100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 5 InterviewSolution

201.	Gagandeep has written the following code to display an image in the background of HTML document:<BODY Bg Ground= “Animals.jpeg">but he is not getting the desired output. Help him in identifying correct code from the followinga.<BODY Bg= “Animals.jpeg”>b. <BODY BACK= “Animals.jpeg”>c. <BODY BGIMAGE= “Animals.jpeg”>d. <BODY BACKGROUND= “Animals.jpeg”>
Answer» Correct option is d. <BODY BACKGROUND= “Animals.jpeg”>

Discussion

202.	What will be the angle of diffracting for the first minimum due to Fraunhofer diffraction with sources of light of wavelength `550nm` and slit of width `0.55mm`?A. 0.001 radB. 0.01 radC. 1 radD. 0.1 rad
Answer» Correct Answer - A Using `d sin theta = n lamda`, for `n =1` `sin theta = (lamda)/(d) =(550 xx10^(-9))/(0.55 xx10^(-3))=10^(-3)=0.001 rad`.

Discussion

203.	A block of mass `M` is kept on a smooth surface and touches the two springs as shown in the figure but not attached to the springs. Initially springs are in their natural length. Now, the block is shifted `(l_(0)//2)` from the given position in such a way that it compresses a spring and released. The time - period of oscillation of mass will be A. `(pi)/(2)sqrt((M)/(K))`B. `2pisqrt((M)/(5K))`C. `(3pi)/(2)sqrt((M)/(K))`D. `pisqrt((M)/(2K))`
Answer» Correct Answer - C `T=pi[sqrt((M)/(K))+(1)/(2)sqrt((M)/(K))]=(3pi)/(2)sqrt((M)/(K))`

Discussion

204.	What is the heat source of fluidized bed in tem tech thermal sand reclamation system?(a) Flame heating(b) Ultraviolet heating(c) Infrared heating(d) Conduction heating
Answer» The correct choice is (c) Infrared heating To explain: In tem tech thermal sand reclamation system, overall there are two beds involved in it and in them, shortwave infrared emitters are used in the beds.

Discussion

205.	Up to what accuracy can the temperature of the bed be controlled in tem tech thermal sand reclamation system?(a) ± 1%(b) ± 2%(c) ± 3%(d) ± 4%
Answer» The correct choice is (b) ± 2% Explanation: The temperature of the bed can be controlled to an accuracy of ± 2%, in tem tech thermal sand reclamation system. These fluidized beds essentially need a very accurate temperature control.

Discussion

206.	Light of wavelength `6000 Å` is reflected at nearly normal incidence from a soap films of refractive index 1.4 The least thickness of the fringe then will appear black isA. infinityB. `200 Å`C. `2000 Å`D. `1000 Å`
Answer» Correct Answer - C (c ) Black fringe means destructive interference. For destructive interference, we have `2 mu t cos r = n lambda` Where `mu` is refractive index, `lambda` is wavelength and `t` is thichness `implies t = (n lambda)/(2 mu cos r)` For minima, we have `cos r = 1, n = 1` `implies t = (lambda)/(2 mu) = (6000)/(2 xx 1.4) = 2142 Å` `~~ 2000 Å`

Discussion

207.	What should be the width of each slit to obtain n maxima of double slit pattern within the central maxima of single slit pattern?
Answer» Let 'a' be the width of each slit. Linear separation between n bright fringes, x = nβ = \(\frac{n\lambda D}{d}\) Corresponding angular separation, θ₁ = \(\frac{\mathrm{x}}{D}=\frac{n\lambda}{d}\) Now, the angular width of central maximum in the diffraction pattern of a single slit, θ₂ = \(\frac{2\lambda}{a}\) As θ₁ = θ₂ \(\frac{n\lambda}{d}=\frac{2\lambda}{a}\) \(\therefore\)a = \(\frac{2d}{n}\); where d = separation between slits \(\frac{n\lambda}{d}=\frac{2\lambda}{a}\) \(n=\frac{2d}{a}\) ,where d is separation between slit and a width of slit.

207.

What should be the width of each slit to obtain n maxima of double slit pattern within the central maxima of single slit pattern?

Answer»

Let 'a' be the width of each slit.

Linear separation between n bright fringes, x = nβ = \(\frac{n\lambda D}{d}\)

Corresponding angular separation, θ₁ = \(\frac{\mathrm{x}}{D}=\frac{n\lambda}{d}\)

Now, the angular width of central maximum in the diffraction pattern of a single slit,

θ₂ = \(\frac{2\lambda}{a}\)

As θ₁ = θ₂

\(\frac{n\lambda}{d}=\frac{2\lambda}{a}\)

\(\therefore\)a = \(\frac{2d}{n}\); where d = separation between slits

\(\frac{n\lambda}{d}=\frac{2\lambda}{a}\)

\(n=\frac{2d}{a}\) ,where d is separation between slit and a width of slit.

Discussion

208.	1 – cos2A =(A) sin2A(B) 2sin2A(C) cos2A(D) 2cos2A
Answer» Correct option is: (C) cos²A

Discussion

209.	The number of terms in an A.P. 51, 48, 45, …, 6 is – (A) 14 (B) 16 (C) 17 (D) 18
Answer» Correct answer is (B) 16

Discussion

210.	\( \int_{\pi}^{0} \frac{\sqrt{x^{2}+1}\left(\log \left(x^{2}+1\right)-2 \log x\right)}{x^{4}} d x \)
Answer» Let I = \(\int\cfrac{\sqrt{x^2+1}(log(x^2+1)-2 log x)}{x^4}dx\) \(=\int\cfrac{x\sqrt{1+\frac1{x^2}}log(\frac{x^2+1}{x^2})}{x^4}dx\) (\(\because\) 2log x = log x² and log(x² + 1) - 2logx = log(x² + 1) - logx² = log(\(\frac{x^2+1}{x^2}\))) \(=\int\cfrac{\sqrt{1+\frac1{x^2}}log(1+\frac1{x^2})}{x^3}dx\) Let 1 + \(\frac1{x^2}\) = t ⇒ \(\frac{-2}{x^3}dx = dt\) ⇒ \(\frac{dx}{x^3} = -\frac{dt}2\) \(\therefore\) I = \(-\frac12\int\sqrt t\) log t dt \(=-\frac12[log t\int \sqrt t dt-\int(\frac{d}{dt}log t\int\sqrt t dt)dt]\) \(=-\frac12[\frac23log t t^{3/2}- \int\frac1t\times\frac23t^{3/2}dt]\) \(=-\frac12[\frac23t^{3/2}log t - \frac23\int t^{1/2}dt]\) \(= -\frac12(\frac23t^{3/2}log t - \frac23\times\frac23t^{3/2})\) \(=-\frac12[\frac23(1+\frac1{x^2})^{3/2}log(1+\frac1{x^2}-\frac49(1+\frac1{x^2})^{3/2}]\) (By putting t = 1 + \(\frac1{x^2}\))

210.

