(a^2-b^2\right)+2 x(a c+b d)+\left(c^2-d^2\right)\) is having two different real roots.

Given,

\(\beta x+y+9=0 \text {. }\)

Slope of line \(1\left(m_1\right)=-\frac{a}{b}=\frac{-\beta}{1}=-\beta\)

And

\(-3 x+y-4=0\)

Slope of line \(2\left(m_2\right)=3\)

We know that, \(\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)

\(\Rightarrow \tan 45^{\circ}=1=\left|\frac{-\beta-3}{1-3 \beta}\right| \)

\(\Rightarrow \frac{-(\beta+3)}{1-3 \beta}=\pm 1\)

So either \(\frac{-(\beta+3)}{1-3 \beta}=1\) or \(\frac{-(\beta+3)}{1-3 \beta}=-1\)

First, consider \(\frac{-(\beta+3)}{1-3 \beta}=1\)

\(\Rightarrow-(\beta+3)=1-3 \beta\)

\(\Rightarrow-\beta-3=1-3 \beta\)

\(\Rightarrow 2 \beta=4\)

\(\Rightarrow \beta=2\)

Also \(\frac{-(\beta+3)}{1-3 \beta}=-1\)

\(\Rightarrow-(\beta+3)=-1+3 \beta\)

\(\Rightarrow-4 \beta=2\)

\(\Rightarrow \beta=-\frac{1}{2}\)

So \(\beta=2\) or \(\frac{-1}{2}\)

Cost of the Tomato last year = Rs.10

So, Cost of the Tomato this year = 10 + 20% of 10 = Rs.(10 + 2) = Rs.12

Cost of the Potato last year = Rs.5

Let the cost of Potato this year be x

Therefore, cost of 5 Potato and 3 Tomato this year = 5x + (3 × 12) = 5x + 36 ….. (i)

cost of 5 Potato and 3 Tomato last year = 5 × 5) + (3 × 10 = 55 …….. (ii)

This year the cost is Rs.11 more

From (i) and (ii), we get

5x + 36 = 55 + 11

⇒ 5x = (66 – 36)

⇒ x = 6

Increase in the cost of Potato = (6 – 5) = Rs. 1

Percentage increase = \(\frac{1}{5}\) × 100% = 20%

Hence, the correct answer is 20%

Given:

Total number of \(\mathrm{t}\)-shirts \(=150\)

Lost in transit \(=40 \%\)

Formula used:

\(\mathrm{SP}=[\mathrm{CP}+(\mathrm{CP} \times\) profit \(\%)] \mathrm{SP}=[\mathrm{CP}-(\mathrm{CP} \times\) loss \(\%)]\)

\(\mathrm{SP}=\) selling price \(\mathrm{CP}=\) cost price

\(\mathrm{CP}=\) cost price

Let \(\mathrm{CP}\) be Rs. \(20 \mathrm{x}\)

According to the question,

\(40 \%\) of the \(\mathrm{t}\)-shirts are lost in transit

So, only \(150 \times 60 \%=90 \mathrm{t}\)-shirts can be sold

\(45 \mathrm{t}\)-shirts at \(20 \%\) profit

\(\Rightarrow 45 \times 20 x \times \frac{6}{5} \)

\(\Rightarrow 45 \times 24 x \)

\(\Rightarrow 1080 x \)

\(45 \text { t-shirt at } 5 \% \text { loss } \)

\(\Rightarrow 45 \times 20 x \times \frac{19}{20} \)

\(\Rightarrow 45 \times 19 x \)

\(\Rightarrow 855 x \)

\(1080 x+855 x=7740 \)

\(\Rightarrow 1935 x=7740\)

\(\Rightarrow x=4 \)

\(\Rightarrow 20 x=80 \)

\(\Rightarrow \text { CP }=\text { Rs. } 80\)

Given:

\(x+y+x y=3 \)

\(y+z+y z=8 \)

