100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 103 InterviewSolution

5101.	If α, β are the roots of ax2 + bx + c = 0 and −α, γ are the roots of a1x2 + b1x + c1 = 0 then β, γ are roots of Ax2 + x + C = 0, where C-1 =
Answer» \(\because\) α and ß are roots of ax² + bx + c = 0 \(\therefore\) α + ß = -b/a and α ß = c/a Also, -α & \(\gamma\) are roots of a₁x² + b₁x + c₁ = 0 \(\therefore\) -α + \(\gamma\) = \(\frac{-b_1}{a_1}\) and -α\(\gamma\) = c₁/a₁ Again, ß and \(\gamma\) are roots of Ax² + x + c = 0 \(\therefore\) ß + \(\gamma\) = -1/A ⇒ A = -(ß + \(\gamma\)) and ß\(\gamma\) = C/A \(\therefore\) C = AB\(\gamma\) = -(ß + \(\gamma\))ß\(\gamma\) (\(\because\) A = -(ß + \(\gamma\))) \(\therefore\) C^-1 = \(\frac{-1}{ß\gamma(ß+\gamma)}\)

5101.

If α, β are the roots of ax2 + bx + c = 0 and −α, γ are the roots of a1x2 + b1x + c1 = 0 then β, γ are roots of Ax2 + x + C = 0, where C-1 =

Answer»

\(\because\) α and ß are roots of ax² + bx + c = 0

\(\therefore\) α + ß = -b/a and α ß = c/a

Also, -α & \(\gamma\) are roots of a₁x² + b₁x + c₁ = 0

\(\therefore\) -α + \(\gamma\) = \(\frac{-b_1}{a_1}\) and -α\(\gamma\) = c₁/a₁

Again, ß and \(\gamma\) are roots of Ax² + x + c = 0

\(\therefore\) ß + \(\gamma\) = -1/A ⇒ A = -(ß + \(\gamma\))

and ß\(\gamma\) = C/A

\(\therefore\) C = AB\(\gamma\) = -(ß + \(\gamma\))ß\(\gamma\) (\(\because\) A = -(ß + \(\gamma\)))

\(\therefore\) C^-1 = \(\frac{-1}{ß\gamma(ß+\gamma)}\)

Discussion

5102.	Find value of ' \( K \) ' such that \( x^{2}+2(K-1) x+k+5=0 \) (a) has \( x=2 \) lying inside the roots (b) has one root less than \( 2 \& \) other greater than 2. (c) has roots of opposite signs. (d) has roots such that 5 lies outside the roots. (e) has both roots greater than 5. (f) has exactly one root in \( (1,3) \).
Answer» Given quadratic equation is \(x^2 + 2(k - 1)x + (k + 5) = 0\) By comparing with ax² + bx + c = 0, we obtain \(a = 1>0, b=2(k - 1)\) & \(c = k + 5\) Let α and \(\beta \) are roots of quadratic equation . Let \(f(x) = x^2 + 2(k - 1)x + k + 5\). (a) Given that \(\alpha < 2 < \beta\) Required condition are (i) \(D < 0\) (ii) \(a f(2) < 0\) (i) \(D = b^2 - 4ac > 0\) ⇒ \(4(k - 1)^2 - 4(k + 5) > 0\) ⇒ \(4k^2 - 8k + 4 -4k - 20 > 0\) ⇒ \(4k^2- 12k - 16> 0\) ⇒ \(k^2 - 3k - 4 > 0\) ⇒ \((k - 4)(k + 1) > 0\) ⇒ \(k < -1 \;or\; k> 4\) .....(1) (ii) \(a f(2) < 0\) ⇒ \(1.(4 + 4k - 4 + k + 5) < 0\) ⇒ \(5k + 1 < 0\) ⇒ \(k < \frac{-1}5\) ......(2) \(\therefore\) From (1) & (2), we get \(k< - 1\) ⇒ \(k \in (-\infty, -1)\) (b) Given that \(\alpha < 2 < \beta\) same as question part (a). \(\therefore\) \(k \in (-\infty, -1)\). (c) Quadratic equation has roots of opposite sign. \(\therefore\) Product of roots = negative < 0. ⇒ \(\frac ca < 0\) ⇒ \(\frac{k + 5}1 < 0\) ⇒ \(k< - 5\) ⇒ \(k \in(-\infty, - 5)\) (d) Case I: \(\alpha, \beta < 5\) Required conditions are (i) \(D \ge 0\) (ii) \(a f(5) > 0\) (iii) \(5 > \frac{-b}{2a}\) (i) \(D \ge 0\) ⇒ \(4k^2 - 12k - 16 \ge 0\) ⇒ \(k^2 - 3k - 4 \ge 0\) ⇒ \((k - 4) (k + 1) \ge 0\) ⇒ \(k \le -1 \; or\; k \ge 4\) ....(1) (ii) \(a f(5) > 0\) ⇒ \(1(25 + 10k - 10 + 10) > 0\) ⇒ \(25 + 10k > 0\) ⇒ \(k > \frac{-25}{10}\) ⇒ \(k > \frac{-5}2\) .....(2) (iii) \(5 > \frac{-b}{2a}\) ⇒ \(5 > \frac{-2(k - 1)}{2(1)}\) ⇒ \(5 > 1 - k\) ⇒ \(k > - 4\) ....(3) From (1), (2) & (3), we obtain \(k\in(\frac{-5}2 , -1] \cup [4,\infty)\) .....(4) Case II: \(\alpha, \beta < 5\) Required conditions are (i) \(D \ge 0\) (ii) \(a f(5) > 0\) (iii) \(5 > \frac{-b}{2a}\) From (i), we get \(k \le -1\;or\; k\ge 4 \) .....(1b) From (ii), we get \(k > \frac{-5}2\) .....(2b) From (iii), we get \(k < -4\) ......(3b) From(1b), (2b), (3b), we obtain \(x\in \phi\) ....(5) \(\therefore\) From (4) & (5), we get \(x\in (\frac{-5}2, -1] \cup [4, \infty)\). (e) has both roots greater than 5 is solution of part (d) case II. \(\therefore\) This case is not possible because solution set is empty set. \(\therefore\) Both roots of given quadratic equation never be greater than 5. (f) Quadratic equation has exactly one root in (1, 3). \(\therefore\) \(f(1) f(3) < 0\) ⇒ \((1 + 2k - 2 + 5 + k) (9 + 6k - 6 + k + 5) < 0\) ⇒ \((3k + 4) (7k + 8) < 0\) ⇒ \(\frac{-4}3 < k < \frac{-8}7\) ⇒ \(k \in (\frac{-4}3, \frac{-8}7)\).

5102.

Find value of ' \( K \) ' such that \( x^{2}+2(K-1) x+k+5=0 \) (a) has \( x=2 \) lying inside the roots (b) has one root less than \( 2 \& \) other greater than 2. (c) has roots of opposite signs. (d) has roots such that 5 lies outside the roots. (e) has both roots greater than 5. (f) has exactly one root in \( (1,3) \).

Answer»

Given quadratic equation is

\(x^2 + 2(k - 1)x + (k + 5) = 0\)

By comparing with ax² + bx + c = 0, we obtain

\(a = 1>0, b=2(k - 1)\) & \(c = k + 5\)

Let α and \(\beta \) are roots of quadratic equation .

Let \(f(x) = x^2 + 2(k - 1)x + k + 5\).

(a) Given that \(\alpha < 2 < \beta\)

Required condition are

(i) \(D < 0\)

(ii) \(a f(2) < 0\)

(i) \(D = b^2 - 4ac > 0\)

⇒ \(4(k - 1)^2 - 4(k + 5) > 0\)

⇒ \(4k^2 - 8k + 4 -4k - 20 > 0\)

⇒ \(4k^2- 12k - 16> 0\)

⇒ \(k^2 - 3k - 4 > 0\)

⇒ \((k - 4)(k + 1) > 0\)

⇒ \(k < -1 \;or\; k> 4\) .....(1)

(ii) \(a f(2) < 0\)

⇒ \(1.(4 + 4k - 4 + k + 5) < 0\)

⇒ \(5k + 1 < 0\)

⇒ \(k < \frac{-1}5\) ......(2)

\(\therefore\) From (1) & (2), we get

\(k< - 1\)

⇒ \(k \in (-\infty, -1)\)

(b) Given that \(\alpha < 2 < \beta\)

same as question part (a).

\(\therefore\) \(k \in (-\infty, -1)\).