\( \int_{\pi}^{0} \frac{\sqrt{x^{2}+1}\left(\log \left(x^{2}+1\right)-2 \log x\right)}{x^{4}} d x \)

Answer»

Let I = \(\int\cfrac{\sqrt{x^2+1}(log(x^2+1)-2 log x)}{x^4}dx\)

\(=\int\cfrac{x\sqrt{1+\frac1{x^2}}log(\frac{x^2+1}{x^2})}{x^4}dx\)

(\(\because\) 2log x = log x² and log(x² + 1) - 2logx = log(x² + 1) - logx² = log(\(\frac{x^2+1}{x^2}\)))

\(=\int\cfrac{\sqrt{1+\frac1{x^2}}log(1+\frac1{x^2})}{x^3}dx\)

Let 1 + \(\frac1{x^2}\) = t

⇒ \(\frac{-2}{x^3}dx = dt\)

⇒ \(\frac{dx}{x^3} = -\frac{dt}2\)

\(\therefore\) I = \(-\frac12\int\sqrt t\) log t dt

\(=-\frac12[log t\int \sqrt t dt-\int(\frac{d}{dt}log t\int\sqrt t dt)dt]\)

\(=-\frac12[\frac23log t t^{3/2}- \int\frac1t\times\frac23t^{3/2}dt]\)

\(=-\frac12[\frac23t^{3/2}log t - \frac23\int t^{1/2}dt]\)

\(= -\frac12(\frac23t^{3/2}log t - \frac23\times\frac23t^{3/2})\)

\(=-\frac12[\frac23(1+\frac1{x^2})^{3/2}log(1+\frac1{x^2}-\frac49(1+\frac1{x^2})^{3/2}]\)

(By putting t = 1 + \(\frac1{x^2}\))

Discussion

211.	The number of odd numbers between 0 and 50 is – (A) 25 (B) 26 (C) 24 (D) 27
Answer» Correct answer is (A) 25

Discussion

212.	In right ∆ABC, ∠B=90°, AB = √15cm, BC=1Cm and AC = 4cm then tan A will be equal to –(A) √15/4(B) √15/1(C) 1/√15(D) 4/√15
Answer» Correct answer is (C) 1/√15

Discussion

213.	find the domain and range of f(x)=x²-y/x-2
Answer» \(f(x) = \frac {x^2 - 4}{x - 2}\) Domain = \(R -\{2\}\) (∵ f(x) is not defined at x = 2) \(f(x) = \frac {x^2 - 4}{x - 2} = \frac{(x-2)(x+ 2)}{x - 2} = x+ 2 , x\ne 2\) \(\therefore f(x) \ne 4\) Range = \(R - \{4\}\).

213.

find the domain and range of f(x)=x²-y/x-2

Answer»

\(f(x) = \frac {x^2 - 4}{x - 2}\)

Domain = \(R -\{2\}\) (∵ f(x) is not defined at x = 2)

\(f(x) = \frac {x^2 - 4}{x - 2} = \frac{(x-2)(x+ 2)}{x - 2} = x+ 2 , x\ne 2\)

\(\therefore f(x) \ne 4\)

Range = \(R - \{4\}\).

Discussion

214.	Evaluate : `(cos^(2)(45^(@) + theta)+ cos^(2) (45^(@) -theta))/(tan(60^(@) + theta) xx tan(30^(@) - theta))+(cot30^(@) + sin 90^(@))xx (tan60^(@) -sec0^(@))`
Answer» `(cos^(2)(45^(@) + theta)+ cos^(2) (45^(@) -theta))/(tan(60^(@) + theta) xx tan(30^(@) - theta))+(cot30^(@) + sin 90^(@))xx (tan60^(@) -sec0^(@))` `(cos^(2)(45^(@) + theta)+ cos^(2)(45^(@) - theta))/( tan(60^(@) + theta) xx cot(60^(@) + theta)) + (sqrt(3)+1)(sqrt(3)-1)` ` = 1+2=3`

Discussion

215.	sin2A =(A) 2tanA/1+tan2A(B) cos2A – sin2A(C) 2cos2A(D) 2sin2A
Answer» Correct option is: (A) \(\frac{2tanA}{1+tan^2A}\)

Discussion

216.	Evaluate \( \int_{0}^{2}\|4 x-5\| d x \).
Answer» \(\int\limits_0^2\|4x - 5\|dx\) \(\because\) \|4x - 5\| = \(\begin{cases}4x-5&; \quad x\geq\frac54\\-(4x-5) &;x\leq\frac54\end{cases}\) \(\therefore\) \(\int\limits_0^2\|4x-5\|dx=\int\limits_0^{5/4}-(4x-5)dx+\int\limits_{5/4}^2(4x-5)dx\) \(=[-\frac{4x^2}2+5x]_0^{5/4}+[\frac{4x^2}2-5x]_{5/4}^2\) \(=(-2\times(5/4)^2+5\times5/4)-0)+(2\times2^2-5\times2\) \(-2\times(5/4)^2+5\times5/4)\) \(=2\times\frac{25}4-4\times\frac{25}{16}+8-10\) \(=\frac{25}2-\frac{25}4-2\) \(\frac{25}4-2\) = \(\frac{25-8}4=\frac{17}4\).

216.

Evaluate \( \int_{0}^{2}|4 x-5| d x \).

Answer»

\(\int\limits_0^2|4x - 5|dx\)

\(\because\) |4x - 5| = \(\begin{cases}4x-5&; \quad x\geq\frac54\\-(4x-5) &;x\leq\frac54\end{cases}\)

\(\therefore\) \(\int\limits_0^2|4x-5|dx=\int\limits_0^{5/4}-(4x-5)dx+\int\limits_{5/4}^2(4x-5)dx\)

\(=[-\frac{4x^2}2+5x]_0^{5/4}+[\frac{4x^2}2-5x]_{5/4}^2\)

\(=(-2\times(5/4)^2+5\times5/4)-0)+(2\times2^2-5\times2\) \(-2\times(5/4)^2+5\times5/4)\)

\(=2\times\frac{25}4-4\times\frac{25}{16}+8-10\)

\(=\frac{25}2-\frac{25}4-2\)

\(\frac{25}4-2\) = \(\frac{25-8}4=\frac{17}4\).

Discussion

217.	Solvepx + qy = p - qqx - py = p + q
Answer» According to the question we have Px+qy=p-q ---------(1) qx-py=p+q ---------(2) Multiplying (1) with p and (2) with q we get, p²x+pqy=p²-pq -----------(3) q²x-pqy=pq+q² -----------(4) Adding (3) with (4) we get, p²x+q²x=p²+q² or, x(p²+q²)=p²+q² or, x=1 Putting in (1) we get, p.1+qy=p-q or, qy=p-q-p or, qy=-q or, y= -1 ∴, x=1 and y= -1

217.

Solvepx + qy = p - qqx - py = p + q

Answer»

According to the question we have

Px+qy=p-q ---------(1)
qx-py=p+q ---------(2)
Multiplying (1) with p and (2) with q we get,
p²x+pqy=p²-pq -----------(3)
q²x-pqy=pq+q² -----------(4)
Adding (3) with (4) we get,
p²x+q²x=p²+q²
or, x(p²+q²)=p²+q²
or, x=1
Putting in (1) we get,
p.1+qy=p-q
or, qy=p-q-p
or, qy=-q
or, y= -1
∴, x=1 and y= -1

Discussion

218.	The value of cot15. Cot75 .cot60. Cot30 +tan45 is equal to
Answer» cot(90° – 75) cot(90° – 60°) cot 45° cot 60° cot 75° = tan 75° tan 60° (1) cot 60° cot 75° = 1

218.