\(x+z+x z=15\)

\(x+y+x y=3\)

Adding 1 on both sides of the equation

\(1+x+y+x y=4\)

\(\Rightarrow(1+x)+y(1+x)=4\)

\(\Rightarrow(1+x)(1+y)=4\quad...(1)\)

Similarly performing the same on other two equations,

\(y+z+y z=8\)

\(1+y+z+y z=9\)

\(\Rightarrow(1+y)+z(1+y)=9\)

\(\Rightarrow(1+y)(1+z)=9\quad....(2)\)

\(x+z+x z=15\)

\(1+x+z+x z=16\)

\(\Rightarrow(1+x)+z(1+x)=16\)

\(\Rightarrow(1+x)(1+z)=16\quad....(3)\)

Multiplying equations \((1),(2)\) and \((3)\)

\({[(1+x)(1+y)(1+z)] 2=[4 \times 9 \times 16]} \)

\(\Rightarrow[(1+x)(1+y)(1+z)]=[2 \times 3 \times 4]\quad....(4)\)

From equation (1) and (4)

(1+z)=6

\(\Rightarrow z=5\)

Similarly, from equation (2) and (4)

\((1+x)=\frac{8}{3} \)

\(\Rightarrow x=\frac{5}{3}\)

Similarly, from equation (3) and (4)

\((1+y)=\frac{3}{2} \)\)

\(\therefore x+y+z=\frac{5}{3}+\frac{1}{2}+5=\frac{43}{6}\)

Speed of Kuldeep on first day of the month \(=\frac{\text { Distance Covered }}{\text { Time Taken }}\)

Let speed on the first day be \(\mathrm{S}_1\) and time be \(\mathrm{T}_1\)

Speed on the last day be \(\mathrm{S}_2\) and time be \(\mathrm{T}_2\).

\(\Rightarrow \mathrm{S}_1=\frac{D}{T_1}\)\(\Rightarrow \mathrm{D}=\mathrm{S}_1 \times \mathrm{T}_1\)

\(\Rightarrow \mathrm{D}=10 \times\left(\frac{6}{60}\right)\)\(\Rightarrow \mathrm{D}=\frac{60}{60}\)

\(\Rightarrow \mathrm{D}=1 \mathrm{~km}\)

Now, Speed of Kuldeep on last day of the month \(=\frac{\text { Distance Covered }}{\text { Time Taken }}\)

\(\Rightarrow S_2=\frac{D}{T_2}\)\(\Rightarrow S_2=\frac{1}{\left(\frac{4}{60}\right)}\)

\(\Rightarrow S_2=\frac{60}{4}\)\(\Rightarrow S_2=15 \mathrm{~km} / \mathrm{h}\)

Now, the improvement in Kuldeep's speed \(=\left(\mathrm{S}_2-\mathrm{S}_1\right)\)

= 15 – 10= 5 km/h

So, Increase in Kuldeep's speed in \(\mathrm{m} / \mathrm{s}=\left(S_2-S_1\right) \times\left(\frac{5}{18}\right)\)

\(=5 \times\left(\frac{5}{18}\right)\)\(=\frac{25}{18} \mathrm{~m} / \mathrm{s}\)

\(\therefore\) Kuldeep increased his speed by \(\frac{25}{18} \mathrm{~m} / \mathrm{s}\) in one month.

Rs. 645 and Rs. 625Given;

Total sum = Rs. 1301

Amount of A after 7 years = Amount of \(B\) after 9 years

Rate of Interest \(=4 \%\) per annum

Concept Used:

C.I \(=P\left[\{1+(R / 100)\}^n-1\right]\)

\(R =\) Rate of Interest

\(T =\) Time

Calculation:

Let the sum of \(A\) and \(B\) be Rs. \(x\) and Rs. \((1301-x)\) respectively.