(c) Quadratic equation has roots of opposite sign.

\(\therefore\) Product of roots = negative < 0.

⇒ \(\frac ca < 0\)

⇒ \(\frac{k + 5}1 < 0\)

⇒ \(k< - 5\)

⇒ \(k \in(-\infty, - 5)\)

(d) Case I: \(\alpha, \beta < 5\)

Required conditions are

(i) \(D \ge 0\)

(ii) \(a f(5) > 0\)

(iii) \(5 > \frac{-b}{2a}\)

(i) \(D \ge 0\)

⇒ \(4k^2 - 12k - 16 \ge 0\)

⇒ \(k^2 - 3k - 4 \ge 0\)

⇒ \((k - 4) (k + 1) \ge 0\)

⇒ \(k \le -1 \; or\; k \ge 4\) ....(1)

(ii) \(a f(5) > 0\)

⇒ \(1(25 + 10k - 10 + 10) > 0\)

⇒ \(25 + 10k > 0\)

⇒ \(k > \frac{-25}{10}\)

⇒ \(k > \frac{-5}2\) .....(2)

(iii) \(5 > \frac{-b}{2a}\)

⇒ \(5 > \frac{-2(k - 1)}{2(1)}\)

⇒ \(5 > 1 - k\)

⇒ \(k > - 4\) ....(3)

From (1), (2) & (3), we obtain

\(k\in(\frac{-5}2 , -1] \cup [4,\infty)\) .....(4)

Case II: \(\alpha, \beta < 5\)

Required conditions are

(i) \(D \ge 0\)

(ii) \(a f(5) > 0\)

(iii) \(5 > \frac{-b}{2a}\)

From (i), we get

\(k \le -1\;or\; k\ge 4 \) .....(1b)

From (ii), we get

\(k > \frac{-5}2\) .....(2b)

From (iii), we get

\(k < -4\) ......(3b)

From(1b), (2b), (3b), we obtain

\(x\in \phi\) ....(5)

\(\therefore\) From (4) & (5), we get

\(x\in (\frac{-5}2, -1] \cup [4, \infty)\).

(e) has both roots greater than 5 is solution of part (d) case II.

\(\therefore\) This case is not possible because solution set is empty set.

\(\therefore\) Both roots of given quadratic equation never be greater than 5.

(f) Quadratic equation has exactly one root in (1, 3).

\(\therefore\) \(f(1) f(3) < 0\)

⇒ \((1 + 2k - 2 + 5 + k) (9 + 6k - 6 + k + 5) < 0\)

⇒ \((3k + 4) (7k + 8) < 0\)

⇒ \(\frac{-4}3 < k < \frac{-8}7\)

⇒ \(k \in (\frac{-4}3, \frac{-8}7)\).

Discussion

5103.	If an integer p is chosen at random in the interval `0le ple5,` then the probality that the roots of the equation `x^(2)+px+(p)/(4)+(1)/(2)=0` are real is -A. ` (4)/(5)`B. ` (2)/(3)`C. ` (3)/(5)`D. None of these
Answer» Correct Answer - B `DrArrP^(2)-((p)/(4)+(1)/(2))ge0` `(p-2)*(p+1)ge0` `pge-1"or" pge0` In `0leple5,` possible values of p are 2,3,4,5 Thus , probability `=(4)/(6)=(2)/(3)`

Discussion

5104.	If α, β are the zeros of the quadratic plynimial p(x) = x2 - (k + 6)x + 2(2k - 1) then the value of k,if α + β = \(\frac12\)αβ, is(a) -7(b) 7(c) -3(d) 3
Answer» Correct option is: (d) 3

Discussion

5105.	The roots α and β of the quadratic equation x2 - 5x + 3(k - 1) = 0 are such that αβ = 1. Find the value k.
Answer» α + β = 5 --- (1) α - β = 1 --- (2) Solving (1) and (2), we get α = 3 and β = 2 Also αβ = 6 Or 3(k - 1) = 6 k - 1 = 2 k = 3

5105.

The roots α and β of the quadratic equation x2 - 5x + 3(k - 1) = 0 are such that αβ = 1. Find the value k.

Answer»

α + β = 5 --- (1)

α - β = 1 --- (2)

Solving (1) and (2), we get

α = 3 and β = 2

Also αβ = 6

Or 3(k - 1) = 6

k - 1 = 2

k = 3

Discussion

5106.	If Sin(A+B)=1 and Cos(A-B)=root 3/2, A>B find the acute angles A & B
Answer» We have Sin (A+B)=1 (A+B)=sin^-1 1 We know sin^-11 = 90 A+B=90 --------(1) Cos (A-B) =√3/2 A-B=cos^-1 √3/2 We know cos^-1 √3/2=30 A-B=30 -----------(2) Solve (1) & (2) A+B=90 A-B=30 add 2A=120 A=60 & B=30 Sin (A+B) = 1 and Cos (A-B) = √3/2 Sin (A+B) = 1 Sin (A+B) = Sin 90^o A + B = 90^o A = 90^o - B ............. (i) Cos (A-B) = √3/2 Cos (A-B) = Cos 30^o A - B = 30^o A = 30^o + B .............(ii) In equation (i) and (ii), 90^o - B = 30^o + B 90^o - 30^o = B + B 60^o = 2B B = 60^o/2 = 30^o Putting the value of B is eq (i) A = 90^o - B = 90^o - 30^o = 60^o Hence, A = 60^o and B = 30^o

5106.

If Sin(A+B)=1 and Cos(A-B)=root 3/2, A>B find the acute angles A & B

Answer»

We have Sin (A+B)=1

(A+B)=sin^-1 1

We know sin^-11 = 90

A+B=90 --------(1)

Cos (A-B) =√3/2

A-B=cos^-1 √3/2

We know cos^-1 √3/2=30

A-B=30 -----------(2)

Solve (1) & (2)

A+B=90

A-B=30 add

2A=120

A=60 & B=30

Sin (A+B) = 1 and Cos (A-B) = √3/2

Sin (A+B) = 1

Sin (A+B) = Sin 90^o

A + B = 90^o

A = 90^o - B ............. (i)

Cos (A-B) = √3/2

Cos (A-B) = Cos 30^o

A - B = 30^o

A = 30^o + B .............(ii)

In equation (i) and (ii),

90^o - B = 30^o + B

90^o - 30^o = B + B

60^o = 2B

B = 60^o/2 = 30^o

Putting the value of B is eq (i)

A = 90^o - B

= 90^o - 30^o

= 60^o

Hence,

A = 60^o and B = 30^o

Discussion

5107.	calculate the current in a circuit if 500 c charge passes through it in 10 minutes
Answer» Charge (Q) = 500 C. Time(t) = 10 minutes. = 10 × 60 = 600 seconds. Using the Formula, Current = Charge/Time. ⇒ I = Q/t ∴ I = 500/600 ⇒ I = 5/6 ∴ I = 0.83 A.

5107.

calculate the current in a circuit if 500 c charge passes through it in 10 minutes

Answer»

Charge (Q) = 500 C.

Time(t) = 10 minutes.

= 10 × 60

= 600 seconds.

Using the Formula,

Current = Charge/Time.

⇒ I = Q/t

∴ I = 500/600

⇒ I = 5/6

∴ I = 0.83 A.

Discussion

5108.	class : 100-120 120-140 140-160 160-180 180-200frequency :10 p 10 q 5 = 60
Answer» By the problem 10+p+10+q+5=60 =>q =35-p Class . Mid value. Freq 100-120. 110 10 120-140 130 p 140-160 . 150. 10 160- 180 . 170. 35 -p 180-200 . 190. 5 Given mean = 145 11010+130p+15010+170(35-p)+1905=14560 =>50p= 1100+1500+5950+950-8700 =>50p=800 =>p =16

Discussion

5109.	48x2 - 13x -1 = 0
Answer» 48x² - 13x -1 = 0 48x² - 16x + 3x - 1 = 0 16x (3x - 1) + 1 (3x - 1) = 0 (16x + 1) (3x - 1) = 0 (16x +1) = 0 or (3x-1) = 0 16x = -1 or 3x = 1 x = -1/16 or x = 1/3 x = -1/16, 1/3

5109.