The value of cot15. Cot75 .cot60. Cot30 +tan45 is equal to

Answer»

cot(90° – 75) cot(90° – 60°) cot 45° cot 60° cot 75°

= tan 75° tan 60° (1) cot 60° cot 75°

= 1

Discussion

219.	If sina=cosa find the value of 2tana + cos*2a
Answer» SinA = CosA We know that Sin 45 = Cos 45 So, A =45 So, 2tan A + Cos²A = 2tan 45 + cos² 45 = 2(1) +(1/√2)² = 2+1/2 = 4+1/2 = 5/2

219.

If sina=cosa find the value of 2tana + cos*2a

Answer»

SinA = CosA
We know that Sin 45 = Cos 45
So, A =45
So,

2tan A + Cos²A = 2tan 45 + cos² 45
= 2(1) +(1/√2)² = 2+1/2 = 4+1/2 = 5/2

Discussion

220.	solve 1≤\|x-2\|≤3
Answer» x−2<0 x<2 1≤−x+2≤3 1−2≤−x+2−2≤3−2 −1≤−x≤1 1≥x≥−1 x∈[−1,1] x−2>0 x>2 1≤x−2≤3 3≤x≤5 x∈[3,5] x∈[−1,1]U[3,5]

Discussion

221.	The minimum value of z = 3x – 2y subject to constraints 2x + y ≤ 4, x ≥ 0, y ≥ 0 is (A) 6 (B) -6 (C) -8 (D) -12
Answer» Correct answer is (C) -8

Discussion

222.	The minimum value of P = 6x + 16y subject to the constraints x ≤ 40, y ≥ 20 and x, y ≥ 0 is (a) 240 (b) 320 (c) 0 (d) None of these
Answer» (c) 0 The minimum value of P = 6x + 16y subject to the constraints x ≤ 40, y ≥ 20 and x, y ≥ 0 is 0.

222.

The minimum value of P = 6x + 16y subject to the constraints x ≤ 40, y ≥ 20 and x, y ≥ 0 is (a) 240 (b) 320 (c) 0 (d) None of these

Answer»

(c) 0

The minimum value of P = 6x + 16y subject to the constraints x ≤ 40, y ≥ 20 and x, y ≥ 0 is 0.

Discussion

223.	The maximum value of P = 40x + 50y subject to the constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0 and y ≥ 0 is (a) 120 (b) 230 (c) 200 (d) None of these
Answer» (b) 230 The maximum value of P = 40x + 50y subject to the constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0 and y ≥ 0 is 230.

223.

The maximum value of P = 40x + 50y subject to the constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0 and y ≥ 0 is (a) 120 (b) 230 (c) 200 (d) None of these

Answer»

(b) 230

The maximum value of P = 40x + 50y subject to the constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0 and y ≥ 0 is 230.

Discussion

224.	A person invests Rs. 20,000 at 20 %, Rs. 150 shares at a premium of Rs. 50. The income from these shares in Rs. is (a) 1,000 (b) 3,000 (c) 1,500 (d) 2,000
Answer» (b) Rs. 3,000 A person invests Rs. 20,000 at 20 %, Rs. 150 shares at a premium of Rs. 50. The income from these shares in Rs. is Rs. 3,000.

224.

A person invests Rs. 20,000 at 20 %, Rs. 150 shares at a premium of Rs. 50. The income from these shares in Rs. is (a) 1,000 (b) 3,000 (c) 1,500 (d) 2,000

Answer»

(b) Rs. 3,000

A person invests Rs. 20,000 at 20 %, Rs. 150 shares at a premium of Rs. 50. The income from these shares in Rs. is Rs. 3,000.

Discussion

225.	Find the value of k if x + k is a factor of the polynomial x3 + kx2 - 2x + k + 5.
Answer» Let p(x) = x³+ kx²− 2x + k + 4 As x + k is a factor ∴ p(−k) = 0 p(−k) = −k³+ k³+ 2k + k + 4 ⇒ 3k + 4 As 3k + 4 = 0 ∴k = \(\frac{-4}{3}\)

225.

Find the value of k if x + k is a factor of the polynomial x3 + kx2 - 2x + k + 5.

Answer»

Let p(x) = x³+ kx²− 2x + k + 4

As x + k is a factor

∴ p(−k) = 0

p(−k) = −k³+ k³+ 2k + k + 4

⇒ 3k + 4

As 3k + 4 = 0

∴k = \(\frac{-4}{3}\)

Discussion

226.	The total investment made in buying Rs. X shares at a premium of 25 % is Rs. 125 X. The number of shares bought is (a) 500 (b) 250 (c) 125 (d) 100
Answer» (d) 100 The total investment made in buying Rs. X shares at a premium of 25 % is Rs. 125 X. The number of shares bought is 100.

226.

The total investment made in buying Rs. X shares at a premium of 25 % is Rs. 125 X. The number of shares bought is (a) 500 (b) 250 (c) 125 (d) 100

Answer»

(d) 100

The total investment made in buying Rs. X shares at a premium of 25 % is Rs. 125 X. The number of shares bought is 100.

Discussion

227.	Reduce the given Rational expressions to its lowest form (i) (x3a - 8)/(x2a + 2xa + 4) (ii) (10x3 - 25x2 + 4x - 10)/(-4 - 10x2 )
Answer» (i) \(\frac{x^{3a}-8}{x^{2a}+2x^a+4}=\frac{(x^a)^3-2^3}{x^{2a}+2x^a+4}\) \(=\frac {(x^a-2)(x^{2a}+2x^a+4)}{x^{2a}+2x^a+4}\) (∵ a³ - b³ = (a - b)(a² + ab + b²) and here a = x^a and b = 2) = x^a - 2 (ii) \(\frac{10x^3-25x^2+4x-10}{-4-10x^2}\) \(=\frac{5x^2(2x-5)+2(2x-5)}{-2(2+5x^2)}\) \(=\frac{(2x-5)(2+5x^2)}{-2(2+5x^2)}\) \(= \frac{2x-5}{-2}\) \(=\frac{5-2x}{2}\)

227.

Reduce the given Rational expressions to its lowest form (i) (x3a - 8)/(x2a + 2xa + 4) (ii) (10x3 - 25x2 + 4x - 10)/(-4 - 10x2 )

Answer»

(i) \(\frac{x^{3a}-8}{x^{2a}+2x^a+4}=\frac{(x^a)^3-2^3}{x^{2a}+2x^a+4}\)

\(=\frac {(x^a-2)(x^{2a}+2x^a+4)}{x^{2a}+2x^a+4}\) (∵ a³ - b³ = (a - b)(a² + ab + b²) and here a = x^a and b = 2)

= x^a - 2

(ii) \(\frac{10x^3-25x^2+4x-10}{-4-10x^2}\)

\(=\frac{5x^2(2x-5)+2(2x-5)}{-2(2+5x^2)}\)

\(=\frac{(2x-5)(2+5x^2)}{-2(2+5x^2)}\)

\(= \frac{2x-5}{-2}\)

\(=\frac{5-2x}{2}\)

Discussion

228.	Find the value of "a" in the polynomial 2a + 2xa + 5a + 10 if (a + x) is one of its factors.
Answer» If a + x is a factor of polynomial P(x) then x = -a is a root of P(x). ∴ P(-a) = 0 ⇒ 2a + 2(-a)a + 5a + 10 = 0 ⇒ -2a² + 7a + 10 = 0 ⇒ \(a = \frac{-7 \pm \sqrt{49 + 80}}{-4}\) \(= \frac{7\pm \sqrt{129}}{4}\)

228.