\(\Rightarrow x \{1+(\frac{4} { 100})\}^7=(1301- x )\{1+(\frac{4} { 100})\} 9\)

\(\Rightarrow \frac{x} {(1301- x )}=\{1+(\frac{4} { 100})\}^2\)

\(\Rightarrow \frac{x} {(1301- x )}=(\frac{26} { 25})^2\)

\(\Rightarrow 625 x=676(1301-x)\)

\(\Rightarrow 1301 x=676 \times 1301\)

Sum of \(A=x=676\).

Sum of \(B=(1301-x)=1301-676\)

\(\Rightarrow\) Sum of \(B=625\).

\(\therefore\) The sum of \(A\) and \(B\) initially was Rs. 676 and Rs. 625.

Given:

Time taken by A to complete \(20 \%\) work \(=4\) days

Time taken by \(B\) to complete \(33.33 \%\) of work \(=10\) days

They worked together for 9 days

C completed the remaining work in 6 days

As we know,

Work \(=\) Time \(\times\) Efficiency

Time taken by A to complete \(100 \%\) work \(=4 \times 5=20\) days

Time taken by B to complete \(100 \%\) work \(=10 \times 3=30\) days

L.C.M of 20 and \(30=60=\) Total work

Efficiency of \(A=\frac{60}{20}\)

\(=3\) units/day

Efficiency of \(B=\frac{60}{30}\)

\(=2\) units/day

Worked done by \(A\) and \(B\) together in 9 days \(=(3+2) \times 9\) \(=45\) units

Remaining work \(=(60-45)\) units

\(=15\) units

C can complete the remaining work in 6 days

Efficiency of \(C=\frac{15}{6}=2.5\) units/day

Total efficiency of \(B\) and \(C=(2+2.5)\) units/day

\(=4.5\) units / day

Now, \(75 \%\) of total work \(=75 \%\) of 60 units

\(=45\) units

Required time \(=\frac{45}{4.5}\) days

\(=10\) days

\(\therefore B\) and \(C\) can complete \(75 \%\) work in 10 days.

Hence, the correct answer is 10.

Given:

\(f(x)=\sqrt{20-x^2}\)

It is defined only when \(20-x^2 \geq 0\)

So, \(y \geq 0 \quad\)........(i)

Let, \(y=f(x)\)

\(\Rightarrow y=\sqrt{20-x^2}\)

Squaring both sides, we get,

\(y^2=20-x^2\)

\(\Rightarrow x^2=20-y^2\)

\(\Rightarrow x=\sqrt{20-y^2}\)

It is defined only when \(20-y^2 \geq 0\)

\(\Rightarrow y^2 \leq 20\)

\(\Rightarrow y^2-20 \leq 0 \)

\(\Rightarrow(y-2 \sqrt{5})(y+2 \sqrt{5}) \leq 0\)

\(\Rightarrow y \in[-2 \sqrt{5}, 2 \sqrt{5}]\)

From equations (i) and (ii), we get: \(y \in[0,2 \sqrt{5}]\)

\(\log _a(a b)=x \)\( \Rightarrow \frac{\log a b}{\log a}=x \)

\( \Rightarrow \frac{\log a+\log b}{\log a}=x \)

\( \Rightarrow 1+\frac{\log b}{\log a}=x \)

\( \Rightarrow \frac{\log b}{\log a}=x-1 \)

\( \Rightarrow \frac{\log a}{\log b}=\frac{1}{x-1} \)

\( \Rightarrow 1+\frac{\log a}{\log b}=1+\frac{1}{x-1} \)

\( \Rightarrow \frac{\log b}{\log b}+\frac{\log a}{\log b}=\frac{x}{x-1} \)

\( \Rightarrow \frac{\log b+\log a}{\log b}=\frac{x}{x-1} \)

\( \Rightarrow \frac{\log (a b)}{\log b}=\frac{x}{x-1} \)

\( \Rightarrow \log_b (a b)=\frac{x}{x-1}\)