48x2 - 13x -1 = 0

Answer»

48x² - 13x -1 = 0

48x² - 16x + 3x - 1 = 0

16x (3x - 1) + 1 (3x - 1) = 0

(16x + 1) (3x - 1) = 0

(16x +1) = 0 or (3x-1) = 0

16x = -1 or 3x = 1

x = -1/16 or x = 1/3

x = -1/16, 1/3

Discussion

5110.	in the given figure two triangles ABC and DBC lie on same base of BC such that PQ parallel BA and PR parallel BD.prove that QR parallel AD
Answer» I think you are looking for this question: Click here: -> https://bit.ly/2EuPdLI

Discussion

5111.	Three pipes A, B, and C can fill a tank in 12 hours, 15 hours, and 20 hours respectively. If pipe A is opened all the time and pipe B and C are open for one hour each alternately. Pipe B opened first then pipe C opened alternatively. At what time the tank will be full?1. 9 hours2. 8 hours3. 6 hours4. 7 hours5. None of these
Answer» Correct Answer - Option 4 : 7 hours Given: Pipe A can fill a tank in = 12 hours Pipe B can fill a tank in = 15 hours Pipe C can fill a tank in = 20 hours Calculation: Let the total capacity of the tank be LCM (12, 15, 20) = 60 litres Now, Tank filled by A in 1 hour = 60/12 ⇒ 5 litre/hour Tank filled by B in 1 hour = 60/15 ⇒ 4 litre/hour Tank filled by C in 1 hour = 60/20 ⇒ 3 litre/hour Now, According to the question, For first-hour pipe A and B is opened = (5 + 4) litre ⇒ 9 litre For second-hour pipe A and C is opened = (5 + 3) litre ⇒ 8 For third-hour pipe A and B is opened = (5 + 4) litre ⇒ 9 litre For fourth-hour pipe A and C is opened = (5 + 3) litre ⇒ 8 Similarly, So…. On Now, We conclude that – 2 hours = 17 litre × 3 × 3 6 hours = 51 litre Now, The remaining work = (60 – 51) ⇒ 9 litre Again 7^th hour A and B are opened and fill the remaining tank ⇒ 9/9 ⇒ 1 hour Now, The tank fills in = (6hour + 1 hour) ⇒ 7 hours ∴ The tank will be fill in 7 hours

5111.

Three pipes A, B, and C can fill a tank in 12 hours, 15 hours, and 20 hours respectively. If pipe A is opened all the time and pipe B and C are open for one hour each alternately. Pipe B opened first then pipe C opened alternatively. At what time the tank will be full?1. 9 hours2. 8 hours3. 6 hours4. 7 hours5. None of these

Answer» Correct Answer - Option 4 : 7 hours

Given:

Pipe A can fill a tank in = 12 hours

Pipe B can fill a tank in = 15 hours

Pipe C can fill a tank in = 20 hours

Calculation:

Let the total capacity of the tank be LCM (12, 15, 20) = 60 litres

Now,

Tank filled by A in 1 hour = 60/12

⇒ 5 litre/hour

Tank filled by B in 1 hour = 60/15

⇒ 4 litre/hour

Tank filled by C in 1 hour = 60/20

⇒ 3 litre/hour

Now,

According to the question,

For first-hour pipe A and B is opened = (5 + 4) litre

⇒ 9 litre

For second-hour pipe A and C is opened = (5 + 3) litre

⇒ 8

For third-hour pipe A and B is opened = (5 + 4) litre

⇒ 9 litre

For fourth-hour pipe A and C is opened = (5 + 3) litre

⇒ 8

Similarly, So…. On

Now,

We conclude that –

2 hours = 17 litre

× 3 × 3

6 hours = 51 litre

Now,

The remaining work = (60 – 51)

⇒ 9 litre

Again 7^th hour A and B are opened and fill the remaining tank

⇒ 9/9

⇒ 1 hour

Now,

The tank fills in = (6hour + 1 hour)

⇒ 7 hours

∴ The tank will be fill in 7 hours

Discussion

5112.	Three pipes A, B and C can fill a tank in 12 hours, 15 hours, and 20 hours respectively. If pipe A is opened all the time and pipe B and C are open for one hour each, alternatively, with pipe A. In how much time the tank will be full?1. 9 hours2. 8 hours3. 6 hours4. 7 hours5. None of these
Answer» Correct Answer - Option 4 : 7 hours Given: Pipe A can fill a tank in = 12 hours Pipe B can fill a tank in = 15 hours Pipe C can fill a tank in = 20 hours Calculation: Let the total capacity of tank be LCM (12, 15, 20) = 60 litres Now, Tank filled by A in 1 hour = 60/12 = 5 litres Tank filled by B in 1 hour = 60/15 = 4 litres Tank empty by C in 1 hour = 60/20 = 3 litres Now, According to the question, For first hour pipe A and B is opened = (5 + 4) litres = 9 litres For second hour pipe A and C is opened = (5 + 3) litres = 8 litres Hence, for every 2 hours, volume filled in the tank = (9 + 8) litres = 17 litres So, in (2 × 3) = 6 hours,volume of the tank filled = 3 × 17 = 51 litres The remaining volume to be filled = (60 – 51) = 9 litres Again in the 7^th hour, A and B are opened and fill the remaining tank in time = 9/9 = 1 hour Now, the time taken to fill the tank = 6 hour + 1 hour = 7 hours ∴ The tank will be fill in 7 hours

5112.

Three pipes A, B and C can fill a tank in 12 hours, 15 hours, and 20 hours respectively. If pipe A is opened all the time and pipe B and C are open for one hour each, alternatively, with pipe A. In how much time the tank will be full?1. 9 hours2. 8 hours3. 6 hours4. 7 hours5. None of these

Answer» Correct Answer - Option 4 : 7 hours

Given:

Pipe A can fill a tank in = 12 hours

Pipe B can fill a tank in = 15 hours

Pipe C can fill a tank in = 20 hours

Calculation:

Let the total capacity of tank be LCM (12, 15, 20) = 60 litres

Now,

Tank filled by A in 1 hour = 60/12 = 5 litres

Tank filled by B in 1 hour = 60/15 = 4 litres

Tank empty by C in 1 hour = 60/20 = 3 litres

Now,

According to the question,

For first hour pipe A and B is opened = (5 + 4) litres = 9 litres

For second hour pipe A and C is opened = (5 + 3) litres = 8 litres

Hence, for every 2 hours, volume filled in the tank = (9 + 8) litres = 17 litres

So, in (2 × 3) = 6 hours,volume of the tank filled = 3 × 17 = 51 litres

The remaining volume to be filled = (60 – 51) = 9 litres

Again in the 7^th hour, A and B are opened and fill the remaining tank in time = 9/9 = 1 hour

Now, the time taken to fill the tank = 6 hour + 1 hour = 7 hours

∴ The tank will be fill in 7 hours

Discussion

5113.	Two pipes A and B can fill a tank in 15 hours and 18 hours, respectively. Both pipes are opened simultaneously to fill the tank. In how many hours will the empty tank be filled?1. \(8 \dfrac{2}{11}\)2. \(7 \dfrac{2}{11}\)3. \(9 \dfrac{2}{11}\)4. \(10 \dfrac{2}{11}\)
Answer» Correct Answer - Option 1 : \(8 \dfrac{2}{11}\) Given: Pipe A can fill in 15 hours. Pipe B can fill in 18 hours. Calculation: Tank filled by Pipe A in 1 hour ⇒ 1/15 Tank filled by Pipe B in 1 hour ⇒ 1/18 Tank filled by both Pipes in 1 hour ⇒ (1/15) + (1/18) ⇒ 11/90 Time taken By both pipes to fill the tank ⇒ 90/11 \(\Rightarrow 8\frac{2}{{11}}\;{\rm{hours}}\) ∴ Time taken by both pipes is \(8\frac{2}{{11}}\;{\bf{hours}}\).

5113.

Two pipes A and B can fill a tank in 15 hours and 18 hours, respectively. Both pipes are opened simultaneously to fill the tank. In how many hours will the empty tank be filled?1. \(8 \dfrac{2}{11}\)2. \(7 \dfrac{2}{11}\)3. \(9 \dfrac{2}{11}\)4. \(10 \dfrac{2}{11}\)

Answer» Correct Answer - Option 1 : \(8 \dfrac{2}{11}\)

Given:

Pipe A can fill in 15 hours.

Pipe B can fill in 18 hours.