Find the value of "a" in the polynomial 2a + 2xa + 5a + 10 if (a + x) is one of its factors.

Answer»

If a + x is a factor of polynomial P(x)

then x = -a is a root of P(x).

∴ P(-a) = 0

⇒ 2a + 2(-a)a + 5a + 10 = 0

⇒ -2a² + 7a + 10 = 0

⇒ \(a = \frac{-7 \pm \sqrt{49 + 80}}{-4}\)

\(= \frac{7\pm \sqrt{129}}{4}\)

Discussion

229.	Ramesh opens a Saving Bank Account on 16.06.2007 with a deposit of Rs. 700. He deposited Rs. 1,500 on 07.07.2007. find the amount on which he would receive the interest at the end of July, 2007 (a) Rs. 700 (b) Rs. 1,500 (c) Rs. 2,200 (d) Rs. 800
Answer» Correct answer is (c) Rs. 2,200

229.

Ramesh opens a Saving Bank Account on 16.06.2007 with a deposit of Rs. 700. He deposited Rs. 1,500 on 07.07.2007. find the amount on which he would receive the interest at the end of July, 2007 (a) Rs. 700 (b) Rs. 1,500 (c) Rs. 2,200 (d) Rs. 800

Answer»

Correct answer is

(c) Rs. 2,200

Discussion

230.	Indian GST model has ____ rate structure (a) 3 (b) 4 (c) 5 (d) 6
Answer» (b) 4 Indian GST model has 4 rate structure.

230.

Indian GST model has ____ rate structure (a) 3 (b) 4 (c) 5 (d) 6

Answer»

(b) 4

Indian GST model has 4 rate structure.

Discussion

231.	Mother, Father and Son line up at random for a family picture. E: Son on one end F: Father in middlethen P(E/F) is(a) 0 (b) 1 (c) 1/2 (d) 1/3
Answer» (b) 1 E: Son on one end F: Father in middle then P(E/F) is 1.

231.

Mother, Father and Son line up at random for a family picture. E: Son on one end F: Father in middlethen P(E/F) is(a) 0 (b) 1 (c) 1/2 (d) 1/3

Answer»

(b) 1

E: Son on one end

F: Father in middle

then P(E/F) is 1.

Discussion

232.	Four years ago, the ratio of half the age of A to 1/3 of the age of B is 8 ∶ 7. Also half of the present age of B exceeds 1/3 of present age of A by 11. Find the present age of A?1. 462. 363. 664. 565. 48
Answer» Correct Answer - Option 2 : 36 Given: Four years ago, the ratio of half the age of A to 1/3 of the age of B = 8 ∶ 7 Half of the present age of B exceeds 1/3 of present age of A by 11 Calculations: Let the present ages of A and B be a and b respectively. Four years ago A’s age = (a – 4) Four years ago B’s age = (b – 4) Four years ago, the ratio of half the age of A to 1/3 of the age of B = 8 ∶ 7 \( \Rightarrow \;\frac{{\frac{1}{2}\; \times\; \left( {a\; - \;4} \right)}}{{\frac{1}{3}\; \times\; \left( {b\; - \;4} \right)}} = \frac{8}{7}\) ⇒ 21a – 16b = 20 ----(1) Half the present age of B exceeds 1/3 of present age of A by 11 ⇒ b/2 = a/3 + 11 ⇒ 2a – 3b = -66 ----(2) On solving equation (1) and (2) ⇒ a = 36 year and b = 46 years ∴ The present age of A is 36

232.

Four years ago, the ratio of half the age of A to 1/3 of the age of B is 8 ∶ 7. Also half of the present age of B exceeds 1/3 of present age of A by 11. Find the present age of A?1. 462. 363. 664. 565. 48

Answer» Correct Answer - Option 2 : 36

Given:

Four years ago, the ratio of half the age of A to 1/3 of the age of B = 8 ∶ 7

Half of the present age of B exceeds 1/3 of present age of A by 11

Calculations:

Let the present ages of A and B be a and b respectively.

Four years ago A’s age = (a – 4)

Four years ago B’s age = (b – 4)

Four years ago, the ratio of half the age of A to 1/3 of the age of B = 8 ∶ 7

\( \Rightarrow \;\frac{{\frac{1}{2}\; \times\; \left( {a\; - \;4} \right)}}{{\frac{1}{3}\; \times\; \left( {b\; - \;4} \right)}} = \frac{8}{7}\)

⇒ 21a – 16b = 20 ----(1)

Half the present age of B exceeds 1/3 of present age of A by 11

⇒ b/2 = a/3 + 11

⇒ 2a – 3b = -66 ----(2)

On solving equation (1) and (2)

⇒ a = 36 year and b = 46 years

∴ The present age of A is 36

Discussion

233.	Three years ago the ratio of ages of Gargi and Supriya is 2 ∶ 1, also the present age of Gargi is 27 years more than Supriya. What is the sum of their present ages?1. 892. 683. 874. 935. 105
Answer» Correct Answer - Option 3 : 87 Given∶ Three years ago the ratio of ages of Gargi and Supriya is 2 ∶ 1 The present age of Gargi is 27 years more than Supriya. Calculations∶ Let the present ages of Gargi and Supriya be G and S respectively. According to question, \(\frac{{{\rm{G}} - 3}}{{{\rm{S}} - 3}} = \frac{2}{1}\) ⇒ 2S - G = 3 ---(1) Also, G = 27 + S ⇒ G -S = 27 ---(2) On solving the equations (1) and (2), We get, G = 57 and S = 30 ⇒ present age of Gargi = 57 years and the present age of Supriya = 15 years ⇒ Sum of their present ages will be; 57 + 30 = 87 years.

233.

Three years ago the ratio of ages of Gargi and Supriya is 2 ∶ 1, also the present age of Gargi is 27 years more than Supriya. What is the sum of their present ages?1. 892. 683. 874. 935. 105

Answer» Correct Answer - Option 3 : 87

Given∶

Three years ago the ratio of ages of Gargi and Supriya is 2 ∶ 1

The present age of Gargi is 27 years more than Supriya.

Calculations∶

Let the present ages of Gargi and Supriya be G and S respectively.

According to question, \(\frac{{{\rm{G}} - 3}}{{{\rm{S}} - 3}} = \frac{2}{1}\)

⇒ 2S - G = 3 ---(1)

Also, G = 27 + S

⇒ G -S = 27 ---(2)

On solving the equations (1) and (2), We get, G = 57 and S = 30

⇒ present age of Gargi = 57 years and the present age of Supriya = 15 years

⇒ Sum of their present ages will be; 57 + 30 = 87 years.

Discussion

234.	Ghanshyam borrow a sum of Rs. 13,360 at 8.75 % per annum compound interest and paid back in two instalments. What was the amount of each instalment1. 75692. 84513. 70024. 70215. 1520
Answer» Correct Answer - Option 1 : 7569 Answer : 7569 Given : Net amount = 13360 Rs. Rate % = 8.75% = 7/80 Formula used : \(amount = \frac{x}{{\left( {1 + r\;\% } \right)}} + \frac{x}{{{{\left( {1 + R\% } \right)}^2}}} \ldots \ldots \ldots \frac{x}{{{{\left( {1 + r\% } \right)}^n}}}\) Calculation : \(13360 = x/(18.75\% ) + x/{(18.75)^2}\) X=7569 Rs.

234.

Ghanshyam borrow a sum of Rs. 13,360 at 8.75 % per annum compound interest and paid back in two instalments. What was the amount of each instalment1. 75692. 84513. 70024. 70215. 1520

Answer» Correct Answer - Option 1 : 7569

Answer : 7569

Given : Net amount = 13360 Rs.

Rate % = 8.75% = 7/80

Formula used : \(amount = \frac{x}{{\left( {1 + r\;\% } \right)}} + \frac{x}{{{{\left( {1 + R\% } \right)}^2}}} \ldots \ldots \ldots \frac{x}{{{{\left( {1 + r\% } \right)}^n}}}\)

Calculation :

\(13360 = x/(18.75\% ) + x/{(18.75)^2}\)

X=7569 Rs.

Discussion

235.	A fan is sold for Rs. 900 cash payment of Rs. 200 down payment followed by two equal monthly installments of each Rs. 375. The annual rate of interest is ____ (approximately) (a) 25 % (b) 30 % (c) 54 % (d) 59 %
Answer» (d) 59% A fan is sold for Rs. 900 cash payment of Rs. 200 down payment followed by two equal monthly installments of each Rs. 375. The annual rate of interest is 59% (approximately).

235.

A fan is sold for Rs. 900 cash payment of Rs. 200 down payment followed by two equal monthly installments of each Rs. 375. The annual rate of interest is ____ (approximately) (a) 25 % (b) 30 % (c) 54 % (d) 59 %

Answer»

(d) 59%

A fan is sold for Rs. 900 cash payment of Rs. 200 down payment followed by two equal monthly installments of each Rs. 375. The annual rate of interest is 59% (approximately).

Discussion

236.	Ratio of Nidhi’s age and Amit’ age is 2 ∶ 3. Nidhi's age was 16 years 4 years ago. What is the present age of Amit?1. 30 years2. 32 years3. 33 years4. 35 years5. 28 years
Answer» Correct Answer - Option 1 : 30 years Given: Ratio of Nidhi’s age and Amit’s age is 2 ∶ 3. Concept Used: Ratio concept used. Calculation: Let the present of Nidhi and Amit be 2x and 3x. Present of Nidhi = (16 + 4)years = 20 years So, 2x = 20 ⇒ x = 10 So, present age of Amit = 3x = 3 × 10 = 30 years ∴ Present age of Amit is 30 years old.

236.

Ratio of Nidhi’s age and Amit’ age is 2 ∶ 3. Nidhi's age was 16 years 4 years ago. What is the present age of Amit?1. 30 years2. 32 years3. 33 years4. 35 years5. 28 years

Answer» Correct Answer - Option 1 : 30 years

Given:

Ratio of Nidhi’s age and Amit’s age is 2 ∶ 3.

Concept Used:

Ratio concept used.

Calculation:

Let the present of Nidhi and Amit be 2x and 3x.

Present of Nidhi = (16 + 4)years = 20 years

So, 2x = 20

⇒ x = 10

So, present age of Amit = 3x = 3 × 10 = 30 years

∴ Present age of Amit is 30 years old.

Discussion

237.	Which is the right sequence of a stages of Internationalization a. Domestic, Transnational, Global, International, Multinational b. Domestic, International, Multinational, Global, Transnational c. Domestic, Multinatinal, International, Transnational, Global d. Domestic, Internatinal, Transnational, Multinational, Global
Answer» b. Domestic, International, Multinational, Global, Transnational

Discussion

238.	Rohan purchased a mobile Rs. 5000 cash or Rs. 500 down payment followed by 4 equal instalments. If the rate 25 % per annum at simple interest calculate monthly instalment and find the nearest integer value. 1. 11822. 17203. 15224. 11555. 2102
Answer» Correct Answer - Option 1 : 1182 Given : After down payment 5000 – 500= 4500 Rate % = 25 % Total instalment = 4 Formula used : \(instalment = \frac{{100 \times amount}}{{\begin{array}{{20}{c}} {100 \times n + \;R \times \frac{{n\left( {n - 1} \right)}}{2}}\\ {} \end{array}}}\) calculation :* \( = (4500×25×4)/(12×100) + 4500 = 4875 \) Instalment \(\frac{{100 \times \;4875}}{{\begin{array}{*{20}{c}} {100 \times 4 + \;25 \times 4/12 \times \frac{3}{2} \times }\\ {} \end{array}}} = {\rm{ }}{\bf{1181}}.{\bf{8}}\;\)

238.

Rohan purchased a mobile Rs. 5000 cash or Rs. 500 down payment followed by 4 equal instalments. If the rate 25 % per annum at simple interest calculate monthly instalment and find the nearest integer value. 1. 11822. 17203. 15224. 11555. 2102

Answer» Correct Answer - Option 1 : 1182

Given : After down payment 5000 – 500= 4500

Rate % = 25 %

Total instalment = 4

Formula used :

\(instalment = \frac{{100 \times amount}}{{\begin{array}{*{20}{c}} {100 \times n + \;R \times \frac{{n\left( {n - 1} \right)}}{2}}\\ {} \end{array}}}\)

calculation :

\( = (4500×25×4)/(12×100) + 4500 = 4875 \)

Instalment

\(\frac{{100 \times \;4875}}{{\begin{array}{*{20}{c}} {100 \times 4 + \;25 \times 4/12 \times \frac{3}{2} \times }\\ {} \end{array}}} = {\rm{ }}{\bf{1181}}.{\bf{8}}\;\)

Discussion

239.	The Theory of Absolute Cost Advantage is given by a. David Ricardo b. Adam Smithc. F W Taylor d. Ohlin and Heckscher
Answer» The Theory of Absolute Cost Advantage is given by Adam Smith.

Discussion

240.	The total ages of Amit, Sumit and Puneet is 85 years. Five years ago the ratio of their ages was 2 : 3 : 5. What is the present age of Sumit?1. 21 years2. 26 years3. 27 years4. 23 years5. None of these
Answer» Correct Answer - Option 2 : 26 years Solution: Given: Total ages of Amit, Sumit and Puneet = 85 years Ratio of Amit, Sumit and Puneet ages five years ago = 2 : 3 : 5 Calculation: Let five years ago the ages of Amit, Sumit and Puneet is 2x, 3x and 5x respectively. Then, their present ages is (2x + 5) + (3x + 5) + (5x + 5) = 85 ⇒ 2x + 5 + 3x + 5 + 5x + 5 = 85 ⇒ 10x + 15 = 85 ⇒ 10x = 85 – 15 ⇒ 10x = 70 ⇒ x = 7 Present age of Sumit = 3x + 5 ∴ 26 years

240.

The total ages of Amit, Sumit and Puneet is 85 years. Five years ago the ratio of their ages was 2 : 3 : 5. What is the present age of Sumit?1. 21 years2. 26 years3. 27 years4. 23 years5. None of these

Answer» Correct Answer - Option 2 : 26 years

Solution:

Given:

Total ages of Amit, Sumit and Puneet = 85 years

Ratio of Amit, Sumit and Puneet ages five years ago = 2 : 3 : 5

Calculation:

Let five years ago the ages of Amit, Sumit and Puneet is 2x, 3x and 5x respectively.