Calculation:

Tank filled by Pipe A in 1 hour

⇒ 1/15

Tank filled by Pipe B in 1 hour

⇒ 1/18

Tank filled by both Pipes in 1 hour

⇒ (1/15) + (1/18)

⇒ 11/90

Time taken By both pipes to fill the tank

⇒ 90/11

\(\Rightarrow 8\frac{2}{{11}}\;{\rm{hours}}\)

∴ Time taken by both pipes is \(8\frac{2}{{11}}\;{\bf{hours}}\).

Discussion

5114.	Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours, then what percentage of the tank will remain unfilled?1. 17%2. 50%3. 33%4. 25%
Answer» Correct Answer - Option 4 : 25% Given: Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours Concept Used: Pipe and cistern Calculation: Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively To find the total capacity take lcm of three Lcm of 15,12 and 10 is 60 Total capacity = 60 unit which is 100% Efficiency of Pipe A = 60/15 = 4 Efficiency of Pipe B = 60/12 = 5 Efficiency of Pipe C = 60/10 = 6 As per the question, All three pipes were opened for 3 hours ⇒ (4 + 5 + 6) × 3 ⇒ 45 unit work done Remaining work ⇒ 60 - 45 = (15/60) × 100 = 25%

5114.

Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours, then what percentage of the tank will remain unfilled?1. 17%2. 50%3. 33%4. 25%

Answer» Correct Answer - Option 4 : 25%

Given:

Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours

Concept Used:

Pipe and cistern

Calculation:

Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively

To find the total capacity take lcm of three

Lcm of 15,12 and 10 is 60

Total capacity = 60 unit which is 100%

Efficiency of Pipe A = 60/15 = 4

Efficiency of Pipe B = 60/12 = 5

Efficiency of Pipe C = 60/10 = 6

As per the question,

All three pipes were opened for 3 hours

⇒ (4 + 5 + 6) × 3

⇒ 45 unit work done

Remaining work

⇒ 60 - 45 = (15/60) × 100 = 25%

Discussion

5115.	A train passes a station that is 270 m long in 90 sec. at a speed of 108 km / hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds
Answer» Correct Answer - Option 3 : 81 seconds Given: Length of the station = 270 m Time taken on crossing the station = 90 sec Speed = 108 km / h Concept: Crossing a bird feeder = crossing its own length Crossing a station = Crossing the combined length of train and station. Calculations: Speed of train = 108 km / h = 108 × 5 / 18 ⇒ 30 m / s Let the length of the train be x metres. Then, (x + 270) / 30 = 90 ⇒ x + 270 = 90 × 30 ⇒ x + 270 = 2700 ⇒ x = 2430 So, time taken by the train to cross the bird feeder ⇒ 2430 / 30 ⇒ 81 s The time taken by the train to cross the bird feeder is 81 seconds.

5115.

A train passes a station that is 270 m long in 90 sec. at a speed of 108 km / hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds

Answer» Correct Answer - Option 3 : 81 seconds

Given:

Length of the station = 270 m

Time taken on crossing the station = 90 sec

Speed = 108 km / h

Concept:

Crossing a bird feeder = crossing its own length

Crossing a station = Crossing the combined length of train and station.

Calculations:

Speed of train = 108 km / h = 108 × 5 / 18

⇒ 30 m / s

Let the length of the train be x metres.

Then, (x + 270) / 30 = 90

⇒ x + 270 = 90 × 30

⇒ x + 270 = 2700

⇒ x = 2430

So, time taken by the train to cross the bird feeder

⇒ 2430 / 30

⇒ 81 s

The time taken by the train to cross the bird feeder is 81 seconds.

Discussion

5116.	A train passes a station that is 270 m long in 90 sec. at a speed of 108 km/hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds5. 90 seconds
Answer» Correct Answer - Option 3 : 81 seconds Given: Length of the station = 270 m Time taken on crossing the station = 90 sec Speed = 108 km/h Calculations: Speed of train = 108 km/h = 108 × 5/18 ⇒ 30 m/s Let the length of the train be x metres. Then, (x + 270)/30 = 90 ⇒ x + 270 = 90 × 30 ⇒ x + 270 = 2700 ⇒ x = 2430 So, time taken by the train to cross the bird feeder ⇒ 2430 /30 ⇒ 81 s The time taken by the train to cross the bird feeder is 81 seconds.

5116.

A train passes a station that is 270 m long in 90 sec. at a speed of 108 km/hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds5. 90 seconds

Answer» Correct Answer - Option 3 : 81 seconds

Given:

Length of the station = 270 m

Time taken on crossing the station = 90 sec

Speed = 108 km/h

Calculations:

Speed of train = 108 km/h = 108 × 5/18

⇒ 30 m/s

Let the length of the train be x metres.

Then, (x + 270)/30 = 90

⇒ x + 270 = 90 × 30

⇒ x + 270 = 2700

⇒ x = 2430

So, time taken by the train to cross the bird feeder

⇒ 2430 /30

⇒ 81 s

The time taken by the train to cross the bird feeder is 81 seconds.

Discussion

5117.	Solve the equations and select the correct option:3x + 5y = 132x + y = 41. x > y2. x < y3. x = y4. x ≥ Y
Answer» Correct Answer - Option 2 : x < y Given: 3x + 5y = 12 2x + y = 4 Calculation: 3x + 5y = 13 ….(1) 2x + y = 4 ….(2) Multipying equation (2) by 5 ⇒ 10x + 5y = 20 ….(3) Subtracting equation 2 from 3 ⇒ 10x + 5y = 20 -3x + 5y = 13 ⇒ 7x = 7 ⇒ x = 1 Replacing value of x in equation 1 ⇒ 3 + 5y = 13 ⇒ 5y = 10 ⇒ y = 2 ∴ x < y Common mistake: While multiplying and adding or subtracting terms changes the constant value should be changed

5117.

Solve the equations and select the correct option:3x + 5y = 132x + y = 41. x > y2. x < y3. x = y4. x ≥ Y

Answer» Correct Answer - Option 2 : x < y

Given:

3x + 5y = 12

2x + y = 4

Calculation:

3x + 5y = 13 ….(1)

2x + y = 4 ….(2)

Multipying equation (2) by 5

⇒ 10x + 5y = 20 ….(3)

Subtracting equation 2 from 3

⇒ 10x + 5y = 20

-3x + 5y = 13

⇒ 7x = 7

⇒ x = 1

Replacing value of x in equation 1

⇒ 3 + 5y = 13

⇒ 5y = 10

⇒ y = 2

∴ x < y

Common mistake:

While multiplying and adding or subtracting terms changes the constant value should be changed

Discussion

5118.	Find the wrong term in the following number series.1150, 1342, 1527, 1705, 1876, 20411. 13422. 15273. 18764. 11505. 2041
Answer» Correct Answer - Option 5 : 2041 Given: 1150, 1342, 1527, 1705, 1876, 2041 Calculation: 1342 – 1150 = 192 1527 – 1342 = 185 1705 – 1527 = 178 1876 – 1705 = 171 2040 – 1876 = 164 ∴ The wrong term is 2041

5118.

Find the wrong term in the following number series.1150, 1342, 1527, 1705, 1876, 20411. 13422. 15273. 18764. 11505. 2041

Answer» Correct Answer - Option 5 : 2041

Given:

1150, 1342, 1527, 1705, 1876, 2041

Calculation:

1342 – 1150 = 192

1527 – 1342 = 185

1705 – 1527 = 178

1876 – 1705 = 171

2040 – 1876 = 164

∴ The wrong term is 2041

Discussion

5119.	The curve of angle of incidence versus angle of deviaton wshown has been plotted for prism. Q. The value of refractive index of the prism used isA. `sqrt(3)`B. `sqrt(2)`C. `sqrt(3)//sqrt(2)`D. `2//sqrt(3)`
Answer» `delta=i+e-A` `delta_(min)=60^(@)` when `i=e` `therefore60^(@)=2i-A=2(60^(@))-A` `thereforeA=60^(@)` `thereforemu=(sin((A+delta_(min))/(2)))/(sin((A)/(2)))=(sin((60+60)/(2)))/(sin((60)/(2)))=sqrt(3)`

Discussion

5120.	Assertion`:-` In general, heat of neutralization of `CH_(3)COOH` by `NaOH` is less than heat of neutralization of `H_(2)SO_(4)` by `NaOH`. Reason`:-` `H_(2)SO_(4)` is a di basic acid while `CH_(3)COOH` is monobasic acid.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False
Answer» Correct Answer - B