Then, their present ages is (2x + 5) + (3x + 5) + (5x + 5) = 85

⇒ 2x + 5 + 3x + 5 + 5x + 5 = 85

⇒ 10x + 15 = 85

⇒ 10x = 85 – 15

⇒ 10x = 70

⇒ x = 7

Present age of Sumit = 3x + 5

∴ 26 years

Discussion

241.	Subsidiaries consider regional environment for policy / Strategy formulation is known as a. Polycentric Approach b. Regiocentric Approach c. Ethnocentric Approach d. Geocentric Approach
Answer» Subsidiaries consider regional environment for policy / Strategy formulation is known as Regiocentric Approach.

Discussion

242.	According to this theory the holdings of a country’s treasure primarily in the form of gold constituted its wealth. a. Gold Theory b. Ricardo Theory c. Mercantilism d. Hecksher Theory
Answer» c. Mercantilism

Discussion

243.	In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.182431497211033ABCDE What will come in place of ‘E’?1. 1412. 873. 1224. 1255. None of these
Answer» Correct Answer - Option 4 : 125 The pattern of the original number series is ⇒ 18 ⇒ 18 + (2² + 2) = 24 ⇒ 24 + (3² - 2) = 31 ⇒ 31 + (4² + 2) = 49 ⇒ 49 + (5² - 2) = 72 ⇒ 72 + (6² + 2) = 110 According the pattern of the above series the number series starting from the given number would be as: ⇒ 33 ⇒ A = 33 + (2² + 2) = 39 ⇒ B = 39 + (3² – 2) = 46 ⇒ C = 46 + (4² + 2) = 64 ⇒ D = 64 + (5² – 2) = 87 ⇒ E = 87 + (6² + 2) = 125 ∴ The required number in place of E would be 125.

243.

In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.182431497211033ABCDE What will come in place of ‘E’?1. 1412. 873. 1224. 1255. None of these

Answer» Correct Answer - Option 4 : 125

The pattern of the original number series is

⇒ 18

⇒ 18 + (2² + 2) = 24

⇒ 24 + (3² - 2) = 31

⇒ 31 + (4² + 2) = 49

⇒ 49 + (5² - 2) = 72

⇒ 72 + (6² + 2) = 110

According the pattern of the above series the number series starting from the given number would be as:

⇒ 33

⇒ A = 33 + (2² + 2) = 39

⇒ B = 39 + (3² – 2) = 46

⇒ C = 46 + (4² + 2) = 64

⇒ D = 64 + (5² – 2) = 87

⇒ E = 87 + (6² + 2) = 125

∴ The required number in place of E would be 125.

Discussion

244.	In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.42649012416220823ABCDE What will come in place of ‘C’?1. 1052. 1013. 654. 1045. None of these
Answer» Correct Answer - Option 1 : 105 The pattern of the original number series is ⇒ 42 ⇒ 42 + (11 × 2) = 64 ⇒ 64 + (13 × 2) = 90 ⇒ 90 + (17 × 2) = 124 ⇒ 124 + (19 × 2) = 162 ⇒ 162 + (23 × 2) = 208 According the pattern of the above series the number series starting from the given number would be as: ⇒ 23 ⇒ A = 23 + (11 × 2) = 45 ⇒ B = 45 + (13 × 2) = 71 ⇒ C = 71 + (17 × 2) = 105 ⇒ D = 105 + (19 × 2) = 143 ⇒ E = 143 + (23 × 2) = 189 ∴ The required number in place of ‘C’ would be 105.

244.

In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.42649012416220823ABCDE What will come in place of ‘C’?1. 1052. 1013. 654. 1045. None of these

Answer» Correct Answer - Option 1 : 105

The pattern of the original number series is

⇒ 42

⇒ 42 + (11 × 2) = 64

⇒ 64 + (13 × 2) = 90

⇒ 90 + (17 × 2) = 124

⇒ 124 + (19 × 2) = 162

⇒ 162 + (23 × 2) = 208

According the pattern of the above series the number series starting from the given number would be as:

⇒ 23

⇒ A = 23 + (11 × 2) = 45

⇒ B = 45 + (13 × 2) = 71

⇒ C = 71 + (17 × 2) = 105

⇒ D = 105 + (19 × 2) = 143

⇒ E = 143 + (23 × 2) = 189

∴ The required number in place of ‘C’ would be 105.

Discussion

245.	In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct options.11111212014721133683ABCDEWhat will come in place of ‘D’?1. 1812. 1873. 1834. 1895. None of these
Answer» Correct Answer - Option 3 : 183 The pattern of the original number series is ⇒ 111 ⇒ 111 + 1³ = 112 ⇒ 112 + 2³ = 120 ⇒ 120 + 3³ = 147 ⇒ 147 + 4³ = 211 ⇒ 211 + 5³ = 336 According the pattern of the above series the number series starting from the given number would be as: ⇒ 83 ⇒ A = 83 + 1³ = 84 ⇒ B = 84 + 2³ = 92 ⇒ C = 92 + 3³ = 119 ⇒ D = 119 + 4³ = 183 ⇒ E = 183 + 5³ = 308 ∴ The required number in place of ‘D’ would be 183.

245.

In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct options.11111212014721133683ABCDEWhat will come in place of ‘D’?1. 1812. 1873. 1834. 1895. None of these

Answer» Correct Answer - Option 3 : 183

The pattern of the original number series is

⇒ 111

⇒ 111 + 1³ = 112

⇒ 112 + 2³ = 120

⇒ 120 + 3³ = 147

⇒ 147 + 4³ = 211

⇒ 211 + 5³ = 336

According the pattern of the above series the number series starting from the given number would be as:

⇒ 83

⇒ A = 83 + 1³ = 84

⇒ B = 84 + 2³ = 92

⇒ C = 92 + 3³ = 119

⇒ D = 119 + 4³ = 183

⇒ E = 183 + 5³ = 308

∴ The required number in place of ‘D’ would be 183.

Discussion

246.	In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.85667469480336013ABCDE What will come in place of ‘D’?1. 4852. 7253. 4834. 7205. None of these
Answer» Correct Answer - Option 2 : 725 The pattern of the original number series is ⇒ 8 ⇒ 8 × 7 = 56 ⇒ 56 + 11 = 67 ⇒ 67 × 7 = 469 ⇒ 469 + 11 = 480 ⇒ 480 × 7 = 3360 According the pattern of the above series the number series starting from the given number would be as: ⇒ 13 ⇒ A = 13 × 7 = 91 ⇒ B = 91 + 11 = 102 ⇒ C = 102 × 7 = 714 ⇒ D = 714 + 11 = 725 ⇒ E = 725 × 7 = 5045 ∴ The required number in place of D would be 725.

246.

In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.85667469480336013ABCDE What will come in place of ‘D’?1. 4852. 7253. 4834. 7205. None of these

Answer» Correct Answer - Option 2 : 725

The pattern of the original number series is

⇒ 8

⇒ 8 × 7 = 56

⇒ 56 + 11 = 67

⇒ 67 × 7 = 469

⇒ 469 + 11 = 480

⇒ 480 × 7 = 3360

According the pattern of the above series the number series starting from the given number would be as:

⇒ 13

⇒ A = 13 × 7 = 91

⇒ B = 91 + 11 = 102

⇒ C = 102 × 7 = 714

⇒ D = 714 + 11 = 725

⇒ E = 725 × 7 = 5045

∴ The required number in place of D would be 725.

Discussion

247.	In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.151209607680614404915207ABCDE What will come in place of ‘C’?1. 76802. 22953. 44814. 34845. 3584
Answer» Correct Answer - Option 5 : 3584 The pattern of the original number series is ⇒ 15 × 8 = 120 ⇒ 120 × 8 = 960 ⇒ 960 × 8 = 7680 ⇒ 7680 × 8 = 61440 ⇒ 61440 × 8 = 491520 According the pattern of the above series the number series starting from the given number 7 would be as: ⇒ A = 7 × 8 = 56 ⇒ B = 56 × 8 = 448 ⇒ C = 448 × 8 = 3584 ∴ The required number in place of C would be 3584.

247.

In the following question, a number series is given, after the number series, a number and then A, B, C, D and E are given. Complete the number series starting from the given number based on the pattern of the original number series and choose correct option.151209607680614404915207ABCDE What will come in place of ‘C’?1. 76802. 22953. 44814. 34845. 3584

Answer» Correct Answer - Option 5 : 3584

The pattern of the original number series is

⇒ 15 × 8 = 120

⇒ 120 × 8 = 960

⇒ 960 × 8 = 7680

⇒ 7680 × 8 = 61440

⇒ 61440 × 8 = 491520

According the pattern of the above series the number series starting from the given number 7 would be as:

⇒ A = 7 × 8 = 56

⇒ B = 56 × 8 = 448

⇒ C = 448 × 8 = 3584

∴ The required number in place of C would be 3584.

Discussion

248.	A tank takes 24 hours to filled but due to leakage it takes 6 hours more to full it. How much time leakage will take to empty tank? If tank is full.1. 100 hrs.2. 120 hrs.3. 150 hrs.4. 125 hrs.5. 170 hrs.
Answer» Correct Answer - Option 2 : 120 hrs. Given: Time to fill tank = 24 hours due to leakage it takes 6 hours more to full. Calculations: Total work is done in 1 hour, ⇒ \(\frac{1}{{24}}\; - \;\frac{1}{X}\; = \;\frac{1}{{30}}\) ⇒ \(\frac{1}{{24}}\; - \;\frac{1}{{30}}\; = \frac{1}{X}\) ⇒ \(\frac{{30\; - \;24}}{{720}}\; = \;\frac{1}{x}\;\) ⇒ \(\frac{6}{{720}}\; = \;\frac{1}{x}\) ⇒ \(\frac{1}{{120}} = \;\frac{1}{x}\) ∴ It will take 120 hours to empty the tank.

248.

A tank takes 24 hours to filled but due to leakage it takes 6 hours more to full it. How much time leakage will take to empty tank? If tank is full.1. 100 hrs.2. 120 hrs.3. 150 hrs.4. 125 hrs.5. 170 hrs.

Answer» Correct Answer - Option 2 : 120 hrs.

Given:

Time to fill tank = 24 hours

due to leakage it takes 6 hours more to full.

Calculations:

Total work is done in 1 hour,

⇒ \(\frac{1}{{24}}\; - \;\frac{1}{X}\; = \;\frac{1}{{30}}\)

⇒ \(\frac{1}{{24}}\; - \;\frac{1}{{30}}\; = \frac{1}{X}\)

⇒ \(\frac{{30\; - \;24}}{{720}}\; = \;\frac{1}{x}\;\)

⇒ \(\frac{6}{{720}}\; = \;\frac{1}{x}\)

⇒ \(\frac{1}{{120}} = \;\frac{1}{x}\)

∴ It will take 120 hours to empty the tank.

Discussion

249.	If 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4, then θ = ______1. 3π/42. π/33. π/44. π/6
Answer» Correct Answer - Option 2 : π/3 Concept: The sum of an infinite G.P. is S = \(\rm a\over 1-r\) Where a is the first term of the series and r is the common ratio Calculation: Given 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4 First term of the G.P is a = 1 and r = sin θ ∴ \(\rm 1\over 1 - \sin θ\) = 2√3 + 4 1 - sin θ = \(1\over2\sqrt3 + 4\) 1 - sin θ = \({1\over2\sqrt3 + 4} \times {2\sqrt3 - 4\over 2\sqrt3 - 4}\) 1 - sin θ = \( {4-2\sqrt3 \over 16-12}\) 1 - sin θ = \( 1-{\sqrt3 \over 2}\) sin θ = \( {\sqrt3 \over 2}\) ⇒ θ = \(\boldsymbol{\pi\over3}\)

249.

If 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4, then θ = ______1. 3π/42. π/33. π/44. π/6

Answer» Correct Answer - Option 2 : π/3

Concept:

The sum of an infinite G.P. is

S = \(\rm a\over 1-r\)

Where a is the first term of the series and r is the common ratio

Calculation:

Given 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4

First term of the G.P is a = 1 and r = sin θ

∴ \(\rm 1\over 1 - \sin θ\) = 2√3 + 4

1 - sin θ = \(1\over2\sqrt3 + 4\)

1 - sin θ = \({1\over2\sqrt3 + 4} \times {2\sqrt3 - 4\over 2\sqrt3 - 4}\)

1 - sin θ = \( {4-2\sqrt3 \over 16-12}\)

1 - sin θ = \( 1-{\sqrt3 \over 2}\)

sin θ = \( {\sqrt3 \over 2}\)

⇒ θ = \(\boldsymbol{\pi\over3}\)

Discussion

250.	Two vessels A and B contains spirit and water in the ratio 1 : 2 and 5 : 11 respectively. These contents are mixed in the ratio 3 : 4. What is the ratio of spirit and water in the new mixture?1. 3 : 162. 16 : 33. 9 : 194. 19 : 9
Answer» Correct Answer - Option 3 : 9 : 19 Given: Two vessels A and B contains spirit and water in the ratio 1 : 2 and 5 : 11 respectively. There contents are mixed in the ratio 3 : 4. Calculation: Let the ratio of spirit in the new mixture be 'x' Also, the spirit in vessel A be 1/3 and in vessel B be 5/16 ⇒ (x - 5/16)/(1/3 - x) = 3/4 ⇒ (16x - 5)/4 = (1 - 3x) ⇒ 16x - 5 = 4 - 12x ⇒ 28x = 9 ⇒ x = 9/28 ⇒ x = 9/(9 + 19) ∴ Required ratio of spirit and water in the new mixture = 9 : 19

250.

Two vessels A and B contains spirit and water in the ratio 1 : 2 and 5 : 11 respectively. These contents are mixed in the ratio 3 : 4. What is the ratio of spirit and water in the new mixture?1. 3 : 162. 16 : 33. 9 : 194. 19 : 9

Answer» Correct Answer - Option 3 : 9 : 19

Given:

Two vessels A and B contains spirit and water in the ratio 1 : 2 and 5 : 11 respectively. There contents are mixed in the ratio 3 : 4.