Discussion

5121.	Choose the correct code in which all the molecules have dative `pi` bond. I `- P_(4)O_(10)` , II `- (SiH_(3))_(3) N`. , III `-P_(4)O_(6)` , IV `-BF_(3)`A. I, II, III, IVB. II & IVC. I, III, & IVD. I, II & IV
Answer» Correct Answer - D Conceptual

Discussion

5122.	Which of the following displacement occur:A. `Zn+2H^(+)toZn^(2+)+H_(2)uarr`B. `Fe+2Ag^(+)toFe^(2+)+Agdarr`C. `Cu+Fe^(2+)toCu^(2+)+Fedarr`D. `Zn+Pb^(2+)toZn^(2+)+Pbdarr`
Answer» Correct Answer - A::B::D

Discussion

5123.	Select the correct statement(s) with respect to the `p pi-dpi` dative bond.A. In `(Ph_3Si)_2O`, the Si-O-Si group is nearly linearB. Silanols such as `(CH_3)_3` SiOH are stronger protonic acids then their carbon analogsC. Trisilysl phosphine `(H_3Si)_3` P is planarD. `CIO_4^-` does not polymerise at all
Answer» Correct Answer - ABD (A)Steric repulsions of bulkier groups and `ppi-dpi` dative bonding favour for a linear Si-O-Si group. (B)Due to stabilization of the conjugate base anion by O `(ppi)to Si(dpi)`bonding motion. ( C)In is pyramidal because `ppi-dpi` bonding is not effective due to the bigger size of phosphorus atom (D)Most effective due to small size of chlorine.

Discussion

5124.	Assertion:- `NO_(3)^(-)` is planar while `NH_(3)` is pyramidal Reason:- N in `NO_(3)^(-)` is `sp^(3)` and in `NH_(3)` it is `sp^(3)` hybridised with one ione pair.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False.
Answer» Correct Answer - A

Discussion

5125.	List of Basicity or protocity of an acid.
Answer» It is number of H⁺ ions furnished by a molecule of an acid. An acid may be classified according to its basicity. Thus we may have, (i) Mono basic or Mono protic acids like HCl, HNO₃, CH₃COOH, HCN etc. (ii) Dibasic or Diprotic acids like, H₂SO₄, H₂CO₃, H₂SO₃, H₂S etc. (iii) Tribasic or Triprotic acids like H₃PO₄ ,H₃AsO₄ etc.

5125.

List of Basicity or protocity of an acid.

Answer»

It is number of H⁺ ions furnished by a molecule of an acid. An acid may be classified according to its basicity. Thus we may have,

(i) Mono basic or Mono protic acids like HCl, HNO₃, CH₃COOH, HCN etc.

(ii) Dibasic or Diprotic acids like, H₂SO₄, H₂CO₃, H₂SO₃, H₂S etc.

(iii) Tribasic or Triprotic acids like H₃PO₄ ,H₃AsO₄ etc.

Discussion

5126.	What is Arrhenius Acid? Give its Examples.
Answer» Arrhenius Acid : Substance which gives H⁺ ion on dissolving in water (H⁺ donor) Ex: HNO₃, HClO₄, HCl, HI, HBr, H₂SO₄, H₃PO₄ etc.

5126.

What is Arrhenius Acid? Give its Examples.

Answer»

Arrhenius Acid : Substance which gives H⁺ ion on dissolving in water (H⁺ donor)

Ex: HNO₃, HClO₄, HCl, HI, HBr, H₂SO₄, H₃PO₄ etc.

Discussion

5127.	The basicity of phosphorous acid is : (A) 1 (B) 2 (C) 3 (D) 4
Answer» Correct option: (B) 2 Explanation: Phosphorous acid has two replaceable H⁺ ions.

5127.

The basicity of phosphorous acid is : (A) 1 (B) 2 (C) 3 (D) 4

Answer»

Correct option: (B) 2

Explanation:

Phosphorous acid has two replaceable H⁺ ions.

Discussion

5128.	The characteristics of an acid is : (A) turns blue litmus to red. (B) turns phenolphthalein pink from colourless. (C) decompose carbonates (D) oxy compounds of non-metals
Answer» Correct option: (A) turns blue litmus to red. Explanation: Statement (A) indicates characteristic of acid.

5128.

The characteristics of an acid is : (A) turns blue litmus to red. (B) turns phenolphthalein pink from colourless. (C) decompose carbonates (D) oxy compounds of non-metals

Answer»

Correct option: (A) turns blue litmus to red.

Explanation:

Statement (A) indicates characteristic of acid.

Discussion

5129.	Select the suitable reason (s) for higher strength of an acid or base : (A) higher value of Ka or Kb (B) higher extent of ionisation (C) (A) and (B) both (D) Larger number of replaceable H atoms.
Answer» Correct option: (C) (A) and (B) both Explanation: K_a or K_b and degree of ionisation are the measure of strength of an acid or base.

5129.

Select the suitable reason (s) for higher strength of an acid or base : (A) higher value of Ka or Kb (B) higher extent of ionisation (C) (A) and (B) both (D) Larger number of replaceable H atoms.

Answer»

Correct option: (C) (A) and (B) both

Explanation:

K_a or K_b and degree of ionisation are the measure of strength of an acid or base.

Discussion

5130.	Arrhenius theory of acid-base is not applicable in : (A) aqueous solution (B) in presence of water (C) non-aqueous solutions (D) none of the above
Answer» Correct option: (C) non-aqueous solutions Explanation: Since Arrhenius theory is only applicable to aqueous medium.

5130.

Arrhenius theory of acid-base is not applicable in : (A) aqueous solution (B) in presence of water (C) non-aqueous solutions (D) none of the above

Answer»

Correct option: (C) non-aqueous solutions

Explanation:

Since Arrhenius theory is only applicable to aqueous medium.

Discussion

5131.	What are the types of Arrhenius acids ?
Answer» Types of Arrhenius Acid are : (1) Mono basic acid : Gives single H⁺ per molecule e.g HA (HNO₃, HClO₄, HCl) (2) Di basic acid : Gives two H⁺ per molecule e.g H₂A(H₂SO₄, H₃PO₃) (3) Tri basic acid : Gives three H⁺ per molecule e.g H₃A(H₃PO₄)

5131.

What are the types of Arrhenius acids ?

Answer»

Types of Arrhenius Acid are :

(1) Mono basic acid : Gives single H⁺ per molecule e.g HA (HNO₃, HClO₄, HCl)

(2) Di basic acid : Gives two H⁺ per molecule e.g H₂A(H₂SO₄, H₃PO₃)

(3) Tri basic acid : Gives three H⁺ per molecule e.g H₃A(H₃PO₄)

Discussion

5132.	Define Arrhenius base.
Answer» Arrhenius base : Any substance which releases OH^– (hydroxyl) ion in water (OH^– ion donor) OH^– ion is present also in hydrated form of H₃O₂^-, H₇O₄^-, H₅O₃^- First group elements (except Li.) form strong bases

5132.

Define Arrhenius base.

Answer»

Arrhenius base : Any substance which releases OH^– (hydroxyl) ion in water (OH^– ion donor)

OH^– ion is present also in hydrated form of H₃O₂^-, H₇O₄^-, H₅O₃^-
First group elements (except Li.) form strong bases

Discussion

5133.	Which of the following statements is true in aqueous medium?A. `CH_(2)CH_(2)S^(-)` is a stronger base and more nucleophilic than `CH_(3)CH_(2)O^(-)`B. `CH_(3)CH_(2)S` is a stronger base but is less nucleophilic than `CH_(3)CH_(2)O^(-)`C. `CH_(3)CH_(2)S^(-)` is a weaker base but is more nucleophilic than `CH_(3)CH_(2)O^(-)`D. `CH_(3)CH_(2)S^(-)` is a weaker base and less nucleophilic than `CH_(3)CH_(2)O^(-)`
Answer» Correct Answer - C In protic solvent `CH_(3)CH_(2)S^(-)` is more nucleophilic than `CH_(3)CH_(2)O`:

Discussion

5134.	Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would she observe? A. The solution turned red. B. Yellow precipitate was formed. C. White precipitate was formed. D. The reaction mixture became hot.
Answer» Correct option is B. Yellow precipitate is formed The relation between lead nitrate and potassium iodide is an example of precipitation reaction. Pb (No₃)₂ + 2 KI \(\longrightarrow\) PbI₂ + 2KNO₃ (yellow ppt) The yellow ppt of load iodide can be recovered by filtration. Correct option is B. Yellow precipitate is formed

5134.

Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would she observe? A. The solution turned red. B. Yellow precipitate was formed. C. White precipitate was formed. D. The reaction mixture became hot.

Answer»

Correct option is B. Yellow precipitate is formed

The relation between lead nitrate and potassium iodide is an example of precipitation reaction.

Pb (No₃)₂ + 2 KI \(\longrightarrow\) PbI₂ + 2KNO₃ (yellow ppt)

The yellow ppt of load iodide can be recovered by filtration.

Correct option is B. Yellow precipitate is formed

Discussion

5135.	The average speed of nitrogen molecule is v. if the temperature is doubled and nitrogen molecule dissociate into nitrogen atoms completely then new average speed becoems.A. vB. `sqrt(2)v`C. `2v`D. 4v
Answer» `v_(mg)=sqrt((8)/(pi)(RT)/(M))` `N_(2)(g)rarr2N(g)` `((v_(avg))N)/((v_(avg))N_(2))sqrt((28xx2T)/(14xxT))=sqrt(4)=2` `(v_(avg))_(N)=2v`

Discussion

5136.	What kind of acid does not completely dissociate?
Answer» A weak acid is one that does not dissociate completely in solution; this means that a weak acid does not donate all of its hydrogen ions (H⁺) in a solution.

Discussion

5137.	A photoelectric plate is initially exposed to a spectrum of hydrogen gas excited to second energy level. Late when the same photoelectric plate is exposed to a spectrum of some unknown hydrogen like gas, excited to second energy level. It is found that the de-Broglie wavelength of the photoelectrons now ejected has increased `sqrt(6.1)` times. For this new gas difference of energies of first Lyman line and Balmer series limit is found to be two times the ionization energy of the hydrogen atom is ground state. Detect the atom and determine the work function of the photoelectric plate.
Answer» For minimum de-Broglie wavelength of electron, the momentum and hence `KE` of electron must be maximum. Case-I. When hydrogen spectra is used, for maximum energy of photoelectron `13.6[(1)/(1^(2))-(1)/(2^(2))]-phi=(1)/(2)mv_(1)^(2) rArr (1)/(2)mv_(1)^(2)=(3)/(4)xx13.6eV-phi` For unknown gas of atomic no.`z` `13.6z^(2)[(1)/(1^(2))-(1)/(2^(2))]-phi=(1)/(2)mv_(2)^(2) rArr (1)/(2)mv_(2)^(2)=z^(2)(3)/(4)xx13.6ev-phi` Comparing the de-Broglie wavelengths `(lambda_(1))/(lambda_(2))=(h//mv_(1))/(h//mv_(2))=((v_(2))/(v_(1)))` `rArr ((lambda_(1))/(lambda_(2)))^(2)=((3)/(4)xx13.6-phi)/((3)/(4)xx13.6z^(2)-phi)=(1)/(6.1)`....(`1`) For the unknown gas Energy of first lvman line-Energy of Balmer series limit `=2xx(13.6)eV` `(13.6)z^(2)((3)/(4))-13.6z^(2)((1)/(4))=2(13.6)` `z^(2)//2=2`, `z=2` `:. ` Unknown gas is helium Now from equation (`1`) `(10.2-phi)/(40.8-phi)=(1)/(6.1)` `rArr phi=4.2eV`

Discussion

5138.	Study the following table and answer the questions based on it.Production of various models of mobile phones of a company (in lakhs) per year in given years. ABCD200675494556200848605778201089596387201257797492What was the average production (in lakhs) of mobile phones in 2012?1. 63.752. 54.253. 61.54. 75.5
Answer» Correct Answer - Option 4 : 75.5 According to the given information : Production of mobile phones in 2012 = 57 + 79 + 74 + 92 = 302. Number of models = 4 (A, B, C, D) Average = (Total Production) / (Number of Models) Average production = 302 / 4 = 75.5. Hence, the correct answer is "75.5".

5138.

Study the following table and answer the questions based on it.Production of various models of mobile phones of a company (in lakhs) per year in given years. ABCD200675494556200848605778201089596387201257797492What was the average production (in lakhs) of mobile phones in 2012?1. 63.752. 54.253. 61.54. 75.5

Answer» Correct Answer - Option 4 : 75.5

According to the given information :

Production of mobile phones in 2012 = 57 + 79 + 74 + 92 = 302.

Number of models = 4 (A, B, C, D)

Average = (Total Production) / (Number of Models)

Average production = 302 / 4 = 75.5.

Hence, the correct answer is "75.5".

Discussion

5139.	The diameter of a cylindrical piston is 5 cm. The stroke length of the piston is 10 cm. Find the displacement volume, if π = 4.1. 125 cm32. 1000 cm33. 500 cm34. 250 cm3
Answer» Correct Answer - Option 4 : 250 cm³ Concept: Displacement volume: It is the volume covered while going from T.D.C to B.D.C. It is given by: \(\frac{\pi}{4}d^2L\) where d = diameter of the cylinder and L = stroke length of the piston. Calculation: Given: d = 5 cm, L = 10 cm Displacement volume: \(\frac{\pi}{4}d^2L\) \(\frac{4}{4}\times 5^2\times 10=250\,cm^3\) The relation mentioned in the question was ambiguous so we have made some assumptions as of now until the question gets clear.

5139.

The diameter of a cylindrical piston is 5 cm. The stroke length of the piston is 10 cm. Find the displacement volume, if π = 4.1. 125 cm32. 1000 cm33. 500 cm34. 250 cm3

Answer» Correct Answer - Option 4 : 250 cm³

Concept:

Displacement volume:

It is the volume covered while going from T.D.C to B.D.C. It is given by:

\(\frac{\pi}{4}d^2L\)

where d = diameter of the cylinder and L = stroke length of the piston.

Calculation:

Given:

d = 5 cm, L = 10 cm

Displacement volume:

\(\frac{\pi}{4}d^2L\)

\(\frac{4}{4}\times 5^2\times 10=250\,cm^3\)

The relation mentioned in the question was ambiguous so we have made some assumptions as of now until the question gets clear.

Discussion

5140.	How many associate banks were merged with the State Bank of India in the year 2017?1. 42. 53. 64. 7
Answer» Correct Answer - Option 2 : 5 The correct answer is 5. 5 associate banks were merged with the State Bank of India in the year 2017. These banks were State Bank of Bikaner and Jaipur (SBBJ), State Bank of Travancore (SBT), State Bank of Patiala (SBP), State Bank of Mysore (SBM), State Bank of Hyderabad (SBH). The merger was approved by the cabinet in June 2016 to make it a global-sized bank. The government order read “the entire undertaking of SBBJ, SBM, SBT, SBP and SBH shall stand transferred to and vested in the State Bank of India from April 01, 2017.” State Bank of Saurashtra was merged with SBI in 2008. State Bank of Indore was merged with it in 2010. State Bank of India Headquarters- Mumbai, Maharashtra Chairperson- Dinesh Kumar Khara

5140.

How many associate banks were merged with the State Bank of India in the year 2017?1. 42. 53. 64. 7

Answer» Correct Answer - Option 2 : 5

The correct answer is 5.

5 associate banks were merged with the State Bank of India in the year 2017.
- These banks were
  - State Bank of Bikaner and Jaipur (SBBJ),
  - State Bank of Travancore (SBT),
  - State Bank of Patiala (SBP),
  - State Bank of Mysore (SBM),
  - State Bank of Hyderabad (SBH).

The merger was approved by the cabinet in June 2016 to make it a global-sized bank.
- The government order read “the entire undertaking of SBBJ, SBM, SBT, SBP and SBH shall stand transferred to and vested in the State Bank of India from April 01, 2017.”
- State Bank of Saurashtra was merged with SBI in 2008.
- State Bank of Indore was merged with it in 2010.
State Bank of India
- Headquarters- Mumbai, Maharashtra
- Chairperson- Dinesh Kumar Khara

Discussion

5141.	Which of the following banks is known as the first development bank of India?1. IDBI2. IFCI3. ICICI4. SFC
Answer» Correct Answer - Option 2 : IFCI IFCI IFCI, previously Industrial Finance Corporation of India, is a Non-Banking Finance Company in the public sector. Established in 1948 as a statutory corporation, IFCI is currently a company listed on BSE and NSE. IFCI has seven subsidiaries and one associate

5141.