Calculation:

Let the ratio of spirit in the new mixture be 'x'

Also, the spirit in vessel A be 1/3 and in vessel B be 5/16

⇒ (x - 5/16)/(1/3 - x) = 3/4

⇒ (16x - 5)/4 = (1 - 3x)

⇒ 16x - 5 = 4 - 12x

⇒ 28x = 9

⇒ x = 9/28

⇒ x = 9/(9 + 19)

∴ Required ratio of spirit and water in the new mixture = 9 : 19

Discussion

Explore topic-wise InterviewSolutions in .

What will be the angle of diffracting for the first minimum due to Fraunhofer diffraction with sources of light of wavelength `550nm` and slit of width `0.55mm`?A. 0.001 radB. 0.01 radC. 1 radD. 0.1 rad

What is the heat source of fluidized bed in tem tech thermal sand reclamation system?(a) Flame heating(b) Ultraviolet heating(c) Infrared heating(d) Conduction heating

Up to what accuracy can the temperature of the bed be controlled in tem tech thermal sand reclamation system?(a) ± 1%(b) ± 2%(c) ± 3%(d) ± 4%

Light of wavelength `6000 Å` is reflected at nearly normal incidence from a soap films of refractive index 1.4 The least thickness of the fringe then will appear black isA. infinityB. `200 Å`C. `2000 Å`D. `1000 Å`

What should be the width of each slit to obtain n maxima of double slit pattern within the central maxima of single slit pattern?

1 – cos2A =(A) sin2A(B) 2sin2A(C) cos2A(D) 2cos2A

The number of terms in an A.P. 51, 48, 45, …, 6 is – (A) 14 (B) 16 (C) 17 (D) 18

\( \int_{\pi}^{0} \frac{\sqrt{x^{2}+1}\left(\log \left(x^{2}+1\right)-2 \log x\right)}{x^{4}} d x \)

The number of odd numbers between 0 and 50 is – (A) 25 (B) 26 (C) 24 (D) 27

In right ∆ABC, ∠B=90°, AB = √15cm, BC=1Cm and AC = 4cm then tan A will be equal to –(A) √15/4(B) √15/1(C) 1/√15(D) 4/√15

find the domain and range of f(x)=x²-y/x-2

Evaluate : `(cos^(2)(45^(@) + theta)+ cos^(2) (45^(@) -theta))/(tan(60^(@) + theta) xx tan(30^(@) - theta))+(cot30^(@) + sin 90^(@))xx (tan60^(@) -sec0^(@))`

sin2A =(A) 2tanA/1+tan2A(B) cos2A – sin2A(C) 2cos2A(D) 2sin2A

Evaluate \( \int_{0}^{2}|4 x-5| d x \).

Solvepx + qy = p - qqx - py = p + q

The value of cot15. Cot75 .cot60. Cot30 +tan45 is equal to

If sina=cosa find the value of 2tana + cos*2a

solve 1≤|x-2|≤3

The minimum value of z = 3x – 2y subject to constraints 2x + y ≤ 4, x ≥ 0, y ≥ 0 is (A) 6 (B) -6 (C) -8 (D) -12

The minimum value of P = 6x + 16y subject to the constraints x ≤ 40, y ≥ 20 and x, y ≥ 0 is (a) 240 (b) 320 (c) 0 (d) None of these

The maximum value of P = 40x + 50y subject to the constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0 and y ≥ 0 is (a) 120 (b) 230 (c) 200 (d) None of these

A person invests Rs. 20,000 at 20 %, Rs. 150 shares at a premium of Rs. 50. The income from these shares in Rs. is (a) 1,000 (b) 3,000 (c) 1,500 (d) 2,000

Find the value of k if x + k is a factor of the polynomial x3 + kx2 - 2x + k + 5.

The total investment made in buying Rs. X shares at a premium of 25 % is Rs. 125 X. The number of shares bought is (a) 500 (b) 250 (c) 125 (d) 100

Reduce the given Rational expressions to its lowest form (i) (x3a - 8)/(x2a + 2xa + 4) (ii) (10x3 - 25x2 + 4x - 10)/(-4 - 10x2 )

Find the value of "a" in the polynomial 2a + 2xa + 5a + 10 if (a + x) is one of its factors.

Ramesh opens a Saving Bank Account on 16.06.2007 with a deposit of Rs. 700. He deposited Rs. 1,500 on 07.07.2007. find the amount on which he would receive the interest at the end of July, 2007 (a) Rs. 700 (b) Rs. 1,500 (c) Rs. 2,200 (d) Rs. 800

Indian GST model has ____ rate structure (a) 3 (b) 4 (c) 5 (d) 6

Mother, Father and Son line up at random for a family picture. E: Son on one end F: Father in middlethen P(E/F) is(a) 0 (b) 1 (c) 1/2 (d) 1/3

Four years ago, the ratio of half the age of A to 1/3 of the age of B is 8 ∶ 7. Also half of the present age of B exceeds 1/3 of present age of A by 11. Find the present age of A?1. 462. 363. 664. 565. 48

Three years ago the ratio of ages of Gargi and Supriya is 2 ∶ 1, also the present age of Gargi is 27 years more than Supriya. What is the sum of their present ages?1. 892. 683. 874. 935. 105

Ghanshyam borrow a sum of Rs. 13,360 at 8.75 % per annum compound interest and paid back in two instalments. What was the amount of each instalment1. 75692. 84513. 70024. 70215. 1520

A fan is sold for Rs. 900 cash payment of Rs. 200 down payment followed by two equal monthly installments of each Rs. 375. The annual rate of interest is ____ (approximately) (a) 25 % (b) 30 % (c) 54 % (d) 59 %

Ratio of Nidhi’s age and Amit’ age is 2 ∶ 3. Nidhi's age was 16 years 4 years ago. What is the present age of Amit?1. 30 years2. 32 years3. 33 years4. 35 years5. 28 years

Rohan purchased a mobile Rs. 5000 cash or Rs. 500 down payment followed by 4 equal instalments. If the rate 25 % per annum at simple interest calculate monthly instalment and find the nearest integer value. 1. 11822. 17203. 15224. 11555. 2102

The Theory of Absolute Cost Advantage is given by a. David Ricardo b. Adam Smithc. F W Taylor d. Ohlin and Heckscher

The total ages of Amit, Sumit and Puneet is 85 years. Five years ago the ratio of their ages was 2 : 3 : 5. What is the present age of Sumit?1. 21 years2. 26 years3. 27 years4. 23 years5. None of these

Subsidiaries consider regional environment for policy / Strategy formulation is known as a. Polycentric Approach b. Regiocentric Approach c. Ethnocentric Approach d. Geocentric Approach

According to this theory the holdings of a country’s treasure primarily in the form of gold constituted its wealth. a. Gold Theory b. Ricardo Theory c. Mercantilism d. Hecksher Theory

A tank takes 24 hours to filled but due to leakage it takes 6 hours more to full it. How much time leakage will take to empty tank? If tank is full.1. 100 hrs.2. 120 hrs.3. 150 hrs.4. 125 hrs.5. 170 hrs.

If 1 + sin θ + sin2 θ + ... upto ∞ = 2√3 + 4, then θ = ______1. 3π/42. π/33. π/44. π/6

Two vessels A and B contains spirit and water in the ratio 1 : 2 and 5 : 11 respectively. These contents are mixed in the ratio 3 : 4. What is the ratio of spirit and water in the new mixture?1. 3 : 162. 16 : 33. 9 : 194. 19 : 9