Which of the following banks is known as the first development bank of India?1. IDBI2. IFCI3. ICICI4. SFC

Answer» Correct Answer - Option 2 : IFCI

IFCI

IFCI, previously Industrial Finance Corporation of India, is a Non-Banking Finance Company in the public sector.
Established in 1948 as a statutory corporation, IFCI is currently a company listed on BSE and NSE.
IFCI has seven subsidiaries and one associate

Discussion

5142.	Tesla is a measure of1. Magnetic flux density 2. Electric flux density3. Magnetic potential 4. Electric potential
Answer» Correct Answer - Option 1 : Magnetic flux density Tesla (T) is a measure of magnetic flux density. Electrical potential measures in volts (V). Magnetic potential measures in Amperes (A).

5142.

Tesla is a measure of1. Magnetic flux density 2. Electric flux density3. Magnetic potential 4. Electric potential

Answer» Correct Answer - Option 1 : Magnetic flux density

Tesla (T) is a measure of magnetic flux density.

Electrical potential measures in volts (V).

Magnetic potential measures in Amperes (A).

Discussion

5143.	Which of the following statements may not be considered as a code of ethics for professional engineers?1. Engineers shall hold paramount the safety, health, and welfare of the public2. Engineers shall extend help to poor workers3. Engineers shall perform services only in areas of their competence4. Engineers shall issue public statements only in an objective and truthful manner.
Answer» Correct Answer - Option 2 : Engineers shall extend help to poor workers The correct answer is Engineers shall extend help to poor workers. Code of Ethics refers to a set of guidelines to bring about acceptable behaviors in members of a particular group, association or profession. The salient values envisaged in the code of ethics are: Allegiance to the various ideals enshrined in the preamble to the Constitution Apolitical functioning Good governance for the betterment of the people to be the primary goal of civil service Duty to act objectively and impartially Accountability and transparency in decision-making Maintenance of highest ethical standards Merit to be the criteria in the selection of civil servants consistent, however, with the cultural, ethnic and other diversities of the nation Provision of healthy and congenial work environment Communication, consultation, and cooperation in the performance of functions i.e. participation of all levels of personnel in management.

5143.

Which of the following statements may not be considered as a code of ethics for professional engineers?1. Engineers shall hold paramount the safety, health, and welfare of the public2. Engineers shall extend help to poor workers3. Engineers shall perform services only in areas of their competence4. Engineers shall issue public statements only in an objective and truthful manner.

Answer» Correct Answer - Option 2 : Engineers shall extend help to poor workers

The correct answer is Engineers shall extend help to poor workers.

Code of Ethics refers to a set of guidelines to bring about acceptable behaviors in members of a particular group, association or profession.

The salient values envisaged in the code of ethics are:
- Allegiance to the various ideals enshrined in the preamble to the Constitution
- Apolitical functioning
- Good governance for the betterment of the people to be the primary goal of civil service
- Duty to act objectively and impartially
- Accountability and transparency in decision-making
- Maintenance of highest ethical standards
- Merit to be the criteria in the selection of civil servants consistent, however, with the cultural, ethnic and other diversities of the nation
- Provision of healthy and congenial work environment
- Communication, consultation, and cooperation in the performance of functions i.e. participation of all levels of personnel in management.

Discussion

5144.	Mechanical efficiency of a gas turbine as compared to internal combustion reciprocating engine is (a) higher (b) lower (c) same (d) un-predictable.
Answer» Correct option is (b) lower

Discussion

5145.	The air standard efficiency of closed gas turbine cycle is given by (rp = pressure ratio for the compressor and turbine)(a) η = 1 – \(\cfrac{1}{(r_p)^{γ-1}}\)(b) η = 1 – ( rp) γ − 1(c) η = 1 – \(\left(\cfrac{1}{r_p}\right)^{γ-1/γ}\)(d) η = (rp)γ-1/γ - 1
Answer» (d) η = (rp)^γ-1/γ - 1

Discussion

5146.	With the increase in pressure ratio thermal efficiency of a simple gas turbine plant with fixed turbine inlet temperature (a) decreases (b) increases (c) first increases and then decreases (d) first decreases and then increases.
Answer» (b) increases

Discussion

5147.	The process of abstracting steam at a certain section of the turbine and subsequently using it for heating feed water supplied to the boiler is called (a) Reheating (b) Regeneration (c) Bleeding (d) Binary vapour cycle
Answer» (b) Regeneration Regeneration is the process of bleeding steam and using it for feed water heating.

5147.

The process of abstracting steam at a certain section of the turbine and subsequently using it for heating feed water supplied to the boiler is called (a) Reheating (b) Regeneration (c) Bleeding (d) Binary vapour cycle

Answer»

(b) Regeneration

Regeneration is the process of bleeding steam and using it for feed water heating.

Discussion

5148.	An inward flow reaction turbine has an external diameter of 1 m and its breadth at inlet is 250 mm. If the velocity of flow at inlet is 2 m/s and 10% of the area of flow is blocked by blade thickness, the weight of water passing through the turbine will be nearly (a) 10 kN/s (b) 14 kN/s (c) 18 kN/s (d) 22 kN/s
Answer» (b) 14 kN/s d₁ = 1 m b₁ = 0.25 m V_F1 = 2 m/s k = 0.9 = coefficient of blade thickness Q = k⋅πd₁b₁VF₁ Q = 0.9 × π × 1 × 0.25 × 2 = 1.413 m³/sec Weight of water passing through turbine = ρgQ = 1000 × 9.81 × 1.413 = 14 kN/s

5148.

An inward flow reaction turbine has an external diameter of 1 m and its breadth at inlet is 250 mm. If the velocity of flow at inlet is 2 m/s and 10% of the area of flow is blocked by blade thickness, the weight of water passing through the turbine will be nearly (a) 10 kN/s (b) 14 kN/s (c) 18 kN/s (d) 22 kN/s

Answer»

(b) 14 kN/s

d₁ = 1 m

b₁ = 0.25 m

V_F1 = 2 m/s

k = 0.9 = coefficient of blade thickness

Q = k⋅πd₁b₁VF₁

Q = 0.9 × π × 1 × 0.25 × 2 = 1.413 m³/sec

Weight of water passing through turbine = ρgQ = 1000 × 9.81 × 1.413 = 14 kN/s

Discussion

5149.	With which religion are the Dilwara temples associated ?1. Buddhism2. Jainism3. Hinduism & Jainism4. Hinduism
Answer» Correct Answer - Option 2 : Jainism The correct answer is Jainism. The Dilwara Temples or Delvada Temples are a group of svetambara Jain temples located about 2+1⁄2 kilometers from the Mount Abu settlement in Sirohi District, Rajasthan's only hill station. Dilwara temples are associated with the Jainism religion. The earliest were built by Vimal Shah and supposedly designed or at least financed by Vastupala, Jain minister of Dholka. They date between the 11th and 16th centuries, forming some of the most famous monuments in the style of Māru-Gurjara architecture, famous for their use of very pure white marble and intricate marble carvings. They are managed by Seth Shri Kalyanji Anandji, Sirohi and are a pilgrimage place for Jains, and a significant general tourist attraction. Although Jains built many temples at other places in Rajasthan, the Dilwara temples are believed to be the most impressive. Jain architecture developed largely as an offshoot of Hindu and Buddhist styles. It cannot be accredited with a style of its own. Compared to the number of Hindu temples in India, Jain temples are few and spaced out. The regional styles of Jain temple architecture are easily distinguishable in different parts of the country. Early Jain temple architecture was mostly rock-cut and bricks were hardly used. In later years, however, brick temples were constructed on a large scale. At the same time, they also deviated from Hindu and Buddhist sites to build on their own. The majority of Jain temples in India consist of three core building elements: Image chamber Garbhagriha Hall Mandapa Porch

5149.

With which religion are the Dilwara temples associated ?1. Buddhism2. Jainism3. Hinduism & Jainism4. Hinduism

Answer» Correct Answer - Option 2 : Jainism

The correct answer is Jainism.

The Dilwara Temples or Delvada Temples are a group of svetambara Jain temples located about 2+1⁄2 kilometers from the Mount Abu settlement in Sirohi District, Rajasthan's only hill station.
- Dilwara temples are associated with the Jainism religion.
The earliest were built by Vimal Shah and supposedly designed or at least financed by Vastupala, Jain minister of Dholka.
They date between the 11th and 16th centuries, forming some of the most famous monuments in the style of Māru-Gurjara architecture, famous for their use of very pure white marble and intricate marble carvings.
They are managed by Seth Shri Kalyanji Anandji, Sirohi and are a pilgrimage place for Jains, and a significant general tourist attraction.
Although Jains built many temples at other places in Rajasthan, the Dilwara temples are believed to be the most impressive.

Jain architecture developed largely as an offshoot of Hindu and Buddhist styles.
- It cannot be accredited with a style of its own. Compared to the number of Hindu temples in India, Jain temples are few and spaced out.
- The regional styles of Jain temple architecture are easily distinguishable in different parts of the country.
Early Jain temple architecture was mostly rock-cut and bricks were hardly used. In later years, however, brick temples were constructed on a large scale.
- At the same time, they also deviated from Hindu and Buddhist sites to build on their own.
The majority of Jain temples in India consist of three core building elements:
- Image chamber Garbhagriha
- Hall Mandapa
- Porch

Discussion

5150.	If mf is the mass of fuel supplied per kg of air in one second, then the mass of gases leaving the nozzle of turbojet will be(a) (1 – mf ) kg/s(b) \(\cfrac{1}{(1+m_f)}kg/s\) (c) (1 + mf ) kg/s(d) \(\cfrac{1}{(1-m_f)}kg/s\)
Answer» (c) (1 + mf ) kg/s For turbojet engine, total mass of gases coming out will be sum of fuel and air supplied. Hence, mass flow = (1 + mf ) kg/s

5150.

If mf is the mass of fuel supplied per kg of air in one second, then the mass of gases leaving the nozzle of turbojet will be(a) (1 – mf ) kg/s(b) \(\cfrac{1}{(1+m_f)}kg/s\) (c) (1 + mf ) kg/s(d) \(\cfrac{1}{(1-m_f)}kg/s\)

Answer»

(c) (1 + mf ) kg/s

For turbojet engine, total mass of gases coming out will be sum of fuel and air supplied.

Hence, mass flow = (1 + mf ) kg/s

Discussion

Explore topic-wise InterviewSolutions in .

If α, β are the roots of ax2 + bx + c = 0 and −α, γ are the roots of a1x2 + b1x + c1 = 0 then β, γ are roots of Ax2 + x + C = 0, where C-1 =

If an integer p is chosen at random in the interval `0le ple5,` then the probality that the roots of the equation `x^(2)+px+(p)/(4)+(1)/(2)=0` are real is -A. ` (4)/(5)`B. ` (2)/(3)`C. ` (3)/(5)`D. None of these

If α, β are the zeros of the quadratic plynimial p(x) = x2 - (k + 6)x + 2(2k - 1) then the value of k,if α + β = \(\frac12\)αβ, is(a) -7(b) 7(c) -3(d) 3

The roots α and β of the quadratic equation x2 - 5x + 3(k - 1) = 0 are such that αβ = 1. Find the value k.

If Sin(A+B)=1 and Cos(A-B)=root 3/2, A&gt;B find the acute angles A &amp; B

calculate the current in a circuit if 500 c charge passes through it in 10 minutes

class : 100-120 120-140 140-160 160-180 180-200frequency :10 p 10 q 5 = 60

48x2 - 13x -1 = 0

in the given figure two triangles ABC and DBC lie on same base of BC such that PQ parallel BA and PR parallel BD.prove that QR parallel AD

Three pipes A, B and C can fill a tank in 12 hours, 15 hours, and 20 hours respectively. If pipe A is opened all the time and pipe B and C are open for one hour each, alternatively, with pipe A. In how much time the tank will be full?1. 9 hours2. 8 hours3. 6 hours4. 7 hours5. None of these

Two pipes A and B can fill a tank in 15 hours and 18 hours, respectively. Both pipes are opened simultaneously to fill the tank. In how many hours will the empty tank be filled?1. \(8 \dfrac{2}{11}\)2. \(7 \dfrac{2}{11}\)3. \(9 \dfrac{2}{11}\)4. \(10 \dfrac{2}{11}\)

Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours, then what percentage of the tank will remain unfilled?1. 17%2. 50%3. 33%4. 25%

A train passes a station that is 270 m long in 90 sec. at a speed of 108 km / hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds

A train passes a station that is 270 m long in 90 sec. at a speed of 108 km/hr. The time taken by the train to pass a bird feeder is:1. 45 seconds2. 60 seconds3. 81 seconds4. 55 seconds5. 90 seconds

Solve the equations and select the correct option:3x + 5y = 132x + y = 41. x > y2. x < y3. x = y4. x ≥ Y

Find the wrong term in the following number series.1150, 1342, 1527, 1705, 1876, 20411. 13422. 15273. 18764. 11505. 2041

The curve of angle of incidence versus angle of deviaton wshown has been plotted for prism. Q. The value of refractive index of the prism used isA. `sqrt(3)`B. `sqrt(2)`C. `sqrt(3)//sqrt(2)`D. `2//sqrt(3)`

Choose the correct code in which all the molecules have dative `pi` bond. I `- P_(4)O_(10)` , II `- (SiH_(3))_(3) N`. , III `-P_(4)O_(6)` , IV `-BF_(3)`A. I, II, III, IVB. II & IVC. I, III, & IVD. I, II & IV

Which of the following displacement occur:A. `Zn+2H^(+)toZn^(2+)+H_(2)uarr`B. `Fe+2Ag^(+)toFe^(2+)+Agdarr`C. `Cu+Fe^(2+)toCu^(2+)+Fedarr`D. `Zn+Pb^(2+)toZn^(2+)+Pbdarr`

List of Basicity or protocity of an acid.

What is Arrhenius Acid? Give its Examples.

The basicity of phosphorous acid is : (A) 1 (B) 2 (C) 3 (D) 4

The characteristics of an acid is : (A) turns blue litmus to red. (B) turns phenolphthalein pink from colourless. (C) decompose carbonates (D) oxy compounds of non-metals

Select the suitable reason (s) for higher strength of an acid or base : (A) higher value of Ka or Kb (B) higher extent of ionisation (C) (A) and (B) both (D) Larger number of replaceable H atoms.

Arrhenius theory of acid-base is not applicable in : (A) aqueous solution (B) in presence of water (C) non-aqueous solutions (D) none of the above

What are the types of Arrhenius acids ?

Define Arrhenius base.

Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would she observe? A. The solution turned red. B. Yellow precipitate was formed. C. White precipitate was formed. D. The reaction mixture became hot.

The average speed of nitrogen molecule is v. if the temperature is doubled and nitrogen molecule dissociate into nitrogen atoms completely then new average speed becoems.A. vB. `sqrt(2)v`C. `2v`D. 4v

What kind of acid does not completely dissociate?

The diameter of a cylindrical piston is 5 cm. The stroke length of the piston is 10 cm. Find the displacement volume, if π = 4.1. 125 cm32. 1000 cm33. 500 cm34. 250 cm3

How many associate banks were merged with the State Bank of India in the year 2017?1. 42. 53. 64. 7

Which of the following banks is known as the first development bank of India?1. IDBI2. IFCI3. ICICI4. SFC

Tesla is a measure of1. Magnetic flux density 2. Electric flux density3. Magnetic potential 4. Electric potential

Mechanical efficiency of a gas turbine as compared to internal combustion reciprocating engine is (a) higher (b) lower (c) same (d) un-predictable.

The air standard efficiency of closed gas turbine cycle is given by (rp = pressure ratio for the compressor and turbine)(a) η = 1 – \(\cfrac{1}{(r_p)^{γ-1}}\)(b) η = 1 – ( rp) γ − 1(c) η = 1 – \(\left(\cfrac{1}{r_p}\right)^{γ-1/γ}\)(d) η = (rp)γ-1/γ - 1

With the increase in pressure ratio thermal efficiency of a simple gas turbine plant with fixed turbine inlet temperature (a) decreases (b) increases (c) first increases and then decreases (d) first decreases and then increases.

The process of abstracting steam at a certain section of the turbine and subsequently using it for heating feed water supplied to the boiler is called (a) Reheating (b) Regeneration (c) Bleeding (d) Binary vapour cycle

With which religion are the Dilwara temples associated ?1. Buddhism2. Jainism3. Hinduism & Jainism4. Hinduism

If mf is the mass of fuel supplied per kg of air in one second, then the mass of gases leaving the nozzle of turbojet will be(a) (1 – mf ) kg/s(b) \(\cfrac{1}{(1+m_f)}kg/s\) (c) (1 + mf ) kg/s(d) \(\cfrac{1}{(1-m_f)}kg/s\)

If Sin(A+B)=1 and Cos(A-B)=root 3/2, A>B find the acute angles A & B